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REESE LIBRARY 
 
 OF THR 
 
 UNIVERSITY OF CALIFORNIA. 
 
 Received 
 
 CUS^ 
 Accessions No..'~F&.&..& Shelf No 
 
V 
 
THE 
 
 PROGRESSIVE 
 
 PRACTICAL ARITHMETIC, 
 
 CONTAINING 
 
 THE THEORY OF NUMBERS, IX CONNECTION WITH CONCISE ANALYTIC 
 
 AND SYNTHETIC METHODS OF SOLUTION, AND DESIGNED 
 
 AS A COMPLETE TEXT-BOOK ON THIS SCIENCE, 
 
 FOR 
 
 COMMON SCHOOLS AND ACADEMIES. 
 
 BY 
 
 HORATIO N. ROBINSON, LL.D., 
 
 AUTHOR OF A WORK ON ALGEBJft*J ' ^^METiiTIS^SfcSlf^OMETRY, SURVEYING ATO 
 NAVIGATION, ASTRONMjfc.' fUirBBBNTIAL ' JUffl l&TJRpR^L CALCULUS, ETC. 
 
 UNIVERSITY 
 
 IVISON & PHINNEY, 48 & 50 WALKER ST. 
 CHICAGO: S. C. GRIGGS & CO. 
 
 CINCINNATI : MOORE, WILSTACH, KEYS & CO. PHILADELPHIA : SOWER, BARNES * CO. 
 
 BOSTON: BROWN, TAGGARD AND CHASE. ST. LOUIS: KEITH AND WOODS. 
 
 NEW ORLEANS : BLOOMFIELD, STEELE & CO. BUFFALO : PHINNEY & CO. 
 
 1859. 
 
Entered, according to Act of Congress, in the year 1858, by 
 HORATIO N. ROBINSON, LL.D., 
 
 la the Clerk's Office of the District Court of the United States for the Northern 
 District of New Yorl 
 
 ELECTROTYPE!) AT THE 
 BOSTON STEREOTYPE FOUNDRY. 
 
UNIVERSITY 
 
 PROGRESS and improvement characterize almost every art and 
 science ; and within the last few years the science of Arithmetic has 
 received many important additions and improvements, which have 
 appeared from time to time successively in the different treatises pub- 
 lished upon this subject. 
 
 In the preparation of this work it has been the author's aim to com- 
 bine, and to present hi one harmonious whole, all these modern im- 
 provements, as well as to introduce some new methods and practical 
 operations not found in other works of the same grade ; in short, to 
 present the subject of Arithmetic to the pupil more as a science than 
 an art ; to teach him methods of thought, and how to reason, rather 
 than what to do; to give unity, system, and practical utility to the 
 science and art of computation. 
 
 The author believes that both teacher and pupil should have the 
 privilege, as well as the benefit, of performing at least a part of the 
 thinking and the labor necessary to the study of Arithmetic ; hence 
 the present work has not "been encumbered with the multiplicity of 
 "notes," "suggestions," and superfluous operations so common to 
 most Practical Arithmetics of the present day, and which prevent the 
 cultivation of that self-reliance, that clearness of thought, and that 
 vigor of intellect, which always characterize the truly educated mind. 
 
 The author claims for this treatise improvement upon, if not superi- 
 ority over, others of the kind in the following particulars, viz. : In 
 the mechanical and typographical style of the tcork ; the open and 
 attractive page ; the progressive and scientific arrangement of the 
 subjects ; clearness and conciseness of definitions ; fullness and accu- 
 racy in the new and improved methods of operations and analyses; 
 brevity and perspicuity of rules; and in the very large number of 
 
 (iii) 
 
IV PREFACE. 
 
 examples prepared and arranged with special reference to their prac- 
 tical utility, and their adaptation to the real business of active life. 
 The answers to a part of the examples have been omitted, that the 
 learner may acquire the discipline resulting from verifying the opera- 
 tions. 
 
 Particular attention is invited to improvements in the subjects of 
 Common Divisors, Multiples, Tractions, Percentage, Interest, Propor- 
 tion, Analysis, Alligation, and the Roots, as it is believed these 
 articles contain some practical features not common to other authors 
 upon these subjects. 
 
 It is not claimed that this is a perfect work, for perfection is impossi- 
 ble ; but no effort has been spared to present a clear, scientific, com- 
 prehensive, and complete system, sufficiently full for the business man 
 and the scholar ; not encumbered with unnecessary theories, and yet 
 combining and systemizing real improvements of a practical and 
 useful nature. How nearly this end has been attained the intelligent 
 and experienced teacher and educator must determine. 
 
 In conclusion, the author takes this opportunity of acknowledging 
 his obligations to those teachers from whom he has received valuable 
 hints and suggestions, many of which have been incorporated into 
 this work. He desires also to make particular acknowledgment for 
 the valuable services rendered in the preparation of this work by Mr. 
 D. W. Fish, of Rochester, N. Y., who mostly prepared the Primary 
 and Intellectual Arithmetics of this series, and Mr. J. H. French, 
 of Syracuse, N. Y., Superintendent of the New York State Map and 
 Atlas Survey. These gentlemen have had long and successful experi- 
 ence as teachers, and to them the public are indebted for the excellent 
 plan and superior arrangement of this treatise, and for many of the 
 new and practical features which it contains. 
 
CONTENTS. 
 
 SIMPLE NUMBERS. 
 
 Page 
 
 Definitions, 9 
 
 Roman Notation, 10 
 
 Table of Roman Notation, , . 11 
 
 Arabic Notation, 12 
 
 Numeration Table, 17 
 
 Laws and Rules for Notation and Numeration, 18 
 
 Addition, 22 
 
 Subtraction 32 
 
 Multiplication, 41 
 
 Contractions, 48 
 
 Division 53 
 
 Contractions, 63 
 
 Applications of preceding Rules, 67 
 
 General Principles of Division, 72 
 
 Exact Divisors, 74 
 
 Prime Numbers, 75 
 
 Factoring Numbers, 75 
 
 Cancellation, 77 
 
 Greatest Common Divisor, 81 
 
 Multiples, 87 
 
 Classification of Numbers, 92 
 
 COMMON FRACTIONS. 
 
 Definitions, &c., 94 
 
 General Principles of Fractions, 97 
 
 Reduction of Fractions, , 97 
 
 * (v) 
 
VI CONTENTS. 
 
 Page 
 
 Addition of Fractions, 104 
 
 Subtraction of Fractions, 106 
 
 Multiplication of Fractions, 109 
 
 Division of Fractions, 114 
 
 Promiscuous Examples, 120 
 
 DECIMALS. 
 
 Decimal Notation and Numeration, 124 
 
 Reduction of Decimals, 129 
 
 Addition of Decimals, 132 
 
 Subtraction of Decimals, 134 
 
 Multiplication of Decimals, 135 
 
 Division of Decimals, 136 
 
 DECIMAL CURRENCY. 
 
 Notation and Numeration of Decimal Currency, 139 
 
 Reduction of Decimal Currency, 141 
 
 Addition of Decimal Currency, 142 
 
 Subtraction of Decimal Currency, 143 
 
 Multiplication of Decimal Currency, 144 
 
 Division of Decimal Currency, 145 
 
 Additional Applications, 147 
 
 When the Price is an Aliquot Part of a Dollar, 147 
 
 To find the Cost of a Quantity, 148 
 
 To find the Price of One, 149 
 
 To find the Quantity 149 
 
 Articles sold by the 100 or 1000, 150 
 
 Articles sold by the Ton, 151 
 
 Bills / 152 
 
 Promiscuous Examples 154 
 
 COMPOUND NUMBERS. 
 
 Reduction, 158 
 
 Definitions, &c 7 158 
 
 English Money, 159 
 
 Troy Weight 161 
 
CONTENTS. Vll 
 
 Page 
 
 Apothecaries' Weight, 162 
 
 Avoirdupois Weight 163 
 
 Long Measure, 166 
 
 Surveyors' Long Measure, 168 
 
 Square Measure, 169 
 
 Surveyors' Square Measure, 172 
 
 Cubic Measure, 173 
 
 Liquid Measure, 175 
 
 Dry Measure, 176 
 
 Time, 178 
 
 Circular Measure, 180 
 
 Counting ; Paper ; Books, &c., 181 
 
 Reduction of Denominate Fractions, 183 
 
 Addition of Compound Numbers, 190 
 
 Addition of Denominate Fractions, 193 
 
 Subtraction 194 
 
 To find the Difference in Dates, 196 
 
 Table, 197 
 
 Subtraction of Denominate Fractions, 198 
 
 Multiplication of Compound Numbers, 199 
 
 Division, 201 
 
 Longitude and Tune 203 
 
 Duodecimals 206 
 
 Promiscuous Examples, 210 
 
 PERCENTAGE. 
 
 Definitions, &c., 213 
 
 Commission and Brokerage, 218 
 
 Stocks 222 
 
 Profit and Loss, 225 
 
 Insurance, 231 
 
 Taxes, 232 
 
 Custom House Business, 235 
 
 Simple Interest, 239 
 
 Partial Payments or Indorsements, 245 
 
 Problems in Interest, 251 
 
Vlll CONTENTS. 
 
 Page 
 
 Compound Interest 254 
 
 Discount, 257 
 
 Banking, 260 
 
 Exchange, 264 
 
 Equation of Payments, 269 
 
 RATIO AND PROPORTION. 
 
 Ratio 277 
 
 Proportion, 280 
 
 Simple Proportion 281 
 
 Compound Proportion, 287 
 
 Partnership, 292 
 
 Analysis, 296 
 
 Alligation Medial, 305 
 
 Alligation Alternate, 306 
 
 Involution, . 311 
 
 Evolution, 312 
 
 Square Root, 313 
 
 Cube Root, 320 
 
 Arithmetical Progression, 326 
 
 Geometrical Progression, 329 
 
 Promiscuous Examples, 332 
 
PRACTICAL ARITHMETIC. 
 
 DEFINITIONS. 
 
 1. Quantity is any thing that can be increased, diminished, 
 or measured. 
 
 2. Mathematics is the science of quantity. 
 
 3. A Unit is one, or a single thing. 
 
 4. A Number is a unit, or a collection of units. 
 o. An Integer is a whole number. 
 
 6. The Unit of a Number is one of the same kind or 
 name as the number. Thus, the unit of 23 is 1 ; of 23 dollars, 
 1 dollar ; of 23 feet, 1 foot. 
 
 7. Like Numbers have the same kind of unit. Thus, 74, 
 16, and 250 ; 7 dollars and 62 dollars ; 19 pounds, 320 pounds, 
 and 86 pounds ; 4 feet 6 inches, and 17 feet 9 inches. 
 
 8. An Abstract Number is a number used without refer- 
 ence to any particular thing or quantity. Thus, 17 ; 365 ; 8540. 
 
 9. A Concrete Number is a number used with reference 
 to some particular thing or quantity. Thus, 17 dollars ; 365 
 days ; 8540 men. 
 
 NOTES. 1. The unit of an abstract number is 1, and is called Unity. 
 2. Concrete numbers are, by some, called Denominate Numbers. 
 Denomination means the name of the unit of a concrete number. 
 
 10. Arithmetic is the Science of numbers, and the Art of 
 computation. 
 
 11. A Sign is a character indicating an operation to be 
 performed. 
 
 1. A Rule is a prescribed method of performing an op- 
 eration. 
 
 Define quantity. Mathematics. A unit. A number. An integer. 
 
 The unit of a number. Like numbers. An abstract number. A 
 
 concrete number. The unit of an abstract number. Denominate 
 numbers. Arithmetic. A sign, or symbol. A rule. 
 
10 SIMPLE NUMBERS. 
 
 NOTATION AND NUMERATION. 
 
 13. Notation is a method of writing or expressing numbers 
 by characters ; and, 
 
 14:. Numeration is a method of reading numbers expressed 
 by characters. 
 
 15. Two systems of notation are in general use the 
 Roman and the Arabic. 
 
 NOTE. The Roman Notation is supposed to have been first used 
 by the Romans ; hence its name. The Arabic Notation was intro- 
 duced into Europe by the Arabs, by whom it was supposed to have 
 been invented. But investigations have shown that it was adopted by 
 them about 600 years ago, and that it has been in use among the Hin- 
 doos more than 2000 years. From this latter fact it is sometimes 
 called the Indian Notation. 
 
 THE ROMAN NOTATION 
 
 1C. Employs seven capital letters to express numbers, 
 thus : 
 
 Letters, I V X L C D M 
 
 Values, one, five, ten, fifty, h 5* d . hn Sj 
 
 one 
 hundred, hundred, thousand. 
 
 17. The Roman notation is founded upon five principles, 
 as follows : 
 
 1st. Repeating a letter repeats its value. Thus, II repre- 
 sents two, XX twenty, CCC three hundred. 
 
 2d. If a letter of any value be placed after one of greater 
 value, its value is to be united to that of the greater. Thus, 
 XI represents eleven, LX sixty, DC six hundred. 
 
 3d. If a letter of any value be placed before one of greater 
 value, its value is to be taken from that of the greater. Thus, 
 IX represents nine, XL forty, CD four hundred. 
 
 Define notation. Numeration. What systems of notation are now 
 in general use ? From what are their names derived ? What ar> 
 to express numbers in the Roman notation ? What is the value of each ? 
 What is the first principle of combination ? Second ? Third ? 
 
NOTATION AND NUMERATION. 11 
 
 4th. If a letter of any value be placed between two letters, 
 each of greater value, its value is to be taken from the united 
 value of the other two. Thus, XIV represents fourteen, 
 XXIX twenty-nine, XCIV ninety-four. 
 
 5th. A bar or dash placed over a letter increases its value 
 one thousand fold. Thus, V signifies five, and V five thou- 
 sand ; L fifty, and L fifty thousand. 
 
 TABLE OF ROMAN NOTATION. 
 
 I is One. XX is Twenty. 
 
 II " Two. XXI " Twenty-one. 
 
 III " Three. XXX " Thirty. 
 
 IV " Four. XL " Forty. 
 V " Five. L " Fifty. 
 
 VI " Six. LX " Sixty. 
 
 VII " Seven. LXX " Seventy. 
 
 VIII " Eight. LXXX " Eighty. 
 
 IX " Nine. XC Ninety. 
 
 X Ten. C " One hundred. 
 
 XI " Eleven. CC " Two hundred. 
 
 XII " Twelve. . D " Five hundred. 
 
 XIII " Thirteen. DC " Six hundred. 
 
 XIV " Fourteen. M " One thousand, [dred. 
 XV " Fifteen. MC " One thousand one hun- 
 
 XVI " Sixteen. MM " Two thousand. 
 
 XVII " Seventeen. ^X " Ten thousand. 
 
 XVIII " Eighteen. ~C " One hundred thousand. 
 
 XIX " Nineteen. M " One million. 
 
 NOTE. The system of Roman notation is not well adapted to the 
 purposes of numerical calculation ; it is principally confined to the 
 numbering of chapters and sections of books, public documents, &c. 
 
 Express the following numbers by letters : 
 
 1. Eleven. Ans. XL 
 
 2. Fifteen. Ans. 
 
 Fourth ? Fifth ? Repeat the table. What is the value of LVTI ? 
 
 CLxxm? xcvin^_ CDXXXII? xcix- DCXIX? 
 
 VMDCCXLIX ? MDXXVCDLXXXIX ? To what uses is the 
 Roman notation now principally confined ? 
 
12 SIMPLE NUMBERS. 
 
 3. Twenty-five. 
 
 4. Thirty-nine. 
 
 5. Forty-eight. 
 
 6. Seventy-seven. 
 
 7. One hundred fifty-nine. 
 
 8. Five hundred ninety-four. 
 
 9. One thousand five hundred thirty-eight. 
 
 10. One thousand nine hundred ten. 
 
 11. Express the present year. 
 
 THE ARABIC NOTATION 
 
 18. Employs ten Characters or figures to express numbers. 
 Thus, 
 
 Figures, 0123456789 
 
 Nimcs and } naught one, two, three, four, five, six, seven, eight, nine. 
 
 C or 
 values, $ ciphcr> 
 
 19. The first character is called naught, because it has no 
 value of its own. The other nine characters are called signif- 
 icant figures, because each has a value of its own. 
 
 SO. The significant figures are also called Digits, a word de- 
 rived from the Latin term digitus, which signifies finger. 
 
 21. The naught or cipher is also called nothing, and zero. 
 
 The ten Arabic characters are the Alphabet of Arithmetic, 
 and by combining them according to certain principles, all 
 numbers can be expressed. We will now examine the most 
 important of these principles.* 
 
 %%. Each of the nine digits has a value of its own ; hence 
 any number not greater than 9 can be expressed by one 
 figure. 
 
 * Fractional and decimal notation, and the notation of compound numbers, will be 
 discussed in their appropriate places. 
 
 What are used to express numbers in the Arabic notation ? What 
 is the value of each ? What general name is given to the significant 
 figures ? Why ? Numbers less than ten, how expressed ? 
 
NOTATION AND NUMERATION. 13 
 
 S3. As we have no single character to represent ten, we 
 express it by writing the unit, 1, at the left of the cipher, ? 
 thus, 10. In the same manner we represent 
 
 2 tens, 3 tens, 4 tens, 5 tens, 6 tens, 7 tens, 8 tens, 9 tens, 
 
 or or or or or or or or 
 
 twenty, thirty, forty, fifty, sixty, seventy, eighty, ninety, 
 
 20; 30; 40; 50; 60; 70; 80; 90. 
 
 24. When a number is expressed by two figures, the right 
 hand figure is called units, and the left hand figure tens. 
 
 We express the numbers between 10 and 20 by writing 
 the 1 in the place of tens, with each of the digits respectively 
 in the place of units. Thus, 
 
 eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, nineteen. 
 
 11, 12, 13, 14, 15, 16, 17, 18, 19. 
 
 In like manner we express the numbers between 20 and 
 30, between 30 and 40, and between any two successive tens. 
 Thus, 21, 22, 23, 24, 25, 26, 27, 28, 29, 34, 47, 56, 72, 93. 
 The greatest number that can be expressed by two figures 
 is 99. 
 
 SJt5. We express one hundred by writing the unit, 1, at the 
 left hand of two ciphers, or the number 10 at the left hand of 
 one cipher; thus, 100. In like manner we write two hun- 
 dred, three hundred, &c., to nine hundred. Thus, 
 
 one two three four five six seven eight nine 
 hundred, hundred, hundred, hundred, hundred, hundred, hundred, hundred, hundred, 
 
 100, 200, 300, 400, 500, 600, 700, 800, 900. 
 
 26. When a number is expressed by three figures, the 
 right hand figure is called units, the second figure tens, and 
 the left hand figure hundreds. 
 
 As the ciphers have, of themselves, no value, but are always 
 used to denote the absence of value in the places they occupy, 
 
 Tens, how expressed ? The right hand figure called what ? Left 
 hand figure, what ? What is the greatest number that can be expressed 
 by two figures ? One hundred, how expressed ? When numbers are 
 expressed by three figures, what names are given to each 5 
 
14 SIMPLE NUMBERS. 
 
 we express tens and units with hundreds, by writing, in place 
 of the ciphers, the numbers representing the tens and units. 
 To express one hundred fifty we write 1 hundred, 5 tens, and 
 units; thus, 150. To express seven hundred ninety-two, 
 we write 7 hundreds, 9 tens, and 2 units ; thus, 
 
 I I 1 
 
 792 
 
 The greatest number that can be expressed by three figures 
 is 999. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Write one hundred twenty-five. 
 
 2. Write four hundred eighty-three. 
 
 3. Write seven hundred sixteen. 
 
 4. Express by figures nine hundred. 
 
 5. Express by figures two hundred ninety. 
 
 6. Write eight hundred nine. 
 
 7. Write five hundred five. 
 
 8. Write five hundred fifty-seven. 
 
 ST. We express one thousand by writing the unit, 1, at 
 the left hand of three ciphers, the number 10 at the left hand 
 of two ciphers, or the number 100 at the left hand of one 
 cipher ; thus, 1000. In the same manner we write two 
 thousand, three thousand, &c., to nine thousand ; thus, 
 
 one two three four five six seven eight nine 
 
 thousand, thousand, thousand, thousand, thousand, thousand, thousand, thousand, thousand. 
 
 1000, 2000, 3000, 4000, 5000, 6000, 7000, 8000, 9000. 
 
 S8. When a number is expressed by four figures, the 
 places, commencing at the right hand, are units, tens, hundreds, 
 thousands. 
 
 Use of the cipher, what ? Greatest number that can be expressed by 
 three figures ? One thousand, how expressed ? How many figures 
 used ? Names of each ? 
 
NOTATION AND NUMERATION. 15 
 
 To express hundreds, tens, and units with thousands, we 
 write in each place the figure indicating the number we wish 
 to express in that place. To write four thousand two hun- 
 dred sixty-nine, we write 4 in the place of thousands, 2 in the 
 place of hundreds, 6 in the place of tens, and 9 in the place of 
 units; thus, 
 
 I I 1 I 
 
 4t z o v 
 
 The greatest number that can be expressed by/owr figures 
 f)999. 
 
 EXAMPLES FOR PRACTICE. 
 
 Express the following numbers by figures : 
 
 1. One thousand two hundred. 
 
 2. Five thousand one hundred sixty. 
 
 3. Three thousand seven hundred forty-one. 
 
 4. Eight thousand fifty-six. 
 
 5. Two thousand ninety. 
 
 6. Seven thousand nine. 
 
 7. One thousand one. 
 
 8. Nine thousand four hundred twenty-seven. 
 
 9. Four thousand thirty-five. 
 
 10. One thousand nine hundred four. 
 Head the following numbers : 
 
 11. 76; 128; 405; 910; 116; 3416; 1025. 
 
 12. 2100; 5047; 7009; 4670; 3997; 1001. 
 
 Next to thousands come tens of thousands, and next 
 to these come hundreds of thousands, as tens and hundreds 
 come in their order after units. Ten thousand is expressed 
 by removing the unit, 1, one place to the left of the place 
 
 Greatest number expressed by four figures r Tens of thousands, how 
 expressed ? Hundreds of thousands ? 
 
16 SIMPLE NUMBERS. 
 
 of thousands, or by writing it at the left hand of four ci- 
 phers; thus, 10000; and one hundred thousand is expressed 
 by removing the unit, 1, still one place further to the left, or 
 by writing it at the left hand of five ciphers ; thus, 100000. 
 "We can express thousands, tens of thousands, and hundreds of 
 thousands in one number, in the same manner as we express 
 units, tens, and hundreds in one number. To express five 
 hundred twenty-one thousand eight hundred three, we write 5 in 
 the sixth place, counting from units, 2 in the fifth place, 1 in 
 the fourth place, 8 in the third place, in the second place, 
 (because there are no tens,) and 3 in the place of units; 
 thus, 
 
 II S J I I 1 1 
 
 5 21803 
 
 The greatest number that can be expressed by five figures 
 is 99999 ; and by six figures, 999999. 
 
 EXAMPLES FOB PRACTICE. 
 
 Write the following numbers in figures : 
 
 1. Twenty thousand. 
 
 2. Forty-seven thousand. 
 
 3. Eighteen thousand one hundred. 
 
 4. Twelve thousand three hundred fifty. 
 
 5. Thirty-nine thousand five hundred twenty-two. 
 
 6. Fifteen thousand two hundred six. 
 
 7. Eleven thousand twenty-four. 
 
 8. Forty thousand ten. 
 
 9. Sixty thousand six hundred. 
 
 10. Two hundred twenty thousand. 
 
 11. One hundred fifty-six thousand. 
 
 12. Eight hundred forty thousand three hundred. 
 
 Greatest number expressed by five figures ? Six figures ? 
 
NOTATION AND NUMERATION. 17 
 
 13. Five hundred one thousand nine hundred sixty -four. 
 
 14. One hundred thousand one hundred. 
 
 15. Three hundred thirteen thousand three hundred thir- 
 teen. 
 
 16. Seven hundred eighteen thousand four. 
 
 17. One hundred thousand ten. 
 Read the following numbers : 
 
 18. 5006; 12304; 96071; 5470; 203410. 
 
 19. 36741; 400560; 13061; 49000; 100010. 
 
 20. 200200; 75620; 90402; 218094; 100101. 
 
 For convenience in reading large numbers, we may point 
 them off, by commas, into periods of three figures each, count- 
 ing from the right hand or unit figure. This pointing enables 
 us to read the hundreds, tens, and units in each period with 
 facility. 
 
 3O. Next above hundreds of thousands we have, succes- 
 sively, units, tens, and hundreds of millions, and then follow 
 units, tens, and hundreds of each higher name, as seen in the 
 following 
 
 NUMERATION TABLE. 
 
 I | | j . f . | 
 
 CM CM ^*-( CM tt_i tj-j ft i ti_. Ci_i 
 
 000 000000 
 
 r~^ 
 
 4 
 
 98,765,432,109,876,556,789,012,345 
 
 ninth eighth seventli sixth fifth fourth third second first 
 period, period, period, period, period, period, period, period, period. 
 
 How may figures be pointed off? One million, how expressed? 
 Next period above millions, what ? Give the name of each successive 
 period. 
 
 B* 
 
18 SIMPLE NUMBERS. 
 
 NOTE. This is called the French method of pointing off the peri- 
 ods, and is the one in general use in this country. 
 
 31. Figures occupying different places in a number, as 
 units, tens, hundreds, &c., are said to express different orders 
 of units. 
 
 Simple units are called units of the first order. 
 Tens " " " " " second " 
 
 Hundreds " " " third " 
 
 Thousands " " fourth 
 
 Tens of thousands " " " " " fifth " 
 
 and so on. Thus, 452 contains 4 units of the third order, 5 
 units of the second order, and 2 units of the first order. 
 1,030,600 contains 1 unit of the seventh order, (millions,) 3 
 units of the fifth order, (tens of thousands,) and 6 units of the 
 third order, (hundreds.) 
 
 EXAMPLES FOR PRACTICE. 
 
 Write and read the following numbers : 
 
 1. One unit of the third order, four of the second. 
 
 2. Three units of the fifth order, two of the third, one of the 
 first. 
 
 3. Eight units of the fourth order, five of the second. 
 
 4. Two units of the seventh order, nine of the sixth, four 
 of the third, one of the second, seven of the first. 
 
 5. Three units of the sixth order, four of the second. 
 
 6. Nine units of the eighth order, six of the seventh, three 
 of the fifth, seven of the fourth, nine of the first. 
 
 7. Four units of the tenth order, six of the eighth, four of 
 the seventh, two of the sixth, one of the third, five of the sec- 
 ond. 
 
 8. Eight units of the twelfth order, four of the eleventh, six 
 of the tenth, nine of the seventh, three of the sixth, five of the 
 fifth, two of the third, eight of the first. 
 
 Units of different orders are what ? 
 
NOTATION AND NUMERATION. 19 
 
 
 
 32. From the foregoing explanations and illustrations, we 
 derive several important principles, which we will now pre- 
 sent. 
 
 1st. Figures have two values, Simple and Local. 
 
 The Simple Value of a figure is its value when taken alone ; 
 thus, 2, 5, 8. 
 
 The Local Value of a figure is its value when used with an- 
 other figure or figures in the same number ; thus, in 842 the 
 simple values of the several figures are 8, 4, and 2 ; but the 
 local value of the 8 is 800 ; of the 4 is 4 tens, or 40 ; and of 
 the 2 is 2 units. 
 
 NOTE. When a figure occupies units' place, its simple and local 
 values are the same. 
 
 2d. A digit or figure, if used in the second place, expresses 
 tens ; in the third place, hundreds ; in the fourth place, thou- 
 sands ; and so on. 
 
 3d. As 10 units make 1 ten, 10 tens 1 hundred, 10 hun- 
 dreds 1 thousand, and 10 units of any order, or in any place, 
 make one unit of the next higher order, or in the next place 
 at the left, we readily see that the Arabic method of notation 
 is based upon the following 
 
 TWO GENERAL LAWS. 
 
 I. Numbers increase from right to left, and decrease from 
 left to right, in a tenfold ratio. 
 
 II. Every removal of a figure or number one place to the 
 left, increases its local value tenfold ; and every removal of a 
 figure or number one place to the right, diminishes its local 
 value tenfold. 
 
 Thus, 
 
 6 is 6 units. 
 
 60 is 10 times 6 units. 
 
 600 is 10 times 6 tens. 
 
 6000 is 10 times 6 hundreds. 
 
 60000 is 10 times 6 thousands. 
 
 First principle derived ? What is the simple value of a figure ? Local ? 
 Second principle ? Third ? First law of Arabic notation ? Second ? 
 
20 SIMPLE NUMBERS. 
 
 4th. The local value of a figure depends upon its place from 
 units of the first order, not upon the value of the figures at the 
 right of it. Thus, in 425 and 400, the value of the 4 is the 
 same in both numbers, being 4 units of the third order, or 4 
 hundred. 
 
 NOTE. Care should be taken not to mistake the local value of a 
 figure for the value of the whole number. For, although the value of 
 the 4 (hundreds) is the same in the two numbers, 425 and 400, the value 
 of the whole of the first number is greater than that of the second. 
 
 5th. Every period contains three figures, (units, tens, and 
 hundreds,) except the left hand period, which sometimes con- 
 tains only one or two figures, (units, or units and tens.) 
 
 33. As we have now analyzed all the principles upon 
 which the writing and reading of whole numbers depend, we 
 will present these principles in the form of rules. 
 
 RULE FOR NOTATION. 
 
 L Beginning at the left hand, write the figures belonging to 
 the highest period. 
 
 II. Write the hundreds, tens, and units of each successive 
 period in their order, placing a cipher wherever an order of 
 units is omitted. 
 
 RULE FOR NUMERATION. 
 
 L Separate the number into periods of three figures each, 
 commencing at the right hand. 
 
 II. Beginning at the left hand, read each period separately, 
 and give the name to each period, except the last, or period 
 of units. 
 
 34. Until the pupil can write numbers readily, it may be 
 well for him to write several periods of ciphers, point them off, 
 over each period write its name, thus, 
 
 Trillions, Billions, Millions, Thousands, Units. 
 
 000 , 000, 000, 000 , 000 
 
 Fourth principle ? What caution is given ? Fifth principle ? Rule 
 for notation ? Numeration ? 
 
NOTATION AND NUMERATION 
 
 and then write the given numbers underneWBf ifftlef? J^fe-S , 
 priate places. Vv^d . - BI 
 
 x^vsf r * p 
 
 EXERCISES IN NOTATION AND NUMERATION. 
 
 Express the following numbers by figures : 
 
 1. Four hundred thirty-six. 
 
 2. Seven thousand one hundred sixty-four. 
 
 3. Twenty-six thousand twenty-six. 
 
 4. Fourteen thousand two hundred eighty. 
 
 5. One hundred seventy-six thousand. 
 
 6. Four hundred fifty thousand thirty-nine. 
 
 7. Ninety-five million. 
 
 8. Four hundred thirty-three million eight hundred sixteen 
 thousand one hundred forty-nine. 
 
 9. Nine hundred thousand ninety. 
 
 10. Ten million ten thousand ten hundred ten. 
 
 11. Sixty-one billion five million. 
 
 12. Five trillion eighty billion nine million one. 
 
 Point off, numerate, and read the following numbers : 
 
 13. 8240. 
 14. 400900. 
 15. 308. 
 16. 60720. 
 
 17. 1010. 
 18. 57468139. 
 19. 5628. 
 20. 850026800. 
 
 21. 370005. 
 22. 9400706342. 
 23. 38429526. 
 24. 74268113759. 
 
 25. Write seven million thirty-six. 
 
 26. Write five hundred sixty-three thousand four. 
 
 27. Write one million ninety-six thousand. 
 
 28. Numerate and read 9004082501. 
 
 29. Numerate and read 2584503962047. 
 
 30. A certain number contains 3 units of the seventh order, 
 6 of the fifth, 4 of the fourth, 1 of the third, 5 of the second, 
 and 2 of the first ; what is the number ? 
 
 31 . What orders of units are contained in the number 290648 ? 
 
 32. What orders of units are contained in the number 
 1037050? 
 
22 SIMPLE NUMBERS. 
 
 ADDITION. 
 
 MENTAL EXERCISES. 
 
 35. 1. Henry gave 5 dollars for a vest, and 7 dollars for 
 a coat ; how much did he pay for both ? 
 
 ANALYSIS. He gave as many dollars as 5 dollars and 7 dollars, 
 which are 12 dollars. Therefore he paid 12 dollars for both. 
 
 2. A farmer sold a pig for 3 dollars, and a calf for 8 dol- 
 lars ; how much did he receive for both ? 
 
 3. A drover bought 5 sheep of one man, 9 of another, and 
 3 of another ; how many did he buy in all ? 
 
 4. How many are 2 and 6? 2 and 7 ? 2 and 9 ? 2 and 8 ? 
 2 and 10 ? 
 
 5. How many are 4 and 5 ? 4 and 8 ? 4 and 7 ? 4 and 9 ? 
 .6. How many are 6 and 4? 6 and 6 ? 6 and 9 ? 6 and 7 ? 
 
 7. How many are 7 and 7 ? 7 and 6 ? 7 and 8 ? 7 and 10 ? 
 7 and 9 ? 
 
 8. How many are 5 and 4 and 6 ? 7 and 3 and 8 ? 6 and 9 
 and 5? 
 
 3G. From the preceding operations we perceive that 
 Addition is the process of uniting several numbers of the 
 same kind into one equivalent number. 
 
 37. The Sum or Amount is the result obtained by the 
 process of addition. 
 
 38. The sign, -(-, is called plus, which signifies more. 
 When placed between two numbers, it denotes that they are 
 to be added ; thus, 6 -\- 4, shows that 6 and 4 are to be added. 
 
 39. The sign, =, is called the sign of equality. When 
 placed between two numbers, or sets of numbers, it signifies 
 that they are equal to each other; thus, the expression 
 6 + 4 = 10, is read 6 phis 4 is equal to 10, and denotes that 
 the numbers 6 and 4, taken together, equal the number 10. 
 
 Define addition. The sum or amount ? Sign of addition ? Of 
 equality ? 
 
ADDITION. 
 
 23 
 
 ADDITION TABLE. 
 
 1- 
 
 - 1= 2 
 
 2+ 1= 3 
 
 3+ 1= 4 
 
 4+ 1= 5 
 
 1- 
 
 - 2= 3 
 
 2+ 2= 4 
 
 3+ 2= 5 
 
 4+ 2= 6 
 
 1- 
 
 - 3= 4 
 
 2- 
 
 - 3 =: 5 
 
 3+ 3= 6 
 
 4+ 3= 7 
 
 1- 
 
 - 4= 5 
 
 2- 
 
 - 4= 6 
 
 3+ 4= 7 
 
 4+ 4= 8 
 
 1- 
 
 - 5= 6 
 
 2- 
 
 - 5= 7 
 
 3+ 5= 8 
 
 4- 
 
 - 5= 9 
 
 1- 
 
 - 6= 7 
 
 2- 
 
 - 6= 8 
 
 3+ 6= 9 
 
 4J 
 
 -6 = 10 
 
 1- 
 
 - 7= 8 
 
 2- 
 
 - 7= 9 
 
 3+ 7 = 10 
 
 4- 
 
 - 7 = 11 
 
 1- 
 
 - 8= 9 
 
 2- 
 
 -8 = 10 
 
 3- 
 
 - 8 = 11 
 
 4- 
 
 - 8 = 12 
 
 1+ 9 = 10 
 
 2+ 9 = 11 
 
 3- 
 
 -9 = 12 
 
 4- 
 
 - 9 = 13 
 
 1 + 10 = 11 
 
 2 + 10 = 12 
 
 3- 
 
 -10 = 13 
 
 4- 
 
 -10 = 14 
 
 1 + 11 = 12 
 
 2 + 11 1= 13 
 
 3- 
 
 -11 = 14 
 
 4- 
 
 -11 = 15 
 
 1 + 12 = 13 
 
 2 + 12 = 14 
 
 3- 
 
 -12 = 15 
 
 4 + 12 = 16 
 
 5- 
 
 - 1= 6 
 
 6+ 1= 7 
 
 7+ 1= 8 
 
 8+ 1= 9 
 
 5- 
 
 - 2= 7 
 
 6+ 2 = -8 
 
 7+ 2= 9 
 
 8+ 2 = 10 
 
 5- 
 
 - 3= 8 
 
 6+ 3= 9 
 
 7+ 3 = 10 
 
 8+ 3 = 11 
 
 5- 
 
 - 4= 9 
 
 6+ 4 = 10 
 
 7+ 4 = 11 
 
 8+ 4=12 
 
 5+ 5 = 10 
 
 6+ 5 = 11 
 
 7+ 5 = 12 
 
 8+ 5 = 13 
 
 5- 
 
 - 6 = 11 
 
 6+ 6 = 12 
 
 7+ 6 = 13 
 
 8+ 6 = 14 
 
 5- 
 
 -7 = 12 
 
 6+ 7 = 13 
 
 7+ 7 = 14 
 
 8+ 7 = 15 
 
 5- 
 
 -8 = 13 
 
 6- 
 
 -8 = 14 
 
 7+ 8 = 15 
 
 8+ 8 = 16 
 
 5- 
 
 -9 = 14 
 
 6- 
 
 -9 = 15 
 
 7+ 9 = 16 
 
 8+ 9 = 17 
 
 5- 
 
 -10 = 15 
 
 6- 
 
 -10 = 16 
 
 7 + 10 = 17 
 
 8 + 10 = 18 
 
 5 + 11 = 16 
 
 6- 
 
 -11 = 17 
 
 7 + 11 = 18 
 
 8 + 11 = 19 
 
 5 + 12 = 17 
 
 6- 
 
 -12 = 18 
 
 7 + 12 = 19 
 
 8 + 12 = 20 
 
 9+ 1 = 10 
 
 10+ 1 = 11 
 
 11- 
 
 -1 = 12 
 
 12- 
 
 1- 1 = 13 
 
 9+ 2 = 11 
 
 10+ 2 = 12 
 
 11- 
 
 - 2 = 13 
 
 12- 
 
 -2 = 14 
 
 9+ 3 = 12 
 
 10- 
 
 -3 = 13 
 
 11- 
 
 -3 = 14 
 
 12- 
 
 -3 = 15 
 
 9- 
 
 -4 = 13 
 
 10- 
 
 -4 = 14 
 
 11- 
 
 - 4 = 15 
 
 12- 
 
 - 4 = 16 
 
 9- 
 
 -5 = 14 
 
 10- 
 
 -5 = 15 
 
 11- 
 
 -5 = 16 
 
 12- 
 
 - 5 = 17 
 
 9- 
 
 - 6 = 15 
 
 10+ 6 = 16 
 
 11+ 6 = 17 
 
 12- 
 
 - 6 = 18 
 
 9- 
 
 -7 = 16 
 
 10+ 7 = 17 
 
 11- 
 
 -7 = 18 
 
 12+ 7 = 19 
 
 9- 
 
 -8 = 17 
 
 10+ 8 = 18 
 
 11- 
 
 -8 = 19 
 
 12+ 8 = 20 
 
 9- 
 
 -9 = 18 
 
 10- 
 
 -9=19 
 
 11- 
 
 - 9 = 20 
 
 12+ 9 = 21 
 
 9- 
 
 -10 = 19 
 
 10- 
 
 - 10 = 20 
 
 11- 
 
 -10 = 21 
 
 12 + 10 = 22 
 
 9 + 11 = 20 
 
 10- 
 
 -11=21 
 
 11- 
 
 -11=22 
 
 12 + 11=23 
 
 9 + 12 = 21 
 
 10- 
 
 -12 = 22 
 
 11- 
 
 -12 = 23 
 
 12 + 12 = 24 
 
24 SIMPLE NUMBERS. 
 
 CASE I. 
 
 40. When the amount of each column is less 
 than 10. 
 
 1. A farmer sold some hay for 102 dollars, six cows for 
 162 dollars, and a horse for 125 dollars ; how much did he re- 
 ceive for all ? 
 
 OPEKATION. ANALYSIS. We arrange the numbers so 
 
 ,3^ that units of like order shall stand in the 
 
 same column. We then add the columns 
 
 102 separately, for convenience commencing at 
 
 162 the right hand, and write each result under 
 
 J25 the column added. Thus, we have 5 and 2 
 
 and 2 are 9, the sum of the units ; 2 and 6 
 
 Amount, 389 are 8, the sum of the tens ; 1 and 1 and 1 
 
 are 3, the sum of the hundreds. Hence, the 
 
 entire amount is 3 hundreds 8 tens and 9 units, or 389, the Answer. 
 
 EXAMPLES FOR PRACTICE. 
 
 (2.) (3.) (4.) (5.) 
 
 pounds. rods. cents. days. 
 
 132 245 312 437 
 
 243 321 243 140 
 
 324 132 412 321 
 Ans. 699 
 
 6. What is the sum of 144, 321, and 232 ? Ans. 697. 
 
 7. What is the amount of 122, 333, and 401 ? Ans. 856. 
 
 8. What is the sum of 42, 103, 321, and 32 ? Ans. 498. 
 
 9. A drover bought three droves of sheep. The first con- 
 tained 230, the second 425, and the third 340 ; how many 
 sheep did he buy in all? Ans. 995. 
 
 CASE II. 
 
 41. When the amount of any column equals or 
 exceeds 10. 
 
 1. A merchant pays 725 dollars a year for the rent of a 
 Case I is what ? Give explanation. Case n is what ? 
 
ADDITION. 25 
 
 store, 475 dollars for a clerk, and 367 dollars for other ex- 
 penses ; what is the amount of his expenses ? 
 
 OPERATION. ANALYSIS. Arranging the num- 
 
 -3 ^ bers as in Case I, we first add the 
 
 ||| column of units, and find the sum 
 
 725 to be 17 units, which is 1 ten and 
 
 475 7 units. We write the 7 units in 
 
 gg7 the place of units, and the 1 ten in 
 
 - the place of tens. The sum of the 
 
 Sum of the units, 17 figures in the column of tens is 15 
 
 Sum of the tens, 15 tens, which is 1 hundred, and 5 
 
 Sum of the hundreds, 14 tens. We write the 5 tens in the 
 
 l 
 
 Total amount, 1567 
 
 the place of hundreds. We next 
 
 add the column of hundreds, and find the sum to be 14 hundreds, 
 which is 1 thousand and 4 hundreds. We write the 4 hundreds in 
 the place of hundreds, and 1 thousand in the place of thousands. 
 Lastly, by uniting the sum of the units with the sums of the tens 
 and hundreds, we find the total amount to be 1 thousand 5 hundreds 
 6 tens and 7 units, or 1567. 
 
 This example may be performed by another method, which 
 is the common one in practice. Thus : 
 
 OPERATION. ANALYSIS. Arranging the numbers as before, we 
 
 725 add the first column and find the sum to be 17 units ; 
 
 475 writing the 7 units under the column of units, we add 
 
 ggy the 1 ten to the column of tens, and find the sum to be 
 
 16 tens ; writing the 6 tens under the column tens, we 
 
 1567 add the 1 hundred to the column of hundreds, and find 
 
 the sum to be 15 hundreds ; as this is the last column, 
 
 we write down its amount, 15 ; and we have the icJwle amount, 1567, 
 
 as before. 
 
 NOTES. 1. Units of the same order are written in the same column ; 
 and when the sum in any column is 10 or more than 10, it produces 
 one or more units of a higher order, which must be added to the next 
 column. This process is sometimes called " carrying the tens." 
 
 2. In adding, learn to pronounce the partial results witHout naming 
 the numbers separately ; thus, instead of saying 7 and 5 are 12, and 
 5 are 17, simply pronounce the results, 7, 12, 17, &c. 
 
 Give explanation. Second explanation. What is meant by carry- 
 ing the tens ? 
 
26 SIMPLE NUMBERS. 
 
 . From the preceding examples and illustrations we 
 deduce the following 
 
 RULE. I. Write the numbers to be added so that all the units 
 of the same order shall stand in the same column ; that is, units 
 under units, tens under tens, fyc. 
 
 II. Commencing at units, add each column separately, and 
 write the sum underneath, if it be less than ten. 
 
 III. If the sum of any column be ten or more than ten, write 
 the unit figure only, and add the ten or tens to the next column. 
 
 IV. Write the entire sum of the last column. 
 
 PROOF. 1st. Begin with the right hand or unit column, and 
 add the figures in each column in an opposite direction from 
 that in which they were first added ; if the two results agree, 
 the work is supposed to be right. Or, 
 
 2d. Separate the numbers added into two sets, by a hori- 
 zontal line ; find the sum of each set separately ; add these 
 sums, and if the amount be the same as that first obtained, the 
 work is presumed to be correct. 
 
 NOTE. By the methods of proof here given, the numbers are united 
 in new combinations, which render it almost impossible for two pre- 
 cisely similar mistakes to occur. 
 
 The first method is the one commonly used in business. 
 
 EXAMPLES FOR PRACTICE. 
 
 (2.) 
 
 miles. 
 
 (3.) 
 
 inches. 
 
 (4.) 
 
 tons. 
 
 (5.) 
 
 feet. 
 
 (6.) 
 
 bushels. 
 
 24 
 
 321 
 
 427 
 
 1342 
 
 3420 
 
 48 
 
 479 
 
 321 
 
 7306 
 
 7021 
 
 96 
 
 165 
 
 903 
 
 5254 
 
 327 
 
 82 
 
 327 
 
 278 
 
 8629 
 
 97 
 
 250 
 
 1292 
 
 1929 
 
 22531 
 
 10865 
 
 - Rule, first step ? Second ? Third ? Fourth ? Proof, first method ? 
 Second ? Upon what principle are these methods of proof founded ? 
 
ADDITION. 27 
 
 (7.) (8.) (9.) (10.) 
 
 hours. years. gallons. rods. 
 
 347 7104 3462 47637 
 
 506 3762 863 3418 
 
 218 9325 479 703 
 
 312 4316 84 26471 
 
 424 2739 57 84 
 
 11. 42 + 64 + 98 + 70 + 37 = how many ? Am. 311. 
 
 12. 312 + 425 + 107 + 391 + 76 = how many ? 
 
 Am. 1311. 
 
 13. 1476 + 375 -f 891 + 66 + 80 = how many ? 
 
 Ans. 2888. 
 
 14. 37042 + 1379 + 809 + 127 + 40 = how many ? 
 
 Ans. 39397. 
 
 15. What is the sum of one thousand six hundred fifty-six, 
 eight hundred nine, three hundred ten, and ninety-four ? 
 
 Ans. 2869. 
 
 16. Add forty-two thousand two hundred twenty, ten thou- 
 sand one hundred five, four thousand seventy-five, and five 
 hundred seven. Ans. 56907. 
 
 17. Add two hundred ten thousand four hundred, one hun- 
 dred thousand five hundred ten, ninety thousand six hundred 
 eleven, forty-two hundred twenty-five, and eight hundred 
 ten. Ans. 406556. 
 
 18. What is the sum of the following numbers : seventy- 
 five, one thousand ninety-five, six thousand four hundred thir- 
 ty-five, two hundred sixty-seven thousand, one thousand four 
 hundred fifty-five, twenty-seven million eighteen, two hundred 
 seventy million twenty-seven thousand ? Ans. 297303078. 
 
 19. A man on a journey traveled the first day 37 miles, 
 the second 33 miles, the third 40 miles, and the fourth 35 miles ; 
 how far did he travel in the four days ? 
 
 20. A wine merchant has in one cask 75 gallons, in another 
 65, in a third 57, in a fourth 83, in a fifth 74, and in a sixth 
 67 ; how many gallons has he in all? Ans. 421. 
 
28 SIMPLE NUMBERS. 
 
 21. An estate is to be shared equally by four heirs, and 
 the portion to each heir is to be 3754 dollars ; what is the 
 amount of the estate? Ans. 15016 dollars. 
 
 22. How many men in an army consisting of 52714 in- 
 fantry, 5110 cavalry, 6250 dragoons, 3927 light-horse, 928 
 artillery, 250 sappers, and 406 miners ? 
 
 23. A merchant deposited 56 dollars in a bank on Monday, 
 74 on Tuesday, 120 on Wednesday, 96 on Thursday, 170 on 
 Friday, and 50 on Saturday ; how much did he deposit during 
 the week ? 
 
 24. A merchant bought at public sale 746 yards of broad- 
 cloth, 650 yards of muslin, 2100 yards of flannel, and 250 
 yards of silk ; how many yards in all ? 
 
 25. Five persons deposited money in the same bank ; the 
 first, 5897 dollars; the second, 12980 dollars; the third, 
 65973 dollars ; the fourth, 37345 dollars ; and the fifth as 
 much as the first and second together ; how many dollars did 
 they all deposit ? Ans. 141072 dollars. 
 
 26. A man willed his estate to his wife, two sons, and four 
 daughters ; to his daughters he gave 2630 dollars apiece, to 
 his sons, each 4647 dollars, and to his wife 3595 dollars; 
 how much was his estate ? Ans. 23409 dollars. 
 
 (27.) (28.) (29.) (30.) (31.) 
 
 476 908 126 443 180 
 
 390 371 324 298 976 
 
 915 569 503 876 209 
 
 207 245 891 569 314 
 
 841 703 736 137 563 
 
 632 421 517 910 842 
 
 234 127 143 347 175 
 
 143 354 274 256 224 
 
 536 781 531 324 135 
 
 245 436 275 463 253 
 
ADDITION. 
 
 29 
 
 (32.) 
 3120 
 8417 
 2645 
 9016 
 1857 
 4532 
 
 (33.) 
 14903 
 32087 
 63124 
 51037 
 86710 
 39425 
 
 (34.) 
 40371 
 56108 
 92436 
 15682 
 23761 
 58619 
 
 (35.) 
 
 43916 
 
 16782 
 
 58468 
 
 20375 
 
 3741 
 
 897 
 
 (36.) 
 2981043 
 7126459 
 9412767 
 7891234 
 109126 
 84172 
 72120 
 27676921 
 
 (37.) 
 
 1278976 
 
 7654301 
 
 876120 
 
 723456 
 
 31309 
 
 4871 
 
 978 
 
 (38.) 
 
 416785413 
 
 915123460 
 
 31810213 
 
 7367985 
 
 654321 
 
 37853 
 
 2685 
 
 1371781930 
 
 39. Add 8635, 2190, 7421, 5063, 2196, and 1245. 
 
 Ans. 26750. 
 
 40. Add 246034, 29865, 47321, 58653, 64210, 5376, 9821, 
 and 340. Ans. 461620. 
 
 41. Add 27104, 32540, 10758, 6256, 704321, 730491, 
 2787316, and 2749104. 
 
 42. Add 1, 37, 29504, 6790312, 18757421, and 265. 
 
 Ans. 25577540. 
 
 43. Add 56263, 211964, 56321, 18536, 4340, 279, and 73. 
 
 Ans. 347776. 
 
 44. Add 3742 bushels, 493 bushels, 927 bushels, 643 bush- 
 els, and 953 bushels. 
 
 45. Add 7346 acres, 9387 acres, 8756 acres, 8394 acres, 
 and 32724 acres. Ans. 66607. 
 
 46. Henry received at one time 15 apples, at another 115, 
 and at another 19 ; how many did he receive ? 
 
 c* 
 
30 SIMPLE NUMBERS. 
 
 47. A man commenced farming at the west, and raised, the 
 first year, 724 bushels of corn ; the second year, 3498 bushels ; 
 the third year, 9872 bushels; the fourth year, 9964 bushels; 
 the fifth year, 11078 bushels; how many bushels did he raise 
 in the five years ? Ans. 35136 bushels. 
 
 48. A has 3648 dollars, B has 7035 dollars, C has 429 
 dollars more than A and B together, and D has as many dol- 
 lars as all the rest ; how many dollars has D ? How many- 
 have all? Ans. D has 43590 dollars. 
 
 49. A man bought three houses and lots for 15780 dollars, 
 and sold them so as to gain 695 dollars on each lot ; for how 
 much did he sell them ? Ans. 17865 dollars. 
 
 50. At the battle of Waterloo, which took place June 18th, 
 1815, the estimated loss of the French was 40000 men; of 
 the Prussians, 38000 ; of the Belgians, 8000 ; of the Hano- 
 verians, 3500 ; and of the English, 12000 ; what was the entire 
 loss of life in this battle ? 
 
 51. The expenditures for educational purposes in New 
 England for the year 1850 were as follows : Maine, 380623 
 dollars ; New Hampshire, 221146 dollars ; Vermont, 246604 
 dollars ; Massachusetts, 1424873 dollars ; Rhode Island, 
 136729 dollars; and Connecticut, 430826 dollars; what was 
 the total expenditure ? Ans. 2840801 dollars. 
 
 52. The eastern continent contains 31000000 square 
 miles ; the western continent, 13750000 ; Australia, Green- 
 land, and other islands, 5250000 ; what is the entire area of 
 the land surface of the globe ? 
 
 53. The population of New York, in 1850, was 515547; 
 Boston, 136881; Philadelphia, 340045; Chicago, 29963; 
 St. Louis, 77860; New Orleans, 116375; what was the en- 
 tire population of these cities ? Ans. 1216671. 
 
 54. The population of the globe is estimated as follows : 
 North America, 39257819; South America, 18373188; Eu- 
 rope, 265368216 ; Asia, 630671661 ; Africa, 61688779 ; 
 Oceanica, 23444082 ; what is the total population of the 
 globe according to this estimate ? Ans. 1038803745. 
 
ADDITION. 
 
 31 
 
 55. The railroad distance from New York to Albany is 144 
 miles ; from Albany to Buffalo, 298 ; from Buffalo to Cleveland, 
 183 ; from Cleveland to Toledo, 109 ; from Toledo to Spring- 
 field, 365; and from Springfield to St. Louis, 95 miles; what 
 is the distance from New York to St. Louis ? 
 
 56. A man owns farms valued at 56800 dollars ; city lots 
 valued at 86760 dollars ; a house worth 12500 dollars, and 
 other property to the amount of 6785 dollars ; what is the 
 entire value of his property? Ans. 162845 dollars. 
 
 (57.) 
 
 15038 
 
 7404 
 
 34971 
 
 30359 
 
 6293 
 
 2875 
 
 16660 
 
 64934 
 
 80901 
 
 7444 
 
 57068 
 
 17255 
 
 32543 
 
 40022 
 
 56063 
 
 33860 
 
 17548 
 
 28944 
 
 16147 
 
 38556 
 
 234882 
 
 3-9058 
 
 152526 
 
 179122 
 
 7626 
 
 1218099 
 
 (58.) 
 
 26881 
 
 12173 
 
 39665 
 
 33249 
 
 6318 
 
 4318 
 
 34705 
 
 80597 
 
 95299 
 
 8624 
 
 53806 
 
 18647 
 
 41609 
 
 35077 
 
 46880 
 
 41842 
 
 26876 
 
 36642 
 
 29997 
 
 44305 
 
 262083 
 
 39744 
 
 169220 
 
 198568 
 
 8735 
 
 1395860 
 
 (59.) 
 41919 
 19577 
 74736 
 66768 
 12673 
 7193 
 51365 
 
 155497 
 
 183134 
 16845 
 
 111139 
 35902 
 82182 
 75153 
 
 132936 
 82939 
 44424 
 65586 
 52839 
 83211 
 
 522294 
 78861 
 
 353428 
 
 386214 
 17005 
 
 (60.) 
 
 93808 
 
 41371 
 
 110525 
 
 102936 
 
 17087 
 
 13251 
 
 112110 
 
 220619 
 
 225255 
 
 68940 
 
 176974 
 
 86590 
 
 149162 
 
 109355 
 
 283910 
 
 112511 
 
 72908 
 
 157672 
 
 86160 
 
 119557 
 
 839398 
 
 117787 
 
 471842 
 
 571778 
 
 41735 
 
82 SIMPLE NUMBERS, 
 
 SUBTEACTION. 
 
 MENTAL EXERCISES. 
 
 43. 1. A farmer, having 14 cows, sold 6 of them; how 
 many had he left ? 
 
 ANALYSIS. He had as many left as 14 cows less 6 cows, which 
 are 8 cows. Therefore, he had 8 cows left. 
 
 2. Stephen, having 9 marbles, lost 4 of them ; how many 
 had he left? 
 
 3. If a man earn 10 dollars a week, and spend 6 dollars for 
 provisions, how many dollars has he left ? 
 
 4. A merchant, having 16 barrels of flour, sells 9 of them ; 
 how many has he left ? 
 
 5. Charles had 18 cents, and gave 10 of them for a book ; 
 how many had he left ? 
 
 6. James is 17 years old, and his sister, Julia, is o years 
 younger ; how old is Julia ? 
 
 7. A grocer, having 20 boxes of lemons, sold 1 1 boxes ; 
 how many boxes had he left ? 
 
 8. From a cistern containing 25 barrels of water, 15 bar- 
 rels leaked out ; how many barrels remained ? 
 
 9. Paid 16 dollars for a coat, and 7 dollars for a vest; 
 how much more did the coat cost than the vest ? 
 
 10. How many are 18 less 5 ? 17 less 8 ? 12 less 7 ? 
 
 11. How many are 20 less 14 ? 18 less 12 ? 19 less 11 ? 
 
 12. How many are 11 less 3? 16 less 11? 19 less 8? 
 20 less 9 ? 22 less 20 ? 
 
 44. Subtraction is the process of taking one number from 
 another equal to or greater than itself. 
 
 45. The Minuend is the number to be subtracted from. 
 
 46. The Subtrahend is the number to be subtracted. 
 
 Define subtraction. Minuend. Subtrahend. 
 
SUBTRACTION. 
 
 33 
 
 47. The Difference or Remainder is the result obtained 
 by the process of subtraction. 
 
 NOTE. The minuend and subtrahend must be like numbers ; thus, 
 5 dollars from 9 dollars leave 4 dollars ; 5 apples from 9 apples leave 4 
 apples ; but it would be absurd to say 5 apples from 9 dollars, or 5 
 dollars from 9 apples. 
 
 48. The sign, , is called minus, which signifies less. 
 "When placed between two numbers, it denotes that the one 
 after it is to be taken from the one before it. Thus, 8 6 = 2 
 is read 8 minus 6 equals 2, and denotes that 6, the subtrahend, 
 taken from 8, the minuend, equals 2, the remainder. 
 
 SUBTRACTION TABLE. 
 
 1 I 
 
 2 2= 
 
 3 3= 
 
 4 4= 
 
 2 1= 1 
 
 3 2= 1 
 
 4 3= 1 
 
 5 4= 1 
 
 3 1 2 
 
 4 2= 2 
 
 5 3= 2 
 
 6 4= 2 
 
 4 1 3 
 
 5 2= 3 
 
 6 3= 3 
 
 7 4= 3 
 
 5 1 = 4 
 
 6 2= 4 
 
 7 3= 4 
 
 8 4= 4 
 
 6 1 = 5 
 
 7 2= 5 
 
 8 3= 5 
 
 9 4= 5 
 
 7 I 6 
 
 8 2= 6 
 
 9 3= 6 
 
 10 4= 6 
 
 8^7 
 HZ / 
 
 9 2= 7 
 
 10 3= 7 
 
 11 4= 7 
 
 9 = 8 
 
 10 2= 8 
 
 11 3= 8 
 
 12 4= 8 
 
 10 = 9 
 
 11 2= 9 
 
 12 3= 9 
 
 13 4= 9 
 
 11 =10 
 
 12 2 = 10 
 
 13 3 = 10 
 
 14 4 = 10 
 
 12 =11 
 
 13 2=11 
 
 14 3 = 11 
 
 15 4 = 11 
 
 13 =12 
 
 14 2=12 
 
 15 3 = 12 
 
 16 4 = 12 
 
 5 5= 
 
 6 6= 
 
 7 7= 
 
 8 8= 
 
 6 5= 1 
 
 7 6= 1 
 
 8 7= 1 
 
 9 8= 1 
 
 7 5= 2 
 
 8 6= 2 
 
 9 7= 2 
 
 10 8= 2 
 
 8 5 = 3 
 
 9 6= 3 
 
 10 7= 3 
 
 H_ 8= 3 
 
 9 5= 4 
 
 10 6= 4 
 
 11 7= 4 
 
 12 8= 4 
 
 105= 5 
 
 11 6= 5 
 
 12 7= 5 
 
 138= 5 
 
 11 5 = 6 
 
 12 6= 6 
 
 13 _ 7= 6 
 
 148= 6 
 
 12 5= 7 
 
 13 6= 7 
 
 14 7 = 7 
 
 158= 7 
 
 135= 8 
 
 146= 8 
 
 157= 8 
 
 168= 8 
 
 145= 9 
 
 156= 9 
 
 16 7= 9 
 
 17 8= 9 
 
 15 5 = 10 
 
 16 6 = 10 
 
 177 = 10 
 
 188 = 10 
 
 16 5 = 11 
 
 17 6 = 11 
 
 18 7 = 11 
 
 19 8 = 11 
 
 175 = 12 
 
 18 6 = 12 
 
 19 7 = 12 
 
 20 8 = 12 
 
 Difference or remainder. What is the sign of subtraction ? 
 
34 
 
 SIMPLE NUMBERS. 
 
 9 9 
 
 10 10= 
 
 11 11= 
 
 12 12= 
 
 10 9= 1 
 
 11 10= 1 
 
 12 11= 1 
 
 13 12= 1 
 
 11 9 2 
 
 12 10= 2 
 
 13 11= 2 
 
 14 12= 2 
 
 12 9 = 3 
 
 13 10= 3 
 
 14 11= 3 
 
 15 12= 3 
 
 13 9 4 
 
 14 10= 4 
 
 15 11 4 
 
 16 12= 4 
 
 14 9= 5 
 
 15 10= 5 
 
 16 11= 5 
 
 17 12= 5 
 
 15 9= 6 
 
 16 10= 6 
 
 17 11= 6 
 
 18 12= 6 
 
 16 9= 7 
 
 17 10= 7 
 
 18 11= 7 
 
 19 12= 7 
 
 17 9= 8 
 
 18 10= 8 
 
 19 11= 8 
 
 20 12= 8 
 
 18 9= 9 
 
 19 10= 9 
 
 2011= 9 
 
 21 12= 9 
 
 19 9=10 
 
 2010 = 10 
 
 21 11 = 10 
 
 22 12 = 10 
 
 20 9 = 11 
 
 21 10 = 11 
 
 22 11 = 11 
 
 23 12 = 11 
 
 21 9=12 
 
 2210 = 12 
 
 23 11 = 12 
 
 24 12 = 12 
 
 CASE I. 
 
 49. When no figure in the subtrahend is greater 
 than the corresponding figure in the minuend. 
 
 1. From 574 take 323. 
 
 OPERATION. 
 Minuend, 574 
 
 Subtrahend, 323 
 
 Remainder, 
 
 251 
 
 ANALYSIS. We write the less num- 
 ber under the greater, with units under 
 units, tens under tens, &c., and draw 
 a line underneath. Then, beginning at 
 the right hand, we subtract separately 
 each figure of the subtrahend from the 
 figure above it in the minuend. Thus, 3 from 4 leaves 1 , which is the 
 difference of the units ; 2 from 7 leaves 5, the difference of the tens ; 
 3 from 5 leaves 2, the difference of the hundreds. Hence, we have 
 for the whole difference, 2 hundreds 5 tens and 1 unit, or 251. 
 
 EXAMPLES FOR PRACTICE. 
 
 
 (20 
 
 (3.) 
 
 (4.) 
 
 (5.) 
 
 Minuend, 
 
 876 
 
 676 
 
 367 
 
 925 
 
 Subtrahend, 
 
 334 
 
 415 
 
 152 
 
 213 
 
 Remainder, 
 
 542 
 
 
 
 
 Oaso 1 is what ? Give explanation. 
 
SUBTRACTION. 
 
 35 
 
 From 
 Take 
 
 (6.) 
 876 
 523 
 
 (7.) 
 732 
 522 
 
 (8.) 
 
 987 
 
 782 
 
 (9.) 
 498 
 178 
 
 Remainders. 
 
 10. From 3276 take 2143. 1133. 
 
 11. From 7634 take 3132. 4502. 
 
 12. From 41763 take 11521. 30242. 
 
 13. From 18346 take 5215. 13131. 
 
 14. From 397631 take 175321. 222310. 
 
 15. Subtract 47321 from 69524. 22203. 
 
 16. Subtract 16330 from 48673. 32343. 
 
 17. Subtract 291352 from 895752. 604400. 
 
 18. Subtract 84321 from 397562. 313241. 
 
 19. A farmer paid 645 dollars for a span of horses and 
 a carriage, and sold them for 522 dollars ; how much did he 
 lose? 
 
 20. A man bought a mill for 3724 dollars, and sold it for 
 4856 dollars ; how much did he gain ? Ans. 1132 dollars. 
 
 21. A drover bought 1566 sheep, and sold 435 of them; 
 how many had he left? Ans. 1131 sheep. 
 
 22. A piece of land was sold for 2945 dollars, which was 832 
 dollars more than it cost ; what did it cost ? 
 
 23. A gentleman willed to his son 15768 dollars, and to 
 his daughter 4537 dollars ; how much more did he will to the 
 son than to the daughter ? Ans. 11231 dollars. 
 
 24. A merchant sold goods to the amount of 6742 dollars, 
 and by so doing gained 2540 dollars ; what did the goods cost 
 him ? 
 
 25. If I borrow 15475 dollars of a person, and pay him 
 9050 dollars, how much do I still owe him ? 
 
 26. In 1850 the white population of the United States was 
 19,553,068, and the slave population 3,204,313 ; how much 
 was the difference? 
 
 27. The population of Great Britain in 1851 was 20,936,468, 
 and of England alone, 16,921,888; what was the difference? 
 
36 SIMPLE NUMBERS. 
 
 CASE II. 
 
 5O. When any figure in the subtrahend is greater 
 than the corresponding figure in the minuend. 
 1. From 846 take 359. t 
 
 OPERATION. ANALYSIS. In this example we 
 
 (7) (13) (16) cannot take 9 units from 6 units. 
 
 Minuend, 846 From the 4 tens we take 1 ten, which 
 
 Subtrahend, 359 equals 10 units, and add to the 6 
 
 Remainder ~4 8 7~ un ^ s > making 16 units ; 9 units from 
 
 16 units leave 7 units, which we 
 
 write in the remainder in units' place. As we have taken 1 ten 
 from the 4 tens, 3 tens only are left. We cannot take 5 tens from 
 3 tens ; so from the 8 hundreds we take 1 hundred, which equals 10 
 tens, and add to the 3 tens, making 13 tens; 5 tens from 13 
 tens leave 8 tens, which we write in the remainder in tens' place. 
 As we have taken 1 hundred from the 8 hundreds, 7 hundreds only 
 are left ; 3 hundreds from 7 hundreds leave 4 hundreds, which we 
 write in the remainder in hundreds' place, and we have the whole 
 remainder, 487. 
 
 NOTE. The numbers written over the minuend are used simply to 
 explain more clearly the method of subtracting ; in practice the pro- 
 cess should be performed mentally, and these numbers omitted. 
 
 The following method is more in accordance with prac- 
 tice. 
 
 OPERATION. ANALYSIS. Since we cannot take 9 units from 6 
 l^s units, we add 10 units to 6 units, making 16 units ; 
 
 Jsl I 9 units from 16 units leave 7 units. But as we have 
 
 added 10 units, or 1 ten, to the minuend, we shall 
 359 have a remainder 1 ten too large,*to avoid which, we 
 
 add 1 ten to the 5 tens in the subtrahend, making 6 
 tens. We can not take 6 tens from 4 tens ; so we add 
 10 tens to 4, making 14 tens ; 6 tens from 14 tens 
 leave 8 tens. Now, having added 10 tens, or 1 hundred, to the 
 minuend, we shall have a remainder 1 hundred too large, unless we 
 add 1 hundred to the 3 hundreds in the subtrahend, making 4 hun- 
 dreds ; 4 hundreds from 8 hundreds leave 4 hundreds, and we have 
 for the total remainder, 487, the same as before. 
 
 Case n is what ? Give explanation. Second explanation, 
 
SUBTRACTION. 
 
 37 
 
 NOTE. The process of adding 10 to the minuend is sometimes called 
 borrowing 10, and that of adding 1 to the next figure of the subtrahend, 
 carrying one. 
 
 51. From the preceding examples and .illustrations we 
 have the following general 
 
 RULE. I. Write the less number under the greater, placing 
 units of the same order under each other. 
 
 II. Begin at the right hand, and take each figure of the sub- 
 trahend from the figure above it, and write the result under- 
 neath. 
 
 III. If any figure in the subtrahend be greater than the cor- 
 responding figure above it, add 10 to that upper figure be- 
 fore subtracting, and then add 1 to the next left hand figure of 
 the subtrahend. 
 
 PROOF. Add the remainder to the subtrahend, and if their 
 sum be equal to the minuend, the work is supposed to be right. 
 
 EXAMPLES FOR PRACTICE. 
 
 
 (2.) 
 
 (3.) 
 
 (*) 
 
 (5.) 
 
 Minuend, 
 
 873 
 
 7432 
 
 1969 
 
 8146 
 
 Subtrahend, 
 
 538 
 
 6711 
 
 1408 
 
 4377 
 
 Remainder, 
 
 335 
 
 
 
 
 
 (6.) 
 
 (7.) 
 
 (8.) 
 
 (9.) 
 
 
 gallons. 
 
 bushels. 
 
 miles. 
 
 days. 
 
 From 
 
 3176 
 
 9076 
 
 7320 
 
 5097 
 
 Take 
 
 2907 
 
 4567 
 
 3871 
 
 3809 
 
 
 (10.) 
 
 (11.) 
 
 (12.) 
 
 (13.) 
 
 
 dollars. 
 
 rods. 
 
 acres. 
 
 feet. 
 
 From 
 
 76377 
 
 67777 
 
 900076 
 
 767340 
 
 Take 
 
 45761 
 
 46699 
 
 899934 
 
 5039 
 
 
 
 
 What do we mean hy borrowing 10 ? By carrying ? Kule, first step ? 
 Second? Third? Proof? 
 
38 
 
 SIMPLE NUMBERS. 
 
 14. 479 382 = how many ? Ans. 97. 
 
 15. 6593 1807 = how many? Ans. 4786. 
 
 16. 17380 3417 = how many ? Ans. 13963. 
 
 17. 80014 r 43190 = how many ? Ans. 36824. 
 
 18. 282731 90756 how many? Ans. 191975. 
 
 19. From 234100 take 9970. 
 
 20. From 345673 take 124799. 
 
 21. From 4367676 take 256569. Ans. 4111107. 
 
 22. From 3467310 take 987631. Ans. 2479679. 
 
 23. From 941000 take 5007. Ans. 935993. 
 
 24. From 1970000 take 1361111. Ans. 608889. 
 
 25. From 290017 take 108045. 
 
 26. Take 3077097 from 7045 676. Ans. 3968579. 
 
 27. Take 9999999 from 60000000. Ans. 50000001. 
 
 28. Take 220202 from 4040053. Ans. 3819851. 
 
 29. Take 2199077 from 3000001. Am. 800924. 
 
 30. Take 377776 from 8000800. Ans. 7623024. 
 
 31. Take 501300347 from 1030810040. 
 
 32. Subtract nineteen thousand nineteen from twenty thou- 
 sand ten. Ans. 991. 
 
 33. From one million nine thousand six take twenty thou- 
 sand four hundred. Ans. 988606. 
 
 34. What is the difference between two million seven 
 thousand eighteen, and one hundred five thousand seven- 
 teen? Ans. 1902001. 
 
 35. What is the difference between thirteen thousand thir- 
 teen, and eleven hundred eleven? Am. 11902. 
 
 36. From fifty million take five hundred five. 
 
 37. A merchant bought 5875 bushels of wheat, and sold 
 2976 bushels; how many bushels remained in his posses- 
 sion ? Am. 2899 bushels. 
 
 38. The Declaration of Independence was published July 
 4, 1776 ; how many years to July 4, 1860 ? 
 
 39. Massachusetts contains 7800 square miles, and New 
 Hampshire 9491 square miles ; which is the larger, and how 
 many square miles ? 
 
SUBTRACTION. 39 
 
 40. Sir Isaac Newton was born in the year 1642, and died 
 in 1727 ; to what age did he live ? 
 
 41. Mont Blanc, in Europe, is 15,680 feet high, and Chim- 
 borazo, in America, is 21,427 feet ; what is the difference in 
 their heights ? Am. 5747 feet. 
 
 42. La Place, the celebrated mathematician and philoso- 
 pher, was born in 1749, and died in 1827 ; how long did he 
 live? 
 
 43. If a company enter into business with a capital of 
 12750 dollars, and at the end of the year have 27194 dollars, 
 how much will they have gained ? Ans. 14444 dollars. 
 
 44. The population of the United States in 1800 was 
 5,305,925, and in 1850, 23,387,125 ; what was the increase in 
 50 years? Ans. 18,081,200. 
 
 45. The population of the United States by the census 
 of 1850 was 23,387,125; it is estimated that in 1860 it wjll 
 be 31,095,535 ; what is the estimated increase for the 10 
 years? Ans. 7,708,410. 
 
 46. According to the same rate of increase, the population 
 of the United States in 1870 will be 40,617,708 ; what will 
 be the increase of population from 1850 ? Ans. 17,230,583. 
 
 47. According to the same rate of increase, in 1900 the 
 population will be 101553377 ; what will be the increase from 
 1850? Ans. 78166252. 
 
 EXAMPLES COMBINING ADDITION AND SUBTRACTION. 
 
 1. A merchant gave his note for 5200 dollars. He 
 paid at one time 2500 dollars, and at another 175 dollars; 
 what remained due ? Ans. 2525 dollars. 
 
 2. A traveler who was 1300 miles from home, traveled 
 homeward 235 miles in one week, in the next 275 miles, in the 
 next 325 miles, and in the next 280 miles ; how far had he 
 still to go before he would reach home? Ans. 185 miles. 
 
 3. A man deposited in bank 8752 dollars ; he drew out at 
 one time 4234 dollars, at another 1700 dollars, at another 962 
 
40 SIMPLE NUMBERS. 
 
 dollars, and at another 49 dollars ; how much had he remain- 
 ing in bank ? Ans. 1807 dollars. 
 
 4. A man bought a farm for 4765 dollars, and paid 750 
 dollars for fencing and other improvements ; he then sold it for 
 384 dollars less than it cost him ; how much did he receive 
 for it? Ans. 5131 dollars. 
 
 5. A forwarding merchant had in his warehouse 7520 bar- 
 rels of flour; he shipped at one time 1224 barrels, at another 
 time 1500 barrels, and at another time 1805 barrels; how 
 many barrels remained ? 
 
 6. A had 450 sheep, B had 175 more than A, and C 
 had as many as A and B together minus 114; how many 
 sheep had C ? Ans. 961 sheep. 
 
 7. A farmer raised 1575 bushels of wheat, and 900 bushels 
 of corn. He sold 807 bushels of wheat, and 391 bushels of 
 corn to A, and the remainder to B ; how much of each did 
 he sell to B ? Ans. 768 bushels of wheat, and 509 of corn. 
 
 8. A man traveled 6784 miles ; 2324 miles by railroad, 
 1570 miles in a stage coach, 450 miles on horseback, 175 
 miles on foot, and the remainder by steamboat ; how many 
 miles did he travel by steamboat ? Ans. 2265 miles. 
 
 9. Three persons bought a hotel valued at 35680 dollars. 
 The first agreed to pay 7375 dollars, the second agreed to 
 pay twice as much, and the third the remainder ; how much 
 was the third to pay ? Ans. 13555 dollars. 
 
 10. Borrowed of my neighbor at one time 750 dollars, at 
 another time 379 dollars, and at another 450 dollars. Having 
 paid him 1000 dollars, how much do I still owe him? 
 
 Ans. 579 dollars. 
 
 11. A man worth 6709 dollars, received a legacy of 3000 
 dollars. He spent 4379 dollars in traveling; how much had 
 he left ? 
 
 12. In 1850 the number of white males in the United 
 States was 10026402, and of white females 9526666; of 
 these, 8786968 males, and 8525565 females were native 
 born; how many of both were foreign born? Ans. 2240535. 
 
MULTIPLICATION. 41 
 
 MULTIPLICATION. 
 
 MENTAL EXERCISES. 
 
 1. What will 4 pounds of sugar cost, at 8 cents a 
 pound ? 
 
 ANALYSIS. Four pounds will cost as much as the price, 8 cents 
 taken 4 times ; thus, 8 -j- 8 + 8 -\- 8 32. But instead of adding, 
 we may say, since one pound costs 8 cents, 4 pounds will cost 4 
 times 8 cents, or 32 cents. 
 
 2. If a ream of paper cost 3 dollars, what will 2 reams 
 cost? 
 
 3. At 7 cents a quart, what will 4 quarts of cherries 
 cost? 
 
 4. At 12 dollars a ton, what will 3 tons of hay cost? 4 
 tons? 5 tons? 
 
 5. There are 7 days in 1 week ; how many days in 6 weeks ? 
 in 8 weeks ? 
 
 6. What will 9 chairs cost, at 10 shillings apiece ? 
 
 7. If Henry earn 12 dollars in 1 month, how much can he 
 earn in 5 months ? in 7 months ? in 9 months ? 
 
 8. What will 11 dozen of eggs cost, at 9 cents a dozen? at 
 10 cents? at 12 cents? 
 
 9. When flour is 7 dollars a barrel, how much must be 
 paid for 7 barrels ? for 9 barrels ? for 12 barrels ? 
 
 10. At 9 dollars a week, what will 4 weeks' board cost ? 
 7 weeks' ? 9 weeks' ? 
 
 11. If I deposit 12 dollars in a savings bank every month, 
 how many dollars will I deposit in 6 months ? in 8 months ? 
 in 9 months ? 
 
 12. At 9 cents a foot, what will 4 feet of lead pipe cost ? 
 7 feet? 10 feet? 
 
 13. When hay is 8 dollars a ton, how much will 3 tons 
 cost ? 4 tons ? 7 tons ? 9 tons ? 11 tons ? 
 
 D* 
 
42 
 
 SIMPLE NUMBERS. 
 
 14. What will be the cost of 11 barrels of apples, at 2 dol- 
 lars a barrel ? at 3 dollars ? 
 
 15. At 1 cents a pound, what will 9 pounds of sugar cost ? 
 11 pounds ? 12 pounds ? 
 
 54. Multiplication is the process of taking one of two 
 given numbers as many times as there are units in the other. 
 
 55. The Multiplicand is the number to be taken. 
 
 56. The Multiplier is the number which shows how 
 many times the multiplicand is to be taken. 
 
 57. The Product is the result obtained by the process of 
 multiplication. 
 
 58. The Factors are the multiplicand and multiplier. 
 
 NOTES. 1. Factors are producers, and the multiplicand and mul- 
 tiplier are called factors because they produce the product. 
 
 2. Multiplication is a short method of performing addition when 
 the numbers to be added are equal. 
 
 59. The sign, X> placed between two numbers, denotes 
 that they are to be multiplied together ; thus 9 X 6 = 54, is 
 read 9 times 6 equals 54. 
 
 MULTIPLICATION TABLE. 
 
 IX 1 = 1 
 
 2X 1= 2 
 
 3X 1 3 
 
 4X 1= 4 
 
 IX 2 = 2 
 
 2X 2 = 4 
 
 3X 2= 6 
 
 4X 2= 8 
 
 X 3= 3 
 
 2X 3 = 6 
 
 3X 3= 9 
 
 4x 3 = 12 
 
 X 4= 4 
 
 2X 4= 8 
 
 3X 4 = 12 
 
 4 X 4 = 16 
 
 X 5= 5 
 
 2X 5 = 10 
 
 3X 5 = 15 
 
 4x 5 = 20 
 
 X 6= 6 
 
 2X 6 = 12 
 
 3X 6=18 
 
 4x 6 = 24 
 
 X 7 = 7 
 
 2X 7 = 14 
 
 3X 7 = 21 
 
 4x 7 = 28 
 
 X 8= 8 
 
 2X 8 = 16 
 
 3X 8 = 24 
 
 4X 8 = 32 
 
 IX 9= 9 
 
 2X 9 = 18 
 
 3X 9 = 27 
 
 4 X 9 = 36 
 
 1 X 10 10 
 
 2 X 10 = 20 
 
 3X 10 = 30 
 
 4X 10 = 40 
 
 1 X 11 = 11 
 
 2 X 11 22 
 
 3 X 11 = 33 
 
 4 X 11=44 
 
 IX 12=12 
 
 2X 12 = 24 
 
 3 X 12 = 36 
 
 4 X 12 = 48 
 
 Define multiplication. Multiplicand. Multiplier. Product. Fac- 
 tors. Multiplication is a short method of what ? What is the sign of 
 multiplication ? 
 
MULTIPLICATION. 
 
 43 
 
 5X 1= 5 
 
 6X 1= 6 
 
 7X 1= 7 
 
 8X 1= 8 
 
 5X 2 = 10 
 
 6X 2 = 12 
 
 7X 2 = 14 
 
 8X 2 = 16 
 
 5X 3 = 15 
 
 6X 3 = 18 
 
 7X 3 = 21 
 
 8 X 3 = 24 
 
 5 X 4 = 20 
 
 6X 4 = 24 
 
 7 X 4 = 28 
 
 8X 4 = 32 
 
 5X 5 = 25 
 
 6X 5 = 30 
 
 7X 5 = 35 
 
 8X 5 = 40 
 
 5 X 6 = 30 
 
 6 X 6 = 36 
 
 7X 6 = 42 
 
 8X 6 = 48 
 
 5X 7 = 35 
 
 6 X 7 = 42 
 
 7X 7 = 49 
 
 8X 7 = 56 
 
 5X 8 = 40 
 
 6 X 8 = 48 
 
 7X 8 = 56 
 
 8 X 8 = 64 
 
 5X 9 = 45 
 
 6X 9 = 54 
 
 7X 9 = 63 
 
 8X 9 = 72 
 
 5 X 10 = 50 
 
 6X 10 = 60 
 
 7X 10 = 70 
 
 8 X 10 = 80 
 
 5X 11 = 55 
 
 6 X 11 = 66 
 
 7X 11 = 77 
 
 8 X 11 = 88 
 
 5 X 12 = 60 
 
 6X 12 = 72 
 
 7X 12 = 84 
 
 8X 12 = 96 
 
 9X 1= 9 
 
 10 X 1= 10 
 
 11 X 1= 11 
 
 12 X 1= 12 
 
 9X 2= 18 
 
 10 X 2 = 20 
 
 11 X 2= 22 
 
 12 X 2= 24 
 
 9X 3= 27 
 
 10 X 3= 30 
 
 11 X 3= 33 
 
 12 X 3= 36 
 
 9 X 4 = 36 
 
 10 X 4 = 40 
 
 11 X 4= 44 
 
 12 X 4= 48 
 
 9 X 5 = 45 
 
 10 X 5 = 50 
 
 11 X 5 55 
 
 12 X 5= 60 
 
 9 X 6 = 54 
 
 10 X 6= 60 
 
 11 X 6= 66 
 
 12 X 6= 72 
 
 9X 7= 63 
 
 10 X 7 = 70 
 
 11 X 7= 77 
 
 12 X 7= 84 
 
 9X 8= 72 
 
 10 X 8 = 80 
 
 11 X 8= 88 
 
 12 X 8= 96 
 
 9 X 9= 81 
 
 10 X 9= 90 
 
 11 X 9= 99 
 
 12 X 9 = 108 
 
 9 X 10 = 90 
 
 10X10 = 100 
 
 11 X10 = 110 
 
 12X10 = 120 
 
 9X 11= 99 
 
 10x11 = 110 
 
 11 XH = 121 
 
 12X11 = 132 
 
 9X 12 = 108 
 
 10X12 = 120 
 
 11 X12 = 132 
 
 12X12 = 144 
 
 OPERATION. 
 
 CASE I. 
 
 6O. When the multiplier consists of one figure. 
 1. Multiply 374 by 6. 
 
 ANALYSIS. In this example it is 
 Lri required to take 374 six times. If we 
 l|| take the units of each order 6 times, 
 we shall take the entire number 6 
 times. Therefore, writing the multi- 
 plier under the unit figure of the mul- 
 tiplicand, we proceed as follows: 6 
 times 4 units are 24 units ; 6 times 7 
 tens are 42 tens ; 6 times 3 hundreds 
 are 18 hundreds; and adding these 
 partial products, we obtain the entire 
 product, 2244. 
 
 Multiplicand, 
 Multiplier, 
 
 units, 
 tens, 
 hundreds, 
 
 Product, 
 
 374 
 6 
 
 24 
 42 
 18 
 
 2244 
 
 Case I is what ? Give explanation. 
 
44 SIMPLE NUMBERS. 
 
 The operation in this example may be performed in another 
 way, which is the one in common use. 
 
 OPERATION. ANALYSIS. Writing the numbers as before, we 
 374 begin at the right hand or unit figure, and say: 6 
 times 4 units are 24 units, which is 2 tens and 4 
 units; write the 4 units in the product in units' 
 place, and reserve the 2 tens to add to the next prod- 
 uct ; 6 times 7 tens are 42 tens, and the two tens re- 
 served in the last product added, are 44 tens, which is 4 hundreds 
 and 4 tens ; write the 4 tens in the product in tens' place, and reserve 
 the 4 hundreds to add to the next product ; 6 times 3 hundreds are 
 18 hundreds, and 4 hundreds added are 22 hundreds, which, being 
 written in the product in the places of hundreds and thousands, 
 gives, for the entire product, 2244. 
 
