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LIBRARY
University of California.
GIFT OF
^' r
/:
t •
: .7 ^
^
/
/
r
* W
ECLECTIC EDUCATIONAL SERIES,
RAY'S
NEW HIGHER
ARITHMETIC
A Bevised Edition of the Higher Arithmetic
BY
JOSEPH RAY, M. D.
'♦
LaU Ptv/eaaor in Woodvxxrd ColUfie,
IVERSITY
VAN ANTWERP, BRAGG & CO.
CINCINNATI AND NEW YORK.
Ray's Mathematical Series.
ARITHMETIC.
Bay's New Primary Arithmetic.
Ray's New Intellectual Arithmetic.
Ray's New Practical Arithmetic.
Bay's New Higrher Arithmetic.
TWOBOOK SERIES.
Bay's New Elementary Arithmetic.
Bay's New Practical Arithmetic.
AliGEBRA.
Bay's New Elementary Algrebra.
Bay's New Higrher Algrebra.
HIGHER MATHEMATICS
Bay's Plane and Solid Oeometry.
Bay's Oeometry and Trigronometry.
Bay's Analytic Oeometry.
Ray's Elements of Astronomy.
Bay's Surveyingr and Navigration.
Bay's Differential and Integrral Calculus.
Copyright,
1880,
BY Van Antwerp, Braoo & Co.
PREFACE.
Ray*s Higher Arithmetic was published nearly twentyfive
years ago. Since its publication it has had a more extensive circii
iation than any other similar treatise issued in this country. To
adapt it more perfectly to the wants of the present and future, it has
been carefully revised.
It has been the aim of the revision to make Ray's New Higher
Arithmetic thoroughly practical, useful, and teachable. To this
end the greatest care has been given to securing concise definitions
and explanations, and, at the same time, the systematic and thorough
presentation of each subject. The pupil is taught to think for him*
self correctly, and to attain his results by the shortest and best
methods. Special attention is given to modem business transactions,
and all obsolete matter has been discarded.
Almost every chapter of the book has been entirely rewritten,
without materially changing the general plan of the former edition,
although much new, and some original matter has been introduced.
Many of the original exercises are retained.
Particular attention is called to the rational treatment of the
Arithmetical Signs, to the prominence given to the Metric System,
and to the comprehensive, yet practical, presentation of Percentage
and its various Applications. The method of combining the algebraic
and geometric processes in explaining square and cube root will com
mend itself to teachers. The chapter on Mensuration is unusually'
full and varied, and contains a vast amount of useful information.
(Hi)
11 I 90S
IV PREFACE.
The Topical Outlines for Keview will prove invaluable to both
teachers and pupils in aiding them to analyze and to classify their
arithmetical knowledge and to put it together so as to gain a com
prehensive view of it as a whole.
Principles and Formulas are copiously interspersed as summaries,
to enable pupils to work intelligently.
The work, owing to its practical character, logical exactness, and
condensation of matter, will be found peculiarly adapted to the wants
of classes in High Schools, Academies, Normal Schools, Commercial
Schools and Colleges, as well ad to private students.
The publishers take this opportunity of expressing their obligations
to J. M. Gbeenwood, a. M., Superintendent of Public Schools, Kan
sas City, Mo., who had the work of revision in charge, and also to
Rev. Db. U. Jesse Knisely, of Newcomerstown, Ohio, for his
valuable assistance in revising the final proofsheets.
Cincinnati, Jvly^ 1880.
CONTENTS.
PAGE
I. Introduction 9
II. Numeration and Notation 15
III. Addition 23
IV. Subtraction 27
Business Tenns and Explanations 29
V. Multiplication 31
When the multiplier does not exceed 12 . . . .32
When the multiplier exceeds 12 . . . . . .34
Business Terms and Explanations 36
Ck)ntractions in Multiplication 38
VI. Division 43
Long Division 45
Short Division 47
Contractions in Division 49
Arithmetical Signs 50
General Principles 52
Contractions in Multiplication and Division ... 53
VII. Properties OP Numbers 59
Factoring 61
Greatest Common Divisor 64
Least Common Multiple 68
Some Properties of the Number Nine ^ . . . 70
Cancellation 72
VIII. Common Fractions 75
Numeration and Notation of Fractions .... 77
Reduction of Fractions 78
Common Denominator 82
Addition of Fractions 85
Subtraction of Fractions 86
Multiplication of Fractions r / ^'
vi CONTENTS.
FAOB
Division of Fractionfl 90
The G. C. D. of Fractions 92
The L. C. M. of Fractions 94
IX. Decimai. Fractions 99
Numeration and Notation of Decimals 100
Eeduction of Decimals 103
Addition of Decimals 105
Subtraction of Decimals 106
Multiplication of Decimals . . . . . . 108
Division of Decimals Ill
X. Circulating Decimals 115
Beduction of Circulates 118
Addition of Circulates 120
Subtraction of Circulates 121
Multiplication of Circulates ^ . 122
Division of Circulates 123
XI. Compound Denominate Numbers 125
Measures of Value 125
Measures of Weight 130
Measures of Extension 133
Measures of Capacity 139
Angular Measure 142
Measure of Time 143
Comparison of Time and Longitude .... 146
Miscellaneous Tables 146
The Metric System 147
Mea.sure of Length 149
Measure of Surface 149
Measure of Capacity 150
Measure of Weight . . * 150
Table of Comparative Values 152
Reduction of Comi>ound Numbers 154
Addition of Compound Numbers 160
Subtraction of Compound NumberH ..... 163
Multiplication of Compound Numbers .... 165
Division of Compound Numbers 167
Longitude and Time 169
Aliquot Parts 172
CONTENTS. vii
PAOC
XII. Ratio 175
XIII. Proportion 177
Simple Proportion 178
Compound Proportion 184
XIV. Percentage 188
Additional Formulas . 197
Applications of Percentage 197
XV. Percentage. — AppLiCATiONa (Without Time,) . . 199
I. Profit and Loss 199
II. Stocks and Bonds 204
ni. Premium and Discount 208
IV. Commission and Brokerage 213
V. Stock Investments 220
VI. Insurance 228
VII. Taxes 232
VIII. United States Revenue 236
XVI. Percentage. — Applications. ( Wilh Time.) . . . 242
I. Interest 242
Common Method 245
Method by Aliquot Parts 246
Six Per Cent Methods 246
Promissory Notes 254
Annual Interest 259
II. Partial Payments 261
U.S. Rule 261
Connecticut Rule 264
Vermont Rule 265
Mercantile Rule 266
in. True Discount 266
IV. Bank Discount . 268
v. Exchange 278
Domestic Exchange 279
Foreign Exchange 281
Arbitration of Exchange 283
VI. Equation of Payments 286
VII. Settlement of Accounts 292
Account Sales 296
Storage Accounts ^7
viii CONTENTS.
PAGE
vni. Compound Interest . . . . . . . 298
IX. Annuities 308 /
Contingent Annuities .317
Personal Insurance 322
XVn. Partnership 327
Bankruptcy 331
XVIII. Alligation . 333
Alligation Medial 333
Alligation Alternate 334
XIX. Involution 342
XX. Evolution 347
Extraction of the Square Root 349
Extraction of the Cube Root 354
Extraction of Any Root 359
Homer's Method 360
Applications of Square Root and Cube Root . . . 363
Parallel Lines and Similar Figures 366
XXI. Series 369
Arithmetical Progression 369
Geometrical Progression 373
XXII. Mensuration 378
Lines 378
Angles 379
Surfaces 379
Areas . . . • 382
Solids 390
Miscellaneous Measurements 395
Masons' and Bricklayers' Work .... 395
Gauging 396
Lumber Measure 398
To Measure Grain or Hay 398
XXIII. Miscellaneous Exercises 401
OF THE
UNIVERSITY
RAT'S
HIGHER ARITHMETIC.
I. mTEODUOnOK
Article 1. A definition is a concise description of any
object of thought, and must be of such a nature as to dis
tinguish the object described from all other objects.
2. Quantity is any thing which can be increased or
diminished; it embraces number and magnitude. Number
answers the question, **How many?" Magnitude, **How
much?"
3. Science is knowledge properly classified.
4. The primary truths of a science are called Prin
ciples.
6. Art is the practical application of a principle or the
principles of science.
6. Mathematics is the science of quantity.
7. The elementary branches of mathematics are Arith
metic, Algebra, and Geometry.
8. Arithmetic is the introductory branch of the science
of numbers. Arithmetic as a science is composed of defini
(9)
10 BA Y'S HIGHER ARITHMETIC,
tions, principles, and processes of calculation ; as an art, it
teaches how to apply numbers to theoretical and practical
purposes.
9. A Proposition is the statement of a principle, or of
something proposed to be done.
10. Propositions are of two kinds, demonstrable and
indemonstrable.
Demonstrable propositions can be proved by the aid of
reason. Indemonstrable propositions can not be made simpler
by any attempt at proof.
11. An Axiom is a selfevident truth.
12. A Theorem is a truth to be proved.
13. A Problem is a question proposed for solution.
14. Axioms, theorems, and problems are propositions.
15. A process of reasoning, proving the truth of a prop
osition, is called a Demonstration.
16. A Solution of a problem is an expressed statement
showing how the result is obtained.
17. The terra Operation, as used in this book, is applied
to illustrations of solutions.
18. A Rule is a general direction for solving all prob
lems of a particular kind.
19. A Formula is the expression of a general rule or
principle in algebraic language; that is, by symbols.
20. A Unit is one thing, or one. One tiling is a con
crete unit ; one is an abstract unit.
21. Nuniber is the expression of a definite quantity.
Numbers are either abstract or concrete. An abstra^;^ num
ber is one in which the kind of unit is not named ; a concrete
number is one in which the kind of unit is named. Con
crete numbers are also called Denominate Numbers.
INTRODUCTION. H
22. Kumbers are also divided into Integral, Fractional,
and Mixed.
An Integral number , or Integer, is a whole number; a
Fractimud nmnber is an expression for one or more of the
equal parts of a divided whole; a Mixed number is an
Integer and Fraction united.
23. A Sign is a character used to show a relation among
numbers, or that an operation is to be performed.
24. The signs most used in Arithmetic are
+  X ^ v/
()
• .
•• \ J ••••
25. The sign of Addition is [+], and is called plus. The
numbers between which it is placed are to be added.
Thus, 3 + 5 equals 8.
Plus is described as a perpendicular cross, in which the bisecting
lines are equal.
26. The sign of Subtraction is [ — ], and is called minus.
When placed between two numbers, the one that follows it
is to be taken from the one that precedes it. Thus, 7 — 4
equals 8.
Minus is described as a short horizontal line.
Plus and Minus are Latin words. I^us means more; minus means
less.
Michael Steifel, a German mathematician, first introduced {
and — in a work published in 1544.
27. The sign of Multiplication is [ X ]> and is read mul
iiplied by, or tim^s. Thus, 4 X 5 is to be read, 4 multiplied
by d, or 4 times 5.
The sign is described as an oblique cross.
William Oughtred, an Englishman, born in 1573, first introduced
the sign of multiplication.
28. The sign of Division is [ ^ ], and is read divided by.
When placed between two numbers, the one on the left is
12 BAY'S HIGHER ARITHMETIC.
to be divided by the one on the right. Thus, 20 =4
equals 5.
The sign is described as a short horizontal line and two dots;
one dot directly above the middle of the line, and the other jnst
beneath the middle of it.
Dr. John Pell, an English analyst, born in 1610, introduced the
sign of division.
29. The Radical »ign, [ l/ ]» indicates that some root is
to be found. Thus, /36 indicates that the square root of
36 is required; ^125, that the cube root of 125 is to be
found; and v^625 indicates that the fourth root of 625 is
to be extracted.
The root to be found is shown by the small figure placed between
the branches of the Eadical sign. The figure is called the index.
30. The signs, +> — > X, r, i/, are symbols of
operation.
31. The sign of Equality is [=], two short horizontal
parallel lines, and is read eqvxds or is equal to, and sig
nifies that the quantities between which it is placed are
equal. Thus, 3 + 5=9 — 1. This is called an equaJthOUy
because the quantity 3 + 5 is equal to 9 — 1.
32. Ratio is the relation which one number bears to
another of the same kind. The sign of Ratio is [ : ]. Ratio
is expressed thus, 6 : 3 = f = 2, and is read, the ratio of
6 to 3 = 2, or is 2.
The sign of ratio may be described as the sign of division with
the line omitted. It has the same force as the sign of division, and
is used in place of it by the French.
33. Proportion is an equality of ratios. The sign of
Proportion is [::], and is used thus, 3: 6:; 4: 8; this
may be read, 3 is to 6 as 4 is to 8; another reading, the
ratio of 3 to 6 is equal to the ratio of 4 to 8.
INTRODUCTION.
13
34. The signs [(), ], are signs of Aggregation — the
first is the Parenihem, the second the Yinadum, They are
used for the same purpose; thus, 24 — (8 f 7), or 24 —
8  7, means that the sum of 8 + 7 is to be subtracted
from 24. The numbers within the parenthesis, or under
the vinculum, are considered as one quantity.
36. The dots [. . . .], used to guide the eye from words
at the left to the right, are called Leaders, or the sign of
Gontimuitiony and are read, and so on.
36. The sign of Dedvetion is [.'.], and is read therefore^
hence, or consequently.
37. The signs, =, :, : :, ( ), , . . . ., .•.,
are symbols of relation.
38. Arithmetic depends upon this primary proposition:
that any number may be increased or diminished. ** In
creased" comprehends Addition, Multiplication, and Invo
lution; "decreased," Subtraction, Division, and Evolution.
39. The fundamental operations of Arithmetic in the
order of their arrangement, are : Numeration and Notation,
Addition, Subtraction, Multiplication, and Division.
Topical Outline.
Introduction.
1. Definition.
2. Quantity.
3. Science.
4. Principles.
5. Art.
6. Mathematics.
7. Proposition.
8. Demonstration.
9. Solution.
10. Operation.
11. Rule.
12. Formula.
13. Unit.
14. Number.
15. Sign. .
16. Signs most used.
17. Primary Proposition.
18. Fundamental Operations.
14
HAY'S HIGHER ARITHMETIC.
Topical Outline of Arithmetic.
Preliminary Definitions...
1. Definition.
2. Quantity.
3. Science.
4. Mathematics.
5. Proposition.
6. Tlieorem.
7. Axiom.
8. Demonstration.
9. Solution.
10. Rule.
l^ 11. Sign.
1. Definitions.
1. As a Science... ^ 2. Classification of Numbers
^ ( Numeration
1.
3. Operations. ^
and
, Notation.
2. Addition.
3. Subtraction.
4. Multiplication.
5. Division.
6. Involution.
7. Evolution.
{Abstract
Ck)ncrete.
' Integral.
2.  Fractional.
. Mixed,
g / Simple.
v^ I Compound.
r 1. Terms often used.
2. As an Art < 2. Signs.
^ 3. Applications.
/'I. Problem.
2. Operation
3. Solution.
4. Principle.
5. Formula.
6. Rule.
^ 7. Proof.
f 1. To Integers.
2: To Fractions.
3. To Compound Numbers,
4. To Ratio and Proportion.
5. To Percentage.
6. To Alligation.
7. To Progression.
8. To Involution and Evolution,
l^ 9. To Mensuration.
n. IsTUMERATION AKD NOTATIOK
40. Kiuneration is the method of reading numbers.
Notation is the method of writing numbers. Numbers
are expressed in three ways ; viz., by words, letters, and figures.
41. The first nine numbers are each represented by a
single figure, thus:
I 23456789
one. two. three, four. five. six. seven, eight, nine.
All other numbers are represented by combinations of
these and another figure, 0, called zero, naught, or cipher.
Kemabk. — The cipher, 0, is used to indicate no value. The other
figures are called significant figures, because they indicate some value.
42. The number next higher than 9 is named ten, and
is written with two figures, thus, 10 : in which the cipher, 0,
merely serves to show that the unit, 1, on its left, is different
from the unit, 1, standing alone, which represents a single
thing, while this, 10, represents a single group of ten things.
The nine numbers succeeding ten are written and named
as follows:
II 12 13 14 15 16
eleven, twelve. thirteen, fourteen, fifteen. sixteen.
17 18 19
seventeen, eighteen, nineteen.
In each of these, the 1 on the left represents a group of ten
tilings, while the figure on the right expresses the units or
single things additional, required to make up the number.
Kemabk. — The words eleven and tivelve are supposed to be derived
from the Saxon, meaning one left after ten, and two left after ten. The
words thirteen, fourteen, etc., are contractions of three and ten, ftmr and
te»,ete. ^^^j
16 HAY'S HIGHER ARITHMETIC,
The next number above nineteen {nine and fen), is ten
and ten^ or two groups of few, written 20, and called twenty.
The next numbers are twentyoney 21 ; twentytwo, 22 ; etc. ,
up to three tern, or thirty, 30 ; forty, 40 ; fifty, 50 ; sixty, 60 ;
seventy, 70 ; eighty, 80 ; ninety, 90.
The highest number that can be written with two figures
is 99, called ninetynine; that is, nine tens and nine units.
The next higher number is 9 tens and ten, or ten tens,
which is called one hundred, and written with three figures,
100; in which the two ciphers merely show that the unit on
their left is neither a single thing, 1, nor a group of ten
things, 10, but a group of ten tens, being a unit of a higher
order than either of those already known.
In like manner, 200, 300, etc., express two hundreds, three
hundreds, and so on, up to ten hundreds, called a thousand,
and written with four figures, 1000, being a unit of a still
higher order.
43. The Order of a figure is the place it occupies in a
number.
From what has been said, it is clear that a figure
in the 1st place, with no others to the right of it, expresses
units or single things; but standing on the left of another
figure, that is, in the 2d place, expresses groups of tens;
and standing at the left of two figures, or in the 3d place,
expresses tens of tens, or hundreds; and in the 4th place,
expresses tens of hundreds or Hwusands, Hence, counting
from the right hand.
The order of Units is in the 1st place, 1
The order of Tens is in the 2d place, 10
The order of Hundreds is in the 3d place, 100
The order of ThotLsands is in the 4th place, 1000
By this arrangement, the same figure has differerd values
according to the place, or order, in which it stands. Thus, 3
in the first place is 3 units; in the second place 3 tens, or
thirty; in the third place 3 hundreds; and so on.
NUMERATION AND NOTATION.
17
44i The word Units may be used in naming all the orders,
as follows:
Simple units are called Units of the Ist order.
Tens " " Units of the 2d order.
Hundreds " " Uniis of the Sd order.
Thousands " " Units ^ the 4th order.
etc. etc.
45. The following table shows the place and name of
each order up to the fifteenth.
Table op Orders.
15th. 14th. 13lh. 12th. 11th. 10th. 9th. 8th. 7th. 6th. 5th. 4th. 8d. 2d. Ist
ac
o
00
•
•
•
•
•
•
00
^3
•
•
•
OD
a
•
00
G
•
a
03
•
•
OD
c
o
■
o
•
oo
a
o
O
•
oo
a
o
OQ
S
o
00
S
C3
00
f^
•4H
«44
p^
«4H
9
O
H
.^4
o
00
O
00
O
00
^3
00
O
OQ
a
o
n3
2
o
3
c
^
O
oo
S
C
1
o
10
21
«iH
'V
o
TS
TJ
p
o
'V
00
a
c
a
pq
S
00
a
1—^
C
P
W
00
a
s
p
w
oo
C
H
00
■♦»
46. For convenience in reading and writing numbers,
orders are divided into groups of three each, and each
group is called a period. The following table shows the
grouping of 'the first fifteen orders into five periods :
Table of Periods.
00
a
o
o S
PQ
CO
C
O
O Z3
OQ
G
a
w
6 5 4
00
00 ^
a 'S
3 2 1
ac
'V
0)
C
P
w
00
OQ *J
(K
6th Period. 4th Period.
H. A. 2.
9 8 7
3d Period.
6 5 4
2d Period.
3 2 1
Ist Period.
18 HA Y'S HIGHER ARITHMETIC.
47. It will be observed that each period is composed of
units, tens, and hundreds of the same denomination.
48. List of the Periods, according to the common or
French method of Numeration.
First Period, Units.
Second " Thousands.
Third " Millions.
Fourth " Billions.
Fifth " Trillions.
Sixth Period, Quadrillions.
Seventh " Quiniillions.
Eighth '' Sextillions.
Ninth " Septillions.
Tenth " Ckitillions.
The next twelve periods are, Nonillions, Decillions, Undecillions,
Duodecillions, Tredecillions, Quatuordecillions, Quindecillions,
Sexdecillions, Septendecillions, Octodecillions, Novendecillions,
Vigintillions.
Principles. — 1. Ten units of any order always make one
of ike next higher order,
2. Removing a significant figure one place to the left increases
its value tenfold; one pkice to the right, decreases iti value ten
fold,
3. Vojcant orders in a number are filled with ciphers.
Problem. — Express in words the number which is repre
sented by 608921045.
Solution. — The number, as divided into periods, is 608*921*045;
and is read six hundred and eight million nine hundred and
twentyone thousand and fortyfive.
Bule for Numeration. — 1. Begin at the right , and point
the number into periods of three figures each.
2. Commence at the left, and read in succession each period
with its name.
Remark. — Numbers may also be read by merely naming each
figure wiih the name of the place in which it stands. This method,
however, is rarely used except in teaching beginners. Thus, the
numbers expressed by the figures 205, may be read two hundred and
five, or two hundreds no tens and five units.
NUMERATION AND NOTATION. 19
Examples in Numeration.
7
4053
204026
4300201
40
7009
500050
29347283
85
12345
730003
45004024
278
70500
1375482
343827544
1345 165247 6030564 830070320
832045682327825000000321
8007006005004003002001000000
60030020090080070050060030070
504630!209J02'80OV0^24d703l2505O7
Problem. — Express in figures the number four million
twenty thousand three, hundred and seven. 4020307.
SoiiUTiON. — Write 4 in miUyms period; place a dot after it to
separate it from the next period : then write 20 in thoiisands period ;
pXace another dot: then write 307 in units period. This gives
4*20'307. As there are but <wo places in the thousands period, a
cipher must he put before 20 to complete its orders, and the number
correcHy written, is 4020307.
Note. — Every period, except the highest, must have three figures ;
and if any period is not mentioned in the given number, supply its
place with three ciphers.
Bule for Notation. — Begin at (lie lefty and write each
period in its proper place— filling ilie vacant orders with ciphers.
Proof. — ^Apply to the number, as written, the rule for
Numeration, and see if it agrees with the number given.
1. Seventyfive.
2. One hundred and thirtyfour.
3. Two hundred and four.
4. Three hundred and seventy.
5. One thousand two hundred
and thirtyfour.
Examples m Notation.
6. Nine thousand and seven.
7. Forty thousand five hundred
and sixtythree.
8. Ninety thousand and nine.
9. Two hundred and seven thou
sand four hundred and one.
20
BAY'S mo HER ARITHMETIC.
10. Six hundred and forty thou
sand and forty.
11. Seven hundred thousand and
seven.
12. One million four hundred
and twentyone thousand six
hundred and eightyfive.
13. Seven million and seventy.
14. Ten million one hundred
thousand and ten.
15. Sixty million seven hundred
and five thousand.
16. Eight hundred and seven
million forty thousand and
thirtyone.
17. Two billion and twenty mill
ion.
18. Nineteen quadrillion twenty
trillion and five hundred
billion.
19. Ten quadrillion four hun
dred and three trillion
ninety billion and six hun
dred million.
20. Eighty octillion sixty sex
tillion three hundred and
twentyfive quintillion and
thirtythree billion.
21. Nine hundred decillion sev
enty nonillion six octillion
forty septillion fifty quad
rillion two hundred and
four trillion ten million
forty thousand and sixty.
English Method of Numeration.
49. In the English Method of Numeration six figures
make a period. The first period is wnife, the second millions^
the third biUions, the fourth triUionSy etc.
The following table illustrates this method :
§
H
>w.
OQ
H
OD
^ .2
./^
m
CD
H
OD
G
CO
S3
O
a
H
oo
u
C
CO
C
c
p
CO
a
.13
00
a
CO
a
00
O
S3
H
CO
D
CO »J
r; "fh
OQ
w H p tn
H
00
CO
S
O
S3
H
00
00
00 M
§ C
HP
00
0)
w
H
OD
a
00
a
a
00
O
00
P
00 ■»
432109 876543 210987 654321
By this system the twelve figures at the right are read, two
hundred and ten thousand nine hundred and eightyseven
NUMERATION AND NOTATION
21
million six hundred and fiftyfour thousand three hundred
and twentyone. By the French method they would be read,
two hundred and ten billion nine hundred and eightyseven
million six hundred and fiftyfour thousand three hundred
and twentyone.
Roman Notation.
50. In the Roman Notation, numbers are represented by
seven letters. The letter I represents one; V, five; X, ten;
L, fifty; C, (yrve hundred; D, five hundred; and M, one thou
sand. The other numbers are represented according to the
following principles :
1st. Every time a letter is repeated, its value is repeated.
Thus, n denotes tujo; XX denotes tu^enty.
2d. Where a letter of less value is placed before one of
greater value, the less is taken from the greater; thus, TV
denotes fi)ur,
3d. Where a letter of less value is placed after one of
greater value, the less is added to the greater; thus, XI
denotes eleven,
4th. Where a letter of less value stands between two letters
of greater value, it is taken from the following letter,
not added to the preceding; thus, XIV denotes fourteen,
not mieen,
5th. A bar [ — 1 placed over a letter increases its value a
thousand times. Thus, V denotes five thoumnd; M denotes
one million.
»
Roman Table.
I One.
II Two.
Ill Three.
IV Four.
V . ^ Five.
VI Six.
IX Nine.
X Ten.
XI Eleven.
XIV Fourteen.
XV Fifteen.
XVI Sixteen.
XIX Nineteen.
XX Twenty.
XXI Twentyone.
XXX Thirty.
22
RAY'S HIGHER ARITHMETIC
XL
L
LX
XC
C
cccc
D
.OMAN TaBT.E.
(Continued.)
Forty.
DC ....
Six hundred.
Fifty.
DCC ....
Seven hundred
Sixty.
DCCC ....
Eight hundred
Ninety.
DCCCC . . .
Nine hundred.
One hundred.
M ....
One thousand.
Four hundred.
MM ....
Two thous,and.
Five hundred.
MDCCOLXXXI
1881.
1. Definitions.
Topical Outline.
Numeration and Notation.
2. Methods.
1. Arabic.
1. Characters.....
1. Names.
2. Values.
{i
1. Significant
Zero.
Simple.
2. Terms.
f 1. Order.
I 2. Periods, i ^' ^"^^^^
2. Roman... g ^
oS
3. Principles.
4. Rules.
1. Names.
LocaL
Frencl
I 2. English.
2. Values...
1. Alone.
2. Repeated.
3. Preceding.
4. Following.
5. Between.
6. Line Above.
3. Ordinary Language.
ni. ADDITION.
51. Addition is the process of uniting two or more
like numbers into one equivalent number.
Sum or Amoimt is the result of Addition.
62. Since a number is a collection of units of the same
kind, two or more numbers can he united into one sum, only
when their units are of the same kind. Two apples and 3
apples are 5 apples; but 2 apples and 3 peaches can not be
united into one number, either of apples or of peaches.
Nevertheless, numbers of different names may be added together,
if they can be brought under a common denomination.
Principles.— 1. Only like numbers can be added.
2. The sum is equal to dU the units of aU Hie parts.
3. The sum is the same in kind as the numbers added.
4. Units of the same order, and only such, can he added
direcUy.
5. The sum is the same in whatever suecession the numbers
are added.
Bemark. — Like numbers, similar numbers, and numbers of the same
kind are those having the same unit.
Problem. —What is the sum of 639, 82, and 543?
Solution. — ^Writing the numbers as in the margin, operation.
Bay, 3, 5, 14 units, which are 1 ten and 4 units. Write 63 9
the 4 units beneath, and carry the 1 ten to the next 82
column. Then 1,,5, 13, 16 tens, which are 6 tens to be 5 43
written beneath, and 1 hundred to be carried to the 12 64
next column. Lastly, 1, 6, 12 hundreds, which is set
beneath, there being no other columns to carry to or add.
Demonstration. — 1. Figures of the same order are written in the
(23)
24 RAY'S HIGHER ARITHMETIC.
same column for ccnveniencej since none but units of the same name
can be added. (Art. 62.)
2. Commence at the right to add, so that when the sum of any
column is greater than nine, the tens may be carried to the next
column, and, thereby, units of the same name added together.
3. Carry one for every ten, since ten units of each order make one
unit of the order next higher.
Rule for Adding Simple Numbers. — 1. Wriie the num
bers to be addedy so that figures of the same order mmj stand
in the same column, and draw a line directly beneath.
2. Begin at Hie right and add each column separately^
writing the units under the column added, and carrying the
tens, if any, tx) Hie next column. At the last column ivrite the
last whole amount.
Methods of Proof. — 1. Add the figures downward
instead of upward; or
2. Separate the numbers into two or more divisions;
find the sum of the numbers in each division, and then
add the several sums together; or
3. Commence at the left; add each column separately;
place each sum under that previously obtained, but extend
ing one figure further to the right, and then add them
together.
In each of these methods the result should be the same
as when the numbers are added upward.
Note. — For proof by casting out the 9's, see Art. 105.
Examples for Practice.
Find the sum,
1. Of 76767 ; 7654 ; 50121 ; 775. Ans. 135317.
2. Of 97674; 686; 7676; 9017. Ans. 115053.
3. Of 971; 7430; 97476; 76734. Ans. 182611.
4. Of 999; 3400; 73; 47; 452; 11000; 193; 97; 9903;
42 ; and 5100. Ans. 31306.
5. Of four hundred and three ; 5025 ; sixty thousand and
seven; eightyseven thousand; two thousand and ninety;
and 100. 154625.
6. Of 20050; three hundred and seventy thousand two
hundred ; four million and five ; two million ninety thousand
seven hundred and eighty; one hundred thousand and
seventy; 98002; seven million five thousand and one; and
70070. 13754178.
7. Of 609505 ; 90070 ; 90300420 ; 9890655 ; 789 ; 37599 ;
19962401; 5278; 2109350; 41236; 722; 8764; 29753; and
370247. 123456789.
8. Of two hundred thousand two hundred; three hundred
million six thousand and thirty; seventy million seventy
thousand and seventy ; nine hundred and four million nine
thousand and forty; eighty thousand; ninety mUlion nine
thousand; six hundred thousand and sixty; five thousand
seven hundred. 1364980100.
In each of the 7 following examples, find the sum of the
consecutive numbers from A to B, including these numbers :
A.
B.
9.
119
131.
.4*18. 1625.
10.
987
1001.
Am. 14910.
11.
3267
3281.
Arts. 49110.
12.
4197
4211.
Am. 63060.
13.
5397
5416.
Am. 108130.
14.
7815
7831.
Am. 132991.
15.
31989
32028.
Am. 1280340.
16. Paid for cofiee, $375; for tea, $280; for sugar, $564;
for molasses, $119; and for spices, $75: what did the whole
amount to? $1413.
17. I bought three pieces of cloth: the first cost $87; the
second, $25 more than the first; and the third, $47 more
than the second: what did all cost? $358.
18. A man bought three bales of cotton. The first cost
H. A. 8.
26 ^A Y'S HIGHER ARITHMETIC
$325; the second cost $16 more than the first; and the
third, as much as both the others: what sum was paid for
the three bales? $1332.
19. A has $75; B has $19 more than A; C has as
much as A and B, and $23 more ; and D has as much as
A, B, and C together: what sum do they all possess?
$722.
• _
Suggestions. — ^Two things are of the greatest importance in
arithmetical operations, — absolute accuracy and rapidity. The
figures should always be plain and legible. Frequent exercises in
adding long columns of figures are recommended ; also, practice in
grouping numbers at sight into iem and iwerUies is a useful exercise.
Accountants usually resort to artifices in Addition to save time
and extra lahor, such as writing the number to be carried in small
figures beneath the column to which it belongs, also writing the
whole amount of each column separately, ete.
Topical Outline.
Addition.
1. Definitions.
2. Sign.
8. Principles.
' 1. Writing the Numbers.
2. Drawing Line Beneath.
3. Addiug, Reducing, etc
4. Operation..
5. Rule.
6. Methods of Proof.
IV. SUBTEAOTIOR
53. Subtraction is the process of finding the difierence
between two numbers of the same kind.
The larger number is the Minuend; the less, the Subtra
hend; the number left, the Difference or Remainder,
Minuend means to be diminished; subtrahend, to be sub
tracted,
64. Subtraction is the reverse of Addition, and since
none but numbers of the same kind can be added together
(Art. 62), it follows, therefore, that a number can be sub
tracted only from another of the same kind : 2 cents can
not be taken from 5 apples^ nor 3 cows from 8 horses.
Principles. — 1. Tlie minuend and subtrahend must be of the
same kind,
2. The difference is the same in hind as the minuend and
subtrahend,
3. The difference equals the minuend minus the subtrahend.
4. The minuend equals the difference plus the subtrahend,
5. The subtrahend equals the minuend minus the difference.
Problem. — From 827 dollars take 534 dollars.
OPERATION.
Solution. — After writing figures of the same $8 2 7
order in the same column, say, 4 units from 7 units 534
leave 3 units. Then, as 3 tens can not be taken $293 Bern,
from 2 tens, add 10 tens to the 2 tens, which make
12 tens, and 3 tens from 12 tens leave 9 tens. To compensate for
the 10 tens added to the 2 tens, add one hundred (10 tens) to the 5
hundreds, and say, 6 hundreds from 8 hundreds leaves 2 hundreds;
and the whole remainder is 2 hundreds 9 tens and 3 units, or 293.
DSMONSTRATIOK. — 1. 6inoe a number can be subtracted only from
28 BAY'S HIOHEE ARITHMETIC.
another of the Rame kind (Art. 64), figures of the same name are
placed in the same column to be convenient to each other, the
less number being placed below as a matter of custom.
2. Commence at the right to subtract,, so that if any figure is
greater than the one above it, the upper may be increased by 10,
and the next in the subtrahend increased by 1, or, as some prefer, the
next in the minuend decreased by 1.
Bule for Subtracting Simple Numbers. — 1. TFrife ihe
less number under the greater, placing units urtder units, tens
under tens, etc., and draw a line directly beneath.
2. Begin at the rigid, subtract each figure from iJie one above
it, placing the remainder beneatli.
3. If any figure exceeds the one above it, add ten to the
upper, subtract the lower from the sum, increa^ by 1 the units
if the next order in Hie subtraJiend, and proceed as before.
Proof. — Add the remainder to the less number. If the
work be correct, the sum will be equal to the greater.
Note. — For proof by casting out the 9's, see Art. 105.
Examples for Practice.
1. From 30020037 take 50009.
OPERATION.
Eemark. — When there are no figures in 30020037
the lower number to correspond with those 50009
in the upper, consider the vacant places 29 9 700 2 8 .Re?ii.
occupied by zeros.
2. From 79685 take 30253.  Ans. 49432.
3. From 1145906 take 39876. Ans. 1106030.
4. From 2900000 take 777888. Am. 2122112.
5. From 71086540 take 64179730. Ans. 6906810.
6. From 101067800 take 100259063. Ans. 808737.
7. How many years from the discovery of America in
1492, to the Declaration in 1776 ? 284 years.
8. A farm that coat $7253, was sold at a loss of $395 ;
for how much was it sold? $6858.
SUBTRACTION. 29
9. The difference of two numbers is 19034, and the
greater is 75421 : what is the less ? 66387.
10. How many times can the number 285 be subtracted
from 1425? 5 times.
11. Which is the nearer number to 920736; 18l6045 or
25427? Neither. Why?
12. From a tract of land containing 10000 acres, the
owner sold to A 4750 acres; and to B 875 acres less than
A : how many acres had he left ? 1375 acres.
13. A, B, C, D, are 4 places in order in a straight line.
From A to D is 1463 miles ; from A to C, 728 miles ; and
from B to D, 1317 miles. How fer is it from A to B, from
B toC, and from C to D?
A to B, 146 miles ; B to C, 582 miles ; C to D, 735 miles.
BUSINESS TERMS AND EXPLANATIONS.
55. Bookkeeping is the science and art of recording
business transactions.
56. Business records are called Accounts, and are kept
in bo<^ called Account Books. The books mostly used
are the Daybooh and Ledger. In the Daybook are recorded
the daily transactions in business, and the Ledger is used
to classify and arrange the results of all transactions under
distinct heads.
67. Each account has two sides : Dr. — Debits, and Cr. —
Credits. Sums a person oioes are his Debits; sums owing
to him are his Credits. The difference between the sum
of the Debits and the sum of the Credits, is called the
Balance. Debits are preceded by **To," and Credits by
'* By."
58. Finding the difference between the sum of the
Debits and the sum of the Credits, and writing it under
the less side as Balance, is called Balancing.
30
HAY'S HIGHER ARITHMETia
Practical Exercises.
Dr.
JAMES CRAIG.
Cr.
1878.
1878.
July 4
.. 6
» 10
H 25
» 81
To Merchandise . .
„ Interest ....
„ Sundries . . .
,, Merchandise . .
Ditto . . .
560 50
24 90
870 60
320 10
125 40
July 5
M 11
„ 26
„ 31
By Cash
„ BiUs Payable . .
„ Sundries . . .
„ Cash
" Balance . . . •
550 50
890 70
310 SO
100 00
49 50
1878.
1901 50
190150
Aug. 1
To Balance ....
49 50
Balance the following account:
Dr.
THOMAS BALDWIN.
Cr.
1879.
1879.
Jan. 3
» 16
„ 25
,. 31
To Merchandise . .
„ Sundries . . .
„ Cash .....
„ Merchandise . .
810 30
580 20
381 25
60 75
Jan. 7
„ 20
„ 31
By Sundries . • .
„ Cash
„ Merchandise . .
,, Balance ....
1
1000 00
300 00
225 20
1879.
Feb. 1
To Balance ....
Topical Outline.
Subtraction.
1. Definition.
2. Terms
3. Sign.
4. Principles.
5. Operation
6. Rule.
7. Proof.
8. Applications.
1. Minuend.
2. Subtrahend.
. 3. Diflference or Remainder.
1. Writing the Numbers.
2. Drawing Line Beneath.
3. Subtracting and Writing Difference.
Y. MULTIPLIOATION.
69. 1. Multiplication is taking one number as many
times as there are units in another ; or,
2. Multiplication is a short method of adding numbers
that are equal.
60. The number to be taken, is called the Multiplicand;
the other number, the Multiplier; and the result obtained,
the Product. The Multiplicand and Multiplier are together
called Factors (makers), because they make the Product.
PROBLEM.rHow many trees in 3 rows, each containing
42 trees?
. Solution. — Since 3 rows contain 3 times oPEBATioir.
as many trees as one row, take 42 three First row, 42 trees
times. This may be done by writing 42 Second row, 4 2 trees,
three times, and then adding. This gives Third row, 42 trees.
126 trees for the whole number of trees. 126 trees.
Instead,* however, of writing 42 three
times, write it <mee; then placing under it 42 trees,
the figure 3, the number of tiirus it is to be 3
taken, say, 3 times 2 are 6, and 3 times 4 126 trees,
are 12. This process is MvUipHcatioTi,
Principles. — 1. The multiplicand may be eiOier concrete
or abstra4±,
2. The multiplier mu^st always be an abstract number.
3. The product is the same in kind as the multiplicand.
4. The product is the same, tvhichever factor is taken as the
multiplier.
5. Tlie partial products are the same in kind as the midti
plicand.
6. The sum of the partial products is equal to the total product.
SA Y'S HIGHER ASITHMETIU.
MOLTTPLICATION TaBLE.
61. Multiplication is divided into two cases:
1, When the multiptter does wit exceed 12,
2. When the midlipHer exeeedt 12.
CASE I.
62. When the multiplier does not exceed 13.
Problem. — At the rat« of 53 miles an hour, how far
will a railroad car run in four hours?
Solution. — Here saj, 4 times 3 (units) are 12 operatioh,
(units); write the 2 in units' place, and carry the 63 miles.
1 (ten); then, 4 times 5 are 20, and 1 carried 4
makes 21 (tens), and tlie work is complete. 212 miles.
DehOnstratiOM. — The multiplier being written under the mul
tiplicand for convenience, begin with units, so that if the product
should contain tens, they may be carried to the ten«; and ho on for
each successive order.
MULTIPLICATION. 88
Since every figure of the multiplicand is multiplied, therefore,
the whole multiplicand is multiplied.
Bule. — 1. Write the mvUiplicandy and place the multiplier
under it, so thai units of tJie same order shall stand in the same
column, and draw a line beneath,
2. Begin with units; multiply each figure <^ the midtiplicand
by the multiplier y carrying as in Addition,
Proof. — Separate the multiplier into any two parts;
multiply by these separately. The sum of the products
must be equal to the first product.
Examples for Practice.
1. 195X3. Am. 585.
2. 3823X4. Ans. 15292.
3. 8765 X 5. Ans. 43825.
4. 98374 X 6. Ans. 590244.
5. 64382 X 7. Am. 450674.
6. 58765X8. Am, 470120.
7. 837941 X 9. Am. 7541469.
8. 645703 X 10. Am. 6457030.
9. 407649 X H. Am. 4484139.
10. If 4 men can perform a certain piece of work in 15
days, how long will it require 1 man?
Solution. — One man must work four times as long as four men.
4 X 15 days = 60 days.
11. How many pages in a halfdozen books, each con
taining 336 pages? 2016 pages.
12. How far can an ocean steamer travel in a week, at
the rate of 245 miles a day? 1715 miles.
13. What is the yearly expense of a cottonmill, if
$32053 are paid out every month ? $384^36.
14. A receives from his business an average of $45 a
day. He pays three clerks $3; three, $9; and three, $12
a week ; other expenses amount to $4 a day ; what are his
profits for one week? $174.
84 JfiA yS HIGHEB ABITHMETia
CASE II.
63. When the multiplier exceeds 12.
Problem.— Multiply 246 by 235.
Solution. — First multiply hy 6 operation.
(units), and place the first figure of 2 4 6
the product, 1230, under the 5 (units). 235
Then multiply by 3 (tens), and place 12 3 product by 5
the first figure of the product, 738, 738 product by 3
under the 3 (tens). Lastly, multiply 49 2 product by 2
by 2 (hundreds), and place the first STSTo product by 235
figure of the product, 492, under the 2
(hundreds). Then add these several products for the entire product
DEMONSrtiATiON.— The of the first product, 1230, is unU» (Art.
62). The 8 of the second product, 738, is tens, because 3 (tens) times
6 = 6 times 3 (tens) = 18 (tens) ; giving 8 (tens) to be written in the
tens' column. The 2 of the third product, 492, is hundreds, because 2
(hundreds) times 6 = 6 times 2 (hundreds) = 12 (hundreds), giving
2 (hundreds) to be written in the hundreds' column. The right
hand figure of each product being in its proper column, the other
figures will fall in their proper columns; and each line being the
product of the multiplicand by a part of the multiplier, their sum
will be the product by aU the parts or the whole of the multiplier.
Rule. — 1. Write the mvUiplier under the multiplicand,
placing fijurea (^ the tame order in the same cdumuy and draw
a line beneath.
2. Multiply each figure of the mvUiplieand by each figure of
Hie mxdtiplier successively ; first by the units^ figure, then by the
ten^ figure, etc.; phdng the righthand fi^pire of each product
under that fi>gure of the multiplier whidi produces it, then draw
a line beneath,
3. Add the several partial products together; their sum will be
the rexpdred produ^i.
Methods of Proof.— 1. Multiply the multiplier by the
multiplicand; this product must be the same as the first
product.
MULTIPLICATION.
85
2. The same as when the multiplier does not exceed 12.
Note. — For proof by casting out the 9*8, see Art, 106.
Bemark. — Although it is custom
ary to use the figures of the multi
plier in regular order beginning with
units, it will give the same product
to use them in any order, observing
that the righthand figure of ea4:h parlicU
product must he pku^ under the figure
of the multiplier which produces it.
OPERATION.
246
235
7 38 product by 30
492 product by 200
1230 product by 6
67 810 product by 235
Examples for Practice.
1. 7198X216.
2. 8862 X 189.
3. 7575X7575.
4. 15607X3094.
5. 93186X4455.
6. 135790X24680.
7. 3523725 X 2583.
8. 4687319 X 1987.
9. 9264397X9584.
10. 9507340X7071.
11. 1644405X7749.
12. 1389294X8900.
13. 2778588X9867.
14. 204265X562402.
Am. 1554768.
Am. 1674918.
Am. 57380625.
Am. 48288058.
Am. 415143630.
Am. 3351297200.
Am. 9101781675.
Am. 9313702853.
Am. 88789980848.
Am. 67226401140.
Am. 12742494345.
Am. 12364716600.
Am. 27416327796.
Am. 114879044530.
Practical Problems.
1. In a mile are 63360 inches: how many inches are
there in the circumference of the earth at the equator if
the distance be 25000 miles ? 1584000000 inches.
2. The flow of the Mississippi at Memphis is about 434000
cubic feet a second: required the weight of water passing
that point in one day of 86400 seconds, if a cubic foot of
water weigh 62 pounds? 2324851200000 pounds.
36 BAY'S HIGHER ARITHMETIC.
3. John Sexton sold 25625 bushels of wheat, at $1.20 a
bushel, and received in payment 320 acres of land, valued
at $50 an acre ; 60 head of horses, valued at $65 a head ;
10 town lots, worth $150 each ; and the remainder in
money : how much money did he receive ? $9350,
4. K light comes from the sun to the earth in 495
seconds, what is the distance from the earth to the sun,
light moving 192500 miles a second ? 95287500 miles.
5. If 3702754400 cubic feet of solid matter is deposited
in the Gulf of Mexico by the Mississippi every year, what
is the deposit for 6000 years? 22216526400000 cu. ft.
6. The area of Missouri is 65350 square miles: how
many acres are there in the State, allowing 640 acr^ to
each square mile? 41824000 acres.
7. In the United States, at the close of 1878, there were
81841 miles of railroad : if the average cost of building be
$50000 a mile, what has been the total cost of building
the railroads in this country? $4092050000.
8. The number of pounds of tobacco produced in this
country in 1870 was 260000000. If this were manufact
ured into plugs one inch wide and six inches long, and
four plugs weigh a pound, what would be the length in
inches of the entire crop ? 6240000000 inches.
BUSINESS TERMS AND EXPLANATIONS.
64. A Bill is an account of goods sold or delivered,
services rendered, or work done. Usually the price or value
is annexed to each article, and the date of purchase given.
It is customary to write the total amount off to the rigtt,
and not directly under the column of amounts added.
66. A Receipt is a written acknowledgment of pay
ment. The common form consists in signing the . name
after the, words ** Received Payment" written at the foot
of the bill. .. . .
MVLTIPLICA TION.
m
\, Joseph Allen bought of Seth Ward, at Springfield,
HI, Jan. 2, 1879, 30 barrels of flour, at $3.60 a barrel;
48 barrels of mess pork, at $16.25 a barrel; 16 boxes of
candles, at $3.50 a box; 23 barrels of molasses, at $28.75
a barrel; and 64 sacks of coffee, at $47.50 a sack. Place
the purchases in bill form.
1879.
Solution.
Joseph Allen,
Springfield, III., Jan. 2, 1879.
Bought of Seth Ward.
Jan.
2
>i
2
i>
2
j»"
2
>•
2
To 30 bl. flour.
@ S 8.60 a bl.
It
48
i>
mess pork, „ .16.25
ft
„ 16 boxes candles, „ 3.50 „ box
„ 23 bl. molasses, ,, 28.75 „ bl.
„ 64 sacks coffee, „
47.50 „ sack
1 108
00
780
00,
56
00
661
25
8040
00
^15
■
25
2. At St. Louis, March 1, 1879, Chester Snyder bought
of Thomas Glenn, 4 lb. of tea, at 40 ct. ; 21 lb. of butter,
at 21 ct.; 58 lb. of bacon, at 13 ct.; 16 lb. of lard, at 9
ct.; 30 lb. of cheese, at 12 ct.; 4 lb. of raisins, at 20 ct.;
and 9 doz. of eggs, at 15 ct. . Place these purchases in the
form of a receipted bill? $20.74.
66. A Statement of Account is a written form reur
dered to a customer, showing his debits and credits as they
appear on the books. The following is an example :
1880.
John SMrm,
Cincinnati, Feb. 2, 1880.
171 Account with Van Antwerp, Bragg & Co.
Jau.
2
10
29
81.
To 525 McGuffey's Revised First Readers, @ 16c.
„ 50 Ray's New Higher Arithmetics, „ 75c.
Cr.
By Cash ,
: ,, Merchandise
84
37
50
■75
»»
20
12
121
32
50
■75

. 988
JSf
38 BAY'S HIGHER ARITHMETIC.
3. James Wilson & Co. bought of the Alleghany Coal
Co., March 2, 1880, five hundred tons of coal, at $2.75 a
ton, and sold the same Company during the month, as
follows: March 3d, 14 barrels of flour, at $6.55 a barrel;
March 10th, 6123 pounds of sugar, at 8c. a pound; they
also paid them on account, on March 15th, cash, $687.50.
Make out a statement of account in behalf of the Alleghany
Coal Co. under date of April 1, 1880. $105.96.
CJONTRACnONS IN MULTIPLICATION.
CASE I.
67. When the multiplier is a composite number.
A Composite Number is the product of two or more
whole numbers, each greater than 1, called its facUyrs.
Thus, 10 is a composite number, whose factors are 2 and
5; and 30 is one whose factors are 2, 3, and 5.
Problem. — At 7 cents a piece, what will 6 melons cost?
Analysis. — Three times 2 times operation.
are 6 times. Hence, it is the same 7 cents, cost of 1 melon,
to take 2 times 7, and then take 2
this product 3 times, as to take 6 14 cents, cost of 2 melons,
times 7. The same may be shown 3
of any other composite number. J^ cents, cost of 6 melons.
Bule. — Separate the multiplier into two or more factors.
Multiply first by one of the factorSy then this product by another
factor, and so on till each factor has been used as a muUipiier.
The last prodiLct wiU be the result required.
Examples for Practice.
1. At the rate of 37 miles a day, how fer will a man
walk in 28 days ? 1036 miles.
MULTIPLICATION. 89
2. Sound moves about 1130 feet per second: how fer
will it move in 54 seconds? 61020 feet.
3. If an engine travel at an average speed of 25
miles an hour, how far can it travel in a week, or 168
hours? 4200 miles.
CASE II.
68. When the multiplier is 1 with ciphers annexed,
as 10, 100, 1000, etc.
Demonstration. — By the principles of Notation (Art 48),
placing one cipher on the right of a number, changes the units
into tens, the tens into hundreds, and so on, and, therefore, mtiUiplies
the number by 10.
Annexing tvoo ciphers to a number changes the units into hun
dreds, the tens into thousands, and so on, and, therefore, multiplies
the number by 100. Annexing three ciphers multiplies the number
by 1000, etc.
Rule. — Anr\£x to the mtdtiplicand as many ciphers as there
are in tfie mvltiplier; the resvlt will he the required product.
Examples for Practice.
1. Multiply 743 by 10. Ans. 7430.
2. Multiply 375 by 100. Ans. 37500.
3. Multiply 207 by 1000. Ans. 207000.
CASE III.
69. When ciphers are on the right in one or both
factors.
Problem.— Find the product of 5400 by 130.
OPERATION.
5400
Solution.— Find the product of 54 by 13, 130
and then annex three ciphers ; that is, as 16 2
many as there are on the right in both the 5J
factors. 702000
40 BAY'S JSIGHEM ARITHMETIC.
Analysis. — Since 13 times 54 = 702, it follows that 13 times 54
hundreds (5400) = 702 hundreds (70200) ; and 130 times 5400= 10
times 13 times 5400 = 10 times 70200 = 702000.
Bule. — Multiply as if there wei'C no ciphers on the right in
the numbers; then annex to the product as many ciphers as
there are on tJie right in boHi the factors.
Examples for Practice.
1. 15460 X 3200. Ans. 494720(K).
2. 30700 X 5904000. Ans. 181252800000.
CASE IV.
70. When the multiplier is a little less or a little
greater than 10, 100, 1000, etc.
Problem.— Multiply 3046 by 997.
Analysis. — Since 997 is equal to 1000 operation.
diminished by 3, to multiply by it is the same • 3 4 6
as to multiply by 1000 (that is, to annex 3 99 7
ciphers) and by 3, and take the difference of 3046000
the products ; and the same can be shown in 913 8
any similar case. 3036862
Note. — Where the number is a little greater than 10, 100, 1000,
etc., the two products must be added,
Bnle. — Annex to the mtdtiplicand as many ciphers as there
are figures m the multiplier; multiply the multiplicand by the
difference between Hie multiplier and 100, 1000, etc, and add
or subtract the smaller result as the multiplier is greater or less
than 100, 1000, etc.
Examples for Practice.
1. 7023 X 99. Ans. 695277.
2. 16642 X 996. Ans. 16575432.
3. 372051 X 1002. Am. 372795102.
MULTIPLICATION. 41
CASE V.
71. When one part taken as units, in the multi
plier, is a flBkctor of another part so taken.
Problem.— Multiply 387295 by 216324.
Solution. — Commence with the 3 of operation.
the multiplier, and obtain the first partial 3 8 7 2 9 5
product, 1161885 ; then multiply this prod 216324
uct by 8, which gives the product of the 11618 8 5
multiplicand by 24 at once (since 8 times 9 2 9 5 8
3 times any number make 24 times it). 83655720
Set the righthand figure under the right 83781 203580
hand hgure 4 of the multiplier in use.
Multiply the second partial product by 9, which gives the product
of the multiplicand by 216 (since 9 times 24 times a number make
216 times that number). Set the righthand figure of this partial
product under the 6 of the multiplicand ; and, finally, add to obtain
the total product.
Rule. — 1. Multiply the mvUiplieand by some figure or fibres
cf the mvUiplierf which are a factor of one or more parts of the
mvUiplier.
2. Multiply this partial product by a factor of some other
figure or figures of tJie multiplier, and write Uie righthand
figure thus obtained under the tnghthand fi^re of the mtdtiplier
thus used,
3. Continue thus untU the entire multiplier is used, and then
odd the partial products.
Examples for Practice.
1. 38057 X 48618. Ans, 185025522B.
2. 267388 X 14982. Ans, 4006007016.
3. 481063 X 63721. Ans. 30653815423.
4. 66917 X 849612. Ans. 56853486204.
5. 102735 X 273162. Ans. 28063298070.
6. 536712 X 729981. Ans. 391789562472.
H. A. 4.
42
RAY'S HIGHER ARITHMETIC
Topical Outline.
1. Definitions.
2. Tenniu.
8. Sign.
4. Principles.
5. Opeiation.
6. Rule.
7. Proof.
8. Applications.
9. Contractions.
Multiplication.
1. Multiplicand.
2. Multiplier.
3. Partial Product
4. Product
1. Writing Numbers.
2. Drawing Line Beneath.
%. Finding Partial Products.
4. Drawing a Line Beneath Partial Products.
5. Adding the Partial Products.
YL DIYISIOK
•
72. 1. Division is the process of finding how many
times one number is contained in another ; or,
2. Division is a short method of making several sub
tractions of the same number.
3. Division is also an operation in which are given the
product of two &ctors, and one of the factors, to find the
other factor.
73. The product is the Dividend; the given factor is
the Divisor; and the required factor is the Quotient. The
Remainder is the number which is sometimes left after
dividing.
Note. — Dividend signifies to be divided. Quotient is derived from
the Latin word quoties, which signifies haw often.
Problem. — ^How many times is 24 cents contained in
73 cents?
Solution. — Twentyfour cents operation.
from 73 cents leaves 49 cents; 24 73 cents,
cents from 49 cents leaves 25 cents ; 24
24 cents from 25 cents leaves 1 cent. 49 cents remaining.
Here, 24 cents is taken 3 times 24
from {out of) 73 cents, and 1 cent ^ cents remaining,
remains ; hence, 24 cents is eontatTied 24
in 73 cents 3 times, with a remainder "J ^^^ remaining,
of 1 cent.
74. The divisor and quotient in Division, correspond to
the factors in Multiplication, and the dividend corresponds to
the product. Thus :
Factors / 5 X 3 = 1 5  p^^^^^,^
I 3X5 = 15 i
Dividend I * ' > Divisors and Quotients.
ll5!3 = 5i
(43)
44 BAY'S HIGHER ARITHMETIC.
75, There are three methods of expressing division ;
thus,
12^3, J^, or 3)12.
Each indicates that 12 is to , be divided by 3.
Principles. — 1. Wlien. iJie dividend and divisor are . lUce
numbers, the quotient is abstract.
2. When the divisor is an abstract number, the quotient is
like the dividend,
3. The remainder is like tlie dividend.
4. Tlie dividend is equal to the produet of the quotient by
the divisor, plus the remainder.
, 76. Multiplication is a short method of making several
additions of the same number ; Division is a short method
of making several subtractions of the same number ; hence,
Division is the reverse of Multiplication.
77. All problems in Division are divided into two
classes :
li To find the number of equal parts of a nujnber.
2. To divide a number into equal parts.
78. Two methods are employed in solving problems in
Division : Long Division, when the work is written in full
in solving the problem; and Short Division, when the
result only is writteil, the work being performed in the mind.
The following illustrates the methods :
Problem. — Divide 820 by 5.
LONG DIVISION. SHORT DIVISION.
5)820(164 Quotient. 5)820
i_ 164 Quotient.
3 2 tens.
30 Both operations are performed on the same
2 units. principle. In the first, the subtraction is writ
20 ten ; in the second, it is performed mentally.
DIVISION. 45
LONG DIVISION.
Problem. — Divide $4225 equally among 13 men.
Solution. — As 13 is not contained operation.
in 4 (thousandR)i therefore, the quo
tient has no thousands. Next, take 42 Sc «^ J "c « J
2 5 ^^ p ^ a
(hundreds) as a partial dividend; 13 Susi^S .eSs
is contained in it 3 (hundreds) times ; 13)4225)326
after multiplying and subtracting, there 3 9 h undreds,
are 3 hundreds left. Then bring down 3 2
2 tens, and 32 tens is the next partial 2 6 t ens,
dividend. In this, 13 is contained 2 6 5
(tens) times, with a remainder of 6 tens. 65^ units.
Lastly, bringing down the 5 units, 13
is contained in 65 (units) exactly 5
(units) times. The entire quotient is 3 hundreds 2 tens and 5 units.
This may be further shown by separating the dividend into parts,
each exactly divisible by 13, as follows:
DIVISOR. DIVIDEND. QUOTIENT.
13)3900 + 260 + 65(300+2045
3900
+ 260
+ 260
+ 65
+ 65
Biile for Long Dmsion. — 1. Draw curved lines on tke
right and left of tke dividend, placing the divisor on the left.
2. Find how ofte^i the divisor is contained in the lefthand
figure, or figures, of the dividend, and write the number in the
quotient at the right of (he dividend,
3. Multiply the divisor by this quotient figure, and write the
produ/^ under that part of the dividetul from which it was
obtained,
4. Subtract this product from the figures above it ; to the re
mainder bring down the next figure of tJie dividend, and divide
as before, until all the figures of tShC dividend are brought
down.
46
RAY'S HIOHEB ARITHMETia
b. If at any time after a figure is brought dovm, the number
ihui formed is too smaU to contain the divisor, a cipher must
be placed in the qv/otient, and anoUier figure brought down, after
which divide as before,
6. ^ there is a final remainder after the last division, place
the divisor under it and annex it to the quotient.
Proof. — Multiply the Divisor by the Quotient, and to
this product add the Remainder, if any ; the sum is equal
to the Dividend when the work is correct.
Notes. — 1. The product must never be greater than the partial
dividend from which it is to be subtracted; if so, the quotient
figure is too large, and must be diminished.
2. The remainder after each subtraction must be less than the
divisor ; if not, the last quotient figure is too smaUy and must be
increased.
3. The order of each quotient figure is the same as the lowest
order in the partial dividend from which it was derived*
Examples for Practice.
1.
1004835 : 33.
2.
5484888 : 67.
3.
4326422 : 961.
4.
1457924651 : 1204.
5.
65358547823 : 2789.
6.
33333333333 : 5299.
7.
245379633477 : 1263.
8.
555555555555 : 123456.
9.
555555555555 : 654321.
Ans. 30449^
Ans, 81864
Aw5. 4502
Ans, 1210900^1
Ans, 23434402^Vf
Ans. 6290495^V
Ans. 194283161j4f
Ans, 4500028^^:^^
Ans. 849056ff
In the following, multiply A by itself, also B by itself
divide the difference of the products by the sum of A and B.
Ans. 909.
4ns, 9158.
Ans. 104672.
A.
B.
10.
2856
3765.
11.
33698
42856.
12.
47932
152604
A.
B.
13.
4986
5369.
14.
3973
4308.
15.
23798
59635.
16.
47329
65931.
DIVISION. 47
In the following, multiply A by itself, also B by itself:
divide the difference of the products by the difference of
A and B.
Am. 10355.
Am. 8281.
Am. 83433.
Am. 113260.
17. K 25 acres produce 1825 bushels of wheat, how much
is that per acre ? 73 bushels.
18. How many times 1024 in 1048576 ? 1024 times.
19. How many sacks, each containing 55 pounds, can be
filled with 2035 pounds of flour? 37 sacks.
20. How many pages in a book of 7359 lines, each page
containing 37 lines? Id8f pages.
21. In what time will a vat of 10878 gallons be filled, at
the rate of 37 gallons an hour? 294 hours.
22. In what time will a vat of 3354 gallons be emptied,
at the rate of 43 gallons an hour? 78 hours.
23. The product of two numbers is 212492745; one is
1035; what is the other? 205307.
24. What number multiplied by 109, with 98 added to
the product, will give 106700? 978.
SHORT DIVISION.
Problem. — ^How often is 2 cents contained in 652 cents?
Solution. — ^Two in 6 (hundreds) is contained 3 oferatiok.
(hundreds) times; 2 in 5 (tens) is contained 2 2 )652
(tens) times, with a remainder of 1 (ten) ; lastly, 1 326
(ten) prefixed to 2 makes 12, and 2 in 12 (units)
is contained 6 times, making the entire quotient 326.
Remarks. — Commence at the left to divide, so that if there ia a
Temainder it may be carried to the next lower order.
48 J^A Y'S HIOffEB ARITHMETIC.
By the operation of the rule, the dividend is separated into parts
corresponding to the different orders. Having found the number
of times the divisor is contained in each of these parts, the sum of
these must give the number of times the divisor is contained in the
whole dividend. Analyze the preceding dividend thus :
652 = 600 + 40 + 12
2 in 600 is contained 300 times.
2 in 40 is contained 20 times.
2 in 1 2 is contained 6 times.
Hence, 2 in 652 is contained 326 times.
Rule for Short Division. — 1. Write the divisor on the
left of the dividend with a curved line between them, and draw
a line directly beneath the dividend. Begin at die left, divide
Buecemvely each figure or figures of the dividend by the divisor ,
and set the quotient beneath.
2. Whenever a remainder occurs, prefix it to Uie figure in
the next lower order, and divide as before,
3. If the figure, except the first, in any order does not oon
tain the divisor, place a cipher beneath it, prefix U to the figure
in the next lower order, and divide as before.
4. If there is a remainder after dividing the last figure,
place the divisor under it and annex it to the quxMent,
Proof. — ^The same as in Long Division.
Examples for Practice.
1. Divide 512653 by 5. Am. 102530f.
2. Divide 534959 by 7. Ans. 76422f
3. Divide 986028 by 8. Ans. 123253f
4. Divide 986974 by 11. Ans. 897241^.
5. At $6 a head, how many sheep can be bought for
$222 ? 37 sheep,
6. At $5 a barrel, how many barrels of flour can be
bought for $895? 179 barrels.
DIVISION. 48
OONTBACTIONS IN DIVISION.
CASE I.
79. When the divisor is a composite number.
This ca«e presente no difficulty except when remainders
occur.
Problem. — ^Divide 217 by 15.
Solution. — 15 = 3X5, hence 217^3 = 72 and 1 remainder;
72 r 5 =14 and 2 remainder. Dividing 217 by 3, the quotient is
72 ihreeiy and 1 unU remainder. Dividing by 5, the quotient is 14
(ffleen»)y and a remainder of 2 threes; hence the quotient is 14,
and the tnia remainder is 2X^ + 1 = 7*
Bule. — 1. Divide (he dividend by one factor of the divisor y cmd
divide this qiwtiefni by another fadoTy and so on^ tiU each fouior
has been used; the last quotient vM be the required result.
2. MvUiply each remainder by aU of the divisors preceding
(he one which produced it. The sum of the products, plus the
first remainder, toiU be the true remainder,
Remark. — ^This rule is not much used.
CASE II.
80. When the divisor is 1 with ciphers annexed.
This case presents no difficulty. Proceed thus:
Problem.— Divide 23543 by 100.
OPERATION.
SoLxmoN.—l 100 ) 235143
235^
Rule. — Out off as many figures in the dividend' as there are
ciphers in the divisor; the figures cut off vnU be the remainder,
«nd the other figure or figures ike quotient.
H. A. 5.
50 BAY'S HIGHER ARITHMETIC.
CASE III.
81. When ciphers are on the right of the divisor.
Problem.— Divide 3846 by 400.
Solution. — To divide by 400 is the operation.
same as to divide by 100 and then by 4 4 100)38 46
(Art. 79). Dividing by 100 gives 38, and  9 Quotient,
46 remainder (Art. 80); then, dividing 200 + 46 = 246, Kem.
by 4 gives 9, and 2 remainder: the true
remainder is 2 X 100 + 46 = 246 (Art. 79).
Bule. — 1. Out off the ciphers at the right of the dm8(yry
and as many figures from the right of the dividend.
2. Divide the refmaining part of the dividend by the remain
ing part of the divisor.
3. Annex to the remainder the fibres cut off, and thus
obtain the true remmnder.
AKITHMEnCAL SIGNS.
82. If a number be multiplied, it is simply repeated as
many times as there are units in the multiplier; if a num
ber be divided, it is simply decreased by the divisor as many
times as there are units in the quotient. It is thus evident,
that Addition and Subtraction are the fundamental concep
tions in all the operations of Arithmetic; and, hence, all
numbers may be classified as follows :
1. Numbers to be added; or, positive numbers.
2. Numbers to be subtracted; or, negative numbers.
83. Positive numbers are distinguished by the sign +»
negative numbers by the sign — ; thus, + 8 is a positive 8,
and — 8 a negative 8.
Eemark. — When a number is preceded by no sign, as, for
example, the number 4 in the first of the following exercises, it is
to be considered positive.
ARITHMETICAL SIGN& 61
84. The signs X and 7 do not show whether their remtUs
are to be added or to be subtracted ; they simply show what
operations are to be performed on the positive or negative
numbers which they follow.
Thus, in the statement, f 12 — ^ X 2) the sign X shows that 5 is
to be taken twice, but it does not show what is to be done with the
resulting 10 ; that is shown by the — . We are to take two 5's from
12. So, in 18 + 9 H 3, the sign f shows that 9 is to be divided by
3 ; what is to be done with the quotient, is shown by the + before
the 9.
86. In every such numerical statement, the + or the —
must be understood to affect the whole restttt of the opera
tions indicated between it and the next + or — , or between it
and the close of the expression.
Thus, in 5 + 7X2X9 — 2X6, the f indicates the addition of
126, not of 7 only ; and the — indicates the subtraction of 12. The
same meaning is conveyed by5f(7X2X9) — (2X6).
86. When the signs X and + occur in succession, they
are to have their particular effects in the exact order of
their occurrence.
Thus, we would indicate by 96 ? 12 X 4, that the operator is first
to divide by 12, and then multiply the quotient by 4. The result
intended is 32, not 2; if the latter were intended, we Rhould write
96 7 (12 X 4). Usage has been divided on this point, however.
Remabk.— It will be observed that in no case can the sigti X
or ^ affect any number before the preceding f or — , or beyond the
following + or — .
Exercises.
1. 4X3 + 7X2 — 9X3+6X4 — 3X3 = ?
SoLXjnoN. + 4X3 = 12, 7X2 = 14, 9X3 = — 27, 6X4 = 24,
— 3X3 = — 9. Grouping and adding according to the signs, we
have, 12 + 14 4 24 = 50 ; and — 27 — 9 = — 36. Therefore, 50 — 36
=14, Ans,
2. 2X2 — 1X2 — 2X2 — 5X3— 5X3 — 4 X 2 —
4x2 — 8X3 — 5X2 — 9X3 — 7X2 — 12X4 — 7X
2^?
52 RAY'S HIGHER ARITHMETIC.
80JLirA0N.+ 4— 2 — 4 — 15 — 15 — 8 — 8 — 24— 10— 27 — 14
':8 — 14. Grouping and adding, we have, 4 — 189= — 185, An^
3. 21^3X7 — 1x141 X 4^2 + 18 ^3x6i
■2x2) + (4— 2 + 6 — 7)X4x6v8 = ? 59.
Bemabk. — ^Whenever several numbers are included within the
marks of parenthesis, brackets, or yinculum, the^are regarded as
one number. Note the advantage of this in example 3.
4. 16x4^8—7 + 48^16—3—7x4x0x9X16 +
24X6r48 — 4X9M2 = ? 1.
5. (16M6x96^8 — 7 — 5 + 3)X[(27^9)^8—
l]+(91M3x7— 45— 3)X9=? 9.
GENERAL PBINCIPLES.
87. The following are the General Principles of Mul
tiplication and Division.
Principle I. — MvUiplying either factor of a produd, muUi'
plies ihe product by the same number.
Thus, 5X4 = 20, and 5X4X2 = 40, whence 20X2=40.
n. — Dividing either factor of a product, divides the produd
by the same number.
Thus, 5X4 = 20, and 5X4f2 = 10, whence 20^^2 = 10.
ni. — MvUiplying one factor of a product, and dividing the
other factor by the same number, does not alter the product.
Thus, 6X4 = 24, and 6X2X4 ^2 = 24, whence 6X2X2 =
24.
TV .— Multiplying the dividend, or dividing the divisor, by
any number, multiplies the quotient by that number.
If 24 be the dividend and 6 the divisor, then 4 is the quotient;
hence 24X256 = 8, and 24 i (6 h 2) = 8.
CONTRACTIONS. 68
V. — Dividing the dividend^ or muUiplying (ke divisor, bg
any number, divides the quotient by that nund)er.
Thus, if 24 be the divideod and 6 the diyisor, then 24 i 2 = 12;
and 1256 = 2; whence 24 ^ (6 X 2) = 2. Therefore, 412 = 2.
VL — Multiplying or dividing both dividend and divisor by
the same number^ does not change the quotient.
Thus, 24X2 = 48, and 6X2 = 12; consequently, 48112 = 4,
and 24^6 = 4^
CONTBACnONS IN MULTIPLICATION AND DIVISION.
CASE I.
88. To multiply by any simple part of 100, 1000, eto.
Note. — ^Let the pupil study carefully the following table of
equivalent parts:
Pabtb of 100. Parts of 1000.
12^ = 1 of 100. 125 =t of 1000.
16 = 1 of 100. 166 = I of 1000.
25 = i of 100. 250 = i of 1000.
33i = i of 100. 333i = i of 1000.
37i = I of 100. 375 = I of 1000.
62^ = I of 100. 625 = I of 1000.
66f = I of 100. 666 = i of 1000.
75 = I of 100. 750 = I of 1000.
87^ = J of 100. 875 = J of 1000.
Pboblem. — ^Multiply 246 by 87^. operation.
24600
SoLTJTiON. — Since 87^ is J of 100, 7
annex two ciphers to ^ the multiplicand, g \ 1 72200
which multiplies it by 100, and then . — TTmrZ
♦«L » r ^1 IX 4n«. 21625
take } of the result.
Bule. — Multiply by 100, 1000, efc., and take such a part
of the residt as the multiplier is of 100, 1000, etc.
64 I^A Y'S HIOHEB ARITHMETia
Examples for Practice.
1. 422X33^. Am. 14066f.
2, 6564 X 62f Am. 410250.
8. 10724 X 16*. Am. 178733.
CASE II.
89. To multiply by any number whose digits are
all alike.
Problem.— Multiply 592643 by 66666.
Solution.— Multiply 592643 by 99999 (Art. 70), the product is
59263707357 ; take } of this product, since 6 is  of 9 ; the result
is 39509138238.
Bnle. — Multiply as if the digits were ffs, and take such a
part of the product as the digit is of 9.
Examples for Practice.
1. 451402 X 3333. Am. 1504522866.
2. 281257 X 555555. Am. 156253732635.
3. 630224 X 4444000. Am. 2800715456000.
case III.
90. To divide by a number ending in any simple
part of lOOy lOOOy etc.
Problem.— Divide 6903141128 by 21875.
Solution. — Multiply both by 8 and 4 successively. The divisor
becomes 700000, and the dividend 220900516096, while the quotient
remains the same. (Art. 87, vi.) Performing the division as in
Art. 81, the quotient is 315572, and remainder 116096. The remain
der heing a part of the dividend, has been made too large by the
multiplication by 8 and 4, and is, therefore^ reduced to its true
dimensions by dividing by 8 and 4. This gives 3628 for the true
remainder.
CONTRACTION& 66
Bole. — Multiply both dividend and diviMr by sudi a number
08 will convert the final figures of the divisor into ciphers, and
then divide the firnner product by the latter.
Notes. — 1. If there be » remainder, it should be divided by the
multiplier, to get the true remainder.
2. The multiplier is 3, 4, 6, etc., according as the final portion
of the divisor is thirds, fourths, sixths, etc., of 100, 1000.
Examples fob Practige.
1. 300521761 — 225. Ans. 1335652^
2. 1510337264 T 43750. Ans. 34521f^
3. 2250071236171406250. Ans. 160003^^^
4. 620712480 r 20833 J. Quoe. 29794. Rem. 4146
5. 742851692 r2916. QuoL 254692. Bern. 2^
General Problems.
Note. — Let the pupil make a special problem under each general
problem, and solve it.
1. When the separate cost of several things is given,
how is the entire cost found?
2. When the sum of two numbers, and one of them, are
given, how is the other found?
3. When the less of two numbers and the difference
between them are given, how is the greater found ?
4. When the greater of two numbers and the difference
between them are given, how is the less found?
5. When the, cost of one article is given, how do you
find the cost of any number at the same price?
6. If the total cost of a given number of articles of
equal value is stated, how do you find the value of one
article ?
7. When the divisor and quotient are given, how do you
find the dividend?
66 BA Y'S HIGHER ARITHMETIC.
8. How do you divide a number into parts, each contain
ing a certain number of units ?
9. How do you divide a number into a given number of
equal parts?
10. K the product of two numbers, and one of them, are
given, how do you find the other?
11. If the dividend and quotient are given, how do you
find the divisor?
12. If you have the product of three numbers, and two
of them are given, how do you find the third?
13. K the divisor, quotient, and remainder are given.
Low do you find the dividend ?
14. K the dividend, quotient, and remainder are given,
how do you find the divisor?
MlBOELLANEOUS EXERCISES.
1. A grqper gave 153 biarrels of flour, worth $6 a bar
rel, for 64 barrels of sugar : what did the sugar cost per
barrel ? $17.
2. When the divisor is 36, quotient 217, and remainder
25, what is the dividend ? 7620.
3. What number besides 41 will divide 4879 without a
remainder? 119.
4. Of what number is 103 both the divisor and the
quotient? 10609.
5. What is the nearest number to 53815, that can be
divided by 375 without a remainder? 54000.
6. A fiirmer bought 25 acres of land for $2675 : what did
19 acres of it cost? $2033.
7. I bought 15 horses, at $75 a head : at how much per
head must I sell them to gain $210 ? $89.
8. A locomotive has 391 miles to run in 11 hours : after
running 139 miles in 4 hours, at what rate per hour must
the remaining distance be run? 36 miles.
MISCELLANEO US EJ^RCISES. 57
9. A merchant bought 235 yards of cloth, at $5 per
yard ; after reserving 12 yards, what will he gain by selling
the remainder at $7 per yard ? $386.
10. A grocer bought 135 barrels of pork for $2295 ; he
sold 83 barrels at the same rate at which he purchased, and
the remainder at an advance of $2 per barrel : how much
did he gain ? $104.
11. A drover bought 5 horses, at $75 each, and 12 at
$68 each; he sold them all at $73 each: what did he
gain? $50.
At what price per head must he have sold them to have
gamed $118? $77.
12. A merchant bought 3 pieces of cloth of equal length,
at $4 a yard; he gained $24 on the whole, by selling 2
pieces for $240: how many yards were there in each
piece? 18 yards.
13. If 18 men can do a piece of work in 15 days, in
how many days will one man do it? 270 days.
14. If 13 men can build a wall in 15 days, in how many
days can it be done if 8 men leave? 39 days.
15. If 14 men can perform a job of work in 24 days, in
how many days can they perform it with the assistance of
7 more men? 16 days.
16. A company of 45 men have provisions for 30 days :
how many men must depart, that the provisions may last
the remainder 50 days? 18 men.
17. A horse worth $85, and 3 cows at $18 each, were
exchanged for 14 sheep and $41 in money: at how much
each were the sheep valued? $7.
18. A drover bought an equal number of sheep and hogs
for $1482 : he gave $7 for a sheep, and $6 for a hog :
what number of each did he buy? 114.
19. A trader bought a lot of horses and oxen for $1260;
the horses cost $50, and the oxen $17, a head; there were
twice as many oxen as horses: how many were there of
each? 15 horses and 30 oxen.
58
RAY'S HIGHER ARITHMETia
20. In a lot of silver change, worth 1050 cents, one
seventh of the value is in 25cent pieces; the rest is made
up of 10cent, 5cent, and 3cent pieces, of each an equal
number: how many of each coin are there?
Of 25cent pieces, 6; of the others, 50 each.
21. A speculator had 140 acres *of land, which he might
have sold at $210 an acre, and gained $6800; but after
holding, he sold at a loss of $5600: how much an acre
did the land cost him, and how much an acre did he sell
it for? $165, cost; and $125, sold for.
Topical Outline.
DrVTBION.
1. Definitions.
2. Terms
8. Sign.
4. Principles.
( 1. Dividend.
J 2. Divisor.
& Opttration ^
3. Quotient.
4. Remainder.
rl. Writing the Numbers.
2. Drawing Curved Lines.
3. Finding Quotient Figure.
4. Multiplying Divisor and Writing Product
5. Drawing Line.
6. Subtracting.
7. Annexing Lower Order.
8. Repeating the Process from 8.
^9. Writing Remainder.
6 Rules.. / 1^^ Division.
I Short Division.
7 Prool
8. Applications.
d. Contractions / Division.
I Multiplication and Division.
f 1. .Of Numbers... / ^^^^"f^
10. Arithmetical Signs. \ ^ Negative.
I 2. Of Operation... /Multiplying.
I Dividing.
11. General Principles.
12. Applications.
Tn. PEOPEETIES OF NUMBEES.
DEFINITIONS.
91. 1. The Properties of Numbers are those qualities
which belong to them.
2. Kumbers are classified (1), as Integral, Fractional,
and Mixed (Art. 22) ; (2), as Abstract and Concrete (Art.
21); (3), Prime and Composite; (4), Even and Odd; (5),
Perfect and Imperfect.
3. An integer is a whole number; as, 1, 2, 3, etc.
4. Integers are divided into two classes— prime numbers
and composite numbers.
5. A prime number is one that can be exactly divided
by no other whole number but itself and unity, (1) ; as,
1, 2, 3, 5, 7, 11, etc.
6. A composite number is one that can be exactly
divided by some other whole number besides itself and
unity; as, 4, 6, 8, 9, 10, etc.
Remark. — Every compoFiite number is the product of two or
more other numbers, called its /acUna (Art. 60).
7. Two . numbers are prime to each other, when unity is
the only number that will exactly divide both; as, 4 and 5.
Kemabk. — Two prime numbers are always prime to each other:
sometimes, also, two composite numbers ; as, 4 and 9.
8 An even number is one which can be divided by 2
without a remainder ; as, 2, 4, 6, 8, etc.
9. An odd number is one which can not be divided by
2 without a remainder; as, 1, 3, 5, 7, etc.
Remabe. — All even numbers except 2 are composite : the odd
numbers are partly prime and partly composite.
