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Hill Collection Early American Mathematics Books Digitized by tlie Internet Arcliive in 2008 witli funding from IVIicrosoft Corporation littp://www.arcliive.org/details/commonsclioolaritOOIiagaricli COMMON SCHOOL ARITHMETIC BY D. B. HAGAR, PRINCIPAL OF STATE NORMAL SCHOOL, SALEM, MASS. PHILADELPHIA: COWPERTHWAIT cS: COMPANY. HAGAR'S Mathematical Series. I. Primary Lessons in Numbers. II. Elementary Arithmetic. III. Common School Arithmetic. IV. Elementary Algebra. (In press.) To be followed by other Books of a Complete Series. FOR TEACHERS. A Manual of Dict.a.tion Problems and Key to the Common School Arithmetic. Containing carefully prepared Dictation Problems and Topical Reviews, and all answers omitted in the Common School Arithmetic, with Solutions of Difficult Problems .... Price, Si.oo Dictation Problems (separately) .... "50 Foi-wanied, postpaid, on receipt of the Price. Entered., accordiitf; to Act of Coiif^ress, in the year 137 1, by DANIFJ. E. HAGAR and HENRY n. MAGI.ATHLIM In the Office of the Librarian of Congress, at Washington. Westcott & Thohson, Sherman & Co. Stereolyptrs and EUclrotyfers. PItilada. Printers, Philada, INTRODUCTION. This Common School Arithmetic is designed to be a com- plete manual for learners Avho may be prepared to advance beyond the first lessons in Numbers. It has been constructed with a view to the most rapid and thorough progress of the pupil by the use of the least number of books possible, and by the greatest economy of time. It combines mental and written exercises in a practical sys- tem. All obsolete and valueless material and all merely puzzling problems have been excluded, but no pains have been spared to embody valuable modern methods of computa- tion and topics having direct relation to business as it is transacted at the present day. The work is sufficiently comprehensive to render the use of a higher arithmetic quite unnecessary. It is ample enough in its range of subjects and exercises to qualify the learner for a skillful and i:)rompt solution of all ordinary problems of a commercial character, and at the same time to subserve the purposes of mental discipline. The Primary Lessons in Numbers and the Elementary Arithmetic, of this series, it is believed, form a valuable com- •])endious course sufficient for a majority of pupils. The Primary Lessons and the Common School Arithmetic like- wise form a two-book course, but full and complete. SUGGESTIONS TO TEACHEKS. A TEACHER should never undertake a recitation in Arith- metic without a full understanding of the subject of the lesson. Preparation should be made for the elucidation of difficulties, and for making plain the way of the learner, whenever re- quired, by happily-chosen, familiar illustrations. In forming classes, pupils of the same attainments, as nearly as possible, should be placed together. Vary the exercises in a class, so as to secure animation and interest and retain the attention of each member during the entire recitation. Tolerate no indefinite answering of questions. Require all principles and rules to be recited exactly, and all forms of so- lution to be logically and concisely expressed. Let all answers to Test Questions be definite and prompt. Do not overlook the importance of mental arithmetic. The plan of this book is to combine mental and written exercises, and to require a reason for every process. The proficiency of the learner should be often tested with problems not found in the book. Require, as an occasional test, the formation of problems and their solution without regard to rules. Seek, above all, to make the arithmetical exercise useful in the cultivation of the invaluable habit of self-reliance. En- deavor to give such a practical character to the instruction in the science of Numbers, that the knowledge acquired may be found readily available in the many computntions required in business. To attain (his object, good Ixjoks are aids, but can never perform the duties or assume the responsibilities of the 'eacher. 4 CONTENTS. SECTION PAGiS I. Pkeliminary Defi- nitions " ii. nubieration and no- TATION 9 Orders of Units 11 Decimal System 12 III. Addition 17 IV. Subtraction 24 V. Review Problems 30 VI. Multiplication 33 VII. Division 41 VIII. Review Problems 49 IX. Factoring 52 X. Divisors and Fac- tors 56 Greatest Common Divisors 58 XI. Multiples 60 Least Common Mul- tiples 62 XII. Factors in Divis- ion 64 Division by Factors... 65 Cancellation 66 XIII. Analysis 68 XIV. Review Problems 70 XV. Common Fractions 72 XVI. Reduction of Frac- tions 75 Higlier or Lower Terms 75 Common Denomina- tor 79 XVII. Addition of Frac- tions 82 XVIII. Subtraction OF Frac- tions 83 XIX. Multiplication of Fractions 85 Compound Frac- tions 88 XX. Division of Frac- tions 90 Complex Fractions... 93 1* SECTION PAQE XXI. Relation of Num- bers 95 XXII. Review Problems 98 XXIII. Decimal Fractions 102 XXIV. Reduction of Deci- mals 106 XXV. Addition and Sub- traction OF Dec- I3IALS Ill XXVI. Multiplication of Decimals 113 XXVII. Division of Deci- mals 115 XXVIII. United States Money 118 XXIX. Reduction of U. S. Money 120 XXX. Computations 122 XXXI. Business Methods... 124 XXXII. Bills and Ac- counts 128 XXXIII. Review Problems... 132 XXXIV. Denominate Num- bers 133 Measures of Exten- sion 133 Measures of Capa- city 139 Measures of Weight 141 Measures of Time,.. 144 Miscellaneous Measures 147 XXXV. Compound Numbers 149 Reduction De- scending 150 Reduction Ascend- ing 152 XXXVI. Addition of Com- pound Numbers. 157 XXXVII. Subtraction of Compound Nttm- bers.... 160 Difference of Dates. 161 6 CONTENTS. SECTION PAGE XXXVIII. Multiplication of Compound Num- bers 162 XXXIX, Division of Com- pound Numbers. 161 Longitude and Time 165 XL. Aliquot Parts 168 XLI. MEASURE3IENTS 170 XLII. Review Problems.. 174 XLin. Percentage 176 XLIV. Profit and Loss 182 XLV. Commission 185 XLVI. Insurance 187 XLVII. Review Problems.. 189 XLVIII. Simple Interest 192 XLIX. Partial Payments. 202 L. Present Worth 206 Commercial Dis- couiit 206 True Discount 207 LI. Banking - 208 LII. Annual Interest... 213 LIII. Compound Inter- est 215 LIV. Review Problems.. 217 LV. Ratio 219 LVI. Proportion 221 Simple Proportion. 223 Compound Pro- portion 225 LVII. Distributive Pro- portion 229 LVIII. Partnership 2.30 LIX. Average of Pay- ments 235 Debit and Credit Account 239 Cash Balances 241 LX. Investments 212 Corporate Stocks 213 Government Secur- ities 243 LXI. Exchange 247 Domestic Ex- change 248 Foreign Exchange.. 250 LXII. General Taxes 254 LXIIl. National TAxes.... 257 LXIV. Review Problems. 258 SECTION PAGB LXV. Involution 262 Powers 263 Process of Involu- tion 263 LXVI. Evolution 264 INlethod for Square Root 265 Method for Cube Root 271 LXVII. Mensuration 278 Polygons 278 Right - angled Tri- angles 282 Circles 283 Volumes 284 Similar Figures 287 LXVIII. Boards and Timber... 290 Squared Timber 290 Round Timber 291 LXIX. Stone and Brick Work 293 LXX. Grain and Hay 295 LXXI. Gauging 296 LXXII. Metric Sy'Stem 298 Linear Measures 299 Surface Measures 299 Cubic Measui-es 299 Liquid and Dry Meas- ures 300 Weights 300 Computations 302 LXXIII. Series or Progres- sion 304 Arithmetical Pro- gression 301 Geometrical Progres- sion 306 LXXIV. Life Interests 308 Carlisle Tables 309 LXXV. Problems for Anal- ysis 311 LXXVI. General Review 321 APPENDIX. Roman Notation 323 Contractions 326 Duodecimals. 328 Accurate Interest 3;i0 Examination Problems 331 Common School Arithmetic. SECTION I. PRELIMIJfAET BEFIKITIOKS. 'RTICLE 1. — A Unit is one, or a single thing of any kind. 3. A Number is a unit, or a collection of units. Thus, one, two, three, four, five, are numbers. 3. The Unit of a number is one of the collection forming that number. Thus, one is the unit of six, one book is tlie unit of six books. 4. An Integer is a number formed wholly of entire units. Thus, three, five, six, nine, are integers. Integers are also called integral or vohoh numbers. 6, Similar Numbers are those which have the same unit. Thus, three yards and five yards are similar numbers. 6. Dissimilar Numbers are those which do not have the same unit. Thus, three yards and three books are dissimilar numbers. 7. A Concrete Number is one that names the kind of unit numbered. Thus, five bushels, in which the kind of unit is named, is a concrete number. 7 8 PRELIMINARY DEFINITIONS. 8. An Abstract Number is one that does not name the kind of unit numbered. Thus, five, in which the kind of unit is not named, is an abstract number. 9. Arithmetic is the science of numbers and the art of com- puting by them. 10. A Solution in Arithmetic is the process of answering a question which requires computation. 11. A Proof of a solution is the process of testing its cor- rectness. 12. A Problem is a question for solution. 13. A Principle is a general truth. 14. A Rule is U concise statement of the method of solving a problem. 15. An Example is a problem which is used to illustrate a principle or rule. 16. An Exercise is a problem which is intended to render knowledge familiar by drill or practice. EXEItCISES. 1. How many units in one? In one dollar? Three is a col- lection of how many units ? 2. What is the unit of two books? Of four? Of five pounds ? Of seven houses ? 3. Are two cents and five cents similar or dissimilar numbers ? Why are three men and five books dissimilar numbers ? 4. Is four yards a concrete or an abstract number ? Three ? Two boys ? 5. Why is one mile the unit of four miles ? Why is one the unit of six ? 6. Why is two houses a concrete number ? Why is four an abstract number ? 7. Of the two numbers, two miles and ten miles, what is the unit ? 8. Of the two numbers, seven dollars and nine dollars, what is the unit? NUMERATION AND NOTATION. SECTION 11. JfVMEEATIOK AJfD JTOTATIOJ^. 17. The Naming of numbers is performed by means of a small number of words. A single thing is named one; one and one is named two; one and one and one is named three; and so we have the separate names, One, two, three, four, five, six, seven, eight, nine, ten. 18. Ten, by being regarded as forming a set or collection of units, may be treated as a single thing, or as a unit equal to ten ones. One and ten, two and ten, three and ten, four and ten, etc., by change of form, give the familiar names, Eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, nineteen. Two tens, three tens, four tens, etc., by change of form, give the names, Twenty, thirty, forty, fifty, sixty, seventy, eighty, ninety. Twenty and one, twenty and two, etc. ; thirty and one, thirty and two, etc., to ninety and nine, by change of form, give the names, Twenty-one, twentij-two, etc.; thirty-one, thirty-two, etc., to ninety-nine. 19. One Hundred is the name given to a collection of ten tens. One hundred and one hundred, two hundred and one hun- dred, etc., form Two hundred, three hundred, etc., to nine hundred. FIGURES. 20. Fi.g-ures are the characters commonly used to represent numbers. They are as follows — PRINTED, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. WRITTEN, ^/fJ-Z^^/cP/ NAMED, Zero, one, two, three, four, five, six, seven, eight, nine. 10 NUMERATION AND NOTATION. The figure is sometimes called a Cipher, or Naught, be- cause when written alone it expresses no value, or the absence of number; and the figures 1, 2, 3, 4, 5, 6, 7, 8, 9 are called Numerals, or Significant Figures, because each expresses as many ones as are denoted by its name. Numbers greater than nine are expressed by repeating or combining two or more of the ten figures. 21. Exact Tens are written with the figure expressing the number of tens at the left of 0, which marks the absence of ones ; and tens and ones are wi'itten with the figure expressing the tens at the left of the figure expressing the ones. Thus, Ten, or 1 ten. Eleven, or 1 ten and 1 one. Twelve, or 1 ten and 2 ones. Thirteen, or 1 ten and 3 ones. Twenty, or 2 tens. Twenty-one, or 2 tens and 1 one. Thirty, or 3 tens, and so on to ninety-nine, or 9 tens and 9 ones. 22. Exact Hundreds are written with the figure expressing the hundreds at the left of two zeros. Hundreds, tens and ones are expressed together in a number by writing the figure expressing the tens at the left of the figure expressing the ones, and the figure expressing the hun- dreds at the left of that expressing the tens. Thus, One hundred, or 1 hundred tens ones, is written 100. Two hundred, or 2 hundreds tens ones, is written 200. Four hundred ten, or 4 hundreds 1 ten ones, is written ^10. Five hundred six, or 5 hundreds tens 6 ones, is written 506. Nine hundred seventy-eight, or 9 hundreds 7 tens 8 ones, is written 978. 23. Numeration is the method of naming numbers, and of reading numbers expressed by figures. 24. Notation is the method of writing numbers, or of express- ing numbers by fitnires. is written 10, ii 11, (( 12, li 13, 11 20, (( 21, « 30, NUMERATION AND NOTATION. U WRITIEN EXERCISES. Write in figures — 1. One hundred thirty-six; two hundred thirteen. 2. Four hundred forty-four ; one hundred eleven. 3. Three hundred twenty-five ; five hundred ten. 4. Six hundred seventeen ; two hundred twenty. 5. Seven hundred five ; eight hundred fifteen. 6. Nine hundred nine ; seven hundred four. 7. Five hundred ; eight hundred seventy-one. 8. Six hundred ; three hundred eighty. 9. Five hundred twenty-two ; nine hundred ninety-nine. 10. Seven hundred one ; three hundred twenty-five. ORDERS AND PERIODS OF UNITS. 25. Orders of Units are denoted by the successive figures used in expressing a number. Thus, in 365, the 5, which expresses 5 ones, denotes units of the First Order; the 6, which expresses 6 tens, denotes units of the Second Order; and the 3, which expresses 3 hundreds, denotes units of the Third Order. 26. In naming numbers, the first three orders of units are regarded as forming a group, called the Class, or Period, of Units, having ones, tens and hundreds. Thus, 425 forms a period composed of 425 units. 27. Ten hundreds form One Thousand; ten thousands form One Ten-Thousand; and ten ten-thousands form One Hundred- Thousand. These three orders of units form a group, called the Period of Thousands, having ones, tens and hundreds. Thus, 363425 is composed of 363 thousands 425 units, or of two periods, and is read tliree hundred sixty-three thousands four hundred twenty- five. 406007 is composed of 406 thousands 007 units, or of two periods, and is read four hundred six thousands seven. In like manner are formed and read other periods. 12 NVMERATION AND NOTATION. THE DECIMAL SYSTEM. 28. Simple or Primary Units are the units expressed by a numeral when written alone, or, by the order of ones in the period of units, when in a collection. In writing numbers, the order of simple units may be marked by placing a point (.), called the Decimal Point, at the right of the units' period ; and the different periods may be separated by a Comma (,). Thus, 215 thousands 463 units may be written, 215,463. Units are understood to be Primary Units when not other- wise indicated by the expression or its connection. 29. The Names of the Orders of Units, and the Names of the Periods, are given in the following TABLE. 6th Period. 5th Period. 4th Period. 3d Period. 2d Period. 1st Period. '~t^ .— ^ ,— -s .^— V ^-— V '^"^ f 1 •S NAMES OF J 3 a a o s !3 PERIODS. o a o •3 § i 3 u •^ J3 3 V G? tH n § EH t3 (M ^ ^ «H <« «M O O o o o o ORDERS OF UNITS. W t< O H E^ O Weho Who WtHO Weho 420, 73 5, 80 0, 612, 309, 25 4. The number expressed is Four hundred hceniy quadrillions, seven hundred thirty-Jive trillions, eight hundred billions, six hundred twelve iniillions, three hundred nine thousands, tivo hundred Jiftij-four units. In reading numbers the name of the units' period is not usually given, since when omitted it is readily understood. NUMERATION AND NOTATION. 18 30. The Periods above Quadrillious, in order, are — Quiniillions, SexUllions, Septillions, Octillions, Nonillions, De- cillions, Undecillions, Duodecillions, I'redecillions, Quatuordecil- lions, QuindecUlions, SexdecilUons, Septendecillions, Octodecil- lions, Novendecillions, Vigintillions, etc. 31. The Scale of numbers is the arrangement of their units. In the ordinary system of notation, or that which has been explained, the scale is ten, because the units are so arranged that ten ones are one ten, ten tens are one hundred, ten hundreds are one thousand, etc. 32. The method of expressing numbers by ten figures is termed the Arabic Notation, from its having been introduced into Europe by the Arabs. 33. The method of expressing numbers by the scale of ten is termed the Decimal System, from the Latin decern, which sig- nifies ten; and the scale often is, for the same reason, termed the Decimal Scale. Principles of Numeration and Notation. 34. — 1. Ten ^(n^ts of any order in a number are ahvays equal to one unit of the next higher order. 2. The same figure represents invariably the same number of units. 3. The name and value of the units represented by a figure in a number are always those of its order in that number. 4. The absence of units in any order in a number is marJced by a cipher. 5. The order of simple units in a number may be known by having the decimal point expressed or understood at the right of that order. EXMUCISES IN NUMERATION. 35. --Ex. 1. Write and read 56073402. Solution.— 56073402, separated into periods, is 56,073,402, or 56 mil- lions 73 thousands 402 units, and is read fifty-six millions seventy-three thousands four hundred two. 2. Write and read 735467005. Write and read 93606121. 2 14 NUMERATION AND NOTATION. 36. Rule for Numeration. — Beginning with the lowest order of units, separate the figures of the number into periods of three figures each. In reading begin at the left; read the hundreds , tens and ones of each period, and give the name of each period, except the last, after its ones. Write and read- 1. 3U. 2. 1780. Z. 23U- 4. 16110. 5. 70008. 6. 134-020. 7. 68110. 8. 89000. 9. 143211. 10. 456104. 11. 215779. FJtOBLJEMS. 12. 132401. 13. 3000835. 14. 92416512. 16. 732534902. 16. 7324768291. 17. 44444444444. 18. 56073014211597. 19. 313134405678012. 20. 14132486879011326. 21. 59444632132007955. 22. 3567890038531900210. EXERCISES IN NOTATION. 37. — Ex. 1. Write in figures twenty-two millions four hun- dred six thousands. Solution. — Writing 22 for tlie tens and ones of 22,406,000 millions, 40G for the hundreds, tens and ones of thousands, 000 for the absence of hundreds, tens and ones of units, gives 22,406,000 as the required expression. 2. Write in figures three hundred sixty-five millions nine hundred twenty-five thousands seven hundred seventy-five. 8. Write in figures nine hundred thirty-two thousands four hundrefl forty-seven. 4. Write in figures four hundred eighteen millions eight hundred sixty-three thousands two hundred three. NUMERATION AXD NOTATION. 16 38. Rule for Notation.— Wi'ite the figures representing the hundreds, tens and ones of each period uv their order. Mark hij a cipher any order in the number which has no units given. PROBLJEMS. Write in figures — 1. Three hundred fourteen ; four hundred ten. 2. Five hundred six ; nine hundred seventy-seven. 3. Sixteen thousand ninety-one ; twenty -five thousand one hundred. 4. One hundred eighty-three thousand ; two hundred nine thousand ninety-nine. 5. Nine thousands seven hundreds three tens four ones. 6. Four millions six ; ten millions ; five hundred five millions, 7. Five hundred thousands four hundred six ; one hundred one thousands one hundred one. 8. Thirty-seven millions one hundred seventy-one thousands eleven. 