 Gl. The unit value of a number is not changed by re- 
 peating the number. As the multiplier always expresses 
 times, the product must have the same unit value as the mul- 
 tiplicand. But, since the product of any two numbers will be 
 the same, whichever factor is taken as a multiplier, either 
 factor may be taken for the multiplier or multiplicand. 
 
 NOTE. In multiplying, learn to pronounce the partial results, as in 
 addition, without naming the numbers separately; thus, in the last 
 example, instead of saying 6 times 4 are 24, 6 times 7 are 42 and 2 to 
 carry are 44, 6 times 3 are 18 and 4 to carry are 22, pronounce only 
 the results, 24, 44, 22, performing the operations mentally. This will 
 greatly facilitate the process of multiplying. 
 
 
 EXAMPLES 
 
 FOB PRACTICE. 
 
 
 Multiplicand, 
 
 (2.) 
 7324 
 
 (3.) 
 
 6812 
 
 (4.) 
 34651 
 
 Multiplier, 
 Product, 
 
 4 
 
 6 
 
 5 
 
 29296 
 
 40872 
 
 173255 
 
 (5.) 
 
 82456 
 
 (6.) 
 92714 
 
 (?) 
 28093 
 
 (8.) 
 46247 
 
 3 
 
 7 
 
 8 
 
 9 
 
 
 Second explanation. Repeating a number has what effect on the 
 unit value ? The product must be of the same kind as what ? 
 
MULTIPLICATION. 45 
 
 9. Multiply 32746 by 5. Ans. 163730. 
 
 10. Multiply 840371 by 7. Ans. 5882597. 
 
 11. Multiply 137629 by 8. Ans. 1101032. 
 
 12. Multiply 93762 by 3. Ans. 281286. 
 
 13. Multiply 543272 by 4. Ans. 2173088. 
 
 14. Multiply 703164 by 9. Ans. 6328476. 
 
 15. What will be the cost of 344 cords of wood at 4 dol- 
 lars a cord? Ans. 1376. 
 
 16. How much will an army of 7856 men receive in one 
 week, if each man receive 6 dollars ? Ans. 47136 dollars. 
 
 17. In one day are 86400 seconds ; how many seconds in 
 7 days ? Ans. 604800 seconds. 
 
 18. What will 7640 bushels of wheat cost, at 9 shillings a 
 bushel ? Ans. 68760 shillings. 
 
 19. At 5 dollars an acre, what will 2487 acres of land 
 cost? Ans. 12435 dollars. 
 
 20. In one mile are 5280 feet ; how many feet in 8 miles ? 
 
 Ans. 42240 feet. 
 
 CASE II. 
 
 62. When the multiplier consists of two or more 
 figures. 
 
 1. Multiply 746 by 23. 
 
 OPERATION. ANALYSIS. Writ- 
 
 Multiplicand, 746 in g tne multiplicand 
 
 Multiplier, 23 and multiplier as in 
 
 Case I, we first mul- 
 
 2238 $gSr* tiply each figure in the 
 1492 2o|{gf a t n h |. mul - multiplicand by the 
 
 Product, 17158 {g3SEf'- t figure of the mul- 
 
 tiplier, precisely as m 
 
 Case I. We then multiply by the 2 tens. 2 tens times 6 units, or 6 
 times 2 tens, are 12 tens, equal to 1 hundred, and 2 tens ; we place the 
 2 tens under the tens figure in the product already obtained, and add 
 the 1 hundred to the next hundreds produced. 2 tens times 4 tens 
 are 8 hundreds, and the 1 hundred of the last product added are 9 
 hundreds ; we write the 9 in hundreds' place in the product. 2 tens 
 
 Case II is what ? Give explanation. 
 
46 SIMPLE NUMBERS. 
 
 times 7 hundreds are 14 thousands, equal to 1 ten thousand and 4 
 thousands, which we write in their appropriate places in the product. 
 Then adding the two products,, we have the entire product, 17158. 
 
 NOTES. 1. When the multiplier contains two or more figures, the 
 several results obtained by multiplying by each figure are called partial 
 products. 
 
 2. When there are ciphers between the significant figures of the 
 multiplier, pass over them, and multiply by the significant figures only. 
 
 B3. From the preceding examples and illustrations we 
 deduce the following general 
 
 RULE. I. Write the multiplier under the multiplicand, placing 
 units of the same order under each other. 
 
 II. Multiply the multiplicand by each figure of the multi- 
 plier successively, beginning with the unit figure, and write the 
 first figure of each partial product under the figure of the mul- 
 tiplier used, writing down and carrying as in addition. 
 
 III. If there are partial products, add them, and their sum 
 will be the product required. 
 
 G4:. PROOF. 1. Multiply the multiplier by the multipli- 
 cand, and if the product is the same as the first result, the 
 work is correct. Or, 
 
 2. Multiply the multiplicand by the multiplier diminished 
 by 1, and to the product add the multiplicand ; if the sum be 
 the same as the product by the whole of the multiplier, the 
 work is correct. 
 
 Multiply 
 By 
 
 Ans. 
 
 EXAMPLES 
 
 (2.) 
 4732 
 36 
 
 FOK PRACTICE. 
 
 (3.) 
 
 8721 
 47 
 
 (4.) 
 17605 
 204 
 
 28392 
 14196 
 
 61047 
 34884 
 
 70420 
 35210 
 
 170352 
 
 409887 
 
 3591420 
 
 What are partial products ? When there are ciphers in the multi- 
 plier, how proceed ? llule, first step ? Second ? Third ? Proof, 
 first method ? Second ? 
 
MULTIPLICATION. 47 
 
 (5.) (6.) 
 
 7648 81092 
 
 328 194 
 
 8. How many yards of linen in 759 pieces, each piece con- 
 taining 25 yards ? Am. 18975 yards. 
 
 9. Sound is known to travel about 1142 feet in a second of 
 time ; how far will it travel in 69 seconds ? 
 
 10. A man bought 36 city lots, at 475 dollars each ; how 
 mucft did they all cost him ? Ans. 17100 dollars. 
 
 11. What would be the value of 867 shares of railroad 
 stock, at 97 dollars a share ? Ans. 84099 dollars. 
 
 12. How many pages in 3475 books, if there be 362 
 pages in each book ? Ans. 1257950 pages. 
 
 13. In a garrison of 4507 men, each man receives annually 
 208 dollars ; how much do they all receive ? 
 
 14. Multiply 7198 by 216. Ans. 1554768. 
 
 15. Multiply 31416 by 175. Ans. 5497800. 
 
 16. Multiply 7071 by 556. Ans. 3931476. 
 
 17. Multiply 75649 by 579. Ans. 43800771. 
 
 18. Multiply 15607 by 3094. Ans. 48288058. 
 
 19. Multiply 79094451 by 76095. Ans. 6018692248845. 
 
 20. Multiply live hundred forty thousand six hundred nine, 
 by seventeen hundred fifty. Ans. 946065750. 
 
 21. Multiply four million twenty-five thousand three hun- 
 dred ten, by seventy-five thousand forty-six. 
 
 Ans. 302083414260. 
 
 22. Multiply eight hundred seventy-seven million five hun- 
 dred ten thousand eight hundred sixty-four, by five hundred 
 forty-five thousand three hundred fifty-seven. 
 
 Ans. 478556692258448. 
 
 23. If one mile of railroad require 116 tons of iron, worth 
 65 dollars a ton, what will be the cost of sufficient iron to 
 construct a road 128 miles in length? Ans. 965120 dollars. 
 
48 SIMPLE NUMBERS. 
 
 CONTRACTIONS. 
 CASE I. 
 
 When the multiplier is a composite number. 
 
 A Composite Number is one that may be produced by 
 multiplying together two or more numbers ; thus, 18 is a com- 
 posite number, since 6X3 = 18; or, 9 X 2 = 18 ; or, 3 X 
 3X 2 = 18. 
 
 66. The Component Factors of a number are the sev- 
 eral numbers which, multiplied together, produce the given 
 number ; thus, the component factors of 20 are 10 and 2, 
 (10 X 2 = 20 ;) or, 4 and 5, (4 X 5 = 20 ;) or, 2 and 2 and 
 5, (2 X 2 X 5 = 20.) 
 
 NOTE. The pupil must not confound the factors with the parts of a 
 number. Thus, the factors of which 12 is composed, are 4 and 3, 
 (4X3=12 ;) while the parts of which 12 is composed are 8 and 4, 
 (8 -f- 4 = 12,) or 10 and 2, (10 + 2 = 12.) The factors are multiplied, 
 while the parts are added, to produce the number. 
 
 1. What will 32 horses cost, at 174 dollars apiece ? 
 
 OPERATION. ANALYSIS. The fac- 
 
 Multiplicand, 174 COSt of 1 llOFSC. tors of 32 are 4 and 
 
 1st factor, 4 8 - If we multiply the 
 
 cost of 1 horse by 4, 
 
 696 cost of 4 horses. we obtain the cost of 4 
 *2d factor, 8 horses ; and by multi- 
 
 Product, 5568 cost of 32 horses. ft cost _ f . 4 
 
 horses by 8, we obtain 
 
 the cost of 8 times 4 horses, or 32 horses, the number bought. 
 67'. Hence we have the following 
 
 RULE. I. Separate the composite number into two or more 
 factors. 
 
 II. Multiply the multiplicand by one of these factors, and 
 
 What are contractions? Case I is what? Define a composite 
 number. Component factors. What caution is given ? Give ex- 
 planation. Rule, first step ? Second ? 
 
MULTIPLICATION. 49 
 
 that product "by another, and so on until all the factors have 
 been used successively ; the last product will be the product re- 
 quired, 
 
 NOTE. The product of any number of factors will be the same in 
 whatever order they are multiplied. Thus, 4 X 3 X 5 = 60, and 
 5X 4X 3 = 60. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Multiply 3472 by 48 = 6 X 8. Ans. 166656. 
 
 3. Multiply 14761 by 64 = 8 X 8. 
 
 4. Multiply 87034 by 81 = 3 X 3 X 9. Ans. 7049754. 
 
 5. Multiply 47326 by 120 = 6 X 5 X 4. 
 
 6. Multiply 60315 by 96. Ans. 5790240. 
 
 7. Multiply 291042 by 125. Ans. 36380250. 
 
 8. If a vessel sail 436 miles in 1 day, how far will she sail 
 in 56 days ? Ans. 24416 miles. 
 
 9. How much will 72 acres of land cost, at 124 dollars an 
 acre ? Ans. 8928 dollars. 
 
 10. There are 5280 feet in a mile; how many feet in 84 
 miles ? Ans. 443520 feet. 
 
 11. What will 120 yoke of cattle cost, at 125 dollars a 
 yoke ? 
 
 CASE II. 
 
 68. When the multiplier is 10, 100, 1000, &c. 
 
 If we annex a cipher to the multiplicand, each figure is re- 
 moved one place toward the left, and consequently the value of 
 the whole number is increased tenfold, (33.) If two ciphers 
 are annexed, each figure is removed two places toward the 
 left, and the value of the number is increased one hundred 
 fold ; and every additional cipher increases the value tenfold. 
 
 69. Hence the following 
 
 RULE. Annex as many ciphers to the multiplicand as there 
 are ciphers in the multiplier ; the number so formed will he 
 the product required. 
 
 Case II is what ? Give explanation. Rule ? 
 E . 
 
50 SIMPLE NUMBERS. 
 
 EXAMPLES HOR PRACTICE. 
 
 1. Multiply 347 by 10. Ans. 3470. 
 
 2. Multiply 4731 by 100. Ans. 473100. 
 
 3. Multiply 13071 by 1000. 
 
 4. Multiply 89017 by 10000. 
 
 5. If 1 acre of land cost 36 dollars, what will 10 acres 
 cost? Ans. 3 60 dollars. 
 
 6. If 1 bushel of corn cost 65 cents, what will 1000 bushels 
 cost ? Ans. 65000 cents. 
 
 CASE III. 
 
 7O. When there are ciphers at the right hand of 
 one or both of the factors. 
 
 1. Multiply 1200 by 60. 
 
 OPERATION. ANALYSIS. Both multiplicand and 
 
 Multiplicand, 1200 multiplier may be resolved into their 
 
 Multi lier 60 component factors ; 1200 into 12 and 
 
 100, and 60 into 6 and 10. If these 
 
 Product, 72000 several factors be multiplied together 
 
 they will produce the same product as 
 
 the given numbers, (67.) Thus, 12 X 6 = 72, and 72 X 100 = 
 7200, and 7200 X 10 = 72000, which is the same result as in the 
 operation. Hence the following 
 
 RULE. Multiply the significant figures of the multiplicand 
 by those of the multiplier, and to the product annex as many 
 ciphers as there are ciphers on the right of both factors. 
 
 EXAMPLES FOR PRACTICE. 
 
 (2.) (3.) 
 
 Multiply 4720 10340000 
 
 By 340000 105000 
 
 1888 5170 
 
 1416 1034 
 
 1604800000 1085700000000 
 
 Case III is what ? Give explanation. Rule. 
 
MULTIPLICATION. 51 
 
 4. Multiply 70340 by 800400. Am. 56300136000. 
 
 5. Multiply 3400900 by 207000. Ans. 703986300000. 
 
 6. Multiply 634003000 by 40020. Ans. 25372800060000. 
 
 7. Multiply 10203070 by 50302000. 
 
 Ans. 513234827140000. 
 
 8. Multiply 30090800 by 600080. Ans. 18056887264000. 
 
 9. Multiply eighty million seven thousand six hundred, by 
 eight million seven hundred sixty. Ans. 640121605776000. 
 
 10. Multiply fifty million ten thousand seventy, by sixty- 
 four thousand. Ans. 3200644480000. 
 
 11. Multiply ten million three hundred fifty thousand one 
 hundred, by eighty thousand nine hundred. 
 
 Ans. 837323090000. 
 
 12. There are 296 members of Congress, and each one re- 
 ceives a salary of 3000 dollars a year ; how much do they all 
 receive ? 
 
 EXAMPLES COMBINING ADDITION, SUBTRACTION, AND 
 MULTIPLICATION. 
 
 1. Bought 45 cords of wood at 4 dollars a cord, and 9 loads 
 of hay at 13 dollars a load; what was the cost of the wood 
 and hay ? Ans. 297 dollars. 
 
 2. A merchant bought 6 hogsheads of sugar at 31 dollars 
 a hogshead, and sold it for 39 dollars a hogshead ; how much 
 did he gain? 
 
 3. Bought 288 barrels of flour for 1875 dollars, and sold 
 the same for 9 dollars a barrel ; how much was the gain ? 
 
 Ans. Ill dollars. 
 
 4. If a young man receive 500 dollars a year salary and 
 pay 240 dollars for board, 125 dollars for clothing, 75 dollars 
 for books, and 50 dollars for other expenses, how much will 
 he have left at the end of the year? Ans. 10 dollars. 
 
 5. A farmer sold 184 bushels of wheat at 2 dollars a 
 bushel, for which he received 67 yards of cloth at 4 dollars a 
 yard, and the balance in groceries ; how much did his gro- 
 ceries cost him ? 
 
52 SIMPLE NUMBERS. 
 
 6. A sold a farm of 320 acres at 36 dollars an acre ; B 
 sold one of 244 acres at 48 dollars an acre ; \vhich received 
 the greater sum, and how much? Ans. B, 192 dollars. 
 
 7. Two persons start from the same point and travel in 
 opposite directions, one at the rate of 35 miles a day, and the 
 other 29 miles a day ; how far apart will they be in 16 days? 
 
 Ans. 1024 miles. 
 
 8. A merchant tailor bought 14 bales of cloth, each bale 
 containing 26 pieces, and each piece 43 yards; how many 
 yards of cloth did he buy? Ans. 15652 yards. 
 
 9. If a man have an income of 3700 dollars a year, and his 
 daily expenses be 4 dollars ; what will he save in a year, or 
 365 days ? Ans. 2240 dollars. 
 
 10. A man sold three houses ; for the first he received 
 2475 dollars, for the second 840 dollars less than he received 
 for the first, and for the third as much as for the other two ; 
 how much did he receive for the three ? Ans. 8220 dollars. 
 
 11. A man sets out to travel from Albany to Buffalo, a 
 distance of 336 miles, and walks 28 miles a day for 10 days ; 
 how far is he from Buffalo ? 
 
 12. Mr. C bought 14 cows at 23 dollars each, 7 horses at 
 96 dollars each, 34 oxen at 57 dollars each, and 300 sheep at 
 2 dollars each ; he sold the whole for 3842 dollars ; how 
 much did he gain ? Ans. 310 dollars. 
 
 13. A drover bought 164 head of cattle at 36 dollars a 
 head, and 850 sheep at 3 dollars a head ; how much did he 
 pay for all ? 
 
 14. A banker has an income of 14760 dollars a year; he 
 pays 1575 dollars for house rent, and four times as much for 
 family expenses ; how much does he save annually ? 
 
 Ans. 6885 dollars. 
 
 15. A flour merchant bought 936 barrels of flour at 9 dol- 
 lars a barrel; he sold 480 barrels at 10 dollars a barrel, and 
 the remainder at 8 dollars a barrel ; how much did he gain or 
 lose ? Ans. Gained 24 dollars. 
 
DIVISION. 53 
 
 DIVISION. 
 
 MENTAL EXERCISES. 
 
 71 . 1. How many hats, at 4 dollars apiece, can be bought 
 
 for 20 dollars ? 
 
 
 
 ANALYSIS. Since 4 dollars will buy one hat, 20 dollars will buy 
 as many hats as 4 is contained times in 20, which is 5 times. There- 
 fore, 5 hats, at 4 dollars apiece, can be bought for 20 dollars. 
 
 2. A man gave 1 6 dollars for 8 barrels of apples ; what 
 was the cost of each barrel ? 
 
 3. If 1 cord of wood cost 3 dollars, how many cords can 
 be bought for 15 dollars ? 
 
 4. At 6 shillings a bushel, how many bushels of corn can 
 be bought for 24 shillings ? 
 
 5. When flour is 6 dollars a barrel, how many barrels can 
 be bought for 30 dollars ? 
 
 6. If a man can dig 7 rods of ditch in a day, how many 
 days will it take him to dig 28 rods ? 
 
 7. If an orchard contain 56 trees, and 7 trees in a row, 
 how many rows are there ? 
 
 8. Bought 6 barrels of flour for 42 dollars ; what was the 
 cost of 1 barrel ? 
 
 9. If a farmer divide 21 bushels of potatoes equally 
 among 7 laborers, how many bushels will each receive ? 
 
 10. How many oranges can be bought for 27 cents, at 3 
 cents each? 
 
 11. A farmer paid 35 dollars for sheep, at 5 dollars apiece ; 
 how many did he buy ? 
 
 12. How many times 4 in 28 ? in 16 ? in 36 ? 
 
 13. How many times 8 in 40 ? in 56 ? in 64 ? 
 
 14. How many times 9 in 36 ? in 63 ? in 81 ? 
 
 15. How many times 7 in 49 ? in 70 ? in 84 ? 
 
 E* 
 
54 
 
 SIMPLE NUMBERS. 
 
 7!3. Division is the process of finding how many times 
 one number is contained in another. 
 
 73. The Dividend is the number to be divided. 
 
 74. The Divisor is the number to divide by. 
 
 75. The Quotient is the result obtained by the process of 
 division, and shows how many times the divisor is contained 
 in the dividend. 
 
 NOTES. 1. When the dividend does not contain the divisor an exact 
 number of times, the part of the dividend left is called the remainder, 
 and it must be less than the divisor. 
 
 2. As the remainder is always a part of the dividend, it is always 
 of the same name or kind. 
 
 3. When there is no remainder the division is said to be complete. 
 
 7O. The sign, -j-, placed between two numbers, denotes 
 division, and shows that the number on the left is to be divided 
 by the number on the right. Thus, 20 -f- 4 = 5, is read, 20 
 divided by 4 is equal to 5. 
 
 Division is also indicated by writing the dividend above, and 
 
 12 
 the divisor below a short horizontal line ; thus, = 4, shows 
 
 that 12 divided by 3 equals 4. 
 
 DIVISION TABLE. 
 
 1- 
 
 
 -| 
 
 2- 
 
 - 2= 1 
 
 3- 
 
 - 3= 1 
 
 4- 
 
 - 4= 1 
 
 2- 
 
 
 = 2 
 
 4- 
 
 - 2= 2 
 
 6- 
 
 -3=2 
 
 8- 
 
 - 4= 2 
 
 3- 
 
 
 3 
 
 6- 
 
 - 2= 3 
 
 9- 
 
 - 3= 3 
 
 12- 
 
 - 4= 3 
 
 4- 
 
 
 = 4 
 
 8- 
 
 - 2= 4 
 
 12- 
 
 - 3= 4 
 
 16- 
 
 - 4 4 
 
 5- 
 
 
 5 
 
 10- 
 
 - 2= 5 
 
 15- 
 
 - 3= 5 
 
 20- 
 
 -45 
 
 6- 
 
 
 = 6 
 
 12- 
 
 -2=6 
 
 18- 
 
 - 3= 6 
 
 24- 
 
 -46 
 
 7- 
 
 -1=7 
 
 14- 
 
 - 2= 7 
 
 21- 
 
 - 3= 7 
 
 28- 
 
 - 4 7 
 
 8- 
 
 -1=8 
 
 16- 
 
 - 2= 8 
 
 24- 
 
 - 3= 8 
 
 32- 
 
 - 4= 8 
 
 9- 
 
 -1=9 
 
 18- 
 
 -2=9 
 
 27- 
 
 - 3= 9 
 
 36- 
 
 - 4= 9 
 
 10- 
 
 -1 = 10 
 
 20- 
 
 -2 = 10 
 
 30- 
 
 -3 = 10 
 
 40- 
 
 - 4 = 10 
 
 11- 
 
 - 1 = 11 
 
 22- 
 
 - 2 = 11 
 
 33- 
 
 - 3 = 11 
 
 44- 
 
 - 4 = 11 
 
 12- 
 
 - 1 = 12 
 
 24- 
 
 -2 = 12 
 
 36- 
 
 -3 = 12 
 
 48- 
 
 - 4 = 12 
 
 Define division. Dividend. Divisor. Quotient. Remainder. 
 What is complete division ? What is the sign of division. 
 
DIVISION. 
 
 55 
 
 5- 
 
 - 5= 1 
 
 6- 
 
 - 6= 1 
 
 7- 
 
 - 7= 1 
 
 8- 
 
 - 8= 1 
 
 10- 
 
 - 5 2 
 
 12- 
 
 -6=2 
 
 14- 
 
 -7=2 
 
 16- 
 
 - 8= 2 
 
 15- 
 
 - 5 3 
 
 18- 
 
 - 6= 3 
 
 21- 
 
 -7=3 
 
 24- 
 
 -8=3 
 
 20- 
 
 - 5= 4 
 
 24- 
 
 -6=4 
 
 28- 
 
 -7=4 
 
 32- 
 
 - 8= 4 
 
 25- 
 
 Q mz o 
 
 30- 
 
 - 6= 5 
 
 35- 
 
 -7=5 
 
 40- 
 
 -8=5 
 
 30- 
 
 - 5= 6 
 
 36- 
 
 -6=6 
 
 42- 
 
 -7=6 
 
 48- 
 
 - 8= 6 
 
 35 
 
 - 5= 7 
 
 42- 
 
 -6=7 
 
 49- 
 
 -7=7 
 
 56- 
 
 -8=7 
 
 40- 
 
 - 5 8 
 
 48- 
 
 - 6= 8 
 
 56- 
 
 -7=8 
 
 64- 
 
 - 8= 8 
 
 45- 
 
 - 5= 9 
 
 54- 
 
 -6=9 
 
 63- 
 
 -7=9 
 
 72- 
 
 - 8= 9 
 
 50- 
 
 -5 = 10 
 
 60- 
 
 -6 = 10 
 
 70- 
 
 - 7 = 10 
 
 80- 
 
 - 8 = 10 
 
 55- 
 
 - 5 = 11 
 
 66- 
 
 - 6 = 11 
 
 77- 
 
 - 7 = 11 
 
 88- 
 
 - 8 = 11 
 
 60- 
 
 -5 = 12 
 
 72- 
 
 - 6 = 12 
 
 84- 
 
 -7 = 12 
 
 96- 
 
 -8 = 12 
 
 9- 
 
 - 9= 1 
 
 10- 
 
 -10= 1 
 
 11- 
 
 -11= 1 
 
 12- 
 
 -12= 1 
 
 18- 
 
 - 9= 2 
 
 20- 
 
 -10= 2 
 
 22- 
 
 -11= 2 
 
 24- 
 
 -12= 2 
 
 27- 
 
 -9=3 
 
 30- 
 
 -10= 3 
 
 33- 
 
 -11= 3 
 
 36- 
 
 -12= 3 
 
 36- 
 
 -9=4 
 
 40- 
 
 -10= 4 
 
 44- 
 
 -11= 4 
 
 48- 
 
 -12= 4 
 
 45- 
 
 - 9= 5 
 
 50- 
 
 -10= 5 
 
 55- 
 
 -11= 5 
 
 60- 
 
 -12= 5 
 
 54- 
 
 - 9= 6 
 
 60- 
 
 -10= 6 
 
 66- 
 
 -11= 6 
 
 72- 
 
 -12= 6 
 
 63- 
 
 - 9= 7 
 
 70- 
 
 -10= 7 
 
 77- 
 
 -11= 7 
 
 84- 
 
 -12= 7 
 
 72- 
 
 - 9= 8 
 
 80- 
 
 -10= 8 
 
 88- 
 
 -11= 8 
 
 96- 
 
 -12= 8 
 
 81- 
 
 _ 9= 9 
 
 90- 
 
 -10= 9 
 
 99- 
 
 -11= 9 
 
 108- 
 
 -12= 9 
 
 90- 
 
 -9 = 10 
 
 100- 
 
 -10 = 10 
 
 110- 
 
 -11 = 10 
 
 120- 
 
 -12=10 
 
 99- 
 
 - 9 = 11 
 
 110- 
 
 -10 = 11 
 
 121- 
 
 -11 = 11 
 
 132- 
 
 -12=11 
 
 108- 
 
 - 9 = 12 
 
 120- 
 
 -10 = 12 
 
 132- 
 
 -11 = 12 
 
 144- 
 
 -12=12 
 
 CASE I. 
 
 77. When the divisor consists of one figure. 
 1. How many times is 4 contained in 848 ? 
 
 OPERATION. 
 
 ANALYSIS. After writing the divisor 
 on the left of the dividend, with a line 
 between them, we begin at the left hand 
 and say : 4 is contained in 8 hundreds, 
 2 hundreds times, and write 2 in hun- 
 dreds' place in the quotient; then 4 is 
 contained in 4 tens 1 ten times, and write the 1 in tens' place in the 
 quotient ; then 4 is contained in 8 units 2 units times ; and writing thH 
 2 in units' place in the quotient, we have the entire quotient, 212. 
 
 Divisor, 
 Quotient, 
 
 Dividend, 
 4)848 
 
 212 
 
 Case I is what ? Give first explanation. 
 
56 SIMPLE NUMBERS. 
 
 2. How many times is 4 contained in 2884 ? 
 
 OPERATION. ANALYSIS. As we cannot divide 2 thousands by 
 4^2884 ^ we take the 2 thousands and the 8 hundreds to- 
 gether, and say, 4 is contained in 28 hundreds 7 hun- 
 721 dreds times, which we write in hundreds' place in 
 the quotient ; then 4 is contained in 8 tens 2 tens 
 times, which we write in tens' place hi the quotient ; and 4 is con- 
 tained in 4 units 1 unit, time, which we write in units' place in the 
 quotient, and we have the entire quotient, 721. 
 
 3. How many times is 6 contained in 1824 ? 
 
 OPERATION. ANALYSIS. Beginning as in the last example, we 
 
 6)1824 sa y> 6 i s contained in 18 hundreds 3 hundreds times, 
 
 which we write in hundreds' place in the quotient ; 
 
 then 6 is contained in 2 tens no times, and we write 
 
 a cipher in tens' place in the quotient ; and taking the 2 tens and 4 
 
 units together, 6 is contained in 24 units 4 units times, which we 
 
 write in units' place in the quotient, and we have 304 for the entire 
 
 quotient. 
 
 4. How many times is 4 contained in 943 ? 
 
 OPERATION. ANALYSIS. Here 4 is contained in 9 
 
 4)943 hundreds 2 hundreds times, and 1 hundred 
 
 over, which, united to the 4 tens, makes 
 
 235 ... 3 Rem. 14 tens ; 4 i n 14 tens, 3 tens times and 2 
 
 tens over, which, united to the 3 units, 
 
 make 23 units ; 4 in 23 units 5 units times and 3 units over. The 
 3 which is left after performing the division, should be divided by 4 ; 
 but the method of doing it cannot be explained until we reach 
 Fractions ; so we merely indicate the division by placing the divisor 
 under the dividend, thus, f . The entire quotient is written 235|, 
 which may be read, two hundred thirty-five and three divided by 
 four, or, two hundred thirty-five and a remainder of three. 
 
 From the foregoing examples and illustrations, we deduce 
 the following 
 
 RULE. I. Write the divisor at the left of the dividend, with 
 a line between them. 
 
 Second. Third. Rule, first step? 
 
DIVISION. 
 
 ^e divi- 
 
 II. Beginning at the left hand, divi( 
 dividend by the divisor, and write the 
 dend. 
 
 III. If there be a remainder after dividing any figure, re- 
 gard it as prefixed to the figure of the next lower order in the 
 dividend, and divide as before. 
 
 IV. Should any figure or part of the dividend be less than 
 the divisor, write a cipher in the quotient, and prefix the num- 
 ber to the figure of the next lower order in the dividend, and 
 divide as before. 
 
 V. If there be a remainder after dividing the last figure, 
 place it over the divisor at the right hand of the quotient. 
 
 PROOF. Multiply the divisor and quotient together, and to 
 the product add the remainder, if any ; if the result be equal 
 to the dividend, the work is correct. 
 
 NOTES. 1 . This method of proof depends on the fact that division is 
 the reverse of multiplication. The dividend answers to the product, the 
 divisor to one of the factors, and the quotient to the othe)\ 
 
 2. In multiplication the two factors are given, to find the product : 
 in division, the product and one of the factors are given to find the 
 other factor. 
 
 llTT 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Divide 7824 by 6. 
 
 OPERATION. 
 Divisor. 6)7824 Dividend. 
 
 1304 Quotient. 
 
 (2.) 
 
 4)65432 
 
 (3.) 
 5)89135 
 
 PROOF. 
 
 1304 Quotient. 
 6 Divisor. 
 
 7824 Dividend. 
 
 6)178932 
 
 (5.) 
 
 7)4708935 
 
 (6.) 
 8)1462376 
 
 (7.) 
 9)7468542 
 
 Second step? Third? Fourth? Fifth? Proof? How does divis- 
 ion differ from multiplication ? 
 
58 
 
 SIMPLE NUMBERS. 
 
 8. Divide 3102455 by 5. 
 
 9. Divide 1762891 by 4. 
 
 10. Divide 546215747 by 11, 
 
 11. Divide 30179624 by 12. 
 
 12. Divide 9254671 by 9. 
 
 13. Divide 7341563 by 7. 
 
 Quotients. 
 
 Quotients. 
 620491. 
 440722|. 
 49655977. 
 2514968^. 
 1028296. 
 Hem. 
 
 14. Divide 3179632 by 5. 
 
 15. Divide 19038716 by 8. 
 
 16. Divide 84201763 by 9. 
 
 17. Divide 2947691 by 12. 
 
 18. Divide 42084796 by 6. 
 
 Sums of quotients and remainders, 20680083. 28. 
 
 19. Divide 47645 dollars equally among 5 men; how 
 much will each receive ? Am. 9529 dollars. 
 
 20. In one week are 7 days; how many weeks in 17675 
 days? Ans. 2525 weeks. 
 
 21. How many barrels of flour, at 6 dollars a barrel, can be 
 bought for 6756 dollars? Ans. 1126 barrels. 
 
 22. Twelve things make a dozen ; how many dozen in 
 46216464? Ans. 3851372 dozen. 
 
 23. How many barrels of flour can be made from 347560 
 bushels of wheat, if it take 5 bushels to make one barrel ? 
 
 Ans. 69512 barrels. 
 
 24. If there be 3240622 acres of land in 11 townships, 
 how many acres in each township ? 
 
 25. A gentleman left his estate, worth 38470 dollars, to be 
 shared equally by his wife and 4 children ; how much did 
 each receive ? Ans. 7694 dollars. 
 
 CASE II. 
 
 78. When the divisor consists of two or more figures. 
 
 NOTE. To illustrate more clearly the method of operation, we will 
 first take an example usually performed by Short Division. 
 
 Case II is what ? 
 
DIVISION. 59 
 
 1. How many times is 8 contained in 2528 ? 
 OPERATION. ANALYSIS. As 8 is not contained in 2 thou- 
 
 8 } 2528 ( 316 sands, we take 2 and 5 as one number, and con- 
 9 A sider hoAv many times 8 is contained in this 
 
 partial dividend, 25 hundreds, and find that it 
 12 is contained 3 hundreds times, and a remainder. 
 
 8 To find this remainder, we multiply the divisor, 
 
 8, by the quotient figure, 3 hundreds, and sub- 
 tract the product, 24 hundreds, from the par- 
 tial dividend, 25 hundreds, and there remains 
 1 hundred. To this remainder we bring down 
 the 2 tens of the dividend, and consider the 12 tens a second partial 
 dividend. Then, 8 is contained in 12 tens 1 ten time and a remain- 
 der ; 8 multiplied by 1 ten produces 8 tens, which, subtracted from, 
 12 tens, leave 4 tens. To this remainder we bring down the 8 units, 
 and consider the 48 units the third partial dividend. Then, 8 is con- 
 tained in 48 units 6 units times. Multiplying and subtracting as 
 before, we find that nothing remains, and we have for the entire 
 quotient, 316. 
 
 2. How many times is 23 contained in 4807 ? 
 
 OPERATION. ANALYSIS. We first find how 
 
 Divisor. Divid'd. Quotient. many times 23 is contained in 48, 
 
 23 ) 4807 ( 209 the first partial dividend, and place 
 
 46 the result in the quotient on the 
 
 ~TTT right of the dividend. We then 
 
 multiply the divisor, 23, by the 
 
 quotient figure, 2, and subtract the 
 
 product, 46, from the part of the 
 
 dividend used, and to the remainder bring down the next figure of 
 the dividend, which is 0, making 20, for the second partial dividend. 
 Then, since 23 is contained in 20 no times, we place a cipher in the 
 quotient, and bring down the next figure of the dividend, making a 
 third partial dividend, 207 ; 23 is contained in 207, 9 times ; multi- 
 plying and subtracting as before, nothing remains, and we have for 
 the entire quotient, 209. 
 
 NOTES. 1 . When the process of dividing is performed mentally, and 
 the results only are written, as in Case I, the operation is termed Short 
 Division. 
 
 2. When the whole process of division is written, the operation is 
 termed Long Division. 
 
 Give first explanation. Second. What is long division ? What is 
 short division r When is each used r 
 
60 SIMPLE NUMBERS. 
 
 3. Short Division is generally used when the divisor is a number 
 that will allow the process of dividing to be performed mentally. 
 
 From the preceding illustrations we derive the following 
 general 
 
 RULE. I. Write the divisor at the left of the dividend, as 
 in short division. 
 
 II. Divide the least number of the left hand figures in the 
 dividend that will contain the divisor one or more times ', and 
 place the quotient at the right of the dividend, with a tine be- 
 tween them. 
 
 III. Multiply the divisor by this quotient figure, subtract 
 the product from the partial dividend used, and to the remain- 
 der bring down the next figure of the dividend. 
 
 IV. Divide as before, until all the figures of the dividend 
 have been brought down and divided. 
 
 V. If any partial dividend will not contain the divisor, 
 place a cipher in the quotient, and bring down the next figure 
 of the dividend, and divide as before. 
 
 VI. If there be a remainder after dividing all the figures of 
 the dividend, it must be written in the quotient, with the divi- 
 sor underneath. 
 
 NOTES. 1. If any remainder be equal to, or greater than the divisor, 
 the quotient figure is too small, and must be increased. 
 
 2. If the product of the divisor by the quotient figure be greater 
 than the partial dividend, the quotient figure is too large, and must be 
 diminished. 
 
 79. PROOF. 1. The same as in short division. Or, 
 
 2. Subtract the remainder, if any, from the dividend, and 
 divide the difference by the quotient ; if the result be the same 
 as the given divisor, the work is correct. 
 
 80. The operations in long division consist of five prin- 
 cipal steps, viz. : 
 
 1st. Write down the numbers. 
 
 Rule, first step ? Second ? Third ? Fourth ? Fifth ? Sixth ? First 
 direction ? Second ? Proof? Recapitulate the steps in their order. 
 
DIVISION. 61 
 
 2d. Find how many times. 
 
 3d. Multiply. 
 
 4th. Subtract. 
 
 5th. Bring down another figure. 
 
 EXAMPLES FOR PRACTICE. 
 
 3. Find how many times 36 is contained in 11798. 
 
 OPERATION. PROOF BY MULTIPLICATION. 
 
 Dividend. 
 
 Divisor. 36) 1179#'(327 Quotient. 327 Quotient. 
 
 108J' 36 Divisor. 
 
 9S17 
 
 72 
 
 2^ 
 
 252 26 Remainder. 
 
 26 Remainder. 11798 Dividend. 
 
 4. Find how many times 82 is contained in 89634. 
 
 OPERATION. PROOF BY DIVISION. 
 
 82 ) 8964-( 1093 89634 Dividend. 
 
 82 8 Remainder. 
 
 76^1 Quotient. 1093 ) 89626 ( 82 Divisor. 
 
 738 8744 
 
 ~254 2186 
 
 246 2186 
 
 5. Find how many times 154 is contained in 32740. 
 
 6. Divide 32572 by 34. Ans. 958. 
 
 7. Divide 1554768 by 216. Ans. 7198. 
 
 8. Divide 5497800 by 175. Ans. 31416. 
 
 9. Divide 3931476 by 556. Ans. 7071. 
 10. Divide 10983588 by 132. Ans. 83209. 
 
 F 
 
62 SIMPLE NUMBERS. 
 
 11. Divide 73484248 by 19. Am. 3867592. 
 
 12. Divide 8121918 by 21. Am. 386758. 
 
 13. Divide 10557312 by 16. Ans. 659832. 
 
 14. Divide 93840 by 63. Rem. 33. 
 
 15. Divide 352417 by 29. Rem. 9. 
 
 16. Divide 51846734 by 102. Hem. 32. 
 
 17. Divide 1457924651 by 1204. Rem. 1051. 
 
 18. Divide 729386 by 731. Rem. 579. 
 
 19. Divide 4843167 by 3605. Rem. 1652. 
 
 20. Divide 49816657 by 9101. Rem. 6884. 
 
 21. Divide 75867308 by 10115. Rem. 4808. 
 
 Quotients. Rem. 
 
 22. Divide 28101418481 by 1107, 25385201. 974. 
 
 23. Divide 65358547823 by 2789. 23434402. 645. 
 
 24. Divide 102030405060 by 123456. 826451. 70404. 
 
 25. Divide 48659910 by 54001. 901. 5009. 
 
 26. Divide 2331883961 by 6739549. 346. 7. 
 
 27. A railroad cost one million eight hundred fifty thousand 
 four hundred dollars, and was divided into eighteen thousand 
 five hundred and four shares ; what was the value of each 
 share? Ans. 100 dollars. 
 
 28. If a tax of seventy-two million three hundred twenty 
 thousand sixty dollars be equally assessed on ten thousand 
 seven hundred thirty-five towns, what amount of tax must 
 each town pay ? Ans. 6736 T VV^ dollars. 
 
 29. In 1850 there were in the United States 213 college 
 libraries, containing 942321 volumes ; what would be the 
 average number of volumes to each library ? 
 
 Ans. 442 4 ? f ^ vols. 
 
 30. The number of post offices in the United States in 
 1853 was 22320, and the entire revenue of the post office 
 department was 5937120 dollars; what was the average 
 revenue of each office? Ans. 266 dollars. 
 
DIVISION. 63 
 
 CONTRACTIONS. 
 
 CASE I. 
 
 81. When the divisor is a composite number. 
 
 1. If 3270 dollars be divided equally among 30 men, how 
 many dollars will each receive ? 
 
 OPERATION. ANALYSIS. If 3270 dollars be divided 
 
 5)3270 equally among 30 men, each man will receive 
 
 as many dollars as 30 is contained times in 
 
 6)654 3270 dollars. 30 may be resolved into the 
 
 109 Ans. factors 5 and 6 ; and we may suppose the 30 
 
 men divided into 5 groups of 6 men each ; 
 
 dividing the 3270 dollars by 5, the number of groups, we have 
 
 654, the number of dollars to be given to each group ; and dividing 
 
 the 654 dollars by 6, the number of men in each group, we have 
 
 109, the number of dollars that each man will receive. Hence, 
 
 RULE. Divide the dividend by one of the factors, and the 
 quotient thus obtained by another, and so on if there be more 
 than two factors, until every factor has been made a divisor. 
 The last quotient will be the quotient required. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Divide 3690 by 15 = 3 X 5. Ans. 246. 
 
 3. Divide 3528 by 24 = 4 X 6. Ans. 147. 
 
 4. Divide 7280 by 35 = 5 X 7. Ans. 208. 
 
 5. Divide 6228 by 36 = 6 X 6. Ans. 173. 
 
 6. Divide 33642 by 27 = 3 X 9. Ans. 1246. 
 
 7. Divide 153160 by 56 7 X 8. Ans. 2735. 
 
 8. Divide 15625 by 125 = 5 X 5 X 5. Ans. 125. 
 
 82. To find the true remainder. 
 
 1. Divide 1143 by 64, using the factors 2, 8, and 4, and find 
 the true remainder. 
 
 What are contractions ? Case I is what ? Give explanation. Rule. 
 
64 SIMPLE NUMBERS. 
 
 OPERATION. ANALYSIS. Divid- 
 
 2)1143 ing 1143 by 2, we 
 
 Q 7777 -, have a quotient of 
 
 8)571 1 rem. .,... ,^ . . 
 
 L 571, and a remainder 
 
 4)71- 3X2=6 " of 1 undivided, which, 
 
 17 q v ft V 9 Aft a being a P art of the 
 
 - given dividend, must 
 
 55 true rem. also be a part of the 
 
 true remainder. The 
 
 571 being a quotient arising from dividing by 2, its units are 
 
 2 times as great in value as the units of the given dividend, 1 143. 
 Dividing the 571 by 8, we have a quotient of 71, and a remainder 
 of 3 undivided. As this 3 is a part of the 571, it must be multiplied 
 by 2 to change it to the same kind of units as the 1. This makes a 
 true remainder of 6 arising from dividing by 8. Dividing the 7 1 by 
 4, we have a quotient of 17, and a remainder of 3 undivided. This 
 
 3 is a part of the 71, the units of which are 8 times as great in value 
 as those of the 571, and the units of the 571 are 2 times as great 
 in value as those of the given dividend, 1143 ; therefore, to change 
 this last remainder, 3, to units of the same value as the dividend, 
 we multiply it by 8 and 2, and obtain a true remainder of 48 arising 
 from dividing by 4. Adding the three partial remainders, we obtain 
 55, the true remainder. Hence, 
 
 RULE. I. Multiply each partial remainder by all the pre- 
 ceding divisors. 
 
 II. Add the several products, and the sum will be the true 
 remainder. 
 
 EXAMPLES FOR PRACTICE. 
 
 
 
 
 
 
 
 
 
 
 
 Hem, 
 
 2. 
 
 Divide 
 
 34712 
 
 by 
 
 42 
 
 6 
 
 X 
 
 7. 
 
 20. 
 
 3. 
 
 Divide 
 
 401376 
 
 by 
 
 64 
 
 8 
 
 X 
 
 8 
 
 . 
 
 
 32. 
 
 4. 
 
 Divide 
 
 139074 
 
 by 
 
 72 
 
 = 3 
 
 X 
 
 4 
 
 X 
 
 6. 
 
 42. 
 
 5. 
 
 Divide 
 
 9078126 
 
 by 
 
 90 
 
 = 3 
 
 X 
 
 5 
 
 X 
 
 6. 
 
 6. 
 
 6. 
 
 Divide 
 
 18730627 
 
 by 
 
 120 
 
 = 4 
 
 X 
 
 5 
 
 X 
 
 6. 
 
 67. 
 
 7. 
 
 Divide 
 
 7360479 
 
 by 
 
 96 
 
 n 
 
 X 
 
 6 
 
 X 
 
 8. 
 
 63. 
 
 8. 
 
 Divide 
 
 24726300 
 
 by 
 
 70 
 
 2 
 
 X 
 
 5 
 
 X 
 
 7. 
 
 60. 
 
 9. 
 
 Divide 
 
 5010207 
 
 by 
 
 84 
 
 rr 
 
 X 
 
 2 
 
 X 
 
 6. 
 
 15. 
 
 Explain the process of finding the true remainder when dividing by 
 the factors of a composite number. 
 
DIVISION. 65 
 
 CASE II. 
 
 83. When the divisor is 10, 100, 1000, &c. 
 
 1. Divide 374 acres of land equally among 10 men ; how 
 many acres will each have ? 
 
 OPERATION. ANALYSIS. Since we have shown, 
 
 110)3714 ^at ^ remove a figure one place 
 
 toward the left by annexing a cipher 
 
 Quotient. 37 4 Rem. increases its value tenfold, or multi- 
 
 or, 37 T 4 <y acres. plies it by 10, (68,) so, on the con- 
 
 trary, by cutting off or taking away 
 
 the right hand figure of a number, each of the other figures is removed 
 one place toward the right, and, consequently, the value of each is 
 diminished tenfold, or divided by 10, (32.) 
 
 For similar reasons, if we cut off two figures, we divide by 
 100, if three, we divide by 1000, and so on. Hence the 
 
 RULE. From the right hand of the dividend cut off as 
 many figures as there are ciphers in the divisor. Under the 
 figures so cut off, place the divisor, and the whole will form the 
 quotient. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Divide 4760 by 10. 
 
 3. Divide 362078 by 100. 
 
 4. Divide 1306321 by 1000. 
 
 5. Divide 9760347 by 10000. 
 
 6. Divide 2037160310 by 100000. 
 
 CASE III. 
 
 84. When there are ciphers on the right hand of 
 the divisor. 
 
 1. Divide 437661 by 800. 
 
 OPERATION. ANALYSIS. In this example we 
 
 8 1 00) 437 6 1 61 resolve 800 into the factors 8 and 
 
 100, and divide first by 100, by cut- 
 
 547 - - - 61 Rem. t i ng O fl- two right hand figures of the 
 
 Case II is what? Give explanation. Rule. Case in is what 
 Give explanation. 
 
66 SIMPLE NUMBERS. 
 
 dividend, (83,) and we have a quotient of 4376, and a remainder of 
 61. We next divide by 8, and obtain 547 for a quotient; and the 
 entire quotient is 
 
 2. Divide 34716 by 900. 
 
 OPERATION. ANALYSIS. Dividing 
 
 9|00)347|16 as in the last example, we 
 
 have a quotient of 38, and 
 38 Quotient. 5, 2d rem. ^ remainders> 16 and 
 
 5 X 100 + 16 = 516, true rem. 5, Multiplying 5, the 
 
 38, Ans. last remainder, by 100, 
 
 the preceding divisor, and 
 
 adding 16, the first remainder, (82,) we have 516 for the true re- 
 mainder. But this remainder consists of the last remainder, 5, pre- 
 fixed to the figures 16, cut off from the dividend. Hence, 
 
 85. When there is a remainder after dividing by the sig- 
 nificant figures, it must be prefixed to the figures cut off from 
 the dividend to give the true remainder ; if there be no other 
 remainder, the figures cut off from the dividend will be the 
 true remainder. 
 
 EXAMPLES FOR PRACTICE. 
 
 Quotients. Bern. 
 
 3. Divide 34716 by 900. 38 516 
 
 4. Divide 1047634 . by 2400. 436 1234 
 
 5. Divide 47321046 by 45000. 1051 26 
 
 6. Divide 2037903176 by 140000. 63176 
 
 7. Divide 976031425 'by 92000. 3425 
 
 8. Divide 80013176321 by 700000. 376321 
 
 9. Divide 19070367428 by 4160000. 4584 927428 
 
 10. Divide 379025644319 by 554000000. 535644319 
 
 11. The circumference of the earth at the equator is 24898 
 miles. How many hours would a train of cars require to travel 
 that distance, going at the rate of 50 miles an hour ? 
 
 Ans. 497 f*. 
 
 12. The sum of 350000 dollars is paid to an army of 14000 
 men ; what does each man receive ? Ans. 25 dollars. 
 
 How is the true remainder found ? 
 
PROMISCUOUS EXAMPLES. 67 
 
 EXAMPLES IN THE PRECEDING RULES. 
 
 1. George Washington was born in 1732, and lived 67 
 years ; in what year did he die ? Ans. in 1799. 
 
 2. How many dollars a day must a man spend, to use an 
 income of 1095 dollars a year ? Ans. 3 dollars. 
 
 3. If I give 141 dollars for a piece of cloth containing 47 
 yards, for how much must I sell it in order to gain one dollar 
 a yard ? Ans. 188 dollars. 
 
 4. A speculator who owned 500 acres, 17 acres, 98 acres, 
 and 121 acres of land, sold 325 acres ; how many acres had 
 he left? Ans. 411 acres. 
 
 5. A dealer sold a cargo of salt for 2300 dollars, and gained 
 625 dollars ; what did the cargo cost him ? 
 
 Ans. 1675 dollars. 
 
 6. If a man earn 60 dollars a month, and spend 45 dol- 
 lars in the same time, how long will it take him to save 900 
 dollars from his earnings ? 
 
 7. If 9 persons use a barrel of flour in 87 days, how many 
 days will a barrel last 1 person at the same rate ? 
 
 Ans. 783 days. 
 
 8. The first of three numbers is 4, the second is 8 times 
 the first, and the third is 9 times the second ; what is their 
 sum? Ans. 324. 
 
 9. If 2, 2, and 7 are three factors of 364, what is the 
 other factor? Ans. 13. 
 
 10. A man has 3 farms ; the first contains 78 acres, the 
 second 104 acres, and the third as many acres as both the 
 others ; how many acres in the 3 farms ? 
 
 11. If the expenses of a boy at school are 90 dollars for 
 board, 30 dollars for clothes, 12 dollars for tuition, 5 dollars 
 for books, and 7 dollars for pocket money, what would be the 
 expenses of 27 boys at the same rate ? Ans. 3888 dollars. 
 
 12. Four children inherited 2250 dollars each; but one 
 dying, the remaining three inherited the whole ; what was the 
 share of each? Ans. 3000 dollars. 
 
68 SIMPLE NUMBERS. 
 
 13. Two men travel in opposite directions, one at the rate 
 of 35 miles a day, and the other at the rate of 40 miles a day ; 
 how far apart are they at the end of 6 days ? 
 
 14. Two men travel in the same direction, one at the rate 
 of 35 miles a day, and the other at the rate of 40 miles a 
 day ; how far apart are they at the end of 6 days ? 
 
 15. A man was 45 years old, and he had been married 19 
 years ; how old was he when married? Ans. 26 years. 
 
 16. Upon how many acres of ground can the entire popu- 
 lation of the globe stand, supposing that 25000 persons can 
 stand upon one acre, and that the population is 1000000000 ? 
 
 Ans. 40000 acres. 
 
 17. Add 384, 1562, 25, and- 946 ; subtract 2723 from the 
 sum ; divide the remainder by 97 ; and multiply the quotient 
 by 142 ; what is the result ? Ans. 284. 
 
 18. How many steps of 3 feet each would a man take in 
 walking a mile, or 5280 feet? Ans. 1760 steps. 
 
 19. A man purchased a house for 2375 dollars, and ex- 
 pended 340 dollars in repairs ; he then sold it for railroad 
 stock worth 867 dollars, and 235 acres of western land val- 
 ued at 8 dollars an acre ; how much did he gain by the trade ? 
 
 Ans. 32 dollars. 
 
 20. The salary of a clergyman is 800 dollars a year, and 
 his yearly expenses are 450 dollars; if he be worth 1350 dol- 
 lars now, in how many years will he be worth 4500 dollars ? 
 
 Ans. 9 years. 
 
 21. How many bushels of oats at 40 cents a bushel, must 
 be given for 1600 bushels of wheat at 75 cents a bushel ? 
 
 Ans. 3000 bushels. 
 
 22. Bought 325 loads of wheat, each load containing 50 
 bushels, at 2 dollars a bushel ; what did the wheat cost ? 
 
 23. If you deposit 225 cents each week in a savings bank, 
 and take out 75 cents a week, how many cents will you have 
 left at the end of the year ? Ans. 7800 cents. 
 
 24. The product of two numbers is 31383450, and one of 
 the numbers is 4050 ; what is the other number ? 
 
PROMISCUOUS EXAMPLES. 69 
 
 25. The Illinois Central Railroad is 700 miles long, and 
 cost 31647000 dollars ; what did it cost per mile ? 
 
 Ans. 45210 dollars. 
 
 26. What number is that, which being divided by 7, the 
 quotient multiplied by 3, the product divided by 5, and this 
 quotient increased by 40, the sum will be 100 ? Ans. 700. 
 
 27. How many cows at 27 dollars apiece, must be given 
 for 54 tons of hay at 17 dollars a ton ? 
 
 28. A mechanic receives 56 dollars for 26 days' work, and 
 spends 2 dollars a day for the whole time ; how many dollars 
 has he left ? Ans. 4 dollars. 
 
 29. If 7 men can build a house in 98 days, how long would 
 it take one man to build it ? Ans. 686 days. 
 
 30. The number of school houses in the State of New 
 York, in 1855, was 11,137 ; suppose their cash value to have 
 been 5,301,212 dollars, what would be the average value? 
 
 Ans. 476 dollars. 
 
 31. A thief, traveling 5 miles an hour, has 48 miles the 
 start of an officer who is pursuing him at the rate of 1 1 miles 
 an hour ; in how many hours will the officer overtake him ? 
 
 Ans. 8 hours. 
 
 32. The average beat of the pulse of a man at middle age 
 is about 4500 times in an hour ; how many times does it beat 
 in 24 hours ? Ans. 108000 times. 
 
 33. How many years from the discovery of America, in 
 1492, to the year 1900 ? 
 
 34. According to the census, Maine has 31766 square 
 miles; New Hampshire, 9280; Vermont, 10212; Massachu- 
 setts, 7800; Rhode Island, 1306; Connecticut, 4674; and 
 New York, 47000 ; how many more square miles has all 
 New England than New York? 
 
 35. What is the remainder after dividing 62530000 by 
 87900? Ans. 33100. 
 
 36. A pound of cotton has been spun into a thread 8 miles 
 in length ; allowing 235 pounds for waste, how many pounds 
 will it take to spin a thread to reach round the earth, suppos- 
 ing the distance to be 25000 miles? Ans. 3360 pounds. 
 
70 SIMPLE NUMBERS. 
 
 37. John has 8546 dollars, which is 342 dollars less than 
 4 times as much as Charles has ; how many dollars has 
 Charles ? Am. 2222 dollars. 
 
 38. The quotient of one number divided by another is 37, 
 the divisor 245, and the remainder 230; what is the divi- 
 dend? Ans. 9295. 
 
 39. What number multiplied by 72084 will produce 
 5190048? Ans. 72. 
 
 40. There are two numbers, the greater of which is 73 
 times 109, and their difference is 17 times 28 ; what is the less 
 number? Ans. 7481. 
 
 41. The sum of two numbers is 360, and the less is 114 ; 
 what is the product of the two numbers ? Ans. 28044. 
 
 42. What number added to 2473248 makes 2568754? 
 
 Ans. 95506. 
 
 43. A farmer sold 35 bushels of wheat at 2 dollars a bush- 
 el, and 18 cords of wood at 3 dollars a cord; he received 9 
 yards of cloth at 4 dollars a yard, and the balance in money ; 
 how many dollars did he receive ? Ans. 88 dollars. 
 
 44. A cistern whose capacity is 840 gallons has two pipes ; 
 through one pipe 60 gallons run into it in an hour, and through 
 the other 39 gallons run out in the same time ; in how many 
 hours will the cistern be filled ? Ans. 40 hours. 
 
 45. Two men start from the same place, and travel in the 
 same direction ; one travels 32 miles a day, and the other 46; 
 how far will they be apart at the end of 15 days ? 
 
 Ans. 210 miles. 
 
 46. Henry resides 1 mile from school, and George 2 miles ; 
 if they each attend school 295 days, how many more miles 
 will George travel in that time than Henry ? Ans 590 miles. 
 
 47. If the remainder be 3252, and the minuend 4248, what 
 is the subtrahend ? Ans. 996. 
 
 48. How many times will a coach wheel 16 feet in circum- 
 ference revolve in going 160 miles, there being 5280 feet in a 
 mile ? Ans. 52800 times. 
 
PROMISCUOUS EXAMPLES. 71 
 
 49. A salt dealer has 3524 barrels in store; how many 
 barrels will he have left after selling 75 barrels to one man, 
 149 to another, 854 to another, and 287 to another? 
 
 Am. 2159 barrels. 
 
 50. How many yards of muslin will it take to make dresses 
 for a family of 7 girls, each girl having 3 dresses, and each 
 dress averaging 9 yards? Ans. 189 yards. 
 
 51. If a man can do a piece of work in 855 days, how 
 many men can do it in 45 days? Ans. 19 men. 
 
 52. A man, traveling 27 miles a day, starts from Boston 
 to meet another man, traveling 45 miles a day, from New York, 
 216 miles distant; in how many days will they meet ? 
 
 Ans. 3 days. 
 
 53. The population of New York in 1855 was 3466212, 
 of which 1124211 attended church regularly; how many 
 persons in the state were not regular attendants at church ? 
 
 Ans. 2342001 persons. 
 
 54. A farmer receives 684 dollars a year for produce from 
 his farm, and his expenses are 375 dollars a year ; how many 
 dollars will he save in five years ? 
 
 55. The salt manufacturer at Syracuse pays 58 cents for 
 wood to boil one barrel of salt, 10 cents for boiling, 5 cents to 
 the state for the brine, 28 cents for the packing barrel, and 3 
 cents for packing and weighing, and receives 125 cents from 
 the purchaser ; how many cents does he make on a barrel ? 
 
 Ans. 21 cents. 
 
 56. A company of 15 persons purchase a township of 
 western land for 286000 dollars, of which sum one man pays 
 6000 dollars, and the others the remainder, in equal amounts ; 
 how much does each of the others pay ? Ans. 20000 dollars. 
 
 57. If 256 be multiplied by 25, the product diminished by 
 625, and the remainder divided by 35, what will be the quo- 
 tient? Ans. 165. 
 
 58. Two men start from different places, distant 189 miles, 
 and travel toward each other ; one goes 4 miles, and the other 
 5 miles an hour ; in how many hours will they meet ? 
 
72 SIMPLE NUMBERS. 
 
 GENERAL PRINCIPLES OF DIVISION. 
 
 8G. The quotient in Division depends upon the relative 
 values of the dividend and divisor. Hence any change in the 
 value of either dividend or divisor must produce a change in 
 the value of the quotient. But some changes may be produced 
 upon both dividend and divisor, at the same time, that 
 will not affect the quotient. The laws which govern these 
 changes are called General Principles of Division, which we 
 will now examine. 
 
 I. 54-^-9 = 6. 
 
 If we multiply the dividend by 3, we have 
 54 x 3 -^ 9 = 162 -- 9 = 18, 
 
 and 18 equals the quotient, 6, multiplied by 3. Hence, 
 Multiplying the dividend by any number, multiplies the quotient 
 by the same number. 
 
 II. Using the same example, 54 -7- 9 = 6. 
 If we divide the dividend by 3 we have 
 
 V-T-9 = 18-r-9= 2, 
 
 and 2 = the quotient, 6, divided by 3. Hence, Dividing the 
 dividend by any number^ divides the quotient by the same 
 number. 
 
 III. If we multiply the divisor by 3, we have 
 
 54 _L. 9 X 3 = 54 ~ 27 2, 
 
 and 2 = the quotient, 6, divided by 3. Hence, Multiplying 
 the divisor by any number, divides the quotient by the same 
 number. 
 
 IV. If we divide the divisor by 3, we have 
 
 5* -T-| = 54-^8 = 18, 
 
 Upon what does the value of the quotient depend ? What is the 
 first general principle of division? Second? Third? Fourth? 
 
GENERAL PRINCIPLES OF DIVISION. 73 
 
 and 18 r= the quotient, 6, multiplied by 3. Hence, Dividing 
 the divisor by any number, multiplies the quotient by the same 
 number. 
 
 V. If we multiply both dividend and divisor by 3, we have 
 
 54 X 3 -^ 9 X 3 = 162 -^- 27 6. 
 
 Hence, Multiplying both dividend and divisor ty ike same num- 
 ber, does not alter the value of the quotient. 
 
 VI. If we divide both dividend and divisor by 3, we have 
 
 jyt^f =18 -f.3=6. 
 
 Hence, Dividing both dividend and divisor by the same num- 
 ber, does not alter the value of the quotient. 
 
 87. These six examples illustrate all the different changes 
 we ever have occasion to make upon the dividend and divisor 
 in practical arithmetic. The principles upon which these 
 changes are based may be stated as follows : 
 
 PRIN. I. Multiplying the dividend multiplies the quotient ; 
 and dividing the dividend divides the quotient. (8G. I and II.) 
 
 PRIN. II. Multiplying the divisor divides the quotient , and 
 dividing the divisor multiplies the quotient. (8G. Ill and IV.) 
 
 PRIN. III. Multiplying or dividing both dividend and 
 divisor by the same number, does not alter the quotient. (8G. 
 V and VI.) 
 
 88. These three principles may be embraced in one 
 
 GENERAL LAW. 
 
 A- change in the dividend produces a LIKE change in the 
 quotient; but a change, in the divisor produces an OPPOSITE 
 change in the quotient. 
 
 NOTE. If a number be multiplied and the product divided by the 
 same number, the quotient will be equal to the number multiplied. 
 Thus, 15 X 4 = 60, and 60 -f- 4 = 15. 
 
 Fifth ? Sixth ? Into how many general principles can these be con- 
 densed ? What is the first ? Second ? Third ? In what general law 
 are these embraced ? 
 
74 PROPERTIES OF NUMBERS. 
 
 EXACT DIVISORS. 
 
 80o An Exact Divisor of a number is one that gives 
 a whole number for a quotient. 
 
 As it is frequently desirable to know if a number has an exact 
 divisor, we will present a few directions that will be of assistance, 
 particularly in finding exact divisots of large numbers. 
 
 NOTE. A number whose unit figure is 0, 2, 4, 6, or 8 is called an 
 Even Number. And a number whose unit figure is 1, 3, 5, 7, or 9, is 
 called an Odd Number. 
 
 2 is an exact divisor of all even numbers. 
 
 4 is an exact divisor when it will exactly divide the tens 
 and units of a number. Thus, 4 is an exact divisor of 268, 
 756, 1284. 
 
 5 is an exact divisor of every number whose unit figure is 
 or 5. Thus, 5 is an exact divisor of 20, 955, and 2840. 
 
 8 is an exact divisor when it will exactly divide the hun- 
 dreds, tens, and units of a number. Thus, 8 is an exact 
 divisor of 1728, 5280, and 213560. 
 
 9 is an exact divisor when it will exactly divide the sum of 
 the digits of a number. Thus, in 2486790, the sum of the 
 digits 2 + 4+8 + 6 + 7+9 + = 36, and 36 -J- 9 4. 
 
 10 is an exact divisor when occupies units' place. 
 100 when 00 occupy the places of units and tens. 
 
 1000 when 000 occupy the places of units, tens, and hun- 
 dreds, &c. 
 
 A composite number is an exact divisor of any number, 
 when all its factors are exact divisors of the same number. 
 Thus, 2, 2, and 3 are exact divisors of 1 2 ; and so also are 4 
 (=2X2) and 6 (= 2 X 3). 
 
 An even number is not an exact divisor of an odd number. 
 
 If an odd number is an exact divisor of an even number, 
 
 What is an exact divisor ? "What is an even number ? An odd num- 
 ber ? When is 2 an exact divisor ? 4 ? 5 ? 9 ? 10 ? 100 ? 1000 ? 
 When is a composite number an exact divisor ? An even number is 
 not an exact divisor of what ? An odd number is an exact divisor of 
 what? 
 
FACTORING NUMBERS. 
 
 75 
 
 twice that odd number is also an exact divisor of the even 
 number. Thus, 7 is an exact divisor of 42 ; so also is 7 X 2, 
 or 14. 
 
 PRIME NUMBERS. 
 
 9O. A Prime Number is one that can not be resolved 
 or separated into two or more integral factors. 
 
 For reference, and to aid in determining the prime factors 
 of composite numbers, we give the following : 
 
 TABLE OF PRIME NUMBERS FROM 1 TO 1000. 
 
 1 
 
 59 
 
 139 
 
 233 
 
 337 
 
 439 
 
 557 
 
 653 
 
 769 
 
 883 
 
 2 
 
 61 
 
 149 
 
 239 
 
 347 
 
 443 
 
 563 
 
 659 
 
 773 
 
 .887 
 
 3 
 
 67 
 
 151 
 
 241 
 
 349 
 
 449 
 
 569 
 
 661 
 
 787 
 
 907 
 
 5 
 
 71. 
 
 157 
 
 251 
 
 353 
 
 457 
 
 571 
 
 673 
 
 797 
 
 911 
 
 7 
 
 73 
 
 163 
 
 257 
 
 359 
 
 461 
 
 577 
 
 677 
 
 809 
 
 919 
 
 11 
 
 79 
 
 167 
 
 263 
 
 367 
 
 463 
 
 587 
 
 683 
 
 811 
 
 929 
 
 13 
 
 83 
 
 173 
 
 269 
 
 373 
 
 467 
 
 593 
 
 691 
 
 821 
 
 937 
 
 17 
 
 89 
 
 179 
 
 271 
 
 379 
 
 479 
 
 599 
 
 701 
 
 823 
 
 941 
 
 19 
 
 97 
 
 181 
 
 277 
 
 383 
 
 487 
 
 601 
 
 709 
 
 827 
 
 947 
 
 23 
 
 101 
 
 191 
 
 281 
 
 389 
 
 491 
 
 607 
 
 719 
 
 829 
 
 953 
 
 29 
 
 103 
 
 193 
 
 283 
 
 397 
 
 499 
 
 613 
 
 727 
 
 839 
 
 967 
 
 31 
 
 107 
 
 197 
 
 293 
 
 401 
 
 503 
 
 617 
 
 733 
 
 853 
 
 971 
 
 37 
 
 109 
 
 199 
 
 307 
 
 409 
 
 509 
 
 619 
 
 739 
 
 857 
 
 977 
 
 41 
 
 113 
 
 211 
 
 311 
 
 419 
 
 521 
 
 631 
 
 743 
 
 859 
 
 983 
 
 43 
 
 127 
 
 223 
 
 313 
 
 421 
 
 523 
 
 641 
 
 751 
 
 863 
 
 991 
 
 47 
 
 131 
 
 227 
 
 317 
 
 431 
 
 541 
 
 643 
 
 757 
 
 877 
 
 997 
 
 53 
 
 137 
 
 229 
 
 331 
 
 433 
 
 547 
 
 647 
 
 761 
 
 881 
 
 
 FACTORING NUMBERS. 
 
 CASE I. 
 
 91. To resolve any composite number into its 
 prime factors. 
 
 Whai is a prime number ? In factoring numbers, Case I is what ? 
 
76 PROPERTIES OP NUMBERS. 
 
 1. What are the prime factors of 2772 ? 
 OPERATION. ANALYSIS. We divide the given number by 
 
 2 
 2 
 3 
 3 
 7 
 11 
 
 2772 2, the least prime factor, and the result by 2; 
 this gives an odd number for a quotient, divisible 
 by the prime factor, 3, and the quotient resulting 
 from this division is also divisible by 3. The 
 next quotient, 77, we divide by its least prime 
 factor, 7, and we obtain the quotient 11 ; this be- 
 ing a prime number, the division can not be car- 
 ried further. The divisors and last quotient, 2, 
 2, 3, 3, 7, and 11 are all the prime factors of the 
 
 given number, 2772. Hence the 
 
 RULE. Divide the given number by any prime factor ; di~ 
 vide the quotient in the same manner, and so continue the 
 division until the quotient is a prime number. The several 
 divisors and the last quotient will he the prime factors required. 
 
 PROOF. The product of all the prime factors will be the 
 given number. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What are the prime factors of 1 1 40 ? Ans. 2 } 2, 3, 5, 19* 
 
 3. What are the prime factors of 29925 ? 
 
 4. What are the prime factors of 2431 ? 
 
 5. Find the prime factors of 12673. 
 
 6. Find the prime factors of 2310. 
 
 7. Find the prime factors of 2205. 
 
 8. What are the prime factors of 13981 ? 
 
 CASE II. 
 
 92. To resolve a number into all the different sets 
 of factors possible. 
 
 1. In 36 how many sets of factors, and what are they? 
 Give explanation. Rule. Proof. Caac II is what ? 
 
CANCELLATION. 77 
 
 OPERATION. ANALYSIS. Writing the 36 at 
 
 '2 X 18 th k^ f ^ e s *& n = we arran g e 
 
 3 NX 12 a ^ the different sets of factors into 
 
 4. y o which it can be resolved under 
 each other, as shown in the opera- 
 
 36 rr<i tion, and we find that 36 can be 
 
 2X2X9 resolved into 8 sets of factors. 
 2X3X6 
 3X3X4 
 2X2X3X3 
 
 EXAMPLES FOR PRACTICE. 
 
 2. How many sets of factors in the number 24? What 
 are they ? Ans. 6 sets. 
 
 3. In 125 how many sets of factors ? What are they ? 
 
 Ans. 2 sets. 
 
 4. In 40 how many sets of factors, and what are they ? 
 
 Ans. 6 sets. 
 
 5. In 72 how many sets of factors, and what are they ? 
 
 Ans. 16 sets. 
 
 CANCELLATION. 
 
 93. Cancellation is the process of rejecting equal factors 
 from numbers sustaining to each other the relation of dividend 
 and divisor. 
 
 It has been shown (7T) that the dividend is equal to the 
 product of the divisor multiplied by the quotient. Hence, if 
 the dividend can be resolved into two factors, one of which is 
 the divisor, the other factor will be the quotient. 
 
 1. Divide 63 by 7. 
 
 OPERATION. ANALYSIS. We see in 
 
 Divisor, #)# x 9 Dividend. this example that 63 is 
 
 composed of the factors 7 
 
 9 Quotient. an( J g ? an ^ j-jjat fa e factor 
 
 7 is equal to the divisor. 
 
 Therefore we reject the factor 7, and the remaining factor, 9, is the 
 quotient. 
 
 Give explanation. What is cancellation ? Upon what principle is 
 it based ? Give first explanation. 
 
78 PROPERTIES OF NUMBERS. 
 
 Whenever the dividend and divisor are each composite 
 numbers, the factors common to both may first be rejected 
 without altering the final result. (87 5 Prin. III.) 
 
 2. "What is the quotient of 24 times 56 divided by 7 times 
 48? 
 
 OPERATION. ANALYSIS. 
 
 24 X 56 4 X X * X $ We first in- 
 
 4, A.US. dicate the op- 
 
 7x48 # X X $ eration to be 
 
 performed by 
 
 writing the numbers which constitute the dividend above a line, and 
 those which constitute the divisor below it. Instead of multiplying 
 24 by 56, in the dividend, we resolve 24 into the factors 4 and 6, 
 and 56 into the factors 7 and 8 ; and 48 in the divisor into the fac- 
 tors 6 and 8. We next cancel the factors 6, 7, and 8, which are 
 common to the dividend and divisor, and we have left the factor 4 
 in the dividend, which is the quotient. 
 
 NOTE. When all the factors or numbers in the dividend are can- 
 celed, 1 should be retained. 
 
 95. If any two numbers, one in the dividend and one in 
 the divisor, contain a common factor, we may reject that factor. 
 
 3. In 54 times 77, how many times 63 ? 
 
 OPERATION. ANALYSIS. In this example we see that 9 will 
 
 6 11 divide 54 and 63 ; so we reject 9 as a factor of 54, 
 
 ^ , #tf an( ^ reta ^ n ^ e f actor 6, and also as a factor of 63, 
 
 ft X /iff and retam the f ac tor 7. Again, 7 will divide 7 in 
 
 0$ the divisor, and 77 in the dividend. Dividing 
 
 ^ both numbers by 7, 1 will be retained in the 
 
 divisor, and 11 in the dividend. Finally, the 
 
 product of 6 X 1 1 = 66, the quotient. 
 
 4. Divide 25 X 16 X 12 by 10 X 4 X 6 X 7. 
 
 OPERATION. 
 
 54 $ 
 
 ANALYSIS. In 
 this, as in the pre- 
 ceding example, we 
 
 ^0X^X0X7 7 
 
 tors that are com- 
 mon to both divi- 
 dend and divisor, 
 
 Give second explanation. 
 
CANCELLATION. 79 
 
 and we have remaining the factor 7 in the divisor, and the factors 5 and 
 4 in the dividend. Completing the work, we have ^- = 2, Ans. 
 
 From the preceding examples and illustrations we derive 
 the following 
 
 RULE. I. Write the numbers composing the dividend above 
 a horizontal line, and the numbers composing the divisor 
 below it. 
 
 II. Cancel all the factors common to both dividend and 
 divisor. 
 
 III. Divide the product of the remaining factors of the div- 
 idend by the product of the remaining factors of the divisor y 
 and the result will be the quotient. 
 
 NOTES. 1. Rejecting a factor from any number is dividing the number 
 by that factor. 
 
 2. When a factor is canceled, the unit, 1, is supposed to take its 
 place. 
 
 3. One factor in the dividend -will cancel only one equal factor in the 
 divisor. 
 
 4. If all the factors or numbers of the divisor are canceled, the 
 product of the remaining factors of the dividend will be the quotient. 
 
 5. By many it is thought more convenient to write the factors of 
 the dividend on the right of a vertical line, and the factors of the divisor 
 on the left. 
 
 EXAMPLES FOR -PRACTICE. 
 
 1. What is the quotient of 16 X 5 X 4 divided by 20 X 8 ? 
 
 FIRST OPERATION. SECOND OPERATION. 
 
 2 0, 
 
 = 2, 
 
 2, Ans. 
 2. Divide the product of 120 X 44 X 6 X 7 by 72 X 33 X 14. 
 
 Rule, first step ? Second ? Third ? What is the effect of rejecting 
 a factor ? What is the quotient when all the factors in the divisor are 
 canceled ? 
 
80 PROPERTIES OF NUMBERS. 
 
 FIRST OPERATION. 
 
 _ 
 
 --- 
 
 3 
 
 SECOND OPERATION. 
 
 ,10 
 
 "U 
 
 3. Divide the product of 33 X 35 X 28 by 11 X 15 X 14. 
 
 Am. 14. 
 
 4. What is the quotient of 21 X 11 X 26 divided by 14 X 
 13? Am. 33. 
 
 5. Divide the product of the numbers 48, 72, 28, and 5, by 
 the product of the numbers 84, 15, 7, and 6, and give the 
 result. Ans. 9^-. 
 
 '6. Divide 140 X 39 X 13 X 7 by 30 X 7 X 26 X 21. 
 
 Ans. 4^. 
 
 7. What is the quotient of 66 X 9 X 18 X 5 divided by 
 22 X 6 X 40 ? Ans. 10J. 
 
 8. Divide the product of 200 X 36 X 30 X 21 by 270 X 
 40 X 15 X 14. Ans. 2. 
 
 9. Multiply 240 by 56, and divide the product by 60 mul- 
 tiplied by 28. Ans. 8. 
 
 10. The product of the numbers 18, 6, 4, and 42 is to be 
 divided by the product of the numbers 4, 9, 3, 7, and 6 ; what 
 is the result ? Ans. 4. 
 
 11. How many tons of hay, at 12 dollars a ton, must be 
 given for 30 cords of wood, at 4 dollars a cord ? Ans. 10 tons. 
 
GREATEST COMMON DIVISOR. 81 
 
 12. How many firkins of butter, each containing 56 pounds, 
 at 13 cents a pound, must be given for 4 barrels of sugar, each 
 containing 182 pounds, at 6 cents a pound ? Am. 6 firkins. 
 
 13. A tailor bought 5 pieces of cloth, each piece containing 
 24 yards, at 3 dollars a yard. How many suits of clothes, at 
 18 dollars a suit, must be made from the cloth to pay for it? 
 
 Am. 20 suits. 
 
 14. How many days' work, at 75 cents a day, will pay for 
 115 bushels of corn, at 50 cents a bushel? Ans. 76fdays. 
 
 GREATEST COMMON DIVISOR. 
 
 9O. A Common Divisor of two or more numbers is a 
 number that will exactly divide each of them. 
 
 97. The Greatest Common Divisor of two or more num- 
 bers is the greatest number that will exactly divide each of 
 them. 
 
 Numbers prime to each other are such as have no common 
 divisor. 
 
 NOTE. A common divisor is sometimes called a Common Measure ; 
 and the greatest common divisor, the Greatest Common Measure. 
 
 CASE I. 
 
 98. "When the numbers are readily factored. 
 
 1. What is the greatest common divisor of 6 and 10 ? 
 
 Am. 2. 
 
 OPERATION. ANALYSIS. We readily find by inspection 
 6 . . 1 that 2 will divide both the given numbers ; 
 
 hence 2 is a common divisor ; and since the 
 
 ** ' _ quotients 3 and 5 have no common factor, but 
 
 are prime to each other, the common divisor, 
 2, must be the greatest common divisor. 
 
 2. What is the greatest common divisor of 42, 63, and 105 ? 
 
 What is a common divisor ? The greatest common divisor ? A 
 common measure ? The greatest common measure ? What is Case I ? 
 Give analysis. 
 
82 PROPERTIES OF NUMBERS. 
 
 OPERATION. ANALYSIS. We observe that 3 
 
 14.. 21 .. 35 
 
 42 . . 63 . . 105 wn l exactly divide each of the given 
 
 numbers, and that 7 will exactly 
 divide each of the resulting quo- 
 2 . . 3 . . 5 tients. Hence, each of the given 
 
 -~^-^ numbers can be exactly divided by 3 
 
 ** X 21, Ans. times 7 ; and these numbers must be 
 
 component factors of the greatest 
 
 common divisor. Now, if there were any other component factor of 
 the greatest common divisor, the quotients, 2, 3, 5, would be exactly 
 divisible by it. But these quotients are prime to each other. Hence 
 3 and 7 are all the component factors of the greatest common divisor 
 sought. 
 
 3. What is the greatest common divisor of 28, 140, and 280 ? 
 
 OPERATION. ANALYSIS, We first divide by 4 ; 
 
 28 . . 1 40 . . 280 tnen tne quotients by 7. The re- 
 
 ; suiting quotients, 1, 5, and 10, are 
 
 ' " * _ prime to each other. Hence 4 and 
 
 1 5 . . 10 7 are all the component factors of 
 
 the greatest common divisor. 
 
 4 X 7 = 28, Ans. 
 
 From these examples and analyses we derive the following 
 
 RULE. I. Write the numbers in a line, with a vertical line 
 at the left, and divide by any factor common to all the numbers. 
 
 II. Divide the quotients in like manner, and continue the 
 division till a set of quotients is obtained that have no common 
 factor. 
 
 III. Multiply all the divisors together, and the product will 
 be the greatest common divisor sought. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. What is the greatest common divisor of 12, 36, 60, 72 ? 
 
 Ans. 12. 
 
 2. What is the greatest common divisor of 18, 24, 30, 36, 
 42? Ans. 6. 
 
 Rule, first step ? Second ? Third ? 
 
GREATEST COMMON DIVISOR. 83 
 
 3. What is the greatest common divisor of 72, 120, 240, 
 384? Ans. 24. 
 
 4. What is the greatest common divisor of 36, 126, 72, 
 216? Ans. 18. 
 
 ' 5. What is the greatest common divisor of 42 and 112 ? 
 
 Ans. 14. 
 
 6. What is the greatest common divisor of 32, 80, and 
 256? Ans. 16. 
 
 7. What is the greatest common divisor of 210, 280, 350, 
 630, and 840 ? Ans. 70. 
 
 8. What is the greatest common divisor of 300, 525, 225, 
 and 375 ? Ans. 75. 
 
 9. What is the greatest common divisor of 252, 630, 1134, 
 and 1386? Ans. 126. 
 
 10. What is the greatest common divisor of 96 and 544 ? 
 
 Ans. 32. 
 
 11. What is the greatest common divisor of 468 and 1184? 
 
 Ans. 4. 
 
 12. What is the greatest common divisor of 200, 625, and 
 150? Ans. 25. 
 
 CASE II. 
 
 99. When the numbers can not be readily factored. 
 
 As the analysis of the method under this case depends upon 
 three properties of numbers which have not been introduced, 
 we present them in this place. 
 
 I. An exact divisor divides any number of times its dividend. 
 
 II. A common divisor of two numbers is an exact divisor 
 of their sum. 
 
 III. A common divisor of two numbers is an exact divisor 
 of their difference. 
 
 What is Case II ? What is the first principle upon which it is 
 founded ? Second ? Third ? 
 
84 
 
 PEOPEBTIES OP NUMBERS. 
 
 84 
 70 
 
 OPEE 
 
 2 
 2 
 
 ATION. 
 
 203 
 168 
 
 35 
 
 14 
 14 
 
 2 
 2 
 
 28 
 
 7, 
 
 
 
 
 
 Ans. 
 
 1. What is the greatest common divisor of 84 and 203 ? 
 
 ANALYSIS. We draw two vertical 
 lines, and place the larger number on 
 the right, and the smaller number on 
 the left, one line lower down. We 
 then divide 203, the larger number, by 
 84, the smaller, and write 2, the quo- 
 tient, between the verticals, the prod- 
 uct, 168, opposite, under the greater 
 number, and the remainder, 35, below. 
 We next divide 84 by this remainder, 
 
 writing the quotient, 2, between the verticals, the product, 70, on the 
 left, and the new remainder, 14, below the 70. We again divide the 
 last divisor, 35, by 14, and obtain 2 for a quotient, 28 for a product, 
 and 7 for a remainder, all of which we write in the same order as in 
 the former steps. Finally, dividing the last divisor, 14, by the last 
 remainder, 7, and we have no remainder. 7, the last divisor, is the 
 greatest common divisor of the given numbers. 
 
 In order to show that the last divisor in such a process is 
 the greatest common divisor, we will first trace the work in the 
 reverse order, as indicated by the arrow line below. 
 
 OPERATION. 7 divides the 14, as proved 
 
 by the last division; it will 
 also divide two times 14, or 28, 
 (I.) Now, as 7 divides both 
 itself and 28, it will divide 35, 
 their sum, (II.) It will also 
 divide 2 times 35, or 70, (I ;) 
 and since it is a common 
 divisor of 70 and 14, it must 
 divide their sum, 84, which 
 is one of the given numbers, 
 (II.) It wiU also divide 2 
 times 84, or 168, (I;) and 
 
 since it is a common divisor of 168 and 35, it must divide their 
 Bum, 203, the larger number, (II.) Hence 7 is a common divisor 
 of the given numbers. 
 
 Again, tracing the work in the direct order, as indicated below, we 
 
 203 
 
 168 
 
 35 
 
 28 
 
 Give analysis. 
 
84 
 70 
 14 
 
 GREATEST COMMON DIYISOE. 85 
 
 know that the greatest common divisor, whatever it be, must divide 
 ^ 2 times 84, or 168, (I.) Then 
 
 203 since it will divide both 168 
 
 and 203, it must divide their 
 difference, 35, (HI.) It will 
 also divide 2 times 35, or 70, 
 (I ;) and as it will divide both 
 35 70 and 84, it must divide their 
 
 difference, 14, (III.) It will 
 ~ also divide 2 times 14 or 28, 
 
 J (I;) and as it will divide both 
 
 28 and 35, it must divide their 
 difference, 7, (III;) hence, it 
 cannot be greater than 7. 
 Thus we have shown, 
 
 1st. That 7 is a common divisor of the given numbers. 
 2d. That their greatest common divisor, whatever it be, 
 cannot be greater than 7. Hence it must be 7. 
 
 From this example and analysis, we derive the following 
 
 RULE. I. Draw two verticals, and write the two numbers, 
 one on each side, the greater number one line above the less. 
 
 II. Divide the greater number by the less, writing the quo- 
 tient between the verticals, the product under the dividend, and 
 the remainder below. 
 
 III. Divide the less number by the remainder, the last divisor 
 by the last remainder, and so on, till nothing remains. The 
 last divisor will be the greatest common divisor sought. 
 
