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MATHEMATICS
OF ACCOUNTING
BY
ARTHUR B. CURTIS, B.C.S., C.P.A.
AND
JOHN H. COOPER, B. Accts., C.P.A.
THIRD EDITION
PRENTICE-HALL, INC.
Englewood Cliffs
PRENTJOT^ALL ACCOUNTING SERIES
//. A. Finney, Editor
COPYRIGHT, 1925, 1934, 1947, BY
PRENTICE-HALL, INC.
ENGLEWOOD CLIFFS, N J.
ALL RIGHTS RESERVED. NO PART OF THIS BOOK MAY BE
REPRODUCED IN ANY FORM, BY MIMEOGRAPH OR ANY OTHER
MEANS, WITHOUT PERMISSION IN WRITING FROM THE
PUBLISHERS.
First Printing
. . .June, 1925
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Fourth Flint ing .
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Eighth Printing
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Revised Edition
First "» ' : .•
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Secon . 1 ' . • . *
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Third !'
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Thud Edition
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Thirteenth Printing, June, 1957
Preface
It is now somewhat more than twenty years since the publica-
tion of the first edition of Mathematics of Accounting. Those
familiar with the original edition, and the revised edition ten years
later, will find that in the present revision the sequence of subjects
has been considerably changed; the treatment of some subjects
has been amplified; certain other subjects have been added. Some
problems have been changed; new problems and review problems
have been included.
Subjects new to Part 1 are: Factors and Multiples, Business
Insurance, and Payroll Records and Procedure. The chapter on
Graphs has been expanded to include Index Numbers.
Part 2 has been enlarged to include chapters on the following:
Permutations and Combinations, Probability, Probability and
Mortality, Life Annuities, Net Premiums, and Valuation of Life
Insurance Policies.
Algebraic formulas have been changed to conform to standard
usage, while arithmetical substitutions and detailed solutions have
been retained. Those desiring to work entirely from the tables in
the appendix will find table references in the detailed solutions.
Grateful recognition is here given to those who have taught
from the previous editions and who have responded with sugges-
tions for this revision,
ARTHUR B. CURTIS
JOHN H. COOPER
Contents
PART I
2SAPTEB PAGB
1. FUNDAMENTAL PROCESSES AND SHORT METHODS FOR THE
ACCOUNTANT 1
Addition; Drill tables; Streamline addition; Drill table; Combinations
whose sum is 10; Drill table; Adding where the same number is repeated
many times; Drill table; Group addition; Drill table; Addition of two
columns at a time; Drill table; Recording addition by columns; Practical
applications; Subtraction; Avoid errors; Difference between a given minu-
end and several subtrahends; Balancing an account; Complement method;
Subtracting on an adding machine; Practical problems; Multiplication;
Contractions in multiplication; To multiply by factors of the multiplier;
To multiply when a part of the multiplier is a factor or multiple of another
part; To multiply a number of two figures by 11 ; To multiply any number
by 11; Multiplying by 25; Multiplying by 15; Multiplying numbers ending
with ciphers; Multiplication by numbers near 100, and by numbers near
1,000; Multiplication of two numbers each near 100, 1,000, and so forth;
Multiplying by numbers a little larger than 100; Multiplication of two
numbers each a little more than 100; Cross multiplication; To cross-multi-
ply a number of three digits by a number of two digits; To cross-multiply
a number of three digits by another number of three digits; Preparation of
a table of multiples of a number; Division; To divide by 25, 50, or 125;
Abbreviated division; Use of tables in division; Reciprocals in division.
2. CHECKING COMPUTATIONS 29
Methods; Rough check; Absolute check; Check numbers obtained by
casting out the nines: Verification of addition; Verification of subtraction;
Verification of multiplication; Verification of division; Verification of
division where there is a remainder. Check numbers obtained by casting
out the elevens: Verification of addition; Verification of subtraction;
Verification of multiplication; Verification of division. Check number
thirteen.
3. FACTORS AND MULTIPLES 35
Factors; Tests of divisibility; Greatest common divisor; Least common
multiple; Cancellation.
4. COMMON FRACTIONS 39
Terms explained; Reduction of fractions; Principle; Mixed numbers; To
change a mixed number to an improper fraction; Addition and subtraction
of fractions; Multiplication of fractions; Division of fractions; To find the
product of any two mixed numbers ending in J ; To multiply a mixed num-
ber by a mixed number; Decimal fractions; Addition and subtraction; Mul-
tiplication; Division; To abbreviate decimal multiplication when a given
number of decimal places is required; Division of decimals; To change a
decimal fraction to an equivalent common fraction; To change a common
fraction to a decimal; Aliquot parts; The use of aliquot parts; Multiplica-
tion by aliquot parts; Division by aliquot parts.
vi CONTENTS
CHAPTEB PAGH
5. PERCENTAGE 53
Relation between percentage and common decimal fractions; Applica-
tions; Definitions; Fundamental processes; Computations; Daily record
of departmental sales ; Per cent of returned sales by departments ; Clerk's
per cent of average sales; Per cent of income by source; Per cent of expense;
Per cent of increase or decrease; Operating statistics; Budgeting; Profits
based on sales; Marking goods; Commissions.
6. COMMERCIAL DISCOUNTS 71
Cash discount; Trade discount; Single discount equivalent to a series;
To find the net price; Transportation charges on discount invoices;
Anticipation.
7. SIMPLE INTEREST 77
Definition; Short method of calculating; Sixty-day method; Method
using aliquot parts; The cancellation method; Dollars-times-days method,
6%; Interchanging principal and time; Exact or accurate interest; Accu-
mulation of simple interest; Symbols; Simple amount; Rate; Time; Present
worth; Comparison of simple amount and simple present worth; True
discount.
8. BANK DISCOUNT 87
Definition; Counting time; Finding the difference between dates by use
of a table; Proceeds; To find bank discount and proceeds; To find the face
of a note when the proceeds, time, and rate of discount are given.
9. PARTIAL PAYMENTS 91
Partial payments on debts; Methods; United States Rule; Merchants' Rule.
10. BUSINESS INSURANCE 95
Kinds of insurance; Policy; Fire insurance; Form of policy; Rates; To find
the premium; Agent's commission; Cancellation of policies; Coinsurance;
Use and occupancy insurance; Group life insurance; Health insurance;
Workmen's compensation insurance.
11. PAYROLL RECORDS AND PROCEDURE 107
Requirements; Payroll procedure; Timebooks; Time-clock cards; Deduc-
tions; Withholding exemptions; Payroll sheets; Piecework system; Pay
checks; Pay envelopes and receipts; Coin sheet and currency memorandum.
12. AVERAGE 123
Simple average; Moving averages; Progressive average; Periodic average;
Weighted average.
13. AVERAGING DATES OF INVOICES 131
Definition; Use; Term of credit; Average due date; Focal date; Methods;
Rule for product method.
14. EQUATION OF ACCOUNTS, OR COMPOUND AVERAGE . . . 135
Definition; Rule for the product method; When to date forward or
backward.
CONTENTS vii
CHAPTER PAGL
15. ACCOUNT CURRENT 139
Definition; Methods.
16. STORAGE 141
Definition; Running account.
17. INVENTORIES 143
Valuation of inventories; Cost or market, whichever is lower; Average cost
method; "First-ill, first-out" method of inventory; "Last-in, first-out"
method of inventory; Merchandise turnover; Number of turnovers; Per
cent of mark-down to net cost; Computation of inventory by the retail
method; Determining the ratio of cost to retail.
18. GROSS PROFIT COMPUTATIONS 155
Gross profit; Rate per cent of gross profit; Procedure; Uses; Cost of goods
sold; Rate per cent of cost of sales; Fire Losses; Use of gross profit test in
verification of taxpayer's inventory; Installment sales of personal property;
Computation of gross profit ; Reserve for unearned gross profit ; Bad debts ;
Deferring income; its effect on tax.
19. ANALYSIS OF STATEMENTS 169
Financial and operating ratios; Costs, expenses, and profits; Ratio of gross
profit to net sales; Ratio of operating profit to net sales; Ratio of net profit
to net sales; Ratio of operating profit to total capital employed; Ratio of
net profit to net worth; Earnings on common stockholders' investments;
Working capital ratio ; Sources of capital; Manner in which capital is invest-
ed; Turnover of total capital employed; Turnover of inventories; Turnover
of accounts receivable; Turnover of fixed property investment.
20. PARTNERSHIP 181
Definition; Mathematical calculations; Goodwill; Profit-sharing agreements;
Lack of agreement; Losses; Arbitrary ratio; Ratio of investment; Division
of profits by first deducting interest on capital; Profits insufficient to cover
interest on investment; Adjustments of capital contribution; Profit sharing
in ratio of average investment; Liquidation of partnership; Methods;
Total distribution; Periodic distribution.
21. GOODWILL 201
Definition; Basis of valuation; Earning power determined from profit and
loss statements; Methods of valuing goodwill; Case illustrations; Valuation
by appraisal; Valuation by number of years' purchase price of net profits;
Valuation on basis of excess of profits over interest on net assets; Basis of
stock allotment; Common stock only; Preferred stock for net assets; Bonds
preferred stock, and common stock.
22. BUSINESS FINANCE 213
Stock rights; Sale of stock and rights, federal income tax; Working capital;
Cumulative voting; Book value of shares of stock; Profits distribution.
23. PUBLIC FINANCE AND TAXATION 223
Governmental functions; Purposes of taxes; Appropriations; Kinds of
taxes. Property Tax: Determination of tax rate; To find the amount
of tax.
Viii CONTENTS
CHAPTER PAGE
24. FUNDAMENTALS OF ALGEBRA ' . . . . 229
Explanation ; Symbols and terms ; Positive and negative numbers ; Addition
of positive and negative numbers; The coefficient; Parentheses, brackets,
and braces; Subtraction; Multiplication; Division.
25. EQUATIONS 235
Simple equations; Fractions; Clearing of complex fractions; Simultaneous
equations with two or more unknowns; Arithmetical solution of problems
containing unknown quantities.
26. LOGARITHMS 249
Uses of logarithms; Exponents; Parts of a logarithm; Characteristic; Posi-
tive characteristic; Negative characteristic; Mantissa; How to use a table
of logarithms; To find a number when the logarithm is given; To find a
number whose mantissa is not in the table; Rules for computation by loga-
rithms; Multiplication by logarithms; Division by logarithms; Powers of
numbers; Process with a negative characteristic; Roots of numbers, Process
with negative characteristics; The slide rule; Use of slide rule; Accuracy of
calculations made by the slide rule; Theory of the slide rule; How to learn to
use the slide rule; Reading the slide rule; Construction of model slide rule;
Multiplication on the slide rule; Division on the slide rule.
27. GRAPHS AND INDEX NUMBERS 265
Charts and graphs; Circle chart; Comparison of circles; Bar chart; Line or
curve chart; Rules for coordinate charts; Logarithmic chart; Ratio charts.
Index Numbers: Uses of index numbers; Index numbers; Economic position
of agriculture; Construction of index numbers; Composite price indexes;
Weighted index numbers ; Farm evaluation on the basis of crop production
index; Computation of the crop production index.
28. PROGRESSION 285
Definition ; Increasing series ; Decreasing series ; Arithmetical progression ;
Symbols; Relation of elements. Increasing Series: To find the number of
terms ; To find the first term ; To find the last term ; To find the common dif-
ference; To find the sum. Decreasing Series : To find the number of terms;
To find the first term; To find the last term; To find the common difference;
To find the sum. Geometrical Progression : Elements ; Increasing series ; To
find the first term; To find the last term; To find the sum; To find the
ratio; Decreasing series; To find the first term; To find the last term; To
find the sum; To find the ratio; Progression problems solved by the use of
logarithms.
29. FOREIGN EXCHANGE 293
Foreign trade; Rate of exchange; Par of exchange; Current rate of
exchange; Six classes of problems; Conversion of one monetary unit into
terms of another; Conversion of decimals of one monetary unit into mone-
tary units of a smaller denomination; Interest on foreign exchange; To find
the value of a time bill of exchange; Foreign exchange accounts; Averaging
accounts in foreign exchange; Conversion of foreign branch accounts.
PART II
30. COMPOUND INTEREST 311
Compound interest; Compound interest method; Actuarial science; Sym-
bols: Principal; Time; Rate; Ratio of increase; Compound amount tables;
CONTENTS •*
CHAPTER ' PAGE
30. COMPOUND INTEREST (Cont.}
Calculation of compound amount; Compound amount of given principal;
Compound interest; Results of frequent conversions of interest; Nominal
and effective rates; Effective interest; Compound present worth; Com-
pound discount; Rate; Time; Compound amount for fractional part of con-
version period.
31. ORDINARY ANNUITIES 327
Definition ; Kinds of annuities ; Rent of an annuity ; Amount of an ordinary
annuity ; Analysis of compound interest ; Relation of compound interest and
annuities ; Procedure in computing the amount of an annuity ; Semiannual
or quarterly basis; Rent of ordinary annuity; Use of effective interest in
annuities; Sinking fund contributions; Present value of an ordinary
annuity; Amortization; Computation of the rents or periodic payments of
the present value of an ordinary annuity ; Payment of debt by installments ;
Computation of the term of an annuity; Use of effective rate in annuities;
Computation of the rate of an annuity; Selection of rate by calculation of
amounts of annuities; Solution of annuity problem with limited data.
32. SPECIAL ANNUITIES 349
Annuity due; To find the amount of an annuity due; Present value of an
annuity due; Comparison of present value of an ordinary annuity and that
of an annuity due ; To find the present value of an annuity due ; Rents of the
amount of an annuity due; Rent of the present value of an annuity due;
Effective interest on annuity due; Deferred annuity; Perpetuity; Perpetui-
ties payable at intervals longer than a year.
33. BOND AND BOND INTEREST VALUATION 367
Definitions; Bonds sold at par; Bonds purchased at a discount or at a pre-
mium; Price and rate of yield; Use of bond tables; Bond table, first form;
Bond table, second form; Interpolating in bond tables; Bond values com-
puted without tables; Bonds sold at a discount; Bonds sold at a premium;
Values of bonds between interest dates ; Interest accrued between interest
dates ; Bond discount or premium between interest dates ; Illustration of the
practical process of calculating the value of a bond bought at a discount ;
Theoretical procedure illustrated; Illustration of the practical process of
calculating the value of a bond bought at a premium; Bonds bought on
a yield basis ; Bonds to be redeemed above par ; Serial redemption bonds ;
Frequency of redemption periods; Alternative solution; Bonds redeemed
by other than equal annual payments; Bonds redeemable from a fund;
Effective rate of interest on bonds ; Effective rate of interest on bonds sold
at a premium; Effective rate on bonds sold at a discount; Computation
when bond table is not available ; Approximation by averages ; Amortization
of discount, premium, or discount and expense on serial redemption bonds;
Bonds outstanding method ; Scientific method.
34. ASSET VALUATION ACCOUNTS 40£
Asset valuation; Depreciation; Depletion; Depreciation methods; Straight-
line method; Working-hours or unit-product method; Sum-of-digits
method; Sinking-fund method; Annuity method of depreciation; Fixed-
percentage-of-diminishing-value method; Composite life; Depletion; Cal-
culation of depletion; Capitalized cost; Perpetuity providing for ordinary
annual expenses and for replacement of asset; Capitalization of a wasting
jsset.
x CONTENTS
CHAPTER PAGE
35. BUILDING AND LOAN ASSOCIATIONS 425
Control; Classes of stock; Withdrawal of funds; Plans of organization;
Terminating plan; Serial plan; Dayton or Ohio plan; To find the time
required for stock to mature (rate of interest given) ; To find the effective ,
rate of interest on money invested in installment shares.
36. PERMUTATIONS AND COMBINATIONS 439
Permutation; Number of ways of doing two or more things together;
Combinations.
37. PROBABILITY 445
Probability; Permutations and combinations in probability; Compound
events; Independent events; Mutually exclusive events; Empirical
probability.
38. PROBABILITY AND MORTALITY 453
Life insurance; Mortality table; Notation; Probability of living; Proba-
bility of dying; Joint life probabilities.
39. LIFE ANNUITIES 459
Factors involved; Pure endowment; Life annuity; Commutation columns;
Life annuities due; Use of commutation table; Deferred annuity; Deferred
We annuity due; Temporary life annuities; Temporary annuities due; Life
annuities with payments m times a year; Forborne temporary annuity due.
40. NET PREMIUMS 469
Net single premium; Annual premiums; Term insurance; Annual premium
for term insurance; Net single premium for endowment insurance; Annual
premium for endowment insurance.
41. VALUATION OF LIFE INSURANCE POLICIES 473
Mortality and the level premium; Policy reserves; Interest and the pre-
mium; Loading; Dividends and net cost; Terminal reserves; Retrospec-
tive method; Transformation; Reserve valuation for limited payment life
insurance; Preliminary term valuation.
APPENDIX APPENDIXES
I. Practical Business Measurements 481
II. Tables of Weights, Measures, and Values 489
III. Tables . . . . • 497
INDEX 540
PARTI
CHAPTER(lp
Fundamental Processes and Short
Methods for the Accountant
Addition. Addition is the process of combining numbers of the
same denomination. Quantities of such unlike measures as dollars
and yards cannot be added; but quantities like yards, feet, &i\dinches
can be changed to like numbers and then added. Like numbers are
numbers that express the same kind of units. The sum is the
number resulting from adding two or more like numbers, and the
addends are the different numbers to be added.
Addition is the most fundamental of all numerical operations.
It is essential that the clerk, the businessman, arid the accountant
be able to add with precision and rapidity. The ability to recog-
nize the sums of numbers instantly is acquired by constant practice
and careful study.
Drill tables. Practice adding the columns of numbers in the
following table until you can complete the operation in twenty-five
seconds, without error. State sums only; that is, do not repeat the
numbers to be added.
581654596729489*
1?1<*322. 747l'6462
796375943959848 •
49^125432845817
786689278876379
Practice stating the sums of the following columns of numbers
until you can do all of them correctly in less than two and a half
minutes.
34234335576523538
22233323323322232
[Drill iable continued on next page.]
3
FUNDAMENTAL PROCESSES AND SHORT METHODS
567232-72847446332
431222222-32,422222
668.9 7. 68796754 6' 635
45849635345484568
4432123. 2232212323
67457576886346396
54226414413331321
676988798995-79998
95886487865577776
8,28633567 £.5 5 375464
76747589645695966
44314644346^55551
78879657689727445
3 4 5 5365467652 3< '4 .4 5
Streamline addition. Omit unnecessary words: that is, do not
name the number to be added; name only the sum.
6 In the example at the left, a common way of adding would be (com-
7 mencing at the top): 5 and 7 are 12, 12 and 8 are 20, 20 and 4 are 24,
8 arid so forth. Instead of adding in this manner, proceed to the answer
4 by saying (mentally), " 12, 20, 24, 27, 29, 38."
3
2
Drill table.
3 4267 5 .1835
2 5 9 8 9. 3 4 . 7 2 8 '
7377268671
40 6. 9586963
5933872499
6842625746,
1584397537
Combinations whose sum is 10. Combinations of two or more
numbers whose sum is 10 are of frequent occurrence. When these
combinations are recognized, addition may be shortened by adding
such combinations as 10.
FUNDAMENTAL PROCESSES AND SHORT METHODS 5
4 In this example, the addition may be performed as follows: (com-
7, mencing at the top) 14, 16, 21, 30, 40.
3 Or, it may be added in this manner: 11, 21, 30, 40.
2 Do not try to form combinations. Unless they are instantly recog-
5 nized, add the numbers in the regular manner.
9
5)
3
J?
40
Drill table.
7
2
1
5
6
4
9
8
2
3
5
6
4
,9
8
2
7
4
,5
fr.
2
7.'
1
4<
8
2.
5
3
8
2
7
3
1
8
7
2
5
4
5
2
3
)
\
6
!)
lj
7,
2/
9
4
£
3^
1
5
9"
8
3
7
4*
2.
9
7'
3
6
3
t
2
5
2
7
4
5"
1.
7.
5
5
9
Adding where the same number is repeated many times. In
obtaining averages, in adding statistics, and in other work involv-
ing addition, often the same number is repeated many times. Use
multiplication to save time in adding.
724 In this example, 7 occurs four times and 6 occurs three times in the
785 third column. The sum of the third column may be found as follows:
773 Carried 4
748 4X7 28
696 3X6 18
687 «o
679 5U
5092
-7
^
The work is actually performed mentally thus: 4 (28) 32, (18)
50. Where the columns are long, a side calculation may be
necessary.
Drilftable. . . .. , , ,
68
63
64
,67
59
54
57
•IS
284
273
281
311
314-
321
318
W
V4
33
32
31
34
32
33
36
.86
.75
86
.29
36
.75
.95
W
23,
23.
24.
25
26
31
32,
33
56
95
72
31
54
72
69
47
47.
39
38.
45
39
42.
38,.
56
85
64,
58.
95
74
56
4l.
72,
72
69
68
67
71.
53
37
48
95
83
44
,93
59
•-1 0 CLA
FUNDAMENTAL PROCESSES AND SHORT METHODS
Group addition. The most practical method of adding is to
group or combine two or more figures mentally, and to name
results only.
5
4
7
3
8
6
1
J2
36
Mental Steps
10 19
14
36
Drill tabl
t -v
/
Instead of saying, "5 and
4 are 9, and 7 are 16, and 3
are 19, and 8 are 27, and 6
are 33, and 1 are 34, arid 2 are
30," simply think, ''9, 19,
33, 36."
ib
t
6 -'4 8 5 3" 7 ; 0* 3 8 6 4 5 7 3j 8 7 4 2 9
3652487632576393847
7' 3 6 2 9 8- 4 6 7 4 5 7 6 2 4- • 6 9 7 4
Z 6 3. 5875325973846827
8 2 4 .8 3 7 4 6. 5 4, 9 7 6 3 5 8 7 3 6
386247 8 635863476248
5737358237634862359
437427864-7648395821
55998636777753421 73
Addition of two columns at a time. Two columns of figures may
he added at the same time, as shown in the following illustration:
Mental titeps
56 Tens
Units
Tens
Explanation
Units
28(7) 76 (2) 84
43 (3) 124 (4) 127
_21 (5) 147 (6) 148
148
(1) 56 and 20 = 76 (2) 76 and 8 = 84
(3) 84 and 40 = 124 (4) 124 and 3 = 127
(5) 127 and 20 = 147 (6) 147 and 1 = 148
Drill table.
s
^ ^
79 82 24 37 65 39, 28 28.
48 84 33 44 81 58 39 59-
81 95 46 53 42 48 23 86
1£ 83 52 66 73 73 37 63
Recording addition by columns. ^Accountants are subject to
interruptions, but the time required to re-add a column of figures
for the purpose of picking up the carrying figure may be saved if
the total of each column is recorded separately. The separate
column totals are also convenient to use in checking the work ; for
instance, if in a final summary of additions there is an error of
$100.00, the hundreds' columns of the subtotals may be verified
quickly without the necessity of re-adding all the columns.
FUNDAMENTAL PROCESSES AND SHOR1 METHODS 7
Fxample — Method 1 Example — Method 2 Example— Method 3
4572 4572 4572
39S6 3986 39S6
2173 2173 2173
5911 5911 5911
2765 2765 2765
4937 4937 4937
24 24 24
32 ^ 34 34
40 43 43
20 24 24
24344 24344
Explanation 1. Add each column separately, setting the sums one place to
the left, as in the example. After the last column has been added, add the
individual sums in regular order; that is, from right to left.
Explanation 2. In Method 2, a little time is saved by adding to each column
the number carried from the column at the right.
Explanation 3. Method 3 differs from Method 2 in the writing of the columns'
sums. It is somewhat easier to write the sums one below the other. This
cannot be done in Method 1 because carrying figures are not used, and another
step is required to complete the answer: that is, finding the grand total of the
units, tens, hundreds, and so forth.
A modification of the third method is useful in adding columns of dollars
and cents.
$ 644 22 The total, $4,062.08, is obtained by adding each column sepa-
821 94 rately as explained under Method 3. The computation will
314 26 appear as follows, the purpose of the horizontal lines being to
712 84 separate cents from dollars, and hundreds from thousands.
976.54
592 28
$4~062~08
Sum of the first column 28
Bum of the second column, 28 plus 2, carrying number . . . 30
Sum of the third column, 19 plus 3, carrying number 22
Sum of the fourth column, 24 plus 2, carrying number . . 26
Sum of the fifth column, 38 plus 2, carrying number 40
As there are no more columns, write the carrying number 4
The total, $4,062.08, is obtained by reading the numbers at the right, com-
mencing at the bottom.
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1395
2764
8351
6248
5347
4586
Problems
'V?* 6'
36S8 $367
4932, 421.
7A63; 281 .
2898 • 633
6598 855.
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98
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$786
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822
753
629,
367.
521
57
56
86
75
43
.54
8 FUNDAMENTAL PROCESSES AND SHORT METHODS
Practice Problems
Average weekly earnings from payroll reports.
^ ^f
2.
<<• fc
\t
4.
VK- a
<# 0.7 b.
- 17;'
8.
^33 20
35 72
30
85
33.20
28
28
37
61
22
.53
23.25
25.13
29 88
21.
.99
35 72
28
24
29
.97
35
62
35 25
37.41
39 24
35
31
42 28
22
61
31
.36
16
.22
32 18
31 65
33 47
28
89
31.56
21
46
21
.34
31
.91
37 17
31.40
35.29
27
.89
29.82
23
.91
33,
.34
14
.16
35 99
22 93
23.22
19
.11*
37.33
22
41
31,
.78
27
.14
36 66
32.16
33.49
19.
.15
28 97
17
99
34
73
31
62
35 96
26 37
28 34
18.
.96.
54.61
25
21
26
74
33
84
37 37
36 52*
33.64
30
73'
39 06
33.
25
22
88
30
89
18.17
32 05 ,
34 60
30.
70
28.29
28
72.
28
09
39
04
29 64
32 58.
26 49
18.
,95
26 87
19
49-
28
20
15
76
15 82
23 60
28.81
33,
,45'
27.01
18.
64
20
80
20
87^
29 87
23 44
37.92
31
(TO
39.52
37.
12
37
53
38
72
27 84
36 37
41.54
30
41"
41.86
31.
04
28
94
37.
16 ,
36 83
Tabulation of advertising lineage.
7, ^\9. ,
10.
^tQLlj
12.
' "-I?-,
14.
J> "^5.^
16.
26,228
29,207
22,107-
14,849
10,049
57,104
13,022
10,755
13,818
17,588
1 5,977- •
'11,966
14,745
71,075
15,223
16,850
27,122-
28,267
39,082"
36,021
8,562
119,035
17,058-
15,573
17,077
15,095
9,644
7,888
5,575
28,857
18,048-
10,259
32,094-
36,072
23,449»
19,634
12,376
39,190
26,174
19,635
32,936
32,835
18,930
15,033
2,175
16,085
28,169
24,572
21,499
18,116
46,520
43,778
7,531
15,484
14,949
15,057
20,655
24,094
25,140
19,271
8,650
28,192
24,478 '
19,445
22,338
10,365
8,015 4
13,412
15,530
14,711
22,175
24,493
13,412
60,475
38,795
93,323
23,680
22,865
23,680
37,335
Addition of dollars and cents, irregular items.
f t.
'jVfl-S is. 0\j$r4
20. *|21£t)
22.
13p.
10
86
.35
80
45.
40
$6
.35
1
,955.05
12.
65
52
.67
44
.82
34
20
*' 52*
67-
531.03
10
57
44
00
37
45
28
66
42
57
442 . 85
50.
05
208
.33
127
.29
135
68
208
33
2
,148 74
1,275.
48
394
.68
4,151
.36
945
21
878
52k
30
,149 39
260,
73
64
.72
24.94-
2
.72
111
56
112 64
7,
81
6
.29
.72
118,
.61
,6
46
509.74
78,
.13
27
.33
71
.97
.75
44
.77
27.02
2
.50
.62
12
.09
32
.49
.69
153 07
111
82
27
.65
.35'
8
.58
3
07
1
,512.34
29
.53
7
.33
160
.31-
33
55
70
63
1
,002.90
54
.53:
16
.29
45
15:
8
45
5
37
282 51
15
.36-
4
.59
128
.60
24
.85
35
,15
66 66
147
62*
111
52
41
.51.
138
.34
9.98
146.43
5
27
59
.68
46
.43
92
.54
214
34
641.51
FUNDAMENTAL PROCESSES AND SHORT METHODS
Practical applications. In the following problems will be found
examples of business records requiring addition for the completion
of the record.
Problem 1
In this problem, cash register tapes provided the source of the entries on
Form 1. As the sales were registered, the classification was imprinted on the
tape. At the end of the day, the classified items appearing on the tape were
accumulated on Form 1, and the totals transferred to Form 2. At the end ot
the week, Form 2 was added; at the end of the month, the weekly totals were
accumulated to monthly totals. Thus, sales for the month were analyzed by
departments or classes.
Add the columns on Form 1 (Saturday's sales), transfer the totals to Form 2,
and find the total sales for the week.
Form 1
Candy
Cigars
Soda
Drugs
Own
Remedies
Patent
Medicines
Toilet
Articles
X3?
.10
.10
.45
75
1.25
.35
1.25
.25
.15
1.64
.45
.50
1.15
.80
.15
.20
.10
.15
1 50
.80
.45--
.25
.45
.75*
1.25
.89
2.65
.75*
.50
.10
1 50
.90
33
.75
90
25.
.50
U75
1.65
2,35
1.85
W
/•'"
1,0'
*<\
4 . />, ~
£ v <-
Iff
Form 2
Day
Candy
Cigars
Soda
Drugs
Own
Heine-
•t •
Patent
Medi-
Toilet
Articles
Totals
^
.^
dies
cines
IVlon.
12 65
19 15
T
3 95
27.63
Vis
/9 85
5.00
^4(
Tues.
8 50
16.10
6 80
33 98
2.47
12.20
3 65
..S^ 70
Wed.
11 25
8 75
4 50
15.20
1 75
2 55
10.45
...$*.'£>"
Thurs.
9 65
4 25
2.55
7 65
2 85
4 86
4.63
..2&.ifv
Fri.
10 35
5.55
3.75
12.84
3.68
5.49
3.85
...j*»\5f J
Sat.
..4<<;:<?
I'^O
(•&
1^1$
47/4
£•£'£
7':Ty
" jryjir
1>*C
s.^.
&A-
(&-.n
*"*'
ai..o
^.^
^,lf
Form 2 is self-proving — that is, the sum of the daily totals must equal the
sum of the departmental or classification totals.
Problem 2
The "peg board'7 is used for accumulating numbers having to do with many
kinds of information. The numbers are entered on narrow forms which are
attached to the "peg board." The forms are held in place and cross extension
as well as "footings" are thus permitted.
10 FUNDAMENTAL PROCESSES AND SHORT METHODS
In the following example, this arrangement is used to accumulate total
departmental sales made by a salesman.
Salesman
R. F.
Salesman
R. F.
Salesman
R. F.
Salesman
R. F.
Salesman
u. F. rotaz »Srtte«
Dept.
,f>qte 4/2
/j^ 4/3
JCtyte 4/4
£)it^ 4/5
7^% y6
138 &7
.587 . 23
;347 58
£o f 0 1
f 637 82
1
645.39
321.69
123 63
563 85
495 71 ..... .
2
362 45
847.86
219 23
149 27
826 45
3
472 31
123.45
547 81
462 38
718 26
4
45 97
671.17
359 34
326 49
534 58
5
273.14
372 45
135 67
857 62
149 17
6
928,. 63
436 49
569 81
318 48
529 32
7
7 *V l_J VJ *^\ld
"\ trfi S^i
i *i i vl
. A , jj" LV
, j. i'
V^ L*™ ^?M
']/) **^ •
J.H r ^
> » • l
(a) Find the total of eacli day's sales.
(b) Find the total sales for each department.
The answer in the lower right corner proves the work.
Subtraction. Subtraction is the process of finding the differ-
ence between two like numbers. The minuend is the number to be
diminished, and the subtrahend is the number to be taken from the
minuend. ""-
Addition and subtraction are closely related. Subtraction by
adding is the method used by the expert cashier and by money
changers. The " making change" method of subtraction consists
in adding to the amount of the purchase enough to make the sum
equal to the amount tendered in payment.
Example
Y buys groceries to the value of $1.34 and gives the cashier two one-dollar
bills in payment. How much change should he receive?
FUNDAMENTAL PROCESSES AND SHORT METHODS 11
Solution
The cashier in making change may return to Y a penny, a nickel, a dime,
and a half dollar, saying: "$1.34, 35, 40, 50, $2.00," which means $1.34 + .01
= $1.35; $1.35 + .05 = $1.40; $1.40 + .10 = $1.50; and $1.50 + .50 = $2.00.
Other coins than those mentioned may be returned by the cashier, but it is
customary to make change in the largest coins possible.
Exercise
As the cashier, make change, using the largest denominations possible,
assuming the following purchases were made and two one-dollar bills were offered
in payment.
1. $1.44 5. $1.64 9. $1.17 13. $1.43
2. 1.67 6. 1.32 10. 1.29 14. 1.38
3. 1.27 7. 1.82 11. 1.54 16. 1.49
4. 1.41 8. 1.1 1 12. 1.56 16. 1.05
Avoid errors. Many errors in subtraction are made in borrow-
ing from the next higher order. When that order is reached, it is
not uncommon to overlook the fact that borrowing has taken place.
Errors of this kind can be avoided by changing subtraction to the
process of addition; that is, by adding to the subtrahend the num-
ber required to make the subtrahend equal to the minuend.
Explanation. Instead of thinking, "7 from 16 is 9," think, "7 + 9 «= 16."
Write the 9. Add 1, the digit carried over, to the
8, making 9. 9 -f 8 = 17. Write 8, and add 1, Example
the digit carried over, to 1, making 2. 2 + 0 = 2. Minuend 8276
Write 0. 3 + 5 = 8. Write 5. Answer: 5,089. Subtrahend 3187
Difference 5089
Problems
1. 9574 2. 7436 3. 6175 4. 8147 5. 6328 6. 5317
5886 3569 2897 4368 2549 3428
Difference between a given minuend and several subtrahends.
In instances similar to the following example, the final result can
be found in one operation by the application of the foregoing
method of subtraction.
Example
From a fund of $3,456, the following disbursements were made: $594, $375,
and $286. What was the balance left in the fund?
Explanation. Write the problem as shown in the solution. Begin at the
right, and add the units' column of subtrahends, (6 + 5 + 4), adding (and
setting down) enough (in this instance, 1) to make the units' figure
of the sum the same as the units' figure of the minuend. Add the
tens' column of the subtrahends, including the carrying figure, . ?
(I +8 + 7 + 9), adding (and setting down) enough (in this in- 594
stance, 0) to make the tens' figure of the sum equal the tens' figure 375
of the minuend. Add the hundreds' column of the subtrahends, 286
including the carrying figure, (2 + 2 + 3 + 5), adding (and setting $2,201
12 FUNDAMENTAL PROCESSES AND SHORT METHODS
down) enough (in this instance, 2) to make the hundreds' figure of the minuend.
To the carrying figure, 1 , add enough (in this case. 2) to make the thousands'
figure of the minuend; set down 2.
Problems
1. $1,562 2. $2,756 28 3. $5,987 4. $4,875 6. $2,975
437 52770 235 365 762
122 7 55 789 1,529 194
254 528 75 1,526 284 275
Balancing an account. In most cases, inspection will tell which
side of the account is the greater in amount. Add the larger side,
and put the same footing on the smaller side, leaving space for the
balance; then add from the top downward, Mipply'mg the figures
necessary to make the column total equal to the footing previously
placed there.
Example
Debits Credits
$ 1,956.18 $ 134.26
3,452 75 258 19
289.34 764 83
5,726.31 2,375 94
_ Balance, 7,891 36 *
jjMgjjS $U2424.58
Explanation. The balance, $7,891.36, was found as follows: Inspection
showed the debit side to be the larger in amount. It was therefore added, and
the footing of the account, $11,424.58, was placed under both debit and credit
columns. The first order of the credits — that is, the cents — addsjto 22. Insert
6 to make 28. With 2, the digit carried over, the second order, the dimes, adds
to 22. Insert 3 to make 25. The third order, the dollars, with the digit carried
over, adds to 23. Insert 1 to make 24. The fourth order, the tens of dollars,
with the digit carried over, adds to 23. Insert 9 to make 32. The fifth order,
the hundreds of dollars, with the digit carried over, adds to 16. Insert 8 to
make 24. The sixth order, the thousands of dollars, with the digit carried over,
adds to 4. Insert 7 to make 11.
Problems
1. Debits Credits 2. Debits Credits 3. Debits Credits
$856.73 $298.56 $725 14 $1,356.17 $3,586.28 $ 591.18
34596 264.39 23951 69135 192.75 2,751.26
298.85 6.15 64.28 256.38 38472 185.35
142 31 .............. .............. 75.19 265 54 ..................
^Complement method. The complement of a number is the
difference between that number and the unit of a next higher order
Thus, the complement of 6 is 4; the complement of 8 is 2; and the
complement of 68 is 32.
If, in subtracting a number less than 10 from a given number,
Its complement is added, the result will be 10 too large. If two
FUNDAMENTAL PROCESSES AND SHORT METHODS 13
complements are added, the result will be 20 too large ; and if three
complements are added, the result will be 30 too large.
To find the sum of a column containing numbers to be sub-
tracted, add the complements of the subtractive items, and from
the sum of each order deduct as many tens as there are subtractive
items in the order.
Example
A practical application of the complement method of subtraction is that of
finding the net increase in a statistical record such as the following:
Sales
Sales
Increase
Dept.
This Mo.
Last Mo.
Decrease*
1
$ 427 95
$ 346 29
$ 81 66
2
515 86
457 75
58 11
3
395 57
385.86
9.71
4
402.75
416 87
14 12*
$1,742.13
$1,606.77
$135.36
Solution
The difference between the sales this month and the sales last month for
each department is shown as an increase or a decrease. The difference between
the total sales this month and the total sales last month is $135.36. To prove
that the departmental increases and decrease are correct, add the third column,
beginning at the top and adding downward, using the complement each time
on the last number. Thus, 8 and 8 are 16; write 6, and drop the 10, as one
complement was added and the answer is 10 too large. 14 and 9 are 23; write 3
and carry 10, dropping one 10. 19 and 6 are 25; write 5 and carry 1, again
dropping one 10. 14 and 9 are 23; write 13, dropping one 10 as before.
Example Solution
Find the net increase of the fol- In this problem there are four
lowing items: items showing decreases; therefore,
each time a complement is added, the
Increase final result will be 10 too large, and
Decrease* in this case, the final result will be 40
15 60 too large, so 40 is deducted each time.
4 51* Begin at the top and add downward:
17 20 9 (comp.), 15, 23 (comp. was 8), 30,
61 96 31, 37, 46, 51, subtract 40, write 1
29 00 and carry 1.
8 62* Now the next column. 7(6andl),
124 20 12, 14, 23, 27, 29, 33, 35, 38, 41, 45,
59 40 54, 60, subtract 40, write 0 and
89 83* carry 2. Next column, 7, 13, 20, 21,
199.30 30, 32, 36, 45, 46, 55, 62, 64, 68, 70,
113 79* subtract 40, write 0 and carry 3.
132 46 Adding the tens: 4 (I and 3 car-
34.99 ried), 5, 11, 13, 15, 20, 22, 31, 40, 43,
122 . 65 46, 48, but subtract 20 as only two
580.01 complements were used, write 8 and
carry 2, The complement 10 may be
14 FUNDAMENTAL PROCESSES AND SHORT METHODS
added each time there is no item, making the answer 68, then subtract 40, leaving
28 as before. Remember, subtract as many 10's as there are complements added.
Finally the hundreds' column. There are but five items in this column;
therefore, with the 2 carried, proceed as follows: 3, 4, 13, 15, subtract 10 (only one
complement was added) and write 5. Answer: 580.01.
Problems
The items to be subtracted are marked (*) in Problems 1 and 2.
1. $58 10
19 66
45 55
77 28
9 01*
16.11
14 12*
2. $122 65
175 50
89 88*
17 20
1 48
8 62*
36 95
3. $48.75 Gain 4. $20 25 Gain
3 1.25 Gain
3 . 20 Loss
65.50 Gain
15. 25 Loss
16. 38 Gain
26 65 Gain
4. 50 Loss
41 50 Gain
28 45 Gain
38 47 Gain
12 34 Loss
49 82 Gain
Subtracting on an adding machine. If increase or decrease
columns^ are being verified on an adding machine that does not have
a direct subtraction device, add the complements of the numbers
to be subtracted.
To subtract $219.48, set 780.52 on the keyboard and strike all
nines to the left of the number; and to subtract $102.79, set 897.21
and strike all nines to the left of the number. Striking of the
nines eliminates from the totalizers the number 1 that would
otherwise be included in the answer.
Practical problems. In the following problems, both addition
and subtraction have to be performed in order to complete the
records.
Problem 1
This problem illustrates a section of a twelve-month moving-average schedule
used in cost accounting and other cumulative work. Assuming that twelve
months covers a cycle of business changes due to seasonal variations, and so
forth, the moving twelve months' total provides a fairly reliable amount for
comparative purposes.
The earliest month's results are subtracted from the twelve months' total
and the current month's results are added, making a current twelve-month
accumulation. The record is self-proving.
Total, 12/31/43.. J
Dept. 1
^125,275.93
' Dept. 2
$56,472.29 !
Dept. 3
$4,207.23 !
Dept. 4 Total
$7,200.49
Deduct Jan., 1943
9,495.79
4,907.63
368.80
502.50
Add Jan., 1944...
8,805.67
4,480.25
358.79
588.79
12 mos. totals
Deduct Feb., 1943
8,933.07
4,093.19
293.67
496.68
Add Feb., 1944...
9,033.48
4,123.97
235.80
517.90 ....
12 mos. totals
Deduct Mar., 1943
10,854.92
4,837.07
331.04
480.09
Add Mar., 1944..
8,588.37
4,001.18
334.17
521.72
12 mos. totals. . .
FUNDAMENTAL PROCESSES AND SHORT METHODS 15
Problem 2
From the following sales record, find the increase or decrease in sales by
departments.
COMPARATIVE SALES RECORD
Dept. No.
1
February, 2nd Year
$ 7,134.95
February, 1st Year
$ 6,834.79
2
6,225 19
5,764 87
3
7,934 97
8,375 16
4
6,354 76
5,986 35
5
3,695.15
3,756 89
6
. . 9,767 98
9,475 18
7
8,567 39
8,467 . 57
8
5,607 18
4,865 84
9
11,365 39
10,785 65
14,572 86
13,764 16
Increase or
Decrease f
Total .
Problem 3
A daily business record may be prepared from cash register totals and other
information. With the aid of the amounts given, complete the record for the
day. Some of the sections contain items that are needed to complete other
sections.
Cash Receipts
Rec'd. on Acc't. $234 56
Other Receipts . 59 32
Cash Sales 497.85
Total Receipts
Cash on Hand
Opening Balance $250 . 75
Receipts
Total
Paid Out
Closing balance
Accounts Payable
Bal. for'd $315.20
Invoices Today. 262.35
Total
Paid Today.... 136.57
Balance
Sales
Cash Sales . . . $
Credit Sales.. 152.35
Total Sales
Bank Account
Bal. for'd.... $2,872.63
Today's Dep __
Total _
Today's Cks.. 175.32
Balance
Cash Sales Summary
Total for'd... $2,542.75
Today's Cash
Sales „
Total to for-
ward „
Problem 4
Cash Paid Out
For Stock $ 85 42
For Expenses . . 19 56
Personal 27.50
Deposit 652.80
Total
Accounts Receivable
Hal. for'd $481.52
Credit Sales
Total
Rec'd. on Acct
Balance
Credit Sales Summary
Total for'd $638.47
Today's Credit
Sales
Total to forward
In the following table of Gross Profits by Departments, add the Goods on
Hand, March 1, 1st Year, to the Purchases for the Year, and from this sum
subtract the Goods on Hand, March 1, 2nd Year. This gives the Cost of Goods
Sold. The operation should be performed without transferring any of the
figures. Use the complements of the numbers in the column Goods on Hand,
March 1, 2nd Year.
16 FUNDAMENTAL PROCESSES AND SHORT METHODS
The difference between the Cost of Goods Sold and the Sales will give the
Profit or Loss.
To verify the work, add all the columns, and deal with the totals in the same
way as with the figures for the departments. The difference between the total
Cost of Goods Sold column and the Sales column should equal the difference
between the Profit and the Loss columns, showing the Net Profit of the ten
departments for the year.
GROSS PROFITS BY DEPARTMENTS
Goods Goods
On Hand Purchases On Hand Cost of
March 1 , for the March 1 , Goods
Dept. 1st Year Year 2nd Year Sold Sales Profit Loss
\ $3,475 86 $ 9,846 37 $2,347 11 $12,678 92 .
2 1,357.10 6,72540 1,47586.. 6,18890
3 3,276 84 10,326 85 3,827 84 8,297 63 .. .
4 5,475.90 11,176 98 5,874 13 13,586 47 . ..
5 4,276 83 9,798 34 4,207 16 10,508 92 . ..
6 3,785 47 8,376.41 3,648 10 8,756 13 ...... .......
7 2,98617 9,38657 3,01474 8,964 85 ... .. . __ .
8 3,275.83 8,724 18 2,817 56 9,575 34 . .. ...
9 2,976 95 9,543 34 2,734 15 10,789 18
10 3,532 25 10,217 60 3,375 89 .................. 12,756 84 ...... .. .__„_.
Footings -. 1LI1 _^ -.--
Multiplication. Multiplication is a short process of addition;
that is, a number is to be taken as an addend a given number of
times.
How many bushels of grain are in three bins each containing
146 bu.?
A ddition Multiplication
146 146
146 __3
146 438
438
Multiplication involves three numbers, the multiplicand (the
number to be repeated, 146) ; the multiplier (the number showing
the number of repetitions, 3) ; and the product (the number show-
ing the result, 438).
The multiplicand and the product are always like numbers.
146 bushels multiplied by 3 equals 438 bushels.
Problems
1. What is the cost of 640 acres of land at $42.50 an acre?
2. How many minutes are there in an ordinary year?
3. A barrel of flour contains 196 pounds. What is the weight of flour produced
in one day by a mill that produces 375 barrels?
FUNDAMENTAL PROCESSES AND SHORT METHODS 17
4. Sound travels about 1,120 feet in a second. How far will it travel in
15 seconds?
5. How many peaches are in 12 crates, if there are 84 peaches in each crate?
Accuracy and speed in multiplication depend largely upon a
thorough mastery of the multiplication tables. Tables previously
learned should be reviewed. Continue with frequent drills 011
combinations up to 25 X 25. The following table of multiples
from 12 X 12 to 25 X 25 is given for reference and drill. Tables
of multiples prepared in this manner facilitate the work of pay roll
extension, inventory extension, billing, and so forth.
TABLE OF MULTIPLES
J4^ 13
14
16
16
17
18
19
20
21
22
23
24
/-2&
12(144^)156
168
180
192
204
216
228
240
252
264
276
288
(300
13 TSG 169
182
195
208
221
234
247
260
273
286
299
312
325
14 168 182
196
210
224
238
252
266
280
294
308
322
336
350
15 180 195
210
225
240
255
270
;2S5
300
315
330
345
360
375
16 192 208
224
240
256
272
288
304
320
336
352
368
384
400
17 204 221
238
255
272
289
31)6
323
340
357
374
391
408
425
18 216 234
252
270
288
306 i
(323f
>342
360
378
396
414
432
450
19 228 247
266
285
304
323
342"
361
380
399
418
437
456
475
20 240 260
280
300
320
340
360
380
400
420
440
460
480
500
21 252 273
294
315
336
357
378
399
420
441
462
483
504
525
22 264 286
308
330
352
374
396
418
440
462
484
506
528
550
23 276 299
322
345
368
391
414
437
460
483
506
529
552
575
24 288 312
336
360
384
408
432
456
480
504
528
552
576
600.
25^300 )*25
350
375
400
425
450
475
500
525
550
575
600
625
Contractions in multiplication. Contractions in multiplication
may often be made by observing the peculiarities of the multiplier
and the multiplicand and calling into use factors, multiples,
complements, supplements, reciprocals, aliquots, and the like.
To multiply by factors of the multiplier. The ordinary method
and the shorter method of multiplying by factors are shown in the
following example. Observe that in the ordinary method there
are two multiplications and an addition, while in the shorter
method there are only two multiplications.
Multiply 567 by 27.
Ordinary Method
567
27
3969
1134
15309
Example
Solution
Shorter Method
567 27 = 9 X 3
9
5103
3
15309
18 FUNDAMENTAL PROCESSES AND SHORT METHODS
Problems
Multiply:
1. 4,584 by 64. 3. 1,459 by 35. 5. 8,756 by 42.
2. 8,359 by 54. 4. 2,684 by 27. 6. 6,123 by 45.
To multiply when a part of the multiplier is a factor or multiple
of another part.
Example
Multiply 34,768 by 4H8.
Solution
34768
__488
27HT44 product by 8
16688640 product of 60 times product by 8
16966784
Problems
Multiply:
1. 45,692 by 549. 3. 21,347 by 497. 6. 84,123 by 248.
2. 49,871 by 648. 4. 33,546 by 355. 6. 13,456 by 153.
To multiply a number of two figures by 11. Observation of the
ordinary method shows that, in the answer, the sum of the two
digits is written between the two digits.
Ordinary Method Shorter Method
54 54
11 Jl
54 594
54
594
When the sum of the two digits is 10 or more, 1 must be carried
to the digit at the left; for example, 64 X 11 = 704, and 93 X 11
- 1,023.
To multiply any number by 11. Observation of the ordinary
method shows that, in the answer, the units' digit of the multipli-
cand is the units' digit of the product; that the tens' digit of the
product is the sum of the units' digit and the tens' digit of the
multiplicand; that the hundreds' digit of the product is the sum of
the tens' digit and the hundreds' digit of the multiplicand; and so
on. When the sum of two digits is 10 or more, 1 must be carried.
Ordinary Method Shorter Method
8937 8937
11 _11
8937 98307
8937
98307
FUNDAMENTAL PROCESSES AND SHORT METHODS 19
Multiplying by 26. Annex two ciphers to the multiplicand,
and divide by 4.
Example
Multiply 7,562 by 25.
Solution
4)756200
"189050
Problems
Multiply each of the following by 25:
1. 3,874,; 6 2. 3,948. 3. 7,981. 4. 5,426.
Multiplying by 15. Annex a cipher to the multiplicand, and
increase the result by one-half of the multiplicand.
Example
Multiply 8,435 by 15.
Solution
84350
42175
126525
Problems
Multiply each of the following by 15:
1. 7,432. 2. 8,397. 3. 3,926. 4. 9,536.
Multiplying numbers ending with ciphers. Multiply the sig-
nificant figures in each number, and to the product annex as many
ciphers as there are final ciphers in both the multiplier and the
multiplicand.
Example
Multiply 756,000 by 4,200.
Solution
756
42
31752cvoOo ( '
Annex five ciphers. Answer: 3,175,200,000.
Problems
Multiply:
1. 325,000 by 2,300. 3. 24,100 by 4,200.
2. 370 by 480. 4, 8,300 by 2,100.
Multiplication by numbers near 100, as 98, 97, 96, and so forth,
and by numbers near 1,000, as 997, 996, and so forth. This
method is of value in finding the net proceeds of some amount less
2%, 3%, and so forth, and also in many other situations.
20 FUNDAMENTAL PROCESSES AND SHORT METHODS
Example
Multiply 3,247 by 97.
Solution
Multiply the number by 100, and subtract 3 times the number.
324,700 = 3,247 X 100
9,741 = 3,247 X 3
314,959 = 3,247 X 97
Multiplication by a number near 1,000 is accomplished in the
name manner by multiplying by 1,000 instead of by 100.
Problems
Multiply:
1. 2,450 by 98. 2. 7,318 by 97. 3. 5,438 by 90. 4. 8,752 by 95.
Multiplication of two numbers each near 100, 1,000, and so
forth. Products of numbers in this class may be calculated
mentally.
Example
Multiply 90 by 98.
Explanation. Stop 1. Multiply the complements of the two numbers, and
if the product occupies units' place only, prefix a cipher.
Result, 08. Solution
Step 2. Subtract the complement of one number Complement
from the other number, and write the result at the left CH\ 4
of the result in Step 1. The complement of either mini- ()s 2
her subtracted from the other number leaves the same ()4()s
remainder; as, 96 — 2 or 98 — 4 each equals 94. Answer:
9,408.
Example
Multiply 92 by 88.
Solution
(Complement
92 8
88 1 2
8096
Explanation. The product of the complements is 90, the last two figures of
the answer. 88 — 8 or 92 — 12 = 80, the lirst two figures of the answer.
Answer: 8,090.
Example
Multiply 996 by 988.
Solution
Complement
996 4
988 12
984,048
FUNDAMENTAL PROCESSES AND SHORT METHODS 21
Explanation. When numbers near 1,000 are multiplied, ciphers are prefixed
to the product of the complements, so that the product occupies three places.
Problems
Multiply:
1. 97 by 96. 2. 88 by 98. 3. 995 by 992. 4. 997 by 994.
Multiplying by numbers a little larger than 100, as 101, 102,
and so forth. Annex two ciphers to the multiplicand, and to this
add the product of the multiplicand and the units' figure of the
multiplier. Annex three ciphers for multipliers over 1,000.
Example
Multiply 3,475 by 104.
Solution
347500
13900 (4 X 3,475)
361400
Problems
Multiply :
1. 2,875 by 102. 2. 3,490 by 105. 3. 2,972 by 1,004. 4. 4,508 by 1,006.
Multiplication of two numbers each a little more than 100. To
the sum of the numbers (omitting one digit in the hundreds'
column), annex two ciphers, and add the product of the supple-
ments (excess over 100).
Example
Multiply 112 by 113.
Solution
112
113
12500 (sum of numbers, with one digit in the hundreds* column omitted)
156 (product of supplements, 12 X 13)
1265(5
Explanation. In instances similar to the foregoing, a knowledge of the
multiplication tables to 20 X 20 makes mental results possible, and is invaluable
in inventory and other extensions.
Problems
Multiply:
1.114 by 112. 2. 106 by 108. 3. 116 by 111. 4. 118 by 115.
Cross multiplication. When the multiplicand and the multi-
plier are each numbers of two figures, the work may easily be kept
in mind and the partial products added without being written
down.
22 FUNDAMENTAL PROCESSES AND SHORT METHODS
Example
Multiply 47 by 38.
Solution Graphic Solution
47
38 3;
1786
* o i -
A *«
Explanation. 8X7 = 56. Write 6, carry 5. (8 X 4) + (3 X 7) + 5 = 58,
Write 8, carry 5. (3 X 4) -f 5 - 17. Write 17. Answer: 1,7<S6.
Problems
Multiply:
1. 53 by 20. 2. 48 by 57. 3. 74 by 32. * 4. 65 by 28
To cross-multiply a number of three digits by a number of two
digits. A three-digit number may be multiplied by a two-digit
number in a manner similar to that of multiplying a two-digit
number by a two-digit number.
Example
Multiply 346 by 28.
Solution
346
28
9688
Explanation. 8X6 = 48. Write 8, carry 4. 4 (carried) + (8 X 4) -f
(6 X 2) = 48. Write 8, carry 4. 4 (carried) -f (8 X 3) 4- (4 X 2) = 36. Write
6, carry 3. 3 (carried) + (2 X 3) - 9. Write 9. Answer: 9,(>XX.
A graphic presentation of the steps required appears as follows:
42 3
34il>
2J
Problems
1. 324 X 28 4. 428 X 34 7. 2X9 X 85 10. 693 X 42
2. 543 X 42 5. 51(5 X 26 8. 356 X 48 11. 384 X 56
3. 658 X 56 6. 513 X 76 9. 7X5 X 34 12. 473 X 65
To cross-multiply a number of three digits by another number
of three digits. Comparison of the graphic presentation with that
above shows that the first three steps arc the same, the next three
are new, and the final three are the same.
Example
Multiply 428 by 356.
Solution Graphic Solution
,1 23 465 78
428 42* 4 M 4st/S \2 8 ±28
356 3 5 <} 3
152,368
23 465 78 9.
V 4sixS \Z 8 428
A, 3"^ >TM> $56
FUNDAMENTAL PROCESSES AND SHORT METHODS 23
Explanation. 6 X 8 = 48. Write 8, carry 4. 4 (carried) + (6 X 2) +
(8X5) = 56. Write 6, carry 5. 5 (carried) + (6 X 4) + (8 X 3) + (2 X 5)
= 63. Write 3, carry 6. 6 (carried) + (5 X 4) + (2 X 3) = 32. Write 2,
carry 3. 3 (carried) + (3 X 4) = 15. Write 15. Answer: 152,368.
Problems
1. 124 X 251 6. 832 X 425 11. 436 X 579
2. 262 X 158 7. 639 X 256 12. S32 X 656
3. 328 X 245 8. 819 X 325 13. 295 X 638
4. 638 X 256 9. 677 X 283 14. 767 X 842
6. 784 X 364 10. 518 X 824 15. 698 X 476
Preparation of a table of multiples of a number. It is not
uncommon to have to use the same number many times in making
calculations, especially in cost accounting. A saving of time and
increased accuracy in the work arc achieved if a table of multiples
of the number is constructed. Suppose that you have to perform
a number of multiplications in which 32(5,834 is one of the factors.
A table of multiples may be constructed with an adding machine
by locking the repeat key. Sub-total after each pull of the handle.
The sub-totals should check with the product column shown below.
If the table is prepared by repeated additions, and not with an
adding machine, the 10th product should be computed, as it will
verify all, unless there are compensating errors in the work.
TABLE OF MULTIPLES
Multiplier Product
1 326,834
2 (326,834 + 326,834) . 653,668
3 (653,608 -f 326,834) 980,502
4 (980,502 -f 326,834) . ... 1 ,307,336
5 (1,307,336 + 326,834) . . 1 ,634,170
6 (1,634,170 + 326,834) . 1,961,004
7 (1,96 1,004 + 326,834) . . 2,2X7,838
8 (2,2X7,838 + 326,834) 2,614,672
9 (2,614,672 + 326,834) 2,94 1 ,506
Verification
10(2,941,506 + 326,834) 3,268,340
Example
Multiply 326,834 by 5,249.
Solution
2941506 = 9 times 326,834
1307336 = 4 times 326,834
653668 = 2 times 326,834
1634170 = 5 times 326,834
1715551666 - product
If the table is prepared without toe use of an adding machine, proceed as
outlined on the next page.
24 FUNDAMENTAL PROCESSES AND SHORT METHODS
1. Write 326,834 near the bottom of a slip of paper or a card.
2. Start the table by writing 326,834. Place the slip or card just above thig
number, thus:
r
326,834
1. 326,834
2
3 ...
3. Add the two numbers, placing the sum, 653,668, on line 2. This is
two times the number.
4. Move the slip or card down one line and add again, placing the sum,
980,502, on line 3, forming three times the number.
5. Continue moving the slip or card down one line each time and adding.
6. When 9 times the number is obtained, check the accuracy of the work by
repeating the process once more. The result should be ten times the number.
Problems
Set up a table of multiples of 245,386, and multiply 245,386 by the following
numbers:
1. 2,465 2. 3,542 3. 2,498 4. 5,347 6. 6,173
Division. Division is the process of finding how many times
one number is contained in another number. The dividend is the
number to be divided, the divisor is the number by which we divide,
and the quotient is the number showing how many times the divi-
dend contains the divisor.
The remainder is a number less than the divisor, and results
when the dividend does not contain the divisor exactly. It is an
undivided portion of the dividend.
Short division is the method used when the products of the
divisor and the digits of the quotient are omitted.
Example Solution
Divide 3,476 by 2. 2)3476
1738
Long division is the method used when the work is written in full.
Example Solution
Divide 5,839 by 24. ?4)5839(243
48
103
96
79
72
7
To divide by 25, 60, or 125. The work of division can be
lessened by making the operation one of multiplication.
FUNDAMENTAL PROCESSES AND SHORT METHODS 25
Example Solution
Divide 1,400 by 25. 14 X 4 = 56.
Explanation. Divide 1,400 by 100 by dropping the zeros. But, 100 ia
4 times the actual divisor, therefore the quotient 14 is £ of the actual quotient,
so 14 X 4 or 56 is the actual quotient.
In a similar manner, 1,400 divided by 50 is 2S; and 14,000 divided by 125
is 112. (Note: Further reference to this method is given under the subject of
division by aliquot parts of 100.)
Abbreviated division. Instead of writing the product and
then subtracting, the product of each digit of the divisor is sub-
tracted mentally, using the " making change" method, and only
the remainder is written.
3285
2347708756
667
1995
1236
66
Use of tables in division. If a number of divisions are to be
made witli the same divisor, it is advantageous to set up a table of
multiples of the divisor.
Example
Assume that 328 is to be used a number of times as a divisor, and that one
of the dividends is 587,954, a table of multiples could be set up thus:
TABLE OF MULTIPLES
Multiplier Product
\ . . .... 328
2 . 656
3 . . 984
4 ..... 1,312
5 ... ... ... 1,640
6. ... .. . .. 1,968
7 . 2,296
8 .. . .. 2,624
9 2,952
Explanation. Inspection shows the first digit in the quotient to be 1 . The
second partial dividend is 2,599. The table of
multiples shows the largest product contained Solution
therein to be 2,296, opposite 7. The third 328)5S7954(1792fft
partial dividend is 3,035, and the table of multi- 328
pies shows the largest product contained therein 2599
to be 2,952, opposite 9. The fourth partial 2296
dividend is 834, and the largest product con-
tained therein is 656, opposite 2. The remainder
is 178. The fraction ttf may be reduced to
, or it may be changed to a decimal. 8<*4 178 __ 89
656 328 164
178
26 FUNDAMENTAL PROCESSES AND SHORT METHODS
Division in this manner is rapid, as no time is lost through
selection of a quotient so large that when the product is found it
exceeds the dividend, necessitating another trial.
Problems
Divide the following numbers by 144 after setting up a table of multiples
of 144:
1. 374,825. 2. 628,256. 3. 496,287.
Reciprocals in division. The reciprocal of any number is
found by dividing 1 by the number. The reciprocal of 5 is 1 -r- 5,
or .2, and the reciprocal of 25 is 1 -f- 25, or .04.
The quotient in a division may be found by multiplying the
dividend by the reciprocal of the divisor. Hence, in instances in
which it is necessary to find what per cent each item is of the total
of the items, the use of the reciprocal of the divisor will save time
and provide a check on these computations.
To find what per cent each item is of the total of the items:
(a) Divide 1 by the total of the items to obtain the reciprocal
of the total.
(6) Using the result obtained in (a) as a fixed multiplier, mul-
tiply each of the individual items, and the respective results
obtained will be the per cents which the individual items are of the
total sum.
Example
Find the per cent that each department's monthly expense is of the total
monthly expense.
Department Expense
A $ 600 00
B 500 00
C 1,200 00
D 700 00
E . 1,000 00
Total .. $4,000 00
Solution
Divide 1 by 4,000 to obtain the reciprocal, .00025. Multiply the expense
of each department by this reciprocal, and the product will be the per cent that
the department's expense is of the total expense.
Department Expense Reciprocal Per Cent
A $ 600 00 X .00025 = 15 %
Crf
/O
B 500.00 X .00025 = 12i
C 1,20000 X 00025 = 30%
D 70000 X .00025 = 17i%
E 1,000 00 X .00025 = 25 %
Total $4,000 00 100%
FUNDAMENTAL PROCESSES AND SHORT METHODS 27
The foregoing method of calculating the rate per cent lias a
great many applications in an accountant's work. Another illus-
tration is given — that of calculating the per cent that each item in
a profit and loss statement is of net sales.
QUALITY MEAT MARKET
PROFIT AND Loss STATEMENT FOK THE YEAR
Detail Amount Per Cent
Net sales $20,000 00 100 00
Cost of merchandise sold 15,712.00 78.56
Gross profit $4,288 .00 ~~2~l~. 44
Expenses N
Salaries and wages $2,266 00 11 .33
Advertising 22 00 .11
Wrappings 172 00 .86
Refrigeration 210 00 1 .05
Heat, light, and power 54 00 .27
Telephone 54 00 .27
Rent 33S 00 1.69
Interest 146 00 .73
Depreciation of store equipment 152 00 .76
Repairs to store equipment 44 00 .22
Insurance 10 00 .05
Taxes 42 00 .21
Losses from had debts 38 00 .19
Other expenses 284 00 1 42
Total expenses _3»832 °° 19I1(?
Net profit $ _456 00 ~ 2J28
Explanation. The foregoing is a simple statement, and the per cents can
be determined mentally if each item is divided by the amount of net sales. For
the purpose of illustration, however, find the reciprocal of $20,000.00, which is
.00005 (1 -T- 20,000); then multiply each item by this reciprocal, and the results
will be as shown in the per cent column.
Problems
1. The floor space occupied by Z Manufacturing Company was as follows:
Service Department X 600 sq. ft.
Service Department Y . . . 1,100 sq. ft.
Service Department Z . . . . . 550 sq. ft.
Producing Department A . . 2,000 sq. ft.
Producing Department B 1 ,568 sq. ft.
Producing Department C . 2,234 sq. ft.
Sales Department . 600 sq. ft.
Administrative Offices .... . 550 sq. ft.
97262 sq. ft.
The Building and Maintenance Expense account shows a total of $2,982.50.
What amount of this expense should be distributed to each of the departments?
18 FUNDAMENTAL PROCESSES AND SHORT METHODS
2. In the following tabulation, find the per cent that each department's floor
space is of the total floor space:
Sq. Ft. Per Cent
Floor Space of Total
Dept. 1 2,456
Dept. 2 1,014
Dept. 3 875
Dept. 4 1,252
Dept. 5 74S _._...
0,345 JOOJXJ
3. Calculate the per cent that each item is of net sales.
THE FOOD MART
PROFIT AND Loss STATEMENT
Net Sales $35,00010000%
Cost of Merchandise Sold 27,909 .
Gross Profit $7,031 .... ....
Expenses
Salaries and Wages $4,080 .... ....
Advertising 28
Wrappings .... 200 .
Refrigeration . . 30S
Heat, Light, and Power 100
Telephone SI
Rent . 410
Interest 203
Depreciation of Store Equipment 147
Repairs to Store Equipment . . 45 .... ...
Insurance . . 21
Taxes 39
Losses from Bad Debts 119
Other Expenses 490
Total Expenses . 0,349 " T7
Net Profit $T,2S2 ~ ...
CHAPTER^}
Checking Computations
Methods. Addition may be checked by adding the second
time, adding from the bottom to the top if the first addition was
from the top to the bottom. This is preferable to performing the
work in the same way the second time, as a mistake once made is
likely to be repeated.
Subtraction may be checked by adding the subtrahend and the
remainder. The sum should equal the minuend.
Multiplication may be checked by interchanging the multiplier
and the multiplicand and inuliiph inn again.
Division may be checked by multiplying the divisor and the
quotient, adding to this product any remainder. The answer
should equal the dividend.
Rough check. Rough check is an approximate check and is
often used to locate large errors. It is also used in determining
approximate results. It is especially useful in checking misplace-
ment of the decimal point in multiplication and division of decimal
fractions.
A rough check of addition may be made as follows:
Example Check
54,S92 55
36,071 36
53,784 54
21,342 21
_76,854 J77
242,943 243
If the required result is thousands, disregard the three columns
at the right, except to increase the fourth column sum by one if the
digit in the third column is 5 or more. The check shows the
answer to be approximately 243,000.
Absolute check. There is no such thing as an absolute check
because there are always possibilities of offsetting errors, but the
use of several methods of checking computations makes the prob-
ability of error so slight that one may rely on the result as correct.
Check numbers obtained by casting out the nines. A simple
and easily remembered check is that of casting out the nines. Add
29
30
CHECKING COMPUTATIONS
the digits of the number, divide the sum by nine, and use the
remainder, which is called "the excess/' as the check number. In
the number 4,875, the sum of the digits is 24, and 24 divided by 9
equals 2 with an excess of 6.
Verification of addition.
Explanation. The sum of the digits of 8,342 is 17 (S + 3 -f 4 + 2).
out 9 and set down 8. If a number contains a 9, skip
it in adding the digits; thus, in 8,907, 8 + 6 + 7
equals 21 . Cast out the nines and set down the excess,
3. Find the check number of each line in the same
way. Add the check numbers, and cast the nines
out of their sum. Find the check number of the sum
of the column being verified. The final check number
in each case is 5.
Problems
Add, and verify by casting out the nines:
1.
2487
3156
29S2
4750
8928
2.
7452
8129
5758
2253
70S5
3.
4501
2765
4567
8250
2435
Example
8342
8967
8378
9276
8431
43394—5
Cast
8
3
8
6
_7
32—5
4.
1231
4567
1085
3426
7531
Verification of subtraction.
Example
7856 8
213S 5
5718 3
Explanation. 7,856 checks 8, and 2,138 checks 5. 8-5 = 3, and 5,718
i' hecks 3.
Problems
Subtract, and verify by casting out the nines:
1.
7496
2831
2.
7428
1956
3.
4751
3286
4.
8237
5129
Verification of multiplication.
Example
482 5
376 J7
181232—8 35—8
Explanation. 482 checks 5, and 376 checks 7.
and the product, 181,232, also checks 8.
7 X 5 - 35. 35 checks 8,
CHECKING COMPUTATIONS 31
Problems
Multiply, and verify by casting out the nines:
1. 2. 3. 4.
456 412 832 765
287 654 254 414
Verification of division. Division may be verified by multi-
plication; that is, the product of the quotient and the divisor
should equal the dividend. Apply the same principle in verifying
with check numbers.
Example
13)76492(5884
65 Explanation. 76,492
77^ checks 1. 13 checks 4.
104 5,884 checks 7. 4X7 =
— 28, and 28 checks 1,
J/*J which is also the check
_ number of the dividend.
52
52
Problems
Divide, and verify by casting out the nines:
1. 11,550 by 42. 2. 60,882 by 73. 3. 11,049 by 127. 4. 9,854 by 26.
Verification of division where there is a remainder. The check
number of the remainder added to the product of the check number
of the quotient and the check number of the divisor should equal
the check number of the dividend.
Example Explanation. Step 1 : The
32)75892(2371 remainder, 20, checks 2. The
Q4 quotient, 2,371, checks 4.
—- The divisor, 32, checks 5. 2 +
1 *® (4X5) = 22, and 22 checks 4.
_ Step 2: The dividend,
229 75,892, checks 4.
224 step 1 and Step 2 should
52 produce the same check num-
32 ber.
Problems
Divide, and verify by casting out the nines:
1. 34,765 by 52. 2. 29,878 by 87. 3. 95,763 by 26. 4. 8,476 by 4J
Check numbers obtained by casting out the elevens. Because
casting out nines does not reveal errors in computations if two
32 CHECKING COMPUTATIONS
digits have been transposed, some persons prefer to use eleven as a
check number.
Begin with the left-hand digit of the first number, and subtract
it from the digit to its immediate right. If the digit to the right
is smaller, add eleven before subtracting. Using the remainder as
a new digit, subtract it from the third digit from the left, first
adding eleven if necessary. Use this remainder as a new digit,
and subtract it from the fourth digit from the left, first adding
eleven if necessary. Continue in this manner until all the digits
in the number have been used. The final remainder is the check
number of the number.
Another method of checking results by means of the number
eleven is to use alternate digits. From the sum of the first, third,
fifth, etc., digits (beginning at units' place) subtract the sum of the
second, fourth, sixth, etc., digits. If the subtraction cannot be
performed, eleven is first added to the sum of the odd digits, and
the sum of the even digits is subtracted, the remainder being the
check number.
Verification of addition.
Explanation. Begin at the left with the number 4,324. 4 from 14 (3 + 11)
= 10. 10 from 13 (2 + 11) = 3. 3 from 4 = 1, the Example
check number of 4,324.
Take the second number, 8,6cS9. 8 from 17 , {
(6 + 11) = 9. 9 from 19 (8 + 1 1) = 10. 10 from 20 8b89 10
(9 + 11) = 10, the check number of 8,689. 6327 2
Check all the numbers in the same manner. Add ^>A 7 n
the check numbers. The sum of the check numbers _
checks 1, and the sum of the numbers checks 1. 31791 — 1 23 — I
Problems
Add, and verify by casting out the elevens:
1. 2. 3. 4.
3789 2450 9755 8307
5462 1279 8256 7165
9581 2075 3851 2693
3998 2754 8632 2198
5314 9287 6311 5183
Verification of subtraction.
Example
7453 6
1289 2
6164 4
Explanation. 7,453 checks 6. 1,289 checks 2. 6-2 = 4 and 6,164
checks 4.
CHECKING COMPUTATIONS 33
Problems
Subtract, and verify by casting out the elevens:
1. 2. 3. 4.
8795 3465 7985 3079
1560 2134 5698 1002
Verification of multiplication.
Example
584 1
256 3
149504 3
Explanation. 584 checks 1 . 256 checks 3. 3X1=3, and 149,504 checks 3
Problems
Multiply, and verify by casting out the elevens:
1. 2. 3. 4.
346 4289 7437 287
275 324 2856 J6
Verification of division.
Example 1 Example 2
24)89784(374JI 31)75893(2448
72 62
177 138
168 m
98 149
96 124
24 253
24 248
~5
Explanation 1. 89,784 checks 2. 24 checks 2. 3,741 checks 1. 2X1=2,
the check number of the dividend.
Explanation 2. 75,893 checks 4. 31 checks 9. 2,448 checks 6. The
remainder checks 5. 5 + (9 X 6) = 59. 59 checks 4, the same check number
as that of the dividend.
Problems
Divide, and verify by casting out the elevens:
1. 80,925 by 83. 2. 124,392 by 142. 3. 25,874 by 49. 4. 28,769 by 135.
Check number thirteen. If thirteen is used as a check number,
transpositions and shif tings of figures are readily detected. How-
ever, in checking by 13, it is necessary actually to divide by 13.
34 CHECKING COMPUTATIONS
TABLE OF MULTIPLES
1
13
6
78
2
26
7
91
3 ...
. 39
8
104
4
... 52
9
117
5
65
10
130
All the dividing is done mentally.
Example
Cast out 13 from 247,563.
Explanation. Begin with the two left-hand digits. 24 checks 11. 11, with
the next digit, 7, is 1 1 7, and 1 1 7 checks 0. Use the next two digits. 56 checks 4.
4 with the next digit is 43, and 43 checks 4.
The verification of addition, subtraction, multiplication, and
division is performed in the same manner as with 9 and 11. The
difference is in the method of arriving at the check number, as has
been outlined.
Problems
1. Add, and verify by check number 13:
24875
32986
79840
80475
13048
93476
2. Subtract, and verify by check number 13:
84756
21348
3. Multiply, and verify by check number 13:
4875
259
4. Divide, and verify by check number 13:
975,648
348
CHAPTER^)
Factors and Multiples
Factors. The factors of a number are the integers whose prod-
uct is the number. Thus, the factors of 6 are 2 and 3, the factors
of 18 are 3 and 6, or 2 and 9. A prime factor is a prime number,
that is, a number not exactly divisible by any number except
itself arid 1.
Factoring is the process of separating a number into its factors.
Example Solution
What are the prime factors of 315? 3)315 The prime factors of
3)105 315 are, therefore,
5) 35 3X3X5X7.
7
Example Solution
What are the factors of 315? 9)315 The factors of 315 are,
7) 35 therefore, 9X7X5.
5
Factoring is important for its assistance in the solution of
problems in fractions, practical measurements, percentage, and all
problems in which cancellation is used. One use of factors was
given on page 17, "to multiply by factors of the multiplier," and
another on page 18, "to multiply when a part of the multiplier is
a factor or multiple of another part."
Tests of divisibility. To be able to factor a number quickly,
one must become thoroughly familiar with the tests of divisibility.
A number is divisible by :
1. Two, if is is an even number or if it ends in zero.
2. Three, if the sum of its digits is divisible by 3. Thus, 41754
is divisible by 3 because the sum of the digits is 21, and 21 is
divisible by 3.
3. Four, if the two right-hand figures are zeros, or if they
express a number divisible by 4. Thus, 13724 is divisible by 4
because 24 is divisible by 4.
4. Five, if the units' figure is either a zero or a 5.
5. Six, if it is an even number the sum of whose digits is divisible
by 3. Thus, 846, 918, and 54252 are divisible by 6,
35
36 FACTORS AND MULTIPLES
6. Eight, if the three right-hand digits are zeros, or if they
express a number divisible by 8. Thus, 2000 and 5624 are divisi-
ble by 8.
7. Nine, if the sum of its digits is divisible by 9.
8. Ten, if the right-hand figure is zero.
(There is no simple method of testing divisibility by 7.)
Greatest common divisor. A common divisor of two or more
numbers is a number that evenly divides each of them. Thus, a
common divisor of 16 and 24 is 4.
The greatest common divisor of two or more numbers is the
greatest number that will evenly divide each of them. It is the
product of all their common factors.
Example Solution
Find the greatest common divisor 3)36 63 54
of 36, 63, and 54. 3^2 ~2l~~18
~T~T~6
Since 4, 7, and 6 have no common factors, the G. C. D. is 3 X 3 = 9.
A practical application of the principles involved in finding
the G. C. D. is in reducing common fractions to their lowest terms.
Problems
Find the G. C. D. of the following:
1. 64, 160, 320, 640 3. 32, 48, 12<S
2. 36, 54, 90 4. 81, 729, 2187
5. X, K, and Z own land on a new street. X has 600 feet frontage, Y has
720 feet, and Z has 900 feet. If they wish to cut this land into lots of equal
width, how wide will the lots be, and how many will each have?
6. If you have three coils of steel cable measuring, respectively, 2205, 2940,
and 4704 feet, and wish to cut the whole quantity into pieces of the greatest equal
length possible without waste or splices, what will be the length of each piece?
How many lengths will be cut from each coil?
Least common multiple. A common multiple of two or more
numbers is a number that is evenly divisible by each of them.
Thus, 24 is a common multiple of 3 and 8.
The least common multiple of two or more numbers is the
least number that is evenly divisible by each of them. Thus, 12
is the L. C. M. of 4 and 6.
Example
What is the L. C. M. of 12, 28, 30, 42, and 64?
FACTORS AND MULTIPLES 37
Solution
2)12 28 30 42 64
2) 6 14 15 21 32
3)" 3 7~15 21 16
7) 1 7 5 7 16
f 1 5 1 16
2X2X3X7X5X16 = 6,720
Explanation. Notice that any number not divisible by the factor is brought
down, and the process is repeated as long as at least two of the numbers have a
common factor. Finally, the L. C. M. is the product of the factors and the
numbers having no common factor.
Problems
Find the L. 0. M. of the following:
1. 6, IS, 30, 42 3. 45, 63, 72, 99
2. 16, 24, 64, 96 4. 14, 35, 42, 28
Cancellation. Certain compulations involving division can be
shortened by removing or cancelling equal factors from both divi-
dend and divisor.
Example
If 32 units of product sell for $57.60, what will 18 units of the same product
sell for at the same rate?
Solution
3.60
9 UM
? = 32.40
w
Problems
Using cancellation, divide:
1. 27 X 48 X 96 X 38 2. 8 X 12 X 15 X 6
19 X 16 X 9 X 2 5 X 4 X 3 X 18
3. If 15 tons of coal cost $258.00, how much will 25 tons cost at the same
rate?
4. A ship's provisions will last 36 men for 216 days. How long will they
last 124 men?
CHAPTER!^
Common Fractions
Terms explained. A unit is a single quantity by which another
quantity of the same kind is measured: 1 foot is the unit of 5 feet;
1 barrel is the unit of 10 barrels; 1 acre is the unit of 40 acres, and
so forth.
These integral units are often divided into equal parts known as
fractional units, as i ft., ^ bbL, ^ A., and so forth.
A fraction is an expression for one or more of the equal parts of
a unit, as •£ ft., f ft., § bbl., % A., and so forth.
The number above the line in the expression of a fraction is
called the numerator', the number below the line is called the
denominator.
The denominator indicates the number (and hence the size) of
parts into which the unit is divided.
The numerator indicates the number of these parts taken.
A proper fraction expresses less than a unit, or its numerator is
less than its denominator; as, $, f, $, and so forth.
An improper fraction is a fraction whose numerator is equal to
or greater than its denominator; as, f, f, f, and so forth.
A mixed number is a number expressed by a whole number and
a fraction; as, 2l, 3£, 16f, and so forth.
Reduction of fractions. Reduction is the process of changing
the numerator and the denominator of a fraction without changing
the value of the fraction.
A fraction is reduced to higher terms when the numerator and
the denominator are expressed in larger numbers.
A fraction is reduced to lower terms when the numerator and
the denominator are expressed in smaller numbers, and it is reduced
to its lowest terms when there is no common divisor of its numerator
and denominator.
Principle. Multiplying or dividing both numerator and
denominator of a fraction by the same number does not change
the value of the fraction. Thus, £-£ may be reduced to the equiva-
lent fraction % by dividing both terms by 4. The fraction iri has
been reduced to lower terms. Again, i~f may be reduced to the
equivalent fraction f by dividing both terms by 8. Here the
39
40 COMMON FRACTIONS
fraction if has been reduced to lowest terms, since 2 and 3 do not
have a common divisor.
Conversely, $ may be changed to an equivalent fraction whose
denominator is 24 by multiplying both terms by 8 (obtained by
dividing 24 by 3), or £f\ Thus, the fraction f has been reduced
to a higher given denominator.
Mixed numbers. It is sometimes desirable to change a mixed
number to an improper fraction, or, conversely, to change an
improper fraction to a mixed number.
To change a mixed number to an improper fraction. Multiply
the whole number by the denominator of the fraction, add the
numerator, and place the sum over the denominator, thus, 3^ is
Y, 4| is V, and 6* is V>.
To change an improper fraction to a whole or a mixed number,
divide the numerator by the denominator; thus V2 is 4, f is l£,
Vis 1| or U, and V-is4f.
Problems
1. Reduce to lowest terms: A, A, A, H, iff, U, *t, M, if, li
2. Change to equivalent fractions having denominators as indicated:
i to Hths i to 15ths t to 25ths
1 to Oths i to 24ths A to 4Sths
$ to 20ths | to 24ths | to 32nds
i to Sths | to 36ths -fy to 36ths.
3. Reduce to equivalent fractions whose denominators are 24: TV I, I, I,
it.
4. Change to improper fractions: 4t, 3j, li, 7i, 8|, 6i, 3|, 5|, 5f, 9|.
5. Change to whole or mixed numbers: V, ¥, V, ¥, H, V, f , V, ff, ¥•
6. Is the number of fractional units increased or decreased when we reduce
A to •}? Is the size of the fractional unit increased or decreased when we reduce
A to t?
Addition and subtraction of fractions. Similar fractions are
fractions that have a common denominator. Only similar frac-
tions can be added or subtracted.
To add fractions, reduce the fractions to similar fractions
having a common denominator and add the numerators.
To subtract fractions, reduce the fractions to similar fractions
having a common denominator and subtract the numerators.
Example Solution
Add: i, |, and
12
6
8
JJ
17
COMMON FRACTIONS 41
Explanation. Inspection shows that 12 is the least common denominator.
i is A. £ is A, and i is A- Adding the^umerators of the similar fractions
gives 17, and fj is 1 A- /,
Example <' - Solution
Subtract: ^ t = ft
I ~ A ( ft — A - A
Multiplication of fractions, (a) To multiply a fraction by a
whole number, multiply the numerator or divide the denominator
of the fraction by the whole number.
Example Solution
Multiply 6 X A- 6 X A = f 8 = 2i
or
12 -f- 6 = 2, and -J - 2i
(6) To multiply a whole number by a fraction, multiply the
whole number by the numerator of the fraction and write the
product over the denominator. Cancel when possible.
Example Solution
• > Find f of 35. I X 35 = Y - 14
or
' 7
2 X M = 14
» X 1
(c) To multiply a fraction by a fraction, multiply the numer-
ators to obtain the numerator of the answer, and multiply the
denominators to obtain the denominator of the answer. Cancel
when possible.
Example Solution
Find io,' It. £xi&=28 = &
or
5
__
3 x n 8
8
(d) To multiply a mixed number by a mixed number, reduce
each mixed number to an improper fraction and proceed as in (c).
Example Solution
Find the product of: i X V = W
3i X4i
Find the product of: 4
42
COMMON FRACTIONS
Problems
Find:
1. 9 X A 3. | of 35
2. 24 X 1 4. A of 16
Division of fractions, (a) To divide a fraction by a whole
number, divide the numerator or multiply the denominator by the
whole number.
6. | of H
6. £ of f f
7. 3i X 4*
8. 12f X 8 i
Example
Divide H by ,5.
Solution
25 -r- 5 = 5 Answer:
5
_
2X x
or
5
2S
(fr) To divide any quantity — a whole number, a mixed number,
or a fraction, by a fraction, invert the divisor and multiply.
Example Solution
Divide 8 by f . 4
Example
Divide 16| by £.
Example
Divide* by*.
1X2"
Solution
13 3
W X 0
TxT
2
Solution
30
_ 1
" K)2
Problems
Divide:
a. M by 3
b. -H »>y 9
c. 8 by |
d. 9 by ?
e. 16j by
f. 1SJ l)y
g. 3i by li
h. 9j by 3i
1. How many pieces of wire each Sj inches long can be cut from 40 feet
of wire?
2. If 1 of a ton of coal costs $12.75, what is the cost of one ton?
3. How many sash weights each weighing 2-J- pounds can be cast from 120
pounds of pig iron, if i of the quantity of pig iron is wasted in the casting oper-
ation?
4. A room is 18f feet long and 14^ feet wide. The width of the room is
what part of the length of the room?
5. A carpenter has a board that is 20 feet long, but it is -J- longer than he
needs. How long a board does he need?
6. What is the cost of 7£ tons of coal at $14| a ton?
COMMON FRACTIONS
43
7. A house and lot are valued at $6,600. If the lot is worth f as much as
the house, what is the value of each?
8. If a man can earn $2f a day, how long will it take him to earn $46-f ?
9. A table is 20 feet long. How many people can be seated on the two
sides if you allow if feet for each person?
10. Henry's time book shows that his working time for one week was as
follows: Monday, 7^ hours; Tuesday, 8-J hours; Wednesday, S hours; Thurdsay,
9^ hours; Friday, 8i hours; Saturday, Of hours.
He is paid straight time for 8 hours or less and time and a half for hours in
excess of S each day other than Saturday, when he receives double-time pay for
hours worked. How much did he earn at $| an hour?
11. The shipping clerk reported that he dispatched 320 packages averaging
2Sj pounds each. What was the total weight of packages dispatched?
12. A cubic foot of water weighs 62i pounds, and there are approximately
7^ gallons to the cubic foot. Estimate the weight of water that a 10-gallon keg
will contain.
To find the product of any two mixed numbers ending in £.
(a) Wlien the sum of the whole numbers is an even number. To
the product of the whole numbers, add one-half of their sum, and
annex i.
Multiply 24 £ by
192
16
208 1
Example
Solution
(8 X 24)
(i of the sum of 24 and 8)
(-J annexed)
Multiply:
1. Si by 4i
2. 12£ by 8i.
Problems
3. 28iby 12J.
4. H>i by 14J.
6. 18ihy 18i.
6. 10i by 34 i.
(b) When the sum of the whole numbers is an odd number. To
the product of the whole numbers, add one-half of their sum, less
1, and annex f.
Multiply 151 by 6l.
Example
Solution
90 (6 X 15)
_10_ (i of 15 + 6 - 1)
100| (I annexed)
44 COMMON FRACTIONS
Problems
Multiply:
1. 18i by 5i 3. 38£ by 5i 6. 23* by
2. l4 by 7£. 4. 13i by 8|. 6. 19£ by
To multiply a mixed number by a mixed number.
Example
Multiply 524i by 27f
14148 6 = common denominator of fractions
1741 4]
13^ 3 } = numerators of changed fractions
_ * lj
14336^ f = li
Explanation. Multiply 524 by 27, obtaining the first part of the answer,
14,148. Next, take i of 524, obtaining 174|. Then take £ of 27, obtaining 13^.
Finally, take -g- of ^, obtaining £. Add the four partial products, and the com-
plete product is 14,336^.
Problems
Multiply:
1. 247| by 39i 3. 59| by 15|. 6. 181f by 6f .
2. 849i by 28i 4. 176f by 34f . 6. 56£ by 12|.
Decimal fractions. A decimal fraction is a fraction whose
denominator is some power of ten, indicated by a decimal point
placed just to the right of the units' place. Thus, .1 is TO~, .05 is
Tthr, and .25 is tVV or i.
Addition and subtraction. To add or to subtract decimals,
write the numbers so that the decimal points fall vertically and
proceed as in whole numbers.
Example Solution
Add: .01, 4.72, 78.25, and .005. .01
4.72
78 25
.005
82.985
Example Solution
Subtract: 47.02
47.02 - .92 _ 92
46.10
Problems
1. Add: 25.679, .0356, 2.78, and .017.
2. Add: 136.2, 28.348, .004, and 1.356.
COMMON FRACTIONS 45
3. Subtract: 13.48 from 27.049.
4. Subtract: .003 from .47.
Multiplication. To multiply decimal fractions, multiply as in
whole numbers and point off as many decimal places in the product
as there are places in both multiplicand and multiplier.
Example Solution
Multiply 3.06 X .8. 3 06
.8
2.448
Explanation. Since there are 3 decimal places in both the multiplicand and
the multiplier, point off three decimal places in the product.
Example Solution
Multiply: 23.8564
23.8564 by 6.72 6 72
477128
1669948
1431384
160315008
Explanation. As there are 6 decimal places in the multiplicand and the
multiplier, point off six decimal places in the product. The answer is 160.315008.
Rough check: 24 X 7 = 168.
Division. Proceed as with whole numbers, annexing zeros to
the dividend if necessary. The number of decimal places in the
quotient must equal the number in the dividend minus the number
in the divisor.
Example Solution
Divide : . 24)54 . 864(228 . 6
54.864 by .24 6 8
2 06
144
0
Explanation. Divide by writing the remainders only. The quotient is 2286.
As there are three decimal places in the dividend and two decimal places in the
divisor, point off one decimal place in the quotient. The answer is, therefore.
228.6.
Example Solution
Divide: 49.099
256.7894 by 5.23 5 . 23)256.78940
47 58
5194
4870
163
46 COMMON FRACTIONS
Explanation. Predetermine the placing of the decimal. As there are two
decimals in the divisor, place the decimal point over the third decimal place in
the dividend. Place the first figure of the quotient over the last figure of the
partial dividend. One zero has been annexed to the dividend in order to obtain
a quotient to three decimals. Rough check: 49 X 5 = 245.
Problems
Multiply: Divide:
1. 34.278 X 1.45 6. 5.8769 by 1.34
2. 395.264 X .035 7. .0084 by 1.5
3. 74.26 by .00423 8. 45.87 by .0056
4. .056 by .083 9. 8.45 by 25.3
6. 18.42 X .045 10. 956 by 4.87
To abbreviate decimal multiplication when a given number of
decimal places is required. It is a waste of time to carry out
decimal multiplication to a denomination smaller than that in
which the data are expressed; often it is unnecessary to carry it
beyond the third or fourth decimal.
Example
Multiply 4.7892 by 3.1705, and obtain the answer correct to four decimal
places.
Solution
4
5
7892
6713
= multiplicand
= multiplier reversed
14~
^3676
= 4.7892 X 3.
,4789
2
=4. 7892 X .1
,3352
4
i
= 4. 7802 X 07
, 287
3
%
2 =4 7B02 X 006
23
9
a
3
00 = 4.7302 X 0005
% 0
15.
,2128
0
Explanation. The multiplier, 3.1765, is written in the reverse order, 56713,
the units' digit being placed under the lowest order of the multiplicand that is
desired in the product — ten thousandths. Multiply by each digit of the reversed
multiplier, beginning with that digit of the multiplicand which stands directly
above the digit of the multiplier used, taking care to include the digit carried
over from the multiplication of the one (or two) rejected digits at the right.
Example
Multiply 4.7869347 by 7.25, and obtain the product correct to three decimal
places.
Solution
4 786 9347
527
33 508 3 =4 7869 X 7
.957 2 =4 7860 X 2
239 0 = 4.7800 X 5
34 704 5
COMMON FRACTIONS 47
Problems
Multiply:
1. 5.987654 by 3.147, obtaining the product correct to the 4th decimal.
2. 3.590f by 14.57, obtaining the product correct to the 3rd decimal.
3. 184.2Sy by 3.145, obtaining the product correct to the 4th decimal.
4. 44.187542 by 6.2434, obtaining the product correct to the 3rd decimal.
Division of decimals. Division of decimals may often be
abbreviated, especially when the divisor is given to a greater
number of decimal places than are contained in the dividend, and
when only three or four decimal places are essential in the quotient.
Example
Divide 4.39876 by 2.4871934, and obtain the quotient correct to three decimal
places.
Kohdion
Ordinary Method Abbreviated Method
2 4871934)4 39S 7600 (1J76S 2.487 2W)4 398 7(1 768
~2 487 1934 ' 2_487 2
1 911 56660 I"911 5
1_741_0353S L741J?
~T70~53l220 ~T70~5
149231604 149 2
19 8975472 MM)
^1 4020688 T~4
Explanation. Observation of the ordinary method shows that the third
decimal place in the quotient is not affected by the digit in the third decimal
place in the divisor (except through the digits carried).
Since the units' digit of the divisor is contained in the units' digit of the
dividend, the first digit in the quotient is in the units' place, and as three decimal
places are required, the quotient will contain four digits. Therefore, the last
four digits of the divisor will not affect the quotient, except through the digits
carried over.
The first four digits of the divisor, 2.487, are contained once in 4.398. Multi-
plication of that part of the divisor used, by the quotient digit (including the
digit carried over from the one or two following digits — in this case considering
the 9 as a unit and adding it to the 1, making 2) gives 2487 2, and this result
deducted from the previous dividend leaves 1911 5 for the new dividend.
Cancel the right-hand digit, 7, of the divisor, and divide 1911 by 248, obtain-
ing the quotient 7. Multiplying the divisor by 7 (and including the carrying
digit) gives 1741 0, and subtracting leaves a new dividend of 170 5.
Cancel another digit, 8, of the divisor, and divide by 24. This is contained
6 times in 170. The product (including the digit carried over) is 149 2, and
this product subtracted leaves a new dividend of 21 3.
48 COMMON FRACTIONS
Cancel another digit, 4, of the divisor. Divide 21 by 2, using the carried
digit; the result is 8. The new product is 19 9, and this product subtracted
from 21 3 leaves a remainder of 1 4.
Example
Divide 8.47 by 31.76983476, and obtain the quotient correct to three decimal
places.
Solution
31.769 83476)8 4700 (.260
6 3540 =".31.760 (8) X .2, or 6.3539(6). Use 6.3540.
2 1160
1J90G2 = 31.76 (.95) X .06, or 1.9001(8). Use 1.9062.
"2098
1906 = 31.7 (GO) X .006, or .1906(1). Use .1906.
"192
Problems
Divide:
1. 4.3954 by 37.265872, obtaining the quotient correct to the 3rd decimal.
2. 65.157 by 4.4976348, obtaining the quotient correct to the 4th decimal.
3. 1.297648 by 15.782643, obtaining the quotient correct to the 3rd decimal.
4. 3.489765 by .28765431, obtaining the quotient correct to the 3rd decimal
To change a decimal fraction to an equivalent common fraction.
Write the denominator of the decimal, omit the decimal point, and
reduce to lowest terms. Thus, to reduce to common fractions in
lowest terms or to mixed numbers :
.75 = AV = * -025 = T?8ir = A
6.25 = 6T% = 6£ 4.125 = 4^- = 4j
To change a common fraction to a decimal. A common frac-
tion may be regarded as an indicated division. Thus: -§- may be
regarded as 2 -5- 5; therefore, f expressed as a decimal is .4; simi-
larly, | is .14$, f is .375, and TV is .4375.
Aliquot parts. An aliquot part of any number is a number that
is contained in it an integral number of times. Thus, 5, 10, 20,
and 50 are aliquot parts of 100; that is, 5 = ^V of 100, 10 = yV
of 100, and so forth.
The use of aliquot parts. As a means of saving time in multi-
plication and in division, it is useful to know the decimal equiva-
lents of common fractions, or, conversely, to know the common
fraction equivalents of decimal fractions. Aliquot parts are of
value in addition and subtraction if an adding machine or a calcu-
lating machine is used, because machines are not adapted for
general work involving common fractions.
COMMON FRACTIONS 49
TABLE OF ALIQUOT PARTS OF 1
Common
Decimal
Common
Decimal
Fraction
Equivalent
Fraction
Equivakn
*
.50
*
.11*
*
.33*
A
.10
i
.66|
A
.09^
.25
.08*
I
.75
A
.4lf
*
.20
TS
.58*
*
.16*
.91*
*
.83*
5
.06*
*
. 14y
A
.06*
2
28-r
A
.18*
I
.42*
A
.31*
4
57*
Tff
.43f
I
71?
IT
.56*
*
.85*
t?
.681
i
.12*
ri
.93*
3
37*
.04
f
. 62*
A
.03*
ff
X7i
A
-09|
The fractions in the above table can be extended as decimals
as far as the work demands.
Problems
Express the following aa decimal fractions; non-terminating fractions should
be carried to the sixth decimal place and the common fraction annexed:
21 1 3 5 11
31126 2
5 1 1 15 1 1
•ff TIT T Tff TT TO"
812 13 1
V T8 ^ TT T W
54 \\ 5 5 3
T T I (T ~S <5 "Saf
Multiplication by aliquot parts.
Example
Find 16*% of $475.34.
Solution
6)$475_34
" $79.22
Explanation. Since .16* equals i, find -g- of $475.34.
Example
Find the cost of 256 units at 37^ each.
Solution
256 X | X $1 = $96
Explanation. 37i?f is | of $1. Therefore, 256 X f X $1 = $96.
50
COMMON FRACTIONS
Problems
Extend the following items mentally:
1. 72
@ .T2*
9.
18
(W,
.33*
17.
64 @
.25
25.
72
@ .83*
2. 45
@ -lit
10.
39
(r/>
.6ft*
18.
27 <fo
.22f
26.
32
(«> .87*
3. 24
© .08*
11.
55
($
.09ix
19.
32 ©
.18f
27.
36
& .411
4. 36
® .50
12.
16
©
.75
20.
96 &
•03i
28.
27
^ .44f
6. 15
@ .06f
13.
49
(n\
.28}
21.
48 (a,
.56i
29.
12
(«; .75
6. 75
fe .93*
14.
32
4
.43*
22.
60 (",
.58^
30.
14
@. .07|
7. 48
@ .16*
15.
28
«!>
.57|
23.
48 (<i,
.37*
31.
18
(tn .16*
8. 32
@ .06*
16.
24
<&
.62*
24.
35 («,
•1472
32.
16
^ .87 ^
Division by aliquot parts. It is diffirult to divide a number by
a mixed number. If the divisor is an aliquot part, the quotient
may be found by multiplication, as follows:
Example
Divide 4,875 by 16*.
Explanation. Sinco 16* is Jr of 100, divide 4,875 by J of 100, Solution
or ^ir2-. This is the same as multiplying by n5#. Therefore, divide 48 75
by 100 by pointing off two decimal places from the right, and multi- 6
ply the result by 6. The answer is 292.50, or 292^. 292 ~50
Example
The production cost of 1,250 units is $3,170.
unit.
Find the cost per
Kxplanation. 1,250 is i of 10,000. Divide S3, 170 by 10,000
by pointing off 4 decimal places from the right; then multiply the
result by 8. The cost per unit is found to be $2.536.
Solution
3170
s
2~5360
Divide:
1. 1, 342 by Hi
2. 2,578 by 12i
Problems
3. 3, 126 by 33i-
4. 384 by 25.
5. 158 by 6i.
6. 4,275*by 14f
Problems
1. A manufacturer pays dividends amounting to fs-
his capital. If the
How many cords
dividends amount to $37,500, what is the capital?
2. A fuel dealer had 36 cords of wood and sold | of it.
did he sell?
3. If a merchant buys an article for $12^ and sells it for $16, the profit is
what fraction of the selling price? What fraction of the cost price?
4. A crate containing 10 dozen oranges cost $4.50. If they are sold at the
rate of 65 cents a dozen, but i dozen are spoiled, the profit is what fraction of the
selling price?
5. A man has $37^ and spends $12^. What fraction of his money does
he keep?
FRACTIONS 51
6. A factory normally employed 48 men. During a dull period 16 received
temporary lay-offs. What fraction of the force continued to work?
7. The last reading of a gas meter was 67,324 cu. ft.; the previous reading
was 64,815 cu. ft. At $1.45 a thousand cubic feet, find the amount of the gas hill.
8. An investment of $18,000 produces an annual income of $720. At the
same rate, what should an investment of $25,000 produce?
9. Tires costing §18.75 were installed when the speedometer registered
18,985 miles. The four tires were replaced \\hen the speedometer registered
34,652 miles. SI. 00 was allowed for each old tire. What was the average tire
cost per mile, correct to the nearest tenth of a mill?
10. An excavation 8 feet in depth required the removal of 5,328 cu. ft. of
earth and rock. The average depth of earth was 5 ft., and the cost of earth
removal was §1-} a cu. yd. The remainder was lock and cost $4g- a cu. yd. for
removal. What was the cost of making the excavation?
CHAPTER
Percentage
Relation between percentage and common and decimal frac-
tions. Percentage is a continuation of the subject of fractions.
It is the process of computing by hundredths, but instead of the
term hundredths, the Latin expression per cent is used. The sign-
(%) generally replaces the words per cent, thus, 5%, 10%, and so
forth.
Any per cent may be expressed either as a common fraction or
as a decimal, thus:
Common Fraction Decimally
1%. - TOO 01
5% . TiU 05
100%
300%..
*%.. . ,,2)() or 10;)0 .00* or .005
•°5- Too "r .0,000 °005
Care should be taken in writing per cents. Do not write both
the sign and the decimal point; thus, 2% and .02 are the same,
but 2% and .02 % are widely different, since the first is equivalent
to ro and the second to Woo-
Applications. Percentage admits of applications in many
fields. Business operations are guided by carefully prepared
statistics, and the relationships of items in statistics are often more
clearly reflected when they are expressed in terms of percentage.
There are numerous problems involving percentage besides those
having to do with financial considerations, such as finding the per
cent of increase or decrease in volume; per cent of shrinkage of
material; per cent of waste in manufacturing operations; per cent
of yield of crops.
Definitions. The base is the number or quantity represented
by 100%. The base may be, for example, total sales, total
53
To 0"
12J 125
H)0 °l 1000
1 00
100
1
MOO
100
3.
1
100 or 1006
TO* ( . 5
100 OI 10,000
54 PERCENTAGE
expenses, the face value of a note, the par value of a bond, pounds
of material used, capacity, and so forth.
The rate is the number of hundredth*, or the per cent. The
rate may be, for example, 6% or 25%, which are written decimally
as .06 and .25.
The percentage is the product of the base and the rate. The
percentage may be, for example, the interest cost of a sum of money,
the departmental portions of an expense item, the increase in
pounds of material used, and the like.
Fundamental processes. In percentage and its application,
three fundamental mathematical principles are involved, namely:
(1) to find a given per cent of a number; (2) to find what per cent
one number is of another; and (3) to find a number when a certain
per cent of it is known.
Computations. Computations in percentage are based on
these principles.
Principle 1. The percentage is the product of the base and the
rate.
Base X Kate = Percentage
Example
6% interest on $500 is $30. (500 X .00 = 30)
Problems
In the following, convert the per cent either to a common fraction or to a
decimal fraction, whichever is the easier.
Find:
1. 25% of 5,280 ft, 6. 2f % of 180 11)8.
2. 10% of 846 Ibs. 7. £' 'c of 240 pil.
3. 16f % of 24 bu. 8. -£% of $5,000.
4. 37i% of $60. 9. 20% of 95 yds.
5. 80% of 120 pp. 10. 14f ' ,', of 42 in.
11. If an expense item of $16.00 is reduced 6-f%, what will be the amount of
this item after the reduction?
12. A commission of 12-J% was earned on a $240 sale. What was the
commission?
13. A sample of grain showed 2$% weed seed. How many bushels of weed
seed are in 600 bushels of this grain?
14. An item sells for 40 cents. What will be the selling price after a reduction
of 15%?
16. Anticipated requirements for copper will exceed the manufacturer's stock
by 35%. If 185 pounds are on hand, how many pounds will have to be
purchased?
Principle 2. The rate may be found by dividing the percentage
by the base.
Percentage -*• Base = Rate.
PERCENTAGE 55
Example
$30 -*• $500 = .06 or 6%.
Problems
In the following find what per cent of:
1. 72 is 24 6. 12 is 20
2. 60 is 50 7. 64 is S
3. 180 is 120 8. 90 is 10
4. 360 is 90 9. 150 is 25
6. 50 is 20 10. 125 is 25
11. Last year's taxes on a house were $520. This year's taxes were $640
What per cent were this year's taxes of last year's taxes?
12. A pile of lumber contained 4,500 feet, and 3,300 feet were used. What
per cent remained?
13. Wages are increased from §1.50 an hour to $1.75 an hour. Find the
per cent of increase.
14. A new style of packaging reduced the shipping weight from 130 Ibs. to
121 Ibs. What was the per cent of saving in shipping weight?
15. The inspector rejected 5 items out of 140 produocd. What was the
per cent of rejects?
Principle 3. The ba.se may be found by dividing the percentage
hy the rate.
1'eiccntagc 4- Rate == Base.
Example
30 -5- .06 = 500.
Problems
Find the number of which:
1. 25 is 20% 6. SO is 43%
2. 125 is 16|% 7. 374 is 17%
3. 240 is 75% 8. 375 is \%
4. 48 is i% 9. 4i is J%
6. 72isl2i% 10. 20 is 40 %
11. The fire insurance premium on a house was $22.50. The house was
insured for 80% of its value at £%• Find the value of the house.
12. Sales increased each year over .the preceding year as follows: 15% the
second year, 20% the third year, and 25 % the fourth year. If the fourth year's
sales were $21,562.50, what were the first year's sales?
13. A bankrupt can pay his creditors 72 cents on the dollar. If his assets
are $13,475.28, what are his liabilities?
14. The gross income of a rental property is $1,HOO a year. P^xpenses are
$500. If the net income is a return of 6^% on the investment, find the value
of the property.
PERCENTAGE
16. One workman completes a unit in 7^ hours. Another workman com-
nlfctes a similar unit in 5f hours. The first workman took what per cent more
time than the second workman to complete the unit?
Miscellaneous Problems
1. A machine that cost $50 was marked up 30%. What was the marked
price?
2. After a clerk's salary was increased (>i%, he received $S50 a year. What
was his former salary?
3. A 4-apartment building cost $18,000. Repairs average l^r% of the cost;
taxes, 2^-%; insurance on 90% valuation, f%; other expenses amount tc
$114.25. What should the annual rental income be in order to return the
owner 8% on his investment? What should be the average monthly rental of
each apartment?
4. A product shrinks 10% in processing. TIow many pounds of raw material
will be required to process 252 pounds of finished product?
5. A creditor received $037.73 from a bankrupt estate paying 68 cents on tlu
dollar. What was the creditor's loss on the account?
Daily record of departmental sales. The following tabulation
is designed to show the total daily sales by departments, and the
total sales for the week, both by departments and for the business
as a whole. After Saturday's sales have been entered, the total
departmental sales for the week may be found and also the per cent
that each department's sales is of total sales. The per cent that
each day's sales is of total sales for the week is also obtainable.
DAILY RKCORD OF DKPAUTM i:\TAL SALES
Dcpt. Mon. Tues. \Vc<L Thurs. Fri. Sat. Total I'crCcnt
A $475. SO $275 S3 $329 SO $424 S3 $3S7 92 $412 15 . ...
B 324. IS 174 82 274 19 2S5 27 304.14 319 2S ..
C 456.19 259 SO 179 SO 25S 24 2SO 39 30574...
1) 421 40 20S 75 142.50 2SO 22 17S 90 200 57
K 175 00 125 34 150 S5 210 05 102 50 1S7 50 . . ..
Total ._ . . _. ~~ _--""""-- _ - r~ ltMM)0'p
pcr - ~-
Cent 100 00%
Problem
Prepare a form similar to the above, enter the sales in the proper columns,
and find: (a) the total sales for each day in all departments (add downward);
(b) the total sales for each department for the week (add across); (c) in two
ways, the total sales in all departments for the entire week; (d) the per cent of
grand total sales made each day (total for each day divided by the grand total) ;
(e) the per cent of grand total sales made in each department (total of each
department divided by the grand total).
Per cent of returned sales by departments. In some lines of
business it is important to keep a close check on the volume of
PERCENTAGE
57
returned sales. This may be done advantageously by means
of per cents derived from tabulated results.
Problem
Prepare a form similar to the following, enter the data, anil find: (a) the net
%aies for each department, and the net sales for all the departments; (6) the
per cent of returned sales in each department, and the total per cent of returned
sales.
SALES AND
SALES BY DKPARTMENTS
Dept.
A
Saks
$ 24,Sf>3 95
Returned Net
Sales Sales
$ 756 S2
Per Cent
of Sales
Returned
B
C
D
110,356 SO
53,76S 21
16,135 40
1 ,32S 95
975 32
62S 74 ...
E
Total
9,356 24
256 4S .
Clerk's per cent of average sales. As a measure of efficiency,
the following tabulation may bo made for a department, and each
['lerk's weekly or monthly sales compared with the average weekly
or monthly sales.
MONTHLY SALES OF CLERKS— DEPT. A
Clerk's Monthly l*er Cent of
Number Sales Average
I $2,756 SO ................
1,954 36 ..............
2,075 S3 .............
2,634 S7 .........
2,315 62
2
3
4
5
Total
Average
100 00%
Problem
Prepare a form similar to the above, enter the data, and find: (a) the total
monthly sales; (b) the average monthly sales per clerk; and (c) what per rent
*ach clerk's sales are of the average sales per clerk.
Per cent of income by source. In accounting for the income
3f a public service enterprise, it is desirable to show the per cent of
income from each source when the company's activities are varied.
Problems
^ 1. In the following tabulation of gross earnings of a public utility corporation,
ind what per cent the earnings from each source are of the total gross earnings.
58 PERCENTAGE
Source Gross Earnings Per Cent
Electric light and power $15,817,324 00
Electric and stearn railroads . 0,763,656 00
City railways and bus lines . 4,248,824 00
Gas 3,191,720 00
Heat . 672,394 00
Bridges 589,691 00
Ice - 2,54,670 00
Water 88,303 00
Miscellaneous 21,816 JX) ....__.._.
S3 1 ,64'S,39SJ)0 JOQ QQ < ;,
2. In the following tabulation of the revenue from transportation of an inter-
urban railway, find what per cent each iten: of revenue is of the total revenue.
RKVKNUU FROM TRANSPORTATION
Source Amount 1'er Cent
Passengers $657,855 00 .
Baggage 550 00 .....
Parlor and chair cars. .. . . . 9,894.00
Special cars . 2500
Mail 1,500 00
Express 21,96200
Milk . . 1,666 00
Freight . 264,214 00
Miscellaneous. ... . 26900
^^7,935^00 H)0l)0%
Per cent of expense. Items of operating expenses and their
relation to total expenses are more easily compared if expressed in
terms of percentage.
Problems
1. In the following report of an interurban railway company, find what per
cent each group of expenses is of total operating expenses.
OPERATING KXPKNSKS
Item Amount Per Cent
Way and structures $228,690 00
Equipment 9S,979 00
Power 105,890 00
Conducting transportation . . . 249,427 00
Traffic 52,82300
General and miscellaneous 141,560 00
Transportation for investment (credit) . . 8,40300
100 00%
2. In the following statement of the operating expenses of a restaurant for
a period of one month, find what per cent each item of expense is of total oper-
ating expeo^es.
PERCENTAGE 59
OPERATING EXPENSES
Item Amount Per Cent
Superintendent's labor $ 75 (X)
General labor 1,776 00
Extra labor . . 160 00
Supplies . . 200 00
Electricity . 58 00
Fuel ... 75 00
Laundry. . 103 00
Ice . . . 22 00
Repairs and renewals — equipment ... 110 00
Meals to employees . . 340 00
Music \ . 75 00
Miscellaneous . 66 00
Total $;V)60 00 100 00%
Percent of increase or decrease. Percentage is often employed
to find the relation between numbers; that is, to find how much
larger or smaller one number is than another.
Problems
1. In the following departmental sales tabulation, find: (a) the increase or
the decrease in monthly sales by departments; (h) each department's per cent of
increase or decrease (divide increase or decrease in each department by that
department's monthly sales for This Month Last Year).
Per Cent Per Cent
Increase Decrease Increase Decrease
This Month
This Month
Dept.
This Year
Last Year
A
$2,973 69
$2,795. 84
B
1,426 S3
1,S52.1S
C
3,752 89
3,565 62
D
2,5S1 28
2,67S 15
E
2,076 S2
1,825 38
Total
2. In the following condensed balance sheet of a municipal railway, find the
increase or decrease for each item, and also the per cent of increase or decrease.
.4 sscts
Capital Assets
Current Assets
Deferred Assets
This Year
. $ 7,912,526
2,174,925
132,124
Laxt Year
37,610,139
2,241,395
132,125
Increas( , Per Cent
Decrease] Inc., De,c.\
Total Assets
$10,219,575
$9,983,659
Liabilities, Reserves, and
Surplus
Funded Debt
. $ 3,992000
$4,192000
Current Liabilities ....
Reserves ...
Surplus
269,720
1,568,469
4,389,386
343,126
1,615,743
3,832,790
Total Liabilities, etc
. $10,219,575
39,983,659
60
PERCENTAGE
3. In the following tabulation of advertising expenditures and direct sales
resulting therefrom, compute the totals, the increase, and the per cent of increase.
Jan
Feb. ...
Mar.. .
Apr. . . .
May. . . .
June . .
July
Aug .
Sept
Get... .
Nov .
Dec
This
Advertising
$2,238 00
2,154.00
2,435 86
2,425 46
2,293.12
2,035 76
none
none
none
2,212 56
7X5 24
none
Year
Sales
$4,251.44
7,461 60
8,773 84
7,292.12
8,709.04
8,412 28
7,3X3.46
7,656 80
8,227.84
4,298 70
5,260 X4
5,6X3 96
Last
Advertising
$
1,769 64
1,787.96
1,769 53
1,840.26
1,831 70
1,825.49
none
none
none
1,142 04
1,306 26
none
Year
Sales
$ 3,762.00
5,067 16
5,232 48
7,818 00
4,867 20
4,673 12
5,0X3 20
4,454 56
4,650 88
4,976 40
2,6X2 00
3,542 XO
Total
Year ago $10,607.52 $37,650.77
Increase
% Increase.. . . . ... .
4. In the following statement, find the increase or decrease of revenues and
expenses and the per cent of increase or decrease:
Increase,
This Year Last Year Decrease^ Per Cent
Railway operating revenue $X66,197 $970,060
Other operating revenue X/21X 7,X2() ... ....
Total operating revenue $X74,415 SOTTJSSO .... " ..T
Railway operating expense:
Way and structures $ 91,3X0 $ 85,569
Equipment 64,249 61 ,866 .....
Power 108,313 114,906 ....
Conducting transportation 196,259 21 1,144
Traffic 9,496 10,157
General and miscellaneous .... 12S,S49 12X,XX7 .
Depreciation 17,324 44,645
Taxes (except income taxes) . 26,1X5 29,840 ... .
Total $642,055 «GS7^014 .7.71 ~" _ j
Operating income $232,360 $29Q~,X66 ^_^_^ ~"
Non-operating income:
Interest funded securities 2,579 5,105
Interest unfunded securities. ... 7,765 6,328
Total 1^0»344 $ 1],433 ~ ...............
Gross income $242,704 $302,299 ~
Deductions from gross income:
Interest $160,318 $161,402
Miscellaneous 3,216 3,257 ................
Total $163,534 $164^659 ^~TZ^ .~
Net income $ 79,170 $137,640
PERCENTAGE
61
Operating statistics. The operations of a public utility engaged
in transportation afford an excellent opportunity for the presenta-
tion of statistics for managerial control. The following problem
has been derived from the report of such an enterprise.
Problem
From the following data, ascertain the required answers.
SECTION OF INCOME
Income
Operating revenue:
Railway operating revenue
Coach operating revenue
Total operating revenue. . . .
Non-operating income
Total revenue from all sources
Operating expenses:
Railway operating expenses .
Coach operating expenses .
Total operating expenses. .
Net revenue from all sources
STATEMENT
Thin Year
. . $ 22,413,089
818,328
184,273
Last Year
$ 21,678,900
5jJL282
$ 21, 730,188
141,707
$ 23,410,290 $ 21,871,955
. $ 10,572,497
____ 780,558
$J7T359,055
$ 0,057,235
$ 15,383,494
41,701
»JM_25J95
$ 73,440,700
Railway revenue car-miles .52,803, 1 1 1 48,248,330
Coach revenue coach-miles. . .. 3,529,795 157,540
Railway revenue car-hours 5,092, 1 90 5,207, 1 70
Railway revenue passengers 357,920,108 340,1 10,298
Railway transfer passengers . .1 23,3 1 0,520 1 1 1 ,445,9 1 2
Railway total passengers ... 481,230,094 457,502,210
Coach revenue passengers . . 10,504,723 978,782
Coach transfer passengers . . 387,228
Coach total passengers 10,951 ,951 978,782
Total revenue and transfer passengers 492,188,045 458,540,992
Railway operating revenue per car-mile (cents) ...
Coach operating revenue per coach-mile (cents) . . , . . .. .
Railway operating expenses per car-mile (cents) . . ..
Coach operating expenses per coach-mile (cents) .. .......... .. ..
Railway operating revenue per car-hour ($ and
cents) -
Railway operating expenses per car-hour ($ and
cents)
Ratio of transfer passengers to revenue passengers
— railway (per cent)
Ratio of transfer passengers to revenue passengers
— coach (per cent)
Railway revenue passengers per car-mile operated .
Railway transfer passengers per car-mile operated
Total railway passengers per car-mile operated
Coach revenue passengers per coach-mile operated
Coach transfer passengers per coach-mile operated
62 PERCENTAGE
Statistics (Continued) This Year Last Year
Total coach passengers per coach-mile operated
Ratio of railway operating expenses to railway oper-
ating revenue (per cent)
Ratio of coach operating expenses to coach operating
revenue (per cent)
Budgeting. Percentage is also applied in budgeting, as shown
by the following example from hotel accounting.
Example
Among the several items of the budget is, China and Glassware, $3,500, to
be distributed to four departments on the basis of the previous year's expense
for this item in the four departments, as follows:
Department Per Cent
Rooms 1 1 29
Restaurant 5529
Coffee Shop 14 86
Beverages . . . 1 8 56
Total 106~00%
Solution
Department Per Cent Budget
Rooms 1 1 29 $ 395 00
Restaurant 55 29 1,935.00
Coffee Shop 14 86 520 00
Beverages J^L^L. 650 00
Total WJO% S^XTOO
Problems
1. The following year it was found that the actual disbursements for China
and Glassware amounted to $2,280.74, and other farts were as given in the
tabulation below. Compute the per cent for the distribution of the budgeted
amount for the next year, and the per cent that the expense of China and Glass-
ware is of the income for each department.
China and Per Cent Per Cent
Gross Glassware of of
Department Income Expense Expense Income
Rooms $141,857 50 $ 269 53 . .
Restaurant 59,626.90 1,252 16
Coffee Shop 33,587.45 335 87
Beverages 9,061 .65 423 18
$244,133 50 $2,280.74 100 00%
2. The following budget is that of an estimated operating statement.
Per Cent of
Net sales: Total Sales
Class A $2,000,000
Class B 200,000
Class C 250,000
Class D 50,000
$2,500,000 UK) 00%
PERCENTAGE 63
Per Cent of
Sales
Production costs:
Class A $1,200,000
Class B 140,000
Class C 162,500
Class D _40,000 ii^iz^;
$1,542,500
Ptr Cent
Gross margin $ 957,500
Selling:
Sales administration $ 50,000
General sales department expense. . . . 12,500
Special promotion, etc 12,500
District operating expense 400,000
Advertising A 100,000
Advertising]* 12,500
Advertising C 25,000
Selling cost $ 612,500 .... ___..
Net margin $_ 345^00 ~
Calculate: (a) the per cent of net sales in each class, as compared with total net
sales; (b) the per cent of pioduction cost in each class, based on sales of each
class; (c) the per cent that selling cost is of total net sales; (d) the per cent that
the net margin is of total net sales.
3. The following is the budget for the Water Department of a municipality.
Find the per cent that each budget expenditure is of the total for the department.
Amount Per Cent
Pump station and filter plant salaries .... $17,300.00
Office salaries and expenses 4,600.00
Chemicals, filter plant 1,000 00
Power — pump station and filter plant 15,000 00
Light, heat, and supplies 3,000 00
Water service 3,000 00
Meters and installation 6,000 00
Water main extensions and fire hydrants . 3,000 00 .
Motor truck repairs 150 00
Interest on outstanding warrants 4,246 00
Total $^296JO MjjjO%
4. Compute the increase or decrease and the per cent of increase or decrease
in the following comparative budget.
Public Buildings This Last % %
and Utilities Year Year Inc. Dec. Inc. Dec.
City hall engineers and jani-
, tors $ 5,060 $ 4,284
City hall fuel and supplies ... 4,000 2,216
City hall maintenance and re-
pairs 1,000 850
64
PERCENTAGE
Public Buildings This Last
and Utilities Year Year
Detention hospital repairs. . . $ 300 $ 400
Detention hospital light and
fuel 900 700
Park light and fuel 1 ,200 1 ,050
Septic tank electric power. ... 600 600
Septic tank repairs 100 100
Incinerator fuel and light. . . . 1,000 1,000
Electric lighting — streets,
alleys 14,500 14,000
Interest on warrants 2,2H4 2,170
Contingent fund 5,000. 4,036
Detention hospital insurance 433
Library insurance ... . 281
Park insurance 104
$35,1)44 $32~224
Inc. Dec.
%
Inc.
%
Dec.
Profits based on sales. In the income statement, it is custo-
mary to base all percentage calculations on sales. With sales
equalling 100%, cost of sales, overhead, and net profit are expressed
as per cents of sales. Overhead expenses are those incurred in
operating a business — such as salaries and wages, rent, heat, light and
power, depreciation, taxes, insurance, advertising, telephone, post-
age, and so forth. In marking goods bought for resale, these
expenses must be taken into consideration. A few items of over-
head expense do not fluctuate, but many of them have a fairly
constant ratio to gross sales. The merchant determines the ratio
of overhead expenses to sales from his own experience and that of
others engaged in similar businesses. This per cent of cost of doing
business plus the per cent of profit decided upon deducted from
100% determines the per cent which the cost of goods plus freight
and drayage bears to the selling price.
Sales
= 100%
Invoice Price plus Freight and
75%
Cartage
Overhead Profit
15% 10%
Cost of Sales
75%
Gross Profit on Sales
25%
Example
If overhead charges amount to 15% of sales, and a profit of 10% on sales is
desired, what is the selling price of an article with an invoice cost of $21.00 and
freight and cartage of $1.50?
PERCENTAGE 65
Solution
15% + 10% = 25%
100% - 25% - 75%
$21.00 + $1.50 = $22.50, the cost.
$22.50 -T- 75% = $30.00, the selling price.
Verification
25% of $30.00 = $7.50, the overhead and profit.
$30.00 - $7.50 = $22.50, the cost.
Problems
1. An article that cost $15.00 was sold for $20.00. What is the profit per
cent on the selling price?
2. With an overhead expense of 20%, what per cent of profit on sales is
made by selling for $1.50 articles that have an invoice cost of $1.00?
3. What is the per cent of gross profit on sales in Problem 2?
4. How much must the article in Problem 2 be reduced to sell at cost?
What per cent is this of the marked price?
5. A merchant sold an article for $12.00 and made a profit of 12^% on the
selling price. What was his profit in dollars?
6. Find the per cent of reduction of marked price to produce cost.
Cost Marked Price Per Cent Reduction
a. $ 20 $ 25
b. 2 50 2 75
c. 1 00 1 20
d. 03 .05
e. 3 00 6 00
/. .40 .50
g. 09 .12
h. 15.00 25 00
7. Find the per cent of profit on the selling price.
Per Cent Profit on
Cost Selling Price Selling Price
a. $ 1.00 $ 1 20
6. 10.00 15.00
c. .60 .75
d. 3.50 7.00
«. 6.00 8 00
/. 150 00 175 00
g. 16.00 24 00
h. 75 00 125 00 ....
8. The factory price of an automobile is $1,300. Freight charges from
factory to dealer are $65.00. If the dealer's overhead is 20% and he expects a
net profit of 15% on sales, what should be the selling price of the automobile?
9. A furniture dealer bought a shipment of 20 chairs at $30.00 each. He
marked them to sell at a profit of 40% on cost. The entire shipment was sold
66 PERCENTAGE
in the fall clearance sale at 25% reduction from marked price. What was the
profit or loss?
10. Complete the following:
% on
Cost Selling Price % on Cost Selling Price
a. $ 4 00 $ 6 00
b. 15.00 25 00
c. .16 .20
d. .04 .08
e. .OS .10
/. 5 00 7 00 ..
g. 500 00 750 00
h. 24.00 32 00
11. The invoice price of an article is $12.00. Freight is 75 cents. It costs
18% to do business and you desire a net profit of 10% on sales. What is the
selling price of the article?
12. If the invoice cost is $28.00, freight $2.00, overhead 25%, net profit on
sales 15%, what is the selling price?
13. A stock of merchandise valued at $8,750.00 was damaged by fire and
water. The loss was estimated to be 25%. Find the value of the damaged
merchandise.
14. A merchant's overhead, or cost of doing business, is 22-f %. He desires
to make a net profit of 7ir%. What will be the selling price of an item that cost
this merchant $4.90?
16 Merchandise is bought for $3.50 less 25% and sold at $3.50 net. What
is the rate per cent of profit?
16. A tea and coffee merchant blends a 40 j£ tea with a 70 £ tea in the ratio
of 2 to 1. If the blend is sold at 65^ a pound, what is the rate per cent of profit
on cost?
17. A chair manufacturer finds the cost of material in a certain type chair
to be $7.50. Manufacturing cost (labor and overhead) is $14.80. Selling and
administrative expenses are 20% of sales. What is the manufacturer's price
for this chair if he desires to net 10% on the selling price?
18. A clerk was ordered to mark a lot of suits so as to make a profit of 20%
after allowing 5% discount for cash. By mistake he marked the suits $24.75
each, which resulted in a loss to the clothier of 8%. At what price should the
suits have been marked?
Marking goods. Merchants frequently indicate the cost price
and the selling price on each article. The buyer may use the cost
price marking for comparison with current quotations ; slow sellers
may be checked for desirability of reducing the selling price ; inven-
tory of stock may be taken at cost; and, under special systems of
accounting, a daily record of cost of sales is achieved.
In order to conceal the cost price from the customer, a set of
symbols is used, interpretation of which depends upon a knowledge
of the key to the letters or characters. Any word or phrase of ten
PERCENTAGE 67
letters or any ten arbitrary characters may be used as the key.
An extra letter or character is used to prevent repetition of a letter,
and this extra letter or character is called a repeater. The repeater
makes it more difficult for a stranger to decipher the marks. The
word or phrase used must not contain the same letter twice.
Otherwise the same letter will represent two different numbers.
If the cost and the selling price are both written on the same
tag, the cost price is usually written below, and the selling price
above, a horizontal line. If both cost and selling price are marked,
a separate key is used for each.
Example
Use the word "blacksmith" with repeater "w" as the selling key, and the
phrase "pay us often" with repeater "x" as the cost key, and mark an article
to sell at $6.50 with a cost of $4.25.
Solution
b 1 a c k s m i t h Repeater
1234567890 w
pay u soften Repeater
1234567890 x
S.kh
U.as
Example
With the same keys, mark an article to sell at $9.55, with a cost of $7.00.
Solution
T.kw
F.nx
The following are examples of key words and phrases:
Blacksmith Buy for cash
Charleston Black horse
Buckingham Cash profit
Republican Pay us often
Authorizes Our last key
Problems
1. Using "Charleston" as the key word and "x" as the repeater, indicate
the following costs:
a. $5.56 d. 86.62 g. $6.20 j. $1.44 m. $15.00
b. $6.50 e. $7.50 h. $5.00 k. $26.50 n. $2.35
c. $2.45 /. $12.50 t. $.25 I. $12.47 o. $1.60
2. Use as the cost key "pay us often" and repeater "w," as the selling key
"authorizes" and repeater "x," and show markings for the following:
68 PERCENTAGE
Coat
Selling Price
a.
' $ 2.25
$ 3.50
b.
1.15
1.50
c.
1.25
1 65
d.
23.50
29.50
e.
.65
.90
3. If the selling key is "Bridgepost" with repeater "w," and the cost key
is " Cumberland" with repeater "x," write in figures the prices given in the
following:
B.dt Bg.ow
a' Oud e' Cu.dx
R.pg It.ww
U.xd J' Ux.dx
Be.ot
^
' Cb.dx g' libd
I.tw R.dt
a' U.ed *' C.rd
Commissions. The commission business in this country is
largely the result of our industrial and commercial development.
Economic conditions demand that there shall he agents who shall
represent either the buyer or the seller. The compensation paid
the agent for his services is called a commission. The principles
of percentage apply in commission.
The person who transacts business for another is the agent, and
the one for whom the business is transacted is the principal. The
fee, usually a per cent of the dollar volume of the transaction, is
the commission.
\ . , V^ ^ Problems
1. An agent sells oil for $3,475.00 at 3i% commission. What is the amount
of the commission?
2. A merchant buys goods through an agent at a cost of $275.00. The
agent charges 2^% commission. What is the total cost of the goods to the
merchant?
3. An agent sells a consignment of merchandise for $1,824, retaining his
commission of 3%. How much does he remit to his principal?
4. If $302.75 was charged for selling $8,650.00 of merchandise, what was
the rate of commission?
5. A realtor's fee for selling a house and lot WHS $150.00. If the rate was
2%, what was the amount received by the principal?
6. An agent's commissions for one week were $216.80. If his sales were
$10,840.00, what rate did he charge?
7. The invoice price on a shipment of merchandise was $1,283.38, i
agent's commission. If the agent's rate was 3%, what was the commission?
8. The proceeds of a sale received by the principal were $828.78. The
commission deducted by the agent was $43.62. What was the rate?
9. Find the net proceeds of the following:
PERCENTAGE 69
ACCOUNT SALES
Boston, Mass., Oct. 5t 19 —
Sold for Account of
Friends Milling Co., Friendsville, Minn.
By Puritan Brokerage Company
19-
Aug.
4
350 bbls. Flour (o\ $4.51
24
175bbls. Flour @ 4.4S
Sept.
r>
320 bbls. Flour (8l 4.50
19
60 bbls. Flour @ 4.53
Oct.
1
30 bbls. Flour @ 4.52
Total Sales
Charges
Aug.
Oct.
1
4
1
5
Freight $420.60
Cartage 26 50
Storage 22 90
Commission (fll 3%
Total Charges
Net Proceeds
10. The manufacturing cost of a certain type machine is $$0f.W. The
manufacturer wishes to catalog this machine at a list price that will net a profit
of 25% on sales after allowing a dealer's discount of 25% and agent's commission
of 16-f %. Find the catalog list price.
CHAPTER^
Commercial Discounts
Cash discount. Cash or time discount is a deduction for
immediate payment, or for payment within a definite time. The
deduction is a certain per cent of the invoice.
The expression "Terms: 2/10, 1/30, n/60" means that 2% of
the invoice price may be deducted by the purchaser if payment is
made within 10 days of the date of the invoice, that 1 % may be
deducted if the invoice is paid within 30 days from the date of the
invoice, and that the invoice is due in 60 days without discount.
In some cases notice is given to the effect that interest at a specified
rate will be charged after the due date.
The acceptance of a cash discount is usually of ndv:ml:igo to
the purchaser. The following table indicates the annual interest
rates to which the usual cash discounts are equivalent:
i% 10 days, net 30 days = 9 per cent a year
1% 10 days, net 30 days =18 per rent a year
li% 10 days, net 30 days = 27 per cent a year
2% 10 days, net 30 days = 30 per cent a year
2% 10 days, net 60 days — 14.4 per cent a year
2% 30 days, net 4 months = 8 per cent a year
The rate per cent a year is calculated by taking the number of
days between the discount date of payment and the end of the
credit period, dividing the number of days in a year (360) by this
number^ and multiplying the quotient by the rate of discount
under consideration.
X Rate of Discount = Equivalent Annual Interest Rate
Number of Days Between
Discount Date and
End of Credit Period
Problems
1. Find the equivalent annual interest rate for the following terms:
2% 30 days, net 60 days
3% 10 days, net 30 days
3% 30 days, net 60 days
3% 10 days, net 4 months
72 COMMERCIAL DISCOUNTS
2. To pay an invoice of $1,500, with terms 2/10, n/30, the purchaser bor-
rowed the money at 6% in order to take advantage of the 2% discount. What
benefit did he secure by borrowing the money?
3. A merchant was able to obtain 5% discount on an invoice of $720 by
borrowing the money at the bank for 90 days at 6% interest. Plow much was
lie able to save?
Trade discount. Mercantile or trade discounts are reductions
from list prices, or from the amount of the invoice without regard
to time of payment. By offering different rates of trade discounts
to wholesalers and retailers, the manufacturer can send the same
catalog to both classes of customers. Revised discount sheets are
issued as prices fluctuate, hut the same catalog may be used a year
or more because the list prices are fixed.
Rules of percentage are applied in commercial discounts:
Invoice price = base
Per cent of discount = rate
Discount = percentage
Several discounts are sometimes given. These are known as
chain discounts or a series of discounts.
The order in which the discounts are deducted will not affect
the result; thus, a selling price stated as list price "less 10%, 20%,
and 5%" is the same as a selling price stated as list price "less 5%,
20%, and 10%." This is shown in the following example, in
which $100.00 is used as the base:
Example
$100.00 X .10 .. . . $1000 $100.00 X .05 $500
$100.00 - $10.00 . $90 00 $100.00 - $5.00 $95 00
$ 90.00 X .20 . SIS 00 $ 95.00 X .20 $19 00
$ 90.00 - $1S.OO $72 00 $ 95.00 - $19.00 $76.00
$ 72.00 X .05 . .. .$3.60 $ 76.00 X .10 $7.60
$ 72.00 - $3.60 . . . . $68 40 $ 76.00 - $7.60 $68 40
$100.00 - $68.40 $31.60 $100.00 - $68.40 $31.60
The total discount in each case is $31.60.
The dollar amount of discount determined from a series of
rates is not the same as the amount of discount determined from a
single rate equal to the sum of the series of rates. The sum of the
series of rates is 35%] 35% of $100.00 is $35.00, whereas the correct
discount is $31.60.
