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I 
 
 AN ALGEBRAIC ARITHMETIC 
 
 » 
 
^^n^g^ 
 
AN ALGEBRAIC ARITHMETIC 
 
 AN EXPOSITION OF THE THEORY 
 AND PRACTICE OF 
 
 ADVANCED ARITHMETIC 
 
 BASED ON THE ALGEBRAIC EQUATION 
 
 BY 
 
 S. E. COLEMAN, B.S. 
 
 WILLIAM WHITING FELLOW AT HARVARD UNIVERSITY 
 
 FORMERLY INSTRUCTOR IN MATHEMATICS IN THE OAKLAND 
 
 HIGH SCHOOL, OAKLAND, CALIFORNIA 
 
 Wetjj gorit 
 THE MACMILLAN COMPANY 
 
 LONDON : MACMILLAN A CO., Ltd. 
 1898 
 
 All rights reserved 
 / *■' OF THE 
 
 UNIVERSITY 
 
 OF 
 
 £^LlFORfi:^ 
 

 Copyright, 1897, 
 By the MACMILLAN COMPANY. 
 
 Vodnooti ^xt»» 
 
 J. S. Cuihinj? A Co. - Rerwick k Smith 
 Norwood MasB. U.S.A. 
 
PREFACE 
 
 This arithmetic is not ojffered to the public as a refine- 
 ment or super-refinement of the methods of existing text- 
 books on the subject. It is a new departure. 
 
 For a number of years arithmetics have been under- 
 going a progressive change. Voluminous works, in which 
 the isolated treatment of related topics and the multi- 
 plicity of detail relating to business arithmetic completely 
 obscured the unity of the science, have by degrees given 
 place to more compendious works. The change has, 
 however, been little more than a process of successive 
 elimination. To the former plethora has succeeded an 
 ever-increasing leanness, until at last the skeleton of the 
 subject stands revealed indeed, for it alone remains. The 
 arithmetic of to-day is merely a compilation of examples, 
 classified and miscellaneous, with illustrative solutions 
 accompanied by brief explanatory notes and a few defini- 
 tions. The task of infusing a living, rational principle 
 into these dry bones is left entirely to the teacher. 
 
 A second change in the mathematics of the grammar 
 school, contemporaneous with that above mentioned, has 
 been the introduction of elementary geometry and, more 
 recently, of elementary algebra. The situation is best 
 described by saying that the latter subjects have partially 
 
 173093 
 
VI PREFACE 
 
 superseded arithmetic, since the whole time devoted to 
 mathematics has remained substantially the same. 
 
 The reason for these changes is not far to seek. The 
 mathematics are chiefly valuable as a factor in education 
 in that they afford a means of developing the reasoning 
 powers of the child; and as arithmetic, in spite of the 
 numerous attempts to improve the text-books and the 
 methods of teaching, persisted in remaining little more 
 than a collection of rule-of-thumb methods for turning 
 out "answers," the progressive teacher naturally turned 
 to other branches of mathematics which embodied a logi- 
 cally coherent science not yet perverted by "practical 
 applications." 
 
 Much was gained by so doing. Geometry, which in 
 its complete and rigidly demonstrative form is a fairly 
 difficult subject for the high school, was found to contain 
 a large number of facts that could be established by 
 simple yet fairly conclusive reasoning. And not only is 
 the method by which these facts are acquired in their 
 logical relation of the highest value in developing the 
 reasoning powers of the child, but the facts themselves 
 possess a value far higher than that of mere utility. For 
 example, it may possibly prove of use to some member of 
 a large grammar-school class to know the empirical rule 
 by which the contents of a cask or the number of feet of 
 lumber in a round log are determined ; but it is important 
 that all should know, and should be able to give some 
 simple explanation of the fact, that similar surfaces are 
 to each other as the squares and similar solids as the 
 cubes of their like dimensions. These are universal 
 truths, depending on the nature of space, by which all 
 
PREFACE Vll 
 
 physical existence is conditioned; and a knowledge of 
 them is therefore an essential part of a complete educa- 
 tion; but this, from its very nature, can be true of no 
 empirical rule. 
 
 The introduction of elementary algebra into the schools 
 is of more doubtful value for two principal reasons. 
 The conceptions of geometry can be represented by fig- 
 ures and objects, and are therefore readily grasped by 
 the child, but those of algebra can be so represented only 
 to a very limited degree; and, for the most part, are 
 abstractions the nature of which the child comprehends 
 with difficulty. Moreover, the time that can be allotted 
 to the subject in the grammar school is barely sufficient 
 to carry the pupil over the essentially uninteresting 
 details of algebraic manipulation that necessarily precede 
 any but the simplest applications of the science. 
 
 But while the wisdom of introducing elementary algebra 
 into the schools may, for these and minor reasons, be 
 seriously questioned, the experience of several years as 
 a teacher has led the author to the conclusion that the 
 application of certain algebraic conceptions to arithmetic 
 would contribute largely toward the rational presentation 
 of the subject, thus increasing its disciplinary value, and 
 at the same time preparing the way for a natural transi- 
 tion to the algebra of the high school. 
 
 These conceptions are the use of letters as the general 
 representatives of (positive) numbers and of the equation 
 to express their relations. 
 
 Both ideas are introduced into the first chapter, and 
 developed, so far as the purpose demands, in the second. 
 The comparatively large amount of space given to prob- 
 
Vlll PREFACE 
 
 lems is due to the fact that they afford at once the most 
 interesting introduction to the subject and the best means 
 of explaining the significance of the equation and its 
 transformation.* 
 
 Among the numerous and important applications of 
 these ideas throughout the book, may be noted the 
 following : 
 
 The cases of percentage are reduced to three (Art. 32), 
 and all are shown to be contained in the single equation 
 p = hr. 
 
 All the applications of percentage not involving time 
 are shown to be merely special applications of the three 
 percentage formulas. (For example, see table under 
 Profit and Loss, Art. 33.) 
 
 All the formulas of simple interest are derived from 
 the two: i=prt and a=p-\-i-j these being obtained 
 directly from the definitions. 
 
 The interest formulas are shown to be special develop- 
 ments of the earlier percentage formulas (Art 48). 
 
 The principles of proportion are rigidly demonstrated, 
 affording a simple and elegant illustration of the method 
 of algebraic proof, freely used in the chapter on men- 
 suration. 
 
 * The attention of teachers is called to the fact that an equation 
 is not a quantity, but an expression of relation, and therefore can- 
 not be operated upon, in the usual sense of the word. Operations 
 are not performed upon an equation, but upon its members. The 
 abbreviated and, to beginners, highly misleading forms of state- 
 ment, " Multiply the equation by 3," "Subtract 10 from the equa- 
 tion," and the lik«, should be studiously avoided. For a similar 
 reason, the word "transpose" should not be used. A term is 
 transposed by addition or subtraction, and the specific operation 
 should always be named. 
 
PREFACE IX 
 
 The algebraic and geometrical explanation of evolution 
 are combined in one, the algebraic symbols being the 
 natural method of expressing the geometrical relations. 
 Thus each set of ideas confirms the other. 
 
 The chapter on mensuration embodies the principles 
 enunciated in the early part of this preface as fully as 
 the limited time generally allotted to this subject permits. 
 In schools where a more extended course in elementary 
 geometry is given, it will afford a convenient opportunity 
 for a review of the most important results of the course. 
 The author takes pleasure in acknowledging his indebted- 
 ness to Hill's Lessons in Geometry for many valuable 
 suggestions in the preparation of this chapter, and recom- 
 mends the book as being admirably adapted to the needs 
 of grammar schools. 
 
 Among the features of the book not resulting from the 
 algebraic method of treatment, the author would call 
 attention to the use of the article to mark the logical 
 divisions of the subject; to the rational explanation of 
 the application of simple and compound proportion to the 
 solution of problems;"* and to the treatment of partial 
 payments. The latter subject is placed after compound 
 interest, where the effect of the two usual methods of 
 applying partial payments can be intelligibly discussed, 
 and the manner in which interest is compounded by the 
 United States Rule is fully explained. 
 
 It is believed that the examples, which for the most 
 part have been compiled from various sources, present a 
 
 * The author has never seen anything on this point in any text- 
 book, except variations of the rule of thumb : More requires more, 
 and less requires less. 
 
X PREFACE 
 
 fairly extensive and varied application of the subject 
 matter. 
 
 The author will receive with pleasure any suggestions 
 or criticisms that will be of assistance in the improvement 
 of later editions of this work. 
 
 S. E. COLEMAN. 
 
 Cambbidgb, Mass., Aug. 11, 1897. 
 
CONTENTS 
 
 CHAPTER I 
 INTRODUCTION 
 
 PAOB 
 
 The Usb of Letters to denote Numbebs .... 1 
 
 The Equation . 4 
 
 Definitions 6 
 
 Axioms 7 
 
 The Solution of Equations 7 
 
 The Solution of Pboblems 9 
 
 Directions for the Solution of Problems ... 12 
 
 CHAPTER II 
 
 ADDITION. SUBTRACTION. MULTIPLICATION. DIVISION 
 
 Inverse Processes 15 
 
 Addition 16 
 
 Subtraction 17 
 
 Multiplication 20 
 
 Division 27 
 
 CHAPTER III 
 PERCENTAGE AND ITS APPLICATIONS 
 
 Percentage 31 
 
 Percentage Formulas 37 
 
 xi 
 
XU CONTENTS 
 
 PAQB 
 
 Profit and Loss 38 
 
 Commission and Brokerage 43 
 
 Commercial Discount 46 
 
 Insurance 48 
 
 Taxes 50 
 
 Duties 62 
 
 CHAPTER IV 
 
 APPLICATIONS OF PERCENTAGE INVOLVING TIME 
 
 Simple Interest 57 
 
 Accurate Interest 62 
 
 Problems in Interest 63 
 
 Interest Formulas 68 
 
 Present Worth and True Discount 70 
 
 Bank Discount 72 
 
 Annual Interest 75 
 
 Compound Interest 76 
 
 Partial Payments . . 78 
 
 CHAPTER V 
 PROPORTION. PARTNERSHIP. AVERAGE OF PAYMENTS 
 
 Ratio 84 
 
 Proportion 85 
 
 Problems in Simple Proportion 89 
 
 Compound Proportion 92 
 
 Problems in Compound Proportion 94 
 
 Partnership . .97 
 
 AyBRAOB of Payments . . 99 
 
CONTENTS XUl 
 
 CHAPTER VI 
 INVOLUTION AND EVOLUTION 
 
 PAOE 
 
 Involution 101 
 
 Evolution 102 
 
 The Square of the Sum of Two Numbers . . . 103 
 
 Square Root 105 
 
 The Cube of the Sum of Two Numbers . . . .110 
 
 Cube Root 112 
 
 CHAPTER VII 
 MENSURATION 
 
 Lines 117 
 
 Angles 118 
 
 Plane Figures 118 
 
 Area of Parallelograms 120 
 
 Area of Triangles 121 
 
 Area of Polygons . 124 
 
 The Circle 126 
 
 Similar Plane Figures 131 
 
 Solids 132 
 
 Prisms and Cylinders 132 
 
 Pyramids and Cones 135 
 
 The Sphere 139 
 
 Similar Solids 141 
 
ALGEBRAIC ARITHMETIC 
 
 CHAPTER I 
 
 INTRODUCTION 
 
 1. The Use of Letters to denote Numbers. It is often 
 necessary to speak of something whicli is true not only 
 of one number or set of numbers, but of all numbers or 
 of all similar sets of numbers. For example, the sum of 
 9 and 6 is 15, and their difference is 3. If we add this 
 sum and difference, we get 18, which is twice the greater 
 of the two numbers. If we subtract the difference from 
 the sum, we get 12, which is twice the smaller of the two 
 numbers. The same relation is true of any two numbers, 
 and may be expressed in the general statement : 
 
 If the sum and difference of two numbers he added, the 
 result is twice the greater of the numbers; if the difference 
 be taken from the sum, the result is twice the smaller of the 
 numbers. 
 
 This may be expressed more briefly by the aid of 
 signs; thus: 
 
 sum of two numbers + difference of the numbers 
 
 = twice the larger number. 
 
 sum of two numbers — difference of the numbers 
 
 = twice the smaller number. 
 
 B 1 
 
2 ALGEBRAIC ARITHMETIC 
 
 The statement may be still further shortened by using 
 letters to represent the numbers. Thus, let g stand for 
 the greater number, I for the less, s for their sum, and 
 d for their difference. We then have 
 
 s-^d = 2g, 
 and s — d = 2l; 
 
 in which, it must be remembered, a and b denote any 
 two numbers, provided only that a is greater than b. 
 From the above equations it is clear that 
 
 9='-ii (1) 
 
 and ' = '-^- (2) 
 
 These equations express the fact that the larger of two 
 numbers is equal to one-half the sum found by adding the 
 sum and the difference of the numbers, and the smaller 
 number is equal to one-half the remainder found by sub- 
 tracting the difference of the numbers from their sum. 
 These equations are, in fact, a very convenient state- 
 ment of the rule for finding any two numbers when 
 their sum and difference are given. 
 
 Note. Rules stated in the form of equations are called for- 
 mulas. 
 
 The following chapter will afford numerous illustra- 
 tions of the advantage of using letters to denote numbers 
 when we are studying their general properties. 
 
 2. Operations to be performed with numbers denoted 
 by letters are indicated by the usual signs of arithmetic, 
 in the same manner as when the numbers are expressed 
 
INTRODUCTION 8 
 
 with figures ; with the exception that the product of two 
 numbers is indicated by writing the letters together, the 
 sign of multiplication being omitted. 
 
 Thus the sum of any two numbers a and b is indicated 
 by a -f 6, their difference by a — 6, their product by a6, 
 
 and their quotient by a ^ 6, or by -• 
 
 b 
 
 In each case the numerical value of the result can be 
 found only when the values of a and b are given. 
 
 Note 1. The word numerical relates to particular numbers, 
 that is, to numbers expressed by figures ; the word literal, to num- 
 bers expressed by letters. Thus 12 is a numerical quantity ; a, b, 
 c, etc. , are literal quantities. 
 
 Note 2. The sign of multiplication cannot be omitted between 
 numerical factors, but is omitted between a numerical and a literal 
 factor. Thus 6 x a x 6 is written Bab. (Omit "times" in read- 
 ing.) 
 
 EXAMPLES 1 
 
 If a = 4, 6 = 1, c = 3, and d = 2, find the numerical 
 values of 
 
 1. a 4- 6. 6. ac — 2bd. 
 
 2. a — &4-C. 7. 4:Cd-\-ab. 
 
 3. 5a-2d. 8. Sa-5b-\-2acd. 
 
 4. 12b-2a. 9. abc -^ abd + bed. 
 
 5. a + 86 — 6d 10. 2a-r-d + 4c. 
 
 If a = 6, 6 = 5, c = 2, and d = 0, find the values of 
 
 11. 2ab-{-b — cd. ^^ ^4.^. 
 
 r ' c b 
 
 12. a -T- c X 6 — ac. ^ r , o ^ 
 
 -_ 2 ac — 6 -f 3 d 
 
 13. 6 a6 -J- 5 c. • 7aH-a6c * 
 
4 ALGEBRAIC ARITHMETIC 
 
 3. The Equation. The statement that two numbers or 
 two sets of numbers are equal is called an equation. 
 
 Equations are used in Art. 1, and the values of the 
 letters in the examples of Art. 2 are given by means 
 of equations. 
 
 The part of an equation on the left of the sign of 
 equality is called the left side, left member, or first mem- 
 ber of the equation; that on the right, the right side, 
 right member, or second member. 
 
 Many problems can be most easily solved by the use 
 of letters to denote the numbers to be found, and equa- 
 tions to express the relations that exist between these 
 numbers and the given numbers of the problem. How 
 this is done will be shown by the following examples : 
 
 Ex. 1. If 5 be added to 3 times a certain number, 
 the result is 29. Eind the number. 
 
 The problem may be stated more briefly thus : 
 
 3 times a certain number 4- 5 = 29 ; 
 
 or, if we let a stand for the number, it may be stated 
 still more briefly by the equation 
 
 3a + 5 = 29. (1) 
 
 Now if from this equation we can find the value of a, 
 that is, the number that a represents, this value will be 
 the answer to the problem. Let us try to do this. 
 
 Subtracting 5 from both members of the equation, 
 
 we get 
 
 3a = 24. (2) 
 
 Dividing the sides of this equation by 3, we have 
 
 a = 8. (3) 
 
THE EQUATION O 
 
 To prove the result, replace a in equation (1) by its 
 value. This gives 
 
 3x8+5 = 29, 
 29 = 29. 
 
 The equation is said to be satisfied by a = 8 ; which 
 means that when 8 is substituted for a, the equation is 
 true. It is evident that it would not be satisfied by any 
 other value of a. 
 
 Ex. 2. The sum of two numbers is 38, and their 
 difference is 8. What are the numbers ? 
 
 The answer can be written down at once by substi- 
 tuting s = 38, and (Z = 8 in formulas (1) and (2), Art. 1. 
 The pupil should carefully compare this method of solu- 
 tion with the following : 
 
 Let us denote the greater of the numbers by x ; then, 
 since the smaller number is 8 less than the larger, it will 
 be denoted by a; — 8. 
 
 The problem states that 
 
 the larger number + the smaller number = 38. 
 Hence x-\-{x-^) = 38, (1) 
 
 or a; -h a; - 8 = 38, 
 
 or 2a;-8 = 38. 
 
 Add 8 to both sides of the equation ; then 
 
 2a; = 46. 
 Divide both sides by 2 ; then 
 
 a; = 23 = larger number. 
 Hence a; — 8 = 16 = smaller number. 
 
6 ALGEBBAIC ARITHMETIC 
 
 Proof. Substituting these values in equation (1), we 
 have 
 
 23 + 15 = 38, 
 
 38 = 38. 
 
 Note. It is necessary to notice the punctuation after the equa- 
 tions in the solution of a problem ; for the equations always occur 
 as parts of sentences, and the punctuation helps to make the mean- 
 ing clear, just as in the case of any other kind of sentence. 
 
 4. Definitions. The figures, letters, and signs used in 
 arithmetic are called symbols. 
 
 Any combination of symbols denoting a number is 
 called an expression. If it contains letters, it is called 
 an algebraic expression. 
 
 The members of an equation are expressions. 
 
 The parts of an expression which are separated from 
 each other by the signs of addition or subtraction are 
 called the terms of the expression. 
 
 Thus the expression 2ab ~c-\- 5 has three terms ; 
 5(iocy has one term. 
 
 A term may consist of two or more factors. Thus the 
 term 5 axy contains four factors. 
 
 If the factors of a product are separated into groups 
 in any way, either group of factors is called the coefficient 
 of the other group. 
 
 Thus in the term 5 axy^ 5 is the coefficient of axy, 5 a 
 is the coefficient of xy, 5 ay is the coefficient of x, etc. 
 If a term has a numerical factor, it is generally spoken 
 of as the coefficient of the term. 
 
 Terms containing the same literal factors are called 
 like terms. 
 
AXIOMS 7 
 
 Thus 5 ahx and 9 dbx are like terms ; 3 ab and 7 cdx 
 are unlike terms. 
 
 5. Axioms. We have seen that some problems can be 
 stated in the form of equations in which a letter stands 
 for the answer ; and that the value of the letter which 
 satisfies the equation is the answer to the problem. In 
 solving such equations, frequent use is made of the fol- 
 lowing simple truths, or axioms : 
 
 Ax. 1. If equal numbers are added to equal numbers, 
 the sums are equal. 
 
 Ax. 2. If equal numbers are subtracted from equal 
 numbers, the remainders are equal. 
 
 Ax. 3. If equal numbers are multiplied by equal num- 
 bers, the products are equal. 
 
 Ax. 4. If equal numbers are divided by equal num- 
 bers, the quotients are equal. 
 
 Thus, if a = 5 and c = d, 
 
 then a-\- c=b-}-d by Ax. 1. 
 
 a — c = b — d by Ax. 2. 
 
 ac = bd by Ax. 3. 
 
 and a-i-c = b-i-d by Ax. 4. 
 
 The four axioms may be summed up in the statement : 
 Equal numbers will still remain equal numbers after they 
 have been increased) diminished, multiplied, or divided by 
 equal numbers. 
 
 6. The Solution of Equations. Let us now look again 
 at the solution of Ex. 1, Art. 3. The algebraic (or sym- 
 
a ALGEBRAIC ARITHMETIC 
 
 bolical) statement of the problem is 3 a + 5 = 29, in 
 which a stands for the answer. 
 
 Since the members of this equation are equal numbers, 
 if we subtract 5 from each of them, the remainders will 
 be equal by Ax. 2. This gives 
 
 3a + 5-5 = 29-5, 
 or 3a = 24; 
 
 that is, we form an equation out of the equal remainders. 
 This could not have been done if we had subtracted 
 more from one member than from the other, for in that 
 case the remainders would have been unequal, and the 
 equation would have been destroyed. 
 
 We wish to obtain a alone in the left member of the 
 equation. We can now do this by dividing that side by 
 the coefficient of a; but since we must preserve the 
 equality of the members, we divide both by 3, and obtain 
 a = 8. In this operation we use Ax. 4. 
 
 Since, in solving the original equation, we have made 
 use of only those operations which do not destroy the 
 equality of its members, we know that the last equation 
 is true. It therefore gives us the required value of a. 
 
 Exercise. Find what axioms have been used in the 
 solution of Ex. 2, Art. 3. 
 
 Ex. 1. Find the value of x if 
 
 2 a; 4- 5 = 15 — a;. 
 
 Since we wish to obtain x alone in the left side, and 
 only numerical quantities in the other, we must get rid 
 of the X in the right side and the 5 in the left. The x 
 
THE SOLUTION OF EQUATIONS 9 
 
 will disappear from the right side if we add x to it, since 
 15 — a; -f a; = 15. Hence, adding x to both members to 
 preserve their equality, we have 
 
 2a;-f-5-}-a; = 15 — aj + a; by Ax. 1. 
 
 or 3a; + 5 = 15. 
 
 Subtract 5 from both sides ; then 
 
 3a; = 10 by Ax. 2. 
 
 Divide both sides by 3 ; then 
 
 a; = JJ>_ = 3 J by Ax. 4. 
 
 Note. This value of x satisfies not only the given equation, 
 but also all the equations derived from it ; that is, x has the same 
 value throughout the solution, which must be the case in the 
 solution of any equation. 
 
 
 
 EXAMPLES 
 
 2 
 
 
 Solve the following equations : 
 
 
 
 1. 
 
 3a; 4-4: 
 
 = a; + 10. 
 
 1 7 
 
 
 2. 
 
 4a;4-4: 
 
 = a; + 7. «• 
 
 S^x 
 
 
 3. 
 4. 
 
 5. 
 
 5a;-5: 
 
 a; + 4 = 
 
 14 = 1. 
 
 X 
 
 = 20-2a;. ^ 
 2(5 -a;). 
 
 8. 
 
 a; + |a; 
 5^4 
 
 = 10. 
 
 :1. 
 
 7. The Solution of Problems. 
 
 Ex. 1. What number is that whose half added to 16 
 gives 25 ? 
 
 Let X denote the number. 
 
 Then ^ will denote half the number, and ^ + 16 will 
 2 ^ 
 
 denote the half added to 16. 
 
10 ALGEBRAIC AKITHMETIC 
 
 But the problem states that this is 25 ; hence 
 ^ + 16 = 25. 
 
 Subtract 16 from both sides ; then 
 
 1 = 9 by Ax. 2. 
 
 Multiply both sides by 2 ; then 
 
 a; = 18 by Ax. 3. 
 
 Proof : J^ + 16 = 25, 
 
 25 = 25. 
 
 Ex. 2. A man having $92 spent a part of it, and then 
 had 3 times as much as he had spent. How much did 
 he spend? 
 
 Let X be the number of dollars he spent. 
 Then 92 — a; will be the number of dollars he had left. 
 But the problem tells us that this is 3 times as much 
 as he spent. Hence we have the equation 
 
 3a; = 92-a;. (1) 
 
 Add X to both sides ; then 
 
 4a; = 92 by Ax. 1. 
 
 Divide both sides by 4 ; then 
 
 a; = 23 by Ax. 4. 
 
 Hence the man spent % 23. 
 
 Remarks. In problems involving concrete numbers, 
 like the last, it is not necessary to express the kind of 
 unit in the equation. Thus, in this problem, we do not 
 write 
 
 ^3a;=$92~$aj; 
 
THE SOLUTION OF PROBLEMS 11 
 
 for, though the statement is correct, it is not so simple 
 as when made without the sign. 
 
 The members of the equation are to be regarded as 
 abstract numbers, denoting the number of times the con- 
 crete unit is contained in the quantities to be compared. 
 Thus the members of (1) denote the number of times $ 1 
 is contained in the sum of money the man had left. 
 
 The quantities to be compared must be of the same 
 kind, and must be measured by the same unit. For 
 example, we cannot compare a sum of money with a dis- 
 tance, nor can we compare two sums of money when 
 one is measured in dollars and the other in cents or in 
 dimes. 
 
 Such statements as 
 
 100^= $1, 
 
 16 oz. = 1 lb., 
 
 are not equations at all in the sense in which we shall 
 use the word in this book. The first of these statements 
 means that the two sums are equal in value; the second, 
 that the two weights are equal ; but in neither of them 
 are the two numbers equal. Equations, as we shall use 
 them, will always mean that the two members are equal 
 numbers. 
 
 Ex. 3. A can do a piece of work in 10 days, but A 
 and B working together can do it in 6 days. In how 
 many days can B do it alone ? 
 
 Let X = the number of days it would take B to do the 
 work alone. 
 
 Then - = the part of the work he can do in one day. 
 
12 ^ ALGEBRAIC ARITHMETIC 
 
 From the problem we know that A can do ^ of the 
 work in one day ; and A and B together, ^ of the work 
 in one day. 
 
 H+ro (^) 
 
 Multiply both sides by 30 x, the L. C. M. of the denom- 
 inators; then 
 
 20f = 30^ + ^ by Ax. 3. 
 
 6 « 10 
 
 or 5x = S0-{-3x. 
 
 Subtract 3 x from both sides ; then 
 
 2a;=30 by Ax. 2. 
 
 Divide both sides by the coefficient of a?; then 
 
 ■ a; = 15 by Ax. 4. 
 
 Hence B can do the work in 15 days. 
 
 Proof by substitution : Keplace x in (1) by 16 ; then 
 
 Proof by analysis: Since A can do the work in 10 
 days, in one day he can do J^ of it ; since B can do the 
 work in 15 days, in one day he can do yV of it. Hence 
 A and B working together can do ^ 4- J^-, or ^, of it in 
 one day, or the whole piece of work in 6 days. 
 
 8. From the examples of Art. 3 and Art. 7, the follow- 
 ing directions for the solution of similar problems may 
 be deduced : 
 
 I. Denote the required number by some letter (it is 
 customary to use x). This is called the unknown quantity. 
 
PROBLEMS 13 
 
 II. If there are other numbers in the problem that 
 depend on the unknown quantity, find expressions for 
 them in terms * of the unknown quantity. 
 
 III. Write these expressions in the form of an equa- 
 tion which expresses in symbolic form the conditions 
 of the problem. 
 
 IV. Clear the equation of fractions, if there are any, 
 by multiplying both members by the /. c. m. of the 
 denominators. 
 
 V. By addition or subtraction remove all terms con- 
 taining the. unknown quantity to one side of the equation, 
 and all other terms to the other side. 
 
 VI. After adding together the terms containing the 
 unknown quantity, divide the members of the equation 
 by its coefficient. This gives the answer. 
 
 EXAMPLES 3 
 
 1. John is 3 times as old as James, and the sum of 
 their ages is 16 years. What is the age of each ? 
 
 2. A boy bought a top and a ball for 24 cents, paying 
 5 times as much for the ball as for the top. What did 
 he pay for each ? 
 
 3. Ida's sister gave her some money, and her brother 
 gave her twice as much. After spending 12 cents, she 
 had 18 cents left. How much was given her by each ? 
 
 * A number is said to be expressed in terms of another number 
 when the expression for it contains the letter that represents the 
 other number. Thus in Ex. 2, Art. 3, the smaller number, cc — 8, 
 is expressed in terms of the larger number, x ; and in Ex. 2, 
 Art. 7, the number of dollars the man had left, 3aj, is expressed in 
 terms of the number of dollars he spent, x. 
 
14 ALGEBRAIC ARITHMETIC 
 
 4. The sum of two numbers is 50, and their difference 
 is 18. Find them. 
 
 5. The sum of three numbers is 126. The second is 
 twice the first, and the third is equal to the sum of the 
 other two. What are the numbers ? 
 
 6. A boy, after spending half his money, earned 14 
 cents, and then had 30 cents. How much had he at 
 first? 
 
 7. A and B together can do a piece of work in 8 da., 
 and A working alone can do it in 20 da. In what time 
 can B do it ? 
 
 \/s. Fred has 3 times as many marbles as Harry, lack- 
 ing 2 ; and both together have 26. How many has each ? 
 
 9. The sum of two numbers is 62, and the greater is 
 3 less than 4 times the smaller. Find the numbers. 
 
 10. A father is 6 years more than 4 times as old as 
 his son, and the sum of their ages is 71 years. Find 
 the age of each. 
 
 11. If ^ of a certain number be subtracted from f of 
 it, the remainder will be 8. What is the number ? 
 
 12. Divide 42 into two parts, such that one part shall 
 be I of the other. 
 
 13. One of two apple trees bore f as many apples as 
 the other, and both yielded 21 bu. How many bushels 
 did each yield ? ^ 
 
 14. A lad having 45 cents bought an equal number 
 of pears, oranges, and bananas ; the pears being 3 cents 
 each, the oranges 4 cents, and the bananas 2 cents. How 
 many of each did he buy ? 
 
