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LIBRARY 

University of California. 

GIFT OF- 




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* 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. 

TWO-BOOK 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 twenty-five 
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 proof-sheets. 

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 self-evident 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 twenty-oney 21 ; twenty-two, 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 ninety-nine; 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 






«4-4 


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 
twenty-one thousand and forty-five. 

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. Seventy-five. 

2. One hundred and thirty-four. 

3. Two hundred and four. 

4. Three hundred and seventy. 

5. One thousand two hundred 
and thirty-four. 



Examples m Notation. 

6. Nine thousand and seven. 

7. Forty thousand five hundred 
and sixty-three. 

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 twenty-one thousand six 
hundred and eighty-five. 

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 
thirty-one. 

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 
twenty-five quintillion and 
thirty-three 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 eighty-seven 



NUMERATION AND NOTATION 



21 



million six hundred and fifty-four thousand three hundred 
and twenty-one. By the French method they would be read, 
two hundred and ten billion nine hundred and eighty-seven 
million six hundred and fifty-four thousand three hundred 
and twenty-one. 

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 Twenty-one. 

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; eighty-seven 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. Book-keeping 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 Day-booh and Ledger. In the Day-book 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.-r-How 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 m-ultiptter 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 half-dozen 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 cotton-mill, 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 

DEMONS-rtiATiON.— 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 right-hand 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 right-hand 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 right-hand 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 right-hand 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 right-hand 
figure thus obtained under the tnght-hand 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. — Twenty-four 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+204-5 
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 left-hand 
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. 849056f||||f 



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? Id8|f 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 by5-f(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 — 1x14-1 X 4-^2 + 18 -^3x6-i- 
■2x2) + (4— 2 + 6 — 7)X4x6-v-8 = ? 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 + 
24X6-r-48 — 4X9-M2 = ? 1. 

5. (16-M6x96-^8 — 7 — 5 + 3)X[(27-^-9)-^8— 
l]+(91-M3x7— 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 5X4-f-2 = 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 24X2-5-6 = 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 12-5-6 = 2; whence 24 -^ (6 X 2) = 2. Therefore, 4-1-2 = 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, 48-1-12 = 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. 34521||f|^ 



3. 22500712361-7-1406250. Ans. 160003^^^ 

4. 620712480 -r- 20833 J. Quoe. 29794. Rem. 4146| 

5. 742851692 -r-2916|. 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 25-cent pieces; the rest is made 
up of 10-cent, 5-cent, and 3-cent pieces, of each an equal 
number: how many of each coin are there? 

Of 25-cent 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 /acUn-a (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 right-hand 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 right-hand 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 right-hand 
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 9999-f 6 

768945=J 8000 = 8X 1000 = 8X (9994-l) = 8X 999+8 

900 = 9X 100 = 9X (994-l) = 9X 99-|-9 
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 twenty-ninths. Eight twenty-firsts. 
Nine forty-seconds. Nineteen ninety-thirds. Thirteen one- 
hundredths. Twenty-four one-hundred-and-fifteentbs. 



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. ^. 


8--m- 




Ans. f f . 


4. m- 




Ant. ■^. 


9-HH- 




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 ninety-ninths. Ans. |^, -J^. 

2. Reduce |^, f , and ^ to sixty-thirds. 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. 5|H. 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. i|9A. ^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^, -J-J. 

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. A-i^off Am, m 



3. -H-^off Ans. ii^ 

4. A-i^of 4. Ans, ^ 

5. H — liV- ^ns. H 
6- A-i^. . Am. ^ 



7. A — H- Am. ^^ 

8. ii-A' 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 23|f 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 fi-H = 
}J. 30 — 23 = 7. Therefore the answer is 1]\. 

9. 12|— lOJf. Ans. l\i. 

10. 12|| — 9f|. Am. 3^. 

11. 5|f — 2^. Ans. 3^. 

12. 7^— Sf Am. ^^. 

13. 15 — f Am. l4. 

14. 18 — 5|. ^n«. 12f. 

15. I of 2| — 3|f. ^n«. ||. 

16. ^ — \ of If ^rw. l|. 

17. J^ of 4J — ^ of 3^ = what? Am. 13f . 

18. ll| + 4--9H = 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. 6|X41. 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 f-f-f = }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. 6-M. 

5. 21 ^t1^. 



Am. -^ 

Am. 4s 

Am. -^ 

Am. 9 

Am. ^^ 
Am. 1-J-, 

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«. 4|H- 



40f • 73 • ^' 

12 ^A-^-i^ A-nit ^ 



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. 4|f. 