Single discount equivalent to a series.
First method. To find the single discount that is equivalent to
a series of <li>(ioimfs71TuE^^ from 100%.
Use the remainder as a new base. Multiply it by the second dis-
count, and deduct the product. Use this remainder as & jiew base.
COMMERCIAL DISCOUNTS 73
Compute each discount successively, ^proceeding as before. The
difference between 100% and the, last result will Lii Uic-.iiijLi^
discount.
Example
What single discount is equivalent to a series of discounts of 20%, 10%,
and 8i%?
Solution
100% -20% 80%
80% X 10% . .... 8%
80% - 8%.. . ... . . 72%
72% X8i%.. . ... 6%
72% - 6%.. . . 60%
100% - 66% .. . . ... 34%
Explanation. 100 % represents the invoice price. 20%, or I of 100%, equals
20%, which subtracted from 100%, leaves 80%; 10%, or TV of 80%, equals 8%,
which subtracted from 80% leaves 72%; S-J-%, or rV of 72%, equals 6%, which
subtracted from 72% leaves 66 %. 100%, the invoice price, less 66%, the
selling price, leaves 34%, the single discount.
Second method. To find the single discount that is equivalent
to a series of discounts, subtract each single discount from 100%
and find the product of the remainders. Subtract the final prod-
uct from 100%, and the remainder is the single discount equiva-
lent to the series of discounts.
Example
What single discount is equivalent to the series 20%, 10%, and 5%?
Solution
100% 100% 100%
20% v J0% 5%
80% ' 90% "95%
.80 X .90 X .95 = .684, or OS.4%
100% — 68.4% = 31.6%, the single discount
A short method. To find the single discount that is equivalent
to any two discounts, subtract the product of the discounts from
the sum of the discounts.
Example
What single discount is equivalent to discounts of 20% and 20%?
Solution
20% + 20% = 40%
20% X 20% = 4%
40% - 4% = 36%
To find the net price. To find the net price, the list price and
discounts being given: Reduce the discount series to a single dis-
74 COMMERCIAL DISCOUNTS
count^multiply the invoice price by this single discount, and deduct
the result from the invgicejjrice.
Example
What is the net price of an invoice of $600.00, less 30%, 20%, and 10%?
Solution
100% 100% 100%
30% 20% 10%
70% 80% 90%
.70 X .SO X .90 504, or 50.4%
100% - 50.4% 49.6%, the rate of discount
$600.00X49.0% .. . . $297.60, the discount
$600.00 - $297.60 $302.40, the net price
If the amount of discount is not desired, the net price may be
found as follows:
$600.00 X 50.4% = $302.40
Problems
1. In each of the following, calculate by the short method the single dis-
count that is equivalent to the scries:
(a) 10% and 5%. (c) 40% and 5%. (c) 35% and 10%.
(b) 20% and 5%. (d) 15% and 10%. (/) 30% and 20%.
2. In each of the following, find the net price:
(a) $350.00, less 10%, 10%, and 5%. (c) $480.00, less 20%, 10%, and 5%.
(6) $500.00, less 33£%, 5%, and 2|%. (d) $1 ,200.00, less 5%, 2i%, and 1 %.
(c) $900.00, less 50%, 20%, and 5%.
3. The list price of an invoice is $750.00, with discounts of 10%, 5%, and
2t%. The terms of the invoice are: 2/10, 1/30, and n/60. What amount will
he necessary to pay the invoice: (a) within the 10-day period; (6) within the
30-day period?
4. B purchases merchandise listed at $3,500.00, less 20% and 25%. He
sells this merchandise at the same list price, less 15%, 10%, and 5%. Does he
gain or lose, and what amount?
5. A dealer offers merchandise at a list price of $5,000.00, less discounts of
25 %, 10%, and 10%. Another dealer offers the same merchandise at a list price
of $4,800.00, less discounts of 20%, 15%, and 5%. Which is the better offer,
and by what amount?
6. Which is the better offer, and by what amount, on an invoice of $425.00:
(a) 30%, 20%, and 10%; or (b) a single discount of 50%?
7. The list price of an item is $24.00. If bought at that price less 33^% and
10%, and then sold at the same list price less 20% and 5%, what is the profit?
8. The net cost of an invoice of merchandise was $1,200.00. What was
the list price, if the cost was 25% and 20% off list?
COMMERCIAL DISCOUNTS 75
^-9. If the list price is $400.00, and the net price is $380.00, \\hat is the single
rate of discount?
10. What single discount is equivalent to 25 %, 20%, and 12j%?
Transportation charges c^j^s^^ynvoices. In some cases
transportation charges are paid by the seller; in other cases, by the
purchaser. If the purchaser is to pay the transportation charge,
_aiicTas" a matter of convenience the'seTIeFpreprfys Tf , TJie~ seller adds
the jcharge to the invoice. The purchaser is not entitled to cash
discount on the added charge.
If_a shipment is made "freight allowed/' the discount should
be figured after the deduction for freiglit; otherwise, it would be
equivalent to taking discount on the transportation charge.
Problems
1. An invoice of books amounts to $4.85, and parcel post charges are 79 cents,
a total of $5.64. If terms are 2/10, what is the discount if paid within the 10-day
period?
2. Complete the following invoice:
6 doz. Items @ $6.65 ..............
24 doz. Articles @ .45 ..............
Girdoz. Items @ Al\ .............
3 only Items @ 2.34 __.... _
Less 15%
Less freiglit allowance ..............
525 Ibs. @ .45| cwt. ..............
Net " 7ZTZI
If the terms of the above invoice are 1/10, n/30, what will be the discount if
paid within 10 days?
3. An invoice for paper, freight allowed, amounted to $1,754.50. The freight
bill paid by the purchaser was $238.54. If 2% discount was allowed for pay-
ment within 10 days, what was the amount of the check?
Anticipation. In retail business, invoices for purchase^ often
have dating terms, The terms may be 2/10, 90 days extra. IfHie
merchandise is received within 10 days and checked by the receiv-
ing department, the purchaser will deduct 2% cash discount, and
an anticipation discount on the balance computed at 6% (usually)
for 90 days, which is equivalent to an additional discount of l£%.
Another case is that of spring purchases of fall merchandis*
invoiced 2/10, November 1 dating. An invoice with these terms
may be discounted 2% if paid before November 11, and if paid
July 15 would be subject to anticipation discount for 119 days at
the customary rate, say, 6%.
76 COMMERCIAL DISCOUNTS
Example
An invoice billed April 2, for $3,250.75, terms 2/10, Nov. 1 dating, freight
allowed, was paid April 28. Freight paid by the purchaser was $132.48.
What was the amount of the check?
Solution
Invoice $3;250.75
Less freight 132 48
$3,118~27
Less discount, 2% 62 37
$3,055 90
Less anticipation, 6% for 197 days 100 34
Amount of check ^M^Oi5
Problems
1. An invoice for $21.25 dated Dec. 28, terms 2/10, Feb. 26, was paid Jan. 7.
What was the amount of the check?
2» What is the anticipation on an invoice for $475.50, dated June 10, terms
2/10^ 90 days extra, if paid June 25?
3. Find the amount earru-J by paying an invoice for $1,275.25, dated July 12,
terms 2/10, Oct. 1 dating, on July 2<S.
CHAPTEI&T^J
Simple Interest
Definition. Interest, as commonly defined, is a payment for
the use of borrowed money or credit. This payment depends upon
the rate per cent charged and upon the length of time for which
interest is calculated. The sum loaned or the amount of credit
used is the principal. The number of hundredths of the principal
that is taken is the rate, which is usually expressed as a per cent.
The principal, with the interest added, is called the amount.
Short method of calculating. There are a great many methods
of computing interest, each of them possessing more or less merit.
However, with the accountant the chief consideration is not how
many methods there are, but rather how accurately and how
quickly he can solve a problem in interest.
The computation of the product of principal, rate, and time is
the shortest method when the time is full years or fractional parts
of a year, such as ^-, ^, -J, any number of lOths, and so forth; other-
wise, the operation may be shortened by taking advantage of
aliquot parts, multiples and fractions, cancellation, and so forth.
The following principles and methods of computing interest are
quick and accurate when the rate is £%.
Sixty-day method. To find the interest at 6% for:
6 days, point off 3 additional places to the left of the decimal point in the
principal.
60 days, point off 2 additional places to the left of the decimal point in the
principal.
600 days, point off 1 additional place to the left of the decimal point in the
principal.
For 6,000 days, the interest will be the same as the principal.
Example
Find the interest on $256.75 for 6 days at 6%.
Solution
Pointing off 3 places to the left of the decimal point in the principal gives
25675, «r 26£.
Example
Find the interest on $345.65 for 36 days at 6%.
77
78 SIMPLE INTEREST
Solution
Point off 3 additional places to the left of the decimal point in the principal,
and multiply by 6. The answer is $2.07.
For rates other than 6%, see adjustments on page 80.
Problems
Find the interest at 6% on:
1. $180.00 for 60 days. 6. $26.50 for 18 days.
2. $150.00 for 54 days. 7. $752.25 for 6 days.
3. $262.50 for 24 days. 8. $15.80 for 54 days.
4. $32.75 for 36 days. 9. $75.40 for 30 days.
6. $65.50 for 12 days. 10. $12.85 for 24 days.
Method using aliquot parts.
Example
Find the interest on $275.84 for 124 days at 6%.
Solution
$2 75 84 = interest for 60 days
2 75 84 = interest for 60 days
IS 38 = interest for 4 days
$5~|70 06 = interest for 124 days
Explanation. Pointing off 2 decimals, as indicated by the vertical line, gives
the interest for 60 days. Double this to find the interest for 120rlays. Four
days' interest is -fa of 60 days' interest. The sum, $5.70, is the interest for
124 days.
Example
Find the interest on $754.90 for 137 days at 6%.
$ 7'
7
1
Solution
54 90 = interest for 60 days
54 90 = interest for 60 days
50 98 = interest for 12 days
62 90 = interest for 5 days
.68 = interest for 137 days
Explanation. Pointing off 2 decimal places gives the interest for 60 days.
Double this to find the interest for 120 days. Twelve days is ^ of 60 days;
therefore, the interest for 12 days is £ of 60 days' interest, or $1.5098. Five
days is T* of 60 days, and the interest is -£2 of $7.5490, or $0.629. The sum,
$17.24, is the interest for 137 days.
After a little practice, any number of days can be resolved into
6- or 60-day periods and easy fractions thereof.
Example
Find the interest on $247.64 for 8 days at 6%.
SIMPLE INTEREST 79
Solution
$1247.64 = interest for 6 days
|§82.54 = interest for 2 days
$|33§.18 = interest for 8 days
Explanation. To find the interest for 6 days, point off 3 decimals, as indi-
cated by the vertical line. Two days' interest is 4 of 6 days' interest. The
inswer is, therefore, 33^.
For rates other than 6%, see adjustments on page 80.
Problems
Find the interest at 6% on:
1. S286.75 for 9 days. 6. $175.82 for 34 days.
2. $189.22 for 8 days. 7. $38.95 for 19 days.
3. $256.35 for 27 days. 8. $47.56 for 17 days.
4. $178.56 for 39 days. 9. $29.10 for 2 days.
6. $38.29 for 40 days. 10. $1,286.75 for 21 days.
The cancellation method. The cancellation method may be
ised to advantage in many interest calculations, especially in those
laving fractional rates and rates other than 6%.
Example
Find the interest on $750.00 for 45 days at 5%.
Solution
125 15 .01
4 0
Explanation. Writing below the line 12 times 30, instead of 360 days, facili-
,ates cancellation.
Problems
Find the interest, by the cancellation method, on:
1. $840.00 for 12 days at 2%. 6. $284.00 for 34 days at 6%.
2. $320.00 for 15 days at 4%. 7. $368.00 for 56 days at 5%.
3. $160.80 for 16 days at 5%. 8. $775.14 for 79 days at 5%.
4. $275.75 for 74 days at 6%. 9. $250.00 for 91 days at 6%.
6. $112.50 for 85 days at 4%. 10. $500.00 for 102 days at 4%.
j^ Example
Find the interest on $345.75 for 96 days at 4i%.
Solution
Z
345.75 X 00 X .09 31.1175 00 oort
12X20X2 - -g- = *3'889'
80 SIMPLE INTEREST
Problems
Find the interest, by the cancellation method, on:
1. $360.80 for 3H days at 4£%. 3. $1,000.00 for 40 days at 5j%.
2. $312.32 for 45 days at 4|%. 4. $1,600.00 for 75 days at 4i%.
Dollars-times-days method, 6%. This method is rapid, and
is particularly valuable when a calculating machine is used. It is
a modification of the cancellation method, where 6% and 360 days
anxiwD .of tlie factors. Tims:
$ X Days X .00
300"
6,000
Assume that there are no other items that can be cancelled. The
number of dollars is multiplied by the number of days, and the
product divided by 6,000. Any number may be divided by 6,000
by pointing off 3 decimals, and dividing the resultant number by 6.
Example
Find the interest on $256.50 for 2S days at 6%.
Solution
Multiply the number of dollars by the number of days, point off 3 decimal
places in addition to the number of decimal places in the principal, then divide by 6.
$256 50
28
6)7 182 00
1 . 1 97 or $1.20, the interest
This method may be used for any rate by adding to or subtract-
ing from the interest computed at 6%, the fractional part thereof
that the specified rate is greater or less than the 6% rate.
For 8%, increase the interest by ^ of the amount computed at 6%.
For 7%, increase the interest by ^ of the amount computed at 6%.
For 5%, decrease the interest by i of the amount computed at 6%.
For 4%, decrease the interest by ^ of the amount computed at 6%.
For 4^%, decrease the interest by -j- of the amount computed at 6%.
The above adjustments may be used with any of the 6%
methods in solutions in which the rate is more or less than 6%.
Problems
Find the interest on the following:
1. $275.12 fop 73 days at 5%. 4. $138.42 for 28 days at 4£%.
2. S132.S& for 28 days at 8%. 5. $276.95 for 17 days at 8%.
3. 15280.60 for 70 days at 4%. 6. $640.64 for 56 days at 7%.
SIMPLE INTEREST 81
Interchanging principal and time. Under the 60-day method,
the computations may often be shortened by interchanging the
principal and the time.
Example
Find the interest on $6,000.00 for 31 days at 6%.
Solution
Interchanging the principal and the time, the problem becomes that of
finding the interest on $31.00 for 6,000 days. Apply the 6%, 60-day method,
and the interest is found to be $31.00, since the interest is equal to the principal
when the rate is 6% and the time is 6,000 days.
Problems
Find the interest on the following:
1. $2,400.00 for 23 days at 6%. 4. $3,000.00 for 193 days at 6%.
2. $3,600.00 for 7 days at 6%. 6. $4,500.00 for 3S days at Q%.
3. $6,000.00 for 156 days at 6%. 6. $4,200.00 for 41 days at 6%.
Exact or accurate interest. Exact or accurate interest is that
which is obtained when a year is taken as 365 days. For full
years, all methods of computing interest give the same result- a
certain per cent of the principal; hence the results differ only when
fractional parts of a year are used.
Example
Find the exact interest on $1,200.00 for 93 days at 6%.
Solution
The cancellation method previously explained is the method used, as it i&
probably the most practical.
240
K«»X 93 XJ>6 _ 1,339.20 _
cttltf — T o "~~ 'ff'AO.OtJ
MT^f' » «J
73
Problems
Find the exact interest on:
1. $750.00 for 45 days at 6%. 2. $1,200.00 for 68 days at 7%.
3. $1,600.00 for 73 days at 6i%.
Accumulation of simple interest. Simple interest accumulates
in like amount each period, if the principal and rate are unchanged.
Symbols. Let i equal the rate of interest, n the number of
periods, and P the principal. Then accumulation of simple
interest on any sum of money, for any number of periods, may be
found as follows:
82 SIMPLE INTEREST
Example
Find the simple interest on $100.00 for 5 years at 6%.
Algebraic Formula Arithmetical Substitution
P(l x in) = Interest. 100(1 X .06 X 5) = 30.
Solution
1 X .06 = .06, interest on 1 for 1 year at 6%
.06 X 5 = .30, interest on 1 for 5 years at 6%
.30 X 100 =- $30.00, interest on $100.00 for 5 years at 6%
TABLE OF SIMPLE INTEREST
(1)
(2)
(3)
(4)
(5)
Total Int.
End of
Interest Due
at End of
Sum Due it
Year
Principal
Each Year
Each Year
End of Year
1
$100 00
$6.00
$ 6.00
$106 00
2
100 00
6.00
12 00
112.00
3
100.00
6.00
18 00
118 00
4
100.00
6 00
24 00
124 00
5
100.00
6 00
30.00
130 00
Problems
1. A man borrows $500.00 for 9 years at 4%. What amount of interest will
he pay during this period? Write the formula and solution, as shown in the
example above.
2. What is the amount of interest due on $300.00 at the end of 10 years if
the rate is 7%? Write the formula and solution, as shown in the example
above.
3. Construct a table in columnar form, similar to the table above (omitting
column 5), for $400.00 invested for 5 years at 6%.
4. What is the interest accumulation on a debt of $4,270.00 for 8 years at
5% simple interest?
Simple amount. The simple amount is found by adding to the
principal the totaljimple interest. It is the amount due at the
end of the stated period.
Example
What is the amount of $100.00 for 5 years at 6%?
Algebraic Formula Arithmetical Substitution
P + [P(l X in)] - Amount. 100 + [100(1 X .06 X 5)] = 130.
Solution
1 X .06 = .06, interest on 1 for 1 year at 6%
.06 X 5 = .30, interest on 1 for 5 years at 6%
100 X .30 = 30.00, interest on 100 for 5 years at 6%
100 + 30.00 = $130.00, amount of $100.00 for 5 years at 6%
The simple amount is shown in column 5 of the table above;
hence the construction of a table is omitted here.
SIMPLE INTEREST 83
Problems
Write formulas and solutions for the following:
1. The amount of a $200.00 note due in 6 years; interest, 5%.
2. The amount due in 6 years on $530.00 at 6%.
3. The amount due on a note for $750.00 with 4 % interest. No interest has
been paid during a period of 4 years.
Rate. To find the rate, when the principal, interest, and time
are given, divide the given interest by the interest on the principal
at 1% for the given time.
Example
At what rate will $100.00 produce $24.00 in 4 years?
Algebraic Formula Arithmetical Substitution
Interest _ . , . . 24 fl
= Rate °f lnterest'
100(1 X .01 X 4)
Solution
1 X .01 = .01, interest on 1 at 1% for 1 year
.01 X 4 = .04, interest on 1 at 1 % for 4 years
100 X .04 = 4.00, interest on 100 at 1 % for 4 years
24 -f- 4 = 6, or 6%, the rate of interest
Problems
Write formulas and solutions for each of the following:
Principal
1. $ 400 00
Interest
$ 48.00
Time
3 years
Rate
2. 2,000.00
500.00
5 years
3. 800.00
336 00
6 years
4. 300 00
126.00
7 years
5. 150.00
40.50
6 years
Time. To find the time, when the principal, interest, and rate
of interest are given, divide the interest by the product of the
principal and the given rate for one year.
Example
In what time will $100.00 invested at 6% produce $24.00 interest?
Algebraic Formula Arithmetical Substitution
Interest _. 24
Time.
X in) 100(1 X .06 X 1)
Solution
1 X .06 = .06, interest on 1 at 6% for 1 year
100 X .06 * 6, interest on 100 at 6% for 1 year
24 -r 6 = 4, the number of years
84 SIMPLE INTEREST
Problems
Principal
Interest Rate
1. $1,000 00
$240 00 6%
2. 750 00
90 00 4%
3. 300 00
81 00 4i%
Time
Present worth. The present worth of a debt, due at some
future time, without interest, is the sum which must be invested
now in order to produce the specified amount at the end of the
period. Thus, since $1 invested for 5 years at 6% will amount to
$1.30, the amount that must be invested now at 6% to produce
$1 at the end of 5 years is 1.00/1.30, or $.7692.
To find the present worth of a sum, multiply the sum by the
present worth of $1.00 for the given time.
Example
What is the present worth of a note for $100.00, due in 5 years, without
interest, money being worth 6%?
Algebraic Formula Arithmetical Substitution
( ~ — ] = Present worth. 100 (-- /t * ^ \ = 76.92.
\1 -f (1 X m)J \1 + (1 X .00 X 5)/
Solution
1 X .06 = .06, interest for 1 year on 1
.06 X 5 = ..30, interest for 5 years on 1
1 + .30 = 1.30, amount of 1 for 5 years
1 -T- 1 .30 = .7692, present worth of 1 for 5 years
100 X .7692 = $76.92, present worth of $100.00 for 5 years
Verification
$76.92 X .06 = $4.6154, interest for 1 year on present worth
$4.6154 X 5 = $23.08, interest for 5 years on present worth
$76.92 + $23.08 = $100.00, amount due in 5 years
TABLE OF PRESENT WORTH
P
(1)
Years
1
2
3
4
5
(2)
Principal
$100 00
100 00
100.00
100.00
100.00
(3)
Divided by
Amount of $1
$1.06
1.12
1.18
1.24
1 30
(4)
Equals
Present Worth
$94 34
89 29
84.74
80 65
76 92
(5)
Principal Minus
Present Worth
Equals Discount
$ 5 66
10.71
15 26
19 35
23 08
Comparison of simple amount and simple present worth. The
following comparative chart is presented to illustrate the accumu-
SIMPLE INTEREST
85
lation of simple interest on a sum and on the present worth of the
same sum:
130
124
118
112
106
AMOUNT KXH
100
100
94.34
90-
89.28
84.74
80.65
r» 80-
PRRSKNT
76.92
WORTH
70-
60-
^~ '
" — — .
'— — ~__^.^
~~~ "^
r-^
, -—1
"""""'I
0-
BASIS OP
AMOUNT
BASIS OF
PRESENT WORTH
Jan. 1, Dec. 31. Dec. 31. Dec. 31. Dec. 31. Dec. 31.
IstYr. letYr. 2nd Yr. 3rd Yr. 4th Yr. 6th Yr.
Figure 1.
The amount starts at $100.00, and accumulates to $130.00 in
5 years. The present worth starts at $76.92, and accumulates to
$100.00. The rate of interest is the same in each case, 6%.
Problems
1. What is the present value of a 6-year note for $650.00, without interest,
if money is worth 5%? Write the formula and solution, as shown in the example
on/page 84.
' 2. A note for S3, 500. 00, without interest, is due in 5 years. What is its
present value, money being worth 6%? Write the formula and solution, as
shown in the example on page 84.
3/ Construct a table in columnar form, similar to the table on page 84,
(omitting column 5). Use $1.00 as the principal, 4 years as the time, and 5%
aaoihe interest rate.
4. Construct a comparative chart showing the difference in value of the
amount and the present worth of $400.00 due in 8 years, interest at 5%.
True discount. True discount is the difference between the
sum due and its present worth computed on a simple interest
basis. (See page 84.)
Example
Find the true discount on a debt of $100.00 due in 5 years, without interest,
money being worth 6%.
86 SIMPLE INTEREST
Solution
The present worth"of the debt is the sum shown in the solution on page 84,
or $76.92.
$100.00 - $76.92 = $23.08, the true discount
Refer to column 5 of the Table of Present Worth, page 84, for
the method of showing the true discount in columnar form.
Problems
1. What is the true discount on a debt of $750.00 due in 3 years, money being
worth 4%? Write the formula and solution.
2. Find the difference between the present value and the face value of a
non-interest-bearing note for $500.00, due in 4 years, money being worth 6%.
Write the formula and solution.
3. Construct a table in columnar form, similar to the table on page 84, for
$1.00 at 4% for 6 years.
4. Find the difference between the true discount and the simple interest on
$650.00 for 8 years at 4%.
CHAPTER 8
Bank Discount
Definition. Bank discount is a deduction made from the
amount due at maturity on a note or draft, in consideration of its
being converted into cash before maturity. If the note does not
bear interest, its face value is the amount due at maturity. If the
note does bear interest, the amount due at maturity is the face
value plus interest on the face value for the period and at the rate
specified in the note.
In bank discount, the time is the period from the date of dis-
count to the date of maturity of the note. The date of maturity
of a note is the day 011 which it is due. Notes due a given number
of days after date mature after the exact number of days have
elapsed. Notes due a given number of months after date mature
on the same date so many months hence, except notes made on tho
31st and falling due in a 30-day month, which mature on the 30th,
and notes made on the 29th, 30th, or 31st of some month and
falling due in February, which mature on the last day of February.
Example
A note due 30 days after January 31, will mature on March 2; but if the note
is due in one month, it will mature on the last day of the succeeding month, or
February 28. If the year should be a leap year, the maturity dates would be
March 1 and February 29.
Counting time. In counting time, the usual method is to
count the first succeeding day as one day. To illustrate, if a note
is given on January 15 for 10 days, the 16th is counted as the first,
the 17th as the second, the 18th as the third, and January 25 as the
tenth day.
Finding the difference between dates by use of a table. By
numbering the days of the year, a calendar may be made for
determining the number of days between any two dates. A
portion of such a table, and the use made of it, are illustrated on
page 88.
87
2
306
3
307
4
308
5
309
6
310
7
311
8
312
9
313
10
314
88 BANK DISCOUNT
May 1 121 Nov. 1 305
2 122
3 123
4 124
5 125
6 126
7 127
8 128
9 129
10 130
The number of days between May 4 and November 9 is found
as follows :
The table shows that November 9 is the 313th day of the year
The table shows that May 4 is the 124th clay of the year
Therefore, the difference is ISO days, the time required
Another form of table is one that shows the number of days
from any day of any month to the corresponding day of any other
month not more than one year later.
Jan. Feb. Mar. Apr. May June July Aug. Kept. Oct. Nov. Dec.
January ... 365 31 59 90 120 151 181 212 243 273 304 334
February . 334 305 28 59 89 120 150 181 212 242 273 303
March .. 306 337 365 31 61 92 122 153 184 214 245 275
April 275 306 334 365 30 61 91 122 153 183 214 244
May .... 245 276 304 335 365 31 61 92 123 153 184 214
June 214 245 273 304 334 365 30 61 92 122 153 183
July 184 215 243 274 304 335 365 31 62 92 123 153
August 153 184 212 243 273 304 334 365 31 61 92 122
September . 122 153 181 212 242 273 303 334 365 30 61 91
October. .. 92 123 151 182 212 243 273 304 335 365 31 61
November.. 61 92 120 151 181 212 242 273 304 334 365 30
December... 31 62 90 121 151 182 212 243 274 304 335 365
Example
A note due August 17 was discounted June 10. What was the term of
discount?
Solution
From the table, June 10 to August 10 61 days
August 10 to August 17 7 days
Total 68 days
Banks do not compute time uniformly, but the methods given
here are in common use.
NOTE: Tables similar to those above may also be used to
good advantage in computing the unexpired time of insurance
policies.
Proceeds. The proceeds of a note is the difference between the
amount due at maturity and the bank discount.
BANK DISCOUNT 89
To find bank discount and proceeds.
Compute the bank discount as simple interest on the amount
due at maturity for the unexpired time (term of discount). Deduct
the bank discount from the value of the note at maturity to obtain
the proceeds.
Example 1
Find the bank discount and the proceeds if a non-interest-bearing note for
$420.00 due in 90 days is discounted at 0^.
Solution
Face of note ... .' .............................. $420.00
Bank discount, 90 days ...................... (5 30
Proceeds .............................. 1413.70
Example 2
A note for $780.00 dated May 5, payable in 6 months with interest at 6%,
is discounted at 6% on August 3. Find the bank discount and the proceeds.
Solution
Face of note ................................ $780 00
Interest for 6 months at f)% ....................... 23 40
Value of note at maturity ..... ........ $803 40
Bank discount on $803.40 for 94 days ;it 6% ... 12 59
Proceeds ....................................... $790~si
To find the face of a note when the proceeds, time, and rate
of discount are given. Divide the proceeds of the note by the
proceeds of $1.00 for the given rate and time.
Example
For what sum must a non-interest-bearing note be drawn, due in 90 days,
so that when it is discounted at a bank at 6% per annum, the proceeds will bo
$537.40?
Solution
$0.015 = bank discount on $1.00 for 90 days
$1.00 - $0.015 - $0.985, the proceeds of $1.00
$537.40 -T- $0.985 = 545.58 times, or $545.58
Problems
Find the barik discount and the proceeds on:
J) A note for $750.00, due May 30 without interest, and discounted April 16
at 6%.
@ A note for $1,200.00, due December 4 without interest, and discounted
Oct. 29 at 6%.
(§/ A note for $1,500.00, dated October 8 and due in 4 months with interest
at 6%, discounted December 1 at 6%.
90 BANK DISCOUNT
AT) A note for $800.00, dated September 9 and due in G months with interest
at T%, discounted November 11 at 6%.
{€) A note for $250.00, dated July 11 and due in 90 days with interest at
5i%, discounted September 1 at 6%.
$443.03 wiis received a.s the proceeds of a 90-day note discounted at 6%.
the face of ttie note?
(ft For what sum must a 60-day note be drawn in order that the proceeds
will be $600.00 when the note is discounted at 6%?
(K\ Find the date of maturity, the term of discount, the bank discount, and
the proceeds of a 60-day note for $750.00, dated July 8 and discounted July 17
at 5%.
(§) Find the date of maturity and the term of discount of a 90-djty sight
draft, dated May 14, accepted May 17, and discounted June 10.
(iCy Find the date of maturity, the term of discount, the bank discount, and
the proceeds of a note for $050.00, dated Nov. 30, due in 3 months, and dis-
counted Jan. 5 at 0%.
CHAPTER 9
Partial Payments
Part payments on debts. A debtor may by agreement make
equal or unequal payments on the principal at regular or irregular
intervals. Any partial payment of a note or draft should be
recorded on the back of the note or draft.
Methods. There are two methods of applying payments of
principal and interest to the reduction of an interest-bearing debt.
The method adopted by the Supreme Court of the United States
is termed the "United States Rule"; the other method, which has
been widely adopted by businessmen, is termed the "Merchants'
Rule."
United States Rule. The United States Rule is now a law in
many of the states, having been made so either by statute or by
court decision.
The court holds that when a part payment is made on an
interest-bearing debt, the payment must first be used to discharge
the accumulated interest, and what remains of the payment is then
applied in cancellation of the principal. If the payment is smaller
than the accumulated interest, no cancellation takes place, and the
previous principal continues to draw interest until the accumulated
payments exceed the accumulated interest.
Example
An interest-bearing note for $1,800.00 dated March J, 1944, had the following
indorsements:
September 27, 1944 .... $500 00
March 15, 1945 2500
June 1, 1945 700 00
How much was due September 1, 1945?
Solution
Yr. Mo. Da. Yrs. Moz. Days
Date of note. . . 1944—3 1
First payment, $500.00 .... 1944— 9—27 6 26
Second payment, $25.00 .. . 1945-3 15 5 18
Third payment, $700.00 1945—6 1 2 16
Settlement 1945—9 1 3 0
I 6 0
91
S>2 PARTIAL PAYMENTS
Explanation. The time is found by successive subtractions of the first date
from the second, the second from the third, and so on. The sum of the different
times is equal to the time between the date of the note and the date of settlement.
Face of note, March 1, 1944 $1,80000
Interest on $1,800.00 at 6% from March 1 to Sept. 27, 6 months and
26 days 61 SO
Amount due Sept. 27, 1944 $1, SGI SO
Deduct payment ,500 00
Balance due Sept. 27, 1944... ^1~36T~SO
Interest on $1,361.80 at 6% from Sept. 27 to March 15, 5 months and
18 days, $38.13. As this inteicst is larger in amount than the pay-
ment made at March 15, the interest is not added and the payment
is not deducted.
Interest on $1,361.80 at 6% from Sept. 27 to June 1, 1945, 8 months
and 4 days 55 38
Amount due June 1, 1945 $1,417.18
Deduct sum of payments: March 15 $ 25 00
June 1 700 00 725 00
Balance due June 1, 1945 $ (>92~Ts
Interest on $692.18 at 6% from June 1 to Sept. I, 1945, 3 months • . _ 10 38
Balance due September 1, 1945 $ 702 56
Problems
1. A note for $1,650.00 was dated May 20, 1944. The interest was 6% from
date, and the following payments were indorsed:
Sept. 8, 1944 $ 45 00
Dec. 14, 1944 20 00
Feb. 26, 1945 5000
July 5, 1945 90 00
Nov. 14, 1945 25000
What amount was due December 17, 1945?
2, A note for $1,200.00 was dated June 20, 1941, The interest was 6% from
date, and the following payments were indorsed:
Oct. 2, 1941 $120 40
February S, 1942 .. 2950
May 23, 1942 5640
December 11, 1942 .. 38875
What amount was due January 23, 1943?
3. A note for $1,000.00 was dated April 10, 1938. The interest was 7% from
date, and the following payments were made:
November 10, 1939 $ 80 50
July5, 1940 10000
January 10, 1941 45080
October 1, 1943 . 50000
What amount was due January 1, 1944?
PARTIAL PAYMENTS 93
Merchants' Rule. Find the amount of the debt (principal and
interest) to the date of final settlement, or if the debt runs for more
than one year, find the amount to the end of the first year. Deduct
from this the sum of all the payments and interest on same to the
date of settlement, or to the end of the year. The remainder will
be the amount due at the date of settlement, or at the beginning
of the next year.
Example
For purposes of comparison, the same problem will be used here as was used
to illustrate the United States Kule.
Solution
Face of note, March 1 , 1942 $1,SOO 00
Interest, 1 year at (>% to March 1, 1943 I OS 00
$I,90S 00
Deduct:
First payment, September 27, 1942 $500 00
Interest at 6% to March 1, 1943, 5 months
and 4 days. ^12 S3 512 S3
Balance due at begi lining of second year $1,395 17
Interest on $1,395. 17 at (>%, March 1 to Sept. 1,
1943, 0 months 41 SO
*l~437l)3
Deduct:
Second payment, March 15, 1943 $ 25 00
Interest at 6% from March 15 to Sept. 1,
1943, 5 months and 16 days 09
Third payment, June 1, 1943 700.00
Interest at (>';< from June 1 to Sept. 1, 1943,
3 months __10 50 730 19
Balance due ~~ $ 700 ~~S4
The difference of $1.72 between the balance as computed by
the Merchants' Rule and the balance as computed by the United
States Rule is small, but a much greater difference will occur when
the time is long and the amount large.
It is usual to compute the balance due on notes of one year or
less by the Merchants' Rule.
Problems
1. A note for $950.00 with interest at 6% was dated Feb. 3, 1943, and had
the following indorsements:
March 1, 1943 $150.00 July 8, 1943 $300.00
June 3, 1943 96 . 00 December 20, 1 943 .... 250 . 00
What amount was due January 17, 1944?
2. A note for $791.84 with interest at 6% was dated December 14, 1942, and
bore the following indorsements;
94 PARTIAL PAYMENTS
January 3, 1943 $100 00 July 29, 1943 $324 00
March 16, 1943 240.00 Augusts, 1943 20.00
What amount was due November 14, 1943?
3. A note for $1,200.00 dated April 1, 1942, bore interest at 7% and had the
following indorsements:
April 12, 1942 $161 08 July 28, 1942 J 17 90
July 19, 1942 224 14 January 29, 1943 100 25
What amount was due April 1, 1943?
CHAPTER 10
Business Insurance
Kinds of insurance. There are at least 21 kinds of insurance
applicable to the ordinary business being done in big cities and as
many as 150 kinds of insurance covering all branches of human
endeavor.
Policy. An insurance policy is a written contract. The con-
sideration given for the protection promised consists of a premium
to be paid in money and the fulfillment by the insured of acts of
commission and omission according to the terms and conditions
set forth in the policy.
Fire insurance. Fire insurance is guaranty of indemnity for
loss or damage to property by fire. Insurance companies are
liable for loss or damage resulting from the use of water or chemi-
cals used in extinguishing the fire and from smoke. A fire loss is
predicated on the sound value at the time the loss is sustained and
not at the time the insurance is written.
Form of policy. With but a few exceptions, fire insurance
companies use a State standard-form policy made mandatory by
the State in which they operate. The New York State standard
form of policy is the one that is generally used, as it embraces
nearly all that is contained in other forms. The form attached to
the policy is known as a rider. £The rider form directly applies the
insurance to fit the facts and conditions of the particular risk. It
also amends the standard form, which is not a contract until com-
pleted by descriptions and amendments,
Rates. Probably no phase of insurance interests the business-
man more than his insurance rate. Independent rating bureaus
operate in different parts of the country. Their business is to
inspect and to measure the hazards in terms of rates. Rate
schedules are compiled for this purpose. The charges are in the
nature of penalties for hazards.
Example
A particular building has been inspected and surveyed by the rater. The
degree of municipal and local protection has been measured. This establishes
the basic rate. Assume the basic rate to be .40, which is a charge commensurate
with the degree of protection and covers all general hazards that cannot be
95
96 BUSINESS INSURANCE
segregated and measured. The better the city protection, the lower will be the
basic rate.
Basic rate 40
Area: 15,SOO sq. ft 04
(The standard unit area is 1 ,000 sq. ft., and an additional
charge is made for larger areas.)
Parapet wall deficiency 04
Skylights not standard construction 02
Metal stacks through roof . .... 06
Outside wood cornices, loading docks, and wooden conveyor. .06
Gallery decks used for storage ... 03
Occupancy hazard (woodworking mill) 92
Shavings allowed to accumulate 05
No cans for collecting waste. . . 05
No drip pans under machines 05
Floor oil-soaked . . 05
Total 177
Credit for open finish (inside walls) . . ... OS
Building rate unexposed 1 .69
Kxposuro:
From buildings No. 2 and No. 5 at IS ft 34
From building No. 6 at 15 ft .02
From office at 23 ft .05
From buildings No. 9 and No. 10 . . 07
Exposure charge .4S
Total building rate 217
If this assured would have the parapet wall brought up to the standard
requirements, his rate would be reduced .04. By having the shavings removed
daily, and by installing waste cans and drip pans under the machines, the rate
could be reduced another .15. As a matter of fact, the owner makes his own
rate — the rater simply measures the hazards in terms of rates.
To find the premium. Insurance companies charge a certain
number of cents or dollars for insuring each $100.00 worth of
property. Thus, the insurance rate in the foregoing example is
$2.17 for each hundred dollars of insurance carried. If the build-
ing is valued at $05,000.00 and is insured for full value, the amount
of the premium would be computed as follows:
$2 17 the rate per $100.00 of insurance.
X 6 50 the number of hundred dollars of insurance purchased.
$1,410.50 the premium, or cost of the insurance for one year.
Agent's commission. Local agents of the fire insurance
companies are located in almost every city and town. They act
as the representatives of the companies, >oliciiing the business
and collecting the premiums. For this service they receive a
certain per cent of the premiums.
BUSINESS INSURANCE 97
Example
A store building valued at $10,000.00 was insured for 80% of its value, the
rate being SI. 25 a hundred. What was thf agent's commission if he received
15% of the premium?
Solution
80% of $10,000.00 = $8,000.00, the insured value.
SO X SI. 25 = $100.00, the premium.
15% of $100.00 = $15.00, the agent's commission.
Cancellation of policies. Both the insurance company and the
insured have the right to cancel an insurance policy at any time.
When the policy is canceled by the insurance company, the por-
tion of the premium to be repaid to the insured is determined pro
rata.
Example
On April 10, the owner of a building insured his property for one year. The
premium was $36.00. On October 20, the policy was canceled by the insurance
company. What rebate did the insured receive for the unexpircd term?
Solution
The time from April 10 to October 20 is 103 days, expired term of the policy.
(See page 88 for table of number of days between dates.)
ill of $30.00 is $19.04, amount of premium earned.
$36.00 - $19.04 = $16.96, amount of premium returned.
When the policy is canceled by the insured, the amount of
premium returned is determined by the "short rate/' The short
rate is an arbitrary per cent fixed by the insurance companies, and
is shown by a table like the one at the top of page 98.
Example
On May 2, a one-year policy was written on a shop. The premium was
$38.75. On September 26, the policy was canceled at the request of the insured.
What rebate did the insured receive?
Solution
From May 2 to September 26 is 147 days. The table shows that 60% of
the premium is to be retained when the policy has been in force 150 days, which
is the number of days next higher than 147. Then 40% of the premium will be
returned
$38.75 X .40 = $15.50, the return premium.
Coinsurance. This is a form of insurance in which the person
who insures his property agrees to carry insurance equal to a
certain percentage of the valuation of the property. If he fails
to carry that percentage with an insurance company, he (the
98 BUSINESS INSURANCE
SHORT RATE CANCELLATION TABLE
Period exceeding 20 days and not exceeding 25 days, to be the rate of 25 days,
and so on up to one year.
Policy in Force
Per Cent
of A nnual
J'rem.
Pi
rticy
in Force
Per Cent
of Annual
Prem.
1 day
... 2%
55
days
.. 29%
2 days
.-* 4%
60
days
or 2
months. .
.. 30%
3 days
... 5%
65
days
.. 33%
4 days
... 6%
70
days
.. 36%
5 days
... 7%
75
days
. 37%
6 days
... s%
SO
days
. . 3S%
7 days
. .. 9%
85
days
.. 39%
S days
... 9%
90
days
or 3
months
40%
9 days
. .. 10%
105
days
.. 46%
lOdays
, .. 10%
120
days
or 4
months
. 50%
1 1 days
... 1 1 %
135
days
.. 56%
12 days
, . . . 1 1 %
150
days
or 5
months
. 60%
13 days
.. 12%
165
days
.. 66%
14 days
.... 13%
ISO
davs
or 6
months . .
. 70%
1 5 days
. ... 13%
195
days
.. 73%
1 6 days ,
.... H%
210
days
or 7
months
75%
17 days
.... 15%
225
davs
.. 7S%
IS days
.... 16%
240
davs
or S
months .
.. so%
19 days
. ... 16%
255
davs
.. S3%
20 days
.... 17%
270
days
or 9
months ....
... S5%
25 days
. ... 19%
2S5
days
.. ss%
30 days or 1 month ,
20%
300
days
or 1<
I) months
00%
35 days
... 23%
315
<lays
, .. 93%
40 days
25%
330
days
or 1
1 months
. 95%
45 days
• • 27%
345
days
.. 9S%
50 days
. - 2S%
360
days
or 1
2 months
. 100%
insured) becomes a coinsurer on the loss, in the ratio which his
lack of insurance bears to the amount he should have carried.
Illustration of SOC/0 coinsurance clause:
Case 1. Value of building and contents $75,000
Assured should carry S0% of value or 60,000
Insurance actually carried , . . 45,000
Loss by (ire 10,000
Paid by insurance company, 75% of loss, or. . 7,500
Assured must bear 25% of loss, or 2,500
Insurance carried was only 75% of what assured should have
carried to comply with the 80 % clause.
Case 2. Value of property $10,000
Insurance required S,000
Insurance carried 9,000
Losses up to $9,000 Paid in full
BUSINESS INSURANCE 99
Case 3. Value of property $10,000
Insurance required 8,000
Insurance carried S,000
Losses exceeding $S,000
Face of policy, $8,000, is paid.
Case 4. Value of property .. . $10,000
Insurance required 8,000
Insurance carried ... 5,000
Losses exceeding $8,000
Face of policy, $5,000, is paid.
Ixjsses under $8,000
Paid in the proportion that $5,000 bears to $8,000, or
f of the loss.
Problems
1. \Vh;it premium must be paid on a policy for $3,7(10 at $1.50 a hundred?
2. A house worth $12,000 is insured for J of its value for three years at $2.35
a hundred. How much is the premium?
3. An agent wrote a policy of $4,500 on a store building at a, rate of 85 cents
If the agent's commission >\as 15%, what was the amount of his commission?
4. Find the amount paid by the insurance company under the 80% coinsur-
ance clause in the following:
(«) (&) M (d)
Value of property $50,000 $75,000 $100,000 $200,000
Insuiance carried 40,000 00,000 80,000 80,000
Lossbyhrc J 0,000 45,000 40,000 40,000
Paid by insurance company .. .
5. You are presented the following tornado insurance plan and arc asked to
select one of the four policies and to decide upon whether to insure for one or
three years. In your opinion, what policy should be taken and for how long
a term? The sound value of the property to be insured is $1,242,000.
(^insurance
(I) None
(2) 50%
(3) 80%
(4) 90%
Amount of '
Insurance
$ 200,000
021,000
993,600
1.117.800
One- Year Three- Year
Rate Premium /fate Premium
20 50
.102
255
.0749 . ..
. 1872
.0678
1695
Compute the premiums at the one-year rate and at the three-year rate.
Find the average yearly premium on each policy at the three-year rate, and
make comparisons in order to determine which policy to accept.
6. A one-year policy on a dwelling was dated June 5. The premium was
$42.50. On October 1, the policy was canceled at the request of the insured.
Find the amount of return premium.
Use and occupancy insurance. This kind of insurance is pro-
tection against loss due to interruption of business by fire or
tornado. It is insurance against a loss that is suffered on account
of destruction of the property.
100 BUSINESS INSURANCE
The insurance recovery or indemnity is the profit that would
have been made if business had not been interrupted and, in
addition, the total of expenses that must continue during suspen-
sion of business. A business that is not profitable may be so
insured in order to recover the continuing expenses.
Generally speaking, use and occupancy insurance insures gross
profits plus the salaries of key employees kept on the payroll
account. The policy excepts payroll (other than that of the key
employees), heat, light, power, and expenses of maintaining prop-
erties not destroyed (such as taxes, depreciation, and maintenance
thereof). These items can be picked up by analyzing the running
expense accounts. In no event does the policy pay expenses
required to be insured unless it is proved that they continue after
the fire.
Coinsurance clauses are also applicable in use and occupancy
insurance.
A simple procedure to arrive at use and occupancy value for
the past twelve months is as follows:
Total Sales
Deduct:
Cost of Merchandise
(Opening Inventory + Purchases — Closing
Inventory)
Ordinary Labor Payroll ...
Light, Heat, and Power . . . .
Total deductions
Actual 100% use and occupancy value for the period
The foregoing procedure is predicated on the assumption that
all expenses other than ordinary payroll and light, heat, and power
will continue at the same cost as if the business were operating.
A more exact method is one considered in the light of a problem
in arithmetic or algebra, as follows:
Let x = Use and Occupancy Insurable Interest Each Day
a = Expenses That Do Not Continue During Suspension of Business
6 = Selling Price of Merchandise
c = Cost of Merchandise
d — Number of Working Days in the Month
Then:
Example
The expenses of a business for a given month were determined as follows:
BUSINESS INSURANCE 101
Part of Expense Part of Expense
That Must That Will Xot
Continue During Continue During
Item Total Suspension Suspension
Payroll $45,000
Salaries and Wages of Key Em-
ployees Who Must Be Retained. $20,000
Salaries and Wages of Employees
Not Retained ... * .. $25,000
Power . 750 — 750
Heat and Light 525 225 300
Leasehold . . 1,200 1,200
Advertising 1 ,725 725 1 ,000
Taxes .... . 950 050
Insurance . 1,375 500 S75
Interest ... . . 525 525
Other Expenses 1,950 S75 1,075
#54,000 $25,000 $29,000
Find the estimated amount of insurance to he carried for each day of the
month if sales are estimated to be $1 SI ,500 and cost of merchandise sold $109,500.
Average number of working days each month is 25.
Solution
b - c - a _ 181,500 - 109,500 - 29,000 _
d ~X 25 -1,7-0
Therefore, on the basis of estimates, $1,720 is the amount of insurance to be
carried for each day in the month.
The same result may be obtained in the following manner, using the esti-
mates given:
Sales for the Month $1 SI ,500
Less: Cost of Sales 109,500
dross Profit 72,000
Total Expenses . 54,000
Net Profit 1S,000
Add: Expenses That Must Continue During Suspension 25,000
Use and Occupancy Value for the Month . $ 43,000
43,000 4- 25 = 1,720 ~~ '
Problems
1. The gross profit of a business was $200,000 after charging raw materials
and payroll into manufacturing cost, but excluding light, heat, and power.
Continuing payroll of key men was fixed at $20,000. If the policy contained
the 80% clause, what was the required amount of insurance?
2. An audit of the expense accounts of the company insured in Problem I
showed that items that would not have to be continued after the fire totaled
$60,000. What amount would be collectible for a twelve-month period, other
facts being as stated in Problem 1?
3. Assume that it takes 15 months to rebuild the plant. How much insur-
ance would be collectible?
102 BUSINESS INSURANCE
4. If a manufacturer on a Sept. 30 fiscal-year basis had a fire on April 1
and it is shown by previous experience that the following six months are the
most profitable — in fact, that 66$% of the net earnings are made in that period —
would the adjustment take this into consideration, or would it be made on an
average for the year?
6. If the conditions in Problem 4 were reversed, what earnings would the
adjustment reflect?
6. Compute the use and occupancy value from the following data: Beginning
inventory, $ 1 70,4X2.00; ending inventory, $171 ,72 1 .77 , manufacturing — including
raw materials, labor, light, heat, and power, maintenance, depreciation, adminis-
tration, insurance, tuxes, interest, advertising, and all other expenses, $3,409,-
058.42. Fixed charges that arc included in the foregoing and that are expected
to continue are: administrative salaries, $35,200; interest, $4,X(K); taxes, 89,901 .22;
dues and pledges, $0,150; credit information, $235, insurance, $7, 91 X. 49; salaries
of office, supervisors, and foremen that will have to be retained, $237,075;
miscellaneous expenses, $42,395.02. Sales were $3, 551, 70S. SI.
Group life insurance. While this type of insurance is a part of
the subject of life insurance, it is presented in this chapter because
it is a common form of business insurance. The principles of life
insurance are presented in another chapter.
Group life insurance affords employees of a business with
ordinary life insurance at low cost so long as they are employed by
the particular employer, as the employer pays a part of the pre-
mium. The operation of this type of insurance is best explained
by an example of an actual plan.
Group Life Plan
1. Eligibility. The following plan of group life insurance is
offered to all present employees of the company who will have
completed six months or more of continuous service on November
11, 19 — , and to all new employees after they have been with the
company for six months,
2. Amounts of insurance. The amount of insurance available
to each employee under age 65, nearest birthday, will be based on
annual earnings as follows:
Class Annual Earnings Life Insurance
1. Less than $1 ,200 \ $1,000
2. $1 ,200 but loss than $2,200 1 ,500
3. $2,200, but less than $2,800 2,000
4. $2,SOO, but less than $3,200 2,500
5. $3,200, but less than $3,800 3,000
6. $3,800, but less than $4,200 3,500
7. $4,200, but less than $4,800 . ... 4,000
8. $4,800, but less than $5,200.. . ... 4,500
9. $5,200 and over 5,000
BUSINESS INSURANCE 103
3. Cost of insurance. The monthly cost of the insurance will
be based on the employee's insurance age on each anniversary date
of the plan, as shown in the following schedule :
Em ploycc 's Mon thly
Attained Age on Policy Contribution per
Anniversary Each Year $1,000 of Insurance
Age 44 and under $0 . 70
Ages 45 to 54, inclusive 1 . 00
Ages 55 to 59, inclusive 1 . 50
Age GO and over 1 .SO
Problems
1. Employee Y is 42 years of age and his earning classification is Class 5.
What is the monthly deduction for his insurance?
2. If Y were 14 years older, what would be the monthly deduction?
3. B is 40 years of age and earns $3,000 a year. How much insurance is
available to him, and what will be his monthly contribution?
4. Company A" insures each of its employees for $1,000. Under age 50 the
cost to the employee is 60 cents a month; at age 50 or over, the cost is $1.00 a
month. There are 54 employees, classified as follows:
Age Number
18 . 1
22 . . . . 6
25 . 10
29 .... 4
30 7
45 . ... 12
47 . . X
52 .. ... 2
56 . . 3
5S i
What is the amount of the monthly payroll deduction?
5. The manual shoWvS the cost of group insurance on a monthly basis to be
is follows:
Age Premium
18 .. . . . $ 51
22 ... 53
25 . 54
29 . . . . . 55
30 . .55
45 ... . . .80
47 90
52 1 26
56 1.71
58 • 2.00
With the number in each age group being that given in Problem 4, what is
the amount of insurance premium that is borne by Company XI
104 BUSINESS INSURANCE
Health insurance. Some plans are contributory and others
non-contributory. In either case, the benefits are much the same;
but in contributory plans the employee pays a part of the cost in
the form of a monthly premium deducted from wages, while in the
non-contributory plans the cost is borne by the employer. Few
businesses have their own insurance departments, most of the
plans being handled by insurance companies under a group plan.
Incapacities include sickness and non-occupational accidents
(occupational accidents being covered by Workmen's Compensa-
tion Insurance), but the employer usually reserves the right to
withhold benefits if the incapacity is the result of the employee's
misconduct or negligence.
The following examples are illustrative of the many ways in
which the factor of service is employed to favor the veteran
worker.
Example 1
Amount and Duration of
Length of Service Disability Itenejits
Under 2 years Such practice as the company may establish
2 but less than 5 years . . . Full pay 4 weeks, half-pay 9 weeks
f> but less than 10 years . Full pay 13 weeks, half-pay 13 weeks
10 years and over Full pay 13 weeks, half-pay 39 weeks
Example 2
Amount and Duration of
Length of Serriee Disability Benefits
1 but less than 10 years 50% of wages
10 but less than 30 years 75% of wages
30 years and over 100% of wages
Maximum: 26 weeks in 3 years
Example 3
Amount and Duration of
Length of Xermce Disability Benefits
6 months but less than 1 year 35% of wages; maximum: $14.00
per week, for 0 weeks
1 but less than 2 years . 50' 'v of wages; maximum: $20.00
per week, for 13 weeks
2 but less than 3 years. . 00% of wages; maximum: $24.00
per week, for 13 weeks
3 but less than 4 years 70% of wages; maximum: $28.00
per week, for 26 weeks
4 years and over 75% of wages; maximum: $30.00
per week for 26 weeks.
Problems
1. A was insured under the plan in Example 1. He was employed for 3 years
and became incapacitated for a period of 6 weeks. His average weekly wage
was $35.80. What amount of disability benefit, was he entitled to receive?
BUSINESS INSURANCE 105
2. B was insured under the plan in Example 2. He had been with the same
employer for 12 years. Two years ago he drew compensation for 8 weeks, and
last year for 12 weeks. This year he was again incapacitated for a period of
8 weeks. If his average weekly wage was $45.00, what amount of disability
benefit was he entitled to this year?
3. C was employed by an employer using the plan in Example 3, and had
worked for this employer for a period of 6 years. He became incapacitated
when receiving a weekly salary of $60.00, and was unemployed for 10 weeks.
What was the amount of compensation paid?
Workmen's compensation insurance. This type of insurance
is financial protection against loss of time for the wage-earning
group, resulting from accident and occupational sickness while on
duty. The cost is levied on the employer in the form of a premium
on the payroll classified according to the hazard of occupation. A
few states have their own Workmen's Compensation Insurance
Departments, but in most states the insurance is carried by the
insurance companies specializing in this type of insurance, gener-
ally referred to as casualty insurance companies.
Problems
Find the cost of workmen's compensation insurance on payrolls divided into
four classifications with respective rates as follows:
1. $239,530.39 (m .(HI per 0
75,535.62 © .519 per C
241,327.85 © .081 per C
99,791.48 © .586 per C
2. $272,584.07 ® .611 per C
91,856.68 © .5 19 per C
292,258.87 © .081 per C
148,735.42 © .586 per C
3. $254,248.83 © .581 per C
79,950.31 © .548 per C
272,368.08 © .085 per C
105,553.36 @ .564 per C
4. A deposit of $100.00 was made on a public liability policy. The payroll
audit was as follows:
$381,839.77 © .052 per C
294,212.15 © .026 per C
138,631.05 © .026 per C
What amount of additional premium was due on completion of the payroll
audit?
CHAPTER 11
Payroll Records and Procedure
Requirements. The term payroll records has gained new
significance since enactment of the Social Security Act and
more recently the Current Tax Payment Act of 1943. Formerly
each business handled its payroll system in accordance with its
own particular needs. Now payroll systems are becoming more or
less standardized in so far as certain information must be provided
in order to meet the tax requirements.
Requirements at the time of wage payments are that the
employer must deduct the taxes, both Federal Old Age Benefit Tax
and Withholding Tax. Tax legislation has not dictated the form
of records to be kept, but regulations have stipulated that certain
information must be available, and that a statement shall be fur-
nished the employee on or before January 31 of the succeeding
calendar year, or, on the day on which the last payment of wages
is made where employment is terminated before the close of the
calendar year. Records needed are best determined from the
reports required.
For operational purposes many employers give a pay statement
with each wage payment. The pay detail can be shown on a stub
attached to the pay check, on a duplicate of the pay check, on the
cash pay envelope, or on a separate slip.
Payroll procedure. Payroll procedure involves, first of all, the
production of the time .card, which the individual employee either
fills out or else inserts in the time recorder at stated times and
which therefore contains the basic information for other records.
The following forms contain information transferred from the
time card: the payroll summary sheet, the pay check or the pay
envelope (if a pay envelope is used, a pay receipt is also required),
and the individual employee's historical earning record.
Tjmebooks. Another method that is still employed to quite
an extent entails the work of timekeepers who keep time books.
The pencil or pen records that these timekeepers turn in show that
individual employees work a certain number of hours and fractions
of hours on particular work.
Time-clock cards. Time-clock cards provide a written record
of the time each employee is on duty. The employee's time card
107
108
PAYROLL RECORDS AND PROCEDURE
may be either a job ticket covering a single job, or a daily time
report recording all jobs worked upon, during the day. Attendance
records or the In-and-Out clock cards are the controls on total
time. A time-clock card must be prepared for each individual
employee for each day or each pay period, according to the system
•ifoppinif
time, anc
when coi
work reco
number oi
At the er
elapsed ti
from the
and check(
on the att
NO. 138 PAY
NAME Ed Wolper-3
ENDING 8/28/46
DEDUCTIONS
r o A B .'J
INC TAX 2 Of
imprinted with tl
the pay period d
number, employe
and possibly othei
ing data. It show
identification of tl
the employee, sta
ie day or
ate, clock
e's name,
" identify-
s the date,
ie job and
rting and
"j^to
2/-
TOTAL t 4 ^
12—
1
»755
2759
*7s?
5749
2
5803
3
4
21130
5
a 12oi
*12°3
51202
£ 12oo
No.
NAME
DAILY COST CA
138 R;
I Ed Wolper
RD
kTF / '*•
6
s!228
21224
*1229
51227
21226
7
8
TIME IMPRINTS
ii'-Mrn
QUANTITY
JOB NO
COST
9
^
10
*432
2435
*432
5438
£433
CO
11
AUG 23 164
4-
u.
2
27S-
+*
12
AUG 23 160
CO
13
AUG 23 160
1
to
+
+31
ff
14
AUG 23 15 1
IS
16
AUG 23 149
/
u.
<.
TO,
ff
?JM
g
7*
ft
Jf
ef
AUG 23 140
to
TO
OTAL DCC
Nt
AUG. 23 139
'3
u.
t
S9,
/O
UCTK
AUG 23 126
to
PAY
AUG 23 120
7
*•
+
t//
77
:ime, rate, elapsed
I amount earned
npleted. Piece-
rds also show the
f pieces produced,
id of the day the
Line is computed
clock registration
ed with that shown j
endance records.
AUG 23 113
CO
AUG 23 112
<*
u.
,.
7^2
/ 32
AUG. 23 100
to
AUG 23 100
+
«*.
j
AUG 23
96
v>
AUG 23
95
7
u.
,-r
216
77
AUG 23
88
w
AUG 23
87
7
u.
<*
,17
77
AUG 23
80
v>
TOTALS
72
792
DATE r/** FOREMAN 0 K&&*~~~/
i and Out Clock Card and Daily Cost Card
The illustration is that of the weekly In-and-Out clock card
from which the payroll is prepared and the daily cost card used
for cost accounting purposes. The In-and-Out clock card illus-
trated shows the exact minute of entering and leaving. Some
clocks register this time in tenths of an hour instead of the exact
time. The daily cost card shows hours and tenths, beginning at
the bottom and reading toward the tog, so arranged to facilitate
computation of elapsed .timA. The closing hour, 16.4, is 24 minutes
past 4 o'clock.
PAYROLL RECORDS AND PROCEDURE 109
The In-and-Out card shows 8 hours' elapsed time on Monday,
the factory hours being from 8 A. M. to 12 M., and 12:30 P. M. to
4:30 P. M. Of course, it is impossible to check in and out on the
specified hour, and a certain tolerance is allowed. Different com-
panies have varying rules regarding tardy registrations. Wage
and Hour inspectors object to too early registration, and more
than 15 or 20 minutes early is likely to be counted as overtime.
The daily cost card shows 7.2 hours' productive time; therefore,
the difference of .8 of an hour is nonproductive time. A recon-
ciliation can be made showing where the .8 of an hour was not
utilized on productive work.
. 1 Between Jobs 837 and 266
. 1 Between Jobs 266 and 490
. 1 Between Jobs 722 and 61 1
. 1 At noon (Starting time being 12,6 instead of 12,5.)
. 1 Between Jobs 598 and 701
.2 Between Jobs 701 and 431
. 1 At close of day (Finishing time being 16.4 instead of 16.5.)
.8 Total lost or nonproductive time
Deductions. Fixed or standard deductions, such as group
insurance, employees' benefit association dues, hospital service
dues, and so forth, can be entered at the time the caid is made up.
Federal Old Age Benefit Tax and Withholding Tax cannot be
entered until earnings are computed. ^
At the present time, F. 0. A. B. Tax is t% of earnings. A
portion of the Withholding Tax schedule effective January 1, 1946,
that for a weekly payroll, is presented for use with the problems.
Withholding exemptions. For income tax computations, the
personal exemption is on a per capita basis; therefore, withholding
exemptions are on a pcr-ciapita-basis as follows:
A single person is allowed one exemption.
Husband and wife have two exemptions: if both are working, either spouse
may take both exemptions or each may take one; if one is not working, the
other may take both exemptions.
One exemption may be taken for each dependent (a person whose income is
less than 3#00 a year, who is closely related to the taxpayer, and for whom
the taxpayer provides more than one-half the support).
The number of exemptions claimed determines the proper
column to be used in the wage-bracket tables in determining the
tax to be withheld.
Employees' names on time cards and payrolls are marked to
indicate the number of withholding exemptions claimed. Find
the employees' earnings in the two columns at the left; where this
line intersects the exemption column, the amount of tax to be with-
held is shown.
If the payroll period with respect Co an employee in WEEKLY
At g*
••** I tZ-Z
And me number of wttMioldIn0 exemption* claimed t»—
il'II
12 ____
13 . .
$14 ____
$11.
12
13-
15
% of, 1
»~« $0 $0
$2.001 .10 o
2.1Of 3Q O
2 3O .50 O
2.5O ,7O| O
$0
O
O
$0
o
o
o
o
$0
O
O
8
if;::
1 2O
1 30
t20
21
...
8:::
$21
$22
170
1.80
20O
220
2.40
.20
.40
.50
$25
26
27
28
(29
$30
25O
27O
29O
300
320
.70
.90
1 OO
1 20
140
33
$34
$31
$3?
$33
H4
$35
60
5 70
590
340
3 60
370
390
4.10
2 20
.,„
.20
.40
$35
f 36
$37
$40
6 10
62O
t4O
60
680
420
4 4O
4 60
480
4 90
§4O
60
280
290
3 10
1-30
$4O
$41
f 42
$43
$44
$41
$42
143
$44
$45
5 1O
5 3O
5 40
330
3 40
3 60
380
400
1 40
1 60
1 8O
200
2.10
X|
$47
$48
$49
$46
$47
$48
$49
$5O
780
8OO
820
23O
.50
.60
.80
1 OO
1.2O
1!
$54
$51
$52
$53
$54
$55
18
6 BO
7OO
720
7 30
7 50
SCO
6 2O
5 30
550
5 TO
§20
30
IPS
380
1.30
1 bO
1 7O
1 80
20O
$55
$56
$57
ill
$56
$57
$58
$59
$60
9VO
90O
10 1O
10 3O
1O5O
$60
$62
(66
$68
$62
$64
$66
168
*/0
1O7O
11 1O
11 5O
Ulg
770
79O
8 10
580
too
70
6 3O
6.50
400
4 20
4 40
450
4.70
2 ?0
2 40
250
~ TO
2.90
.40
.50
.70
.38
1
2
2 3O
27O
.50
.80
13
(74
(76
$78
12 GO
13 OO
13 40
13 TO
14 10
1060
11 OC
11 4O
11 TO
12.1O
1O.1C
48O
52O
5 50
590
€20
300
330
3 TO
4.OO
4.40
1 20
1 5O
1.9O
2 2O
25O
$«O
$82
(84
14 5O
138
1560
16 OO
1050
1O9O
11 20
J1
50
«• -
47O
118
57O
6-10
* .
290
3 2O
3 6O
»1OO
*105
$UO.
$115
$120.
!$105
*110.
$115.
.'$130
,$125.
18 SO
1950
20 4O
21 4O
165O
1750
18 4O
19 4O
20 JO
12 4O
1330
14.30
290
380
SJ8
6-30
$; t
$1 .
• •
Si ft $•
:«,) . $-
21 2O
22 2O
23 1O
24 10
2500
1920
20 2O
21 1O
2200
23.00
IS*
£i§
2100
ns
18 __
1900
17X»
11 2O
12 1O
13 1O
14 OO
15-OO
9-20
10 10
11 1O
12 CO
12^90
is
11?
7JOL
.
$••«
s. •«)
SirtO
$: w
$20O and over.
28 4O
30 3O
32 2O
26 4Ol
28 3Ol
-yi ?n!
24 4O|
26 30
9* **)'
'Ki >'
-
22.40
24 30
26 2O
28 OO
2990
18 4O
2O 3O
22 1O
24 .OO
25.90
16,40 14 4Oi
18 2O 16 2O
2O 1O 18 lO
22.00 20.00
23-90' 21 SO]
1O3O
12 2O
14 1O
12, 1C
14.0C
*5 ac
1» percent off tt»*
over S2OO
72.801
18-80! 16.8C
PAYROLL RECORDS AND PROCEDURE
111
Example
Brown's earnings are $29.00 for the week, and his number of withholding
exemptions is 2. What is the amount of Withholding Tax?
Solution
In the columns at the left find the bracket $29.00 to $30.00, and follow across
to the intersection of the column headed "2." The tax is found to be $1.40.
Problems
1-2. Complete the following weekly time cards. Notice that the daily
recordings are made across the card instead of vertically, as in the illustration on
page 10S. Make the extension of elapsed time at the extreme right of the card.
1.
PAY ENDING 10/25
Nt. 24
NAME Garret Knlgbt - 3
IN
OUT
IN
OUT
IN
OUT
M 7 57
M1201
M1259
M4 05
TUSOI
TU1202
TU1258
TU402
W 7 53
W1202
W1257
W 3 59
TH?59
TH1201
TH 1 01
TH404
FR75«
FR1203
FR1256
FR402
' '
SA758
SA1210
FOA.
WHD
O £"
DEDUCTIONS
r • ,
TAX
MP<: ^'J/« ^ >UT J'> "/)
Z~J Q~ 2S-
TOTALS
I '
t
TOTAL DFt>iJCTlON«»
TOTA
L *
NET PAY
$
PAY
NO. 25 ENDING Sept. 24
NAME Alice Vharton - 1
IN
OUT
IN
OUT
IN
OUT
M 8 2»
M 1203
M1259
M 4 4C
TU 8 25
TU1205
TU1287
TU4 60
W 8 28
WH M
W12S5
W4 46
TH 832
TH1201
TH 1255
TH4 55
FR 8 17
FRl2oz
FR 1 01
FR 449
SA 830
SA1210
roj
WHC
INS
TOT
DEDUCTIONS
i a
MRS • -*~* • AMT
TV
TOTAL PAY
TOTAI OfOUCT'GMt
/(.
M *
NET MY
ft
112
PAYROLL RECORDS AND PROCEDURE
3-4. Complete the following semi-monthly time cards. A section of
Withholding Tax schedule for semi-monthly pay periods is given to enable
to compute the tax:
A section of the
you
$54 to $56 SI . 50
56 to 58 . . . 1 80
68 to 60 2 20
3.
PAY
NO. 16 ENDING Oct. 31
NAME H«rry Burr - 2
DATE
IN
OUT
IN
OUT
IN
OUT
16
^
M 7 32
M1204
M 12^9
M 4 02
*/>
TU 749
TUl2ot
TU1253
TU 4 03
y»
W 7«9
W1204
w!2».
w 4 oo
*/*>
TM7 SB
TH1203
M12M
T«4 04
^
PR 8 oo
r*12oo
rnl23»
r* 401
%
3A79*
M
« 105
M
&
%
M 7 3*
Ml2ot
M lOO
M 4<»*
%
TU 745
T012C3
TU 1259
TU 4 04
%
w 7M
w 12o.
w 12 IT
w 4o«
%
TH 7 54
«12o<
Wl24«
Tn4 OX
%
rn 7 49
r« 12oa
rni2M
fA4°l
%
SA 7 59
W 1.3
%
%
W7W
M 12oi
M 1257
M 4 03
M 4'«
M 9 34
FO/
WHC
0
DEDUCTIONS
HRS « .— ^*r- AMT
TAX
TOTAL P
TOTAL otoucTior
NET PAY
H
c
j^
TOT
IL f
$
PAYROLL RECORDS AND PROCEDURE
113
The following problem contains overtime, which is to be computed at
time and one-half. At the bottom of the time card, enter the regular hours on
the first line and the overtime hours on the second line. You will notice that
f;he rate has already been adjusted to one and one-half times the regular rate.
PAY
NO. 14 ENDING Sept. 30
NAME Richard Knight - 2
DATE
IN
OUT
IN
OUT
IN
OUT
16
FR 7 50
FR1205
FR1230
FR 4 35
X
X
x>
M 7 58
M1202
M1229
M 4 33
>20
TU 7 57
TU1200
TU1229
TU4 30
TU 4 59
TU 7 02
X
W 7 49
W1203
W 1228
W 4 38
X
TH 8 13
TH1207
TH1229
TH 4 37
%
FR 749
FR1201
FR1230
FR 4 39
%
x>
,%
M 7 59
M 12 00
M1230
M 4 34
X
TU 8 00
TU1202
TU1230
TU4 30
^
W 7 58
W1203
W1228
W 4 37
%
TH748
TH1205
TH1230
TH4 38
%
FR 7 50
FR1200
FR 1232
FR 4 22
%
rot
WHC
INS
TOT
SEDUCTIONS
A
HRS 9 —
f//
TAX
TOTAL PA
TOTAL DEDUCTION
NIT PAY
>— -
ILt
Payroll sheets. Whether payment is made by cash or check,
listings of payments to employees are made on payroll sheets for
record purposes, and the hours worked, rate of pay, gross earnings,
deductions, and net pay are shown for each employee.
Piecework system. Quantity of work produced rather than
number of hours worked is the basis of earnings under a piecework
system. Under this system there is an incentive for the skilled
worker to produce more, thereby increasing his earnings. The
payroll is designed to record the number of pieces produced rather
than the number of hours worked; hence, no provision need be
made for overtime.
114
PAYROLL RECORDS AND PROCEDURE
1
if
gl,
o>
3
o
0
a I"
Q> O
"E °
S °
o^
Oo
c
"8
a
I
CQ
•<
d
Thu
v
Exemption
Name
fe|**1
G^CJ
^
. 0)
d
PC
0
c;S
O
•£^«s
£-S9
<»? CO
PAYROLL RECORDS AND PROCEDURE
115
£
03
P,
•o «
-g %
•S 2
a;
3 S
O
d
'a ^
00
d^
pq
6
i
H
W
o
aj
COrH
!2 CO
CSJ
S
(E
'h
^•r
S
116
PAYROLL RECORDS AND PROCEDURE
O T3
£"0
Ss"*
-.2
O +•*
<D '
2
o3
^ o
pt
0) ^
•* 2
(D >>
s s.
.S3
SO)
O
2 *
03 .2
"-P x
o5 ^
o ^
73 fl
S2
So
.s
'5
art g-
C5 QJ CO
"Id
No.
TIME SH
and
PAYROLL SU
Q
PQ
« s
'
H
tl
vo
CO
^ M
H K
CO
CM
CM
rH^
-P
0-H
O 6
& CO
PQPn
<H;
to
00
IOLO
CM CM
rH rH
$$
£*
PAYROLL RECORDS AND PROCEDURE
117
tt
c
^c
s
o
c
O
CO
OT
ss
•^
o
H
I-
JM
Z X
j^oh
<OS^K
S
00
l«3
*S?
*>£
<*
^01
^14
fclfclsl
(Slw-V
ci
118
PAYROLL RECORDS AND PROCEDURE
Pay checks. After the payroll has been figured, the next opera-
tion is the writing of checks. Two types of payroll checks are
illustrated, the stub portions on this page, and the check por-
tions on the following page.
Form A
BOONE COMPANY,
BOONEVILLE, MO.
Statement of Employee's Earnings and Payroll Deductions
RATE
TIME WORKhD
AMOUNT
EARNED
OTHFR
COMPEN-
SATION
TOTAL
AMOUNT
TAXABLE
DEDUCTIONS
Days
Hours
Standard
Hours
Tax
Federal
Old AKC
Tax
Miscel
1.00
5
40
40
40.00
4.00
44.00
2.10
.44
1.51
This is your statement of earnings and receipt for deductions as required by law.
Save it carefully as it is the basis of any claim for Unemployment Insurance or
Old Age Pension.
Form A is the typo produced on a bookkeeping machine
writing from the time cards and making the payroll chock, pay-
roll summary sheet, and earnings record in a single operation.
Form B
Pay-Roll R«milUoc« Voucher
Employ** S. S.
Acct No .
341
Herman Shultz - 3
SEP 15 19
Hours nr\
Worked Reg fc*U Overtime
Amount Earned
90.00
••*«*».
Fed. Old-Age Tax
.90
Inc Tax
3.70
Gr. Ins
1.00
Misc
5.30
TXHIltuBlll.
10.90
•rttaeiMhtt
79.10
DETACH AND RETAIN THIS VOUCHER
Marter Industries, Inc.
PAYROLL RECORDS AND PROCEDURE
119
Form A (Continued)
BOONE COMPANY,
BOONEVILLE, MO.
PAYROLL CHECK
i
I DATE
AUG19
CHECK NO
56793
PAY TO THE ORDER OF
D.C. DANBURY 364 09 1898 EXACTLY $ 39.95
Not Good for Over Two Hundred Dollar*
\ FIRST NATIONAL BANK
| BOONEVILLE, MO.
89-11
Thit chick not valid ttnle$* prettnted for payment within 60 day* from date ofiuue.
Treasurer
Form B (Continued)
:-~ • • • us-" ' * . • !~ ' :.. It : •: 'i. • ' " :«: .iTrrnSn'S'i? .I:.::' ui- -jtii:!" i it " . '
DATE SEP 15 19
Master Industries, Inc. No. 2317
JlwnujaflwiviA JloUonaf JKan£ Lecenter, Minn.
Lecenter, Minn.
PAY Seventy-nine and 10/100 $ 79.10
TO THE
OF Herman Shultz - 3
PAY CHECK
Master Industries, Inc.
BY.
120
PAYROLL RECORDS AND PROCEDURE
Form B is the type of check produced on a typewriter from data
on the payroll summary sheet. Three operations are required;
preparation of the payroll summary sheet, writing of checks,
and, finally, posting to employees' earning records.
Pay envelopes and receipts. Some factories pay their employ-
ees by cash instead of by check. In such cases, pay envelopes and
pay receipts are used.
Employ**'*
Nam*
PAY-ROLL RECEIPT
Lee Spence — 2
Employ**'* .Q1 c7flnet Company .
S S. Acd. N«.191-57-8055 rwi No.l^f
R.<*iv.d From Ace Sales Company
Earning, to Au«U9t 20 19
Hour.
Worked. Regular *U Ov*rtim*
Amount Earned
Commission
$ 21.00
$ 15.00
Total
36.00
DEDUCTIONS:
F.d.ral Old-Ag*
Ann'ty Tax @ ^% * -
Inc. T« _
$ 2.60
Group Int..
$ .50
Total Deduction* * 3*46
N*k Amount H*r*with $ 3g>54
The flap, printed with the remittance data, when signed by
the employee becomes a receipt for wages. The face of the enve-
lope, printed the same as the flap, contains a carbon copy of the
data and is the employee's permanent record.
Coin sheet and currency memorandum. Where employees are
paid in cash, each employee receiving an envelope containing the
exact amount of his net earnings, it is necessary to prepare a
currency and coin sheet in order to have the correct number of
units of each denomination.
PAYROLL RECORDS AND PROCEDURE
121
^
J/
Name
Curiency and Coin Sheet
Net
Earnings
Hi
21
2/3
S+
Currency
20
10
Coin
50
25
10
05
01
A currency memorandum,
made up from the foregoing, is
taken to the bank to enable
the paying teller to make up
the amount of money required
in different denominations.
PAY ROLL
Pitt "
FOR _. -_
Currency
•tUAM
"«i^""'"
«v.
?"• ^
/<r^
00
I0*i *
/o
o o
JLS
5 • *^
JC
0 0
i».
fiil^^
so
O *
H.1v^ -/
X
oc
X
/
CO
nimM 4r
f*
Nickel. V^
/o
Pennle* </
*/
TOTAL
X/J
r*/
Problems
Rule currency and coin sheets and complete them for Problems 1, 2, and
3, pages 114, 115, and 116.
CHAPTER/12/
Average
Simple average. The simple average of a group of items is
determined by adding the items to be averaged and dividing the
sum by the number of items added.
Example
From the following statistics, find the average rate per kilowatt hour for
electrical energy:
New England States 2 «SSf*
South Atlantic States 2 77^
Atlantic States 2 I Of4
North Central States . 1 88?
Pacific Northwest . . 181^
Solution
2.88 + 2.77 + 2.19 + 1.88 -f 1.81 = 11.53
11.53 -r- 5 = 2.306
Explanation. The number of items to be added is 5, and the sum is 11.53$£.
1 1.53 divided by 5 equals 2.306, therefore, 2.306^ is the average rate per kilowatt
hour.
Problems
1. The following delivery record shows the number of deliveries made each
day by the live trucks of the delivery department:
TRUCKS
DAY No. 1 No. 2 No. 3 No. 4 No. 5 TOTAL AVKUAUK
Monday 242 320 271 141 243
Tuesday 217 328 393 182 218 J / L
Wednesday .. 256 290 296 120 325 ,„. .. /..
Thursday . . 302 289 344 149 297
Friday 293 306 301 216 218
Saturday . . 317 365 423 227 303
Total .... I '
Average. . . «2.7/
(a) Calculate the total number of deliveries made by each truck, and the
daily averages for the week (vertical columns).
(6) Calculate the total number of deliveries made each day, and the average
number of deliveries per truck (horizontal lines).
123
124 AVERAGE
2. The monthly output of motor cars and trucks for one year was as follows:
January 231,728
February 323,796
March 413,327
April 410,104
May 425,783
June 396,796
July 392,076
August 461,298
September 415,285
October . . 397,096
November 256,936
December ... . . 233,135
Total .... . ..: 2%
What was the average monthly output for the year?
3. The sales of five clerks on a certain day were as follows:
A $356 80
It 438.90
C 395 60
1) 410.85
E . 440.90
Total ...
(a) Find the average sales.
(6) Which clerks sold above the average?
(c) Which clerks sold below the average?
Moving averages. Moving averages are a series of simple
averages of statistics applicable to groups of an equal number
of time units, each successive group excluding the data for the
first time unit of the preceding group and including the data for the
time unit immediately following those of the preceding group.
For example, a yearly moving average, by months, may begin with
an initial group including the data applicable to the twelve months
of 1943. The next group would omit the data applicable to Janu-
ary, 1943, and include the data applicable to the remaining
eleven months of 1943 and that applicable to the month of January,
1944.
Example
The labor costs in a certain manufacturing plant for the first six months of
1943 were as follows:
January $3,363. 17
February 3,644. 15
March 4,472.90
April 3,209.20
May 3,415.40
June 4,152.05
AVERAGE 125
The labor costs for the next two months were:
July ............................................. $3,824.06
August .......................................... 4,015.25
What has been the average labor cost for each six months since January I,
1943?
Solution
The labor cost for the period from January 1 to June 30 is the sum of the
labor costs for each of the six months, or $22,256.87. The average for the period
is $22,256.87 -^ 6, or $3,709.48.
The average for the period from February 1 to July 31 is computed as follows:
Total: January 1 to June 30 ....................... $22,256.87
Deduct: January labor cost ........................ 3,363 17
$T8~893770
Add: July labor cost .............................. 3>8?4_L()6
122,717 76
$22,717.76 -^ 6 = $3,786.29
The average for the period from March 1 to August 31 is calculated in the
same manner:
Total: February 1 to July 31 ....................... $22,717.76
Deduct: February labor cost ....................... 3,644 15
$19,0731)7
Add: August labor cost .......................... _ 4,015 25
$23,088.86 -T- 6 = $3,848.14 - -____-
Comparison of these averages, $3,709.48, $3,786.29, and $3,848.14, shows an
increase for each period.
In permanent records, these averages should be tabulated.
Moving Increase or
194S Labor Cost Average Decrease^
January- June .............. $22,256 87 $3,709 48 $ ..........
February- July ............. 22,717 76 3,786 29 76.81
March-August ............. 23,088 . 86 3,848 .14 61 . 85
t Indicate decreases by means of daggers.
Further comparisons, based on the figures of prior periods,
may be made in succeeding years. A column may be annexed
to show the increase or decrease of the average of each six months'
period compared with the simple average for the preceding year.
Another column may be used to show the increase or decrease in
the moving average for the current six months' period compared
with the moving average for the same period of the preceding
year.
126 AVERAGE
Problems
1. Below are stated the labor costs, for the succeeding months, of the com-
pany in the preceding example; compute the moving averages.
September $4,275. 60
October 3,981 28
November 4,013 75
December 4,010 80
2. Using the averages obtained in Problem 1, complete the tabular record
for the year 1943.
3. Find the simple average for the year 1943.
Progressive average. The method of progressive average is
cumulative. The results of the latest period are added to the
total previously computed, and the amount is divided by the
previous divisor plus 1.
Example
Department A sales were: January, $5,364.00; February, $4,872.00; March,
$5,024.00. Department B sales were: January, $2,561.00; February, $2,325.00;
March, $2,753.00. Find the progressive monthly averages.
Solution
SALES RECORD
Dept. Jan. Feb. Total Aver. March Total Aver. April
A 5,364 4,872 10,236 5,118 5,024 15,260 5,087
B 2,561 2,325 4,886 2,443 2,753 7,639 2,546 etc.
Explanation. Department A sales for January and February total $10,236.00.
$10,236.00 -h 2 = $5,118.00, the average for the two months. $10,236.00 +
$5,024.00 =» $15,260.00, the total sales for the three months. $15,260.00 -5- 3
= $5,087.00, the average for the three months. The record for the year would
be completed in this manner.
The totals and averages of Department B are computed in the same way.
Problem
Using the above record and the following information, complete the record
('or the six months' period.
Department A sales:
April $5,986.00
May 6,125.00
June 6,398.00
Department B sales:
April 2,482.00
May 2,593.00
June 2,715.00
Periodic average. Periodic average is simple average applied
for several periods to statistics applicable to the same unit of time.
AVERAGE 1*7
It may be used to show a variation in expenses, earnings, sales,
and so forth.
Example
EXPENSES
Month 1943 1942 1941 1940 Total Average
January $478 60 $392 cS5 $429 65 $356 00 $1,657. 10 $414 28
February 462 37 529 83 531 .33 535 35 2,058.88 514 72
March 347 92 629 89 432 45 567 89 1,978 15 494 54
Explanation. The expenses for January for the four years are totaled; the
total, $1,657.10, divided by 4, the number of years shown, equals $414.28, the
average monthly expense for January. The other averages are calculated in
the same manner.
Problem
The output of a factory for the first quarter of the years 1943, 1942, 1941,
and 1940 is shown in the following table:
Month 1943 1942 1941 1,940 Total Average
January 231,728 238,908 309,544 240,592
February 323,796 304,735 364,180 283,577
March ... . 413,327 394,513 434,470 374,425
Compute the periodic average.
Weighted average. Weighted averages take into account not
merely the number of units to be averaged, but also the value of
each unit.
The average-price method of pricing requisitions in cost
accounting is illustrated in the following example.
Example
A stock record shows the following receipts:
4,800 Ibs. © 20^
3,000 Ibs. @ 18£
4,000 Ibs. @2\i
What is the average price per pound for the month?
Solution
4,800 Ibs. @ 20^ = $ 960 00
3,000 Ibs. @ 18£ = 540.00
4,000 Ibs. @ 21 £ = 840 00
11^800 Ibs. = $2,340" 00
2,340 -r 11,800 = 19.83, or 19.83^ per pound, the average price.
Example
A manufactured product is composed of four ingredients, the relative pro-
portions and costs per pound being as follows:
128 AVERAGE
Material Pounds Price per Pound
A I *l 50
B".. . 3 75
C 4 1 25
D 2 2 00
It was found in the second year that owing to price fluctuations, the raw
material costs had increased as follows:
Material Per Cent
A . 50
B ..... 100
C 10
D . 25
What was the average per cent of increase in the cost of raw material com-
posing the finished product?
Solution
Per Cent
Cost Total
Material
A
B ..
C . .
D.
$12.75 $4 50
4.50 -I- 12.75 = 35.29%, the weighted average per cent
Problems
1. A stock card shows the following receipts:
3,000 Ibs. © 2S^
2,000 Ibs. <$ 27ff
1,500 Ibs. (a) 29jS
4,000 Ibs. fe 30^
What is the average price per pound?
2. X owns the following securities:
$3,000 Power Corp. Si's
$1,000 Alabama Company 6's
10 shares Northern Power Co.,
7% Preferred Stock, Par Value,
$100.00 a share
$2,000 Union Depot Co. 5's
What is the average rate of interest earned on X's investment, assuming; that
the securities were bought at par?
3. A product is manufactured from the following materials used in the relative
oroportions given:
Cost
Total
Price
Increased
Pounds
per Lb.
Cost
Increase
Cost
1
$1.50
$ 1 50
50
$ .75
3
.75
2 25
100
2.25
4
1.25
5.00
10
.50
2
2.00
4 00
25
1 00
AVERAGE 129
Material
A .
B.
C .
D
If the raw materials used advance in price at the following rates, what will
be the average per cent of increase in the cost of the finished product?
A .. .. 25% C . 33i%
B 30% D 50%
Pounds
4
Price
per Lb.
30ff
... . 6
.3
. 2
25^
75^
CHAPTER 13
Averaging Dates of Invoices
Definition. Averaging dates of invoices is the process of
finding the date when several invoices due at different daces may
he paid in one amount, without loss of interest to either debtor or
creditor. This date is called the equated date of payment.
Use. The process of averaging the dates of invoices is most
frequently used in bankruptcy settlements, where claims when
filed with a trustee must show the average due date of the items
if interest is to be obtained on overdue amounts. In general, the
equated date is important in the settlement of bills of long stand-
ing, and in the fixing of the date of a note in settlement of invoices.
Term of credit. A term of credit is the time elapsing between
the date of a bill and the date on which it becomes due; as, "Bill
purchased January 10, Term of Credit 10 days." The due date
would be January 20.
Average due date. The average due date is the date on
which settlement of the complete account should be made by pay-
ment of the amount of the invoices, without charge for interest
on overdue items or allowance for discount on prepaid items.
Focal date. The focal date is an assumed date of settlement
with which the due dates of the several items may be compared, to
determine the equated date of payment.
Any elate ma}' be used as the focal date, and the final result
will be the same. In the interest method, any rate per cent may
be used, and the result will be the same. However, 6% is usually
used, as the computations are then less complicated.
In all calculations, use the nearest dollar. For example, for
S115.29, use SI 15.00; and for $101.84, use $162.00.
When several bills are sold, some of which have a term of credit,
first find the due date of those with a term of credit, arid then find
the equated date of the several bills.
With respect to bills with terms of credit, the due date of such
bills, rather than the invoice date, is used in computing the equated
date.
Do not use fractions of a day in determining the average date.
Methods. There are two methods in common use: the Prod-
uct Method, and the Interest Method.
131
132 AVERAGING DATES OF INVOICES
Rule for product method. Use as the focal date the last
of the month preceding the first item. Multiply each item by the
number of days intervening between the assumed date and the
due date of the item, and divide the sum of the several products
by the sum of the account. Count forward from the assumed date
the number of days obtained in the quotient. The result will be
the average due date.
Example
Find the date at which the following bills of merchandise may be paid in one
amount without loss to either party: Due January 1, $150.00; February 14,
$200.00; April 20, $155.00; June 15, $200.00.
Solution by Product Method
(Focal date, Dec. 31)
Due January 1 $150 X 1 = 150
Due February 14 200 X 45 = 9,000
Due April 20 155X110 = 17,050
Due June 15 J200 X 166 = 33,200
705 59,400
59,400 4- 705 = 84 days.
84 days after December 31 is March 25.
Explanation. For convenience, assume December 31 as the date of settle-
ment. On the first bill, which is due January 1, there would be interest for
1 day. On the second bill there would be interest from December 31 to Febru-
ary 14, or 45 days, which is equivalent to interest on $9,000.00 for 1 day. On
the third bill there would be interest from December 31 to April 20, or 110 days,
which is equivalent to interest on $17,050.00 for 1 day. On the fourth bill
there would be interest from December 31 to June 15, or 166 days, which is
equivalent to interest on $33,200.00 for 1 day.
If all the bills were paid December 31, the debtor would be entitled to interest
on $59,400.00 for 1 day, or interest on $705.00, the amount of the account, for
84 days. It is evident that the bills could be paid at a time 84 days later than
December 31, or March 25, without loss to either party.
Verification
The interest on $150.00 for 83 days is $2.08
The interest on $200.00 for 39 days is 1.30
Total gain of interest to debtor $3J38
The interest on $155.00 for 26 days is ITT67
The interest on $200.00 for 82 days is 2.72
Total gain of interest to creditor $3.39
The gain of interest to the debtor is on all bills paid after they are due.
The gain of interest to the creditor is on all bills paid before they are due.
These two results should be equal, or within a few cents of the same amount.
The reason for a little discrepancy is the fraction of a day which is disregarded
in determining the due date of the account.
AVERAGING DATES OF INVOICES 133
Solution by Interest Method
January 1 $150 00 for 1 day = $0 03, interest
February 14 200 00 for 45 days = 1 . 50, interest
April 20 155 00 for 110 days = 2 84, interest
June 15 200.00 for 166 days = 5 53, interest
Total interest $9.90
Interest on $705.00 for 1 day is $0.1175.
$9.90 •*- $0.1175 = 84, or 84 days.
Explanation. Assume December 31, the last day of the month preceding
the first item, to be the date of settlement. If the amount of the account,
$705.00, is paid December 31 , there will be a loss of interest to the debtor of $9.90.
The interest on $705.00 for 1 day at 6% is $0.1175. It will take a principal of
$705.00 as many days to produce $9.90 as the number of times that $0.1175 is
contained in $9.90, or 84 days, the same result as was obtained by the product
method.
Problems
1. Several invoices mature as follows:
April 12 $260 00 August 18 $120 00
May 25 500.00 September 2 300 00
At what date may the foregoing invoices be paid in one amount without loss
to either party?
2. Calculate the average due date of the following invoices:
June 10 $400 00 August 15 $250 00
July 27 100 00 September 22 300 . 00
3. Calculate the average due date of the following invoices:
May 8 $275 00 on 60 days' credit
May 24 150 00 on 2 months' credit
June 10 300 . 00 on 90 days' credit
July 1 250 00 on 30 days' credit
CHAPTER 14
Equation of Accounts, or Compound Average
Definition. Equation of accounts, or compound average, is
the process of finding the date when the balance of an account
having both debits and credits can be paid without loss to either
debtor or creditor. With respect to bills with terms of credit, the
due date of the bill rather than the invoice date is used in comput-
ing the equated date. Credits other than for cash (such as non-
interest-bearing notes) are extended to the due date thereof.
Summarizing briefly, the date to be used for each debit and credit
is the date when the item has a cash value of the amount shown in
the entry.
Rule for the product method. After finding the date that each
item has a cash value, use the last day of the month preceding the
earliest date as the focal date for both sides of the account. Find
the number of days between the focal date and the due date of
each item; multiply each item by the number of days intervening
between the focal date and the due date of the item. Find the
sum of the products on both the debit and the credit sides of the
account. Divide the difference between the sums of the debit
and the credit products by the balance of the account. The
quotient will be the number of days between the focal date and
the average date of the account.
When to date forward or backward. The average date is
forward from the focal date when the balance of the account and
the excess of the products are on the same side (both debits or both
credits) ; if they are on opposite sides, the average date is backward
from the focal date.
Example
At what date may the balance of the following account be paid without loss
of interest to either party?
Debits Credits
July 1, Mdse. 30 days $250.00 Aug. 15, Cash $400.00
July 26, Mdse. 30 days 425 00 Sept. 10, Cash 300.00
Aug. 15, Mdse. 60 days 320.00 Sept. 20, Cash 150.00
Aug. 30, Mdse. 60 days 500 . 00
135
136 EQUATION OF ACCOUNTS, OR COMPOUND AVERAGE
Solution by Product Method
Since the earliest date is July 31, the assumed focal date would be June 30.
However, since July 31 is an end-of-month date, this date is used, as each multi-
plier is 31 less than it would be if June 30 were used.
My 31, $ 250 00 X 0 = 00,000 Aug. 15, $400 00 X 15 = 6,000
Aug. 25, 425 00 X 25 = 10,625 Sept. 10, 300 00 X 41 = 12,300
Oct. 14, 320 00 X 75 = 24,000 Sept. 20, 150.00 X 51 = 7,650
Oct. 29, 500 00 X 90 = 45,000
$1,495.00 79^625 $850 00 25^950
Debit side $1,495 00 $79,625.00
Credit side 850.00 25,950 00
$~645.00 $53~675.00
$53,675.00 4- $645.00 = 83.
The equated date is, therefore, 83 days after July 31, or October 22.
Explanation. First find the due date of each item. For convenience, assume
July 31, the earliest due date, as the day of settlement for ail the items on each
side of the account. Proceed as in the process of averaging dates, which was
described in the preceding chapter. With July 31 used as the focal date, there
is a loss of interest on the total debits equivalent to the interest on $79,625.00 for
1 day, and a gain of interest on the total credits equivalent to the interest on
$25,950.00 for 1 day; or a net loss of interest equivalent to the interest on
$53,675.00 for I day, which is equal to the interest on $645.00 for 83 days. It is
evident that the date when there would be no loss of interest to either party
must be 83 days after July 31, or October 22.
Solution by Interest Method
Debits
July 31, 0 days' interest on $ 250 00 = $ .00
Aug. 25, 25 days' interest on 425 00 = 1 77
Oct. 14, 75 days' interest on 320 00 = 4 00
Oct. 29, 90 days' interest on 500 00 = 7 50
$1,495766 $13727
Credits
Aug. 15, 15 days' interest on $ 400.00 = $ 1 00
Sept. 10, 41 days' interest on 300 00 - 2 05
Sept. 20, 51 days' interest on 150 00 = 1 28
$~850.00 $"4.33
Dr. $1,495 00 Interest, $13 27
Cr. 850_00 Interest, 4 33
6000)_ 645^00 $ 8.94
$ 1075 interest for one day.
$8.94 + .1075 = 83.
The equated date of payment is, therefore, 83 days after July 31, or
October 22.
Explanation. The explanation of the product method is applicable to the
interest method. The variance is in finding interest on each item and dividing
the interest balance by the interest for 1 day on the balance of the account.
EQUATION OF ACCOUNTS, OR COMPOUND AVERAGE 137
Problems
Find the equated date in each of the following:
1.
Debits
June 10, Mdse
Aug. 20, Mdse
Oct. 30, Mdse
2.
Debits
May 3, Mdse. 60 days.
June 15, Mdse. 60 days.
July 20, Mdse. 30 days
Aug. 27, Mdse. 60 days
3.
Debits
Mar. 1, Mdse. 30 days
Mar. 20, Mdse. 2 mos
Apr. 5, Mdse. 60 days
Credits
$500 00 July 5, Cash $300 00
100 00 Aug. 10, Cash 150 00
250 00 Sept. 25, Cash 200 00
Credits
$300 00 June 20, Cash $150.00
250 00 July 1, Note, 30 days with-
175.00 out int 200.00
225.00 Aug. 10, Note, 60 days, int.,
6% 300.00
Credits
$225.00 Mar. 31, Cash
300 00 Apr. 15, Cash
150.00 May 10, Casli
$150.00
100 00
200.00
CHAPTER 15
Account Current
Definition. An account current is a transcript of the ledger
account. It should show the dates on which sales were made,
the term of credit for each item, cash payments, and, if settlements
were made by note, the date and other details of each note.
Methods. Two methods are used in finding the amount due:
the Interest Method, and the Product Method.
Example
Find the balance due January 1 on the following ledger account, which bears
interest at 6%.
J. B. JOHNSON
Dr. Cr.
Sept. 1, Balance $1,200 00 Oct. 1, Cash $1,000 00
Sept. 20, Mdse. 30 days ... 400 00 Nov. 10, Cash 200.00
Oct. 30, Mdse. 30 days ... 520 00 Dec. 3, Cash 400.00
Nov. 25, Mdse. 30 days 350.00 Dec. 15, Note 10 days 300.00
Solution by Interest Method
J. B. JOHNSON
Dr. Cr.
Date Due Amount Days Interest Date Amount Days Interest
Sept. 1 $1,200 00 122 $24 40 Oct. 1 $1,000 00 92 $15 33
Oct. 20 400 00 72 4.87 Nov. 10 200 00 52 1 73
Nov. 29 52000 33 2.86 Dec. 3 40000 29 1.93
Dec. 25 350_00 7 .41 Dec. 25 300 00 7 .35
$2~470 00 $32.54 $1,900.00 $19.34
1,900 00 19 34
$ 570 00 -f- $13.20 = $583.20
Explanation. The number of days opposite each entry is the actual number
of days from the date of the item to January 1, the date which is taken as the
focal date.
Solution by Product Method
Dr.
Cr.
Date Due
Amount
Days Product
Date
Amount
Days Product
Sept.
1
$1,200
X
122
= $146,400
Oct.
1
$1,000
X
92
= $ 92,000
Oct.
20
400
X
73
= 29,200
Nov.
10
200
X
52
10,400
Nov.
29
520
X
33
17,161
Dec.
3
400
X
29
= 11,600
Dec.
25
350
X
7
2,450
Dec.
25
300
X
7
2,100
$2,470
$195,210
$1,900
$116,100
1,900
116,100
$ 570 $ 79,110
The intent on $79,110 for 1 day = $ 13.20
$570.00 + $13.20 = $583.20
139
140 ACCOUNT CURRENT
In some instances it is more convenient to find the equated due
date, and then calculate the interest on the balance of the account
from that date to the date of settlement.
Problems
1. Find the amount that will settle the following account Sept. 10, interest
at 6%.
Dr. Cr.
Mar. 15, Mdse. 4 mos $450 00 July 5, Cash $400 00
Mar. 30, Mdse. 60 days .... 375 00 July 30, Cash 375 00
Apr. 18, Mdse. 30 days .... 700 00 Aug. 15, Cash 690.00
May 15, Mdse. 4 mos 62000 Sept. 5, Cash 61500
May 30, Mdse. 4 mos 410 00
2. Find the amount that will settle the following account on June 1, interest
at 6%.
Dr. Cr.
Jan. 4, Mdse. 30 days $500 00 Feb. 20, Cash $300 . 00
Jan. 30, Mdse. 30 days 200 00 Feb. 28, Note, 60 days with
Feb. 5, Mdse. 30 days 600 00 interest at 6% ... 300 00
Mar. 1, Mdse. 30 days 400.00 Mar. 20, Cash 15000
3. A borrowed $10,000.00 from a bank on January 2, giving a note secured
by a mortgage for building a home, due in one year, with interest at 6%. From
time to time the bank advanced him money to pay contractors' estimates.
Before maturity the bank had actually advanced $9,000.00, as follows:
January 31 $3,000 00
March 15 3,00000
April 15 1,500.00
May 15 1,500.00
On June 1, the following year, the maker of the note desires to pay it. (a)
How should interest be computed? (6) What amount is due June 1?
CHAPTER 16
Storage
Definition. Storage is the charge made by a warehouse or
depositary for the storing of goods until they are required for use
or for transportation to some other point.
Running account. When goods are being received and
delivered, the storage company keeps a running account, showing
the dates at which goods are received and delivered, together with
details of the number of packages, barrels, and so forth. Storage
is charged for the average number of days for which one package,
barrel, or box has remained in storage. The average number of
days is divided by 30 to reduce the average number of days to
months, or by 7 to reduce the average number of days to weeks, as
the case may be ; then the number of months or weeks is multiplied
by the price per month or per week.
Example
The following is a memorandum of the quantity of salt stored with a storage
company at 4^ per barrel per term of 30 days' average storage.
Time in
Equivalent
Date
Receipts
Deliveries
Balance
Storage
for 1 Day
June 4
120 bbi.
120 bbl.
28 days
3,360 bbl.
July 2
20 bbl.
100 bbl.
18 days
1,800 bbl.
July 20
100 bbl.
200 bbl.
10 days
2,000 bbl.
July 30
50 bbl.
150 bbl.
11 days
1,650 bbl.
Aug. 10
100 bbl.
50 bbl.
15 days
750 bbl.
Aug. 25
50 bbl.
Obbl.
9,560 bbl.
Explanation. 9,560 bbl. for 1 day are equivalent to 1 barrel for 9,560 days,
and 9,560 divided by 30 (the number of days per term) equals 318-f terms. In
some cases a full month's storage is charged for any part of a month that goods
remain in storage; in other cases, 15 days or less are called one-half of a month,
and any period of over 15 days is counted as a whole month. 318f terms would
be charged for as 319 terms, and 319 X .04 = 12.76. Therefore, $12.76 is the
storage charge.
141
142 STORAGE
Problems
the following memoranda, compute storage at 4ff per barrel per term
of 30 days' average storage:
Received Delivered
Feb. 10 ........... 300 bbl. Feb. 20 .. . 150 bbl.
Feb. 19 .......... 150 bbl. Mar. 5 ....... 200 bbl.
Mar. 12 ......... 500 bbl. Mar. 15 ....... 400 bbl.
v Mar. 30 . . . . 300 bbi. Apr. 14 ..... 300 bbl.
; 2/)A grower stored 5,000 bushels of potatoes at 5^ per cwt., the term being
30 days' average storage. The following is a memorandum of the transactions
that occurred. Compute the amount of storage.
Received Delivered
Sept. 1 .......... 2,500 bushels Nov. 4 .......... 500 bushels
Sept. 10 ......... 1,500 bushels Dec. 10 ......... 600 bushels
Oct. 5 ........... 1,000 bushels Jan. 15 .......... 750 bushels
Feb. 1 ........ 1,500 bushels
Mar. 18. . 750 bushels
Apr. 2 ........ 000 bushels
CHAPTER 17
Inventories
Valuation of inventories. The bases of inventory valuation
most commonly used by business concerns are: (a) cost; and (b)
cost or market, whichever is lower. However, the average cost
method is used in some instances — the tobacco industry, for
example — and market value as a basis is used in grain and cotton
inventories and in inventories of dealers in securities.
Cost or market, whichever is lower. In valuing inventories
at cost or market, whichever is lower, a comparison of inventory
totals at the two values is not sufficient. It is necessary to consider
each item or group of similar items purchased at the same price,
and to make the extension at the cost or market price, whichever
is lower. *^f\
& f
Example
One hundred tons of sugar (200,000 Ibs.) were purchased at 7|4 a pound, and
later, 50 tons (100,000 Ibs.) were purchased at 6^ a pound. The entire 150 tons
were on hand at the close of the year, at which time the market value of sugar
was 6^^ a pound. Compute the inventory at: (a) cost; and (b) at cost or market,
whichever is lower.
Solution
(a) 200,000 Ibs. @ .07 .............................. $14,000
100,000 Ibs. @ .06 ............................... 6,000
Inventory at cost .................. §?0,000
(b) 200,000 Ibs. @ .065
100,000 Ibs. @ .06
_
Inventory at cost or market, whichever is lower . . $19,000
Problems
A
(jj.* Given the following inventory of a retail shop for children's clothing, toys,
and so forth (correct as to quantities and values), state the amount which should
be shown on a balance sheet as merchandise inventory, adopting the method of
valuing inventory at cost or market, whichever is lower.
* C. P. A., Maryland.
143
144
INVENTORIES
Value Per Unit
Item
150 knit towels $
16 crepe cle chine carriage sets 10 00
125 lingerie and pongee hats 2 .00
85 rubber bibs with sleeves
240 creepers 2
200 spring coats 9
50 spring coats 17
8 play yards 6
8 desks used in office 55
140 shirts
200 boys' wash suits 0
125 bloomers 1
5 cribs 21
12 electric trains 1
Total
)St
Market
38
$ 0 35
00
12.50
.00
1.75
.50
.50
05
1 98
50
10 00
50
18 50
00
5.75
.00
60 00
.75
.79
20
5 98
85
1 cSO
00
19 98
50
1.50
Total
Value
Cost
Market
$ 57.00
$ 52 50
160 00
200 00
250 00
218.75
42 50
42.50
492 00
475 20
1,900 00
2,000.00
875 00
925 00
48 00
46.00
440 00
480 00
105 00
110.60
1,240 00
1,196 00
231 25
225 00
105 00
99 90
18 00
18 00
$5,963 75 $6,089 45
( 2.*/ You are called in by the X. Y. Z. Clothing Company to advise them on
the-mlculation of their inventory. They have always followed the policy of
cost or market, whichever is lower. You are informed that the inventory will
be used for the tax return, as well as for the annual report to stockholders.
Value per Unit
Item Cost Market
13 suits, grade A $60 00 $55 00
12 suits, grade B 40 00 37 50
77 suits, grade C 30 00 30 00
7 suits, grade D 20.00 22 00
24 overcoats, grade 1 75 00 80 00
5 overcoats, grade 2 40 00 45 00
10 overcoats, grade 3 30 00 25 00
6 topcoats, grade X 20 00 17 . 50
9 topcoats, grade Y 15 00 12 50
18 topcoats, grade Z 10.00 11 .00
Total
Total
Cost
I 780 00
480 00
510 00
140 00
1,800 00
200 00
300 00
120 00
135 00
180 00
Value
Market
$ 715 00
450 00
510 00
154 00
1,920 00
225 00
250 00
105 00
112.50
198.00
$4,645 00 $4,639 50
Which total figure would you advise the company to use for: (a) tax reports;
(b) annual report to stockholders?
Average cost method. The general rule that the average cost-
method of valuing inventories will not be accepted for income tax
purposes is subject to certain exceptions. In the tobacco industry,
for example, tobacco is bought from the producer, usually in small
quantities and at greatly varying prices. Different grades of
tobacco are mixed and stored in hogsheads, and it is practically
impossible to determine the exact cost of any particular hogshead.
The inventory is therefore averaged monthly, according to grades,
as follows:
First method. From the inventory of each grade at the begin-
ning of the month is subtracted the amount of tobacco of that
*C. P. A., Wisconsin.
INVENTORIES
14*
grade used, leaving so many pounds costing so many dollars; to
this is added the tobacco of that grade purchased during the
month, and a new average is determined. This is the inventory
for the close of the month, and is consequently the opening inven-
tory of the next month.
Example
STSCK CARD
RECEIVED
ISSUED
BALANCE
Date
Quan-
tity
Rate
Amount
Date
Quan-
tity
Rate
Quan-
tity
Rate
Amount
6-29
9-30
12-10
100,000
80,000
125,000
$1 00
1.10
.95
$100,000
88,000
118,750
9-1
12-5
12-18/
80,000
30,000
20,000
$1 00
1 08
1 08
100,000
20,000
100,000
70,000
195,000
175,000
$1.00
1 00
1 08
1.08
$100,000
20,000
108,000
75,600
194,350
172,750
Inv't
12-31
175,000
$0 987
$172,750
Explanation. It will be noticed that the receipt of 125,000 at .95 on Dec. 10
was extended into the balance column in quantity and amount only, and that
the issuance on Dec. 18 was made at the rate established on the first of the
month. This is the method that is used when receipts are frequent, as it saves
the time that would be required to compute a new rate after each receipt, and
establishes a standard rate of issuance for the month.
Second method. When receipts are not frequent and are large
in amount, a new average price is computed upon the entry of
each receipt.
Example
STOCK CARD
RECEIVED
ISSUED
BALANCE
Date
Quan-
tity
Rate
Amount
Date
Quan-
tity
Rate
Quan-
tity
Rate
Amount
6-29
9-30
12-10
100,000
80,000
125,000
$1.00
1.10
.95
$100,000
88,000
118,750
9-1
12-5
12-18
80,000
30,000
20,000
$1.00
1.08
99i
100,000
20,000
100,000
70,000
195,000
175,000
$1.00
1 00
1.08
1.08
.99^
.99i
$100,000
20,000
108,000
75,600
194,350
174,430
Inv't
12-31
175,000
$0.99^
$174,430
146
INVENTORIES
^~\ Problems
fy. Rule two stock cards as in the preceding example, and enter the following
data. Compute the balances: (a) by the first method; and (6) by the second
method.
Received
July 5 80,000 units @ $0.90
Aug. 15. ... 20,000 units @ 1 .00
Sept. 1 30,000 units @ 1.10
.— Dec. 8 20,000 units @ 1 .20
Issued
Aug. 1 50,000 units
Dec. 2 20,000 units
Dec. 20 30,000 units
2. Complete the following stock ledger card, using average-price method.
Stores Ledger
Actual Receipt Price
./£"
.20
Average Price
J5
/7o<f
Name {ffludshs^A
' 5 7/7&t& part No. 5& "A 7&
Minimum /#? "
Maximum 3O0 Location jL3
Drawing No.
Unit
REFERENCE
QUANTITY
ON HAND
Date Number
Remarks
Received
Issued
Quantity
Value
I
Z
OCT 4 4523
OCT 10 34567
OCT 12 35$54
OCT 13 38765
OCT 15 -39458
OCT 16 4587
OCT 19 40156
100
50
5
10
3
12
10
1
2
3
—
J^
5
6
F
7
8
6
9
01
10
U
Zl
81
11
12
13
"First-in, first-out" method of inventory. Where the same
merchandise has been purchased at various prices during the year,
and the goods on hand cannot be identified with specific invoices,
the amount on hand at the end of the year may be inventoried
at the latest purchase price. If, however, the quantity on hand is
greater than the amount purchased at the last price, the balance
may be inventoried at the next to the last purchase price, and so on.
This method is termed "first-in, first-out method" of inventory.
Example
Inventory, December 31 275,000 units
Invoices :
November 10 125,000 units @ $0.95 per C
September 5 80,000 units @ 1.10 per G
June 10 70,000 units @ 1.00 per G
How should the foregoing inventory be valued?
INVENTORIES 147
Solution
125,000 units @ $0.95 per C ........................ $1,187.50
80,000 units @ 1.10 per C ........................ 880.00
70,000 units @ 1.00 per C ........................ 700.00
275,000 units inventoried ........................... $2,767 . 50
"Last-in, first-out" method of inventory. Under the "last-in,
first-out" method inventories are valued at the cost of goods
earliest acquired, and in computing profits from sales the cost of
goods last acquired is used. This method will show smaller
profits when prices are rising and larger profits when prices are
falling than the afirst-in, first-out" method. Businesses which
use raw materials or other goods includable in inventory, which are
subject to sharp price fluctuations; businesses in which the value of
inventory is large compared with other assets and sales; and
businesses in which production consumes an extended period are
most likely to benefit from the use of this method. (Consult the
Internal Revenue Code relative to the requirements incident to
adoption and use of this method.)
Example
A has an opening inventory of 10 units at 10 cents a unit, and during the
year he makes purchases of 10 units as follows:
January
1 @ .11 =«
11
April
2 @ .12 =
?4
July
3 @ .13 =
39
October
4 @ .14 =
56
10 f
730
His closing inventory shows 15 units. What is the value of the closing
inventory?
Solution
10 @ .10 = 1.00
1 @ .11 (Jan.) = .11
2 @ .12 (Apr.) = .24
_2 @ .13 (July) = .26
Totals 15 1.61
Problem
Value the closing inventory, using the " last-in, first-out'* method.
Opening inventory: 50 units at $1.00
Production:
First quarter: 50 units at $1.50
Second quarter: 100 units at $1.75
Third quarter: 50 units at $2.00
Fourth quarter: 100 units at $2.25
Closing inventory: 150 units
148 INVENTORIES
Merchandise turnover. The number of times that the valu«
of the inventory is contained in the cost of sales is the merchandise
turnover.
The final inventory should not be used in computing turnover,
unless it represents a normal inventory for the fiscal period, or is
the first inventory that has been taken.
If a perpetual inventory system is in use, the monthly inven-
tories should be added to the inventory at the beginning of the
period, and the sum divided by the number of months in the fiscal
period plus one. In a year there would thus be thirteen inven-
tories— the one at January 1, and the twelve inventories at the
ends of the months. When a perpetual inventory is not used,
add the inventory at the beginning of the fiscal period to the one
at the close of the period; then divide by two. The quotient will
be the estimated average inventory for the period. If semiannual
inventories are taken, use three inventories and divide by three.
If quarterly inventories are taken, use five inventories and divide
by five.
FORMULA
Cost of Sales -f- A verage Inventory at Cost — Rate of Turnover
Example
A department store found the average inventory of Department A for the
fiscal period to be $30,000. The cost of sales for the same period was $120,000.
$120,000 -^- $30,000 = 4, the rate of turnover.
An estimated inventory at the end of any period may be
obtained by dividing the sales for the period by 100% plus the
per cent of gross profit based on cost, and deducting the quotient
from the total of purchases and first of period inventory. A more
complete discussion of the gross profit test is given in Chapter 18.
Example
In the above example, assume that in Department A the total cost of mer>
chandise was $150,000, that the sales were $144,000, and that the average profits
were 20%. Using the per cent of gross profits to determine the average inven
tory, the solution would be as follows:
$144,000 (sales) -i- 120% = $120,000, the cost of sales.
$150,000 (total cost of goods) - $120,000
(cost of sales) = $30,000, the estimated inventory.
$120,000 (cost of sales) -f- $30,000
(inventory) = 4, the rate of turnover.
Number of turnovers. The number of turnovers varies in
different lines of business. Records show turnovers varying from
1 to more than 20, depending on the kind of business. It is possi-
INVENTORIES 149
blc to make a larger profit by several turnovers with a small
mark-up* than by 1 or 2 turnovers with a large mark-up. Limited
capital and frequent turnovers can produce a profit equal to that
produced by a greater capital turned fewer times a year. If a
merchant turns $1 eight times in the course of a year, he has used
i of the capital that would be required if the rate of turnover
were 1.
Example 1 .
A merchant had a rate of mark-up of 50%, with a turnover of 1. He found
that by using a rate of mark-up of 30% he had a turnover of 2. If his former
sales were $300,000 annually, how much were his gross profits increased, pro-
vided he continued to use the same investment in merchandise?
$300,000 + 150% = $200,000, cost of sales.
$300,000 - $200,000 = $100,000, gross profits.
Under the new policy he turns the $200,000 twice, the equivalent of $400,000
annually.
$400,000 at 30% = $120,000, profits.
$120,000 - $100,000 = $20,000, increased profits due to lowering
the rate of mark-up and increasing
the rate of turnover.
Example 2
What investment in merchandise would he required under the new policy
to make the same amount of profits that was made under the old policy?
$100,000 -h 30% = $333,333.33, cost of goods sold to make profits of $100,000.
Since there were 2 turnovers, the cost of goods sold was twice the amount of
the a.verage inventory. Therefore:
$333,333.33 -T- 2 = $166,666.67, the average inventory.
Hence, the merchant could make the same amount of gross profits with an
investment $33,333.33 smaller than that required under his old policy.
Problems
1. A rate of mark-up of 30% results in 2 turnovers of an average inventory
of $30,000. If the expenses of conducting the business are $8,000, what is the
net profit?
2. A rate of mark-up of 20% results in 3 turnovers of an average inventory
of $30,000. If expenses remain at $8,000, what is the net profit?
3. The Cost of sales in Department B was $42,000. The average inventory
was $12,000. What was the number of turnovers?
4. A merchant's sales amounted to $42.000. His average inventory was
$10,000, and the average rate of mark-up was 40%. Find the number of
turnovers.
* "Mark-up," as used in this text, refers to the addition made to the cost jf
merchandise to produce the selling price.
150 INVENTORIES
6. Commodity X, with a rate of mark-up of 40%, had a turnover of 2. With
a rate of mark-up of 30%, it had a turnover of 3. If prior sales were $56,000,
find the sales and the increase in gross profit with the 30% rate of mark-up.
6. A rate of mark-up of 35 % results in a turnover of 2 and in sales amounting
to $540,000. A rate of mark-up of 20 % results in a turnover of 4. How muck
less capital under the latter plan is required to make as much profit as under the
former plan?
7.* On January 1, a concern dealing in a single commodity had an inventory
of merchandise which cost $20,000. The goods were marked to sell at 125%
of cost, and all subsequent purchases during the six months ending June 30
were marked at the same rate. The selling price of the inventory at June 30
was $24,000. Purchases and sales by months were:
Purchases Sales
(Cost) (Selling Price)
January $ 8,000 $ 9,000
February 9,000 9,500
March 14,000 12,000
April 16,000 18,000
May.... . .. . 13,000 22,000
June 10,000 18,000
(a) Compute estimated inventories at cost price at the end of each of the
six months.
(b) Compute the rate of turnover for the six months' period, using (1) the
January 1 and June 30 inventories; (2) ail the inventories.
(c) State which method gives the more accurate results.
Per cent of mark-down to net cost. If an item costs $1 and
is marked $1.25, in order to sell the item for cost the price must be
reduced 25 i. The marked price is the base when prices are
reduced. 25 jS is £ of $1.25. £ = 20%.
An item costs $1 and is marked $1.50. 50^ reduction is -V oJ
$1.50, or 33£%.
Problems
Calculate the per cent of mark-down for each of the following items:
Marked Per Cent of Mark-down
Item Cost Price to Produce Cost
A $ 2 00 $ 2 50 ...
H 1 00 1.25
(y . .35 .40
7) 80 1 00
E 15.00 25 00 .
F 3.50 4 00 .......
6' 20 30
H 5.00 7 00
/ 08 .10
J 2.00 4 00
K 4000 7500
L 12.00 18.00
1 American Institute Examination.
INVENTORIES
151
Computation of inventory by the retail method. The need for
frequent inventories has led many department stores to adopt the
"retail method " of computing inventories. The accuracy of the
inventory by this method depends upon the care exercised in
recording the mark-ups and the mark-downs of merchandise
prices, and the classification of merchandise into departments
and groups and sub-classes within the departments. In addition
to the usual records showing sales (at selling price only), records
are kept which show the opening inventory and purchases at
cost and at retail (or selling) prices. An estimated inventory may
be prepared from such records in the following manner.
INVENTORY COMPUTATION
Cost Retail
Inventory, beginning of period $ 6,000 $ 8,000
Purchases during the period, including freight
and cartage 74,000 1 1 1 ,200
Totals $<XO,000 $Tl9,200
% Mark-on = $39,000 ~ $119,200 or 32.XS59%.)~
. .. 104,200
Inventory at retail . . $ 157)00
Estimated inventory = $15,000 - ($15,000 X 32.XX59%) = $10,007".
The foregoing illustration does not take into consideration
changes in selling price after the original mark-up. Price changes
must be dealt with, and the retail mercantile business has terms
for these changes that are not generally understood; therefore, to
prevent any • '• .• :• • ' • •'• j the following diagram is presented
and the terms explained.
Cost Plus Original Mark-Up
Original Retail (Selling Price)
Original mark-up. The amount by which the original retail
Erice of an article exceeds the cost is the original mark-up.
152 INVENTORIES
Additional mark-up. An amount that increases the original
retail price is an additional mark-up.
Additional mark-up cancellation. A reduction in the additional
mark-up is an additional mark-up cancellation, and the amount
cannot exceed the amount of the additional mark-up.
Net mark-up. The sum of additional mark-ups minus the sum
of additional mark-up cancellations is the net mark-up.
Mark-downs. Deductions from the original retail price to
establish a new but lower retail price are mark-downs.
Mark-down cancellations. A reduction of the amount of a
mark-down is a mark-down cancellation. Mark-down cancella-
tions cannot exceed the total mark-clowns. It is evident that the
retail price of merchandise is increased when the mark-down is
reduced, but such an increase is not to be considered as an addi-
tional mark-up.
Net mark-down. The difference between the sum of the mark-
downs and the sum of the mark-down cancellations is the net
mark-down.
Mark-on. The difference between cost and the original retail
plus the net mark-up is the mark-on.
To illustrate the terms, let the following transactions be
assumed.
INVENTORIES
153
Cost of Article
$1.00
Original
Mark-Up
Original Retail (or Selling) Price
$1.00 -f .50 = $1. 50 _
$1.50
Addn'l
Mark-Up
1st Adjusted Retail Price
$1.50 -f -25 = $1.75
2nd Adjusted Retail Price
$1.75 - .10 - $1.05
3rd Adjusted Retail Price
$1.65 - .15 = $1.50
4th Adjusted Retail Price
_ $1^.50^-^.1^= $1.35
5th Adjusted Retail Price
$1.35 - .35 = $1.00
$1.00
e
25*
6th Adjusted Retail Price
$1.00 + .25 = $1.25
a. Additional mark-up
cancellation
Net mark-up = 15^
Mark-on = 65^
b. Additional mark-up
cancellation
Net mark-up = 0
Mark-on =50^
c. Mark-down
d. Mark-down
e. Mark-down cancellation
Net mark-down =
.15 + .35 - .25 = .25
Determining the ratio of cost to retail. In determining the ratio
of cost to retail, it is customary to include additional mark-ups,
and additional mark-up cancellations hut to exclude mark-downs
and mark-down cancellations. To illustrate, let us assume the
following facts :
Inventory at beginning of month:
Cost $30,000.00
Retail 43,000 00
Purchases :
Cost 46,000 00
Retail 55,000.00
Returned purchases:
154
INVENTORIES
Cost 1,000 00
Retail 1,50000
Additional mark-ups 5,500 00
Additional mark-up cancellations 2,000 00
Mark-downs 6,000 00
Mark-down cancellations 1,000 00
Sales at retail 71,000.00
Compute the inventory by the retail method.
Solution
Cost Retail
Inventory $30,000 00 $ 43,000 00
Purchases 46,000 00 55,000 00
$76,000 "(JO $ 98,000 00
Deduct: Returned purchases 1,000 00 1,500 00
$75,000.00 $"90,500*00
Additional mark-ups less cancellations thereof 3,500 00
$100,000~00
$100,000 - $75,000 = $25,000.
$25,000 ~ $100,000 = 25%.
Mark-downs less mark-down cancellations 5,000 00
$ 95,00(fOO
Sales at retail __71_,000_ 00
End-of-month inventory at retail value $ 24,000 . 00
$24,000 X 25% = $6,000.
$24,000 - $6,000 = $18,000, the cost value of the inventory.
Problems
1. From the records kept for Department B, the following information
obtained:
Cost
Inventory at Beginning of Month $15,000 00
Purchases 36,00000
Returned Purchases 500.00
Additional Mark-Dps
Additional Mark-Up Cancellations
Sales '.
Retail
$25,000 00
54,000 00
700 00
2,000 00
1,000 00
60,000 00
Calculate by the retail method of inventory the cost of the book inventory
at the end of the month.
2.* In a certain department of a large dry-goods house, the purchases for
one year were $30,000. They were in the first place marked up for selling pur-
poses to $45,000. Later, additional mark-ups amounting to $2,000 were made,
and mark-downs aggregating $5,000 were also recorded. At the end of the
fiscal period there were found to be on hand goods of a marked selling value of
$10,000. State how you would arrive at their inventory value for the purpose
of closing the books, and calculate the amount. Explain fully.
* American Institute Examination.
CHAPTER 18
Gross Profit Computations
Gross profit. The gross profit represents the margin between
the sales and the cost of goods sold, and when expressed as a per
cent of sales indicates to one who is familiar with trade practice
whether a sufficient margin of profit is being made. Use of the per
cent of gross profit to check the correctness of the value set upon
the inventory is called the gross profit test of inventory.
Rate per cent of gross profit. The gross profit test is based
on the supposition that in normal times and under normal condi-
tions, any business will produce approximately the same per cent
of gross profit on sales in any one period of time as in any other
corresponding period of time.
Procedure. Statements of the gross profit and sales for each
of several prior periods should be obtained. The gross profit for
any one period divided by the sales for the same period gives the
rate of gross profit for that period, based on sales. Disregard any
per cent that is abnormal. Add the remaining per cents, and
divide by the number added. The quotient is the average per
cent of gross profit in prior periods.
Uses. The per cent of gross profit may be used in two ways:
first, to prove inventories; and second, to compute the estimated
inventory when it is impossible or impracticable to take a physical
inventory.
Example
Assume that the average gross profit for the past five years has been 40%
of sales, and that an audit of the books shows that the inventory, taken prior
to the beginning of the audit, and valued at $100,000, seems smaller than it
should be, while the previous inventory and purchases amounted to $400,000.
The sales for the period are $400,000. Show by comparative statement the
possibility of error.
Solution
In the following set-up, both the average and the current per cents and
results are shown. As the profit in prior years has been 40% of sales, the cost
of goods sold has been 60% of sales. 60% of $400,000 (sales) = $240,000, cost
of sales.
155
156
GROSS PROFIT COMPUTATIONS
Sales
Cost of Sales
First of year inven-
tory and purchases $400,000
Less: Current inv't. . J00,000
Gross profit
CURRENT YEAR
Actual Currtn*
Amounts Per Cent
$400,000 100
CURRENT YEAR IN TERMS
OF AN AVERAGE YEAR
Test Average
Amounts Per Cent
$400,000 100
300,000
__
25%
J40,000
$160,000
60
40 <
If the sales are correct, the cost of sales is $60,000 too high, unless the rate
has really changed. This discrepancy may be caused by any of the following: the
volume of sales may be incorrectly stated; the current inventory may be errone-
ous, and the cost of sales affected thereby; or there may be an abnormal increase
in the cost of merchandise purchased, when compared with the 5-year average.
The accountant should determine the reason for the discrepancy.
Cost of goods sold. The average rate per cent of gross profit,
applied to the sales for the current period, will give the estimated
gross profit for the current period. Deduction of the estimated
gross profit from the sales gives the estimated cost of goods sold.
This procedure may be reduced to a formula as follows :
AVERAGE FOR PRIOR PERIODS
1. Sales — Cost of sales = Gross profit.
2. Gross profit -5- Sales = Per cent of gross profit (based on sales).
APPLICATION TO CURRENT PERIOD
3. Sales X Per cent of gross profit (prior periods) = Estimated gross profit.
4. Sales — Estimated gross profit (current period) = Estimated cost of sales.
Example
Sales
First period $400,000
Second period 450,000
Third period 350,000
Fourth period 100,000
Cost Gross
of Sales Profit
$300,000 $100,000
340,000 110,DOO
260,000 90,000
What was the cost of sales during the fourth period?
S'olution
AVERAGE FOR PRIOR PERIODS
First period, $100,000 ^ $400,000 - 25.00%
Second period, 110,000 + 450,000 = 24 44%
Third period, 90,000 -J- 350,000 = 25.71%
75.15%
75% "5- 3 = 25%, the average rate of gross profit.
GROSS PROFIT COMPUTATIONS 157
APPLICATION TO CURRENT PERIOD
$100,000 X 25% = $25,000, estimated gross profit.
$100,000 - $25,000 = $75,000, estimated cost of sales.
Rate per cent of cost of sales. If the rate of profit lias been
based on cost price instead of on selling price, the cost of sales may
be tested by the following computations :
AVERAGE FOR PRIOR PERIODS
1 . Sales — Cost of sales - Gross profit.
2. dross profit -f- Cost of sates — Per cent of gross profit (based on cost of sates)
APPLICATION TO CURRENT PERIOD
3. Sales -r- (100% -f- Per cent of gross profit) = Cost of sales.
Example
Sales Cost of Sales
First period $400,000 $300,000
Second period 450,000 340,000
Third period 350,000 260,000
Fourth period 100,000
What was the cost of saleh for the last period?
Solution
AVERAGE FOR PRIOR PERIODS
First period, $100,000 4- $300,000 = 33 33%
Second period, 110,000-5- 340,000= 3235%
Third period, 90,000 -r 260,000 = _34_61_%
100729%
100% 4- 3 = 33^%, average per cent of gross profit.
APPLICATION TO CURRENT PERIOD
$100,000 -f- 1.33i (1 + .33i) - $75,000, cost of sales.
$100,000 - $75,000 = $25,000, gross profit.
It follows that if the cost of sales can be found, any element
(inventory at beginning of period, purchases, closing inventory,
and so forth) which goes to make up the cost of sales can be found,
provided the other elements of the costs are given.
Fire losses. Insurance companies are generally willing to
settle inventory losses resulting from fire on the basis of values
determined by the gross profit method.
Example
The insurance company agrees that the following facts are to be the basis
of its reimbursement to the insured for his fire losses:
Average gross profit for 4 years, 40% of sales.
Sales for this period to date of fire, $50,000.
Cost of goods available for sale, $300,000.
158 GROSS PROFIT COMPUTATIONS
Solution
$50,000 (sales) X 40% (rate of gross profit) = $20,000, gross profit.
$50,000 (sales) - $20,000 (gross profit) = $30,000, cost of goods sold.
$300,000 (goods available for sale) - $30,000
(cost of goods sold) = $270,000, estimated inventory
at date of fire.
Use of gross profit test in verification of taxpayer's inventory.
Assessors make use of the gross profit test to determine the approxi-
mate inventory and to check the item of inventory in the schedule
filed by the taxpayer, since assessment dates seldom coincide with
closing dates. The following forms have been given to the tax-
payer to fill out, the date of assessment being May 1.
FOR MERCHANTS
1. Book value of last inventory of stock of merchandise
2. Add purchases since last inventory to May 1
3. Add in-freight and cartage paid since last inventory to May 1
4. Total of above three items
Deduct from above total net result of following two items:
5. Amount of net sales from date of last inventory to
May 1
6. Less. Gross profit on sales estimated at %
(Previous near % may be used where actual % is
unknown.)
7. Net inventory of merchandise on May 1 (Item 4 less Item 6)
FOR MANUFACTURERS
1. Book value of raw materials, finished goods, and work-in-
process at last inventory. Date
2. Acid purchases of raw materials and finished goods since last
inventory to May 1
3. Add amount paid for in-freight and cartage from last inventory
to May 1
4. Add amount paid for labor and manufacturing expenses from
last inventory to May 1
5. Total of above four items
Deduct from above total the net result of the following two items:
6. Amount of net sales from date of last inventory to
May 1
7. Less: Gross profit on sales estimated at %
(Previous year % may be used where actual % is not known.)
8. Net value of raw materials, goods-in-process, and finished
goods on May 1 (Item 5 — Item 7)
Problems
1. From the figures in the following tabulation, calculate the per cent of gross
profit for each year, and by means of the average per cent of gross profit calculate
the inventory at the end of the first half of the fifth year.
GROSS PROFIT COMPUTATIONS
159
Sales
First year $120,000
Second year . . . 150,000
Third year 165,000
Fourth year . ... 180,000
Fifth year (6 mo.)... 95,000
Opening
Purchases Inventory
$ 90,000
100,000 10,000
110,000 12,000
122,000 10,000
62,000 11,000
Closing
Inventory
$10,000
12,000
10,000
11,000
Per Cent of
Gross Profit
2. From the following facts, find the inventory as of December 31:
Inventory, January 15 following, $16,578.50.
Sales, December 31 to January 15, $2,890.00.
$765 of the above sales shipped and invoiced before December 31.
Purchases, December 31 to January 15, at cost, $1,256.50.
Average gross profit, 25 % of cost.
3. The average gross profit of the X. Company for the past three years has
been 45% of the sales. During the fourth year the sales amounted to $159.500.
Goods were purchased to the amount of $105,000. Returned purchases totaled
$5,000 for the period. Freight paid on purchases was $6,000. The inventory
at the beginning of the period was $40.000. Current market prices are 10%
above the purchase prices for the year. Find the cost of replacing the goods at
the end of the year.
4. On April 30, the board of managers of the Ames Mercantile Company
removed the superintendent on the general suspicion that his books misrepre-
sented the true financial condition of the business. Prepare a statement showing
the nature and the probable extent of the misrepresentations; also an approxi-
mate statement of income and profit and loss for the four months ending April 30.
The following is a trial balance taken from the books, April 30:
Capital Stock
Furniture and Fixtures
Inventory, January 1
Cash
Accounts Payable
Accounts Receivable
Loans Payable .
Sales
Purchases . . . ....
Salaries, Salesmen
Advertising . . . ....
Salaries, Office
Rent
Interest
Insurance, January 1 to December 31
Stationery and Printing
Reserve for Depreciation of Furniture & Fix-
tures
Surplus, January 1
$ 10,000
128,600
15,450
24,600
40,700
2,200
1,650
1,100
400
200
999
105
$ 75,000
39,000
10,000
51,000
2,710
48,294
S226J004 $226,004
160 GROSS PROFIT COMPUTATIONS
An analysis of the Purchases, Sales, and Inventory accounts revealed the
following:
Opening Closing
Purchases Hales Inv't Inv't
First year $122,000 $153,750 $101,000 $100,000
Second year 123,000 153,170 100,000 102,000
Third year 121,000 154,722 102,000 128,600
5.* The hooks of a concern recently burned out contained evidence of pur-
chases, including inventory, to the amount of $200,000, and sales of $40,800,
since the last closing. Upon investigation, however, the auditor ascertained
that a sale of merchandise had been made just prior to the fire, and not recorded
in the books, at an advance of two-fifths over cost less a 10% cash discount; the
profit on the transaction was $31,928. The past history of the business indi-
cated an average gross profit of 50% on cost of goods sold.
(a) What amount should be claimed as fire loss?
(6) What rate of gross profit do the transactions finally yield?
6.f The store and stock of the Diamond Jewelry Company was destroyed by
fire on November 1. The safe was opened, and the books were recovered intact.
The trial balance taken off was as follows:
Cash in Bank . .... $ 1,000
Accounts Receivable . . 10,000
Accounts Payable ...... $ 30,000
Merchandise Purchases 90,000
Furniture and Fixtures 7,500
Sales 110,000
General Expense 18,000
Insurance 1 ,500
Salaries 5,500
Real Estate— Store Lot 50,000
Store Building 35,000
Capital Stock 50,000
Surplus 28,500
$218,500 $218,500
The average gross profit as shown by the books and accepted by the insurance
companies was 40% of sales. The insurance adjuster agreed to pay 75% of
the book value of furniture and fixtures, 90% of the book value of the store
building, and the entire loss on merchandise stock.