(59)
60 RA Y^S HIQHEB ARITHMETIC.
10. A perfect number is one which is equal to the sum
of aU its divisors; as, 6 = 1 + 2+3; 28 = 1 + 2 + 4 +
7 + 14.
11. An imperfect number is one not equal to the sum
of all its divisors. Imperfect numbers are Abuniard or Z>e
fective: Abundant when the number is less than the sum
of the divisors ; as, 18, less than 1 + 2 + 3 + 6 + 9; and
Defective when the number is greater than the sum; as,
16, greater than 1+2 + 4 + 8.
12. A diYisor of a number, is a number that will exactly
divide it.
13. One number is divisible by another when it contains
that other without a remainder ; 8 is divisible by 2.
14. A multiple of a number is the product obtained by
taking it a certain number of times ; 15 is a multiple of
5, being equal to 5 taken 3 times; hence,
1st. A multiple of a number can always be divided by U
vrithoid a remainder,
2d. Every multiple is a corrvposile number.
15. Since every composite number is the product of
factors, each fiictor must divide it exactly; hence, every
lactor of a number is a divisor of it.
16. A prime &ctor of a number is a prime number that
will exactly divide it : 5 is a prime &ctor of 20 ; while 4 is
a factor of 20, not a prime factor ; hence,
1st. The prime factors of a number are aU the prime numr
bers that 'unll exactly divide it.
Example. — 1, 2, 3, and 5 are the prime factors of 30.
2d. Every composite number is equal to the product of aU
its prime factors.
Example. — All the prime factors of 15 are 1, 3, and 5 ; and 1 X
3 X 5 = 15.
17. Any &ctor of a number is called an aliquot part
of it.
Example. — 1, 2, 3, 4, and 6, are aliquot parts of 12.
\
FACTORING. 61
FACTORING.
92. Factoring is resolving composite numbers into fac
tors ; it depends on the following principles and propositions.
Pkenceple 1. — A f(uioT of a number is a factor of any
mvUiple of thai number.
Demonstration. — Since 6 = 2X3, therefore, any maltiple of
6 = 2 X 3 X some number ; hence, every factor of 6 is also a factor
of the maltiple. The same May be proved of the multiple of any
composite number.
Principle 2. — A factor of any two numbers is also a fador
of their Bum.
Demonstbation. — Since each of the numbers contains the factor
a certain number of times, their sum must contain it as often as
both the numbers ; 2, which is a factor of 6 and 10, must be a factor
of their sum, for 6 is 3 ticos, and 10 is 5 twos, and their sum is 3
twos + 5 twos =S twos,
93. From these principles are derived the six following
propositions :
Prop. I. — Every number ending udih 0, 2, 4, Q, or S, is
dimsible hy 2.
Demonstration. — Every number ending with a 0, is either 10 or
some number of tens; and since 10 is divisible by 2, therefore, by
Principle 1st, Art. 92, any number of tens is divisible by 2.
Again, any number ending with 2, 4, 6, or 8, may be considered
as a certain number of tens plus the figure in the units' place ; and
since each of the two parts of the number is divisible by 2, there
fore, by Principle 2d, Art. 92, the number itself is divisible by 2;
tKus, 36 = 30 + 6 = 3 tens f 6 ; each part is divisible by 2 ; hence,
36 is divisible by 2.
Conversely, No number is divisible by 2, unless it ends uHh
0, 2, 4, 6, or 8.
Prop. II. — A number is divisible by 4, when the number
denoted by its two righthand digits is divisible by 4.
62 BA Y' S HIGHER ARITHMETIC.
Demonsteation.— ^Since 100 is divisible by 4, any namber of
hundreds will be divisible by 4 (Art. 92, Principle 1st) ; and any
number consisting of more than two places may be regarded as a
certain number of hundreds plus the number expressed by the
digits in tens' and units' places (thus, 884 is equal to 3 hundreds
+ 84) ; then, If the latter part (84) is divisible by 4, both parts,
or the number itself, will be divisible by 4 (Art. 02, Prin. 2d).
Conversely, No number is divisiUe by 4, unless the number
denoted by its two righthand digits is divisible by 4.
Prop. III. — A number ending in or 5 is divisible by 5.
Demonstration. — ^Ten is divisible by 5, and every number of two
or more figures is a certain number of tens, plus the righthand
digit ; if this is 5, both parts of the number are divisible by 5, and,
hence, the number itself is divisible by 5 (Art. 92, Prin. 2d).
Conversely, No number is divisible by 5, unless it ends in
or 5.
Prop. IV. — Every number ending in 0, 00, etc., is divis
ible by 10, 100, etc.
Demonstration. — If the number ends in 0, it is either 10 or a
multiple of 10 ; if it ends in 00, it is either 100, or a multiple of 100,
and so on ; hence, by Prin. 1st, Art. 92, the proposition is true.
Prop. V. — A composite number is divisible by ike product
of any two or Tnore of its prime factors.
Demonstration. — Since 2 X 3 X 5 = 30, it follows that 2X3
taken 5 times, makes 30 ; hence, 30 contains 2 X 3 (6) exactly 5
times. In like manner, 30 contains 3X5 (1^) exactly 2 times, and
2X5 (10), exactly 3 times.
Hence, if any even number is divisible by S, it is also
divisible by 6.
Demonstration. — An even number is divisible by 2 ; and if also
by 3, it must be divisible by their product 2X3, or 6.
Prop. VI. — Every prime number, except 2 and 5, ends
with 1, 3, 7, or 9.
Demonstration. — This is in consequence of Props. I. and HI.
FACTO RING.
63
Prop. VII.— Jny integer w divisible by 9 or by S, if the
sum of its digits be thus divmble.
94. To find the prime fkotors of a composite
number.
Problem. — ^Find the prime factors of 42.
SoLunoN.^42 is divisible by 2, and 21 is
divisible by 3 or 7, which is found by trial;
hence, the prime factors of 42 are 2, 3, 7, .'. 2 X
3X7=42.
OPERATION.
2)42
3)21
Bule. — Divide tJie given nuniber by any prime number that
wiU eaxicUy divide it; divide Hie quotient in like manner, and
so continue until Hie quotient is a prime number; the several
divism's and the last quotient are Hie prime fcuiors.
Kemabks. — 1. Divide first by the smallest prime factor.
2. The least divisor of any number is a prime number ; for, if it
were a composite number, its factors, which are less than itself,
would also be divisors (Art. 92), and then it would not be the
least divisor. Therefore, the prime factors of any number may be
found by dividing it first by the least number that will exactly
divide it, then dividing this quotient in like manner, and so on.
3. Since 1 is the factor of every number, either prime or com
posite, it is not usually specified as a factor.
Find the prime factors of:
1. 45. Ans. 3, 3, 5.
2. 54. Ans, 2, 3, 3, 3.
3. 72. Ans, 2, 2, 2, 3, 3.
4.
75.
Ans, 3, 5, 5
5.
96.
Ans. 2, 2, 2, 2, 2, 3
6.
98.
Ans, 2, 7, 7
7. Factor 210.
8. Factor 1155.
9. Factor 10010.
10. Factor 36414.
11. Factor 58425.
Ans, 2, 3, 5, 7.
Ans, 3, 5, 7, 11.
Am, 2, 5, 7, 11, 13.
Ans, 2, 3, 3, 7, 17, 17.
Ans, 3, 5, 5, 19, 41.
64 ^^ y^ HIOHEB ABITHMETia
95. The prime factors common to several numbers may
be found by resolving each into its prime factors, then
taking the prime factors alike in all.
Find the prime factors common to:
1. 42 and 98. Am. 2, 7.
2. 45 and 105. Am. 3, 5.
3. 90 and 210. Am. 2, 3, 5.
4. 210 and 315. Am. 3, 5, 7.
96. To find all the divisors of any composite ntun
ber.
Any composite number is divisible, not only by each of
its prime factors, but also by the product of any two or
more of them (Art. 93, Prop. V.) ; thus,
42 = 2 X 3 X 7 ; and all its divisors are 2, 3, 7, and
2 X 3, 2 X 7, and 3 X 7; or, 2, 3, 7, 6, 14, 21. Hence,
Bule. — Resolve Hie number into iU prime fadorSy and then
form from these factors all the different products of which they
wiU admit; the prims factors and their products wiU be aU Hie
divisors of the given number.
Find all the divisors:
1. Of 70. Am. 2, 5, 7, and 10, 14, 35.
2. Of 196. Am. 2, 7, and 4, 14, 28, 49, 98.
3. Of 231. Am. 3, 7, 11, and 21, 33, 77.
4. Of 496 ; and name the properties of 496.
GREATEST COMMON DIVISOR.
97. A common divisor (C. D.) of two or more num
bers, is a number that exactly divides each of them.
98. The greatest common divisor (G. C. D.) of two
or more numbers is the greatest number that exactly divides
each of them.
GREATEST COMMON DIVI80B. 65
Principles. — 1. Every prime fadot cf a number u a
dimor of that number.
2. Every product of two or more prime factora of a number^
is a divisor of that number,
3. Every number is equal to the continued product of all its
prime factors.
4. A divisor of a number is a divisor of any number of
times that number,
5. A common divisor of two or more numbers is a divisor
of their sum, and also of their difference.
6. The product of all the prime factors, common to two or
more numbers, is their greatest common divisor.
7. The greatest commen divisor of tujo numbers, is a divisor
<f their difference.
To Find the Qreatist Common Divisob.
CASE I.
99. By simple Motoring.
Problem.— Find the G. C. D. of 30 and 105.
OPERATIOK.
30 = 2X3X5. I 3X6 = 15, G.C.D.
105 = 3X5X7. ) '
Demonstration. — ^The product 3X5 is a divisor of both the
numbers, since each contains it, and it is their greatest common
divisor, since it contains all the factors common to both.
PROBLEM.^Find the G. C. D. of 36, 63, 144, and 324.
OPERATION.
36, 63, 144, 324
Solution.^ 3
3
12, 21, 48, 108
4, 7, 16, 36
..3X3 = 9, G. C. D.
Bale. — Resolve the given numbers into their prime factors,
and take the product of the faetors common to all the numbers.
H. A. 6.
66
BAY','^ mOHER ARITHMETIC.
Find the greatest common divisor:
1. Of 30 and 42.
2. Of 42 and 70.
3. Of 63 and 105.
4. Of 66 and 165.
5. Of 90 and 150.
6. Of 60 and 84.
7. Of 90 and 225.
8. Of 112 and 140.
9. Of 30, 45, and 75.
10. Of 84, 126, and 210.
11. Of 16, 40, 88, and 96.
12. Of 21, 42, 63, and 126.
^na. 2X 3 = 6.
Am. 2x7 = 14.
Am. 3 X 7 = 21.
^rw. 3x11=33.
Am. 2 k 3 X 5 = 30.
Am. 2 X 2 X 3 = 12.
Am. 3 X 3 X 5 = 45.
Am. 2 X 2 X 7 = 28.
Am. 3 X 5 = 15.
Am. 2 X3X7 = 42.
Am. 2X2X2=8.
Am. 3 X 7 = 21.
CASE II.
100. By successive divisions.
Problem.— Find the G. C. D. of 348 and 1024.
OPERATION.
348) 1024(2
696
328)348(1
328
20 ) 3 2 8 (16
20
128
120
Demonstbation.— If 348
will divide 1024, it is the
G. C. D. ; but it will not
divide it.
If 328 (Art 98, Prin. 5,)
will divide 348, it is the
G. C. D. of 328, 348, and
1024 ; but it will not divide
348 and 1024 exactly.
If 20 will divide 328 (by
same process of reasoning),
it is the G. C. D.; but there
is a remainder of 8; hence,
if 8 will divide 20, it is the
G. C. D. ; but there is also a remainder of 4.
Now, 4 divides 8 without a remainder. Therefore, 4 is the
greatest number that will divide 4, 8, 20, 328, 348, and 1024, and is
the G. C. D.
_8)20(_2^
16
G. C. D. 4)8
OBEATE8T COMMON DIVISOR. 67
Bule. — Divide tiie greater number by the less, and the
divisor by the remainder, and 90 on; always dividing the last
divisor by the last remainder, till withing remains; the hut
divisor vriU be the greatest ccmvMm divisor sovght.
Note. — A condensed form of operation condensed opekation.
may be used after the pupils are familiar «.« 1024 2
with the preceding process. «««
Bemabk. — To find the greatest common
divisor of more than two numbers, find the ^ " ^^^ ^ "
G. C. D. of any two ; then of that G. C. D., 1^ 8 2
and any one of the remaining numbers, 4 8 2
and so on for all of the numbers ; the last
C. D. will be the G. C. D. of all the numbers.
Remark. — If in any case it be obvious that one of the numbers has
a prime factor not found in the other, that factor may be suppressed
by division before applying the rule. Thus, let the two numbers be
715 and 11011. It is plain that the prime 5 divides the first but not
the second ; and since .that prime can be no factor of any common
divisor of the two, their G. C. D. is the same as the G, C. D. of 143
and 11011.
Find the greatest common divisor of:
1. 85 and 120. Ans. 5.
2. 91 and 133. Ans, 7.
3. 357 and 525. Ans. 21.
4. 425 and 493. Ans. 17.
5. 324 and 1161. Ans. 27.
6. 589 and 899. Ans. 31.
7. 597 and 897. ■ Ans. 3.
8. 825 and 1287. Ans. 33.
9. 423 and 2313. Ans. 9.
10. 18607 and 24587. Ans. 23.
11. 105, 231, and 1001. Ans. 7.
12. 165, 231, and 385. Am. 11.
13. 816, 1360, 2040, and 4080. Ans. 136.
14. 1274, 2002, 2366, 7007, and 13013. Ans. 91.
68 RAY' 8 HIGHER ARITHMETIC.
LEAST COMMON MULTIPLE.
101. A common multiple (C. M.) of two or more
numbers, is a number that can be divided by eaufii of them
without a remainder.
102. The least common multiple (L. G. M.) of two
or more numbers, is the lead number that is divisible by
each of them without a remainder.
Principles. — 1. A miiUiple of a number is divisiUe by
that number.
2. A multiple of a number mmt contain aU of the prime
factors of that number.
3. A commxm multiple of two or mjore numbers is divisible
by each of those numbers.
4. A common multiple of two or mme numbers contains aU
of the prme factors of ecLch of (hose numbers.
5. The hast common multiple of turn or more numbers must
contain all of the prim/e fadtms of each <f those numbers, and
no other factors. ^
6. If two or more numbers are prims to each other, their
continued product is their least common multiple.
To Find the Least Common Multiplk
CASE I.
103. By factoring the numbers separately.
Problem. — ^Find the L. C. M. of 10, 12, and 15.
Solution. — Resolve each operatioit.
number into its prime f ac 10 = 2X5
tors. A multiple of 10 con 1^=2X2X3
tains the prime factors 2 and 15=3X5
5 ; of 12, the prime factors .'. L. C. M. = 2 X 2 X 3X5 = 60.
2, 2, and 3 ; of 15, the prime
factors 3 and 5. But the L. C. M. of 10, 12, and 15 must contain
LEAST COMMON MULTIPLK 69
all of the different prime factors of these numbers, and no other
factors ; hence, the L. C. M. = 2 X 2 X3 X 5 = 60 (Art. 102, Prin.
2 and 5).
Bule. — Resolve each number into its prime factors, and
dien take the continued jproduct of aU the different prime factorSy
using ecuih factor the greatest number of times it occurs in any
one of the given numbers.
Remabks. — 1. Each factor must be taken in the least common mul
tiple the greatest number of times it occurs in either of the numbers.
In the preceding solution, 2 must be taken twice, because it occurs
twice in 12, the number containing it most.
2. To avoid mistakes, after resolving the numbers into their
prime factors, strike out the needless factors.
Find the L. C. M. of:
1. 8, 10, 15. Ans. 120.
2. 6, 9, 12. Ans. 36.
3. 12, 18, 24. Ans. 72.
4. 8, 14, 21, 28. Ans. 168.
5. 10, 16, 20, 30. Ans. 60.
6. 15, 30, 70, 105. Ans. 210.
CASE II.
104. By dividing the numbers suooessively by their
common primes.
Problem.— Find the L. C. M. of 10, 20, 25, and 30.
Solution. — Write the operation.
numbers as in the margin. 2
Strike out 10, because it is *.
;p 20
25
30
10
25
15
2
5
3
contained in 20 and 80.
Next, divide 20 and 30 by
the prime factor 2; write 2X5X2X5X3 = 300 L. CM.
the quotients 10 and 15,
and the undivided number 25 in a line beneath. Divide these num*
bers by the common prime factor 5. The three quotients — 2, 5, 3,
are prime to one another ; whence, the L. C. M. is the product of the
divisors 2, 5, and the quotients 2, 5, 3. By division, all needless
factors are suppressed.
70 RAY'S HIGHER ARITHMETia
Bule. — 1. WrUe the numbers in a harizontxd line; strike
out any number that will exactly divide any of the others; divide
by any prime number that will divide tuHp or more of them
uriUumt a remainder; write the quotients and undivided num^
hers in a line beneath,
2. Proceed with this line as before, and continue the opera
tion tUl no number greater than 1 unU exactly divide two or
nurre of the numbers.
3. Multiply together the divisors and the numbers in the last
line; their product will be the least common multiple required.
Remark. — Prime factors not obvious may be found by Art lOO.
Find the least common multiple of:
1. 6, 9, 20. Ans. 180.
2. 15, 20, 30. Ans. 60.
3. 7, 11, 13, 5. Ans. 5005.
4. 35, 45, 63, 70. ' Ans. 630.
5. 8, 15, 20, 25, 30. Ans. 600.
6. 30, 45, 48, 80, 120, 135. Ans. 2160.
7. 174, 485, 4611, 14065, 15423. Ans. 4472670,
8. 498, 85988, 235803, 490546. Am. 244291908.
9. 2183, 2479, 3953. Ans. 146261.
10. 1271, 2573, 3403. Ans. 105493.
SOME PROPERTIES OF THE NUMBER NINR
105. Addition, Subtraction, Multiplication, and Division
may be proved by ** casting out the 9's." To cast the 9's
out of any number, is to divide the sum of the digits by 9,
and find the excess.
Problem.— Find the excess of 9's in 768945.
Explanation. — Begin at the left, thus : 7 + 6 are 13 ; drop the
9; 4 + 8 are 12; drop the 9; 3 + 4 + 5 are 12; drop the 9; the
excess is 3. The 9 in the number was not counted.
CASTING OUT NINEi
Principle. — Any number divided by 9, wiU
remainder as the sum of its digits divided by 9.
ILLUSTRATION.
700000 = 7 X 100000 = 7 X (99999 4 1) =7 X 99999 + 7
60000 = 6X 10000 = 6X (9999 + l) = 6X 9999f 6
768945=J 8000 = 8X 1000 = 8X (9994l) = 8X 999+8
900 = 9X 100 = 9X (994l) = 9X 999
40 = 4X 10 = 4X (9 + l) = 4X 9 + 4
5 = 5X 1= =6
Whence, 7X99999 + 6X9999 + 8X999 + 9X99 + 4X9 + 7 +
6 + 8+9 + 4 + 5 = 768945.
Solution. — An examination of the above shows that 768945 has
been separated into multiples of 9, and the sum of the digits com
posing the number ; the same may be shown of any other number.
There qan be no remainder in the multiples, except in the sum of
the digits.
Proof op Addition. — The sum of the excess of ffs in the
several numbers must equal the excess of ffs in their sum.
Illustration. — The excesses in the num
bers are 8, 2, 4, and 3, and the excess in the
sum of these excesses is 8. The excess in the
sum of the numbers is 8, the two excesses
being the same, as they ought to be when the
work is correct.
operation.
7352
5834
6241
7302
26729
8
2
4
3^
8
Proof of Subtraction. — The excess of 9*s in the minuend
must equal the sum of the excess of 9*s in the subtrahend and
remainder.
Illustration. — As the min
uend is the sum of the subtra
hend and remainder, the reason
of this proof is seen from that of
Addition.
operation.
Minuend, 7 6 4
Subtrahend, 1234
Eemainder, 6 40 6
8
1
7
Proof of Multiplication. — Find the excess of 9*8 in ths
factors and in the product, TJie excess of 9*8 in the product
of the excesses of the factorSy should equal the excess in the product
of the factors themselves.
72 R^ Y'S HIGHER ABITHMETIC.
IiiLuaTRATiON. — Multiply 835 by 76; opebation.
the product iB 63460. The excess in the 835X76 = 63460
multiplicand is 7, in the multiplier 4, 835, excess =7
and in the product 1; the two former 76, " =4
multiplied, give 28 ; and the excess in 28 7X^=28, " =1
is also 1, as it should be. 63460, '' =1
Proof op Division. — Find the excess of 9'« in each of the
terms. To the excess of ffs in the product of the excesses in the
divisor and quotient^ add ike excels in the remainder; the excess
in the sum shotdd equal the excess in the dividend.
OPERATION.
Ilhtstbation.— Divide 8915 by 25; the quo 35 6, excess 5
tient is 356, and the remainder, 15. The excess 2 5, ** 7
of 9's in the divisor is 7 ; in the quotient, 5 ; 35
their product is 35, the excess of which is 8.
The excess in the remainder is 6. 6 f 8 = 14,
of which the excess is 5. The excess of 9'b in
the dividend is also 5. ^ ^
3 5, « 8
15, " 6
14,
8915,
CANCELLATION.
106. Oanoellation is the process of crossing out equal
factors from dividend and divisor.
The sign of Cancellation is an oblique line drawn
across a figure ; thus, ^, ^, ^.
Principles. — 1. Canceling a factor in any number^ divides
Hie number by that factor.
2. Canceling a factor in both dividend and divisoTy does not
change the quotient. (Art. 129, III.)
Problem. — Multiply 75, 153, and 28 together, and divide
hy the product of 63 and 36.
CA NCELLA TION. 73
SoiiUnoN. — Indicate the operations operation.
as in the margin. Cancel 4 out of 25 17 7
28 and 36, leaving 7 above and 9 7^5 v T^il v9il
below. Cancel this 7 out of the divi /f ^ A APP A^p^
dend and out of the 63 in the divisor, P P X p p
leaving 9 below. Cancel a 9 out of 9 9
the divisor and out of 153 in the 3
dividend, leaving 17 above. Cancel 2 5x17
3 out of 9 and 75, leaving 25 above ^ = 1 4 1 J
and 3 below. No further canceling is
possible; the factors remaining in the dividend are 25 and 17,
whose product, 425, divided by the 3 in the divisor, gives 141}.
Bule for Cancellation. — 1. Indicate Hie mulHplicatwtis
which prodtice the dividend, and those, if any, vMch produce the
divisor.
2. Gancd equal factors from dividend and divisor ; mvUiply
together the factors remaining in the dividend, and divide the
product by the product of the factors left in the divisor.
Note. — If no factor remains in the divisor, the product of the
factors remaining in the dividend will be the quotient ; if only one
factor is left in the dividend, it will be the answer.
Examples for Practice.
1. How many cows, worth $24 each, can I get for 9
horsQ3, worth $80 each? 30 cows.
2. I exchanged 8 barrels of molasses, each containing 33
gallons, at 40 cents a gallon, for 10 chests of tea, each
containing 24 pounds: how much a pound did the tea
cost me? 44 cents.
3. How many bales of cotton, of 400 pounds each, at 12
cents a pound, are equivalent to 6 hogsheads of sugar, 900
pounds each, at 8 cents a pound? 9 bales.
4. Divide 15 X 24 X 112 X 40 X 10 by 25 X 36 X 56
X90. ^.
H. A. 7.
74
BAY'S HIQHEB ARITHMETIC.
Topical Outline.
1. Definitions..
2. Factoring
3. G. C. D.
4. L. C. M..
Properties of Numbers.
Properties.
Numbers Glassiiiect.
Integer
Fraction.
L Mixed.
Diyisor; Divisible ; •Multiple ; Factors;
L Prime Factors; Aliquot Parts.
r 1. Definition.
2. Principles.
3. Propositions.
4. Operations.
5. Rules.
6. Applications.
Prime.
Composite.
Even.
Odd.
Perfect.
Imperfect
1. Definitions....
2. Principles.
3. Operations....;
I 2.
Common Divisor.
Greatest Common Divisor.
r Case I J^"l®
\ Applications.
Case II... / Rule.
I Applications.
1. Definitions.....  ^' Common Multiple.
2. Principles.
3. Operations
Least Common Multiple.
r Case I / ^^^e
\ Applications.
Case II... /Rule
\ Applications.
). Some Properties of the No. 9 / ^' I'rinciple.
1 2. Application to...
6. Cancell:>t;on... ,
1. Definition.
2. Sign, /.
3. Prin<'iples.
4. Operation.
5. Rule.
6. Applications.
1. Addition.
2. Subtraction.
3. Multiplicatiou
4. Division.
vm. coMMOi^ FRAcrriONs.
PEFINinONS.
107. A Fraction is an expression for one or more of
the equal parts of a divided whole.
108. Fractions are divided into two classes; viz., com
mon fractions and decimal fractions.
100. A Common Fraction is expressed by two num
bers, one above and one below a horizontal line; thus, J,
which is read tioo thirds.
HO. The Denominator is the number below the line.
It shows the number of parts into which the whole is
divided, and thus the size of the parts.
111. The Numerator is the number above the line. It
shows how many of the parts are taken.
Note. — The denominator denominatesj or names, the parts; the
numerator numbers the parts.
112. The Terms of a fraction are the numerator and
the denominator.
IiiiiUSTRATiON. — The expression i,four fifths, shows that the whole
is divided into five equal parts, and that four of those parts are
taken. 5 is the denominator, 4 is the numerator, and the terms of
the fraction are 4 and 5.
113. Every fraction implies: 1. That a number is di
vided ; 2. That the parts are equal ; 3. That one or more
of the parts are taken.
114. There are two ways of considering a fraction whose
numerator is greater than 1. Four fifths may be 4 fifths
of one thins:, or 1 fifth of four things ; therefore,
^ (75)
76 RAY'S HIGHER ARITHMETIC.
The numerator of a fraction may be regarded as showing
the number of units to be divided; the denominator, the
number of parts into which the numerator is to be divided ;
the fraction itself being the value of one of those parts.
Hence, a fraction may be considered as an indicated
ditrision (Art. 75) in which,
1. The dividend is the numerator,
2. The divisor is the denominaior.
3. The quotient is Hie fraction itself.
116. The Value of a fraction is its relation to a unit.
Ue. Fractions are divided into classes with respect to
their value and form.
(1). As to value, into Proper, Improper, and Mixed.
(2). As to form, into Simple, Complex, and Compound.
117. A Proper Fraction is one whose numerator is less
than its denominator; as, \.
118. An Improper Fraction is one whose numerator is
equal to, or greater than, its denominator; as, f.
119. A Mixed Number is a number composed of an
integer and a fraction ; as, 3f .
120. A Simple Fraction is a single fraction whose
terms are integral; as, , , f.
121. A Complex Fraction is one which has one or
1 1
both of its terms fractional ; as, ^r, ^, or ^.
a 5
122. A Compound Fraction is a fraction of a fraction ;
as, \ of .
123. An Integer may be expressed as a fraction by
writing 1 under it as a denominator ; thus, ^, which is read
seven ones.
COMMON FBACTION& 77
124. The Beciprocal of a number is 1 divided by that
number ; thus, the reciprocal of 5 is .
125. Similar Fractions are those that have the same
denominator ; as, f and f .
126. Dissimilar Fractions are those that have unlike
denominators; as, f and f.
Remark. — The word "fraction" is from the Latin, fimngo, I
break, and literally means a broken number. In mathematics,
however, the word " fraction,'' as a general term, means simply the
Indicated quotient of a required division.
NUMEBATION AND NOTATION OF FRACTIONa
127. Numeration of Fractions is the art of reading
fractional numbers.
128. Notation of Fractions is the art of writing frac
tional numbers.
Bule for Beading Common Fractions. — Bead the num
ber of. parts taken as expressed by the nttmerator, and then the
size of the parts as expressed by the denominator.
Example. — J is read seven ninths,
Kemabk. — Seven ninths (J), signifies 7 ninths of one, or i of 7,
or 7 divided by 9.
Bule for Writing Common Fractions. — Write the num
ber of parts ; plaee a horizontal line below it, under which unite
the number whicfi indicates the size of the parts.
Fractions to be written in figures:
Seven eighths. Four elevenths. Five thirteenths. One
seventeenth. Three twentyninths. Eight twentyfirsts.
Nine fortyseconds. Nineteen ninetythirds. Thirteen one
hundredths. Twentyfour onehundredandfifteentbs.
78 RA Y' 8 HIGHER ARITHMETIC
129. Since a fetction is an indicated division (Art. 114) ;
therefore,
Principles. — ^I. A Fradion is mvltipliedf
1st. By multiplying the numerator.
2d. By dividing the denominator,
n. A Fraction is divided,
1st. By dividing the numerator. '
2d. By multiplying the denominator.
HI. The value of a Fraction is not changed,
1st. By multiplying both terms by the same number.
2d. By dividing both terms by Uie same number.
Remark. — The proof of I is found in Art. 87, Principle IV ; the
proof of U is in Principle V ; and of III, in Principle VI.
REDUCTION OF FRACTIONS.
130. Beduction of Fractions consists in changing their
form without altering their value.
CASE I.
131. To reduce a fraction to its lowest terms.
Remarks. — 1. Reducing a fraction to lower terms, is changing it
to an equivalent fraction whose terms are smaller numbers.
2. A fraction is in its lowettt terms when the numerator and de
nominator are prime to each other; as, f, but not J.
Problem. — Reduc3 f^ to its lowest terms.
Solution. — Dividing both terms by the first operation.
common factor 2, the result is § ; dividing 2 ) fg = JJ
this by 5 (129, iii\ the result is f , which 5 ) ^ = , Ans.
can not be reduced lower.
Or, dividing at once by 10, the greatest second operation.
common divisor of both terms, the result is 10 ) ^ = , Ans.
f , as before. i
EEDUCTION OF FRACTIONS. 79
Bule. — Reject aU fcustors common to both terms of thefraetwn
Or, divide both terms of the fraction by their greatest common
divism'.
Reduce to their lowest terms :
1 H
Ans. f.
6. Mf
Ans. ff.
2. If.
Ans. f .
7. Ml.
Ans. \^.
3. iV».
Am. ^.
8m
Ans. f f .
4. m
Ant. ■^.
9HH
Ans. i.
5.,m
Ans. ^.
10. t'W^
>
Ans. ^.
Express
the following in their simplest
forms :
11. 923
: 1491.
Ans. ^.
13. 2261
=4123.
Ans. ^.
12. 890
: 1691.
Ana. ^.
14. 6160
^ 40480.
Ans. :^.
CASE II.
132. To reduce a fraction to higher terms.
Remabk. — Reducing a fraction to higher terms, is changing it
to an equivalent fraction whose terms are larger numbers.
Pkoblem. — Reduce f to fortieths.
SoiitmoN. — Divide 40 by 8, the quo operation.
tient is 5 ; multiply both terms of the 40^8 = 5
given fraction, f , by 5 (129, m), and the . __ 5X5 __. o, ^^^
result is fj, the equivalent fraction re 8X5
quired.
Rule. — Divide ihe required denominator by the denominator
of the given fraction ; multiply both terms of the given fraction
by this quotient ; the result is the equivalent fraction required.
1. Reduce ^ and ^ to ninetyninths. Ans. ^, J^.
2. Reduce ^, f , and ^ to sixtythirds. Ans. ff > fi» A*
3. Reduce ^, ^, and ^ to equivalent fractions having
6783 for a denominator. Ans. 4^» iHi* iHi'
80 JiA rs HIGHEB ARITHMETIC.
CASE III.
133. To reduce a whole or mixed number to an
improper fraction.
Problem. — Reduce 3f to an improper fraction ; to fourths.
Solution. — In 1 (unit), there are 4 fourths; in 3 (units), there
are 3 times 4 fourths, = 12 fourths : and 12 fourths { 3 fourths =
15 fourths.
Bule. — Multiply together tJie whole number and the denom
inator of the fraction: to the product add the numerator , and
vrrite the sum over the denominator,
1. In $7f , how many eighths of a dollar? Ans, ^.
2. In 19f gallons, how many fourths? Ans. ^.
3. In 13f^ hours, how many sixtieths? Ans. ^^,
Reduce to improper fractions:
4. 11. Ans. ^.
5. 15^. Ans. ^.
6. 127. Ans. 4^.
7. 109^. Ans. ifp.
8. 5H. Ans. VtV^.
9. 13fJ. Ans. i^fi.
Bemark. — To reduce a whole number to a fraction having a
given denominator, is a special case under the preceding.
Problem. — ^Reduce 8 to a fraction whose denominator is 7.
Solution. — Since  equals one, 8 equals 8 times J, or ^.
CASE IV.
' 134. To reduce an improper firaetion to a whole
or mixed number.
Problem. — Reduce ^ of a dollar to dollars.
Solution. — Since 5 fifths make 1 dollar, there will be as many
dollars in 13 fifths as 5 fifths are contained times in 13 fifths ; that
is, 2 dollars.
DEDUCTION OF FRACTIONS. 81
BiQe. — Divide the numerator by the denominator; the quotient
vnU be the wJwle or mixed number,
Kebiark. — If there be a fraction in the answer, reduce it to its
lowest terms.
1. In ^ of a dollar, how many dollars? $4.
2. In i^l^ of a bushel, how many bushels? 34J bu.
3. In ^^ of an hour, how many hours? I^tV ^o^"^*
Reduce to whole or mixed numbers:
4. f. Ans. 1.
5. i9A. ^ns, 35.
6. ^^. Am. 88.
7. HP ^^' 105fV.
8. 4l^ ^^' 327^.
9. j^p. An*, 509^.
CASE V.
135. To reduce compound to simple fractions.
Problem. — Reduce f of ^ to a simple fraction.
Solution. — J oi ^ = ^j; I oi ^ operation.
= JV; andfof ^ = 2X_5 = li. i oi ^ = ^^ = 1^, Am,
^' * ^ 28 28 * ^ 4X7 28
Bule. — Multiply the numerators together for the numerator,
and the denominators together for the denominator of the frac
Hon, canceling common faetors if they occur in both terms,
Bemark. — Whole or mixed numbers must be reduced to im
proper fractions before applying the rule.
Problem. — ^Reduce f of ^ of ^ to a simple fraction.
OPERATION.
Solution. — ^Indicate the work, ^
and employ cancellation, as ^X^X7
shown in the accompanying 3X10X12 ^'W.
operation. n a
82 RAY'S HIGHER ARITHMETIC
Reduce to simple
fractions :
1. 4 of i off
.Ins. \,
2. 1 of f of 2f
i4ns. .
3. i of II of 2.
^ns. 2.
4. i of i of 3i.
Am, \\.
5. f of 1 of ^ of 8f .
Am. 3f
6. \ of 1 of f of f of
4.
4n8. .
7. A of f of tV
ofH
of 7i.
ui^lS. J.
8. if of A of A
ofM
of 1t^5.
^ns. ^^.
COMMON DENOMINATOR
136. A common denominator of two or more fi*actions,
is a denominator by which they express like parts of a unit.
137. The least common denominator (L. C. D.) of
two or more fractions, is the least denominator by which
they can express like parts of a unit.
Principles — 1. Ordy a common multiple of different de
nominators can become a common denominator,
2. Only a least common mvUiple can become a least common
dcno7ninator.
CASE VI.
138. To reduce fractions to equivalent fractions
having a common denominator.
Problem. — Reduce ^, f , and f to a common denominator.
OPERATIOK.
Solution.— Since 2X^X4 = 24, 24 is a 1X3X4 ^12
common multiple of all the denominators. 2X^X4 24
The terms of ^ must be multiplied by 3 X 4 ; 2X2X4 _ 16
the terms of , by 2 X 4; and the terms of , by 3X2X4 24
2X3. The values of the fractions are un 3X2X3 _ 1_8
altered. (l29, m.) 4X2X3~~24
V
s
REDUCTION OF FR ACTIONS. 88
Bule. — Multiply both terms of eadi fraction by the d/enommoir
tors of the other fractions.
Remark. — Since the denominator of each new fraction is the
product of the ^ame numbers — viz., all the denominators of the
given fractions — it is unnecessary to find this product more than
once. The operation* is generally performed as in the following
example :
Pboblem. — Beduce ^, f , and f to a common denominator.
OPERATION.
2X^X7 = 70, common denominator.
1X5X7 = 35, first numerator. i = fi)
3X2X7 = 42, second numerator. } = ^ C Ana.
6X2X5 = 60, third numerator. ? = ?S 3
Note. — Mixed numbers and compound fractions must first be
reduced to simple fractions ; the lowest terms are preferable.
Reduce to a common denominator:
1. h I h ^rui. ii I*, H
2. h h h ^ris. ^, Y^, ^.
3. 1, h I ^^. Hh 1%, m
4. h h h h ^^. ni m^ uh m
5. I, i of a^, I of . Am. 1^, J^, ff
6. loff f off,iofioff of2f. An8.m>m^m'
"Remark. — When the terms of the fractions are small, and one
denominator is a multiple of the others, reduce the fractions to a
common denominator, by multiplying both terms of each by such a
number as will render its denominator the same as the largest de
nominator. This number wUl be found by dividing the largest denomina
tor by the denominator of the fraction to be redv/xd.
Problem. — Reduce ^ and f to a common denominator.
OPERATION.
Solution. — The largest denominator, 6, is a 1X2 2
multiple of 3; therefore, if we multiply both gw 2^^^
terms of J by 6 divided by 3, which is 2, it is 5 ^
reduced to ^. "^ =~q
84 R^ Y'S HIGHER ABITHMETia
Reduce to a common denominator:
1. \y , and f. Ans. f, f, .
2. I, I and ■^. Am. ^, f , ^.
3. f, t, A and H ' ^w«. i, it, ^, H.
CASE VII.
139. To reduoe firaotions of different denominators,
to equivalent firactions having the least conunon de
nominator.
Problem. — Reduce f , ^, and ^ to equivalent fractions,
having the least common denominator.
OPEKATION.
Solution. — Find the L. C. M. of 8, 9, 5 5X9 4 5.
and 24, which is 72 ; divide 72 by the given "g = oTTg ^^"70
denominators 8, 9, and 24, respectively ; fj 7 V 8 6 6
multiply both terms of each fraction by —= w = ^
the quotient obtained by dividing 72 by ^ q \/ ^ 07
its denominator ; the L. C. M. is the de — = — _ — = —
2424V372
nominator of the equivalent fractions. In "^
practice it is not necessary to multiply each denominator in form,
Bule. — 1. Find the L. C. M, of the denominators of the
given fractions, for the L. C D.
2. Divide this L. G. D. by the demmiinator of each fraction,
and midtiply the numerator by the quoti&nt,
3. Write the 'product after each multiplication as a fmrner
ator above the L. C. D.
Remabk.— ^All expressions should be in the simplest form.
Reduce to the least common denominator :
1. h h f ^»« A' A» tI
^' T* s* inr» J* Ans, j^, jj^, J^, JJ.
3 ^, I, H. An». H, M. H
4 I. I. A. M ^n» ■$. f. I. f
6 i, A. H. A ^n« H. J^. M. M
6. If, 3, and ^ of 3f ^««. J^, iyj^, J.
ADDITION OF FB ACTIONS. 85
ADDITION OF FRACTIONa
140. Addition of Fractions is the process of uniting
two or more fractional numbers in one sum.
Bemabk. — As integers to be added must express Hhe units (Art.
52), BO fractions to be added must express like pcais of like units.
Problem. — ^What is the sum of f, , and j^?
Solution. — Beducing the given fractions to operation.
equivalent fractions having a common denom f = H i = if
inator, we have if, Jf , and JJ. Since these ^^ = JJ
are now of the same kind, they can be added H + M + H = i J
by adding their numerators. Their sum is $}=lff,^7M.
H=iM.
Bnle. — Meduce {he ftactions to a common denominator^ add
their numerators, and vnrite the sum over tlie common denomr
inator.
Kemares. — 1. Each fractional expression should be in its sim
plest form before applying the rule.
2. Mixed numbers and fractions may be added separately and
their sums united.
S. After adding, reduce the sum to its lowest terms.
Examples for Practice.
1. i, I, and ■^. Ans, f
2. I f , I, and ^. An^. 2^
3. 1 and 2. Ans. 4j\
4. 2^, af, and 4f . Ans, lOf^
5 T%» A» A» a^id ^' ^^ Hi
6. H, 2i, 3i, and 4^. . Am. 11^
7. I of f , and ^ of  of 2^. Ans, 1^%
^' i + 4f + ih + U' ^^« Wt
^ l + H + H + M + l* ^^. ^3^
10. ^ of 96J4 f of H of &J. Ans. 59^
11 i+i+n+i^+m+m ^m. dm
86 BAY'S HIOHEE ABITHMETIC.
SUBTRACTION OF FRACTIONS.
141. Subtraction of Fractions is the process of finding
the difference between two fractional numbers.
Remabk. — In subtraction of integers, the numbers must be of
like units (Art. 54) ; in subtraction of fractions, the minuend and
subtrahend must express like parts of like units.
Problem. — ^Find the difference between f and ^.
Solution. — Reducing the given frac operahok.
tions to equivalent fractions having a { =}^
common denominator, we have J = fj, ^^ = f
and A = M; the difference is A = ^. }^ — J4 = ^ = ^, ^iw.
Bule. — Redv^ce the fractions to a common denominator^ and
write the difference of their numerators over the common denowr
inator,
Remabk. — Before applying the rule, the fractions should be in
their simpi st form. The difference should be reduced to its lowest
terms.
Examples for Practice.
1. i — T^. ^ns. H
2. Ai^off Am, m
3. H^off Ans. ii^
4. Ai^of 4. Ans, ^
5. H — liV ^ns. H
6 Ai^. . Am. ^
7. A — H Am. ^^
8. iiA' Am. ^
Remark. — When the mixed numbers are small, reduce them to
improper fractions before subtracting; if they are not small, sub
tract whole numbers and fractions separately, and then unite the
results. Thus,
MULTIPLICATION OF FRACTIONS. 87
Problem. — Subtract 23f from 31 J.
SoiiUnoN. — Beducing the fractions to a operation.
common denominator, } = \\, Bat ff can 31} H + li=H
not be taken from H. Take a unit, \i, 2 3j; iJ — ii = i
from the integer of the minuend and add it 7 U, Am.
to«. Then ii+if = H, and fiH =
}J. 30 — 23 = 7. Therefore the answer is 1]\.
9. 12— lOJf. Ans. l\i.
10. 12 — 9f. Am. 3^.
11. 5f — 2^. Ans. 3^.
12. 7^— Sf Am. ^^.
13. 15 — f Am. l4.
14. 18 — 5. ^n«. 12f.
15. I of 2 — 3f. ^n«. .
16. ^ — \ of If ^rw. l.
17. J^ of 4J — ^ of 3^ = what? Am. 13f .
18. ll + 49H = wbat? ^rw. lOHf.
19. A man owned ^ of a ship, and sold  of his share :
how much had he left? Am. f^.
20. After selling ^^ of f + ^ of f of a farm, what part
of it remains? Am. ^.
21. 3i + 4f  5^ + 16  7H + 10  14f, is equal
to what? Am. 6f^.
22.5l — 2i + \^ — ^ + 3^+Si 16i, is equal
to what? Am. .
23. 1 — I of f — I of f = what? Am. ^.
MULTIPLICATION OF FRACTIONS.
142. Multiplication of Fractions is finding the product
when either or when each of the factors is a fractional num
ber. There are three cases:
1. To multiply a fruction by an integer,
2. To mvUiply an integer by a fraction.
3. To mvUiply one fraction by another.
«iriH
88
BA Y*S HIGHER ARITHMETIC.
Note. — Since any whole number may be expressed in the form
of a fraction, the first and second cases are special cases of the third.
Problem. — Multiply f by f.
Solution. — Once f is J. \ times f is J of operaxion.
f = A i tim^ h ^^^^ is 5 times /5 = i. f X f = iJ, 47i«.
Problem. — ^Multiply \ by 6.
Solution. — Six times 3 fourths is 18
fourths. Beducing to its simplest form,
we have 4.
Problem. — Multiply 8 by f .
Solution. — One fifth times 8 is f , 3
fifths times 8 is 3 times =^*. Re
ducing to a mixed number, we have 4^.
Note. — The three operations are alike, and from them we may
derive the rule.
operation.
iXf = ^ = 4l,^7w.
OPERATION.
fX = V = 4f,^t«.
or,8X«=¥ = 4f
Bule. — MvUiply {lie numerators together for the numercdor
of Hie product, and the denominators for the denominator of the
product
Kemark. — ^Whole numbers may be expressed in the form of
fractions.
1. tt X 12.
2. H X 18.
3. If X 24.
Examples for Practice.
Ans. 9^.
Ans, 8 J.
Ans. 14J.
4. T^ X 28.
5. H X 30.
6. 3f X 5.
Ans. 15f.
Ans. 26.
Ans. 18 J.
OPERATIONS.
3f 3 = V
Remark. — In multiplying a
mixed number by a whole num
ber, multiply the whole number
and the fraction separately, and
add the products; or, reduce
the mixed number to an im " "*'
proper fraction, and multiply it ; as, in the last example.
15
3*
¥X5 =^
^^ = 18}, ^ns.
MULTIPLICATION OF FEACTIONS
89
7. 45 X f
8. 50 X H
9. 25 X .
10. 32x2f.
AuB. 35.
Am. Z^.
Am. 18.
Am. 76.
11. 28 X 3.
12. if X h
13. li X H.
14. H X H.
;ln«. 102.
Am. ^.
4n«. if.
15. What will 3i yards of cloth cost, at $4^ per yard ?
teKMAKK. — ^In finding the product of two mixed numberfl, it is
generally best to reduce them to improper fractions ; thus,
41 = 1; 3i = Jjf ; $JX ¥== V^ = $l 5, An*.
OPERATION.
SoiiTJnoN. — The operation may be performed without
reducing to improper fractions ; thus, 3 yards will cost
$131, And 1 of a yard will cost \ of $41 = $11; henoe,
the whole will cost $15.
$41
131
ii
16. 6X41. Am. 30.
17. 4 X 2. Am. l^.
Ann, $15
18. 12f X3^. Am. 40^.
19. 7ii X 3^. Am. 2&i.
Am. 4.
Am. 7h
Am. 4^.
Am. 94.
Am. 5%.
Am. 4.
Ans. 49.
20. Multiply i of 8 by i of 10.
21. Multiply I of 5f by f of aj.
22. Multiply f of  of 5f by f of 3.
23. Multiply 5, 4^, 2^, and f of 4f .
24. Multiply , f , ^, i of 2^, and ^ of 3^.
25. Multiply , i, ^, 3^, and 3.
26. Multiply 3i, 4, 5,  of ^s^^and 6.
27. At i of a dollar per yard, what will 25 yards of
cloth cost? $21.
28. A quantity of provisions will last 25 men 12f days:
how long will the same last one man ? 31 8 days.
29. At 3i cents a yard, what will 2 yards of tape
cost? ' 9 cents.
30. What must be paid for f of f of a lot of groceries
that cost $18f ? $7^.
31. K owns I of a ship, and sells f of his share to L:
what part has he left? fs*
H. A. 8.
90 RAY'S HIGHER ARITHMETIC.
DIVISION OF FEACriONa
143. Division of Fractions is finding the quotient
when the dividend or divisor is fractional, or when both
are fractional. There are three cases:
1. To divide a fradion by an integer.
2. Tb divide an integer by a fra^stwn.
3. To divide one fraction by another.
Note. — Since any whole number may be expreesed in the form of
a fraction, the first and second cases reduce to the third case.
Problem.— Divide f by ^^
Solution. — } is contained in 1, seven operation.
times ; } is contained in J, J of 7 = J times ; "*f = fXJ = H
I is contained in 4, 5X J=V times; f is
contained in f , J of */■ == ii times. It will be seen that the terms of
the dividend have been miUtipliedf and that the terms of the divisor
have exchanged places. Writing the terms thus is called " inverting
the terms of the divisor," or simply, " inverting the divisor."
Problem. — ^Divide 3 by .
OPERATION.
Solution. — J is contained in fff = }Xf = ^ = 7i, Ans.
1, five times ; J is contained in Or, 3^1 = ^ = 7 J.
3, three times 5 times = 15 times ;
f is contained in 3, J of 15 times = ^^ = 7^ times.
Problem. — Divide f by 2.
OPERATION.
Solution. — ^Two is contained ^^f = ^Xi = A = ?r ^^^^
in 1, J times ; 2 is contained in Or, ^ s 2 = J.
I, J of f = iV times ; 2 is con
tained in f , 4 times ^^ = ^ = ^ times.
Note. — From these solutions we may derive the following rule.
Bule. — Multiply the dividend by the divisixr vnth its termn
inverted.
DIVISION OF FRACUONS.
91
Bkmakk. — The terms of the divisor are inyerted because the
solution requires it. The same may be shown by a different solu
tion, as below.
Problem. — Divide  by f.
Solution. — Bednce both' dividend and
divisor to a common denominator. The
quotient of H "^ H ^^ the same as 10 r 27
= ij. The same result is obtained by
multiplying the dividend by the divisor, with its terms inverted ;
thus, i X I = i?.
OFERATIOK*
Beharb:. — Mixed nnmbers mnst be rednced to improper frac
tions. Use cancellation when applicable.
Examples for Pbacticb.
1. A^3.
3. 1^8.
4. 6M.
5. 21 ^t1^.
Am. ^
Am. 4s
Am. ^
Am. 9
Am. ^^
Am. 1J,
Am. 261
8 *iy«.
10. lf^5.
11. ^^i.
12. \^^^.
13. ^^^^.
14. 54if^25.
15. Divide 1^ by \ of f of 7.
16. Divide ^ of 3^ of 3^ by ^ of \^.
17. Dividef of^byjof .
18. Divide I of 3 by H of 7.
19. Divide i of I X ih by ^ of 3^.
20. Divide ^ of 5^^ by  of ^ of 3^.
21. Divide ^ of ^ of ^ by f of ^ of f
22. Divide 1 times 4 by 1^ times 3.
23. Divide ^ by f of 8} times j^ of 3^.
24. Divide ^ of f of 27^ by  of ^^ of 5^.
25. What is 2 X I of 19^^(4^ X tV ^^ ^)?
Am. •^.
Am, fl
Am. ^jf.
Am. 12^.
Am. lOf
Ans. H.
Am. 2\.
Am. f .
Am. ^.
Am. 2J.
A718. If.
Am. f .
4n3. If.
Am. j^.
J.w«. If.
Aii8. 34^.
^ns. 3^.
92
BA Y'S HIGHER ARITHMETIC.
9
144. To reduce complex to simple fractions.
Problem. — Reduce — to a simple fraction.
OPERATION.
Explanation.— The mixed If =^V 2 J
numbers are reduced to im V "^ i = V X 4 = If = f f> ^"»
proper fractions^ and the
numerator is divided by the denominator. (Art. 114.)
Itule. — Divide the numerator by the denominator' , as in
division of fractions.
Eeduce to simple fractions:
1.
f
2 y
2f
^ ii
Am. ■^.
Ant. .
Ans. If.
4.
^* 18
62
6.
16^
Am, 1\,
Ans, W,
Ans, 3.
Bemabk. — Complex fractions may be multiplied or divided, by
reducing them to simple fractions. The operation may often be
shortened by cancellation.
7. ^ X ^. Ans. iH
31 ^
25 "^ lOi* ^* ^'
Q ^ V ^ Ann 650
94 244 ' ^sWt"'
2 • 12^
10. r? 4 ~. ^n«. 4H
40f • 73 • ^'
12 ^A^i^ Anit ^
THE GREATEST COMMON DIVISOE OF FRACTIONS.
145. The gpreatest common divisor of two or more
fractions is the greatest fraction that will exactly divide
each of them.
O, a D. OF FRACTIONS. 98
One fraction is divisible by another when th^ numerator
of the divisor is a factor of the numerator of the dividend,
and the denominator of the divisor is a multiple of the
denominator of the dividend.
Thus, ^5 is divisible by ,», ; for A = i! ; H "^ A = ^•
The greatest common divisor of two or more fractions,
must be that fraction whose numerator is the G. C. D. of
all the numerators, and whose denominator is the L. C. M.
of the denominators.
Thus, the G. C. D. of ^ and J is ^fy.
Problem. — ^Find the G. C. D. of , f , and j^.
Solution. — Since 5 and 7 are both operation.
prime numbers, 1 is the G. C. D. of 1 = G. C. D. of 5, 25, 7
all the numerators ; 96 is the L. C. 9 6 = L. C. M. of 8, 32, 12
M. of 8, 32, and 12 ; therefore, the G. ^j, Ans.
C. D. of the fraction is ^5.
Bule. — Fhid the G, C. D, of the numerators of the fractionSy
and divide it by the L, C, M, of their denominators,
Eemark. — The fractions should be in their simplest forms before
the rule is applied.
Fmd the greatest common divisor:
1. Of 83^ and 268f. Am. 2^.
2. Of 14^ and 95f . Am, ^.
3. Of 59i and 735. Am. 2.
4. Of 23^2^ and 213. Am. 2.
5. Of 418 and 1772. Am. .
6. Of 261 and 652. Am. 4f.
7. Of 44t, 546, and 3160. Am, 4f
8. A farmer sells 137^ bushels of yellow com, 478^ bushels
of white corn, and 2093f bushels of mixed corn : required
the size of the largest sacks that can be used in shipping, so
as to keep the com from being mixed ; also the number of
sacks for each kind. 3 bushels ; 44, 153, and 670.
94 HA Y'S HIQBER ARITHMETIC,
9. A owris a tract of land, the sides of which ate 134f ,
128^, and 115^ feet long: how many rails of the greatest
length possible will be needed to fence it in straight lines,
the fence to be 6 rails high, and the rails to lap 6 inches at
each end? 354 rails.
THE LEAST COMMON MULTIPLE OF FRACTIONS.
146, The least common multiple of two or more
^'actions is the least number that each of them will divide
exactly.
Note. — The G. C. D. of several fractions must be a fraction, but
the L. C. M. of several fractions may be an integer or a fraction.
A fraction is a mtdtiple of a given fraction when its
numerator is a multiple, and its denominator is a divisor, of
the corresponding terms of the given fraction.
Illustration. — j^ is a multiple of y^. 8 is a multiple of 2, and
11 is a divisor of 33 ; hence, ■A:JA = AX^^ = 12. The same
result is otherwise obtained ; thus, ^ = f J, and JJ t ^ = 12.
A fraction is a comnum multiple of two or more given
fractions when its numerator is a common multiple of the
numerators of the given fractions, and its denominator is a
common divisor of the denominators of the given fractions.
A fraction is the least common multiple of two or more
fractions when its numerator is the least common multiple
of the given numerators, and its denominator is the greatest
common divisor of the given denominators.
Problem. — ^Find the L. C. M. of i, f , and f .
Solution. — The L. C. M. of operation.
the numerators is 15. The G. L. C. M. of 1, 3, 5 = 3 X ^ = 1 5
C. D. of the denominators is 1 ; G. C. D. of 3, 4, 6 = 1
therefore, the L. C. M. of the •"• V, Ans.
fractions is ^, or 15.
Z. a M. OF FBACTJON& 95
Bule. — Divide the L. C. M. of Oie mmwraJUm by the O. C.
D. of ike denominators.
Kemabk. — ^The fractions must be in their simplest forms before
the rule is applied.
Find the least common multiple :
1 Of I, f , I, I, and f Am. 60.
2. Of 4}, 6f , 5, and 10^. Am. 472^.
3. Of 4> 4» A» 5f » and 12. Ans. 350.
4. A can walk around an island in 14^ hours ; B, in 9^^
hours ; C, in 16 hours ; and D, in 25 hours. If they start
from the same point, and at th^ same time, how many hours
afier starting till they are all together again ? 100 hours.
Promiscuous Exercises.
Note to Teachers. — All problems marked thus [*], are to be
solved mentally by the class. In the solution of such problems, the
following is earnestly recommended :
1. The teacher will read the problem slowly and distinctly, and
not repeat it.
2. The pupil designated by the teacher, will then give the answer
to the question.
3. Some pupil, or pupils, will now reproduce the question in the
exact language in which it was first given to the class.
4. The pupil, or pupils, called upon by the teacher, will give a
short, logical analysis of the problem.
1. What is the sum of 3, 4^, 5^, f of , and  of ^
of I? isu
2. The sum of 1^ and — is equal to how many times
their difference? ^ 5 times.
3. What i8(2f + ofl_M)=l^? 6.
96 SAT^ S nrOHEB ABITBMETia
4. Eeduce i^^LM and 5 x (100 ?^ + 14) to
their simplest forms. 16 and 26^^.
5. What is i of 5^ — ^ of 3f ? ^.
6. What is If X AAf X ifl X If equal to? ^.
7.* X^X^? tV
8. Ixli^x^ix^ix^^i? Vft.
32354 ^^
9. (2 + i)^(3 + _ ^i^^,^
(2i)X(43f) ^
10. ii^lililiizil^ what? 4M.
4iX4il ^
11. Add I of I of I, i X I of 1 J, and \. ^.
12. 4 of V' of what number, diminished by — 3& — ,
leaves ff ? ^.
13.* James's money equals f of Charles's money ; and J
of James's money + ^^^ equals Charles's money : how much
has each? James, $36; Charles, $60.
14.* A leaves L for N at the same time that B leaves N
for L. The two places are exactly 109 miles apart: A
travels 1\ miles per hour, and B, 8J miles per hour; in
how many hours will they meet, and how fer will each have
traveled? 6ff hours. A, 51^ miles; B, 57^ miles.
15. What number multiplied by •§■ of f of 3^ will
produce 2}? 2f.
16. What, divided by 1, gives 14f ? 23.
17. What, added to 14f , gives 29 ? ISyf^.
18.* I spend ^ of my income in board, ^ of it in clothes,
and save $60 a year : what is my income ? $216.
19.* Divide 51 into two such parts that J of the first is
equal to f of the second. 27 and 24.
20.* \ is what part of ? .
COMMON FRACTIONS.
21. Divide ^ of 3f by jj of 7; ^d ^ of
of ^ of 54^.
22. Multiply ^ of 2^ by ^ of 19^; and
of 14^ by y\ of ^ of 13.
23. (iMi+i♦H + H!l)^f =wbat?
^. J 24 * A bequeathed ^ of his estate to his (
rest to his younger, who received $526 less thj
What was the estate?
25. Find the sum, difference, and product
also, the quotient of their sum by the differen
Sum 5^, diff. Iff, prod. 8fJ
26.* A cargo is worth 7 times the ship : wl
4Jf I cargo is j^ of the ship and cargo ?
27. By what must the sum of ^^^, ^^
multiplied to produce 1000?
2S, Multiply the sum of all the divisors of
I ing 1, by the number of its prime fitctors ex<
divide by 149^,
and I 29. ^ is what part of 10^? Reduce 1
^'' ^rCHarle^' 1^ 30..Multiply 1, 14f, ?!, 1, 1*, and 6.
^/ that B lea^^^ 5i' ^' 4' 7V 2'
i ^^ ^^AfiS aP^' 31. J of I of what number equals 9?
y ., per bour;^ 32.* A .63gallon cask is  fuU: 9^ gaUont
b\ ^^^ ^eacb b*^ off, how full wiU it be?
^ bo^ ^ g 57^ 0»^ . 33. If a person going 3f miles per hoi
^ tJUl^' i of ^ I I journey in 14f hours, how long would he be,
by t ^^ ^ ^^ I 5i mUes per hour?
" '■ I 34. A man buys 32f pounds of coffee, a
24}' ^^ pound : if he had got it 4 cents a pound cheaj
29^ t ^ \k\si cl<>*^ I more pounds would he have received ?
I \)OSS^9 e I**.' 1 35. Henry spent j^ of his money and then
\jicoTJCi^ ' f tbe ^ i ^^ tlkeii lost f of all his money, and had in
uirts tb»^ » ^1^^ , ' than at first. How much had he at first?
^^ f H. A, 9.
98
BAY'S HIGHER ARITHMETIC
Topical Outline.
Common Fractions.
1. Definition.
2. ClasKS.
& Terms..
' 1. Ab to Kinds
< 1. Common.
* \2. Decimal.
( 1. Proper.
2. As to Value
• • 2. Improper.
i .3. Mixed.
r 1. Simple.
L 3. As to Form.
• ' 2. Complex.
.8. Compound.
f 1. Numerator..
2. Denominator.
8. Similar.
L 4. Dissimilar.
4. Principles.
5. Reduction.
1. Cases....
2. Principles.
1. Lowest Tprms.
2. Higher Terms.
8. Mired Numbers to Improper Fractions.
4. Improper Fractions to Mixed Numbers.
5. Compound to Simple Fractions.
6. Common Denominator.
7. licast Common Denominator.
1^ 8. Rules.
6. Practical Applications..
1. Addition
2. Subtraction.
8. Multiplication.
4. Division
Is.
18.
{^
18.
rl. Deflni
J 2. Prlnd
\%, Rule,
I. Definition.
Principles.
Rule.
1. Definition.
Principles.
Rule.
I. Definition.
Principles.
Rule.
1. Definition.
Principles.
5. Divisors, Multiples, etc
IX DECIMAL FRAOnOIirS.
147. A Decimal Fraction is a fraction whose denoni
iuator is 10, or some product of 10, expressed by 1 with
ciphers annexed.
Bemask 1. — A decimal fraction is also defined as &Jraetion who9t
denominator is some power cf 10. By the ^^potoer" of a quantity, is
usually understood, either that quantity itself, or the product
arising from taking only that quantity a certain number of times
as a factor. Thus, 9 = 3 X 3» or the aeeond power of 3.
Remark 2. — Since decimal fractions form only one of the daeees
(Art. 108) under the term fixutUmSy the general principles relat
ing to common fractions relate also to decimals,
148. The orders of integers decrease from left to right
in a tenfold ratio (Art. 48). The orders may be continued
from the place of units toward the right by the same law
of decrease.
149. The places at the right of imits are called decimal
places, and decimal fractions when so written, without a
denominator expressed, are called decimals.
160. The decimal point, or separatriz, is a dot [ . ]
placed at the left of decimals to distinguish them from
integers.
Thus, ^ is written .1
TrAnr " " .001
From this, it is evident that,
The denominojlxyr of any decimal is 1 with as many cyphers
annexed as there are places in the dednud.
151. A pure decimal consists of decimal places only ;
as, .325
' (99)
100 I^AY'S HIGHER ARITHMETIC.
152. A mixed decimal consists of a whole number and
a decimal written together; as, 3.25
Remark. — A mixed decimal may be read as an improper frac
tion, since ^^^H^
153. A com.plex decimal has a common fraction in its
righthand place; as, .033^
154. From the general law of notation (Arts. 48 and
148) may be derived the following principles :
Principle I. — If, in any decimal^ ilie point be moved to
the right, Hie decimal is multiplied by 10 as often as the point is
removed one place.
Illustration. — If, in the decimal .032, we move the point one
place to the right, we have .32. The first has three decimal places,
and represents thousandths; while the second has two places, and
represents hundredths, (Art; 129, Prin. i.)
Principle II. — if, in any decimjol, the point be moved to the
left, the decivwl is divided by 10 a^ often as the point is removed
one place.
Illustration. — If, in the decimal .35 we move the point one
place to the left, we have .035. The first represents hundredths;
the second, thrascmdihs, while the numerator is not changed. (Art.
129, Prin. n.)
Principle III. — Decimal ciphers may be annexed to, or
omitted, from, the right of any number without altering its valve.
Illustration.— .5 is equal to .500; for i^ox ISo — i^ The
reverse may be shown in the same way. (Art. 129, Prin. in.)
NUMERATION AND NOTATION OF DECIMALS.
155. Since .6 = ^; .06 = y^ ; and .006 = y^, any
figure expresses tenths, hundredths, or thomandths, according
as it is in the 1st, 2d, or 3d decimal place; hence, these
places are named respectively the tenil\s\ the hundredtiis\ the
DECIMAL FRACTIONS.
thousmidih^ place ; other places are named in the same way,
as seen in the following table:
Table op Becdul Orderb.
1st plac«
.2 .... read 2 Tenths.
2d "
.08 ... .
' 8 Hundredths.
3d "
.005 . . .
' 5 Thoueandths.
4th "
.0007 . . .
' 7 Tenthousandths.
6th "
.00003 . .
' 3 Hundredthousandths.
6lh "
.000001 . .
• 1 Millionth.
7lh ■■
.0000009 .
* 9 Tenmillionths.
8th "
.00000004.
' 4 Hundredmillionths.
9th '■
.000000006
' 6 BiUionths.
Note:.— The names of the decimal orders are derired from the
names of the ordera of whole numbera. Tlie table may therefore be
extended to trilliontha, quadrilliooths, etc.
Problem.— Eead the decimal .0325
SoLDTiciN. — The numerator is 325; the denomination u ten
IhousandtliB eince there are four decimal places. It is read three
liundred and twentyfive tenthousandths.
Bule. — Bead the number expressfd in tAc d^dmeU ptacet a*
tiie numerator, give it the denomination exj^tased by Uie riglit
hand fyure.
Examples to be Read.
102
BA Y'S HIOHER ARITHMETIC.
5.
.7200
13.
2030.0
6.
.5060
14.
40.68031
7.
1.008
15.
200.002
8.
9.00J
16.
.0900001
9.
105.01
17.
61.010001
10.
.0003
18.
31.0200703
11.
00.100
19.
.000302501
12.
180.010
20.
.03672113
Exercises in Notation.
156. The numerator is written as a simple number; the
denomination is then expressed by the use of the decimal
point, and, if necessary, by the use of ciphers in vacant
places.
Problem. — Write eightythree thousand and one billionths.
Explanation. — First write the numerator, operation.
83001. If the point were placed immediately .000083001
at the left of the 8, the denominator would be
hundredthousandths ; it is necessary to fill four places with ciphers
so that the final figure may be in billionths' place.
Biile. — WrriQ ihe numerator as a vMe number; then place
Vie decimal point m (hat ihe rigidhand figure shall be of ihe
same name as ihe decimal.
Examples to be Written.
1. Five tenths.
2. Twentytwo hundredths.
3. One hundred and four thou
sandths.
4. Two units and one hun
dredth.
5. One thousand six hundred
and five tenthousandths.
6. Eightyseven hundredthou
sandths.
7. Twentynine and one half
tenmi Ilionths.
SEDUCTION OF DECIMALS,
103
8. Nineteen million and one
billionths.
9. Seventy thousand and forty
two units and sixteen hun
dredths.
10. Two thousand units and
fiftysix and one third mill
ionths.
11. Four hundred and twenty
one tenths.
12. Six thousand hundredths.
13. Fortyeight thousand three
hundred and five thou
sandths.
14. Eight units and one half a
hundredth.
15. Thirtythree million ten
millionths.
16. Four hundred thousandths.
17. Four hundredthousandths.
18. One unit and one half a
billionth.
19. Sixtysix thousand and three
millionths.
20* Sixtysix million and three
thousandths.
21. Thirtyfour and one third
tenths.
REDUCTION OF DEaMALS.
167. Beduction of decimals is changing their form
vdthout altering their value.
CASE I.
158. To reduce a decimal to a common fraction.
Problem. — Beduce .24 to a common fraction.
Solution. — .24 is equal to ^j which, operation.
reduced, is A. . 2 4 = ^^^ == ,<^, Aiiz.
Problem. — ^Reduce .12^ to a common fraction.
operation.
Solution.— Write 12J as a 221
numerator, and under it place . 1 2 J =? —  =.^= ^^ = J, Am,
100 as a denominator. Re
duce the complex fraction according to Art. 144.
Bule. — FHte tfte decimal as a common fraction; then re
duce the fraction to its lowest terms.
104 BAY'S HIOHER ARITHMETIC,
Bemakk. — If the decimal contains many decimal places, an
approximate value is sometimes used. For example, 3.14159 =
nearly 3.
Reduce to common fractions .
1.
.25625
Am. T^ff.
9.
ll.Of
Am. lliV
2.
.15234375
Am. ^^.
10.
.390625
Am. .
3.
2.125
Arvi. 2\.
11.
.19441
Am. ^j.
4.
19.01750
Am. 19j^.
12.
.24
Am. \i.
5.
16.00^
Am. l&^hf.
13.
.33i
Am. \.
6.
360.028f
Am. 350"^.
14.
.66$
Am. \.
7.
.6666661
Am. ^.
15.
.25
Ans. \.
8.
.003125
CAS
16.
E II.
.75
Am. \.
169. To reduce common fractions to decimals.
Problem. — Reduce ^ to a decimal.
OPERATION.
8oLiiTioN.The fraction  = J of 7. 7 = 8)7.000
7.0, ^ of 7.0 = .8, with 6 tenths remaining ; .6. .87 5,. Ans,
=r.'60, \ of .60 =.07, with 4 hundredths re
maining; .04 = .040, J of .040 = .005. The answer is .875.
Note. — Another form of solution may be obtained by multiply
ing both terms of the fraction by 1000; dividing both terms of the
resulting fraction by the first denominator and writing the answer
as a decimal. Thus, J = }^^, ^J = x^^\ = .875. Both solutions
depend on Art. 129, Prin. ni.
. 'Rvle.— Annex ciphers to the numerator and divide it by the
denominator. Then point off as many decimal places in Hie
qmtient as there are ciphers annexed.
Bemark. — Any fraction in its lowest terms haying in its de
nominator any prime factor other than 2 and 5, can not be reduced
exactly to a decimal. Thus, ^j = .08333 {. The sign + is used at
ADDITION OF DECIMALS, 105
the end of a decimal to indicate that the result is less than the true
quotient. The sign — is also sometimes used to Indicate that the
last figure is too great Thus, \ = .1428 f, or, by abbreviating, \
= .143—.
Reduce to decimals:
1. . Am, ,1b
2. \. Am, .125
3. ^V Am. .05
4. If. Am, .46875
5. ^^. Am, .005625
6. ^. Am, .8
7. ^^. Am, .495
8. ^j. Am, .078125
9. ^. ^««. .05078125
10. y^. ^«5. .0009765625
Note. — The rule converts a mixed number into a mixed decimal,
and a complex into a pure decimal ; thus, 9f = 9.375, since f = .375 ;
and .263^5 = .2612, since ^ = .12.
11. l^ Am. 16.5
12. 42^ Am. 42.1875
13. .015^ Am, .01525
14. lOl.Olf Am, 101.0175
15. 751.19^ Am, 75119.0375
16. 2.00^ Am, 2.00003125
ADDITION OF DECIMALS.
160. Addition of Decimals is finding the sum of two
or more decimals.
Note. — Complex decimals, if there are any, must be made pure,
aa far, at least, as the decimal places extend in the other numbers.
Vi..
Problem.— Add 23.8 and 17^ and .0256 and .41.
operation.
Solution. — Write the numbers 2 3.8
as in the operation, and add as in 17i = 17.5
simple addition. Write the deci .0256
mal point in the sum to the left of .412^= .4166f
tenths. 41 .74221, Ana,
106 BAY'S HIOHEB ARITHMETIC.
Bule. — 1. Wriie Ihe numbers so that figures of the scam
order shaU stand in the same column.
2. Then add as in simple numbers, and put ilie decimal point
to the left of tenths.
Bemark. — ^The proof of each fundamental operation in decimals
is the same as in simple numbers.
1. Find the sum of 1 + .9475 1.9475
2. Of 1.33 added to 2.66^ 4.
3. Of 14.034, 25, .000062^, .0034 39.0374625
4. Of 83 thousandths, 2101 hundredths, 25 tenths, and
94i^ units. 118.093
5. Of .16f, .37^, 5, 3.4, .OOOJ 8.9805«j
6. Of 4 units, 4 tenths, 4 hundredths. 4.44
7. Of .Hi + .6666f + .2222221 1.
8. Of .14f, .018f, 920, .0139^^.. 920.1754
9. Of 16.008J, .00741, .2f, .00019042 16.299768199
10. Of 675 thousandths, 2 millionths, 64, 3.489107, and
.00089407 68.29000307
11. Of four times 4.067 and .000^ 16,272
12. Of 216.86301, 48.1057, .029, 1.3, 1000. 1266.29771
13. Add 35 units, 35 tenths, 35 hundredths, 35 thou
sandths. 38.885
14. Add ten thousand and one millionths; four hundred
thousandths ; 96 hundredths ; fortyseven million sixty thou
sand and eight billionths. 1.017101008
SUBTKACTION OF DEaMALS.
161. Subtraction of DecimalB is finding the difference
between two decimals.
Problem. — From 6.8 subtract 2.057
Solution. — Write the numherR so that units of operation.
the same order stand in the same column ; sup 6 . 8
pose ciphers to be annexed to the 8, and subtract 2. 057
as in whole numbers. 4.743, Ans.
SUBTRACTION OF DECIMALS.
107
Problem.— From 13.256f subtract 6.77^
ExPLAKATiON. — In this example,
the complex decimals are carried out
by division to the same place, and the
common fractions treated by Art. 141.
OPERATION.
ia.256i
6.77i = 6.773t
6.483f, Am.
Bule. — 1. WrSe the subtrahend beneath the mnvfind so Viat
uiiUs of the same order stand in the same column.
2. Subtract as in simple numbers, and write the decimal
pohd as in addition of decimals,
NoT£s. — 1. If eitlier or both of the given decimals be complex,
proceed as directed in the second problem.
2. If the minuend has not as many decimal places as the subtra
hendy annex decimal ciphers to it^ or suppose them to be annexed,
until the deficiency is supplied.
Examples for Practice.
1. Subtract 8.00717 from 19.54
2. 3 thousandths from 3000.
3. 72.0001 from 72.01
4. Subtract .93^ from 1.169^
5. How much is 19 — 8.999^?
6. How much less is .04^ than .4?
7. How much is .65007 — ^?
8. What is 2f — 1^ in decimals ?
9. Subteaet 1 from 1.684
10. f of a millionth from .000
11. 1^ hundredths from 49f tenths.
12. 10000 thousandths from 10 units.
13. 24^ tenths from 3701 thousandths.
14. 1^ units from 1875 thousandths.
15. f of a hundredth from ^ of a tenth,
16.^64^ hundredths from 100 units.
11.58283
2999.997
.0099
.238^
lO.OOOf
.35f
.15007
.95
.684
.000443H
4.9225
0.
1.251
0.
0.
99.351
i»
108 HA rS HIOHEB ARITHMETIC.
MULTIPLICATION OF DECIMALS.
162. Multiplication of Decimals is finding the product
when either or when each of the factors is a decimal.
Principle. — T/ie numbtr of decimal places in the product
equaU Uie number of decimal places in both factors.
. Problem. ^Multiply 2.56 by .184
OPERATION.
Solution.— 2.56 = fJJ, and .184 := ^y^. 2.5 6
Now, f JJ X iVW  AVijW; that 18, the product . .184
of hundredths by thousandths is hundredth 10 2 4
thousandths. This requires five places of 20 48
decimals, or as many as are found in both 25 6
factors. .47104, Ann.
Bule. — 1. Multiply as in whole numbers,
2. Point off as many decimal places in ffie product as there
are decimal places in the two faxstors.
Remarks. — 1. If the product does not contain as many places as
the fac(tors, prefix ciphers till it does contain as many.
2. Ciphers to the right of the product are omitted after pointing.
Examples for Practice.
1. IX.l .1
2. 16 X .03^ .b^
3. .OlX.ll .0015
4. .J080 X 80. 6.4
5. 37.5 X 82 3093.75
6. 64.01 X .32 . 20.4832
7. 48000x73.. 3504000.
8. 64.66X18. 1164.
9. .56JX.O33V .0172H
10.. 738X120.4 98855.2
MULTIPLICATION OF DECIMALS,
109
11. .0001 X 1.006
12. 34 units X. 193
13. 27 tenths X A\
14. 43. 7004 X. 008
15. 21.0375 X 4.44
16. 9300.701 X 251.
17. 430.0126X4000.
18. .059 X .059 X .059
19. 42 units X 42 tenths.
20. 21 hundredths X 600.
21. 7100 X i of a miUionth.
22. 26 millions X 26 millionths.
23. 2700 hundredths X 60 tenths.
24. 6.3029 X .03275
25. 135.027 X 1.00327
.0001006
6.562
1.134
.3496032
93.5
2334475.951
1720050.4
.000205379
176.4
13.5
.0008875
676.
162.
.206419975
135.46853829
163. Oughtred's Method for abbreviating multiplica
tion, may be used when the product of two decimals is
riequired for a definite number of decimal places less than
is found in both Actors.
Problem.— Multiply.3.8640372 by 1.2479603, retaming
only seven decimal places in the product.
Explanation. — It is evident that we need
regard only that portion of each partial
product which affects the figures in and above
the seventh decimal place.
Beginning with the highest figure of the
multiplier, we obtain the first partial product.
Taking the second figure of the multiplier, we
carry each figure of the partial product one
place to the right, so that figures of the same
order shall be in Uie same column. This
product is carried out one place further than
is required, so as to secure accuracy m the
seventh place, and we draw a perpendicular
line to separate this portion. The product of
OPERATION.
3 6 9 7 4 2
3.8640372
1.2479603
38640372
7728074
1545614
270482
34776
2318
11
4.8221650
4
8
G
3
4
5
,04 by the righthand
110 MAY'S HIGHER ARITHMETIC.
figure in the seventh place, would extend to the ninth place of
decimals ; so we may reject the la»t figure, and commence with the 7.
With each succeeding figure of the multiplier, we commence to
multiply at that figure of the multiplicand which will produce a
product in the eighth place. It is also convenient to place each
figure of the multiplier directly oVer the first figure of the multipli
cand taken. In multiplying hy .007, we have 7X3 = 21 ; but, if we
had been expressing the complete work, we should have 5 to carry
to this place ; the corrected product is therefore 21  ^ = 26. The
product from the last figure, 3, is carried two places to the right. In
the total product, the eighth decimal is dropped ; but the seventh
decimal figure is corrected by the amount carried.
Bule. — 1. Multiply only mck figures as shall produce one
more than the required nuwher of decimal places.
2. Begin wiih Vie highest order of the multiplier; under the
rightrhand figure of each partial product, place the riglUrhand
figure of the succeeding one. In obtaining su/h rightliand
figure, let that number he added which would he carried from
multiplying the figure of the next lower order.
3. Add the partial produds, and ryect the righthand figure.
Kemarks.— 1. It will be found convenient to write the multiplier
in a reverse order, with its uniti^ figure under that decimal figure of
the multiplicand whose order is next lower than the lowest required.
Thus, in the fourth example, the 8 would be written under the 1.
2. In carrying the tens for what is left out on the right, carry also
one ten for each 5 of units in the omitted part ; thus, 1 ten for 5 or
14 units, 3 tens for 25 or 26 units, etc. Make the same correction
for the final figure rejected in the product.
ExAMi^^Es FOR Practice.
1. Multiply 27.653 by 9.157, preserving three decimal
places. 253.219
2. Multiply 43.2071 by 3.14159, preserving four decimal
places. 135.7390
3. Multiply 3.62741 by 1.6432, preserving four decimal
places. 5.9606
DIVISION OF DECIMALS. Ill
4. 9.012 X 48.75, preserving one place. 439.3
5. 4.804136 X .010759, preserving six places. .051688
6. Slij^ X 26, preserving three places. 21813.475
7. 702.61 X 1.258^, preserving three places. 884.020
8. 849.931 X .0424444, preserving three places. 36.075
9. 880.695 X 131.72 true to units. 116005
10. .025381 X .004907, preserving five places. .00012
11. 64.01082 X .03537, preserving six places. 2.264063
12. 1380. 37^ X .234f, preserving two places. 82416
DIVISION OF DEajdALS.
164. Division of Decimals is the process of finding the
quotient when either or when each term is a decimal.
165. Since the dividend corresponds to the product in
multiplication (Art. 73), and the decimal places in the divi
dend are as many as in both factors (Art. 162, Prin.),
we derive the following principles :
Principles. — 1. TAe dividend mud contain as many deci
mal places as the divisor ; and when both have the same number,
Hie quotient is an integer,
2. The quotierd must contain as many decimal places as
th£ number of those in the dividend exceeds the number of those
in Hue divisor.
Problem.— Divide .50312 by .19
OPERATION.
.19). 50312(2. 648, Am.