CHAPTER II 
 
 ADDITION, SUBTRACTION, MULTIPLICATION, DIVISION 
 
 9. We shall now study a little more fully the way in 
 which operations are performed upon numbers denoted 
 by letters. 
 
 There are four fundamental operations, or processes, 
 by certain combinations of which all the problems of 
 arithmetic are solved. These are addition, subtraction, 
 multiplication, and division. You have already learned 
 that subtraction is the inverse of addition, and division 
 the inverse of multiplication; by which is meant that 
 subtraction itii-does what addition does, and division un- 
 does what multiplication does. 
 
 Hence, if to any number I add any other number, and 
 afterwards subtract the same number, I shall have left 
 the first number unchanged, since the two operations 
 exactly cancel each other. 
 
 Thus, if a and h are any two numbers, 
 
 a + b-b = a. (1) 
 
 Again, if I multiply any number by any other number, 
 then divide the product by the same number, the quotient 
 will be the first number. If the division be performed 
 first, then the multiplication, the result will still be the 
 first number. 
 
 16 
 
16 ALGEBRAIC ARITHMETIC 
 
 Thus, a X 6 -J- 6 = a ^ 6 X 6 = a, (2) 
 
 ab a L 
 or — = - X 6 = a. 
 
 
 
 On account of this inverse relation, every fact in 
 addition gives one or more corresponding facts in sub- 
 traction, and similarly for multiplication and division. 
 
 Thus, since 5 + 7 = 12, it follows that 
 
 (7 _|. 5) _ 5 = 7^ or 12 - 5 = 7, 
 and that (5 + 7) - 7 = 5, or 12 - 7 = 5; 
 and since 6 x 8 = 48, it follows that 
 
 (6 X 8) -- 8 = 6, or 48 -^ 8 = 6, 
 and that (8 x 6) -^ 6 = 8, or 48 -r- 6 = 8. 
 
 ADDITION 
 
 10. If we have a group of a things and a second group 
 of h things, and if we form a single group from these 
 by putting the two groups together, we shall have as 
 many things in the single group thus formed as there 
 were at first in both the groups. It is clear that this 
 will be true whether we put the first group with the 
 second or the second with the first; that is, the sum 
 of the things is not changed by the way in which they 
 are brought together. 
 
 This fact or law is expressed symbolically thus : 
 
 a + b=:b + a\ (1) 
 
 and briefly in words thus : 
 
 Additions may he performed in any order. 
 
 Thus, for any three numbers a, 6, and c, 
 a+6-fc=a+cH-6=64-a + c=a+(6H-c) = (6-|-a)-|-c, etc. 
 
SUBTRACTION 17 
 
 11. If I take a marbles from a box twice, and after- 
 wards take the same number of marbles 3 times, I have 
 in all taken a marbles (2 -f 3) times, or 5 times ; which 
 makes 5 a marbles. 
 
 Thus the number I take the first time is 
 
 a -{-a, or 2 a, 
 and the second time, a-\- a-\-a, or 3 a. 
 
 Hence I take in all, a-\-a-\-a-\-a-{-ay or 5a. 
 Hence 2a-f3a = (2 + 3)a = 5a. 
 
 If I take a things m times, and again n times, I take 
 in all a things (m -f- n) times, or {m + n) a things. 
 
 Hence ma -{- na = {m -\- n) a. (1) 
 
 From the formula we have the rule : To add terms 
 having a common factor, write the common factor with a 
 coefficient equal to the sum of the coefficients of the terms 
 added. 
 
 Ex. 1. 3a6 + 5a6 + a?> = 9a6. 
 
 Ex. 2. 4ac-f-66c=(4a + 66)c. 
 
 Exercise. Show that the above equations are true 
 when a = l, 6 = 2, c = 3; when a = 4, 6 = 3, c = 2; 
 when a = h = c = ^. 
 
 Are they true for all values of the letters ? 
 
 SUBTRACTION 
 
 12. If there are a apples in one basket and h in 
 another, and I take away c of them, the number remain- 
 ing will be a -f 6 — c. This result does not show whether 
 I take the apples partly from each basket or all from 
 
18 ALGEBRAIC ARITHMETIC 
 
 one. It merely indicates that the whole number of 
 apples, a -h 6, has been diminished by c. 
 
 If I take them all from the first basket, the number 
 remaining in it will be a — c, and the whole number of 
 apples remaining will be a — c + 6. Similarly, if I take 
 them all from the second basket, the whole number re- 
 maining will be a-\-(b — c)j or 6 — c + a. 
 
 The whole number remaining will be the same which- 
 ever way I take the c apples ; hence 
 
 a-f-6 — <? = a — cH-6 = 6 — c-fa. (1) 
 
 From this we have the law : Subtractions may be per- 
 formed in any order. 
 
 Note. This law is limited to the case where the minuend is at 
 least as large as the subtrahend. 
 
 13. From articles 10 and 12 it is clear that 
 
 a+(6 + c)=fl + 6 + (?, (1) 
 
 (H-(6-c)=a-f 6-(?. (2) 
 
 Hence a poA'enthesis which is preceded by the sign of 
 addition may be removed from an expression without affect- 
 ing its value. 
 
 14. The expression a—(b-{-c) means that from a we 
 are to subtract the sum of b and c. We shall evidently 
 obtain the same result by first subtracting b from a, then 
 subtracting c from the remainder •, hence 
 
 a-(b-\-c)=a-b-c. (1) 
 
 The expression a — (b — c) means that from a we are 
 to subtract the difference between b and c ; hence if we 
 
SUBTRACTION 19 
 
 subtract b from a, we subtract c units too much. Hence, 
 to obtain the correct result, we must add c to the re- 
 mainder. That is, 
 
 a-(b-c)=a-b + c. (2) 
 
 From (1) and (2) it follows that a parenthesis luhich 
 IS preceded by the sign of subtraction may be removed 
 from an expression if all the -f- signs within the paren- 
 thesis be changed to — signs, and all the — signs to -\- 
 signs. 
 
 Ex. a-(c-2& + 5)=a-c + 26-5. 
 
 EXAMPLES 4 
 
 If a = 6, 6 = 5, c = 4, d = 2, and e = 1, find the 
 values of the following expressions (1) by substituting 
 in the given expressions, then performing the indicated 
 operations, (2) by removing parentheses, combining like 
 terms, then substituting the values of the letters. The 
 results should agree. 
 
 1. 3ad+(2c-ae). 8. 3a-(26-a + 5). 
 
 2. bde-{a-2d). 9. Zad-{ad-b)-\-2b-c. 
 
 4. 7cd+(8a-5cd + 9). 11. 20e-[3a-(26 + a)]. 
 
 5. 5a&4-[6c-(3c-5e)]. 12. 2 6c - (a6 - 6c) + 2 a6. 
 
 6 ^—(p — c) . 13 a4-(2c — ci) e 
 
 d ' c a 
 
20 ALGEBRAIC ARITHMETIC 
 
 MULTIPLICATION 
 
 15. For the case where thfe multiplier is an integer, 
 multiplication is defined as the process of taking one 
 number as many times as there are units in another 
 number. 
 
 Thus 3x5 = 5 + 5 + 5 
 
 (the 5 being taken as many times as there are units in 
 the multiplier 3), 
 and 4a = a + a + a + a. 
 
 This definition fails when the multiplier is a mixed 
 number or a fraction, for we cannot take anything a 
 fraction of a time. A fractional multiplier does not 
 indicate how many times the multiplicand is to be taken, 
 but what part of it. 
 
 Thus f X a means that 3 of the 4 equal parts of a 
 are to be taken. The multiplier itself is 3 of the 4 equal 
 parts of unity. Hence we have the following definition 
 of multiplication, which holds for any value of the mul- 
 tiplier : 
 
 Multiplication is the process of doing to the multipli- 
 cand what was done to unity to obtain the multiplier. 
 
 Numerical examples will make the meaning clear. 
 
 Multiply 10 by |. 
 
 To obtain the multiplier, 1 was divided into 5 equal 
 parts, and two of these parts were taken. Hence divide 
 10 into 5 equal parts and take two of them. The result 
 is 4. 
 
 Multiply 5 by 3. 
 
 The multiplier is three ones (1 + 1 + 1). 
 
 Hence the product is three fives {b-^-b-^- 6). 
 
MULTIPLICATION 21 
 
 16. The number of dots in the figure, counted by rows, 
 
 is three fives ; counted by columns, it is five 
 threes. The number of dots must he the same 
 • • • • • whichever way they are counted; hence 
 3x5 = 5x3. 
 The same reasoning holds for the product of any two 
 integers. Thus if there are a rows of 6 dots each, the 
 whole number of dots, counted by rows, will be ah ; and 
 if counted by columns it will be ha. 
 Hence ab = ba. (1) 
 
 This is a law of multiplication, which, expressed in 
 words, is : The order in which factors are taken does not 
 affect the value of the product ; or, more briefly : 
 The factors of a product may he taken in any order. 
 
 Note. It should be observed that, in the proof of this law, we 
 have assumed that a and h are integers. The law holds good when 
 either or both the factors are fractions ; but the proof is different 
 and rather more difficult. 
 
 The law holds for any number of factors; thus for 
 three factors a, h, and c, 
 
 ahc = ach = cha = a(hc), etc. (2) 
 
 17. It follows directly from the law stated in the pre- 
 ceding article that a numher is multiplied hy multiplying 
 any one of its factors. 
 
 Thus, if we wish to multiply the number ahc by any 
 number x, we know from this law that 
 
 (abc) X = (ax) bc = a (bx) c = ab (ex). (1) 
 
 Exercise. Show that 30 is multiplied by 7 by multi- 
 plying any one of its prime factors by 7. 
 
22 ALGEBRAIC ARITHMETIC 
 
 18. To multiply 24 by 2 we multiply the 4 units by 
 2 and the 2 tens by 2 ; that is, 
 
 2 X 24 = 2(20 + 4)= 2 X 20 + 2 X 4. 
 
 We may separate the multiplicand into parts in any 
 way, and multiply it by multiplying each of those parts. 
 For example, 
 
 2x24 = 2(12 + 7 + 5)= 2x12 4-2x7 + 2x5. 
 
 The same fact is true of any number. 
 
 Thus, o (/fi 4- /i) = a/n + an. (1) 
 
 Hence the law : A number is multiplied by multiplying 
 each of its parts (terms). 
 
 Note. The parts of a number are not its factors. A number 
 is produced from its parts by addition, not by multiplication. 
 
 Ex.1. 5(3a6 + 2)=15a6 + 10. 
 
 Ex.2. 2a(b + cd)=2ab-\-2acd. 
 
 Ex. 3. 3b(cd-{-S^-\-^e) = 3bcd + 9a-^2e. 
 
 Exercise. Show that these equations are true when 
 a = 1, 6 = 2, c = 5, and d = 3 ; also when a = 6, 6 = 1, 
 c = 2, and d = 3. Are they true for other values of the 
 letters ? 
 
 19. Not only may the multiplicand be separated into 
 parts, but the multiplier may be also. This is done in 
 finding a numerical product when the multiplier consists 
 of more than one figure. 
 
MULTIPLICATION 23 
 
 Thus the operation of finding 
 
 42 X 35, when expressed 
 
 ily, IS 
 
 35 = 
 
 30 + 5 
 
 42 = 
 
 40 + 2 
 
 2x 5 = 
 
 10 
 
 2x30 = 
 
 60 
 
 40 X 5 = 
 
 200 
 
 40x30 = 
 
 1200 
 
 42 X 35 = 1470 
 
 If we should separate the factors into parts in any way, 
 and should multiply each part of the multiplicand by 
 each part of the multiplier, the sum of these partial 
 products would be the product of the factors. 
 
 Exercise. Find the product of 35 and 42 after separ 
 rating the factors into the parts : 35 = 12 + 20 + 3, and 
 
 42 = 20 + 22. 
 
 Note. The sign of multiplication is omitted between paren- 
 theses, and between a parenthesis and a factor. 
 
 Ex.1. (a + 6)(c + d)=a(c + (r)+6(c + c?) 
 = ac + ac^ + 6c + bd. 
 
 Ex.2. 2a(6 + 4c)=2a6 + 8ac. 
 
 Ex. 3. (3a + 6)(c + 5e)=3a(c + 5e)+&(c + 5e) 
 = 3 ac + 15 ae + 6c + 5 be. 
 
 Exercise. Show that the above results are true when 
 a = 1, 6 = c = 2, c? = 4, and e = 3. 
 
 Give the letters a different set of values, and show that 
 the results are true for those values. 
 
24 ALGEBRAIC ARITHMETIC 
 
 20. The expression 4 (8 — 3) means that the difference 
 between 8 and 3 is to be taken 4 times. 
 
 Hence 4(8 - 3)= 4 x 5 = 20. 
 
 The result can be found differently as follows : If we 
 take 4x8, every time we have taken 8 instead of 5 we 
 have taken 3 too many. Hence we have taken 3 too 
 many 4 times, or 4 x 3 too many in all ; and the result 
 will be correct if we subtract that number. 
 
 Hence 4(8 - 3)= 4 x 8 - 4 x 3. 
 
 The same reasoning holds for any numbers ; hence, in 
 general, 
 
 a(b — c)=ab — ac. (1) 
 
 Hence the law given in Art. 18 may be extended so as 
 to read : An expression is multiplied by multiplying each 
 of its terms, whether they are to be added or subtracted. 
 
 21.* To multiply (c — c^) by (a — b), first take (c — d) 
 a times, which gives a(c — d), or ac — ad. This result is 
 too large, for in taking the multiplicand a times instead 
 of (a — b) times, we have taken it b times too many. 
 Hence we must subtract 6 (c — d), or (be — bd). 
 
 Hence {a — b){c — d)= a{c — d)— b{c — d) 
 = a{c — d) — (bc — bd) 
 = ac — ad — bc-{- bd. 
 
 From this example we may deduce the following rule 
 for the multiplication of algebraic quantities: Multiply 
 each term of the multiplicand by each term of the multiplier. 
 
 » This article may be omitted at the discretion of the teacher. 
 
MULTIPLICATION 25 
 
 When the two terms of a product have both + or both — 
 before them, put -f- before their product; when one has + 
 and the other —, put — before their product. In using the 
 first terms of the expressions, which have no sign, apply the 
 rule as if they had the -f- sign. 
 
 The rule for the signs may be briefly stated: Like 
 signs give -\- and unlike signs give — . 
 
 Ex. {2a-b)(Sa-2b)=2a(Sa-2b)-b(Sa-2b) 
 = 6aa-4:ab-Sab-{-2bb 
 = 6aa-7ab-\-2bb. 
 
 Exercise. Show that this result is true when a = 3 
 and 6 = 2; when a = 6 and 6 = 5. 
 
 22. When a factor is to be taken more than once in a 
 product, instead of repeating the factor the required 
 number of times, it is written only once with a small 
 figure to the right and a little above it. This figure 
 shows how many times the factor is to be repeated, and 
 is called an exponent. 
 
 Thus, 52 = 5 X 5, 23 = 2 X 2 X 2. The answer to the 
 example at the end of the last article would be written 
 
 6a^-7ab-h2b^ 
 
 It should be noticed that a coefficient and an exponent 
 have very different meanings. 
 
 Thus 3x6 = 6+6 + 6 = 18; 
 
 63 = 6x6x6 = 216. 
 4:a = a-\-a-\-a + a; 
 
26 ALGEBRAIC ARITHMETIC 
 
 Ex. 1. 2ax5a^ = 2x5aaa = 10a\ 
 Ex. 2. 3a(a-4:b)=3a'-12ab. 
 Ex. 3. (a-\-by=(a + b)(a + b)=a(a-{-b)+b(a + b) 
 = a:'-{-ab-\-ab + b^ = a^-{-2ab-{- Jy". 
 
 Exercise. Show that the result of Ex. 3 is true when 
 a = 20 and 6 = 4; when a = 40 and 6 = 3; when a = 4 
 and 6 = 3. 
 
 EXAMPLES 5 
 Multiply : 
 
 1. 2a by 4a. 7. a + 6 by 3. 
 
 2. 3 a by 5 cd 8. 3 a — 6 by 5. 
 
 3. ^o?' by a6. 9. o?-\-a by a. 
 
 4. 2a62 by 10a26. 10. 2a2-a by a\ 
 
 5. 4a62c by 2ca?. 11. a2 + 2a-2 by 3a. 
 
 6. Ice by ^o?y. 12. 6c-|-ca + «6 by a6c. 
 
 If a = 3, 6 = 2, c = 1, and d==5, find the numerical 
 values of : 
 
 13. 2a^c. 19. a(d-c-b). 
 
 14. 5a62d 20. Sb^id'-ac). 
 
 15. a^b^c'd^ 21. (a -\- b)(d - c). 
 
 16. 6a6'' + 3c*. 22. (a2 + 2 6)(a - 6). 
 
 17. 5a^c-b\ 23. (a + 6)*. 
 
 18. 3(a« + 6). 24. (2 6 + <«)'. 
 
DIVISION 27 
 
 Simplify the following by removing parentheses and 
 combining like terms : 
 
 25. 2(a-b)+S(a-\-b). 28. 7 a(b - c)-2b(a- c). 
 
 26. Sa(b-\-c) — (ab-\-2ac). 29. |(6 - 2c)- |(c- 2&).\/' 
 
 27. c(a + b)-c(a-b). 30. 2[3a& - 4a(c - 2 6)]. 
 
 Find: 
 
 31. (x + yy. 37.* (a -by. 
 
 32. (2a-^by. 38. (2a;-32/)2. 
 
 33. (a2 + c)2. 39. (a^ - cy. 
 
 34. (a + 6)(3a + 26). 40. (a-b)(2a-b). 
 
 35. (a-6)(2a + 5&). 41. (2a - ft^^^g^ _ 252^. 
 
 36. (a + b)(a-2b). 42. (a + 6)^. 
 
 DIVISION 
 
 23. We have already referred to division as the inverse 
 of multiplication (Art. 9). It is the process by which, 
 when the product of two factors (the dividend) and one 
 of the factors (the divisor) are given, the other factor 
 (the quotient) is found. 
 
 In consequence of this relation between the two pro- 
 cesses, it is easy to derive the laws and rules of division 
 from the corresponding laws and rules of multiplication. 
 We shall proceed to do this. 
 
 24. Since a number is multiplied by multiplying any 
 one of its factors (Art. 17), it follows that a number is 
 divided by dividing any one of its factors. 
 
 Thus (abc)-^d=-^xbc = ax^xc = abx^- (1) 
 d a a 
 
 * This and the following are to be taken or omitted with Art. 21. 
 
28 ALGEBRAIC ARITHMETIC 
 
 It will be seen from this that the result is the same 
 whether the division is performed before or after any or 
 all of the multiplications ; hence : 
 
 Divisions may he performed in any order. 
 
 Ex. 1. Show that 336 is divided by 2 by dividing any 
 one of its factors 4, 6, and 14, by 2. 
 
 Ex. 2. Show that 42 is divided by 3 by dividing any 
 one of its prime factors by 3. 
 
 25. Since a number is multiplied by multiplying each 
 of its parts (Art. 18), it follows that a number is divided 
 by dividing each of its parts. 
 
 .Thus (6 + c)-^/7i=- + -. (1) 
 
 ^ ^ mm ^ ^ 
 
 This law is employed in every numerical example in 
 division. Thus the steps in the process of dividing 762 
 by 3 are as follows : 
 
 762 = 600 4- 150 + 12, 
 3)762 = :&oo+^f^ + ¥ 
 = 200 + 50 + 4 
 = 254. 
 
 If the method of solution seems unfamiliar, it is 
 because we are accustomed to perform the separate steps 
 mentally, and to put down only the result. 
 
 26. It follows from Art. 20, that an expression is 
 divided by dividing each of its terms, whether they are to be 
 added or subtracted. 
 
 Thus {a-b)-^c = ^-^' (1) 
 
PROBLEMS 29 
 
 EXAMPLES 6 
 Divide : 
 
 1. 15a by 5a. 7. 12a^ + 15a^b by 3a. 
 
 2. 12a2by4a. 8. 18 a^ft - 12 ac by 6 a. 
 
 3. Sab by 26. 9. 5x^y — xy^ by 7 xy. 
 
 4. 6xy by 2a;2/2. lo. a^b-^-Sbcd by 3a. 
 
 5. 30 a^dc by 3 ac. 11. (a + &)^ by a + &. 
 
 6. a'b'cd' by b'd. 12. 15 a^d^ + 5 a'b^ - 3 a*b by 5 a^ft. 
 
 PROBLEMS 
 Note. See Art. 8 for directions. 
 
 13. Three boys, counting their money, found they had 
 190 cents. The second had twice as many cents as the 
 first, and the third as many as both the others, plus 4 
 cents. How many cents had each ? 
 
 14. A cistern filled with water has two faucets, one of 
 which will empty it in 5 hr., the other in 20 hr. How 
 long will it take both to empty it ? 
 
 15. If 12 be added to the half of a certain number, 
 the sum will be 20. Find the number. 
 
 16. A farmer divided 52 apples among 3 boys in such 
 a manner that B had ^ as many as A, and C had 2 less 
 than } as many as A. How many had each ? 
 
 17. The whole number of hands employed in a factory 
 is 1000. There are twice as many boys as men, and 11 
 times as many women as boys. How many of each are 
 there ? 
 
30 ALGEBRAIC ARITHMETIC 
 
 18. A and B invest equal amounts in trade. A gains 
 $ 1260, and B loses ^ 870 ; A's money is now double B's. 
 What sum did each invest ? 
 
 19. Divide 100 into two parts such that twice one part 
 is equal to 3 times the other. 
 
 20. The sum of two numbers is 36, and their differ- 
 ence is half the greater. Find them. 
 
 21. A man of 40 has a son 10 yr. old. In how many 
 years will the father be 3 times as old as the son ? 
 
 22. A father's age is 3 times that of his son, and in 
 10 yr. it will be twice as great. How old are they ? 
 
 23. A has $15 more than B, B has $5 less than C, 
 and they have $ 65 in all. How much has each ? 
 
 24. In a regiment containing 1200 men, there were 3 
 times as many cavalry as artillery less 20, and 92 more 
 infantry than cavalry. How many of each ? 
 
 25. What are the ages of three brothers, whose united 
 ages are 48 years, and their birthdays 2 years apart ? 
 
 26. The difference of the squares of two consecutive 
 numbers is 15. What are the numbers ? 
 
 27. At the time of marriage, a man was twice as old 
 as his wife ; but 18 years later his age was | times hers. 
 Required their ages on the wedding day. 
 
CHAPTER III 
 
 PERCENTAGE AND ITS APPLICATIONS 
 
 PERCENTAGE 
 
 27. Three closely related operations are frequently- 
 employed in commercial, or business arithmetic; namely: 
 
 I. To find a certain part of a number. 
 II. To find what part one number is of another. 
 III. To find a number when a certain part of it is given. 
 
 Ex. 1. What is I of 50? 
 
 What part of 50 is 30 ? 
 
 What is the number of which 30 is f ? 
 
 Ex. 2. A man had 75 sheep, and he sold f of them. 
 How many did he sell ? 
 
 A man had 75 sheep, and he sold 50 of them. What 
 part of his sheep did he sell ? 
 
 A man sold 50 sheep, which was J of all he had. How 
 many had he at first ? 
 
 It will now be seen what is meant by saying that these 
 operations are closely related. The three questions in 
 each of the examples involve the same three numbers, 
 of which two are given and the third required ; and any 
 one of the three can be found if the other two are given. 
 
 28. It is customary in business to express the frac- 
 tion that one number is of another in hundredths, even 
 
 31 
 
32 ALGEBRAIC ARITHJVIETIC 
 
 when the fraction can be readily reduced to lower 
 terms. 
 
 Thus4is^of 8;3is^of30; lis^of40.. 
 
 In stating problems, the denominator 100 is omitted, 
 and the phrase per centy which means hundredths, is used 
 instead. 
 
 The sign % means per cent. 
 
 The following expressions exhibit the different ways 
 of denoting a fractional part : 
 
 i=^ = .50 =50 per cent = 50%. 
 2 100 ^ ^ 
 
 i = ^ = .125 = 12|. per cent = 12^%. 
 
 1 i 
 200 == 100 " .005 = \ per cent = i%. 
 
 5 = 1?^ = 1.25 = 125 per cent = 125%. 
 
 1 = 152 = 1.00 = 100 per cent = 100%. 
 
 Note. It should be remembered that 100% of a number is 
 once the number, or the number itself. 
 
 EXAMPLES 7 (Oral) 
 
 Name the corresponding fractions in lowest terms : 
 
 2% 15% 40% 80% 100% \% 
 
 4% 20% 45% 85% 120% i% 
 
 6% 25% 60% 90% 125% y\% 
 
 10% 35% 75% 95% 175% 12^% 
 
PERCENTAGE 33 
 
 EXAMPLES 8 
 
 Express as fractions in the lowest terms. The results 
 should be memorized : 
 
 6i% 
 
 8W 
 12i% 
 
 16|% 
 33J% 
 
 
 62i% 
 66f% 
 83i% 
 
 
 87i% 
 2*% 
 
 
 BXAMPT.KS 9 
 
 
 
 Express as 
 
 per cent. 
 
 Memor: 
 
 ize the first four columns : 
 
 
 * 
 
 TJHF 
 
 A 
 A 
 
 « 
 
 
 29. In computations, per cent is expressed as a com- 
 mon fraction (in lowest terms) or as a decimal, according 
 as the one form or the other is the more convenient. 
 
 Ex. 1. What is 20% of 85? (Case I, Art. 27.) 
 
 20% of a number is ^ of it ; and \ of 85 is 17. 
 
 Ex. 2. What per cent of 30 is 18 ? (Case 11.) 
 
 18 is H of 30 ; and ^ is f, or 60%. 
 
 Ex. 3. 8 is 48% of what number ? (Case III.) 
 
 Since 48 is 8% of the number, 1% of it is |- of 48, 
 
 or 6; and 100% of the number is 100 x 6, or 600. Or, 
 Since 8%, or -^, of the number is 48, -^ of it is ^ of 
 
 48, or 24 ; and |f of the number is 25 x 24, or 600. 
 
 Ex. 4. What number diminished by 5% of itself is 
 38 ? (Case III.) 
 
 100% - 5%, or 95%, of the number is 38. 95% = |^. 
 Hence the number is 20 times ^ of 38, or 40. 
 
84 ALGEBRAIC ARITHMETIC 
 
 EXAMPLES 10 (Oral) 
 Find 
 
 1. 5% of 120. 5. 8% of 300 sheep. 
 
 2. 121% of 72. 6. 6i% of f 320. 
 
 3. 25% of 96. 7. 81% of 24 men. 
 
 4. 33i% of 66. 8. 75% of 300 bu. 
 
 What per cent of 
 
 9. 40 is 15? 13. 72 rd. are 18 rd.? 
 
 10. 12^ is 2|? 14. 1 T. are 2 cwt.? 
 
 11. 1 da. are 16 hr.? 15. 1 gal. are 3 pt.? 
 
 12. lib. are 2oz.? 16. ^480 are ^24? 
 
 What is the number of which 
 
 17. 30 is 20%? 21. 96 is 133J%? 
 
 18. 84 is 12%? 22. 55 is 125%? 
 
 19. 5 is i%? 23. 72 is 66|%? 
 
 20. 16 is 32%? 24. 15 is 16f%? 
 
 25. A farmer had 150 sheep, and sold 16f % of them. 
 How many did he sell ? 
 
 26. A boy increasing his money by 25% of itself has 
 $ 1. What had he at first ? 
 
 27. A gi'ocer bought 150 dozen eggs, and found 
 of them bad or broken. How many were salable ? 
 
 28. What number increased by 8J% of itself is 130 ? 
 
 29. A clerk has his salary increased 12^%, and he 
 then gets $ 18 per week. What was his salary before 
 the increase? 
 
PERCENTAGE 36 
 
 30. A man sold a horse for $100 at 20% above cost. 
 
 Find the cost. 
 
 31. In a school of 75 pupils 3 were absent. What per 
 cent was absent? 
 
 32. John has 36 cents, which is 37|-% of what his 
 brother has. How much has his brother? 
 
 33. A clerk spends 88% of his salary and saves $144. 
 What is his salary ? 
 
 34. 37^% of a stock of goods valued at $1200 was 
 destroyed by fire. What was the loss? 
 
 30. Percentage includes all operations in which a per 
 cent of a number is given or required. 
 
 The number of which the per cent is taken is called 
 the base. 
 
 The per cent, when expressed decimally or as a com- 
 mon fraction, is usually called the rate per cent, or simply 
 the rate. 
 
 The percentage is the result obtained by taking a cer- 
 tain per cent of the base. It is, therefore, a product^ of 
 which the factors are the base and the rate. 
 
 Note. It will be seen, from the two definitions of percentage, 
 that the word is used (1) to name a class of operations, (2) to 
 name the result of an operation. 
 
 31. If we use the initial letters of the words base, rate, 
 and percentage to denote the numbers called by these 
 names, we can easily express algebraically the relations 
 that these numbers bear to one another. The equations 
 expressing these relations are percentage formulas. 
 
36 ALGEBRAIC ARITHMETIC 
 
 From the definition of percentage, we know that 
 
 p = bt: (1) 
 
 Ex. 1. A man invests $1265, and gains 12% on his 
 investment. How much does he gain ? 
 
 6 = $1265, r=.12, p^? 
 $1265 
 
 .\2 
 
 $151.80= p. 
 
 Divide the members of (1) by h ; then 
 
 f = .or. = |. (2) 
 
 Here we have given a product (percentage) and one of 
 the factors (base) to find the other factor (rate per cent). 
 
 Ex. 2. A merchant owes $15,120," of which he can 
 pay only $9828. What per cent of his debts can he 
 pay? 
 
 p = $9828, 6 = $15120, r = ? 
 