Draft journal entries to include the account against the insurance companies.
Installment sales of personal property. The large increase in
sales of personal property on the installment plan, and the option
that the government allows a taxpayer coming within the defini-
tion of an installment dealer to return his gross income from sales
on the installment basis, are indications of the growing importance
of this subject.
The installment plan of selling was devised for the purpose of
* American Institute Examination,
t C. P. A., Oklahoma.
GROSS PROFIT COMPUTATIONS
161
stimulating sales, whereas the installment basis of reporting income
was devised for the purpose of deferring from year to year the
income to be realized from installment sales, with a view to the
possible effect that this deferment might have upon the amount of
federal income tax to be paid. It is, of course, essential that the
latest Federal Income Tax Law be observed.
Computation of gross profit. The gross profit to be reported
may be ascertained by taking that proportion of the total cash
collections received in the taxable year from installment sales
(such collections being allocated to the year against whose sales
they apply) which the annual gross profit to be realized on the
total installment sales made during each year bears to the gross
contract price of all such sales made during that particular year.
Example
The books of the Model Credit Company, selling merchandise on the install'
ment plan, show the following:
First
Year
Sales $ 80,000
Cost of Sales
Inventory (old) $ 45,000
Purchases _55>000
$100,000
Less: Inventory (new) . . 40,000
Cost of sales $150,000
Gross profit $ 20,000 ^ ^
Collections were made in the fourth year on each year's contracts as follows:
First Second Third Fourth
Year Year Year Year
$1,600 $4,800 $25,000 $70,000
What was the gross profit to be reported for the fourth year?
Second
Third
Fourth
Year
Year
Year
$110,000
$130,000
$90,000
$ 40,000
75,000
$ 50,000
90,000
$48,000
50,000
$fl 5,000
50,000
$140,000
48,000
$98,000
40,000
$"G57000
$ 92,000
$58,000
$45,000
$ 38,000
$32,000
Solution
Per Cent of Gross Profit
First year, $20,000 (gross profit) •*- $ 80,000 (sales)
Second year, 45,000 " " -4- 110,000 "
Third year, 38,000 " " + 130,000 "
Fourth year, 32,000 " " ^ 90,000 "
Profit on Collections in Fourth Year
Collected on first-year contracts, $ 1,600 X 25.00%
4,800 X 40.91%
25,000 X 29.23%
" second-year "
" " third-year
" fourth-year " 70,000 X 35.55% =
Gross profit realized in the fourth year. . , ,
= 25.00%
= 40 91%
= 29.23%
= 35.55%
$ 400 00
1,963 68
7,307.50
24,855^.00
$34,526718
162 GROSS PROFIT COMPUTATIONS
Reserve lor unearned gross profit. The gross income to be
realized on installment sales is credited to " Reserve for Unearned
Gross Profit/7 and at this time this account is debited with the
gross profit on collections. The balance of the account represents
gross profit on installment sales contracts remaining unpaid at the
date of closing.
Example
The books of the X.Y.Z. Company, selling merchandise on the installment
plan, show the following:
First Second Third Fourth
Year Year Year Year
Sales $89,257 99 $111,825 86 $137,012 32 $97,912 26
Gross profits 29,962~89 48,068 37 ~38j28~63 39,168~71
Collections during the
fourth year on each
year's accts 1,635.35 4,832 00 25,182. 14 69,927 92
What amount should be ci edited to Reserve for Unearned Gross Profit to
represent deferred income for the fourth year? What amount should be debited
to Reserve for Unearned Gross Profit to represent income realized from the
first, second, third, and fourth years' collections received in the fourth year?
Solution
(a) Per Cent of Gross Profit
First year $29,962.89 ^ $ 89,257.99 = 33 57%
Second year 48,068.37 -*• 11 1,825.86 - 42 98%
Third year 38,128.63 -;- 137,012.32 = 27 83%
Fourth year 39,168.71 4- 97,912.26 = 40 00%
Profit on Collections
First-year accounts $ 1,635.35 X 33.57% = $ 548 99
Second-year accounts. . . . 4,832.00 X 42.98% = 2,076 79
Third-year accounts 25,182.14 X 27.83% = 7,008 19
Fourth-year accounts. . . . 69,927.92 X 40.00% = 27,971 . 17
Gross profit realized in 4th yr $37,605^14
Journal entries
Installment Sales Contracts $ 97,912.26
Cost of Sales $ 58,743 55
Reserve for Unearned Gross Profit. . . 39,168 71
Cash $101,577.41
Installment Sales Contracts $101,577.41
Reserve for Unearned Gross Profit.*. ... $ 37,605. 14
Realized Gross Profit on Installment
Sales $ 37,605. 14
Bad debts. The bad debts written off during the year should
be allocated by years, and a charge should be made to Reserve
for Unearned Gross Profit for the percentage of gross profit in
each year's write-off, and to Profit and Loss (Bad Debts) for the
remainder, the entire credit^ being made to Installment Sales
Contracts.
GROSS PROFIT COMPUTATIONS 163
Example
During the fourth year, bad accounts were written off as follows:
First-Yr. Accts. Second-Yr. Accts. Third-Yr. Accis. Fourth-Yr. Accts.
$67 65 $141 05 $65 62 $126 25
What amount should be charged to these accounts: Profit and Loss (Bad
Debts), and Reserve for Unearned Gross Profit?
Solution
Unrealized
Profit Remainder
$ 67.65 X 33.57% $ 22 71 $ 44 94
141.05X42.98% 60.62 cSO 43
65.62X27.83% 1826 47.36
126.25 X 40.00% . . 50 50 75 75
$152.09 $248 48
Reserve for Unearned Gross Profit $152 00
Profit and Loss (Had Debts) . 248 48
Installment Sales Contracts $400 57
Problems
1. The X.Y.Z. Company's books for the 5th year showed:
Sales .... . $128,642 60
Gross profit 42,975 12
Collections were made in the fifth year on each year's contracts as follows:
1st Yr. 2nd Yr. 3rd Yr. 4th Yr. 5th Yr.
$230 60 $1,590 31 $9,326 80 $21,256 30 $82,327 58
Calculate: (a) the per cent of gross profit for the fifth year; (b) the amount
to be credited to Reserve for Unearned Gross Profit; (c) the amount to be debited
to Reserve for Unearned Gross Profit. Use the rates given in the solution on
page 162 for the first four years.
2. The analysis of bad debts written off during tire 5th year was:
Ist-Yr.Acct. 2nd-Yr.Acct. 3rd- Yr. Acct. 4th-Yr. Acct. 5th-Yr. Acct.
$8 35 $209 75 $910.40 $150 80 $470 62
What amounts should be charged to Reserve for Unearned Gross Profit and
to Profit and Loss, respectively?
3. Results for the 6th year:
Sales $140,695 39
Gross profit 54,541 07
Collections were made in the sixth year on each year's contracts as follows:
1st Yr. 2nd Yr. 3rd Yr. 4th Yr. 5th Yr. 6th Yr.
$62.70 $492.54 $2,798.30 $4,689.30 $2,657 80 $90,275.89
(a) Calculate the per cent of gross profit for the 6th year.
(b) Calculate for the 6th year the gross profit on collections made.
4. Accounts receivable were written off as follows:
1st Yr. 2nd Yr. 3rd Yr. 4th Yr. 5th Yr. 6th Yr.
$52.83 $31.50 $51.10 $150.00 $163.82 $108.28
164 GROSS PROFIT COMPUTATIONS
Compute the charges to be made to Profit and Loss (Bad Debts) and to
Reserve for Unearned Gross Profit.
5.* The "A & B" Company is engaged in the business of retailing musical
merchandise. The majority of the sales consist of installment sales of pianos
and talking machines, on which the initial payment is less than 25% of the sales
price and the balance is payable in monthly installments over a period of three
to five years. The company was incorporated and began business on January 1.
The following schedules are submitted on the various classes of merchandise:
SALES
Piano Install- Machine Install- Other
ment Sales ment Sales Mdse Sales
First year $148,650 00 $92,475 00 $38,337 60
Second year 163,520 00 88,535 00 39,543 50
Third year 180,400 00 94,256.00 40,731 . 15
PURCHASES
First year 106,322 37 67,432. 18 27,108 88
Second year 120,987 41 55,116 92 27,224 35
Third year 140,125 25 60,013.22 27,469 33
INVENTORIES
First year 20,103 14 10,248 31 8,323 64
Second year 32,105 86 15,012 83 15,299 41
Third year 39,29444 18,14477 13,52131
Attention is called to the fact that "Other Merchandise Sales" are sales for
cash, or credit sales other than installment sales.
No adjustments to Deferred Income account are made until the end of the
year. Additions to this income are made at the end of the year on the basis
of the balance due on the current year's installment sales, and deductions are
made on the basis of cash received during the current year on installment sales
of previous years. On December 31 of the third year, the unpaid balances on
third-year piano installment sales amount to $110,425.50, and on third-year
machine installment sales to $60,475.00 — exclusive of accrued interest. The
following amounts were received during the third year on installment sales of
previous years:
On first-year piano installment sales $30,285.00
On second-year piano installment sales 42,413 00
On first-year machine installment sales 25,386 . 00
On second-year machine installment sales 26,285.00
The above amounts are exclusive of interest, which is credited direct to
interest revenue.
Fractional percentages may be disregarded in the computation of ratios —
over -J- of 1 % should be added, and less than \ of 1 % should be dropped.
On first-year machine installment sales, uncollectible balances amounting to
$399.00 were charged on the books of the company to expense, and credited to
installment sales contracts.
Federal income taxes paid in the third year were charged to surplus.
* C. P. A., Michigan.
GROSS PROFIT COMPUTATIONS
165
Depreciation is calculated at the following rates:
Buildings 2% Furniture and Fixtures ... 10%
Auto Trucks 25%
The following is a copy of the trial balance as of December 31, end of thiid
year, before closing and before apportionment of deferred income on installment
sales :
Cash $ 15,327.48
Notes Receivable . 2,000.00
Accounts Receivable . 20,842 . 1 1
Installment Sales Contracts . . .. 205,418.50
Inventories .. 62,418.10
War Bonds 5,000.00
Real Estate 10,00000
Buildings . 40,00000
Furniture and Fixtures . . 4,500 00
Auto Trucks 3,000 . 00
Notes Payable $ 50,000 . 00
Accounts Payable 13,458 25
Deferred Income on Installment Sales 83,245 70
Reserve for Depreciation, Buildings 1,600 00
Reserve for Depreciation, Fur. <fc Fix 900 00
Reserve for Depreciation, Trucks ... 1 ,500 00
Capital Stock . ... 150,000 . 00
Surplus 75,556.21
Sales 315,387.15
Piano Rentals 1 ,785 00
Interest on Installment Sales . . 2,035 23
Interest on War Bonds 237 50
Cash Discounts on Purchases . . . 2,452.07
Purchases 227,607.80
Salaries, Officers 14,000 00
Salaries, Store 8,10146
Light and Heat 717.68
Advertising 4,01571
Truck Expense ... 508.53
Sundry Store Expense 2,239 17
Salaries, Office 2,020 00
Traveling Expense .... 648 50
Postage 472 30
Telephone and Telegraph 441 40
Insurance 1 ,309 06
Real Estate and Personal Property Taxes 2,029 69
Bad Debts, Accounts Receivable 709 66
Bad Debts (first year machine install-
ment sales) 399 00
Repairs, Sundry 365. 68
Donations 20000
Cash Discounts on Sales 444 48
State Franchise Tax 187.80
Capital Stock Tax 233.00
Interest Paid 3,000 00
$698,157.11 $698,157.11
166 GROSS PROFIT COMPUTATIONS
You are asked to give: (a) the net taxable income (for federal tax purposes)
for the third year; (6) a balance sheet of the "A & B" Company as of January 1,
beginning of fourth year.
Deferring income; its effect on tax. The statement was made
in the second paragraph of this subject that the installment basis
of accounting defers income with a view to the possible effect that
deferment may have on the amount of federal income tax to be
paid. Since the income is deferred, the tax is deferred (not saved).
The amount of profit realized and to be realized from the sales
of a particular year, if not taxed in that particular year, will be
taxed eventually, and the saving of tax results from a possible
reduction in the rate of tax or from the spreading of taxable income
over several years. If it is anticipated that the rate of tax will be
increased, it may not be wise to defer the income.
Second, a change from the accrual to the installment basis
results in double taxation, for Section 44 (c) of the Internal
Revenue Code provides as follows: "If a taxpayer entitled to the
benefits of subsection (a) elects for any taxable year to report
his net income on the installment basis, then in computing his
income for the year of change or any subsequent year, amounts
actually received during any such year on account of sales or other
dispositions of property made in any prior year shall not be
excluded/'
The amount of gross income which may be deferred on install-
ment sales is governed by :
(1) The terms of sale;
(2) Annual increase, if any, in sales;
(3) Per cent of year's sales collected in the current year; and
(4) Fluctuation of gross profits.
Example
Assume the terms of sale to be 10% down, and 10% a month; the annual
increase in sales to be $10,000; the per cent of year's sales collected, and the
sales throughout the year, to be uniform; and the per cent of gross profit to be
fixed.
Gross Profit
Year Sales on Sales
First $50,000 30%
Second 60,000 30%
Third 70,000 30%
Fourth 80,000 30%
Fifth 90,000 30%
Since it has been assumed that the sales are uniform throughout the year
and that collections are met promptly, the second year's business may be analyzed
as follows:
GROSS PROFIT COMPUTATIONS
167
Down Payments
Jan 10% of $ 5,000
Feb 10% of
Mar 10% of
Apr 10% of
May 10% of
of
of
June 10'
July 10<
Aug 10% of
Sept 10% of
Get 10% of
Nov 10% of
Dec 10% of _
Year's sales $60,000
J )own payments
Install, payments
Total payments
Ratio of payments to sales: $37,500 -r
5,000
5,000
5,000
5,000
5,000
5,000
5,000
5,000
5,000
5,000
5,000
Installment Payments
500
500 10% of $ 5,000 = $ 500
500 10% of 10,000 = 1,000
500 10% of 15,000 = 1,500
500 10% of 20,000 = 2,000
500 10% of 25,000 = 2,500
500 10% of 30,000 = 3,000
500 10% of 35,000 = 3,500
500 10% of 40,000= 4,000
500 10% of 45,000= 4,500
500 10% of 45,000 = 4,500
500 10% of 45,000= 4,500
. $ 6,000
. 31,500
. $37^00
$60,000 = 62.5%.
$31,500
A comparison of the income to be reported on the accrual basis
and on the installment basis may be made as follows:
ACCRUAL BASIS
Sales
Second
Year
. $60,000
30%
. . 1S.OOO
Third
Year
$70,000
30%
21,000
BASIS
$22,500
43,750
Fourth
Year
$SO,()0()
30%
24,000
$26,250
50,000
Fifth
Year
$90,000
30
27,000
$30,000
56,250
Gross profit ( %)
Gross profit ($)
INSTALLMENT
Collections :
Ist-year accounts $18,750
2nd-year accounts 37.500
3rd-year accounts
4th-year accounts
5th-year accounts . . .
Gross income to be reported:
30 % of Ist-year coll
30 % of 2nd-year coll
$ 5,625
. 11,250
$ 6,750
13,125
$ 7,875
15,000
$ 9,000
16,875
30 % of 3rd-year coll
30 % of 4th-year coll . . .
30 % of 5th-year coll
Total income reported
. . $16,875
$19,875
$22,875
$25,875
Income deferred . . .
. . $ 1,125
$ 1,125
$ 1,125
$ 1,125
It may be observed from the foregoing analysis that with an
annual increase of $10,000 in sales, and with a constant gross profit
ratio of 30%, the amount of income deferred from year to year is
$1,125.
With an annual increase of $20,000 in sales, and other cond>
168 GROSS PROFIT COMPUTATIONS
tions the same, the amount of income deferred would be $2,250
(2 X $1,125).
Problems
1. Assume the terms of sale to be 10% down and 5% a month, the annual
increase in sales $10,000, the per cent of year's sales collected and the sales
throughout the year uniform, and the per cent of gross profit fixed.
Gross Profit
Year Sales on Sales
First $50,000 30%
Second 60,000 30%
Third 70,000 30%
Fourth 80,000 30%
Fifth 90,000 30%
Show the amount of income deferred when the installment basis is used.
2. If the terms of payment were 5% down and 5% a month, and other con-
ditions were the same as in Problem 1, what would be the amount of income
deferred each year?
CHAPTER 19
Analysis of Statements
Financial and operating ratios. An analysis of the financial
and the operating ratios of a business means a study of the relation-
ships that are expressed in the statistics presented. Well-known
and commonly used ratios are those of expenses and earnings to
sales, and of earnings on capital employed. Other ratios, relation-
ships, and turnovers that are indicators of the condition of a busi-
ness should also be considered.
A summary of financial and operating ratios, relationships, and
turnovers would include the following:
(1) Ratio of costs and expenses to net sales.
(2) Ratio of gross profit to net sales.
(3) Ratio of operating profit to net sales.
(4) Ratio of net profit to net sales.
(5) Ratio of operating profit to total capital employed.
(6) Ratio of net profit to net worth.
(7) Earnings on common stockholders' investments.
(8) Working capital ratio.
(9) Sources of capital.
(10) Manner in which capital is invested.
(11) Turnover of total capital employed.
(12) Turnover of inventories.
(13) Turnover of accounts receivable.
(14) Turnover of fixed property investment.
There«are many other ratios which are important measures of
efficiency, but of which only brief mention can be made in this
chapter. Depending on the type of business being analyzed,
these other ratios might include the labor turnover, the unit of
output per operative, the average wage per man, the average wage
per hour, and other statistics.
Costs, expenses, and profits. Costs, expenses, and profits
should be expressed as per cents of money values and, where possi-
ble, should be expressed in terms of dollars per production unit,
such as the ton, pound, yard, or gallon. The per cents, compared
with those of previous years, show whether sales prices have been
169
170 ANALYSIS OF STATEMENTS
adjusted proportionately to cOvSts of production and distribution.
The unit prices supplement the per cents and afford a direct
comparison.
Ratio of gross profit to net sales. The ratio of gross profit to
net sales is an indication of the spread between the cost of pro-
duction and the selling price. The gross profit must be as large
as possible, for out of it must come the expenses of selling, adminis-
tration, finance, and other charges, before a net return is realized
on capital.
Ratio of operating profit to net sales. The ratio of operating
profit to net sales expresses the basic relationship between profits
and sales. Operating profits represent the gain before the deduc-
tion of federal taxes, interest on borrowed money, and extra-
ordinary losses, but do not include miscellaneous income not
attributable to ordinary operations.
Ratio of net profit to net sales. The ratio of net profit to net
sales indicates the margin of profit on the selling price. The
rapidity of stock turnover, and the capital invested in accounts
receivable, in inventory, and in plant, should be considered with
this ratio.
Ratio of operating profit to total capital employed. The ratio
of operating profit to total capital employed forms a ready basis for
a comparison of the operating results of a business or of several
plants under a single control. Capital employed includes plant,
inventories, accounts receivable, cash balances, and so forth,
regardless of the source of such capital, and is readily determined
by referring to the asset side of the balance sheet.
Ratio of net profit to net worth. The ratio of net profit to net
worth expresses the measure of earnings available to the stock-
holders or proprietors, and is the final indicator of the success or
failure of any business.
Earnings on common stockholders' investments. The earn-
ings on common stockholders' investments are based on the
stockholders' share of the net profit, in relation to their interest in
the net worth of the business. There are two ways in which these
earnings may be stated: (a) as a per cent of the amount of such
investments; and (6) in dollars earned per share outstanding.
Example
The following profit and loss statement, together with certain other facts,
is presented to illustrate items 1-7 in the summary on page 169. The numbers
in parentheses refer to the numbered ratios in the summary.
ANALYSIS OF STATEMENTS 171
BLANK MERCANTILE COMPANY
PROFIT AND Loss STATEMENT
FOR THE TWELVE MONTHS' PERIOD ENDED DECEMBER 31, 19 — .
Sales:
Gross Sales ........... $693,004 . 10
Less: Sales Rebates and
Allowances ... $ 870.64
Prepaid Freight . . ___ 200.25
^ 1,070^89
-- - -
^
Net Sales ........... -- - - $691j933 21 1C0.00%
Cost of Sales:
Inventory, beginning of
year .......... $107,278 46
Purchases .......... $624,225 28
Freight ............ 16,271 98
$640,497"26
Less: Purchase Rebates
and Allowances 630 81
639,866 45
$747,144 91
Inventory, end of year. 124,814 04
Cost of Sales ........ ~~~ 622,330^87 89 94 (1)
Gross Profit .......... $ 69,602~34 10"~06% (2)
Delivery Expenses:
Salaries of Drivers ..... $ 3,414 34
Dep'n on Equipment . . 2,839 57
Auto Repairs ....... 1,562 53
Gasoline and Oil ..... 1,479 27
Drivers' Expenses ..... 1 19 40
Drayage .............. 66 84
Total ............... $ 9,481.95 1.37 (1)
Selling Expenses:
Salesmen's Salaries ..... $ 11,812 50
Salesmen's Expenses. . . 1,942.06
Advertising ........... 844 . 32
Telephone and Tele-
graph .......... 642.57
Total ............... $ 15,241.45 2.20 (1)
172
ANALYSIS OF STATEMENTS
General Expenses:
Salaries $ 8,722 33
Expenses 613 36
Executive Salaries 3,600 00
Taxes (other than fed-
eral) 1,906 23
Insurance 1,723 46
Depreciation 1,259 54
Light, Heat, and Water 829 49
Printing and Stationery 444.50
Postage 408.52
Collections 219 76
Repairs 115 91
Storage 22 69
Miscellaneous 380 50
Total " 20,246 29 2 93 (1)
Total Expense " 44,969 69 6.50% (1)
Net Operating Profit $ 24,632 65 3 56% (8}
Additions to Income:
Discount on Purchases. $ 9,565 86
Interest Earned 563 32
Bad Debts Recovered . 102 53
Total _10,23JL71 1.48
$ 34,864.36 5.04%
Deductions from Income:
Discount on Sales $ 4,771 . 92
Interest Paid for Money
Borrowed 4,373 16
Interest Paid on Build-
ing Contract 3,010 . 00
Bad Debts Reserve .... 1,283 . 91
Donations 162 00
Total " 13,600.99 1 97
Net Profit $ 21,263.37 3 07% (4)
Supplemental
Total Capital Used (see Balance Sheet, below) $276,317.34
Ratio of Profit to Capital 7 . 69% (5)
Net Worth (beginning of year) 124,252.36
Ratio of Profit to Net Worth 17 . 11 (6)
Common Stock Outstanding 1 13,400 .00
Number of Shares ($50.00 par value) 2,268
Per cent earned 18 . 75 (7)
Dollars earned per share 9 . 38
Working capital ratio. This ratio is probably the best-known
measure applied to financial statements, because more than any
other it has been stressed by bankers and businessmen. It is
computed by dividing the amount of the current assets by the
amount of the current liabilities. If the quotient is 2, the current
assets are said to be in a "2 to 1 " ratio; that is, in a ratio of $2 of
current assets to each $1 of current liabilities.
ANALYSIS OF STATEMENTS 173
What the working capital ratio should be depends upon differ-
ences in types of business, location, and other factors, the effect of
which is to vary somewhat the proportions involved. While some
lines of trade may be expected to maintain a 2-to-l ratio, others
may necessitate a proportion as high as 10 to 1.
The rapidity with which receivables and inventory are turned is
a factor bearing on the adequacy of the working capital ratio.
With respect to accounts receivable, there is a range of turnover
from 3 days in some of the retail chain stores to 80 or 90 days in
coal and heavy manufacturing industries. The turnover of
inventories is most rapid in such industries as slaughtering and
meat packing, retail chain stores, chemical products, and iron and
steel, while the turnover of inventories is found to be slow in such
industries as tobacco products, machinery manufacturing, leather
products, and rubber goods.
Example
The following balance sheet is presented to illustrate the working capital
ratio. It will also be referred to in later paragraphs, where the computation of
other ratios is discussed.
BLANK MERCANTILE COMPANY
BALANCE SHEET
DECEMBER 31, 19 — .
Assets
Current:
Cash in Banks $ 13,598.85
Cash on Hand 4,113.24 $ 17,712 09
Accounts Receivable — Customers $ 64,832 57
Accounts Receivable — Others . . . 647 92
Notes Receivable — Customers 5,329 91
Notes Receivable— Others 227.31
Securities 1,274 34
Accrued Interest 32 98
Railroad Claims i3_J6.
$"72^438779
Less: Reserve for Bad Debts . 1,890.06 70,548 73
Merchandise Inventory . 124,814 04
Total $213,074.86
Fixed:
Land $ 3,450.00
Warehouse Building $ 50,373.48
Warehouse Equipment 545 . 77
Delivery Equipment 14,090 . 39
Furniture and Fixtures 2,488 . 85
$ 67,498 49
Less: Accumulated Depreciation 9,152.48 58,346.01
Total $ 61,796.01
174 ANALYSIS OF STATEMENTS
Deferred Charges:
Prepaid Insurance $ 1,298. 13
Prepaid Interest 148 34
Total 1,446 47
$276,317 34
Liabilities
Current:
Payroll $ 1,131.77
Accounts Payable 16,177.08
Notes Payable— Banks 50,000 00
Notes Payable— Others 17,600 00
Notes Payable— Stockholders 11,700 00
Accrued Taxes 1,575 17
Accrued Interest — Notes 1 ,393 92
Accrued Interest— Contracts 3,010.00
Total $102,587,94
Fixed:
Warehouse Contract for Deed . 43,000 00
Net Worth:
Capital Stock— Common $11 3,400 00
Surplus 17,329 40 130,729 40
$276,317 34
In the foregoing balance sheet, the current assets are stated at $213,074.86,
and the current liabilities are stated at $102,587.94.
213,074.86 -T- 102,587.94 = 2.077.
The ratio of working capital is, therefore, 2.077.
Sources of capital. The sources of capital may be stated in a
general way under four heading's, as follows:
(1) Short-term borrowings and credits.
(2) Long-term borrowings and credits.
(3) Stockholders' investments.
(4) Surplus (earnings left in the business).
Summarizing the liability section of the foregoing balance
sheet and dividing each section total by the total of all sections,
the ratio of capital supplied by each source is as shown in the
right-hand column of the following tabulation :
Amount Per Cent
Current Liabilities $102,587.94 37. 13
Fixed Liabilities 43,000.00 15.56
Capital Stock— Common 113,400.00 41 .04
Surplus 17,329.40 6.27
1276,317.34 100. 00
ANALYSIS OF STATEMENTS 175
Manner in which capital is invested. The manner in which
the capital is employed in the business is shown by a summary
of the asset sections.
Amount Per Cent
Current Assets $213,074.86 77. 12
Fixed Assets 61,796.01 22.36
Deferred Charges 1,446.47 .£2
$276,317.34 100.00
Turnover of total capital employed. This item expresses the
relation of the net sales to the total capital employed. The aver-
age capital employed throughout the year should be used, but, in
the absence of monthly statements, the capital at the beginning
of the year and the capital at the end of the year should be added
and divided by two to give an estimate of the average capital
employed. In arriving at this average, investments not employed
in operations should be eliminated from the total assets, for, as a
rule, they represent a surplus not required in the conduct of the
business. Income from such investments should be eliminated
from the statement of earnings before the ratio is computed.
Total assets at beginning of year $246,351 . 89
Total assets at end of year 276,317 . 34
2)$522,669~23
Average capital employed (securities not eliminated,
as the amount was negligible) $261,334.61
The turnover of total capital employed is, therefore :
$691,933.21 (net sales) -f- $261,334.61 (average capital) = 2.64.
Turnover of inventories. The subject of inventory turn-
over was presented in Chapter 17.
The rate of turnover is computed as follows:
$622,330.87 (cost of sales) •*- $112,131.69 (average inventory) = 5.55.
Turnover of accounts receivable. The normal credit period,
whether it be 30, 60, or 90 days, is compared with the average
number of days' sales uncollected obtained from the following
formula, as a means of judging the efficiency of the collection
department :
A Counts receivable at end of fiscal period _. . . _ . ,
— — - — - — : r-; X Days in fiscal period
Sales for fiscal period
— Average number of days' sales uncollected.
The Accounts Receivable account showed $64,832.57 of out-
standing accounts at the close of the fiscal period. The sales for
the fiscal period of 12 months amounted to $691,933.21, and the
176
ANALYSIS OF STATEMENTS
average term of credit granted at time of sale was 30 days. The
average number of days' sales represented in standing accounts is
computed as follows:
84,832.57
691,933.21
X 365 = 34.
If the average number of days' sales uncollected is greater than
the average term of credit, the presence of overdue accounts is
indicated. This is true of the example just given.
Turnover of fixed property investment. This turnover ex-
presses the relationship between the volume of business done and
the capital invested in plant and equipment. Large investments
in property and equipment increase the expense burden through
charges for depreciation, insurance, taxes, and so forth, and may
make a favorable or an unfavorable operating statement, depend-
ing on the volume of business handled.
The number of dollars of sales for each dollar of fixed property
investment is calculated as follows :
$691,933.21 (net sales) -r- $58,346.01 (net fixed property investment) = 11.86.
Problems
1. From the balance sheets and supplemental information, determine the
ratios named, following the balance sheets.
This Year Last Year
Current Assets $215,003 48 $213,074.86
Fixed Assets— Net 57,535 04 61,796.01
Deferred Charges _M93.59 1,446.47
Total $2737732.11 $276,317734
Liabilities
Current Liabilities $ 86,229 . 30 $102,587 . 94
Fixed Liabilities 38,000 00 43,000 00
Total Liabilities $124,229730 $145,587794
Net Worth
Capital Stock $124,300.00 $114,300.00
Surplus 25,202.81 16,429.40
Total Net Worth $149,502.81 $130,729.40
Total $273/732. 11 $276,317.34'
Annual Sales i688,f67798 $691,933^21
Annual Expense 47,340.74 44,969.69
Ratios
Current Ratio
Worth to Debt
Worth to Fixed Assets
Sales to Fixed Assets
Sales to Current Debt
Sales to Worth
Expense to Sales (%)
ANALYSIS OF STATEMENTS
177
2. The United Manufacturing Company's card in the credit file of the Second
National Bank contained the data for the year ended January 31, 1944, and
from their balance sheet and profit and loss statement you have entered the
comparative figures for the year ended January 31, 1945. Compute the com-
parative ratios for 1945.
COMPARATIVE RATIOS
COMPARATIVE FINANCIAL STATEMENTS
1/31 1/31
_ 1944 1945 19 19 19
ASSE
TQ I/Si t/31
TS 1944 1945 19 19 19
FIXED ASSETS TO
TANGIBLE NET WORTH
24.8
CASH
3,206
1,862
ACCOUNTS RECEIVABLE
45,199
42,267
CURRENT DEBT TO
TANGIBLE NET WORTH
48.0
NOTES TRADE AO
EPT RECV.
INVENTORIES
89,342
83,218
NET WORKING CAPITAL REP
BY FUNDED DEBTS
NET SALES TO
INVENTORY
4.3
NET WORKING CAPITAL REP
BY INVENTORY
107.7
TOTA
CURRENT f
168,709
127^349
DUE FROM AFFILIA
FE OR SUBS'Y
INVENTORY COVERED BY
CURRENT DEBT
61.3
LAND BUILDINGS
MACHINERY.flXTl
t£S
L_ 28,244
49,248
AVERAGE COLLECTION PERIOD
42.5
NOTES ACC'TS (Of TORS. PARTNERS)
2L716
2,156
TURNOVER OF
TANGIBLE VET WORTH
3.4
TURNOVER OF
NET WORKING CAPITAL
4.7
TO'
U. ASSETS
168,709
179.754
NET PROFITS ON NET SALES
.52
LIABILITIES
NET PROFITS ON
TANGIBLE NET WORTH
1.8
ACCEPT , NOTES PAYABLE
NET PROFITS ON
NET WORKING CAPITAL
2.4
PAYABLE AFFILIATE OR SUBS'Y
— AtCnilALS —
' '
'
CURRENT ASSETS
TO CURRENT DEBT
2 5
Due Officers
_8,130
.5,321
TOTAL DEBT INCLUDING N W
TANGIBLE NET WORTH
148.0
TOTAL CURRENT
54,725
55,825
SALES
388,553
394,774
'MORTGAGES
EXPENSES
Deferred Bank Loan
8,280
NET PROFIT
2,048
1,664
WORKING CAPITAL
83,022
71,523
TANGIBLE NET WORTH
113,963
114,649
-~
FIXED ASSETS
26,244
49,244
CAPITAL STOCK
103,^00
los'.Voo
FUNDED DEBT
8,280
SURPLUS
10,883
12.548
, ,_i,
3. From the data given in the following balance sheet and profit and loss
statement, together with the supplemental data, compute the fourteen financial
and operating ratios relationships, and turnovers outlined in the preceding
sections of this chapter.
178 ANAL/SIS OF STATEMENTS
BLANK MERCANTILE COMPANY
BALANCE SHEET
DECEMBER 31, 19 —
Assets
Current:
Cash in Banks $ 13,771 58
Cash on Hand 3,616 34 $ 17,387.92
Accounts Receivable — Customers . . $ 59,424 48
Accounts Receivable— Others 704 30
Notes Receivable— Customers. .. 3,746 76
Notes Receivable— Others 272. 19
Securities 994 64
Accrued Interest 52 30
Railroad Claims 50J)5
$~65,245~62
Less: Reserve for Bad Debts 3,852.57 61,393 05
Merchandise Inventory 136,222 51
Total $215,003 48
Fixed:
Land $ 3,450.00
Warehouse Building $ 50,180 55
Warehouse Equipment 545 77
Delivery Equipment . ... 14,090 39
Furniture and Fixtures 2,503 85
$ 67,320 56
Less: Accumulated Depreciation . 13,235 52 54,085 04 57,53504
Deferred Charges:
Prepaid Insurance 1,193 59
Total $273J32"1T
Liabilities
Current:
Payroll $ 1,116.17
Accounts Payable 13,325 . 73
Notes Payable— Banks 24,000 00
Notes Payable— Others 14,500 00
Notes Payable— Stockholders. . . 26,70000
Accrued Taxes 1,641 97
Accrued Interest— Notes 2,285 33
Accrued Interest — Contracts.. . 2,660.00
Total $86,229.20
Fixed:
Warehouse Contract for Deed . 38,000 . 00
Net Worth:
Capital Stock— Common $123,400.00
Surplus 26,102.91 149,502.91
Total $273,732.11
ANALYSIS OF STATEMENTS
BLANK MERCANTILE COMPANY
179
PROFIT AND Loss STATEMENT
FOR THE YEAR ENDED DECEMBER 31, 19 — .
Sales:
Gross Sales $689,361.43
Less: Sales Rebates and Al-
lowances $ 1,059 89
Prepaid Freight 133 56
__1,193.45
Net Sales $688,167.98 100.00%
Cost of Sales:
Inventory, beginning of year $124,814.04
Purchases.... $611,332.45
Freight __iy 84_6^
$626,517713
Less: Pur. Rebates and Al-
lowances 1,392 74
~ 625,124.39
$749,938.43
Inventory, end of year 136,222 51
Cost of Sales ' 613,715 92 %
Gross Profit $ 74,452.06 %
Delivery Expenses:
Salaries of Drivers $ 3,874 . 27
Dep'n on Equipment 2,818 . 08
Auto Repairs 1,430.61
Gasoline and Oil 1,231 29
Drivers' Expenses 125 35
Drayage 52 91
Total " $ 9,532.51 %
Selling Expenses:
Salesmen's Salaries $ 12,300 00
Salesmen's Expenses 2,015 78
Advertising 1,357 . 83
Telephone and Telegraph ... 536 . 2 1
Total 16,209.82 %
180
ANALYSIS OF STATEMENTS
General Expenses :
Salaries
Expenses
Executive Salaries
Taxes (other than federal) .
Insurance
Depreciation
Light, Heat, and Water . . .
Printing and Stationery. . .
Postage
Collections
Repairs
Storage
Miscellaneous
Total
Total Expense
Net Operating Profit
Additions to Income:
Discount on Purchases . .
Interest Earned
Bad Debts Recovered . .
Total
Deductions from Income:
Discount on Sales
Interest Paid for Money
Borrowed
Interest Paid on Building
Contract
Bad Debts Reserve
Donations
Total
Net Profit
8,797 50
265 43
4,175 00
2,069 17
1,937 82
1,264 96
826 33
516 70
486 85
238 65
106 47
18 29
895 24
$ 21,598.41
47,340 74
$ 27,111 32
$ 9,759 20
1,348 60
10 65
11,118.45
$ 38,229.77
$ 4,523.98
4,443.87
2,660 00
3,446 80
269 20
15,343 85 ...
$ 22,885.92 ...
Supplemental
Total Capital Employed (see Balance Sheet) $
Ratio o^f Profit to Capital
Net Worth (beginning of year) 130,729 . 40
Ratio of Profit to Net Worth
Common Stock Outstanding (see Balance Sheet)
Number of Shares ($50.00 per value) . ....
Per Cent Earned
Dollars Earned Per Share
CHAPTER 20
Partnership
Definition. A partnership association is defined by Chancellor
Kent as follows: " A contract of two or more competent persons to
place their money, effects, labor, and skill, or some or all of them,
in lawful commerce and business, and to divide tiie profits and
bear the losses in certain proportions."
Mathematical calculations. The most important mathemati-
cal calculations in partnership accounting are concerned with :
(1) Division of profits.
(2) Division of assets upon liquidation.
(3) Calculation of goodwill.
Goodwill. The calculation of goodwill also has to be considered
in connection with the other types of business organizations-
namely, individual proprietorship and corporation — -when changes
in ownership, reorganizations, consolidations, and so forth, are
made; see Chapter 21.
Profit-sharing agreements. Profits may be shared in many
ways. A few of the most common methods of profit distribution
are:
(1) Arbitrary ratios.
(2) In the ratio of capital invested at organization of business.
(3) In the ratio of capital accounts at the beginning or at the
end of each period.
(4) In the ratio of average investments.
(5) Part of the profits may be distributed as salaries or as
interest on capital invested, and the remainder in some other
ratio.
(6) If the investment is less than the amount agreed upon,
interest is charged o™ the shortage ; and if the investment is more
than the amount agreed upon, interest is credited on the excess;
the resulting profit or loss is then distributed in a ratio agreed upon.
Lack of agreement. If the partners have failed to include
in their articles of co-partnership an agreement as to the method
by which profits are to be distributed, the law provides that the
181
T82 PARTNERSHIP
profits shall be divided equally, regardless of the ratio of the part-
ners7 respective investments.
Losses. If losses are incurred and no provision has been made
for their distribution, the profit-sharing ratio governs.
Arbitrary ratio.
Example
A and B are partners. A has $3,000.00 invested, while B has $2,500.00
invested. A is to receive f of the profits, and B is to receive £. The profits
for the year are $2,400.00. What is each partner's share?
Solution
Net profits $2,400 00
A's share, f of $2,400.00 1,600 00
B's share, i of $2,400.00 800.00
Problems
A and B were partners. Gain or loss was to be divided f and f , respectively.
A invested $3,500.00, and B invested $2,400.00. During the year, A withdrew
$500.00, and B withdrew $700.00. At the end of the year the books showed the
following assets and liabilities:
Cash on Hand and in Bank $8,000 00
Inventory of Merchandise 7,500 00
Notes Receivable 790 00
Accounts Receivable 840 . 00
Notes Payable . . 4,70000
Accounts Payable 7,24000
(a) What has been the gain or loss? (6) What is each partner's net capital
at the end of the year?
Ratio of investment.
Example
January 1 , A's investment $10,000.00
January 1 , #'s investment 6,000 00
January 1, C's investment 4,000 00
Total ' $20,000700
December 31, Profits $ 4,000 00
Profits are to be shared in the ratio of investments at the beginning of the
year.
Solution
Investment Ratio Profits Shares
A $10,000 i£ $4,000 $2,000
B 6,000 A 4,000 1,200
C 4,000 A 4,000 800
$20,000 |S $4,000
Explanation. Add the beginning-of-year investments of each of the partners,
and take for the numerator of the fraction representing each partner's share his
investment at the beginning of the year, and for the denominator the total
PARTNERSHIP 183
capital. Using these fractions, calculate the fractional parts of the net profit
or lose, and these will be the partners' shares.
Problems
In each of the following, show the division of net profit or net loss, which
is to be calculated in the ratio of investments:
INVESTMENTS
A B C NET PROFIT NET Loss
1. $4,000 $4,000 $2,000 $2,500
2. 5,000 3,000 1,500 $1,200
3. 6,000 7,500 2,500 2,000
4. 2,000 3,500 1,500 1,400
5. 3,500 2,500 1,000 750
Division of profits by first deducting interest on capital.
Example
January 1, A's investment $10,000
January 1, B's investment 6,000
January 1 , f"s investment 4,000
December 31, Net profits 4,000
By agreement, each partner is to receive 5% interest on his investment
(this interest to be deducted from total profits), and the balance of the profits is
to be distributed equally.
Solution
A's investment, $10,000, X .05 $ 500, interest
ZTs investment, 6,000, X .05 300, interest
C's investment, 4,000, X .05 200, interest
Total $l7XXJ
Net profits, $4,000 - $1,000 - $3,000, to be divided equally. $3,000 -r- 3
= $1,000, each partner's share after interest is deducted.
Interest Profit Total Credit
A $500 $1,000 $1,500
B 300 1,000 1,300
C 200 1,000 1,200
Total $4,000
Problems
Show the division of profits in each of the following:
RATE OF INT.
INVESTMENTS NET ON BALANCE TO
ABC PROFITS INVESTMENT BE DIVIDED
1. $ 8,000 $ 4,250 $ 3,700 $4,000 5% Equally
2. 9,750 3,500 10,000 6,000 6% Equally
3. 4,725 5,300 5,250 5,300 6% Equally
4. 12,000 6,000 4,000 4,500 4% *, T, I
5. 20,000 10,000 5,000 5,000 6% i, i, i
Profits insufficient to cover interest on investment. If it is
agreed that each partner is to be credited with interest on his
184 PARTNERSHIP
invevStment, the interest must be credited to each partner, even
though the total profits are not large enough to cover the credit.
Any over-distribution incurred by the distribution of the interest
should be divided among the partners in accordance with the agree-
ment as to the division of profits. The same rule applies where
there is a loss before interest is credited.
Example
January 1, A's investment $10,000
January 1, B's investment 6,000
January 1^ C"s investment 4,000
December 31, Business profits 700
By agreement, each partner is to receive 5% interest on his investment, and
the profits are to be shared equally.
Solution
A's investment, $10,000, X .05 $ 500, interest
£'s investment, 6,000, X .05 300, interest
C"s investment, 4,000, X .05 200, interest
Total interest to be credited $ 1 ,000
Profits earned 700
Net loss $ 300
Since the loss is to be shared equally, each partner's loss is $100.
Credit Debit Net
Interest Loss Credit
A $500 $100 $400
II 300 100 200
(J 200 100 100
Total $700
Problems
Find the net credit or debit to each partner in each of the following:
INVESTMENTS PROFIT OR Loss INTEREST BALANCE TO
ABC BEFORE INTEREST RATE BE DIVIDED
1. $ 8,000 $ 8,000 $ 4,000 Profit, $ 800 6% Equally
2. 5,000 7,000 2,000 Profit, 140 6% A, A, A
3. 3,800 4,200 5,000 Loss, 200 6% Equally
4. 10,000 7,500 5,000 Profit, 2,150 6% f , $, i
6. 15,000 15,000 10,000 Profit, 2,000 6% Equally
Adjustments of capital contribution. If the partners do not
invest the agreed amounts, adjustments may be made, provided the
contract so states. Partners may be charged with interest en
the amount of the shortage of their investment from the agreed
amount, and may be credited with interest on the excess of their
investment over the agreed amount. These adjustments should
be mad(t before the profits for the period are prorated. If interest
adjustments result in an over-distribution of profits, the amount
PARTNERSHIP 185
over-distributed is divided in the ratio of the division of profits,
unless otherwise agreed.
Example
Agreed to Invest Invested
January 1, A $10,000 $12,000
January 1, B 6,000 5,000
January 1, C 4,000 2,000
December 31, Profits for the year. ... 3,100
By agreement, A is to be allowed 5% interest on his excess investment, and
B and C are to be charged 5% interest on their shortages. After these adjust-
ments have been made, profits are to be divided equally.
Solution
A's excess, $2,000, X .05 $100, interest
7?'s shortage, $1,000, X .05 50, interest
C"s shortage, $2,000, X .05 100, interest
Charge to #'s account $ 50
Charge to C's account TOO $150
Credit to A's account 100
Net amount of interest $ 50
The net amount of interest, $50, is added to net profits.
Profits before distribution:
Net profits $3,100
Add net interest 50
Total $3,150
$3,150 -T- 3 = $1,050, each partner's share after interest adjustment.
Afs* profits $1,050
Add interest 100 $1,150, total credit of A
B's -J- profits $1,050
Less interest 50 1,000, net credit of B
C's -J profits $1,050
Less interest 100 950, net credit of C
Total profits $3,100
Problems
1. Prepare a statement of profit distribution from the following facts:
A B C D
Agreed investment $6,000 $6,000 $8,000 $4,000
Investment 7,000 6,000 6,000 2,500
Profit ratio after adjustment of
6% interest on excess or de-
ficiency of investment 25% 25% 33J% 16|%
Net profits before adjustments for interest, $6,500.
2. Prepare a statement of profit distribution from the following facts:
X Y Z
Agreed investment $5,000 $4,500 $4,500
Investment 4,000 4,000 6,000
186 PARTNERSHIP
Profits to be shared equally after adjustments for 6% interest. Profits before
adjustments for interest, $600.
3. The capital of a certain organization was to be $40,000.00, of which A and R
were to contribute one-half each, A to receive 55% of the profits and B to receive
45%. A, being short of funds, invested only $15,000.00, and, the firm being
short of capital, 11 put in the balance until A could make up his shortage, with
the provision that he be allowed 6% interest on the excess of his investment
over the agreed amount. The profits for the year were $12,000.00. Show
distribution of profits.
Profit sharing in ratio of average investment.
First method. Multiply the original investment by the number
of days or months during which the amount was in the business
without change. The product may be termed Day-Dollars or,
Month-Dollars. The ratio of any product to the total of the prod-
ucts is the average capital ratio for that partner.
When the capital is changed, either by additional investment
or by withdrawal, the changed capital is multiplied by the number
of days or months to find its value in day-dollars or month-dollars,
and for each change a new calculation is made. The ratio of the
total of the day-dollars or month-dollars for each partner to the
sum of the day-dollars or month-dollars for all the partners gives
the ratio of each partner's investment to the total investment.
Example
A
Debit ('red it
Feb. 1 $1 ,000 Jan. 1 $10,000
June 1 1,500 May 1 4,000
Nov. 1 500 July 1 1,000
H
July 1 $1,000 Jan. 1 $ 6,000
Dec. 1 1,000 Aug. 1 4,000
Oct. 1 2,000
Net profits of the business for the year were $4,530.
Solution
A
MONTHS IN MONTH-
INVESTED BUSINESS DOLLARS
Jan. 1, $10,000 X 1 month $10,000
Feb. 1, 9,000 X 3 months 27,000
May 1, 13,000 X 1 month 13,000
June 1, 11,500 X 1 month 11,500
July I, 12,500 X 4 months 50,000
Nov. 1, 12,000 X 2 months 24,000
A's month-dollars investment $135,500
PARlNtRSHIP 187
R
Jan. 1, S 6,000 X 6 months $36,000
July 1, 5,000 X 1 month 5,000
Aug. 1, 9,000 X 2 months 18,000
Oct. 1, 1 1,000 X 2 months 22,000
Dec. 1, 10,000 X 1 month 10,000
B's month-dollars investment $ 01 ,000
Total month-dollars investment $226,500
A's share of profits, ty~~y, <>f $4,530 $ 2,710
^^O,OUU
01 000
#'s share of profits, ~>,.--rrnn of $4,530. . 1,820
22b"3°° $ 4,530
If the average investment is desired, it can be found by dividing the month-
dollars by 12, as:
A's month-dollars, S135,500 -~ 12 . $1 1,201 67
B's month-dollars, SOI, 000 4- 12 7,583 33
Total average monthly investment $1X,S75 00
The ratios of the average monthly investments are the same as the ratioti
of the month-dollars investments.
Second method. Multiply each investment by the number
of months from the date made until the end of the period; find
the sum of the products obtained. Likewise, multiply each with-
drawal by the number of months from the date withdrawn until
the end of the period; find the sum of the products obtained.
Deduct the sum of the withdrawal products from the sum of the
investment products; the result for each partner should be the same
as the month-dollars obtained by the first method.
The example under the first method is used in the following
solution.
Solution
A
TIME TO
INVESTMENTS END OF MONTH-
Date Amount YEAR DOLLARS
Jan. 1 $10,000 X 12 months = $120,000
May 1 4,000 X 8 months = 32,000
July 1 1,000 X 6 months = 6,000
~ $158,000
WITHDRAWALS
Feb. 1 $ 1,000 X 11 months = $ 11,000
June 1 1,500 X 7 months = 10,500
Nov. 1 500 X 2 months = 1,000
22,500
A's month-dollars $135,500
188 PARTNERSHIP
B
INVESTMENTS
Jan. 1 $ 6,000 X 12 months =* $ 72,000
Aug. 1 4,000 X 5 months = 20,000
Oct. 1 2,000 X 3 months = 6,000
* 98,000
WITHDRAWALS
July 1 $ 1,000 X 6 months = $ 6,000
Dec. 1 1,000 X 1 month = 1 ,000
~~ 7,000
#'s month-dollars $ 91,000
The distribution of the profits is the same as in the preceding example.
Problems
1. A, #, and C began business January 1. Their accounts for the year
appear as follows:
A
Jan. 1 $ 7,500
July 1 2,500
n
Mayl $4,000 Jan. 1 $10,000
C
Oct. 1 $7,000 Jan. I $10,000
Aug. I 3,000
Their profits for the year were $3,310. Determine the share of each partner,
if profits were divided on the basis of average investment.
2. Ames and Brown engaged in the hardware business, and at the end of
the first year their books showed a profit of $2,35701. They had agreed to
share profits and losses equally, after allowing 6% interest on average invest-
ment. Their investments and withdrawals for the year were:
Ames
July 1 $ 500 Jan. 1 $3,000
Sept. 15 1,500
Brown
Sept. 15 $1,000 Jan. 1 $2,500
July 1 250
Determine the net capital of each partner at the end of the year.
3. C. H. John and C. B. Arthur formed a partnership. John invested
$15,000, but four months later withdrew $3,000. Arthur invested $10,000, and
eight months later withdrew $2,000. Interest at 6% was to be credited on
average investment; the remainder of the profits was to be distributed in pro-
portion to original investments. The first year's profits, before interest adjust-
ment, were $2,500. What was the net capital of each partner at the beginning
of the second year?
Liquidation of partnership. Because of the nature of the
association, a partnership must necessarily be terminated on or
PARTNERSHIP 189
before the death of any one of the partners. It is not necessary to
discuss here the various causes of dissolution, but only the problems
met with at the time of settlement. The purpose of the formation
of a partnership is the making of profits, and the division of losses
is governed by the same general rule as the division of profits.
Profits should be credited and losses should be charged before any
division of assets is made. If this rule were not followed, an unfair
distribution of capital would result.
When dissolution is accompanied by liquidation, each of
the partners has an equal obligation to share in the work. But
since it does not usually require the time of all the partners, any
one of the partners, or an outsider, may liquidate the business.
In liquidation, profits or losses must first be divided in the profit
or loss ratio, and the remaining capital should then be shared by
the partners in the capital ratio.
In insolvency, partners must share losses in the profit and loss
ratio, and not in the capital ratio. This may at times result in a
deficit in capital for some one or more of the partners. Each
partner with a deficit should contribute to the firm the amount of
his deficit. But if he is totally unable to pay into the firm ai)>
portion of his deficit, the remaining partners must bear this loss in
the profit and loss ratio.
The governing profit and loss ratio, when a partner is unable to
pay, should be stated in fractions, of which the numerators are
the profit-and-loss-sharing per cents of the partners with credit
balances, and the denominators are the sums thereof. It is evident
that it is incorrect to compute the test loss division by multiplying
the loss by the profit and loss per cents, since the full amount of the
test loss would not be distributed.
Methods. Liquidation may be accomplished in two ways:
(1) All the assets may be converted , all the liabilities paid, the
profits or losses distributed, and all the capital divided at one time.
(2) A periodic distribution of the capital may be made before
all the assets are converted.
Total distribution. The first method of liquidation does not
involve any very difficult calculations.
Example
From the following figures, show the amount of capital distributed to each
partner at dissolution:
ABC
Capital balances before conversion of
assets $10,000 $6,000 $4,000
Profit ratio 40% 40% 20%
Assets converted into cash $30,000
Liabilities to be paid 14,000
190
PARTNERSHIP
Solution
Assets Liabilities Net Assets
$30,000 - $14,000 = $16,000
Total Investment, $20,000, less Net Assets, $16,000 = Loss, $4,000
Capital balances before conversion of assets. $10,000
Distribution of loss _L>600
Balances $~S7,400
Cash distributed <S,400
B
(40%)
$6,000
1,600
$4;400
4,400
C
(20%}
$4,000
_J400
$3^00
3,200
Total
(100%)
$20,000
16,000
Periodic distribution. Periodic distribution may result from
either of two causes:
(1) The desire of the partners to reduce the capital of the firm,
or to completely dissolve the firm, even though it is still solvent.
(2) Forced liquidation.
As the assets are converted into cash, and the debts are paid,
the balance of cash should be distributed periodically to the part-
ners. This should be done in such a way as to reduce the accounts
to the profit and loss ratio existing among the partners. The
distribution is made on the assumption that all book assets may be
a total loss until converted into cash.
The following example illustrates the adjustment of capital
ratios to profit and loss ratios.
Example
From the following data, show the periodic distribution of the cash collected:
A B C
Capital balances before conver-
sion of assets $10,000 $6,000 $4,000
Profit ratio 40% 40% 20%
First period:
Net loss $ 1 ,000
Cash collected . . . 9,000
Assets unrealized . . 10,000
Second period:
Net loss.. .... 1,000
Cash collected . . . 5,000
Assets unrealized . . 4,000
Third period:
Cash collected 2,000
All other assets uncollectible.
PARTNERSHIP
191
Solution
A
1. Capital balances before conversion of
assets $10,000
2. Distribution of loss . 400
3. Balance after distiibution of loss . $ 9,600
For the purpose of making a test, it will
be assumed that the unrealized assets
will never be realized.
4. Test loss in profit and loss ratio . . .
5. After the test loss has been deducted,
the remaining amounts will show the
proper distribution of the cash bal-
ance (3-4)
6. Balance at the end of the first period (3-5) $ 4,000
7. Net loss for second period . . .... 400
8. Balance after distribution of loss $ 3,600
The balances of the accounts are now in
the profit and loss ratio.
9. Distribution of cash
10. Balance at end of second period
1 1. Net loss for third period
12. Balance after distribution of loss
13. Cash distribution
$6,000
400
$5,600
$4,000
200
$3,800
Total
$20,000
1,000
$19,000
(4,000) (4,000) (2,000) (10,000)
5,600
1,600
$4,000
400
$3^600
1^800
$27)00
J200
$1^800
9,000
$io,ooo
1,000
$~9,6o6
2,000
2,000
1,000
1 period
$"1,600
SOO
$1,600
800
$ 800
400
n of loss
$ " SOO
SOO
$ SOO
SOO
$ 400
400
5,000
47000
2,000
2,000
__ 2££2
The following example illustrates the adjustment of capital
to the profit and loss ratio, where a deficiency of one partner is
involved.
Example
Show how each period's cash should be distributed in the following:
A B C
Capital balances before conver-
sion of assets . $10,000 $8,000 $2,000
Profits to be shaied equally.
First period:
Net loss
Cash to be distributed
Second period:
Net loss
Cash to be distributed . .
Third period:
Remaining assets sold for
Solution
A
Capital balances before conversion of assets $10,000
First period's loss distributed 500
Balance after distribution of loss
Test loss of amount of the remaining assets
It will be observed from the test loss that
C's possible loss is $2,000 greater than his
capital. If the test loss should become
$1,500
8,000
1,500
3,000
4,000
B
$8,000
500
C
$2,000
500
Total
$20,000
1,500
$ 9,500 $7,500 $1,500 $18,500
(3,500) (3,500) (3,500) (10,500)
192
PARTNERSHIP
(1,000) (1,000) 2,000
an actual loss, C will owe the firm $2,000,
and if C should be unable to pay in this
$2,000, A and B would be required to
bear this additional loss. To provide
against this contingency, a further test
loss charge of $2,000 is made against A
and#
When the sum of the two test losses, $4,500
($3,500 + $1,000), is deducted from A's
investment of $9,500, it can be seen that
A should receive $5,000; it can also be
seen that the sum of B's test losses
deducted from his investment gives the
amount of cash which is payable to him.
Cash distribution
Balance of capital undistributed
Second period's loss distributed
Balance after distribution of loss
Test loss of unrealized assets
What applied above applies again here. C"s
account shows a possible loss, and the
amount must be distributed as a test loss
to be taken up by the other partners.
Test loss for C's s
Cash distribution
Balance undistril
Third period's ne
Cash on hand . . .
Cash distributed
The following example illustrates a return of investments and
loans of partners which is complicated by the accounting principle
that " loans must be paid before capital is returned to partners. "
$J5,qoq
$~4,500
,500
$" 4,000
(2,000)
$3,000
$4,500
500
$4,000
(2,000)
$1,500
500
$1,000
(2,000)
$ 8,000
$107566
1,500
$ 9,000
(6,000)
7's account
(500) (500) 1 000
tion
1 ,500 1 ,500
3 000
tributed
$~~2,566 $2^500 $1,000
$ 6660
j net loss
.... 667 667 666
2 000
ted
$ 1,833 $1,833 $ 334
1.833 1.833 334
$ 4,066
4.000
Partners
A.
B.
C.
D.
Example
Capital Accounts Loan Accounts Profit Ratio
$33,000
28,500
18,000
10,500
$90,000
$10,500
10,000
21,000
|8,500
$60,000
The partners have decided upon a dissolution, and after paying all their
liabilities, they have:
Cash $ 30,000
Other assets 110,000
Net loss 10,000
How should the cash be distributed among the partners, assuming that no
member of the firm has private property with which to repay a capital account
that has been reduced by losses to a debit balance?
PARTNERSHIP
195
Profit and loss ratio ....
Capital balances before conver-
sion of assets
Loss distributed
Balance of capital
Test loss of assets
Possible deficiency of capital
Solution
A
40%
B
30%
$33,000
4,000
$28,500
3,000
C
20%
$18,000
2,000
D
10%
$10,500
1,000
Total
100%
$90,000
10,000
$29,000 $25,500 $16,000
(44,000) (33,000) (22,000)
$ 9,500 $80,000
(11, 000) (110,000)
$15,000 $ 7,500 $ 6,000 $ 1,500 $30,000
By applying the possible loss of unrealized assets against capital accounts,
it is found that the possible loss is greater than the capital invested.
In practice, as long as the firm has assets and owes each partner money on a
loan account, a partner will generally not pay cash into the firm, since the firm
would have to pay it back immediately. The problem states that none of the
members of the firm has money other than that invested in the firm. As the
shortages are debts due the firm, and as the partners have loan accounts, these
loan accounts will undoubtedly be used as a set-off. Therefore, the shortages
will be deducted as follows:
A B C D Total
Loans by partners $10,500 $10,000 $21,000 $18,500 $60,000
Less possible shortages J 5,000 7,500 6,000 1^500 30,000
Kach partner's standing in the
business after distribution of
the test loss ($4,500) $2,500 $15,000 $17,000 $30,000
As A's loan is not enough to take up the possible shortage, and as the problem
states that A has no other property, it is necessary to distribute A's test shortage
to the other partners.
A B C D Total
Ratio of distribution £ J $# H
Balances after distribution of test
loss ($4,500) $2,500 $15,000 $17,000 $30,000
A's test shortage distributed . 4,500 02,250) (U>00) J750)
Balances . 0 $~250 $13^>o6" $16,250 $30"^00
Cash distribution 0 250 13,500 16,250 30,000
The accounts of the partnership now stand:
Net assets $110,000
A's capital
^'s capital
^"s capital
/)'s capital
A's loan
B'sloan $10,000
Less payment 250
C'a loan $21,000
Less payment 13,500
D'sloan $18^500
Less payment 16,250
29,000
25,500
16,000
9,500
10,500
9,750
7,500
2,250
$110,000 $110,000
194
PARTNERSHIP
Problems
1. Ay B, and C decided to dissolve partnership. On the basis of the following
facts, show the proper distribution for each period:
ABC
Capital $8,000 $4,000 $6,000
Ratios 444% 22f% 33i%
First period:
Net loss $4,000
Assets uncollected 9,000
Cash. 5,000
Final period:
Assets converted 7,000
Losses 2,000
2. Show the periodic casli distribution based on the following facts:
Capital Loans Ratio
X $22,000 $2,000 50%
Y 8,000 3,000 33^%
Z 4,000 1,000 16f%
First period:
Cash $ 4,000
Assets uncoliected 36,000
Second period:
Cash 10,000
Assets uncollected 24,000
Loss 2,000
Third period:
Cash 21,000
Loss 3,000
C. P. A. Problems
1.* The capital of a partnership is contributed as follows:
A $90,000
B . . . . 45,000
C 15,000
The partnership agreement provides for profit sharing in the following ratios:
A 50%
B 30%
C 20%
The partners' salaries are as follows:
A . $5,000
B 3,000
C 2,000
At the end of the first year's business, C dies. The books are closed, and
the net assets of the business are shown to be $152,500. A and B liquidate the
affairs of the partnership, and distribute the surplus as follows:
First distribution $42,410.20
Second distribution 74,622 . 30
Final distribution 31,967.50
* C. P. A., Maryland.
PARTNERSHIP
195
Prepare a statement of the partners' accounts, showing how the distribution
of assets should be made and how the losses should be apportioned.
2.* A, B, C, and D enter into partnership with a capital of $100,000. A
invests $40,000; B, $30,000; f , $20,000; and /), $10,000. They are to share
profits or losses in the following proportions: .1, 35%; B, 28%; C, 22%; and
/), 15%. They are also to receive stipulated salaries chargeable to the business.
At the end of six months, there is a loss of $S,000, and meantime the partners
have drawn against prospective profits as follows: .1, $400; B, $600; C, $600;
and /), $400.
They dissolve partnership, and agree to distribute the proceeds of firm assetvS
monthly as realized. C and D enter other businesses, and A and B remain to
wind up the firm's affairs, it being stipulated that from all moneys collected
and paid over to C and /), a commission of 5< '0 be deducted and divided equally
between A and B for their services in liquidating the partnership.
The realization anil liquidation lasts four months, and the transactions are
as follows:
Kxpwscs and
First month
Second month
Third month
Fourth month
Prepare partners' accounts, showing the amount payable monthly to each
partner.
3.f A, B, (', and D formed a personal-service partnership, the clientele of
the firm being personal clients of the respective partners.
All fees received and all expenses were pooled by the firm, and the partnership
agreement stated that the net earnings for the year were to be shared as follows:
Losses on
Idealization,
A .wets
Liabilities
Exclusive of
Realized
Liquidated
Commissions
$ 30,190
$ 7,900
$ 400
50,300
0,100
750
20,010
3,SO()
340
9,500
2,200
110
$110,000
$20,000
$1,600
B
C
D
40%
16*%
10%
On August 31, as a result of a dispute, a supplementary agreement covering
the remainder of the year was made between the partners. This agreement
provided that the distribution of net earnings was to be made on the basis of
the above percentages, except that in the distribution of the net earnings for
the last four months of the year, so far as C and D were concerned, a net earning
was to be assumed on the basis of payment by the clients of A and B of gross
fees of $175,000 and $250,000, respectively, instead of the amounts actually
received from those clients.
The deficiency in A's gross fees was to be charged to him, and the excess in
Z?'s gross fees credited to him.
* C. P. A., New York.
t American Institute Examination.
196 PARTNERSHIP
No adjustment for expenses was to be applicable to either the deficiency or
the excess.
The net income from January 1 to August 31 was $75,000.
From September 1 to December 31, the following gross fees were received:
From clients of A $110,000
From clients of B 290,000
From clients of C 15,000
From clients of D 25,000
The operating expenses for the last four months were $55,000.
Determine the total net income of each partner for the year, taking into
account the supplementary agreement.
4.* On January 1, 19 — , Adams, Burk, and Oline became partners in the
operation of a dry goods business in Scranton, Pa.
At December 31 of the same year, the trial balance of the partnership, before
any adjustments were made, was as follows:
Adams, capital $ 50,000
Burk, capital 30,000
Cline, capital 20,000
Inventory of merchandise, January 1 . . $125,000
Accounts receivable, customers. . . 75,000
Accounts receivable, employees . . 3,000
Cash 6,000
Notes payable 60,000
Accounts payable 15,000
Sales 500,000
Purchases, including freight 323,000
Salaries and store expenses. ... . 125,000
Bad debts written off . 2,500
Interest paid on notes payable . 6,000
Salary to Mr. Adams 2,500
Salary to Mr. Burk 4,000
Salary to Mr. Cline 3,000
$675,000 $675,000
Prepare a balance sheet as of December 31, a profit and loss statement for
the year ended the same date, and a statement of the partners' accounts after
the following adjustments have been made:
Interest to be credited on partners' capital at 6 % per annum.
Mr. Adams owns the store, which the partnership occupies under an agree-
ment providing for an annual rent of $10,000 payable in monthly installments
in advance. No rent has been paid during the year. The year's rent should
therefore be credited to Adams, together with $325 interest on unpaid monthly
installments.
Of the interest paid on notes payable, $2,000 applies to the period subsequent
to December 31; accrued taxes, $1,000; accrued wages, $1,500. A reserve of
$1,500 is required to cover possible losses from doubtful accounts.
Ten per cent of the profits, if any, after the foregoing adjustments have
been made, is to be credited to " Bonuses to department managers and salesmen."
* C. P. A., Pennsylvania.
PARTNERSHIP 197
The remaining profits or losses are to be apportioned to the partners as
follows:
Mr. Adams 40%
Mr. Burk 33i%
Mr. Ciine 26$%
5.* A partnership composed of two members divides its profits equally, after
all items of income and expense for each calendar year have been determined.
One of the items of income is interest on partners' withdrawals, which is calcu-
lated and charged to each partner at the end of the year. By agreement, the
interest calculation is made on the partners' average monthly balances as shown
by the books. Partner A's account for the calendar year 19—, before interest
is charged to him, is found to be as follows:
Debits Credits
January 1, 19—, Balance $ 1,080 21 $
January account 6,000 00 550.00
February account 2,500 00 550 00
March account 3,052 74 550.00
April account 13,009 81 9,550 00
May account 5 . 45 550 00
June account 1,15420 55000
July account 1,500 00 550 00
August account 1 ,500 00 550 00
September account 500 00 550 00
October account 1 ,000 00 4,050 00
November account 1 ,014 10 550.00
December account 1,000 00 550 00
Show a statement of the interest which partner A should be charged at
December 31, 19 — ; simple interest, 6% per annum.
6.f A and B are in partnership. A receives two-thirds and B one-third of
the profits. On November 30, 1933, the Profit and Loss account (after interest
on capital has been charged at 5%), shows a profit of $6,000. On December 1,
1932, the start of the year under audit, A had a capital of $10,000.00 in the
business, and during the year he has drawn out $4,500.00. B on the same date
had a capital of $8,000.00, and during the year has drawn out $1,000.00.
Make up the two capital accounts as they should appear on November 30,
1933.
7.f A, By and C formed a partnership. A agreed to furnish $5,000, B and C
each $3,500. A was to manage the business, and was to receive one-half of
the profits; B and C were each to receive one-quarter. A supplied merchandise
valued at $4,250, but no additional cash. B turned over to A, as manager,
$4,500 cash, and C turned over $2,750. The business was conducted by A for
some time, but exact books were not kept. While manager, A purchased addi-
tional merchandise amounting in all to $37,500, and made sales amounting to
$50,000. The cash received and paid out for the partnership was not kept
separate from A's personal cash. B took over the management to straighten
out the affairs. He found accounts receivable amounting to $10,000. Of these
he collected $2,250. The remaining merchandise he sold for $250. These
* C. P. A., North Carolina.
fC. P. A., Indiana.
198 PARTNERSHIP
receipts he deposited to the firm's credit in the bank. The balance of accounts
receivable proved worthless. The outstanding accounts payable amounted to
$1,000, of which $750 had been incurred in purchasing merchandise, while $250
represented expenses. B paid these accounts.
A presented receipted claims, showing that during his management he had
paid other expenses of $1,200. By mutual agreement, B was held to be entitled
to $50 on account of interest on excess capital contributed, and A and C were
each charged $37.50 for shortage of contributed capital.
(a) Prepare the Trading and Profit and Loss accounts and the accounts of
each of the partners, including the final adjustments to be made at the close of
the partnership.
(6) Show how the above final adjustments would be modified if A proved to
have no assets or obligations other than those of the partnership.
8.* A and B, who are partners in a trading firm, decide to admit C as from
January 1, 1934.
They make an agreement with (\ as follows:
C is unable to contribute any tangible assets as his capital investment, but
agrees to allow his share of the profits to be credited to his capital account until
lie shall have one-fifth interest. (i is to share profits and losses to the extent
of one-fifth.
C is to receive a salary of $30,000 per annum, payable monthly, in addition
to his share of the profits.
The balance sheet of A and B at December 31, 1933, is as follows:
A sse ts Liabilities
Cash $ 1,500 Accounts Payable $ 8,000
Accounts Receivable 10,000 Capital Accounts:
Merchandise 7,500 .1 $10,000
Furniture and Fixtures 1,500 B _5»?99
Goodwill 2,500 ~~ 15,000
$23,000 $23,000
During the six months ended June 30, 1934, the business has sustained unusual
losses, and it is decided to dissolve the partnership.
The balance sheet at that date is as follows:
A ssets Lia bilitws
Cash $ 500 Accounts Payable $12,500
Accounts Receivable 12,500 Capital Accounts:
Merchandise 5,000 A $10,000
Furniture and Fixtures 1 ,500 B 5,000
Goodwill 2,500 15,000
Deficit: Being loss on trading
for 6 mos 5,500
$27,500 $27^00
Accounts receivable were sold for $9,000, the buyer assuming all responsi-
bility for collection and loss, if any.
Merchandise realized $6,500, and furniture and fixtures $500.
You are asked to make an examination of the accounts from January 1,
and to prepare statements showing the realization of assets, the adjustment of
the partnership accounts, and the distribution of funds,
* American Institute Examination,
PARTNERSHIP 199
In your examination, you find that C has not drawn his salary for four
months, and that B has advanced to the partnership $2,500 as a temporary loan.
You find that these liabilities are included in the sum of $12,500 shown as accounts
payable.
C is ascertained to have no assets.
9.* A, B, and C were in partnership, A's capital being $90,000, 7?'s $50,000
and C's $50,000. By agreement, the profits were to be shared in the following
ratio: A, 60%; B, 15%; (7, 25%. During the year, C withdrew $10,000. Net
losses on the business during the year were $1 5,000, and it was decided to liquidate.
It is uncertain how much the assets will ultimately yield, although none of them
is known to be bad. The partners therefore mutually agree that as the assets
are liquidated, distribution of cash on hand shall be made monthly in such a
manner as to avoid, so far as feasible, the possibility of one partner's being paid
cash which he might later have to repay to another. Collections are made as
follows: May, $15,000, June, §13,000; July, $52,000. After this no more can
be collected. Show the partners1 accounts, indicating how the cash is distributed
in each installment; the essential feature in the distribution is the observance
of the agreement given above.
10. f Brown, Green, and Black engage in a soliciting business under an
agreement that Brown is to receive a salary of $200 per month, (Jreen a salary
of $150 per month, and Black a salary of $100 per month; that the earnings are
to be determined at any time at the request of any partner; and that the profits
of the business are to be divided on the basis of the amount of business secured
by each.
The partnership is in business nine (9) months, and the business record for
that period is as follows:
Brown's business $4,500.00
Green's business 2,SOO.OO
Slack's business 3,000 00
Net profits of the business amount to $5,026.50.
The partners then decide to rescind the agreement as to salaries, and to
divide the profits on the basis of business secured individually, treating all
salaries drawn as advances.
Drawings:
Brown $1,000 00
Green . . . ... 1,200 00
Black 900 00
You find that the following errors have occurred during the nine months:
Office furniture charged to expense $ 65 00
Accts. rec. (Green's business) worthless 210 00
Cash advanced by Black — credited to his account as busi-
ness secured 400 00
Items not paid nor entered in the books:
Brown's salary $200.00
Green's salary 150.00
Advertising 27 50
Clerk hire 130.00
* American Institute Examination,
t C. P. A., Indiana.
200 PARTNERSHIP
Telephone 6 00
Rent 50 00
Stationery and supplies — exp 15 00
Show the journal entries necessary to readjust the accounts. Make up a
statement of the Profit and Loss account, showing all corrections and the dis-
tribution of the profits.
11.* Brown and Green entered into a joint venture.
On May 1, 19—, they purchased 5,000 tons of coal in Philadelphia at $4 per
ton, f.o.b., for which they gave notes on May 10 for one-half at 3 months and
for the other half at 6 months. The coal was shipped to Mexico City on May 15,
the freight, and so forth, amounting to $5,000.
A joint banking account was opened on May 10, each party contributing
$G,000.
The freight was paid by check on May 20, and on May 25 a check was drawn
for $1,000 for charges at Mexico City.
The coal was sold at $7 per ton, and the proceeds used to purchase a cargo
of timber, which was shipped to Philadelphia. Freight and other charges
thereon, amounting to $3,750, were paid by check June 30.
During July, four-fifths of the timber was sold for $32,000. This amount
was received and paid into the joint account August 2.
In order to close the transaction, Brown agreed to take over the remaining
one-fifth at cost price, including freight and charges, and he paid a check for
this into the joint account August 10.
The first note fell due and was paid August 13, and on the same day the
other note was paid under discount at the rate of 4% per annum.
Prepare accounts showing the results of the foregoing transactions; disregard
interest on capital contributions.
* American Institute Kxamination.
CHAPTER 21
Goodwill
Definition. Goodwill is an intangible asset, and may be
defined in general terms as the value of any benefits or advantages
which may accrue to a business from its being soundly established,
bearing a good reputation, having a favorable location, and so
forth. It results in the earning of a higher rate of net income than
that of less fortunate concerns in the same line of business.
Basis of valuation. When two or more businesses are consoli-
dated or merged, the payment made for each business depends
upon :
(1) The value of the net assets of each business.
(2) The earning power of each business.
A committee should be formed, consisting of members from
each of the businesses being consolidated or merged (proprietor-
ship, firm, or corporation); this committee should have the assist-
ance of an appraiser and an accountant in the preparation of a
report dealing with the net assets and the earning power.
The report should contain a balance sheet of each business,
stating the values at which it is proposed to take over the assets,
and stating the liabilities to be assumed.
The value of the fixed assets and of the inventory should be
determined by the appraiser. The accountant, after making
an audit, should submit the other balance sheet items.
Earning power determined from profit and loss statements.
The following points should receive consideration when earning
power is being determined from profit and loss statements :
(1) Number of years included. The value of goodwill depends
to some extent on whether profits have been uniform year after
year, or have steadily increased or decreased, or have fluctuated
from year to year. Therefore, in order to show the trend of pro-
fits, it is necessary to have profit and loss statements for several
years. A statement of average profits is insufficient, as it does
not show the trend.
(2) Adjustments to correct profits. Adjustments may be neces-
sary to correct errors, such as:
201
202 GOODWILL
(a) Wrong classification of capital and revenue expenditures.
(6) Omission of provision for depreciation, bad debts, and so
forth.
(c) Inadequate provision for repairs.
(d) Anticipation of profits on consignments and sales for
future delivery.
(3) Uniformity of methods.
(a) If the methods of computing the manufacturing costs
are not uniform, the cost statements should be revissd
and put on a uniform basis.
(6) The depreciation charges should be analyzed as to
method and rate. If different methods and rates have
been used, adjustments should be made so that the
charges will have been calculated on a uniform basis.
(c) There may be a wide difference in the management
salaries paid by the consolidating companies for the same
services. The salaries should be adjusted. In a single
proprietorship or partnership, salaries may not have been
paid or credited; in that case they should be included at
an arbitrary figure.
(d) If, iii a partnership, interest on capital has been charged
as an expense, the entries should be reversed and the item
of interest on capital thus eliminated.
(4) Eliminations. Eliminations may have to be made for
extraordinary and non-operating profits or losses.
Methods of valuing goodwill. Goodwill may be valued on the
basis of:
(1) An appraisal of goodwill.
(2) A number of years' purchase price of the net profits.
(3) A number of years1 purchase price of excess profits over
interest on net assets.
Capitalization of profits in excess of interest on net assets is
usually calculated as follows:
Net assets $100,000.00
Profits 10,000 00
Interest on net assets © 6% 6,000 00
Excess of profits over interest 4,000 00
Excess capitalized at 20% (4,000 + .20) 20,000 00
Case illustrations. The following four cases of goodwill valua-
tion, taken from reports of consolidations, show how goodwill has
been valued in practice.
GOODWILL 203
Case 1. The goodwill of the consolidating units was fixed at the
sum of the profits for the two preceding years, plus an additional
10%.
Case 2. The goodwill was based on the total profits for the five
years preceding, less five years' interest on the net worth.
Case 3. The goodwill was the average annual earnings for the
four years preceding consolidation, less the following deductions:
(a) Profits on favorable contracts about to expire.
(6) $100,000 for the estimated value of services rendered by
the retiring president.
(c) 6% interest on actual capital invested.
The remainder was capitalized on a 10% basis.
Case 4. From the net profits of each company the following
items were deducted:
(a) 7% on capital actually employed.
(fr) li% on sales.
(f) 2% depreciation on brick buildings,
(rf) 4% depreciation on frame buildings.
(e) 8% depreciation on machinery.
The remainder was capitalized at 20%, or 5 times the amount of
such earnings in excess of 7% on capital and other deductions
agreed upon.
Valuation by appraisal. There is no particular problem in the
calculation of the value of goodwill by appraisal. It may be
appraised by a disinterested party; or, more often, it is the amount
on which the vendor and the vendee agree. They usually appraise
the net assets, and agree that the purchase price shall be a certain
amount in excess of the value of the net assets. This excess is the
payment for goodwill.
Valuation by number of years' purchase price of net profits.
The goodwill may be estimated at so many years' purchase price
of the net or gross profits of any one year, or at so many years'
purchase price of the average profits of a number of years.
Example
The consideration of the sale of a business, as agreed to between the parties,
is four years' purchase price of the average profits for the preceding three years,
plus the net value of the assets.
Net value of assets $100,000
Profits of preceding three years:
1st year $20,000
2nd year 15,000
3rd year 28,000
What is the selling price of the business?
204 GOODWILL
Solution
Net value of assets $100,000
Profits of preceding three years:
1st year $20,000
2nd year 15,000
3rd year 28,000
$63,000
$63,000 -s- 3 = $21,000, average profits for three
years.
$21,000 X 4 (goodwill) 84,000
Selling price $184,000
Valuation on basis of excess of profits over interest on net
assets. The value of goodwill is calculated under this method by,
first, deducting from the average profits a fair return of interest on
the capital invested, and, second, by multiplying the remainder of
the profits, or the excess, by an agreed number of years' purchase
price.
Example
A agrees to buy a certain business, and to pay for it in cash. He agrees to
give dollar for dollar of the value of the net assets, plus a six years7 purchase
price of the excess of the profits over the interest on capital at 6%. Net assets
are valued at $100,000, and average profits are $18,000. What is the purchase
price of the business, including goodwill?
Solution
Net assets $100,000
Profits, average $18,000
$100,000 X .06 _°j9°0
Excess profits «T2JOOO
$12,000 X 6 (goodwill) 72,000
Purchase price $172^)00
It would be more favorable to the seller to determine the value by using a
higher rate of interest and capitalizing the excess profits at this rate; thus:
Net assets $100,000
Profits, average $18,000
$100,000 X .08 8,000
Excess profits $10,000
$10,000 -T- .08 (goodwill) 125,000
Purchase price $225,000
$225,000 - $172,000 « $53,000, advantage to the seller.
In the foregoing example, the goodwill represents the capitali-
zation of that portion of the profits which is not attributable to the
net tangible assets. The rate to be used depends largely on the
kind of business under consideration. In some lines of business
the per cent may be as low as 6% or 8%; in others it may be 10%;
and in still others 15%, or even 20%.
GOODWILL
205
Basis of stock allotment. Since most phases of the calculation
of the value of goodwill are found in consolidations, an example of
consolidation is given. The matter of stock allotment is included,
because when an agreement has been reached as to the valuation
of the assets and as to the earning power of each of the businesses,
the next question to decide is the method of making payment.
The following three typical methods will be presented:
(1) Payment entirely in common stock.
(2) Payment in preferred stock for the net assets; payment in
common stock for the goodwill.
(3) Payment in bonds for the fixed assets, or for an agreed
percentage thereof; payment in preferred stock for the balance of
the net assets; payment in common stock for the goodwill.
In the allotment of securities, the fundamental rule is to dis-
tribute them in such a manner that, if the income of the consolida-
tion is the same as the combined income of the several businesses,
oach of the old businesses, or the former owners or stockholders
thereof, will receive the same net income as before the consolidation.
To illustrate how this principle would operate under each of the
three methods outlined, assume that three companies are to be
consolidated on the basis of the following statements :
Net assets $40,000
Average earnings 4,000
Rate of income on net
assets 10%
H
$60,000
12,000
20
Vo
C
$120,000
20,000
10-1%
Total
$220,000
30,000
Common stock only. When only common stock is to be issued,
it must be issued in the ratio of the net earnings if the income of the
consolidation is to be distributed in the ratio in which the com-
panies contributed earnings. To determine the amount of stock
which is to be issued, capitalize the earnings by dividing the income
of each company by a rate of income agreed upon. Thus, if it is
agreed that the rate be 10%, the distribution of common stock is
made as follows :
A B C Total
Stock to he issued:
A: $4,000 -T- .10 $40,000
B: $12,000 -5- .10 $120,000
C: $20,000 -T- .10 $200,000
Total $360,000
Less net assets trans-
ferred 40,000 60,000 ^20,000 _220,000
Goodwill 0 $ 60,000 $~80^KK) $140,000
206
GOODWILL
Ten per cent was chosen as the basic rate, because it was the
lowest rate earned by any one of the three companies.
If the profits of the consolidation amount to $36,000, it will be
possible to pay a 10% dividend, which would be distributed as
follows :
A: 10% of $40,000 .......................... $ 4,000
B: 10% of $120,000 ............................... 12,000
C: 10% of $200,000 ........................ __20i0f)0
$36jKJO
This is an equitable division, so far as profits are concerned.
However, it is objectionable because it gives each old company an
interest in the assets which is proportionate to the profits earned
before the consolidation, instead of an interest proportionate to the
assets contributed. This might work a hardship in case of
liquidation.
Net Assets
A ..................... $ 40,000
B ..................... 60,000
C ..................... 120,000
$220,000
Goodwill
0
$60,000
^ 80,000
$140,000
Total
$ 40,000
120,000
200,000
$360,000
Fraction
Assume that after a number of years it is decided to liquidate
the consolidated company, and that in the meantime, all of the
profits have been paid out as dividends. The goodwill lias no
realizable value, so there is $220,000 to be distributed as follows :
Former stockholders of A :
Former stockholders of B:
Former stockholders of C:
of $220,000
of 220,000
fSHrof 220,000
$ 24,444.45
.... 73,333 33
. 122,222 22
$220,000 00
lose, and the former
The former stockholders of A would
stockholders of B and C would profit.
Former
Stockholders Assets
of Company Contributed
A $ 40,000
B 60,000
C J 20,000
$220,000
Preferred stock for net assets. In order to avoid giving an
advantage to one or more companies at the expense of the others,
it is advisable to issue preferred stock for the net assets, and
common stock for the goodwill. The goodwill should be allotted
to the several companies in the ratio of the excess of the profits
contributed over the dividends on the preferred stock.
Liquidating
Dividend
Gain
Loss
$ 24,444.45
$15,555.55
73,333 33
$13,333 33
122,222 22
2,222 22
$220,000 00
$15,555 55
$15,555 55
GOODWILL 207
Assume that in the above illustration 6% stock, preferred as to
assets, is to be issued for the net assets, and that common stock is
to be issued for the goodwill.
ABC Total
Earnings $4,000 $12,000 $20,000 $3l>,000
Less dividends on preferred
stock:
A: 6 9; of $ 40,000 2,400
B. 6% of 60,000 3,600
C: 6% of 120,000 7,200
Excess earnings $1,600 $ S,400 $12,SOO
Common stock should be issued in the ratio of the excess
earnings. If five years' purchase of the excess profits were agreed
upon, the distribution of stock would be:
.4 B r Total
Preferred stock $40,000 $60,000 $120,000 $220,000
Common stock S,()00 42,000 64,000 1 14,000
Assuming profits of $,36,000 as before, the distribution of
dividends would be:
Profits $36,000
Preferred dividends: 6% of $220,000 13,200
Balance available for common stock dividends $22, SCO
Then, $22,800 4- $114,000 - 20%, the rate per cent which
could be paid on the common stock.
A B C
Preferred dividends:
6% of $ 40,000 $2,400
6% of 60,000 $ 3,600
6% of 120,000 ... $ 7,200
Common dividends:
20% of $ X,000 . . 1,600
20% of 42,000 8,400
20% of 64,000 _12'800
Total dividends . $4,000 JJM2JOOO $20,000
These dividends are in each case equal to the profits contrib-
uted to the consolidation by the several companies.
It is important to note that goodwill should be based on the
profits contributed minus the profits to be returned as preferred
dividends, and not on the total profits.
Bonds, preferred stock, and common stock. If bonds are
issued for a percentage of the net assets, preferred stock for the
remaining net assets, and common stock for the goodwill, the good-
will should be based on the profits turned in minus the bond
interest and the preferred dividends.
208 GOODWILL
Assume that 5% bonds are to be issued for 80% of the net
assets, 6% preferred stock for the remaining net assets, and
common stock for the goodwill, which is to be computed by capita-
lizing at 20% the earnings of each company in excess of bond
interest and preferred dividends to be paid to former stockholders.
The issues of the three classes of securities would be computed as
follows :
.4 B C Total
Bonds:
A: 80% of $ 40,000 $32,000
J3:80%of 60,000 .... $48,000
C: 80% of 120,000 .... $96,000
Total bonds $176,000
Preferred Stock:
yl:20%of $ 40,000 8,000
B: 20% of 60,000 .... 12,000
C:20%of 120,000. .. 24,000
Total preferred stock.. .. 44,000
Common Stock:
A: Earnings $ 4,000
B: Bond interest $1 ,600
Pfd. dividend 480 2,080
Excess ' $ 1,920
$1,920 -T- .20 " ~ ^ $ 9,600
B: Earnings $12,000
Bond interest $2,400
Pfd. dividend 720 _3, 1 20
Excess $ 8,880
$8,880 4- .20 ' $44,400
C: Earnings $20,000
Bond interest $4,800
Pfd. dividend 1 ,440 6,240
Excess " *^f760
$13,760 -*- .20 "" $68,800
Total common stock. .. . $122,800
With profits of $36,000 before allowance for bond interest and
preferred dividends, the former stockholders would receive interest
and dividends as follows:
A B C Total
Bond interest:
5% of $32,000 $1,600
5% of 48,000 . $ 2,400
5% of 96,000 . $4,800
Total $ 8,800
Preferred dividends:
6% of $ 8,000 480
6% of 12,000 720
6% of 24,000 1,440
Total 2,640
GOODWILL 209
Common dividends:
20% of $ 9,600 $1,920
20% of 44,400 $ 8,880
20% of 68,800 $13,760
Total $24,560
Total distribution f4^ IH£22 $20,000 $36,000
Conclusion. The illustrations given are merely indicative of
the principles to be borne in mind in the distribution of stock and
other securities; they cannot be accepted as procedures to be
invariably followed, for several reasons.
First, in the illustrations, the profits of the consolidation are
assumed to be the same as the combined profits of the separate
companies before they were consolidated. However, consolidations
are usually made with the object of increasing profits; hence the
question is raised as to how the additional profits should be divided.
Should the preferred stock be participating or non-participating?
Second, the question of control involves the matter of the
voting rights of the several classes of stock.
These and other considerations would tend to cause modifica-
tions in the methods described, but the illustrations serve to
indicate the basic principles which must be followed in security
allotment in order that the stockholders of the several consolidating
companies may preserve their interests in the assets and earnings
of the consolidation.
Problems
1. A, B, and C are about to consolidate. The following data are presented;
A B (J Total
Net Assets $250,000 $150,000 $600,000 $1,000,000
Average Profits 50,000 15,000 150,000 215,000
Interest Hate 10%
Profit Rate 20% 10% 25%
Prepare tabulations showing the stock distribution:
(a) Preferred stock for the net assets, and common stock for the goodwill.
(6) Show possible disadvantage of issuing only common stock.
2. Using the data in Problem 1, show the security allotment if 5% bonds
are issued for 80% of the net assets, 6% preferred stock for the remainder, and
common stock for the goodwill, which is to be based on excess earnings capitalized
at 15%.
3. A, B, and C call upon you to draw up plans for their consolidation. They
submit the following information:
Assets A B C
Plants $3.50,000 $200,000 $180,000
Materials 100,000 20,000 20,000
Accounts Receivable 80,000 60,000 40,000
Cash 20,000 10,000 10,000
210
GOODWILL
Liabilities ABC
Accounts Payable $ 70,000 $ 30,000 $ 20,000
Capital 350,000 200,000 100,000
Surplus 130,000 60,000 130,000
Average income 30,000 35,000 40,000
Upon your recommendation, the consolidation will issue: (a) 6% bonds for
the fixed assets; (b) 7% preferred stock for the remaining net assets; (c) common
stock for the goodwill, which is to be based on excess profits capitalized at 10%.
Assuming that the consolidation will have net profits amounting to $105,000,
prepare statements showing the allotment of securities and the distribution of
profits.
4.* The net worth and profits of three companies are as follows:
X
Capital $100,000
Profits 50,000
r
Capital 200,000
Profits 50,000
Z
Capital 250,000
Profits 50,000
(a) Give your theory of how a consolidation should be made.
(b) Show the respective interests of A', of Y, and of Z in the consolidated
company, using a factor of 6% to represent the normal value of money.
5.f A has agreed to sell to B the goodwill of the X. Y. Company on the basis
of three years' profits of the business, which arc to be determined by you, on
sound principles of accounting and as accurately as possible, from the follow1 ng
statement handed you by A. You are required to compute the value of the
goodwill, but are not expected to take into account any considerations except
those presented by the statement.
Credits 1st Year
Sales (selling prices substantially uni-
form throughout period) $038,400
Estimated value of construction work
performed arid charged to property 110,000
Appreciation of real estate upon re-
valuation by experts
Profit on sale of Bethlehem Steel Co.
stock
Inventory at end of period:
Production material at cost 72,000
Finished goods at selling prices 76,500
$8967900
2nd Year
$602,500
77,600
80,000
103,100
114,000
3rd Year
$ 564,000
154,000
85,000
106,600
150,000
$977~200 $1,059,600
* C. P. A., Michigan.
t American Institute Examination.
GOODWILL
211
Debits
Production materials purchased $233,000
Production labor 50,850
Production expense (including de-
preciation) 66,750
Helling expenses 52,500
Interest 96,000
Cost of construction work 74,600
Inventory at beginning of period:
Production material at cost 51,400
Finished goods at selling prices 54,900
ioso^dod
Balance, being profit claimed by A. .. $216,900
6.* In the preceding problem, does the basis used for arriving at the value
of the goodwill— three years' piofits- -appear to you to be reasonable in view
of the facts disclosed to you? If not, what advice would you offer upon the
question if A or It were your client?
7.* .1 and H are partners in business, and have the following statement:
$252,400
61,400
$ 220,300
60,900
69,300
55,650
94,000
49,000
70,300
62,800
98,500
86,000
72,000
76,500
$730,250
$246,950
103,100
114,000
$ 815,900
$ 243,700
Store ...
Accounts Receivable . . .
Cash
Furniture and Fixtures .
Merchandise
Miscellaneous Equipment
$15,000
12,000
9,000
2,S()0
37,000
4,200
$80,000
Accounts Payable $10,000
Bills Payable 5,000
-t's Capital 30,000
#'s Capital 35,000
$80,000
C is admitted as a special partner, under the following arrangement: C is to
contribute $30,000, and is to be entitled to one-third of the profits for 1 year.
Before the contribution is made, the following changes are to be made in the
books: store to be marked down 5%; allowance for doubtful accounts to be
created, amounting to 2 ',<',; merchandise to be revalued at $35,000; furniture
and fixtures to be revalued at $2,500. At the end of the year, the goodwill is
to be fixed at 3 times the net profits for the year in excess of $20,000, this good-
will to be set up on the books and the corresponding credit to be to A and II
equally. A, B, and C are each to draw $3,000 in cash, and the remaining profits
are to be carried to their capital accounts.
During the year, the following transactions took place:
Merchandise bought on credit $240,000
Cash purchases 25,000
Cash sales 125,000
Sales on credit 175,000
Accounts payable paid (face, $245,000; discount, 2%) . . . 240,100
Accounts receivable collected (face, $170,000; all net ex-
cept $50,000, on which 2% was allowed) 169,000
Buying expenses, paid cash .... 1,500
Selling expenses, paid cash 21,000
Delivery expenses, paid cash 9,000
Management expenses, paid cash 4,500
Miscellaneous expenses, paid cash 3,000
Interest on notes payable, paid cash 250
* American Institute Examination.
212 GOODWILL
The partners each withdrew $3,000 cash, as agreed.
When the books were closed for the purpose of determining the profits and
goodwill, the following were agreed upon:
Value of merchandise on hand $60,000
Depreciation on store 285
Additional allowance for doubtful debts 165
Furniture and fixtures written down 200
The goodwill having been estimated and duly entered, C then contributes
enough cash to make his capital account equal one-third of the total capital.
Prepare statements showing how the accounts are to be adjusted, and pre-
pare the balance sheet after the final adjustment.
CHAPTER 22
Business Finance
Stock rights. Corporations, in undertaking to secure addi-
tional capital, not infrequently offer additional stock to their
stockholders at a price below the prevailing market quotation of
the outstanding shares.
'Phis privilege of subscribing has value as long as the market
price of the old stock remains higher than the offering price of the
new stock; and if the stockholders prefer not to exercise the rights,
they may sell them in the market for whatever they will bring.
Example
A corporation has a capital stock of $100,000, divided into 1,000 common
shares of $100 par value. The entire amount is outstanding, and the market
quotation is $150. Finding that $50,000 additional capital is needed, the
directors decide to offer to the stockholders 500 shares of new common stock
at $125. They accordingly announce on August 1 that stockholders of record
as of September 1 will have the privilege of subscribing for the new issue in the
proportion of one share of the new stock for every two shares of the old stock
held on the latter date. The subscriptions are payable on or before October 1
following, and transferable warrants for the rights are to be issued as soon as
practicable after September 1. What is the value of a right?
Explanation. According to the conditions of this offer, every holder of two
of the old shares at the close of business on September 1 will be entitled to
subscribe to one of the new shares. He will therefore come into possession of
two " rights/' as that term is used on the New York Stock Exchange, or one
right for every old share held. (On some stock exchanges the term right indi-
cates the privilege of subscribing to one share of the new issue.)
Trading in the rights will begin following the declaration of the directors
on August 1, and will continue until October 1. Until the warrants are in the
hands of the stockholders — during the period from August 1 to September 1 —
the trading will be on a "when issued" basis; that is, delivery and payment for
the rights will be made when the warrants are available. During this time the
stock will sell "rights-on"; that is, the market value of the shares will include
the value of the rights.
With the delivery of the warrants on September 1, the stock will sell
"ex-rights," its price no longer including the value of the rights. With the
issuance of the warrants, and until October 1, trading in the rights will be for
immediate delivery and payment; that is, delivery and payment the day after
the sale is made.
Should a holder of two shares exercise his privilege of subscription, he would
own:
213
214 BUSINESS FINANCE
2 shares @ $150 $300
!_ share © 125 125
3 shares @ 141.67 $425
It will be noticed that the difference between the market price of the old
stock and the average price of the three shares is $8.33, the value of a right; also
that the difference between the market price of the old stock and the offering
price of the new stock is $25.00, or three times the value of a right.
Therefore, the following formula may be used:
Formula
Market price — Offering price
- = Value of a right.
Number of rights to purchase 1 share -f- 1
Substitution
150-125 25
-2+-l-=8->or*U3-
During the second period — that is, while the stock is quoted ex-rights — the
value of the rights may be ascertained as follows:
Formula
Market price — Offering price
' 1,1
Number of rights to purchase I share
It will be found that the market price of the rights during the second period
will tend to coincide with this value. Any appreciable difference opens an
opportunity for a profit.
The foregoing discussion and example apply only to values on
the market. The profit or loss resulting from the sale of rights,
and the profit or loss from the sale of stock acquired by the exercise
of rights, are governed by Section 29.22(a)-8, Regulations 111.
Sale of stock and rights, federal income tax. Ordinarily,
a stockholder derives no taxable income from the receipt of rights
to subscribe for stock, nor from the exercise of such rights, but if he
sells the rights instead of exercising them, he may derive taxable
income, or sustain a loss.
The following rule is stated in Sec. 29.22(a)-8, Regulations 111.
"(1) If the shareholder does not exercise, but sells, his rights
to subscribe, the cost or other basis, properly adjusted, of the stock
in respect of which the rights are acquired shall be apportioned
between the rights and the stock in proportion to the respective
values thereof at the time the rights are issued, and the basis for
determining gain or loss from the sale of a right on one hand or a
share of stock on the other will be the quotient of the cost or other
basis, properly adjusted, assigned to the rights or the stock,
divided, as the case may be, by the number of rights acquired or
by the number of shares held."
BUSINESS FINANCE 215
Example
A purchased 100 shares of stock at $125.00 a share, and in the following
year the corporation increased its capital by 20%. A, therefore, received 100
rights, entitling him to subscribe to 20 additional shares of stock; the subscription
price was $100.00 a share. Assume that at the time that the rights were issued
the stock had a fair market value of $120.00 a share, and that the rights had a
fair market value of $3.00 each. If, instead of subscribing for the additional
shares, A sold the rights at $4.00 each, his taxable gain would be computed as
follows:
100 shares® $125.00... . $12,500.00, cost of stock in respect of
which rights were issued
100 shares ® $120.00. . $12,000.00, market value of old stock
100 rights @ $3.00 . . $300.00, market value of rights
12 000
T9^nri °^ 12>^ $12,195.12, cost of old stock apportioned
onn ^° surn stock after issuance of rights
3 °- of 12,500 $304.88, cost of old stock apportioned
12,300 to rights
100 rights @ $4.00 . . . $400.00, sales price of rights
$400.00 - $304.88 $95.12, profit on sale of rights
For the purpose of determining the gain or loss from the subsequent sale of
the stock in respect of which the rights were issued, the adjusted cost of each
share is $121.95— that is, $12,195.12 -f- 100.
Rule 2 of Sec. 29.22(a)-8, Regulations 111, states:
"(2) If the shareholder exercises his rights to subscribe, the
basis for determining gain or loss from a subsequent sale of a share
of the stock in respect of which the rights were acquired shall be
determined as in paragraph (1). The basis for determining gain
or loss from a subsequent sale of a share of the stock obtained
through exercising the rights shall be determined by dividing the
part of the cost or oth^r basis, properly adjusted, of the old shares
assigned to the rights, plus the subscription price of the new shares,
by the number of new shares acquired."
Example
A purchased 100 shares of stock at $125.00 a share, and in the following year
the corporation increased its capital by 20%. A, therefore, received 100 rights
entitling him to subscribe to 20 additional shares of stock; the subscription price
was $100.00 a share. Assume that at the time that the rights were issued the
stock had a fair market value of $120.00 a share, and that the rights had a fair
market value of $3.00 each. A exercised his rights to subscribe, and later sold
for $140.00 a share 10 of the 20 shares thus acquired. The profit is computed
as follows:
Cost of old stock apportioned to rights in accordance
with the computation in the example under Rule 1 ... $ 304 88
Subscription price of 20 shares at $100.00 a share 2,000.00
Basis for determining gain or loss from sale of shares
acquired by exercise of rights $2,304.88
216 BUSINESS FINANCE
$2,304.88 -T- 20 = $115.24, basis for determining gain or loss from sale of
each share of stock acquired by exercise of rights.
Proceeds of sale:
10 shares <& $140.00 $1,400 00
Cost of stock sold :
10 shares @ $115.24 1,152 40
Profit $ 247 "60
The basis for determining the gain or loss from the subsequent sale of the
remaining 10 shares of stock acquired on subscription is $115.24 a share; and the
basis for determining the gain or loss on the stock in respect of which the rights
were issued is $121.95 a share — that is, $12,195.12 -T- 100, as in the example
under Rule 1.
Problems
1. A company has a capital stock of $1,000,000, divided into 10,000 common
shares of $100 par value. The entire amount is outstanding, and the market
quotation is $150 a share. Finding that $500,000 of additional capital is needed,
the directors decide to offer to the stockholders 5,000 shares of new common
stock at par. What is the approximate market value of a right if each stock-
holder may subscribe for one share of new stock for every two shares of the old
stock held?
2. A corporation offered, at $100 a share, one share of its new stock for each
six shares held. The stock was selling at $185 a share when the offer was
announced. What was the approximate market value of a right?
3. W owned 100 shares of Purity Baking Common that cost him $1.3,400.
Later, he received rights to subscribe to additional stock, but since he did not
care to increase his investment, he sold the rights at 3f less commission, receiving
therefor $357.30. At the date when the stock was quoted ex-rights, the average
market values were:
Stock 124i
Rights 3f
(a) What was W's loss on the sale of the rights?
(6) What was the carrying value of the stock?
4. Smith owned 100 shares of common stock in the W. Corporation, which
offered rights to subscribe to new common stock at $100 a share, the basis of
the offering being one share for each five shares held. The average market
values on the date when the stock sold ex-rights were:
Stock 150.50
Rights 11.8125
Smith later sold his rights at $14.50.
(a) if Smith paid $120 a share for the original 100 shares, what is his profit
on the sale of the rights?
(6) What is the carrying value of the 100 shares?
Working capital. One of the most difficult problems for any-
one entering a new business is to know how much money will be
required to finance the enterprise until the receipts will equal or
BUSINESS FINANCE
217
exceed the disbursements. While this is strictly a question of
finance, the accountant is often called upon to deal with it.
Example
A manufacturer gives you the following data, and requests that you estimate
the amount of working capital required to finance the making and selling of an
article:
Selling price, each $100
Cost to make, each 60
Selling expenses, each 20
Overhead, each 10
Net profit, each 10
Sales, first month . . 50 articles
" second month 100
" third month 150
" fourth month 200
" each month thereafter 200 "
All the sales are installment sales, the payments being $10 per month.
Assume arbitrarily that the complete cost of $90 on each article is incurred at
the time that the sale is made.
What will be the largest amount of capital required, and in which month
will it be required?
Solution
Working
Total Receipts Deficiency Capital
Each Month Each Month Required
$ 500
1,500
3,000
5,000
7,000
9,000
11,000
13,000
15,000
17,000
Months
First ...
Second .
Third .
Fourth
Fifth...
Sixth ..
Seventh
Eighth
Ninth
Tenth .
Eleventh
Twelfth.
Total Costs
Each Month
$ 4,500
9,000
13,500
18,000
18,000
18,000
18,000
18,000
18,000
18,000
18,000
18,000
18,500
19,500
Deficiency
Each Month
$ 4,000
7,500
10,500
13,000
11,000
9,000
7,000
5,000
3,000
1,000
500f
l,500f
f Receipts from collections are more than the costs for the month.
$ 4,000
11,500
22,000
35,000
46,000
55,000
62,000
67,000
70,000
71,000
70,500
69,000
The above table shows in the last column the amount of working capital
required to finance the business by months. The greatest amount required is
found to be $71,000 in the tenth month. Thereafter, the collections are greater
than the costs.
Problems
1. A company is about to be formed for the purpose of manufacturing a
specialty. After careful investigation, the following estimates have been made:
Selling price, each $75
Cost to make, each 35
Selling and administration expense 14
Net profit 26
218 BUSINESS FINANCE
First month 30 machine*,
Second " 70
Third " 180 "
Fourth " 200 "
Each month thereafter 225 "
The terms of payment are $15 down, and $5 per month. What is the greatest
amount of working capital that will be required, and in which month will this
amount be needed?
2. The X. Company plans to sell on the installment basis, direct from factory
to consumer. Their product is a specialty retailing at $100, payable $10 with
order and balance in nine equal installments.
Cost to manufacture:
Material 40%
Labor 35%
Burden 25%
Helling expense 15% of sale?
Administration expense 4% of sales-1
Other expense 1 % of sales
Labor cost is expected to increase 14y%, which will decrease the profit 35%.
Estimated sales:
First month 50 machines
Second " . 100
Third " 150
Fourth " 200
Each month thereafter 200 "
Assuming that all the expenses of a sale are paid during the month in which
the sale is made, prepare a schedule showing the essential facts, and the amount
of working capital needed monthly. '
3.* On the basis of the following facts, determine, by months, the cash
requirements of an installment dealer for the first year's operations:
1. Cost of article $50 00
2. Sales price 90 00
3. Selling expense 15 00
4. Overhead 15 00
5. Profit 10.00
6. Sales for the first month were 100 articles
7. Sales for the second month were 200 articles
8. Sales for subsequent months were 300 articles per month
9. Merchandise paid for on the month following the sale
10. Expenses paid during the month of sale
11. Payments are received at the rate of $10 down and $10 per
month; assume that no irregularities are experienced
4.t The A. B. Company acquired the right to sell musical instruments in a
given territory. They request you:
* C. P. A., Wisconsin,
t C. P. A., Pennsylvania.
BUSINESS FINANCE 219
(a) To state how much capital will be required to carry on the business
during the first year.
(6) To demonstrate by computation how your estimates would work out
during the first six months.
Assume that the sales for the first year will total $180,000 from the sale of
instruments, and $24,000 from service work (respectively, $15,000 and $2,000
monthly). The overhead and direct selling costs are estimated at $30,000 for
the year. This amount includes all expenses except the cost of instruments
sold and parts used in service, the latter being estimated at $12,000.
The instruments are purchased on 60 days' credit, at a discount of 30% from
the price at which they are sold to the customer. Twenty per cent of the instru-
ments are sold for cash, and 80% on lease contracts. When they are sold on
lease contracts, 25 % is required as a down payment, and the balance in 12
months. All payments are made to the A. B. Company. The leases are dis-
counted at the bank, the charges by the bank being added to the price charged
the customer. The service charges are billed and payable in 30 days.
Cumulative voting. Cumulative voting is a method whereby
each shareholder is entitled to cast a number of votes equal to the
product of the number of shares which he holds and the number of
directors to be elected. The shareholder may cast all his votes for
any one or more of the directors to be elected, or may distribute his
votes in any way that he desires. Thus, the minority stock-
holders, by combining their votes, may elect a representative on
the board of directors.
Example
A corporation has an outstanding capital stock of $100,000, composed of
1,000 shares of common stock with a par value of $100 each. The stockholders
are to elect seven directors at the annual meeting. Calculate the least number
of shares required to elect three of the directors, provided that the cumulative
method of voting is used.
Formula Substitution
Explanation.
a = Number of shares outstanding
6 = Number of directors to be elected
c — Number of directors minority desires to elect
x = Required number of shares
1,000 X 7 = 7,000, total votes of 1,000 shares
376 x 7 = 2,632, total votes of 376 shares
2,632 -r- 3 = 877, the number of votes each of the three
directors would receive if the holders of the
376 shares of stock cast all their votes for
three directors
7,000 - 2,632 = 4,368, balance of votes
4,368 -T- 5 = 873, the largest number of votes the remaining
stockholders could cast for five directors
220 BUSINESS FINANCE
By the same method of calculation, it will be found that the owners of 375
shares would have 875 votes for each of three directors, while the remaining
stockholders would have 875 votes for each of five directors. This would give
a tie vote, and neither side could elect the desired number.
Problems
1. In a corporation which uses the cumulative method of voting, how many
of the seven directors can you safely seek to elect, if you own 1,501 out of the
4,000 voting shares?
2. Seven directors are to be elected by the X. Company, which has a voting
capital of 5,000 shares. How many shares are necessary to elect four directors
under the cumulative plan?
Book value of shares of stock. A corporation is owned by the
stockholders, whose evidences of ownership are shares of stock.
The hook value of a share of stock is equal to the quotient of the
net worth divided by the number of shares of stock outstanding.
Formula
Assets — Liabilities
-vr— - — -- = Book value per share
Number of shares
Example
A corporation has assets of $340,000 and liabilities of $120,000. It has a
capital stock of $200,000. If the shares have a par value of $100 each, what
is their book value?
Solution
$340,000 - $120,000 $220,00000
$220,000 + 2,000 shares 110.00
Problems
1. Find the book value of a share of stock in each of the following companies
Par Value
Assets Liabilities Capital of Shares
(a) $350,829.75 $134,082.47 $100,000 $100
(6) $575,850.00 $190,260.75 25,000 shares No Par
(c) $1,322,080 35 $110,809.20 $1,000,000 $50
2. A company has assets of $385,915.28 and liabilities of $158,910.75. If
there are 5,000 shares outstanding, each with a par value of $50, what is the
book value of each share?
Profits distribution. The distribution of profits in the part-
nership type of business organization was discussed under the
subject of partnership (Chapter 20). The distribution of the
profits of corporations is illustrated in the following examples.
Example 1
At the end of a certain year, a corporation had outstanding 2,250 shares
of common stock, par value $100. A 2% dividend was declared. Net earnings
BUSINESS FINANCE 221
were $53,320.84. What was the amount of the dividend, and what amount
of the year's profits, after the declaration of the dividend, remained in surplus?
Solution
2,250 shares @ $100 $225,000, capital stock
2% of $225,000 $4,500, dividend
$53,320.84 - $4,500 $48,820.84, credit to surplus
Example 2
A company had the following number of outstanding shares, each with a
par value of $50: common, 858,860 shares; 6% cumulative preferred, 291,047
shares; 5% non-cumulative and non-participating preferred, 28,849 shares.
The company declared a 6% dividend on the common stock. What amount
of profits was distributed to each class of stock? What amount of profits was
necessary to cover the dividends?
Solution
Capital Dividends
858,860 shares common @ $50 $42,943,000 00
6 % of $42,943,000 $2,576,580 00
291,047 shares 6% cumulative preferred (m $50. 14,552,350.00
6% of $14,552,350 873,141 .00
28,849 shares 5% non-cumulative preferred (a\
$50 1,442,450.00
ri% of $1,442,450 _ 72,122.50
Profits necessary to cover dividends $3,521,843.50
Problems
1. Calculate the amount of dividends on the following stocks; each class has
a par value of $100:
182,260 shares common @ 6%;
57,633 shares 7% cumulative preferred.
2. A company earned $2,850,460.03. It paid from this 8% dividends on
$12,379,850 preferred stock outstanding. If the preferred stock was non-
participating, what per cent was earned for the common stock, of which there
was $10,600,000 outstanding? Par value in each case was $50 a share.
3. The outstanding stock of a corporation consists of:
Preferred A stock, without par value, series 1, cumulative divi-
dends $7 per share 10,000 shares
Participating preference stock, without par value, cumulative
dividends $8 per share 16,301 shares
Preferred R stock, without par value, non-cumulative dividends
$3.50 per share 6,659 shares
Common stock without par value 198, 145 shares
The participating preference stock entitles the holder to receive, among other
things, participating dividends equal share for share to any dividends paid from
time to time on the common stock.
If, in the course of a year, 50^ is paid on each share of common stock, what
will be the total amount of dividends paid?
CHAPTER 23
Public Finance and Taxation
Governmental functions. Governmental functions are divided
among three major classes of governmental units — the federal
government, the forty-eight state governments, and thousands of
local government bodies, including counties, towns, townships,
cities, villages, and such special units as drainage, levee, irrigation,
and park districts.
Purposes of taxes. Federal government taxes are used to pay
the army and navy, the salaries of governmental personnel, pen-
sions, and other goverment expenses such as education, highways,
economic development, social welfare, and so forth. Expenses of
the federal government in 1943 amounted to more than 78 billion
dollars; most of this was for national defense, which cost 72 billions.
The budget included a billion dollars for aid to agriculture and
almost 2 billions for interest on the public debt.
State government taxes are used to pay the salaries of state
government personnel, the support of schools, universities, and
asylums, and for sundry other state expenses. The largest single
item in state expenditures for the 48 states during the year 1942
was that for operation and maintenance, which in that year
amounted to $4,083,877,000; of this amount, $1,030,117,000 was
for operation and maintenance of schools.
County taxes are used to pay the salaries of county employees,
the cost of roads, charities, and miscellaneous other county
expenses.
City taxes are used to pay the salaries of city employees, police
and fire protection, support of schools, and other city expenses.
Town taxes are used to pay the salaries of town employees,
support of schools, and other town expenses.
Taxes levied by special units are for the purpose of paying for
and maintaining the special units.
Appropriations. Funds deemed to be necessary for the conduct
of government are set up by the respective governing bodies as
appropriations. The size of the appropriations may result from
modifications in the quantity and/or quality of governmental
activities and from changes in commodity prices and wage and
223
224
PUBLIC FINANCE AND TAXATION
^ TAX REVENUES |
| M THE U. S. |
W/////////M^^^
Courtesy of Minnesota Taxpayers' Assn.
salary levels. To meet these appropriations, taxes are levied upon
persons and property.
Kinds of taxes. Commodity taxes may be limited in their
scope, applying to particular commodities, such as taxes on
tobacco, liquors, and so forth; or they may be general, applying to
the manufacture and sale of all or most commodities, such as sales
taxes, manufacturers' excise taxes, and so forth.
PUBLIC FINANCE AND TAXATION 825
Highway taxes are best known as the motor vehicle tax or
license, and the motor vehicle fuel tax known as the gasoline tax.
The general property tax and the special property taxes are the
major sources of revenue for state and local governments. The
tax rate imposed on the assessed value of a piece of property is a
total of a series of separate tax rates imposed by local and state
governments. Real property is listed and valued by the assessor,
while personal property is usually listed by the taxpayer, who sets
his own value on the articles he lists, although the assessor may
change the values if investigation proves them to be incorrect.
Taxes on business consist of state bank taxes and state taxes on
insurance companies, railroads, and public service enterprises such
as telephone and light and power companies. There are also state
taxes on business in general in the form of license fees and franchise
taxes.
Income taxes are levied on the income of both persons and
business. The first income tax was a federal tax, but now most of
the states have income tax laws. The excess profits tax (repealed
in 1946) is a tax on the excess profits of a corporation and is a
federal tax.
Death taxes on transfer of property of a deceased person to his
beneficiaries or heirs are levied by the federal government and by
most of the state governments. One form is the federal estate
tax, another is the state inheritance tax. To prevent distribution
of large holdings of property in order to escape estate and inherit-
ance taxes, the gift tax was enacted by the federal government.
Several of the states also have gift taxes.
The chart on page 224 summarizes the general sources of tax
revenues.
Income, inheritance, estate, and gift taxes are complex and
constitute complete studies in themselves; also, rates and regula-
tions are frequently undergoing changes. Therefore, these sub-
jects will not be presented in this text.
Property Tax
Determination of tax rate. The amount of money needed
divided by the assessed valuation of the property determines the
rate to be levied. For example :
State tax:
Budget S 3,750,000
Assessed value of property in state 1 ,250,000,000
3,750,000 -v- 1,250,000,000 = .30%, State rate
226 PUBLIC FINANCE AND TAXATION
County tax:
Budget $ 90,000
Assessed value of property in county . . 4,500,000
90,000 + 4,500,000 = 2 00%, County rate
City tax:
Budget 75,000
Assessed value of property in city 2,500,000
75,000 -T- 2,500,000 = <LOO%, Citv rate
5.30%, Total rate
To find the amount of tax. Multiply the assessed valuation by
the rate. Tax rates may be expressed: as so many mills on the
dollar; as a certain rate per cent; or as dollars on the thousand.
Example
Property is assessed at $8,500; the tax rate is $40.60 a thousand. Find
the tax.
Solution
$ 40 60, rate on each thousand
<S 5 , number of thousands assessed
$345.10, the tax.
Problems
1. What is the total tax rate on the following?
State:
Budget $ 5,500,000
Assessed valuation 2,500,000,000
County :
Budget 72,000
Assessed valuation 6,000,000
City:
Budget 125,000
Assessed valuation 3,750,000
School District:
Budget 35,000
Assessed valuation 1,750,000
2. In a certain town the tax rate on $1,000 was as follows:
State $ 1 .20
County 30 00
Town:
Library $1 . 24
Revenue 91
Road and Bridge 2 74
Road Drag 91
5.80
School District 1 .00
Total rate
If the assessed value of X Company's property in this town was $93,960,
what was the amount of property tax?
PUBLIC FINANCE AND TAXATION 227
3. If property is assessed at $6,880 and the tax rate is $100,30 a thousand,
what is the tax?
4. If the tax rate is $100.30 a thousand, what is the rate per cent? What is
this rate in mills on the dollar?
5. A tax rate of $99.20 a thousand is made up of the following rates:
State:
Debt $ 7 33
Road, Bridge, and Soldiers' Relief 1 . 10
School 1 .23
Teachers' Retirement .04
County :
Revenue 27.09
City:
Revenue 61.41
School District 1 .00
Total ...~.H
Property valued at $75,000 is assessed at $ of its value. What per cent of
the total tax applies to each division?
6. In a certain city taxes are due January 1, hut may be paid in two install-
ments, the first on or before May 31 and the second on or before Oct. 31.
If the first half is not paid on or before May 31, the following penalties will
attach: During June, 3%; July, 4%; August, 5%; September, 6%; October, 7%.
During November and December, the penalty is 8% computed on any
amount unpaid.
The second one-half cannot be paid until the first one-half has been paid.
If the tax rate is $96.50 a thousand, find the total tax paid on the following:
(a) Property with assessed value of $35,000, both installments paid on
August 10.
(b) Property with assessed value of $7,500, the iirst installment paid July 15
and the second installment paid December 15.
7.* A city with an assessed valuation of $1,000,000 and estimated receipts
for current expenses from miscellaneous sources of $50,000 and of $2,000 from
sinking fund investments has submitted to you the following budget of expendi-
tures for the year:
Mayor and other Commissioners $20,000
Water Department 15,000
Bond Interest 5,000
Fire Department 20,000
Police Department 21,000
Health Department 15,000
Retirement of Bonds 10,000
Street Department 18,000
General Government 25,000
What tax levy must be made to provide the necessary revenue?
8.f The assessed valuation of the taxable property of the State of W., as
* C. P. A., North Carolina.
t Adapted from C. P. A. pjxamination.
PUBLIC FINANCE AND TAXATION
determined by the Tax Commission for a certain year, was $4,068,268,534.
What would a citizen whose property was assessed at $507,374 have to pay
for each of the following purposes, and what would be the amount of his total
tax bill?
Purposes — Total Amount to Be Raised
Interest on Certificates of Indebtedness $ 199,339.42
Free High Schools 175,000.00
State Graded Schools 200,000.00
Highway Improvements 1,700,000.00
General Purposes 100 . 00
In addition to the above, the following mill taxes are assessed:
University f mill
Normal Schools . . \ mill
Common Schools ^ mill
CHAPTER 24
Fundamentals of Algebra
Explanation. The work of an accountant is complicated in
many particulars and requires technical calculations. It is
extremely difficult to make some of these calculations by arithme-
tic. On the other hand, if the fundamentals of algebra are under-
stood, the calculations may be made with comparative ease. Only
the more common and more useful principles of algebra will be
discussed in this chapter.
Symbols and terms. In algebra, the letters of the alphabet
are usually employed as symbols to represent numbers.
The following signs have the same meaning in algebra as in
arithmetic :
-f is read 'plus"
— is read 'minus"
X is read 'times" or "multiplied by"
-f- is read 'divided by"
= is read 'equals"
An exponent is a number or symbol written at the right of
another number or symbol and a little above it, to show how many
times the latter is to be used as a factor and to indicate its power.
For example, "25*" = 25 X 25 X 25 = 15,625. When no expo-
nent is indicated, 1 is understood to be the exponent.
An equation is an expression of equality between two numbers.
An axiom is a statement which is admitted to be true without
any proof. Algebraic operations make use of the following axioms :
(1) The equality of both sides of an equation is not destroyed
by the addition of the same number to both sides.
(2) The equality of both sides of an equation is not destroyed
by the subtraction of the same number from both sides.
(3) The equality of both sides of an equation is not destroyed
by the multiplication of both sides by the same number.
(4) The equality of both sides of an equation is not destroyed
by the division of both sides by the same number.
Positive and negative numbers. A positive or negative state
of any concrete magnitude may be expressed without reference to
the unit; thus, numbers that are greater than zero are positive, and
numbers that are less than zero are negative.
229
230 FUNDAMENTALS OF ALGEBRA
A number which has a " +," or positive, sign prefixed to it is
called a positive number; thus, +5. If a " — " sign precedes the
unit, it is called a negative number, and is written thus, —5.
Addition of positive and negative numbers. When two or
more positive and negative numbers are combined into a single
number, the result is called the sum of the numbers.
Example
The sum of +6 and -4 is +2
1 +4a and -2a is +2a
' +2 and -6 is -4
* +3a and —la is — 4a
' -4 and -3 is -7
1 — 3a and — 4a is — 7a
From the foregoing, the following rules may be derived :
(1) To add two numbers or terms of different signs, subtract
the smaller number or term from the larger, and prefix the sign of
the larger.
(2) To add two negative numbers, add their absolute values
and prefix the negative sign.
Example
Add +4a, — 3a, +6a. The problem may be restated thus: (+4a) + (— 3a)
Solution
+ 4a
+ 6a
+ 10a
- 3a
The addition of positive quantities is made in the same way as
in ordinary addition.
The addition of a positive and a negative number is equivalent
to deducting the smaller number from the larger, and retaining the
sign of the larger.
Problems
Find the sum of each of the following:
1. +4, -3, +7, -2, +6, +4, -8, -6.
2. -4, -8, +6, +5, -4, -3, -2, -7.
3. +1, -4, -6, +2, +7, +6, +4, +§.
The coefficient. The number or letter put before a mathemati-
cal quantity, known or unknown, to show how often it is to be
taken, is called the coefficient. In adding terms which are multiples
FUNDAMENTALS OF ALGEBRA 231
of the same letter, add the coefficients of these terms, and prefix
the proper sign. Thus, +6a, +76, — 5a, +66 becomes:
+6a + 76
-5a + 66
Added + a +136
It is convenient to arrange the terms in columns, so that like
terms stand in the same column.
Problems
Find the sum of each of the following:
1. +4a, +36, -4a, -26.
2. +6a, +46, -6c, +3a, -46, +4r.
3. +4*, -4y, +4*, -3*, +3//, +2*.
6. -
6. +4z, +3z, -Go;, +2z, -12*,
7. +4a, +36, +3c, -4c, +46, -a.
8. —a, —6, +c, +6, — c, +a, — r.
9. +c, +z, -z, +y, -2c, + 6x, -
10. +
Hereafter, when a term is not preceded by a positive or a nega-
tive sign, it is to be understood as a positive term.
Parentheses, brackets, and braces. Parentheses, brackets,
and braces are used to indicate that the part inclosed is to be
employed as a single term or as a single unit.
Since the same rules apply to all signs of aggregation, in the
explanation given hereafter, only parentheses will be mentioned.
RULES, (a) When a term in parentheses is preceded by a
"+," or positive, sign, the parentheses may be removed without
any change in the signs of the inclosed terms.
(6) If a term in parentheses is preceded by a " —," or
negative, sign, when the parentheses are removed it is necessary to
change each of the positive and the negative signs of the terms
inclosed.
Examples
(a - 6) + (c - d) - (e -/) = a - 6 + c - d - e +f
(4a + 36 - c) - (d + 4* - /) = 4a + 36 - c - d - 4e + /
Wherever possible, like terms should be combined ; as :
(3a - 36 + 4c) - (2a - 3c) = a - 36 + 7c
If several algebraic expressions are inclosed one within the
other by inclosure signs, such as parentheses or brackets, eliminate
the innermost pair of inclosure signs first*
232 FUNDAMENTALS OF ALGEBRA
Example 1
a + [b - (c - d)] = a + [b - c + d]
-a + 6-c + d.
Example 2
a + b - [- (b + c) + (c - d)] = a + b - [-b - c + c - d\
=a+6+6+c-c+d
= a -f 26 -f d.
Problems
Simplify the following by removing all signs of aggregation:
1. x 4- y + (x + y) - (2x + 2y).
2. 3a + (6 - 4c) - (4a - 6 - c).
3. 42 + (37 + 6) - (40 + 20).
4. 30 - [20 + (3 + 4) - (4 + 2)].
6. (3x - 4y) - (6* -f 2t/) + (4x - 3*/).
6. -3a + [46 - (6c + la) - 56 + 6c].
7. -362 -f 4a2 - (262 - 4a2) + 4a.
8. a + 6 + c + d - (2a + 26 - 2c + 2c).
9. -x - y - (z - x) - y - z(x + y + z).
10. (8a - 46) + (3c + d) - (3o -f 6 4- c).
Subtraction. Subtraction is the process of determining one ot
two numbers when their sum and one of the numbers are given.
The minuend is the sum of the two numbers.
The subtrahend is the number to be deducted.
The remainder is the required number.
To subtract, change the sign of the subtrahend and add the
subtrahend and the minuend.
Example 1
Subtract 8x from 4:r.
Solution
&x from 4r = (+4s) - (+8&)
Removing parentheses: = 4x — Sx
= -4s
Example 2
Deduct — 8# from — 4#.
Solution
-Sx from -4z = (-4z) - (-Sx)
Removing parentheses: = — 4x + Sx
Example 3
Deduct — Sx from 4x.
-8x from 4x = (+4x) - (-8x)
Removing parentheses: = 4x + Sx
= +12a?
FUNDAMENTALS OF ALGEBRA 233
To subtract algebraic expressions having two or more terms,
change the sign of each term of the subtrahend and proceed as in
addition.
Example
Subtract 3a - 5c - 3d from 7c - 16a - 2d.
Solution
The changing of the signs of the subtrahend is usually done mentally; thus.
-16a + 7c - 2d
3a - 5c - 3d
Oa-h 12c + d
Problems
Subtract:
1. MX from 28z.
2. I3x from 3x.
3. IGafrom -18a.
4. — 6a from 18a.
5. a -f 2a from 8a — 3a.
6. 4a - 36 from 3a -f- 46.
7. 16a - 146 + 3c from -13a -f 26 - 4c.
8. - 16a + 46 + 2c from 18a - 146 -f- 3c.
9. 3cs - 562; from -6cs + 76z.
10. 3p + 4g from 4p — 4g.
Multiplication. When two or more numbers are multiplied,
the result is called the product of the numbers.
When two numbers with like signs, either positive or negative,
are multiplied, the product is positive.
Examples
(+a) X (+6) - +ab
(-a) X (-6) = +ab
(-4) X(-4) = 4-16
When two numbers with unlike signs are multiplied, the
product is negative.
Examples
(+a) X (-6) = -a6
(+4) X(-3) = -12
The exponent to be used in the product is equal to the sum of
the exponents appearing in the multiplier and the multiplicand.
Examples
(0) * X (a)1 = a2
(a4) X (-a3) = -a7
(-3a2) X C-2a3) = +6a6
234 FUNDAMENTALS OF ALGEBRA
Problems
Multiply the following:
1. a2 by a2. 7. 3a + 100 by 6.
2. a3 by a6. 8. 3(a + b).
3. -a3 by a4. 9. 1,000 + s - 6 by 6.
4. ab by db. 10. (100 - c) - 6 by 4.
6. a2 by fc2. 11. [20,000 - (2,000 + z)] by 12£.
6. a + 6 + c by 5. 12. [40,000 - (T - B - 2,000)] by 12.
Division. Division is the process of finding one of two numbers
when the other number and the product of the two numbers are
given. When the dividend and the divisor have like signs, either
positive or negative, the quotient is a positive number.
Examples
ab -f- a — b
— ab -. 6 = a
-14 ^--7 = 2
When the dividend and the divisor have unlike signs, the
quotient is a negative number.
Examples
—ab -T- a = — b
ab -T- —b= —a
-10 ^ 5 = -2
10 -T- 5 = 2
The exponent to be used in the quotient is equal to the differ-
ence between the exponent in the dividend and the exponent in the
divisor.
Examples
a7 -7- a2 = a8
— ab2 -f- b = -ab
8a664 -r- -2ab3 = -4a66
206 -f- 202 = 20'
Problems
Divide the following:
1. 63 by 9. 7. 5a2 - 606 by 5.
2. -40 by 8. 8. 50a10 - 25a15 by 5.
3. -125 by -5. 9. 20,000 - (2,000 - 5x) by 5.
4. 48a by 12a. 10. 1,600 - (200 + 4x) by 4.
6. a26 by a. 11. GOa28 - 30a24 + 15a20 by 15a6.
6. 9a2 by 3a. 12. 72a8 - 18a10 - 54a6 by 9a2.
CHAPTER 25
Equations
Simple equations. An equation is an expression of equality
between two magnitudes or operations. The members of the
equation are separated by the sign of equality, " =," which means
"is equal to." Either member of an equation may contain
numerals, letters, or both. A simple equation is an equation of
the first degree, and contains but one unknown.
Example
Simple equation: x = 50
Or: 10* = 500
If the same number is added to or subtracted from both sides
of an equation, the equality is not destroyed.
Examples
Simple equation: x = 50
Adding 10 to each side: x + 10 = 60
Simple equation : lOx = 500
Subtracting 10 from each side: lOz — 10 = 490
If both sides of an equation are multiplied by or divided by f he
same number, the equality is not destroyed.
Example 1
3z = 15
Multiplied by: 5
15* = 75
Simplifying, or dividing by 15: x = 5
Example 2
4)20s « 40
Divided by 4: 5* = 10
Simplifying, or dividing by 5: x = 2
Any term of either member of an equation may be transposed
from one side to the other by changing the sign of the term, and the
235
236
EQUATIONS
equality of both sides of the equation is not destroyed. This
operation is equivalent to either adding the same quantity to, or
subtracting the same quantity from, both sides of the equation.
Simple equation:
Transposing the " — 10" to
the right side of the equa-
tion, and changing the
sign:
Or:
Example 1
lOz - 10 = 490
lOz
lOx
490 + 10
500
Simple equation:
Transposing:
Or:
Example 2
x + 5 = 15
x = 15 — 5
x = 10
Example 3
Solve the equation, lOx — 14 = 6x -\- 2.
The equation:
Transposing "§x" to the
left side, and changing
the sign to "-":
Transposing " — 14" to
the right side, and
changing the sign:
Uniting similar terms:
Dividing by 4:
Adding:
Left side
lOx =40
-14 = -14
26
Solution
lOz - 14 = 6z + 2
- 6z - 14 = +2
lOz — 6z = 14 + 2
4z = 16
x = 4
Verification
Adding:
Right side
6z = 24
2=2
26
Therefore the two sides of the equation are equal, and x = 4.
Problems
In the following equations, solve for the unknown quantities, and check your
results:
1. 40a; - 20 = 10s + 10.
2. 10z + 5 = 25.
3. 8a + 8 = 2a + 32.
4. 16 + 5a = 8a + 1.
5. 496 - 4 = 376 + 8.
6. 2x + 6(4z - 1) = 98.
7. 39 + 4(a + 6) = 8a - 1.
8. 2,000 - (5* + 500) = 2,500.
9. 86 - 5(46 + 3) = 1 - 4(26 - 7).
10. 18y - (lOy - 8) = 20y - (6y + 4).
EQUATIONS 237
Example
How can 90 be divided into two parts in such a way that one part will be
four times the other?
Solution
Let: x — the smaller part
Then: 4z = the larger part
Adding: 5x = 90
Dividing each member of
the equation by 5: x = 18, the smaller part
4x = 72, the larger part
Example
How many dimes and cents are there in $2.40, if there are 60 coins in all?
Solution
Let: x = the number of dimes
Then: 60 — x — the number of cents
Therefore: 10z = the value of the dimes
But: 60 — x — the value of the cents
Simplifying: lOz -f 60 - x = 240
Transposing: 9z = 180
x = 20, or there are 20 dimes
60 — x = 40, or there are 40 cents
Verification
20 dimes = $2.00
40 cents = .40
60 coins = $2.40
Example
The P. Q. Company wrote off depreciation on its building, which cost $50,000,
at the rate of 2% of the original cost per annum. This amount was included in
General Administration Expense account, and constituted Y$ of that account.
The purchases cost two and one-half times the old inventory. The value
of the old inventory was twice the amount of the selling expense. The Belling
and General Administration Expense accounts each equalled 10% of the sales.
The new inventory was valued at an amount equal to the selling expense.
The interest and discount costs were £ of the selling expense. Set up a profit
and loss statement showing the net profit from operations.
Solution
$50,000 X 2% = $1,000, Depreciation
$1,000, Depreciation = -j^ of General Adm. Expense
Multiplying by 10: $1,000 X 10 = $10,000 = General Adm. Expense
$10,000 = Selling Expense
Selling Expense = ytr of Sales
Multiplying by 10: $10,000 X 10 = $100,000, or Sales
2 X $10,000 = $20,000, or Old Inventory
2^ X $20,000 = $50,000, Purchases
i of $10,000 « $2,000, Interest and Discount
238 EQUATIONS
THE P. Q. COMPANY
PROFIT AND Loss STATEMENT
DECEMBER 31, 19 — .
Sales $100,000
Cost of Sales:
Old Inventory $20,000
Purchases 50,000
Total $70,000
Less New Inventory 10,000 60,000
Gross Profit $"40^000
Selling Expense $10,000
General Administrative Expense 10,000 20,000
Operating Profit $ 20,000
Interest and Discount 2,000
Net Profit $ 18,000
Problems
1. How can 640 he divided into two parts, in such a way that one part will
be seven times the other?
2. How many quarters and cents are there in $4.00, if there are 40 coins
in all?
3. A sum of money, $10.00, is made up of dimes and quarters, the total
number of coins being 88. How many dimes and quarters are there?
4. In a certain dairy, the ice cream contains 14% cream. If the mixture
is to be made from coffee cream of 20% butter fat and milk which tests 4%,
what portion of each will have to be used in a mixture of 100 Ibs.?
5. How many pounds of coffee worth 25^ per pound must a grocer mix
with other coffee worth 42^ per pound to make a mixture worth 34^ per pound?