Solution. — The division is per 38
formed as in integers. The quo 12 3
tient 18 pointed according to 114
Principle 2. The quotient must 9 1
have 6 — 2 = 3 places. 7*6
152
152
112 RAY'S HIGHER ARITHMETIC.
Proof. — By expressing the decimals as common fractions
we have;
miifi ^ iV^ = imVir X W =!fU = 2.648
Problem.— Divide .36 by .008
Solution. — The dividend has a less num operation.
ber of decimal places than the divisor. .008 ) . 360
Annex one cipher, making the number 45, Am,
equal. The quotient is an integer.
Problem.— Divide .002^ by .06
OPERATION.
Solution.— Reducing the . 002 1 = . 002475
mixed decimals to equiva .06 J =.0 66
lent pure decimals, we have .066). 002475 (.037 5, ilf».
.002475 and .066. Dividing, 198
we find one more decimal 4 9 5
place necessary to make the 4 6 2
division exact ; and, pointing 330
by Principle 2, we have .0375 • 3 30
Bule. — Divide as in whole numbers, and point off as matiy
decimal places in the quotient as those in the dividend exceed
tJwse in tJw divisor.
Note. — When the division is not exact, annex ciphers to the
dividend, and carry the work as far as may be necessary.
Examples for Practice.
1.
63 : 4000.
.01575
2.
3.15 : 375.
.0084
3.
1.008 : 18.
.056
4.
4096 : .0S2
128000.
5.
9.7 : 97000.
.0001
6.
.9^.00075
1200.
7. 13^78.12^ .1664
8. 12.9^8.256 1.5625
9. 81.209671.28 63.445
10. 1MOO. .01
11. 10.1 M7. .59412—
12. .001^100. .00001
DIVISION OF DECIMALS, 113
13. 12755 f 81632. .15625
14. 2401 T 21.4375 112.
15. 21.13212^.916 23.07
16. 36.72672^.5025 73.088
17. 2483.25^5.15625 481.6
18. 142.0281^9.2376 15.375
19. .08^^.121 .66f = §.
20. .0001— .01 .01
21. 95.3 ^. 264 360.984848 f
22. 1000 ^ .001 1000000.
23. Ten ~ 1 tenth. 100.
24. .000001^.01 .0001
25. .00001 MOOO. .00000001
26. 16.275 4. 41664 39.0625
27. 1 tenmillionth f 1 hundredth. .00001
• 166^ Oughtred's Method. — ^If the quotient is not re
quired to contain figures below a certain denomination, the
work may sometimes be abridged.
Problem.— Divide 84.27 by 1.27395607, securing a quo
tient true to four places of decimals.
OPERATION.
Solution.— since 1.2)f';{}>jj^) 84.27000 ( 66.1483 —
the divisor is greater 7643 736
than 1 and less than 2, the quotient 7 8 3 2 6 4
will contain six places, — two of in 7 6 43 7 4
tegers and four of decimals. The i ggQQ
highest denomination of the divisor, 12 7 40
multiplied by the lowest denomina g j ^ q
tion of the quotient, would obtain a 50 9 6
figure in the fourth place. We take ~10^54~
one place more as in multiplication 1 1 Q
(Art. 163), and also cut off two figures ^r
of the divisor, since these can not affect „ q
the quotient above the fourth place.
After obtaining the first figure of the quotient, we drop one right
hand figure of the divisor for each figure obtained. To prevent
errors, we cancel the figure before each division.
H. A. 10.
114
BAY^S HIQHER ABITHMETIC.
Bule. — Fini the figure of the dividend thcd would remit firom
imdtiplying a unit in the highest devwminajtion of ihe dmsor by a
unit of the lowed denomination required in the quotient. Take
one more figure of the dividend to secure aceimwy. Oat off any
figures of ihe divisor not needed for the abbreviated dividend.
Divide as usual until the fibres remaining in the dividend
are all divided. At each subsequent division, drop a figure
from the divisor , carrying the number necessary from the produd
of ihe figure omitted.
Continue until ihe divisor is reduced to two figures.
Bemabk. — In the quotient, 5 units of an omitted order may be
taken as 1 unit of the next higher ordjer.
Examples for Practice.
1. 1000 r .98, preserving two places.
2. 6215.75 f .99^, preserving three places.
3. 28012 = .993, preserving two places.
4. 52546.35 r.99f, preserving three places.
5. 4840 ^ .9875, preserving two places.
6. 2= 1.4142136, preserving seven places.
7. 9.869604401^3.14159265, preserving eight places.
3.14159265
1020.41
6246.985
28209.47
62678.045
4901.27
1.4142135
Topical Outline.
Decimal Fractions.
Definitions.
Decimal Point
{Pure.
Mixed.
Complex.
Principles.
Numeration, Rule.
Notation, Rule.
Reduction
In.
I
Dec. to a Com. Fraction.
Com. Fraction to a Dec
Addition, Rule.
Subtraction, Rule.
Multiplication, Rule.
Abbreviated Multiplication.
Diyisiou, Rule.
AbbrvviHtod J>lvli»iou.
X. OmOULATrKG DECIMAiS.
107. In reduciDg common fractions to decimals, the
process, in some cases, does not terminate. Thb gives rise
to Circulating Decimals.
Principle I. — Ij any prime facUyn other Hian 2 and 5 are
found in Hie denominator of a fraction in Us lowed termSy the
resulting decimal vnU be interminate.
Demonstration. — If the fraction is in its lowest terms, the
nnmerator and denominator are prime to each other (131, Bem. 2).
In the process of reduction, the numerator is multiplied by 10. By
this means the factors 2 and 5 may be introduced into the numera
tor as many times as necessary; but no others are introduce<).
Therefore, if any factors other than 2 and 5 are found in the denom
inator, the division can not be made complete, and the resulting
decimal will be interminate.
Thus, ^2^ = 2X2X2X2X2X2X6 ^^ '^^375
and, ^ = 2X2X2X2 = ^^^^
l>«t, 61> = 2X^^ = 11^^^+
It is evident that the first will terminate if the numerator be mul
tiplied six times by 10, carrying the decimal to the sixth place. In
the same way we reduce ^ to sl decimal containing four places.
But since the factor 3 is found in the denominator of 7^, the frac
tion can not be exactly reduced, though the numerator be multi
plied by any power of 10.
•
Principle II. — Every interminate decimal arising from the
reducti&n of a comnum fra^ction will, if the division be carried
far evuyagh, contain the same figure, or set of figures, repeated
in the same order.
(115)
116 HAY'S mOHER ARITHMETIC.
Demonstration. — Each of the remainders must be less than the
denominator which is used as the divisor (Art. 78, Note 2). If the
division be carried far enough, some remainder must be found equal
to some remainder already found, and the subsequent figures in the
quotient must be similar to the figures found from the former
remainder.
Thus, in reducing , we find the decimal .142857, and then have
the remainder 1, the number we started with; if we annex a cipher,
we shall get 1 for the next figure of the quotient, 4 for the next, etc.
168. 1. In terminate decimals, on this account, have
received the name of OircuUding or Reimrring Decimals,
2. A Circulate or Circulating Decimal has one or
more figures constantly repeated in the same order.
3. A Bepetend is the figure or set of figures repeated,
and it is expressed by placing a dot over the first and last
figure ; thus, f = .142857 ; if there be one figure repeated,
the dot is placed over it, thus, f = .6666 + = .6
4. A Pure Circulate has no figures but the repetend ; as,
.5 and .124
5. A Mixed Circulate has other figures before the
repetend; as, .2083 and .31247
6. A Simple Bepetend has one figure; as, .4
7. A Compound Bepetend has two or more figures ; as,
M
8. A Perfect Bepetend is one which contains as many
decimal places as there are units in the denominator, less 1 ;
thus,  = .i42857
9. Similar Bepetends begin and end at the same deci
• • • •
mal place; ap, .427 and .536
10. Dissimilar Bepetends begin or end at diflferent deci
• • • •
mal places; as, .205 and .312468
CIRCULATINO DECIMALS. 117
11. Conterminous Bepetends end at the same place;
as, .50397 and .42618
12. Cooriginou8 Bepetends begin at the same place;
• • •
as, .5 and .124
169. Any terminate decimal may be considered a circu
• • •
late, its repe tend being ciphers; as, .35 = .350 = .350000
Any simple repetend may be made compound, and any
compound repetend still more compound, by taking in one
• • •
or more of the succeeding repetends; as, .3 = .33333, and
.0562 = .056262, and .257 = .257257257
Kemarks. — 1. When a repetend is thus enlarged, be careful to
take in no part of a repetend without taking the whole of it ; thus, if
we take in 2 figures iwthe last example, the result, .25725, would be
incorrect, for the next figure understood being 7, shows that 25725
is not repeated.
2. A repetend may be made to begin at any lower place by carry
ing its dots forward, each the same distance ; thus, .5 = .555, and
.294i = .29414, and 5.1836 = 5.183683
3. Dissimilar repetends can be made similar, by carrying the dots
forward till they all begin at the same place as the one farthest
from the decimal point.
4. Similar repetends may be made conterminous by enlarging the
repetends until they all contain the same number of figures. This
number will be the least common multiple of the numbers of figures
in the given repetends.
For, suppose one of the repetends to have 2, another 3, another
4, and the last 6 figures ; in enlarging the first, figures must be taken
in, 2 at a time, and in the others, 3, 4, and 6 at a time.
170. Circulating decimals originate, as has been already
shown, in changing some common fractions to decimals.
Then, having given a circulate, it can always be changed to
an equivalent common fraction.
171. Circulating decimals may be added, subtracted,
multiplied, and divided as other fractions.
118 RAY'S HIGHER ARITHMETia
CASE I.
172. To reduce a pure circulate to a common
Araction.
Problem. — Change .53 to a common fraction.
OPERAnON.
Solution. — Removing the 100 times the repetend = 5 3.5 3 ,
decimal point one place to the Once the repetend = : .5 3
right, multiplies the repetend /. 99 times the repetend = 5 3.
hy 10 ; two places, by 100, and Once the repetend = JJ, Aw,
so on. Then, multiplying by
100, and subtracting the repetend from the product, removes all of
that part to the right of the decimal point, and, dividing 53 by 99,
we have the common fraction whidi produced the given repetend.
« ■
Problem. — Change .456 to a common fraction.
OPERATION.
1000 times the repetend = 4 5 6.4 5 6
Once the repetend =• .4 56
999 times the repetend = 4 5 6. 1
.*. once the repetend = fff, Aw.
Problem. — Change 25.6 to a common fraction.
OPERATION.
Carry the dot forward thus : 2 5.6 = 2 6.6 2 6
1000 times the repetend = 6 2 5.6 2 5
Once the repetend = .6 25
999 times the repetend = 6 2 5.
.*. once the repetend = ff J
Whence the 25.625 = 25 f }{, Aw,
Note. — From these solutions the following rule is derived.
Bule. — WriU the repetend for ihe numerator , omitting Hie
decimal point and the dote, and for the denominator write cw
many ff8 as there are fibres in the repetend^ and reduce the
fraction to Ue lowed terms.
CZBCULAUNG DECIMALS. 119
CASE II.
173. To reduce a mixed circulate to a common
fraction.
Problem. — Change .82143Y to a common fraction.
OPEBATION.
8 21 itii
Omitting the decimal point, we have : .821437= ^^^ =
821X999 + 437 ^ 821(1000 — 1)4437
1000X999 "" 999000 ^
821000 — 821 + 437 820616 102677
999000 999000 124875
Or, briefly, 821437 — 821 102677
, Ans,
Ang
999000 124876'
Problem. — Change .048 to a common fittction.
OPERATION.
Omitting the decimal point, we have :
4ft 4 8 4(10 — 1^
8
100 100 ' 900 900 ' 900
9 00 ■^900"' 9 00 ~"2 26' ^'
Or, briefly, 48 — 4 44 11
900 ""900 226' ****
Note. — ^The following rule is derived from the preceding solu
tions.
Rule. 1. For ihe nvmeftcAor, subtract the part which precedes
ihe repetend from ihe whole expression, both quantities being con
siiered as units.
2. For ihe dentyminatory write as inany 9'« as there are fibres
in the repetend, and annex as many ciphers as there are decimal
figures before each rqpeiend.
120
RAY'S HIGHER ARITHMETIC.
Reduce to common fractions:
1.
.3
2.
.05
3.
.123
4.
2.63
5.
.31
6.
.0216
7.
48.1
i
8.
1.001
tV
9.
.138
TtS'
10.
.2083
2A
11.
85.7142
H
12.
.063492
lis
13.
.4476190
48iV.
14.
.09027
85f
ttV
ADDITION OF CIECULATES.
174. Addition of Circulates is the process of finding
the sum of two or more circulates. Similar circulates only
can be added.
Problem.— Add .256, 5.3472, 24.815, and .9098
OPERATION.
.25666666,^6
5.8472727272
2 4.8158i58158
.9098000000
31.3295552097
Solution. — Make the circulates similar.
The first column of figures which would
appear, if the circulates were continued, is
the same as the first figures of the repe
tends, 6, 7, 1, 0, whose sum, 14, gives 1 to be
carried to the righthand column. Since
the last six figures in each number make a repetend, the last six
figures of the sum also make a repetend.
Bule. — Make the repetends dmilar, if iJiey be rwt 8o; add,
and point off as in ordinary decimalsy increasing the rigkt4iand
column by the amount, if any, which nnyuM be carried to it if
the circulates were continued; then make a repetend in the sum,
similar to those above.
Bemark. — In finding the amount to be carried to the righthand
column, it may be necessary, sometimes, to use the two gu<lceediDg
figures in each repetend.
SUBTBACTION OF CIRCULATES. 121
RxAAfPTJRS FOB PRACTICE.
1. Add .453, .068, .327, .946 1.796
2. Add 3.04, 6.456, 23.38, .248 33.1334
3. Add .25, .104, .61, and .5635 1.536
4. Add i.03, .257, 5.04, 28.0445245 34.37
5. Add .6, .138, .05, .0972, .0416 1.
6. Add 9.21107, .65, 5.004, 3.5622 18.45
7. Add .2045, .09, and .25 .54
8. Add 5.0770, .24, and 7.124943 12.4
9. Add 3.4884, 1.637, 130.81, .066 136.00
SUBTRACTION OF CIRCULATES.
176. Subtraction of Circulates is the process of find
ing the difference between two circulates. The two circu
lates must be similar^
Problem.— Subtract 9.3i56 from 12.902i
OPERATION.
Solution. — Prepare the numbers for sub 1 2.9 0212121
traction. If the circulates were continued, the 9.3 1 5 6 1 5 6 1
next figure in the subtrahend (5) would be 3.5 8 6 5 5 5 9
larger than the one above it (2) ; therefore,
carry 1 to the righthand figure of the subtrahend.
Rule. — Make ike repetends similar^ if they be not so ; mtbtract
and point off as m ordinary decimals, carrying 1, Jwwever,
to the righthand figure of the subtrahendy if on continuing the
circulates it be found necessary ; then make a repetend in the
remainder, simUar to those above,
Bemark. — It may be necessary to observe more than one of the
succeeding figures in the circulates, to ascertain whether 1 is to be
carried to the righthand figure of the subtrahend or not.
H. A. 11.
122 BAY'S HIGHER ARITHMETIC
Examples for Practice.
1. Subtract .0074 from .26 .259
2. Subtract 9.09 from 15.35465 6.25
3. Subtract 4.5i from 18.23673 13.72
4. Subtract 37.0128 fit)m 100.73 63.7i
5. Subtract 8.27 from 10.0563 1.7836290
6. Subtract 190.476 from 199.6428571 9.16
7. Subtract 13.637 from 104.i 90.503776
MULTIPLICATION OF CLBCULATES.
176. Mtiltiplication of drculates is the process of
finding the product when either or when each of the factors
is a circulate.
Problem.— Multiply .3754 by 17.43
Solution. — ^In forming the partial operation.
products, carry to the righthand figures .3754
of each respectively, the numbers 1, 3, 0, 1 7.4 3 = 1 7.4 1
arising from the multiplication of the il601777
figures that do not appear. The repetend 2628lili
of the multiplier being equal to J, J of 3.7 5 4 4 4 4 4
the multiplicand is 125148, whose figures 1 2 5 i 4 8
are set down under those of the multipli . .
cand from which they were obtained. 6.5 4 5 2 4 8 1 Aw^
Point the several products, carry them forward until their repetends
are similar, and add for answer.
Bule. — 1. Ij the multiplier contain a repetend^ change Uto a
common fraction.
2. Hien multiply as in mvltiplieation of decinuds, and add
to the rightlMnd figure of each partial product the amount
necessary if the repetend were repeated,
3. Make the partial products similar, and find their sum.
DIVISION OF CIRCIJLATE& 123
Examples for Practice.
1. 4.735 X 7.349 34.800liS
2. .07067 X .9432 .066665
3. 714.32X3.456 2469.173814
4. 16.204X32.75 530.810446
f). 19. 072 X. 2083 3.97348
6. 8.7543X4.7157 17.7045082
7. 1.256784X6.42081 8.069583206
DIVISION OF CIRCULATES.
177. Division of Circulates is the process of finding
the quotient when either or when each of the terms is a
circulate.
Problem. — ^Divide .154 by .2
OPERATION.
.i54 = HI, .2 = f
Rule. — Change the terms to common fractions; then divide
(18 in division of fractions, and rediLce the result to a repetend,
Kemark. — This is the easiest method of solving problems in di
yision of circulates. The terms may be made similar, however, and
the division performed without changing the circulates to common
fractions.
Examples for Practice.
1. .*75r.i 6.8i
2. 51.49i ^ 17. 3.028
3. 681.5598879^94. 7.2506371
4. 90.5203749 ^ 6.754 13.40i
5. 11.068735402 ~ .245 45.13
6. 9.5^30663997 ^ 6.217 1.53
7. 3.500691358024^7.684 .45
124
BAY'S HIGHER ARITHMETIC.
Topical Outline.
Circulating Decimai^.
1. Principles.
2. Definitions —
1. Circulate.
2. Repetend.
3. Pure Circulate.
4. Mixed Circulate.
5. Simple Repetend.
6. Compound Repetend.
7. Perfect Repetend.
8. Similar Repetends.
9. Dissimilar Repetends.
10. Conterminous Rei)etends.
, 11. Cooriginous Repetends.
3. Reduction / Case I.
lease II.
{
{
Definition.
4. Addition \ Rule.
Applications.
Definition.
5. Subtraction..... } Rule.
Applications.
Definition.
6. Multiplication \ Rule.
Applications.
Definition.
7. Division \ Rule.
Applications.
XL COMPOUND DENOMESTATE
. NUMBERS.
178. 1. A Measure is a standard unit used in estimating
quantity. Standard units are fixed by law or custom.
2. A quantity is measured by finding how many times it
contains the unit.
3. Denomination is the name of a unit of measure of a
concrete number.
4. A Denominate Number is a concrete number
which expresses a particular kind of quantity; as, 3 feet, 7
pounds.
5. A Ck>mpound Denominate Number is one expression
of a quantity by different denomincttiom under one kind of
measure; as, 5 yards, 2 feet, and 8 inches.
6. All measures of denominate numbers may be embraced
under the following divisions: Valuer Weight, Extenshn,
and Time.
MEASUBES OF VALUE.
179. 1. Value is the worth of one thing as compared
with another.
2. Value is of three kinds: Intrinsic^ Commercial, and
Nominal.
3. The Intrinsic Value of any thing is measured by the
amount of labor and skill required to make it useful.
4. The Commercial Value of any thing is its purchasing
power, exchangeability, or its worth in market.
(125)
126 RAY'S HIGHER ARITHMETIC.
5. The Nominal Value is the name value of any thing.
6. Value is estimated among civilized people by its price
in money.
7. Money is a standard of value, and is the medium of
exchange; it is usually stamped metal, called coins, and
printed bills or notes, called paper money.
8. The money of a country is its Currency. Currency
is national or foreign.
United States Money.
180. United States Money is the legal currency of the
United States. It is based upon the decimal system; that
is, ten units of a lower order make one of the next higher.
The Dollar is the unit. The same unit is the standard
of Canada, the Sandwich Islands, and Liberia.
Table.
10 mills, marked m., make 1 cent, marked ct.
10 cents ** 1 dime, " d.
10 dimes " 1 dollar, " $.
10 dollars " 1 eagle, ** e.
Note. — The cent and miUy which are yj^ and j^un of a dollar,
derive their names from the Latin centum and mt//e, meaning a hun
dred and o thousand ; the dime^ which is j\y of a dollar, is from the
French word disrwe, meaning Un,
Bemabks. — 1. United States money was established, by act of
Congress, in 1786. The first money coined, by the authority of the
United States, was in 1793. The coins first made were copper cents.
In 1794 silver dollars were made. Gold eagles were made in 1795;
gold dollars, in 1849. Gold and silver are now both legally standard.
The trad^ dollar was minted for Asiatic commerce
2. The coins of the United States are classed a^ brmze, nickd, silver.
MEASURES OF VALUE.
127
and gold. The name, value, composition, and weight of each coin
are shown in the following table :
Table.
COIN.
VALUE.
COMPOSITION.
1
WEIGHT.
BBONZS.
One cent
Icent
95 parts copper, 5 parts tin & sine
48 grains Troy.
NICKEL.
3cent piece.
5cent piece.
Scents.
5 cents.
75 parts copper, 25 parts nickel.
75 •* ** 25 " **
30 grains Troy.
77.16 •♦
SILVER.
Dime.
Quarter dollar.
Half dollar.
Dollar.
10 cents.
25 cents.
50 cents.
100 cents.
90 parts silver, 10 parts copper.
90 •* ** 10 ** ••
90 •♦ •• 10 " ••
90, •• •• 10 ** ••
2.5 grams.
6.25 ••
12.5 ••
412.5 grains Troy.
GOIiD.
Dollar.
Quarter eagle.
Three dollar.
Half eagle.
Eagle.
Double eagle.
100 cents.
2>^ dollars.
3 dollars.
5 dollars.
10 dollars.
20 dollars.
90 parts gold, 10 parts copper.
90 " " 10 ••
90 " " 10 "
90 •* " 10 •'
90 •* •• 10 ••
90 *• •• 10 "
25.8 grains Troy.
64.5
77.4 •*
129
258 •♦ *'
516 " ••
3. A deviation in weight of ^ a grain to each piece, is allowed by
law in the coinage of Double Eagles and Eagles ; of ^ of a grain in
the other gold pieces ; of IJ grains in all silver pieces ; of 3 grains
in the fivecent piece ; and of 2 grains in the smaller pieces.
4. The mill is not coined. It is used only in calculations.
181. In reading U. S. Money, name (he dollars and all
higher denominaiiona together as dollars^ the dimes and cents as
cents, and die next figure, if there he one, as mills ;
Or, name the whole number as dollars, and the rest as a
decimal of a dollar.
Thus, $9,124 is read 9 dollars 12 ct. 4 mills, or 9 dollars 124
thousandths of a dollar.
128 BAY'S HIOHER ABITHMETIQ.
English or Sterling Money.
182. English or Sterling Money is the currency of
the British Empire.
The pound sterling (worth $4.8665 in U. S. money) is
the unit, and is represented by the sovereign and the £1
banknote.
Table.
4 farthings, marked qr., make 1 penny, marked d.
12 pence " 1 shilling, " s.
20 shillings " 1 pound, ^ " £.
Equivalent Table.
& 8. d. qr.
1 = 20 = 240 = 960.
1 = 12 = 48.
1 = 4.
Bebcarks. — 1. The abbreviations, JS, s., d., q., are the initials
of the Latin words l\hra^ 8olidariu8f denariuSy qwtdixinSf signifying,
respectively, pound, shilling, penny, and quarter.
2. The coins are gold, silver, and copper. The gold coins are the
sovereign ( = £1), half sovereign ( = 108.), guinea ( = 21s.), and
halfguinea ( = lOs. 6d.) The silver coins are the crown ( = 58.),
the half crown ( =2s. 6d.), the florin ( = 28.), the shilling, and the
sixpenny, fourpenny, and three penny pieces. The penny, half
penny, and farthing are the capper coins. The guinea, halfguinea,
crown, and halfcrown are no longer coined, but some of them are
in circulation. •
3. The standard for gold coins is J} pure gold and ^ alloy ; the
standard for silver is f J pure silver and ^ copper ; pence and
halfpence are pure copper.
French Money.
188. French Money is the legal currency of France.
Its denominations are decimal.
MEASURES OF VALUE 129
The franc (worth 19.3 cents in U. S. money) is the unit
The franc is also used in Switzerland and Belgium, and.
under other names, in Italy, Spain, and Greece.
Table.
10 millimes, marked m., make 1 centime, marked c.
10 centimes " 1 decime, ** d.
10 decimes " 1 franc, " fr.
Equivalent Table.
1 fr. = 10 d. == 100 c. = 1000 m.
1 d. = 10 c. = 100 m.
1 c. = 10 m.
Remark.— The coins are of gold, silver, and bronze. The gM
coins are 100, 60, 20, 10, and 5franc pieces ; they are .9 pure gold.
The 20franc piece weighs 99.55 gr. The siktr coins are 6, 2, and
1 franc pieces, and 50 and 20centime pieces ; they are now .835
pure silver. The brome coins are 10, 5, 2, and 1centime pieces.
The decime is not used in practice. All sums are given in francs
and centimes, or hundredths.
German Money.
184. German Money is the legal currency of the Grer
man Empire.
The mark, or reichsmark (worth 23.8 cents in U. S.
money), is the unit. The only other denomination is the
pfennig (penny).
Table.
100 pfennige, marked Pf., make 1 mark, marked EM.
Remabk. — The coins are of gold, silver, and copper. The gold
coins are of the value of 40, 20, 10, and 5 marks; the silver coins, 3,
2, and 1mark pieces, and 50, 20, and 10 pfennige. The coppa, 2 and
1 pfennig pieces. Gold and silver are .9 fine.
130 BAY'S HIGHER ARITHMETIC.
MEASURES OF WEIGHT.
185. Weight is the measure of the force called gravity,
which draws bodies toward the center of the earth.
The standard unit of weight in the United States is the
Troy pound of the Mint.
188. Three kinds of weight are in use, — 2Voy Weight ^
ApotJiecariea^ Weight, and Avoirdupois Weight.
Troy Weight
187. Troy Weight is used in weighing gold, silver,
platinum, and jewels. Formerly it was used in philosoph
ical and chemical works.
Table.
24 grains, marked gr., make 1 pennyweight, marked pwt.
20 pwt. ** 1 ounce, " oz.
12 oz. ** 1 pound, ft).
Equivalent Table.
ft). oz. pwt. gr.
1 = 12 = 240 = 5760.
1 = 20 = 480.
1 = 24.
Hem ARK. — The Troy pound is equal to the weight of 22.7944
cubic inches of pure water at its maximum density, the barometer
being at 30 inches. The standard pound weight is identical with the
Troy pound of Great Britain.
Apothecaries' Weight.
188. Apothecaries' Weight is used by physicians and
apothecaries in prescribing and mixing dry medicines.
Medicines are bought and sold by Avoirdupois Weight
MEASURES OF WEIGHT. 131
Table.
20 grains, marked gr., make 1 scruple, marked 9.
3 9 *' 1 dram, ** 3.
83 "1 ounce, ** 5.
12 § ** 1 pound, " lb.
Equivalent Tablr
ft). §. 5. 9. gr.
1 = 12 = 96 = 288 = 5760.
1 = 8 = 24 = 480.
1 = 3 = 60.
1 = 20.
Kemark. — The pound, ounce, and grain of this weight are the
same as those of Troy weight ; the pound in each contains 12 oz. =
5760 gr.
Avoirdupois or Commercial Weight.
189. Avoirdupois or Commercial Weight is used for
vreighing all ordinary articles.
Table.
16 ounces, marked oz., make 1 pound, marked lb.
25 lb. ** 1 quarter, *' qr.
4 qr. "1 hundredweight, ** cwt.
20 cwt. '' 1 ton, '* T.
Equivalent Table.
T. cwt. qr. lb. oz.
1 = 20 =^ 80 = 2000 = 32000.
1 = 4 = 100 = 1600.
1 =r 25 = 400.
1 = 16.
132 JiA yS HlOffJSIi AMITHMETia
Kemabks. — 1. In Great Britain, the qr. = 28 lb., the cwt. = 112
lb., the ton = 2240 lb. These values are used at the United States
customhouses in invoices of English goods, and are still used in
some lines of trade, such as coal and iron.
2. Among other weights sometimes mentioned in books, are : 1
stone, horseman's weight, = 14 lb.; 1 stone of butcher's meat = 8 lb.;
1 clove of wool == 7 lb.
3. The lb. avoirdupois is equal to the weight of 27.7274 cu. in. of
distilled water at 62° (Fahr.) ; or 27.7015 cu. in. at its maximum
density, the barometer at 30 inches. For ordinary purposes, 1 cubic
foot of water can be taken 62^ lb. avoirdupois.
4. The terms gross and net are used in this weight. Gross weight
is the weight of the goods, together with the box, cask, or whatever
contains them. Net weight is the weight of the goods alone.
5. The word avoirdupois is from the French avoirs, du, poisj signify
ing goods of weight.
6. The ounce is often divided into halves and quarters in weigh
ing. The sixteenth of an ounce is called a dram.
COMPARISON OP WEIGHTS.
190. The pound Avoirdupois weighs 7,000 grains Troy,
and the Troy pound weighs 5,760 grains, hence there are
1,240 grains more in the Avoirdupois pound than in the
Troy pound.
The following table exhibits the relation between certain
denominations of Avoirdupois, Troy, and Apothecaries^
Weight.
Avoirdupois.
Troy.
Apotheci
iries'.
1 lb. —
1t¥t «>.
—
1*
lb.
1 oz. —
•H4 oz.
—
m
S
1 ft).
1 lb.
1 oz.
1
5
1 gr.
1
gf
1 pwt.
i
3
1 pwt.
H
a.
MEASUBES OF EXTENSION. 133
Bemabk. — In addition to the foregoing, the following, called
Diamond Weighty is used in weighing diamonds and other precious
stones.
Table.
16 parts make 1 carat grain = .792 Troy grains.
'4 carat grains " 1 carat =3.168 "
Note. — ^This carat is entirely different from the assay carat, which
has reference to i\i% fineness of gold. The mass of gold is considered
as divided into twentyfour parts, called carats, and is said to be so
many carats fine, according to the number of twentyfourths of pure
gold which it contains.
MEASURES OF EXTENSION.
191. 1. Extension is that property of matter by which
it occupies space. It may have one or more of the three
dimensions,— length, breadth, and thickness.
2. A lin6 has only one dimension, — length.
3. A sur&ce has two dimensions, length and breadth.
4. A solid or volume has three dimensions, — length,
breadth, and thickness.
192. Measures of Extension embrace :
^ Long Measure.
1. Linear Measure.
Chain Measure.
Mariners' Measure.
Cloth Measure.
2. Superficial Measure.  ^^'^ Measure.
(Surveyors Measure.
3. Solid Measure. T Liquid Measure.
4. Measures of Capacity. < Apothecaries Measure.
6. Angular Measure. vDry Measure.
134 BA Y'S HIGHER ARITHMETIC.
Long or Linear Measure.
193. Linear Measure is used in measuring distances, or
length, in any direction.
The standard unit for all measures of extension is
the yard, which is identical with the Imperial yard of
Great Britain.
Table.
12 inches, marked in., make 1 foot, marked ft.
3 ft. ** 1 yard, '' yd.
5^ yd. or 16^ ft. *' 1 rod, " rd.
320 rd. ** 1 mile, " mi.
Equivalent Table.
mi. rd. yd. ft. in.
1 = 320 = 1760 = 5280 = 63360.
1 = 5^ = 16^ = 198.
1 = 3 = 36.
1 = 12.
Bemares. — 1. The standard yard of the United States was
obtained from England in 1856. It is of bronze, and of due length
at 59.8° Fahr. A copy of the former standard is deposited at
each state capital : this was about jjhrj ^^ ^^ i^^^h too long.
2. The rod is sometimes called perch or pole. The furlong, equal to
40 rods, is seldom used.
3. The inch may be divided into halves, fourths, eighths, etc., or
into tenths, hundredths, etc.
4. The following measures are sometimes used :
12 lines
make 1 inch.
3 barleycorns
1 "
3 inches
1 palm.
4 inches
1 hand.
9 inches
1 span.
18 inches
1 cubit.
3 feet
1 pace.
MEASURES OF EXTENSION. 135
194. Chain Measure is used by surveyors in measuring
land, laying out roads, establishing boundaries, etc.
Table.
7.92 inches, marked in., make 1 link, marked li.
100 li. "1 chain, ** ch.
80 ch. "1 mile, ** mi.
Equivalent Table.
mi. ch. li. in.
1 = 80 = 8000 = 63360.
1 = 100 = 792.
1 = 7.92.
Kemarks. — 1. The surveyors* chain, or Gunter's chain, is 4 rods,
or 66 feet in length. Since it consists of 100 links, the chains and
links may he written as integers and hundredths ; thus, 2 chains 56
links are written 2.56 ch.
2. The engineers' chain is 100 feet long, and consists of 100 links.
3. The engineers' leveling rod is used for measuring vertical dis
taqces. It is divided into feet, tenths, and hundredths, and, hy
means of a vernier, may be read to thousandths.
196. Mariners' Measxire is used in measuring the depth
of the sea, and also distances on its surface.
Table.
6 feet make 1 fathom.
720 feet " 1 cablelength.
Kemarks. — 1. A nautical mile is «ne minute of longitude, meas
ured on the equator at the level of the sea. It is equal to 1.152
statute miles. 60 nautical miles = 1 degree on the equator, or 69.16
statute miles. A league is equal to 3 nautical miles, or 3.458 statute
miles.
2. Depths at sea are measured in fathoms ; distances are usually
measured in nautical miles.
136 liAY'S mOHER ARITHMETIC.
196. Cloth Measure is used in measuring drygoods.
The standard yard is the same as in Linear Measure, but
is divided into halves, qtuirterSf eighthsy sixteenthSy etc., in
place of feet and inches.
Rebiarks. — 1. There was formerly a recognized tahle for Cloth
Measure, but it is now obsolete. The denomiuations were as follows :
2i inches, marked in., make 1 nail, marked na.
4 na. or 9 in. " 1 quarter, " qr.
4 qr. " 1 yard, " yd.
2. At the customhouse, the yard is divided decimally.
Superficial or Surface Measure.
197. 1. Superficial Measure is used in estimating the
numerical value of surfaces ; such as, land, weatherboard
ing, plastering, paving, etc.
2. A surfJEtce has length and breadth, but not thickness.
3. The area of a surface is its numerical value; or the
number of times it contains the measuring unit,
4. A superficial iinit is an assumed unit of measure Vor
surfaces.
Usually the square, whose side is the linear unit, is the
unit of measxire; as, the square inch, square foot, square
yard,
5. A Beetangle may be defined as
a surface bounded by four straight lines
forming four square corners; as either of
the figures A, B.
6. When the four sides are equal the
rectangle is called a square; as the
figure C.
7. The area of a rectangle is equal to its length mtdtiplied by
its breadth.
MEASUEES OF EXTENSION,
137
Explanation. — Take a rectangle 4 inches long
by 3 inches wide. If upon each of the inches in
the length, a square inch he conceived to stand,
there will be a row of 4 square inches, extending
the whole length of the rectangle, and reaching 1
inch of its width. Ap the rectangle contains as many such rows as
there are inches in its width, its area must be equal to the number
of square inches in a row (4) multiplied by the number of rows (3),
= 12 square inches. This statement (7), as commonly understood,
can present no exception to Prin. 2, Art. 60.
Table.
144 square inches (sq. in.) make 1 square foot, marked sq. ft.
9 sq. ft. ** 1 square yard, ** sq. yd.
30J sq. yd. " 1 square rod,
leOsq. rd. .r . ___.
" 1 acre,
((
<(
sq. rd.
A.
Equivalent Table.
A. sq. rd. sq. yd. sq. ft.
1 = 160 = 4840 = 43560
1 = 30i = 272i
1 = 9
1
sq. in.
6272640.
39204.
1296.
144.
Note. — ^The following, though now seldom used, are often found
in records of calculations :
40 perches (P.), or sq. rds., make 1 rood, marked E.
4 roods " 1 acre, " A.
198. Surveyors' Measure is a kind of superficial
measure, which is used chiefly in government surveys.
Table.
625 square links (sq. li.) make 1 square rod, sq. rd.
16 so. rd. ** 1 sauare chain, so. ch.
16 sq. rd.
10 sq. ch.
640 A.
36 sq. mi. (6 miles square)
H. A. 12.
((
((
((
1 square chain, sq. ch.
1 acre, A.
1 square mile, sq. mi.
1 township, , JX^^^
//
' or T^!£ \
UNIVERSITY I
MAY'S HIGHEB ABITHMETja
Eqdivalent Table.
Tp. Bq. mi. A. sq. ch. 8q. rd. sq. li.
1 = 36 = 23040 = 230400 ^ 368.6400 = 2304000000.
1 = 640 ^ 6400 = 102400 = 64000000.
1 = 10 = 160 =. 100000.
1 = 16 = 10000.
1 ^ 625.
Solid Measure.
IBS. A Solid has length, breadth, and thickness.
Solid Meaaure \a used in estimating the Ciontenia or vdwme
of solids.
A Cube is a solid, bounded by six equal squares, called
facei. Its length, breadth, and thickness are all equal.
Remark. — The size or name of anj cube, like that oE a square,
depends upon i(s side, as cubic iuch, cubic foot, cubic yard.
ExPLANATios. — It each side of a cube
is 1 inch iong, it ia calied a cubic inch; it
each side is 3 feet (I yard) long, aa repre
sented ID the figure, it ia a cubic or solid
Wiien the base o£ a cube is 1 square yard,
it contains3X3=9 square feet; and I foot
high on this base, contains 9 solid feet; 2
teet high contains 9 X 2 = 18 Boiid feet ; 3 teel high
= 27 solid tceL Also it may be shown that 1 solid
contains 12 X 12 X 12 = 1728 solid or eMe inches.
The unit by which all solids are measured is a culje,
whose side is a linear inch, foot, etc, and their size or
solidity will be the number of times they contain this unit.
Bemark.— The simplest solid ia the lecbmyuiir solid, which is
bounded by sin rectangles, called its far.ee, each opposite pair being
equal, and perpendicular lo the other tour; as, for example, the
MEASURES OF EXTENSION, 139
ordinary form of a brick or a box of soap. If the length, breadth,
and thickness are. the same, the faces are squares, and the solid is
a cabe.
Table.
1728 cubic inches (cu. in.) make 1 cubic foot, cu. ft.
27 cu. fk. ** 1 cubic yard, cu. yd.
Equivalent Table.
cu. yd. cu. ft. cu, in.
1 = 27 = 46656.
1 = 1728.
Hemakes. — 1. A perch of stone is a mass 16\ ft. long, IJ ft. wide,
and 1 ft. high, and contains 24 cu. ft.
2. Earth, rockexcavations, and embankments are estimated by
the cubic yard.
3. Bound timber will lose ^ in being sawed, hence 50 cubic ft. of
round timber is said to be equal to 40 cubic ft. of hewn timber,
which is a ton,
4. Firewood is usually measured by the cord. A pile of wood 4
ft. high, 4 ft. wide, and 8 ft. long, contains 128 cubic feet or one cord.
One foot in length of this pile, or 16 cu. ft., is called a cord foot,
5. Planks and scantling are estimated by board measure. In this
measure, 1 reduced footy 1 ft. long, 1 ft. wide, and 1 in. thick, contains
12 X 12 X 1 = 144 cu. in. All planks and scantling less than an
inch thick, are reckoned at that thickness ; but, if more than an
inch thick, allowance must be made for the excess.
Measures of Capacity.
200. Capacity means room for things.
Measures of Capacity are divided into Measures of
Liquids and Measures of Dry Substances,
201. Liquid Measure is used in measuring liquids, and
in estimating the capacities of cisterns, reservoirs, etc.
The gallon, which contains 231 cu. in., is the unit of
measure in liquids.
140 BA Y'S HIGHER ARITHMETIC.
Note. — This gallon of 231 cubic inches was the standard in
England at the time of Queen Anne. The present imperial gallon
of England contains 10 lb. of water at 62° Fahr., or 277.274 cubic
inches.
Table.
4 gills, marked gi., make 1 pint, marked pt.
2 pt. "1 quart, ** qt.
4 qt. " 1 gallon, ** gal.
Equivalent Table.
gal. qt. pt. gi.
1 = 4 = 8 = 32.
1=2= 8.
1=4.
Note. — Sometimes the barrel is estimated at SIJ gal., and the
hogshead at 63 gal.; but usually each package of this description
is gauged separately.
202. Apothecaries' Fluid Measure is used for meas
uring all liquids that enter into the composition of medical
prescriptions.
Table.
60 minims, marked nt, make 1 fluid drachm, marked f^,
8 fs ** 1 fluid ounce, " f§.
16 ^ * 1 pint, " O.
8 ** 1 gallon, " cong.
Equivalent Table.
cong. O. f§. f^. Vl,
1 r= 8 ==r 128 = 1024 = 61440.
1 = 16 = 128 = 7680.
1 = 8 = 480.
1 = 60.
MEASURES OF EXTENSION. 141
Notes. — 1. Cong, is an abbreviation for eongioLriumy the Latin for
gallon ; O. is the initial of octam, the Latin for one eighth^ the pint
being one eighth of a gallon.
2. For ordinary purposes, 1 teacup = 2 wineglasses = 8 table
spoons = 32 teaspoons = 4 f§.
203. Dry Measure b used for measuring grain, fruit,
vegetables, coal, salt, etc.
The Winchester bushel is the unit; it was formerly used
in England, and so called from the town where the standard
was kept. It is 8 in. deep, and 18^ in. in diameter,
and contains 2150.42 cu. in., or 77.6274 lb. av. of distilled
water at maximum density, the barometer at 30 inches.
Note. — ^This bushel was discarded by Great Britain in 1826, and
the imperial bushel substituted; the latter contains 2218.192 cu. in.,
or eighty pounds avoirdupois of distilled water.
Table.
2 pints, marked pt, make 1 quart, marked qt.
8 qt. ** 1 peck, " pk.
4 pk. "1 bushel, ** bu.
Equivalent Table.
bu. pk. qt. pt
1 = 4 = 32 = 64.
1 = 8 = 16.
1=2.
Kemabks. — 1. 4qt. or J peck = 1 dry gal. = 268 8 cu. in. nearly.
2. The qaarter is still used in England for measuring wheats of
which it holds eight bushels, or 480 pounds avoirdupois.
3. When articles usually measured by the above table are sold
by weight, the bushel is taken as the unit. The following table gives
the legal weight of a bushel of various articles in avoirdupois
pounds :
142
BAY'S HIGHER AMITHMETia
Table.
ARTICLES.
LB.
60
80
56
56
32
60
56
50
60
EXCEPTIONS.
Beans.
Coal.
Corn (Indian).
Flax Seed.
Oats.
Potatoes (Irish).
Rye.
Salt.
Wheat.
Me., 64 ; N. Y., 62.
rOhio, 70 of cannel ; Ind., 70 mined out of
( the state; Ky., 76 of, anthracite.
N. Y., 58 ; Cal., 52 ; Arizona, 54.
N. Y. and N. J., 55 ; Kan., 54.
r Md., 26 ; Me., N. H., N. J., Pa., 30 ; Neb., 34 ;
\ Montana, 35; Oregon and Wash., 36.
Ohio, 58 ; Wash., 50.
La., 32; Cal., 54.
Mass., 70 ; Pa., coarse, 85 ; ground, 70 ; fine^
62 ; Ky. and 111., fine, 55 ; Mich., 56 ;
Col. and Dak., 80.
Comparative Table of Measures.
cu. in. gal. cu. in. qt. cu. in. pt.
Liquid Measure, 231 57f 28J
Dry Measure (J pk.), 268 67^ 33f
cu. in.gi.
Angular or Circular Measure.
201. A plane angle is the difference of direction of two
straight lines which meet at a point.
Explanation. — Thus, the two lines AB and AC
meet at the point A, called the apex. The lines AB
and AC are the sides of the angle, and the difference
in direction, or the opening of the lines, is the angU
itself,
. Angular Measxire is used to measure angles, directions,
latitude, and longitude, in navigation, astronomy, etc.
A circle is a plane surface bounded by a line, all the
points of which are equally distant from a point within.
■
MEASURE OF TIME. 143
ExPLANATtONS. — The bounding lino I>
ADBEA is a evrcumfertnce. Every point y^
of this line is at the same distance from /
the point C, which is called the center, I
The cvrde is the area included within the a I _
circumference. Any straight line drawn \ /
from the center to the circumference is \ /
called a radius; thus, CD and CB are ^^^^^ ^^y^
radii. Any part of the circumference, as E
AEB or AD, is an are. A straight line,
like AB, drawn through the center, and having its ends in the cir
cumference, is a diameter; it divides the circle into two equal parts.
Notes. — 1. Every circumference contains 360 d^^rees ; and, the
apex of an angle being taken as the center of a circle, the angle is
measured by the number of degrees in the arc included by the sides
of the angle.
2. The angle formed by two lines perpendicular to each other, as
the radii AC and DC in the above figure, is a right angU, and is
measured by the fourth part of a circumference, 90°, called a quadrant.
Table.
60 seconds, marked ", make 1 minute, marked. '.
6(K " 1 degree, *' ®.
360° " 1 circumference, '* c
Equivalent Table.
1 == 360 = 21600 = 1296000.
1 = 60 = 3600.
1 = 60.
Note. — The twelfth part of a circumference, or 30°, is called a
siffn.
• MEASURE OF TIME.
205. 1. Time is a measured portion of duration.
2. A Year is the time of the revolution of the earth
144
RAY'S HIGHER ARITHMETIC,
around the sun ; a Day is the time of the revolution of the
earth on its axis.
3. The Solar Day is the interval of time between two
successive passages of the sun over the same meridian.
4. The Mean Solar Day is the mean, or average, length
of all the solar days in the year. Its duration is 24 hours,
and it is the unit of Time Measure.
5. The Civil Day, used for ordinary purposes, commences
at midnight and closes at the next midnight.
6. The Astronomical Day commences at noon and closes
at the next noon.
Table.
60 seconds, marked sec., make 1 minute, marked rain.
1 hour, ** hr.
1 day, ** da.
1 week, ** wk.
1 month, " men.
1 year, ** yr.
1 common year.
1 leap year.
1 century, marked cen.
60 min.
24 hr.
7 da.
4 wk.
12 calendar mon.
365 da.
366 da.
100 yr.
Note.— 1 Solar year = 365 da. 5 hr. 48 min. 46.05 sec. = 365 J da.,
nearly.
Equivalent Table.
vr. mo.
wk.
I = 12 = 52 =
1 =
da.
hr. min. Rec.
(365 —
1366 —
8760 — 525600 — 31536000
8784 — 527040 — 31622400
7 —
168 — 10060 — 604800
1 —
24 — 1440 — 86400
1 — 60  3600,
1 — 60,
MEASUBE OF TIME. 145
Note. — The ancients were unable to find accurately the number
of days in a year. They had 10, afterward 12, calendar months,
corresponding to the revolutions of the moon around the earth. In
the time of Julius Caesar the year contained 365J days ; instead of
taking account of the J of a day every year, the common or civil
year was reckoned 365 da^s, and every 4th year a day was inserted
(called the intercalary day), making the year then have 366 days.
The extra day was introduced by repeating the 24th of February,
which, with the Komans, was called the sixth day before the kalends of
March. The years containing this day twice, were on this account
called bissextile, which means having two sixths. By us they are gen
erally called leap years.
But 3651^ days (365 days and 6 hours) are a little longer than the
true year, which is 365 days 5 hours 48 minutes 46.05 seconds. The
difference, 11 minutes 13.95 seconds, though small, produced, in a
long course of years, a sensible error, which was corrected by
Gregory XIII., who, in 1582, suppressed the 10 days that had been
gained, by decreeing that the 5th of October should be the 15th.
206. 'To prevent difficulty in future, it has been decided
to adopt the following rule.
Hule for Leap Years. — Every year that is divisible by 4
is a leap year, unless it ends with two ciphers; in which case
it must be divisible by 400 to be a leap year.
Illustration.— Thus, 1832, 1648, 1600, and 2000 are leap years ;
but 1857, 1700, 1800, 1918, are not.
Notes. — 1. The Gregorian calendar was adopted in England in 1752.
The error then being 11 days. Parliament declared the 3d of September
to be the 14th, and at the same time made the year begih January
1st, instead of March 25th. Kussia, and all other countries of the
Greek Church, still use the Julian calendar; consequently their
dates (Old Style) are now 12 days later tlian ours (New Style), The
error in the Gregorian calendar is small, amounting to a day in
3600 years.
2. The year formerly began with March instead of January ; con
sequently, September, October, November, and December were the 7th,
8th, 9th, and 10th months, as their names indicate; being derived
from the Latin numerals Septem (7), Octo (8), Novem (9), Decern (10).
H. A. 13.
146 RAY'S HIGHER ARITHMETIC,
COMPARISON OF TIME AND LONGITUDE.
207. The longitude of a place is its distance in degrees,
minutes, and seconds, east or west of an established meridian.
•
Note. — The difference of longitude of two places on the same side of
the e!<tablislied meridian, is found by subtracting the Igj^s longitude
from the greater ; but, of two places on opposite sides of the meridian,
the difference of longitude is found by adding the longitude of one to
the longitude of the other.
The circumference of the earth, like other circles, is
divided into 360 equal parts, called degrees of longitude.
The sun appears to pass entirely round the earth, 360°,
once in 24 hours, one day; and in 1 hour it passes over
15°. (360° f 24 = 15°.)
As 15° equal 900', and 1 hour equals 60 minutes of
timey therefore, the sun in 1 minute of time passes over 15'
of a degree. (900' ^ 60 = 15'.)
As 15' equal 900", and 1 minute of time equals 60 seconds
of tiine, therefore, in 1 second of time the sum passes over
15" of a degree. (900" ^ 60 = 15".)
Table for Comparing Longitude and Time.
15° of longitude ^= 1 hour of time.
15' of longitude = 1 min. of time.
15" of longitude = 1 sec. of time.
Note. — If one place has greater east or less west longitude than
another, its time must be later; and, conversely, if one place has
later time than another, it must have greater east or less west
longitude.
MISCELLANEOUS TABLES.
208. The words folio, quarto, octavo, etc., used in speak
ing of books, show how many leaves a sheet of paper makes.
THE METRIC SYSTEM.
147
A sheet folded into
2 leaves,
called a folio, ]
make
s 4
4 "
" a quarto or 4to,
8
8 "
** an octavo or 8vo,
16
12 "
** a duodecimo or 12mo,
24
16 
" a 16mo,
32
32 ''
** a 32mo,
64
Also, . 24 sheets of paper make 1 quire.
20 quires ** 1 ream.
2 reams " 1 bundle.
5 bundles ** 1 bale.
pages.
12 things
make 1 dozen.
12 dozens or 144
things
1 gross.
12 grass or
144 dozens
1 great gross.
20 things
1 score.
56 lb.
1 firkin of butter.
100 lb.
1 quintal of fish.
196 lb.
1 bbl. of flour.
200 lb.
1 bbl. of pork.
THE METRIC SYSTEM.
Historical.
The Metrio System is an outgrowth of the French Revolution of
1789. At that time there was a general disposition to break away
from old customs ; and the revolutionists contended that every thing
needed remodeling. A commission was appointed to determine
an invariable standard for all measures of length, area^ solidity,
capacity, and weight. After due deliberation, an accurate survey
was made of that portion of the terrestrial meridian through
Paris, between Dunkirk, France, and Barcelona, Spain ; and from
this, the distance on that meridian from the equator to the pole
was computed. The quadrant thus obtained was divided into ten
million equal parts ; one part was called a metefi\ and is the hose
of the system. From it all measures are derived.
148
BAY'S HIGHER ARITHMETIC.
In 1795 the Metric System was adopted in France. It is now
used in nearly all civilized countries. It was author^
ized by an act of Congress in the United States in 1866.
209. The Metric System is a decimal system
of weights and measures.
The meter is the primary unit upon which
the system is based, and is also the imit of
length. It is 39.37043 inches long,* which is
very nearly one tenmillionth part of the distance
on the earth's surface from the equator to the
pole, as measured on the meridian through Paris.
o •
0)
00
CO
•
CO
4
tf
^
O
Kemabk. — The stoTicfcird meter, a bar of platinum, is
kept among the archives in Paris. Duplicates of this
bar have been furnished to the United States.
210. The names of the lAmer denominations
in each measure of the Metric System are
formed by prefixing the Laiin numerals, dec*
(.1), ceaii (.01), and imJUi (.001) to the unit of
that measure ; those of the higJier denominations,
by prefixing the Greek numerals, deka (10),
hddo (100), kUo (1000), and mipna (10000), to
the same unit.
These prefixes may be grouped about the unit
of measure, showing the decimal arrangement
of the system, as follows:
rmiUi = .001
Lower Denominations. \ centi = .01
I deci = .1
Unit of Measure =
deka =
hekto =
kilo =
Higher Denominations.
1.
10.
100.
1000.
myria= 10000.
THE METRIC SYSTEM. 149
211. The units of the various measures, to which these
prefixes are attached, are as follows:
. The Meter, which is the unit of Length.
The Ar, " ** " " " Surface.
The Liter, ** '' '' " " Capacity.
The Gram, '* '* *• '* " Weight.
Bemabk. — The name of each denomination thus derived, imme
diately shows its relation to the unit of measure. Thus, a cenii
tneter is one onehundredth of a meter; a kilogram is a thousand
grams ; a hehtdiier is one hundred liters, etc.
Measure of Length.
212. The Meter is the unit of Length, and is the de
nomination used in all ordinary measurements.
Table.
10 millimeters, marked mm. , make 1 centimeter, marked cm.
10 centimeters
10 decimeters
10 meters
10 dekameters
10 hektometers
10 kilometers
Hemareb. — 1. The figure on page 148 shows the exact length
of the decimeter, and its subdivisions the centimeter and millimeter.
2. The centimeter and millimeter are most often used in meas
uring very short distances ; and the kilometer, in measuring roads
and long distances.
Measure of Surface.
213. The Ar (po. ar) is the unit of Land Measure ; it
is a square, each side of which is 10 meters (1 dekameter)
in length, and hence its area is one square dekameter.
1 decimeter.
" dm.
1 meter.
" m.
~1 dekameter.
*' Dm.
1 hektometer.
" Hm.
1 kilometer.
" Km.
1 myriameter.
" Mm.
150 EAY'S HIGHER ARITHMETIC,
Table.
100 centars, marked ca., make 1 ar, marked a.
100 ars *' 1 hektar, *' Ha.
Kemare. — The iquare meter (marked m^) and its subdivisions are
used for measuring small surfaces.
Measure of Capacity.
214. The Liter (p^o. le'ter) is the unit of Capacity. It
is equal in volume to a cube whose edge is a decimeter ;
that is, one tenth of a meter.
Table.
10 milliliters, marked ml., make 1 centiliter, marked cl.
10 centiliters " 1 deciliter, ** dl.
10 deciliters " 1 liter, '* 1.
10 liters " 1 dekaliter, '' Dl.
10 dekaliters '* 1 hektoliter, '' HI.
Bemarks. — 1. This measure is used for liquids and for dry sub
stances. The denominations most used are the liter and hektoliter ;
the former in measuring milk, wine, etc., in moderate quantities, and
the latter in measuring grain, fruit, etc., in large quantities,
2. Instead of the milliliter and the kiloliter, it is customary to use
the cubic centimeter and the cubic meter (marked m^), which ar4
their equivalents.
3. For measuring wood the ster (pro. stdr) is used. It is a cubic
meter in volume.
Measure of Weight.
215. The Gram {pro, gram) is the imit of Weight. It
was determined by the weight of a cubic centimeter of dis
tilled water, at the temperature of melting ice.
THE METRIC SYSTEM. 151
Table.
10 milligrams, marked mg., make 1 centigram, marked eg.
10 centigrams
* 1 decigram.
dg.
10 decigrams
* 1 gram.
g
10 grams
* 1 dekagram.
Dg
10 dekagrams
* 1 hektogram,
Hg.
10 hektograms
' 1 kilogram.
Kg.
10 kilograms
* 1 myriagram.
Mg.
10 myriagrams,
or 100
kilograms *
* 1 quintal.
. ^
10 quintals, or
1000
(( (
* 1 metric ton,
M.T.
Bemarks. — 1. The gram, kilogram {pro, klKogram), and metric
ton are the weights commonly used.
2. The gram is used in all cases where great exactness is required ;
such as, mixing medicines, weighing the precious metals, jewels,
letters, etc.
3. The kilogram, or, as it is commonly abbreviated, the "kilo,"
is used in weighing coarse articles, such as groceries, etc.
4. The metric ton is used in weighing hay and heavy articles
generally.
216. Since, in the Metric System, 10, 100, 1000, etc.,
units of a lower denomination make a unit of the higher
denomination, the following principles are derived:
Principles. — 1. A number u redvjced to a lower denom
inaiion by removing ihe decimal point as many places to the
RIGHT as there are ciphers in the muUiplier,
2. A number is reduced to a higher denomination by
remxmng the decimal point as many places to the left as
there are ciphers in the divisor.
Illustrations. — Thus, 15.03 m. is read 15 meters and 3 centi
meters ; or, 15 and 3 hundredths meters. Again, 15.03 meters = 1.503
dekameters = .1503 hektometer = 150.3 decimeters = 1503 centi
meters. As will be seen, the reduction is effected by changing the
decimal point in precisely the same manner as in United Skites Money.
152
BA rS HIGHER ARITHMETIC.
217. The following table presents the legal and approx
imate values of those denominations of the Metric System
which are in common use.
Table.
DENOMINATION.
LEGAL VALUE.
1
APPBOX. VALUE.
Meter.
39.37 inches.
3 ft 3 inches.
Centimeter.
.3937 inch.
J inch.
Millimeter.
.03937 inch.
■it inc^
Kilometer.
.62137 mile.
1 mile.
Ar.
119.6 sq. yards.
4 sq. rods.
Hektar.
2.471 acres.
2i acres.
Square Meter.
1.196 sq. yards.
10 sq. feet.
Liter.
1.0567 quarts.
1 quart
Hektoliter.
2.8375 bushels.
2 bu. 3} pecks.
Cubic Centimeter.
.061 cu. inch.
■jJj cu. inch.
Cubic Meter.
1.308 cu. yards.
35J cu. feet.
Ster.
.2759 cord.
\ cord.
Gram.
15.432 grains troy.
lb\ grains.
Kilogram.
2.2046 pounds av.
2J pounds.
Metric Ton.
2204.6 pounds av.
1 T. 204 pounds.
Topical Outline.
Compound Numbers.
1. Preliminary Definitions.
2. Value.
8. Weight.
ri. Definitions.
2. United States and Canadian Money.
 3. English Money.
4. French Money.
5. German Money.
1. Troy Weight
2. Apothecaries' Weight
.3. Avoirdupois Weight
TOPICAL OUTLINE.
153
Compound Numbers. — {Concluded,)
1. Linear... .
4. Extension. .
1. Long Measure.
2. Chain Measure.
3. Mariners' Measure.
.4. Cloth Measure.
„ . , „ f 1. Surface Measure.
2. Superficial Measure.... <
3. Solid Measure.
4. Measures of Capacity. / 1 Liqui<i
2. Surveyors' Measure.
5. Angular Measure.
5. Time.
6. Comparison of Time and Longitude.
7. Miscellaneous Tables.
12.
Dry.
8. Metric System.
1. Historical.
2. Terms..
1. How Derived.
2. Lower Denominations...
3. Higher Denominations..
1. Meter.
2. Ar.
3. Liter.
L 4 Gram.
1. Ivcngth.
2. Surface.
3. Capacity.
I 4. Weight.
5. Principles.
6. Table of Legal and Approximate Values.
3. Units...
4. Measures.
154 BAY'S HIGHER ARITHMETIC,
REDUCTION OF COMPOUND NUMBEES.
218. Reduction of Compound Numbers is the process
of changing them to equivalent numbers of a different de
nomination.
Reduction takes place in two ways:
From a higlier denomination to a lower,
From a lower denomination to a higher.
Principles. — 1. Reduction from a higher denomination to
a lower, is performed by midtiplication.
2. Reduction from a lower denomination to a higher,, is per
formed by division.
Problem.— Reduce 18 bushels to pints.
OPERATION.
Solution. — Since 1 bu. = 4 pk., 18 bu. i g \y^
= 18 times 4 pk. =: 72 pk., and since 1 pk. 4
=r 8 qt., 72 pk. = 72 times 8 qt. = 576 qt.; y^ 1^
and since 1 qt. = 2 pt., 576 qt. =: 576 times o
2 pt. = 1152 pt. Or, since 1 bu. = 64 pt., ^ . « .
multiply 64 pt. by 18, which gives 1152 pt. o
as before. This is sometimes called Beduo
lUm De^cendiny. 1 8 bu. = 1 1 5 2 Jt
Problem. — Reduce 236 inches to yards.
Solution. — Since 12 inches = 1 ft., operation.
236 inches will be as many feet as 12 in. 12)236 in.
is contained times in 286 in., which is 3)192 ft.
19 ft., and since 3 ft. = 1 yd., l^ ft will ^ ^
be as many yd. as 3 ft. is contained times 2 3 6 in =64 vd
in 19 ft., which is 6 yd. Or, since 1
yd. =r 36 in., divide 236 in. by 36 in., which gives 6f yd., as before.
This is sometimes called Beductlon Ascending.
Note.— In the last example, instead of dividing 236 in. by 36 in.
the unit of value of yards, since 1 inch is equal to ^^ yards, 236
inches = 236 X iV = W yd. = ^ yd. The operation by division
is generally more convenient.
REDUCTION OF COMPOUND NUMBERS. 155
Remark. — Reduction Descending diminishes the size^ and, there
fore, increases the number of units given ; while Reduction Ascend
ing increases the ^ize, and, therefore, diminishes the number of units
given. This is further evident from the fact, that the multipliers
in Reduction Descending are larger than 1 ; but in Reduction As
cending smaller than 1.
Problem. — Reduce f gallons to pints.
Solution.— Multiply by 4 to operation.
reduce gal. to qt. ; then by 2 to _3 X ^ X ^ = 3 pt.
reduce qt. to pt. Indicate the g
operation, and cancel. . gal. = 3 pt.
Problem. — ^Reduce 5f gr. to S.
OPERATION.
Solution. — Al ?
though this is Ee ^ 1^ y ivi^ J 5
duction Ascending, ^7&^' y 2038 84
we use Prin. 1, in . ^
multiplying by the successive
unit values, ^V> h and }. ^j g^'=TiO'
Problem. — Reduce 9.375 acres to square rods.
operation.
9.3 7 5
160
562500
9375
1500.00 sq. rd.
9.3 7 5 A. = 1 5 sq. rd.
Problem. — Reduce 2000 seconds to hours.
OPERATION.
2 sec. = I hr.
156 JiA y » S HIGHER ARITHMETIC.
Problem. — Reduce 1238.73 hektograms to grams.
OPERATION.
123 8.7 3X100 = 123873 grams.
Problem. — How many yards in 880 meters?
OPERATION.
3 9.3 7 in. X 8 8
12X3
= 9 6 2.3 7 7 + yd.
Bemark. — Abstract factors can not produce a concrete result;
sometimes^ however, in the steps of an indicated solution, where the
change of denomination is very obvious, the abbreviations m&y be
omitted until the result is written.
From the preceding exercises, the following rules are
derived :
219. For reducing from higher to lower denomina
tions.
Bule. — 1. Multiply {he highest denomination given, by that
number of (he next lower which makes a unit of the higher,
2. Add to the prodiict the number, if any, of the Uywer
denomination,
3. Proceed in like manner vdth the remit thus obtained,
tUl the whole is reduced fo the required denomination,
220. For reducing from lower to higher denomina
tions.
Bule. — 1. Divide the given quantity by that number of its
own denomination which makes a unit of the next higher,
2. Proceed in like manner with the quotient thus obtained,
till (he wJwle is reduced to (he required denomination,
3. The last quotient, with the several remainders, if any,
annexed, will be the answer.
Note. — In the Metric System the operations are performed by
removing the point to tlie right or to the left
RED UCTION OP COMPO UND NUMBERS. 1 57
Examples for Practice.
1. How many square rods in a rectangular field 18.22
chains long by 4.76 eh. wide? 1387.6352 sq. rd.
2. Beduce 16.02 chains to miles. .20025 mi.
3. How madly bushels of wheat would it take to. fill 750
hektoliters? 2128^ bu.
4. Reduce 35.781 sq. yd. to sq. in. 46372.176 sq. m.
5. Eeduce 10240 sq. rd. to sq. ch. 640 sq. ch.
6. How many perches of masonry in a rectangular
solid wall 40 ft. long by 7^ ft. high, and 2 ft. average
thickness ? 82f P.
7. How many ounces troy in the Brazilian Emperor's
diamond, which weighs 1680 carats? 11.088 oz.
8. Reduce 75 pwt. to 3. 30 3.
9. Reduce f gr. to §. ^hs l
10. Reduce 18f 3 to oz. av. 2^ oz.
11. Reduce 96 oz. av. to oz. troy. 87^ oz. troy.
12. How many gal. in a tank 3 ft. long by 2\ ft. wide
and 1 ft. deep ? 75^ gal.
13. How many bushels in a bin 9.3 ft. long by 3 ft.
wide and 2\ ft. deep? 61 bu., nearly.
14. How many sters in 75 cords of wood? 271.837+ s.
15. Reduce 2\ years to seconds. 70956000 sec.
16. Fortynine hours is what part of a week? ^f wk.
17. Reduce 90.12 kUoliters to liters. 90120 1.
18. Reduce 25" to the decimal of a degree. .00694°
19. Reduce 192 sq. in. to sq. yd. ^j sq. yd.
20. Reduce 6 cu. yd. to cu. in. 311040 cu. in.
21. Reduce $117.14 to mills. 117140 mills.
22. Reduce 6.19 cents to dollars. $.0619
23. Reduce 1600 mills to dollars. $1.60
24. Reduce $5f to mills. 5375 mills.
25. Reduce 12 lb. av. to lb. troy. 14^^ lb.
26. How many grams in 6.45 quintals? 645000 g.
27. Reduce .216 gr. to oz. troy. .00045 oz. troy.
158 RAY'S HIGHER ARITHMETIC.
28. Reduce 47.3084 sq. mi. to sq. rd. 4844380.16 sq. rd.
29. Reduce 4j 9 to lb. ^^ lb.
30. Reduce 7^ oz. av. to cwt. yjg^ cwt.
31. Reduce 99 yd. to miles. yf^ mi.
32. How many acres in a rectangle 24^^ rd. long by
16.02 rd. wide? 2.4530625 a^res.
33. How many cubic yards in a box 6J ft. long by 2^ ft.
wide and 3 ft. high? 1^ cu. yd.
34. Reduce 169 ars to square meters. 16900 m^.
35. Reduce 2^ f^ to n^. 1200 n^.
36. If a piece of gold is ^ pure, how many carats fine
is it ? 204 carats.
37. In 18 carat gold, what part is pure and what part
alloy? ff pure, and ^ alloy.
38. How many square meters of matting are required to
cover a floor, the dimensions of which are 6 m., \\ dm.
by 5 m., 3 cm.? 30.9345 m^.
39. How many cords of wood in a pile 120 ft. long,
6 J ft. wide, and 8f ft. high ? 53^ C.
40. How many sq. ft. in the four sides of a room 21^ ft.
long, 16^ ft. wide, and 13 ft. high ? 988 sq. ft.
41. What will be the cost of 27 T. 18 cwt. 3 qr., 15 lb.
12 oz. of potash, at $48.20 a ton? $1346.97—.
42. What is the value of a pile of wood 16 m., 1 dm.,
5 cm. long, 1 m., 2 dm., 2 cm. wide, and 1 m., 6 dm.,
8 cm. high, at $2.30 a ster? $76.13+
43. What is the cost of a field 173 rods long and 84 rods
wide, at $25.60 an acre? $2325.12
44. If an open court contain 160 sq. rd. 85 sq. in.;
how many stones, each 5 inches square, will be required to
pave it ? 250909 stones.
45. A lady had a grassplot 20 meters long and 15
meters wide; after reserving two plots, one 2 meters
square and the other 3 meters square, she paid 51 cents
a square meter to have it paved with stones : what did the
paving cost? $146.37
BED UCTION OF COMPO UND NUMBERS. 1 59
46. A cubic yard of lead weighs 19,128 lb.: what is the
weight of a block 5 ft. 3^ in. long, 3 ft. 2 in. wide, and
1 ft. 8 in. thick? 9 T. 17 cwt. 7 lb. 11.37 oz.
47. A lady bought a dozen silver spoons, weighing 3 oz.
4 pwt. 9 gr., at $2.20 an oz., and a gold chain weighing
13 pwt., at ?1J a pwt.: required the total cost of the spoons
and chain. $23,331
48. A wagonbed is 10 ft. long, 3^ ft. wide, and 1 ft.
deep, inside measure : how many bushels of corn will it hold,
deducting one half for cobs? 22 bu. 4 qt. 1.5 — pt.
49. K a man weigh 160 lb. avoirdupois, what will he
weigh by troy weight ? 194 lb. 5 oz. 6 pwt. 16 gr.
50. The forewheel of a wagon is 13 ft. 6 in. in circum
ference, and the hind wheel 18 ft. 4 in.: how many more
revolutions will the forewheel make than the hind one in
50 miles? 5155.55+ revolutions.
51. An apothecary bought 5 ft. 10 §. of quinine, at
$2.20 an ounce, and sold it in doses of 9 gr., at 10 cents
a dose: how much did he gain? $219.33^
52. How many steps must a man take in walking from
Kansas City to St. Louis, if the distance be 275 miles, and
each step, 2 ft. 9 in.? 528000 steps.
53. The area of Missouri is 65350 sq. mi.: how many
hektars does it contain? 16925940.91+ Ha.
54. A schoolroom is 36 ft. long, 24 ft. wide, and 14
ft. high; required the number of gallons of air it will
contain? 90484.36+ gal.
55. Allowing 8 shingles to the square foot, how many
shingles will be required to cover the roof of a barn
which is 60 feet long, and 15 feet fiom the comb to the
eaves? 14400 shingles.
56. A boy goes to bed 30 minutes later, and gets up
40 minutes earlier than his roommate : how much time
does he gain over his roommate for work and study
in the two years 1884 and 1885, deducting Sundays
only? 731 hours, 30 min.
160 HAY'S HIGHER ARITHMETIC.
ADDITION OF COMPOUND NUMBEKS.
221. Compoand Numbers may be added, subtracted,
multiplied, and divided. The priuciples ujx)n which these
operations are performed are the same as in Simple Num
bers, with this variation ; namely, that in Simple Numbers
ten units of a lower denomination make one of the next
higher, while in Compound Numbers the scales vary.
Addition of Compound Numbers is the process of find
ing the sum of two or more similar Compound Numbers.
Problem. — Add 3 bu. 2\ pk.; 1 pk. 1^ pt.; 5 qt. 1 pt;
2 bu. 1^ qt; and .125 pt.
Solution. — Eeduce the frac
tion in each number to lower
denominations, and write units
of the same kind in the same
column. The righthand column,
when added, gives 3j\ pt. = 1 qt.
Wt P^m write the 1/j and add
the 1 qt. with the next column, 6 1 1^= AtiB,
making 9 qt. = 1 pk. 1 qt.; write
the 1 qt. and carry the 1 pk. to the next column, making 4 pk. = 1 bu.;
as there are no pk. left, write down a cipher and carry 1 bu. to the
next column, making 6 bu.
Problem.— Add 2 rd. 9 ft. 1\ in.; 13 ft. 5.78 in.; 4 rd.
11 ft. 6 in.; 1 rd. lOf ft.; 6 rd. 14 ft. 6 in.
Solution. — The numbers are prepared,
written, and added, as in the last ex
ample; the answer is 16 rd. 9^ ft. 9.655
in. The \ foot is then reduced to 6
inches, and added to the 9.655 in., making
15.655 in. = l ft. 3.655 in. Write the
3.655 in., and carry the 1 ft., which gives
16 rd. 10 ft. 3.655 in. for the final answer.
OPERATION.
bu.
pk.
qt.
pt.
3
2
2
:
 3 bu.
2ipk.
1
u=
— 1 pk. \\ pt.
5
1 :
— 5 qt.
1 pt.
2
1
i
2bu.
Hqt.
\'
.125
pt
U^ III,
OPERATION.
rd. ft.
in.
2 9
7.25
13
5.78
4 11
6
1 10
8
6 14
6.625
16 91
9.655
but \ ft. =
3 6.
16 10
3.655
ADDITION OF COMPOUND NUMBERS. 161
Rule. — 1. Write the numbers to be added, placing units of
the same denomination in the same column.
2. Begin with Hie lowest denomination^ add the numbers,
and divide^ their sum by the number of units of this denom
ination whidi make a unit of the next higher,
3. Write the remainder under the column added, and carry
the quotierd to the next column.
4. Proceed in the same manner with all (he columns to Hie
last, under which write its entire sum,
Eemark. — The proof of each fundamental operation in Com
pound Numbers is the same as in Simple Numbers.
Examples for Practice.
1. Add f mi.; 146^ rd.; 10 mi. 14 rd. 7 fl. 6 in.; 209.6
rd.; 37 rd. 16 ft. 2^ in.; 1 mi. 12 ft. 8.726 in.
12 mi. 180 rd. 9 ft. 4.633 in.
2. Add 6.19 yd.; 2 yd. 2 ft. 9f in.; 1 ft. 4.54 in.; 10 yd.
2.376 ft.; f yd.; 1 ft.;  in. 21 yd. 2 ft. 3.517 in.
3. Add 3 yd. 2 qr. 3 na. 1^ in.; 1 qr. 2 na.; 6 yd. 1 na.
2.175 in.; 1.63 yd.;  qr.; f na. 12 yd. 1 na. 0.755 in.
4. If the volume of the earth is 1; Mercury, .06; Venus,
.957; Mars, .14; Jupiter, 1414.2; Saturn, 734.8; Uranus,
82; Neptune, 110.6; the Sun, 1407124; and the Moon,
.018, what is the volume of all? 1409467.775
5. James bought a balloon for 9 francs and 76 centimes,
a ball for 68 centimes, a hoop for one franc and 37 cen
times, and gave to the poor 2 francs and 65 centimes, and
had 3 francs and 4 centimes left. How much money did
he have at first? 17^ francs.
6. Add 15 sq. yd. 5 sq. ft. 87 sq. in.; 16^ sq. yd.; 10 sq.
yd. 7.22 sq. ft.; 4 sq. ft. 121.6 sq. in. ; ^ sq. yd.
43 sq. yd. 7 sq. ft. 37.78 sq. in.
7. Add 101 A. 98.35 sq. rd.; 66 A. 74^ sq. rd.; 20 A.; 12
A. 113 sq. rd.; 5 A. 13.33^ sq. rd. 205 A. 139.18^ sq. rd.
H. A. 14.
162 RAY'S HIGHER ARITHMETIC.
8. Add 23 cu yd. 14 cu. ft. 1216 cu. in.; 41 cu. yd. 6
cu. ft. 642.132 cu. in.; 9 cu. yd. 25.065 cu. ft.; ^ ca yd.
75 cu. yd. 4 cu. ft. 1279.252 cu. in.
9. Add I C; I cu. ft.; 1000 cu. in.
107 cu. ft. 1072 cu. in.
10. Add 2'lb. troy, 6f oz.; If lb.; 12.68 pwt; 11 oz. 13
pwt. 19^ gr.; I K). II oz.;  pwt.
5 lb. troy, 9 oz. 9 pwt. 2. 85 J gr.
11. Add 83 14.6 gr.; 4.18 g; l^z\ 23 29 18 gr.; 1§
12 gr.; 19. 1 lb. 2 g 4 3 1 9.
12. Add y\ T.; 9 cwt. 1 qr. 22 lb.; 3.06 qr.; 4 T. 8.764
cwt; 3 qr. 6 lb.; ^ cwt. 5 T. 6 cwt. 2 qr. 14^ lb.
13. Add .3 lb. av.; f oz. 5^ oz.
14. Add 6 gal. 3^ qt.; 2 gal. 1 qt. .83 pt. ; 1 gal. 2 qt.
^ pt.; I gal.; I qt.;  pt. 11 gal. 2 qt. .11 pt.
15. Add 4 gal. .75 pt.; 10 gal. 3 qt. IJ pt.; 8 gal.  pt.;
5.64 gal.; 2.3 qt.; 1.27 pt; ^ pt. 29 gal. 2 qt. .05f pt.
16. Add 1 bu. ^ pk.; ^^ bu.; 3 pk. 5 qt. 1^ pt.; 9 bu.
3.28 pk.; 7 qt. 1.16 pt.; ^ pk. 12 bu. 3 pk. .46^f pt.
17. Add I bu.;  pk.; ^ qt.;  pt. 2 pk. f pt.
18. Add 6 f3 2 f3 25 nt ; 2J f3; 7 fs 42 nt ; 1 f3 2 fs;
3f3 6f5 51 rn,. 14 f3 7 f^ 38 nt.
19. Add  wk.; ^ da.; ^ hr.; \ min. ; ^ sec.
4 da. 30 min. 30J sec.
20. Add 3.26 yr. (365 da. each); 118 da. 5 hr. 42 min.
37J sec; 63.4 da.; 7 hr.; 1 yr. 62 da. 19 hr. 24f min.;
Y^ da. 4 yr. 340 da. 1 hr. 14 min. 55^ sec.
21. Add 27° 14' 55.24"; 9° 18^"; 1° 15^; 116° 44'
23.8" 154° 14' 57.29"
22. Add ^\ \ct.\ \ m. 50 ct. 2f m.
23. Add 3 dollars 7 m.; 5 dollars 20 ct.; 100 dollars 2 ct.
6 m.; 19 dollars \ ct. $127 23 ct. 4 m.
24. Add £21 6s. 3d.; £5 17s.; £9.085; 16s. 7id.;
£^. £37 10s. 8.15d.
25. Add I A.; f sq. rd.; i sq. ft.
107 sq. rd. 12 sq. yd. 6 sq. ft. 34^ sq. in.
SUBTRACTION OF COMPOUND NUMBERS. 163
SUBTRACTION OF COMPOUND NUMBERS.
222. (Subtraction of Compound Numbers is the
process of finding the difierence between two similar Com
pound Numbers.
Problem. — From 9 yd. 1 ft. 6^ in. take 1 yd. 2.45 ft.
Solution. — Change the J in. to a decima], operation.
making the minuend 9 yd. 1 ft. 6.5 in.; reduce yd. ft. in.
.45 ft. to inches, making the subtrahend 1 yd. 2 9 1 6.5
ft 5.4 in. The first term of the difference is 1.1 1 2 5.4
in. To subtract the 2 ft., increase the minuend 7 2 1.1 Am,
term by 3 feet, and the next term of the subtra
hend by the equivalent, 1 yard. Taking 2 ft. from 4 ft. we have a
remainder 2 ft, and 2 yd. from 9 yd. leaves 7 yd., making the answer
7 yd. 2 ft 1.1 in.
Problem. — ^From 2 sq. rd. 1 sq. ft. take 1 sq. rd. 30 sq.
yd. 2 sq. ft.
Solution. — Write the numbers as before. operation.
If the 1 sq. ft. be increased by a whole sq. sq.rd. sq.yd. sq. ft.
yd. and the next higher part of the subtra 2 1
hend by the same amount, we shall have to 1 30 2
make an inconvenient reduction of the first Ans. IJ sq.ft.
remainder as itself a minuend. Hence, we
make the convenient addition of J sq. yd., and, subtracting 2 sq. ft.
from Z\ sq. ft, we have 1\ sq. ft. Then, giving the same increase to
the 30 sq. yd., we proceed as in the former case, increasing the upper
by one of the next higher, and, having no remainder higher than
feety the answer is, simply, \\ sq. ft = 1 sq. ft 36 sq. in.
Hule. — 1. Place the subtrahend under tlie minuend^ so that
numbers of the same denomination stand in the same column.