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:-J-A = 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^^,^ 

(2-i)X(4-3f) ^ 

10. ii^lililiizil^ what? 4M. 

4iX4i-l ^ 

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}? 2|f. 

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 .63-gallon 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 
right-hand 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 Ten-thousandths. 


6th " 


.00003 . . 


' 3 Hundred-thousandths. 


6lh " 


.000001 . . 


• 1 Millionth. 


7lh ■■ 


.0000009 . 


* 9 Ten-millionths. 


8th " 


.00000004. 


' 4 Hundred-millionths. 


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 twenty-five ten-thousandths. 

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 eighty-three 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 
hundred-thousandths ; 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 rigid-hand figure shall be of ihe 
same name as ihe decimal. 



Examples to be Written. 



1. Five tenths. 

2. Twenty-two hundredths. 

3. One hundred and four thou- 
sandths. 

4. Two units and one hun- 

dredth. 



5. One thousand six hundred 

and five ten-thousandths. 

6. Eighty-seven hundred-thou- 
sandths. 

7. Twenty-nine and one half 

ten-mi 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 
fifty-six and one third mill- 
ionths. 

11. Four hundred and twenty- 
one tenths. 

12. Six thousand hundredths. 

13. Forty-eight thousand three 
hundred and five thou- 
sandths. 



14. Eight units and one half a 
hundredth. 

15. Thirty-three million ten- 
millionths. 

16. Four hundred thousandths. 

17. Four hundred-thousandths. 

18. One unit and one half a 
billionth. 

19. Sixty-six thousand and three 

millionths. 
20* Sixty-six million and three 

thousandths. 
21. Thirty-four 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 ; forty-seven 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. 64.66|X18. 1164. 

9. .56JX.O33V .0172-H 
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 right-hand 



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 right-liand 
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 right-hand 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, A-m, 
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. 81.2096-7-1.28 63.445 

10. 1-MOO. .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 ten-millionth -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. Co-originou8 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)4-437 
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 right-hand 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 right-hand 
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 right-hand 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 right-hand 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 right-hand 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 right-hand 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 right-lMnd 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. .*75-r-.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. Co-originous 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. 

3-cent piece. 
5-cent 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 five-cent 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 
bank-note. 

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 
half-guinea ( = lOs. 6d.) The silver coins are the crown ( = 58.), 
the half crown ( =2s. 6d.), the florin ( = 28.), the shilling, and the 
six-penny, four-penny, and three- penny pieces. The penny, half- 
penny, and farthing are the capper coins. The guinea, half-guinea, 
crown, and half-crown 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 
half-pence 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 5-franc pieces ; they are .9 pure gold. 
The 20-franc piece weighs 99.55 gr. The siktr coins are 6, 2, and 
1 -franc pieces, and 50 and 20-centime pieces ; they are now .835 
pure silver. The brome coins are 10, 5, 2, and 1-centime 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 1-mark 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 hundred-weight, ** 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 
custom-houses 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 twenty-four parts, called carats, and is said to be so 
many carats fine, according to the number of twenty-fourths 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 cable-length. 

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 dry-goods. 
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 custom-house, 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, weather-board- 
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, rock-excavations, 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. Fire-wood 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 tea-cup = 2 wine-glasses = 8 table- 
spoons = 32 tea-spoons = 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 ten-millionth 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 one-hundredth 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 (p-o. 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, klKo-gram), 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)19-2 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. Forty-nine 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. yj-g^ 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 grass-plot 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 wagon-bed 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 fore-wheel 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 fore-wheel 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 school-room 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 fi-om the comb to the 
eaves? 14400 shingles. 

56. A boy goes to bed 30 minutes later, and gets up 
40 minutes earlier than his room-mate : how much time 
does he gain over his room-mate 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 right-hand 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. 3|d.; £5 17|s.; £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 -^* sq-ro* 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 twenty-four hours? 

Noon; 1037.4 statute miles per hour. 

4. What is the relative time between Mobile and Chi- 
cago? Chicago time 1 min. 49|f 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 steam-ship 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 fore-wheel 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 sea-water 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 horse-power 
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 court-yard 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 X-3 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. Thirty-six men of a ship's crew die, which is 42f % 
of the whole : what was her crew ? 84 men. 

12. A stock-farmer 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. 

$377-1- 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? 5f|f 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 20000-f-450-=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 semi-annually. 

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. = St-ock. 

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? $1-12J, 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 Commission-Merchant, 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 commission-merchant 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 packing-house 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 bank-notes : 
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 delivei-y 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 rolling-mill 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 North-western 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. 