The total quantity desired is 50 Ibs.
6. Barnes has $6,000 invested in 5% bonds. How much must he invest
in 8% stock to make his average net income 6%?
7. An estate of $33,120 was to be divided among the mother, three sons,
and three daughters. The mother was to receive four times as much as each
son, and each son three times as much as each daughter. How much was each
to receive?
8. In the making of candy, a mixture of 75% sugar at 5jz£ per pound and 25%
corn syrup at 2£ per pound is used. If the price of sugar advances to 7£ per
pound, what must be the ratio of sugar and corn syrup to be used, if the cost of
production is to remain the same?
9. A vinegar manufacturer makes various grades of vinegar, ranging from
50% to 100% pure cider vinegar. How much 100% pure cider vinegar must
be mixed with 63% pure to make 100 gallons of 75% pure?
10. How much milk with a butter fat test of 3.5% must be mixed with milk
testing 4.65% to fill 75 quart bottles which are to test 3.95%?
EQUATIONS 239
Fractions. In many accounting problems it is necessary to use
simple fractions in algebraic form, such as " yV of x" or "^ times
o
x 4ot
a," which when simplified are written T^ or -=-•
10 o
Example
Divide 129 into two parts, in such a way that | of one part will equal |- of
the other.
Solution
Let: x — one part
Then: 129 — x = the other part
2 v 3
The problem states that: ~ = - (129 — x)
9 8
The common denominator of the
fractions, as seen by inspec-
tion, is 72, and by the process
of changing the fractions to a
common denominator, the re-
suit is: ^ = | (129 - x)
Multiplying each side of the
equation by 72, to clear the
equation of the fractions, the
result is: HKT = 27(129 - x)
Eliminating the parentheses: Hu; = 3,483 — 27 x
Transposing: 43.r = 3,483
Dividing by 43: x = 81
129 - x = 48
Verification
| of 81 = 18
f of 48 = 18
Example
The superintendent of a certain plant was hired under a contract which
provided that he was to receive 10% of the net profits of the business as a salary,
after his salary had been deducted as an expense. The profits for the year
were $13,200. Compute the superintendent's salary.
Solution
Let: s = the amount of salary
The problem is stated: * = TV($ 13,200 - s)
Clearing of fractions by multiplication: 10s = $13,200 — s
Transposing: 11s = $13,200
Dividing by 11: s = $1,200
Verification
Net Profits $13,200
Less Salary 1,200
$12,000
Dividing by 10 * 1,200
240 EQUATIONS
Clearing of complex fractions. To clear an equation of a
complex fraction, multiply the opposite term of the equation by
the denominator of the complex fraction, and solve the resulting
equation by the usual methods.
Example
Find the value of (1.06)10 in the following equation:
i__l_
(1.06) 10 = $7 3600871
Solution
STEPS
1
1 -
10
'- = $7.3600871
.06
Multiplying right-hand
term by .06: 1 - 7j-^yr0 = 7.3600871 X .06 (1)
Clearing: 1 - * ' = .4416052 (£)
Transposing and changing
signs: • = 1 - .4416052 (8)
Clearing: ~~ = .5583947 (4)
Changing and dividing: (1.06)10 = 1 -f- .5583947 (5)
Clearing: (1.06) I0 = 1.790847 (6)
Problems
1. One-half of a certain number exceeds one-sixth of the same number by 8.
What is the number?
2. Find the value of (1.05)10 in the following complex fraction:
I
1 -
(L°5)1° = 7.72173.
.05
3. Find the value of (1.03)15 in the following complex fraction:
— ^^—^ = 18.598913.
.Uo
4. Find the value of (1.04)20 in the following:
^ — = .0735818
" .04
EQUATIONS S41
6. Find the value of (1.005)48 in the following:
= 42.5803178.
Simultaneous equations with two or more unknowns. When
each of two equations contains two or more unknown quantities,
and every equation containing those unknowns may be satisfied by
the same set of values for the unknown quantities, the equations
are said to be simultaneous.
The value of the unknown quantities in two or more simul-
taneous equations may sometimes be found by combining the
equations into a single equation containing only one unknown
quantity.
This combining may be done in several different ways, and is
known as elimination.
The method of elimination by addition and subtraction is
probably the most simple, and will therefore be the one used here.
It is often necessary to multiply one, or sometimes both, of the
equations by a number that will make the terms that contain one
of the unknowns in each equation of equal absolute value. Substi-
tute known values where possible.
Add or subtract the resulting equations, and the sum or
remainder will be an equation containing one unknown less than
the previous equations.
The chief difficulty in the practical application of these rules is
the expression of the unknowns in the form of equations. It seems
advisable to make a written statement of each condition, equation,
or unknown, and also a similar statement of each of the knowns.
After each statement, a symbol or letter should be used to
represent each unknown or known. In algebra, the letters "x"
"y," and "z" are commonly used, but it seems to be more practical
to use the initial letter of the name of the thing whose value is to
be found.
Example 1
Carol has five times as much money as Mary. Together they have $60.
How much money has each?
Statement of equations:
M = the number of dollars belonging to Mary
C = 5M, or five times as much as belongs to Mary
M + C = $60
Solution
Substitution of 5M for C: M + 5M = 60
Combining: 6M = 60
Dividing by 6: M = 10
C = 5M, or $50
242 EQUATIONS
Example 2
It cost $98.50 to manufacture and sell a certain article. The cost of the
labor was equal to the cost of the material used. The cost of the overhead was
$9.50 more than the expenses. The overhead and expenses totaled $2.50 more
than the material. Find the cost of each item.
Statement of equations:
STEPS
M = Cost of Material (1)
L = Cost of Labor (2)
O = Cost of Overhead (5)
K = Cost of Expenses (4)
M -f- L + O -f- E = $98.50 (5)
Solution
M = L (6)
() = E + $9.50 (7)
M = 0 + E - $2.50 (8)
M = E + $9.50 -f E - $2.50 (9)
M = 2E -f $7.00 (10)
M = 2E -f $7.00 in terms of E (11)
L = 2K -f $7.00 " " " " (12)
O = E + $9.50 " " " " (13)
E = E " " " " (14)
Adding (11), (12), (IS), and (14): M-\-L + 0 + E = 6E + $23.50
Substituting: $98.50 = 6# + $23.50 (16)
Transposing: QE = $98.50 - $23.50 (J7)
Dividing by 6: E = $12.50
Substituting in (/I), (^), (/3), and (^), the value of each item may be
found.
Example 3
In the following simultaneous equations, solve for the values of a and b:
STEPS
10a - 66 = 38 (1)
14a + 86 = 4 (£)
Multiplying (1) by 7: 70a - 426 = 266 (3)
Multiplying (2) by 5? 70a -f 406 = ^0 (4)
Subtracting (4) from (3) : ^826 = 246 (6)
Dividing by 82: —6 = 3 (6)
Or: 6 = -3 (7)
Substituting the value of 6 in (1) : lOo + 18 = 38 (8)
Transposing 18 in step (8): 10a = 20 (9)
Dividing by 10: a = 2 (10)
Example 4
The following are the condensed balance sheets of three companies who wish
to know their true worth as of December 31 :
EQUATIONS
243
Company
A
Assets, exclusive of intercompany in-
vestments $200,000 00
investment in Company B (50%) . . 350,000 00
" Company C (20%) .... 250,000 00
" Company C (20%). ..
" Company A (10%) ...
S800JOOO 00~
Company
B
Company
C
$500,000 00 $300,000.00
100,000 00
____ 100,000 00
$600,000 00 $400~000 00
Capital Stock $500,000 00
Surplus 300,000 00
$400,000 00 $300,000 00
200,000 00 100,000.00
As each company owns stock in each of the others, there are three unknown
quantities. The true worth may be found as follows:
Company A owns . .
Company B owns
Company C owns . .
Let:
Let:
Let:
Solution
SUMMARY OF OWNKRSIIII'
Company A
Net Assets
$200,000 00
500,000 00
.. 300,000 00 10%
A — Net Worth of Company A
B = Net Worth of Company B
C = Net Worth of Company C
Company B
50 %
Statement of equations :
A = 200,000 +
B = 500,000 +
C = 300,000 +
+ 1C
Company C
20%
20%
Solution
Adding:
Multiplying (4) by 10:
Multiplying (2) by 5:
Subtracting (6) from (5):
Multiplying (3) by 7:
Subtracting :
Multiplying (9) by 10:
Dividing by 97:
Using (5) and substituting
the value of — rg-A :
Transposing:
Using (2) and substituting
the value of — £0:
Transposing:
STEPB
A -£# -
ir - 200,000.00
(1)
/y -
1C = 500,000 00
(£)
AA +
C = 300,000 00
(3)
xVA 4- ir# +
1C - 1,000,000.00
(4)
9A + 5£ +
6CY = 10,000,000.00
(*)
+ SB —
C = 2,500,000 00
(6)
9A +
76' = 7,500,000.00
(7)
xV^ +
7Cy = 2,100,000.00
(8)
= 5,400,000 00
(9^
97A
= 54,000,000 00
(10)
A
= 556,701 03
(ID
-55,670.10 + C = 300,000 . 00
(12)
C= 355,670.10
(IS)
J5 - 71,134
.02 = 500,000.00
(14)
B - 571,134.02
U6)
244 EQUATIONS
Verification
ABC Total
Assets, exclusive of inter-
company investments .. $200,000.00 $500,000.00 $300,000.00
A owns: £ of B 285,567.01
iofC 71,134.02
Total worth of A $556,701703 $ 556,701 . 03
B owns: i of C 71jL134^02
Total worth of B $571,134 02 571,134.02
C owns: A of A 55>670J°
Total worth of C $355,67010 355,670J 0
Total value of the three
companies $1,483,505 15
Problems
1. For a certain piece of advertising, the total cost of printing, envelopes,
and postage is $14.50. The envelopes cost $1.50 more than the postage. The
cost of printing is $0.50 more than the combined costs of postage and envelopes.
Find the separate costs.
2. At the end of the year, the books of the Blank Company showed a net
profit of $12,247.50. The treasurer was to receive 12^-% of the net profits as a
bonus. The treasurer defaulted, and it was found that his account was short
$4,847.55. Show: (a) the true profit; and (b) the treasurer's account balance.
3. A man has $7,000 invested at 5%. How much must he invest at 65-% to
make his total income equal to 6% on his total investment?
4. In a gallon of a certain kind of paint there are equal parts of pigment
and oil. How much oil must be added to a gallon of this paint to make a paint
of $ pigment and % oil?
6. A merchant made 25% on his capital the first year, excluding his salary.
He withdrew $1,800 for his personal expenses, and had $9,153.13 left. Find
the amount of his investment.
6. A certain candy contains 40% corn syrup and 60% sugar. The syrup
costs 2ff per pound and the sugar 5^ per pound. If sugar advances to 6^ and the
cost of the syrup is unchanged, in what proportions must the ingredients of
the mixture be used in order to keep the cost the same?
7. A and B were partners, and agreed to share profits in proportion to the
capital invested. The profits were $2,000. A owned a f interest plus $400,
and his share of the profits was $900. What was the value of the business and
the capital of each partner?
8. The audited statements of a company show the following: The cash is
$2,400 more than the expenses. The accounts receivable equal twice the amount
of cash less $8,000. The cash and the accounts receivable together exceed the
expenses by $10,000. Find the amount of cash, accounts receivable, and
expenses.
9. The following is the balance sheet of the B. Company:
EQUATIONS
245
Assets
Cash $ 200
Accounts Receivable 2,500
Plant and Equipment 18,000
Officers' Accounts:
President Smith 3,000
Treasurer Brown 4,500
$28,200
Liabilities
Accounts Payable $ 5,000
Capital Stock 10,000
Surplus or Net Profits 13,200
$28,200
By agreement, the president was to receive, in lieu of salary, 15% of the net
profits, and the treasurer was to receive 10% of the net profits. Both the
deductions were to be included in expenses.
The treasurer, who was not bonded, has disappeared, and you are now
requested to state the amount of Smith's and of Brown's accounts and also the
true amount to be credited to surplus.
10. You are requested to find the value, for consolidated purposes, of the
following balance sheets, all as of the same date:
Assets, other than stock
Stock in B Company
A
. .. $400,000
60,000
B
$200,000
C
$200,000
20,000
Stock in C Company . . . .
60,000
20,000
Deficit
40000
$520,000
$260,000
$220,000
Liabilities
$200 000
$160000
$ 40000
Capital Stock
300,000
100,000
100 000
Surplus
20,000
80,000
$520,000
$260,000
$220,000
The investments in stock were at par and cost, there being neither surplus nor
deficit at the date of purchase.
11. Three companies agree to consolidate, and each agrees to accept its
pro rata share in the capital stock of the new corporation, D. Corporation D i»
formed with 500,000 shares of no par value stock.
Total Assets .............. $ 750,000
Total Liabilities
Capital Stock
Surplus
125,000
500,000
125,000
Total Assets .............. $1,000,000 Total Liabilities .......... $ 375,000
Capital Stock ............ 375,000
Surplus .................. 250,000
$1,000,000
$1,000,000
Total Assets $1,750,000 Total Liabilities $ 550,000
Capital Stock 1,000,000
Surplus 200,000
$1775(^000 $1,750,000
246 EQUATIONS
STOCK OWNERSHIP
ABC
A owned 15% 15%
Current at $50,000 $150,000
B owned 15% 10%
Current at $75,000 $ 37,500
C owned 5 % 5 %
Current at $25,000 $30,000
What percentage of the capital stock of Corporation D will each incorporalor
receive, arid what will be the book value of each company's interest in Corpo-
ration D?
Arithmetical solution of problems containing unknown quanti-
ties. Home accountants prefer to solve problems containing
unknown quantities by an arithmetical rather than an algebraic
method. The arithmetical method consists of making estimates
of the unknown quantities on the basis of quantities which are
known. A second test is made, based upon the results of the first
estimate. Succeeding tests or approximations are then made until
the correct value is ascertained.
A peculiarity of this method is that mistakes made in the
computations will always be eliminated, and the final computation
will always be correct. An error in computation may necessitate
a, greater number of approximations, but in the end it will be
eliminated in the process of solution.
Example
You are requested to find the value, for consolidated balance sheet purposes,
of the following corporations. The separate balance sheets show the value of
each corporation as of the same date.
AMES COMPANY
Assets $200,000 Liabilities . . . . $100,000
Stock of Brown Co. (par) 30,000 Capital Stock . . . 150,000
Stock of Coulter Co. (par) 30,000 Surplus 10,000
$260,000 $2fyyJOQ
BROWN COMPANY
Assets $100,000 Liabilities $ 80,000
Stock of Coulter Co. (par) 10,000 Capital Stock 50,000
Deficit 20,000
$130,000 jiU30,QOQ
COULTER COMPANY
AssetvS $100,000 Liabilities . $ 20,000
Stock of Brown Co. (par) 10,000 Capital Stock 50,000
Surplus 40,000
$110,000 sno,ojo
The holding company purchased the stock of the subsidiaries, paying book
value therefor. Book value was par, for the subsidiaries had neither surplus
nor deficit at the date of acquisition by the holding company.
£
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EQUATIONS
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248 EQUATIONS
Problems
1. A corporation wishes to create an insurance fund equal to 25% of the
net profits after deduction of the insurance fund and of the manager's bonus of
10% of the profits, the bonus and the insurance fund both to be considered as
expenses. The profits are -147,250. Find the amount of the bonus, the insur-
ance fund, arid the balance to be carried to surplus.
2. The assets and liabilities of two companies are as follows:
Smith Jones
Company Company
Assets, exclusive of intercompany investments $360,000 $320,000
Due from Jones Company . . . 20,000
Due from Smith Company SO, 000
Deficit . . . . 100,000 160,000
i$4SO,000 $560,000
Liabilities, exclusive of intercompany invest-
ments ... . . \ . . *400,000 $540,000
Due to Jones Company SO, 000
Due to Smith Company . ... 20,000
84X0,000 #560,000
Both companies having failed, you are required to state how many cents on
the dollar each firm can pay.
3. Three companies, M, Ar, and O, agree to consolidate. Their balance sheets
show assets as follows:
M: Other assets, $ 42,5,000 + 10%.V -h 15%0
N: Other assets, 462,500 + 15',',J/ + I5%0
O: Other assets, 1,145,000 + V2%</0M + 5%N
What are the assets of each company?
CHAPTER 26
Logarithms
Uses of logarithms. It is not the purpose of this chapter to
explain how logarithms are derived, but to make as clear as possible
the simple use of logarithms by the accountant. The accountant
desires to know how to make a particular calculation accurately
and in the least possible time. Logarithms are an exceedingly
valuable tool for this purpose and have many practical applications.
The use of logarithms greatly simplifies the multiplying and
dividing of numbers, the raising of numbers to powers, and the find-
ing of the roots of numbers. Logarithms reduce the multiplying
and dividing of numbers to problems of addition and subtraction;
the finding of the power of a number to a problem, in multipli-
cation; and the extraction of a root to a problem in division.
Exponents. Logarithms are exponents; that is, the logarithm
of a number is the exponent indicating the power to which a base
must be raised to produce the number. The base of the common
system of logarithms is 10. Therefore:
The logarithm of 100 is 2, because 10 must be raised to the 2nd power to
produce 100.
The logarithm of 1,000 is 3, because 10 must be raised to the 3rd power to
produce 1,000.
The logarithm of 10,000 is 4, because 10 must be raised to the 4th power to
produce 10,000.
The logarithm of 100,000 is 5, because 10 must be raised to the 5th power to
produce 100,000.
Obviously, the log of a number between 10 and 100 will be
something between 1 and 2; for instance, the log of 50 is 1. (59897.
And obviously, the log of a number between 100 and 1000 will bo
something between 2 and 3; for instance, the log of 625 is 2.79588.
A table may now be made as follows :
Number Log
10 ... I. 00000
50 . 1 69897
100 . . 2 00000
625 . 2 79588
1000 3 00000
10000 4 00000
100000 5.00000
249
250 LOGARITHMS
Parts of a logarithm. The part of a logarithm at the left of the
decimal point is called the characteristic; the part at the right of the
decimal point is called the mantissa.
If the number is 10 or more hut less than 100, the characteristic is 1.
If the number is 100 or more but less than 1000, the characteristic is 2.
If the number is 1000 or moic but less than 10000, the characteristic is 3.
Numbers which are less than 1 have negative characteristics.
The nature of positive and negative characteristics is indicated
below:
Characteristic
Number of Loq
100,000 5
10,000 . 4
1 ,000 . 3
100 . 2
10 . 1
1. . 0
1 . -1
01.. -2
.001 -3
.0001 ... -4
From the foregoing, the following rules for the determination
of the characteristic may be developed:
Logs for the number 1 and all numbers in excess thereof have
positive characteristics; the characteristic is one less than the
number of figures at the left of the decimal point in the number.
Logs for numbers less than 1 have negative characteristics; the
characteristic is one more than the number of zeros between the
decimal point and the first significant figure at the right thereof.
The usual way of writing the logarithm of a number is as
follows:
log US = 2.07 1SS2
Sometimes a sign, ";//," is used to separate a number from its
logarithm; the number is then read: 118 nl (the number whose
logarithm is) 2.071882; or reversed, it is read: 2.071882 In
(the logarithm of the number) 118.
Characteristic. The characteristic, as has been stated, is the
part of the number at the left of the decimal point of the logarithm.
The characteristic of the logarithm of each number from 1 to 9
inclusive is 0.; from 10 to 99 inclusive, it is 1.; from 100 to 999
inclusive, 2. ; and so forth. Also, the characteristic of each number
from .1 to .9 is — 1.; from .01 to .09 inclusive, — 2.; and so forth.
LOGARITHMS 251
Examples
The logarithm of 5. has a characteristic of 0.
...... 25. " " " *' 1.
" 490. " " " " 2.
The logarithm of 370. has a characteri.stic of 2.
41 4. " " " " 0.
« 0 " " " " 0.
.
" " " .3 " " " " -1. or 9. (mantissa) -10
.49 " " " " -1. or 9. " -10
Positive characteristic. From the foregoing list of numbers
and of the characteristics of their logarithms, it will be seen that
the characteristic for the log of 41)0. is the same as that for the log
of 370. Also, it is the same for the, log of 4. as for (he log of 2.7.
It is not the value of the digits in the number, but the number of
digits at the left of the decimal point in the number, that gives the
value to the positive characteristic.
Negative characteristic. If a number is less than 1., its log has
a negative characteristic; that is, the log of .3 or of .49 has a nega-
tive characteristic of —1., written as 1., and the log of .01 or of
.0245 has a negative characteristic! of — 2.
Examples
The log of .().-, 2 (>99()
The lojr of .(MM).") 4 0990
The log of .000005 0 0990
If a given quantity is added to any number, and from the sum
is subtracted the same quantity, the result is the same as the,
original number; that is, if 10 be added to 4, making 14, and from
that sum 10 be subtracted, the result, is still 4, although the form
in which it is written is different; as, 4 — 4 + 10 — 10. Instead
of writing a logarithm with a negative characteristic as, 1.1950
in some calculations it is found to be more convenient to indicate
the negative characteristic; in the following manner: 9.1959 — 10.
Or it may be convenient in some cases to use a larger number;
thus, 1.1959 may be represented as 29.1959 - 30, or as 0.1959 - 7.
Any number may be added, provided the same number is used as a
negative quantity. The change in form does not change the value.
A characteristic may be either positive or negative, but a
mantissa is always positive. The small dash over the characteristic
is intended to serve as a reminder that only the characteristic is a
minus quantity, while the mantissa is invariably a plus quantity.
Mantissa. The mantissa of the log of a number is a group of
figures which stands for or represents the sequence of the digits of
the number.
252 LOGARITHMS
The mantissa of the log of 125. is shown by the table to be
.096910, which is also the mantissa of the logs of 12.5, 1.25, and
125,000. The mantissa is determined by the sequence of the digits
of a number, and the, characteristic is used to show the correct
placing of the decimal point.
Examples of Numbers and the Mantissas of Their Logarithms
Tho sequence of dibits in 125. is indicated by the numti^a, .090910
" " " 1250. " " <: " li .090910
" " " " " 125000. " " " " <k .090910
" " " 12.5 " 4k .090910
" " " " " 1.25 4l <4 " '4 <v .()9r>910
« n „ ,2- „ u ., t< .. .()()(;()!()
« u n i. i, 4-- u n ,4 4. „ .(J70(i94
t- a u ^25 *» .• <k «< - .795SSO
u u 4* ()9()< « .« «< - « .0(i;j7,ss
" " " " 1. " " " " *' .000000
Examples of Numbers and Their Complete Logarithms
iatic and ,\>
loK 1235. - 3.09107
" 123.5 ---- 2.09107
" 123.") - 1.09107
1. 23.') 0 09107
" .123.") - 1.09107
.0123.') - 2.09107
" .001235 -.-. 3.09 107
The table given in Appendix III of this book is called a ,sv.r-
pluce table. 'This means that the mantissas as given are accurate,
to the sixth place. However, it does not necessarily follow that by
the use of a six-place table calculations can be performed accurately
to the sixth place. Tables of logarithms accurate to six, eight, or
even ten places are sometimes used, but the six-place table is
sufficient for ordinary purposes. If the accountant ha> many
computations involving large numbers, he should procure a more
extended table.
How to use a table of logarithms.
For numbers of one siynijicant figure. The table shows only the
mantissa of each logarithm. If it is desired to find the mantissa of
a number such as 2, 20, 200, 2,000, or of any number whose only
significant figure is 2, it is necessary to turn to the table and in the
column at the left, headed "Ay run down the line until the number
2(K) is reached. To the right of this number, in the column headed
'()/' the mantissa .301030 is found. This is the mantissa for the,
logs of 2, 20, 200, or .2, .02, .002, and so forth. The mantissa for
LOGARITHMS 253
any other number of one significant figure may be found in a
similar manner. In order to obtain the complete logarithm of a
number it is necessary to supply the characteristic. The charac-
teristic of the log of 2 is (). Hence, the complete logarithm of 2 is
0.30 1030.
For numbers of two significant fit/nres. If it is desired to find
the logarithm of a number containing two significant figures as,
17, 170, or 1.7 it is necosary to look in the column at the left of
the table, headed "A'/' and run down the column until 170 is
reached. In the column to the rin;ht of 170, headed "0," the
mantissa .230441) is found. Keep in mind that this is the mantissa,
for the logs of 17, 170, 1,700, and so forth.
For numbeis of three snjnijicnnt figures. Assume that the
logarithm of I IX i> desired. In the left-hand column of the table,
headed lk A," iind t he number 1 1 S. To t he riirht of t his number, in
the column headed "0," the mantissa .071 SS2 is given. This, of
course, is the mantissa for the logs of 1 IS, 1.1S, 1 1 ,<SOO, and so
forth. In the foregoing illustrations, the mantissa only was found.
By the rules previously given, the characteristic of the log of IIS
is ascertained to be 2. Therefore, t he logarithm of the number
I IS is 2.071SS2.
For numbers of four sn/nijieont Jujures. To illustrate: It/ is
required to find the logarithm of 1,(>4S. Find the number 1(>4 at
the left of the table, in the column headed " A V On the horizontal
line to the right of 104, in the column headed "S," the mantissa
found is .2K>Of>7. The characteristic for the log of 1,04X is 3.
The complete logarithm of 1,048 is 3.210957.
Interpolation for numbers of Jive or more si(/nijie(int Ji(/ure,s.
The logarithm of a number of five or more significant, figures may
be found by the process of interpolation. This method is based
upon the assumption that the differences of the mantissas are
proportional to the differences of the numbers given. This pro-
portion is not strictly exact, for the difference's really grow smaller
as the mantissas themselves grow larger. However, the results
obtained deviate only slightly from the true results, and are
sufficiently accurate for most purposes.
Example
Find the logarithm of 131,525.
Solution
In the column headed "/)/* on the line horizontal to the number 131 in the
column headed ''A'," the difference between the mantissa of the log of 1,315 and
the mantissa of the log of the next higher number, 1,310, is given as 330; this is
more correctly stated as .000330. The excess of 131,525 over 131, .500 is T2A of
the difference between 131. ,500 and 131,600. Therefore, multiply .000330 by
254 LOGARITHMS
) obtain the fractional part of the difference in the mantissa; it is .0000825.
The mantissa for 131,500 is .1 1X926. To this add the .OOOO.S25, which will give
the mantissa for 131,525, .1 1900S5. Stated again:
The log of 131,600. has a mantissa of .1 19256
" " " 131,500. u " " " .11S926
The difTerence of 100 = the difTerence of .000330
131,525 - 131,500 = 25
The difTerence of 25 in the numbers requires that a proportional part of the
difference in the mantissas, .000330, be added to the mantissa oi the log of the
smaller number.
iVo of .000330 = .OOOOS25
.1 1X920 1- .0000X25 - .1 1900X5
.11900X5 is approximately the correct mantissa. The characteristic for the
log of 131,525 is 5. Therefore, the complete log of 131,525 is 5.1 1900X5.
Problems
Find the logarithms of the following (express your answers in the form:
"log 1 IX - 2.071XX2").
1. 4
6. 5
11. 1.127
16. .S237X
2. 20
7. X2
12. 1,275
17. .03264
3. 30
S. 775
13. 1,4S2
18. .0003X2
4. .00
9. S27
14. 739.S2
19. 1.00375
5. 25
10. S.37
15. 6X,439
20. 2 4S7(>5
To find a number when the logarithm is given. If the above
process of finding the logarithm from a number is reversed, the
number can easily be found from the logarithm. This process is
called finding the antilogarithm. It is necessary first to find the
digits of the number, and this must be done from the mantissa.
If the mantissa can be found in the table, take the digits corre-
sponding to the mantissa, and point off these digits decimally as
indicated by the characteristic.
Example
Find the number whose logarithm is 0.2S1033.
Solution
In the table, the mantissa, .2S1033, is found in the vertical column headed
"0," opposite the number 191. This indicates that the sequence of digits,
ignoring possible initial or final zeros, is 191. Using the characteristic for the
correct placing of the decimal, the number is found to be 1.91.
To find a number whose mantissa is not in the table. If the
mantissa is not given in the table, it is necessary to reverse the
process of interpolation.
LOGARITHMS 255
Example
Find the number \vho>e logarithm is 5.11900S5.
In the table, the mantissa for the number nearest the one given, and less
than it, is found to be .11S92(>. Subtracting this from the mantissa given in
the problem, the difference is .OOOOS25. Use this as the numerator, and the
diffeience, .000330 (indicated in the table in the column at the right, headed " /)")
as the denominator. Reduce this fraction .OOOOS25 .0003300; the result is .25,
or 25 100 of the difference between 1,315 and 1,31(>. Ar.ncx this amount to
1,31*5) and the sequence of digits is found to be 131,525, which, when pointed
off as indicated by the characteiistic, ghes the result, 131,525.
Problems
Find the numbers lepresented by the1 tollo\\ing logarithms:
1. l.()5X011 6. 1J)9M)70 11. S.I 720 19- 10 16. 4.037540
2. 2.71IS07 7. 2.<»M7.s4 12. 19.003S91 -20 17. 2.0S2()S5
3. 0.0X1241 8. 3.C>2107(> 13. 3.17(i()<)l -4 18. <>. 403794 - 7
4. 9.67209S - 10 9. 5.(>3,s9XS 14. (U321XS - 9 19. 3.390791
5. 2.707570 10. I.S()()23<> 15. 7.30304S 20. 1.S44334
Rules for computation by logarithms.
RULE 1. To multiply numbers, add their logarithms; the sum
is the logarithm of tin* product.
RrLK 2. To divide numbers, subtract the logarithm of the
divisor from the logarithm of the dividend; the remainder is the
logarithm of the quotient.
Rt'LK 3. To obtain a power of a number, multiply the loga-
rithm of the number by the exponent of the power sought; the
product is the logarithm of the power of the number.
RTLK 4. To obtain a root of a number, divide the logarithm
of the number by the index of the root sought; the quotient is the
logarithm of the root of the number.
Multiplication by logarithms.
Example
Multiply H35 by 22, using the method of Rule I.
Solution
log 635 2 802774
log 22 1 342423
log of product . ... 4.145197
In the table, the mantissa .145190 is found to correspond to the digits 1,397.
The characteristic 4 indicates that the product has five digits to the left of the
decimal point, hence, the product is 13,970.
256 LOGARITHMS
Problems
Multiply:
1. 25 by 25 8. 1.43 by .032 15. 145.3 by 6.296
2. 42 by 37 9. 1,480 by .138 16. .003 by .002
3. 240 by 381 10. 92.7 by 8.75 17. 100.05 by 100.25
4. 762 by 431 11. 3.39 by 8.92 18. 47.2 by 200
5. 42.5 by 49.2 12. 9.293 by 48.67 19. .999 by 647.2
6. 34.7 by 1.42 13. 143.9 by 1.478 20. 3.8 by 4.9
7. 1,430 by .249 14. 1.278 by 3.84
Division by logarithms. Rule 2 gives the procedure for divi-
sion by means of logarithms.
Division of a greater number by a lesser number.
Example
Divide S75 by 37.
Solution
log 875 ... 2 94200S
log 37 I 56X202
log quotient 1 373X06
In the table, the mantissa, .373X06, coi respond-; to the dibits 23,648. The
characteristic indicates that the quotient has t\\v digits to the left of the decimal
point. Pointed oil, the an>\ver is 23.648.
Division of a lesser number by a greater number. At times it is
necessary to divide a smaller number by a larger one; this, of
course, produces a quotient which is less than 1, and requires the
operation of subtraction of a greater logarithm from a lesser one.
To do this, change the form of the logarithm of the minuend; that
is, add 10 (or a multiple of 10) to the characteristic of the minuend,
and write -10 (or a multiple of -U), the same multiple to be used in
each case) after the mantissa.
Example
Divide 269 by 239,000.
Solution
log 269 - 2.429752, or 12 429752 - 10
Deduct log 239,000, or ... 5 378398
log quotient. ... 7 0.51354 — 10
Changing 7.051354 — 10 to its simple form, it becomes 3.051354. By
reference to the logarithm table of mantissas, and by applying the rule for
pointing off numbers by the characteristic, the quotient is found to be .0011255
(correct to five significant figures).
The above method is also applicable to the subtraction of a
negative logarithm from a positive one.
LOGARITHMS
257
Example
Divide 14,200 by .000191.
Solution
log 14.200 = 4.152288, or 14. 152288 - 10
log .000191 = 4.281033, or 6 281033 - 10
log quotient 7.871255 - 0
By finding the antilog of 7.871255, the resulting quotient is ascertained to
be 74,345,500 (correct to five significant places).
Problems
Divide:
1. 128 by 64 8. 2.486 by 3.45
2. 2,160 by 150 9. .6843 by 89
3. 344 by 8 10. 9.278 by 12.43
11. 6 by 2 *
12. 89 by 47
13. 2.1 by4S
14. 3.875 by 23S.7
Rule
4. 93 by .31
5. 649.4 by 24.3
6. 8.42 by 2.48
7. .3472 by 124
Powers of numbers.
the powers of numbers.
15. 1.425 by 892.7
16. 147.25 by 9,276
17. .03 by 6,000
18. .01 25 by 3,427
19. 1 .005 by 927.8
20. 2.4255 by 384.275
21. 6,497.S by 2.S74
gives the procedure for finding
Example
Find the fourth power of 26.
Solution
log 26 . 1 414973
Multiply by the exponent of the power . . . . 4
log power . 5 659892
In the table, the mantissa .659892 represents the sequence of digits 456,975.
The chin acteristic 5 indicates that the product has six digits to the left of the
decimal point. Hence, the fourth power of 26 is 456,975 (correct to five signifi-
cant places).
Process with a negative characteristic.
Example
Find the fifth power of .025.
Solution
log .025 = 2.397940, or
Multiply by the exponent of the power. .
41.9S9700 - 50, changed
The mantissa .9897
Pointed off bv the characteristic
Find the value of:
1. 5th power of 25
2. 4th power of 35
3. 5th power of 2
4. 7th power of 1 .25
6. 4th power of 2.47
6. 3rd power of 575
7. 9th power of 4
Problems
8. 12th power of 7
9. 121
10. 142
11. 9,200'
12. 18*
13. 1472
14. .01 52
8 397940 - 10
— J?
4 1 989700 - 50
9 9897
976562
00000000976562
16. 8.921
16. 1461
17. .07*
18. .97*
19. 303
20. 274
21. 9B
258 LOGARITHMS
Roots of numbers. Rule 4 gives the procedure for finding
the roots of numbers.
Example
Find the cube root of 875.
Solution
log 875 2.942008
Divide by index of the root 3)2_M200^
log quotient 0.980669
In the table, the mantissa .980669 corresponds to the sequence of digits
95,646, and the characteristic indicates that the number is 9.5646.
Process with negative characteristics. In stating the equiva-
lent of the negative characteristic of a logarithm, care must be
taken to see that the right-hand number, or minus quantity, is
exactly divisible by the index of the root with a quotient of 10 or a
multiple thereof.
Example
Find the 4th root of .125.
The negative characteristic of the complete logarithm of .125 may be stated
in several different ways: as 1.096910; as 3.096910 - 4, as 7.096910 - 8, as
19.096910 - 20; or as 39.096910 - 40.
In this problem, in order to avoid any complication, it is best to have the
right-hand number of the characteristic exactly divisible by 4 (the index of the
required root), with a quotient of 10.
Solution
log .125 I 09691C
Or, log .125 39 096910 - 40
Divide by index of root 4)_39J^969i°J7 i9
log quotient 9 774228 - 10
Or 1.774228
The mantissa .774228 corresponds to the succession of digits 594,604 (accurate
fo five places). Pointed off by the characteristic, —1, the 4th root of .125 is
594604.
Problems
Find the value of the following to five significant places:
1. Square root of 64. 8. 10th root of 10.
2. Square root of 97. 9. \Xl25.
3. Square root of .64. 10. tyl. _
4. Cube root of 81. 11. ^89.
6. Cube root of .081. 12. ^/89~27. 19. 506 X
6. 6th root of 49. 13. V-9643. 20. 3.872 X
7. 7th root of 750. 14. ^2980. 21. -v/225 X
The slide rule. This is an old device, yet it is used to only a
limited extent by accountants in general. It has been described as
"logarithms on a stick." In its simple form, this mechanical
LOGARITHMS 259
device consists of a grooved base rule into which a slide rule is
fitted. A third part is the runner. The graduations on the upper
part of the base and slide are called "upper scale/' while those on
the lower part of the base and slide are called "lower scale. "
These graduations are for different purposes, and because of the
different requirements they are graduated differently.
Use of slide rule. The slide rule is used either to check figures
or for original calculations. It may be used to check or compute
any operation involving multiplication, division, raising to powers,
or extracting of roots. It has a great many applications in busi-
ness, although it is generally considered as a device used only by
engineers.
The slide rule should appeal to the accountant as a device
which may be carried in the pocket or the brief case; its weight
is negligible.
Accuracy of calculations made by the slide rule. It is possible,
after having attained proficiency in the handling of the slide rule,
to obtain results in which the margin of error will not be more than
one-quarter of 1%. This is satisfactory for most business prob-
lems. As in logarithms, when the slide rule is used the digits are
taken from the left to the right of a number, regardless of the value
of the number. The slide rule is as nearly accurate in the calcula-
tion of decimal numbers as it is in the calculation of whole numbers
of large denominations. It will give correct answers of two places,
and if careful computations have been made, a three-place solution
of a good degree of accuracy may be expected. If extreme care is
used in the computation of a problem, an answer of four places may
be had with a fair degree of accuracy. Of course, a long and care-
fully graded slide rule will give better results than either a short or
a carelessly graded rule.
Theory of the slide rule. The theory of the slide rule is indi-
cated very roughly in the following simple example and illustration.
Example
Find the sum of 4 and 6.
RULE1
01234 5678 9 10 11 12
01234 56789 10 11 12
RULE 2
Figure 2.
Solution
Above are two ordinary rulers, set opposite each other. To find the sum of
4 and 6, perform the following steps:
260 LOGARITHMS
(1) Set the "0" on Rule 1 over the "4" on Rule 2.
(2) Observe the figure "6" on Rule 1.
(3) On Rule 2, immediately below the "6" on Rule 1, will be found "10"—
the sum of 4 and 6.
The process of subtraction may be shown as follows.
Example
Find the difference between 9 and 4.
RULE 1
0
1
2
3
4
5
678
9
10 11 12
0 1
2345
6
7
8
9
10
11 12
RULE
2
Figure 3.
Solution
(1) Locate the subtrahend "9" on Rule 2.
(2) On Rule 1, locate the minuend "4," and place it immediately over bhe
subtrahend "9" on Rule 2.
(3) On Rule 2, the number immediately below the "0" on Rule 1 will be the
remainder, which in this case is 5.
The foregoing examples show that addition and subtraction
of small numbers can be performed on two ordinary rulers. The
principle of the slide rule is similar.
On the common slide rule, the graduations are made according
to logarithms. Hence, if any two numbers on the slide rule are
added, the result obtained is the sum of two logarithms.
The addition of the logarithms of numbers results in multipli-
cation of the numbers; and the subtraction of the logarithms of
numbers results in division of the numbers.
How to learn to use the slide rule. Practice is without doubt
the only efficient method of learning how to use the slide rule. If
possible, a slide rule should be obtained for use in this chapter.
If, however, a slide rule is not available, a cardboard model may be
made for practice. Care must be used to have the markings as
accurate as possible, in order to obtain fair results. A model such
as the following can be used very conveniently for the practice
material.
i ra 13
i i i i i i i i i 1 1
4
5 |6
7 |8
9
i i I I I 1 1 1 1 I " j
1 !2 !3
4
5 6
7 8
9
Figure 4.
LOGARITHMS 261
Reading the slide rule. In reading the numbers, go over the
rule from left to right, as follows: 1, 2, 3 ... 10; then, beginning
at 1 again and calling it 10, read 10, 20, 30 ... 100; then, again
beginning at 1 and calling it 100, read 100, 200, 300 ... 1,000.
It is possible to do this because the mantissa for 10 is the same as
that for 100, 1,000, etc.
It will be noticed in Figure 4 that the spaces decrease from left
to right. These decreases should correspond exactly to the differ-
ences between the logarithms from 1 to 10. Assume that you
divide your model rule into 1,000 equal parts. Then, since log
2 = .301, the 2 would be placed at the 301st graduation. Log
3 = .477; therefore 3 would be placed at the 477th graduation.
Log 4 = .602; therefore 4 would be placed at the 602nd gradua-
tion. Similarly, 5 would be placed at the 698th; 6 at the 778th;
7 at the 845th; 8 at the 903rd; and 9 at the 954th.
Marks can be put in to show the mantissas for the logarithms
of 1.5, 2.5, 3.5, and so on, but they will not be half the distance
between the previous graduations, because log 1.5 is 0.176, and this
is not half the difference between log 1 and log 2.
It will be noticed that the distance from 1 to 2 is divided into
10 divisions. These are read from left to right, like telephone
numbers, thus: one-one, one-two, one-three, and so forth, to
one-nine; then 2. These are understood as 1.1, 1.2, 1.3, and so
forth. Consulting the table of logarithms for the logs of 110,
120, 130, and so forth, we find that the marks will be placed at the
following graduations: 41, 79, 113, 146, 176, 204, 230, 255, and
278.
Construction of model slide rule. Obtain a piece of cardboard
12^ inches long and 1 inch in width. Rule a line midway along the
full length, and mark it off in graduations of one-eighth inch. This
will make a measure with 100 graduations instead of 1,000, but for
the purpose of this work it will be satisfactory.
Since log 2 = .301, mark 2 at 30.1
log 3 = .477, mark 3 at 47.7
log 4 = .602, mark 4 at 00.2
log 5 = .698, mark 5 at 69.8
log 6 = .778, mark 6 at 77.8
log 7 = .845, mark 7 at 84.5
log 8 = .903, mark 8 at 90.3
log 9 = .954, mark 9 at 95.4
Between 1 and 2 are 1.1, 1.2, 1.3, .., 1.9, as previously
explained.
262
LOGARITHMS
Since log 1.10 = .041, mark at 4.1
log 1 .20 = .079, mark at 7.9
log 1.30 = .113, mark at 11.3
log 1.40 = .146, mark at 14.6
log 1.50 = .176, mark at 17.6
log 1 .60 = .204, mark at 20.4
log 1.70 = .230, mark at 23,0
log l.SO = .255, mark at 25.5
log 1.90 = .278, mark at 27.8
For closer graduations, you will find that you can make 10
Indentations with a sharp pin in one-eighth inch space. By care-
fully counting the points placed, you can use the full logarithm
and make a fairly accurate slide rule.
Having completed the graduations, cut the cardboard length-
wise on the medial line. This will give two pieces with measure-
ments exactly the same. These two measures may now be used as
Rule 1 and Rule 2 in the following simple problems.
Multiplication on the slide rule. Add the logarithms of the
numbers to be multiplied.
Example
Multiply 2 by 3.
RULE 1
1 2
3
4
5 6
7
8
91
1 2 3 4
5 6
7 8
9 1
RULE
2
Figure 6.
Solution
(1) Locate "2" on Rule 2.
(2) Place "1" on Rule 1 over "2" on Rule 2.
(3) Locate "3" on Rule 1.
(4) Read the number on Rule 2 immediately below, which is "6."
Problems
Multiply:
1. 2 by 4. 2. 3 by 4. 3. 2 by 5. 4. 3 by 2. 6. 4 by 2. 6. 3 by 3.
If the problem is of such a nature that the rules cannot be
operated by placing the left-hand "1" on Rule 1 over the number
on Rule 2, it is necessary to use the right-hand "I" on Rule 1.
Example
Multiply 4 by 5.
1
2 345
678
91
RULE 1
BULE2
1 2
3
456
7 8 91
Figure 6,
LOGARITHMS 263
Solution
(1) Locate "4" on Rule 2.
(2) Place " 1 " on Rule 1 over "4" on Rule 2, so that the "5" on Rule I is
over Rule 2. In this case it is necessary to use the " 1 " on the right-hand side
of Rule 1.
(3) Read the number immediately under "5" on Rule 1, which is "2."
"2" may be either 2., .2, .02, 20, 200, or any other number in which the left-hand
digit is "2." In this case the number can be determined by inspection to be 20.
Problems
Multiply:
1. 5 by S. 2. 4 by 5. 3. 3 by 10. 4. 20 by 20. 5. 40 by 5. 6. 2 by 30.
Division on the slide rule. Subtraction of logarithms results
in the division of the numbers.
Example
Divide 9 by (>.
RULE 1
1 2
i
3
4
5 G 7 8 0 1
1 ' ' ' ' ' 2 3
4 5
6
7 8 U 1
RULE 2
Figure 7.
Solution
(1) Locate "9" on Rule 2.
(2) Place "6" on Rule 1 over "9" on Rule 2.
(3) Read on Rule 2 immediately below " 1 " on Rule 1. As the " 1 " is over
the 5th graduation on Rule 2, the digits will be 15, and by inspection the answer
is determined to be 1.5.
Problems
Divide:
1. 6 by 2. 4. 6 by 4. 7. 35 by 7. 10. 95 by 5.
2. X by 4. 5. 9 by 4. 8. 25 by 5. 11. IS by 12.
3. 5 by 2. 6. 64 by 8. 9. 12 by 4. 12. 30 by IS.
Another type of problem is that of multiplying two or more
numbers and dividing their product by another number.
Example
8X9-5-4 = ?
Solution
(1) Set Rule 1 so that the " 1 " on the right is over "8" on Rule 2.
(2) Read the number immediately under "9" on Rule 1, which is "72."
(3) Set "4" on Rule 1 over "72" on Rule 2.
(4) Read the number on Rule 2 immediately under "1" on Rule 1, and the
answer is found to be 18.
Problems
Solve the following by the use of the slide rule:
1. 16 X 6 -r- 32. 3. 35 X 35 -f- 5. 6. 37 X 19 -i- 13. 7. 8 X 14 -h 12.
2. 20 X 40 + 8. 4. 45 X 12 -i- 8. 6. 44 X 34 ~ 27. 8. 4 X 7 -^ 200.
The following problems illustrate some of the practical appli-
cations of the slide rule.
264 LOGARITHMS
Problems
1. Payroll calculation. "Brown's time rani for a particular clay showed the
following:
Time
Job No.
II
M
12
3
30
21
2
10
32
50
45
1
30
S~
00
If Brown was paid 54^ an hour, what was the labor cost chargeable to each
job?
2. Prorating expend . The power cost of a small plant is to be distributed
to the departments on the basis of horsepower hours, as follows:
Dept. A . 45 horsepower
Dept. B.. 35 horsepower
Dept. C . . . . 90 horsepower
Dept. 1) .... . 20 horsepower
Dept. 10. .. . .5 horsepower
Dept. K ... 5 horsepower
Total 200 horsepower
The total power cost was $450. What was the cost of power in each depart-
ment ?
3. The air fare from .V to }" is $40. The traveler goes over three divisions
of airway, respectively 380, 230, and 190 miles in length. Find the amount of
the (are to be apportioned to each division.
4. An article that cost $25 is sold for $50 less 20%- Find the per cent of
gain on the cost. Find the per cent of gain on the selling price.
6. (liven:
Sales . . $500
Cost of (ioods Sold 300
Selling Expenses . . 75
(ieneral Expenses . . 50
Profit
What per cent of the sales is each item?
6. Work the following:
% on Selling
(^ost Selling Price, Price
(a) $ 5 00 20 ...
(6) 8 00 40 . ... ...
(c) 12 00 25 .. ...
(d) 20 00 20
(e) 10 00 30
7. The list price of an article is $25, less 10% and 5%. Find the net cost
8. Find the interest on $600 at 5% for 3 years 6 months.
9. A field is 40 rods wide and 80 rods long. How many acres does it contain?
10. Find the cost of SO items at $1.50 a dozen.
CHAPTER 27
Graphs and Index Numbers
Charts and graphs. Charts and graphs arc becoming increas-
ingly popular as a means of presenting the results of accounting
and mathematical computations. Accountants, credit men, pro-
duction managers, sales managers, advertising men, and general
business executives are realizing more and more how greatly
graphic charts may help them in their work. The reason is
obvious. Long rows of figures must be thoroughly studied if the
relations between quantities are to be grasped. This is a tedious
task. On the other hand, pages of valuable data may be presented
on a simple chart that will convey more real information than the
most elaborately written report. It is necessary, however, to
distinguish between important and unimportant data. Fur-
thermore, a method of presentation must be chosen that will
convey a correct impression, for it is quite possible to prepare
misleading charts from correct data. Two important points
must, therefore, be borne in mind:
(1) The selection of the data;
(2) The selection of the design.
Circle chart. This type of chart is used extensively for popular
presentation, and is designed to exhibit the true proportions of the
component parts of a group total. It is adapted to such purposes
as exhibiting the distribution of disbursements, the sources of
receipts, and the allocation of appropriations in government
finance.
The circle with sectors, however, is not so desirable a form of
presentation as the bar chart, described in later paragraphs, since it
does not possess the same degree of flexibility. It is impossible,
for instance, in a profit and loss analysis, to exhibit a loss. More-
over, it does not always permit a convenient arrangement of cap-
tions, which must sometimes be written in at an angle. Another
disadvantage is that the figures are not easily compared. For
these reasons, it is probably best to limit the circle chart to the
illustration of facts which are not intended to be compared from
266
GRAPHS AND INDEX NUMBERS
period to period. However, the sector method is so widely used
that it is perhaps better understood generally than any other.
The circle is segmented on the basis of 100°, not 360°, as geo-
graphic circles are segmented. This is because the chart circle is
designed to exhibit a percentage scale.
Example
Distribution of the expense dollar.
Figure 8. Circle Chart.
Comparison of circles. Comparisons in magnitude are some-
times made by presenting circles of different sizes. The objection
to this method is the resulting confusion in the mind of the reader
as to whether area or diameter is used as the basis of measurement.
It is impossible to estimate accurately the difference between
the diameters of two circles by merely looking at them. When
comparative diameters are being estimated, the circles themselves
have to be subordinated in the mind of the reader while the diam-
eters are visualized. Charts were devised because of their ease of
comprehension, and their purpose is defeated when conflicting
GRAPHS AND INDEX NUMBERS 267
metal processes are involved. For this reason, circles of different
sizes should never be used for comparative purposes.
The same criticism applies to squares and cubes. A cube
whose edge is twice that of another will possess eight times the
cubic content and four times the outside area of the smaller
one, and unless the basis of measurement is carefully explained, the
comparison may easily be misleading.
Problems
1. Using the figures in the following condensed operating statement, prepare
a circle chart showing the distribution of the "sales dollar."
Sales . . $66,734.49
Plant Operating F,xpenses .. .... . 21,464.91
Payroll . .. ... 18,055 22
Taxes . 1,055.07
Depreciation 14,252.63
Depletion . 4,667.20
Net Profit . . . . 7,239 46
2. Prepare a circle chart to illustrate the following accounts receivable
analysis.
Current to 60 days old ... 47 08%
90 days old . . 11.69%
120 days old 6 62%
4 to 6 months old . . 10.65%
6 months to 1 year old 8 54%
Over 1 year old 15.42%
3. Prepare a circle chart to illustrate the distribution of the sales dollar.
Raw materials , . . $ 55
Wages and salaries . . 25
Direct taxes . .07
Selling, advertising, and miscellaneous expense 05
Reinvestment in the business . . .03
Wear and tear on equipment ... .02|
Dividends 02£
Bar chart. The bar chart, like the circle chart, is designed to
exhibit the true proportions of the component parts of a group
total.
Bars used in charting may consist of single heavy lines, or they
may be widened into rectangles. Three ways of presenting data
by means of bars are in common use :
(1) Comparisons are made by presenting a series of bars of
different lengths, each bar representing a different magnitude (see
Figure 9).
(2) A single bar is subdivided into component parts (see
Figure 10).
268 GRAPHS AND INDEX NUMBERS
(3) A combination of (1) and (2) may be employed (see Figure
11).
Where color is not used, bars or parts of a bar may be differenti-
ated by crosshatching, and also by the use of solid black and white,
1st Year - $6,803,407 mmmmmmmmmmmmmmmmmm
2nd Year - 7,008,564
3rd Year - 7,602,939
4th Year - 8,411,776
5th Year - 10,122,473
6th Year - 12,089,857
Figure 9. Bar Chart Showing Sales for 6-year Period.
Crosshatching consists of the fine parallel lines drawn across the
face of the bar at various angles, and sometimes crossed into small
rectangles.
In making comparisons by means of parallel bars, it is essential
that the bars be of the same width, in order that measurement by
length be not confused with that of area.
^ s.
AL.r,br~$J.<5u,uuu
Cost of Sales
$135,000
Expenses,
$35,000
Profit,
$ iu.ua
Figure 10. Bar Chart Showing Net Profit on Sales.
Bars may be placed in either horizontal or vertical position.
In the bar method of charting, figures may be placed at the
*ide or in the bat and decimal points kept in line; it is thus easy to
foot the figures representing the various components and to verify
the total (Figure 11).
Because of its decimal divisions, a millimeter scale is
convenient in constructing these charts.
Problems
i. Chart the following data, using vertical bars.
Period Amount
t 40,000
2 50,000
3 60,000
4 75,000
5 80,000
6 90,000
7 95,000
8 100,000
9 105,000
10 110,000
GRAPHS AND INDEX NUMBERS
269
2. Chart the following data, using the single
bar subdivided into component parts.
Sales
81,000,000
Cost of Sales
550,000
General Expenses
200,000
Selling Expenses
150,000
Net Profit
100,000
3. Chart the following,
using horizontal bars.
Period
Amount
]
2,931
2
9,052
3
13,541
4
16,403
5
20,919
6
28,063
7
36,365
s
41,393
9
49,404
10
56,615
4. Chait the following similarly to Figure 12.
CiUOWTH IN NUMliKH OF EMPLOYEES
A'O.
Year Employees
1st
2nd
3rd
4th
5th
6th
6,587
6,893
7,205
7,581
7,810
8,104
No.
No.
Male
Female
1,859
4,728
2,OS 1
4,812
2,270
4,935
2,317
5,264
2,168
5,642
2,278
5,826
Line or curve chart. The Hue or
curve chart is probably adaptable to
a greater variety of uses than any other
type of graphic presentation. It is used
particularly to exhibit trends and fluc-
tuations in data, the abnormal condi-
tions being shown by unusual " peaks7'
and "valleys."
On line charts, the scale is indicated
by vertical and/or horizontal rulings.
If the coordinate type of ruling is used,
the lines in each direction are spaced
an equal distance apart, horizontal lines
marking the vertical scale, and vertical
lines marking the horizontal scale.
EARNINGS RETAINED AND
INVESTED IN BUSINESS
STOCK DIVIDEND (RETAINED
IN THE BUSINESS)
GASH DIVIDENDS
TAXES
DEPRECIATION
INTEREST
MATERIAL AND ALL
OTHER EXPENSES
WAGES, SALARIES
AND COMMISSIONS
$580,252
fOTiwj
fMMifttf'OWtfw
:%2W$&
S?. 102,456 !:
S 894 168
^840.618^
115,440,450
<
$2W
48,458
^$50,035,903
Figure 11.
270
GRAPHS AND INDEX NUMBERS
All rectilinear charts have two axes — the horizontal, called
the z-axis, and the vertical, called the ?/-axis.
Rules for coordinate charts. Custom has established certain
rules governing the construction of coordinate charts, which must
he observed if they are to be plotted in accordance with good
usage.
(1) The zero line should always appear, or attention should be
specifically called to its omission.
IstYr. 2nclYr. 3rd Yr. 4th Yr. 6th Yr. 6th Yr.
Figure 12. Bar Chart Showing Annual Gross and Net Income.
(2) The time element should always be expressed by the hori-
zontal scale, and magnitude by the vertical scale.
(3) The curves should be sharply distinguished from the ruling.
(4) Figures and lettering should be so placed that they are
read from the bottom or from the right-hand side.
(5) Exact data should be inserted at the top of the chart, the
figures in each case appearing immediately above the correspond-
ing point on the curve.
(6) The figures of the vertical scale should be placed on the
left. In wide charts they may be repeated on the right.
(7) The horizontal scale should read from left to right, and the
vertical scale from bottom to top.
GRAPHS AND INDEX NUMBERS
271
It is considered good practice to make the zero line heavier
than the other coordinate rulings. In percentage charts the 100%
line is also accentuated by heavier ruling.
Example
The earnings of a corporation over a period of years were as follows:
1st yea
2nd ye
3rd yet
4th yei
5th ye;
6th yei
7th yei
8th yet
9th yei
10th y<
50
45
40
86
80
1
325
20
15
10
5
n
r .
ar . .
ir
ir .
vr
ir
ir
ir
ir
^ar
$39,202,000
37,555,000
28,621,000
30,438,000
28,693,000
27,319,000
28,358,000
35,941,000
31,772,000
34,249,000
—
—
—
~
—
^
^
—
—
\
\
\
/
/
\
X
^^"
\
\
A
rera
£C
/
/
"•
"
\
-"'
"-^
^
—
—
^,
'
_
—
—
—
|
YEARS
Figure 13.
NOTE: Limited space does not permit insertion of exact figures (Rule 5).
Problems
1. Using the following data, prepare a line chart showing the corporation's
earnings and the dividends paid over a period of years.
272 GRAPHS AND INDEX NUMBERS
Year Net Income Dividends Paid
1st $28,154,431 $16,354,000
2nd . 35,422,514 10,360,632
3rd . . 40,129,417 16,369,400
4th . 2S,6S4,916 16,404,509
5th 31,54S,606 17,47S,459
6th 32,070,274 1S,209,281
7th 30,61S,77X 20,639,196
Sth 32,600, 1 50 20,662,854
9th 44,552,482 20,662,854
I Oth 35,419,903 20,943,094
11th . . 35,657,410 22,609,650
2.* From the information given in the following table, prepare a line graph
of the Bonded Debt Limit and the Net Bonded Indebtedness of the City of X
for the 15-year period. (Scale, 1 in. = $2,000,000.)
Year
1st
2nd
3rd
4th
5th
6th
7th
8th
Oth
10th
llth
12th
13th
14th
15th
Logarithmic chart. The common logarithms of the
table, being the expressions of numbers in terms of the power of 10,
are particularly adapted to the presentation of percentage relation-
ships. This fact has led to the construction of a chart in which
percentage relationships are revealed by a comparison of different
sets of data plotted in terms of numerical magnitude.
The logarithmic chart is a variation of the rectilinear type,
The ruling differs from that of the customary coordinate chart
in that the data lines representing the scale are not evenly spaced,
but conform to certain proportions expressed by the first ten
numbers in a table of common logarithms.
Figure 14 illustrates the method of laying out such a scale.
The first column of figures is purely for drafting purposes, and
Assessed
Bonded
Net Bonded
Valuation
Debt Limit
Indebtedness
$442,932,255
$\ 5,887,062
•1:510,698,500
460,548,763
18,272,142
9,321,050
486,424,005
20,804,104
10,577,500
496,342,170
23,291,794
11,101,500
505,713,510
23,919,607
11,921,000
521,239,125
24,702,675
14,730,750
539,457,120
25,491,759
16,566,000
574,020,559
20,307,724
16,534,750
588,556,266
27,289,865
18,254,800
075,011,540
28,988,846
22,030,250
681,198,160
30,588,436
23,965,500
677,070,755
27,750,500
24,800,000
725,603,037
29,033,300
27,403,300
755,229,851
30,773,800
25,023,500
810,509,504
33,974,550
25,744,500
* O. P. A., Wisconsin.
GRAPHS AND INDEX NUMBERS
273
consists of the first ten figures of a logarithmic table, only the first
two decimal digits being used in each case.
The condensed table of logarithms of numbers, page 274, will
be of assistance in preparing a logarithmic chart.
In the vertical scale, each horizontal line is spaced the distance
from the base of the scale which represents the proportion that
its scale number bears to 100; that is, the first line above the base
line is thirty one-hundredths of the total height of the scale.
100 10
05 9
30 8
S4 7
77 C
ea 5
€0 4
/'
.2
X 0
03*7 d
/
X
1
^
30 ^
/
/
/
X
igita
LO«8. (
P ^
1
.8
2
0 .4
Y. a
3
7 .6
via
4
0 .6
5
9 .1
6
7 .J
7
4.f
8 9
0.951.
Figure 14. Logarithmic Chart.
the next line is forty-seven one-hundredths, and so on, the top
line representing 100. A ruler having a 100-millimeter scale
may be conveniently used for making these horizontal lines.
The second vertical column represents the numerical magnitude
wale, and is numbered from 1 to 10, or some multiple thereof, the
ruled lines being spaced according to the logarithms of these
numbers. The result is that when data are plotted on such a chart
in terms of numerical magnitude, the relationships shown between
the various groups of data plotted will be correct percentage
relationships. This does not hold true where data are plotted
numerically on an ordinary coordinate chart; where the curves
274 GRAPHS AND INDEX NUMBERS
represent widely differing magnitudes, an attempted comparison
of the fluctuations in the data will be misleading. In order that
data plotted on a coordinate chart may present correct percentage
relationships, the lines must represent a percentage, not a numeri-
cal, scale.
LOGARITHMS OF NUMBERS
Num- Loga- Num- Loga- Num- Loga- Num- Loga-
ber rithrn bcr rithrn her rithm ber rithm
1
0.00
10
1 00
100
2 00
1,000
3 00
2
0.30
20
1.30
200
2.30
2,000
3 30
3
0.47
30
1.47
300
2 47
3,000
3 47
4
0.60
40
1.60
400
2 60
4,000
3 60
6
0.69
50
1.69
500
2 69
5,000
3 69
6
0.77
60
1.77
600
2 77
6,000
3 77
7
0.84
70
1 84
700
2.84
7,000
3 84
8
0 90
80
1 90
800
2 90
8,000
3 90
9
0.95
90
1 95
900
2 95
9,000
3 95
10,000 4.00
In comparing on a logarithmic scale data of widely differing
magnitudes, it is necessary to use inoia than one grouping of
logarithmic rulings. Each such grouping is called a cycle, because
it represents 10, or some multiple thereof. For instance, if data
represented by figures in the hundreds group and data running
into the thousands were to be compared, it would be necessary to
use two cycles; if the figures ran into the ten-thousands group, it
would be necessary to use three cycles.
It will be noticed that the base line in a logarithmic chart is
numbered 1, instead of 0. This is because in the tables the log of
1 is .0. Therefore, in a logarithmic chart there is no zero line.
To illustrate the use of a full logarithmic chart, a simple
example in multiplication may be cited. Applying the principle
that in multiplication the logarithm of the product of two numbers
is the sum of the logarithms of the numbers, the addition may be
performed graphically on a logarithmic chart. By doubling the
distance of any number represented by a digit, the square of the
digit is obtained. Thus, the distance from 1 to 9 is twice the dis-
tance from 1 to 3, because 9 is the square of 3.
Since the scales of both axes are the same, lines projected at
right angles from identically numbered points on both axes will
complete a square, the diagonal of which is 45°, to the right of any
point on the base of the rr-axis; for instance, any horizontal line
which intersects this diagonal will, if projected downward from
the intersecting point to the base line of the x-axis, and at right
GRAPHS AND INDEX NUMBERS
275
angles thereto, record on the .r-axis digit scale the product of the
two numbers representing the starting points of the diagonal and
the horizontal lines.
Figure 14 illustrates a computation of this kind, and affords
a mechanical demonstration'of the logarithmic principle referred to.
A diagonal line is projected upward at an angle of 45° from the
10,000
9,000
8,000
7.000
6.000
6,000
4,000
8,000
2.000
tooo
900
800
700
600
600
400
800
200
100
Jan.
Z
Z
z
Feb. Mar.
Figure 15.
Apr.
Ratio Chart.
May
June
digit 2 on the x-axis. The point at which it intersects the hori-
zontal line numbered 4 on the y-scale is directly above digit 8 on
the x-scale, and 2 times 4 is 8. Likewise, the sum of the distances 1
to 2 on the z-axis, and 1 to 4 on the 7/-axis, will, if laid off by a
pair of dividers, arrive at point 8 on either scale.
276
GRAPHS AND INDEX NUMBERS
Ratio charts. Figure 15 illustrates a two-cycle chart. Only
the horizontal lines are ruled in accordance with the logarithmic
scale. The vertical lines are evenly spaced. This is called a
semi-logarithmic or ratio-ruled chart, and is the kind most com-
monly used.
The use of the full logarithmic chart, with the ratio ruling
in both directions, is rare, and is confined chiefly to mathematical
demonstrations.
The ratio chart has its limitations. While it exhibits correct
percentage relationships, it does not record correct numerical
magnitudes, and it should bo used only where percentage relation-
ships are desired. But because relationships between fluctuating
data are more easily grasped in terms of percentages than in purely
numerical terms, the semi-logarithmic, chart has a wide field of
usefulness.
Where the percentage of increase or decrease of one item, such
as sales, is to be compared with the percentage of increase or
decrease of some other item, such as expenses, the result is best
shown in the ratio or semi-logarithmic; chart.
January
February
March
. April
May
Juno
Example*
dross 1'ro fit
Sales
Cost
(40',)
Expense
Net Profit
$ f>,400
$ 3,240
* 2,100
« 2,000
S 160
4,500
2,700
1,SOO
1,500
300
8,000
4.SOO
3,200
2,200
1,000
10,000
('),()()()
4,000
2,500
1,500
7,000
4,200
2,SO()
2,100
700
0,000
3,600
2,400
1,500
900
$40,000
$24,540
SI 0,360
$11,800
$4,560
Problems
1. Prepare a ratio chart to illustrate the following
statements :
condensed operating
Sales
1st Year 2nd Year 3rd Year 4th Year 5th Year 6th Year
. $61,960.29 $74,401.38 $80 598 00 $72 887 60 $79 647 14 $65 315 48
Tost of Sales. ,
Gross Profit. .
Expenses
. 27,745.59 32,967.89 48,935.10 41,660.83 48,945.54 33,414.53
. 34,214.70 41,433.49 31,662.90 31,226.77 30,701.60 31,900.95
. 31,152.64 33,30886 2864073 3058032 3034226 29 736 O9
Net Profit . .
. 3,062.06 8,124.63 3,022.17 646.45 359.34 2,164.93
* Charted in Figure 15.
GRAPHS AND INDEX NUMBERS
27T
2. Prepare a ratio chart to illustrate the following comparative income
accounts of a public utility company:
Power-Opr.,
Conducting;
Mainte-
Transporta-
Gross
nance and
tion, and
Fixed
Net
Yoar
Earnings
Renewals
General
Tnxos
Charges
Income
1st ^
$22,147^)74
$37001,198
$ 8,027,973
$17591,523
$ 8,827,988
$ 5(0,7081
2nd
23,282,408
3,492,301
9,097,001
1,059,518
8,901,120
72,312
3rd
24,210,592
3,030,088
9,081,213
1,000,230
9,321,559
538,490
4th
23,901.398
3,r>9 1,209
8,825,005
1,808,951
9,531,232
201,311
5th
21,315,455
3,017,318
8,077,405
1,783,510
9,022,031
581,501
<>th.
27,279,517
4,091,928
9,382,587
1,812,511
9,01 1,908
2,377,553
7th
29,720,927
4,459,039
10,723,912
2,100,709
9,573,522
2,803,085
8th
31,704,127
4,755,004
13,355,575
2,128,8)9
9,029,553
1,534,810
9th
30,039,520
4,955,124
17,287,117
2,315,750
9,735,052
1,715,877
10th
39,400,341
5,905,409
20,028,504
2,00 1,253
9,823,110
382,005
llth
12,911,010
8,500,400
19,874,309
2,798,821
9,870,158
1,807,292
12th
43,235,972
8,500, 100
20,407,117
2,580.001
9,853,177
1,829,277
131h
45,552,031
8,500,400
22, 179,553
2,095,708
10,010,370
1,800,000
14th
40,215,488
8,500, 100
22,078,890
2,700,903
10,104,921
1,810,305
f Loss.
3. Prepare a ratio chart to show the following expenses:
Salaiies
Kent .....
Ciedit Losses .
Heat ... .
Light . .
Taxes . .
Shipping and Receiving .
Depreciation .
Miscellaneous Expenses .
Power ..... . ..
Freight .
Delivery Expense . ,
Insurance . . .
HINT. Kearrangc items fioiu highest to lowest per cent.
3
5
25
1 0
5
5
1
Index Numbers
Uses of index numbers. Current newspapers, magazines,
bulletins, and books contain much economic information expressed
in terms of index numbers. Following are examples from the
Monthly Review of Agricultural and Business Conditions issued by
the Federal Reserve Bank:
"From June 1939 to March 1942 prices rose 30 per cent, according to the
index of wholesale prices compiled by the U. 8. Bureau of Labor
x/o
GRAPHS AND INDEX NUMBERS
" Industrial production reached a peak in February (1945) when the index
stood at 236 per cent of the 1935-1939 average."
"The index of prices paid by farmers was unchanged at 173 for the fourth
consecutive month according to data from the Department of Agriculture."
(July 30, 1945)
Index numbers may be used in the form of a chart similar to
the following, which appeared in a recent issue of the Chicago
Tribune:
INDEX
1939-1
225
200
175
150
125
100
75
DO latest Figure for August, 1945
^.
Pric
b
es Recei
y Farmer
^d
/
s —
/
y.
..-«••*
••••*
^^1
..*"'
N
rices Pai
y Farmer
d
b
s
1939 1940
1942 1943 1944
1945
Index numbers. An index number is a number that is used for
measuring trends in prices or in other movements which can be
quantitatively expressed by means of statistical data. Among
those more frequently used in business are the following:
Production indexes. Automobile production, building con-
struction, electric power output, steel production fponrningr of
capacity).
Trade indexes. Carloadings, department store sales, check
clearings, inventories (of manufacturers, wholesalers, and retailers),
value of imports and exports.
Financial indexes. Prices of stocks and bonds, business fail-
ures, new capital issues, commercial bank loans, prices of basic
commodities, prices of agricultural products, foreign exchange
rates.
Miscellaneous. Wage rates, employment, financial situation of
the government, foreign affairs.
Most of the indexes listed in the foregoing are expressed
statistically ; therefore, their trends can be measured mathematically.
The nature of the particular business primarily determines the
indexes most useful to it. To answer the question, "How is
business?" recourse may be had for purposes of comparison to
some index of general business conditions, for example, to the
Federal Reserve index of industrial production, a national sum-
mary charted for a period of years : /
GRAPHS AND INDEX NUMBERS
279
INDUSTRIAL PRODUCTION
PER PHYSICAL VOLUME SEASONALLY ADJUSTED, PER
CENT 1935-39 -loo CENT
IDU
240
220
200
180
160
140
120
100
80
60
/\
i^
260
240
220
200
180
160
140
120
100
80
60
r
^ -
-
/
-
-
s
-
-
s
-
-
f
-
-
y
' \
f
**
-
\
,/
-
1937 1938 1939 1940 1941 1942 1943
1944
or possibly to indexes somewhat more restrictive as to locality and
type of business, such as the following from the Ninth Federal
Reserve District Monthly Review of July 30, 1945:
/riONS- 1935 -1939 = 100
,/ uric
May
,/ unc
J 'unc
1,94»
1945
1944
1V43
237
201
20S
176
224
207
201
173
I S3
172
153
146
1 73
163
152
137
1 02
150
149
143
HO
100
102
12S
114
120
114
109
1 39
203
143
145
1S5
ISO
177
17S
134
143
151
229
233
228
Bank Debits— 93 Cities . . .
Hank Debits — Fanning Centers
City Dept. Store Sales
City Dept. Store Stocks
Country Dept. Store Sales
Country Lumber Sales ...
Miscellaneous Carloadings
Total Carloadings (excl. Misc.)
Farm Prices — Minn, (unadj.)
Employment — Minn, (nnadj. 1936 = 100)
Minnesota Payrolls (unadj. 1936 - 100)
Economic position of agriculture. Agriculture creates a stream
of income that exceeds that of any other industry. This income
flows through various channels into the industrial and commercial
life of the nation. Probably as many or more people are engaged
in the processing and handling of agricultural commodities from
the producer to the consumer than are engaged in the actual
production processes. Since agriculture and other industries are
customers of one another, any enhancement of the buying power'
of one operates to the advantage of the other. With this back-
ground, the industry of agriculture is chosen to furnish much of the
material in this chapter.
Suppose we desire to compare the price of wheat for different
years. The July 15, 1937-1941 average price of wheat was
280 GRAPHS AND INDEX NUMBERS
73 cents a bushel, and on July 15, 1945, the price was $1.48 a
bushel. We can say that wheat has gone up 75 cents a bushel,
but a better comparison is to show what per cent the 1945 price is
of the 1937-1941 average.
1.48 -1- .73 = 2.03, approximately
2.03 X 100 - 203%
Thus the 1945 price of wheat is 203% of the 1937 1941 average.
Expressed as an index number, it is 203 when the base is t41937-
1941 average price equals 100."
Construction of index numbers. In the study of price fluctua-
tions, the first thing to determine is the original base price. This
may be: (a) the price of a commodity on a certain date; (/>) the
average price of a commodity during a certain year; or (r) the
average price of a commodity during several years. Whichever
is used, that price is assigned a numerical value of 100.
Other index numbers for following years or periods are obtained
by multiplying the price for the years or periods by 100 and divid-
ing the product by the original base price. Thus, the index
numbers of the prices paid to Minnesota wheat growers from June
15, 1939 to June 15, 1944, were as follows:
PRICE RECEIVED BY MINNESOTA FARMERS
(Fifteenth-of-Month Comparison)
Arrr. Price Index \ 'umber or
All Wheat (hu.) per Bushel Jtelutire Price
1930-1939 average ......... S 79 100
1 940 ..... . . 75 95
J941 ....... . S5 10S
1942... 1 01 12S
1943 ..... I 27 161
1944 (0 mos.). 1 48 1X7
Index numbers or relative prices may be found for all farm
products for which average prices are known. The base period
taken for making calculations is arbitrarily selected, depending on
the period for which comparisons are wanted.
Problems
Given the following data, compute the index numbers:
1. Corn (bu.):
1930-1939 average .............................. $ .52
1940 ...................... ..... 48
1941 ............. .............. 53
1942 ......................... 68
1943 ......................................... 90
1944 (6 mos.) ................................... 1.01
GRAPHS AND INDEX NUMBERS 281
2. Potatoes (bu.):
1930-1939 avenge $ 50
1940 ... ... . 50
1941 . 4S
1942 .... 93
1943 ... . ..1.27
1944(6 mos.) .. 1.12
3. Hogs (lOOllxs.):
1930-1939 average $ 6 78
1940 . 5 20
1941 . 9 04
1942 13 10
1943 13 70
1944 (6 mos.) . . . ... 13 25
4. If the average farm price of A\lieat used as the base was S,5 cents a bushel
(as in 1941), what is the index number for 1944, when the price was S1.4S cents
a bushel?
Composite price indexes. It is commonly desired to express
the price of a group of commodities or of all commodities in a single
number or series. Thus, we say that the index of prices received
by farmers in mid- April (1945) was 203, calculated from the
average base price of 1910 1914. The index 203 was obtained by
taking the average price for all farm products for each of the two
periods of comparison. At the same time, the prices paid by
farmers for things used in farm production and family maintenance,
including interest and taxes, was 173, calculated from the average
base price of 1910 1914. The ratio of the indexes of prices
received to prices paid is the so-called parity ratio established by
law which designates the years 1910 1914 as the base period. This
ratio averaged 100 in the base years. It was 117 in mid-April
(1945): 203 -r 173 - 1.17, and 1.17 X 100 - 117. At the bottom
of the depression in 1932, the parity ratio was 01. The term
parity, as applied to the price of an agricultural commodity, is that
price which will give to the commodity a purchasing power equiva-
lent to the average purchasing power of the commodity in the base
period, 1910-1914.
Weighted index numbers. An index number of prices which
will satisfactorily measure changes in the price level must include a
considerable number of representative commodities, and these
commodities must be weighted in accordance with their importance
in trade and industry. The number of items may be 30, or the
number may be 500 or even more. For purposes of illustration, we
shall use three principal agricultural commodities: wheat, corn-
and cotton.
282 GRAPHS AND INDEX NUMBERS
Suppose we desire to compute the average rise in the price of ths
three commodities from 1939 to 1942, making due allowance for
the total quantity of each commodity produced, so that the index
of prices found will give us information on the increase in buying
power accruing to the producers as a result of the rise in prices.
We obtain the production for the base period (1939) and the
average prices for the years 1939 and 1942 from the Yearbook of the
Department of Agriculture for 1943 and set up the material as
follows:
Wheat (lorn Cotton
Total
Wheat
(bu.)
(lorn
(bu.)
Cotton
(500-lb. bale a)
Total production in
1939 . .
741,180
,000
2,5X0,912
,000
1
1
,817
,000
Average price
in
1939
7S/-
a
bu.
50
*t
al
>u.
S45
4
5 a 1
>ale
Average price
in
1942.
SI 26
a
bu.
85
5<£ a bu.
$94
0
5 a bale
Value at 1939 prices $57N, 120,400 $1,405,958,010 $ 537,082,050 $2,581,101,000
Value at 1942 prices $933,880,800 $2,206,079,760 $1,118,509,050 $4,259,075,610
Price index: $4,259,075,010 -r- $2,581,101,000 = 1.05
1.05 X 10()'(', = 105%
Average rise in price: 165% — 100% = 65%
Explanation. The values of the commodities produced in 1939 were calcu-
lated at the average prices in both 1939 and 1942, and the total value of all three
for each of the two years was found. The aggregate value for 1942 was then
divided by the aggregate value for 1939. This method of constructing an index
number of prices is recommended because it does not require the statistics of
current production. (It is frequently impossible to obtain such statistics
promptly.)
The calculation procedure of the foregoing example may he
stated in terms of a formula for weighted aggregative price index
as:
where :
p represents the price of a commodity;
q represents the quantity of a commodity;
o represents the base period, the period from which the price changes are
measured ;
n represents the given period, the period being compared with the base;
2) is the symbol of summation, or addition.
In the illustration for 1942, the index number formula is:
P — . ^'^21939
There is no short cut. The q0 in the numerator and the q0 in the
denominator may not be canceled. The quantity of each com-
GRAPHS AND INDEX NUMBERS 283
modity included in the index must be multiplied by its respective
price in both the base year and the given year and the several
products added, as indicated in the formula and illustrated in the
example.
. Problems
1. Corn and barley are two principal feed grains used in fattening hogs for
market. From the following figures calculate the weighted index for 1942, using
1939 as the base year.
Hogs (Ibs.) Corn (bu.) Burlcy (bu.)
Total production in 1939 17,081,824,000 2,580,912,000 278,163,000
Average price in 1939 (cents) 7 74 56 8 40 5
Average price in 1942 (cents) 13 04 85 5 59 4
2. Calculate the individual price indexes for hogs, corn, and barley. Did the
price of hogs keep pace with the price of corn? How about the price of hogs
and the price of barley? If both feeds were available, which would be the more
profitable to use, assuming that each has the same feed value for pork production?
3. Find the weighted index for the three following fresh fruits sold in large
quantities for home canning.
Apples (bu.) Peaches (bu.) Pears (bu.)
Total production in 1939 . 139,379,000 64,222,000 29,279,000
Average price in 1939 .... $ .64 $ S2 $ 70
Average price in 1 942 .... $ I 38 $1 49 $ 1 . 57
4. In the following tabulation are given the index numbers for the average
price paid by farmers and the average price received by farmers for the years
1932 to 1942, inclusive, based on 1910-1914 averages, the years chosen for
computation of parity prices, at which time parity was 100.
Prices Paid Prices Received
1932 . 124 62
1933 120 81
1934 . . 129 103
1935 130 107
1936 128 125
1937 134 105
1938 127 93
1939 125 96
1940 126 103
1941 133 J42
1942 . . 151 175
Calculate the parity price indexes for the years 1935, 1940, and 1942.
Farm evaluation on the basis of crop production index. The
farm owner, the banker or financial agent making farm loans, arid
other interested persons may partially evaluate a farm in terms of
crop production. Other factors, such as building improvements,
fencing, drainage, location relative to roads, rural electrification
line, and so forth, are also to be considered in fixing the total value.
The crop production index is a comparison of the yield per acre
284 GRAPHS AND INDEX NUMBERS
of all crops on a given farm with the average yield per acre of all
crops on a number of farms in the same locality — community,
township, or county. A crop production index of 92 indicates that.
the yield of crops on a farm with this index is 92% of the average
for the locality; a farm with a crop production index of 105 has a
yield 5% greater than the average.
Computation of the crop production index. Assume that a farm
has a cropping system as follows :
50 acres of corn yielding 40 bushels per acre
50 acres of oats yielding 35 bushels per acre
50 acres of alfalfa yielding 2^ tons per acre
and that the average yields for the county in which this farm is
located are :
Corn . . ........... 30 bushels per acre
Oats . . . ....... . 25 bushels per acre
Alfalfa ............................ 3 tons per acre
The crop production index is computed as follows:
50 X 40 = 2,000, the number of bushels of corn produced
50 X 35 = 1,750, the number of bushels of oats produced
50 X 2 J = 125, the number of tons of alfalfa produced
150 acres
Next, find how many acres would be required to produce each
crop using the county averages.
2,000 bu. of corn -r 30 1m. per acre would require 66^ acres for corn
1,750 bu. of oats -f- 25 bu. per acre would require 70 acres for oats
125 tons of alfalfa -r- 3 tons per acre would require 41 £ acres
acres required
Since 178^ acres would be needed to produce what this farm produced, on the
basis of county averages, the crop production index is 118.9.
178i
Irn X 10° = 118'9
loO
Problem
Ten acres of a 160-acre farm were used for farmstead and pasture, and the
150 acres in cultivation produced the following crops, as compared with the
county average:
Number of Yield per County Average
Crop Acres Acre Yield per Acre
Corn .............. 40 45 bu. 42^ bu.
Oats ............... 40 50 bu. 48 bu.
Alfalfa ............. 30 2 tons 2.^ tons
Wheat .............. 40 25 bu. 20 bu.
150
Compute the crop production index for this farm.
CHAPTER 28
Progression
Definition. A progression is a series of numbers, each term of
which is obtained from the preceding or following term by a fixed
law; as, 2, 4, 6, 8, and so forth; or, 2, 4, 8, 16, and so forth.
Increasing series. A progression, each term of which is
greater than the preceding term, is known as an increasing or
ascending series.
Decreasing series. A progression, each term of which is less
than the preceding term, is known as a decreasing or descending
series.
Arithmetical progression. When each term of a progression
differs from the preceding or following term by a common differ-
ence, the progression is said to be arithmetical.
Symbols. The five elements of an arithmetical progression
are represented by certain well-established symbols:
Number of terms ... . . . . n
First term . . . a
Last term . . I
Common difference . . d
Sum of the terms . s
Relation of elements. The five elements whose symbols are
indicated above are so related that if any three of them are given,
the remaining two may be found.
Increasing Series
The formulas used in connection with increasing series, and the
methods of solution, are shown below; in each case the values of the
terms have been taken as follows:
Number of terms . 12
First term ... .2
Last term 35
Common difference 3
Sum 222
To find the number of terms.
Algebraic Formula Arithmetical Substitution
I — a 35 — 2
h 1 = H. '- 0 h 1 = number of terms.
d .5
985
286 PROGRESSION
Solution
Simplifying the numerator: 35 — 2 = 33
Dividing by denominator: 33 -r- 3 = 11
Adding: 11 + 1 = 12, the number of terms
To find the first term.
Algebraic Formula Arithmetical Substitution
I - (n - \)d = a. 35 - (12 - 1)3 = first term.
Solution
Removing parentheses: 12 — 1 = 11
Multiplying by 3: 1 1 X 3 = 33
Subtracting: 35 — 33 = 2, the first term
To find the last term.
Algebraic Formula Arithmetical Substitution
a + (n - l)d = 1. 2 + (12 - 1)3 = last term,
Solution
Removing parentheses: 12 — 1 = 11
Multiplying by 3: 11 X 3 = 33
Adding: 2 + 33 = 35, the last term
To find the common difference.
Algebraic Formula Arithmetical Substitution
I- a 35 - 2 ....
= a. -— — - = common difference.
Solution
Simplifying the numerator: 35 — 2 = 33
Simplifying the denominator: 12 — 1 = 11
Dividing: 33 -7-11 =3, the common difference
To find the sum.
Algebraic Formula Arithmetical Substitution
\ (a + 1) - 8. ~ (2 + 35) = sum.
Solution
Adding the terms in parentheses: 2 + 35 = 37
12
Multiplying by the fraction: 37 X — = 222, the sum
A
Decreasing Series
The decreasing series will be illustrated in the same manner a*»
the increasing series, with the values of the terms taken as follows:
PROGRESSION 287
Number of terms 8
First term 7
Last term —21
Common difference . . . —4
Sum -56
To find the number of terms.
Algebraic Formula Arithmetical Substitution
J-a,! (-21) -7, , . f.
1_ i = n< h 1 = number of terms.
a —4
Solution
Simplifying the numerator: ( — 21) — 7 = —28
Dividing the denominator: (—28) -j- (—4) = 7
Adding: 7+1=8, the number of terms
To find the first term.
Algebraic Formula Arithmetical Substitution
I - (n - })d = a. -21 - [(8 - l)(-4)] = first term.
Solution
Removing the parentheses: 8 — 1=7
Multiplying: 7 X (-4) = -28
Simplifying: (-21) - (-28) = (-21) +28
And: -21 + 28 = 7
To find the last term.
Algebraic Formula Arithmetical Substitution
a + (n - \)d = I. 7 + 1(8 - l)(-4)] = last term.
Solution
Removing parentheses: 8—1=7
Multiplying by -4: 7 X (-4) = -28
Subtracting: 7 - 28 = -21
To find the common difference.
Algebraic Formula Arithmetical Substitution
I- a (-21) - 7
— — - = a. — - — — — = common difference,
n — 1 o — 1
Solution
Simplifying the numerator: (-21) - 7 = -28
Simplifying the denominator: 8 — 1 = 7
Dividing: -28 ^ +7 = -4
To find the sum.
Algebraic Formula Arithmetical Substitution
I (a + I) = s. I 7 + (-21) = sum.
6 A \ \
288 PROGRESSION
Solution
o
Simplifying the fraction: 9 = ^
Adding: 7 + (-21) = -14
Multiplying: 4 X (-14) - -56
Problems
1. Given a — 2, n = 6, / = 12; find d and s.
2. Find the sum of all the even numbers from 10 to SO inclusive.
3. The first term of a progression is 6, the number ot terms is 15, the common
difference is 7; find the last term and the sum.
4. Given n = 12, / = — 17, s = -72, find d and a.
5. The first term is 6, the last term is 1S1, and the common difference is 7;
find the number of terms.
6. Given I = 57, n = 23, a = —9; find d and s.
7. A building is to be leased for a term of 21 years. The first yeai's rental
is to be $10,000.00, equal increases in rent are to be made each year, and the
rental for the twenty-first year is to be $30,000.00. Find: (n) the difference in
each year's rental; (b) the total rental that will be paid during the period of
21 years.
8. A deposited $25.00 in his savings account on January 1, and on the first
of each month thereafter deposited $5.00 more than the previous month. How
many dollars did he deposit December 1, and what was the amount of the
accumulated deposits? Do not take interest into consideration in solving this
problem.
9. A punch board has 50 numbers in each section (numbers 1 to 50). A
person pays the amount of the number punched. If the board has four sections,
what will be the amount of revenue derived from the board?
10. A man invests his savings in the shares of a building and loan association,
depositing $240.00 the first year. At the beginning of the second year he is
credited with $16.80 interest, and deposits $223.20. At the beginning of the
third year he is credited with $33.00 interest, and deposits $206.40. What is
his credit at the end of 10 years, and how much cash has he paid in?
11. A bond issue of $40,000.00 bearing interest at 4% is to be retired in
10 equal annual installments. What amount of interest will be paid during the
life of the issue?
12. An employee started work for a company at $1,200.00 for the first year,
with a guaranteed yearly increase of $100.00. What was his salary 12 years
later? How much had the company paid him in the course of the 12 years?
Geometrical Progression
A geometrical progression is one in which any term after the
first is obtained by multiplying the preceding term by a fixed
number known as the ratio.
PROGRESSION 289
Elements. In a geometrical progression, there are five ele-
ments so related that any three being given, the others may be
found. These five elements, together with their symbols, are:
Number of terms n
First term a
Last term I
Sum of the series s
Ratio r
Increasing series. As in an arithmetical progression, the
formulas and solutions in a geometrical progression are based on
one set of terms, and the solutions are given as though the required
term in the example in question were lacking. In some cases, two
different formulas may be used.
Number of terms . . . 6
First term ... 3
Last term . . 96
Sum of terms . . . 189
Ratio of increase. . . ... 2
To find the first term.
Algebraic Formula Arithmetical Substitution
1 % r ff
(r)«-i = «• (2)° 1 = firsttmn-
Solution
Simplifying the exponent: (2)6"1 = (2Vr>
Finding the power: (2)5 = 32
Dividing: 90 -r- 32 = 3, the first term
To find the last term.
Algebraic Formula Arithmetical Substitution
a(r)"-1 - 1. 3(2)c-1 = last term.
Solution
Simplifying the exponent: (2)6"1 = (2^
Finding the power: (2)5 = 32
Simplifying: 3 X 32 = 96, the last term
To find the sum.
Algebraic Formula Arithmetical Substitution
IT - a (96 X 2) - 3
—l =s. -2-—-
Solution
Simplifying: (96 X 2) = 192
Subtracting: 192 - 3 = ISO
Simplifying the denominator: 2 — 1 = 1
Dividing: 189 -M = 189, the sum
290
PROGRESSION
To find the ratio.
Algebraic Formula
'l
A rithmetical Substitution
.
= ratio.
Simplifying the exponent:
Simplifying the fraction:
Extracting the 5th root:
Decreasing series.
Number of terms.
First term
Last term
Ratio of decrease
Sum of terms
6 -
96 -r-
5
Solution
1 = 5
3 = 32
>2 = 2, the ratio
6
96
3
- i
. 189
To find the first term.
Algebraic Formula
I
A rithmetical Substitution
3
(i)6"1
. = first term.
Simplifying the denominator: (i)6~
Dividing: 3 -f- ^
To find the last term.
Algebraic Formula
a(r)»-' = L
Simplifying the exponents: 6 — 1
Finding the power: (£)5
Multiplying: 96 X A
To find the sum.
Algebraic Formula
Ir - a
Solution
i = -Jr
r - 1
= s.
= 96, the first term
A rith metical Substitution
96(i)b~l = last term.
Solution
= 5
= A
= 3, the last term
Arithmetical Substitution
(3 X i) - 96
- 1
= sum.
Multiplying:
Subtracting:
Simplifying the denominator:
Dividing:
To find the ratio.
Algebraic Formula
3 X i = f
f- 96 = -
-J- — l = — ^
-j -- ^ = 189, the sum
Arithmetical Substitution
'"'/I
"V96 "
PROGRESSION 291
Solution
Simplifying the fraction: <nr = ^2"
Simplifying the radical: \/~&? = i> the ratio
Progression problems solved by the use of logarithms. The
use of logarithms replaces laborious calculations in solving prob-
lems in progression.
Example
What is the average yearly rate of increase if $100.00 placed at interest for
10 years produces $179.08?
Algebraic Formula Arithmetical Substitution
n\ ~Vd^^nd~ __ _ 10/ 179.08 _
K ~ \Vaiue at beginning *' Kate " \ lOCUX)
Solution
179.0S -4- 100.00 - 1.790S
log 1.790S = 0.253047
log 0.253047 -r- 10 = log 0.025305
In the table of logarithms, it is found that log 0.025305 stands for 1.00.
1.06 - 1.00 = .06, or 6%
Problems
1. A machine costing $27,500.00 is found to be worth only $2,750.00 at the
end of 12 years. Find the fixed percentage of diminishing value.
2. A building cost $80,000.00. At the end of each year the owners deducted
10 fo from its carrying value as estimated at the beginning of the year. What
is the estimated value at the end of 10 years? NOTE: The value at the end
of the 10th year is the value at the beginning of the llth year; therefore:
Value - $80,000 X (.9)11-1.
3. An asset that cost $15,000.00 has been written down 3% of the decreasing
balance each year for 10 years. At the end of the 10th year, what is its value
as shown by the books?
4. If the population of a city increases in 5 years from 150,000 to 175,000,
find the average rate of yearly increase.
5. If a savings bank pays 5% compounded annually, what will be the amount
of $200.00 at the end of 6 years?
6. There are seven terms in a geometric progression of which the first term
is 2 and the last term is 1.458. Find r and 5.
7. Find a and s in a geometric progression where r = 2, a = 9, and I = 256.
8. Find r and s in a geometric progression where a = 1 7, I = 459, and n = 4.
9. If the bacteria in milk double every two hours, how many times will the
number be multiplied in 24 hours?
CHAPTER 29
Foreign Exchange
Foreign trade. The foreign trade connections of American
concerns make it necessary for the accountant to be acquainted
with the basic principles of exchange values.
Rate of exchange. Theoretically, the rate of exchange between
any two countries is the ratio between the values of the amounts
of metal in their standard monetary units. This theoretical rate
is commonly known as "par of exchange/' but because of the many
economic factors in foreign business, another rate, the "current
rate/' is usually used.
Par of exchange. A country that stands ready to redeem all of
its obligations in gold, upon demand and without restriction, is said
to be "on the gold standard." Conditions at this writing (April
1946) are chaotic, and all countries are "off the gold standard."
However, coinages of world monetary systems are based on actual
or theoretical monetary units containing gold or silver of a legally
established weight and fineness. The mint par rate of exchange
between any two countries is the ratio between the amounts of
gold, if the countries are on a gold basis, or of silver, if they are on a
silver basis, contained in their standard monetary units. If one
country is on a gold basis and another is on a silver basis, reference
must be made to the market prices of the two kinds of bullion.
If a foreign coin contains 516.4058 grains of fine silver, and the
market price of silver in terms of gold in this country is 60 cents for
5 ounces (480 grains), the rate of exchange is: (516.4058 -v- 480)
X .60 = .646. The silver par of exchange is not fixed because the
market price of silver in terms of gold money is continually chang-
ing. Calculation of the par rate of exchange of the English sover-
eign and the United States gold dollar is shown in the following
example.
Example
Find the par of exchange of the Knglish sovereign and the United States gold
dollar.
293
294 FOREIGN EXCHANGE
Solution
Weight of English sovereign piece 123 2744700 grains
Deduct alloy, Sj- % 10 272H725
Net weight of gold . 1 13 001,5975
*Weight of United States gold dollar 15 23X0952 grains
Deduct alloy, 10% . . 1 523S095
Net weight of gold . . . .13 7142857
The par of exchange is 1 13.0015975 -5- 13.7142S57 - S 239099S
A comparative table showing the values of foreign coins
is issued at frequent intervals by the United States Treasury
Department.
Current 'rate of exchange. The current rate of exchange,
under normal conditions, will fluctuate between the gold import
and the gold export point, slightly below or above par. This
fluctuation is caused chiefly by demand and supply.
Under abnormal conditions, the rates are quite different from
par, owing to a number of causes, some of which are:
(1) Suspension of the gold standard in the country where the
rate is quoted, or in the country whose currency is quoted.
(2) The extent to which the currency of either country is depre-
ciated as a result of management of the currency or inflation.
(3) Lack of confidence in the stability of foreign government.
(4) Issuance of a large amount of paper money by the foreign
country.
(5) Shrinkage of gold and other legal reserves.
Sterling quotations are usually made on " demand," but may
be made on 30-, GO-, or 90-day drafts. Where time is a factor in
calculations, three days of grace are allowed, interest being calcu-
lated on the basis of 3(>5 days to the year for sterling drafts.
Six classes of problems. The mathematics of foreign exchange
may be resolved into six classes of problems:
(1) Conversion of one monetary unit into terms of another.
(2) Interest on foreign exchange.
(3) Bankers' and brokers' problems of valuing time bills of
exchange.
(4) Finding the value of an account as a whole.
(5) Averaging accounts of foreign exchange bearing interest.
(6) Foreign branch house accounting.
Conversion of one monetary unit into terms of another. The
necessity of converting one monetary unit into terms of another
* By proclamation of the President, the weight of the gold dollar \ras fixed at
15<i5f grains nine-tenths fine on January 31, 1934.
FOREIGN EXCHANGE 295
results from the simple purchase of letters of credit, travelers*
checks, and cable transfers on a foreign country. These transfers
are calculated at the quoted or current rate of exchange. This
current rate is the rate prevailing for that particular class of
transfer on the date of purchase.
To make the calculations, reduce the amount of foreign money
to the term of the monetary unit to which the exchange rate is
applicable, that is, in units and a decimal fraction thereof, as,
1*150 7s. 6d. = £150.375. Multiply by the conversion rate and
add the charges for the transfer.
Example
What is the cost of a letter of credit on London for £150 7s. 6d., if the bank
charges \( 'v for its service? Assume that the current rate of exchange is $4.0").
Solution
C150 000
7s. - -Jo- of £1 :*50
<>d. = 2To ()f M 025
£150 375
Multiply l»y rate of exchange 4 05
Dollar value at current rate . . . . (109 02
Service charge of 2 ' ( •* 05
Total cost . . (>12 07
Conversion of decimals of one monetary unit into monetary
units of a smaller denomination. This is simply a matter of
reduction of decimals of denominate numbers.
Example
What value of London exchange may be purchased with $1,000, if the rate
is S4.40i?
Solution
$1,000 ~ S4 40375 - 224 02(>9, the number of pounds sterling
.02o9 X 20 = .53Ss.; therefore, no shillings
.r>3S X 12 = (5.450d.
Answer: €224 Os. 6d.
Problems
Find the cost in dollars of each of the following:
1. €1,200 16s. at S3.7.H.
2. £8,240 5s. Sd. at $3.90; commission, 1%.
3. Calculate how much exchange can be purchased with the following. $750
on England, rate, S3. 845.
Interest on foreign exchange. To find the interest on foreign
exchange :
(a) Reduce the amount of foreign money to the term of the
monetary unit to which the exchange rate is applicable.
296 FOREIGN EXCHANGE
(b) Compute the interest in the terms of that monetary
unit.
(c) Reduce the interest to its proper exchange units.
Example
Find the interest on £200 5s. fxl. for 63 days at 6%.
Solution
£ .... . £200 00
5s. - | of £1 (decimally) . 25
(3d. - Vo of €1 (decimally) . 025
£200 275
Interest for GO days (point ofT 2 places) . C 2 0027500
Interest for 3 days (2*0 of GO days' int.) . 1001375
t 2 102SS75
The above amount is ordinary interest and is
changed to exact interest by deducting ^ 02SS066
C 2 0740S09
Reducing the decimal of a pound to shillings, .0740S09 X 20 = 1.4X1G1S
shillings. Reducing the decimal of a shilling in pence, .4S101S X 12 = 5.779410,
or practically 0 pence. Answer: £2 Is. (kl.
Problems
1. Find the interest on £250 10s. (kl. for 93 days at 6%.
2. Find the interest on I' 1,000 15s. lOd. for 03 day; at 5<;)t
To find the value of a time bill of exchange. The four elements
entering into the determination of the value of a time hill of
exchange are:
(1) Current rate of demand exchange.
(2) Interest on money (exact time).
(3) Stamp charges.
(4) Commission of home or foreign banks.
Example
If the demand rate of exchange is $4.05000, discount rate 4ro, stamps $$( [,,
and commission \%, what is the 60-day rate?
Solution
Demand rate . . S4 05000
Deduct:
Interest at 4% for ^W of a year 02790
Stumps, ?V% -.-. 00203
Commission, -\-% . . . 01013
04012
SO-day rate S4 00988
FOREIGN EXCHANGE 297
Problem
Determine the proceeds of a 60-day draft on London for £200, if the following
conditions prevail :
Demand rate on London $3.90
Interest rate . . 7%
Stamp charges . -^rc
Commission charged by Knglish hank }-%
Commission charged by American bank j %
Foreign exchange accounts. Kates of exchange fluctuate;
therefore, it is necessary to establish a method of keeping records
of transactions to show current values.
Procedure: (a) Extend the items of debits and credits in foreign
currency at the rates of exchange stated for each item. (In
practice, this would be performed when the entry is recorded.)
(b) Strike a balance of the foreign currency, and convert this
balance at the current rate of exchange. Insert this result among
the dollar items.
(c} Find the difference between the total debit dollars and the
total credit dollars. A debit difference indicates a loss on exchange,
while a credit difference indicates a profit on exchange.
Example
State the balance of the following account in both foreign and domestic
Currency, and show the profit or loss on exchange, exclusive of interest charges
and credits. The current price of exchange is quoted at .$4.51 on the last day
of the month.
DAVIS & COMPANY- -BANKKKS, LONDON
Debits:
Jan. 1 Remittance, sight bill, C 1,000 at $4.4(>j[
Jan. 10 Remittance, sight bill, £500 10s. at $4.49i
Credits:
Jan. S Drafts drawn, Cl 00 at $4.47
Jan. 24 Cable, £1,100 10s. (id. at $4.52$
Solution
Foreign Rale Domestic
Debits:
Jan. 1 Remittance . C 1,000 Os. $4 4(>g $4,463 75
Jan. 10 Remittance 500 10s. 4 495 2,249 75
Profit on exchange 60 27
£1,500 10s. $6,779 77
Credits:
Jan. S Drafts C 100 Os. Od. $4 47 $ 447 00
Jan. 24 Cable . . 1,100 10s. 6<i. 4 525 4,979 88
Balance 299 19s. 6d. 4 51 1,352 89
£1,500 10s. Od. $6,779 77
298 FOREIGN EXCHANGE
Explanation. The balance of the account is found by first deducting the
smaller side of the account in foreign money from the larger side. In the above
case, the smaller side amounts to £1,200 10s. 6d., and the larger side to £1,500
10s.; the balance is therefore £299 19s. 6d. Second, convert this balance into
domestic money at the current rate of exchange, $4.51; the balance is found to
be $1,352.89.
The profit or loss on exchange is the difference between the total debit and
the total credit of the domestic money columns.
There might be a question, in determining the correct valuation of the balance
on hand, as to whether the £299 1 9s. 6d. should have been valued at the last
buying price of $4.49i, or at $4.51, the current market price. This is purely a
question of accounting and finance. The usual practice is to take the current
rate on the last day of the month, when given; otherwise, the cost of the exchange
on hand, if that can be determined by inspection, will be acceptable.
Problems
1. Find the balance of the following account in both foreign and domestic
currency, and the profit or loss on exchange, exclusive of interest:
Debits:
Aug. 1 Balance £ 500 10s. 4d. © $4 . 85
Aug. 10 Remittance 11,000 Os. Od. @ 3 90
Aug. 20 Collection 200 Os. 6d. («} 3 875
Aug. 25 Discounts 196 5s. 6il. (a) 3 88
Credits:
Aug. 3 Cable £ 200 Os. Od. (m $3 88
Aug. 7 Sight draft 400 10s. Od. (m 3 S75
Aug. 9 Sight draft 300 Os. Od. © 3 89
Aug. 15 Cable 2,000 15s. 6d. (m 3 87i
Aug. 20 Cable 400 Os. Od. (5) 3 87i
Demand rate on Aug. 31, $3.90.
2.* A dealer in foreign exchange finds from his books that he has had tht
following transactions in London exchange during a particular month:
Exchange bought in local market :
Jan. 1 30-day bill, payable in London, £300 («), $4.75
Jan. 15 Hill due in London, at sight, £2,500 @ $4.76
Exchange sold in local market:
Jan. 5 Bill due in London, at sight, £1,000 @ $4.77
Jan. 20 Cable transfer, £2,000 @ $4.78
Foreign correspondent's draft honored and paid:
Jan. 20 Bill at 30 days after sight, accepted December 21, £500 @ $4.78
State how the balance of the account stands at the close of the month, and
how much profit or loss has been derived from the transactions. (At January 31,
the rate for cable transfers is $4. SO.)
Is the profit or loss so seated final?
Averaging accounts in foreign exchange. In general, trans-
actions in foreign exchange accounts are large, and involve the
* American Institute Examination.
FOREIGN EXCHANGE
299
holding of considerable sums of money; because of this, interest
is a very important factor. If the interest has not been taken care
of at the time of the transaction, it must necessarily be considered
later.
The principle of averaging accounts, as explained previously in
this text, will be applied here.
Example
Find the average due date of the following account; also the amount due
June 1 following, including interest at 6%:
Debits
Jan. 2
Jan. 31
Mar. 16 .
Credits
£600 Feb. 15 . £500
300 Apr. 1 200
100
Solution
Debits Days Day-£ Total
Jan. 2
£ 600
2
£ I 200
Jan. 31
300
31
9300
Mar. 16
100
75
7500
Feb. 15
£1,000
Credits
C 500
Days
46
Day-£
£23 000
£18,000
Total
Apr. 1
200
91
18,200
Balance. £
£ 700
300
£41,200
£23.200
Using the last day of the month preceding the first item as the focal date,
23,200 -f- 300 = 77 days. As the balances of the £'s and the Day-£'s are on
opposite sides, the due date will be counted backward from the focal date;
that is, £300 should have been paid 77 days before December 31, or on October 15,
in order that neither party should lose interest.
From October 15 of one year to June 1 of the next year is 229 days. £300 for
229 days at 6% is:
229
300 X .06 X - I = £11.29315
ouO
.29315 X 20 = 5.8630s.
.863 X 12 = 10.356d.
Answer:
Principal due October 15 last
Interest to June 1
Total due June 1
£3000s. Od.
1 1 5s. IQd.
£311 5s7"iOcT
Problems
1. Find the average due date of the following account, arid the amount due
on July 1, including interest at 5%:
300 FOREIGN EXCHANGE
Del/its Credits
June 1
£200 Os. Od.
June 4 . .
£400 15s 9d
June 4
300 10s. 6d.
June 12 .
400 Os. Od.
June 6
. . . 400 Os. Od.
June 24 ....
500 1 5s. Od.
June 12
. . 600 15s. 3d.
June 30
600 Os Od
June 25
700 Os. Od.
June 28 500 Os. Od.
2. Bond, who is located in New York, has an account with Waite in London.
Waite engages an accountant to prepare from the following data a statement to
be mailed to Bond:
Debits Credits
May 12 . £ 650 June 10 £ 400
May 30. . . .217 June 30 400
June 12 . . 240 July 1, Balance 557
July 1 . J250
£1,357 £1,357
Find the average due date of the account, and the interest at 5% to July 1,
using 365 days to the year.
Conversion of foreign branch accounts.* The conversion of
foreign branch accounts requires the application of different rates
of exchange.
Current asset and current liability values should be converted
at the rate prevailing as of the date of the balance sheet.
Fixed asset values should be converted at the rate prevailing
at the time of purchase, or at the average rate for purchases during
a fiscal period. If there have been no changes in fixed assets
during the fiscal period, the fixed assets should be valued at the
same rate as in the preceding period. Differences in values of
fixed assets from one period to another, due to fluctuations in
exchange, should not be allowed to affect the results of the fiscal
period.
Remittances should be converted at the actual rate prevailing
on the day of the transmittal of the money.
Revenue items should be converted at the average rate for the
period.
The Controlling or Adjustment account and the Old Inventory
account should be converted at the rate established on the head
office books at the end of the preceding fiscal period.
The following example illustrates the above principles of
conversion.
Example
A corporation having its head office in Boston had a branch office in London.
The trial balances of the head office and of the branch office on December 31 were*
* See page 294, " Current rate of exchange. "
FOREIGN EXCHANGE
301
BOSTON OFFICE
TRIAL BALANCE
Cash .. . .
Branch Account
Remittances
Expenses .
Income
Capital Stock
Surplus
$ 100,000
876,000
$ 130,500
40,000
60,000
800,000
25,500
$f,0l6,006 $1,016,000
LONDON BRANCH OFFICE
TRIAL BALANCE
Cash
Remittances
Customers .
Inventory, December 31
Expenses
Income
Creditors
Boston OiTicc Control
£ 15,000
30,000
130,000
50,000
10,000
£ 25,000
10,000
_ 200,000
£235,000 £235^000
An analysis of the Remittance account showc
June 1, from London Branch
Sept. 1, from London Branch
Dec. 1, from London Branch .
Current rate
Average rate
£10,000 © $4 42
10,000 & 4 28
10,000 © 4 35
$4 40
4 30
From the foregoing data, prepare:
(a) Statement of conversion.
(h) Consolidated trial balance working paper.
(c) Consolidated profit and loss statement.
(d) Consolidated balance sheet.
Solution
BRANCH OFFICE
Cash .
Remittances
Customers
Inventory
Expenses
Income . . .
Creditors
Boston Control
Profit on Exchange
STATEMENT OF CONVERSION
] Bounds
Pounds
Rate
Dollars
Dollars
15,000
$4 40
66,000
30,000
4 35
130,500
130,000
4 40
572,000
. 50,000
4 40
220,000
10,000
4 30
43,000
25,000
4 30
107,500
10,000
4 40
44,000
200,000
4 38
876^000
4.000
2357000
235,000
F,03 1,500
1^031,500
302
FOREIGN EXCHANGE
cv
j-<
;43
o
fl
8 '
O
g
& S
o sfi
d
r<
cp O
o
O
CJ
d
a ^
W
fi
o
a
C5
3
£
0 g
W o
^
3
fci
PQ
S
S
o
O
o
o
^
^3
o
5 fl
i 2
)«
Remittances
Expenses at
Income at B
a
O
W
31
'§•£
da
Remittances
Customers .
>
1
c
£
c
HH
Expenses at
w
+0
c3
<D
1
HH
Creditors . . .
G
O
0
1
a
0
1
OH
FOREIGN EXCHANGE 303
CONSOLIDATED PROFIT AND Loss STATEMENT
Income:
Boston Office ....................... $ 60,000
London Office ......... 107,500
Total ............ . $167,500
Expenses :
Boston Office $ 40,000
London Office .... 43,000
Total . _ 83,000
Operating Income . $ 84,500
Profit on Exchange t . ___ 4,000
Net Profit to Surplus ....... $jS8j500
f Some may question the advisability of including the Profit on Exchange in the
Not Profits. The sotting up of an account "Reserve for Fluctuation of Exchange"
is sometimes advooatod. This is purely an accounting problem, however, and will
not bo discussed here.
CONSOLIDATED BALANCE SHEET
.4 ssvts
Cash:
Boston . . ... $100,000
London .... . . 66,000 $166,000
Customers . 572,000
Inventory ................. 220,000
$958,000
Liabilities and Capitol
Creditors . . . . . . $ 44,000
Capital:
Capital Stock .... $800,000
Surplus:
Balance, Jan. 1 . ... $25,500
Net Profit ........... 88,500 114,000 914,000
In the above example, no fixed assets were stated. When, as
sometimes happens, this item appears in the Branch Trial Balance,
other difficulties are encountered.
To overcome these difficulties, it is necessary, in making a
statement of conversion, to divide the account on the Head Office
books, "Branch Control," into "Branch Control — Fixed Assets,"
and "Branch Control— Current Assets." The rate at the time of
purchase should be used for the conversion of the Branch Control —
Fixed Assets account. By converting at this rate, the value of the
fixed assets is found in terms of the monetary unit of the country
in which the Head Office is located. By deducting the value of the
fixed assets, or the Branch Control — Fixed Assets, from the
Branch Control account, the value of the Branch Control —
Current Assets is found.
The following example illustrates this point:
304
FOREIGN EXCHANGE
Example
A. United States company doing business in London through its London
Branch, received a trial balance of the branch on December 31, as follows:
LONDON BRANCH
Cash
Remittances
Customers
Inventory (new)
Expenses
Plant
Creditors .
Income from Sales
New York Control ...
£ 5,000
25,000
50,000
25,000
25,000
100,000
£ 50,000
50,000
130,000
£230,000
€230,000
NEW YORK OFFICE
Cash
London Brand
Remittances
Expenses .
Capital Stock
Surplus . .
* 100,000
617,500
$118,875
50,125
500,000
148,750
$767,625 $7077625
£5,000 © $4.75i
£5,000 © $4.75f
The following remittances were received:
£5,000 (m $4.75
£5,000 © $4.75i
£5,000 © $4.76
The current rate of exchange was $4.76^.
The average rate of exchange was $4.75^.
The fixed rate of exchange for the plant was $4.74^.
From the information given, show: (a) a statement of conversion; (b) a
balance sheet of the London Branch; and (c) a consolidated balance sheet of the
New York Office. Show also: (d) the accounts on the books of the New York
Office in which the profit is taken up.
Solution
STATEMENT OF CONVERSION
Cash
£ 5,000
@ $4 7 6 3- $
23,825 00
Remittances
25,000
© 4 75|r
118,875 00
Customers
50,000
© 4 76^
238,250 00
Inventory. .
25,000
© 4 76£
119,125 00
Expenses. .
25,000
© 4 75|
118,812 50
Plant . .
100,000
© 4 74£
474,500 00
Creditors
£ 50,000 © 4 76^
<1
i 238,250 00
Income from Mdse
50,000 © 4 75|
237,625 00
New York Control
(Plant Acct.) ....
100,000 (a\ 4 74£
474,500 00
New York Control
(Current Acct.) .
30,000 © 4. 70f
143,000 00
Profit on Exchange
12 50
£230,000 £230,000
$1,093,387 50 $1,093,387 50
FOREIGN EXCHANGE
305
CONVERSION OF BRANCH CONTROL ACCOUNT
Plant and Fixed Assets
London Branch Control
Deduct:
Value of Plant £100,000
Rate of exchange for fixed assets 4.74^
Value in dollars
New York Control £130,000
Less Plant Account J^i000
Current Assets at Branch £ 30^000
The rate is found to be 143,000 -^ 30,000 = 4.76f .
BALANCE SHEET OF LONDON BRANCH
Assets
$617,500
474,500
$143,000
143,000
Cash
Customers
Inventory
Plant
Liabilities
Creditors
Capital
Plant
Current (Schedule A)
( 1r edits
Balance
Income
Profit on Kxchangc
€ 5,000 Ca\ $4 76i $ 23,825
50,000 (a, 4.704- 238,250
25,000 (a\ 4.764 1 1 9, 1 25
100,000 (n> 4.74i 474,500
£180,000 $855,700
50,000 (m $4.7(>i 238,250
£130,000 $617,450
100,000 Cm $4 74£ $474,500
30,000 @ 4 76i 142,950 617,450
SCHEDULE A
Current Account
£30,000 (m $4 76| $143,000 00
. . 50,000 <& 4 75i 237,625.00
£80,000
12.50
$380.637.50
Debits
Expenses
£25000 (<
7* $4 75j-
$118812 50
Remittances
25,000 (j
l\ 4 . 754-
118,875 00
£50,000
237,687 50
Balance
.... £30,000
$142,950 00
The rate is found to be 142,950 -r- 30,000 = 4.76^.
NEW YORK OFFICE
LONDON CONTROL ACCOUNT
$474,500 00 Remittances
143,000 00 Expenses
237,625 00 Ii§il I Plant Acct
' I Current Acct
p i f Plant Acct
' I Current Acct
Income from Mdse
Profit on Exchange
12 50
$855,137 50
$118,875.00
118,812.50
474,500.00
142,950.00
$855,137.50
306 FOREIGN EXCHANGE
PROFIT AND Loss ACCOUNT
Expenses of Branch $1 18,812 50 Income of Branch . . $237,625.00
Expenses of Home Office. 50,125 00 Profit on Exchange of
Profit to Surplus 68,700 00 Branch 12 50
$237L637^5Q $237^637 50
NEW YORK BALANCE SHEET
Assets
Current:
Cash:
New York $100,000
London 23,825 $123,825
Customers — London 238,250
Inventory— London 119,125 $481 ,200
Fixed:
Plant— London 474,500 $955,700
Liabilities
Current :
Creditors—London 238,250
$717,450
Capital:
Capital Stock $500,000
Surplus $148,750
Net Profits 68,700 217,450
"jm 7,450
A balance sheet of the London Branch is next set up, but this balance sheet
needs no explanation.
In the above solution, note particularly the schedule of the current account
and the method of finding the rate.
Various methods of taking up the profits of the branch on the Head Office
books are used. Probably no explanation need be made, except as to the credit
to the Profit and Loss account of the $12.50 profit on exchange. This might
have been credited to an account called "Reserve for Fluctuation of Exchange."
Problems
1.* On December 31, the trial balance on the books of the London. Office of
the A. Rubber Company is as follows:
£ s. d. £ s. d.
Estate Purchase 3,000 0 0
Estate Development 8,000 0 0
Estate Produce Stock, Mar. 1 . . . . 600 0 0
Cash at Bank, London 800 0 0
Estate Manager, Jan. 1 928 12 8
Remittance to Estate Manager . . . 1,000 0 0
London Office Expenses 400 0 0
Share Capital 12,000 00
Creditors 1,900 00
Profit and Loss Balance 828 12 8
£14,728"! 2^ £l4^728"l2~8
* C. P. A. Examination.
FOREIGN EXCHANGE 307
After the above balances have been taken out, the accounts to December 31
are received from the estate manager, as follows (the dollar to be taken at 4s. 4d.) :
Balance, Jan. 1 ................. $ 4,280
Remittances from London .......... 8,600
Rebates . . ... 131
Sale of Produce . . .. 2,000
Profit on Rice ... . . 249
Expenditures on Development . . . . $ 9,000
Expenditures on Purchase of New Land. . 2,800
Expenditures on Upkeep of Estate ... . 1,640
Balance Carried Forward ............... 1,820
115,266
The produce unsold at December 31 was valued by the manager at $5,500.
You are required to construct the Revenue account and the balance sheet for
presentation to the shareholders.
2. Change into dollars and cents the following items of a London Brunch,
and show the new value of the Head Office account:
Fixed assets ..... . £6,000 Rates of exchange:
Inventory (new) . 500 Current ............. $4 78-g-
Cash ... 1,000 Average remittance ...... 4 62fV
Profit of period ... 1,500 Opening rate .......... 4 86fV
Head Office account . . . 5,000 Average rate of year 4 60T7<5
Remittance from Head Office. . 1,000 Balance on Head Office
books .................. $27,200
PART IB
CHAPTER 30
Compound Interest
Compound interest. In computing simple interest (Chapter
7), it has been seen that the principal remains constant. In com-
puting compound interest, the principal is increased by the
additions of interest at stated intervals. The total amount
accumulated at the end of some given time is the compound
amount, and the difference between the compound amount and
the original principal is the compound interest. The principal of
compound interest is logical, for if the periodic interest were paid
to the lender, he would have this additional principal available
for investment during the following period, and so on to the end of
the last period.
Compound interest method. The compound interest method
is the most accurate and scientific means of finding the true value
of an investment. For this reason it is essential that accountants,
and also investors in general, be familiar with it. The method is
based on the foregoing assumption that all accumulations of
interest become a part of the investment at the end of each interest
period.
Actuarial science. Actuarial science is the mathematical sci-
ence based upon compound interest and upon insurance probabili-
ties (see Chapter 38). It deals with the investment of funds, and
also with the mortality tables used by insurance companies (see
Table 7, page 536, in the Appendix). The actuary uses tables for
the greater part of his work : yet he must also have a knowledge of
the fundamentals of his science. The accountant's interest in
actuarial science is to give the best service to his clients by being
able to compute investment values, prepare schedules of amortiza-
tion, set up sinking fund accounts, and so forth.
Symbols. In the choice of the symbols used in this text, the
attempt has been made to select those which are most commonly
accepted.
The following symbols are given at this time for reference :
1 A unit of value, as $1, or the basis of any unit of value.
t The rate of interest for a single period.
j The nominal annual rate, if interest is compounded more often
than once each year.
311
COMPOUND INTEREST
n The number of periods.
r The periodic ratio of increase, or (1 + t).
m Frequency of periods during year.
s The compound amount of 1 .
v The present worth of 1.
D The compound discount on 1 or the quantity of discount.
7 The compound interest on 1 or the quantity of interest.
an The present value of an annuity of 1 . (A is also used.)
R Rent, or periodic payment of an annuity.
sn Amount of an annuity of 1.
P The principal.
S Any amount.
Principal. The principal is a sum of money or element of
value for which interest is paid, or on which interest computations
are based; 1 or $1 will be used in most of the calculations in this
work.
Time. The time which an investment has to run is commonly
stated in years and months, but in compound interest computa-
tions it is better to state the time as a number of periods of equal
duration. Thus, if a problem calls for "four years and six months,
interest compounded semiannually," the time should be stated as
nine periods of six months each.
Example
Number of
Frequency of Periods, or
Time Compoundijig Value of n
1 year Annually 1
1 year . ... . Semiannually 2
1 year . Quarterly 4
1 year. . . . . Monthly 12
3 years, 6 months . . . . Semiannually 7
3 years, 6 months . Quarterly 14
5 years . . . Annually 5
5 years Monthly 60
Rate. The rate is the measure of interest on the investment
or principal. It may be indicated in different ways; for example,
as .06, 6%, or yjj-g-. The rate is generally stated as so much a
year, but when the period of compounding is of any length other
than a year, it is necessary to restate the rate as so much per
period.
If the interest period is a half-year, it is necessary to divide the
stated annual rate by two, and multiply the time in years by
two; if the interest is compounded quarterly, it is necessary to
divide the annual rate by four, and multiply the time in years by
four.
COMPOUND INTEREST
Example
313
Time
1 year
Rate
6%
1
Frequency of 1
Compounding
Annually
Number of
°eriods, or
Value of n
1
Rate,
or Value
ofi
6%
1 year ....
1 year. .
6%
6%
Semiannually
Quarterly
2
4
v /O
3%
l4r%
1 year
2 years
2 years
. .. .6%
. . . 6%
6%
Monthly
Semiannually
Monthly
12
4
24
1 2 /O
*%
3%
i%
3 years, 6 months
3 years, 6 months .
3 years, 6 months. . .
. . 8%
. • • 4%
. ... 4%
Semiannually
Semiannually
Quarterly
7
7
14
4 /C/
4%
2%
1%
Ratio of increase. From every investment, the investor
expects to receive the amount of his investment plus interest. He
buys bonds, stocks, or other investments with the expectation of
receiving an increased amount in return. If the total investment
be multiplied by 1 plus the interest rate expressed decimally — as,
for example, 1.06- — the result will be the amount of his investment
at the end of one interest period. The 1 plus the interest rate is
called the ratio of increase.
Again, if the principal, $1, is placed at interest at 6% for 1
year, it will be worth $1.06 at the end of the year. The amount to
be added to the $1 is $0.06. The ratio of increase is expressed as
(1 + .06), or 1.06; if the symbols previously given were used, the
symbol would be (1 + 2), or r.
Compound amount tables. A compound amount table is a
compilation of the value of 1 for various numbers of periods at
various rates per cent. It is constructed by making the successive
multiplications (1.06) l, (1.06)2, (1.06)3, (1.06)4, and so forth.
A compound amount table is given in Table 2, Appendix III,
page 512.
When a compound amount table is available, many calcula-
tions may be eliminated. However, if one is not available, or if
the factors to be used are not given in the table, the desired amount
may be obtained by one of the methods described below.
Calculation of compound amount. The following methods of
calculation are given to show the different means of arriving at the
value of $1 for a given time.
First method. The first method shows each step taken to find
the value at the end of each period.
Example
find the value of $1 at 6% compound interest for 8 years.
314 COMPOUND INTEREST
Solution
1
.00
X
1.06
= 1.06
for
1
period
1
.06
X
1.06
=
.1236
for
2
periods
1
.1236
X
1.06
=
.191016
for
3
periods
1
.191016
X
1.06
=
.262477
for
4
periods
1
.262477
X
1.06
=
.338225
for
5
periods
1
.338225
X
1.06
=
.418519
for
6
periods
1
.418519
X
1.06
=
.503630
for
7
periods
1.503630
X
1.06
=
.593848
for
X
periods
It can be seen that the compound amount, 1.593848, is the sum
of the investment and the compound interest.
The above method of arriving at the compound amount
becomes very laborious when there are many periods.
Second method. *In this method, multiplication is performed by
using powers of numbers.
Example
Find the value of (1 + i)12 or (1.06)12.
Solution
ALGEBRAIC INCREASE
(1 + i) X (I + i) = (1 + i)2
(1 + i)2 X (1 + i)2 = (1 + i)4
(1 + i)4 X (1 + i)4 = (1 + i)8
(1 + i)8 X (1 + i)4 = (1 + *)12
ARITHMETICAL INCREASE
Exponents Powers of Katio of
Added Increase Multiplied
(1.06) = 1.06 for 1 period
(1.06) = 1.06
(1.06)2 =1.1236 for 2 periods
(1.Q6)2 = 1.1236
(1.06)4 = 1 .262477 for 4 periods
(1.06)4 = 1.262477
(1.06)8 - 1.593848 for 8 periods
(1.06)4 = 1.262477
(1.06)12 = 2.012196 for 12 periods
The above principle may be applied for any power of a number.
If the compound interest table does not extend to a sufficient
number of periods, the required value may be found by this
method, as illustrated in the following example:
Example
Find the compound amount of $1 at 6% for 80 years.
COMPOUND INTEREST 315
Solution
Assume that we referred to the compound interest table, and found that the
highest value shown at 6% was for 20 years, and was 3.2071355. The calcu-
lation for 80 years could be made as follows:
Powers of
Ratio of
Exponents Increase
Added Multiplied
(LOG)20 = 3.2071355
(LOG)20 = 3.2071355
(LOG)40 = 10.285718
0_. OG^° = 10.285718
(LOG)80 = 105.795994
Third method. The calculation by the third method is made by
the use of logarithms.
Example
Find the value of (LOG)80.
Solution
log LOG... . ... .0 0253059
Multiply by exponent 80
log of 80th power of 1 .06 ~~~2 0244720
Antilog 2.024472 . .105 80
Problems
Work the following problems by the methods indicated, and check your
results by referring to the compound interest table. Show the complete work,
as in the foregoing solutions.
Find the compound amount of:
1. (1.04)4 by Method 1. 7. (LOG)24 by Method 2.
2. (L03)6 by Method 1. 8. (L02)40 by Method 3.
3. (LOG)8 by Method 1. 9. (1.005)60 by Method 3.
4. (1.02)6 by Method 2. 10. (1.04)75 by Method 3.
6. (1.005)30 by Method 2. 11. (1.04) 60 by Method 3.
6. (1.03)20 by Method 2. 12. (1.05)80 by Method 3.
Compound amount of given principal. To compute the com-
pound amount of any principal, apply the following procedure.
Procedure: (a) Compute the compound amount of 1 for the
number of periods at the given rate, (1 + i}n, or s.
(b) Multiply the compound amount of 1 by the number of
dollars in the investment, P(l + i)n = S.
Example
What will be the compound amount of $100 placed at interest at 6% for
4 years, interest compounded annually?
Formula Arithmetical Substitution
P(l + i> = S lOO(LOG)4 = $126.25
316 COMPOUND INTEREST
Extended Solution
1.00 X 1.06 = 1.06; or, amount of 1 for 1 year = (1.06)1
1.06 X 1.06 = 1.1236; or, amount of 1 for 2 years = (1.06)2
1.1236 X 1.06 = 1.1910; or, amount of 1 for 3 years = (1.06)*
1.1910 X 1.06 = 1.2625; or, amount of 1 for 4 years = (1.06)4
$100 X 1.2625 = $126.25; or, compound amount of $100 for 4 years
Hereafter, instead of the compound amount of 1 at the end of
each year being found as above, the solutions will be shortened by
the use of the value of the expression (1 + i)n, or s, as in the follow-
ing. Find the value of s4 at 6%.
Contracted Solution
(1.06)4 = 1.2625, compound amount of 1 for 4 years
$100 X 1.2625 = $126.25, compound amount of $100 for 4 years
Use the compound interest table given in Table 2, Appendix
III, page 512.
Problems
Construct formulas and write contracted solutions for the following:
Principal Rate Compounded Years
1.
$ 600
.00
6%
Annually
4
2.
$ 400
.00
3%
Annually
5
3.
$ 600
.00
5%
Annually
3
4.
$1,000
00
4%
Annually
12
6.
$ 256
25
6%
Annually
10
6.
$1,247
.50
3%
Annually
6
7.
$3,847
.50
4%
Annually
8
8.
$1,472
25
3i%
Annually
4
9.
$2,442
.50
7%
Annually
10
10.
$8,247
.50
9%
Annually
20
Compound interest. Since an investment placed at interest
for a definite time at a fixed rate will produce a given amount, the
difference between this amount and the original investment will bo
the increase, or compound interest. To compute the compound
interest, it is therefore necessary to find the amount at the end of
the time and to deduct the principal from it.
Procedure: (a) Determine the compound amount of 1 for the
number of periods at the given rate, (1 + i)n = s.
(6) Find the compound interest on 1 by deducting 1 from the
compound amount of 1, s — 1 = I.
(c) Multiply the compound interest on 1 by the number of
dollars in the investment, P(s — 1) = /.
Example
Find the compound interest on $100.00 for 4 years at 6%.
Formula Arithmetical Substitution
P(s - 1) = / 100(1.2625 - 1) = $26.25
COMPOUND INTEREST 317
Solution
(1.06)4 = 1.2625, compound amount of 1 for 4 years
1.2625 — 1 = .2625, compound interest on 1 for 4 years
$100 X .2625 = $26.25, compound interest on $100 for 4 years
TABLE OF ANALYSIS OF COMPOUND INTEREST
(1)
End of
Period
1
(2)
Principal
$1 00
(3)
Simple
Interest
$ 06
(4)
Interest
on Interest
$
(5)
Compound
Interest
$ 06
(6)
Compound
Amount
$1 06
2
3
4
1 00
1 00
1.00
.12
.18
.24
.0036
.011016
.022477
.1236
.191016
.262477
1 1236
1 191016
1.262477
Problems
1. Find the compound interest on $500 for 5 years at 3%.
2. Find the compound interest on $650 for 6 years at 4%.
3. Find the compound interest on $2,560 for 4 years at 6%.
4. Construct a table of analysis of the compound interest on $800 at 4%
for 4 years.
5. Construct a table showing the complete analysis of $447.20 at 5% com-
pound interest for 4 years.
Results of frequent conversions of interest. Compound inter-
est is usually stated as a certain rate per annum, but if the interest
is to be compounded more often than once each year, the total
accumulation will be greater than the accumulation of (1 + i)n
times the principal, where i is used as the annual rate and n as the
number of years.
A study of the following will show the difference between the
amount of $100 at 6% interest, compounded monthly for 10
years, and the amount of $100 at 6% interest compounded annually
for 10 years.
100 X (1.005)120 (compounded monthly for 10 years). . . . $181 94
100 X (1.06)10 (compounded annually for 10 years). . . . 179.08
Difference caused by frequent conversions $ 2 86
Nominal and effective rates. Nominal, as the word implies, is
defined as "in name only." In illustration (a), above, 6% is the
nominal rate. Effective interest is the interest actually received
by the investor, and is based upon the amount invested and upon
1 year as the period of time. In illustration (a), $6.17 is the
amount of interest received in 1 year on an investment of $100; in
effect, this is 6.17% on the investment, or an effective rate of
6.17%.
In order to distinguish between nominal and effective rates, the
following symbols are used:
318 COMPOUND INTEREST
i = the effective annual rate
j = the nominal annual rate
m = the number of conversions each year
The procedure in calculating the effective rate, the nominal
rate being given, is as follows:
Procedure: (a) Find the number of conversion periods each
year; (6) find the rate per period; (c) determine the compound
amount of 1 for the number of periods found in (a) and at the rate
per period found in (6) ; (d) deduct 1 from the compound amount
of 1.
Example
What is the annual effective rate if the nominal rate is 6%, compounded
quarterly?
Formula A rithmctical Substitution
Solution
1X4 = 4, number of periods
.06 -r- 4 == .015, rate per period
1 X 1.015 = 1.015, compound amount of 1 for
1 period at 1.5%
1.015 X 1.015 = 1.030225, compound amount of 1 for
2 periods at 1.5%
1.030225 X 1.030225 = 1.061364, compound amount of 1 for
4 periods at 1.5%
1.061364 - 1 = .061364, or 6.1364%, annual effective
rate
The procedure in calculating the nominal rate, the effective
rate being given, is as follows.
Procedure: (a) Determine the log of 1 plus the effective rate,
or the ratio of increase.
(6) Divide the log found in (a) by the number of periods of
compounding.
(c) Find the antilog of the quotient of (b), to determine 1 plus
the nominal rate.
(d) Deduct 1 to determine the nominal rate.
Example
An insurance cormoany receives 6% effective interest on a certain investment*
what is the nominal rale per annum, if interest is compounded quarterly?
Formula Arithmetical Substitution
KX(1 + t) - l]mn = Nominal rate [^(1.06) - 1]4 = .058788
COMPOUND INTEREST 319
Solution
log (1.06) = 0.0253059
0.0253059 -4- 4 = 0.0063265, log of ratio of increase
antilog 0.0063265 = 1.014672, ratio of increase
1.014672 - 1 = .014672, rate of increase for 1 quarter
.014672 X 4 = .058788, or 5.8788%, nominal rate per annum
NOTE. v^T-OG) is sometimes written (1.06)*.
Effective interest. While calculations of the effective rate and
of the nominal rate are always based on a unit period of 1 year,
most investments are for periods of more than 1 year. The added
feature of a number of years may be included in the calculation by
either of two methods :
(1) Find the effective rate for the year, and use the actual
number of years for the period of investment.
(2) Find the rate for one period, and also the number of
periods, and use these results as the values of i and n. See page
312 under "rate."
The use of the first, or effective interest method, is advanta-
geous in some annuity computations. However, because of its
simplicity the second method is the one most commonly used; it
will be employed hereafter unless it is necessary to use the first
method for explanatory purposes.
First method.
Procedure : (a) Calculate the effective rate per year.
(6) Find the compound amount for the number of years.
(c) Multiply the compound amount by the principal in dollars.
Example
What amount will be due in 5 years if $200 is placed at interest at 5%, com-
pounded quarterly?
Solution
( ! + V ) - 1 = .050945, effective rate for 1 year
(1. 050945) B = 1.282037, compound amount for 5 years
1.282037 X $200 = $256.41, compound amount of $200 for 5 years
Second method.
Procedure : (a) Find the total number of periods.
(6) Find the rate per period.
(c) Determine the compound amount of 1 for the number of
periods at the rate found in (a).
This method is preferable when interest tables are available.
320 COMPOUND INTEREST
Example
What amount will be due in 5 years if $200 is placed at interest at 5%, com-
pounded quarterly?
Solution
The number of interest periods is 4 X 5, or 20. The rate of interest per
period is .05 -f- 4, or .0125.
(1.0125)20 = 1.282037, compound amount for 20 periods at
1.25%
1.282037 X $200 = $256.41, compound amount of $200 for 5 years
at 5%, compounded quarterly
Problems
1. Find the compound amounts of the following:
(a) $1,500 at 2% compounded quarterly for 5 years.
(6) $450.25 at 4% compounded semiannually for 8 years.
(c) $1,250 at 3% compounded quarterly for 10 years.
2. Find the effective rate equivalent to :
(a) 4% compounded quarterly. (c) 8% compounded quarterly.