 IV. If more than two numbers be given, first find the greatest 
 common divisor of two of them, and then of this divisor and one 
 of the remaining members, and so on to the last ; the last common 
 divisor found will be the greatest common divisor of all the 
 given numbers. 
 
 NOTES. 1. When more than two numbers are given, it is better to 
 begin with the least two. 
 
 2. If at any point in the operation a prime number occur as a re- 
 mainder, it must be a common divisor, or the given numbers have no 
 common divisor. 
 
 Rule, first step ? Second ? Third ? Fourth ? What relation have 
 numbers when their difference is a prime number ? 
 
86 
 
 PROPERTIES OF NUMBERS. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. What is the greatest common divisor of 221 and 5512 ? 
 
 OPERATION. 
 
 
 
 5512 
 
 221 
 
 2 
 
 442 
 
 
 
 1092 
 
 
 4 
 
 884 
 
 208 
 
 1 
 
 208 
 
 Am. 13 
 
 1 
 
 13 
 
 
 
 78 
 
 
 6 
 
 78 
 
 
 
 ~*0 
 
 2. Find the greatest common divisor of 154 and 210. 
 
 Ans. 14. 
 
 3. What is the greatest common divisor of 316 and 664? 
 
 Ans. 4. 
 
 4. What is the greatest common divisor of 679 and 1869 ? 
 
 Ans. 7. 
 
 5. What is the greatest common divisor of 917 and 1495 ? 
 
 Ans. 1. 
 
 6. What is the greatest common divisor of 1313 and 4108 ? 
 
 Ans. 13. 
 
 7. What is the greatest common divisor of 1649 and 5423 ? 
 
 Ans. 17. 
 
 The following examples may be solved by either of the fore- 
 going methods. 
 
 8. John has 35 pennies, and Charles 50 : how shall they 
 arrange them in parcels, so that each boy shall have the same 
 number in each parcel ? Ans. 5 in each parcel. 
 
 9. A speculator has 3 fields, the first containing 18, the 
 second 24, and the third 40 acres, which he wishes to divide 
 into the largest possible lots having the same number of acres 
 in each ; how many acres in each lot ? Ans. 2 acres. 
 
MULTIPLES. 87 
 
 10. A farmer had 231 bushels of wheat, and 273 bushels 
 of oats, which he wished to put into the least number of bins 
 containing the same number of bushels, without mixing the 
 two kinds ; what number of bushels must each bin hold ? 
 
 Ans. 21. 
 
 11. A village street is 332 rods long; A owns 124 rods 
 front, B 116 rods, and C 92 rods; they agree to divide their 
 land into equal lots of the largest size that will allow each 
 one to form an exact number of lots ; what will be the width 
 of the lots ? Ans. 4 rods. 
 
 12. The Erie Railroad has 3 switches, or side tracks, of the 
 following lengths: 3013, 2231, and 2047 feet; what is the 
 length of the longest rail that will exactly lay the track on 
 each switch ? Ans. 23 feet. 
 
 13. A forwarding merchant has 2722 bushels of wheat, 
 1822 bushels of corn, and 1226 bushels of beans, which he 
 wishes to forward, in the fewest bags of equal size that will 
 exactly hold either kind of grain ; how many bags will it 
 take? Ans. 2885. 
 
 14. A has 120 dollars, B 240 dollars, and C 384 dollars ; 
 they agree to purchase cows, at the highest price per head that 
 will allow each man to invest all his money; how many 
 cows can each man purchase? Ans. A 5, B 10, and C 16. 
 
 MULTIPLES. 
 
 100. A Multiple is a number exactly divisible by a 
 given number ; thus, 20 is a multiple of 4. 
 
 101. A Common Multiple is a number exactly divisible 
 by two or more given numbers ; thus, 20 is a common multiple 
 of 2, 4, 5, and 10. 
 
 1O2; The Least Common Multiple is the least number 
 exactly divisible by two or more given numbers ; thus, 24 is 
 the least common multiple of 3, 4, 6, and 8. 
 
 What is a multiple ? A common multiple ? The least common 
 multiple ? 
 
PROPERTIES OF NUMBERS. 
 
 103. From the definition (1OO) it is evident that the 
 product of two or more numbers, or any number of times their 
 product, must be a common multiple of the numbers. Hence, 
 A common multiple of two or more numbers may be found by 
 multiplying the given numbers together. 
 
 104. To find the least common multiple. 
 
 FIRST METHOD. 
 
 From the nature of prime numbers we derive the follow- 
 ing principles : 
 
 I. If a number exactly contain another, it will contain all 
 the prime factors of that number. 
 
 II. If a number exactly contain two or more numbers, it 
 will also contain all the prime factors of those numbers. 
 
 III. The least number that will exactly contain all the 
 prime factors of two or more numbers, is the least common 
 multiple of those numbers. 
 
 1. Find the least common multiple of 30, 42, 66, and 78. 
 
 OPERATION. ANALYSIS. The 
 
 30 = 2 X 3 X 5 number cannot be 
 
 42 = 2 X 3 X 7 * ess ^an *8, s * nce 
 
 66 = 2X3XH ** must contain 78 ; 
 
 rro _ OWQ\X1Q hence it HlUSt COn- 
 
 2X6X1 tain the factors of 
 
 2X3X13X11X7X5 = 
 
 Q 
 ^ X X ! 
 
 We here have all the prime factors of 78, and also all the factors of 
 66, except the factor 11. Annexing 11 to the series of factors, 
 
 2X 3X 13 X 11, 
 
 and we have all the prime factors of 78 and 66, and also all the 
 factors of 42 except the factor 7. Annexing 7 to the series of factors, 
 
 2 X 3 X 13 X 11 X 7, 
 and we have all the prime factors of 78, 66, and 42, and also all the 
 
 How can a common multiple of two or more numbers be found ? 
 First principle derived from prime numbers ? Second ? Third ? 
 Give analysis. 
 
LEAST COMMON MULTIPLE. 89 
 
 factors of 30 except the factor 5. Annexing 5 to the series of factors, 
 2XSX 13 X 11 X7X5, 
 
 and we have all the prime factors of each of the given numbers ; and 
 hence the product of the series of factors is a common multiple of 
 the given numbers, (II.) And as no factor of this series can be 
 omitted without omitting a factor of one of the given numbers, the 
 product of the series is the least common multiple of the given 
 numbers, (III.) 
 
 From this example and analysis we deduce the following 
 
 RULE. I. Resolve the given numbers into their prime factors. 
 
 II. Take all the prime factors of the largest number, and 
 such prime factors of the other numbers as are not found in the 
 largest number, and their product will be the least common 
 multiple. 
 
 NOTE. When a prime factor is repeated in any of the given numbers, 
 it must be used as many times, as a factor of the multiple, as the 
 greatest number of times it appears in any of the given numbers. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Find the least common multiple of 7, 35, and 98. 
 
 Ans. 490. 
 
 3. Find the least common multiple of 24, 42, and 17. 
 
 Ans. 2856. 
 
 4. What is the least common multiple of 4, 9, 6, 8 ? 
 
 Ans. 72. 
 
 5. What is the least common multiple of 8, 15, 77, 385 ? 
 
 Ans. 9240. 
 
 6. What is the least common multiple of 10, 45, 75, 90 ? 
 
 Ans. 450. 
 
 7. What is the least common multiple of 12, 15, 18, 35 ? 
 
 Ans. 1260. 
 
 Rule, first step ? Second ? What caution is given ? 
 H* 
 
90 PROPERTIES OP NUMBERS. 
 
 SECOND METHOD. 
 
 1O5. 1. What is the least common multiple of 4, 6, 9, 
 and 12 ? 
 
 OPERATION. ANALYSIS. We first write 
 
 4 . . 6 . . 9 . . 12 ti ie given numbers in a series, 
 
 with a vertical line at the left. 
 Since 2 is a factor of some of 
 
 2. .3. .9.. 6 
 
 . . 9 . . 3 the gi ven numbers, it must be 
 
 a factor of the least common 
 
 3 multiple sought. Dividing as 
 
 2X2X3X3=: 36, Ans. an y of the numbers ^as are 
 
 divisible by 2, we write the 
 
 quotients and the undivided number, 9, in a line underneath. We 
 now perceive that some of the numbers in the second line contain 
 the factor 2 ; h^nce the least common multiple must contain another 
 2, and we again divide by 2, omitting to write down any quotient 
 when it is 1. We next divide by 3 for a like reason, and still again 
 by 3. By this process we have transferred all the factors of each 
 of the numbers to the left of the vertical ; and their product, 36, 
 must be the least common multiple sought, (1OA, III.) 
 
 2. What is the least common multiple of 10, 12, 15, and 75 ? 
 OPERATION. ANALYSIS. Weread- 
 
 2,5 
 
 2,3 
 
 5 
 
 10 . . 12 . . 15 . . 75 ily see that 2 and 5 are 
 
 among the factors of the 
 given numbers, and must 
 be factors of the least 
 
 6.. 3.. 15 
 
 - common multiple : hence 
 v V H V /) SOD Avt* 
 & A o A ^ - ouu, sns. we diyide ey num ber 
 
 that is divisible by either of these factors or by their product ; thus, 
 we divide 10 by both 2 and 5 ; 12 by 2 ; 15 by 5 ; and 75 by 5. 
 We next divide the second line in like manner by 2 and 3 ; and 
 afterwards the third line by 5. By this process we collect the 
 factors of the given numbers into groups ; and the product of the 
 factors at the left of the vertical is the least common multiple sought. 
 
 3. What is the least common multiple of 6, 15, 35, 42, 
 and 70? 
 
 Give explanation. 
 
3,7 
 
 2,5 
 
 LEAST COMMON MULTIPLE. 91 
 
 OPERATION. ANALYSIS. In this oper- 
 
 15 . 42 . . 70 ation we omit the 6 and 35, 
 
 because they are exactly con- 
 
 o . . 2 . . 10 tained in some of the other 
 
 3X7X2X5=r 210, Am. g iven numbers; thus, 6 is 
 
 contained in 42, and 35 in 
 
 70 ; and whatever will contain 42 and 70 must contain 6 and 35. 
 Hence we have only to find the least common multiple of the re- 
 maining numbers, 15, 42, and 70. 
 
 From these examples we derive the following 
 
 RULE. I. Write the numbers in a line, omitting any of the 
 smaller numbers that are factors of the larger, and draw a 
 vertical line at the left. 
 
 II. Divide by any prime factor, or factors, that may be con- 
 tained in one or more of the given numbers, and write the quotients 
 and undivided numbers in a line underneath, omitting the 1's/ 
 
 III. In like manner divide the quotients and undivided num- 
 bers, and continue the process till all the factors of the given 
 numbers have been transferred to the left of the vertical. Then 
 multiply these factors together, and their product will be the least 
 common multiple required. 
 
 EXAMPLES FOR PRACTICE. 
 
 4. What is the least common multiple of 12, 15, 42, and 
 60? Ans. 420. 
 
 5. What is the least common multiple of 21, 35, and 42 ? 
 
 Ans. 210. 
 
 6. What is the least common multiple of 25, 60, 100, and 
 125? . Ans. 1500. 
 
 7. What is the least common multiple of 16, 40, 96, and 
 105? Ans. 3360. 
 
 8. What is the least common multiple of 4, 16, 20, 48, 60, 
 and 72? Ans. 720. 
 
 9. What is the least common multiple of 84, 100, 224, and 
 300? Ans. 16800. 
 
 Rule, first step ? Second ? Third ? 
 
92 PROPERTIES OP NUMBERS. 
 
 10. What is the least common multiple of 270, 189, 297, 
 243? Ans. 187110. 
 
 11. What is the least common multiple of 1, 2, 3, 4, 5, 6, 7, 
 8,9? Ans. 2520. 
 
 12. What is the smallest sum of money for which I could 
 purchase an exact number of books, at 5 dollars, or 3 dollars, 
 or 4 dollars, or 6 dollars each ? Ans. 60 dollars. 
 
 13. A farmer has 3 teams; the first can draw 12 barrels 
 of flour, the second 15 barrels, and the third 18 barrels ; 
 what is the smallest number of barrels that will make full 
 loads for any of the teams ? Ans. 180. 
 
 14. What is the smallest sum of money with which I can 
 purchase cows at $30 each, oxen at $55 each, or horses at 
 $105 each? Ans. $2310. 
 
 15. A can shear 41 sheep in a day, B 63, and C 54 ; what 
 is the number of sheep in the smallest flock that would furnish 
 exact days' labor for each of them shearing alone ? 
 
 Ans. 15498. 
 
 16. A servant being ordered to lay out equal sums in the 
 purchase of chickens, ducks, and turkeys, and to expend as 
 little money as possible, agreed to forfeit 5 cents for every fowl 
 purchased more than was necessary to obey orders. In the 
 market he found chickens at 12 cents, ducks at 30 cents, and 
 turkeys at two prices, 75 cents and 90 cents, of which he im- 
 prudently took the cheaper ; how much did he thereby for- 
 feit? Ans. 80 cents. 
 
 CLASSIFICATION OF NUMBERS. 
 
 Numbers may be classified as follows : 
 
 106. I. As Even and Odd. 
 
 107. II. As Prime and Composite. 
 
 What is the first classification of numbers ? What is an even num- 
 ber ? An odd number ? Second classification ? A prime number ? 
 A composite number ? 
 
CLASSIFICATION OP NUMBERS. 93 
 
 108. m. As Integral and Fractional. 
 
 An Integral Number, or Integer, expresses whole things. 
 Thus, 281 ; 78 boys; 1000 books. 
 
 A Fractional Number, or Fraction, expresses equal parts 
 of a thing. Thus, half a dollar; three-fourths of an hour; 
 seven-eighths of a mile. 
 
 109. IV. As Abstract and Concrete. 
 HO. V. As Simple and Compound. 
 
 A Simple Number is either an abstract number, or a 
 concrete number of but one denomination. Thus, 48, 926 ; 
 48 dollars, 926 miles. 
 
 A Compound Number is a concrete number whose value is 
 expressed in two or more different denominations. Thus, 32 
 dollars 15 cents ; 15 days 4 hours 25 minutes ; 7 miles 82 
 rods 9 feet 6 inches. 
 
 111. VI. As Like and Unlike. 
 
 Like Numbers are numbers of the same unit value. 
 
 If simple numbers, they must be all abstract, as 6, 62, 487 ; 
 or all of one and the same denomination, as 5 apples, 62 ap- 
 ples, 487 apples; and, if compound numbers, they must be 
 used to express the same kind of quantity, as time, distance, 
 &c. Thus, 4 weeks 3 days 16 hours; 1 week 6 days 9 
 hours ; 5 miles 40 rods ; 2 miles 100 rods. 
 
 Unlike Numbers are numbers of different unit values. Thus, 
 75, 140 dollars, and 28 miles ; 4 hours 30 minutes, and 5 
 bushels 1 peck. 
 
 What is the third classification ? What is an integral number ? A 
 fractional number ? What is the fourth classification ? An abstract 
 number ? A concrete number ? What is the fifth classification ? A 
 simple number ? A compound number ? Sixth classification ? What 
 are like numbers ? Unlike numbers ? 
 
94 FRACTIONS. 
 
 FRACTIONS. 
 
 DEFINITIONS, NOTATION, AND NUMERATION. 
 
 . If a unit be divided into 2 equal parts, one of the 
 parts is called one half. 
 
 If a unit be divided into 3 equal parts, one of the parts is 
 called one third, two of the parts two thirds. 
 
 If a unit be divided into 4 equal parts, one of the parts is 
 called one fourth, two of the parts two fourths, three of the 
 parts three fourths. 
 
 If a unit be divided into 5 equal parts, one of the parts is 
 called one fifth, two of the parts two fifths, three of the parts 
 three fifths, &c. 
 
 The parts are expressed by figures ; thus, 
 
 One half is written 
 One third " 
 
 Two thirds " 
 One fourth " 
 
 Two fourths " f 
 
 Three fourths " 
 
 One fifth is written 
 
 Two fifths 
 
 One seventh " 
 
 Three eighths " 
 
 Five ninths |- 
 
 Eight tenths " T 8 <y 
 
 Hence we see that the parts into which a unit is divided take 
 their name, and their value, from the number of equal parts 
 into which the unit is divided. Thus, if we divide an orange 
 into 2 equal parts, the parts are called halves ; if into 3 equal 
 parts, thirds ; if into 4 equal parts, fourths, &c. ; and each 
 third is less in value than each half, and each fourth less than 
 each third; and the greater the number of parts, the less 
 their value. 
 
 When a unit is divided into any number of equal parts, one 
 or more of these parts is a fractional part of the whole number, 
 and is called & fraction. Hence 
 
 118. A Fraction is one or more of the equal parts of a 
 unit. 
 
 Define a fraction. 
 
DEFINITIONS, NOTATION, AND NUMERATION. 95 
 
 . To write a fraction, two integers are required, one 
 to express the number of parts into which the whole number 
 is divided, and the other to express the number of these parts 
 taken. Thus, if one dollar be divided into 4 equal parts, 
 the parts are called fourths, and three of these parts are 
 called three fourths of a dollar. This three fourths may be 
 written 
 
 3 the number of parts taken. 
 
 4 the number of parts into which the dollar is divided. 
 
 115. The Denominator is the number below the line. 
 It denominates or names the parts ; and 
 
 It shows how many parts are equal to a unit. 
 
 116. The Numerator is the number above the line. 
 It numerates or numbers the parts ; and 
 
 It shows how many parts are taken or expressed by 
 the fraction. 
 
 117. The Terms of a fraction are the numerator and de- 
 nominator, taken together. 
 
 118. Fractions indicate division, the numerator answering 
 to the dividend, and the denominator to the divisor. Hence, 
 
 11O. The Value of a fraction is the quotient of the nu- 
 merator divided by the denominator. 
 
 13O. To analyze a fraction is to designate and describe 
 its numerator and denominator. Thus, f is analyzed as fol- 
 lows : 
 
 4 is the denominator, and shows that the unit is divided 
 into 4 equal parts ; it is the divisor. 
 
 3 is the numerator, and shows that 3 parts are taken ; it is 
 the dividend, or integer divided. 
 
 3 and 4 are the terms, considered as dividend and divisor. 
 
 The value of the fraction is the quotient of 3 -^- 4, or J. 
 
 How many numbers are required to write a fraction ? Why ? De- 
 fine the denominator. The numerator. What are the terms of a frac- 
 tion ? The value ? What is the analysis of a fraction ? 
 
96 FRACTIONS. 
 
 EXAMPLES FOR PRACTICE. 
 
 Express the following fractions by figures : - 
 
 1. Seven eighths. 
 
 2. Three twenty-fifths. 
 
 3. Nine one hundredths. 
 
 4. Sixteen thirtieths. 
 
 5. Thirty-one one hundred eighteenths. 
 
 6. Seventy-five ninety-sixths. 
 
 7. Two hundred fifty-four four hundred forty-thirds. 
 
 8. Eight nine hundred twenty-firsts. 
 
 9. One thousand two hundred thirty-two seventy-five thou- 
 sand six hundredths. 
 
 10. Nine hundred six two hundred forty-three thousand 
 eighty-seconds. 
 
 Read and analyze the following fractions : 
 
 11. A; iV; A; *i; If; rfe; A- 5 ^ ill- 
 
 12. T 9 <&; m 
 
 TUTTIJ5 
 
 Fractions are distinguished as Proper and Improper. 
 
 A Proper Fraction is one whose numerator is less than its 
 denominator ; its value is less than the unit, 1. Thus, fy, T 5 g, 
 A f are P r P er fractions. 
 
 An Improper Fraction is one whose numerator equals or 
 exceeds its denominator; its value is never less than the 
 unit, 1. Thus, -f, , J 4 5 -, -^ 5 -, |-g, J F 8 ^- are improper fractions. 
 
 122. A Mixed Number is a number expressed by an in- 
 teger and a fraction; thus, 4^-, 17|, 9 T 3 ^ are mixed numbers. 
 
 12*S. Since fractions indicate division, all changes in the 
 terms of a fraction will affect the value of that fraction according 
 to the laws of division ; and we have only to modify the lan- 
 guage of the General Principles of Division (8T) by substi- 
 tuting the words numerator, denominator, find fraction, or value 
 
 "What is a proper fraction ? An improper fraction ? A mixed 
 number ? What do fractions indicate ? 
 
REDUCTION. 97 
 
 of the fraction, for the words dividend, divisor, and quotient, 
 respectively, and we shall have the following 
 
 GENERAL PRINCIPLES OF FRACTIONS. 
 
 124* PRIN. I. Multiplying the numerator multiplies the 
 fraction, and dividing the numerator divides the fraction. 
 
 PRIN. II. Multiplying the denominator divides the fraction, 
 and dividing the denominator multiplies the fraction. 
 
 PRIN. III. Multiplying or dividing both terms of the frac- 
 tion by the same number does not alter the value of the fraction. 
 
 These three principles may be embraced in one 
 
 GENERAL LAW. 
 
 12&. A change in the NUMERATOR produces a LIKE change 
 in the value of the fraction ; but a change in the DENOMINA- 
 TOR produces an OPPOSITE change in the value of the fraction. 
 
 KEDUCTION. 
 
 CASE I. 
 
 126. To reduce fractions to their lowest terms. 
 
 A fraction is in its lowest terms when its numerator and de- 
 nominator are prime to each other ; that is, when both terms 
 have no common divisor. 
 
 1. Reduce the fraction to its lowest terms. 
 
 FIRST OPERATION. ANALYSIS. Dividing both 
 
 || if =. ^, Ans. terms of a fraction by the same 
 
 number does not alter the value 
 
 of the fraction or quotient, (124, III;) hence, we divide both 
 terms of |f, by 2, both terms of the result, ff, by 2, and both terms 
 of this result by 3. As the terms of f are prime to each other, the 
 lowest terms of || are |. We have, in effect, canceled all the fac- 
 tors common to the numerator and denominator. 
 
 First general principle ? Second ? Third ? General law ? What 
 is meant by reduction of fractions ? Case I is what r What is 
 meant by lowest terms? Give analysis. 
 
 I 
 
98 FRACTIONS. 
 
 SECOND OPERATION. In this operation we have divided 
 
 12} 48 _ 4- Ans. both terms of the fraction by their 
 
 greatest common divisor, (97 1 ,) and 
 thus performed the reduction at a single division. Hence the 
 
 RULE. Cancel or reject all factors common to loth numera- 
 tor and denominator. Or, 
 
 Divide both terms by their greatest common divisor. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce ||J to its lowest terms. Ans. . 
 
 3. Reduce f f to its lowest terms. Ans. f . 
 
 4. Reduce f f to its lo wes t terms. Ans. f . 
 
 5. Reduce | to its lowest terms. 
 
 6. Reduce sVfe*? to its lowest terms. 
 
 7. Reduce f^ T to its lowest terms. 
 
 8. Reduce -jtffg to its lowest terms. 
 
 9. Reduce $f $ to its lowest terms. Ans. J. 
 
 10. Reduce |f J$ to its lowest terms. Ans. J. 
 
 11. Reduce /$; to its lowest terms. Ans. . 
 
 12. Express in its simplest form the quotient of 441 divided 
 by 462. Ans. f. 
 
 13. Express in its simplest form the quotient of 189 di- 
 vided by 273. Ans. T V 
 
 14. Express in its simplest form the quotient of 1344 di- 
 vided by 1536. Ans. . 
 
 CASE II. 
 
 137. To reduce an improper fraction to a whole 
 or mixed number. 
 
 1. Reduce 3- to a whole or mixed number. 
 
 OPERATION. ANALYSIS. Since 
 
 324-M5 = 21 T \ = 21f, Ans. 15 fifteenths equal 
 
 1,324 fifteenths are 
 equal to as many times 1 as 15 is contained times in 324, which is 
 times. Or, since the numerator is a dividend and the denom- 
 
 \ 
 
 Rule. Case II is what ? Give explanation. 
 
REDUCTION. 99 
 
 inator a divisor, (118,) we reduce the fraction to an equivalent 
 whole or mixed number, by dividing the numerator, 324, by the 
 denominator, 15. Hence the 
 
 RULE. Divide the numerator by the denominator. 
 
 NOTES. 1. When the denominator is an exact divisor of the numer- 
 ator, the result will be a whole number. 
 
 2. In all answers containing fractions, reduce the fractions to their 
 lowest terms. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. In J^ 3 - of a week, how many weeks? Ans. If-. 
 
 3. In ^^- of a bushel, how many bushels ? Ans. 23f . 
 
 4. In A 6 1 of a dollar, how many dollars ? 
 
 5. In - 8 T 7 ^ of a pound, how many pounds ? Ans. 54. 
 
 6. Reduce -ftp to a mixed number. 
 
 7. Reduce Zffi to a whole number. 
 
 8. Change J^f 2 - to a mixed number. Ans. 18. 
 
 9. Change ^ f - 1 to a mixed number. 
 
 10. Change ^%/ fl to a mixed number. Ans. 1053f f. 
 
 11. Change ai^aa to a whole number. Jtws. 7032. 
 
 CASE III. 
 
 128. To reduce a whole number to a fraction hav- 
 ing a given denominator. 
 
 1. Reduce 46 yards to fourths. 
 
 OPERATION. ANALYSIS. Since in 1 yard there are 4 fourths, 
 
 4g in 46 yards there are 46 times 4 fourths, which are 
 
 4 184 fourths r= ^4. I n practice we multiply 46, 
 
 the number of yards, by 4, the given denominator, 
 A f ? ^ ns - and taking the product, 184, for the numerator of a 
 fraction, and the given denominator, 4, for the de- 
 nominator, we have ^^-. Hence we have the 
 
 RULE. Multiply the whole number by the given denominator ; 
 take the product for a numerator, under which write the given 
 denominator. 
 
 Rule. Case HI is what ? Give explanation. Rule. 
 
100 FRACTIONS. 
 
 NOTE. A whole number is reduced to a fractional form by writing 
 1 under it for a denominator ; thus, 9 - 3.. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce 25 bushels to eighths-of a bushel. Ans. ^ &. 
 
 3. Reduce G3 gallons to fourths of a gallon. Ans. 5 ^ 2 . 
 
 4. Reduce 140 pounds to sixteenths of a pound. 
 
 5. In 56 dollars, how many tenths of a dollar ? Ans. - 5 T 6 a -. 
 
 6. Reduce 94 to a fraction whose denominator is 9. 
 
 7. Reduce 180 to seventy-fifths. 
 
 8. Change 42 to the form of a fraction. Ans, ^Ji. 
 
 9. Change 247 to the form of a fraction. 
 
 10. Change 347 to a fraction whose denominator shall 
 be 14. Ans. fja. 
 
 CASE IV. 
 
 129. To reduce a mixed number to an improper 
 fraction. 
 
 1. In 5| dollars, how many eighths of a dollar ? 
 OPERATION. 
 
 5 a ANALYSIS. Since in 1 dollar there are 8 eighths, 
 
 Q in 5 dollars there are 5 times 8 eighths, or 40 
 
 eighths, and 40 eighths -f- 3 eighths := 43 eighths, 
 or 4g 3 -. From this operation we derive the following 
 
 RULE. Multiply the whole number by the denominator of 
 the fraction ; to the product add the numerator, and under the 
 sum write the denominator. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. In 4 dollars, how many half dollars ? .Ans. f . 
 
 3. In 7 If weeks, how many sevenths of a week ? 
 
 4. In 341 J acres, how many fourths? Ans. 
 
 5. Change 12 T 7 2 years to twelfths. 
 
 G. Change SGy 9 ^ to an improper fraction. Ans. 
 
 7. Reduce 21 fa to an improper fraction. Ans. 
 
 8. Reduce 225i| to an improper fraction. Ans. - 
 
 Case IV is what ? Give explanation. Rule. 
 
REDUCTION. 101 
 
 9. In 96^% how many one hundred twentieths ? 
 
 10. In 1297 B \, how many eighty-fourths ? Ans. 
 
 11. What improper fraction will express 400f $ ? 
 
 CASE v. 
 13O. To reduce a fraction to a given denominator. 
 
 As fractions may be reduced to lower terms by division, 
 they may also be reduced to higher terms by multiplication ; 
 and all higher terms must be multiples of the lowest terms. 
 (103.) 
 
 1. Reduce to a fraction whose denominator is 20. 
 
 OPERATION. ANALYSIS. "We first divide 20, the 
 
 20 _i_ 4 5 required denominator, by 4, the denomi- 
 
 nator of the given fraction, to ascertain 
 
 _^ X 5 __ ^ ^ n ^ if it be a multiple of this term, 4. The 
 
 4X5 division shows that it is a multiple, and 
 
 that 5 is the factor which must be em- 
 
 ployed to produce this multiple of 4. We therefore multiply both 
 terms of f by 5, (124,) and obtain -^f, the desired result. Hence the 
 
 RULE. Divide the required denominator by the denominator 
 of the given fraction, and multiply both terms of the fraction by 
 the quotient. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce f to a fraction whose denominator is 15. 
 
 Ans. -ft. 
 
 3. Reduce f to a fraction whose denominator is 35. 
 
 4. Reduce Jf to a fraction whose denominator is 51. 
 
 Ans. ff. 
 
 5. Reduce f to a fraction whose denominator is 150. 
 
 6. Reduce | to a fraction whose denominator is 3488. 
 
 Ans. 
 
 7. Reduce to a fraction whose denominator is 1000. 
 
 Case V is what ? How are fractions reduced to higher terms ? 
 What are all higher terms ? Give analysis. Rule. 
 
 T* 
 
102 FRACTIONS. 
 
 CASE VI. 
 
 131. To reduce two or more fractions to a com- 
 mon denominator. 
 
 A Common Denominator is a denominator common to two 
 or more fractions. 
 
 1. Reduce f and f to a common denominator. 
 
 OPERATION. ANALYSIS. We multiply the terms of the 
 
 3><5 first fraction by the denominator of the second, 
 
 = $ and the terms of the second fraction by the 
 
 denominator of the first, (124.) This must re- 
 
 2 <x ^ duce each fraction to the same denominator, 
 
 8 for each new denominator will be the product 
 
 5X4 of the given denominators. Hence the 
 
 RULE. Multiply the terms of each fraction by the denomina- 
 tors of all the other fractions. 
 
 NOTE. Mixed numbers must first be reduced to improper fractions. 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce , , and J to a common denominator. 
 
 Ans. ,Jf,. 
 
 3. Reduce f and | to a common denominator. 
 
 > Ans. 5, f|. 
 
 4. Reduce f, -f^ and to a common denominator. 
 
 Ans. HI, SJg, m- 
 
 5. Reduce f , , f , and ^ to a common denominator. 
 
 Ans. Mi, f4J, Ml, MI- 
 G. Reduce ^5, , and f to a common denominator. 
 
 Ans. mifoh- 
 
 7. Reduce f , 2, J, and to a common denominator. 
 
 A*"- -HI, ? iif' Aft- 
 
 8. Reduce 1 J, ^ , and 4 to a common denominator. 
 
 Ans. 
 
 C.ise TI ic wlmt ? "What is a common denominator ? Give analysis. 
 
EEDUCTION. 103 
 
 CASE VII. 
 
 To reduce fractions to the least common de- 
 nominator. 
 
 The Least Common Denominator of two or more fractions 
 is the least denominator to which they can all be reduced, and 
 it must be the least common multiple of the lowest denom- 
 inators. 
 
 1. Eeduce , f, and T 5 2 to the least common denominator. 
 OPERATION. ANALYSIS. We first find 
 
 g -^2 the least common multiple 
 
 " . ' g 
 
 " ' 
 
 2, 3 .. ^ 
 
 Q' a 4.9 ^ t ^ ie S^ ven denominators, 
 
 III " ' which is 24. This must be 
 
 2X3X2X2 24 the least common denom- 
 
 i 4 ^ inator to which the frac- 
 
 3 _9 I Ans. tions can be reduced. (III.) 
 
 f_ __ ^J j We then multiply the terms 
 
 of each fraction by such a 
 
 number as will reduce the fraction to the denominator, 24. Re- 
 ducing each fraction to this denominator, by Case V, we have the 
 answer. 
 
 $ince the common denominator is already determined, it is 
 only necessary to multiply the numerators by the multipliers. 
 Hence the following 
 
 KULE. I. Find the least common multiple of the given de- 
 nominators, for the least common denominator. 
 
 II. Divide this common denominator by each of the given 
 denominators, and multiply each numerator by the correspond- 
 ing quotient. The products will be the new numerators. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce 2 \, ^, f g, and T \ to their least common de- 
 nominator. Ans. T V 2 , T^ T*> rfir- 
 
 3. Reduce , ^-, T 3 F , ? 2 T to their least common denominator. 
 
 iSI, M 
 
 What is Case VLT ? What must be the least common denominator ? 
 Give, analysis. Rule, first step. Second. 
 
104 FRACTIONS. 
 
 4. Reduce f, 2^, f, and 6 to their least common denomi- 
 nator. Ans. 3&Wft>Mf,W- 
 
 5. Reduce 5j, 2, and If to their least common denomina- 
 tor. Ans. - 4 /-> J/-, -V-- 
 
 6. Reduce T 9 T , f , f , and to their least common denomi- 
 nator. Ans. I?!, ?4, Ml, 4ft- 
 
 7. Reduce j- , , f , 2|, and ^- to their least common de- 
 nominator. Ans. -Jf f, TS> TS ret, i%- 
 
 8. Change f, T \, 3f, 9, and J- to equivalent fractions hav- 
 ing the least common denominator. 
 
 9. Change , If, |, J|-,and 6 to equivalent fractions hav- 
 ing the least common denominator. 
 
 10. Change 2 r 7 a , |J, 4, If, |, and | to equivalent frac- 
 tions having the least common denominator. 
 
 11. Reduce ^, , , and j 7 ^ to a common denominator. 
 
 12. Reduce , ^, 2, and ^ to a common denominator. 
 
 13. Reduce -f, T 7 ^, f , and 3-1 to equivalent fractions hav- 
 ing a common denominator. Ans. |, , -, |. 
 
 14. Change / T , f , and |^ to equivalent fractions having a 
 common denominator. Ans. &%%, ^o^ ? ^^ < 
 
 15. Change ^ T , 7J-, , and 5 to equivalent fractions haV- 
 ing a common denominator. Ans. f |, ^ 5 -, f, - 3 F 3 & -. 
 
 16. Change ^ 6^, T 9 ^, 7, f, and 1^- to equivalent fractions 
 having a common denominator. 
 
 ADDITION. 
 
 133. 1. What is the sum of , f , f , and ? 
 
 OPERATION. ANALYSIS. Since the 
 
 -yi 2, ^rzs. ^ ven ^actions have a 
 
 common denominator, 8, 
 their sum may be found by adding their numerators, 1, 3, 5, and 
 7, and placing the sum, 16, over the common denominator. We 
 thus obtain ^ = 2, the required sum. 
 
 2. Add ^, A, ^ A, and T V Jws. 2J. 
 
 3. Add 
 
 Give first explanation. 
 
ADDITION. 
 
 4. What is the sum of T 
 
 A 
 
 5. What is the sum o 
 
 6. What is the sum o 
 
 , A H> 
 Jfo 
 JJ, 
 
 105 
 
 IVEBSIT' 
 
 our K 
 ltvn^ 
 
 and f Jf ? 
 
 134. 1. What is the sum off and f? 
 
 OPERATION. ANALYSIS. In 
 
 whole numbers 
 we can add like 
 
 numbers only, or those having the same unit value ; so in fractions 
 we can add the numerators when they have a common denominator, 
 but not otherwise. As | and -f have not a common denominator, 
 we first reduce them to a common denominator, and then add the nu- 
 merators, 27 -|- 10 == 37, the same as whole numbers, and place the 
 sum over the common denominator. Hence the following 
 
 RULE. I. When necessary, reduce the fractions to a com- 
 mon or to their least common denominator. 
 
 II. Add the numerators, and place the sum over the common 
 denominator. 
 
 NOTE. If the amount be an improper fraction, reduce it to a whole 
 or a mixed number. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Add f to f. Ans. f |. 
 
 3. Addf to|i. Ans. 1 
 
 4. Add f , |, f and T V. Ans. 
 
 5. Add if, Sfc and ^-. Ans. 
 
 6. Add ^, A, /y, and A. Ans. f. 
 
 7. Add fj, if *, ||, 3, and f . ^w. 3 
 
 8. Add f , J, f, t, $, f J, f , and &. -4n. 
 
 9. Add 7J-, of, and 10}. 
 
 OPERATION. 
 f + f+ f 1| 
 7 _L 5 _l_ 10 ^^ 22 
 
 Ans. 23 } 
 
 ANALYSIS. The sum of the frac- 
 tions^,f,andf is 1^; the sum of 
 ^ e integers, 7, 5, and 10, is 22 ; 
 and the sum of both fractions 
 an d integers is 23}J. Hence, 
 
 Give second explanation ? Rule, first step. Second. 
 
106 
 
 FRACTIONS. 
 
 To add mixed numbers, add the fractions and integers sep- 
 arately, and then add their sums. 
 
 NOTE. If the mixed numbers are small, they may be reduced to im- 
 proper fractions, and then added after the usual method. 
 
 10. What is the sum of 14|, 3 r % If, and ? Ans. 21JJ. 
 
 11. What is the sum of }, l/^-, lOf , and 5 ? Ans. 18/ . 
 
 12. What is the sum of 17f , 18&, and 26^ ? 
 
 13. What is the sum of &, ^ 1 , 3, and % ? 
 
 14. What is the sum of 125f-, 327 &, and 25? Ans. 478/ T . 
 
 15. What is the sum of JJ&, |J, 1^, , and Jg$ ? 
 
 16. What is the sum of 3 T 9 ^, 2f, 40f, and 10 T V ? 
 
 17. Bought 3 pieces of cloth containing 125}, 96}, and 
 48| yards ; how many yards in the 3 pieces ? 
 
 18. If it take 5 yards of cloth for a coat, 3 yards for a 
 pair of pantaloons, and } of a yard for a vest, how many yards 
 will it take for all ? Ans. 9-^. 
 
 19. A farmer divides his farm into 5 fields; the first con- 
 tains 26^3- acres, the second 40j-f acres, the third 51 f- acres, 
 the fourth 59f acres, and the fifth 62 acres ; how many acres 
 in the farm? Ans. 241 if. 
 
 20. A speculator bought 175f bushels of wheat for 205 
 dollars, 325f- bushels of barley for 296f dollars, 270|f bush- 
 els of corn for 200^ dollars, and 437^ bushels of oats for 
 156J | dollars ; how many bushels of grain did he buy, and how 
 much did he pay for the whole ? ^ m f 1209^ 9 F bushels. 
 
 * 1 850f dollars. 
 
 SUBTRACTION. 
 
 135. 1. From' A take A- 
 
 OPERATION. ANALYSIS. Since the given 
 
 J^ -fa -fa ^ Ans. fractions have a common denom- 
 
 inator, 10, we find the difference 
 by subtracting 3, the less numerator, from 7, the greater, and write 
 
 How aro mixorl mimlvr- \\:\t\- -.1 ? (Jive note. 
 
SUBTRACTION. 107 
 
 the % remainder, 4, over the common denominator, 10. "We thus 
 
 obtain T ^- = f , the required difference. 
 
 2. From f take f . Ans. . 
 
 3. From f J take T . Ans. . 
 
 4. From f take / T . 
 
 5. From f take f . 
 
 6. From -/^ take T y F . 
 
 7. From iff take JJf. Ans. ft. 
 
 136. 1. From | take f . 
 
 OPERATION. ANALYSIS. 
 
 f - 1 = f - f ft = 3J? 3V 3J > = ft = A, ^- As ^ whole 
 
 numbers, we 
 
 can subtract like numbers only, or those having the same unit value, 
 so, we can subtract fractions only when they have a common de- 
 nominator. As f and f have not a common denominator, we first 
 reduce them to a common denominator, and then subtract the 
 less numerator, 30, from the greater, 32, and write the difference, 2, 
 over the common denominator, 36. We thus obtain ^ = T ^, the 
 required difference. Hence the following 
 
 RULE. I. When necessary, reduce the fractions to a 
 common denominator. 
 
 II. Subtract the numerator of the subtrahend from the 
 numerator of the minuend, and place the difference over the 
 common denominator. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. 
 
 From take f . 
 
 Ans. T 5 ? . 
 
 3. 
 
 From f take f . 
 
 -4ws. 9 <y. 
 
 4. 
 
 Subtract T 4 T from |. 
 
 -4?15. y 1 ^. 
 
 5. 
 
 Subtract -^ from T 8 ^. 
 
 -4^5. 4^' 
 
 6. 
 
 Subtract ^ from |f . 
 
 Ans. ^ 5 . 
 
 7. 
 
 Subtract ^5% from f . 
 
 ^^s- yWff- 
 
 8. 
 
 What is the difference between 9 
 
 ^ and 2f ? 
 
 Give explanations. Riile, first step. Second. 
 
108 FRACTIONS. 
 
 OPERATION. ANALYSIS. We first reduce the frac- 
 
 9 = 9^y tional parts, J and f , to a common denom- 
 
 g3 2-C inator, 12. Since we cannot take ^ from 
 
 ^, we add 1 r= ^| to T %, which makes ^-f , 
 
 6/2- ^715. and T % from if leaves T ^. We now add 1 
 
 to the 2 in the subtrahend, (5O,) and say 
 
 3 from 9 leaves 6. We thus obtain 6 T 7 ^, the difference required. 
 
 Hence, to subtract mixed numbers, we may reduce the 
 fractional parts to a common denominator, and then subtract 
 the fractional and integral parts separately. Or, 
 
 We may reduce the mixed numbers to improper fractions, 
 and subtract the less from the greater by the usual method. 
 
 9. From 8 take 3}. Ans. 4|f. 
 
 10. From 2o take 9 T V Ans. 16 T \. 
 
 11. From 4f take f. 
 
 12. Subtract 1| from 6. 
 
 13. Subtract 120^ from 450. Ans. 330 1|. 
 
 14. Subtract T 4 ^\ from 3 T V Ans. 3 3 4 T \. 
 
 15. Find the difference between 49 and 75. 
 
 16. Find the difference between 227| and 196f. 
 
 17. From a cask of wine containing 31 gallons, 17f gal- 
 lons were drawn ; how many gallons remained ? Ans. 13$. 
 
 18. A farmer, having 450^ acres of land, sold 304| acres ; 
 how many acres had he left ? Ans. 145i. 
 
 19. If flour be bought for 6 dollars per barrel, and sold 
 for 7 dollars, what will be the gain per barrel ? 
 
 20. From the sum of f and 3 take the difference of 4 
 and 5. Ans. 3|f. 
 
 21. A man, having 25} dollars, paid 6 dollars for coal, 2 
 dollars for dry goods, and j of a dollar for a pound of tea ; 
 how much had he left ? Ans. 
 
 22. What number added to 2f will make 7 ? Ans. 
 
 23. What fraction added to | will make ? Ans. 
 
 In how many ways may mixed numbers be subtracted ? What arc 
 they? 
 
MULTIPLICATION. 109 
 
 24. A gentleman, having 2000 dollars to divide among his 
 three sons, gave to the first 91 2 dollars, to the second 545 
 dollars, and to the third the remainder ; how much did 
 the third receive ? Am. $542 T V 
 
 25. Bought a quantity of coal for 136-j- 9 ^ dollars, and of 
 lumber for 350f dollars. I sold the coal for 184| dollars, and 
 the lumber for 41 6 J dollars. How much was my whole gain? 
 
 Ans. 
 
 MULTIPLICATION. 
 
 CASE I. 
 
 137. To multiply a fraction by an integer. 
 
 1. If 1 yard of cloth cost f of a dollar, how much will 5 
 yards cost ? 
 
 OPERATION. ANALYSIS. Since 1 yard cost 
 
 3 NX 5 - 15 33 ^ ns 3 fourths of a dollar, 5 yards 
 
 will cost 5 times 3 fourths of a 
 
 dollar, or 15 fourths, equal to 3f dollars. A fraction is multiplied 
 by multiplying its numerator, (124.) 
 
 2. If 1 gallon of molasses cost /^ of a dollar, how much 
 will 5 gallons cost ? 
 
 OPERATION. ANALYSIS. Since 5, the 
 
 T/Q- X 5 = - = 1^ Ans. multiplier, is a factor of 20, the 
 
 denominator, of the multipli- 
 cand, we perform the multiplication by dividing the denominator, 
 20, by the multiplier, 5, and we have -, equal to If dollars. A 
 fraction is multiplied by dividing its denominator, (124.) Hence, 
 
 Multiplying a fraction consists in multiplying its numerator, 
 or dividing its denominator. 
 
 NOTE. Always divide the denominator when it is exactly divisible 
 by the multiplier. 
 
 EXAMPLES FOR PRACTICE. 
 
 3. Multiply ^ by 5. Ans. \f = 2f 
 
 4. Multiply T 3 T by 7. Ans. 1|. 
 
 Case I is what ? Give explanations. Deduction. 
 J 
 
110 FRACTIONS. 
 
 5. Multiply T 9 by 12. Ans. 7f . 
 
 6. Multiply / T by 63. Ans. 15. 
 
 7. Multiply 5 by 9. 
 
 OPERATION. ANALYSIS. In multiply- 
 
 5,1 ing a mixed number, we first 
 
 9" Q r multiply the fractional part, 
 
 5 1 __ 11 and then the integer, and 
 
 4 JLJL s/ 9 9 9 _ ^Q j add the two products ; or we 
 
 45 reduce the mixed number to 
 
 4 q an improper fraction, and 
 
 * then multiply it. 
 
 8. Multiply 7f by 12. Ans. 9 If 
 
 9. Multiply T y T by 8. ^n. 5& 3 T . 
 
 10. Multiply T fe by 51. Ans. 2. 
 
 11. Multiply 15| by 16. Ans. 250. 
 
 12. Multiply Jjy. by 22. ^ws. 16|. 
 
 13. If a man earn 8 T 9 ^ dollars a week, how many dollars 
 will he earn in 12 weeks ? 
 
 14. What will 9 yards of silk cost at ^ of a dollar per 
 yard ? 
 
 15. What will 27 bushels of barley cost at of a dollar 
 per bushel ? Ans. 23| dollars. 
 
 CASE II. 
 
 138. To multiply an integer by a fraction. 
 
 1. At 75 dollars an acre, how much will f of an acre of 
 land cost ? 
 
 FIRST OPERATION. ANALYSIS. 3 fifths of an 
 
 5 ) 75 price of an acre. acre will cost three times as 
 
 much as 1 fifth of an acre. 
 
 15 cost of of an acre. Dividing 75 dollars by 5, we 
 
 have 15 dollars, the cost of 
 
 Ans 45 " " i of an acre > which we mul - 
 
 tiply by 3, and obtain 45 
 
 dollars, the cost of f of an acre. 
 
 Explain the process of multiplying mixed numbers. What is Case 
 II ? Give first explanation. 
 
MULTIPLICATION. 
 
 Ill 
 
 Or, multiplying the price 
 of 1 acre by 3, we have the 
 cost of 3 acres ; and as % 
 of 3 acres is the same as 
 -| of 1 acre, we divide the 
 cost of 3 acres by 5, and 
 we have the cost of ^ of an 
 acre, the same as in the first 
 operation. Hence, 
 
 Multiplying by a fraction consists in multiplying by the nu- 
 merator and dividing by the denominator of the multiplier. 
 
 SECOND OPERATION. 
 75 price of 1 acre. 
 
 3 
 
 5 ) 225 cost of 3 acres. 
 AnS. 45 " " f of an acre. 
 
 15 
 
 NOTE. By using the vertical line and 
 cancellation, we shall shorten, and com- 
 bine both operations in one. 
 
 45, Ans. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Multiply 3 by f . 
 
 3. Multiply 100 by T 9 . 
 
 4. Multiply 105 by & 
 5 Multiply 19 by f 
 6. Multiply 24 by 6|. 
 
 OPERATION. 
 
 24 
 
 = o 
 
 144 
 159, Ans. 
 
 Or,* 
 
 7. Multiply 42 by 9. 
 
 8. Multiply 80 by 14-^. 
 
 9. Multiply 156 by f. 
 
 10. At 8 dollars a bushel, what will 
 seed cost ? 
 
 Ans. H. 
 
 Ans. 64f 
 
 Ans. 85. 
 
 Ans. 5|f. 
 
 ANALYSIS. We 
 multiply by the in- 
 teger and fraction 
 separately,and add 
 the products ; or, 
 reduce the mixed 
 number to an im- 
 proper fraction, 
 and then multiply by it. 
 
 Ans. 409. 
 Ans. 1165. 
 Ans. 108. 
 ^ of a bushel of clover 
 
 Give second explanation. Note. Deduction. 
 
112 FRACTIONS. 
 
 11. If a man travel 36 miles a day, how many miles will 
 he travel in lOf days ? Am. 384 miles. 
 
 12. If a village lot be worth 450 dollars, what is T 7 ? of it 
 worth? Ans. 262^- dollars. 
 
 13. At 16 dollars a ton, what is the cost of 2J tons of hay ? 
 
 CASE III. 
 
 139. To multiply a fraction by a fraction. 
 1. At f of a dollar per bushel, how much will f of a bushel 
 of corn cost ? 
 
 OPERATION. ANALYSIS. 
 
 1st step, | _i_ 4 A, cost of of a bushel. Since Ibush- 
 
 2d step, A X 3 = A, " " f " " " 
 
 Or, $ 
 
 o 
 
 $ cost f times f of a dollar, or 3 times 
 
 of f of a dollar. Dividing f of a 
 
 3 _ dollar by 4, we have T 2 ^, the cost of 
 
 -I _ , .| w - \ of a bushel. A fraction is di- 
 
 vided by multiplying its denomina- 
 
 tor, (124.) Multiplying the cost of \ of a bushel by 3, we have -% 
 of a dollar, the cost of f of a bushel. It will readily be seen that we 
 have multiplied together the two numerators, 2 and 3, for a new 
 numerator, and the two denominators, 3 and 4, for a new denom- 
 inator, as shown in the whole work of the operation. Hence, for 
 multiplication of fractions, we have this general 
 
 RULE. I. Reduce all integers and mixed numbers to 
 improper fractions. 
 
 II. Multiply together the numerators for a new numerator, 
 and the denominators for a new denominator. 
 
 NOTE. Cancel all factors common to numerators and denominators. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Multiply f by J. Ans. ^. 
 
 3 Multiply I by \ . Ans. T V 
 
 4. Multiply by f f . Ans. f^. 
 
 5. Multiply 4 by f . Ans. 3f . 
 
 "What is Case III ? Give explanation. Rule, first step ? Second ? 
 What shall be done with common factors ? 
 
MULTIPLICATION. 113 
 
 6. What is the product of y 9 ^, f , f , and ? Ans. %. 
 
 7. What is the product of If, f , 2, and 5 ? ^ ws . llff 
 
 8. What is the product of of -&, of f of , and f of 
 
 . 
 
 O 
 
 OPERATION. Or, 
 
 6 # * 7~~30' 
 
 NOTE. Fractions with the word o/ between them 
 are sometimes called compound fractions. The word 
 of is simply an equivalent for the sign of multiplica- 
 tion, and signifies that the numbers between which \ _~- 
 it is placed are to be multiplied together. ^U | I -g- . 
 
 9. Multiply T \ of 2 by of 7. Ans. Hf. 
 
 10. Multiply f of 16 by ^ of 26f. 
 
 11. What is the product of 3, ^ of , and f oi 
 
 12. What is the value of 2 times f of f of 1 ? Jws. 2. 
 
 13. What is the value of of of If times f of 8 ? 
 
 14. What is the product of 12^ multiplied by 5 times 6f ? 
 
 1-5. At f of a dollar per yard, what will f of a yard of 
 cloth cost ? Ans. ^ of a dollar. 
 
 16. If a man own ^ of a vessel, and sell of his share, 
 what part of the whole vessel will he sell ? 
 
 17. When oats are worth of a dollar per bushel, what is 
 f of a bushel worth ? 
 
 18. What will 7f pounds of tea cost, at f of a dollar per 
 pound ? Ans. 4^- dollars. 
 
 19. What is the product of 9f by 4 ? 
 
 9f 
 
 42 23 2 
 
 39 product by 4. Or, 9? X4| = - X = 46. 
 
 6f f. 
 
 46 " 4. 
 
 What does " o/"' signify when placed between two fractions ? What 
 is a compound fraction ? 
 * 
 
114 FRACTIONS. 
 
 To multiply mixed numbers together we may either mul- 
 tiply by the integer and fractional part separately, and then 
 add their products ; or, we may reduce both numbers to 
 improper fractions, and then multiply as in the foregoing rule. 
 
 20. Multiply 12 by 8. Ans. 108|. 
 
 21. What cost 6 1 cords of wood, at 2f dollars a cord ? 
 
 22. What cost } of 2 tons of hay, at 11-& dollars a ton ? 
 
 Ans. $21 T 3 F . 
 
 23. What will 8f cords of wood cost, at 2f dollars per 
 cord? Ans. 22^ dollars. 
 
 24. What must be paid for f of 6 tons of coal, at of 7 
 dollars per ton ? 
 
 25. A man owning $ of a farm, sold of his share ; what 
 part of the whole farm had he left ? Ans. f. 
 
 26. Bought a horse for 125J dollars, and sold him for of 
 what he cost ; how much was the loss ? Ans. $25^. 
 
 27. A owned f of 123| acres of land, and sold f of his 
 share ; how many acres did he sell ? Ans. 49 T 8 ^. 
 
 28. If a family consume 1 barrels of flour a month, how 
 many barrels will five such families consume in 4 T 9 TI months ? 
 
 DIVISION. 
 
 CASE I. 
 
 14O. To divide a fraction by an integer. 
 
 1. If my horse eat T 9 <y of a ton of hay in 3 months, what 
 part of a ton will last him 1 month ? 
 
 OPERATION. ANALYSIS. If he eat T 9 7 of a ton in 
 
 ^ -L- 3 = T ^j, Ans. 3 months, in 1 month he will eat ^ of 
 
 T 9 7 of a ton, or T 9 7 divided by 3. Since 
 
 a fraction is divided by dividing its numerator, (124,) we divide 
 the numerator of the fraction, - t %, by 3, and we have ^, the answer. 
 
 2. If 3 yards of ribbon cost f of a dollar, what will 1 yard 
 cost? 
 
 Case I is what ? Give first explanation. 
 
DIVISION. 115 
 
 OPERATION. ANALYSIS. Here we cannot exactly 
 
 5 _._ 3 __ 5 ^ ng divide the numerator by 3 ; but, since a 
 
 fraction is divided by multiplying the 
 
 denominator, (124,) we multiply the denominator of the fraction, 
 , by 3, and we have T \, the required result. Hence, 
 
 Dividing a fraction consists in dividing its numerator, or 
 multiplying its denominator. 
 
 NOTE. We divide the numerator when it is exactly divisible by the 
 divisor ; otherwise we multiply the denominator. 
 
 EXAMPLES FOR PRACTICE. 
 
 3. Divide f by 2. Ans. f 
 
 4. Divide ? 9 T by 3. Ans. f. 
 
 5. Divide || by 5. Ans. ff. 
 
 6. Divide T W by 25. 
 
 7. Divide tf by 14. Ans. fo. 
 
 8. Divide |i by 21. Ans. VV 
 
 9. If 6 pounds of sugar cost f of a dollar, how much will 
 1 pound cost ? 
 
 10. At 7 dollars a barrel, what part of a barrel of flour can 
 be bought for f of a dollar ? Ans. . 
 
 11. If a yard of cloth cost 5 dollars, what part of a yard 
 can be bought for f of a dollar ? Ans. ^ . 
 
 12. If 9 bushels of barley cost 1\ dollars, how much will 1 
 bushel cost ? 
 
 OPERATION. NoTE . We reduce the mixed number 
 
 71 - _3_6_ to an improper fraction, and divide as 
 
 4- + $=%, Ans. before ' 
 
 13. If 12 barrels of flour cost 76f dollars, how much will 
 1 barrel cost ? 
 
 OPERATION. ANALYSIS. Here we first divide as in 
 
 12 ) 76^ simple numbers, and we have a remainder 
 
 of 4^. We reduce this remainder to an 
 
 6|-, Ans. improper fraction, %., which we divide (as 
 
 in Ex. 1,) and annex the result, f , to the partial quotient, 6, and 
 
 WG have 6^, the required result. 
 
 Give second explanation. Deduction. 
 
116 FRACTIONS. 
 
 14. How many times will 16 j gallons of cider fill a vessel 
 that holds 3 gallons ? Ans. 5&. 
 
 15. If 9 men consume f of 9| pounds of meat in a day, 
 how much does each man consume ? Ans. f of a pound. 
 
 16. A man paid $99f for 4 cows; how much was that 
 apiece? Ans. $24ff. 
 
 CASE II. 
 
 141. To divide an integer by a fraction. 
 
 1. At f of a dollar a yard, how many yards of cloth can be 
 bought for 12 dollars? 
 
 FIRST OPERATION. ANALYSIS. As many yards as f of a 
 
 12 dollar, the price of 1 yard, is contained 
 
 A times in 12 dollars. Integers cannot be di- 
 
 vided by fourths, because they are not of 
 3 ) 48 the same denomination. Reducing 12 dol- 
 
 16 yards ^ ars to fourths by multiplying, we have 48 
 
 fourths ; and 3 fourths is contained in 48 
 fourths 16 times, the required number of yards. 
 
 SECOND OPERATION. ANALYSIS. Here we divide the integer 
 3 \ 12 by the numerator of the fraction, and mul- 
 
 tiply the quotient by the denominator, 
 which produces the same result as in the 
 first operation. Hence, 
 
 16 yards. 
 
 Dividing by a fraction consists in multiplying by the denom- 
 inator, and dividing by the numerator of the divisor. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Divide 18 by f . 
 3. Divide 63 by T V 
 4. Divide 42 by f 
 5. Divide 120 by & 
 6. Divide 316 by ^. 
 
 Ans. 48. 
 Ans. 117. 
 Ans. 49. 
 Ans. 205f 
 Ans. 877. 
 
 Case II is what ? Give first explanation. Second. Deduction. 
 
DIVISION. 117 
 
 7. How many bushels of oats, worth f of a dollar per bushel, 
 will pay for of a barrel of flour, worth 9 dollars a barrel ? 
 
 Am. 15. 
 
 8. If f of an acre of land sell for 21 dollars, what will an 
 acre sell for at the same rate ? Am. $49. 
 
 9. When potatoes are worth of a dollar a bushel, and 
 corn f of a dollar a bushel, how many bushels of potatoes are 
 equal in value to 16 bushels of corn ? Am. 22. 
 
 10. If a man can chop 2f cords of wood in a day, in how 
 many days can he chop 22 cords ? 
 
 OPEKATION. 
 
 21 = -^ 
 
 22 ANALYSIS. We reduce the mixed number 
 
 to an improper fraction, and then divide the 
 integer in the same manner as by a proper 
 
 11 )88 fraction. 
 
 Am. 8 days. 
 
 11. Divide 75 by 13f. Am. 
 
 12. Divide 149 by 24. Am. 
 
 13. A farmer distributed 15 bushels of corn among some 
 poor persons, giving them If bushels apiece; among how 
 many persons did he divide it ? 
 
 14. Divide f of 320 by % of 9. Am. 25 f. 
 
 15. Bought of 7 cords of wood for J of $32 ; how much 
 did 1 cord cost ? Am. $3f 
 
 16. A father divided 183 acres of land equally among his 
 sons, giving them 45 J acres apiece ; how many sons had he ? 
 
 Am. 4. 
 
 CASE III. 
 
 14. To divide a fraction by a fraction. 
 
 1. How many pounds of tea can be bought for of a dol- 
 lar, at of a dollar a pound ? 
 
 How divide by a mixed number ? Case in is what ? 
 
118 FRACTIONS. 
 
 OPERATION. ANALYSIS. As 
 
 First step, f 1 X 3 = f man Y pounds as f 
 
 Second step, f J -~ 2 = JJ = If- of a dollar is con - 
 
 11 2*11 "0 11 tained times in 11 
 
 Whole work, ~ ^ ~ = ~: X ~= =lf,^.of a dollar. 1 is 
 
 '12 3 **V 8 contained in ji }J 
 
 times, and ^ is con- 
 
 tained in || 3 times as many times as 1, or 3 times \%, which is f f 
 times, which is the number of pounds that could be bought at J- of 
 a dollar per pound ; but f is contained but -| as many times as ^, 
 and || divided by 2 gives ||, equal to 1-| times, or the number of 
 pounds that can be bought at f of a dollar per pound. 
 
 We see in the operation that we have multiplied the dividend by 
 the denominator of the divisor, and divided the result by the numer- 
 ator of the divisor, which is in accordance with 14O for dividing a 
 fraction. Hence, by inverting the terms of the divisor, the two 
 fractions will stand in such relation to each other that we can mul- 
 tiply together the two upper numbers for the numerator of the quo- 
 tient, and the two lower numbers for the denominator, as shown in 
 the operation. For division of fractions, we have this general 
 
 RULE. I. Reduce integers and mixed numbers to improper 
 fractions. 
 
 II. Invert the terms of the divisor, and proceed as in multi- 
 plication. 
 
 NOTES. 1. The dividend and divisor may be reduced to a common 
 denominator, and the numerator of the dividend be divided by the nu- 
 merator of the divisor ; this will give the same result as the rule. 
 
 2. Apply cancellation where practicable. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. 
 
 Divide | by J. 
 
 ^t/zs. 1^-. 
 
 3. 
 
 Divide f by . 
 
 Ans. 3^. 
 
 4. 
 
 Divide -f- by T 9 jj. 
 
 .te. t-g. 
 
 5. 
 
 Divide % by /j. 
 
 ^4^5. 
 
 G. 
 
 Divide | by f }. 
 
 Ans. -jjG. 
 
 7. 
 
 How many times is f contained in f ? 
 
 -4ws. IjV 
 
 8. 
 
 How many times is y contained in 1 ? 
 
 ^tws. 3f. 
 
 Rule, first step. Second. What other method is mentioned ? 
 
DIVISION. 119 
 
 9. How many times is T 7 5 contained in ^ ? Ans. 2f . 
 
 10. How many times is ^ contained in % ? 
 
 11. How many times is of contained in f of 2 ? 
 
 12. What is the quotient of T 9 <y of 4, divided by of 3| ? 
 
 13. What is the quotient of of I of 36 divided by 1J 
 times f ? 
 
 01 
 
 14. What is the value of ? 
 
 OPERATION. 
 
 3_ J __7 t 35__# $ This example 
 
 - ~^"~^ ~ ~7 X == f , -4w5. i s only another 
 
 4f -V- * 8 ^ 5 form for ex- 
 
 pressing divis- 
 
 ion of fractions ; it is sometimes called a complex fraction, and the 
 process of performing the division is called reducing a complex frac- 
 tion to a simple one. 
 
 We simply reduce the upper number or dividend to an improper 
 fraction, and the lower number, or divisor, to an improper fraction, 
 and then divide as before. 
 
 15. What is the value of - ? Ans. f f . 
 
 16. What is the value of -? Ans. 20. 
 
 17. What is the value of -^-P Ans. f^ 
 
 4 f 
 
 18. What is the value of ? *? Ans. I. 
 
 * 
 
 P p r 
 
 19. What is the value of - ~ ? Ans. . 
 
 o 
 
 20. If a horse eat f of a bushel of oats in a day, in how 
 many days will he eat 5 bushels? Ans. 14. 
 
 21. If a man spend If dollars per month for tobacco, in 
 what time will he spend lOf dollars? Ans. 6| months. 
 
 What is a complex fraction ? 
 
120 
 
 FRACTIONS. 
 
 22. How many times will 4f gallons of camphene fill a 
 vessel that holds of | of 1 gallon ? Am. 10J-. 
 
 23. If 14 acres of meadow land produce 32 f tons of hay, 
 how many tons will 5 acres produce ? Jras. 112. 
 
 24 If 2 yards of silk cost $3, how much less than $17 
 will 9 yards cost ? Ans. $2|. 
 
 25. If f of a yard of cloth cost ft of a dollar, how much 
 will 1 yard cost ? 
 
 26. A man, having $10, gave of his money for clover 
 seed at $3 a bushel ; how much did he buy? Ans. 2 bush. 
 
 27. How many tons of hay can be purchased for $11 9ft, 
 at $9f per ton? Ans. 
 
 PROMISCUOUS EXAMPLES. 
 
 1. Reduce ^, , f, and to equivalent fractions whose de- 
 nominators shall be 24. Ans. f , f , 2 \, ^. 
 
 2. Change i to an equivalent fraction having 91 for its 
 denominator Ans. ff. 
 
 3. Find the least common denominator of f, If, of f, 2, 
 * of * of 1ft. 
 
 4. Add 4i f, $ of 1|, 3, and JJ. 
 
 5. Find the difference between f of 6ft and f of 4 r \. 
 
 ^45. Ifjjf. 
 
 6. The less of two numbers is 475 6 1, and their difference 
 is 128f ; what is the greater number? Ans. 4885/ B . 
 
 7. What is the difference between the continued products of 
 3,i,S,4f,and3I,S,4,f? Ans. 3&. 
 
 4 2J- 
 
 8. Reduce the fractions and to their simplest form. 
 
 9. What number multiplied by f will produce 18257 ? 
 
 Ans. 304.']. 
 
 10. A farmer had - of his sheep in one pasture, ^ in an- 
 other, and the remainder, which were 77, in a third pasture ; 
 how many sheep had he ? Ans. 140. 
 
 11. What will 7f cords of wood cost at of 9;, dollars per 
 cord? Ans. $24. 
 
PROMISCUOUS EXAMPLES. 121 
 
 12. At J of a dollar per bushel, how many bushels of apples 
 can be bought for 5|- dollars? 
 
 13. Paid $1837f for 7350^ bushels of oats ; how much was 
 that per bushel ? Ans. % of a dollar. 
 
 14. If 235J- acres of land cost $4725|, how much will 628 
 acres cost? Ans. $12601. 
 
 15. A man, owning f of an iron foundery, sold of his share 
 for$540; what was the value of the foundery ? Ans. $4055 1. 
 
 16. f of |- of what number is - - less than 14f ? 
 
 L4 T<J Ans. 27. 
 
 17. A merchant bought 4 cords of wood at $3 per cord, 
 and paid for it in cloth at f of a dollar per yard ; how many 
 yards were required to pay for the wood ? 
 
 18. How many yards of cloth, j of a yard wide, will line 
 20 yards, 1^ yards wide ? Ans. 34. 
 
 19. If the dividend be , and the quotient ^ ff , what is the 
 divisor? 
 
 20. If the sum of two fractions be |, and one of them be 
 /yj what is the other ? Ans. -/^. 
 
 21. If the smaller of two fractions be f ^, and their differ- 
 ence g*j, what is the greater ? Ans. f . 
 
 22. If 3 1 pounds of sugar cost 33 cents, how much must be 
 paid for 65 pounds ? 
 
 23. If 324 bushels of barley can be had for 259 bushels 
 of corn, how much barley can be had for 2000 bushels of 
 corn? Ans. 2500 bushels. 
 
 24. A certain sum of money is to be divided among 5 per- 
 sons ; A is to have J, B , C T V, D ^ and E the remainder, 
 which is 20 dollars ; what is the whole sum to be divided ? 
 
 Ans. $50. 
 
 25. What number, diminished by the difference between J 
 and -| of itself, leaves a remainder of 34 ? Ans. 40. 
 
 26. If f of a farm be valued at $1728, what is the value of 
 the whole ? 
 
122 FRACTIONS. 
 
 27. Bought 320 sheep at $2- per head ; afterward bought 
 435 at $1 per head ; then sold f of the whole number at $1J 
 per head, and the remainder at $2; did I gain or lose, and 
 how much ? Ans. Lost $44. 
 
 28. If 5 be added to both terms of the fraction , will its 
 value be increased or diminished? Ans. Increased T . 
 
 29. If 5 be added to both terms of the fraction f , will its 
 value be increased or diminished ? Ans. Diminished 5 . 
 
 30. How many times can a bottle holding % of of a gal- 
 lon, be filled from a demijohn containing f of If gallons ? 
 
 Ans. 7. 
 
 31. Bought J of 7 cords of wood for | of $32 ; how much 
 did 1 cord cost ? 
 
 32. Purchased 728 pounds of candles at 1 6| cents a pound ; 
 had they been purchased for 3|- cents less a pound, how many 
 pounds could have been purchased for the same money ? - 
 
 Ans. 953|. 
 
 33. What number, divided by If, will give a quotient of 
 9? Ans. 12 jj. 
 
 34. The product of two numbers is 6, and one of them is 
 1846; what is the other? Ans. 5 | 5 . 
 
 35. A stone mason worked 11 days, and after paying his 
 board and other expenses with f of his earnings, he had $20 
 left ; how much did he receive a day ? 
 
 36. If f of 4 tons of coal cost $5, what will J of 2 tons 
 cost? Ans. $5. 
 
 37. In an orchard f of the trees are apple trees, T V peach 
 trees, and the remainder are pear trees, which are 20 more than 
 of the whole ; how many trees in the orchard ? Ans. 800. 
 
 38. A man gave 6| pounds of butter, at 12 cents a pound, 
 for of a gallon of oil ; how much was the oil worth a gal- 
 lon ? Ans. $1. 
 
 39. A gentleman, having 271^- acres of land, sold of it, 
 and gave f of it to his son ; what was the value of the re- 
 mainder, at $57| per acre ? Ans. $4577 B 3 ^. 
 
PROMISCUOUS EXAMPLES. 123 
 
 40. A horse and wagon cost $270 ; the horse cost 1 times 
 as much as the wagon; what was the cost of the wagon? 
 
 41. What number taken from 2 times 12J will leave 
 20? Ans. 11. 
 
 42. A merchant bought a cargo of flour for $2173^, and 
 sold it for f of the cost, thereby losing of a dollar per bar- 
 rel ; how many barrels did he purchase ? Ans. 126. 
 
 43. A and B can do a piece of work in 14 days ; A can do 
 J as much as B ; in how many days can each do it ? 
 
 Ans. A, 6 days ; B, 8 days. 
 
 44. How many yards of cloth f of a yard wide, are equal 
 to 12 yards f of a yard wide ? Ans. 11 J. 
 
 45. A, B, and C can do a piece of work in 5 days ; B and 
 C can do it in 8 days ; in what time can A do it ? 
 
 46. A man put his money into 4 packages ; in the first he 
 put f , in th second , in the third ^, and in the fourth the re- 
 mainder, which was $24 more than T L of the ^v hole ; how much 
 money had he? Ans. $720. 
 
 47. If $7|- will buy 3 cords of wood, how many cords can 
 be bought for $10 ? Ans. 4J'. 
 
 48. How many times is of f of 27 contained in % of ^ of 
 42? 
 
 49. A boy lost J- of his kite string, and then added 30 feet, 
 when it was just f of its original length ; what was the length 
 at first ? Ans. 100 feet. 
 
 50. Bought f of a box of candles, and having used of 
 them, sold the remainder for f of a dollar; how much would 
 a box cost at the same rate ? Ans. $5-ff. 
 
 51. A post stands in the mud, in the water, and 21 feet 
 above the water ; what is its length ? 
 
 52. A father left his eldest son f of his estate, his youngest 
 son f- of the remainder, and his daughter the remainder, who 
 received $1723f less than the youngest son; what was the 
 value of the estate ? Ans. $21114f. 
 
124 DECIMALS. 
 
 DECIMAL FRACTIONS. 
 
 Decimal Fractions are fractions which have for 
 their denominator 10, 100, 1000, or 1 with any number of 
 ciphers annexed. 
 
 NOTES. 1. The word decimal is derived from the Latin decent, 
 which signifies ten. 
 
 2. Decimal fractions are commonly called decimals. 
 
 3. Since ro z= iVo ' To" z= TWo"' &c -> tne denominators of decimal 
 fractions increase and decrease in a tenfold ratio, the same as simple 
 numbers. 
 
 DECIMAL NOTATION AND NUMERATION. 
 
 144:. Common Fractions are the common divisions of a 
 unit into any number of equal parts, as into halves, fifths, 
 twenty-fourths, &c. 
 
 Decimal Fractions are the decimal divisions of a unit, thus : 
 A unit is divided into ten equal parts, called tenths ; each of 
 these tenths is divided into ten other equal parts called hun- 
 dredths ; each of these hundredths into ten other equal parts, 
 called thousandths; and so on. Since the denominators of 
 decimal fractions increase and decrease by the scale of 10, the 
 same as simple numbers, in writing decimals the denomina- 
 tors may be omitted. 
 
 In simple numbers, the unit, 1, is the starting point of 
 notation and numeration ; and so also is it in decimals. We 
 extend the scale of notation to the left of units' place in 
 writing integers, and to the right of units' place in writing 
 decimals. Thus, the first place at the left of units is tens, 
 and the first place at the right of units is tenths ; the second 
 place at the left is hundreds, and the second place at the 
 right is hundredths ; the third place at the left is thousands, 
 and the third place at the right is thousandths ; and so on. 
 
 What are decimal fractions? How do they differ from common 
 fractions ? How are they written ? 
 
NOTATION AND NUMERATION. 125 
 
 The Decimal Point is a period ( . ), which must always be 
 placed before or at the left hand of the decimal. Thus, 
 T 6 u is expressed .6 
 
 " '54 
 
 .279 
 
 NOTE. The decimal point is also called the Separatrix. This is a 
 correct name for it only when it stands between the integral and deci- 
 mal parts of the same number. 
 
 .5 is 5 tenths, which = -^ of 5 units ; 
 
 .05 is 5 hundredths, " = T V of 5 tenths ; 
 
 .005 is 5 thousandths, " = T \j of 5 hundredths. 
 
 And universally, the value of a figure in any decimal place 
 is y 1 ^ the value of the same figure in the next left hand place. 
 
 The relation of decimals and integers to each other is clear- 
 ly shown by the following 
 
 NUMERATION TABLE. 
 
 5 
 s.3 
 
 SlS 
 
 4753.62418695 
 
 Integers. Decimals. 
 
 By examining this table we see that 
 
 Tenths are expressed by one figure. 
 
 Hundredths " " " two figures. 
 
 Thousandths " " " three 
 
 Ten thousandths " " four " 
 
 And any order of decimals by one figure less than the corre- 
 sponding order of integers. 
 
 14<>. Since the denominator of tenths is 10, of hun- 
 
 What is the decimal point ? What is it sometimes called ? "What is 
 the value of a figure in any decimal place ? 
 
 K* 
 
126 DECIMALS. 
 
 dredths 100, of thousands 1000, and so on, a decimal may be 
 expressed by writing the numerator only ; but in this case 
 the numerator or decimal must always contain as many 
 decimal places as are equal to the number of ciphers in the 
 denominator ; and the denominator of a decimal will always 
 be the unit, 1, with as many ciphers annexed as are equal to 
 the number of figures in the decimal or numerator. 
 The decimal point must never be omitted. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Express in figures thirty-eight hundredths. 
 
 2. Write seven tenths. 
 
 3. Write three hundred twenty-five thousandths. 
 
 4. Write four hundredths. Ans. .04. 
 
 5. Write sixteen thousandths. 
 
 6. Write seventy-four hundred-thousandths. Ans. .00074. 
 
 7. Write seven hundred forty-five millionths. 
 
 8. Write four thousand two hundred thirty-two ten-thou- 
 sandths. 
 
 9. Write five hundred thousand millionths. 
 10. Read the following decimals : 
 
 .05 .681 .9034 .19248 
 
 .24 .024 .0005 .001385 
 
 .672 .8471 .100248 .1000087 
 
 NOTE. To read a decimal, we first numerate from left to right, and 
 the name of the right hand figure is the name of the denominator. We 
 then numerate from right to left, as in whole numbers, to read the 
 numerator. 
 
 A mixed number is a number consisting of integers 
 and decimals; thus, 71.406 consists of the integral part, 71, 
 and the decimal part, .406; it is read the same as 7 
 71 and 406 thousandths. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Write eighteen, and twenty-seven thousandths. 
 
 2. Write four hundred, and nineteen ten-millionths. 
 
 How many decimal places must there be to express any decimal ? 
 
KOTATION AND NUMERATION. 127 
 
 3. Write fifty-four, and fifty-four millionths. 
 
 4. Eighty-one, and 1 ten-thousandth. 
 
 5. One hundred, and 67 ten-thousandths. 
 
 6. Read the following numbers : 
 
 18.027 100.0067 400.0000019 
 
 81.0001 54.000054 3.03 
 
 75.075 9.2806 40.40404 
 
 147. From the foregoing explanations and illustrations 
 we derive the following important 
 
 PRINCIPLES OF DECIMAL NOTATION AND NUMERATION. 
 
 1. The value of any decimal figure depends upon its place 
 from the decimal point : thus .3 is ten times .03. 
 
 2. Prefixing a cipher to a decimal decreases its value the 
 same as dividing it by ten ; thus, .03 is T ^ the value of .3. 
 
 3. Annexing a cipher to a decimal does not alter its value, 
 since it does not change the place of the significant figures of 
 the decimal ; thus, T 6 ff , or .6, is the same as T 6 n % or .60. 
 
 4. Decimals increase from right to left, and decrease from 
 left to right, in a tenfold ratio ; and therefore they may be 
 added, subtracted, multiplied, and divided the same as whole 
 numbers. 
 
 5. The denominator of a decimal, though never expressed, 
 is always the unit, 1, with as many ciphers annexed as there 
 are figures in the decimal. 
 
 6. To read decimals requires two numerations ; first, from 
 units, to find the name of the denominator, and second, towards 
 units, to find the value of the numerator. 
 
 148. Having analyzed all the principles upon which the 
 writing and reading of decimals depend, we will now present 
 these principles in the form of rules. 
 
 RULE FOR DECIMAL NOTATION. 
 
 I. Write the decimal the same as a whole number, placing 
 
 "What is the first principle of decimal notation ? Second ? Third ? 
 Fourth ? Fifth ? Sixth ? Kule for notation, first step ? 
 
128 DECIMALS. 
 
 ciphers where necessary to give each significant figure its true 
 local value. 
 
 II. Place the decimal point before the first figure. 
 
 RULE FOR DECIMAL NUMERATION. 
 
 I. Numerate from the decimal point 9 to determine the de- 
 nominator. 
 
 II. Numerate towards the decimal pointy to determine the 
 numerator. 
 
 III. Read the decimal as a whole number, giving it the name 
 or denomination of the right hand figure. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Write 425 million ths. 
 
 2. Write six thousand ten-thousandths. 
 
 3. Write one thousand eight hundred fifty-nine hundred- 
 thousandths. 
 
 4. Write 260 thousand 8 billionths. 
 
 5. Read the following decimals : 
 
 .6321 .748243 .2962999 
 
 .5400027 .60000000 .00000006 
 
 6. Write five hundred two, and one thousand six millionths. 
 
 7. Write thirty-one, and two ten-millionths. 
 
 8. Write eleven thousand, and eleven hundred-thousandths. 
 
 9. Write nine million, and nine billionths. 
 
 10. Write one hundred two tenths. Am. 10.2. 
 
 11. Write one hundred twenty-four thousand three hun- 
 dred fifteen thousandths. 
 
 12. Write seven hundred thousandths. 
 
 13. Write seven hundred-thousandths. 
 
 14. Read the following numbers : 
 
 12.36 9.052 62.9999 
 
 142.847 32.004 1858.4583 
 
 1.02 4.0005 27.00045 
 
 Second ? Rule for numeration, first step ? Second ? Third ? 
 
REDUCTION. 129 
 
 REDUCTION. 
 CASE I. 
 
 149. To reduce decimals to a common denomina- 
 tor. 
 
 1. Reduce .5, .375, 3.25401, and 46.13 to their least com- 
 mon decimal denominator. 
 
 OPEEATION. ANALYSIS. A common denominator must con- 
 
 .50000 ta ^ n as man y decimal places as is equal to the 
 
 37500 greatest number of decimal figures in any of the 
 
 3*25401 given decimals. We find that the third number 
 
 ' contains five decimal places, and hence 100000 
 
 " must be a common denominator. As annexing 
 
 ciphers to decimals does not alter their value,(l 44:., 3)we give to each 
 
 number five decimal places by annexing ciphers, and thus reduce the 
 
 given decimals to a common denominator. Hence, 
 
 RULE. Give to each number the same number of decimal 
 places, by annexing ciphers. 
 
 NOTES. 1. If the numbers be reduced to the denominator of that 
 one of the given numbers having the greatest number of decimal places, 
 they will have their least common decimal denominator. 
 
 2. A whole number may readily be reduced to decimals by placing 
 the decimal point after units, and annexing ciphers ; one cipher re- 
 ducing it to tenths, two ciphers to hundredths, three ciphers to thou- 
 sandths, and so on. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce .17, 24.6, .0003, 84, and 721.8000271 to their 
 least common denominator. 
 
 3. Reduce 7 tenths, 24 thousandths, 187 millionths, 5 hun- 
 dred millionths, and 10845 hundredths to their least common 
 denominator. 
 
 4. Reduce to their least common denominator the following 
 decimals: 1000.001, 841.78, 2.6004, 90.000009, and 6000. 
 
 What is meant by the reduction of decimals ? Case I is what ? 
 Give explanation. Rule. 
 
130 DECIMALS. 
 
 CASE II. 
 
 150. To reduce a decimal to a common fraction. 
 
 1. Reduce .75 to its equivalent common fraction. 
 
 OPERATION. ANALYSIS. We omit the decimal point, 
 
 ^75 _. _7 5 a. supply the proper denominator to the deci- 
 mal, and then reduce the common fraction 
 thus formed to its lowest terms. Hence, 
 
 RULE. Omit the decimal point, and supply the proper 
 denominator. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce .125 to a common fraction. Ans. . 
 
 3. Reduce .16 to a common fraction. Ans. -/$. 
 
 4. Reduce .655 to a common fraction. Ans. $^. 
 
 5. Reduce .9375 to a common fraction. Ans. -ff . 
 
 6. Reduce .0008 to a common fraction. Ans. T2 Ttf. 
 
 CASE III. 
 
 151. To reduce a common fraction to a decimal. 
 
 1. Reduce f- to its equivalent decimal. 
 
 FIRST OPERATION. ANALYSIS. We first annex 
 
 f. = J $$ = T Vfr = .75, Ans. tlie same number of ciphers 
 
 to both terms of the fraction ; 
 
 SECOND OPERATION. this d e s not alter its value. 
 
 \ o f\f) We then divide both resulting 
 
 terms by 4, the significant fig- 
 .75 ure of the denominator, to ob- 
 
 tain the decimal denominator, 
 
 100. Then the fraction is changed to tEe decimal form by omitting 
 the denominator. If the intermediate steps be omitted, the true 
 result may be obtained as in the second operation. 
 
 2. Reduce T X 6 to its equivalent decimal. 
 
 Case II is what? Give explanation. Rule. Case III is what? 
 Explain first operation. Second. 
 
REDUCTION. 131 
 
 THIRD OPERATION. ANALYSIS. Dividing as in the former 
 
 16) 1.0000 example, we obtain a quotient of 3 fig- 
 
 ures, 625. But since we annexed 4 
 
 .Ub2o, Ans. ciphers, there must be 4 places in the 
 
 required decimal; hence we prefix 1 cipher. This is made still 
 
 plainer by the following operation ; thus, 
 
 A = iWftfr = T*fcfo = -0625. 
 From these illustrations we derive the following 
 
 RULE. I. Annex ciphers to the numerator, and divide by 
 the denominator. 
 
 II. Point off as many decimal places in the result as are 
 equal to the number of ciphers annexed. 
 
 NOTE. A common fraction can be reduced to an exact decimal when 
 its lowest denominator contains only the prime factors 2 and 5, and 
 not otherwise. 
 
 EXAMPLES FOR PRACTICE. 
 
 3. Reduce to a decimal. Ans. .625. 
 
 4. Reduce f to a decimal. 
 
 5. Reduce J-f to a decimal. Ans. .9375. 
 
 6. Reduce J to a decimal. 
 
 7. Reduce ^ to a decimal. Ans. .08. 
 
 8. Reduce -fy to a decimal. Ans. .046875. 
 
 9. Reduce f to a decimal. 
 
 10. Reduce F 3 ^ to a decimal. 
 
 11. Reduce B g^ to a decimal. Ans. .00375. 
 
 12. Reduce T F to a decimal. Ans. .008. 
 
 13. Reduce to a decimal. Ans. .33333+. 
 
 NOTE. The sign, -f-, in the answer indicates that there is still a 
 remainder. 
 
 14. Reduce f to a decimal. Ans. .513513+. 
 
 NOTE. The answers to the last two examples are called repeating 
 decimals ; and the figure 3 in the 13th example, and the figures 513 in 
 the 14th, are called repetends, because they are repeated, or occur in 
 regular order. 
 
 Third operation. Rule, first step ? Second ? When can a common 
 fraction be reduced to an exact decimal ? 
 
132 
 
 DECIMALS. 
 
 ADDITION. 
 
 20.074 ? 
 