9. Two hundred forty-nine millions ; seventeen billions nine millions. 10. Ninety-three thousands one hundred eighty-six. 11. One hundred fifty-two .millions four hundred twenty-five thousands three hundred thirty-three. 12. Seven hundred fifty -five trillions one hundred six bil- lions four hundred fifteen millions one hundred five units. 13. One quintillion twenty-five quadrillions one hundred fif- teen trillions seven billions eight hundred eighty-eight millions five hundred fifty units. 14. Eight hundred eighty-eight quintillions six thousand six hundred six. 15. Three hundred thirty-seven billions four hundred forty- nine millions two thousands three hundred eleven. 16. Five decillions one hundred six nonillions eight octil- lions four septillions one hundred nineteen sextiilions six hun- dred seventy-nine quintillions four quadrillions three hundred fifteen trillions seven hundred twenty billions forty-six millions three thousands one. 16 NUMERATION AND NOTATION. TEST QUESTIONS. 39. — 1. "What is a Unit ? A number ? Name some number. What is the unit of a number ? Give an illustration of the unit of a number. What is an integer ? 2. What are Similar Numbers? Name two. AVhat are dissimilar numbers ? Name two. What is a concrete number ? What is an ab- stract number ? Give an illustration of a concrete number. Of an ab- stract number. 3. What is Arithmetic? A solution? A proof ? A problem? A principle ? A rule ? An example ? An exercise ? 4. How is the Naming of numbers performed ? Give the names of the first ten numbers. How may ten be regarded ? What numbers do we get by combining ten with each of the first nine numbers, and chang- ing their form? Two tens, three tens, etc., by change of form give what numbers ? Twenty and one, twenty and two, etc. ? 5. What are Figures ? What does written alone express ? What is called? What does each of the other nine figures express when written alone? What are they called? How are exact tens written? How are the numbers between the tens written ? How are exact hun- dreds written ? How are hundreds, tens and ones expressed together in a number? 6. What is Numeration? What is notation? How are orders of units denoted in expressing a number? 7. What are Simple Units, or units of the first order ? Units of the second order? Units of the third order? How many simple units does 2 of tlie first order express ? 2 of the second order ? 2 of the third order ? 8. How many orders compose a Period ? What is the name of the first period ? Of the second ? Name the orders of each. How many units in the period of units equal one unit in the period of thousands? Give tlie names of the first six periods in their order. Name in order ])erio(ls higher than the period of quadrillions. How may the order of simple units be marked? How may tlie diflerent periods be sepa- rated ? 9. What is the Scale of numbers? Wiiat is the scale in the ordinary system? Wliat is the method of expressing numbers by figures called? Of expressing nimibcrs by tlie scale of ten? What is the scale of ten termed ? 10. Recite the Principles of numeration and notation. The rule for numeration. The rule for notation. ADDITION. 17 SECTION III. ADDITION. 40. — Ex. 1. If you have 5 dollars and your brother has 4, how many dollars have both ? 2. If James has 6 books and John has 7, how many have both ? 3. How many cents are 9 cents and 3 cents ? 4. Add by 2's from 1 to 21. Solution.— 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21. 5. Add by 3's from 1 to 22. Add by 3's from 22 to 52. 6. Add by 4's from 2 to 34. Add by 4's from- 34 to 62. 7. Add by 5's from 3 to 33. Add by 5's from 33 to 68. 8. Add by 6's from 4 to 34. Add by 6's from 34 to 70. 9. Add by 7's from 5 to 40. Add by 7's from 40 to 75. 10. Add by 8's from 6 to 46. Add by 8's from 46 to 94. 11. Add by 9's from 7 to 52. Add by 9's from. 52 to 88. 12. What is the sum of 8 dollars and 7 dollars ? 13. What is the unit of 8 dollars and 7 dollars ? 14. Why are 8 dollars and 7 dollars similar numbers ? What is the unit of their sum ? 15. Are 9 books and 8 dollars similar or dissimilar numbers? 16. Why cannot 9 books and 8 dollars be united into one number ? Because 9 books and 8 dollars are neither 17 books nor 17 dollars. 17. Only what kind of numbers, then, can be united so as to form one number ? 18. In my garden there are 5 roses upon one bush, 7 upon another and 2 upon another. How many roses are there in all ? 19. How many ones are 5, 7 and 2? 7, 5 and 2? 7, 2 and 5 ? 2, 7 and 5 ? 2, 5 and 7 ? 20. When the same numbers are added in different orders, is the result changed ? 21. 4, 3, 5, 1 and 2 are how many ? 22. 8, 1, 9, 3, 2 and 5 are what number? IS ADDITION. DEFINITIONS. 4-1. Addition is the process of uniting two or more numbers to find their sum. 42. The Sum is the result of the addition. It contains as many ones as all the numbers added. 43. A Si^ is a mark used for abbreviating an expression. 44. The Sign of Addition is +> and is called plus. When placed between two numbers, it means that they are to be added. 45. The Sign of Equality is =, and is read equals or equal When placed between arithmetical expressions it de- notes that they are equal. Thus, 8 + 5 = 13, is read, eight plus five equaU thirteen, and means that the sum of eight and five is thirteen. Principles of Addition. 46. — 1. Only similar numbers can he added. 2. The sum and the numbers added must be similar. 3. The sum of numbers will be the same in whatever order they are added. CJ^STS I. When the Sum of all the Units of each Order is Less than Ten. 47.— Ex. 1. What is the sum of 3142, 2320 and 516? r SI42 Solution. — For convenience in adding, write JS, umbers j ^g^O ^^^^ numbers so that figures representing units of added, J _ ^ „ the same order stand in the same cohimn. ^ Add the ones, tens, hundreds, and thousands Sum, 5978 separately. 6 ones + ones + 2 ones are 8 ones, which write for the ones of the sum. 1 ten -f- 2 tens + 4 tens are 7 tens, wliich write for the tens of the sum. T) hundreds + 3 hundreds + 1 liundrcd are 9 iiundreds, wliich write for the hundreds of the sum. 2 thousands + 3 thousands are 5 thousands, which write for the thou- sands of the sum. Hence, the sum is 5 thousands 9 hundreds 7 tens 8 ones, or 5978. ADDITION. ■ite and add — (2.) 61 25 (3.) 187 12 (4.) 42 55 (5.) 423 354 19 6. What is the sum of 13, 173 and 202? Ans. 388. 7. How many dollars are 101 dollars, 65 dollars and 113 dollars ? 8. How many books are 341 books, 113 books and 202 books? Ans. 656. C^SE II. Wlien the Suiu of all the Units of auy Order is Greater than Ten. 48.— Ex. 1. What is the sum of 5591, 1428 and 2335? /Solution. — Write the numbers as in the pre- OOJl ceding case. , , , X 1428 Begin to add with the ones, so that wlien the ' I 2335 sura of any of the orders of units is greater than r» ~r\(r>r i nine, its tens may be conveniently added with bum, y3o4. ^, -^ i- ,, ^ 1 • 1 1 the units of the next liigher order. 5 ones + 8 ones + 1 one are 14 ones, or 1 ten and 4 ones. Write tlie 4 ones for the ones of tlie sum, and reserve the 1 ten to add with the tens. 1 ten + 3 tens -f 2 tens + 9 tens are 15 tens, or 1 hundred, and tens. Write the 5 tens for the tens of the sura, and reserve the 1 hundred to add with the hundreds. ' 1 hundred + 3 hundreds + 4 hundreds + 5 hundreds are 13 hundreds, or 1 tliousand and 3 hundreds. Write the 3 lumdreds for the liun- dreds of the sum, and reserve the 1 thousand to add witli the thousands. 1 thousand + 2 thousands + 1 thousand + 5 thousands are 9 thou- sands, wliich we write for the thousands of the sum. Hence, the sum is 9 thousands 3 hundreds 5 tens 4 ones, or 9354. The explanation may be abbreviated by naming only results. Thus, Five, thirteen, fourteen ; write 4 and reserve 1. One, four, six, fifteen ; write 5 and reserve 1. One, four, eight, thirteen; write 3 and reserve 1. One, three, four, nine; write 9. Ans. 9354. Proof. — Tlie correctness of the sohition may be proved by reviewing the work carefully, or by adding the numbers downward. If the work is correct, the result in each case v/ill equal the first result, since the sum of numbers must be the same in whatever order they may be added. (Art. 46—3.) 20 ADDITION. Add and prove — C-^-) (3.) (4.) (5.) (6.) 1683 467 1062 114 7703 456 305 3457 590 4104 312 587 2004 309 3492 7. What is the sum of 615 + 3045 + 5000 ? 8. How many are 3600, 7240 and 797 ? 9. A farmer raised from one field 1213 bushels of wheat ; from a second, 1308 bushels ; from a third, 2230 bushels ; and from a fourth, 244 bushels. How much wheat did he raise ? Ans. 4995 bushels. 40. Rule for Addition.— TFrJ?5e tlte nimvhei^s so that all figures of the same order shall standi in the same column, and draiv a line under them. Begin at the right, add the units of each order sepa- rately, and lurite the sum, if less than ten, under the column added. If the sum is ten or more, ivrite the figure standing for its ones, and add its tens with the units of the next higher order. Write thewhole sum of the units of the highest order. PROOF.— Add the numhers a second time, in a, different order. If the xvorh is correct, the result will he the same by both methods. rjtOJiljEMS. Add and pr- ove — (1.) (2.) (3.) (4.) (5.) 4306 172 3334 146 4455 4507 903 678 504 777 8421 129 105 810 91 6. What is the sum of 563, 491 and 708 ? Ans. 1762. 7. What is the sum of 423, 567 and 385? 8. Find the sum of 785, 584, 175 and 145. Ans. 1689. ADDITION. 21 9. Find the sum of 2385, 385 and 500. Ans. 3270. 10. Find the sum of 1603, 495, 1708 aud 793. 11. How many are 11063, 4461, 1030 and 309? 12. How many are 64004, 15300, 1008 and 488 ? 13. How many are 3560, 1246, 8556 and 2451 ? 14. How many are 2063 + 950 + 805 + 470 ? Ans. 4288. 15. How many are 1623, 2045, 705 and 3435? 16. How many are 4000, 7891, 1631, 500 aud 19? 17. 600 + 419 + 6663 + 1000 + 317 = what number? 18. 1798 + 4444 + 6666 + 1333 + 122 = how many ? 19. 20120 + 3290 + 3167 + 532 + 499 = how many ? 20. Add 138935, 113467 and 12506. Ans. 264908. 21. Add 38394, 22957 and 601826. Ans. 663177. 22. Add 2352 yards, 3800 yards and 1785 yards. 23. How many dollars are 13177 dollars, 7346 dollars aud 1275 dollars ? 24. Upon a platform car there are three rough blocks of stone. One of them weighs 3470 pounds; another, 5362 pounds ; and the third, 2567 pounds. What is the weight of the three ? Ans. 11399 pounds. 25. Six bales of cotton contain, respectively, 470, 480, 45d, 500, 512 and 475 pounds. What is their entire weight ? 26. If four boxes weigh, respectively, 1618, 1450, 963 and 1341 pounds, what is their entire weight? 27. A merchant bought at one time 513 barrels of flour ; at another, 763 barrels ; and at another, 1 347 barrels. How many barrels did he buy in all ? Ans. 2623. 2!J ADDITION. 28. A gardener took from one tree 348 pears ; from another, 316 ; from another, 159 ; and from another, 96. How many did he take from the whole ? Ans. 919. 29. In a certain school there are in the first section 31 pupils ; in the second, 39 ; in the third, 42 ; in the fourth, 47 ; and in the fifth, 64. How many pupils are there in the five sections? Ans. 223. 30. Thompson has in his farm 150 acres ; Reed in his farm, 317 acres ; Howland in his farm, 72 acres ; and two others have each 875 acres. How many acres have they all ? 31. A large work consists of five volumes : there are 612 pages in the first volume, 709 in the second, 691 in the third, and 1357 in the remaining two together. How many pages in the entire work ? Ans. 3369. 32. The distance from New York to Chicago by railroad is 911 miles ; from Chicago to Omaha, 401 miles ; from Omaha to Ogden, 1101 miles; from Ogden to Sacramento, 743 miles; from Sacramento to San Francisco, 117 miles. What is the whole distance from New York to San Francisco ? 33. A lumber-dealer has 37270 feet of boards in one yard, 9536 in another, 45098 in vessels, and 8876 at a mill. How many feet of boards has he in all these places ? 34. I bought a house for 6236 dollars ; I paid 869 dollars for repairs, and 300 dollars for painting it. For what price must it be sold to gain 325 dollars ? Am. 7730 dollars. 35. The sailing distance from New OHeans to Charleston is 1167 miles; from Charleston to Norfolk, 431 miles; from Norfolk to New York, 308 miles ; and from New York to Bos- ton, 356 miles. How many miles must a steamer sail, that ^hall touch at all these places, in going from New Orleans to Boston? Ans. 2262. 36. The mariner's compass was invented in China 1120 years before Christ; America was discovered by Columbus 1492 years after Christ ; and steam was first applied by Fulton to ])ropclling boats 315 years after the discovery of America. How many years after the invention of the mariner's compass was steam first applied to propelling boats ? ADDITION. 23 Copy, add and prove — (37.) (38.) (39.) (40.) (41.) 131 600 180 4811 11319 256 176 316 1141 4121 702 31 701 7891 3006 91 45 9 123 723 61 39 114 416 345 42. Alaska contains 577,390 square miles; California, 188,981; and Oregon, 95,274. How many square miles do the three con- tain? Ans. 861,645. 43. In 1868 there were employed in the United States, as cultivators of the land, 3,219,495 persons ; as day-laborers, 960,000 ; as servants, 560,000 ; as mechanics and manufac- turers, 480,905 ; and as merchants, 123,564. How many per- sons were employed in all of the occupations named? 44. The area of America is 15,977,000 square miles; of Europe, 3,954,000 square miles; of Asia, 15,742,000; of Africa, 11,734,000; and of Oceanica, 3,700,000. What is the area of the whole ? Ans. 51,107,000 square miles. 45. In 1870 the population of America was 85,950,000; of Europe, 300,390,000; of Asia, 795,000,000; of Africa, 174,240,000; and of Oceanica, 30,102,000. What was the population of the whole ? TEST QUESTIONS. 50.— 1. What is Addition? The sum or amount? A sign? The sign of addition ? The sign of equality ? Illustrate the use of the sign of addition. Of the sign of equality. 2. What are Principles of addition ? Why cannot 5 dollars and 4 books be added ? Why can 4 dollars and 5 dollars be added ? Sho-w that the sum of numbers is the same in whatever order they are added. 3. What is the Rule for addition ? Why are the numbers in addition •nritten so that all figures of the same order shall stand in the same column? Why begin at the right to add? When the sum of any column is ten or more, why add its tens with units of the next higher order? Give the proof of nddition. The renson for the proof. 24 SUBTRACTION. SECTION IV. SUBTRACTION. 51. — Ex. 1. If John has 7 books, liow many more must he obtain to have 12? 2. In my purse there "were 11 dollars ; 6 dollars have since been taken out. How many remain ? 3. William is 17 years old, and his brother is 8. "NYhat is the difference in their ages? 4. Subtract by 2's from 21 to 1. Solution.— 19, 17, 15, 13, 11, 9, 7, 5, 3, 1. 5. Subtract by 3's from 22 to 1. By 3's from 52 to 22. 6. Subtract by 4's from 32 to 4. By 4's from 60 to 32. 7. Subtract by 5's from 33 to 3. By 5's from 68 to 33. 8. Subtract by 6's from 34 to 4. By 6's from 70 to 34. 9. Subtract by 7's from 40 to 5. By 7's from 75 to 40. 10. Subtract by 8's from 46 to 6. By 8's from 94 to 46. 11. Subtract by 9's from 52 to 7. By 9's from 88 to 52. 12. What is the unit of 12 and 7? What is the unit of the difference of 12 and 7 ? 13. What is the unit of the difference of 13 dollars and 9 dollars ? 14. Are 13 dollars and 9 dollars and their difference similar or dissimilar numbers ? 15. Why can you not subtract 9 books from 13 dollars? Because the numbers liave no common unit. 16. Only what kind of a number, therefore, can be taken from another number ? 17. If you take 7 cents from 16 cents, what luimber will be the difference ? What number added to 7 will make 16 ? DEFINITIONS. 52. Subtraction is the process of taking one number from another. 53. The Difference is the result obtained by the subtraction. SUBTRACTION. 26 54. The Minuend is the number from which the subtraction is to be made. 55. The Subtrahend is the number which is to be subtracted. The subtrahend cannot be greater than the minuend. When the sub- trahend is equal to the minuend, the difference is 0. 5G. The Sig'u of Subtraction is — , and is called minus. When placed between two numbers it denotes that the one on the right is to be taken from that on the left. Thus, 13 — 5 is read, thirteen minus five, and means that five is to be taken from thirteen. Principles of Subtraction. 57. — 1. Only similar numbers can be subtracted. 2. The minuend, subtrahend and difference mxist be similar numbers. 3. The sum of the difference and subtrahend must equal the minuend. C^SE I. Wlieu the Units of the Subtrahend do not Exceed those of the same Orders in the Minuend. 58.— Ex. 1. From 569 subtract 235. Minuend, 569 Solution. — For convenience in subtracting, Subtrahend 235 write the figures of the subtrahend under figures representing the same orders in the minuend. Difference, 004 Subtract each order of units separately: 9 ones — 5 ones are 4 ones, which write for the ones of the difference. 6 tens — 3 tens are 3 tens, which write for the ten^< of the difference. 5 hundreds — 2 hundreds are 3 hundreds, which writs for the hun- dreds of the difference. Hence, the difference is 3 hundreds 3 tens 4 units, or 334. (2.) (3.) (4.) (5.) From 496 873 98 3318 dollars, Subtract 365 651 54 208 dollars. In Ex. 5, as there are no units of the thousands' order in the subtra- hend, we proceed as if the subtrahend had been written 0208. .3 26 SUBTRACTION. 6. A person born in the year 1819 was how old in 1871 ? 7. Johnson had 765 dollars, and paid away 351 dollars. How many dollars had he left ? Ans. 414. 8. What is the difference between 3467 and 1352 ? 9. What is the difference between 25674 and 5473 ? 10. Andrew Holt bought a plantation for 14630 dollars, and sold it for 15750 dollars. How many dollars did he gain by the operation ? Ans. 1120. ca.se II. T^licn the Units of one or more Orders of the Subtrahend Ex- ceed those of the Same Orders of the Minuend. 59 —Ex. 1. From 8662 take 4734. 7.16.5.12. Solution. — Write the numbers as in the pre- Mlnuend, 8G62 ceding case. Subtrahend, 4734. Since 4 ones cannot be subtracted from 2 nQQQ ones, take 1 ten from the 6 tens, leaving 5 tens; JJ ' and add 10 ones, which are equal to the 1 ten, to the 2 ones, making 12 ones ; 12 ones — 4 ones are 8 ones, which write for the ones of the difference. 5 tens — 3 tens are 2 tens, which write for the tens of the difference. Since 7 hundreds cannot be subtracted from 6 hundreds, take 1 thou- sand from the 8 thousands, leaving 7 thousands, and add the 10 hundreds that equal the 1 thousand to the 6 hundreds, making 16 hundreds; 16 hundreds — 7 hundreds are 9 hundreds, which write for the hundreds of the differeiice. 7 thousands — 4 thousands are 3 thousands, which write for the thou- sands of the difference. Hence, the diflcrence is 3 thousands 9 hundreds 2 tens 8 units, or 3928. The explanation may be abbreviated thus : 2 + 10 are 12 ; 12 — 4 are 8, which write. 6^1 are 5 ; 5 — 3 are 2, which write. 6 + 10 are 16 ; 16 — 7 are 9, which write. 8 — 1 are 7 ; 7 — 4 are 3, which write. Ans. 3928. "We begin at the right to subtract, so tliat when any figure of the sub- trahend denotes a greater number tlian tlie corresponding figure of the minuend, 1 may be taken from tlie ne.