(b) 7% compounded semiannually. (d) 6% compounded monthly.
3. Construct a table, similar to the one on page 317, for 4 years at 3 %, interest
to be compounded semiannually.
Compound present worth. Sometimes it is desired to find
what principal placed at interest now will amount to a certain sum
at a definite future time.
The present worth of a sum which is due at the end of a certain
number of periods is a smaller sum which, if put at compound
interest at a given rate, will amount to the known sum in the given
time.
The ratio of increase employed in accumulating 1 to a com-
pound amount is the same as the ratio of increase employed in
accumulating a present worth to 1. To illustrate:
Compound Amount Present Worth
Basis of calculation: $1 .00 Present worth of $1 : $ . 792094
Multiplying: 1.06 Multiplying: 1 06
$1.06 $ .839619
Multiplying: 1.06 Multiplying: 1 06
$1 . 1236 $ .889996
Multiplying: 1.06 Multiplying: 1.06
$1.191016 $ .943396
Multiplying: 1.06 Multiplying: 1.06
Compound amount: $1 .262477 Basis of calculation: $1.000000
1 + 1.262477 = .792094, present worth
or
1 -T- .792094 = 1.262477, compound amount
COMPOUND INTEREST 321
Procedure: (a) Compute the present worth by dividing 1 by the
compound amount of 1, 1 -5- s = t»n.
(6) Multiply the present worth of 1 by the number of dollars
to be produced, S X vn = P.
Example
What amount of money, invested at compound interest at 6% for 4 years,
will produce $100?
Formula Arithmetical Substitution
S X vn = P 100 X .7921 = $79.21
Solution
(1.06)4 = 1.262477, compound amount of 1 for 4 years at 6%
1 -7- 1.262477 = .7921, compound present worth of 1 for 4 years
at 6%
$100 X .7921 = $79.21, compound present worth of $100 for
4 years at 6%
Verification
79.21 X 1.06 = $ 83.96, compound amount for 1 year
83.96 X 1.06 = $ 89.00, compound amount for 2 years
89.00 X 1.06 = $ 94.34, compound amount for 3 years
94.34 X 1.06 = $100.00, compound amount for 4 years
(1) (2)
(3)
(4)
End of
Compound Amount
Present
Period Principal
(Inverted Order)
Worth of 1
$1.00 -f-
$1.262477
$ .792094
1 1 00 ^
1.191016
.839619
2 1.00-7-
1 . 1236
.889996
3 1.00 -T-
1.06
943396
4 1.00 -^
1 00
1.000000
Problems
Set up formulas, solutions, and verifications for the following:
1. What amount placed in the bank at 4%, interest compounded semi-
annually, will accumulate to $2,000 in 5 years?
2. What principal will have to be placed at interest at 3i%, compounded
semiannually, to accumulate to $3,000 in 3 years?
3. What amount of money will have to be placed on deposit to cancel a debt
of $2,375.50 due in 5 years without interest, if the amount deposited is to be
credited with interest at 4%, compounded quarterly?
4. Construct a table, similar to the one above, for $1 afe 4% for 5 years,
interest compounded semiannually.
Compound discount. The compound discount is the difference
between 1 and the present worth of 1.
Procedure : (a) Calculate the compound discount by deducting
from 1 the present worth of 1, (1 — vn) = D.
322 COMPOUND INTEREST
(6) Multiply the compound discount on 1 by the number of
dollars, S(l - vn) = D.
Example
What is the compound discount on $100 due in 4 years, if money can be
invested at 6%, interest compounded annually?
Formula Arithmetical Substitution
S(l - vn) = D 100(1 - .792094) = $20.79
Solution
(1.06)4 = 1.262477, compound amount of 1 for 4 years
at 6%
1 -f- 1.262477 = .792094, compound present worth of 1 due at
the end of 4 years at 6%
1 — .792094 = .207906, compound discount on 1 due at the
end of 4 years at 6 %
$100 X .207906 = $20.79, compound discount on $100 due at the
end of 4 years at 6 %
Problems
Construct formulas and write solutions for the following:
1. Find the compound discount on $500 due in 4 years, money being worth
5%.
2. Compute the compound discount on $600 due in 5 years, money being
worth
3. Required, the compound discount on $800 due in 10 years, money being
worth 4%.
Rate. The rate of interest may be computed if the principal,
the amount, and the time are known. The computation invoh es
the use of a principle illustrated in the chapter on logarithms (see
page 256).
Procedure : (a) Calculate the compound amount of 1 by dividing
the given compound amount by the principal.
(6) Determine the log of the compound amount of 1.
(c) Divide the log of the compound amount of 1 by the number
of periods.
(d) Determine the ratio of increase by finding the antilog of (c).
(e) Deduct 1 from the ratio of increase.
Example
If $100 amounts to $126 25 in 4 years, what is the rate of interest?
Formula Arithmetical Substitution
/Amount , . 4/126.25 ,
COMPOUND INTEREST 323
Solution
126.25 -T- 100 — 1.2625, compound amount of 1 for 4 years
at the unknown rate
log 1.2625 = 0.101231
0.101231 -T- 4 = 0.025307
antiiog of 0.025307 = 1.06, ratio of increase
1.06 - 1 = .06, or 6%, the rate
Verification
1.00 X 1.06 = 1.06
1.06 X 1.06 = 1.1236
1.1236 X 1.06 = 1.1910
1.1910 X 1.06 -= 1.2625
$100.00 X 1.2625 - $126.25
Problems
1. If $100 amounts to $130.70 in 5 years, what is the rate of interest?
2. If $1,000 amounts to $1,127.16 in 2 years, what is the rate of interest, com-
pounded monthly? Set up the formula, the solution, and the verification.
3. Compute the annual rate of interest for each of the following:
(a)
$ 100
$ 133.82
5 years
(W
200
310 59
10 years
(c)
80
212.26
20 years
(d)
1,000
2,830.75
25 years
(e)
40
68.10
18 years
4. At what nominal rate of interest per annum will $200 amount to $268.78
in 5 years, if interest is converted semiannually?
6. At what rate of interest will any principal double itself in 10 years?
Time. By applying the principles of logarithms, the time may
l)e computed if the principal, the amount, and the rate are given.
Procedure : (a) Determine the compound amount of 1 by divid-
ing the compound amount by the principal.
(6) Determine the log of the compound amount of 1.
(c) Determine the log of the ratio of increase.
(d} Divide (6) by (c), to determine the time in periods.
Example
If $100 placed at interest at 6%, compounded annually, amounts to $126.25,
what is the time of the investment?
Formula Arithmetical Substitution
. , Amount . f 126.25
log of (1 + t) log of (1.06)
(b)
1,000
3,207.14
6%
(c)
200
533.17
4%
(d)
40
62.32
3%
(e)
500
1,621.70
4%
CO
300
609.84
6%
(g)
100
200.00
5%
324 COMPOUND INTEREST
Solution
126.25 -4- 100 = 1.2625, compound amount of 1 at 6% for
n periods
log 1.2625 = 0.101231
log 1.06 = 0.025306
0.101231 -*- 0.025306 = 4, time in periods, or 4 years
Problems
Compute the time in each of the following:
Principal Amount Rate Convertible
(a) $ 100 $ 240.66 5% Annually
Semiannually
Compound amount for fractional part of conversion period.
In the problems thus far, the time contained an exact number of
conversion intervals. How shall compound interest be computed
when there is a fractional part of an interest period, for example, if
the time is 4 years, 2 months, interest at 6% convertible semi-
annually?
In actual practice, simple interest is customarily used for
fractions of an interest period. In the example above: (a) com-
pound interest would be computed for 8 years at 3%; (6) simple
interest would be computed on the amount found in (a) at 3 % for
2 months; and (c) the sum of the answers found in (a) and (6)
would be the amount due.
Problems
1. Find the compound amount of $250 for 3 years and 3 months, interest at
5% converted annually.
2. Find the compound amount of $1,575 for 4 years and 3 months, interest at
5% converted semiannually.
3. Find the present value of $2,750 due in 2 years, 8 months, if the interest
rate is 6%, compounded semiannually.
Review Problems
1. How much money will have accumulated after 8 years if $500 is invested
now at 4% converted quarterly?
2. Calculate the value of vb at 3% by dividing 1 by (1.03)5 as shown by the
compound amount table. Compute the value of t;6 at 3% by logarithms.
3. If the annual interest rate is 5%, what is the corresponding rate of
discount? f HINT: d =* l .• J
COMPOUND INTEREST 325
4. How long will it take money to double itself when invested at 3 % con-
verted annually?
5. What nominal rate, convertible monthly, is equivalent to 4% a year
effective?
6. Find the present value of a debt of $750 due in 5 years, if the current
rate of interest is 6% convertible monthly.
7. If you can get 4% converted annually, how much will you need to invest
to accumulate $7,500 in 15 years?
8. Find the present value of $1,250 due in 4 years, 6 months, with interest
at 4% convertible semiannually.
9. At age 42, a man has $6,725 to invest. What interest rate, converted
annually, must he receive in order for the investment to accumulate to $12,500
at age 60?
10. Calculate by means of logarithms the present value of $382.45 due in
5 years without interest, if money is worth 4% effective.
CHAPTER 31
Ordinary Annuities
Definition. An annuity is a series of equal payments made at
equal intervals of time. Examples of annuities are: premiums on
life insurance, interest payments on bonds or mortgages, rentals of
property, pensions, sinking fund payments, regular preferred stock
dividends, and so forth.
The word annuity suggests annual payments, but the broad
meaning of the term is a series of equal payments made at equal
stated intervals, whether these intervals are one year, six months,
three months, or any other period of time.
Kinds of annuities. Annuities are of two kinds :
(1) Ordinary annuity. This is a series of payments where each
periodical payment is made at the end of a period.
(2) Annuity due. This is a series of payments where each
periodical payment is made at the begining of the period.
Ordinary annuities will be discussed in this chapter and annui-
ties due in the next chapter.
Rent of an annuity. The periodic payments are known as
"rents," arid the single periodic payment is represented by the
symbol R.
Amount of an ordinary annuity. The amount of an annuity of
1 for any number of periods (n) is represented by sn (read "s sub n ").
This symbol used in conduction with the rate of interest becomes
sn|t ; for example, s^5% represents the amount of an annuity of 1 a
period, for 10 periods, at 5%. The amount of an ordinary annuity
is the sum at the end of the term of all the periodic payments, plus
the interest on all payments.
Analysis of compound interest. To understand annuities, it is
necessary to know how compound interest tables are built up.
Take (1 + i)n = (LOG)4. This amount can be found in the 6%
column of a compound interest table, the fourth number from the
top. An analysis shows that it is composed of three elements: a
principal of 1; an annual addition of .06 simple interest; and
"interest on interest" of .02247696. A further analysis may be
made as shown on the next page.
327
328
ORDINARY ANNUITIES
End of End of End of End of
Initial 1st 2nd 3rd 4th
Payment Period Period Period Period Total
Invested $1.00 $1.00
1st annual interest
on 1 .06
Interest on .06.... .0036 .0036 0036
Interest on 1st
.0036 .000216 000216
Interest on 2nd
.0036 .000216
Interest on .000216 .00001296 .07146096
2nd annual interest
on 1 .06
Interest on 06 . . .0036 0036
Interest on .0036 . .000216 .067416
3rd annual interest
on 1 .06
Interest on .06 .... . 0036 . 0636
4th annual interest
on 1 .06 .06
Total 1 26247696
When the above tabulation is summarized, three distinct parts
appear:
1 . The principal $1 . 00
2. The four equal annual amounts of simple interest,
$.06 24
3. The accumulations of " interest on interest" 02247696
Total $1 26247696
This may be further reduced to :
1. Principal $1 00
2. Compound interest 26247696
3. Compound amount $f 26247696
Relation of compound interest and annuities. In the above
table of analysis, it can be seen that there is a series of payments of
$.06 each at regular stated intervals of 1 year, and that compound
interest is calculated on each of these $.06 payments until the end
of the fourth year. Therefore, .26247696 is the amount of an
annuity of .06 for 4 years at 6%.
An annuity of 1 may be computed from this result as follows:
.26247696 -r- 6 = .04374616, amount of an annuity of .01 for
4 years at 6%
.04374616 X 100 = 4.374616, amount of an annuity of 1 for 4 years
at 6%
Procedure in computing the amount of an annuity. From the
foregoing discussion may be derived the following general pro-
ORDINARY ANNUITIES 3*9
cedure for computing the amount of an annuity for any given
number of periods at any stated interest rate :
Procedure : (a) Calculate the compound amount at the periodic
rate and for the number of periods given, or obtain the compound
amount from a compound amount table, (1 + i)n = s.
(b) Deduct 1 from the compound amount, 5 — 1 = 7.
(c) Divide the compound interest on 1 by the rate per cent
expressed decimally, to obtain the amount of the annuity of 1,
(d) Multiply the amount of the annuity of 1 by the number of
dollars of each annuity rent, R X s-^ — S.
Example
It is desired to find the amount of an ordinary annuity of $100 for 5 years
at 6%.
Solution
1.338226 = compound amount of 1 at 6%, or (1.06)5
1.338226 - 1 = .338226, compound interest on 1 at 6%
.338226 -s- .06 = 5.6371, amount of annuity of 1
$100 X 5.6371 = $563.71, amount of annuity of $100 for 5 years
at 6%
From the above, the following may be derived:
(1 + i)», or (1.06)5 = 1.338226, compound amount, or s.
(1 + i)n - 1, or (1.06)5 - 1 = .338226, compound interest, or 7.
-. > or -'- — — = 5.6371, amount of an annuity of 1, or s-\ .
Rsn^ = S or $100 X 5.6371 = $563.71
Verification
Rent paid at end of first year $100.00
Interest at 6% on $100 $ 6. 00
Rent paid at end of second year 1 00 . 00 106 . 00
Amount of annuity for 2 years $206.00
Interest at 6% on $206 $ 12.36
Rent paid at end of third year ... . . 100 . 00 1 12 36
Amount of annuity for 3 years . $318.36
Interest at 6% on $318.36. ... . ... $ 19. 10
Rent paid at end of fourth year . . 100.00 11910
Amount of annuity for 4 years $437 . 46
Interest at 6% on $437.46 .. $2625
Rent paid at end of fifth year. . .... 100.00 126.25
Total $563.71
Semiannual or quarterly basis. If the rents are payable every
six months, or every three months, and the interest is to be com-
pounded on the same dates, the method of using the rate per period
330
ORDINARY ANNUITIES
and the time in periods is preferable to the method of finding the
effective interest for 1 year and using the time in years. Only the
former method will be illustrated.
Procedure: (a) Find the nominal rate per period, —
(6) Find the number of periods, mn.
(j\mn
1 H — j .
(d) Determine the compound interest on 1 for the number of
(j \mn
1 H — ) — 1 = /.
(e) Divide the compound interest found in (d) by the nominal
j
rate per cent per period, / -f- — = s-|t.
(/) Multiply the amount of the annuity of 1 by the number of
dollars of each periodic rent, Ksr^.
Example
What will be the amount of an ordinary annuity of 8 rents of $50 each,
payable every 6 months, interest at 6% per year, compounded semiannually?
Formula Arithmetical Substitution
Rs-t = ,Sf
Solution
4X2 = 8, number of periods
.06 4- 2 = .03, rate per cent per period
(1.03)8 = 1.26677, 1 at compound interest for 8 periods
1.26677 - 1 = .26677, compound interest on 1 for 8 periods
.26677 -T- .03 = 8.8923, amount of annuity of 1 for 8 periods, or s^3%.*
$50 X 8.8923 = $444.62, amount of annuity of $50 for 8 periods
Problems
1. Find the amount of an ordinary annuity of:
(a) $200 at 5% for 4 years, rents and interest payable annually.
(b) $120 " 4% 6 '
(c) $250 " 3% 20 " semiannually.
(d) $250 " 6% 10 ' ' ' u tt
(e) $500 " 4% 30 ' ' " annually.
(/) $100 " 6% 40 ' '
2. Construct a table of analysis of the amount of an annuity of 1 for 5 years
at 4%.
3. If for 5 years $250 is deposited at the end of every six months in a bank
paying 3^%, interest converted semiannually, what will be the amount credited
to the account at the end of the term? Set up a schedule showing: (a) number
The value of «^J% may be found in Table 4, page 527.
ORDINARY ANNUITIES 331
of periods; (6) amount deposited; (c) interest each period; (d) amount to be
added to the account; (e) balance of the account each period.
4. A purchases a house, and agrees to pay $60 each month for 1 year. If
money is worth 6%, interest compounded monthly, what sum paid in one
amount at the end of the year would be the equivalent of A's total monthly
payments?
Rent of an ordinary annuity. Frequently the amount of an
ordinary annuity is known and it is desired to find the periodic
rent, as in problems of sinking funds.
Procedure : (a) Compute the amount of the annuity of 1 for the
given number of periods at the given rate per period, s .
(6) Divide the number of dollars of the required amount by the
amount of an annuity of 1. The result will be the rent, or periodic
payment, P -r- sn-{ = R.
Example
What should be the amoant of each equal annual payment into a fund which,
in 4 years at 6%, interest compounded annually, is to amount to $1,000?
Solution
P •*- s-,. = R 1000 + slv = R
n \ 416/0
In Table 4 it will be found that
SV|6% = 4-374616, amount of annuity of 1 for 4 years
at 6%
$1,000 -T- 4.374616 = $228.59, rent required to accumulate $1,000
at the end of 4 years
Verification
Rent at end of first year $ 228 59
Interest on $228.59 for 1 year at 6%. 13 72
Rent at end of second year .... 228 59
Amount of annuity at the end of 2 years $ 470 90
Interest on $470.90 for 1 year at 6% * 28.25
Rent at end of third year 228 59
Amount of annuity at the end of 3 years $ 727 74
Interest on $727.74 for 1 year at 6% 43 66
Rent at end of fourth year 228 60
Amount of annuity at the end of 4 years at 6 % .... $1 ,000 00
TABLE OF AMOUNT OF ANNUITY
(1)
End of
Period
1
(2)
Rent
$228 59
(3)
Interest
Accumulation
$
(4)
Addition to
Principal
$228 59
(5)
Total
Amount
$ 228 59
2
3
4
228 59
228 59
228.60
13 72
28.25
43.66
242 31
256.84
272.26
470.90
727.74
1,000.00
332 ORDINARY ANNUITIES
Problems
1. A savings bank pays 2%, interest compounded quarterly. How much
must be deposited at the end of each quarter in order to accumulate $400 at the
end of 2 years? Prepare a table for verification.
2. A company owes $600, due in 4 years. How much must be set aside
semiannually at 4%, interest compounded semiannually, to accumulate to the
amount of the debt at maturity? Prepare formula, solution, and table.
3. A company issued bonds for $30,000, due in 10 years. Interest is at 5%,
compounded quarterly. How much must the company set aside every three
months in order to be able to meet the payments on the bonds when they become
due?
4. A company has a debt of $20,000, due at the end of 10 years. Money is
worth 5 %, interest compounded annually. How much must be set aside annually ,
to accumulate to the amount of the debt?
5. At the age of 30, Y decides that he ought to deposit in the bank, every
three months, an amount which will have accumulated to $25,000 by the time
he is 55. The bank allows him 4%, interest compounded quarterly. What is
the amount of F's quarterly deposits?
Use of effective interest in annuities. Very often the rents are
paid annually, and the interest is compounded semiannually or
quarterly; when such is the case, the effective late of interest must
be used.
If the interest is compounded more or less frequently than the
rents are paid, it is necessary to convert the nominal interest rate
to the effective interest rate applicable to the rent periods.
Procedure: (a) Calculate the effective periodic interest on the
(j\m
1 + — I — 1.
m)
(fe) Calculate the amount of an annuity of 1, using the effective
rate per period; the number of periods corresponds to the number
01 rents, s~,..
7 nil
(c) Multiply the amount of an annuity of 1, found in (6), by
the number of dollars of each rent, Rs^. = 8.
; nit
Example
What will be the amount of an annuity, the annual payments of which are
$100 for 4 years at 6%, interest compounded quarterly?
Solution
The solution to this example will be stated in two parts
(1) Calculation of the effective rate per year.
(2) Calculation of the amount of the annuity of $100.
ORDINARY ANNUITIES 333
PART 1
Formula Arithmetical Substitution
I1 + ") ~ l = Effective rate ( 1 + ^J - 1 = .0613635
.06 -f- 4 = .015, rate per period
1 H- .015 = 1.015, ratio of increase
(1.015)4 = 1.0613635, compound amount of 1 for 1 year
1.0613635 - 1 = .0613635, effective rate per year
PART 2
Use the effective rate found in Part 1, and proceed as in the examples previ-
ously given.
Formula Arithmetical Substitution
(1.0613635)4 = 1.2689855, compound amount of 1 for 4
periods at 0 1 363,") ' [ *
1.2689855 - 1 = .2689855, compound interest on 1 for 4
periods at 6.13635%
.2689855 -f- .0613635 = 4.383477, amount of annuity of 1 for 4
periods at • .•>• ••>•">
$100 X 4.383477 = $438.35, amount of annuity of $100 for
4 years
Verification
First year:
Rent at end of year ....................... $100 00
Second year:
$100 X .0613635 (effective rate) ............. $ 614
Rent .................................... 100 00 106 14
Amount of annuity for 2 years ............. $206. 14
Third year:
$206.14 X .0613635 .................. $1265
Rent ................................ JOO 00 H2^ 65
Amount of annuity for 3 years ...... . $318.79
Fourth year:
$318.79 X .0613635 ......... $ 19 56
Rent ......................... 100 00 119 56
Amount of annuity for 4 years . ...... $438 35
Problems
1. Barlow has a 5-year annuity for which the payments, made at the end
of each year, are $300 each. Interest is at 4%, compounded semiannually.
What is the amount of the annuity? Prepare proof of answer.
2. Ware desires to know how much he will have in the savings bank at the
end of 25 years if he deposits $150 at the end of each six months. The bank pays
4%, and the interest is compounded at the end of each quarterly period.
* The compound amount of 1 for 4 periods at 6.13635% is the same as the com-
pound amount of 1 for 16 periods at 1£%, and (1.015)16 is readily found in the com-
pound amount table given in Appendix III.
334 ORDINARY ANNUITIES
3. Deposits of $500 are made at the end of each year for 20 years. If the
bank credits the account with quarterly interest at 4% (nominal rate), what
will be the amount of the accumulation?
Sinking fund contributions. A sinking fund produced by
equal periodic payments accumulating at compound interest is one
type of annuity. The principles applicable to annuities will be
further illustrated with special reference to sinking funds.
To find the rent of a sinking fund, divide the number of dollars
required in the total fund by the amount of the annuity of 1 for the
specified number of periods at the given rate, P -f- s^. = R.
Example
A company borrowed $2,500 for 5 years, and established a sinking fund to
provide for the payment of the debt. The contributions to the funcl were to
be made at the end of each year. If money is worth 6%, what should be the
amount of each annual contribution?
Solution
P + s . = R $2,500 -5- s,~ = $443.49
nit 0,6/0
In Table 4 it will be found that
s -, „ = 5.63709, amount of ordinary annuity of 1 for
6 |fi /o
5 periods at 6%
$2,500 -5- 5.63709 = $443.49, contribution to sinking fund.
TABLE OF ACCUMULATION OF SINKING FUND CONTRIBUTIONS
(1)
(2)
(3)
(4)
(5)
End of
Yearly
Total
Period
Contribution
Interest
Increase
Fund
1
$ 443 49
$...-
$ 443 49
$ 443 49
2
443 49
26 61
470 10
913 59
3
443 49
54 82
498 31
1,411 90
4
443 49
84 71
528 20
1,940 10
5
443 49
116 41
559 90
2,500 00
$2,217.45 $282 55 $2,500 00
Problems
1. A company establishes a sinking fund to provide for the payment of a
debt of $8,000 maturing in 4 years. The contributions to the fund are to be
made at the end of each six months. Interest at 4% is to be compounded
semiannually. What must be the amount of each semiannual contribution?
Construct a table, as in the example above.
2. A debt of $30,000 is due in 4 years. A sinking fund is to be established,
and contributions are to be made at the end of each six months. What must
be the amount of each semiannual contribution, if interest at 4 % is compounded
semiannually? Construct a table, as in the example above.
3. A has an obligation of $8,000 maturing in 3 years. How much must he
set aside each month at 6%, interest compounded monthly, in order to be able
to pay the debt when due?
ORDINARY ANNUITIES 335
Present value of an ordinary annuity. The present value of an
ordinary annuity ^presented by the symbol a-, is the sum which,
if put at compound interest, will produce the periodic rents of the
annuity contract as they become due.
First method. Procedure: Find the present value of each peri-
odic rent separately ; add the present values of these periodic rents ;
their sum is the present value of the annuity.
Example
Assume that it is desired to find the value, at the beginning of the first period,
of a series of four annual periodic payments of Off each. Respective payments
are to be made at the end of each year.
PRESENT VALUE AT DATE OF CONTRACT
1st rent, payable at end of 1st year ... .06 X - = .00 X .94339 = 056003
2nd rent, payable at end of 2nd year... .00 X 77-7^, = -06 X .SS999 = .053399
3rd rent, payable at end of 3rd year. . . .00 X 7^7.77, = -00 X .83962 = .050377
4th rent, payable at end of 4th year. . . .00 X 7777,77-4 = -OG X .79209 = .047525
( 1 .Uo) — -
Present value of an annuity of .00 = 207004
Second method. It will he noted that the present value at 6%
of an annuity of 4 rents of 6^ each is the same as the compound
discount on 1 for 4 periods at 6%. The computation of the com-
pound discount on 1 for 4 periods at the rate of 6 % is as follows :
(l.OO)4 = 1.202477, compound amount of 1 for 4 periods at 0%
1 -r- 1.202477 = .792094, present value of 1 for 4 periods at 0%
1 - .792094 = .207904, compound discount on 1 for 4 periods at 6%
Substituting symbols for figures, it is apparent that the present
value of an annuity for n periods, the rents of which when stated
in cents are the same as z, is equal to the compound discount on 1
for n periods at the rate of i.
In the example just given, the basis of calculation is the periodic
payment of 6^, and this is stated as "an annuity of 6^.M How-
ever, the basis most frequently used, or the common basis, is 1,
and is expressed as "an annuity of 1." (Annuity tables are built
on this basis.) Therefore, in order to find the present value of an
annuity of 1, divide the compound discount by the rate per cent
expressed decimally. Using the figures given above, the calcula-
tion would be: .207904 -f- .06 = 3.465105, or the amount of an
annuity of 1 for 4 periods at 6%.
336 ORDINARY ANNUITIES
Procedure: (a) Calculate the compound discount on 1 for the
required number of periods and at the required rate per cent,
1 - t;n = D.
(b) Divide the compound discount by the given rate per cent,
expressed decimally. The quotient will be the present value of an
annuity of 1, D -5- i = a~}..
(c) Multiply the present value of the annuity of 1 by the num-
ber of dollars of each rent, R X a-^ = A.
Example
The terms of an annuity contract call for the payment of $100 at the end
of each year for 4 years. If money is worth 6%, interest compounded annually,
what is the present value of the annuity contract at the beginning of the first
year?
Formula Arithmetical Substitution
1
Ra , = A 100
(LOG)4
= $346.51
.06
Solution
(1.06)4 = 1.262477, compound amount of 1 for 4 years at
<)%
1 -r- 1.262477 = .792094, present value of 1 for 4 years at 6%
1 - .792094 = .207906, compound discount on 1 for 4 years
at 6%
.207906 4- .06 = 3.4651, present value of an annuity of 1 for 4
years at 6%, or the value of «^6^.*
3.4651 X $100 = $346.51, present value of an annuity of $100
Verification
Beginning of first year:
Present value of contract $346 51
End of first year:
Deduct:
Rent $100.00
Interest on $346.51 at 6% 20 79
Reduction in value of annuity contract 79 21
Present value $267~30
End of second year:
Deduct:
Rent $100 00
Interest on $267.30 __ 16 04
Reduction in value of annuity contract .... 83 . 96
Present value $183 34
* The value of a4{e% may be obtained directly from Table 5, page 532.
ORDINARY ANNUITIES 337
End of third year:
Deduct:
Rent $100.00
Interest on $183.34 11.00
Reduction in value of annuity contract. 89.00
Present value $ 94 34
End of fourth year:
Deduct:
Rent $100 00
Interest on $94.34 5 66
Reduction in value of annuity contract ... 94 34
$~(TOO
Amortization. Payments made on the principal, as shown in
the above example, are known as amortization payments. Amor-
tization is the gradual repayment of the principal through the
operation of the two opposing forces of compound interest and
periodic payments. Compound interest increases the principal,
while the payments reduce it. In the verification above, it can
be seen that the excess of each payment over the interest for
the period is the amount by which the principal is reduced. The
amount of this reduction is the amortization.
Problems
1. Find the present value of each of the following ordinary annuities:
Rents Paid Interest Years
(a) $400 Annually 3% 6
(6) $225 Annually 47o 10
(c) $350 Annually 4% 10
(d) $255 Semiannually 6% 3
(e) $340 Semiannually 7% 3
2. By the terms of an annuity contract, $500 is to be paid at the end of each
six months for 4 years. Money is worth 6%, interest compounded Semiannually.
(a) Find the present value of the annuity. (6) Submit solution and verification
showing the applications each year of the payments as to interest, amortization
of principal, and new principal against the present value of the annuity.
3. What is the value at the beginning of the first period of an annuity of $300
payable at the end of each year for 10 years, money being worth 4%, interest
compounded Semiannually? Prepare columnar table.
4. A man purchases a house for $1,800 cash and sixteen notes of $400 each,
without interest, one due at the end of each six months until all the notes are
paid. If money is worth 4%, interest compounded serniannuaily, what is the
cash value of the property?
5. What is the cash value of a contract which calls for the payment of $50 at
the end of each month for 5 years, if money is worth 6%, interest convertible
monthly?
6. A disability insurance contract provides that the insured may choose
one of the following options: (a) $50 a month, payable at the end of each month.
338 ORDINARY ANNUITIES
for 48 months; (6) $500 cash, and $50 a month for 36 months; (c) $100 a month
for 12 months, and $50 a month for the ensuing 26 months. If money is worth
6%, interest compounded semiannually, which is the best option?
Computation of the rents or periodic payments of the present
value of an ordinary annuity. If the present value of an annuity,
the rate per cent, and the time are given, the rents or periodic
payments may be calculated as follows:
Procedure: (a) Determine the present value of an annuity of 1
for the required number of periods at the given rate per period,
or a ..
n i
(b) Divide the given present value of the annuity by the present
value of the annuity of 1 found in (a), A -f- a — R.
Example
What annual rent will be produced by an ordinary annuity the present
value of which is $346,51, if there are four rents, and money is worth 6 %?
Formula Arithmetical Substitution
-i-fi. ^^-=$100.00.
1 - 1_
(1.06)_4
.06 ~~
Solution
(LOG)4 = 1.262477, compound amount of 1 for 4 years
at 6%
1 -f- 1.2G2477 = .792004, present value of 1 for 4 years at 6%
1 — .792094 = .207900, compound discount on 1 for 4 years
at 6%
.207906 -T- .06 = 3.4651, present value of an annuity of 1 for
4 years at 6%, or value of a .*
$346.51 -T- 3.4651 = $100, rent of annuity
See verification shown on pages 336 337.
Problems
1. If the present value of an annuity contract is $6,000, what amount must
be paid at the end of each year for 10 years to cancel the obligation, money being
worth 4%, interest compounded annually?
2. An annuity contract is worth $12,000 at the present date. If the time to
maturity is 10 years, and money is worth 3%, interest compounded semiannually,
what periodic payment must be made at the end of each six months to cancel
the contract in 10 years?
3. The present value of a 12-year annuity contract is $8,000. If money is
worth 4%, interest convertible quarterly, what amount must be paid at the
end of each quarter to cancel the contract in 12 years?
* This value is readily found in Table 5, page 532.
ORDINARY ANNUITIES 339
Payment of debt by installments. In many contracts it is
agreed that the principal of the debt together with the interest are
to be paid in equal periodic payments; each payment is to cancel
the interest due to date, and the balance is to be applied toward the
repayment of the principal.
The procedure, formula, and solution are similar to those given
on page 338.
Example
If Smith borrows $1,000 from Jones at 0%, and agrees to cancel the debt
(principal and interest) in five equal annual payments, what will be the amount
of each payment?
The formula for "rent of present value of annuity," given on page 338, is
applicable.
TABLE OF INSTALLMENT PAYMENTS
(1)
End of
Period
1
(2)
Payment
Made
$ 237 30
(3)
To Cancel
Interest
$ (>0 00
(4)
Amortization
of Principal
$177 39
(5)
Balance of
Principal
$1,000 00
2
237 39
49 3(5
18S 03
822 61
3
237 39
3S 07
199 32
634 58
4
237 39
2tt 11
211 28
435 26
5
Totals
237 39
13 44
223 95
223 98
$1,1 SO 95
SING 98
$999 97
$ 03
A slight error, such as that in the above table, is likely to occur
in many problems in installment payments; it is caused by the fact
that the computations are not carried to a sufficient number of
decimal places. In such cases the difference should be corrected in
the last payment. In the table, the fifth payment should be
$237.42, instead of $237.39. This would leave column (5) with no
remainder, and column (4) would have a total of $1,000.
Problems
1. A contracts for the purchase of a house, and agrees to pay for it in install-
ments of equal amounts over a period of 10 years. The cash value of the house
is $10,000. If money is worth 5%, interest convertible quarterly, what should
he the amount of each payment?
2. What annuity, payable quarterly for 20 years, would be required to repay
a loan of $12,840, the nominal rate of interest being 4% per annum?
3. A debt of $3,500, with interest at 5%, compounded semiannually, will be
discharged, principal and interest, by equal payments at the end of each six
months for 10 years. Determine the amount of each payment.
4. A house cost $15,000. The purchaser paid $3,000 cash, and agreed to
pay the balance in equal quarterly payments, principal and interest, over a
period of 8^r years. If money is worth 6%. interest convertible quarterly, what
is the amount of each payment?
340 ORDINARY ANNUITIES
6.* A company purchased machinery on December 1, at a cost of $40,000.
Twenty-five per cent of the cost was paid in cash, and the balance is to be paid
in 60 monthly installments of equal amount. The monthly payments are to
be represented by notes, payable on the first day of each month and secured
by chattel mortgage. Interest at 6% per annum, or £ of 1% per month, is to
be included in the notes. Compute the amount of each note, taking the com-
pound interest on $1 as .348850 for the given time and rate.
Computation of the term of an annuity. When the rate per
cent, the amount of the annuity, and the size of each periodic pay-
ment are stated, it is possible to calculate the number of periodic
payments to be made.
In calculating the time, it is necessary to resort to equations
and logarithms.
Procedure: (a) Divide the amount of the annuity by the num-
ber of dollars in one of the periodic payments, to find the amount
of the annuity of 1 at the given rate.
(b) Multiply the amount of the annuity of 1 by the rate per
period expressed decimally, to find the compound interest on 1 for
the unknown time.
(c) Add 1 to the compound interest on 1 to find the compound
amount of 1.
(d) Determine the log of the compound amount of 1 found
in (c).
(e) Determine the log of 1 plus the rate per cent for the period
expressed decimally; that is, the ratio of increase.
(/) Divide the log of the compound amount of 1, (d), by the log
of the ratio of increase, (e), to find the number of periods, or the
number of periodic payments.
Example
The amount of an annuity is $1,318.08, the rent is $100 each year, and the
rate is 6%. Find the term.
Formula
/Amount of annuity vy \1
XJ _ Term
log (Ratio of increase)
Arithmetical Substitution
log (1.06)
Solution
10
Dividing by rent: 1,318.08 -s- 100 = 13.1808
Multiplying by rate: 13.1808 X .06 = .790848
* C. P. A., Ohio.
ORDINARY ANNUITIES 341
Adding 1: 1 + .790848 = 1.790848
log of: 1.790848 = 0.253059
log of: 1.06 = 0.025306
Dividing: 0.253059 -r- 0.025306 = 10
As the rents are to be paid annually, the annuity term will be 10 years.
If the payments are to be made more often than once each year,
the rate should be reduced to a rate per period.
Example
A purchases a house for $2,629.02, and agrees to pay $500 down and $50 at
the end of each month. Each payment is to cancel the interest to date, and
the balance is to apply against the principal. The debt bears 6% interest.
How many months will it take A to pay off the debt?
Reducing the rate to the rate per period, .06 -f- 12 = .005.
Formula Arithmetical Substitution
log
"*
-: = Term
X .005
50
48
log (l + i) log 1.005
Solution
Dividing: 2,129.02 ~ 50 = 42.5S04
Multiplying: 42.5804 X .005 - .212902
Subtracting: I - .212902 = .787098
Dividing: 1 -f- .787098 = 1.270489
log of: 1.270489 = 0.1039709
log of: 1.005 = 0.0021661
Dividing: 0.1039709 ~ 0.0021661 = 48, approx.
Hence 48 monthly payments will be necessary to pay off the debt.
Problems
1. L. Miller purchased a house and lot for $7,800. He agreed to pay $2,800
cash, and $50 at the end of each month until the debt should be paid. How
many months will it take Miller to pay the debt, if each payment is to cancel
the interest due and the balance is to apply against the principal? The interest
rate is 6%.
2. B sells his residence for $10,500. lie receives a down payment of $2,500.
The balance is to be paid on the basis of $75 each month. Each monthly pay-
ment is to cancel the interest first, and the balance is to apply against the princi-
pal. The contract states that interest is at the rate of 6% per annum. How
long will it take the buyer to pay off the debt?
3. Cole buys a farm for $18,000, and agrees to pay $6,000 down, the balance
to draw interest at 6% until paid. Any payments made are to apply on interest
due to date, and if the payments exceed the interest due, the balance is to be
applied to the reduction of the principal. Cole desires to know how long it
will take him to pay for the farm if he makes equal quarterly payments of $400.
Use of effective rate in annuities. If interest is compounded
more frequently than the rents are paid, or vice versa, it ie neces-
342 ORDINARY ANNUITIES
sary to reduce the rate of interest to an effective rate for a perioc
corresponding to the periods of the rent payments.
Procedure: (a) Calculate the effective rate of interest for OIK
rent period.
(6.-1) I f the amount of the annuity is known, use the procedun
previously given.
(6.-2) If the present value of the annuity is known, use the
procedure previously ^iven.
Example
The rents of $100 each are to he paid annually, the amount is $318.64, anc
the interest rate is 6%, compounded semiannually. Find the term.
PART 1
Formula A rith me tic a I Su bstit at ion
f 1 + -M - 1 = KfTective late M + '^j - 1 = .0609
Solution
( ^ ~^~ 9 / ~ 1-0609, effective ratio of increase for 1 year
1 .0009 - I - .0609, effective rate
PART 2
As the problem states the amount of the annuity, it may be solved by pro
cedure (6.- 1).
Form ula + 1 rith mctical Substit ution
/_-! ___ o
.
log (Ratio of increase) log 1.0609
Solution
Dividing: 31S.64 -5- 100 = 3.1864
Multiplying: 3.1864 X .0609 = .1940517
Adding 1: 1 + .1940517 = 1.1940517
log of: 1.1940517 = 0.077022
log of: 1.0609 = 0.025674
Dividing: log 0.077022 -*- log 0.025674 = 3
The result, 3, indicates that there are three annual payments of $100 each.
I Aerification
First year:
Rent ....................... $100 00
Second year:
$100 X .0609 .......................... $ 6 09
Rent .................................. 100 00 106 09
$206 09
Third year:
$206.09 X .0609 ................ . $ 1 2 55
Rent ............ 100 00 112 55
Amount ................. ~~~ ~~ $318 64
ORDINARY ANNUITIES 343
Problems
1. Compute the number of periods in each of the following:
Amount of an
Ordinary Annuity Rents Rate
(a) $ cS 0191 $ 1 00 44%
(6) 663.29 100 00 4%
(c) 9,549.11 1,00000 5%
(d) 1,061.82 200 00 3%
2. Smith desires to repay his debt of $5,000 by paying $500 at the end ot
each year. If money is worth 5%, interest convertible semiannually, how many
payments will be have to make?
3. A has a debt of $10,570. lie makes a payment of $1,000 at the end of
each year. Money is worth 6%, interest convertible quarterly. How many
full periodic payments will A have to make in order to cancel the debt?
4. To repay a loan of $1,000 bearing interest at 4%, convertible quarterly,
Adams makes a payment of $100 at the end of each six months. How long
will it take him to cancel the debt?
5. A building is puicluised for $15,000. Payment is to be made in install-
ments of $1,500 at the end of each six months' period. Interest is at 0%, com-
pounded quarterly. How many payments will be required to cancel the debt?
Computation of the rate of an annuity. The mathematical
theory of an annuity deals with five elements term, rent, rate of
interest, present value of the annuity, and amount of the annuity.
Tf the rate and any other three elements are given, the missing
element may be found. However, as the rate is used twice in
annuity calculations, it can be only approximated if it is not given.
To obtain an approximate rate by inspection, find a trial rate,
and test it to see whether, when it is used with the given component
parts, it produces an amount equivalent to or nearly equivalent to
the amount of the annuity. If the trial rate proves to be near the
required rate, use it as a basis, and select another rate such that of
the two rates chosen one will be more and the other less than the
required rate. Proceed to select the approximate rate by the proc-
ess of interpolation.
The test rates chosen may be found by means of an annuity
table, or by the calculation of the amount of an annuity of 1.
Selection of rates by use of an annuity table.
Procedure: (a) Divide the given amount of the annuity by the
annuity rent, to find the amount of an annuity of 1.
(b) Choose from the annuity table the two amounts nearest to
the amount found in (a).
(c) Proceed by interpolation to find the approximate required
rate.
344
ORDINARY ANNUITIES
Example
4>*Uff(f/lC
An ordinary annuity the amount of which is $1,099.62 has fhe annual rents
of $200 each. What is the rate of the annuity?
SECTION OF ANNUITY TABLE
Periods 3? %
4(/o
4i%
5f/o
0%
1
1
.00000
1
00000
1
00000
1
00000
1
00000
2
2
.03500
2
.04000
2
04500
2
05000
2
06000
3
3
.10622
3
.12160
3
13702
3
15250
3
1 S360
4
4
.21494
4
24646
4
.27819
4
31012
4
37461
5
5
.36246
5
.41632
5.47071
5
.52563
5
63709
$1,099.62 -f- $200
Solution
$5.4981, amount of an annuity of $1 for 5 periods
The second step is to choose from the annuity table, horizontally to the right
of the fifth period, the two amounts nearest to $5.49X1. By inspection, the
amount $5.47071, in the 4-J-% column, is found to he the nearest one under
$5.4981, and $5.52563, in the 5% column, is found to be the nearest one over
$5.4981. Using these rates and amounts as a basis, the following may be
derived :
Amount of a $1 annuity at 5% $5 52563
Amount of a $1 annuity at 4-j% 5 47071
Difference in amount caused by \r/0 difference in interest
rate $ 05492
Amount of a $1 annuity at unknown rate $5 49S1
Less amount at \\ % 5 4707 1
Difference $ 02739
of i% — .2493%, or approximately 1 %
4i% + •£% = 4f %, the approximate rate
Selection of rate by calculation of amounts of annuities. If
no annuity table is at hand, it is necessary to obtain the two basic
rates by the calculation of the amounts of the annuities, using
estimated rates.
Procedure: (a) As this is a five-payment annuity and each pay-
ment is $200, the total paid in will be $1,000. Deducting this
$1,000 from the amount, $1,099.62, the interest is found to be
$99.62.
(6) The payments will draw interest thus :
$200 for 4 years
200 for 3 years
200 for 2 years
200 for 1 year
200 for 0 years
10 years
ORDINARY ANNUITIES 34$
Hence we have the equivalent of $200 for 10 years.
Simple interest on $200 for 10 years at 5% is $100, or a little
more than the interest in the problem when compound interest is
disregarded. Thus, it can be seen that the interest is probably less
than 5%, but as the compounding of interest is infrequent, the
variation will be small. Hence it is advisable to try the rate of 5%.
First trial rate. The solution by the 5% mte is as follows:
Solution
(1.05)6 = 1.276281, compound amount of 1 at 5% for 5
periods
1.27G2S1 — 1 = .270281, compound interest on 1 at 5% for 5
periods
.2702X1 -r- .05 = 5.52503, amount of annuity of 1 at 5% for 5
periods
$200 X 5.52503 = $1,105.13, amount of annuity of $200 at 5% for
5 periods
This is found to be a little more than the required amount; therefore the next
trial rate should be less than 5%.
Second trial rate. A trial rate of 4^% will be used.
Solution
(1.045)5 = 1.2401819, compound amount of 1 at 4.5% for
5 years
1.2401X19 - 1 = .2401X19, compound interest on 1 at 4.5% for
5 years
2401X19 -r- .045 = 5.470709, amount of annuity of 1 at 4.5% for
5 years
$200 X 5.470709 = $1,094.14, amount of annuity of $200 at 4.5%
for 5 years
This is found to be smaller than the amount in the problem, $1,099.02.
Interpolate between the rates used in the first and second trials, as follows*
Interpolation of A mounts
Amount of annuity at 5% $1,105. 13
Amount of annuity at 4^% 1,094 14
Difference in amount caused by \% difference in rate. . $ 10 99
Amount of annuity at unknown rate $1,099 62
Amount of annuity at 4^% 1,094. 14
Difference in amount. . . $ 5 48
of i% = approximately -J-%
+ i% = 4|-%, the required rate
346 ORDINARY ANNUITIES
The approximate rate is found from the present value of annui-
ties in exactly the same manner as from the amounts of annuities,
except that the formula for the present value is used instead of the
formula for the amounts.
Problems
1. Calculate the rates in each of the following:
Amount of
No.
Annuity
Rents
Periods
(a)
$ 5 58
$ 1 00
5
(b)
232 76
10 00
15
(c)
4,486 52
100 00
21
(d)
4,358 54
125 00
20
(e)
1,729 46
137 50
10
(/)
347 50
20 00
12
(g) 6,684 00 200 00 20
2. Calculate the rates in each of the following:
] ^resent Value
No. of Annuity Rents Periods
(a) $ 4 45 $ 1 00 5
(b) 77 22 10 00 10
(c) 196 90 25 50 10
(d) 5,018 75 500 00 15
(e ) 657 . 42 20 00 36 monthly payments.
(/) 1,200 00 100 00 20
(g) 7,500 00 500 00 25
Solution of annuity problem with limited data. It is not
uncommon to find in examinations in accounting aproblem followed
by a list of numerical values of certain terms, from which the
candidate must calculate certain other values needed for solution
of the problem. The purpose of this is to test the candidate's
knowledge of the relationships between actuarial terms.
Problems
1.* A company is issuing $100,000 of 4%, 20-year bonds, which are to be
paid at maturity by means of a sinking fund into which annual deposits are to
be made. The board of directors wishes to assume that this fund will earn
5ri% f°r the first 5 years, 5% for the next 5 years, and 4% for the last 10 years,
What is the annual deposit required?
Given:
£*% 5% 4%
tf, 5 581 5 526 5 416
»S10 12.875 12.578 12006
(1 -M)6 1.307 1.276 1 217
(l+O10 1.708 1.629 1.480
1 American Institute Examination.
ORDINARY ANNUITIES 347
2.* A city, with its fiscal year ending April 30, prepares its budget and
makes its tax levy for the subsequent fiscal year during March, taxes being
payable on or after November 1.
A bond election was held in June, 1942, and bonds of $1,000,000 were issued
dated August 1, 1942, due in 20 years. A sinking fund was to be provided, calcu-
lated on a basis of 4%, interest compounded annually.
An audit having been made as of April 30, 1943, the balance of $409,588.25
in the sinking fund is found to differ from the actuarial requirements.
Calculate the correct amount which should have been in the fund, and
ascertain the annual adjustment that the city must thereafter make to be able
to meet the bonds at maturity; the difference is to be spread over the subsequent
levies, and not provided for in the next levy only.
Assume that 4% interest will be earned in the future, that all taxes will be
collected in full by the end of the fiscal year, and that a deposit of the correct
amount is to be made in the sinking fund annually on April 30.
Given, at 4%:
r8 = .7306902 (1 + i)8 = 1.3685690
r9 = .7205867 (1 -ft)9 = 1.4233118
v10 = .6755642 (1 + i)10 = 1.4802443
(1 -f r)10 = 2.1068492
(1 + i)20 = 2.1911231
(1 + i)21 = 2.2787681
Review Problems
1. Amos Brown sets aside $400 at the end of each year to provide a fund
for his daughter's college expenses. If he invests the money at 3% effective,
compounded annually, what will be the amount at the end of 10 years?
2. Ben Told invests $200 at the end of each year. At the end of the fifth
year he has accumulated $1,040.81. Write the equation whose solution will
give the rate of interest. Solve and check your answer as nearly us possible
from the s 1 table. Interest convertible annually.
3. If in Problem 1 the interest realized had been 2^%, what would have
been the amount?
4. An, annuity of $1.00 a year amounted to $8.00 in 7 years. What was
the effective rate of interest?
5. If money can be invested at 3% effective, how many full years will be
necessary to accumulate a fund of at least $2,000 from $100 set aside at the
end of each year?
6. Find the amount of an annuity of $1,000 a year paid in four quaiterly
installments of $250 for 6 years if the interest rate is 4% effective.
7. How long will it take to accumulate $1,500 by depositing $20 at the
end of each month if the bank pays 2% effective? Give your answer to the
nearest month.
8. If you pay a paving tax of $52.17 at the end of each year for 10 years and
the rate of interest is 5%, what is the actual tax for the paving?
9. What is the present value of an annuity of $1,800 a year in monthly
installments for 10 years if money is worth 4% effective?
' American Institute Examination.
348 ORDINARY ANNUITIES
10. How long will it take to pay for a house and lot priced at $6,000 if you
pay $1,000 down and $800 at the end of each year until full payment is made,
assuming interest to be 6% effective?
11. If $750 invested at the end of each year for 6 years amounts to $4,912.62,
what is the rate of interest?
12. An annuity of $100 a year for H years amounts to $900. Find the effective
rate of interest, convertible annually.
13. Find the amount of an annuity of $500 for 10 years: (a) with effective
rate 4%; (b) with a nominal rate of 4%, converted quarterly.
14. George Smith deposits $50 in a savings bank at the end of each throe
months. The bank pays 2% convertible semiannually. What will be the
amount to Smith's credit at the end of 5 years?
16. Find the present value of an annuity of $500 a year for 10 years if money
is worth 4% effective.
16. A realtor offers a house for $4,000 cash and $1,000 a year for six years
without interest. A buyer desires to pay cash. If money is worth 6% effective,
what should be the cash price of the house?
17. Find the present value of an annuity of $2,000 a year for 5 years if money
is worth 4%, converted quarterly.
18. A realtor offers a house for $1,500 cash and $50 a month for 10 years,
without interest. If money is worth 6% effective, what is the equivalent cash
price?
19. What is the present value of an annuity of $840 a year in quarterly
installments for six years: (a) if money is worth 4% nominal, convertible quar-
terly; (6) if money is worth 4% nominal, convertible semiannually?
20. How many years will it take to accumulate $785 if $100 is invested at
the end of each year at 4^%?
CHAPTER 32
Special Annuities
Annuity due. An annuity due is an annuity the periodic pay-
ments of which are made at the begining of each period. A com-
parison of the following tables will show the difference between an
annuity due and an ordinary annuity; it will be remembered that
the payments of an ordinary annuity are made at the end of eacli
period.
(1)
(2)
Hogi lining of 1st >r. Contract made
2nd
3rd
4th
5th
Contract ends
(3)
Rents Payable
in Ordinary
Annuity
None
1st payment
2nd
3rd
4th
(4)
Rents Payable
in Annuity
Due
1st payment
2nd
3rd "
4th "
None
(D
End of
Period
1
2
3
4
5
(2)_
4- Period
impound
Amount
00
(3) (4)^
Amount of 5-Period
Ordinary (Compound
Annuiti/ Amount
1 00 00
(«r>)
A mount of
Annuiti/
Due '
1 06
06
2 06
06
2 1S36
.1236
3 1836
.1236
3 374616
191016
4 374616
191016
4 637093
262477
5 637093
262477
5 637093
Amount of an ordinary annuity for 5 periods. . . 5 637093
Deduct 1 000000
Amount of an annuity due for 4 periods . 4 637093
To find the amount of an annuity due. From the above tabu-
lations it can be seen that the amount of an annuity due of 1 for 4
periods is 1 less than the amount of an ordinary annuity for 5
periods.
349
350
SPECIAL ANNUITIES
Totfd Value,
First
Period
1
Second Third
Period Period
Fourth
Period
End of Last
Period
1 262477
1
1 191016
1
1 1236
1
1 06
ANALYSIS OF THE AMOUNT OF AN ANNUITY DUE
Assumed Periods
Rent, beginning of 1st yr
" 2nd "
" " " 3rd "
" 4th "
Knd of 4th year, annuity due 4 637093
As shown above, each periodic rent draws interest from the
begining of the year in which it is deposited, and continues to
draw interest until the due date. The sum of these periodic
amounts is the amount of the annuity due.
Since each payment of an annuity due is made at the beginning
of a period, the final result is that the amount of an annuity due
exceeds the amount of an ordinary annuity by the interest on the
Amount of the ordinary annuity for one period.
The amount of an annuity due, represented by the symbol S',
*iay be found in either of two ways:
(1) By adding the interest for one period to the amount of an
ordinary annuity for the number of periods specified.
(2) By finding the amount of an ordinary annuity for one pay-
ment more than the number of payments specified, and then
deducting one payment or rent from the total.
First method. Procedure: (a) Compute the amount of an
ordinary annuity at the given rate per cent and for the given rents.
(6) Multiply the amount found in («) by 1 plus the rate of
interest per period expressed decimally.
Example
Annuity payments of $100 are to be made at the beginning of each year for
4 years. Money is worth 6%. What is the amount of the annuity?
Formula
A rith mctical Substitution
Solution
(1.06)4 = 1.262477, compound amount of 1 at 6% for
4 periods
1.262477 - 1 = .262477, compound interest on 1 at 6% for
4 periods
.262477 -T- .06 = 4.374616, amount of an ordinary annuity of
1 nt 6% for 4 periods, or $4 6%.*
The value of «4 fi% may be found in Table 4, page 529.
SPECIAL ANNUITIES 351
100 X 4.374616 = 437.4616, amount of an ordinary annuity of
100 at 6Vo for 4 periods
$437.4616 X 1.06 = §463.71, amount of an annuity due of $100
} 'erijicatwn
Beginning of 1st period:
Contract made.
Rent ... ................... $100.00 $100 00
Beginning of 2nd period:
Rent . ............... 100 00
Interest on $100 at 6 % ............ 6 00 106.00
New principal ................. $206 00
Beginning of 3rd period:
Rent . . . . ................ $100 00
Interest on $206 at (>% ... . 12 36 112 36
New principal ....... .... $318.36
Beginning of 4th period:
Rent ........ . $100 00
Interest on $31S.36 at 6% J9 10 JJO 10
New principal . . $437 46
End of 4th period:
Interest on $437.46 . . 26 25
Amount due . .... $463 71
Second method. Procedure: (a) Determine the amount of an
ordinary annuity of 1 for the required number of periods plus 1.
(ft) Deduct 1 from the result found in (a).
(r) Multiply the difference found in (ft) by the number of
dollars in each periodic rent, and this product will be the amount
of the annuity due.
Formula
A rithmctical Substitution
Solution
(1.06)* = 1.338225, compound amount of 1 for 5 periods
at 6%
1.338225 - 1 = .338225, compound interest on 1 for 5 periods
at 6%
.338225 -v- .06 = 5.63709, amount of an ordinary annuity of 1 for
5 periods at 6 %, or s^% from Table 4, page 529.
5.63709 - 1 = 4.63709, amount of an annuity due of 1 for 4
periods at 6%
$100 X 4.63709 = $463.71, amount of an annuity due of $100 for
4 years at 6%
352 SPECIAL ANNUITIES
Problems
Prepare the formula, solution, and verification for each of the following:
1. An annuity contract calls for the payment of $1,000 at the beginning of
each year for 5 years. Money is worth 6%, interest compounded annually.
What is the amount of the annuity at the end of the fifth year?
2. For 5 years a man deposits in the bank $150 on the first of each quarter.
The bank allows him 4% interest, compounded quarterly. What will be the
amount of his savings in the bank at the end of the 5-year period?
3. The X.Y.Z. Company deeds a house and lot to Smith. In return, Smith,
is to deposit with the company, over a period of 10 years, $200 at the beginning
of each six months. The rate of interest is to be 6%, compounded semiannually.
What will be the amount of the accumulation at the end of the 10-year period?
Present value of an annuity due. The present value of an
annuity due, represented by the symbol A', is the present or actual
cash value, at the date of the first payment, of all the payments to
be made under the annuity contract. The following table shows
the present value of an annuity due for which payments of 1 are to
be made for 4 years; interest is calculated at 6%, compounded
annually :
Rent, beginning of first period.
Total.
Value at
Beginning
of Period
1 000000
First
Period
I
Second
Period
Third
Period
Fourth
Period
043390
1.
8X9996
1.
.839619
1
3 673011
Comparison of present value of an ordinary annuity and that
of an annuity due.
Example
the terms of an annuity
be made.
Example
Under the terms of an annuity due, four annual payments of $1 each aie to
made. Interest is at 6%. Find the present value of the annuity.
TABLE OF COMPARISON OF PRESENT VALUE OF AN ORDINARY
ANNUITY AND AN ANNUITY DUE
(I)
(2)
(3)
(4)
Four Rents,
Three Rents,
Number of
Ordinary
Ordinary
Annuity Due
Rents
Annuity
Annuity
of Four Rents
1
.943396
943396
1 000000
2
1 S33392
1 833392
1 943396
3
2 673011
2 673011
2 833392
4
3 465105
3 673011
SPECIAL ANNUITIES 353
To find the present value of an annuity due. This may be
done in two ways:
(1) If a number in column 2 of the above table is multiplied by
1 plus the rate per cent, (1.06), the product will be the number in
column 4 corresponding to the same number of rents; thus:
3.465105 X 1.06 = 3.673011, the present value of an annuity duo
at 6% for 4 periods.
(2) If 1 is added to the present value of an ordinary annuity,
the sum will be the present value of an annuity due of one morv
rent than the ordinary annuity ; thus :
Present value of an ordinary annuity of 3 rents 2.673011
Adding 1 1 000000
Present value of an annuity due for 4 periods 3 67301 1
First method. Procedure : (a) Compute the present value of an
ordinary annuity, using the given number of dollars in the rents,
the given rate per cent, and the given number of periods.
(6) Multiply the present value found in (a) by the rate per cent
per period plus 1.
Example
Under the terms of an annuity due, four annual payments of $100 each are
to be made. Money is worth 6%, interest compounded annually. Find the
present value of the annuity.
Formula
(Ran])(\ + i) = A'
Arithmetical Substitution
L i \
100 1 ^6-J(1.06) = $367.30.
Solution
(1.06)4 = 1.262477, compound amount of 1 at 6% for
4 periods
1 -f- 1.262477 = .7920937, present value of 1 at 6% for 4
periods
1 - .7920937 = .2079063, compound discount on 1 at 6% for
4 periods
.2079063 -f- .06 = 3.465105, present value of an ordinary annuity
of 1, or o^, from Table 5, page 532.
100 X 3.465105 = 346.5105, present value of an ordinary annuity
of 100
$346.5105 X 1,00 - $367.30, present value of an annuity due of
$100
354 SPECIAL ANNUITIES
Verification
Beginning of first period:
Present value $367 . 30
Rent deducted $100.00 100.00
New principal $267 . 30
Beginning of second period:
Rent 100.00
Less interest on $267.30 16 04
Balance to apply on principal 83 . 96
New principal $183734
Beginning of third period:
Rent $100.00
Less interest on $183.34 11 00
Balance to apply on principal 89 00
New principal $94 . 34
Beginning of fourth period:
Rent $100 00
Less interest on $94.34 5 66
Balance to apply on principal 94 34
Second method. Procedure: (a) Compute the present value of
an ordinary annuity of 1 at the required rate and for one less than
the required number of periods.
(6) To the present value of the annuity found in (a)- add 1, and
the result is the present value of an annuity due of 1 for the
required number of periods.
(c) Multiply the present value of an annuity due of 1 by the
number of dollars in each rent.
Formula
R(a--\. + 1) = A'
Arithmetical Substitution
A L_ \
100 \ y" + 7 = $367-30-
Solution
(1.06)4-1, or (1.06)3 = 1.191016, compound amount of 1 for 3
periods at 6%
1 ^ 1.191016 = .839619, present value of 1 for 3 periods
at 6%
1 - .839619 = .160381, compound discount on 1 for 3
periods at 6%
.160381 -4- .06 = 2.6730, present value of an ordinary annuity
of 1 for 3 periods at 6%, or a^|ft% from
Table 5, page 532.
2.6730 + 1 = 3.6730, present value of an annuity due of
1 for 4 periods at 6%
$100 X 3.6730 = $367.30, present value of an annuity due
of $100 for 4 years at 6% '
SPECIAL ANNUITIES 355
Problems
1. What is the present value of an annuity due in which $100 payments
are to be made on the first day of each six months for 10 years, if money is worth
5%, interest compounded semiannually? Prepare formula, solution, and
verification.
2. The rents of an annuity due are $500 each, and are payable semiannually
for 10 years. If money is worth 6%, interest compounded semiannually, what
is the value of the annuity at the date of the payment of the first rent? Prepare
formula, solution, and verification.
3. A contract provides for the payment of $150 on the first of each quarter
for a period of 10 years. Interest is 4%, compounded quarterly. What is the
present value of the contract?
4. What is the present value of a contract which calls for the payment of
$50 on the first of each month for a period of 10 years, interest to be computed
monthly at 6% per annum?
5.* A is considering two propositions for the investment of $75,000 belonging
to an estate. The first proposition offers him six 7% notes maturing as follows:
On July 1 , 1943 $ 5,000
On July 1, 1945 . 5,000
On July 1, 1947 . 5,000
On July 1, 1949 5,000
On July 1, 1950.. . . . 5,000
On July 1, 1951 .... .... . . 50,000
Total $75,000
The second proposition offeis him two 5% notes, maturing as follows:
On July 1, 1946 $25,000
On July 1,1951 60,000
Total $85,000
In each case the loan is adequately secured, and the interest is payable
semiannually; each proposition is offered to A for $75,000 in cash on July 1,
1941.
A requests you to determine which proposition is the better one for him to
accept. State your findings, and demonstrate the correctness of your answer.
Given: The present value of 1, ten periods hence at 3^%, is .708919.
Rents of the amount of an annuity due. The rents may be
found by the following procedure :
Procedure: (a) Use the Second Method (see page 351) to com-
pute the amount of an annuity due of 1 for the required number of
periods and at the required rate per period.
(6) Divide the given amount of the annuity due by the amount
of an annuity due of 1, as computed in (a), to find the rent.
* Adapted from 0. P. A., Illinois.
356 SPECIAL ANNUITIES
Example
The rents of an annuity due are paid at the beginning of each yeai for 4 years.
At the end of the fourth year the amount of the annuity is $100. Interest is
at 6%, compounded annually. Compute the rents.
Formula
P
= R'
A rith mrtica I tiuhstit u tion
100
.06
Solution
(1.06)6 = 1.338225, compound amount of 1 for 5 period'*
at 6%
1.338225 — 1 = .338225, compound interest on 1 for 5 periods
at 6%
,338225 -T- .06 = 5.6371, amount of an ordinary annuity of 1 for 5
periods at 6%, or s^6% from Table 4, page 529
5.6371 — 1 = 4.6371, amount of an annuity due of 1 for 4 periods
at 6%
$100 -r- 4.6371 = $21.57, rent of the amount of an annuity due of
$100 for 4 years at 6%
Verification
Beginning of first year:
Rent $ 21 57
Beginning of second year:
Rent $21.57
Interest on $21.57 at 6% L29 22 86
Amount $ 44 43
Beginning of third year:
Rent $21.57
Interest on $44.43 at 6% 2.67 24J24
Amount $~68767
Beginning of fourth year:
Rent $21 57
Interest on $68.67 at 6% 4. 12 25 69
Amount $ 94 36
End of fourth year:
Interest on $94.36 at 6% 5 66
$100 02
Problems
1. What are the rents of an annuity due which amount to $4,762.40 in
10 years, if money is worth 6% per annum? Prepare formula, solution, and
verification.
2. Mr. Ames desires to know how much he must deposit in the bank on the
first day of each six months' period for 10 years, in order that at the end of the
SPECIAL ANNUITIES 357
tenth year he may have $12,000 accumulated. The bank pays 4%, interest
compounded semiannually. What is the semiannual deposit required? Pre-
pare formula, solution, and verification.
3. Brown wishes to save $8,000 by making equal deposits on the first of each
quarter for 10 years. His bank allows him 4%, interest compounded quarterly.
What is the quarterly deposit required?
4. A wishes to have $5,000 saved at the end of 3 years. He decides to make
equal deposits on the first of each month. His bank allows him 6%, interest
compounded monthly. What is the monthly deposit required?
Rent of the present value of an annuity due. The rent of the
present value of an annuity due may be found by the following
procedure:
Procedure: (a) Use the Second Method (see page 354) to com-
pute the present value of an annuity due of 1 for the required
number of periods and at the given rate per cent.
(6) Divide the given present value of the annuity due by the
present value of an annuity of 1, as computed in (a).
Example
A man owes $5,000. He wishes to know the size of each of ten equal annual
payments which will exactly cancel the debt, and pay the accrued interest due
on the date of each payment. The first payment is to be made at once. Money
is worth 0% per annum.
Form ula
P
A rithmetical Substitution
i .o«)Vo '
Solution
(1.06)9 = 1.6S9479, compound amount of 1 for 9 periods
at 6%
1 -5- 1.689479 = .591898, present value of I for 9 periods at 6%
1 — .591898 = .408102, compound discount on 1 for 9 periods
at 6%
.408102 -T- .06 = 6.8017, present value of an ordinary annuity for
9 periods at 6%, or a9;6% from Table 5,
page 532.
6.8017 + 1 = 7.8017, present value of an annuity due for
10 periods at 6%
$5,000 -*- 7.8017 === $640.89, present value of an annuity due of
$5,000
358 SPECIAL ANNUITIES
Problems
1. Give the formula, solution, and verification for each of the following:
Debt to Be
Retired Payments Rents Per Cent Compounded
(a) $1,000 00 4 ........ 6 Annually
(6) 2,7(32 50 8 ........ 4 Semiannually
(c) 4,875 40 16 ........ 6 Quarterly
2. A purchases a farm for $15,000, with interest at 6%. He desires to pay
'or it in twelve equal annual payments, the first payment to be made immediately.
What is the amount of each payment necessary to cancel the debt and interest?
Effective interest on an annuity due. When the interest is
compounded more often than the payments are made, the effective
rate should be found for a period corresponding to the period of an
annuity payment.
Procedure: (a) Reduce the given interest rate to an effective
rate for a period corresponding to the period of an annuity pay-
ment.
(b) Using as the rate for each period the effective rate found in
(a), proceed by following the instructions given in preceding pages
of this chapter.
Example
What is the amount of an annuity due, the annual payments of which are
$100 for 3 years, and the interest on which is 0%, compounded quarterly?
PAHT 1
Formula Arithmetical Substitution
+ -
-) - l = Effective rule ( 1 + -'^-J - 1 = .0613635
Solution to Part 1
.06 -r 4 = .015, quarterly rate
1 H- .015 = 1.015, ratio of increase for 1 quarter
(1.015)4 = 1.0613635, compound amount of 1 for 4 periods
1.0613635 — 1 = .0613635, effective interest for an annuity pay-
* ment period
PART 2
Formula
((I _i_ f\«+i __ i \
1 1 j «= Amount of annuity due
An
1W(»
Arithmetical Substitution
. 0613635)' - 1
-0613635
SPECIAL ANNUITIES 359
Solution to Part 2
Substituting for i the rate found in Part 1, the solution is:
(1.0G13635)4 = 1.2689855, compound amount of 1 for 4
periods at '• ' >•'••{"»
1.2689855 — 1 = .2689855, compound interest on 1 for 4
periods at i'».i.{« ».{.">• t
.2689855 -J- .0613635 = 4.3834, amount of an ordinary annuity of
1 for 4 periods at 6.13635%
4.3834 - 1 = 3.3834, amount of an annuity due for 3
periods at •'» 1 3« ),{,"> /
$100 X 3.3834 - $338.34, amount of annuity due of $100
for 3 years
Problems
1. What will be the amount at the end of 4 years of an annuity the rents of
which are $1,000 payable at the end of each year, and the nominal rate 4%,
interest compounded semiannually? Give formula, solution, and verification.
2. A company desires to know how much will be accumulated at the end
of 5 years if $5,000 is placed in a sinking fund at the end of each year. The
fund bears a nominal rate of 6%, interest compounded semiannually. Give
verification.
3. The R-M Company desires to accumulate a sinking fund of $50,000 to
meet a bond issue of that amount which is due in 10 years. What will be the
annual payments made on the last day of each year, if the fund is so placed
that it will bear an annual rate of 6%, interest to be compounded semiannually?
Construct columnar table showing payments, interest, addition to fund, arid
amount of fund.
4. The R.P.S. Company has a bond issue of $50,000 coming due at the end
of 5 years. The directors desire to know how much money must be placed in
the bank at the end of each year in order that this fund may exactly cover the
bond issue at the end of the required time. The bank pays 4%, interest com-
pounded quarterly. Construct a columnar table, as in the preceding problem.
5. The annual rents of an ordinary annuity are $500, the time is 6 years, and
money is worth 6%, interest compounded serniannually; what is the present
value? Construct a columnar table.
6. What is the present value of a contract in which a company agrees to
pay $500 at the beginning of each year for 5 years, interest to be allowed at the
rate of 5% per annum, compounded quarterly? Construct a columnar table.
Deferred annuity. A deferred annuity is an annuity in which
a number of periods are to expire before the periodic payments or
rents are to begin.
The amount of a deferred annuity is the same as the amount of
one which is not deferred, since no payments are made until after
the time of deferment has expired.
The present value of a deferred annuity is the value at the
beginning of the period of deferment.
There are two methods of computing the present value of a
360 SPECIAL ANNUITIES
deferred annuity. In each of these two distinct operations are
necessary.
First method. Procedure: (a) Determine the present value of
an ordinary annuity of 1 at the given rate and for the number of
periods corresponding to the number of rents.
(6) Multiply the present value of 1 found in (a) by the number
of dollars in each rent.
(c) Multiply the present value of the annuity found in (6) by
the present value of 1 for the number of deferred periods.
Example
Find the present value of an annuity contract which calls for four equal
annual payments of $100 each, the first rent to be paid at the end of the seventh
year. Interest is to be calculated at 6%.
An analysis shows that this is an ordinary annuity for 4 years, deferred for
6 years.
Formula
Let: n = the number of rents
m = the number of deferred periods
Ran\t - vrn = Present value of deferred annuity.
A rithmetical Substit ution
100
1 (1.06V
.06
P-l -
L(1.06)'J
$244.28.
Short Solution, Part 1
1 -r- 1.262477 = .7920937, present value of 1 due at the end of
4 years at 6%
1 - .7920937 = .2079063, compound discount on 1 due at the
end of 4 years at 6 %
.2079063 -f- .06 = 3.465105, present value of annuity of 1 for 4
years at 6 %, or a^6% from Table 5, page 532
$100 X 3.465105 = $346.51, present value of annuity of $100 for
4 years
Short Solution, Part 2
(1.06)6 = 1.418519
1 -^ 1 .418519 = .7049605, present value of 1 due at the end of
6 years at 6%
$346.51 X .704605 = $244.28, present value of deferred annuity
Verification, Part 1
Present time:
Present value of deferred annuity $244 28
Multiply by (1.06)6 1 418519
End of deferred period:
Value at end of deferred period $346.51
SPECIAL ANNUITIES 361
End of 1st period:
Rent $100 00
Interest on $346.51 at 6% 20 79
Amortization 79 . 21
Balance $267~30
End of 2nd period:
Rent $100 00
Interest on $267.30 at 6 % 1 6.04
Amortization 83 96
Balance $183.34
End of 3rd period:
Rent $100 00
Interest on $183.34 at 6% 1 1 00
Amortization 89 00
Balance $ "94" 34
End of 4th period:
Rent $100 00
Interest on $94.34 at 6 % 5 66
Amortization 94 34
Second method. Procedure: (a) Compute the present value of
an ordinary annuity of 1 at the given rate per cent, for a number of
periods equal to the sum of the periods of the annuity and deferred
periods, or n + m periods.
(b) Compute the present value of an ordinary annuity of 1, at
the given rate and for a number of periods equal to the deferred
periods, or m periods.
(c) Find the difference between the present values found in (a)
and (6).
(d) Multiply the difference found in (c) by a number equal to
the number of dollars in each rent.
Formula
K(<ln+m\i — amlt) — Present value of deferred annuity.
A rith m ctica I S u bstit ution
1
100
1
= $244.28.
Solution
(LOG)10 = 1.790847, compound amount of 1 at 6%
for 10 periods
1 -v- 1.790847 = .5583948, present value of 1 at 6% for 10
periods
1 — .5583948 = .4416052, compound discount on 1 at 6%
for 10 periods
362 SPECIAL ANNUITIES
.4416052 -f- .06 = 7.360087, present value of annuity of 1 at
6% for 10 periods, or a10 6% from
Table 5, page 532.
(1.06)6 = 1.418519, compound amount of 1 at 6%
for 6 periods
1 -T- 1.418519 = .7049605, present value of 1 at 6% for
6 periods
1 - .7049605 = .2950395, compound discount on 1 at 6%
for 6 periods
.2950395 4- .06 = 4.917324, present value of annuity of 1 at
6% for 6 periods, or a6l6% from Table 5,
page 532.
7.360087 - 4.917324 = 2.442763, difference in present value of
annuities of 1 for 10 periods and 6 periods
$100 X 2.442763 =* $244.28, present value of deferred annuity
of $100
Probably most of the difficulties in the solution of problems
such as the above arise from failure to make a complete analysis
before beginning the work. Problems of this type may be analyzed
and solved in various ways. Be sure to see each problem in all its
parts before attempting to calculate the amounts.
Problems
Find the present value of each of the following deferred annuities. (NOTE:
If interest is to be compounded semianrmally, quarterly, or monthly, this same
condition usually prevails during the period of deferment.)
Number of Years
Payments
Payments Made
Kate
Rents
Deferred
1.
$100
Annually
5%
5
5
2.
$500
Annually
4%
6
4
3.
$250
Semiannually
4%
10
5
4.
$200
Quarterly
6%
16
3
6.
$100
Quarterly
6%
12
5
6.
$ 50
Monthly
6%
48
3
7. What is the present value of an annuity contract in which the A.B.
Company agrees to pay to Mr. Ladd a monthly installment of $40 for 10 years,
the first payment to be made 10 years hence? Money is worth 6%, interest
convertible monthly during the annuity period, and annually during the deferred
period.
8. A company is issuing $100,000 of 4%, 20-year bonds, which it wishes to
pay at maturity by means of a sinking fund in which equal annual deposits are
to be made. The board of directors wishes to assume that this fund will earn
5% interest for the first 10 years, and 4% for the last 10 years. What is the
annual deposit required?
Given:
5% 4c/o
810 12.578 12 006
(1 -f t)10 1 629 1 .480
SPECIAL ANNUITIES 363
9. A lease has 5 years to run at $1,200 a year, with an extension for a further
> years at SI, 500 a year. The payments are due at the end of each year. If
noney is worth 59o, what should be the sum paid now in lieu of the 10 years'
•ent?
10.* On December 31, 1943, A is indebted to B in the following amounts:
$1,500, due December 31, 1944, without interest
$3,500, due December 31 , 1946, with interest at the rate of 6% pay-
able annually
$5,000, due December 31, 1948, with interest at the rate of 6% from
December 31, 1943, not payable until maturity of note but to be
compounded annually
$6,000, due December 31, 1949, with interest at the rate of 5% pay-
able annually
On this date (December 31, 1943), .4 learns that on December 31, 1947, he
vill fall heir to $200,000, and he arranges with B to cancel the four notes in
exchange for one note due in 4 years. It is then agreed that the new note shall
nclude interest to maturity calculated at 5%, compounded annually, and that B
ihall not lose by the exchange.
What will be the amount of the new note?
Given at 5%:
i'1 = .9523X10 (1 -hi)1 = 1.0500000
v2 = .9070295 (1 -f i)2 = 1.1025000
i'3 = .S63X376 (1 + i)3 - 1.1576230
vb = .7S35262 (1 + i)4 - 1.2155062
Given at 6%:
(1 + i)3 = 1.1910160
(1 + i)b = 1.3382256
11.* On January 1, 1938, A leased a building to B for the period ending
3ecember 31, 1952, at an annual rental of $7,000 payable annually in advance.
Subject to this lease, A leased the same property to C on January 1, 1944, for a
,erm of 50 years at an annual rental of $10,000, payable annually in advance,
7 to receive the rental of $7,000 payable by B during the remainder of B's lease.
?or this lease C paid to A an additional $1,500 as a bonus.
Omitting all consideration of income tax questions, how should the various
tccounts appear on C"s books if he calculates interest on the investment at 6%
>er annum?
Given at 6%:
(1 _f i)» = 1.689479 v» = .591899
(1 -|- {)io = 1.790848 v10 = .558395
(1 + i)41 = 10.902861 v41 = .091719
12. f The Belgian pre-armistice debt to the United States amounted to
>1 7 1,800, 000. The settlement provided that no interest was to be charged on
his part of the war debt and that graduated payments on account of principal
vere to be made, totaling $9,400,000, by June 15, 1931, the balance to be payable
it the rate of $2,900,000 per annum for 56 years.
Assuming an interest rate of 3% per annum, calculate the loss to the United
States by the waiving of interest calculated at June 15, 1931.
* Adapted from American Institute Examination,
f American Institute Examination.
364 SPECIAL ANNUITIES
The present wlue of $1.00 at 3% due in 56 years is $0.1910361, and in 56
years $1.00 at 3% will amount to $5.2346131.
13. A note, the face value of which is $1,000, bears interest at the rate of
8% per annum, and is payable in monthly installments of $25, including interest.
It is desired to discount this note at the bank, so that the bank shall have an
effective rate of 12% per annum. What is the amount of the discount to be
deducted by the bank?
Perpetuity. A perpetuity is defined as a series of periodic pay-
ments which are to run indefinitely. This form of annual pay-
ment is most often found as the effect of the establishment of an
endowment fund, the rents of which are to be used for a special
purpose. As the endowment fund is never to be returned, the
amount has no meaning, but its present value is the value of the
periodic payment divided by the prevailing rate of interest.
The present value of a perpetuity is denoted by aw.
Procedure: Divide the periodic rent by the current rate
per cent expressed decimally, to find the present value of the
perpetuity.
Example 1
It is desired to establish a fund the annual income from which, at 6%, will
be $300. Find the present value of the perpetuity.
Formula Arithmetical Substitution
*f -A. ^ - W.OOO
Solution
$300 -5- .06 = $5,000, the amount of the endowment fund
Example 2
Find the present value of a perpetuity of $50 a month at 6% nominal con-
vertible monthly.
Solution
S - $10'000
Example 3
What is the present value of a perpetuity of $500 a year, if money is worth
4% convertible semiannually?
Solution
- $12'376-24
Perpetuities payable at intervals longer than a year. In this
case the annual interest on the principal will accumulate during
the interval to equal the payment due at the end of that time.
SPECIAL ANNUITIES 365
Example
What is the present value of a perpetuity of $1,000 every 5 years, if interest
is at 4% a year?
Solution
$1,000 1
•04
= $25,000 X 0.1846271 = $4,615.68
Problems
1. If a farm produces a net annual income of $3,600, what is its present value
if money is worth 5%?
2. What is the present value of a perpetuity of $3,500 payable every 5 years,
if money is worth 4% convertible semiannually?
3. Find the present value of a perpetuity of $250 a year if money is worth 6 %.
4. What is the present value of a perpetuity of $(>00 a year if money is worth
4% convertible quarterly?
5. Find the annual rent of a perpetuity whose present value is $15,000, if in-
terest is 5% a year.
6. Find the present value of a perpetuity of $10,000, payable every 5 years,
if money is worth 4^% a year.
Review Problems
1. Brown sets aside $500 at the beginning of each year to provide for his
daughter's college education. If he invests the money at 3% effective, what
will be the amount following the 10th payment?
2. Find the present value of a premium of $27.42, assuming that the insured
will live to pay 20 premiums, money being worth 4% effective.
3. White bought a house, agreeing to pay $1,000 down and $500 each six
months until he paid $8,000. If money is worth 5% effective, what should be
the cash price of the house?
4. Find the present value of an annuity of $600 a year for 10 years, if
money is worth 4% effective: (a) deferred 5 years, (b) deferred 8 years.
5. If you deposit $150 in a savings bank at the beginning of each quarter
and the bank pays 3% nominal convertible quarterly, how much will you have
to your credit at the end of 5 years?
6. Cole, at age 22, takes a 20-payment life insurance policy of $1,000 on
which the premium is $27.42 payable at the beginning of each year. If he
should die at the end of 10 years just before the eleventh premium is due, by
how much would his estate be increased by having taken the insurance instead
of having put the premiums into a savings bank paying 2% effective?
7. Find the present value of an annuity of $1,000 a year to be paid in
quarterly installments for 12 years, deferred for 5 years, if money is worth 3%
effective.
8. An insurance premium of $64.50 is payable semiannually in advance for
20 years. If interest is at 3 % convertible semiannually, find the amount of the
payments at the end of 20 years.
366 SPECIAL ANNUITIES
9. Attached to a $100 bond are coupons worth $2.50 each six months for
the next 20 years. What is the present value of these coupons, assuming money
to be worth 4% effective?
10. How long will it take to accumulate $10,000 by investing $500 each six
months at 3% convertible semiannually?
11. What rate of interest compounded monthly is equivalent to 6% com-
pounded quarterly?
12. Find the annual payment required to accumulate $3,000 in 10 years,
when money is worth 3% convertible semiannually.
13. If it takes an orchard 6 years to reach profitable maturity, and for a
period of 20 years it is expected to yield a net income of $3,500 a year, what is
the cash value of the orchard if money is worth 4% effective?
14. Find the present value of an annuity due of $250 a year payable annually
for 6 years, if money is worth 5%.
15. Find the present value of an annuity of $75 a month for 8 years, 6 months,
if money is worth 6 % nominal converted monthly.
CHAPTER 33
Bond and Bond Interest Valuation
Definitions. A bond is a promise to pay a specified sum of
money at a de terminable future date. It differs from a note, in
general, in that it is usually a long term obligation. Bonds are
generally issued by a city, state, nation, or corporation, and seldom
by individuals. The sum written in the body of the instrument is
known as the face, par, or nominal par, and is the amount on which
interest is calculated. Bonds usually have a par of $100, $500, or
$1,000.
Interest payments may be made annually, semiannually, or
quarterly. These interest payments are known as nominal inter-
est, or cash interest.
The rate per cent earned on the actual money invested is
termed the effective or investment interest. The cash rate and
the effective rate are not the same unless the bond is purchased at
par. Quotations are sometimes made "on a basis/7 which means
at an effective rate on the money invested. The price of a bond
will usually be either above or below par; because the cash rate is
either above or below the effective rate.
A bond is said to be "redeemed" when it is bought back by the
company which issued it. If the market price of a bond is more
than the par value, the bond is said to be above par; or if less, the
bond is said to be below par. Bonds sold above par are said to be
sold " at a premium," and if sold below par, they are said to be sold
"at a discount."
Bonds are usually sold on the open market for whatever they
will bring. Whether they bring more or less than par depends
upon:
(1) The cash or coupon rate of interest.
(2) The redemption price.
(3) The current rate of interest.
(4) The length of time until maturity.
(5) The character of the security.
Only the first four of the above can be mathematically con-
sidered.
368 BOND AND BOND INTEREST VALUATION
Bonds sold at par. It is apparent that if a man desires to buy
a certain bond which has a face value of $1 ,000, and this bond bears
a cash rate of interest exactly the same as that which his money is
worth on the market for other securities of the same general class,
he will be willing to pay $1,000 for the bond. In such a case, the
interest that he will receive from this bond will be equivalent to
the interest that he would receive from any other investment of the
same general class. But if he considered the purchase of another
bond of $1,000, which had a cash rate of interest lower than that
of other investments of the same general class, it is apparent that
he would not pay a full $1,000 for this bond. And again, if there
were on the market securities of the same general class paying a
higher rate per thousand, he would be willing to pay more than tln»
par value for a bond of this class.
This may be illustrated graphically, as follows:
5%
4%
^ \
Cash or Bond
Rate of Interest
Par of Bond
1. Cash or bond rate of interest has a tendency to lift the price of the bond
above par.
2. Effective rate of interest has a tendency to bring the price of the bond
below par.
a. This is an illustration of an equal pull of cash and effective interest.
If all other factors were equal, the result would be a price at par.
6. If the cash rate is larger than the effective rate, the price should be
above par.
c. If the effective rate is larger than the cash rate, the pull is below par.
Bonds purchased at a discount or at a premium. If a bond is
purchased at a discount, the effective rate io higher than the cash
rate, for two reasons: (a) the investment is less than par; and (6)
the investor makes additional income equal to the difference
between the cost and the par to be collected at maturity. On the
other hand, if a bond is purchased at a premium, the effective rate
is lower than the cash rate, for two reasons : (a) the investment is
BOND AND BOND INTEREST VALUATION
369
more than par; and (6) the investor loses the difference between
the price paid and the par to be collected at maturity.
Price and rate of yield. In business, the usual questions
which arise with regard to bonds are: "What price should be paid
for bonds if a certain investment rate of interest is desired by the
investor?" and "What rate will a bond, purchased at a certain
price, produce?"
Use of bond tables. In the calculation of the cost, bond tablevS
are generally used. They are reproduced below in order that the
student may familiarize himself with their use.
Bond table, first form. The following is a very simple form of
bond table, in which only values, effective rates, and nominal rates
are shown:
TABLE OF BOND VALUES
FIVE YEARS, INTEHEST PAYABLE SEMIANNUALLY
.
Xowinal Rate
Effective Rate
3
3i
4
4i
5
6
7
"4
89 20
91 30
93 52
95 08
97 84
102.16
106.48
5 )i
SS.70
90 85
93 00
95.10
97 31
101.61
105 92
ST
XX 20
90 34
92 49
94.63
96 78
101.07
105 37
54
87 . 70
89 84
91 98
94.12
96 26
100.53
104 81
6
87 20
89 34
91.47
93.60
95 73
100.00
104 27
Example
What is the purchase price of a 5-year, 5% bond (interest payable .semi-
it nnually), bought on a (}'/0 basis?
Referring to the table, the answer is found to be $95 73.
Problems
1. A $1,000 bond due in 5 years bears interest at 4%, payable semiannually,
and is bought on a 5^-% basis. What is the purchase price, or value?
2. A $1,000 bond due in 5 years bears interest at 6%, payable semiannually,
and is purchased on a 5^% basis. State the value of the bond.
3. What price can be paid for city bonds of $1,000 each, due in 5 years,
interest at 5%, payable semiannually, bought on a 6% basis?
4. If John Jones bought one 4^%, $1,000 bond for $946.30, what per cent
may he expect on his investment?
5. The coupon rate is 4jr%; the number of bonds is 5; the purchase price is
$4,680.00. What per cent is made on the investment?
Bond table, second form. The most common form of bond
table shows the "Price" as indexed between the "Nominal" and
the "Effective" rates. A new and very efficient form of bond
370 BOND AND BOND INTEREST VALUATION
table is that used by Johnson, Stone, Cross, and Kircher in their
volume, Yields of Bonds and Stocks (New York, Prentice-Hall,
Inc., enlarged edition, 1938). In this form of table the "Effec-
tive" rate is indexed between the " Nominal" rate and the "Price."
An additional feature is the showing of the cash rate of return.
A page from Yields of Bonds and Stocks is reproduced on page 371.
Example
What will be the rate of yield of a bond purchased at $97.25, if it bears 5%
interest, payable semiannually, and matures in 5 years?
Procedure: (a) Turn to the 5% table.
(6) Find the column headed 5 years.
(c) In the column at the left, find the price $97.25.
(d) Find the rate at the intersection of the year column and the price line.
In this case it is 5.639%.
Interpolating in bond tables. It frequently happens that a
given yield or a given price is not listed in a bond table. In such
cases, however, the desired price or yield rate can be obtained by
interpolation.
Example
What price can an investor pay for a bond due in 5 years, if he wishes to
obtain a yield of 6% and the bond bears interest at 5%, payable semiannually?
Procedure: (a) Turn to the 5% table.
(b) Find the column headed 5 years.
(c) Select from this column two effective rates, one just above and the other
just below the rate desired.
(d) In the price column at the left, select the prices corresponding to the
rates selected in (c).
(e) The rate of yield is obtained by interpolating between the yields of the
prices found in (c).
Solution
When yield = 6.057, the price is $95.50
" " = 6.000, " " "
" " = 5.996, " " " 95.75
6.057 - 5.996 = 0.061
6.057 - 6.000 = 0.057
$95.75 - $95.50 = $.25
Since a difference of 0.061 in the rate of yield means a difference of 25^ in the
price, a difference of 0.057 in the rate of yield will mean a difference of fy of .25;
hence the price sought is fy X .25 greater than $95.50. That is:
|| X .25 = .2336
$95.50 + .2336 = $95.7336 (Answer)
Example
What is the rate of yield on a 5% bond due in 4 years, interest payable
semiannually, if the bond is purchased at 96£?
BOND AND BOND INTEREST VALUATION
371
TABLE OF YIELDS OF 6% BOND
YIELDS IN PER CENT PER ANNUM, CORRECT TO THE NEAREST FIVE TEN-
THOUSANDTHS OP 1%, INTEREST PAYABLE SEMIANNUALLY
Price
3
Years
3J
Years
4
Years
4J
Years
5
Years
5}
Years
6
Years
Current
Income
94
7.262
6.961
6.736
6.561
6 422
6 308
6 213
5.319
94}
7.164
6.876
6.661
6.494
6.361
6.252
6.161
5.305
94*
7.067
6.792
6.586
6.427
6.299
6 19,5
6.109
5.291
94|
6.970
6.708
6.512
6.360
6.238
6.139
6.057
5.277
95
6 873
6.624
6.438
6.293
6.178
6 083
6 005
5.263
95}
6.776
6.540
6.364
6.227
6.117
6 028
5.953
5 249
96^
6 680
6.457
6.290
6.160
6.057
5.972
5 902
5.236
951
6.584
6.374
6.216
6.094
5.996
5.917
5.850
5.222
96
6.489
6.291
6.143
6.028
5.936
5.861
5 799
5.208
96}
6.394
6.209
6 070
5.962
5.876
5 806
5.748
5 195
96 J
6 299
6.126
5 997
5.897
5.817
5.751
5 697
5 181
961
6.204
6.044
5.924
5.832
5.757
5.697
5 646
5 168
97
6.110
5.962
5.852
5.766
5 698
5.642
5 596
5 155
971
6.016
5.881
5.780
5.701
5 639
5.588
5 545
5 141
97 i
5 922
5.800
5.708
5.637
5.580
5.533
5 495
5 128
97J
5.828
5.718
5.636
5.572
5.521
5.479
5.445
5 115
98
5.735
5.638
5.565
5.508
5.463
5 425
5 395
5 102
98}
5 642
5 557
5 493
5.444
5.404
5 372
5 345
5 089
98*
5.550
5.477
5 422
5.380
5 346
5 318
5 295
5 076
981
5.457
5.397
5 351
5.316
5.288
5.265
5 . 246
5.063
99
5.365
5.317
5 281
5.252
5.230
5.211
5 196
5 051
99}
5.274
5.237
5 210
5.189
5.172
5.158
5 147
5 038
99|
5 182
5.158
5 140
5.126
5.115
5.105
5 098
5 025
991
5.091
5.079
5.070
5.063
5.057
5.053
5 049
5 013
100
5.000
5.000
5.000
5.000
5.000
5 000
5 000
5 000
100}
4.909
4.921
4 930
4.937
4 943
4.948
4 951
4 988
100|
4.819
4.843
4.861
4.875
4.886
4.895
4.903
4.975
1001
4.729
4.765
4.792
4.813
4.829
4.843
4.854
4 963
101
4.639
4.687
4.723
4 751
4.773
4.791
4.806
4 950
1011
4.550
4.609
4.654
4.689
4.716
4 739
4.758
4 938
10U
4.460
4.532
4.585
4 627
4.660
4.687
4.710
4 926
1011
4.371
4.454
4.517
4.565
4.604
4.636
4.662
4 914
102
4.282
4 377
4.449
4.504
4.548
4.584
4.615
4 902
102}
4.194
4.301
4 381
4.443
4.493
4.533
4 567
4 890
102^
4.106
4.224
4 313
4 382
4.437
4.482
4.520
4.878
1021
4.018
4.148
4.245
4.321
4.382
4.431
4.472
4.866
103
3.930
4 072
4.178
4.260
4.326
4.380
4.425
4.854
103}
3.843
3 996
4 111
4 200
4.271
4.330
4 378
4.843
103]
3.755
3.920
4.044
4.140
4.216
4.279
4.331
4 831
1031
3.669
3.845
3.977
4.079
4.162
4,229
4.285
4.819
* From Johnson, Stone, Cross, and Kircher, Yields of Bonds and Stocks. New
York: Prentice-Hall, Inc., enlarged edition, 1938.
372 BOND AND BOND INTEREST VALUATION
Solution
When price = $96.75, the yield is 5.924
" " = 96.875, " " "
" = 97.000, " " " 5.852
$97.00 - $96.75 = $.25
$97.00 - 96.875 = $.125
5.924 - 5.852 = 0.072
Since a difference of 25^ in the price means a difference of 0.072 in the rate
of yield, a difference of 12^ in the price will mean a difference in the rate of
yield of i-~ of 0.072; hence the sought for rate of yield is 12^/25 X 0.072 less
than 5.924. That is:
Yield sought = 5.924 - (ljj X 0.072 J
= 5.924 - 0.031)
= 5.888% (Answer)
Problems
By means of the foregoing bond table, supply the missing factor i
the following:
in each of
1.
Par of
Bonds
$ 1,000
Nominal
Price Rate
$ 977 50 5%
Effective Interest
Rate Years Payable
3 Semiannually
2.
1,000
fro/
6 (/ 4
3.
5,000
5,087.50
5%
34
4.
4,000
fyO/
4^ c/ 5
5.
10,000
9,812.50
5%
5i
6.
2.000
5%
STJT % 6
Bond values computed without tables. If no bond tables are
available, the values of bonds sold at a discount or at a premium
may be found by either of two methods of calculation.
Bonds sold at a discount. The discount on bonds is the dif-
ference between the cost (which is below par) and the par value.
Let C = the par value of the bond;
C" = the redemption price when more than par;
r = the rate named in the bond, that is, the coupon or cash interest;
i - the rate of yield expected ;
n = the number of years to maturity;
P = the cost of the bond.
Two methods of calculation are given.
First method. Procedure : (a) Compute the present value of the
par of the bond at the investment rate of interest and for the
number of periods between the date of purchase and the maturity
of the bond, Cvn.
BOND AND BOND INTEREST VALUATION 373
(6) Compute at the effective interest rate the present value of
an annuity for a number of periods equal to the number of bond-
interest or cash-interest payments, the rents of the annuity to be
of the same amount as the periodic bond-interest or cash-interest
payments, rCa~\it
(c) Add the present value of the par of the bond found in (a),
and the present value of the annuity found in (6). The sum will
be the present value of the bond.
Example
What will be the cost of a 5% bond for SI 00, maturing in 4 years, bought so
as to produce 6% effective interest, payable semiannually?
Formula
G> + r(1<rnt = P
A rith mctical 8 ubst it ation
100 1 ' .J + 2.50 \ ,;" = . 196.49.
Xolutioti, Part 1
Finding the present value of the par of the bond:
(1.03)8 = 1. 20(5770 1, compound amount of 1 at 3% for
<S periods
1 -T- 1.2007701 = .7S94092, present value of 1 at 3% for 8 periods
$100 X .7894092 = $78.94, piescnt value of $100 at 3% for 8
periods
Solution, Part '2
Finding the present value of an annuity the rents of which, $2.50, are the
aiime as the periodic interest payments on the bond:
.7S94092 = present value of 1 , as found above
1 - .7894092 = .2105908, compound discount on 1 at 3% for
8 periods
.2105908 -r- .03 = 7.01909, present value of an annuity of 1
$2.50 X 7.01909 = $17.549, present value of an annuity of $2.50
Solution, Part 3
Adding:
$78.94 + $17.55 = $90.49, cost of bond
A convenient and condensed statement showing, for each
period, the carrying value, the accumulation of discount, the
coupon interest, and the effective interest may be made as follows.
374
BOND AND BOND INTEREST VALUATION
End of
Period
1
$2 89
$2 50
2 ....
.... 2 91
2 50
3 ...
. . . 2 92
2 50
4
2 93
2 50
5
2 94
2 50
6
2 96
2 50
7
. . 2 97
2 50
8
2 99
2 50
Effective Coupon Accumulation Carrying
Interest Interest of Discounts Value
$ 96 49
$ 39 96 88
.41 97 29
.42 97 71
.43 98 14
.44 98 58
.46 99 04
47 99 51
.49 100 00
Second method. The theory of this method is that the $2.50
interest received will offset $2.50 of the $3.00 expected; therefore,
the cost of the bond will be the par value of the bond less the
present value at the effective interest rate of an annuity of $.50.
Procedure: (a) Compute the present value of an annuity of 1 at
the effective interest rate for a number of periods equal to the
number of periods that the bond has yet to run, a^.
(b) Calculate the difference between the number of dollars of
income per period at the desired rate and at the cash rate, using as
the basis in both cases the par of the bond, Ci — Cr.
(c) Multiply the present value of the annuity found in (a), by
the difference found in (6); the result will be the discount on the
par value of the bond, (Ci — Cr) • ani.
(d) Deduct the discount found in (c) from the par value of the
bond, and the result will be the price.
Formula
C - [(Ci - Cr) - «nl] = P
Arithmetical Substitution
/. 1 '
100 -
(3.00 - 2.50) '
(1.03T
".03 >
= $96.49
Solution, Part 1
(1.03)8 = 1.2667701, compound amount of 1 at 3% for
8 periods
1 -f- 1.2667701 - .7894092, present value of 1 at- 3% for 8 periods
1 - .7894092 = .2105908, compound discount on 1 at 3% for
8 periods
$.2105908 + .03 = $7.01969, present value of an annuity of 1
Solution, Part 2
3.00 — 2.50 = .50, difference between effective and cash
interest for 1 period
$7.01969 X .50 = $3.51, discount on bond
BOND AND BOND INTEREST VALUATION 375
Solution, Part 3
$100 - $3.51 = $96.49, cost of bond
Verification
First Period :
Cost of bond $ 96. 49
Interest at 3% on $9C .49 $2.89
Coupon interest 2 . 50
Accumulation of discount 39
$ 961J8
Second Period:
Interest at 3% on $96.88 $2 91
Coupon interest 2 50
Accumulation of discount 41
$ 97729
Third Period:
Interest at 3% on $97.29 $2.92
Coupon interest !J. 50
Accumulation of discount 42
$ 97" 71
Fourth Period:
Interest at 3 % on $97.71 $2 93
Coupon interest 2 . 50
Accumulation of discount 43
$ 98714
Fifth Period:
Interest at 3% on $9S.14 $2.94
Coupon interest 2 50
Accumulation of discount .44
$~9~8~58
Sixth Period:
Interest at 3% on $98.58 $2 96
Coupon interest 2 50
Accumulation of discount 46
$ 99~04
Seventh Period:
Interest at 3% on $99.04 $2.97
Coupon interest 2 . 50
Accumulation of discount .47
$~99~5I
Eighth Period:
Interest at 3 % on $99.51 . $2 . 99
Coupon interest . . 2 50
Accumulation of discount _ .49
Par of bond $100 00
Problems
1. What should be the purchase price of a $1,000, 5-year, 5% bond (interest
payable semiannually), bought so that it will produce 6£%? Prove your work
by means of the table.
376 BOND AND BOND INTEREST VALUATION
2. If money is worth 6%, interest payable semiannually, what should be the
purchase price of five $100 bonds, bearing a cash rate of 5%, and having 5 years
to run? Prove your answer by means of the table.
3. Construct in columnar form a table showing the carrying value, the cash
interest, the effective interest, and the amortization of a 5-year, 6% bond of
$500, bought on a 7% basis (interest payable semiannually).
4. A $500 bond, maturing in 6 years and bearing interest at 6%, payable
semiannually, is bought on an 8% basis. Construct a columnar table, as in
problem 3.
6. Show in columnar form the carrying value, the cash interest, the effective
interest, and the accumulation of discount for a 4-year bond, the par value of
which is $1,000. The bond bears 5% interest, payable semiannually; the effec-
tive rate is 6%, convertible semiannually.
Bonds sold at a premium. As stated previously, if the effective
rate is less than the coupon rate, the bond will sell at a premium,
which means that it will be priced above par. When bonds sell at
a premium, part of the money received for each coupon is used to
cancel part of the premium paid for the bond.
As in the case of discount on bonds, two methods of calculating
the price of ,a bond sold at a premium are in common use.
First method. Procedure: The procedure here is the same as
the procedure for the first method of finding the price of a bond
purchased at a discount.
Example
What price should be paid for a $100, 0%, 4-year bond, in order that the
investor may realize 5% on his investment? Interest coupons are payable
semiannually.
Formula
O + Cr - a-]t = P
Arithmetical fiubstit ution
- $103.58.
/
Solution, Part 1
Finding the present value of the face of the bond:
(1.025)8 = 1.2184029, compound amount of 1 at 2^% for
8 periods
1 -T- 1.2184029 - .8207466, present value of 1 at 2£% for 8
periods
$100 X .8207460 = $82.07, present value of face of bond
Solution, Part 2
Finding the present value of the annuity of $3:
1 - .8207466 = .1792534, compound discount on 1 at 2i% for
8 periods
BOND AND BOND INTEREST VALUATION
.1792534 -T- .025 = 7.1701372, present value of an annuity of 1 at
2i9o for 8 periods
?3 X 7.1701372 = $21.51, present value of an annuity of $3
377
Solution, Part 3
$82.07 + $21.51 = $103.58, cost of bond
COLUMNAR TABLE SHOWING VERIFICATION
Effective
Interest
(Coupon
Interest
A tnortization
Carrying
Value
$2 59
S3 00
S 41
$103 58
103 17
2 58
3 00
42
102 75
. . 2 57
3 00
43
102 32
2 56
3 00
44
101 88
2 55
3 00
45
101.43
2 54
3 00
46
100 97
2 52
251
3 00
3 00
.48
49
100.49
100 00
Adding :
End of
Period
\
2
3 .
4
/
8
Second method. Procedure: (a) Compute the present value of
an annuity of 1 at the investment rate of interest and for a number
of periods equal to the number of unexpired periods of the bond,
a -, .
n\i
(h) From the number of dollars of one cash interest subtract
the number of dollars found by multiplying the par of the bond by
the effective rate of interest, Cr — Ci.
(c) Multiply the present value of the annuity found in (a) by
the number of dollars of the excess of the investment interest
found in (6). The result obtained is the premium on the bond,
(Cr - Ci) • a^,
(d) Add the par value of the bond and the premium found in
(c) to obtain the present value of the bond at a premium.
Formula
[(Cr - Ci) - oj + C = P
A rith metical S ubstit ution
i_\
(L025^8 I
(3 00 - 2.50)
+ 100 = $103.58
. -025 /J
Solution , Part 1
(1.025)8 = 1.2184029, compound amount of 1 at
for
for 8
8 periods
1 -f- 1.2184029 =-- .8207466, present value of 1 at
periods
1 - .8207466 = .1792534, compound discount on 1 at 2£% for
8 periods
378 BOND AND BOND INTEREST VALUATION
.1792534 -T- .025 = 7.1701372, present value of an annuity of 1 at
2-3-% for 8 periods
$.50 X 7.1701372 = $3.58, present value of an annuity of $.50
Solution, Part 2
Adding: $100 + $3.5S = $103.58
Verification
First Period:
Cost of bond ............................... $103 . 58
Coupon interest ......................... S3 00
Less 2i% interest on $103.58 ................. 2 59
Amortization of premium .................... .41
$103. 17
Second Period:
Coupon interest .......................... $3 00
Less 2^% interest on $103.1 7 ................. J2J5S
Amortization of premium .................... .42
$102.75
Third Period:
Coupon interest ............................. $3 00
Less 2i% interest on $102.75 ................ 2 57
Amortization of premium ................... 43
$102.32
Fourth Period:
Coupon interest ........................... $3 00
Less 2i% interest on $102.32 ................. 2 56
Amortization of premium ................... .44
Fifth Period:
Coupon interest ............................. $3 00
Less 2^% interest on $101 .88 ................ 2 55
Amortization of premium .................... 45
$FoT43
Sixth Period:
Coupon interest ............................ $3 00
Less 2£ % interest on $101 .43 .............. 2 54
Amortization of premium ................... 46
$100.97
Seventh Period:
Coupon interest ............................. $3 00
Less 2|% interest on $100.97 ................ 2 52
Amortization of premium .................... .48
$100 49
Eighth Period:
Coupon interest ............................. $3 00
Less 2i% interest on $100.49 ................. 2.51
Amortization of premium ................... .49
Par of bond ................................ $100.00
BOND AND BOND INTEREST VALUATION 379
Problems
Fill in the price in each of the following; the interest is payable semiannually:
1,
Face of
Bond
$ 100
Time to
Run
5 years
Cash
Interest
6%
Effective
Interest Price
7% $
9,.
1,000
10 years
5%
6%
3
5,000
15 years
5i%
6%
4,
2,000
12 years
6%
5^%
ft.
3,000
6 years
6^%
4i%
6.
5,000
20 years
&S°/n
5%
In problems 7, 8, 9, and 10, set up a columnar table showing: (a) the number
of periods; (b) the effective interest; (c) the coupon interest; (d) the amortization
of premium or discount; and (e) the carrying value. The interest is payable
semiannually.
Price
Face of
Time to
Cash or
Effective
Bond Run, Years
Coupon Interest Interest
1.
$ 100
4
5%
6%
8.
3,000
3i
6%
5%
9.
7,500
4
fiv("o
5^ %
10.
5,000
5
4%
5%
11.* A $10,000, 5% coupon bond is bought on a 4% basis. It is due 1-j years
hence, and interest is payable semiannually. Find the cost of the bond.
12. What is the difference in the purchase price of two $1,000, 20-year bonds,
bought to yield 6%, if one of the bonds has a semiannual coupon of $25, while
the other has a semiannual coupon of $35?
13. Davis died on April 1, 1933. His estate contained five $1,000 bonds
of the X.Y.Z. Company, bearing 6% interest, payable July 1st and January 1st.
The bonds were due on July 1, 1938, and were inventoried at 104^. On July 1,
1933, the trustee purchased five more of the same bonds on a 5% basis. Com-
pute the price paid by the trustee for the bonds. Assume the value of $1, due
after ten periods at 2jr%, to be $.781198402.
14. Find the price for a $1,000 bond bearing interest at 5^%, payable May 1
and November 1, maturing May 1, 1953, if bought on May 1, 1942 at a price
to yield the purchaser 5%.
15. Suppose the bond described in Problem 14 were bought to yield the
purchaser 6%. Find the price.
Values of bonds between interest dates. Heretofore, in finding
the values of bonds we have used even periods in the calculation
of the interest. However, if it is desired to find the value of a bond
at a date other than an interest date, additional computations are
necessary. Two factors must be taken into account: (1) accrued
interest at the cash rate must be computed for the fraction of an
* C. P. A. Examination.
380 BOND AND BOND INTEREST VALUATION
interest period elapsed; and (2) the amortization of the premium
or the accumulation of the discount must be computed for the
fraction of an interest period elapsed.
Interest accrued between interest dates. Finding the amount
of the accrued interest for a fractional part of a period is a simple
computation which requires no explanation. The accrued interest
must always be considered when the exact value of an investment
is being determined.
Bond discount or premium between interest dates. For prac-
tical purposes, the amortization of the premium or the accumula-
tion of the discount between interest dates may be calculated on a
proportional basis, by means of interpolation. This method gives
a fair degree of accuracy. The amount may be readily found if
bond tables are used.
Illustration of the practical process of calculating the value of a
bond bought at a discount.
Procedure: (a) By the use of bond tables, or of formulas pre-
viously given, determine the value of the bond at the interest date
just preceding the purchase date, and the value of the bond at the
interest date just subsequent to the purchase elate.
(6) Calculate the discount for one period by finding the dif-
ference between the values determined in (a).
(c) Calculate the part of the discount which is in the same pro-
portion to the discount for one period, found in (fr), as the fractional
part of the period which has elapsed is to the total period.
(rf) Add the discount for the fractional period, found in (c), to
the value of the bond at the interest date just preceding the pur-
chase date.
(e) Determine the accrued interest, at the rate specified in the
bond, on the par of the bond for the time expired since the last
interest date.
(/) Add the accrued interest found in (e) to the amount found
in (d) ; the sum is the value of the bond, with interest.
Example
On March 1, 1944, what was the value of a $1,000, 4^-% bond, due January
1, 1949? Interest coupons are payable January 1 and July 1, and money is
worth 6 %, interest compounded semiannually .
Solution
Value 9 periods before maturity, at 6% $941 60
Value 10 periods before maturity, at 6% 936 02
Accumulation of discount during 1 period $ 5 58
BOND AND BOND INTEREST VALUATION 381
Interpolation
Accumulation of discount during 1 period .............. $ 5.5S
As two months of the six months' period have elapsed, the
simple proportional part of the discount accumulated
is i of $5.58, or $1.86.
Value 10 periods before maturity, . . 036 02
Add 2 months' accumulation of discount .... . I 86
Value 9 periods and 4 months before maturity . . $937.88
Add accrued portion of next interest coupons 7 50
Total value of bond, with interest .............. . $945 38
Theoretical procedure illustrated. As the discount accumu-
lates at the rate of 3% for one whole period, and as exactly one
third of a period has expired, the accumulation may be expressed
by a fraction:
-
(1.03J-1
When the 3rd root, of (1.03) is extracted, the fraction becomes:
(1.009002) - 1
(1.03) - 1"" X '
009902
Simplifying, X 5.5S ............ $,842
Accumulation of discount by interpolation (as above) . 1 . SO
Accumulation of discount by theoretical process 1 842
Error by interpolation in the practical process of calculation $ 018
Illustration of the practical process of calculating the value of a
bond bought at a premium.
Procedure: (a) Determine from bond tables the value of the
bond at the interest date just preceding the purchase date, and the
value of the bond at the first interest date subsequent to the pur-
chase date.
(6) Determine the premium for one period by finding the dif-
ference between the values found in (a).
(c) Calculate the part of the premium which is in the same
proportion to the premium for one period as the expired part of the
period is to the whole period.
(d) Deduct the premium found in (c) from the value of the
bond at the interest date just preceding the purchase date.
(e) Determine the accrued interest on the par of the bond for
the expired fractional part of the period.
(/) Add the accrued interest found in (e) to the amount found
in (d)] the sum is the purchase price of the bond, with accrued
interest.
382 BOND AND BOND INTEREST VALUATION
Example
On May 1, 1943, what should have been paid for a $1,000, 6% bond, due
January 1, 1947, if interest coupons were payable January 1 and July 1, and
money was worth 5%, interest compounded semiannually?
Solution
Value 8 periods before maturity, at 5% . .... $1,03585
Value 7 periods before maturity, at 5% . . . . 1,031 75
Amortization of premium during 1 period . . $ 4 10
Interpolation
Amortization of premium during 1 period $ 4 10
As 4 months of the 8th period have expired, the
proportional part of the 6 months' period is
two-thirds. Therefore, the amortization
which has taken place is f of $4.10, or $2.73.
Value 8 periods before maturity $1,035 85
Deduct 4 months' amortization of premium 2 73
Value 7 periods and 2 months before maturity $1,033 12
Interest at 0% for 1 period . $30 00
As 4 months have expired since any interest was
paid, there is due £, or -f , of $30 . 20 00
Value of bond, with accrued interest $1,053 12
Bonds bought on a yield basis. For bonds bought on a strictly
yield basis, the following procedure may be used :
To the price of the bond at the last preceding interest date, add
interest thereon at the effective (yield) rate for the expired portion
of the period during which the purchase is made.
Example
A $1,000 bond maturing October 1, 1952 with interest at 6% payable April 1
and October 1 was bought on July 1, 1942 at a price to yield the investor 5%.
What price was paid for the bond?
Solution
(a) To obtain the price of the bond on April 1, 1942, which was the last
interest payment date prior to the date of purchase, follow the procedure on
page 376. The price of the bond on April 1, 1942 is found to be $1,080.92.
(b) To determine the price of the bond on July 1, 1942, compute the elapsed
time from April 1 to July 1 , which is 3 months or one-half a period. Then,
$1,080.92 X (1.025)* = $1,094.35
Problems
1. A $100 bond bears 5% interest, payable semiannually, and is due in
5 years and 4 months. What price, plus accrued interest, should an investoi
pay for the bond if he wishes his investment to produce 4%?
BOND AND BOND INTEREST VALUATION 383
2. A $1,000, 5% bond is due in 6 years and 3 months, and interest is payable
semiannually. The effective interest rate is 6%. What is the value of the
bond, with accrued interest?
3. A $1,000, 5-g-% bond, with interest payable annually, is redeemable at
par in 20 years and 4 months, and is bought on a 6% basis. Find the purchase
price.
4. A $1,000, 5% bond, with interest payable semiannually, is redeemable
at par in 8 years and 5 months, and is bought on a 4-g-% basis. Find the cash
price.
6. Five $1,000, 6% bonds, with interest payable semiannually, are due in
4 years and 4 months, and are purchased on a 4^% basis. Find the value of
the bonds.
Bonds to be redeemed above par. A form of bond which is
redeemable (usually at the option of the maker) at a premium is a
callable bond.
To find the value of a bond redeemable above par at the option
of the maker, before it is due, it is necessary to determine the value:
(1) On the assumption that the bond will be redeemed at the
optional redemption date, and at the optional price.
(2) On the assumption that the maker will not redeem the
bond until maturity, and that it will then be redeemed at the par
value.
After the two prices have been found, the purchaser should pay
the lower of the two prices.
Example
A $1,000, 6% bond, with interest payable semiannually, is due in 15 years,
but the maker has the option of redeeming it at the end of 10 years at 105. What
price should an investor pay for the bond, if he purchases it on a 5% basis?
(1) Calculation on the assumption that the bond will be redeemed at the
end of 10 years at 105:
Formula
CV + Cr - an]l = P
A rithmctical S ubstitution
\(1.025)2V
/I -
1,0501 „- -~^}+ 30 I (^-l = $1,108.45.
Solution, Part 1
(1.025)20 = 1.638616, compound amount of 1 at 2£% for
20 periods
1 -T- 1.638616 = .61027, present value of 1 at 2^% for 20 periods
$1,050 X .61027 = $640.78, present value of redemption price of
bond
384 BOND AND BOND INTEREST VALUATION
Solution, Part 2
.61027 = present value of 1 at 2£% for 20 periods
1 - .61027 = .38973, compound discount on 1 at 2i% for
20 periods
.38973 -T- .025 = 15.5892, present value of annuity of 1
$30 X 15.5892 = $467.67, present value of annuity of $30
Solution, Part 8
$640.78 + $467.67 = $1,108.45, value based on optional redemption
(2) Calculation on the assumption that the bond will not be paid until
maturity:
Formula
Cv» + Cr • a-,, = P
Arithmetical Substitution
Solution, Part 1
(1.025)30 = 2.097567, compound amount of 1 at 2j-% for
30 periods
1 -f- 2.097567 = .476742, present value of 1 at 2i% for 30
periods
$1,000 X .476742 = $476.74, present value of par of bond
Solution, Part 2
.476742 = present value of 1 at 2i% for 30 periods
1 - .476742 = .523258, compound discount on 1 at 2^% for
30 periods
.523258 -i- .025 = 20.93029, present value of an annuity of 1
$30 X 20.93029 = $627.91, present value of an annuity of $30
Solution, Part 8
$476.74 -f $627.91 = $1,104.65, price of bond based on par
A comparison of the two results shows:
Value based on 15-year redemption price ............ $1,108 45
Value based on 20-year redemption price ........... 1,104.65
Therefore, the purchaser should pay the lower price, or $1,104.65.
Problems
1. A $1,000, 5^% bond, with interest payable semiannually, matures iiv
10 years, but the company has the option of redemption at the end of 5 year&
at 104. What price should an investor pay for the bond, if he purchases it on
a 5% basis?
2. A $1,000, 5% bond, with interest payable semiannually, matures in
20 years, but the company has the option of redeeming the bond at the end of
BOND AND BOND INTEREST VALUATION 385
15 years by paying a bonus or premium of 10%. What price should an investor
pay for five of these bonds, if he purchases them on a 4% basis?
3. A $1,000, 5^% bond, with interest payable annually, is redeemable in
5 years with a bonus of 10%. What price should be paid for this bond by a
purchaser who desires to realize 6% on his investment? (v* at 6% = .7473.)
Construct a table of verification.
Prepare the formula, solution, and verification for each of the following:
Effective Coupon Amount Redeemable Time Coupons
Interest Interest of Bond Value to Run Payable
4. 5% 6% $ 100 104 6 years Annually
6. 6% 8% 1,000 105 4i " Annually
6. 5% 6% 500 108 4^ " Semiannually
7. 6% 4% 500 110 5i " Semiannually
Serial redemption bonds. Many public and private corpora-
tions desire to pay off a portion of their bonds each year instead of
setting up a sinking fund. Serial redemption bonds may be
redeemed in equal or unequal periodic amounts.
If the bonds are not redeemed in equal periodic amounts, it is
difficult to derive a formula or plan by which computations may be
shortened or systematized. But if the redemptions are to be made
in equal amounts and at regular periodic dates, formulas and solu-
tions for finding the value of such an issue may be derived.
For the purpose of finding the present value of bonds to be
redeemed in a series, it is well to analyze the issue into its com-
ponent parts, and to calculate the value of each component part
separately. The following example will illustrate this point :
Example
What is the present value of a bond issue of $10,000 bearing 5% interest,
payable annually? These bonds are to be redeemed serially, $2,000 each year.
Money is worth 6%, effective interest.
ANALYSIS OF THE CALCULATION OF THE VALUE OF A SERIAL
REDEMPTION BOND
Present Present
Multiplied Value of Value of
by Present Annuity a Series of
First Second Third Fourth Fifth Value of of Annuities
Period Period Period Period Period Annuity of 1 Principal of Interestt
Principal $2,000 $2,000 $2,000 $2,000 $2,000 $4212363 $8,424.73
Interest payments 100 94339 $ 94 34
100 100 1 83339 183.34
100 100 100 2 67301 267.30
100 100 100 100 3.46511 346.51
100 100 100 100 100 4.21236 421 24
$8.'424.73 $1.312 73
Summary
Present value of annuity of $2,000 at 6% $8,424.73
Present value of 5 annuities of $100 each (the number
of rents varies from 1 to 5, as shown above) at 6% 1,312. 73
Present value of the serial redemption bonds $9,737.46
386 BOND AND BOND INTEREST VALUATION
A study of the above analysis shows two general divisions :
(1) The calculation of the present value of an annuity, the rents
of which are the same as the amounts of the bonds which are
redeemed annually.
(2) The calculation of the present value of a series of annuities,
the rents of which are the same as the periodic interest payments
on each series of bonds.
The first part needs no explanation, since the computation is
similar to that of the present value of an ordinary annuity.
The sum of the values of the second part may be found by cal-
culating the value of each separate annuity and then adding these
values.
Let C = the par value of the bond;
R — annual amount redeemed;
r = the rate named in the bond, that is, the coupon or cash rate;
i = the rate of yield expected;
n = the number of years to maturity;
N = the number of equal redemptions to retire the issue;
P = the cost of the bond.
Procedure, Part 1 : (a) Find the present value of an annuity of
1 at the effective rate per cent for as many periods as the bond has
serial payments, a^..
(b) Multiply the present value of the annuity of 1 found in (a)
by the number of dollars to be paid each period on the principal
of the bonds, Ra-{t*
Procedure, Part 2 : (a) Use the present value of an annuity of 1
found in Part 1 (a).
(b) From the number of annuities subtract the present value
of an annuity of 1 found in (a), N — a- .
(c) Divide the annuity discount found in (6) by the effective
rate to obtain the cumulative present worth of the annuities,
N - a-,.
nit.
i
(d) Multiply the average annuity price of 1 by the number of
dollars in each interest payment or rent. The result will be the
value of a series of annuities, Cr ( = — — )•
Procedure, Part 3 : Add the result found in 1 (b) to that found
in 2 (d).
Formula
t + Cr|
BOND AND BOND INTEREST VALUATION 387
Arithmetical Substitution
Solution, Part 1
(1.06)5 = 1.3382256, compound amount of 1 at 6%
for 5 periods
1 -f- 1.33S2256 = .7472582, present value of 1 at 6% for
5 periods
1 - .7472582 - .2527418, compound discount on 1 at 6%
for 5 periods
.2527418 -r- .06 = 4.2123638, present value of annuity of 1 at
6 % for 5 periods (See Table 5, page 532.)
$2,000 X 4.2123638 = $8,424.73, present value of annuity of $2,000
Solution, Part 2
4.2123638 = present value of annuity of 1 at 6% for 5 periods
5 — 4.2123638 = .7876362, annuity discount on 5 annuities
.7876302 -r- .06 = 13.12727, cumulative present value of annuities
$100 X 13.1273 = $1,312.73, cumulative present value of annuity
of $100
Solution, Part 3
$8,424.73 -f $1,312.73 = $0,737.46, sum of present value of annuities
Problems
1. Compute the purchase price of $5,000 of serial bonds, issued January 1,
1934, with 5% interest, payable annually, and dated to mature in five equal
annual installments. Money is worth 5^%.
2. Verify the solution of the above problem by setting up a columnar table
showing: (a) date of maturity; (b) bonds outstanding; (c) coupon interest each
year; (d) effective interest each year; (c) accumulation of discount; (/) carrying
value.
3. Compute the purchase price of $50,000 of serial bonds, issued January 1,
1934, bearing 5% interest, coupons due annually. These bonds were to mature
serially in equal annual payments, beginning January 1, 1935, and each year
thereafter for 10 years. Money was worth 4%.
4. Prepare a columnar table for Problem 3.
5. A $20,000 serial bond issue, with interest at 5^%, payable semiannually,
is redeemable in ten equal semiannual installments. Money is worth 5%, con-
vertible semiannually. Set up a columnar table similar to that in Problem 2.
Frequency of redemption periods. In the example on page
385, bonds were redeemed at each interest date. It is more usual,
however, to find that the interest is payable semiannually, while
the bond redemptions occur once a year.
388
BOND AND BOND INTEREST VALUATION
P-( Pk Pu,
BOND AND BOND INTEREST VALUATION 389
Example
A $6,000 issue of serial bonds is to be redeemed in equal installments of
$2,000, on January 1 of each year. The coupons are at the rate of 5%, payable
semiannually. If money is worth 6%, interest convertible semiannually, what
is the present value of the issue? (For the purpose of illustration, short-term
bonds are used.)
A study of the analysis on page 388 shows three divisions:
(1) The calculation to find the present value of an annuity of
$2,000, the annual payment on the bond issue, at the effective
annual rate of 6.09%.
(2) The calculation to find the present value of an annuity of
$25, the amount of the interest payment due at the end of each
year.
(3) The calculation to find the present value of the series or
six rents of $25 each at 3% semiannual interest.
It should be noticed that one of the interest payments of $25
due at the end of each year is separated from the other interest
payments. This separation is made to reduce the remaining inter-
est payments to a series of annuities of regularly increasing terms.
The computation may be shortened if the annual bond redemp-
tion payment of $2,000 and the annual interest payment of $25 are
combined. It is then necessary to find the present value of an
annuity the rents of which are $2,025 at 6.09% per annum. The
formula and solution are derived by a method similar to that
explained on page 386.
Let g represent the value of one coupon on the periodic cash interest on one
bond.
A' /"')
Formula
(R + g) • an{t + g
in which o-|f in the first part of the formula is calculated at the effective annual
rate.
A rithmetical Substitution
/, i_\ /'"(w1
2025 (L0609)3 +25\ -°3
2'°25\ -;0609~~/ + 25r ~^3~
Solution, Part 1
(1.03)2 = 1.0609, effective ratio of increase
1.0609 - 1 = .0609, effective rate per annum
(1.0609)3 = 1.19405228, compound amount of I at 6.09%
for 3 periods
390 BOND AND BOND INTEREST VALUATION
1 -4- 1.19405228 = .8374843, present value of 1 at 6.09% for
3 periods
1 - .8374843 = .1625157, compound discount on 1 at 6.09%
for 3 periods
.1625157 -f- .0609 = 2,668568, present value of an annuity of 1 at
6.09% for 3 periods
$2,025 X 2.668568 = $5,403.85, present value of annuity of $2,025
at 6.09% for 3 periods
Solution, Part 2
(1.03)° = 1.19405228, compound amount of 1 at 3% for
6 periods
1 -T- 1.19405228 = .8374843, present value of 1 at 3% for 6 periods
1 - .8374843 = .1625157, compound discount on 1 at 3% for
6 periods
.1625157 -T- .03 = 5.4171914, present value of an annuity of 1 at
3% for 6 periods
6 - 5.4171914 = .5828086, annuity discount on 6 rents
.5828086 -5- .03 = 19.42695, present value of 6 rents of 1 at 3%
for 6 periods
$25 X 19.42695 = $485.67, present value of 6 rents of $25 each
Solution, Part 3
$5,403.85 + $485.67 = $5,889.52, value of serial bond
Alternative solution. The following method of solving the pre-
ceding example is preferred when the number of redemptions are
few. The procedure is that of rinding the present value of the
respective payments of interest and principal.
July 1 Interest $ 150 X .970873 = $ 145 63
Jan. 1 Int. and Principal 2,150 X .942595 = 2,026 58
July 1 Interest 100 X .915141 - 91 51
Jan. 1 Int. and Principal 2,100 X .888487 = 1,865 82
July 1 Interest 50 X .862608 = 43 13
Jan. 1 Int. and Principal 2,050 X .837484 = 1,71 6 _84
Total = $5,889jy.
Bonds redeemed by other than equal annual payments. If the
bonds are to be redeemed in any other way than by equal annual
payments beginning at the end of the first year, an analysis must
be made of the component parts, and computations made for each
part separately.
Example
What is the value of an issue of 5% serial redemption bonds for $10,000, if
$2,000 is to be redeemed at the end of 6 years and $2,000 at the end of each year
thereafter until all the bonds have been redeemed? Interest coupons are payable
semiannually. Money is worth 6%, interest converted serniannually.
BOND AND BOND INTEREST VALUATION
391
ANALYSIS OF A SERIES OF SERIAL REDEMPTION BONDS, WITH
INTEREST PAYMENTS
SEMIANNUAL PERIODS
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1ST
SERIES
25 25 25 25 2
25 25 25 25 2
25 25 25 25 25
25 25 25 25 25
25s-
IN* <N <N <N fri
-N25
25\
2ND
SERIES
25 25 25 25 2
25 25 25 25 2
25 25 25 25 25
25 25 25 25 25
25
25
25 25 \25
25 25 25\
3RD
SERIES
25 25 25 25 2
25 25 25 25 2
25 25 25 25 25
25 25 25 25 25
25
25
25 25 25 25\25
25 25 25 25 25 x
4m
SKKIKH
25 25 25 25 2
25 25 25 25 2
25 25 25 25 25
25 25 25 25 25
25 25 25 25 25 25 25
25 25 25 25 25 25 25
OTH
SERIES
TOTALS
25 25 25 25 2
25 25 25 25 2
250 250 250 250 25
25 25 25 25 25
250 250 250 250 250
25
25
25 25 25 25 25 25 25 25 25
25 25 25 25 25 25 25 25 25
From the above analysis there are found to be four component
parts in the example :
(1) An annuity of $2,000, the payments of which are to begin
at the end of the sixth year and are to be made at the end of each
year thereafter, at an effective interest rate of 6%, convertible
0.09%
annually. This series of annuities
semiaimually, or
deferred for 5 years.
(2) An annuity, the rents of which are $25, payable at the end
of the sixth year, and annually thereafter, at 6.09%.
(3) A series of ten annuities of unequal length, the rents of
which are $25, payable at the end of each half-year. Each annuity
is for 1 period longer than the preceding one. The rate of effective
interest is 3%. This series of annuities is deferred for 5 years.
(4) A series of ten annuities of equal length, the rents of which
are $25, payable at the end of each period, with interest at 3%.
Formulas will not be given for this example, which does not
involve anything new, but only the combining of certain principles
already explained.
To shorten the calculation, (1) and (2) above can be combined,
making a deferred annuity the rents of which are $2,025 for 5
periods at 6.09%.
Procedure: (a) Find the present value at the beginning of the
sixth year of (1) and (2) combined.
(6) Find the present value of (3) at the beginning of the sixth
year.
(c) Multiply the sum of the present values found in (a) and (6)
by the present worth of 1 for ten periods at 3%. The result is the
392 BOND AND BOND INTEREST VALUATION
present value at the beginning of the first period of all the bonds
and interest payments falling due after the end of the fifth year.
(d) Calculate the present value of (4) ; this result is the value
at the beginning of the first year.
(e) Add the results found in (c) and (d). Their sum is the
present value of the series of serial redemption bonds, with all the
interest payments.
Solution, Part 1
NOTE: The compound amount and present value of (1. 0609) 5 are the same
as the compound amount and present value of (1.03)10.
(1.0609)5 = 1.3439164, compound amount of 1 for 5
periods at 6.09%
1 -T- 1.3439164 = 7440939, present value of 1 for 5 periods at
6.09%
1 - .7440939 = .2559061, compound discount on 1 for 5
periods at 6.09%
.2559001 -T- .0609 = 4.2020707, present value of an annuity of 1
for 5 periods at 6.09%
$2,025 X 4.2020707 = $8,509.19, value of an annuity of $2,025 for
5 periods at 6.09%
Solution, Part 2
.7440939 = present value of I for 10 periods at 3% (same
as 1 for 5 periods at 6.09%)
1 — .7440939 = .2559061, compound discount on 1 for 10 periods
at 3%
.2559061 -r- .03 = S.530202, present value of an annuity of I for
10 periods at 3%
10 - 8.530202 = 1.469798, annuity discount on the series of 10
annuities of 1
1.469798 -r- .03 = 48.9932, present value of the series of 10 annui-
ties of 1
$25 X 48.9932 = $1,224.83, present value of the series of 10
annuities of $25 each
Solution, Part 8
$8,509.19 + $1,224.83 = $9,734.02, amount of present value at the
beginning of the sixth year
.7440939 = present value of 1 for 10 periods at 3%
$9,734.02 X .7440939 = $7,243.02, present value of serial bonds
and of the series of 10 annuities
Solution, Part 4
8.530202 = present value of an annuity of 1 for 10 periods
at 3%
$25 X 10 = $250, sum of the 10 payments at the end of
each period
$250 X 8.530202 = $2,132.55, present value of the annuity of
$250 for 10 periods at 3%
BOND AND BOND INTEREST VALUATION
393
Solution, Part 5
$7,243.02 + $2,132.55 = $9,375.57, present value of serial bonds,
with all interest payments
VERIFICATION OF CALCULATION OF THE VALUE OF A SERIES
OF SERIAL REDEMPTION BONDS
Knd of
Period
1..
2
3
4 .
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20..
Bonds Effective Coupon
Redeemed Interest Interest
$2,000.00
2,000 00
2,000 00
2,000 00
2,000.00
$281.27
$250 00
282.20
250 00
283.17
250 00
284.17
250 00
285 19
250 00
286 . 25
250 00
287. S3
250.00
288.45
250.00
289.61
250.00
290.80
250 00
292 02
250 00
293.28
250 00
234.58
200 00
235 62
200 00
176.69
150 00
177.49
150 00
118.31
100 . 00
118 86
100 00
59 43
50 00
59.71
50 00
Amortization
Bonds Less
of Discount
Discount
$9,375.57
$31 27
9,406 84
32 20
9,439 04
33.17
9,472.21
34 17
9,506 38
35.19
9,541 57
36 25
9,577 82
37.33
9,615.15
38 45
9,653.60
39 61
9,693 21
40 80
9,734.01
42 02
9,776.03
43 28
7,819 31
34 58
7,854.89
35 62
5,889 51
26 69
5,916 20
27 49
3,943 69
IS 31
3,962 00
18 86
1,980 86
9 43
1,990 29
9 71
0 00
Problems
1. The State Highway Department of Michigan desires to know the value
of a series of road bonds which it is about to issue. The bonds will have a pal-
value of $200,000, and will bear 5% interest, payable semiannually. They are
to be redeemed serially, in installments of $40,000. The first redemption pay-
ment is to be made one year from the date of issue, and the other payments
are to be made annually thereafter. Money is worth 6%, interest compounded
semiannually. Find the value of the bonds, and draw up a table of analysis
as a proof of your solution.
2. A corporation wishes to float an issue of serial bonds for $100,000. These
bonds are to be redeemed in yearly installments of $20,000, the first redemption
payment to be made at the beginning of the sixth year. The interest rate is 5%,
payable semiannually, and the effective rate is 4^%, interest convertible semi-
annually. Draw up a table of analysis. From the analysis, prepare the formula
and solution.
3. An issue of serial bonds bearing 4% interest, payable semiannually, is to
be redeemed serially in installments of $4,000. The first redemption payment
is to be made at the end of the fifth year, and the other payments are to be
made annually thereafter. At what price must $20,000 of these serial bonds
be purchased in order to net the purchaser 5% annual effective interest?
394 BOND AND BOND INTEREST VALUATION
4. An issue of $50,000 of bonds bearing interest at 5%, payable semiannually,,
is sold to produce 5^% effective interest, convertible semiannually. The bonds
are to be retired serially, as follows:
$10,000 at 104 in 6 years
10,000 at 103 in 7 years
10,000 at 102 in 8 years
10,000 at 101 in 9 years
10,000 at par in 10 years
Set up a schedule in columnar form, showing the book value, cash interest,
effective interest, amortization of discount, and par value of bonds outstanding
each year.
Bonds redeemable from a fund. Frequently, bonds are
redeemed periodically from a fund; that is, as soon as money is put
into the fund, or when money becomes available at the end of an
interest period, it is at once used to redeem outstanding bonds.
No difficulty would be encountered in making the computations
necessary in this system of redemption, except for the fact that
bonds are usually issued in denominations of $100, $500, or $1,000,
and the payments into the fund may exceed the expenditures from
the fund for the redemption of bonds. A residue would then be
left in the fund, and this residue should earn interest.