 $9828 ^ gg^gg^ 
 $15120 /"^ 
 
 Divide the members of (1) by r ; then 
 
 (3) 
 
 Ex. 3. What number increased by 18% of itself is 
 2950 ? 
 2950 is 100% + 18% or 118% of the number; hence 
 ly = 2950, r = 1.18, 6 = ? 
 
 6 = ???? = 2500. 
 1.18 
 
PERCENTAGE 37 
 
 Note. In some problems, as in this one, one of the given num- 
 bers is not directly stated ; but must be found from the conditions 
 of the problem before the formula can be applied. 
 
 32. Percentage Formulas. The three cases of percent- 
 age and their formulas are : 
 
 Case I. To find a given per cent of a number. 
 
 p = br. (1) 
 
 Case II. To find what per cent one number is of another. 
 
 -f- (2) 
 
 Case III. To find a number when a certain per cent of 
 it is gvven. 
 
 *=^- (3) 
 
 EXAMPLES 11 
 
 1. A merchant failing was able to pay his creditors but 
 ^0. He owes A $ 3500, B $ 1200, C $ 1134, D $ 650. 
 
 What will each receive ? 
 
 2. A person whose annual income is f 450 pays $ 125 
 for board, $ 140 for clothing, f 25 for books, and $ 30 for 
 sundries. What per cent of his income is each item, and 
 what per cent remains ? 
 
 3. The deaths in a certain city, during the year, are 
 980, which is 3^% of the population. What is the popu- 
 lation ? 
 
 4. Sold cloth for $3.50 per yard, which was 70% of 
 its cost. What was the cost per yard ? 
 
38 ALGEBRAIC ARITHMETIC 
 
 5. A merchant failing owes $3500; his property is 
 valued at $ 2100. What per cent of his indebtedness can 
 he pay ? 
 
 6. A shepherd lost 12% of a flock of sheep by disease, 
 and then had 2200. How many were in the flock at 
 first? 
 
 7. Sold a house and lot, which cost me $ 1450.75, at a 
 gain of 15%. What was the gain ? 
 
 8. A man spent in one year $ 2150, which was 5|% 
 of what he had. How much had he ? 
 
 9. A man having $5800 worth of hay lost $870 
 worth by fire. What per cent of the whole was the part 
 lost ? 
 
 10. A tailor, after using 75% of a piece of cloth, had 
 9 J yards left. How many yards were in the whole piece ? 
 
 11. A man drew 25% of his bank deposits, and spent 
 33 J % of the money thus drawn in the purchase of a 
 horse worth $250. How much money had he in the 
 bank at first ? 
 
 12. A man owning f of a cotton-mill, sold 35% of his 
 share for $ 24,640. What part of the whole mill did he 
 still own, and what was its value ? 
 
 PROFIT AND LOSS 
 
 33. Gains, losses, and selling price (S. P.) are always 
 reckoned as a per cent of the cost; in other words, they 
 are percentages computed on the cost as base. 
 
 The following table shows what quantities are denoted 
 
PROFIT AND LOSS 39 
 
 by the letters of the percentage formulas (Art. 32) in the 
 various problems that occur in Profit and Loss : 
 
 Table 
 
 b = cost, 
 
 ' r = rate of gain, p = profit. (1) 
 
 r = rate of loss, p = loss. (2) 
 
 r = 1 + rate of gain, ) ^ ^ p (3) 
 
 ■;1 
 
 r = 1 — rate of loss, y ' ' (4) 
 
 34. Ex. 1. A. man sells a farm for $2081.25, gaining 
 11%. What did the farm cost him ? 
 
 (Case III, Art. 32 and (3) of Table.) 
 
 p = S. P. = $ 2081.25 (= 111 % of cost). 
 
 r = l.ll, 6 = cost = ? 
 
 6 = $ 2081.25 -^ 1.11 = $ 1875. Ans. 
 
 Ex. 2. At what price must goods that cost $3.50 per 
 yard be sold to lose 20% ? 
 
 (Case I, and (4) of Table.) 
 
 The S. P. will be 100% - 20%, or 80% of the cost. 
 b = $3.50, r = 1 - .2 = .8, p = S. P. = ? 
 jp = $ 3.50 X .8 = $ 2.80. Ans. 
 
 Ex. 3. Find the gain per cent on a horse sold for $ 72 
 at a gain of $ 9.50. (Case II, and (1) of Table.) 
 
 6 = $ 72 - $ 9.50 = $ 62.50 = cost. 
 
 p = $9.50, r = ? 
 
 r = $ 9.60 -J- $ 62.50 = .162 = 16^%. Ana. 
 
40 ALGEBRAIC ARITHMETIC 
 
 EXAMPLES 12 (Oral) 
 
 1. Bought a COW for $ 40, and sold her for 20% above 
 cost. What did I receive for her ? 
 
 2. A watch that cost $25 was sold at a loss of 10%. 
 What was the loss, and the selling price ? 
 
 3. A tailor bought cloth at $ 6 a 3^ard, and wished to 
 sell it at a gain of 16|%. At what price must he sell it? 
 
 4. A merchant sells silk at a profit of $ 1,50 per yard, 
 which is 37i% gain. What did it cost, and what is the 
 selling price ? 
 
 5. A watch was sold for $ 34, at a gain of 6J%. What 
 was the cost ? 
 
 6. A dealer lost 12^% on a reaper by selling it for 
 $ 56. For what should he have sold it to gain 12^% ? 
 
 7. Sold melons for f .40 that cost $ .30. What was 
 the gain per cent ? 
 
 8. What per cent is gained on an article bought for 
 $3 and sold for $5? 
 
 9. If corn selling for 21 cents a bushel yields a profit 
 of 50%, what did it cost ? 
 
 EXAMPLES 13 
 
 4. A man offers a farm, for which he gave $ 3450, 
 for 20% less than its cost. What is his price ? 
 
 Note. Computations are simplified by expressing the per cent 
 as a common fraction when it is an aliquot part of 100. 
 
 5. For how much per barrel must I sell flour costing 
 $4.50 per barrel to gain 16|% ? 
 
 SuGOESTioN. To $4. 60 add ^ of it. 
 
PROFIT AND LOSS 41 
 
 6. Sold a cargo of wheat for $ 16,000, at a profit of 
 25%. What was the cost of the cargo ? 
 
 7. A merchant made a profit of $ 156 by selling a 
 quantity of silks at a gain of 12%. What was the cost 
 of the silks, and for how much were they sold ? 
 
 8. A merchant marked a piece of carpeting 25% 
 more than it cost him, but, anxious to effect a sale, and 
 supposing he would still gain 5 % , sold it at a discount of 
 20% from his marked price. Did he gain or lose ? 
 
 Suggestion. S. P. = 80 % of marked price ; marked price = 125 % 
 of cost. S. P. = ? % of cost ? 
 
 9. Sold a lot of books for ^480, and lost 20%,. For 
 what should I have sold them to gain 20% ? 
 
 10. A man bought a pair of horses for $450, which 
 was 25% less than their real value, and sold them for 
 25% more than their real value. What was his gain? 
 
 Suggestion. The real value is the base in both operations. 
 
 11. A merchant pays $6840 for a stock of goods, and 
 sells them at an advance of 26|^% on the purchase price. 
 After deducting $ 375 for expenses, what is his gain ? 
 
 12. A dealer bought 108 bbl. of apples at $4.62^, and 
 sold them so as to gain $ 114.88^. What was his gain 
 per cent ? 
 
 13. My goods are marked to sell at retail at 40% 
 above cost. I furnish my wholesale customers at 12% 
 discount from the retail price. What per cent profit do 
 I make on goods sold at wholesale ? 
 
 Suggestion. 88% of 140% of cost, or ^^^ x 140% of cost - 
 wholesale S. P. 
 
42 ALGEBRAIC ARITHMETIC 
 
 14. At what price must shovels that cost $ 1.12 each 
 be marked in order to abate 5% (of marked price), and 
 yet make 25% profit ? 
 
 16. By selling coffee at 18 cents per pound, I make a 
 profit of 20%. For how much must I sell it to make a 
 profit of 16|% ? 
 
 16. Bought land at ^ 60 an acre. How much must I 
 ask an acre, that I may deduct 25% from my asking 
 price, and still make 20 % on the cost ? 
 
 17. Find the loss per cent on goods sold for $425.98, 
 at a loss of $134.52. 
 
 18. Sold goods for $ 3.50 less than cost, and lost 14%. 
 What per cent should I have gained by selling for $2.75 
 above cost ? 
 
 19. Two sets of furniture were sold for $35 each. 
 On one there was a gain of 16f % ; on the other a loss of 
 16|%. Was there a gain or a loss on both, and how 
 much per cent ? 
 
 20. A hardware merchant bought three dozen agate 
 basins at the rate of 3 for $ 5, and sold them at a gain of 
 $ 10 on the whole. What was the average selling price 
 of each, and what was the gain per cent ? 
 
 21. I bought a horse of Mr. A. for 15% less than it 
 cost him, and sold it for 30% more than I paid for it. 
 I gained $ 15 in the transaction. How much did the 
 horse cost me ? How much did it cost Mr. A. ? For 
 what did I sell it ? 
 
 22. If tea, when sold at a loss of 25%, brings $1.25 
 per pound, what would be the gain or loss per cent if 
 sold for $1.60 per pound? 
 
COMMISSION AND BROKERAGE 48 
 
 COMMISSION AND BROKERAGE 
 
 35. A person who buys and sells goods or lands, col- 
 lects debts, or transacts other business of like nature for 
 another person is called a commission merchant or agent. 
 
 The pay received for such services is called commission. 
 It is usually a percentage on the money paid for property 
 bought ; on the money received for property sold ; on the 
 money collected. 
 
 A broker is a person who buys and sells stocks, bonds, 
 bills of exchange, etc., for a commission, which is called 
 brokerage. 
 
 The money that remains from a sale after the com- 
 mission and other expenses have been paid is called the 
 net proceeds. 
 
 Table 
 
 f am't of sale, \ f '' = ^^*® ^^ com., p = com. (1) 
 
 "" I money collected, j \ r = 1 — rate of com., p = proceeds. (2) 
 
 b = am't of purchase, r = 1 -f rate of com., p = remit. (3) 
 
 36. Ex. 1. Find the commission on the sale at auction 
 of a house and the furniture for $9346.80 at 6i%. 
 
 6 = $ 9346.80, r = .06J, p = ? 
 
 p = $ 9346.80 X .0625 = $ 584.175 com. 
 
 Ex. 2. Find the net proceeds of the above sale. 
 b = $ 9346.80, r = 1 - .06^ = .93}, p = ? 
 p = $ 9346.80 X .93| = $ 8762.625 proceeds. 
 Or, from Ex. 1, 
 
 $9346.80 - $ 584.176 = $8762.625 proceeds. 
 
44 ALGEBRAIC ARITHMETIC 
 
 Ex. 3. I send ^3120 to a commission merchant to 
 buy flour at 4% commission. Find cost of flour and 
 commission. 
 
 The remittance includes the investment +4% of it; 
 hence is 104% of the investment. 
 
 p = $3120, r = 1 + .04 = 1.04, 6 = ? 
 6 = $ 3120 -f- 1.04 = $ 3000. Ans. 
 
 EXAMPLES 14 
 
 4. Find the commission on the sale of a farm for 
 $13,750, at 2f%. 
 
 5. A commission merchant sells 225 bbl. of potatoes 
 at $ 3.25 per bbl., and 316 bbl. of apples at |4.50 per bbl. 
 What is his commission at 4^% ? 
 
 6. A dealer sends his agent in Havana $6720.80, 
 with which to purchase fruits, after deducting his com- 
 mission of 5%. What sum did the agent invest, and 
 what was his commission? 
 
 7. If $63 is paid for collecting a debt of $1260, 
 what is the rate of commission ? 
 
 8. An architect charges |% for plans and specifica- 
 tions, and 1^% for superintending the construction of 
 a building which cost $32,000. What is his fee? 
 
 9. My agent has purchased goods for me to the 
 amount of $ 12.50, for which he charges a commission 
 of If %. What sum must I remit to pay for goods and 
 commission ? 
 
 10. Sent to my agent in Cincinnati $ 765 to purchase 
 bacon, after deducting his commission of 2%. What is 
 his commission, and what does he expend for bacon ? 
 
COMMISSION AND BROKERAGE 45 
 
 11. A grocer sends $2490 to a commission merchant 
 to buy sugar at 3|% commission. If he pays 8 cents 
 a pound for the sugar, for what must the grocer sell the 
 whole to gain 16|% on the whole cost, and at how much 
 per pound? 
 
 12. A collector collected rents at 3% commission and 
 received f 87.60 for his services. What sum of money 
 did he collect? 
 
 13. I pay $275 for a lot and build on it a house cost- 
 ing $ 1720, which my agent rents for $ 25 a month, charg- 
 ing 5% commission. What per cent do I make a year 
 on the money invested ? 
 
 14. Find the commission on the sale of 100 bales of 
 cotton, averaging 480 lb. to the bale, at $18 per cwt., 
 the commission being 5%. 
 
 15. An agent sells 450 tons of hay at $ 13 a ton, com- 
 mission 5%, and with the proceeds buys wool at 22^ 
 cents per pound, commission 4%. What is his whole 
 commission, and how many pounds of wool does he buy ? 
 
 16. An agent in Boston received 28,000 lb. of cotton, 
 which he sold at $ .12^ per lb. He paid $ 45.86 freight 
 and cartage, and after retaining his commission, he re- 
 mits $3252.89 as the net proceeds of the sale. What 
 was the rate of his commission ? 
 
 17. A collector remits $1890 to his principal after 
 deducting his commission of 10%. What was the 
 amount collected? 
 
 18. A farm was sold for $9384 at a commission of 
 1%. Find commission and proceeds of sale. 
 
46 ALGEBRAIC ARITHMETIC 
 
 19. Kemitted to a stockbroker $ 10,650, to be invested 
 in stocks, after deducting J% brokerage. What amount 
 of stock did he purchase? 
 
 20. A broker received ^45,337 to invest in bonds, 
 after deducting a commission of |^%. What amount 
 did he invest, and what was his commission? 
 
 COMMERCIAL DISCOUNT 
 
 37. Manufacturers and wholesale dealers avoid the 
 inconvenience and expense of issuing price-lists of their 
 goods with every change in their market value by deter- 
 mining upon a fixed list price for every article (largely in 
 excess of its true value), from which they give their cus- 
 tomers certain discounts, determined by current market 
 prices. 
 
 Goods are frequently subject to two or more discounts 
 (the last generally being for cash payment) ; and in such 
 cases each discount is reckoned by itself on the sum 
 remaining after subtracting the preceding discounts. 
 
 In stating commercial discount, the sign % is usually 
 omitted. 
 
 38. Find the cost of a bill of goods amounting to $ 800 
 at 20 and 5 off, and 5 off for cash. 
 
 5 )$ 800 = list price of goods. 
 160 =20% discount. 
 20 )$ 640 
 
 32 =5% discount. 
 
 20 )$ 608 
 
 30.40 = 5% off for cash. 
 $577.60 = cost of the goods. 
 
UNIVERSITY 
 
 OF 
 
 *-^ COMMERCIAL DISCOUNT 47 
 
 Or as follows : 
 
 5 X ?P X ^0 
 
 EXAMPLES 15 
 
 1. Bought goods to the amount of $650 at 10 off, 
 and 5 off for cash. What was the cost ? 
 
 2. Find the cost of a bill of goods marked at $450 
 at 40% off, and 5% off for cash. 
 
 3. By getting a discount of 20, and 10 off for cash, I 
 pay $ 1080 for a bill of goods. What was the list price ? 
 What single discount would give the same reduction ? 
 
 4. For what must I sell goods which were sold me 
 for $ 830, list price, at 30, 10, and 5 off, to gain 20% ? 
 
 5. Find the amount of a bill of $ 1560, discounts 
 being 40, 25, and 5. Find the single equivalent discount. 
 
 6. Sold a bill of goods marked at $ 250 for 30, and 
 5 off. How much more did I receive than if I had 
 given a discount of 35% ? 
 
 7. Paid $ 655 for a bill of goods after a discount of 
 16|%. AVhat was the invoice price ? 
 
 8. Find the cash value of a bill of cloth amounting 
 to $425.50 at a discount of 10%, and 5% off for cash. 
 Find the equivalent single discount. 
 
 9. Find the cost of a stove listed at $ 25, discounts 
 being 10 and 7^. 
 
 10. I paid $ 1.50 for a book after a discount of 25%, 
 and 16| off. What was its marked price ? 
 
48 ALGEBRAIC ARITHMETIC 
 
 INSURANCE 
 
 39. Insurance is a guaranty to pay a certain sum of 
 money in case of loss or damage. It is classed as insur- 
 ance on. property and insurance on life. 
 
 That on property is called fire insurance, if against loss 
 by fire; marine insurance, if against loss at sea; stock 
 insurance, if against the loss of cattle, horses, etc. 
 
 The sum paid for obtaining the insurance is called the 
 premium, and the written contract is called the policy. 
 
 The premium is a certain per cent of the sum insured, 
 and is paid in advance. In life insurance it is generally 
 paid annually. 
 
 Fire-insurance companies rarely insure property for 
 more than two-thirds of its value, and in no case pay for 
 more than the value of the property destroyed, whatever 
 may be the face of the policy. 
 
 EXAMPLES 16 
 
 1 . What is the premium for insuring goods for $ 14,500, 
 at 1^% ? 
 
 2. A house worth $15,000 is insured for | of its 
 value, at |%. What is the premium ? 
 
 3. A ship valued at $40,000 is insured for | of its 
 value, at 1^%, and its cargo, valued at $36,000, at f %. 
 What is the cost of insurance ? 
 
 4. A merchant paid $ 1450 premium for the insurance 
 of a cargo of cotton, the rate of insurance being 2|%. 
 For what sum was the cargo insured ? 
 
 5. If it cost $ 93.50 to insure a store for ^t)f its value, 
 at 1|%, what was the store worth? 
 
INSURANCE 4y 
 
 6. A merchant pays $ 50 for an insurance of $ 32,500 
 on a shipment of goods from New York to St. Louis. 
 What is the rate of insurance ? 
 
 7. A house valued at $ 1200 had been insured for | 
 of its value for 3 years, at 1% per annum. During the 
 third year it was destroyed by fire. What was the 
 actual loss to the owner, no allowance being made for 
 interest ? 
 
 SuGGBSTiON. The difference between the amount of the insur- 
 ance and the premium for the three years is what he gets from the 
 insurance. 
 
 8. A merchant has his store and goods insured for 
 $ 5500 at |-% premium. What is the cost to him ? If 
 the store and goods are destroyed, what sum does the 
 insurance company lose ? 
 
 9. An insurance company loses $3528 by the wreck 
 of a carload of flour which it had insured for $3600. 
 What was the rate of insurance ? 
 
 10. A merchant insures a cargo of goods for $ 81,800, 
 which sum includes the value of the goods and the pre- 
 mium at 2|%. What is the premium, and the value 
 of the goods ? 
 
 Suggestion. The premium is always computed on the amount 
 of insurance; hence in this case the base is $81,800. 
 
 11. A merchant ships $ 31,360 worth of wheat from 
 Chicago to Buffalo. For what must he get it insured at 
 2% so as to cover both the value of the wheat and the 
 premium paid for its insurance ? 
 
 Suggestion. 6 = amount of insurance = $31,360 + premium. 
 Hence $31,360 is what per cent of 6 ? 
 
50 ALGEBRAIC ARITHMETIC 
 
 12. A merchant shipped a cargo of flour worth ^3597 
 from New York to Liverpool. For what must he insure 
 it, at 3J%, to cover the value of the flour and the pre- 
 mium ? 
 
 13. I insure my life for $8000, paying $19.80 per 
 $ 1000 per year. What do I pay the company if I live 
 20 years after insurance ? 
 
 14. The annual premium on a life insurance at 2\% 
 is $ 126. What is the amount of the insurance ? 
 
 TAXES 
 
 40. A tax is a sum of money assessed upon the inhabi- 
 tants of a town, district, county, or state, or upon their 
 property, to meet some public expense, such as the sup- 
 port of the schools, or of the government, or the building 
 of public works. 
 
 A tax assessed, without regard to property, upon every 
 male citizen within certain age limits (fixed by law) is 
 called a poll tax, or capitation tax. A person so assessed 
 is called a poll. 
 
 A property tax is assessed at a certain per cent on the 
 estimated, or assessed, value of taxable property. 
 
 Taxable property is of two kinds : (1) Real estate, or 
 fixed property; as houses and lands; (2) Personal, or 
 movable property; as furniture, merchandise, cattle, 
 money, etc. 
 
 EXAMPLES 17 
 
 1. What sum must be assessed to raise $83,600 net, 
 after deducting the cost of collection at 5% ? 
 
 Remark. The cost of collection is 5% of the amount collected. 
 See Art. 35, second paragraph, and (2) of the Table. 
 
TAXES 51 
 
 2. In a certain district, a school-house is to be built 
 at a cost of $ 18,500. What amount must be assessed to 
 cover this and the collector's fees at 3% ? 
 
 3. A county builds a bridge for $ 4410. The property- 
 is valued at $1,000,000. What is the tax per $100, 
 including the cost of collection at 2% ? 
 
 4. In a certain town a tax of $5000 is to be assessed. 
 There are 500 polls, each assessed $ .75, and the valua- 
 tion of the taxable property is $ 370,000. What will be 
 the rate of property tax, and how much will be A's tax, 
 whose property is valued at $7500, and who pays for 
 2 polls? 
 
 Suggestion. Subtract the amount to be raised by poll tax from 
 the whole sum to be assessed ; and find the per cent that the 
 remainder is of the value of the taxable property. This is the rate 
 of taxation. 
 
 5. A tax of $ 11,384, besides cost of collection at 3J%, 
 is to be raised in a certain town. There are 760 polls 
 assessed at $ 1.25 each, and the personal property is 
 valued at $124,000, and the real estate at $350,000. 
 Find the tax rate, and find a person's tax whose real 
 estate is valued at $6750 and personal property at 
 $ 2500, and who pays for 3 polls. 
 
 6. In the above town, how much is B's tax on $ 15,000 
 real estate, $ 2750 personal property, and 2 polls ? 
 
 7. What is C's tax on $9786 and 1 poll ? 
 
 8» How much taxes will a person pay whose property 
 is assessed at $ 7500, if he pays f % town tax, ^% state 
 tax, 1 J mills on a dollar school tax ? 
 
52 ALGEBKAIC AUlTHMEriC 
 
 9. I buy a lot for $ 400 and build a house on it for 
 $2000. I pay an insurance on the house of |^% on J of 
 its value, and a tax on the whole of 14 mills on a dollar, 
 the property valuation being | of the cost. For how 
 much must I rent the house per month to realize 10% a 
 year on my money ? 
 
 10. A tax of $ 56,000, including cost of collecting, is to 
 be raised in a city on a property valuation of $ 22,400,000. 
 Assuming that the uncollectible tax will be 10% of the 
 tax assessed, what will be the tax rate expressed in mills 
 on a dollar ? 
 
 11. In the above city, how much is A's tax on $ 27,500 ? 
 
 DUTIES 
 
 41. The taxes levied on imported goods are called cus- 
 toms or duties. 
 
 Duties are of two kinds : specific and ad valorem. 
 
 A specific duty is a tax on goods according to weight, 
 number, or measure, without regard to value. 
 
 An ad valorem duty is a percentage of the cost of goods 
 in the country from which they are imported. 
 
 Many articles are subject to both kinds of duty. 
 
 Gross weight is the weight of goods including the boxes 
 or other packing material. 
 
 Net weight is the weight after deducting the weight of 
 the packing material. 
 
 Specific duties are calculated on the net weight of 
 goods. All custom-house weights are long-ton weights. 
 
 The following list is taken from two successive tariffs 
 of the United States. The new superseded the old July 
 24,1897: 
 
DUTIES 
 
 53 
 
 ABTI0LK8 
 
 Old Kate of Duty 
 
 N::\v Rate of Duty 
 
 Alcoholic perfumery 
 
 $2 per gal. and 50% 
 
 $.60 per lb. and 45% 
 
 Earthen and crock- 
 
 
 
 ery ware .... 
 
 30% 
 
 55% 
 
 Glass, cut, engraved, 
 
 
 
 or painted . . . 
 
 35% 
 
 60% 
 
 Tin plate . . . . 
 
 U ct. per lb. 
 
 1| ct. per lb. 
 
 Machinery .... 
 
 35% 
 
 45% 
 
 Cigars 
 
 $4 per lb. and 25% 
 
 $4.50 per lb. and 25% 
 
 Horses valued at 
 
 
 
 $150 or less . . 
 
 20% 
 
 $30 per head 
 
 Wheat 
 
 20% 
 
 25 ct. per bu. 
 
 Cotton clothing, 
 
 
 
 ready-made . . . 
 
 40% 
 
 50% 
 
 Cotton hosiery, val- 
 
 
 
 ued at not more 
 
 
 
 than $1 per doz. 
 
 
 
 pairs 
 
 50% 
 
 $.50 per doz. and 15% 
 
 Shirts and drawers 
 
 
 
 valued at not more 
 
 
 
 than $1.50 per doz. 
 
 60% 
 
 $ .60 per doz. and 15% 
 
 Collars and cuffs of 
 
 
 
 linen 
 
 30ct. perdoz. and3% 
 
 40 ct. per doz. and 20% 
 
 Laces and embroid- 
 
 
 
 eries of linen . . 
 
 50% 
 
 60% 
 
 Silk velvets . . . 
 
 $1.50 per lb. 
 
 $1.50 per lb. and 15% 
 
 Lead pencils . . . 
 
 50% 
 
 45 ct. per gross and 25% 
 
 EXAMPLES 18 (Oral) 
 
 1. Which is the higher duty on horses valued at 
 $ 150 ? Less than $ 150 ? 
 
 2. For what value of wheat are the two duties on that 
 article equal ? Which is the higher, and by how much, 
 when wheat is worth $ .60 ? 
 
54 ALGEBRAIC ARITHMETIC 
 
 3. What is the difference between the new and old 
 duties on $1000 worth of cut glass? On $25,000 
 worth of machinery? 
 
 4. What is the cost per gross of lead pencils on which 
 the two rates of duty are equal ? Which is the greater 
 for pencils worth more than that? For pencils worth 
 less? 
 
 5. What was the old duty on $ 5000 worth of ready- 
 made clothing ? What is the new ? 
 
 6. What is the new duty on 100 lb. of perfumery 
 worth $ 2 per pound ? 
 
 7. What is the new duty on a dozen collars valued at 
 $ 1.20 per doz. ? 
 
 8. What is the difference between the old duty and 
 the new on a ton of tin plate ? 
 
 EXAMPLES 19 
 
 Find the (new) duty : 
 
 1. On 1000 boxes of cigars, each containing 100 cigars, 
 invoiced at $ 7.25 per box. Net weight 12 lb. per 1000. 
 
 2. On 12 gross lead pencils at $1.00 per gross. 
 
 20 gross lead pencils at $ 2.25 per gross. 
 5 gross lead pencils at $ 5.00 per gross. 
 
 3. On machinery invoiced at $ 26,500. 
 
 4. On 150 yd. silk velvet at $ 1.75 per yd. Net weight 
 75 1b. 
 
 5. On 15 doz. shirts at $ 1.50 per doz. 
 
 20 doz. linen collars at $1.10 per doz. 
 50 yd, linen lace at 5 ^ per yd. 
 7 doz. cotton hose at $ .90 per doz. 
 
MISCELLANEOUS EXAMPLES ,65 
 
 EXAMPLES 20 (Miscellaneous) 
 
 1. A man had $5420 in bank. He drew out 15% of 
 it, then 20% of the remainder, and afterwards deposited 
 121% of what he had drawn. How much had he then 
 in bank ? 
 
 2. If a man owning 45% of a steamboat sells 16|% 
 of his share for $ 5860, what is the value of the whole 
 boat? 
 
 3. A man sold two houses at $2500 each; on one he 
 gained 20%, on the other he lost 20%. What was his 
 loss on the two sales ? 
 
 4. A man bought a piece of property which after- 
 wards increased in value each year at the rate of 25% 
 on the value of the previous year, for 4 years ; and was 
 then worth $16,000. What did it cost ? 
 
 5. After deducting 6^% commission and $132 for 
 storage, my agent sends me $ 23,654.25 as the net pro- 
 ceeds of a consignment of pork and flour. What was the 
 amount of the sale ? 
 
 6. After taking out 15% of the grain in a bin, there 
 remained 40 bu. 3^ pk. How many bushels were there 
 at first ? 
 
 7. The profits of a farm in 2 years were $3485, and 
 the profits of the second year were 5 % greater than those 
 of the first year. What were the profits of each year ? 
 
 8. If |- of a farm is sold for what f of it cost, what is 
 the gain per cent ? 
 
 9. What is the cost of goods sold for $47,649, at a 
 profit of 16}% ? 
 
56, ALGEBRAIC ARITHMETIC 
 
 10. A broker receives $ 7125 to invest in cotton, after 
 deducting his commission of 2|^%. How many pounds 
 of cotton can he buy at 11^ cents a pound ? 
 
 11. Sold a farm for $14,700, and lost 12%. What 
 per cent should I have gained by selling it for $ 21,000 ? 
 
 12. I buy a house for $6500 and spend $500 for 
 repairs. I rent it for $ 77.50 a month, out of which I 
 pay a yearly insurance of |% on -f- of its whole cost, 
 including repairs, and a yearly tax of 1% on J of the 
 same. What per cent of income a year do I realize on 
 the whole cost ? 
 
 13. For what sum must a policy be made out to cover 
 the insurance on property worth $2100, at |^% ? 
 
 14. I bought a lot of coffee at 12^ per pound. Allow- 
 ing that the coffee will fall 5% short in weighing it out, 
 and that 10% of the sales will be in bad debts, for how 
 much per pound must I sell it to make a clear gain of 
 14% on the cost? 
 