\t higher order of the minuend. Subtrahend, 4734 Proof.— To test tlie solution, we add the Difference 39'^ 8 difference and subtrahend, and since the result - — is equal to the minuend, the work is known to Minu&ixd, 8662 be correct. (Art. 57-3.) SUBTRACTIOX. 27 Solve and prove — (2.) (3.) (4.) (5.) From 4107 6483 1922 5r6'5' pounds, 8vhiTad 3058 938 1415 6718 pounds. 6. What is the difference between 1634 and 1329 ? 7. I bought goods for 9363 dollars, and sold them for 10382 dollars. How much did I gain by the transaction ? Ans. 1019 dollars. 60. Rule for Subtraction.— Tr77Y<^ the subtrahend under the minuend, to that figures of the same order shall stand in the same column. Begin at the right, subtract tJi e units of each order of the subtrahend from units of the same order of the minuend, if possible, and write thejiifference beneath. When the units of any order of the minuend are less than those of the same order of the subtrahend, increase their 7iumber by adding ten, the value of a unit tahen from the next higher order of the minu- end, and subtract; then consider the units of that higher order of the minuend one less. PROOF. — Add the differeitce and subtrahend. If the ivorh is connect, the result will equal the lyvhvuend. PROBIjEMS. Solve and prove — (1.) (2.) (3.) (4.) JVom 4592 9863 34061 46603 soldiers, Take 3683 416 12341 928 soldiers. 5. How many are 69351 less 40402? Ans. 28949. 6. How many are 32443 less 965 ? 7. How many are 17630 less 8542? Ans. 9088. 8. What is the difference between 92334 and 13403? 9. What is the difference between 46304 and 37413? 10. How much more is 32045 than 862? Ans. 31183. 11. How much less is 40920 th^u 91605? Ans. 50685, 28 SUBTRACTION. 12. 36841 — 27777 = what number? Ans. 9064. 13. 81710 — 70809 = what number ? Ans. 10901. 14. What is the difference between 4000 and 1321 ? 3.9.9.10. Solution. — As there are no units expres.=ed Minuend, 4^000 in the three lower orders of the minuend, re- Subtrahend 132 1 duce one of the four thousands to hundreds, then one of the hundreds to tens, and then one Difference, 2b 7 J ^^ ^j^g ^^^^ ^^ ^^^^_ r^^^ minuend will then be 3 thousands 9 hundreds 9 tens 10 ones, from which the subtrahend can easily be taken. Then, 10 ones — 1 one are 9 ones, which write for the ones of the difference. 9 tens — 2 tens are 7 tens, which write for the tens of the difference. 9 hundreds — 3 hundreds are 6 hundreds, which write for the hundreds of the difference. 3 thousands — 1 thousand are 2 thou- sands, which write for the thousands of the difference. Hence, the answer is 2679. Solve and prove — . (15.) (16.) (17.)- (18.) From 34-00 m^n, 5002 8003 20000 hn^Yida, Take 2306 " 2151 7004 4242 " 19. Subtract 199 from 1000. Ans. 801. 20. Subtract 3003 from 30030. An&. Ti^ll. 21. How many are 19000 men less 567 men? 22. How many yards are 31005 yards less 1777 yards? 23. How many tons are 51067 tons less 45075 tons ? 24. From J 00000 take 16898. Ans. 83102. Find the difference — 25. Between 36512 and 18735. Ans. Villi. 26. Between 12701 and 4110. 27. Between 7125 and 3647. Ans. 3478. 28. Between 38217 and 9548. Ans. 28669. 29. Between 1000000 and 107701. 30. Between 500807 and 480705. Am. 20102. 31. If a man should buy a farm for 8000 dollars, and should make a payment of 5925 dollars upon it, how many dollars would he still owe? Ans. 2075. 32. The Pilgrims landed at Plymouth in the year 1620. SUBTRACTION. 29 How long was it from that time to the Declaration of Indepen- dence, in the year 1776? 33. A merchant bought 63100 bushels of corn, and sold 17734 bushels. How many bushels had he left ? 34. I bought a piece of property for 95565 dollars, and sold it for 100000 dollars. What was the gain ? Ans. 4435 dollars. 35. In a certain election A received 25316 votes and B 21925 votes. What was A's majority ? ^ns. 3391. 36. By selling a farm for 7535 dollars I gained 1675 dollars. What v.-as its cost ? 37. At a certain election the successful candidate received 25316 votes, which was 3191 votes more than were given to the rival candidate. How many votes did the latter receive ? 38. The population of a town is 2763 ; ten years ago it was 2595. What was the increase in the ten years ? Ans. 168. 39. A mill sawed 427563 feet of boards one year and 385895 the year before. How many more feet did it saw in the one year than in the other? Ans. 41668. 40. St. Peter's Church, Rome, has standing-room for 54000 persons, and the New York Academy for 1326 persons. How many more persons can stand at the same time in the one than in the other ? 41. There are 41823360 seeds in a bushel of timotliy, and 16400960 seeds in a bushel of clover. How many more seeds in a bushel of the one than in a bushel of the other ? TEST QUESTIONS. 61. — 1. What is Subtraction? The difference? The minuend? The subtrahend? When the subtrahend and minuend are equal, what is their diflerence ? What is the sign of subtraction ? Ilhistrate its use. 2. What are the Principles of subtraction? Why cannot 5 apples be subtracted from 8 dollars ? If the minuend and subtrahend express dollars, what will the difference express? If the minuend is 8, what must be the sum of the subtrahend and difference? 3. What is the Rule for subtraction ? Why is the subtrahend written under the minuend so that figures shall stand in the same column ? Why begin at the right to subtract? .What is the method of proof? When the subtrahend and difference are given, how may the minuend be found ? 30 BE VIEW PROBLEMS. SECTION V. ■ REVIEW PROBLEMS. MENTAL, EXEJtCISES. 62, — Ex. 1. How many tens and ones in 96? In 78? 2. How many hundreds, tens and ones in 365? In 431? In 987? 3. How many thousands and units in 55625? In 107308? 4. John has 14 dollars, William 10 dollars and Henry 9 dollars. How many dollars have they all ? 5. A farmer has 26 cows in one field, 15 in another and 23 in a third. How many cows has he in the three fields ? Solution. — He has as many cows in the three fields as the sum of 26, 15 and 23. 26 + 10 = 36 ; 36 + 5 = 41 ; 41 + 20 = 61 ; 61 + 3 = 64. Therefore he has 64 cows in the three fields. 6. A man paid 18 dollars fiar a coat, 33 dollars for an over- sack and 9 dollars for pantaloons. How much did he pay for all? 7. A drayman moved 41 boxes on one day, 17 on another day, 34 on a third day and 23 on a fourth day. How many boxes did he move in all? 8. A certain orchard contains 28 apple trees, 18 pear trees, 32 peach trees and 1 1 cherry trees. How many trees does it contain altogether? 9. If I had 9 dollars more I should have 27 dollars. How much money have I? 10. John had 48 cents, but has since spent 29 cents. How many cents has he remaining ? Solution.— He has as many cents remaining as the difference between 48 and 29. 48 — 9 = 39 ; 39 — 20 = 19. Therefore he has 19 cents remaining. 11. Robert is 15 years old, and his father is 53 years old. What is the difiercnce in their ages ? 12. Susan is 13 years old. In how many years will she be 42 years old? 13. AVillio gathered 27 nuts in the forenoon, and in the after- REVIEW PROBLEMS. 31 noon as many more less 19. How many did he gather in the afternoon ? 14. A drover bought 16 sheep at one time and 112 at another ; he then sold 18. How many had he left ? 15. I had 54 cents in my purse, but took out 15 cents at one time and 16 cents at another. How many cents then remained in the purse ? WRITTEN EXERCISES. 63. — Ex. 1. A man paid 700 dollars for a span of horses, 450 dollars for a carriage and 175 dollars for a double harness. What sum did he pay for the whole? Ans. 1325 dollars. 2. What is the sum of one million five hundred seventy-five thousand three hundred twenty-two, plus four hundred nineteen thousand three hundred sixty-five, plus one hundred thirty-two thousand three hundred fifty-five ? 3. I bought some property for 5390 dollars, and sold it for 6585 dollars. How much did I gain ? Ans. 1195 dollars. 4. If the smaller of two numbers is 5390 and their diflference 1195, what is the larger number? 5. Mount St. Elias is 17900 feet in height and Mount Wash- ington 6284 feet. How much higher is Mount St. Elias than Mount Washington? Ans. 11616 feet. 6. If the minuend is four hundred seventeen thousand seven hundred twelve, and the subtrahend is one hundred thousand ninety-three, what is the difference ? Ans. 317619. 7. A man's salary is 1200 dollars, and his income from other sources is 655 dollars. His expenses are 850 dollars. How much is his net income ? Solution. — His income must be 1200 dollars. the sum of his salary and his income 655 " from other sources. 1200 + 65-5 = To^r J n ^^'^^ dollars. Income, looo dollars. tj- ^. • ^ u i • • ' His net income must be his income Kvjoenses, 850 " less his expenses. 1855 dollars — Net Income, 1005 dollars. 850 dollars = 1005 dollars. His net income, therefore, is 1005 dollars. 32 REVIEW PROBLEMS. 8. A merchant bought goods to the amount of 7835 dollars. He then sold a part for 9563 dollars, and the balance for 361 dollars. How much did he gain by the transaction ? 9. The great bell of Moscow weighs 432000 pounds, and the bell of City Hall, New York, weighs 22300 pounds. What is the difference in their weight ? 10. A grain-dealer had 9867 bushels of corn. He sold 3479 bushels at one time and 5372 bushels at another time. How many bushels had he then left ? Solution.— If he sold 3479 34-79 bushels. 9867 bushels. bushels at one time and 5372 5372 " 8851 " bushels at another time, he oozr -f ^ 1 1 -I r\-i /:> i i i sold in all 3479 bushels -|- 5372 8851 bushels. 10 lb bushels. , , , 00-11,1.1 bushels, or 880I bushels. If he had 9867 bushels, and sold 8851 bushels, he had left 9867 bush- els — 8851 bushels, or 1016 bushels. Ans. 1016 bushels. 11. Kaspar sold 516 barrels of flour at one time, 419 bar^jels at another time, and had 316 remaining. How many barrels had he at first ? Ans. 1251. 12. A man had 16745 acres of land. To A he sold 5304 acres and to B 6319 acres. How much had he then unsold? 13. Andrew Jones bought a piece of land for 18068 dollars. He sold a part for 9563 dollars, and the remainder for 9385 dollars. How much did he gain by the operation ? 14. A man had 15675 dollars ; afterward he expended 5000 for a house and 1225 for furniture, and then earned 963 dollars. How much money had he then ? Aiis. 10413 dollars. 15. The sailing distance from New York to Cape Horn is 7232 miles ; from Cape Horn to Canton, 8840 miles ; from Canton to Cape Good Hope, 7000 miles ; and from Cape Good Hope to New York, 6790 miles. The sailing distance from Canton to San Francisco is 6090 miles, and the distance by railroad from San Francisco to New York, 3273 miles. How much shorter is the route from Canton to New York, by way of San Francisco, than by cither Cape Good Hope or Cape Horn? Aixs. 4427 miles shorter than by Ca]>e Good Hope; 6709 miles shorter tlian by Cape Horn. MULTIPLICATION. 83 SECTION VI. MUL TIPLICA TlOJf. 64. — Ex. 1. If a boy take from a library 4 books each day for 3 days, how many times does he take 4 books ? How many books does he take in all ? 2. Four boys have each 6 cents. How many cents have they in all ? How many times must 6 cents be taken to make the number they have in all ? 3. What is the difference between 6 cents -f- 6 cents + 6 cents + 6 cents and 4 times 6 cents ? 4. How many times 9 cents will 3 pencils cost at 9 cents each? 5. Give tilt 2's from one 2 to six 2's. Solution. — One 2 is 2, two 2's are 4, three 2's -are 6, four 2's are 8, five 2's are 10, and six 2's are 12. 6. Give the 2's from seven 2's to twelve 2's. 7. Give the 3's from three 3's to twelve 3's. 8. Give the 4's from five 4's to twelve 4's. 9. Give the 5's from five 5's to twelve 5's. 10. Give the 6's from six 6's to twelve 6's. 11. Give the 7's from seven 7's to twelve 7's. 12. Give the 8's from eight 8's to twelve 8's. 13. Give the 9's from nine 9's to twelve 9's. 14. Give the lO's from ten lO's to twelve lO's. 15. Give the ll's from eleven ll's to twelve ll's. 16. How many are 5 times 7? 4 times 8 dollars? Is the number which denotes the times another number is taken ab- stract or concrete ? 17. When 7 dollars are taken 6 times, what is the resuh ? To which of the given numbers is the result similar ? 18. Is the result the same when we take 8 five times as when we take 5 eight times ? 19. How many are 10 times 6? 10 times 9? 20. How many are 11 times 5? 11 times 10? 21. How many are 12 times 12? 12 times 11 ? 34 MULTIPLICATION. DEFINITIONS. 65. Multiplication is the process of taking one of two num- bers as many times as there are units in the other. 66. The Multiplicand is the number taken or multiplied. 67. The Multiplier is the number which shows how many- times the multiplicand is to be taken. 68. The Product is the result of the multiplication. 69. The multiplier and multiplicand are called Factors of the product. When the factors are more than two, the multiplication is called Con- tinued Multiplication, and the result a Continued Product. 70. The Si^ of Multiplication is X , and is read, times, or multiplied by. When placed between two numbCTs, it denotes that they are to be multiplied together. Thus, 8 X 5 is read, e'ght multiplifd hy five, ov five times eight. Principles of Multiplication. 71. — 1. Tlie multiplier is always a7i abstract number. 2. The product and the multiplicand are similar numbers. 3. In* finding the product of tivo factors, either, abstractly con- sidered, may be used as the multiplier. ca.se I. Wlien the Multiplier consists of but One Order of Units. 72.— Ex. 1. Multiply 3427 by 6. Multiplicand, 34^^7 Solution. — For convcnicnoo write the Multiplier 6 multiplier under the multiplicand, and begin Product, ' 20562 -tf;-gl!t to multiply ' 6 times 7 ones are 42 ones, or 4 tens and 2 ones. Write the 2 ones for the ones of the product, and reserve tlie 4 tens to add to the product of the tens. () times 2 tens are 12 tens, and 12 tens + 4 tens are 16 tens, or 1 hun- dred, and 6 tens. Write the 6 tens for the tens of the product, and re.'^e^vl■ tin- 1 luuidred to add to llic product of tiie hundreds. M ITL TIPLICA TION. 35 6 times 4 hundreds are 24 hundreds, and 24 hundreds + 1 liundred are 25 hundreds, or 2 thousands and 5 hundreds. Write the 5 hundreds for the hundreds of the product, and reserve the 2 thousands to add to the product of the thousands. 6 times 3 thousands are 18 thousands, and 18 thousands + 2 thousands are 20 thousands, or 2 ten-thousands and thousands, which write. Hence, the product is 2 ten-thousands thousands 5 hundreds 6 tens 2 ones, or 20562. We begin at the right to multiply, so that when the product exceeds nine, we may add its tens to the next product. The explanation may be abbreviated thus : 6 times 7 are 42 ; write the 2 and reserve the 4. 6 times 2 are 12, and 4 are 16 ; write the 6 and reserve the 1. 6 times 4 are 24, and 1 is 25 ; write the 5 and reserve the 2. 6 times 3 are 18, and 2 are 20, which write. Ans. 20562. The correctness of the work may be tested or proved by carefully re- viewing the whole process. Solve and prove — (2.) (3.) (4.) (5.) MuUiphj 645 916 men. 1125 3246 yards. By _J^ 5 _J _3 6. Multiply 30874 by 6. Ans. 185244. 7. What is the product of 73121 by 7? Ans. 511847. 8. What will 8 flirms cost at 13035 dollars each? 9. How many square miles in 9 townships, if each -contain 24375 square miles ? ^ns. 219375. CASE II. When the Multiplier consists of More than One Order of Units. 73— Ex. 1. Multiply 962 by 34. Multiplicand 962 Solution.— Since 34 is equal to 3 tens + 4 "' ' ones, 34 times 962 must equal 4 times 962 -\- Multiplier, J^ 3 t^„g times 962. Partial j 3848 4 times 962, or the product of 962 by the Products 1 2886 '^"^^ °^ t^^^ multiplier, is 3848, the first par- 7 Qc)iynQ ^^^^ product. 1 rodtict, J^/U^ 3 tgj^g ^ 2 ones are 6 tens, which write for the tens of the second partial product. 3 tens X 6 tens are 18 tens of tens, or 18 hundreds, or 1 thousand and 8 hundreds; write the 8 hun- dreds for the hundreds of the partial product, and reserve the 1 thou- sand for the thousands. 3 tens X 9 hundreds are 27 tens of hundreds, 36 MUL TIPLICA TION. or 27 thousands, and 27 thousands -|- 1 thousand are 28 thousands, or 2 ten-thousands and 8 thousands, which write in the partial product, making the second partial product 2886 tens, or 28860. The entire product, or the sum of the partial products, 3848 -\- 28860, is 32708. The explanation may be abbreviated, thus : 962 X 4 is 3848 units. 3 times 2 are 6, which write for the tens. 3 times 6 are 18 ; write the 8 for the hundreds, and reserve the 1 for the thousands. 3 times 9 are 27, and 1 are 28, which write, making 2886 tens. Adding the partial products, we have 32708, the answer required. Proof. — Prove the solution by reversing the order of the factors in the multiplication, taking the 34 for the multipli- cand and the 962 for the multiplier. The result being the same as at first obtained, the work is presumed to be correct, since the product of two factors is the same, whichever is taken as the multiplier. (Art. 71 — 3.) 32708 Solve and prove — (2.) (3.) (4.) (5.) Multiply 8125 813 3104 614 By _^ j55 16 207 4298 In Ex. 5 we omit to multiply by the tens, since 1228 the product of any number multiplied by is 0. 6. What is the product of 763 by 305? Am. 232715. 7. What is the product of 706 by 408 ? 8. What is the product of 403 by 62 ? Ans. 24986. C^SE III, When either Factor has One or More Ciphers on the Right. 74.— Ex. 1.- Multiply 465 by 100. Multiplicand, 465 Solution. — Since 10 units of any order Multiplier, 100 ^^^ always equal to 1 of the next higher r, J , //^r:nr} ^^'^^^ (Art. 34—1), the writing of a cipher on Jrodud, 40000 ^j^^ ^j^j^j ^^ ^ number, whk'h removes its figures each an order to the left, must multiply it by 10. In like manner, the writing of two ciphers on the right of a number nnist multiply it by 100; the writing of three ciphers must multiply it by 1000, etc. MULTIPLICATION. 37 Hence, to 465 X 1> which is 465, we annex two ciphers, and have 100 times the number, or 46500, the answer requii'ed. 2. Multiply 465 by 500. Solution.— Since 500 is 100 times 5, 500 Multiplicand, 465 times 465 must be 100 times 5 times 465. Multiplier, 500 5 times 465 is 2325, and 100 times as Product, 232500 """"^^ ^' ^^f """^ ^'^"^ "?^'^' ''""'^''^' or 232500, the answer required. 3. Multiply 4650 by 500. Solution. — 5 times 4650 is 5 times Multiplicand, 4650 455 with a cipher annexed, or 23250; MuUijilier, 500 and 100 times as much is 23250 with two Product, 2325000 "P^^!"' annexed, or 2325000, the answer required. Multiply — 4. 516 by 10. Am. 5160. 5. 1302 by 100. 6. 95 by 1000. 7. 254 by 600. 8. 75 by 300. Am. 22500. How many are — 9. 40 times 560? Am. 22400. 10. 80 times 3400? 11. 200 times 500? ^ns. 100000. 12. 120 times 4110? 13. 1000 times 1000? 14. If you can travel 12 miles in one hour, how far, at the same rate, can you travel in 100 hours? 15. What will 45 casks of molasses cost at 5*0 dollars each ? 16. If there are 640 acres in a square mile, how many acres are there in 150 square miles ? Ans. 96000. 75. Rule for Multiplication.— 7/ the -multiplier consists of one order of units, multiply each order of the multi- plicand, beginning at the right, by the multiplier. Write the units of each result in the product, and re- serve the tens, if any, to be added to the next result. If the Tnultiplier consists of more than one order of units, multiply each order of the inultiplicand by each order of the multiplier, ivrite the right-hand figure of each partial product under the order of the multiplier used, and add the partial products. 