 1. What Is the sum of 3.703, 621.57, .672, and 
 
 OPERATION. 
 3.703 
 
 621.57 
 
 .672 
 20.0074 
 
 645.9524 
 
 ANALYSIS. We write the numbers so that fig- 
 ures of like orders of units shall stand in the same 
 columns ; that is, units under units, tenths under 
 tenths, hundredths under hundredths, &c. This 
 brings the decimal points directly under each 
 other. Commencing at the right hand, we add 
 each column separately, and carry as in whole 
 numbers, and in the result we place a decimal 
 point between units and tenths, or directly under the decimal point 
 in the numbers added. From this example we derive the following 
 
 RULE. I. Write the numbers so that the decimal points 
 shall stand directly under each other. 
 
 II. Add as in ivhole numbers, and place the decimal point, 
 in the result, directly under the points in the numbers added. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Add 
 
 Sum, 
 
 .199 
 
 2.7569 
 
 .25 
 
 .654 
 
 3.8599 
 
 3. Add 
 
 4.015 
 6.75 
 
 27.38203 
 375.01 
 2.5 
 
 Amount, 415.65703 
 
 4. Add 1152.01, 14.11018, 152348.21, 9.000083. 
 
 Am. 153523.330263. 
 
 5. Add 37.03, 0.521, .9, 1000, 4000.0004. 
 
 Ans. 5038.4514. 
 
 6. What is the sum of twenty-six, and twenty-six hun- 
 dredths ; seven tenths ; six, and eighty-three thousandths ; 
 four, and four thousandths ? Ans. 37.047. 
 
 Explain the operation of addition of decimals. Give rule, first step. 
 Second. 
 
ADDITION. 133 
 
 7. What is the sum of thirty-six, and fifteen thousandths ; 
 three hundred, and six hundred five ten-thousandths ; five, 
 and three millionths ; sixty, and eighty-seven ten-millionths ? 
 
 Ans. 401.0755117. 
 
 8. "What is the sura of fifty-four, and thirty-four hun- 
 dredths ; one, and nine ten-thousandths ; three, and two hun- 
 dred seven millionths; twenty-three thousandths; eight, and 
 nine tenths ; four, and one hundred thirty-five thousandths ? 
 
 Ans. 71.399107. 
 
 9. How many yards in three pieces of cloth, the first piece 
 containing 18.375 yards, the second piece 41.625 yards, and 
 the third piece 35.5 yards? 
 
 10. A's farm contains 61.843 acres, B's contains 143.75 
 acres, C's 218.4375 acres, and D's 21.9 acres; how many 
 acres in the four farms ? 
 
 11. My farm consists of 7 fields, containing 12} acres, 18f 
 acres, 9 acres, 24| acres, 4|| acres, 8^ acres, and 15^ acres 
 respectively ; how many acres in my farm ? 
 
 NOTE. Reduce the common fractions to decimals before adding. 
 
 Ans. 93.6375. 
 
 12. A grocer has 2 barrels of A sugar, 5| barrels of B 
 sugar, 3| barrels of C sugar, 3.0642 barrels of crushed 
 sugar, and 8.925 barrels of pulverized sugar ; how many bar- 
 rels of sugar has he ? Ans. 23.8642. 
 
 13. A tailor made 3 suits of clothes ; for the first suit he 
 used 2 yards of broadcloth, 3^ yards of cassimere, and 
 yards of satin ; for the second suit 2.25 yards of broadcloth, 
 2.875 yards of cassimere, and 1 yard of satin ; and for the 
 third suit 5-^ yards of broadcloth, and 1 yards of satin. 
 How many yards of each kind of goods did he use ? How 
 many yards of all ? Ans. to last, 18.375. 
 
 L 
 
134 DECIMALS. 
 
 SUBTRACTION. 
 
 153. 1. From 91.73 take 2.18. ANALYSIS. In each of these 
 
 three examples, we write the 
 
 OPERATION, subtrahend under the minu- 
 
 91-73 end, placing units under 
 
 2.18 units, tenths under tenths, 
 
 Ans. 89.55 ^ c * Commencin g at the 
 
 right hand, we subtract as 
 
 2. From 2.9185 take 1.42. *? wholc mbers > d f 
 
 the remainders we place the 
 
 OPERATION. decimal points directly under 
 
 2.9185 those in the numbers above. 
 
 j ^2 In the second example, the 
 
 number of decimal places in 
 
 Ans. 1.4985 t h e mmue nd is greater than 
 
 the number in the subtra- 
 
 3. From 12465 take 95.58746. hend] and in the third exam . 
 
 OPERATION. pie the number is less. In 
 
 124.65 koth cases, we reduce both 
 
 or "OVA a mmuend and subtrahend to 
 
 VO.tJOl 1O . . , 
 
 the same number of decimal 
 
 Ans. 29.06254 places, by annexing ciphers; 
 
 or we suppose the ciphers to 
 be annexed, before performing the subtraction. Hence the 
 
 RULE. I. Write the numbers so that the decimal points 
 shall stand directly under each other, 
 
 II. Subtract as in whole numbers, and place the decimal 
 point in the result directly under the points in the given numbers. 
 
 4. Find the difference between 714 and .91 6. Ans. 713.084. 
 
 5. How much greater is 2 than .298 ? Ans. 1.702. 
 
 6. From 21.004 take 75 hundredths. 
 
 7. From 10.0302 take 2 ten-thousandths. Ans. 10.03. 
 
 8. From 900 take .009. Ans. 899.991. 
 
 9. From two thousand take two thousandths. 
 
 10. From one take one millionth. Ans. .999999. 
 
 Explain subtraction of fractions. Give the rule, first step. Second. 
 
MULTIPLICATION. 135 
 
 11. From four hundred twenty-seven thousandths take 
 four hundred twenty-seven millionths. Ans. .426573. 
 
 12. A man owned thirty-four hundredths of a township of 
 land, and sold thirty-four thousandths of the township ; how 
 much did he still own ? Ans. .306. 
 
 MULTIPLICATION. 
 
 154. 1. What is the product of .35 multiplied by .5 ? 
 
 OPERATION. ANALYSIS. We perform the multiplication the 
 
 .35 same as in whole numbers, and the only difficulty 
 
 5 we meet with is in pointing off the decimal places 
 
 in the product. To determine how many places to 
 
 .175, Ans. p i n t O ff, we may reduce the decimals to common 
 fractions ; thus, .35 n= T 3 ^ and .5 =. T 6 ^. Perform- 
 ing the multiplication, and we have ^ X -& = ^jfe, and this 
 product, expressed decimally, is .175. Here we see that the prod- 
 uct contains as many decimal places as are contained in both mul- 
 tiplicand and multiplier. Hence the following 
 
 RULE. Multiply as in whole numbers, and from the right 
 hand of the product point off as many figures for decimals as 
 there are decimal places in both factors. 
 
 NOTES. 1. If there be not as many figures in the product as there 
 are decimals in both factors, supply the deficiency by prefixing ciphers. 
 
 2. To multiply a decimal by 10, 100, 1000, &c., remove the point as 
 many places to the right as there are ciphers on the right of the multi- 
 plier." 
 
 EXAMPLES. 
 
 2. Multiply 1.245 by .27. Ans. .33615. 
 
 3. Multiply 79.347 by 23.15. Ans. 1836.88305. 
 
 4. Multiply 350 by .7853. 
 
 5. Multiply one tenth by one tenth. Ans. .01. 
 
 6. Multiply 25 by twenty-five hundredths. Ans. 6.25. 
 
 Explain multiplication of decimals. Give rule. If the product have 
 less decimal places than both factors, how proceed ? How multiply by 
 10, 100, 1000, &c. ? 
 
136 DECIMALS. 
 
 7. Multiply .132 by .241. Ans. .031812. 
 
 8. Multiply 24.35 by 10. 
 
 9. Multiply .006 by 1000. Ans. 6. 
 
 10. Multiply .23 by .009. Ans. .00207. 
 
 11. Multiply sixty-four thousandths by thirteen million ths. 
 
 Ans. .000000832. 
 
 12. Multiply eighty-seven ten-thousandths by three hun- 
 dred fifty-two hundred-thousandths. 
 
 13. Multiply one million by one millionth. Ans. 1. 
 
 14. Multiply sixteen thousand by sixteen ten-thousandths. 
 
 Ans. 25.6. 
 
 15. If a cord of wood be worth 2.37 bushels of wheat, how 
 many bushels of wheat must be given for 9.58 cords of wood ? 
 
 Ans. 22.7046 bushels. 
 
 DIVISION. 
 
 155. 1. What is the quotient of .175 divided by .5 ? 
 
 OPERATION. ANALYSIS. We perform the division the same as 
 .5 ) .175 in whole numbers, and the only difficulty we meet 
 - with is in pointing off the decimal places in the quo- 
 Ans. 35 tient. To determine how many places to point off, 
 we may reduce the decimals to common fractions; thus, .175 = 
 an( ^ -5 iV Performing the division, and we have 
 
 175 5 M$ 10 35 
 
 VX _ -.!. 
 
 1000 10 1000 100 
 
 and this quotient, expressed decimally, is .35. Here we see that the 
 dividend contains as many decimal places as are contained in both 
 divisor and quotient. Hence the following 
 
 RULE. Divide as in whole numbers, and from the right 
 hand of the quotient point off as many places for decimals 
 as the decimal places in the dividend exceed those in the 
 divisor. 
 
 Explain division of decimals. Give rule. 
 
DIVISION. 137 
 
 NOTES. 1. If the number of figures in the quotient be less than the 
 excess of the decimal places in the dividend over those in the divisor, 
 the deficiency must be supplied by prefixing ciphers. 
 
 2. If there be a remainder after dividing the dividend, annex ciphers, 
 and continue the division: the ciphers annexed are decimals of the 
 dividend. 
 
 3. The dividend must always contain at least as many decimal places 
 as the divisor, before commencing the division. 
 
 4. In most business transactions, the division is considered suffi- 
 ciently exact when the quotient is carried to 4 decimal places, unless 
 great accuracy is required. 
 
 5. To divide by 10, 100, 1000, &c., remove the decimal point as 
 many places to the left as there are ciphers on the right hand of the 
 divisor. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Divide .675 by .15. Am. 4.5. 
 
 3. Divide .288 by 3.6. Ans. .08. 
 
 4. Divide 81.6 by 2.5. Ans. 32.64. 
 
 5. Divide 2.3421 by 21.1. 
 
 6. Divide 2.3421 by .211. 
 
 7. Divide 8.297496 by .153. Ans. 54.232. 
 
 8. Divide 12 by .7854. 
 
 9. Divide 3 by 3 ; divide 3 by .3 ; 3 by .03 ; 30 by .03. 
 
 10. Divide 15.34 by 2.7. 
 
 11. Divide .1 by .7. Ans. .142857+. 
 
 12. Divide 45.30 by .015. Ans. 3020. 
 
 13. Divide .003753 by 625.5. Ans. .000006. 
 
 14. Divide 9. by 450. Ans. .02. 
 
 15. Divide 2.39015 by .007. Ans. 341.45. 
 
 16. Divide fifteen, and eight hundred seventy-five thou- 
 sandths,by twenty-five ten-thousandths. Ans. 6350. 
 
 17. Divide 365 by 100. 
 
 18. Divide 785.4 by 1000. Ans. .7854. 
 
 19. Divide one thousand by one thousandth. 
 
 Ans. 1000000. 
 
 When are ciphers prefixed to the quotient ? If there be a remainder, 
 how proceed ? If the dividend have less decimal places than the divi- 
 sor, how proceed ? How divide by 10, 100, 1000, &c. ? 
 
 L* 
 
138 DECIMALS. 
 
 PROMISCUOUS EXAMPLES. 
 
 1. Add six hundred, and twenty -five thousandths ; four 
 tenths ; seven, and sixty-two ten-thousandths ; three, and fifty- 
 eight millionths ; ninety-two, and seven hundredths. 
 
 Ans. 702.501258. 
 
 2. What is the sum of 81.003 + 5000.4+5.0008 + 
 73.87563 + 1000 + 25 + 3.000548 + .0315 ? 
 
 3. From eighty-seven take eighty-seven thousandths. 
 
 4. What is the difference between nine million and nine 
 millionths? Ans. 8999999.999991. 
 
 5. Multiply .365 by .15. Ans. .05475. 
 
 6. Multiply three thousandths by four hundredths. 
 
 7. If one acre produce 42.57 bushels of corn, how many 
 bushels will 18.73 acres produce ? Ans. 797.3361. 
 
 8. Divide .125 by 8000. Ans. .000015625. 
 
 9. Divide .7744 by .1936. 
 
 10. Divide 27.1 by 100000. Ans. .000271. 
 
 11. If 6.35 acres produce 70.6755 bushels of wheat, what 
 does one acre produce ? Ans. 11.13 bushels. 
 
 12. Reduce .625 to a common fraction. Ans. f. 
 
 13. Express 26.875 by an integer and a common fraction. 
 
 Ans. 26. 
 
 14. Reduce T f ? to a decimal fraction. Ans. .016. 
 
 15. Reduce to a decimal fraction. Ans. .5. 
 
 16. How many times will .5 of 1.75 be contained in .25 of 
 17? Ans. 5. 
 
 17. What will be the cost of 3| bales of cloth, each bale 
 containing 36.75 yards, at .85 dollars per yard ? 
 
 18. Traveling at the rate of 4f miles an hour, how many 
 hour* will a man require to travel 56.925 miles. 
 
 Ans. 12$ hours. 
 
NOTATION AND NUMERATION. 139 
 
 DECIMAL CURRENCY. 
 
 156. Coin is money stamped, and has a given value es- 
 tablished by law. 
 
 157. Currency is coin, bank bills, treasury notes, &c., in 
 circulation as a medium of trade. 
 
 158 . A Decimal Currency is a currency whose denom- 
 inations increase and decrease in a tenfold ratio. 
 
 NOTE. The currency of the United States is decimal currency, and 
 is sometimes called Federal Money , it was adopted by Congress in 1786. 
 
 NOTATION AND NUMERATION. 
 
 The gold coins of the United States are the double eagle, 
 half eagle, and quarter eagle, three dollar piece, and dollar. 
 
 The silver coins are the dollar, half and quarter dollar, dime 
 and half dime, and three cent piece. 
 
 The nickel coin is the cent. 
 
 NOTES. 1. The following pieces of gold are in use, but are not legal 
 coin, viz. ; the fifty dollar piece, and the half and quarter dollar pieces. 
 
 2. The copper cent and half cent, though still in circulation, are no 
 longer coined. 
 
 3. The mill is used only in computation ; it is not a coin. 
 
 TABLE. 
 
 10 mills (m.) make 1 cent, . . . c. 
 
 10 cents " 1 dime, . . . d. 
 
 10 dimes " 1 dollar, . . . $. 
 
 10 dollars " 1 eagle, . . . E. 
 
 UNIT EQUIVALENTS. 
 Mills. Cents. 
 
 10 = 1 Dimes. 
 
 100 =10 = 1 Dollarg . 
 
 1000 = 100 =10 = I Eagle . 
 
 10000 = 1000 = 100 = 10 = 1 
 
 NOTE. The character $ is supposed to be a contraction of U. S., 
 (United States,") the U being placed upon the S. 
 
 "What is coin? Currency? Decimal currency? Federal money? 
 What are the gold coins of U. S. ? Silver ? Copper ? What are the 
 denominations of U. S. currency ? What is the sign of dollars ? From 
 what derived ? 
 
140 DECIMAL CURRENCY. 
 
 159* The dollar is the unit of United States money; 
 dimes, cents, and mills are fractions of a dollar, and are sepa- 
 rated from the dollar by the decimal point ; thus, two dollars 
 one dime two cents five mills, are written $2.125. 
 
 By examining the table, we see that the dime is a tenth part 
 of the unit, or dollar ; the cent a tenth part of the dime or a 
 hundredth part of the dollar ; and the mill a tenth part of the 
 cent, a hundredth part of the dime, or a thousandth part of the 
 dollar. Hence the denominations of decimal currency increase 
 and decrease the same as decimal fractions, and are expressed 
 according to the same decimal system of notation ; and they 
 may be added, subtracted, multiplied, and divided in the same 
 manner as decimals. 
 
 Dimes are not read as dimes, but the two places of dimes 
 and cents are appropriated to cents ; thus, 1 dollar 3 dimes 
 2 cents, or $1.32, are read one dollar thirty-two cents ; hence, 
 
 When the number of cents is less than 1 0, we write a cipher 
 before it in the place of dimes. 
 
 NOTE. The half cent is frequently written as 5 mills ; thus, 24<| cents, 
 written $.245. 
 
 1OO. Business men frequently write cents as common 
 fractions of a dollar ; thus, three dollars thirteen cents are 
 written SSy 1 ^, and read, three and thirteen hundredths dollars. 
 In business transactions, when the final result of a computation 
 contains 5 mills or more, they are called one cent, and when 
 less than 5, they are rejected. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Write four dollars five cents. Ans. $4.05. 
 
 2. Write two dollars nine cents. 
 
 3. Write ten dollars ten cents. 
 
 4. Write eight dollars seven mills. Ans. $8.007. 
 
 What is the unit of U. S. currency ? What is the general law of 
 increase and decrease ? In practice, how many decimal places are given 
 to cents ? In business transactions, how are cents frequently written ? 
 What is done if the mills exceed 5 ? If less than 5 ? 
 
REDUCTION. 141 
 
 5. Write sixty-four cents. Ans. $0.64. 
 
 6. Write three cents two mills. 
 
 7. Write one hundred dollars one cent one mill. 
 
 8. Read $7.93 ; $8.02 ; $6.542. 
 
 9. Read $5.272; $100.025; $17.005. 
 
 10. Read $16.205; $215.081; $1000.011; $4.002. 
 
 REDUCTION. 
 
 161. By examining the table of Decimal Currency, we see 
 that 10 mills make one cent, and 100 cents, or 1000 mills, 
 make one dollar ; hence, 
 
 To change dollars to cents, multiply by 100 ; that is, annex 
 two ciphers. 
 
 To change dollars to mills, annex three ciphers. 
 To change cents to mills, annex one cipher. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Change $792 to cents. Ans. 79200 cents. 
 
 2. Change $36 to cents. 
 
 3. Reduce $5248 to cents. 
 
 4. In 6.25 dollars how many cents ? Ans. 625 cents. 
 
 NOTE. To change dollars and cents to cents, or dollars, cents, and 
 mills to mills, remove the decimal point and the sign, $. 
 
 5. Change $63.045 to mills. Ans. 63045 mills. 
 
 6. Change 16 cents to mills. 
 
 7. Reduce $3.008 to mills. 
 
 8. In 89 cents how many mills ? 
 
 162* Conversely, 
 
 To change cents to dollars, divide by 100 ; that is, point off 
 two figures from the right. 
 
 To change mills to dollars, point off three figures. 
 To change mills to cents, point off one figure. 
 
 How are dollars changed to cents ? to mills ? How are cents changed 
 to mills? How are cents changed to dollars ? Mills to dollars ? to cents? 
 
142 DECIMAL CURRENCY. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Change 875 cents to dollars. Ans. $8.75. 
 
 2. Change 1504 cents to dollars. 
 
 3. In 13875 cents how many dollars ? 
 
 4. In 16525 mills how many dollars? 
 
 5. Reduce 524 mills to cents. 
 
 6. Reduce 6524 mills to dollars. 
 
 ADDITION. 
 
 1G3. 1. A man bought a cow for 21 dollars 50 cents, a 
 horse for 125 dollars 37^ cents, a harness for 46 dollars 75 cents, 
 and a carriage for 210 dollars ; how much did he pay for all ? 
 
 OPERATION. 
 
 $ 21 50 ANALYSIS. Writing dollars under dol- 
 
 125*375 ^ ais ' cents un( l er cents, &c., so that the 
 decimal points shall stand under each 
 other, we add and point off as in addition 
 of decimals. Hence the following 
 
 Ans. $403.625 
 
 RULE. I. Write dollars under dollars, cents under cents, fyc. 
 II. Add as in simple numbers, and place the point in the 
 amount as in addition of decimals. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the sum of 50 dollars 7 cents, 1000 dollars 75 
 cents, 60 dollars 3 mills, 18 cents 4 mills, 1 dollar 1 cent, and 
 25 dollars 45 cents 8 mills ? Ans. $1137.475. 
 
 3. Add 364 dollars 54 cents 1 mill, 486 dollars 6 cents, 93 
 dollars 9 mills, 1742 dollars 80 cents, 3 dollars 27 cents 6 
 mills. Ans. $2689.686. 
 
 4. Add 92 cents, 10 cents 4 mills, 35 cents 7 mills, 18 cents 
 6 mills, 44 cents 4 mills, 12 cents, and 99 cents. Ans. $3.126. 
 
 Explain the process of addition of decimal cxirrency . Rule, first step. 
 Second. 
 
SUBTRACTION. 143 
 
 5. A farmer receives 89 dollars 74 cents for wheat, 13 dol- 
 lars 3 cents for corn, 6 dollars 374- cents for potatoes, and 19 
 dollars 62^ cents for oats; what does he receive for the 
 whole? Am. $128.77. 
 
 6. A lady bought a dress for 9 dollars 17 cents, trimmings 
 for 87^ cents, a paper of pins for 6^ cents, some tape for 4 
 cents, some thread for 8 cents, and a comb for 1 1 cents ; what 
 did she pay for all ? Am. $10.3375. 
 
 7. Paid for building a house $2175.75, for painting the 
 same $240.37=1, for furniture $605.40, for carpets $140.12; 
 what was the cost of the house and furnishing ? 
 
 8. Bought a ton of coal for $6.08, a barrel of sugar for 
 $26.625, a box of tea for $16, and a barrel of flour for $7.40 ; 
 what was the cost of all ? 
 
 9. A merchant bought goods to the amount of $7425.50 ; 
 he paid for duties on the same $253.96, and for freight 
 $170.09 ; what was the entire cost of the goods ? 
 
 10. I bought a hat for $3.62, a pair of shoes for $1|, an 
 umbrella for $lf , a pair of gloves for $.62|, and a cane for 
 $.87 ; what was the cost of all my purchases ? Am. $8.25. 
 
 SUBTRACTION. 
 
 164. 1. A man, having $327.50, paid out $186.75 for 
 a horse ; how much had he left ? 
 
 OPERATION. ANALYSIS. Writing the less number un- 
 
 $327.50 der the greater, dollars under dollars, cents 
 
 186 75 under cents, &c., we subtract and point off 
 
 in the result as in subtraction of decimals. 
 
 Am. $140.75 Hence the following 
 
 RULE. I. Write the subtrahend under the minuend, dollars 
 under dollars, cents under cents, fyc. 
 
 11. Subtract as in simple numbers, and place the point in 
 the remainder, as in subtraction of decimals. 
 
 Explain the process of subtraction. Give rule, first step. Second. 
 
144 DECIMAL CURRENCY. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. From $365 dollars 5 mills take 267 dollars 1 cent 8 
 mills. Ans. $97.987. 
 
 3. From 50 dollars take 50 cents. Ans. $49.50. 
 
 4. From 100 dollars take 1 mill. Ans. $99.999. 
 
 5. From 1000 dollars take 3 cents 7 mills. 
 
 6. A man bought a farm for $1575.24, and sold it for 
 $1834.16; what did he gain ? Ans. $258.92. 
 
 7. Sold a horse for 145 dollars 27 cents, which is 37 dol- 
 lars 69 cents more than he cost me ; what did he cost me ? 
 
 8. A merchant bought flour for $5.62 J a barrel, and sold 
 it for $6.84 a barrel ; how much did he gain on a barrel ? 
 
 9. A gentleman, having $14725, gave $3560 for a store, 
 and $7015.87^- for goods ; how much money had he left ? 
 
 10. A lady bought a silk dress for $13J, a bonnet for $5^, a 
 pair of gaiters for $lf , and a fan for $| ; she paid to the shop- 
 keeper a twenty dollar bill and a five dollar bill ; how much 
 change should he return to her? Ans. $3.75. 
 
 NOTE. Reduce the fractions of a dollar to cents and mills. 
 
 11. A gentleman bought a pair of horses for $480, a har- 
 ness for $80.50, and a carriage for $200 less than he paid for 
 both horses and harness ; what was the cost of the carriage? 
 
 Ans. $360.50. 
 
 MULTIPLICATION. 
 
 165. 1. If a barrel of flour cost $6.375, what will 85 
 barrels cost ? 
 
 OPERATION. 
 
 $6.375 ANALYSIS. We multiply as in simple 
 
 85 numbers, always regarding the multiplier 
 as an abstract number, and point off' from 
 the right hand of the result, as in multipli- 
 51000 cation of decimals. Hence the following 
 
 Ans. $541.875 
 
 Give analysis for multiplication in decimal currency. 
 
DIVISION. 145 
 
 RULE. Multiply as in simple numbers, and place the point 
 in the product, as in multiplication of decimals. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. If a cord of wood be worth $4.275, what will 300 cords 
 be worth? 4ns. $1282.50. 
 
 3 What will 175 barrels of apples cost, at $2.45 per bar- 
 re! ? An*. $428.75. 
 
 4. What will 800 barrels of salt cost, at $1.28 per barrel? 
 
 5. A grocer bought 372 pounds of cheese at $.15 a pound, 
 434 pounds of coffee at $.12| a pound, and 16 bushels of pota- 
 toes at $.33 a bushel ; what did the whole cost ? 
 
 6. A boy, being sent to purchase groceries, bought 3 pounds 
 of tea at 56 cents a pound, 15 pounds of rice at 7 cents a 
 pound, 27 pounds of sugar at 8 cents a pound ; he gave the 
 grocer 5 dollars ; how much change ought he to receive ? 
 
 7. A farmer sold 125 bushels of oats at $.37^- a bushel, 
 and received in payment 75 pounds of sugar at $.09 a pound, 
 12 pounds of tea at $.60 a pound, and the remainder in cash; 
 how much cash did he receive ? Ans. $32.92. 
 
 8. A man* bought 150 acres of land for $3975 ; he after- 
 ward sold 80 acres of it at $32.50 an acre, and the remainder 
 at $34.25 an acre ; how much did he gain by the transaction ? 
 
 Ans. $1022.50. 
 
 i 
 DIVISION. 
 
 166. 1. If 125 barrels of flour cost $850, how much 
 will 1 barrel cost ? 
 
 OPERATION. ANALYSIS. We divide as in 
 
 125 ) $850.00 ( $6.80, Ans. simple numbers, and as there 
 
 is a remainder after dividing 
 the dollars, we reduce the div- 
 idend to cents, by annexing two 
 ciphers, and continue the di- 
 vision. Hence the following 
 
 Rule. Give rule for division in decimal currency. 
 M 
 
146 DECIMAL CURRENCY. 
 
 RULE. Divide as in simple numbers, and place the point in 
 the quotient, as in division of decimals. 
 
 NOTES. 1. In business transactions it is never necessary to carry 
 the division further than to mills in the quotient. 
 
 2. If the dividend will not contain the divisor an exact number of 
 tbnes, ciphers may be annexed, and the division continued as in divis- 
 ion of decimals. In this case it is always safe to reduce the dividend 
 to mills, or to 3 more decimal places than the divisor contains, be- 
 fore commencing the division. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. If 33 gallons of oil cost $41.25, what is the cost per gal- 
 lon ? Am. $1.25. 
 
 3. If 27 yards of broadcloth cost $94.50, what will 1 yard 
 cost? 
 
 4. If 64 gallons of wine cost $136j what will 1 gallon cost? 
 
 Ans. $2.125. 
 
 5. At 12 cents apiece, how many pine-apples can be bought 
 for $1.32? Ans. 11. 
 
 6. If 1 pound of tea cost 54 cents, how many pounds can 
 be bought for $405 ? 
 
 7. If a man earn $180 in a year, how mucji does he earn 
 a month ? 
 
 8. If 100 acres of land cost $2847.50, what will 1 acre 
 cost? Ans.' $28.475. 
 
 9. What cost 1 pound of beef, if 894 pounds cost $80.46? 
 
 Ans. $.09. 
 
 10. A farmer sells 120 bushels of wheat at $1.12 a bushel, 
 for which he receives 27 barrels of flour ; what does the flour 
 cost him a barrel ? 
 
 11. A man bought 4 yards of cloth at $3.20 a yard, and 
 37 pounds of sugar at $.08 a pound ; he paid $6.80 in cash, 
 and the remainder in butter at $.16 a pound ; how many pounds 
 of butter did it take ? Ans. 56 pounds. 
 
 12. A man bought an equal number of calves and sheep, 
 paying $166.75 for them ; for the calves he paid $4.50 a head, 
 and for the sheep $2.75 a head ; how many did he buy of each 
 kind? Ans. 23. 
 
OF 
 
 APPLICATIONS. 
 
 13. If 154 pounds of sugar cost $18.48, W 
 cost? 
 
 14. A merchant bought 14 boxes of tea for $560; it being 
 damaged he was obliged to lose $106.75 on the cost of it ; 
 how much did he receive a box ? Ans. $32.37. 
 
 ADDITIONAL APPLICATIONS. 
 
 CASE I. 
 
 167. To find* the cost of any number or quantity, 
 when the price of a unit is an aliquot part of one dollar. 
 
 168. An Aliquot Part of a number is such a part as will 
 exactly divide that number; thus, 2, 3, and 7^ are aliquot 
 parts of 15. 
 
 NOTE. An aliquot part may be a whole or a mixed number, while a 
 factor must be a whole number. 
 
 ALIQUOT PARTS OF ONE DOLLAR. 
 
 50 cents = 
 33 cents = 
 25 cents = 
 20 cents = 
 16f cents = 
 
 of 1 dollar, 
 
 of 1 dollar, 
 
 of 1 dollar, 
 
 of 1 dollar, 
 
 of 1 dollar. 
 
 121 cents = of 1 dollar. 
 10 cents fa of 1 dollar. 
 
 8 cents = fa of 1 dollar. 
 
 6 cents = -fa of 1 dollar. 
 
 5 cents = ^ of 1 dollar. 
 
 1. What will be the cost of 3784 yards of flannel, at 25 
 cents a yard ? 
 
 OPERATION. ANALYSIS. If the price were $1 a yard, 
 
 4 ^ 3784 ^ e cos ^ wou ^ be as many dollars as there are 
 
 yards. But since the price is \ of a dollar a 
 
 yard, the whole cost will be \ as many dollars 
 
 as there are yards ; or, } of 3784 = 3784 -^ 4 = $946. Hence the 
 
 RULE. Take such a fractional part of the given number as 
 the price is part of one dollar. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What cost 963 bushels of oats, at 33 cents per bushel? 
 
 Ans. $321. 
 
 Case I is what ? What is an aliquot part of a dollar ? Give ex- 
 planation. Rule. 
 
148 DECIMAL CURRENCY. 
 
 3. What cost 478 yards of delaine, at 50 cents per yard ? 
 
 4. What cost 4266 yards of sheeting, at 8 cents a yard? 
 
 Ans. $355.50. 
 
 5. What cost 1250 bushels of apples, at 12 i cents per 
 feushel? Ans. $156.25. 
 
 6. What cost 3126 spools of thread, at 6| cents per spool? 
 
 Ans. $195.375. 
 
 7. At 16 cents per dozen, what cost 1935 dozen of eggs? 
 
 . Ans. 322.50. 
 
 8. What cost 56480 yards of calico, at 12 per yard ? 
 
 9. At 20 cents each what will be the cost of 1275 salt 
 barrels? Ans. $255. 
 
 CASE II. 
 
 169. The price of one and the quantity being given, 
 to find the cost. 
 
 1. How much will 9 barrels of flour cost, at $6.25 per 
 barrel ? 
 
 OPERATION. ANALYSIS. Since one barrel cost $6.25,9 
 
 $6.25 barrels will cost 9 times $6.25, and $6.25 X 
 9 = $56.25. Hence 
 
 Ans. $56.25 
 
 RULE. Multiply the price of one by the quantity. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. If a pound of beef cost 9 cents, what will 864 pounds 
 cost? Ans. $77.76. 
 
 3. What cost 87 acres of government land, at $1.25 per 
 acre? 
 
 4. What cost 400 barrels of salt, at $1.45 per barrel ? 
 
 Ans, $580. 
 
 5 What cost 16 chests of tea, each chest containing 52 
 pounds, at 44 cents per pound ? 
 
 Case n is what ? Give explanation. Rule. 
 
APPLICATIONS. 149 
 
 CASE III. 
 
 170 . The cost and the quantity being given, to find 
 the price of one. 
 
 1. If 30 bushels of corn cost $20.70, what will 1 bushel 
 cost? 
 
 OPERATION. ANALYSIS. If 30 bushels cost $20.70, 1 
 
 310 ) $210 70 bushel will cost -fa of $20.70; and $20.70 -f- 
 
 30 = $.69. Hence, 
 $.69 
 
 RULE. Divide the cost by the quantity. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. If 25 acres of land cost $175, what will 1 acre cost? 
 
 3. If 48 yards of broadcloth cost $200, what will 1 yard 
 cost? Am. $4.1 6. 
 
 4. If 96 tons of hay cost $1200, what will 1 ton cost? 
 
 5. If 10 Unabridged Dictionaries cost $56.25, what will 1 
 cost? Ans. $5.62. 
 
 6. Bought 18 pounds of tea for $11.70 ; what was the price 
 per pound ? Ans. $.65. 
 
 7. If 53 pounds of butter cost $10.07, what will 1 pound 
 cost? 
 
 8. A merchant bought 800 barrels of salt for $1016 ; what 
 did it cost him per barrel ? 
 
 9. If 343 sheep cost $874.65, what will 1 sheep cost ? 
 
 Ans. $2.55. 
 
 10. If board for a family be $684.37 for 1 year, how much 
 is it per day? Ans. $1.87. 
 
 CASE IV. 
 
 171. The price of one and the cost of a quantity 
 being given, to find the quantity. 
 
 1. At $6 a barrel for flour, how many barrels can be bought 
 
 for $840 ? 
 
 Case III is what ? Give explanation. Rule. Case IY is what ? 
 
150 DECIMAL CURRENCY. 
 
 OPERATION. ANALYSIS. Since $6 will buy 1 barrel 
 
 6 ) 840 of flour > $840 will buy as many barrels 
 
 . ~T7I as there are dollars, or as many barrels as 
 
 Am. ] reis. $6 is contained times in $840; 840-^-6 
 
 z= 140 barrels. Hence, 
 
 RULE. Divide the cost of the quantity by the price of one. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. How many dozen of eggs can be bought for $5.55, if one 
 dozen cost $.15 ? Am. 37 dozen. 
 
 3. At $12 a ton, how many tons of hay can be bought for 
 $216? Ans. 18 tons. 
 
 4. How many bushels of wheat can be bought for $2178.75, 
 if 1 bushel cost $1.25 ? Ans. 1743 bushels. 
 
 5. A dairyman expends $643.50 in buying cows at $19 
 apiece ; how many cows does he buy ? Ans. 33 cows. 
 
 6. At $.45 per gallon, how many gallons of molasses can 
 be bought for $52.65 ? 
 
 7. A drover bought horses at $264 a pair; how many 
 horses did he buy for $6336 ? 
 
 8. At $65 a ton, how many tons of railroad iron can be 
 bought for $117715? Ans. 1811 tons. 
 
 CASE V. 
 
 172. To find the cost of articles sold by the 100, 
 1000, &c. 
 
 1. What cost 475 feet of timber, at $5.24 per 100 feet ? 
 
 FIRST OPERATION. 
 
 $5 24 ANALYSIS. If the price were $5.24 per 
 
 475 f oot > the cost f 475 feet would be 475 X 
 
 $5.24 $2489. But since $5.24 is the 
 
 2620 price of 100 feet, $2489 is 100 times the true 
 
 3668 value. Therefore, to obtain the true value, 
 
 2096 we divide $2489 by 100, which we may do 
 
 i \ ^QOAA b y cuttin off two % ures from the ri S ht and 
 100 ) $2489.00 the result is $24.89. Or, 
 
 Ans. $24.89 
 Give explanation. Rule. Case V is what ? Give first explanation. 
 
APPLICATIONS. 151 
 
 SECOND OPERATION. ANALYSIS. Since 1 foot costs T ^, or. 01, 
 $5.24 of> $5.24, 475 feet will cost f|, or 4.75 times 
 
 4 75 $5.24, which is $24.89. 
 
 NOTE. For the same reasons, when the price 
 
 o fi , is per thmisand, we divide the product by 1000, 
 
 or, which is more convenient in practice, we re- 
 
 2096 duce the given quantity to thousands and deci- 
 
 mals of a thousand, by pointing off three figures 
 
 $24.8900 from the right hand. Hence the 
 
 RULE. I. Reduce the given quantity to hundreds and deci- 
 mals of a hundred, or to thousands and decimals of a thousand. 
 
 II. Multiply the price by the quantity, and point off in the 
 result as in multiplication of decimals. 
 
 NOTE. The letter C is used to indicate hundreds, and Mto indicate 
 thousands. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What will 42650 bricks cost, at $4.50 per M ? 
 
 Ans. $191.925. 
 
 3. What is the freight on 2489 pounds from Boston to New 
 York, at $.85 per 100 pounds ? Ans. $21.156+. 
 
 4. What will 7842 feet of pine boards cost, at $17.25 
 perM? Ans. $135.274+. 
 
 5. What cost 2348 pine-apples, at $12 per 100 ? 
 
 6. A broom maker bought 1728 broom-handles, at $3 per 
 1000 ; how much did they cost him ? 
 
 7. What is the cost of 2400 feet of boards, at $7 perM; 
 865 feet of scantling, at $5.40 per M; and 1256 feet of lath, at 
 $.80 per C? Ans. $31.519. 
 
 8. What will be te cost of 1476 pounds of beef, at $4.374- 
 per hundred pounds ? 
 
 CASE VI. 
 
 173. -To find the cost of articles sold by the ton of 
 2000 pounds. 
 
 1 . How much will 2376 pounds of hay cost, at $9.50 per ton ? 
 Give second explanation. Rule, first step. Second. Case VI is what ? 
 
152 DECIMAL CURRENCY. 
 
 OPERATION. ANALYSIS. Since 1 ton, or 2000 pounds, cost 
 
 2 ) $9.50 $9.50, 1000 pounds, or -| ton, will cost ^ of $9.50, 
 
 ^ or $9.50 -f- 2 = $4,75. One pound will cost 
 
 T J rg -, or .001, of $4.75, and 2376 pounds will 
 
 2.376 cost f , or 2.376 times $4.75, which is $1 1.286. 
 
 $11.28600 Hence > 
 
 RULE. I. Divide the price of 1 ton by 2, and the quotient 
 will be the price of 1000 pounds. 
 
 II. Multiply this quotient by the given number of pounds 
 expressed as thousandths, as in Case V. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. At $7 a ton, what will 1495 pounds of hay cost ? 
 
 Ans. $5.2325. 
 
 3. At $8.75 a ton, what cost 325 pounds of hay ? 
 
 Ans. $1.421+. 
 
 4. What is the cost of 3142 pounds of plaster, at $3.84 per 
 ton? Ans. $6.032+. 
 
 5. What is the cost of 1848 pounds of coal, at $5.60 per 
 ton? 
 
 6. Bought 125 sacks of guano, each sack containing 148 
 pounds, at $18 a ton; what was the cost? 
 
 7. What must be paid for transporting 31640 pounds of 
 railroad iron from Philadelphia to Richmond, at $3.05 per 
 ton? Ans. $48.251. 
 
 BILLS. 
 
 17" 4:. A Bill, in business transactions, is a written state- 
 ment of articles bought or sold, together with the prices of 
 each, and the whole cost. 
 
 Find the cost of the several articles, and the amount or 
 footing of the following bills. , 
 
 Give explanation. Rule. What is a bill ? Explain the manner of 
 making out a bill. 
 
BILLS. 153 
 
 Mr. JOHK RICE, NEW YoKK ' June 20 > 1859 ' 
 
 Bo't. of BALDWIN & SHERWOOD, 
 
 7 yds. Broadcloth, fa) $3.60 
 
 9 " Satinet, " 1.12 
 
 12 Vesting, .90 
 
 24 " Cassimere, " 1.37 
 
 32 " Flannel, .65 _ 
 
 , , D $99.925 
 
 See d Payment, BALDWIN & SHERWOOD . 
 
 (2-) 
 
 DANIEL CHAPMAN & Co., BosTON > Jan ' ^ 1860 ' 
 
 ^o'^. O/PALMEK & BROTHER. 
 
 67 pairs Calf Boots, (a) $3.75 
 108 " Thick " " 2.62 
 75 " Gaiters, " 1.12 
 27 " Buskins, " .86 
 35 Slippers, " .70 
 50 Rubbers, 1.04 _ 
 
 $717.93 
 Jtec a. Jraymentj 
 
 PALMER & BROTHER, 
 
 By GEO. BAKER. 
 
 (3.) 
 
 G. B. GEANNIS, CHARLESTON, Sept. 6, 1859. 
 
 Bo't. of STEWART & HAMMOND, 
 325 Ibs. A. Sugar, (a) $.07 
 148 " B. " .06^ 
 
 286 Rice, " .05 
 
 95 O. J. Coffee, " .12^- 
 50 boxes Oranges, " 2.75 
 75 " Lemons, " 3.62^ 
 12 " Raisins, " 2.85 ___ 
 
 501.75 
 Payment, by note at 4 mo. 
 
 STEWART & HAMMOND. 
 
154 DECIMAL CURRENCY. 
 
 Messrs. OSBORN & EATON, ST " LouIS > Oct ' 15 > 1858 ' 
 Bo't. of ROB'T. H. CARTER & Co., 
 20000 feet Pine Boards fa> $15 per M. 
 
 7500 
 
 Plank, 
 
 t( 
 
 9.50 
 
 10750 
 
 Scantling, 
 
 u 
 
 6.25 
 
 3960 
 
 Timber, 
 
 (i 
 
 2.62 
 
 5287 
 
 a 
 
 a 
 
 3.00 " 
 
 $464.6935 
 Hec'd. Payment, 
 
 KOB'T. H. CARTER & Co. 
 
 (5.) 
 
 Mr. J. C. SMITH, CINCINNATI, May 3, 1861. 
 
 Bo't. of SILAS JOHNSON, 
 
 25 Ibs. Coffee Sugar, 
 
 5 " Y. H. Tea, 
 
 26 " Mackerel, 
 4 gal. Molasses, 
 
 46 yds. Sheeting, 
 
 30 Bleached Shirting, 
 
 6 skeins Sewing Silk, ' 
 4 doz. Buttons, 
 
 Chgd. in a | c . 
 
 PROMISCUOUS EXAMPLES. 
 
 1. What will 62.75 tons of potash cost, at $124.3^ per ton ? 
 
 ~Ans. $7802.9625. 
 
 2. "What cost 15 pounds of butter, at $.17 a pound? 
 
 Aw. $2.55. 
 
 3. A cargo of corn, containing 2250 bushels, was sold for 
 $1406.25 ; what did it sell for per bushel ? Aw. $ . 
 
PROMISCUOUS EXAMPLES. 155 
 
 4. If 12 yards of cloth cost $48.96, what will one yard 
 cost ? 
 
 5. A traveled 325 miles by railroad, and C traveled .45 of 
 that distance ; how far did C travel ? Ans. 146.25 miles. 
 
 6. If 36.5 bushels of corn grow on one acre, how ma^iy 
 acres will produce 657 bushels ? Ans. 18 acres. 
 
 7. Bought a horse for $105, a yoke of oxen for $125, 4 
 cows at $35 apiece, and sold them all for $400 ; how much 
 was gained or lost in the transaction ? 
 
 8. A man bought 28 tons of hay at $19 a ton, and sold it 
 at $15 a ton ; how much did he lose ? Ans. $112. 
 
 9. If a man travel 4f miles an hour, in how many hours 
 can he travel 34 miles ? Ans. 7.5 hours. 
 
 10. At $.31 per bushel, how many bushels of potatoes 
 can be bought for $9 ? Ans. 28.8 bushels. 
 
 11. If a man's income be $2000 a year, and his expenses 
 $3.50 a day, what will he save at the end of a year, or 365 
 days? 
 
 1 2. A merchant deposits in a bank, at one time, $687.25, 
 and at another, $943.64 ; if he draw out $875.29, how much 
 will remain in the bank ? 
 
 13. Bought 288 barrels of flour for $1728, and sold one 
 half the quantity for the same price I gave for it, and the other 
 half for $8 per barrel ; how much did I receive for the whole ? 
 
 Ans. $2016. 
 
 14. What will eight hundred seventy-five thousandths of a 
 cord of wood cost, at $3.75 per cord ? Ans. $3.281+. 
 
 15. A drover bought cattle at $46.56 per head, and sold 
 them at $65.42 per head, and thereby gained $3526.82 ; how 
 many cattle did he buy ? Ans. 187. 
 
 16. If 36.48 yards of cloth cost $54.72, what will 14.25 
 yards cost? Ans. $21.375. 
 
 17. A house cost $3548, which is 4 times as much as the 
 furniture cost ; what did the furniture cost ? Ans. $887. 
 
 18. How many bushels of onions at $.82 per bushel, can 
 be bought for $112.34? 
 
156 DECIMAL CURRENCY. 
 
 19. If 46 tons of iron cost $3461.50, what will 5 tons cost? 
 
 20. A gentleman left his widow one third of his property, 
 worth $24000, and the remainder was to be divided equally 
 among 5 children ; how much was the portion of each child ? 
 
 Ans. $3200. 
 
 21. A man purchased one lot, containing 1 60 acres of land, at 
 $1.25 per acre ; and another lot, containing 80 acres, at $5 per 
 acre ; he sold them both at $2.50 per acre ; what did he gain 
 or lose in the transaction ? 
 
 22. A druggist bought 54 gallons of oil for $72.90, and 
 lost 6 gallons of it by leakage. He sold the remainder at 
 $1.70 per gallon; how much did he gain ? Ans. $8.70. 
 
 23. A miller bought 122J- bushels of wheat of one man, 
 and 75 bushels of another, at $.93J per bushel. He sold 60 
 bushels at a profit of $12.50 ; if he sell the remainder at 
 $.81 per bushel, what will be his entire gain or loss ? 
 
 Ans. $4.718+ loss. 
 
 24. A laborer receives $1.40 per day, and spends $.75 for 
 his support ; how much does he save in a week ? 
 
 25. How many pounds of butter, at $.16 per pound, must 
 be given for 39 yards of sheeting, at $.08 a yard? 
 
 Ans. 19 pounds. 
 
 26. What cost 23487 feet of hemlock boards, at $4.50 per 
 1000 feet? Ans. $105.6915. 
 
 27. A man has an income of $1200 a year; how much 
 must he spend per day to use it all ? 
 
 28. Bought 28 firkins of butter, each containing 5 6 pounds, 
 at $.17 per pound ; what was the whole cost ? 
 
 29. A merchant bought 1 6 bales of cotton cloth, each bale 
 containing 13 pieces, and each piece 26 yards, at $.07 per 
 yard ; what did the whole cost ? Ans. $378.56. 
 
 30. What cost 4868 bricks, at $4.75 per M ? 
 
 31. A farmer sold 27 bushels of potatoes, at $.33 per 
 bushel ; 28 bushels of oats, at $.25 per bushel ; and 19 bush- 
 els of corn, at $.50 per bushel ; what did he receive for the 
 whole? Ans. $25.50. 
 
PROMISCUOUS EXAMPLES. 157 
 
 32. John runs 32 rods in a minute, and Henry pursues 
 him at the rate of 44 rods in a minute; how long will it take 
 Henry to overtake John, if John have 8 minutes the start ? 
 
 Am. 21^ minutes. 
 
 33. If 4 barrels of flour cost $32.3, what will 7 barrels 
 cost? Am. $51. 
 
 34. If .875 of a ton of coal cost $5.635, what will 9 tons 
 cost? Am. $59.57. 
 
 35. For the first three years of business, a trader gained 
 $1200.25 a year; for the next three, he gained $1800.62 a 
 year, and f#r the next two he lost $950.87 a year ; supposing 
 his capital at the beginning of trade to have been $5000, what 
 was he worth at the end of the eighth year ? Am. $12100.82. 
 
 36. What will be the cost of 18640 feet of timber, at $4.50 
 per 100 ? Ans. $838.80. 
 
 2i 
 
 37. Reduce 57 to a decimal fraction. Ans. .78125. 
 
 6 % 
 
 38. What will 1375 pounds of potash cost, at $96.40 per 
 ton? Ans. $66.275. 
 
 39. Reduce .5625 to a common fraction. Ans. T 9 ^. 
 
 40. Reduce ^-, .62^, .37^, f , to decimals, and find their 
 sum. Ans. 1.464375. 
 
 41. A man's account at a store stands thus : 
 
 Dr. Cr. 
 
 $4.745 $2.76 
 
 2.62 1.245 
 
 1.27 .62 
 
 * .45 3.45 
 
 5.28 1.87J 
 
 What is due the merchant ? Ans. $4.41. 
 
 42. A gardener sold, from his garden, 120 bunches of on- 
 ions at $.12^- a bunch, 18 bushels of potatoes at $.62 per 
 bushel, 47 heads of cabbage at $.07 a head, 6 dozen cucum- 
 bers at $.18 a dozen; he expended $1.50 in spading, $1.27 
 for fertilizers, $1.87 for seeds, $2.30 in planting and hoeing; 
 what were the profits of his garden ? Ans. $23.68. 
 
 N 
 
158 REDUCTION. 
 
 REDUCTION. 
 
 175* A Compound Number is a concrete number whose 
 value is expressed in two or more different denominations. 
 (110.) 
 
 176. Reduction is the process of changing a number from 
 one denomination to another without altering its value. 
 
 Reduction is of two kinds, Descending and Ascending. 
 
 177. Reduction Descending is changing a number of one 
 denomination to another denomination of less unit value ; thus, 
 $1 = 10 dimes = 100 cents = 1000 mills. 
 
 178. Reduction Ascending is changing a number of one 
 denomination to another denomination of greater unit value ; 
 thus, 1000 mills = 100 cents = 10 dimes =$1. 
 
 179. A Scale is a series of numbers, descending or as- 
 cending, used in operations upon compound numbers. 
 
 NOTE. In simple numbers and decimals, the scale is uniformly 10 ; 
 in compound numbers the scales are varying. 
 
 CURRENCY. 
 
 180. I. UNITED STATES MONEY. 
 
 TABLE. 
 
 10 mills (m.) make 1 cent, ct. 
 
 10 cents " 1 dime, d. 
 
 10 dimes " 1 dollar, $. 
 
 10 dollars " 1 eagle, E. 
 
 UNIT EQUIVALENTS. 
 
 ct. m. 
 
 ,1. 1 ~ 10 
 
 $ 1 = 10 = 100 
 
 E. 1 = 10 = 100 = 1000 
 
 1 = 10 = 100 = 1000 = 10000 
 SCALE uniformly 10. 
 
 NOTE. This table is given here for the purpose of presenting all the 
 compound numbers together ; but Decimal Currency has been previ- 
 ously treated of so fully, that additional exercises are not considered 
 necessary. 
 
 AVhat. is a compound number? Reduction i* what? Reduction 
 descending ? Reduction ascending ? A scale ? Scale of simple num- 
 bers and decimals ? Scale of compound numbers ? Repeat the table 
 of U. S. money. 
 
COMPOUND NUMBERS. 159 
 
 II. ENGLISH MONEY. 
 English Money is the currency of Great Britain. 
 
 TABLE. 
 
 4 farthings (far. or qr.) make 1 penny, d. 
 
 12 pence " 1 shilling, s. 
 
 20 shillings " 1 pound or sovereign,. ., or sov. 
 
 UNIT EQUIVALENTS. 
 
 d. far. 
 
 14 
 
 ,orsov. 1 = 12 = 48 
 
 1 20 = 240 = 960 
 .SCALE ascending, 4, 12, 20; descending, 20, 12, 4. 
 
 NOTES. 1. The equivalents of each unit in all lower denominations 
 may easily be learned, when but few denominations and small numbers 
 are used. 
 
 2. Farthings are generally expressed as fractions of a penny ; thus, 
 1 far., sometimes called 1 quarter, (qr.) =d. ; 3 far. = f d. 
 
 CASE I. 
 
 18S. To perform reduction descending. 
 1. Reduce 21 18 s. 10 d. 2 far. to farthings. 
 
 OPERATION. ANALYSIS. We multiply 
 
 21 18s. 10 d. 2 far. 21 by 20, because there 
 20 are 20 times as many shil- 
 
 lings as pounds, and add 
 18 shillings to the product, 
 making 438 shillings in 21 
 
 5266 d. ^ s< e next mu hiply 438 
 
 M shillings by 12, because 
 
 there are 12 times as many 
 
 Ans. 21066 far. pence as shillings, and add 
 
 10 pence to the product, 
 
 making 5266 pence in 21 18s. 10 d. Lastly, we multiply the 5266 
 pence by 4, because there are 4 times as many farthings as pence, 
 and add 2 farthings to the product; and we obtain 21066 farthings 
 in the given number. Hence the following 
 
 RULE. I. MULTIPLY the highest denomination of the given 
 number by that number of the scale which ivill reduce it to 
 the next lower denomination, and add to the product the given 
 number, if any, of that lower denomination. 
 
 Repeat the table of English money. The scale. Case I is what r 
 Give explanation. Rule. 
 
160 REDUCTION. 
 
 II. Proceed in the same manner with the results obtained in 
 each lower denomination, until the reduction is brought to the 
 denomination required. 
 
 CASE II. 
 
 183. To perform reduction ascending. 
 
 I. Reduce 21066 farthings to pounds. 
 
 OPERATION. ANALYSIS. We first divide 
 
 4) 21066 far. the 21066 far. by 4, because 
 
 there are \ as many pence as 
 
 12)5266d. + 2far. farthings, and we find that 
 
 2|0 ) 43|8 s. + 10 d. 21066 far. 5266 d. + a re- 
 
 o7 f _l_ i Q mainder of 2 far. We next 
 
 21 . +- 18 s. ^^ 5266 d> by 12> because 
 
 Ans. 21 18 s. 10 d. 2 far. there are ^ as many sh ini ng8 
 
 as pence, and we find that 5266 d. z= 438 s. -j- 10 d. Lastly, we 
 divide the 438 s. by 20, because there are -^ as many pounds as 
 shillings, and we find that 438 s. = 21 + 18 s. The last quotient 
 with the several remainders annexed in the order of the succeeding 
 denominations gives the answer 21 18 s. 10 d. 2 far. Hence the 
 following 
 
 RULE. I. DIVIDE the given number by that number of the 
 scale which will reduce it to the next higher denomination. 
 
 II. Divide the quotient by the next higher number in the 
 scale ; and so proceed to the highest denomination required. 
 The last quotient, with the several remainders annexed in a 
 reversed order, will be the answer. 
 
 NOTE. Reduction descending and reduction ascending mutually 
 prove each other. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. In 14194 farthings how many pounds? 
 
 2. In 14 15 s. 8 d. 2 far. how many farthings ? 
 
 3. Reduce 15 19 s. 11 d. 3 far. to farthings. 
 
 4. In 15359 farthings how many pounds? 
 
 5. In 46 sov. 12s. 2 d. how many pence ? 
 
 6. In 11186 pence how many sovereigns? 
 
 Case II is what ? Give explanation. Rule. Proof, how performed ? 
 
COMPOUND NUMBERS. 
 
 161 
 
 WEIGHTS. 
 
 184:. Weight is a measure of the quantity of matter a 
 body contains, determined according to some fixed standard. 
 Three scales of weight are used in the United States 
 and Great Britain, namely, Troy, Apothecaries', and Avoir- 
 dupois. 
 
 I. TROY WEIGHT. 
 
 185. Troy Weight is used in weighing gold, silver, and 
 jewels ; in philosophical experiments, &c. 
 
 TABLE. 
 
 24 grains (gr.) make 1 pennyweight,, .pwt. or dwt. 
 
 20 pennyweights " 1 ounce, oz. 
 
 12 ounces " 1 pound, Ib. 
 
 UNIT EQUIVALENTS. 
 
 pwt. gr. 
 
 1 = 24 
 
 i b . 1 20 = 480 
 1 = 12 = 240 = 5760 
 SCALE ascending, 24, 20, 12; descending, 12, 20, 24. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. How many grains in 14 Ib. ' 
 10 oz. 18 pwt. 22 gr.? 
 
 OPERATION. 
 
 14 Ib. 10 oz. 18 pwt. 22 gr. 
 12 
 
 178 oz. 
 20 
 
 3578 pwt. 
 24 
 
 14334 
 7156 
 
 2. How many pounds in 
 85894 grains ? 
 
 OPERATION. 
 
 24 ) 85894 gr. 
 20 ) 3578 pwt. + 22 gr. 
 12 )178 oz. + 15 pwt. 
 14 Ib. + 10oz. 
 
 Ans. 141b. lOoz. 18 pwt. 
 22 gr. 
 
 85894 gr., Ans. 
 
 3. In 5 Ib. 7 oz. 12 pwt. 9 gr., how many grains ? 
 
 4. In 43457 grains how many pounds ? 
 
 Define weight. Troy weight. Repeat the table. Give the scale. 
 
162 
 
 REDUCTION. 
 
 5. Reduce 41760 grains to pounds. Ans. 7 Ib. 3 oz. 
 
 6. A miner had 14 Ib. 10 oz. 18 pwt. of gold dust; how 
 much was it worth at $.75 a pwt. ? Ans. $2683.50. 
 
 7. How many spoons, each weighing 2 oz. 15 pwt., can be 
 made from 5 Ib. 6 oz. of silver ? Ans. 24. 
 
 8. A goldsmith manufactured 1 Ib. 1 pwt. 16 grs. of gold into 
 rings, each weighing 4 pwt. 20 gr. ; he sold the rings for $1.25 
 apiece ; how much did he receive for them ? Ans. $62.50. 
 
 II. APOTHECARIES' WEIGHT. 
 
 186. Apothecaries' Weight is used by apothecaries and 
 physicians in compounding medicines ; but medicines are 
 bought and sold by avoirdupois weight. 
 
 TABLE. 
 
 20 grains (gr.) make 1 scruple, sc. or 9- 
 
 3 scruples " 1 dram, dr. or 3 
 
 8 drams 
 12 ounces 
 
 1 ounce, oz. or 
 
 1 pound, Ib. or 
 
 UNIT EQUIVALENTS. 
 
 so. gr. 
 
 dr. ^1 = 20 
 
 oz 1 = 3 = 60 
 
 lb . 1 8 = 24 480 
 1 = 12 = 96 = 288 == 5760 
 SCALE ascending, 20, 3, 8, 12; descending, 12, 8, 3, 20. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. How many gr. in 12 ib 
 85 35 19 15 gr.? 
 
 OPERATION. 
 
 12 ib 835 19 15 gr. 
 
 12 s 
 
 2. How many Ib in 73175 
 gr.? 
 
 OPERATION. 
 
 2|0) 7317)5 gr. 
 3)36589 + 15gr. 
 
 152 5 
 
 8 
 
 8 ) 12193 + 19 
 12 ) 152 + 3 3 
 
 
 
 1219 3 
 3 
 
 12tb + 8 
 
 3658 B 
 20 
 
 Ans. 12lb8 3 3 19 15 gr. 
 
 73175 gr., Ans. 
 
 
 Define apothecaries' weight. Repeat the table. Give the scale. 
 
COMPOUND NUMBERS. 163 
 
 3. In 1 Gib. 11 oz. 7 dr. 2 sc. 19 gr., how many grains? 
 
 4. Reduce 47 ft 6 4 5 to scruples. Am. 13692' sc. 
 
 5. How many pounds of medicine would a physician use in 
 one year, or 365 days, if he averaged daily 5 prescriptions 
 of 20 grains each ? Ans. 6 ib. 4 1 ). 
 
 III. AVOIRDUPOIS WEIGHT. 
 
 187. Avoirdupois Weight is used for all the ordinary pur- 
 poses of weighing. 
 
 TABLE. 
 
 16 drams (dr.) make 1 ounce, oz. 
 
 16 ounces " 1 pound, Ib. 
 
 100 Ib. "1 hundred weight, . cwt. 
 / 20 cwt., or 2000 Ibs., " 1 ton, T. 
 
 UNIT EQUIVALENTS. 
 
 oz. dr. 
 
 Ib. 1 = 16 
 
 cwt . 1 16 = 256 
 
 T 1 = 100 = 1600 = 25600 
 
 l' = 20 = 2000 = 32000 = 512000 
 SCALE ascending, 16, 16, 100, 20 ; descending, 20, 100, 16, 16. 
 
 NOTE. The long or gross ton, hundred weight, and quarter were 
 formerly in common use ; but they are now seldom used except in 
 estimating English goods at the U. S. custom-houses, and in freighting 
 and wholesaling coal from the Pennsylvania mines. 
 
 LONG TON TABLE. 
 
 28 Ib. make 1 quarter, marked .qr. 
 
 4 qr. = 112 Ib. " 1 hundred weight, " cwt. 
 
 20 cwt. = 2240 Ib. 1 ton, T. 
 
 SCALE ascending, 28, 4, 20; descending, 20, 4, 28. 
 
 The following denominations are also in use. 
 56 pounds make 1 firkin of butter. 
 
 100 
 
 
 
 1 quintal of dried salt fish. 
 
 
 100 " 
 
 M 
 
 1 cask of raisins. 
 
 
 196 
 
 u 
 
 1 barrel of flour. 
 
 
 200 " 
 
 
 
 1 
 
 < 
 
 beef, pork, or fish. 
 
 
 280 " 
 
 
 
 1 
 
 < 
 
 salt at the N. Y. State 
 
 salt works. 
 
 56 " 
 
 
 
 1 bu 
 
 shel 
 
 
 a 
 
 32 " 
 
 M 
 
 1 
 
 < 
 
 oats. 
 
 
 48 
 
 " 
 
 1 
 
 i 
 
 barley. 
 
 
 56 
 
 M 
 
 1 
 
 1 " corn or rye. 
 
 
 60 
 
 ( 
 
 1 " wheat. 
 
 
 Define avoirdupois weight. Repeat the table. Give the scale. The 
 long ton table. What other denominations are in use ? "What is the 
 value of each ? 
 
164 REDUCTION. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. In 25 T. 15 cwt. 70 Ib. 
 how many pounds ? 
 
 OPERATION. 
 
 25 T. 15 cwt. 70 Ib. 
 
 20 
 
 .5 
 
 100 
 
 2. In 51570 pounds how 
 many tons ? 
 
 OPERATION. 
 
 100) 51570 Ib. 
 
 2|0)51|5cwt. + 701b. 
 
 515 CWt. O K m I i r 
 
 25 P. -(-15 cwt. 
 Ans. 25 T. 15 cwt. 70 Ib. 
 
 51570 Ib., Ans. 
 
 3. Reduce 3 T. 14 cwt. 74 Ib. 12 oz. 15 dr. to drams. 
 
 4. Reduce 1913551 drams to tons. 
 
 5. A tobacconist bought 3 T. 15 cwt. 20 Ib. of tobacco, at 
 22 cents a pound; how much did it cost him ? Ans. $1654.40. 
 
 6. How much will 115 pounds of hay cost, at $10 per ton ? 
 
 7. A grocer bought 10 barrels of sugar, each weighing 
 
 2 cwt. 17 Ib., at 6 cents a pound; 5 barrels, each weighing 
 
 3 cwt. 6 Ib., at 7 cents a pound ; he sold the whole at an 
 average price of 8 cents a pound ; how much was his whole 
 gain? Ans. $51.05. 
 
 8. Paid $360 for 2 tons of cheese, and retailed it for 12 
 cents a pound ; how much was my whole gain ? Ans. $140. 
 
 9. If a person buy 10 T. 6 cwt. 3 qr. 14 Ib. of English iron, 
 by the long ton weight, at 6 cents a pound, and sell the same 
 at $130 per short ton, how much will he gain ? Ans. $115.85. 
 
 10. A farmer sold 2 loads of corn, weighing 2352 Ibs. each, 
 at $.90 per bu. ; what did he receive ? Ans. $75.60. 
 
 11. How many pounds in 300 barrels of flour ? Ans. 58800. 
 
 12. A grocer bought 3 barrels of salt at $1.25 per barrel, 
 and retailed it at J of a cent per pound ? what did he gain ? 
 
 Ans. $2.55. 
 
 STANDARD OF WEIGHT. 
 
 188. In the year 1834 the U. S. government adopted a 
 uniform standard of weights and measures, for the use of the 
 custom houses, and the other branches of business connected with 
 the general government. Most of the States which have adopt- 
 ed any standards have taken those of the general government. 
 
COMPOUND NUMBERS. 1G5 
 
 189. The United States standard unit of weight is the 
 Troy pound of the mint, which is the same as the imperial 
 standard pound of Great Britain, and is determined as fol- 
 lows : A <3ubic inch of distilled water in a vacuum, weighed 
 by brass weights, also in a vacuum, at a temperature of 62 
 Fahrenheit's thermometer, is equal to 252.724 grains, of which 
 the standard Troy pound contains 5760. 
 
 190. The U. S. Avoirdupois pound is determined from 
 the standard Troy .pound, and contains 7000 Troy grains. 
 Hence, the Troy pound is f $ = |Ai O f an avoirdupois 
 pound. But the Troy ounce contains -^f^ = 480 grains, and 
 the avoirdupois ounce ^-^p == 437.5 grains ; and an ounce Troy 
 is 480 437.5 =. 42.5 grains greater than an ounce avoirdu- 
 pois. The pound, ounce, and grain, Apothecaries' weight, 
 are the same as the like denominations in Troy weight, the only 
 difference in the two tables being in the divisions of the ounce. 
 
 191. COMPARATIVE TABLE OF WEIGHTS. 
 
 Troy. Apothecaries'. Avoirdupois. 
 
 1 pound = 5760 grains, = 5760 grains, = 7000 grains. 
 1 ounce = 480 " = 480 " 437.5 " 
 
 175 pounds, 175 pounds, = 144 pounds. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. An apothecary bought 5 Ib. 10 oz. of rhubarb, by 
 avoirdupois weight, at 50 cents an ounce, and retailed it at 
 12 cents a dram apothecaries' weight ; how much did he gain? 
 
 Ans. $33.75. 
 
 2. Change 424 drams apothecaries' weight to Troy weight. 
 
 Ans. 4 Ib. 5 oz. 
 
 3. Change 20 Ib. 8 oz. 12 pwt. Troy weight to avoirdu- 
 pois weight. Ans. 17/ 7 3- Ib. 
 
 4. Bought by avoirdupois weight 20 Ib. of opium, at 40 
 cents an ounce, and sold the same by Troy weight at 50 cents 
 an ounce; how much was gained or lost? Ans. $17.83^. 
 
 What is the II. S. standard of weight ? How obtained ? How is 
 the avoirdupois pound determined ? How is the apothecaries' pound 
 determined ? What are the values of the denominations of Troy, avoir- 
 dupois, and apothecaries' weight? 
 
166 REDUCTION. 
 
 MEASURES OF EXTENSION. 
 
 11)2. Extension has three dimensions length, breadth, 
 and thickness. 
 
 A Line has only one dimension length. 
 
 A Surface or Area has two dimensions length and breadth. 
 
 A Solid or Body has three dimensions length, breadth, and 
 thickness. 
 
 I. LONG MEASURE. 
 
 193. Long Measure, also called Linear Measure, is used 
 in measuring lines or distances. 
 
 TABLE. 
 
 12 inches (in.) make 1 foot, ft. 
 
 3 feet " 1 yard, yd. 
 
 5\ yd., or 16^ ft., " 1 rod rd. 
 
 40 rods " 1 furlong, .... fur. 
 
 8 furlongs, or 320 rd., " 1 statute mile,, .mi. 
 
 UNIT EQUIVALENTS. 
 
 ft. in. 
 
 yd. 1 = 12 
 
 rd. '1 3 = 36 
 
 fur . 1 = 51 = 16* = 198 
 
 mi 1 = 40 220" = 666" = 7920 
 1 = 8 = 320 1760 = 5280 = 63360 
 SCALE ascending, 12, 3, 5J, 40, 8; descending, 8,40, =>, 3, J2. 
 
 The following denominations are also in use : 
 
 3 barleycorns make 1 inch, 5 ll , se( | b ? shoemakers in measuring 
 
 ' ( the length of the foot. 
 
 4 inches " 1 hand 5 used in . measm ' hl t he height of 
 
 ' ( horses directly over the fore feet. 
 6 feet " 1 fathom, used in measuring 1 depths at sou, 
 
 1.15 statute miles 1 geographic mile, J " 8Cfl in measuring dis- 
 
 ' ^ tances at sea. 
 3 geographic " " 1 league. 
 
 60 " > 1 , < of latitude on a meridian or of 
 
 69| statute ^ Degree ^ ] ong j tude on the equator. 
 S60 degrees " the circumference of the earth. 
 
 TTow many dimensions has extension? Define a line. Surf?! <<> or 
 ;in a. A solid or body. Define long measure. What are the d> 
 
 The value of each. What other denominations arc us: d ? 
 
COMPOUND NUMBERS. 
 
 107 
 
 NOTES. 1. For the purpose of measuring cloth and other goods sold 
 by the yard, the yard is divided into halves, fourths, eighths, and six- 
 teenths. The old table of cloth measure is practically obsolete. 
 
 2. The geographic mile is ^V ofy|?r r Ti^To ^ ^ e distance round 
 the center of the earth. It is a small fraction more than 1.15 statute 
 miles. 
 
 3. The length of a degree of latitude varies, being 68.72 miles at the 
 equator, 68.9 to 69.05 miles in middle latitudes, and 69.30 to 69.34 miles 
 in the polar regions. The mean or average length is as stated in the 
 table. A degree of longitude is greatest at the equator, where it is 
 69.16 miles, and it gradually decreases toward the poles, where it is 0. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. In 2 mi. 4 fur. 32 rd. 
 2 yd. how many inches ? 
 
 OPERATION. 
 
 2 mi. 4 fur. 32 rd. 2yd. 
 8 
 
 20fiir. 
 40 
 
 ~832rd. 
 
 5* 
 416 
 4162 
 
 4578yd. 
 3 
 
 13734ft. 
 12 
 
 2. In 164808 inches how 
 many miles? 
 
 OPERATION. 
 12) 164808 in. 
 3)13734 ft. 
 
 54)4578 yd. 
 _2j 2 
 
 Tl ) 9156 
 
 4|0)83|2rd. + f yd. = 2 yd. 
 8)20fur.-f-32rd. 
 ~~2 mi. + 4 fur. 
 Ans. 2 mi. 4 fur. 32 rd. 2 yd. 
 
 164808 in., Ans. 
 
 3. The diameter of the earth being 7912 miles, how many 
 inches is it? Ans. 501304320 inches. 
 
 4. In 168474 feet how many miles ? 
 
 5. In 31 mi. 7 fur. 10 rd. 3 yd., how many feet ? 
 
 6. If the greatest depth of the Atlantic telegraphic cable 
 from Newfoundland to Ireland be 2500 fathoms, how many 
 miles is it? Ans. 2 mi. 6 fur. 29 rd. 1 ft. 
 
168 
 
 REDUCTION. 
 
 7. If this cable be 2200 miles in length, and cost 10 cents 
 a foot, what was its whole cost? Ans. $1161600. 
 
 8. A pond of water measures 4 fathoms 3 feet 8 inches in 
 depth ; how many inches deep is it ? Ans. 332. 
 
 9. How many times will the driving wheels of a locomo- 
 tive turn round in going from Albany to Boston, a distance of 
 200 miles, supposing the wheels to be 18 ft. 4 inches in cir- 
 cumference? Ans. 57600 times. 
 
 10. If a vessel sail 120 leagues in a day, how many stat- 
 ute miles does she sail ? Ans. 414. 
 
 11. How many inches high is a horse that measures 14J- 
 hands? Ans. 58. 
 
 SURVEYORS' LONG MEASURE. 
 
 194. A Gunter's Chain, used by land surveyors, is 4 rods 
 or 66 feet long, and consists of 100 links. 
 
 TABLE. 
 
 * 
 
 7.92 inches (in.) make 1 link, 1. 
 
 25 links " 1 rod, rd. 
 
 4 rods, or 66 feet, " 1 chain., ch. 
 
 80 chains " 1 mile, . . mi. 
 
 UNIT EQUIVALENTS. 
 
 1. in. 
 
 rd. 1 = 7.92 
 
 ch . 1 = 25 i= 198 
 
 mi. 1 4 = 100 = 792 
 1 = 80 = 320 = 8000 = 63360 
 
 SCALE ascending, 7.92, 25, 4, 80 ; descending, 80, 4, 25, 7.92. 
 
 NOTE. Rods are seldom used in chain measure, distances being 
 taken in chains and links. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. In 3 mi. 51 ch. 73 1. how many links? 
 
 2. Reduce 29173 1. to miles. 
 
 3. A certain field, enclosed by a board fence, is 17 ch. 31 1. 
 long, and 12 ch. 87 1. wide ; how many feet long is the fence 
 which encloses it ? Ans. 3983.76ft. 
 
 Repeat the table of surveyors' long measure. Give the scale. 
 
COMPOUND NUMBERS. 
 
 169 
 
 12 in. 
 
 1ft. 
 
 II. SQUARE MEASURE. 
 
 A Square is a figure having four equal sides, and 
 four equal angles or corners. 
 
 1 square foot is a figure having four 
 sides of 1 ft. or 12 in. each, as shown 
 in the diagram. Its contents are 12 
 X 12 ^=: 144 square inches. Hence 
 
 The contents or area of a square, or 
 of any other figure having a uniform 
 length and a uniform breadth, is found 
 by multiplying the length by the breadth. 
 Thus, a square foot is 12 in. long and 12 in. wide, and the con- 
 tents are 12 X 12 = 144 square inches. A board 20 in. long 
 and 10 in. wide, is a rectangle, containing 20 X 10 = 200 
 square inches. 
 
 1OO . Square Measure is used in computing areas or sur- 
 faces ; as of land, boards, painting, plastering, paving, &c. 
 
 TABLE. 
 
 144 square inches (sq. in.) make 1 square foot, marked sq. ft. 
 9 square feet 1 square vard, " so. vd. 
 
 12 in. = 1 ft. 
 
 y square feet 
 301 square yards 
 40 square rods 
 
 4 roods 
 640 acres 
 
 1 7 
 
 square yard, 
 i square rod, 
 1 rood, 
 1 acre, 
 1 square mile, 
 
 R. 
 
 UNIT EQUIVALENTS 
 
 sq. yd. 
 
 1 = 
 
 30*- = 
 1210 = 
 
 272| = 
 10390 = 
 
 sq. yd. 
 sq. rcl. 
 
 R. 
 
 A. 
 
 " sq. mi. 
 
 sq. in. 
 
 144 
 1296 
 39204 
 1568 160 
 
 sq.mi. 1= 4= 160= 4840= 43560= 6272640 
 1 = 640 = 2560 = 102400 = 3097600 = 27878400 = 4014489600 
 SCALE ascending, 144, 9, 301, 40, 4, 640; descending, 640, 4, 
 40, 301, 9, 144. 
 
 Define a square. How is the area of a square or any rectangular 
 figure found ? For what is square measure used ? Repeat the table. 
 Give the scale. 
 
170 REDUCTION. 
 
 Artificers estimate their work as follows : 
 
 By the square foot : glazing and stone-cutting. 
 
 By the square yard : painting, plastering, paving, ceiling, and 
 paper-hanging. 
 
 By the square of 100 feet: flooring, partitioning, roofing, slating, 
 and tiling. 
 
 Brick-laying is estimated by the thousand bricks; also by the 
 square yard, and the square of 100 feet. 
 
 NOTES. 1. In estimating the painting of moldings, cornices, &c., the 
 measuring-line is carried into all the moldings and cornices. 
 
 2. In estimating brick-laying by the square yard or the square of 
 100 feet, the work is understood to be 1-^ bricks, or 12 inches, thick. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. In 10 A. 1 R. 25 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in. 
 how many square inches ? 
 
 OPERATION. 
 
 10 A. 1 R. 25 sq. rd. 16 sq.yd. 4 sq. ft. 136 sq. in. 
 
 4 
 
 41 R. 
 40 
 
 1665sq.rd. 
 30j 
 
 416J 
 49966 
 
 50382J sq. yd. 
 _ 9_ 
 
 453444i sq.ft. 
 144 
 
 36 = sq. ft, 
 1813912 with 136 sq.in. 
 1813776 
 453444 
 
 65296108 sq. in., Ans. 
 2. In 65296108 sq. in. how many acres? 
 
 How do artisans estimate work ? 
 
 
COMPOUND NUMBERS. 171 
 
 OPERATION. 
 
 144 ) 65296108 sq. in. 
 
 9 ) 453445 sq. ft. + 28 sq. in. 
 30f l 50382 sq. yd. + 7 sq. ft. 
 
 121 ) 201 528 fourths sq. yd. 
 
 4|0)166|5 sq. rd. + ^=15} sq. yd. 
 4)41R.-f-25sq.rd. 
 
 10 A. + 1 R. 
 
 Ans. 10 A. 1 R. 25 sq. rd. lof sq. yd. 7 sq. ft. 28 sq. in. 
 
 C 10 A. 1 R. 25 sq. rd. 15 sq. yd. 7 sq. ft. 28 sq. in. 
 
 Or 3 6 sq. ft. 108 sq. in. 
 
 Or 10 A. 1 R. 25 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in. 
 
 ANALYSIS. Dividing by the numbers in the ascending scale, and 
 arranging the remainders according to their order in a line below, 
 we find the square yards a mixed number, 15-|. But % of a sq. yd. 
 | of 9 sq. ft. = 6| sq. ft. ; and f of a sq. ft. = f of 144 sq. in. = 
 108 sq. in. Therefore ^ sq. yd. := 6 sq. ft. 108 sq. in. ; and adding 
 108 sq.in. to 28 sq. in. we have 136 sq. in., and 6 sq. ft. to 7 sq.ft. we 
 have 13 sq. ft. = 1 sq. yd. 4 sq. ft., and writing the 4 sq. ft. in the 
 result, and adding 1 sq. yd. to 15 sq. yd. we have for the reduced 
 result, 10 A. 1 R. 25 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in. 
 
 3. Reduce 87 A. 2 R. 38 sq. rd. 7 sq. yd. 1 sq. ft. 100 sq. 
 in. to square inches. 4ns. 550355068 sq. in. 
 
 4. Reduce 550355068 square inches to acres. 
 
 5. A field 100 rods long and 30 rods wide contains how 
 many acres ? Ans. 18 A. 3 R. 
 
 6. How many rods of fence will enclose a farm a mile 
 square? 4ns. 1280 rods. 
 
 7. How much additional fence will divide it into four equal 
 square fields ? 4ns. 640 rods. 
 
 8. How many acres of land in Boston, at $1 a square foot, 
 will $100000 purchase ? 
 
 Ans. 2 A. 1 R. 7 sq. rd. 9 sq. yd. 3J s.q. ft. 
 
 9. How many yards of carpeting, 1 yd. wide, will be required 
 to carpet a room 18 ft. long and 16 ft. wide ? 4ns. 32| yd. 
 
172 REDUCTION. 
 
 10. What would be the cost of plastering a room 18 ft. long, 
 16 ft. wide, and 9 ft. high, at 22 cts. a sq. yd. ? Ans. $22.44. 
 
 11. What will be the expense of .slating a roof 40 feet 
 long and each of the two sides 20 feet wide, at $10 per 
 square? Ans. $160. 
 
 SURVEYORS' SQUARE MEASURE. 
 
 1OT. This measure is used by surveyors in computing the 
 area or contents of land. 
 
 TABLE. 
 
 625 square links (sq. 1.) make 1 pole, P. 
 
 16 poles " 1 square chain,, .sq. ch. 
 
 10 square chains " 1 acre, A. 
 
 640 acres " 1 square mile,. ..sq. mi. 
 
 36 square miles (6 miles square) " 1 township, Tp. 
 
 UNIT EQUIVALENTS. 
 
 P. sq. 1. 
 
 sq. Ch. 1 = 625 
 
 A. 1 = 16 ~ 1000 
 
 sq. mi. 1 = 10 = 160 10000 
 
 Tp. 1 = 640 = 6100 = 102400 =. 64000000 
 1 = 36 = 23040 =. 230400 = 3686400 = 2304000000 
 SCALE ascending, 625, 16, 10, 640, 36; descending, 36, 640, 
 10, 16, 625. 
 
 NOTES. 1. A square mile of land is also called a section. 
 
 2. Canal and railroad engineers commonly use an engineers' chain, 
 which consists of 100 links, each 1 foot long. 
 
 3. The contents of land are commonly estimated in square miles, 
 acres, and hundredths ; the denomination, rood, is fast going into dis- 
 use. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. How many poles in a township of land ? 
 
 2. Reduce 3686400 P. to sq. mi. 
 
 3. In 94 A. 7 sq. ch. 12 P. 118 sq. 1. how many square 
 links? 
 
 4. What will be the cost of a farm containing 4550000 
 square links, at $50 per acre ? Ans. $2275. 
 
 Repeat the table of surveyors' square measure. Give the scale. 
 
COMPOUND IS UMBERS. 173 
 
 III. CUBIC MEASURE. 
 
 198. A Cube is a solid, or body, 
 having six equal square sides, or 
 faces. If each side of a cube be 1 
 yard, or 3 feet, 1 foot in thickness 
 of this cube will contain 3X3X1 
 = 9 cubic feet, and the whole cube will 
 contain 3 X 3 X 3 27 cubic feet. 
 
 olt. =1 yd. O 
 
 A solid, or body, may have the 
 
 three dimensions all alike or all different. A body 4 ft. long, 
 3 ft. wide, and 2 ft, thick contains 4 X 3 X 2 = 24 cubic or 
 solid feet. Hence we see that 
 
 The cubic or solid contents of a body are found by multiply- 
 ing the length, breadth, and thickness together. 
 
 199. Cubic Measure, also called Solid Measure, is used 
 in estimating the contents of solids, or bodies ; as timber, wood, 
 stone, &c. 
 
 TABLE. 
 
 1728 cubic inches (cu. in.) make 1 cubic foot, cu. ft. 
 
 27 cubic feet " 1 cubic yard, cu. yd. 
 
 16 cubic feet " 1 cord foot, cd. ft. 
 
 8 cord feet, or ? (t l d f d ^ 
 128 cubic feet, $ 
 
 rt/l ,./,, .. - ( perch of stone ? -^ , 
 
 24} cubic feet 1 | g masonrV) \ Pch. 
 
 SCALE ascending, 1728, 27. The other numbers are not in a 
 regular scale, but are merely so many times 1 foot. The unit 
 equivalents, being fractional, are consequently omitted. 
 
 NOTES. 1. A cubic yard of earth is called a load. 
 
 2. Railroad and transportation companies estimate light freight by 
 the space it occupies in cubic feet, and heavy freight by weight. 
 
 3. A pile of wood 8 feet long, 4 feet wide, and 4 feet high, contains 
 1 cord ; and a cord foot is 1 foot in length of such a pile. 
 
 4. A perch of stone or of masonry is 16 feet long, 1^ feet wide, and 
 1 foot high. 
 
 Define a cube. How are the contents of a cube or rectangular 
 solid found? For what is cubic measure used? Repeat the table. 
 Give the scale. How is railroad freight estimated ? What is under- 
 stood by a cord foot ? By a perch of stone or masonry ? 
 
174 REDUCTION. 
 
 5. Joiners, bricklayers, and masons make no allowance for windows, 
 doors, &c. Bricklayers and masons, in estimating their work by 
 cubic measure, make no allowance for the corners of the walls of 
 houses, cellars, &c., but estimate their work by the girt, that is, the 
 entire length of the wall on the outside. 
 
 6. Engineers, in making estimates for excavations and embankments, 
 take the dimensions with a line or measure divided into feet and deci- 
 mals of a foot. The estimates are made in feet and decimals, and the 
 results are reduced to cubic yards. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. In 125 cu. ft. 840 cu. in. how many cu. in. ? Ans. 216840. 
 
 2. Reduce 5224 cubic feet to cords. Ans. 40|f. 
 
 3. In a solid, 3 ft. 2 in. long, 2 ft. 2 in. wide, and 1 ft. 8 in. 
 thick, how many cubic inches? Ans. 19760. 
 
 4. How many small cubes, 1 inch on each edge, can be 
 sawed from a cube 6 feet on each edge, allowing no waste for 
 sawing? Ans. 373248. 
 
 5. In a pile of wood 60 feet long, 20 feet wide, and 15 feet 
 high, how many cords ? Ans. 140-|. 
 
 6. How many cubic feet in a load of wood 10 feet long, 3| 
 feet wide, and 3 feet high ? Ans. 113| cu. ft. 
 
 7. If a load of wood be 12 feet long and 3 feet wide, how 
 high must it be to make a cord ? Ans. 3f ft. high. 
 
 8. The gray limestone of Central New York weighs 175 
 pounds a cubic foot. What is the weight of one solid yard ? 
 
 Ans. 2 T. 7 cwt. 25 Ib. 
 