 15. An agent sells for Johnson & Co. 3500 lb. of butter 
 at 20^ per pound, and 2580 lb. of cheese at 9^ per pound, 
 at a commission of 5%. He invests the balance in dry 
 goods, after deducting his commission of 2J% for pur- 
 chasing. How many dollars' worth of goods do Johnson 
 & Co. receive? What is the entire commission of the 
 agent ? 
 
CHAPTER IV 
 
 APPLICATIONS OF PERCENTAGE INVOLVING TIME 
 
 42. The money paid for the use of money is called 
 interest. It is always a percentage of the sum loaned. 
 
 The sum loaned is called the principal. 
 
 The rate per cent of the principal paid for its use for 
 a certain time is called the rate of interest. It is under- 
 stood to be for a year unless otherwise specified. 
 
 The sum of the principal and the interest is called the 
 amount. It is the sum that the borrower must pay back 
 to cancel his debt. 
 
 In computing interest for a fraction of a year, it is 
 customary to reckon each month as ^^ of a year, and a 
 day as -^^ of a month. 
 
 SIMPLE INTEREST 
 
 43. Ex. 1. What is the interest on $100 for 2^ yr. 
 at8%? 
 
 The interest for 1 yr. is S% of $100, or $8, and for 
 21 yr. is 21 X $ 8, or $ 20. Or, 
 
 Since the interest is 8% of the principal for 1 yr., for 
 21 yr. it will be 2i- x 8% or 20% of the principal; and 
 20% of $100 = $20. 
 
 Ex. 2. Find the interest and amount of $200 for 3 yr. 
 at 5%. 
 
 67 
 
58 ALGEBRAIC ARlTIiMETIC 
 
 The interest is 15% of the principal, or $30; the 
 amount is $ 200 + $ 30 = $ 230. 
 
 EXAMPLES 21 (Oral) 
 
 rind the interest and amount of : 
 
 3. $ 100, at 6%, for 1 yr. ; 2^ yr. ; 3 yr. 4 mo. 
 
 4. f 500, at 5%, for 6 mo. ; 2 yr. ; 2^ yr. 
 
 5. $50, at 12%, for 1 mo.; 1^ yr. 
 
 6. $1000, at 3%, for 1 yr. ; 2 yr. 4 mo. 
 
 7. $40, at 6%, for 2 mo.; 6 mo.; 10 mo. 
 
 8. $ 5 at 10%, for 2 yr. ; 3| yr. 
 
 9. $ 10 at 6%, for 1 mo. ; 9 mo. 
 
 10. $ 300 for 6 mo., at 6% ; at 8%. 
 
 11. $60 for 8 mo., at 6%; at 12%. 
 
 12. $ 200 for 3 mo., at 1% a month. 
 
 13. $ 250 for 2| yr., at 4% ; at 10%. 
 
 14. $ 6 for 7 mo., at 1% a month. 
 
 44. To compute interest at 6 per cent. Keckoning a 
 month SiS ^ oi Si year and a day as ^^ of a month, the 
 interest, at 6%, 
 
 for 1 yr. = .06 of the principal ; 
 for 2 mo. = .01 of the principal ; 
 for 1 mo. = .005 of the principal; 
 for 6 da. = .001 of the principal ; 
 . for 1 da. = .000^ of the principal. 
 
SIMPLE INTEREST 59 
 
 Hence to find the decimal fraction of the principal that 
 the interest, at 6%, for any given time is, take 6 times the 
 number of year's and ^ the number of months as hundredths^ 
 and \ the number of days as thousandths. 
 
 Ex. 1. Find the interest of ^375.50 for 3 yr. 5 mo. 
 21 da., at 6%. 
 The interest for 
 
 3 yr. = 3 X .06 = .18 of the principal ; 
 
 5 mo. = 5 X .005 = .025 of the principal ; 
 
 21 da. = 21 X .000^ = .0035 of the principal; 
 
 .2085 of the principal. 
 
 Or, following the rule exactly, 
 
 ^375.5 
 (3 X .06) = .18 -2085 
 
 (I X .01) = .025 7510 
 
 (Vx. 001) =.0035 300 
 
 .2085 19 
 
 $78.29 interest. 
 
 Note. In forming the multiplier, the operations indicated in 
 parentheses should be performed mentally, only the results being 
 set down. 
 
 EXAMPLES 22 
 
 Find the interest and amount, at 6%, of : 
 
 2. $ 760, for 1 yr. 9 mo. 27 da. 
 
 3. $ 179.50, for 1 yr. 1 mo. 8 da. 
 
 4. $ 325, for 2 yr. 11 mo. 6 da. 
 
 6. $ 758.75, for 3 yr. 2 mo. 16 da. 
 
60 ALGEBRAIC ARITHMETIC 
 
 6. $ 1024.25, for 2 yr. 3 mo. 22 da. 
 
 7. $ 584.50, for 1 yr. 2 mo. 14 da. 
 
 8. $ 725.84, for 1 yr. 3 mo. 11 da. 
 
 9. $ 387.95, for 3 yr. 7 mo. 24 da. 
 10. $ 42.20, for 24 da. 
 
 45. To find the years, months, and days between two 
 dates, add mentally to the earlier date first the years, 
 then the months, then the days necessary to obtain the 
 later date, in each case recording only the result. 
 
 Ex. Find the time from Jan. 26, 1895, to June 8, 1897. 
 
 From Jan. 26, 1895, to Jan. 26, 1897, 2 yr. ; to May 26, 
 4 mo. ; to June 8, 12 da. (4 in May and 8 in June). 
 Time: 2 yr. 4 mo. 12 da. 
 
 Note. Observe that (1) the last day is counted, the first is not ; 
 (2) where in counting the days we pass from one month to the 
 next, the whole number of days in the former is taken as thirty 
 for any month of the year. 
 
 46. For rates other than 6 per cent, the multiplier is most 
 readily found by the six per cent method, as follows : 
 
 First find the multiplier for the given time, at 6% ; 
 then 
 
 for 3% take ^ of it ; for 7% add ^ ; 
 
 for 4% subtract J of it ; for 8% add |; 
 
 for 5% subtract ^ ; for 9% add ^, etc. 
 
 For rates higher than 10%, it is easier to form a 12% 
 multiplier with the months as hundredths, and ^ the 
 days as thousandths. 
 
SIMPLE INTEREST 61 
 
 EXAMPLES 23 
 
 Find the interest on : 
 
 1. $ 721.56, for 1 yr. 4 mo. 10 da., at 6%. 
 
 2. $ 54.75, for 3 yr. 24 da., at 5%. 
 
 3. $ 1000, for 11 mo. 18 da., at 7%. 
 
 4. $3046, for 7 mo. 26 da., at 8%. 
 
 5. $ 1821.50, from April 1 to Nov. 12, at 6%. 
 
 6. $ 700, from Jan. 15 to Aug. 1, at 10%. 
 
 7. $316.84, from Oct. 20, 1895, to March 10, 1897, 
 at 7%. 
 
 8. $127.36, from Dec. 12, 1893, to July 3, 1895, 
 
 at 4^%. 
 
 Find the amount of : 
 
 9. $ 3146, for 2 yr. 3 mo. 10 da., at 7%. 
 
 10. $ 1008.80, for 10 mo. 16 da., at 6^%. 
 
 11. $ 2000, for 15 da., at 12^%. 
 
 12. $ 137.60, for 127 da., at 10%. 
 Note. Count 30 da. to a month. 
 
 13. $ 1671.64, from June 1, 1894, to April 1, 1896, 
 at 7%. 
 
 14. $ 250, from June 5, 1896, to Feb. 14, 1897, at 8%. 
 
 15 . $ 340.50, from May 25, 1895, to Sept. 9, 1897, at 9 % . 
 
 16. $25, for 93 da., at 12%. 
 
 17. $ 145.20, for 1 yr. 11 mo. 29 da., at 7%. 
 
62 ALGEBRAIC ARITHMETIC 
 
 18. $ 450, for 3 yr. 2 mo. 21 da., at 8%. 
 
 19. A man engaged in business was making 12^% 
 annually on his capital of $ 16,840. He quit his business, 
 and loaned his money at T-|-%. What did he lose in 
 2 yr. 3 mo. 18 da. by the change ? 
 
 20. A speculator borrowed $9675, at 6%, April 15, 
 1894, with which he purchased flour at $ 6.25 per bbl. 
 May 10, 1895, he sold the flour at $ 7| per bbl., cash. 
 What did he gain by the transaction ? 
 
 21. A man borrows $1000 at 10% interest, and with 
 it buys a note for $ 1100, maturing in 5 mo., but which 
 not being paid when due, runs 1 yr. 6 mo. beyond matu- 
 rity, drawing interest at 6% after maturity. How much 
 does he gain ? 
 
 47. Accurate Interest. The common method of com- 
 puting interest is accurate for whole years; but is not 
 accurate for months, since no month is exactly -^^ of a 
 year ; nor for days, since a day is reckoned as -g^-^ of a 
 year. 
 
 To compute accurate interest, find the interest for 
 years by the common method ; and for any fraction of a 
 year, take as many 365ths of a year's interest as there 
 are days. 
 
 Note. The number of days in each month can be remembered 
 from the following : 
 
 " Thirty days hath September, 
 April, June, and November : 
 All the rest have thirty-one, 
 Except the second month alone, 
 Which has but twenty-eight, in fine, 
 Till leap-year gives it twenty-nine." 
 
SIMPLE INTEREST 63 
 
 Ex. 1. Find the accurate interest on $535 from July 
 25 to Oct. 3, at 6%. 
 
 Time : July, 6 da. 
 Aug., 31 
 
 Sept., 30 $535x6x70 ^ ^ ^^^ ^^^^ 
 
 Oct., 3 10 X 365 
 
 70 da. 
 
 EXAMPLES 24 
 
 2. Find the interest in Ex. 1 by the common method. 
 
 3. Find the exact interest on 3 United States bonds, of 
 $ 1000 each, at 6%, from May 1 to Oct. 15. 
 
 4. What is the exact interest on a $ 500 United States 
 bond, at 5%, from Nov. 1 to April 10? 
 
 Find the exact interest on : 
 
 5. $375, from June 12, 1896, to Dec. 14, 1897, at 7%. 
 
 6. $ 5760, from Nov. 8, 1896, to March 1, 1897, at 6%. 
 
 7. 1 12,085, from Sept. 4, to Dec. 17, at 5%. 
 
 8. $ 1250, from April 1, to Dec. 7, at 6%. 
 
 9. What is the difference between the exact interest 
 for 90 da. on $1,000,000 of 6% bonds and the interest 
 reckoned on the basis of 360 da. to the year ? 
 
 PKOBLEMS IN INTEREST 
 
 48. Interest Formulas. Since the interest and the 
 amount in any problem in interest are percentages of the 
 principal, the relation that these quantities bear to one 
 
64 ALGEBRAIC ARITHMETIC 
 
 another can easily be expressed by the formulas of 
 Art. 32. 
 
 Thus, applying (1) Art. 32 to Ex. 2, Art. 43, we see 
 that: 
 
 To find the interest y 6 = $ 200, r = .15. 
 
 Hence p (the interest) = $ 200 x 0.15 = $ 30. 
 
 To find the amount, & = $ 200, r = 1.15. 
 
 Hence p (the amount) = $ 200 x 1.15 = f 230. 
 
 It will be seen that, in finding the interest, r is the 
 product of two factors ; namely, the rate of interest and 
 the time. We shall obtain much more useful formulas 
 by denoting each of these factors by a letter, and by 
 always using the same letters to denote the same ele- 
 ments. For this purpose we shall use the initials of the 
 names of the elements. 
 
 Thus, p = the principal (base), 
 
 r = the rate of interest (per annum, unless other- 
 wise specified), 
 
 t = the time (expressed in the same denomina- 
 tion as that for which the rate is given), 
 
 i = the interest, 
 
 a = the amount =:p-\-i. 
 
 Note. It should be observed thatp is not here a percentage of 
 some number, as in the preceding formulas. It is the base, of 
 which i and a are percentages. 
 
 49. The formula expressing the rule by which we 
 have computed the interest in all preceding examples is 
 
 '' = prt-, (1) 
 
 in which the product rt is the multiplier (Art. 46). 
 
PROBLEMS 65 
 
 This equation involves four quantities, any one of 
 which can be found if the other three are given. For 
 we have only to solve the equation for the unknown 
 quantity ; then substitute the given values of the other 
 quantities, and perform the indicated operations. 
 
 Ex. 1. What sum of money will gain $84 interest in 
 2 yr., at 7% ? 
 
 1= $84, r = .07, t = 2j p = ? 
 
 Solve (1) for p by dividing its sides by rtf and inter- 
 change the sides ; then 
 
 p-if (2) 
 
 Hence ^ = ^^^=$600. Ans, 
 
 .14 
 
 NoTB. The value of p is the answer to the question : 84 is 14 % 
 of what number ? (Case III. Art. 32.) 
 
 Ex. 2. At what rate will $300 gain $ 60 in 4 years ? 
 p = $ 300, t = $ 60, ^ = 4, r = ? 
 Divide the members of (1) by pt ; then 
 
 r = l. (3) 
 
 Analysis. The interest on $ 300 for 4 yr. is the same as the 
 interest on |1200 for one year, the rate remaining the same. 
 Hence the rate is yf^ ^ , or 5 %. 
 
 Note. The t in all the interest formulas is really an abstract 
 multiplier; its value is the number of years. The product of 
 dollars and years is, of course, impossible. 
 
66 ALGEBRAIC ARITHMETIC 
 
 Second solution and analysis : 
 15 
 
 $300 x^ "^ 
 
 The interest for 1 yr. is \ the interest for 4 yr., or 
 $ 15. Hence the rate is -^^-^^ or 5%. 
 
 Ex. 3. In what time will % 500 gain $ 60, at 4% ? 
 
 p = $500, i = $60, r = .04, t=? 
 Solving (1) for ty we have 
 
 Hence t = ^,,f ^^^^^ = ^ = 3. 3 yr. Ans. 
 
 $500x0.04 20 ^ 
 
 Analysis. For 1 yr. the interest is % 500 x 0.04, or $20. Hence 
 to gain % 60 it will take as many years as $ 20 is contained times 
 in $60, or 3 yr. 
 
 60. From the definition of amount we have the for- 
 mula 
 
 a=p + i. (1) 
 
 Hence, replacing i by its value prt, 
 
 a=p-\-prf, (2) 
 
 or a=p{l + rf), (3) 
 
 Equation (3) expresses the fact that the amount may 
 be obtained directly from the principal by adding 1 to 
 the multiplier by which the interest is obtained. 
 
 Ex. 1. Find the amount of $ 250 for S^ yr., at 6%. 
 The interest is 15% of the principal ; hence the amount 
 
PROBLEMS 67 
 
 is 115% of it. The multiplier for the interest is .15; 
 for the amount it is 1 + .15, or 1.15. 
 
 a = 1 250 X 1.15 = $ 287.5. (Case I.) 
 
 Ex. 2. What principal will amount to $267.90 in 
 
 2yr.,at7%? 
 
 a = $267.90, r = .07, t = 2, p=? 
 Solve (3) for p by dividing its sides by (1 + rt) ; then 
 
 Hence p = f|^ = »|^ = $ 235. (Case III.) 
 
 Subtract p from both sides of (2) ; then 
 a—p=: prtf 
 or prt=a—p. (6) 
 
 Divide both sides of (5) hy pt-, then 
 
 . = ^. (6) 
 
 But a—p = i. 
 
 Hence, substituting, r = — : 
 
 pt 
 
 which is the same as (3) Art. 49. 
 
 Divide the members of (5) by pr ; then 
 
 pp 
 
 (7) 
 
68 ALGEBRAIC ARITHMETIC 
 
 Replacing a—phy t, 
 
 < = -; 
 
 which is the same as (4) Art. 49. 
 
 51. Interest Formulas. From the last two articles we 
 
 obtain the following set of interest formulas. References 
 are given to the corresponding percentage formulas. 
 
 Given the Principal, Rate, and Time ; to find the Interest. 
 
 / = pri. (Case I.) (1) 
 
 Given the Principal, Rate, and Time; to find the Amount. 
 
 Given the Interest (or Amount), Principal, and Time; to 
 find the Rate. 
 
 /^ . ''*_ \ (Case II.) (3) 
 
 Given the Interest, Rate, and Time ; to find the Principal. 
 p=^' (Casein.) (4) 
 
 Given the Amount, Rate, and Time ; to find the Principal. 
 
 P = Yf7i (Casein.) (5) 
 
 Given the Interest (or Amount), Principal, and Rate; to 
 find the Time. 
 
 t = J-, i = ^^zJL. (None.) (6) 
 
 pr pp 
 
PROBLEMS 69 
 
 EXAMPLES 25 (Oral) 
 Find : 
 
 1. Sum that will gain $ 20 in 2 yr., at 5%. 
 
 2. Sum that will amount to $ 228 in 2 yr., at 7%. 
 
 3. Rate at which $ 400 will gain $ 84 in 3 yr. 
 
 4. Time in which $ 200 will gain ^ 56 at 7%. 
 
 5. Rate at which $ 120 will gain $ 60 in 10 yr. 
 
 6. Sum that will amount to $ 350 in 15 yr., at 5%. 
 
 7. Time in which $ 1000 will gain $ 250 at 5%. 
 
 8. Sum that will gain $ 900 in 3 yr., at 3%. 
 
 9. Rate at which $ 5 will gain ^ 1 in 3 yr. 
 
 10. Sum that will amount to $ 260 in 3 yr. 9 mo., at 8%. 
 
 11. Time in which $ 100 will gain $ 15, at 6%. 
 
 12. Rate at which $ 50 will gain ^ 1.50 in 6 mo. 
 
 13. Sum that will gain $ 40 in 6 mo., at 8%. 
 
 14. Rate at which $ 200 will gain $ 25 in 2^ yr. 
 
 15. Time in which $ 75 will gain $ 5, at 4%. 
 
 16. Time in which ^150 will gain $ 21, at 8%. 
 
 17. Rate at which $ 400 will amount to ^ 460 in 2 J yr. 
 
 18. Time it will take $ 700 to amount to $ 749, at 7%. 
 
 19. Rate at which any sum will double itself in 10 yr. ; 
 in 8 yr. 4 mo. ; in 16 yr. ; in 20 yr. 
 
 20. Time it will take money to double itself, at 5%; at 
 6%; at 8%. 
 
70 ALGEBRAIC ARITHMETIC 
 
 EXAMPLES 26 
 Find: 
 
 1. The principal tliat will gain ^ 213 in 5 yr. 10 mo. 
 20 da., at 7%. 
 
 2. Sum that will amount to $1028 in 4 mo. 24 da., 
 at 7%. 
 
 3. Time in which $ 1301.64 will amount to f 1522.92, 
 at 5%. 
 
 4. Kate at which $ 1350 will amount to $ 1539 in 2 yr. 
 4 mo. 
 
 5. Sum that will amount to $ 761.44 in 3 yr. 4 mo. 
 24 da., at 5%. 
 
 6. Rate at which $ 1500 will gain $ 252 in 2 yr. 4 mo. 
 24 da. 
 
 Note. ( = 2f yr. When the months and days are not expres- 
 sible as a fraction with small terms, it is simpler to form a 1% mul- 
 tiplier for the given time, and solve by the second set of formulas (3). 
 Solve Ex. 6 by both methods. 
 
 7. Time in which $ 175.12 will gain $ 6.43, at 6%. 
 
 8. Kate at which $2085 will gain $68.11 in 5 mo. 
 18 da. 
 
 9. Sum that will gain $ 173.97 in 4 yr. 4 mo., at 69^. 
 
 10. Sum that will amount to $ 1596 in 2 yr. 6 mo., 
 at 5J%. 
 
 PRESENT WORTH AND TRUE DISCOUNT 
 
 52. The present worth of any debt is the sum or prin- 
 cipal which at the current rate of interest will amount to 
 that debt when it becomes due. 
 
PRESENT WORTH AND TRUE DISCOUNT 71 
 
 The difference between the amount of the debt and its 
 present worth is called the true discount. 
 
 Problems in present worth and true discount are solved 
 by formula (5) ; in which a is the amount of the debt, 
 andp its present worth (base). 
 
 Ex. 1. A merchant buys a bill of goods for $ 700 on 3 
 months' time. What is the present worth of the debt, 
 money being worth 6% ? 
 
 P = 
 
 iI55L_ = iI22=$ 689.66. Ans. 
 
 l-^rt 1+ 0.015 1.015 
 
 EXAMPLES 27 
 
 2. What is the present worth and discount of a debt 
 of $ 1000 due in 1 yr. 6 mo., the current rate of interest 
 being 6% ? 
 
 3. A merchant buys goods for $4200 on 4 mo. credit, 
 but is offered a discount of 3% for cash. If money is 
 worth ^% a month, what is the difference ? 
 
 4. Bought a house and lot for $19,500 cash, and sold 
 .them for $22,000, payable ^ in cash and the remainder 
 in 1 yr. 6 mo. How much did I gain, computing dis- 
 count at 6% ? 
 
 5. A merchant holds two notes, one for $356.25, due 
 Dec. 1, 1897, and the other for $497.50, due Feb. 1, 1898. 
 What would be due him in cash on both notes Sept. 15, 
 
 1897, discounting at 6% ? 
 
 6. Which is the more profitable, to buy coal at $8.75 
 per ton on 6 mo. credit, or at $8.60 on 2 mo. credit, 
 money being worth 7% ? 
 
72 ALGEBRAIC ARITHMETIC 
 
 7. What sum must I put at interest at 8% to liquidate 
 a debt of $ 2500 due 3 yr. hence ? 
 
 8. Bought a house for $ 7500, payable in 4 mo., and 
 sold it for $7500 cash. If money is worth 6%, what 
 did I gain? 
 
 9. Find the difference between the interest and true 
 discount of $270 for 9 mo., at 8%. 
 
 BANK DISCOUNT 
 53. Promissory Notes. 
 
 $ 150.00. Cambridge, Mass., July 28, 1897. 
 
 Ninety days after date, I promise to pay Charles Bond 
 One Hundred and Fifty Dollars, value received. 
 
 John Brainard. 
 
 $2000.00. Berkeley, Cal., May 15, 1896. 
 
 Sixty days after date, I promise to pay to the order 
 
 of Frank Barnes Two Thousand Dollars, value received, 
 
 with interest at 6%. 
 
 W. B. Slack. 
 
 The above are examples of promissory notes ; so called 
 because they contain a promise to pay a certain sum, at 
 a specified time, for value received. 
 
 The person who signs a note is called the maker; 
 the person to whom or to whose order it is to be paid 
 is called the payee. The sum named in a note is called 
 its face. 
 
 In most states a note is not legally due till three days 
 after the expiration of the time specified in the note. 
 
BANK DISCOUNT 73 
 
 These are called days of grace. They are counted in 
 by bankers in discounting notes. A note is said to 
 mature on the last day of, grace. 
 
 To find the date of maturity when the time is ex- 
 pressed in days, count forward from the date of the note, 
 the specified number of days plus the days of grace, 
 reckoning the actual number of days in the months 
 passed over. When the time is expressed in months, 
 calendar months are always to be understood. 
 
 A note, like the second given above, made payable to 
 the order of the payee, or one made payable to hearer, 
 is a negotiable note ; that is, it can be bought and sold. 
 
 54. Bank Discount. If the holder of a negotiable note 
 sells it to a bank, he will receive the amount of the note 
 at maturity minus a percentage of that sum, called the 
 bank discount, which is computed at a certain per cent 
 per month or per annum. 
 
 The sum received for the note is called the proceeds 
 or avails of the note. 
 
 Note. Banks reckon 12 mo. of 30 da. each, or 360 da. to a 
 year; and count the actual number of days in a given time. 
 
 Ex. 1. Find the discount and proceeds of the first note 
 in the last article, if discounted at a bank Aug. 31, at 1 % 
 per month. 
 
 The term of discount = 93 da. — (3 da. in July 4- 31 
 da. in Aug.) = 59 da. 
 
 Discount = interest on $ 150 f or.59 da. at 1 % per mo. 
 
 = $ 150 x0.019| = f 2.95. 
 Proceeds = $ 150 - $ 2.95 -= $ 147.05. 
 
74 ALGEBRAIC ARITHMETIC 
 
 Ex. 2. Find the discount and proceeds of the second 
 note, discounted on the day it was made, at 6%. 
 
 Interest on $ 2000 for 2 mo. 3 da. = f 2000 x. 0105 = ^21. 
 Amount of note at maturity = $ 2021. 
 Discount = interest on $ 2021 for 2 mo. 3 da. = $ 21.22. 
 Proceeds = $ 2021 - $ 21.22 = ^ 1999.78. 
 
 Note. The present worth of the note at 6% discount is, of 
 course, its face ; and the true discount on the amount at maturity 
 is $21. The excess of the bank discount above the true discount 
 is equal to the interest on the true discount for the given time 
 ($21x.0105 = $.22). 
 
 EXAMPLES 28 
 
 Find the date of maturity, the term of discount, and 
 the proceeds of the following : 
 
 3. $ OST^^j^. Boston, July 27, 1897. 
 
 Three months after date, I promise to pay to the order 
 of M. Levering Nine Hundred Fifty-seven and rf^ Dol- 
 lars, value received. 
 
 Discounted Aug. 10, at 8%. T. J. Jennings. 
 
 4. $916^^. Glendalb, Cal., Feb. 5, 1896. 
 
 Two months after date, we jointly and severally 
 promise to pay C. R. Crowley, or order, Nine Hundred 
 Sixteen and y^^ Dollars, value received, with interest 
 at 8%. 
 
 Discounted Feb. 21, at 10%. James Little. 
 
 T. B. Long. 
 
ANNUAL INTEREST 76 
 
 5. $700.00. New York, April 10, 1897. 
 Four months after date, I promise to pay to the order 
 
 of Edward Brill Seven Hundred Dollars, value received. 
 Discounted June 10, at 8%. ^ ^ Gorden. 
 
 Write the following in the form of promissory notes, 
 and find the proceeds : 
 
 6. Note of $650, given Jan. 8, 1897, payable 60 da. 
 after date; discounted Feb. 1, at 1% per month. 
 
 7. Note of $1840, given July 5, 1897, payable in 
 30 da. ; discounted July 8, at 1 % per month. 
 
 8. Note of $2550, given May 3, 1897, payable in 
 3 mo., with interest at 6% ; discounted May 3, at 6%. 
 
 Note. The note bears interest for 3 mo. 3 da. ; but the term 
 of discount is 95 da., which is 3 mo. 5 da., as banks reckon time. 
 
 9. Note of $56.25, given July 29, 1897, payable in 
 6 mo., with interest at 10% ; discounted Oct. 1, at 1% 
 per month. 
 
 ANNUAL INTEREST 
 
 55. If a note reads " with interest payable annually,'^ 
 or " with annual interest," the interest is due at the end 
 of each year, and thereafter draws simple interest until 
 paid. Interest so computed is called annual interest. 
 
 Ex. 1. 
 
 $ 1000.00. Chicago, Jan. 13, 1897. 
 
 Three and one-half years from date, I promise to pay 
 Henry Ames, or bearer. One Thousand Dollars, for value 
 received, with interest at 6%, payable annually. 
 
 M. J. Clarkson. 
 
76 ALGEBRAIC ARITHMETIC 
 
 Find the amount of the note at maturity (not counting 
 days of grace), no interest having been paid. 
 
 The simple interest on the principal for 3^ yr. = $ 210. 
 The annual interest on the principal = $ 60. 
 
 The first year's interest remains unpaid 2^ yr. ; the 
 second, li yr. ; the third, i yr. This is equivalent to 
 the use of $ 60 for (2^ + 1 1- + i) yr., or 4^ yr. 
 
 Interest on $ 60 for ^ yr. = $ 16.20. 
 
 Total interest = $ 210 + 1 16.20 = $ 226.20. 
 
 Amount due at maturity = $ 1000 + $ 226.20 = $ 1226.20. 
 
 EXAMPLES 29 
 
 Find the annual interest of : 
 
 2. $8000for 5yr., at6%. 
 
 3. $ 1500 for 4 yr., at 7%. 
 
 4. $3500 for 10 yr., at 8%. 
 
 5. $675for9iyr., at8%. 
 
 6. $800 for 4 yr., at 7%. 
 
 COMPOUND INTEREST 
 
 56. In compound interest, the interest, when due, is 
 added to the principal, thus forming a new principal 
 for the next period. The interest may be compounded 
 with the principal annually, semiannually, or quarterly, 
 according to agreement. Annual periods are understood 
 unless otherwise stated. 
 
COMPOUND INTEREST 77 
 
 Ex. 1. What is the compound interest of $750 for 
 1 yr. 3 mo., at 8% per annum, payable semiannually ? 
 
 $ 750 1st principal. 
 1.04 
 
 $ 780 amount at end of J yr., 2d principal. 
 1.04 
 
 f 811.20 amount at end of 1 yr., 3d principal. 
 1.02 
 
 $ 827.42 amount in 1 yr. 3 mo. 
 750.00 
 
 $ 77.42 compound interest. 
 
 EXAMPLES 30 
 
 2. Find the simple interest, annual interest, and com- 
 pound interest of $2500 for 6 yr., at 6%. 
 
 3. Find the amount of $350 in 3 yr., at 7% com- 
 pound interest. 
 
 Find the compound interest on : 
 
 4. $ 1200, for 3 yr., at 5%, payable annually. 
 
 5. $864.50, for 4 yr., at 8%, payable annually. 
 
 6. $ 680, for 2 yr., at 7%, payable semiannually. 
 
 7. $460, for 1 yr. 5 mo. 18 da., at 6 %, payable quar- 
 terly. 
 
 8. $ 1250, for 3 yr. 7 mo. 18 da., at 5 %, payable semi- 
 annually. 
 