$.06-j-.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 joint-stock companies, or 
mutual companies. 

Remakks. — 1. In joint-stock 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^^, re-insured $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; re-insured 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 ; re-insured | 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\% ; re-insured 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 %; re-insured 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%; re-insured $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 forty-nine 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 PoU-Tax, 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 property-tax. 

Formula. — Taxable Property X -Ka^ = Amount of Tax, 

Note.— If there be a poll-tax, the sum produced by it should be 
added to the property-tax, 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 poll-tax 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 poll-tax of 
$1.25 is levied, what should be the rate of taxation? 

3-95^% = 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 poll-tax 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 Pi-operty. 
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 poll-tax of $1, and a property-tax 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 custom-houses 
established at ports of entry and ports of delivery. The principal 
custom-house 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 custom-house. The ton, for custom-house 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 cologne-water: 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 dry-goods 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? $.62|f 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 pig-iron, American manufacture, to be 
of equal quality with Scotch pig-iron : 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 marble-cutter 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:— Commission-Merchant, 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, 
Joint-Stock and Mutual Companies. 

2. Three Cases. 

1. Definitions :— Tax, State and Government, 

Direct and Indirect, Poll-tax, 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 half-yearly, 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 43-J-f^, 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 50-f- ($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 .'. $840-5-$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 semi-annual 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, $490. 84-j-$. 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 Twenty-five 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 "iron-clad" 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 semi-annually: 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, twenty-five 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 twenty-nine 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 
bank-bills. 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 " bank-bills," 
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 bank-notes 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 Twenty-five. 
$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 Fifty-three 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 ninety-two -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 seventy-two -^^ 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 eighty-six 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, twenty-eight 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 thirty-seven -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 eighty-five -^^ 
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, 

$4.20-5-$. 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 ? 25|ff , 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, tm-ies 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; 

$4.20-i-$.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. 

$2240-5-1.10 = 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 right-hand 
members together for a dividend, and dividing the product by the 
product of the left-hand 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 right-hand qvmitity of each 
equation and tlie lefb-lw/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 left-hand 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 right-hand quantity in one 
of the same denomination as the left-hand 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 20-f- 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 PAYMEm-S. 

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 left-hand, 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 right-hand, 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 
$244-f $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 semi-annually 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 half-year 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 ^semi-annually. $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 semi-annually? 

$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 
half-year 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 fifty-Jive. 



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 5-290 


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. -27-22 7926 


1.3159 3178 


1.36H8 618:< 


8 


M7I6 5938 


1.2184 0290 


1.2867 70(« 


1.3168 0»)4 


1.3685 6905 


1.4-221 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 18-26 


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 


1.9-225 01.32 


2.1068 4918 


2.3078 6031 


20 


1.4859 4740 


1.6386 1644 


1.8061 11-23 


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 


2.03-27 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 


2.65-23 35-24 


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 


2.1-222 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.-26-20 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 


2.8-209 9520 


3.4606 9589 


4.-2412 5799 


5.1927 8391 


6.3516 1-548 


43 


2.3431 89:^6 


2,8915 2008 


3.5645 1677 


4.:te97 0-202 


5.40(« 95-27 


6.6374 3818 


44 


2.3900 5:n4 


2.9638 0808 


3.6714 5-227 


4.54:« 4160 


5.6165 1506 


6.9)61 2290 


45 


2.4378 5421 


3.0379 0328 


3.7815 9584 


4.70-23 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 28-24 


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 


9.08-26 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 


5.98-27 i;i27 


7.68<>5 8S71 


9.8638 6463 


53 


2.856:1 3475 


3.7013 9016 


4.7904 1247 


6.1921 08-24 


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 fifty-five. 