 9. A cellar wall, 32 ft. by 24 ft., is 6 ft, high and 1 ft. thick. 
 How much did it cost at $1.25 a perch? Ans. $50.909+ 
 
 10. How much did it cost to dig the same cellar, at 15 
 cents a cubic yard ? AM. $25.60. 
 
 11. My sleeping room is 10 ft. long, 9 ft. wide, and 8 ft. high. 
 If I breathe 10 cu. ft. of air in one minute, in how long a time will 
 I breathe as much air as the room contains ? Ans. 72 min. 
 
 12. In a school room 30 ft. long, 20 ft. wide, and 10 ft. high, 
 with 50 persons breathing each 10 cu. ft, of air in one minute, 
 in how long a time will they breathe as much as the room 
 contains? Ans. 12 min. 
 
 How are excavations and embankments measured ? 
 
COMPOUND NUMBERS. 175 
 
 MEASURES OF CAPACITY. 
 
 I. LIQUID MEASURE. 
 
 194. Liquid Measure, also called Wine Measure, is used 
 in measuring liquids ; as liquors, molasses, water, &c. 
 
 TABLE. 
 
 4 gills (gi.) make 1 pint, pt. 
 
 2 pints " 1 quart, qt. 
 
 4 quarts " 1 gallon, gal. 
 
 31J gallons 1 barrel, . .. ..bbl. 
 
 2 barrels, or 63 gal. " 1 hogshead,, .hhd. 
 
 UNIT EQUIVALENTS. 
 
 pt. gi. 
 
 qt. 1 4 
 
 g al. 1=2= 8 
 
 bbi. 1 4 = 8 = 32 
 
 hhd . 1 = 3U 126 = 252 = 1008 
 
 1 = 2 = 63 = 252 504 = 2016 
 
 SCALE ascending, 4, 2, 4, 31, 2 ; descending, 2, 31, 4, 2, 4. 
 
 The following denominations are also in use : 
 
 36 gallons make 1 barrel of beer. 
 
 54 " or 1^ barrels " 1 hogshead " " 
 
 42 " " 1 tierce. 
 
 2 hogsheads, or 120 gallons, " 1 pipe or butt. 
 
 2 pipes or 4 hogsheads, " 1 tun. 
 
 NOTES. 1. The denominations, barrel and hogshead, are used in es- 
 timating the capacity of cisterns, reservoirs, vats, &c. 
 
 2. The tierce, hogshead, pipe, butt, and tun are the names of casks, 
 and do not express any fixed or definite measures. They are usually 
 gauged, and have their capacities in gallons marked on them. 
 
 3. Ale or beer measure, formerly used in measuring beer, ale, and 
 milk, is almost entirely discarded. 
 
 What is liquid measure ? Repeat the table. Give the scale. What 
 other denominations are sometimes used ? How are the capacities of 
 cisterns, reservoirs, &c., reckoned ? Of large casks ? 
 
176 
 
 EEDUCTION. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. In 2 hhd. 1 bar. 30 gal. 2 
 qt. 1 pt. 3 gi. how many gills ? 
 
 OPERATION. 
 2 hhd. 1 bar. 30 gal. 2 qt. 
 
 2 
 
 5bbl. 
 
 [lpt.3gi. 
 
 185 
 
 187^ gal. 
 4 
 
 752 qt. 
 _2 
 
 1505pt. 
 4 
 
 2. In 6023 gi. how many 
 hhds. ? 
 
 OPERATION. 
 
 4 ) 6023 gi. 
 2 ) 1505 pt. + 3 gi. 
 4 ) 752 qt. + 1 pt. 
 
 188 gal. 
 
 j 2 
 
 63 )376 
 
 [gal. 
 
 2 hhd. + 1 bar. 
 Ans. 2 hhd. 1 bar. 30 gal. 
 
 1 pt. 3 gi. 
 
 But J- gal. = 2 qt., making 
 the Ans. 2 hhd. 1 bar. 30 gal. 
 
 2 qt. 1 pt. 3 gi. 
 
 6023 gi., Ans. 
 
 3. Reduce 3 hogsheads to gills. 
 
 4. Reduce 6048 gills to hogsheads. 
 
 5. In 13 hhd. 15 gal. 1 qt. how many pints? 
 6.. In 6674 pints how many hogsheads ? 
 
 7. What will be the cost of a hogshead of wine, at 6 cents 
 a gill? Ans. $120.96. 
 
 8. A grocer bought 10 barrels of cider, at $2 a barrel; 
 after converting it into vinegar, he retailed it all at 5 cents a 
 quart ; how much was his whole gain ? Ans. $43. 
 
 9. At 6 cents a pint, how much molasses can be bought for 
 $3.84? Ans. 8 gal. 
 
 10. How many demijohns, that will contain 2 gal. 2 qt. 1 pt. 
 each, can be filled from a hogshead of wine ? Ans. 24. 
 
 II. DRY MEASURE. 
 
 UOo. Dry Measure is used in measuring articles not 
 liquid, as grain, fruit, salt, roots, ashes, &c. 
 
 What is dry measure ? 
 
COMPOUND NUMBERS. 177 
 
 TABLE. 
 
 2 pints (pt.) make 1 quart, .......... qt. 
 
 8 quarts " 1 peck, ......... pk. 
 
 4 pecks " 1 bushel,. bu. or bush. 
 
 UNIT EQUIVALENTS. 
 qt. pt. 
 
 pk. 1 = 2 
 bu . 1 = 8 = 16 
 1 = 4 = 3 = 64 
 
 SCALE ascending, 2, 8, 4 ; descending, 4, 8, 2. 
 
 NOTE. In England, 8 bu. of 70 Ibs. each are called a quarter, used in 
 measuring grain. The weight of the English quarter is of a long ton. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. In 49 bu. 3 pk. 7 qt. 1 pt. how many pints ? 
 
 2. In 3199 pt. how many bushels ? 
 
 3. Reduce 1 bu. 1 pk. 1 qt. 1 pt. to pints. 
 
 4. Reduce 83 pints to bushels. 
 
 5. An innkeeper bought a load of 50 bushels of oats at 65 
 cents a bushel, and retailed them at 25 cents a peck ; how 
 much did he make on the load ? Ans. $17.50. 
 
 STANDARD OF EXTENSION. 
 
 2O2* The U. S. standard unit of measures of extension, 
 whether linear, superficial, or solid, is the yard of 3 feet, or 36 
 inches, and is the same as the imperial standard yard of 
 Great Britain. It is determined as follows : The rod of a 
 pendulum vibrating seconds of mean time, in the latitude of 
 London, in a vacuum, at the level of the sea, is divided into 
 391393 equal parts, and 360000 of these parts are 36 inches, 
 or 1 standard yard. Hence, such a pendulum rod is 39.1393 
 inches long, and the standard yard is f f ^$$$ of the length of 
 the pendulum rod. 
 
 2O 3. The U. S. standard unit of liquid measure is the old 
 English wine gallon, of 231 cubic inches, which is equal to 
 8.339 pounds avoirdupois of distilled water at its maximum 
 density, that is, at the temperature of 39.83 Fahrenheit, the 
 barometer at 30 inches. 
 
 Repeat the table. What is a quarter ? What is the U. S. standard 
 unit of measurement of extension ? How is it determined ? What is 
 the U.S. standard unit of liquid measure ? 
 
178 REDUCTION. 
 
 2O 4. The U. S. standard unit of dry measure is the Brit- 
 ish Winchester bushel, which is 18 inches in diameter and 8 
 inches deep, and contains 2150.42 cubic inches, equal to 
 77.6274 pounds avoirdupois of distilled water, at its maximum 
 density. A gallon, dry measure, contains 268.8 cubic inches. 
 
 NOTE. 1. The wine and dry measures of the same denomination 
 are of different capacities. The exact and the relative size of each may 
 be readily seen by the following 
 
 2O5. COMPARATIVE TABLE OF MEASURES OF CAPACITY. 
 
 Cu. in. in Cu. in. in Cu. in. in Cu. in. in 
 one gallon. one quart. one pint. one gill. 
 
 Wine measure, 231 57| 28J 7, 
 
 Dry measure, (\ pk.,) 268f 67j 33-f 
 
 2. The beer gallon of 282 inches is retained in use in a few places 
 only by custom. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. A fruit dealer bought a bushel of strawberries, dry 
 measure, and sold them by wine measure ; how many quarts 
 did he gain ? Am. 5f quarts. 
 
 2. A grocer bought 40 quarts of milk by beer measure, and 
 sold it by wine measure ; how many quarts did he gain ? 
 
 Ans. 8f |- quarts. 
 
 3. A bushel, or 32 quarts, dry measure, contains how many 
 more cubic inches than 32 quarts wine measure ? 
 
 Ans. 302f cu. in. 
 TIME. 
 
 2O6. Time is used in measuring periods of duration, as 
 years, days, minutes, &c. 
 
 TABLE. 
 
 60 seconds (sec.) make 1 minute, min. 
 
 60 minutes " 1 hour, h. 
 
 24 hours 
 7 days 
 
 365 days 
 
 366 days 
 
 12 calendar months 
 
 1 day, da. 
 
 1 week, wk. 
 
 1 common year,. . .yr. 
 
 1 leap year, yr. 
 
 1 year, yr. 
 
 100 years " 1 century, C. 
 
 What is the IT. S. standard unit of dry measure ? How is it ob- 
 taiiK d r 'What is tlu: relative si/o of the wine and the dry gallon? 
 What is the ,si/e of a beer gallon ? What is time ? Repeat the table. 
 
COMPOUND NUMBERS. 
 
 179 
 
 wk. 
 1 : 
 
 yr. mo. 
 1 = 12 : 
 
 UNIT EQUIVALENTS. 
 
 min. sec. 
 
 b. 1 = 60 
 
 da. 1 = 60 3600 
 
 I 24 = 1440 = 86400 
 
 7 = 168 = 10080 = 604800 
 
 365 = 8760 = 525600 = 31536000 
 
 366 =i 8784 = 527040 = 31622400 
 
 SCALE ascending, 60, 60, 24, 7, 4; descending, 4, 7, 24, 60, 60. 
 The calendar year is divided as follows : 
 
 No. of mo. 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 10 
 11 
 12 
 
 Season. 
 
 Winter, 
 
 
 
 Spring, 
 
 Summer, 
 u 
 
 Autumn, 
 
 Winter, 
 
 Names. 
 
 Abbreviations. 
 
 January, 
 
 Jan. 
 
 February, 
 
 Feb. 
 
 March, 
 
 Mar. 
 
 April, 
 
 Apr. 
 
 rfay, 
 
 
 June, 
 
 Jun. 
 
 July, 
 
 
 
 August, 
 
 Aug. 
 
 (T September, 
 
 Sept. 
 
 < October, 
 
 Oct. 
 
 d November, 
 
 Nov. 
 
 December, 
 
 Dec. 
 
 No. of days. 
 
 31 
 
 28 or 29 
 
 31 
 
 30 
 
 31 
 
 30 
 
 31 
 
 31 
 
 30 
 
 31 
 
 30 
 
 31 
 
 365 or 366 
 
 NOTES. 1. The exact length of a solar year is 365 da. 5 h. 48 min. 46 
 sec. ; but for convenience it is reckoned 11 min. 14 sec. more than this, 
 or 365 da. 6 h. = 365^ da. This ^ day in 4 years makes one day, 
 which, every fourth, bissextile, or leap year, is added to the shortest 
 month, giving it 29 days. The leap years are exactly divisible by 4, 
 as 1856, 1860, 1864. The number of days in each calendar month 
 may be easily remembered by committing the following lines : 
 
 " Thirty days hath September, 
 April, June, and November ; 
 All the rest have thirty-one, 
 Save February, which alone 
 Hath twenty-eight; and one day more 
 We add to it one year in four." 
 
 2. In most business transactions 30 days are called 1 month. 
 EXAMPLES FOR PRACTICE. 
 
 1. Reduce 365 da. 5 h. 48 min. 46 sec. to seconds. 
 
 2. Reduce 31556926 seconds to days. 
 
 Give the scale. What is the length of each of the calendar months ? 
 What is the exact length of a solar year r Explain the use of bissextile 
 or leap year. What is the length of a month in business transactions ? 
 
180 REDUCTION. 
 
 3. In 5 wk. 1 da. 1 h. 1 min. 1 sec. how many seconds ? 
 
 4. In 3114061 seconds how many weeks? 
 
 5. How many times does a clock pendulum, 3 ft. 3 in. long, 
 beating seconds, vibrate in one day? Ans. 86400. 
 
 6. If a man take 1 step a yard long in a second, in how 
 lono- a time will he walk 10 miles ? Ans. 4 h. 53 min. 20 sec. 
 
 O 
 
 7. In a lunar month of 29 da. 12 h. 44 min. 3 sec. how 
 many seconds? Ans. 2551443. 
 
 8. How much time will a person gain in 40 years, by rising 
 45 minutes earlier every day ? Ans. 456 da. 13 h. 30 min. 
 
 CIRCULAR MEASURE. 
 
 SOT. Circular Measure, or Circular Motion, is used prin- 
 cipally in surveying, navigation, astronomy, and geography, 
 for reckoning latitude and longitude, determining locations of 
 places and vessels, and computing difference of time. 
 
 Every circle, great or small, is divisible into the same num- 
 ber of equal parts, as quarters, called quadrants, twelfths, 
 called signs, 360ths, called degrees, &c. Consequently the 
 parts of different circles, although having the same names, are 
 of different lengths. 
 
 TABLE. 
 
 60 seconds (") make 1 minute, . . . '. 
 
 60 minutes " 1 degree, . . . . 
 
 30 degrees " 1 sign, S. 
 
 12 signs, or 360, " 1 circle, C. 
 
 UNIT EQUIVALENTS. 
 
 { _- fo 
 
 s. 1 = GO = 3600 
 
 C. 1 = 30 = 1800 = 108000 
 1 = 12 = 360 = 21600 = 1296000 
 SCALE ascending, 60, 60, 30, 12; descending, 12, 30, 60, 60. 
 
 NOTES. 1. Minutes of the earth's circumference are called geo- 
 graphic or nautical miles. 
 
 2. The denomination, signs, is confined exclusively to Astronomy. 
 
 Define circular measure. How are circles divided ? Repeat the 
 table. Give the scale. What is a geographic mile? What is a 
 feign.? 
 
COMPOUND NUMBERS. 181 
 
 3. Degrees are not strictly divisions of a circle, but of the space 
 about a point in any plane. 
 
 4. 90 make a quadrant, or right angle, and 60 a sextant, or ^ of a 
 circle. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Reduce 10 S. 10 10' 10" to seconds. 
 
 2. Reduce 1116610'' to signs. 
 
 3. How many degrees in 11400 geographic or nautical 
 miles? Ans. 190. 
 
 4. If 1 degree of the earth's circumference is 69 statute 
 miles, how many statute miles in 11400 geographic miles, or 
 190 degrees? Ans. 13148. 
 
 5. How many minutes, or nautical miles, in the circum- 
 ference of the earth? Ans. 21 GOO 7 or mi. 
 
 6. A ship during 4 days' storm at sea changed her longitude 
 397 geographical miles ; how many degrees and minutes did 
 she change ? Ans. 6 37'. 
 
 IN COUNTING. 
 
 12 units or things. . . .make. . . .1 dozen. 
 
 12 dozen " 1 gross. 
 
 12 gross " 1 great gross. 
 
 20 units " 1 score. 
 
 3O9. PAPER. 
 
 24 sheets ........ make ........ 1 quire. 
 
 20 quires " 1 ream. 
 
 2 reams " 1 bundle. 
 
 5 bundles " 1 bale. 
 
 S1O. BOOKS. 
 
 The terms folio, quarto, octavo, duodecimo, Sec., indicate 
 the number of leaves into which a sheet of paper is folded. 
 A sheet folded in 2 leaves is called a folio. 
 
 A sheet folded in 4 leaves 
 A sheet folded in 8 leaves 
 A sheet folded in 12 leaves 
 A sheet folded in 16 leaves 
 A sheet folded in 1 8 leaves 
 A sheet folded in 24 leaves 
 A sheet folded in 32 leaves 
 
 a quarto, or 4to. 
 
 an octavo, or 8vo. 
 
 a 12mo. 
 
 a 16mo. 
 
 an ISmo. 
 
 a 24mo. 
 
 a 32mo. 
 
 What is a degree ? Repeat the table for counting. For reckoning 
 paper. For indicating the size of books. 
 
182 REDUCTION. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. If in Birmingham, England, 150 million Gillott pens are 
 manufactured annually, how many great gross will they make ? 
 
 Ans. 86805 great gross 6 gross 8 dozen. 
 
 2. In 100000 sheets of paper, how many bales ? 
 
 Ans. 20 bales 4 bundles 6 quires 16 sheets. 
 
 3. What is the age of a man 4 score and 10 years old ? 
 
 4. How many printed pages, 2 pages to each leaf, will 
 there be in an octavo book, having 8 fully printed sheets ? 
 
 Ans. 128 pages. 
 
 5. How large a book will ten 32mo. sheets make, if every 
 page be printed ? Ans. 640 pages. 
 
 PROMISCUOUS EXAMPLES IN REDUCTION. 
 
 1. How many suits of clothes, each containing 6 yd. 3 qr., 
 can be cut from 333 yards of cloth ? Ans. 48. 
 
 2. A man bought a gold chain, weighing 1 oz. 15 pwt., at 
 seven dimes a pennyweight; what did it cost? Ans. $24.50. 
 
 3. A physician, having 2 Ib 3 55 19 10 gr. of medicine, 
 dealt it out in prescriptions averaging 15 grains each ; how 
 many prescriptions did it make ? Ans. 886. 
 
 4. A man bought 1 T. 11 cwt. 12 Ibs. of hay, at 1 cents 
 a pound ; what did it cost ? Ans. $38.90. 
 
 5. What will be the cost of a load of oats weighing 1456 
 pounds, at 37 cents per bushel? Ans. $17.0625. 
 
 6. If one bushel of wheat will make 45 pounds of flour, how 
 many barrels will 1000 bushels make ? Ans. 229-bbl. 116 Ib. 
 
 7. A load of wheat weighing 2430 pounds is worth how 
 much, at $1.20 a bushel? Ans. $48.60. 
 
 8. Paid $12.50 for a barrel of beef; how much was that 
 per pound ? Ans. 6 cents. 
 
 9. If a silver dollar measure one inch in diameter, how 
 many dollars, laid side by side on the equator, would reach 
 round the earth ? Am. 1573862400. 
 
 10. In 10 mi. 7 fur. 4 ch. 70 1., how many links ? 
 
 Ans. 87470 links. 
 
DENOMINATE FRACTIONS. 183 
 
 11. What is the value of a city lot, 25 feet wide and 100 
 feet long, if every square inch is worth one cent? Ans. $3600. 
 
 12. How many cords of wood can be piled in a shed 50 ft. 
 long, 25 ft. wide, and 10 ft. high ? Ans. 97 Cd. 5 cd. ft. 4 cu. ft. 
 
 13. A cistern 10 feet square and 10 feet deep, will hold 
 how many hogsheads of water? Ans. 118 hhd. 46^ gal. 
 
 14. A bin 8 feet long, 5 feet wide, and 4J feet high, will 
 hold how many bushels of grain ? Ans. 144 T 9 j bu. 
 
 15. How many seconds less in every Autumn than in 
 either Spring or Summer? Ans. 86400 sec. 
 
 16. If a person could travel at the rate of a second of dis- 
 tance in a second of time, how much time would he require to 
 travel round the earth? Ans. 15 days. 
 
 17. How many yards of carpeting, 1 yd. wide, will be re- 
 quired to carpet a room 20 ft. long and 18 ft. wide ? Ans. 40. 
 
 18. A printer calls for 4 reams 10 quires and 10 sheets of 
 paper to print a book ; how many sheets does he call for ? 
 
 Ans. 2170. 
 
 19. How many times will a wheel, 16 ft. 6 in. in circumfer- 
 ence, turn round in running 42 miles ? Ans. 13440. 
 
 20. How many days, working 10 hours a day, will it re- 
 quire for a person to count $10000, at the rate of one cent 
 each second ? Ans. 27 da. 7h. 46 min. 40 sec. 
 
 21. A town, 6 miles long and 4^ miles wide, is equal to 
 how many farms of 80 acres each ? Ans. 216. 
 
 22. At $21.75 per rod, what will be the cost of grading 
 10 mi. 176 rds. of road ? Ans. $73428. 
 
 REDUCTION OF DENOMINATE FRACTIONS. 
 
 CASE I. 
 
 211. To reduce a denominate fraction from a 
 greater to a less unit. 
 
 1. Reduce -fa of a bushel to the fraction of a pint. 
 Case I is what ? 
 
184 REDUCTION. 
 
 OPERATION. 
 
 J x X f X f =V> -^ 5 - ANALYSIS. To reduce bushels 
 
 Q to pints, we must multiply by 4, 
 
 ,- ' 8, and 2, the numbers in the 
 
 $0 1 scale. And since the given num- 
 
 4 ber is a fraction of a bushel, we 
 
 <g indicate the process as in multi- 
 
 g plication of fractions, and after 
 canceling, obtain -f, the Answer.. 
 
 4 = pt., Ans. Hence, 
 
 RULE. Multiply the fraction of the higher denomination by 
 the numbers in the scale successively, between the given and the 
 required denominations. 
 
 NOTE. Cancellation may be applied wherever practicable. 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce T ^\7Q- of a to the fraction of a penny. 
 
 Ans. % d. 
 
 3. Reduce T^^ of a week to the fraction of a minute. 
 
 Ans. -j 7 ^ min. 
 
 4. What part of a gill is ^W of a hosghead ? Ans. ^ gi. 
 
 5. What fraction of a grain is ^ of an ounce ? Ans. gr. 
 
 6. Reduce yuutjuzjiy f a m ^ e t the fraction of an inch. 
 
 Ans. T^fr^s in. 
 
 7. Reduce f of of 2 pounds to the fraction of an ounce 
 Troy. Ans. f oz. 
 
 8. Reduce ^<y of a hogshead to the fraction of a pint. 
 
 Ans. ff pt. 
 
 9. Reduce y^^ of an acre to the fraction of a rod. 
 
 Ans. $ rd. 
 
 CASE II. 
 
 212. To reduce a denominate fraction from a less 
 to a greater unit. 
 
 1. Reduce | of a pint to the fraction of a bushel. 
 
 Give explanation. Rule. Case II is what ? 
 
DENOMINATE FRACTIONS. 185 
 
 OPERATION. 
 
 ANALYSIS. To reduce 
 _^1_1J . pints to bushels, we must 
 
 5 2~84~80' ^ divide by 2 ' 8 ' and 4 ' the 
 
 numbers of the scale. And 
 
 Or, 5 
 
 2 
 8 
 
 80 
 
 1 bu., Am. 
 
 since the given number of 
 pints is a fraction,, we indi- 
 cate the process, as in divis- 
 ion of fractions, and cancel- 
 ing, obtain -fa, the Answer* 
 
 RULE. Divide the fraction of the lower denomination by the 
 numbers in the scale, successively, between the given and the 
 required denomination. 
 
 NOTE. The operation will frequently be shortened by cancellation. 
 EXAMPLES FOR PRACTICE. 
 
 2. What part of a rod is ^ of a foot ? Ans. y^ rd. 
 
 3. What part of a pound is -f of a dram ? Ans. T / F ^ lb. 
 
 4. Reduce of a cent to the fraction of an eagle. 
 
 Ans. Yifa, E. 
 
 5. A hand is of a foot ; what fraction is that of a mile? 
 
 6. Reduce ^ of 2 pwt. to the fraction of a pound. Ans. ^| w lb. 
 
 7. How much less is f of a pint than of a hogshead ? 
 
 Ans. tJJhhd. 
 
 8. In f of an inch what fraction of a mile ? Ans. TTr5 ^ ff <j- mi. 
 
 9. f of an ounce Troy is f of what fraction of 2 pounds ? 
 10. f of an ounce is of what fraction of 2 pounds Troy? 
 
 CASE III. 
 
 S13. To reduce a denominate fraction to integers 
 of lower denominations. 
 
 1. What is the value of f of a hogshead of wine ? 
 
 Give explanation. Rule. Case LTI is what ? 
 
186 
 
 REDUCTION. 
 
 OPERATION. 
 
 f hhd. X 63 = -2|JL gal. = 39 J gal. 
 
 f gal. X 4 = -tf qt. = If qt ; f qt. X 2 = f pt, ~ 1 pt. 
 
 Ans. 39 gal. 1 qt. 1 pint. 
 
 ANALYSIS, f hhd. | of 63 gal., or 39| gal. ; and f gal. f of 
 4 qt., or If qt. ; and f qt. = of 2 pt., or 1 pt. Hence, 
 
 RULE. I. Multiply the fraction by that number in the scale, 
 which will reduce it to the next lower denomination, and if the 
 result be an improper fraction, reduce it to a whole or mixed 
 number. 
 
 II. Proceed with the fractional part, if any, as before, 
 until reduced to the denominations required. 
 
 III. The units of the several denominations, arranged in 
 their order, will be the required result. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce f- of a month to lower denominations. 
 
 Ans. 17 da. 3 h. 25 min. 42f- sec. 
 
 3. What is the value of f of a ? Ans. 8s. 6 d. 3f far. 
 
 4. What is the value off of a bushel ? 
 
 5. Reduce f of 15 cwt. to its equivalent value. 
 
 Ans. 12 cwt. 85 Ibs. 11 oz. 6f dr. 
 
 6. Reduce f of f of a pound avoirdupois to integers. 
 
 Ans. 4oz. llff dr. 
 
 7. What is the value of f of an acre ? Ans. 3 R. 13 P. 
 
 8. Reduce f of a day to its value in integers. 
 
 Ans. 16 h. 36 min. 55^ sec. 
 
 9. What is the value of f of a pound Troy ? 
 
 10. What is the value of of f>i tons? Ans. 4 T. 5 cwt. 5 5$ Ib. 
 
 1 1. What is the value of of 3 acres ? Ans. 1 A. 1 R. 20 P. 
 
 CASE IV. 
 
 SI 4. To reduce a compound number to a fraction 
 of a higher denomination. 
 
 1. What part of a week is 5 da. 14 h. 24 min. ? 
 Give explanation. Rule. Case IV is what ? 
 
DENOMINATE FRACTIONS. 187 
 
 OPERATION. ANALYSIS. To find 
 
 5 da. 14 h. 24 min. = 8064 min. what part one compound 
 
 1 wk. = 10080 min. number is of another, 
 
 8964 4 w k An* they must be reduced to 
 
 TOO B a- 5 w the same denomination. 
 
 In 5 da. 14 h. 24 min. there are 8064 minutes, and in 1 week there 
 are 10080 minutes. Since 1 minute is yo^o ^ a wee k 8064 min- 
 utes is iVoVo I of a week. Hence, 
 
 RULE. Reduce the given number to its lowest denomination 
 for the numerator, and a unit of the required denomination 
 to the same denomination for the denominator of the required 
 fraction. 
 
 NOTE. If the given number contain a fraction, the denominator of 
 this fraction must be regarded as the lowest denomination. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What part of a mi. is 6 fur. 26 rd. 3 yd. 2 ft. ? Ans. mi. 
 3; What fraction of a is 13 s. 7 d. 3 far. ? 
 
 4. Reduce 10 oz. lOpwt. 10 gr. to the fraction of a pound 
 Troy. Ans. %$% Ib. 
 
 5. Reduce 2 cd. ft. 8 en. ft. to the fraction of a cord. 
 
 Ans. T \ Cd. 
 
 6. Reduce 1 bbl. 1 gal. 1 qt. 1 pt. 1 gi. to the fraction of a 
 hogshead. Ans. J hhd. 
 
 7. What part of 2 rods is 4 yards 1 feet? Ans. -/^. 
 
 8. Reduce If pecks to the fraction of a bushel. Ans. f bu. 
 
 9. What part of 9 feet square are 9 square feet ? 
 
 10. From a piece of cloth containing 8 yd. 3 qr. a tailor cut 
 2 yd. 2 qr. ; what part of the whole piece did he take ? Ans. f. 
 
 CASE v. 
 
 215. To reduce a denominate decimal to integers 
 of lower denominations. 
 
 1 . Reduce .78125 of a pound Troy to integers of lower de- 
 nominations. 
 
 Give explanation. Rule. Case V is what ? 
 
188 REDUCTION. 
 
 OPERATION. ANALYSIS. We first multiply 
 
 .78125 Ib. bv 12 to reduce the given number 
 
 12 from pounds to ounces, and the 
 
 result is 9 ounces and the decimal 
 
 9.37500 oz. .375 O f an oz. We then multiply 
 
 20 this decimal by 20 to reduce it to 
 
 7.50000 pwt. pennyweights, and get 7 pwt. and 
 
 24 .5 of a pwt. This last decimal we 
 
 multiply by 24, to reduce it to 
 
 12.0000 gr. grains, and the result is 12 gr. 
 
 9 oz. 7 pwt. 12 gr., Ans. f* e the answer is 9 oz ' 7 ? wt 
 
 RULE. I. Multiply the given decimal by that number in the 
 scale which will reduce it to the next lower denomination, and 
 point off as in multiplication of decimals. 
 
 II. Proceed with the decimal part of the product in the same 
 manner until reduced to the required denominations. The in- 
 tegers at the left will be the answer required. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the value of .217 ? Ans. 13' 1.2". 
 
 3. What is the value of .659 of a week ? 
 
 Ans. 4 da. 14 h. 42 min. 43.2 sec. 
 
 4. Reduce .578125 of a bushel to integers of lower denom- 
 inations. Ans. 2 pk. 2 qt. 1 pt. 
 
 5. Reduce .125 bbl. to integers of lower denominations. 
 
 Ans. 3 gal. 3 qt. 1 pt. 2 gi. 
 
 6. What is the value of .628125 ? 
 
 7. What is the value of .22 of a hogshead of molasses ? 
 
 Ans. 13 gal. 3 qts. 3.52 gi. 
 
 8. What is the value of .67 of a league ? 
 
 Ans. 2 mi. 3 rd. 1 yd. 3$ in. 
 
 9. What is the value of .42857 of a month ? 
 
 Ans. 12 da. 20 h. 34 min. 13 sec. 
 10. What is the value of .78875 of a long ton ? 
 
 Ans. 15 cwt. 3 qr. 2 Ib. 12.8 oz. 
 
 Give explanation. Rule. 
 
DENOMINATE FRACTIONS. 189 
 
 11. What is the value of 5.88125 acres ? Ans. 5 A. 3 R. 21 P. 
 
 12. Reduce .0055 T. to pounds. Ans. 11 Ib. 
 
 13. Reduce .034375. of a bundle of paper to its value in 
 lower denominations. Ans. 1 quire 9 sheets. 
 
 CASE VI. 
 
 216. To reduce a compound number to a decimal 
 of a higher denomination. 
 
 1. Reduce 3 pk. 2 qt. to the decimal of a bushel. 
 
 OPERATION. ANALYSIS. Since 8 quarts make 
 
 2.00 qt. 1 peck, and 4 pecks 1 bushel, there 
 
 ~~7~~~ will be | as many pecks as quarts 
 
 (183), and J as many bushels as 
 
 8 
 
 .8125 bu., Ans. pecks. 
 
 Or we may reduce 3 pk. 2 qt. to 
 
 Or, 3 pk. 2 qt. = 56 qt. the fract i on O f a bushel (as in 214), 
 1 bu. = 32 qt. an d we aav e |_| O f a bushel, which, 
 ff = .8125 bu., Ans. reduced to a decimal, equals .8125. 
 Hence the 
 
 RULE. Divide the lowest denomination given by that num- 
 ber in the scale which will reduce it to the next higher, and an- 
 nex the quotient as a decimal to that higher. Proceed in the 
 same manner until the whole is reduced to the denomination 
 required. Or, 
 
 Reduce the given number to a fraction of the required de- 
 nomination, and reduce this fraction to a decimal. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Reduce 3 qt. 1 pt. 1 gi. to the decimal of a gallon. 
 
 Ans. .90625 gal. 
 
 3. Reduce 10 oz. 13 pwt. 9 gr. to the decimal of a pound 
 Troy. Ans. .8890625 Ib. 
 
 4. Reduce 1.2 pints to the decimal of a hogshead. 
 
 Ans. .00238 + hhd. 
 
 5. What part of a bushel is 3 pk. 1,12 qt. ? Ans. .785 bu. 
 
 Case VI is what ? Give explanations. Rule. 
 
190 ADDITION. 
 
 6. What part of an acre is 3 R. 12.56 P. ? 
 
 7. Reduce 17 yd. 1 ft. 6 in. to the decimal of a mile. 
 
 Ans. .00994318 + mi. 
 
 8. Reduce .32 of a pint to the decimal of a bushel. 
 
 Ans. .005 bu. 
 
 9. Reduce 4| feet to the decimal of a fathom. 
 
 Ans. .8125 fathom. 
 
 10. Reduce 150 sheets of paper to the decimal of a ream. 
 
 Ans. .3125 Rm. 
 
 11. Reduce 47.04 Ib. of flour to the decimal of a barrel. 
 
 12. Reduce .33 of a foot to the decimal of a mile. 
 
 13. Reduce 5 h. 36 min. 57 T 6 <y sec. to the decimal of a day. 
 
 ADDITION. 
 
 1. A miner sold at one time 10 Ib. 4oz. 16 pwt. 8 gr. 
 of gold ; at another time, 2 Ib. 9 oz. 3 pwt. ; at another, 1 1 oz. 
 20 gr. ; and at another, 25 Ib. 16 pwt. 23 gr. ; how much did 
 he sell in all ? 
 
 OPERATION. ANALYSIS. Arranging the num- 
 
 ib. oz. pwt. gr. hers in columns, placing units of the 
 
 10 4 16 8 same denomination under each oth- 
 
 2930 er > we fi rs t a dd the units in the 
 
 11 20 right hand column, or lowest de- 
 
 o 1 - f\ IP OQ nomination, and find the amount to 
 
 zo u ID zo - , . , . i n 
 
 be 51 grains, which is equal to 2 
 
 Ans. 39 1. 17 3 pwt. 3gr. We write the 3 gr. under 
 
 the column of grains, and add the 2 
 
 pwt. to the column of pwt. We find the amount of the second col- 
 umn to be 37 pwt., which is equal to 1 oz. 17 pwt. Writing the 17 
 pwt. under the column of pwt., we add the 1 oz. to the next column. 
 Adding this column in the same manner as the preceding ones, we 
 find the amount to be 25 oz., equal to 2 Ib. 1 oz. Placing the 1 oz. 
 under the column of oz., we add the 2 Ib. to the column of Ib. 
 Adding the last column, we find the amount to be 39 Ib. Hence 
 the following 
 
 What is addition of compound numbers ? Give explanation. 
 
COMPOUND NUMBERS. 191 
 
 RULE. I. Write the numbers so that those of the same unit 
 value will stand in the same column. 
 
 II. Beginning at the right hand, add each denomination as 
 in simple numbers, carrying to each succeeding denomination 
 one for as many units as it takes of the denomination added ', to 
 make one of the next higher denomination. 
 
 EXAMPLES FOR PRACTICE. 
 
 (2.) (3.) 
 
 . s. d. Ib. . 3. 9. gr. 
 
 48 13 8 12 8 7 2 15 
 
 51 6 4 10 4 1 10 
 
 67 11 3 15 00 2 1 19 
 
 76 18 10 11 6 12 
 
 244 10 1 13 4 4 2 00 
 
 (4.) (5.) 
 
 T. cwt. Ib. oz. dr. bu. pk. qt. pt. 
 
 4 7 18 4 10 1371 
 
 15 98 15 5 3220 
 
 3 9 10 6 15 161 
 
 10 15 04 17 051 
 
 9 12 42 11 2 45 2 4 
 
 6. What is the sum of 4 mi. 3 fur. 30 rd. 2 yd. 1 ft. 10 in., 
 5 mi. 6 fur. 18 rd. 1 yd. 2 ft. 6 in., 10 mi. 4 fur. 25 rd. 2 yd. 
 2 ft. 11 in., and 6 fur. 28 rd. 4 yd. 2 ft. 1 in. ? 
 
 7. Find the sum of 197 sq. yd. 4 sq.ft. 1 104 sq. in., 122 
 sq. yd. 2 sq. ft. 27 j sq. in., 5 sq. yd. 8 sq. ft. 2| sq. in., and 237 
 sq. yd. 7 sq. ft. 128 sq. in. ? 
 
 Ans. 563 sq. yd. 4 sq. ft. 118.825 sq. in. 
 
 NOTE. When common fractions occur, they should be reduced to a 
 common denominator, to decimals, or to integers of a lower denomi- 
 nation, and added according to the usual method. 
 
 Give the Rule. 
 
192 ADDITION. 
 
 
 
 A. 
 26 
 19 
 456 
 
 R. P. 
 
 3 28 
 2 38 
 2 20 
 
 (8.) 
 sq. yd. 
 25 
 30 
 16 
 
 sq. ft. sq. in. 
 8 125 
 7 150 
 6 98 
 
 503 
 
 1 8 
 
 12Q-) 5 85 
 (i) = *(*) 
 
 503 
 
 1 8 
 
 13 
 
 1 13 
 
 mi. fur. 
 1 7 
 
 (9.) 
 rd. yd. ft. 
 30 4 2 
 
 in. 
 11 
 
 hhd. 
 27 
 
 (10.) 
 
 gal. qt. 
 65 3 
 
 pt. 
 
 2 
 
 3 
 
 4 
 
 00 
 
 2 1 
 
 10 
 
 112 
 
 60 
 
 2 
 
 3 
 
 10 
 
 7 
 
 25 
 
 1 2 
 
 11 
 
 50 
 421 
 14 
 
 29 
 00 
 39 
 
 
 2 
 1 
 
 1 
 3 
 2 
 
 16 
 
 bu. 
 23 
 
 3 16 
 
 (11.) 
 
 pk. qt. 
 3 7 
 
 4 
 
 pt. 
 1 
 
 2 
 
 yr. 
 
 25 
 
 (12.) 
 da. h. 
 300 19 
 
 min. 
 54 
 
 sec. 
 35 
 
 34 
 
 2 
 
 
 
 1 
 
 21 
 
 40 
 
 12 
 
 40 
 
 24 
 
 42 
 
 3 
 
 5 
 
 
 
 3 
 
 112 
 
 14 
 
 15 
 
 17 
 
 51 
 
 1 
 
 4 
 
 1 
 
 6 
 
 19 
 
 11 
 
 45 
 
 59 
 
 23 
 11 
 
 
 3 
 
 3 
 4 
 
 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 57 
 
 109 11 37 16 
 
 bundles 1 ream 15 quires 
 bundles 1 ream 10 quires 
 13 sheets, how much does 
 
 13. If a printer 
 ' sheets of paper, 
 ' sheets, and the 
 
 one day use 4 
 the next day 3 
 next 2 bundles 
 
 10 
 
 he use in the three days ? 
 
 Am. 2 bales 1 ream 6 quires 19 sheets. 
 14. A tailor used, in one year, 2 gross 5 doz. 10 buttons, 
 another year 3 gross 7 doz. 9, and another year 4 gross 6 
 doz. 11 ; how many did he use in the three years? 
 
 Ans. 10 gross 8 doz. 6. 
 
COMPOUND NUMBERS. 193 
 
 15. A ship, leaving New York, sailed east the first day 3 
 45' 50"; the second day, 4 50' 10'' ; the third, 2 10' 55"; 
 the fourth, 2 39" ; how far was she then east from the place 
 of starting? Arts. 12 47' 34". 
 
 16. A man, in digging a cellar, removed 127 cu. yd. 20 cu. 
 ft. of earth ; in digging a drain, 6 cu. yd. 25 cu. ft. ; and in 
 digging a cistern, 17 cu. yds. 18 cu. ft. ; what was the amount 
 of earth removed, and what the cost at 16 cents a cu. yd.? 
 
 Ans. 152 cu. yds. ; $24.37. 
 
 17. A farmer received 80 cents a bushel for 4 loads of 
 corn, weighing as follows : 2564, 2713, 3000, and 3109 Ibs. ; 
 how much did he receive for the whole? Ans. $162.657-(- 
 
 18. A druggist sold for medicine, in three years, at an aver- 
 age price of 9 cents a gill, the following amounts of brandy, 
 viz. : 1 bbl. 4 gal. 1 pt. ; 30 gal. 2 qt, 1 gi. ; 2 bbl. 15 gal. ; 
 how much did he receive for the whole? Ans. $415.17. 
 
 218. To add denominate fractions. 
 
 1. Add |- of a mile to -J of a furlong. 
 
 OPERATION. ANALYSIS. We find the 
 
 mi. 6 fur. 26 rd. 11 ft. value of each fraction in in- 
 \ f _ 13 rd 5-L ft tegers of less denominations 
 
 (213), and then add their 
 Ans. 7 fur. 00 values as in compound num- 
 
 Or, J fur. -j- 8 = 2 V m>- bers ( 217 } ' 
 
 A mi. + 1 mi. = & mi. = 7 fur. . % we , . "* r . e<Juce the , 
 
 given fractions to fractions of 
 
 the same denomination (212), then add them, and find the value 
 of their sum in lower denominations (213). 
 
 2. Add f of a rod to J of a foot. Ans. 13 ft. 1 in. 
 
 3. What is the sum of of a mile, f of a furlong, and of 
 a rod ? Ans. 7 fur. 27 rd. 8 ft. 3 in. 
 
 4. What is the sum of f of a pound and f of a shilling ? 
 
 Ans. 13 s. 10 d. 2f qr. 
 
 5. What is the sum of | of a ton and ^ of 1 cwt. ? 
 
 Ans. 12 cwt. 42 Ib. 13f oz. 
 
 Give explanation of the process of adding denominate fractions. 
 Q 
 
194 SUBTRACTION. 
 
 6. What is the sum of |- of a day added to ^ an hour ? 
 
 Ans. 9 h. 30 min. 
 
 7. What is the sum of of a week, f of a day, and of 
 an hour ? Ans. 1 da. 22 h. 15 min. 
 
 8. Add f of a hhd. to f of a gal. 
 
 9. What is the sum of f- of a cwt., 8f lb., and 3 T 9 ^ oz. by 
 long ton table ? Ans. 73 lb. 1 oz. 3fi dr. 
 
 10. What is the sum of f of a mile, f of a yard, and f- of 
 a foot ? 
 
 11. Sold 4 village lots ; the first contained of of an 
 acre ; the second, 60 J rods ; the third, f of an acre ; and the 
 fourth, f of of an acre ; how much land in the four lots ? 
 
 Ans. 3 R. 26 P. 126 T 4 T 5 ? sq. ft. 
 
 12. A farmer sold three loads of hay ; the first weighed 
 1 T., the second, 1-^ T., and the third, 18f cwt. ; what was 
 the aggregate weight of the three loads ? 
 
 Ans. 3 T. 5 cwt. 91 lb. 10 oz. 
 
 SUBTRACTION. 
 
 219. 1. If a druggist buy 25 gal. 2 qt. 1 pt. 1 gi. of 
 wine, and sell 18 gal. 3qt. 1 pt. 2 gi., how much has he left? 
 
 OPERATION. ANALYSIS. Writing the subtrahend 
 
 gal. qt. .pt. gi. under the minuend, placing units of the 
 
 25 2 1 1 same denomination under each other, ' 
 
 18 3 2 we begin at the right hand, or lowest 
 
 A ~7> Q 3~ denomination ; since we cannot take 
 
 2 gi. from 1 gi., we add 1 pt. or 4 gi. to 
 
 1 gi., making 5 gi. ; and taking 2 gi. from 5 gi., we write the remain- 
 der, 3 gi., underneath the column of gills. Having added 1 pt. or 
 4 gi. to the minuend, we now add 1 pt. to the pt. in the subtra- 
 hend, making 1 pt. ; and 1 pt. from 1 pt. leaves pt., which we write 
 in the remainder. Next, as we cannot take 3 qt. from 2 qt., we add 
 1 gal. or 4 qt. to 2 qt., making 6 qt, and taking 3 qt. from 6 qt., we 
 write the remainder, 3 qt., under the denomination of quarts. Add- 
 ing 1 gal. to 18 gal., we subtract 19 gal. from 25 gal., as in simple 
 
 What is subtraction of compound numbers ? Give explanation. 
 
COMPOUND NUMBERS. 
 
 195 
 
 numbers, and write the remainder, 6 gal., under the column of gal- 
 lons. Hence the following 
 
 RULE. I. Write the subtrahend under the minuend, so that 
 units of the same denomination shall stand under each other. 
 
 II. Beginning at the right hand, subtract each denomination 
 separately, as in simple numbers. 
 
 III. If the number of any denomination in the subtrahend 
 exceed that of the same denomination in the minuend, add to 
 the number in the minuend as many units as make one of the 
 next higher denomination, and then subtract ; in this case add 
 1 to the next higher denomination of the subtrahend before 
 subtracting. Proceed in the same manner with each denomi- 
 nation. 
 
 EXAMPLES FOR PRACTICE. 
 
 (2.) (3.) 
 
 lb. oz. pwt. gr. A. B. P. 
 
 From 18 6 10 14 25 2 16.9 
 
 Take 10 5 4 6 19 3 25.14 
 
 Hem. 
 
 8168 
 
 5 2 31.76 
 
 
 W 
 
 
 
 
 (5.) 
 
 
 
 T. 
 
 cwt. 
 
 lb. 
 
 y r - 
 
 da. 
 
 h. 
 
 min. 
 
 sec. 
 
 14 
 
 11 
 
 69f 
 
 38 
 
 187 
 
 16 
 
 45 
 
 50 
 
 10 
 
 12 
 
 98-| 
 
 17 
 
 190 
 
 20 
 
 50 
 
 40 
 
 20 361 19 55 10 
 
 6. A Boston merchant bought English goods to the amount 
 of 4327 13s. 7d., and he paid 1374 10s. 11} d.; how 
 much did he then owe ? 
 
 7. From 300 miles take 198 mi. 7 fur. 25 rd. 2yd. 1ft. 
 10 in. Ans. 101 mi. 14 rd. 2 yd. 2 ft. 8 in. 
 
 8. What is the difference in the longitude of two places, 
 one 75 20' 30" west, and the other 71 19' 35" west? 
 
 Ans. 4 55". 
 
 9. From 10 ft 7 4 3 1 9 15 gr. take 3ft82329 
 18 gr. Ans. 6 ft 11 1 3 1 9 17 gr. 
 
 Give the Rule. 
 
196 SUBTRACTION. 
 
 10. The apparent periodic revolution of the sun is made in 
 365 da. 6 h. 9 min. 9 sec., and that of the moon in 29 da. 12 h. 
 44 mill. 3 sec. ; what is the difference ? 
 
 Ans. 335 da. 15 h. 25 min. 6 sec. 
 
 11. A man, having a hogshead of wine, drank, on an aver- 
 age, for five years, including two leap years, one gill of wine 
 a day ; how much remained ? Ans. 5 gal. 3 qt. 1 pt. 1 gi. 
 
 12. A section of land containing 640 acres is owned by 
 four men ; the first owns 196 A. 2 R. 16 P. ; the second, 200 
 A. 1 R. ; the third, 177 A. 36 P. ; how much does the fourth 
 own ? Ans. 65 A. 3 R. 7.75 P. 
 
 13. From a pile of wood containing 75 J Cd. was sold 
 at one time 16 Cd. 5 cd. ft.; at another, 24 Cd. 6cd.ft. 12 
 cu. ft. ; at another, 27 Cd. 112 cu. ft. ; how much remained in 
 the pile ? Ans. 6 Cd. 3 cd. ft. 4 cu. ft. 
 
 14. If from a hogshead of molasses 10 gal. 1 qt. 1 pt, be 
 drawn at one time, 15 gal. 1 pt. at another, and 14 gal. 3 qt. 
 at another, how much will remain ? 
 
 22O. To find the difference in dates. 
 
 1. What length of time elapsed from the discovery of 
 America by Columbus, Oct. 14, 1492, to the Declaration of 
 Independence, July 4, 1776 ? 
 
 FIRST OPERATION. ANALYSIS. We place the earlier date 
 
 yr. mo. da. unde the later, writing first on the left 
 
 177b 7 the number of the year from the Chris- 
 
 1492 10 14 tian era, next the number of the month, 
 
 2o o TT counting January as the first month, and 
 
 next the number of the day from the 
 
 first day of the month. Instead of the number of the year, month, 
 and day, some use the number of years, months, and days that 
 
 SECOND OPERATION. JmVe el P Sed sinCG tnC C<nristian era tnus : 
 
 vr> * wt "" fla * ' instead of saying July is the 7th month, 
 
 1775 6 3 we sa y ^ months and 3 days have 
 
 1491 13 elapsed, and instead of saying October 
 
 is the 10th month, we say 9 months and 
 
 8 20 13 days have elapsed. 
 
 How is the difference of dates found ? 
 
COMPOUND NUMBERS. 
 
 197 
 
 Both methods will obtain the same result ; the former is generally 
 used. 
 
 NOTES. 1. When hours are to be obtained, we reckon from 12 at 
 night, and if minutes and seconds, we write them still at the right of 
 hours. 
 
 2. In finding the time between two dates, or in computing interest, 
 12 months are considered a year, and 30 days a month. 
 
 When the exact number of days is required for any period 
 not exceeding one ordinary year, it may be readily found by 
 the following 
 
 TABLE, 
 
 Showing the number of days from any day of one month to the same day 
 of any other month within one year. 
 
 FROM ANY 
 DAY OP 
 
 TO THE SAME DAY OF THE NEXT. 
 
 Jan. 
 
 Feb. 
 
 Mar. 
 
 Apr. 
 
 May. 
 
 June 
 
 July 
 
 Aug. 
 
 Sept. 
 
 Oct. 
 
 Nov. 
 
 Dec. 
 
 January . . . 
 February . . 
 March .... 
 April .... 
 
 365 
 334 
 306 
 275 
 245 
 214 
 '184 
 153 
 122 
 92 
 61 
 31 
 
 31 
 365 
 337 
 306 
 276 
 245 
 215 
 184 
 153 
 123 
 92 
 62 
 
 59 
 28 
 365 
 334 
 304 
 273 
 243 
 212 
 181 
 151 
 120 
 90 
 
 90 
 59 
 31 
 365 
 335 
 304 
 274 
 243 
 212 
 182 
 151 
 121 
 
 120 
 89 
 61 
 30 
 365 
 334 
 304 
 273 
 242 
 212 
 181 
 151 
 
 151 
 
 120 
 92 
 61 
 31 
 365 
 335 
 304 
 273 
 243 
 212 
 182 
 
 181 
 150 
 122 
 91 
 61 
 30 
 365 
 334 
 303 
 273 
 242 
 212 
 
 212 
 181 
 153 
 122 
 92 
 61 
 31 
 365 
 334 
 304 
 273 
 243 
 
 243 
 212 
 184 
 153 
 123 
 92 
 62 
 31 
 365 
 335 
 304 
 274 
 
 273 
 242 
 214 
 183 
 153 
 122 
 92 
 61 
 30 
 365 
 334 
 304 
 
 304 
 273 
 245 
 214 
 184 
 153 
 123 
 92 
 61 
 31 
 365 
 335 
 
 334 
 303 
 275 
 244 
 214 
 183 
 153 
 122 
 91 
 61 
 30 
 365 
 
 May 
 
 June . . 
 
 July 
 
 August . . . 
 September . 
 October. . . . 
 November . 
 December. . 
 
 If the days of the different months are not the same, the 
 number of days of difference should be added when the earlier 
 day belongs to the month from which we reckon, and subtracted 
 when it belongs to the month to which we find the time. If 
 the 29th of February is to be included in the time computed, 
 one day must be added to the result. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. George Washington was born Feb. 22, 1732, and died 
 Dec. 14 1799 ; what was his age ? Ans. 67 yr. 9 mo. 22 da. 
 
 How can the number of days, if less than a year, be obtained ? 
 Q* 
 
198 SUBTRACTION. 
 
 3. How much time has elapsed since the declaration of 
 independence of the United States ? 
 
 4. How many years, months, and days from your birthday 
 to this date ; or what is your age ? 
 
 5. How long from the battle of Bunker Hill, June 17, 1775, 
 to the battle of Waterloo, June 18, 1815 ? Am. 40 yr. 1 da. 
 
 6. What length of time will elapse from 20 minutes past 
 2 o'clock, P. M., June 24, 1856, to 10 minutes before 9 o'clock, 
 A. M., January 3, 1861 ? Ans. 4 yr. 6 mo. 8 da. 18 h. 30 min. 
 
 7. How many days from any day of April to the same day 
 of August? of December? of February? 
 
 8. How many days from the 6th of November to the 15th 
 of April? Ans. 160 days. 
 
 9. How many days from the 20th of August to the 15th 
 of the following June ? Ans. 299 days. 
 
 To subtract denominate fractions. 
 
 1. From -f of an oz. take % of a pwt. 
 
 OPERATION. ANALYSIS. We per- 
 
 il oz. =7 pwt. 12 gr. form the same reduc- 
 
 i p-wt. 21 gr. tions as in addition of 
 
 denominate fractions, 
 6 pwt. 15 gr., Ans. ( 218 ^ and then sub- 
 
 ' tract the less value from 
 
 Or, fo.X20=r^pwt. the greater. 
 
 Y I- */. pwt. = 6 pwt. 15 gr. 
 
 2. What is the difference between ^ rod and f of a foot ? 
 
 Ans. 7ft. 6 in. 
 
 3. From take of f of a shilling. 
 
 4. From f of a league take ^ of a mile. 
 
 Ans. 1 mi. 2 fur. 16rd. 
 
 5. From 8-ft- cwt. take 1 qr. 2f Ib. 
 
 Ans. 8 cwt, 2 qr. 14 Ib. 5 oz. 15^\ dr. 
 
 6. From J- of a week take of a day. 
 
 Ans. 1 da. 4 h. 48 min. 
 
 Give explanation of the process of subtracting denominate fractions. 
 
COMPOUND NUMBERS. 199 
 
 7. Two persons, A and B, start from two places 120 miles 
 apart, and travel toward each other ; after A travels f , and 
 B f , of the distance, how far are they apart ? 
 
 Ans. 41. mi. 7 fur. 9 rd. 8 ft. 7 in. 
 
 8. From a cask of brandy containing 96 gallons, - leaked 
 out, and f of the remainder was sold ; how much still remained 
 in the cask ? Ans. 25 gal. 2 qt. 3 gi. 
 
 MULTIPLICATION. 
 
 1. A farmer has 8 fields, each containing 4 A. 2 R. 
 27 P. ; how much land in all ? 
 
 OPERATION. ANALYSIS. In 8 fields are 8 times as much 
 
 A. R. P. land as in 1 field. We write the multiplier 
 
 4 2 27 under the lowest denomination of the mul~ 
 
 8 tiplicand, and proceed thus; 8 times 27 P- 
 
 - 77~ are 216 P., equal to 5 R. 16 P. ; and we 
 
 write the 16 P. under the number multiplied. 
 
 Then 8 times 2 R. are 16 R., and 5 R. added make 21 R., equal to 
 
 4 A. 1 R. ; and we write the 1 R. under the number multiplied. 
 
 Again, 8 times 4 A are 32 A., and 4 A. added make 36 A., which we > 
 
 write under the same denomination in the multiplicand, and the 
 
 work is done. Hence, 
 
 RULE. I. Write the multiplier under the lowest denomina- 
 tion of the multiplicand. 
 
 II. Multiply as in simple numbers, and carry as in addi- 
 tion of compound numbers. 
 
 EXAMPLES FOR PRACTICE. 
 
 (2.) 
 
 bu. pk. qt. pt. 
 
 4251 
 
 (3.) 
 
 mi. fur. rd. ft. 
 
 9 4 20 13 
 
 2 
 
 6 
 
 9130 
 
 57 3 4 12 
 
 Multiplication of compound numbers, how performed ? Rule. 
 
200 MULTIPLICATION. 
 
 (4.) (5.) 
 
 . s. d. Ib. oz. pwt. gr. 
 
 5 18 4 3 4 22 
 
 4 7 
 
 (6.) (7.) 
 
 T. cwt. Ib. oz. 
 
 14 16 48 12 13 10' 35" 
 
 11 9 
 
 8. In 6 barrels of grain, each containing 2 bu. 3 pk. 5 qt., 
 how many bushels ? Am. 17 bu. 1 pk. 6 qt. 
 
 9. If a druggist deal out 3 Ib 4 1 3 2 9 16 gr. of med- 
 icine a day, how much will he deal out in 6 days ? 
 
 10. If a man travel 29 mi. 3 fur. 30 rd. loft, in 1 day, 
 how far will he travel in 8 days ? 
 
 11. If a woodchopper can cut 3 Cd. 48 cu. ft, of wood in 1 
 day, how many cords can he cut in 12 days ? Ans. 404 Cd. 
 
 12. What is the weight of 48 loads of hay, each weighing 
 1 T. 3 cwt. 50 Ib. ? 
 
 OPERATION. ANALYSIS. When the multi- 
 
 T - cwt - lb - plier is large, and a composite 
 
 1 3 50 number, we may multiply by one 
 
 6 of the factors, and that product 
 
 "^ \ by the other. Multiplying the 
 
 / 1 00 weight of 6 loads. . , , ,, .. , , , 
 
 weight of 1 load by 6, we obtain 
 
 the weight of 6 loads, and the 
 
 56 8 00 weight of 48 loads. weight of 6 loads multiplied by 
 
 8, gives the weight of 48 loads. 
 
 13. If 1 acre of land produce 45 bu. 3 pk. 6 qt. 1 pt. of 
 corn, how much will 64 acres produce ? Ans. 2941 bu. 
 
 14. How much will 120 yards of cloth cost, at 1 9 s. 8 d. 
 per yard ? 
 
 15. If $80 will buy 4 A. 3 R. 26 P. 20 sq. yd. 3 sq. ft. of 
 land, how much will $4800 buy ? Ans. 295 A. 10 sq.yd. 
 
 16. If a load of coal by the long ton weigh 1 T. 6 cwt. 2 qr. 
 26 lb. 10 oz., what will be the weight of 73 loads ? 
 
 Ans. 97 T. 11 cwt. 3 qr. 11 lb. 10 oz. 
 
COMPOUND NUMBERS. 201 
 
 17. The sun, on an average, changes his longitude 59' 8.33" 
 per day ; how much will be the change in 365 days ? 
 
 18. If 1 pt. 3 gi. of wine fill 1 bottle, how much will be re- 
 quired to fill a great gross of bottles of the same capacity ? 
 
 DIVISION, 
 
 223. 1. If 4 acres of land produce 102 bu. 3 pk. 2 qt. of 
 
 wheat, how much will 1 acre produce ? 
 
 OPERATION. ANALYSIS. One acre will produce ^ 
 
 pt. bu. P k. qt. pts. as muc h as 4 acres. Writing the divi- 
 
 4 ) 102 3 2 _ sor on the left of the dividend, we divide 
 
 25261 ^ ku. by 4, and we obtain a quotient of 
 
 25 bu., and a remainder of 2 bu. We 
 
 write the 25 bu. under the denomination of bushels, and reduce the 
 2 bu. to pecks, making 8 pk., and the 3 pk. of the dividend added 
 makes 11 pk. Dividing 11 pk. by 4, we obtain a quotient of 2 pk. 
 and a remainder of 3 pk. ; writing the 2 pk. under the order of 
 pecks, we next reduce 3 pk. to quarts, adding the 2 qt. of the 
 dividend, making 26 qt., which divided by 4 gives a quotient of 6 qt. 
 and a remainder of 2 qt. Writing the 6 qt. under the order of 
 quarts, and reducing the remainder, 2 qt., to pints, we have 4 pt., 
 which divided by 4 gives a quotient of 1 pt, which we write under 
 the order of pints, and the work is done. 
 
 2. A farmer put 132 bu. OPERATION. 
 
 1 pk. of apples into 46 barrels ; bn - P k - 
 
 how many bu. did he put into 46 ) 132 
 a barrel? _?_ 
 
 40 
 
 4 
 
 When the divisor is large, and 
 not a composite number, we di- ( P 
 
 vide by long division, as shown 
 in the operation. From these 23 
 
 examples we derive the 
 
 184(4qt. 
 
 1^1 Am. 2bu.3pk.4qt. 
 
 Explain the process of dividing compound numbers. 
 
202 DIVISION. 
 
 RULE. I. Divide the highest denomination as in simple 
 numbers, and each succeeding denomination in the same man- 
 ner, if there be no remainder. 
 
 II. If there be a remainder after dividing any denomina- 
 tion, reduce it to the next lower denomination, adding in the 
 given number of that denomination, if any, and divide as be- 
 fore. 
 
 III. The several partial quotients will be the quotient re- 
 quired. 
 
 NOTES. 1. When the divisor is large and is a composite number, 
 we may shorten the work by dividing by the factors. 
 
 2. When the divisor and dividend are both compound numbers, they 
 must both be reduced to the same denomination before dividing, and 
 then the process is the same as in simple numbers. 
 
 EXAMPLES FOR PRACTICE. 
 
 (3.) (4.) 
 
 . s. d. T. cwt. Ib. 
 
 5 ) 25 8 4 7 ) 45 15 25 
 
 518 6 10 75 
 
 (5.) (6.) 
 
 wk. da. h. min. 
 
 4 )3 5 22 _ 00 10 ) 25 42' 40" 
 
 ~~6 17 30 2 34 16 
 
 7. Bought 6 large silver spoons, which weighed 11 oz. 3 pwt. ; 
 what was the weight of each spoon ? 
 
 8. A man traveled by railroad 1 000 miles in one day ; 
 what was the average rate per hour ? 
 
 Ans. 41 mi. 5 fur. 13 rd. 5 ft. 6 in. 
 
 9. If a family use 10 bbl. of flour in a year, what is the 
 average amount each day ? Ans. 5 Ib. 5 oz. 144& dr. 
 
 10. The aggregate weight of 123 hogsheads of sugar is 
 57 T. 19 cwt. 42 Ib. 14 oz. ; what is the average weight per 
 hogshead ? Ans. 9 cwt. 42 Ib. 10 oz. 
 
 11. How many times are 5 10 s. 10 d. contained in 537 
 10s. 10d.? _ Ans. 97. 
 
 Give the rule. When the divisor is a composite number, how may 
 \ve proceed? When the divisor and dividend are" both compound 
 numbers, how proceed ? 
 
COMPOUND NUMBERS. 203 
 
 12. A cellar 50 ft. long, 30 ft. wide, and 6 ft. deep was ex- 
 cavated by 5 men in 6 days ; how many cubic yards did each 
 man excavate daily? Arts. 11 cu. yd. 3 cu.ft. 
 
 13. If a town 5 miles square be divided equally into 150 
 farms, what will be the size of each farm ? 
 
 Ans. 106 A. 2 R. 26 P. 20 sq. yd. 1 sq.ft. 72 sq. in. 
 
 14. How many times are 4 bu. 3 pk. 2 qt. contained in 
 336 bu. 3pk. 4qt.? Ans. 70. 
 
 15. A merchant tailor bought 4 pieces of cloth, each con- 
 taining 60 yd. 2.25 qr. ; after selling of the whole, he made 
 up the remainder into suits containing 9 yd. 2 qr. each ; how 
 many suits did he make ? Ans. 17. 
 
 LONGITUDE AND TIME. c 
 
 SS54. Every circle is supposed to be divided into 360 
 equal parts, called degrees* 
 
 Since the sun appears to pass from east to west round the 
 earth, or through 360, once in every 24 hours, it will pass 
 through ^ of 360, or 15 of the distance, in 1 hour ; and 1 of 
 distance in y 1 ^ of 1 hour, or 4 minutes ; and V of distance in 
 fa of 4 minutes, or 4 seconds. 
 
 TABLE OF LONGITUDE AND TIME. 
 
 360 of longitude = 24 hours, or 1 day of time. 
 
 15 " " =1 hour " 
 
 1 " " = 4 minutes " " 
 
 1' " " =4 seconds " " 
 
 CASE I. 
 
 225. To find the difference of time between two 
 places, when their longitudes are given. 
 
 1. The longitude of Boston is 71 3', and of Chicago 87 
 30 ; ; what is the difference of time between these two places ? 
 
 Explain how distance is measured by time. Repeat the table of 
 longitude and time. Case I is what r 
 
204 LONGITUDE AND TIME. 
 
 OPERATION. ANALYSIS. By subtraction of 
 
 87 30' compound numbers we first find 
 
 yj g/ the difference of longitude be- 
 
 tween the two places, which is 
 
 16 27' 16 o 27/. Since 1 of longitude 
 
 ' 4 makes a difference of 4 minutes 
 
 , , Z '. 7Z" ., of time, and 1' of longitude a 
 
 1 h. o ram. 48 sec., Ans. ,._ , . 
 
 difference of 4 seconds of time, 
 
 we multiply 16 27'. the difference in longitude, by 4, and we obtain 
 the difference of time in minutes and seconds, which, reduced to 
 higher denominations, gives 1 h. 5 min. 48 sec., the difference in 
 time. Hence the 
 
 HULE. Multiply the difference of longitude in degrees and 
 minutes by 4, and the product will be the difference of time in 
 minutes and seconds, which may be reduced to hours. 
 
 NOTE. If one place be in east, and the other in west longitude, the 
 difference of longitude is found by adding them, and if the sum be 
 greater than 180, it must be subtracted from 360. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. New York is 74 1' and Cincinnati 84 24' west longi- 
 tude ; what is the difference of time ? Ans. 41 min. 32 sec. 
 
 3. The Cape of Good Hope is 18" 28' east, and the Sand- 
 wich Islands 155 J west longitude; what is the difference of 
 time ? Ans. 11 h. 33 min. 52 sec. 
 
 4. Washington is 77 V west, and St. Petersburg 30 
 19' east longitude ; what is their difference of time ? 
 
 Ans. 7 h. 8 min. 20 sec. 
 
 5. If Pekin is 118 east, and San Francisco 122 west 
 longitude, what is their difference of time ? 
 
 6. If a message be sent by telegraph without any loss of 
 time, at 12 M. from London, 0' longitude, to Washington, 
 77 1' west, what is the time of its receipt at Washington ? 
 
 NOTE. Since the sun appears to move from east to west, when it is 
 exactly 12 o'clock at one place, it will be past 12 o'clock at all places 
 east, and before 12 at all places west. Hence, knowing the difference 
 of time between two places, and the exact time at one of them, the 
 exact time at the other will be found by adding their difference to Jhe 
 given time, if it be east, and by subtracting if it be west. 
 
 Ans. 6 h. 51 min. 56 sec., A. M. 
 Give explanation. Rule. 
 
COMPOUND NUMBERS 
 
 7. A steamer arrives at Halifax, 63 3 1 
 P. M. ; the fact is telegraphed to St. 
 
 without loss of time ; what is the time of itF^^eipt at St. 
 Louis ? Ans. 2 h. 13 min. 24 sec., P. M. 
 
 8. If, at a presidential election, the voting begin at sunrise 
 and end at sunset, how much sooner will the polls open and 
 close at Eastport, Me., 67 west, than at Astoria, Oregon, 124 
 west ? Ans. 3 h. 48 min. 
 
 9. When it was 1 o'clock, A. M., on the first day of Jan- 
 uary, 1859, at Bangor, Me., 68 47' west, what was the 
 time at the city of Mexico, 99 5' west? 
 
 Ans. Dec. 31, 1858, 58 min. 48 sec. past 10, P. M. 
 
 CASE II. 
 
 226. To find the difference of longitude between 
 two places, when the difference of time is known. 
 
 1. If the difference of time between New York and Cincin- 
 nati be 41 min. 32 sec., what is the difference of longitude ? 
 
 OPERATION. ANALYSIS. Since 4 minutes of time 
 
 min. sec. make a difference of 1 of longitude, and 
 
 4 ) 41 32 4 seconds of time, a difference of I' of 
 
 ~~ ~~ . longitude, there will be 1 as many de- 
 
 ' 7 s ' grees of longitude as there are minutes 
 
 of time, and \ as many minutes of longitude as there are seconds of 
 
 time. Hence, 
 
 RULE. Reduce the difference of time to minutes and sec- 
 onds, and then divide by 4 ; the quotient will be the difference 
 in longitude, in degrees and minutes. 
 
 2. What is the difference of longitude between the Cape 
 of Good Hope and the Sandwich Islands, if the difference of 
 time be 11 h. 33 min. 52 sec. ? Ans. 173 28'. 
 
 3. What is the difference of longitude between Washington 
 and St. Petersburg, if their difference of time be 7 h. 8 min. 
 20 sec.? Ans. 107 20'. 
 
 Case II is what ? Give explanation. Rule. 
 R 
 
206 DUODECIMALS. 
 
 4. When it is half past 4, P. M., at St. Petersburg, 30 19' 
 east, it is 32 min. 36 sec. past 8, A. M., at New Orleans, west; 
 what is the difference of longitude ? Am. 119 J 21' 
 
 5. The longitude of New York is 74 1' west. A sea cap- 
 tain leaving that port for Canton, with New York time, finds 
 that his chronometer constantly loses time. What is his longi- 
 tude when it has lost 4 hours ? 8 h. 40 min. ? 13 h. 25 min. ? 
 
 Ans. 14 1' west; 55 59' east; 127 14' east. 
 
 6. When the days are of equal length, and it is noon on 
 the 1st meridian, on what meridian is it then sunrise ? sun- 
 set ? midnight ? Ans. 90 west ; 90 east ; 180" east or west. 
 
 DUODECIMALS. 
 
 Duodecimals are the divisions and subdivisions of 
 a unit, resulting from continually dividing by 12, as 1, y 1 ^, T , 
 TrVs? & c ' ^ n practice, duodecimals are applied to the meas- 
 urement of extension, the foot being taken as the unit. 
 
 If the foot be divided into 12 equal parts, the parts are 
 called inches, or primes ; the inches divided by 12 give sec- 
 onds; the seconds divided by 12 give thirds; the thirds di- 
 vided by 12 give fourths ; and so on. 
 
 From these divisions of a foot it follows that 
 
 I 7 (inch or prime) ......... j s T L. o f a foot. 
 
 I" (second) or T ^ of ^, ..... ^ O f a foot. 
 
 V" (third) or T V of T V of T V, . . T7 V F of a foot, &c. 
 
 TABLE. 
 
 12 fourths, marked (""), make 1 third ........ marked V" 
 
 12 thirds i second, " 1" 
 
 12 seconds i prime, or inch, " V 
 
 12 primes, or inches, " 1 foot, ft. 
 SCALE uniformly 12. 
 
 The marks ', ", '", "", are called indices. 
 
 What are duodecimals ? To what applied ? Explain the divisions 
 of the foot. Repeat the table. 
 
COMPOUND NUMBERS. 207 
 
 NOTE. Duodecimals are really common fractions, and can always 
 be treated as such ; but usually their denominators are not expressed, 
 and they are treated as compound numbers. 
 
 ADDITION AND SUBTRACTION OF DUODECIMALS. 
 
 328. We add and subtract duodecimals the same as other 
 compound numbers. 
 
 EXAMPLES. 
 
 1. Add 13 ft. 4' 8", 10 ft. 6' 7", 145 ft. 9' II 7 '. - 
 
 Ans. 169ft. 9' 2". 
 
 2. Add 179ft. 11' 4", 245ft. 1' 4", 3ft. 9' 9". 
 
 Ans. 428ft. 10' 5". 
 
 3. From 25ft. 6' 3" take 14ft. 9' 8". Ans. 10ft. 8' 1". 
 
 4. From a board 15 ft. 7' 6" in length, 3 ft. 8' 11" were 
 sawed off; what was the length of the piece left? 
 
 Ans. 11 ft. 10' 7". 
 
 MULTIPLICATION OF DUODECIMALS. 
 
 229* Length multiplied by breadth gives surface, and 
 surface multiplied by thickness gives solid contents (108). 
 
 1. How many square feet in a board 11 feet 8 inches long 
 and 2 feet 7 inches wide ? 
 
 OPERATION. ANALYSIS. We first multiply by the 7'. 
 
 lift. 8' 7 twelfths times 8 twelfths equals 56 one 
 
 2 7' hundred forty-fourths, which equals 4 
 
 twelfths and 8 one hundred forty-fourths. 
 
 6 ft - We write the 8 144ths marked with two 
 
 23 4' indices to the rigRt, and add the 4 12ths 
 
 30ft 1' 8" ^ ^ e nex ^ P r duct. 7' times 11 equals 
 
 77', which added to 4' equals 81', equal to 
 
 6 feet and 9'. We write the 9' under the 
 
 inches, or 12ths, and the 6 under the feet, or units. 2 times 8' 
 equals 16', or 1 foot and 4'. We write the 4' under the 9', and 
 add the 1 foot to the next product. 2 times 1 1 feet are 22 feet, and 
 1 foot added make 23 feet, which we write under the 6 feet. Add- 
 
 How are duodecimals added and subtracted ? Give analysis of ex- 
 ample 1. 
 
208 DUODECIMALS. 
 
 ing these partial products, and we have 30 ft. V and 8" for the 
 entire product. 
 
 It will be seen from the above that the number of indices to every 
 product of any two factors is equal to the sum of the indices of those 
 factors ; thus 7' X & z= 56" ; 4' X 5"' = 20""'. Hence the 
 
 KULE. I. Write the several terms of the multiplier under 
 the corresponding terms of the multiplicand. 
 
 II. Multiply each term of the multiplicand by each term of 
 the multiplier, beginning with the lowest term in each, and call 
 the product of any two denominations the denomination denoted 
 by the sum of their indices, carrying I for every 12. 
 
 III. Add the partial products, carrying 1 for every 12; 
 their sum will be the required answer. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. How many square feet in a board 13ft. 9' long and 11' 
 wid e? Ans. 12ft. 7' 3". 
 
 3. How many square feet in a stock of 4 boards, each 11 ft. 
 9' long and 1 ft. 3' wide ? Ans. 58 ft. 9'. 
 
 4. How many square yards of plastering on the walls of a 
 room 12ft. 11' square, and 9ft. 3' high, allowing for two win- 
 dows and one door, each 6 ft. 2' high and 2 ft. 4' wide ? 
 
 Ans. 48 sq. yd. 2 ft. 9'. 
 
 5. How many solid feet in a mow of hay 30 ft. 4' long, 
 25ft. 6' wide, and 12ft. 5' high? Ans. 9604ft. 3' 6". 
 
 6. How many cords in a pile of wood 18ft. 6' long, 12ft. 
 wide, and 5 ft. 6' high ? Ans. 9 cords 69 ft. 
 
 7. How many cubic yards of earth must be removed in 
 digging a cellar 36 ft. 10' long, 22 ft. 3' wide, and 5 ft. 2' deep ? 
 
 Ans. 156cu.yd. 22ft. 3' 1" . 
 
 8. What would it cost to plaster a wall 32 ft, 8' long and 
 9 ft. high, at 17 cents per square yard ? Ans. $5.55 . 
 
 9. How many yards of carpeting, 27' wide, will be re- 
 quired to cover a floor 48 ft. long and 33 ft. 9' wide ? 
 
 Ans. 240 yards. 
 
 Give the rule. 
 
COMPOUND NUMBERS. 209 
 
 DIVISION OF DUODECIMALS. 
 
 23 O. 1. A flagstone, 3 ft, 9' wide, has a surface of 20 ft. 
 II 7 3" ; what is its length ? 
 
 OPERATION. ANALYSIS. We divide 
 
 3ft. 9' ) 20ft. 11' 3" ( 5 ft. 7'. the surface by the width 
 
 18 9' to obtain the length. The 
 
 divisor is something more 
 
 than 3 ft., and to obtain 
 
 2 2' 3 the first quotient figure, we 
 
 consider how many times 
 
 3ft. and something more is contained in nearly 21ft. (20ft. 11'); 
 we estimate it to be o times, and multiplying the divisor by this 
 quotient figure, we have 18ft. 9', which, subtracted from 20ft. 11', 
 leaves 2 ft. 2', to which we bring down 3", the last term of the divi- 
 dend. We next seek how many times the divisor is contained in 
 this remainder, and find by trial the quotient 7' ; multiplying the 
 divisor by this figure, we obtain 2 ft. 2' 3", and there is no remain- 
 der. Hence the 
 
 RULE. I. Write the divisor on the left hand of the dividend, 
 as in simple numbers. 
 
 II. Find the first term of the quotient either by dividing the 
 first term of the dividend by the first term of the divisor, or by 
 dividing the first two terms of the dividend by the first two 
 terms of the divisor ; multiply the divisor by this term of the 
 quotient, subtract the product from the corresponding terms of 
 the dividend, and to the remainder bring down another term of 
 the dividend. 
 
 III. Proceed in like manner till there is no remainder, or 
 till a quotient has been obtained sufficiently exact. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Divide 44 ft. 5' 4" by 16ft. 8'. Ans. 2 ft. 8'. 
 
 3. The square contents of a walk are 184 ft. 3', and the 
 length is 40 ft. 11' 4" ; what is the width? Ans. 4ft. 6'. 
 
 4. A blanket whose square contents are 14 ft. 6', is to be 
 lined with cloth 2 ft. 7' wide ; how much in length will be re- 
 quired ? 
 
 Give analysis of example 1. Rule. 
 
210 PROMISCUOUS EXAMPLES. 
 
 5. A block of granite contains 64 ft. 2' 5" ; its width is 
 2 ft. 6', and its thickness 3 ft. 7' ; what is its length ? 
 
 NOTE. Since the solid contents are the product of the three dimen- 
 sions, we divide the solid contents by any two dimensions or by their 
 product, to obtain the other dimension. 
 
 Ans. 7 ft. 2'. 
 
 PROMISCUOUS EXAMPLES. 
 
 1. In 115200 grains Troy, how many pounds? 
 
 2. In 365 da. 5 h. 48 niin. 46 sec., how many seconds ? 
 
 Ans. 31556928. 
 
 3. A man wishes to ship 1560 bushels of potatoes in bar- 
 rels containing 3 bu. 1 pk. each ; how many barrels will be 
 required ? Ans, 480. 
 
 4. Reduce 295218 inches to miles. 
 
 5. Reduce 456575 grains to pounds, apothecaries' weight. 
 
 Ans. 79 ib 3 15 IB 15 gr. 
 
 6. How many sheets in 3 reams of paper ? 
 
 7. What is the value of 4 piles of wood, each 20 ft. long, 6 ft. 
 wide, and 10 ft. high, at $3.25 per cord ? Ans. $121.87. 
 
 8. How many bottles, each holding 1 qt. 1 gi., can be filled 
 from a barrel of cider? Ans. 112. 
 
 9. At $26.40 per sq. rd. for land, what will be the cost of a 
 village lot 8 rd. long, and 4 rd. wide ? Ans. $980.10. 
 
 10. Divide 259 A. 1 R. 10 P. of land into 36 equal lots. 
 
 Ans. 7 A. 32J- P. 
 
 1 1 . How many times can a box holding 4 bu. 3 pk. 2 qt. be 
 filled from 336 bu. 3 pk. 4qt.? Ans. 70. 
 
 1 2. What is the value of .875 of a gallon ? 
 
 13. What part of a mile is 2 fur. 36 rd. 2 yd. ? Ans. &. 
 
 14. What part of 2 days is 13 h. 26 min. 24 sec. ? 
 
 15. From 26 A. 2 R. of land, 5 A. 3 R. were sold ; what 
 part of the whole piece remained unsold ? Ans. -f^. 
 
 ] 0. What is the difference between f of a pound sterling 
 and 5 pence? Ans. 11 s. 6 d. 
 
 1 7. What is the sum of | of a yard, | of a foot, and | of 
 an inch ? Ans. 7 inches. 
 
PROMISCUOUS EXAMPLES. 211 
 
 18. Reduce 3 cwt. Iqr. 7 Ib. of coal to the decimal of a 
 long ton. Ans. .165625. 
 
 19. Benjamin Franklin was born Jan. 18, 1706, and George 
 Washington Feb. 22, 1732 j how much older was Franklin 
 than Washington ? Ans. 26 yr. 1 mo. 4 da. 
 
 20. The longitude of Boston is 71 4' west, and that of 
 Chicago 87 30' west; when it is 12 M. at Boston, what is the 
 time in Chicago ? Ans. 10 h. 58 min. 16 sec. A. M. 
 
 21. If the difference of time between New York and New 
 Orleans be 1 h. 4 sec., what is the difference in longitude ? 
 
 Ans. 15 1'. 
 
 22. Add of a mile, of a furlong, and r 3 ? of a rod to- 
 gether. Ans. 5 fur. 33 rd. 8 ft. 3 in. 
 
 23. If a bushel of barley cost $.80, what will 20 bu. 3 pk. 
 6 qt. cost? Ans. $16.75. 
 
 24. What is the value of .875 of a gross ? Ans. 10 doz. 
 
 25. How many acres hi a field 56^- rods long, and 24.6 
 rods wide ? Ans. 8 A. 2 R. 29.9 P. 
 
 26. How many perches of masonry in the wall of a cellar 
 which is 20 feet square on the inside, 8 feet high, and 1 feet 
 in thickness ? Ans. 44.6+. 
 
 27. A, B. and C rent a farm, and agree to work it upon 
 shares ; they raise 640 bu. 3 pk. of grain, which they divide 
 as follows : one fourth is given for the rent ; of the remainder 
 A takes 10 bu. more than one third, after which B takes one 
 half of the remainder less 7 bushels, and C has what is left ; 
 how much is C's share ? Ans. 161 bu. 3 pk. 6 qt. 
 
 28. What is the value in Troy weight of 13 Ib. 8 oz. 11.4 dr. 
 avoirdupois weight ? Ans, 16 Ib. 5 oz. 10 pwt. 11.7 -f- gr- 
 
 29. . If 154 bu. 1 pk. 6 qt. cost $173.74, how much will 1.5 
 bushels cost? Ans. $1.687+. 
 
 30. What is the value of .0125 of a ton ? Ans. 25 Ibs. 
 
 31. What fraction of 3 bushels is T V of 2bu. 3 pk. ? 
 
 Ans. T 7 /^' 
 
 32. How many wine gallons in a water tank 4 feet long, 
 3 feet wide, and 1 ft. 8 in. deep ? Ans. 174^. 
 
212 PROMISCUOUS EXAMPLES. 
 
 33. How many bushels will a bin contain that is 7 feet 
 square, and 6 ft, 8 in. deep ? Ans. 301.336 + bu. 
 
 34. How much, must be paid for lathing and plastering 
 overhead a room 36 feet long and 20 feet wide, at 26 cents a 
 square yard ? 
 
 35. How many shingles will it take to cover the roof of a 
 building 46 feet long, each of the two sides of the roof being 
 20 feet wide, allowing each shingle to be 4 inches wide, and 
 to lie 5 inches to the weather ? Ans. 13248. 
 
 36. John Young was born at a quarter before 4 o'clock, A. 
 M., Sept. 4, 1836; what will be his age at half past 6 o'clock, 
 P. M., April 20, 1864 ? Ans. 27 yr. 7 mo. 16 da. 14 h. 45 min. 
 
 37. How many cubic yards of earth were removed in dig- 
 ging a cellar 28 ft. 9' long, 22 ft. 8' wide, and 7 ft. 6' deep ? 
 
 Ans. ISl-^ cu. yd. 
 
 38. What will 30 bu. 54 Ib. of wheat cost, at $1.37 per 
 bushel? Ans. $42.4875. 
 
 39. How many square yards of carpeting will it take to 
 .cover a floor 24 ft. 8' long and 18 ft. 6' wide ? Ans. 50f. 
 
 40. What is the cost of 54 bu. 8 Ib. of barley, at 84 cents 
 per bushel ? Ans. $45.50. 
 
 41. What is the depth of a lot that has 120 feet front, and 
 contains 18720 square feet? 
 
 42. How many steps of 30 inches each must a person 
 take in walking 21 miles? 
 
 43. How long will it require one of the heavenly bodies to 
 move through a quadrant, if it move at the rate of 3' 12" 
 per minute ? Ans. 1 da. 4 h. 7 min. 30 sec. 
 
 44. How many times will a wheel, 9 ft. 2 in. in circum- 
 ference, turn round in going 65 miles ? 
 
 45. If a man buy 10 bushels of chestnuts, at $5.00 per 
 bushel, dry measure, and sell the same at 22 cents per quart, 
 liquid measure, how much is his gain? Ans. $31.92. 
 
 46. What will it cost to build a wall 240 feet long, 6 feet 
 high, and 3 feet thick, at $3.25 per 1000 bricks, each brick 
 being 8 inches long, 4 inches wide, and 2 inches thick ? 
 
 Ans. $379.08. 
 
 
PERCENTAGE. 213 
 
 PERCENTAGE. 
 
 Per cent, is a term derived from the Latin words per 
 centum, and signifies by the hundred, or hundredths, that is, a cer- 
 tain number of parts of each one hundred parts, of whatever de- 
 nomination. Thus, by 5 per cent, is meant 5 cents of every 100 
 cents, $5 of every $100, 5 bushels of every 100 bushels, &c. 
 Therefore, 5 per cent, equals 5 hundredths =. .05 = T ^^ = -$. 
 8 per cent, equals 8 hundredths = .08 = T ^ =: ^. 
 
 232. Percentage is such a part of a number as is indi- 
 cated by the per cent. 
 
 233. The Base of percentage is the number on which 
 the percentage is computed". 
 
 334:. Since per cent, is any number of hundredths, it is 
 usually expressed in the form of a decimal; but it may be 
 expressed either as a decimal or a common fraction, as in the 
 
 following 
 
 TABLE. 
 