38 MUL TIP Lie A TION. If either factor has one or more ciphers on the right, multiply without regard to these ciphers, and annex to the result as inany ciphers as are on the right of both the factors. PROOF.— Revieiv the ivorh, or reverse the order of the factors and inultiply. If the work is correct, the result will he the same hy both methods. PROBLEMS. Multiply and prove — 1. 216 by 8. Arts. 1728. 2. 405 by 9. 3. 1315 by 6. Am. 7890. 4. 116 by 1000. 5. 413 by 70. Ans. 28910. 6. 555 by 4. 7. 4163 by 7. Am. 29141. 8. 3162 by 11. 9. 51003 by 90. 10. 18300 by 18. 11. 738 by 235. 12. 756 by 72. 13. 3216 by 5. 14. 248 by 19. 15. 160 by 30. 16. 365 by 37. 17. 1040 by 11. 18. 4561 by 603. 19. 11140 by 13. 20. 5704 by 974. 21. 4402 by 222. 22. 4561 by 4005. Am. 54432. Am. 4712. Am. 13505. Am. 2750283. Am. 5555696. 23. How many are 312 times 4144? 24. HoAV many are 999 times 345 ? Am. 1292928. 345000 = 345 X 1000 345 = 345 X 1 999 Solution. — Since 1000 times any number, less once the number, must be 999 times tlie number, we here abridge the solution by taking the 344655 = 345 X rnnhi()licaiul 1000 times, or once too many, by annexing three ciphers, and tbL'ii subtracting the multiplier. This method of abridgment applies whenever the nniltiijlier is 1 less than 100, 1000, 10000, etc. 25. 4573 X 99 = what number? Am. AbT121. 26. 13i6 X 999 = what number? 27. 1230 X 9999 = what number? Am. 12298770, 28. What is the product of 1036 by 990? MVL TIPLICA TION. cM 29. What is the product of 4455 by 105 ? 445500 = 4455 X 100 Solution.— Here the solution may Qc^a>>^ r //^^v ^ ^'^ abridged by writing the product of " '^^ 4455 by the 1 hundred at once, by 4(>777o =4455 X 105 annexing two ciphers, and writing under it the product of 4455 by the 5 ones, and then adding the two partial products. 30. What is the product of 6307 by 1003 ? 31. If 5362 feet of boards can be sawed in a mill in one day, how many feet can be sawed in it in 313 days ? Ans. 1678306. 32. At 125 dollars a month, how many dollars can be earned in 12 months? Ans. 1500. 33. What will 3158 tons of coal cost at 8 dollars per ton? Q 7 r o Solution. — At 8 dollars a ton, 3158 tons of coal will cost o 3158 times 8 dollars, which is equal to 8 times 3158 dollars, or 25264 dollars. 25264 8 dollars is the true multiplicand, but since 8 times 3158 gives the same product as 3158 times 8, we, for convenience, in the solution consider both factors as abstract numbers, and make the smaller factor the nuiltiplier. 34. What will 8344 yards of cloth cost at 6 dollars a yard ? Ans. 50064 dollars. 35. How many oranges in 47 boxes when each box contains 279 oranges? ^ns. 13113. 36. Two factors are 7312 and 7000. What is their product? Ans. 51184000. 37. There were 6 drawers in a desk, 8 compartments in each drawer, and 87 dollars in each compartment. How many dollars did the desk contain ? Solution. — Since in the desk 87 dollars. there were 6 drawers, and each had 48 No. of compartments. « compartments, there were in the desk 6 times 8, or 48 compartments. ^"^^ Since there were 48 compart- 348 ' ments in the desk, and 87 dollars 417 6 dollars ^" each, the desk contained 48 times 87 dollars, gr 4176 dollars. 40 MULTIPLICATION. 38. What is the continued product of the factors 6, 8 and 87? 39. If 15 pounds of hay are required by 1 horse for 1 day, . how many pounds are required by 5 horses for 30 days ? 40. How many gallons in 1025 casks, if each cask contain 63 gallons ? Ans. 64575, 41. In a bushel of rye are 888390 seeds. How many seeds are there in 25 bushels ? ■ Am. 22209750. 42. If the w^eight of a cubic foot of white-oak wood is 43 pounds, what is the weight of 128 cubic feet? 43. What is the continued product of the factors 12, 420 and 310? Am. 1562400. 44. How many yards of cloth in 32 bales, each bale having 121 pieces, and each piece 31 yards? Am. 120032. 45. What is the population of a State containing 8320 square miles, if each square mile has 91 inhabitants ? 46. If Pennsylvania -'ontain 46000 square miles, what will be its population at 75 persons to a square mile ? Ans. 3450000 persons. 47. Sound moves 1142 feet in a second. How far will it move in one hour, or 3600 seconds? Am. 4111200 feet. 48. If a railroad, 1035 miles in length, should obtain government aid to the amount of 52400 dollars per mile, what would be the entire amount received ? TEST QUESTIONS. <6. — 1. What is MtTLTiPLicATioN ? The product? The factors of a product? The multiplicand? The multiplier? What does the sign of multiplication denote when written between two numbers? 2. What are the Principles of multiplication ? Show that the mul- tiplicand and product are similar numbers. Why must the multiplier be regarded always as an abstract number ? Show that the product is the same in whatever order its factors are used. 3. Recite the Rule for multiplication. Give the proof. Why, ui multiplication, do you begin at the right to multiply? When the nnil- tiplier consists of more than one order of units, how many partial prod- ucts may there be ? What will be the unit of the partial product when the multiplier is a number of simple units? When the multiplier is a number of tens ? When there are partial products, how do you obtain the entire product? DIVISION. 41 SECTION VII. Dirisio.Y. 77. — Ex. 1. How many barrels, holding 3 bushels each, wiil be required to hold 30 bushels of apples ? 2. If a farmer should raise 21 bushels of potatoes, and should wish to put them into barrels holding 3 bushels each, how many barrels would be required ? 3. If you have 28 apples, and wish to distribute them equally among 7 boys, how many can you give to each boy ? 4. How many times are 9 bushels contained in 36 bushels ? What is the product of 9 bushels by 4 ? . 5. How can you show that 9 bushels are contained in 36 bushels 4 times ? 6. When you distribute 35 apples equally among 7 boys, do you find how many times 7 boys are contained in 35 apples, or do you find one of the 7 equal parts of 35 ? 7. When you find how many times 9 cents are contained in 72 cents, is the result a concrete or an abstract number ? 8. When you find one of the 9 equal parts of 72 cents, is the result a concrete or an abstract number ? 9. How many times are 9 dollars contained in 29 dollars, and how many dollars remain ? 4 * 42 DIVISION. 10. If you have 47 peaches to distribute among 7 boys, how many entii'e peaches can you give to each, and how many peaches will remain ? 11. 7 times 6 peaches, and 5 peaches, are how many peaches ? DEFINITIONS. 78. Division is the process of finding how many times one number is contained in another ; or. Division is the process of separating one of two numbers into as many equal jiarts as there are units in the other. 79. The Dividend is the number to be divided. 80. The Divisor is the number by which to divide. 81. The Quotient is the result obtained by the division. 82. The Remainder is a part of the dividend remaining un- divided. 83. The Si^ of Division is -=-, and is read, divided hi/. The dividend is placed at the left of the sign, and the divisor at the right of it. Thus, 40 -H 8 is read, forty divided by eight. Division is sometimes denoted by placing the dividend over the divisor, with a line between them. Thus, ^;^ is read, sixteen divided by four. Division is also denoted by a curved line, ), with the divisor on the left and the dividend on the right. Thus, 5)10 is read, ten divided by five. 84. A Parenthesis ( ), enclosing two or more numbers, or a Vinculnm, , drawn over them, denotes that the expression is to be treated as a single number. Thus, (16 + 4) -f- 5, or li5^f^-4- 5, denotes that the sum of 16 and 4, or 20, is to be divided by 5. 85. The Xamcs of the equal parts into which a number may be divided differ according to their number. A half of a number is one of two equal parts into which it is divided. DIVISION. 43 A third of a number is one of the three equal parts into which it is divided. A fourth of a number is one of the Jour equal parts into which it is divided. In like manner we have the names fifths, sixths, sevenths, eighths, ninths, tenths, twentieths, thirti/ fourths, forty-sixths, etc. Halves, thirds, fourths, etc., are expressed by writing the number denoting the name of the parts, as a divisor, under a line, and the number denoting the number of the parts repre- sented, as a dividend, above the line. Thus, ^ signifies 1 divided by 2, or 1 half of 1, and is read, one half. f signifies 2 divided by 3, or 2 thirds of 1, and is read, two thirds. 86. A Fraction is a number which represents one or more of the equal parts into which a unit, or one, is divided. Tims, I, expressing 3 of the four equal parts of 1, is a fraction. Principles of Division. 87. — 1. Division is the reverse of multiplication. 2. The quotient must he an abstract number when the divisor and dividend are similar numbers. 3. The quotient must be a concrete number and the divisor an abstract number when the divisor and dividend are dissimilar numbers. 4. The remainder and dividend must be similar mimbers. 5. The dividend is equal to the product of the integer of the quotient multiplied by the divisor, jylus the remainder. ca.se; I. Short Division. 88.— Ex. 1. Divide 4313 by 4. Divisor, 4^)4-313 Dividend. Solution.— For convenience we . i . first divide the liighest order of 107 8 J Quotient. ^^itg^ 4 jg contained in 4 thcu- 4 sands 1 thousand times. Write 1 , (D-i a Tk I- i" the thousands' order in the 4-313 Proof q^^^^j^^^ 4 is not contained in 3 hundreds any number of hundred times. 44 DIVISION. Write in the huiulrods' order in the quotient, Jind unite the 3 hundreds with the 1 ten, making 31 tens. 4 is contained in 31 tens 7 tens times, with a remainder 3 tens. "Write 7 in the tens' order in tlie quotient, and unite the 3 tens with tlie 3 ones, making 33 ones. ■ 4 is contained in 33 ones 8 times, with a remainder 1. Write 8 in the ones' order in tlie quotient, and the remainder 1, with 4, the divisor, under it as a part of the quotient. The required quotient is 107S|. Prove the correctness of the solution by multiplying the integer of the quotient by the divisor, and adding the remainder. For, (Art. 87 — oj the dividend must be equal to the integer of tlie quotient multiplied by the divisor, plus the remainder. The explanation of the solution may be abridged thus : 4 in 4, 1 ; 4 in 3, 0; 4 in 31, 7 ; 4 in 33, 8, with 1 as a remainder. Ans. 1078|^. Division is called Short Division when in the solution only the divisor, dividend and quotient are written. Solve and prove — (2.) (3.) (4.) (5.) (6.) 4)940 6)672 5 )6570 7)847 8)968 (7.) (8.) (9.) (10.) (11.) 8 )4162 2 )1931 6 )6753 5) 47235 9)817 12. If 6 boys share equally 1386 apples, liow many -will each boy have? Ans. 231. C^SE II. Long Division. 89.— Ex. 1. Divide 16013 by 5, or find one fifth of 16013. Divisor. Dividend. Quotiont. ^, - • . ^ • i • i , _j Solution. — 6 is not contained in 1 ten- OyluUlo(o/C(Jr.^- thousand any number of ten-thousand ^Q times ; hence, we unite the 1 ten-thousand with the 6 thousands, making 16 thousands. 5 is contained in 16 thousands 3 thou- sands times, with a remainder. Write 3 in ]^g the thousands' order of the quotient. 5X3 ■tQ thousands -= 15 thousands, which, written iHider the 16 tiiousands and subtracted, 3 Tirm. leaves 1 tiiousand. Unite the 1 thousand with the hundreds, making 10 hundreds. 10 10 DIVISION. 45 6 is contained in 10 hundreds 2 hundred tiraos. Write 2 in the hun- dreds' order of the quotient. 5X2 hundreds =- 10 liundreds, wliich, written under the 10 liundreds and subtracted, leaves no remainder. is not contained in 1 ten any number of tens times. Write in the tens' order of the quotient, and unite with the 1 ten the 3 ones, making 13 ones. 5 is contained in 13 ones 2 ones times, with a remainder. Write 2 in the ones' order in the quotient. 5X2 ones = 10 ones, which, written under the 13 ones and subtracted, leaves 3 remainder. W^rite the remainder 3 over the divisor 5 as a part of the quotient, which gives 3202|, the result required. 3202-- 5 16010 3 16013 Proof. — Prove the solution by multiplying the integer of the quotient by the divisor, and adding the remainder, which gives the dividend ; hence, the work is correct. Division is called Long Division when each process of the solutiou is written. Solve and prove — (2.) (3.) 8)3368( 1 1)235 4-( (6.) (7.) 21)640( 15)915( (4.) 9)3706( (8.) 25)806( (5.) 12M004( (9.) 7)1803( Am. 220f. 10. Find one eighth of 1765. 11. When a steamer sails 3018 miles in 14 days, what is her progress per day ? A^is. 21 5 j\ miles, 12. At the rate of 27 miles per hour, in how many hours will a train of cars run 2563 miles? Ans. 94ff . C^SE III. When the Divisor is 1, with One or More Ciphers on the Rig-ht. 90. It has been shoAvn that a number is multiplied by 10 by removing each figure one order to the left, which is done by annexing one cipher ; by 100, by removing each figure two orders to the left, or annexing txvo ciphers, etc. (Art. 74.) 4(i DIVISION. Hence, since division is the reverse of multiplication (Art. 87), To divide by 10, remove each figure one order to the right by removing the decimal point one order to the left. To divide by 100, remove each figure two orders to the right by removing the decimal point two orders to the left, etc. Thus, 2563 -=- 10 is 256.3, or 256yV 2563 -- 100 is 25.63, or 2b^i^. 2563 -- 1000 is 2.563, or 2^W The decimal point in the result separates the integer of the quotient from the fractional part. The first order on the right of the decimal point is tenths ; the second order, hundredths ; the third, thousandths. Ex. 1. Divide 67325 by 1000. 67325 -4- 1000 = Solution. — Remove the decimal point in /?'y ' many horses can he purchase? 35. It is proposed to divide a ti'act of land containing 72000 acres into farms of 320 acres each. How many farms will it make? Ans. 225. 36. In an orchard there are 44520 trees in 212 equal rows. How many trees are there in each row ? 37. If the dividend is 126072 and the divisor 612, what is the quotient ? 38. The height of a mountain in Asia is 28176 feet, and the height of Mount Washington is 6284 feet. How mauy times as high as the latter is the former? Ans. 4|-|^. TEST QUESTIONS. 92. — 1. What is Division? The dividend? The divisor? The quotient ? "What in division corresponds to the factors of tlic product in muhiplication ? What is the product? What, then, in division maybe regarded as the factors of the dividend? 2. What is the Sign of division ? When written between two num- bers what does it denote? In what other ways* may division be denoted '/ 3. Recite the Principles of division. Sliow that division is the re- verse of multiplication. Show when the quotient will be an abstract number. When the quotient will be a concrete number. 4. What is a Rkmainder in division ? Why are the remainder and dividend similar numbers? How may the remainder be changed to a fractional part of the quotient ? 5. Recite tlie Rule for division. How does long division clifier from short division ? What is the reason for removing the decimal point to the left in dividing by 10, 100, etc. ? 0. What is the Pkoof of division ? The reason for it ? Since multi- plication and division are the reverse of each other, how may multipli- cation be proved ? Sliow that the product divided by the multiplier gives the multiplicand. REVIEW. 49 SECTION VIII. REVIEW PROBLEMS. MENTAL EXERCISES. 93.— Ex. 1. The factors of a product are 11 and 12. What is the product? 2. The product of 15 by 9 is what number? 3. If a boy can earn 18 dollars in one month, how many dollars can he earn in 10 months? • 4. At 14 dollars a ton, what will 13 tons of hay cost? Solution. — If one ton cost 14 dollars, 13 tons will cost 13 times 14 dollars, or 182 dollars. The multiplication may be conveniently per- formed thus : 13 times 14 = 3 times 14 + 10 times 14 ; 3 times 14 = 42 ; 10 times 14 = 140 ; 42 + 140 = 182. 5. At 16 dollars each, what will 12 garments cost? 6. If 14 men can do a piece of work in 21 days, in how many days can one man do it ? 7. John lives 5 miles from a certain place ; Andrew, 7 miles farther away ; and Benjamin, 8 times as far as Andrew. How far from the place does Benjamin live ? 8. How many lengths of 12 feet each are there in a fence which is 132 feet long? 9. A and B start from points 121 miles apart, and travel toward each other, A at the rate of 5 miles per hour, and B at the rate of 6 miles per hour. In how many hours will they meet ? 10. When flour is worth 8 dollars a barrel, how many barrels of flour will pay for 24 tons, of coal at 5 dollars a ton ? 11. If 10 men can do a piece of work in 6 days, in how many days can 4 men do it ? Solution. — If 10 men can do a piece of work in 6 days, 1 man can do it in 10 times 6 days, which are 60 days ; and 4 men can do it in one fourth of 60 days, which is 15 days. 12. When 12 tons of coal at 7 dollars a ton pay for 21 pairs of boots, how much are the boots worth a pair ? 13. If 8 cords of wood will buy 32 pairs of shoes, how many cords will buy 40 pairs of shoes ? 50 BEri£!W. 14, I sold 5 dozen of eggs at the rate of 4 for 7 cents, and received 5 cents in money, and the balance in coffee at 25 cents a pound. How many pounds of coffee did I receive ? WMITTEN EXERCISES. 94. — Ex. 1. If in making mortar, 6 bushels of sand are re- quired for each cask of lime used, how many bushels of sand will be required for 97 casks of lime ? Ans. 582. 2. How many loads, each contaiuiug 1345 bricks, are there in a pile containing 286485 bricks ? 3. If the front and rear walls of a house each require for their construction 31G50 bricks, and the other two walls each require 43400 bricks, how many bricks are required for the four walls? ' Aiis. 150100. 4. The factors of a product are 8043 and 405. What is that product? Ahs. 1232415. 5. The product of two factors is 9225 ; onfe of the factors is 45. What is the other factor? 45)9225(205 90_ "225 225 PoM'TioN. — Since a product is the result ob- tained by multiplying one of two factors by the otlier, the quotient obtained l)y dividing the prod- uct by one of the factors must be the other factor. 9225 -f- 45 ^ 205, the factor required. HE VIEW. 61 6. If a product is 9225 and the multiplicand 205, what is the multiplier? 7. If a man has 15000 dollars, and should spend enough of it to pay for a farm of 80 acres at 78 dollars per acre, how much would he have left ? 8. If a merchant should purchase 1011 barrels of flour at 12 dollars a barrel, and pay down 7919 dollars, how much would he then owe for the flour? A^is. 4213 dollars. 9. The dividend is 15750 and the divisor 25. What is the quotient, or the other factor of the dividend ? Aiis. 630. 10. What is the average of 16, 22 and 28? Solution. — The average of two numbers is one half of the sum of those numbers ; the average of th7-ee numbers is one third of their sum, etc. The sum of 16, 22 and 28 is 66, and one third of 66 is 22. 11. The elevation of the northern lakes above the sea is as follows : Superior, 627 feet ; Michigan, 587 feet ; Huron, 574 feet ; and Ontario, 282 feet. What is their average elevation above the sea? Ans. 517|- feet. 12. What is the value of (14 + 6) + 16 X 2 — (^63 — 19~+4) + 45 ^ 9 — 3 ? iU + 6') + 16X,?