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.'2-250 4:iO 


1.2597 1-20 


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 


1.99-2.) 6-26 


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 


2.59-37 425 


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 8-25 


4.1772 482 


16 


2.1828 746 


2.5401 517 


2.9521 638 


3.4-259 426 


3.970B 059 


4.5949 730 


17 


2.2920 183 


2.6927 7-28 


3.1588 152 


3.7000 181 


4.3-276 :«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 


5.4-274 3-26 


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 


1-2.1721 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 1-29 


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 


9.3-25!) 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 5-20 


11.4-239 4-22 


15.9681 718 


22.2512 -250 


30.9126 805 


37 


6.0814 069 


8.6360 871 


1-2.-2-236 181 


17.-2456 256 


24.'i538 im 


34.0039 486 


38 


6.38->4 773 


9.1542 524 


13.0792 714 


18.6-252 756 


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 


4-2 


7.7615 876 


11.5570 327 


17.1442 568 


25.3;i94 819 


37.3175 .m 


54.7636 992 


4} 


8.1496 669 


1-2.2504 546 


18.3443 548 


27.3666 404 


40.6761 098 


60.-2400 692 


44 


8.5571 503 


1-2.9854 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 


1-2.6428 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 


38.61-21 509 


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 4-25 



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 semi-annually. $3597.67 

4. If $1000 is deposited for a child, at birth, and draws 
7% compound interest, payable semi-annually, 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 semi-annually. $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 semi-annually. $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 semi-annually. 

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 semi-annually? 

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 semi-annually? 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 semi-annually ? 

$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, life-estates, 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, half-year, 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 semi-annually, allowing 5^ interest, 
payable annually? $32400. 

Suggestion. — Here tlie yearly payment is $1620, by allowing 5^ 
interest on the half-yearly 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 semi-annually. $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 
semi-annually. $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 semi-annually? 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 fifty-jwe. 



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 


11.061-241 


10.200744 


8.771540 


23 


14.856842 


13.488574 


1*2.303379 


11.272187 


10.371059 


8.883218 


24 


15.246963 


13.798642 


1-2.550353 


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 


1-2.277674 


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 


1-2.854000 


11.586934 


9.606575 


35 


18.664613 


16.374194 


14.498246 


1-2.917672 


11.654568 


9.644159 


36 


18.908?82 


16.546852 


14.620987 


13.085-208 


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 


13.45-2449 


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 


12.164-267 


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 


13.86-21-24 


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 


6-261 


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 


1-2.037 


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 


2.5-22 


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 


0.3-JOal 


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 life-eBtate or widow's dower 
(which Ib a life-efltate in one third of her husband's real estate): 
Estimate the value of the property in which the life-estate is held; the yearly 
interest of this sum, at an a(p eed rate, will be a life-annuity, whose value for 
the given aye and rate will be the value of the life-estate, 

2 The reversion of a life-annuity, life-estate, or dower is found 
by deducting its value from the value of the property. 



Examples for Practice. 

1. What must be paid for a life-annuity of $650 a year, 
by a person aged 72 yr., int. 7%? $3355.30 

2. What is the life-estate and reversion in $25000, age 
55 yr., int. 6%? Life-estate, $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 life-annuity 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, semi-annually, 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 $7619.04|f 

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 fi-om 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 78-i- 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 degi-eea, 

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 iff|J. 


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 



60X4H-42 
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= 604-4 

16384= 60 2X4+2(60X42) + 4» 

24576 = 608 + 2(60»X4)4-3(60X4M 

64» = 262144— 60 » + 3(602 X4) + 3(60X4 2)4-4 ». 

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 factoi-s 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 20-inch 

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 right-hand 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 left-hand 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 right-hand 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 right-liand 
column for a dividend; cut off the right-hand figure from tJie 
last number of the previous column, two ri^it-hand 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 right-hand 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. 



RiOHT-ANOLED TRIANGLES. 




380. A Right-angled 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 
right-angled 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 right-angle4 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 + 264-26 + 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 = 12-1-48 + 192-1- 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 + 12-f 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 ; twenty-second 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 straight-edge 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 right-angled when it has one right angle ; 
it is acute-angled 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 half-work? 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 fire-place 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 
fire-place 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 twenty-four sectors. 

Remarks. — 1. The hexagon is com|K)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 twenty-four 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 
twenty-four triangles forming such a poly- 
gon would differ very little from one of 
the twenty-four 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 right-angled 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. ? 

2290.221-1- 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 square-edged 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 square-edged 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 wagon-bed 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 hay-mow 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 10-acre 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 hour-hand and 
minute-hand 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 minute-hand as 
far from 8 as the hour-hand 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 minute-hand mid- 
way between 12 and the hour-hand? when is the hour-hand midway 
between 4 and the minute-hand? 

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? $2.295-f . 

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 re-paid 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 60-day 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 semi-annually, 
and due in 20 yr. are bought so as to yield 89^ payable semi-annually : 
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 semi-annually, 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 -R-4 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 minute-hand, 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 right-angled 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? 33.0232-f 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? 

.80449-1- 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, 6.9615242-f rd. 

Space within, 1.9537115-f- 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, $17.87747-1- 






408 JiA Y'S HIQHEIt ARITHMETIC, 

103. A 12-inch 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 water-trough, 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. 



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