Begin at the lowest denomination y andy subtracting the parts
successively from right to lefty write the remainders beneath,
2. Jf any number in the subtrahend be greater than that
of the same denomination in the minuendy increase the upper
by a unit, or sudi other quantity of the next higher denmnina
164 BA Y'S JnOHkB ARITHMEIia
tion as vnU render the subtraction possible, and give an equal
increase to the next higher term of the subtrahend.
Remark. — The increase required at any part of the minuend is,
commonly, a unit of the next higher denomination. In a few
instances it will be convenient to use more, and, if less be required,
the tables will show what fraction is most convenient. Sometimes
it is an advantage to alter the form of one of the given quantities
before subtracting.
Examples for Practice.
1. Subtract  mi. from 144.86 rd. 16.86 rd.
2. Subtract 1.35. yd. from 4 yd. 2 qr. 1 na. If in.
3 yd. 1 qr. f in.
3. Subtract 2 sq. rd. 24 sq. yd. 91 sq. in. from 5 sq. rd.
16 sq. yd. 6 sq. ft. 2 sq. rd. 22 sq. yd. 8 sq. ft. 41 sq. in.
4. Subtract 384 A. 43.92 sq. rd. from 1.305 sq. mi.
450 A. 148.08 sq. rd.
5. Subtract 13 cu. yd. 25 cu. ft. 1204.9 cu. in. from 20 cu.
yd. 4 cu. ft. 1000 cu. in. 6 cu. yd. 5 cu. ft. 1523.1 cu. in.
6. Subtract 9.362 oz. troy from 1 lb. 15 pwt. 4 gr.
3 oz. 7 pwt. 22.24 gr.
7. Subtract $ 3 from ^ g. 3 3 1 9 15^ gr.
8. Subtract 56 T. 9 cwt. 1 qr. 23 lb. from 75.004 T.
18 T. 10 cwt. 2 qr. 10 lb.
9. Subtract ^ lb. troy from ^ lb. avoirdupois. 0.
10. Subtract 12 gal. 1 qt. 3 gills from 31 gal. 1 pt.
18 gal. 3 qt. 3 gi.
11. Subtract .0625 bu. from 3 pk. 5 qt. 1 pt.
3 pk. 3 qt. 1 pt.
12. Subtract 1 f§ 4 fs 38 nt from 4 f^ 2 fs.
2"f3 5f3 22 nt.
13. Subtract 275 da. 9 br. 12 min. 59 sec. from 2.4816 yr.
(allowing 365 J days to the year.)
1 yr. 265 da. 18 hr. 29 min. 21.16 sec.
MULTIPLICATION OF COMPOUND NUMBERS. 166
14. Find the difference of time between Sept. 22d, 1855,
and July 1st, 1856. 9 mon. 9 da.
15. Find the difference of time between December Slst,
1814, and April Ist, 1822. 7 yr. 3 mon.
16. Subtract 43° 18' 57.18" from a quadrant.
46° 41' 2.82"
17. Subtract 161° 34' 11.8" from 180°. 18° 25' 48.2"
18. Subtract ^ ct. from $^. 8 ct. 4J m.
19. Subtract 5 dollars 43 ct. 2\ m; from 12 dollars 6 ct.
8^ m. $6.635f
20. Subtract £9 18s. e^d. from £20. £10 Is. 5id.
21. From \ A. 10 sq. in. take 79 sq. rd. 30 sq. yd. 2 sq.
ft. 30 sq. in. 16 sq. in.
22. From 3 sq. rd. 1 sq. ft. 1 sq. in. take 1 sq. rd. 30 sq.
yd. 1 sq. ft. 140 sq. in. 1 sq. rd. 1 sq. ft. 41 sq. in.
23. From 3 rd. 2 in. take 2 rd. 5 yd. 1 ft. 4 in. 4 in.
24. From 7 mi. 1 in. take 4 mi. 319 rd. 16 ft. 3 in.
2 mi. 4 in.
25. From 13 A. 3 sq. rd. 5 sq. ft. take 11 A. 30 sq. yd. 8 sq
ft. 40 sq. in. 2 A. 1 sq. rd. 30 sq. yd. 1 sq. ft. 32 sq. in.
26. From 18 A. 3 sq. ft. 3 sq. in. take 15 A. 3 sq. rd. 30 sq.
yd. 1 sq. ft. 142 sq. in. 2 A. 156 sq. rd. 3 sq. ft. 41 sq. in.
MULTIPLICATION OF COMPOUND NUMBERS.
223. Compound Multiplication is the process of mul
tiplying a Compound Number by an Abstract Number.
Problem. — Multiply 9 hr. 14 min. 8.17 sec. by 10.
SoLUTiOK. — Ten times 8.17 sec. =
81.7 sec. = 1 min. 21.7 sec. Write 21.7
sec. and carry 1 min. to the 140 min.
obtained by the next multiplication.
This gives 141 min. = 2 hr. 21 min. "^ ^ " ^ ^ ^ ^'^
Write 21 min. and carry 2 hr. This gives 92 hr. = 3 da. 20 hr.
OPERATION.
da. hr. min. sec.
9 14 8.1 7
10
166 RA Y'S HIGHER ARITHMETIC.
Problem.— Multiply 12 A. 148 sq. rd. 28 sq. yd. by 84.
Solution.— Since 84 = 7X12, multiply operation.
by one of these factors, and this product ^* sqro* sq. yd.
by the other ; the last product is the one 12 148 28 1
required. The same result can be obtained 7
by multiplying by 84 at once; performing 9 8 2 17 J
the work separately, at one side, and trans 1 2
ferring the results. 1086 30 24
Rule. — 1. Write the mvltiplier under the lowest denomina
tion of the multiplicand.
2. Multiply the lowest denomination first, and divide the
product by the number of units of this denomination which
make a unit of the next higher; write the remainder under the
denomination multipliedy and carry tJie quotient to Hie product
of the next high^ denomination,
3. Proceed in like manner with all the denominations, vrriting
the entire product at the last.
Examples for Practice.
1. Multiply 7 rd. 10 ft. 5 in. by 6. 45 rd. 13 ft.
2. Multiply 1 mi. 14 rd. S\ ft. by 97.
101 mi. 126 rd. 8 ft. 3 in.
3. Multiply 5 sq. yd. 8 sq. ft. 106 sq. in. by 13.
77 sq. yd. 5 sq. ft. 82 sq. in.
4. Multiply 41 A. 146.1087 sq. rd. by 9.046
379 A. 23.4593+ sq. rd.
5. Multiply 10 cu. yd. 3 cu. ft. 428.15 cu. in. by 67.
678 cu. yd. 1 cu. ft. 1038.05 cu. in.
6. Multiply 7 oz. 16 pwt. 5f gr. by 174.
113 K). 3 oz. 5 pwt. 16^ gr.
7. Multiply 2 3 1 9 13 gr. by 20. 6 § 3 5.
8. Multiply 16 cwt. 1 qr. 7.88 lb. by 11.
8 T. 19 cwt. 2 qr. 11.68 lb.
DIVISION OF COMPOUND NUMBERS. 167
9. Multiply 5 gal. 3 qt. 1 pt. 2 gills by 35.108
208 gal. 1 qt. 1 pt. 2.52 gills.
10. Multiply 26 bu. 2 pk. 7 qt. .37 pt. by 10.
267 bu. 7 qt. 1.7 pt.
11. Multiply 3 fs 48 nt by 12. 5 f? 5 fs 36 nt.
12. Multiply 18 da. 9 hr. 42 min. 29.3 sec. by 16yV.
306 da. 4 hr. 25 min. 2 sec, nearly.
13. Multiply £215 16s. 2\di. by 75. £16185 14s. d.
14. Multiply 10° 28' A2^" by 2.754 28° 51' 27.765"
DIVISION OF COMPOUND NUMBERS.
224. Division of Compound Numbers is the process
of dividing when the dividend is a Compound NTumber.
The divisor may be Simple or Compound, hence there
are two cases:
1. To divide a Compound Number into a number of equal
parts,
2. To divide one Compound Number by another of the name
kind.
Note. — Problems under the second case are solved by reducing
both Compound Numbers to the same denomination, and then
dividing as in simple division.
Problem. — ^Divide 5 cwt. 3 qr. 24 lb. 14f oz. of sugar
equally among 4 men.
SoiiUTiON. — 4 into 5 cwt. gives a opekation.
quotient 1 cwt., with a remainder 1 cwt., cwt. qr. lb. oz.
= 4 qr., to be carried to 3 qr., making 4)5 3 24 14f
7 qr.; 4 into 7 qr. gives 1 qr., with 3 qr., 1 1 2 4 1 5 1
— 75 lb., to be carried to 24 lb., = 99 lb.;
4 into 99 lb. gives 24 lb., with 3 lb., = 48 oz., to be carried to 14 oz.,
making 62 oz.; 4 into 62f oz. gives 15^1 oz., and the operation is
complete.
168 RAY'S HIGHER ARITHMETIC.
Problem. — If $42 purchase 67 bu. 2 pk. 5 qt. If pL
of meal, how much will $1 purchase?
OPERA.TION.
Solution. — Since 42 = 6 X 7, divide bu. pk. qt. pt.
first by one of these factors, and the 6)67 2 5 If
resulting quotient by the other ; the 7)11 1 \l\
last quotient will be the one required. \ 2 3 1 jyt
Rule. — 1. Write the quantity to be divided in the order
of its denominations, beginning with tJie highest; place the
divisor on the Uft,
2. Begin with Hie highest denomination, divide eacl% number
separately, and write Hie gv^otient beneatJi.
S, If a remmnder occur after any division, reduce it to the
next lower denomination, and, before dividing, add to it the
number of its denomiyiation.
Examples for Practice.
1. Divide 16 mi. 109 rd. by 7. 2 mi. 107 rd.
2. Divide 37 rd. 14 ft. 11.28 in. by 18.
2rd. 1ft. 8.96 in.
3. Divide 675 C. 114.66 cu. ft. by 83.
8 C. 18.3453+ cu. ft.
4. Divide 10 sq. rd. 29 sq. yd. 5 sq. ft. 94 sq. in. by 17.
19 sq. yd. 4 sq. ft. 119^ sq. in.
5. Divide 6 sq. mi. 35 sq. rd. by 22^. 170 A. 108 sq. rd.
6. Divide 1245 cu. yd. 24 cu. ft. 1627 cu. in. by 11.303
110 cu. yd. 6 cu. ft. 338.4+ cu. in.
7. Divide 3 § 7 3 18 gr. by 12. 2319 l^ gr.
8. Divide 600 T. 7 cwt. 86 lb. by 29.06
20 T. 13 cwt. 20 lb. 14 oz. 12+ dr.
9. Divide 312 gal. 2 qt. 1 pt. 3.36 gills by 72.
4 gal. 1 qt. 1.79+ gills.
10. Divide 19302 bu. by 6.215
3105 bu. 2 pk. 6 qt. 1.5+ pt.
LONGITUDE AND TIME. 169
11. Divide 76 yr. 108 da. 2 hr. 3* min. 26.18 sec. by 45.
1 yr. 254 da. 27 min. 31.25— sec.
12. Divide 152° 46' 2" by 9. 16° 58' 26".
225. liongitude and Time give rise to two cases:
1. To find the difference of longitude between two places
when the difference of time is given,
2, To find ihe difference of time when their longitudes are given.
Problem. — The difference of time between two places is
4 hr. 18 min. 26 sec: what is their difference of longitude?
Solution. — Every hour of time corre
j X 1 KO f 1 ^ J • * OPERATION.
spends to 15° of longitude; every minute ,
t ^' ^ ic/ * 1 x J J l»r. mm. sec.
of time to 15' of longitude ; every second .
of time to 15'''' of longitude (Art. 207).  e
Hence, multiplying the hours in the — ~
• 6436"^ 30 '^^
diflference of time by 15 will give the
degrees in the difference of longitude,
multiplying the minutes of time by 15 will give minutes ('') of
longitude, and multiplying the seconds of time by 15 will give
seconds ('''') of longitude.
Problem. — The difference of longitude between two
places is 81° 39' 22": what is their difference of time?
SoLunoN.^15 into 81°
gives 5 (marked hr.), and 6° operation.
to be carried. Instead of ^ ^)^l° 39^ 22''
multiplying 6 by 60, add ^ hr. 2 6 min. 3 7 ^^^ sec.
ing the 39^, and then divid
ing, proceed thus ; 15 into 6° is the same as 15 into 6 X ^ =
6 V ^ 4
— S — = 24^^, and as 15 into 39^ gives 2' for a quotient and 9^
15
remainder, the whole quotient is 26'' (marked min,), and remainder
9^ = 9 X 6(/^, which, divided by 15, gives ?X?Pl =, 3^//^ ^hich
15
with 1^, obtained by dividing 22^^ by 15, gives 37xV^ (marked sec.).
The ordinary mode of dividing will give the same result, and may
be used if preferred.
H. A. 15.
170
RAY'S HIGHER ARITHMETIC,
226. From these solutions we may obtain the following
rules :
CASE I.
Rule. — Multiply ihe difference of time by 15, according to
the rule for MvUipHccUion of Compound Numbers, and mark
the product ° ' " instead of hr. min. sec.
CASE II.
Rule. — Divide the difference of longitude by 15, a/xording
to the rule for Dvvimn of Coinpound Numbers, an^ mark the
quotient hr. min. sec, instead of ° '
ff
Note. — The following table of longitudes, as given in the records
of the U. S. Coast Survey, is to be used for reference in the solution
of exercises. "W." indicates longitude West, and "E." longitude
East of the meridian of Greenwich, England.
Table op Longitudes.
Pl.ACE.
LONOrrUDE.
Portland, Me., ....
o
70
t
15
H
18 W.
Boston, Mass., ....
71
3
50 ••
New Haven, Conn., .
72
55
45 "
New York City, . . .
74
24 "
Philadelphia, Pa., .
75
9
3 *•
Baltimore, Md., . . .
76
86
59 *•
Washington, D. C, .
77
36 "
Richmond, Va., . . .
77
26
4 "
Charleston, S. C , . .
79
55
49 •'
Pittsburgh, Pa., . . .
80
2
••
Savannah, Ga., . . .
81
5
26 "
Detroit, Mich., . . .
&3
3
"
Cincinnati, O., ...
84
29
45 ••
Ixjuisville, Ky., . . .
85
25
"
Indianapolis, Ind., .
86
6
"
Nashville, Tenn,, . .
86
49
"
Chicago, 111
87
3..
"
Mobile, Ala., . . .
88
2
28 "
Madison, Wis., . . .
89
24
3 ••
New Orleans, La., . .
90
3
28 "
St Louis, Mo., ....
90
12
H "
Minneapolis, Minn.,
93
11
8 "
PLACE.
Des Moines, Iowa, . . .
Omaha, Neb.,
Austin, Tex.,
Denver, Col
Salt Lake City, Utah, .
San Francisco, Cal., . .
Sitka, Alaska,
St. Helena Island, . . .
Reykjavik, Iceland, . .
Rio Janeiro, Brazil, . .
St. Johns, N. F., ....
Honolulu, Sandwich Is.
Greenwich, Eng., ....
Paris, France,
Rome, Italy
Berlin, German Em p., .
Vienna, Austria
Constantinople,Tu rkey ,
St. Petersburg, Russia, .
Bombay, India,
Pekin, China
Sydney, Australia, . . .
LONGITUDE.
o /
93 37
95 56
97 44
104 59
111 53
122 27
135 19
5 42
22
43 20
52 43
157 52
2 20
12 28
13 23
16 20
28 59
30 16
72 48
116 26
151 11
It
16 W.
14 "
12 "
33 "
47 "
49
42
E
"
**
"
"
••
"
•*
•*
(<
(I
(«
<4
LONGITUDE AND TIME. 171
Examples for Practice.
1. It is six o'clock A. M. at New York ; what is the time
at Cincinnati? 18 min. 2.6 sec. after 5 o'clock A. M.
2. The difference of time between Springfield, HI., and
Philadelphia being 58 min. 1^^ sec, what is the longitude
of Springfield ?  89° 39' 20" W.
3. At what hour must a man start, and how fast would
he have to travel, at the equator, so that it would be noon
for him for twentyfour hours?
Noon; 1037.4 statute miles per hour.
4. What is the relative time between Mobile and Chi
cago? Chicago time 1 min. 49f sec. faster.
5. A man travels from Hali&x to St. Louis ; on arriving,
his watch shows 9 A. M. Halifax time. The time in St.
Louis being 13 min. 32^ sec. after 7 o'clock A. M., what
is the longitude of Halifax? 63° 35' 18" W.
6. Noon occurs 46 min. 58 sec. sooner at Detroit than
at Galveston, Texas: what is the longitude of the latter
place? 94° 47' 30" W.
7. When it is five minutes after four o'clock on Sunday
morning at Honolulu, what is the hour and day of the week
at Sydney, Australia?
41 min. 12 sec. after 12 o'clock A. M., Monday.
8. What is the difference in time between St. Petersburg
and New Orleans? • 8 hr. 1 min. 17 sec.
9. When it is one o'clock P. M. at Rome, it is 54 min.
34 sec. after 6 o'clock A. M. at Buffalo, N. Y.: what is
the longitude of the latter? 78° 53' 30" W.
10. When it is six o'clock P. M. at St. Helena, what
is the time at San Francisco?
12 min. 56^ sec. after 10 o'clock A. M.
11. A ship's chronometer, set at Greenwich, points to 4
hr. 43 min. 12 sec. P. M.: the sun being on the meridian,
what is the ship's longitude? 70° 48' W.
172
BA F'^y HIOHER ARITHMETIC.
ALIQUOT PARTS.
227. All aliquot part is an exact divisor of a number.
Aliquot parts may be used to advantage in finding a
product when either or when each of the factors is a
Compound Number.
Problem. — Find the cost of 28 A. 145 sq. rd. \b\ sq. yd.
of land, at $16 per acre.
Solution. — Multiply $16, the operation.
price of 1 A., by 28 ; the product, $ 1 6
$448, is the price of 28 A. 145 '2 8
sq. rd. is made up of 120 sq. rd., 12 8
20 sq. rd., and 5 sq. rd. 120 sq. 3 2
rd. = f of an acre ; hence take f
of $16, the price per acre, to find
the cost of 120 sq. rd. 20 sq. rd.
= J of 120 sq. rd., and cost \ as
much. 5 sq. rd. = J of 20 sq. rd.,
and cost \ as much. 15 sq. yd.
= ^ of 1 sq. rd., or ^ of 5 sq. rd.,
and cost y^y as much as the latter. Add the cost of the several
aliquot parts to the cost of 28 acres. The result is the cost of the
entire tract of land.
120 sq. rd.= f
2 sq. rd. == \
5 sq. rd. = J
15Jsq.yd. = ^5
$448
12.
2.
.50
.0 5
$462.55
Problem. — A man travels 3 mi. 20 rd. 5 yd. in 1 hr.;
how far will he go in 6 da. 9 hr. 18 min. 45 sec. (12 hr.
to a day)?
OPERATION.
rd. yd.
Solution. — This
example is solved
like the preceding,
except that here
the multiplications
and divisions are
performed on a
Compound instead
of a Simple Number.
mi.
3
6 da. =8X9 hr.
1 5 min. = J of 1 "
3 " =1 of 15 min.
4 5 sec. = J of 3 min.
249
20
80
5
9
27
188
1
220
225
2J
245
U
49
i
•
12
lA
«
ALIQUOT PARTS,
173
Note. — In all questions in aliquot parts, one of the numbers
indicates a raie, and the other is a Compound Number whose txi/ue
at this rate is to be found.
Bule for Aliquot Parts. — Multiply the number indicating
the rate by tlie number of thai denomination for whose unit tlie
rate is given y and separate the numbers of the otlier denomina
tions into parts whose values can be obtahted directly by a
simple division or muUiplieation of one of the preceding values.
Add tfiese different values; the residt will be the entire value
required.
Notes. — 1. Sometimes one of the values may be obtained by
adding or subtracting two pi^eceding values instead of by multiplying
or dividing.
2. Aliquot parts are generally used in examples involving U. S.
money, and the following table should be memorized for future use.
Aliquot Parts of 100.
5
10
tV
= iV
1
TTT
12i
16
1
1
20 ^\
25 =i
50 =i
Kemark. — The following multiples of aliquot parts of 100, are
often used: 18f=A, 37^=1, 40 = , 60 = f, 62^ = f, 75 = i.
Examples for Practice.
1. If a man travel 2 mi. 105 rd. 6^ ft. in 1 hour, how
far can he travel in 30 hr. 29 min. 52 sec.?
71 mi. 12 rd. i ft. ^ in.
2. An old record says that 694 A. 1 R 22 P. of land,
at $11.52 per acre, brought $8009.344 ; what is the error in
the calculation ? $10 too much.
3. At $15.46 an oz., what will be the cost of 7 lb. 8 oz.
16 pwt. 11 gr. of gold? $1435.04
1 74 BA Y' jS HIOHUR ARITHMETIC.
4. What is the cost of 88 gal. 3 qt. 1 pt. of vinegar, at
374^ ct. a gallon? $33.33—
5. If the heart should beat 97920 times in each day, how
many times would it beat in 8 da. 5 hr. 25 min. 30 sec?
805494 times.
6. At a cost of $8190.50 per mile over the plain, and at a
rate of $84480 per mile of tunnel, what is the cost of a rail
way 17 mi. 150 rd. plain, and 70 rd. tunnel?
$161557.80 nearly.
7. What is the value of 20 T. 1 cwt. 13 lb. of sugar, at
$3f per cwt.? $1504.2375
8. If £3 6s. silver weigh 1 lb. troy, how much will 17 tb.
11 oz. 16 pwt. 9 gr. be worth? £59 7s. +
9. K a steamship could make 16 mi. 67f rd. in 1 hr.,
how far could it go in 24 da. 22 hr. 56 min. 12 sec. ?
9709 mi. 29.09 rd.
Topical Outline.
Operations with Compound Numbers.
1. Definition.
1. Reduction J 2. nnsps f ^ From Higher to Lower.
From Lower to Higher.
f 1. Dennitioii.
■J 2. Cases i^'
Is. Rules. ^^•
2. Addition / 1 Definition.
\ 2. Rule.
3. SubtracUon / 1 Definition.
\ 2. Rule.
4. Multiplication / 1 Definition.
(2.
Rule.
{1. Definition.
2. Cases.
3. Rule.
rCase I.— Rule.
6. Longitude and Time. I Case II.— Rule.
(. Table of Longitudes.
Definition.
7. Aliquot Parts \ 2. Rule.
Table.
I 3.
Xn. RATIO.
DEFINITIONS.
228. 1. Batio is a Latin word, signifying rdation or eon
nectuni; in Arithmetic, it is tJie measure of the rdation af
one number to another of the same kindy expressed by their
quotient,
2. A Ratio is found by dividing the first number by the
second ; as, the ratio of 8 to 4 is 2. The ratio is abstract.
8. The Sign of Batio is the colon (:), which is the sign
of division, with the horizontal line omitted ; thus, 6 : 4
si^ifies the ratio of 6 to 4 = .
4. Each number is called a term of the ratio, and both
together a couplet or ratio. The first term of a ratio is
the antecedent, which means going before; the second term
is the consequent, which mesms foUotuing.
5. A Simple Batio is a single ratio consisting of two
terms ; as, 3 : 4 = .
6. A Compound Batio is the product of two or more
simple ratios ; as, < ' I = J— 
^ 15:8/ ' "
5
8'
7. The Beciprocal of a Batio is 1 divided by the ratio,
or the ratio inverted ; thus, the reciprocal of 2 : 3, or , is
8. Inverse Batio is the quotient of the consequent di
vided by the antecedent; thus,  is the inverse ratio of
4 to 5.
9. The Value of the Batio depends upon the relative
size of the terms.
(175)
176 RAY'S HIGHER ARITHMETIC.
220. From the preceding definitions the following prin
ciples are derived;
^ ^ ^ , Antecedent
Principles. — 1. licdw = — •
Conseqtient
2. Antecedent = Gomequent X fi^w).
_ , Antecedent
3. Consequent = — ^7;
Hence, by Art. 87 :
1. The Ratio is multiplied by multiplying t/w Antecedent or
dividing the Consequent.
2. The Ratio is divided by dividing the Antecedent or mul
tiplying the Consequent.
3. The Ratio is not changed by multiplying or dividing
both terms by the same number.
General Law.— J.ny change in the Antecedent produces a
like change in the Ratio, but any clw/nge in tJw Consequent
produces an opposite cfiange in Hie Ratio.
Problem. — What is the ratio of 15 to 36 ?
OPERATION.
15 ; 36 = Ji = ^3.
Bnle. — Divide the Antecedent by the Consequent.
Examples for Practice.
1. What is the ratio of 2 ft. 6 in. to 3 yd. 1 ft. 10 in.? j\.
2. What is the ratio of 4 mi. 260 rd. to 1 mi.
96 rd. ? f^f .
3. What is the ratio of 13 A. 145 sq. rd. : 6 A. 90 sq.
rd.? ff,
4. What is the ratio of 3 lb. 10 oz. 6 pwt. 10^ gr. : 2 lb.
14i pwt? i,^jft^.
PBOPaRTION. 177
5. What is the ratio of 10 gal. 1.54 pt. : 7 gal. 2 qt.
.98 pt? 1^.
6. What is the ratio of 56 bu. 2 pk. 1 qt. : 35 bu. 3 pk.
6.055 qt.? ft^f.
7. K the antecedent is 7 and the ratio J^, what is the
consequent ? 4f .
8. If the consequent is ^ and the ratio f , what is the
antecedent ? ^j,
9. What is the ratio of a yard to a meter, and of a meter
to a yard? mm*. IMMU
10. What is the ratio of a pound avoirdupois to a pound
troy ? m
11. Find the difference between the compound ratios
12. Find the difference between the ratio 4 : 1\ and the
inverse ratio. WW 
13. If the consequent is 6^, and the ratio is 2^, what is
the antecedent, and what is the inverse ratio of the two
numbers? Antecedent 14f, inverse ratio ^.
XIII. PEOPORTIOl^.
DEFINITIONS.
230. i. Proportion is an equality of ratios.
Thus, 4 : 6 : : 8 : 12 is a proportion, and is read ^ is to Q as 8
18 to 12.
2. The Sign of Proportion is the double colon ( ; : ).
Note. — It is the same in effect as the sign of equality, which is
sometimes used in its place.
3. The two ratios compared are called couplets. The
first couplet is composed of the first and second terms, and
the second couplet of the third and fourth terms.
178 JRAY'S HIOHEB ARITHMETIC.
4. Since each ratio has an antecedent and consequent,
every proportion has two antecedents and two consequents,
the 1st and 3d terms being the antecedents, and the 2d and
4th the consequents.
5. The first and last terms of a proportion are called the
extremes; the middle terms, the means. All the terms
are called proportionals, and the last term is said to be a
fourth proportional to the other three in their order.
6. When three numbers are proportional, the second num
ber is a mean proportional between the other two.
Thus, 4:6 : : 6:9; six is a mean proportional between 4 and 9.
7. Proportion is either Simple or Compound: Simple
when both ratios are simple ; Compound when one or both
ratios are compound.
Principles. — 1. In every proportion the product of the
means is equal to the produd of Hie extremes,
2. The product of the extremes divided by either mean, will
give the other mean.
3. Tlie product of (he means divided by either extreme, will
give Hie other extreme.
SIMPLE PROPORTION.
231. 1. Simple Proportion is an expression of equality
between two simple ratios,
2. It is employed when three terms are given and we wish
to find the fourth. Two of the three terms are alike, and
the other is of the same kind as the fourth which is to be
found.
3. All proportions must be true according to Principle 1,
which is the test. Principles 2 and .3 indicate methods of
finding the wanting term.
SIMPLE PMOFORTION. 179
4. The Statement is the proper arrangemeut of the
terms of the proportion.
Problem. — If 6 horses cost $300, what will 15 horses
cost?
STATEMENT.
6 horfles : 15 horses :: $300 : ($ ).
OPERATION.
(« O /\ /\ \y 1 E
~ =$750. Or,($300X15)^6 = $750,^n«.
• Solution. — Since 6 horses and 15 horses maybe compared, they
form the first couplet; also, $300 and $ — may be compared, as they
are of the same unit of value.
Notes. — 1. To find the missing extreme, we use Prin. 3.
2. To prove the proportion, we use Prin. 1. Thus, $750 X 6 ^=
$300 X 15.
Problem. — If 15 men do a piece of work in 9f da., how
long will 36 men be in doing the same?
STATEMENT.
Solution.— Since 36 men will ^^^ ^^^ ^^ ^^
require less time than 15 men to 36*15"94*M
do the same work, the answer
should be less than 9f da.; make a operation.
decreasing ratio, Jf , and multiply 9 f = ^ da.
the remaining quantity by it. ^ X M = ^ <!*•> ^^•
Bnle. — 1. For the third term, vrrUe that number which is
of the 8ame denominatimi as the number required,
2. For the second term, unite the greater of the two re
maining numbers, when the fourth term is to he greater than
the third; and the less, when the fourth term is to be less
Vmn the third,
3. Divide the product of the second and third terms by the
first; the qmtient will be Hie fourth term, or number required.
180 BA Y'S HIGHER ARITHMETIC,
Examples FOR Practice.
Note.— Problems marked with an asterisk are to be solved
mentally.
1.* If I walk 10^ mi. in 3 hr., how far will I go in 10
hr., at the same rate? 35 mi.
2. If the forewheel of a carriage is 8 ft. 2 in. in cir
cumference, and turns round 670 times, how often will the
hind wheel, which is 11 ft. 8 in. in circumference, turn
round in going the same distance? 469 times.
3. If a horse trot 3 mi. in 8 min. 15 sec, how far caji
he trot in an hour, at the same rate? 21^ mi.
4. What is a servant's wages for 3 wk. 5 da., at $1.75
per week? $6.50
5. What should be paid for a barrel of powder, containing
132 lb., if 15 lb. are sold for «5.43? $47.85
6. A body of soldiers are 42 in rank when they are 24
in file : if they were 36 in rank, how many in file would
there be? 28.
7. If a pulse beats 28 times in 16 sec, how many times
does it beat in a minute? 105 times.
8. If a cane 3 ft. 4 in. long, held upright, casts a shadow
2 ft. 1 in. long, how high is a tree whose shadow at the
same time is 25 ft. 9 in.? 41 ft. 2 in.
9. If a farm of 160 A. rents for $450, how much should
be charged for one of 840 A ? $2362.50.
10. A grocer has a false gallon, containing 3 qt. 1^ pt.:
what is the worth of the liquor that he sells for $240, and
what is his gain by the cheat? $225, and $15 gain.
11. If he uses 14f oz. for a pound, how much does he
cheat by selling sugar for $27.52? $2.15
12. An equatorial degree is 365000 ft.: how many ft. in
80° 24' 37" of the same? 29349751^^ ft.
13. If a pendulum beats 5000 times a day, how often
does it beat in 2 hr. 20 min. 5 sec ? 486:^^ times.
SIMPLE PROPORTION. 181
14.* If it takes 108 days, of 8^ hr. , to do a piece of work,
how many days of 6 hr. would it take ? 136 days.
15. A man borrows $1750, and keeps it 1 yr. 8 mon.:
how long should he lend $1200 to compensate for the
favor? 2 yr. 5 mon. 5 da.
1^ A garrison has food to last 9 mon., giving each man
1 lb. 2 oz. a day : what should be a man's daily allowance,
to make the same food last 1 yr. 8 mon.? 8^^ oz.
17. A garrison of 560 men have provisions to last during
a siege, at the irate of 1 lb. 4 oz. a day per man ; if the
daily allowance is reduced to 14 oz.' per man, how large a
reinforcement could be received? 240 men.
18. A shadow of a cloud moves 400 ft. in 18f sec: what
was the wind's velocity per hour? 14^ mi.
19. If 1 ft), troy of English standard silver is worth £3
6s., what is 1 lb. av. worth? £4 2^d.
20. If I go a journey in 12f days, at 40 mi. a day, how
long would it take me at 29f mi. a day? 17 da.
21.* If  of a ship is worth $6000, what is the whole
of it worth ? . $10800.
22. If A, worth $5840, is taxed $78.14, what is B worth,
who is taxed $256.01 ? $19133.59—
23.* What are 4 lb. 6 oz. of butter worth, at 28 ct. a
lb.? $1.22i
24. If I gain $160.29 in 2 yr. 3 mon., what would I
gain in 5 yr. 6 mon., at that rate? $391.82
25. If I gain $92.54 on $1156.75 worth of sugar, how
much must I sell to gain $67.32? $841.50 worth.
26. If coffee costing $255 is now worth $318.75, what
did $1285.20 worth cost? $1028.16
27. A has cloth at $3.25 a yd., and B has flour at $5.50
a barrel. K, in trading, A puts his cloth at $3.62^, what
should B charge for his flour? $6.13y\
28.* If a boat is rowed at the rate of 6 miles an hour,
and is driven 44 feet in 9 strokes of the oar, how many
strokes are made in a minute? 108 strokes.
182 BAY'S HIGHER ARITHMETIC.
29. If I gain $7.75 by trading with $100, how much
ought I to gain on $847.56? $65.6859
30. What is a pile of wood, 15 ft. long, 10\ ft. high,
and 12 ft. wide, worth, at $4.25 a cord? $62.75
Eemark. — In Fahrenheit's thermometer, the freezing point of
water is marked 32°, and the boiling point 212° : in the Centigrade,
the freezing point is 0°, and the boiling point 100°: in Keaumer's,
the freezing point is 0°, and the boiling point 80°.
31. From the above data, find the value of a degree of
each thermometer in the degrees of the other two.
V F.=^° R. = F C; l"" C. =1° F. =
°R.; l°R.=li°C. = 2i°F.
32. Convert 108° F. to degrees of the other two ther
mometers. 33^° R. and 42f ° C.
33. Convert 25° R. to degrees of the other two thermom
eters. 31i° C. and 88^° F.
34. Convert 46° C. to degrees of the other two ther
mometers. 36^° R. and 114° F.
Kemabks. — 1. In the working of machinery, it is ascertained that
the available power is to the weight overcome, inversely as the diatanees they
pass over in the same tim£,
2. Inverse variation exists between two numbers when one in
creases as the other decreases.
3. The available power is taken  of the whole power, \ being
allowed for friction and other impediments.
• 35. If the whole power applied is 180 lb. and moves 4 ft. ,
how far will it lift a weight of 960 lb. ? 6 in.
36. If 512 lb. be lifted 1 ft. 3 in. by a power moving
6 ft. 8 in. , what is the power ? 144 lb.
37. A lifts a weight of 1410 lb. by a wheel and axle; for
every 3 ft. of rope that passes through his hands the weight
rises 4^ in.: what power does he exert? 270 lb.
38. A man weighing 198 lb. lets himself down 54 ft. with
a uniform motion, by a wheel and axle : if the weight at the
hook rises 12 ft., how much is it? 594 lb.
SIMPLE PEOPORTION, 183
39. Two bodies free to move, attract each other with
forces that vary inversely as their weights. If the weights
are 9 lb. and 4 lb., and the smaller is attracted 10 ft., how
far will the larger be attracted ? 4 ft. 5^ in.
40. Suppose the earth and moon to approach each other
in obedience to this law, their weights being 49147 and 123
respectively, how many miles would the moon move while
the earth moved 250 miles ? 99892+ mi.
Can the three following questions be solved by pro
portion?
41. If 3 men mow 5 A. of grass in a day, how many
men will mow 13^ A. in a day ?
42.* If 6 men build a wall in 7 da., how long would 10
men be in doing the same?
43.* If I gain 15 cents each, by selling books at $4.80 a
doz., what is my gain on each at $5.40 a doz.?
44. A clock which loses 5 minutes a day, was set right
at 6 in the morning of January 1st : what will be the right
time when that clock points to 11 on the 15th?
11 min. 17.35+ sec. past noon.
45. If water begin and continue running at the rate of
80 gal. an hour, into a cellar 12 ft. long, 8 ft. wide, and 6
ft. deep, while it soaks away at the rate of 35 gal. an hour,
in what time will the cellar be full? 95.75+ hr.
46. Take the proportion of 4 : 9 : : 252 : a fourth term.
If the third and fourth terms each be increased by 7, while
the first remains unchanged, what multiplier is needed by
the second to make a proportion? fl^.
47. Prove that there is no number which can be added
to each term of 6 : 3 : : 18 : 9 so that the resulting num
bers shall stand in proportion.
48. A certain number has been divided by one more than
itself, giving a quotient jL: what is the number? \.
49. If 48 lb. of seawater contain 1^ lb. of salt, how
much fresh water must be added to these 48 lb. so that 40
lb. of the mixture shall contain \ lb. of salt? 72 lb.
184 BAY'S HIGHER ARITHMETIC.
CJOMPOUND PROPORTION.
232. Compound Proportion is an expression of equality
between two ratios when either or when each ratio is Com
pound.
Problem. — If 3 men mow 8 A. of grass in 4 da., how
long would 10 men be in mowing 86 A.?
STATEMENT.
1 men : 3 men 1 ^ , r \ a
Solution. — Since « *  ^ fi A \ ' * v / ^^*
the denomination
of the required term operation.
is days, make the 9
third term 4 da. In T g
forming the first and 3X^^X^ ,, \, j .
second terms, con ^ ^ ^ ="%==<> T cla. Ans,
sider each denom A r A p
ination separately ; *^ r
10 men can do the same amount of work in less time than 3 men ;
hence, the first ratio is, 10 men : 3 men, the less number being the
second term. Since it takes 4 da. to mow 8 A., it will take a greater
number of days to mow 36 A., and the second ratio is,'8 A. : 36 A.,
the gii'eater number being the second term. Then dividing the con
tinued product of the means by that of the extremes (Art. 230,
Prin. 3), after cancellation, we have 5f da., the required term.
Rule. — 1. For the third term, write that nwnher which is
of the same denomination as the number required,
2. Arrange each pair of numbers having the sam£ denom
ination in the compound ratio, as if, with the third term, Hiey
formed a simple proportion,
3. Divide the product of the numbers in the second and third
tenns by the product of the nwinbers in the first term: the quotient
wiU be the required term.
233. Problems in Compound Proportion are readily
solved by separating all the quantities involved into ttoo
causes and two effects.
COMPOUND PROPORTION. 185
Problem. — If 6 men, in 10 days of 9 hr. each, build
25 rd. of fence, how many hours a day must 8 men work
to build 48 rd. in 12 days?
Solution. — 6 men 10 da. and 9 hr. constitute the first cause,
whose effect is 25 rd. ; 8 men 12 da. and ( ) hr. constitute the second
cause, whose effect is 48 rd. Hence,
STATEMENT.
8 men. ^
: 12 da. >:: 25 rd. : 48 rd.
( ) hr. J
OPERATION.
? 5
Biile of Cause and Effect. — 1. Separate all the quan
tities contained in the question into two causes and Hmr effects,
2. Write, for the first tefi^m of a propoHion, all the qudn
tities that constitute the first cause ; for the second term, aU tJiat
constitute the second cause; for the third, all that comtittUe tJie
effect of the first cause ; and for the fourth, all that constitute the
effect of the second coMse,
3. The required quantity may be indicated by a bracket,
and found by Art. 230, Principles.
Note. — The two causes must be exactly alike in the numbei' and
kind of their terms ; and so must the two effects.
Examples for Practice.
1. If 18 pipes, each delivering 6 gal. per minute, fill a
cistern in 2 hr. 16 min., how many pipes, each delivering
20 gal. per minute, will fill a cistern 7^ times as large as
the first, in 3 hr. 24 min. ? 27 pipes.
H. A. 16.
186 JiA Y'S HIGHER ARITHMETIC.
2. The use of $100 for 1 year is worth $8 : what is the
use of $4500 for 2 yr. 8 mon. worth ? $960.
3. If 12 men mow 25 A. of grass in 2 da. of 10^ hr.,
how many hours a day must 14 men work to mow an 80 A.
field in 6 days ? . 9f hr.
4. If 4 horses draw a raiboad car 9 miles an hour, how
many miles an hour can a steam engine of 150 horsepower
drive a train of 12 such cars, the locomotive and tender
being counted 3 cars ? 22^ mi. per hr.
5. If 12 men, working 20 days 10 hours a day, mow
247.114 hektars of timothy, how many men in 30 days,
working 8 hours a day, will mow 1976912 centars of
timothy of the same quality ? * 8 men.
6. If the use of $3750 for 8 mon. is worth $68.75,
what sum is that whose use for 2 yr. 4 mon. is worth
$250? $3896.10+
7. If the use of $1500 for 3 yr. 8 mon. 25 da. is worth
$336.25, what is the use of $100 for 1 yr. worth? $6.
8. A garrison of 1800 men has provisions to last 4^
months, at the rate of 1 lb. 4 oz. a day to each : how long
will 5 times as much last 3500 men, at the rate of 12 oz.
per day to each man? 1 yr. 7^ months.
9. What sum of money is that whose use for 3 yr., at
the rate of $4^ for every hundred, is worth as much as the
use of $540 for 1 yr. 8 mon., at the rate of $7 for every
hundred? $466. 66f
10. A man has a bin 7 ft. long by 2 ft. wide, and 2 ffc.
deep, which contains 28 bu. of corn : how deep must he
make another, which is to be 18 ft. long by 1^ ft. wide, in
order to contain 120 bu. ? 4^ ft.
11. If it require 4500 bricks, 8 in. long by 4 in wide, to
pave a courtyard 40 ft. long by 25 ft. wide, how many
tiles, 10 in. square, will be needed to pave a hall 75 ft.
long by 16 ft. wide? 1728 tiles.
12. If 150000 bricks are used for a house whose walls
average 1^ ft. thick, 30 ft. high, and 216 ft. long, how
COMPOUND PBOPORTION.
187
many will build one with walls 2 ft. thick, 24 ft. high, and
324 ft. long? 240000 bricks.
13. If 240 panes of glass 18 in. long, 10 in. wide, glaze a
house, how many panes 16 in. long by 12 in. wide will
glaze a row of 6 such houses? 1350 panes.
14. If it require 800 reams of paper to publish 5000 volumes
of a duodecimo book containing 320 pages, how many reams
will be needed to publish 24000 copies of a book, octavo
size, of 550 pages ? 9900 reams.
15. If 15 men cut 480 sters of wood in 10 days, of 8
hours each: how many boys will it take to cut 1152 sters
of wood, only ^ as hard, in 16 days, of 6 hours each, pro
vided that while working a boy can do only f as much as
a man, and that \ of the boys are idle at a time throughout
the work ? 24 boys.
Topical Outline.
Ratio and Proportion.
1. Ratio
2. Proportion.
1. Definitions.
2. Principles.
3. General Law.
1. Definitions.
2. Principles.
3. Kinds
4. Statement
5. Rules.
I
1. Simple.
2. Compound.
XIY. PEEOEI^rrAGE,
DEFINITIONS.
234. 1. Percentage is a term applied to all calcula
tions in which 100 is the basis of comparison; it is also
used to denote the result arising from taking so many
hundredths of a given number.
2. Per Cent is derived from the Latin phrase pet
centumy which means by or on the hundred,
3. The Sign of Per Cent is ^; the expression 4% =;
yI^, is read "4 per cent'* equals yj^; or, decimally, .04
4. The elements in Percentage are the BasCy the Bate,
the Percentage^ and the Amount or Difference,
5. The Base is the number on which the Percentage is
estimated.
6. The Bate is the number of hundredths to be taken.
7. The Percentage is the result arising from taking
that part of the Base expressed by the Rate.
8. The Amount is the Base phis the Percentage.
9. The DifTerenoe is the Base minus the Percentage.
NOTATION.
236. It is convenient to use the following notation:
1. Base =B.
2. Rate =R.
3. Percentage = P.
A f Amount = A.
1 Difference = D.
(ISS)
PERCENTAGE. 189
Remark. — All problems in Percentage refer to two or more of
the above terms. Owing to the relations existing among these
terms, any two of them being given the others can be found. These
relations give rise to the following cases :
CASE I.
236. Given the base and the rate, to find the per
centage.
Principle. — The percentage of any number is the rnrne^
part of that number as the given rate is of 100^.
Problem. — If I have 160 sheep, and sell 35% of them:
bow many do I sell?
OPERATIONS.
1. 160 sheep X3 5 = 56 sheep, Am.
>^
2. 100/o = 160 sheep.
Ifo = 1.60 sheep.
.. 35^ = 1.60X35 = 56 sheep, Ans.
3. 1 6 X iV = ^ ^ sheep, Ans. (Art. 218, Bern.)
Suggestion.— 3 5^ = yVtt = sV
Solution. — Take .36 of the base ; the result is the percentage.
Formula.— B X B = P.
Rule 1. — Multiply the hose by the rate ^ expressed deci
mxdly; the product is the percentage.
Rule 2. — Find that part of the base which the rate % is
qfim.
Examples for Practice.
1. Find 62^^ of 1664 men. 1040 men.
2. Find 35% of f yV
3. Find 9f % of 48 mi. 256 rd. 4 mi. 184 rd.
190 BAY'S HIGHER ARITHMETIC.
4. Fiud 11^% of $3283.47 $364.83
5. Find ZZ\% of 127 gal. 3 qt. 1 pt. 42 gal. 2 qt. 1 pt
6. Find 98^ of 14 cwt. 2 qr. 20 lb.
14 cwt. 1 qr. 15f lb.
7. Find 40% of 6 hr. 28 min. 15 sec.
2 hr. 35 min. 18 sec.
8. Find 104^ of 75 A. 75 sq. rd. 78 A. 78 sq. rd.
9. Find 15^ of a book of 576 pages. 90 pages.
10. Find 56^% of 144 cattle. 81 cattle.
11. Find 16^ of 1932 hogs. 322 hogs.
12. Find 1000% of $5.43f $54.37^
13. Find ^% of f sV
14. What part is 25% of a farm? \,
15. What part of a quantity is 18f% of it; 31^%;
37i%; 43f%; 56^%; 62%; 68f%; 81^%; 83J%; 87^%;
93% ? A, 1^, f , tV. A» I, ii, «, h h H of it.
16. How much is 100% of a quantity; 125% of it;
250%; 675%; 1000%; 9437^%?
1 time, 1J,2^, 6f, 10, 94f times the quantity.
17.* A man owning f of a ship, sold 40% of his share :
what part of the ship did he sell, and what part did he
still own ? ^ sold ; ^ left.
18.* A owed B a sum of money; at one time he paid
him 40% of it; afterward he paid him 25% of what he
owed ; and finally he paid him 20% of what he then owed :
how much does he still owe? ^^ of it.
19. Out of a cask containing 47 gal. 2 qt. 1 pt., leaked
6f%: how much was that? 3 gal. If pt.
20. A has an income of $1200 a year; he pays 23% of
it for board; 10f% for clothing; 6f% for books; ^% for
newspapers; 12^% for other expenses: how much does he
pay for each item, and how much does he save at the end
of the year? $276, bd.; $124.80, cL; $81, bks.; $7, npr.
$154.50, other ex.; $556.70 saved.
21. Find 10% of 20% of $13.50 27 ct.
22. Find 40% of 15% of 75% of $133.33^ $6.
PERCENTAGE. IM
23. A man coDtracts to supply dressed stone for a court
house for $119449, if the rough stone costs him 16 ct. a
cu. ft.; but if he can get it for 15 ct. a cu. ft., he will
deduct Z^o from his bill; how many cu. ft. would be
needed, and what does he charge for dressing a cu. ft. ?
358347 cu. ft., and 17^ ct. a cu. ft.
24. 48% of brandy is alcohol ; how much alcohol does a
man swallow in 40 years, if he drinks a gill of brandy 3
times a day? 657 gal. 1 qt. 1 pt. 2.4 gills.
25. A had $1200; he gave 30% to a son, 20% of the
remainder to his daughter, and so divided the rest among
four brothers that each after the first had $12 less than the
preceding : how much did the last receive ? $150.
26. What number increased by 20% of 3.5, diminished
by 12% of 9.6, gives 3i? 4.
CASE II.
237. Given tbe base and the percentage, to find
the rate*
Pkinciple. — The rate equals Uie number of hundredths Uiat
the percentage is of the base.
Problem. — ^What per cent of 45 is 9?
Solution. — 9 is J of 45 ; but  of any operation.
number ia equal to 20^ of that number ; ^^ := ^ ^n^ 20^, Arts.
hence, 9 is 20^ of 45
P
Formula. — — = R.
Bnle 1. — Divide the percentage by the base; the qvMient is
the rate;
Bnle 2. — Find that part of 100^ that the percentage is
of the base.
192
RAY'S HIQHER ARITHMETIC.
Examples for Practice.
1. 15 ct. is how many ^ of $2?
2. 2 yd. 2 ft. 3 in. is how many ^ of 4 rd.?
3. 3 gal. 3 qt. is what % of 31^ gal.?
4. f is how many ^ of .  ?
'5. I of I of 4^ is what % of 1^?
6,
^ is how many % of f ?
3 ^ / 10
7. $5.12 is what % of $640?
8. $3.20 is what % of $2000?
9. 750 men is what % of 12000 men?
10. 3 qt. 1^ pt. is what ^ of 5 gal. 2J qt.?
11. A's money is 50^ more than B's; then B's money is
222%
16%
2%
25%
how many ^ less than A's ?
12. What % of a number is 8^ of 35^ of it?
13. What % of a number is 21% of 2^% of it?
14. What ^ of a number is 40^ of 62% of it?
15. 12^ of $75 is what % of $108?
16. U. S. standard gold and silver are 9 parts pure to 1
part alloy: what % of alloy is that? 10%
17. What ^ of a meter is a yard? 91 ^;fgj %
18. How many ^ of a township 6 miles square, does a
man own who has 9000 acres ? ^^Tt% •
19. How many ^ of a quantity is 40^ of 25^ of it ?
also, 16% of 37^% of it? also, 4^^ of 120% of it? also,
2% of 80% of 66^ of it? also, % of 36% of 75% of it?
also, ^% of 22^% of 96% of it?
10, 6, 5, ItV 7^^, Wt7%
20. 30% of the whole of an article is how many % of f
of it? 45%.
21. 25% of f of an article is how many % of f of it?
22. How many % of his time does a man rest, who sleeps
7 hr. out of every 24 ? 29^% .
PERCENTAOK 193
CASE III.
238. Given the rate and the percentage, to find
the base.
Principle. — The base bears the same ratio to the percerdxige
that 100% does to the rate.
Problem. — 95 is 5% of what number?
OPERATION.
100/^ = H* = 19X100 = 1900, Ana.
Or, 95!. 05=1900, Am.
Formula. — ~ = B.
Bnle 1. — Divide the percentage by the rate, and then mul
tiply the qtu>tient by 100; the product is the base.
Bnle 2. — Divide the percentage by the rate expressed deci
mally ; the quotient is the base.
Examples for Practice.
1. $3.80 is 5^ of what sum? $76.
2. ^ is 80^ of what number? ■^.
3. 16 is 1^% of what number? 1066.
4. 31^ ct. is 15f^ of what? $2.
5. $10.75 is 3\fo of what? $322.50
6. 162 men is 4% of how many men? 3375 men.
7. $19.20 is ^fo of what? $3200.
8. $189.80 is 104% of what? $182.50
9. 16 gal. 1 pt. is 6^ of what? 262 gal. 2 qt.
10. 10 mi. 316 rd. is 75% of what? . 14 mi. 208 rd.
H. A. 17. ^
194 RAY'S HIGHER ARITHMETIC.
11. Thirtysix men of a ship's crew die, which is 42f %
of the whole : what was her crew ? 84 men.
12. A stockfarmer sells 144 sheep, which is 12^ % of his
flock: how many sheep had he? 1125 sheep.
13. A merchant sells 35% of his stock for $6000: what
is it all worth at that rate? $17142.86
14. I shot 12 pigeons, which was 2§% of the flock: how
many pigeons escaped ? 438 pigeons.
15. A, owing B, hands him a $10 bill, and says, ** there is
6J^ of your money:" what was the debt? $160.
16. $25 is 62^% of A's money, and 41f^ of B's: how
much has each? A $40, B $60.
17. A found $5, which was 13J% of what he had before:
how much had he then ? $42.50
18. I drew 48% of my funds in bank, to pay a note of
$150: how much had I left? $162.50
19. A farmer gave his daughter at her marriage 65 A.
106 sq. rd. of land, which was 3^ of his farm : how much
land did he own? 2188 A. 120 sq. rd.
20. A pays $13 a month for board, which is 20% of his
salary : what is his salary ? $780 a year.
21. Paid 40 ct. for putting in 25 bu. of coal, which was
llf ^ of its cost: what did it cost a bu. ? 14 ct.
22. 81 men is 5% of 60% of what? 2700 men.
23. A, owning 60% of a ship, sells 7^% of his share for
$2500: what is the ship worth? $55555. 55f
24. A father, having a basket of apples, took out 33^%
of them ; of these, he gave 37^^ to his son, who gave 20^
of his share to his sister, who thus got 2 apples: how many
apples were in the basket at first ? 80 apples.
25. B lost three dollars, which was 31J^ of what he
had left: how much had he at first? $12.60
26. Bought 8000 bu. of wheat, which was bl^% of my
whole stock: how much had I before? 6000 bu.
27. If 32^ of 75% of 800% of a number is 1539, what
is that number? • 801^.
PER CENTA OE. 195
CASE IV.
239. Given the rate and the amount or the differ
ence, to find the base.
Principle. — The base is equal to ihe anumnt divided by 1
plus the rate, or the difference divided by 1 minus tlie raJte,
Problem. — ^A rents a house for $377, which is an advance
of 16^ on the rent of last year: what amount did he pay
last year?
OPERATION.
$3771 1.16 = $325.00, Ans.
Or, 10 0^ = rental last year ;
1 6 Jl^ = increase this year ;
hence, 1 OO^o f 1 6^o = 1 16 % =$377;
l^o = $3.25;
.. 100^ = 3.25X100 = $325.00, ^ns.
Problem. — John has $136, which is 20^ less than
Joseph's money: how many dollars has Joseph?
OPERATION.
$136^(1 — .2)=$170.00, Am.
Or, 1 J^ = Joseph's money ;
8 ^0 = John's money = $136;
l^o = $1.7;
.. 100^c)=$1.7X 100 = $170.00, ^w^.
Or, 80%=i = $136; ^=$34; and  = $34X 5 = $170.00, Ans.
(A^(l+K))
F0RMULA.~B=j)^(j_j^^
Bule. — Divide ike sum by 1 plus (he rate, or divide the
difference by 1 minus ihe rate; ihe quotiefiit wUl be ihe base*
196 RAY'S HIGHER ARITHMETIC.
Examples for Practice.
•
1. $4.80 is 33^^ more than what? $3.60
2. f is 50% more than what? 
3. 96 da. is 100^ more than what? 48 da.
4. 2576 bu. is 60% less than what? 6440 bu.
5. 87^ ct. is 87^% less than what? $7.
6. 42 mi. 60 rd. is 55^ less than what ? 93 mi. 240 rd.
7. 2 lb. 9^ oz. is 50% less than what number of
pounds? 5ff lb.
8. ^ is 99^ less than what? 155.
9. $920.93f is 337% more than what ? $210.50
10. $4358.06^ is 233^% more than what? $1307.41^
11. In 64^ gal. of alcohol, the water is 7^^ of the
spirit: how many gal. of each? 60 gal. sp., 4^ gal. w.
12.* A coat cost $32; the trimmings cost 70% less, and
the making 50^ less, than the cloth: what did each cost?
Cloth $17.77, trimmings $5.33^, making $8.88
13.* If a bushel of wheat make 39^ lb. of flour, and the
cost of grinding be 4%, how many barrels of flour can a
farmer get for 80 bu. of wheat? 15^ barrels.
14.* How many eagles, each containing 9 pwt. 16.2 gr. of
pure gold, can I get for 455.6538 oz. pure gold at the mint,
allowing 1^% for expense of coinage? 928 eagles.
15. 2047 is 10% of 110% less than what number? 2300.
16. 4246 is 6% of 50^ of 466% more than what
number ? 3725.
17. A drew out of bank 40% of 50% of 60% of 70%
of his money, and had left $1557.20 : how much had he at
first? $1700.
18. I gave away 42^% of my money, and had left $2 :
what had I at first? $3.50
19. In a school, 5% of the pupils are always absent, and
the attendance is 570: how many on the roll, and how
many absent? 600, enrolled; 30, absent.
APPLICATIONS OF PERCENTAGE. 197
20. A man dying, left 33^% of his property to his wife,
60% of the remainder to his son, 75% of the remainder to
his daughter, and the balance, ?500, to a servant : what was
the whole property, and each share ?
Property, $7500 ; wife had »2500 ;
son, $3000 ; daughter, $1500.
21. In a company of 87, the children are 37^^ of the
women, who are 44J% of the men: how many of each?
54 men, 24 women, 9 children.
22.* Our stock decreased 33^%, and again 20^ ; then it
rose 20%, and again 33 J^ ; we have thus lost $66: what
was the stock worth at first? $450.
23.* A brewery is worth 4% less than a tannery, and
the tannery 16% more than a boat; the owner of the
boat has traded it for 75^ of the brewery, losing thus
$103 : what is the tannery worth ? $725.
ADDITIONAL FORMULAS.
240. The following additional formulas, derived from
preceding data, may also be employed to advantage :
1. By definition, A = B + P ; also, D = B — P.
r A
2. From Case IV. <
A = B + (BxR) = B + P.
APPLICATIONS OF PERCENTAGE.
241. The Applications of Percentage may be divided
into two classes, those in which time is not an essential
element, and those in which it is an essential element, as
follows:
198
BAY'S HIOHEB ARITHMETIC,
Those in which Time is not an
Essential Element
Those in which Time is an
Essential Element...
I
\
1. Profit and Loss.
2. Stocks and Dividends.
3. Premium and Discount.
4. Commission and Brokerage,
5. Stock Investments.
6. Insurance.
7. Taxes.
8. United States Revenue.
1. Simple Interest.
2. Partial Payments.
3. True Discount.
4. Bank Discount.
5. Exchange.
6. Equation of Payments.
7. Settlement of Accounts.
8. Compound Interest.
9. Annuities.
Note.— These topics will be presented in the order in which they
stand.
Topical Outline.
Percentage.
1. Definitions.
2. Notation.
Ok x>aSCS.M... «
u.
m.
IV.
1^
I 3.
I 3.
i 3.
I 3.
1. Principle.
Formula.
Rules.
1. Principle.
Formula.
Rules.
1. Principle.
Formula.
Rules.
1. Principle.
Formula.
Rule.
4. Additional Formulas.
„ ,, , ^ . ( Time not an Element
5. Applications of Percentage. { *_.
*^*^ I Time an Element.
XY. PEECENTAGE.APPLIOATIOI^rS.
I. PROFIT AND LOSS.
DEFINITIONS.
242. 1. Profit and Loss are commercial terms, and pre
suppose a cost price.
2. The Cost is the price paid for any thing.
3. The Selling Price is the price received for whatever
is sold.
4. Profit is the excess of the Selling Price above the Cost.
5. Loss is the excess of the Cost above the Selling Price.
243. There are four cases of Profit or Loss, solved like
the four corresponding cases of Percentage.
The cost corresponds to the Base; the per cent of profit
or loss, to the Bate; the profit or loss, to the Percentage;
the cost plus the profit, or the selling price, to the Amount;
and the cost minus the loss, or the selling price, to the
Difference.
CASE I.
244. Given the cost and the rate, to find the profit
or loss.
Problem. — Having invested 84800,
my rate of profit is 1S% : what is my $4800
profit? .13
Solution.— Since the coRt is $4800, and 14400
the rate of profit is 13^; the profit is 13^ ^^^^
of $4800, which is $624. $ 6 2 4.0 Profit.
(199)
200 BAY'S HIGHER ARITHMETIC.
Examples for Practice.
1. If a man invests $1450 so as to gain 14^%, what is
his profit? $210.25
2. I bought $1760 worth of grain, and sold it so as to
make 26^% profit : what did I receive for it? $2222.
3. If a man invests $42540, and loses 11^ of his capital :
to what does his loss amount, and how much money has
he left ? $4963, loss ; $37577, left.
4. A man buys 576 sheep, at $10 a head. K his flock
increases 21^ per cent, and he sells it at the same rate per
head, how much money does he receive ? $7000.
5. The cost of publishing a book is 50 ct. a copy ; if the
expense of sale be 10% of this, and the profit 25% : what
does it sell for by the copy? 68f ct.
6. A began business with $5000: the 1st year he gained
14f %, which he added to his capital ; the 2d year he gained
8^, which he added to his capital; the 3d year he lost
12^, and quit: how much better off was he than when he
started? $452.92
7. A bought a farm of government land, at $1.25 an
acre; it cost him 160% to fence it, 160% to break it up,
80% for seed, 100% to plant it, 100^ to harvest it, 112%
for threshing, 100^ for transportation ; each acre produced
35 bu. of wheat, which he sold at 70 ct. a bushel: how
much did he gain on every acre above all expenses the first
year? $13.10
8. For what must I sell a horse, that cost me $150, to
gain 35% ? $202.50
9. Bought hams at 8 ct. a lb.; the wastage is 10% : how
must I sell them to gain 30^ ? llf ct. a lb.
10. I started in business with $10000, and gained 20%
the first year, and added it to what I had ; the 2d year I
gained 20%, and added it to my capital; the 3d year I
gained 20% : what had I then? $17280.
PROFIT AND LOSS. 201
11. I bought a cask of brandy, containing 46 gal., at
$2.50 per gal.; if 6 gal. leak out, how must I sell the rest,
so as to gain 25% ? • $3.59 per gal.
CASE II.
246. Given the cost and the profit or loss, to find
the rate.
Problem. — ^A man bought part of a mine for $45000,
and sold it for $165000 : how many per cent profit did he
make?
OPERATION.
$165000 — $45000 = $120000 profit.
100^o=$46000;
1^^=$450; and 1 20000f450=266}/c, Am.
Solution. — Here the profit is $120000, which, compared with
$45000, the cost, is WW = i that is, f of 100^^ = 266f ^c
Examples for Practice.
1. If I buy at $1 and sell at $4, how many per cent do I
gain? 300%.
2. If I buy at $4 and sell at $1, how many per cent do
Hose? 75%.
3. If I sell f of an article for what the whole cost me,
how many per cent do I gain ? 80^ .
4.* Paid $125 for a horse, and traded him for another,
giving 60^ additional money. For the second horse I re
ceived a third and $25 ; I then sold the third horse for $150 :
what was my per cent of profit or loss? 12^^ loss.
5.* A man bought a farm for $1635, which depreciated in
value 25%. Selling out, he invested the proceeds so as to
make 33^^ profit : what was his per cent of profit or loss on
the entire transaction ?
202 RAY'S HIGHER ARITHMETIC.
6. It cost me $1536 to raise my wheat crop : if I sell it
for $1728, what per cent profit is that per bushel? 12^%.
7. If I pay for a lb. of sugar, and get a lb. troy, what
^ do I lose, and what % does the grocer gain by the
cheat? ll\% loss; 21^% gain.
8. A, having failed, pays B $1750 instead of $2500,
which he owed him: what % does B lose? 30^.
9. An article has lost 20^ by wastage, and is sold for
40^ above cost: what is the gain per cent? 12%.
10. If my retail profit is 33 J^, and I sell at whole
sale for 10% less than at retail, what is my wholesale
profit? 20%,
11. Bought a lot of glass; lost 15% by breakage: at
what ^0 above cost must I sell the remainder, to clear 20^
on the whole ? 41^^ .
12. If a bushel of com is worth 35 ct. and makes 2^ gal.
of whisky, which sells at $1.14 a gal., what is the profit
of the distiller who pays a tax of 90 ct. a gallon ? ^t^% •
13. I had a horse worth $80 ; sold him for $90 ; bought
him back for $100: what % profit or loss? 12 J^ loss.
CASE III.
246. Given the profit or loss and the rate, to
find the cost.
Problem. — By selling a lot for 34f % more than I gave,
my profit is $423.50: what did it cost me?
Solution. — Since 34f ^ = $423.50, operation.
\(fo = $423.50 f 34f, = $12.32 ; and 3 4 J ^^ = $ 4 2 3.5 ;
100^0, or the whole cost, = 100 times 1 ^ = $ 1 2.3 2
$12.32=11232, as in Case III of 1 OO/o =$1 2 32, ^?i8.
Percentage.
Bemark. — After the cost is known, the profit or loss may be
added to it, or subtracted from it, to get the selling price (amount
or difference).
''\ xi r( A r: "y"
■^ OK THC
PROFIT AND LOI^, UNIVERSlT^foi
Examples for Practice.
1. How large sales must I make in a year, at a profit of
8^ , to clear $2000 ? $25000.
2. I lost $50 by selling sugar at 22^^ below cost : what
was the cost? $222. 22
3. If I sell tea at 13^% profit, I make 10 ct. a lb.: how
much a pound did I give? 75 ct.
4. I lost a 2 dollar gold coin, which was 1\% of all I
had: how much had I? $35.
5. A and B each lost $5, which was 2J^ of A's and
3^% of B's money: which had the most money, and how
much ? A had $30 more than B.
6. I gained this year $2400, which is 120^ of my gain
last year, and that is 44^ of my gain the year before :
what were my profits the two previous years ?
$2000 last year; $4500 year before.
7. The dogs killed 40 of my sheep, which was A\% of my
flock: how many had I left? 920 sheep.
CASE IV.
247. Given the selling price (amount or difference)
and the rate, to find the cost.
Problem. — Sold goods for $25.80, by which I gained
T\% • what was the cost?
Solution.— The cost is 100^ ; operation.
the $25.80 being l\(fo more, is 10 0^ = cost price ;
107i/o ; then, l<fo = $25.80 ^ 107 J 1 7 i /o = $ 2 5.8 ;
— 24 ct., and 100^, or the cost, = .*. 1^=24 ct.
100 times 24 ct. = $24; as in 100^o=$24, Ans,
Case rV of Percentage.
Remabk. — After the cost is found, the difference between it and
the selling price (amount or difference) will be the profit or loss.
204 RAY'S HIGHER ARITHMETia
Examples for Practice.
1. Sold cloth at $3.85 a yard; my profit was 10^ : how
much a yard did I pay? $3.50
2. Gold pens, sold at $5 apiece, yield a profit of 33^%:
what did they cost apiece? $3.75
3. Sold out for $952.82 and lost 12% : what was the cost,
and what would I have received if I had sold out at a profit
of \2% ? $1082.75, and $1212.68
4. Sold my horse at 40% profit; with the proceeds I
bought another, and sold him for $238, losing 20% : what
did each horse cost me ? $212.50 for 1st, $297.50 for 2d.
5. Sold flour at an advance of 13^^ ; invested the
proceeds in flour again, and sold this lot at a profit of
24^, realizing $3952.50: how much did each lot cost me?
1st lot, $2812.50; 2d lot, $3187.50
6. An invoice of goods purchased in New York, cost me
8% for transportation, and I sold them at a gain of 16f^
on their total cost on delivery, realizing $1260: at what
were they invoiced? $1000.
7. For 6 years my property increased each year, on the
previous, 100^, and became worth $100000: what was it
worth at first? $1562.50
IT. STOCKS AND BONDS.
DEFINITIONS.
248. 1. A Company is an association of persons united
for the transaction of business.
2. A company is called a Corporation when authorized
by law to transact business as one person.
Corporations are regulated by general laws or special
acts, called Charters.
STOCKS AND BONDS. 205
3. A Charter is the law which defines the powers, rights,
and legal obligations of a corporation.
4. Stock is the capital of the corporation invested in
business. Those owning the stock are Stockholders.
5. Stock is divided into Shares, usually of $50 or $100
each.
6. Scrip or Certificates of Stock are the papers issued
bj a corporation to the stockholders. Each stockholder is
entitled to certificates showing the number of shares that he
holds.
7. Stocks is a general term applied to bonds, state and
national, and to the certificates of stock belonging to
corporations.
8. A Bond is a written or printed obligation, under seal,
securing the payment of a certain sum of money at or
before a specified time.
Bonds bear a fixed rate of interest, which is usually payable
either annually or semiannually.
The principal classes of bonds are government, state, city,
county, and railroad.
9. An Assessment is a sum of money required of the
stockholders in proportion to their amounts of stock.
Bemark. — Usually, in the formation of a company, the stock
subscribed is not all paid for at once ; but assesamenta are made as the
needs of the business require. The stock is then said to be paid for
in installments. Other assessments may be made to meet losses or to
extend the business.
10. A Dividend is a sum of money to be paid to the
stockholders in proportion to their amounts of stock.
Remark. — The gross earnings of a company are its total receipts in
the transaction of the business ; the net earnings are what is left of the
receipts after deducting all expenses. The dividends are paid out
of the net earnings.
206 liA Y'S HIGHER ABITHMETIC.
249, Problems involving dividends and assessments give
rise to four cases, solved like the four corresponding cases
of Percentage.
The quantities involved are, the Stock, the Bate, and
the Dividend or Assessment.
The stock corresponds to the Base; the dividend or
assessment, to the Percentage; the stock plus the dividend,
to the Amount; and the stock minus the assessment, to
the Difference,
CASE I.
250. Given the stock and the rate, to find the
dividend or assessment.
FoBMULA. — Stock X Baie = Dividend or AsseasmenL
Examples for Practice.
1. I own 18 shares, of $50 each, in the City Insurance
Co., which haj3 declared a dividend of 7^% : what do I
receive? $67.50
2. I own 147 shares of railroad stock ($50 each), on
which I am entitled to a dividend of 5%, payable in stock:
how many additional shares do I receive ?
7 shares, and $17.50 toward another share.
3. The Western Stage Co. declares a dividend of 4J per
cent: if their whole stock is $150000, how much is dis
tributed to the stockholders ? $6750.
4. The Cincinnati Gas Co. declares a dividend of 18% :
what do I get on 50 shares ($100 each) ? $900.
5. A railroad company, whose stock account is $4256000,
declared a dividend of 3^^ : what sum was distributed
among the stockholders? $148960.
6. A telegraph company, with a capital of $75000, de
clares a dividend of 7%, and has $6500 surplus: what has
it earned? $11750.
STOCKS AND BONDS 207
7. I own 24 shares of stock ($25 each) in a fuel company,
which declares a dividend of 6% ; I take my dividend in
coal, at 8 ct. a bu. : how much do I get ? 450 bu.
CASE II.
251. Given the stock and dividend or assessment,
to find the rate.
~, Dividerid or Assessment ,
Formula. — ; = Rale.
Stock
Examples for Practice.
1. My dividend on 72 shares of bank stock ($50 each) is
$324: what was the rate of dividend? 9^.
2. A turnpike company, whose stock is $225000, earns
during the year $16384.50: what rate of dividend can it
declare? 7^, and $684.50 surplus.
3. The receipts of a certain canal company, whose stock
is $3650000, amount in one year to $256484; the out
lay is $79383: what rate of dividend can the company
declare? ^%y and $12851 surplus.
4. I own 500 shares ($100) in a stock company. K I
have to pay $250 on an assessment, what is the rate? ^%.
CASE III.
252. Given the dividend or assessment and the
rate, to find the stock.
_ Dividend or Assessment ^i ^i
Formula. = Stock.
Rate
Examples for Practice.
1. An insurance company earns $18000, and declares a
15^ dividend : what is its stock account? $120000.
208 RAY'S HIGHER ARITHMETIC.
2 A man gets $94.50 as a 7^ dividend: how many
shares of stock ($50 each) has he? 27 shares.
3 Received 5 shares ($50 each), and $26 of another
share, as an S% dividend on stock: how many shares
had I? 69 shares.
CASE IV.
253. Given the rate and the stock plus the divi
dend, or the stock minus the assessment, to find the
stock.
Stock f Div idend
Formulas. ~ /S^odk = J ^ 1 f i^ote.
Stock — As sessment
l — Bate.
Examples for Practice.
1. Received 10^ stock dividend, and then had 102 shares
($50 each), and $15 of another share: how many shares had
I before the dividend? 93 shares.
2. Having received two dividends in stock, one of 5%,
another of 8^, my stock has increased to 567 shares: how
many had I at first? 500 shares.
III. PREMIUM AND DISCOUNT.
254. 1. Premium, Discount, and Par are mercantile
terms applied to money, stocks, bonds, drafts, etc.
2. Drafts, Bills of Exchange, or Checks are written
orders for the payment of money at some definite place and
time.
3. The Par Value of money, stocks, drafts, etc., is the
nominal value on their face.
4. The Market Value is the sum for which they sell.
PREMIUM AND DISCOUNT. 209
5. Discount is the excess of the par value of money,
stocks, drafts, etc., over their market value.
6. Fremiuin is the excess of their market value over
their par value.
7. Bate of Premium or Bate of Discount is the rate
per cent the premium or discount is of the face.
255. Problems involving premium or discount give rise
to four cases, corresponding to those of Percentage.
The quantities involved are the Par Valvs, the Rate,
the Premium or Discount, and the Market Value,
The par value corresponds to the Base; the premium or
discount, to the Percentage ; and the market value, to the
Amourd or Difference,
CASE I.
256. Given the par value and the rate, to find the
premium or discount.
FOBMUIAS. — \ _ .
( Discount =
Par Value X i?«te.
Par Value X Rotie.
Note. — If the result is a premium, it must be added to the par to
get the market value ; if it is a discount, it must be mbtracted.
Examples for Practice.
1. Bought 54 shares of railroad stock ($100 each) at 4%
discount: find the discount and cost.
Dis., «216; cost, $5184,
2. Buy 18 shares of stock ($100 each) at 8^ discount:
find the discount and cost. $144, and $1656.
3. Sell the same at 4^^ premium : find the premium, the
price, and the gain. $81, $1881, and $225.
4. Bought 62 shares of railroad stock ($50 each) at 28^
premium : what did they cost? $3968.
H. A. 18.
210 BAY'S HIGHER ABITHMETIC.
5. What is the cost of 47 shares of railroad stock ($50
each) at 30% discount? $1645.
6. Bought $150 in gold, at f% premium: what is the
premium and cost? $112J, and $151. 12^
7. Sold a draft on New "Sork of $2568.45, at ^^ prem
ium : what do I get for it ? $2581.29
8. §old $425 uncurrent money, at 3% discount: what did
I get, and lose?  $412.25, and $12.75
9. What is a $5 note worth, at 6^ discount? $4.70
10. Exchanged 32 shares of bank stock ($50 each), 5^
premium, for 40 shares of railroad stock ($50 each), 10^ dis
count, and paid the difference in cash : what was it ? $120.
11. Bought 98 shares of stock ($50 each), at 15% dis
count ; gave in payment a bill of exchange on New Orleans
for $4000, at f % premium, and the balance in cash : how
much cash did I pay ? $140.
12. Bought 56 shares of turnpike stock ($50 each), at
69%; sold them at 76^% : what did I gain? $210.
13. Bought telegraph stock at 106%; sold it at 91%:
what was my loss on 84 shares ($50 each) ? $630.
14. What is the diflerence between a draft on Philadel
phia of $8651.40, at 1 J^ premium, and one on New Orleans
for the same amount, at ^% discount? $151.40
CASE II.
257. Qiven the face and the discount or premium,
to find the rate.
Formulas. — Bate= "
Discount or Premium
Par Vctlv£.
Difference between Ma rket and Par Value
Par Value.
\
Notes. — 1. If the par and the market value are known, take their
difference for the discount or premium.
2. If the rate of profit or loss is required, the market value or cost is
the standard of comparison, not the faxic.
PREMIUM AND DISCOUNT. 211
Examples for Pbactice.
1. Paid $2401.30 for a draft of $2360 on New York:
what was the rate of premium? li%«
2. Bought 112 shares of raibroad stock ($50 each) for
$3640 : what was the rate of discount ? 35^ .
3. If the stock in the last example yields 8^ dividend,
what is my rate of profit? 12j\%.
4. I sell the same stock for $5936 : what rate of premium
is that? what rate of profit? 6%; 63^^.
5. If I count my dividend as part of the profit, what is
my rate of profit ? <^^ A^ •
6. Exchanged 12 Ohio bonds ($1000 each), 1% premium,
for 280 shares of railroad stock ($50 each) : what rate . of
discount were the latter? 8^^.
7. Gave $266. 66 of notes, A% discount, for $250 of gold :
what rate of premium was the gold? 2%.
8. Bought 58 shares of mining stock ($50 each), at 40^
premium, and gave in payment a draft on Boston for $4000 :
what rate of premium was the draft? li%
9. Keceived $4.60 for an uncurrent $5 note: what was
the rate of discount? 8^.
10. Paid $2508.03 for 26 shares of stock ($100 each), and
brokerage, $25.03: what is the rate of discount? ^\%
CASE III. *
258. Given the discount or premium and the rate,
to find the face.
.Discount or Pi'emium
FoRMUiA. — Par Value =
Bate.
Notes. — 1. After the face is obtained, add to it the premium, or
subtract the discount, to get the market value or cost.
2. If the profit or loss is given, and the rate per cent of the face
corresponding to it, work by Case III, Percentage.
212 BA Y'S HIOHEB ARITHMETia
Examples fOr Practice.
1. Paid 36 ct. premium for gold \% above par: how
much gold was there? $48.
2. Took stock at par ; sold it for 2\fo discount, and lost
$117: how many shares ($50 each) had I? 104 shares.
3. The discount, at 7^^, on stocks, was $93.75: how
many shares ($50 each) were sold ? 25 shares.
4. Buy stock at 4^% premium; sell at 8J^ premium;
profit, $345 : how many shares ($100 each) ? 92 shares.
5. Buy stocks at 14^ discount; sell at 3^^ premium;
profit, $192.50: how many shares ($50 each)? 22 shares.
6. The premium on a draft, at ^^, was $10.36: what
was the face? $1184.
7. Buy stocks at 6% discount ; sell at 42^ discount ; loss,
$666 : how many shares ($50 each) ? 37 shares.
8. Bought stock at 10% discount, which rose to 5^
premium, and sold for cash; paying a debt of $33, I
invested the balance in stock at 2% premium," which, at
par, left me $11 less than at first: how much money had
I at first? $148.50
CASE IV.
259. Given the market value and the rate, to find
the par value.
Market Value
FOBMUI.AS.— Par Value =^ J 1 + ^'« «/ Premium.
' Market Value
1 — Rate of Discount.
Notes. — 1. After the face is known, take the difference between it
and the market value, to find the discount or premium.
2. Bear in mind that the rate of premium or discount, and the
rate of profit or loss, are entirely different things; the former is
referred to the par value or face, as a standard of comparison, the
latter to the market value or cost.
COMMISSION AND BROKERAGE. 213
Examples for Practice.
1. What is the face of a draft on Baltimore costing
$2861.45, at \\<fo premium? $2819.16
2. Invested $1591 in stocks, at 2^6^ discount ; how many
shares ($50 each) did I buy ? 43 shares.
3. Bought a draft on New Orleans, &t i% discount, for
$6398.30: what was its face? $6430.45
4. Notes at 65% discount, 2% brokerage, cost $881.79:
what is their &ce? $2470.
5. Exchanged 17 railroad bonds ($500 each) 25% below
par, for bank stock at 6:^% premium: how many shares
($100 each) did I get? 60 shares.
6. How much gold, at f % premium, will pay a check for
$7567 ? $7520.
7. How much silver, at 1\% discount, can be bought for
$3172.64 of currency ? $3212.80
8. How large a draft, at \^ premium, is worth 54 city
bonds ($100 each), at 12% discount? $4740.15
9. Exchanged 72 Ohio State bonds ($1000 each), at 6\fo
premium, for Indiana bonds ($500 each), at 2% premium :
how many of the latter did I get ? 150 bonds.
IV. COMMISSION AND BROKERAGE.
DEFINITIONS.
260. 1. A CommissionMerchant, Agent, or Factor
is a person who sells property, makes investments, collects
debts, or transacts other business for another.
2. The Principal is the person for whom the commission
merchant transacts the business.
8. Commission is the percentage paid to the commission
merchant for doing the business.
214 RAY'S mOHER ARITHMETIC.
4. A Consignment is a quantity of merchandise sent to
a commission merchant to be sold.
5. The person sending the merchandise is the Consignor
or Shipper, and the commission merchant is the Consignee.
When living at a distance from his principal, the consignee
is spoken of as the Correspondent.
6. The Net Proceeds is the sum left after all charges
have been paid.
7. A Guaranty is a promise to answer for the payment
of some debt, or the performance of some duty in the case
of the failure of another person, who, in the first instance,
is liable. Guaranties are of two kinds : of payment^ and of
collection.