— (63—19 + 4) +45-^9 — 3 = 20+ 16 X !2 — 40 + 45^9 — 3 = 20 + 32 — 40 + 5—3; =57-43; =14 Solution. — Combining first the numbers in the parentheses, we have 20 for the value of (14 + 6), and 40 for the vahie of (63 — r9 + 4). Combining the numbers affected by the signs of multiplication and those by the sign of division, we have 32 for the value of 16 X 2, and 5 for 45 -=- 9. Then, combining the numbers as indicated by the signs of ad- dition and subtraction, we have 14 as the value required. 13. What is the value of 4 + 6 X 5 — 16 -- 8 — (4 X 2) ? 14. Henry has 7375 dollars, which is 7 times as much as Daniel has, lacking 780 dollars. How many dollars has Daniel? Ans. 1165. 15. Tv.o candidates at an election received in the aggregate 15653 votes. If one of them received 783 votes more than the other, how manv votes did the other receive ? 52 ' FACTORING. SECTION IX. FACTORS AKD DIVISORS. 95. — Ex. 1. Of what two integers is 6 the product ? 2. What two integers multiplied together produce 15? What two produce 21 ? 3. What integers are factors of 6 ? Of 15 ? Of 21 ? 4. Of what three integers is 30 the continued product ? 5. Of what sets of two integers is 30 the product ? 30 = 2 X 15, or 3 X 10, or 5 X 6. 6. Of what sets of integers greater than 1 is 24 the product ? 7. Name some numbers that are the product of integers greater than 1. 8. What are the smallest integers greater than 1 that will divide 21 without a remainder ? 30 without a remainder ? 9. Give the sets of integers greater than 1 which, when multiplied together, will produce 30. 10. Of what number are 3 and 5 the factors? 2, 3 and 5 the factors ? DEFINITIONS. 96. The Factors of a number are the integers which being multiplied together will produce that number. Thus, 2, 3 and 5 are the factors of 30 ; for 2 X 3 X 5 = 30. 97. A Prime Number is an integer that has no factor except itself and 1. Thus, 1, 3, 5, 7 and 11 are prime numbers. 98. A Composite Number is an integer that has other fac- tors besides itself and 1. Thus, 4 and 6 are composite numbers, since 4 = 2X2; and 6 =: 2 X 3. 99. A Prime Factor is a factor which is a prime number. The prime factor 1 is not commonly mentioned, since it is a factor of every integer. Numbers are said to be vtutiially prime, or prime to each other, when they have no common factor exce«t 1. FACTORING. 58 100. Factoring is the process of finding the factors of com- posite numbers. The number of times a number is taken as a factor may be denoted by writing a small figure, called an Exponent, at the right and above the figure or figures of the factor. Thus, 3^ =; 3 X 3, and denotes that 3 is taken twice as a factor. 11^ = 11 X 11 X 11, iind denotes that 11 is taken 3 times as a factor. 101. An Exact Divisor of a number is any integer which will divide the number without a remainder. Thus, 1, 2, 3, 4, G and 12 are each exact divisors of 12. The Exact Divisors of a number are called, also. Divisors, or Measures, of that number, and must be factors of it. A number is said to be divisible by its exact divisors. Thus, 12 is divisible by its exact divisors 1, 2, 3, 4, 6 and 12. 102. Any number is divisible by 2 when its right-hand' figure is 0, 2, 4, 6, or 8. For such numbers are composed of some exact number of twos. Numbers divisible by 2 are Even Numbers, and all others are Odd Numbers. 103. A number is divisible by 4 if its tens and ones are divisible by 4. For 4 is an exact divisor of 100, and of any number of hundreds; hence, if the tens and ones of a number are divisible by 4, the number itself must be. Thus, 648 and 7312 are each divisible by 4. 104. A number is divisible by 5 if its right-hand figure is or 5. A number whose right-hand figure is is an exact number of tens; a number whose right-hand figure is 5 is an exact number of tens plus 5. 5 is an exact divisor of 5 or 10 ; hence, any exact number of tens, or any exact number of tens plus 5, is divisible by 5. Thus, 70 and 75 are each divisible by 5. 105. A number is divisible by 3 or 9 when the sum of the ones represented by its figures is divisible by 3 or 9. Take, for example, the number 7542 ; 7542 = 7000 + 500 + 40 + 2 ; and 7000 = 7 times 999 + 7 ; 500 = 5 times 99 + 5 ; 40 = 4 times 9 + 4 ; and 2 = 2; where the figures expressing the number of each order plus 5* 54 FACTORING. the exact number of times 9 are the figures of the given number. Now, 7 times 999, 5 times 99 and 4 times 9, being each divisible by 9 and by 3, if the sum of the ones represented by the figures of the number are divisible by 9 or by 3, the number itself is thus divisible. Thus, 7542 and 9765 are each divisible by 9 and by 3. 106. Principles. — 1. Every number is equal to the product of alt iLs prime factors. 2. A number is divisible by all its j^rime factors, and by ah- the products of two or more of them, and is divisible by no other numbers. WRITTEN EXERCISES. 10?. — Ex. 1. What are the prime factors of 70? Solution. — Since the right-hand -^^^ figure is 0, we can divide by the prime g)QQ numbers 2 and 5. (Arts. 102 and 104.) Dividing by these prime numbers gives for a quotient 7, which is also Proof, J? X <5 X 7 = 70 prime. Hence, the prime factors of 70 are 2, 5 and 7. 7 2. What are all the factors or divisors of 66 ? 2)66 2y 3, 11 Solution.— Since every prime Q)'^g S y^ 2 = 6 factor of a number, and every — -ii ^ ^ = $>^ product of two or more of these ^-'- 1 1 \( Q '9'?? prime factors, is an exact divisor of the number, and no other 11 /\ o y^ /C = ob numbers can be exact divisors of tliat number (Art. 106 — 2), the prime factors 2, 3 and 11, and the prod- ucts of 3 by 2, 11 by 2, 11 by 3, and 11 by 2 times 3, must be all the fac- tors or exact divisors of 66. Hence, 2, 3, 11, 6, 22, 33 and 66 ar£ the factors and divisors required. 3. What are the prime factors of 84? Av.<<. 2\ -S and 7. 4. What are all the factors or divisors of 56 ? Ans. 2\ 7, 4, 8, 14, 28, 56. 108. Rule for Factoring.— D/f^fZe the given ninnber hy any of its prim,e factors greater than 1. Divide the quotient, if composite, in lilce vnanner, and so proceed FA CTORING. 55 jontil the last quotient is a, priDie iiiunhcr. The last quotient and the several divisors will be the prime factors of tJie ninnber. Tlie prime factors of a given number, and the various products of these factors, by talcing two together, three together, etc., are all the different fac- tors or exact divisors of that number. 1^ no JUL EMS. What are the prime factors of — Ans. 3^ 11. Ans. 3, 7, 11. Ans. 2, 3, 5, 37. Ans. 3, 7, 17, 19. What are all the different factors or divisors of — 13. 70? Ans. 2, 5, 7, 10,14, 16. 42? Ans. 2, 3, fi, 7, 14, 21 and 42. 17. 63? 18.105? ^n.s. 3, 5, 7, 15, 21, 35 and 105. 1. 75? Avs. 3, 5, 5. 7. 99? 2. 144? Ans. 2^ 3^ 8. 231? 3. 116? 9. 875? 4. 340? Ans. 2\ 5, 17. 10. 1110? 5. 180? 11. 4004? 6. 3809 ? Ans. 13, 293. 12. 6783? 35 and 70. 14. 30? 15. 56? Ans. 2,4,7,8,14,28 and 56. 19. How many of the different factors or divisors of 100 are prime, and how many are composite ? Ans. Three are prime and gix are composite. 20. How many of the different factors or divisors of 210 are prime, and how many are composite ? TEST QUESTTONS. 109. — 1. What are the Factors of a number? What is a prime number? Name some number that is the product of two prime num- bers. What is a composite number ? Give an example of a composite number. 2. Wliat is a Prime Factor ? Name a prime factor of 30. A com- posite factor of 30. When are two numbers prime to each other? 56 FA CTORING. 3. What is an Exact Divisor of a number ? How do even numbers differ from odd numbers ? Of what are divisors or measures of a num- ber factors ? 4. By what is a number said to be Divisible ? How can you know that a number is divisible by 2 ? That a number is divisible by 4 ? By 5? By 9 or 3? 5. What is Factoring ? How do you denote the number of times a factor is taken ? How do you find all the factors of a number ? 6. What are Principles of factoring? Show that 18 is the product of all its prime factors. Show that 18 is divisible by the various prod- ucts of its prime factors. 7. What is the Rule for factoring ? In factoring an even number why do you divide by 2 ? In factoring numbers why do you divide by their prime factors in succession ? SECTION X. commojY divisors. 110. — Ex. 1. What numbers are exact divisors of 15? Of 20? 2. What number is an exact divisor of both 15 and 20? 3. What is the unit of 15? Of 20? Of the exact divisor of both 15 and 20? 4. What numbers are exact divisors of 33 dollars ? 5. What number is an exact divisor of both 22 dollars and 33 dollars ? 6. What is the unit of 22 dollars ? Of 33 dollars ? Of the exact divisors of both 22 dollars and 33 dollars ? 7. What are the exact divisors common to 12 and 18? 8. What prime factors have 12 and 18 in common? ^Yliat is the product of those factors ? {). What is the greatest exact divisor common to 12 and 18? To 8 and 20? 10. What exact divisor is common to 5 and 3 times 5? 11. What exact divisors arc common to G and 7 times 6? 12. AVhat exact divisors are common to and 42 ? To 6 and 42 and their sum? To 6 and 42 and their difference? FACTORING. 57 DEFINITIONS. 111. A Common Dirisor, or Commou Measure, of two or more numbers is any exact divisor (Art. 101) of each of tho.se numbers. Thus, 2 is a common divisor of 8, 10 and 18. Only similar numbers can have a common divisor, since one number to measure another must have the same unit. 112. The Greatest Common Divisor, or Greatest Common Measure, of two or more numbers is the greatest exact divisor of each of them. Thus, 6 is the greatest common divisor of 12, 24 and 42. Numbers that are prime to each other have no common divisor, for they can have no common factor greater than 1. 113. Principles. — 1. The greatest common divisor of two or onore numbers is the product of all their common prime factors. 2. A divisor of a number is a divisor of any integral number of times that number. 3. A common divisor of two or more numbers is also a divisor of their sum and of their difference. C^SE I. Common Divisor. 114. — Ex. 1. What is a common divisor of 14 and 20. Solution. — 2 is an exact divisor of any number whose ^J j-A-i '^^ right-hand figure is 4 or (Art. 102); hence 2 is an 2' ]^Q exact divisor of both 14 and 20, and therefore a com- mon divisor of the given numbers. 2. What are all the common divisors of 45 and 90 ? 45 = 3^3X5 3)45, 90 Solution. — By fac- 90=^3X3X5X2 ^r '?)7^ '^n Coring we find the com- or, oy io, ou ^^^ ^^-^^^ factors of 9 x^ o-_ n 5j 5, 10 4:5 and 90 to be 3, 3 \, ~~i ^ and 5 ; lience 3, 3 and 5 ' are common divisors of 3X3X5 = 45 45and90. 58 FACTORING. Since every product of two or more of the prime factors must be a divisor, the various products that can be formed by 3, 3 and 5 must likewise be common divisors of the given numbers. These products are 3 X 3, or 9 ; 5 X 3, or 15 ; and 5 X 3 X 3, or 45. Hence, 3, 5, 9, 15 and 45 are the common divisors required. 3. What is a common divisor of 27 aud 39? Ans. 3. 4. What are all the common divisors of 45 and 75 ? Ans. 3, 5 and 15. 115. Rule for finding Common Divisors.— i^/^cZ tlie prime fac- tors of the given nwrvibers. All coimnon factors are eoTmyion divisors, and all the various products of those factors are all the coininon divisors. PROBLEMS. 1. What is a common divisor of 25 and 35? Ans. 5. 2. What are all the different common divisors of 27 and 81 ? 3. What are the common composite factors or divisors of 24, 72 and 84? J ws. 4, 6 and 12. 4. What are all the common divisors of 105, 210, 315. Arts. 3, 5, 7, 15, 21, 35 and 105. C^SK II. Greatest Common Divisors. 116. — Ex. 1. What is the greatest common divisor of 12, 18 and 78 ? 12 = ^ X t? X ^ Solution. — By factor- /;? = ^ X "? X ^ ing we find the prime fac- tors common to 12, 18 and 78 — 2 X 3X 13 78 are 2 and 3. Greatest Com. Div. ^^2X3 = 6 Since the product of p. all the common prime fac- ' io\i(0 1 o lyo ^^^^ ^^ *^''^ greatest com- ? 1 mon divisor (Art. 113), 3) 6y 9, 39 2 X 3, or 6, must be that ">, 3, 13 ^^^^^•'^'■- Or, dividing by 2, we take out that common factor. Dividing the resulting quotients by 3, we take out that common factor and obtain quotients that are prime to each other. Hence 2 and 3 are all the factors common to the given numbers, and their product, 2 X 3, or 6, must be the greatest common divisor required. FACTORING. 69 2. What is the greatest common divisor of 91 and 133 ? 91) 1S3( 1 Solution. — Since any number is the great- Q -I est divisor of itself, if 91 be a diviisor of 133, it must be the greatest common divisor of 42)91(2 91 and 133. We find that it is not a divisor 84- of 133, since on trial 42 remains. 7 ) /^/^P ^^ '^'^ ^® ^ divisor of 91, it is also a divisor of Z) 133, which is once 91, plus 42. (Art. 113—2.) It _T is not a divisor of 91, since on trial 7 remains. If 7 is a divisor of 42, it must be also a divisor of 91, which is twice 42, plus 7. On trial it is found to be a divisor of 42. Now, 7 is the greatest divisor of 7 and 42 ; hence, 7 is the greatest common divisor of 42 and 91, and therefore 7 is the greatest common divisor of 91 and 133. 3. What is the greatest common divisor of 84 and 132? 117. Rules for finding the Greatest Common Divisor.— i. Find the prime factors common to the given numbers, and the product of those factors will he the greatest coin- ?non divisor required. Or, 2. Divide the greater' number by the less, and if there be a remainder , divide the divisor by it, and so continue to divide the last divisor by the last remain- der till an exact divisor is found. That divisor will he the greatest convmon divisor of the two numbers. 3. If more than two numbers are given, find first the greatest common divisor of two of theiiv, and then the greatest common divisor of that divisor and another of the numbers, and so on till all the numbers have been used. The last common d,ivisor will he the greatest common divisor of all the numbers. Pn01iZ,EMS. What is the greatest common divisor of — 1. 32, 48 and 80? Ans. 16. ! 4. 308 and 630? Ans. 14. 2. 75 and 165? | 5. 91 and 117? 3. 72 and 168 ? Ans. 24. I 6. 21, 30, 39 and 81 ? Ans. 3. 60 FACTORING. 7. I have three rooms, the first of which is 16 feet wide ; the second, 20 feet; and the third, 24 feet. What must be the width of carpeting which will exactly fit each room ? Ans. 4 feet. 8. There is a garden 84 feet wide and 1068 feet long. What must be the length of the longest rails that will, without cutting, exactly enclose it ? Ans. 12 feet. 9. Find the greatest common divisor of 99 bushels, 261 bushels and 504 bushels. Ans. 9 bushels. 10. James has 66 dollars, Edward has 77 dollars, and Arthur has 264 dollars. If they should purchase fleur at the highest price per barrel that would allow each to exactly use his money, how much would the flour cost per barrel ? Ans. 11 dollars. SECTION XI. MULTIPLES. 118.— Ex. 1. What number is 7 times 6 ? 5 times 6 ? 2. What number is some integral number of times 6 ? 3. What prime factors have 6 and 7 times 6 in common ? 6 and 5 times 6 ? 4. What prime factors have 6 and 42 in common ? 6 and 30? 5. What two numbers are contained in 6 an exact number of times ? Of what two numbers is 6 an exact number of times ? 6. Of what two numbers is 15 an exact number of times? Of what number are those two numbers prime factors? 7. What numbers from 5 to 30 contain 5 an integral num- ber of times ? 8. What numbers from 6 to 24 contain botli 2 and 3 an in- tegral number of times ? 9. What is the least number that contains both 2 and 3 an integral number of times? 10. What is the least number that is an integral number of times 3 and 4? What are the prime factors of 3 and 4? Of 12? FA CTOEING. 61 11, What is the least number that is an integral number of times 10 and 6? What are the prime factors of 10 and 6 which when taken the least number of times will form those numbers ? What are the prime factors of 30 ? DEFINITIONS. 119. A Multiple of a number is any integral number of times that number. Thus, 10, which is twice 5, is a multiple of 5. 120. A Coiinnon Multiple of two or more numbers is any number which is an integral number of times each of them. Thus, 12, which is 3 times 4, is a common multiple of 3 and 4. 121o The Least Commou Multiple of two or more numbers is the least number which is aii integral number of times each of them. Thus, 6 is the least common multiple of 2 and 3. Only similar numbers can have a common multiple, since numbers must be similar to be factors of the same product. 122. Principles. — 1. A multiple of a number contains all the prime factors of that number. 2, A common muUiple of two or more numbers contains all the prime factors of those numbers. 3. The least common nniltiple of two or more numbers is the least number that contains all the prime factors of those numbers. C^SE I. Common Multiples. 123. Ex. 1. — Find a common multiple of 9 and 11. ^ -f Ky Q QQ Solution. — Since a common multiple of the given numbers must be a number which contains those numbers as factors, their product, or 99, is a common multiple. 2. Find a common multiple of 5, 7 and 9. Ans. 315. 124. Rule for finding a Common Multiple.— Multipli/ the given niiTYibers together, and the product will be a common multiple of those numbers. 62 FACTORING. PROBLEMS. 1. What is a common multijjle of 3, 5 and 6? Ans. 90. 2. Find a common multiple of 11 and 13. 3. What is a common multiple of 7, 10 and 25 ? Ans. 1750. C^?k.SE II. Least Common Multiples. 125. Ex. 1. — Find the least common multiple of 4, 12 and 30. j^--- 2 y^ 2 Solution. — A multiple of 4 must -iq) ^V $?y "? contain its prime factors 2 and 2; a '*' ^ s/ =6'^ of 30 must contain the additional prime factor 5. 2, 2, 3 and 5 are all the prime factors of the numbers; hence, the product of these factors, or 60, is their least common multiple. 2. Find the least common multiple of 42, 49 and 70. 7 )A2 A9 70 Solution.— By division we « \~^ ^y — Tn ^^^^ ''"^ ^^^"^ prime factor 7, ■^ > / , -t 1/ common to the given numbers, 3, 7, 5 and liave left the factors 6, 7 and 10. 7 y^ 2 y^ 3 y^ 7 y^ 5 ^ 147 We take out the prime fac- tor 2, common to 6 and 10, and have left the factors 3, 7 and 5, which have no factor common to any two of them. Hence, 7, 2, 3, 7 and 5 are all the prime factors of the numbers, and their product, 1470, is the least common multiple required. 3. Find the least common multiple of 8, 11 and 15. Ans. 1320. 126. Rules for finding the Least Common Multiple.— 1. Find the prime ftvctora of tlie glueii nwDibers, ami the product of the different prime factors, eacJi factor being ta^hen the greatest numher of times it occurs in any of the numbers, will be the least comm^ou multiple. Or, 2. Place the given numbers in a horizontal line; FACTORING. 63 divide hy any prime ninnher that is a factor^ of two or iy%ore of them, and write the quotients and un- divided numbers heloiv. Divide these, if possible, in lihe manner, and so continue until the quotients and undivided numbers are prime to each other. TJie product of the divisors and the numbers remaining in the last horizontal line will be the least common multiple. PROBLEMS, What is the least commou multiple of — 1. 24, 15 and 16? Ans. 240. 2. 42 and 56 ? Ans. 168. 3. 9, 11 and 48? 4. 13 and 29? .4ns. 377. 5. 18, 27 and 30 ? 6. 60, 50 and 35? Am. 2100. 7. Find the least commou multiple of 40, 36, 32, 30, 28, 24, 20 and 18. 