 Decimals. Common Fractions. Lowest Terms. 
 
 1 
 
 per 
 
 cent. 
 
 m 
 
 .01 
 
 n= 
 
 ToT 
 
 = 
 
 Too 
 
 2 
 
 per 
 
 cent. 
 
 M 
 
 .02 
 
 " 
 
 ToT 
 
 u 
 
 -fa 
 
 4 
 
 per 
 
 cent. 
 
 II 
 
 .04 
 
 " 
 
 ToT 
 
 u 
 
 A 
 
 5 
 
 per 
 
 cent. 
 
 u 
 
 .05 
 
 " 
 
 ToT 
 
 " 
 
 i 
 
 6 
 
 per 
 
 cent. 
 
 U 
 
 .06 
 
 " 
 
 6 
 
 TTO" 
 
 M 
 
 
 7 
 
 per 
 
 cent. 
 
 11 
 
 .07 
 
 " 
 
 To~o 
 
 " 
 
 ToT 
 
 8 
 
 per 
 
 cent. 
 
 " 
 
 .08 
 
 " 
 
 ToT 
 
 u 
 
 "^T 
 
 10 
 
 per 
 
 cent. 
 
 u 
 
 .10 
 
 u 
 
 TTo 
 
 " 
 
 TO" 
 
 16 
 
 per 
 
 cent. 
 
 II 
 
 .16 
 
 11 
 
 ITO 
 
 (t 
 
 ^ 
 
 20 
 
 per 
 
 cent. 
 
 u 
 
 .20 
 
 (I 
 
 T 2 A 
 
 " 
 
 i 
 
 25 
 
 per 
 
 cent. 
 
 " 
 
 .25 
 
 u 
 
 TrV 
 
 M 
 
 f 
 
 50 
 
 per 
 
 cent. 
 
 II 
 
 .50 
 
 u 
 
 TT 
 
 M 
 
 
 100 
 
 per 
 
 cent. 
 
 M 
 
 1.00 
 
 u 
 
 100 
 too 
 
 " 
 
 1 
 
 125 
 
 per 
 
 cent. 
 
 " 
 
 1.25 
 
 (i 
 
 m 
 
 M 
 
 f 
 
 4 
 
 per cent. " 
 
 .005 
 
 u 
 
 1000 
 
 U 
 
 
 f 
 
 per 
 
 cent. 
 
 " 
 
 .0075 
 
 u 
 
 To~o o o 
 
 (( 
 
 ToT 
 
 12* 
 
 per 
 
 cent. 
 
 || 
 
 .125 
 
 u 
 
 ToTT 
 
 u 
 
 A 
 
 16* 
 
 per 
 
 cent. 
 
 " 
 
 .1625 
 
 u 
 
 T 
 
 u 
 
 
 AMiat is meant by per cent. ? From what is the term derived ? 
 "What is percentage ? What is the base of percentage ? How is per 
 cent, expressed? 
 
214 PERCENTAGE. 
 
 EXAMPLES FOR PRACTICE. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Express decimally 3 per cent. ; 6 per cent. ; 9 per cent. ; 
 14 per cent. ; 24 per cent. ; 40 per cent. ; 112 per cent. ; 150 
 
 per cent. 
 
 r cent. 
 
 2. Express decimally 6J per cent, ; 8 j per cent, ; 33 J per 
 nt. ; 7 per cent. ; lOf per cent. ; 9| per cent, ; 103^- per 
 
 cent. ; 225 per cent. 
 
 3. Express decimally per cent. ; f per cent. ; per cent. ; 
 f per cent. ; per cent. ; 1| per cent. ; 2| per cent. ; 4 per 
 cent.; 5f per cent.; 7| per cent.; 12 per cent.; 25 f per 
 cent. 
 
 4. Express in the form of common fractions, in their lowest 
 terms, 6 per cent. ; 8 per cent. ; l2 per cent. ; 14 per cent. ; 
 18 J per cent. ; 21| per cent, ; 31^ per cent. ; 37 per cent. ; 
 40 f per cent. ; 112 per cent. ; 225 per cent. 
 
 CASE I. 
 
 23>. To find the percentage of any number. 
 
 1. A man, having $125, lost 4 per cent, of it; how many 
 dollars did he lose ? ' 
 
 OPERATION. 
 
 $125 ANALYSIS. Since 4 per cent, is 1^0-= .04, he lost 
 
 .04 of $125, or $125 X .04 = $5. Or, 4 per cent, 
 is ^ Jg., and -^ of $125 = $5. Hence the 
 
 RULE. Multiply the given number or quantity by the rate 
 per cent, expressed decimally, and point off as in decimals. Or, 
 
 Take such a part of the given number as the number ex- 
 pressing the rate is part of 100. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is 6 per cent, of $320 ? Ans. $19.20. 
 
 3. What is 8 per cent, of $327.25 ? Ans. $26.18. 
 
 Case I is what ? Give explanation. Rule. 
 
PERCENTAGE. 215 
 
 4. What is 7% per cent, of $56.75 ? Ans. $4.11-^. 
 
 5. What is 12J- per cent, of 2450 pounds ? 
 
 Ans. 306.25 pounds. 
 
 6. What is 6f per cent, of 19072 bushels? 
 
 Ans. 1287.36 bushels. 
 
 7. What is 33 per cent, of 846 gallons ? 
 
 Ans. 282 gallons. 
 
 8. What is 9f per cent, of 275 miles? Ans. 26.95 miles- 
 '9. What is 15 per cent, of 450 sheep ? 
 
 10. What is 50 per cent, of 1240 men ? 
 
 11. What is 105 per cent, of $5760 ? Ans. $6048. 
 
 12. What is 175 per cent, of $12967 ? 
 
 13. What is 25 per cent, of f ? 
 
 25 per cent, equals T 2 ^ = \, and J X \ = A Ans. 
 
 14. What is 15 per cent, of f ? Ans. T ^. 
 
 15. What is 2 per cent, of 6| ? Ans. . 
 
 16. What is 33 per cent, of jk? Ans. &. 
 
 17. What is 84 per cent, of 7? Ans. 6^. 
 
 18. Find per cent, of $40.80 Ans. $.306. 
 
 19. Find If per cent, of $15.60 Ans. $.26. 
 
 20. A farmer, having 760 sheep, kept 25 per cent, of them, 
 and sold the remainder ; how many did he sell ? 
 
 21. A man has a capital of $24500; he invests 18 per 
 cent, of it in bank stock, 30 per cent, of it in railroad stocks, 
 and the remainder in bonds and mortgages ; how much does 
 he invest in bonds and mortgages? Ans. $12740. 
 
 22. A speculator bought 1576 barrels of apples, and upon 
 opening them he found 12 per cent, of them spoiled; how 
 many barrels did he lose ? 
 
 23. Two men engaged in trade, each wfth $2760. One of 
 them gained 33^ per cent, of his capital, and the other gained 
 75 per cent, ; how much more did the one gain than the other ? 
 
 Ans. $1150. 
 
 24. A man, owning f of an iron foundery, sold 35 per cent, 
 of his share ; what part of the whole did he sell, and what 
 part did he still own ? Ans. He still owned f . 
 
216 PERCENTAGE. 
 
 25. A owed B $575.40 ; he paid at one time 40 per cent, 
 of the debt ; afterward he paid 25 per cent, of the remainder ; 
 and at another time 12 per cent, of what he owed after the 
 second payment ; how much of the debt did he still owe ? 
 
 Am. $226.56f. 
 
 CASE II. 
 
 236. To find what per cent, one number is of an- 
 other. 
 
 1. A man, having $125, lost $5 ; what per cent, of his 
 money did he lose ? 
 
 OPERATION. ANALYSIS. We multi- 
 
 5 -7- 125 := .04 4 per cent. ply the base by the rate 
 
 Or, P er cent - to obtain the 
 
 ^ = & = .04 = 4 per cent. Percentage ( 235 ) ; con- 
 
 versely, we divide the per- 
 centage by the base to obtain the rate per cent. Or, since $125 is 
 100 per cent, of his money, $5 is T |^, equal to -^ of 100 per cent, 
 which is 4 per cent. Hence the 
 
 RULE. Divide the percentage by the base, and the quotient 
 will be the rate per cent, expressed decimally. Or, 
 
 Take such a part of 100 as the percentage is part of the 
 
 base. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What per cent, of $450 is $90 ? Ans. 20. 
 
 3. What per cent, of $1400 is $175? Ans. 12. 
 
 4. What per cent, of $750 is $165 ? 
 
 f>. What per cent, of $240 is $13.20 ? Ans. 5. 
 
 C. What per cent, of $2 is 15 cents ? 
 
 7. What per cent, of 6 bushels 1 peck is 4 bushels 2 pecks 
 6 quarts ? Ans. 75 per cent. 
 
 8. What per cent, of 1 5 pounds is 5 pounds 10 ounces 
 avoirdupois weight ? Ans. 37J- per cent. 
 
 9. What per cent, of 250 head of cattle is 40 head ? 
 
 Case II is what ? Give explanation. Rule. 
 
PERCENTAGE. 217 
 
 10. From a hogshead of sugar containing 760 pounds, 100 
 pounds were sold at one time, and 90 pounds at another; what 
 per cent, of the whole was sold ? 
 
 11. A man, having 600 acres of land, sold o f jt a t one 
 time, and ^ of the remainder at another time ; what per cent, 
 remained unsold ? Ans. 50 per cent. 
 
 CASE III. 
 
 237. To find a number when a certain per cent, of 
 it is given. 
 
 1. A man lost $5, which was 4 per cent, of all the money 
 he had ; how much had he at first ? 
 
 OPERATION. ANALYSIS. We are here required to 
 
 $5 -f- .04 .$125. find the base, of which $5 is the per- 
 
 Or, centage. Now, percentage equals base 
 
 -5- X 100 in $125 multiplied by the rate per cent; con- 
 
 versely, base equals percentage divided 
 
 by rate per cent. Or, $5 is 4 per cent, of all he had ; \ of $5, or ^, 
 equals 1 per cent, of all he had, and 100 times % equals 100 per 
 cent., or all he had. Hence the 
 
 RULE. Divide the percentage by the rate per cent., ex- 
 pressed decimally, and the quotient will be the base, or number 
 required. Or, 
 
 Take as many times 100 as the percentage is times the rate 
 per cent. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. 16 is 8 per cent, of what number ? Ans. 200. 
 
 3. 42 is 7 per cent, of what number ? 
 
 4. 75 is 12i per cent, of what number? Ans. 600. 
 
 5. 33 is 2J per cent, of what number ? Ans. 1200. 
 
 6. $281.25 is 374 per cent, of what sum of money ? 
 
 Ans. $750. 
 
 7. A farmer sold 50 sheep, which was 20 per cent, of his 
 whole flock ; how many sheep had he at first ? 
 
 Case III is what ? Give explanation. Rule. 
 S 
 
218 PERCENTAGE. 
 
 8. I loaned a man a certain sum of money ; at one time 
 he paid me $59.75, which was 12 per cent, of the whole sum 
 loaned to him ; how much did I loan him ? 
 
 9. A merchant invested $975 in dry goods, which was 15 
 per cent, of his entire capital ; what was the amount of his 
 capital? Ans. $6500. 
 
 10. If a man, owning 40 per cent, of an iron foundery, sell 
 25 per cent, of his share for $1246.50, what is the value of 
 the whole foundery ? -Ans. $12465. 
 
 11. A' merchant pays $75 a month for clerk hire, which is 
 25 per cent, of his entire profits ; how much are his profits for 
 one year, after paying his clerk hire ? Ans. $2700. 
 
 12. A produce buyer, having a quantity of corn, bought 
 2000 bushels more, and he found that this purchase was 40 
 per cent, of his whole stock ; how much had he before he 
 bought this last lot ? Ans. 3000 bushels. 
 
 COMMISSION AND BROKERAGE. 
 
 338. An Agent, Factor, or Broker, is a person who trans- 
 acts business for another, or buys and sells money, stocks, 
 notes, &c. 
 
 339. Commission is the percentage, or compensation 
 allowed an agent, factor, or commission merchant, for buying 
 and selling goods or produce, collecting money, and transact- 
 ing other business. 
 
 24:0. Brokerage is the fee, or allowance paid to a broker 
 or dealer in money, stocks, or bills of exchange, for making 
 exchanges of money, buying and selling stocks, negotiating 
 bills of exchange, or transacting other like business. 
 
 NOTE. The rates of commission and brokerage are not regulated by 
 law, but are usually reckoned at a certain per cent, upon the money 
 employed in the transaction. 
 
 Define an agent, factor, or broker. What is meant by commission ? 
 Brokerage ? 
 

 COMMISSION AND BROKERAGE. 219 
 
 CASE I. 
 
 . To find the commission or brokerage 011 any 
 sum of money. 
 
 1. A commission merchant sells butter and cheese to the 
 amount of $1540 ; what is his commission at 5 per cent. ? 
 
 OPERATION. ANALYSIS. 
 
 1540 X .05 = $77, Am. Since the com- 
 
 Or, yfo = 2-V , and ^ X 1540 = $77. mission on & 1 is 
 
 5 cents or .05 of 
 
 a dollar, on $1540 it is $1540 X .05 = $77. Or, since 5 per cent 
 is T |^=:^ of the sum received, the commission is -fa of $1540 
 = $77. Hence the 
 
 RULE. Multiply the given sum by the rate per cent, ex- 
 pressed decimally, and the result will be the commission or bro- 
 kerage. Or, 
 
 Take such a part of the given sum as the number expressing 
 the per cent, is part of 100. 
 
 EXAMPLES FOB PRACTICE. 
 
 2. A commission merchant sells goods to the amount of 
 $6756 ; what is his commission at 2 per cent.? Ans. $135.12. 
 
 3. What commission must be paid for collecting $17380, 
 at 3 per cent. ? Ans. $608.30. 
 
 4. An agent in Chicago purchased 4700 bushels of wheat, 
 at 75 cents a bushel ; what was his commission at 1 per cent, 
 on the purchase money ? 
 
 5. A broker in New York exchanged $25875 on the Suf- 
 folk Bank, Boston, at per cent. ; how much brokerage did 
 he receive ? Ans. $64.6875. 
 
 6. An auctioneer sold at auction a house for $3284, and 
 the furniture for $2176.50 ; what did his fees amount to at 
 2 per cent. ? 
 
 7. A broker negotiates a bill of exchange of $2890 for % 
 per cent, commission ; how much is his brokerage ? 
 
 Ans. $23.12. 
 
 Case I is what ? Give explanation. Rule. 
 
220 PERCENTAGE. 
 
 8. An agent buys for a manufacturing company 26750 
 pounds of wool, at 32 cents a pound, and receives a commis- 
 sion of 2 J per cent. ; what amount does he receive ? 
 
 Ans. $235.40. 
 
 9. If I sell 400 bales of cotton, each weighing 570 pounds, 
 at 9 cents a pound, and receive a commission of 2 per cent., 
 how much do I make by the transaction ? Ans. $461.70. 
 
 10. A commission merchant in New Orleans sells 450 bar- 
 rels of flour at $7.60 a barrel ; 38 firkins of butter, each con- 
 taining 56 pounds, at 25 cents a pound ; and 105 cheeses, each 
 weighing 48 pounds, at 9 cents a pound ; how much is his 
 commission for selling, at 5J per cent. ? Ans. $243.133. 
 
 11. A lawyer collected a note of $950, and charged 6 per. 
 cent, commission ; what was his fee, and what the sum to be 
 remitted ? Ans. Fee, $61.75 ; remitted, $888.25. 
 
 12. An insurance agent's fees are 6 per cent, on all sums 
 received for the company, and 4 per cent, additional on all 
 sums remaining, at the end of the year, after the losses are 
 paid ; he receives, during the year, $30456.50, and pays losses 
 to the amount of $19814.15; how much commission does he 
 receive during the year ? Ans. $2253.084. 
 
 CASE II. 
 
 4. To find the commission or brokerage, when 
 it is to be deducted from the given sum, and the bal- 
 ance invested. 
 
 1. A merchant sends his agent $1260 with which to buy 
 merchandise, after deducting his commission of 5 per cent. ; 
 what is the sum invested, and how much is the commission ? 
 
 OPERATION. 
 
 $1260 ~ 1.05 = $1200, inwKio'i. 
 
 $1260 $1200 = $60, commission. 
 
 Or, i^ + T&T = &H $ 1260 -Mu- = $ 1200 > iuvested; 
 
 And $1260 $1200 = $60, commission. 
 
 Cusc II is what? (Jive explanation, llulc. 
 
COMMISSION AND BROKERAGE. 221 
 
 ANALYSIS. Since the commission is 5 per cent., the agent must 
 receive $1.05 for every $1 he expends ; he can invest as many 
 dollars as $1.05 is contained times in $1260, which is $1200; and 
 the difference between the given sum and the sum invested is his 
 commission. 
 
 Or, the money expended is -J$$ of itself, the commission is T -$ of 
 this sum, and the commission added to the sum expended is -j-jj-J of 
 the whole sum. Since $1260 is ff = fi$1260 -^-|i = $1200, 
 the sum expended; and $1260 $1200=: $60 the commission. 
 Hence the 
 
 RULE. I. Divide the given amount by 1 increased by the rate 
 per cent, of commission, and the quotient is the sum invested. 
 
 II. Subtract the investment from the given amount, and the 
 remainder is the commission. 
 
 EXAMPLES FOK PRACTICE. 
 
 2. A man sends $3246.20 to his agent in Boston, request- 
 ing him to lay it out in shoes, after deducting his commission 
 of 2 per cent; how much is his commission? Ans. $63.65. 
 
 3. What amount of stock can be bought for $9682, and al- 
 low 3 per cent, brokerage ? Ans. $9400. 
 
 4. A flour merchant sent $10246.50 to his agent at Chica- 
 go, to invest in flour, after deducting his commission of 3^- 
 per cent. ; how many barrels of flour could he buy at $5.50 
 per barrel? Ans. 1800 barrels. 
 
 5. An agent receives a remittance of $4908, with which to 
 purchase grain, at a commission of 4^- per cent. ; what will be 
 the amount of the purchase ? 
 
 6. Remitted $603.75 to my agent in New York, for the pur- 
 chase of merchandise, agent's commission being 5 per cent. ; 
 what amount of broadcloth at $5 per yard should I receive ? 
 
 Ans. 115 yds. 
 
 7. A commission merchant receives $9376.158, with or- 
 ders to purchase grain ; his commission is 3 per cent., and he 
 charges 1^ per cent, additional for guaranteeing its delivery at 
 a specified time ; how much will he pay out, and what are 
 Ms fees ? Ans. Fees, $403.758. 
 
222 PEP.CENTAGE. 
 
 8. A real estate broker, whose stated commission is 1J 
 per cent., receives $13842.07, to be used in the purchase of 
 city lots ; how much does he invest, and what is his commis- 
 sion ? Ans. $13604 invested; $238.07 commission. 
 
 9. A broker received $10650, to be invested in stocks after 
 deducting per cent, for brokerage ; what amount of stock 
 did he purchase ? 
 
 STOCKS. 
 
 24:3. A Corporation is a body authorized by a general 
 law, or by a special charter, to transact business as a single 
 individual. 
 
 244. A Charter is the legal act of incorporation, and de- 
 fines the powers and obligations of the incorporated body. 
 
 24:5. A Firm is the name under which an unincorporated 
 company transacts business. 
 
 246. Capital or Stock is the property or labor of an indi- 
 vidual, corporation, company, or firm; it receives different 
 names, as Bank Stock, Railroad Stock, Government Stock, &c. 
 
 247. A Share is one of the equal parts into which the 
 stock is divided. 
 
 248. Stockholders are the owners of the shares. 
 
 249. The Nominal or Par Value of stock is its first cost, 
 or original valuation. 
 
 NOTE. The original value of a share varies in different companies. 
 A share of bank, insurance, railroad, or like stock is usually $100. 
 
 250. Stock is At Par when it sells for its first cost, or 
 original valuation ; 
 
 2>1. Above Par, at a premium, or in advance, when it 
 sells for more than its original cost ; and 
 
 252. Below Par, or at a discount, when it sells for less 
 than its original cost. 
 
 Define a corporation. A charter. A firm. Capital or stock. Shares. 
 Stockholders. Par value. At par. Above par. Below par. 
 
STOCKS. 223 
 
 253. The Market or Real Value of stock is what it will 
 bring per share in money. 
 
 S54r. A Dividend is a sum paid to stockholders from the 
 profits of the business of the company. 
 
 255. An Assessment is a sum required of stockholders to 
 meet the losses or expenses of the business of the company. 
 
 2oG. Premium or advance, and discount on stock, divi- 
 dends, and assessments, are computed at a certain per cent, 
 upon the original value of the shares of the stock. 
 
 CASE I. 
 
 257. To find the value of stock when at an ad- 
 vance, or at a discount. 
 
 1. What will $3240 of bank stock cost, at 8 per cent, ad- 
 vance ? 
 
 OPERATION. ANALYSIS. Since 
 
 $1 -|-.08 == $1.08 $1 of the stock at 
 
 $3240 X $1-08 = $3499.20, Am. P ar value wil1 cost 
 
 $1 plus the premi- 
 um, or $1.08, $3240 of the same stock will cost 3240 X $1.08 in 
 $3499.20. If the stock w.ere 8 per cent, below par, $1 minus the 
 discount, or $1.00 $.08 = $.92, would show what $1 of the stock 
 would cost. Hence the 
 
 EULE. Multiply the par value of the stock by the number 
 indicating the price of $1 of the same stock, and the product 
 will be the real value. 
 
 NOTE. In all examples relating to stocks, $100 is considered the 
 par value of a share of stock, unless otherwise stated. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. If the stock of an insurance company sell at 5 per cent. 
 
 k below par, what will $1200 of the stock cost? Ans. $1140. 
 3. What is the market value of 35 shares of New York 
 Central Railroad stock, at 15 per cent, below par ? 
 Market value. A dividend. An assessment. Case I is what? Give 
 explanation. Rule. 
 
224 PERCENTAGE. 
 
 4. What must be paid for 48 shares of Panama Railroad 
 stock, at a premium of 5^ per cent., if the par value be $150 
 per share? A/is. $7596. 
 
 5. What costs $5364 stock in the Minnesota copper mines, 
 at 9 per cent, above par ? 
 
 6. A man purchased $6275 stock in the Pennsylvania Coal 
 Company, and sold the same at a discount of 12 per cent.; 
 what was his loss ? Ans. $753. 
 
 7. What must be paid for 125 shares of United States 
 stock, at 4J per cent, premium, the par value being $1000 per 
 share? Ans. $130937.50. 
 
 8. Bought 42 shares of Illinois Central Railroad stock, at 
 14 per cent, discount, and sold the same at an advance of 12 
 per cent. ; how much did I gain ? Ans. $1113. 
 
 9. What is the market value of 175 shares of stock in the 
 Suffolk Bank, at J per cent, advance ? Ans. $17631.25. 
 
 10. Bought 75 shares of stock in the Bank of New Orleans, 
 of $50 each, at 3 per cent, discount, and sold it at 2 per cent, 
 advance; what was my gain ? Ans. $196.875. 
 
 11. B exchanged 28 shares of bank stock, of $50 each, 
 worth 7 per cent, premium, for 25 shares of railroad stock, of 
 $100 each, at 12 per cent, discount, and paid the difference 
 in cash ; how much cash did he pay ? Ans. $689.50. 
 
 CASE II. 
 
 358. To find how much stock may be purchased for 
 a given sum. 
 
 1. How many shares of bank stock, at 3 per cent, advance, 
 may be bought for $5150 ? 
 
 OPERATION. ANALYSIS. Since the stock 
 
 $5150 -^- 1.03 = $5000 = is at 3 per cent, advance, $1 
 
 50 shares Ans. ^ stock at par will cost $1.03 ; 
 
 and if we divide $5150, the 
 
 whole sum to be expended, by $1.03, the cost of $1 of stock, the 
 quotient must be the amount of stock purchased. Hence the 
 
 Case n is what ? Give explanation. 
 
PROFIT AND LOSS. 225 
 
 RULE. Divide the given sum by the cost of $1 of stock, 
 and the quotient will be the nominal amount of stock purchased. 
 
 2. How many shares of railroad stock, at 5 per cent, ad- 
 vance, can be purchased for $6300 ? Ans. 60 shares. 
 
 3. I invested $6187.50, in Ocean Telegraph stock, at 10 
 per cent, discount ; how much stock did I purchase ? 
 
 Ans. $6875. 
 
 4. I sent my agent $53500 to be invested in Illinois Cen- 
 tral Railroad stock, which sold at 7 per cent, advance ; what 
 amount did he purchase ? Ans. $50000. 
 
 5. Sold 50 shares of stock in a Pittsburg ferry company, 
 at 8 per cent, discount, and received $1150; what is the par 
 value of 1 share ? Ans. $25. 
 
 PROFIT AND LOSS. 
 
 259. Profit and Loss are commercial terms, used to ex- 
 press the gain or loss in business transactions, which is usually 
 
 ' reckoned at a certain per cent, on the prime or first cost of 
 articles. 
 
 CASE I. 
 
 260. To find the amount of profit or loss, when the 
 cost and the gain or loss per cent, are given. 
 
 1. A man bought a horse for $135, and afterward sold him 
 for 20 per cent, more than he gave ; how much did he gain ? 
 OPERATION. ANALYSIS. Since $1 
 
 $135 X .20= $27, Ans. S ains 2 <> cents, or 20 per 
 
 i <ttiQ*vi _ <97 cent, $135 will gain $135 
 = *; $ o X * - 27. 
 
 > since 20 
 
 per cent, equals -$ = ^, the whole gain will be ^ of the cost. 
 Hence the following 
 
 RULE. Multiply the cost by the rate per cent, expressed 
 decimally. Or, 
 
 Take such part of the cost as the rate per cent, is part of 100. 
 
 Rule. What is meant by profit and loss ? Case I is what ? Give 
 explanation. Rule. 
 
226 PERCENTAGE. 
 
 EXAMPLES FOB PRACTICE. 
 
 2. A grocer bought a hogshead of sugar for $84.80, and sold 
 it at 12J- per cent, profit ; what was his gain ? 
 
 3. A miller bought 500 bushels of wheat at $1.15 a bushel, 
 and he sold the flour at 16 per cent, advance on the cost of 
 the wheat; what was his gain? Ans. $95.834-. 
 
 4. Bought 76 cords of wood at $3.62 a cord, and sold it 
 so as to gain 26 per cent. ; what did I make ? 
 
 5. A hatter bought 40 hats at $1.75 apiece, and sold them 
 at a loss of 14f- per cent. ; what was his whole loss ? 
 
 6. A grocer bought 3 barrels of sugar, each containing 230 
 pounds, at 8 cents a pound, and sold it at 18 T 2 T per cent, profit ; 
 what was his whole gain, and what the selling price per pound ? 
 
 Ans. Whole gain, $10.35 ; price per pound, 9 cents. 
 
 7. A sloop, freighted with 3840 bushels of corn, encoun- 
 tered a storm, when it was found necessary to throw 37^ per 
 cent, of her cargo overboard ; what was the loss, at 62 cents 
 a bushel ? Ans. $900 loss. 
 
 8. A gentleman bought a store and contents for $4720 ; he 
 sold the same for 12J- per cent, less than he gave, arid then 
 lost 15 per cent, of the remainder in bad debts ; what was his 
 entire loss ? Ans. $1209.50. 
 
 9. A man commenced business with $3000 capital ; the 
 first year he gained 22 per cent., which he added to his capi- 
 tal ; the second year he gained 30 per cent, on the whole sum, 
 which gain he also put into his business ; the third year he 
 lost 1 6 per cent, of his entire capital ; how much did he make 
 in the 3 years ? Ans. $981.25. 
 
 CASE II. 
 
 261. To find the gain or loss per cent., when the 
 cost and selling price are given. 
 
 1. Bought wool at 32 cents a pound, and sold it for 40 cents 
 a pound ; what per cent, was gained ? 
 
 Case II is what ? Give explanation. Rule. 
 
PROFIT AND LOSS. 227 
 
 OPERATION. 
 
 40 _ 32 8 ; 8 -j- 32 = & .25, ^W5. 
 6 r , 40 32 = 8 ; 8 ^- 32 = -fa = ; X 100 = 25 per cent. 
 
 ANALYSIS. Since the gain on 32 cents is 40 32 =: 8 cents, the 
 whole gain is -% =. -J- of the purchase money ; and ^ reduced to a 
 decimal is 25 hundredths, equal to 25 per cent. Or, if the gain were 
 equal to the purchase money, it would be 100 per cent. ; but since 
 the gain is -^ = of the purchase money, it will be \ of 100 per 
 cent., equal to 25 per cent. Hence the following 
 
 RULE. Make the difference between the purchase and selling 
 prices the numerator, and the purchase price the denominator ; 
 reduce to a decimal, and the result will be the per cent. Or, 
 
 Take such a part of 100 as the gain or loss is part of the 
 purchase price. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. A man bought a pair of horses for $275, and sold them 
 for $330 ; what per cent, did he gain ? Ans. 20 per cent. 
 
 3. If a merchant buy cloth at $.60 a yard, and sell it for 
 $.75 a yard, what does he gain per cent. ? 
 
 4. A speculator bought 108 barrels of flour at $4.62 a 
 barrel, and sold it so as to gain $114.88^-; what per cent, 
 profit did he make ? Ans. 23 per cent. 
 
 5. Bought sugar at 8 cents a pound, and sold it for 9 cents 
 a pound ; what per cent, was gained ? 
 
 6. A drover bought 150 head of cattle for $42 per head, 
 and sold them for $5400 ; what was his loss per cent. ? 
 
 Ans. 14f per cent. 
 
 7. If I sell for $15 what cost me $25, what do I lose per 
 cent. ? Ans. 40 per cent. 
 
 8. Bought paper at $2 per ream, and sold it at 25 cents 
 a quire; what was the gain per cent. ? Ans. 150 per cent. 
 
 9. If I sell of an article for f of its cost, what is gained 
 per cent. ? Ans. 50 per cent. 
 
 10. If ^ of an article be sold for what of it cost, what is 
 the loss per cent. ? Ans. 37 per cent. 
 
228 PERCENTAGE. 
 
 11. If I sell 3 pecks of clover-seed for what one bushel 
 cost me, what per cent, do I gain ? Ans. 33 per cent. 
 
 12. A, having a debt against B, agreed to take $.871 on 
 the dollar ; what per cent, did A lose ? 
 
 13. A grocer bought 7 cwt. 20 Ib. of sugar, at 7 cents a 
 pound, and sold 3 cwt. 42 Ib. at 8 cents, and the remainder at 
 8^- cents ; what was his gain per cent. ? Ans. 18^ per cent. 
 
 14. Bought 2 hogsheads of wine, at $1.25 a gallon, and 
 sold the same at $1.60 ; what was the whole gain, and what 
 the gain per cent. ? Ans. Gain 28 per cent. 
 
 15. A grain dealer bought corn at $.55 a bushel and sold 
 it at $.66, and wheat for $1.10, and sold it for $1.37^-; upon 
 which did he make the greater per cent. ? 
 
 Ans. o per cent., upon the wheat. 
 
 CASE III. 
 
 263. To find the selling price, when the cost and 
 the gain or loss per cent, are given. 
 
 1. Bought a horse for $136 ; for how much must he be sold 
 to gain 25 per cent. ? 
 
 OPERATION. ANALYSIS. Since $1 of cost 
 
 $1 + .25 =z $1.25. sells for $1.25, $136 of cost will 
 
 $1.25 X 136 = $170, Ans. sel1 for 136 timcs $!- 25 > which 
 
 Or i -4- 2 5 J.25 _ 5 equals $170, the selling price. 
 
 ~ r ' slnce the cost is *> and 
 
 will be l-f = } of the cost, or 
 
 } of $136 = $170. If the horse had been sold at a loss of 25 per 
 cent., then $1 of cost would have sold for $1 minus .25, or $.75, 
 &c. Hence, 
 
 RULE. Multiply $1 increased by the gain or diminished Inj 
 the loss per cent, by the number denoting the cost. Or, 
 
 Take such a part of the cost as is equal to -f{}$ increased or 
 diminished by the gain or loss per cent. 
 
 Case ILI is what ? Give explanation. Rule. 
 
 
PKOFIT AND LOSS. 229 
 
 EXAMPLES FOR PRACTICE. 
 
 2. If 12 hundred weight of sugar cost $140, how must 
 it be sold per pound to gain 25 per cent. ? Ans. 14 cents. 
 
 o. Bought a hogshead of molasses for 30 cents a gallon, 
 and paid 16f per cent, on the prime cost, for freight and cart- 
 age ; how much must it sell for, per gallon, to gain 33^- per 
 cent, on the whole cost ? Ans. $.46. 
 
 4. For what price must I sell coffee that cost 10^ cents a 
 pound, to gain 17 per cent.? 
 
 5. If I am compelled to sell damaged goods at a loss of 15 
 per cent., how should I mark goods that cost me $.62^- ? 
 
 $1.20? $3.87^- ? Ans. $.53; $1.02; $3.29f. 
 
 6. A man, wishing to raise some money, offers his house 
 and lot, which cost him $3240, for 18 per cent, less than cost ; 
 what is the price ? % 
 
 7. C bought a farm of 120 acres, at $28 an acre, paid 
 $480 for fencing, and then sold it for 12^ per cent, advance 
 on the whole cost ; what was his whole gain, and what did he 
 receive an acre ? Ans. $480 gain ; $36 an acre. 
 
 8. Bought a cask of brandy, containing 52 gallons, at 
 $2.60 per gallon ; if 7 gallons leak out, how must the remain- 
 der be sold per gallon, to gain 37J- per cent, on the cost of the 
 whole? Ans. $4.13. 
 
 9. A merchant bought 15 pieces of broadcloth, each piece 
 containing 23^ yards, for $840, and sold it so as to gain 18J 
 per cent. ; how much did he receive a yard ? 
 
 CASE IV. 
 
 263. To find the cost, when the selling price and 
 the gain or loss per cent, are given. 
 
 tl. A merchant sold cloth for $4.80 a yard, and by so doing 
 lade 33^ per cent. ; how much did it cost ? 
 OPERATION. 
 $1 _j_ .331. = $l.33i ; $4.80 -r-'l.33 = $3.60, Ans. 
 Or, $4.80 | of the cost ; $4.80 -=- = $3.60. 
 T Case IV is what? 
 
230 PERCENTAGE. 
 
 ANALYSIS. Since the gain is 33^- per cent, of the cost, $1 of the 
 cost, increased by 33|- per cent., will be what $1 of cost sold for : 
 therefore there will be as many dollars of cost, as 1.33^ is con- 
 tained times in $4.80, or $3.60. Or, since he gained 33 per cent. 
 = of the cost, $4.80 is of the cost ; $4.80 ~- = $3.60. 
 
 NOTE. If the rate per cent, be loss, we subtract it from $1, instead 
 of adding it. Hence the following 
 
 RULE. Divide the selling price by $1 increased by the gain 
 or diminished by the loss per cent., expressed decimally, or in 
 the form of a common fraction, and the quotient will be the 
 cost. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. By selling sugar at 8 cents a pound, a merchant lost 20 
 per cent. ; what did the sugar cost him ? Ans. 10 cents. 
 
 3. Sold flour for $6.12 per barrel, and lost 12 per cent. ; 
 what^was the cost ? Ans. $7.00. 
 
 4. A grocer, by selling tea at $.96 a pound, gains 28 per 
 cent. ; how much did it cost him ? Ans. $.75. 
 
 '5. Sold a quantity of flour for $1881, which was 18J per 
 cent, more than it cost ; how much did it cost ? 
 
 6. Sold 25 barrels of apples for $69.75, and made 24 per 
 cent. ; how much did they cost per barrel ? 
 
 7. Sold 9 cwt. of sugar at $8J per cwt., and thereby lost 
 12 per cent. ; how much was the whole cost? 
 
 8. Having used a carriage six months, I sold it for $96, 
 which was 20 per cent, below cost ; what would I have received 
 had I sold it for 15 per cent. above cost? Ans. $138. 
 
 9. B sells a pair of horses to C, and gains 12 per cent. ; 
 C sells them to D for $570, and by so doing gains 18f per 
 cent. ; how much did the horses cost B ? Ans. $426.66f . 
 
 10. A grocer sold 4 barrels of sugar for $24 each ; on 2 
 barrels he gained 20 per cent., and on the other 2 he lost 20 
 per cent. ; did he gain or lose on the whole ? Ans. Lost $4. 
 
 11. A person sold out his interest in business for $4900, 
 which was 40 per cent, iriore than 3 times as much as he began 
 with ; how much did he begin with ? Ans. $1166.66|. 
 
 Give explanation. Rule. 
 
INSURANCE. 231 
 
 INSURANCE. 
 
 264. Insurance on property is security guaranteed by 
 one party to another, for a stipulated sum, against the loss of 
 that property by fire, navigation, or any other casualty. 
 
 265. The Insurer or Underwriter is the party taking the 
 risk. 
 
 266. The Insured is the party protected. 
 
 267. The Policy is the written contract between the 
 parties. 
 
 268. The Premium is the sum paid by the insured to the 
 insurer, and is estimated at a certain rate per cent, of the 
 amount insured, which rate varies according to the degree of 
 hazard, or class of risk. 
 
 XOTE. As a security against fraud, most insurance companies take 
 risks at not more than two thirds the full value of the property 
 insured. 
 
 269* To find the premium when the rate of insur- 
 ance and the amount insured are given. 
 
 1. What must I pay annually for insuring my house to the 
 amount of $3250, at 1 per cent, premium ? 
 
 OPERATION. ANALYSIS. We 
 
 $3250 X -01 or .0125 = $40.625. multiply the amount 
 
 Or, 1 per ct. = jfo = & ; insured, $3250, by 
 
 $3250 X A = &40.62A. the ra * e ! ** P ? r 
 
 cent., and the result, 
 
 $40.625, is the premium. Or, the rate, \\ per cent, is -^=.-fa of 
 the amount insured, and -fa of $3250 is $40.62^. Hence the 
 
 RULE. Multiply the amount insured by the rate per cent., 
 and the product will be the premium. Or, 
 
 Take such a part of the amount insured as the rate is part 
 of 100. 
 
 Define insurance. Insurer, or underwriter. Policy. Premium. 
 To what amount can property usually be insured ? Give analysis of 
 example 1. Rule. 
 
232 PERCENTAGE. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the premium on a policy for $750, at 4 per 
 cent.? Ans. $30. 
 
 3. What premium must be paid for $4572.80 insurance, at 
 2 percent.? Ans. $114.32. 
 
 4. A house and furniture, valued at $5700, are insured at 
 If per cent. ; what is the premium ? Ans. $99.75. 
 
 5. A vessel and cargo, valued at $28400, are insured at 3 
 per cent.; what is the premium? Ans. $994. 
 
 6. A woolen factory and contents, valued at $55800, are 
 insured at 2% per cent. ; if destroyed by fire, what would be 
 the actual loss of the company ? Ans. $54237.60. 
 
 7. What must be paid to insure a steamboat and cargo 
 from Pittsburg to New Orleans, valued at $47500, at f of 1 
 percent.? Ans. $356.25. 
 
 8. A gentleman has a house, insured for $8000, and the fur- 
 niture for $4000, at 2f per cent. ; what premium must he 
 pay? Ans. $285. 
 
 9. A cargo of 4000 bushels of wheat, worth $1.20 a bushel, 
 is insured at J of 1^- per cent, on f of its value ; if the cargo be 
 lost, how much will the owner of the wheat lose ? Ans. $1636. 
 
 10. What will it cost to insure a factory valued at $21000, 
 at | per cent. ; and the machinery valued at $15400, at | per 
 cent.? Ans. $264.25. 
 
 TAXES. 
 
 270. A Tax is a sum of money assessed on the person 
 or property of an individual, for public purposes. 
 
 271. When a tax is assessed on property, it is apportioned 
 at a certain per cent, on the estimated value. 
 
 When assessed on the person, it is apportioned equally 
 among the male citizens liable to assessment, and is called a 
 poll tax. Each person so assessed is called a poll. 
 
 What is a tax r How is a tax on property apportioned ? On the 
 person, how ? 
 

 TAXES. 233 
 
 S72J. Property is of two kinds real estate, and personal 
 property. 
 
 S73. Real Estate consists of immovable property, such 
 as lands, houses, &c. 
 
 74. Personal Property consists of movable property, 
 such a,s money, notes, furniture, cattle, tools, &c. 
 
 175. An Inventory is a written list of articles of proper- 
 ty, with their value. 
 
 S76 . Before taxes are assessed, a complete inventory of all 
 the taxable property upon which the tax is to be levied must 
 be made. If the assessment include a poll tax, then a complete 
 list of taxable polls must also be made out. 
 
 I. A tax of $3165 is to be assessed on a certain town; 
 the valuation of the taxable property, as shown by the as- 
 sessment roll, is $600,000, and there are 220 polls to be as- 
 sessed 75 cents each ; what will be the tax on a dollar, and 
 how much will be A's tax, whose property is valued at $3750, 
 and who pays for 3 polls? 
 
 OPERATION. 
 
 $.75 X 220 = $165, amount assessed on the polls. 
 
 $3165 $165 = $3000, amount to be assessed on the property. 
 
 $3000 -J- $600,000 = .005, tax on $1. 
 
 $3750 X -005 = $18.75, A's tax on property. 
 
 $.75 X 3 = $2.25, A's tax on 3 polls. 
 
 $18.75 4- $2.25 = $21, amount of A's tax. 
 
 Hence the following 
 
 RULE. I. Find the amount of poll tax, if any,* and subtract 
 this sum from the whole amount of tax to be assessed. 
 
 II. Divide the sum to be raised on property, by the whole 
 amount of taxable property, and the quotient will be the per 
 cent., or the tax on one dollar. 
 
 III. Multiply each man's taxable property by the per cent., 
 or the tax on $1, and to the product add his poll tax, if any ; 
 the result will be the whole amount of his tax. 
 
 What is real estate ? Personal property ? An inventory ? Explain 
 the process of levying a state or other tax. Rule. 
 
234 
 
 PERCENTAGE. 
 
 NOTE. Having found the tax on $1, or the per cent., which in the 
 preceding example we find to be 5 mills, or ^ per cent., the operation 
 of assessing taxes may be greatly facilitated by finding the tax 011 $2, 
 $3, &c., to $10, and then on $20, $30, &c., to $100, and arranging 
 the numbers as in the following 
 
 TABLE. 
 
 Prop. 
 
 Tax. 
 
 Prop. 
 
 Tax. 
 
 Prop. 
 
 Tax. 
 
 Prop. 
 
 Tax. 
 
 $1 gives 
 
 $.005 
 
 $10 
 
 $.05 
 
 $100 
 
 $ .50 
 
 $1000 
 
 $5.00 
 
 2 
 
 .01 
 
 20 
 
 .10 
 
 200 
 
 1.00 
 
 2000 
 
 10. 
 
 3 
 
 .015 
 
 30 
 
 .15 
 
 300 
 
 1.50 
 
 3000 
 
 15. 
 
 4 " 
 
 .02 
 
 40 
 
 .20 
 
 400 
 
 2.00 
 
 4000 
 
 20. 
 
 5 
 
 .025 
 
 50 
 
 .25 
 
 500 
 
 2.50 
 
 5000 
 
 25. 
 
 6 
 
 .03 
 
 60 
 
 .30 
 
 600 
 
 3.00 
 
 6000 
 
 30. 
 
 7 
 
 .035 
 
 70 
 
 .35 
 
 700 
 
 3.50 
 
 7000 
 
 35. 
 
 8 
 
 .04 
 
 80 
 
 .40 
 
 800 
 
 4.00 
 
 8000 
 
 40. 
 
 9 " 
 
 .045 
 
 90 
 
 .45 
 
 900 
 
 4.50 
 
 9000 
 
 45. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. According to the conditions of the last example, how 
 much would be a person's tax whose property was assessed 
 at $3845, and who paid for 2 polls ? 
 
 Finding the amount from the table, 
 
 The tax on $3000 is . . 
 
 " " " 800 . . , . " 
 
 5 . . . " 
 2 polls " 
 
 $15.00 
 4.00 
 .20 
 .025 
 1.50 
 
 Total tax is $20.725 
 
 3. How much would be Ws tax, who was assessed for 1 
 poll, and on property valued at $5390 ? Am. $27.70. 
 
 4. A tax of $9190.50 is to be assessed on a certain village ; 
 the property is valued at $1400000, and there are 2981 polls, 
 to be taxed 50 cents each ; what is the assessment on a dollar ? 
 what is C's tax, his property being assessed at $12450, and he 
 paying for 2 polls? Am. $.005J on $1 ; $(''.). 17 i, C's lax. 
 
 5. What is the tax of a non-resident, having property in 
 the same village valued at $5375 ? Ans. $29.5625. 
 
 Explain the table and its use. 
 
CUSTOM HOUSE BUSINESS. 235 
 
 6. A mining corporation, consisting of 30 persons, are 
 taxed $4342.75 ; their property is assessed for $188000, and 
 each poll is assessed 62^- cents ; what per cent, is their tax, 
 and ho\v much must he pay whose share is assessed for $2500, 
 and who pays for 1 poll ? Ans. 2 T 3 ,y per cent.; $58.125. 
 
 7. In a certain county, containing 25482 taxable inhab- 
 itants, a tax of $103294.60 is assessed for town, county, and 
 state purposes ; a part of this sum is raised by a tax of 30 
 cents on each poll ; the entire valuation of property on the as- 
 sessment roll is $38260000 ; what per cent, is the tax, and how 
 much will a person's tax be who pays for 3 polls, and whose 
 property is valued at $9470 ? Ans. to last, $24.575. 
 
 8. The number of polls in a certain school district is 225, 
 and the taxable property $1246093.75 ; it is proposed to build 
 a union school house at an expense of $10000; if the poll tax 
 be $1.25 a poll, and the cost of collecting be 2^- per cent., what 
 will be the tax on a dollar, and how much will be E's tax, who 
 pays for 1 poll, and has property to the amount of $11500 ? 
 
 Ans. $.008, tax on $1 ; $93.25, E's tax. 
 
 9. In a certain district the school was supported by a rate- 
 bill ; the teacher's wages amounted to $300, the fuel and other 
 expenses to $50.50 ; the public money received was $92, and 
 the whole number of days' attendance was 2585 ; A sent 2 
 pupils 118 days each ; how much was his rate-bill ? Ans. $2.36. 
 
 CUSTOM HOUSE BUSINESS. 
 
 S77. Duties, or Customs, are taxes levied by government 
 on imported goods, for the support of government and the 
 protection of home industry. 
 
 278. A Custom House is an office established by govern- 
 ment for the transaction of business relating to duties. 
 
 It is lawful to introduce merchandise into a country only 
 
 Define duties. A custom house. 
 
236 PERCENTAGE. 
 
 at points where custom houses are established. A seaport 
 town, having- a custom house, is called a port of entry. To 
 carry on foreign commerce secretly, without paying the duties 
 imposed by law, is smuggling. 
 
 27O. Tonnage is a tax levied upon the vessel, independent 
 of its cargo, for the privilege of coming into a port of entry. 
 Its amount is regulated by the size of the vessel. 
 
 28O. Revenue is the income to government from duties 
 and tonnage. 
 
 Duties are of two kinds ad valorem and specific. 
 . Ad Valorem Duty is a percentage computed on the 
 market value of merchandise in the country from which it is 
 imported. 
 
 383. Specific Duty is a sum computed on the weight or 
 measure of goods, without regard to their cost. 
 
 CASE I. 
 
 384. To compute ad valorem duties. 
 
 1. What is the duty, at 8 per cent., on a quantity of silk 
 which cost, in Florence, $6850 ? 
 
 OPERATION. ANALYSIS. Since the duty is ad valorem, 
 
 $6850 we multiply the cost, $6850, by .08, the 
 
 .08 given rate per cent., and we have $548, the 
 
 Ans. Hence the 
 $548.00, Ans. 
 
 RULE. Multiply the cost of the goods ly the given rate per 
 cent., and the product will be the required duty. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the duty, at 6 per cent., on a shipment of spring 
 hats, bought in Paris for $8565 ? Ans. $513.90. 
 
 3. John Jones imported, from Havana, 50 hhds. of W. I. 
 molasses, which were invoiced at 36 cents per gallon ; what 
 was the duty at 15 per cent. ? Ans. $170.10. 
 
 Define a port of entry. Smuggling. Tonnage. Revenue. Ad 
 valorem duty. Specific duty. What is Case I ? Give explanation. 
 Rule. 
 

 CUSTOM HOUSE BUSINESS. 287 
 
 4. What is the duty, at 12 per cent., on 175 bags of Java 
 coffee, each containing 115 Ibs., valued at 15 cents per pound ? 
 
 Ans. $362.25. 
 
 CASE II. 
 
 285. To compute specific duties. 
 
 It is the design of government to tax only so much of the 
 merchandise as will be available to the importer in market ; 
 consequently, the law directs that certain allowances be made 
 before specific duties are computed. 
 
 286. Gross Weight is the weight of the goods together 
 with the package or covering which contains them. 
 
 287. Net Weight is what remains after all allowances 
 have been made. 
 
 288. Draft, or Tret, is an allowance for waste, and is to be 
 deducted always from the gross weight or measure. 
 
 DKAFT BY WEIGHT. 
 
 On 112 Ib 1 lb. 
 
 Above 112 lb. and not exceeding 224 lb., 2 lb. 
 
 " 224 lb. " " " 336 lb., 3 lb. 
 
 336 lb. " " " 1120 lb., 4 lb. 
 
 " 1120 lb. " " " 2016 lb., 7 lb. 
 
 2016 lb 9 lb. 
 
 By law and usage no greater allowance than 9 lb. can be 
 made for draft on one box or package ; and sometimes several 
 pieces are weighed together for one draft. 
 
 289. Leakage is an allowance of 2 per cent, on all 
 liquors in casks paying duty by the gallon. 
 
 290. Breakage is an allowance of 10 per cent, on ale, 
 beer, and porter, in bottles ; and of 5 per cent, on all other 
 liquors in bottles. 
 
 NOTE. Leakage and breakage are the principal kinds of draft 
 allowed for liquors. 
 
 291. Tare is an allowance for the weight of the pack- 
 
 What is Case II ? Define gross weight. Net weight. Draft. Leak- 
 age. Breakage. Tare. 
 
Zoo PERCENTAGE. 
 
 age or covering that contains the goods. It is sometimes 
 reckoned by the piece, but is generally a certain per cent, of 
 that part of the gross weight which remains after the deduc- 
 tion of draft. 
 
 To compute ad valorem duties, we have the following 
 
 RULE. I. When the merchandise is subject to leakage or 
 breakage; Deduct the legal alloivance, and compute the duty 
 on the remainder. 
 
 II. When the merchandise is subject to both draft and 
 tare; First deduct the draft, and on the remainder compute 
 the tare, and deduct this also ; then compute the duty on the 
 net weight or measure. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. What is the duty, at 2 cents per pound, on 6 barrels of 
 sugar, weighing 250 pounds apiece, allowing draft for each 
 barrel separately, and tare at 15 per cent. ? Ans. $25.194. 
 
 2. What is the duty on 50 casks of nails, weighing each 
 450 Ibs. gross, at 1^ cents per pound, draft being allowed for 
 each cask, and tare at 8 per cent. ? Ans. $256.45. 
 
 3. Imported 200 boxes of figs, weighing 60 Ib. each ; draft 
 was reckoned on each parcel of 10 boxes, and tare at 9 Ib. per 
 box. What was the duty at 4 cents per Ib. ? Ans. $404.80. 
 
 4. What is the duty, at 14 cents per pound, on 6 barrels 
 of indigo, weighing 450 Ib. each, the legal rate of tare being 3 
 percent.? Ans. $363.40+. 
 
 5. Required the duty, at 2 1 cents per Ib., on 4 casks of shot, 
 weighing 200 Ib. each, legal tare being 3 per cent. Ans. $1 7.285. 
 
 6. Required the duty on 3 pipes of port wine, gross esti- 
 mate as follows: first pipe, 170 gallons; second, 176 gal- 
 lons ; third, 180 gallons ; duty, 15 cents per gallon. 
 
 Ans. $77.322. 
 
 7. What is the duty on 250 dozen bottles of porter, at 5 
 cents per bottle ? Ans. $135. 
 
 Give rule for computing ad valorem duties. 
 
SIMPLE INTEREST. 
 
 239 
 
 SIMPLE INTEREST. 
 
 292. Interest is a sum paid for the use of money. 
 
 293. Principal is the sum for the use of which interest 
 is paid. 
 
 294. Rate per cent, per annum is the sum per cent, paid 
 for the use of $100 annually. 
 
 NOTE. The rate per cent, is commonly expressed decimally, as hun- 
 dredths (231). 
 
 290. Amount is the sum of the principal and interest. 
 
 296. Simple Interest is the sum paid for the use of the 
 principal only, during the whole time of the loan or credit. 
 
 297. Legal Interest is the rate per cent, established by 
 law. It varies in different States, as follows : 
 
 Alabama, 8 per cent. 
 
 Arkansas, 6 
 
 Connecticut, G 
 
 Delaware, 6 
 
 Dist. of Columbia, ... 6 
 
 Florida, 8 
 
 Georgia, 8 
 
 Illinois, 6 
 
 Indiana, 6 
 
 Iowa, 7 
 
 Kentucky, 6 
 
 Louisiana, 5 
 
 Maine, 6 
 
 Maryland, 6 
 
 Massachusetts, 6 
 
 Michigan, , , 7 
 
 Mississippi, . 
 
 Missouri, 6 
 
 New Hampshire, 6 
 
 New Jersey, 6 
 
 New York, 7 
 
 North Carolina, 6 
 
 Ohio, .6 
 
 Pennsylvania, 6 
 
 Rhode Island, 6 
 
 South Carolina, 7 
 
 Tennessee, 6 
 
 Texas, 8 
 
 ! United States (debts), 6 
 
 | Vermont, 6 
 
 I Virginia, 6 
 
 i Wisconsin, 7 
 
 . 8 per cent. 
 
 NOTE. When the rate per cent, is not specified, in accounts, notes, 
 mortgages, contracts, &c., the legal rate is always understood. 
 
 29S. Usury is illegal interest, or a greater per cent, than 
 the legal rate. 
 
 CASE i. 
 
 299. To find the interest on any sum, at any rate 
 per cent., for years and months. 
 
 Define interest. Principal. Rate per cent, per annum. Amount, 
 What is simple interest ? Legal interest ? Usury ? Case I ? 
 
240 
 
 PERCENTAGE. 
 
 In percentage, any per cent, of any given number is so 
 many hundredths of that number ; but in interest, any rate per 
 cent, is confined to 1 year, and the per cent, to be obtained 
 of any given number is greater than the rate per cent, per 
 annum if the time be more than 1 year, and less than the rate 
 per cent, per annum if the time be less than 1 year. Thus, 
 the interest on any sum, at any rate per cent., for 3 years 6 
 months, is 31 times the interest on the same sum for 1 year ; 
 and the interest for 3 months is of the interest for 1 year. 
 
 1. What is the interest on $75.19 for 3 years 6 months, at 
 6 per cent. ? 
 
 OPERATION. 
 
 875.19 
 .06 
 
 $4.5114 
 
 22557 
 135342 
 
 ANALYSIS. The interest on $75.19, for 1 yr., 
 at 6 per cent, is .06 of the principal, or $4.51 14, 
 and the interest for 3 yr. 6 mo. is 3fy =. 31 times 
 the interest for 1 yr., or $4.5114 X 3, which is 
 $15.789 +i the Ans. Hence, the following 
 
 $15.7899, Ans. 
 
 RULE. I. Multiply the principal by the rate per cent., and 
 the product ivill be the interest for 1 year. 
 
 II. Multiply this product by the time in years and fractions 
 of a year, and the result will be the required interest. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the interest of $150 for 3 years, at 4 per cent. ? 
 
 Ans. $18. 
 
 3. What is the interest of $328 for 2 years, at 7 pur cent. ? 
 
 4. What is the interest of $125 for 1 year 6 months, at 6 
 percent.? Ans. Si 1.25. 
 
 5. What is the interest of $200 for 3 years 10 months, at 
 7 per cent ? Ans.' $5:1(5(1 + . 
 
 6. What is the interest of $76.50 for 2 years 2 months, at 
 5 per cent. ? Ans. $8.2<s7 -f- . 
 
 Explain the difference between percentage and interest. (iive 
 analysis. Rule. 
 
SIMPLE INTEREST. 241 
 
 7. What is the interest of $1270.25 for 11 months, at 7 
 per cent. ? Ans. $81.89 -f . 
 
 8. What is the interest of $2569.75 for 4 years 6 months, 
 at 6 per cent. ? 
 
 9. What is the interest of $1500.60 for 2 years 4 months, 
 at 6J- per cent. ? Ans. $218.8375. 
 
 10. What is the amount of $26.84 for 2 years 6 months, at 
 5 per cent. ? Ans. $30.195. 
 
 11. What is the amount of $450 for 5 years, at 7 per cent. ? 
 
 12. What is the interest of $4562.09 for 3 years 3 months, 
 at 3 per cent. ? Ans $444.80 -f- . 
 
 13. What is the amount of $3050 for 4 years 8 months, at 
 5 per cent. ? Ans. $3701.25. 
 
 14. What is the interest of $5000 for 9 months, at 8 per 
 cent, ? Ans. $300. 
 
 15. If a person borrow $375 at 7 per cent,, how much will 
 be due the lender at the end of 2 yr. 6 mo. ? 
 
 16. What is the interest paid on a loan of $1374.74, at 6 
 per cent., made January 1, 1856, and called in January 1, 
 18GO? Ans. $329.937 + . 
 
 17. If a note of $605.70 given May 20, 1858, on interest at 
 8 percent., be taken up May 20, 1861, what amount will then 
 be due if no interest has been paid? Ans. $751.068. 
 
 CASE II. 
 
 3OO. To find the interest on any sum, for any 
 time, at any rate per cent. 
 
 The analysis of our rule is based upon the following 
 
 Obvious Relations between Time and Interest. 
 
 I. The interest on any sum, for 1 year, at 1 per cent., is 
 .01 of that sum, and is equal to the principal with the separatrix 
 removed two places to the left. 
 
 II. A month being -fa of a year, -fa of the interest on any 
 sum for 1 year is the interest for 1 month. 
 
 What is Case II ? Give the first relation between time and interest. 
 Second. 
 
 U 
 
242 PERCENTAGE. 
 
 III. The interest on any sum for 3 days is ^y = -^ = .1 
 of the interest for 1 month, and any number of days may 
 readily be reduced to tenths of a month by dividing by 3. 
 
 IV. The interest on any sum, for 1 month, multiplied by 
 any given time expressed in months and tenths of a month, 
 will produce the required interest. 
 
 1. What is the interest on $724.68 for 2 yr. 5 mo. 19 da., 
 at 7 per cent. ? 
 
 OPERATION. ANALYSIS. We remove 
 
 2yr. 5 mo. 19 da. = 29. 6^ mo. the separatrix in the given 
 10 \ <fc7O/icQ principal two places to the 
 
 12) $7-2468 left, and we have $7.2468, 
 
 $.6039 the interest on the given sum 
 
 29.6 for 1 year at 1 per cent. 
 
 (300 L). Dividing this by 
 12, we have $.6039, the inter- 
 est for 1 month, at 1 per cent. 
 54351 (II.) Multiplying this 
 
 12078 quotient by 29.6^, the time 
 
 expressed in months and deci- 
 malg of a month> (m IV ^ 
 
 we have $17.89557, the in- 
 
 $125.26899, Ans. terest on the given sum for 
 
 the given time, at 1 per cent. 
 
 (IV.). And multiplying this product by 7 (7 times 1 per cent.), 
 we have $125.268 -f-, the interest on the given principal, for the 
 given time, at the given rate per cent. Hence, 
 
 RULE. I. Remove the separatrix in the given principal 
 two places to the left ; the result will be the interest for 1 year, 
 at 1 per cent. 
 
 II. Divide this interest byl2', the result will be the interest 
 for 1 month, at 1 per cent. 
 
 III. Multiply this interest by the given time expressed in 
 months and tenths of a month ; the result will be the interest 
 for the given time, at 1 per cent. 
 
 IV. Multiply this interest by the given rate ; the product 
 will be the interest required. 
 
 Give the third. Fourth. Give analysis. Rule. 
 
SIMPLE INTEREST. 243 
 
 CONTRACTIONS. After removing the scparatrix in the principal 
 two places to the left, the result may be regarded either as the in- 
 terest on the given principal for 12 months at 1 per cent., or for'l 
 month at 12 per cent. If we regard it as for 1 month at 12 per 
 cent., and if the given rate be an aliquot part of 12 per cent., the 
 interest on the given principal for 1 month may readily be found by 
 taking such an aliquot part of the interest for 1 month as the given 
 rate is part of 12 per cent. Thus, 
 
 To find the interest for 1 month at 6 per cent., remove the sep- 
 aratrix two places to the left, and divide by 2. 
 
 To find it at 3 per cent, proceed as before, and divide by 4 ; at 4 
 per cent., divide by 3 ; at 2 per cent., divide by 6, &c. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the interest of $100 for 7 years 7 months, at 6 
 per cent. ? Am. $45.50. 
 
 3. What is the amount of $47.50 for 4 years 1 month, at 9 
 percent? Am. $64.956 + . 
 
 4. What is the amount of $2000 for 3 months, at 7 per 
 cent? Am. $2035. 
 
 5. What is the interest of $250 for 1 year 10 months and 
 15 days, at 6 per cent. ? Am. $28.12^. 
 
 6. What is the interest of $36.75 for 2 years 4 months and 
 12 days, at 7 per cent. ? Am. $6.088 + . 
 
 7. What is the amount of $84 for 5 years 5 months and 9 
 days, at 5 per cent. ? 
 
 8. What is the interest of $51.10 for 10 months and 3 
 days, at 4 per cent. ? 
 
 9. What is the interest of $175.40 for 15 months and 8 
 days, at 10 per cent. ? Am. $22.31 -f- . 
 
 10. What is the amount of $1500 for 6 months and 24 
 days, at 7 per cent. ? Am. $1563.75. 
 
 11. What is the amount of $84.25 for 1 year 5 months 
 and 10 days, at 6 per cent. ? 
 
 12. What is the interest of $25 for 3 years 6 months and 
 20 days, at 6 per cent. ? Am. $5.33^. 
 
 13. What is the interest of $112.50 for 3 months and 1 
 day, at 9 J- per cent. ? Am. $2.70 +. 
 
 What contractions are given ? 
 
244 PERCENTAGE. 
 
 14 What is the interest of $408 for 20 days, at 6 per 
 cent.? Am. $1.36. 
 
 15. What is the interest of $500 for 22 days, at 7 per 
 cent. ? 
 
 16. What is the amount of $4500 for 10 days, at 10 per 
 cent.? Ans. $4512.50. 
 
 17. What is the amount of $1000 for 1 month 5 days, at 
 6} per cent. ? Ans. $1006.56. 
 
 18. Find the interest of $973.68 for 7 months 9 days, at 
 4J- per cent. 
 
 19. If I borrow $275 at 7 per cent., how much will I owe 
 at the end of 4 months 25 days ? 
 
 20. A person bought a piece of property for $2870, and 
 agreed to pay for it in 1 year and 6 months, with 6J- per cent, 
 interest; what amount did he pay ? Ans. $3149.825. 
 
 21. In settling with a merchant, I gave my note for $97.75, 
 due in 11 months, at 5 per cent.; what must be paid when 
 the note falls due? Ans. $102.23 -f. 
 
 22. How much is the interest on a note of $384.50 in 2 
 years 8 months and 4 days, at 8 per cent. ? 
 
 23. What is the interest of $97.86 from May 17, 1850, to 
 December 19, 1857, at 7 per cent. ? Ans. $51.98 +. 
 
 24. Find the interest of $35.61, from Nov. 11, 1857, to 
 Dec. 15, 1859, at 6 per cent. Ans. $4.474. 
 
 25. Required the interest of $50 from Sept. 4, 1848, to 
 Jan. 1, 1860, at 3 per cent. 
 
 26. Required the amount of $387.20, from Jan. 1 to Oct. 
 20, 1859, at 7 per cent. Ans. $408.957 +. 
 
 27. A man, owning a furnace, sold it for $6000 ; the terms 
 were, $2000 in cash on delivery, $3000 in 9 months, and the 
 remainder in 1 year 6 months, with 7 per cent, interest; 
 what was the whole amount paid ? Ans. $6262.50. 
 
 28. Wm. Gallup bought bills of dry goods of Geo. Bliss 
 & Co., of New York, as follows, viz.: Jan. 10, 1858, $350; 
 April 15, 1858, $150 ; and Sept. 20, 1858, $550.50 ; he bought 
 on time, paying legal interest ; what was the whole amount 
 of his indebtedness Jan. 1, 1859 ? Ans. $1092.66 +. 
 
PARTIAL PAYMENTS. 
 PARTIAL PAYMENTS OR IN 
 
 OO1. A Partial Payment is payment' 
 bond, or other obligation ; when the amount 
 written on the back of the obligation, it becomes a receipt, and 
 is called an Indorsement. 
 
 $2000. SPRINGFIELD, MASS., Jan. 4, 1857. 
 
 1. For value received I promise to pay James Parish, or 
 order, two thousand dollars, one year after date, with interest. 
 
 GEORGE JONES. 
 On this note were indorsed the following payments : 
 
 Feb. 19, 1858, $400 
 
 June 29, 1859, $1000 
 
 Nov. 14, 1859, $520 
 
 What remained due Dec. 24, 1860? 
 
 OPERATION. 
 
 Principal on interest from Jan. 4, 1857, $2000 
 
 Interest to Feb. 19, 1858, 1 yr. 1 mo. 15 da., 135 
 
 Amount, $2135 
 
 Payment Feb. 19, 1858, 400 
 
 Remainder for a new principal, $1735 
 
 Interest from Feb. 19, 1858, to June 29, 1859, 1 yr. 
 
 4 mo. 10 da., 141.69 
 
 Amount, $1876.69 
 
 Payment June 29, 1859, 1000. 
 
 Remainder for a new principal, $876.69 
 
 Interest from June 29, 1859, to Nov. 14, 1859, 4 mo. 
 
 15 da., 19.725 
 
 Amount, $896.415 
 
 Payment Nov. 14, 1859, 520. 
 
 Remainder for a new principal, $376.415 
 
 Interest from Nov. 14, 1859, to Dec. 24, 1860, 1 yr. 
 
 1 mo. 10 da., 25.09 
 
 Remains due Dec. 24, 1860, $401. 505 -f- 
 
 What is meant by partial payment ? By an indorsement ? 
 
 U* 
 
24G PERCENTAGE. 
 
 $475.50. NEW YORK, May 1, 1855. 
 
 2. For value received, we jointly and severally promise to 
 pay Mason & Bro., or order, four hundred seventy-five dol- 
 lars fifty cents, nine months after date, with interest. 
 
 JONES, SMITH & Co. 
 
 The following indorsements were made on this note : 
 
 Dec. 25, 1855, received, $50 
 
 July 10, 1856, 15.75 
 
 Sept. 1, 1857, 25.50 
 
 June 14, 1858, " 104 
 
 How much was due April 15, 1859 ? 
 
 OPERATION. 
 
 Principal on interest from May 1, 1855, $475.50 
 
 Interest to Dec. 25, 1855, 7 mo. 24 da., 21.63 
 
 Amount, $497.13 
 
 Payment Dec. 25, 1855, 50. 
 
 Remainder for a new principal, $447.14 
 
 Interest from Dec. 25, 1855, to June 14, 1858, 2 yr. 
 
 5 mo. 19 da., 77.29 
 
 Amount, $524.42 
 
 Payment July 10, 1856, less than interest") 
 
 then due, > $15.75 
 
 Payment Sept. 1, 1857, J 25.50 
 
 Their sum less than interest then due, . . . $41.25 
 
 Payment June 14, 1858, 104. 
 
 Their sum exceeds the interest then due, $145.25 
 
 Remainder for a new principal, $379.17 
 
 Interest from June 14, 1858, to April 15, 1859, 10 mo. 
 
 1 da., 22.19 
 
 Balance due April 15, 1859, $401.36 -f- 
 
 These examples have been wrought according to the method 
 prescribed by the Supreme Court of the U. S., and are suf- 
 ficient to illustrate the following 
 
PARTIAL PAYMENTS. 247 
 
 UNITED STATES RULE. 
 
 I. Find the amount of the given principal to the time of the 
 first payment, and if this payment exceed the interest then due 
 subtract it from the amount obtained, and treat the remainder 
 as a new principal. 
 
 II. But if the interest be greater than any payment, cast the 
 interest on the same principal to a time when the sum of the 
 payments shall equal or exceed the interest due ; subtracting the 
 sum of the payments from the amount of the principal, the re- 
 mainder will form a new principal, on which interest is to be 
 computed as before. 
 
 SAN FRANCISCO, June 20, 1858. 
 
 3. Three years after date we promise to pay Ross & 
 Wade, or order, five hundred fourteen and -f^ dollars, for 
 value received, with 10 per cent, interest. WILDER & BRO. 
 
 On this note were indorsed the following payments : Nov. 
 12, 1858, $105.50 ; March 20, 1860, $200 ; July 10, 1860, 
 $75.60. How much remains due on the note at the time of 
 its maturity? Ans. $242.12-+. 
 
 CHARLESTON, May 7, 1859. 
 
 4. For value received, I promise to pay George Babcock 
 three thousand dollars, on demand, with 7 per cent, interest. 
 
 JOHN MAY. 
 
 On this note were indorsed the following payments : 
 
 Sept. 10, 1859, received $25 
 
 Jan. 1, 1860, " 500 
 
 Oct. 25, 1860, " 75 
 
 April 4, 1861, " 1500 
 
 How much was due Feb. 20, 1862 ? Ans. $1344.35 + . 
 
 Give the United States Court rule for computing interest where 
 partial payments have been made. 
 
248 PERCENTAGE. 
 
 NEW ORLEANS, Aug. 3, 1850. 
 
 5. One year after date I promise to pay George Bailey, or 
 order, nine hundred twelve y 7 ^ dollars, with 5 per cent, in- 
 terest, for value received. JAMES POWELL. 
 
 The note was not paid when due, but was settled Sept. 15, 
 1853, one payment of $250 having been made Jan. 1, 1852, 
 and another of $316.75, May 4, 1853. How much was due 
 at the time of settlement ? Ans. $467.53 . 
 
 S184.56. CINCINNATI, April 2, 1860. 
 
 6. Four months after date I promise to pay J. Ernst & 
 Co. one hundred eighty-four dollars fifty-six cents, for value 
 received. S. ANDERSON. 
 
 The note was settled Aug. 26, 1862, one payment of $50 
 having been made May 6, 1861. How much was due, legal 
 interest being 6 per cent. ? Ans. $154.188 -f- . 
 
 NOTE. A note is on interest after it becomes due, if it contain no 
 mention of interest. 
 
 7. Mr. B. gave a mortgage on his farm for $6000, dated 
 Oct. 1, 1851, to be paid in 6 years, with 8 per cent, interest. 
 Three months from date he paid $500 ; Sept. 10, 1852, $1126 ; 
 March 31, 1854, $2000 ; and Aug. 10, 1854, $876.50. How 
 much was due at the expiration of the time ? Ans. $3284.84 -f- 
 
 SO2. The United States rule for partial payments has 
 been adopted by nearly all the States of the Union ; the only 
 prominent exceptions are Connecticut, Vermont, and New 
 Hampshire. 
 
 CONNECTICUT RULE. 
 
 I. Payments made one year or more from the time the in- 
 terest commenced, or from another payment, and payments less 
 than the interest due, are treated according to the United States 
 rule. 
 
 Give Connecticut rule for partial payments. 
 
PARTIAL PAYMENTS. 249 
 
 
 
 II. Payments exceeding the interest due, and made within 
 one year from the time interest commenced, or from a former 
 payment, shall draw interest for the balance of the year, pro- 
 vided the interval does not extend beyond the settlement, and the 
 amount must be subtracted from the amount of the principal for 
 one year ; the remainder will be the new principal. 
 
 III. If the year extend beyond the settlement, then find the 
 amount of the payment to the day of settlement, and subtract it 
 from the amount of the principal to that day ; the remainder 
 will be the sum due. 
 
 WOODSTOCK, CT., Jan. 1, 1858. 
 
 1. For value received, I promise to pay Henry Bowen, or 
 order, four hundred sixty dollars, on demand, with interest. 
 
 JAMES MARSHALL. 
 
 On this note are indorsed the following payments : April 
 16, 1858, $148 ; March 11, 1860, $75 ; Sept. 21, 1860, $56. 
 How much was due Dec. 11, 1860 ? Ans. $238.14+. 
 
 8O"5, A note containing a promise to pay interest an- 
 nually is not considered in law a contract for any thing more 
 than simple interest on the principal. For partial payments 
 on such notes, the following is the 
 
 VERMONT EULE. 
 
 I. Find the amount of the principal from the time interest 
 commenced to the time of settlement. 
 
 II. Find the amount of each payment from the time it was 
 made to the time of settlement. 
 
 III. Subtract the sum of the amounts of the payments from 
 the amount of the principal, and the remainder will be the 
 sum due. 
 
 RUTLAND, April 11, 1856. 
 
 $600. 
 
 1. For value received, I promise to pay Amos Getting, or 
 order, six hundred dollars on demand, with interest annually. 
 
 JOHN BROWN. 
 
 Give the Connecticut rule for partial payments. The Vermont rule. 
 
250 PERCENTAGE. 
 
 On this note were indorsed the following payments : Aug. 
 10, 1856, $156; Feb. 12, 1857, $200; June 1, 1858, $185. 
 What was due Jan. 1, 1859 ? Am. $105.50+. 
 
 SO4:. In New Hampshire interest is allowed on the an- 
 nual interest if not paid when due, in the nature of damages 
 for its detention ; and if payments are made before one year's 
 interest has occurred, interest must be allowed on such pay- 
 ments for the balance of the year. Hence the following 
 
 NEW HAMPSHIRE RULE. 
 
 I. Find the amount of the principal for one year, and de- 
 duct from it the amount of each payment of that year, from the 
 time it was made up to the end of the year ; the remainder will 
 be a new principal, with which proceed as before. 
 
 II. If the settlement occur less than a year from the last an- 
 nual term of interest, make the last term of interest a part of a 
 year, accordingly. 
 
 575 - KEENE, N. H., Aug. 4, 1858. 
 
 1. For value received, I promise to pay George Cooper, or 
 order, five hundred seventy-five dollars, on demand, with in- 
 terest annually. DAVID GREENMAN. 
 
 On this note were indorsed the following payments : Nov. 
 4, 1858, $64; Dec. 13, 1859, $48; March 16, 1860, $248; 
 Sept. 28, 1860, $60. What was due on the note Nov. 4, 
 1860? Ans. $215.3^. 
 
 3O>. When no payment whatever is made, upon a note 
 promising annual interest, till the day of settlement, in New 
 Hampshire the following is the 
 
 COURT RULE. 
 
 Compute separately tJie interest on the principal from the 
 time the note is given to the time of settlement, and the interest 
 on each year's interest from the time it should be paid to the 
 time of settlement. Th.e sum of the interests thus obtained, 
 added to the principal, will be the sum due. 
 
 The New Hampshire rule. The New Hampshire court rule. 
 
PARTIAL PAYMENTS. 251 
 
 KEENE, N. H., Feb. 2f, 1855. 
 
 1. Three years after date, I promise to pay James Clark, 
 or order, five hundred dollars, for value received, with interest 
 annually till paid. JOHN S. BRIGG-S. 
 
 What is due on the above note, Aug. 2, 1859 ? Ans. $649.40. 
 
 PROBLEMS IN INTEREST. 
 
 SO6. In examples of interest there are five parts involved, 
 the Principal, the Rate, the Time, the" Interest, and the 
 Amount. If any three of these be given, the others may be 
 obtained. 
 
 CASE I. 
 
 3O7. The time, rate per cent., and interest being 
 given, to find the principal. 
 
 1. What principal in 2 years, at 6 per cent., will gain 
 $31.80 interest ? 
 
 OPERATION. ANALYSIS., Since $1, in 
 
 S.I 2, interest of $1 in 2 years at 6 per cent. 2 years, at 6 per cent., will 
 
 $31.80 -I- .12 = $265, Ans. S ain $- 12 interest, the prin- 
 cipal that will gain $31.80, 
 
 at the same rate and time, must be as many dollars as $.12 is con- 
 tained times in $31.80; dividing, we obtain $265, the required 
 principal. Hence, 
 
 RULE. Divide the given interest by the interest of $1 for 
 the given time and rate, and the quotient will be the principal. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What principal, at 6 per cent., will gain $28.12 in 6 
 years 3 months ? Ans. $75. 
 
 3. What sum, put at interest for 4 months 18 days, at 4 
 per cent., will gain $9.20 ? Ans. $600. 
 
 4. What sum of money, invested at 7 per cent., will pay 
 me an annual income of $1260 ? Ans. $18000. 
 
 5. What sum must be invested in real estate, yielding 10 
 per cent, profit in rents, to produce an income of $3370 ? 
 Ans. $33700^ 
 
 How many parts are considered in examples in interest ? What 
 are they ? What is Case I ? Give analysis. Rule. 
 
252 PERCENTAGE. 
 
 CASE II. 
 
 308. The time, rate per cent., and amount being 
 given, to find the principal. 
 
 1. What principal in 2 years 6 months, at 7 per cent, 
 will amount to $88.125? 
 
 OPERATION. ANALYSIS. 
 
 $1.175 Amt. of $1 in 2 years 6 months, at 7 per cent. Smce $*' m 
 
 $88.125 -f- 1.175 = $75, Ans. 
 
 months, at 7 
 
 per cent., will amount to $1.175, the principal that will amount to 
 $88.125, at the same rate and time, must be as many dollars as 
 $1.175 is contained times in $88.125 ; dividing, we obtain $75, the 
 required principal. Hence the 
 
 RULE. Divide the given amount by the amount of $1 for 
 the given time and rate, Ofid the quotient will be the principal 
 required. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What principal, at 6 per cent., will amount to $655.20 
 in 8 months ? Ans. $630. 
 
 3. What principal, at 5 per cent., will amount to $106.855 
 in 5 years 5 months and 9 days ? Ans. $84. 
 
 4. What sum, put at interest, at 5 per cent, for 8 years 
 5 months, will amount to $1897.545 ? Ans. $1297.09+. 
 
 5. What sum, at 7 per cent, will amount to $221.075 in 3 
 years 4 months ? Ans. $179.25. 
 
 6. What is the interest of that sum, for 11 years 8 days, at 
 10 per cent, which will at the given rate and time amount to 
 $857.54? Ans. $460.04. 
 
 CASE III. 
 
 309. The principal, time, and interest being given, 
 to find the rate per cent. 
 
 1. I lent $450 for 3 years, and received for interest $67.50 ; 
 what was the rate per cent ? 
 
 Give case IE. Analysis. Rule. Case IH. 
 
PROBLEMS IN INTEREST. 253 
 
 OPERATION. ANALYSIS. Since at 
 
 $ 4.50 1 per cent. $450, in 3 
 
 3 years, will gain $13.50 
 
 interest, the rate per 
 $1 3.0 0, int. of $450 for 3 years at 1 per cent. , . , ,, 
 
 cent, at which the same 
 
 $67.50 -f- 13.50 = 5 percent., Ans. principal, in the same 
 
 time, will gain $67.50, 
 
 must be equal to the number of times $13.50 is contained in $67.50; 
 dividing, we obtain 5, the required rate per cent. Hence the 
 
 RULE. Divide the given interest by the interest on the prin- 
 cipal for the given time at 1 per cent., and the quotient will be 
 the rate per cent, required. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. If I pay $45 interest for the use of $500 3 years, 
 what is the rate per cent. ? Ans. 3. 
 
 3. The interest of $180 for 1 year 2 months 6 days is 
 $12.78 ; what is the rate per cent. ? Ans. 6. 
 
 4. A man invests $2000 in bank stock, and receives a 
 semi-annual dividend of $75 ; what is the rate per cent. ? 
 
 5. At what per cent, must $1000 be loaned for 3 years 3 
 months and 29 days, to gain $183.18 ? Ans. 5. 
 
 6. A man builds a block of stores at a cost of $2163:0, and 
 receives for them an annual rent of $2596.80 ; what per cent, 
 does he receive on the investment ? Ans. 12. 
 
 CASE IV. 
 
 310. Principal, interest, and rate per cent, being 
 given, to find the time. 
 
 1. In what time will $360 gain $86.40 interest, at 6 per 
 
 cent. ? 
 
 OPERATION. ANALYSIS. Sinee in 
 
 $ 360 1 year $360, at 6 per 
 
 .06 cent., will gain $21.60, 
 
 the number of years in 
 $21.60 Intereetof$360inlyearat6 P ercent. wMch ^ game princi _ 
 
 $86.40H- 21.60 = 4 years, Ans. pal, at the same rate, 
 
 will gain $86.40, will be 
 
 Analysis. Rule. Case IV. Analysis. 
 V 
 
254 PERCENTAGE. 
 
 as many as $21.60 is contained times in $86.40 ; dividing, we ob- 
 tain 4 years, the required time. Hence the 
 
 RULE. Divide the given interest by the interest on the prin- 
 cipal for 1 year, and the quotient will be the time required in 
 years and decimals. 
 
 NOTE. The decimal part of the quotient, if any, may be reduced to 
 months and days (by 2O9). 
 
 EXAMPLES FOR PRACTICE. 
 
 2. The interest of $325 at 6 per cent, is $58.50 ; what is 
 the time ? Ans. 3 years. 
 
 3. B loaned $1600 at 6 per cent, until it amounted to 
 $2000 ; what was the time ? Ans. 4 years 2 months. 
 
 4. How long must $204 be on interest at 7 per cent., to 
 amount to $217.09 ? Ans. 11 months. 
 
 5. Engaging in business, I borrowed $750 of a friend at 6 
 per cent, and kept it until it amounted to $942 ; how long did 
 I retain it ? Ans. 4 years 3 months 6 days. 
 
 6. How long will it take $200 to double itself at 6 per cent, 
 simple interest ? Ans. 1 6 years 8 months. 
 
 7. In what time will $675 double itself at 5 per cent. ? 
 
 NOTE. The time in years in which any sum will double itself may 
 be found by dividing 100 by the rate per cent. 
 
 COMPOUND INTEREST. 
 
 311. Compound Interest is interest on both principal and 
 interest, when the interest is not paid when due. 
 
 NOTE. The simple interest may be added to the principal annually, 
 semi-annually, or quarterly, as the parties may agree ; but the taking 
 of compound interest is not legal. 
 
 1. What is the compound interest of $200, for 3 years, at 
 6 per cent. ? 
 
 Rule. In what time will any sum double itself at interest ? What 
 is compound interest ? 
 
COMPOUND INTEREST. 255 
 
 OPERATION. 
 
 $200 Principal for 1st year. 
 
 X .06 z= 12 Interest for 1st year. 
 
 $212 Principal for 2d year. 
 $212 X -06 = 12.72 Interest for 2d year. 
 
 $224.72 Principal for 3d year. 
 $224.72 X -06 = 13.483 Interest for 3d year. 
 
 $238.203 Amount for 3 years. 
 200.000 Given principal. 
 
 $38.203 Compound interest. 
 
 RULE. I. Find the amount of the given principal at the 
 given rate for one year, and make it the principal for the 
 second year. 
 
 II. Find the amount of this new principal, and make it the 
 principal for the third year, and so continue to do for the given 
 number of years. 
 
 III. Subtract the given principal from the last amount, and 
 the remainder will be the compound interest. 
 
 NOTES. 1. "When the interest is payable semi- annually or quar- 
 terly, find the amount of the given principal for the first interval, and 
 make it the principal for the second interval, proceeding in all respects 
 as when the interest is payable yearly. 
 
 2. When the time contains years, months, and days, find the amount 
 for the years, upon which compute the interest for the months and 
 days, and add it to the last amount, before subtracting. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the compound interest of $500 for 2 years at 7 
 per cent. ? Ans. $72.45. 
 
 3. What is the amount of $312 for 3 years, at 6 per cent, 
 compound interest ? Ans. $371.59-)-. 
 
 4. What is the compound interest of $250 for 2 years, 
 payable semi-annually, at 6 per cent.? Ans. $31.37-}-. 
 
 5. What will $450 amount to in 1 year, at 7 per cent, com- 
 pound interest, payable quarterly ? Ans. $482.33. 
 
 6. What is the compound interest of $236 for 4 years 7 
 months and 6 days, at 6 per cent. ? Ans. $72.66-}-. 
 
 Explain operation. Give rule. 
 
256 
 
 PERCENTAGE. 
 
 7. What is the amount of $700 for 3 years 9 months and 
 24 days, at 7 per cent, compound interest ? Ans. $906.55+- 
 
 A more expeditious method of computing compound interest 
 than the preceding, is by means of the following 
 
 TABLE, 
 
 Showing the amount of$l, or 1, at 3, 4, 5, 6, and 7 per cent., compound 
 interest, for any number of years, from 1 to 20. 
 