8. In a Guaranty of Payment, the guarantor makes an
absolute agreement that the instrument shall be paid at
maturity.
9. The usual form of a guaranty, written on the back of a
note or bill, is:
*^For value received^ I her dry guaranty the payment of the
vrithin. John Saunders."
10. A Broker is a person who deals in money, bills of
credit, stocks, or real estate, etc.
11. The commission paid to a broker is called Brokerage.
261. Commission and Brokerage involve four cases,
corresponding to those of Percentage.
The quantities involved are the Arrwurd Bought or SM^
the Rate of Gommimon or Brokerage^ the Commission or Bro
kerage, and the Cost or Net Proceeds.
The amount of sale or purchase corresponds to the Base;
the commission or brokerage, to the Percentage; the cost, to
the Amount; and the net proceeds, to the Difference.
COMMISSION AND BROKERAOE. 215
CASE I.
262. Given the amount of sale, purchase, or col
lection and the rate, to find the commission.
FoBMUiiA. — Amount of Sale or Purchase X R<^ = Commission,
Examples for Practice.
1. I collect for A $268.40, and have 5% commission:
what does A get? $254.98
2. I sell for B 650 barrels of flour, at $7.50 a barrel, 28
barrels of whisky, 35 gal. each, at $1.25 a gal.: what is
my commission, at 2^%? $137.25
3. Received on commission 25 hhd. sugar (36547 lb.), of
which I sold 10 hhd. (16875 lb.), at 6 ct. a lb., and 6 hhd.
(8246 lb.) at 5 ct. a lb., and the rest at 5^ ct. a lb.: what
is my commission, at 3%? $61.60
4. A lawyer charged 8^ for collecting a note of $648.75:
what are his fee and the net proceeds?
$51.90, and $596.85
5. A lawyer, having a debt of $1346.50 to collect, com
promises by taking 80%, and charges 5% for his fee: what
are his fee, and the net proceeds? $53.86, and $1023.34
6. Bought for C, a carriage for $950, a pair of horses for
$575, and harness for $120; paid charges for keeping, pack
ing, shipping, etc., $18.25; freight, $36.50: what was my
commission, at 3^^ , and what was the whole amount of my
bill? $54.83, and $1754.58
7. An architect charges 3^^ for designing and superin
tending a building, which cost $27814.60: to what does his
fee amount? $973.51
8. A factor has 2^ commission, and 3^% for guar
antying payment: if the sales are. $6231.25, what does
he get? $389.45
216 RAY'S HIGHER ARITHMETIC,
9. Sold 500000 lb. of pork at 5^ ct. a lb.: what is my
commission at 1 J% ? $343.75
10. An architect charges \\% for plans and specifications,
and 2J^ for superintending: what does he make, if the
building costs $14902.50? ' $614.73
11. A sells a house and lot for me at $3850, and charges
\% brokerage: what is his fee? $24.06+
12. I have a lot of tobacco on commission, and sell it
through a broker for $4642.85: my commission is 2^%, the
brokerage IJ^^: what do I pay the broker, and what do I
keep? I keep $63.84; brokerage, $52.23
CASE II.
263. Given the commission and the amount of the
sale, purchase, or collection, to find the rate.
_ Commission _
FoBMULA.— —IT, ;; — ; — = Bo^e.
Amount of oale or riirchase
Examples for Practice.
1. An auctioneer's commission for selling a lot was $50,
and the sum paid the owner was $1200 : what was the rate
of commission ? ^ 4%.
2. A commissionmerchant sells 800 barrels of flour, at
$6.43f a barrel, and remits the net proceeds, $5021.25:
what is his rate of commission? ^i%'
3. The cost of a building was^l9017.92, including the
architect's commission, which was $553.92: what rate did
the architect charge? S%.
4. Bought flour for A; my whole bill was $5802.57, in
cluding charges, $76.85, and commission, $148.72: find the
rate of commission. 2f %.
5. Charged $52.50 for collecting a debt of $1050: wha(
was my rate of commission ? 5%^
COMMISSION AND BROKERAOK 217
6. An agent gets $169.20 for selling property for 88460:
what was his rate of brokerage? 2%.
7. My commission for selling books was $6.92, and the net
proceeds, $62.28: what rate did I charge? 10^.
8. Paid $38.40 for selling goods worth $6400: what was
the rate of brokerage? %.
9. Paid a broker $24.16, and retained as my part of the
commission $42.28, for selling a consignment at $2416: what
was the rate of brokerage, and my rate of commission?
Brok. 1%; com. 2f %•
CASE III.
264. Given the commission and rate, to find the
sum on which commission is charged.
FoRBiuui. = AmourU of Sale or Purchase.
Rale
Note. — After finding the sum on which commission is charged,
subtract the commission to find the net proceeds, or add it to find
the whole cost, as the case may be.
Examples for Practice.
1. My commissions in 1 year, at 2^%, are $3500: what
were the sales, and the whole net proceeds?
$140000, and $136500.
2. An insurance agent's income is $1733.45, being 10^
on the sums received for the company: what were the com
pan/s net receipts? $15601.05
3. A packinghouse charged 1^% commission, and cleared
52376.15, after paying out $1206.75 for all expenses of
packing : how many pounds of pork were packed, if it cost
4\ ct. a pound ? 5308000 lb.
4. Paid $64.05 for selling coffee, which was ^^ broker
age: what are the net proceeds? $7255.95
H. A. 19.
218 BAY'S mOHER ARITHMETIC.
5. An agent purchased, according to order, 10400 bushels
of wheat ; his commission, at l^'^o > was $156, and charges
for storage, shipping, and freight, $527.10: what did he pay
a bushel, and what was the whole cost?
$1.20 a bu., and $13163.10, whole cost.
6. Received produce on commission, at 2^%; my surplus
commission, after paying ^ brokerage, is $107.03: what
was the amount of the sale, the brokerage, and'^net pro
ceeds? Sale, $6116; brok., $30.58; pro., $5978.39
CASE IV.
265.* Given the rate of commission and the net
proceeds or the whole cost, to find the sum on which
commission is charged.
FOBMULAS. —
: Amovmi of Sale or Purchase.
(1 + Bate)
I Net Proceeds . ^ _ , « ,
7i dIZT ^^ Amount of Sale or Purchase.
Bale)
Note. — After the sum on which commission is charged is known,
find the commission by subtraction.
Examples for Praotice.
1. A lawyer collects a debt for a client, takes 4^ for his
fee, and remits the balance, $207.60: what was the debt
and the fee? $216.25, and $8.65
2. Sent $1000 to buy a carriage, commission 2^% : what
must the carriage cost? $975.61
Suggestion.— lOO^c + 2\fo — 102J^ = $1000 ; find Ifoy then 100^.
3. A buys per order a lot of coffee; charges, $56.85;
commission, 1\%; the whole cost is $539.61: what did the
coffee cost? $476.80
COMMISSION AND BROKERAGE. 219
4. Buy sugar at 2J^ commission, arid 2^% for guaran
teeing payment: if the whole cost is $1500, what was the
cost of the sugar? $1431.98
5. Sold 2000 hams (20672 lb.); commission, 2^^, guar
anty, 2^, net proceeds due consignor, $2448.34: what did
the hams sell for a lb.? 12^ ct.
6. Sold cotton on commission, at 5^; invested the net
proceeds in sugar; commission, 2%; my whole commission
was $210: what was the value of the cotton and sugar?
Cotton, $3060; sugar, $2850.
Suggestion. — Note carefully the different processes required here
for commission in buying and commission in selling.
7. Sold flour at 3^% commission; invested f of its
value in coffee, at 1^^ commission; remitted the balance,
$432.50 : what was the value of the flour, the coffee, and
my commissions ? Flour, $1500; coffee, $1000, 1st
com., $52.50, 2d com., $15.
8. Sold a consignment of pork, and invested the proceeds
in brandy, after deducting my commissions, 4% for selling,
and l\^o for buying. The brandy cost $2304.00 : what
did the pork sell for, and what were my commissions ?
Pork, $2430; 1st com., $97.20; 2d com., $28.80
9. Sold 1400 barrels of flour, at $6.20 a barrel; invested
the proceeds in sugar, as per order, reserving my commis
sions, A% for selling and 1^ for buying, and the ex
pense of shipping, $34.16: how much did I invest in
sugar? $8176.
10. An agent sold my com; and, after reserving his com
mission, invested all the proceeds in corn at the same price ;
his commission, buying and selling, was 3%, and his whole
charge $12 : for what was the corn first sold ? $206.
11. My agent sold my flour at 4% commission ; increas
ing the proceeds by $4.20, I ordered the purchase of wheat
at 2^ commission; after which, wheat declining 3J^, my
whole loss was $5: what was the flour worth? $53.
220 BAY'S HIGHER ARITHMETIC.
V. STOCK INVESTMENTS.
DEFINITIONS.
266. 1. A Stock Exchange is an association of brokers
and dealers in stocks, bonds, and other securities.
Remarks. — 1. The name "Stock Exchange" is also applied to
the building in which the association meets to transact business.
2. New York city is the commercial center of the United States,
and the transactions of the New York Stock Exchange, as tele
graphed throughout the country, determine the market value of
nearly all stocks sold.
2. United States Government Bonds are of two kinds, —
coupon and registered.
Kemarks. — 1. Coupon bonds may be transferred like banknotes :
the interest is represented by certificates, called coupons, printed at
the bottom of the bond, which may be presented for payment when
due.
2. Regialei'ed bonds are recorded in the name of the owner in the
U. S. Treasurer*s office, and the interest is sent directly to the owner.
Registered bonds must be indorsed, and the record must be changed,
to effect a transfer.
3. The United States also issues legal tender notes, known as
" Greenbacks," which are payable in coin on demand, and bear no
interest.
3. The various kinds of United States bonds are dis
tinguished, 1st. By the rate of interest; 2d. By the date
at which they are payable. Most of the bonds are payable
in coin ; a few are payable in currency.
Example. — Thus, "U. S. 4J*s, 1891," means bonds bearing in
terest at ^fc1 and payable in 1891. " U. S. cur. 6's, 1899," means
bonds bearing interest at 6^, and payable in currency in 1899.
Quotations of the principal bonds are given in the leading daily
papers.
4. Bonds are also issued by the several states, by cities
and towns, by counties, and by corporations.
STOCK INVESTMENTS. 221
Bemarks. — Legitimate stock transactions involve the following
terms and abbreviations, which need explanation :
1. A person who anticipates a decline, and contracts to deliver
stocks at a future day, at a fixed price which is lower than the
present market price, expecting to buy in the interval at a still
lower price, is said to sell short,
Sliort sales are also made for cash, deliverable on the same day,
or in the regvJar way, where the certificates are delivered the day
after the sale. In these cases the seller borrows the stock from a
third party, advancing security equivalent to the market price, and
waits a decline; he buys at what he considers the lowest point,
returns his borrowed stock, and reclaims his security.
2. A person who buys stock in anticipation of a rise is said to
buy long.
3. Those who sell short are interested, of course, in forcing
the market price down, and are called hears; while those who
buy long endeavor to force the market price up, and are known
as htdh.
4. The following are the principal abbreviations met with in
stock quotations: c, means coupon; r., registered; prefd, or pf., pre
Jerredf applied to stock which has advantages over common stock
of the same company in the way of dividends, etc.; xd., wiihovi
dividendj meaning that the buyer is not entitled to the dividend
about to be declared ; c, cash; s3, sSO, s60, seller's option., three, thirty,
or sixty days, as the case may be, means that the seller has the privi
lege of closing the transaction at any time within the specified
limit; bS, b30, b60, buyer'' s option, three ctoys, etc., giving the buyer
the privilege; be., between calls, means that the price was fixed
between the calls of the whole list of stocks, which takes place in
the New York Exchange twice a day ; opg., for deliveiy at the opening
of the books of transfer'.
6. The usual rate of brokerage is ifo on the par value of the
stock, either for a purchase or a sale.
267. The quantities involved in problems in stock in
vestments are: the Amount Invested, the Market Value, the
Rate of Dividend or Interest paid on the par value of the
stock, the Rate of Inemne on the amount invested, and the
Income itself.
These quantities give rise to five eases, all of which may
be solved by the principles of Percentage.
.222 MAY'S HIGHER ARITHMETia
NOTATION.
268. The following notation may' be adopted to ad
vantage in the formulas :
Amount Invested = A. I.
Market Value = M. V.
Rate of Dividend or Interest = R D.
Rate of Income = R. I.
Income = I.
CASE I.
269. Given the amount invested, the market value,
and the rate of dividend or interest, to find the in
come.
A. I.
Formula.— rrr X ^ D. = ^
M. V.
Problem. — A person invests $5652.50 in Mutual In
surance Company stock at 95 cents : what will be his
income if .the stock pay 10% dividend annually?
OPERATION.
$5652.50^.95=$59 5 0^ par value of stock purchased.
$5950X.1=$595= income.
Or, 9 5 ^c = J? of the par value = $5652.50
h<fo = ^ " •* " " =$ 297.50
10% = A " " " " =$ 595, ^ns.
Solution. — As many dollars* worth of stock can be bought as
$.95 is contained times in $5652.50, which iri 5950 times. Therefore
$5950 is the par value of the stock purchased ; and 10^ dividend on
$5950 is $595, the income on the investment.
Examples for Practice.
1. A invests $28000 in Lake Shore Kailroad stock, at
70^. If the stock yields 8^ annually, what is the amount
of his income? $5200.
STOCK INVESTMENTS, 223
2. B invests $100962 in U. S. cur. 6's, 1899, at 106^%.
If gold is at I premium, what does the government save
by paying him his interest in greenbacks? $7.11
3. If I invest $10200 in Tennessee 6's, new, at 30^,
what is my annual income ? $2040.
4. A broker invested $36000 in quicksilver preferred
stock, at 40^ : if the stock pays 4^, what is the income
derived? $3600.
5. Which is the better investment, stock paying 6^^
dividend, at a market value of 106^%, or stock paying 4^%
dividend, at 104^% ? The former, lffff^%.
6. Which is the more profitable, to invest $10000 in 6%
stock purchased at 75%, or in 5^ stock purchased at 60%,
allowing brokerage ^%? 5^ stock is $31.74+ better.
CASE II.
270. Given the amount invested, the market valuer
and the income, to find the rate of dividend or in
terest.
A. I.
Formula. — I. ^ ^/ ' = R. D.
M. V.
Problem. — Invested $10132.50 in railroad stock at 105%,
which pays me annually $965 : what is the rate of dividend
on the stock?
Solution.— By Art. 259, $10132.50 ^ 1.05 = $9650 = par value
of the stock ; and by Art. 251, $965 4 $9650 = .1, or 10^, Am,
Examples for Practice.
1. A has a farm, valued at $46000, which pays him 5%
on the investment. Through a broker, who charges $56.50
for his services, he exchanges it for insurance stock at 9^
premium, and this increases his annual income by $1072:
what dividend does the stock pay? 8%.
224 RAY'S HIGHER ARTTHMETTC.
2. What dividend must stock pay, in order that my rate
of income on an investment of $64968.75 shall be 4^%,
provided the stock can be bought at 103^% ? 4\%,
3. My investment was $9850, my income is $500, and
the market value of the stock 108^^%, brokerage \% : what
is the rate of dividend ? 5^^ .
t 4. The sale of my farm cost me $500, but I gave the
proceeds to a broker, allowing him ^%, to purchase railroad
stock then in market at 102% ; the farm paid a 5^ income,
equal to $2075, but the stock will pay $2025 more : what is
the rate of dividend? * 10^%.
5. Howard has at order $122400, and can allow broker
age ^%,and buy insurance stock at 101^, yielding 4^%;
but if he send to the broker $100 more for investment, and
buy rollingmill stock at 103^^ , the income will only be half
80 large : what rate does the higher stock pay ? 2^^ .
CASE III.
271. Oiven the income* rate of dividend, and mar
ket value, to find the amount invested.
FORMUIiA. — =r^ X M. V. — A. I.
Problem. — If U. S. bonds,. pa3dng 5% interest, are sell
ing at 108^%, how much must be invested to secure an
annual income of $2000 ?
OPERATION.
$2000^$. 05 = 40000; $40000 X 1.085 = $43400, Ans.
Solutions. — 1. To produce an income of $2000 it will require as
many dollar's worth of stock at par as 5 ct. is contained times in
$2000, which is 40000 ; and, at lOSJ^c, it will require $40000 X 1.08^
= $43400.
2. $1 of stock will give 5 cents income, and $2000 income will
require $40000 worth of stock at par. $40000 of stock, at lOSJ^,
will cost $40000 X 1.085 = $43400.
STOCK INVESTMENTS. 225
Examples for Practice.
1. What amount is invested by A, whose canal stock,
yielding 4^ , brings an income of $300, but sells in market
for n% ? «6900.
2. If I invest all my money in 5^ furnace stock, salable
at 75%, my income will be $180: how much must I borrow
to make an investment in 6^ state stock, selling at 102^ ,
to have that income? $360.
3. If railroad stock be yielding 6^, and is 20% below par,
how much would have to be invested to bring an income
of $390? $5200.
4. A banker owns 2^% stocks, at 10% below par, and
3% stocks, at 15% below par. The income from the former
is 66^ more than from the ^atter, and the investment in
the latter is $11400 less thi»,ii in the former: required the
whole investment Mid income. $31800, and $960.
5. Howard M. Holden sold $21600 U. S. 4's, 1907, reg
istered, at 99f^, and immediately invested a sufficient
amount of the proceeds in Illinois Central Railroad stock,
at 80%, which pays an annual dividend of 6^; he receives
$840 from the railroad investment ; with the remainder of
his money he bought a farm at $30 an acre: required the
amount invested in railroad stock, and the number of acres
in the farm? $11200 in R. R. stock; 342^ acres.
6. W. T. Baird, through his broker, invested a certain
sum of money in Philadelphia 6's, at 115i^, and three
times as much in Union Pacific 7's, at 89%, brokerage
^ % in both cases : how much was invested in each kind of
stock if his annual income is $9920?
$34800 in Phila. 6'8 ; $104400 in U. P. 7's.
7. Thomas Reed, bought 6% mining stock at 114^^, and
4^ furnace stock at 112% , brokerage ^^; the latter cost him
$430 more than the former, but yielded the same income:
what did each cost him? Mining, $920; furnace, $1350.
226 HA Y' S HIQHEB ARITHMETIC,
CASE IV.
272. Oiyen the market value and the rate of divi
dend or interest, to find the rate of income.
R. D. „ ^
Formula. — rr^7= R. I.
M. V.
Problem. — What per cent of his money will a man
realize in bu3dng ^% stock at 80%?
OPERATION.
$.0 6^$.8 = /^ = 5% = 7J/c.
Solution. — ^The expenditure of 80 ct. buys a dollar's worth of
Btock, giving an income of 6 ct. The rate per cent of income on
investment is $.06 4 $.80 = .07 J, or 7J^o.
Examples for Practice.
1. What is the rate of income on Pacific Mail 6's, bought
at 30^? 20%.
2. What is the rate of income on Union Pacific 6's, bought
at 110%? ^hfor
3. Which is the better investment: U. S. new 4's, regis
tered, at 99f ^ ; or U. S. new 4^'s, coupons, at 106^ ?
The latter is ^^f, better.
4. Thomas Sparkler has an opportunity of investing
$30000 in Northwestern preferred stock, at 76%, which
pays an annual premium of 5^; in Panama stock, at 125^,
which pays a premium annually of 8^%; or he can lend
his money, on safe security, at 6^% per annum. Prove
which is the best investment for Mr. Sparkler.
The Panama stock.
*
5. Thomas Jackson bought 500 shares of Adams Express
stock, at 105^^, and paid \\fo brokerage : what is the rate
of income on his investment per annum if the annual divi
dend is 8%? 7^^%.
STOCK INVESTMENTS. 227
CASE V.
273. Giyen the rate of income and the rate of
dividend or interest, to find the market value.
Formula. — ' = M. V.
XV. I.
Problem.— What must I pay for Lake Shore 6's ($100 a
share), that the investment may yield 10^ ?
OPERATION.
$.06j.l=3% = 60^o = $60 ashare.
Solution.— If bought for $100, or par, it will yield 6^ ; to yield
\(fo it must be bought for $100 X 6 = $600 ; to yield 10^ it must be
bought for T^ of $600 = $60.
Examples for Practice.
1. What must I pay for Chicago, Burlington & Quincy
Railroad stock that bears 6^, that my annual income on
the investment may yield 5^ ? 120^.
2. Which is the best permanent investment : 4's at 70^ ,
5's at 80%, 6's at 90%, or lO's at 120^ ? Why?
3. *rhe rate of income being 7% on the investment, and
the dividend rate 4%, what is the market value of $3430 of
the stock? $1960.
4. In a mutual insurance company one capitalist has an
investment paying 8^: what is the premium on the stock,
the dividend being 9^ ? 12^.
5. Suppose 10% state stock 20% better in market than
4% railroad stock; if A's income be $500 from each, how
much money has he 'paid for each, the whole investment
bringing 6^% ? $11250, railroad ; $5400, state.
6. At what figure must be government 5 per cent's to
make my purchase pay 9%? 55f^.
228 BA Y' S. HIOHER ARITHMETIC.
VL INSURANCE.
DEFINITIONS.
274. 1. Insurance is indemnity against loss or damage.
2. There are two kinds of insurance, viz.: Property In
surance and Personal Insurance.
3. Under Property Insurance the two most important
divisions are : Fire Insurance and Marine Insurance.
4. Fire Insurance is indemnity against loss by fire.
5. Marine Insurance is indemnity against the dangers
of navigation.
Notes. — 1. Transit Insurance is applied to risks which are taken
when property is transferred by railroad, or by railroad and water
routes combined.
2. There are several minor forms of property insurance, also,
such as Live Stock Insurance, Steam Boiler Insurance, Plate Glass In
surance, etc., the special purposes of which are indicated by their
names.
6. Personal Insurance is of three kinds, viz.: Life In
surance. Accident Insurance, and Health Insurance. Personal
insurance will be discussed in another chapter.
7. The Insurer or Underwriter is the party or company
that undertakes to pay in case of loss.
8. The Risk is the particular danger against which the
insurer undertakes.
9. The Insured is the party protected against loss.
10. A Contract is an agreement between two or more
competent parties, based on a sufficient consideration, each
promising to do or not to do some particular thing possi
ble to be done, which thing is not enjoined nor prohibited
by law.
INSURANCE. 229
11. The Primary Elements of a contract are: the
Parties, the Consideration^ the Subject Matter, the Consent
of the Parties, and the Tbne.
12. The written contract between the two parties in in
surance is called a Policy.
13. The Premium is the sum paid for insurance. It is
a certain per cent of the amount insured.
14. The Rate varies with the nature of the risk.
15. The Amount or Valuation is the sum for which the
premium is paid.
Notes. — 1. Whoever owns or has an interest in property, may
insure it to the full amount of his interest or liability.
2. Only the (ictual loss can be recovered by the insured, whether
there be one or several insurers.
3. Usually property is insured for about two thirds of its value.
16. Insurance business is usually transacted by incorpor
ated companies.
17. These companies are either jointstock companies, or
mutual companies.
Remakks. — 1. In jointstock companies the capital is owned by
individuals who are the stockholders. They share the profits and
losses.
2. In mutual companies the profits and losses are divided among
the insured.
276. The operations in insurance are included under the
principles of Percentage.
The quantities involved are, the Amount insured, the
Per Cent of premium, and the Premium,
The amount corresponds to the Base, and the premium, to
the Percentage.
CASE I.
276. Given the rate of insurance and the amoun
insured, to find the premium.
FomnTLA.— Amount Insured X Ka^ = Premium,
1^
230 J^A Y'S HIQHER ARITHMETIC.
Examples for Practice.
1. Insured f of a vessel worth $24000, and f of its cargo
worth $36000, the former at 2^%, the latter at 1J%: what
is the premium ? $607.50
2. Insured a house for $2500, and furniture for $600, at
3^^: what is the premium ? $18.60
3. What is the premium on a cargo of railroad iron worth
$28000, at 1%? $490.
4. Insured goods invoiced at $32760, for three months, at
^%: what is the premium? $262.08
5. My house is permanently insured for $1800, by a de
posit of 10 annual premiums, the rate per year being ^:
how much did I deposit, and if, on terminating the insur
ance, I receive my deposit less 5%, how much do I get?
$135 deposited; $128.25 received.
6. A shipment^ of pork, costing $1275, is insured at
4^, the policy costing 75 cents: what does the insurance
cost? $7.83
7. An insurance company having a risk of $25000, at
•j^^, reinsured $10000, at ^, with another office, and
$5000, at 1^0 y with another: how much premium did it
clear above what it paid ? $95.
CASE II.
277. Oiyen the amount insured and premium, to
find the rate of insurance.
^ Premium r, .
Formula. ^ = Raie,
Amount Insured
Examples for Practice.
1. Paid $19.20 for insuring  of a house, worth $4800
what was the rate? f^
INSURANCE. 231
2. Paid $234, including cost of policy, $1.50, for insuring
a cargo worth $18600: what was the rate? li%'
3. Bought books in England for $2468 ; insured them for
the voyage for $46.92, including the cost of the policy, $2.50:
what was the rate? ^\%
4. A vessel is insured for $42000; $18000 at 2^,
$15000 at 3f%, and the rest at 4%: what is the rate on
the whole $42000 ? ^%.
5. I took a risk of $45000; reinsured at the same
rate, $10000 each, in three offices, and $5000 in another;
my share of the premium was $262.50: what was the
rate? 2%.
6. I took a risk at l^fo ; reinsured  of it at 2^ , and
\ of it at 2^% : what rate of insurance do I get on what
is left? ^%.
CASE III.
278. Oiven the premium and rate of insurance, to
find the amount insured.
_, PreTniwm, i . r ■»
Formula. — = Amou7U Insureds
Bate,
Examples FOR Practice.
1. Paid $118 for insuring, at f^ : what was the amount
insured ? $14750.
2. Paid $411.37^ for insuring goods, at 1^^ : what was
their value? $27425.
3. Paid $42.30 for insuring f of my house, at ^^% : what
is the house worth ? $7520.
4. Took a risk at 2\% ; reinsured f of it at 2^%;
my share of the premium was $197.13: how large was the
risk ? $26284.
5. Took a risk at If %; reinsured half of it at the same
rate, and ^ of it at 1^^ ; my share of the premium was
$58.11 : how large was the risk? $19370.
232 BA Y'S HIQHER ARITHMETIC,
6. Took a risk at 2%; reinsured $10000 of it at 1\%,
and $8000 at ^\%\ niy share of the premium was $207.50:
what sum was insured ? $28000.
7. The Mutual Fire Insurance Company insured a build
ing and its stock for  of its value, charging If^. The
Union Insurance Company relieved them of \ of the risk,
at 1^%. The building and stock being destroyed by fire,
the Union lost fortynine thousand dollars less than the
Mutual: what amount of money did the owners of the
building and stock lose? $51750.
VII. TAXES.
DEFINITIONS.
270. 1. A Tax is a sum of money levied on persons, or
on persons and property, for public use.
2. Taxes in this country are: (1.) /Stofe and Local Taxes;
(2.) Taoces for the National GovemTnent
3. Taxes are further classified as Direct and Indirect.
4. A Direct Tax is one which is levied on the person or
property of the individual.
5. An Indirect Tax is a tax levied on articles of con
sumption, for which each person pays in proportion to the
quantity or number of such articles consumed.
6. A PoUTax, or Capitation Tax, is a direct tax levied
on each male citizen liable to taxation.
7. A Property Tax is a direct tax levied on property.
Note. — In legal works properly is treated of under two heads ;
viz , Real Property^ or Real Estate^ including houses and lands ; and
Personal Property, including money, bonds, cattle, horses, furniture, —
in short, all kinds of movable property..
TAXES. 233
8. State and Local Taxes are generally direct^ while the
United States Taxes are indirect,
9. An Assessor is a public officer elected or appointed to
prepare the Assessment RoU.
10. An Assessment Boll is a list of the names of the
taxable inhabitants living in the district assessed, and the
valuation of each one's property.
11. The Collector is the public officer who receives the
taxes.
Note. — In some states all the taxes are collected by the counties ;
in others, the towns collect; while in others the collections are
made by separate collectors. Generally a number of different taxes
are aggregated, — such as state, county, road, school, etc.
280. The quantities involved in problems under tax
ation are, the Assessed Value of the property, the Rate of
Taocatioriy the Taa, and the Amount Left after taxation.
They require the application of the four cases of Percent
age, the assessed value corresponding to the Base; the tax,
to the Percentage; and the amount left after taxation, to
the Difference,
CASE I.
281. Given the taxable property and the rate, to
find the propertytax.
Formula. — Taxable Property X Ka^ = Amount of Tax,
Note.— If there be a polltax, the sum produced by it should be
added to the propertytax, to give the whole tax.
Examples for Practice.
1. The taxable property of a county is $486250, and the
rate of taxation is 78 ct. on $100 ; that is, ^% : what is
the tax to be raised ? $3792.75
H. A. 20.
234
BAY'S HIQHEJR ABITHMETIC,
Kemark. — The rate of taxation being usually small, is expressed
most conveniently as so many cents on $100, or as so many mills
on $1.
2. A's property is assessed at $3800 ; the rate of taxation
is 96 ct. on $100 {^^fo)  what is his whole tax, if he
pays a polltax of $1 ? $37.48
Remarks.^— 1. In making out bills for taxes, a table is used, con
taining the units, tens, hundreds, thousands, etc., of property, with
the corresponding tax opposite each.
2. To find the tax on any sum by the table, take out the tax on
each figure of the sum, and add the results. In this table, the rate
is l\<fo, or 125 ct on $100.
Tax Table.
Prop.
Tax.
Prop.
Tax.
Prop.
Tax.
Prop.
Tax.
Prop.
Tax.
$1
.0125
$10
.125
$100
$1.25
$1000
$12.50
$10000
$125
2
.025
20
.25
200
2.50
2000
25.
20000
250
3
.0375
30
.375
300
3.75
3000
37.50
30000
375
4
.05
40
.50
400
5.
4000
50.
40000
500
5
.0625
50
.625
500
6.25
5000
62.50
50000
625
6
.075
60
.75
600
7.50
6000
75.
60000
750
7
.0875
70
.875
700
8.75
7000
87.50
70000
875
8
.10
80
1.
800
10.
8000
100.
80000
1000
9
.1125
90
1.125
900
11.25
9000
112.50
90000
1125
3. What will be the tax by the table, on property assessed
at $25349 ?
Solution.— The tax for $20000 is $250; for $5000 is 62.50; for
$300 is $3.75; for $40 is .50; for $9 is .1125; which, added, give
$316.86, the tax on $25349.
4. Find the tax for $6815.30 $85.19
5. What is the tax on property assessed at $10424.50,
and two polls, at $1.50 each? $133.31
6. A's property is assessed at $251350, and B's at $25135.
What is the difference in their taxes? $2827.69—
TA2LES. 235
CASE II.
282. Given the taxable property and the tax, to
find the rate.
Tax ^
FoEMUIiA.— 7= 77— =r = Bxiie.
laxable Froperly
Examples for Practice.
1. Property assessed at $2604, pays 819.53 tax : what is
the rate of taxation ? % — 75 ct. on $100.
2. The taxable property in a town of 1742 polls, is
$6814320; a tax of $66913 is proposed: if a polltax of
$1.25 is levied, what should be the rate of taxation?
395^% = 95 ct. on $100.
3. An estate of $350000 pays a tax of $5670 : what is the
rate of taxation ? lf^% = $1.62 on $100.
4. A's tax is $50.46 ; he pays a polltax of $1.50,
and owns $8704 taxable property ; what is the rate of tax
ation? ^fo = 56i ct. on $100.
CASE III.
283. Given the tax and the rate, to find the assessed
value of the property.
Formula. —  — = Taxable Pioperty.
Bdte
Note.— If any part of the tax arises from polls, it should be first
deducted from the given tax.
Examples for Practice.
1. What is the assessed value of property taxed $66.96,
at If % ? $3720.
236 RAY'S HIGHER ARITHMETIC.
2. A corporation pays $564.42 tax, at the rate of x*oV^,
or 46 ct. on $100 : find its capital. «122700.
3. A is taxed $71.61 more than B ; the rate is 1^%, or
$1.32 on $100: how much is A assessed more than B?
$5425.
4. A tax of $4000 is raised in a town containing 1024
polls, by a polltax of $1, and a propertytax of ^^^0 (24
ct. on $100) : what is the value of the taxable property in
it ? $1240000.
5. A's income is 16% of his capital; he is taxed 2^% of
his income,. and pays $26.04: what is his capital ? $6510.
CASE IV.
284. Given the amount left after payment of tax
and the rate, to find the assessed value of the prop
erty.
T, AmouTvt Left after Payment m n tl
FoBMUiiA. — = Taxable Property,
1 — Bate (fo.
Examples for Practice.
1. A pays a tax of l^V^ ($1.35 on $100) on his capital,
and has left $125127.66: what was his capital, and his
tax? Capital, $126840; tax, $1712.34
2. Sold a lot for $7599, which covered its cost and 2^
beside, paid for tax : what was the cost ? $7450.
VIII. UNITED STATES REVENUE.
DEFINITIONS.
285. 1. The Ee venue of the United States arises from
the Internal Revenue, from the OustomSy and from Sales of
PiMic Lands.
UNITED STATES BEVENUE. 237
2. The Internal Revenue is derived from taxes on spirits,
tobacco, fermented liquors, banks, and from the sale of
stamps, etc.
3. The Customs or Duties are taxes imposed by Gov
ernment on imported goods. They are of two kinds, —
ad valorem and specific,
4. Ad Valorem Duties are levied at a certain per cent
on the cost of the goods as shown by the invoice.
5. Specific Duties are certain sums collected on each
gallon, bushel, yard, ton, or pound, whatever may be the
cost of the article.
Note. — Problems involving specific duty only are not solved, of
course, by the principles of Percentage.
6. An Invoice is a detailed statement of the quantities
and prices of goods purchased.
7. A Tariff is a schedule of the rates of duties, as fixed
by law.
Notes. — 1. The collection of duties is made at the customhouses
established at ports of entry and ports of delivery. The principal
customhouse officers are collectors, naval officers, surveyors, and
appraisers.
2. Ihre is an allowance for the weight of whatever contains the
goods. Duty is collected only on the quantities passed through
the customhouse. The ton, for customhouse purposes, consists of
20 cwt., of 112 lb. each.
3. On some classes of goods, both specific and ad valorem duties
are collected. The cost price, if given in foreign money, must be
changed to United States currency,
286. Problems in United States Customs, where the
duty is wholly or in part ad valorem, are solved by the
principles of Percentage.
The quantities involved are: the Invoice Price corre
sponding to the Base; the Duty, corresponding to the Per
centage ; and the Total Cost of the importation, corresponding
to the Amount.
238 RAY'S HIGHER ARITHMETIC.
CASE I.
287. Given the invoice price and the rate, to find
the duty.
FoBMUiiA. — Invoke Price ^(^ Rale = Duty.
Examples for Practice.
1. Import 24 trunks, at $5.65 each, and 3 doz. leather
satchels, at $2.25 each; the rate is 35^ ad valorem: what
is the duty? $75.81
2. What is the duty on 45 casks of wine, of 36 gal. each,
invoiced at $1.25 a gal., at 40 ct. a gal. specific duty ? $648.
3. There is a specific duty of $3 per gallon, and an ad
valorem duty of 50% on colognewater: what is the total
amount of duty paid on 25 gallons, invoiced at $16.50 per
gallon? $281.25
4. What is the total duty on 36 boxes of sugar, each
weighing 6 cwt. 2 qr. 18 lb., invoiced at 2^ ct. per lb., the
specific duty being 2 ct. per lb., and the ad valorem duty
25^? $r04.97
5. What is the duty on 575 yards of broadcloth, weighing
1154 lb., invoiced at $2.56 per yd., the specific duty being
50 ct. per lb. , and the ad valorem duty 35^ ? If the freight,
charges, and losses amount to $160.80, how much a yard
must I charge to gain 15% ?
Duty, $1092.20; price per yard, $5.45
6. A merchant imported a ton of manilla, invoiced at
5 ct. per lb.; he paid a specific duty per ton, which, on this
shipment, was equivalent to an ad valorem duty of 22^^ :
what is the specific duty ? $25 per ton.
7. Received a shipment of 3724 lb. of wool, invoiced at
23 ct. per lb.; the duty is 10 ct. T)er lb., and 11^ ad valorem,
less 10^: what is the total amount of duty? $419.96
UNITED STATES REVENUE. 239
8. A drygoods merchant imports 1120 yards of dress
goods, \\ yd. wide, invoiced at 23 ct. a sq. yd.; there is a
specific duty of 8 ct. per sq. yd., and an ad valorem duty
of 40^: what must he charge per yard, cloth measure, to
clear 25^ on the whole? $.62f per yd.
CASE II.
288. Given the invoice price and the duty^ to find
the rate.
IhUy _
FoBMUiA.—  ; — : — frr = Bate,
Invoice Pnce
Examples for Practice.
1. If goods invoiced at $3684.50 pay a duty of $1473.80,
what is the rate of duty? 40%.
2. If laces invoiced at $7618.75, cost, when landed,
$10285.31J, what is the rate of duty? 35^.
3. Forty hhd. (63 gal. each) of molasses, invoiced at 52
ct. a gallon, pay $453.60 duty. The specific duty is 5 ct. a
gallon: what is the additional ad valorem duty? 25%.
CASE III.
289. Given the duty and the rate, to find the in
voice price.
Formula. — ^ = Invoice Price,
Bate
Examples for Practice.
1. Paid 8575.80 duty on watches, at 25%: at what were
they invoiced, and what did they cost me in store?
Invoiced, $2303.20; cost, $2879.
2. The duty on 1800 yards of silk was $2970, at 60%
240 BA Y'S HIGHER ABITHMETIC.
%
ad valorem : what was the invoice price per yard, and what
must I charge per yard to clear 20% ?
Invoiced at $2.75; sell at 85.28 per yd.
3. The duty on 15 gross, qt. bottles of porter, at a tax of
35 ct. a gallon, was $151.20: if this were equivalent to an
ad valorem duty of 20^% on the entire purchase, how
many bottles were allowed for a gallon, and at how much
per bottle must the whole be sold to clear 20% ?
5 bottles to gal. ; 50 ct. per bottle.
CASE IV.
* 290. Given the entire cost and the rate, to find the
invoice price.
^ Whofe Cost y. . „ .
Formula. ==:Invovce Frice.
l + BcUe
Examples for Practice.
1. 1000 boxes (100 each) pf cigars, weighing 1200 lb., net,
cost in store $13675. There is a specific duty of $2.50 per
lb., an ad valorem duty of 25%, and an internal revenue
tax of 60 ct. a box : freight and charges amount to $75 ;
find the invoice price per thousand cigars. $80.
2. Supposing No. 1 pigiron, American manufacture, to be
of equal quality with Scotch pigiron : at what price must
the latter be invoiced, to compete in our markets, if Amer
ican iron sells for $45 a ton ; freight and charges amounting
to $10 a ton, and the specific duty being equivalent, in this
instance, to an ad valorem duty of 25% ? $28 a ton.
3. A marblecutter imports a block of marble 6 ft. 6 in.
long, 3 ft. wide, 2 ft. 9f in. thick ; the whole cost to him
being $130 ; he pays a specific duty of 50 ct. per cu. ft. and
an ad valorem duty of 20^ : freight and charges being
$20.80, what was the invoice price per cu. ft.? $1.25
TOPICAL OUTLINE.
241
Topical Outline.
Applications op Percentage.
(WUhmt Time.)
1. Profit and Loss.
2. Stocks and Bonds..
3. Premium and Discount.
4. Commission and Brokerage.
1. Definitions :— Cost, Selling Price, Profit,
Loss.
2. Four Cases.
1. Definitions :— Company, Corporation, Char
ter, Stock, Shares, Scrip, Bond, Assess
ment, Dividend.
2. Four Cases.
1. Definitions :— Drafts, Par Value, Market
Value, Discount, Premium, Rate.
2. Four Cases.
•
1. Definitions:— CommissionMerchant, Prin
cipal, Commission, Consignment, Con
signor, Consignee, Correspondent, Net
Proceeds, Guaranty, Broker, Brokerage.
2. Four Cases.
5. Stock Investments •<
6. Insurance ^
7. Taxes «
1. Definitions :— Stock Exchange, Government
Bonds, State Bonds, etc.
2. Notation.
3. Five Cases.
1. Definitions:— Fire Insurance, Marine In
surance, Personal Insurance, Insurer,
Risk, Insured, Contract, Primary Ele
ments, Policy, Premium, Rate, Amount,
JointStock and Mutual Companies.
2. Three Cases.
1. Definitions :— Tax, State and Government,
Direct and Indirect, Polltax, Property
tax, Assessor, Assessment Roll, Collector.
2. Four Cases.
8. United States Revenue.
1. Definitions:— Internal Revenue, Customs,
Ad Valorem Duties, Specific Duties, In
voice, Tariff, Tare.
2. Four Cases.
H. A. 21.
XYI. PEEOEETAGE APPLIOATIOI^S.
I. INTEEEST.
DEFINITIONS.
291. 1. Interest is money charged for the use of money.
Note. — The profits accruing at regular periods on permanent,
investments, such as dividends or rents, are called interest, since
they are the increase of capital, unaided by labor.
2. The Principal is the sum of money on which interest
is charged.
Note. — The principal is either a sum loaned ; money invested to
secure an income ; or a debt, which not being paid when due, is
allowed by agreement or by law to draw interest.
3. The Rate of Interest is the number of per cent the
yearly interest is of the principal.
4. The AiAount is the sum of the principal and interest.
5. Interest is Payable at regular intervals, yearly, Judf
yearly, or quarterly , as may be agreed : if there is no agree
ment, it is understood to be yearly.
Note. — If interest is payable halfyearly, or quarterly, the rate
is still the rate per annumy or rate per year. In short loans, the rate
per month is generally given ; but the rate per year, being 12 times
the rate per month, is easily found ; thus, 2^ a month = 24^ a
year.
6. The Legal Rate is the highest rate allowed by the
law.
7. If interest be charged at a rate higher than the law
allows, it is called Usury; and, in some states, the person
offending is subject to a penalty.
(242)
Rehake:. — The per cent of interest that ia legal in the different
Btales and territories, is exhibited in the following table.
Alabama
Ariaona 
Arkansa*
California
Canada
Colorado 
Connecticut
Dakota 
Delaware^
District Columbia.
Florida
Georgia
Illinois
Indiaaa
Kauatis
Keutuckjr
Louisiana .*.
Maine
Maryland
Massachuaetts
Michigan
Minnesota
Mississippi
Missouri
Montana
Nebraska
Nevada
New Hampshire..
New Jersey
New MeiicO"
New York
North CaroliuD~.
Ohio
Oregon
Pennsylvania
Rhode Island
South Carolina...
Tennessee
Texas
United States
Utah
Vermont
Virginia
Washington Tei
ritory,
West Virginia.
Wisconsin
Wyoming
10^
12)S
Any.
Any.
10^
Note. — When the per i
lot mentioned in the
t, the first column gives the per cet
collected by law. If stipulated in the note, a per cer
high a« that in the second column may be collected.
8. Interest is either SimpU or Cbrnpowiwi.
9. Simple Interest is interest which, ev<^n
244 RAY'S HIGHER ARITHMETIC.
when due, is not convertible into principal, and therefore
can not accumulate in the hands of the debtor by drawing
interest itself, however long it may be retained.
Note. — Compound Interest is interest which, not being paid when
due, is convertible into principal, and from that time draws interest
itself and accumulates in the hands of the debtor, according to the
time it is retained. It will be treated of in a separate chapter.
292. Simple Interest differs from the applications of
Percentage in Chapter XV, by taking time as an element
in the calculation, which they do not.
293. The five quantities embraced in questions of in
terest are: the Principal^ the Interedy the Ratey the Time,
and the Amourd, — or the sum of the principal and interest.
Any three of these being given, the others may be found.
They give rise to five cases.
The principal corresponds to the Bdse^ and the interest
to the Percentage.
NOTATION.
294. The following notation can be adopted to ad*
vantage in the formulas : «
Principal = P.
Eate =R
Interest = I.
Amount = A.
Time = T.
CASE I.
295. Given the principal, the rate, and the time,
to find the interest and the amount.
Principle. — The irdered is equal to the continued product
of the principal, rate, and time.
Formulas. , p _i_ t a
INTEREST. 245
PXBXT = I.
{
KoTE. — The time is expressed in years or parts of years, or both.
COMMON METHOD.
Problem. — What is the interest of $320 for 3 yr. 5 mo.
18 da., at 4% ?
OPERATION.
$320X.04X3A = $44.37J, Int.
Solution.— The interest of $320 for 1 year, at 4^, is $12.80; and
for 3/y years, it is 3^ times as much as it is for 1 year, or $12.80 X
3 A = $44.37i.
Remark. — Unless otherwise specified, 30 days are considered to
make a month; hence, 5 mo. 18 da. = ^ of a year. T r 
General Rule. — 1. Multiply the principal by the rate, and
that product by the time expressed in years; the product is the.
intered, ^
2. Add the principal and interest, to find the amount.
^
Examples fob Practioe.
Find the simple interest of:
1. $178.63 for 2 yr. 5 mo. 26 da., at 7^. $31.12
2. $6084.25 for 1 yr. 3 mo., at 4^. $342.24
3. $64.30 for 1 yr. 10 mo. 14 da., at 9^. $10.83
4. $1052.80 for 28 da., at 10^. $8.19
5. $419.10 for 8 mo. 16 da., at 6^. $17.88
6. $1461.85 for 6 yr. 7 mo. 4 da., at 10^. $964.01
7. $2601.50 for 72 da., at 1^%. $39.02
8. $8722.43 for 5 yr., at 6%. $2878.40
9. $326.50 for 1 mo. 8 da., at 8^. $2.76
10. $1106.70 for 4 yr. 1 mo. 1 da., at 6%. $271.33
11. $10000 for 1 da., at 6%. $1.67
1 yr. =
21.2880
3 yr. = 3
63.864
6 mo. = J
10.644
1 mo. = i
1.774
1 8 da. = A
1.064
1 da. = ^
.059
246 RAY'S HIGHER ARITHMETIC.
METHOD BY ALIQUOT PARTS.
296. Many persons prefer computing interest by the
method of Aliquot Parts. The following illustrates it :
Problem. — What is the simple interest of $354.80 for
3 yr. 7 mo. 19 da., at 6%?
Solution. — The yearly in operation.
terest, being 6^ of the prin $354.80
cipal, is found, by Case I of .06
Percentage; the interest for 3
yr. 7 mo. 19 da. is then ob
tained by aliquot parts; each
item of interest is carried no
lower than mills, the next
figure being neglected if less
than 5 ; but if 5 or over, it is $77.41 Ans,
counted 1 mill.
SIX PER CENT METHODS.
FiBST Method.
297. The six per cent method possesses many advantages,
and is readily and easily applied, as the following will show :
At 6^ per annum, the interest on $1
for 1 year is 6 ct., or .06 of the principal.
u 2 mo. " 1 " " .01 *' " "
it I it a ^ it it QQ5 a it u
« 6 da. = ^ mo. " yV *' " ^^ " " "
« 1 " = ^ *' '' ^V « « .OOOi " "
Problem. — What is the interest of $560 for 3 yr. 8 mo.
12 da., at 6%?
Solution. — The interest on opebation.
$lfor 1 yr., at 6 ^o, is 6 ct., and $5 60X.222 = $124.32, Ana.
for 3 yr. is 18 ct.; the interest
on $1 for 8 mo. is 4 ct., and the interest for 12 da. is 2 mills ; henco,
the interest for the entire time is $.222 : therefore, the interest on
$560 for 3 yr. 8 mo. 12 da. is equal to $.222 X 660 = $124.32
INTEREST, 247
Per Cent Bule. — 1. Tofe «ic cents J(yr every year, J a
cent for every motvthf and i of a miU for every day; their mm
18 the interest on $1, at 6%, /or Hie given Utne.
2. MuUiply the interest on $1 for the given time by the prinr
cipal; the prodtust is the interest required.
Remarks. — 1. To find the interest at any other rate than 6^, in
crease or decrease the interest at 6^ by such a part of it as the given
rate is greater or less than 6^.
2. After finding the interest at 6^o, observe that the interest at
5^ == interest at 6^ — J of itself. And the interest
at 4J^ = int. at 6^ — J of it.
at 4 ^ = int. at 6% — J of it.
at 3 fo = i int. at 6^.
at 2 ^ = J int at 6^.
at 1 J^ = J int. at 6^.
at 1 fo=i int. at 6^.
at 7 ^ = int. at 6^  J of it.
at 7ifo = " Qfc f J of it
at 8 fo= " 6^0 + J of it.
&t 9 ^0= " 6^0 + J of it.
at 12fof 18^, 24^ = 2, 3, 4 times interest at 6^.
at 5^, 10^, 15^, 20^ =xV» h h i interest at 6^, after moving
the point one figure to the right.
3. Or, having the interest at 6^, multiply by the given rate and
divide by 6.
Second Method.
Problem. — ^What is the simple interest of $354.80 for 3
yr. 7 mo. 19 da., at 6%?
Solution. — The interest of any opebation.
sum ($354.80), at 6^, equals the ($354.80^2 ) X .436 J =
interest of half that sum ($177.40), $77.4055 3, Am.
at 12^. But 12^ a year equals
1^ a month, and for 3 yr. 7 mo., or 43 mo., the rate is 43^, and for
19 da., which is \^ of a month, the rate is iM] hence, the rate for
the whole time is 43Jf^, and 43 J J^ of the principal will be the
interest. To get 43J§^o, multiply by 43JJ hundredths = .43 J =
.436J.
Remark. — In the multiplier .436 J, the hundredths (43) are the
number of months, and the thousandths (6J) are J of the days
(19 da.), in the given time, 3 yr. 7 mo. 19 da.
248 BAY'S mo HER ARITHMETTC,
Rule. — Reduce the years, if any, to m(wii/w, and vrrite the
whole rmmber of months as decimal hundredtlis ; after which,
'place ^ of die days, if any, as thousandths; multiply half the
principal by this number, the product is the interest.
Note. — In applying the rule, when the number of days is 1 or 2,
place a cipher to the right of the months, and write the } or  ;
otherwise, they will not stand in the thousandths' place : thus, if
the time is 1 yr. 4 mo. 1 da., the multiplier is .160J
Bemark. — Exact or Accurate Interest requires that the common
year should be 365, and leap year 366 days ; hence, exaA;t interest is
T^y less for common years, aud ^ less for leap years than the ordi
nary interest for 360 days.
Examples for Practice.
Find the simple interest of:
1. $1532.45 for 9 yr. 2 mo. 7 da., at 12%. $1689.27
2. $78084.50 for 2 yr. 4 mo. 29 da., at 18^. $33927.72
3. $512.60 for 8 mo. 18 da., at 7%. $25.72
4. $1363.20 for 39 da., at 1\% a month. $22.15
5. $402.50 for 100 da., at 2^ a month. $26.83
6. $6919.32 for 7 yr. 6 mo., at 6%. $3113.69
7. $990.73 for 9 mo. 19 da., at 7^. $55.67
8. $4642.68 for 5 mo. 17 da., at 15^. $323.05
9. $13024 for 9 mo. 13 da., at 10^. $1023.83
10. $615.38 for 4 yr. 11 mo. 6 da., at 20%. $607.17
11. $2066.19 for 3 yr. 6 mo. 2 da., at 30^. $2172.94
12. $92.55 for 3 mo. 22 da., at 5%. , $1.44
Find the amount of:
13. $757.35 for 117 da., at 1J% a month. $801.65
14. $1883 for 1 yr. 4 mo. 21 da., at 6%. $2040.23
15. $262.70 for 53 da., at 1% a month $267.34
16. $584.48 for 133 da., at 7^%. $600.67
17. $392.28 for 71 da., at 2J% a month. $415.49
INTEREST, < 249
18. Find the interest of $7302.85 for 365 da., at 6%,
counting 360 da. to a year. $444.26
19. If I borrow $1000000 in New York, at 7%, and lend
it at Ifc in Ohio, what do I gain m 180 da.? $479.45
Note. — In New York 365 days are counted a year instead of 360.
20. Find the interest of $5064.30 for 7 mo. 12 da., at 7%,
in New York. $218.45
21. If I borrow $12500 at 6%, and lend it at 10^, what
do I gain in 3 yr. 4 mo. 4 da.? $1672.22
22. If $4603.15 is loaned July 17, 1881, at 7%, what is
due March 8, 1883? $5130.34
Note. — When the time is to be found by subtraction, and the days
in the subtrahend exceed those in the minuend, disregard days in
subtracting, and take the months in the minuend one less than the
actual count. Having found the years and months by subtraction,
find the days by actual count, beginning in the month preceding the
later of the two dates. Thus, in the above example, we find 1 yr.
7 «io. (not 8), and count from Februcui'y 17th to March 8th 19 da.
23. Find the interest, at 8%, of $13682.45, borrowed from
a minor 13 yr. 2 mo. 10 da. old, and retained till he is of
age (21 years). $8543.93
24. In one year a broker loans $876459.50 for 63 da., at
\\^o a mo., and pays 6% on $106525.20 deposits: what is
his gain? $21216.96
25. What is a broker's gain in 1 yr., on $100, deposited
at 6%, and loaned 11 times for 33 da., at the rate of 2%
a month? $18.20
26. Find the simple interest of £493 16s. 8d. for 1 yr.
8 mo.; at 6^. £49 7s. 8d.
Note. — In England the actual number of days is counted.
27. Find the simple interest of £24 18s. 9d. for 10 mo.,
at 6%. £1 4s. 11 Jd.
28. Of £25 for 1 yr. 9 mo., at 5%. £2 3s. 9d.
29. Of £648 15s. 6d. from June 2 to November 25, at
5%. £15 12s. lOd.
250 CRAY'S HIGHER ARITHMETIC
CASE II.
298. Given the principal, the rate, and the interest,
to find the time.
F0RMU1.A.— ~ .^ ~ = T.
Problem. — John Thomas loaned 8480, at 5%, till the in
terest was 8150 ; required the time.
Solution. — The inter operation.
est on $480 for 1 yr., at $1 50f ($480X. 06) = 6^ years.
5^, is $24. If the prin
cipal produce $24 interest in 1 year, it will require as many years
to produce $150 interest as $24 is contained times in $150, which is
6J years, or 6 yr. 3 mo. From the preceding, the following rule is
derived :
Briile. — Divide the given interest by Hue interest of the prin
cvpal for one year; the quotient is the time,
Kemark. — If the principal and amount are given, take their
difference for the interest.
Examples for Practice.
In what time will:
1. $1200 amount to $1800, at 10^ ? 5 yr.
2. $415.50 to $470.90, at 10^? 1 yr. 4 mo.
3. $3703.92 to $4122.15, at 8^ ? 1 yr. 4 mo. 28 da.
Note — A part of a day, not being recognized in interest, is
omitted in the answer, but must not be omitted in the proof,
4. In what time will any sum, as $100, double itself by
simple interest, at 4^, 6, 7, 9, 10, 12, 20, 25, 30% ?
22f, 16?, 13i, IH, 10, 8i, 5, 4, 3^ yr.
INTEREST. 251
5. In what time will any sum treble itself by simple in
terest, at 4, 10, 12% ? 50, 20, 16 yr.
6. How long must I keep on deposit 81374.50, at 10%,
to pay a debt of $1480.78? 9 mo. 8 da.
7. How long will it take $3642.08 to amount to 84007.54,
at 12% ? 10 mo. 1 da.
8. How long would it take $175.12 to produce $6.43
interest, at 6% ? 7 mo. 10 da.
9. How long would it take $415.38 to produce $10.69
interest in New York, at 7^ ? 134 days.
CASE III.
299. Given the principal, interest, and time, to find
the rate.
Formula. — = ::r^^ B.
PXlfe^XT
Problem. —At what rate per cent will $4800 gain $840
interest in 2\ years?
Solution. — The operation.
interest of $4800, Int. of $4800 at 1^ for 2J yr. = $120;
at ] ^, for 2} yr. is .'. $8405$l 20 = 7; or,
$120 ; hence, the 1^X7 = 7^, Amr
rate is as many
times 1^ as $120 is contained times in $840, which is 7 times; or,
7^. Hence the rule.
Bule. — Divide (he given inierest by Hie interest of the prin
cipal for the given time, at 1%,
Examples for Practice.
1. At what rate per annum will any sum treble itself, at
simple interest, in 5, 10, 15, 20, 26, 30 years, respectively ?
40, 20, 13i, 10, 8, 6%.
252 RAY'S mo HER ARITHMETIC.
2. At what rate of interest per annum will any gum
quadruple itself, at simple interest, in 6, 12, 18, 24, and 30
years, respectively? 50, 25, 16f, 12^, 10^.
3. What is the rate of interest when $35000 yields an
income of 8175 a month? 6%.
4. What is the rate of interest when $29200 produces
$6.40 a day? 8^.
5. What is the rate of interest when $12624.80 draws
$315.62 interest quarterly. 10%.
6. Find the rate when stock, bought at 40% discount,
yields a semiannual dividend of 5%? 16f % per annum.
7. A house that cost $8250, rents for $750 a year; the
insurance is 3^^, and the repairs ^%, every year: what
rate of interest does it pay? 8% — .
CASE IV.
900. Given the interest, rate, and time, to find the
principal.
Problem. — A man receives $490.84 interest annually on
a mortgage, at 7^: what is the amount of the mortgage?
SoLUTiON.r—Since $.07 is opebation.
the interest of $1 for one year, $490. 84j$. 07 = 7012
$490.84 is the interest of as $1X7012 = $7012, Ans.
many dollars as $.07 is con Or, 7^=$490.84
tained times in $490.84, which 1^=$70.12
is 7012; therefore, $7012 is 100 /o = $7012.
the amount of the mortgage.
Or, thus: the time being 1 year, $490.84 is 7^ of the principal;
1^ of it is j of $490.84, which is $70.12, and hence 100^ of it, or the
whole principal, is $7012.
BrUle. — Multiply ihe rate by the time, and divide the intereM
by ihe product; the quotient will he ihe principal.
f
INTEREST. 253
Examples for Practice.
What principal will produce:
1. J1500 a year, at 6%? $26000.
2. $1830 in 2 yr. 6 mo., at 5% ? $14640.
3. $45 a mo., at 9^? $6000.
4. $17 in 68 da., at 1% a mouth? $750.
5. $656.25 in 9 mo., at 3^%? $25000.
6. $86.15 in 9 mo. 11 da., at 10^? $1103.70
7. $313.24 in 112 da., at 7%? $14383.47
8. $146.05 in 7 mo. 14 da., at 6^? $3912.05
9. $58.78 in 1 yr. 3 mo. 20 da., at 4% ? $1125.57
10. $79.12 in 5 mo. 25 da., at. 7% in N. Y.? $2357.46
CASE V.
901. Given the amount, rate, and time, to find the
principal.
FoBMULA.— A i ( 1 + R X T ) = p.
Problem. — What is the par value of a bond which, in
8 yr. 8 mo., at 6^, will amount to $15200?
OPERATION.
Solution.— There are as $1 5200^$l. 52 = 10000;
many dollars in the princi $1 X 10000 = $! 0000, Ans.
pal as $1.52 is contained Or, 100^ = bond;
times in $15200, which is 1 00^ + 52 ^o = $l 5200;
10000 ; therefore, $10000 is 1^ = $100
the par value of the bond. 100 ^ = $1000 0, Am.
Or, the amount is \%^ of
the principal ; hence, \^i of the principal is $15200 ; and yj^ of the
principal is $100, and JJJ of the principal is $10000,
Hole. — Divide ihe amount by the amount of $1 for fJte
given time and raJte; ihe quotient unll be ihe principal.
254 BA Y'S mOHER ARITHMETIC.
Examples for Practice.
1. What principal in 2 yr. 3 mo. 12 da., at 6^, will
amount to $1367.84 ? $1203.03
2. What principal in 10 mo. 26 da., will amount to
$2718.96, at 10% interest? $2493.19
3. What principal, at 4^^, will amount to $4613.36 in
3 yr. 1 mo. 7 da.? $4048.14
4. What principal, at 7%, will amount to $562.07 in 79
da. (365 da. to a year) ? $553.68
PKOMISSOBY NOTEa
DEFINITIONS.
802. 1. A Promissory Note is a written promise by
one party to pay a specified sum to another.
2. The Face of a note is the sum promised to be paid.
3. The Maker is the party who binds himself to pay the
note by signing his name to it.
4. The Payee is the party to whom the payment is
promised.
Note. — The maker and the payee are called the wigiival parties to
a note.
5. The Holder of a note is the person who owns it: he
may or may not be the payee.
6. The Indorser of a note is the person who writes his
name across the back of it.
7. A Time Note is one in which a specified time is set
for payment.
Eemark. — There are various forms used in time notes, the prin
cipal ones being as follows?
PR0MI8S0BY NOTEIS. 255
1. Oedinaby Fokm.
<{450.50 Boston, Mass., June 30, 1880.
Three months after date, I promise to pay Thomas Ford,
or order, Four Hundred and Fifty, and 3^, Dollars, for value
received, vdth interest,
Edward E. Morris.
2. Joint Note.
$350. Denver, Col., Jan. 2, 1881.
Twelve months after date, we, or either of us, promise to
pay Frank R. Hairis, or bearer. Three Hundred and Fifty
Dollars, for value received, with interest from dale.
James West.
Daniel Tate.
Bemarks. — 1. This note is called a "joint note," because James
West and Daniel Tate are jointly liable for its payment.
2. If the note read, " We jointly and severally promise to pay,"
etc., it would then be called a joint and several note, and the makers
would be both jointly and singly liable for its payment.
3. Principal and Surety Note.
$125.00 St. Louis, Mo., Nov. 20, 1880.
Ninety days after date, I promise to pay James Miller, or
order, One Hundred and Twentyfive Dollars, for value received,
negotiable and payable without defalcation or discount.
Surety, James Miller. H. A. White.
Bemare. — The maker of this note, H. A. White, is the piirhcipal,
and the note is made payable to the order of the surety, James Miller,
who becomes security for its payment, indorsing it on the back to
the order of Mr. Whitens creditor.
256 BAY'S HIGHER ARITHMETIC,
8. A Demand Note is one in which no time is specified,
and the payment must be made whenever demanded by the
holder. The following is an example ;
Demand Note.
$1100.00 Cincinnati, O., Mar. 18, 1881.
For value received, I promise to pay David Swinton, or
ord/ir, on demand, Eleven Hundred Dollars, witli interest.
Henry Rudolph.
9. A promissory note is negotiable — that is, transferable
to another party by indorsement, — when the words **or
order," or ** or bearer," follow the name of the payee, as in
the above examples ; otherwise, the note is not negotiable.
10. If the note is payable to "bearer," no indorsement is
required on transferring it to another party, and the maker
nnlv IS rpsnonsihlft for its nflvmftnt.
only is responsible for its payment
11. If the note is payable to "order," the payee and each
holder in his turn must, on transferring it, indorse the note
by writing his name on its back, thus becoming liable for its
payment, in case the maker fails to meet it when due.
Bemarks. — 1. An indorser may free himself from responsibility
for payment if he accompany his signature with the words, " with
out recourse ;" in which case his indorsement simply signifies the
transfer of ownership.
2. If the indorser simply writes his name on the back of the note,
it is called a " blank " indorsement ; but if he precedes his signature
by the words, " Pay to the order of John Smith," for example, it is
called a " special " indorsement, and John Smith must indorse the
note before it can legally pass to a third holder.
12. It is essential to a valid promissory note, that it
contain the words "value received," and that the sum of
money to be paid should be written in words.
PROMISSOBY NOTES. 257
fr
13. If a note contains the words " with interest," it draws
interest from date, and, if no rate is mentioned, the legal
rate prevails.
Kemabks. — 1. The face of such a note is the sum mentioned with
its interest from date to the day of payment.
2. If a note does not contain the words " with interest," and is
not paid when due, it draws interest from the day of maturity, at
the legal rate, till paid.
14. A note matures on the day that it is legally due.
However, in many of the states, three days, called Days of
GracCy are allowed for the payment of a note after it is
nominally due.
Bemabks. — 1. The day before "grace" begins, the note is nomi
nally due ; it is legally due on the last day of grace.
2. The days of grace are rejected by writing " without grace " in
the note.
3. Notes falling due on Sunday or on a legal holiday, are due tho
day before or the day after, according to the special statute of each
state or territory.
4. If a note is payable " on demand," it is legally due when pre
sented.
15. When a note in bank is not paid at maturity, it goes
to protest — ^that is, a written notice of this fact, made out
in legal form by a notary public, is served on the indorsers,
or security.
Kemaiik. — Some of the states require that certain words shall be
inserted in every negotiable note in addition to the usual forms.
For instance, in Pennsylvania, the words " without defalcation " are
required; in New Jersey, "without defalcation or discount;" and
in Missouri, the statute requires the insertion of " negotiable and
payable without defalcation or discount." In Indiana, in order to
evade certain provisions of the law, the following words are usually
inserted, "without any relief from valuation or appraisement laws."
This constitutes what is known as the "ironclad" note.
303. If a note be payable a certain time ** after date,"
proceed thus to find the day it is legally due:
H. A. 22.
258
RA Y'S HIGHER ARITHMETIC.
Bule. — Add to the date of ihe note, Hie number of years and
months to elapse before payment ; if this gives the day of a month
higher than that month contains, take tlie last day in that month;
then, count the number of days mentioned in the note and 3 more:
this wUl give the day the note is legally due ; but if it is a Sunday
or a national holiday, it mud be paid the day previous,
Remarks.^1. When counting the days, do not reckon the one from
which the counting begins. (For exceptions, see Art. 313, 16,Kem.2.)
2. The months mentioned in a note are calendar months. Hence,
a 3 mo. note would run longer at one time than at another ; one
dated Jan. 1st, will run 93 days ; one dated Oct. 1st, will run 95
days : to avoid this irregularity, the time of short notes is generally
given in days instead of months ; as, 30, 60, 90 days, etc.
3. When the time is given in days, it is convenient to use the
following table in determining the day of maturity of a note :
Table
Showing the Number of Days from any Day of:
•
365
31
59
90
120
151
181
212
243
273
304
334
•
334
365
28
59
89
120
150
181
212
242
273
303
p
•
306
>
•
275
p
•
a
s
o
•
a
>
•><
•
153
184
212
243
•
122
153
181
212
9
•
92
123
151
182
212
243
273
304
335
365
31
61
<
•
61
92
120
151
181
212
•
31
62
90
121
151
182
To the
same day
of next
245
276
304
335
365
31
61
214
245
273
304
334
365
30
61
92
122
153
183
1.84
215
243
274
304
335
365
31
62
92
123
153
Jan.
337
365
31
61
92
306
334
365
30
61
Feb.
Mar.
April.
273
304
334
365
31
61
92
122
242
273
303
334
365
30
61
91
Mav.
June.
122
153
184
214
245
275
91
242 , 212
273 243
July.
122
153
183
214
?!44
92
123
153
184
214
Aug.
304
334
3H5
30
274
304
335
365
Sept.
Oct
Nov.
Dec.
Kemark. — In leap years, if the last day of February be included
in the time, 1 day must be added to the number obtained from the
table.
.1
ANNUAL INTEREST. 259
Examples for Practice.
Find the maturity and amount of each of the following
notes:
$560.60 Chattanooga, Tenn., June 3, 1872.
1. For value received, sixty days after date, I promise to
pay Madison Wilcox five hundred and sixty ^ dollars,
with interest at 7^. James Daily.
$567.47—
8430.00 ToPEKA, Kan., Jan. 30, 1879.
2. Six months after date, I promise to pay Christine Ladd
four hundred and thirty dollars, for value received, with
interest at 12^ per annum. Amos Lyle.
$456.23
$4650.80 St. Louis, Mo., August 10, 1875.
3. For value received, three months after date, I promise
to pay Oliver Davis, or order, four thousand six hundred
and fifty ^^ doyars, with interest at the rate of 10^ per
annum, negotiable and payable without defalcation or dis
count. Milton Moore.
Surety, Oliver Davis. $4770.95 —
ANNUAL INTEREST.
304. Annual Interest is interest on the principal, and
on each annual interest after it is due.
Annual interest is legal in some of the states.
Explanation.— If Charles Heydon gives a note and agrees to pay
the interest annually, but fails to do so, and lets the note run for
three years before settlement, the first year's interest would draw
interest for two years, and the second year's interest would draw
interest for one year. The last year's interest would be paid at the
time of settlement, or as it fell due.
260 HA Y'S HIOHEB ARITHMETIC.
Problem.— Find the amount of $10500 for 4 yr. 6 mo.,
interest 6^, payable annually.
OPERATION.
Int. of $ 1 5 for 4 J yr., at 6^ = $ 2 8 3 5;
" " $630 " 8 " " " =$ 302.40
Total annual interest, =$3137.40
Amount =$10600 + $3 13 7. 40 = $ 136 3 7. 4
Solution.— The interest of $10500 for 1 yr., at 6^, is $630, and
for 4J yr. is $2835. The first anniual interest draws interest 3J yr.,
the second 2^, the third 1 J, and the fourth \ yr. This is the same
as one annual interest for (3 J + 2^ + 1^1^ = 8) eight years. But
the interest of $630, for 8 yr., at 6^, is $302.40; therefore, the
amount = $10500 + $2835 + $302.40 = $13637.40
Bule.— 1. Find the interest on ihe 'principal for the entire
time, and on each yearns interest till the time of settleinent.
2. The sum of these interests is the interest required.
Note. — In Ohio and several other states, interest on unpaid annual
interest is calculated at the legal rate only.
Examples for Practice.
1. Find the amount of a note for $1500, interest 6%,
payable annually, given Sept. 3, 1870, and not paid till
March 1, 1874. $1838.71
2. A gentleman holds six $1000 railroad bonds, due in
3 years, interest 6^, payable semiannually: no interest
having been paid, what amount is owing him when the
bonds mature? $7161.
3. $2500.00 St. Paul, Minn., Jan. 11, 1869.
For value received, I promise to pay Morgan Stuart,
or order, twentyfive hundred dollars, with interest at 7%,
payable annually. Leonard Douglas.
What was due on this note March 17, 1873, if the interest
was paid the first two years? $2898.825
PARTIAL PAYMENTS, 261
II. PARTIAL PAYMENTS.
905. A Partial Payment is a 'payment in part of a
note or other obligation.
Whenever a partial payment is made, the holder should
write the date and amount of the payment on the back of
the note. This is called an Indorsement.
908. The following decision by Chancellor Kent, ''John
son's Chancery Rep., Vol. I, p. 17," has been adopted by the
Supreme Court of the United States, and, with few excep
tions, by the several states of the Union, and is known as
the ** United States Rule;"
XJ. S. Bule. — 1. ** The rule for casting interest when par
tial payments have been made, is to apply the payment, in Hie
first place, to the disdiarge of the interest due,
2. ^^If the payment exceeds the interest, the surplus goes
towards discharging the principal, and the subsequent interest
is to be computed on the balance of principal remaining due,
3. **ijr the payment be less than ike interest, the suiplus of
interest must not be taken to augment the principal; but in
terest continues on the former principal until Hie period when
the payments, taken together, exceed the inta^est due, and then
the surplus is to be applied towards discharging the principal,
and intei'est is to be computed on the balxmce as aforesaid,
dKyi. This rule is based upon the following principles:
Principles. — 1. Payments must be applied first to the dis
diarge of interest due, and the balance, if any, toward paying
the principal,
2. Interest must, in no case, become part of the principal.
3. Interest or payment must not draw interest,
Eemarks. — 1. It is worthy of remark, that the whole aim and
tenor of legislative enactments and judicial decisions on questions of
262 EAY'S HIGHER ARITHMETIC,
interest, have been to favor the debtor, by disallowing compound
interest, and yet ihu veiy rule fails to secure the end in view, and
really maintains and enforces the principle of compound interest in
a most objectionable shape ; for it makes interest due (not every year as
compound interest ordinarily does), bvi as often as a payment is made; by
which it happens that the closer the paymerUs are together, the greajler the
loss of the debtor, who thus suffers a penalty for his very promptness.
To illustrate, suppose the note to be for $2000, drawing interest
at 6^, and the debtor pays every month $10, which just meets the
interest then due ; at the end of the year he would still owe $2000.
But if he had invested the $10 each month, at 6^, he would have
had, at the end of the year, $123.30 available for payment, while the
debt would have increased only $120, being a difference of $3.30 in
his favor, and leaving his debt $1996.70, instead of $2000.
2. To find the difference of time between two dates on the note,
reckon by years and months as far as possible, and then count the
davs.
Problem.
^850. Cincinnati, April 29, 1880.
Ninety days after date, I promise to pay Stephen
Ludlow, or order, eight hundred and fifty dollars, with in
terest; value received. Charles K. Taylor.
Indoi'sements,— Oct, 13, 1880, $40 ; June 9, 1881, $32 ; Aug. 21, 1881,
$125 ; Bee. 1, 1881, $10 ; March 16, 1882, $80.
What was due Nov. 11, 1882?
Solution.— Interest on principal ($850) from April 29
to Oct. 13, 1880, 5 mo. 14 da., at 6^ per annum, $23.233 .
850.
Whole sum due Oct. 13, 1880, . . „ . . 873.233
Payment to be deducted, 40.
Balance due* Oct. 13, 1880, ..... 833.233
Interest on balance ($833,233) from Oct. 13, 1880, to
June 9, 1881, being 7 mo. 27 da., .... 32.913
Payment not enough to meet the interest, . . . 32. '
Surplus interest not paid June 9, 1881, .... MS
Interest on former principal ($833,233) from June 9,
1881, to Aug. 21, 1881, being 2 mo. 12 da., . . 9.999
Whole interest due Aug. 21, 1881, 10.912
PARTIAL PAYMENTS. 263
(Brought forward) Int. due Aug. 21, 1881, . . 10.912
833.233
Whole sum due Aug. 21, 1881, 844.145
Payment to be deducted, 125.
Balance due Aug. 21, 1881, 719.145
Interest on the above balance ($719,145) from Aug. 21,
1881, to Dec. 1, 1881, being 3 mo. 10 da., . . 11.986
Payment not enough to meet the interest, . , .10.
Surplus interest not paid Dec. 1, 1881, .... 1.986
Interest on former principal ($719,145) from Dec. 1,
1881, to March 16, 1882, being 3 mo. 15 da., . . 12.585
Whole interest due March 16, 1882, .... 14.571
719.145
Whole sum due March 16, 1882, . . . . . 733.716
Payment to be deducted, ...... c 80.
Balancedue March 16, 1882, 653.716
Interest on balance ($653,716) from March 16, 1882, to
Nov. 11, 1882, being 7 mo. 26 da., .... 25.713
Balance due on settlement, Nov. 11, 1882, . . . $679.43
•
Examples for Practice.
1. $304^^. Chicago, March 10, 1882.
For value received, six months after date, I promise to
pay G. Riley, or order, three hundred and four ^^ dollars.
No payments. H. McMakin.
What was due Nov. 3. 1883? 8325.63
2. $429^^. Indianapolis, April 13, 1873.
On demand, I promise to pay W. Morgan, or order,
four hundred and twentynine 3^ dollars, value received,
with interest. r Wilson.
Indorsed : Oct. 2, 1873, $10 ; Dec. 8, 1873, $60 ; July 17, 1874,
$200.
What was due Jan. 1, 1875? $195.06
264 RAY'S HIGHER ARITHMETIC.
3. $1750. New York, Nov. 22, 1872.
For value received, two years after date, I promise to
pay to the order of Spencer & Ward, seventeen hundred
and fifty dollars, with interest at 7 per cent.'
Jacob Winston.
Indorsed: Nov. 25, 1874, $500; July 18, 1875, $50; Sept. 1, 1875,
$600 ; Dec. 28, 1875, $75.
What was due Feb. 10, 1876? $879.71
4. A note of $312 given April 1, 1872, 8% from date,
was settled July 1, 1874, the exact sum due being $304.98.
Indorsed: April 1, 1873, $30.96; Oct 1, 1873,$ ; April 1,
1874, $20.40
Restore the lost figures of the second payment. $11.08
5. There have been two equal annual payments on a 6^
note for $175, given two years ago this day. The balance is
$154.40: what was each payment? $20.50
6. A merchant gives his note, 10% from date, foi*
$2442.04: what sum paid annually will have discharged
the whole at the end of 5 years? $644,204
308. In Connecticut and Vermont the laws provide the
following special rules relative to partial payments:
Connecticut Bule. — 1. Gompvle ike interest to the time of
the first payment, if that be oive year or more from, ihe time
tJiat interest commenced; add it to ihe prindpai, and deduct
the payment from the sum total.
2. Jff' there be after payments made, compute the interest on
Hie balance due to ihe next payment, and then deduct the pay
ment as above; and in like manner from one payment to
another, till all the payments are absorbed: Provided, ihe time
between one payment and another be one year or more.
3. But if any payment be made before one yearns interest
hath accrued, then compute ihe interest on ihe sum then due
on the obligation, for one year, add it to ihe principal, and
compute ihe interest on ihe sum paid, from tJie time it vxis
PARTIAL PAYMENTS. 265
jpaici, wp to the end of ike year; add it to 1}ie mm paid^ and
deduct that mm from the principal and interest added as above.
(See Note.)
4. ijf' any payment be made of a less sum tliait the interest
arisen at tfie time of such paymenty no interest is to be com
putedy bvt only on the principal mm, for any period.
Note. — Jff a year does not extend beyond the time of payment; but.
if it doesj then find the amount of the principal renuiining unpaid up to
the time of settiementj likevjise the amount of the payment or payments
from the time they were paid to the time of settlement, and deduct the sum
of these several amounts from the amount of the principal.
What is due on the 2d and 3d of the preceding notes, by
the Connecticut rule? 2d, $194.88 ; 3d, $877.95
Vermont Bule. — 1. On all notes, etc., payable with in
terest, partial payments are applied, first, to liquidate the
interest that has accrued at the time of such payments; and,
secondly f to the extinguishment of the principal.
2. When notes, etc., are drawn wiHi interest payable
ANNUALLY, the annual interests that remain unpaid are svbjed
to simple interest from the time they become due to the time
of final settlement.
3. Partial payments, made after annual interest begins
to accrue, also draw simple interest from the time such
payments are made to the end of the year; and their amounts
are then applied, first, to liquidate the simple interest that has
accrued from the unpaid annual interests; secondly, to Uqui
date tJie annual interests that have become due; and, thirdly,
to the extinguishment of Hie principal.
$1480. Woodstock, Vt., April 12, 1879.
For value received, I promise to pay John Jay, or order,
fourteen hundred and eighty dollars, with interest annually.
James Brown.
Indorsed : July 25, 1879, $40 ; May 20, 1880, $50 ; June 3, 1 881, $350.
What was due April 12, 1882? $1291.95
H. A. 23.
266 RAY'S HIGHER ARITHMETIC,
909. Business men generally settle notes and accounts
payable within a year, by the Mercantile Rule.
Mercantile Bule. — 1. Fiad the amount of the principal
from the date of the note to the date of 8ettlem£nt
2. Find the amount of eojoh payment from its date to the
date of setttement
3. From the amount of the principal svhtract the sum of the
amounts of the paym£nt8,
NoTR — In applying this rule, the time should be found for the
exact number of days.
Examples for Practice.
1. $950.00 New York, Jan. 25, 1876.
For value received, nine months after date, I promise
to pay Peter Finley nine hundred and fifty dollars, with 7%
interest. Thomas Hunter.
The following payments were made on this note: March
2, 1876, $225; May 5, 1876, $174.19; June 29, 1876,
$187.50; Aug. 1, 1876, $79.15
Required — the balance at settlement. $312.57
2. A note for $600 was given June 12, 1865, 6% interest,
and the following indorsements were made : Aug. 12, 1865,
$100; Nov. 12, 1865, $250; Jan. 12, 1866, $120: what was
due Feb. 12, 1866, counting by months instead of days ?
$146.65
III. TRUE DISCOUNT.
DEFINITIONS.
310. 1. Discount on a debt payable by agreement at
some future time, is a deduction made for ** cash," or present
payment.