40, 36, 32, 30, 28, 24-, 20, 18 Solution. -Write 40 as a factor of the answer. The 4^-> X 5 X ^ X 7 = 10080 largest factor common to 40 and 3G is 4; write 9, the remaining factor of 36, as a factor of the answer. The largest factor common to 40 and 32 is 8 ; write 4, the remaining factor of 32, as a factor of the answer. The largest factor common to 40 and 30 is 10; and 3, the remaining foctor of 30, is found in 9. The largest factor common to 40 and 28 is 4; write 7, the remaining factor of 28, as a factor of the answer. The largest factor common to 40 and 24 is 8 ; and 3, the remaining factor of 24, is found in 9. Tlie factors of 20 are found in 40, and the factors of 18 in 36. The continued product of the factors of the answer is 10080. 8. Find the least common multiple of 8, 10, 11, 90 and 132. 9. What is the least common multiple of 7, 16, 21 and 28? 10. Find the smallest number that will exactly contain 9, 15, 18 and 20. Ans. 180. 11. What is the least number of cents with which you may purchase either slates at 18 cents or arithmetics at 63 cents each? Am. 126. 64 FACTORING. 12. What is the smallest sum of money for which I can purchase calves at 17 dollars each, yearlings at 34 dollars each, or cows at 68 dollars each ? Ans. 68 dollars. 13. What is the least number of acres that can be exactly divided into lots of 12 acres, 15 acres or 16 acres each? Ans. 240. TEST QUESTIONS. 127. — 1. What is a Common Divisor of two or more numbers ? Why cannot 4 dollars and 6 yards have a common divisor ? What is the greatest common divisor of two or more numbers ? What numbers have no common divisor ? 2. What is the Principle in relation to the greatest common divisor of numbers? Of what is the divisor of a number a divisor? Of what is a common divisor of two or more numbers also a divisor? 3. What is the Rule for finding the common divisor of two or more numbers ? For finding the greatest common divisor of two or more numbers ? 4. What is a Multiple of a number? A common multiple of two or more numbers? The least common multiple of two or more numbers? 5. What is the Principle in relation to a multiple of a number? In relation to a multiple of two or more numbers ? In relation to the least common multiple of two or more numbers? 6. What is the Rule for finding a common multiple? For finding the least common multiple of two or more numbers? When numbers are prime to each other, how is their least common multiple found ? SECTION XII. FACTORS r.W DiriSIOJ^. 128. The Value of a quotient in division depends upon the relative values of dividend and divisor. Hence, Any change in the factor. i of the dividend or divisor must affect the value of the quotient. Thus, 24-- 6 = 4; and (24X2) -^6 = 8, or, 24 -- (6 -- 2) = 8; also, (24 -f- 2) --6 =2, or, 24 -f- (6 X 2) = 2. FACTORING. 66 The same change in the factors of both dividend and divisor does not affect the value of the quotient. Thus, (24 -- 2) -- (6 4- 2) = 4 ; or, (24 X 2) -^ (6 X 2) = 4. 129. General Principles of Division. — 1. Multiplying the dividend, or dividing the divisor, multiplies the quotient. 2. Dividing the dividend, or vndtiplying the divisor, divides tJie quotie)U. 3. Dividing or multijohjing both dividend and divisor by the same number does not change the quotient. C^SK I. Division by Factors. 130. Ex. 1. Divide 9702 by 21, using the factors of 21. Solution. — The factors of 21 are 3 and 7. Since 21 3)9702 times a number is 7 times 3 times the number, one twenty- _, s „„g . first of a number must be one seventh of one third of that J- number. 462 One third of 9702, the dividend, is 3234, and one seventh of 3234 is 462. 2. Divide 4677 by 45, using factors. 5)A677 Solution. — 45 is equal to 5 X 9- , -• Dividing 4677 by 5, we have 935/t>cs, 9 J 9oO , ^ ones = ^ ^^^^ 2 ones as a remainder. 103, 8 fives =■ 40 Dividing by 9, we have 103 forty-fives, _ --, . , "T^n and 8 fives as a remainder. True Remainder, 4-" rt^^ ^ ^ 4.- 1 • ^ • o Tlie first })artial remainder is 2 ones, or 2, and the second partial remainder 8 fives, or 40 ; hence, 2 + 40, or 42, is the whole or true remainder, and 103|| is the quotient required. The factors 5, 3 and 3 could have been used with the same result. 3. Divide 825 by 86, using factors. Ans. 22f|. 131. Rule for Dividing by Factors.— i^//^,rZ any convenient set of factors of the rlivisor ; divide tlie dividend hy one of these factors, and the quotient thus obtained by another, and so on. till all the factor's are used. If there be onr or more reri%a.ind;ers, multiply each. by the divisors preceding the one that produced it, and add the prod^ucts and the remainder, if any , from the first division. The sum ivill be the true remainder. 6* 66 FACTORING PnOBLEMS. Divide, using factors — 1. 2954 by 14. Ans. 211. 2. 3728 by 28. Ans. 133f 3. 8316 by 27. 4. 88763 by 32. Ans. 2773f|. 5. 47839 by 42. Ans. 1139^^ 6. 11630 by 81. 7. 2520 by 105. Aiis. 24. 8. 196473 by 72. Am. 2728f|. 9. Divide 94596 by 2300, using the tactors 23 and 100. ogg SoLTJTiox. — Dividing by 100, we have 23 \ 00J94-o\ 96 ( 4-1^^ 945 hundreds, and 96 ones as a remainder. 92 Dividing by 23, we have 41 twenty-three ^f- hundreds, and 2 hundreds as a remain- der. ^^ 2 hundreds + 96 ones = 296, tlie true ^ remainder, and '^1-^^^% is the quotient required. In the computation we denote, for brevity, the factoring of tlie divisor by cutting off the two ciphers by a mark, and in the division of the divi- dend by 100 we set off the remainder by the same mark. 10. Divide 782967 by 3700, using the factors 37 and 100. A71S. 21imi- 11. Divide 46370 by 90, using the foctors 9 and 10. 12. What is the quotient of 345600 divided by 5000 ? Am. 69^%%. 13. Divide 16632 by 5148, using the factors 11, 18 and 26. 14. If 700 barrels of apples cost 2100 dollars, how much will one barrel cost? ^1??^. 3 dollars. 15. If 63 bushels of wheat make one load, how many full loads can be made from 1937 bushels, and how many bushels will remain ? Ans. 30 full loads, and 47 bushels over. CASK II. Cancellation. 132. Cancellation is the process of shortening computations by striking out equal factors from the dividend and divisor, and using only the remaining factors. FA CTORING. G7 133.— Ex. 1. Divide 11 X 3 X 2 by 11 X 2. 1 1 Solution. — Indicate the ti- X 3 XS- 1X3X1 division by writing the divi- y— T^ — ~ 1 sy -I — ~ ^ ''^"^ °^'^^' ^^^^ divisor. TTXlt 1X1 DWiAc botli dividend and ^ 1 divisor by the factors 11 and 2, by cancelling those common factors in both, whicli does not change tlio quotient. (Art. 129—3.) When a factor is cancelled, 1 remains, and if not written is understood. 2. Divide 15 X 6 X 7 by 10 X 18. Solution. — Cancelling the factor 5, com- mon to both dividend and divisor, we have . c, 1 in the dividend 3 in place of 15, and in the ~ divisor 2 in place of 10. We next cancel 3X6, or 18, common to both dividend and divisor, and have left -|, or Z\. 3. Divide 11 X 9 X 8 by 11 X 3 X 2. Ans. 12. 134. Rule for Cancellation.— C«-?^ceZ Uv the dividend and divisor all factors common to both, and then divide the product of the remaining factors of the dividend by the product pf the remaining factors of the diviso?\ PHOB T.EMS. 1. Divide 35 X 8 X 3 by 12 X 7 X 5. 2. Divide 42 X 15 X 80 by 75 X 24. Ans. 28. 3. Divide 96 X 63 X 5 by 72 X 35. 4. Di^dde 108 X 77 X 2 by 18 X 26 X H. Ans. 'J 1 :r 5. Divide 65 X 16 X 33 by 26 X 22 X 8. 6. How many are (300 X 45 X 6) -- (150 X 30 X 18) ? 7. How many barrels of beef, at 21 dollars a barrel, are worth as much as 14 tons of coal, at 6 dollars a ton ? Ans. 4. 8. How many bushels of corn, at 90 cents a bushel, will pay ibr 120 yards of cloth, at 15 cents a yard ? 9. I exchanged 90 bushels of potatoes, at 75 cents a bushel, for tubs of butter containing 54 pounds each, at 25 cents a pound. How many tubs of butter did I receive ? Ans. 5. 68 FA CTORING. SECTION XIII. AXALYSIS. 135. — Ex. 1. If 24 bushels of wheat cost 72 dollars, what v.ill 5 bushels cost? Solution. — If 24 bushels cost 72 dollars, 1 bushel will cost one twenty-fourth of 72 dollars, which is 3 dollars. If 1 bushel cost 3 dollars, 5 bushels will cost 5 times 3 dollars, which are lo dollars. 2. If 15 men can earn 75 dollars in a given time, how much can 7 men earn in the same time ? 3. When 7 hats cost 35 dollars, how much will 15 hats cost? 4. If 7 hats cost 35 dollars, how many hats will cost 75 dollars ? Solution. — If 7 hats cost 35 dollars, 1 hat will cost \ of 35 dollars, which is 5 dollars, and as many hats will cost 75 dollars as 5 dollars are contained times in 75 dollars, which are 15. 5. If 4 persons require 48 dollars' worth of provisions in a certain time, how many persons Avill require 108 dollars' worth in the same time ? 6. When 13 bushels of corn are worth as much as 39 bushels of oats, how many bushels of corn are worth as much as 27 bushels of oats ? 7. If 17 men can earn as much in one day as 51 boys, who are each paid 50 cents a day, how much can 1 man earn in a day? 8. If 9 barrels of flour are worth as much as 36 cords of wood, how many cords of wood are worth as much as 7 barrels of flour? 9. When 28 cords of wood are worth as much as 7 barrels of flour, how many cords of wood are worth as much as barrels of flour ? DEFINITION. 136. Analysis, in Arithmetic, i.s the process of stating, in regular order, the reasons for each step in the solution of a problem whose conditions require several computations. FA CTOBING. 69 WRITTEN JSXIDRCISES. Ex. 1. If 12 horses cost 2100 dollars, liow much will 21 horses cost at the same rate ? 12)2100 dollars. Soi^ution by Anai^ysis.-K 12 — - — - horses cost 2100 dollar.s, 1 horse will -^ ' ^ cost one twelfth of 2100 dollars, which 21 is 175 dollars. 1 ly n If 1 horse cost 175 dollars, 21 horses ^r-^ will cost 21 times 175 dollars, whicli are 3675 dollars. 367 5 dollars. Solution by CANCf:Li,ATioN. — If Or, 12 horses cost 2100 dollars, 1 horse 525 7 will cost one twelfth of 2100 dol- jnirin-y ~)-j- lars, and 21 horses will cost 21 =^S075 times as much. Indicating the work, ^^ and cancelling, we have 3675, the ^ same result. 2. When 45 tons of coal cost 270 dollars, what will 60 tons cost? Ans. 360 dollars. 3. If 180 quarts of oats be sufficient for 30 horses for a cer- tain time, how many quarts will be sufficient for 63 horses for the same time ? 4. If 15 men can do a piece of work in 54 days, in what time can 9 men do the same ? 5. What time will 48 men require to mow a field that 10 men can mow in 32 days ? Ajas. 6f days. 6. A cistern can be filled in 265 minutes by 5 equal pipes running into it. In what time could it be filled by 25 such pipes? Ans. 53^ minutes. . TEST QUESTIONS. 137. — 1. Upon what does the Value of a quotient depend? What effect upon the quotient is produced hy multiplying the dividend or dividing the divisor ? What effect by dividing the dividend or multiply- ing the divisor? What changes in both dividend and divisor do not affect the value of the quotient ? 2. How do you divide by using Factors? If there be one or more remainders, liow do you find the true remainder ? 3. What is Cancellation ? The rule for cancellation ? Analysis ? 70 REVIEW PROBLEMS. - SECTION XIV. REVIEW PROBLEMS. MENTAL EXERCISES. 138. — Ex. 1. Of what numbers are 2, 3 and 11 the prime factors ? 2. What are all the exact divisors of 66 ? 3. Name the composite numbers from 2 to 30. From 30 to 45. 4. Five 20's are how many 4's ? Seven 12's are how many 14's? 5. I sold 9 oranges at 8 cents apiece, and spent the money for melons at 18 cents each. Hoav many melons did I pur- chase ? 6. What common factors have 18 and 42? 7. What is the greatest number that will exactly divide 27 and 63 ? 8. When cheese is 15 cents a pound, and butter 35 cents, what is the least number of pounds of cheese that can be ex- changed for an exact number of pounds of butter ? 9. When bvitter is 45 cents a pound, how many pounds of sugar, at 18 cents a pound, will cost as much as 2 pounds of butter ? '^ 10. If 5 men can perform a piece of work in 14 days, in what time can 7 men do it ? 11. If 8 men can do a piece of work in 15 days, how many men can do it in 12 days ? 12. When coffee is 35 cents a pound, how many pounds of sugar, at 14 cents a pound, will cost as much as 2 pounds of coffee ? 13. If 6 men can do a piece of work in 14 days, how many men can do it in 7 days ? 14. I sold 11 pears at 6 cents each, and bought with the money, oranges at 3 cents each. How many oranges did I buy? -REVIEW PROBLEMS. 71 WRITTEN EXERCISES. 139.— Ex. 1. What are the prime factors of 56? 2. Which of the numbers 31, 98 and 101 are prime numbers ? 3. Find the greatest common divisor of 4165 and 686. 4. Which of the numbers 700, 575 and 335 have 25 for an exact divisor ? 5. Find all the exact divisors of 90. 6. What is the least common multiple of 8, 11 and 13? Ans. 1144. 7. What exponent will denote the number of times that 5 is taken as a factor in 3125 ? 8. I have 14 bushels of oats, 22 bushels of rye and 24 bushels of corn. What is the capacity of the largest sacks, of equal size, into which the whole may be put without mixing ? Ans. 2 bushels. 9. Three men start at the same time and place to walk in the same direction round a circle. A can make the circuit in 5 hours, B in 8 hours, and C in 10 hours. In what time after they start will they all be together again at the point of starting? Ans. 40 hours. 10. If 1350 men can build a bridge in 30 days, in what time can 1500 men build it? 11. When the cost of 42 pounds of rice is 294 cents, what is the cost of 28 pounds ? Ans. 196 cents 12. How many tons of ii'on can be bought for 95285 dollars, if 50 tons can be bought for 4250 dollars? Ans. 1121. 13. If 27 men can earn 2187 dollars in one month, how many men can earn 5832 dollars in the same time ? Ans. Tl. 14. If 18 men can sow a field of oats in 12 days, how long will it take 48 men to sow a field of the same size ? 15. How many pounds of tea, at 80 cents a pound, are equal in value to 6 bushels of wheat, Avorth 280 cents a bushel ? 16. I exchanged 15 pieces of cloth, each containing 30 yards worth 14 cents per yard, for 21 barrels of apples, each con- taining 3 bushels. How much did the apples cost me per bushel ? Ans. 100 cents. 72 FRACTIONS. SECTION XV. COMMO.Y FR. i CTIOjYS. 140. — Ex. 1. If an apple be cut into two equal parts, what part of the apple will one of the pieces be ? 2. What is one half of an apple, or one half of anything ? One third of an apple, or one third of anything ? 3. One of the three equal parts of an apple is what part of the whole ? Two of the three equal parts are what part of the whole ? 4. What is one fourth of anything? Two fourths? Three fourths ? 5. How many halves in a single thing or unit? How many thirds ? How many fourths ? 6. What is one fifth of a unit? One sixth ? One seventh? One eighth ? 7. What are two fifths of anything? Three fiftlis? Four fifths? 8. Which is the greater — a half or a third ? A third or r^ fourth ? A third or a fiftli ? 9. How are halves expressed by figures? Thirds? Fourths? Fifths? 10. What does "I" signify ? What does f signify ? FRACTIONS. 73 11. In f , what expresses the number of parts into which a unit has been divided ? What expresses the number of parts taken ? DEFINITIONS. 141. A Fraction is a number which expresses one or more of the equal parts into which a unit is divided. Thus, one half, two thirds, five fourths, etc., are fractions. 142. The Unit of the Fraction is 'the unit, or whole thing, which is considered as di^♦ded into parts. Thus, tlie unit of tiie fraction of an apple is one apple ; the unit of the fraction of a dollar is one dollar, etc. When no particular unit is named, the abstract unit 1 is understood. Thus, halves, thirds, etc., are understood to be halves, thirds, etc., of 1. 143. A Fractional Unit is one of the equal parts of the unit of the fraction. Thus, one half, one third, etc., is the fractional unit of halves, thirds, etc. 144. The Penominator of a fraction is the number which denominates or names the parts of the unit. Thus, four is the denominator of f . 145. The T^mnerator of a fraction is the number which numerates or numbers the fractional units taken. Thus, three is the numerator of |. 14(>. The Terms of a fraction are its numerator and denom- inator. 147. A CoKimwi Fraction is an expression of any number of parts of a unit, written by placing the numerator above the denomin*ator, with a line between them. Thus, f, wliich is an expression of two fifths, is a common fraction. 148. A Proper Fraction is one whose numerator is less than its denominator ; and an Improper Fraction is one whose nume- rator is not less than its denominator. Thus, f, f, etc., are proper fractions, and f, f, etc., are improper frac- tions. 74 FRA CTIONS. 119. A Mixed Number is a number expressed by an integer and a fraction ; as, 5f . 150. Similar Fractions are such as have the same denomina- tor ; as, I, I, etc. 151. Dissimilar Fractions are such as have different denomi- nators ; as, f , |, etc. 152. An Integer may be expressed fractionally by writing it as a numerator, with 1 as its denominator. Thus, 3 may be expressed f, and be rea^three ones, or three. 153. Fractions are read by pronouncing the number in the numerator, and then naming the parts denoted by the denom- inator. Thus, I is read one half; f, tivo thirds ; ^j, three twenty-firsts, etc. 151. Fractions may not only bo regarded as a number of parts of a unit, but as another method of indicating division. Thus, 4 may be regarded as f of 1 ; -|- of 3 ; or 3 divided by 5. 155. Principles. — 1. The value of a fraction is the quotient obtained hij dividing the numerator by the denominator. 2. The value of a fraction is less than 1 rvhen the numerator is less than the denominator. 3. The value of a fraction equals or exceeds 1 ivhen the nume- rator equals or exceeds the denominator. EXERCISES. 156. Name the kind of fraction, and read — 1. s 4. ft •J* 7. 1^. 10. ||. 5. 1 .? T4- 8. i^. 11. ^. ^. 6. ToTF- 9. 19t. 12. 36^ 3. Write in figures — 13. Nine tenths. 14. Seven eighths. 15. Eleven thirteenths 19. Six thousand one nineteenths. 20. How may 3 be expressed fractionally? 21. Is the value of ^ greater or less than 1 ? Of j%? 16. 21 seventy-seconds. 17. Three one-hundredths. IS. 1 7 t wo-hundrcd-twentieths. FRACTIONS. 76 SECTION XVI. REDUCTIOJV OF FRACTIOKS. 157, Reduction of Fractions is the process of changing their form of expression without changing their value. C^SE I, Fractions Reduced to Larger or Smaller Terms. 158. — Ex. 1. One third of an apple is how many sixths of an apple ? 2. What is \ expressed in terms twice as large ? 3. Express \ in terms twice as large, f in terms 3 times as large. 4. Two sixths of an apple are how many thirds of an apple? What is "I expressed in terms one half as large ? 5. Two eighths are how many fourths? Express -^ in terms one third as large. 6. Ten fifteenths are how many fifths ? Express -^ in terms one sixth as large. DEFINITIONS. 159. A fraction is reduced to Larger, or Higher, Terms when expressed in an equivalent fraction with larger terms. 160. A fraction is reduced to Smaller, or Lower, Terms when expressed in an equivalent fraction with smaller terms. 161. A fraction is in its Lowest Terms when expressed in terms which are prime to each other. 162. Principle. — Multiplying or dividing both terms of a fraction by the same number does not change its value. For, in the one case, as the number of parts is increased, their size is diminished ; and, in the other case, as the number of parts is dimin-* ished, their size is increased. Thus ■^=^. -^^^ = -^-<9 6^3 _2_^ 76 FRA CTIONS. WJtITTEN EXEJtCTSES. 