 9. $ 790, for 9 mo. 27 da., at 8%, payable quarterly. 
 
78 ALGEBRAIC ARITHMETIC 
 
 10. What sum placed at simple interest for 3 yr. 10 mo. 
 18 da., at 7 %, will amount to the same as $ 1500 placed 
 at compound interest for the same time, and at the 
 same rate, payable semiannually ? 
 
 11. How much must a father, at the birth of his son, 
 set apart for his benefit, so that with the interest at 7%, 
 compounded semiannually, it may amount to $10,000 
 when his son shall become 21 years of age ? 
 
 PARTIAL PAYMENTS 
 
 57. When partial payments are made on notes or other 
 obligations bearing interest, they may be applied in either 
 of two ways ; namely : 
 
 (1) To the debt of principal, leaving the interest un- 
 paid till the time of final settlement. 
 
 (2) To the debt of interest first, and the remainder, if 
 any, to the principal. 
 
 There are other methods which are formed by various 
 combinations of these two. 
 
 It will be seen from the following articles that by the 
 first method the debt draws simple interest ; by the sec- 
 ond, compound interest. 
 
 Note. All interest is in effect compounded when it is paid, 
 since it allows the lender to loan it again and so draw interest on 
 interest, while, If not paid, the debtor has the use of the interest 
 money without paying interest on it. 
 
 The acknowledgment of a partial payment, stating 
 the time and amount of the same, is written on the back 
 of the note j and is called an indorsement. 
 
PARTIAL PAYMENTS 79 
 
 58. The first method mentioned in the last article is 
 commonly used when partial payments are made on mer- 
 cantile accounts which are past due, and on notes con- 
 taining the words " with interest " and running for a year 
 or less. It is called the Merchants' Rule. 
 
 Ex. 1. What is due Oct. 1, 1897, on a note of $ 750, 
 with interest at 6%, dated June 1, 1897, and bearing the 
 following indorsements : July 1, ^ 100 ; Aug. 19, $ 250 ? 
 
 Interest on 1st principal for 1 mo. =$ 750 x .005 = $ 3.75 
 
 Interest on 2d principal for 1 mo. 18 da. = 650 x .008 = 5.20 
 Interest on 3d principal for 1 mo. 12 da. = 400 x .007 = 2.80 
 
 Total interest $11.75 
 
 Amount due Oct. 1 = $400 + $11.75 = $411.75. 
 
 The same result is obtained by the following method 
 of procedure, which is the usual way of stating 
 
 The Merchants' Rule 
 
 Find the amount of the note or debt from its date to the 
 time of settlement. 
 
 Find the amount of each payment from its date to the 
 time of settlement. 
 
 Subtract the sum of the amounts of the payments from 
 the amount of the note or debt. 
 
 Thus, in the above example, 
 Amount of $ 750 for 4 mo. is . . . $ 765.00 
 
 Amount of $ 100 for 3 mo. is . . . $ 101.50 
 Amount of $ 250 for 1 mo. 12 da. is 251.75 353.25 
 
 Amount due S 411.75 
 
80 ALGEBRAIC ARITHMETIC 
 
 EXAMPLES 31 
 
 Write out the following in proper form on paper, 
 placing the indorsements on the back, and solve by the 
 Merchants^ Eule : 
 
 2. Face, ^1500. Date, Jan. 1, 1895. Interest, 6%. 
 Indorsements: Aug. 7, 1895, $500; Dec. 7, 1895, $500. 
 What is due Jan. 1, 1896? 
 
 3. Face, $480. Date, March 3, 1894. Interest, 7%. 
 Indorsements: Sept. 3, 1894, $196.80; March 3, 1895, 
 $ 214. Sept. 3, 1895, paid the amount due. Find it. 
 
 4. Face, $1000. Date, July 20, 1894. Interest, 8%. 
 Indorsements : March 5, 1895, $ 50 ; July 5, 1895, $ 450. 
 What was still due on the date of last payment ? 
 
 5. Face, $1230. Date, Jan. 1, 1896. Interest, 51%. 
 Indorsements : March 1, 1896, $ 98 ; June 7, 1896, $ 500 ; 
 Sept. 20, 1896, $290; Dec. 10, 1896, $100. What is 
 due Jan. 1, 1897 ? 
 
 6. Face, $800. Date, March 1, 1896. Interest, 10%. 
 Indorsements: Aug. 10, 1896, $200; Sept. 1, 1896, $50; 
 Jan. 1, 1897, $ 15. What was due March 1, 1897 ? 
 
 59. The second method of applying partial payments, 
 mentioned in Art. 57, is generally employed in the case 
 of interest-bearing notes that run for more than a year ; 
 but is also frequently used when the time is less than a 
 year. 
 
 Under the application of this method, three cases may 
 arise ; namely, a payment may be (1) equal to, (2) greater 
 than, or (3) less than, the interest accumulated at the 
 time of the payment. 
 
PARTIAL PAYMENTS 81 
 
 In the first case, the payment just cancels the interest, 
 and the principal, or interest-bearing debt, remains un- 
 changed. The debtor, in the end, pays just as much as 
 if such payment had been deferred until he was able to 
 make a payment large enough to diminish the principal; 
 and, meanwhile, he loses the use of the payment. 
 
 In the second case, the principal is diminished ; hence 
 the total interest on the debt is diminished by such 
 payment. 
 
 In the third case, if the unpaid balance of the interest 
 were added to the principal, the interest-bearing debt 
 would be increased. This would increase the total in- 
 terest ; and besides losing the use of such payment, the 
 debtor would actually have more to pay, in the end, than 
 if he had kept the money till he was able to make a 
 sufficiently large payment to reduce the principal. 
 
 It was to prevent such manifest injustice to the debtor 
 that the Supreme Court of the United States adopted the 
 following rule : 
 
 The United States Rule 
 
 Find the amount of the principal to the time when the 
 payment, or the sum of the payments, equals or exceeds 
 the interest. 
 
 From this amount deduct the payment or sum of the 
 payments. 
 
 Consider the remainder as a new principal, and proceed 
 as before. 
 
 Ex. 1. A note of $500, dated Feb. 1, 1895, and bear- 
 ing interest at 6%, is indorsed as follows: May 1, 1895, 
 o 
 
82 ALGEBRAIC ARITHMETIC 
 
 ^40; Nov. 14, 1895, ^8; April 1, 1896, $18; May 1, 
 1896, $30. What was due Sept. 16, 1896 ? 
 
 $ 500.00 1st principal. 
 
 7.50 interest to May 1. 
 $507.50 
 
 40.00 1st payment. 
 $ 467.50 2d principal. 
 
 15.04 interest to Nov. 14. 
 
 10.67 interest to April 1. 
 $493.21 
 
 26.00 2d and 3d payments. 
 $467.21 3d principal. 
 
 2.34 interest to May 4. 
 $ 469.55 
 
 30.00 4tli payment. 
 
 $ 439.55 4tli principal. 
 
 9.89 interest to Sept. 16. 
 $ 449.44 amount due. 
 
 Note. It will be seen from the note to Art. 57 that by the 
 United States Rule the interest is compounded as often as a pay- 
 ment is made which equals or exceeds the unpaid interest. 
 
 EXAMPLES 32 
 
 2 to 6 inclusive. Solve Ex. 2 to 6 inclusive of the last 
 article by the United States Rule, and compare the results 
 with those obtained by the Merchants' Rule. Account 
 for the difference in the results. 
 
 7. What was due Aug. 5, 1896, on a note for $ 2500, 
 with interest at 7%, dated Aug. 5, 1895, and bearing the 
 
PARTIAL PAYMENTS 83 
 
 following indorsements : Jan. 1, 1896, $ 500 ; March 10, 
 
 1896, $ 750 ? 
 
 8. A note for $ 16,500, dated May 20, 1896, and bear- 
 ing interest at 7%, is indorsed as follows : Sept. 1, 1896, 
 $ 25 ; Oct. 14, 1896, $ 150 ; March 20, 1897, $ 45 ; July 5, 
 
 1897, $ 300. Find the amount due Nov. 11, 1897. 
 
 9. Find the amount due Jan. 1, 1897, on a note for 
 $497.39, with interest at 6%, dated March 15, 1894, and 
 indorsed as follows: Nov. 3, 1894, $57.50; June 15, 
 1895, $ 22.25 ; Aug. 1, 1895, $ 125 ; Sept. 15, 1895, $ 175. 
 
 10. A note for f 10,000 runs 4 yr., at 8% interest, on 
 which were made quarterly payments of $500. What 
 was the amount due at the time of settlement ? 
 
 11. On a note for $1000, at 6% interest, payments 
 were made as follows: in 1 yr., $50; in 1 yr. 6 mo., 
 $250; in 2 yr., $224; in 2 yr. 8 mo., $20; in 2 yr. 
 10 mo., $ 110. Find the amount due at the end of 4 yr. 
 
CHAPTER V 
 
 PROPORTION. PARTNERSHIP. AVERAGE OF 
 PAYMENTS 
 
 60. Ratio. The relative magnitude of two numbers, 
 measured by the quotient of the first divided by the 
 second, is called their ratio. 
 
 Thus the ratio of 12 to 3 is 4 ; of 9 da. to 4 da. is 2\ ; 
 of 3 pt. to 1 gal., is f . 
 
 Concrete numbers of different kinds can have no ratio 
 to one another. For example, we cannot compare feet 
 and pounds with respect to their magnitude. Moreover, 
 concrete numbers of the same kind must be expressed 
 in the same unit before their ratio can be taken. 
 
 A ratio is always an abstract number, and may be ex- 
 pressed as a rate per cent (Case II, Art. 32). Thus, in 
 percentage, the rate is the ratio of the percentage to 
 the base. 
 
 The ratio of any two numbers a and b is expressed 
 
 by the notation a : 6 or - ; and a is called the first term 
 
 of the ratio, or the antecedent; and 6, the second term, 
 or the consequent. 
 
 The product of two or more ratios is called a compound 
 ratio. 
 
 Thus the ratio compounded of the ratios 3 : 4 and 5 : 7 
 is 16 : 28 ; since J X ^^ = if . 
 
 84 
 
PROPORTION 85 
 
 PROPORTION 
 
 61. A statement of the equality of two ratios is called 
 a proportion, and is expressed in three ways ; thus : 
 
 
 t = A, 
 
 4 
 
 : 8 = 6 : 12, 
 
 4 
 
 : 8 : : 6 : 12. 
 
 The last is the usual notation, and is read "4 is to 8 
 as 6 is to 12." 
 
 The four terms of a proportion are said to be propor- 
 tional or in proportion. 
 
 Thus 4, 8, 6, 12, are proportional. 
 
 The first ratio of a proportion is called the first couplet ; 
 the second ratio, the second couplet. 
 
 The first and fourth terms of a proportion are called 
 extremes ; and the second and third terms, means. 
 
 62. Denote any four proportional numbers by the 
 letters a, 6, c, d; then 
 
 a:b:: c:d, (1) 
 
 Multiply the sides of this equation by bd ; then 
 
 ad=bc. (3) 
 
 Hence the law : 
 
 (i.) The product of the extremes of any proportion is 
 equal to the product of the means. 
 
86 ALGEBBAIC ARITHMETIC 
 
 If equation (2) or (3) be solved for each of the quan- 
 tities in succession, we obtain 
 
 6c , be x.s 
 
 a = -, (^=-; (4) 
 
 a a 
 
 , ad ^ ad .f.. 
 
 6 = —, c = — . (5) 
 
 c 
 
 From (4) we have the law : 
 
 (ii.) The product of the means divided by either extreme 
 will give the other extreme. 
 
 From (5) we have the law : 
 
 (iii.) The product of the extremes divided by either mean 
 will give the other mean. 
 
 From (i.) it follows that a proportion is verified, or 
 proved, by showing that the product of the extremes is 
 equal to the product of the means. 
 
 From (ii.) and (iii.) it follows that if three terms of a 
 proportion are given, the fourth term can be found. 
 
 EXAMPLES 33 
 
 Verify the following proportions : 
 
 1. 12 : 1728 : : 1 : 144. 
 
 2. 27.03 : 9.01 :: 16.05 : 5.35. 
 
 3. i:*::f:A. 
 
 Find the value of x in each of the following propor- 
 tions : 
 
 4. 8: 62:: 20: a;. 6. «: 20 : : 120 : 50. 
 6. 12:a;::l:144. 7. 80:4::a;:i. 
 
PROPORTION 87 
 
 8. 2.5: 62.5:: 5: a;. 12. | : a; : : J : 59.0625. 
 
 9. 175.35:a;::i:f 13. tV^I^-^^-I- 
 
 10. 4|:a;::9f :27^. 14. a; : 38J : : 8| : 76^. 
 
 11. ic : 9.01 :: 16.05 : 5.35. 15. 7.5 : 18 : : a; : 7tV 
 
 63. If four numbers a, h, c, c?, are proportional, that 
 
 is, if 
 
 a : 6 : : c : d, (1) 
 
 then it is also true that 
 
 6 : a : : d : c, (2) 
 
 a.ci'.hidj (3) 
 
 c'.a.'.d'.h. (4) 
 
 This is easily proved ; for, from (1), we know that 
 
 ad = be, (Art. 62.) (5) 
 
 or be = ad. (6) 
 
 Divide the sides of (6) by ac ; then 
 
 - = - , or b: a: :d: c. 
 a c 
 
 Similarly, dividing the sides of (6) by ab gives (4); 
 and dividing the sides of (5) by cd gives (3). 
 
 Four other proportions can be obtained from the four 
 given by interchanging the couplets. 
 
 Exercise. Express the proportionality of the num- 
 bers 5, 15, 7, 21, in as many ways as possible. 
 
 64. When any substance is sold at a fixed price per 
 pound, the cost of any amount of it is so related to its 
 
88 ALGEBRAIC ARITHMETIC 
 
 weight that when the weight is doubled the cost is also 
 doubled, when the weight is halved the cost is also 
 halved, and so on. This relation is expressed by saying 
 that the cost and the weight are proportional. 
 
 For example, if coffee is 20 cents per lb., 3 lb. will 
 cost 60 cents, and 5 lb. will cost 100 cents ; and 
 
 3 lb. ^ 20 ct. 
 
 61b. 100 ct.' 
 
 or 3 lb. : 5 lb. : : 20 ct. : 100 ct. 
 
 And, in general, if a pounds of the coffee cost f h and 
 c pounds of it cost $ d, then 
 
 alb. :clb. ::$6:$c?; 
 
 which is the symbolical statement of the fact that the 
 ratio of any two values of the cost is equal to the ratio 
 of the corresponding weights. 
 
 Definition. One quantity is said to be proportional 
 to another when the two are so related that the ratio of 
 any two values of the one is equal to the ratio of the 
 corresponding values of the other. 
 
 It should be observed that in the proportion, corre- 
 sponding values are both antecedents or both consequents. 
 
 65. Definition. The reciprocal of a number is 1 di- 
 vided by that number. Hence the reciprocal of a fraction 
 is the fraction inverted, and the reciprocal of a ratio is 
 the ratio formed by interchanging its antecedent and 
 consequent. 
 
 If the number of men to do a given piece of work be 
 doubled, the time required to do the work will be halved j 
 
PROBLEMS 89 
 
 if 3 times as many men work, the time required will be 
 ^ as long ; if only \ as many men work, it will take 4 
 times as long; and so on. This relation is expressed 
 by saying that the time required to do the work is in- 
 versely proportional to the number of men working. 
 
 For example, if 1 man can do a piece of work in 48 da., 
 4 men can do it in J of 48 da., or 12 da., and 6 men can 
 do it in ^ of 48 da., or 8 da. 
 
 . ^ 4 men 8 da. 
 
 Also, = — -— — t 
 
 ' 6 men 12 da. 
 
 or 4 men : 6 men : : 8 da. : 12 da. ; 
 
 from which it will be seen that the ratio of the two num- 
 bers of men is equal to the reciprocal of the ratio of the 
 corresponding numbers of days. 
 
 Definition. One quantity is said to be inversely 
 proportional to another when the ratio of any two values 
 of the one is equal to the reciprocal of the ratio of the 
 corresponding values of the other. 
 
 Observe that in the case of inverse proportionality, 
 corresponding values are both extremes or both means. 
 
 PROBLEMS IN SIMPLE PROPORTION 
 
 66. A statement of the equality of two simple ratios 
 is called a simple proportion. 
 
 Problems involving two pairs of quantities, propor- 
 tional or inversely proportional, three of which quantities 
 are given, can be solved by simple proportion. 
 
 Ex. 1. If 20 lb. of sugar cost $1.20, what will 45 lb. 
 cost ? 
 
90 ALGEBRAIC AEITHMETIC 
 
 Let $ X denote the cost of 45 lb. of sugar j then, sincej^ T 
 the cost is proportional to the weight, \ ^ i^ 
 
 201b. :451b.:: $1.20 :$a;. ^* i "^N, 
 
 Hence $ a; = ^^ x $ 1.20 = $ 2.70. ^ns. 
 201b. 
 
 Note. In concrete problems the product of the extremes or of 
 the means will be the product of two concrete numbers, and this 
 has no meaning. A multiplier is necessarily an abstract number. 
 Hence in the above example we cannot multiply $ 1.20 by 45 lb. ; 
 but we can multiply it by the ratio of 45 lb. to 20 lb. , for all ratios 
 are abstract numbers (Art. 60). 
 
 Since the ratio of 45 lb. to 20 lb. is the same as the ratio of 45 
 to 20, it is unnecessary to retain concrete denominations in the 
 solution, except in the case of the number of the same kind as 
 the answer. Hence we may proceed as follows : 
 
 .06 
 
 The required quantity may be taken as any one of the four 
 terms of the proportion, but it is customary to write it as the 
 fourth. 
 
 Solution by Analysis. If 20 lb. of sugar cost $1.20, 1 lb. 
 will cost -^^ of $1.20, or 6^, and 45 lb. will cost 45 times 6)*, or 
 $2.70. 
 
 Ex. 2. In how many days can 12 men do a piece of 
 work that 60 men can do in 8 da. ? 
 
 The number of men and the number of days are in- 
 versely proportional; hence 
 
 12 men : 60 men : : 8 da. : a: da. 
 
PROBLEMS 91 
 
 Solution by Analysis. Since it takes 60 men 8 da., it will 
 take 1 man 60 times 8 da., and it will take 12 men ^^ of 60 x 8 da., 
 Q^ 60x8 da. 
 12 
 
 EXAMPLES 34 
 
 Note. All problems in proportion can be solved by analysis. 
 The learner should become familiar with both methods of solution. 
 
 3. If 20 yd. of cloth cost $ 180, find the cost of 45 yd. 
 
 4. If 18 bu. of wheat make 4 bbl. of flour, how many 
 barrels will 200 bu. make ? 
 
 5. How many men will be required to build 32 rods 
 of wall in the same time that 5 men can build 10 rods ? 
 
 6. If 5 sheep can be bought for ^ 20.75, how many 
 sheep can be bought for $ 398.40 ? 
 
 7. When 10 bbl. of flour cost $ 112.50, what will be 
 the cost of 476 bbl. of flour ? 
 
 8. If a train runs 30 mi. in 50 min., in what time will 
 it run 260 mi. ? 
 
 9. If a horse travels 12 mi. in 1 hr. 36 min., how far 
 at the same rate will he travel in 15 hr. ? 
 
 10. How many days will 12 men require to do a piece 
 of work that 95 men can do in 7^ da. ? 
 
 11. If f of an acre of land cost $60, what will 45f 
 acres cost? 
 
 12. If by selling $5000 worth of dry goods a mer- 
 chant gains $ 456.25, what amount must he sell to gain 
 $ 1000 ? 
 
92 ALGEBRAIC ARITHMETIC 
 
 13. If a pasture will feed 120 horses 81 da., how many- 
 horses will it feed 108 da. ? 
 
 14. If a business yields $ 700 profits in 1 yr. 8 mo., in 
 what time will it yield f 1050 profits at the same rate ? 
 
 15. If it takes a train 2^ hr. to go a certain distance at 
 the rate of 27 mi. an hour, how long will it take to go the 
 same distance at the rate of 21 mi. an hour ? 
 
 16. If 15 men can build a wall in 6 da., how many 
 men would be required to build it in 4^ da.? 
 
 17. If a quantity of provisions is sufficient to support 
 225 men 25 da., how many days will it support 75 men ? 
 
 18. If 12 men earn $ 78 in 4 da., how many men will 
 earn $ 58 J in the same time at the same wages ? 
 
 COMPOUND PROPORTION 
 
 67. A statement of the equality of two compound 
 ratios, or of a compound ratio and a simple one, is called 
 a compound proportion. 
 
 For example, the equation 
 
 may be expressed as a compound proportion ; thus : 
 
 3:4 
 
 2 
 
 ;J}::9:30; 
 
 in which form the product of the ratios written one above 
 the other is understood. By taking the product, the pro- 
 portion is reduced to a simple one. , Thus the above 
 
 becomes 
 
 6 : 20 : : 9 : 30. 
 
COMPOUND PROPORTION 93 
 
 From the meaning of a compound proportion, it follows 
 that the product of the extremes is equal to the product 
 of the means, as in the case of a simple proportion; 
 hence a missing term is found in the same way. 
 
 Ex. 1. Find the value of x in the proportion 
 
 3 
 
 The proportion means that 
 
 5 X 3 : 6 X 7 : : 10 : aj. 
 
 "Vr-^- 
 
 Hence ^^ 6x7x10 ^^3 
 
 5x3 
 
 68. The amount of work done by a given number of 
 men is proportional to the time, and the amount of work 
 done in a given time is proportional to the number of 
 men. If both the time and the number of men vary, the 
 amount of work is proportional to their product. 
 
 ^or example, 6 men in 4 da. can do 24 times as much 
 work as 1 man in 1 da. 
 
 Hence, if 1 man can dig 2 rd. of ditch in 1 da., 6 men 
 in 4 da. can dig 6x4x2 rd., or 48 rd., and 5 men in 
 3 da. can dig 5 x 3 x 2 rd., or 30 rd. 
 
 ., 6men 4da._ 48rd. 
 
 ^^'''' 5 men ^3 da. ~ 30 rd.' 
 
 6 men: 5 men) .^ , ^^ , 
 or . , o ^ ^ : : 48 rd. : 30 rd. ; 
 
 4 da. : 3 da. ) 
 
 that is, the ratio of the two amounts of work is equal to 
 the product of the ratios of the corresponding numbers 
 of men and days. 
 
94 ALGEBRAIC ARITHMETIC 
 
 Definition. One quantity is said to be proportional 
 to the product of two or more other quantities when the 
 ratio of any two values of that quantity is equal to the 
 product of the ratios of the corresponding values of 
 the others. 
 
 PROBLEMS IN COMPOUND PROPORTION 
 
 69. Ex. 1. If 18 men build 126 rd. of wall in 60 da., 
 how many rods will 6 men build in 110 da. ? 
 Symbolical statement : 
 
 18 men : 6 men ] ^ o^ i i 
 
 60 da. :110da.r = ^^^^^- = ^^^- 
 
 Hence g. rd. = ^^ >< ,^f xj^^ ^^- = 77 rd. 
 18 X 60 
 
 Solution by Analysis. One man in 60 da. will build — of 
 
 126 
 126 rd., or — rd. : 6 men in the same time will build 6 times as 
 18 
 
 many rods, or — rd. ; in 1 da. the 6 men will build — of 
 
 ''is 60 
 
 126ji6 ^ 126x_6 j,^ ^^^ .j^ j^Q ^^ ^^ ^.j^ ^^.j^ j^Q 
 
 18 18 X 60 ' 
 
 times as many rods as in 1 da., or — -^ — ^-- — rd. 
 
 18 X 60 
 
 Ex. 2. If 18 men build 126 rd. of wall in 60 da., how 
 many men will it take to build 77 rd. in 110 da. ? 
 
 126 rd. : 77 rd. ) .^ 
 
 ■fiA J ar\A > : : 18 men : a; men. 
 
 110 da. : 60 da. ) 
 
 xj«„«« r. r«^« 77 X 60 X 18 men ^ _^^ 
 
 Hence x men = — — — --— = b men. 
 
 126 X 110 
 
PROBLEMS 95 
 
 Explanation op the Method. In problems of this class all 
 the numbers occur in like pairs, except one which is of the same 
 kind as the answer. Take this as the third term of the proportion ; 
 then the fourth term, when found, is the answer. Consider each 
 of the pairs of numbers separately, forming a first couplet from 
 each, as in simple proportion. 
 
 In solving by analysis, begin with the number like the answer, and 
 consider the effect upon it of the given change in each of the other 
 numbers, separately. Thus, in the analysis of Ex. ] , we considered 
 the effect upon the number of rods caused first by the change in 
 the number of men from 18 to 6 (the time remaining unchanged), 
 then by the change in the number of days from 60 to 110 (the num- 
 ber of men remaining unchanged). In each case we first reason 
 to 1 of the number that is changed. 
 
 EXAMPLES 35 
 
 3. If 8 men earn $ 320 in 8 da., how much will 12 men 
 earn in 4 da. ? 
 
 4. If it costs $ 41.25 to pave a sidewalk 5 ft. wide and 
 75 ft. long, what will it cost to pave a similar walk 8 ft. 
 wide and 566 ft. long ? 
 
 5. If 16 horses consume 48 bu. of oats in 12 da., how 
 many bushels will 20 horses consume in 8 wk. ? 
 
 6. What sum of money will gain $300 in 8 mo., if 
 $ 800 gain | 70 in 15 mo. ? 
 
 7. If 10 men can cut 46 cords of wood in 18 da., work- 
 ing 10 hr. a day, how many cords can 40 men cut in 
 24 da., working 9 hr. a day ? 
 
 8. What is the cost of 36^ yd. of cloth IJ yd. wide, if 
 ^ yd., If yd. wide, cost $ 3.37^ ? 
 
96 ALGEBRAIC ARITHMETIC 
 
 9. A contractor employs 45 men to complete a work 
 in 3 mo. What additional number of men must lie 
 employ to complete the work in 2^ mo. ? 
 
 10. How many days will 21 men require to dig a ditch 
 80 ft. long, 3 ft. wide, and 8 ft. deep, if 7 men can dig a 
 ditch 60 ft. long, 8 ft. wide, and 6 ft. deep in 12 da. ? 
 
 11. When the shadow of a post 10 ft. 6 in. high is 
 12 ft. 3 in. long, what is the length of shadow of a post 
 8 ft. 9 in. high ? 
 
 12. The shadow of a post 16 ft. 3 in. high is 5 ft. 5 in. 
 long. What height of post will give a shadow 3 ft. 4 in. 
 long? 
 
 13. If a vat 16 ft. long, 7 ft. wide, and 15 ft. deep holds 
 384 bbl., how many barrels will a vat 17^ ft. long, lOJ ft. 
 wide, and 13 ft. deep hold ? 
 
 14. What is the weight of a block of granite 8 ft. long, 
 
 4 ft. wide, and 10 in. thick, if a similar block 10 ft. long, 
 
 5 ft. wide, and 16 in. thick weighs 5200 lb. ? 
 
 15. If it costs $15 to carry 20 tons IJ mi., what will 
 it cost at the same rate to carry 400 tons ^ mi. ? 
 
 16. If 6 laborers can dig a ditch 34 yd. long in 10 da., 
 how many days will 20 laborers require to dig a similar 
 ditch 170 yd. long? 
 
 17. If a man walk 192 mi. in 6 da., walking 8 hr. a 
 day, how far can he walk in 18 da., walking 6 hr. a day ? 
 
PARTNERSHIP 97 
 
 PARTNERSHIP 
 
 70. The association of two or more persons for the 
 purpose of carrying on business is called partnership. 
 
 The persons thus associated are called partners, and 
 together they form a company or firm. 
 
 The money or property invested is called the capital or 
 stock. 
 
 The money and property of all kinds belonging to a 
 company, including the amounts due it, are called its 
 resources or assets ; its debts are called liabilities. 
 
 The profits and losses of a company are usually divided 
 among the partners proportionally to the capital of each, 
 if all invest for the same time ; and proportionally to the 
 product of capital and time if the times are different. 
 
 Problems in partnership are therefore solved by the 
 same methods as other problems in proportion. 
 
 EXAMPLES 36 
 
 1. A and B form a partnership. A furnishes $400 
 capital and B $600. They gain $250. What is the 
 profit of each ? 
 
 Suggestion. (1) The whole capital is ^ 1000 ; hence 
 $ 1000 : $ 400 : : $ 250 : A's share. 
 $ 1000 : $ 600 : : $ 250 : B's share. 
 
 (2) Each partner receives the same fraction (or per cent) of the 
 whole gain that his capital is of the whole capital. 
 
 (3) The gain of each partner is the same fraction (or per cent) 
 of his capital that the whole gain is of the whole capital. 
 
 2. A, B, and C traded in company. A put in $ 8000 ; 
 B, $4500; and C, $3500. Their profits were $6400. 
 What is each partner's share of the profits ? 
 
98 ALGEBRAIC ARITHMETIC 
 
 3. A and B, in trading for 3 yr., make a profit of 
 $4800. A invested | as much stock as B. What is 
 each man's share of the profits ? 
 
 4. Brooks & Co. fail in business; their liabilities 
 amount to $22,000; their resources to $8800. Tliey 
 owe A $4275, and B $2175.50. What will each of 
 these creditors receive? 
 
 5. Four persons engage in manufacturing, and invest 
 jointly $ 22,500. At the end of a certain time A's share 
 of the gain is $2000; B's, $2800.75; C's, $1685.25; 
 and D's, $ 1014. How much capital did each put in ? 
 
 6. Three partners, A, B, and C, furnish capital as 
 follows : A, $ 500 for 2 mo. ; B, $ 400 for 3 mo. ; C, 
 $ 200 for 4 mo. They gain $ 600. What is each part- 
 ner's share ? 
 