Example
The X.Y.Z. Company issues $100,000 of bonds, par value $100, and sets up
a sinking fund for their periodic redemption. The bond interest and the sinking
fund interest are each 6%. The bonds arc to run for 5 years, with interest
payable semiannually, and are to be kept alive in the treasury. Sinking fund
payments are to be made semiannually. Interest is to be paid by the trustee
on the balance remaining in the fund. Show: (a) the periodic payments into
the sinking fund; (b) the interest accrued periodically on bonds redeemed; (c) the
total amount that must be invested in the sinking fund each period; (d) the
amount of bonds purchased periodically for the sinking fund; (e) the cash balance
held in the treasury; and (/) the interest on the cash balance held in the treasury.
Let: i = the effective interest per period.
And: n = the number of periods.
Formula Arithmetical Substitution
= R — — - — X $100,000 = $11,723.05
\J~ (1.03V0
~~ .03
From Table 6, page 534, the value of is found to be .1172305,
1 "" (L03T10
.03
and .1172305 X $100,000 = $11.723.05.
BOND AND BOND INTEREST VALUATION
395
The remaining part of the solution can be derived from the following table:
Periodic
Payment
Balance
from
Preceding
Period
Total
Amount
Available
Bond
Interest
Bonds
Re-
deemed
Cash
Balance
Interest
on Cash
Balance
Bonds
Out-
standing
$100,000
$11,723
05
$11,723
.05
$3,000
$ 8,700
$23
05
$
.69
91,300
11,723
05
$23 74
11,746
79
2,739
9,000
7
79
23
82,300
11,723
05
8 02
11,731
07
2,469
9,200
62
07
1
86
73,100
11,723
05
63.93
11,786
98
2,193
9,500
93
98
2
82
63,600
11,723
05
96 80
11,819
85
1,908
9,900
11
85
.36
53,700
11,723
05
12 21
11,735
26
1,611
10,100
24
26
.73
43,600
11,723
05
24 99
11,748
.04
1,308
10,400
40
01
1
.20
33,200
11,723
05
41 24
11,764
29
996
10,700
68
29
2
.05
22,500
11,723
05
70 34
11,793
39
675
11,100
18.
39
,55
11,400
11,723
06
18.94
11,742.
00
342
11,400
In the above example, the first payment into the fund is
$11,723.05. The payment on the interest will be $3,000, leaving
$8,723.05 to be used for the redemption of bonds. As the bonds
issued are of $100 denomination, only $8,700 of this fund can be
used for the redemption of bonds, leaving a balance of $23.05 cash
in the hands of the trustee.
In order to verify the solution of a problem of this kind, it is
necessary to charge the trustee with interest at the sinking fund
rate on the balance remaining in his hands. The interest on
$23.05, the cash balance in the trustee's hands, for 6 months at
3% is $.69. This interest added to the cash balance gives the
balance from the preceding period.
Problems
1. A $200,000 bond issue, maturing in 8 years, bears interest at 6%, payable
semiannually. The par value of the bonds is $1,000. A sinking fund is set up,
and the trustee is to purchase bonds semiannually at par and keep them alive
in the treasury. Money is worth 6%, interest convertible semiannually. Pre-
pare a table, showing: (a) the semiannual periodic payments to the sinking fund;
(b) the interest accrued on the bonds; (c) the par value of the bonds purchased
semiannually; (d) the cash balance in the hands of the trustee; and (e) the interest
on the cash balance.
2. An issue of $200,000 of 6%, 10-year bonds is floated by a corporation.
The par value is $100, arid the interest coupons are payable semiannually. The
bonds are to be redeemed semiannually, and the sinking fund payments neces-
sary for the redemption of the principal and the payment of the interest are
placed in the hands of the trustee. The sinking fund rate is 6%. Prepare a
table, as in Problem 1.
Effective rate of interest on bonds. One question which is
very important to the investor is, ''What rate of interest will I
receive on this bond?" or "What will be the effective rate on the
396 BOND AND BOND INTEREST VALUATION
money in vested ?" The investor knows the amount of each inter-
est coupon, but the coupon rate is based on par value and not on
the amount of money which has been invested.
Because of the fact that the unknown effective interest rate
must be used more than once in the algebraic formula for the cal -
culation of the price of a bond, the absolute effective rate is difficult
to find.
However, two methods which give a close approximation may
be used. The rate may be found :
(1) By the use of bond tables, or test rates, and interpolation.
(2) By formulas specially constructed to give approximately
the required rate. Most formulas, and sometimes a method of
averages, will give the rate accurately enough for ordinary purposes.
Effective rate on bonds sold at a premium.
Example
If a $100, 3-year, 6% bond, bearing semiannual interest coupons, is bought
at $105.38, what rate of interest will be realized on the investment?
SECTION OF BOND TABLE
CASH INTEREST PAYABLE SEMIANNUALLY
Effective
Interest
Rate
3%
$"2T /o
4% 4\%
5%
6%
7%
3 75
$97 89
$99 30
$100
70
$102
.11
$103
.52
$106
33
$109 14
3 80
97.75
99.16
100
56
101
97
103
.37
106
18
108 99
3.875
97.54
98 95
100
35
101
75
103
16
105
96
108.77
3 90
97 48
98 88
100,
,28
101
.68
103
09
105
89
108 70
4.00
97.20
98 60
100
00
101
.40
102
80
105
.60
108 40
4 10
96 92
98 32
99
72
101
12
102
52
105
31
108 11
4.12i
96.86
98.25
99
65
101
05
102
.45
105
24
108 04
4.20
96.65
98.05
99
44
100
84
102
.23
105
.02
107 82
4.25
96.51
97.91
99,
,30
100
.70
102
.09
104
.88
107 67
Solution
In the 6% column of the section of the bond table given above will be found,
opposite 4%, the price of $105.60, and opposite 4.1%, the price of $105.31.
The rate is therefore somewhere between 4% and 4.1%. A more exact approxi-
mation may be determined as follows:
Interpolation
Value on a basis of 4% $105 60
Value on a basis of 4.1 % 105 31
Difference in value caused by a difference in rate of .1 %. . $ .29
Value on a basis of 4% $105 60
Price paid 105.38
Difference between price paid and value on a basis of 4 % . $ .22
If the difference between the price paid and the value on a basis of 4% is
$.22, and the difference in the value caused by a difference of .1% in the rate
is $.29, the rate earned will be ff of .1 % greater than 4%, or 4,070%.
BOND AND BOND INTEREST VALUATION 397
Effective rate on bonds sold at a discount.
Example
What will be the rate of income on an investment in a 4%, semiannual, 3-year
bond bought at $99.35?
Solution
By reference to the table above, the computation may be made as follows:
Value of a 4% bond at 4.2% effective interest . . . $99 44
Value of a 4% bond at 4.25 % effective interest 99 30
Difference in value caused by a difference in rate of .05 % $ 1 4
Value on a basis of 4.2% $99 44
Price of bond 99 35
Difference $ .09
A of -05% 032%
4.20% + .032% 4.232%
Hence, the approximate rate is 4.232%.
Computation when bond table is not available. If no bond
table is available, an approximate rate may be computed by the
use of test rates, but care must be exercised in the choice of the
rates, which must be as close to the actual rate as it is possible to
estimate.
Example
If a $100, 5%, 5-year bond, with interest payable semiannually, is bought at
$97.31, what rate of interest will be realized on the investment?
Solution
The first step is to find the cost at an estimated effective rate. Assume this
rate to be 5^%. By the formula, previously given, for finding the purchase
price of bonds, this bond at 5.5% is worth $97.84. As this price is higher than
the price paid, the rate is too low, and it is necessary to try a higher rate. Make
a second trial at 5.75%. The purchase price is then found to be $96.78. As
one price is slightly above and the other slightly below the actual price paid, the
approximate rate can be found by interpolation.
Interpolation
Value on a basis of 5.50% $97 84
Value on a basis of 5.75% 96_Z?
Difference in value caused by a difference in rate of ^ % . . $ 1 . 06
Value on a basis of 5.50 % $97^*84
Price paid for bond 97 31
Difference between price at 5.50% and price paid $ .53
Since the difference between the price paid and the price at 5.50% is $.53,
and the difference between the value on a 5.50% basis and the value on a 5.75%
basis is $1.06, the rate earned will be approximately 5.50% plus y5^ of the
difference between 5.50% and 5.75%, or 5.625%.
Approximation by averages. A fair approximation may be
made by the use of averages. In the above example, the rate
would be found as shown on the next page.
398
BOND AND BOND INTEREST VALUATION
Semiannual yield $ 2. 50
Total gain on redemption, $100 —
$97.31 = $2 69
Average gain on redemption ($2.69 -f-
10) .269
Average yield per period . ... $ 2 769
Capital invested at beginning of 10th
period before maturity $ 97 31
Capital invested at beginning of last
period before maturity ($100 - .269) 99 731
Total $197041
Dividing by 2, to find average capital $ 9S 5205
Average yield -f- average capital . Rate
2.769 -T- 9S.5205 .. . ... 02811
Multiplying by 2 . 05622, or 5 622%
Correct yield . 0562468, or 5.625%
Problems
In each of the following, find the effective rate by two methods: (1) by the
use of bond tables, or test rates, and interpolation; (2) by the use of averages.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Purchase
Nominal
Price
Rate
$ 92 00
4%
1,015 00
4%
$9,750 00
4%
545 00
5%
925 00
5%
983 75
5%
9,732 50
5J%
5,201 50
5£%
96 85
6%
73 55
6%
Par Time in Interest
Value
$ 100 00
1,000 00
10,000 00
500 00
1,000 00
1,000 00
10,000 00
5,000 00
100 00
100 00
Effective Kate
Fears Payable Method I Metho
3
9
29
8
15
18
5
9
4
6
Scmianiuuilly
Amortization of discount, premium, or discount and expense
on serial redemption bonds. The calculation of the amortization
of bond discount, bond premium, or bond discount and expense,
when bonds are to be redeemed serially, or in unequal amounts, is
a problem which requires particular attention, since the distribu-
tion over the period of years must be equitable. The two methods
which are most commonly used are :
(1) The bonds outstanding method.
(2) The scientific or effective-interest method.
Bonds outstanding method. It would be incorrect to write off
the discount or premium, or the discount and expense, on serial
redemption bonds or on a bond issue which has no regular redemp-
tion period, by the straight-line method. In some cases it is
difficult to calculate the portion to be amortized by the scientific
method. Because of the ease of the calculations and the fair
BOND AND BOND INTEREST VALUATION
399
degree of accuracy which it affords, the bonds outstanding method
is the one most commonly used.
Procedure : (a) Find the sum of the bonds outstanding during
each period of the life of the bond issue.
(6) Use the sum found in (a) as the denominator of a series of
fractions, and use the sum of the bonds outstanding each period as
successive numerators. The sum of all these fractions, of course,
will in every problem be equal to 1.
(c) Multiply the total bond discount or the premium, or the
total bond discount arid expense, by the appropriate fraction, to
obtain the portion of discount or premium to be amortized each
period.
Example
An issue of ten bonds of $1,000 each, bearing 5% interest, payable semi-
annually, is to be redeemed as follows: $3,000 at the end of the sixth year; $3,000
at the end of the eighth year; and $2,000 at the end of each year thereafter.
The bonds are sold at a discount of $400. Compute an equitable amortization
ot the discount over the life of the bonds.
Solution
Periods of On e-h a If Bonds
Amortization
Discount
Bonds Less
Year Each
Outstanding
Fraction
Written Off
on Bonds
Discount
$10,000 00
$400 00
$9,600 00
1
10,000 00
10/160
$ 25 00
375 00
9,625 00
2
10,000.00
10/160
25 00
350 00
9,650.00
3
10,000 00
10/160
25 00
325 00
9,675 00
4
10,000 00
10/160
25 00
300 00
9,700 00
5
10,000 00
10/160
25 00
275 00
9,725 00
6
10,000 00
10/160
25 00
250 00
9,750.00
7
10,000 00
10/160
25 00
225.00
9,775 00
8
10,000.00
10/160
25 00
200.00
9,800.00
9
10,000.00
10/160
25 00
175.00
9,825 00
10
10,000 00
10/160
25 00
150 00
9,850 00
11
10,000 00
10/160
25 00
125 00
9,875.00
12
7,000 00
10/160
25 00
100 00
6,900 00
13
7,000 00
7/160
17 50
82 50
6,917 50
14
7,000 00
7/160
17 50
65 00
6,935 00
15
7,000 00
7/160
17 50
47 50
6,952 50
16
4,000 00
7/160
17 50
30 00
3,970 00
17
4,000 00
4/160
10 00
20 00
3,980 00
18
2,000.00
4/160
10 00
10 00
1,990 00
19
2,000 00
2/160
5.00
5 00
1,995 00
20
000 00
2/160
5.00
0.00
000 00
160/160
400.00
400 BOND AND BOND INTEREST VALUATION
Scientific method. To find by the scientific method the
amount of discount or premium to be amortized on an issue of
serial redemption bonds, it is necessary to find the approximate
effective rate of interest which these bonds will bear; and to find
the approximate effective rate it is necessary to use other approxi-
mations.
Procedure: (a) Determine the average life of the bonds, in
periods.
(6) Determine, by the use of a bond table or by annuity calcu-
lations, the approximate effective interest rate for one bond having
a life of the same number of periods as the average life found in (a).
(c) From a bond table or by annuity calculations, find the
value of one bond at each of the annual redemption dates, at the
approximate effective rate found in (6).
(d) By using the values found in (e), find the total value of the
bonds redeemed at each redemption date.
(e) Add the values found in (d).
(/) Compare the sum found in (e) with the actual price received
for the bonds.
(</) By the same process, determine another approximate rate.
(h) By interpolation, determine the error, using the approxi-
mate rates found above.
The computation of the periodic amortization is comparatively
simple when the cost of the serial redemption bonds, the nominal
or coupon rate, and the effective rate of interest are known. Val-
uation of each member of the series is equally simple. Refer to the
table on page 402, which shows the amortization of discount for a
series of serial redemption bonds. The periodic amortization is the
difference between the effective income and the actual cash income.
It may be observed that the difficulty of the whole calculation is
the determination of the effective rate.
The example that was previously given under the bonds out-
standing method would be solved by the scientific method as
follows :
Solution
(a)
Bonds
Maturing In
Product
3
6 years
18
3
8 "
24
2
9 "
18
2
10 "
20
10
80
80 -r 10 = 8, or an average life of 8 years
9,600 -T- 100 = $96, the average sales price per hundred
BOND AND BOND INTEREST VALUATION 401
(6) Refer to a bond table which shows the values at different effective rates
of a 5% bond maturing in 8 years, with interest payable semiannually. The
price nearest to 96 is found to be 96.02. This amount is opposite the effective
rate of 5f%.
(c) The next step is to find the value at 5f % of all the bonds in the series.
To do this, refer to the bond table which shows a cash rate of 5% and an effective
rate of 5f%.
(d) The results will be:
Par Value Years Value at
of Bonds to Run <5f% Total Value
$3,000 00
6
$96.85
$2,905 50
3,000 00
8
96 02
2,880.60
2,000 00
9
95 63
1,912 60
2,000.00
10
95.27
1,905 40
(e) Value of series $9,604 10
(/) As the sales price of $9,600 for the bonds is below the price based on
a 5f % rate, the effective rate is a little larger than 5f%. A test at 5|% would
result in the following:
Par Value
Years
Value at
of Bonds
to Rim
*f%
Total Value
$3,000 00
6
$96 24
$2,887 20
3,000 00
8
95 24
2,857 20
2,000 00
9
94 79
1,895 80
2,000 00
10
94 36
1,887 20
Value of series 19,527 40
(h)
Interposition
Value of series at 5f % $9,604 10
Value of series at 5f % 9,52740
Difference caused by a difference in rate of -g-% $ 76. 70
Value of series at 5g % $9^604710
Sales price ... ... . 9,600 JX)
Difference . . $ 47IO
410/7670 of fc% . ... "" 1)06682
Trial rate ~™^T625™
Add fractional rate found .006682
Effective rate ITJI6J16^
Effective rate to be used semiannually (5.631682
-7-2) 2.815841%
402
BOND AND BOND INTEREST VALUATION
TABLE SHOWING AMORTIZATION OF DISCOUNT FOR SERIES
OF SERIAL REDEMPTION BONDS
End of Bonds
Effective Rate,
Coupon Rate,
Amortization
Bonds Less
Period Redeemed
2.815841 %
2.5%
of Discount
Discount
$9,600 00
1
$270.32
$250.00
$20 32
9,620 32
2
270 89
250 00
20 89
9,641 21
3
271 48
250 00
21 48
9,662.69
4
272 09
250 00
22 09
9,684 78
5
272.71
250 00
22 71
9,707 49
6
273 35
250 00
23 35
9,730 84
7
274 01
250 00
24 01
9,754 85
8
274 68
250 00
24.68
9,779 53
9
275 38
250.00
25 38
9,804 91
10
276 09
250 00
26.09
9,831 00
11
276 83
250 00
26 83
9,857 83
12 $3,000 00
277 58
250 00
27 58
6,885 41
13
193 88
175.00
18.88
6,904 29
14
194.41
175.00
19 41
6,923 70
15
194 96
175.00
19 96
6,943 66
16 3,000 00
195.52
175.00
20 52
3,964 18
17
111.63
100 00
11.63
3,975 81
18 2,000 00
111 95
100 00
11 95
1,987 76
19
55 97
50.00
5 97
1,993 73
20 2,000 00
56 14
50 00
6 14
.13*
* Error caused by
approximation
and by the use
of bond tables
having only
places.
Problems
1.* On January 1, 1934, a corporation floated a bond issue of $300,000 to be
retired serially over a period of 8 years as follows :
Dec. 31, 1934
Dec. 31, 1935
Dec. 31, 1936
Dec. 31, 1937
$10,000 Dec. 31, 1938 $ 30,000
15,000 Dec. 31, 1939 . 35,000
20,000 Dec. 31, 1940 40,000
25,000 Dec. 31, 1941 125,000
The discount and expense of issuing the bonds amounted to $33,000.
Draft a schedule, showing how much of such bond discount and interest you
would claim as a deduction from gross income for federal income tax purposes
for each of the years 1934 to 1941, inclusive.
2.* A city wishes to buy new fire equipment. The cost will be $500,000,
and the equipment will have an estimated life of 10 years, and no salvage value.
It is necessary to issue bonds to pay for this purchase, although, at the present
time, interest rates are high — 6 %, payable annually.
How would you suggest that these bonds be issued, and what will be the
annual cost to the taxpayers?
It is expected that a sinking fund would not earn more than an average
of 3%.
' American Institute Examination.
BOND AND BOND INTEREST VALUATION 403
The bonds will be issued in denominations of $100 and multiples thereof.
Given:
(LOG)9 = 1.679479 (1.06)10 = 1.790848
(1.03)9 = 1.304773 (1.03)10 = 1.343916
3.* A series of 5% bonds totalling $100,000, dated January 1, 1934, is redeem-
able at par by ten annual payments of $10,000 each, beginning December 31,
1944. What equal annual payments to a sinking fund are required to be pro-
vided on a 4% basis in order to pay off the bonds as they mature?
The first payment to the sinking-fund trustees is to be made on December 3 1 ,
1934, and the further payments are to be made annually thereafter.
What is the status of the sinking fund on December 31, 1943, 1944, and
1945?
Given at 4%:
(1 + t)10 = 1.48024428 (1 + i)20 = 2.19112312
v™ = .67556417 v20 = .45638695
Given at 5%:
(1 + t)10 = 1.6288946 (1 + i)20 = 2.6532977
i'10 = .6139133 v20 = .3768895
4. On June 1, a corporation sold a $3,000,000 issue of 6%, first mortgage,
20-year bonds at a discount of $300,000. According to the terms of sale these
bonds were to be retired at the rate of $150,000 each year, the purchases to be
made in the open market. The first retirement in the amount of $75,000 was
to be made on March 1 following the date of issue, and $75,000 was to be retired
each six months thereafter.
Any premium paid or discount received on bonds purchased for retirement
was to }>e added or deducted, whichever the case might be, to that year's portion
of discount, and was to be amortized as shown by the schedule of amortization.
Prepare a schedule, showing the amortization of bond discount by the bonds
outstanding method.
5. An issue of $175,000 of 6^% bonds was sold for 95, the amount of the
discount being $8,750. Other expenses pertaining to the issue of the bonds
amounted to $596.67, making the total of bond discount and expense $9,346.67.
These bonds are dated March 30, 1944, arid the due dates are as follows:
$ 3,500 due April 15, 1944 $ 12,000 due April 15, 1949
5,000 due April 15, 1945 14,500 due April 15, 1950
7,000 due April 15, 1946 15,000 due April 15, 1951
8,000 due April 15, 1947 100,000 due April 15, 1952
10,000 due April 15, 1948
Prepare a schedule, showing the annual charge for amortization of bond
discount and expense as of the close of each year, December 31.
6.f On April 1, 1934, Southern Railway Equipment Gold 4^'s, due serially
on each successive coupon date in October and April between October 1, 1934,
up to, and including, April 1, 1948, were sold to yield 4f %. The Bank of
* American Institute Examination.
t Moore, Justin H., Handbook of Financial Mathematics. New York, Prentice-
Hall, Inc., 1929.
404 BOND AND BOND INTEREST VALUATION
Montrose purchased 2 bonds due April 1, 1937, 4 due October 1, 1938, 1 due
October 1, 1942, 7 due April 1, 1943, and 1 due April 1, 1945.
(a) What was the total price paid? (b) Set up a schedule showing the
investor's situation in regard to the book value of this investment at the begin-
ning of each six-month period until the final maturity.
7.* On November 1, 1933, an investor purchased 20 bonds, each with a par
value of $1,000, and maturing as follows:
Number
of Bonds
May 1, 1934 1
Nov. 1, 1934 . 3
May 1, 1935 ... 2
May 1, 1936 5
May 1, 1937 9
The price paid for the 20 bonds was $20,417.11. The coupon rate is 6%,
payable May 1 and November 1.
(a) Determine the rate of yield, (b) Set up a schedule showing the investor's
situation in regard to the book value of this investment at the beginning of each
six-month period until the final maturity.
Review Problems
1. A serial issue of $20,000 in denominations of $1,000 with interest at 4%
has maturity of $1,000 at each interest date, March 1 arid September 1. Find
the price at the date of issue, March 1, to yield the purchaser 3^%.
2. If, in Problem 1, the first maturity were to occur 2 years following the
date of issue, and then each six months thereafter, find the price at the date of
issue, March 1, that would yield the purchaser 3^ -%.
3. A $1,000 bond with interest at 4y%, maturing April 1, 1954 and redeem-
able at 102 on any interest date after January 1, 1945, was sold on July 21, 1942
on a 5% basis. What price was paid for it?
4. A loan of $5,000 at 5% payable scmianiiually is repayable on each interest
date in installments of $1,250. Find the purchase price to yield the investor 4%.
5. A loan of $5,000 with interest at 6% payable semiannually is to run 5 years
and then be redeemed in annual installments of $1,000. What is the purchase
price to yield 5% convertible semiannually?
* Mooro, Justin H., Handbook of Financial Mathematics New York, Prentice-
Hall, Inc., 1929.
CHAPTER 34
Asset Valuation Accounts
Asset valuation. At the moment that an asset is purchased
for use, it is said to be worth cost. When this asset can no longer
be used for the purpose for which it was purchased, and can be
sold for little or nothing, it is said to be worth scrap value. The
difference between the scrap value and the cost is the depreciation.
Depreciation. Depreciation is the decline in value of a physi-
cal asset caused by wear, tear, action of the elements, obsolescence,
supersession, or inadequacy and resulting in an impairment of
operating effectiveness.
Depletion. Depletion is the progressive extinction of a wasting
asset by a reduction in the quantity. A coal mine is depleted
year by year as the coal is mined.
Depreciation methods. There are many methods of calculat-
ing the periodic depreciation charge. Probably the following are
the most serviceable:
(1) Straight-line method.
(2) Working-hours method, or unit-product method.
(3) Sum-of-digits method.
(4) Sinking-fund method.
(5) Annuity method.
(6) Fixed-percentage-of-diminishing-value method.
The accountant should know how to calculate depreciation by
each method, should be able to discuss the advantages and dis-
advantages of each, and should be able to formulate tables of
comparison showing the results of each.
Straight -line method. This is the simplest method, and is the
one most commonly used.
Procedure: (a) Find the difference between the cost and the
scrap value.
(6) Divide the difference found in (a) by the number of periods
which the asset is expected to be of service. The result is the
depreciation charge per period.
Example
What will be the depreciation charge and the asset valuation at the end of
each year for an asset costing $1,000, and having an estimated life of 10 years
and an estimated scrap value of $100?
405
406
ASSET VALUATION ACCOUNTS
Formula
Cost — Scrap = Depreciation
Depreciation -f- Number of years = Annual charge
A rithmetical Substitution
$1,000 - $100 - $900
$900 4- 10 = $90
TABLE OF DEPRECIATION
Periodic
Depreciation
Years Charge
1
2
3
4
5
6
7
8
9
10
$90
90
90
90
90
90
90
90
90
90
Accumulated
Depreciation
Reserve
$ 90
180
270
360
450
540
630
720
810
900
Asset
Value
$1,000
910
820
730
f>4()
550
460
370
280
190
100
Problems
1. Set up a table showing the annual depreciation, carrying value, and
accumulated depreciation for an asset costing $3,500, and having a life of 10
years and an estimated scrap value of $500.
2. Prepare tables showing by the straight-line method the depreciation on
the following machines:
Assets Cost
Lathes $5,000
Milling machines 3,000
Power equipment 4,100
Furniture 600
Estimated
tier a p Yalw
$600
400
200
150
Estimated
Life in
Years
10
12
10
15
3. The following fixed assets belong to the Western Hardware Company:
Asset Cost
Buildings .. $100,000
Machinery 70,000
Tools 20,000
Patterns 10,000
Estimated
Scrap Value
$35,000
25,000
5,000
none
Estimated
Life in Years
20
15
10
8
Compute the amount of annual depreciation by the straight-line method.
4. Complete the following schedule of fixed assets and depreciation, for the
purpose of supporting an income tax return (allow six months' average or
additions) :
ASSET VALUATION ACCOUNTS
407
Previous
Current
Cost
Additions
Kate
Reserve
Depreciation
. $75,000 00
none
none
none
none
s
92,519 61
SI, 046 91
3%
$ 9,461 92
$
ngs
35,654 IS
1,126 19
5%
12,319 14
1 tools
51,252 19
9,217.62
10%
11,463 21
B
3,469 52
417 51
10%
1,121 44
- .
icks
. 3,219 52
1,750 19
25%
1,749 32
712 92
$""."...."./
none
$"" "...
5',o
12S 34
».."..."..".
$
Asset
Land
Brick buildings
Wooden buildings
Machinery and tools-
Office furniture
Automobile trucks
Spur track
Working -hours or unit-product method. This mot hod is based
on the number of hours which the asset is in use, or on the number
of units produced.
Example
A certain one-purpose machine which costs $1,000, and has no scrap value,
lias been installed in a factory. A machine of this class produces 10,000 units
of product during its life. Assuming that the annual production is as given
below, set up a table shoeing the depreciation to be written oil each year.
First year
1,000 units
Fifth year
1,000 units
Second year
2,000 "
Sixth year
1,200 "
Third year
1,SOO "
Seventh year
1,200 "
Fourth year
1,000 "
Eighth year .
. 800 "
Solution
TABLE OF DEPRECIATION
I'm!
Periodic
A ecu mutated
Fraction
Depreciation
Depreciation
Asset
Year
of Cost
Charge,
R( serve
Value
$1,000
[
1,000/10,000
$100
$ 100
900
2
2,000/10,000
200
300
700
3
1,800/10,000
ISO
480
520
4
1,000/10,000
100
580
420
5
1,000/10,000
100
080
320
6
1,200/10,000
120
SOO
200
7
1,200/10,000
120
920
80
8
800/10,000
80
1,000
0
Problems
1. Show in appropriate form the yearly depreciation, accumulated deprecia-
tion, and asset value of a machine which cost $7,400, and which will have a scrap
value of $200. Assume that machines of this class have a working-hour average
life of 24,000 hours; also assume that the machine will be run as follows:
First year 2,000 hours Sixth year 2,000 hours
Second year 2,000 " Seventh year 3,000 "
Third year 1,800 " Eighth year 3,000 "
Fourth year 2,600 " Ninth year . 3,000 "
Fifth year 2,800 " Tenth year 1,800 "
2. An aircraft motor costing $7,700, and having an estimated scrap value
of $1,000, is assumed to have a useful life of 2,000 hours. If this motor is oper-
408 ASSET VALUATION ACCOUNTS
ated 350 hours during a certain month, what should be the charge for depreciation
in that month?
Sum-of -digits method. Those who believe th it the deprecia-
tion charge should be large during the early years of the useful life
of the asset, will find the surn-of-digits method of value.
Procedure: (a) Find the sum of the digits, or numbers repre-
senting the periods of useful life of the asset. Use this sum as the
denominator of certain fractions.
(6) Use the same digits or numbers in inverse order as the
numerators of these fractions.
(c) Compute the periodic depreciation by multiplying the
total depreciation by the fractions obtained in (a) and (6).
Example
An asset is valued at $1,000, and has a scrap value of $100. What should
be the depreciation charges if the asset is to be written down in 9 years by the
sum-of-digits method?
Solution
TABLE OF DEPRECIATION
Year
1
2
3
4
5
0
7
8
9
45 45/45 $900
The denominator of the fractions used in the second column is found by
adding the first column. The numerators are the same numbers taken in
inverse order.
Problem
Compute by the sum-of-digits method the depreciation charges, the asset
valuation, and the depreciation provision for each year on each of the following
machines :
Life of Asset,
Asset Cost Scrap Years
Power lathe $1,300 $200 10
Hack saw 350 38 12
Turret lathe . 3,000 200 7
Boiler. . . ... 2,700 180 8
Power equipment . 800 50 5
Delivery equipment . . . 2,000 500 5
Present the information by means of table?
Periodic
Accumulated
Fractional
Depreciation
Depreciation
Part
Charge
Reserve
Atawt Value
$1,000
9/45
$180
$180
820
8/45
160
340
660
7/45
140
480
520
6/45
120
600
400
5/45
100
700
300
4/45
80
780
220
3/45
b'O
840
160
2/45
40
880
120
1/45
20
900
100
ASSET VALUATION ACCOUNTS 409
Sinking-fund method. This method is based on the assump-
tion that a fund will be provided to replace the asset at the expira-
tion of its life. Seldom is such a fund provided, but the method
can be applied without actually accumulating the fund. The
procedure is the same as though the fund were actually created.
Procedure : (a) Find the amount of the total depreciation of the
asset by deducting the scrap value from the cost.
(6) Divide the total depreciation found in (a) by the amount
of an ordinary annuity of 1 at the sinking fund interest rate, for the
number of periods of the life of the asset, s^t-. This will give the
periodic sum to be placed in the sinking fund.
(c) To the periodic sum found in (6), add a sum equal to the
interest on the sinking fund for the period. This will give the
periodic charge to depreciation, and the credit to reserve for
depreciation.
Example
An asset costs $1,000, and has a scrap value of $100 at the end of 10 years.
Determine the periodic depreciation charge by the sinking-fund method, on a
(>% interest basis.
Formula Arithmetical Substitution
Cost - Scrap 0 . .. . ., 1,000-100
— — = Periodic deposit /,~^^r;,r~ , = $(>8.28.
to sinking fund. (l^^l
* .06
The formula and substitution just shown give only the first periodic charge
to depreciation. Each periodic charge thereafter is an amount equal to the
sum of the first periodic payment and the interest on the accumulated depreci-
ation reserve. Table II, on page 410, shows the periodic charges to depreciation
and the credits to the reserve account for each of the 10 years.
The depreciation entries are independent of the fund entries. If a sinking
fund were provided for the above example, the entries for the fund would be as
shown in Table I:
TABLE I— ENTRIES TO THE SINKING FUND
Debit to Credit to Credit to Accumulation
Year
Sinking Fund
Cash
Interest
of Fund
1
$ 68 28
$ 68.28
$.-.
$ 68 28
2
72 38
68.28
4.10
140 66
3
76 72
68 28
8.44
217.38
4
81.32
68 28
13.04
298 70
5
86.20
68 28
17 92
384.90
6
91.37
68 28
23 09
476.27
7
96.86
68.28
28 58
573.13
8
102.67
68.28
34.39
675.80
9
108 83
68.28
40 55
784.63
10
115 37
68.29
47.08
900.00
$900.00 $682.81 $217.19
410 ASSET VALUATION ACCOUNTS
TABLE II— DEPRECIATION ENTRIES BY THE SINKING-FUND
METHOD OF DEPRECIATION
Depreciation Accumulated
End of
Year
Charge and
Reserve Credit
Depreciation
Reserve
Asset
Value
$1,000 00
1
$ 68 28
$ 68 28
931 72
2
72 38
140 66
859 34
3
76 72
217 38
782 62
4
81 32
298 70
701 30
5
86 20
384 90
615 10
6
91 37
476 27
523 7£
7
96 86
573.13
426 87
8
102 67
675 80
324 20
9
108 83
784 63
215 37
10
115 37
900 00
100.00
Problems
1. Set up sinking fund depreciation tables for the following assets, using 5%
as the sinking fund rate.
Asset Cost Scrap Life
Office furniture . $ 500 $ 125 8 years
Factory furniture 1,000 100 10 years
Machinery .. . 15,000 3,000 8 years
Delivery equipment . 4,000 500 5 years
2.* The owner of an unimproved building site who is desirous of developing
it so that it will produce an income, receives from a proposed lessee a proposition
relative to the erection of a building at a cost of $100,000. In calculating the
annual expenses, which are to be made the basis of rentals, the owner assumes
a life of 50 years for the proposed building, and calculates by the straight-line
method that the depreciation charge should be $2,000 per year. The prospective
lessee contends that this depreciation charge is too large, that, as depreciation
charged into expense does not represent actual expenditures, an amount of cash
equal to the depreciation charge should be set aside annually and invested in
interest-bearing securities, and the interest obtained annually reinvested. He
demonstrates by calculation that an annual depreciation charge of $477.68 so
handled will at an interest rate of 5% amount to $100,000 at the end of 50 years,
and hence argues that this amount rather than $2,000 should be taken into
consideration in determining the annual rental.
You are asked by the owner to give your opinion as to which of the two
methods should be used, and why. Give your answer.
Annuity method of depreciation. The theory applied in this
method is that the depreciation charge should include, in addition
to the amount credited to the reserve, interest on the carrying
value of the asset.
The investment in property is regarded, first, as the amount of
scrap value which draws interest, and second, as an investment in
* C. P. A., Wisconsin.
ASSET VALUATION ACCOUNTS 411
an annuity to be reduced by equal periodic amounts. The interest
on the scrap value plus the equal periodic reduction of the invest-
ment is the charge to depreciation, offset by a credit to interest
computed on the diminishing value of the property, and a credit to
the reserve account for the balance. This charge to depreciation
is the same each period during the life of the property. The theory
of an investment in an annuity is that the annuity is to be reduced
by equal periodic payments, and as the credits to interest wUl
decrease, the credits to the reserve must correspondingly increase.
Procedure: (a) Find the difference between the cost and the
scrap value.
(6) Divide the difference found in (a) by the present value of
an annuity of 1.
(c) Calculate the interest on the scrap value for one period at
the given rate per cent.
(d) Determine the sum of (6) and (c). This sum will be the
periodic charge to depreciation.
Example
Calculate by the annuity method the annual charge to depreciation for an
asset valued at $1,000, with a scrap value of $100, which is to be written off in
10 years on a 6% basis.
Formula
[ Cost — Scrap 1 . . . ._ . .. .
I „__ _j_ (h(.ra,p x i) = Periodic charge.
L ««I* J
A rith metica I Kiibstit ut io n
(LOG)1
Solution, Par I 1
$1,000 — $100 = $900, sum to be depreciated
(1.06)™ = 1.7908477, compound amount of 1 for 10
periods at 6%
1 -r- 1.790S477 - .5583948, present value of 1 for 10 periods
at 6%
1 - .5583948 = .4416052, compound discount on 1 for 10
periods at 6%
.4416052 -f- .06 = 7.360087, present value of an annuity of 1 for
10 years at 6%
$900 •*- 7.360087 = $122.28, rent of the present value of annuity
412 ASSET VALUATION ACCOUNTS
Solution, Part 2
$100 X .06 *= $6.00, interest on scrap value
Solution, Part 3
$122.28 -h $6.00 = $128.28, periodic charge to depreciation
In the above example, the $900 represents the present value of the sum to
be spread over the life of the asset, and the $100 represents the scrap value. In
the following tables, the fifth column always contains the carrying value of the
annuity, plus $100. The two tables are given to show the similarity between an
annuity in which an investment was made and equal annual rents withdrawn,
and the annuity method of depreciation.
TABLE OF REDUCTION OF AN ANNUITY
End of Rents Credits to Amortization of Present Value of
Period Withdrawn Interest Investment Annuity, Plus $100
$1,000 00
1 $ 128 28 $ 60 00 $ 68.28 931 72
2 128 28 55 90 72 38 859 34
3 128 28 51 56 76 72 782 62
4 128 28 46 96 81.32 701 50
5 128 28 42.08 86 20 615 10
6 128.28 36.91 91 37 523 73
7 12828 31.42 9686 42687
8 128 28 25.61 102 67 324 20
9 128 28 19.45 10S 83 215.37
10 128 29 12 92 115 37 100.00
£f,282 81 $382 81 $900. 00
TABLE OF REDUCTION OF THE VALUE OF AN ASSET
End of Depreciation Credits to Credits to Value of
Period
1
2
3
4 '
5
6
7
8
9
10
$1,282.81 $382.81 $900.00
Problem
The Acme Manufacturing Company, believing that the annuity method of
depreciation is the correct one, desires that you construct tables for the following
machines (one table for each) :
Charge
Interest
Reserve
Asset
$1,000 00
128.28
$ 60 00
$ 68 28
931 72
128 28
55 90
72 38
859 34
128 28
51.56
76.72
782 62
128 28
46.96
81 32
701 50
128 28
42 08
86 20
615 10
128 28
36 91
91 37
523 73
128 28
31.42
96 86
426 87
128.28
25 61
102 67
324 20
128.28
19 45
108 83
215 37
128 29
12.92
115 37
100.00
ASSET VALUATION ACCOUNTS 413
Interest
Assets Cost Scrap Life Rate
Lathes .................... $5,000 $ 500 10 years 5%
Milling machines ........... 4,500 1,000 8 years 5%
Grinders ................... 1 ,200 200 5 years 5%
Motors .................. 2,000 200 0 years 5%
Fixed-percentage-of-diminishing-value method. By this
method a uniform rate on diminishing value gives the amount to
be charged to depreciation each year. As the book value declines
each year, the percentage of book value declines similarly.
The difficulty encountered in this method is that of finding the
rite per cent to be used in the calculation of the charge.
Procedure: (a) Divide the scrap value by the cost.
(6) Extract the root, the index of which corresponds to the
number of periods of depreciation to be taken on the life of the
asset, of the quotient obtained in (a).
(c) Deduct from 1 the result obtained in (b), to find the rate
per cent to be used.
(d) Multiply the net asset value or the carrying value of the
asset at the beginning of each period by the rate found in (c), to
obtain the depreciation charge for each period.
Example
What will be the depreciation charges for an asset valued at $1,000, with a
scrap value of $100, which is to he written off in 10 years by the fixed-percentage-
of-diminishing- value method9
The following are the formula and solution for the calculation of the rate:
Formula Arithmetical Substitution
n/Sc7ap value t 10/ 100 on ™0/w
- V lvalue " '• ' ~ Vl,0 = 20-S072%'
10
,000
Solution, Part 1
100 -T- 1,000 = .1
log .1 = 1.000000
Changed, T.OOOOOO = 9.000000 - 10
9.000000 - 10 -h 10 = .900000 - 1
Changed = 1.900000
The antilog of T.900000 = .79432K
1 - .794328 = .205672, or 20.567%
Solution, Part 2
$1,000 X 20.567% = $205.67, first depreciation charge
$1,000 - $205.67 = $794.33, new asset value
$794.33 X 20.567% = $163.37, second depreciation charge
his process is continued for each of the 10 years.
414
ASSET VALUATION ACCOUNTS
TABLE OF DEPRECIATION
(Rate, 20.567%)
Periodic
Accumulated
Depreciation
Depreciation
Asset
Year
Charge
Reserve
Value
$1,000 00
1
$205 67
$205 67
794 33
2
163.37
369 04
630 96
3
129.77
498 81
501 19
4
103 08
601 89
398 11
5
81 88
683 77
316 23
6
05 04
748 81
251 19
7
51 66
SOO 47
199 53
8
41 04
841 51
158 49
9
32 60
874 11
125 89
10
25 89
900 00
100 00
Problems
1. Construct a comparative columnar table showing the periodic depreciatior
charges computed by the straight-line, sum-of-digits, sinking-fund, annuity, and
fixed-percentage-of-diminishing-value methods for an asset costing $10,000, and
having a probable life of 10 years and a scrap value of $1,500. Use an interest
rate of 4% per annum.
2. An asset costing $2,000 has a life of 5 years. It has no scrap value, but
for the purposes of calculation, use $1. Money is worth 5%. Construct com
partitive columnar tables showing the carrying value and the annual depreciation
charge computed by the straight-line, sum-of-digits, sinking-fund, annuity, and
fixed-percentage-of-diminishing-value methods.
3. A businessman, having heard much about correct depreciation but under-
standing little of the methods of calculation used, calls on you to explain to him
by means of comparison the five most important methods. As an illustration,
use an asset costing $4,000, with a scrap value of $500 and a life of 5 years, and
an interest rate of 6%.
Composite life. Often the depreciable assets of a business have
wearing values (cost less scrap value) and teiins of effective life
that vary widely, yet it is desirable to ascertain the life of the plant
as a whole, as when bonds secured by a mortgage on buildings and
equipment are issued. The bonds should not be issued for a term
of years exceeding the composite life of the plant; and, for a margin
of safety, the term of the bonds should be considerably shorter than
the composite life of the plant.
If interest is not a factor, the composite life is found by dividing
the total wearing value by the total depreciation.
Wearing Annual
Cost 8 crap Value Charge
$45,000 $5,000 $40,000 $ 800
12,000 2,000 10,000 400
30,000 5,000 25,000 1,000
12,000 2,000 10,000 500
Unit Life
Bldg. (Brick) 50
Bldg. (Frame) 25
Heavy Machinery 25
Boiler 20
85,000 -^ 2,700
$85,000
31.5, approximate years
$2,700
ASSET VALUATION ACCOUNTS 415
If interest is a factor, as when depreciation is computed on the
sinking fund basis, the annual charge for depreciation is the annual
rent, the accumulation of which should equal the wearing value.
Making use of the data from above and using 5% as the interest
rate, we have:
Unit Life Wearing Value Annual Charge
Bldg. (Brick) 50 $40,000 $ 191 .07
Bldg. (Frame) 25 10,000 209 . 53
Heavy Machinery 25 25,000 523 81
Boiler 20 J0,000 J*02 .43
$85,000 $17226784
1,226.84 -T- 85,000 = 1.4433%
The per cent is low because the major asset lias a 50-year life,
and because over long periods the interest is also a large factor.
To find the composite life, let r denote the rate of depreciation
and i the interest rate. Then,
loK(l+i)
loi
For the foregoing problem
log ( 1 + i)
log 1.05
__ log 4.464 _ .649724
loig'LOS ~~ .02U89
.649724 4- .021189 = 30.66 years.
Problems
1. A company's plant consists of: (a) Buildings: cost, $100,000; life, 40 yean»»
scrap value, $10,000. (6) Engine: cost, $40,000; life, 25 years; scrap valu*»-
$5,000. (c) Boiler: cost, $12,000; life, 20 years; scrap value, $2,000. (d) Elec-
trical equipment: cost, $7,500; life, 15 years; scrap value, $1,500. Compute the
charge for depreciation by the sinking fund method, on a 4% basis, and find the
composite life of the plant.
2. Find the composite life of a plant consisting of the following:
Item
A
B
C
D
Interest at 5%.
Depreciation Problems from C. P. A. Examinations
1.* A manufacturing company has a factory building which cost, with its
equipment, $100,000. The company has set up a depreciation reserve of $30,000.
* C. P. A., Michigan.
Cost Scrap Value Life
$ 6,000
$ 200
10 years
3,500
500
15 years
12,000
1,000
20 years
15,000
2,500
12 years
416 ASSET VALUATION ACCOUNTS
An appraisal made shows the replacement cost to be $160,000, and the depreci-
ated sound value to be $130,000.
(a) Prepare the necessary entries to give effect to the appraisal figures.
(6) How would you treat the item of depreciation on the increased values,
for the purpose of determining costs?
2.* A machine which cost $1,200 has been used for 5 years, and has depreci-
ated annually 10%. The latter amount has been credited to the Reserve
account.
(a) At the end of the first 5 years, the machine is traded for a new one which
costs $1,500; an allowance of $300 is made on the old machine, the balance being
paid in cash. Prepare the necessary entries to take care of this transaction.
(6) Assume that the machine is traded for one costing $1, 700, and that an
allowance of $700 is made, the balance being paid in cash. Prepare the necessary
entries.
3.f A manufacturing plant, operating to the date of negotiations relative
to its disposition, was acquired by a newly formed corporation, the price being
based on the present sound values, which were stated as follows:
Present
tiound Value Age
Machinery $116,500 4^- years
26,300 4 years
217,300 2\ years
16,750 2 years
57,550 I year
Equipment $ 13,300 6 years
1 1 ,650 2 years
27,660 1 year
Buildings: A . $285,700 12 years
A . 15,000 5i years
A . 16,600 1 year
B . 525,000 5 years
The estimated life of the machinery is 10 years from the date of original
purchase; of the equipment, 15 years from the (late of purchase; of buildings A,
30 years; and of building #, 45 years.
It is desired to set up the assets on the books at present reproductive values,
with a corresponding depreciation reserve to bring the net book value to the
amount of the "sound values " given above. Compute the " reproductive value "
and the depreciation reserve, and give the future annual depreciation provision,
all on the basis of a uniform rate each year until the book value is extinguished.
It may be assumed for the purposes of your answer that the assets will have
no salvage value.
4.f The City Dairy Company bottles and distributes milk. Its sales average
40,000 Ibs. per day. It operates three pasteurizers, each of which has a capacity
of 2,500 Ibs. per hour.
These machines cost $1,200 apiece, installed, and they have been in use for
3 years. At the time that they were installed, their life was estimated at 15
*C. P. A., Michigan.
| American Institute Examination.
J C. P. A., Wisconsin.
ASSET VALUATION ACCOUNTS 417
years, and their salvage value at the end of that period at $50 each. Experience
has shown that repair and maintenance charges on these machines will average
3 % of their cost per year.
The Smith Dairy Machinery Company manufactures a new type of machine
which is guaranteed to have a productive capacity of 12,000 Ibs. per hour, and
to save, in comparison with the old type, 80% of the cost of live steam and
refrigeration used in pasteurization. The life of these machines is estimated at
10 years, and their salvage value at the end of that period at $100 each. The
manufacturers offer to install them ready to operate at $8,500 each, and to
remove the old machines and make an allowance for their estimated scrap value.
Their guarantee also provides for replacement of broken or defective parts, and
for complete maintenance to keep the machines in good working order for 1 year.
Subsequent repair and maintenance charges may be assumed to average 5%
annually of the machines' original cost.
You are called upon to make a special examination of the accounts, with a
view to determining whether it would be advantageous from a profit and loss
standpoint to install the new type of machine. Your examination discloses that
the present cost of live steam and refrigeration used in pasteurization averages
44^ per 1,000 Ibs. of milk.
Required: (a) Assuming future production to average 5% increase over the
sales given above, state the conclusions that you would report to your client,
and show your method of arriving at them.
(6) Assuming that the manufacturer's proposition has been accepted, draft
the entries which you consider should be made to record the changes.
5.* The Plant and Equipment and Reserve for Depreciation accounts, pre-
sented below, represent the transactions of the A. Company for the year 1944,
as recorded by the bookkeeper, from January 1 of that year.
PLANT AND EQUIPMENT
1944 1944
Jan. 1 Balance $500,000 Jan. 31 Screw-cutting lathe. $ 150
17 Planer 2,000 Apr. 17 Steam engine 300
Mar. 21 Bolt machine . 1,250 Sept. 30 Steel and lumber. . . 400
Apr. 16 Crane 3,000 Dec. 31 Balance 524,650
May 3 Electrical equipment
for crane . . . . 1,200
27 Roof of machine shop 3,500
June 3 Lathe belting . . . 750
Aug. 20 Wm. Smith, Con-
tractor 10,000
Dec. 31 Machine shop. . .. 3,800
«525,500 $525,500
RESERVE FOR DEPRECIATION— PLANT AND EQUIPMENT
1944 1944
Dec. 31 Balance $175,000 Jan. 1 Balance $125,000
Dec. 31 Depreciation at 10%
per annum 50,000
$175,000 $175,000
1 Adapted from American Institute Examination.
418 ASSET VALUATION ACCOUNTS
The following is a description of the transactions; you are required to make
any entries that you deem necessary to correct the accounts, giving reasons
therefor, and setting up corrected accounts.
The balances at the beginning are assumed to be correct.
Planer, $2,000, is a standard machine, purchased new.
Bolt machine was made in company's own shop. The SI, 250 represents
cost of castings, $500, and direct labor, $750. The machine shop pay roll was
$20,000 ($15,000 direct, and $5,000 indirect) during the year; castings and parts
purchased were $17,000; general supplies were $4,000; rent was $2,500; light
heat, and power were $3,500.
Crane and equipment, $4,200, are standard machinery, purchased new.
Roof of machine shop was destroyed by weight of snow during the winter.
Belting for all equipment, amounting to $25,000, was charged to plant and
equipment when the plant was opened, and has not siiu-e been depreciated.
William Smith is engaged in erecting an addition to the plant buildings.
$10,000 is the first payment on the uncompleted work.
Machine shop, $3,800, represents the cost of making tools, setting machines,
•ind installing new machinery, as follows:
Tool making $1,000
Setting machines for special work . 1,800
Installing planer . . . 300
Installing bolt machine* ... . . . 200
Installing crane . . 500
$3,800
Screw-cutting lathe— cost, 1937, $2,000.
Steam engine—cost, 1934, $15,000.
Steel and lumber, $400, represents salvage from machine shop roof.
Prior to December 31, 1943, a separate account was kept for land and
buildings.
Ten per cent per annum depreciation on plant and equipment has been
written oil.
6.* A machine costing $81 is estimated to have a life of 4 years and a residual
value of $16. Prepare a statement showing the annual charge for depreciation
computed by each of the following methods: (a) straight-line; (b) constant-
pereentage-of-diminishmg-value; (r) annuity. (For convenience in arithmetical
calculation, assume the rate of interest to be 10%.)
Depletion. The provision for the extinction of wasting assets,
such as mines, timber lands, or gravel pits, is called a provision for
depletion.
We shall deal with two classes of problems; namely, the deter-
mination of the amount of depletion each year, and the capitaliza-
tion of the wasting asset.
Calculation of depletion. The amount of depletion usually
stands in the same proportion to the total cost of the wasting asset
as the units of product removed stand to the total units of product
that it was estimated the asset would produce when new; but if the
1 American Institute Examination.
ASSET VALUATION ACCOUNTS 419
property is leased and it is not possible to remove all the product
before the lease expires, it is plain that depletion should be based
on the quantity to be extracted during the period of the lease.
Problems
1. The estimated recoverable tonnage in a coal mine was placed at 2,214,363
tons. The value of "Coal Lands" was $90,443.62. Compute the depletion
charge per ton of coal mined.
2. A tract of timber was valued at $25,965.86, and its footage was estimated
at 17,228,000. The following year the timber cut was 5,184,336 feet. Compute
the rate of depletion per thousand feet, and the depletion charge for the year.
3. A mining property was valued at $50,000, and the estimated recoverable
tonnage placed at 1,000,\)00 tons.
During the next 7 years, tonnage was removed as follows:
First year 50,000 tons
Second year . 00,000 tons
Third year 70,000 tons
Fourth year . . . X0,00l) tons
Fifth year . 80,000 tons
Sixth year 80,000 tons
Seventh year . 80,000 tons
Prospecting and development toward the close of the seventh year cost
$25,000, and resulted in an estimated recoverable tonnage of 2,000,000 tons.
Appreciation due to discovery was placed on the books at $60,000.
The tonnage removed during the eighth and ninth years was 100,000 tons
and 120,000 tons, respectively.
Calculate the annual charges for depletion.
[Suggestion: Set up two accounts, Mining Property (Cost), and Mining
Property (Discovery Value), and credit these accounts with the proper depletion
charges each year. This problem is illustrative of the complications that arise
when there is more than one valuation on a particular property.]
Capitalized cost. Capitalization is the cost of an indefinite
number of renewals of anything. The value is found as a per-
petuity, page 364. Capitalized cost is the cost of the renewals
plus the original cost. When an endowment fund provides for the
periodic replacement of a useful memorial, such as a building, a
bridge, and so forth, the amount of the fund is the capitalized cost
of the memorial.
Procedure: (a) Divide the cost of the asset by the given rate
per cent expressed decimally, to find the capitalization.
(6) Divide the capitalization found in (a) by the amount of an
ordinary annuity of 1 at the given rate per cent and for the time,
expressed in periods, which represents the life of the asset. (This
will give the amount of an endowment fund necessary to replace
the asset periodically at the end of each term of years.)
420 ASSET VALUATION ACCOUNTS
(c) Add the cost of the asset to the amount of the endowment
fund found in (6). The sum of these two items is the amount
necessary to provide for the first cost and for the replacement fund.
Example
It is desired to set aside a fund which will provide for the erection of a memorial
costing $2,000, and for the replacement of the memorial at the end of each 5 years.
Money is worth 6%. Find the amount of the fund.
Formula Arithmetical Substitution
Cost 2,000
t 0(>
— + Cost = Capitalized cost. ' + 2,000 - $7,913.21.
8 i (1.0o)° — 1
.06
Solution
2,000 -f- .()() - 33,333.33, capitalization of asset
(I. (Mi)5 - 1.33S225(), compound amount of 1 at 6%
for 5 year*
1.338225(5 - I - .33S225(i, compound interest on 1 at 6%
for 5 years
.33H2250 -i- .0<> = 5.(>37093, amount of ordinary annuity of
1 at (>% for 5 years
33,333.33 •*• 5.037093 = 5,913.21, sum available for investment at
compound interest
$5,913.21 -f $2,000 = $7,913.21, capitalized cost
Verification
$2,000 — present cost of memorial
$7,913.21 - $2,000 - $5,913.21, sum available for investment
at compound interest
$5,913.21 X 1.3382250 = $7,913.21, total fund 5 years hence
Problems
1. The cost of the Davis Memorial Building was $150,000. It is estimated
that the building will last 50 years. Money is worth 4%. Calculate the
capitalized cost of the building.
2. A philanthropist desires to provide a fund for the erection of a college
building costing $200,000, and for the replacement of the building at the end
of each 50 years. Money is worth 4%. Calculate the amount of the fund,
assuming that repairs and renewals necessary during the 50-year periods are
not to be paid for out of the fund.
Perpetuity providing for ordinary annual expenses and for
replacement of asset. Many endowments provide for ordinary
annual expenses as well as for the replacement of the asset.
Procedure: (a) Divide the cost of the asset by the given rate
per cent expressed decimally, to find the capitalization.
(6) Divide the capitalization of the asset by the amount of an
ordinary annuity of 1 at the given rate per cent for a number of
ASSET VALUATION ACCOUNTS 421
periods, expressed in years, equal to the life of the asset. This
will give the replacement fund.
(c) Divide the annual upkeep by the given rate per cent, to
find the capitalization of the upkeep fund.
(d) Add the original cost of the asset, the replacement fund,
and the upkeep fund to find the total fund necessary.
Example
The Rutledge Home cost $75,000. The annual expenses of running the
institution, including repairs and upkeep, are estimated at $5,000. If the life
of the building is 50 years, and money is worth 4%, what is the amount of the
fund necessary to pnnide for the perpetuity of the home?
Formula
Cost
, r^ A . o
— - + Cost + . = Sum.
,s i
A rithmcticnl Substitution
75,000
.04
Solution, J'art 1
$75,000 -5- .04 - $1,S75,000, capitalization
$1,S75,000 ~ $15200708 = $12,281.03, building replacement fund
Solution, I* art 2
$75,000 = cost of building
Solution, Part ft
$5,000 -f- .04 = $125,000, upkeep fund
Summary
Building fund $ 12,281.03
Cost of building 75,000 00
Upkeep fund 125,000 00
Total fund necessary . $212,281 63
Problems
1. The original cost of a public library is $100,000, the estimated life of the
building is 50 years, and the annual cost of upkeep is $12,000. Money is worth
4%. Calculate the amount of the fund necessary to provide for the perpetuity
of the library.
2. The cost of a memorial is $55,000, its life is 25 years, and its upkeep is
$2,400 a year. Money is worth 4%. Calculate the amount of the fund neces-
sary to provide for building the memorial, keeping it in repair, and replacing it
at the end of each 25-year period.
422 ASSET VALUATION ACCOUNTS
3. A charitable institution is to be built and maintained by a trust com-
mittee. It is estimated that the building will cost SI 00,000, arid that its life
will be 60 years. The estimated annual income and costs are as follows:
Income :
Donations $4,000
Expenses :
Miscellaneous service charges . . . . . 1,000
Heat ... 2,000
Matron ... . 1,800
Help . 3,000
Food . . . 5,000
Medical attention . . . . 1,200
Incidentals ... . . . 1,000
Find the value of the endowment at 5%, interest convertible annually.
Capitalization of a wasting asset. Investment in a wasting
asset such as a mine or timberlands should yield not only interest
on the investment, but additional income to provide funds for the
restoration of the capital originally invested. A sinking fund,
(•ailed a redemption fund, is created for the purpose of restoring the
original capital. The investment rate will usually be higher than
the rate which can be earned on the redemption fund.
To find the fair market value at which a wasting asset may be
capitalized, divide the estimated annual average income by the
sum of : (a) the rent of an ordinary annuity of 1 at the redemption
fund rate for the length of time estimated to deplete the asset
(this to provide the annual sinking fund payment to restore the
investment), and (6) a fair annual rate of income (this to provide a
return to the investor).
Example
A mine produces an average annual operating income of $10,000. At tli<*
present rate of depletion, it is estimated that the mine will last 8 years. If
the sinking fund will earn 4%, and the owners are to receive a dividend of 6%.
what should be the capitalized value?
Formula
Operating profit
h Dividend rate
Capitalized value
Arithmetical Substitution
.04
ASSET VALUATION ACCOUNTS
Solution
(1.04)8 = 1.3685691, compound amount of 1 for 8
periods at 4%
1.3685691 - 1 « .3685691, compound interest on 1 for 8
periods at 4%
.3685691 ^ .04 = 9.214226, amount of annuity of 1 for 8
periods at 4%
1 •*- 9.214226 - .1085278, the rent of an annuity that will
amount to I
.1085278 + .06 = .1685278, rent of annuity plus the dividend
rate
$10,000 -v- .1685278 = $59,337.39, capitalized value
423
TABLE OF
Annual
Years Income
\ $10,000
2 10,000
3 10,000
4 10,000
5 10,000
6 10,000
7 10,000
8 10,000
CAPITALIZATION OF
6% on Sinking
Capitalized Fund
Value Payments
$3,560 24 $6,439.76
3,560 24 6,439 76
3,560 24 6,439 76
3,560 24 0,439 76
3,560 24 6,439 76
3,560 24 6,439 76
3,560.24 6,439 76
3,560.24 6,439 76
WASTING
Interest
on Sinking
Fund
$ .
ASSET
Sinking Fund
Accumulations
$ 6,439 76
13,137 11
20,102 35
27,346 20
34,879 81
42,714 76
50,863 11
59,337.39
257 59
525 48
804 09
1,093 85
1,395 19
1,708 59
2,034 52
Problems
1. Find the capitalized value of a coal mine which will produce a net income
of $20,000 a year for 30 years; the annual income rate is 6%, and a sinking fund
is to be accumulated at 4%.
2. A tract of timber will yield an annual revenue of $20,000 for 20 years.
If annual dividends are declared at 5%, and payments are made annually into
a sinking fund which bears 4% interest, what is the value of the timber rights?
3. A gravel pit is estimated to contain 3,500,000 cubic yards of gravel. This
pit is leased at a royalty of 10^ per cubic yard of gravel extracted, and the average
annual output is 150,000 cubic yards. If a 6% dividend is paid on the stock,
and a fund equal to the capital stock is accumulated at 4%, what should be the
capitalized value of the property?
Review Problems
1. A machine that cost $1,000 is estimated to have a life of 10 years and a
scrap value of $200. Compute the annual depreciation charge by:
(a) The straight line method.
(6) The fixed percentage of diminishing value method.
(c) The sinking fund method, using 6% effective interest
(d) The annuity method, using 6% effective interest.
2. A mine with a net annual yield of $75,000 will be exhausted in 15 years
at the present rate of output. What is the mine worth, on a 5% basis?
424 ASSET VALUATION ACCOUNTS
3. A certain make of bench drill costs $17.50 and lasts 3 years. How much
can be paid for a better grade of drill that will last 6 years, money being worth 4 %?
(HINT: Capitalized costs must be equal. Solve for x.)
4. A roof made of one material will cost $300 and last for 20 years. If made
of another type of material, it will last for the life of the building, which is esti-
mated to be 75 years. How much can one afford to pay for the permanent type
of roof if money is worth 5%?
6. Calculate the fixed percentage to be written off each year for an asset
costing $2,400, estimated life 6 years and scrap value $400.
CHAPTER 35
Building and Loan Associations
Control. Building and loan associations are organized under
state laws, and in most cases are under the direct supervision of
the state banking department. The banking department requires
semiannual or annual reports, and the associations are subject to
special examinations from time to time by the state bank exam-
iners. In addition to this, annual audits are usually made by
committees of stockholders, and in a great many cases independent
audits are made by certified public accountants.
Classes of stock. The amount of capital stock of a building
and loan association is fixed by the charter, and the minimum
capital and the par value per share are stated at the beginning of
the charter. The classes of stock issued are installment stock and
fully-paid stock.
Installment stock. Ownership of this class of stock is generally
evidenced by a pass book, in which the weekly or monthly pay-
ments are recorded. The monthly payments are usually $1 for
each share with a maturity value of $200, or 50ji for each share with
a maturity value of $100. In some organizations, if it is desired
to mature the shares in a shorter time, double payments may be
made. Shares of this class participate in all the earnings of the
association.
When the monthly installments, called dues, and the profits or
dividends credited to the stockholder equal the face value of the
shares, the shares are matured or paid-up. The amount may then
be withdrawn, or fully-paid shares may be issued.
If the stockholder, also called a member, has borrowed from the
association, the maturing of his stock effects a reduction in his
indebtedness.
Fully-paid stock. Ownership of this class of stock is evidenced
by a stock certificate, and the stockholder usually receives a given
rate of interest, although in some cases stockholders participate in
earnings in the same manner as holders of installment stock.
Withdrawal of funds. In most cases, funds deposited in an
association must be left for at least six months. After that time
any member may withdraw all or a part of his funds, together with
the dividends credited and not already paid. In the event of with-
drawal before the maturity of the shares, some associations retain
425
426 BUILDING AND LOAN ASSOCIATIONS
a membership or withdrawal fee of 2% of the par value of each
share, and issue membership certificates which are transferable
upon the books of the association. The member may retain this
certificate, or may assign it to someone else, as he sees fit. When
another account is opened in the association, either by the member
or by anyone else holding his certificate of membership, credit is
given for the amount which the certificate represents.
Other associations permit the withdrawal of the face amount
of the deposits, and allow the holder interest for the equated time.
The rate of interest is fixed by the association, arid is usually lower
than the per cent earned by the shares. This difference in rates
results in a profit to the association. Such profit on withdrawals
is added to the other profits, and distributed to the shares remain-
ing in the association.
Plans of organization. There are three principal plans upon
which building and loan associations are organized : the terminating
plan; the serial plan; and the permanent or perpetual plan, also
called the Dayton or Ohio plan.
Terminating plan. The life of the association is limited under
the terminating plan to any number of years that may be agreed
upon. When the specified number of years elapses, the association
goes out of existence, or a new one may be formed for another
period of time. On becoming a member subsequent to the date
of issue of the stock, the purchaser pays the book value of the share
or shares purchased. The book value consists of back dues, plus
dividends that have accumulated to the credit of the stock since
the date of original issue.
The amount of the monthly payments necessary to retire the
stock is determined in the same manner as is the rent of the present
worth of an annuity.
Determination of the amount to be paid monthly under the termi-
nating plan.
Procedure : (a) Compute the interest rate per period by dividing
the annual rate by the number of times that conversion takes place.
(6) Compute the number of periods by multiplying the time
in years by the number of times that the interest is converted
annually.
(c) Compute the present value of an annuity of 1, using the
periodic rate found in (a) and the number of periods found in (6).
(d) Divide the par value of one share by the present value of
an annuity of 1 found in (c).
Example
What should be the monthly payment per share, each share having a par
BUILDING AND LOAN ASSOCIATIONS 427
value of $100, if it is desired to mature the shares in 5 years, money being worth
6%?
Solution
p
— = Monthly payment
«„;»"- 51 .7256
100 -5- 51.7256 = 1.933
.*. $1.93 = Monthly payment.
By tliis plan, the association would be dissolved at the end of the 5-year
period.
Serial plan. Under this plan, stock is issued at specified dates.
Each issue constitutes a new series, and shares in the profits in
proportion to the length of time that the series is outstanding.
The distribution of profits under this plan is similar to the distri-
bution in a partnership using the average investment method.
New members may pay all back payments to the beginning of the
current series plus a charge for interest on these payments, or they
may wait for a new series to begin; a new series may begin annually,
semiannually, quarterly, or monthly, depending on the demand for
shares. As soon as a new series is opened, issuance of shares in the
previous series is stopped.
Distribution of profits. When several series of stock, maturing
at as many different dates, have been issued, the distribution of
profits is a complicated matter. Of the many methods of deter-
mining the proper profit distribution, the most familiar are the
partnership method and Dexter's Method, commonly called
"Dexter's Rule." The purpose of these methods is to secure an
equitable distribution of profits among the members of the associa-
tion, upon the basis of the amounts that they have contributed
against the face value of the shares registered in their names, and
upon the basis of the time that each dollar paid by the association
members has been in the possession of the association.
Partnership method. The name indicates the substance of
this method, and an example and solution are given to show its
application.
Example
The Alpha Building and Loan Association issued six series of shares, as
follows:
Series
1
Date Number of Shares
Jan. 1 1942 500
2
3
July 1
Jan. 1
1942
1943
500
400
4
5
July 1
Jan. 1
1943
1944
300
400
6
July 1
1944
400
428
BUILDING AND LOAN ASSOCIATIONS
The dues in each series were $1 a share, payable monthly. The net profits
from interest, fines, and so forth, less the operating expenses for the half-year
ended Dec. 31, 1944, were $965.22, and the undivided profits on July 1, 1944,
were $4,272.78.
The status of the shares on July 1, 1944, was as follows:
Series
1
2
3
4
5
Date of Issue Shares
Jan. 1, 1942 500
July , 1942 500
1943 400
1943 300
1944 400
Number of Paid per Profit per Value per
Jan.
July
Jan.
Share
Share
Share
$30
$4 12
$34 12
24
2.66
26 66
18
1.51
19 51
12
.69
12 69
6
.19
6 19
Distribute the profits fur the half-year ending December 31, 1944, by the
partnership method.
Solution
On December 31, 1944, the amount paid on each share of each series was
as follows:
Paid per Paid per
Series Share Series Share
\ $36 4 .... $18
2 30 5 .12
3 24 6 6
Dues of $1 have been paid at the beginning of the month on each of the
500 shares in the first series for 36 months. The average time is found to be
18^ months. That is, $1 was invested for 36 months; $1 for 35 months; $1 for
34 months; and so forth. The average, 18^ months, is the sum of the first term
and the last term, divided by 2; thus, - - — = 18^. Solving for series 2, 3, 4, 5,
2i
and 6, we have 15^-, 12^-, 9^, 63, and 3^, respectively, as the average time for
each series.
As there are 500 shares in the first series, and each share has paid $36, the
capital of the first series is $18,000. $18,000 for 18^ months equals $333,000 for
1 month. Solving for each series, we have the following:
Series Dollars for 1 Month
2 30 X 15* " X 500
232 500
3 24 X 12^ " X 400
120,000
4 18X9^ " X 300
51,300
5 12 X 6^ " X 400
31 200
6 6 X 3^ " X 400
8 400
Undivided profits, July 1, 1944
$776,400
$ 4,272 78
Earnings, July 1, 1944, to Dec. 31,
1944 965 22
Profits available for distribution
$ 5,238.00
The profits available for distribution, $5,238, are prorated among the series
in the proportion that the equated capital of each series bears to the total equated
capital, as follows:
BUILDING AND LOAN ASSOCIATIONS
429
Series
o QQQ
1 ' >4 of $5,238 = $2,246 59 for 500 shares, or $4.49 per share
" = 1,568 57 " 500 " " 3.14 "
" = 809 58 " 400 " " 2.02 "
" = 346.10 " 300 " " 1.15 "
" = 210 49 " 400 " " .53 " "
" = 56_67 " 400 " " .14 " "
$5,238~00
2 ^^
3 lj.200 lt
7,764
4 513 f<
7,764
5 312 •<
5 7,764
A
6
H4
7,764
Dexter' s rule for distribution of profits. This method is a modi-
fication of the partnership method. Its principles are illustrated
in the following solution, which is based on the example given
under the partnership method.
Solution
To the capital of each series on July 1, 1944, as shown by the column "Value
per Share" in the previous tabulation, should be added the contribution of $1 per
share for each of the six months of the current half-year, computed by the
average method.
1 paid
July
1
=
$1
for 6
months, or $ 6
for
moil
1 "
Aug.
1
=
1
1
5
a
< 5
n
i
1 "
Sept.
1
=
1
1
4
kl
' 4
(i
i
1 "
Oct.
1
=
1
1
3
n
' 3
it
1
1 "
Nov.
1
=
1
1
2
u
< 2
tl
<
1 "
Dec.
1
=
i
1 1
month, ' 1
" i
$21 " I "
$21 -T- 6 = $3.50, average for 0 months
Add $3.50 to the value of each share, and multiply by the number of shares,
to find the capital value of the series.
Series Shares
Dollars per Share
Capital per
Scries
1
500
$37
62
$18,810
00
2
500
30
16
15,080
00
3
400
23
01
9,204
00
4
300
16
19
4,857
00
5
400
9
69
3,876
00
6
400
3
50
1,400
00
Total capital
$53,227
66
The previous distributions of profits are unchanged. To these are added
the profits of the current period, computed on the basis of the present earning
capitals ; in other words, the capital $53,227 earned a profit of $965.22, or 1 .81 34 %.
430 BUILDING AND LOAN ASSOCIATIONS
Hence, the profit for each series and for each share in a series is calculated as
follows :
Profit
Profit per
per
Share for
Series
Capital
Rate
Series
Shares
Last Period
1
$18,810
.018134
$341.10
500
$0.68
2
15,080
.018134
273.46
500
.55
3
9,204
.018134
166 91
400
.42
4
4,857
.018134
88 08
300
.29
5
3,876
.018134
70 29
400
.18
6
1,400
.018134
25 38
400
.06
$965~22
Adding the profit per share for the last period to the value of the share as
of July 1, 1944, and including the monthly payments for the period, gives the
value per share as of December 31, 1944.
COMPARISON OF PARTNERSHIP PLAN AND DEXTER'S RULE
PARTNERSHIP PLAN DEXTER'S RULE
Profit per
Profit per
Profit per
Profit per
Share to
Share to
Share for
Share for
Difference
Series
July 1,1944
Dec. SI, 1944
Last 6 Mos.
Last 6 Mos.
Inc.
Dec.
1
$4 12
$4 49
$0 37
$0.68
$0.31
2
2 66
3.14
.48
.55
.07
3
1 51
2.02
.51
.42
$0.11
4
.69
1.15
.46
.29
.17
5
.19
.53
.34
.18
.16
6
.00
.14
.14
.06
.08
The above comparison shows that the partnership plan favors
the newer series at the expense of the old. This is unjust, because
it was really the old series that produced the profits. Dexter's
Rule is thus a more equitable method of calculating the distribu-
tion of profits in a building and loan association using the serial
plan.
Problems
1. A building and loan association issued a new series each year, as follows:
Number of
Series
Date of Issue
Shares
1
Jan. 1, 1941
300
2
Jan. 1, 1942
400
3
Jan. 1, 1943
500
4
Jan. 1, 1944
300
The dues were $1 per month per share. The profits to the end of the fourth
year were $4,800. Find the value of one share in each series at the end of the
fourth year, using the partnership method.
2. The first series of a certain building and loan association is 3 years old and
has 1,000 shares; the second series is 2 years old and has 500 sharas; and the
third series is 1 year old and has 400 shares. The net assets are $60,650.00
Payments were $1 per month per share.
BUILDING AND LOAN ASSOCIATIONS 431
By the partnership method, compute: (a) net profits; (b) profit for 1 share
in each series; (c) value of 1 share in each series.
3. By Dexter *s Rule, find the value as of June 30, 1943, of 1 share in each
series :
STATEMENT, DECEMBER 31, 1942
Number of Paid per Profit per Value per
Series Date Issued Shares Share Share Share
1 Jan. 1, 1941 500 $24 $3 32 $27 32
2 July I, 1941 400 18 I 89 19 89
3 Jan. 1, 1942 300 12 .86 12 86
4 July 1, 1942 400 6 23 6 23
The fifth scries was issued January 1, 1943, and comprised 300 shares. The
dues in each series were $1 per month per share. The profits for the six months
ended June 30, 1943, after all expenses had been deducted, amounted to $648.75.
4. The sixth of the above series was issued July 1, 1943, and comprised
400 shares. The net profits for the six months ended December 31, 1943, were
$698.75. Find the value of 1 share in each series as of December 31, 1943.
Withdrawal value. If 20 shares of the third series in the
example on page 427 are withdrawn on October 1, 1944, what is
their withdrawal value, assuming that the association allows 5%
interest on shares withdrawn before maturity?
Each share of the third series lias a paid up value on October
1, 1944, of $21; hence the 20 shares are worth $420. To this
amount add 5% interest for the equated time, 11 months,
( - — ]> or $19.25; this gives a withdrawal value of $439.25.
The book value of each share of this series is $22.51 ($21 in each
payment, plus $1.51 profits as of July 1, 1944). Twenty shares at
$22.51 gives a book value of $450.20. The difference, $10.95, is
the profit on the transaction, and makes the divisible profits for
the half-year $976.17 ($965.22 + $10.95).
Problems
(Based on example, page 427.)
1. Find the withdrawal value of 10 shares of the second series, withdrawn
December 31, 1944, interest allowed at 4%.
2. Find the withdrawal value of 20 shares of the fourth series, withdrawn
November 1, 1944, interest allowed at 5%.
3. Find the withdrawal value of 10 shares of the third series, withdrawn
July 1, 1944, assuming that the association permits withdrawals at book value,
but retains a membership fee of 2% of the par value of each share.
Dayton or Ohio plan. The plan most commonly used by
building and loan associations is the Dayton or Ohio plan. Use of
this plan eliminates the uncertainty as to the time of the maturity
of a loan, thus giving the borrower a definite contract. Under
432 BUILDING AND LOAN ASSOCIATIONS
other plans a successful association would soon mature its stock,
while an unsuccessful one would greatly prolong the time during
which the borrower would have to continue his interest payments.
Furthermore, under the Dayton plan the borrower does not have-
to own stock. Each payment that the borrower makes does two
things: first, it pays the interest; second, it reduces the principal.
The same idea is applied in the Federal Farm Loan Act, and is an
application of the subject " Payment of debt by installments,"
discussed on page 339. The monthly payment is found by the
formula used under the terminating plan.
Problems
1. Find the monthly payment which will cover both principal and interest
of a $4,000 loan at 5% for 10 years.
2. Construct a schedule for the first 2 years.
3. Construct a schedule for the last 2 years, showing the final amortization
of the debt.
To find the time required for stock to mature (rate of interest
given). Such a problem is simply that of finding the time that it
takes an annuity of annual rents payable in twelve monthly install-
ments to accumulate to a certain amount.
Procedure: (a) Divide the maturity value by the number of
dollars in the periodic payment.
(6) Multiply the quotient found in (a) by the periodic rate pel-
cent.
(c) Determine the logarithm of 1 plus the product found in (6).
(d) Determine the logarithm of 1 plus the periodic rale per
cent.
(e) Divide the logarithm found in (c) by the logarithm found
in (d).
Example
The Washington Building and Loan Association yields the investor a nominal
rate of 7%, convertible monthly. What is the time required for payments
of $1 a month to mature $100?
For nnda
. I / Maturity value .\ 1
log 1 + ( u - - - - X i )
L \Perionic payment / J
^r~~Ti — . — ^ ^ Term.
log (1 + i)
A rith metica I S ubstitut ion
log [l + (^ X .0058333)1
log 1.0058333 -79 periods.
BUILDING AND LOAN ASSOCIATIONS 433
Solution
Dividing: 100 -r 1 = 100 (1)
Multiplying: 100 X .0058333 = .58333 (#)
Adding 1: 1 + .58333 = 1.58333 (3)
log: 1.58333 = 0.199572 (4)
log: 1.0058333 = 0.002520 (5)
Dividing: log 0.199572 -f- log 0.002526 = 79 (upprox.) (6)
Problems
1. If the nominal rate of interest in the above example had been 5%, what
would have been the time of maturity?
2. If the monthly payment is 50^ and the nominal rate is 5%, convertible
monthly, what time is required to mature a $100 share?
3. If payments of 50 £ a month on a $100 share earn for the investor an
effective rate of 6%, convertible annually, in what time will the stock mature?
To find the effective rate of interest on money invested in
installment shares. This is a problem similar to that of finding
the effective rate earned on an annuity.
Example
On July 1, 1937, the Zenith Building and Loan Association issued $100 par
value stock on which monthly payments of $1 per share were to be made. The
stock matured on January 1, 1944, by the association's accepting 2^ on the
SOth payment of $1. What was the effective rate of interest earned?
To find the effective rate, it is necessary to use estimated rates, as explained
under the heading "Computation of the rate of an annuity," page 343. The
formula for the first estimated rate is as follows:
Formula Arithmetical Substitution
„ Amount.
A, + OJL- .\
\ i /
.005660
Solution
First trial rate, 6.8%:
.068 ~ 12 = .005f, monthly rate
1.005| to 79th power (by logs) = 1.5625808, compound amount
1 .5625808 - 1 = .5625808, compound interest
.5625808 -j- .005f = $99.28, amount of annuity
The first trial rate is found to be too small.
Second trial rate, 7.2%:
.072 -T- 12 = .006, monthly rate
1.006 to 79th power (by logs) = 1.6041402, compound amount
1.6041402 - 1 = .6041402, compound interest
.6041402 -T- .006 = $100.69, amount of annuity
The second trial rate is found to be too large.
434 BUILDING AND LOAN ASSOCIATIONS
Interpolation
Value at 7.2% ... $100 69
Value at 6.8% J)9 2$
Difference of .4% -S~l.il
$1.41 -T- 4 = .35, the difference represented by .1 %
$100 — .02, excess of last payment $99 9<S
Value of 79 payments at unknown rate 99 98
Value of 79 payments at 6.8% 9928
Excess over rate of 6.8% . .. $"~70
Difference in value represented by difference of .1 % .3f-
-70 -f- .35 = 2, or .2% to be added to 6.8%, or 7%
Problems
1. If, in the foregoing example, the stock had been matured by the associ-
ation's acceptance of the full amount of the 70th payment, what would have
been the effective rate earned?
2. Payments of 50^ a month mature $100 in 10 years, 4 months, without its
being necessary for any part of the 125th payment to be made. What is the
effective rate of interest earned?
Classified Problems on Building and Loan Associations
Distribution of profits to shareholders.
1. Five shares of stock in a building and loan association had a book value
of $215.80 at the beginning of a 6 months' period. The dues of $5 per month
for the next 6 months, payable in advance, were paid when due. What is the
average book value for the period that should be used in the distribution of
profits to these 5 shares?
2. B subscribed for 20 shares in a building and loan association, and because
his subscription was made at a time between dividend dates, he had to pay $20
in dues each month for 4 months. What was the book value of his payments?
3. Twenty shares of stock had a book value of $800 at the beginning of a
6 months' period. The shareholder became delinquent for 3 months. At the
beginning of the fourth month he paid $80, and then paid $20 each month for
the next 2 months. What was the average book value of his 20 shares? If his
delinquency had been covered by fines, and he had therefore been allowed full
participation in profits, what would have been the book value?
4. Ten shares of stock in the X. Building and Loan Association had a book
value of $365.80 on July 1, 1943. Dues of $1 per month per share were paid
for the next 6 months. On December 31, 1943, the average book value of
holdings in the association was $126,178.36. The association reported a net
gain for the 6 months amounting to $4,116.80. Find: (a) the rate per cent
earned; (6) the dividend on the 10 shares, December 31, 1943; and (c) the book
value of the 10 shares, December 31, 1943.
5. Adams paid $70 in advance for 1 share of paid-up stock in the Ames
Building and Loan Association. The maturity value of the shares was placed
at $100. The per cents of earnings for the 5 succeeding semiannual periods were
4.6%, 3.9%, 4.2%, 5.1%, and 4.9%, respectively. What was the book value
of Adams' share at the beginning of the sixth period?
BUILDING AND LOAN ASSOCIATIONS 435
6. The Garfield Building and Loan Association issued shares at the beginning:
of each month. Smith subscribed for 40 shares just 1 month before the end
of a 6 months' period. He paid for these shares at the rate of $40 a month.
What was the average book value of these 40 shares that was used in the dis-
tribution of profits for the 6 months' period? How much was Smith entitled to
receive as dividends if the association showed a net gain of $2,631.25, and a book
value of $85,160.72?
Shares issued in series.
1. B has 20 shares of $100 each in each of 2 series of the Capital City Building
and Loan Association. Twenty shares are of a series which is ending its second
6 months' period, and 20 shares are of a series which is ending its first 6 months'
period. B has paid his dues of $20 a month on each group of shares. The
association's rate of profit for the 6 months' period just ended is 5^%. What
are the dividends on each group of shares?
Withdrawal values.
1. Lee paid $20 a month, for 54 months, on 20 shares of stock in the Midway
Building and Loan Association. When the 55th payment was due, he withdrew
his money for the withdrawal value. The association allowed him the sum of
his payments, and simple interest at 5% a year. The value of the stock had
been accumulating at 6%, interest convertible monthly. What was the book
value of the stock? What was the withdrawal value? How much profit was
retained by the association?
2. If, in Problem 1, the interest paid on withdrawals had been calculated at
4%, what would have been the difference between the book value and the with-
drawal value?
3. If, in Problem 1, the stock had been accumulating at 7%, interest con-
vertible monthly, and simple interest at 5% was paid on withdrawals, what
would have been the difference between the book value and the withdrawal
value?
4. A stockholder who has been making payments at the rate of $10 a month
for 85 months, withdraws at the date of the <H6th payment. The stock has been
accumulating at tiie rate of 5g-%. The association allows 4% simple interest
on the payments withdrawn. What is the difference between the book value
and the withdrawal value if the stock has a maturity value of $200 a share?
The interest rate from the borrower's standpoint
1. A certain building and loan association operating on a 7 % nominal interest
basis will finance the building of your house. If you make payments of $1 a
month on each $100 share, your loan will mature with the 79th monthly payment.
On the other hand, however, the Mutual Life Insurance Company, through its
financial agents in this city, is offering loans on real estate at 5-Jr%. Suppose
that you pay the 5i% interest monthly in advance, and invest the difference
between this amount and the $1 per share payable to the building and loan
association in a sinking fund at 4% interest, payable monthly, will this prove
to be a better proposition at the end of 78 months?
2. If, in Problem 1, the difference in the monthly payments were placed in
a savings bank at 4%, interest convertible semiannually, would the insurance
company's proposition be the better one?
436 BUILDING AND LOAN ASSOCIATIONS
The interest rate from the borrower's standpoint when the interest and
dues are considered together as a single sum for the payment of interest and
principal.
1. Jones borrowed $1,000 from an association operating on the basis of a 6%
nominal interest rate, convertible monthly. Each month he paid $5 dues and
$5 interest. His stock matured at the end of 11 years and 3 months, after
Jones had made the monthly payment of dues and interest required at that
time. What effective rate of interest has Jones paid on his loan? (NOTE. — It
will be necessary to use an estimated rate, and to solve by interpolation.)
2. An association is operating on a 7% basis. Smith borrows $1,200, paying
$7 interest and $12 dues each month. The 79th payment is $7 interest and
$2.40 dues. This is the final payment, and matures the stock. What is the
effective interest rate?
Review.
1.* At the end of its fourth year, the Thrifty Building and Loan Association
has 500 shares in the first series, 400 in the second, 1,000 in the third, and 800 in
the fourth, and its total net earnings for all years amount to $8,364.
(a) Compute the rate of earnings under the simple interest partnership plan.
(6) Prepare a share statement which includes the profit per series, the book
value per series, the book value per share, and such other details as may be
necessary.
(c) Compute the book value of A/'s 50 shares of stock in the first series.
(d) Compute the withdrawal value of A^'s 10 shares in the second series,
assuming that 4% interest is allowed on withdrawals.
2. The Ypsilanti Building and Loan Association desires that you present to
them a statement that they may use to inform their customers as to the com-
parative cost and returns of their loans compared with loans of a similar nature
obtained from other sources. Bank loans in Ypsilanti, on good, salable property,
may be obtained by the payment of 7% semiannual interest, and an equal
semiannual reduction of the principal. The building and loan association will
lend on good, salable property, on condition that the borrower will pay, for
each $1 ,000 borrowed, $5 each month as a repayment of the loan, and $6 interest
each month. These payments are to continue until the $5 and the cumulative
interest shall be a sum sufficient to repay the loan. Interest is allowed on the
repayment of the loan at 8%, compounded semiannually, January and July.
Eight per cent simple interest is allowed on each monthly payment until the
date of compounding.
3. Distribute a profit of $2,940 on the following series on the partnership
plan.
Paid in Total Paid in Average Total for Profit
Series Per Share Shares Per Series Aro. *\fonths One Month Per Series
1 $60 100 30 5
2 48 200 24 5
3 36 300 18.5
4 24 250 12.5
5 12 400 6.5
$2,940.00
* C. P A., Pennsylvania.
BUILDING AND LOAN ASSOCIATIONS 437
4. How many years will be required to mature stock with a par value of
$100 a share if monthly payments of $1.55 are made regularly, interest at 6%?
5. If stock with a par value of $100 a share matures in 7 years, payments
being $1 a share monthly, what effective rate of interest is earned on the
investment?
6. A building and loan association organized on the serial plan issued 500
shares of Series A stock, par value $100 a share, on the first day of its fiscal
year, and 500 shares quarterly thereafter for a period of two years. The first
of each month, payments of 50 cents a share were made on this stock. At the
end of two years it was found that profits for the last quarter-year available for
distribution to shareholders amounted to $720. Find the profit per share for
the quarter on the basis of the equated capitals of the respective series.
7. If in Problem 6 dividends have been credited at the rate of 30 cents a share
each quarter, find the profits per share if the $729 were to be distributed on the
basis of earning capital.
CHAPTER 36
Permutations and Combinations
Permutation. A permutation is each arrangement which can
be made by using all or part of a number of tilings. The "number
of permutations of n things taken r at a time," represented by the
symbol „/%, is the number of arrangements of r tilings that can be
formed from n things. Thus, using the three letters a, 6, and c
taken two at a time, the permutations are ab, ac, 6a, 6c, ca, and cb.
Using all of them at the same time, the permutations are abc} ach,
bac, bca, cao, arid cba.
Since nPr is used to denote the number of permutations of n
things taken r at a time1, its value is determined as follows:
For first place: any one of n things may l>o chosen;
For second place: any one of the remaining, or n — 1, things may he chosen;
For third place: any one of the remaining, or n — 2, things may be chosen;
For fourth place: any one of the remaining, or n — 3, things may be chosen;
and so forth;
For the last or
rth place: there remains a choice of n — (r — 1) or n — r + 1 things,
Therefore, J\ = H(H - l)(w - 2)0* - 3) - • • (n - r + 1).
Take the three letters a, fr, and c two at a time. For the fh>fc
place, there is a choice of 3 letters; for the second place, there is a
choice of 3 — 1 letters; therefore, 3/^2 = 3(3 — 1) = 0, the number
of permutations of three letters taken two at a time, as shown in the
first paragraph.
Using the three letters a, &, and c all at the same time, r - n.
Therefore, the symbol may be expressed nPn when all of the n
things are taken at once, and
nPn — n(n — \)(n — 2)(n — 3) • • • (until n factors are used).
The symbol n!, called "factorial n" denotes the product of all
integers from n to 1 inclusive, and the expression is abbreviated to
nPn = n\. Solving the foregoing example, we have 3/^3 = 3 • 2 • 1
= 6, the number of permutations of the three letters taken three
at a time, as shown in the first paragraph.
NOTE: To avoid confusing the multiplication sign (X) and
the sign for quantity (x), use is made of the • placed above the line
of writing, and is read "times" or "multiplied by." Also, the
430
440 PERMUTATIONS AND COMBINATIONS
. . . placed on the line of writing indicates omission of the "in-
between" factors.
Example 1
Determine the number of three-letter code words that can be made from the
letters of the word bunch, not repeating a letter in any word.
Solution
The answer is the number of arrangements that can be made from five
objects (the letters of the word bunch) taken three at a time.
Formula
rfr = n(n - 1) • • • (n - r + 1)
Arithmetical S institution
J>z = 5(5 - 1)(5 - 2) = 5 • 4 • 3 = 60
Example 2
Determine the number of five-letter code words obtainable in the foregoing
example.
Solution
The answer is the number of arrangements that can be made from five
objects taken all at the same time, or J*n = til] and, since n\ is the product of all
the integers from n to 1 , we have
n\ = n(n — l)(n — 2)(w — 3) (ft — 4), or five factors in all.
Arithmctica I ft ubst it utio n
J\ = 5 • 4 • 3 • 2 • 1 = 120
Notice thht the last factor (n — r + 1) is one more than the
difference be6ween the number of things, n, and the number of
places, r. 1 hus, if n is 7, and r is 5, the last factor is 3 ; also, the
number of factors will be equal to r. So we may write the formula
nPr = n(n - i)(n — 2) • • • (until r factors are used).
Example 3
Five perrons enter a doctor's waiting room in which there are seven vacant
chairs. In how many ways can they take their places?
Solution
Formula
nPr — n(n — \}(n — 2) • • • (until r factors are used)
Arithmetical Substitution
-jP, = 7(7 - 1)(7 - 2) (7 - 3) (7 - 4) (five factors)
= 7 • 6 • 5 • 4 - 3 = 2,520
n\
The formula as used above may be expressed as nPr = -: -—TV
(n - r) !
Using the data in Example 3, we have:
PERMUTATIONS AND COMBINATIONS) 441
_7-0-:.fi:4.;8. 1^ = 2,520
Notice that the last two factors cancel the two below the line,
and that the remaining factors are the same as in the preceding
solution.
Using the three letters a, b, and c, two at a time, we have*
n\ _ 3!^ _ 3-2- 1 _ ,
" r ~ (n - r}\ " (3"-~2)"l 1 " '
In the foregoing illustrations, the objects were distinct, that is,
there were no repetitions of objects. If the objects are not
distinct, the formula is altered to read:
/',= -WU
alblcl
where, of the n things, there are a alike, b alike, and c alike.
Example 1
How many permutations may be made of the letters of the word Illinois?
Solution
There are eight letters in the word Illinois, but three are i's and two are I'n.
Then we have:
8! 40,320
3! -2! 12
3,360
Example 2
How many permutations may be made of the letters of the word Indianolaf
Solution
There are nine letters in the word Indianola, but two are a's, two are I'H, and
two are n's. Then we have:
9! 362,880
2!- 21-21- -*--«•»»
Number of ways of doing two or more things together. If a
certain tiling can be done in m ways, and a second thing can be
done in n ways, the two things can be done in succession in m • n
ways, or mn ways. The principle can be extended to find the
number of ways of doing three or more things together, as m • n • p •
. . . ways.
442 PERMUTATIONS AND COMBINATIONS
Example
In how many ways can two positions, the one that of bookkeeper and the
other that of stenographer, be filled when there are five applicants for the position
of bookkeeper and three applicants for that of stenographer?
Solution
Assuming that all applicants are qualified, there are five ways of filling the
position of bookkeeper, and for each of these there is a choice of three stenogra-
phers; hence, the two positions can be filled in 5-3 = 15 ways.
For each of the m ways of filling the position of bookkeeper there are n ways
of filling the position of stenographer; that is, there are n ways of filling both
positions for each way of filling the position of bookkeeper. Therefore, there
are in all mn ways of filling the two positions together.
Problems
1. If three dice are thrown together, in how many ways can they fall?
2. There are eight vacant seats to be filled by five persons. In how many
ways can they take their places?
3. How many five-place numbers can be made from the digits 1, 2, 3, 4, f>
0, and 7?
4. What is the number of permutations of the letters (a) of the word Indiana;
(b) of the word Illiopolis?
6. Using three letters at a time, how many permutations can be formed
with the letters abed?
6. (a) How many permutations of the letters abcde can be formed four at
a time? (b) Five at a time? (c) Three at a time?
7. How many permutations may be made of six objects taken: (a) six at a
time? (6) five at a time? (r) two at a time?
8. Given the numbers 2, 3, 4, 5, and 6. How many four-phictj numbers
can be formed therefrom?
9. The Greek alphabet contains 24 letters. If no repetition of letters arc
allowed, how many three-letter fraternities can be named therefrom?
10. A signal man has five flags, no two of which are alike, (a) How many
different signals can be made by placing them in a row using all five of them
each time? (6) How many by using three at a time?
Combinations. A combination is a set or selection of r things
out of a total of n things without reference to the order within the
selection; therefore, ab and ba are the same combination. The
"number of combinations that can be made from a total of n
things taken r at a time" is denoted by the symbol nCr. Since
each combination can be arranged in more than one way, the
number of permutations is denoted by r!, and the total number of
permutations for all combinations is rlnCr. The total number of
permutations of n things taken r at a time is denoted by the symbol
nPr, as in previous paragraphs.
PERMUTATIONS AND COMBINATIONS 443
Therefore, r!nCr = «Pr
and r
(n-r)!
Substituting and dividing by r!,
??:
Example
Find the number of combinations \\hich can be made with the four letters
a, 6, c, and d taken three at a time.
/Solution
= -_ ^. =
4/3 3!(4-3J! 3-2-1
These combinations are: abc, abd, acrf, and bed. Notice in
permutations that the three letters a, b, and c form six permuta-
tions, abc, acbj bac, bca, cab, and cba, but there is only one combina-
tion abc, since all others are merely a rearrangement of the same
letters. The addition of the fourth letter makes possible three
more com! )inations.
Example
How many lines ran be drawn connecting seven points, no throe of which are
in the same straight line?
Solution
If we let the points be represented by the letters a, b, c,, d, c, /, arid g, and
any line connecting two of them by the symbol ah, nc, and so on, we find that
ab and ha is the same line, that ac and ca is another line, and so forth; therefore,
the problem is that of finding the number of combinations of seven objects
taken two at a time.
,, _ 7! __ _ 7-6 _
72 ~ 2!(7 - 2)! ~ 21 ~ Zl
NOTE: The factors from o to 1 above the line cancel the same factors below
the line, leaving the factors indicated in the solution.
If r is large and the difference between n and r is small, the following formula
will save considerable work: nC!r = nCn r.
Example
Find the value of 25^23
Solution
25^23 = 26^25-23 = 26^'% == = 300
Problems
1. How many combinations can be made with the five letters a, 6, c, d, and
e taken three at a time?
444 PERMUTATIONS AND COMBINATIONS
2. If twelve members of an association are available for committee assign*
ments, how many different committees of four each can be selected?
3. You have nine friends that you wish to invite to dinner parties of four
guests each. How many dinner parties can you have without having the same
company of four twice?
4. A committee consisting of two men and one woman is to be formed from
a party of five men and four women. In how many ways can the committee be
chosen?
CHAPTER 37
Probability
Probability. One of the principal applications of permutations
and combinations is found in the theory of probability. The
probabilities for the occurrence of one or more events, in cases in
which it is possible to count the number of equally likely ways in
which the event can happen or fail, are known as priori probabilities.
Counting of some sort is the background of probability. For
example, if a coin is tossed, the chances are even between heads
and tails. If a die is thrown, the chances of throwing any one of
the numbers 1 to 6 is 1 in G, as there are six surfaces numbered
from 1 to 6. The chance of drawing an ace from a well-shuffled
pack of 52 cards is evidently 4 in 52.
If an event can occur in m ways and fail in n ways, and if each
of these ways is equally likely, then the probability of its occurring is
m
m + n
and the probability of failure ir» an event is
n
m + n
Example
Compute the probability of throwing a 3 in the first throw of a die.
Solution
m 1 1
P =
m + n 1+5 0
Example
Compute the probability of failing to throw a 3 in the first throw of a die-
Solution
m + n 1+5 6
The mathematician computes the ratio of successes to the
total number of ways in which the event can occur. For example,
445
446 PROBABILITY
he would say that the chances of throwing a 4 in one toss of a die
is "one chance in six." The average person usually calculates the
ratio of the successes to the failures, and would say "the odds are
five to one against throwing a 4."
Problems
1. A bag contains ten black balls and fifteen white ones. What is the
probability that a ball drawn at random will be black?
2. A box contains six times as many black balls as white ones and one ball
is drawn at random. What is the probability that the ball drawn will be black?
3. If you are to win a prize valued at $12.00 by throwing an ace in a single
throw with a die, what is the value of your expectation?
Permutations and combinations in probability. The following
examples will illustrate the statement made at the beginning of
this chapter.
Example
What is the probability of obtaining a 6 if two dice are tossed?
Solution
Under permutations we learned that a succession of acts can be performed
together in as many ways as the result of their continued product. Since each
die has 6 faces, the two dice can fall in 0 X 6, or 36, ways. In these 36 ways,
the sum 6 can appear in any one of the following 5 ways: 5 and 1 , 1 and 5, 4 and 12,
2 and 4, 3 and 3. Therefore, the probability of throwing a 6 is /V
Example
If two cards are drawn from a complete deck of 52 cards, what is the proba-
bility that both are hearts?
Solution
First is the determination of the number of combinations of 52 objects taken
2 at a time.
n\ 52'
C - - ___ ^ - 1 Q9fi
" P"»!(n -r)I " 21(52-2)1 " '
Second is the determination of the number of selections of two hearts.
nr _--
2!(13 - 2)!
Therefore, the probability of selecting two hearts is T^l «- or -pr-
Example
Two prizes are offered in a lottery of 20 tickets. What is your probability
of winning a prize if you hold five tickets?
Solution
First, determine the number of ways in which five tickets can be selected.
PROBABILITY 447
"Cr = 81(20-5)! = 15>5°4
If you hold the two prize tickets, the remaining tickets may be any three
of the remaining 18, so the number of selections containing both prizes is
18!
nCr = = 816
Next, determine the number of selections containing the first prize and not
the second.
nCr = C = 3'°6°
The number of selections containing the second prize and not the first is
evidently the same, 3,060.
Therefore, the probability of winning a prize is
816 + (2 X 3,060) = 6,936 = 17
15,504 ~ 15,504 ~~ 38
Example
In the foregoing example, what is the probability that you will not win ?,
prize?
Solution
As two of the tickets are winners, 18 are not winners, and this is the number
from which five tickets must be selected, then,
18! 8,568
n r 51(18 - 5)!
A.S before,
The probability that there will be no winner is
8,568 _ 21
15,504 " 38
Checking the answer to the preceding problem, we have:
i _ 2 1 _ 17
Problems
1. A complete deck of cards numbers 52 and is made up of 13 cards in each
of the four suits. If four cards are drawn, find the following probabilities:
(a) That all are hearts;
(b) That there is one card of each suit;
(c) That there are two diamonds and two clubs.
2. (a) In a single throw of two dice, what is the probability of throwing a
ten? (b) What would be the probability in a single throw of three dice?
448 PROBABILITY
3. If you toss six coins, what is the probability that there are four heads
and two tails?
4. The cash drawer contains five ten-dollar bills, six five-dollar bills, and
seven one-dollar bills. How many different sums may be formed with three
bills taken out at random?
Compound events. The j oint occurrence of two or more simpler
events in connection with one another is called a compound event.
If two or more events occur without influencing one another, they
are said to be independent ; but if any one of them does affect the
occurrence of the others, they are said to be dependent. When the
occurrence of any one of the events excludes the occurrence of any
other on that occasion, the events are said to be mutually exclusive.
Independent events. The probability that n independent
events will happen favorably on a given occasion (when all of them
are in question) is the product of their separate probabilities. If
the separate probabilities that an event can occur favorably are
represented by p\, p*, . . . pn, and P equals the probability that
all these events will happen together at a given trial, then,
P = Pl X P2 X ' • • Pn
Example
What is the probability that 2, 3, and 4 are thrown in succession with a die?
Solution
The probability of getting 2 is ^-, that of getting 3 is -g-, and that of getting 4
is #. As these are the probabilities of independent events, the joint probability
will be | X i X i = TTTT.
Example
What is the probability of getting 2, 3, and 4 in one throw with throe dice?
Solution
Consider the three dice as being thrown separately. Then there are 3 chances
in 6 of getting the first number, 2 chances in 6 of getting the second number, and
one chance in 6 of getting the third number. Since these events are all inde-
pendent of one another, the joint probability will be P = f X f X i = -£$•
Example
From a box containing 5 brown marbles and 4 green marbles, 3 marbles are
drawn. What is the probability that all three will be brown?
Solution
Consider each drawing an independent event. Since there are at first
9 marbles and 5 are brown, the probability on the first drawing will be -J. If a
brown marble has been drawn, then on the second drawing the probability will
be f-. Now, if 2 brown marbles have been drawn, on the third drawing the
probability will be y. Since these three probabilities have been independent
events, the joint probability is
PROBABILITY 449
P=t xix* = A
Mutually exclusive events. Let the number of mutually exclu-
sive events be represented by pb p2, . . . pn. The probability
that some one of these events will occur is equal to their sum;
therefore, P = p\ + p2 + * * ' pn-
Example
Three horses are entered in a race. Snowball's chances of winning are -J-,
Thunderbolt's chances are ^-, and Fleetwind's is -J-. What is the probability that
the race will be a tie?
Solution
The winning of the race by Snowball, Thunderbolt, or Fleetwind forms a
set of mutually exclusive events, since only one can be the winner. The proba-
bility that one of them wins the race is
Since the race may be a tie, that probability is
Example
If my chance of completing a certain engagement is -^, and your chance of
completing it is ^, what is the probability that the engagement will be completed
if we both work independently of one another?
Solution
If we work together to complete the engagement,
P. = i x * = A
If I complete the engagement and you fail to complete it,
Pt = t X (1 - *) = TV
If you complete it and I fail,
p, = £ X (1 - t) = A
The sum of pi + p* + p$ = •& + T2" + A = f, the probability that the
engagement will be completed by one or the other of us.
This result can be checked by assuming that both will fail:
From a box containing 5 brown marbles and 4 green marbles, 2 marbles are
drawn at random. What is the probability that both are of the same color?
Solution
The probability that the 2 are brown is determined as shown on the next page.
450 PROBABILITY
5' -.0 .nrt ^5^-36; g.i
2!(5_2)! ^ 2!(9-2)! ""' 36 18
The probability that the 2 are green is determined in the same manner.
= 6 and '
21(4-2)1 21(9-2)1 ""' 36 18
These two events are mutually exclusive; hence, the probability that both
marbles are of the same color is
TF + T8" = ¥
The probability that there will be one of each color is
5X4 = 5X4 = 5
~9T 36 ~ 9
2!(9 - 2)!
Problems
1. A and B engage in a game of checkers. The probability that A will wiii
a game is f and that B will win a second game is -5-. What is the probability
that both win?
2. From a bowl containing 5 red marbles and 6 white ones, 4 marbles are
drawn at random. What is the probability that they are all white?
3. In Problem 2, what is the probability that the 4 marbles drawn at random
are red?
4. In Problem 2, determine the probability that of the 4 marbles drawn, 2 aro
red and 2 are white.
6. If three cards arc drawn from an ordinary pack of playing cards, what
is the probability that all three will be spades?
6. In Problem 5, what is the probability that all three cards are black?
7. What is the probability that all three will be of the same suit?
8. If the cards are replaced after each drawing, what is the probability that
the first three are hearts?
9. At an election, 500 of the registered voters in the precinct cast their
ballots. Two hundred voted in favor of a certain amendment and 300 voted
against it. If five voters are chosen at random, what is the probability that
they all voted for the amendment?
Empirical probability. The practical application of probabili-
ties leads to a consideration of probabilities which are derived
from experience. These are called empirical probabilities. The
connection between probability and statistics, a subject devoted
to the analysis and interpretation of data, is found in empirical
probability.
Example
The following table gives the average daily sales of 92 market gardeners in a
certain public market.
PROBABILITY
451
Average
Daily
Sales
2 50-
7 49
7 50-
12.49
12 50-
17 49
17 50-
22.49
22 50-
27.49
27 50-
32.49
32 50-
37 49
37 50-
42 49
42 50-
47 49
47.50-
52 49
Num-
ber of
Garden-
2
8
27
21
16
3
11
2
1
1
ers
What is the probability of a person engaging in market gardening in this
market of having average daily sales of less than $17.50?
Solution
The table shows that of 92 gardeners, the number who make less than $17.50
average daily sales is 2 + 8 + 27 = 37. The required probability is, therefore,
|r£, expressed decimally as 0.402.
Problems
1. A study of market gardening showed that 100 gardeners had marketed
in a particular market as follows:
Number of
Years Gardeners
1 to 5 . . . . 31
6 to 10 . . 22
11 to 15 10
16 to 20 19
21 to 25 . . 6
26 to 30 . ... 3
31 to 35 ... 5
36 to 40 ... 3
41 to 45 . \_
100
What is the probability of a gardener being in this market for 20 years or less?
2. A survey among farmers to determine their average income disclosed tho
following facts:
Per Cent of
Income Range
Negative income (Loss)
0 to $ 500
$ 500 to $ 1,000
$ 1,000 to $ 1,500
$ 1,500 to $ 2,000
$ 2,000 to $ 3,000
$ 3,000 to $ 4,000
$ 4,000 to $ 5,000
$ 5,000 to $ 7,500 .
$ 7,500 to $10,000..
$10,000 and over
Farmers
4.55
12.53
21.89
17.01
14.69
13 31
7.66
3.25
3.17
0.93
1.01
100.00
452
PROBABILITY
Based on the above survey, what is C's probability of making $2,000 to
$3,000 a year if he engages in farming?
3. A manufacturer of electric light bulbs made a test with 200 bulbs of
uniform design. The results were tabulated as follows:
Life in
Hours
400
to
500
500
to
600
600
to
700
700
to
800
800
to
900
900
to
1,000
1,000
to
1,100
1,100
to
1,200
1,200
to
1,300
1,300
to
1,400
1,400
to
1,500
1,500
to
1,600
Number of
Bulbs
2
4
7
16
23
27
37
31
24
18
8
3
What is the probability that a bulb will burn out in less than 800 hours,
based on the experience of the foregoing test?
4. If three new bulbs are placed in operation at the same time, what is the
probability that all of them will last 1,200 hours?
HINT: Cube of a single probability.
5. Find the probability that a bulb will be "alive" between 900 and 1,300
hours.
CHAPTER 38
Probability and Mortality
Life insurance. Life insuiance is based upon probabilities
determined by the actual study of large collections of mortality
statistics. If an event has happened m times in n possible cases
(where n is a large number), then, in the absence of further knowl-
171
edge, it may be assumed for many practical purposes that — is the
n
best estimate of the probability of the event and that confidence
771
in this estimate may increase as n increases. The fraction — is
n
called the frequency ratio.
According to the American Experience Table of Mortality (see
page 536), of 69,804 men living at age 50, the number living ten
years later will be 57,917. The probability that a man aged 50
will live ten years is taken to be
57,917
69^04
Mortality table. Application of the theory of probability is
made in the study of problems involving the duration of human
life, such as life insurance, life annuities, pensions, and so forth.
Tables that show the number of deaths expected to occur during a
given age are used in the solution of these problems. Census
records and vital statistics gathered by governmental agencies are
the basis of some mortality tables. Others are based upon the
records of life insurance companies. Results based upon mortality
tables are applicable only to large groups of individuals.
The table on page 454 is taken from the American Experience
Table. (The entire table is given in Table 7, in the Appendix.)
Column (1) is the age column and contains the age of 100,000
people or their survivors.
Column (2) indicates the number of people living at the begin-
ning of the year designated on the same line in column (1). The
table starts with 100,000 people alive at age 10 and at age 11 shows
that only 99,251 have survived. The number that have died in
the interval, 749, is shown in column (3).
453
454
Column
Column
(4) is
(5) is
PROBABILITY AND MORTALITY
749
100,000
:« ">251
'" 100,000
= .00749 at age 10, and so on for each age.
= .99251 at age 10, and so on for each age.
AMERICAN EXPERIENCE TABLE OF MORTALITY
(1)
(2)
(3)
(4)
(5)
Age
Number
Living
Number of
Deaths
Yearly
Probability
of Dying
Yearly
Probability
of Living
X
I,
d,
V*
Px
10
11
12
13
14
15
16
17
18
19
20
100,000
99,251
98,505
97,762
97,022
96,285
95,550
94,818
94,089
93,362
92,637
749
746
743
740
737
735
732
729
727
725
723
0.007490
0 007516
0.007543
0.007569
0.007596
0 007634
0 007661
0.007688
0 007727
0.007765
0 007805
0 992510
0.992484
0.992457
0.992431
0.992404
0.992366
0.992339
0.992312
0.992273
0.992235
0 992195
30
85,441
720
0.008427
0 991573
50
69,804
962
0.013781
0.986219
70
38,569
2,391
0.061993
0.938007
90
847
385
0.454545
0.545455
95
3
3
1.000000
0.000000
Notation. For convenience, the letter I is used to designate
entries in the "living" column. Subscripts appended to the I
denote specific entries in this column; thus, /i6 indicates the
number living at age 15.
It is customary to refer to the age as x, that is, any age ; there-
fore, lx would apply to any entry in column (2).
The number dying is denoted by d, which with the subscript, as
in di5, indicates the number dying between the age indicated and
the next. Since x stands for any year, dx indicates any entry in
column (3).
PROBABILITY AND MORTALITY 455
The probability of a person dying is expressed by q. With the
subscript, as in #15, it denotes the probability of a person of the
age indicated dying before reaching the following age. Similarly,
qx stands for the probability of a person aged x dying before reach-
ing age x + 1. This probability is shown in column (4).
The probability of living is denoted by the letter py and pi6
denotes the probability that a person aged 15 will live to become
age 16. This probability is shown by the table, column (5), to be
.992366. The symbol px denotes the probability that a person
aged x will live to age x + 1.
For convenience, the symbols already explained and others
based on them are shown in the following summation :
x = a person or a life aged x years
lx = the number of persons living at age x
lx+i = the number living at age x -f- 1
/,+n = the number living at age x -\- n
dx = the number of persons dying in the age interval x to x + 1
dx+i = the number dying in the age interval x + 1 to x + 2
px = the probability that a person of age x will live one year
qx = the probability that a person of age x will die within one year
npx = the probability that a person of age £ will live at least n years
\nQx — the probability that a person of age x will not live n years
n\q* = the probability that a person of age x will die within one year after reach-
ing the age of x + n
Pxy = the probability that two persons, of ages x and ?/, respectively, will live
at least a year
r.pxy — the probability that two persons, of ages x and y, respectively, will live
at least n years
Probability of living. On page 453 was shown the probability
r *j
that a man aged 50 will live ten years is taken to be r Q' = .8297.
O«_/,
The probability that a person aged 50 will survive 10 years is
equal to the ratio between the number living at age 60 and the
number living at age 50. Expressed as a formula,
_
10^50 — 7~~
tso
The probability that a person aged x will live to age x + 1 is
equal to the ratio of the number of people living at age x + 1 to the
number living at age x, or
Example
What is the probabilitv that a person of age 30 will live at least a year?
456 PROBABILITY AND MORTALITY
Solution
_ U, _ /a, _ 84721 _
P* ~ IT ~ Hi ~ 85441 ~ -991573
The probability that a person will live longer than one year or
n years is expressed by the formula
Example
Determine the probability that a person age 40 will live to be 60.
Solution
/GO _ 57917 _
/To " 78106 ~ '741
Probability of dying. The probability that a person aged x will
die within a year is expressed by
Example
What is the probability that a person aged 40 will die within one year?
Solution
The probability that a person aged x will not live to age x + n
is ascertained by the following formula:
What is the probability that a person aged 30 will not live to age 40?
Solution
, 'so — '30+10
MSO = ~ ;
£30
_ 85441 - 78106
~ 8544 J
= .085849
PROBABILITY AND MORTALITY 457
Since the sum of the probability of living and the probability
of dying equals 1, or certainty, the foregoing example may be
solved as follows :
|lO#30 = 1 — IOPZO = 1—7—
/30
_ 781Q6
85441
= .085849
The probability that a person aged x will die within one year
after reaching age x + n is ascertained by the following formula :
Example
What is the probability that a person aged 30 will die within one year after
reaching age 40?
Solution
85441
Joint life probabilities. The probability that two persons (x)
and (y) will survive at least a year is denoted by pzy. As the
probabilities of life, or of death, of two or more persons are assumed
to be independent of each other, it follows that
Example
A husband and wife are aged 37 and 32, respectively. What is the proba-
bility that both will be alive at the end of 20 years?
Solution
The probability that the husband will be alive at the end of 20 years it?
WtM, or 0.7729
The probability that the wife will be alive at the end of 20 years is
, or 0.8076
Since the probability of two separate and distinct events is the
product of the probabilities of each event, the probability that
both will be alive at the end of 20 years is
0.7729 • 0.8076 = 0.6242
458 PROBABILITY AND MORTALITY
NOTE : Since (x) is used in life insurance to denote the age of
a person, it is confusing to use it to represent times, or multiplied
by; therefore, multiplication is indicated by the period placed above
the line. The formula may also be written
—
Pxy —
Problems
1. What is the probability that a child aged 12 will die between the ages
15 and 16?
2. A father and son are aged 35 years and 13 years, respectively. Find
the probability that both will be living on the son's twenty-first birthday.
3. What is the probability that a man aged 35 will die within 5 years?
What is the probability that he will die in the year after he reaches age 40?
4. A husband and wife are aged 42 and 40, respectively. The husband
has purchased an annuity, the first payment to be on his 65th birthday. What
is the probability that both will be living at that time?
5. What is the probability that neither will be living when the first payment
of the annuity described in Problem 4 is due?
6. If the annuity described in Problem 4 is payable for 20 years, what is the
probability that both husband and wife will be living when the twentieth pay-
ment is due?
7. A husband and wife are aged 26 and 24, respectively, at the date of their
marriage. What is the probability that they will live to celebrate their golden
wedding anniversary?
8. Y is aged 40 and Z is aged 35. Calculate the following:
(a) That Y will survive the first year but Z will not.
(6) That both will survive one year.
(c) That Z will survive the first year but Y will not.
(d) That both will survive 15 years.
(e) That both will die during the first year.
9. Find the probability that a person aged 50 will live to age 70.
10. Find the probability that a person aged 25 will not live to age 35. What
is the probability that this person will die between the ages of 35 and 36?
CHAPTER 39
Life Annuities
Factors involved. In Chapter 30 it is shown that the present
value of a sum of money payable n years in the future depends
upon the rate of interest which can be earned.
If the payment of this sum of money at a future time is contin-
gent on some person being alive at such future time, the present
value depends upon the rate of interest and also upon the probabil-
ity that the person will be living. For example, if two equally
good insurable risks aged 25 and 65, respectively, are to receive
$1,000 each upon attaining ages 35 and 75, respectively, the present
value of the promised payment to the person aged 25 would be
relatively much greater than to the person aged 65.
Pure endowment. A pure endowment contract promises to
pay to the holder thereof a definite sum of money if he is living at
the end of a specified period, but nothing to his beneficiaries if he
fails to survive this period.
The present value of an n-year pure endowment of 1 to a person
now aged x is expressed by the symbol nEX9 which is equivalent to
the present value of 1 to be received at the end of n years, multi-
plied by the probability npx that a person aged x will survive n
years.
If nEx denotes the present value of an n years' pure endowment
to a person of age xy we have
TJ1 I f I T) T~) * \ n
n-I-J x I/I i "\ I £ x /~
OH
Example
A person aged 20 is to receive $5,000 upon attaining age 25. Find the
present value of the probability, interest at 8^%.
Solution
*25
5,000^*0 = 5,000 (L03*)5
459
460 LIFE ANNUITIES
The probability that the person will receive the money is
- j— = .8419732, the present value of 1 for 5 years at 3i%
(1.03-jf)
The present value to the person aged 20 is, then,
$5,000 X .8419732 X .9610846 = $4,046.04
Problems
1. A girl aged 10 is to receive $5,000 upon attaining age 18. Find the present
value of the inheritance, interest at 3^%.
2. A persor, aged 20, is to receive $10,000 upon reaching age 30. Find the
present value of his expectation on the basis of 3^% interest and the American
Experience Table of Mortality.
3. Find the present value of a pure endowment of $2,000 to a person aged 30
payable if he reaches the age of 60, on a 3-g-% basis.
Life annuity. A series of periodical payments during the con-
tinuance of one or more lives constitutes a life annuity. The
simplest form of a life annuity to a person age x is the payment of
1 at the end of each year so long as the person now aged x lives.
Such an annuity consists of the sum of pure endowments of 1 each
year. The symbol for a life annuity is ax\ therefore, ax = iEx +
2EX + 3#* • • • + nEx • • • to table limit.
Substituting these values gives :
vlI+i + v*lx+2 + vHf+3 ' ' ' + vnlx+n
ax = — ±— -- ± ----- — — -- ---- -- • • • to table limit.
Lx
Example
Find the value of a life annuity of $1,000 a year to a person now aged 90,
interest at 3-j%.
.035)-4/94) + ((1.035)- V)
Present values are found in Table 3. Values of ko, ki, and so forth, are
found in Table 7.
Substituting all the indicated values and solving, we have
ago = .8738
$1,000 X .8738 = $873.80
Problem
A life pension of $500 a year, payable at the end of each year, is granted to a
person now aged 91. What is the present value of this pension, interest at 3^%?
Commutation columns. In the examples and in the preceding
problem, the computations are not particulary arduous, because the
LIFE ANNUITIES 461
of the annuitant made it necessary to make only a few com-
putations. But, in cases where the annuitant is younger — for
example, age 20 — it is evident that a great amount of work would
be required in order to solve the problem. Much of this compu-
tation may be eliminated by the use of tables called ' ' commutation
columns" (see Table 8).
The first column of this table, the Dx column, has been con-
structed of the products of similar ?/s and Z's and the product
denoted by the letter D. The computation was therefore reduced
to the addition of the values found in the Dx column. To save
time in adding these values, another commutation column was
formed, containing the sums of all the Z>'s from any particular
value of Dx to the table limit. This is the Nx column of Table 8.
Therefore, the work is materially reduced by using the tables
and the formula :
The solution to the example on page 460 now becomes :
A^t AWi 33.47
~D, '=~7>;r ^ 38:3047 =
$1,000 X .8738 = 1873.80
Using the commutation tables, the present value of the pure
endowment on page 459 may be found by the formula
Substituting the values for the example on page 459, we have
»*«• - P£l - "S(>9207
40556.2
and
$5,000 X .809207 - $4,046.04
From the foregoing formula it may be found that 1 at age a
will purchase an n-year pure endowment of
Dz
Dx+n
Problems
1. What is the present value of a life annuity of $3,000 to a person aged 30,
interest at 3i%?
2. Find the value of a life annuity of $2.500 at 3^% to a person aged 35.
462 LIFE ANNUITIES
Life annuities due. The principles of annuities apply in life
insurance. The preceding illustration was that of a life annuity
where the payment was made at the end of each year. When the
payments are to be made at the beginning of each year, the life
annuity is a life annuity due. Actuaries use the symbol a, to
represent the present value of an annuity; and, since an annuity
due differs from an ordinary annuity by an additional payment
made at the beginning of the period, the present value of an annuity
due is 1 + ax, and the symbol becomes
a, = 1 + ax
using a different type "a" from that used in the ordinary life
annuity. Since the different type "a" is somewhat difficult to
make, the regular a may be used and distinguished by a bar
over it, thus,
tin = 1 -f- ®x
N +i
Use of commutation table. Since ax = —~- > the life annuity
due formula may be written as
&x=sl+~157
arid if for 1 we substitute -rr' we have
which is equivalent to
the formula for the present value of a life annuity due of 1 payable
to a person aged x. The values may be obtained from the
commutation table.
Example
Find d3o
. N,
596804 . . A U1
a3o = OQ44Q o from ttie table
a3o =» 19.6054, also shown in the table in the 1 — az column.
LIFE ANNUITIES 463
Problems
(Use the commutation table.)
1. Find #20.
2. Find D35.
3. If x = 75, find Dx.
4. Find Nxitx = 22.
6. When Dx = 25630.1, what is the value of *•?
6. When Nx = 20S510, what is the value of j?
7. At what age does Ar, = 157,255?
8. At what age does Dx = 19S7.87?
9. What is the difference in value between <740 and a40?
10. Find (1) a30; (2) a3o.
Deferred annuity. When the first payment under a life annu-
ity is to be made after the lapse of a specified number of years
(contingent upon the annuitant (x) being alive), instead of being
made a year after the payment of the single premium, the annuity
is deferred.
Since under an ordinary annuity the first payment is made at
the end of one year, then if an annuity is deferred n years, the first
payment is made at the end of n + I years; but an annuity provid-
ing for the first payment at the end of n years is deferred n — I years,
for the annuity is entered upon at the end of n — 1 years, and the
first payment is not made until one year later; and it is a deferred
life annuity due.
The present value of a life annuity of $1.00 deferred for n years
is expressed by the symbol
but in n years the annuitant's age will be x + n, and the value of
the annuity will be a,+n; and, since it is desired to find the value of
this annuity now, we discount it by multiplying ax+n by the regular
present-value symbol, vn.
However, three factors are to be considered as follows :
(a) The value of the life annuity, ax+n;
(6) The present value of 1 in n years, vn\
(c) The probability that the person aged x will be living n years from now, npx.
Therefore, the formula becomes :
and, making substitutions so that the solution may be obtained
from the commutation table, we have
n|a* = ~~DT
464 LIFE ANNUITIES
Example
What single premium will a person aged 30 have to pay to obtain a life
annuity of $2,500, so that he will receive his first annuity payment at the end
of his 46th year?
Solution
D
From the commutation tables, it is found :
253745
"|flso ~ 1577l7>
= 16.08669
$2,500 X 16.08669 - $40,216.73
Deferred life annuity due. The first payment of an annuity
due would be made 1 year before that of an ordinary deferred
annuity; therefore, the deferred life annuity due is the equivalent
of an ordinary life annuity deferred for n — 1 years, and the
formula is
, .
nax = n-i\ax or nax =
Example
Y is aged 55, and he desires to purchase a life annuity of $2,500, the first
payment to be made at age 65. What is the single premium payment?
Solution
and
_ 4S616.4
~ 9733.40
= 4.994cS
$2,500 X 4.994S = $12,487.00
Problems
1. A child 15 years of age is to receive $2,400 a year for life, the first payment
to be made at age 21. Calculate the value of this annuity at 3^%.
2. What single premium payment will a person aged 35 have to pay to
obtain a life annuity of $3.000 from which he will receive his first annuity pay-
ment at age 60?
LIFE ANNUITIES 465
Temporary life annuities. A temporary life annuity continues
for n years, contingent on the annuitant living that long; hence, it
is not an annuity certain. The symbol for a temporary life annu-
ity is axn\j and the formula is
'"' D,
Example
Find the present value of a life annuity of $2,000 for 25 years, to a person
aged 40.
Solution
= 324440 - 43343J
19727.4"""
- 14.24906
$2,000 X 14.24906 - $28,498.12
Temporary annuities due. The present worth of a temporary
life annuity due, also termed an immediate temporary annuity, is
equivalent to the difference between the present worth of a whole
life annuity due and a deferred life annuity due, and may be
expressed as
Uxn\ ~ <*•* — n|«x
Substituting values, the formula for use with commutation
tables becomes
Nx - Nx+n
Example
Y buys a temporary life annuity of $1,200 for his widowed mother aged 50.
Payments are to begin at once and to continue until age 75. What is the present
value of this annuity due?
Solution
"' />*
#50 25 ~ fT—
_ 181663 - 11728.9
12498.6
= 13.59625
$1,200 X 13.59625 = $16,315.50
466 LIFE ANNUITIES
Problems
1. Find the present value of a temporary life annuity of $1,500 for 5 years
co a person aged 65.
2. Find the values of a20ioi, nl6-M[j and r/3510>
3. Find the values of io|«2o, 2o|«i6, and io|^35.
Life annuities with payments m times a year. Annuity con-
tracts often provide that payments shall be made more frequently
than once a year, such as quarterly or monthly, the latter being
more common. For an annuity payable m times a year, the
symbol ar(m) is used to denote its present value, and the formula
used to determine the value when the payments are made at the
end of the period is
, w - 1
«x(w) = fix H — o —
2m
Example
Find the present value of a life annuity of $(500 a year payable monthly, the
first installment to be paid in one month, for a person 35 yeais of age.
Solution
o
2-rii
Substituting values,
«35(12) = 035 + U
_ #36 _ 482326
035 ~ AT* ~ 24^447
= 17.0143
U = -45S3
17.6143 + .4583 - 1S.0726
$600 X 18.0726 - $10,843.56
If the payments are made at the beginning of the period, they
constitute an annuity due, and its present value will be:
2m
For a deferred life annuity of 1 a year, payable in in install-
ments a year, the present value is
For a temporary life annuity for n years, payable in m install-
ments a year, the present value is
LIFE ANNUITIES 467
Example
The value of a temporary life annuity of $180 a year payable annually for
15 years to a person aged 45 is $1,873.96.
What is the present value of an annuity of the same annual rent if paid in
monthly installments of $15 ea^h, the first payment one month hence?
Solution
$l80a46fri = $1,873.96
a^I10'4109 11
" 1B^ = §=1 |^ .46606
Therefore,
a46<12)16l- = 10.4109 + fl
= 10.4109 + .2447 - 10.6556
$180(10.6556) - $1918.01
Problems
1. Find the present value of a pension of $75 a month payable at the end
of each 3 months to a pensioner aged 65.
2. A corporation executive aged 5S is to be retired at age 65. During retire-
ment he will receive $3,600 a year payable in monthly installments. Find the
present value of this retirement allowance on a 3^% basis.
3. A life annuity contract provided for a payment of $750 a year for 15
years, the first payment to be made at age 60. At age 60 the annuitant desired
monthly payments. Find the amount of the monthly payments.
4. A widow was to receive $1,800 a year for life in annual payments, the
first payment to be made one year after her husband's death. When the first
payment was due, the widow was 65 and asked that the payments be made
monthly. What amount should she receive monthly?
Forborne temporary annuity due. A forborne temporary
annuity due is created when a person who is entitled to a life
annuity due of 1 a year forbears to draw it and agrees that the
unpaid installments are to accumulate as pure endowments until
he is aged x + n.
On page 465 the present value of a temporary annuity due was
found by the formula
and on page 461 it is given that 1 at age x will purchase an n-year
pure endowment of
Then the present value would buy a pure endowment equal to
468 LIFE ANNUITIES
7) AT — AT A7 — N
D^^N^N^ or N* Nl"
Dx+n
Problems
1. Find the amount at age 60 of a forborne temporary annuity due of 1 a year
that is to be accumulated for a person now aged 40.
2. A man was to receive a life annuity of $1200 a year, the first payment to
be made one year mter his 60th birthday. At that time he was still employed
at a good salary, and so decided to postpone the beginning of the annuity for
5 years. What yearly sum will he receive, the first payment to begin on his
66th birthday? (Use American Experience Table of Mortality and 83- %, and
treat the postponed annuity as a forborne annuity.)
CHAPTER 40
Net Premiums
Net single premium. The net single premium is equal to the
present value of the benefit influenced by rates of mortality and
interest.
The net single premium for a whole life policy (a policy pay-
able at death only) is denoted by Ax. Solution of a problem of
this type is simplified by use of the commutation columns Mx and
Ac, thus:
A -^
x ~ Df
Example
Find the net single premium for $3,000 whole life insurance on a person
aged 24.
Solution
A~< = ^
$3,000 X .303644 = $910.93
Annual premiums. Life insurance premiums are most fre-
quently paid in equal annual payments, but they may be paid
semiannually, quarterly, or monthly, and, in the case of industrial
insurance, weekly. Rates other than annual are greater in pro-
portion than annual rates, for they include interest and additional
overhead or administrative costs.
On an ordinary life policy, payments continue throughout the
life of the insured. On a limited payment life policy, the premium
payments are limited to a certain number of years, such as 20
years on a 20-payment life policy.
The net annual premium is the annual payment made at the
beginning of each policy year, the sum thereof being the equivalent
of the net single premium. The annual premiums constitute an
annuity due payable by the policy holder to the insurance com-
pany. Px is the symbol used for the net annual premium and,
using commutation columns,
469
470 NET PREMIUMS
Example
Find the net annual premium for an ordinary life policy of $1,000 issued to
a person aged 30.
Solution
_3/3o_ 10259.0 _
1 3° ~ ~N^ ~ 596804 --017189
$1,000 X .017189 = $17.19
If the premium-paying period is limited to a certain number of
years, such as 10 years in a 10-payment life policy, then the pay-
ments are equivalent, interest and mortality considered, to the
single net premium. nPx is the symbol used for the net annual
premium for an n-payment life policy to a person aged x, and, in
terms of commutation columns:
Example
Find the net annual premium for a 20-payment lift1 policy for $2000 issued
to a person aged 45.
Solution
MK 7192.S1 7192.S1
20 46 #45 - A',* 253745 - 4S616 205129
$2000- .03506 = $70.12
Term insurance. Other than group life insurance, term insur-
ance is the lowest-cost life insurance obtainable. The term may
be one year, five years, or ten years, and so forth, and the face value
of the policy is payable in the event of death within the stated term.
The net single premium for term insurance may be ascertained
from commutation columns, using the formula
Example
Find the net single premium for a 10-year term insurance of $5,000 at age 2£
Solution
11()31.1 - 9094.96
37673.6
$5.000 X .067318 = $336.59
37673.6
= .0673,8
NET PREMIUMS 471
Annual premium for term insurance. The payments of pre-
mium constitute an annuity due for a definite term, and the net
annual premium may be determined from commutation columns,
using the formula
Example
Find the net annual premium for a 10-year term insurance of $3,000 at Age 20.
Solution
* 20 io I ~ "T7 \r
A -jo — A ao
_ 13207.3 - 10259.0
9S4400 - 59(>S04
" So = -007701
$3,000 X .007761 = $23.38
Net single premium for endowment insurance. An endow-
ment policy provides for payment of the face value of the policy
at the end of the stated period if the insured he living, or to the
named beneficiary or beneficiaries should death occur before the
end of the stated period.
Endowment insurance may be considered as term insurance of
1 for n years plus an n-year pure endowment of 1, and the net
single premium may be found from commutation columns by the
use of the following formula:
Mx — Mjcm -h />>,+„
A.n\ = -f)~ •
Example
Find the net single premium on a 10-year endowment policy for $5,000 at
rtgc 30.
Solution
_ Mao — A/40 + (>w
^30 10*1 ~~ ~~ ~~ TV
^30
_ 10259.0 - 8088.91 + 19727.4
30440.*
_ 21897.49 _
~ 1^04408 ~
$5,000 X .719346 = $3,596.73
Annual premium for endowment insurance. The net annual
premium for r years for an n-year endowment insurance of 1 may
be found from commutation columns, using the formula
472 NET PREMIUMS
M, - M1+n + L
Example
Find the net annual premium on a 15-year endowment policy for $10,000
purchased at age 40.
Solution
__ ~ A/55 + £>55
40 161 ~ vr \T~
N 40 — A/55
^ 8Q88-01 - 5510.54 + 9733.40
344167 - 124876"
$10,000 X .056143 = $561.43
Miscellaneous Problems
1. Find the net single premium for a whole life insurance of $1,000 at
age 30.
2. What is the increase in the net single premium for a whole life insurance
of $2,000 from age 25 to 26?
3. What is the present value of a life annuity of $1,000 a year at age 25?
4. Find the net annual premium for an ordinary life policy of $1,000 at
age 20.
5. Find the net annual premium for a 20-payment life policy for $10,000
at age 30.
6. What is the net annual premium for a 10-payment life policy for $2,000
at age 60?
7. What is tne net single premium for 10-year term insurance of $5,000 at
age 20V
8. What is the net annual premium on a 10-year endowment policy for
$5,000 at age 45?
9. Find the net annual premium for a 20-payment endowment at age 65
for $5,000 if the insured is 45 at date of issue.
10. Find the difference in annual premiums between a 20-payment life
policy and a 20-year endowment policy, each for $5,000, issued at age 30.
CHAPTER 41
Valuation of Life Insurance Policies
Mortality and the level premium. If the cost of insurance on
a group of men were to be met each year by payment into a fund
of just the amount necessary to meet that year's death loss, the
amount would be low at first, and finally prohibitive, hence the
necessity of a level premium. In order to have a level premium,
an excess over current death losses is collected during the early
years to bear the burden of later years when losses exceed the
premium income.
This excess premium is known as the " reserve." When cal-
culated on the basis prescribed by law, it is called the " legal
reserve/' and "legal reserve" companies are referred to as "old
line" companies.
Policy reserves. To show the meaning of insurance reserves,
a simple illustration is given.
Assume that an ordinary life policy for $1,000 is purchased at
age 20. The net annual premium, calculated from Table 8.
would be
$1,000 X .013847 = $13.85
Term insurance for one year at the same age would be :
1st year:
. C20 351.07 -_„.
A20 = yr- = = .00754
DM 40556.2
$1,000 X .00754 = $7.54
3th year:
C25 293.55
A25 = ^ = 3767^6 = -00779
$1,000 X .00779 = $7.79
and so on.