163. — Ex. 1. Reduce f to forty-seconds. 4 X g 24_ Solution. — Since 42, the required denominator, is 6 7X5 42 times as large as 7, the given denominator, and, since multiplying both terms of a fraction by the same immber does not cliange its value, (Art. 162,) we multiply both terms of ^ by 6, which gives If, the fraction required. 2. Change f to an equivalent fraction whose numerator is 20. 3. Reduce |4 to its lowest terms. 24^2 12 12^3 _ 4 Solution.— Since dividing both terms of 43-^2 21' 21-^3 7 a fraction by the same number does not change the value of the fraction, we divide both terms of |f by 2, or cancel that factor in each term, which gives, as the fraction in lower terms, \\. Dividing both terms of if by 3 gives 4, which, since the terms are prime to each other, is the result required. 24^6 4 Second Solution. — Since 6 is the greatest common 42-^6 7 divisor of the terms of ff, we can obtain the lowest terms of the fraction by dividing both terms by that divisor, which gives f, the same result as at first obtained. 4. Reduce |^ to its lowest terms. 5. Change ff to its lowest terms. 164;. Rules for Reduction of Fractions to Higher or the Lowest Terms.— i. To reduce a fraction to higher temns, multi- ply both terms of the fraction hy such a number as will give the required term. 2. To reduce a fraction to its lowest terms, cancel in both terms all common factors, or divide both terms by their greatest common divisor. rnOBLEJTS. 1. Reduce ^ to forty-ninths. Ans. ff. 2. Reduce f and ^ to sixteenths. A)is. y^, \^. 3. Reduce f and | to twenty-seventlis. 4. Reduce ^, ^ and .^-f to one-hundred-fifths. FEA CTIONS. 77 Reduce to lowest terms — 5. 11^. . Ans. f. TS- 8. ^. Am. If. 9. Hi. Ans. If. 10 1702 ^^- 188G- 7 ^82 J „ „ 13 11. In what lower terms can ^ be expressed? J„, 1 2 8 6 4 orirl 2 C^SE II. Integers or Mixed Numbers Reduced to luiproper Fractions. 165. — Ex. 1. How many fourths in 1 orange? In 3 oranges ? In 6 oranges ? 2. How many thirds in 1 apple ? In 4 apples ? In 7 apples ? 3. How many fifths in 1 ? In 3 ? In 6 ? 4. How many sixths of a cake are 2 cakes ? Are 2^ cakes ? 3-|- cakes ? 3f cakes ? 5. How many sevenths in 1 ? In 1\1 In 5f ? WRITTEN EXERCISES. 166. — Ex. 1. Reduce 29 to sixths. 29 6 sixths Solution. — Since in 1 there are 6 sixths, in ■fn,, o/r//)? = ^~-^ 29 there must be 29 times 6 sixths, which are J-|^. 2. Reduce 18f to an equivalent improper fraction. r r 18~ = 18-\ — ; Solution. — Since in 1 there are |, in 18 there must be 18 times f, or J-f^ ; i|^ XS=—; ^ -f ^ = ^ and f are ^\ the fraction required. 3. Reduce 19 to fifths. Ans. ¥ 4. Change 41f to an improper fraction. A ns. 33 1 167. Rule for Reduction of Integers or Mixed Numbers to Improper Fractions.— MiUfiply the integer hy the given denomi- nator, and, if there he a fractional part, ad,d its nu- merator to the product. Tlie result ivritten over the given denominator will he the required frjoiction. 7* 78 FRACTIONS. PROBLEMS. Reduce to equivalent improper fractions — 1. 104f. Am. ^^. 2. 28f 3. 19i§i-. ^*^- W- 5. 42f. 6. 23^. Ans. ^-H- 7. 28H- 4. 1444. 8- 115tI?- ^^• 14 3 7 8 125 9. Change 36 to thirteenths and 41 to fourteenths. A n't 4_68 5J4 CASE III. Improper Fractions Reduced to Integers or Mixed Numbers. lg8,_Ex. 1. In four fourths of an orange, how many oranges ? In twelve fourths of an orange ? 2. How many apples in three thirds of an apple ? In twelve thirds of an apple? 3. How many ones in f ? In f ? In | ? In ^2 ? In \^ ? 4. How many cakes in i/ of a cake ? In ^3 of a cake ? WRITTEN EXERCISES. 169. — Ex. 1. Reduce J-f^ to an e(iuivalent integer or mixed number. ~ := XQ'y -^ g ^ 18~ Solution.— Since 9 ninths equal one, 167 ^ ^ ninths must equal as many ones as 9 is contained times in 167, or 18| times. Hence, ip ^^ 18|. 2. Reduce ^/ to an equivalent integer or mixed number. 170. Rule for Reduction of Improper Fractions to integers or Mixed Yi\xm\s^vs.— Divide the numerator by the djenoirhinv^tor . PROBLEMS. Reduce to an integer or mixed number — 1. ^|i. Ans. 129. 3. V^. Am. 2|i 4. ifi. Am. 151. 5. H^. 6. "tV- ^'i«- 28|i. 7. What is the value of ^^ dollars? Am. 31 doHan 8. How many miles are ^^^ miles ? FEA CTIONS. 79 C^SE IV, Dissimilar Fractions Reduced to Similar Fractions. 171. — Ex. 1. How many sixths of an apple is 1 third of au apple ? Are 2 thirds of an apple ? 2. Express f and ^ each as sixths. 3. Express \ and f each as sixteenths. 4. What is a common multiple of the denominators of \ and f ? 5. Express \, f and f each as twelfths. What is tlie least common multiple of the denominators of ^, f and ^ ? DEFINITIONS. 172. Fractions have a Common Denominator when their de- nominators are alike. 173. Fractions have the Least Common Denominator when their denominators are the smallest that they can have in common. 174. Fractions are said to be reduced to a common denomi- nator when they are changed to equivalent fractions with denominators alike. 175. Principles, — 1. A common denominator of hvo or more fractions is a covimon. multiple of their denominators. 2. The least common denominator of two or more fractions is the least common multiple of their denominators. WJRITTJEN EXERCISES. 176. — Ex. 1. Reduce f and ^ to equivalent fractions having a common denominator. 5^^ 15 Solution.— Since the denominator of 27 i'^ '^ times tlie ^ ^7 denominator of f, we multiply both terms of ~ by 3, which J_^ gives as its equivalent \^. Hence, W and 2:V are the fractions ' required. 2. Reduce ^4*, |- and -^ to similar fractions. Ans. f|, If and ^. 80 FHA CTIONS. 3. Reduce f and i to similar fractions. ^ ^ Solution. — Multiplying both terms of J by o, the denoni- 4 ~0 inator of \, we have if ; and multiplying both terms of i by Jl __ _i_ 4, the denominator of |, we have v.'V- Hence, | and ^ ^ ^| ^ ^^ and 2*jy, which are similar fractions. 4. Reduce -|^, f and \^ to equivalent fractions having the least common denominator. iy.12 _ 13_ 2x12' 24 5X3 15 Solution. — The least common multiple of the de- nommators 2, 8 and 12 is 24; hence, 24 is the least common denominator. \ reduced to twenty-fourths is || ; f is \\ ; and {\ is 8XS 24 11X2 22 ja^a ^~2l ^1 > lience, if, \\ and || are the fractions required. 5. Reduce yV, xi ^^^^ It *° ^^® least common denominator. J ,, c. 5 8 8 .5 5 177. Rules for Reducing Fractions to a Common Denominator. — 1. Multiply hotlh terms of oue or more of the frac- tions by any muinher that will malce the denomiivators alike. Or, 2. Multiply both temns of each fraction by the de- nomiivators of the other fractions. 178. Rule for Reducing Fractions to the Least Common Denom- inator. — Find, the least common multiple of all the denominators for the least coi^vmoii denominator, and multiply both terms of each fraction by such a num- ber as ivill reduce it to that denominatoj\ FJtOBT^EMS. Reduce to equivalent fractions having a common denomina- tor — 1 1 i 1 ^„ which, reduced to its lowest terms, ;= JLi ^* 2 5* 2. Reduce .35 to an equivalent common fraction. 240. Rule for Reduction of Decimals to Common Fractions.— Oinit the decimal point, ivj%te the denominator under the given numerator, and, if necessary , reduce the fractioji to its lowest terms. PMOBIjEMS. Reduce to common fractions in their lowest terms — 5. .08. A7i8.^. 6. 1.06. 7. .096. Ans. ^. 8. .006943. 9. Reduce 1.06 to a mixed number. Ans. 1-^. 10. Reduce 503.1875 to a mixed number. Ans. 503^. 11. Reduce .16f to an equivalent common fraction, 1fil-^-J.-lO_l S0LUTi0N.-.16f is equal to ^J, '^^ 3~ 100 ~ 100~ S00~~ 6 which, reduced (Art. 211), is \, the fraction required. 12. Reduce .lO/y ^^^ ^^ equivalent common fraction. 13. Reduce 5.3f to an equivalent mixed number. 14. Reduce .43-||- to an equivalent common fraction. A 'it 9 43 3 -^'''*'- 99 00' Cj^lSE III. 1. .0075. Ans. 4-3 0- 2. .375. 3. .03125. Ans. -^. 4. .3216. ^n-s. m. Common Fractions Rcdnced to Decimals. 241. — Ex. 1. In 1 half of a melon, how mauy tenths of a melon ? In 1 fifth of a melon ? 2. How many tenths in -I? Hundredths in |? 3. How many hundredths in ^? In |^? In f ? REDUCTION OF DECIMALS. 109 WRITTEN EXERCISES. 242. — Ex. 1. Reduce f to an equivalent decimal. 8)3.000 Solution.— f is i of 3. 3 is equal to 30 tenths ; i of 30 a^ry [~ tenths is 3 tenths, with 6 tenths remaining. 6 tenths are 60 hundredtlis; \ of 60 hundredths is 7 Iiundredths, with 4 hundredths remaining. 4 hundredths are 40 thousandths ; | of 40 thousandths is 5 thousandtlis. 3 tenths + 7 Iiundredths + 5 tliousaudths = .375. Hence, | is equiva- lent to .375. 2. Reduce ^-^ to an equivalent decimal. 243. Rule for Reduction of Common Fractions to Decimals.— JSe- duce the nuineTator to tciitlis, hundredtlis, etc., hy an- nexing ciphers ; divide the result by the denominator, and point off as many orders for decimals in the quotient, as there were ciphers anneHced. FROHIEMS. Reduce to decimals — 1. 2^. An8. .36. I 5. ||. Am. .95. 6. i 2. f. 3. |. Ans. 2.5. 7. -|t|. Am. .3216. Ans. .0078125. 1 128* 4. ^. An8. .09375. 9. Reduce 3f to a mixed decimal. oS_ ^i^ -^ ly r Solution. — 3|^3 and \. |is.75; hence, 4 4 ' 3f is 3 and .75, or 3.75. 10. Reduce 47y\ to a mixed decimal. Ans. 47.1875. 11. Reduce j^ to a decimal. 12. Reduce l6o|f to a mixed decimal. Am. 100.96. 13. Reduce .151 to a pure decimal. 14. Reduce l.OOf to a mixed decimal. 15. Reduce 503^ to a mixed decimal. 16. Reduce f to a complex decimal of two orders. o )^ nn Solution. — | is | of 2. 2 is equal to 20 tenths ; | of 20 — '- tenths is 6 tenths, with 2 tenths remaining. .6" 6"- 2 tenths are 20 hundredths; i of 20 hundredths is 6 hundredths, with 2 hundredths remaining. 10 110 BEBVCTION OF DECIMALS. Here the continual recurring of the same figure in the result, with a like remainder, shows that j has no equivalent pure decimal. In such cases, when the reduction has been carried to any desirable extent, we may express the remainder in the form of a common fraction, and the result will be a complex decimal ; or we may use the sign -|- to indicate the incompleteness of the result, as .66 + for .66|. 17. Reduce ^ to a complex decimal of four orders. 18. Reduce -|^ to a decimal of four orders. Ans. .1111 -]-. 19. Reduce y\ to a decimal of four orders. 20. Reduce ^^ to a complex decimal of five orders. 21. Reduce q-qVo ^o ^ decimal of six orders. Ans. .000444+. 22. Reduce -f-^-f to a mixed decimal of five orders. . TEST QUESTIONS. 244. — 1. "What is a Decimal Fraction? Why is it called decimal? How is a decimal, when written, distinguished from an integer? What is the denominator of a decimal ? 2. What do the figures at the right of the Decimal Point express ? From what are the names of decimal orders derived ? To what do the orders tenths, hundredths, thousandths, etc., correspond ? Of what does a pure decimal consist ? A mixed decimal ? A complex decimal? 3. How many units of any order in a decimal are equal to one of the order next higher? Why may integers and decimals form one expres- sion ? What is the rule for reading decimals ? For writing decimals ? 4. How are decimals reduced to a common denominator? Why does annexing a cipher to a decimal not change its value ? What is the rule for reducing decimals to common fractions ? 5. What is the Kule for reduction of common fractions to decimals ? How do you proceed when the common fraction has no equivalent pure decimal ? 6. What is a Number? What are figures? Wh;it is numeration? Notation ? What is tlie scale of numbers ? What is tlie scale of the ordinary system? What is termed the decimal system? (Art. 33.) 7. What is an Integer? A unit of an integer? (Arts. 3, 4.) A frac- tion? A fractional unit ? The unit of a fraction ? (Arts. 141, 143.) A decimal fraction ? How is a common fraction expressed by figures ? How ia a decimal expressed by figures ? How do integers and decimals cor- respond in expression ? (Art. 228.) REDUCTION OF DECIMALS. Ill SECTION XXV. ADDITIOJf AJfD SUBTRACTIOJ^ OF DECIMALS. 245. — Ex. 1. How many tenths are 5 tenths and 4 tenths? 2. How many tenths are -^ and y\ ? .4 and .3 ? 3. How many hundredths are y^^^ and y^-j^ ? .43 and .8 ? 4. How many hundredths are -^-^ less y^-^ ? .51 less .8 ? 246. Principle. — Decimals which are similar may he added or subtracted like integers. WRITTEN EXERCISES. 247.— Ex. 1. Add 13.634, 35.423 and 8.56. 13.6 3 A Solution. — Writing the numbers so that all the figures op: /(j)o of the same order stand in the same column, and adding as in addition of integers, gives 57.617, the sum required. The 8.56 is made similar to the other decimals by an- 8.560 57 .6 17 nexing a cipher. 2. From 963.75 subtract 585.125. 963.750 Solution. — Writing the numbers so that figures of the /- o/r i pp: same order stand in the same column, and subtracting as in subtraction of integers, gives 378.625, the difierence re- 378.625 qnired. The cipher annexed to the minuend is usually understood, and the subtraction performed in the same manner as if it were written. 3. What is the sum of 145.07, 3.476 and 11.05? 4. What is the difference between 56.77 and 7.899 ? 248. Rule for Addition and Subtraction of Decimals. — Write the Tiumhers so that figures of the same order shall he iiv'the same coluimi. Add or subtract in the same Tnanner as if the num- hers were integers, and place the decimal point at the left of the order of tenths in the result. 112 REDUCTION OF DECIMALS. FltOBXEMS. 1. What is the sum of .89, .269, 15.2 and .2? Ans. 16.559. 2. What is the sum of 11.35, 19, 3.41 and 100.678? 3. What is the difference between .9173 and .2138 ? Ans. .7035. 4. AVhat is the difference between 407 and 91.713? Ans. 315.287. 5. Required the value of 270.2 less 75.4075. 6. 450 + 376.004 + 1.08 + .76 + .05 = what number? Ans. 827.894. 7. .001 — .00099 = what number? Ans. .00001. 8. 100 — .10101 = what number ? Ans. 99.89899. 9. What is the sum of ten thousand one hundred one thou- sandths, ninety-nine, and eighty-nine thousand eight hundred ninety-nine hundred thousandths ? 10. What is the sum of 98.75 miles, 100.3655 miles and 15.7875 miles? 11. If 41.674 cubic feet of oak wood, or 64.693 cubic feet of white pine wood, will Aveigh a ton, how many more cubic feet are there in a ton of pine than in a ton of oak ? Am. 23.019 cubic feet. 12. Add seventy -five hundredths; eight, and sixty-seven hundredths ; seven, and three hundred fifty-five thousandths ; and thirty-one, and seven hundred thirty-five thousandths. Am. 48.51. 13. From three millions three, take three, and three mil- lionths. 14. In one field there are 31.175 acres ; in another, 9.1825 acres ; and in a third, 25.75 acres. How many acres in the three ? 15. What part of 1 must be added to 9.999999 to make 10? Ans. One millionth part. 16. From the sum of one and five hundredths, eighteen and thirty-five ten-thousandths, and four hundred four millionths, subtract eleven and ninety-nine ten-millionths. Am. 8.0538941. MULTIPLICATION OF DECIMALS. 113 section xxvi. multiplicjltiojY of decimals. 249. — Ex. 1. How many tenths of a dollar are 3 times 3 tenths of a dollar ? 2. How many tenths are 3 times 3% ? 4 times y^ ? 3. How many hundredths of a dollar are 7 times 6 hun- dredths of a dollar ? 4. How many hundredths are 7 times j^ ? 6 times .08 ? 5. How many hundredths in j^ X yo '• -6 X .4 ? 6. How many thousandths in jwo X 8 ? y^^^ X 1% ? 7. How many ones are 10 times y\ ? 10 times .6 ? 8. How many tenths are 10 times yf ^ ? 10 times .05 ? 9. How many hundredths are 9 times yto? How many tenths are 3 times .9 ? 250. Principles. — 1. Each removal of the decimal point one order to the right makes the value of an expr&ssion ten-fold. For, by the removal, each figure is made to express units of the next higher order. Thus, 16.4 = sixteen and four tenths ; and 164. = one hundred sixty- four, or ten times 16.4. 2. Each removal of the decimal point one order to the left makes the value of an expression one-tenth as large as before. For, by the removal, each figure is made to express units of the next lower order. Thus, 164. = one hundred sixty-four ; and 16.4 = sixteen and four- tenths, which is one tenth as much as 164. WRITTJEK EXERCISES. 251.— Ex. 1. Multiply .49 by 6. .A9 Solution. — 6 times 9 hundredths are 54 hundredths, or 5 6 tenths and 4 hundredths. Write 4 for the hundredths of the result, and reserve the 5 ^.94- tenths. 6 times 4 tenths are 24 tenths ; 24 tenths and 5 tenths are 29 tenths, or 2 ones and 9 tenths. Write 2 ones 9 tenths in the result, which gives 2.94, the product required. 114 MULTIPLICATION OF DECIMALS. 2. Multiply 7.6 by .06. 7" Q Solution. — .00 is the same as y^^j of 6 ; hence, .06 times 7.6 /^^P is the same as j^^ of 6 times 7.6. — 6 times 7.6 are 45.6, and j^u of 6 times 7.6 is j^q of 45.6, .4-56 which, found by removing the decimal point two orders to the left, is .456. Hence, 7.6 multiplied by .06 is .456. Or, the solution may be explained as follows : 6 hundredtlis times 6 tenths is 36 thousandths, or 3 hundredths and G thousandths. Write 6 in the thousandths' order in the result, and re- serve tlie 3 hundredths. 6 hundredtlis times 7 is 42 hundredths ; 42 hundredths and 3 hun- dredtlis are 45 liundredtlis, or 4 tenths and 5 hundredths, which we write in the result, and have, as before, .456. By the first explanation it appears that in decimals, as in common fractions (Art. 195), Multiplying by a fraction is the same as multiplying by its numerator and dividing the result by its denominator. By that process, when the factors are decimals. The number of decimal orders in the jiroduct is made as many as there are in both factors. 3. Multiply 1.704 by .35. . Ans. .5964. 4. Multiply .0051 by 51. Ans. .2601. 252. Rules for Multiplication of Decimals.— i. Multiply as in integers, and place the decimal point in the pro- duct, so that it shall have as many decimal orders as are contained in both factors. Or, 2. If the multiplier is a decimal, multiply hy its numerator and divide hy its denominator. PROBL EMS. Multiply — 1. .125 by .025. Ans. .003125. 2. 8.25 by 4.5. 3. 958 by .34. Ans. 325.72. 4. 500.83 by 121. Ans. 60600.43. 6. 4.6337 by 100. ^»s. 463.37. 7. 4.6337 by 1000. 8. 3.007 by '.36. Ans. 1.08252. 9. .285 by .003. 10. 3.84062 by 70000. 5. 1.007 by .0041. I Ans. 268843.4. DIVISION OF DECIMALS. 115 11. What is the product of 9.688 by .2|? Ans. 2.6642. 12. What will 56 pounds of coffee cost, at .37|- of a dollar per pound ? 13. If a box must have a capacity of 24.958 cubic feet to contain a ton of anthracite coal, what must be the capacity of a box that will contain 4.5 tons ? 14. What is the weight of 128 cubic feet of common soil; if the weight of a cubic foot is 137.125 pounds ? Ans. 17552 pounds. SECTION XXVII. DIVISION OF DECIMALS. 253. — Ex. 1. In 9 tenths of a dollar hoAV many times 3 tenths? 2. How many times 3 tenths is -^^ ? Is .9 ? 3. In 42 hundredths of a dollar how many times 6 hun- dredths of a dollar ? 4. How many times is .8 contained in .48 ? .3 in .06 ? 5. What is 6 ^ 10 ? .6 - 10 ? .6 ^.6 ? .06 ^ .6 ? WniTTEN EXERCISES. 254— Ex. 1. Divide 1.345 by 5. 5) 1.34-5 Solution.— 1.345 is 1345 thousandths; \ of 1345 thoii- .269 S'indths is 269 thousandths, or .269. 2. Divide 1.46 by .25. 25)146.00(5.84 Solution.-.25 is the same as fi^, or ^^^ X 25. 19^ First, divide 1.46 by j^q-, by multiplying by ■ 100, which is done by removing the decimal '^-''^ point in the dividend two orders to the right, 200 making the dividend 146. 146 divided by 25 is inn 5, with 21 ones remaining. 21 ones are 210 . tenths, which divided by 25 is 8 tenths, with 10 tenths remaining. 