It arises from the consideration of the present tvorth of the
debt.
TRUE DISCOUNT, 267
2. Present Worth is that sum of money which, at a
given rate of interest, will amount to the same as the debt
at its maturity.
3. True Discount, then, is the difference between the
present worth and the whole debt.
Remarks. — 1. That is, it is the simple interest on the present
worthy from the day of discount until the day of maturity.
2. Discount on a debt must be carefully distinguished from Com
mercial Discount, which is simply a deduction from the regular
price or value of an article ; the latter is usually expressed as such
a "per cent off."
311. The different cases of true discount may be solved
like those of simple interest: the present worth correspond
ing to the Principal; the discount, to the Interest; and the
face of the debt, to the Amount, The following case is
the only one much used :
312. Given the face, time, and rate, to find the
present worth and true discount.
Problem. — Find the present worth and discount of
$5101.75, due in 1 yr. 9 mo. 19 da., at
OPERATION.
Amount of $ 1 for 1 yr. 9 mo. 1 9 da., at6^ = $1.108J,
and $5101. 7 5 .$1.108^ = 4603. 775;
$1X4603.775 = $4603.77 5, present worth;
$5101.75— $4603. 775 = $497. 97 5, true discount.
Solution. — The amount of $1 for 1 yr. 9 mo. 19 da., at 6^, is
$1.108^, and $5101.75 is the amount of as many dollars as $1.108 J is
contained times in $5101.75, which is 4603.775 times ; therefore, the
present worth is $4603.775, and the true discount is $497,975
Bule. — 1. Divide the debt by the ainourd of %\ for the given
time and raie ; the quotient is the present tuorih.
2. The difference bet>ween the debt and the preset worth is
the true discount,
OF THE '^
UNiVERS/T
\^
268 BAY'S mOHER ARITHMETIC.
Examples for Practice.
1. Find the true discount on a debt for $5034.15 due in
3 yr. 5 mo, 20 da., without grace, at 7%. . $984.33
2. What is the present worth of a note for $2500, bearing
6% interest, and payable in 2 yr. 6 mo. 15 da., discounted
at 8%.. $2394.10
IV. BANK DISCOUNT.
DEFINITIONS.
313. 1. A Bank is an institution authorized by law to
deal in money.
2. The three chief provinces of banks are : the receiv
ing of deposits, the loaning of money, and the issuing of
bankbills. Any or all of these provinces may be exercised
by the same bank.
Bemarks. — 1. A Bank of Deposit is one which takes charge of
money, stocks, etc., belonging to its customers.
Money so intrusted is called a deposity and the customers are
known as deposUors.
2. A Bank of Discount is one which loans money by discounting
notes, drafts, etc.
3. A Bank of Issue is one which issues notes, called " bankbills,"
that circulate as monev.
. 3. The banks of the. United States may be divided into
two general classes, — ^National Banks and Private Banks.
Kemarks. — 1. National Banks are organized under special legis
lation of Congress. They must conform to certain restrictions as to
the number of stockholders, amount of capital stock, reserve, circu
lation, etc. In return, they have privileges which give them certain
advantages over Private Banks : they are banks of issue, of discount,
and of deposit.
2. Private Banks are unable to issue their own banknotes to
advantage, owing to the heavy tax on their circulation ; they are,
therefore, confined to the provinces of deposit and discount.
BANK DISCOUNT,
269
4. A Check is a written order on a bank by a depositor
for money. The following is a usual form:
No. 1032.
Cincinnati, Nw, 27, 1851.
Pay to.
Bank of Cincinnati,
Rvfui B, Shaffer,
OR Order
Four Hundred and Twentyfive.
$425^
j^ Dollars.
Oeorge Potter HoUister,
Kemabks. — 1. Before this check ifl paid, it must be indorsed by
Kuf us B. Shaffer. He may either draw the money from the bank
itself, or he may transfer the check to another party, who must also
indorse it, as in the case of a promissory note. (See Art. 302, 11.)
2. If the words or Bearer are used in place of or Orders no indorse
ment is necessary, and any one holding the check may draw the
money for it.
5. A Draft is a written order of one person or company
upon another for the payment of money. The following is
a customary form:
$1453^.
New Orleans, La., July 25, 1880.
At sight.
Pay to the
order of.
First Ndtional Bank.
Fourteen Hundred Fiftythree and.
j^ Dollars
Value Received and Charge to Acxx)unt of
To John R. Williams & Co,,
No. 2136. Memphis, Tenn,
Robert James.
270 HA Y'S HIGHER ARITHMETIC,
6. Drafts may be divided into two classes, Sight Drafts
and Time Drafts,
7. A Sight Draft is one payable "at sight." (See the
form above.)
8. A Time Draft is payable a specified time "after
sight," or ** afl^r date."
9. The signer of the draft is the maker or drawer.
10. The one to whom the draft is addressed, and who is
requested to pay it, is the drawee.
11. The one to whom the money is ordered to be paid, is
the payee.
12. The one who has possession of the draft, is called
the owner or holder; when he sells it, and becomes an
indorser, he is liable for its payment.
13. The Indorsement of a draft is the writing upon the
back of it, by which the payee transfers the payment to
another.
Bemarks. — 1. A special indorsement is an order to pay the draft to
a particular person named, who is called the indorsee^ as " Pay to
F. H. Lee. — W. Harris," and no one but the indorsee can collect the
bill. Drafts are usually collected through banks.
2. When the indorsemeM is in blanks the payee merely writes his
name on the back, and any one who has lawful possession of the
draft can collect it.
3. If the drawee promises to pay a draft at maturity^ he writes
across the face the word "Accepted," with the date, and signs his
name, thus: "Accepted, July 11, 1881. — H. Morton." The ncceptor
is first responsible for payment, and the draft is called an acceptance.
4. A draft, like a promissory note, may be payable "to order,"
or " bearer," and is subject to protest in case the payment or accept
ance is refused.
5. A. time draft is universally entitled to the three days grace:
but, in about half of the states, no grace is allowed on sight drafts.
14. When a bank loans money, it discounts the time notes^
BANK DISCO UNT. 271
drafts, etc., offered by the borrower at a rate of discount
agreed upon, but not in accordance with the principles of
true discount.
15. Bank Discount is simple interest on the face of a
note, calculated from the day of discount to the day of
maturity, and paid in advance.
16. The Proceeds of a note is the amount which remains
after deducting the discount from the face.
Bemabks. — 1. Sitice the face of every note is a debt due at a future
time, its cost ought to be the present worth of that debt, and the
bank discount should be the same as the true discount. As it is,
the former is greater than the latter ; for, hank discount ia interest on
the face of the note, while true discount is the interest on the present worihf
which is always less than the face. Hence, their difference is the interest
of the difference between the present worth and face ; that is, the
interest of the true discount, (See Art. 310, 3.)
2. In discounting notes, the three days of grace are always taken
into account ; and in Delaware, the District of Columbia, Maryland,
Missouri, and Pennsylvania, the day of discount and the day ofmaivr
rity are both included in the time.
3. If the borrower is not the maker of the notes or drafts,
he must be the last indorser, or holder of them.
314. Problems in bank discount are solved like those of
simple interest. The face of the note corresponds to the
Prindpcd; the bank discount, to the Interest; the proceeds,
to the Prindpcd less the Interest; and the term of discount,
to the Time,
CASE I.
315. Given fhe flEkce of the note, the rate, and time,
to find the discount and proceeds.
Problem. — What is the bank discount of $770 for 90
days, at 6%?
272 BAY'S HIGHER ARITHMETIC.
OPERATION.
Int. of $1 for 93 da., at 6 ^^ =$.0 1 55 = Rate of Discount.
$770X. 0155 = 111. 935, Discount.
$770 — $11.935 = $758.065, Proceeds.
Solution.— Find the interest of $770 for 93 da., at 6^; this is
$11,935, which is the bank discount. The proceeds is the difference
between $770 and $11,935, or $758,065
pRTNCiPLES. — 1. The interest on the sum discounted for Uie
given time (plus the tiiree days of grace) at the given rate per
cent, is Hie bank discount,
2. The proceeds is equal to Uie sum discounted minus ilie
discount.
Bule. — 1» Find the interest on the sum discounted for three
days more Hum the given time, at the given rate; it is the
discount
2. Subtract the discount from Hie sum discounted, and the
remainder is the proceeds,
Kemarks. — 1. As with promissory notes, the three days of grace
are not counted on a draft bearing the words " without grace."
2. The following problems present, each, two dates, — one showing
when the note is nominally due, and the other when it is legally due.
Thus, in an example which reads, " Due May Yg," the first is the
nominal, and the second the legal date.
Examples for Practice.
Find the day of maturity, the time to run, and the pro
ceeds of the following notes :
1. $792.50 QumcY, III., Jan. 3, 1870.
Six months after date, I promise to pay to the order
of Jones Brothers seven hundred and ninetytwo j^ dollars,
at the First National Bank, value received.
Discounted Feb. 18, at 6/o. ALBERT L. ToDD.
Due July ^; 138 da. to run; proceeds, J774.27
BANK DISCOUNT. 273
2. 81962yV(r N^^ Y^»^» '^"^^ 26, 1879.
Value received, four months after date, I promise to
pay B. Thoms, or order, one tliousand nine hundred sixty
two ^jj dollars, at the Chemical National Bank.
Discounted Aug. 26, at 7^.
Due Nov. 2 6/^^; 95 da. to run; pro. $1926.70
3. $2672^. Philadelphia, March 10, 1872.
Nine months after date, for value received, I promise
to pay Edward H. King, or order, two thousand six hun
dred seventytwo ^^ dollars, without defalcation.
T.. * J T 1 in * /!^ Jeremiah Barton.
Discounted July 19, at 6^.
Due Dec. lyiaJ 1^8 da. to run; pro. $2606.27
4. $3886. St. Louis, Jan. 31, 1879.
One month after date, we jointly and severally promise
to pay C. McKnight, or order, three thousand eight hun
dred eightysix dollars, value received, negotiable and pay
able, and without defalcation or discount.
T. Monroe,
I. Foster.
Discounted Jan. 31, at 1J% a month.
Due Feb. ^s/^ Mar.; 32 da. to run; pro. $3823.82
Kemark.— A note, drawn by two or more persons, "jointly and
severally," may be collected of either of the makers, but if the words
"jointly and severally " are not used, it can only be collected of the
makers as a firm or company.
5. $2850. Austin, Tex., April 11, 1879.
For value received, eight months after date, we promise
to pay Henry Hopper, or order, twentyeight hundred and
fifty dollars, with interest from date, at six per cent per
^°^^"" HaNNA & TUTTLE.
Discounted June 15, at 6^.
Due Dec. ^^,4; 182 da. to run; pro. $2875.47
274 RA Y'S HIGHER ABITHMETIC.
6. $737yV(r Boston, Feb. 14, 1880.
Value received, two months after date, I promise to pay
to J. K. Eaton, or order, seven hundred thirtyseven f^^
dollars, at the Suffolk National Bank.
William Allen.
Discounted Feb. 23, at 10^^.
Due April ^^/xi\ ^4 da. to run; pro. $726.34
7. $4085^. New Orleans, Nov. 20, 1875.
Value received, six months after date, I promise to pay
John A. Westcott, or order, four thousand eightyfive ^^
dollars, at the Planters' National Bank.
E. Waterman.
Discounted Dec. 31, 1875, at 5^.
Due May 20/33, 1^76; 144 da. to run; pro. J4003.50
CASE II.
316. Given the proceeds, time, and the rate of dis
count, to find the flEU^e of the note.
Problem. — For what sum must a 60 da. note be drawn,
to yield $1000, when discounted, at 6% per annum?
OPERATION.
The bank discount of $1 for 6 3 da., at 6^=$. 0105
Proceeds of $1 = $.9895; $1000 J.989 5 = $1 1 0. 6 1, the
face of the note.
Solution. — For every $1 in the face of the note, the proceeds, by
Case I, is $.9895 ; hence, there must be as many dollars in the face
as tliis sum, $.9895, is contained times in the given proceeds, $1000;
this gives $1010.61 for the face of the note.
Hule. — Divide the given proceeds by the proceeds of $1 for
Uie given time and rate; or, divide the given discount by the
discount of $1 for the given time and rate; the quotient w
equal to the face of the note.
BANK DISCOUNT. 275
Examples for Practice.
1. Find the face of a 30 da. note, which yields $1650,
when discounted at 1^^ a mo. $1677.68
2. The face of a 60 da. note, which, discounted at 6%
per annum, will yield $800. $808.49
3. The face of a 90 da. note, bought for $22.75 less than
its face, discounted at 7^. $1258.06
4. The face of a 4 mo. note, which, discounted at 1 ^ a
month, yields $3375. $3519.29
5. The face of a 6 mo. note, which, discounted at 10^
per annum, yields $4850. $5109.75
6. The face of a 60 da. note, discounted at 2% a month,
to pay $768.25 $801.93
7. The face of a 40 da. note, which, discounted at 8^,
yields $2072.60 . $2092.60
8. The face of a 30 da. and 90 da. note, to net $1000
when discounted at 6%.
$1005.53 at 30 da.; $1015.74 at 90 da.
CASE III.
317. Given the rate of bank discount, to find the
corresponding rate of interest.
Problem. — What is the rate of interest when a 60 da.
note is discounted at 2^ a month?
OPERATION.
Assume $ 1 as the face of the note.
Then, 6 3 days = time.
$4.20 = discount.
$95.80= proceeds.
$.00175 = interest of $ 1, at 1 ^, for the given time.
And $95. 80X. 00175 = $. 1676 5, interest of proceeds, at 1 ^cy
for the given time ; then,
$4.205$. 16765 = 25^/^, rate.
276 BAY'S lUOHER ARITHMETia
Solution. — The discount of $100 for 63 days, at 2^ a month, is
$4.20, and the proceeds is $95.80 The interest of $1 for the given
time, at \(fo, is $.00175, and of $95.80 is $.16765 To find the rate,
we proceed thus ; $4.20 ^ $.16765 = 25^^.
Bule. — 1. Find ike discount and proceeds of $100 or $1
for the time the note runs,
2. Divide the discount by Hie interest of the proceeds^ at 1^^
for the same time; the quotierd is ike rale.
Examples for Practice.
What is the rate of interest :
1. When a 30 da. note is discounted at 1, 1^, 1\, 2% a
month ? 12^11, 15^/ IS^^A^ 24,2^% per annum.
2. When a 60 da. note is discounted at 6, 8, 10^ per
annum ? ^ iWa > ^A\> ^^A%^ P^^ annum.
3. When a 90 da. note is discounted at 2, 2^, 3% a
month ? 25ff , 32^^, 39f^^ per annum.
4. When a note running 1 yr. without grace, is discounted
at 5, 6, 7, 8, 9, 10, 12^ ? 5^, G^f , 74, 8^, 9, 11^, 13^%.
5. My note, which will be legally due in 1 yr. 4 mo. 20
da., is discounted by a banker, at 8%: what rate of interest
does he receive? 9^.
Remark.— It may seem unnecessary to regard the time the note
has to run, in determining the rate of interest ; but, a comparison
of examples 1 and 3, shows that a 90 da. note, discounted at 2^ a
month, yields a higher rate of interest than a 30 da. note of the same
face, discounted at 2^ a month. The discount, at the same rate, on
all notes of the same face, tmies as the time to run, and if in each
case, it was referred to the mme prindpcUy the rate of interest would
be the same ; but when the discount becomes Inrgerj the proceeds or
principal to which it is referred, becomes smaller j and therefore the
rate of interest corresponding to any rate of discount increases %oith the time
the note has to run. Hence, the profit of the discounter is greater
proportionally on long notes than on short ones, at the same rate.
BANK DISCOUNT. 277
CASE IV.
318. Given the rate of interest, to find the corre
sponding rate of discount.
Problem. — What is the rate of discount when a 60 day
note yields 2% interest a month?
OPEBATION.
Assume $ 1 as the face of the note.
Then, 6 3 days = the time ;
$4.20= the interest;
and $104.20 = the amount ;
also, $.18235 = interest of amount, at 1 ^, for the given time;
$4.20i$.18235 = 23^2V> rate per cent.
Bemark. — Note the difference between this solution and that
under the preceding case. In that, the interest was found on the
proceeds ; in this, it is computed on the amount.
Kule.— 1. Find the intered and Uie amount of $100 or $1
for the given time,
2. Divide the interest by the interest of the amount at 1^
for Hie given time; the quotient is (he corresponding rate of
discount.
Examples for Practice.
What rates of discount :
1. On 30 da. notes, yield 10, 15, 20^ interest?
2. On 60 da. notes, yield 6, 8, 10% interest?
3. On 90 da. notes, yield 1, 2, 4% a month interest?
lliWr. 22 W. 42114 %•
4. On notes running 1 yr. without grace, yield 5, 6, 7, 8,
9, 10^ interest? 4^, 5, 6^, 7^, ^M, ^^fc
278 BAY'S HIGHER ARITHMETIC.
*
V. EXCHANGE.
DEFINITIONS.
319. 1. Exchange is the method of paying a debt in a
distant place by the transfer of a credit.
Bemabk. — Th'is method of transacting business is adopted as a
convenience: it avoids the danger and expense of sending the
money itself.
2. The payment is effected by means of a B\J(l of Exchxtn^e,
3. A Bill of Exchange is a written order on a person
or company in a distant place for the payment of money.
Bemark. — The term includes both drafts and checks.
4. Exchange is of two kinds: Domestic, or Inland, and
Foreign.
5. Domestic Exchange treats of drafts payable in the
country where they are made. (See page 269 for form.)
6. Foreign Exchange treats of drafts made in one
country and payable in another. ^
7. A foreign bill of exchange is usually drawn in trip
licate, called a Set of Exchange; the different copies,
termed respectively the first, second, and third of exchange,
are then sent by different mails, that miscarriage or delay
may be avoided. When one is paid, the others are void.
The following is a common form:
Foreign Bill op Exchange.
1. Exchange for New York, July 2, 1878.
£1000. Thirty days after sight of this First of
Exchange (Second and Third of same tenor and date unpaid),
pay to the order of James S. RoUinjs One Thxmsand Pounds
Sterling, value received, and charge to amount of
To John Brown & Co., J. S. Chick.
No. 1250. Liverpool, England.
)
EXCHANGE. 279
8. The Bate of Exchange is a rate per cent of the face
of the draft.
9. The Course of Exchange is the current price paid
in one place for bills of exchange on another.
10. The Par of Exchange is the estimated value of the
monetary unit of one country compared with that of another
country, and is either Intrhmc or (JommerdaL
11. Intrinsic Par of Exchange is based on the com
parative weight and purity of coins.
12. Commercial Par of Exchange is based on com
mercial usage, or the price of coins in different countries.
DOMESTIC EXCHANGE.
320. Where time is involved, problems in Domestic
Exchange are solved in accordance with the principles of
Bank Discount.
Examples for Practice.
1. What is paid for $3805.40 sight exchange on Boston,
at% premium? $3824.43
2. What is the cost of a check on St. Louis for $1505.40,
at \fo discount? $1501.64
3. What must be the face of a sight draft to cost $2000,
at 1^ premium? $1987.58
4. What must be the face of a sight draft to cost $4681.25,
at l\fo discount? $4740.51
5. What will a 30 day draft on New Orleans for $7216.85
cost, at 1% discount, interest 6% ?
OPERATION.
$1— $.00f = $.9962 5, rate of exchange.
$.0055 = bank discount of $ 1 for 3 3 da.
$.99075 = cost of exchange for $ 1.
$7 216.85X.99075 = $7150.09 4+, Ana.
280 BAY'S HIGHER ARITHMETIC.
Solution. — From the rate of exchange subtract the bank dis
count of $1 for 33 days, at 6^; the remainder is the cost of exchange
for $1. Then multiply the face of the draft by the cost of exchange
for $1, which gives $7150.094, the cost of the draft.
6. What will a 60 day draft on New York for $12692.50
cost, at \^o premium, interest 6%? $12654.42
7. What must be the face of an 18 day draft, costing
$5264.15, at \% premium, interest 6^ ? $5256.27
8. What must be the face of a 21 day draft, costing
$6836.75, at \% discount, interest 6%? $6925.04
9. A commission merchant in St. Louis sold 5560 lb. of
bacon, at \\\ ct. a lb.; his commission is 2J^, and the
course of exchange 98^^: what is the amount of the draft
that he remits to his consignor? $632.91
10. A grain merchant in Toledo sold 11875 bu. of corn,
at 40 cents a bushel; deducting 3% as commission, he pur
chased a 60 day draft with the proceeds, at 2% premium;
required the face of the draft? $4564.14
11. Sold lumber to the amount of $20312.50, charging
\\% commission, and remitted the proceeds to my consignor
by draft; required the face of the draft, exchange \% dis
count? $20108.35
12. If a 45 day draft for $5500 costs $5538.50, find the
rate of exchange.
OPEKATION.
The bank discount of $5500 for 48 days, at 6 ^=$44.
Then $5538.50 + $44 — $5500=$82. 50 premium;
$82.50
And ^_ ^^ = . 1 5 = 1 J ^, rate of exchange.
13. A father sent to his son, at school, a draft for $250,
at 3 mo., interest 6%, paying $244.25 for it; find the rate
of exchange. \% discount.
14. An agent owing his principal $1011.84, bought a
draft with this sum and remitted it ; the principal received
$992 : find the rate of exchange. 2^ premium.
EXCHANGE.
281
FOREIGN EXCHANGE.
321. 1. In Foreign Exchange it is necessary to find the
value of money in one country in terms of the monetary unit
of another country. We reduce foreign coins by comparison
with U. S. money, to find their value.
Remark. By an Act of Congress, March 3, 1873, the Director
of the Mint is authorized to estimate annually the values of the
standard coins, in circulation, of the various nations of the world.
In compliance with this law, the Secretary of the Treasury issued
the following estimate of values of foreign coins, January 1, 1884:
COUNTRY.
Austria
Bolivia
Brazil
Chili
Cuba
Denmark, Norway, Sweden...
Egypt
France, Belgium, Switz
Great Britain
German Empire
India..
Japan
Mexico
Netherlands
Portugal
Russia
Tripoli
Turkey
MONETARY UNIT.
Florin ,
Boliviano
Milreis of 1000 reis...
Peso
Peso
Crown
Piaster
Franc
Pound sterling
Mark
Rupee of 16 annas
Yen (silver)
Dollar
Florin
Milreis of 1000 reis...
Rouble of 100 copecks.
Mahbub of 20 piasters.
Piaster
VALUE IN
U. 8. MONEY.
.806
.546
.912
.932
.268
.049
.193
4.866J
.238
.383
.869
.875
.402
1.08
.645
.727
.044
Note. — The Drachma of Greece, the Lira of Italy, the Peseta (100
centimes) of Spain, and the Bolivar of Venezuela, are of the same
value as the Franc. The Dollar , of the same value as our own, is the
standard in the British Possessions, N. A., Liberia, and the Sandwich
Islands. The Peso of Ecuador and the United States of Colombia,
and the Sol of Peru, are the same in value as the Boliviano.
H. A. 24.
282 HA Y' S HIGHER ARITHMETIC.
2. Bills of Exchange on England, Ireland, and Scotland
are bought and sold without reference to the jxir of exchange,
3. It is customary in foreign exchange to deal in drafts
on the various commercial centers. London exchange is the
most common, and is received in almost all parts of the
civilized world.
Kemark. — London exchange is quoted in the newspapers in
several grades : as, " Prime bankers' sterling bills," or bills upon
banks of the highest standing ; " good bankers," next in rank ;
" prime " and " good commercial," or bills on merchants, etc. The
prices quoted depend upon the standing of the drawee and upon
the demand for the several classes of bills. Usually a double quo
tation, one for 60 day bills, and the other for 3 day bills, is given
for each class.
Examples for Practice.
1. Find the cost of a bill of exchange on London, at 3
days sight, for £530 12s., exchange being quoted at $4.88
OPERATION.
£530 12s. = £530.6
$4.88X530.6 = 12589.328
2. Find the cost of a bill of exchange on London, at
sight, for £625 10s. lOd., when exchange is $4.87 pound
sterling. $3046.39
3. What is the cost of a bill on Paris for 1485 francs,
$1—5.15 francs. $288.35
4. What is the cost of a bill on Amsterdam for 4800
guilders, quoted at 41^ cents, brokerage J^ ?
OPERATION.
$.415X4800 = $1992.
$1992X.00i = $9.9 6, brokerage.
$1992 + $9.96 = $2001.9 6, cost of bill.
5. What will a draft in St. Petersburg on New York for
$5000 cost, if a rouble be worth $.74? 6756§f roubles.
ARBITRATION OF EXCHANGE. 283
6. What will a draft on Charleston for $4500 cost at Eio
Janeiro (milreis = $.54), discount 2% ? 8166. 66f milreis.
7. A gentleman sold a 60 day draft, which was drawn on
Amsterdam for 1000 guilders; he discounted it at 6%, and
brokerage was \%\ what did he get for it, a guilder being
valued at 40^ cents? $397.77 \
AEBITRATION OF EXCHANGE.
DEFINITIONS.
322. 1. Owing to the constant variation of exchange, it
is sometimes advantageous to draw through an intermediate
point, or points, in place of drawing directly. This is called
Circular Exchange.
2. The process of finding the proportional exchange be
tween two places by means of Circular Exchange, is called
the Arbitration of Exchange.
3. Simple Arbitration is finding the proportional ex
change when there is but one intermediate point.
4. Compound Arbitration is finding the proportional
exchange through two or more intermediate points.
Problem. — ^A merchant wishes to remit $2240 to Lisbon :
is it more profitable to buy a bill directly on Lisbon, at 1
milreis = $1.10; or to remit through London and Paris, at
£1 = $4.88 —  25.20 francs, to Lisbon, 1 milreis = 5.5 francs ?
OPERATION.
$224051.10 = 2036.364 milreis, direct exchange.
( ) milreis =$ 2 2 4 0.
$4.88 = 25.20 francs.
5 . 5 francs = 1 milreis.
2240X25.20X1
= 2103.129 milreis, circular exchange.
4.88X5.5
Hence, the gain is 6 6 .7 6 5 milreis in buying circular exchange.
284 BA Y'S HIGHER ARITHMETIC.
Solution.— By direct exchange $2240 will buy 2036.364 milrels.
By circular exchange we have a set of equations, all of whose
members are known except one. Multiplying the righthand
members together for a dividend, and dividing the product by the
product of the lefthand members, the quotient is the required mem
ber, which, in this case, is 2103.129 milreis. The diflference in favor
of the circular method is 66.765 milreis.
Problem. — A merchant in Memphis wishes to remit
$8400 to New York; exchange on Chicago \&\\% premium,
between Chicago and Detroit 1% discount, and between
Detroit and New York \% discount: what is the value of
the draft in New York, if sent through Chicago and Detroit ?
OPERATION.
($ ) N. Y. = $ 8 4 0, Memphis.
$ 1 . 1 i Mem. = %\ Chicago.
$.9 9 Chi. = $1, Detroit.
$.99i Det. = $l, N. Y.
8400X1X1X1 =$8401.46, .In^
1.015X.99X.995
Solution.— By the problem, $1,015 in Memphis = $!• in Chicago ;
$.99 in Chicago = $1 in Detroit; and $.995 in Detroit = $1 in New
York. Hence, multiplying and dividing, as in the preceding
problem, we obtain the answer, $8401.46
Biile 1. — 1. Form a series of equationSy expressing the con
ditions of the question; i/te first containing the quantity given
equal to an unknown number of the qiumtity requiredy and all
arranged in suck a way thai the righthand qvmitity of each
equation and tlie lefblw/nd quantity in the equation next foUow
ingy shjJl he of the same denominationy and also the righthand
quantity of die la^t and the lefthand quantity of the first,
2. Cancel equal factors on opposite sides y and divide Hie
product of Hie quantities in the column which is complete by
t/ie product of those in the other column. The quotient will be
tlie quantity required.
ARBITRATION OF EXCHANGE. 285
Rule 2. — Reduce Hie given quardity to iJie denomination
vrith which it is compared; reduce this restdt to the denominor
tion with which it is compared; and so on, until the required
denomination is reached^ indicating the operations by imriting
as midtipliers the proper unit values. The compound fraction
thus obtained, reduced to its simplest form, uiU be the amount
of the required denomination.
Notes. — 1. The first is sometimes called the chain rule, because
each equation and the one following, as well as the last and first,
are connected as in a chain, having the righthand quantity in one
of the same denomination as the lefthand quantity of the next.
2. In applying this rule to exchange, if any currency is at a
premium or a discount, its unit value in the currency with which
it is compared, must be multiplied by such a mixed number, or
decimal, as will increase or diminish it accordingly.
3. If commission or brokerage is charged at any point of the
circuit, for effecting the exchange on the next point, the sum to be
transmitted must be diminished accordingly, by multiplying it by
the proper number of decimal hundredths.
Examples for Practice.
1. A, of Galveston, has $6000 to pay in New York. The
direct exchange is ^^ premium; but exchange on New
Orleans is ^% premium, and from New Orleans to New
York is J% discount. By circular exchange, how much
will pay his debt, and what is his gain ?
$5999.96; $30.04 gain.
2. A merchant of St. Louis wishes to remit $7165.80 to
Baltimore. Exchange on Baltimore is ^% premium; but,
on New Orleans it is ^% premium ; from New Orleans to
Havana, % discount; from Havana to Baltimore, \^
discount. What will be the value in Baltimore by each
method, and how much better is the circular ?
Direct, $7147.93; cir., $7183.77; gain, $36.84
286 BAY'S HIOHEE ARITHMETIC.
3. A Louisville merchant has $10000 due him in Charles
ton. Exchange on Charleston is \% premium. Instead of
drawing directly, he advises his debtor to remit to his agent
in New York at ^% premium, on whom he immediately
draws at 12 da., and sells the bill at f % premium, interest
off at 6^. What does he realize, and what gain over the
direct exchange? Realizes «10029.94; gain, $17.44
Suggestion. — Interest at 6^ per annum is \(fo a month, or J^
for 15 days, which diminishes the rate of premium to \<fo*
4. A Cincinnati manufacturer receives, April 18th, an
account of sales from New Orleans; net proceeds $5284.67,
due June Yy. He advises his agent to discount the debt at
6%, and invest the proceeds in a 7 day bill on New York, in
terest off at 6%, at \% discount, and remit it to Cincinnati.
The agent does this, April 26. The bill reaches Cincinnati
May 3, and is sold 2it \% premium. What is the proceeds,
and how much greater than if a bill had been drawn May 3,
on New Orleans, due June 7, sold at J^^ premium, and
interest off at 6% ? Proceeds, $5296.10; gain, $35.65
5. A merchant in Louisville wishes to pay $10000, which
he owes in Berlin. He can buy a bill of exchange in Louis
ville on Berlin at the rate of $.96 for 4 reichmarks; or he is
offered a circular bill through London and Paris, brokerage
at each. place \%, at the following rates: £1=$4.90 =
25.38 francs, and 5 francs = 4 reichmarks. What is the
difference in the cost? $81.35
VI. EQUATION OP PAYMENTS.
DEFINITIONS.
323. 1. Equation of Payments is the process of find
ing ihe time when two or more debts, due at different times,
may be paid without loss to either debtor or creditor.
EQ UA TION OF PA YMENTS. 287
2. The time of payment is called the Equated Time.
3. The Term of Credit is the time to elapse before a
debt is due.
4. The Average Term of Credit is the time to elapse
before several debts, due at different dates, may be paid
at once.
5. To Average a^ Account is to find the equitable
time of payment of the balance.
6. Settling or Closing an Account is finding how much
is due between the parties at a particular time. It is some
times called Striking a Balance.
7. A Focal Date is a date from which we begin to
reckon in averaging an account.
324. Equation of Payments is based upon the following
principles :
Principles. — 1. Tluit any sum of money paid before it is
duCy is balanced by keeping an equal sum of money for an
equal time after it is due,
2. Thai the interest is the measure for the use of any sum
of money for any time,
Kemare. — The first statement is not strictly correct, although it
has the sanction of able writers, and is very generally accepted.
For short periods, it will make no appreciable difference.
Problem. — A buys an invoice of $250 on 3 mo. credit;
$800 on 2 mo.; and $1000 on 4 mo.; and gives his note for
the whole amount ; how long should the note run ?
OPEBATION.
Debts. Terms. Equivalents.
250 X 3 = 750
800 X 2 = 1600
1000 X 4 = 4000
2050 6350
♦•* 6350 _ o A \
time to run = mo. = 3 mo. 3 da., Ans.
2050 '
288 HA y'/S' HIGHER ARITHMETia
Solution. — The use of $1 for 6350 months is balanced by the use
of $2050 for 3 months and 3 days. Or, the interest of $1 for 6350
months equals the interest of $2050, 3 months 3 days, at the same
rate.
Problem. — ^I buy of a wholesale dealer, at 3 mo. credit,
as follows: Jan. 7, an invoice of $600; Jan. 11, $240; Jan.
13, $400; Jan. 20, $320; Jan. 28, $1200; I give a note at
3 mo. for the whole amount : when is it dated ?
OPERATION.
$600X = $1 for days.
$240X 4 = $1 " 960 '"
$400X 6=$1 " 2400 "
$320X13 = $1 " 4160 "
$1200 X 2 1 = $ 1 *' 25200 "
$2760 $1 " 32720 "
Whence 327 20f 27 60== 11.8+ days; hence the time is 12
days, nearly, after Jan. 7 ; that is, Jan. 1 9.
Solution. — Start at the date of the first purchase, and proceed as
in the preceding solution.
325. When the terms of credit begin at the same date,
we have the following rule :
Rule. — Mtdtiply eadi debt by its term of credit, and divide
tJw mm of the products by Hie sum of the debts; the quotierU
tvill be the equaled time.
326. If the account has credits as well as debits, it is
called a compound eqmdion, and may be averaged on the
same principle as the simple equation.
The following problem will illustrate the difference between
simple and compound equations, as it involves both debits
and credits:
Problem. — What is the balance in the following account,
and when is it due?
EQUATION OF PAYMEmS.
A in aa^t «rUh B.
SoLirnoN. — Start with March 1, the earliest day upon which a
payment is made or falls due. Find tlie producL'', botli on the
I>r. and Cr. side of the account, using the dates when the itemii
are due, as cash. The balance due, $625, ie found by subtracting
the amount of the credits from the amount of the debits. The
balance of products, 5S050, divided by the balance of items, 629,
determines the mean time to be 93 days after March 1 ^ June 2.
Bule. — 1. FiTtd when each item is due, and take the earlUel
or laie^ date ae the focal date.
2. Find the difference betioeen tiie focal date and the re
maining dates, and multiply each item, hj its corre^mnding
difference.
3. Find the difference bdween the Dr. and Or. Hems, and
dUo between the Jh. and Or. produda, and divide tiie difference
of the prodvela by the difference of the items. Add t!te quotient
to the focal date '^ it be the first date, or subtract if it be the last
date; the restdt in either case vnll be the equated tinie,
4. Jj' the two balances be on oppodte sides of the aecowit, the
quotient mitgt be eubiraded from the focal date if the first, or
added if it be the last date.
Remark. — The solutions thus far have been by the prodml method.
The same reiult will be oblained by dividing the interest of the
product balance for 1 day by the inlerest of the item balance for 1
day. This depends upon the principle that if both dividend and
divisor are multiplied or divided by the same number the quotient
is nnchBn){ed.
290
RAY'S BJOHER ABITHMETia
Examples for Practice.
What is the mean time of the following invoices:
1. A to B, Dk.
1877.
May
June
»t
July
Aug;
Dol.
Ct.
When
due
Days after
15
1
10
20
1
15
To Invoice at ^
months.
Sum total.
800
700
900
600
500
1000
*

4500
Products.
Oct. 30, 1877.
Bemabk. — Start with Sept. 15, the day the first debt is due.
2. Dr. E in accH current with F. Cb.
Feb.
Mar.
A^r.
4
20
1
5
To Invoice 3 mo.
I< «« 9 <(
It tt Q 4t
It <l O (t
550
260
150
325
—
P'.'od.
May
June
July
8
28
3
1
By cash,
'*^ remit, June 5,
" note, 1 munth,
>« "2 "
tl50
420
.340
170
—
Prod,
3.
E owes F $205, due Feb. 26.
H, Wright to Mason & Giles, Db.
1876.
Feb.
it
Mar:
Apr.
May
June
1
20
10
8
10
15
" 3
" 3
" cash,
" 3
" 3

Dol.
Ct.
WTien
due
Days after
AprUS.
lOnths.
900
«(
700
•(
600
500
t(
900
u
400
Sum total.
4000
Prodtiets.
Due June 13, 1876,
Bemark. — Start with April 8, the time the first payment is due.
4. Dr. . A in acc^t current with B. Cb.
Mar.
Apr.
June
May
19
20
15
10
To Invoice cash,
tt «t «<
<« t« <«
U <« M
$901)
800
700
600
—
Prod.
Feb.
Mar.
June
July
20
5
20
10
By cash,
" remit, Mar. 15,
'• cash,
4t ••
MOO
300
2on
500
—
Prod,
A owes B a balance of $1600, due April 23.
J
EQ UA TION OF PA YMENTS,
291
Kemark. — Start with Feb. 20, and date the remittance, March 15,
the day it is received.
5. Dr.
C in qjccH current with D,
Cr.
Jan.
Feb.
(I
Apr.
4
3
15
2
To invoice 2 mo.
1 "
2 '•
cash,
<<
«250
140
450
100
Prod.
Mar.
Apr.
May
June
10
21
4
20
16
By cash,
" note, 2 mo.
" remit., May 25,
" accep. 16 da. sight
$350
200
240
120
500
Prod.
C is Cr. $470, due Aug. 12.
6. I owe $912, due Oct. 16, and $500, due Dec. 20. K
I pay the first, Oct. 1, 15 days before due, when should I
pay the last? Jan. 16, next, 27 days after due.
7. Oct. 3, I had two accounts, amounting to $375, one
due Dec. 6, and the other Nov. 6, but equated for Nov. 16:
what was each in amount? $250, $125.
8. A owes $840, due Oct. 3 ; he pays $400, July 1 ; $200,
Aug. 1 ; when will the balance be due?
April 30, next year.
9. I owe $3200, Oct. 25 ; I pay $400, Sept. 15 ; $800,
Sept. 30: when should the balance be paid? Nov. 12.
10. An account of $2500 is due Sept. 16 ; $500 are paid
Aug. 1 ; $500, Aug. 11 ; $500, Aug. 21 : when will the
balance be due? Nov. 9.
11. Exchange the five following notes for six others, each
for the same amount, and payable at equal intervals: one
of $1200, due in 41 days; one of $1500, due in 72 days;
one of $2050, due in 80 days; one of $1320, due in 110
days; one of $1730, due in 125 days; total, $7800.
The notes are $1300 each, and run 25, 50,
75, 100, 125, 150 days respectively.
12. Burt owed in two accounts $487 ; neither was to draw
interest till after due, — one standing a year, arid the othei two
years. He paid both in 1 yr. 5 mo., finding the true dis
count of the second, at 6%, exactly equal to the interest of
the first : what difference of time would the common rule
have made? 3 days*
292
RAY'S HIGHER ARITHMETIC.
yil. SETTLEMENT OF ACCOUNTS.
DEFINITIONS.
327. 1. An Account is a written statement of debit and
credit transactions between two parties, giving the date,
quantity, and price of each item.
2. An Account Current is a written statement of the
business transactions between two parties, for a definite time.
328. In settling an account, the parties may wish to
find:
(1). When the balance is equitably due.
(2); Wliot mm J at a given time, should be paid to balance
the accotint
The first process is Averaging the Account (Art. 324) ;* the
second is Finding the Cask Balance.
Dr.
Henry Armor in accH with City Bank.
Cr.
1875.
Dol.
a.
1874.
Dol.
C*.
Jan.
5
To check,
300
Dec.
31
By balance old account,
500
it
20
.t tt
600
1875.
4(
21
ti «
100
Jan.
7
" casb.
50
<t
27
tt tt
850
,
ti
15
tt tt '
400
(t
31
** . *'
75
tt
24
tt tt
1000
Explanation. — The two parties to this account are Henry
Armor and City Bank. The lefthand, or Dr. side, shows, with
their dates, the sums paid by the bank on the checks of Henry
Armor, for which he is their debtor. The righthand, or Cr. side,
shows, with their dates, the sums deposited in the bank by Henry
Armor, for which he is their creditor.
329. Generally, in an account current, each item draws
interest from its date to the day of settlement.
SETTLEMENT OF ACCOUNTS.
293
Problem. — Find the interest due, and balance this ac
count:
F. H, WUlis in accH with E. S, Kennedy.
Dr. Cte
1879.
Jan.
6
t»
13
ti
m
ti
2.
<4
28
To check.
It
Dol.
a.
1878.
170
Dec.
31
480
1879.
96
Jan.
7
500
t.
20
50
it
30
By balance forward,
" cash deposit,
tt It K
<t •« •«
ct.
Interest to Jan. 31, at 6 per cent.
Db.
$170, 6fcy 25 da
480, " 18
96, " 15
5 0, " 6
50, " 3
OPERATION.
Cr.
it
u
ti
u
$1290
= $.708
= 1.44
= .24
= .50
r= .025
$2,913
$325, 6%, 31 da =$1,679
= 3.20
.32
.04
8 0*0,
17 5,
240,
(t
u
u
24
11
1
((
u
u
$1540
$1296
$244
$2.913
$2.33
/. Interest due = $2. 33; and cash balance due Willis is
$244f $2.33= $246.33, Ans.
Solution.— The interest is found on each item from its date to
the day of settlement; then the sum of the items and the sum
of the interests are found on each side. The diflference between
the sums of the interests is equal to the interest due, which,
added to the diflference between the sums of the items, gives the
balance due.
Bule.— 1. Find the number of days to dapse between (h^
date when each item is due and die date of settlement
2. Find the mm of the items on Hie Dr. side, and then add
to each item its interest^ if tJie item is dii^ before the date of
settlement, or »id)tract it if due after the date of settlement; do
the same loith Or. side.
3. The difference between the anumnts on Hie two sides of
the aecouni is Hie cash balance.
294
BAY'S HIGHER ARITHMETIC.
1. Find the equated time and cash balance of the follow
ing account, July 7, 1876 ; also April 30, 1876, interest
^ per annum :
Henry Hammond.
Db.
Or.
1876.
Jan.
4
To Mdse, at 3 mo.
1900.10
Feb.
1
ii ti 44 ^ 44
400.00
44
IS
4t 44 44 O 44
700.50
Mar.
7
44 44 (4 A 44
600.40
April
May
8
•• Cash,
500.20
10
'♦ Mdse, at 30 da.
400.00
June
1
' 1 mo.
100.60
Equated time. May 15, 1876.
Cash balance July 7, 1876, $3633.62
Cash balance April 30, 1876, $3592.80
Kemarks. — 1. When, in forming the several products, the cents
are. 50 or less, reject them; more than 50, increase the dollars bj 1.
2. Tlie cash balance is the sum that Henry Hammond will be
required to pay in settling his account in full at any given date.
2. Find the equated time and cash balance of the follow
ing account, Oct. 4, 1876, money being worth 10^ per
annum :
William Smith.
Dr.
Cr.
1876.
Dol.
Ct.
1876.
Dol.
a.
Jan.
1
To Mdse,
800
00
Jan.
10
By Cash,
400
00
41
16
at 30 da.
180
30
44
28
44 44
200
00
Feb.
14
44 44 44 QQ 44
40()
60
Feb.
15
" Bills Rec. at 60 da.
180
30
Mar.
25
4. (1
500
00
4.
28
'• Cash ,
100
00
April
1
" Cash,
800
00
Mar.
30
" Bills Rec. at 90 da.
450
00
May
7
" Mdse,
600
00
April
May
14
" Cash,
400
60
«i
21
at 60 da.
700
00
1
44 «4
500
00
June
10
4. 44
200
00
.4
15
'• Bills Rec. at 1 mo.
680
OO
44
15
" 90 da.
20(M)
00
June
16
♦' Cash,
300
00
July
12
** *fc
500
00
July
19
44 44
700
ft)
Aug.
4
•* 2 mo.
1000
10
Aug.
10
" Bills Rec. at 20 da.
200
00
Sept.
1
'* Cash,
150
00
Oct.
3
4* ftk
1100
00
Equated time, June 18, 1876; cash balance, $2389.70
Kemark. — All written obligations, of whatever form, for which
a certain amount is to be received, "are called BiUs Receivable.
SETTLEMENT OF ACCOUNTS.
295
3. Find the equated time and the cash balance of
account, March 31, 1876, interest 10% per annum.
Dr.
Gewge Cummings,
Cr.
1876.
Dol.
a.
1875.
Dol J
Ct.
Jan.
2
To Cash,
800
00
Dec.
1876.
1
By Mdse, at 1 mo. '
583
00
«
21
(i K
194
00
Feb.
»
.t tt 14 •(
40
00
Mar.
4
41 t(
150
00
Mar.
30
«l 41 tt 4t
130
00
Equated time Feb. 11, 1876; cash balance, $110.48
Find the interest due, and balance the following ac
counts :
A, L. Morris in acc'l with T, J. Fisher & Co.
4. Dr. Cr.
1871.
Dol.
Ct.
1870.
Dol.
a.
Jan.
13
To check,
350
Dec.
31
By bal. flrom old acc't.
813
64
44
22
44 44
275
1871.
Feb.
25
It It
100
Feb.
4
" cash.
120
May
1
it 44
400
Mar.
17
41 It
500
June
23
• 4 44
108
25
May
31
44 U
84
50
Interest to June 30, at 6 per cent.
Int. due Morris, $13.34; bal. due Morris, S298.23
6. Dr.
Wm. White in accH with Beach & Beiiy.
Cr.
1875.
Dot.
Ct.
1875.
Dol.
Ct.
July
O
««
To check.
212
50
June
80
By bal. from old acc't,
1102
50
4t "^
20
66
July
6
" cash deposit,
50
Aug.
7
2:»
44
15
44 4 44
95
t»
25
300
Aug.
9
44 t4 tt
168
75
Sept.
5
110
Sept.
18
44 4t 44
32
.t
11
46
40
Oct.
3
44 44 4t
79
90
44
27
454
25
Interest to Oct. 12, at 10 per cent.
Int. due White, $19.68; bal. due White, $123.68
296
JIAY'S HIGHER ARITHMETIC.
ACCOUNT SALES.
3dO. 1. An Account Sales is a written statement made
by an agent or consignee to his employer or consignor, of
the quantity and price of goods sold, the charges, and the
net proceeds.
2. Quaranty is a charge made to secure the owner
against loss when the goods are sold on credit.
3. Storage is a charge made for keeping goods, and is usu
ally reckoned by the week or month on each piece or article.
4. In Account Sales the charges for freight, commission,
etc., are the Debits, and the proceeds of sales are the
Credits; the Net Proceeds is the difference between the
sums of the credits and debits.
331. Account Sales are averaged by the following rule :
Bnle. — 1. Average the sales alone; this result wUl he the
date to be given to the commission and guaranty,
2. Make the sales the Or, side, and the charges the Dr.
side, and find the eguaied time for paying the net proceeds.
1. Find the equated time of paying the proceeds on the
following account of Charles Maynard:
Charles Maynard^s Consignment,
1876.
Aug.
12
ti
14
«
24
tt
29
((
3i»
«
31
1876.
Aug.
10
«i
31
ti
31
(1
31
By J. Barnes, at 10 da.^
" Cash
*' Bills Receivable, at 30 da.
" Cash
•• (4eorge Hand ,
** Bills Beceivable. at 20 da.,
<4
(t
((
t<
ti
CHARGES.
To Cash paid, Freight .....$ 75.00
" " •' Storage 10.00
" " " Insurance >6 per cent 10.26
•* " " Commission 2^ per cent „ 205.23
Net proceeds due Maynard.
Dol,
50
60
800
00
850
00
210
00
4900
00
1400
00
8210
800
7910
Ct.
60
51
09
Equated time Sept. 4, 1876.
SETTLEMENT OF ACCOUNTS.
297
2. Make an account sales, and find the net proceeds and
the time the balance is due in the following:
William Thomas sold on account of B. F. Jonas 2000
bu. wheat July 8, 1876, for $2112.50; July 11, 300 bu.
wheat, and took a 20 day note for $362.50 ; paid freight
July 6, 1876, $150.00; July 11, storage, ?6; drayage, $5;
insurance, $4; commission, at 2^%, $61.87; loss and gain
for his net gain, $11.57 ^
Net proceeds, $2236.56; due July 12, 1876.
3.
Md^e. Co. "5."
1870.
July
18
tt
'24
it
30
1876.
July
15
15
30
30
30
30
By Note, at 20 da
ti >t «i 25 '*
" Cash J!!!"!!!!!!!!i!!!!Z!!!!""!!!l!""!!!!iZ!!!!"Z!!!!;!!!
CHARGES.
To Cash paid, Freight f 33.00
" *' " Drayage 4.00
•* " " Inaurunce 1.50
•• *• *• Storage 3.00
" " " Commission, at 2'4 per cent 5.29
C. V. Oanies's uet proceeds 62.73
1109.52
Less our \i net loss 3.93
fl05.59
Find C. V. Cames's net proceeds if paid Jan. 1, 1877,
money being worth 10^ per annum ; also the equated time.
Equated time Aug. 19, 1876; net proceeds, $65.03
STORAGE AC(X)UNTS.
332. Storage Accounts are similar to bank accounts,
one side showing how many barrels, packages, etc., are
received, and at what times; the other side showing how
many have been delivered, and at what times.
Storage is generally charged at so much per month of
30 days on each barrel, package, etc.
298
BAY'S HIGHER ARITHMETIC,
1. Storage to Jan. 31, at 5 ct. a bbl. per month.
RECEIVED.
DELIVERED.
1876.
EU.
Balance on hand.
Day$.
Products.
1876.
Bbl.
January
2
5
7
10
14
17
20
24
28
200
150
30
120
80
150
75
60
200
January
10
13
17
20
25
27
30
31
110
90
20
115
140
72
100
Storage, $19.25; bbl. on hand, 418.
2. Storage to Feb. 20, at 5 ct a bbl. per month.
RECEn^ED.
DELIVERED.
Bbl.
Balance on hand.
Bayi,
Products.
Bbl.
January
31
418
February
5
100
February
4
9
12
16
250
120
lUO
30
10
12
14
18
20
80
220
140
90
288
Storage, $15.85; bbl. on hand, 0.
VIII. COMPOUND' INTEREST.
DEFINITIONS.
333. 1. Compound Interest is interest computed both
upon the principal and upon each accrued interest as
additional principal.
2. Annual Interest is the gain of a principal whose yearly
interests have become debts at simple interest; but, distinct
from this, Com])Oiuid Interest is Hie whole gain of a prhicipal,
increased ai the end of each interval by all the interest drawi
during that interval.
3. The final amount in Compound Interest is called tho
compound amount.
COMPOUND INTERJSST. 299
Example.— Let the principal be $1000, the rate per cent 6. The
first year's interest is $60. If this be added as a new debt, the prin
cipal will become $1060, and the second year's interest $63.60 ; in
like manner, the third principal is $1123.60, and a third year's
interest $67,416; then, the whole amount is $1191.016, and the
whole gain, $191,016
Bemark. — If the above debt be, at the same rate, on annvxd intaest
(Art. 304), the whole amount will be $1190.80, and the whole gain
$190.80; the difference is the interest of $3.60 (an interest upon an
interest debt) for one year, $.216
Compound Interest has four cases.
CASE I.
334. Given the principal, rate, and time, to find the
compound interest and amount.
Problem. — Find the compound amount of $1000, in 4
years, at 2^ per annum
Solution. — Multiplying the principal, $1000, by 1.02, the num
ber expressing the amount of $1 for a year, we have the first year's
amount, $1020. Continuing the use of the factor 1.02 until the
fourth product is obtained, we have for the required amount,
$1082.43216 The same numerical result would have been obtained
by taking 1.02 four times as a factor, and multiplying the product
by 1000.
Remark. — Compound Interest may be payable semiannually or
quarterly, and in such cases the computation is made by a multi
[>lication similar to the last.
Example. — Let the debt be $1000, and let the interest be com
pounded at 2^0 quarterly. In one year there are four intervals,
and, as seen in the last process, the year's amount is $1082.43216
The real gain on each dollar is $.0824+, or, a fraction over S^jf(.
According to the usual form of statement, this debt 1s compounding
300 JRA Y'S HIGHER ARITHMETIC.
" at 8^ per annum^ payable qiMiierly.^^ But this must not be under
stood as an exact statement of the real gain ; for, when the quarterly
rate is 2^, the annual rate exceeds 8^, and when the annual rate is
exactly 8^, the quarterly rate is 1.943^, nearly.
Bemark. — To ascertain the true rate for a smaller interval when
the yearly rate is given, requires to separate the yearly multiplier
into equal factors. Thus the true halfyear rate, when the annual
rate is 21^, is found by separating 1.21 into two equal factors, 1.10
X 1.10; and the quarterly rate, when the annual is 8^, is found by
separating 1.08 into four factors, each nearly 1.01943 It is true th^t
this process is rarely demanded, and that it is very tedious when
'the intervals are small; but, in a proj>er place, it will be a useful
exercise. (See Art. 388).
BiQe. — Find the amount of Hie principal for the firtt in
tervaly at the rate for that intervaly and in like manner treat
the whole debt, at Hie end of each interval, as a principal at
maple interest through the following interval or part of an
interval; (he result will be the compound amount. To find the
compound interest, deduct the orig^inal debt from the compound
amount.
Examples for Practice.
1. Find the compound amount and interest of S3850, for
4 yr. 7 mo. 16 da., at 5%, payable annually.
64826.59, and $976.59
2. The compound interest of $13062.50, for 1 yr. 10 mo.
12 da., at 8%, payable quarterly. $2082.25
3. The compound amount of SIOOO, for 3 yr., at 10%,
payable ^semiannually. $1340. 10
4. What sum, at simple interest, 6^, for 2 jr. d mo. 27
da., amounts to the same as $2000, at compound interest,
for the same time and rate, payable semiannually?
$2016.03
5. What is the difference between the annual and the
compound interest of $5000 in 6 years, 6% per annum?
$22,596
L
COMPOUND INTEEEST. 301
6. Required the amount of $1000, at compound interest,
21%, for 2 yr. 6 mo. $1617.83
Behabks. — 1. In obtaining the answer to the last problem, the
compound amount at the end of the second full interval is treated
as a sum at simple interest for 6 months. But, calculated at a true
halfyear rate, the amount is only $1610.51, and this sum continued
at the same true rate for the remaining half year will amount to the
same sum which the debt would have reached in a full interval ;
for,
$1464.10 X 1.21 = $1771.561 ; and
$1464.10, for 2 intervals, at 10% comp. int. =
$1464.10 X 1.10 X 1.10 = $1771.561
2. Let the student carefully note that the interest drawn during
a year may be considered as the mm of interests compounded
through smaller intervals, at smaller rates. Thus, 6^ a year may
be regarded as the sum of interest compounded through quarterly
intervals at 1.467^, through monthly intervals at .487^, or daily
at .016^, approximately. Suppose 7 months have passed since
interest began. The year may be taken as a period of 12 intervals,
and the interest as having been compounded through 7 of them, at
.487^. If the amount then reached be continued at compound
interest through the other 5 intervals, the amount will be the
same as that of the debt continued to the end of the year at the
full rate.
3. In this view, the statement may be made, general, that the worth
of the debt at any point in a year, is the pHndpalj whichy compounding ai
a true partial rate for the remaining fraction of a year^ wiU amount to the
same sum as the debt continued through that year at the annual rate.
This is in strict accordance with the results obtained by Algebra,
but in the common calculations of Arithmetic, the interest is added
to the debt, and the rate divided, only according to the statements
*' payable annually," "payable quarterly," etc.
CALCULATION BY TABLES.
836. When the intervals are many, the actual multipli
cations become laborious; and, therefore, tables of compound
interest have been prepared to shorten the work.
302
RAY'S HIGHER ARITHMETIC.
Amownl of $1 at Compound Interest in any nujnber of yean, not
exceeding fiftyJive.
Yrs.
2 per cent.
2H per cent.
3 per cent.
3>6 per cent.
4 per cent.
4^ per cent.
1
1.0200 0000
1.0250 0000
1.0300 0000
1.0350 0000
1.0400 0000
1.0450 0000
2
1.0«M (XMX)
1.0506 2500
1.0609 0000
1.0712 2500
1.0816 0000
1.0920 2500
3
1.U612 0800
1.0768 9062
1.0927 27rt»
1.1087 1787
1.1248 6400
1.1411 6612
4
1.0824 ;{216
1.1U88 1289
1.1255 0881
1.1475 2300
1.1698 .5856
1.1925 i860
5
l.KMO 8080
1.1314 0821
1.1692 7407
1.1876 8631
1.2166 5290
1.^61 8194
6
1.1261 6242
1.1596 9342
1.1940 5230
L'lm 5533
1.2653 1902
1.3022 6012
7
l.HSfl 8W7
l.lb86 8575
1.2298 7387
1. 2722 7926
1.3159 3178
1.36H8 618:<
8
M7I6 5938
1.2184 0290
1.2867 70(«
1.3168 0»)4
1.3685 6905
1.4221 0061
9
1.1950 9257
1.2488 6297
1.3047 7318
1.3628 9735
1.4'2:« 11»1
1.4860 9514
10
1.21t>9 9442
1.2800 8454
1.3439 1638
1.4105 9876
1.4802 4428
1,5529 6942
11
1.24.^ 7431
1.3120 8666
1.3842 3;«7
1.4599 6972
1.5394 5406
1.6228 5305
12
1.2682 4179
1.3448 8882
1.4257 6089
1.5110 6866
1.6010 3222
1.6958 8143
13
1.2936 0663
1.3785 1104
1.4685 a371
1.56.39 56JI6
1.6650 7351
1.7721 9610
14
1.3194 7876
1.4129 7382
1.5125 8972
1.6186 9452
1.7316 7645
1.8519 4492
15
1.3458 6834
1.4482 9817
1.5579 6742
1.6753 4883
1.8009 4351
1.9352 8244
16
1.3727 8570
1.4845 0562
1.6047 0644
l.7:»9 mn
1.8729 8125
2.0223 7015
17
1.4002 4142
1.5216 1826
1.6528 4763
1.7946 7555
1.9479 «1050
2.1133 7681
18
1.4*282 4625
1.5596 5»72
1.7024 3306
1.8574 8920
2.0258 1652
2.2084 7877
19
1.4568 1117
1.5986 5019
1.7535 0605
1.9225 01.32
2.1068 4918
2.3078 6031
20
1.4859 4740
1.6386 1644
1.8061 1123
1.9897 8886
2.1911 2:il4
2.4117 1402
21
1.5156 6&34
1.6795 8185
1.8602 9457
2.0591 3147
2.2787 6807
2.5202 4116
22
1.5459 7967
1.7215 7140
1.9161 0341
2.1315 1158
2.3699 1879
2.6336 5201
23
1.5768 9926
1.7646 1068
1.9735 8651
2.2061 1448
2.4647 1555
2.7521 6635
24
1.6084 3725
1.8087 2595
2.0327 9411
2.2833 2849
2.56;*3 W17
2.8760 1383
25
1.6406 0599
1.8539 4410
2.0937 7793
2.3632 4498
2.6658 36:»
3.0054 3446
26
1.6734 1811
1.9002 9270
2.1565 9127
2.4459 5856
2.7724 6979
8.1406 7901
27
1.7068 8B48
1.9478 0UO2
2.2?12 8901
2.5315 6711
2.88:«6858
3.2820 0956
28
1.7410 '2421
1.9964 9502
2.2b. 9 2768
2.6201 7196
2.9987 0832
3.4296 9999
29
1.77)S 4469
2.(M64 07;»
•2.3565 6551
2.7118 7798
3.1 lb6 5145
8.5840 3649
30
J.811! fil58
2.0975 6758
2.4272 6247
2.80B7 9370
8.2433 9751
8.7453 1813
31
1.8475 88S2
2.1500 0677
2.5000 8035
2.9050 3148
8.3731 3341
8.9138 5745
32
1.8845 4059
2.'2U37 5694
2.5750 8276
3.0067 0759
3.5080 5875
4.0899 8101
33
1.9222 3140
2.2588 5086
2.6523 3524
3.1119 4235
3.6483 8110
4.2740 3018
34
1.9606 7603
2.3153 2213
2.7319 0530
3.2208 ma
8.7943 16:{4
4.4663 6154
35
1.9998 8956
2.3732 0519
2.8138 6245
3.3.3a5 9045
3.9460 8899
4.6673 4781
36
2.0898 8734
2.4.325 3532
2.8982 7833
8.4502 6611
4.1039 3255
4.8773 7846
37
2.0SU6 85«)9
2.49:» 4870
2.9852 2668
3.5710 2543
4.2680 8986
5.0968 6049
38
2.1222 9879
2.5556 8242
3.0747 8348
3.6960 1132
4.4.388 1:H5
5.:K62 1921
39
2.1647 4477
2.6195 7448
3.1670 2898
3.8253 7171
4.6163 6599
5.5658 991(8
40
2.2080 3966
2.6850 6384
3.2620 3779
3.9592 5972
4.8010 2063
5.8163 6454
41
2.2522 0046
2.7521 9(M3
3.3598 9893
4.0978 3381
4.9930 6145
6.0781 0094
42
2.2972 4447
2.8209 9520
3.4606 9589
4.2412 5799
5.1927 8391
6.3516 1548
43
2.3431 89:^6
2,8915 2008
3.5645 1677
4.:te97 0202
5.40(« 9527
6.6374 3818
44
2.3900 5:n4
2.9638 0808
3.6714 5227
4.54:« 4160
5.6165 1506
6.9)61 2290
45
2.4378 5421
3.0379 0328
3.7815 9584
4.7023 5655
5.8411 7568
7.2482 4843
46
2.4866 J 129
3.1138 50S6
3.8950 4:^72
4.8669 4110
6.0748 2271
7.5744 1961
47
2.5363 4351
3.1916 9713
4.0118 9503
5.0372 84<>4
6.3178 1562
7.9152 6849
48
2.5870 7039
3.27I4 8956
4.1322 5188
5.2i:t5 8898
6.57<6 2824
8.2714 5557
49
2.6:188 1179
S.mi 7680
4.2562 1944
6.3960 6459
6.83:« 4'J:J7
8.6436 7107
50
2.6915 8803
3.4371 0872
4.3839 0602
5.5849 2686
7.1066 83:35
9.0826 3627
51
2.7454 1979
3.5230 3644
4.5154 2320
5.78tW 9930
7.3909 5068
9.4391 0490
52
2.800:J 2»19
3.6111 1235
4.6508 8590
5.9827 i;i27
7.68<>5 8S71
9.8638 6463
53
2.856:1 3475
3.7013 9016
4.7904 1247
6.1921 0824
7.9940 5226
10.3077 %t&\
54
2.01 :M 6144
3.79:J9 2491
4.9;M1 2485
6.4(188 3202
8.3138 1435
10.7715 8677
55
2.9717 3067
3.8887 7303
5.0821 4iSo9
6.6331 4114
8.6463 6692
11.2563 0817
Svhtract $1 from the Amount in this Table to find tJte Interest.
COMPOUND INTEREST.
303
Amownl of $1 at Compound Interest in any number of years, not
exceeding fiftyfive.
Yrs.
5 per ceut.
6 per cent.
7 per cent.
8 per cent.
9 per cent.
10 per cent.
1
i.(eooooo
1.0600 000
1.0700 000
1.0800 000
1.0900 000
l.KXXl (XXI
2
1.1023 000
i.vim 000
1.1449 000
1.1664 0(N)
1.1881 UN)
1.210UIHXJ
3
1.1576 "^50
1.1910 160
l.'2250 4:iO
1.2597 120
l.29.)0 '2JH)
i.:»io 0(>o
4
1.2155 063
1.2624 770
1.3107 9«i0
1:MH 890
1.4115 816
1.4641 000
5
^ 1.2762 816
1.3382 256
1.4025 517
1.4693 281
1. 5:^86 240
1.6105 100
6
1.3400 936
1.4185 191
1.5007 :m
1.5868 743
1.6771 001
1.7715 610
7
1.4071 004
1.5036 30:j
1.5038 481
l.(i057 815
1.7i:« 243
1.8*280 JttJl
1.9487 171
8
1.4774 554
1.7181 862
1.S509 3112
1.992.) 626
2.14:i5 888
9
1.5513 282
1.6894 790
1.8384 592
l.mn) 046
•2.1718 9;«
2.:i579 477
10
1.6288 tMO
1.7908 477
1.9671 514
2.1589 250
2.3673 637
2.5937 425
11
1.7103 394
1.8982 986
2.1048 5*20
2.3316 300
2.5801 264
2.8531 167
12
1.7958 563
2.0121 965
2.2o21 916
2.5181 701
'2.8126 648
'AASM '284
13
1.8856 491
2.1329 28;t
2.4098 450
2.7196 2:^7
3.0658 046
3.45'22 712
14
1.9799 316
2.2609O40
2.5783 .342
2.9371 936
3.:i417 270
3.7974 9b;j
15
2.0789 282
2.3965 582
2.7590 315
3.1721 601
3.fr4'24 825
4.1772 482
16
2.1828 746
2.5401 517
2.9521 638
3.4259 426
3.970B 059
4.5949 730
17
2.2920 183
2.6927 728
3.1588 152
3.7000 181
4.3276 :«4
5.0544 703
18
2.4066 192
2.8543 :<92
3.3799 323
3.9960 195
4.7171 204
5.5599 173
19
2.5269 502
3.a»5 995
3.6165 275
4.3157 Oil
5.1416 613
6.1159 000
20
2.6532 977
3.2071 355
3.8696 845
4.6609 571
5.6044 108
6.7275 000
21
2.7859 626
3.3995 636
4.1403 621
5.0338 337
6.1088 077
7.4002 499
23
2.92)2 607
3.6035 374
4.4304 017
6.4:i65 404
6.6581) WW
8.1402 749
23
3.0715 2:«
3.8197 497
4 7405 299
5.8714 637
7.'2678 745
8.9543 024
24
3.2250 993
4.0489 346
5.072J 670
6.3411 807
7.9110 832
9.8497 :r27
25
3.3863 519
4.2918 707
5.4274 326
6.WM 752
8.6'230 807
10.8347 059
26
3.5556 727
4.549;j 830
5.8073 529
7.3963 632
9.3991 579
11.9181 765
27
8.7334 56:j
4.8223 459
6.2138 676
7.9880 615
10.2450 821
13.1099 942
28
3.9201 291
5.1116 867
6.&488 3^
8.6271 0&4
11.1671 395
14.4200 936
29
4.1161 350
5.4183 879
7.1142 571
9.3172 749
12.1721 821
Id.mW 930
30
4.3219 424
5.74:M 912
7.6122 550
10.0626 569
13.2676 785
17.4494 im
31
4.^380 393
6.0881 006
8.1451 129
10.8676 694
14.4617 695
19.1943 425
32
4.7649 415
6.45J3 867
8.7152 708
11.7:J70 830
15.76:« 288
21.1137 768
33
5.00:il 885
6.8405 899
9.325!) 398
12.6760 496
17. 1 8*20 284
23.'2251 544
34
5.25:» 480
7.2510 2>3
9.9781 13.)
13.6901 336
18.7284 109
25.5476 699
35
5.5160 IM
7.6860 868
10.6765 815
14.7853 443
20.4139 679
28.1024 369
36
5.7918 161
8.1472 520
11.4239 422
15.9681 718
22.2512 250
30.9126 805
37
6.0814 069
8.6360 871
12.2236 181
17.2456 256
24.'i538 im
34.0039 486
38
6.38>4 773
9.1542 524
13.0792 714
18.6252 756
26.4366 805
37.4043 4:M
39
6 7047 512
9.70t« 075
13.9^8 '204
20.1152 977
28.8159 817
41.1447 778
40
7.0399 887
10.2857 179
14.9744 578
21.7.M5 215
31.4094 200
45.2592 556
41
7.3919 882
10.9028 610
16.0226 690
23.4624 &32
34.2362 679
49.7851 811
42
7.7615 876
11.5570 327
17.1442 568
25.3;i94 819
37.3175 .m
54.7636 992
4}
8.1496 669
12.2504 546
18.3443 548
27.3666 404
40.6761 098
60.2400 692
44
8.5571 503
12.9854 819
19.6284 596
29.5559 717
44.;Wi9 697
66.264U 761
45
8.9850 078
13.7646 108
21.0021 518
31.9204 494
48.3272 861
72.8904 m7
46
9.4342 582
14.5904 875
22.4?26 234
34.4740 853
52.6767 419
80.1795 321
47
9.9059 711
15.4659 167
24.0457 070
37.2320 122
57.4176 486
88.1974 8.5:1
48
10.4012 697
16.3938 717
25.?2S9 065
40.2105 7:J1
62.5852 370
97.0172 :«J8
49
10.9213 331
17.3775 WO
27.5299 300
43.4274 190
68.2179 08:^
106.7189 572
50
11.4673 998
18.4201 543
29.4570 251
46.9016 125
74.;«75 '201
1 17.3908 5'29
51
12.0407 698
19.5233 635
31.5190 168
50.65.37 415
81.0496 969
128.1299 382
52
12.6428 063
20.6968 853
:«.?25:t 480
54.7060 408
88.3441 696
142.04'29 320
53
13.2749 487
2I.9:«6 9*5
:i6.0861 224
59.0825 241
96.2951 449
15«.'2472 252
54
13.9383 961
2:J.2550 204
38.6121 509
6:{.80!)1 '260
104.9617 079
171.8719 477
55
14.6356 300
24.650:j 216
41.3150 015
68.9138 561
114.4082 610
lb9.0591 425
Subtract $1 from the Amount in this Table to find ifie Interest,
304 BAY'S HIGHER ARITHMETIC.
How to use the table in finding the Compound Amount:
1. Observe at what intervals interest is payahky and also Uie
rate 'per interval.
2. If die number of full intervals can be found in tlie year
column, note Hie sum corresponding to it in Hie column under
Hie proper rate; multiply Hiis sum, or its amount for any re
maining fraction of an interval, by the principal,
S. If the number of intervals be not found in the table,
separate Hie whole time into periods whidi are each udtJiin ihe
limits of ihe table; find Hie amount of the principal for one
of them, make Hwi a principal for Hie next, and so mi, till the
whole time has been taken into the calculation.
Examples for Practice.
1. Find the compound amount of $750 for 17 yr., at 6%,
payable annually. $2019.58
2. Of $5428 for 33 yr., 5% annually. $27157.31
3. The compound interest of $1800 for 14 yr., at 8%,
payable semiannually. $3597.67
4. If $1000 is deposited for a child, at birth, and draws
7% compound interest, payable semiannually, till it is of
age (21*yr.), what will be the amount? $4241.26
5. Find the amount of $9401.50, at compound interest for
19 yr. 4 mo., 9%, payable semiannually. $51576.68
6. Find the compound amount of $1000 for 100 yr., at
10%, payable annually. $13780612.34
7. The compound interest of $3600 for 15 yr. , at 8^ ,
payable quarterly. $8211.71
8. The compound interest of $4000 for 40 yr., at 5%,
payable semiannually. $24838.27
9. The compound interest of $1200 for 27 yr. 11 mo. 4 da.,
at 12^, payable quarterly. $31404.74
COMPOUND INTEREST. 305
CASE II.
336. Given the principal, rate, and compound in
terest or amount, to find the time.
Problem. — Find the time in which $750 will amount to
$2000, the interest being 8 %, payable semiannually.
Solution. — Since a compound amount is found by multiplying
the principal by the amount of $1, we here reverse that process, and
say: $2000 i 750 = $2.66666666, the amount of $1, at 4^^. The
number next lower, in the 4^ column, is $2.66583633, the amount
for 25 intervals, and is less than $2.66666666 by $.00083033
Since the amount for 25 intervals will, according to the table,
gain $.10663343 in 1 interval, it will gain $.00083303 in such a frac
tion of an interval as the latter sum is of the former; .00083033;
.10663346 = j^, nearly; hence, the required period is 25 jVr inter
vals of 6 mo., or, 12 yr. 6 mo. 1 da., j4w«.
Bule. — 1. Divide {he amount by the principal.
2. If die quotient he found in the table under the given rate,
ihe years opposite wHl he Hie required number of intervals ; but
if not found eocacUyy in Hie table, take tJie number next less,
noting its deficiency, its number of years, and its gain during
a full interval.
3. Divide the deficiency by the interval gain, and annex the
qux)tient to Hie number of full intervals ; the result will be ihe
required Hine.
Examples for Practice.
In what time, at compound interest, will:
1. $8000 amount to $12000, at 6%?
6 yr. 11 mo. 15 da.
2. $5200 amount to $6508, 6^, payable semiannually?
3 yr. 9 mo. 16 da.
H. A. 26.
306 RAY'S HIGHER ARITHMETIC.
3. jil2500 gain $5500, 10%, payable quarterly?
3 yr. 8 mo. 9 da.
4. U gain »1, at 6, 8, 10%?
11 yr. 10 mo. 21 da.; 9 yr. 2 da.; 7 yr. 3 mo. 5 da.
5. $9862.50 amount to $22576.15, 12%, payable semi
annually? 7 yr. 1 mo. 7 da.
CASE III.
337. Giyen the principal, the compound interest or
amount, and the time, to find the rate.
Problem.— At what rate will $1000 amount to $2411.714
in 20 years?
Solution.— Dividing $2411.714 by 1000, we have $2.411714,
which, in the table, corresponds to the amount of $1 for the time,
at ^(fo^
Rule. — Divide Hie amount by ihe principal; search in (ke
table, opposite ihe given number of full intervals^ for the eicact
q^wtient or ilie number nearest in value; if the time contain
also a part of an interval, find the amount of the tabular sum
for tJiat time, before comparing with the quotient; the rate per
cent at the head of the column will be the exact, or ihe approx
imate rate.
Examples for Practice.
At what rate, by compound interest,
1. Will $1000 amount to $1593.85 in 8 yr.? 6%.
2. $3600 amount to $9932.51 in 15 yr.? 7%.
3. $13200 amount to 48049.58, in 26 yr. 5 mo. 21 da. ?
5%.
4. $2813.50 amount to $13276.03, in 17 yr. 7 mo. 14 da.,
interest payable semiannually? 9%.
COMPOUND INTEEEST. 307
5. $7652.18 gain $17198.67, interest payable quarterly,
in 11 yr. 11 mo. 3 da.? 10^.
6. Any sum double itself in ^0, 15, 20 yr. ?
1st, between 7% and 8^; 2d, nearly 5^;
3d, little over ^%.
CASE IV.
338. Given the compound interest or amount, the
time, and the rate, to find the principal.
Problem. — ^What principal will yield 831086.78 com
pound interest in 40 yr., at 8^?
Solution.— In 40 yr. $1 will gain $20.7245215, at the given rate;
the required principal must contain as many dollars as this interest
is contained times in the given interest; $31086.78^20.7245215 =
$1500, ^718.
Bule. — Divide Hie given interest or amount by the interest
or amount of $1 for the given time and at ike given rate; the
quotient will he the required principal.
Kemark. — If the amount be due at some future time, the prin
cipal is the present worth at compound interest, and the difference
between the amount and present worth is the compound discount.
Examples for Pra^ctice.
What principal, at compound interest,
1. Will yield $52669.93 in 25 yr., 6^? $16000.
2. Will gain $1625.75 in 6 yr. 2 mo., 7^, payable semi
annually? $3075.
3. WiU yield $3598.61 in 3 yr. 6 mo. 9 da., 10^, payable
quarterly ? $8640.
4. Will yield $31005.76 in 9^ yr., at 8%, payable semi
annually? $28012.63
308 RAY'S HIGHER ARITHMETIC.
5. Will amount to $27062.85 in 7 yr., at 4% ?
$20565.54
6. What is the present worth of $14625.70, due in 5 yr.
9 mo., at 6^ compound interest, payable semiannually ?
$10409.77
7. What is the compound discoimt on $8767.78, due in
12 yr. 8 mo. 25 da., 5%? $4058.87
IX. ANNUITIES.*
DEFINITIONS.
339. 1. An Annuity is a sum of money payable at
yearly or other regular intervals.
{1. Perpetual or Limited.
2. Certain or Contingent.
3. Immediate or Deferred.
2. A Perpetual Annuity, or a Perpetuity, is one that
continues forever.
3. A Limited Annuity ceases at a certain time.
4. A Certain Annuity begins and ends at fixed times.
5. A Contingent Annuity begins or ends with the hap
pening of a contingent event — as the birth or the death of a
person.
6. An Immediate Annuity is one that begins at once.
7. A Deferred Annuity, or an Annuity in Reversion,
is one that does not begin immediately ; the term of the re
version may be definite or contingent.
* Since the problems in annuities may be classed usder the
Applications of Percentage, the subject is presented here, instead
of being placed after Progression ; however, those who prefer may
omit this chapter until after Progression has been studied.
ANNUITIES. 309
8. An annuity is Forborne or in Arrears if not paid
when due.
9. The Forborne or Final Value of an annuity is the
amount of the whole accumulated debt and interest, at the
time the annuity ceases.
10. The Present Value of an annuity is that sum, which,
put at interest for the given time and at the given rate,
will amount to the final value.
11. The value of a deferred annuity at the time it com
mences, may be called its Initial Value ; its Present Value
is the present worth of its mitial value, at an assumed rate
of interest.
12. The rules for annuities are of great importance ; their
practical applications include leases, lifeestates, rents, dowers,
pensions, reversions, salaries, life insurance, etc.
Kemark. — An annuity begins, not at the time the first payment
is made, but one interval before ; if an annuity begin now, its first
payment will be a year, halfyear, or quarter of a year hence, accord
ing to the interval named.
CASE I.
«
340. Given the payment, the interval,* and the
rate, to find the initial value of a perpetuity.
Problem. — What is the initial value of a perpetual lease
of $250 a year, allowing 6^ interest ?
OPERATIOK.
Solution. — The initial value must $ 1
be the principal, which, at 6^, yields . 06
$250 interest every year ; it is found, .06)250.0000
by Art 300, $4166.66 1, Am.
Bule. — Divide tke given payment by the interest of $1 for
one interval at ike proposed rate.
310 RA Y'S HIGHER ARITHMETIC,
Examples for Practice.
1. What is the initial value of a perpetual leasehold of
$300 a year, allowing 6^ interest? $5000.
2. What must I pay for a perpetual lease of $756.40 a
year, to secure 8% interest? $9455.
3. Ground rents on perpetual lease, yield an income of
$15642.90 a year: what is the present value of the estate,
aUowing 7% interest? $223470.
4.^ What is the initial value of a perpetual leasehold of
$1600 a year, payable semiannually, allowing 5^ interest,
payable annually? $32400.
Suggestion. — Here tlie yearly payment is $1620, by allowing 5^
interest on the halfyearly payment first made.
5. What is the initial value of a perpetual leasehold of
$2500 a year, payable quarterly, interest 6% payable semi
annually; 6% payable annually: 6% payable quarterly?
$41979.161; $42604.16; $41666. 66f
CASE II.
341. To find the present value of a deferred per
petuity, when the payment, the interval, the rate,
and the time the perpetuity is deferred are known.
Problem. — Find the present value of a perpetuity of
$250 a year, deferred 8 yr., allowing 6% interest.
Solution. — Initial value of perpetuity of $250 a year, by last
rule = $4166.66§ The present value of $4166.66f , due 8 yr. hence,
at 6^ compound interest, = $4166.66§ ^ 1.5938481 (Art. 335 ). Use
the contracted method, reserving 3 decimal places ; the quotient,
$2614.22, is the present value of the perpetuity.
Bule. — Find the initial value of tJie perpetuity by tlie last
rule; then find the present worili of this mm for the time the
ANNUITIES. 311
'perpetuity is deferred, by Case IV of Compound Interest ; iJm
wiU be the present vakie required.
Examples for Practice.
1. Find the present value of a perpetuity of $780 a year,
to commence in 12 yr., int. 5%. $8686.66
2. Of a perpetual lease of $160 a year, to commence iu
3 yr. 4 mo., int. 7%. $1823.28
3. Of the reversion of a perpetuity of $540 a year, de*
ferred 10 yr., int. 6%. $5025.55
4. Of an estate which, in 5 yr. , is to pay $325 a year for
ever: int. 8%, payable semiannually. $2690.67
5. Of a perpetuity of $1000 a year, payable quarterly, to
commence in 9 yr. 10 mo. 18 da., int. 10^, payable semi
annually. $3858.88
CASE III.