 Yrs. 
 
 3 per cent. 
 
 4 per cent. 
 
 5 per cent. 
 
 6 per cent. 
 
 7 per cent. 
 
 1 
 
 1.030,000 
 
 1.040,000 
 
 1.050,000 
 
 1.060,000 
 
 1.07,000 
 
 2 
 
 1.060,900 
 
 1.081,600 
 
 1.102,500 
 
 1.123,600 
 
 1.14,490 
 
 3 
 
 1.092,727 
 
 1.124,864 
 
 1.157,625 
 
 1.191,016 
 
 1.22,504 
 
 4 
 
 1.125,509 
 
 1.169,859 
 
 1.215,506 
 
 1.262,477 
 
 1.31,079 
 
 5 
 
 1.159,274 
 
 1.216,653 
 
 1.276,282 
 
 1.338,226 
 
 1.40,255 
 
 6 
 
 1.194,052 
 
 1.265,319 
 
 1.340,096 
 
 1.418,519 
 
 1.50,073 
 
 7 
 
 1.229,874 
 
 1.315,932 
 
 1.407,100 
 
 1.503,630 
 
 1.60,578 
 
 8 
 
 1.266,770 
 
 1.368,569 
 
 1.477,455 
 
 1.593,848 
 
 1.71,818 
 
 9 
 
 1.304,773 
 
 1.423,312 
 
 1.551,328 
 
 1.689,479 
 
 1.83,845 
 
 10 
 
 1.343,916 
 
 1.480,244 
 
 1.628,895 
 
 1.790,848 
 
 1.96,715 
 
 11 
 
 1.384,234 
 
 1.539,454 
 
 1.710,339 
 
 1.898,299 
 
 2.10,485 
 
 12 
 
 1.425,761 
 
 1.601,032 
 
 1.795,856 
 
 2.012,196 
 
 2.25,219 
 
 13 
 
 1.468,534 
 
 1.665,074 
 
 1.885,649 
 
 2.132,928 
 
 2.40,984 
 
 14 
 
 1.512.590 
 
 1.731,676 
 
 1.979,932 
 
 2.260,904 
 
 2/57,853 
 
 15 
 
 1.557,967 
 
 1.800,944 
 
 2.078,928 
 
 2.396,558 
 
 2.75,903 
 
 16 
 
 1.604,706 
 
 1.872,981 
 
 2.182,875 
 
 2.540,352 
 
 2.95,216 
 
 17 
 
 1.652,848 
 
 1.947,900 
 
 2.292,018 
 
 2.692,773 
 
 3.15,881 
 
 18 
 
 1.702,433 
 
 2.025,817 
 
 2.406,619 
 
 2.854,339 
 
 3.37,293 
 
 19 
 
 1.753,506 
 
 2.106,849 
 
 2.526,950 
 
 3.025,600 
 
 3.61,652 
 
 20 
 
 1.806,111 
 
 2.191,12312.653,298 
 
 3.207,135 3.86,968 
 
 8. What is the amount of $800 for 6 years, at 7 per cent. 
 
 OPERATION. 
 
 From the table $1.50073 Amount of $1 for the time, 
 800 Principal. 
 
 $1200.58400, Ans. 
 
 9. What is the compound interest of $120 for 15 years, at 
 5 per cent.? Ans. $129.47+. 
 
 Of what use is the table in computing compound interest ? 
 
DISCOUNT. 257 
 
 10. What is the amount of $.10 for 20 years, at 7 per 
 cent.? Ans. $.38696. 
 
 DISCOUNT. 
 
 81'2. Discount is an abatement or allowance made for 
 the payment of a debt before it is due. 
 
 313. The Present Worth of a debt, payable at a future 
 time without interest, is such a sum as, being put at legal in- 
 terest, will amount to the given debt when it becomes due. 
 
 1. A owes B $321, payable in 1 year; what is the pres- 
 ent worth of the debt, the use of money being worth 7 per 
 cent. ? 
 
 OPERATION. ANALYSIS. The 
 
 Am'tof $1, 1.07) $321 ($300, Present value, amount of $1 for 
 
 321 ! Y ear is $1-07; 
 
 therefore the pres- 
 
 $321 Given sum or debt. ent WQrth of eyery 
 
 300 P resen t rth. $1.07 of the given 
 
 $21 Discount. debt is * l * and 
 
 the present worth 
 
 of $321 will be as many dollars as $1.07 is contained times in $321. 
 $321 -^ 1.07 = $300, Ans. Hence the following 
 
 RULE. I. Divide the given sum or debt by the amount of 
 $1 for the given rate and time, and the quotient will be the pres- 
 ent worth of the debt. 9 
 
 11. Subtract the present worth from the given sum or debt) 
 and the remainder will be the discount. 
 
 NOTE. The terms present worth, discount, and debt, are equivalent 
 to principal, interest, and amount. Hence, when the time, rate per 
 cent., and amount are given, the principal may be found by (3O8) 
 and the interest by subtracting the principal from the amount. 
 
 EXAMPLES FOB PRACTICE. 
 
 2. What is the present worth of $180, payable in 3 years 
 4 months, discounting at 6 per cent. ? Ans. $150. 
 
 Define discount. Present worth. Give analysis. Rule. 
 
258 PERCENTAGE. 
 
 3. What is the present worth of a note for $1315.389, due 
 in 2 years 6 months, at 7 per cent.? Ans. $1119.48. 
 
 4. What is the present worth of a note for $866.038, due 
 in 3 years 6 months and 6 days, when money is worth 8 per 
 cent.? What the discount ? Ans. $190.15-)-, discount. 
 
 5. What is the present worth of a debt for $1005, on which 
 $475 is to be paid in 10 months, and the remainder in 1 year 3 
 months, the rate of interest being 6 per cent. ? 
 
 NOTE. When payments are to be made at different times without 
 interest, find the present worth of each payment separately, and take 
 their sum. 
 
 Ans. $945.40+. 
 
 6. I hold a note against C for $529.925, due Sept. 1, 1859 ; 
 what must I discount for the payment of it to-day, Feb. 7, 
 1859, money being worth 6 per 'cent. ? Ans. $17.425. 
 
 7. A man was offered $3675 in cash for his house, or 
 $4235 in 3 years, without interest; he accepted the latter 
 offer ; how much did he lose, money being worth 7 per cent. ? 
 
 Ans. $175. 
 
 8. A man, having a span of horses for sale, offered them 
 for $480 cash in hand, or a note of $550 due in 1 year 8 
 months, without interest ; the buyer accepted the latter offer ; 
 did the seller gain or lose by his offer, and how much, interest 
 being 6 per cent. ? Ans. Seller lost $20. 
 
 9. What must be discounted for the present payment of a 
 debt of $2637.72, of which $517.50 is to be paid in 6 months, 
 $793.75 in 10 months, and the remainder in 1 year 6 months, 
 the use of money being worth 7 per cent. ? Ans. $187.29 +. 
 
 10. What is the difference between the interest and discount 
 of $130, due 10 months hence, at 10 per cent, ? Ans. $.83. 
 
 PROMISCUOUS EXAMPLES IN PERCENTAGE. 
 
 1. A merchant bought sugar in New York at 0] rents per 
 pound ; (lie. wastage by transportation and retailing was 5 per 
 cent., and the interest on the first cost to the time of sale was 
 2 per cent. ; how much must he ask per pound to gain 25 per 
 cent. ? Ans. 8 cents. 
 
PROMISCUOUS EXAMPLES. 259 
 
 2. A person purchased 2 lots of land for $200 each, and , ^ 
 sold one at 40 per cent, more than cost, and the other at 20 
 per cent, less ; how much did he gain ? Ans. $40. 
 
 3. Sold goods to the amount of $425, on 6 months' credit, 
 which was 25 more than the goods cost; what was the true 
 profit, money being worth 6 per cent. ? Ans. $12.62 -(-. 
 
 4. Bought cotton cloth at 13 cents a yard, on 8 months' 
 credit, and sold it the same day at 12 cents cask; how much * 
 did I gain or lose per cent., money being worth 6 per cent. ? 
 
 Ans. Lost 4 per cent. 
 
 5. A farmer sold a pair of horses for $150 each; on one 
 he gained 25 per cent., on the other he lost 25 per cent. ; did 
 he gain or lose on both, and how much ? Ans. Lost $20. 
 
 6. A man invested of all he was worth in the coal trade, 
 and at the end of 2 years 8 months sold out his entire interest 
 for $3100, which was a yearly gain of 9 per cent, on the 
 money invested ; how much was he worth when he commenced 
 trade? Ans. $3750. 
 
 7. In how many years will a man, paying interest at 7 per 
 cent, on a debt for land, pay the face of the debt in interest ? 
 
 Ans. 1 4f years. 
 
 8. Two persons engaged in trade ; A furnished f of the 
 capital, and B f; and at the end of 3 'years 4 months they 
 found they had made a clear profit of $5000, which was 124- 
 per cent, per annum on the money Invested ; how much cap- 
 ital did each furnish ? Ans. A, $7500 ; B, $4500. 
 
 9. Bought $500 worth of dry goods, and $800 worth of 
 groceries ; on the dry goods I lost 20 per cent., but on the 
 groceries I gained 15 per cent. ; did I gain or lose on the 
 whole investment, and how much ? Ans. Gained $20. 
 
 10. What amount of accounts must an attorney collect, in 
 order to pay over $1100, and retain 8^- per cent, for collect- 
 ing? Ans. $1200. 
 
 11. A merchant sold goods to the amount of $667, to be 
 paid in 8 months ; the same goods cost him $600 one year 
 previous to the sale of them ; money being worth 6 per cent., 
 what was his true gain ? Ans. $5.346 -(-. 
 
260 PERCENTAGE. 
 
 12. A nurseryman sold trees at $18 per hundred, and 
 cleared ^ of his receipts ; what per cent, profit did he make ? 
 
 Ans. 50 per cent. 
 
 13. If | of an article be sold for what f- of it cost, what is 
 the gain per cent. ? . Ans. 40|. 
 
 14. A lumber merchant sells a lot of lumber, which he has 
 had on hand 6 months, on 10 months' credit, at an advance of 
 30 per cent, on the first cost ; if he is paying 5 per cent, inter- 
 est on capital, what are his profits per cent. ? Ans. 21 J-J-. 
 
 15. A person, owning | of a piece of property, sold 20 per 
 cent, of his share ; what part did he then own ? Ans. ^. 
 
 16. A speculator, having money in the bank, drew 60 per 
 cent, of it, and expended 40 per cent, of 50 per cent, of this for 
 728 bushels of wheat, at $1.50 per bushel ; how much was 
 left in the bank ? Ans. $3640. 
 
 17. I wish to line the carpet of a room, that is 6 yards long 
 and 5 yards wide, with duck yard wide ; how many yards of 
 lining must I purchase, if it will shrink 4 per cent, in length, 
 and 5 per cent, in width ? Ans. 43f f . 
 
 18. A's money is 28 per cent, more than B's ; how many 
 per cent, is B's less than A's ? Ans. 21. 
 
 19. A capitalist invested -| of his money in railroad stock, 
 which depreciated 5 per cent, in value ; the remaining f he in- 
 vested in bank stock, which, at the end of 1 year, had gained 
 $1200, which was 12 per cent, of the investment ; what was the 
 whole amount of his capital, and what was his entire loss or 
 gain ? Ans. $25000, capital ; $450, gain. 
 
 20. C's money is to D's as 2 to 3 ; if % of C's money be 
 put at interest for 3 years 9 months, at 10 per cent., it will 
 amount to $1933.25 ; how much money has each ? 
 
 Ans. C, $2812 ; D, $4218. 
 
 BANKING. 
 314. A Bank is a corporation chartered by law for the 
 
 purpose of receiving and loaning money, and furnishing a 
 paper circulation. 
 
 What is a bank? 
 
BANKING. 261 
 
 315. A Promissory Note is a written or printed engage- 
 ment to pay a certain sum, either on demand or at a specified 
 time. 
 
 31G. Bank Notes, or Bank Bills, are the notes made 
 and issued by banks to circulate as money. They are payable 
 in specie at the banks. 
 
 317. The Face of a note is the sum made payable 'by 
 the note. 
 
 318. Days of Grace are the three days usually allowed 
 by law for the payment of a note after the expiration of the 
 time specified in the note. 
 
 319. The Maturity of a note is the expiration of the 
 days of grace ; a note is due at maturity. 
 
 320. Notes may contain a promise of interest, which will 
 be reckoned from the date of the note, unless some other time 
 be specified. 
 
 The transaction of borrowing money at banks is conducted 
 in accordance with the following custom : the borrower pre- 
 sents a note, either made or indorsed by himself, payable at 
 a specified time, and receives for it a sum equal to the face, 
 less the interest for the time the note has to run. The amount 
 thus withheld by the bank is in consideration of advancing 
 money on the note prior to its maturity. 
 
 321. Bank Discount is an allowance made to a bank for 
 the payment of a note before it becomes due. 
 
 322. The Proceeds of a note is the sum received for it 
 when discounted, and is equal to the face of the note less the 
 discount. 
 
 CASE I. 
 
 323. Given the face of a note to find the proceeds. 
 The law of custom at banks makes the discount of a note 
 
 Define a promissory note. Bank notes. The face of a note. Days 
 of grace. The maturity of a note. Explain the process of discounting 
 a note at a bank. Define bank discount. The proceeds of a note. 
 What is Case I ? 
 
262 PERCENTAGE. 
 
 equal to the simple interest at the legal rate for the time spe- 
 cified in the note. Hence the 
 
 RULE. I. Compute the interest on the face of the note for 
 three days more than the specified time ; the result will be the 
 discount. 
 
 II. Subtract the discount from the face of the note, and the 
 remainder will be the proceeds. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. "What is the discount, and what the proceeds, of a note 
 for $450, at 60 days, discounted at a bank at 6 per cent. ? 
 
 Ans. Discount, $4.725 ; proceeds, $445.275. 
 
 2. What are the proceeds of a note for $368, at 90 days, 
 discounted at the Bank of New York ? Ans. $361.345 -[-. 
 
 3. What shall I receive on my note for $475.50, at 60 
 days, if discounted at the Crescent City Bank, New Orleans ? 
 
 Ans. $471.33+. 
 
 4. What are the proceeds of a note for $10000, at 90 days, 
 discounted at the Philadelphia Bank ? Ans. $9845. 
 
 5. Paid, in cash, $240 for a lot of merchandise. Sold it 
 the same day, receiving a note for $250 at 60 days, which I 
 got discounted at the Hartford Bank. What did I make by 
 this speculation ? Ans. $7.37. 
 
 6. A note for $360.76, drawn at 90 days, is discounted at 
 the Vermont Bank. Find the proceeds. Ans. $355.1 68 -f-. 
 
 7. Wishing to borrow $530 of a western bank which is 
 discounting paper at 8 per cent., I give my note for $536.75, 
 payable in 60 days. How much do I need to make up the 
 required amount ? Ans. $.7(>-ir>. 
 
 NOTES. 1. To indicate the maturity of a note or draft, a- vortical 
 line ( | ) is used, with the day at which the note is nominally due on 
 the left, and the date of maturity on the right ; thus, Jan. ' \ 10 . 
 
 2. When a note is on interest, payable at a future specified time, the 
 amount is the face of the note, or the sum made payable, and must be 
 made the basis of discount. 
 
 Give rule. 
 
BANKING. 
 
 Find the maturity, term of discount, and proceeds of the 
 
 following notes : 
 
 $500. BOSTON, Jan. 4, 1859. 
 
 8. Three months after date, I promise to pay to the order of 
 John Brown & Co. five hundred dollars, at the Suffolk Bank, 
 value received. JAMES BARKER. 
 
 Discounted March 2. ( Due, April 4 | 7 . 
 
 Am. < Term of discount, 36 da. 
 
 ( Proceeds, $497. 
 
 $750. ST. Louis, June 12, 1859. 
 
 9. Six months after date, I promise to pay Thomas Lee, or 
 order, seven hundred fifty dollars, with interest, value re- 
 ceived. BYRON QUINBY. 
 
 Discounted at a broker's, Nov. 15, at 10 per cent. 
 
 (Due, Dec."| 1B . 
 
 Ans. I Term of discount, 30 da. 
 
 (Proceeds, $766.434+. 
 
 CASE II. 
 
 324. Given the proceeds of a note, to find the 
 face. 
 
 1. I wish to borrow $400 at a bank. For what sum must 
 I draw my note, payable in 60 days, so that when discounted 
 at 6 per cent. I shall receive the desired amount ? 
 
 OPERATION. ANALYSIS. $400 is the 
 
 $1.0000 proceeds of a certain note, 
 
 .0105 disc, on $1 for 63 da. the face of which we are 
 
 required to find. We first 
 
 $ .9895 = proceeds of $1. obtain the proceeds of $1 
 
 $400 H- -9895 = $404.244 = by the last case, and then 
 
 face of the required note. divide the given proceeds, 
 
 $400, by this sum ; for, as many times as the proceeds of $1 is con- 
 tained in the given proceeds, so many dollars must be the face of 
 the required note. Hence the 
 
 Give Case II. Analysis. 
 
264 PERCENTAGE. 
 
 RULE. Divide the proceeds by the proceeds of $1 for the 
 time and rate mentioned, and the quotient will be the face of 
 the note. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the face of a note at 60 days, which yields 
 $680 when discounted at a New Haven bank ? 
 
 Ans. $687.215+. 
 
 3. What is the face of a note at 90 days, of which the pro- 
 ceeds are $1000 when discounted at a Louisiana bank ? 
 
 Ans. $1013.085+. 
 
 4. Wishing to borrow $500 at a bank, for what sura must 
 my note be drawn, at 30 days, to obtain the required amount, 
 discount being at 7 per cent. ? Ans. $503.22 +. 
 
 5. James Hopkins buys merchandise of me in New York, 
 at cash price, to the amount of $1256. Not having money, 
 he gives his note in payment, drawn at 6 months. What 
 must be the face of the note ? Ans. $1302.341 +. 
 
 EXCHANGE. 
 
 Exchange is a method of remitting money from one 
 place to another, or of making payments by written orders. 
 
 !'3G. A Bill of Exchange is a written request or order 
 upon one person to pay a certain sum to another person, or to 
 his order, at a specified time. 
 
 SS7. A Sight Draft or Bill is one requiring payment to 
 be made " nt sight," which means, at the time of its presenta- 
 tion to the person ordered to pay. In other bills, the time 
 specified is usually a certain number of days "after sight." 
 
 There are always three parties, and usually four, to a trans- 
 action in exchange : 
 
 328. The Drawer or Maker is the person who signs the 
 order or bill. 
 
 the rule. Define exchange. A bill of exchange. A si^lit draft. 
 The drawer. 
 
EXCHANGE. 265 
 
 329. The Drawee is the person to whom the order is 
 addressed. 
 
 330. The Payee is the person to whom the money is or- 
 dered to be paid. 
 
 331. The Buyer or Remitter is the person who purchases 
 the bill. He may be himself the payee, or the bill may be 
 drawn in favor of any other person. 
 
 332. The Indorsement of a bill is the writing upon its 
 back, by which the payee relinquishes his title, and transfers 
 the payment to another. The payee may indorse in blank by 
 writing his name only, which makes the bill payable to the 
 bearer, and consequently transferable like a bank note ; or he 
 may accompany his signature by a special order to pay to 
 another person, who in his turn may transfer the title in like 
 manner. Indorsers become separately responsible for the 
 amount of the bill, in case the drawee fails to make payment. 
 A bill made payable to the bearer is transferable without in- 
 dorsement. 
 
 333. The Acceptance of a bill is the promise which the 
 drawee makes when the bill is presented to him to pay it at 
 maturity ; this obligation is usually acknowledged by writing 
 the word " Accepted," with his signature, across the face of 
 the bill. 
 
 Three days of grace are usually allowed for the payment of 
 a bill of exchange after the time specified has expired. But in New 
 York State no grace is allowed on sight drafts. 
 
 From these definitions, the use of a bill of exchange in mon- 
 etary transactions is readily perceived. If a man wishes to 
 make a remittance to a creditor, agent, or any other person 
 residing at a distance, instead of transporting specie, which is 
 attended with expense and risk, or sending bank notes, which 
 are liable to be uncurrent at a distance from the banks that 
 issued them, he remits a bill of exchange, purchased at a bank 
 or elsewhere, and made payable to the proper person in or 
 
 The drawee. The payee. The buyer. An indorsement. An 
 acceptance. What of grace on bills of exchange ? 
 
 W 
 
266 PERCENTAGE. 
 
 near the place where he resides. Thus a man by paying 
 Boston funds in Boston, may put New York funds into the 
 hands of his New York agent. 
 
 334. The Course of Exchange is the variation of the 
 cost of sight bills from their par value, as affected by the rela- 
 tive conditions of trade and commercial credit at the two places 
 between which exchange is made. It may be either at a pre- 
 mium or discount, and is rated at a certain per cent, on the 
 face of the bill. Bills payable a specified time after sight are 
 subject to discount, like notes of hand, for the term of credit 
 given. Hence their value in the money market is affected by 
 both "the course of exchange and the discount for time. 
 
 335. Foreign Exchange relates to remittances made be- 
 tween different countries. 
 
 336. Domestic or Inland Exchange relates to remit- 
 tances made between different places in the same country. 
 
 An inland bill of exchange is commonly called a Draft. 
 In this work we shall treat only of Inland Exchange. 
 
 CASE I. 
 
 337. To find the cost of a draft. 
 
 $500. SYRACUSE, May 7, 1859. 
 
 1. At sight, pay to James Clark, or order, five hundred 
 dollars, value received, and charge the same to our account. 
 To M. SMITH & Co. 
 
 Messrs. BROWN & FOSTER, ) 
 
 Baltimore. 
 
 What is the cost of the above draft, the rate of exchange 
 being 1 per cent, premium? 
 
 OPERATION. ANALYSIS. Since ex- 
 
 $500 X 1 .01 5 = $507.50, Ans. chan ? e is at ^ ST, Cent ; 
 
 premium, each dollar ot 
 
 the draft will cost $1.015 ; and to find the whole cost of the draft, 
 
 How is exchange conducted ? Explain course of exchange. For- 
 eign exchange. Inland exchange. Define a draft. What is Case I 
 Give analysis. 
 
EXCHANGE. 267 
 
 we multiply its face, $500, by 1.015, and obtain $507.50, the re- 
 quired Ans. 
 
 $480. BOSTON, June 12, 1859. 
 
 2. Thirty days after sight, pay to John Otis, or bearer, four 
 hundred eighty dollars, value received, and charge the same 
 to account of AMOS TRENCHARD. 
 
 To JOHN STILES & Co., 
 New York. 
 
 What is the cost of the above draft, exchange being at a 
 premium of 3 per cent. ? 
 
 OPERATION. ANALYSIS. Since 
 
 $1.0000 time is allowed, the 
 
 .0055 z=z discount for 33 days. draft must suffer dis- 
 
 "TTr count in the sale. The 
 
 $ .9945=: proceeds of $1. ,. 
 
 discount of $1, at the 
 
 .Uo z= rate of exchange. i i L T p 
 legal rate in Boston, for 
 
 $1.0245= cost of $1 of the draft. the specified time, al- 
 
 $480 X 1.0245 = $491.76, Ans. lo g grace, is $.0055, 
 
 which, subtracted from 
 
 $1, gives $.9945, the cost of $1 of the dr.aft, provided sight ex- 
 change were at par ; but sight exchange being at premium, we add 
 the rate, .03, to .9945, and obtain $1.0245, the actual cost of $1. 
 Then, multiplying $480 by 1.0245, we obtain $491.76, the Ans. 
 From these examples we derive the following 
 
 RULE. I. For sight drafts. Multiply the face of the draft 
 by 1 plus the rate when exchange is at a premium, and by 
 1 minus the rate when exchange is at a discount. 
 
 II. For drafts payable after sight. Find the proceeds of $1 
 at bank discount for the specified time, at the legal rate where 
 the draft is purchased; then add the rate of exchange when 
 at a premium, or subtract it when at a discount, and multiply 
 the face of the draft by this result. 
 
 EXAMPLES FOR PRACTICE. 
 
 3. A merchant in Cincinnati wishes to remit $1000 by 
 
 Give analysis. Rule I; IE. 
 
268 PERCENTAGE. 
 
 \ 
 
 draft to his agent in New York ; what will the bill cost, ex- 
 change being at 3 per cent, premium ? Am. $1030. 
 
 4. What will be the cost in Rochester of a draft on Albany 
 for $400, payable at sight, exchange being at f per cent, pre- 
 mium? Ans. $403. 
 
 5. A merchant in St. Louis orders goods from New York, 
 to the amount of $530, which amount he remits by draft, ex- 
 change being at 2f per cent, premium. If he pays $20 for 
 transportation, what will the goods cost him in St. Louis ? 
 
 Ans. $564.575. 
 
 6. What will be the cost, in Detroit, of a draft on Boston 
 for $800, payable 60 days after sight, exchange being at a pre- 
 mium of 2 per cent. ? Ans. $806.20. 
 
 7. A man in Philadelphia purchased a draft on Chicago for 
 $420, payable 30 days after sight ; what did it cost him, the 
 rate of exchange being 1^ per cent, discount? Ans. $411.30. 
 
 8. A merchant in Portland receives from his agent 320 
 barrels of flour, purchased in Chicago at $10 per barrel ; in 
 payment for which he remits a draft on Chicago, at 2 per 
 cent, discount. The transportation of his flour cost $312. 
 What must he sell it for per barrel to gain $400 ? Ans. $12. 
 
 CASE II. 
 
 338. To find the face of a draft which a given sum 
 will purchase. 
 
 1. A man in Indiana paid $369.72 for a draft on Boston, 
 drawn at 30 days ; what was the face of the draft, exchange 
 being at 3 J per cent, premium ? 
 
 OPERATION. ANALYSIS. We find, 
 
 $369.72 1.027 = $360, Ans. b ^ Case T ' that a draft 
 
 for $1 will cost $1.027; 
 
 hence the draft that will cost $369.72 must he for as many dollars as 
 1.027 is contained times in $369.72; dividing, we obtain 813(30, the 
 Ans. From this example and analysis we derive the following 
 
 What is Case II ? (Jive analysis. 
 
EQUATION OF PAYMENTS. 269 
 
 RULE. Divide the given cost by the cost of a draft for $1, 
 at the given rate of exchange ; the quotient will be the face of 
 the required draft. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What draft may be purchased for $243. GO, exchange 
 being at 1 per cent, premium? Ans. $240. 
 
 3. What draft may be purchased for $79.20, exchange be- 
 ing at 1 per cent, discount ? Ans. $80. 
 
 4. An agent in Pittsburg holding $282.66, due his em- 
 ployer in New Haven, is directed to make the remittance 
 by draft, drawn at 60 days. What will be the face of the 
 draft, exchange being at 2 per cent, premium ? Ans. $280. 
 
 5. An emigrant from Bangor takes $240 in bank bills to 
 St. Paul, Min., and there pays per cent, brokerage in ex- 
 change for current money. What would he have saved by 
 purchasing in Bangor a draft on St. Paul, drawn at 30 days, 
 exchange being at 1 per cent, discount ? Ans. $5.60. 
 
 6. A Philadelphia manufacturer is informed by his agent in 
 Buffalo that $3600 is due him on the sale of some property. 
 He instructs the agent to remit by a draft payable in 60 days 
 after sight, exchange being at f per cent, premium. The agent, 
 by mistake, remits a sight draft, which, when received in Phila- 
 delphia, is accepted, and paid after the expiration of the three 
 days of grace. If the manufacturer immediately puts this 
 money at interest at the legal rate, will he gain or lose by the 
 blunder of his agent ? Ans. He will lose $8.24-|-. 
 
 EQUATION OF PAYMENTS. 
 
 339. Equation of Payments is the process of finding the 
 mean or equitable time of payment of several sums, due at 
 different times without interest. 
 
 340. The Term of Credit is the time to elapse before a 
 debt becomes due. 
 
 Bole. Define equation of payments. Term of credit. 
 W* 
 
270 EQUATION OF PAYMENTS. 
 
 5141 . The Average Terra of Credit is the time to elapse 
 before several debts, due at different times, may all be paid 
 at once, without loss to debtor or creditor. 
 
 342. The Equated Time is the date at which the several 
 debts may be canceled by one payment. 
 
 CASE I. 
 
 343. When all the terms of credit begin at the 
 same date. 
 
 1. On the first day of January I find that I owe Mr. Smith 
 8 dollars, to be paid in 5 months, 10 dollars to be paid in 2 
 months, and 12 dollars to be paid in 10 months ; at what time 
 may I pay the whole amount ? 
 
 OPERATION. 
 $ 8 X 5 = 40 
 10 X 2 = 20 
 12 X 10 = 120 
 
 30 180 -7- 30 = 6 mo., average time of credit. 
 
 Jan. 1. -f- 6 mo. = July 1, equated time of payment. 
 ANALYSIS. The whole amount to be paid, as seen above, is $30 : 
 and we are to find how long it shall be withheld, or what term of 
 credit it shall have, as an equivalent for the various terms of credit 
 on the different items. Now, the value of credit on any sum is meas- 
 ured by the product of the money and time. And we say, the credit 
 on $8 for 5 mo. = the credit on $40 for 1 mo., because 8 X 5 = 40 
 X 1. In the same manner, we have, the credit on $10 for 2 mo.z=: 
 the credit on $20 for 1 mo. ; and the credit on $12 for 10mo.:= 
 the credit on $120 for 1 mo. Hence, by addition, the value of the 
 several terms of credit on their respective sums equals a credit of 1 
 month on $180; and this equals a credit of 6 months on $30, be- 
 cause 
 
 30X6z=180X 1. 
 
 RULE. I. Multiply each payment by its term of credit, and 
 divide the sum of tie products In/ the sum of the payments ; the 
 quotient will be the average term of credit. 
 
 Average term of credit. Equated time. Give Case I. Analysis. 
 Bule. 
 
AVERAGING CREDITS. 271 
 
 II. Add the average term of credit to the date at which all 
 the credits begin, and the result will be the equated time of 
 payment. 
 
 NOTES. 1. The periods of time used as multipliers must all be of 
 the same denomination, and the quotient will be of the same denomi- 
 nation as the terms of credit ; if these be months, and there be a re- 
 mainder after the division, continue the division to days by reduction, 
 always taking the nearest unit in the last result. 
 
 2. The several rules in equation of payments are based upon the 
 principle of bank discount ; for they imply that the discount of a sum 
 paid before it is due equals the interest of the same amount paid after 
 it is due. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. On the 25th of September a trader bought merchandise, 
 as follows : $700 on 20 days' credit ; $400 on 30 days' credit ; 
 $700 on 40 days' credit : what was the average term of credit, 
 and what the equated time of payment ? 
 
 A ( Average credit, 30 days. 
 
 (. Equated time of payment, Oct. 25. 
 
 3. On July 1 a merchant gave notes, as follows : the first 
 for $250, due in 4 months ; the second for $750, due in 2 
 months ; the third for $500, due in 7 months : at what time 
 may they all be paid in one sum ? Ans. Nov. 1. 
 
 4. A farmer bought a cow, and agreed to pay $1 on Mon- 
 day, $2 on Tuesday, $3 on Wednesday, and so on for a week ; 
 desirous afterward to avoid the Sunday payment, he offered to 
 pay the whole at one time : on what day of the week would 
 this payment come ? Ans. Friday. 
 
 5. Jan. 1, I find myself indebted to John Kennedy in sums 
 as follows : $650 due in 4 months ; $725 due in 8 mouths ; and 
 $500 due in 12 months : at what date may I settle by giving 
 my note on interest for the whole amount ? Ans. Aug. 20. 
 
 CASE II. 
 
 344. When the terms of credit begin at different 
 dates, and the account has only one side. 
 
 345. An Account is the statement or record of mercantile 
 transactions in business form. 
 
 Give Case II. Define an account. 
 
272 
 
 EQUATION OF PAYMENTS. 
 
 3<1G. The Items of an account may be sums due at the 
 date of the transaction, or on credit for a specified time. 
 
 An account may have both a debit and a credit side, the 
 former marked Dr., the latter Cr. Suppose A and B have 
 dealings in which there is an interchange of money or prop- 
 erty ; A keeps the account, heading it with B's name ; the Dr. 
 side of the account shows what B has received from A ; the 
 Cr. side shows what he has parted with to A. 
 
 347. The Balance of account is the difference of the two 
 sides, and may be in favor of either party. 
 
 If, in the transactions, one party has received nothing from 
 the other, the balance is simply the whole amount, and the 
 account has but one side. Bills of purchase are of this class. 
 
 NOTE. Book accounts bear interest after the expiration of the term 
 of credit, and notes after they become due. 
 
 348. To Average an Account is to find the mean or 
 equitable time of payment of the balance. 
 
 349. A Focal Date is a date to which all the others are 
 compared in averaging an account. 
 
 1. When does the amount of the following bill become due, 
 by averaging ? 
 J. C. SMITH, 
 
 1859. To C. E. BORDEN, Dr. 
 
 June 1. To Cash, $450 
 
 " 12. " Mdse. on 4 mos., 500 
 
 Aug. 16. " Mdse., 250 
 
 FIRST OPERATION. 
 
 SECOND OPERATION. 
 
 Due. 
 
 da. 
 
 Items. 
 
 Prod. 
 
 June 1 
 
 
 
 450 
 
 
 Oct. 12 
 
 133 
 
 500 
 
 66500 
 
 Aug. 16 
 
 70 
 
 250 
 
 19000 
 
 
 1200 
 
 85500 
 
 Due. 
 
 da. 
 
 Items. 
 
 Prod. 
 
 June 1 
 
 133 
 
 450 
 
 59850 
 
 Oct. 12 
 
 
 
 500 
 
 
 Aug. 16 
 
 57 
 
 250 
 
 14250 
 
 
 1200 
 
 74100 
 
 85500 -f- 1200 = 71 da. 
 A ^ 71 da. after June 1, 
 Ans '\orAug. 11. 
 
 74100-^1200 = 62 da. 
 
 < 62 da. before Oct. 12, 
 Ans '\ or Aug. 11. 
 
 Define items. Balance. To average an account. A focal date. 
 

 AVERAGING ACCOUNTS. 273 
 
 ANALYSIS. By reference to the example, it will be seen that the 
 items are due June 1, Oct. 12, and Aug. 16, as shown in the two 
 operations. In the first operation we use the earliest date, June 1, 
 as a focal date, and find the difference in days between this date and 
 each of the others, regard being had to the number of days in cal- 
 endar months. From June 1 to Oct. 12 is 133 da. ; from June 1 to 
 Aug. 16 is 76 da. Hence the first item has no credit from June 1, 
 the second item has 133 days' credit from June 1, and the third 
 item has 76 days' credit from June 1, as appears in the column 
 marked da. After this we proceed precisely as in Case I, and find 
 the average credit, 71 da., and the equated time, Aug. 11. 
 
 In the second operation, the latest date, Oct. 12, is taken for a 
 focal date ; the work is explained thus : Suppose the account to be 
 settled Oct. 12. At that time the first item has been due 133 days, 
 and must therefore draw interest for this time. But interest on 
 $450 for 133 days = the interest on $59850 for 1 da. The second 
 item draws no interest, because it falls due Oct. 12. The third item 
 must draw interest 57 days. But interest on $250 for 57 days = 
 the interest on $14250 for 1 day. Taking the sum of the products, 
 we find the whole amount of interest due on the account, at Oct. 12, 
 equals the interest on $74100 for 1 day; and this, by division, is 
 found to be equal to the interest on $1200 for 62 days, which time 
 is the average term of interest. Hence the account would be settled 
 Oct. 12, by paying $1200 with interest on the same for 62 days. This 
 shows that 1200 has been due 62 days ; that is, it falls due Aug. 11, 
 without interest. Hence the following 
 
 RULE. I. Find the time at which each item becomes due, 
 by adding to the date of each transaction the term of credit, if 
 any be specified, and write these dates in a column. 
 
 II. Assume either the earliest or the latest date for a focal 
 date, and find the difference in days between the focal date and 
 each of the other dates, and write the results in a second column. 
 
 III. Write the items of the account in a third column, and 
 multiply each sum by the corresponding number of days in the 
 preceding column, writing the products in a final column. 
 
 IV. Divide the sum of the products by the sum of the items. 
 The quotient will be the average term of credit when the 
 
 Give analysis. Rule. 
 
274 EQUATION OP PAYMENTS. 
 
 earliest date is the focal date, or the average term of interest 
 when the latest date is the focal date ; in either case always 
 reckon from the focal date toward the other dates, to find the 
 equated time of payment. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. JOHN BROWN, 
 
 1859o To JAMES GREIGG, Dr. 
 
 Jan. 1. To 50 yds. Broadcloth, fa) $3.00, . . . $150 
 " 16. 2000 " Calico, " .10, . . 200 
 Feb. 4. 75 Carpeting, " 1.33, . . 100 
 March 3. 400 " Oil Cloth, " .40,... 160 
 If James Greigg wishes to settle the above bill by giving 
 his note, from what date shall the note draw interest ? 
 
 Ans. Jan. 28. 
 
 3. ABRAM RUSSEL, 
 
 1859, To WYNKOOP & BRO., Dr. 
 
 March 1. To Cash, . . , . . e . . > , $300 
 
 April 4. " Mdse., . . ... . <, .... 240 
 
 June 18. " " on 2 mo., . , c * . 100 
 Aug. 8. " Cash, . . * o , . o . o . . . . . . 400 
 
 "What is the equated time of payment of the above account ? 
 
 Ans. May 26. 
 
 4. JOHN OTIS, 
 
 1858. To JAMES LADD, Dr. 
 
 June 1. To 500 bu. Wheat, fa) $1.20, ..,,.. $600 
 
 " 12. " 200 " " 1.50, 300 
 
 " 15. " 640 " " " 1.30, ..... o 832 
 
 " 25. <fc 760 " " 1.00,, .<*. 760 
 
 " 30. " 500 " 1.50, , . . , .. 750 
 
 When is the whole amount of the above bill due, per 
 
 average ? Ans. June 18. 
 
 5. My expenditures in building a Louse, in the year 1856, 
 were as follows : Jim. 1C, $536.78 ; Feb. 20, $425.36 ; March 4, 
 $259.25 ; April 24, $786.36. If at the last date I agree to 
 
AVERAGING ACCOUNTS. 
 
 275 
 
 sell the house for exactly what it cost, with reference to interest 
 on the money expended, and take the purchaser's note for the 
 amount, what shall be the face of the note, and what its 
 date ? Am f Face, $2007.75. 
 
 " \ Date, March 8, 1856. 
 6. THOMAS WHITING, 
 1859. To ISRAEL PALMER, Dr. 
 
 Jan. L To 60 bbls. Flour, (3> $7.00, $420 
 
 " 28. 90 bu. Wheat, 1.50, ... . 135 
 Mar. 15. " 300 bbls. Flour, " 6.00, . . . . 1800 
 If credit of 3 months be given to each item, when will the 
 above account become due ? Ans. May 27. 
 
 CASE in. 
 
 When the terms of credit begin at different 
 times, and the account has both a debt and a credit 
 side. 
 
 1. Average the following account. 
 
 DAVID WARE. 
 Dr. Cr. 
 
 185 
 June 
 
 3. 
 1 
 
 To Mdse 
 
 400 
 
 00 
 
 1851 
 July 
 
 3. 
 4 
 
 By Mdse 
 
 200 
 
 Of) 
 
 
 
 Ifi 
 
 " Draft 3 mo 
 
 800 
 
 00 
 
 Au * 
 
 <>0 
 
 " Cash 
 
 150 
 
 00 
 
 Oct. 
 
 20 
 
 " Cash, 
 
 250 
 
 00 
 
 Sept. 
 
 20 
 
 
 500 
 
 00 
 
 Dr. 
 
 OPERATION. 
 
 Cr. 
 
 Foral ) 
 date. \ 
 
 Due 
 
 da. 
 
 Items. 
 
 Prod. 
 
 Due 
 
 July 4 
 Aug. 20 
 Sept. 20 1 
 
 da. 
 
 Items. 
 
 Prod. 
 
 June 1 
 Sept. 19 
 Oct. 20 
 
 141 
 31 
 
 
 400 
 
 800 
 250 
 
 56400 
 24800 
 
 108 
 
 61 
 
 1 30 
 
 200 
 150 
 500 
 
 21600 
 9150 
 15000 
 
 Balances. 
 
 1450 
 850 
 
 81200 
 45750 
 
 
 850 45750 
 
 
 600 
 
 35450 
 
 35450 -7- 600 =. 59 da., average term of interest. 
 Oct. 20 59 da. = Aug. 22, balance due. 
 
 What is Case III r Explain operation. 
 
276 
 
 EQUATION OP PAYMENTS. 
 
 ANALYSIS. In the above operation we have written the dates, 
 showing when the items become due on either side of the ac- 
 count, adding 3 days' grace to the time allowed to the draft. The 
 latest date, Oct. 20, is assumed as the focal date for both sides, and 
 the two columns marked da. show the difference in days between 
 each date and the focal date. The products are obtained as in the 
 last case, and a balance is struck between the items charged and the 
 products. These balances, being on the Dr. side, show that David 
 Ware, on the day of the focal date, Oct. 20, owes $600 with interest 
 on $35450 for 1 day. By division, this interest is found to be equal 
 to the interest on $600 for 59 days. The balance, $600, therefore, 
 has been due 59 days. Reckoning back from Oct. 12, we find the 
 date when the balance fell due, Aug. 22. Hence the following 
 
 RULE. I. Find the time when each item of the account is 
 due ; and write the dates, in two columns, on the sides of the 
 account to which they respectively belong. 
 
 II. Use either the earliest or the latest of these dates as the 
 focal date for both sides, and Jind the products as in the last 
 case. 
 
 III. Divide the balance of the products by the balance of the 
 account; the quotient will be the interval of time, which must 
 be reckoned from the focal date TOWARD the other dates when 
 both balances are on the same side of the account, but FROM 
 the other dates when the balances are on opposite sides of the 
 account. 
 
 2. What is the balance of the following account, and when 
 is it due ? 
 
 JOHN WILSON. 
 
 Dr. 
 
 Cr. 
 
 185 
 
 9. 
 
 
 
 
 185 
 
 9. 
 
 
 
 
 Jan. 
 
 1 
 
 To Mdse. . 
 
 448 
 
 00 
 
 Jan. 
 
 20 
 
 By Am't bro't forward 
 
 560 
 
 00 
 
 Feb. 
 
 4 
 
 " Cash. . 
 
 364 
 
 00 
 
 Feb. 
 
 16 
 
 
 264 
 
 00 
 
 
 20 
 
 
 
 232 
 
 00 
 
 a 
 
 25 
 
 " Cash 
 
 900 
 
 00 
 
 A 
 
 Balance, $680. 
 Due March 13. 
 
 3. If the following account be settled by giving a note, what 
 shall be the face of the note, and what its dale? 
 
 Give analysis. Rule. 
 
RATIO. 
 
 277 
 
 Dr. 
 
 ISAAC FOSTER. 
 
 Cr. 
 
 185 
 
 
 
 
 
 185* 
 
 J. 
 
 
 
 
 -J an 
 
 1 
 
 To Mdse. on 3 mo. 
 
 145 
 
 86 
 
 May 
 
 11 
 
 By Cash 
 
 11 
 
 00 
 
 " 
 
 12 
 
 a 5 
 
 37 
 
 48 
 
 July 
 
 12 
 
 a 
 
 15 
 
 00 
 
 June 
 
 Aug. 
 
 3 
 4 
 
 .< 2 
 
 12 
 66 
 
 25 
 
 48 
 
 Oct. 
 
 12 
 
 
 82 
 
 00 
 
 Ans. 
 
 $154.07, face of note. 
 Mar. 26, 1858, date. 
 
 RATIO. 
 
 l> t . Ratio is the comparison with each other of two num- 
 bers of the same kind. It is/of two kinds arithmetical and 
 geometrical. 
 
 352. Arithmetical Ratio is the difference of the two 
 numbers. 
 
 353. Geometrical Ratio is the quotient of one number 
 divided by the other. 
 
 354. When we use the word ratio alone, it implies geo- 
 metrical ratio, and is expressed by the quotient arising from 
 dividing one number by the other. Thus, the ratio of 4 to 8 
 is 2, of 10 to 5 is , &c. 
 
 355. Ratio is indicated in two ways. 
 
 1st. By placing two points between the numbers compared, 
 writing the divisor before and the dividend after the points. 
 Thus, the ratio of 5 to 7 is written 5 : 7 ; the ratio of 9 to 
 4 is written 9 : 4. 
 
 2d. In the form of a fraction ; thus, the ratio of 9 to 3 is J ; 
 the ratio of 4 to 6 is f . 
 
 356. The Terms are the two numbers compared. 
 
 357. The Antecedent is the first term. 
 
 358. The Consequent is the second term. 
 
 359. No comparison of two numbers can be fully ex- 
 plained but by instituting another comparison ; thus, the com- 
 
 NOTE. It is thought best to omit the questions at the bottom of the pa^es. in the 
 remaining pfH of this work, leaving the teacher to use such as may be deemed ap- 
 propriate. 
 
278 RATIO. 
 
 parison or relation of 4 to 8 cannot be fully expressed by 2, 
 nor of 8 to 4 by J. If the question were asked, what relation 
 4 bears to 8, or 8 to 4, in respect to magnitude, the answer 2, 
 or , would not be complete nor correct. But if we make 
 unity the standard of comparison, and use it as one of the 
 terms in illustrating the relation of the two numbers, and 
 say that the ratio or relation of 4 to 8 is the same as 1 to 2, 
 or the ratio of 8 to 4 is the same as 1 to %, unity in both cases 
 being the standard of comparison, then the whole meaning is 
 conveyed. 
 
 36O. A Direct Ratio arises from diivding the consequent 
 by the antecedent. 
 
 261. An Inverse or Reciprocal Ratio is obtained by di- 
 viding the antecedent by the consequent. Thus, the direct 
 ratio of 5 to 15 is -^ 5 - = 3 ; and the inverse ratio of 5 to 15 is 
 
 *==* 
 
 362. A Simple Ratio consists of a single couplet ; as 
 3 : 12. 
 
 363. A Compound Ratio is the product of two or more 
 simple ratios. Thus, the compound ratio formed from the 
 simple ratios of 3 : 6 and 8 : 2 is f X f = 3 X 8 : 6 X 2 = 
 
 it=*. 
 
 364:. In comparing numbers with each other, they must 
 be of the same kind, and of the same denomination. 
 
 365. The ratio of two fractions is obtained by dividing 
 the second by the first ; or by reducing them to a common de- 
 nominator, when they are to each other as their numerators. 
 Thus, the ratio of -fr : f is f -^ & = f f 2, which is the 
 same as the ratio of the numerator 3 to the numerator 6 of 
 the equivalent fractions -ft- and T V 
 
 Since the antecedent is a divisor and the consequent a divi- 
 dend, any change in either or both terms will be governed by 
 the general principles of division, (87. ) We have only to 
 substitute the terms antecedent, consequent, and ratio, for divi- 
 sor, dividend, and quotient, and these principles become 
 
RATIO. 279 
 
 GENERAL PRINCIPLES OF RATIO. 
 
 PRIN. I. Multiplying the consequent multiplies the ratio ; 
 dividing the consequent divides the ratio. 
 
 PRIN. II. Multiplying the antecedent divides the ratio ; di- 
 viding the antecedent multiplies the ratio. 
 
 PRIN. III. Multiplying or dividing both antecedent and con- 
 sequent by the same number does not alter the ratio. 
 
 These three principles may be embraced in one 
 
 GENERAL LAW. 
 
 A change in the consequent produces a LIKE change in the 
 ratio ; but a change in the antecedent produces an OPPOSITE 
 change in the ratio. 
 
 36O. Since the ratio of two numbers is equal to the con- 
 sequent divided by the antecedent, it follows, that 
 
 1. The antecedent is equal to the consequent divided by 
 the ratio ; and that, 
 
 2. The consequent is equal to the antecedent multiplied by 
 the ratio. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. What part of 9 is 3? 
 
 f = i; or, 9 : 3 as 1 : , that is, 9 has the same ratio to 3 that 1 
 b#s to \. 
 
 2. What part of 20 is 5 ? Ans. f 
 
 3. What part of 36 is 4? Ans. . 
 
 4. What part of 7 is 49 ? Ans. 7 times. 
 
 5. What is the ratio of 16 to 88 ? Ans. 5%. 
 
 6. What is the ratio of 6 to 8? Ans. |. 
 
 7. What is the ratio of 6J- to 78 ? Ans. 12. 
 
 8. What is the ratio of 16 to 66 ? Ans. 4|. 
 
 9. What is the ratio of to f ? Ans. f . 
 
 10. What is the ratio of f to T 4 F ? Ans. f. 
 
 11. What is the ratio of 3^ to 16|? Ans. 5. 
 
 12. What is the ratio of 3 gal. to 2 qt. 1 pt. ? Ans. ^. 
 
280 PROPORTION. 
 
 13. What is the ratio of 6.3 s to 8 s. 6 d. ? Ans. Iff. 
 
 14. What is the ratio of 5.6 to .56 ? Ans. T V 
 
 15. What is the ratio of 19 Ibs. 5 oz. 8 pwts. to 25 Ibs. 11 
 oz. 4 pwts. ? Ans. !-. 
 
 16. What is the inverse ratio of 12 to 16? Ans. }. 
 
 17. What is the inverse ratio of f to J ? Ans. T 9 . 
 
 18. What is the inverse ratio of 5J to 17? Ans. ^. 
 
 19. If the consequent be 16 and the ratio 2f, what is the 
 antecedent ? Ans. 7. 
 
 20. If the antecedent be 14.5 and the ratio 3, what is the 
 consequent ? Ans. 43.5. 
 
 21. If the consequent be J and the ratio , what is the an- 
 tecedent? Ans. 1. 
 
 22. If the antecedent be f and the ratio , what is the 
 consequent ? Ans. ? 
 
 PROPORTION. 
 
 367. Proportion is an equality of ratios. Thus, the ratios 
 6 : 4 and 12:8, each being equal to f , form a proportion. 
 
 368. Proportion is indicated in two ways. 
 
 1st. By a double colon placed between the two ratios ; thus, 
 
 2 : 5 : : 4 : 10. 
 
 2d. By the sign of equality placed between the two ratios ; 
 thus, 2 : 5 = 4 : 10. 
 
 369. Since each ratio consists of two terms, every pro- 
 portion must consist of at least four terms. 
 
 370. The Extremes are the first and fourth terms. 
 
 371. The Means are the second and third terms. 
 
 372. Three numbers maybe in proportion when the first 
 is to the second as the second is to the third. Thus, the num- 
 bers 3, 9, and 27 are in proportion since 3 : 9 : : 9 : 27, the 
 ratio of each couplet being 3. 
 
 In such a proportion the second term is said to be a mean 
 proportional between the other two. 
 
 373. In every proportion the product of the extremes is 
 equal to the product of the means. Thus, in the proportion 
 
 3 : 5 : : 6 : 10 we have 3 X 10 = 5 X <>. 
 
SIMPLE PROPORTION. 281 
 
 Four numbers that are proportional in the direct 
 order are proportional by inversion, and also by alternation, or 
 by inverting the means. Thus, the proportion 2 : 3 : : 6 : 9, 
 by inversion becomes 3 : 2 : : 9 : 6, and by alternation 2:6:: 
 3:9. 
 
 375. From the preceding principles and illustrations, it 
 follows that, any three terms of a proportion being given, the 
 fourth may readily be found by the following 
 
 RULE. I. Divide the product of the extremes by one of the 
 means, and the quotient will be the other mean. Or, 
 
 II. Divide the product of the means by one of the extremes, 
 and the quotient will be the other extreme. 
 
 EXAMPLES FOR PRACTICE, 
 
 Find the term not given in each of the following proportions* 
 
 1. 48 : 20 : : ( ) : 50. Ans. 120. 
 
 2. 42 : 70 : : 3 : ( ). Ans. 5. 
 
 3. ( ) : 30 : : 20 : 100. Ans. 6. 
 
 4. 1 : ( ) ::7 : 84. Ans. 12. 
 
 5. 48yd. :( ):: $67.25 : $201.75. Ans. 144yd. 
 
 6. 3 Ib. 12 oz. : ( ) : : $3.50 : $10.50. Ans. 11 Ib. 4 oz. 
 
 7. ( ) : $38.25 : : 8 bu. 2 pk. : 76 bu. 2 pk. Ans. $4.25. 
 
 8. 4: 381 :: ( ) : 76. Ans. 8. 
 
 9. ( ) : 12 : : | : If . Ans. 7. 
 10. A:( )::J:f. Ans. f. 
 
 SIMPLE PROPORTION. 
 
 376. Simple Proportion is an equality of two simple 
 ratios, and consists of four terms, any three of which being 
 given, the fourth may readily be found. 
 
 377. Every question in simple proportion involves the 
 principle of cause and effect. 
 
 378. Causes may be regarded as action, of whatever 
 kind, the producer, the consumer, men, animals, time, distance, 
 weight, goods bought or sold, money at interest, &c. 
 
 379. Effects may be regarded as whatever is accom- 
 
 x* 
 
282 PROPORTION. 
 
 plished by action of any kind, the thing produced or consumed, 
 money paid, &c. 
 
 38. Causes and effects are of two kinds simple and 
 compound. 
 
 381. A Simple Cause, or Effect, contains but one element ; 
 as goods purchased or sold, and the money paid or received 
 for them. 
 
 383. A Compound Cause, or Effect, is the product of two 
 or more elements ; as men at work taken in connection with 
 time, and the result produced by them taken in connection 
 with dimensions, length and breadth, &c. 
 
 383. Causes and effects that admit of computation, that 
 is, involve the idea of quantity, may be represented by num- 
 bers, which will have the same relation to each other as the 
 things they represent. And since it is a principle of philoso- 
 phy that like causes produce like effects, and that effects are 
 always in proportion to their causes, we have the following 
 proportions : 
 
 1st Cause : 2d Cause : : 1st Effect : 2d Effect. 
 Or, 1st Effect : 2d Effect : : 1st Cause : 2d Cause; 
 
 in which the two causes, or the two effects forming one coup- 
 let, must be like numbers, and of the same denomination. 
 
 Considering all the terms of the proportion as abstract num- 
 bers, we may say that 
 
 1st Cause : 1st Effect : : 2d Cause : 2d Effect, 
 which will produce the same numerical result. 
 
 But as ratio is the result of comparing two numbers or 
 things of the same kind (364), the first form is regarded as 
 the most natural and philosophical. 
 
 384. Simple causes and simple effects give rise to simple 
 ratios ; compound causes and compound effects to compound 
 ratios. 
 
 38*. 1. If 5 tons of coal cost $30, what will 3 tons cost ? 
 
 NOTE. The required term will be denoted by a ( ), and designated 
 "blank." 
 
SIMPLE PROPORTION. 
 
 283 
 
 STATEMENT. 
 
 tons. tons. $ $ 
 
 5 : 3 : : 30 : ( ) 
 1st cause. 2d cause. 1st effect. 2d effect. 
 
 OPERATION. 
 5 V ( ) = 3 X 30 
 
 o v 2/i6 
 o ^ w_ 
 
 ANALYSIS. In this 
 
 example an effect is 
 required, and 5 tons 
 must have the same 
 ratio to 3 tons, as 
 $30, the cost of 5 
 tons, to (blank) dol- 
 
 tons. 
 
 bar. 
 
 15 
 
 1st cause. 
 
 STATEMENT. 
 
 bar. $ 
 
 ( ) : : 90 
 2d cause. 1st effect. 
 OPERATION. 
 
 $ 
 
 30 
 2d effect. 
 
 00 
 
 $0 
 
 Since the product of the extremes is equal to the product of the 
 means (373), and the product of the means divided by one of the 
 extremes will give the other; (blank) dollars will be equal to the 
 product of 3 X 30 divided by 5, which is $18, Ans. 
 
 2. If 15 barrels of flour cost $90, how many barrels can be 
 
 bought for $30 ? 
 
 ANALYSIS. In this ex- 
 ample a cause is required, 
 and the statement may be 
 read thus: If 15 barrels 
 cost $90, how many or 
 (blank) barrels will cost 
 $30? The product of the 
 extremes, 30 X 15, di- 
 ( ) = 5 bar., Ans. vided by the given mean, 
 
 90, will give the required 
 
 term, 5, as shown in the operation. Hence we deduce the following 
 
 RULE. I. Arrange the terms in the statement so that the 
 causes shall compose one couplet, and the effects the other, put- 
 ting ( ) in the place of the required term. 
 
 II. If the required term be an extreme, divide the product 
 of the means by the given extreme ; if the required term be a 
 mean, divide the product of the extremes by the given mean. 
 
 NOTES. 1. If the terms of any couplet be of different denominations, 
 they must be reduced to the same unit value. 
 
 2. If the odd term be a compound number, it must be reduced to its 
 lowest unit. 
 
 3. If the divisor and dividend contain one or more factors common 
 to both, they should be canceled. If any of the terms of a proportion 
 contain mixed numbers, they should first be changed to improper frac- 
 tions, or the fractional part to a decimal. 
 
 4. "When the vertical line is used, the divisor and the required term 
 are written on the left, and the terms of the dividend on the right. 
 
284 PROPORTION. 
 
 386. We will now give another method of solving ques- 
 tions in simple proportion, without making the statement, and 
 which may be used, by those who prefer it, to the one already 
 given. We will term it the 
 
 SECOND METHOD. 
 
 Every question which properly belongs to simple propor- 
 tion must contain four numbers, at least three of which must 
 be given (3 TO). Of the three given numbers, one must 
 always be of the same denomination as the required number. 
 The remaining two will be like numbers, and bear the same 
 relation to each other that the third does to the required num- 
 ber; in other words, the ratio of the third to the required 
 number will be the same as the ratio of the other two num- 
 bers. 
 
 Regarding the third or odd term as the antecedent of the sec- 
 ond couplet of a proportion, we find the consequent or required 
 term by multiplying the antecedent by the ratio (366). 
 
 By comparing the two like numbers, in any given question, 
 with the third, we may readily determine whether the answer, 
 or required term, will be greater or less than the third term ; 
 if greater, then the ratio will be greater than 1, and the two 
 like numbers may be arranged in the form of an improper 
 fraction as a multiplier ; if the answer, or required term, is to 
 be less than the third term, then the ratio will be less tfran 1, 
 and the two like numbers may be arranged in the form of a 
 proper fraction, as a multiplier. 
 
 1. If 4 cords of wood cost $12, what will 20 cords cost? 
 
 OPERATION. ANALYSIS. It will 
 
 ^^3 x 20 De readily seen in this 
 
 1 X -^, written = $60. example, that 4 cords 
 
 ^ and 20 cords are the 
 
 like terms, and that 
 
 $12 is the third term, and of the same denomination as the answer 
 or required term. 
 
 If 4 cords cost $12, will 20 cords cost more, or less, than 4 cords? 
 evidently more : then the answer or required term will be greater 
 
SIMPLE PRO PORTION. 285 
 
 than.the third term, and the ratio greater than 1. The ratio of 4 
 cords to 20' cords is ^, or 5 ; hence the ratio of $12 to the answer 
 must be 5, and the answer will be -^ or 5 times $12, which is $60. 
 
 2. If 12 yards of cloth cost $48, what will 4 yards cost ? 
 
 OPERATION. ANALYSIS. In this example we 
 
 48 X T 4 2 = $16, Ans. see that 12 yards and 4 yards are 
 the like terms and $48 the third 
 term, and of the same denomination as the required answer. 
 
 If 12 yards cost $48, will 4 yards cost more or less than 12 yards? 
 less : then the ratio will be less than 1, and the multiplier a proper 
 fraction. The ratio of 12 yards to 4 yards is T ^ ; hence the ratio of 
 $48 to the answer is T \, and the answer will be T ^ times $48, which 
 is $16. Hence the following 
 
 RULE. I. With the two given numbers, which are of the 
 same name or kind, form a ratio greater or less than 1, accord- 
 ing as the answer is to be greater or less than the third given 
 number. 
 
 II. Multiply the third number by this ratio, and the product 
 will be the required number or answer. 
 
 NOTE. 1. Mixed numbers should first be reduced to improper frac- 
 tions, and the ratio of the fractions found according to (365). 
 
 2. Reductions and cancellation may be applied as in the first method. 
 
 The following examples may be solved by either of the 
 foregoing methods. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. If 48 cords of wood cost $120, how much will 20 cords 
 cost? Ans. $50. 
 
 2. If 6 bushels of corn cost $4.75, how much will 75 bush- 
 els cost ? Ans. $59.37. 
 
 3. If 8 yards of cloth cost $3, how many yards can be 
 bought for $50 ? Ans. 114f yds. 
 
 4. If 12 horses consume 42 bushels of oats in 3 weeks, how 
 many bushels will 20 horses consume in the same ^time ? 
 
 $. If 7 pounds of sugar cost 75 cents, how many pounds 
 can be bought for $9 ? Ans. 84 Ibs. 
 
 6. What will 11 Ib. 4 oz. of tea cost, if 3 Ib. 12 oz. cost 
 $3.50? Ans. $10.50. 
 
286 SIMPLE PROPORTION. 
 
 7. If a staff 3 ft. 8 in. long cast a shadow 1 ft. 6 in., what 
 is the height of a steeple that casts a shadow 75 feet at the 
 same time ? Ans. 183 ft, 4 in. 
 
 8. At $2.75 for 14 pounds of sugar, what will be the cost 
 of 100 pounds? Ans. $19.64?. 
 
 9. How many bushels of wheat can be bought for $51.06, 
 if 12 bushels can be bought for $13.32 ? 
 
 10. What will be the cost of 28J- gallons of molasses, if 15 
 hogsheads cost $236.25 ? Ans. $7.12J. 
 
 11. If 7 barrels of flour are sufficient for a family 6 months, 
 how many barrels will they require for 11 months? 
 
 12. At the rate of 9 yards for 512 s., how many yards of 
 cloth can be bought for 44 16s.? Ans. 72 yds. 
 
 13. An insolvent debtor fails for $7560, of which he is 
 able to pay only $3100 ; how much will A receive, whose 
 claim is $756? Ans. $310. 
 
 14. If 2 pounds of sugar cost 25 cents, and 8 pounds of 
 sugar are worth 5 pounds of coffee, what will 100 pounds of 
 coffee cost ? 4ns. $20. ' 
 
 15. If the moon move 13 10' 35'' in 1 day, in what time 
 will it perform one revolution ? 
 
 16. If 8J bushels of corn cost $4.20, what will be the cost 
 of 13 bushels at the same rate ? Ans. $6.48. 
 
 17. If 1 f yards of cotton cloth cost 6 pence, how many 
 yards can be bought for 10 6 s. 8 d. ? Ans. 694f yds. 
 
 18. If 12J- cwt. of iron cost $42, how much will 48 1 cwt. 
 cost? Ans. $163.50-1-. 
 
 19. What quantity of tobacco can be bought for $317.23, 
 if 8 J Ibs. cost $lf ? Ans. 15 cwt. 22.7+ Ibs. 
 
 20. If 15| bushels of clover seed cost $156|, how much 
 can be bought for $95.75 ? Ans. 9 bu. 2 pk. 2 qt, 
 
 21. If | of a barrel of cider cost $fy, how much will $ of 
 a barrel cost ? . * Ans. $ T 9 T . 
 
 22. If a piece of land of a certain length, and 4 rods in 
 breadth, contain of an acre, how much would there be if it 
 were llf rods wide ? Ans. 2 A. 28 rods. 
 
 23. If 12 cwt. of iron cost $42|, what will 48$ cwt. cost? 
 
SDIPLE PROPORTION. 287 
 
 24. A grocer has a false balance, by which 1 pound will 
 weigh but 12 oz. ; what is the real value of a barrel of sugar 
 that he sells for $28 ? Ans. $21. 
 
 25. A butcher in selling meat sells 14| oz. for a pound ; 
 how much does he cheat a customer, who buys of him to the 
 amount of $;jo ? Ans. $2.46 -f- . 
 
 26. If a man clear $750 by his business in 1 yr. 6 mo., how 
 much would he gain in 3 yr. 9 mo. at the same rate ? 
 
 27. If a certain business yield $350 net profits in 10 mo., 
 in what time would the same business yield $1050 profits ? 
 
 28. B and C have each a farm ; B's farm is worth $25 
 an acre, and C's $30J- ; if in trading B values his land at $28 
 an acre, what value should C put upon his ? Ans. $34.16. 
 
 29. If I borrow $500, and keep it 1 yr. 4 irio., for how long 
 time should I lend $240 as an eouivalent for the favor ? 
 
 Ans. 2 yr. 9 mo. 10 da. 
 
 COMPOUND PROPORTION. 
 
 387. Compound Proportion embraces that class of ques- 
 tions in which the causes, or the effects, or both, are compound. 
 
 The required term may be a cause, or a single element of a 
 cause ; or it may be an effect, or a single element of an effect* 
 
 1. If 16 horses consume 128 bushels of oats in 50 days, 
 how many bushels will 5 horses consume in 90 days ? 
 
 i 
 
 STATEMENT. 
 1st causo. 2d cause. 1st effect. 2cl effect 
 
 ll- 
 
 Or, 16 X 50 : 5 X 90 : : 128 : ( ) 
 
 OPERATION. ANALYSTS. In this ex- 
 
 $ X 00 9 X ;f$$ 8 ample the required term 
 
 72 bu. i g the second effect; and 
 
 X$ X $0 the question may be read, 
 
 If 16 horses in 50 days 
 
 consume 128 bushels of oats, 5 horses in 90 days will consume 
 how many, or (blank) bushels ? 
 
 NOTE. These questions are most readily performed by cancellation. 
 
288 PROPORTION. 
 
 2. If $480 gain $84 interest in 30 months, what sum will 
 gain $21 in 15 months? 
 
 STATEMENT. 
 
 1st cause. 2d cause. 1st effect. 2d effect. 
 
 84 21 
 
 * * "*- 
 
 (30 (15 
 
 OPERATION. ANALYSIS. The re- 
 
 $0 2 X &X. quired term in this ex- 
 
 $240, Ans. ample is an element of 
 $^ X <!$ the second cause ; and 
 
 the question may be 
 
 read, If $480 in 30 months gain $84, what principal in 15 months 
 will gain $21 ? 
 
 3. If 7 men dig a ditch 60 feet long, 8 feet wide, and 6 feet 
 deep, in 12 days, what length of ditch can 21 men dig in 2f 
 days, if it be 3 feet wide and 8 feet deep ? 
 
 STATEMENT. 
 
 7 121 (60 ( ( ) 
 
 12 : I 2f : : { 8 : < 3 
 
 (6(8 
 Or, 7 X 12 : 21 X f : : 60 X 8 X 6 : ( ) X 3 X 8 
 
 OPERATION. ANALYSIS. In 
 
 2X X 8 X 00 5 X $ X # 2 this example the 
 
 T" 80 ft., Ans. required term is 
 the length of the 
 
 Or, 
 
 3 ditch, and is an element of the 
 
 ^ second effect. The question, 
 
 as stated, will read thus : if 7 
 
 00 5 men, in 12 days, dig a ditch GO 
 
 $ feet long, 8 feet wide, and 6 
 
 /2 feet deep, 21 men, in 2| days, 
 
 will dig a ditch how many, or 
 
 (blank) feet long, 3 feet wide, 
 
 ( ) 80 ft., Ans. and 8 feet deep ? 
 
 Hence we have the following 
 
 HULK. I. Of tic f/iven terms, select tloxc which constitute 
 ihc anuses, and those which constitute the effects, ami arrange 
 them in couplets, putting ( ) in place of the required term. 
 
COMPOUND PROPORTION. 289 
 
 II. Then, if the blank term ( ) occur in either of the ex- 
 tremes, make the product of the means a dividend, and the 
 product of the extremes a divisor ; but if the blank term occur 
 in either mean, make the product of the extremes a dividend^ 
 and the product of the means a divisor. 
 
 NOTES. 1. The causes must be exactly alike in the number and kind 
 of their terms ; the same is true of the effects. 
 
 2. The same preparation of the terms by reduction is to be observed 
 as in simple proportion. 
 
 888. We will now solve an example according to the 
 Second Method given in Simple Proportion. 
 
 1. If 18 men can build 42 rods of wall in 16 days, how 
 many men can build 28 rods in 8 days ? 
 
 OPERATION. ANALYSIS. We see in 
 
 $$ 4 ^0 2 tnis example that all the 
 
 Xffi X X = 24 men. terms appear in couplets, 
 
 4$ except one, which is 18 
 
 men, and that is of the same kind as the required answer. 
 
 Since compound proportion is made up of two or more simple 
 proportions, if this third or odd term be multiplied by the compound 
 ratio, or by the simple ratio of each couplet successively, the prod- 
 uct will be the required term. 
 
 By comparing the terms of each couplet with the third term we 
 may readily determine whether the answer, or term sought, will be 
 greater or less than the third term ; if greater, then the ratio will be 
 greater than 1, and the multiplier an improper fraction ; if less, the 
 ratio will be less than 1, and the multiplier a proper fraction. 
 
 First we will compare the terms composing the first couplet, 42 
 rods and 28 rods, with the third term, 18 men. If 42 rods require 
 18 men, how many men will 28 rods require? less men ; hence the 
 ratio is less than 1, and the multiplier a proper fraction, |f ; next, 
 if 16 days require 18 men, how many men will 8 days require ? 
 more men ; hence the ratio is greater than 1, and the multiplier an 
 improper fraction, ^ 6 -. Regarding the third term as the antecedent 
 of a couplet, the consequent being the term sought, if we multiply 
 this third term by the simple ratios, or by their product, we shall 
 have the required term or answer, thus : 18 X ff X Y 24, as 
 shown in the operation. 
 
 2. 5 compositors, in 16 days, of 14 hours each, can compose 
 20 sheets of 24 pages in each sheet, 50 lines in a page, and 
 
290 COMPOUND PROPORTION. 
 
 40 letters in a line ; in how many days, of 7 hours each, will 
 10 compositors compose a volume to be printed in the same 
 letter, containing 40 sheets, 16 pages in a sheet, 60 lines in a 
 page, and 50 letters in a line ? Ans. 32 days. 
 
 OPERATION, 
 days. comp. hours, sheets, pages, lines. lottos. 
 
 16X A X -V 4 X f& X |} X f % X 1% = 32 days. 
 
 BY CANCELLATION. ANALYSIS. The required term or an- 
 16 swer is to be in days ; and we see that 
 
 / all the terms appear in pairs or couplets, 
 
 jj except the 16 days, which is of the same 
 
 kind as the answer sought. 
 
 40 We will proceed to compare the terms 
 
 of each couplet with the 16 days. First, 
 if 5 compositors require 16 days, how 
 many days will 10 compositors require ? 
 less days; hence the multiplier is the 
 32 days, Ans. proper fraction T 5 ^, and we have 16 X yV 
 Next, if 14 hours a day require 16 days, 
 
 how many days will 7 hours a day require ? more days ; hence the 
 multiplier is the improper fraction ^, and we have 16 X ^ X V' 
 Next, if 20 sheets require 16 days, how many days will 40 sheets 
 require ? move days ; hence the multiplier is the improper fraction 
 $, and we have 16 X Y% X ty X ff Pursuing the same method 
 with the other couplets, we obtain the result as shown in the opera- 
 tion. Hence we have the following 
 
 RULE. I. Of the terms composing each couplet form a 
 ratio greater or less than 1, in the same, manner as if the 
 answer depended on those two and the third or odd term. 
 
 II. Multiply the third or odd term by these ratios successively, 
 and the product will be the answer sought. 
 
 NOTE. By the odd term is meant the one that is of the same kind 
 as the answer. 
 
 The following examples may be solved by either of the 
 given methods. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. If 16 horses consume 128 bushels of oats in 50 days, 
 how many bushels will 5 horses consume in 90 days ? 
 
COMPOUND PROPORTION. 291 
 
 2. If a man travel 120 miles in 3 days when the days are 
 12 hours long, in how many days of 10 hours each will he 
 require to travel 360 miles ? Ans. lOf days. 
 
 3. If 6 laborers dig a ditch 34 yards long in 10 days, how 
 many yards can 20 laborers dig in 15 days? Ans. 170 yds. 
 
 4. If 450 tiles, each 12 inches square, will pave a cellar, 
 how many tiles that are 9 inches long and 8 inches wide will 
 pave the same ? Ans. 900. 
 
 5. If it require 1200 yards of cloth f- wide to clothe 500 
 men, how many yards which is wide will it take to clothe 
 960 men ? Ans. 3291f yds. 
 
 6. If 8 men will mow 36 acres of grass in 9 days, of 9 
 hours each day, how many men will be required to mow 48 
 acres in 12 days, working 12 hours each day ? Ans. 6 men. 
 
 7. If 4 men, in 2 days, mow 6| acres of grass by work- 
 ing 8 hours a day, how many acres will 15 men mow in 3j 
 days by working 9 hours a day? Ans. 40}^ acres. 
 
 8. If, by traveling 6 hours a day at the rate of 4 miles 
 an hour, a man perform a journey of 540 miles in 20 days, in 
 how many clays, traveling 9 hours a day at the rate of 4 
 miles an hour, will he travel 600 miles ? Ans. 14f days. 
 
 9. If 21 yards of cloth If yards wide cost $3.37, what 
 cost 361 yards, 11 yards wide? Ans. $52.79 +. 
 
 10. If 5 men reap 52.2 acres in 6 days, how many men 
 will reap 417.6 acres in 12 days ? Ans. 20 men. 
 
 11. If 6 men dig a cellar 22.5 feet long, 17.3 feet wide, and 
 10.25 feet deep, in 2.5 days, of 12.3 hours, in how many days, 
 of 8.2 hours, will 9 men take to dig another, measuring 45 
 feet long, 34.6 wide, and 12.3 deep? Ans. 12 days. 
 
 12. If 54 men can build a fort in 244 days, working 12 
 hours each day, in how many days will 75 men do the same, 
 when they work but 101 hours each day ? Ans. 21 days. 
 
 13. If 24 men dig a trench 33f yards long, 5-| wide, and 
 3J- deep, in 189 days, working 14 hours each day, how many 
 hours per day must 217 men work, to dig a trench 23 1 yards 
 long, 3 wide, and 2 deep, in 5^ days ? Ans. 16 hours. 
 
292 PARTNERSHIP. 
 
 PARTNERSHIP. 
 
 389. Partnership is a relation established between two 
 or more persons in trade, by which they agree to share the 
 profits and losses of business. 
 
 390. The Partners are the individuals thus associated. 
 
 391. Capital, or Stock, is the money or property invested 
 in trade. 
 
 392. A Dividend is the profit to be divided. 
 
 393. An Assessment is a tax to meet losses sustained. 
 
 CASE I. 
 
 394. To find each partner's share of the profit or 
 loss, when their capital is employed for equal periods of 
 time. 
 
 1. A and B engage in trade; A furnishes $300, and B 
 $400 of the capital $ they gain $182; what is each one's 
 share of the profit ? 
 
 OPERATION. ANALYSIS. Since 
 
 $300 the whole capital 
 
 $4.00 employed is $300 
 
 + $400 = $700, it 
 
 $700, whole stock. i s evident that A 
 
 f -g.o .3, A's share of the stock. furnishes f -- =z f 
 
 4- 4 B g of the capital, and 
 
 $182 X ?' = $78, A's share of the gain. B *#f =/ f . the 
 
 capital. And since 
 
 $182X* = *104,B-. - each man's share of 
 
 the profit or loss will have the same ratio to the whole profit or 
 loss that his share of the stock has to the whole stock, A will have 
 -^ of the entire profit, and B ^, as shown in the operation. 
 
 We may also regard the whole capital as the first cause, 
 and each man's share of the capital as the second cause, the 
 whole profit or loss as the^rs^ effect, and each man's share of the 
 profit or loss as the second effect, and solve by proportion thus : 
 
PARTNERSHIP. 
 
 1st cause. 2<1 cause. 
 
 $700 : $300 
 $700 : 
 
 #00 
 ( ) 
 
 B , s profit . 
 
 ( ) <J> I O, A'S profit. 
 
 Hence we have the following 
 
 RULE. Multiply the whole profit or loss by the ratio of each 
 man's share of the capital to the whole capital. Or, 
 
 The whole capital is to each man's share of the capital as the 
 whole profit or loss is to each man's share of the profit or loss. 
 
 .2 Three men trade in company; A furnishes $8000, B 
 $12000, and C 20000 of the capital; their gain is $1680; 
 what is each man's share ? 
 
 Ans. A's $336; B's $504; C's $840. 
 
 3. Three persons purchased a house for $2800, of which A 
 paid $1200, B $1000, and C $600 ; they rented it for $224 
 a year ; how much of the rent should each receive ? 
 
 4. A man failed in business for $20000, and his available 
 means amounted to only $13654 ; how much will two of his 
 creditors respectively receive, to one of whom he owes $3060, 
 and to the other $1530 ? Ans. $2089.062 ; $1044.531. 
 
 5. Four men hired a coach for $13, to convey them to 
 their respective homes, which were at distances from the place 
 of starting as follows : A's 16 miles, B's 24 miles, C's 28 
 miles, and D's 36 miles; what ought each to pay? 
 
 . 
 S ' 
 
 ( A $2. C. $3.50. 
 '(B$3. D$4.50. 
 
 6. A captain, mate, and 12 sailors took a prize of $2240, 
 of which the captain took 14 shares, the mate 6 shares, and 
 each sailor 1 share ; how much did each receive ? 
 
 7. A cargo of corn, valued at $3475.60, was entirely lost ; 
 of it belonged to A, of it to B, and the remainder to C ; 
 how much was the loss of each, there being an insurance of 
 $2512? Ans. $120.45, A's. $240.90, B's. $602.35, C's. 
 
 Y* 
 
294 PARTNERSHIP. 
 
 8. Three persons engaged in the lumber trade ; two of the 
 persons furnished the capital, and the third managed the busi- 
 ness; they gained $2571.24, of which C received $6 as often 
 as D $4, and E had -^ as much as the other two for taking care 
 of the business ; how much was each one's share of the gain ? 
 
 .Am. $1285.62, C's. $857.08, D's. $428.54, E's. 
 
 9. Four persons engage in the coal trade; D puts in 
 $3042 capital ; they gain $7500, of which A takes $2000, B 
 $2800.75, and C $1685.25 ; how much capital did A, B, and 
 C put in, and how much is D's share of the gain ? 
 
 A (A, $6000. C, $5055.75. 
 s ' I B, $8402.25. D's gain, $1014. 
 
 CASE II. 
 
 395. To find each partner's share of the profit or 
 loss when their capital is employed for unequal periods 
 of time. 
 
 It is evident that the respective shares of profit and loss will 
 depend upon two conditions, viz. : the amount of capital in- 
 vested by each, and the time it is employed. 
 
 1. Two persons form a partnership; A puts in $450 for 
 7 months, and B $300 for 9 months ; they lose $156 ; how 
 much is each man's share of the loss ? 
 
 OPERATION. ANALYSIS. The 
 
 $450 X 7 = $3150, A's capital for 1 mo. use of $450 capital 
 
 $300 X 9 = $2700, B' 8 for 7 months is the 
 
 ~ same as the use of 
 
 $5800, entire 7 timeg $ 450> or 
 
 f J$& = ^ A's share of the entire capital. $ 3150 for l month 5 
 
 nn = A B's and f $ 300 for 9 
 
 ii w v 7 ouu , months is the same 
 
 $156 X -ft = $84, A'S loss. as the uge of 9 timeg 
 
 $156 X A = $72, B's $300, or $2700 for 
 
 1 month. The en- 
 tire capital for 1 month is equivalent to $3150 + $2700= $r>sr>(). 
 If the loss, $156, be divided between the two partners, according to 
 Case I, the results will be the loss of each as shown in the operation. 
 
PARTNERSHIP. 295 
 
 Examples of this kind may also be solved by proportion as in 
 Case I, the causes being compounded of capital and time ; thus, 
 
 $5850 : $3150 : : $156 : ( ) 
 $5850 : $2700 : : $156 : ( ) 
 
 ( ) 
 
 ( ) = $84, A's loss. ( ) = $72, B's loss. 
 
 Hence the following 
 
 RULE. Multiply each man's capital by the time it is em- 
 ployed in trade, and add the products. Then multiply the 
 entire profit or loss by the ratio of each product to the sum of 
 the products, and the results will be the respective shares of 
 profit or loss of each partner. Or, 
 
 Multiply each man's capital by the time it is employed in 
 trade, and regard each product as his capital, and the sum of 
 the products as the entire capital, and solve by proportion, as in 
 Case I. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Three persons traded together; B put in $250 for 6 
 months, C $275 for 8 months, and D $450 for 4 months; 
 they gained $795 ; how much was each man's share of the 
 gain? 
 
 3. Two merchants form a partnership for 18 months. A at 
 first put in $1000, and at the end of 8 months he put in $600 
 more ; B at first put in $1500, but at the end of 4 months he 
 drew out $300 ; at the expiration of the time they found that 
 they had gained $1394.64; how much was each man's share 
 of the gain? Ans. A's $715.20 ; B's $679.44. 
 
 4. Three men took a field of grain to harvest and thresh 
 for I of the crop ; A furnished 4 hands 5 days, B 3 hands 
 6 days, and C 6 hands 4 days ; the whole crop amounted to 
 372 bushels ; how much was each one's share ? 
 
 5. William Gallup began trade January 1, 1856, with a 
 capital of $3000, and, succeeding in business, took in M. H. 
 Decker as a partner on the first day of March following, with 
 
296 ANALYSIS. 
 
 a capital of $2000 ; four months after they admit J. New- 
 man as third partner, who put in $1800 capital; they con- 
 tinue their partnership until April 1, 1858, when they find 
 that $4388.80 has been gained since Jan. 1, 1856; how 
 much was each one's share ? ( $2106, Gallup's. 
 
 Ans.\ $1300, Decker's. 
 
 ( $ 982.80, Newman's. 
 
 6. Two persons engage in partnership with a capital of 
 $5600; A's capital was in trade 8 months, and his share of 
 the profits was $560 ; B's capital was in 10 months, and his 
 share of the profits was $800 ; what amount of capital had 
 each in the firm ? Ans. A, $2613.33; B, $2986.66$. 
 
 7. A, B, and C, engage in trade with $1930 capital; A's 
 money was in 3 months, B's 5, and C's 7 ; they gained $117, 
 which was so divided that J- of A's share, was equal to of 
 B's and to of C's ; how much did each put in, and each 
 gain ? r A, $700 capital ; $26 gain. 
 
 Ans. 1 B, $630 $39 " 
 
 ( C, $600 $52 
 
 * 
 
 ANALYSIS. ' 
 
 396. Analysis, in arithmetic, is the process of solving 
 problems independently of set rules, by tracing the relations 
 of the given numbers and the reasons of the separate steps of 
 the operation according to the special conditions of each question. 
 
 397. In solving questions by analysis, we generally reason 
 from the given number to unity, or 1, and then from unity, or 
 1, to the required number. 
 
 398. United States money is reckoned in dollars, dimes, 
 cents,andmills(18O),one dollar being uniformly valued in all 
 the States at 100 cents ; but in most of the States money is 
 sometimes still reckoned in pounds, shillings, and pence. 
 
 NOTE. At the time of the adoption of our decimal currrnry by 
 Congress, in 1786, the colonial currency, or bills of credit, issued by the 
 colonies, had depreciated in value, and this depreciation, being unequal 
 in the different colonies, gave rise to the different values of the State 
 currencies ; and this variation continues wherever the denominations of 
 shillings and pence are in use. 
 
ANALYSIS. 297 
 
 399. In New England, Indiana, \ 
 Illinois, Missouri, Virginia, Kentucky, > $1 = 6 s. 72 d. 
 Tennessee, Mississippi, Texas, ) 
 
 New York, Ohio, Michigan, $1 = 8 s. = 96 d. 
 
 New Jersey, Pennsylvania, Dela-) 
 ware, Maryland, } f 1 = 7s. 6d. = 90 d. 
 
 South Carolina, Georgia, Canada, ) 
 Nova Scotia, J $1 = 5 * = 60 d ' 
 
 EXAMPLES FOR PRACTICE. 
 
 1. What will be the cost of 42 bushels of oats, at 3 shillings 
 per bushel, New England currency ? 
 
 OPERATION. ANALYSIS. Since 
 
 $ 1 bushel costs 3 shil- 
 
 42 X 3= 126s. 
 126-:- 6 =$21 Or, 
 
 lings, 42 bushels will 
 cost 42 times 3s., or 
 
 WI, Ans. 42X3= 126s.; and 
 as 6 s. make 1 dollar 
 
 New England currency, there are as many dollars in 126 s. as 6 is 
 contained times in 126, or $21. 
 