C Comparing these premiums over a period of years, we have:
473
474 VALUATION OF LIFE INSURANCE POLICIES
Ordinary One Year
Age Life Term
20 $13.85 $ 7.54
25 13.85 7.79
30 13 85 8.14
35 13 85 8 64
40 13 85 9.46
45 13 85 10 79
50 13 85 13 32
55 13 85 17 94
60 13 85 25 79
65 13 85 38.77
The excess of the premium on ordinary life over one-year term
insurance is the amount placed in the reserve to be accumulated
for heavier losses which will occur. It will be noticed that between
50 and 55 and from that point on the ordinary life premiums will
be insufficient; therefore, the reserves will be drawn upon to meet
the difference.
Interest and the premium. Reserves are invested in securities
and earn interest which increases the reserves. The amount of
interest earned, therefore, is reflected in lower premiums. If a
company assumes a rate of interest lower than the maximum
permitted under insurance law, the premium is higher and a larger
reserve is accumulated during the early policy years.
Loading. Using the mortality table and an assumed rate of
interest to be earned on the reserve, the actuary arrives at the net
level premium. To this must be added the expense of doing
business, or overhead, which amount is called "loading." The net
premium plus the loading is the premium rate to the purchaser of
insurance.
Expenses are heaviest in the first policy year; therefore, the
plan is modified to permit more of the premium to be used for
expenses, and this is balanced by lowering the amount required
for the reserve. Methods of modification* are not presented in
this text.
Dividends and net cost. Mortality may differ from that shown
by the table; interest may be earned in excess of the rate antici-
pated; the loading may exceed the actual costs. Such savings
result in dividends to policy holders in mutual companies and to
holders of participating policies issued by stock companies. The
net cost of the insurance is the premium paid less these dividend
refunds.
* An extended discussion of these methods will be found in Robert Riegel and
H. J. Loman, Insurance Principles and Practice. New York: Prentice-Hall, Inc.,
rev. ed., 1929.
VALUATION OF LIFE INSURANCE POLICIES 475
Terminal reserves. When a policy its issued, the mathematical
expectation of the future premiums equals the benefit.
As the insured grows older, the value of the future premiums
becomes less and the value of the benefit, conversely, becomes
greater.
The value of the benefit is represented by Ax+n, and the net
annual premium constitutes a life annuity with a value represented
by Px(l + djr+n); therefore, nVx, the terminal reserve, is found by
the formula
»r, = Ax+n -Pt(l + «,,„)
The foregoing method of valuation is called the prospective
method.
Example
Find the terminal reserve of the 20th policy year on an ordinary life policy
of $1,000 issued at age 30.
Solution
1 + 13.5347 = $14.5347
_j|/,._1 0259.0
lm~ A^~ 590804
Substituting values gives:
20^30 = .50849 - (.017190 X 14.5347)
= .25864
$1,000 X .25864 = $258.64
The surrender value is the sum which the insurance company
pays the policy holder upon the surrender and cancellation of the
policy. Whether the amount will be greater or less than an
amount as calculated in the foregoing example is dependent on
averages obtained from the company's records instead of the
theoretical amount so calculated. Policies contain a table showing
the company's contractual surrender value for each $1,000 of
insurance.
Problems
1. Find the terminal reserve of the tenth year on an ordinary life policy of
$3,000 taken at age 20.
2. Find the terminal reserve of the twentieth year on an ordinary life policy
of $5,000 taken at age 30.
476 VALUATION OF LIFE INSURANCE POLICIES
Retrospective method. Under this method the policy value is
found by deducting the accumulated losses from the accumulated
premiums. The formula is :
v - MI Nx " Nz+n - Mx "" Mx+H
n * Nx' DI+n Dx+n
Problems
1. Check the answer to the example, using the retrospective formula.
2. Find the surrender value in Problems 1 and 2 on page 475 by the retro-
spective method.
Transformation. In computing the terminal reserve, the for-
mula used was
and the net single premium at age x, denoted by Ax, was computed
by the formula :
A, = Px(l -f O
Then, substituting for Ax, we have:
which represents the policy value or reserve, for the policy value
is equal to the present value of the difference between the net
premiums for age x + n and age x for the remainder of life.
To express the value of the reserve in terms of annuities, take
the formula:
which denotes the present value of 1 payable at the end of the
year in which a person dies (d being the value, at the beginning of
the year, of the interest for each year on 1), and
p* = nb;-«
which denotes the annual premium Px, for an ordinary life policy
expressed in terms of annuity values. Then substitute the values
of Ax and Px in the formula:
nVx = Ax+n - P,(l -f- ax+n)
Following the algebraic processes of simplifying, the result is
nV, = 1 - \+,aw
VALUATION OF LIFE INSURANCE POLICIES 477
Problems
1. With the aid of the table of life annuities, calculate the terminal reserve
of the twentieth policy year on an ordinary life policy for SI, 000 issued at age 30.
2. Find the terminal reserve the tenth year on an ordinary life policy of
$5,000 issued at age 25.
Reserve valuation for limited payment life insurance. In
determining the reserve valuation for such policies as 10-payment
life, 20-payment life, and so forth, the following principle is funda-
mental : the terminal reserve of the nth policy year equals the net
single premium at the attained age of the insured minus the present
value of the future net premiums. Using m to denote the number
of annual payments, we have
n:mVx = Ax},, - J\(] + rtxfnro_n_,i)
but only when n is less than m. Where n is equal to or greater
than m, the terminal reserve is simply equal to the net single
pi emium.
For endowment insurance, the formula is
nVKxr\ = AKx+n r_n] - PEzr] (1 + <lf + n:r--j=i])
for an r-year endowment.
Problems
1. Find the terminal reserve of the fifteenth policy year for a $2000 20-year
pay-life policy issued at age 30.
2. Find the terminal reserve of the fifteenth policy year for a $3000 20-year
endowment issued at age 45.
Preliminary term valuation. Initial expenses of securing a
policy, such as agents' commissions, medical and inspection fees,
arid other expenses, make it practically impossible to provide any
reserve out of the first year's premium. Under the net level pre-
mium method, the loading is the same each year; therefore,
expenses exceed income, and the deficit must be met from general
funds. To avoid this, the preliminary term valuation is used,
whereby the first year's premium becomes available for expenses
and losses, the policy is renewed at the beginning of the second
year, and policy values begin with that year. Therefore, the
first year is simply term insurance.
Under this plan, assume an ordinary life policy of $1,000 at
age 20, the premium being $16.50.
The net premium for the first year would be :
* 351'07
Z)20 " 46552
$1,000 X .00754 = S7.54
$16.50 - $7.54 = $8.06, the loading for the first year.
478 VALUATION OF LIFE INSURANCE POLICIES
For the second and subsequent years, the net premium will he
the level net premium based upon age 21 :
_ Jf» _ 12916.3
^2l " IvTt " 937843
$1,000 X .01377 = $13.77
$16.50 - 13.77 = $2.73, the loading for each year after the first.
Problems
1. Find the loading for the first and second years on an ordinary life policy
of $2000 at age 30, with an annual premium of $42.60, using the preliminary
term evaluation.
2. Using the preliminary term evaluation, find the loading for the first and
second years on an ordinary life policy of $1000 at age 40, the premium being
$28.80, *
APPENDIXES
APPENDIX PAOfe/
I. Practical Business Measurements 481
II. Tables of Weights, Measures, and Values 489
III. Tables:
1. Logarithms .... ... ... 497
2. Compound amount of 1 : ,s = (1 + z)w 512
3. Present value of 1: vn = ^ - - or vn = (I + f)"n «r)20
(1 + i)n - 1
4. Amount of annuity of 1 : .<? = . 527
i
5. Present value of annuity of 1 : a , — . 530
alt l
1
6. Rent of present value of annuity of 1 : — = . . . 533
«»'* i
7. Amorican Experience Table of Mortality. . . 530
8. Commutation Columns, 3l% 538
479
APPENDIX I
Practical Business Measurements
Practical business measurements. The measurements discussed in
this section are those of practical use, and include measurements of or
pertaining to the following: angles; surfaces; triangles, rectangles, and
other polygons; circles, including circumference, radius, diameter, and
area; area of irregular figures; and solids, such as the sphere, cone, cylin-
der, cube, and prismatoid.
Rectilinear figures. An angle is the difference in the direction of two
lines proceeding from a common point called the vertex.
A right angle is an angle formed by two lines perpendicular to each
other, and is an angle of 90°.
An angle that is less than a right angle is an acute angle, and one that
is greater is an obtuse angle. Acute and obtuse angles are also called
oblique angles.
A surface has two dimensions — length and breadth.
A plane, or a plane surface, is a level surface. A straight edge will fit
on it in any position.
A plane figure is a figure all of whose points lie in the same plane.
A quadrilateral is a plane figure bounded by four straight lines.
Quadrilaterals are of three classes or kinds: the trapezium, which has
four unequal sides, no two of which are parallel ; the trapezoid, which has
two and only two sides parallel; and the parallelogram, which has two
pairs of parallel sides.
The altitude of a quadrilateral having two parallel sides is the perpen-
dicular distance between those sides.
The diagonal of a quadrilateral is the straight line connecting two of
its opposite vertices.
Parallelograms are of three classes or kinds: the rhomboid, which has
one pair of parallel sides greater in length than the other pair, and no
right angles; the rhombus, all of whose four sides are equal; and the
rectangle, whose angles are all right angles.
A square is a rectangle having four equal sides. It is also a rhombus
whose four angles are each 90°.
A triangle is a plane figure bounded by three straight lines. If the
three sides are of equal length, the triangle is called equilateral. If two
sides are of equal length, it is called isosceles. If the three sides are of
different lengths, it is called scalene. If one of the three angles is a right
angle, the triangle is called a right triangle, and the side opposite the
right angle is called the hypotenuse.
Plane figures may be regular or irregular. A regular plane figure has
all its sides and all its angles equal The smallest regular plane figure is
481
482
APPENDIXES
an equilateral triangle; the next, a square; the next, a pentagon; and so
on. Each figure derives its name from the number of its angles or sides-
hexagon, heptagon, octagon, nonagon, decagon, etc.
The perimeter of a plane figure is the sum of the lengths of its sides
The apothem of a polygon is a perpendicular line drawn from the
center of the figure to the middle of a side, the center being the point
within the figure which is equally distant from the middle points of all
the sides.
The altitude of a plane figure is the perpendicular distance from the
highest point above the base to the base or to the base extended.
Circles. A circle is a plane figure bounded by a curved line, called
the circumference, every point of which is equally distant from a point
within called the center.
The diameter of a circle is a straight line drawn through the center
and terminated by the circumference.
The radius of a circle is a straight line drawn from the center to the
circumference, and is equal to one-half the diameter.
Figure 16.
An arc of a circle is any portion of the circumference.
A sector of a circle is bounded by two radii and the intercepted arc.
A chord is the straight line joining the extremities of an arc.
A segment is bounded by an arc and its chord.
A zone is a portion of a circle bounded by two parallel chords.
A tangent to a circle is a straight line having only one point in common
with the curve; it simply touches the circle. A secant enters the figure
from without.
An ellipse is a plane figure bounded by an oval curved line, and has a
long and a short diameter or axis.
Measurement of triangles. It is proved in geometry that the square
erected on the hypotenuse of a right triangle is equal to the sum of the
squares erected on the other two sides. This may be illustrated as in
Figure 16.
PRACTICAL BUSINESS MEASUREMENTS
483
To find the length of the hypotenuse of a right triangle, the lengths of
the other two sides being given, add the squares of the sides forming the
right angle, extract the square root of the sum, and the result will be the
length of the hypotenuse.
To find the length of either of the two sides other than the hypotenuse,
from the square of the hypotenuse subtract the square of the given side,
extract the square root of the remainder, and the result will be the length
of the third side.
To find the area of a triangle, the base and the altitude being given,
multiply the base by one-half the altitude.
To find the area of a triangle when the lengths of the three sides are
given, from half the sum of the three sides, subtract the length of each
side separately. Find the continued product of the three remainders and
the half sum. The square root of the result will be the area.
Measurement of rectangles. The area of a square or of a rectangle
is the product of the length and the breadth.
Either dimension of a rectangle may be found by dividing the area by
the given dimension.
Measurement of quadrilaterals. The area of a trapezium may bo
found by multiplying one-half the sum of the altitudes by the diagonal.
The area of a trapezoid may be found by multiplying the sum of the
parallel sides by one-half the altitude.
Figure 17. Trapezium.
Figure 18. Trapezoid.
The area of a parallelogram may be found by multiplying the base
by the altitude.
Figure 19. Parallelogram.
The area of polygons having equal sides and equal angles may be
found by multiplying the square by one of the equal sides by :
.433, if the figure is a triangle
1.7205, if the figure is a pentagon
2.5981, if the figure is a hexagon
4.8284, if the figure is an octagon
484 APPENDIXES
Measurement of circles. It is shown in geometry that the circum-
ference of a circle bears a fixed ratio to its diameter. This constant ratio
is represented by TT I'pmnoiim-nl "pi")> and is 3.1416.
From this relation the following principles are derived:
The circumference = the diameter X 3.1416
The diameter = the circumference -r 3.1416
The area = the circumference X half the radius
The area of a circle is found by considering the surface to be composed
of an infinite number of isosceles triangles, the bases of which, taken
together, equal the perimeter of the circle. The common altitude of
these triangles constantly approaches the radius of the circle, and will
reach that length when the perimeter consists of very short straight lines;
hence, perimeter (circumference) X i radius = area.
To find the circumference of a circle, multiply the diameter by 3.1416,
or divide the area by one-fourth of the diameter.
To find the diameter of a circle, divide the circumference by 3.1416,
or divide the area by .7854 and extract the square root of the result.
To find the area of a circle, multiply the circumference by one-half
the radius; or, multiply the diameter by one-fourth of the circumference;
or, multiply the square of the diameter by .7854; or, multiply the square
of the radius by 3.1416.
To find the area of an ellipse, multiply the major axis by the minor
axis, and that result by .7854.
To find the area of a sector of a circle, multiply one-half the length of
the arc by the radius; or, take the same part of the area of the circle as the
number of degrees in the arc is of 360°.
To find the area of a segment which is less than a semi-circle, from
the area of a corresponding sector, subtract the area of the triangle
formed by the chord and radii; to find the area of a segment which is
greater than a semi-circle, add the area of the triangle formed by the
chord and radii to the area of a corresponding sector.
To find the area of a zone, from the area of the circle subtract the
areas of the segments not included in the zone.
Problems
1. Harry and George start from the same point, Harry going 4 miles due
west, and George 3 miles due north; how far apart are they?
2. The base of a triangle is 12 inches, and the altitude is 8 inches. What
is the area of the triangle?
3. The three sides of a triangular plot of land are 100 feet, 130 feet, and
150 feet. What is the area of the plot?
4. A rectangular piece of land is 40 rods long and 20 rods wide. What is
the area in square rods?
6. Find the cost of fencing a field 40 rods wide and 55 rods long, if the
fencing costs $2.25 a rod.
PRACTICAL BUSINESS MEASUREMENTS 485
6. A field is in the form of a trapezium, having a diagonal of 90 rods, and
altitudes of 25 rods and 40 rods. What is the area in square rods?
7. One of the parallel sides of a garden is 60 yards long, and the other is
SO yards long. The garden is 52 yards wide. How many square yards does it
contain?
8. Find the area of a parallelogram whose base is 10 feet, and whose alti-
tude is 4 feet.
9. The side of a hexagonal building is 20 feet. What is the floor area?
10. A cylindrical tank is 12 feet in diameter. What must be the length of
a piece of strap iron which is to be used to make a band around the tank, if 1 foot
is allowed for overlapping?
11. The circumference of a circle is 44 feet. What is its diameter?
12. Find the area of the circle in problem 11.
13. The diameters of an ellipse are 00 feet and 40 feet. What is the* area?
14. How much belting will be required to make a belt to run over two pulleys,
each 30 inches in diameter, if the distance between the centers of the pulleys
is 18 feet?
15. If there is a steam pressure of 90 pounds to the square inch, what is the
pressure on a 9-inch piston?
16. If pieces of sod are 12 inches by 14 inches, how many pieces will be
required to sod a lawn 24 feet wide and 2S feet long?
17. Find the cost of painting the four side walls of a room 14 feet long,
10 feet 6 inches wide, and 8 feet high, at 18 cents a square yard, no allowance
being made for openings.
18. A circular walk 5 feet wide is laid around a plot 20 feet in diameter.
What is the cost of the walk at $2.50 a square foot?
19. Find the number of paving blocks required to pave a street one mile
long arid 35 feet wide, if the blocks are one foot long and five inches wide.
20. What part of an acre is a plot of land 78 feet long and 36 feet wide?
Solids. A solid is a magnitude which has length, breadth, and thick-
ness. Solids include the prism, the cylinder, the pyramid, the cone, the
polyhedron, and the sphere.
A prism is a solid whose upper and lower bases are equal and parallel
polygons, and whose sides, or lateral faces, are parallelograms.
A rectangular sold is bounded by six rectangular surfaces.
A cube is a rectangular solid having six square faces.
A triangular prism is a prism whose bases are triangles.
A cylinder is a prism having an infinite number of faces or sides; the
two bases are equal parallel circles.
A pyramid is a solid having for its base a polygon, and for its other
faces three or more triangles which terminate in a common point called
the vertex or apex.
A cone is a pyramid having an infinite number of faces; or, it is a solid
whose base is a circle, and whose convex surface tapers uniformly to a
point called the apeY.
486 APPENDIXES
A polyhedron is a solid bounded by four or more faces.
A sphere is a solid bounded by a curved surface, every point of which
is equally distant from a point within, called the center.
The frustum of a pyramid or of a cone is the solid which remains when
a portion which includes the apex is cut off by a plane parallel to the base.
The axis of a pyramid or of a cone is a straight line that joins the apex
to the center of the base.
The altitude of a pyramid or of a cone is the perpendicular height
from its apex to its base.
The slant height of a pyramid is the distance from the apex to the
midpoint of one side of its base.
The slant height of a cone is the distance from its apex to the circum-
ference of its base.
The diameter of a sphere is a straight line drawn through its center
and terminated at both ends by the surface.
The radius of a sphere is one-half of its diameter.
The circumference of a sphere is the greatest distance around the
sphere.
A hemisphere is one-half of a sphere.
Measurement of solids. To find the contents of a prism or of a
cylinder when the perimeter of the base and the altitude are given,
multiply the area of the base by the altitude.
To find the convex surface of a prism or of a cylinder, multiply the
perimeter of the base by the height.
To find the entire surface of a prism or of a cylinder, add the area of
the bases to the area of the convex surface.
To find the convex surface of a cone, multiply the circumference of the
base by one-half the slant height.
To find the entire surface of a cone, add the area of the base to the
area of the convex surface.
The slant height of a pyramid or of a cone may be found by adding the
square of the altitude to the square of the radius of the base, and extract-
ing the square root of the sum.
To find the volume of a pyramid or of a cone, multiply the area of the
base by one-third the altitude.
The volume of a pyramid is one-third as much as the volume of a
prism that has the same base and altitude.
The volume of a cone is one-third as much as the volume of a cylinder
that has the same base and altitude.
To find the convex surface of a frustum of a pyramid or of a cone,
multiply one-half the sum of the perimeters of the two bases by the slant
height.
To find the entire surface of a frustum of a pyramid or of a cone, add
the area of the two bases to the area of the convex surface.
To find the volume of a frustum of a pyramid or of a cone, find the
product of the areas of the two bases, and extract the square root thereof
This result is the area of a base which is a mean base between the other
two Add the three areas, and multiply by one-third the altitude.
PRACTICAL BUSINESS MEASUREMENTS 487
To find the surface of a sphere, find the area of a great circle of the
sphere, and multiply this area by 4.
To find the volume of a sphere, multiply the convex surface by one-
third the radius.
The volume of a spherical shell (a hollow sphere) is equal to the
volume of the outside sphere minus the volume of the inside sphere.
Problems
1. A cylindrical tank is 12 foot in diameter. If it is filled with water to a,
depth of 6 feet, what is the weight of the water? (1 cu. ft. of water weighs
62.5 pounds.)
2. How many square yards of sheet metal will be required for a smoke-
stack 2 feet in diameter and 12 feet in height, if I inch is allowed for overlapping?
3. Find the cost of painting the entire surface of a cylindrical tank 10 feet
in diameter and 20 icct long, at 10^ per square foot.
4. The boundary lines of the Fort Pombina Airport are marked by cone-
shaped markers; each marker is 3 feet in diameter and has a slant height of
3 feet. If there are 120 of these markers, and 1 inch \\as allowed for over-
lapping, how many square feet of sheet metal were required for their construction?
6. If a freight car is 36 foot long, and S foot 6 inches \\ido, inside measure,
how many hi she-Is of wheat will it contain when filled to a depth of 5 feet?
(A cubic foot is approximately .8 of a bushel.)
6. How many tons of coal will fill a bin 20 feet, by 16 feet, by 8 feet, if there
are 80 cubic feet to a ton?
7. The measurements of a railroad embankment are: length, 400 feet; height,
10 feet; width of base, 14 feet; and width of top, 8 feet. How many cubic yards
of earth will be required?
8. The measurements of a funnel are as follows: larger diameter, 12 inches;
•»rnaller diameter, 1 inch; and slant height, 18 inches. How many square inches
of sheet metal will be required?
9. One of the units of a grain elevator is a concrete cylinder 20 feet in
diameter and 50 feet in height. The bottom is cone-shaped to facilitate tho
drawing off of the grain. The depth of this cone is 5 feet. If the wheat in this
unit is leveled off at the 30-foot mark, how many bushels are in the unit, assuming
that a cubic foot is approximately .8 of a bushel?
10. A bucket is 16 inches wide at the top, and 10 inches wide at the bottom.
The depth is 12 inches. How many gallons of water will the bucket hold?
(231 cu. in. = 1 gal.)
11. What number of square feet of sheet metal will be required to make
XOO pails, each 10 inches dee]), 8 inches in diameter at the bottom, and 11 inches
in diameter at the top? The allowance for seams and for waste in cutting is 10%.
12. How many tiles 1 inch square will be required for the surface of a tiled
dome in the form of a hemispherical surface, if the diameter of the dome is
24 feet?
13. The top of a vat is 9 feet square, and the base is 8 feet square. If the
slant height is 10 feet, what is the capacity of the vat in cubic feet?
488 APPENDIXES
14. How many cubic feet are there in a spherical body whose diameter is
10 feet?
16. The base of a church steeple is in the form of an octagon measuring
6 feet on each side. The slant height of the steeple is 80 feet. What will be the
cost of painting this steeple at 50^ per square yard?
16. A tank is 8 feet long, 6 feet wide, and 3 feet deep. If a cubic foot of
water weighs 62.5 pounds, what is the weight of water in this tank if it is two-
thirds full?
17. If 38 cubic feet of coal weigh a ton, how many tons can bo put into a bin
10 feet long and 8 feet wide, if the coal is leveled off at an average depth of
5 feet?
18. Find the number of square feet of sheet metal required to make 12 gross
of pails, each 14 inches deep, 8 inches in diameter at the bottom, and 11 inches
in diameter at the top, not allowing for seams or waste in cutting.
19. The diagram is that of a cross section of a concrete retaining wall 150 feet
long. Find the number of cubic yards of material necessary to construct such
a wall.
2'
20. If 200 gallons of water flow through a pipe 2 inches in diameter in 4 hours,
how much water will flow through a pipe 4 inches in diameter in the same time?
HINT: The amounts of the liquids are to each other as the squares of the like
dimensions.
21. A cylindrical hot water tank is 5 feet high and 11 inches in diameter.
How many gallons will it contain?
22. A corn crib 32 feet by 10 feet by S feet is filled \\ith oar corn. How
many bushels will it contain if one bushel equals 1^ cubic foot?
23. A barn loft is 36 feet by 24 feet by 8 foet. How many tons of hay will
it hold if it is to be filled with: (a) clover hay weighing one ton for 600 cubic foot;
(6) timothy hay weighing one ton for 500 cubic feet?
24. If a heaped bushel equals 1^- cubic feet, how many bushels of potatoes
may be stored in a bin that is 12 feet by 8 feet by 6 feet?
25. If a cubic foot of steel weighs 484 pounds, what is the weight of a hollow
steel cylinder whose length is 10 feet and the radii of whose outer and innei
circles are 3 feet and 2^ feet, respectively?
APPENDIX II
Tables of Weights, Measures, and Values
Long Measure
U. S. and British Standard
12 inches
3 feet
5-gr yards, or 163- feet
320 rods, or 5,280 feet
,760 yards
40 rods.
8 furlongs
3 miles
foot
yard
rod
mile
mile
furlong
statute mile
league
Metric System
10 millimeters. . . I centimeter
10 centimeters . 1 decimeter
10 decimeters 1 meter
10 meters . 1 dekameter
10 dekameters . 1 hektometer
10 hekto meters . 1 kilometer
10 kilometers . . 1 myriameter
Comparisons of Long Pleasures
1 inch 25.4001 millimeters 1 centimeter . .3937 inch
1 foot 304S01 meter 1 meter. .. 39.37 inches
1 yard 914402 meter I meter. . 3.28083 feet
1 rod 5.029 meters 1 meter. . . 1.09361 1 yards
1 mile 1.00935 kilometers 1 kilometer . . .02137 mile
Square Measure
( r. 8. and British Standard
1 44 square
9 square
30J- square
272^ square
40 square
4 roods
160 square
640 acres. .
43,560 square
4,840 square
inches
feet
yards
feet
rods
rods
feet. . . .
yards .
100 square millimeters
100 square centimeters
100 square decimeters
100 square meters .
100 square dekameters
100 square hekto meters
Metric System
1 square foot
1 square yard
square rod
square rod
rood
acre
acre
square mile
acre
acre
1 square centimeter
1 square decimeter
1 square meter
1 square dekameter
1 square hektometer
1 square kilometer
100 square kilometers 1 square myriameter
489
490
APPENDIXES
Comparisons of Square Measures
1 sq. in 6.452 sq. cm. 1 sq. mm 00155 sq. in.
1 sq. ft 0929 sq. m. 1 sq. cm 155 sq. in.
1 sq. yd 8361 sq. m. 1 sq. m 10.764 sq. ft.
1 sq. rd 25.293 sq. m. 1 sq. m 1.196 sq. yds.
1 sq. mi 2.59 sq. km. 1 sq. km 3861 sq. mi.
1 sq. km 247.11 acres
1 sq. Dm., or 1 are 1,076.41 sq. ft.
100 ares = 1 hektare 2.4711 acres
Solid or Cubic Measure (Volume)
U. S. and British Standard Metric System
1 ,728 cubic inches 1 cubic foot 1 ,000 cubic millimeters ... 1 cu. cm.
27 cubic feet. . 1 cubic yard 1,000 cubic centimeters. . . 1 cu. dm.
128 cubic feet. . cord of wood 1,000 cubic decimeters. ... 1 cu. m.
24f cubic feet . perch of stone 1,000 cubic meters 1 cu. Dm.
2,150.42 cubic inches standard bushel 1,000 cubic dekameters. . . 1 cu. Hm.
231 cubic inches standard gallon 1,000 cubic hektometers. . 1 cu. Km.
40 cubic feet . ton (shipping) 1,000 cubic kilometers ... 1 cu. Mm.
Comparisons of Solid or Cubic Measures (Volume)
1 cu. in 16.3872 cu. cm. 1 cu. cm 061 cu. in.
I cu. ft 02832 cu. in. 1 cu. m 35.314 cu. ft.
1 cu. yd 7646 cu. in. 1 cu. m 1.3079 cu. yds.
1 cu. dm. = 1 liter 61.023 cu. in.
1 liter 1 .05671 liquid quarts
1 liter 9081 dry quart
1 hectoliter or decistore. . . 3.5314 cu. ft. or
2.8375 U. S. Bushels
1 stere, kiloliter, or cu. m . 1.3079 cu. yds. or
28.37 U. S. Bushels
Liquid Measure (Capacity)
U. S. and British Standard Metric System
4 gills 1 pint 10 milliliters 1 centiliter
2 pints 1 quart 10 centiliters 1 deciliter
4 quarts 1 gallon 10 deciliters ... .1 liter
31^ gallons 1 barrel 10 liters 1 dekaliter
2 barrels 1 hogshead 10 dekaliters 1 hektolitei
1 U. S. Gallon. ... 231 cubic inches 10 hektoliters 1 kiloliter
1 British Imperial 10 kiloliters 1 myrialitei
Gallon 277.274 cubic inches
7.4805 U. S. Gallons. . . 1 cubic foot
16 fluid ounces. ... 1 pint
1 fluid ounce 1.805 cubic inches
1 fluid ounce 29.59 cubic centimeters
1.2 U. S. Quarts. ... 1 Imperial Quart
1.2 U. S. Gallons. . . 1 Imperial Gallon
1 gallon gasoline . . 6 pounds (approx.)
1 gallon oil 7^- pounds (approx.)
1 gallon water 8.3 pounds (approx.)
1 liter gasoline. . . 1.59 pounds (approx.)
1 liter gasoline . . . 0.72 kilograms
TABLES OF WEIGHTS, MEASURES, AND VALUES 491
Dry Measure
U '. 8. and British Standard
2 pints 1 quart
8 quarts . . 1 peck
4 pecks 1 bushel
2,150.42 cubic inches I IT. S. Standard Bushel
1.2445 cubic feet 1 U. S. Standard Bushel
2,218.192 cubic inches . 1 British Imperial Bushel
1.2837 cubic feet . 1 British Imperial Bushel
Metric System
[In the Metric System,
the same table is used
for both Liquid Meas-
ure and Dry Measure.]
Comparisons of Liquid and Dry Measures
1 liquid quart.
1 liquid gallon
1 dry quart
1 peck
1 bushel
1 milliliter
1 liter -
28.317 liters
4.543 liteis
3.785 liters.
decimeter
.94636 liter
3.78543 liters
1.1012 liters
S.80982 liters
.35239 hektoliters
.03381 liquid ounce, or
.2705 apothecaries'
dram
61.023 cubic indies
.03531 cubic foot
.2642 U. S. Gallon
2.202 pounds of water
at 62° F.
1 cu. ft.
1 British Imperial Gal.
1 U. S. Gal.
Avoirdupois Measure (Weight)
( Used for weighing all ordinary substances except precious metals, jewels, and
drugs)
U. S. and British Standard
grains 1 dram
16 drams 1 ounce
16 ounces 1 pound
25 pounds . 1 quarter
4 quarters 1 hundredweight
100 pounds 1 hundredweight
20 hundredweight ... 1 ton
2,000 pounds 1 short ton
2,240 pounds 1 long ton
Metric System
10 milligrams . . 1 centigram
10 centigrams 1 decigram
10 decigrams 1 gram
10 grams 1 dekagram
1 0 dekagrams I hektogram
1 0 hektograms 1 kilogram
10 kilograms 1 myriagrani
Troy Measure (Weight)
(Used for weighing gold, silver, and jewels)
24 grains
20 pennyweights .
12 ounces
1 pennyweight
I ounce
1 pound
492 APPENDIXES
Apothecaries' Measure (Weight)
(Used for weighing drugs)
20 grains
. . . . 1 scruple
3 scruples
... 1 dram
8 drams
1 ounce
12 ounces
1 pound
Comparison of Avoirdupois and Troy Measures
1 pound troy 5,760 grains 1 ounce troy 4X0 gniins
1 pound avoirdupois . . . 7,000 grains 1 ounce avoirdupois 437i grains
I karat, or carat 3.2 troy grains
24 karats pure gold
Comparison of Avoirdupois and Troy Measures with Metric Weights
grain 0648 gram ( 15.4324 grains
ounce (avoir.).. 28.3495 grams 1 gram < .03527 ounce (avoir.)
ounce (troy).. . 31.10348 grains ( .03215 ounce (troy)
pound (avoir.) . .45359 kilogram ( 2.20462 pounds
pound (troy) . . .37324 kilogram i kilogram < (avoir.)
( 2.67923 pounds (troy)
.9842 ton of 2,240
1 tonne, or
metric ton
pounds, or 19.68
hundredweight
1.1023 tons of 2,000
pounds
1,000 kilograms 2,204.6 pounds
1.016 metric tens, or
1,016 kilogrims. . 1 ton of 2,240 pounds
Apothecaries* Fluid Measure (Capacity)
60 minims 1 fluid dram
8 fluid drams 1 fluid ounce
16 fluid ounces . I pint
8 pints . . .1 gallon
Comparisons (Approximate Liquid Measure)
Apothecaries' Common Metric
1 minim 1 to 2 drops 0.06 cu. cm.
60 minims, or
1 fluid dram 1 teaspoonful 3.75 cu. cm.
2 fluid drams 1 dessertspoonful 7.50 cu. cm.
4 fluid drains 1 tablespoonful 15.00 cu. cm.
8 fluid drams 1 fluid ounce 28.39 cu. cm.
2 fluid ounces 1 wineglassful 59.20 cu. cm.
4 fluid ounces 1 teacupfui 118.40 cu. cm.
16 fluid ounces 1 pint 473.11 cu. cm
NOTE: Drops are not accurate measures, but for practical purposes it may
be considered that one minim equals one drop of watery liquids and fixed oils,
but two drops of volatile oils and alcoholic liquids, such as tinctures and fluid
extracts.
TABLES OF WEIGHTS, MEASURES, AND VALUES 493
MISCELLANEOUS TABLES
Surveyors1 Long Measure
7.92 inches 1 link
25 links 1 rod
4 rods, or 100 links . 1 chain
80 chains 1 mile
Surveyors' Square Measure
625 square links 1 square rod
16 square rods. . . . . 1 square chain
10 square chains 1 acre
640 acres . . 1 square mile
36 square miles . 1 township
Mariners' Measure
6 feet 1 fathom
120 fathoms 1 cable's length
7^ cable lengths I mile
5,280 feet 1 statute mile
6,080 feet 1 nautical mile, or British Admiralty knot
50.71H feet 1 knot
120 knots, or
1.152-| statute miles 1 nautical or geographical mile
3 geographical miles 1 league
60 geographical miles, or
69.16 statute miles 1 degree of longitude on the equator, or
1 degree of meridian
360 degrees. . . . . . . 1 circumference
NOTE: A knot is properly T^TT °f a marine mile, but current usage makes it
equivalent to a marine mile. Hence, when the speed of vessels at sea is being
measured, a knot is equal to a nautical mile, or 6,086.08 feet, or 2,028.69 yards.
Circular or Angular Measure
60 seconds (60") ... 1 minute (!')
60 minutes (600 • • 1 degree (L°)
30 degrees . . .1 sign
90 degrees 1 right angle or quadrant
360 degrees . . . 1 circumference
NOTE: One degree at the equator is approximately 60 nautical miles.
Counting
12 units or things . . 1 dozen
12 dozen, or 144 units . . 1 gross
12 gross . 1 great gross
20 units ... 1 score
Paper Measure
24 sheets 1 quire
20 quires . . ... . . 1 ream
2 reams . 1 bundle
5 bundles . . .... 1 bale
NOTE: Although a ream contains 480 sheets, 500 sheets are usually sold aa
a ream.
494
APPENDIXES
Books
Books are printed on large sheets of paper, which arc folded into leaves
according to the size of the book. The terms folio, quarto, octavo, and so forth,
as applied to printed books, are based on sheets about IS by 24 inches, or about
half the size now generally used, and indicate the number of leaves into which
each sheet is folded.
A sheet folded in 2 leaves is called a folio
" 4
" X
" 12
" 16
" 24
" 32
a quarto, or 4to
an octavo, or Svo
a 12 mo
a 16 mo
a 24 mo
a 32 mo
and makes 4 pages
a a u a
32
4^
04
Sizes of Paper
Book Papers Bo fid, Ledger, and Writing Papers
25 X 38 38 X 50 14 X 17 20 X 28
3(4 X 41 41 X 61 16 X 21 23 X 31
32 X 44 64 X 44 18 X 23 21 X 32
33 X 44 66 X 44 17 X 28 16 X 42
35X45 45X70 19X24 23 X 36
1 7 X 22 22 X 34
10 mills . .
10 cents
10 dimes
10 dollars
4 farthings
12 pence.
20 shillings. .
MEASURES OF VALUE
United States Money
English Money
A pound sterling = $4.8665 (normal).
1 cent
1 dime
I dollar
1 eagle
1 penny (d.)
1 shilling (s.)
1 pound (£)
French Money
10 milliraes (m.) .
10 centimes. . . .
10 declines
A franc = $0.193 (normal).
1 centime (c.)
1 decime (d.)
1 franc (fr.)
Comparison of Thermometer Scales
To convert from ° F to ° C, subtract 32 from ° F and divide by 1.8.
To convert from ° C to ° F multiply ° C by 1.8 and add 32.
TABLES OF WEIGHTS, MEASURES, AND VALUES 495
Degrees C
-100
- 50
- 40
- 20
- 17 77
- 15
- 10
- 5
0
5
10
15
20
25
30
35
40
45
50
55
00
65
70
75
cSO
85
90
95
100
150
190
200
300
Temperature Equivalents
Degrees F
-148
- 58
- 40
- 4
0
5
14
23
32 ....
41
50
59
OX
77
80
95
104
113
122
131
140
149
158
107
170
185
194
203
212
302
374
392
572
Remarks
Water freezes
Water boils at sea level
(With each 1,000 feet
altitude, boiling point of
water is reduced ap-
proximately 1° C.)
Approximate Weight of Substances
Brick, pressed, host ....
Brick, common, hard
Brick, common, soft
Coal, broken (anlhra ), loose
Coal, broken (bitu.), loose
Cement, concrete, limestone..
Cement, concrete, cinder
Cement, concrete, stone
Cement, concrete, trap rock
Granite
Hemlock, dry
Hickory, dry
Ice ...
Iron, cast
Iron, wrought.
Lbs. Per
Cu. Ft.
150 Lead
125 Limestone, marble1, ordinarily
100 Limestone, marble, piled
52--50 Masonry, granite, dressed
47-52 Masonry, sandstone
148 Sand, pure quartz dry loose
112 Ibs. per struck bu
150 Sand, angular, large and small.
155 Sandstone, dry for building
170 Sandstone, quarried, piled . . .
25 Shales, red or black
53 Shales, quarried, piled . ...
57-0 Slate
450 Soapstone or steatite
485 Steel, heaviest, lowest in carbon
IJbx. Per
C?/. Ft.
709.6
168
96
165
145
112 113
90-106
117
151
86
162
92
175
170
490
496
APPENDIXES
Coal, anthracite egg
Coal, anthracite nut. ... ....
Coal, anthracite stove. .
Coal, bituminous, 111
Coal, bit., Ind. block .
Coal, bit., Iowa lump
Coal, bit., Pittsburgh
Coal, bit., Pocahontas egg and lump
Coal, cannel
Coke, loose. .
Charcoal, hardwood
Charcoal, pine
Peat, dry
WEIGHTS AND MEASURES
Solid Fuels
Lbs. Per Tons Per Cu. Ft.
Cu. Yd. Cu. Yd. Per Ton
1514 .76 36
1536 .77 36
1521 .76 30
1275 .64 42
1161 .58 43
1 256 63 42
. 1255 .63 42
1411 .71 38
1328 .66 49
S70-1026 51 60-65
513 25 19
486 .24 19
1269 .63 42
Anthracite and Pocahontas, approximately 36 cu. ft. for I ton. Othei
bituminous coal, approximately 40^ cu. ft. for 1 ton. Coke, approximately
60-65 cu. ft. for 1 ton.
Bulk Materials
Lbs. Per Tons Per
Cu. Yd. Cu. Yd.
Ashes 1080 52
Asphalt 2700 1 35
Brick, so "t clay . . 2718 1 35
Brick, hard clay. . . . 3397 1 69
Brick, pressed 3806 1 . 90
Bluestone . . 2970 1 .48
Cement, Portland . . 2430 1 .21
Cinders . 1080 .54
Clay, dry . 1701 85
Clay, wet 2970 1.48
Earth, dry, loose .... . 1 890 .94
Earth, dry, shaken . 2214 1 . 10
Earth and sand, dry, loose. 2700 1 35
Earth and sand, dry, rammed . 3240 1 .62
Fire brick . . . 3915 1 95
Fire clay . . . 3510 1 75
Gravel, dry . . 2970 148
Granite 4536 2.26
Lime, quick, shaken . . . 1485 .70
Limestone, loose .. 2592 1.29
Marble, loose 2592 1 . 29
Mud, river 2430 1 .21
Pitch 1863 .93
Rip-rap, limestone 2160 1 .08
Rip-rap, sandstone 2430 1 . 21
Rip-rap, slate 2835 1.41
Sand, dry, loose 2619 1 30
Sand, wet 3186 1 . 5t>
Slag, screenings 2700 1 . 35
Street sweepings 850 . 42
Tar 1674 .83
APPENDIX III-TABLES
Table 1
TABLE OF LOGARITHMS
H.
0
1
2
3
4
5
6
7
8
0
D
100
000000
000434
000808
001301
001734
002166
002598
003029
003161
003891
432
1
4321
4751
5181
5609
o038
tvn.t)
6894
7321
7748
8174
428
2
8600
9026
9451
9876
010300
010724
011147
011570
011993
01241ft
424
3
012S37
013259
013680
014100
4521
4940
5360
5779
6197
6616
420
4
7038
7451
7868
8284
8700
9116
9532
9917
020361
020775
416
105
021189
021003
022016
022428
022841
023252
023664
024075
4480
4896
412
6
5306
5715
6125
6533
6942
7350
7757
8 KM
8571
8978
408
7
9384
97S9
03O1 95
030000
031004
031408
031812
032216
032619
033021
404
8
033424
033826
4227
4628
5029
5130
583O
6230
6629
7028
400
9
7426
7825
8223
8020
9017
9414
9811
040?07
040602
040998
397
110
041393
041787
042182
042576
042969
043362
043755
044148
044540
044932
893
1
r>323
5714
6105
6495
6885
7275
7664
8053
8-142
8830
890
2
9218
9006
9993
0503HO
050766
051153
051538
051921
052309
052694
386
3
O53078
053463
053816
4230
4613
4996
5378
5760
6142
6524
883
4
6905
7286
7666
8016
8426
8805
U185
9563
9942
06032O
379
115
060698
061075
061452
061829
062200
062582
062958
063333
063709
4083
376
6
4458
4832
5206
5580
5953
6326
OCW
7071
7443
7815
373
7
818€
85a7
8928
9298
96(58
070038
070-1O7
070776
071145
071514
370
8
071882
072250
072617
072985
073352
3718
4085
4451
4816
5182
366
9
5547
5912
6276
66-10
7001
7368
7731
8094
8457
8819
363
120
079181
079543
079904
080206
080626
050087
081347
081707
082067
082426
360
1
082785
083144
083503
3S(,l
4219
I57«
4934
52()1
5047
6004
357
2
6360
6716
7071
7126
77K1
8136
81'H)
8.S15
9198
9552
355
3
91)05
090258
05)0611
090903
091315
091067
092018
OU237O
092721
093071
352
4
093422
3772
4122
4471
4820
5169
5518
5806
6215
6562
349
125
6910
7257
7604
7951
8298
8C44
8990
9335
9681
100026
846
6
100371
100715
1010.39
101103
101717
102091
102m
102777
103119
3462
343
7
3W)4
4146
4187
4828
5169
5510
5851
6191
6531
6871
341
8
7210
7519
7888
8227
8565
S908
9211
9579
9910
110253
338
9
110590
110926
111263
111599
111934
112270
112605
112940
113275
3009
335
130
113943
114277
114611
1149H
115278
115611
115943
116276
116608
116940
333
1
7271
7603
793 1
X2fx>
8595
8<)26
9256
9580
9915
12O245
330
2
120574
120903
1212', 1
1215M
121 888
122216
122511
122S71
123198
3525
828
3
3852
4178
4504
4830
5156
5KS1
5806
6131
6156
6781
325
4
7105
7429
7753
8076
839«
8722
9045
9368
9690
130012
323
135
130334
130655
130977
131298
131619
131039
132260
132580
132900
3219
321
6
3539
3858
4177
411)6
4814
5133
5151
57 Ml
6086
6-103
318
7
6721
7037
7354
7671
7987
8303
W>18
MM 1
9219
9564
816
! &
9879
140194
140508
140822
141136
141150
1 1 1 763
142076
142389
142702
314
9
143U15
3327
3639
3951
4263
4574
4Sb5
6196
5607
5818
811
140
146128
146438
146718
117058
147367
147676
147985
148294
148603
148911
309
1
9219
9527
1)835
150142
150149
150756
151(Xi3
151370
151676
151982
807
2
152288
152594
152900
3205
3510
.'',M5
4120
4124
4728
5032
30r»
3
5336
5640
5943
6246
6519
<i852-
7154
7457
7759
8001
803
4
8362
86C4
8965
9266
9567
9868
160168
100409
160769
161068
301
145
161368
161667
161967
162266
162564
162863
3161
3160
3768
4055
299
6
4353
4650
4947
5214
5511
5S38
6134
6130
6726
7022
297
7
7317
7613
7908
8203
8497
8792
90H6
1)380
9674
9968
295
8
170262
170555
170848
171141
171434
171726
172019
172311
172603
172895
2!)3
9
3186
3478
3769
4060
4351
4641
4932
5222
5512
5802
291
150
176091
176381
176670
176959
177248
177536
177825
178113
178401
178689
289
1
8977
9264
9552
9S39
180126
180413
180099
180986
181272
181558
287
2
181844
182129
182415
182700
2985
3270
3555
3839
4123
4407
285
3
4691
4975
5259
5542
5825
6108
6391
6674
6956
7239
283
4
7521
7803
8084
8366
8647
8928
9209
94JX)
9771
190051
281
155
190332
190612
190892
191 in
191451
191730
192010
192289
192567
2846
279
6
3125
3403
3681
3959
4237
4514
4792
5069
r>346
5623
278
7
5900
6176
6453
6729
7005
7281
7556
7832
8107
8382
276
8
8657
8932
9206
9481
9756
200029
200303
200577
200850
201124
274
9
201397
201670
201943
202216
202488
2761
3033
3305
3577
3848
272
N.
0
1
2
3
4
5
6
7
8
9
D
497
498
APPENDIXES
N.
0
1
2
3
4
5
6
7
8
9
D.
160
204120
204391
204663
204934
205204
205475
205746
206016
20628fl
206556
271
1
6826
7006
7365
7634
7904
8173
8441
8710
8979
9247
269
2
9515
0783
210051
210319
210586
210853
211121
211388
211654
211921
267
3
212188
212451
2720
21*86
3252
3518
3783
401!)
4314
4579
266
4
4844
5109
5373
5688
6902
6166
6430
6694
6957
7221
264
165
7484
7747
8010
8273
8536
8798
9060
9323
9585
9846
262
6
220108
220370
220631
220892
221153
221414
221675
221936
222196
222456
261
7
27 Ifi
2976
3236
31!tf5
3755
4015
4274
4533
4792
5051
259
8
5309
5568
5826
6084
6342
6600
6858
71 1'>
7372
7630
2f>8
9
7887
8141-
8400
8657
8913
9170
9426
9682
9938
230193
256
170
2304 19
23070*
230960
231215
231470
231724
231979
232234
232488
232712
205
1
29<>6
3250
3504
3757
4011
4264
4517
4770
5023
5276
2V!
2
6528
5781
6033
6285
6537
6789
7011
7292
7544
7795
2 52
3
8046
8297
8548
87<)9
9049
9299
95 ~jO
9800
240050
240300
2>O
4
240549
240799
241018
241297
211546
241795
242044
242293
2541
2790
249
175
3038
8286
3531
3782
4030
4277
4525
4772
5019
5?f>6
248
6
6513
5759
6006
6252
6499
6745
6991
7237
7482
7728
21<*
7
75)73
821?)
HUM
870')
8951
9198
9413
9687
9932
250176
li J.">
8
250120
250601
250908
2511".!
25 131*5
25HW8
251881
252125
25236S
2610
2.3
9
2853
3090
3338
3580
3822
4064
4306
4548
4790
6031
242 I
180
255273
255514
255755
255996
256237
256477
256718
256958
257198
257439
211
1
7071)
7918
8158
8398
8637
8877
9116
93,";5
9594
9833
239
2
200071
260310
260548
260787
261025
261263
261501
261730
•VH976
262214
238
3
2451
2<J88
2925
3162
3399
3636
3873
4109
43 1 6
4582
237
4
4818
5054
5290
6525
5761
5996
6232
6467
6702
6937
235
185
7172
7400
7611
7875
8110
8314
8578
8812
9046
9279
234
6
9513
9746
99HO
270213
270446
270679
27091-2
271114
271377
271609
233
7
271812
272074
272306
2538
2770
3001
323,t
:-54(U
3690
3927
232
8
4r,8
4, '48')
4620
48r><>
5081
5311
55 1 2
5772
MX)2
(>2:-52
230
9
0162
6692
6<>21
7151
7380
7609
7838
8067
829G
8525
229
190
278754
278982
279211
279 1«9
279667
279895
280123
280351
280578
280806
228
1
281033
281201
281488
281715
281942
282169
2396
2622
284!)
307.1
227
2
3301
3527
3753
:5979
4205
4431
4656
4882
5107
5332
2i6
3
f»p>57
5782
OO07
6232
6456
6681
6905
7130
7354
7578
225
4
7802
8026
8249
8473
8696
8920
9143
9366
95S9
9812
223
195
200035
290257
290480
290702
290925
291117
291360
291591
291813
292034
290
6
2256
2178
20<><J
2920
3141
3.U>3
3584
3804
402".
4216
2211
7
4106
4687
4907
51 27
6347
5567
5787
6007
622(5
(>iio
220
8
<iW)5
6881
7104
7323
7542
7761
7979
81')S
8116
so;-;,')
219
9
8853
9071
8289
9507
9725
9943
300161
300378
300595
300813
218
200
301030
301217
301 164
301681
801898
802114
802331
302547
302761
302980
217
1
3196
3412
3(>'28
3844
40 >9
4275
4491
4706
4921
51. {6
2K.
2
5351
6666
5781
591 >6
6211
6425
6639
6854
7068
7282
215
3
74SH>
7710
79'24
8137
8351
856 1
8778
8991
9'20i
9417
213
4
9630
9843
310056
310268
310181
310693
310906
311118
311330
311542
212
201
811754
311966
2177
2389
2600
2812
3023
8214
3445
3656
211
6
38<>7
4078
42S9
44<H*
4710
4920
5130
5340
6551
57<iO
210
7
5970
6180
6390
6599
680'.)
7018
7227
7430
7646
78:>4
209
8
80(13
8272
8481
8(589
8898
9106
9314
9522
9730
9938
208
9
320146
320354
320562
320769
320977
321184
321391
321598
321805
322012
207
210
322210
322426
322633
322839
823046
323252
323458
323665
323871
324077
206
1
4282
4488
469 1
4890
5105
5310
5516
5721
5026
6131
205
2
<J336
6541
6745
6950
71 .55
7359
7563
7767
7972
8176
201
3
8380
8583
8787
895)1
9191
9398
9601
9805
330008
330211
203
4
330414
330617
330819
831022
331225
331427
331630
331832
2034
2236
202
215
2438
2640
2842
3044
3246
3447
364 Q
3850
4051
4253
202
6
4454
4655
4856
5057
5257
5458
5658
5859
6059
6260
201
7
6460
6660
6860
7060
7260
7459
7659
7858
8058
8257
200
8
8466
8656
8855
9054
9253
9451
9650
9849
340047
340246
199
9
340444
340612
340841
341039
341237
341435
341632
341830
2028
2225
198
N.
0
1
2
3
4
5
6
7
8
9
J>.
TABLE OF LOGARITHMS
499
N.
0
1
2
3
4
5
6
7
8
9
D.
220
342423
342620
342817
843014
34S212
343409
848606
843802
848999
844196
197
1
4392
4589
47S5
4981
5178
5374
5570
5766
6962
6157
196
2
6353
0549
6744
6939
7135
7330
7525
7720
7915
8110
195
3
8305
8500
86^1
8889
9083
9278
9472
9(566
98(50
35005 4
194
4
350248
350442
350636
850829
351023
351216
351410
351603
851796
1089
193
225
2183
2375
2568
2701
2954
3147
8339
3532
3724
8916
103
6
4108
4301
4493
4685
4876
5068
6260
5452
5643
583 4
192
7
6020
6217
6108
(5599
6790
6981
7172
7863
7554
7714
191
8
7935
8125
Ml 6
8506
8696
88S6
9076
0?(>6
9456
9616
190
9
9835
360025
360215
360404
360593
360783
360972
361161
861350
361539
189
230
361728
361917
362105
362294
362482
362671
302859
363048
363236
363424
18ft
1
3612
3800
398.x
4176
4363
45.")!
4739
49'26
5113
5301
18S
2
fit 88
5675
5802
60 49
6*236
6T23
66 1 0
671M5
61)83
71()')
187
3
7356
7542
7729
7915
8101
82S7
8473
865')
88 1 5
!¥):«>
186
4
9216
9401
9587
9772
9958
370143
870328
870513
370698
370883
185
235
371068
371253
371437
371622
371806
1991
2175
2360
2544
2728
184
6
2912
30%
32SO
3161
3617
3831
4015
4 IDS
43.S2
45(>5
184
7
4748
4932
5115
52' >8
5181
5664
5816
6029
6212
(53! )t
183
8
6577
6759
6912
71'2I
7806
7488
7670
7S.V2
NM t
821 (5
182
9
8398
8580
8761
Ml 3
91 '2 4
9806
9487
9668
9819
380030
181
240
380211
380392
380578
380754
880934
381115
381296
381476
381666
881837
181
1
2017
2197
2377
2557
2737
2917
3097
3277
3156
3036
180
2
8815
3995
4174
4353
4533
4712
4891
5070
6219
5428
179
3
5606
5785
51)64
61 4 '2
6321
6l<)<>
6677
(5856
70? 4
7212
178
4
7390
7563
7746
7923
8101
8279
8456
8631
8811
8989
178
245
0166
9343
9520
9608
9875
390051
390228
390405
390582
390759
177
6
39093.")
391112
89128S
3914t)l
391611
1817
11)93
216!)
mi
2521
17(5
7
2697
2873
3048
3'22 1
3100
3575
37.">1
31126
•4101
4277
176
8
4152
4(527
4802
4!>77
5 IT) 2
532(5
6501
5676
r>8.io
6023
175
9
6199
6374
6548
6722
6896
7071
7245
7119
7592
7766
171
250
397940
398114
398287
398461
398634
39880ft
308981
399154
399328
399501
173
1
9674
98 1 7
4(KH)20
400192
400365
400538
400711
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2
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8
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156
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6382
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6602
6848
7003
155
N.
0
1
2
3
4
5
6
7
8
8
1>.
500
APPENDIXES
K.
0
1
2
3
4
5
6
7
8
9
D.
280
447158
447313
447468
447623
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448552
155
1
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2
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451479
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3
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2247
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3165
153
4
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153
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6
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152
7
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151
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9
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290
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148
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300
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478278
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145
1
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144
2
480007
480151
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11.1
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137
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3246
3382
3318
3<>35
130
9
3791
3927
4063
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4335
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4743
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6014
136
320
605150
605280
505421
505557
505693
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500099
506234
506370
136
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6911
7016
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7310
7431
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2
7850
7991
8126
8260
8393
8530
8G(>1
879U
8934
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3
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9337
9471
9600
9740
9874
510009
510143
510277
510411
131
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610679
510813
510947
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1616
1750
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325
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2151
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2418
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208 1
2818
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308 {
133
8
3218
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3484
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4415
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4548
4681
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5079
52 LI
5344
5476
5609
5741
133
8
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6068
6800
6932
7004
132
9
7196
7328
7460
7592
7724
7855
7987
8119
8251
8382
132
330
618514
518646
518777
518909
519010
519171
519303
619434
519506
519697
131
1
9828
9959
520090
620221
520353
520484
520615
620745
520876
521007
131
2
621138
521269
1400
1530
1661
1792
1922
2053
2183
2314
131
3
2444
2575
2705
2835
2966
3096
3226
3350
3 4 SO
3010
130
4
3740
3876
4006
4136
4266
4396
4526
4656
4785
4915
130
335
5045
5174
6304
5434
5563
5693
5822
5951
6081
6210
129
6
6339
6469
6598
6727
6806
6985
7114
7243
7372
7501
129
7
7630
7759
7888
8016
8145
8274
8402
8531
8660
8788
129
8
8917
9045
9174
9302
9 430
9359
9687
9815
9943
530072
128
9
630200
630328
630456
630584
630712
530840
530968
631096
531223
1351
128
K.
0
1
2
3
4
5
6
7
8
9
D.
TABLE OF LOGARITHMS
501
N.
0
1
2
3
4
5
6
7
8
9
D.
340
531479
531607
531734
531862
531990
532117
532245
632372
532500
632627
128
1
2754
2882
3009
3136
3264
3391
3518
3645
3772
3899
127
2
4026
4153
4280
4407
4534
4661
4787
4914
5041
5167
127
3
5294
5421
5547
5671
6800
5927
6053
61 SO
6306
6432
120
4
6568
66S5
6811
6937
7063
7189
7315
7441
7567
7693
126
345
7819
7945
8071
8197
8322
844S
R574
8699
8825
8951
120
6
9076
9202
9327
9452
9578
9703
9829
99.34
540079
540204
25
7
540329
540455
540580
540705
540830
54095.3
541080
641205
1330
1454
25
8
1579
1704
1829
1953
2078
2203
2327
2 152
2570
2701
23
9
2825
2950
3074
3199
3323
3417
3571
3690
3820
3944
•24
350
544008
541192
544316
544440
544564
544688
544812
544936
545000
545183
24
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5431
5555
5678
5802
5925
601')
6172
6296
6419
24
2
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6913
7036
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8635
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9126
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9371
9494
9616
9739
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550106
23
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550228
550351
550473
550595
550717
550840
550962
551084
651206
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22
6
1450
1572
1694
1816
1938
2060
2181
2303
2125
2547
22
7
2668
2790
2911
3033
3155
3276
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3.31!)
3610
3702
21
8
3883
4004
4 1 L'O
4217
4368
4489
4610
4731
48,32
4973
21
9
5094
5215
5330
5457
5578
5699
6820
5940
6061
6182
21
360
556303
556423
556544
556664
556785
556905
557026
557146
557267
557387
120
1
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7627
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8108
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8,3X9
120
2
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9787
120
3
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500.304
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560713
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501101
1221
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1698
1817
1936
2055
2174
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365
2293
2112
2531
2650
2769
2887
3006
3125
3244
3362
110
6
3181
3600
3718
3837
395,3
4074
4192
4311
4429
4518
119
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5021
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9
7026
71 14
7262
7379
7197
7614
7732
7849
7907
8084
118
370
568202
568319
568436
568554
568671
568788
568905
569023
569140
609257
117
1
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9191
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9812
99.39
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570193
570309
670420
117
2
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1912
2038
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2'291
2107
2523
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275,3
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4
2872
2988
3104
3220
3336
3452
3568
3684
3800
3915
110
375
4031
4147
4263
4379
4494
4610
4726
4841
4957
5072
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6
5188
5303
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50.30
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11.3
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7032
7147
7202
7377
11.3
8
7 192
7007
7722
7836
7931
8066
8181
8295
8110
8525
113
9
8639
8754
8808
8983
9097
9212
9320
9441
95.35
900U
114
380
579784
579898
580012
580126
580241
580355
580469
580583
680697
580811
114
1
580925
581039
1153
1267
1381
1495
1608
1722
1H30
1950
111
2
20fi3
2177
2291
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2518
2631
271.3
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30H5
111
3
3199
33 J 2
3120
3539
3052
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3992
4 105
4218
113
4
4331
4414
4557
4670
4783
4896
5009
5122
5235
5348
113
385
5461
5574
5686
5799
5912
6024
6137
6250
6362
6475
113
6
6587
6700
6812
6925
7037
7149
7262
7374
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112
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7823
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8496
8008
8720
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8832
8944
9056
9167
9279
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9615
9720
9838
112
9
9950
590061
590173
690284
590396
590507
590619
690730
590842
590953
112
390
591063
591176
591287
591399
591510
591621
591732
591843
591955
592066
111
1
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22H8
2399
2510
262J
2732
'2843
2954
3004
3175
111
2
3280
3397
3508
3618
3729
3840
3930
4061
4171
4282
111
3
4393
4503
4614
4724
4834
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50.3 r>
5165
5270
6380
110
4
5496
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5717
6827
6937
6047
6157
6267
6377
6487
110
395
6597
6707
6817
6927
7037
7146
7256
7366
7476
7586
110
6
769.3
7805
7914
8024
8134
8213
8",.33
8402
8572
8081
110
7
8791
8900
9009
9119
9228
9337
9116
9.356
9065
9774
109
8
9883
9992
600101
600210
600319
600428
600537
600646
600755
600864
109
9
600973
601082
1191
1299
1408
1517
1625
1734
1843
1951
109
N.
0
1
2
3
4
5
6
7
8
9
D.
502
APPENDIXES
N.
0
1
2
3
4
5
6
7
8
9
D.
400
602060
602169
602277
602386
602494
602603
602711
602819
602928
603036
108
1
3144
3253
3361
3469
3577
3686
3794
3902
4010
4118
las
2
4226
4334
4412
4550
4658
4766
4874
49S2
5080
5197
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3
5305
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5521
5628
5736
5844
5051
6050
6166
6274
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4
6381
6489
6596
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6811
6910
7026
7133
7241
7348
107
405
7455
7562
7609
7777
7884
7991
8098
8205
8312
8410
107
6
8526
8633
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8847
8954
0061
9167
0274
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94feS
107
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9701
9808
0914
610021
610128
610234
610341
610447
610.334
107
8
610600
610767
610873
610979
108(5
1102
1298
140")
1511
1017
100
9
1723
1820
1030
2012
2148
2254
2300
2466
2572
2678
106
410
612784
612890
612996
613102
613207
613313
613110
613525
613C30
613736
106
1
3842
3917
4053
4 159
4204
4370
4475
45M
4(.80
4792
101,
2
4897
5003
5108
5213
5319
5421
5529
5(>'U
5740
5845
101
3
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(51 (X)
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6370
6470
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0790
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101
4
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104
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643847
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1
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4537
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4832
4031
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5220
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5422
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5717
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5913
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6110
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3
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8067
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98
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1056
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2246
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2536
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97
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653213
653309
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653502
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653695
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96
1
4177
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4754
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2
6138
5235
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5427
5523
5610
5715
5810
5906
6002
90
3
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6194
6200
6386
6482
6577
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6709
6864
6960
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4
7056
7152
7247
7343
7438
7534
7629
7725
7820
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96
455
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8107
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8298
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8488
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8870
95
6
8965
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9155
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660391
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660581
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95
8
660865
0960
1055
1150
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1339
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1813
1907
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2096
2191
2286
2380
2475
2569
2663
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1
2
3
4
5
6
7
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TABLE OF LOGARITHMS
503
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1
2
3
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5
6
7
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460
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662852
662947
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9410
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9590
9089
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99(17
670000
670153
93
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670246
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1080
93
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1173
1205
1358
1431
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93
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672467
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92
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86
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86
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5864
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9
6718
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85
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3
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4
0963
1018
1132
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1301
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1554
1639
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84
515
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3154
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84
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5586
6669
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84
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0
1
2
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5
6
7
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504
APPENDIXES
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2
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5417
5494
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76
570
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76
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6630
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6910
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75
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73
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1928
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2078
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22*28
2303
2378
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75
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2978
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3353
75
N.
0
1
2
3
4
5
6
7
8
9
D.
TABLE OF LOGARITHMS
505
N.
0
1
2
3
4
5
6
7
8
9
D.
580
763428
763503
763578
763653
763727
763802
763S77
763952
764027
764101
75
1
4176
4251
4326
4400
4475
4550
4621
4699
4774
4848
75
2
4023
4908
5072
5147
5-221
5-2' >6
5370
5445
5520
5594
75
3
5609
5743
5S18
5892
5966
6041
6115
6190
6264
6388
74
4
6413
6487
6562
6636
6710
6785
6859
6933
7007
7082
74
585
7156
7230
7304
7379
7453
7527
7601
7675
7749
7823
74
6
7808
7972
8046
8120
8194
8268
8312
8416
8400
8564
74
7
8l)38
8712
8786
8860
8934
9008
9082
9156
9230
0303
74
8
9377
9451
9525
9599
9073
9746
9820
9891
0968
770042
74
9
770115
770189
770263
770336
770410
770484
770557
770631
770705
0778
74
590
770852
770926
770999
771073
771146
771220
771293
771367
771440
771514
74
1
K>S7
1661
1731
1808
1881
1955
202S
2102
2175
2248
73
2
9300
2395
2168
2512
2615
I2te8
276'2
2835
2908
2981
73
3
3055
31*28
3201
3274
3348
3121
3404
3.r)'J7
3640
3713
73
4
3786
3860
3933
4006
4079
4152
4225
4208
4371
4444
73
595
4517
4590
4663
4736
4809
4882
4055
5028
5100
5173
73
6
5210
5319
5.192
5465
5538
5010
5683
5756
6820
6902
73
7
5974
6017
6120
6193
0265
6338
6111
6183
6556
6620
73
8
6701
6771
68 1 6
6919
6902
7064
7137
7209
7282
7354
73
9
7127
7199
7572
7614
7717
7769
7862
7931
8006
8079
72
600
778151
778224
778296
778368
778141
778513
778585
778658
778730
778802
72
1
8874
8947
9019
OO'Jl
0163
0236
9308
9380
0452
0524
7''
2
9596
9669
9741
9813
0885
0957
78002!)
780101
780173
780245
72
3
780317
78MS9
780 J61
780533
78CMnOr>
780677
0710
0821
0893
0965
72
4
1037
1109
1181
1253
1324
1396
1468
1510
1612
1684
72
605
1755
1827
1899
1971
2042
2111
21R6
2258
2329
2401
72
6
2173
2544
2616
2688
2759
2831
2902
2974
3046
8117
72
7
3189
3-260
3332
3103
3475
3546
3618
3689
3761
8832
71
8
3904
3975
4016
1118
41S9
4'261
1332
4 103
4475
4546
71
9
4617
4689
4760
4831
4902
4974
5045
6116
5187
5259
71
610
785330
785401
785172
785543
785615
785686
785757
785828
785899
785970
71
1
6041
6112
6183
6'25 1
6325
6396
6167
6538
6600
6680
71
2
6751
6822
6893
6964
7035
7106
7177
7248
7310
7390
71
3
7400
7531
760-2
7673
7714
7815
7885
7956
8027
8098
71
4
8168
8239
8310
8381
8-151
8522
8593
8663
8734
8804
71
615
8875
8946
9016
9087
0157
9228
9299
0360
0440
9510
71
6
9581
%51
9722
9792
0863
9033
790<H)4
700074
700114
790215
70
7
790285
790356
7904'26
7904 96
700567
790637
0707
07 78
08-18
0918
70
8
09SK
1059
1129
1199
1269
1310
1410
1-180
1550
1620
70
9
1691
1761
1831
1901
1971
2041
2111
2181
2252
2322
70
620
792392
792402
792532
702602
792672
792742
792812
792882
702952
793022
70
1
3092
31 62
3*231
3301
3371
3441
3511
3581
3651
S721
70
2
3790
38(50
3930
4000
4070
4139
4209
4279
4349
4418
70
3
4488
4558
4627
4697
4767
4836
4906
4976
5045
5115
70
4
5185
5254
5324
5393
6463
6532
5602
6672
6741
6811
70
625
5880
5949
6019
6088
6158
6227
6297
6306
6436
6505
69
6
6574
6644
6713
6782
6852
6921
6990
7060
7129
71 08
60
7
72f>8
7337
7406
7475
7545
7614
7683
7752
7821
7890
fl»
8
7900
8029
8098
8167
8236
8305
8371
8443
8513
8,182
69
9
fe651
8720
8789
8858
8927
8096
9065
9134
0203
9272
69
630
790341
709409
799478
709517
799616
799685
799754
700823
790892
799961
60
1
800029
80009S
800167
800236
800305
800373
800442
800511
800580
800648
69
2
0717
0786
0854
0923
0992
1061
1129
1198
1266
1335
69
3
140*
1172
1541
1609
1678
1747
1815
1884
1952
2021
60
4
2089
2158
2226
2295
2363
2432
2500
2668
2637
2705
68
635
2774
2842
2910
2079
3047
3116
3184
3252
3321
3389
68
6
3457
3525
3594
3662
3730
3798
3867
3035
4003
4071
68
7
4139
4208
4276
4344
4412
4480
4548
4616
4685
4753
68
8
4821
4889
4957
5025
6093
5161
5229
5297
5365
6433
68
9
6501
5569
5637
5705
6773
6841
6908
6976
6044
0112
68
N.
0
1
2
3
4
5
6
7
8
e
D.
506
APPENDIXES
N.
0
1
2
5
4
5
6
7
8
9
D.
640
806180
806248
806316
806384
806451
806519
806587
806655
806723
806790
68
1
6858
6926
6994
7061
7129
7197
7264
7332
7400
7467
68
2
7535
7603
7670
7738
7806
7873
7941
8008
8076
8143
68
3
8211
8279
8346
8414
8481
8549
8616
8684
8751
8818
67
4
8886
8953
9021
9088
9156
9223
9290
9358
9425
9492
67
645
9560
9627
9604
9762
9829
9896
9964
810031
810098
810165
67
6
810233
810300
810367
810434
810501
810569
810636
0703
0770
0837
67
7
01)04
0071
1039
1106
1173
1240
1307
1374
1441
1508
67
8
1575
1612
1709
1776
1843
1910
1977
2044
2111
2178
67
9
2243
2312
2379
2445
2512
2579
2646
2713
2780
2847
67
650
812013
812080
81 -SO 17
813114
813181
813247
813314
813381
813448
813514
67
1
3581
3618
3714
3781
3818
3914
3981
4018
4114
4181
67
2
4218
4314
4381
4447
4514
45S1
4617
1714
4780
4847
67
3
4913
4980
504(5
5113
5179
5216
5312
5378
5445
5511
60
4
5578
5644
5711
5777
5843
5910
5976
6042
6109
6175
6(5
655
6241
6308
6374
6440
6506
6573
6639
6705
6771
6838
66
6
6!XH
6970
7036
7102
71f»r
7235
7301
7367
7133
7499
66
7
7565
7631
7698
7764
7830
78%
7062
8028
8034
8160
66
8
8220
8202
8358
8124
8490
8Vifi
8622
8688
8754
8820
66
9
8885
8951
9017
9083
9149
9215
9281
9346
9412
9478
66
660
819544
810610
819676
819741
819807
810873
819939
820004
820070
820136
66
1
820P01
820267
8203I-53
820399
820 1 04
820530
820505
0661
0727
0792
66
2
0858
0924
0989
1055
1120
1186
1251
1317
1382
1418
6ti
3
lf> I I
1579
1015
1710
1775
1841
1006
1072
2037
2103
65
4
2168
2233
2299
2364
2430
2495
2560
2626
2691
2756
65
665
2822
2887
2952
3018
3083
3148
3213
3279
3314
3409
65
6
3474
3539
3605
3670
3735
3800
asfio
3930
39%
4061
65
7
412(5
4191
4256
4321
4386
4451
4516
4581
46 lf>
47J1
6,")
8
4776
4841
4906
4971
5036
5101
51 06
5231
5206
5361
6')
0
6426
5491
5556
5621
5686
5751
5815
5SSO
5915
6010
65
670
826075
826140
826204
826269
826334
826399
826464
826528
826593
826658
65
1
6723
6787
6852
6917
6081
7016
7111
7175
7210
7305
65
2
7309
7434
7499
7563
762S
7692
7757
7821
7886
7951
6r>
3
8015
8080
8144
8209
8273
8338
8402
8467
8r>:u
8505
61
4
8CGO
8724
8789
8853
8918
8982
9016
9111
9175
V239
64
675
930 1
9368
9132
9497
9501
9625
9690
9754
9818
9882
64
6
9947
830011
830075
830139
830201
8302(&
830332
830396
8304 (JO
830525
61
7
830589
0653
0717
0781
0845
0909
0973
1037
1102
1166
61
8
1230
129*
1358
1422
1186
1550
1611
1678
1742
1806
61
9
1870
1934
1998
2062
2126
2189
2253
2317
2381
2415
64
680
832509
832573
832637
832700
832764
832828
832802
832056
833020
833083
64
1
3147
3'>11
IV275
3338
3402
3466
3530
359I-!
3657
3721
64
2
3784
3818
3912
3975
4039
4103
4166
4230
4294
4357
61
3
4421
4484
4548
4611
4675
4730
4802
4866
40'29
4993
61
4
5056
5120
5183
5247
5310
5373
5437
5500
5564
5627
63
685
5(591
5751
5817
5881
5941
6007
6071
6134
6197
6261
63
6
6324
6387
645 I
6514
6577
6641
6704
6767
6830
6P»4
63
7
6957
7020
7083
7146
7210
7273
7336
7399
7162
7525
63
8
758K
7652
7715
7778
7811
7904
7967
8030
8093
8156
63
9
8219
8282
8345
8408
8471
8534
8597
8660
8723
8786
63
690
838849
838912
838975
839038
839101
839164
839227
839289
839352
839115
63
1
0178
9541
9604
96(57
9729
9792
9855
9918
9981
840043
63
2
840106
840169
840232
840294
840357
840420
840482
840545
840608
0671
63
3
0733
0796
0859
0921
0984
1046
1109
1172
1234
1297
63
4
1359
1422
1485
1547
1610
1672
1735
1797
1860
1922
63
695
1985
2047
2110
2172
2235
2297
2360
2422
2484
2547
62
6
2609
2672
2734
2796
2859
2921
2983
3046
3108
3170
62
7
3233
8295
8357
3420
3482
3544
3606
3669
3731
3793
62
8
8855
3918
3980
4042
4104
4166
4229
4291
4353
4415
62
9
4477
4539
4601
4664
4726
4788
4850
4912
4974
5036
62
N.
0
1
2
3
4
5
6
7
8
9
D.
TABLE OF LOGARITHMS
507
N.
0
1
2
3
4
5
6
7
8
9
D.
700
846098
845160
845222
8452S4
845346
845408
845470
846532
845594
845656
62
1
5718
5780
5842
5904
5966
6028
6090
6151
6!213
6275
62
2
6337
6399
6401
6523
6585
6646
6708
6770
6832
6894
62
3
6955
7017
7079
7141
7202
7264
7326
7388
7449
7511
62
4
7573
7634
7696
7758
7819
7881
7943
8004
8066
8128
62
705
8189
8251
8312
8374
8435
8497
8659
8620
8682
8743
62
6
8805
880)6
8928
8989
9051
9112
9174
9235
9297
9358
61
7
9419
9481
9542
9604
9665
9720
9788
9819
9911
9972
61
8
850033
850095
850156
850217
850279
850340
850401
850462
850524
850585
61
9
0646
0707
0709
0830
0891
0952
1014
1075
1136
1197
61
710
851258
851320
851381
851442
851503
851564
851625
851686
851747
851809
61
1
1870
1931
1992
2053
2114
2175
2236
2297
2358
2419
61
2
2480
2541
2602
26(53
2724
2785
2840
2907
2LHJS
3029
61
3
3090
3150
3211
3272
3333
3394
3455
3516
3577
S037
61
4
8698
3759
3820
3881
3941
4002
4063
4124
4185
4245
61
715
4306
4367
4428
4488
4549
4610
4670
4731
4792
4852
61
6
4913
4974
5034
6095
5156
5216
5277
6337
5398
6459
61
7
5519
6580
5640
5701
5701
5822
5882
5913
(5003
(5064
61
8
6124
6185
6245
6306
6306
6427
6487
6548
6008
6008
60
9
6729
6789
6850
6910
6970
7031
7091
7152
7212
7272
60
720
857332
857393
857453
857513
857574
857634
857694
857755
857815
857875
60
1
7935
7995
8056
8116
8176
8236
8297
8357
8-117
8477
60
2
8537
8597
8657
8718
8778
bS38
8Sl>8
8958
9018
9078
60
3
9138
9198
9258
9318
9379
9 439
9-199
9559
9019
9679
60
4
9739
9799
9859
9918
9978
860038
860098
860158
860218
860278
GO
725
860338
860398
860458
860518
860578
0637
0697
0757
0817
0877
60
6
0937
OU96
1056
1116
1176
12.i6
1295
1355
1415
1475
60
7
1534
1594
1664
1714
1773
1833
1893
1952
2012
2072
60
8
2131
2191
2251
2310
2,170
2 130
21M>
2519
2608
20(58
60
9
2728
2787
2847
2906
2966
3025
3085
3144
3204
3263
60
730
863323
86.3382
863442
863501
863561
863620
863680
863739
863799
863858
59
1
3917
8977
40IJ6
4096
4155
4214
4274
4333
4392
4452
69
2
4511
4.070
4 030
4689
4748
4808
4867
4920
4985
5045
5»
3
5104
6163
6222
5282
5341
5100
515')
5519
5578
5037
69
4
5696
5755
5814
5874
6933
5992
6051
6110
6169
6228
69
735
6287
6346
6405
6165
6521
6583
6642
6701
6760
6819
59
6
6878
6937
6990
7055
7114
7173
7232
7291
7350
7409
69
7
71(57
752G
75K5
7014
770.'*
7702
7H21
78HO
7939
7998
59
8
8056
8115
8174
8233
8292
8350
8409
8408
8527
8580
69
9
8644
8703
8702
8821
8879
8938
8997
9056
9114
9m
59
740
869232
869290
869349
869408
869166
869525
869584
869642
869701
869760
69
1
9818
9877
9935
9994
870053
870111
870170
870228
870287
870345
69
2
870404
870102
870521
870579
OG38
0696
0755
(J813
0872
O'JUO
68
3
09b9
1017
110(5
1164
1223
1281
1339
1398
1456
1515
68
4
1573
1G31
1690
1748
1800
1865
1923
1981
2040
2098
68
745
2156
2215
2273
2331
2389
2448
2506
2564
2622
2681
68
6
27,19
2797
2855
2913
2972
3030
3088
3146
32O4
3262
68
7
3321
3379
3437
34'JG
3553
3611
3«i9
3727
3785
3844
68
8
3902
3900
4018
4076
4134
4192
4250
4308
4366
4421
58
9
4482
4510
4598
4656
4714
4772
4830
4888
4945
6003
58
750
875061
875119
875177
875235
875293
875351
875409
875466
875524
875582
68
1
5640
5G98
5756
5813
5871
5929
5987
6045
6102
6160
68
2
6218
6276
6333
6391
6419
6507
6564
6622
6680
6737
58
3
6795
6853
6910
6968
7026
7083
7141
7199
7256
7314
68
4
7371
7429
7487
7544
7602
7659
7717
7774
7832
7889
68
755
7947
8004
8062
8119
8177
8234
8292
8349
8407
8464
67
6
8522
8579
8637
8694
8752
8809
8866
8924
8981
9039
57
7
9096
9153
9211
9268
9325
9383
9440
9497
9565
9612
67
8
9669
9726
9784
9841
9898
9956
880013
880070
880127
880185
57
9
880242
880299
880356
880413
880471
880528
0585
0642
0699
0750
67
N.
0
1
2
3
4
6
6
7
8
9
D.
508
APPENDIXES
N.
0
1
2
3
4
6
6
7
8
9
D.
760
880814
880871
880928
880985
881042
881099
881156
881213
881271
881328
57
1
1385
1442
1499
1556
1613
1670
1727
1784
1841
1898
57
2
1955
2012
2069
2126
2183
2240
2297
2354
2411
2468
57
3
2525
2581
2638
2695
2752
2809
2866
2923
2980
3037
57
4
3093
3150
3207
3264
3321
3377
3434
3491
3548
3605
57
765
3661
3718
3775
3832
3888
3945
4002
4059
4115
4172
57
6
4229
4285
4342
4399
4455
4.') 12
4569
4625
4682
4739
57
7
4795
4852
4909
4965
5022
5078
5135
5192
5248
5305
57
8
5361
5418
5474
5531
5587
5644
5700
5757
5813
5870
57
9
5926
5983
6039
6096
6152
6209
6265
6321
6378
6434
56
770
886491
886547
886604
886660
886716
886773
886829
886885
886942
886998
56
1
7054
7111
7167
7223
7280
7336
7392
7449
7505
7561
5(5
2
7617
7674
7730
7786
7842
7898
7955
8011
8067
8123
5<5
3
8179
8236
8202
8348
8404
84fiO
8516
8573
8629
8685
56
4
8741
8797
8853
8909
8965
9021
9077
9134
9190
9246
56
775
9302
9358
9414
9470
9526
9582
9638
9604
9750
0806
56
6
98«£
9918
9974
890030
890086
800141
8901f>7
890253
890.10!)
890365
56
7
890421
890477
890533
0589
00-15
0700
07 50
0812
0868
0024
56
8
0980
1035
1091
1117
1203
1 •>"){>
1314
1870
1426
1482
56
9
1537
1593
164U
1705
1700
1816
1872
1928
1983
2039
56
780
892005
892150
892206
892262
892317
S9237&
892429
892184
892.540
892595
56
1
2651
2707
2702
281H
2873
2929
2! 18.")
3010
3096
3151
56
2
3207
3262
3318
3373
3129
3184
3540
3505
3651
3706
56
3
3702
3817
3873
3928
3<)S4
4039
4094
4150
42(>.>
4261
55
4
4316
4371
4427
4482
4538
4593
4648
4704
4759
4814
55
786
4870
4925
4980
503fi
5091
5146
5201
5257
5312
5367
55
6
6423
5478
5533
5588
5644
5699
5754
5809
5864
5920
65
7
5975
6030
6085
6140
6195
6251
C306
63(U
6416
6471
55
8
6526
6581
6636
66<>2
6747
6802
6857
6912
69f,7
7022
55
9
7077
7132
7187
7242
7297
7332
7407
7462
7517
7372
55
790
897627
897682
897737
897792
897847
897902
897957
80S012
898067
898122
55
1
8176
8231
8286
8341
8396
8451
850(>
8561
8015
8670
55
2
8725
8780
8835
8890
8944
891)0
9054
910!)
9164
9218
55
3
9273
9328
9383
9437
9492
9547
9602
9656
9711
9766
55
4
9821
9875
9930
9985
900039
900094
900149
900203
900258
900312
65
795
900367
900422
900476
900531
0586
0640
0695
0749
0804
0859
55
6
0913
0968
1022
1077
1131
1186
1240
1295
1349
1404
55
7
1458
1513
1567
1622
1(576
1731
1785
1840
1894
1948
54
8
2003
2057
2112
2166
2221
2275
2329
2384
2438
24fl2
64
9
2547
2601
2655
2710
2764
2818
2873
2927
2981
3036
54
800
903090
903144
903199
903253
903307
903361
903416
903470
903524
903578
54
1
3633
3687
3741
3795
38-19
3904
3958
4012
4066
4120
54
2
4174
4229
4283
4337
4391
4445
4499
4553
4007
4661
54
3
4716
4770
4824
4878
4932
4986
5040
5094
5148
5202
54
4
5250
5810
5364
5418
5472
5526
6580
5634
5688
5742
54
805
5796
5850
5904
5958
6012
6066
6119
6173
6227
6281
54
6
6335
6389
6443
6497
6551
6004
6658
6712
6766
6820
51
7
6874
6927
6981
7035
7089
7143
7196
7250
7304
7358
54
8
7411
7465
7519
7573
7626
7680
7734
7787
7841
7895
54
9
7949
8002
8056
8110
8163
8217
8270
8324
8378
8431
54
810
908485
908539
908592
908646
908699
908753
908807
908860
908914
908967
54
1
9021
9074
9128
9181
9235
9289
9342
9396
9449
9503
54
2
9556
9610
9663
9716
9770
9823
9877
9930
9984
910037
53
3
910091
910144
910197
910251
910304
910358
910411
910464
910518
0571
53
4
0624
0678
0731
0784
0838
0891
0944
0998
1051
1104
53
815
1168
1211
1264
1317
1371
1424
1477
1530
1584
1637
53
6
1690
1743
1797
1850
1903
1956
2009
2063
2116
2169
53
7
2222
2275
2328
2381
2435
2488
2541
2594
2647
2700
53
8
2753
2806
2859
2913
2966
3019
3072
3125
3178
3231
53
9
3284
8337
3390
3443
3496
S549
3602
3655
3708
3761
53
K.
0
1
2
3
4
5
6
7
8
9
D.
TABLE OF LOGARITHMS
509
N.
0
1
2
3
4
5
6
7
8
9
D.
820
913814
913867
913920
913973
914026
914079
914132
914184
314237
914290
53
1
4.m
4396
4419
4502
4555
4008
4660
4713
4766
4819
53
2
4872
4925
4977
5030
5083
51 86
5189
5211
5294
6347
63
3
5400
5453
5505
6558
5611
5664
5716
5769
5822
5875
53
4
5927
5080
6033
6085
6188
6191
6243
6296
6349
6401
53
825
6454
6507
6559
6612
6664
6717
6770
6822
6875
6927
53
6
6980
7033
7oar>
7138
7190
7243
7295
7348
7400
7453
53
7
7506
7,">8
7611
7663
7716
7768
7820
7873
7025
7978
52
8
8030
8083
8135
8188
8210
8293
8315
8397
8450
8502
52
9
8555
8607
8659
8712
8764
8816
8869
8921
8973
9026
52
830
919078
919130
919183
919235
919287
919340
919?92
919444
910496
919549
52
1
9001
9653
9706
9758
9810
9862
9914
9967
920019
920071
5'>
2
9201 23
920176
920228
9202SO
920332
920384
920136
920489
0511
0593
52
3
0645
0697
0749
OS01
0853
0906
0958
1010
1062
1114
52
i
1166
1218
1270
1322
1374
1426
1478
1530
1582
1634
52
835
1686
1738
1790
1«I2
1894
1946
1998
2050
2102
2154
52
6
2206
2258
2310
2362
2111
2466
2518
2570
2622
2674
r>2
7
2725
2777
2829
2881
2933
2985
3037
3089
3140
3192
62
8
3214
3296
3348
3399
3451
3503
3555
3607
3658
3710
52
0
3762
3811
3865
3917
3969
4021
4072
4124
4176
4228
52
840
921279
924331
924383
924434
924186
924538
924589
924641
924693
924744
52
1
4706
4818
1800
4(T>1
5003
5054
5106
5157
5201)
5261
52
2
5312
5364
5415
5467
5518
5570
5621
5673
5725
5776
52
3
5828
5879
5931
5982
6034
6085
6137
6188
6240
6291
61
4
6342
6394
6445
6497
6548
6600
6651
6702
6754
6805
61
845
6857
6908
6959
7011
7062
7114
7165
7216
7268
7319
61
6
7370
7422
7473
7524
7576
7627
7678
7730
7781
7832
51
7
7883
7035
7086
8037
808H
8140
8191
8242
8293
8345
51
8
83f>6
8417
8198
8510
8601
8652
8703
8754
8805
8857
51
9
8908
8959
9010
9061
9112
9163
9215
9266
9317
9368
51
850
920410
92') 170
929521
020572
9?9623
929071
929725
929776
929827
929879
51
1
9930
9981
930032
930083
930134
D301H5
930236
930287
930338
930389
51
2
930440
930101
0542
0.712
0613
060 1
0715
0796
0847
0898
51
3
0049
1000
1051
1102
1153
1201
1254
1305
1356
1407
51
4
i ir>8
1509
1500
1610
1661
1712
1763
1814
1865
1915
61
855
1966
2017
2068
2118
2169
2220
2271
?322
2372
2423
51
6
2474
2V21
257r>
2626
2677
2727
277S
282!)
2879
2930
51
7
2981
3031
3082
31 33
3183
323 1
3285
3335
3386
3437
61
8
3487
3538
3589
3639
3690
3740
3791
3841
3892
3913
61
9
3993
4014
4094
4145
4195
4246
4296
4347
4397
4418
51
860
934498
934549
934599
934650
934700
931751
934801
934852
934902
931953
50
1
5003
5O54
5101
5154
5205
5255
5306
5350
5406
5 157
50
2
5507
5558
5608
5658
5709
5759
5809
5860
5910
5960
50
3
roil
6061
6111
6162
6212
6262
6313
6363
6413
6163
50
4
6514
6564
6614
6665
6715
G705
6815
6865
6916
6966
60
865
7016
7066
7117
7167
7217
7267
7317
7367
7418
7468
50
6
7518
7568
7618
7668
7718
7760
781 9
7860
791!)
79(11)
50
7
8019
8069
8110
81 M
821 <>
8269
8320
8370
8120
8470
60
8
8520
8570
8620
8670
8720
8770
8820
8H70
8920
8970
50
9
9020
9070
9120
9170
9220
9270
9320
9369
9419
9169
50
870
OWU9
939560
939619
939669
930719
939769
930819
939869
939918
1)39968
50
1
940018
940068
940118
940168
910218
940267
940317
940367
940417
940167
50
2
0516
0566
0616
0666
0716
0765
0815
0865
0015
0964
50
3
10H
1064
1114
1163
1213
1263
1313
1362
1112
1462
60
4
1511
!.;<>]
1611
1660
1710
1760
1809
1859
1909
1958
50
875
2008
2058
2107
2157
2207
2256
2306
2355
2405
2455
50
6
2504
2554
2603
2653
2702
2752
2801
2851
2901
2950
50
7
3000
3049
3099
3148
3198
3247
3297
3346
3306
3445
49
8
3495
3544
3593
3643
3692
3742
3791
3841
3890
3939
49
9
3989
4038
4088
4137
4186
4236
4285
4335
4384
4433
49
N.
0
1
2
3
4
5
6
7
8
9
D.
510
APPENDIXES
N.
0
1
2
3
4
5
6
7
8
9
D.
880
944483
944532
944581
944631
944680
944729
944779
944828
944877
944927
49
1
4976
5025
5074
5124
5173
5222
5272
5321
5370
5419
49
2
5169
5518
5567
5616
56(>5
5713
5701
5813
5862
5912
49
3
5961
6010
6030
6108
61.37
6207
6256
6305
6354
6403
49
4
6452
6501
6551
6600
6649
6t)')8
6747
6796
6845
685)4
49
885
6913
6992
7011
7000
7140
7180
7238
7287
7336
7385
49
6
7434
7483
7532
7581
76.W
7b7()
7728
7777
7826
7875
49
7
7924
7973
8022
8070
8119
8168
8217
8266
8315
8364
40
8
8413
8462
8511
8500
8609
8057
8706
8755
8804
8853
49
9
8902
8951
8999
9048
9097
9146
9195
92 H
9292
9341
49
890
949390
949139
949488
910536
049585
940634
940683
940731
949780
949829
49
1
9878
9926
5)975
950024
950073
950121
950170
950219
950267
950316
49
2
9503C55
95041 4
950162
0511
0560
0608
0657
0706
0754
0803
49
3
0851
0900
0919
0907
1046
1005
1143
1192
1240
1289
49
4
1338
1386
1435
1183
1532
1580
1629
1677
1726
1775
49
895
1823
1872
1920
1960
2017
2066
2114
2163
2211
2260
48
6
2308
2356
2405
2453
2.302
25,30
2399
2647
2696
2714
48
7
2792
2811
2880
293*
29*6
303 1
30*3
3131
3180
3228
48
8
3276
3325
3373
3121
3(70
3318
3506
3615
3663
3711
48
9
3760
3808
3856
3905
3953
4001
4049
4098
4146
4194
48
900
954243
954291
954339
954387
954135
954484
954532
954580
054628
951677
48
1
4725
4773
4821
486!)
4918
49(>0
.3014
5002
5110
5158
48
2
5207
5255
5303
53.)1
539<)
5147
5405
5543
530*2
5()10
4*
3
5(588
673(5
57H I
5832
5880
59'28
59 7(
002 1
6072
6120
48
4
6168
6216
6265
6313
6361
640!)
6457
6305
6553
6001
48
905
6640
6697
6745
6703
6810
6888
6936
698 1
7032
70*0
48
6
7128
7176
7224
7272
7320
7368
74 16
7464
7512
73.3!)
48
7
7607
7655
7703
7751
7790
7*17
7w<)j
7') 12
7'J'K)
8038
48
8
8086
81. -14
8181
822!)
8-277
8325
837,i
8121
8408
8516
48
9
8561
8612
8059
8707
8755
8803
fe*50
8808
8946
8994
48
910
959041
959089
959137
950185
059232
939280
5)59328
959375
950123
959471
48
1
9518
9.3(>(>
9(514
!)<>()]
9709
9757
9*04
9852
91* )0
9917
4*
2
9995
960012
960000
9601,-!*
960185
5)(>02,-!3
9(>02*1
9()0.-2M
960376
9601 '23
48
3
960171
05 1 *
0566
0613
0(561
0709
0750
0*04
0*31
0899
48
4
0946
0994
1041
10*9
1136
11*1
1231
1279
1326
1374
48
915
1421
1460
1516
1563
1611
1658
1706
1733
ISO!
1848
47
6
18i»5
1913
195JO
2038
20*5
2132
21*0
2227
2275
2322
47
7
2361)
2417
2 464
•2511
255<>
2<>0h
2(>"i3
2701
2748
2795
47
8
28 13
2H90
2937
25)85
3032
307!)
3126
3171
3221
3268
47
9
3316
3363
3410
3457
3504
3552
3599
3646
3(593
3741
47
920
963788
963835
963882
963929
963977
964024
964071
964118
964165
964212
47
1
42(50
4307
4354
4401
4448
44')f>
451-2
15'M>
46.17
4GM
47
2
4731
4778
4*23
4*72
4919
45)06
5013
5061
5108
5155
47
3
5202
5249
62%
5343
53'K)
5437
518 4
5.3.U
5578
5625
47
4
5672
5719
5766
5813
5*00
5907
5954
6001
6018
6095
47
925
6112
6180
6236
62S3
6320
6376
6123
6470
6517
6564
47
6
6611
6(>58
6705
6752
67!)!)
(>M5
6*92
6i)3!>
698(5
7033
47
7
70*0
71'27
7173
7 '2 20
7*267
7314
7361
7408
74.34
7501
47
8
7548
7395
7042
7(5*8
7733
77*?
7*2')
7875
7922
7969
47
9
8016
800'2
8109
8156
8203
*'24!)
8296
8313
8390
8436
47
930
968483
968530
96857(5
96*T>23
96S670
968716
968763
968810
968*56
968903
47
1
8950
8996
90 IH
90590
9136
5J183
92'29
9276
9323
9369
47
2
9416
9163
9.300
95.36
9602
9619
9151)3
9742
9789
9M5
47
3
9882
9928
9975
970021
970068
970114
970101
970207
970254
970300
47
4
9703 17
970393
970440
0186
0533
0579
0026
0672
0719
0765
46
935
0812
0858
0004
0051
0007
1044
1000
1137
1183
1220
46
6
1276
1322
1360
1413
1461
1508
1551
1601
1647
1693
46
7
1740
1786
1832
1870
1025
1071
2018
2064
2110
2157
46
8
2203
2249
2205
2342
2388
2434
2481
2527
2573
2619
46
9
2<J66
2712
2758
2804
2851
2897
2943
2989
3035
3082
46
N,
0
1
2
3
4
5
6
7
8
9
D.
TABLE OF LOGARITHMS
511
N.
0
1
2
3
4
5
6
7
8
9
D.
940
973128
973174
973220
973266
973313
973359
973405
973451
973497
973543
46
1
3590
3636
3682
3728
3774
3820
3866
3913
3959
4005
46
2
4051
4097
4143
4181)
4235
4281
4327
4374
4420
4466
46
3
4512
4558
4604
4650
4696
4742
4788
4834
48SO
4D26
46
4
4972
5018
5064
5110
6156
5202
5248
5294
5340
5386
46
945
5432
5478
5524
5570
5616
5662
5707
5753
5799
5845
46
6
5891
5937
5983
6029
6075
6121
6167
6212
6268
6304
46
7
6350
6396
6442
6488
6533
6579
6625
6671
6717
6763
46
8
6808
6854
6900
6946
6992
7037
7083
7129
7175
7220
46
9
7266
7312
7358
7408
7449
7495
7541
7586
7632
7678
46
950
977724
977769
977815
977861
977906
977952
977998
978013
978089
9781 35
46
1
8181
8226
8272
8317
8363
8409
8454
8500
8546
8591
46
2
8637
8683
8728
8774
8819
8865
8911
8956
9(H)2
9047
46
3
9093
9138
9184
9230
9275
9321
9366
9412
9457
9503
46
4
9548
9594
9639
9685
9730
9776
9821
9867
9912
9958
46
955
980003
980049
980094
980140
980185
980231
980270
980322
980367
980412
45
6
0458
0503
0549
0591
0640
0685
0730
0776
0821
0867
45
7
0912
0957
1003
1048
1093
1139
11S4
1229
1275
1320
4
8
1366
1411
1 456
1501
i:>47
1592
1637
1683
1728
1773
4
9
1819
1864
1909
1954
2000
2045
2090
2135
2181
2226
4
960
982271
982316
982362
982407
982152
9S2497
982543
982588
982633
982678
4
1
2723
2769
2hl4
2859
290 1
29 1<)
2994
3010
3085
SI 30
4
2
3175
3220
3265
3310
3356
3401
3446
3491
3536
35S1
3
3626
3671
3716
3762
3807
.iS")2
3h')7
3942
3987
4032
4
4077
4122
4167
4212
4257
4302
4347
4392
4437
4482
965
4527
4572
4617
4662
4707
4752
4797
4842
4887
4932
6
4977
5022
5067
5112
5157
5202
5247
5292
5337
5382
7
5426
5471
5516
5561
5606
5651
5696
5711
57hfl
5H30
8
5875
5920
5965
6010
6055
(.100
6144
(llcS'J
6°34
C279
9
6324
6369
6413
6458
6503
h;>18
f.593
WW7
(5682
6727
970
986772
986817
986861
986906
986951
98(W6
987010
987085
987130
987175
4
1
7219
7264
7309
7353
7398
7443
7188
75H2
7577
7622
4
2
7666
7711
7756
7800
7845
7HIK)
793 1
7(>7()
N)2^
80(58
4
3
8113
8157
8202
8247
8291
S336
835^1
8425
8170
85 11
4
4
8559
8604
8648
8693
8737
8782
8826
8871
8916
8960
4
975
9005
9049
9094
9138
9183
9227
92 < 2
9316
9361
9405
45
6
94.50
9194
9539
95M
9628
9(572
9717
9761
9806
9S50
44
7
9895
9939
9983
99W2S
9(H)072
<)<)01 J7
990161
990206
990250
99029 4
44
8
90033!)
990383
990428
0172
0516
0561
OG05
0650
0694
0738
44
9
0783
0827
0871
0916
09(>0
1001
1019
1093
1137
1182
44
980
991226
991270
991315
991359
991403
991448
991492
991536
991580
U91625
44
1
1 669
1713
1 758
1802
1846
1890
1935
1979
2023
2067
44
2
2111
2156
2200
2244
2288
2333
2377
2121
2 465
2509
44
3
2554
2598
2612
2686
2730
2774
2819
2863
2907
2951
44
4
2995
3039
3083
3127
3172
3216
3260
3304
3348
8392
44
985
3436
3480
3524
3568
3613
3657
3701
3745
3789
3833
44 J
6
3877
3921
3965
4009
4053
4097
4141
41H5
4229
4278
44 j
7
4317
4361
4405
4449
4193
4537
4581
4625
40W)
4713
44
8
4757
4801
4845
4889
4933
4977
5021
fX)65
5108
5152
44
9
5196
5240
5284
5328
5372
5416
5160
5504
5517
5591
44
990
995635
995679
995723
995767
995811
995854
995898
995942
995986
996030
44
1
6074
6117
6161
(.205
6219
6293
(>337
()'.MO
6124
6468
44
2
6512
6555
6599
6643
6687
6731
6774
(»H1S
6H62
6WO
44
3
6949
6993
7037
7080
7124
7168
7212
7255
7299
7343
44
4
7386
7430
7474
7517
7561
7605
7648
7692
7736
777'J
44
995
7823
7867
7910
7954
7998
8041
8085
8129
8172
8216
44
6
8259
8303
8347
8390
8134
8477
8521
8564
H(X)8
8652
4*
7
8695
8739
8782
8826
8869
8913
8956
9000
9043
9087
44
8
9131
9174
9218
9261
9305
9348
9392
9435
9479
9522
44
9
9565
9609
9652
9696
9739
9783
9826
9870
9913
9957
43
N.
0
1
2
3
4
5
6
7
8
9
D.
512
APPENDIXES
Table 2
COMPOUND AMOUNT OF 1
s = (1 + i)n
n
H%
H%
Jf4%
H%
«%
1
1.0012 5000
.0025 0000
1.0029 1667
1.0033 3333
1.0037 5000
2
1.0025 0156
.0050 0625
1.0058 4184
1.0066 7778
1 0075 1406
3
1.0037 5469
0075 1877
1.0087 7555
1.0100 3337
1.0112 9224
4
1.0050 0938
.0100 3756
1.0117 1781
1 0134 0015
1.0150 8459
5
1.0062 6564
.0125 6266
1.0146 6865
1.0167 7815
1 0188 9115
6
1.0075 2348
.0150 9406
1.0176 2810
1.0201 6740
1 0227 1200
7
1.0087 8288
.0176 3180
1.0205 9618
1.0235 6797
1.0265 4717
8
1.0100 4386
.0201 7588
1 0235 7292
1.0269 7986
1 0303 9672
9
1.0113 0641
.0227 2632
1.0265 5834
1.0304 0313
1.0342 6070
10
1.0125 7055
.0252 U13
1.0295 5247
1.0338 3780
1.0381 3918
11
1.0138 3626
1.0278 4634
1.0325 5533
1.0372 8393
1 0420 3220
12
1.0151 0356
1.0304 1596
0355 6695
1.0407 4154
1 0459 3983
13
1.0163 7244
1.0329 9200
.0385 8736
1.0442 1068
1.0498 6210
14
1.0176 4290
1.0355 7448
.0416 1657
1.0476 9138
1 0537 9908
15
1 0189 1495
1.0381 6341
.0446 5462
1.0511 8369
1.0577 5083
16
1.0201 8860
1 0407 5882
.0477 0153
1 0546 8763
1 0617 1739
17
1.0214 6383
1.0433 6072
.0507 5732
1.0582 0326
1 0656 9883
18
1 0227 4066
1.0459 6912
.0538 2203
1.0617 3060
1 0696 9521
19
1.0240 1909
1.0485 8404
.0568 9568
1.0652 6971
1 0737 0656
20
1.0252 9911
1.0512 0550
.0599 7829
1.0688 2060
1 0777 3296
21
1.0265 8074
1 0538 3352
1.0630 6990
I 0723 8334
1 0817 7446
22
1.0278 6396
1.0564 6810
1 0661 7052
1 0759 5795
1 0858 31 H
23
1.0291 4879
1 0591 0927
1.0692 8018
1 0795 4448
1 0899 0298
24
1 0304 3523
1 0617 5704
1.0723 9891
1 0831 429(5
1 0939 9012
25
1.0317 2327
1.0644 1144
1 0755 2674
1.0867 5344
1.0980 9258
26
1.0330 1293
1.0670 7247
1.0786 0370
1 0903 7595
1 1022 1043
27
1.0343 0419
0697 4015
1 0818 0980
1 0940 1053
1 1063 4372
28
1 0355 9707
0724 1450
1 0849 6508
1 0976 5724
1 1104 9251
29
1.0368 9157
0750 9553
1 0881 2956
1 1013 1609
1 1146 5685
30
1.0381 8768
0777 8327
1.0913 0327
1 1049 8715
1 1188 3682
31
1.0394 8542
1.0804 7773
1.0944 8624
1.1086 7044
1.1230 3245
32
1 0407 8478
1.0831 7892
1.0976 7849
1.1123 6601
1 1272 4383
33
1.0420 8576
1 0858 8687
1 1008 8005
1.1160 7389
1 1314 7099
34
1 0433 8836
1.0886 0159
1 1040 9095
1.1197 9414
1 1357 1401
35
1.0446 9260
1.0913 2309
1.1073 1122
1.1235 2679
1.1399 7293
36
1.0459 9847
1 0940 5140
1.1105 4088
1.1272 7187
1 1442 4783
37
1.0473 0596
1.0967 8653
1.1137 7995
1 1310 2945
1.1485 3876
38
1.0486 1510
1.0995 2850
1.1170 2848
1.1347 9955
1.1528 4578
39
1.0499 2586
1.1022 7732
1.1202 8648
1.1385 8221
1.1571 6895
40
1.0512 3827
1.1050 3301
1.1235 5398
1 . 1423 7748
1.1615 0834
41
1.0525 5232
1.1077 9559
1.1268 3101
1.1461 8541
1.1658 6399
42
1 0538 6801
1.1105 6508
1.1301 1760
1.1500 0603
1.1702 3598
43
1.0551 8535
1.1133 4149
1.1334 1378
1.1538 3938
1.1746 2437
44
1.0565 0433
1.1161 2485
1.1367 1957
1 . 1576 8551
1.1790 2921
46
1.0578 2496
1.1189 1516
1.1400 3500
1.1615 4446
1.1834 5057
46
1.0591 4724
1.1217 1245
1.1433 6010
1.1654 1628
1.1878 8851
47
1.0604 7117
1.1245 1673
1.1466 9490
1.1693 0100
1.1923 4309
48
1.0617 9676
1.1273 2802
1.1500 3943
I 1731 9867
1 1968 1438
49
1 0631 2401
1.1301 4634
1.1533 9371
1 1771 0933
1 2013 0243
60
1.0644 5291
1.1329 7171
1.1567 5778
1 1810 3303
1.2058 0732
COMPOUND AMOUNT OF 1
s = (1 + i)°
513
n
«2%
J2%
Jl'2%
?8%
?3%
1
2
3
4
5
1.0041 6667
1.0083 5069
1.0125 5216
1.0167 7112
1.0210 0767
.0050 0000
.0100 2500
.0150 7513
.0201 5050
.0252 5125
1.0058 3333
1 0117 0069
1.0176 0228
1.0235 3830
1.0295 0894
1.0062 5000
0125 3906
0188 6743
.0252 3535
.0316 4307
1.0066 6667
1.0133 7778
1.0201 3363
1.0269 3452
1.0337 8075
6
7
8
9
10
0252 6187
0295 3379
.0338 2352
1.0381 3111
.0424 5666
1.0303 7751
1.0355 2940
1.0407 0704
1.0459 1058
1.0511 4013
1.0355 1440
1.0415 5490
1.0476 3064
1.0537 4182
1.0598 8865
1.0380 9084
.0445 7891
.0511 0753
.0576 7695
.0642 8743
1.0406 7262
1.0476 1047
1.0545 9451
1.0616 2514
1.0687 0264
11
12
13
14
16
1.0468 0023
1.0511 6180
1 0555 4173
1.0599 3983
1.0643 5625
1.0563 9583
1.0616 7781
1.0669 8620
1.0723 2113
1.0776 8274
1.0660 7133
1.0722 9008
1.0785 4511
1.0848 3662
1 0911 6483
1.0709 3923
1.0776 3260
1.0843 6780
.0911 4510
.0979 6476
1.0758 2732
1.0829 9951
1.0902 1950
1.0974 8763
1 . 1048 0422
16
17
18
19
20
1.0687 9106
1.0732 4436
1.0777 1621
1.0822 0670
1.0867 1589
0830 7115
0884 8651
0939 2894
0993 9858
1048 9558
1.0975 2996
1 1039 3222
1.1103 7182
1 . 1 1 68 4899
1.1233 6395
1048 2704
.1117 3221
.1186 8053
.1256 7229
1.1327 0774
1.1121 6958
1.1195 8404
1.1270 4794
1.1345 6159
1.1421 2533
21
22
23
24
25
1.0912 4387
1 0957 9072
I 1003 5652
1 1049 4134
1 . 1095 4526
.1104 2006
1159 7216
1215 5202
1271 5978
.1327 9558
1.1299 1690
1365 0808
1431 3771
1498 0602
1.1565 1322
1.1397 8716
.1469 1083
.1510 7902
.1612 9202
1.1685 5009
1.1497 3950
1.1574 0443
1.1651 2046
1.1728 8793
1.1807 0718
26
27
28
29
30
1 1141 6836
1 1188 1073
1 1234 7211
1 1281 5358
1.1328 5422
1384 5955
.1441 5185
1498 7261
.1556 2197
1.1614 0008
1 1632 5955
1.1700 4523
1 1768 7049
1837 3557
.1906 4069
1.1758 5353
. 1832 0262
.1905 9763
1.1980 3887
1.2055 2661
1.1885 7857
1.1965 0242
1.2044 7911
1.2125 0897
1.2205 9236
31
32
33
34
36
1.1375 7444
1.1423 1434
1 1470 7398
1 1518 5346
1.1566 5284
1 1672 0708
1.1730 4312
1 1789 0833
1.1848 0288
1.1907 2689
1975 8610
2045 7202
2115 9869
2186 6634
1 2257 7523
1.2130 61 i5
1.2206 4278
1.2282 7180
1 . 2359 4850
1.2436 7318
1.2287 2964
1.2369 2117
1.2451 6731
1.2534 6843
1.2618 2489
36
37
38
39
40
1.1614 7223
1 1663 1170
1 1711 7133
1.1760 5121
1.1809 5143
1 1966 8052
1.2026 6393
1 2086 7725
1 2147 2063
1.2207 9424
1 2329 2559
1 240 1 1765
1 2473 5167
1 2546 2789
1 2619 4655
1.2514 4614
1 2592 6767
1.2671 3810
1.2750 5771
1.2830 2682
1.2702 3705
1.2787 0530
1.2872 3000
1.2958 1153
1.3044 5028
41
42
43
44
46
1 1858 7206
1.1908 1319
1.1957 7491
1.2007 5731
1 2057 6046
1.2268 9821
.2330 3270
.2391 9786
.2453 9385
.2516 2082
1.2693 0791
1.2767 1220
1.2841 5969
1.2916 5062
1.2991 8525
1.2910 4574
1.2991 1477
1.3072 3424
1.3154 0446
1.3236 2573
1 3131 4661
1.3219 0092
1.3307 1360
1.3395 8502
1.3485 1559
46
47
48
49
60
i
1.2107 8446
1.2158 2940
1 2208 9536
1.2259 8242
1.2310 9068
.2578 7892
.2641 6832
.2704 8916
.2768 4161
.2832 2581
1.3067 6383
1.3143 8662
1.3220 5388
1 3297 6586
1.3375 2283
1.3318 9839
1.3402 2276
1.3485 9915
1.3570 2790
1.3655 0932
1.3575 0569
1.3665 5573
1.3756 6610
1.3848 3721
1.3940 6946
514
APPENDIXES
D
»/4%
1%
lVa%
1%%
la/8%
1
l,007,r>
1.01
1.0112 5
1.0125
1.0137 5
2
1.0150 5625
1.0201
1.02262656
1.0251 5625
1 0276 8906
3
1.02266917
1.0303 01
1.0341 3111
1.0379 7070
1 0418 1979
4
1.0303 3919
1.0406 0401
1 0457 6509
1.0509 4531
1 0561 4481
5
1.0380 6673
1.0510 1005
1.0575 2994
1.0640 8215
1.0706 6680
6
1.0438 5224
1.06152015
1.0694 2716
1.07738318
1.0853 8847
7
1.0536 9613
1.0721 3535
1 0814 5821
1.0908 5047
1 1003 1256
8
1 0615 9885
1 0328 6671
1 0936 2462
1.1044 8610
1 1154 4 ISO
9
1.0695 6081
1 0936 8527
1 1051) 278')
1.11829218
1.1307 7918
10
1.0775 8253
1.10462213
1.1183 0958
1.1322 7083
1.1463 2740
11
1.0856 6141
1.11566835
1.13005124
1.1464 2422
1 1C208910
12
1. 09380690
1 126H 2503
1 11367414
1.1607 5452
1 1780 6813
13
1.1020 1015
1 1380 9.12H
1 1565 4078
1.17526395
1 1912 6(r>6
14
1.1102 7553
1 1491 7121
1.16955186
1.1 899 5475
1.2106 8773
15
1.1186 025i>
1.16096896
1.18270932
1.2048 2918
1.2273 3469
16
1.12699211
1.1725 7864
1.19601480
1.21988955
1.2442 1054
17
1.1354 4155
1.18430143
1 20U4 MW7
1.2351 38 17
1.2613 1813
18
1.1439 6039
1 1961 4748
l.:I230 7ti50
1 2505 7739
1 278661.%
19
1.15254009
1 2081 0895
1.2368 36 11
1 2662 0961
1.2962 4316
20
1.1611 8114
1.2201 9004
1.2507 5052
1.2820 3723
1.3140 6630
21
1.1698 9302
1.2323 9194
1 2648 2146
1 2980 6270
1.3321 3402
22
1.1786 6722
1.2447 1586
1.27905071
1.3142 8848
1 35045177
23
1 1875 0723
1.2571 6302
1.2934 4003
1.3307 1709
1.3690 2048
24
l.i iW i:r>3
1.2697 3465
1 3079 9123
1.3173 5105
1 38784151
25
1.2053 8663
1.2824 3200
1.3227 0613
1.3641 9294
1.4069 2738
26
1.2144 2703
1.29525631
1.33758657
1.3812 4535
1 4262 7263
27
1.2235 3523
1 3082 0888
1 3526 3442
1.3985 1092
1 4 158 MhS
28
1.2327 1175
1 3212 9097
1.3678 5156
1 4159 9230
1 4657 (>478
29
1.2419 5709
1 3345 0388
1.3832 3989
1.4336 9221
1.4859 1 «>(),-)
30
1.2512 7176
1.3478 4892
1.3988 0134
1.4516 1336
1.5063 5013
31
1.2606 5630
1.3613 2740
1.4145 3785
1.4697 5853
1 5270 6275
32
1 2701 1122
1 3719 4068
1.4304 5140
1 4881 3031
1 5480 5986
33
1.2796 3706
1 .$880 9009
1 44 65 4398
1 3007 321 1
1 5693 4569
34
1.2892 3131
1 4025 7699
1 4628 1760
1.5255 6629
1.5909 1M1!>
35
1.2989 0359
1,4166 0276
1.4792 7430
1.5440 3587
1.6127 9940
36
1.3086 4537
1.4307 6878
1.4959 1613
1 5639 4382
1.6349 7539
37
1.8184 6021
1.4450 7647
1.5127 4519
1.5834 W12
1.6374 5030
38
1.3283 4866
1 4595 2724
1 5297 6357
I 6032 8678
1 6802 4633
39
1.3383 1128
1.4741 2251
1.5469 7341
l.G'233 2787
1.7033 4971
40
1.3483 4861
1.4888 6373
1.5643 7687
1.643o 1946
1.7267 7077
41
1.3584 6123
1.5037 5237
1.5819 7611
1.6641 6471
1.7505 1387
42
1.3686 4969
1 5187 8989
1.5997 7334
1 6849 6677
1.7745 8343
43
1.3789 1456
1.5339 7779
1.6177 7079
1.7060 2885
1.7989 8396
44
1.3892 5642
1.5493 1757
1.6339 7071
1.7273 5421
1.8237 1999
45
1.3996 7584
1.5648 1075
1.6543 7538
1.7489 4614
1.8487 9614
46
1.4101 7341
1.5804 5885
1.6721) 8710
1 7708 0797
1.8742 1708
47
1.4207 4971
1.5962 6344
1.6918 0821
1.7929 4306
1.89998757
48
1.4314 0533
1.6122 2608
1.7108 4105
1.8153 5485
1.9261 1240
49
1.4421 4087
1.6283 4834
1.7300 8801
1.8380 4679
1.9525 9644
50
1.4529 5693
1.6446 3182
1.7495 5150
1.8610 2237
1.9794 4464
COMPOUND AMOUNT OF 1
s = (14- i)»
515
r
a
ll/2%
1%%
!3/4%
X7/B%
2%
i
2
3
4
5
1.015
1 0302 25
1 0456 7838
1 0(513 6355
1.0772 8400
1.0162 5
1 0327 6106
1.04954648
1.0666 0161
1.0839 3388
1 0175
1 0353 0625
1 0534 2411
1 0718 5903
1 0906 1656
1.0187 5
1.0378 5156
1 0573 1128
1.0771 3587
1.0973 3216
102
1.0404
1.0612 OR
1.0824 321(5
1.1040 8080
6
7
8
9
10
1 0934 4326
1 1098 4491
1 1264 9259
1.1433 8098
1.16054083
1 1015 4781
1.111)1 4796
1.13763899
1.1561 2563
1.1749 1267
1.10970235
1 1291 2215
1 1488 8178
1 1(589 8721
1.1894 4449
1.11790714
1.13886790
1 1602 2167
1.1819 7588
1 2011 3788
1.1261 6242
1.118(58567
1 17165938
1 1950 9257
1.2189 9142
11
12
13
14
15
1 1779 4894
1 1!>.')6 1817
1 21355241
1.2317 5573
1 2502 3207
1.19400500
1.2134 0758
1.2.W1 2545
1 2531 (5374
1.2735 2765
1 2102 5977
1 2314 3931
1 2529 8950
1 274<> H.82
1.2972 2786
1.2207 1546
1 2197 1638
1 2731 4856
1 2')70 2009
1.3213 3922
1 2433 7131
1 2<i82 4179
1 2936 0(5(53
1.3194 787(5
1.3158 toil1,!
16
17
18
19
20
1.26898555
1,28802033
1 3073 1061
1.32(i<> 5075
1.346S 5501
1 2942 22 18
i 3i:>2 5359
1 3366 2b46
1 35S3 4664
1.3801 1977
1 3 1 99 2935
1 34 30 2S 11
1.3()(i,r)31U
1 ;!i)04 4540
1 4147 7820
1 3161 H33
1.37135398
1 3970 6686
1 1232 6187
1 1 199 4803
1.3727 857)
I 4002 1142
1 1282 1(525
1.45(>8 1117
1.1859 1740
21
22
23
24
25
1 3670 5783
1 3875 (-370
1.10837715
1 4295 0281
1 1509 4535
1 10285160
1 1256 4793
1 418H 1471
1 17235795
1 491.2 8377
1.43% 3081
1 4617 2871
1 1<»03 61 1(,
1 5164 4279
1 5J29 8054
1 4771 3155
1 50 IK 3082
1 f>330 4640
I.f><il7 9102
1.5910 7460
1 5156 6684
1 5459 7967
1.57689926
1 6081 3725
1.64060599
26
1 4727 0953
1 5205 <»838
1 5(599 8269
1 (5209 0725
1.6731 1811
27
28
29
30
1 4918 0018
1 51722218
1.53998051
1.5630 8022
1 5153 0810
1 5701 l')36
1 5%9 3868
1 6218 7268
1 5971 5739
1 (>T>4 1290
1 <!538 57(52
1.6828 0013
1.6512 992(5
1 6822 fill 2
1 71380352
1.74593731
1.7068 8648
1.74102121
1.7758 1469
1.81136158
31
32
33
34
35
1.586526-12
1 6103 2432
1 6344 7918
1 6589 9637
1 6838 8132
1.6482 2811
1 (i7.r>0 1182
1 7022 3076
1 72' W 9201
1.7580 0275
1 7122 4913
1 7122 1319
1.7727 0223
1 8037 2152
1.8352 8970
1 7780 7366
1 8120 2379
1.8150 <)<)2l
1.8806 1172
1.9158 7311K
1 8175 8882
1.8815 4059
1 92223110
1 960(5 7(503
1 9998 8955
36
37
38
39
40
1 7091 3954
1 7347 7663
1.76079828
1 7872 1025
1.81401841
1 7865 7030
1 81500207
1 8111 0560
1 8750 8857
1.9055 5875
1 8671 0727
| 9000 8(!89
1 9333 3841
1 9671 7184
2.0015 9734
1 95179582
1 9883 9199
2 0256 7134
2 0(i3(i 5573
2.1023 4928
2 0398 8734
2 080(5 8509
2 l'J22 9879
2 1617 1177
2.2080 3966
41
42
43
44
45
1 8412 2868
1 8688 1712
1 .8968 7982
1 9253 3302
1.9542 1301
1.93652408
1 9679 9260
1 9999 7248
2.0324 7203
2.0654 9970
2 0366 2530
2 0722 (.624
2 1085 3090
2 1H4 3019
2.1829 7322
2.1417 6833
2.18102618
2 2228 37(50
2.201 5 1581
2.3009 7548
2.2522 0046
2.2972 4447
23131 8936
2.3900 5314
2 1378 5421
46
47
48
49
50
1 9835 2621
2.0132 7910
2.0434 7829
2 0741 3046
2.1052 4242
2 01)90 6407
2 1331 7387
2.1678 3794
2 2030 6531
2.2388 6512
2.2211 7728
2 2fJOO 1789
2.2995 9872
2 3398 4170
2.3807 8893
2.3502 3127
23912 9811
2.4391 9120
2.4849 2603
2.5315 1839
2.48661129
2 53(53 4351
2.5870 7039
2.63881179
2.6915 8808
516
APPENDIXES
s = (1 + i)n
n
2'/8%
2*/4%
2%%
2Va%
2%%
1
2
3
4
5
1.0212 5
1.0429 5156
1.0651 1428
1.0877 4796
1,11086201
1.0225
1.0455 0625
1.06903014
1.09308332
1.1176 7769
1.0237 5
1.0480 6406
1.0729 5558
1.0984 3828
1.1245 2619
1.025
1.0506 25
1.0768 9068
1.1038 1289
1.1314 0821
1.0275
1.0557 5625
1,0847 8955
1.1146 2126
1.1452 7334
6
7
8
9
10
1.1344 6844
1.1585 758!)
1.1831 9563
1.2083 3854
1.23401573
1.1428 2544
1.16853901
1.19483114
1.2217 1484
1.2492 0343
1.15123369
1 17857519
1.2065 6(565
J.2U52 2261
1 2645 5915
1 1596 9342
1.18868575
1.21840290
1.21886297
1.28008454
1.1767 6836
1.2091 2949
1.2423 8055
I.'27(i5 4602
1.31 1C 5103
11
12
13
14
15
1.2602 3856
1.2870 1863
1.3143 6778
1 3422 9809
1.3708 2193
1.27731050
1.30604999
133543611
1.36 vl 8343
1.39620680
1 2945 9243
1.3253 3900
1.3568 1580
1.3890 4017
1 4220 2988
1.31208666
1.34488882
1.3785 1101
1.41297382
1.4482 9817
1 3477 2141
1 3847 8378
1.12286533
1.4619 9413
1.5021 9896
16
17
18
19
20
1.3099 5180
1.4297 0087
1.46008202
1.4911 0876
1.5227 9482
1.42762146
1.4597 4294
1.4925 8716
1.5261 7037
1.5605 0920
1.45580309
1.4903 7841
1.5257 7490
1.5620 1205
1.5991 0984
1 48150562
1.5216 1826
1 5596 5872
1.59865019
1.6386 1644
1.5435 0911
1.5859 559 j
1 6295 6973
1.674H 8290
1.7204 2843
21
22
23
24
25
1.5551 5421
1.5882 0124
1.6219 5051
1.6564 1696
1.6916158'?
1 .5956 2066
1.6315 2212
1 (1682 3137
1 7057 6658
1 74-U 4632
1.63708870
1 6759 6955
1.7157 7383
1 7565 2346
1 79b2 4089
1.6795 8185
1.72157140
1.7646 1068
1 .8087 2595
1.8539 4410
1 7677 4021
1.8163 5307
1.86(>3 0278
1.9176 2610
1.9703 6082
26
27
28
29
30
1.7275 6266
1.7642 7336
1.8017 6417
1.8400 5166
1.8791 5276
1.7833 8962
1.8235 1588
1 8615 4499
1 9064 9725
1.9493 9341
1.81094911
1.88467165
1.9291 3261
1.9752 5663
2.0221 6898
1 9002 9270
1.9478 0002
1 9961 9502
'2.0t(i 4 0739
2 0975 6758
2.0245 4575
2 0802 2075
2 1374 2682
2 1962 0000
225660173
31
32
33
34
35
1.9190 8476
1.95986531
2.00151245
2 0440 4458
2.0874 8053
1 9932 5 1 79
2 0381 0303
2.0839 603 4
2.1308 4945
2.1787 9356
2.0701 9549
2.11936263
2.16969749
2.22122781
2.2739 8197
2 1500 0677
2.2037 5694
2,2588 5086
2 3153 2213
237320519
2 31 8(5 5828
2 3824 2138
2 4479 3797
2.5152 5626
2.5844 2581
36
37
38
39
40
2.1318 3949
2.1771 4108
2.2234 0533
2.2706 5269
2.3189 0406
2 2278 1642
2 2779 4229
2 3291 9599
2 3816 0290
2.4351 8S97
2 3279 8904
2.3832 7878
2 4398 8165
2 4978 2884
2.5571 5228
2 4325 3532
2.4933 4870
2.5556 8242
2.6195 7448
2.6850 6384
2.6554 9752
2 7285 2370
2.8035 5810
2 8806 5595
2.9598 7399
41
42
43
44
45
2.3681 8077
2 4185 0462
2.4698 9784
2.5223 8317
2.5759 8381
2 1899 8072
2 5460 0528
2 6032 00 U)
2.66180444
2.72175639
2 6178 8464
2.ti800 5940
2 7437 1081
2 8088 7395
2.8755 8470
2.7521 9043
2.8209 9520
2 8915 2008
2.9638 0808
3.0379 0328
3.0412 7052
3.1249 0546
3 2108 4036
3.2991 3847
3.3898 6478
46
47
48
49
50
2.6307 2347
2.6866 2634
2.7437 1715
2.8020 21U
2.8615 6409
2.7829 9590
2.8456 1331
2.9096 3961
2.9751 0650
3.0420 4640
2.9438 7984
3.013V 9699
3.0853 7466
3.1586 5231
3.2336 7030
3.1138 5086
3.1916 9713
3.2714 8956
3.3532 7680
3.4371 0872
3.4830 8606
3 5788 7093
3.6772 8988
3.7784 1535
3.8823 2177
COMPOUND AMOUNT OF 1
s = (1 + i)«
517
n
3%
3V4%
3Va%
3»/4%
4%
1
2
3
4
5
1.03
1.0609
1.0927 27
J 1*55 0881
1 1592 7407
1.0325
1.0660 5625
1.1007 0308
1.1364 7593
1.1734 1140
1.035
1.0712 25
1.1087 1788
1.1475 2300
1.1876 8631
1.0375
1 .0764 0625
1.1167 7148
1.15S6 5042
1.2020 9981
1.04
1.0816
1.1248 64
1.16985856
1.2166 5290
6
7
P
9
10
1.1940 :230
1 2298 7387
1.2667 7008
1.3047 7318
1.3439 1638
1.2115 4727
1.2509 2255
1.2915 7754
1.33355381
1.3768 9430
1.2292 5533
1 2722 7926
1 3168 0904
1.3628 9735
1.4105 9876
1.2471 7855
1.2939 4774
1.3424 7078
1.39281344
1.4450 4394
1.2653 1902
1.3159 3178
1.3685 6905
1.4233 1181
1.4802 4428
11
12
13
14
15
1.3842 3387
1 4257 6089
1 4685 3371
1.5125 8972
1.5579 6742
1.42164337
1.4678 4678
1 5155 5180
1.5648 0723
1.6156 6347
1.4599 6972
1.51106866
1.5639 5606
1.6186 9152
1.6753 4883
1.49923309
1. 55515133
1.61378387
1.6743 0076
1.7370 8704
1.5394 5406
1.60103222
1 6650 7351
1.7316 7645
1.8009 4351
16
17
18
19
20
1 6047 0644
1 6528 4763
1.7024 3306
1.753.'} 0605
1.8061 1123
1.6681 7253
1 7223 8814
1.77836575
1.8361 6264
1.8958 3792
1.7339 8604
1.7916 7555
1.8574 8920
1.9225 0132
1.9897 8886
1.8022 2781
1.8698 1135
1.9399 2927
2.01267662
2.0881 5200
1.87298125
1.9479 0050
2.0258 1652
2.1068 4918
2.1911 2314
21
22
23
24
25
1.8602 9457
1 9161 03 11
1 9735 b65l
203279111
2 0937 7793
1.95745266
2 0210 6987
2 0867 5464
2 1545 7416
2 2245 9782
20591 3147
2.1315 1158
2.2061 1448
2 2833 2849
2 3632 4498
2.1664 5770
2.2476 9986
2.3319 8860
2.4194 3818
2.51016711
2.2787 6807
2.3699 1879
2.4647 1551
2.5633 041fl
2.6658 3633
26
27
28
29
30
215659127
2.2212 8901
2 2879 2708
2 3565 0551
2 4272 6247
2 2968 9725
2 3715 4611
2 44862167
2 52820188
2 6103 6844
2 4459 5856
2.5315 6711
2.6201 7196
271187798
2.8067 9370
2.6042 9838
2.70195956
2 8032 8305
2,9084 0616
3.0174 7139
2.7724 6978
2.8833 6858
2.9987 0332
3.1186 5145
3.2433 9751
31
32
33
34
35
2 5000 8035
2.5750 8276
2.6523 :r>24
2.73190530
2.8138 6245
2.6952 0541
2.78279959
2 8732 4058
2 9666 2089
3.0630 3607
2.9050 3148
3.00<>7 0759
3.1119 4235
3.2208 6033
3.3335 9045
3.1306 2657
3.2480 2507
3.3698 2601
3.4961 9448
3.6273 0178
3.3731 3341
3.5080 5875
3 6483 8110
3.7943 1631
3.9460 8899
36
37
38
39
40
2 8982 7838
2 9852 2668
3.0747 8348
3.1670 2698
3.2620 3779
3.1625 8475
3 2653 6875
3.3714 9323
3.4810 6676
3.5942 0143
3.4502 6611
3.5710 2543
36960 1132
3 8253 7171
3.9592 5972
3.7633 2559
3.9044 5030
4.0508 6719
4.2027 7471
4.3603 7876
4.1039 3255
4.2680 8986
4.4388 1345
4.6163659')
4.8010 2063
41
42
43
44
45
3 3598 9893
3.4606 9589
3.5645 1677
3.6714 5227
3.78159584
3.71101298
3.8316 2090
3 9561 4858
4.0847 2341
4.2174 7692
4 0978 3381
4.24125799
1.3897 0202
4.5433 4160
4.7023 5855
4.5238 9296
4.6935 3895
4.8695 4666
5.0521 5466
5.2416 1046
4.9930 6145
5.1927 8391
5.4004 9527
5.6165 1508
5.8411 7568
46
47
48
49
60
3.8950 4372
4.0118 9503
4.1322 5188
4.2562 1944
4.3839 0602
4.3545 4492
4.4960 6763
4.6421 8983
4.7930 6100
4.9488 3548
4.86694110
5 0372 8404
5.2135 8898
5.3960 6459
5.5849 2686
5.4381 7085
5 6421 0226
5.8536 8109
6.0731 9413
6.3009 3891
6.0748 2271
6.3178 1562
6.5705 2824
6.a333 4937
7.1066 8335
518
APPENDIXES
s - (14- i)n
n
4V V
437 r
5%
6%%
1
1.0425
1 .045
1.0475
1.05
1.055
2
1.0868 0625
1.092023
1.0972 5025
1.1025
1.1130 2>
3
1.13299552
1.1411 6613
1 1493 759'2
1 1576 25
1.1742 4138
4
1.1811 4783
1.1925 1860
1.2039/128
1.2155 0625
1.2388 240")
5
1,2313 4661
1.2461 8194
1.2611 5991
1.2762 8156
1.3069 6001
6
1.28367884
1.3022 6012
1.32106501
1.3400 9564
1.3788 12S1
7
1.33823519
1.30080183
1.3838 1500
1.4071 0042
1.4546 7910
8
1.3951 1018
1 4221 0061
1.4495 4084
1 4774 5544
1.53408051
9
1.4544 0237
1.4860 9514
1.5184 0031
1.5513 2822
1.6190 9427
10
1.5162 1147
1.5529 6942
1.5905 2133
1.6288 9463
1.7081 4440
11
1.5806 5358
1 0228 5305
1.0060 7423
1.7103 3936
1.8020 1)240
12
1.04783136
1 0458 HI 43
1.7452 1270
1.7958 5633
1.9012 ()74'J
13
1.71786119
1.7721 9010
1 82S1 1037
1 885o 4914
2 0057 7390
14
1.7908 7342
1 8319 4492
1.4141)4361
1 9799 3100
2 11 GO 9140
15
1.8609 8554
1.9352 8244
2 0059 0552
2.0789 2818
2 2324 7649
16
1 9163 3243
2 0223 7015
2.1011 8004
2.1828 7459
2 3552 0270
17
2.02905156
211337081
2 2009 9237
2.2920 1832
2. 4848 02 H
18
2.11528625
2 2084 7877
2 3035 :{431
2 106G 1923
2 0214 01)27
19
2 2051 8591
2 3078 6031
2 1150 5204
2.5209 5020
2.7656 l()9l
20
2.2989 0631
2.41171402
2.5297 6764
2.6532 9771
2.9177 5749
21
2.3966 0983
2.5202 4116
2.64993160
2.7859 6259
307823115
22
2.4984 6575
2 6330 5201
2.7758 0335
2.9252 6072
3 2475 3703
23
2.6040 5054
2.7521 0635
2.9076 5401
3 0715 2376
3.4261 5157
24
2.7153 4819
2.8700 1383
3 0457 6758
3 2250 9994
3.0145 8990
25
2.8307 5049
3.0054 3446
3.1904 4154
3.3863 5494
3.8133 9233
26
2.9510 5739
3.14007901
3.3119 8751
3.5550 7269
4.0231 2893
27
3.0764 7732
3.2820 0966
3.5007 3192
3 7334 5632
4.21110102
28
3.2072 2761
3.4296 9998
3 0070 1068
3 9201 2914
4 4778 4307
29
3.3435 3478
3.5840 3049
3.8111 9998
4.11«1 3500
4.7211 2111
30
3.4856 3501
3.7453 1813
4.0236 5098
4.3219 4238
4.9839 5129
31
3.6337 7450
3.9138 5745
4.2147 8068
4.5380 3949
5 2580 0801
32
3.7882 0992
4.0899 8104
4.4149 8276
4 7619 4147
5 5472 0238
33
3,9492 0884
4 2740 3018
4.0246 9445
5 0031 8854
5.8523 0181
34
4.1170 5021
4 4663 6154
4.8443 6743
5.2533 4797
6.1742 4171
35
4.2920 2485
4.6073 4781
5.0744 7488
5.5160 1537
6.5138 2501
36
4.4744 3590
4.8773 7846
5.3155 1244
5.7918 1614
0.8720 8538
37
4.0645 9943
5.09680019
5 5079 9928
608140094
7 2500 5008
38
4 8628 4491
5.3202 1921
5.8324 7925
0 3854 7729
7.0488 02M
39
5.0695 1581
5.5058 9408
6.1095 2201
6.7047 5115
8.0094 8099
40
5.2849 7024
5.8163 6454
0.3997 2431
7.0399 8871
8.5133 0877
41
5.5005 8147
0.0781 0094
67037 1121
7.3919 8815
8.9815 4076
42
5.7437 3868
6 3510 1548
7.0221 3750
7.7013 8756
9.4755 2550
43
5.9878 4758
6.6374 3818
7.3556 8903
8.1490 66WI-4
9 9966 7940
44
6.24233110
6.9361 2290
7.7050 8426
8.5571 5028
10.5464 9677
45
6.5076 3017
7.2482 4843
8.0710 7576
8.9850 0779
11.1265 5409
46
6.7842 0445
7.5744 1961
8.4544 5186
9.4342 5818
11.7385 1456
47
7.0725 3314
7 9152 6849
8 8560 3832
9 9059 7109
12 3841 3287
48
7.3731 1580
8 2714 5557
9.2767 0014
10.40126965
13.0652 0017
49
7.6864 7J22
8.6436 7107
9 7173 4340
10.9213 3313
13.7838 4948
50
8.0131 4834
9.0326 3627
10.1789 1721
11.4673 9979
14.5419 6120
COMPOUND AMOUNT OF 1
s = (1 + i)n
519
n
6%
*V%%
7%
8%
9%
2
3
4
5
1.06
1.1236
1.1910 16
1.2624 7696
1.3382 2558
1.065
1.1342 25
1.2079 4963
1.2864 6635
1.37008666
1.07
.1449
2250 43
3107 9601
.4025 5173
1.08
1.1664
1 2597 12
1.36048896
1.4693 2808
1.09
1.1881
1.2950 29
1.4115 8161
1.5386 2395
6
7
8
9
10
1.4185 1911
1.5036 3026
1.5938 4807
1 .685)4 7896
1.7908 4770
1 4591 4230
1.55398655
1.6549 9567
1.7625 7039
1.8771 3747
.5007 3035
6057 8148
71 SI 8618
.83S4 5921
1.9671 5136
1 5868 7432
1 7138 2427
1 8509 3021
1 9990 0163
2.1589 2500
1.6771 0011
1 8280 3912
1 9925 626 1
2 1718 9328
2.3673 6367
11
12
13
14
15
1 8982 9856
2.0121 9647
2 1329 2826
2.2609 0396
2.39655819
1 9991 5140
2 12909624
2 2674 875O
2.41487418
2.5718 4101
2.1048 5195
2.2521 9159
2 4098 4500
2.5785 3415
2.7590 3154
2.33163900
2 5181 70 J 2
2 7196 2373
2.9371 9362
3.1721 6911
2.5804 2641
281-266478
3.0658 0101
3 3U7 2703
3.6424 8246
16
17
18
19
20
2.5403 5168
2 6927 7279
2 8543 3915
3 0255 9950
3.2071 3547
2 7390 1067
201704637
3 1066 5438
3 3085 8691
3.52.-S6 4500
2.9521 6375
3.15881521
3.3799 3228
H Hl<>5 2751
3.8696 8446
3.4259 4264
3 7000 1805
3 9960 1950
4.3157 0106
4.6609 5714
3 9703 0588
4.3276 3341
4 7171 204'2
5.1116 6125
5.6044 1077
21
22
23
24
25
3.3995 6360
3 6035 3742
3 8197 4966
4.0489 3464
4.2918 7072
3 7526 8199
3 9966 0632
4 2563 8573
4 5330 5081
4.82769911
4.14056237
4 4304 0174
.7405 2986
5.0723 G695
5.4274 3264
5 0338 3372
5 1365 4041
5 H71 1 6365
63411 8074
6.8484 7520
6.10880771
6.6586 0043
7 2578 7447
791108317
8.6230 8066
26
27
28
29
30
4.5493 8296
4.8223 4594
511168670
5.4183 8790
5.7434 9117
5.1414 9955
5 475(5 9702
583161733
6 2106 7245
6.6143 6616
5.8073 5292
f, 2 1 US 6763
6 6488 3836
7.11425705
7.6122 5504
7 3903 5321
7 9880 6147
8 6271 0639
9 3172 7190
10.06265689
9 3991 5792
10 21508213
11 1671 3952
12 1721 H?OH
13.2676 7847
31
32
33
34
35
6.0881 0064
6 4533 8668
6 8405 8988
7.2510 2528
7.6860 8679
7 0442 9996
7 5021 7946
7 9898 2113
8. 5091 5950
9.0622 5487
8.1451 1290
871527080
9.3253 3975
9.9781 1354
10.67658148
10.8676 6944
1 1 7370 8300
126700 4964
13 6901 3361
14.7853 4429
14.4617 6953
15 7633 2879
17 1820 2838
18.7284 1093
20.4139 6792
36
37
38
39
40
8.1472 5200
8.6360 8712
9.1542 5235
9.7035 0749
10.2857 1794
9.65130143
10 2786 3(503
10.91674737
11.65828595
12 4160 7453
11.42394219
12.22361814
13.0792 7141
13 9918 2011
14.9744 5784
15.9681 7184
17 24562558
18 6252 7563
20.1152 9768
21.7245 2150
22 2512 2503
21 2538 3528
26 4366 8046
28 8159 8170
31.1094 2005
41
42
43
44
45
10.9028 6101
11.55703267
12 2.504 5463
12.9854 8191
13.7646 1083
13 2231 1938
14.0826 2214
li. 9979 9258
15.9728 6209
17.0110 9813
16.0226 6989
17.14425678
18.3443 5475
19.6284 5959
21.0024 5176
23.4624 8322
25 3394 8187
27.3666 4042
29.5559 7166
31.9204 4939
34 2302 6786
37.31753197
40.6761 0()84
44.3369 5973
48.3272 8610
46
47
48
49
50
14.5904 8748
15.4659 1673
16.3938 7173
17.3775 0403
18.4201 5427
18.11681951
19.2944 1278
20.5485 4961
21.8842 0533
23.3066 7868
22.4726 2338
24 0457 0702
25.7289 0651
27.5299 2997
29.4570 2506
34.4740 8534
37.2320 1217
40.2105 7314
43.4274 1899
46.9016 1251
52.6767 4185
5741764862
62,5852 3700
68.2179 0833
74.3575 2008
520
APPENDIXES
Table 3
PRESENT VALUE OF 1
n
Vs%
%%
3/a%
V*%
6/8%
1
2
3
4
5
0.9987 5156
0.9975 0468
0.9962 5936
0.99ft) 1059
0.9937 7337
0.9975 0023
0.99501869
0 9925 3734
0.99006219
0.9875 9321
0.9962 6401
0.9925 4198
0 9888 3385
0 9851 3958
0.9814 5911
0.9950 2488
0 9900 7450
0 9851 4876
0 9802 4752
0.9753 7067
0.9937 8882
0 9876 1622
0 9814 8196
0.9753 8580
0.9693 2750
6
7
8
9
10
0 9925 3270
0.9912 9359
0 9900 5602
09HHH1999
0.9875 8551
0.9851 3038
0 9h2h 7370
0.9WV2 231 1
0.9777 7869
0 9753 4034
0 9777 9238
09741 3936
0.9701 9999
0 !)668 7-121
0.9632 6198
0 9705 1808
0.9056 8963
0.9608 8520
0 9561 0468
0 9513 4794
0.06S3 06S'<
0 9573 23.jf>
0 9513 771.".