 Suggestion. The use of $500 for 2 mo. is equivalent to the 
 use of !| 1000 for 1 mo. ; of $ 400 for 3 mo. to $ 1200 for 1 mo. ; 
 of $200 for 4 mo. to $800 for 1 mo. Hence, divide the profits 
 proportionally to 1000, 1200, and 800. 
 
 7. A, B, and C gain in trade $8000. A furnishes 
 $ 12,000 for 6 mo. ; B, $ 10,000 for 8 mo. j and C, $ 8000 
 for 11 mo. Apportion the gain. 
 
 8. Jan. 1, 1896, three persons began business with 
 $1300 capital furnished by A. March 1, B put in 
 $1000; and Aug. 1, C put in $900. The profits at the 
 end of the year were $ 750. Apportion it. 
 
 9. In a certain firm B has 3 times as much capital as 
 A, and C has ^ as much as the other two. What is each 
 one's share in a loss of $ 786 ? 
 
AVERAGE OF PAYMENTS 99 
 
 10. In a gain of $ 600 A received ^ ; B, ^ ; and C the 
 remainder. If the whole capital was 12 times A^s gain, 
 what was the capital of each ? 
 
 11. Two men receive ^1000 for grading. One fur- 
 nishes 3 teams 20 da., and the other 5 teams 30 da. 
 If the first receives $ 100 for overseeing the work, what 
 does each receive ? 
 
 12. Two men contract to move $ 5316 cu. yd. of gravel 
 at 25 cents a cu. yd., and agree to share the profits in the 
 ratio of 2 to 3. They employ 5 teams 45 da., at $4 
 each per day. What did each make ? 
 
 AVERAGE OF PAYMENTS 
 
 71. Ex. 1. A owes B $ 1200, of which 8 300 is due in 
 
 4 mo., $ 400 in 6 mo., and $ 500 in 12 mo. If he wishes 
 to pay the whole debt at one time, when must he do so in 
 order that neither party shall lose ? 
 
 The loss that is here referred to is the loss of the use 
 of money, which is really loss of interest. 
 
 If A should pay the debt at once, he would lose the 
 use of $300 for 4 mo., $400 for 6 mo., and $500 for 
 12 mo. ; to all of which he is entitled. 
 
 The use of $ 300 for 4 mo. = the use of $ 1 for 1200 mo. 
 The use of 400 for 6 mo. = the use of 1 for 2400 mo. 
 The use of 500 for 12 mo. = the use of 1 for 6000 mo. 
 
 $1200 9600 mo. 
 
 Hence A is entitled to the use of $ 1 for 9600 mo., or 
 to the use of the $ 1200 for ^\^ of 9600 mo., or 8 mo. 
 That is, the whole debt will be due in a single payment 
 in 8 mo. 
 
100 ALGEBRAIC ARITHMETIC 
 
 EXAMPLES 37 
 
 2. On Dec. 1, 1896, a man gave three notes, the first 
 for f 500, payable in 3 mo. ; the second for $ 750, payable 
 in 6 mo. ; and the third for $ 1200, payable in 9 mo. Find 
 the average time of payment. 
 
 3. Bought merchandise Jan. 1, 1895, as follows : f 350 
 on 2 mo., $500 on 3 mo., $700 on 6 mo. What is the 
 average time of payment ? 
 
 4. Find the average date for paying three bills, due 
 as follows: May 31, |100; June 18, $150; July 9, 
 $ 200. (Compute each from May 31.) 
 
 5. If I borrow $ 250 for 8 mo., how long should I lend 
 $ 400 to repay me an equal interest ? 
 
 6. A person owes a debt of $1680, due in 8 mo., of 
 which he pays -g^ in 3 mo., ^ in 5 mo., -g- in 6 mo., and -J- in 
 7 mo. When is the remainder due ? 
 
 7. On a debt of $ 2500, due in 8 mo. from Feb. 1, the 
 following payments were made : May 1, $ 250 ; July 1, 
 $ 300 ; and Sept. 1, $ 500. When is the balance due ? 
 
 8. Dec. 1, 1894, purchased goods to the amount of 
 $1200, on the following terms: 25% payable in cash, 
 30% in 3 mo., 20% in 4 mo., and the balance in 6 mo. 
 Find the average time of payment and the cash value of 
 the goods, computing discount at 7%. 
 
CHAPTER VI 
 INVOLUTION AND EVOLUTION 
 
 72. Involution. Review Art. 22. 
 
 The product of equal factors is called a power of the 
 factor thus repeated. 
 
 The factor taken once is called the first power; the 
 product of two equal factors is called the second power ; 
 of three equal factors, the third power, and so on. 
 
 The second power of a number is also called the square 
 of the number, because it is equal to the area of the 
 square the length of whose side is the given number. 
 For a similar reason the third power of a number is 
 called its cube. 
 
 A number is said to be squared when its second power 
 is taken, and to be cubed when its third power is taken. 
 
 The process of taking any power of a number is called 
 involution. 
 
 EXAMPLES 38 (Oral) 
 Find the indicated power : 
 
 2^ 
 
 62 
 
 V 
 
 1 
 
 1002 
 
 1.22 
 
 22 
 
 1 
 
 32 
 
 1.V 
 
 2^ 
 
 .72 
 
 902 
 
 3« 
 
 .3» 
 
 (1)^ 
 
 .V 
 
 30^ 
 
 2* 
 
 (t)^ 
 
 ikf 
 
 102 
 
 .2* 
 
 .012 
 
 101 
 
102 ALGEBRAIC ARITHMETIC 
 
 73. To find a Power of a Product. Study carefully the 
 following : 
 
 1. 62 = (2x3)* = (2x3)(2x3) = 2x2x3x3 = 22x32 
 = 4x9 = 36. 
 
 2. 103 = (2x5)3=(2 X 5)(2 X 5)(2 X 5) = 2 X 2 X 2 X 5 
 X 5 X 5 = 2^ X 5»= 8 X 125 = 1000. 
 
 3. (3 ahy = (3 ahh) (Sabb) = 3 x 3 x aabbbb = 9 d'b*. 
 
 4. (a^y =:a^ X a^ = aaa x aaa = a% 
 or (a^y = (aaay = aa x aa x aa = a^. 
 
 A product is raised to any power by raising each of its 
 factors to that power. 
 
 (a'd)* 
 
 1 
 
 (2cy 
 
 74. Evolution. The process of taking one of the equal 
 factors of a number is called evolution. It is the inverse 
 of involution. 
 
 One of the equal factors of a number is called a root 
 of the number. One of the two equal factors of a num- 
 ber is called its square root ; one of the three equal fac- 
 tors, its cube root. 
 
 Thus, since 25 is the square of 5, 5 is the square root 
 of 25 ; since 27 is the cube of 3, 3 is the cube root of 27. 
 
 Find: 
 
 EXAMPLES 
 
 3 39 (Oral) 
 
 (oc)^ 
 
 (foa'b'y 
 
 (iaey 
 
 (2aby 
 
 2(aby 
 
 i(aey 
 
 (5aby 
 
 3a(aby 
 
 fabV 
 
THE SQUARE 103 
 
 The square root of a number is indicated by the 
 radical sign (V) placed before it ; the cube root by -^. 
 
 Thus V25 = 5, ^27 = 3. 
 
 The figure placed above the radical sign indicates 
 what root is to be taken, and is called the index of the 
 root. If no index is written, 2 is understood. 
 
 If an expression consists of more than one term or 
 factor, the root of the whole is indicated by placing the 
 radical sign before the expression enclosed in parentheses 
 or placed under a vinculum ; otherwise the sign affects 
 only the term or the factor immediately following. 
 
 Thus Vl6 + 9 = 4 + 9 = 13; V(l^ + 9) = V16 + 9 
 
 EXAMPLES 40 (Oral) 
 Find: 
 
 V49 Vl-21 ^2500 V* 12a-2V4^2 
 
 V400 ^M V1600 VM V25a^+aV96* 
 ^125 V-81 Vl-44: ^\ ^a'b~c 
 ^64 -^.001 ^1 ^ff V9W + 2 6*c 
 
 75. The Square of the Sum of Two Numbers. 
 
 Review Arts. 18 and 19. 
 
 We have learned (Art. 18) that a number is multiplied 
 by multiplying each of its parts, and that for this pur- 
 pose it may be separated into parts, or terms, in any 
 way. 
 
 If both multiplier and multiplicand consist of more 
 than one term, their product is the sum of the partial 
 
104 ALGEBRAIC ARITHMETIC 
 
 products obtained by multiplying each term of the mul- 
 tiplicand by each term of the multiplier (Art. 19). 
 
 If multiplier and multiplicand are equal and are sepa- 
 rated into parts in the same way, the case is like that 
 of Art. 22, Ex. 3, and the exercise. It was there shown 
 that 
 
 {a + bf^a' + Zab + f)'. (1) 
 
 Since a and h may be any two numbers, we have the 
 law: 
 
 The square of the sum of two numbers equals the square 
 of the first yiumher plus twice the product of the numbers 
 plus the square of the second number. 
 
 The operation of squaring a number of two figures is 
 simplest when it is separated into its tens and units. 
 When it is so separated, we shall use t to denote the 
 tens, and w, the units. In this case formula (1) becomes 
 
 (i^uy=e+2iu-\-u\ (2) 
 
 Exercise. Express the meaning of (2) in words. 
 
 76. Illustration of the Formula, (t + uy is the area 
 of the square the length of whose side is t-^u. The 
 square may be divided into four parts, 
 as shown in the figure. Comparing 
 the right member of the formula with 
 the figure, it will be seen that the first 
 term is the area of the largest part; 
 the second term is the area of the 
 two rectangles; the last term is the 
 area of the small square that fills out the corner, and is 
 always the smallest of the terms. 
 
 tu 
 
 ^ 
 
 w« 
 
 t 
 
 
 u 
 
 t* 
 
 <♦* 
 
 s 
 
SQUARE ROOT 105 
 
 Note. The square of any number may be found by the for- 
 mula. For example, 324 = 32 tens + 4 units ; hence t = 320, 
 
 M = 4. 
 
 EXAMPLES 41 J ^ 
 
 Find by the formula : 
 
 1. 561 2. 732. 3. 2082. 4. 3161 
 
 SQUARE ROOT 
 
 77. Find ^784:, 
 
 The problem may be stated thus : Find the side of 
 the square whose area is 784 (sq. in., say). 
 Or thus : Find t and u, when 
 
 t^-{-2tu + u^ = 7S4:. 
 
 Begin by taking the largest value possible for t. This 
 is easily seen to be 20. 
 
 f-\-2tu-^u^ = 7S4:sq, in. 
 Subtract f = 400 sq. in. 
 
 Eemainder = 2tu-i-u^=: 384 sq. in. 
 
 Compare with the figure in the last article. What is 
 the remainder the area of ? The two rectangles and the 
 
 3 i-^u -^ 
 
 small square have one dimension, u, in common. If 
 placed as in the accompanying figure, they form one long 
 rectangle whose dimensions are 2 i + w and u, and whose 
 area therefore is (2t + u)u. 
 
106 
 
 ALGEBRAIC ARITHMETIC 
 
 How is the width of a rectangle found if its length 
 and area are given ? To find the width u, we are obliged 
 to use 2 tj or 40, as the length, since the whole length is 
 as yet unknown. This may give too large a value for u ; 
 if so, we take one less. 
 
 384 --2^ = 384 ^40 = 9+. 
 
 This is too large ; for it gives 
 
 (2^ + u)u = (40 + 9) X 9 = 441, 
 
 and there are only 384 sq. in. Hence take w = 8. 
 
 This gives (2 « -f- w)w = (40 + 8) x 8 = 384. 
 
 Hence V'^^^ = 4^0 + 8 = 48. Arts. 
 
 2t 
 
 FORMULA FOR SQUARE ROOT 
 
 f-^2tu^u^ \t + u 
 
 ^ 202 
 
 ~ 2x20 
 
 OPERATION 
 
 784 I 20 + 8 = 28 
 
 -\-u 
 
 2t-\-u 
 
 
 40 
 _8 
 48 
 
 400 
 
 384 
 384 
 
 2 Hs called the trial divisor. 
 
 2t-\-uiB called the complete divisor. 
 
 The formula for the square may be written 
 
 (f + £/)2=f2-|-(2f + i/)(|. 
 
 (1) 
 
 In this form, the first term of the coefficient of u is the 
 tried divisor. The whole coefficient is the complete divisors- 
 it is the whole length of the addition to the square of the 
 tens. 
 

 SQUARE ROOT 
 
 
 
 EXAMPLES 42 
 
 
 Solve and prove : 
 
 
 
 1. V1156. 
 
 3. VS184. 
 
 5. V324. 
 
 2. V4225. 
 
 4. V841. 
 
 6. V9604. 
 
 107 
 
 78. Let any integer of three figures be separated into 
 its hundreds, tens, and units, and denote these parts 
 by the initial letters ; then the number will be denoted 
 by the expression h + t -{- u. Let us find its square. 
 
 
 
 ^ + 
 
 t 4-w 
 
 
 
 ^ + 
 
 t -\-u 
 
 
 
 hu + 
 
 tu + u^ 
 
 
 ht + t' 
 
 
 tu 
 
 h' + 
 
 ht + 
 
 hu 
 
 
 h^-^2ht + f-{-2hu-\-2tu-\-u^ 
 
 Hence (h-{-t-^ u)^ =h^ -\- f+ u^ + 2ht -\-2hu -i-2 tu. 
 
 All we need observe here is that the square of the 
 number contains the square of each of the figures plus 
 other terms. This is true of any number. 
 
 Thus, the square of 48.7 
 
 .72= .49 
 contains I 8^. = 64. plus other parts ; 
 42 . =16 . 
 
 The square of 12.34 
 
 . 42 = 
 
 contains 
 
 16 
 
 .32 = 
 21 = 04 
 V , =1 
 
 09 
 
 plus other parts. 
 
108 
 
 ALGEBRAIC AKITHMETIO 
 
 It will be seen from this that if the complete square of 
 any number be separated into groups of two figures each, 
 commencing at the decimal point, the number of groups 
 (counting the last figure to the left as a group, if it 
 stands alone) will be equal to the number of figures in 
 the root ; and the square of each figure of the root will lie 
 wholly within the corresponding group.^ 
 
 79. The square root of any number is found as follows : 
 Separate the number into groups, as above directed, and 
 proceed as in Art. 77, always regarding the part of the 
 root already found as so many tens with respect to the 
 next figure of the root. 
 
 Ex. 1. Find the square root of 75076. 
 
 FULL OPERATION 
 
 
 CONTRACTED 
 
 7'50'76 
 
 274 
 
 7'50'76 1 274 
 
 4 
 
 
 4 
 
 2x20 = 40 
 
 350 
 
 
 47 
 
 350 
 
 7 
 
 329 
 
 
 544 
 
 329 
 
 47 
 
 2176 
 
 2x270 = 540 
 
 2176 
 2176 
 
 2176 
 
 4 
 544 
 
 
 
 Explanation. The first trial and complete divisors are ob- 
 tained from the formula precisely as they would be if the given 
 number were 760. That is « = 20 and w = 7. For the second 
 divisors t — 270 and m = 4. 
 
 When the cipher is omitted from the trial divisor, as in tlie con- 
 tracted operation, omit mentally the right-hand figure of the divi- 
 dend in finding the figure of the root. Write the latter, when 
 found, in units' place in the trial divisor, thus completing it. 
 
SQUARE ROOT 
 
 109 
 
 9'41'.57'80 I 30.685 
 9 
 
 Ex. 2. Extract the square root of 941.578. 
 
 Explanation. Complete the last 
 group to the right by the addition of a 
 cipher. Since there is a remainder 
 after using the last group, the root is 
 not exact ; but can be found to as 
 many places as desired by annexing 
 groups of ciphers. 
 
 The first trial divisor, 6, is con- 
 tained times in 4, Place in the 
 root ; and annex to the trial divisor, 
 and the next group to the dividend. 
 
 To find the square root of a fraction, take the square 
 root of its terms separately if they are seen to be perfect 
 squares ; otherwise it is best to reduce to a decimal first, 
 as by so doing evolution is performed but once. 
 
 606 
 
 4157 
 3636 
 
 6128 
 
 52180 
 49024 
 
 61365 
 
 315600 
 306825 
 
 Vi" 
 
 ■g-^* 
 
 TT- 
 
 EXAMPLES 43 
 
 3. V13225. 9. V196.1369. 16. 
 
 4. V11881. 10. V2-251521. 16. 
 
 5. V994009. 11. -y/5S.U0625. 17. 
 
 6. V20506.24. 12. -^17.75. 18. 
 
 7. V2985.5296. 13. V10795.21. 19. 
 
 8. V001225. 14. V^f. 20. 
 
 21. A square field contains 1,016,064 sq. ft. 
 the length of each side ? 
 
 22. A square farm contains 361 acres. Find the length 
 of one side. 
 
 23. A field is 208 rd. long and 13 rd. wide. What is 
 the length of the side of a square field containing an 
 equal area? 
 
 V30i. 
 
 V69f 
 What is 
 
110 ALGEBRAIC ARITHMETIC 
 
 24. If it costs ^ 312 to enclose a field 216 rd. long and 
 24 rd. wide, what will it cost to enclose a square field of 
 equal area with the same kind of fence ? 
 
 25. Find the dimensions of a rectangular field contain- 
 ing 3200 sq. rd., and twice as long as broad. 
 
 26. How many rods of fence will enclose a square 
 field of 4 acres? 
 
 27. How many rods of fence would be required to en- 
 close a field of 4 acres whose length is twice its width ? 
 
 28. What is the difference between the areas of two 
 fields, one of which is 14 rd. square and the other 14 
 sq. rd. ? 
 
 29. An orchard containing 2401 trees has as many rows 
 as there are trees in a row. How many rows has it ? 
 
 80. The Cube of the Sum of Two Numbers. 
 
 a-j-b 
 
 a^b^-2aW^-b^ 
 a»-f2a26-|- a6* 
 
 a3-f.3a26-4-3a62 + 63 
 Hence {a + by = a^-\-Za^b + Zab^ + b^ (1) 
 
 Exercise 1. State the formula in words, calling a 
 "the first number" and b "the second number." 
 
 Solve the following by the formula, and verify by first 
 taking the indicated sum, then cubing : 
 
 (2 + 3)», (3 + 5)3, (i2+9)», (20-f-5)^ 
 
THE CUBE 
 
 111 
 
 The solution is the simplest when the number is sepa- 
 rated into its tens and units. For this case we shall 
 write the formula thus: 
 
 Exercise 2. Express formula (2) in words. 
 
 (2) 
 
 81. Illustration of the Formula. 
 
 / 
 
 7 
 
 ^^6 
 
 
 Tt- 
 
 - 
 
 
 y 
 
 7 
 
 
 
 / 
 
 / 
 
 ^ 
 
 / 
 
 Fig. 1 
 
 Fig. 2 
 
 Fig. 3 
 
 Fig. 4 
 
 {t -f uf is the volume of a cube, the length of whose 
 edge is t-^u. Such a cube can be formed from 8 solids, 
 as follows : A cube whose edge is t, and whose volume 
 is therefore f (Fig. 1); 3 rectangular solids covering 3 
 adjacent faces of the cube and of thickness w, the volume 
 of each being tht (Fig. 2) ; 3 rectangular solids filling the 
 edges, the volume of each of which is tu^ (Fig. 3); a 
 small cube whose edge is u and whose volume is v?, filling 
 the corner (Fig. 4). 
 
 The formula and figures may be applied to any num- 
 ber, if we regard it as being made up of tens and units, 
 as in Art. 76, note. 
 
 EXAMPLES 44 
 
 Solve by the formula : 
 1. 15^ 2. 233. 
 
 68^ 
 
 4. 127». 
 
112 ALGEBRAIC ARITHMETIC 
 
 CUBE ROOT 
 
 82. Find ^46656. 
 
 The problem is to find the edge of a cube whose volume 
 is 46656 (cu. in., say), or to find t and u when 
 
 ^3 _,. 3 ^2^ _|. 3 ^^2 _^ ^3 ^ 4gg5g j.^ i^ (;i^) 
 
 Begin by taking the largest value possible for t. This 
 is 30; hence i^ = 27000. Subtract from the correspond- 
 ing members of (1) ; then 
 
 3 <2|^ 4- 3 tu" + u^ = 19656 cu. in. (2) 
 
 What is the remainder the volume of? (See the 
 figures of the last article.) Observe that the seven addi- 
 tions to the cube of the tens have one dimension, u, in 
 common, and that equation (2) may be written 
 
 (3 «2 + 3 <w + u^u = 19656 cu. in. (3) 
 
 Suppose the seven solids to be laid side by side, forming 
 one solid. The area of its base would be 3 i^ -|- 3 ^it + w^, 
 its height would be u^ and its volume would be the prod- 
 uct of its base and its height, or (3 ^^ + 3 tu + ii?)u. 
 
 How is the height of a rectangular solid found when its 
 volume and the area of its base are given ? To find the 
 height u we are obliged to use 3 1"^ as the area of the base, 
 since the whole area is not yet known. If we find that 
 this gives too large a value for w, we take one less. 
 
 19656 -J- 3 «2 = 19656 -- (3 x 30^) =7+. 
 
 By trial we find this too large ; hence take w = 6. 
 Then 
 
 (3 ^'4-3 <w-f w>= (3 X 302-f 3 X 30 X 6 +6^ X 6=19656. 
 
 Hence -^46656 = 30 + 6 = 36. Ans, 
 
CUBE KOOT 
 
 113 
 
 Formula for Cube Root 
 
 3t' 
 
 + 3tu-\-u^ 
 
 3f + 3tu-^u^ 
 
 3t\ + 3tu^-i-w' 
 3eu-\-3tu^-^u^ 
 
 3 f is called the trial divisor. 
 
 3t^ -{-3tu-\- u^ is called the complete divisor. 
 
 Solution of -y/46656 by the formula : 
 
 46656 I 30 + 6 = 36 
 27000 
 3 X 30^ = 2700 ' 
 3 X 30 X 6 = 540 
 6^= 36 
 
 3276 
 
 19656 
 
 19656 
 
 The formula for the cube may be written 
 
 (f + uf = f3 + (3 f2 4- 3 /^ + u^u. (4) 
 
 The first term in the parenthesis in the right member is 
 the trial divisor; the whole expression within the paren- 
 thesis is the complete divisor. 
 
 1. ^15625. 
 
 2. ^166375. 
 
 EXAMPLES 45 
 
 3. -^10648. 
 
 4. ^912673. 
 
 5. ^42875. 
 
 6. ^474552. 
 
 83. By multiplying the square of h + t -^u (Art. 78) 
 by the first power, the pupil may prove for himself that 
 
 {h -{- 1 -{- uy = h^ + f -\- u^ -\- other terms. 
 
 I 
 
114 ALGEBRAIC ARITHMETIC 
 
 That is, the cube of a number of three figures contains, 
 among other parts, the cube of each of the figures. The 
 same is true of any number. 
 
 Thus the cube of 382.5 
 
 .53 = .125 
 
 23. = 008. 
 
 8» . = 512 . 
 3» . =27 
 
 contains 
 
 plus other parts. 
 
 It will be seen from this that if the complete cube of 
 any number be separated into groups of three figures 
 each, commencing at the decimal point, the number of 
 groups will be equal to the number of figures in the 
 root ; and the cube of each figure of the root will lie wholly 
 within the corresponding group. 
 
 The last group to the left may contain only one or two 
 figures. 
 
 To find the cube root of any number, separate it into 
 groups as above directed, and proceed as in the last 
 article, always regarding the part of the root already 
 found as so many tens with respect to the next figure of 
 the root. 
 
 If the last group to the right of the decimal point is 
 incomplete, it must be completed by annexing ciphers. 
 No such number has an exact cube root. Why not ? 
 
 When a cipher occurs in the root, annex two ciphers 
 to the trial divisor, and another group to the dividend. 
 
 If there is a remainder after the root of the last period 
 is found, the result may be found to as many places as 
 desired by annexing groups of ciphers. 
 
^ 0^7 Si 
 
 UNIVEr^SiTY 
 
 OF 
 
 iLiFORNVh> 
 
 CUBE ROOT 
 
 115 
 
 Ex. 1. ^12812'904. 
 
 12'812'904 [234 
 8 
 
 3 X 202 = 1200 
 
 3 X 20 X 3 = 180 
 
 32= 9 
 
 1389 
 3 X 2302 = 158700 
 3 X 230 X 4 = 2760 
 42= 16 
 
 4812 
 
 4167 
 
 161476 
 
 645904 
 
 645904 
 
 Ex. 2. ^8710.37. 
 
 
 8'710'.370 1 20.57 + 
 8 
 
 3 X 202 = 1200 
 
 3 X 2002 = 120000 
 
 3 X 200 X 5 = 3000 
 
 52= 25 
 
 710 
 710370 
 
 123025 
 
 615125 
 
 3 X 20502 = 12607500 
 
 3 X 2050 X 7 = 43050 
 
 72= 49 
 
 95245000 
 88554195 
 
11. 
 
 </{in- 
 
 12. 
 
 ^Hm- 
 
 13. 
 
 ^2i. 
 
 14. 
 
 ^h 
 
 116 ALGEBRAIC ARITHMETIC 
 
 EXAMPLES 46 
 
 3. ^1030301. 7. ^.091125. 
 
 4. ^4492125. 8. -^.000097336. 
 
 5. ^1045678.375. 9. -^39.4995. 
 
 6. ^4080.659192. 10. ^1250.6894. 
 
 15. What are the dimensions of a cube that has the 
 same volume as a box 2 ft. 8 in. long, 2 ft. 3 in. wide, 
 and 1 ft. 4 in. deep ? 
 
 16. How many square feet in the surface of a cube 
 whose volume is 91125 cu. ft. ? 
 
 17. What is the length of the inner edge of a cubical 
 bin that contains 150 bu. ? (1 bu. contains 2150.42 
 cu. in.) 
 
 18. What is the depth of a cubical cistern that holds 
 200 bbl. of water? (34 gal. = 1 bbl. ; 1 gal. = 231 
 cu. in.) 
 
 19. Find the length of a cubical vessel that will hold 
 4000 gal. of water. 
 
 20. What are the dimensions of a cubical box contain- 
 ing ^ as much as one whose edge is 4 ft. ? 
 
CHAPTER VII 
 MENSURATION 
 
 84. The process of measuring lines, surfaces, and solids 
 is called mensuration. 
 
 Note. All the rules and formulas of mensuration and the facts 
 upon which they depend are proved in geometry. When state- 
 ments are made without explanation in this chapter, it is not because 
 none can be given, but because they cannot be understood without 
 a knowledge of geometry. 
 
 85. Lines. A straight line has the same direction at 
 every point. 
 
 A curved line, or curve, changes its direction at every 
 point. 
 
 Parallel lines have the same direction ; they are every- 
 where equidistant. 
 
 Two straight lines are said to be perpendicular to each 
 other when the angle between them is a right angle 
 (Art. ^Q). 
 
 straight lines — * ■ ■ • — Parallel lines 
 
 Perpendiculftrs 
 Curves 
 
 117 
 
118 ALGEBRAIC ARITHMETIC 
 
 86. Angles. An angle is the difference in the direction 
 of two straight lines. 
 
 The lines are called the sides of the angle; and their 
 point of meeting, its vertex. 
 
 Angles are measured in degrees, a degree being ^^ of 
 the whole angular magnitude about a point. Thus the 
 sum of all the angles that can be drawn with a common 
 vertex at a point is 360°. 
 
 If two lines intersect so as to form four equal angles, 
 each of these angles is a right angle. A right angle is \ 
 of the angular magnitude about a point, and is therefore 
 equal to 90°. 
 
 An obtuse angle is greater than a right angle. 
 
 An acute angle is less than a right angle. 
 
 1 
 
 / 
 
 Acute Right Obtuse 
 
 PLANE FIGURES 
 
 87. A portion of a plane surface bounded by straight 
 lines or curves is called a plane figure. The sum of the 
 lines bounding the figure is called its perimeter. 
 
 88. Polygons. Any plane figure bounded by straight 
 lines is called a polygon. The parts of a polygon are its 
 sides, angles, and vertices. 
 
 A diagonal of a polygon is a straight line joining any 
 two vertices not adjacent. 
 Polygons receive special names according to the num- 
 
PLANE FIGURES 
 
 119 
 
 ber of their sides. A triangle has three sides ; a quadri- 
 lateral, four ; a pentagon, five ; a hexagon, six. 
 
 A regular polygon has equal sides and equal angles. 
 
 Regular Polygons 
 
 Equilateral 
 triangle 
 
 Square 
 
 Pentagon 
 
 Hexagon 
 
 o 
 
 Octagon 
 
 89. Quadrilaterals are classified as follows : 
 
 . parallelo- 
 gram has 
 its oppo- 
 site sides 
 parallel (4 
 classes). 
 
 ^ \ 
 
 A rectangle has four right 
 angles. 
 
 A square is a rectangle whose 
 sides are equal. 
 
 A rhomboid has no right angles. 
 
 A rhombus has four equal sides 
 and no right angles. 
 
 A trapezoid has only two parallel sides. 
 
 A trapezium has no parallel sides. 
 
 7 
 
 ^ 
 
120 
 
 ALGEBRAIC ARITHMETIC 
 
 90. Area of Parallelograms. 
 
 The dimensions of a parallelogram are its base (b) and 
 its altitude (a). 
 
 Any side of a parallelogram may be taken as its base. 
 Its altitude is the perpendicular distance between its base 
 and the opposite side. 
 
 The altitude of a rectangle is equal to the side not 
 taken as base. 
 
 The area (A) of a parallelogram is equal to the product 
 of its base and its altitude. 
 
 Explanation. This is a familiar fact in the case of rectangles ; 
 the common form of statement for this case being that the area of 
 a rectangle is equal to the product of its length and width. 
 