342. Given the rate, the payment, the interval^
and the time to run, to find the present value of an
annuity certain.
Problem. — 1. Find the present value of an immediate
annuity of $250 continuing 8 years, 6% interest.
Solution.
Present value of immediate perpetuity of $250, . . . = $4166.67
Present value of perpetuity of $250, deferred 8 yr., . . = 2614.2 2
Pres. val. of immediate annuity of $250, running 8 yr., = $1552.45
Problem. — 2. The present value of an annuity of $680,
to commence in 7 yr. and continue 10 yr., 5% int.
Solution.
Pres. val. of perpetuity of $680, deferred 7 yr., at 5^o, =$9665.27
Pres. val. of perpetuity of $680, deferred 17 yr., at 6^, = 5933.64
Pres. val. of annuity of $680, deferred 7 yr., to run 10 yr. =$3731.63
312 I^A Y'S HIGHER ARITHMETIC.
Rule. ^i^ind f/ie 'present valv^e of ttoo perpetuities having
the given rate, payment, and interval, one of them commencing
when die annuity commences, and the other when the annuity
ends. The difference betioeen these values will he the present
value of the annuity.
Notes. — 1. This rule applies whether the annuity is immediate or
deferred ; in the latter, the time the annuity is deferred must be
known, and used in getting the values of the perpetuities.
2. By using the initial instead of the present values of the per
petuities, the rule gives the initial value of the deferred annuity,
which may be used in finding its final or forborne value. (Rem. 1,
Case IV.)
Examples for Practice.
1. Find the present value of an annuity of $125, to com
mence in 12 yr. and run 12 yr., int. 7^. $440.83
2. The present value of an immediate annuity of $400,
running 15 yr. 6 mo., int. 8%. $3484.41
3. The present value of an annuity of $826.50, to com
mence in 3 yr. and run 13 yr. 9 mo., int. 6%, payable
semiannually. $6324. 69
4. The present value of an annuity of $60, deferred 12 yr.
and to run 9 yr., int. 4^%. $257.17
5. Sold a lease of $480 a year, payable quarterly, having
8 yr. 9 mo. to run, for $2500: do I gain or lose, int. S%,
payable semiannually? Lose $509.96
CASE IV.
343. Given the payment, the interval, the rate, and
time to run, to find the final or forborne value of an
annuity.
Problem. — Find the final or forborne value of an annuity
of $250, continuing 8 yr., int. 6^.
ANNUITIES. 313
Solution.— The initial value of a perpetuity of $250, at 6^, =
$4166.66§ ; its compound interest, at 6% for 8 yr., = $4166.66J X
.5938481 = $2474.37, the final or forborne value of the annuity.
Bule. — Consider iJie annuity a perpetuity, and find its
initial value by Case J. The compound interest of this sum,
at tJie given rate for the time iJie annuity runs, mil be ike
filial or forborne value.
Notes. — 1. The final or forborne value of an annuity may be
obtained by finding first the initial value, as in Case III, and then
the compound amount for the time the annuity runs.
2. The present value of an annuity can be obtained by finding
first the forborne value, as in this case, and then the present worth
for the time the annuity runs.
Examples for Practice.
1. Find the forborne value of an immediate annuity of
8300, running 18 yr., int. b%. $8439.72
2. A pays $25 a year for tobacco: how much better off
would he have been in 40 yr. if he had invested it at 10^
per annum? $11064.81
3. Find the forborne value of an annuity of $75, to com
mence in 14 yr., and run 9 yr., int. 6^. $861.85
Suggestion. — The 14 yr. is not used.
4. A pays $5 a year for a newspaper: if invested at 9%,
what will his subscription have produced in 50 yr.?
$4075.42
5. An annuity, at simple interest 6%, in 14 yr., amounted
to $116.76 : what would have been the difference had it
been at compound interest 6%? $9.33
6. A boy just 9 yr. old, deposits $35 in a bank : if he
deposit the same each year hereafter, and receive 10%, com
pound interest, what will be the entire amount when he is
of age? $858.29
H. A. 27.
314
HAY'S HIGHER AMITHMETia
Hie present vcdue of $1 per annum in any number of years, not
exceeding fiftyjwe.
Yre.
4 per cent.
6 per cent.
6 per cent.
7 per ceni.
8 per cent.
10 per cent.
1
.961538
.952381
.943396
.934579
.925928
.909091
2
1.H86005
1.859410
1.833393
1.806018
1.783255
1.735537
3
2.775091
2.723248
2.673012
2.624316
2.577097
2.486852
4
3.629895
3.54S9S1
3.465106
3.387211
8.312127
3.169865
5
4.451822
4.329477
4.212864
4.100197
3.992710
8.790787
6
5.242137
5.075692
4.917321
4.766540
4.622880
4.355261
7
6.002055
5.7863TS
5.582381
5.389289
5.206370
4.868419
8
6.732745
6.46;<213
6.209794
5.971299
5.746039
5.334926
9
7.435332
7.107822
6.801692
6.515232
6.246888
5.759024
10
8.110696
7.721735
7.360087
7.023582
6.710081
6.144567
11
8.760477
8.306414
7.88&S75
7.498674
7J38964
6.495061
12
9.385074
8.86:K52
8.383844
7.942688
7.536078
6.813692
13
9.986<M8
9.3D3573
8.852683
8.357651
7.903776
7.103366
14
10.563123
9.898641
9.294984
8.745468
8.244237
7.366687
15
11.118387
10.379658
9.712249
9.107914
8.559479
7.606080
16
11.652296
10.837770
10.106895
9.446649
8.851369
7.823700
17
12.165669
11.274066
10.477260
9.763223
9.121638
8.021553
18
12.659297
11.689587
10.827603
10.050087
9.371887
8.201412
19
13.133939
12.065321
11.158116
10.335595
9.603599
8.364920
20
13.590326
12.462210
11.469921
10.594014
9.818147
8.513564
21
14.029160
12.821153
11.764077
10.835527
10.016803
8.648694
22
14.451115
13.163003
12.041582
11.061241
10.200744
8.771540
23
14.856842
13.488574
1*2.303379
11.272187
10.371059
8.883218
24
15.246963
13.798642
12.550353
11.469334
10.628758
8.984744
25
15.622080
14.093945
12.783356
11.653583
10.674776
9.077040
26
15.982769
14.375185
13.003166
11.825779
10.809978
9.160945
27
16.329586
14.643031
13.210534
11.986709
10.935165
9.287223
28
16.663063
14.898127
13.406164
12.137111
11.051078
9.306567
29
16.98:i715
15.141074
13.fi90721
12.277674
11.158406
9.369606
30
17.2921133
15.872451
13.764831
12.409041
11.257783
9.426914
31
17.588494
15.592811
13.929086
J2.531814
11.349799
9.479013
32
17.873552
15.802677
14.084043
12.646555
11.434999
9.526376
33
18.147646
16.002549
14.230230
12.763790
11.613888
9.569432
34
18.411198
16.192»04
14.368141
12.854000
11.586934
9.606575
35
18.664613
16.374194
14.498246
12.917672
11.654568
9.644159
36
18.908?82
16.546852
14.620987
13.085208
11.717193
9.676506
37
19.142579
16.711287
14.736780
13.117017
11.776179
9.706917
38
19.367864
16.867893
14.846019
13.193473
11.828869
9.732651
39
19.584485
17.017041
14.949075
13.264928
11.878582
9.756056
9.779051
40
19.792774
17.159086
15.046297
13.331709
11.924613
41
19.993052
17.294368
15.138016
13.394120
U. 967235
9.799137
42
20.185627
17.423208
15.224543
13.452449
12.006690
9.817397
43
20 370795
17.545912
15.306173
13.506962
12.043240
9.833998
44
20.548841
17.6627T3
15.383182
13.557908
12.077074
9.849089
45
20.720040
17.774070
15.455832
13.605522
12.108402
9.862806
46
20.884654
17.880067
15.521370
13.650020
12.137400
9.875280
47
21.042936
17.981016
15.589028
13.691608
12.164267
9.886618
48
21.195131
18.077158
15.650027
13.730474
12.189136
9.890926
49
21.3414T2
18.168?22
15.7075?2
13.766799
12.212163
9.906296
50
21.482185
18.255925
15.761861
13.800746
12.233485
9.914814
51
21.617485
18.338977
15.813076
18.832473
12.253227
9.922550
52
21.747582
18.418073
15.861393
13.862124
12.271S06
9.929590
53
21.872675
18.493403
15.906974
13.889836
12.288432
9.935900
54
21.992957
18.565146
15.949976
13.915735
12.304103
9.941817
55
22.106612
18.633472
15.990543
13.939939
12.318614
9.947107
ANNUITIES. 315
CALCULATIONS BY TABLE.
344. By the table on page 314, some interesting and
important cases in annuities can be solved, among which are
the following three :
CASE V.
346. Oiven the rate, time to run, and the present
or final value of an annuity, to find the payment.
Problem. — An immediate annuity running 11 yr., can be
purchased for $6000.: what is the payment, int. 6%?
Solution. — The present value of an immediate annuity of $1 for
11 yr., at 6^o, is $7x886875; $6000 divided by this, gives $760.76, the
payment required.
*
Bule. — Assume $1 for the payment; determine the present
or final value on this suppositiony and divide the given present
or final value by it.
Examples for Practice.
1. How much a year should I pay, to secure $15000 at
the end of 17 yr., int. 7%? $486.38
2. What is the payment of an annuity, deferred 4 yr.,
running 16 yr., and worth $4800, int. 6%? $599.64
CASE VI.
346. Given the payment, the rate, and present value
of an annuity, to And the time it runs.
Problem. — In what time will a debt of $10000, drawing
interest at 6%, be paid by installments of $1000 a year?
Solution. — The $10000 may be considered the present value of
an annuity of $1000 a year at 6^©; but $10000 t 1000 = $10, the
316 BAY'S HIGHER ARITHMETia
present value of an annuity of $1 for the same time and rate ; by
reference to the table, the time corresponding to this present value,
under the head of 6^, is 15 yr.; the balance then due may be thus
found :
Comp. amt. of $10000 for 15 yr., at 6/<, (Art. 336), . = $23965.58
Final val. of annuity $1000 for 15 yr., at 6^o (Art. 343), = 23275.97 ^
Balance due at end of 15 yr $689.61
Bule. — Divide the present value by Hie payment, and look
in the table, under the given rate, for tlie quotierd; the number
of years corresponding to the quotierd or to Hie tabidar number
next less, will be the number of full intervals required,
»
Note. — The difference between the compound amount of the debt,
and the forborne value of the annuity, for that number of intervals,
will be the unpaid balance.
Examples for Practice.
1. In how many years can a debt of $1000000, drawing
interest at 6%, be discharged by a sinking fund of $80000 a
year ? 23 yr., and $60083.43 then unpaid.
2. In how many years can a debt of $30000000, drawing
interest at 5^, be paid by a sinking fund of $2000000?
28 yr., and $798709.00 unpaid.
3. In how many years can a debt of $22000, drawing 7^
interest, be discharged by a sinking fund of $2500 a year ?
14 yr., and $351.53 then unpaid.
4. Let the conditions be the same as those of the illus
trative example, and let each $1000 payment be itself a
year's accumulation of simple interest: what would be the
whole time required to discharge the debt?
15 yr. 8 mo. 19 da.
5. Suppose the national debt $2000000000, and funded at
4%: how many years would be required to pay it off, by a
sinking fund of $100000000 a year?
41 yr., and $3469275 unpaid.
CONTINGENT ANNUITIES. 317
CASE VII.
347. Given the payment, time to run, and present
value of an annuity, to And the rate of interest.
Bule. — Divide Vie preient value by the payment ; Hie quotient
will be the present value of $1 for the given time and rate;
look in the table and opposite Vie given number of years for
t/ie qmtievd or the tabxdar number of nearest value, and at the
head of tlie column wUl be found the rale, or a number as
near the true rale as tJie table can exhibit.
Examples fob Practice.
1. If $9000 is paid for an immediate annuity of $750, to
run 20 yr., what is the rate? About 5^%.
2. If an immediate annuity of $80, running 14 yr., sells
for $650, what is the rate? 8%+.
CONTINGENT ANNUITIES.
DEFINITIONS.
348. 1. Ck>ntingent Annuities comprise Life Annuities,
DowerSf Pensions, etc.
2. The value of such annuities depends upon the erpecta
tion (f life.
3. Exi>ectation of Life is the average number of years
that a person of any age may be expected to live.
4. Tables, called ** Mortality Tables," have been prepared
in England and in this country for the purpose of ascertain
ing how many persons of a given number and of a certain
318
BAY'S HIGHER ARITHMETia
age would die during any one year, and in how many years
the whole number would die.
Remark. — ^These tables, though not absolutely accurate, are
based upon so large a number of observations that their approx
imation is very close. Legal, medical, and scientific authorities
use them in discussing vital statistics, and insurance companies
make them a basis for the transaction of business.
349. The following table differs but slightly from those
prepared in this country:
Carlisle Table
Of Mortality y hosed upon Observations at Carlisle (Eng.), showing the
Rate of Extinction of lOfiOO lives.
*^ ^
•M
*^ ^
«p^
**•* _•
•M
•
o
<
o E
S t
B 1
4>
bo
2g
2 »«
£
u
B t
.
^1
3 3
S ft
3 3
3
3 B
3 W
525 Cfi
52;
52; C«
Jz;
5Z; CO
^
10000
1539
35
5362
55
70
2401
124
1
8461
682
36
5307
66
71
s 2277
134
2
7779
505
37
5251
57
72
2143
146
8
7274
276
38
5194
58
73
1997
156
4
6998
201
39
5136
62
74
1841
166
5
6797
121
40
5075
66
75
1675
160
6
6676
82
41
5009
69
76
1515
156
7
6594
58
42
4940
71
77
1359
146
8
6536
43
43
4869
71
78
1213
132
9
6493
33
44
4798
71
79
1081
128
10
6460
29
45
4727
70
80
953
1J6
11
6431
31
4C
4657
69
81
837
112
12
6400
32
47
4588
67
82
725
102
13
6368
33
48
4521
63
83
623
94
14
6335
35
49
44^
61
84
529
84
15
6300
39
50
4397
59
85
445
78
16
6261
42
51
4338
62
86
367
71
17
6219
43
52
4276
65
87
296
64
18
6176
43
53
4211
68
88
232
51
19
6133
43
M
4143
70
89
181
39
1 20
6090
43
55
4073
73
90
142
37
! 21
6047
42
56
4000
76
91
105
30
22
6005
42
57
3924
82
92
75
21
23
5963
42
58 '
3842
93
93
54
14
24
5921
42
59
3749
106
94
40
10
25
5879
43
60
3643
122
95
30
7
26
5836
43
61
3521
126
96
23
5
27
5793
45
62
3395
127
97
18
4
28
574S
50
63
3268
125
98
14
3
29
5698
56
64
3143
125
99
11
2
30
5642
57
65
3018
124
100
9
2
31
5)85
57
66
2894
123
101
7
2
32
5528
56
67
2771
123
102
5
2
33
5472
55
68
2648
123
103
3
2
34
5417
55
69
2525
124
104
1
1
CONTINOENT ANNUITIES.
319
Table
Showing the wives of AnnuUies on Single Lives, ojceording to ike
Carlisle Table of Mortality.
Age.
4 per ct.
5 per ct.
6 per ct.
7 per ct.
Age.
4 per ct.
5 per ct.
6 per ct.
7perct.
14.28164
12.083
10.439
9.177
' 52
12.25793
11.154
10.208
9.392
1
16.55455
13.995
12.078
10.6(»
53
11.94508
10.892
9.988
9.205
2
17.72616
14.983
12.925
11.342
54
11.62673
10.621
9.761
9.011
3
18.71508
15.821
13.652
11.978
55
11.29961
10..347
9.524
8.807
4
19.23133
16.271
14.042
12.322
56
10.96607
10.063
9.280
8.595
5
19.592(13
16.590
14.325
12.574
57
10.62559
9.771
9.027
8.875
6
19.74502
16.735
14.460
12.698
58
10.28647
9.478
8.772
8.153
7
19.79019
16.790
14.518
12.756
59
9.96331
9.199
8..529
7.940
8
19.76443
16.786
14.626
12.770
60
9.66333
8.940
8.304
7.743
9
19.69114
16.742
14.500
12.7W
61
9.39809
8.712
8.108
7.572
10
J9.98339
16.669
14.448
12.717
62
9.13676
8.487
7.913
7.408
11
19.45857
16.581
14.384
12.669
e:i
8.87150
8.258
7.714
7.229
12
19.334^3
16.494
14.:J21
12.621
04
8.593:iO
8.016
7.502
7.(M2
13
19.20937
16.406
14.257
12.572
65
8.30719
7.765
7.281
6.847
14
19.08182
16.316
14.191
12.522
66
8.00966
7.503
7.049
6.641
15
18.95534
16.227
14.126
12.473
67
7.69980
7.227
6.803
6.421
16
18.83636
16.144
14.067
12.429
68
7.37976
6.941
6.546
6.189
17
18.72111
16.066
14.012
12.389
69
7.04881
6.643
6.277
5.945
18
18.60656
15.987
13.956
12.348
70
6.70936
6.:»6
5.998
5.690
19
18.48649
15.904
13.897
12.305
71
6.35773
6.015
5.704
5.420
20
18.36170
15.817
13.835
12.259
72
6.02548
5.711
5.424
5.162
21
18.23196
15.726
13.769
12.210
7:^
5.72465
5.435
5.170
4.927
22
18.09386
15.628
13.697
12.156
74
5.45812
5.190
4.944
4.719
2)
17.95016
15.525
13.621
12.098
75
5.23901
4.989
4.760
4.549
24
17.80058
15.417
13.541
12.037
76
5.02399
4.792
' 4.579
4.382
25
17.64486
15.303
13.456
11.9?2
77
4.8247:}
4.609
4.410
4.227
26
17.48586
.15.187
13.368
11.901
78
4.62106
4.422
4.238
4.067
27
17.32023
15.065
13.275
11.832
79
4.;»:J45
4.210
4.010
3.883
28
17.15412
14.942
13.182
11.759
80
4.182»9
4.015
3.858
8.713
29
16.99683
14.827
13.096
11.693
81
3.95309
8.799
3.656
8.523
30
16.85215
14.723
18.020
11.636
82
8.74634
3.606
8.474
3.352
31
16.70511
14.617
12.942
11.578
83
8.5.3409
3.406
8.'286
3.174
82
16.55246
14.506
12.860
11.516
84
3.32856
3.211
3.102
2.999
33
16.39072
14.387
12.771
11.448
85
3.11515
3.009
2.909
2.815
84
16.21943
14.260
12.675
11.374
86
2.92831
2.830
2.?39
2.652
85
16.04123
14.127
12.573
11.295
87
2.77593
•2.685
2.599
2.519
36
15.85577
13.987
12.465
11.211
88
2.68337
2.597
2.515
2.439
37
15.66586
13.843
12.354
11.124
89
2.57704
2.495
2.417
2.344
38
15.47129
1.3.695
r2.2:»
11.0:)3
90
2 41621
2.3.39
2.266
2.198
39
15.27184
13.542
12.120
10.939
91
2.:)9835
2.321
2.248
2.180
40
15.07363
13.390
12.002
10.845
92
2.49199
2.412
2.337
2.266
41
14.88314
13.245
11.890
10.757
93
2.59955
2.518
2.440
2.367
42
14.69466
13.101
11.779
10.671
94
2.64976
2.569
2.492
2.419
4:t
14.50529
12.957
11.668
10.585
95
2.67433
2.596
2.522
2.451
44
14.30874
12.806
11.551
10.494
96
2.62779
2.555
2.486
2.420
45
14.10460
12.648
11.428
10.397
97
2.492M
2.428
2.368
2.309
•M
13.88928
12.480
11.296
10.292
98
2.33222
2.278
2.227
2.177
47
13.66208
12.:«)1
11.154
10.178
99
2.08700
2.045
2.004
1.964
48
13.41914
12.107
10.998
10.aV2
100
1.65282
1.624
1.596
1.569
49
13.15312
11.892
10.823
9.908
101
1.210(»
1.192
1.175
1.159
50
12.86902
11.660
10.6.31
9.749
102
0.76183
0.753
0.744
0.7.35
51
12.56581
11.410
10.422
9.573
103
0.3JOal
0.317
0.314
0.312
320 BAY'S HIGHER ARITHMETIC.
CALCULATIONS BY TABLE.
860. The preceding table of life annuities shows the
sum to be paid by a person of any age, to secure an an
nuity of SI during the life of the annuitant.
CASE I.
851. To find the value of a given annuity on the
life of a person whose age is known.
Bule. — Find from the table the value of a life annuity of
$1, for Hie given a^e and rate of interest, and mvltipLy it by
the payinent of Hie given annuity.
Remarks. — 1. To find the value of a lifeeBtate or widow's dower
(which Ib a lifeefltate in one third of her husband's real estate):
Estimate the value of the property in which the lifeestate is held; the yearly
interest of this sum, at an a(p eed rate, will be a lifeannuity, whose value for
the given aye and rate will be the value of the lifeestate,
2 The reversion of a lifeannuity, lifeestate, or dower is found
by deducting its value from the value of the property.
Examples for Practice.
1. What must be paid for a lifeannuity of $650 a year,
by a person aged 72 yr., int. 7%? $3355.30
2. What is the lifeestate and reversion in $25000, age
55 yr., int. 6%? Lifeestate, $14286; rev., $10714.
3. The dower and reversion in $46250, age 21 yr., int.
6%? Dower, $12736.33; rev., $2680.34
CASE II.
352. To find how large a lifeannuity can be pur
chased for a given sum, by a person whose age is
known.
CONTINGENT ANNUITIES. 321
Bule. — Assume $1 a year for Uw annuity; find from the
table its value for tlis given age and rate of interest, and divide
Hie given cost by it; Hie quotient vnU be Hie 'payment required.
Examples for Practice.
How large an annuity can be purchased :
1. For $500, age 26 yr., int 6%?. $37.40
2. For $1200, age 43, int. bfo ? $92.61
3. For $840, age b^, int. 7^? $103.03
CASE III.
853. To find the present value of the reversion
of a given annuity ; that is, what remains of it, after
the death of its possessor, whose age is known.
Hule. — Fiud the present value of Hie annuity during its
whole continua'nee; find its value during tfve given life; their
difference will be Hie value of the reversion.
Note. — It will save work, to consider (he annuity as $1 a year^ then
apply the rule, using the tables in Art. 335 and Art. 350, and multiply
the result by the given payriienL
Examples for Practice.
1. Find the present value of the reversion of a perpetuity
of $500 a year, after the death of a person aged 47, int.
5%. $3849.50
2. Of the reversion of an annuity of $165 a year for 30
yr., after the death of a person 38 yr. old, int. 6%.
$251.76
3. Of the reversion of a lease of $1600 a year, for 40 yr.,
afl«r the death of A, aged 62, int. 7^. $9485.93
322 BA Y'S HIGHER ARITHMETIC,
PERSONAL INSURANCE.
DEFINITIONS.
354. 1. Personal Insurance is of two kinds: (1.) lAje
[nsurance; (2.) Accident Insurance,
2. life Insurance is a contract in which a company
agrees, in consideration of certain premiums received, to pay
a certain sum to the heirs or assigns of the insured at his
death, or to himself if he attains a certain age.
3. Accident Insurance is indemnity against loss by
accidents.
4. The Policies issued by life insurance companies are:
(1.) Tenn Policies; (2.) Ordinary Life Policies; (3.) Joint
Life Policies; (4.) Endovrment Policies; (5.) Reserved Endoiv
ment Policies; (6.) Tontine Savings Fund Policies.
5. The chief policies are, however, the Ordinary Life and
the Endowment.
6. The Ordinary Life Policy secures a certain sum of
money at the death of the insured. Premiums may be paid
annually for life, semiannually, quarterly, or in one pay
ment in advance; or the premiums may be paid in 5, 10, 15,
or 20 annual payments.
7. An Endowment Policy secures to the person insured
a certain sum of money at a specified time, or to his heirs or
assigns if he die before that time.
Remark. — It will be advantageous for the student to examine
an "application" and a "policy" taken from some case of actual
insurance ; by a short study of such papers, the nature of the insur
ance contract will be learned more easily than by any mere verbal
description; additional light may be had from the reports pub
lished by various companies.
355. 1. The following is a condensed table of one of the
leading companies:
PERSONAL INSURANCE.
323
Table.
Annual JFVemwm Rates for an Insurance of flOOO.
LIFE POLICIES.
ENDOWMENT POLICIES.
Payable at death only.
Paynbl
e as indicated, or at death, if prior.
Age.
Annual Payments.
Single
Payment
Age.
In
10 years
In
15 years
In
20 years
For life
20 years
10 years
•
20 to
•
20 to
25
919 89
927 39
$42 56
$326 58
26
$103 91
$66 02
$47 68
26
20 40
27 93
43 37
832 58
26
104 03
66 15
47 82
27
20 93
28 50
44 22
338 83
27
104 16
66 29
47 98
28
21 48
29 09
45 10
345 31
28
104 29
66 44
48 15
29
22 07
29 71
46 02
352 05
29
104 43
66 60
48 33
30
22 70
80 36
46 97
859 05
30
104 58
66 77
48 53
31
23 35
81 03
47 98
366 33
81
104 75
66 96
48 74
32
24 05
81 74
49 02
873 89
82
104 92
67 16
48 97
38
24 78
82 48
50 10
881 73
83
105 11
67 36
49 22
34
25 56
83 26
5122
389 88
84
105 31
67 60
49 49
35
26 38
84 08
52 40
898 34
35
105 53
67 85
49 79
36
27 25
84 93
58 63
407 11
86
105 75
68 12
50 11
87
28 17
85 83
54 91
416 21
87
106 00
68 41
50 47
88
29 15
86 78
56 24
425 64
88
106 28
68 73
50 86
89
30 19
87 78
57 63
435 42
89
106 58
69 09
6130
40
3180
88 &S
59 09
445 55
40
106 90
69 49
 61 78
41
32 47
89 93
60 60
456 04
41
107 26
C9 92
52 31
42
33 72
41*10
62 19
466 89
42
107 65
70 40
62 89
43
35 05
42 84
68 84
478 11
43
108 08
70 92
63 54
44
36 46
43 64
65 57
489 71
44
108 55
71 50
64 25
45
37 97
45 03
67 37
501 69
45
109 07
72 14
55 04
46
39 58
46 50
69 26
514 04
46
109 65
72 86
55 91
47
41 30
48 07
71 25
526 78
47
110 80
73 66
56 89
48
43 13
49 73
73 32
539 88
48
111 01
74 54
57 96
49
45 09
51 50
75 49
553 33
49
111 81
75 51
69 15
50
47 18
53 38
77 77
567 13
50
112 68
76 59
60 45
51
49 40
55 38
80 14
581 24
51
113 64
77 77
61 90
52
5178
57 51
82 63
595 66
52
114 70
79 07
63 48
53
54 31
59 79
85 22
610 36
53
115 86
80 51
65 22
54
57 02
62 22
87 94
625 33
54
117 14
82 09
67 14
55
59 91
64 82
90 79
640 54
55
118 54
83 82
69 24
56
63 00
67 60
93 78
655 99
56
120 09
85 73
57
66 29
70 59
96 91
671 64
57
121 78
87 84
58
69 82
73 78
100 21
687 48
58
123 64
90 15
59
73 60
77 22
103 68
703 49
59
125 70
92 70
60
77 63
80 91
107 35
719 65
60
127 96
95 50
61
81 96
84 88
111 23
735 92
61
130 45
62
86 58
89 16
115 32
752 26
62
133 19
63
91 54
93 76
119 66
768 67
63
186 20
64
96 86
98 73
12A 28
785 10
64
189 52
65
102 55
104 10
129 18
80152
65
148 16
1
J
324 J^A T'S HIGHER ARITHMETia
2. Quantities considered in Life Insurance are:
1. Premium on$1000.
2. The Gain or Loss.
3. Amount of the Policy.
4. Age of the Insured.
^ 5. The Term of years of Insurance.
These quantities give rise to five classes of problems,
but they involve no new principles, and by the aid of the
preceding tables they are easily solved. Simple interest is
intended where interest is mentioned in the following prob
lems:
Examples for PrAlCtice.
1. W. R. Hamilton, aged 40 years, took a life policy for
$5000. Required the annual premium ? 8156.50
2. Conditions as above, how much would he have paid
out in premiums, his death having occurred after he was
53? »2191.
3. Conditions the same, what did the premiums amount
to, interest 6%? $3045.49
4. James Bragg, aged 50 years, took out an endowment
policy for $20000, payable in 10 years, and died after making
6 payments: how much less would he have paid out by
taking a life policy for the same amount, the premium pay
able annually? $7860.
5. Thomas Winn, 28 years of age, took out an endow
ment policy for $10000, payable in 10 years; he died in 18
months: what was the gain, interest on the premiums at
6^, and how much greater would the profit have been had
he taken a life policy, premiums payable annually?
$7789.052 ; $1755.57, profit.
6. P. Darling took out a life policy at the age of 40
years, and died just after making the tenth payment; his
premiums amounted to $3975.10, interest 6^ ; required the
amount of his jjolicy ? $10000.
TOPICAL OUTLINE.
325
7. R C. Storey took out an endowment policy for $10000
for 15 years ; he lived to pay all of the premiums ; but had
he put them, instead, at 6^ interest as they fell due, they
would have amounted to $15426.78: what was his age?
40 years.
8. Allen Wentworth had his life insured at the age of
twenty, on the life plan, for $8000, premium payable annu
ally : how old must he be, that the sum of the premiums
may exceed the policy? 71 years old.
9. T. B. Bullene, aged 40 years, took out a life policy
for $30000, payments to cease in 5 years, the rate being
$9,919 on the $100 ; his death occurred two months after he
had mad§ the third payment: what was gained over and
above the premiums, interest 6%? $20448.
10. , F. M. Harrington took out an endowment policy for
$11000 when he was 42; at its maturity he had paid in
premiums $635.80 more than the face of the policy: what
was the period of the endowment? 20 years.
1. Simple Interest.
Topical Outline.
Applications op Percentage.
{With Time.)
' 1. Definitions:— Interest, Principal, Rate, Amount, Legal
Rate, Usury, Notes Promissory, Face, Payee, In
dorser. Demand Note, Time Note, Principal and
Surety, Maturity, Protest.
f Methods
I
2. Five Cases.
I Rules.
Common.
Aliquots.
Six and Twelve per ct
L Exact Interest.
2. Rule.— Formula.
3. Rule.— Formula.
4. Rule.— Formula,
5. Rule.— Formula.
3. Annual Interest.— Rule
326
BAY'S HIGHER ABITHMETIC.
Topical Outline.— (Oo7i<mw€d.)
Applications of Pekcentage.
{With Time.)
2. Partial Payments.. .
1. Definitions :— Payment, Indorsement
2. Rules. / ^' ^' ^' I^ule.— Principles. Connecticut,
( / Vermont, and Mercantile Rules.
r 1 True Discount / ^* I^finitions:— Present Worth,Discount
*■ I 2.  '
3. Discount... 
2. Bank Discount.
Rule.
1. Definitions :— Bank, Deposit, Issue,
Check, Drafts, Drawee, Payee, In
dorsement, Discount, Proceeds,
Days of Grace, Time to Run;
2. Four Cases, Rules.
4. Exchnnge
1. Definitions .—Domestic and Foreign Exchange, Bill, Set Rate,
Course, Par, Intrinsic, and Commercial.
1. Domestic.
r
2. Kinds.
2. Forei^..,
1. Direct. Table of Values.
2. Circular.
1. Definitions :— Arbitra
tion, Simple and
Compound.
2. Rules.
5. Equation of Payments...
1. Definitions:— Equated Time, Term of Credit,
. Average Term, Averaging Account, Clos
ing Account, Focal Date.
2 Principles.
3. Rules.
6. Settlement of Accounts.
1. Definitions.
2. Kinds..
ri..
I 3. 5
1. Accounts Current; Rule.
Account Sales; Rule.
Storage.
/.
Compound Interest  ^ ^definitions: — Comp. Int, Comp. Amt
I 2. Four Cases, Rules.
8. Annuities. .
9. Personal Insurance
' 1. Definitions :— Perpetual, Limited, Certain, Contingent. Im
mediate, Deferred, Forborne, Final Value, Initial Value,
Present Value.
2. Seven Cases. Rules. Table.
Definitions.
Table.
I 2.
Xyn. PAETK"EBSHIP.
DEFINITIONS.
856. 1. Partnership is the association of two or more
persous in a business of which they are to share the profits
and the losses. The persons associated are called partners;
together they constitute a Fimif Company y or Home.
2. The Capital is the money employed in the business;
the Assets or Besources of a firm are its property, and
opposed to these are its Liabilities or Debts.
3. Partnership has two cases : (1.) When all the shares
of the capital are continued through the same time; (2.)
When the full shares are not continued through the same
time. The first is called Simple Partnersldp ; the second,
Compound PartnerMp,
Principle. — Gains and losses are sliared in proportion to
the sums invested and Hie periods of investment
CASE I.
867. To apportion the gain or loss, when aU of each
partner's stock is employed through the same time.
Problem. — A, B, and C are partners, with $3000, $4000,
and $5000 stock, respectively ; if they gain $5400, what is
each one's share?
OPERATION.
Solution.— The whole 3 ^^^ of $ 5 4 = $ 1 3 5 0, A's share,
stock is $12000, of which 4 ^ " 5 4 0= 1 8 0, B's '*
A owns T^j, B t\, C ^^] ^ ^5^ « 5 400= 2250, C's **
hence, by the principle 12 $5 4 0, whole,
stated, A should have f^
of the gain, or $1350; in like manner, B, $1800; C, $2250.
(327)
328 RAY'S HIGHER ARITHMETIC,
Rule. — Divide the gain or loss among (he partners in pro
poHion to their shares of tJw stock.
Remark. — The division may be made by analysis or by simple
proportion.
Examples for Practice.
1. A and B gain in one year $3600; their store expenses
are $1500. If A's stock is »2500 and B's «1875, how much
does each gain ? A $1200, B 8900.
2. A, B, and C are partners ; A puts in $5000, B 6400,
C $1600. C is allowed $1000 a year for personal attention
to the business ; their store expenses for one year are $800,
and their gain, $7000. Find A's and B*s gain, and Cs
income. A $2000, B $2560, C $1640.
3. A, B, and C form a partnership ; A puts in $24000,
B $28000, C $32000; they lose ^ of their stock by a fire,
but sell the remainder at f more than cost: if all expenses
are $8000, .what is the gain of each ?
A $5714.281, B $6666.66, C $7619.04f
4. A, B, and C are partners; A's stock is $5760, B's,
$7200; their gain is $3920, of which C has $1120: what is
C*s stock, and A's and B*s gain ?
C's stock, $5184; A*s gain, $1244.44; B's gain, $1555.55f
5. A, B, and C are partners; A's stock, $8000; B's,
$12800; C% $15200; A and B together gain $1638 more
than C : what is the gain of each ?
A $2340 ; B $3744 ; C $4446.
6. A, B, and C have a joint capital of $27000 ; none of
them draw from the firm, and when thfey quit A has
$20000; B, $16000; C, $12000: how much did each con
tribute? A, $11250; B, $9000; C, $6750.
7. A, B, C, and D gain 30% on the stock ; A, B, and C
gain $1150; A, B, D, $1650; B, C, D, $1000; A, C, D,
$1600 : what was each man's stock?
A, $2666f ; B, $666J; C, $500; D, $2166.
PABTNERSHIR 329
CASE II.
358. To apportion the gain or loes when the fUll
shares are not continued throagh the same period.
Problem. — A, B, and C are partners; A puts in $2500
for 8 mo.; B, $4000 for 6 mo.; C, $3200 for 10 mo.; their
net gain is $4750: divide the gain.
Solution. — A*s operation.
capital ($2500), $2500X 8 = $20000, A's equivalent,
used 8 months, is $4000 X 6= 24000, B's "
equivalent to 8 X $3200X10= 32000 ,^8 "
$2500, or $20000, $7 6000
used 1 month^'« ^ of $4750 = $1 250, A's share,
capital ($4000), 5 of $4750 = $! 500, B^s "
used 6 months, IS ^^ $47 50 = $2000, C. 
equivalent to 6 X
$4000, or $24000, used 1 month ; Cs capital ($3200), used 10 months,
is equivalent to 10 X $3200, or $32000 used I month. Dividing the
gain ($4750) in proportion to the stock equivalents, $20000, $24000,
$32000, used for the pame time (1 month), the results Will be the
gain of each; A's $1250, B's $1500, Cs $2000.
Bule. — MvMply each partnei^s stock by Hie time it is wsed;
and divide Uie gain of loss in proportion to the products so
obtained.
Examples for Practice.
1. A begins business with $6000 ; at the end of 6 mo. he
takes in B, with $10000 ; 6 mo. after, their gain is $3300 :
what is each share? A's, $1800; B's, $1500.
2. A and B are partners ; A's stock is to B's, as 4 to 5 ;
after 3 mo., A withdraws f of his, and B f of his: divide
their year's gain, $1675. A, $800 ; B, $875.
3. A. B, and C join capitals, which are as ^, ^, \; after
4 mo. , A takes out ^ of his ; aft«r 9 mo. more, their gain is
$1988: divide it. A, $714; B, $728 ; C, $546.
H A. 28.
330 BAY'S HIGHER ARITHMETIC.
4. A and B are partners; A puts in $2500; B, $1500;
after 9 mo., they take in C with $5000; 9 mo. afl;er, their
gain is $3250 : what is each one's gain ?
A's, $1250; B's, $750; Cs, $1250.
5. A and B are partners, each contributing $1000; after
3 months, A withdraws $400, which B advances ; the same
is done aft«r 3 months more ; their year's gain is $800 : what
should each get ? A, $200 ; B, $600.
6. A, B, and C are employed to empty a cistern by two
pumps of the same bore ; A and B go to work first, making
37 and 40 strokes respectively a minute; aft«r 5 minutes,
each makes 5 strokes less a minute ; after 10 minutes, A
gives way to C, who makes 30 strokes a minute until the
cistern is emptied, which was in 22 minutes fiom the start :
divide their pay, $2. A, 46 ct.; B, $1.06; C, 48 ct.
7. A and B are partners; A's capital is $4200; B's,
$5600 : aft«r 4 months, how much must A put in, to entitle
him to ^ the year's gain ? $2100.
8. A and B go into partnership, each with $4500. A
draws out $1500, and B $500, at the end of 3 mo., and
each the same sum at the end of 6 and 9 mo.; at the end
of 1 j£. they quit with $2200 : how must they settle?
B takes $2200, and has t claim on A for $300.
9. A, B, C, and D go in partnership ; A owns 12 shares
of the stock; B, 8 shares; C, 7 shares; D, 3 shares. After
3 mo., A sells 2 shares to B, 1 to C, and 4 to D; 2 jno.
afterward, B sells 1 share to C, and 2 to D ; 4 mo. after
ward, A buys 2 from C and 2 from D. Divide the year's
gain ($18000).
A, $4650; B, $4650; C, $4700; D, $4000.
10. A, with $400; B, with $500; and C, with $300,
joined in business ; at the end of 3 mo. A took out $200 ;
at the end of 4 mo. B drew out $300, and after 4 mo. more,
he drew out $150; at the end of 6 mo. C drew out $100;
at the end of the year they close ; A's gain was $225 : what
was the whole gain ? $675.
BANKRUPTCY. 331
BANKRUPTCY.
DEFINITIONS.
369. 1. Bankruptcy is the inabUity of a person or a
firm to pay indebtedness.
2. A Bankrupt is a pei*son unable to pay his debts.
3. The assets of a bankrupt are usually placed in the
hands of an Assignee, whose duty it is to convert them
into cash, and divide the net proceeds among the creditors
in proportion to their claims.
Bemarks. — 1. This act on the part of a debtor ia called making
an cuisignmeTity and he is said to be able to pay so much on the dollar.
2. All necessary expenses, including assignee's fee (which is
generally a certain rate per cent on the whole amount of property),
must be deducted, before dividing.
4. The amount paid on a dollar can be found by
taking such a part of one dollar as the whole property is of
the whole amount of the debts; each creditor's proportion
may be then found by multiplying his claim by the amount
paid on the dollar.
Note. — Laws in regard to bankruptcy differ in the various
states ; usually a bankrupt who makes an honest assignment is
freed by law of his remaining indebtedness, and is allowed to retain
a homestead of from $500 to $5000 in value, and a small amount of
personal property.
Examples for Practice.
1. A has a lot worth $8000, good notes $2500, and cash
$1500; his debts are $20000: what can he pay on $1, and
what will A receive, whose claim is $4500?
Solution.— $8000 + $2500 + $1500 = $12000, the amount of prop
erty which is J^g^J or f of the whole debts. Hence,  of $1 = (50
ct., the amount paid on $1, and $4500 X .60 = $2700, the sum paid
to A
332
RAY'S HIGHER ARITHMETIC.
2. My assets are $2520 ; I owe A $1200 ; B, $720 ; C,
8600 ; D, $1080 : what does each ^^\.^ and what is paid on
each dollar ?
A, $840 ; B, $504 ; C, $420 ; D, $756 ; 70 ct on $1.
3. A bankrupt's estate is worth $16000; his debts,
$47500; the assignee charges 5% : what is paid on $1, and
what does A get, whose claim is $3650 ?
32 ct on $1, and $1 168.
Topical Outline.
Partnership.
c 1. Definitious.
1. Partnership 2. f ^^e I.— Applications.
\ Case II.— Applications.
f 1. Definitions.
\ 2. Applications.
2. Bankruptcy.
X YIII. ALLIG ATIOl^I .
DEFINITIONS.
860. Alligation is the process of taking quantities of
different values in a combination of average value. It
is of two kinds, Medial and Alternate.
ALLIGATION aMEDIAL.
361. Alligation Medial is the process of finding the
mean or average value of two or more things of different
given values.
Problem. — If 3 lb. of sugar, at 5 ct. a lb, and 2 lb., at
4^ ct. a lb., be mixed with 9 lb., at 6 ct. a lb., what per lb.
is the mixture worth ?
OPERATION.
Solution.— The 3 lb. at 5 ct. Price. Quantity. Cost,
a lb. = 15 ct.; the 2 lb. at 4 ct. 5 ct. X 3 = 1 5 ct
per lb. == 9 ct.; the 9 lb. at 6 ct. 4 J X 2 =9
per lb. = 54 ct.; therefore, the 6 X 9 = 54
whole 14 lb, are worth 78 ct.; 14 ) 7 8 ( 5 f ct.
and 78i 14 = 5f ct. per lb., Ans,
Bule. — Find the vahtes of Hie definite parts , and divide the
sum of Hie values hy the sum of Uie parts.
Examples for Practice.
1. Find the average price of 6 lb. tea, at 80 ct.; 15 lb.,
at 50 ct.; 5 lb., at 60 ct.; 9 lb., at 40 ct. 54 ct. per lb.
2. The average price of 40 hogs, at $8 each ; 30, at $10
each ; 16, at $12.50 each ; 54, at $11.75 each, $10.39 each.
(.1.33)
334 BAY'S HIGHER ARITHMETIC,
3. How fine is a mixture of 5 pwt. of gold, 16 carats fine ;
2 pwt, 18 carats fine; 6 pwt., 20 carats fine; and 1 pwt.
pure gold? 18f carats fine.
4. Find the specific gravity of a compound of 15 lb. of
copper, specific gravity, 7; 8 lb. of zinc, specific gravity,
6J; and \ lb. of silver, specific gravity, lOJ^. 7.445 —
Remarks. — 1. By the specific gravity of a body is usually under
stood, its weight compared with the iveight of an equal bulk of tvater; it
may be numerically expressed as the quotient of the former by the
latter. Thus, a cubic inch of silver weighing 10} times as much as
a cubic inch of water, its specific gravity •■= lOj.
2. To find the specific gravity of a body heavier than water: (1.)
Find its weight in air , (2.) Suspending it by a light thread, find its
weight in water and note the difference; (3.) Divide the first weight
by this difference. For example : if a piece of metal weighs If oz.
in air, but in water only IJ oz., its specific gravity = 1} i (1 J —
IJ) = 7. (See Norton*8 Natural Philosophyy p. 152.)
5. What per cent of alcohol in a mixture of 9 gal., S6%
strong; 12 gal., 92^ strong; 10 gal, 95% strong; and II
gal, 98^ strong? 93%.
6. At a teacher's examination, where the lowest passable
average grade was 50, an applicant received the following
grades: In Orthography, 50; Reading, 25; Writing, 50;
Arithmetic, 60 ; Grammar, 55 ; Geography, 55 : did he suc
ceed, or did he fail ? He failed.
ALLIGATION ALTERNATE.
862. Alligation Alternate is the process of finding the
proportional quantities at given particular prices or values
in a required combination of given average value.
CASE I.
363. To proportion the parts, none of the quanti
ties being limited.
ALLIGATION.
335
Problem.— 1. What relative quantities of sugar, at 9 ct.
a lb. and 5 ct. a lb., must be used for a corapound, at 6 ct.
alb.?
9
Solution. — If you put 1 lb.
at 9 ct. in the mixture to be
sold for 6 ct., you lose 3 ct.;
i£ you put 1 lb. at 5 ct. in the
mixture to be sold at 6 ct.,
you gain 1 ct.; 3 such lb.
gain 3 ct.; the gain and loss would then be equal if 3 lb. at 5 ct. are
mixed with 1 lb. at 9 ct.
OPERATION.
. 3 lb. at 5 ct. = 1 5 ct.
. 1 lb. at 9 ct. = _9 ct.
4 lb. worth 24 ct.
which is ^ = 6 ct. a lb.
Problem. — 2. What relative numbers of hogs, at $3, $5,
$10 per head, can be bought at an average value of $7 per
head?
OPERATION.
Diff. Balance.
Am,
1
1
2
Explanation. — Wr iti ng
the average price 7, and the
particular values 3, 5, 10, 3 4 3 3
as in the margin, we say : 7 5^ _2^ 3 3
3 sold for 7, is a gain of 4, 10 3 4 2 6
which we write opposite ; 5
sold for 7 is a gain of 2 ; 10 sold for 7 is a loss of 3. We wish to
make the gains and losses eqwd; hence, each losing sale must be
balanced by one which gains. To iose 3 foursy will be balanced by
gaining 4 threes^ and a gain of 3 hvos will balance a loss of 2 threes.
To indicate this in the operation, we write the deficiency 3, against
the excess 4; then the excess 4 against the deficiency 3; and in
another column, in the same manner, pair the 3 and 2, writing each
opposite the position which the other has in the colun^.n of differ
ences. The answer might be given in two statements of balance,
thus : for each 3 of the first kind take 4 of the third, and for each 3
of the second kind take 2 of the third. Since each balance column
shows only proportional parts, we may multiply both quantities in any
balance column by any numbet^y fractional or integi^aly and thus the final
answers be varied indefinitely. For example, had the second
balancing column been multiplied by 4, the answer would have
read, 1, 4, 4, instead of 1, 1, 2. The principle just stated is of
great value in removing fractions from the balance columns,
when integral terms are desired.
336 JiA K'aS' higher ARITHMETIC,
Bale. — 1. Write tlie jmrticular values or prices in orders in
a column^ having tlie smallest at the liead; write the averagfe
value in a middle position at the left and separated from iJte
others by a vertical line,
2. In another column to tJie right, and opposite Hie respective
vcdues, place in order the dijffh'ences between tliem and the
average value,
3. Tlien prepare balance columns, giving to each of Hieni two
numbers, one an excess and tlie other a deficiency taken from
the difference column; write eoc/i of these opposite tJie position
which the oilier has in the difference column; so 'proceed until
each number in the difference column lias been balanced with
anoHier; tlien,
The proportional quantity to be taken of each kbid, will be
the sum of Hie nuDibers in a horizontal line to the rigid of its
excess or deficiency.
Note. — The proof of Alligation Alternate is the process of Alli
gation Medial.
Examples for Practige.
1. What relative quantities of tea, worth 25, 27, 30, 32,
and 45 ct. per lb., must be taken for a mixture worth 28 ct.
per lb.? 19, 4, 3, 1, 3 lb. respectively.
Remark. — It is evident that other results raay by obtained by
making the connections differently ; as, 6, 17, 3, 3, 1 lb.; or, 17, 6, 1,
1, 3 lb.
2. What of sugar, at 5, 5^, 6, 7, and 8 ct. per lb., must
be taken for a mixture worth 6f ct. per lb.?
1, 5, 5, 7, 8 lb. respectively; or, 5, 1, 1, 8, 7 lb., etc.
3. What relative quantities of alcohol, 84, 86, 88, 94, and
96% strong, must be taken for a mixture 87^ strong?
10, 7, 3, 1, 3 gal.; or, 7, 10, 1, 3, 1 gal., etc.
4. What of gold and silver, whose specific gravities are
ALLIGATION.
337
19 J and 10, will make a compound whose specific gravity
shall be 16.84 ? 723 lb. silver to 3487 lb. gold.
5. What of silver f pure, and ^ pure, will make a mixt
ure \ pure? 1 lb., f pure; 5 lb., ^ pure.
6. What of pure gold, and 18 carats, and 20 carats fine,
must, be taken to make 22 carat gold ?
1 part 18 carats, 1 part 20 carats, 3 pure.
CASE II.
864. To proportion the parts, one or more of the
quantities, but not the amount of the combination,
being given.
Problem. — How many whole bushels of each of two
kinds of wheat, worth respectively $1.20 and $1.40, per
bushel, will, with 14 bushels, at $1.90 per bushel, make a
combination whose average value is $1.60 per bushel?
Diff. Bal.
OPERATION.
AiiRwerB.
1.60
1.20
.40
3
1.40
.20
3
1.90
.30
4
2
1
19
14
2
17
14
3
15
14
4
13
5
11
14114
6
9
14
7
7
14
8
5
14
9
3
14
10
1
14
Solution. — We find, by Case I, that to have that average value
the parts may, in one balancing, stand 3 of the first to 4 of the third,
and in another, 3 of the second to 2 of the third. By directly com
bining, we obtain the proportions 1, 1, and 2, and as the third
must be 14, we have for one answer 7, 7, 14. But we find the other
answers in the following manner.
The proportion will not be altered if in any balancing column we
multiply both quantities by the same number, hence the answer can
be varied as often as we can multiply or divide the columns, ob
serving the other conditions, which are that the answers shall be
integral, and that the number of 4'8 and 2's taken shall make 14. As
there are more fractions than there are integers between any two
limits, we try fractional multipliers in order to obtain the greatest
number of answers. Observe that the number of 4's taken will not
stand alone in any answer for the third kind of wheat, but will be
H. A. 29.
338 liA Y'S HIGHER ARITHMETIC,
added to some humber of 2's ; the number of d's taken as any
one answer for the first or second kind vyiU not be increased by any
other product ; hence, if we use a frcuUional iuultiplier, it must be
such that its denominator will disappear in multiplying by 3 ; and
this shows that a fractional multiplier will not be convenient unless
it can be expressed as thirds. Therefore, the remaining question is,
How many thirds of 4 with thirds of 2 will make 14? Since 14 =
■^, the question is the same as to ask, How many whole 4*s with whole
2's will make 42 ? It is plain that there can not be more than ten
4*s. We can take J of
1 four and 19 twos, or J of first column and ^ of second.
2 fours " 17 " " f " " " " V " "
Q « (( 1C (C U 3 U t( U it J5 (t it
The answers are now obvious : write 1, 2, 3, etc., parallel with 19,
17, 15, etc., and the 14's in the third row.
Rule. — Find Hie proportional parts, as in Case J, and ob
serve the term or terms in the balance columns , standing oppo
site Hie value or price corresponding to the limited quantity;
tJien find what midtipliers iviU produce tJie given limited quan
tUy in the required place, and of Hiose multipliers v>se only
such as will agree with tlie remaining conditions of the problem.
Examples for Practice.
* 1. How many railroad shares, at 50%, must A buy, who
has 80 shares that cost him 72%, in order to reduce bis
average to 60%? 96 share.^
2. How many bushels of tops, worth respectively 50, 60
and 75 ct. per bu., with 100 bu., at 40 ct. per bu., will mak
a mixture worth 65 ct. a bu.? 2, 2, and 254
3. How much water (0 per cent) will dilute 3 gal. 2 qt.
1 pt. of acid 91 % strong, to 56^ ? 2 gal. 1 qt. ^ pt.
4. A jeweller has 3 pwt. 9 gr. of old gold, 16 carats fine :
how much U. S. gold, 21f carats fine, must he mix with it,
to make it 18 carats fine? 1 pwt. 21 gr.
ALLIGATION.
3e39
5^ How much water, with 3 pt. of alcohol, 96% strong,
and 8 pt., 78^, will make a mixture 60% strong? 4\ pt.
6. I mixed 1 gal. 2 qt. ^ pt. of water with 3 qt. 1 J pt. of
pure acid; the mixture has 15% more acid than desired:
ho^ much water will reduce it to the required strength?
1 gal. 2 qt. 1^ pt.
7. How much lead, specific gravity 11, with ^ oz. copper,
sp. gr. 9, can be put on 12 oz. of cork. sp. gr. ^, so that the
three will just float, that is, have a sp. gr. (1) the same as
water ? 2 lb. 1\ oz.
8. How many shares of stock, at 40%, must A buy, who
hai bought 120 shares, at 74^, 150 shares, at 68^, and
130 shares, at 54^, so that he may sell the whole at 60^,
and gain 20% ? 610 shares.
9. A buys 400 bbl. of flour, at $7.50 each, 640 bbl., at
$7.25, and 960 bbl., at $6.75: how many must he buy at
$5.50, to reduce his average to $6.50 per bbl.? 1120 bbl.
CASE III.
865. To proportion the parts, the amount of the
whole combination being given.
Problem. — If a man pay $16 for each cow, $3 for each
hog, and $2 for each sheep, how many of each kind may he
purchase so as to have 100 animals for $600 ?
Solution.— Pro operation.
ceeding as in Case DifF. Bal. Answers.
I, we find that the
given average re $6
quires 5 of the first
kind with 2 of the 7 13
third, and 10 of the
second kind with 3 of the third ; taken in two parts, 7 are required
in (me balancing, and 13 in another ; these being together 20, which
is contained five times in the required number, 100, if we multiply all
of the terms in the balance columns by 5, we have for one answer
25 sheep, 50 hogs, and 25 cows.
2
3
16
4
3
10
5
2
10
3
12
64
24
25
5
25
38
36
2 6
51
22
27
64
8
28
340 RAY'S HIGHER ARITHMETia
As there are other multipliers affording results within the con
ditionS) we leave the student to find the remaining answers by a
process similar to that shown under Case 11.
Remark. — Suppose that we had to determine 400
how many 7*s and 11*8 would make 400. By 2 22
trial, we find that tm) ll's taken away leave an 3 7 8. ..5 4
exact number of 7*s. It is now unnecessary to 7 7
take single 11*8 or proceed by trial any farther; 9 3 01... 4 3
for, as 378 is an exact number of 7*8, if we take 7 7
away 11*8 and leave 7*s, we must take seven 11*8; j g 2 2 4... 3 2
and thus the law of continuation is obvious : 400
is composed of
. 11*8 2, 9, 16, 23, 30;
With 7'8 54, 43, 32, 21, 10.
Bule. — Proportion the parts as in Case I; Vien, noting the
sums of the balancing columns, find, by trial or by direct di
viston, ivhat multipliers will make those columns together equal
to the given amount of the combination, and of tiiose multipliers
use only such as will agree with the remaining conditions.
Examples for Practice.
1. What quantities of sugar, at 3 ct. per lb. and 7 ct.
per lb., with 2 lb. at 8 ct, and 5 lb. at 4 ct. per lb., will
make 16 lb., worth 6 ct. per lb.?
f lb. at 3 ct, 8^ lb. at 7 ct
2. How many bbl. flour, at $8 and $8.50, with 300 bbl.
at $7.50, and 800 at 87.80, and 400 at $7.65, will make
2000 bbl. at «7.85 a bbl. ?
200 bbl., at «8 ; 300 bbl., at $8.50
'3. What quantities of tea, at 25 ct and 35 ct. a lb., with
14 lb. at 30 ct, and 20 lb. at 50 ct, and 6 lb. at 60 ct,
will make 56 lb. at 40 ct. a lb.?
10 lb. at 25 ct, and 6 lb. at 35 ct
TOPICAL OUTLINE.
341
4. How much copper, specific gravity 7, with silver,
specific gravity 10^, will make 1 tb. troy, of specific gravity
8f ? 7f^ oz. copper, 4Jff oz. silver.
5. How much gold 15 carats fine, 20 carats fine, and
pure, will make a ring 18 carats fine, weigliing 4 pwt.
16 gr.? 2 pwt. 16 gr.; 1 pwt.; 1 pwt.
6. A dealer in stock can buy 100 animals for 8400, at
the following rates, — calves, $9; hogs, $2; lambs, $1 : how
many may he take of each kind?
37 calves, 4 hogs, 59 laml)s;
{pile of nine different answers.)
7. Hiero's crown, sp. gr. 14f , was of gold, sp. gr. 19J,
and silver, sp. gr. 10 J ; it weighed 17 J lb.: how much gold
was in it? lOfJ lb.
Topical Outline.
AlJ^lGATlON.
*
r 1. Definitions.
. Alligation
. 2. Kinds.. .
1. Alligation
Medial...... / 1 DefiniUons.
I 2. Rules.
»
2. Alligation
Alteniatc.. .
1. Definitions.
£ I. Rulel
2. Cases ..J n. Rule.
I III. Rule.
XrX. mYOLTTTlOK
DEFINITIONS.
806. 1. A power of a quantity is either tliat quantity
itself, or the product of a number of factors each equal to
that quantity.
Kemabk. — Regarding unity as a base, we may say, the power of
a quantity is the product arising from taking unity once as a mul
tijilicand, with only the given quantity a certain number of times
as a factor. The power takes its name from the number of times
the quantity is used as factor. Unity is no power of any other
positive number.
2. The root of a power is one of the equal factors which
>roduce the power.
3. Powers are of difierent degrees, named from the number
of times the root is taken to produce the power. The degree
is indicated by a number written to the right of the root,
and a little above ; this index number is called an exponent.
Thus,
5X5, or the 2d jmver of 5, is written 52.
5X5X5, " 3d " " " " 5».
4. The second power of any numl>er is called the square,
because the area of a square is numerically obtained by form
ing a second power.
5. In like manner the €iird power of any number is called
its cube, because the solidity of a cube is numerically ob
tained by forming a third power.
367. To find any power of a number, higher than
the first.
(342)
INVOLUTION. 343
Bole. — Multiply the number by itself ^ and cmUinue the mvlr
tiplicatwn till Hud number has been used as jo/dor as many
times as are indicated by the exponent.
Notes. — 1. The nnmber of multiplications will be one fess than
the exponent, because the root is used twice in the first multiplica
tion, once as multiplicand and once as multiplier.
2. When the K)wer to be obtained is of a high degree, multiply
by some of the pK)wers instead of by the root continually ; thus, to
obtain the 9th power of 2, multiply its 6th power (64) by its 3d
power (8) ; or, its 5th power (32) by its 4th power (16) : the rule
being, that the product of any two or more powers of a number is thai
power whose degree is eqwil to the sum of their degieea,
3. Any power of 1 is 1 ; any power of a number greater' than 1 is
gieaier than the number itself : any power of a number less than 1, is
less than the number itself.
388. From Note 2, 48 X 48 X 4» X 4^ X 4^ = 4^ «; but
the expression on the left is the 5th power of 48; hence,
(48)5 =415. i}^Q,t is, when the exponent of the power required
is a composite number (15), raise tlte root to a power whose
exponent is one of its factors (3), and this resuU to a power
whose exponent is the other factor (5).
Note. — Let the student carefully note the difference between
raising a potver to a power, and mtdliplying together different powers of
the same root ; thus,
23X22=2*.
Here we have multiplied the cube by the square and obtained the
5th power ; but the 5th power is net the square of the cube; this is the
sixth power, and we write
(22)3 = 2«, or (23)2 =23^2 ^2«.
369. Any power of a fraction is equal to that power of
the numerator divided by that power of the denominator.
370. The square of a decimal must contain twice^ and its
cubCy three times as many decimal places as the root, etc.;
344
BA yS HIGHER ARITHMETia
hence, to obtain any power of a decimal, we have the foUow
ing rule:
Rule. — Proceed as if the decimal were a whole number,
and point off in the remit a number of dedmxd places equal
to the number in the root multiplied by the exponent of Hie
power.
Examples for Practice.
Show by involution, that:
1. (5)*
equals
25.
8. ()« equals iffJ.
2. 14*
2744.
9. (.02)* " .000008
3. 6*
7776.
10. (5*)2 •' 390625.
4. 192"
36864.
11. (.046)8 " .000097336
6. 1»">
1.
10 (\\1 n I
^^ W 4'Jiiiti
6. (I)*
tMt
13. 20562 " 4227136.
7. (2i)«
iiH
14. (7.62i)2 " 58. 1406 J
371. Special processes for squaring and cubing
numbers.
Problem.— Find the square of 64.
64
64
PARALLEL OPERATIONS.
6 +4 = 6 tens + 4 units.
60
+ 4
256
3840
60X4H42
602+60X4
4096= 602 + 2(60X4) + 4a
= 3600 + 480+16.
The operations illustrate the following principle :
Principle. — The square of tlie sum of two numbers is equal
to ihe square of the first, plus twice the product of the first by
the second, plus the square of the second. Thus :
^ \  A >, ^ y ^x^
OF THE
INVOLUTrON\ ^p
252 = (22 + 3)2 = 484 + 2 (22 X 3)+ & = 625.
252 = (21 + 4)2 = 441 + 2 (21 X 4) + 16 = 625.
252 = (20 + 5)2 = 400 + 2 (20 X 5) + 25 = 625.
Remark. — The usual application of this principle in Aritkmetie
is, in squaring a number as composed of tens and units. The third
statement above illustrates this ; and, if we represent the tens by t,
the units by u, we have the following statement :
(t + u)*^=l^ + 2tu+u^; or, in common language :
The square <^ any number composed of tens and units is equal to the
square 0/' the tens, + twice the product 0/ the tens by the units, + the squa:i^
oj the units.
Examples for Practice.
Square the following numbers, considering each as the
sum of two quantities, and applying the principle announced
in Art. 371, on page preceding:
1. 19.
361.
4. 40.
1600.
2. 29.
841.
5. 125.
15625.
3. 4.
16.
6. 59.
3481.
Illustration. — Draw a square. From points in the sides, at
equal distances from one of the comers, draw two straight lines
across the figure, each parallel to two sides of the figure. These
two lines will divide the square into four parts, two of them being
squares and two of them rectangles. The base being composed of
two lines, and the square of four parts, we see that
372. The square described on the sum of two lines is
^ual to the sum of the squares described on the lines, plus
twice the rectangle of the lines.
Remark.— Both the principle employed above and the illustra
tion are frequently used in explaining the method for square root
346 BAY'S HIGHER ARITHMETIC.
Problem. — ^Find the cube of 64.
PARALLEL OPEBATIONS.
64«= 4096= 60» + 2(60 X4) + 4»
64= 6044
16384= 60 2X4+2(60X42) + 4»
24576 = 608 + 2(60»X4)43(60X4M
64» = 262144— 60 » + 3(602 X4) + 3(60X4 2)44 ».
The operation illustrates the following principle :
Principle. — The cube of any number composed of two parts,
is equal to Vie cube of the first part, phis three times the square
of the first by tlie second, plus three times tlie first by the square
of the second, plus tfie cube of the second. Thus :
258 = (22 + 3)8 =228 ^ 3 (222 x 3) + 3 (22 X 32) 4 38
= 10648 + 4356 4 594 4 27 = 15625.
Bemark. — The usual application of this principle in Arithmetic
is, in the cubing of a number as composed of tens and units. Bei)re
senting the tens by i, and the units hy u, we have the following state
ment, which the student will express in common language, similar
to that of the principle used in squaring numbers :
{t 4 uy = e» 4 st^u + Sfu^ 4 ti8.
Examples for Practice.
Considering the following numbers as made, each, of two
parts, cube them by the principle just stated :
(1.) 19.
6859.
(4.) 40.
64000.
(2.) 29.
24389.
(5.) 125.
1953125.
(3.) 4.
64.
(6.) 216.
10077696.
XX. EYOLUTIOK
DEFINITIONS.
373. 1. Evolution is the process of finding roots of
numbers.
2. A root of a number is either the number itself or one
of the equal factors which, without any other factor, produce
the number.
Since a number is the first root, as also the first pouer of itself,
no operation is necessary to find either of these ; hence, in evolu
tion, we seek only one of the equal factors which produce a power.
Evolution is the reverse of Involution, and is sometimes called
the Extraction of Hoots.
3. EootSf like powers, are of different degrees, 2d, 3d, 4th,
etc.; the degree of a root is always the same as the degree
of the power to which that root must be raised to produce
the given number.
Thus, the 3d root of 343 is 7, since 7 must be raised to the 3d
power, to produce 343 ; the 5th root of 1024 is 4, since 4 must be
raised to the 5th jjower, to produce 1024.
Since the 2d and 3d powers are called the square and cube, so the
2d and 3d roots are called the square root and cube root.
4. To indicate the root of a number, we use the Radical
Sign (]/), or a fractmial exponmt
The radical sign is placed before the number; the degree
of the root is shown by the small figure between the branches
of the radical sign, called the Index of the root.
Thus, ^18 signifies the cube root of 18 ; ^^9 signifies the 5th
root of 9. The square root is usually indicated without the index
2; thus, 1^10 is the same as f'lO.
(347)
348 HAY'S HIGHER ARITHMETIC.
5. The root of a uiimber raay be indicated by a fractional
exponent whose numerator in 1, a7id whose denominator is the
index of the root to be expressed.
Thus, 1^7 = 7% and 1^5 = 5^ ; fiiniilarly, 4* = 16% the numera
tor indicating a power and the denominator a root.
6. A perfect power is a number whose root can be
exactly expressed in the ordinary notation; as 32, whose
fifth root is 2.
7. An imperfect power is a number whose root can not
be exactly expressed in the ordinary notation ; as 10, whose
square root is 3.1622[
8. The squares and cubes of the first nine numbers are
as follows:
Numbers, 123456789
Squares, 1 4 9 16 25 36 49 64 81
Cubes, 1 8 27 64 125 216 343 512 729
9. The Square Boot of a number is one of the two equal
factors which, without any other factor, produce that num
ber ; thus, 7X7 = 49, and i/49 = 7.
10. The Cube Boot of a number is one of the three
equal factois which, without any other, produce that num
ber; thus, 3 X 3 X 3 = 27, and ^27 = 3.
374. Concerning powers and roots in the ordinary deci
mal notation, we state the following principles:
Principles. — 1. The square of any nwnber has twice as
ynanyy or mie less ilmn twice as many, figures as Hie number
itself lias.
2. 2%cre mil be as many figures in the square root of a
perfect powei* as there are periods of two figures each in the
power, beginning wiHi units, and also a figure in tlie root cor
responding to a part of sucJi period at Uie left in Hue power.
EVOLUTIOK 349
3. The cube of a number has three times as many figures,
or one or two less than Viree times as viant/y as the number
itself has.
4. TJwre wUl be as vmny figures in Hie cube root of a perfect
poivery as there are periods of three figures each in the po^ver,
beginning with units, and also a figure corresponding to any part
of sudi a period at the left hand.
Exercises.
1. Ppove that there will be six figures in the cube of the
greatest integer of two figures.
2. Prove that there will be twelve figures in the fourth
power of the greatest integer of three figures.
EXTRACTION OF THE SQUARE ROOT.
FiEST Explanation.
pROBLE^r. — What is the length of the side of a square
containing 576 sq. in.?
Solution. — The length required will operation.
be expressed by Ihe square root of 576 ; 5 7 6(20
by Principle 2, we know that the root can 4
have no less than two places of figures; 40 400 24 in. Ans,
and since the scjuare of 3 tens is greater, 4 ry^
and that of 2 tens less than 576, the root Jl i 7 ft
must be less than 30 and greater than 20;
hence, 2 is the first figure of the root, and 400 the greatest square of
tens contained in 576. Let the first of the accompanying figures
represent the square whose side is to be found. We see that the
side must be greater than 20, and that the given area exceeds by
176 sq. in. the square whose side is 20 in. long. It is also evident
from the figure that the 176 sq. in. may be regarded as made of three
parts, two of them being rectangles and one a small square ; these
parts are of the same width, and, if that width be ascertained and
350
RAY'S IIIOHEB ARITHMETIC.
20X4=80
20X20=400
16
%
II
X
s
s
added to 20 in., the required side will be found. The two 20inch
rectangles, with the .small square, may be considered as making one
long rectangle of the required
width, as shown in the figure
on the right ; and, as the exact
area of that rectangle is 176
sq. in., if we knew its length,
its true width would be found
by dividing the area by the
length (Art. 197, 7); but we
do know that the length is
greater than 40 in., and hence,
that the width is, in inches,
less than the quotient of 176 by 40; and since, in 176, 40 is
contained more than four times, but not five times, 4 is the
highest number we need try for the width. Now, as the true
length of that rectangle is 40 in. increased by the true widthy the
proper uxiy to try 4, is to add it to 40 and multiply the sum
by 4; thus, 40 in.f 4 in. = 44 in.; and 44 X 4 = 176. This
shows that 4 in. is the width of the rectangle, and hence the required
side is 20 in. f 4 in. = 24 in., Ans,
Remarks. — 1. Since 17 contains 4 as often integrally ^ as 176 con
tains 40, it is convenient to use simply the 17 as dividend with 4 as
divisor, and then annex the quotient to the divisor and to the first
figure of the root.
2. At the first step we ascertained that the whole root was greater
than 2 tens and less than 3 tens ; at the next step we learned that
the units were not equal to 5, and by trial they were found to be 4.
The whole process was a gradual approach to the exact root, — one
figure at a time. It is im()ortant for the student to note that in the
processes of evolution there must be steps of trial. Even the higher
branches will not exempt from all trial work. The most valuable
rules pertaining to such numerical opera
tions, simply narrow the trial by making
the limits obvious. Thus our device above
showed the second part of the root less
than 5; an actual trial showed it to be
exactly 4.
3. If the power had been 58081, we
should have found there were three figures
in the root; here, as in the former caSiS, 4 is found the greatest figure
44
58081(241
180
176
481
481
481
EVOLUTION, 351
which can stand in ten*B place, <ind we may treat the 24 tens exactly
as we treated the 2 tens in the first illustration.
Second Explanation.
We learned in Art. 371 that the square of a number composed of
tens and units is equal to the square of the tens, i)lus twice the
product of the tens by the units, plus the square of the units.
The square of (20 + 4), or 24a, jg 202 + 2 X (20 X 4) + 4}.
Now, if the square of the tens be taken away, there will remain 2 X
(20 X 4) 4 42 = 40 times 4, and 4 times 4, or, simply (40 + 4) X 4.
We see then that if the square of the tens be taken away, the remain
der is a product whose larger factor is the douUe of the tens, increased
by the unitSj the smaller factor being simply the units.
Suppose, then, in seeking the square root of 1764, we have found
the tens of the root to be 4 ; the remainder 164 must be the product
of the units by a factor which is equal to
the sum of twice the tens and once the operation.
units. If we knew the units, that larger 17 6 4(40
factor could be found by doubling the tens 1600 2
and adding the units; if, on the other 80
hand, we knew the larger factor, the units 2
could be found by diject division ; we cfo 32
know that larger factor to be more than 80,
and hence that the units factor is less than the exact number of times
164 contains 80. Therefore, the units figure can not be so great as
3, and the largest we need try is 2. The proper uxty to try 2, is to
add it to 80, and then multiply by 2 ; this being done, we see that,
the product being equal to 164, 2 is the exact number of units, 80 +
2 the larger factor exactly, and 42 the exact root.
Note. — These successive steps showed, that the first figure toas
the root as nearly as tens could express it ; with the second figure
we found the root exactly. Had the power been 1781, the 42
would still have been the true root as far as tens and units could
express it; and at the next step, seeking a figure in tenth^s place,
we would have found the true root, 42.2, as far as expressible by
tenSj unitSy and tenths. Continuing this operation, we find 42.201895
to be the root, ti'ue as far as miUionihs can express it ; so, in any case,
when a figure is correctly found, the true root can not differ from the
whole root obtained, by so much as a unit in the place of that figure.
16 4 4 2, Ans,
164
1
\
352 BAY'S mOHER ARITHMETIC.
375. To extract the square root of a number writ
ten in the decimal notation, as integer, fraction, or
jtnixed number. *
BtQe. — 1. Point off the number into periods of tvx) figures
eachf commencing with units.
2. Find the greatest square in the first period on the left;
place its root on the right, like a quotierd in division; subtract
the square from the period, and to the remainder bring dovm
the next period for a dividend.
3. Double the root already found, as if it were units, and
uTite it on the left for a trial divisor; find how often this is
contained in (he dividend, exclusive of the righthand figure,
annexing the quotient to the root and to the divisor; then mul
tiply the complete divisor by the quotient, subtract the product
from the dividend, and to ike remainder bring down the next
period as before.
4. Double the root as before, place it on the left as a trial
divisor, proceeding as with the former divisor and quotient
figure; continue the operation until the remainder is nothing,
or until the lowest required decimal order of the root has been
obtained.
Notes. — 1. It any product be foand too large, the last figure of the
root is too large.
2. The number of decimal places in the power must be even;
hence the number of decimal periods can be increased only by
annexing ciphers in pairs. Contracted division may be used to find
the lower orders of an imperfect root.
• 3. When a remainder is greater than the previous divisor it does
not follow that the last figure of the root is too small, unless that
remainder be large enough to contain twice the part of the root already
found and 1 more; for this would be the complete divisor, and
would be contained in the remainder if the root were increased by 1.
Hence, the square of any number mtist be increased by a unit more than
twice the number itself, to make the square of the next higher.
Thus, 1252 =15625; simply add 250+1, and find 15876=126«.
EVOLUTION.
353
Examples for Practice.
1. 1/2809.
2. 1/1444.
53.
38.
109.
13625.
8944.9
2490.74
uare root (
7. 1/ 3444736.
8. 1/57600.
1856.
240.
3, 1/11881.
9. 1/16499844.
10. 1/49098049.
11. 1/73005.
12. v''386».
)f 3, true to the
4062.
4. 1/185640625,
5. 1/80012304.
6. 1/6203794.
13. Find the sqi
7007.
270.194
7583.69
7th decimal
place. 1.7320508
14. Find the square root of 9.869604401089358, true to
the 7th decimal place. 3.1415926
15. 1/.030625 X 1/40.96 X l/.00000625 = what? .0028
16. 1/(126)2 X (58)2 X (604)2 = 4414032.
17. 1/12.96 X sq. rt. of ^2^^ = 3.2863
Kemaue. — The remainder, at any point, is equal to the square of
an unknovm part, plus twice the product of that part by a Arnoim part.
The remainder may also be considered as the difference of (too squares^
which is always equal to the product of the mm of the roots by their
difference,
376. The square root of the product of any number of
quantities is equal to the product of their square roots;
thus, i/l6 X .49 = 4 X .7 = 2.8
377. The square root of a common fraction is equal to
the square root of the numerator, divided by the square
root of the denominator.
Remark. — It is advantageous to multiply both terms by what
will render the denominator a square.
Examples for Practice.
1. V>_^=
2. i/34 =
H. A. 30.
.92582+
5.8843+
S. V^^ = ^ nearly.
4. i/272^= 16f
354
JRA Y'S HIGHER ARITHMETIC.
5. i/6f = 2.5298+
6. V^ ifXi^JVXMX l/3"= .45886+
7. 1/123.454321 X .81 = 9.9999
8. 1/1.728 X 4.8 X^ = ^^ X l/^T, written also 2^
1/21.
EXTRACTION OF THE CUBE ROOT.
First Explanation.
Problem. — What is the edge of a cube whose solid inches
number 13824?
OPERATION.
13824(20
8000 4
20^X3 =1200
20 X3X4 = 240
42= 16
1456
5824 24, ^ns.
5824
Solution. — The length
required will be expressed
by the cube root of 13824.
By Prin. 4, we know the
root can have no less than
two places of figures; al:?o,
since the cube of 3 tens is
greater and that of 2 tens is less than 13824, the root must be
less than 30 and more than 20; conse<iuently, 2 is the fii'st figure of
the root, and 8000 is the greatest cube of tens contained in 13824. Let
the first of the accomi»anying figures
represent the cube whose edge is to
be found. We see that it must be
greater than 20 inches, and that the
given solidity exceeds by 5824 cu. in.
the cube whose edge is 20 inches, and
which for convenience we will call A.
(See Fig. 2.) It is also evident from the
second figure, where the separate parts
are shown, that the 5824 cu. in. may be
regarded as made up of seven parts,
three of them being square blocks
( 5) 20 in. long, three of them being rectangular blocks ( 0) of the
same length, and one a small cube ( C). These parts are of the
same thickness, and if that thickness be ascertained and added to 20 in.,
tiie required e<lge will be found. The scpiare blocks and the oblong
blocks, with the small cube, may be considered as standing in line
Fig. 1.
EVOLUTION,
355
ow, as
(Fig. 3) and forming one oblong solid of uniform thickneRs. N
the exact solidity of that solid
is 5824 cu. in., if we knew its
side surface, it8 true thickness
would be found by dividing the
number expressing the solidity
by the number expressing the
surface. But we do know that
side surface to be greater than
3 times 400 sq. in., and hence
the thickness must be less than
the quotient of 5824 by 1200;
and since in 5824, 1200 is con
tained 4 times but not 5 times,
4 is the highest number we need
try for the width ; as the exact
surface of one side is equal to
1200 sq. in., increased by 3
rectangles 20 in. long, and a
small square also, each of a
width equal to the required
thickness, the proper way to
try 4 for that thickness is, to
multiply it by 3 times 20, then
by itself, and, adding the prod
ucts to 1200, multiply the sum
by 4 ; thus, 1200 sq. in. + 80
sq. in. X 3 f 16 sq. in. = 1456 sq. in., and 1456 X 4 = 5824, which
ly
Fig. 2.
y
N \ \ S
3;
B
D
n
Fis. rt.
shows that 4 in. is the true thickness, and 20 in. f 4 in. = the re
quired edge, 24 in., Ans.
Note.— Fig. 1 shows also the complete cube with the section lines
marked.
356 RAY'S HIGHER ARITHMETIC.
Remarks. — 1. Since 3 times the square of 2 tens is equal to 300
times the H<uare of 2, it is allowable to use simply the square of the
first part of the root as units, multiplying by 300 for a trial divisor ;
and so, too, in the second part of the trial work, it will answer to
multiply the first part by the last found figure and by 30.
2. The steps are trial
. , , , OPERATION. i
steps, and as remarked ... ,
under the rule for square 14172488(242
root, our artifices have r
simply narrowed the 2^X300 = 1200 6172
range of the trial. 2X^X30= 240
3. Had the power been ^ ^ = ^^
14172488, there would 1456 _5824
have been three figures
in the root; and here, as 24^X300 = 172800
in the former case, the 24X2X30= 1440
second figure is 4 ; and 2 ^ = 4
348488
348488
we may treat the 24 tens 174244
exactly as we treated the
2 tens in the former illustration.
Second Explanation.
We have seen (page 346) how the cube of a number, of two figures,
is composed ; that, for example,
24 3=8000 + a(20 2X4) + 3(20X4 2) + 4».
Here "we see that if the cube of the tens be taken away, there will
remain
(202X3 + 20X4X3 + 42)X4;
that is, the remainder may be taken as a product of two factors, of
which the smaller is the units, and the larger made up of 3 times the
square of the tens, with 3 times the tens by the units, with also the
square of the units. Suppose, then, in seeking the cube root of 74088,
we find the tens to be 4; the remainder 10088 must be the product of
units by a factor comix)sed of three parts, such as we have described.