 2. What will 180 bushels of wheat cost at 9s. 4d. per 
 bushel, Pennsylvania currency ? 
 
 OPERATION. ANALYSIS. Multi- 
 
 Or, 
 
 plying the number of 
 28 bushels by the price, 
 
 and dividing by the 
 jf _ value of 1 dollar re- 
 
 Ans duced to pence, we 
 we have $224. Or, 
 when the pence in the 
 
 given price is an aliquot part of a shilling, the price may be reduced 
 to an improper fraction for a multiplier, thus : 9 s. 4 d. = 9^ s. = 
 ^ s., the multiplier. The value of the dollar being 7 s. 6 d. = 7-J s. 
 = ^, we divide by ^-, as in the operation. 
 
 3. What will be the cost of 3 hhd. of molasses, at 1 s. 3 d. 
 per quart, Georgia currency ? 
 
298 ANALYSIS. 
 
 OPERATION. 
 
 ANALYSIS. In this example we first 
 reduce 3 hhd. to quarts, by multiplying 
 by 63 and 4, and then multiply by the 
 ^ price, either reduced to pence or tt> an 
 
 _* improper fraction, and divide by the 
 
 405 00 value of 1 dollar reduced to the same 
 denomination as the price. 
 
 $202.50 
 
 4. Sold 9 firkins of butter, each containing 56 lb., at 1 s. 6 d. 
 per pound, and received in payment carpeting at 6 s. 9 d. per 
 yard ; how many yards of carpeting would pay for the butter ? 
 
 OPERATION. ANALYSIS. The operation in this is similar 
 
 to the preceding examples, except that we di- 
 vide the cost of the butter by the price of a, 
 
 , - 
 
 unit of the article received in payment, reduced 
 to the same denominational unit as the price 
 
 112 yd. ^ a un ^ f ^6 ar ticle sold. The result will be 
 the same in whatever currency. 
 
 5. What will 3 casks of rice cost, each weighing 126 
 pounds, at 4 d. per pound, South Carolina currency ? Ans. $27. 
 
 6. How many pounds of tea, at 7 s. per pound, must be 
 given for 28 lb. of butter, at 1 s. 7 d. per pound ? Ans. 6^. 
 
 7. Bought 2 casks of Catawba wine, each containing 72 
 gallons, for $648, and sold it at the rate of 10 s. 6 d. per quart, 
 Ohio currency ; how much was my whole gain ? Ans. $108. 
 
 8. What will be the expense of keeping 2 horses 3 weeks 
 if the expense of keeping 1 horse 1 day be 2 s. 6 d., Canada 
 currency? Ans. $21. 
 
 9. How many days' work, at 6 s. 3 d. per day, must be 
 given for 20 bushels of apples at 3 s. per bushel ? Ans. 9^. 
 
 10. Bought 160 lb. of dried fruit, at 1 s. 6d. a pound, in 
 New York, and sold it for 2 s. a pound in Philadelphia ; how 
 much was my whole gain ? Ans. $12.66. 
 
 11. A merchant exchanged 43 yards of cloth, worth 10s. 
 6 d. per yard, for other cloth worth 8 s. 3 d. per yard ; how 
 many yards did he receive ? Ans. 
 
ANALYSIS. 299 
 
 12. What will be the cost of 300 bushels of wheat at 9 s. 
 4 d. per bushel, Michigan currency ? Ans. $350. 
 
 13. If of f of a ton of coal cost $2f , how much will of 
 6 tons cost ? 
 
 OPERATION. 
 
 ANALYSIS.. Since f of 4 =:| of a ton costs 
 $\ 2 -, 1 ton will cost 28 times -^ of $J, 
 <* $J X -ff 5 and f of 6 tons nr Sf. tons, will 
 cost - 3 times . o 
 
 $16, 
 
 14. If 8 men can build a wall 20 ft. long, 6 ft. high, and 
 4 ft. thick, in 12 days, working 10 hours a day, in how many 
 days can 24 men build a wall 200 ft. long, 8 ft. high, and 6 ft. 
 thick, working 8 hours a day ? 
 
 OPERATION. 
 
 It 1 10 200 10 $ 
 - X X- X- -X X =100 da. 
 I t& 9 20 4 
 
 ANALYSIS. Since 8 men require 12 days of 10 hours each to 
 build the wall, 1 man would require 8 times 12 days of 10 hours 
 each, and 10 times (12 X 8) days of 1 hour each. To build a wall 
 1 ft. long would require -fa as much time as to build a wall 20 ft. 
 long ; to build a wall 1 ft. high would require \ as much time as to 
 build a wall 6 ft. high ; to build a wall 1 ft. thick, J as much time as 
 to build a wall 4 ft. thick. Now, 24 men could build this wall in ^ 
 as many days, by working 1 hour a day, as 1 man could build it, 
 and in | as many days by working 8 hours a day, as by working 1 
 hour a day ; but to build a wall 200 ft. long would require 200 times 
 as many days as to build a wall 1 ft. long ; to build a wall 8 ft. high 
 would require 8 times as many days as to build a wall 1 ft. high ; 
 and to build a wall 6 ft. thick would require 6 times as many days 
 as to build a wall 1 ft. thick. 
 
 15. If 2 pounds of tea are worth 11 pounds of coffee, and 
 3 pounds of coffee are worth 5 pounds of sugar, and 18 pounds 
 of sugar are worth 21 pounds of rice, how many pounds of 
 rice can be purchased with 12 pounds of tea? 
 
300 ANALYSIS. 
 
 OPERATION. ANALYSIS. Since 18 lb. of su- 
 
 gar are equal in value to 21 Ib. of 
 rice, 1 Ib. of sugar is equal to ^ 
 of 21 Ib. of rice, or fl | Ib. of 
 rice, and 5 Ib. of sugar are equal 
 to 5 times ^ Ib. of rice, or ^ Ib. ; 
 
 3 | 385 if 3 Ib. of coffee are equal to 5 Ib. 
 
 Am 128Mb f sugar ' r V lb - of rice > ! lb - of 
 
 coffee is equal to J of ^ lb. of 
 
 rice, or f| lb., and 11 lb. of coffee are equal to 11 times f | lb. of 
 rice, or - 3 j 8 5 - lb. ; if 2 lb. of tea are equal to 11 lb. of coffee, or -^g 5 lb. 
 of rice, 1 lb of tea is equal to J of f lb. of rice, or -^ } D ., and 12 
 lb. of tea are equal to 12 times ^ lb. of rice, or a|-& lb. = 1281 lb. 
 
 16. If 16 horses consume 128 bushels of oats in 50 days, 
 how many bushels will 5 horses consume in 90 days? Ans. 72. 
 
 17. If $10^ w iH buy 4f cords of wood, how many cords 
 can be bought for $24 ? Ans. 11. 
 
 18. Gave 52 barrels of potatoes, each containing 3 bushels, 
 worth 33^ cents a bushel, for 65 yards of cloth ; how much 
 was the cloth worth per yard ? Ans. $.80. 
 
 19. If a staff 3 ft. long cast a shadow 5 ft. in length, what 
 is the height of an object that casts a shadow of 46f ft. at the 
 same time of day? Ans. 28 ft. 
 
 20. Three men hired a pasture for $63 ; A put in 8 sheep 
 7 months, B put in 12 sheep 4 months, and C put in 15 
 sheep 6| months ; how much must each pay ? 
 
 21. If 7 bushels of wheat are worth 10 bushels of rye, 
 and 5 bushels of rye are worth 14 bushels of oats, and 6 
 bushels of oats are worth $3, how many bushels of wheat will 
 $30 buy? Ans. 15. 
 
 22. If $480 gain $84 in 30 months, what capital will gain 
 $21 m 15 months ? Ans. $240. 
 
 23. How many yards of carpeting f of a yard wide are 
 equal to 28 yards J of a yard wide? Ans. 31. 
 
 24. If a footman travel 130 miles in 3 days, when the days 
 are 14 hours long, in how many days of 7 hours each will he 
 travel 390 miles ? Ans. IS. 
 
ANALYSIS. 301 
 
 25. If 6 men can cut 45 cords of wood in 3 days, how 
 many cords can 8 men cut in 9 days ? Ans. 180. 
 
 26. B's age is 1^- times the age of A, and C's is 2^ times 
 the age of both, and the sum of their ages is 93 ; what is the 
 age of each? Ans. A's age, 12 yrs. 
 
 27. If A can do as much work in 3 days as B can do in 
 4 days, and B can do as much in 9 days as C in 12 days, 
 and C as much in 10 days as D in 8, how many days' work 
 done by D are equal to 5 days' done by A? Ans. 8. 
 
 28. The hour and minute hands of a watch are together at 
 12 o'clock, M. ; when will they be exactly together the third 
 time after this ? 
 
 OPERATION. ANALYSIS. Since 
 
 12 X TIT X 3 3 T 3 T h. the minute hand pass- 
 
 Ans. 3h. 16 miri. 21 T 9 T sec., P.M. es the hour hand 11 
 
 times in 12 hours, if 
 
 both are together at 12, the minute hand will pass the hour hand 
 the first time in ^ of 12 hours, or l^j hours ; it will pass the hour 
 hand the second time in T 2 T of 12 hours, and the third time in ^ of 
 12 hours, or 3^ hours, which would occur at 16 min. 21 T ^ sec. 
 past 3 o'clock, P. M. 
 
 29. A flour merchant paid $164 for 20 barrels of flour, 
 giving $9 for first quality, and $7 for second quality ; how 
 many barrels were there of each ? 
 
 OPERATION. ANALYSIS. If all had been 
 
 $9 X 20 $180 ; first quality, he would have paid 
 
 $180 $164 = $16. $ 180 or $ 16 more than he did 
 
 <C) $ 7 $9 . P a y- Every barrel of second 
 
 quality made a difference of $2 
 16-1-2=8 bbl., * **,. I th( f cost . hence there were 
 
 20 8 12 bbl., 1st as man y barrels of second qual- 
 
 ity as $2, the difference in the 
 cost of one barrel, is contained times in $16, &c. 
 
 30. A boy bought a certain number of oranges at the rate 
 of 3 for 4 cents, and as many more at the rate of 5 for 8 cents ; 
 he sold them again at the rate of 3 for 8 cents, and gained on 
 the whole 75 cents ; how many oranges did he buy ? 
 
 z 
 
302 ANALYSIS. 
 
 OPERATION. ANALYSIS. For 
 
 I + t =tt;ft-^ 2 = i!> avera secost. those he bought 
 2| i| i ctgej gain on each . at the rate of 3 for 
 
 75 -f- U = 90, number of oranges. 4 Cents he P aid $ 
 
 of a cent each, and 
 
 for those he bought at the rate of 5 for 8 cents he paid f of a cent 
 each ; and | -f | || cents, what he paid for 1 of each kind, 
 which divided by 2 gives f f cents, the average price of all he bought. 
 He sold them at the rate of 3 for 8 cents, or | cents each ; the dif- 
 ference between the average cost and the price he sold them for, or 
 f f f if = H cents, is the gain on each ; and he bought as 
 many oranges as the gain on one orange is contained times in the 
 whole gain, c. 
 
 31. A man bought 10 bushels of wheat and 25 bushels of 
 corn for $30, and 12 bushels of wheat and 5 bushels of corn 
 for $20 ; how much a bushel did he give for each ? 
 
 OPERATION. ANALYSIS. We may divide or 
 
 w C. multiply either of the expressions 
 
 1st lot, 10 25 $30 by such a number as will render 
 
 2d " 12 5 $20 one of the commodities purchased, 
 
 alike in both expressions. In this 
 1st -L- 5:^:2 5 So i v i i c, , 
 
 example we divide the first by 5 
 
 10 $14 to make the numbers denoting 
 
 lbu,w. := $1.40 the corn alike, (the same result 
 
 1 bu. C. = $ .64 would be produced by multiply- 
 
 ing the second by 5,) and we have 
 
 the cost of 2 bushels of wheat and 5 bushels of corn, equal to $6. 
 Subtracting this from 12 bushels of wheat and 5 bushels of corn, which 
 cost $20, we find the cost of 10 bushels of wheat to be $14 ; there- 
 fore the cost of 1 bushel is ^ of $14, or $1.40. From any one of 
 the expressions containing both wheat and corn, we readily find the 
 cost of 1 bushel of corn to be 64 cents. 
 
 32. A, B, and C agree to build a barn for $270. A and 
 B can do the work in 16 days, B and C in 13 days, and A 
 and C in 11^ days. In how many days can all do it working 
 together ? In how many days can each do it alone ? What 
 part of the pay ought each to receive ? 
 
ANALYSIS. 
 
 303 
 
 OPERATION. 
 what A and B do in 1 day. 
 
 " BandC " " 
 
 " Aandc " " 
 
 = **' what A ' B ' and c do in 
 
 2 days. 
 
 1 -f- 2 -^-, what A, B, and C do in 1 day. 
 -f- -g^- =i | days, time A, B, and C, will do the 
 whole work together. 
 
 -^ = ^5 1 ^ = 20 da., C alone. 
 9 -= 3 ; l-=:26 da,A 
 
 ANALYSIS. Sinco 
 A and B can do the 
 work in 16 days, they 
 can do fa of it in 1 
 day; B and C, in 
 13 or ^- days, they 
 can do ^ of it in 1 
 day; A and C, in llf 
 or Sf. days, they can 
 do fa of it in 1 day. 
 Then A, B, and C, 
 by working 2 days 
 each, can do fa -j- 
 
 X 8$- = -f , the part of the whole C did. 
 X 8f = f, " " " A " 
 
 V 84 = . " " B 
 
 $270 X | = $120, C's share. 
 
 $270Xf = $90, A'S 
 $270 X | = $60, B's 
 
 work, and by work- 
 ing 1 day each they 
 can do -|- of -|-|, or -^ 
 of the work ; and it 
 will take them as 
 many days working 
 together to do the 
 whole work as 9 7 is contained times in 1, or 8-f- days. 
 
 Now, if we take what any two of them do in 1 day from what the 
 three do in 1 day, the remainder will be what the third does ; we 
 thus find that A does -g^, B -$, and C -g^-. 
 
 Next, if we denote the whole w r ork by 1, and divide it by the part 
 each does in 1 day, we have the number of days that it will take 
 each to do it alone, viz. : A 26f days, B 40 days, and C 20 days. 
 And each should receive such a part of $270 as would be ex- 
 pressed by the part he does in 1 day, multiplied by the number of 
 days he works, which will give to A $90, B $60, and C $120. 
 
 33. If 6 oranges and 7 lemons cost 33 cents, and 12 oranges 
 and 10 lemons cost 54 cents, what is the price of 1 of each ? 
 
 Ans. Oranges, 2 cents; lemons, 3 cents. 
 
 34. If an army of 1000 men have provisions for 20 days, 
 at the rate of 18 oz. a day to each man, and they be reinforced 
 by 600 men, upon what allowance per day must each man be 
 put, that the same provisions may last 30 days ? Ans. 7 oz. 
 
 35. There are 54 bushels of grain in 2 bins ; and in one bin 
 are 6 bushels less than as much as there is in the other ; 
 how many bushels in the larger bin ? Ans. 40. 
 
304 ANALYSIS. 
 
 36. The sum of two numbers is 20, and their difference is 
 equal to % of the greater number; what is the greater 
 number? Am. 12. 
 
 37. If A can do as much work in 2 days as C in 3 days, 
 and B as much in 5 days as C in 4 days ; what time will B 
 require to execute a piece of work which A can do in 6 
 weeks? Am. 11^ weeks. 
 
 38. How many yards of cloth, f of a yard wide, will line 
 36 yards 1 yards wide ? Am. 60. 
 
 39. How many sacks of coffee, each containing 104 Ibs , 
 at 10 d. per pound N. Y. currency, will pay for 80 yards of 
 broadcloth at $3 per yard? Am. 24. 
 
 40. A person, being asked the time of day, replied, the time 
 past noon is equal to of the time to midnight ; what was 
 the hour ? Am. 2, P. M. 
 
 41. A market woman bought a number of peaches at the 
 rate of 2 for 1 cent, and as many more at the rate of 3 for 1 
 cent, and sold them at the rate of 5 for 3 cents, gaining 55 
 cents ; how many peaches did she buy ? Ans. 300. 
 
 42. A can build a boat in 18 days, working 10 hours a day, 
 and B can build it in 9 days, working 8 hours a day ; in how 
 many days can both together build it, working 6 hours a day ? 
 
 43. A man, after spending % of his money, and of the 
 remainder, had $10 left ; how much had he at first ? 
 
 44. If 30 men can perform a piece of work in 1 1 days, how 
 many men can accomplish another piece of work, 4 times as 
 large, in of the time ? Ans. 600. 
 
 45. If 16 Ib. of coffee cost $3, how much can be bought 
 for $1.25? Ans. 6lb. 
 
 46. A man engaged to write for 20 days, receiving $2.50 
 for every day he labored, and forfeiting $1 for every day he 
 was idle ; at the end of the time he received $43 ; how many 
 days did he labor ? Ans. 18. 
 
 47. A, B, and C can perform a piece of work in 12 hours ; 
 A and B can do it in 16 hours, and A and C in 18 hours; 
 what part of the work can B and C do in 9^ hours ? Ans. $. 
 
ALLIGATION MEDIAL. 305 
 
 ALLIGATION. 
 
 1OO. Alligation treats of mixing or compounding two or 
 more ingredients of different values. It is of two kinds Alii- 
 gallon Medial and Alligation Alternate. 
 
 4O1 . Alligation Medial is the process of finding the aver- 
 age price or quality of a compound of several simple ingredi- 
 ents whose prices or qualities are known. 
 
 1. A miller mixes 40 bushels of rye worth 80 cents a 
 bushel, and 25 bushels of corn worth 70 cents a bushel, with 
 15 bushels of wheat worth $1.50 a bushel ; what is the value 
 of a bushel of the mixture ? 
 
 OPERATION. ANALYSIS. Since 40 bushels 
 
 80 X 40 = $32.00 of rye at 80 cents a bushel is 
 
 70 X 25 = 1 7.50 worth $32, and 25 bushels of corn 
 
 i *;n \x i ^ . 99 ^ft at 70 cents a bushel is worth 
 
 $17.50, and 15 bushels of wheat 
 
 80 ) 72.00 at $1.50 a bushel is worth $22.50, 
 
 $90 Ans therefore the entire mixture, con- 
 sisting of 80 bushels, is worth 
 
 $72, and one bushel is worth -^ of $72, or 72 -|- 80 = $.90. 
 Hence the following 
 
 RULE. Divide the entire cost or value of the ingredients 
 by the sum of the simples. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. A wine merchant mixes 12 gallons of wine, at $1 per 
 gallon, with 5 gallons of brandy worth $1.50 per gallon, and 
 3 gallons of water of no value ; what is the worth of one gal- 
 lon of the mixture ? Ans. $.975. 
 
 3. An innkeeper mixed 13 gallons of water with 52 gallons 
 of brandy, which cost him $1.25 per gallon ; what is the value 
 of 1 gallon of the mixture, and what his profit on the sale of 
 the whole at> 6 cents per gill ? Ans. $1 a gallon ; $65 profit. 
 
 4. A grocer mixed 10 pounds of sugar at 8 cts. with 12 
 pounds at 9 cts. and 16 pounds at 11 cts., and sold the mixture 
 at 10 cents per pound ; did he gain or lose by the sale, and 
 how much ? Ans. He gained 1 6 cts. 
 
 z* 
 
306 ALLIGATION ALTERNATE. 
 
 5. A grocer bought 7 dozen of eggs at 12 cents a dozen, 
 8 dozen at 10 cents a dozen, 9 dozen at 11 cents a dozen, 
 and 10 dozen at 10 cents a dozen. He sells them so as to 
 make 50 per cent, on the cost ; how much did he receive per 
 dozen? Am. 15f cents. 
 
 6. Bought 4 cheeses, each weighing 50 pounds, at 13 cents 
 a pound; 10, weighing 40 pounds each, at 10 cents a pound; 
 and 24, weighing 25 pounds each, at 7 cents a pound ; I sold 
 the whole at an average price of 9 cents a pound ; how much 
 was my whole gain ? Ans. $6. 
 
 4OS. Alligation Alternate is the process of finding the 
 proportional quantities to be taken of several ingredients, 
 whose prices or qualities are known, to form a mixture of a 
 required price or quality. 
 
 CASE I. 
 
 4LO3. To find the proportional quantity to be used 
 of each ingredient, when the mean price or quality of 
 the mixture is given. 
 
 1. What relative quantities of timothy seed worth $2 a 
 bushel, and clover seed worth $7 a bushel, must be used to 
 form a mixture worth $5 a bushel ? 
 
 OPERATION. ANALYSIS. Since on every in- 
 
 2 
 
 gredient used whose price or qual- 
 
 fr is less tnan tlie mean rate tnere 
 
 7 
 
 will be a gain, and on every in- 
 gredient whose price or quality is greater than the mean rate 
 there will be a loss, and since the gains and losses must be exactly- 
 equal, the relative quantities used of each should be such as repre- 
 sent the unit of value. By selling one bushel of timothy seed worth 
 $2, for $o, there is a gain of $3 ; and to gain $1 would require ^ of 
 a bushel, which we place opposite the 2. By selling one bushel of 
 clover seed worth $7, for $5, there is a loss of $2 ; and to lose $1 
 would require \ of a bushel, which we place opposite tfie 7. 
 
 In every case, to find the unit of value we must divide $1 by the 
 gain or loss per bushel or pound, &c. Hence, if, every time we take 
 ^ of a bushel of timothy seed, we take \ of a bushel of clover seed, 
 the gain and loss will be exactly equal, and \ve shall have \ and \ 
 for the proportional quantities. 
 
ALLIGATION ALTERNATE. 
 
 307 
 
 OPERATION. 
 
 6 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 3 
 
 i 
 
 
 4 
 
 
 4 
 
 4 
 
 
 i 
 
 
 1 
 
 1 
 
 7 
 
 
 1 
 
 
 2 
 
 2 
 
 . 10 
 
 i 
 
 
 3 
 
 
 3 
 
 If we wish to express the proportional numbers in integers, we 
 may reduce these fractions to a common denominator, and use their 
 numerators, since fractions having a common denominator are to 
 each other as their numerators. (365) thus, and are equal to 
 | and f , and the proportional quantities are 2 bushels of timothy 
 seed to 3 bushels of clover seed. 
 
 2. What proportions of teas worth respectively 3, 4, 7 and 
 10 shillings a pound, must be taken to form a mixture worth 
 6 shillings a pound ? 
 
 ANALYSIS. To preserve the 
 equality of gains and losses, we 
 must always compare two prices 
 or simples, one greater and one 
 less than the mean rate, and 
 treat each pair or couplet as a 
 separate example. In the given 
 example we form two couplets, 
 
 and may compare either 3 and 10, 4 and 7, or 3 and 7, 4 and 10. 
 We find that of a Ib. at 3 s. must be taken to gain 1 shilling, 
 
 and \ of a Ib. at 10 s. to lose 1 shilling ; also \ of a Ib. at 4 s. to gain 
 
 1 shilling, and 1 Ib. at 7 s. to lose 1 shilling. These proportional 
 numbers, obtained by comparing the two couplets, are placed in 
 columns 1 and 2. If, now, we reduce the numbers in columns 1 and 
 
 2 to a common denominator, and use their numerators, we obtain 
 the integral numbers in columns 3 and 4, which, being arranged in 
 column 5, give the proportional quantities to be taken of each.* 
 
 It will be seen that in comparing the simples of any couplet, one 
 of which is greater, and the other less than the mean rate, the pro- 
 portional number finally obtained for either term is the difference 
 between the mean rate and the other term. Thus, in comparing 3 
 and 10, the proportional number of the former is 4, which is the 
 difference between 10 and the mean rate 6 ; and the proportional 
 number of the latter is 3, which is the difference between 3 and the 
 mean rate. The same is true of every other couplet. Hence, when 
 the simples and the mean rate are integers, the intermediate steps 
 taken to obtain the final proportional numbers as in columns 1, 2, 3, 
 and 4, may be omitted, and the same results readily found by taking 
 the difference between each simple and the mean rate, and placing 
 it opposite the one with which it is compared. 
 
 * Prof. A. B. Canfield, of Oneida Conference Seminary, N. Y., used this method of 
 Alligation, essentially, in the instruction of his classes as early as 1846, and he was 
 doubtless the author of it. 
 
808 ALLIGATION ALTERNATE. 
 
 From the foregoing examples and analyses we derive the following 
 
 RULE. I. Write the several prices or qualities in a column, 
 and the mean price or quality of the mixture at the left. 
 
 II. Form couplets ly comparing any price or quality Iess 9 
 with one that is greater than the mean rate, placing the part 
 which must be used to gain 1 of the mean rate opposite the less 
 simple, and the part that must be used to lose 1 opposite the 
 greater simple, and do the same for each simple in every couplet. 
 
 III. If the proportional numbers are fractional, they may be 
 reduced to integers, and if two or more stand in the same hori- 
 zontal line, they must be added; the jinal results will be the pro- 
 portional quantities required. 
 
 NOTES. 1. If the numbers in any couplet or column have a com- 
 mon factor, it may be rejected. 
 
 2. We may also multiply the numbers in any couplet or column by 
 any multiplier we choose, without affecting the equality of the gains 
 and losses, and thus obtain an indefinite number of results, any one of 
 which being taken will give a correct final result. 
 
 EXAMPLES FOR PRACTICE. 
 
 3. A grocer has sugars worth 10 cents, 11 cents, and 14 
 cents per pound ; in what proportions may he mix them to 
 form a mixture worth 12 cents per pound? 
 
 Ans. 2 Ibs. at 10 and 11 cts., and 3 Ibs. at 14cts. 
 
 4. What proportions of water at no value, and wine worth 
 $1.20 a gallon, must be used to form, a mixture worth 90 cents 
 a gallon ? Ans. 1 gal. of water to 3 gals, of wine. 
 
 5. A farmer had sheep worth $2, $2, $3, and $4 per 
 head ; what is the smallest number he could sell of each, and 
 realize an average price of $2f per head ? 
 
 Ans. 3 of the 1st kind, 1 each of the 2d and 3d, and 5 of 
 the 4th kind. 
 
 6. What relative quantities of alcohol 80, 84, 87, 94, and 
 96 per cent, strong must be used to form a mixture 90 per 
 cent, strong ? 
 
 Ans. 6 of the first two kinds, four of the 3d, 3 of the 4th, 
 and 16 of the 5th. 
 
ALLIGATION ALTERNATE. 309 
 
 CASE II. 
 
 4O4. When the quantity of one of the simples is 
 limited. 
 
 1. A miller has oats worth 30 cents, corn worth 45 cents, 
 and barley worth 84 cents per bushel ; he desires to form a 
 mixture worth 60 cents per bushel, and which shall contain 40 
 bushels of corn ; how many bushels of oats and barley must 
 he take ? 
 
 OPERATION. ANALYSIS. By 
 
 S30 
 45 
 84 
 
 4 
 
 8 
 5 | 5 
 
 20 ) the same process 
 
 40 ( Ans. as in Case l we 
 50 1 fi . nd the P ro ? or - 
 
 tional quantities 
 of each to, be 4 bushels of oats, 8 of corn, and 10 of barley. But 
 we wish to use 40 bushels of corn, which is 5 times the propor- 
 tional number 8, and to preserve the equality of gain and loss we 
 must take 5 times the proportional quantity of each of the other 
 simples, or 5 X 4 = 20 bushels of oats, and 5 X 10 = 50 bushels 
 of barley. Hence the following 
 
 RULE. Find the proportional quantities as in Case L 
 Divide the given quantity by the proportional quantity of the 
 same ingredient, and multiply each of the other proportional 
 quantities by the quotient thus obtained. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. A merchant has teas worth 40, 60, 75, and 90 cents per 
 pound ; how many pounds of each must he use with 20 pounds 
 at 75 cents, to form a mixture worth 80 cents ? 
 
 Ans. 201bs. each of the first three kinds, and 1301bs. of 
 the fourth. 
 
 3. A farmer bought 24 sheep at $2 a head ; how many 
 must he buy at $3 and $5 a head, that he may sell the whole 
 at an average price of $4 a head, without loss ? 
 
 Ans. 24 at $3, and 72 at $5. 
 
 4. How much alcohol worth 60 cents a gallon, and how 
 much water, must be mixed with 180 gallons of rum worth 
 $1.30 a gallon, that the mixture may be worth 90 cents a 
 gallon ? Ans. 60 gallons each of alcohol and of water. 
 
310 
 
 ALLIGATION ALTERNATE. 
 
 5. How many acres of land worth 35 dollars an acre must 
 be added to a farm of 75 acres, worth $50 an acre, that the 
 average value may be $40 an acre ? Am. 150 acres. 
 
 6. A merchant mixed 80 pounds of sugar worth 6J cents 
 per pound with some worth 8 cents and 10 cents per pound, 
 so that the mixture was worth 7 cents per pound ; how much 
 of each kind did he use ? 
 
 CASE III. 
 
 4:0*5. When the quantity of the whole compound is 
 limited. 
 
 1. A grocer has sugars worth 6 cents, 8 cents, 11 cents, 
 and 13 cents per pound. He wishes to make a mixture of 
 120 pounds worth 10 cents a pound; how many pounds of 
 each kind must he use ? 
 
 ANALYSIS. By Case 
 I we find the propor- 
 tional quantities of each 
 to be 3 Ibs. at 6 cts., 2 
 Ibs. at 7 cts., 3 Ibs at 12 
 cts., and 4 Ibs. at 13 cts. 
 By adding the propor- 
 tional quantities, we find 
 that the mixture would be but 12 Ibs. while the required mixture is 
 120, or 10 times 12. If the whole mixture is to be 10 times as much 
 as the sum of the proportional quantities, then the quantity of each 
 simple used must be 10 times as much as its respective proportion- 
 al, which would require 30 Ibs. at 6 cts., 20 Ibs. at 7 cts., 30 Ibs. at 
 12 cts., and 40 Ibs. at 13 cts. Hence we deduce the following 
 
 RULE. Find the proportional numbers as in Case I. Di- 
 vide the given quantity by the sum of the proportional quan- 
 tities, and multiply each of the proportional quantities by the 
 quotient thus obtained. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. A farmer sold 170 sheep at an average price of 14 
 shillings a head ; for some he received 9 s., for some 12 s., for 
 some 18s., and for others 20s.; how many of each did he 
 sell ? Ans. 60 at 9 s., 40 at 12 s., 20 at 18 s., and 50 at 20 s. 
 
 OPERATION. 
 
 6 
 
 i 
 
 
 3 
 
 
 3 
 
 30 
 
 7 
 
 
 % 
 
 
 2 
 
 2 
 
 20 
 
 12 
 
 
 % 
 
 
 3 
 
 3 
 
 30 
 
 13 
 
 i 
 
 
 4 
 
 
 4 
 
 40 
 
 
 12 
 
 120 
 
INVOLUTION. 311 
 
 3. A jeweler melted together gold 16, 18, 21, and 24 
 carats fine, so as to make a compound of 51 ounces 22 carats 
 fine ; how much of each sort did he take ? Ans. 6 ounces 
 each of the first three, and 33 ounces of the last. 
 
 4. A man bought 210 bushels of oats, corn, and wheat, and 
 paid for the whole $178.50 ; for the oats he paid $, for the 
 corn $-, and for the wheat $1 per bushel ; how many bush- 
 els of each kind did he buy ? Ans. 78 bushels each of oats 
 and corn, and 54 bushels of wheat. 
 
 5. A, B, and C are under a joint contract to furnish 6000 
 bushels of corn, at 48 cts. a bushel ; A's corn is worth 45 cts., 
 B's 51 cts., and C's 54 cts. ; how many bushels must each put 
 into the mixture that the contract may be fulfilled ? 
 
 6. One man and 3 boys received $84 for 56 days' labor; the 
 man received $3 per day, and the boys $, $}, and $lf re- 
 spectively ; how many days did each labor ? Ans. The man 
 16 days, and the boys 24, 4, and 12 days respectively. 
 
 INVOLUTION. 
 
 406. A Power is the product arising from multiplying a 
 number by itself, or repeating it several times as a factor; 
 thus, in 2 X 2 X 2 = 8, the product, 8, is a power of 2. 
 
 407. The Exponent of a power is the number denoting 
 how many times the factor is repeated to produce the power, 
 and is written above and a little to the right of the factor; thus, 
 2 X 2 X 2 is written 2 3 , in which 3 is the exponent. Exponents 
 likewise give names to the powers, as will be seen in the 
 following illustrations : 
 
 3 = 3 l 3, the first power of 3 ; 
 
 3X3 = 3 2 = 9, the second power of 3 
 
 3X3X3 = 33 ==. 27, the third power of 3. 
 
 408. The Square of a number is its second power. 
 4:09. The Cube of a number is its third power. 
 
 41O. Involution is the process of raising a number to a 
 given power. 
 
312 EVOLUTION. 
 
 411. A Perfect Power is a number that can be exactly 
 produced by the involution of some number as a root ; thus, 25 
 and 32 are perfect powers, since 25 = 5 X 5, and 32 = 2 X 
 2X2X2X2. 
 
 I. What is the cube of 15 ? 
 
 OPERATION. ANALYSIS. We multiply 
 
 15 X 15 X 15 = 3375. Ans. l5 b ? 15 ' and the product 
 
 by 15, and obtain 3375, 
 
 which is the 3d power, or cube of 15, since 15 has been taken 3 
 times as a factor. Hence, we have the following 
 
 RULE. Multiply the number by itself as many times, less 1, 
 as there are units in the exponent of the required power. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the square of 25 ? Ans. 625. 
 
 3. What is the square of 135 ? Ans. 18225. 
 
 4. What is the cube of 72 ? Ans. 373248. 
 
 5. What is the 4th power of 24? Ans. 331776. 
 
 6. Raise 7.2 to the third power. Ans. 373.248. 
 
 7. Involve 1.06 to the 4th power. Ans. 1.26247696. 
 
 8. Involve .12 to the 5th power. Ans. .0000248832. 
 
 9. Involve 1.0002 to the 2d power. Ans. 1.00040004. 
 10. What is the cube off? 
 
 OPERATION. 
 
 JL 2 ^2X2X2 2* 8 
 T X 5 X 5 ~" 5~X~5X 5 ~~ 53~~ 125 
 
 It is evident from the above operation, that 
 
 A common fraction may be raised to any power, by raising 
 each of its terms, separately, to the required power. 
 
 II. What is the square of f ? Ans. 9 . 
 
 12. What is the cube of |f ? Ans. f f f }. 
 
 13. Raise 24J to the 2d power. Ans. 612 T %. 
 
 EVOLUTION. 
 
 412. A Boot is a factor repeated to produce a power ; 
 thus, in the expression 5 X 5 X 5 = 125, 5 is the root from 
 which the power, 125, is produced. 
 
SQUARE HOOT. 313 
 
 S I J{. Evolution is the process of extracting the root of a 
 number considered as a power, and is the reverse of Involution. 
 
 414. The Radical Sign is the character, V> which, placed 
 before a number, denotes that its root is to be extracted. 
 
 41t>. The Index of the root is the figure placed above the 
 radical sign, to denote what root is to be taken. When no 
 index is written, the index 2 is always understood. 
 
 416. A Surd is the indicated root of an imperfect power. 
 
 417. Roots are named from the corresponding powers, as 
 will be seen in the following illustrations : 
 
 The square root of 9 is 3, written -\/9 i= 3. 
 The cube root of 27 is 3, written ^27 = 3. 
 The fourth root of 81 is 3, written ^81 = 3. 
 
 418. Any number whatever may be considered a power 
 whose root is to be extracted ; but only the perfect powers can 
 have exact roots. 
 
 SQUARE ROOT. 
 
 419. The Square Root of a number is one of the two 
 
 equal factors that produce the number ; thus the square root 
 
 of 49 is 7, for 7 X 7 = 49. 
 
 420. In extracting the square root, the first thing to be 
 determined is the relative number of places in a given number 
 and its square root. The law governing -this relation is exhib- 
 ited in the following examples : 
 
 Roots. Squares. Roots. Squares. 
 
 Ill 1 
 
 9 81 10 1,00 
 
 99 98,01 100 1,00,00 
 
 999 99,80,01 1000 1,00,00,00 
 
 From these examples we perceive 
 
 1st. That a root consisting of 1 place may have 1 or 2 
 places in the square. 
 
 2d. That in all cases the addition of 1 place to the root 
 adds 2 places to the square. Hence, 
 A A 
 
314 SQUARE ROOT. 
 
 If we point off a number into two-figure periods, commen- 
 cing at the right hand, the number of full periods and the left 
 hand full or partial period will indicate the number of places 
 in the square root ; the highest period answering to the highest 
 figure of the root. 
 
 421. 1. What is the length of one side of a square plat 
 containing an area of 5417 sq. ft. ? 
 
 OPERATION. ANALYSIS. Since the given figure is 
 
 54,17 | 73.6 a square, its side will be the square root 
 
 49 of its area, which we will proceed to com- 
 
 pute. Pointing off the given number, the 
 
 2 periods show that there will be two in- 
 tegral figures, tens and units, in the root. 
 
 146.0 88.00 ^ e tens f ^ e root must be extracted 
 
 1 46 6 87 96 from the first or left hand period, 54 hun- 
 
 dreds. The greatest square in 54 hun- 
 4 dreds is 49 hundreds, the square of 7 tens ; 
 
 we therefore write 7 tens in the root, at 
 the right of the given number. 
 
 Since the entire root is to be the side of a square, let us form a 
 Fig. i. square (Fig. I), the side of which is 70 feet long. 
 
 The area of this square is 70 X 70 = 4900 sq.ft., 
 which we subtract from the given number. This 
 is done in the operation by subtracting the 
 square number, 49, from the first period, 54, 
 and to the remainder bringing down the sec- 
 ond period, making the entire remainder 517. 
 If we now enlarge our square (Fig. I) by the addition of 517 
 square feet, in such a manner as to preserve the square form, its 
 size will be that of the required square. To preserve the square 
 form, the addition must be so made as to extend the square equally 
 in two directions ; it will therefore be composed of 2 oblong figures 
 at the sides, and a little square at the corner (Fig. II). Now, the 
 width of this addition will be the additional length to the side of the 
 square, and consequently the next figure in the root. To find width 
 we divide square contents, or area, by length. But the length of 
 one side of the little square cannot be found till the width of the 
 addition be determined, because it is equal to this width. We will 
 therefore add the lengths of the 2 oblong figures, and the sum will 
 be sufficiently near the whole length to be used as a trial divisor. 
 
SQUARE ROOT. 
 
 315 
 
 Fig. H. 
 
 Fig. III. 
 
 70 
 
 70. 
 
 3 
 
 i 
 ; Two Divisors = 140 
 
 
 Complete Divisor = 143 
 
 Each of the oblong figures is equal in length to the side of the 
 square first formed ; and their united length 
 is 70 + 70 = 140 ft. (Fig. III). This num- 
 ber is obtained in the operation by doubling 
 the 7 and annexing 1 cipher, the result being 
 written at the left of the dividend. Dividing 
 517, the area, by 140, the approximate length, 
 we obtain 3, the probable width of the addi- 
 tion, and second figure of the root. Since 3 is 
 also the side of the little square, we can now 
 find the entire length of the addition, or the complete divisor, which 
 
 is 70 + 70 -f 3 = 143 (Fig. III). 
 This number is found in the oper- 
 ation by adding 3 to the trial di- 
 visor, and writing the result un- 
 derneath. Multiplying the com- 
 plete divisor, 143, by the trial 
 quotient figure, 3, and subtracting 
 the product from the dividend, we 
 obtain another remainder of 88 square feet. With this remainder, 
 for the same reason as before, we must proceed to make a new 
 enlargement ; and we bring down two decimal ciphers, because the 
 next figure of the root, being tenths, its square will be hundredths. 
 The trial divisor to obtain the width of this new enlargement, 
 or the next figure in the root, will be, for the same reason as 
 before, twice 73, the root already found, with one cipher annexed. 
 But since the 7 has already been doubled in the operation, we have 
 only to double the last figure of the complete divisor, 143, and 
 annex a cipher, to obtain the new trial divisor, 146.0. Dividing, we 
 obtain .6 for the trial figure of the root ; then proceeding as before, 
 we obtain 146.'6 for a complete divisor, 87.96 for a product ; and 
 there is still a remainder of .04. Hence, the side of the given 
 square plat is 73.6 feet, nearly. From this example and analysis 
 we deduce the following 
 
 RULE. I. Point off the given number into periods of two 
 figures each, counting from unit's place toward the left and 
 right. 
 
 II. Find the greatest square number in the left hand period, 
 and write its root for the first figure in the root ; subtract the 
 square number from the left hand period, and to the remainder 
 bring down the next period for a dividend. 
 
316 SQUARE HOOT. 
 
 III. At the left of the dividend write twice the first figure of 
 the root, and annex one cipher, for a trial divisor ; divide the 
 dividend by the trial divisor, and write the quotient for a trial 
 figure in the root. 
 
 IV. Add the trial figure of the root to the trial divisor for a 
 complete divisor ; multiply the complete divisor by the trial 
 figure in the root, and subtract the product from the dividend, 
 and to the remainder bring down the next period for a new 
 dividend. 
 
 V. Multiply the last figure of the last complete divisor by 2, 
 and annex one cipher for a new trial divisor, with which pro- 
 ceed as before. 
 
 NOTES. 1. If at any time the product be greater than the dividend, 
 diminish the trial figure of the root, and correct the erroneous work. 
 
 2. If a cipher occur in the root, annex another cipher to the trial 
 divisor, and another period to the dividend, and proceed as before. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. What is the square root of 406457.2516? 
 
 OPERATION. 
 
 40,64,57.25,16 637.54, Ans. 
 36 
 
 Trial divisor, 
 Complete " 
 
 Trial " 
 Complete " 
 
 Trial 
 Complete " 
 
 Trial " 
 Complete " 
 
 120 
 123 
 
 464 
 369 
 
 1260 
 
 1267 
 
 9557 
 8869 
 
 1274.0 
 1274.5 
 
 688.25 
 637.25 
 
 1275.00 
 1275.04 
 
 51.0016 
 ' 51.0016 
 
 NOTES. 3. The decimal points in the work may be omitted, care 
 being taken to point off in the root according to the number of deci- 
 mal periods used. 
 
 4. The pupil will acquire greater facility, and secure greater accura- 
 cy, by keeping units of like order under each other, and each divisor 
 opposite the corresponding dividend, by the use of the lines, as shown 
 in the operation. 
 
 3. What is the square root of 576 ? Ans. 24. 
 
SQUARE ROOT. 317 
 
 4. What is the square root of 6561 ? Am. 81. 
 
 5. What is the square root of 444889 ? Ans. 667. 
 
 6. What is the square root of 994009 ? Ans. 997. 
 
 7. What is the square root of 29855296? Ans. 5464. 
 
 8. What is the square root of 3486784401 ? Ans. 59049. 
 
 9. What is the square root of 54819198225 ? 
 
 NOTE. The cipher in the trial divisor may be omitted, and its place, 
 after division, occupied by the trial root figure, thus forming in suc- 
 cession only complete divisors. 
 
 10. What is the square root of 2 ? 
 
 2. | 1.4142 +, Ans. 
 
 100 
 24 96 
 
 400 
 281 281 
 
 11900 
 2824 11296 
 
 60400 
 28282 56564 
 
 11. Extract the square roots of the following numbers: 
 
 V3 = 1.7320508 + 
 V5 = 2.2360680 -I'- 
 ve = 2.4494897 + 
 
 V7 = 2.6457513 + 
 V8 = 2.8284271 + 
 V10 = 3.1622776 + 
 
 12. What is the square root of .00008836 ? Ans. .0094. 
 
 13. What is the square root of .0043046721 ? Ans. .06561. 
 
 NOTES. 5. The square root of a common fraction may be obtained 
 by extracting the square roots of the numerator and denominator 
 separately, provided the terms are perfect squares ; otherwise, the 
 fraction may first be reduced to a decimal. 
 
 6. Mixed numbers may be reduced to the decimal form before ex- 
 tracting the root ; or, if the denominator of the fraction is a perfect 
 square, to an improper fraction. 
 
 14. Extract the square root of ^ 5 T . Ans. ff. 
 
 15. Extract the square root of ^f. Ans. . 
 
 16. Extract the square root of . , Ans. .81649 + . 
 
 17. Extract the square root of 17f. Ans. 4.168 + . 
 
 AA* 
 
318 
 
 SQUARE ROOT 
 
 APPLICATIONS. 
 
 2. An Angle is the opening between two lines that 
 meet each other ; thus, the two lines, A B and A C, meeting, 
 form an angle at A. 
 
 4123. A Triangle is a figure having three 
 sides and three angles, as A, B, C. 
 
 424. A Right-Angled Triangle is a tri- 
 angle having one right angle, as at C. 
 
 425. The Base is the side on which it 
 stands, as A, C. 
 
 4:26. The Perpendicular is the side 
 forming a right angle with the base, as B, C. 
 
 427. The Hypotenuse is the side opposite the right angle, 
 as A, B. 
 
 4:28. Those examples given below, which relate to trian- 
 gles and circles, may be solved by the use of the two following 
 principles, which are demonstrated in geometry. 
 
 1st. The square of the hypotenuse of a right-angled triangle 
 is equal to the sum of the squares of the other two sides. 
 
 2d. The areas of two circles are to each other as the squares 
 of their radii, diameters, or circumferences. 
 
 1. The two sides of a right-angled triangle are 3 and 4 
 feet ; what is the length of the hypotenuse ? 
 
 ANALYSIS. Squaring 
 
 OPERATION. the two sides and add- 
 
 3 2 = 9, square of one side. in g we find the sum to 
 
 42 = 1 6, square of the other side- ? e 25 > , and s j nce the sum , 
 
 is equal to the square of 
 
 25, square of hypotenuse. the hypotenuse, \ve cx- 
 
 /25 _ 5 ^ nSt tract the square root, and 
 
 ' obtain 5 feet, the hypot- 
 
 enuse. Hence, 
 
 To find the hypotenuse. Add the squares of the two sides, 
 and extract the square root of the sum. 
 
 To find cither of the shorter sides. Subtract the square of 
 flic f liven side from the square of the hypotenuse, and extract the 
 square root of the remainder. 
 
SQUARE ROOT. 319 
 
 EXAMPLES FOR PRACTICE. 
 
 2. If an army of 55225 men be drawn up in the form of a 
 square, how many men will there be on a side ? Ans. 235. 
 
 3. A man has 200 yards of carpeting 1 yards wide ; what 
 is the length of one side of the square room which this carpet 
 will cover ? Ans. 45 feet. 
 
 4. How many rods of fence will be required to inclose 10 
 acres of land in the form of a square ? Ans. 1 60 rods. 
 
 5. The top of a castle is 45 yards high, and the castle is sur- 
 rounded by a ditch 60 yards wide ; required the length of a 
 rope that will reach from the outside of the ditch to the top 
 of the castle. Ans. 75 yards. 
 
 6. Required the height of a May-pole, which being broken 
 39 feet from the top, the end struck the ground 15 feet from 
 the foot. Ans. 75 feet. 
 
 7. A ladder 40 feet long is so placed in a street, that 
 without being moved at the foot, it will reach a window on 
 one side 33 feet, and on the other side 21 feet, from the 
 ground ; what is the breadth of the street ? Ans. 56.65 -f- ft. 
 
 8. A ladder 52 feet long stands close against the side of a 
 building ; how many feet must it be drawn out at the bottom, 
 that the top may be lowered 4 feet ? Ans. 20 feet. 
 
 9. Two men start from one corner of a park one mile 
 square, and travel at the same rate. A goes by the walk 
 around the park, and B takes the diagonal path to the opposite 
 corner, and turns to meet B at the side. How many rods 
 from the corner will the meeting take place ? Ans. 93.7 -\- rods. 
 
 10. A room is 20 feet long, 16 feet wide, and 12 feet high ; 
 what is the distance from one of the lower corners to the op- 
 posite upper corner ? Ans. 28.284271 -f- feet. 
 
 11. It requires 63.39 rods of fence to inclose a circular 
 field of 2 acres ; what length will be required to inclose 3 
 acres in circular form ? Ans. 77.63 rods. 
 
 12. The, radius of a certain circle is 5 feet ; what will be 
 the radius of another circle containing twice the area of the 
 first? Ans. 7.07106 + feet. 
 
820 CUBE ROOT. 
 
 CUBE ROOT. 
 
 4:29. The Cube Koot of a number is one of the three 
 equal factors that produce the number. Thus, the cube root 
 of 27 is 3, since 3 X 3 X 3 = 27. 
 
 4:30 In extracting the cube root, the first thing to be 
 determined is the relative number of places in a cube and its 
 root. The law governing this relation is exhibited in the fol- 
 lowing examples : 
 
 Roots. Cubes. Roots. Cubes. 
 
 Ill 1 
 
 9 729 10 1,000 
 
 99 907,299 100 1,000,000 
 
 999 997,002,999 1000 1,000,000,000 
 
 From these examples, we perceive, 
 
 1st. That a root consisting of 1 place may have from 1 to 
 3 places in the cube. 
 
 2d. That in all cases the addition of 1 place to the root 
 adds three places to the cube. Hence, 
 
 If we point off a number into three-figure periods, com- 
 mencing at the right hand, the number of full periods and the 
 left hand full or partial period will indicate the number of 
 places in the cube root, the highest period answering to the 
 highest figure of the root. 
 
 4131. 1. What is the length of one side of a cubical block 
 containing 413494 solid inches ? 
 
 OPERATION COMMENCED. ANALYSIS. Since the block is a 
 413494 | 74 cube, its side will be the cube root of 
 343 its solid contents, which we will pro- 
 
 com P ute ' Pointin S off the 
 
 14700 70494. T 
 
 given number, the two periods show 
 
 that there will be two figures, tens and 
 
 units, in the root. The tens of the root must be extracted from the 
 first period, 413 thousands. The greatest cube in 413 thousands is 
 343 thousands, the cube of 7 tens ; we therefore write 7 tens in the 
 root at the right of the given number. 
 
CUBE ROOT. 
 
 321 
 
 Since the entire root is to be the side of a cube, let us form a 
 Fig% L cubical block (Fig. I), the side 
 
 of which is 70 inches in length. 
 The contents of this cube are 
 70 X 70 X 70 1= 343,000 solid 
 inches, which we subtract from 
 the given number. This is done 
 in the operation by subtracting 
 the cube number, 343, from the 
 first period, 413, and to the re- 
 mainder bringing down the sec- 
 ond period, making the entire 
 remainder 70494. 
 
 If we now enlarge our cubical 
 block, (Fig. I), by the addition of 70494 solid inches, in such a 
 manner as to preserve the cubical form, its size will be that of 
 the required block. To preserve the cubical form, the addition 
 must be made upon three adjacent sides or faces. The addition 
 will therefore be composed of 3 flat blocks to cover the 3 faces, 
 (Fig. II) ; 3 oblong blocks to fill the vacancies at the edges, 
 (Fig. Ill) ; and 1 small cubical block to fill the vacancy at the cor- 
 ner, (Fig. IV). Now, the thickness of this enlargement will be the 
 additional length of the side of the cube, and, consequently, the 
 second figure in the root. To find thickness, we may divide solid 
 Fi g- II - contents by surface, or area. 
 
 But the area of the 3 oblong 
 blocks and little cube cannot 
 be found till the thickness of 
 the addition be determined, be- 
 cause their common breadth is 
 equal to this thickness. We will 
 therefore find the area of the 
 three flat blocks, which is suffi- 
 ciently near the whole area to be 
 used as a trial divisor. As these 
 are each equal in length and 
 breadth to the side of the cube 
 whose faces they cover, the whole 
 area of the three is 70 X 70 X 
 3 rr: 14700 square inches. This number is obtained in the operation 
 by annexing 2 ciphers to three times the square of 7 ; the result 
 being written at the left hand of the dividend. Dividing, we obtain 
 
322 
 
 CUBE ROOT. 
 
 4, the probable thickness of the addition, and second figure of the 
 Fig IIL root. With this assumed figure, 
 
 we will complete our divisor by 
 adding the area of the 4 blocks, 
 before undetermined. The 3 ob- 
 long blocks are each 70 inches 
 long ; and the little cube, being 
 equal in each of its dimensions 
 to the thickness of the addition, 
 must be 4 inches long. Hence, 
 their united length is 70 -j- 70 
 + 70 + 4 214. This number 
 is obtained in the operation by 
 multiplying the 7 by 3, and an- 
 nexing the 4 to the product, the 
 result being written in column 
 I, on the next line below the 
 trial divisor. Multiplying 214, 
 the length, by 4, the common 
 width, we obtain 856, the area of 
 the four blocks, which added to 
 14700, the trial divisor, makes 
 15556, the complete divisor ; and 
 multiplying this by 4, the second 
 figure in the root, and subtract- 
 ing the product from the divi- 
 dend, we obtain a remainder of 
 8270 solid inches. With this re- 
 mainder, for the same reason as 
 before, we must proceed to make 
 a new enlargement. But since 
 we have already two figures in 
 the root, answering to the two 
 periods of the given number, 
 the next figure of the root must 
 be a decimal ; and we therefore 
 annex to the remainder a period 
 of three decimal ciphers, mak- 
 ing 8270.000 for a new dividend. 
 The trial divisor to obtain the 
 thickness of this second -enhir^e- 
 ment, or the next figure of the root, will be the area of three new flat 
 blocks to cover the three sides of the cube already formed j and this 
 
 OPERATION CONTINUED. 
 
 413494 | 74 
 
 I. IT. 
 
 343 
 
 214 856 
 
 14700 70494 
 15556 62224 
 
 8270.000 
 
 Fig. IV. 
 
CUBE ROOT. 
 
 323 
 
 surface, (Fig. IV,) is composed of 1 face of each of the flat blocks 
 already used, 2 faces of each of the oblong blocks, and 3 faces of 
 the little cube. But we have in the complete divisor, 15556, 1 
 face of each of the flat blocks, oblong blocks, and little cube ; 
 and in the correction of the trial divisor, 856, 1 face of each of 
 the oblong blocks and of the little cube; and in the square of 
 the last root figure, 16, a third face of the little cube. Hence, 16 
 _|_ 856 -}- 15556 = 16428, the significant figures of the new trial 
 
 divisor. This 
 
 OPERATION CONTIN 
 I. II. 
 
 UED. 
 
 413494| 74.5 
 
 number is ob- 
 tained in the 
 operation by 
 adding the 
 square of the 
 last root fig- 
 ure mentally, 
 and combin- 
 ing units of 
 like order, 
 
 343 
 
 214 856 
 
 14700 
 15556 
 
 70494 
 62224 
 
 222.5 111.25 
 
 1642800 
 16539.25 
 
 8270.000 
 8269.625 
 
 
 .375 
 
 thus : 16, 6, and 6 are 28, and we write the unit figure in the new 
 trial divisor ; then 2 to carry, and 5 and 5 are 12, &c. We annex 
 2 ciphers to this trial divisor, as to the former, and dividing, obtain 
 5, the third figure in the root. To complete the second trial di- 
 visor, after the manner of the first, the correction may be found by 
 annexing .5 to 3 times the former figures, 74, and multiplying this 
 number by .5. But as we have, in column I, 3 times 7, with 4 
 annexed, or 214, we need only multiply the last figure, 4, by 3, 
 and annex .5, making 222.5, which multiplied by .5 gives 111.25, 
 the correction required. Then we obtain the complete divisor, 
 16539.25, the product, 8269.625, and the remainder, .375, in the 
 manner shown by the former steps. From this example and analysis 
 we deduce the following 
 
 RULE. I. Point off the given number into periods of three 
 figures each, counting from units' place toward the left and right. 
 
 II. Find the greatest cube that does not exceed the left hand 
 period, and write its root for the first figure in the required 
 root ; subtract the cube from the left hand period, and to the 
 remainder bring down the next period for a dividend. 
 
 III. At the left of the dividend write three times the square 
 of the first figure of the root, and annex two ciphers, for a trial 
 divisor ; divide the dividend by the trial divisor, and write the 
 quotient for a trial figure w the root. 
 
324 CUBE ROOT. 
 
 IV. Annex the trial figure to three times the former figure, 
 and write the result in a column marked I, one line below tlie 
 trial divisor ; multiply this term by the trial figure, and write 
 the product on the same line in a column marked II ; add this 
 term as a correction to the trial divisor, and the result will be 
 the complete divisor. 
 
 V. Multiply the complete divisor by the trial figure, and 
 subtract the product from the dividend, and to the remainder 
 bring down the next period for a new dividend. 
 
 VI. Add the square of the last figure of the root, the last 
 term in column II, and the complete divisor together ^and annex 
 two ciphers, for a new trial divisor ; with which obtain an- 
 other trial figure in the root. 
 
 VII. Multiply the unit figure of the last term in column I 
 by 3, and annex the trial figure of the root for the next term 
 of column I ; multiply this result by the trial figure of the root 
 for the next term of column II ; add this term to the trial 
 divisor for a complete divisor, with which proceed as before. 
 
 NOTES. 1. If at any time the product be greater than the dividend, 
 diminish the trial figure of the root, and correct the erroneous work. 
 
 2. If a cipher occur in the root, annex two more ciphers to the 
 trial divisor, and another period to the dividend ; then proceed as be- 
 fore with column I, annexing both cipher and trial figure. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. What is the cube root of 79.112 ? 
 
 OPERATION. 
 
 79.112 | 4.2928 + , Ans. 
 64. 
 
 122 244 
 
 4400 15112 
 5044 10088 
 
 1269 11421 
 
 529200. 5024000 
 540621 4865589 
 
 12872 25744 
 
 55212300 158411000 
 55238044 110476088 
 
 128768 1030144 
 
 5526379200 47934912000 
 5527409344 44219274752 
 
 3714637248 rem. 
 
CUBE ROOT. 325 
 
 2. What is the cube root of 84604519 ? Am. 439. 
 
 3. What is the cube root of 2357947691 ? Ans. 1331. 
 
 4. What is the cube root of 109 63240788375? Ans. 22215. 
 
 5. What is the cube root of 270671777032189896 ? 
 
 Ans. 646866. 
 
 6. What is the cube root of .091125 ? Ans. .45. 
 
 7. What is the cube root of .000529475129 ? Ans. .0809. 
 
 8. What is the approximate cube of .008649 ? 
 
 Ans. .2052 + . 
 Extract the cube roots of the following numbers : 
 
 = 1.259921+ 
 =. 1.442249+ 
 = 1.587401+ 
 
 = 1.709975+ 
 = 1.817120+ 
 1.912931+ 
 
 APPLICATIONS IN CUBE ROOT. 
 
 1. What is the length of one side of a cistern of cubical 
 form, containing 1331 solid feet? Ans. 11 feet. 
 
 2. The pedestal of a certain monument is a square block of 
 granite, containing 373248 solid inches ; what is the length 
 of one of its sides ? Ans. 6 feet. 
 
 3. A cubical box contains 474552 solid inches ; what is 
 the area of one of its sides ? Ans. 42 J sq. ft. 
 
 4. How much paper will be required to make a cubical 
 box which shall contain f of a solid foot ? Ans. f of a yard. 
 
 5. A man wishes to make a bin to contain 125 bushels, of 
 equal width and depth, and length double the width ; what 
 must be its dimensions ? Ans. Width and depth, 51.223 + 
 inches ; length, 102.446 + inches. 
 
 NOTE. Spheres are to each other as the cubes of their diameters or 
 circumferences. 
 
 6. There are two spheres whose solid contents are to each 
 other as 27 to 343 ; what is the ratio of their diameters ? 
 
 ANALYSIS. Since spheres are to each other as the cubes of their 
 diameters, the diameters will be to each other as the cube roots of 
 the spheres ; and 41/27 = 3, ^/343 7 j hence the diameters required 
 are as 3 to 7. 
 
 BB 
 
326 ARITHMETICAL PROGRESSION. 
 
 7. The diameter of a sphere containing 1 solid foot is 14.9 
 inches ; what is the diameter of a sphere containing 2 solid 
 feet ? Ans. 18.8 inches. 
 
 8. If a cable measuring 4 inches in circumference support 
 a weight of 1800 pounds, what must be the circumference of 
 a cable that will support 3515| pounds? Ans. 5 inches. 
 
 ARITHMETICAL PROGRESSION. 
 
 4L&t3. An Arithmetical Progression, or Series, is a series 
 of numbers increasing or decreasing by a common difference. 
 Thus, 3, 5, 9, 11, &c., is an arithmetical progression with an 
 ascending series, and 13, 10, 7, 4, &c., is an arithmetical pro- 
 gression with a descending series. 
 
 4t$3. The Terms of a series are the numbers of which it 
 is composed. 
 
 434. The Extremes are the first and last terms. 
 
 43*5. The Means are the intermediate terms. 
 
 436. The Common Difference is the difference between 
 any two adjacent terms. 
 
 437. There are Jive parts in an arithmetical series, any 
 three of which being given, the other two may be found. 
 They are as follows : the first term, last term, common differ- 
 ence, number of terms, and sum of all the terms. 
 
 CASE I. 
 
 438. To find the last -term when the first term, 
 common difference, and number of terms are given. 
 
 Let 2 be the first term of an ascending series, and 3 the 
 common difference; then the series will be written, 2, 5, 8, 11, 
 14, or analyzed thus : 2, 2 + 3, 2 + 3 + 3, 2 + 3 + 3 + 3, 
 2 + 3 + 3 + 3 + 3. 
 
 Here we see that, in an ascending series, we obtain the 
 second term by adding the common difference once to the first 
 term ; the third term, by adding the common difference twice 
 to the first term ; and, in general, we obtain any term by 
 
ARITHMETICAL PROGRESSION. 327 
 
 adding the common difference as many times to the first term 
 as there are terms less one. 
 
 NOTE. The analysis for a descending series would be similar. 
 Hence, 
 
 RULE. Multiply the common difference by the number of 
 terms less one, and add the product to the first term, if the 
 series be ascending, and subtract it if the series be descending. 
 
 EXAMPLES. 
 
 1. The first term of an ascending series is 4, the common 
 difference 3, and the number of terms 19; what is the last 
 term? Ans. 58. 
 
 2. What is the 13th term of a descending series whose first 
 term is 75, and common difference 5 ? Ans. 15. 
 
 3. A boy bought 18 hens, paying 2 cents for the first, 5 
 cents for the second, and 8 cents for the third, in arithmetical 
 progression ; what did he pay for the last hen ? 
 
 4. What is the 40th term of the series , -, 1, 1^, &c. ? 
 
 Ans. 10. 
 
 5. A man travels 9 days ; the first day he goes 20 miles, 
 the second 25 miles, increasing 5 miles each day; how far 
 does he travel the last day of his journey ? Ans. 60 miles. 
 
 6. What is the amount of $100, at 7 per cent, for 45 
 years ? $100 + $7 X 45 = $415, Ans. 
 
 CASE II. 
 
 439. To find the common difference when the 
 extremes and number of terms are given. 
 
 Referring to the series, 2, 5, 8, 11, 14, analyzed in 438, 
 we readily see that, by subtracting the first term from any 
 term, we have left the common difference taken as many times 
 as there are terms less one ; thus, by taking away 2 in the 
 fifth term, 2+3 + 3 + 3 + 3, we have 3 taken 4 times. 
 Hence, 
 
 RULE. Divide the difference of the extremes by the number 
 of terms less one. 
 
328 ARITHMETICAL PROGRESSION. 
 
 EXAMPLES. 
 
 1. The first term is 2, the last term is 17, and the number 
 of terms is 6 ; what is the common difference ? Ans. 3. 
 
 2. A man has seven children, whose ages are in arithmetical 
 progression; the youngest is 2 years old, and the eldest 14; 
 what is the common difference of their ages ? Ans. 2 years. 
 
 3. The extremes of an arithmetical series are 1 and 50, 
 and the number of terms is 34 ; what is the common difference ? 
 
 4. An invalid commenced to walk for exercise, increasing 
 the distance daily by a common difference ; the first day he 
 walked 3 miles, and the 14th day 9 miles; how many miles 
 did he walk each day ? 
 
 NOTE. When we have found the common difference we may add it 
 once, twice, &c., to the first term, and we have the series, and conse- 
 quently the means, 
 
 Ans. 3, 3, 4, 4, 5, 5 J, &c. 
 
 CASE III. 
 
 44O. To find the number of terms when the ex- 
 tremes and common difference are given. 
 
 Examining the series, 2, 5, 8, 11. 14, analyzed in 438 5 
 we also see that after taking away the first term from any 
 term, we have left the common difference taken as many 
 times as the number of terms, less 1. Hence, 
 
 RULE. Divide the difference of the extremes by the common 
 difference, and add 1 to the quotient. 
 
 EXAMPLES. 
 
 1. The extremes are 7 and 43, and the common difference 
 is 4 ; what is the number of terms ? Ans. 10. 
 
 2. The first term is 2, the last term is 40, and the common 
 difference is 7 ; what is the number of terms ? Ans. 6. 
 
 3. A laborer agreed to build a fence on the following con- 
 ditions: for the first rod he was to have 6 cents, with an 
 increase of 4 cents on each successive rod ; the last rod came 
 to 220 cents ; how many rods did he build ? Ans. 56 rods. 
 
GEOMETRICAL PROGRESSION. 329 
 
 CASE IV. 
 
 441. To find the sum of all the terms when the 
 extremes and number of terms are given. 
 
 To deduce a rule for finding the sum of all the terms, we 
 will take the series 2, 5, 8, 11, 14, writing it under itself in an 
 inverse order, and add each term ; thus, 
 
 2+5+ 8+ 11 + 14 = 40, once the sum. 
 14 + 11+ 8 + 5+ 2 = 40, 
 1C + 16 + 1C + 1C + 1C = 80, twice the sum. 
 
 Here we perceive that 1C, the sum of the extremes, multi- 
 plied by 5, the number of terms, equals 80, which is twice the 
 sum of the series. Dividing 80 by 2 ogives 40, which is the 
 sum required. Hence, 
 
 RULE. Multiply the sum of the extremes by the number of 
 terms, and divide the product by 2. 
 
 EXAMPLES. 
 
 1. The extremes are 5 and 32, and the number of terms 12 ; 
 what is the sum of all the terms ? Am. 222. 
 
 2. How many strokes does a common clock make in 12 
 hours ? Ans. 78 .strokes. 
 
 3. What debt can be discharged in a year by weekly pay- 
 ments in arithmetical progression, the first bfinjr 824, arid the 
 last si 224? Ans. $32448. 
 
 4. Suppose 100 apples were placed in a line 2 yards apart, 
 and a basket 2 yards from the first apple ; how far would a 
 boy travel to gather them up singly, and return with each 
 separately to the basket ? Ans. 20200 yards. 
 
 GEOMETRICAL PROGRESSION. 
 
 442. A Geometrical Progression is a series of numbers 
 increasing or decreasing by a constant multiplier. 
 
 When the multiplier is greater than a unit, the series is 
 
 BB* 
 
880 GEOMETRICAL PROGRESSION. 
 
 ascending; thus, 2, 6, 18, 54, 162, is an ascending series, in 
 which 3 is the multiplier. 
 
 When the multiplier is less than a unit, the series is descend- 
 ing; thus, 162, 54, 18, 6, 2, is a descending series, in which - 
 is the multiplier. 
 
 443. The Ratio is the constant multiplier. 
 
 444. In every geometrical progression there are five 
 parts to be considered, any three of which being given, the 
 other two may be determined. They are as follows: The first 
 term, last term, ratio, number of terms, and the sum of all the 
 terms. 
 
 The first and last terms are the extremes, and the interme- 
 diate terms are the means. 
 
 CASE I. 
 
 445. To find any term, the first term, the ratio, 
 and number of terms being given. 
 
 The fir.st term is supposed to exist independently of the 
 ratio. Using the ratio once as a factor, we have the second 
 term ; using it twice, or its second power, we have the third 
 term ; using it three times, or its third power, we have the 
 fourth term ; and, in general, the power of the ratio in any 
 term is one less than the number of the term. The ascending 
 series, 2, 6, 18, 54, may be analyzed thus: 2, 2 X 3, 2 X 
 3X3, 2X3X3X3. 
 
 In this illustration we see that 
 
 1st term, 2, is independent of the ratio. 
 
 2d " 6 2 X 3 = the first term into the 1st power of 
 the ratio. 
 
 3d term, 18 = 2 X 3 2 = the first term into the 2d power 
 of the ratio. 
 
 4th term, 54 r= 2 X 3 3 = the first term into the 3d power 
 of the ratio. Hence 
 
 RULE. Multiply the fast term by that power of the ratio 
 denoted by the number of terms less 1. 
 
GEOMETRICAL PROGRESSION. 331 
 
 EXAMPLES. 
 
 1. The first term of a geometrical series is 4, the ratio is 3 ; 
 what is the 9th term ? Ans. 4 X 3 8 26244. 
 
 2. The first term is 1024, the ratio ^, and the number of 
 term 8 ; what is the last term ? Ans. y 1 ^. 
 
 3. A boy bought 9 oranges, agreeing to pay 1 mill for the 
 first orange, 2 mills for the second, and so on ; what did the 
 last orange cost him ? Ans. $2.304. 
 
 4. The first term is 7, the ratio ^-, and the number of terms 
 7 ; what is the last term ? Ans. T ^^T' 
 
 5. What is the amount of $1 at compound interest for 5 
 years, at 7 per cent, per annum ? Ans. $1.40255 + 
 
 NOTE. In the above example the first term is $1, the ratio is $1.07, 
 and the number of terms is 6. 
 
 6. A drover bought 7 oxen, agreeing to pay $3 for the first 
 ox, $9 for the second, $27 for the third, and so on ; what did 
 the last ox cost him ? Ans. $2187. 
 
 CASE n. 
 
 446. To find the sum of all the terms, the ex- 
 tremes and ratio being given. 
 
 If we take the series 2, 8, 32, 128, 512, in which the ratio 
 is 4, multiply each term by the ratio, and add the terms thus 
 multiplied, we shall have 
 
 8 + 32 + 128 + 512 + 2048 = 2728 ={^ Sf e *l ram 
 But 2 + 8 + 32 + 128 + 512 = 682 = j ^ & ce t ^ ttm of a11 
 
 Hence, by subtracting, we get 2048 2= 204G = { ^^^l^ 
 
 1 f *" 
 
 Dividing by 3, theratio less one, 2046-^3 = 682 = 
 
 The subtraction is performed by taking the lower line or 
 series from the upper. All the terms cancel except 2048 and 
 2. Taking their difference, which is 3 times the sum, and di- 
 viding by 3, the ratio less one, we must have the sum of all 
 the terms. Hence 
 
332 PROMISCUOUS EXAMPLES. 
 
 RULE. Multiply the greater extreme by the ratio, subtract 
 the less extreme from the product, and divide the remainder 
 by the ratio less 1. 
 
 NOTE. Let every decreasing series be inverted, and the first term 
 called the last ; then the ratio will be greater than a unit. If the series 
 be infinite, the first term is a cipher. 
 
 EXAMPLES. 
 
 1. The first term is 2, the last term 512, and the ratio 3 ; 
 what is the sum of all the terms ? Am. 767. 
 
 2. The first term is 4, the last term is 262144, and the 
 ratio is 4; what is the sum of the series? Ans. 349524. 
 
 3. The first term of a descending series is 162, the last 
 term 2, and the ratio ; what is the sum ? Ans. 242. 
 
 4. What is the value of , , y^, &c., to infinity ? Ans. . 
 
 NOTE. In the following examples we first find the last term by the 
 Rule under Case I. 
 
 5. What yearly debt can be discharged by monthly pay- 
 ments, the first being $2, the second $6, and the third $18, 
 and so on, in geometrical progression ? Ans. $531440. 
 
 6. If a grain of wheat produce 7 grains, and these be sown 
 the second year, each yielding the same increase, how many 
 bushels will be produced at this rate in 12 years, if 1000 grains 
 make a pint ? Ans. 3604 bu. 2 pk. 1| pt. 
 
 7. Six persons of the Morse family came to this country 
 200 years ago ; suppose that their number has doubled every 
 20 years since, what would be their number now ? 
 
 NOTE. The other cases in Progression will be found in the Higher 
 Arithmetic. 
 
 PROMISCUOUS EXAMPLES. 
 
 1. One half the sum of two numbers is 800, and one half the 
 difference of the same numbers is 200 ; what are the numbers ? 
 
 Ans. 1000 and 600. 
 
 2. What number is that to which, if you add of T \ of itself, 
 the sum will be 61 ? Ans. 55. 
 
 3. What part of a day is 3 h. 21 min. 15 sec. ? Ans. 
 
PROMISCUOUS EXAMPLES. 333 
 
 4. A commission merchant received 70 bags of wheat, each con- 
 taining 3 bu. 3 pk. 3 qt. ; how many bushels did he receive ? 
 
 5. Four men, A, B, C, and D, are in possession of $1100; A 
 has a certain sum, B has twice as much as A, C has $300, and D 
 has $200 more than C ; how many dollars has A ? Ans. $100. 
 
 6. At a certain election, 3000 votes were cast for three candi- 
 dates, A, B, and C ; B had 200 more votes than A, and C, had 800 
 more than B ; how many votes were cast for A ? Ans. 600. 
 
 7. What part of 17 is 31 ? Ans. |f . 
 
 8. The difference between ^ and of a number is 10 ; what is 
 the number? Ans. 560. 
 
 9. A merchant bought a hogshead of rum for $28.35 ; how 
 much water must be added to reduce the first cost to 35 cents per 
 gallon ? Ans. 18 gal. 
 
 10. A and B traded with equal sums of money ; A gained a sum 
 equal to ^ of his stock ; B lost $200, and then he had ^ as much as 
 A ; how much was the original stock of each ? Ans. $500. 
 
 11. A toner sold 17 bushels of barley, and 13 bushels of wheat, 
 for $31.55 ; he received for the wheat 35 cents a bushel more than 
 for the barley ; what was the price of each per bushel ? 
 
 Ans. Barley, $.90; wheat, $1.25. 
 
 12. What is the interval of time between March 20, 21 minutes 
 past 3 o'clock, P. M., and April llth, 5 minutes past 7 o'clock, 
 A. M. ? Ans. 21 da. 15 h. 44 min. 
 
 13. What o'clock is it when the time from noon is T 9 T of the 
 time to midnight ? Ans. 5 o'clock 24 min. P. M. 
 
 14. What is the least number of gallons of wine that can be 
 shipped in hogsheads, tierces, or barrels, just filling the vessels, 
 without deficit or excess ? Ans. 126 gal. 
 
 15. A ferryman has four boats ; one will carry 8 barrels, another 
 9, another 15, and another 16 ; what is the smallest number of bar- 
 rels that will make full freight for either one of the boats ? Ans. 720. 
 
 16. A and B have the same income ; A saves |- of his, but B, 
 by spending $30 a year more than A, at the end of four years finds 
 himself $40 in debt ; what is their income, and how much does 
 each spend a year ? C Income, $160. 
 
 Ans. 1 A spends $140. 
 C B spends $170. 
 
 17. If a load of plaster weighing 1825 pounds cost $2.19, how 
 much is that per ton of 2000 pounds ? Ans. $2.40. 
 
 18. If 2| yards of cloth If yards wide cost $3.37f , what will be 
 the cost of 36^ yards U yards wide ? Ans. $52.79. 
 
 19. I lend my neighbor $200 for 6 months ; how long ought he 
 to lend me $1000 to balance the favor ? Ans. 36 days. 
 
 20. Bought railroad stock to the amount of $2356.80, and found 
 that the sum invested was 40 per cent of what I had left ; what 
 sum had I at first ? Am. $8248.80. 
 
 21. 20 per cent, of -| of a number is what per cent, of ^ of it ? 
 
 Ans. 12. 
 
334 PROMISCUOUS EXAMPLES. 
 
 22. Bought wheat at $1.50 per bushel, corn at $.75 per bushel, 
 and barley at $.60 per bushel ; the wheat cost twice as much as the 
 corn, and the corn twice as much as the barley ; of the sum paid, 
 $243 and J of the whole was for wheat, and $153 and T ^ of the 
 whole was for the corn ; how many bushels of grain did I purchase ? 
 
 Ans. 756. 
 
 23. Divide $630 among 3 persons, so that the second shall have 
 | as much as the first, and the third -| as much as the other two ; 
 what is the share of each ? C 1st, $240. 
 
 Ans. 32d, $180. 
 (3d, $210. 
 
 24. Bought a hogshead of molasses for $28, and 7 gallons 
 leaked out ; at what rate per gallon must the remainder be sold to 
 gain 20 per cent. ? Ans. $.60. 
 
 25. 20 per cent, of -| of a number is how many per cent, of 2 
 times -| of 14 times the number ? Ans. 7^-. 
 
 26. B and C, trading together, find their stock to be worth $3500, 
 of which C owns $2100 ; they have gained 40 per cent, on their first 
 capital; what did each put in ? A $ B, $1500. 
 
 27. If the ridge of a building be 8 feet above the beams, and the 
 building be 32 feet wide, what must be the length of rafters ? 
 
 28. If 12 workmen, in 12 days, working 12 hours a day, can 
 make up 75 yards of cloth, f of a yard wide, into articles of clothing : 
 how many yards, 1 yard wide, can be made up into like articles, by 
 10 men, working 9 days, 8 hours each day ? Ans. 23^. 
 
 29. A grocer sells a farmer 100 pounds of sugar, at 12 cents a 
 pound, and makes a profit of 9 per cent. ; the former sells him 100 
 pounds of beef, at 6 cents a pound, and makes a profit of 10 per 
 cent. ; who gains the most by the trade, and how much? 
 
 Ans. The grocer gains $.536 -}- most. 
 
 30. In 1 yr. 4 mo. $311.50 amounted to $336.42, at simple 
 interest ; what was the rate per cent. ? Ans. 6. 
 
 31. Three persons engage to do a piece of work for $20; A and 
 B estimate that they do $ of it, A and C that they do f of it, and B 
 and C that they do f of it ; according to this estimate, what part of the 
 $20 should each man receive ? Ans. A's, $llf ; B's, $5f ; C's, $2f 
 
 32. Paid $375, at the rate of 2 per cent., for insurance on a 
 cotton factory and the machinery ; for what amount was the policy 
 given? Ans. $15000. 
 
 33. A merchant bought goods in Boston to\the amount of $1000, 
 and gave his note, dated Jan. 1, 1857, on interest after 90 days; 
 six months after the note Vas given he paid\ $560, and 5 months 
 after the first payment he paie^ $406 ; what wasYhie Aug 23, 1859 ? 
 
 Ans. $63.01. 
 
 34. Iff of A's money be equal to f of B's, and f of B's be equal 
 to - of C/s, and f of C's be equal to f of D's, and D has $45 more 
 than C, how much has each ? A $ A, $368 ; C, $360 ; 
 
 Ans ' B, $336 ; D, $405. 
 
Tffep 
 
 PROMISCUOUS EXJBTLHS. w W 
 
 'KSI.TY 
 
 35. A owed B $900, to be paid in S^barsy^ but aylafi. expiration 
 of 9 months A agreed to pay $300 if B V^Ejdk^oiig enough for 
 the balance to compensate for the advancVj^hcw long should B 
 wait after the expiration of the 3 years ? Arts, lo^ mo. 
 
 36. A certain clerk receives $800 a year ; his expenses equal ^ 
 of what he saves ; how much of his salary does he save yearly ? 
 
 37. A merchant sold cloth at $1 per yard, and made 10 per cent, 
 profit ; what would have been his gain or loss had he sold it at $.87-^ 
 per yard ? Ans. Loss, 3| per cent. 
 
 21 9 
 
 38. What is the cube of -~ Ans. fj. 
 
 11 63 
 
 39. What is the cube root of Ans. f . 
 
 40. A miller is required to grind 100 bushels of provender worth 
 50 cents a bushel, from oats worth 20 cents, corn worth 35 cents, 
 rye worth 60 cents, and wheat worth 70 cents per bushel; how 
 many bushels of each must he take ? 
 
 41. A man owesl$6480 to his creditors; his debts are in arith- 
 metical progression,! the least being $40, and the greatest $500 ; 
 required the numberlof creditors aad the common difference between 
 the debts. A 5 24 creditors. 
 
 s ' I $20 difference. 
 
 42. Two ships sail from the same port ; one goes due north 128 
 miles, and the other due east 72 miles ; how far are the ships from 
 each other ? Ans. 146.86 -|- miles. 
 
 43. If 10 pounds of cheese are equal in value to 7 pounds of 
 butter, and 11 pounds of butter to 2 bushels of corn, and 14 bushels 
 of corn to 8 bushels of rye, and 4 bushels of rye to 1 cord of wood ; 
 how many pounds of cheese are equal in value* to 10 cords of wood? 
 
 Ans. 550. 
 
 44. A and B trade until they gain 6 per cent, on their stock ; 
 then f of A's gain is $18 ; now, if A's stock is to B's as -f to ^, how 
 much did each gain, and w r hat was the original stock of each ? 
 
 A 5 A's gain, $45 ; stock, $750. 
 S " I B's " $37.50 ; " $625. 
 
 45. If 20 men, in 21 days, by working 10 hours a day, can dig a 
 trench 30 ft. long, 15 ft. wide, and 12 ft. deep, when the ground is 
 called 3 degrees of hardness, how many men, in 25 days, by work- 
 ing 8 hours a day, can dig another trench 45 ft. long, 16 ft. wide, 
 and 18 ft. deep, when the ground is estimated at 5 degrees of 
 hardness ? Ans. 42. 
 
 46. Wishing to know the height of a certain steeple, I measured 
 the shadow of the same on a horizontal plane, 27^ feet ; I then 
 erected a 10 feet pole on the same plane, and it cast a shadow of 2-| 
 feet ; what was the height of the steeple ? Ans. 1031 ft. 
 
 47. A can do a piece of work in 3 days, B can do 3 times as 
 much in 8 days, and C 5 times as much in 12 days ; in what time 
 can they all do the first piece of work ? Ans. f da. 
 
336 PROMISCUOUS EXAMPLES. 
 
 48. How many building lots, each 75 feet by 125 feet, can be 
 made from 1 A. 1 R. 6 P. 18 sq. yd. ? Ans. 6. 
 
 49. A man bought a house, and as/eed to pay for it $1 on the 
 first day of January, $2 on the first/day of February, $4 on the 
 first day March, and so on, in geometrical progression, through 
 the year ; what will be the cost of the house, and what the average 
 time of payment ? A \ $ 4095 - 
 
 ' s ' I Average time, Nov. 1. 
 
 50. A man sold a rectangular piece of ground, measuring 44 
 chains 32 links long by 36 chains wide; how many acres did it 
 contain ? Ans. 159 A. 2 R. 8.32 P. 
 
 51. What number is that which being increased by its half, its 
 third, and 18 more, will be doubled ? Ans. 108^. 
 
 52. A merchant has 200 Ib. of tea, worth $.62^ per pound, which 
 he will sell at $.56 per pound, provided the purchaser will pay in 
 coffee at 22 cents, which is worth 25 cents per pound ; does the 
 merchant gain or lose by the sale of the tea, and how much per 
 cent. ? Ans. Lost 1H per cent. 
 
 53. A man owes a debt to be paid in 4 equal installments at 4, 
 9, 12, and 20 months, respectively; discount being allowed at 5 per 
 cent., he finds that $750 ready money will pay the debt ; how much 
 did he owe? Ans. $784.74+. 
 
 54. A and B traded upon equal capitals ; A gained a sum equal 
 to -| of his capital, and B a sum equal to -|l of his ; B's gain was 
 $500 less than A's ; what was the capital of each ? Ans. $4000. 
 
 55. I purchase goods in bills as follows: June 4, 1859, $240.75; 
 Aug. 9, 1859, $|137.25; Auk 29, 1859, $65.64; Sept. 4, 1859, 
 $230.36; Nov. 12, 1859, $$3. If the merchant agree to allow 
 credit of 6 mo. qjn each bill, wnen may I settle by paying the whole 
 amount? Ans. Feb. 1, 1860. 
 
 56. A young man inherited a fortune, ^ of which he spent in 3 
 months, and $ of the remainder in 10 months, when he had only 
 $2524 left; how much had he at first? Ans. $5889.33 +. 
 
 57. A man bought a piece of land for $3000, agreeing to pay 7 
 per cent interest, and to pay principaV and interest in 5 equal an- 
 nual installments ; how much was the' annual payment ? 
 
 Ans. $731.67-}-. 
 
 58. A man held three notes, the first for $600, due July 7, 1859 ; 
 the second for $530, due Oct. 4, 1859 ; and the third for $400, due 
 Feb. 20, 1860; he 'made an equitable exchange of these with a 
 speculator for two o,ther notes, one of which was for $730, due Nov. 
 15, 1859; what was the face of the other, and when due? 
 
 A J Face, $800. 
 
 s ' I Due Aug. 29, 1859. 
 
 59. A room is 27 ft. 6 in. by 22 ft. 6 in., and 10 ft. 3 in. above the 
 base-board ; in said room are 2 doors, each 8 ft. by 4 ft. 4 in. ; 2 
 windows, each 6 ft. by 3 ft. 4 in., and a fireplace of 4 ft. by 4 ft. 6 
 in. : what will it cost to plaster the room, at 18 cents per square 
 yard? Ans. $30.328$. 
 

UNIVERSITY OP CALIFORNIA LIBRARY 
 
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