 From any rhomboid or rhombus, a rectangle of the same dimen- 
 sions can be constructed, by cutting off the triangular portion from 
 one end and fitting it on to the other, as shown in the figure. This 
 change in the form of the figure does not change its area, since the 
 figure is composed of the same parts as before, only differently 
 placed. It is clear that the area of the figure is now the product 
 of its two dimensions ; hence it was equal to the product of these 
 dimensions before its form was changed. 
 
 Definition. Figures having the same area are called 
 equivalent figures. 
 
 It follows from what has been said above that paral- 
 lelograms having equal bases and equal altitudes are 
 equivalent. 
 
TRIANGLES 
 
 121 
 
 91. Area of Triangles. 
 
 The dimensions of a triangle are its base and its alti- 
 tude. 
 
 Any side of a triangle may be taken as the base. The 
 altitude is the length of the perpendicular from the 
 base to the opposite vertex. 
 
 The area of a triangle is equal to one-half the product 
 of its base and altitude. 
 
 Explanation. Add to the given triangle an equal triangle 
 inverted. The two together form 
 a parallelogram having the same 
 base and altitude as the given tri- 
 angle. Since the area of the par- 
 allelogram is &a, the area of the 
 triangle is ^ ha. 
 
 It follows that triangles having equal bases and equal 
 altitudes are equivalent. 
 
 4 = ^ 6a 
 
 92. A triangle having one right angle is called a right 
 triangle. 
 
 The side opposite the right angle is called the hypothe- 
 nuse, and the other two sides, the legs. 
 
 If a right triangle be constructed having legs 3 and 
 
 4 units in length respectively, the 
 hypothenuse will be found to be 
 
 5 units long ; hence the sum of the 
 areas of the squares constructed 
 on the legs will be equal to the 
 area of the square constructed on 
 the hypothenuse. It can be proved 
 that this relation is true of any 
 right triangle. That is : 
 
122 ALGEBRAIC ARITHMETIC 
 
 The square of the hypothenuse of a right triangle is 
 equal to the sum of the squares of the other two sides. 
 
 This relation is expressed by the for- 
 mula h^ = 6- -f a^, where a and b are the 
 legs of the right triangle and h is the 
 hypothenuse. Subtract a^ from both 
 sides and interchange members; then 
 f, — y/ h-i — a ^ ^^ = ^^^ ~ ^'^- Take the square root of 
 a = V h'^ — 6^ both sides, and we have b = V/i^ — a\ 
 In the same way we obtain a= -Vh^ — b\ 
 If any two sides of a right triangle are given, the third 
 side can be found from these formulas. 
 
 EXAMPLES 47 
 
 1. The base of a rhombus is 10 ft. 6 in., and its alti- 
 tude 8 ft. What is its area ? 
 
 2. How many acres in a piece of land in the form of 
 a rhomboid, the base being 8.75 chains, and the altitude 
 6 chains ? 
 
 3. Find the area of a triangle whose base is 12 ft. 
 6 in., and altitude 6 ft. 9 in. 
 
 4. What is the cost of a triangular piece of land whose 
 base is 15.48 chains, and altitude 9.67 chains, at $60 
 an acre? 
 
 5. Find the area of the gable end of a house that is 
 28 ft. wide, and the ridge of the roof 15 ft. higher than 
 the foot of the rafters. 
 
 6. What is the base of a triangle whose area is 189 
 sq. ft., and altitude 14 ft. ? 
 
PROBLEMS 128 
 
 7. Find the altitude of a triangle whose area is 20J 
 sq. ft., and base 3 yd. 
 
 8. The legs of a right triangle are 12 in. and 16 in. 
 respectively. What is the length of the hypothenuse ? 
 
 9. The foot of a ladder is 15 ft. from the base of 
 a building, and the top reaches a window 36 ft. above 
 the base. What is the length of the ladder ? 
 
 10. Find the distance diagonally across a floor 30 
 by 40 ft. 
 
 11. What is the length of a path diagonally across 
 a 10-acre square field ? 
 
 12. A room is 20 ft. long, 16 ft. wide, and 12 ft. high. 
 What is the distance from one of the lower corners to 
 the opposite upper corner? 
 
 13. The hypothenuse of a right angle is 35 ft., and 
 one leg 28 ft. Find the other leg. 
 
 14. A ladder 52 feet long stands against the side of 
 a building. How many feet must it be drawn out at the 
 bottom to lower the top 4 ft. ? 
 
 15. Find the diagonal of a cube containing 729 cu. in. 
 
 16. Wliat is the side of a square field whose diagonal 
 is 15 rods ? What is its area ? 
 
 17. A ladder 28 ft, long, placed in a street, reaches 
 the top of a building 18 ft. high on one side, and one 
 15 ft. high on the other. How wide is the street? 
 
 18. Two vessels sail from the same point, one north 
 58 miles, and the other west 72 miles. How far apart 
 are they ? 
 
124 
 
 ALGEBRAIC ARITHMETIC 
 
 93. Area of a Trapezoid. 
 
 The parallel sides of a trapezoid are called the bases ; 
 and the distance between them is the altitude. 
 
 The area of a trapezoid is equal to one-half the product 
 of the altitude and the sum of the bases. 
 
 Explanation. Draw a diagonal. This divides the trapezoid 
 into two triangles whose common altitude is 
 the altitude of the trapezoid. The base of one 
 of the triangles is the lower base of the trap- 
 ezoid, the base of the other is the upper base 
 of the trapezoid. 
 
 The area of the one triangle is } abi, of the 
 other is ^ abz ; the area of the trapezoid is the 
 
 4 = ^0(61 + 62) 
 sum of the areas of the triangles. Hence 
 
 ^ = I a6i + ^ afta = i a(&i + &2). 
 
 Note. The subscripts (1 and 2) used here and in subsequent 
 articles have no numerical signification, but are merely used to 
 distinguish between two values of the same letter. 
 
 94. The area of a trapezium can be found as the sum of 
 the areas of two triangles, if the length of a diagonal and 
 
 of the two perpendiculars from it to the opposite vertices 
 are known. 
 
 The area of any polygon can be found by dividing it 
 into triangles, computing their areas separately, and add- 
 ing the results. 
 
POLYGONS 
 
 126 
 
 95. Area of Regular Polygons. Every regular polygon 
 can be divided into equal triangles 
 having a common vertex at the centre 
 of the polygon. 
 
 The altitude of these triangles is 
 called the apothem of the polygon. 
 
 In the figure the dotted line is the 
 apothem. 
 
 Since the area of the polygon is the sum of the areas 
 of these equal triangles, it follows that : 
 
 The area of a regular polygon is equal to one-half the 
 product of its perimeter (p) and apothem (a). 
 
 EXAMPLES 48 
 
 1. Eind the area of a trapezoid whose bases are 23 ffe. 
 and 11 ft., and the altitude 9 ft. 
 
 2. One side of a quadrilateral field measures 38 rd., 
 the side opposite and parallel to it measures 26 rd., and 
 the distance between the two sides is 10 rd. Find the 
 area. 
 
 3. Find the area of a trapezium whose diagonal is 
 42 ft., and the perpendiculars to this diagonal from the 
 opposite vertices are 16 ft. and 18 ft. 
 
 4. Derive the formula a = - V3 for the altitude (a) of 
 
 an equilateral triangle, the length of whose 
 sides is b. 
 
 5. Find by the formula of Ex. 4 the alti- 
 tude of an equilateral triangle whose sides 
 are 9 in. 
 
126 ALGEBRAIC ARITHMETIC 
 
 6. Find the area of a regular hexagon whose sides are 
 10 in. 
 
 Suggestion. A regular hexagon is divided into six equilateral 
 triangles by its three diagonals passing through the centre. Hence 
 the apothem can be found by the formula of Ex. 4. 
 
 96. The Circle. A circle is a plane figure bounded 
 by a curve, called the circumference, all 
 points of which are at an equal dis- 
 tance from a point within called the 
 centre. 
 
 The radius (r) of a circle is the dis- 
 tance from its centre to its circumfer- 
 ence (c). Its diameter (d) is the distance 
 across it measured through the centre. Hence d = 2r. 
 
 97. The ratio of the circumference of a circle to its diam- 
 eter is the same for all circles, and is generally denoted by 
 
 the Greek letter tt (pronounced pie). That is, - = tt. 
 
 The value of tt is a little less than 3f ; more accu- 
 rately, 3.1416. It is not exactly expressible by any num- 
 ber ; but can be found to as many decimal places as 
 desired. 
 
 From the equations - =7r and d = 2 r we have 
 d 
 
 (? = w(/ = 2 irr, (1) 
 
 </=|, (2) 
 
 . = A. (3) 
 
THE CIRCLE 127 
 
 Hence, if the radius, the diameter, or the circumfer- 
 ence of a circle is given, the other dimensions can be 
 computed. 
 
 98. The Area of a Circle. If a regular polygon of any 
 number of sides be circumscribed about a circle, its 
 apothem will be the radius of the circle. 
 
 Let p denote the perimeter of the polygon, r (radius 
 of circle) its apothem, and Ai its area ; then 
 
 A^==\rp. (Art. 95) 
 
 It is evident that Ai is larger than the area {A) of 
 circle, and that p is larger than the circumference (c) 
 of the circle. 
 
 But the greater the number of sides of the polygon, 
 the more nearly will p be equal to c, and also the more 
 nearly will Ai be equal to A. 
 
 If we should go on increasing the number of sides of 
 the polygon, its area would still be found by the formula 
 Ai = ^ rp, and at the same time we could make p as 
 nearly equal to c and Ai as nearly equal to A as we 
 please. 
 
 Hence it follows that it must at least be very nearly 
 
128 ALGEBRAIC ARITHMETIC 
 
 correct to find the area of the circle by the same formula. 
 It is proved in geometry that it is exactly 
 correct. 
 
 Hence A = ^cr. (1) 
 
 This amounts to regarding the circle as 
 
 composed of a very great (infinite) num- 
 
 - _ _ 2 ber of triangles, whose common altitude 
 
 is the radius of the circle, and the sum 
 
 of whose bases is the circumference. 
 
 Since c = 2 Trr, (1) may be written 
 
 >f = u/^; (2) 
 
 which is the usual formula for finding the area of a circle. 
 It is sometimes convenient to use the formula 
 
 ^=iw^ (3) 
 
 which the pupil may derive for himself from (1) and the 
 equation d — 2r. 
 
 Exercise. Give the meaning of formulas (1), (2), and 
 (3) in words. 
 
 99. Let Ci and Ca denote the circumferences of two 
 circles ; r^ and r^ their radii. 
 
 Then Cj = 2 irri, Cj = 2 ^r^. [Art. 97 (1)] 
 
 Hence, dividing the members of the first equation by 
 the corresponding members of the second, 
 
 £i = ?Z!i = !i; (1) 
 
 Ca 2 ttTj rg ' 
 
 or, in words : The ratio of the ci/rcumferences of two circles 
 
THE CIRCLE 129 
 
 is equal to the ratio of their radii. Or, more briefly : The 
 circumferences of two circles are to each other as their radii. 
 
 Exercise. Prove that the circumferences of two cir- 
 cles are to each other as their diameters. 
 
 100. Let Ai and A2 be the areas of two circles ; rj and 
 r2 their radii. 
 
 Then A^ = ttTj", A^ = Trr/. [Art. 98 (2)] 
 
 That is : The ratio of the areas of two circles is equal to 
 the ratio of the squares of their radii. Or, The areas of two 
 circles are to each other as the squares of their radii. 
 
 Taking the square root of the first and last members 
 of (1), and interchanging them, we have 
 
 In words : The radii of two circles are to each other as 
 the square roots of their areas. 
 
 Note. The relation between (1) and (2) is expressed by say- 
 ing that either is the converse of the other. 
 
 Exercise. (1) Prove from Art. 98 (3) that the areas 
 of two circles are to each other as the squares of their 
 diameters. Prove the converse. 
 
 (2) Prove from Art. 98 (1) that the areas of two circles 
 are to each other as the squares of their circumferences. 
 Prove the converse. 
 
130 ALGEBRAIC ARITHMETIC 
 
 EXAMPLES 49 
 
 1. What is the circumference of a circle whose diame- 
 ter is 20 in. ? 
 
 2. What is the diameter of a tree whose girt is 18 ft. 
 6 in.? 
 
 3. Find the area of a circle whose diameter is 10 ft. 
 
 4. The distance around a circular park is l}j mi. 
 How many acres does it contain? 
 
 5. What is the circumference of a circle whose area 
 is 19.635 sq. ft. ? 
 
 6. What is the side of a square inscribed in a circle 
 whose diameter is 6 rd. ? 
 
 7. The area of a circle is 78.54 sq. ft. Find the side 
 of the inscribed square. 
 
 8. What is the circumference of a circular pond 
 whose radius is 11 rd. ? Its area? 
 
 9. What is the radius of a circle equal in area to a 
 triangle whose base is 13 ft. and altitude 10 ft. ? 
 
 10. A cow is one day tied to the top of -a stake 5 ft. 
 high by a rope 20 ft. long. On the next day she is tied 
 to the bottom of the stake by the same rope. Find the 
 difference in the areas over which she can graze. 
 
 11. What will it cost at $ 2 a rod to fence a circular 
 plot of land containing 1 acre ? 
 
 12. How many times will a carriage wheel 4 ft. in 
 diameter turn round in going 1 mi. ? 
 
 13. A square field contains 31.5 acres. What is the 
 length of its diagonal ? What is the circumference of a 
 circular field of the same area ? 
 
SIMILAR PLANE FIGURES 131 
 
 101. Similar plane figures are plane figures having the 
 same shape ; that is, their corresponding angles are equal 
 and their corresponding lines (like dimensions) are pro- 
 portional. 
 
 Similar figures may be regarded as enlarged or reduced 
 copies of one another. 
 
 All circles are similar figures, and all regular polygons 
 of the same number of sides. 
 
 It is proved in geometry that : 
 
 (i.) Any corresponding lines of similar plane figures are 
 to each other as their other corresponding lines. 
 
 (ii.) The areas of similar plane figures are to each other 
 as the squares of their corresponding lines. 
 
 Conversely, 
 
 (iii.) The corresponding lines of similar plane figures are 
 to each other as the square roots of their areas. 
 
 Note. These general truths, or theorems, were proved in 
 Arts. 99 and 100 for circles. Compare carefully the theorems as 
 given for circles with the more general corresponding theorems of 
 this article. 
 
 EXAMPLES 50 
 
 1 . The length of one side of a triangular field contain- 
 ing 2 A. 80 sq. rd. is 12 chains. Find the area of a field 
 of similar shape whose corresponding side is 48 chains. 
 
 Suggestion. 122 : 482 : : 2.5 A. : x A. [Theorem (ii.)] 
 
 2. The side of a square field containing 18 acres is 
 60 rd. long. Find the side of a square field that contains 
 J as many acres. 
 
132 ALGEBRAIC ARITHMETIC 
 
 3. Two circles are to each other as 9 to 16, the diame- 
 ter of the less being 112 ft. What is the diameter of the 
 greater ? 
 
 SoGGESTiON. 3:4:: 112 : oc. [Theorem (iii.)] 
 
 4. A rectangular field contains 720 sq. rd., and its 
 length is to its breadth as 5 to 4. What are its dimen- 
 sions ? 
 
 Suggestion. Let I — length of field and 6 = its breadth. The 
 area of a rectangle 5 by 4 is 20. Hence 
 
 20 : 720 : : 52 : Z2 ; 20 : 720 : : 4^ : h"^. 
 Solve the proportions for P' and 6^^ then extract the square roots. 
 
 5. It is required to lay out 283 A. 107 sq. rd. of land 
 in the form of a rectangle so that the length shall be 3 
 times the width. Find the dimensions. 
 
 6. A pipe 1.5 in. in diameter fills a cistern in 5 hr. 
 Find the diameter of a pipe that will fill the same cistern 
 in 55.1 min. 
 
 7. If it costs $ 167.70 to enclose a circular field con- 
 taining 17 A. 110 sq. rd., what will it cgst to enclose 
 another \ as large with the same kind of fence? 
 
 8. If 63.39 rd. of fence will enclose a circular field 
 containing 2 A., what length will enclose a circular field 
 of 8 A.? 
 
 SOLIDS 
 
 102. Prisms and Cylinders. The word solid as used in 
 mathematics means a portion of space bounded by sur- 
 faces. It has no reference to what the space may contain. 
 
PRISMS AND CYLINDERS 
 
 133 
 
 A solid whose ends are equal and parallel polygons 
 and whose sides are rectangles is called a right prism. 
 
 The height of a prism is the perpendicular distance 
 between its ends, or bases. 
 
 From the form of their bases prisms are called tri- 
 angular, quadriangular, pentagonal, etc. 
 
 A right prism whose bases are rectangles is called a 
 quadrangular prism, rectangular solid, or parallelopiped. 
 
 A cube is a rectangular solid whose faces are all equal 
 squares. 
 
 Pentagonal 
 Prism 
 
 Cylinder 
 
 Note. The space passed through by a moving surface is called 
 the solid generated by the surface. 
 
 f€E^ 
 
 The solid generated by a rectangle rotating 
 about one of its sides is called a right circular 
 cylinder. 
 
 Note. The word prism is often used for right 
 prism, and cylinder for right circular cylinder. They 
 are so used in what follows. 
 
 103. The area of the lateral surface {S) of a prism or 
 a cylinder is equal to the product of its height (h) and the 
 perimeter (p) of its base. 
 
134 
 
 ALGEBKAIC ARITHMETIC 
 
 The volume (V) of a rectangular solid is equal to the 
 product of its three dimensions. 
 
 The volume of a prism or a cylinder is equal to the 
 product of its height (h) and the area (A) of its base. 
 
 A / 
 
 /-T-A 
 
 V = lbh 
 
 -p 
 
 
 r| . 
 
 kj 
 
 / 
 
 5 = 2irr/» 
 
 S = hp 
 
 K = irr2A 
 
 ¥ = hA 
 
 EXAMPLES 51 
 
 
 1. Find the area of the lateral surface of a prism 
 whose altitude is 7 in., and its base a pentagon, each side 
 of which is 4 in. 
 
 2. What is the entire surface of a cylinder formed 
 by the revolution about one of its sides of a rectangle 
 6 ft. 6 in. long and 4 ft. wide ? 
 
 3. Find the solid contents of a cylinder whose alti- 
 tude is 15 ft., and its radius 1 ft. 3 in. 
 
 4. Find the entire surface of a prism whose base is 
 an equilateral triangle, the perimeter being 18 ft., and 
 the height 15 ft. 
 
 5. Find the contents of a box whose length, width, 
 and depth are, respectively, 4 ft., 3 ft., and 2 ft. 
 
 6. Find its surface. 
 
PYRAMIDS AND CONES 135 
 
 7. Find the number of square feet necessary to make 
 a stove pipe 2^ ft. long and 5 in. in diameter. 
 
 8. Find the amount of tin necessary to make a tin 
 pail cylindrical in form, 6 in. in diameter and 8 in. deep, 
 without a cover. 
 
 9. How many quarts will the pail hold ? 
 
 10. Find the depth of a cylindrical tank that holds 
 20 gal. and is 18 in. in diameter. 
 
 11. A rectangular can is 10 in. square on the bottom 
 and holds 5 gal. How deep is it ? 
 
 12. What is the difference in the number of square 
 feet of lumber necessary to make the sides of a room 
 16 ft. long, 12 ft. wide, and 10 ft. high, and one of cir- 
 cular floor containing the same area and of the same 
 height? 
 
 104. Pyramids and Cones. A regular pyramid is a 
 
 solid whose base is a regular polygon, and whose sides 
 are equal triangles which terminate in a common vertex. 
 
 The common altitude of the triangular sides is called 
 the slant height of the pyramid. 
 
 The solid generated by a right triangle 
 rotating about one of its legs is called a 
 right circular cone. 
 
 The length of the hypothenuse of the 
 generating triangle is the slant height of 
 the cone. 
 
 The height of a pyramid or cone is the perpendicular 
 distance from its vertex to its base. 
 
136 
 
 ALGEBRAIC ARITHMETIC 
 
 The words pyramid and cone are frequently used for 
 regular pyramid and right circular cone respectively. 
 
 Pjrramid 
 
 Frustum 
 
 Cone 
 
 The frustum of a pyramid or of a cone is the part that 
 remains after cutting oif a portion of the top by a plane 
 parallel to the base. 
 
 105. The lateral surface {S) of a pyramid or a cone is 
 
 equal to one-half the product of its slant height (a) and the 
 perimeter (p) of its base. 
 
 This follows directly from the formula of Art. 95, and 
 formula (1) of Art. 98. 
 
 S = ^ap = irar 
 
 The lateral surface of a frustum of a pyramid or a cone 
 
 is equal to one-half the product of the slant height (a) and 
 the sum of the perimeters (pi and j9a) of its bases. 
 
PYRAMIDS AND CONES 
 
 13T 
 
 This follows for the frustum of a pyramid from Art. 
 93. The lateral surface of the frustum of a cone may 
 be regarded as made up of a very great (infinite) number 
 of trapezoids. 
 
 
 S = tra(ri + r^) 
 
 Let Vi be the radius of the lower base of the frustum 
 of a cone, and 7*2 the radius of the upper base. 
 
 Then S = ia(pi-\- p^) = ^ a(2 -rrr^ -\- 2 TrVi) 
 = 1 a X 2 TT (ri 4- ^2) = 7ra(ri + r^). 
 
 The volume of a pyramid or a cone is equal to one-third 
 
 the product of its height (Ii) and the area (A) of its base. 
 (See note to Art. 84.) 
 
 The volume of the frustum of a pyramid or a cone is 
 found as follows : Add the areas (A^ and A2) of the bases 
 and the square root of their product, and multiply this sum 
 by one-third of the height. (See note to Art. 84.) 
 
 For the frustum of a cone, 
 
 V=\h{A,-^A,+ ^A,A,) 
 
 = |/l (Trri^ -f- -nri + -^/irr^Trr^ 
 = \h (irri + irr^^ 4- Trrjra) 
 = i'irh(ri^-\-r^^ + riri). 
 
138 ALGEBRAIC ARITHMETIC 
 
 EXAMPLES 52 
 
 1. Find the lateral surface of a triangular pyramid, 
 the slant height being 16 ft., and each side of the base 
 5 ft. 
 
 2. Find the lateral surface of a cone whose diameter 
 is 17 ft. 6 in., and the slant height 30 ft. 
 
 3. Find the entire surface of a square pyramid whose 
 base is 8 ft. 6 in. square, and its slant height 21 ft 
 
 4. How many cubic feet in the mast of a ship, its 
 height being 50 ft., the circumference at one end 5 ft., 
 and at the other 3 ft. ? 
 
 5. Find how much water can be put into a tin pail 
 10 in. deep, like a frustum of a cone in form, whose bot- 
 tom is 8 in. across, and top 12 in. across. 
 
 6. How many square feet of tin in the pail described 
 in the last example, without cover ? 
 
 7. A conical wood pile is 6 ft. high and 12 ft. in 
 diameter at the base. How many cords are in it? 
 
 8. How many bushels of oats in a conical pile 2 ft. 
 high and 12 ft. around it at the base ? 
 
 9. Find the number of cubic feet enclosed by a barn 
 60 ft. long, 40 ft. wide, and 20 ft. high, with a pyramidal 
 roof 8 ft. high ; all inside measurements. 
 
 10. How many cubic feet of wood are in a log 20 ft. 
 long and 14 in. in diameter ? 
 
 11. At 28 cents per cubic foot, what is the cost of a 
 stone wall 28 in. thick at the base and 18 in. at the top, 
 4 ft. high and 36 id. long? 
 
THE SPHERE 
 
 139 
 
 12. How many cubic feet in a regular eight-sided post 
 10 ft. high, the width of one side being 3 in., and the dis- 
 tance through it 7.24 in. ? 
 
 106. The Sphere. A sphere is a solid bounded by a 
 uniformly curved surface, all points of which are equally 
 distant from a point within called the centre. 
 
 A sphere is generated by a semicircle rotating about its 
 diameter. The radius and the diameter of the generating 
 semicircle are the radius and the diameter, respectively, 
 of the sphere. 
 
 The section of a sphere made by a plane passing 
 through its centre is called a great circle of the sphere. 
 
 107. The surface (S) of a sphere is equal to the lateral 
 surface of the circumscribed cylinder. (Art. 
 84, note.) 
 
 The diameter and the height of the 
 circumscribed cylinder are each equal to 
 the diameter of the sphere ; hence 
 
 ^ = 2 Trr X 2 r = 4 Trr*. 
 
 The area of the surface of a sphere is 
 $ = A irr^, equal to four times the area of its great 
 '^ = f »/* circle. 
 
 108. Let ri and r, be the radii of two spheres. Si 
 and S2 their surfaces. 
 
 Then Si = 4:7rr^% S2 = 4:nri\ 
 
 Hence 
 
 §1 
 S, 
 
 4 7rn^ 
 
 2» 
 
 or 
 
 7rr2 rj 
 S,:S,::n':r,'. 
 
140 ALGEBEAIC ARITHMETIC 
 
 The surfaces of two spheres are to each other as the 
 squares of their radii. 
 
 Conversely, Li = :^i. 
 
 The radii of two spheres are to each other as the square 
 roots of their surfaces. 
 
 Exercise. (1) Prove that the surfaces of two spheres 
 are to each other as the squares of their diameters. 
 Prove the converse. 
 
 (2) Prove that the surfaces of two spheres are to each 
 other as the squares of the circumferences of their great 
 circles. Prove the converse. 
 
 109. The Volume of a Sphere. We have seen that a 
 circle may be regarded as made up of a very great num- 
 ber of triangles having a common vertex at its centre. 
 Similarly, a sphere may be regarded as made up of a 
 very great (infinite) number of pyramids, having their 
 bases in the surface of the sphere and their common ver- 
 tex at its centre. 
 
 The surface of the sphere is the sum of the bases of 
 these pyramids, and its radius is their height. 
 
 Now the volume of a pyramid is the product of its base 
 and one-third its height; hence the volume of a sphere 
 is the product of its surface and one-third its radius. 
 
 Hence, since ^S = 4 irr'^, 
 
 V=^\Sr = ^^'^. (1) 
 
 110. Let r^ and r^ be the radii of two spheres ; Vx and 
 Fj their volumes. 
 
THE SPHERE 141 
 
 Then 
 
 
 F, = i»n', F, = tW. 
 
 Hence 
 
 
 V, tirr/ ri' 
 
 and 
 
 
 
 The volumes 
 
 of 
 
 two spheres are to each other as the cubes 
 
 of their radii. 
 
 
 
 Conversely, 
 
 TJie radii of two spheres are to each other as the cube roots 
 of their volumes. 
 
 Exercise. (1) Prove that the volumes of two spheres 
 are to each other as the cubes of their diameters ; as the 
 cubes of the circumferences of their great circles. 
 
 (2) Prove the converse of each of the above. 
 
 EXAMPLES 53 
 
 1. Find the surface of a sphere whose diameter is 9 in. 
 
 2. Find the volume of a sphere whose diameter is 18 in. 
 
 3. The glass tank of a lamp is spherical in shape and 
 4 in. in diameter on the inside. - How much oil will it 
 hold? 
 
 4. The diameter of the earth is about 8000 mi. Find 
 its surface and volume. 
 
 111. Similar solids are solids having the same form. 
 Their corresponding surfaces are similar, their corre- 
 sponding angles equal, and their corresponding lines pro- 
 portional. 
 
 All spheres are similar solids, and all cubes, 
 
142 ALGEBRAIC ARITHMETIC 
 
 Theorems : 
 
 (i.) Any corresponding lines of similar solids are to each 
 other as their other corresponding lines. 
 
 (ii.) The surfaces of similar figures (plane or solid) are 
 to each other as the squares of their corresponding lines. 
 Conversely, 
 
 (iii.) Tlie corresponding lines of similar figures are to 
 each other as the square roots of their surfaces. 
 
 (iv.) The volumes of similar solids are to each other as 
 the cubes of their corresponding lines. Conversely, 
 
 (v.) The corresponding lines of similar solids are to each 
 other as the cube roots of their volumes. 
 
 Note. These theorems were proved in Arts. 108 and 110 for 
 spheres. Compare the theorems given there with the more general 
 ones of this article. Why are these more general ? 
 
 The reason for the truth of these theorems is that all lines have 
 but one dimension — length; all surfaces are proportional to the 
 product of two dimensions — length and width ; and all volumes 
 to the product of three dimensions — length, width, and thickness. 
 (Compare with Art. 68.) 
 
 This is evident in the case of squares and cubes. If the side of 
 one square is twice that of another, its area is 4 times as great. 
 If the edge of one cube is twice that of another cube, its volume is 
 8 times as great, etc. 
 
 Illustrate the last two statements by drawings. 
 
 EXAMPLES 54 
 
 1. If a marble column 10 in. in diameter contains 27 
 cu. ft., what is the diameter of a column of equal length 
 that contains i81 cu. ft. ? 
 
 2. A ball 4.6 in. in diameter weighs 18 oz. What is 
 
SIMILAR SOLIDS 143 
 
 the weight of another ball of the same density, that is 
 9 in. in diameter ? 
 
 3. Two vessels have the same shape. One is 12 in. 
 deep and holds 7 gal. The other is 7 in. deep; what 
 does it hold ? 
 
 4. A tank was made of 20 sq. ft. of sheet iron, and a 
 tank of the same shape of 30 sq. ft. What is the ratio 
 of their capacities ? 
 
 Suggestion. First find the ratio of their linear dimensions. 
 
 5. What is the ratio of the corresponding edges of 
 two similar rectangular solids whose volumes are respec- 
 tively 2.7 cu. ft. and 1.5 cu. ft. ? 
 