0 9454 6827
0 9395 9580
11
12
13
14
15
0.9863 5257
0 9851 2117
0.9838 9130
0.9826 6297
0 9814 3618
0 9729 0807
0.9701 8187
0 9680 6171
0 96r>6 4759
0.9632 3949
0 9596 6324
0 9560 7795
0.9525 0605
0.9489 4750
091.310224
0 9466 1 189
0 94l90r>34
0<)372 1921
01)325 5616
0.9279 1688
0 9337 .7)80
0 9279 600")
0 0221 9632
0.9164 6840
0.9107 7604
16
17
18
19
20
0.9802 1092
0.9789 8718
0 9777 0498
0.9765 4430
0.9753 2514
0 9608 3740
0.9584 4130
0.95605117
0.9536 6700
0.9512 8878
0 9118 7022
0.9383 5141
093484573
09313 5316
0.9278 7363
0.9233 0037
01)187 0684
0 9141 3616
0.9095 8822
0.9050 629O
0 9051 1905
08994 9710
0.893') 102')
0.8883 5802
0.8828 4027
21
22
23
24
25
0.9741 0750
0.9728 9139
0.9716 7679
0 9701 6371
0.9692 5215
0.9489 1649
09465 5011
0 9441 8964
0.9418 3505
0.9394 8634
09214 0711
0 9209 5353
0 9175 12S6
0.9140 8504
0.9106 7003
0 9005 6010
0 8960 7971
0 81)16 2160
0.8871 8507
0.8827 7181
0.8773 5670
0.8719 0736
0 8664 0170
0 8611 098.-)
0.85576135
26
27
28
29
30
0.9680 4210
0 9668 3355
0.9656 2652
09611 2100
0.9632 1697
0.9371 4318
0 9348 0646
0.9324 7527
0.9301 4990
0.9278 3032
0 9072 6777
0.9038 7823
0 90050135
0.8971 3709
0 8937 8539
0 8783 7901
0 8740 0986
0 8696 6155
0 86r>3 3188
0 8610 2973
0 8504 4606
0 HI 51 6378
0 8399 1432
0 8346 9746
0.8295 1300
31
32
33
34
35
0.9620 1446
0.9(508 1344
0.9590 1392
0 9581 1590
0.9572 1938
0.92551653
0.9232 0851
0.9209 0624
0 91860972
0.9103 1892
0 8904 4622
0 8871 19.V2
0 8838 0525
0.8805 033C
0.8772 1381
0.8567 4600
08521 8358
0 8182 4237
08410 2226
0.8398 2314
0.8243 6075
0.8192 4050
0 8141 5205
0 8090 9520
0.8010 6976
36
37
38
39
40
0 9560 2435
0.9548 3081
0 9">36 3876
0 9524 4820
0.9512 5913
0 91 10 3384
09117 5145
0 9094 8075
0.9072 1272
0.9049 5034
0 873!) 3f>:>r>
08706 7153
0 8674 1871
0.8611 7804
0.8609 4948
0.8356 4492
0 S3 14 8748
0 8273 5073
0 8232 34.").")
0.8191 3886
0.7990 7551
0 7941 1234
0.7891 7997
0 7842 7823
0.7794 0693
41
42
43
44
45
0 9500 7154
0.9488 8543
0.9477 0080
0.9465 1766
0.9453 3599
0.9026 9361
0.9004 4250
0 8981 9701
0 8959 5712
0.8937 2281
0 8577 3298
0 8545 2850
0.8513 3599
08181 5511
0.8449 8671
0.8150 6354
0 81 10 0850
0.8069 7363
0.8029 5884
0.7989 6402
0.7715 6500
0.7697 5493
0.7649 7384
0.7602 2245
0.7555 0057
46
47
48
49
60
0.9441 5579
0.9429 7707
0.94179982
0.9406 2404
0.9394 4973
0.8914 9407
0 8892 7090
0.8870 5326
0 8848 4116
0.8826 3457
0 8418 2985
0 8386 8478
0.8355 5146
0.8324 2985
0.8293 1990
0.7949 8907
0.7910 3390
0.7870 9841
0.7831 8250
0.7792.8607
0.7508 0802
0 7461 4462
0 7415 1018
0.7369 0453
0.7323 2748
j
PRESENT VALUE OF 1
521
n
3/4%
1%
1V8%
iy«%
!3/a%
1
2
3
4
5
0.9925 5583
0.9851 6708
0.9778 3333
0.9705 5417
0.9633 2920
0.9900 9901
0.9802 9605
0.9705 9015
0.9609 8034
0.9514 6569
0.9888 7515
0 9778 7407
0.9669 9537
0.9562 3770
0.9455 9970
0.9876 5432
0.9754 6106
0.9634 1833
0.9515 2428
0.9397 7706
0.9864 3650
0.9730 5C96
0.9598 5890
0.9468 3986
0 9339 9739
6
7
8
9
10
0.9561 5802
0 9490 4022
0.9419 7540
0.93 JO 6318
0.92800315
0.9420 4524
0.9327 1805
0 9234 8322
09143 3982
0 9052 8095
0 9350 8005
0.9246 7743
0 9143 9054
0.9042 1808
0 8941 5880
0.9281 7488
0.9167 1593
0.0053 9845
0.8942 2069
0.8831 8093
092132912
0 9088 3267
0.8965 0571
0.8843 4596
0,87235113
11
12
13
14
15
0 9210 9494
0.9142 3815
09074 3211
0.9006 7733
0.8939 7254
0.8963 2372
0 8874 4923
0 8786 G2(JO
0 8699 C297
0.8613 4947
08842 1142
0 8743 7470
0 8646 4742
0 8550 2835
0.8455 1629
0 8722 7746
0 8615 0860
0.8508 7269
0.8103 6809
0.8299 9318
0.8605 1899
0.8488 4734
0 8373 3400
0.8259 7682
0.8147 7368
16
17
18
19
20
0.8873 1766
0 8807 1231
08741 5(514
0 8676 4878
0.8611 8985
0 8528 2126
08113 7749
0 8360 1731
0 8277 3992
081954447
0.8361 1005
0 82(58 0846
08176 1034
0 8085 1455
0.7995 1995
0.8197 4635
0 8096 2602
0.7996 3064
0.7897 5866
0.7800 0855
0.8037 2250
0 7928 2120
0.7820 6777
0.7714 6020
0.7609 9619
21
22
23
24
25
0.8547 7901
0 8484 1589
0.8421 0014
0.8358 3140
0.8296 0933
08114 3017
0 8033 9621
0 79,14 4179
078756613
0.7797 6844
0.7906 2542
0 7S18 2983
0 7731 3210
0.76453112
0.7560 2583
0 7703 7881
0 7608 6796
0.7514 7453
0.7421 9707
0.7330 3414
0,7506 7472
0.7404 9294
0.7304 4926
0.72054181
0.7107 6874
26
27
28
29
30
0.8234 3358
0.81730380
0 8112 1906
0 8051 8080
0.7991 8690
0 7720 4796
0 7644 0392
0 7568 3557
074934215
0.7419 2292
074761516
0 7392 9806
0.7310 7348
0 7229 4040
0 7148 9780
0.7239 8434
0.7150 4626
0.7062 1853
0 6974 9978
0.6888 8867
07011 2823
0.6916 1847
0.6822 3771
0.67298417
0.6638 5015
31
32
33
34
35
0 7932 3762
0.7873 3262
0.7814 7158
0.77.')C 5418
0.7698 8008
0.73457715
072730411
0 7201 0307
0.7129 733 t
0.7059 1420
0.7069 4467
0 6990 8002
0 6913 0287
06836 1223
0.67600715
0.6803 8387
0 6719 8407
0.6636 8797
0.6554 9429
0.6474 0177
0.6518 5194
0.6459 6985
0.63720821
0.6285 6540
0.6200 3991
36
37
38
39
40
0.7611 4896
0.7584 6051
0 7528 1440
0.7472 1032
0.7416 4796
0.6989 2495
0 6920 0490
0 6851 5337
0.6783 6967
0 6716 5314
0 6684 8667
0.6610 4986
0 6536 9578
0.6464 2352
0.6392 3216
0.6394 0916
O.C315 1522
0.6237 1873
0.6160 1850
0.6084 1334
0.61103000
0.6033 3416
0.5951 5083
0.5870 7850
0.5791 1566
41
42
43
44
46
0.7361 2701
0.7306 4716
0.7252 0809
0.7198 0952
0.7144 5114
0 6650 0311
0.6584 1892
065189992
0 6454 4546
0.6390 5492
0.6321 2080
0.6250 8855
0.6181 3454
0.6112 5789
0.6044 5774
0.6009 0206
0.5934 8352
0.5861 5656
0.5789 2006
0.5717 7290
0.5712 6083
0.5635 1253
0.5558 6933
0.5483 2979
0.5408 9252
46
47
48
49
50
0.7091 3264
0.7038 5374
0.6986 1414
0.6934 1353
0.6882 5165
0.6327 2764
0.6264 6301
0 6202 6041
0.6141 1921
0.6080 3882
0.5977 3324
0.5910 8356
0 5845 0784
0.5780 0528
0.5715 7506
0.5647 1397
0.5577 4219
0.5508 5649
0.5440 5579
0.5373 3905
0.5335 5612
0.6263 1923
0.5191 8050
0.5121 3800
0.5051 9220
522
APPENDIXES
v- = _J—
T e + i »\ x
n
iya%
1%%
!3/4%
178%
2%
1
2
3
4
5
0.9852 2167
0.9706 6175
0.9563 1699
0.9421 8423
0.9282 6033
0.9840 0984
0.9682 7537
0.9527 9249
0.9375 5718
0.9225 6549
0,98280098
0.9658 9777
0.9492 8528
0.9329 5851
0.9169 1254
0.9815 9509
0 9635 2892
0.9457 9526
0.9283 8799
0.91130109
098039216
0.9611 6878
0.91232233
0.9238 4543
0.9057 30C1
6
7
8
9
10
0.9145 4219
0.9010 2679
0.8877 1112
0 8745 9224
0.8616 6723
0.9078 1352
0 8932 9744
0.87901317
0 8649 5791
0.8511 2709
0.9011 4254
0 8856 4378
0.8704 1157
0 8r>->4 4135
0.8407 2860
0 8045 2868
0.8780 6496
0 8619 0426
0.8400 4099
0.8304 6968
0.8879 7138
087056018
0.8534 9037
0.8367 5527
0.8203 4830
11
12
13
14
15
0 8489 3323
0 836.3 8742
0.8240 2702
0.8118 4928
0 7998 5150
0.83751743
0 8241 2539
0 8109 4750
0.7979 8032
0.7852 2048
0 8262 6889
0 8120 5788
0 7()8() 9128
0 7813 6490
0.7708 7459
0 8151 8496
0 80<)1 8156
0 7854 5429
0.7709 9808
0.7568 0793
080126304
0.7884 93 1H
0 7730 32r>3
0.7578 7502
0.7430 1473
16
17
18
19
20
0 7880 3104
0.7763 8526
0.76491159
0.7536 0747
0.7424 7042
0.7726 6468
0 7603 0965
0.7481 5218
0.7361 8911
0.7244 1732
0 7576 1631
0 744.3 8605
0 7317 7990
0.7191 9401
0.7068 2458
0.7428 7895
0 7292 0633
0.7157 8536
0.7026 1139
0.6896 7989
0.7284 4581
0 7141 6256
0.7001 5937
0.6864 3076
0.6729 7133
21
22
23
24
25
0.7314 9795
0 7206 8763
0.7100 3708
0.6995 4392
0.6892 0583
0 7128 3378
0.701 1 3545
06902 1938
0 6791 8267
0.6683 2243
0 6940 6789
0.6827 2028
067097817
0 659 1 3800
0.6480 9632
0.6769 8640
0 6G45 2653
0 6522 9598
0 6402 9053
0.6285 0604
0 6597 7582
0 6468 3904
0 (>341 5592
0 6217 2149
0.6095 3087
26
27
28
29
30
0.6790 2052
0 6689 8574
0 6590 9925
0.6493 5887
0.6397 6243
0 6576 3584
06471 2014
0.6367 7259
0.6265 9049
0.6165 7121
0 6369 4970
0.6259 9479
0 6152 2829
0.6046 4697
0.5942 4764
0 6169 3844
0 0().,5 8375
0.5944 3804
0 5834 9746
0.5727 5824
0 5975 7928
0 5858 6201
0 5743 7455
0.5631 1231
0.5520 7089
31
32
33
34
35
0.6303 0781
0.6209 9292
0.61181568
0 6027 7407
0.5938 6608
0.60671214
0.5970 1071
0 5874 6442
0 5780 7077
0.5688 2732
0.5840 2716
0 5739 8247
0 5011 1053
0.5544 0839
0.5448 7311
0.5622 1668
0 5518 6913
0 5417 1203
0.5317 4187
0.5219 5521
0 5112 4597
0 5306 3330
0.5202 2873
O..r>100 2817
0.5000 2761
36
37
38
39
40
0.5850 8974
0.5764 4309
0.5679 2428
0.5595 3126
0.5512 6232
0.5597 3168
0 5507 8148
05119 7440
0.5333 0814
0.5247 8046
0.5355 0183
0 52(52 9172
0 5172 4002
0.5083 4400
0.4996 0098
0.5123 4867
0.5029 1894
0 1936 6277
0.18457695
0.17565836
0.4902 2315
0.4806 1093
0.4711 8719
0.4619 4822
0 4528 9042
41
42
43
44
45
0.5431 1559
0.5350 8925
0.5271 8153
0.5193 9067
0.5117 1494
0 5163 8914
0.5081 3199
0.5000 0688
0.49201169
0.4841 4434
0 4910 0834
0 4825 6348
0 4742 6386
0.4661 0699
0.4580 9040
0 4669 0391
0.4583 1058
0.4498 7542
0.44159550
0.4334 6798
0.4440 1021
0.4353 0413
0.4267 6875
0.4184 0074
0.4101 9680
46
47
48
49
50
0.5041 5265
0.4967 0212
0.4893 6170
0.4821 2975
0.4750 0468
0.4764 0280
0.4687 8504
0.4612 8909
0.4539 1301
0.4466 5487
0.4502 1170
0.4 1246850
0.4348 5848
0.4273 7934
0.4200 2883
0.4254 9004
0.4176 5894
0.4099 7196
0.4024 2647
0.3950 1984
0.4021 5373
0.3942 6836
0.3865 3761
0.3789 5844
0.3715 2788
PRESENT VALUE OF 1
i
52,
vn =
(1 + i)"
n
2y«%
2V*%
23/8%
2V2%
2%%
1
0.9791 9217
0.9779 9511
0 9768 0098
0.9756 0976
0.9732 3601
2
0.9588 1730
0.9564 7444
0.9541 4015
0.9518 1440
0 9471 8833
3
0.9388 6639
0.9354 2732
0.9320 0503
0.9285 9941
0.9218 3779
4
0.91 <« 3061
0.9148 4335
0 9103 8342
090595064
0.8971 6573
5
0.9002 0133
0.8947 1232
0.88<J2 6342
0.8838 5429
0.8731 5400
6
0.881 1 7009
0.8750 2427
0.8686 3337
0 8622 968V
0.8497 8191
7
0 8031 2861
0 8557 6946
08 484 Sl!>3
0 SI 12 6524
0.8270 4128
8
0.81,11 6878
0 8369 3835
0 8287 9798
0 8207 4657
0 8049 0635
9
0 8275 8264
0.81 S3 2161
0 S0l>:> 70(57
0 8fX)7 2836
0 7833 (538 ">
10
0.8103 6244
0.8005 1013
0.7907 8942
0.7811 9840
0.7623 9791
11
0 7935 0056
0.7828 94 99
0.7721 4388
0.7621 4t78
0.74191)310
12
0 7769 8953
0 7(>.">(i(>748
0 7~»i:> 2H94
0.71355589
0.7221 3440
13
0.7M)S 2206
0 7lh8 11)05
0 7370 1 972
0 7254 2038
0.7028 0720
14
0 71499100
0 7o23 4137
0.71992158
0.7077 2720
0 6839 9728
15
0.7294 8935
0 7102 2628
0.7032 2010
0.6904 6556
0.6656 9078
16
0 7143 1026
0 700 1 6580
0 6869 0609
0 6736 2493
06478 7421
17
0.6994 4701
0 6850 5212
0 «.7<)«> 7053
0.0571 9506
0.6305 3454
18
068189303
0 <>W)9 77(>3
().h.Ml 0467
0.6411 6591
0.6136 5892
19
0.6706 4189
0 6552 3 184
06101 9993
0 6255 2772
0.59723196
20
0.6566 8729
0.64081(147
0.6253 4791
0.6102 7094
0.5812 5057
21
0.6(30 2305
0 6267 1538
0.6108 4045
0 5953 8629
0.5656 9398
22
0 6296 4313
0.6129 2457
0 59<>6 6955
0.5808 6467
0.5505 5375
23
0.61(>5 4K)2
05991 3724
0 58282710
0 5666 9724
0 5358 1874
24
0.6037 1273
0 5862 4668
0.5693 0637
0.5528 7535
0.5214 7809
25
0.5911 5077
0.5733 4639
0.5560 9902
0.5393 9059
0.50752126
26
0 5788 5021
0 5607 2997
05131 9807
0 5262 3472
0.4939 3796
27
0 5668 0559
054839117
0 5305 9(540
0 51 33 9973
0.4807 1821
28
0.5550 1159
0.5363 2388
0.5182 8708
0.5008 7778
0.4678 5227
29
0.5434 6300
0 52 i5 2213
0.5062 6333
0 18H6 6125
0 4553 3068
30
0.5321 5471
0.5129 8008
0.4945 1852
0.4767 4269
0.4431 4421
31
0.5210 8173
0 5016 9201
0.4830 4617
0.4651 1481
0.4312 8391
32
0.51023915
0. 1906 5233
0 4? 1 8 3997
0.4537 7055
0 4197 4103
33
0.4996 2217
0. 1 798 5558
0 4608 9374
044270298
0 4085 0708
34
0.4892 2612
0.4692 9641
0 n(>2 OH6
0.4319 0534
0 3975 7380
35
0.4790 4638
0.4589 6960
0.4397 5722
0.4213 7107
0.3869 3314
36
0.4690 7846
0.4488 7002
0.4295 5529
0.41109372
0.3765 7727
37
0.4593 1796
0.4389 9268
0 1195 JXJ02
0 4010 6705
0 3664 9856
38
0.4497 6055
0 4293 3270
0.4098 5594
0.3912 8492
0 3566 8959
39
0.4404 0201
0.4198 8528
0 4003 4769
0 3817 4139
0.3471 4316
40
0.4312 3819
0.4106 4575
0.3910 6001
0.3724 3062
0.3378 5222
41
0.4222 6506
0.40160954
0.3819 8780
0.3633 4695
0 3288 0995
42
0.4134 7864
0.3927 7216
0.3731 2006
0.3544 8483
0.32M 0968
43
0.4048 7505
0.3841 2925
0.364 \ 6990
0.3458 3886
03114 4495
44
0.3964 5047
0.3756 7653
0 3560 1455
0.3374 0376
0.3031 0944
45
0.3882 0120
0.3674 0981
0.3477 5536
0.3291 7440
0.2949 9702
46
0.3801 2357
0.3593 2500
0.3396 8778
0.3211 4576
0.2871 0172
47
0.3722 1402
0.3514 1809
033180735
0.3133 1294
0.2794 1773
48
0.3644 6906
0.3436 8518
0.3241 0975
0.3056 7116
0.2719 3940
49
0.3568 8524
0.3361 2242
0 3165 9072
0.2982 J576
0.26466122
50
0.3494 5924
0.3287 2608
0.3092 4612
0.2909 4221
0.2575 7783
524
APPENDIXES
1
v»
(1 + i)
n
3%
3*4%
3V2%
33/4%
4%
\
2
3
4
5
0.9708 7379
0.9425 9591
0.9151 4166
0 8884 8705
0.8626 0878
0.9685 2300
0.9380 3681
0.9085 1022
0.87991305
0.8522 1603
0.9661 8357
0.9335 1070
0.9019 4271
0.8714 4223
0.8419 7317
0.9638 5542
0.9290 1727
0.89.51 3834
0 8630 7310
0.8318 7768
0.9615 3846
0 9245 5621
0.8889 9636
0.85480419
0.8219 2711
6
7
8
9
10
0.8374 8426
0.8130 9151
0.7894 0923
0.7064 1673
0.7440 9391
0.8253 9083
0.7994 1000
0 7742 4698
0.7498 7601
0.7262 7216
0.8135 0064
0.7859 9096
0.7594 1156
0.7337 3097
0.7089 1881
0.80180981
0 7728 2874
0 71489517
071797125
O.G920 2048
0.7903 1 153
0.759917M
0 730(5 9021
0 7025 8674
0.6755 G417
11
12
13
14
15
0.7224 2128
0.7013 7988
0.6809 5134
0.6611 1781
0.6418 6195
0.7034 1129
0.6812 7002
0.6598 2568
0.635)0 5635
0.6189 4078
0.6849 4571
0 6617 8330
063940415
061778179
0.5968 9062
0 6670 0769
0 6428 9898
O.()l%(jl67
O.W26I26
0.5756 7639
0.6495 8093
0 0245 9705
0 6005 71 09
0.5774 7508
0.5552 6450
18
17
18
19
20
0 6231 6694
0.6050 1645
O.f>873 9461
0.5702 8603
0.5536 7576
0.5994 5838
0.5805 8923
0 5023 1402
0.54161407
0.5274 7125
0.5767 0591
0.5572 0378
05383 6114
0.5201 5569
0.502o 6588
0 5548 6881
0 5348 1331
0 5154 8271
0. 1968 5080
0.4788 9234
0 5339 0818
0 5133 7325
0.1936281-2
0.1746 4242
0.4563 8095
21
22
23
24
25
0 5375 4928
0.5218 9250
0.50669176
0.49193374
0.4776 0557
0 5108 6804
0.4947 8745
047921302
0.4641 2884
0.14951945
0.4855 7090
0.4691 5063
0.4532 8.363
0.13795713
0.4231 4699
0.4615 8298
0 1448 9925
0.4288 1856
04U3 1910
0.3983 7985
0 4388 3300
0.4219 5539
0 4057 2633
03901 2147
0.3751 KiSO
26
27
28
29
30
0.4636 9478
0 4501 8906
0.4370 7675
0.4243 4636
0.4119 8676
0 4353 6993
0.4216 6579
0.4083 9302
0 3955 3803
0.3830 8768
0.4088 3767
039501224
0.3816 5434
0.3687 4815
0.3562 7811
0.3839 8058
0..-S701 0176
<Ur>(i7 2159
0.3438 3093
0.331 1 0331
0 3606 8923
03168 1657
0.3331 7747
0 3206 5141
0.3083 1867
31
32
33
34
35
0.3999 8715
0.3883 3703
0.3770 2625
0.3660 4490
0.3553 8340
0 3710 2923
0 3593 5035
0.3480 3908
0.3370 8385
0.3264 7346
0.3442 3035
0.33258971
0.3213 4271
0.3104 7GOr>
0.2999 7686
0.3191 2187
0 3078 7940
0 2967 5123
0 '2860 2528
0.2756 8702
0.2964 6026
0.2850 5794
0.27409117
0.2635 5209
0.2534 1547
36
37
38
39
40
0.3450 3243
0.3349 8294
0 3252 2615
0.3157 5355
0.3065 5684
0.3161 9706
0.3062 4413
0 21)66 0448
0.2872 6826
0.2782 2592
0.2898 3272
0.28003161
0 2705 GUM
0.2014 1?50
0.2525 7217
0.2657 2242
0.2561 1800
0 2468 (5072
0.2379 3805
0.2293 3788
0.2436 6872
0 2342 9685
0 2252 8543
0 2166 2061
0 2082 8904
41
42
43
44
45
0.2976 2800
0.2889 5922
0.2805 4294
0.2723 7178
0.2644 3862
0.2694 6820
0 2609 8615
0.2527 7109
0.2448 1462
0.2371 0859
024403137
0 '2:^7 7910
0.2278 0590
0.2201 0-231
0.2126 5924
0.2210 4855
0 2130 5885
0.2053 5793
0.1979 3535
0.1907 8106
0.2002 7793
0.1925 7493
0,1851 6820
0.1780 4635
0.1711 9841
46
47
48
49
50
0.2567 3653
0.2492 5876
0.2419 9880
0.2349 5029
0.2281 0708
0.2296 4512
0.2224 1658
0 2154 1558
0 2086 3494
0.2020 6774
0/2054 6787
0.1985 1968
0 1918 0645
0.1853 2024
0.1790 5337
0.18388536
0.1772 3890
0.1708 3268
0.1646 5800
0.1587 0651
0.1646 1386
0.1582 8256
0.1521 9476
0.14634112
0.1407 1262
PRESENT VALUE OF 1
1
525
n
4V*%
4y2%
4*/4%
5%
6V2%
1
2
3
4
5
0.9592 3201
0 9201 2721
0 8826 1603
0.8466 3408
0.8121 11)02
0.9569 3780
0.9157 2995
0.8762 9660
0.8385 6134
0.8024 5105
0.9546 5394
0.9113 6414
0 8700 3737
0 8305 8460
0.7929 2086
0 9523 8095
0.9070 2948
0.8638 37(50
0 8227 0247
0.7835 2617
0.9478 6730
0 8984 5242
0 8516 136<i
08072 1674
0.7651 3135
6
7
8
9
10
0.7790 1105
0 7472 5281
0.71678926
0 6875 6764
0.6595 3780
0.7678 9574
0.7348 2846
0 7031 8513
0.6729 0443
0.6439 2768
0.7569 6502
0 7226 3964
0.6898 7077
0 <r>s5 8785
0.6287 2349
0.7462 1540
0.7106 8133
0.6768 3936
0 6446 0892
0.0139 1326
0.7252 4583
0 6874 3681
0 65159887
0.61762920
0.5854 3058
11
12
13
14
15
0.6326 4969
0.6068 5822
05821 1819
0 5583 8676
0.5356 2279
0.6161 9874
0 5896 6386
05012 7164
0 5399 7280
0.5167 2044
0 6002 1335
0.5729 9604
0 5470 1293
0 5222 0804
0.4985 2797
0.5846 7929
0.5568 3742
0 5303 2135
0.5050 6795
0.4810 1710
0.5549 1050
0.5259 8152
0.4985 6068
0.4725 6937
0.4479 3305
16
17
18
19
20
0 5137 8685
0 1928 4110
0 4727 4926
0 4531 7650
0.4349 8945
0 4944 6932
0 4731 7639
0.4528 0037
04333 0179
0.41464286
0 4759 2169
0.4543 1051
0.4337 3796
0.41406965
0.3952 9322
0.4581 1152
0. 1362 9669
0.4155 2065
0 3957 3396
0.3768 8948
0.4245 8101)
0. 4024 4653
0.3814 6590
0.3615 7900
0.3427 2890
21
22
23
24
25
0.4172 5607
0.4002 4563
0 3830 2866
0.3682 7689
0.3532 6321
0.3967 8743
0 3797 0089
0 3633 5013
0 3177 0347
0.3327 3060
0.3773 6823
0 3602 5607
0.3439 1987
03283 2116
03131 3624
0.3589 4236
0.3418 4987
0.3255 7131
0.3100 6791
0.2953 0277
0.3248 6158
0.3079 2567
0.2918 7267
0.27665(550
0.2022 3370
26
27
28
29
30
0 3388 6159
0.3250 4709
031179577
0.2990 8467
0.2868 9177
0.3181 0218
0 3046 9137
0 2915 7069
0 2790 1502
0 2670 0002
0.2992 2314
0.2856 5455
0 2727 0124
0 2003 3531
0.2485 3013
0.2812 4073
0 2678 4832
0.2550 9364
0.2429 4632
0.2313 7745
0.2485 6275
0.23560450
0.2233 2181
0.21167944
0.2006 4402
31
32
33
34
35
0.2751 9594
0.2639 7692
0.2532 1527
0.21289235
0.2329 902C
0 2555 0241
0 2 1 H 9991
023397121
0.2238 9589
0.2142 5444
0.2372 6027
0.2265 0145
0 2162 3050
0 2004 2530
0.1970 6473
0.2203 5947
0.2098 6617
0 1998 7254
0 1903 5480
0.1812 9029
0.1901 8390
0.18026910
0.17087119
0.1619 6321
0. 1535 1963
36
37
38
39
40
0.2234 9186
0 2143 8068
0 2056 4094
0.1972 5750
0.1892 1582
0 2050 2817
0 1961 9921
0.1877 5044
0.1796 6549
0.1719 2870
0.1881 2862
0 1795 9772
0.1714 5367
0 1636 7893
0.1562 5673
0.17265741
0.1641 3503
0.1566 0530
0.1491 4797
0.1420 4568
0.14551624
0.1379 3008
0.1307 3941
0.12392362
0.11746314
41
42
43
44
45
0 18150199
0.1741 0263
0.1670 0492
0.1601 9657
0.1536 6577
0 1645 2507
0.1574 4026
0.1506 0054
0.1441 7276
0.1379 6437
0.1491 7110
0 1424 0078
0.1359 4919
0.1297 8443
0.1238 9922
0.1352 8160
0. 1 L'88 3962
0.1227 0440
0.11686133
0.1112 9651
0.11133947
0.1055 3504
0 1000 3322
0.0948 1822
0.0898 7509
46
47
48
49
50
0.1474 0122
0 1413 9206
0.1356 2787
0.13009868
0.1247 9489
0.1320 2332
0.1263 3810
0 12089771
011569158
0.1107 0965
0.1182 8088
0.1129 1731
0.1077 9695
0.1029 0878
0.0982 4228
0.1059 9668
0.10094921
0.0961 4211
0.0915 6391
0.0872 0373
0.0851 8965
0.0807 4849
0.0765 3885
0.0725 4867
0.0687 6652
526
APPENDIXES
n
6%
6»/a%
7%
8%
9%
1
2
3
4
5
0.9433 9623
0.8899 9644
0.8396 1928
0.7920 9366
0.7472 5817
0.9389 6714
0.8816 5928
0.8278 4909
0.7773 2309
0.7298 8084
0.9345 7944
0.8734 3873
0.8162 9788
0.7628 9521
0.7129 8618
0.9259 2593
0.8573 3882
0.7938 3224
0.7350 2985
0.6805 8320
0.9174 3119
0.8416 7999
0.7721 8348
0 7084 2521
0.6499 3139
6
7
8
9
10
0.7049 6054
O.C650 5711
0.6274 1237
0.5918 9846
0.5583 9478
0.6853 3412
0.6435 0621
0.60123119
0.5673 5323
0.5327 2604
0.6663 4222
0.6227 4974
0.5820 0910
0.5439 3374
0.5083 4929
0.6301 6963
0.5834 9040
0.5402 6888
0.5002 4897
0.4631 9349
0.5962 6733
0.5470 3424
0.5018 GG28
0.4604 2778
0.4224 1081
11
12
13
14
15
0.5267 8753
0.4969 6936
0.4f>88 3902
0.4423 0096
0.4172 6506
0 5002 1224
0.4696 8285
0 4410 1676
0.4141 0025
0.3888 2652
0.4750 9280
0.4440 1196
0.4149 6445
0.3878 1724
0.3624 4602
0.4288 8286
0.3971 1376
0 367(5 9792
0 3404 6104
0.31524170
0 3875 3285
0.3555 3473
0 3261 7865
0 2992 4647
0 2745 3804
16
17
18
19
20
0.3936 4628
0 3713 6442
0.3503 4379
0 3305 1301
0.3118 0473
0 3650 9533
034281251
0.3218 8969
0.3022 4384
0.2837 9703
0.3387 3460
0 3165 7439
0.2958 6392
0.2765 0832
0.2584 1900
0.2918 9047
0 2702 6895
0.2502 4903
0.2317 1206
0.2145 4821
0 2518 6976
0.2310 7318
0.2119 9374
0 1944 8967
0.1784 3089
21
22
23
24
25
0.2941 5540
0.2775 0510
0.2617 9726
0.2169 7855
0.2329 9863
0.2664 7608
0.2502 1228
0.2349 4111
0.2206 0198
0.2071 3801
0.2415 1309
0 2257 1317
0.2109 4688
0.1971 4G62
0.1842 4918
0.1986 5575
0.1839 4051
0.1703 1528
0.1576 9934
0.1460 1790
0.1636 9806
(I 1501 8171
0.1377 8139
0 1264 0191
0.1159 6784
26
27
28
29
30
0.2198 1003
0.2073 6795
0 1950 3014
018155674
0.1741 1013
0.1944 9579
0.1826 2515
0.1714 7902
0.1610 1316
0.1511 8607
0.1721 9549
0.16093037
0.1501 0221
0.1405 6282
0.1313 6712
0.1352 0176
0 1251 8682
0.1159 1372
0.1073 2752
0 0993 7733
0.1063 9251
0 0976 0781
0 0895 4845
0 0821 5454
0.0753 7114
31
32
33
34
35
0.1642 5484
0.154!) 5740
0.1461 8622
0.13791153
0.1301 0522
0.1419 5875
0.1332 9460
0.1251 5925
0.1175 2042
0.1103 4781
0.1227 7301
0.1147 4113
0.1072 3470
0.1002 1934
0.0936 6294
0 0920 1605
0 0852 0005
0 0788 8893
0 0730 4531
0.0676 3454
0 0691 4783
0.0634 3838
0 0582 0035
0.0533 9481
0.0489 8607
36
37
38
39
40
0.1227 4077
0 1157 9318
0,10923885
0.1030 5552
0.0972 2219
0.1036 1297
0 0972 8917
0.0913 5134
0.0857 7590
0.0805 4075
0.0875 3546
008180881
0.0764 5686
0.0714 5501
0 0667 8038
0.0626 2458
0 0579 8572
0.0536 9048
0 0497 1341
0.0460 3093
0.0449 4135
004123059
0.0378 2623
0 0347 0296
0.0318 3758
41
42
43
44
45
0.0917 1905
0.0865 2740
0.0816 2962
0.0770 0908
0.0726 5007
0.0756 2512
0.0710 0950
0.0666 7559
0.0626 0619
0.0587 8515
0.0624 1157
0.0583 2857
0.0545 1268
0.0509 4643
0.0476 1349
0.0426 2123
0.0394 6411
0.0365 4084
0.0338 3411
0.0313 2788
0.0292 0879
0.0267 9706
0 0245 8446
0.0225 5455
0.0206 9224
46
47
48
49
50
0.0685 3781
0.0645 5831
0.0609 9840
0.0575 4566
0.0542 8886
0.0551 9733
0.0518 2848
0.0486 6524
0.0456 9506
0.0429 0616
0.0444 9859
0.0415 8747
0.0388 6679
0.0363 2410
0.0339 4776
0.0290 0730
0.0268 5861
0.0248 6908
0.0230 2693
0.0213 2123
0.0189 8371
0.0174 1625
0.0159 7821
0.0146 5891
0.0134 4854
AMOUNT OF ANNUITY OF 1
527
Table 4
AMOUNT OF ANNUITY OF 1
(1 + i)n - 1
"Sli j
n
1
2
3
4
5
1.000 0000
2 005 0000
3.015 0230
4.030 1001
5.050 2506
1%
1 0000000
20100000
3.030 1000
4.060 4010
5.101 0050
1 0000000
2 OT2 5000
3 037 656'-
4.075 6270
5.126 5723
1.000 0000
2.01 5 OlHW
3 043 2230
4.090 9034
5.152 2660
1.000 0000
2 017 5000
3 052 8M3
4.106 2304
5.178 0894
2%
1.0000000
2.020 1HXK)
3 0(50 4<K)0
4 121 6080
5.204 0402
6
7
8
9
10
6.075 5019
7.103 8794
8.141 4088
9.182 1158
10.228 0264
6.152 0151
7.213 5352
8.285 6706
9.368 5273
10.462 2125
6.190 6544
7.268 0376
8.358 8881
9.463 3742
10.581 6664
6.229 5509
7 322 9942
8 432 8391
9.559 3317
10.702 7217
6.268 7060
7 378 4083
8 507 5305
9.(>56 4122
10.825 3993
6.308 1210
7.134 2831
8 5829691
9 751 6'284
10.949 7210
11
12
13
14
15
11 279 1665
12 335 5624
13.397 2402
14.464 2264
15 536 5475
11.566 8347
12.682 5030
13 809 3280
14 947 1213
16 O')0 8955
11.713 9372
12.800 3614
14.021 1159
15.196 3799
16.386 3346
11 863 2625
13041 2114
14 236 8296
15.450 3821
16.682 1378
12.0148139
13.225 1037
14 456 5430
15.709 5325
16.984 4494
12.168 7151
13.412 0897
14 6803313
15.973 9382
17.293 4169
16
17
18
19
20
16 614 2303
17.697 3014
18.785 78~9
19 879 7168
20.979 1154
17 257 8615
18.4304431
19 611 7476
20 810 8930
22.019 0040
17 591 1638
18 811 0534
20016 1915
21 '2<)6 7689
22.562 9785
17.932 3698
19.201 3554
20 489 3757
21.75)6 7164
23.123 6671
18.281 6772
19 601 6066
20.911 6347
22 3 1 1 1 658
23.701 6112
18.639 2853
20.0120710
21.4123121
22.840 5586
24.297 3698
21
22
23
24
25
22 084 01 10
23 194 1311
24 310 4032
26.431 9552
26.559 1150
23239 1910
24 471 5860
25 716 3018
20.973 4648
28 243 1995
23 8450138
25 143 0785
26 437 3(»70
27 788 0840
29.135 4351
24.470 5221
25 837 5799
27.225 1186
28 633 5208
30 063 0236
25.1163894
26 335 92(52
28.0 '20 6549
29 511 0164
31.027 4592
25.783 3172
27 298 9835
28.844 9()32
30.421 8625
32.030 2907
26
27
28
29
30
27 691 9106
28.830 8702
29.974 5220
31 124 3946
32.280 0166
295256315
30.820 8878
32.1290967
33 450 3877
34.784 8915
30. 499 (3280
31.8808734
33 279 3843
31 695 3766
36.129 0688
31 513 9690
32 986 6783
84.481 4787
35 998 7009
37.538 6614
32.570 4397
31.1404224
35.737 8798
37 303 2927
39.017 1503
33.670 9037
35.344 3238
87.051 2103
38.792 2315
40.568 0792
31
32
33
34
35
33.441 4167
34 008 6238
35 781 6669
36 960 5752
38.145 3781
36.132 7404
37.494 0678
38 869 0085
40.257 6986
41.660 2756
37.580 6822
39 050 4407
40.538 5712
42.015 3033
43.570 8696
39.101 7616
40.688 2880
42 298 6123
43.983 0915
45 592 0879
40.699 9504
42.4121996
41.151 4131
45.927 1153
47.730 8398
42.3794108
4 1 227 029(5
46 1 1 1 5702
48.033 80 16
49994 4776
36
37
38
39
40
39.336 1050
40.532 7855
41 735 4494
42.944 1267
44.158 8473
43 076 8784
44 507 6471
45.952 7236
47 412 2308
48.886 3734
45115 5055
46 679 4493
48 262 9424
49.866 2292
51.489 5571
47.275 9692
48.985 1087
50 719 8854
52 480 6837
54.267 8939
49.566 1295
51.433 5368
53.333 623(5
65 2C6 9621
67.234 1339
51 994 3672
54 034 2545
5(5 1 1 1 939(5
58 237 2384
60.401 9832
41
42
43
44
45
45.379 6415
46.606 5397
47.839 5724
49.078 7703
56.324 1642
50.375 2371
51.878 9895
53 397 7794
54.931 7572
56.481 0747
53 133 1765
54 797 3412
56 482 8080
58.188 3369
59.9156911
56.081 9123
57.923 1410
59 791 9881
61.688 8679
63.614 2010
59 235 7312
61 272 3565
63 344 fi228
65.453 1537
67.598 5839
62.610 0228
64.862 2233
67 159 4678
69.502 6571
71.892 7103
46
47
48
49
50
51.575 7850
52 833 6639
54.097 8322
55.368 3214
56.645 1630
58.045 8855
59.626 3443
61.222 6078
62.834 8338
64.463 1822
61.664 6372
63.435 4452
65.228 3882
67.043 7431
68.881 7899
65.568 4140
67.551 9402
69.565 2193
71 608 6976
73.682 8280
69.781 5591
72 002 7364
74.262 7843
76.562 3830
78.902 2247
74.330 5645
76 817 1758
79.353 5193
81.940 5897
84.679 4016
528
APPENDIXES
(1 + i)° - 1
n
2V4%
2V3%
2'/4%
3%
3V2%
4%
1
2
3
4
5
1.0000000
2.022 6000
3.068 0063
4.137 0364
6.230 1197
1.0000000
2.025 0000
3.075 6250
4.152 5156
5.256 3285
1.0000000
2.027 5000
3.083 2563
4.168 0458
5.282 6671
10000000
2 030 0000
3.090 9000
4.1836270
5.309 1358
1.0000000
2.035 0000
3.106 2250
4.214 9429
5.362 4659
1.000 0000
2.040 0000
3.121 6000
4.246 4640
5.416 3226
6
7
8
9
10
6.347 7974
7.490 6228
8.659 1619
9.853 9930
11.075 7078
6.387 7367
7.547 4302
8.736 1159
9.954 5188
11.203 3818
6.427 9404
7.604 7088
8.813 8383
10 056 2188
11.332 7648
6.468 4099
7.662 4622
8.892 3361
10.159 1061
11.463 8793
6.550 1522
7.779 4075
9 Ool 6868
10 368 4958
11.731 3932
6.632 9755
7.898 2945
9 214 220.1
10.582 7953
12.006 1071
11
12
13
14
15
12.324 9113
13.602 2218
14.908 2718
16 243 7079
17.609 1913
12.483 4663
13.795 5530
15.1404418
16 518 9528
17.931 9267
12 6U 4159
13.992 1373
15.3769211
16 799 7864
18.261 7805
12 807 7957
11 1920296
15.617 7905
17.086 3242
18.598 9139
13.141 9919
14.601 9616
16.1130303
17 676 9864
19.295 6809
13.486 3514
15.025 8035
16.626 8377
182919112
20.023 5876
16
17
18
19
20
19.0053981
20.433 0196
21.892 7625
23.385 3497
24.911 5200
19.380 2248
20.8(54 7305
22 386 3187
23 91 6 0074
25.544 6576
10.763 9795
21 307 4889
22 893 4449
24.5230116
26.197 3975
20.156 8813
21.701 5877
23 414 4354
2.') 1168684
26.870 3745
20 971 0297
22 705 0158
24 499 6913
26 357 1805
28.279 6818
21.8245311
23 097 5121
25 6454129
27 671 22' n
29.778 0786
2l
22
23
24
25
26 472 0292
28 067 6499
29 699 1720
31.3674034
33.073 1700
27.183 2711
28 8(i2 8559
30 584 4273
32 349 0380
34.157 7639
27.917 8259
29.08,-) 5662
31 501 9192
33 368 2220
35.285 8481
28 676 4857
30 53(5 7SOH
32 152 8837
34 426 4702
36.459 2643
30 269 4707
32 328 9022
31 1604137
36 606 5282
38.949 8567
31.969 2017
34 247 9f>98
36.617 8886
39.082 6041
41.645 9083
26
27
28
29
30
34.8173163
36.600 7059
38 424 2218
40 28H 7668
42.195 2640
36.011 7080
37.912 0007
39 859 8008
41.8562958
43.902 7032
37.2.56 2089
39.280 7547
41.360 9754
43.498 4022
45.694 6083
38.553 0423
40.709 6335
42 930 922.1
45 218 8:j02
47.575 4157
41.313 1017
43.759 0602
46.290 6273
48 910 79U3
51.622 6773
44.311 7446
47 084 2144
49.967 5830
52.966 2803
56.084 9378
31
32
33
34
35
44.144 6575
46.137 9123
48.1760153
60.259 9756
62.390 8251
46.000 2707
48.150 2775
50351 0345
52.612 8853
64.928 2074
47.951 2100
50.269 8683
52 652 2897
55.100 2277
57.615 4839
50 002 6782
52.502 7585
55.0778113
57.730 1765
60.462 0818
54.429 4710
57 334 5025
60.341 2101
63 453 152 1
66.674 0127
59.328 3353
62.701 4687
firt 209 5274
69 857 9085
73 052 2249
36
37
38
39
40
54.569 6186
56.797 1351
59.075 3774
61.4045733
63.786 1762
57.301 4126
59.733 9479
62.227 2966
64 782 9791
67.402 5535
60.199 9097
62.8:>r> 4072
65.583 9309
68.387 4890
71.2(58 1450
63.275 9443
66.174 2226
69.1594193
72 2H4 2328
75.401 2597
70.007 6032
73 457 8693
77.028 8947
80 724 90(50
84.550 2778
77.598 3139
81.702 2464
85 970 3363
90 409 1497
95.025 5157
41
42
43
44
45
66.221 3652
68.711 3459
71.257 3512
73.860 6416
76.522 5061
70.087 6174
72.839 8078
75.600 8030
78.552 3231
81.516 1312
74.228 0190
77.26') 2895
80 394 1950
83 605 0353
86.904 1738
78.663 2975
82.023 1965
85.483 8923
89.048 4091
92.719 8614
88.509 5375
92.607 3713
96.848 6293
101.238 3313
105.781 6729
99.826 53(s3
104.819 5978
110.0123817
115.412 8770
121.029 3920
46
47
48
49
50
79.244 2624
82.027 2583
84.872 8717
87.782 5113
90.757 6178
84.554 0344
87.667 8853
90.859 5824
94.131 0720
97.484 3488
90.294 0386
93.777 1246
97 355 9056
101.033 2854
104.811 7008
96 501 4572
100.396 5010
104.408 3960
108.540 6479
112.796 8673
110.484 0315
115.350 9726
1203882566
125.601 8456
130.997 9102
126.870 5677
132 945 3904
139.263 2060
145.833 7343
152.667 0837
AMOUNT OF ANNUITY OF 1
(1 + i)n - i
529
S51i
i
n
4Va%
5%
5V2%
6%
7%
8%
1
1.0000000
1.000 0000
1.0000000
1.0000000
1.000 0000
1.0000000
2
2.045 0000
2.050 0000
2.053 0000
2.060 0000
2.070 0000
2.080 0000
3
3.137 0250
3.152 5000
3.168 0250
3.183 6000
3 214 9000
3.246 4000
4
4.278 1911
4 310 1250
4.342 2664
4.374 6160
4.439 9430
4.506 1120
5
5.470 7097
5.525 6313
5.581 0910
5.637 0930
5,750 7390
5.866 6010
6
6.716 8917
6801 9128
6 888 0510
6.975 3185
7.153 2907
7.335 92vO
7
8.0191518
8.1420085
8.206 8938
8.393 8377
8.6540211
8.922 8034
8
9.380 0136
9.549 1089
9.721 5730
9.897 4679
10.259 8026
10.63C 6276
9
10602 1142
11 026 5643
11 2562595
11.491 3160
11.977 9888
12.487 5578
10
12.288 2094
12.577 8925
12.8753538
13.180 7949
13.816 4480
14.486 5625
11
13.841 1788
14.206 7872
14 583 4983
14.971 6426
15.783 5993
16.645 4875
12
15.464 0318
15.917 1265
10.385 5907
168699412
17.888 4513
18.977 1265
13
17.159 9133
17.712 9829
18 '286 7981
18.882 1377
20 140 0429
21.495 2966
14
18.932 1094
19.598 6320
20.292 5720
21.0150659
22.550 4879
24.214 9203
15
20.784 0543
21,578 5636
22.408 0635
23.275 9699
25.129 0220
27.152 1139
16
22 719 3367
23.657 4918
24 641 1400
25 672 5281
27 888 0536
30.324 2830
17
24 741 7069
25 840 3004
20.990 4027
282128798
30.840 2173
33.750 2257
18
20 855 0837
28.132 3847
29.481 2048
30 905 6526
33.999 0325
37.450 2437
19
29.003 5025
30 539 0039
321020711
33.7599917
37 378 9648
41.446 2632
20
31.371 4228
33.005 954 1
34.8083180
30.785 5912
40.995 4923
45.761 9643
21
33 783 1368
35.719 2518
37.786 0755
39.992 7267
44.865 1768
50.422 9214
22
36.303 3780
38 50:> 2144
40 8(54 3097
43 392 2903
49.005 7392
55.456 7552
23
38.937 0300
41 130 4751
44 111 8107
46.995 8277
53 436 1409
60.893 2950
24
41.0891903
44 501 9989
47 537 9983
50.81o 5/74
58.1766708
66.764 7592
25
44.565 2102
47.727 0988
51.152 588ii
54.864 5120
63.249 0377
73.105 9400
26
47.570 6146
51.1134538
54.965 9805
69.156 8827
68.676 4704
79.954 4152
27
50.711 3236
r>4 6091265
58 989 1094
63.705 7657
74.483 8233
87.350 7684
28
53.993 3332
58 102 5828
63 233 5105
68628 1116
80.697 6909
95.338 8298
29
57.423 0332
62.3227119
67.711 3535
73.639 7983
87.3165298
103.965 9362
30
61.007 OG97
66.438 8475
72.435 4780
79.058 1862
94.400 7803
113.283 2111
31
64.7523878
70 760 7899
77.4194293
84.801 6774
102 073 0414
123.345 8680
32
68.666 2452
75.298 8294
82.677 4979
90.889 7780
110 218 1543
134.213 6374
33
72 756 2263
80 063 7708
8H 224 7603
97.343 1647
118.933 4251
145 950 6204
34
77.030 2565
85 066 9594
94.077 1221
104.183 7546
128.258 7648
158.0266701
35
81.4966180
90.320 3074
100.251 3638
111.434 7799
138.236 8784
172.310 8037
36
86.163 9658
95 836 3227
106 765 1888
119.120 8667
i 48.91 3 4598
187.1021480
37
91 Oil 3*43
101.628 1389
113637 2742
127.268 1187
160.337 4020
203.070 3198
38
96.138 2048
107.709 5458
120 887 3243
135901 2058
172 561 0202
220.3159454
39
101.464 4240
1140950231
128.530 1271
145.058 4581
185.640 2916
238.941 2210
40
107.030 3231
120.799 7742
136.605 6141
151.701 9656
199.635 1120
259.056 5187
41
112.846 6876
127 839 7630
14o.ll8 9229
165.047 6836
214.609 5698
280.781 0402
42
118.924 7885
135 23 1 7511
154.100 4636
175.950 5446
230.632 2397
304.243 5234
43
125 276 4040
142.993 3387
163.575 9891
187.507 5772
247.776 4695
329.583 0053
44
131.913 8422
151.143 0056
173.5726685
199.758 0319
266.120 8513
356.949 6457
45
138.849 9651
159.700 1559
184.119 1653
212.743 5138
285.749 3108
386.505 6174
46
146.098 2135
168.685 1637
195.245 7194
226.508 1246
306.751 7626
418.4260668
47
153.672 6331
178.119 4219
206.984 2339
241.0986121
329.224 3860
452.900 1521
48
161.5879016
188 025 3929
219.368 3668
256.564 5288
353.270 0930
490.1321643
49
169.859 3572
198.426 6626
232.433 6270
272.958 4006
378.998 9995
530.342 7374
50
178 503 0283
209 347 9957
246.217 4765
290.335 9046
406.528 9295
673.770 1564
510
APPENDIXES
Table 6
PRESENT VALUE OF ANNUITY OF 1
1 *
1~(T+l)i
n
y*%
1%
1V4%
ll/2%
!3/4%
2%
1
2
3
4
5
0.99.r> 0249
1 .985 0994
2 970 2481
3.950 4957
4.923 8663
0.990 0990
1.970 3951
2.940 9852
3 901 9656
4.853 4312
0.987 6543
1.963 1154
2.926 5337
3.878 0580
4 817 8360
0.985 2217
1.955 8834
2.912 2004
3.854 3847
*.782 6450
0.982 8010
1.948 6988
•2 897 9840
3 830 9425
4.747 8551
0.980 3922
1 941 5609
2.883 8833
3 807 7287
4.713 4595
6
7
8
9
10
5 896 3844
6 862 0740
7.822 9592
8.779 0639
9.730 4119
5.795 4765
6.728 1945
7.651 6778
85660176
9.471 3045
5.746 0099
6 662 7258
7 568 1243
8 462 3150
9.345 5259
5 697 1872
6 598 21 10
7 485 9251
8 300 5173
9.222 1846
5.618 9976
6534 641!
7.405 0530
8 200 4943
9.101 2229
5,601 4309
6 471 9911
7.325 4814
8 102 2367
8.982 5850
11
12
13
14
15
10.677 0267
11 618 9321
12 556 1513
13.488 7078
14.416 C246
10.367 6282
11.255 0775
12 133 7401
13 003 7030
13.865 05?5
10.217 8034
11.079 3120
11 930 1847
12.770 5528
13.600 5459
10.071 1178
10.907 5052
11.731 5322
12.543 3815
13.343 2330
9.927 4918
10.739 5497
11.537 6410
12 322 0059
13 092 8805
9.786 8481
10 575 3412
11 348 3738
12 106 2488
12.849 2035
16
17
18
19
20
15.339 9250
16 258 6319
17.172 7680
18.082 3562
18.987 4192
14.717 8738
15 562 2513
16 398 2686
17.226 0085
18.045 5530
14.420 2923
15 22» 9183
16 029 5489
16.819 3076
17.599 3161
14 131 2641
14 907 0493
15 672 5609
16.420 1684
17.1686388
13.850 4968
14 5'>5 0828
15.326 8627
160460567
16.752 8813
13.577 7093
14.291 8719
14.9U2 0313
15 678 4620
16.351 4333
21
22
23
24
25
19 887 9792
20.784 0590
21 675 6tJ06
22.562 8(562
23.445 6380
18 856 9831
19.660 3793
20.455 8211
21.243 3873
22.023 1557
18 369 6950
19.130 5629
19 882 0374
20.624 2345
21.357 2686
17.900 1367
18.620 8244
19 330 8615
20.030 4054
20.7196112
17.447 5492
18.130 2095
18 801 2476
19 400 6857
20.108 7820
17 Oil 2092
17.658 0482
18 292 2041
18.913 9250
19.523 4505
26
27
28
29
SO
24.324 0179
25.198 0278
?(J.067 6894
26.933 0242
27.794 0540
22.795 2037
23.551) 6076
24 ,*Ki 1132
25.065 7853
25.807 7082
22 081 2530
22 790 2992
23 502 5178
24.200 0176
24.888 9062
21 398 6317
22.067 6175
22 726 7167
23.376 075d
24.015 8380
20.745 7317
21 371 7264
21.9869547
22 591 6017
23.185 8493
20.121 0358
20 706 8978
21.281 2721
21 844 3847
22.396 4556
31
32
33
34
35
28.650 8000
29.503 2836
30.351 5259
31.195 5482
32.035 3713
26.542 2854
27.269 5895
27.989 6926
28.702 6659
29.408 5801
25.569 2901
20 241 2742
26.904 9622
27.560 4564
28 207 8582
24.646 1458
25.267 1387
25.878 9544
26.481 7285
27.075 5946
23.769 8765
24.343 8590
24.907 9695
25.462 3779
26.007 2510
22.937 7015
23.468 3348
23.988 5636
24.498 5917
24.998 6193
36
37
38
39
40
32.871 0162
33.702 5037
34.529 8544
35.353 0890
36.172 2279
30 107 5050
30.799 5099
31.484 6633
32 163 0330
32.834 6861
28 847 2674
29.478 7826
30 102 5013
30.718 5198
31.326 9332
27.660 6843
28 237 1274
28.805 0516
29.364 5829
29.915 8452
26 542 7528
27.009 0446
27.586 2846
28.094 6286
28.594 2296
25.488 8425
25.969 4f>34
26.440 6406
26.902 5888
27.355 4792
41
42
43
44
45
36.987 2914
37.798 2999
38.605 2735
39.408 2324
40.207 1964
33.499 6892
34.158 1081
34.810 0081
35.455 4535
36.094 5084
31.927 8352
32.521 3187
33.107 4753
33.686 3954
34.258 1682
30.458 9608
30.994 0500
31.521 2316
32.040 6222
32.552 3372
29.085 2379
29.567 8014
30.042 0652
30.508 1722
30.966 2626
27.799 4895
28 234 7936
28.661 5623
29.079 9631
29.490 1599
46
47
48
49
50
41.002 1855
41.793 2194
42.580 3178
43.363 5003
44.142 7864
86.727 2361
37.353 6991
37.973 9595
38.588 0787
39.196 1175
34.822 8822
35.380 6244
35.931 4809
36.475 5867
37.012 8757
33.056 4898
33.553 1920
34.042 5537
34.524 6834
34.999 6881
31.416 4743
31.858 9428
32.293 8013
32.721 1806
33.141 2095
29.892 3136
30.286 5820
30.673 1196
31.052 0780
31.423 6059
PRESENT VALUE OF ANNUITY OF 1
i
531
";
^ (1
. + i)n
3i "
i
•
|
n
2J/4%
2Va%
2%%
3%
3Va%
... «
4%
1
0.977 9951
0.975 6098
0.973 2360
0.970 8738
0 966 1830
OJ615385
2
1.934 4696
1.927 4242
1.920 4243
1.913 4697
1.899 6943
1 886 0947
3
2 869 8969
2.856 0236
2.842 2621
2 828 fil!4
2 801 6370
2 775 0910
4
3.784 7402
3.761 5)742
3.739 4279
3.717 0984
3.673 0792
3.629 8952
5
4.679 4525
4.645 8285
4.612 5819
4.579 7072
4.515 0524
4.451 8223
6
5.554 4768
5.508 1254
5.462 3668
5.417 1914
5 328 5530
5242 1369
7
6.410 2463
6.349 3906
6 289 4081
6 230 2830
6 114 5440
6 002 0547
8
7.247 1846
7.170 1372
7.094 3144
7.019 6922
6.873 9555
6.732 7449
9
8.065 7062
7.970 8655
7.877 6783
7.786 1089
7.607 6865
7.135 3316
10
8.866 2164
8.752 0639
8.640 0762
8.530 2028
8.316 6053
8.110 8958
11
9.649 1113
9.514 2087
9 382 0693
9 252 6241
9 001 5510
8 760 4767
12
10.414 7788
10.257 7646
10.104 2037
9.954 0040
9.663 3343
9.385 0738
13
11 163 5979
10.9831850
108070109
10.634 9553
10.302 7385
9.985 6479
14
11.895 9392
11.690 0122
11.491 0081
11.296 0731
10.920 5203
10.563 1229
15
12 612 1655
12.381 3777
12.156 6989
11 937 9351
11.517 4109
11.118 3874
16
13.312 6313
13.055 0027
12 804 5732
12 561 1020
12.094 1168
11.652 2956
17
13 997 6834
13.712 1977
13.435 1077
13.106 1185
12.651 3206
12.165 d689
18
14 667 6611
14 353 3636
11.018 7666
13.753 5131
13.1896817
12.659 2970
19
15.322 8959
14.978 8013
14 646 0016
14.323 7991
13.709 8374
13.133 9394
20
15.963 7124
15.589 1623
15.227 2521
14.877 4749
14.212 4033
13.590 3263
21
16.590 4278
16.181 5486
15.792 9461
154150241
14 697 9742
14.029 1600
22
17.203 3523
16 765 4132
16 343 4999
15.9369166
15 167 1248
14.451 1153
23
17.802 7896
17 332 1105
168793186
16443 (K)84
15 620 4105
14.856 84 17
24
18 389 0362
17.884 9858
17.400 7967
16 935 5421
16.058 3676
15.246 9631
25
18.962 3826
18.424 3764
17.908 3180
17 413 1477
1(U81 5140
15.622 0799
26
19.523 1126
18.9506111
18 402 2559
17 8768424
16.890 3523
15 9H2 7692
27
20.071 5038
19.464 0109
18.8H2 9741
18327 0315
17 285 3645
18 32^5858
28
20. (507 8276
19 Wi4 8887
19.350 8264
18 764 1082
176670189
16.663 0632
29
21.1323198
20.453 5499
19 80(5 1571
19 188 4546
18 035 7670
16983 714<5
30
21.645 3299
20/J30 2926
20.249 3013
19.600 4414
18.392 0454
17.292 0333
31
22.147 0219
21.395 4074
20.680 5852
20 000 4285
18 736 2758
17.588 4936
32
22.637 6742
21.819 1780
21.100 3262
20.388 7655
19.068 8655
17873 5515
33
23 117 5298
22 291 8809
21.508 8333
20 705 7918
19.390 2082
18 147 0457
34
23.586 8262
22.723 7863
21.906 4071
21.131 8367
19.700 6842
18.111 I97H
35
24.045 7958
23.1451573
22.2U3 3403
21.487 2201
20.000 6611
18.661 (5132
36
24.494 6658
23.556 2511
22.6699175
21.8322525
20.290 4938
18.908 2820
37
24.933 6585
23 057 3181
23 03C 4161
22 167 2354
20.570 5254
10 142 57HH
38
25 362 9912
24.348 6030
23.393 1057
22.492 4616
20.841 0874
19.367 8642
39
25.782-3765
24.730 3444
23 740 2488
22.808 2151
21.1024999
19.584 4H48
40
26.193 5222
25.102 7751
24.078 1011
23.114 7720
21.355 0723
19.792 773U
41
26.505 1317
25.466 1220
24 40t) 9110
23.412 4000
21.599 1037
19.9930518
42
26.987 9039
25.820 6068
24.726 9207
23.701 3592
21.834 8828
20.185 (>267
43
27.372 0332
26.166 4457
25 038 3656
23 981 9021
22.062 6887
20 370 7949
44
27.747 7097
26 503 8495
25.341 4751
24.254 2739
22.282 7910
20.548 8413
45
28.115 1195
26.833 0239
25.636 4721
24.518 7125
22.495 4503
20.720 0397
46
28.474 4445
27.1541696
25 923 5738
24.775 4491
22.700 9181
20.884 6536
47
28.825 8626
27.467 4826
26 202 9915
25.024 7078
22.899 4378
21.042 9361
48
29.169 5478
27.773 1537
26.474 9309
25 266 7066
23.091 2443
21.195 1309
49
29.505 6702
28.071 3695
26.739 5922
25 501 6569
23.276 5645
21.341 4720
50
29.834 3963
28.362 3117
26.997 1700
25.729 7640
23.455 6179
21.482 1846
532
APPENDIXES
i
l~
n
4y3%
5%
5'/2%
6%
7%
8%
1
2
3
4
5
0.956 0378
1.872 6678
2.748 9644
3.587 5257
4.389 9767
0.952 3810
1.859 4104
2.723 2480
3.545 9505
4.329 4767
0.947 8673
1 8463197
2 ($97 9334
3 505 1501
4.270 2845
0.943 3962
1 S33 3927
2.073 0120
3 1(5.') 1056
4.2123638
0 934 5794
1.808 0182
2.624 3160
33872113
4.100 1974
0.925 9259
1.783 2648
2 577 0970
3.312 12(58
3.992 7100
6
7
3
9
10
5.157 8725
5.892 7(309
6 505 8861
7.268 7905
7.912 7182
5.075 6923
5.78G 3731
0.4(53 2128
7.107 8217
7.721 7349
4 995 5303
5.682 9671
6 334 5660
G952 1953
7.537 6258
4 9173243
5.5H2 3814
6 U09 7(J38
0.801 6923
7.300 0871
4.7(56 5397
5 389 2894
5.971 2985
65152323
7.023 581C
4.622 8797
5.206 3701
5 746 6389
6 246 887!)
6.710 0814
11
12
13
14
15
8.528 9169
9 118 5808
9.682 8524
10.222 8253
10.739 5457
8.306 4142
8.863 2516
9.393 5730
9 808 6 109
10.379 6580
8 092 5363
8.618 5179
9.1170785
9 589 6479
10.037 5809
7.886 8746
8 383 8439
8.852 6830
9 29 1 9839
9 712 2190
7.498 6744
7.942 6863
8.357 6508
8.745 4680
9.1079140
7.138 9643
7 536 0780
7.903 7759
8.241 2370
8.559 4787
16
17
18
19
20
11.234 0151
11.707 1914
12.159 9918
12.593 2936
13.007 9365
10.837 7696
11.274 0663
11 689 5869
12.085 3209
12.462 2103
10 462 1620
10.861 6086
11.21(5 0745
11. (5076535
11.950 3825
10 1058953
10 477 2597
10.827 (3035
11.158 1165
11.4699212
9 446 6486
9.763 2230
10.059 0809
10.335 5953
10 594 0143
8.851 3692
9.121 6381
9 371 8871
9 603 5992
9.818 1474
21
22
23
24
25
13.404 7239
13.784 4248
14.147 7749
14.495 4784
14.828 2090
12.821 1527
13.1(53 0026
13.488 5739
13.798 6418
14.093 9446
12.275 2441
12 583 1697
12.8750124
13.151 6990
13.413 9327
11.7640766
12041 5817
12.303 3790
12 5503575
12.783 3562
10 835 5273
11.001 2405
11.2721874
11.469 3340
11.653 5832
10 016 8032
10.2007137
10.371 0590
K) 528 7583
10.674 7762
26
27
28
29
30
15.146 6115
15.451 3028
15.742 8735
16.021 8885
.6.288 8885
14.375 1853
14.6130336
14.898 1273
15.141 0736
15.372 4510
13 662 4954
13 898 0999
14 121 4217
14.333 101?
14.533 7452
13 003 1662
13 210 5341
13.406 1643
13 590 7210
13.764 8312
11.8257787
11.986705)1
12 137 1113
12.277 6741
12.40U 0412
10.809 9780
10935 1618
11.051 078.3
11.158 40fiO
11.257 7833
31
32
33
34
35
16.544 3910
16.788 8909
17.022 8621
17.246 7580
17.461 0124
15.592 8105
15.802 6767
16.002 5492
16 192 9040
16.374 1943
14.723 9291
14.904 1982
15.075 0694
15 237 0326
15.390 5522
13.929 0860
14 084 0434
1 1 230 2290
14.368 1411
14.498 2464
12531 8112
126465553
12 753 7900
12 854 0094
12.947 6723
11.349 7994
11 434 9994
11 5138884
11 5869337
11.654 5682
36
37
38
39
40
17.666 0406
17.862 2398
18.049 9902
18.229 6557
18.401 5844
16.546 8517
16.711 2873
16 867 8927
17.017 0407
17.159 0864
15.536 0684
15.673 9985
15 804 7379
15 928 6615
16.046 1247
14.G20 9871
1 1 736 7803
14 8160192
14.949 0747
15 046 2969
13.035 2078
131170166
13.193 4735
13.264 9285
13 331 7089
11.717 1928
11 775 1785
11.828 H690
11 878 5824
11.924 6133
41
42
43
44
46
18.566 1095
18.723 5498
18.874 2103
19.018 3831
19.156 3474
17 294 3680
17.423 2076
17.545 9120
17.662 7733
17.774 0698
16.157 4642
16 2(52 9992
16.363 0324
16 457 8506
16 547 7257
15 138 0159
15.224 5433
15.306 1729
15.383 1820
15.455 8321
13.394 1*204
13.452 4490
13.506 9617
13.557 9081
13.603 5216
11.967 2346
12.006 6987
12.043 2395
12.077 0736
12.108 4015
46
47
48
49
50
19.288 3707
19 414 7088
19.535 6065
19.651 2981
19.762 0078
17.880 0665
17.981 0157
18.077 1578
18 168 7217
18.255 9255
16.632 9154
16.713 6639
16.790 2027
16 862 7514
16.931 5179
15 524 3699
15.589 0282
15.650 0266
15.707 5723
15.761 8606
13 650 0202
13.691 6077
13.730 4744
13.766 7986
13.800 7463
12.137 4088
12.164 2674
12.189 1365
12.212 1634
12.233 4846
RENT OF PRESENT VALUE OF ANNUITY OF 1 533
Table 6
RENT OF PRESENT VALUE OF ANNUITY OF 1
! L
n
*%
1%
ivi%
iy.%
1H%
2%
1
1.0050000
1 010 0000
1 012 5000
1.0150000
1.017 5000
1.020 0000
2
0 503 7531
0507 5121
0.509 394 i
0.511 27 7;>
0.513 1630
0.515 0195
3
0.336 6722
0.340 0221
0.311 7012
0.343 3830
0.345 0675
0 346 7547
4
0.253 1328
0.256 2811
0 257 8610
0 259 4448
0.261 0324
0.262 6238
5
0.203 0100
0.206 0398
0.207 5621
0.209 OS93
0.210 6214
0.212 1584
6
0.169 5955
0 1725184
0 174 0338
0 175 5252
0.1770226
0 178 5258
7
0.145 7285
0.1486283
(U5008S7
0.151 5562
0.1530306
0.151 5120
8
0.127 8289
0 130 6903
0.132 1331
0.133 5840
0.135 0429
0.136 5098
9
0 113 9074
0.116 7101
0.118 1700
0.119 6098
0.121 0581
0.122.5154
10
0.102 7706
0.105 5821
0.107 0031
0.108 4342
0.109 8764
0.111 3265
11
0.093 6590
0 096 4541
0.097 8684
0 099 2938
0.100 7304
0.102 1779
12
0.086 0()64
0 08* 8188
0.090 25cS3
0 091 6800
0.093 1138
0.094 5590
13
0 079 0122
0 082 4148
0.083 8210
0.085 2 10 1
0 086 6728
0.088 1184
14
0074 1361
00769012
0 078 3052
0.079 7233
0.081 1556
0.082 6020
15
0.069 3644
0.072 1238
0.073 52(15
0.074 9444
0.076 3774
0.077 8255
16
0.065 1894
0 067 9446
0.069 3467
0 070 7051
0.072 1996
0.073 6501
17
0.061 5058
0 064 25H1
0 065 6fi02
0 0(37 0797
0.068 51(52
0.069 9698
18
0.058 2317
0 OtiO 9820
0 0(52 3848
0.063 8058
0.065 2449
0.066 7021
19
0 055 3025
0.0580518
0 059 4555
0 000 8785
0.062 3206
0.063 7818
20
0052 600 1
0 055 4153
0 05(5 8204
0.058 2457
0.059 6912
0.061 1567
21
0.050 2816
0 053 0308
0 054 4375
0 055 8655
0 057 3146
0.058 7818
22
0048 1138
0 050 8(>37
0 052 2724
0 053 7033
0.055 1564
0.056 6314
23
0.046 1346
0 048 8858
0 050 2967
0 051 7308
0.053 1880
0.054 6681
24
0 044 320(5
0017 0735
0018 4866
0.049 924 1
0.051 3857
0052 8711
25
0.012 6519
0 045 4068
0 0168225
0.048 2635
0.049 7295
0.051 2204
26
0041 1116
0 04.? 8689
0 045 2873
00167320
0 048 2027
0 049 6992
27
0.039 6856
0 012 4455
0.043 8668
0045 3153
0 046 7908
0 048 2931
28
0.0383617
0.011 1241
0.012 5186
0.044 001 1
0.045 4815
0 046 9897
29
0037 1291
0 039 8950
0 04 1 3223
0.042 7788
0.044 2642
0.045 7781
30
0.035 9789
0.038 748 L
0.010 1785
0.041 6392
0.043 1298
0.044 6499
31
0.034 9030
0.037 6757
0.039 1094
0 040 5743
0.042 0701
0 043 5963
32
0.033 8945
0 036 6709
0.038 1079
0 039 5771
00410781
0042 6106
33
0032 9173
0 035 7274
0.037 1679
0038 6414
0.040 1478
0011 6865
34
0.032 0559
0034 8 KM)
0 036 2839
0037 7619
0.039 2736
0.0408187
35
0.031 2155
0.034 0037
0.035 4511
0.036 9336
0.038 4508
0.040 0022
36
0.030 4219
0033 2143
0 03 1 G653
0.0361524
0.037 6751
0 039 2329
37
0.029 6714
0 032 4680
0.033 9227
0.035 4114
0.036 9426
0 038 5068
38
0.028 9604
0 031 7615
0033 2198
0.034 7161
0.036 2499
0.037 820(1
39
0.028 2861
0031 0916
0 032 5536
0 034 0546
0.035 5940
0037 1711
40
0.027 6455
0 030 4556
0.031 9214
0.033 4271
0.034 9721
0.036 5558
41
0.027 0363
0 029 8510
0.031 3206
0.0328311
0 034 3817
0.035 9719
42
0.026 4562
0 029 2756
0.030 749L
0.032 2643
0.033 8206
0 035 4173
43
0.025 9032
0.0287274
0 030 2047
0.031 7247
0.033 2867
0.031 8899
44
0.025 3751
0.028 2044
0.029 6856
0.031 2104
0 032 7781
0 034 3879
45
0.024 8712
0 027 7050
0.029 1901
0.030 7198
0.032 2932
0.033 9090
4f
0.024 3889
0 027 2278
0.028 7168
0.0302512
0.031 8304
0.033 4534
47
0.023 9273
0 026 771 1
0.0282611
(J.029 8034
0 031 3884
0.033 0179
48
0.023 4850
0.026 3338
0 027 8307
0.029 3750
0 030 9657
0.032 6018
49
0 023 0609
0025 9147
0 027 4156
0 028 9618
0.030 5612
0 032 2040
50
0.022 6538
0.0255127
0.027 0176
0.028 5717
0.030 1739
0.031 8232
1
Note: For Rent of Annuity (1 -j- i)n — 1 .subtract the rate per cent.
Example: Rent of 1 at 1 Vfe% for 25 years is .0482035 — .015 — .0332B35.
534
APPENDIXES
L i
flHll
1-
n
2y4%
2y2%
23/4%
3%
3Va%
4%
1
2
3
4
5
1 022 5000
0.510 9376
0.3184446
0.204 2189
0.213 7002
1 025 0000
0.518 8272
0.350 1372
0.265 8179
0.215 2469
1.027 5000
0.520 7183
0.351 8324
0.267 4206
0.216 7983
1.0300000
0.522 6108
0 353 524
0.269 0271
0.218 3546
1.035 0000
0 526 4005
0.356 9342
0.272 2511
0.221 4814
1.040 0000
0 530 1961
0.360 3485
0.275 4901
0.224 0271
6
7
8
9
10
0.180 0350
0 156 0003
0.137 9846
0.123 9817
0.112 7877
0.181 5500
0.157 495J
0.139 4674
0.1254569
0.114 2588
0.183 0708
0.158 9975
0.140 9580
0.126 9410
0.115 7397
0 184 5975
0.160 5064
0.142 4564
0 128 4330
0.117 2305
0 187 6682
0 163 544f
0.145 4767
0.131 4460
0.120 2414
0.190 7619
0 166 60 W
0.148 5278
0.134 4930
0.1232909
11
12
13
14
15
0.103 6365
0.096 0174
0.089 5769
0.084 0623
0.079 2885
0.105 1060
0.097 4871
0.091 0483
0.085 5365
0.080 7665
0.106 5863
0.098 9687
0.092 5325
0.087 0246
0.082 2592
0 108 0775
0.100 4621
0 094 0295
0.088 5263
0.083 7666
0.111 0920
0.103 4840
0.097 0616
0.091 5707
0.086 8251
0.114 1490
0 106 5522
0.100 1437
0 094 6690
0,089 9411
16
17
18
19
20
0 075 1166
0.071 4404
0.068 1772
0.065 2618
0.062 6421
0.076 5990
0.072 9278
0.069 6701
0 066 7606
0.064 1471
0.078 0971
0 074 4319
0.071 1806
0.068 2780
0.065 6717
0 079 6109
0.075 95'25
0 072 7087
0.069 8139
0.067 2157
0.082 6848
0.079 0431
0 075 8168
0.072 9403
0.070 3611
0 085 8200
0.082 19H5
0 078 9933
00761380
0 073 5818
21
22
23
24
25
0.060 2757
0.058 1282
0.056 1710
0.054 3802
0.052 7860
0.061 7873
0.059 6466
0.057 6964
0 055 9128
0.054 2759
0.063 3194
0 061 1864
0.059 2441
0.05-7 4686
0.055 8400
0.064 8718
0.062 7474
0 060 8139
0 059 0474
0.057 4279
0.068 0366
0 065 9321
0 064 0188
0 062 2728
0.060 6740
0.071 2801
0.069 1988
0 067 3091
0 065 5868
0.064 0120
26
27
28
29
30
0.051 2213
0.0198219
00185253
0.047 3208
0.046 199?
0 052 7688
0 051 3769
0 050 0879
0.0488913
0.047 7776
0 054 3412
0.052 9578
0.051 6774
0 050 4894
0.049 3844
0.055 938S
0.054 5642
0 Oo3 2932
0.0521147
0.051 0193
0 059 2054
0.057 85i!4
0.056 6027
0.055 4454
0.054 3713
0 062 5674
0.061 2385
0 060 0130
0.058 8799
0.057 8301
31
32
33
34
35
0.045 1528
0.044 1742
0.043 2572
0.042 3966
0.041 5873
0.046 7390
0.045 7683
0.044 8594
0.044 0068
0.043 2056
00183545
0 047 3926
0.046 4925
0.045 6488
0.044 8565
0 049 9989
0.049 0466
0 048 1561
0.047 3220
0.016 5393
0 053 3724
0 052 4415
0 051 5724
0.050 7597
0.049 9984
0.056 8554
0 055 9486
0 055 1036
0.051 3148
0.053 5773
36
37
38
39
40
0 040 8252
0.040 1064
0 039 4275
0.038 7854
0,038 1774
0.042 4516
0.041 7409
0.041 0701
0.010 4362
0.039 8362
0 044 1113
0.043 4095
0.0127476
0 042 1226
0.041 5315
0 043 8038
0.045 1116
0 044 4593
0.043 8439
0.043 2624
0.049 2842
0.048 6133
0.047 9821
0 047 3878
0.046 8273
0.052 8869
0 052 2396
0.051 6319
0 051 0608
0.050 5235
41
42
43
44
45
0.037 6009
0.037 0536
0.036 5336
0.036 0390
0.035 5681
0 039 2679
0 038 7288
0 038 2169
0 037 7304
0.037 2C75
0.040 9720
0 040 4418
0.039 9387
0 039 4610
0.039 0069
0.042 7124
0 042 1917
0 041 6981
0 041 2299
0.040 7852
0.046 2982
0 045 7983
0 045 3254
0.044 8777
0.044 4534
0 050 0174
0.049 5402
0049 0899
0.048 6645
0.048 2625
46
47
48
49
50
0.035 1192
0 034 6911
0.034 2823
0.033 8918
0.033 5184
0 036 8268
0.036 4067
0.036 0060
0.035 6235
0.035 2581
0.038 5749
0.038 1636
0 037 7716
0.037 3977
0.037 0409
0 040 3625
0 039 9605
0.039 5778
0.039 2131
0.038 8655
0.044 0511
0 043 6692
0.043 3065
0.042 9617
0.042 6337
0.047 8821
0 047 5219
0.047 1807
0.046 8571
0.046 5502
Note :
PCI Rent of Annuity (1 -f i)n — 1 subtract the rate per cent.
i
* Rent of 1 at 2% for 20 years is .0611567 — .02 — .0411567-
RENT OF PRESENT VALUE OF ANNUITY OF 1
i i
535
1 -
(1 +
n
4V«%
5%
5V2%
6%
7%
8%
1
1 045 0000
J 050 0000
1 055 0000
1 060 (KICK)
1.070 0000
1 080 0000
2
0.533 9976
0.537 8049
0 541 61 SO
0 545 4369
0 553 0918
0 5flO 7lV)2
3
0.363 7734
0 367 2086
0 370 6541
0 374 1098
0381 0517
0.388 0335
4
0.278 7437
0 282 0118
0 285 2945
0 288 5915
0 295 2281
0 301 9208
5
0 227 7916
0.230 9748
0.234 1764
0 237 3964
02138907
0.250 4564
6
0 193 8784
0 197 0175
0 20v> 1790
0 203 3(526
0 209 795S
0 216 3151
7
0.1697015
0 172 8198
0 175 «)644
0 179 1350
0.185 5532
0 1920721
8
0.151 6097
0 154 7218
0 157 8640
0 161 0359
0 167 467S
0171 0148
9
0.137 5745
0 140 6901
0 143 8395
0 147 0222
0 153 4865
0.1600797
10
0.126 378H
0.129 5016
0 132 6678
0.135 8680
0.112 3775
0.149 0295
11
0 117 2t82
0 ] 20 3889
0 \'2t f»707
0 126 7929
0 133 3569
0 1 10 0763
12
0 109 6662
0 112 8254
0 1 16 0292
0 119 2770
0 1 25 9020
0 13269;>0
13
0 103 2754
0.106 4558
0 109 6843
0.112 9601
0 119650!)
0 1265218
14
0 097 8203
0 101 0240
0 lot 2791
0.107 5819
0 114 3149
0 121 2968
15
0 093 1138
0 096 3423
0 099 6256
0.102 9028
0.109 7916
0.1168295
16
00890151
0 092 2049
0 09") 5825
0 098 9521
0 105 8577
0 112 9760
17
0 085 4176
0 088 6999
0 092 0120
0095 4H8
0 102 12.V2
0.109 6291
18
0 OS2 2369
0 085 5 162
0 088 9199
0 0')2 3565
0 099 1126
0 106 7021
19
0079 1073
0 082 7450
0086 1501
0 089 6209
0 096 75:50
0 101 1276
20
0.076 8761
0.080 24 26
0 083 6793
0.0&7 1846
0 094 3929
0.101 8522
21
0 074 6006
0 077 9961
OOH] 4618
0 085 0046
0 092 2890
0 099 83"2
22
0072 5157
0 075 9705
0079 4712
0083 0156
0 090 4058
0 ()«w 0321
23
0 070 6825
0.074 1368
0 077 0690
0 081 2785
0 088 7139
0 096 4222
24
0 068 9870
0 072 4709
0 076 0358
0 079 6790
0 087 1890
0 091 9780
26
0.067 4390
0.070 9525
0 074 5494
0 078 2267
0085 8105
0.093 6788
26
00660214
0.069 5643
0 073 1931
0 076 9041
0084 5610
0.092 5071
27
0064 7195
0 068 2919
0 071 9523
0 07r> 0<)72
0.083 4257
0091 4181
28
0 063 5208
0 067 122',
0 070 81-11
0 071 5926
0 082 3919
0 0!K) 4889
29
0.062 1146
0.066 0455
0 069 7686
0 073 5796
0081 4187
0089(5185
30
0.061 3915
0.065 051 1
0 068 8054
0 072 6189
0 080 58(54
0.088 8274
31
0.0604135
0.064 1321
0 067 9167
0 07 1 7922
0 079 7969
0.088 1073
32
0.059 5632
0063 280 4
0 067 0952
0 071 0023
0 079 0729
0087 4508
33
0 058 7445
0.062 4900
0 006 3347
0 070 2729
0 078 1081
0086 85 1 (i
34
0 057 9819
0 061 7554
0 065 6296
0 0(59 5981
0 077 7%7
0 086 30 1 1
35
0.057 2705
0.061 0717
0 (Mil 9719
0.068 9739
0.077 2310
0 085 8033
36
0.056 6058
0.060 4345
0061 3664
00683918
00767153
0.0853417
37
0.055 9840
0 059 8398
0 063 7999
0 067 8574
0.070 23(511
0 081 9211
38
00554017
0.059 2842
0 063 2722
0 067 3581
0.075 7951
0 084 53K9
39
0.054 8557
0 058 7646
0 062 7799
0 066 8938
0 075 3868
0084 18"jl
40
0.044 3432
0.058 2782
0.062 3203
0.066 4615
0.075 0091
0.083 8602
41
0.053 8616
0.057 8223
0 061 8<X)9
0 066 0589
0 074 6590"
0 083 561 5
42
0.053 4087
0.057 3947
0 061 4893
0 065 ($834
0.074 3359
0 083 2868
43
0 052 9824
0.056 9933
0 061 1134
0 065 3331
0 074 0359
0 083 0311
44
0 052 5807
0 056 6163
0.060 7613
0.065 0061
0 073 7577
0082 8015
45
0.052 2020
0.056 2617
0.060 4313
0.064 7005
0.073 1990
0.082 5873
46
0.051 8447
0.055 9282
0 0601218
0064 4119
0 073 2600
0 082 3899
47
0 051 5073
0 055 6142
O0')98313
0064 1477
0073 0371
0 082 2080
48
0051 1886
0.055 3184
0 059 5585
0.063 8977
0 072 8307
0.082 0103
49
0.050 8872
0 055 0397
0 059 3023
0 063 6636
0 072 6385
0 081 8856
50
0 050 6022
0.054 7767
0.059 0615
0 063 4443
0.072 4599
0.081 7429
1
Note: For Rent of Annuity (1 -f i)n — 1 subtract the rate per cent.
i
.Example : Rent of 1 for 30 years at 3% is .0510193 — .03 = .0210193.
536 APPENDIXES
Table 7
AMERICAN EXPERIENCE TABLE OF MORTALITY
(Based on 100,000 living at age of 10)
Yearly
Yearly
Age
Number
living
Number
dying
proba-
bility of
proba-
bility of
X
1*
dx
dying
living
q*
Px
10
100 000
749
0.007 490
0.992 510
11
99 251
746
0.007 516
0.992 484
12
98 505
743
0.007 543
0.992 457
13
97 762
740
0.007 569
0.992 431
14
97 022
737
0.007 596
0.992 404
15
96 285
735
0.007 634
0.992 366
16
95 550
732
0.007 661
0.992 339
17
94 818
729
0.007 688
0 992 312
18
94 089
727
0.007 727
0.992 273
19
93 362
725
0.007 765
0.992 235
20
92 637
723
0.007 805
0.992 195
21
91 914
722
0 007 855
0.992 145
22
91 192
721
0 007 906
0 992 094
23
90 471
720
0.007 958
0 992 042
24
89 751
719
0.008 Oil
0.991 989
25
89 032
718
0.008 065
0 991 935
26
88 314
718
0.008 130
0 991 870
27
87 596
718
0 008 197
0 991 803
28
86 878
718
0.008 264
0.991 736
29
86 160
719
0.008 345
0.991 655
30
85 441
720
0.008 427
0.991 573
31
84 721
721
0.008 510
0.991 490
32
84 000
723
0.008 607
0 991 393
33
83 277
726
0.008 718
0 991 282
34
82 551
729
0.008 831
0.991 169
35
81 822
732
0 008 946
0 991 054
36
81 090
737
0.009 089
0 990 911
37
80 353
742
0.009 234
0 990 766
38
79 611
749
0.009 408
0.990 592
39
78 862
756
0.009 586
0.990 414
40
78 106
765
0.009 794
0.990 206
41
77 341
774
0.010 008
0 989 992
42
76 567
785
0 010 252
0.989 748
43
75 782
797
0.010 517
0 989 483
44
74 985
812
0.010 829
0.989 171
45
74 173
828
0 Oil 163
0 988 837
46
73 345
848
0.011 562
0 988 438
47
72 497
870
0.012 000
0.988 000
48
71 627
896
0.012 509
0.987 491
49
70 731
927
0.013 106
0.986 894
60
69 804
962
0.013 781
0 986 219
51
68 842
1 Oil
0.014 541
0 985 459
62
67 841
1 044
0.015 389
0.984 611
53
66 797
1 091
0.016 333
0.983 667
54
65 706
1 143
0.017 396
0.982 604
AMERICAN EXPERIENCE TABLE OF MORTALITY
AMERICAN EXPERIENCE TABLE OF MORTALITY
537
Yearly
Yearly
Age
Number
living
Number
dying
proba-
bility of
proba-
bility of
1,
d*
dying
living
q*
P.
55
64 563
1 199
0.018 571
0 981 429
56
63 364
1 260
0.019 885
0.980 115
57
62 104
1 325
0 021 335
0.978 665
58
60 779
1 394
0.022 936
0.977 064
59
59 385
1 468
0.024 720
0.975 280
60
57 917
1 546
0.026 693
0.973 307
61
56 371
1 628
0.028 880
0.971 120
62
54 743
1 713
0 031 292
0.968 708
63
53 030
1 800
0.033 943
0.966 057
64
51 230
1 889
0.036 873
0.963 127
65
49 341
1 980
0.040 i29
0.959 871
66
47 361
2 070
0 013 707
0.956 293
67
45 291
2 158
0 047 647
0.952 353
68
43 133
2 243
0.052 002
0.947 998
69
40 890
2 321
0.056 762
0.943 238
70
38 569
2 391
0.061 993
0.938 007
71
36 178
2 448
0.067 665
0.932 335
72
33 730
2 487
0.073 733
0.926 267
73
31 243
2 505
0.080 178
0.919 822
74
28 738
2 501
G.087 028
0.912 972
75
26 237
2 476
0.094 371
0.905 629
76
23 761
2 431
0.102 311
0.897 689
77
21 330
2 369
0.1 11 064
0.888 93h
78
18 961
2 291
0.120 827
0.879 173
79
16 670
2 196
0.131 734
0.868 266
80
14 474
2 091
0.144 466
0.855 534
81
12 383
1 964
0 158 605
0.841 395
82
10 419
1 816
0.174 297
0.825 703
83
8 603
1 648
0.191 561
0.808 439
84
6 955
1 470
0 211 359
0.788 641
85
5 485
1 292
0.235 552
0.764 448
86
4 193
1 114
0.265 681
0.734 319
87
3 079
933
0 303 020
0 696 980
88
2 146
744
0.346 692
0.65'* 308
89
1 402
555
0.395 863
0.604 137
90
847
385
0.454 545
0.545 455
91
462
246
0.532 466
0.467 ,534
92
216
137
0 634 259
0.365 741
93
79
58
0 734 177
0.265 823
94
21
18
0.857 143
0.142 857
95
3
3
1.000 000
0.000 000
538 APPENDIXES
Table 8
COMMUTATION COLUMNS, 3^ PER CENT
Age
X
Dx
N,
c,
M,
1 -ha,
A*
10
70891.9
1575 535
513.02
17612.9
22.2245
0.24845
11
67981.5
1504 643
493 69
17099 9
22 1331
0.25154
12
65189.0
1436 662
475 08
16606.2
22.0384
0.25474
13
62509.4
1371 473
457 16
16131,1
21.9403
0.25806
14
59938.4
1308 963
439.91
15674.0
21.8385
0.26151
15
57471.6
1249 025
423.88
15234.1
21.7329
0 26508
16
55104.2
1191 553
407.87
14810.2
21.6236
0 26877
17
52832 9
1136 449
392.47
14402.3
21.5102
0.27261
18
50653.9
1083 616
378.15
14009.8
21 3926
0 27659
19
48562.8
1032 962
364.36
13631.7
21.2707
0.28071
20
46556.2
984 400
351 07
13267.3
21.1443
0 28497
21
44630.8
937 843
338 73
12916 3
21.0134
0 28940
22
42782.8
893 213
326 82
12577 5
20 8779
0 29399
23
41009.2
850 430
315 33
12250.7
20 7375
0.29873
24
39307.1
809 421
304 24
11935.4
20.5922
0.30365
25
37673.6
770 113
293 55
11631.1
20.4417
0 30873
26
36106.1
732 440
283 62
11337.6
20 2858
0 31401
27
34601.5
696 334
274 03
11054 0
20.1244
0.31947
28
33157.4
661 732
264 76
10779 9
19.9573
0.32512
29
31771.3
628 575
256 16
10515.2
19.7843
0 33097
30
30440 8
596 804
247 85
10259.0
19.6054
0 33702
31
29163 5
566 363
239 797
10011 2
19.4202
0.34328
32
27937.5
537 199
232 331
9771.38
19 . 2286
0 34976
33
26760.5
509 262
225.406
9539 04
19.0304
0.35646
34
25630.1
482 501
218 683
9313 64
18.8256
0.36339
35
24544 7
456 871
212 157
9094.96
18.6138
0 37055
36
23502.5
432 326
206 383
8882 80
18 3949
0 37795
37
22501 4
408 824
200 757
8676 42
18.1688
0 38560
38
21539.7
386 323
195.798
8475 . 66
17.9354
0.39349
39
20615.5
364 783
190.945
8279.86
17.6946
0.40163
40
19727.4
344 167
186.684
8088.92
17.4461
0 41003
41
18873.6
324 440
182 493
7902.23
17.1901
0 41869
42
18052.9
305 566
178 828
7719 74
16.9262
0 42762
43
17263.6
287 513
175 421
7540.91
16.6543
0 43681
44
16504.4
270 250
172.680
7365.49
16.3744
0 44628
45
15773.6
253 745
170.127
7192.81
16.0867
0.45600
46
15070.0
237 972
168.345
7022 68
15.7911
0 46600
47
14392.1
222 902
166 872
6854.34
15.4878
0.47626
48
13738.5
208 510
166 047
6687.47
15 1770
0 48677
49
13107.9
194 771
165 983
6521.42
14.8591
0.49752
50
12498.6
181 663
166.424
6355 44
14.5346
0.50849
51
11909.6
169 165
167.316
6189.01
14.2041
0 51967
52
11339.5
157 255
168 601
6021.70
13.8679
0 53104
53
10787.4
145 916
170.234
5853.10
13.5264
0 54258
54
10252.4
135 128
172.317
5682.86
13.1801
0 55430
COMMUTATION COLUMNS, 3}$ PER CENT 539
COMMUTATION COLUMNS, 3}-'2' PER CENT
Age
Dx NX Cx MX
1+ax
Ax
55
56
67
58
59
9733.
9229.
8740.
8264.
7801.
40
60
17
44
83
124876
115142
105912.
97172.
88908.
8
6
2
174
177
180.
183
186
646
325
167
140
340
5510
5335
5158
4978.
4795.
54
90
57
40
27
12 8296
12 4753
12 1179
11.7579
11.3958
0 56615
0 57813
0 59022
0 60239
0.61463
60
61
62
63
64
7351.
6913.
6486.
6071
5666.
65
44
75
27
85
81106,4
73754.7
66841.3
60354.5
54283.3
189
192
196
199
201
604
909
117
109
887
4608.
4419.
4226
4030.
3831 .
93
32
41
30
19
11.
10
10.
9,
9.
0324
6683
3043
9410
5791
0.62692
0.63924
0.65155
0.66383
0.67607
65
66
67
68
69
5273.
4890
4518.
4157
3808.
33
55
65
82
32
48616.4
43343.1
38452.5
33933.9
29776.1
204
206
208
208
208
457
522
022
903
858
3629.
3424
3218
3010
2801.
30
84
32
30
40
9.
8.
8.
8.
7.
2193
8626
5097
1615
8187
0 68824
0 70030
0.71223
0 72401
0.73560
70
71
72
73
74
3470.
3145
2833
2535
2253
67
43
42
75
57
25967
22497
19351
16518
13982
7
1
6
2
5
207
205
201
196
189
881
639
851
430
491
2592
2384
2179
1977
1780
54
66
02
17
73
7.
7.
6.
6
6.
4820
1523
8298
5141
2046
0.74698
0 75813
0.76904
0 77972
0 79018
75
76
77
78
79
1987
1739
1508
1295
1100
87
39
63
73
647
11728
9741
8001
6493
5197
9
03
63
00
27
181
171
161
151
140
253
940
889
2646
0891
1591
1409
1238
1076
924
24
99
05
158
894
5
5.
5.
5.
4.
9002
6002
3039
0111
7220
0 80048
0 81062
0 82064
0.83054
0 84032
80
81
82
83
84
923
763
620
494
386
338
234
465
995
641
4096
3173
2410
1789
1294
62
29
05
59
59
128
116
104
91
78
8801
9588
4881
6152
.9565
784
655
538
434
342
805
924
966
478
862
4.
4.
3
3
3.
4368
1577
8843
6154
3483
0 84997
0 85940
0 86865
0 87774
0 88677
85
86
87
88
89
294
217
154
103
65
610
598
383
963
.6231
907
613
395
241
137
95
34
74
36
.398
67
55
45
34
25
0490
8566
1992
82425
09929
263
196
141
95
60
906
857
000
8011
9768
3
2
2
2
0819
8187
5634
3216
0937
0 89578
0 90468
0 91332
0 92149
0 92920
90
91
92
93
94
38
20
9
3
0
.3047
. 18692
.11888
.22236
.827611
71
33
13
4
0
.775
.4700
.2831
.16420
.94184
16
10
5
2
0
82244
385393
588150
285784
685393
35
19
8
3
0
8775
05509
66970
08155
.79576
1
1
1
1
1
8738
6580
4567
2923
1380
0 93664
0 94393
0 95074
0.95630
0 96152
95
0
114232
0
.114232
0
110369
0
110369
1
0000
0.96618
Index
Abbreviated methods (see Short methods)
Accounts:
asset valuation (see Asset valuation
accounts)
balancing, 12
current (see Current account)
foreign exchange, 297-307
averaging, 298-300
branch, conversion of, 300-307
receivable, turnover, 175-176
running, storage, 141-142
Accrual basis, income reported on, 167
Accumulation of simple interest, 81-82
Accurate interest, 81
Actuarial science, 311
Addends, 3
Adding machine, subtraction on, 14
Addition, 3-;10
combinations, sum 10, 4—5
group, 6
of fractions, 40
decimal, 44-45
positive and negative numbers, 230
practical applications, 9-10
recording, by columns. 6-8
same number repeated, 5
streamline, 4
two columns at a time, 6
verification of, 30, 32
Agent, insurance, commission of, 96-97
Agreement, partnership :
lack of, 181-182
profit-sharing, 181-182
Agricultural indexes, 279-280. 283-284
Algebra, 229-234
coefficient, 230-231
division, 234
multiplication, 233-234
parentheses, brackets, braces, 231-232
positive and negative numbers, 229-
230
addition of, 230
subtraction, 232-233
symbols and terms, 229
Aliquot parts:
calculating interest by using, 78-79
defined, 48
division by, 50-51
multiplication by, 49-50
table of, 49
use of, 48
American experience mortality table, 454,
536-537
Amortization:
annuities, 337-338
of discount:
bonds outstanding method, 398-399
scientific method, 400-404
serial redemption bonds, 398-404
Amount :
annuity due, 349-352
of ordinary annuity, 327-331, 344-346
table, 331
simple, 82-83
Analysis:
of compound interest, 327-328
table, 317
of statements (see Statements, analysis
of)
Annuities:
compound interest and, 327-328
deferred, 359-364
due (see Annuities due)
kinds of, 327
life (see Life annuities)
ordinary (see Ordinary annuities)
rent of,' 327
special, 349-366
Annuities due, 349-359
amount of, 349
analysis of, 350-352
denned, 327, 349
effective interest on, 358-359
present value of:
comparison with ordinary annuity,
352
computing, 353-355
rent of, 357-358
Annuity method, depreciation, 410-413
Anticipation, discounts, 75-76
Antilogarithm, finding, 254
Appraisal, valuation by, 203
Appropriations, 223-224
Arithmetical progression (see Progression:
arithmetical)
Arithmetical solution, problems with un-
known quantities, 246-248
Assets :
current, 172-176
perpetuity, ordinary annual expenses
and replacement of, 420-422
replacement of, perpetuity providing
for ordinary annual expenses
and for, 420-422
wasting, capitalization of, 422-424
Asset valuation accounts, 405-424
asset valuation, 405
capitalization, wasting asset, 422-424
composite life, 414-418
540
INDEX
541
Asset valuation accounts (cont.)i
depletion, 405
calculation of, 418-419
defined, 405, 418
depreciation (see Depreciation)
perpetuity, ordinary annual expenses
and replacement of asset, 420-
422
Average, 123-129
approximation by, bonds, 397-398
compound (see Compound average)
cost method, inventories, 144-146
due date, 131
moving, 124-126
periodic, 126-127
progressive. 126
sales, clerk 's per cent of, 57
simple, 123-124
weighted, 127-129
Averaging:
dates of invoices, 131-133
foreign < \rliii w accounts, 298-300
Axiom, d rimed, 'J29
B
Bad debts, gross profit computations,
162-166
Balancing an account, 12
Bank discount, 87--90
counting time, 87
denned, 87
proceeds, 88-89
finding both, 89
finding face of note, proceeds, time,
rate given, 89
table for finding difference between
dates, 87-88
Bar chart, 267-269, 270
Base. 53-54
Bonds, 367-404
allotment, goodwill, 207-209
approximation by averages, 397-398
bought on yield basis, 382- 383
computation without bond table, 397
definitions, 367
discount:
between interest dates, 380
value of bond bought at, 380-381
interest on :
accrued between interest dates, 380
discount or premium between dates,
380
effective rate:
bonds sold at discount, 396-397
bonds sold at premium, 396
values of bonds between dates, 379-
380
premium, between interest dates, 380
price, 369
purchased at discount or premium,
368-369
rate of yield, 369
redeemable from fund, 394-3P5
redeemed above par, 383-385
redemption periods, 387-390
Bonds (cotU):
serial redemption (see Serial redemp-
tion bonds)
sold at discount, 372-376
effective rate of interest on, 396-397
sold at par, 368
sold at premium, 376-379
effective rate of interest on, 396
tables:
first form, 369
interpolating in, 370-372
of values, 369
second form, 369-370
use of, 369
values :
computed without tables, 372
table, 369
yield, rate of, 369
Book value, shares of stock, 220
Braces, in algebra, 231-232
Brackets, in algebra, 231-232
Briggs' table, 272
Buiigeling. 62-64
Building MIK! loan associations, 425-437
classified problems on, 434-437
control, 425
distribution of profits:
Dcxtcr's method, 427, 429, 431
methods compared, 430
partnership method, 427-429
effective rate of interest on money in-
vested in installment shares.
433-434
funds, withdrawal of, 425-426
plans of oig.'ini/Mliori, 426
Dayton or Ohio, 431-432
serial, 427-431
Dexter's method, 427, 429, 431
methods compared, 430
partnership method, 427-429
withdrawal value, 431
terminating, 426
stock, classes of:
fully-paid, 425
installment, 425
time required for stock to mature, 432-
433
Business finance, 213-221
book value, shares of stock, 220
cumulative voting, 219-220
profits distribution, 220-221
stock rights, 213-214
working capital, 216-219
Business insurance, 95-105
agent's commission, 96-97
coinsurance, 97-99, 100
finding premium, 96
fire, 95
group iif«, 102-103 (see also Group life
insurance)
health, 104-105
kinds of, 95
policy, 95
cancellation of, 97, ;J5
form of, 95
rates, 95-96
542
INDEX
Business insurance (cont.):
use and occupancy, 99-102
workmen's compensation, 104, 105
Business measurements, practical, 481-
488
Cancellation, 37
mark-up, 152, 153
method, interest, 79-80
Capital :
contribution, adjustments of, 184-186
how invested, 175
interest on, profits not covering, 183-
184
sources of. 1 74
total employed:
ratio, operating profit to, 170
turnover, 175
working, 210-219
ratio, 169, 172-174
Capitalization of wasting assets, 422-424
Capitalized cost, 419-420
Cash discount, 71-72
Cash payment, 120-121
Casting out elevens, 31-32
Casting out nines, 29-30
Cham discounts, 72
Change making, subtraction method, 10-
11
Characteristic, positive and negative,
250-251
Charts:
bar, 267-269, 270
circle, 265-267
coordinate, rules for, 270-272
curve, 269-270
graphs and, 265
line, 26&-270
].-i: ,-:-: -.'• 272 275
]..•«., -T., -J7.. -J77
Checking computations, 29-34
absolute check, 29
addition, 30, 32
by casting out elevens, 31-32
by casting out nines, 29-30
check number thirteen, 33-34
division, 31, 33
where remainder, 31
methods, 29
multiplication, 30-31, 33
rough check, 29
subtraction, 30, 32-33
Checks, pay, 118-120
Circle chart, 265-267
City taxes, 223
Clerk's per cent of average sales, 57
Clock card, in and out, 108
time, 107-109
Coefficient, defined, 230-231
Coin sheet, 120-121
Coinsurance, 97-99, 100
Combinations, 442-444
in probability, 446-448
C Commercial discounts, 71-76
anticipation, 75—76
cash, 71-72
finding net price, 73-75
invoices, transportation charges on, 75
single, equivalent to series, 72-73
trade, 72
Commissions, percentage, 68-69
Commodity taxes, 224
Common divisor, greatest, 36
Common fractions, 39-51 (see also Frac-
tions: common)
Common multiple, least, 36-37
Commutation columns, table, 460-463
538-539
Complement method, subtraction, 12-14
Complex fractions, clearing of, 240-241
Composite price indexes, 281
Compound average, 135-137
dating forward or backward, 135—136
defined, 135
product method, 135, 136
C /om pound discount, 321-322
Compound interest, 311-325
actuarial science, 311
analysis of, 327-328
table, 317
compound amount:
calculation of, 313-315
fractional part of conversion period,
324
of given principal, 315-316
tables, 313
compound discount, 321—322
computation of, 316-317
conversions of interest, results of, 317
effective rates, 317-320
method. 311
nominal rates, 317-319
present uorth, 320-321
of 1, table, 321
principal, 312
rate, 312-313, 322-323
ratio of increase, 313
relation to annuities, 328
symbols, 311-312
Computations, checking, 29-34 (see also
Checking computations)
Conversion:
foreign exchange (see Foreign ex-
change: conversion)
of interest, frequent, 317
period, compound amount for frac-
tional part of, 324
Coordinate charts, rules for, 270-272
Cost:
capitalized, 419-420
gross profit test:
goods sold, 156-157
sales, rate per cent, 157
or market, inventories, 143-144
to retail, determining ratio of, 153—154
County taxes, 223
Cross multiplication, 21-23 (see also
Multiplication: cros?)
Cumulative voting, 219-220
INDEX
543
Currency memorandum, 120-121
Current account, 139-140
denned, 139
methods, 139-140
Current assets, 172-176
Current Tax Payment Act of 1943, 107
Curve chart, 269-270
D
Oaily cost card, 108
Hates:
forward or backward, 135
of invoices, averaging, 131-133
Dating terms, 75
Dayton or Ohio plan, 431-432
Death taxes, 225
Debts:
bad, gross profit computations, 162-
166
installment payment of, 339-340
Decimal fractions, 44-48
addition of, 41—45
changing:
common fraction to decimal, 48
to equivalent common fraction, 48
defined, 4 4
division of, 45—48
multiplication of, 45
abbreviating, 46-47
relation between percentage and com-
mon and, 53
subtraction of, 44—45
Decrease, per cent of, 59-60
Deductions on payroll records, 109, 118
Deferred annuity, 359-364
Deferred life annuity, 463-464
due, 464
Denominator, 39
Departmental sales, daily record of, 56
Depletion:
calculation of, 418-419
denned, 405, 418
Depreciation:
defined, 405
methods of calculating, 405
annuity, 410-413
f ixed-pe re en tage-of -diminish ing-
value method, 413-414
sinking-fund, 409-410
straight-line, 405-407
sum-of-digits, 408
unit-product, 407-408
MoikiiiK-hou"*. 407-408
tabl--, ini,, 107. 408,409-410,412,414
Dexter 's rule, 427', 429, 431
comparison of partnership plan and,
430
Discount:
amortization of. serial redemption
bonds, 398-404
table, 402
bond, between interest dates, 380
bonds purchased at, 368-369
calculating, 380-381
bonds sold at, 372-376
Discount (cpnt.):
commercial, 71-76 (see also Commer
cial discounts)
compound, 321-322
series of, 72
Distribution of profits, methods of, 427-
431
Dividends, 24
and net cost, life insurance policies, 474
Divisibility, tests of, 35
Division:
abbreviated, 25
by algebra, 234
by aliquot parts, 50-51
by * j ' ' • . 255, 256-257
by .'• .M, '.'-, 24-25
of fractions, 42-43
decimal, 45-46, 47-48
on slide rule, 263-264
reciprocals in, 26-28
short methods, 24-28
use of tables in, 25-26
verification of, 31, 33
where remainder, 31
Divisor, 24
greatest common, 36
Dollars-times-days method, interest, 80
E
Earning power, determining from profit
and loss statements, 201-202
Effective rate of interest:
compound interest, 317-320
on annuities, 332-334, 341-343
on annuities due, 358—359
on bonds, 396-397
Elevens, casting out, 31—32
Empirical probability, 450-452
Endowment:
insurance, 471-472
pure, 459-460
Envelopes, pay, 120
Equation of accounts, 135—137 (see also
Compound average)
Equations, 235-248
defined, 229
fractions, 239
complex, clearing of, 240-241
problems containing unknown quanti-
ties, 246-248
simple, 235-238
simultaneous, two or more unknowns
241-246
Events :
compound, 448
independent, 448
mutually exclusive, 449-450
Exact interest, 81
Expenses, 169
per cent of, 58-59
Exponents:
denned, 229
logarithms, 249
544
INDEX
Factoring, 35
Factors and multiples, 35-37
Federal Farm Loan Act, 432
Federal income tax, sale of stock and
rights, 214-216
Federal Old Age Benefit Tax, 107
Financial indexes, 278
Fire insurance, 95
Fire losses, 157-158
First-in, first-out method, inventory,
146-147
Fixed -percentage -of -dim inishing-valuc
method, depreciation, 413-414
Fixed property investment, turnover of,
176-180
Focal date, 131
Forborne temporary life annuity due,
467-468
Foreign exchange, 293-307
accounts :
averaging, 298-300
branch, conversion of, 300-307
conversion :
branch accounts, 300-307
decimals of one monetary unit into
monetary units of smaller .de-
nomination, 295
one monetary unit into terms of
another, 294-295
interest on, 295-296
par of exchange, 293-294
current, 294
problems, classes of, 294-307
rate of exchange, 293
time bill, value of, 296-297
Fractions, 239
addition of, 40
changing mixed number to, 40
common, 39-51
relation between percentage and
decimal and, 53
complex, clearing of, 240—241
decimal :
adding, 44-45
subtracting, 44—45
division of, 42—43
kinds of, 39
multiplication of, 41—42
two mixed numbers ending in J,
43-44
reduction of, 39-40
subtraction of, 40-41
terms explained, 39
Frequency ratio, 453
Fully-paid stock, 425
G
Geometrical progression (see Piogression:
geometrical)
Gold standard, 293, 294
Goodwill, 201-212
basis of stock allotment, 205
Goodwill, basis of stock allotment (con/.) :
bonas, preferred stock, common
stock, 207-209
common stock only, 205-206
preferred stock for net assets, 206 -
207
basis of valuation, 101
conclusions concerning, 209
defined, 201
determining earning power from profit
and loss statements, 201-202
in partnership, 181
methods of valuing, 202-204
by appraisal, 203
by number of years' purchase price
of net profits, 203-204
examples, 202-203
excess of profits over interest on net
assets, 204
Government, finance of (see Public
finance)
Governmental functions, 223
Graphs, 265-277
Greatest common divisor (G.C.D.), 36
Gross and net income, bar chart, 270
Gross profit computations, 155-168
bad debts, 162-166
cost of goods sold, 156-157
cost of sales, rate per cent, 157
deferring income, effect on tax, 166-168
fire losses, 157-158
installment sales, personal property,
160-161
procedure, 155
test of inventory, 155
in verifying taxpayer's inventory,
,158-160
unearned gross profit, reserve for, 162
uses, 155-156
Group life insurance, 102-103
amount of, 102
cost of, 103
eligibility for, 102
H
Health insurance, 104-105
Highway taxes, 225
I
Improper fractions, 39
In-and-out clock cards, 108-109
Income, per cent of, by source, 57-58
Income tax, 225
federal, sale of stock and rights, 214-
216
Increase, per cent of, 59-60
Index numbers, 278-284
agricultural, 279-280
composite price, 281
construction of, 280-281
crop production, 283-284
nature of, 278-279
production, 278
weighted, 281-284
INDEX
545
Industrial production, index numbers,
278-279
Installment basis, income reported on,
167
Installment selling, 160-161
Installment stock, 425
Insurance:
business (see Business insurance)
endowment, 471-472
life (see Life insurance)
policies (see Life insurance policies)
policies, transformation, 476-477
reserves, 473-474
term, 470, 471
Insurance Principles and Practice, 474
Interest:
and premium, life insurance policies,
474
bond, valuation (see Bonds: interest)
compound (see Compound interest)
effective, use of, in annuities, 332-334
exact or accurate, 81
method (see Interest method)
on capital, deducting, partnership, 183
on foreign exchange, 295-296
on investment, profits not covering,
183-184
simple, 77-86 (see aiso Simple interest)
Interest method:
account current, 139-140
averaging dates of invoices, 131, 133
cancellation, 79-80
Inventories, 143-154
average cost method, 144-146
cost or maikct, 143-144
determining ratio of cost to retail, 153-
151
first-in, first-out method, 146-147
last-in, first-out method, 147
mark-down, per cent of, to net cost, 150
merchandise turnovers, 118
number of, 348-150
retail method, 151-153
turnover, statement analysis, 175
valuation of, 143
Investment:
average, profit sharing in ratio of, 186-
188
ratio of, partnership, 182-183
Invoices:
dates of, averaging (see Averaging
dates of invoices)
discount, transportation charges on, 75
K
Kent, Chancellor, 181
L
Last-in, first-out method, inventory, 147
Least common multiple (L.C.M.), 36-37
Legal reserve, defined, 473
Life annuities, 459-468
commutation table, 460-461, 462-463
deferred, 463-464
due, 464
Life annuities (cont.):
due, 462
use of commutation table, 462-463
nature of, 459, 460
payments m times a year, 466-407
pure endowment, 459-460
table, commutation columns in, 460-
461
temporary, 465
due, 465-466
Life insurance, 453
group, 102-103
limited payment, 477
policies, valuation of, 473-478
Life insurance policies, 473-478
dividends and net cost, 474
interest and premium, 474
loading, 474
mortality and level premium, 473
preliminary term valuation, 477-478
reserves, 473-474
limited payment life insurance, 477
terminal, 475
retrospective method, 476
Line chart, 269-270
Liquidation of partnership, 188-200
by periodic distribution, 190
by total distribution, 189-190
Loading, 474
Logarithm-. 249-264
charts, 272-275
division of, 255, 256-257
exponents, 249
finding number:
when log is given, 254
when mantissa is not in table, 254-
255
multiplication by, 255-256
parts of a logarithm, 250
characteristic, 250-251
mantissa, 251-252
powers of numbers, 225, 257
process with negative characteristic;,
257
roots of numbers, 258
process with negative characteristics,
258
rules for computation of, 255
slide rule (see Slide rule)
table, 249
how to use, 252-253
of numbers, 274
use of, 249
Losses, fire, gross profit test, 157-158
M
Making change, subtraction method, 10-
Mantissa, 251-252
Mark-down, 152, 153
per cent of, to net cost, 150
Market, cost or, inventories, 143-144
Marking goods, percentage, 66-68
Mark-on, 152, 153
Mark-up, 151-152, 153
546
INDEX
Measurements, practical business, 481-
488
Merchandise turnover, 148-150
Merchants' rule, partial payments, 91, 93
Minuend, 10
Mixed number, 39
Mortality:
and level premium, 473
table, 453-454, 536-537
notation, 454-455
Moving average, 124-126
Multiples:
factors arid, 35-37
least common, 36-37
table of, 17, 23-24, 25, 34
Multiplicand, 16
Multiplication:
algebra, 233-234
any number by 11, 18
by aliquot parts, 49-50
by factors of multiplier, 17-18
by 15, 19
by logarithms, 255-256
by numbers a little larger than 100,
1000, etc., 21
by numbers near 100, 1000, etc., 19-20
by 25, 19
contractions in, 17
cross, 21-23
number of throe digits by number of
three digits, 22-23
number of three digits by number of
two digits, 22
number of two figures by 11, 18
numbers ending with ciphers, 19
of fractions, 41-42
decimal, 45. 46-47
mixed number by mixed number, 44
two mixed numbers ending in \,
43-44
of two numbers each a little larger than
100, 1000, etc., 21
of two numbers near 100, 1000, etc.,
20-21
on slide rule, 262-263
part of multiplier factor or multiple* of
another part, 18
table, multiples of number, 23-24
divisicn, 25
multiplication, 23-24
verification of, 30-31, 33
Multiplier, 16
N
Negative and positive numbers, 229-230
Net cost, per cent of mark-down to, 150
Net income, gross and, bar chart, 270
Net premiums, 469-472
annual, 469-470
for endowment insurance, 471-472
for term insurance, 471
single, 469
for endowment insurance, 471
term insurance, 470, 471
Net price, finding, 73-75
Net profit on sales, bar chart, 268
Nines, casting out, 29-30
Nominal rates, compound interest, 317-
319
Notes,"1 V j " • * proceeds, time, and
! . -re given, 89
Numbers, logarithms of, 275
Numerator, 39
Ohio plan, 431-432
Old-line companies, 473
Operating statistics, percentage. 61-62
Ordinary annuities, 327-348
amortization, 337-338
amount of, 327, 328, 331
computing procedure, 328-329
selection of rate by calculation of,
344 346
semiannual or quarterly basis, 329-
331
table, 331
defined, 327
installment payment of debt, 339-340
present value of, 335-337
rents, computation of, 338
problem with limited data, 346-347
rate:
by calculating amounts of annuities.
344
computation, 343-346
effective, use of, 332, 341-343
sinking fund contributions, 334
term of, computation, 340-341
use of effective interest in, 332-334
P
Par:
bonds redeemed above, 383—385
bonds sold at, 368
of exchange, 293-294
Parentheses, in algebra, 231-232
Parity ratio, 281
Partial payments, 91-94
methods, 91-93
merchants' rule, 93
United States rule, 91-92
Partnership, 181-200
agreements, profit-sharing, 181-188
capital contribution, adjusting, 184
186
deducting interest on capital, 183
lack of, 181-182
profits not covering interest on in-
vestment, 183-184
ratio of average investment, 186-188
defined, 181
goodwill, 181
liquidating, 188-200
losses, ratios:
arbitrary, 182
of investment, 182-183
mathematical calculations, 181
method of distributing profits, 427-429
compared with Dexter 's rule, 430
INDEX
547
Partnership (cont.):
profits distribution, 220-221, 427-430
Pay checks, 118-120
Payroll records and procedure:
cash payment, 120-121
coin sheet, currency memorandum,
120-121
deductions, 109, 118
forms, 110, 111, 112, 113, 114, llf>, 116,
117, 118
pay checks, 118-120
pay envelopes and receipts, 120
piecework system, 116
records, 107-116
sheets for, 114
remittance voucher, 118
withholding exemptions, 109
Percentage, 53-69
applications of, 53
average sales, clerk's per cent, f)7
budgeting, 62-64
commissions, 68- 69
computations, 54-56
daily record, departmental sales, 56
decrease or increase, 59-60
definitions, 53-54
expense, 58-59
fundamental processes, 54
income, by source, 57 58
increase or decrease, 59-60
marking goods, (Hi -08
operating statistics, 61-62
profits based on sales, 64-66
returned sales, by departments, 56 57
Periodic average, 126-127
Permutations, 439-442
in probability, 446-448
nature of, 439-441
number of ways of doing two or more
things together, 441-442
Perpetuity:
ordinary annual expenses and replace-
ment of asset, 420-422
paid at longer-than-year intervals, 364—
365
Personal property, installment sales of,
160-161"
Piecework system, payroll records, 116
Policy, insurance, 95
cancellation of, 97
form of, 95
life insurance (see Life insurance;
policies)
Positive and negative numbers, 229-230
Powers of numbers, 225, 257
Premium :
bond, between interest dates 380
bonds purchased at, 368-369
calculating, 381-382
bonds sold at, 376-379
insurance, 96
interest and. 474
level, mortality and, 473
net (see Net premiums)
Present value:
annuity due, 352
Present value, annuity due (cont.):
compared with ordinary annuity.
352
computation of, 352-355
rent of, 357-358
of annuity of 1, table, 530-532
of 1, table. 520-526
ordinary annuity, 335 337, 338
Present worth:
compound, 320-321
of 1, table, 321
simple interest, 84
simple amount and, comparison. 84-
85
Principal, 77, 312
installment payment of, 339 340
interchanging time and, interest, 81
Probability, 445 452
and mortality, 453 458 (w aho
Mortality)
life insurance, 453
combinations in, 446 4 18
compound events, 4-18
empirical, 450 452
events (sec Kvents)
joint life, 457-458
of dying, 450-457
of living, 455-456
permutations in, 416-448
theory of, 445-446
Proceeds, bank discount, 88-89
Product, 16
Production indexes, 278
Product method:
account current, 139 140
averaging dates of invoices, 131—132
compound average, 136
Profit and loss statements, determining
earning power from, 201 202
Profits, 169
based on sales, percentage. 64 (56
gross, computations (see GrosM profit
computations)
partnership, distribution of, 220 221
ratios concerning, 169, 170- 171
Progression, 285-291
arithmetical:
decreasing series, finding values of
terms, 287-288
defined, 285
elements, relation of, 2S.1
increasing series, 28.r> 2S(>
finding values of terms, 285-280
symbols, 285
decreasing series, 28f> 2XS
defined, 285
defined, 285
geometrical, 288-2!) 1
decreasing series nnding values of
terms, 290 291
defined, 288
elements of, 280
increasing 8<-n<>s, finding values of
terms, 289
increasing series 289
problems, use of logarithms, 291
548
INDEX
Progressive average, 126
Proper fractions, 39
Property taxes, 225-228
Public finance:
and taxation, 223-228 (See also Taxes)
appropriations, 223-^224
governmental functions, 223
Q
Quarterly basis, computing annuity, 329-
331
Quotient, 24
R
Rate, 54, 312-313
compound interest, 312-313, 322-323
effective, ordinary annuity, 332, 341-
343
insurance, 95—96
of exchange. 293
of interest, 77
effective, 317-320
nominal, 317-319
of yield, bonds, 369
ordinary annuity, 343-346
Ratios :
average investment, profit sharing in,
186-188
chart, 275, 276-277
cost to retail, determining, 153-154
financial and operating, 169-180 (see
also Statements, analysis of:
ratios)
frequency, 453
geometrical progression, 290-291
of increase, compound interest, 313
parity, 281
partnership:
arbitrary, 182
of investment, 182-183
working capital, 169, 172-174
Receipts, pay, 120
Reciprocals, in division, 26-28
Record, daily, departmental sales, 56
Redemption, serial, bonds (see Bonds:
serial redemption)
Redemption periods, bonds, 387
Reduction of fractions, 39-40
Remainder, 24
Rent:
Aof annuity, 327
present value of, 338
of annuity due, 355-357
present value of, 357-358
Repeater, 67
Reserves :
denned, 473
for unearned gross profit, 162
policy, 473-^74
terminal, 475
Retail, determining ratio of cost to, 153-
154
Rider, insurance policy, 95
Roots of numbers, 225, 258
Running account, 141-142
S
Sales:
average, clerk's per cent, 57
bar charts, 268
departmental, daily record, 56
profits based on, percentage, 64-66
returned, by departments, 56-57
Serial redemption bonds:
amortization of discount, premium, or
discount and expense on, 398
404
nature of, 385
redeemed by other than equal annual
payments, 390
redemption periods, 387-390
value of, analysis of calculation of, 38.3
387, 388
value of series, verification of calcula
tion, 393-394
with interest payments, analysis of,
391-393
Series of discounts, 72
Short methods, 3-28
addition, 3-10 (see also Addition)
calculating simple interest, 77
calculating single discount, equivalent
to any two discounts, 73
division, 21-28 (see also Division)
multiplication, 16-24 (see also Multi-
plication)
subtraction, 10-16 (see also Subtraction)
Signs, arithmetic, 229
Simple amount, 82-83
Simple average, 1 23—1 2 1
Simple equations, 235-238
Simple interest, 77-86
accumulation of, Sl-82
aliquot-parts method, 78
cancellation method, 79-80
defined, 77
doilars-times-davs method, 80
interchanging principal and time, 81
present worth, 84
simple amount and, comparison, 84-
85
rate, 83
short method, 77
simple amount, 82-83
sixty-day method, 77-78
table of,' 82
time, 83-84
true discount, 85-^86
Simultaneous equations, 241-246
Sinking fund contributions, 334
Sinking-fund method, depreciation, 409-
410
Sixty-day method of calculating interest,
77-78
Slide rule:
accuracy of calculations made by, 259
described, 258-259
division on, 263-264
learning to use, 260
model, constructing, 261-262
multiplication on, 262-263
INDEX
549
Slide rule (co nt.):
reading, 261
theory of, 259-260
use of, 259
Social Security Act, 107
Statements, analysis of, 161M80
cost, expenses, profits, 169-170
ratios, 169-170
costs, expenses, to net sales, 169-170
financial and operating, 71
gross profit to net sales, 170
net profit to net sales, 170
net profit to net worth, 170
operating profit to net sales, 170
operating profit to total capital
em ploy od, 170
working capital, 172-174
relationships:
capital, now invested, 175
earnings on comimm stockholders'
investments, 170
sources of capital, 174
turnover :
accounts receivable, 175-176
fixed property investment, 176 180
inventories, 175
total capital employed, 175
State taxes, 225
Statistics, operating, percentage, 61 62
Stock:
allotment, goodwill, 205 209
classes of, building and loan associa-
tions, 425
rights, 213 211
sale of stock and, federal income tax,
214-216
shares of, book value of, 220
time reciiured to mature, 432-433
Stock card, 145
Stockholders, common, earnings on in-
vestments, 170
Storage, 141-142
defined, 141
running account, 141-142
Stores ledger, form, 146
Straight-line method, depreciation, 405-
407
Streamline addition, 4
Subtraction, 10-16
balancing account, 12
by algebra, 232-233
complement method, 12-14
difference between given minuend and
several subtrahends, 11-12
errors, avoiding, 11
of fractions, 40-41
decimal, 44-^45
on adding machine, 14
practical problems, 14—16
verification of, 30, 32-33
Subtrahend, 10
Sum, 3
Sum-of-digits method, depreciation, 408
Symbols:
and terms, algebra, 229
compound interest, 311-312
T
Tables:
aliquot parts, 49
amount of annuity, 331
of 1, 527-529
annuity, section of, 344
asset valuation accounts, 406, 407, 408,
409-410, 412, 414, 423
bond (see Bonds: tables)
Briggs', 272
cancellation, short rate, 98
compound amount, 313
of 1, 512-519
compound interest, analysis of, 317
drill, addition, 3 6
for finding difference between dates, 88
installment payments, 339
life annuities, 462 403
commutation columns in, 160 4(11
logarithms, 249, 252 253, 197 511
%mortality, American experience, 536 -
537
multiples, 17, 23 24, 25, 31
present value of annuity of 1, 530 532
present value of 1, 520-520
present worth of 1, 321
rent, present value annuity of 1, 533 •
r o *•
5Jo
simple interest, 82
sinking fund contributions, 321
use of, in division, 25 20
weights, measures, values, 489 490
Taxes:
deferring income, effect on, 166 168
in U.S., 1943, 221
kinds of, 224-225
public finance and taxation (xrc, Public,
finance)
purposes of, 223
Taxpayer's inventory, gross profit test in,
158-100
Temporary lift1 annuities, -4(55
due, 465-466
forborne, 467 -468
Terminating plan, 12'» 127
Term insurance, -170, 471
Term of credit, 131
Thirteen, check number, 33 3-1
Time:
bill of exchange, value of, 290 297
compound interest, 312, 323 324
counting, bank discount, 87
interchanging principal and, interest,
81
simple interest, 83-84
Timebooks, 107
Time-clock cards, 107-109
forms, 108
Town taxes, 223
Trade:
discount. 72
foreign (see Foreign exchange)
indexes, 278
Transportation charges, on discount in-
voices, 75
550
INDEX
True discount, simple interest. 85-86
Turnover, merchandise, 148-150
U
Unearned gross profit, reserve for, 102
Unit, 39
United States rule, partial payments, 91-
92, 93
Unit-product method, depreciation, 407-
408
Unknown quantities, solution of prob-
lems containing, 246-248
Use and occupancy insurance, 99-102
Valuation:
of asset accounts (see Asset valuation
accounts)
of bonds, 369, 372 (see also Bonds)
of goodwill (see Goodwill: methods ^>f
valuing)
of iuvtntories, 143
Valuation (cont.):
of life insurance policies (see Life in-
surance policies)
Voting, cumulative, 219-220
W
Wasting assets, capitalization of, 422-424
Weighted average, 127-129
Weighted index numbers, 281-284
Withholding exemptions, payroll records,
109
Withholding tax, 107
Working capital, 216-219
ratio, 169, 172-174
W. .• .• method, depreciation
4U7-408
Workmen's compensation, 104, 105
Y
Yield basis, bond bought on,
Yields of Bonds and Stocks, <
382-383
370, 371
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