If we knew the larger factor, the units could be obtained by direct
division ; but we do know that larger factor to be greater than 3 times
the square of 40 ; hence, we know the units must be less than the
quotient of 10088 by 4800, and consequently 2 is the largest figure we
need try for units. The way to try 2, is to compose a larger factor
M
EVOLUTION, 357
after the manner just described ; operation.
hence, multiply 2 by 3 times
40, add 22 or 4, add the sum to
the 4800, and multiply by 2, 42X300 = 4800
which, being done, shows that 4X2X^0= 240
2 is the units figure, and that 2 2 = 4
42 is the root sought. * 5944
74088(42
64
10088
10088
Hem ARK. — Three times the square of the tens is the convenient
trial divisor. This is in most instances a greater part of the com
plete divisor; for example, the least number of tens above otw ten
is 2, and the greatest figure in unit's place can not exceed 9; the
cube of 29 is 24389, the first complete divisor is 1821, the first trial
divisor being 1200, a greater part of it.
378. To extract the cube root of a number written
in the decimal notation as whole number, fraction,
or mixed number.
Bule. — 1. Beginning with units , separate tlie number into
periods of three figures each; the extreme lefi period niay Irnve
hut one or two figures, but the extreme right, whether of units
or decimal orders, must have three places, by the annexing of
ciphers if necessary,
2. Find tJie greatest cube in the highest period, place its root
on the right as a quotient in division, and tlien subtract the
cube from the period, bringing down to the right of the remain
der the next period to complete a dividend,
3. Square Hie root found as if it were units, multiply it by
300, and place the product on the left ojs a trial divisor] find
how often it is contained in the dividend, and place tlie quo
tient figure to the right of the root ; midtiply the quotient by 30
tiines Hie preceding part of the root as units, square also the
quotient, and add tlie two results to the trial divisor; tJien mul
tiply the sum by the quotient, and subtract the product from
the dividend, annexing to the remainder another period as
before.
4. Square Hie wlwle root as before, multiply by 300, pro*
358 JiAY'S HIGHER ARITHMETIC.
deeding as rmth the former trial divisor^ quotient, and addi
tions; continue Uie operaiion until Hie retnainder is notliing, or
until Hie lowest required decimal order of Hie rod has been
obtained.
Notes. — 1. Should any product exceed the dividend, the quotient
figure is too large.
2. If any remainder is larger than the previous divisor, it does
not follow that the last quotient figure is too small, unl&ss the remainder
is ku'ge enough to ccmtain 3 times the square of thai part of the root already
found, with 3 times that part of the root, and 1 more / for this is the proper
divisor if the root is increased by 1.
3. Should decimal periods be required beyond those with which
the operation begins, the operator may annex three ciphers to each
new remainder.
4. When the operator has obtained one more than half the re
quired decimal figures of the root, the last complete divisor and the
last remainder may be used in the manner of contracted division.
879. The cube root of any product is equal to the prod
uct of the cube roots of the factors. Thus,
#^250 X 4 X 648 X 9 = #^125 X 8 X 216 X 27 =
5X2X6X3.
380. The cube root of a common fraction is equal to the
quotient of the cube root of the numerator by the cube root
of the denominator.
Bemarks. — 1. When the terms are not both perfect cubes, multi
ply both by the square of the denominator, or by some smaller factor
which will make the denominator a cube.
2. Reduce mixed numbers to improper fractions, or the fractional
parts of such numbers to decimals.
Examples for Practice.
1. ^512. 8.
2. 1^19683. 27.
3. 1^7301384. 194.
4. f^94818816. 456.
EVOLUTION. 359
5. #"1067462648. 1022.
6. 1^5.088448. 1.72
7. ^22188.041 28.1
8. #"32:65 3.196154
9. #':0079 .1991632
10. #"3.0092 1.443724
11. f':^ .315985
12. #"25. 2.924018
13. #"TT. 2.22398
14. #"17 .87358
15. #"^. .64366
16. #"171.416328875 5.555
17. #'70Tr 19.1393267
1^ "^SIt .2218845
19. #"1 of ^. .6235319
EXTEACrriON OF ANY EOOT.
381. We have seen in the chapter on Involution, that
if a 'power be raised to a power, the new exponent is the
product of the given exponent by the number of times the
power is taken as a factor; that, for example, 2^ raised to
the 4th power, is 2^^*=2^2^ Consequently, reversing
that process, a power may be separated into equal factors,
if the given exponent be a composite number; thus, 2^^
= 24X8^ and consequently \/2^=2^, 0YfV^=2^\ for
2>2 equals either 2^ X 2* X 2*, or 2^ x 2^ x 23 x 2^.
It is important to note the distinction between separating a power
into factors which are powers, and separating that power into equal
factors, having the same roots and exponents. The latter separa
tion would be extracting a root. Thus,
25 — 23X22 because equal to 23+2.
But the square root of 2^ is no< 2 3, nor is the cube root of 2^ equal to 2 2.
If 23 be taken twice as a faxior we have 2^, and V2^ = 2^ ; similarly,
^26"= 22.
382. A root of any required degree may be extracted, by
separating the number denoting tlie degree, into its factors, and
extraditing successively Hie roots denoted by those factors. Thus,
the 9th root is the cube root of the cube root, and the 6th
root the square root of the cube root.
360 RAY'S HIGHER ARITHMETia
HORNER'S METHOD.
383. Homer's Method, named from its inventor, Mr.
W. G. Horner, of Bath, England, may be advantageously
a])plied in extracting any root, especially if the degree of
the root be not a composite number.
Biile for Extracting any Boot. — 1. Make as many
columns as there are units in the index of the root to he
extracted; place the given number at the head of the right
hand column, and ciphers at the head' of the others,
2. Commencing at the right, separate the given number into
periods of as many figures as there are columns ; extract the
required root to within unity, of the lefthand period, for the
1st figure of the root.
3. Write this figure in the 1st column, multiply it then by
itself, and set it in the 2d column; multiply this again by
the same figure, and set it in the Sd column, and so on,
placing the last product in the righthand column, under that
part of the given number from which the figure was derived,
and subtracting it from the figures above it
4. Add the same figure to the 1st column again, multiply
the result by the figure again, adding the product to the 2d
column, and so on, stopping at the neoct to the last column.
5. Repeat this process, leaving off one column at the right
every time, until all the columns have been thus dropped;
then annex one cipher to the number in the 1st column, two
to the number in the 2d column, and so on, to the number in
the last column, to which the next period of figures from the
given number must be brought down,
6. Divide the number in the last column by the number in
the previous column as a trial divisor (making allowance for
completing the divisor) ; this will give the 2d figure of the root,
which must be used precisely as the 1st figure of the root has
been ; o.nd so on, till all the periods have been brought doum.
EVOLUTION.
361
Problem.— Extract the fourth root of 68719476736.
6
6
25
50
126
375
68719476736(512 Jm
625
10
5
75
75
♦500000 '
15201
* 621947
515201
15
5
♦15000
201
515201 '
15403
* 1067466736
1067466736
*200
1
15201
202
♦530G04000
3129368
201
1
15403
203
633733368
•
202
1
♦1560600
4084
203
1
♦2040
2
1564684
2042
Note. — It is convenient to denote by ♦ the place where a column is
dropped; i, e., reached for the last time by the use of the root
figure in hand.
384. The process may often be shortened by thb
Contracted Method. — Obtain one less than half of the
figures required in the root as Hie rule directs; ^len, instead
of annexing dpJiers and bringing down a period to the last
numbers in the columns, leave tlie remainder in Hie rightliand
column for a dividend; cut off the righthand figure from tJie
last number of the previous column, two ri^ithand figures
from the last number in the column before tJiaty and so on,
ahvays cutting off one more figure for every column to the left.
With the number in the rightrhand column and the one
in the previous column, determine Hie next figure of the root,
and use it as directed in the rule, recoUeoting that (he figures
cut off are not used except in carrying the tens they produce,
H. A. 31.
362
EA Y'S HIGHER ARITHMETIC,
This process is continued until the required number of figures
is obtained, observvig Hwt when all Vie figures in the last
number of any column are cut off, thai column will be no
longer used.
Remark. — Add to the 1st column mentally; multiply and add
to the next column in one operation: multiply and subtract from
tlie righthand column in like manner.
Problem. — Extract the cube root of 44.6 to six decimals.
44.600(3.546323
3
9
17600
6
2700
1725000
90
3175
238136
95
367500
12182
100
371716
865
1050
37594^
111
1054
37659
1058
zi^U
%<i>n
Remakk.— The trial divisors
may be known by ending in two
cipliers; the com
iplete diviwrs stand just beneath them. After get
ting 3 figures of the root, contract the operation by last rule.
Examples for Practice.
1. Extract the square root of 15625.
2. Extract the cube root of 68719476736.
3. Extract the fifth root of 14348907.
4. The cube root of 151.
5. 1/97:41
6. t/TM
8. ^/35.2
12.
9. (7782757789696.
10. v" 1367631.
125.
4096.
27.
5.325074
3.1416
1.01943
.83938
2.03848
9.79795897
4.8058955
EVOLUTION.
363
APPLICATIONS OF SQUARE ROOT AND CUBE ROOT.
DEFINITIONS.
386. 1. A Triangle is a figure which has three sides
aud three angles; as, ABC, MNP.
2. The Base of a triangle is that side upon which it is
supposed to stand; as, AB, MN.
3. The Altitude of a triangle is the perpendicular dis
tance from the base to the vertex of the angle opposite;
as, HP.
Remark. — The three angles of a triangle are together eqnal to
180°, or two right angles. The proof of this belongs to Geometry,
but a fair iUvMratlon may be made in the manner indicated above.
Mark the angles of a card or paper triangle, 1 , 2, 3 ; and by two cuts
divide it into three parts. Place the marked angles with their Ver
tices as at O, and it will be seen that the pieces fit a straight ^^^
through O, while the angles cover just twice 90°, or EOD j FOD.
Any angle less than 90°, as HOF, is an ujcule angle; any angle greater
than 90°, as HOE, is an obtuse angle.
4. An Equilateral Triangle is a triangle haying three
equal sides; as, MNP.
Bemark. — ^The angles of an equilateral triangle are 60° each ;
hence, six equilateral triangles can be formed about the same point
as a vertex, each angle at the vertex being measured by the sixth of a
circumference. (Art, 204.)
364
MAY'S HIGHER AEITHMETIC.
RiOHTANOLED TRIANGLES.
380. A Rightangled Triangle is a triangle having
one right angle ; as IGK, where G = 90°.
The side opposite the right angle is called
the hypothenuse ; the other two sides are called
the base and perpendievlar.
It is demonstrated in Geometry that the
square described upon the hypothenuse of a
rightangled triangle is equal to the sum of
the squares described on the other two sides.
Illustration. — A practical proof of this may be made in the
following manner, especially valuable when the triangle has no
equal sides. It will be a useful and entertaining exercise for the
pupil.
Let the triangle be described upon a card, and let it stand upon
the hypothenuse, as AEB does. Make three straight cuts; — one,
perpendicular from A, through the smaller square; one, perpendic
ular from B, through the larger square, and one at right angles
from the end of the second cut. The two squares are thus divided
into five parts, which may be marked, and arranged, as here shown,
in a square equal to one described on the hypothenuse.
Remark. — The perpendicular in an equilateral triangle divides
the base into two equal parts, and also divides the opposite angle into
two which are 30° each. From this it follows that if, in a. right
EVOLUTION. 365
angled triangle, one angle is 30°, the side opposite that angle is half
the hypothenuse ; and, conversely, if one side be half the hypothenuse,
the angle opposite will be 30°.
387. To find the hypothenuse when the other two
sides are given.
Bule. — Add together Hie squares of the base and perpendicular y
and extract the square root of the sum.
388. To find one side when the hypothenuse and
the other side are given.
Bule. — Subtract the square of the given side from the square
of Ike hypothenuse, and extract tJie square root of the differ
ence; or,
Multiply the square root of the sum of the hypothenuse and
side by the square root of their difference.
Representing the three sides by the initial letters h, p, 6, we have
the following
FOBMULAS. — 1. h = v^p^ \ 6*.
3. b = Vh^ — j)2 ; or, b=Vh+pX Vh—p.
Examples for Practice.
1. Find the length of a ladder reaching 12 ft. into the
street, from a window 30 ft. high. 32.81 J ft.
2. What is the diagonal, or line joining the opposite cor
ners, of a square whose side is 10 ft. ? 14.142 + ft.
3. What is saved by following the diagonal instead of the
sides, 69 rd. and 92 rd., of a rectangle? 46 rd.
4. A boat in crossing a river 500 yd. wide, drifted with
the current 360 yd. ; how far did it go? 616 + yd*
366 BAY'S HIGHER ARITHMETia
Bkmark. — Integers expressing the sides of rightangle4 triangles
may be found to anj extent in the following manner : Take any two
unequal numbers ; the mm of their squarei may represent a hypoth
enuse ; the difference of their squares will then stand for one side, and
donMe their prodMct for the remaining side. Thus, from 3 and % form
13, 12, 5; from 4 and 1, form 17, 15, 8; from 5 and 2, form 29, 21, 20.
PARALLEL LINES AND SIMILAR FIGUREa
DEFINITIONS.
889. 1. Parallel Lines are lines which have the same
direction. The shortest distance between two straight par
allels is, at all points, equal to the same perpendicular line.
2. Similar Figures are figures having the same number
of sides, and their like dimensions proportional.
Rebiabks. — 1. Similar figures have their eorreRponding angles
equal.
2. If a line be drawn through any triangle parallel to one of the
sides, the other two sides are divided proportionally, and the triangle
marked ofi*, is similar to the whole triangle. An illustration of this
ha8 already been furnished in solving Ex. 8, Art.. 231.
3. All equilateral triangles are similar; the same is true of all
squares, all circles, all spheres.
3. The areas of similar figures are to each other as the
squares of their like dimensions.
4. The solidities of similar solids are to each other as the
cubes of their like dimensions.
General Exercises in Evolution and its
Applications.
1. One square is 12^ times another : how many times does
the side of the 1st contain the side of the 2d? 3^.
2. The diagonals of two similar rectangles are as 5 to 12 :
how many times does the larger contain the smaller? 5^.
EVOLUTION. 367
3. The lengths of two similar solids are 4 in. and 50 in. ;
the 1st contains 16 eu. in. : wJiat does the 2d contain ?
31250 cu. in.
4. The solidities of two balls are 189 cu. in. and 875 cu.
in. ; the diameter of the 2d is 17^ in. ; find the diameter of
the 1st. lOJ in.
5. In extracting the square root of a perfect power, the
last complete dividend was found 4725: what was the
power? 225625.
6. What number multiplied by ^ of itself makes 504?
42.
7. Separate 91 252 J into three &ctors which are as the
numbers 1, 2^, and 3. 23, 57.5, and 69.
8. What integer multiplied by the next greater, makes
1332 ? 36.
9. The length and breadth of a ceiling are as 6 and 5 ;
if each dimension were one foot longer, the area would be
304 sq. ft. : what are the dimensions? 18 ft., 15 ft.
10. In extracting the cube root of *a perfect integral
power, the operator found the last complete dividend 241984:
what was the power? 2985984.
11. If we cut from a cubical block enough to make each
dimension one inch shorter, it will lose 1657 cubic inches:
what is the solidity? 13824 cu. in.
12. A hall standing east and west, is 46 ft. by 22 ft., and
12^ ft. high : what is the length of the shortest path a fly
can travel, by walls and floor, from a southeast lower comer
to a northwest upper corner ? 57^ ft.
13. How many stakes can be driven down upon a space
15 ft. square, allowing no two to be nearer each other than
1^ ft., and how many allowing no two to be nearer than \\
ft. ? 128, and 180 stakes.
14. What integer is that whose square root is 5 times its
cube root? 15625.
15. If the true annual rate of interest be 10%, what
would be the true rate for each 73 days, if the interest be
368
I^A Y'S HIGHER ABITHMETia
compounded through the year? Prove the result by con
tracted multiplication. (Art. 334, Rem. 2.) 1.924%.
16. If a field be in the form of an equilateral triangle
whose altitude is 4 rods, what would be the cost of fencing
it in, at 75 ct. a rod? $10.39
Topical Outline.
Powers and Roots.
1. Involution.
1. Definitions.
2. Terms.
2. Evolution
. 3. Squaring and
Cubing..
1. Powers.
2. Root
3. Degree.
4. Exponent.
' 1. Algebraic Statements.
2. Numerical Illustrations.
3. Geometrical Illustration.
4. Principles.
1. Definitions.
2. Terms.
1. Root
2. Powers..
3. Radical.
4. Index.
{I
Perfect.
Imperfect
1. First explanation
(Geometrical).
3. Square Root ^ 2. Second explanation
(Algebraic).
3. Rule.
4. Cube Root.
1. First explanation
(Geometrical).
2. Second explanation
(Algebraic).
3. Rule.
5. Roots in General— Horner's Method.
6. Applications.
XXL SERIES.
DEFINITIONS.
390. 1. A Series is any number of quantities having
a fixed order, and related to each other in value according
to a fixed law. These quantities are called Terms; the first
and last are called Extremes, and the others Means.
The Law of a series is a statement by which, from some necessary
number of the terms the others may be computed.
2. There are many different kinds of series. Those usu
ally treated in Arithmetic are distinguished as Arithmetical
and Oeometrical; these series are commonly called Progres
suyiis*
ARITHMETICAL PROGRESSION.
391. 1. An Arithmetical Progression is a series in
which any term differs from the preceding or following
by a fixed number. That fixed number is called the com
mon difference; and the series is Ascending or Descending,
accordingly as the fird term is the least or the greatest.
Thus, 1, 3, 5, 7, 9, is an ascendin;» series, whose common differ
ence is 2; but if it were written in a reverse order (or, if we treated
9 as the first term), the series would be descending.
2. Every Arithmetical Progression may be considered
under the relations of five quantities, such that any three
of them being given, the others may be found. These five are
conveniently represented as follows:
First term, a.
Last term, L
Number of terms, . . . . n.
Common difference, . . . ei.
Sum of all the terms, . . «.
(369)
370 BA Y'S HIGHER ARITHMETIC,
3. These give rise to twenty different cases, but all the
calculations may be made from the principles stated in the
two following cases.
Note. — Some of the problems arising under this subject are, prop
erly, Algebraic exercises. Nothing will be presented here, however,
which is beyond analysis by means of principles and processes exhib
ited in this book. The formulas given are easily understood, and the
student will find it a very simple operation to write the numbers
in place of their corresponding letters, and work according to the
signs. The formulas are presented a^ a (xmvenieiux,
CASE I.
392. One extreme, the common difference, and the
number of terms being given, to find the other ex
treme.
Problem. — Find the 20th term of the arithmetical series
1, 4, 7, 10, etc.
Solution. — Here the series may be considered as made of 20
terms, and we seek the last. The com. diff. is 3, and the terms
are composed thus: 1, 1+3, 1+6, 1+9, etc.; and it is obTioiis
that, as the addiiion of the com. diff. commences in forming the second
lerm, it is taken twice in the third term, three times in the fourth,
and so on; similarly therefore it must be taken 19 times in form
ing the 20th, and the simple operation is, l + (20 — 1)X3 = 58, Ans,
Formula.—^ = a + (n — 1 )rf; or / = a — (n — l)d
Bule. — Multiply the number of terms less one by the com
mon difference^ add the product to Hie given extreme when the
larger is sought, subtract it from the given extreme when the
smaller is sought.
Examples for Practice.
1. Find the 12th term of the series 3, 7, 11, etc. 47.
2. Find the 18th term of the series 100, 96, etc. 32.
3. Find the 64th term of the series 3 J, 5, etc. 145 J.
SERIES. 371
4. Find the 10th term of the series .025, .037, etc. .133
5. Find the 1st term of the series 68, 71, 74, having
19 terms. 20.
6. Find the 1st term of the series 117, 123^, 130, having
6 terms. 97^.
7. Find the first term of the series 18, 12^, 6J, having
365 terms. 2281J.
CASE II.
393. The extremes and the number of terms being
given, to find the sum of the series.
Problem. — What is the sum of 9 terms of the series 1,
4, 7, 10, etc.?
OPERATION.
Explanation. — Writing the series j 1/9 l) v 3 = 25.
in fuU, in the common order, and
also in a reverse order, we have y X ^ = 117 Ans,
Sum = 1 + 4 + 7 4 10 + 13 f 16 4 19 + 22 + 25.
Sum =^ 25 + 22 4 19 4 16 + 13 + 10 + 7 + 4 + 1.
Twice the sum =26 + 26 + 26 + 26426 + 26 + 26 + 26 + 26 = 9
times the sum of the extremes; .*. the sum =^ J of 9X26 = 117, Ans,
If we add a term whose place is a certain distance beyond the
first, to another whose place is equally distant from the last, the
sum will be the same as that of the extremes, and hence, as the
above illustrates, the double of any such series is equal to the
product of the number of terms by the sum of the extremes.
Formula. — % = (a + /) n.
2
Bule. — Multiply ihe sum of ihe extremes by Hie nunxber
of termSy and divide by 2.
Examples for Practice.
1. Find the sum of the arithmetical series whose extremes
are 850 and 0, and number of terms, 57. 24225,
372 RAY'S UIQHER ARITHMETIC.
2. Extremes, 100 and .0001: number of terms, 12345.
617250.61725
3. What is the sum of the arithmetical series 1, 2, 8,
etc., having 10000 terms? 50005000.
4. Of 1, 3, 5, etc., having 1000 terms? 1000000.
5. Of 999, 888, 777, etc., having 9 terms? 4995.
6. Of 4.12, 17.25, 30.38, etc., having 250 terms?
409701.25
7. Whose 5th term is 21; 20th term, 60; number of
terms, 46? 3178.
Examples for Practice.
Kemabk. — ^It is not deemed necessary to formulate a special rule
for each class of examples here introduced. The following are pre
sented as exercises in analysis, each depending on one or more of
the principles above stated.
1. Find the common difference of a series whose extremes
are 8 and 28, and number of terms, 6. 4.
2. Extremes are 4^ and 20f, and number of terms,
14. \\.
3. Insert one arithmetical mean between 8 and 54. 31.
4. Insert five arithmetical means between 6 and 30.
10, 14, 18, 22, 26.
5. Insert two arithmetical means between 4 and 40.
16, 28.
6. Insert four arithmetical means between 2 and 3.
2i, 2, 2f , 2f
7. What is the number of terms in a series whose ex
tremes are 9 and 42, and common difference, 3? 12.
8. Whose extremes are 3 and lOJ, and common differ
ence, f ? 21.
9. In the series 10, 15 ... 500? 99.
10. What principal, on annual interest at 10^, will, in
50 yr., amount to $4927.50? $270.
SERIES, 373
GEOMETRICAL PROGRESSION.
394. 1. A Geometrical Progression is a series in which
any term after the first is the product of the preceding term
by a fixed number. That fixed number is called the ratio;
and the series is ascending or descending accordingly as the
first term is the least or the greatest.
Thus, 1, 3, 9, 27, is an ascending progression, and the com
mon multiplier is the ratio of 3 to 1 (Art. 228, 2), or of any
term to the preceding; considering 27 as the first term, the same
series may be called descending.
2. Any Geometrical Progression may be considered under
the relations of five quantities, of which three must he
known in order to find the others. These five are thus
represented :
First term, a.
Last term, I.
Satio, r.
Number of terms, . . . . n.
Sum of all the terms, . . s.
3. These quantities give rise to 20 different classes of
problems, but all of the necessary calculations depend upon
principles set forth in the following cases.
Note. — Some of the problems arising from these quantities require
such an application of the formulas as can not be understood without
a knowledge of Algebra.
CASE I.
395. From one extreme, the common ratio and
the number of terms, to find the other extreme.
Problem. — Find the 8th term of the series 1, 2, 4,
8, etc.
374 RAY'S HIGHER ARITHMETIC.
Explanation. — Here, 1 being the first statement.
term, and 2 the ratio, we see that the series 2^^ X 1 = 1 2 8, 4?ia.
may be formed thus: 1, 1X2, 1 X 2=*,
1X2', etc., the ratio being raised to its ^ecxmd power in forming
the 3d term, to its ihitd power in forming the 4th; and so, similarly,
the 8th term = lX2^ = 128, Anz.
Formula. — I = ar*i.
Rule. — Consider the given extreme aa the first term, and
mvltiply it hy that power of the ratio whose degree is denoted
by the number of terms less one.
Examples for Practice.
1. Find the last term in the series 64, 32, etc., of 12
terms. g^j.
2. In 2, 5, 12 J, etc., having 6 terms. 195 j^^.
3. In 100, 20, 4, etc., having 9 terms. i&tis 
4. 1st term, 4; common ratio, 3; find the 10th term.
78732.
5. 3d term, 16; common ratio, 6: find the 9th term.
746496.
6. 33d term, 1024; common ratio, f : find the 40th term.
136H.
7. Find the 1st term of the series 90, 180, of 6 terms. 5§.
8 Of j^^y 1^, having 11 terms. xk
CASE II.
396. From one extreme, the ratio, and the number
of terms, to find the sum of the terms.
Problem. — The first term is 3, the ratio 4, the number
of terms 5; required the sum of the series.
(H»ERATI0N.
Explanation.— Writing the 4X3X4* — 3 __ j q g 3 An$.
whole series, we have : 4 — 1 *
SERIES. 375
S=:3 + 12 + 48 + 192 + 768.
Also, 4 S = 12148 + 1921 768 h 3072.
It is evident that the lower line exceeds the upper by the difference
between 3072 and 3; this difference maybe written 4X768 — 3,\)r
4X3X4* — 3, and as this is 3 times the series, we have, once the
series =
4^1
Now, observing the form, note that we have multiplied the last term
by the ratio, then subtracted the first term, and then divided by the
ratio less one. If the series had stood with 768 for first term, and the
multiplier }, we should have had
768 f 192 h 48 h 12 + 3 =S,
192 + 48 + 12f 3 + i = J S;
and thus 768 — } =  of the series \ hence, we can write,
768 — ^X3 ^«« , ,
^^ — = 10 2 3, as before.
Here we have taken the product of the last term by the ratio from the
first term, and have divided by the excess of unity above the ratio.
In either case, therefore, we have illustrated
Bule I. — Find the last tenn and multiply it by the ratio;
then find the difference between this product and the first term,
and divide by the difference between the ratio and unity.
_ qW — a Q a — rl
r — 1 1 — r
The first answer, above given, may take another form, thus :
4X4^ X3 — 3 . ^, 3(4^—1) , ,,
— IS the same as — ; — ', where appear the
4 — 1 4 — 1
ratio 4, the first term 3, and the number of terms 5. This form, often
used when the series is ascending, hai; the following general statement :
g^ a(r* — 1)
r — 1
and corresponds to the following rule :
376 RAY'S HIGHER ARITHMETIC,
Bule II. — Bxme ihe raiio to a 'power denoted by the num
ber of terms y subtract 1, divide Hie remainder by ihe raiio less 1,
and mvltiply Hie qiiotieiit by Hie first term.
Notes. — 1. The amoiiut of a debt at coinix)und interest for a num
ber of complete intervals, is the last term of a geometrical progression,
whose ratio is 1 J the rate per cent. The table (Art. 335) shows
the powers of the ratio. For example, the jjeriod being 4 yr., the
number of terms is five; the first term is the principal, and the power of
the ratio required by Cajfe I, is the fourth.
2. The amount of an annuity at compound interest is conveniently
found by the Formula corresponding to Kule II. The table (Art.
335) is available, and the work very simple. Thus, if the annuity
be $200, the time 40 yr., and the rate 6^, we have a = 200, n = 40,
r = 1.06 Then, writing these values in the Formula, we have :
, $200(1.06*0—1) $200X9.2857179 ^onnr^oon a
Amount = ■■ ^ = — = $30952.39, Ans.
1.06 — 1 .06 ^ ,
3. If the series be a descending one having an infinite number of
terms, the last term is 0, and the product required by Rule I is 0.
Examples for Practice.
1. Find the sum of 6, 12, 24, etc., to 10 terms. 6138.
2. Of 16384, 8192, etc., to 20 terms. 32767fJ.
3. Of I, I, 3^, etc., to 7 terms. liffl
Find the sum of the following infinite geometrical series :
4. Of 1, ^, J, etc. 2.
5. Of f , /^, 3?/^, etc. 1.
6. Of ^, f , ^, etc. 2.
7. Of , 1, f etc. 8 J.
8. Of .36 = .3636, etc. =^%+ yi^, etc. ■^.
9. Of .349206, of 480, of 6. f and ^ and .
10. Find the amount of an annuity of S50, the time being
53 yr., the rate per cent 10. $77623.61
11. Applying the formula used in the last example, to
any case of the same kind, prove the truth of the rule, in
Case IV, of Annuities.
SERIES.
377
12. Calculate a table of amounts of an annuity of $1, for
any number of years from 1 to 6, at 8%.
Remark. — It is not considered neceasary to give special rules for
finding the ratio, and the number of terms when these are unknown ;
80 far as these are admissible here, they involve no principles beyond
what are presented in the matter already given.
Examples for Practice.
1. Find the common ratio: first term, 8; fourth term,
512. 4.
2. First term, 4}; eleventh term, 49375000000. 10.
3. Sixteenth term, 729 ; twentysecond term, 1000000. 3^.
4. Insert 1 geometric mean between 63 and 112. 84.
5. Four geometric means between 6 and 192.
12, 24, 48, 96.
6. Three geometric means between ^^^^^j, and ^.
7. Two geometric means between 14.08 and 3041.28
84.48 and 506.88
Topical Outline.
Series.
Series.
1. Definitions— Terms, Law, Extremes. Means.
' 1. Terms.
1. Arithmetical... .
2. Classes... .
, 2. Geometrical....
2. Cases.
(Formulas.)
' 1. Terms.
2. Cases.
(Formulas.)
H. A. 32.
XXn. MENSUEATIOK
DEFINITIONS.
397. 1. Gteometry is that branch of mathematics which
treats of quantity having extension and form. When a
quantity is so considered, it is called Sfojpwfttide.
2. There are four kinds of Magnitude known to Greom
etry : Lines, Angles, Surfaces, and Solids. A point has posi
tion, but not magnitude.
3. Mensuration is the application of Arithmetic to Geom
etry ; it may be defined also as the art of computing lengths^
areas, and volumes,
LINES.
398. 1. A line is that which has length only.
2. A straight line is the shortest distance between two
points.
3. A broken line is a line made of connected straight
lines of different directions.
4. A curve, or curved line, is a line having no part
straight.
The word "line," used without the qualifying word "curve," is un
derstood to mean a straight line.
5. A horizontal line is a line parallel with the horizon,
or with the water level. (See Art. 389; 1.)
6. A vertical line is a line perpendicular to a horizontal
plane.
(878)
MENSURATION. 379
ANGLES.
399. An angle is the opening, or inclination, of two
lines which meet at a point. (Art. 204.)
Remark. — Angles differing from right angles are called oblUjue
anyles, (See Art. 885, 3, Bern., and Art. 886.)
SURFACES.
Polygons.
400. 1. A surftee is that which has length and breadth
without thickness.
A solid has length, breadth, and thickness.
A line is meant when we speak of the »Ule of a limited surface, or
the ed<fe of a solid; a surface is meant when we speak of the side or
the base of a solid.
2. A Plane is a surface such that any two points in it
can be joined by a straight line which lies wholly in the
surface. The application of a straightedge is the test of a
plane.
3. A plane figure is any portion of a plane bounded by
lines.
4. A polygon is a portion of a plane inclosed by straight
lines ; the perimeter of a polygon is the whole boundary.
5. Area is surface defined in amount. For the numerical
expression of area, a square is the measuring unit. (Art.
197.)
6. A polygon is regular when it has all its sides equal,
and all its angles equal.
7. A polygon having three sides is a trian f^y^
gle ; having four sides, a quadrilateral ; five /y^ \
sides, a pentagon ; six sides, a hexagon, etc. Quadrilateral.
380
BA Y'S HIGHER ARITHMETIC.
The six diagrams following represent regular polygons.
Pentagon.
Hexagon.
Hel>Ugon.
Octagon.
Nooiigon.
Decagon .
8. The diagonal of a polygon is the straight line joining
two angles not adjacent ; as, PN, on the preceding page.
9. The base is the side on which a figure is supposed to
stand.
10. The altitude of a polygon is the perpendicular distance
from the highest point, or one of the highest points, to the
line of the base.
11. The center of a regular polygon is the point within,
equally distant from the middle points of the sides ; the
apothem of such a polygon is the perpendicular line drawn
from the center to the middle of a side ; as, C a center, and
CD an apothem.
Triangles.
401. Triangles are classified with respect* to their angles,
and also with respect to their sides.
Acute Triangles.
Obtuse Triangles.
Scalene.
Isosceles.
Equilateral. Isosceles.
Scalene.
1. A triangle is rightangled when it has one right angle ;
it is acuteangled when each angle is acute; it is obtuse
angled when one angle is obtuse. These three classes may
be named right triangles, acute triangles, obtuse triangles;
the last two classes are sometimes called oblique triangles.
MENSURATION. 881
2. A triangle is scalene when it has no equal sides;
isosceles, when it has two equal sides; and equilateral,
when its three sides are equal.
A right triangle can be scalene, as when the sides are 3, 4, 5 ; or,
isosceles, as when it is one of the halves into which a diagonal di
vides a square. An obtuse triangle can be scalene or isosceles ; an
acute triangle can be scalene, isosceles, or equilateral.
Quadrilaterals.
402. Quadrilaterals are of three classes:
1. A Trapezium is a quadrilateral having no two sides
parallel.
2. A Trapezoid is a quadrilateral having two and only
two sides parallel.
3. A Parallelogram is a quadrilateral having two pairs
of parallel sides. *
Trapezium « Trapezoid. Riiomboid. Rliombus. Rectangle.
403. Parallelograms are of three classes:
1. A Rhomboid is a parallelogram having one pair of
parallel sides greater than the other, and no right angle.
2. A Bhombiis is a parallelogram whose four sides are
equal.
3. A Rectangle is a pai'allelogram whose angles are all
right angles ; when the rectangle has four equal sides, it is
a square.
A square is a rhomhus whose angles are 90° ; it is also the
form of the unit for surface measure. It may properly be
defined, an equilateral rectangle.
382 BA Y'S HIGHER ARITHMETIC.
r G
n
/K
E L
I
AREAS.
Triangles and Quadrilaterals.
404. The general rules depend on the prmciples stated
in the following remarks:
Kemarks. — 1. The area of a rectangle is equal to the product
of its length by its breadth. (Art. 197, Ex.)
2. The diagonal of a rectangle divides it into two equal triangles.
The accompanying figure illustrates this;
EHI = HEF. Observe also that the per
pendicular GL divides the whole into two
rectangles; EGL is half of one of them,
LXjrl the half of the other, and both these
smaller triangles make EGI, which must
therefore be half of the whole; EGI and
EHI have the same base and equal altitudes.
If the triangle EGH be supposed to stand on GH as a 6a»e,
its altitude is EF; the perpendicular which represents the height
is, in such a case, said to fall on the base produced; i. e., extended.
The triangle EGH is equal to the half of GHIL. Any triangle
has an area equal to the product of half the base by the altitude.
Observe also that the trapezoid EFGI is made of two triangles,
EGF and EGI; each of these has the altitude of the trapezoid,
and each has one of the parallel sides for a base. Hence, the area
of each being J its base X the common altitude, the two areas, or
the whole trapezoid, must equal the half of both bases X the altitude.
3. If the piece EGF were taken off and put on the right of EGHI,
the line EF being placed on HI, the whok area would be the same, but
the perimeter would be increased, and the figure would be a rhomboid.
Different quadrilaterals may have equal areas and waeqwd boundaries;
also, they may have the sides in the same order and equal, with unequal
areas. To find accurately the area of a quadrilateral, more must
be known than merely the four sides in order. A regular polygon
has a greater area than any other figure of the same perimeter.
4. When triangles have equal bases their areas are to each other
as their altitudes; the altitudes being equal, their areas are as their
bases. The area of any triangle is equal to half the product of
the perimeter by the radius of the inscribed circle.
MENSUEATION. 383
General Bules.
1. To find the area of a parallelogram.
Bole. — Multiply one of ttoo paraUd sides by the perpendicular
ditiance between them.
n. To find the area of a triangle.
Bole. — Take half Hie product of Hie base by the altitude.
m. To find the area of a trapezoid.
Bole. — MuUiply half the sum of the parallel sides by the
aUiivde.
Note. — ^The following is demonstrated in Geometry :
IV. To find the area of a triangle when the sides
are given. ,
Bule. — Add the three sides together and take half the sum;
from the half sum take the sides separately ; multiply the half sum
and die three remainders together, and extras the square rod of
the product.
Notes. — 1. The area of a trapezium may be found by applying
this rule to the parts when tlie sides are known and the diagonal is
given in length and in special position as between the sides. The
area of any i)olygon may be found by dividing it into triangles and
measuring their bases and altitudes.
2. The area of a rhombus is equal to half the product of ite diag
onaLs; these are at right angles.
Examples for Practice.
1. Find the area of a parallelogram whose base is 9 ft. 4
in. and altitude 2 ft. 5 in. 22 sq. ft. 80 sq. in.
2. Of an oil cloth 42 ft. by 5 ft. 8 in. 26 sq. yd.
384 RAY'S HIGHER ARITHMETIC.
3. How many tiles 8 in. square in a floor 48 ft. by 10 ft.?
1080.
4. Find the area of a triangle whose base is 72 rd. and
altitude 16 rd. 3 A. 96 sq. rd.
5. Base 13 ft. 3 in.; altitude 9 ft. 6 in.
62 sq. ft. 135 sq. in.
6. Sides 1 ft. 10 in.; 2 ft.; 3 ft. 2 in.
1 sq. ft. 102— sq. in.
7. Sides 15 rd.; 18 rd.; 25 rd. 133.66— sq.'rd.
8. What is the area of a trapezoid whose bases are 9 ft,
and 21 ft, and altitude 16 ft.? 240 sq. ft.
9. Bases 43 rd. and 65 rd. ; altitude 27 rd. ?
9 A. 18 sq. rd.
10. What is the area of a figure made up of 3 triangles
whose bases are 10, 12, 16 rd. and altitudes 9, 15, 10^ rd.?
1 A. 59 sq. rd.
11. Whose sides are 10, 12, 14, 16 rd. in order, and dis
tance from the starting point to the opposite corner, 18 rd. ?
1 A. 3.9 — sq. rd.
12. How much wainscoting in a room 25 ft. long, 18 ft.
wide, and 14 ft. 3 in. high, allowing a door 7 ft. 2 in. by 3
ft. 4 in., and two windows, each 5 ft. 8 in. by 3 ft. 6 in.,
and a chimney 6 ft. 4 in. by 5 ft. 6 in.; charging for the
door and windows halfwork? 128^ sq. yd.
13. What is the perimeter of a rhombus, one diagonal be
ing 10 rd., and the area 86.60J sq. rd. ? 40 — rd.
14. Find the cost of flooring and joisting a house of 3
floors, each 48 ft. by 27 ft., deducting from each floor for a
stairway 12 ft. by 8 ft. 3 in., allowing 9 in. rests for the
joists; estimating the flooring and joisting between the walls
at $1.46 a sq. yd., and the joisting in the walls at 76 ct. a
sq. yd. ; each row of rests being measured 48 ft. long by 9
in. wide. $600.78
15. What is the area of a square farm whose diagonal is
20.71 ch. longer than a side? 250 acres.
16. How many sq. yd. of plastering in a room 30 ft;.
MENSURATION. 385
long, 25 ft. wide, and 12 ft. high, deducting 3 windows,
each 8 ft. 2 in. by 5 ft. ; 2 doors each 7 ft. by 3 ft. 6 in. ;
and a fireplace 4 ft. 6 in. by 4 ft. 10 in. ; the sides of the
windows being plastered 15 in. deep? And what will it
cost, at 25 ct a sq. yd.? 215j^ sq. yd.; cost $53.83
17. From a point in the side and 8 ch. from the corner
of a square field containing 40 A., a line is run, cutting off
19i A.: how long is the line? One answer, 20^ ch.
18. How much painting on the sides of a room 20 ft.
long, 14 ft. 6 in. wide, and 10 ft. 4 in. high, deducting a
fireplace 4 ft. 4 in. by 4 ft., and 2 windows each 6 ft. by
3 ft 2 in.? 73^ sq. yd.
19. Find the cost of glazing the windows of a house of 3
stories, at 20 ct. a sq. ft. Each story has 4 windows, 3 ft.
10 in. wide; those in the 1st story are 7 ft. 8 in. high;
those in the 2nd, 6 ft. 10 in. high; in the 3d, 5 ft» 3 in.
high. * $60.56S
Reguiar Polygons and the Circle.
405. Any regular polygon may be divided into equal
isosceles triangles, by lines from the center to the vertices;
the apothem is their common altitude, and the perimeter
the sum of their bases.
406. To find the area of a regular polygon.
Rule. — Multiply the perimeter by Iwlf the apotfiem.
All regular ])olygoiis of the same nuuiber of sides are similar
figures. (Art. 389, Rem. 1.)
407. 1. The Qircle, as already defined (Art. 204), is a
figure bounded by a uniform curve.
2. Any line drawn in a circle, having its ends in the
curve, is called a chord; as AB, BD.
H. A. r..
386 EAY\S HIGHER ARITHMETIC,
3. The portion of the curve which
is cut off by such a line is called an
arc, and the space between the chord
and the arc is called a segment.
Thus, the curve APB is an arc, AB is
the chord of that arc, and these inclose a
segment whose base is AB, and whose height
is OP.
4. If a line be drawn from the middle of a chord to the
center, it will be perpendicular to the chord ; so also, a line
perpendicular at the middle of a chord, will, if extended,
pass through the center, and bisect either of the arcs stand
ing on that chord. Thus, AB is bisected by the perpen
dicular CO, and the arc AP = PB ; so the arc AD = BD.
5. A tangent to a circle as a straight line having only
one point in common with the curve; it simply touches
the circle; a secant enters the figure from without.
If with C as a center, and CO as a radius, a circle were drawn
in the equilateral triangle ABD, the sides would be tangent to
the circle; the circle would be inscribed in the triangle. The circle
of which CB is radius, is circumscribed about ABD.
6. The space inclosed by two radii and an arc, is called
a sector; as, ACP.
The arc of that sector is the same fraction of the whole cir
cumference that the area of the sector is of the whole circle.
Calculations Pertaining to the Circle.
408. The accompanying diagrams present (Fig. 1) a
regular polygon of six sides, (Fig. 2) one of twelve sides,
and (Fig. 3) a circle divided into twentyfour sectors.
Remarks. — 1. The hexagon is comK)sed of six equilateral triangles,
and hence if OB be 1, the side AB= 1, and it is easy to compute the
apolhem, v^ 1 — i — .86602540378
MENSURATION.
387
2. If the distance from center to vertices be unchanged, and a
regular polygon of twelve siden be formed about the same center,
it will differ jess from a circle whose ra
dius is OB, than the hexagon differs from
such a circle. This is evident from the
second figure; and if the polygon be made
of twentyfour sides (the distance from cen
ter to vertices remaining the same), it will
be still nearer the circle in shape and size ;
in the space of the diagram, one of the
twentyfour triangles forming such a poly
gon would differ very little from one of
the twentyfour sectors here shown. The
circle is regarded as composed of an in
finite number of triangles whose common
altitude is the radius and the sum of whose
bases is the circumference. Hence, the area
= J the sum of bases X altitude; or,
Area of circle ~ J circumference X radius.
Area of circle = \ circumference X diameter.
Fig. 2.
3. Since the perimeter of the hexagon is
6, it is easy to compute the next perime
ter shown, 'which is 12 times AP or BP.
The apothem being found above," subtract
it from OP or 1, and obtain .13397459622
the perpendicular of a rightangled tri
angle; then, the base of that triangle be
ing .5, the half of AB, find the hypothe
nuse .517638090205, = PB. Now, if we
treat PB as we treated AB we can find Fig. 3.
the apothem of the second figure, and then.
find one of the 24 sides of another polygon, still more nearly equal to
the circle. If these operations be continued, we shall find results
$th and 9th as follows:
»
Perimeter of polygon of 1536 sides = 6.28318092
Perimeter of polygon of 3072 sides — 6.28318420
If the distance from center to vertices be taken J instead of 1,
the results will be
3.141590 f and 3.141592 f
388 RAY'S HIGHER ARITHMETIC.
Hence, if the circle of diameter 1, be taken as a polygon of 1536
sidefl, and then aa a polygon of 3072 sides, the expressions for
perimeter do not differ at the fourth decimal place. The number
3.1416 is usually given, although by more expeditious methods than
that above illustrated, the calculation has been carried to a great
number of decimal places, of which the following correctly shows
eighteen :
3.141592653589793238
This important ratio, of circumference to diameter, is represented
by the Greek letter tt (pi,),
4. Since circumference == diameter X "^j ^^^ *r®2i = J circum
ference X diameter, we have area =2rX square of diameter. Rep
resenting the circumference by c, area by C, diameter by (/, radius
by R, we have the following formula*^:
C = d7r; G = d^^; C=R'^r.
General Rules.
409. Pertaining to the circle we have the following
general rules:
I. To find the circumference:
1. Multiply the diameter by 3.1415926; or,
2. Divide iJie area hy \ of the diameter; or,
3. Hiiract the square root of 12.56637 tim^s the area,
II. To find the diameter:
1. Divide the circumference by 3.1415926; or,
2. Divide the area 6i/ .785398, and extract the square root,
m. To find the area:
1. Multiply the diameter by ^ of the circumference; or,
2. Multiply Hie square of the diameter by .785398; or,
3. Mtilfiply the square of Hie radius by 3.1415926
MENSUBATION, 389
IV. To find the area of a sector of a cirole :
1. MvlUply ilie arc by one half the radius; or,
2. Take such a fraction of the whole area as the arc is of the
xxHiole circumference,
V. To find the area of a segment less than a semi
circle :
1. Subtract from the area of the sector having the same arc.
Vie area (f the triangle whose base is the base of Hw segment, and
wlwse vertex is tJie center of the eirde ; or,
2. Divide Vie cube of Vie height by twice the base, and increase
the quotient by two thirds cf the product of height and base.
Remark. — Add the triangle to the factor, if the segment be greater
than a semicircle. The second rule gives an approximate result.
Notes. — 1. The side of a square inscribed in a circle is to radius
as V^2 is to 1.
2. The side of an inscribed equilateral triangle Ls to radius as l/ 3
is to 1.
3. If radius be 1, the side of the inscribed regular pentagon is
1.1755; heptagon, .8677; nonagon, .6840; undecagon, .5634
Examples for Practice.
1. What are the circumferences whose diameters are 16,
22i, 72.16, and 452 yd.?
50.265482; 69.900436; 226.6973; 1420 yd.
2. What are the diameters whose circumferences are 56,
182 J, 316.24, and 639 ft.?
17.82539; 58.09; 100.66232; and 203.4 ft.
3. Find the areas of the circles with diameters 10 ft. ;
2 ft. 5 in. ; 13 yd. 1 ft.
78.54 sq. ft. ; 660.52 sq. in. ; 139 sq. yd. 5.637 sq. ft.
4. Whose circumferences are 46 ft. ; 7 ft. 3 in. ; 6 yd.
1 ft. 4 in.
168.386 sq. ft. ; 4 sq. ft. 26.322 sq. in. ; 29.7443 sq. ft;.
390 JiA Y'S HIGHER ARITHMETIC.
5. CircuiD. 47.124 ft., diameter 15 ft. 176.715 sq. ft.
6. If we Baw down through J of the diameter of a round
log uniformly thick, what portion of the log is cut in two ?
.2918
7. What fraction of a round log of uniform thickness ia
the largest squared stick which can be cut out of it ?
.6366
410. 1. A Solid is that which has length, breadth, and
thickness.
A Holid may have plane mirtaces, curved surtnees, or both. A.
aaved tarfaa ia one no part of which ia n >lnne.
2. The taoeB of a solid are the polygons formed by the
intersections of its bounding pLaues; the lines of those inter
sections are called edges.
3. A Prism is a solid having two bases which are parallel
polygons, and faces which are parallelograms.
A prixm is triangular, quadrangular, etc., according to the Rhnpe
of itH base. The first iif the ligiirea here given represents a qmulran
gular prism, theeecond apenbtipnat prism.
4. A right prism is a prism whose &ce8 are rectangles.
5. A Farallelopiped is a prism whose faces are parallel
ograms. Its bounding surfaces are six parallelograms.
The first figure above represents a parallelepiped whose faces are
MENSURATION. 391
6. A Cube 13 a parallelopiped whose feces are squares.
7, A Cylinder is a solid having two bases which are
equal parallel circles, and having an equal diameter in any
parallel plane between them.
. A Pyramid is a solid with only one base, and whose
faces are triangles with a common vertex.
9. A Cone is a solid whose base is a circle, and whose
other surface is convex, terminating above in a point called
the vertex.
10. A &ustum of a pyramid or cone is the solid which
remaios when a portion having the vertex is cut ofi* by a
plane parallel to the base.
11. A Sphere is a solid hounded by
a curved surface, every point of which
is at the same distance from a point
within, called the center. The diamder
of a sphere is a straight line passing
through the center and having its ends
in the surface; the radius is the distjuice
from the center to the surface.
A segnient of n nphere is a portion cut off by one plane, or lielween
two planes; il^ baies nre cirdes, nnd its height is the portion of llie
diameter whicJi iK cut off witli it.
12. The slant heigbt of a 'pyramid is the perpendicular
distance from the vertex to one of the sides of a base ; the
slant height of a cone is the straight line drawn from the
vertex to the circumference of the base.
392 RAY'S HIGHER ARITHMETIC.
13. The altitude of any solid is the perpendicular dis
tance between the planes of its bases, or th^ perpendicular
distance from its highest point to the plane of the base.
14. The Volume of a solid is the number of solid units
it contains ; the assumed unit of measure is a cube. (Art.
199.)
15. Solids are similar when their like lines are proper
tionaly and their corresponding angles e(}ual.
General Rui.es.
I. To find the convex surface of a prism or cyl
inder :
Rule. — Multiply the perimeter of Vie base by tlie altitude,
II. To find the volume of a prism or cylinder :
Rule. — Multiply tlie base by the altitude,
m. To find the convex surface of a pyramid or
cone :
Rule. — Multiply the perimeter of tlie base by one lialf Vie slant
height,
TV. To find the volume of a pyramid or cone:
Rule. — Multiply the base by one third of Vie altitude,
V. To find the convex surface of a Aioistum of a
pyramid or cone: •
Rule. — Multiply Judf the sum of the perimeters of Vie bases
by the slaiii Imgld,
VI. To find the solidity of a Arustum of a pyramid
or cone:
MENSURATION, 393
Rule. — To ike sum of due two hoses add the square root of
iJmr product, and multiply tlie amount by mie Hard of Hie aUi
tude,
VII. To find the surface of a sphere :
Rule. — Multiply Hie circumference by Hie diatnder.
VIII. To find the volume of a sphere :
Rule. — Multiply the cube of Hie diameter by .5235987
Notes. — 1. Similar solids are to each other as the cubes of their
like dimensions.
2 The sphere is regarded as com[K)sed of an infinite number of
cones whose common altitude is the radius, and tlie »«im of whose
bases is the whole surface of the sphere.
3. The cone is regarded as a ])yramid of an infinite number of faces,
and the cylinder as a prism of an infinite number of faces.
Examples for Practice.
*
1. Find the convex surface of a right prism with altitude
11 J in., and sides of base 5^^, 6^, 8^, 10^, 9 in.
450 sq. in.
2. Of a right cylinder whose altitude is If ft., and the
diameter of whose base is 1 ft. 2^ in.
6 sq. ft., 92.6 sq. in,
3. Find the whole surface of a right triangular prism, the
sides of the base 60, 80, and 100 ft. ; altitude 90 ft.
26400 sq. ft.
4. The whole surface of a cylinder ; altitude 28 ft. ; cir
cumference of the base 19 ft. 589.455 sq. ft.
5. Find the convex surface and whole surface of a right
pyramid whose slant height is 391 ft. ; the base 640 ft.
square.
Conv. surf 500480 sq. ft. ; whole surf. 910080 sq. ft.
394 RAY'S HIGHER ARITHMETIC.
6. Of a right cone whose slant height is 66 ft. 8 in. ;
radius of the base 4 ft. 2 in.
125663.706 sq. in. ; 133517.6876 sq. in.
7. Find the solidity of a pyramid whose altitude is 1 ft.
2 in., and whose base is a square 4J in. to a side.
94J cu. in.
8. Whose altitude is 15.24 in., and whose base is a triangle
having each side 1 ft. 316.76 cu. in.
9. What is the solidity of a prism whose bases are squares
9 in. on a side, and whose altitude is 1 ft. 7 in. ?
1539 cu. in.
10. Whose altitude is 6 J ft. , and whose bases are parallel
ograms 2 ft. 10 in. long by 1 ft, 8 in. wide?
30 cu. ft. 1200 cu. in.
11. Whose altitude is 7 in., and whose base is a triangle
with a base of 8 in. and an altitude of 1 ft. ? 336 cu. in.
12. Whose altitude is 4 ft. 4 in., and whose base is a tri
angle with sides of 2, 2 J, and 3 ft. ? 10.75 cu. ft.
13. What is the solidity of a cylinder whose altitude is 10^
in., and the diameter of whose base is 5 in. ? 206.167 cu. in.
14. Find the convex surface of a frustum of a pyramid
with slant height 3 J in., lower base 4 in. square, upper base
2J in. square. 43J sq. in.
15. The convex surface and whole surface pf the frustum
of a cone, the diameters of the bases being 7 in. and 3 in.,
and the slant height 5 in.
Conv. surf. 78.5398 sq. in., whole surf. 124.0929 sq. in.
16. Find the solidity of a frustum of a pyramid whose
altitude is 1 ft. 4J in. ; lower base, lOf in. square ; upper,
4 in. square. 97 4y cu. in.
17. Of a frustum of a cone, the diameters of the bases
being 18 in. and 10 in., and the altitude 16 in.
2530.03 cu. in.
18. What are the surfaces of two spheres whose diameters
are 27 ft. and 10 in. ?
2290.2211 sq. ft. and 314.16 sq. in.
MENSUBATION. 395
19. Find the solidity of a sphere whose diameter is 6 mi.,
and surface 113.097335 sq. mi. 113.097335 cu. mi.
20. Of a sphere whose diameter is 4 ft. 33.5103 cu. ft.
21. Of a sphere whose surface is 40115 sq. mi.
755499i cu. mi.
22. By what must the diameter of a sphere be multiplied
to make the edge of the largest cube which can be cut out
ofit? • .57735
MISCELLANEOUS MEASUKEMENTS.
Masons' and Beicklayers* Work.
411. Masons' work is sometimes measured by the
cubic foot, and sometimes by the perch. The latter is 16^
ft. long, 1^ ft. wide, and 1 ft. deep, and contains 16^ X
1^ X 1 = 24 cu. ft., or 25 cu. ft. nearly.
412. To find the number of perches in a piece of
masonry.
Bule. — Find the solidity of {he wall in cubic feet by the rules
given for mensuration of solids, and divide it by 24f .
Note. — ^Brick work is generally estimated by the thousand bricks ;
the usual size being 8 in. long, 4 in. wide, and 2 in. thick. When
bricks are laid in mortar, an allowance of ^ is made for the mortar.
Examples for Practice.
1. How many perches of 25 cu. ft. in a pile of building
stone 18 ft. long, 8^ ft. wide, and 6 ft. 2 in. high ?
37.74 = 37f perches nearly.
2. Find the cost of laying a wall 20 ft. long, 7 ft. 9 in.
high, and with a mean breadth of 2 ft., at 75 ct. a perch.
$9.39
896 HAY'S HIGHER ARITHMETIC.
3. The cost of a foundation wall 1 ft. 10 in. thick, and
9 ft: 4 in. high^ for a building 36 ft long, 22 ft. 5 in. wide
outside, at $2.75 a perch, aUowmg for 2 doors 4 ft. wide.
$192.98
4. The cost of a brick wall 150 ft. long, 8 ft. 6 in. high,
1 ft. 4 in. thick, at $7 a thousand, allowing ^ for mortar ?
$289.17
5. How many bricks of ordinary size will buiid a square
chimney 86 ft. high, 10 ft. wide at the bottom, and 4 ft. at
the top outside, and 3 ft. wide inside all the way up?
89861+ bricks.
SuQGEsnoN. — ^Find the solidity of the whole chimney, then of the
hollow part; the difference will be the solid part of the chimney.
Gauging.
413. Gtmglng is finding the contents of vessels, in bush
els, gallons, or barrels.
414. To gauge any vessel in the form of a rect
angular solid, cylinder, cone, fhistum of a cone, etc.
Rule. — Fmi the solidity of (he vessel in cubic inches hy Hie
rules already given; this divided by 2150.42, iviU give five con
tents in bushels; by 231, will give it in wine gallons^ which
may be reduced to barrels by dividing Hie uurnber by 31^.
Note. — ^In applying the rule to cylinders, cones, and frustums of
cones, instead of multiplying the square of half the diameter by
3.14159265, and dividing it by 231, multiply the square of the diounieter hy
.0034, which amounts to the same, and is shorter.
Examples for Practice.
1. How many bushels in a bin 8 ft. 3 in. long, 3 ft 5 in.
high, and 2 ft. 10 in. wide? 64.18 bu.
MENSURATION. 397
2. How many wine gallons in a bucket in the form of a
frustum of a cone, the diameters at the top and bottom being
13 in and 10 in., and depth 1^ in. ? 5.4264 gal.
3. How many barrels in a cylindrical cistern 11 ft. 6 in.
deep and 7 ft. 8 in wide? 126.0733 bbl.
4. In a vat in the form of a frustum of a pyramid, 5 ft.
deep, 10 ft. square at top, 9 ft. square at bottom?
107.26 bbL
415. To find the contents in gallons of a cask or
barreL
Bemark. — ^When the staves are straight from the bung to each
end, consider the cask as two frustums of a cone, and calculate its
contents by the last rule ; but when the staves are curved, use this
rule:
Bule. — Add to (he head diameter (inside) tvx) Udrds of (he
difference between the head and bung diameters; but if the staves
are only slightly curved, add six tenths of this difference; this gives
Vie mean diameter; express it in indies ^ square it, multiply it by
the length in inches, anj, this product by .0034: Hie product will
be the contents in wine gallons.
Note. — ^Aiter finding the mean diameter, the contents are found as
if the cask were a cylinder.
Examples for Practice
1. Find the number of gallons in a cask of beer whose
staves are straight from bung to bead, the length being 26
in., the bung diameter 16 inches, and head diameter 13 in.
18.65 gal.
2. In a barrel of whisky, with staves slightly curved,
length 2 ft. 10 in., bung diameter 1 ft. 9 in., head 1 ft. 6 in.
45.32 gal.
3. In a cask of wine with curved staves, length 5 ft. 4 in.,
bung diameter 3 ft. 6 in., head diameter 3 ft. 348.16 gal.
398 BAY'S HIGHER ARITHMETIC.
Lumber Measure.
416. To find the amount of squareedged incli
boards that can be sawed from a round log.
Kemark. — The following is much used by lumbei*men, and is
sufficiently accurate for practical purposes. It is known as Doyle'a
Rule.
Bule. — From the diameter in inches svbtract 4; the square
of the remainder will he the number of square feet of inch boards
yielded by a log 16 feet in lengtJi,
Examples for Practice.
1. Ho^ much squareedged inch lumber can be cut from
a log 32 inches in diameter, and 20 feet long?
OPERATION.
32 — 4 = 28; 28 X 28 X H = ^^^ feet.
Or, J X 28 X 28 = 980 feet.
2. In a log 24 in. in diameter, and 12 tt. long? 300 ft.
8. In a log 25 in. in diameter, and 24 ft. long?
661i ft.
4. In a log 50 in. in diameter, and 12 ft. long?
1587 ft.
To Measure Grain and Hay.
417. Grain is usually estimated by the bushel, and sold
by weight; Hay, by the ton.
Hem ARKS. — 1. The standard bushel contains 2150.4 cubic inches.
A cubic foot is nearly .8 of a bushel.
2. Hay well settled in a mow may be estimated (approximately)
at 550 cubic feet for clover, and 450 cubic feet for timothy, per ton.
MENSURATIOK 399
418. To find the quantity of grain in a wagon or
in a bin:
Rule. — Multiply Hie contents in cubic feet by .8
Kemabks. — 1. If it be com on the cob, deduct one half.
2. For corn not ** shucked," deduct two thirds for cob and shuck.
419. To find the quantity of hay in a stack, rick,
or mow :
Bule. — Divide the cvbical contents in feet by 550 for clover,
or by 450 for timotliy; Hie quotient will be Hie number of tons.
Examples for Practice.
1. How many bushels of shelled com, or corn on the
cob, or corn not shucked, will a wagonbed hold that is 10^
feet long, 3 J feet wide, and 2 feet deep?
58.8 bu.; 29.4 bu. ; 19.6 bu.
2. In a bin 40 feet long, 16 wide, and 10 feet high?
5120 bu.
3. A haymow contains 48000 cubic feet: how many tons
of well settled clover or timothy will it hold?
87^ tons clover ; 106f tons timothy.
Topical Outline.
Mensuration.
1. General Definitions :— Geometry, Magnitude, MensuraUon.
Parallel.
Perpendicular.
I Straight \ Horizontal.
2. Lines . Broken.
Curved.
Vertical.
Diagonal.
400
HAY'S HIGHER ARITHMETia
Mensuration. — (Continued. )
8. Angles }
4. Surfaces.
6. Solids.....
Right
Oblique / Acute.
Obtuse.
1. General Definitions:— Plane, Plane figure. Area, Polygons/
Regular, Perimeter, Similar, Center, Altitude, Base,
Apothem.
f Isosceles.
\ Scalene.
Scalene.
Acute.
2. Triangles... .
(Rules.)
1. Right.
2. Oblique.
Obtuse
{
Isosceles.
EquilateraL
Scalene.
Isosceles.
L Trapezium.
3. Quadrilateral J 2. Trapezoid.
(Rules.) [ 8. Parallelogram.
1. Rhomboid.
2. Rhombus... { Square.
3. Rectangle... j Square.
' 1. Terms:— Circumference, Radius, Chord, Diam
eter, Segment, Sector, Tangent, Secant.
2. Calculations, Value of w.
3. Formulas, Rules.
1. General Definitions :— Solid, Base, Face, Edge, Similar.
4. Circle
2. Prism...
(Rules.)
Triangular. (Right.) _ „ , . ,
Quadl^lar.  ^"^^^T" J^"*
Pentagonal, etc. * \ S )
r Altitude.
3. Pyramid J slant Height.
(Rules.) (Frustum.) ( Convex Surface.
f Altitude.
4 Cone J slant Height
(Rules.) (Frustum.) ( Convex Surface.
5. Cylinder. / Surface.
(Rules.) ^ Solidity.
1. Terms :Radius, Diameter, Seg
ment.
2. Convex Surface, Rule.
. 3. Solidity, Rule.
6. Sphere.
7. General Formulas.
8. Miscellaneous
Applications..
' Masons' and Bricklayers* Work.
Gauging.
Lumber Measure.
. Measuring Grain and Hay.
XXni. MISCELLAZ^TEOUS EXERCISES.
Note.— Nos. 1 to 50 are to be solved mentally.
1. If I gain I ct. apiece by selling eggs at 7 ct. a dozen, how
much apiece will I gain by selling them at 9 ct. a dozen? f^ ct.
2. If I gain ^ ct. apiece by selling apples at 3 for a dime, how
much apiece would I lose by selling them 4 for a dime? \ ct.
3. If I sell potatoes at 37J ct. per bu., my gain is only  of what it
would be, if I charged 45 ct. per bu.: what did they cost me?
26 J ct. per hu.
4. If I sell my oranges for 65 ct., I gain f ct. apiece more than if I
sold them for 50 ct.: how many oranges have I? 40 oranges.
5. If I sell my pears at 5 ct. a dozen, I lose 16 ct.; if I sell them at
8 ct. a dozen, I gain 11 ct.: how many pears have I, and what did
they cost me? 9 dozen at 6 J ct. per dozen.
6. If I sell ^gs at 6 ct. per dozen, I lose f ct. apiece; how much per
dozen must I charge to gain  ct. apiece? 22 ct.
7. One eighth of a dime is what part of 3 ct.? y\.
8. If I lose f of my money, and spend f of the remainder, what
part have I left? J.
9. A's land is \ less in quantity than B's, but ^ better in quality:
how do their farms compare in value? A's = j^ of B's.
10. If f of A's money equals  of B's, what part of B's equals  of
A's? J.
11. I gave A t\ of my money, and B ^^ of the remainder: who got
the most, and what part? B got ^ of it more than A.
12. A is f older than B, and B J older than C: how many times
es age is A's? 2 J.
13. Two thirds of my money equals J of yours; if we \)\xt our
money together, what part of the whole will I own? ■^^,
14. How many thirds in ^? 1J.
15. Reduce  to thirds ; f to ninths ; and f to a fraction, whose nu
merator shall be 8. 21 thirds; 74 ninths;
^ ' ^ ' 13J.
16. What fraction is as much larger than  as f is less than ^? If.
17. After paying out \ and \ of my money, I had left $8 more than
I had spent: what had I at first? $80.
H. A. 34. (401)
402 HA Y'S HIGHER ARITHMETIC.
18. In 12 yr. I shall be J of my present age: how long since was I ^
of my present age? 8f yr.
19. Four times } of a number is 12 less than the number: what is
the number? 108.
20. A man left jfij of his property to his wife, f of the remainder to
his son, and the balance, $4000, to his daughter, what was the estate?
$22000.
21. 1 sold an article for \ more than it cost me, to A, who sold it
for $6, which was \ less than it cost him : what did it cost me? $8.
22. A is J older than B ; their father, who is as old as both of them,
is 50 yr. of age: how old are A and B? A, 27^ yr.; B, 22^ yr.
23. A pole was J under water; the water rose 8 ft., and then there
was as much under water as had been above water before: how long
is the pole? 18 § ft
24. A is f as old as B; if he were 4 yr. older, he would be ^ as old
as B; how old is each? A, 20 yr.; B, 26§ yr.
25. A's money is $4 more than  of B's, and $5 less than  of B's:
how much has each? ' A, $76 • B. $108.
26. Two thirds of A's age is f of B's, and A is SJ yr. the older:
how old is each? A, 31 J yr.; B, 28 yr.
27. If 3 boys do a work in 7 hr., how long will it take a man who
works 4 J times as fast as a boy? 4 J hr.
28. If 6 men can do a work in h\ days, how much time would be
saved by employing 4 more men? 2^ days.
29. A man and 2 boys do a work in 4 hr. : how long would it take
the man alone if he worked equal to 3 boys? 6f hr.
30. A man and a boy can mow a certain field in 8 hr. ; if the boy
rests 3 hr., it takes them 9} hr. ; in what time can each do it?
Man, 13J hr.; boy, 20 hr.
31. Five men were employed to do a work ; two of them failed to
come, by which the work was protracted 4 J days : in what time could
the 5 have done it? 6 days.
32. Three men can do a work in 5 days; in what time can 2 men
and 3 boys do it, allowing 4 men to work equal to 9 boys? 4 J da.
33. A man and a boy mow a 10acre field; how much more does
the man mow than the boy, if 2 men work equal to 5 boys? 4 A.
34. Six men can do a work in 4J days ; after working 2 days, how
many must join them so as to complete it in 3f da. ? 4 men.
35. Eight men can do a certain amount of work in 6 days; after
beginning, how soon must they be joined by 2 more so as to complete
it in 5J days? In 2} days.
MISCELLANEOUS EXERCISES. 403
36. Seven men can build a wall in 5^ days ; if 10 men are employed,
what part of his time can each rest, and the work be done in the same
time? jV
37. Nine men can do a work in 8 J days; how many days may 3
remain away, and yet finish the work in the same time by bringing
5 more with them ? 5/j days.
38. Ten men can dig a trench in 7 J days ; if 4 of them are absent
the first 2J days, how many other men must they then bring with
them to complete the work in the same time? . 2 men.
39. At what times between 6 and 7 o'clock are the hourhand and
minutehand 20 min. apart? 10}f min. after 6, and 54/^ niin. after 6.
40. At what times between 4 and 5 o'clock is the minutehand as
far from 8 as the hourhand is from 3?
32 y\ min. after 4; and 49^^ min. after 4.
41. At what time between 5 and 6 o'clock is the minutehand mid
way between 12 and the hourhand? when is the hourhand midway
between 4 and the minutehand?
13^^ Dciin. after 5; and 36 min. after 5.
42. A, B, and C dine on 8 loaves of bread; A furnishes 5 loaves; B,
3 loaves; C pays the others 8d. for his share: how must A and B
divide the money? A takes 7d.; B, Id.
43. A boat makes 15 mi. an hour down stream, and 10 mi. an hour
up stream: how far can she go and return in 9 hr.? 54 mi.
44. I can pasture 10 horses or 15 cows on my ground ; if I have 9
cows, how many horses can I keep? 4 horses.
45. A's money is 12 ^ of B's, and 16 ^ of C's; B has $100 more
than C: how much has A? $48.
46. Eight men hire a coach ; by getting 6 more passengers, the ex
pense to each is diminished $1}: what do they pay for the coach?
$32f.
47. A company engage a supper; being joined by ^ as many more,
the bill of each is 60 ct. less : what would each have paid if none had
joined them? $2.10
48. By mixing 5 lb. of good sugar with 3 lb. worth 4 ct. a lb. less,
the mixture is worth 8^ ct. a lb. : find the prices of the ingredients.
10 ct. and 6 ct. a lb.
49. By mixing 10 lb. of good sugar with 6 lb. worth only J as
much, the mixture is worth 1 ct. a lb. less than the good sugar: find
the prices of the ingredients and of the mixture.
Ingredients, 8 ct. and 5J ct. ; Mixt., 7 ct. per lb.
50. A and B have the same amount of money ; if A had $20 more,
404 RA Y'S HIGHER ARITHMETIO.
and B $10 less, A would have 2^ times as much as B: what amount
has each? $32^.
51. A and B pay $1.75 for a quart of varnish, and 10 ct. for the
bottle; A contributes $1, B, the rest: they divide the varnish equally,
and A keeps the bottle: which owes the other, and how much?
B owes A 2 J ct
52. How far does a man walk while planting a field of <;orn 285 ft.
square, the rows being 3 ft. apart and 3 ft. from the fences?
5 mi. 6 rd. 6 ft
53. Land worth $1000 an acre, is worth how much a front foot of
90 ft. depth; reserving ^ for streets? $2.295f .
54. I buy stocks at 20 <fo discount, and sell them at 10 <fo premium:
what per cent, do I gain? 37 J <fo.
55. I invest, and sell at a loss of 15 ^; I invest the proceeds again,
and sell at a gain of 15 ^: do I gain or lose on the two speculations,
and how many per cent? Lose 2 J ^.
56. I sell at 8 ^ gain, invest the proceeds, and sell at an advance of
12^ <fc\ invest the proceeds again, and sell at 4 ^ loss, and quit with
$1166 40: what did I start with? $1000.
57. I can insure my house for $2500, at ^ <fo premium annually, or
permanently by paying down 12 annual premiums: which should I
prefer, and how much will I gain by it if money is worth 6 ^ per
annum to me? The latter; gain, $113.33}
58. A owes B $1500, due in 1 yr. 10 mon. He pays him $300 cash,
and a note of 6 mon. for the balance: what is the face of the note,
allowing interest at 6 ^c? $1080.56
59. If I charge 12 ^ per annum compound interest, payable quar
terly, what rate per annum is that? 12yVWWt7 ^•
60. How many square inches in one face of a cube which contains
2571353 cu. in.? 18769 sq. in.
61. What is the side of a cube which contains as many cubic
inches, as there are square inches in its surface? 6 in.
62. The boundaries of a square and circle are each 20 ft; which
is the greater, and how much? Circle; 6.831 sq. ft. nearly.
63. If I pay $1000 for a 5 yr. lease, and $200 for repairs, how much
rent payable quarterly is that equal to, allowing 10 ^ interest?
$307.92 a year.
64. What is the value of a widow's dower in property worth $3000,
lior age l»eingj 40, and interest 5 ^? $669.50
(
MISCELLANEOUS EXERCISES. 405
65. What principal must be loaned Jan. 1, at 9 ^, to be repaid by
5 installments of $200 each, payable on the first day of each of the
five succeeding months? $978.10
66.. After spending 25 ^ of my money, and 25 ^ of the remainder,
I had left $675: what had I at first? $1200.
67. I had a 60day note discounted at 1 ^ a month, and paid $4.80
above true interest: what was the face of the note? $11112.93
68. Invested $10000 ; sold out at a loss of 20^ : how much must I
borrow at 4%, so that, by investing all I have at 18%, I may retrieve
my loss? $4000.
69. If J of an inch of rain fall, how many bbl. will be caught by
a cistern which drains a roof 52 ft. by 38 ft. ? 9.776+ bbl.
70. A father left $20000 to be divided among his 4 sons, aged 6
years, 8 years, 10 years, and 12 years respectively, so that each share,
placed at 4J^ compound interest, should amount to the same when
its possessor became of age (21 yr.) : what were the shares?
$4360.34; $4761.59; $5199.78; $5678.29
71. $30000 of bonds bearing 7% interest, payable semiannually,
and due in 20 yr. are bought so as to yield 89^ payable semiannually :
what is the price ? $27031.08
72. A man wishes to know how many hogs at $9, sheep at $2,
lambs at $1, and calves at $9 per head, can be bought for $400,
having, of the four kinds, 100 animals in all. How many different
answers can be given ? 288 answers.
73. The stocks of 3 partners, A, B, and C, are $350, $220, and $250,
and their gains $112, $88, and $120 respectively ; find the time each
stock was in trade, B's time being 2 mon. longer than A's.
A*8, 8 mon. ; B*8, 10 mon. ; C*s, 12 mon.
74. By discounting a note at 20 9^ per annum, I get 22 W per annum
interest : how long does the note run ? 200 days.
75. A receives $57.90, and B $29.70, from a joint speculation : if A
invested $7.83J more than B, what did each invest ?
A, $16.08} ; B, $8.25
76. A borrows a sum of money at 6^, payable semiannually, and
lends it at 12^^, payable quarterly, and clears $2450.85 a year: what
is the sum? $38485.87
77. Find the sum whose true discount by simple interest for 4 yr.
is $25 more at 6% than at 4% per annum. $449.50
78. I invested $2700 in stock at 25^ discount, which pays 89^
annual dividends : how much must I invest in stock at 4^^ discount
and paying 10^ annual dividends, to secure an equal income?
$2764.80
406 R4 Y'S HIGHER ARITHMETIC.
79. Exchanged $5200 of stock bearing 5^ interest at 69^, for stock
bearing 7% interest at 92%, the interest on each stock having been
just paid : what is my cash gain, if money is worth 6^ to me ?
$216.66J
80. Bought goods on 4 mon. credit; after 7 mon, I sell them for
$1500, 2i% off for cash ; my gain is 15^^, money being worth 6^ :
what did I pay for the goods ? $1252.94
81. The 9th term of a geometric series is 137781, and the 13th
term 11160261 : what is the 4th term? 567.
82. My capital increases every year by the same per cent. ; at the
end of the 3d year it was $13310 ; at the end of the 7th year it was
$19487.171 : what was my original capital, and the rate of gain?
$10000, and 10%.
83. Find the length of a minutehand, whose extreme point moves
4 inches in 3 min. 28 sec. 11.02 — in.
84. Three men own a grindstone, 2 ft. 8 in. in diameter : how many
inches must each grind off to get an equal share, allowing 6 in. waste
for the aperture? 1st, 2.822— in. ; 2d, 3.621+ in. ; 3d, 6.557— in.
85. I sold an article at 20% gain; had it cost me $300 more, I
would have lost 20% : find the cost. $600.
86. A boat goes 16^ miles an hour down stream, and 10 mi. an
hour up stream : if it is 22} hr. longer in coming up than in going
down, how far down did it go? 585 mL
87. Had an article cost 10% less, the number of % gain would
have been 15 more : what was the % gain ? 35%.
88. Bought a check on a suspended bank at 55% ; exchanged it for
railroad bonds at 60%, which bear 7% interest: what rate of interest
do I receive on the amount of money invested? 21j^%.
89. Bought sugar for refinery; 6% is wasted in the process; 30%
becomes molasses, which is sold at 40 per cent less than the same
weight of sugar cost ; at what per cent advance on the first cost must
the clarified sugar be sold, so as to yield a profit of 14% on the invest
ment? 50%.
90. There is coal now on the dock, and coal is running on also,
from a shoot, at a uniform rate. Six men can clear the dock in one
hour, but 11 men can clear it in 20 minutes : how long would it take
4 men ? 5 hr.
91. A distiller sold his whisky, losing 4% ; keeping $18 of the pro
ceeds, he gave the remainder to an agent to buy rye, 8% commission ;
he lost in all $32: what was the whisky worth? $300.
92. A clock gaining 3} min. a day was started right at noon of
the 22d of February, 1804: what was the true time when that clock
MISCELLANEOUS EXERCISES. 407
Rhowed noon a week afterward ; and, if kept going, when did it next
show true time? 35 min. 32.9 sec. after 11 a. M.
True, Sept. 15th, 8^ min. past 5 a. m.
93. The number of square inches in one side of a rightangled tri
angular board is 144, and the base is half the height ; required the
areas of the different triangles which can be marked off by lines par
allel to the baee, at 12, 13, 14, 14^ inches from the smaller end.
36 sq. in. ; 42^ sq. in. ; 49 sq. in. ; 52^^^ sq. in.
94. Suppose a body falls 16 ft. the first second, 48 ft. the next, 80
the next, and so on, constanUy increasing, how far will it have fallen in
4 sec; in 4J sec; in 5 sec? 256 ft.; 324 ft.; 400 ft.
95. A man traveling at a constant increase, is observed to have gone
1 mile the first hour, 3 miles the next, 5 the next, and so on : how far
will he have gone in 6i hr. ? 42 J mi.
96. The number of men in a side rank of a solid body of militia, is
to the number in front as 2 to 3; if the length and breadth be in
creased so as to number each 4 men more, the whole body will contain
2320 men : how many does it now contain ? 1944 men.
97. A grocer at one straight cut took ofi* a segment of a cheese
which had J of the circumference, and weighed 3 lb. : what did the
whole cheese weigh? 33.0232f lb.
98. A wooden wheel of uniform thickness, 4 ft. in diameter, stands
in mud 1 ft. deep: what fraction of the wheel is out of the mud?
.804491 of it.
99. My lot contains 135 sq. rd., and the breadth to length is as 3 to
5 : what is the width of a road which shall extend from one corner
half round the lot, and occupy J of the ground ? 24 J ft.
100. A circular lot 15 rd. in diameter is to have three circular grass
beds just touching each other and the large boundary : what must be
the distance between their centers, and how much ground is left in the
triangular space about the main center?
Distance, 6.9615242f rd.
Space within, 1.9537115f sq. rd.
101. I have an inch board 5 ft. long, 17 in. wide at one end, and 7
in. at the other : how far from the larger end must it be cut straight
across, so that the solidities of the two parts shall be equal ? 2 ft.
102. Four equal circular pieces of uniform thickness, the largest
possible, are to be cut from a circular plate of the same thickness, and
worth $67 : supposing there is no waste, what is the worth of each of
the four, and what is the worth of the outer portion which is left?
Each small circle, $11.49538
Outer portion, $17.877471
408 JiA Y'S HIQHEIt ARITHMETIC,
103. A 12inch ball Ir in the corner where walls and floor are at
right angles : what must be the diameter of another ball which can
touch that ball while both touch the same floor and the same walls?
3.2154 in., or 44.7846 in.
104. A workman had a squared log twice as long as wide or deep ;
he made out of it a watertrough, of sides, ends, and bottom each 8
inches thick, and having 11772 solid inches: what is the capacity of it
in gallons? GS^Sj. gal.
105. How many inch balls can be put in a box which measures,
inside, 10 in. square, and is 5 in. deep? 568 balls.
106. A tin vessel, having a circular mouth 9 in. in diameter, a
bottom 4J in. in diameter, and a depth of 10 in., is \ part full of
water : what is the diameter of a ball which can be put in and just
be covered by the water? 6.1967 in.
Or T . I C
UNIVERSITY
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