 6. What is the. edge of a cube whose entire surface is 
 1050 sq. ft., and what is its volume ? 
 
 7. What must be the inner edge of a cubical bin to 
 hold 1250 bu. of wheat ? 
 
 8. How many gallons will a cistern hold whose 
 depth is 7 ft., the bottom being a circle 7 ft. in diameter 
 and the top 5 ft. in diameter ? 
 
 9. What is the value of a stick of timber 24 ft. long, 
 the larger end being 15 in. square, and the smaller 6 in., 
 at 28 cents a cubic foot ? 
 
 10. The surface of a sphere is the same as that of a 
 cube, the edge of which is 12 in. Find the volume of 
 each. 
 
Examples 1 
 
 2. 
 
 4^, top; 
 
 10. 6f. 
 
 1. 5. 
 
 
 
 20;*, ball. 
 
 11. 18. 
 
 2. 6. 
 
 
 3. 
 
 lOj^, sister ; 
 
 12. 90. 
 
 3. 16. 
 
 
 
 20^, brother. 
 
 13. If 
 
 4. 4. 
 
 
 4. 
 
 16 and 34. 
 
 14. 33,V 
 
 5. 0. 
 
 
 5. 
 
 21, 42, 63. 
 
 
 6. 8. 
 
 , 
 
 6. 
 
 32;^. 
 
 
 7. 28. 
 
 
 7. 
 
 13ida. 
 
 1. 8a2. 
 
 8. 66. 
 
 
 8. 
 
 7 marbles, H ; 
 
 2. 16acd. 
 
 9. 26. 
 
 
 
 19 marbles, F. 
 
 3. 6a8&. 
 
 1©. 16. 
 
 
 9. 
 
 13 and 49. 
 
 4. 20 a^bK 
 
 11. 66. 
 
 
 10. 
 
 13 yr., son ; 
 
 5. Sa&Vrc. 
 
 13. 3. 
 
 
 
 58 yr., father. 
 
 6. S5 cex^y. 
 
 13. 18. 
 
 
 11. 
 
 48. 
 
 7. 3a + 36. 
 
 14. 3. 
 
 
 12. 
 
 18 and 24. 
 
 8. 15a -66. 
 
 15. tVV 
 
 
 13. 
 
 6^ bu. ; 
 14,4 bu. 
 
 9. a^-^a"^. 
 10. 2a5-a*. 
 
 Examples 2 
 
 14. 
 
 I 
 
 11. 3a3 + 6a2-6a. 
 
 1. 3. 
 
 
 
 
 12. a62c2 + a26ca 
 
 2. 1. 
 
 
 
 Examples 4 
 
 + a^b^c. 
 
 8. 3f 
 
 
 
 
 13. 18. 
 
 4. 2. 
 
 
 1. 
 
 38. 
 
 14. 300. 
 
 5. tV 
 
 
 2. 
 
 8. 
 
 15. 900. 
 
 6. 66. 
 
 
 3. 
 
 10. 
 
 16. 76. 
 
 7. 6. 
 
 
 4. 
 
 73. 
 
 17. 29. 
 
 8. ^^: 
 
 
 5. 
 
 163. 
 
 18. 33. 
 
 
 
 6. 
 
 2^. 
 238. 
 
 19. 6. 
 
 Examples 3 1 
 
 7. 
 
 20. 72. 
 
 1. 4yr., 
 
 James ; 
 
 8. 
 
 9. 
 
 21. 20. 
 
 12 yr. 
 
 , John. 
 
 9. 
 
 36. 
 146 
 
 22. 13. 
 
23. 
 
 25. 
 
 12. 
 
 3 a& + 62 _ 3 ^2. 
 
 2. 
 
 27r/o; 
 
 24. 
 
 729. 
 
 13. 
 
 31^, 1st; 
 
 
 3H%; 
 
 25. 
 
 5a + 6. 
 
 
 62^, 2d ; 
 
 
 ^%^ 
 
 26. 
 
 2 a6 + ac. 
 
 
 97^, 3d. 
 
 
 6f%; 
 
 27. 
 
 2 be. 
 
 14. 
 
 4hr. 
 
 
 28f%. 
 
 28. 
 
 6ab-7ac+2bc. 
 
 15. 
 
 16. 
 
 3. 
 
 28000. 
 
 29. 
 
 26-|c. 
 
 16. 
 
 A, 24 apples ; 
 
 4. 
 
 $5. 
 
 30. 
 
 22 a6 - 8 ac. 
 
 
 B, 12 apples ; 
 
 5. 
 
 60%. 
 
 31. 
 
 x^-\-2xy + 2/2. 
 
 
 C, 16 apples. 
 
 6. 
 
 2500 sheep. 
 
 82. 
 
 4a2 + 4a6 4-62. 
 
 17. 
 
 40 men ; 
 
 7. 
 
 $217.61. 
 
 83. 
 
 a* + 2 a^c + c2. 
 
 
 80 boys ; 
 
 8. 
 
 $40000. 
 
 34. 
 
 3a2+5a6+2 62. 
 
 
 880 women. 
 
 9. 
 
 15%. 
 
 35. 
 
 2a2+3a6-5 62. 
 
 18. 
 
 $3000. 
 
 10. 
 
 39 yd. 
 
 36. 
 
 ai-ab-2 b\ 
 
 19. 
 
 40 and 60. 
 
 11. 
 
 $3000. 
 
 37. 
 
 a3_2a6 + 62. 
 
 20. 
 
 12 and 24. 
 
 12. 
 
 52%; 
 
 38. 
 
 4a;2_i2a;y+9y2. 
 
 21. 
 
 6yr. 
 
 
 $45,760. 
 
 39. 
 
 a« - 2 a3c2 + c*. 
 
 22. 
 
 10 yr., son ; 
 
 
 
 40. 
 
 2a2-3a& + 62. 
 
 
 30 yr., father. 
 
 
 Examples 13 
 
 41. 
 
 10a2-9a62 
 
 23. 
 
 A, $30; 
 
 
 
 
 + 2 6*. 
 
 
 B,$15; 
 
 4. 
 
 $2760. 
 
 42. 
 
 a3+3a26+3a62 
 
 
 C, $20. 
 
 5. 
 
 $5.25. 
 
 
 + 68. 
 
 24. 
 
 164 artillery ; 
 
 6. 
 
 $ 12800. 
 
 
 
 
 472 cavalry ; 
 
 7. 
 
 $ 1300 cost. 
 
 
 Examples 6 
 
 
 564 Infantry. 
 
 8. 
 
 Neither. 
 
 
 
 25. 
 
 14 yr. ; 
 
 9. 
 
 $720. 
 
 1. 
 
 3. 
 
 
 16 yr. ; 
 
 10. 
 
 $300. 
 
 2. 
 
 3 a. 
 
 
 18 yr. 
 
 11. 
 
 $1437.60. 
 
 8. 
 
 4 a. 
 
 26. 
 
 7 and 8. 
 
 12. 
 
 23%. 
 
 4. 
 
 3a;y. 
 
 27. 
 
 18 yr.; 
 
 13. 
 
 23i%. 
 
 6. 
 
 10 a6. 
 
 
 36 yr. 
 
 14. 
 
 $1.47. 
 
 6. 
 
 aHd^. 
 
 
 
 15. 
 
 $.17^. 
 
 7. 
 
 4a2 + 6a6. 
 
 
 Examples 11 
 
 16. 
 
 $96. 
 
 8. 
 
 3 a6 - 2 c. 
 
 
 
 17. 
 
 24%. 
 
 9. 
 
 fx-J2/2. 
 
 1. 
 
 A, $1400; 
 
 18. 
 
 11%. 
 
 10. 
 
 a 
 
 
 B, $480; 
 
 19. 
 
 2|%loss. 
 
 
 
 C, $453.60 
 
 20. 
 
 $1.94^; 
 
 11. 
 
 a2 + 2a6 + 62. 
 
 
 D, $260. 
 
 
 16f%. 
 
ANSWERS 
 
 147 
 
 21. $50; 
 $58.82; 
 $65. 
 
 22. 4% loss. 
 
 Examples 14 
 
 4. $378.13. 
 6. $96.90. 
 
 6. $6400.76 invest- 
 
 ment. 
 
 7. 6%. 
 
 8. $600. 
 
 9. $1271.88. 
 
 10. $ 15 com. ; 
 $ 750 inv. 
 
 11. $2905; 
 9|^^perlb. 
 
 12. $2920. 
 
 13. 14f%. 
 
 14. $432. 
 
 15. $506.25; 
 23750 lb. 
 
 16. 5|%. 
 
 17. $2100. 
 
 18. $ 82.11 com. ; 
 $9301.89 pro- 
 ceeds. 
 
 19. $10623.44. 
 
 20. $45111.44; 
 $225.56. 
 
 Examples 15 
 1. $555.75. 
 
 3. 
 
 $256.50. 
 
 $1080; 
 
 7. 
 8. 
 
 9. 
 10. 
 
 $696.11. 
 $666.90 ; 
 
 $3.75. 
 
 $786. 
 $363.80; 
 
 $20.81. 
 $2.40. 
 
 Examples 16 
 
 1. $217.50. 
 
 2. $37.50. 
 
 3. $738. 
 
 4. $58000. 
 
 5. $13600. 
 
 7. $424. 
 
 8. $ 44 prem. ; 
 $5456 loss. 
 
 9. 2o/„. 
 
 10. $1840.50 pre- 
 
 mium ; 
 $79959.50 value. 
 
 11. $32000.^ 
 
 12. $3717.83. 
 
 13. $3168. 
 
 14. $5600. 
 
 Examples 17 
 
 1. $88000. 
 
 2. $19072.16. 
 
 3. $.45. 
 
 4. U%, rate; 
 $95.25, A's tax. 
 
 5. .0228 tax rate ; 
 $214.66. 
 
 7. 
 
 8. 
 
 9. 
 10. 
 11. 
 
 $407.20. 
 
 $224.37. 
 
 $103.13. 
 
 $21.43. 
 
 2 J mills. 
 
 $76.39. 
 
 
 Examples 19 
 
 1. 
 
 $7212.50. 
 
 2. 
 
 $37.15. 
 
 3. 
 
 $11925. 
 
 4. 
 
 $151.88. 
 
 5. 
 
 $30.72. 
 
 
 Examples 20 
 
 1. 
 
 $3902.40. 
 
 2. 
 
 $ 78133.33. 
 
 3. 
 
 $208.33. 
 
 4. 
 
 $6553.60. 
 
 6. 
 
 $25372. 
 
 6. 
 
 48 bu. 
 
 7. 
 
 $1700, Istyr.; 
 
 
 $1785, 2d yr. 
 
 8. 
 
 40f%. 
 
 9. 
 
 $40842. 
 
 10. 
 
 61788.6 lb. 
 
 11. 
 
 25f-% nearly. 
 
 12. 
 
 12%. 
 
 13. 
 
 $2116.94. 
 
 14. 
 
 16^. 
 
 16. 
 
 $863.99. 
 
 
 $68.21. 
 
 Examples 22 
 
 2. $83.22; 
 $ 843.22. 
 
 3. $11.91; 
 $191.41. 
 
148 
 
 ANSWERS 
 
 4. 
 
 $57.20; 
 
 Examples 24 
 
 Examples 28 
 
 
 $382.20. 
 
 2. $6.06. 
 
 3. Mat. Oct. 30 ; 81 
 
 6. 
 
 $146.19; 
 
 3. $82.36. 
 
 da. term of 
 
 
 $904.94. 
 
 4. $10.96. 
 
 dis.; $940.14 
 
 6. 
 
 $ 142.03 ; 
 
 5. $39.55. 
 
 proceeds. 
 
 
 $1166.28. 
 
 6. $106.99. 
 
 4. Apr. 8 ; 46 da. ; 
 
 7. 
 
 $42.28; 
 
 7. $172.17. 
 
 $917.21. 
 
 
 $626.78. 
 
 8. $51.37. 
 
 6. Aug. 13 ; 64 da. ; 
 
 8. 
 
 $55.77; 
 
 9. $205.48. 
 
 $690.04. 
 
 
 $781.61. 
 
 
 6. $641.55. 
 
 9. 
 
 $83.02; 
 
 
 7. $1821.60. 
 
 
 $470.97. 
 
 Examples 26 
 
 8. $2548.53. 
 
 10. 
 
 $.17; 
 
 
 9. $56.69. 
 
 
 $42.37. 
 
 1. $516.71. 
 
 
 
 
 2. $1000. 
 
 Examples 29 
 
 
 Examples 23 
 
 3. 3 yr. 4 mo. 24 da. 
 
 2. $2688. 
 
 1. 
 
 $58.93. 
 
 4. 6%. 
 
 3. $464.10. 
 
 2. 
 
 $ 8.40. 
 
 6. $650.80. 
 
 4. $3808. 
 
 3. 
 
 $67.67. 
 
 6. 7%. 
 
 6. $586.04. 
 
 4. 
 
 $159.75. 
 
 7. 7 mo. 10 da. 
 
 6. $247.52. 
 
 5. 
 
 $67.09. 
 
 8. 7%. 
 
 Examples 30 
 
 6. 
 
 7. 
 
 a 
 
 $38.11. 
 $30.81. 
 $8.93. 
 $3647.61. 
 
 9. $669.12. 
 10. $1403.08. 
 
 2. $900 simple; 
 $ 1035 annual ; 
 
 8. 
 9. 
 
 Examples 27 
 
 $1046.30 comp. 
 3. $428.76. 
 
 10. 
 11. 
 
 $1066.36. 
 $2010.42. 
 
 2. $917.43; 
 
 4. $189.15. 
 6. $311.64. 
 
 12. 
 13. 
 14. 
 15. 
 
 $142.45. 
 $1886.17. 
 $263.83. 
 $410.70. 
 
 $82.57. 
 
 3. $43.65 in favor 
 
 of dis. 
 
 4. $1137.61. 
 
 6. $100.32. 
 
 7. $41.99. 
 
 8. $245.77. 
 
 9. $53.38. 
 
 10. $1540.79. 
 
 11. $2357.79. 
 
 16. 
 
 17. 
 
 $25.78. 
 $165.50. 
 
 5. $838.26. 
 
 6. The first by 5^ 
 
 18. 
 
 $410.73. 
 
 per ton. 
 
 
 19. 
 
 $1936.60. 
 
 7. $2010.13. 
 
 Examples 31 
 
 20. 
 
 $1120.69. 
 
 8. $147.06. 
 
 2. $576. 
 
 21. 
 
 $7.33. 
 
 9. $.92. 
 
 3. $98.33. 
 

 
 ANSWERS 
 
 
 149 
 
 4. $575.34. 
 
 6. 
 
 280 bu. 
 
 9. 
 
 A, $131; 
 
 5. $284.79. 
 
 6. 
 
 $6428.57. 
 
 
 B, $393; 
 
 6. $601.08. 
 
 7. 
 
 220| cd. 
 
 
 C, $262. 
 
 
 8. 
 
 $52.79. 
 
 10. 
 
 A, $1800; 
 
 Examples 32 
 
 9. 
 
 9 men. 
 
 
 B, $600; 
 
 2. $577.38. 
 
 10. 
 
 2|da. 
 
 
 C, $1200. 
 
 3. $99.88. 
 
 11. 
 
 10 ft. 2| in. 
 
 11. 
 
 1st, $357f; 
 
 4. $676.67. 
 
 12. 
 
 10 ft. 
 
 
 2d, $642f. 
 
 5. $285.99. 
 
 13. 
 
 546 bbl. 
 
 12. 
 
 $171.60; 
 
 6. $603.49. 
 
 14. 
 
 2080 lb. 
 
 
 $257.40. 
 
 7. $1386.78. 
 
 15. 
 
 $100. 
 
 
 
 8. $1284.11. 
 
 16. 
 
 15 da. 
 
 
 Examples 37 
 
 9. $162.25. 
 
 17. 
 
 432 mi. 
 
 2. 
 
 June 27, 1897. 
 
 10. $4408.21. 
 
 11. $523.43. 
 
 1. 
 
 Examples 36 
 
 A's, $100; 
 
 3. 
 
 4. 
 
 May 5, 1896. 
 June 23. 
 
 Examples 34 
 
 
 B's, $150. 
 
 5. 
 
 6 mo. 
 
 3. $4.05. 
 
 4. 44|bbl. 
 
 5. 16 men. 
 
 2. 
 
 A's, $3200; 
 B's, $1800; 
 
 C's, $ 1400. 
 
 6. 
 
 7. 
 8. 
 
 5 yr. 20 da. from 
 date of last 
 payment. 
 
 Nov. 26. 
 
 Mar. 7 ; 
 
 $1178.01. 
 
 6. 96 sheep. 
 
 7. $5355. 
 
 8. 7 hr. 13^ min. 
 
 3. 
 4. 
 
 A's, $ 1800 ; 
 B's, $3000. 
 A, $1710; 
 
 9. 112^ mi. 
 
 
 B, $ 870.20. 
 
 
 
 10. 59|da. 
 
 5. 
 
 A, $6000; 
 
 
 Examples 43 
 
 11. $7320. 
 
 
 B, $8402.25; 
 
 3. 
 
 115. 
 
 12. $10958.90. 
 
 
 C, $5055.75; 
 
 4. 
 
 109. 
 
 13. 90 horses. 
 
 
 D, $3042. 
 
 5. 
 
 997. 
 
 14. 2yr. 6mo. 
 
 6. 
 
 A's, $200; 
 
 6. 
 
 143.2. 
 
 15. 3hr. 
 
 
 B's, $240; 
 
 7. 
 
 54.64. 
 
 16. 20 men. 
 
 
 C's, $160. 
 
 8. 
 
 .036. 
 
 17. 75 da. 
 
 7. 
 
 A, $2400; 
 
 9. 
 
 14.0048+. 
 
 18. 9 men. 
 
 
 B, $2666.67; 
 
 10. 
 
 1.5006+ . 
 
 
 
 C, $2933.33. 
 
 11. 
 
 7.625. 
 
 Examples 35 
 
 8. 
 
 A, $388.71 ; 
 
 12. 
 
 4.213+. 
 
 3. $240. 
 
 
 B, $249.17; 
 
 13. 
 
 103.9. 
 
 4. $498.08. 
 
 
 C, $112.12. 
 
 14. 
 
 1.5411 +. 
 
150 
 
 ANSWERS 
 
 16. 
 16. 
 
 If 
 
 17. 
 
 tV^. 
 
 18. 
 
 .91287 + . 
 
 19. 
 
 5.6. 
 
 20. 
 
 8^ 
 
 21. 
 
 1008 ft. 
 
 22. 
 
 240.33 rd. 
 
 23. 
 
 52 rd. 
 
 24. 
 
 $187.20. 
 
 25. 
 
 80 X 40 rd 
 
 26. 
 
 101.2 rd. 
 
 27. 
 
 107.33 rd. 
 
 28. 
 
 182 sq. rd. 
 
 29. 
 
 49 rows. 
 
 3. 
 
 4. 
 
 5. 
 
 6. 
 
 7. 
 
 8. 
 
 9. 
 10. 
 11. 
 12. 
 13. 
 14. 
 15. 
 16. 
 17. 
 18. 
 19. 
 20. 
 
 Examples 46 
 
 , 101. 
 165. 
 101.5. 
 15.98. 
 .45. 
 .046. 
 3.4056. 
 10.77. 
 
 M. 
 
 1.42+. 
 .7936. 
 2 ft. 
 
 12150 sq. ft. 
 5 ft. 8+ in. 
 9 ft. 5.3+ in. 
 8 ft. 1.4 in. 
 3.17 ft. 
 
 Examples 47 
 
 1. 84 sq. ft. 
 
 2. 5|A. 
 
 3. 42j3^ sq. ft. 
 
 4. $449.07. 
 
 5. 210 sq. ft. 
 
 6. 27 ft. 
 
 7. 4^ ft. 
 
 8. 20 in. 
 
 9. 39 ft. 
 
 10. 50 ft. 
 
 11. 56.57 rd. 
 
 12. 28 ft. 3.36 in. 
 
 13. 21ft. 
 
 14. 20 ft. 
 
 15. 15.59. in. 
 
 16. 10.6 rd.; 
 112.5 sq. rd. 
 
 17. 45.08 ft. 
 
 18. 92.45 mi. 
 
 Examples 48 
 
 1. 153 sq. ft. 
 
 2. 2 A. 
 714 sq. ft. 
 Given. 
 7.794+ in. 
 259.8+ sq. in. 
 
 Examples 49 
 
 1. 5 ft. 2.83 in. 
 
 2. 5 ft. 10.67 in. 
 8. 78.54 sq. ft. 
 
 4. 114.59 A. 
 
 5. 15.708 ft. 
 
 8. 4.24 rd. 
 
 7. 7.07+ ft. 
 
 8. 69.12 rd. ; 
 380.13 sq. rd. 
 
 9. 4.55 ft. 
 
 10. 78.54 sq. ft. 
 
 11. $89.68. 
 
 12. 420+ times. 
 
 13. 100.399 sq. rd. 
 
 diagonal ; 
 251.6 rd.circum. 
 
 Examples 50 
 
 1. 40 A. 
 
 2. 34.64+ rd. 
 
 3. 149 ft. 4 in. 
 
 4. 30 rd.; 20 rd. 
 
 5. 369 rd. ; 123 rd. 
 
 6. 3.5 in. 
 
 7. $75. 
 
 8. 126.78 rd. 
 
 Examples 51 
 
 1. 140 sq. in. 
 
 2. 263.89 sq.ft. 
 
 3. 73.63 cu. ft. 
 
 4. 301.177 sq. ft. 
 
 5. 24 cu. ft. 
 
 6. 62 sq. ft. 
 
 7. 3.27^ sq. ft. 
 
 8. 179.07 sq. in. 
 
 9. 3.91 qt. 
 
 10. 18.16 in. 
 
 11. 11.66 in. 
 
 12. 68.80 sq. ft. 
 

 Examples 52 
 
 Examples 53 
 
 4. 
 
 .54433. 
 
 1. 
 
 2. 
 
 3. 
 
 120 sq. ft. 
 824.67 sq. ft. 
 429^ sq. ft. 
 
 1. 254.47 sq. in. 
 
 2. 3053.6+ cu. in. 
 8. .58 qt. 
 
 5. 
 6. 
 
 1.2164+. 
 13.228 ft. edge ; 
 2315.03 cu. ft. 
 
 4. 
 6. 
 6. 
 
 64.99 cu. ft. 
 795.871 cu. in. 
 257 sq. ft. 
 
 4. 201062400 sq. 
 mi.; 
 268083200000 
 
 7. 
 8. 
 
 vol. 
 11 ft. 7 in. 
 1494.257 gal. 
 
 7. 
 8. 
 9. 
 
 1 cd. 98^ cu. ft. 
 6.14 bu. 
 54400 cu. ft. 
 
 cu. mi. 
 Examples 54 
 
 9. 
 10. 
 
 $5.46. 
 
 1 cu. ft. vol. of 
 cube; 
 
 10. 
 
 21.38 cu. ft. 
 
 1. 14.42 in. 
 
 
 1 cu. ft. 659 
 
 11. 
 
 $1276.12. 
 
 2. 9 1b. 
 
 
 cu. in. vol. of 
 
 12. 
 
 3+ cu. ft. 
 
 8. 1.389+ gal. 
 
 
 sphere. 
 
 ^ OF THE 
 
 UNIVERSITY 
 
 OF 
 
NEW AMERICAN EDITION OF 
 
 HALL AND KNIGHT'S ALGEBRA, 
 
 FOR COLLEGES AND SCHOOLS. 
 
 Revised and Enlarged for the Use of American Schools 
 and Colleges. 
 
 By FRANK L. SEVENOAK, A.M., 
 
 Assistant Principal of the Academic Department^ Stevens 
 Institute of Technology. 
 
 Half leather. 12mo. $1.10. 
 
 JAMES LEE LOVE, Instructor of Mathematics, Harvard University, 
 Cambridge, Mass.: — Professor Sevenoak's revision of the Elementary Algebra 
 is an excellent book. I wish I could persuade all the teachers fitting boys for the 
 Lawrence Scientific School to use it. 
 
 VICTOR C. ALDERSON, Professor of Mathematics, Armour Institute, 
 Chicago, 111.: — We have used the English Edition for the past two years in our 
 Scientific Academy. The new edition is superior to the old, and we shall certainly 
 use it. In my opinion it is the best of all the elementary algebras. 
 
 AMERICAN EDITION OF 
 
 ALGEBRA FOR BEGINNERS. 
 
 By H. S. HALL, M.A., and S. R. KNIGHT. 
 
 REVISED BY 
 
 FRANK L. SEVENOAK, A.M., 
 
 Assistant Principal of the Academic Department^ Stevens 
 Institute of Technology. 
 
 16mo. Cloth. 60 cents. 
 
 An edition of this book containing additional chapters on Radicals and 
 the Binomial Theorem will be ready shortly. 
 
 JAMES S. LEWIS, Principal University School, Tacoma, Wash.:— I have 
 examined Hall and Knight's "Algebra for Beginners " as revised by Professor Scv- 
 enoak, and consider it altogether the best book for the purpose intended that I 
 know of. 
 
 MARY McCLUN, Principal Clay School, Fort Wayne, Indiana: — I have 
 
 examined the Algebra quite carefully, and I find it the best I have ever seen. Its 
 greatest value is found in the simple and clear language in which all its definitions 
 are expressed, and in the fact that each new step is so carefully explained. The ex- 
 amples in each chapter are well selected. I wish all teachers who teach Algebra 
 might be able to use the "Algebra for Beginners." 
 
 THE MACMILLAN COMPANY, 
 
 66 FIFTH AVENUE, NEW YORK, 
 
AMERICAN EDITION 
 
 OF 
 
 LOCK'S 
 
 TRIGONOMETRY FOR BEGINNERS, 
 
 WITH TABLES. 
 
 Revised for the Use of Schools and Colleges 
 
 By JOHN ANTHONY MILLER, A.M., 
 
 Professor of Mechanics and Astronomy at the Indiana University, 
 
 8vo. Cloth. $1.10 net. 
 
 IN PREPARA TION. 
 
 AMERICAN EDITION 
 
 OF 
 
 HALL and KNIGHT'S 
 ELEMENTARY TRIGONOMETRY, 
 
 WITH TABLES. 
 
 By H. 8. HALL, M.A., and S. R. KNIGHT, B.A. 
 
 Rerised and Enlarged for the Use of American 
 Schools and Colleges 
 
 By FRANK L. SEVENOAK, A.M., 
 
 Asnttant Principal of the Academic Department, Stevens 
 Institute of Technology. 
 
 THE MACMILLAN COMPANY, 
 
 66 FIFTH AVENUE, NE^W YORK. 
 
ELEMENTARY SOLID GEOMETRY. 
 
 BY 
 
 HENRY DALLAS THOMPSON, D.Sc., Ph.D., 
 
 Professor of Mathematics in Princeton University. 
 
 »mo. Cloth. $1.10, net. 
 
 This is an elementary work on Geometry, brief and interesting, well conceived, 
 d well written. — School of Mines Quarterly. 
 
 THE ELEMENTS OF GEOMETRY. 
 
 By GEORGE CUNNINGHAM EDWARDS, 
 
 Associate Professor of Mathematics in the University of California. 
 
 i6mo. Cloth. $1.10, net. 
 
 PROP. JOHN F. DOWNEY, University of Minnesota : — There is a gain in 
 its being less formal than many of the works on this subject. The arrangement and 
 treatment are such as to develop in the student ability to do geometrical work. The 
 book would furnish the preparation necessary for admission to this University. 
 
 PRIN, F. 0. MOWER, Oak Normal School, Napa, Cal.: — Of the fifty or 
 more English and American editions of Geometry which I have on my shelves, I 
 consider this one of the best, if not the best, of them all. I shall give it a trial in my 
 next class beginning that subject. 
 
 THE MACMILLAN COMPANY, 
 
 66 FIFTH AVENUE, NEW YORK. 
 
MATHEMATICAL TEXT-BOOKS 
 
 SUITABLE 
 
 FOR USE IN PREPARATORY SCHOOLS. 
 
 SELECTED FROM THE LISTS OF 
 
 THE MACMILLAN COMPANY, Publishers. 
 
 ARITHMETIC FOR SCHOOLS. 
 
 By J. B. LOCK, 
 
 Author of " Trigonometry for Beginners" " Elementary Trigonometry," etc^ 
 
 Edited and Arranged for American Schools 
 
 By CHARLOTTE ANGAS SCOTT, D.SC, 
 
 Head of Math. Dept., Bryn Mawr College, Pa. 
 
 16mo. Cloth. 75 cents. 
 
 " Evidently the work of a thoroughly good teacher. The elementary truth, that 
 arithmetic is common sense, is the principle which pervades the whole book, and no 
 process, however simple, is deemed unworthy of clear explanation. Where it seems 
 advantageous, a rule is given after the explanation. . . . Mr. Lock's admirable 
 ' Trigonometry ' and the present work are, to our mind, models of what mathematical 
 school books should be." — ' The Literary World. 
 
 FOR MORE ADVANCED CLASSES. 
 
 \ ARITHMETIC. 
 
 By CHARLES SMITH, M.A., 
 Author of *' Elementary Algebra," "A Treatise on Algebra^* 
 
 AND 
 
 CHARLES L. HARRINGTON, M.A., 
 
 Head Master of Dr. J. Sach's School for Boys, New York. 
 1 6mo. Cloth. 00 cents. 
 
 A thorough and comprehensive High School Arithmetic, containing many good 
 examples and clear, well-arranged explanations. 
 
 There are chapters on Stocks and Bonds, and on Exchange, which arc of more 
 than ordinary value, and there is also a useful collection of miscellaneous examples. 
 
 THE MACMILLAN COMPANY, 
 
 66 FIFTH AVENUE, NEW YORK. 
 
...» ... ..»,~ij,!is s; sSi" E-¥.H: 
 
 AUG 3 191! 
 
 .jyt 24 '919 
 SEP 28 W25 
 
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 60m1,'l6 
 
YB 12539