10 tenths are 100 hun- dredths, which, divided by 25, is 4 hun- dredths. Hence, 1.46 divided by .25 = 5.84. Or, Multiplying both the divisor and dividend by 100, the denominator of the divisor, making the divisor a whole number, and then dividing, we have the quotient 5.84, as before. 116 DIVISION OF DECIMALS. By the first explanation it appears that in decimals, as in common fractions (Art. 206), Dividing by a fraction is performed by multiplying by its de- nominator and dividing the residt by its numerator. By that process in the division of decimals, The number of decimal orders in the quotient is made as many as there are in the dividend, less the number in the divisor. 3. Divide .5964 by .35. Ans. 1.704. 4. Divide .2601 by 51. Ans. .0051. 255. Rules for Division of Decimals.— i. If the divisor is an ijvteger, divide as in integers, and point off as many decimal oi^ders in the quotient as there are decimal orders in the dividend. '2. If the divisor is a decimal, mahe it an integer by lyvoving the decimal point to the right, and move the decii^val point in the dividend as many orders to the right, and then divide. Or, 3. If the divisor is a decimal, multiply the dividend by its denominator and divide the i^esult by the nu- merator. When the divisor is 10, 100, etc., the division may be performed sim- ply by removing the decimal point in the dividend as many orders to the left as there are ciphers in the divisor. PROBLE31S. Divide — 1. .456 by .06. Ans. 7.6. 2. 463.37 by 100. 3. 1.606 by 44. Ans. .0365. 4. 1.08252 by .36. 5. 825.72 by 958. Ans. .34. 6. 463.37 by 1000. 7. 21 by 56. Ans. .375. 8. .1606 by 44. Ans. .00365. 9. 6 by .006. Ans. 1000. 10. 172.8 by .0144. 11. What is the value of 21.17-- .0073? 12. What is the value of .015625 -^ 25 ? Ans, .000625. 13. What is the value of 2.15565 h- 1.05 ? 14. What is the quotient of 3.672 by .81 ? Ans. 4.533+. DIVISION OF DECIMALS. 117 When, as in tlie last problem, the division will not terminate, the sign + may be annexed, to indicate that the quotient is not complete. The quotient is then called an approximate quotient. 15. What is the approximate quotient of 45.5 divided by 2100 ? Ans. .0216+. 16. If you should travel 787.5 miles in 210 hours, at what rate per hour Avould you travel ? 17. If .0001 is a dividend and 1.25 a divisor, what is the quotient ? Ans. .00008. 18. Divide three thousand one hundred tAventy-five mil- lionths, by one hundred twenty-five thousandths. 19. If 375 bushels of potatoes be Avorth as much as 7.5 tons of hay, hoAV many bushels are Avorth as much as 1 ton? Ans. 50. 20. A tract of land containing 125.4 acres was sold for 7586.7 dollars ; Avhat was the price per acre ? 21. How many casks, each containing 31.5 gallons, can be filled from a vat containing 368.25 gallons ? Ans. 11 ; and 21.75 gallons remain. 22. If 16.284 cubic feet of fire-bricks weigh one ton, how many loads, of a ton each, Avill a pile of such bricks, contain- ing 333.822 cubic feet, make? Ans. 20 ; and 8.142 cubic feet remain. TEST QUESTIONS. 256. — 1. What kind of decimals can be added or subtracted? What is the rule for addition or subtraction of decimals ? What does the decimal point in the result of addition or subtraction mark ? 2. In \Vhat simple way may the value of a decimal expression be MULTIPLIED by 10 ? Why does the removal of the decimal point in an expression one order to the right make the value expressed tenfold ? What are the rules for multiplication of decimals ? 3. In what simple way may the value of a decimal expression be biviDED by 10 ? Why does the removal of the decimal point in an ex- pression, one order to the left, make the value expressed one tenth as large ? What are the rules for division of decimals ? When is a quo- tirtit called an approximate quotient? lis UNITED STATES MONEY. UJ^ITED STATES MOXEY. 257. Money is a measure of value used as a medium of trade. 258. Coin, or Specie, is metal stamped, and authorized by government to be used as money. 259. Paper Money consists of Notes, issued by banks and by the Treasury of the United States, as substitutes for coin. Treasury Notes of less face-value than are called Fractional Currencv. 200. Currency is the Coin and Notes in circulation as money. Bidiion is uncoined gold or silver. An Alloy is a baser metal mixed with a finer. A Token is a coin whose intrinsic value is less than that assigned it by law. 261. United States Money is the currency of the United States. UXITED STATES MONET. 119 TABLE. 10 mills (in. J are 1 cent . . .ct. or c. 10 cents " 1 dime d. 10 diines, or 100 cents, " 1 dollar $. 262. The Coins of the United States are made of gokl, silver, nickel and bronze. The Gold Coins are the fifty-dollar piece, double eagle or twenty-dollar piece, eagle or ten-dollar piece, half-eagle, quarter- eagle, three-dollar piece and dollar. The Silver Coins are the dollar, half-dollar, quarter-dollar, dime, half-dime and three-cent piece. The Nickel Coins are the five-cent piece and three-cent piece. The Bronze Coins are the two-cent piece and cent. The gold coins are made of 9 parts of pure gold and 1 part of an alloy consisting of silver and copper. The silver coins are made of 9 parts of pure silver and 1 part of copper. The nickel coins are made of 75 parts of copper and 25 parts of nickel. The bronze coins are made of 95 parts of copper and 5 parts of zinc and tin. 263. Canada Money, or the money of the Dominion of Canada, consists, like United States money, of dollars and cents. Of this money, 100 cents are 1 dollar. The Canada coins are the twenty-cent, ten-cent and five-cent pieces, made of silver, and the cent, made of bronze. 264. The Dollar is the principal unit of United States money. Dimes, cents and mills may be written respectively as tenths, hundredths and thousandths of dollars, and, when decimally expressed, may be separated from dollars by the decimal jioint. 265. Dimes, or tens of cents, are commonly regarded as a number of cents. Thus, 15 dollars 3 dimes 6 cents 7 mills are written $15,367, and read fifteen dollars thirty-six cents seven mills. 266. Any decimal of a dollar less than a cent may be read as a decimal of a cent. Thus, $42.5025 may be read forty-two dollars fifty and twenty-five hun- dredths cents. 120 USITED STATES MONEY. WRITTES EXERCISES. 267. Write and read — 1. $2.45; $17.17; $43.47; $.95; $60. 2. $.05; $.005; $.555; $.606; $6.06. 3. $.0556; $.7585; $.7008; $70.08756. 4. $300.30; $45,303; $90,909; $40.0025. Write— 5. Fifty cents ; five cents five mills. 6. Two dollars twenty-five cents ; thirty dollars four cents. 7. 11 dollars 37 cents; 213 dollars 73 cents. 8. 7 mills ; 31 cents 5 mills ; 14 cents 6 mills. 9. 3 dollars -^^ ; 15 cents 5 mills ; 7 dollars 5 mills. 10. 38 and 5 tenths cents ; 17 dollars 17 cents ; 7 tenths mills. 11. 3 and 3 tenths cents ; 15 and 75 hundredths dollars. 12. 63 dollars 63 cents ; 6 and 5 tenths mills. 13. 600 dollars 6 and 6 tenths cents. SECTION XXIX. REDUCTIOJ^ OF UjYITED STATES MOJ^EY. 268. — Ex. 1. How many mills are 9 cents? Are 11 cents? 2. How many cents are $3 ? Are $5 ? Are $6 ? 3. In $1.15, how many cents? In $2.25 ? 4. In 90 mills, how many cents? In 110 mills? 5. How many mills are 5 dimes ? Are $5 ? 6. How many dollars are 300 cents ? Are 115 cents? Are 500 cents ? Are 5000 mills ? DEFINITIONS. 269. Denomination is the name of the unit expressing a number. Of two denominations, the higher is that which expresses the greater value, and the loicer is that wliich expresses tlie less value. 270. Reduction is the j^rocess of changing a number to an equivalent number of a different denomination. UNITED STATES MONEY. 121 }r KITTEN EXERCISES. 271. — Ex. 1. Reduce 56 ceuts to mills. ^/' V in--^pn Solution. — Since 1 cent is 10 mills, 56 cents must be 56 times 10 mills, or 560 mills. 2. Reduce $43 to cents. /o^ -t nn — /r in SOLUTION. — Since 100 cents are $1, 2546 254-6^ 100^ 25. 4.b , ,, , „ /' , cents must be as many dollars as 100 cents are contained times in 2546 cents, which are 25.46 times. Hence, 2546 cents are $25.46. 7. Reduce 43000 mills to dollars. f^nnn— mnn— /e dividend of that number?. What must the least common multiple of two or more numbers contain? (Art. 122.) COMPOUND NUMBERS. 149 SECTION XXXV. BEDUCTIOJi OF COMPOUjYD A'UMBUES. 358. — Ex. 1. How many inches in 2 feet 6 inches? In" 3 feet 7 inches ? In 2 yards 1 foot ? 2. How many feet in 30 inches ? In 43 inches ? How many yards in 84 inches ? 3. How many gills in 2 quarts 1 pint ? In 1 quart 1 pint 1 gill? In 3 quarts 1 pint? 4. How many quarts in 20 gills? In 13 gills? In 28 gills? 5. How many ounces in 5 pounds 6 ounces? In 3 pounds 8 ounces? In 10 pounds 10 ounces? 6. How many pounds in 86 ounces? In 56 ounces? In 170 ounces? DEFINITIONS. 359. A Simple Denominate Number is a number expressed in units of only one denomination. Thus, 3 yards, and 2 days, are each a simple denominate number. 360. A Compound Denominate Number is a number expressed in units of more than one denomination. Thus, 2 feet 6 inches, 4 days 6 hours, are each a compound denominate number. 361. Reduction of Denominate Numbers is the process of changing them to equivalent numbers of a different denomi- nation. 362. Reduction Descending is the process of changing a number to an equivalent number expressed in units of a lower denomination. 363. Reduction Ascending is the process of changing a number to an equivalent number expressed in units of higher denominations. 364. Principle. — BeducMon descendinc/ is performed by mul- tiplication, and reduction ascending is performed by division. 13* 160 COMPOUND NUMBERS. cj^se; I. Reduction Desceudiug-. 365. — Ex. 1. How many quarts in 3 pecks 7 quarts? In 3 pecks 5 quarts ? In 1 bushel 1 peck ? 2. How many inches in 3 yards 2 feet ? In 4 yards 1 foot ? 3. How many pennyweights in 5 ounces 11 penny^veights ? In 4 ounces 15 pennyweights ? 4. How many days are 9 weeks 5 days ? 5. How many inches in ^ of a yard ? Solution. — Since in 1 yard there are 3 feet, in | of a yard there must be f of 3 feet, or ^^ of 1 foot, whicli equals f of a foot. And since in 1 foot there are 12 inches, in | of a foot there must be f of 12 inches, or 20 inches. Hence f of a yard are 20 inches. 6. How many pounds in | of a ton ? In -J-j^ of a ton ? 7. What part of a quart is -|- of a gallon ? f of a gallon ? WRITTEN EXEBCISKS. 366. — Ex. 1. How many pints are 13 gal. 2 qt. 1 pt. 13 gal. 2 qt. 1 pt. ^ SoLrTioN. — Since 1 gallon is 4 ■^rz" . -» o 7 quarts, 13 gallons must be 13 oz ISO. qt. in lo qal. .-a . -o * a 3 ^ times 4 quarts, or o2 quarts ; and -^ 52 quarts + 2 quarts are 54 quarts. 54 No. qt. in 13 gal. 2 qt. Since 1 quart is 2 pints, 54 Q quarts must be 54 times 2 pints, or 108 pints; and 108 pints -j-1 108 No. pt. in 13 gal. 2 qt. pj^t are 109 pints. Hence, 13 gal. 1 2 qt. 1 pt. are 109 pints. 109 No. pt. in 13 gal. 2 qt. 1 pt. 2. How many yards are 53 mi. 132 rd. 4 yd. ? 3. How many pounds are 5 T. 3 cwt. 15 lb.? 367. Rule for Reduction Descenii\nq.— Multiply the number of the highest dcnmnhiatioii given, by that number of the 7iext lower which equals one of the higher, COMPOUND ^'^UMBEIiS. 151 a7id to the product add the number, if any, of the Joiver denomination. Reduce this result in like manner, and so proceed until the given numher is reduced to the required denomination. PnOBLJEMS. 1. How many square yards are 37 A. 132 sq. rd. ? 2. How many quarts are 308 bu. 1 pk. 6 qt. ? 3. How many quarters are iS% yd. 1 qr. ? 4. How many links are 2 mi. 40 eh. 25 li. ? 5. How many gills in 16 hogsheads? 6. How many ounces in 1 long ton? 7. How many grains are 7 oz. 19 pwt. 13 gr. ? 8. How many square rods in a quarter section of land ? 9. How many square rods in 2 A. 5 sq. eh. 8 sq. rd. ? 10. How many cubic feet are 20 cd. 6 cd. ft. ? 11. How many seconds are 29 d. 12 h. 44 min. 3 sec. ? 12. How many pints are f of a bushel? 9 o a SOLUTIOK.— Since 1 bushel is 4 - hu. = jX4 pk. = - ph. pecks, f of a bushel must be | of 4 pecks, or | of a peck. - p)k. = — X.8qt.=-~ qt. Since 1 peck is 8 quarts, | of a peck must be | of 8 quarts, or y of J- qt. = -r^X 2 pt. = -— p^. a quart. J Since 1 quart is 2 pints, ^-^ of a -J- pt. = i*5- 2^i' quart must be %* of 2 pints, or ip of a pint = 25f pints. 13. How many pounds are y^g- of a ton ? 14. How many gills are f of a hogshead? 15. What part of a second is yiroToTo- ^^ ^ ^^^J ^ 16. Reduce .0525 cwt. to ounces. Solution. — .05^5 art. = .0525 X 100 Ih. = 5.25 Ih. = 5.25 X 16 oz. = 84 oz. 17. Express .09375 of an acre in square rods. 18. Express .7375 of a pound Troy in penny^veights. 152 COMPOUND NUMBERS. 19. Reduce | of a rod to lower integers. Solution.— Since 1 rod is 5| jXS^ =^=3~ = No. of yd. -^^^'^'' '^ o^ ^ ^°^ ™"^^ be f of S 2 16 16 •> ^ b\ yards, or S^V yards. -^X3 =^ = l^ = No. of ft. ^'^'' ^ rV'f/f ' "'^ '^ ^ ^^ 16 16 J J yard must be y^ of 3 feet, or ly\ ^ -^y -17) 60 qS ,1- , . feet. io 16 4 >> bince 1 foot is 12 inches. T? J ,^ . of a foot must be y\ of 12 -rd.=3yd. Ift.S-in. inches, or 3f inches. Hence, f of a rod = 3 yd. 1 ft. 3| in. 20. Reduce ^ of a day to a compound number. 21. Express f of a hogshead as a compound number. 22. Express y\ of a mile as a compound number. 23. Reduce .7375 of a pound Troy to a compound number. Solution.— .757J lb. = .7375 X 12 oz. = 8.85 oz. ,- .85 oz. = .85 X 20 pict. = 17 pwt. Hence, .7375 oi 9.T^o\inA = 8 oz. 17 jnct. 24. Reduce .5625 of a day to a compound number. 25. How many pints are .015625 of a bushel? 26. Reduce .7625 of a degree to lower integers. 27. Express the value of 3.076 cubic yards as a compound number. 28. Express the value of 19.742 acres as a compound number. cAsE II. Reduction Ascending. 368. — Ex. 1. How many pecks in 31 quarts ? In 29 quarts? How many bushels in 40 quarts ? 2. How many yards in 96 inches ? In 60 inches ? 3. How many ounces in 71 pennyweights? In 63 penny- weights ? 4. How many yards in 20 inches ? Solution. — Since 12 inches are 1 foot, there are in 20 inches as many feet as y'j of 20, or J of a foot. Since 3 feet are 1 yard, tliere are in | of a foot as many yards as \ of f, or § of a yard. 5. How many tons in 14 hundred-weight? 6. How many gallons in 5- "f « quart? COMPOUND NUMBERS. 163 WMITTJEN EXERCISES. i 369. — Ex. 1. How many gallons are 109 pints? 2)109 Solution.— Since 2 pints are 1 , quart, there must be one half as 4.JOUf 1 pi. many quarts as pints, or 54 quarts, 13 gal. . . 2 at. with a remainder of 1 pint. Since 4 quarts are 1 gallon, there 10 J pt. = IJ gal ^ qt. 1 pt. ^^^^ y^^ ^^^ foyj.jjj ^j, ^^^^^. gaiio„s as quarts, or 13 gallons, with a remainder of 2 quarts. Hence, 109 pt. = 13 gal. 2 qt. 1 pt. 2. How many miles are 14400 rods? 3. How many tons are 10315 pounds? 370. Ruie for Reduction Ascending. — Divide the given number by that number of its rl enomination jvhich equals one of the next higher, and lujnte tlie remain- der, if any. Divide the quotient in like manner, and so continue until the given number is reduced to the required de- nomination. Tlie last quotient, with the remainders , if any, written in their order froin the highest to the lowest, will be the required result. Reduction Ascending and Reduction Descending, being performed by opposite processes, are proofs of each other. PItOBT.EMS. 1. How many acres are 183073 square yards? 2. How many bushels are 9870 quarts? 3. How many yards are 275 quarters ? 4. How many miles are 20025 links ? 5. How many hogsheads are 32256 gills ? 6. How many tons are 35840 ounces ? 7. How many ounces are 3829 grains ? 8. How many quarter sections of land are 25600 square rods? 154 COMPOUND NUMBERS. 9. How many acres, square chains and square rods are 408 square rods ? 10. How many cords are 2556 cubic feet? 11. How many days are 2551443 seconds? 12. How many bushels are 25f pints ? Solution. — Since 2 pints are 1 quart, ^g~ r= — — No. of pt. tliere must be one half as many quarts 2^g ^ as pints, or %^ of a quart. ■z~-^ 2 = -r~ ~ ^'^- ^f Q^- Since 8 quarts are 1 peck, there must g, g be one eightli as many pecks as quarts, -y^ S = j=No. of 2)k. or I of a peck. S ^ Since 4 pecks are 1 bushel, there must 5 ^' -^^ S ^^ ^J be one fourth as many bushels as pecks, or f of a bushel. 13. What part of a ton is 375 pounds? 14. What part of a hogshead is 1344 gills? 1 5. What part of a day is -^^ of a second ? 16. What decimal of a hundred-weight is 84 ounces? Solution.— 54 nz. = 84 -^ 16, or 5.25 lb. = 5.25 ~ 100, or .0525 cwt. 17. What decimal of an acre is 15 square rods? 18. What decimal of a pound Troy is 177 pennpveights ? 19. Reduce 3 yd. 1 ft. 3f in. to a fraction of a rod. S~ tn. = ~~r of an in. ; -- ^ i^ = -r ; nence, 3- m. = -- ft 4 4 •' 4 16 '4 16 '' lift-=§ of a ft.; jj*S=i: /.e«ce,iji/(. = ^| yd. 3k' y<^- = f »/ « !/* '■ iF f = I '■ '--• ■?i y^- = I '■''• Solution. — Since 12 inches are 1 foot, there must be one twelfth as many feet as inches, or y\ of a foot. Since 3 feet are 1 yard, there must be one third as many yards as feet, or -^-^ of a yard. Since 5^ yards, or V of a yard, arc 1 rod, there must be j-j- as ni;uiy rods as yards, or f of a rod. 20. What fraction of a day is 16 h. 36 min. 55y\ sec. ? 21. Wliat fraction of a hogshead is 39 gal. 8 pt. 3 gi. ? 22. What fraction of a mile is 85 rd. 1 yd. 2 ft. 6 in. ? COMPOUND NUMBERS. 155 23. Reduce 8 oz. 17 pwt. io a decimal of a pound. Solution.— i"i>-(?. = ir -^ 20, or .85 oz. ; S.S5 oz. = S.SS -^ 12, or .7375 lb. 24. Reduce 13 h. 30 rain, to a decimal of a day. 25. Reduce 1 pint to a decimal of a bushel. 26. Reduce 45' 45" to a decimal of a degree. 27. Express 3 cu. yd. 2 cu. ft. 89.856 cu. in. as a mixed deci- mal of a cubic yard. 28. Express 19 A. 118 sq. rd. 21.78 sq. yd. as a mixed deci- mal of an acre. CJ^SE III. One Compound Number Reduced to the Fraction of Another. 371. — Ex. 1. How many feet are 3 yards 2 feet? Are 5 yards 1 foot? 2. What fraction of 16 feet is 1 foot? Is 11 feet? 3. Reduce 3 yards 2 feet to a fraction of 5 yards 1 foot. 4. What fraction of 4 pounds 3 ounces is 2 pounds 5 ounces ? WBITTEIf EXEBCISES. 372.— Ex. 1. Reduce 3 wk. 5 d. to a fraction of 11 wk. 5 d. 1 h. r, , - 7 ^,T / 7 Solution. — Since only similar 3 wk. O d. = 624- fi- r, u J / A ^ ^ numbers can be compared (Art. 11 ivk. 5 d. 1 h. := 1969 h. 214), we reduce each of the given numbers to hours, and have as their equivalent 624 hours and 19G9 hours. Since 1 hour is ^^^-^ of 1969 hours, 624 hours must be y%W of 1969 hours. Hence, 3 wk. 5 d. are j^^-^g of 11 wk. 5 d. 1 h. 2. Reduce 5 mi. 40 rd. to a decimal of 8 mi. 20 rd. 5 mi. 40 rd. = 1640 rd. Solution. — 5 mi. 40 rd. are |f f g = 8 mi. 20 rd. = 2580 rd. j%% of 8 mi. 20 rd. ; and T^y = .635 + . i/^ /n oci Hence, 5 mi. 40 rd. are .635 +of 8 mi. ^^ = -^=.^55+ 20 rd. 2580 129 3. What fraction of 6 gal. 1 qt. 1 pt. is 2 gal. qt. 1 pt. ? 156 COMPOUND NUMBERS. 373. Rules for Reduction of one Cempound Number to the Frac- tion of another. -i. Reduce both of the given nuirvbers to the same denomination ; and then make the nxvmher denoting the part the numerator-, and that denoting the whole the denominator of the fraction required. ^. When the fraction required is a decimal, reduce the comjyion fractioji thus found to a decimal. PItOBZEMS. 1. What fraction of 25° 42' 40" is 7° 42' 48"? 2. "What decimal will express the relation of 5 cwt. 91 lb. to 2 T. 7 cwt. 28 lb. ? 3. From a farm containing 170 A. 16 sq. rd., I sold 37 A. 128 sq. rd. What part of the farm did I sell ? Ajis. |. MISCELLANEOUS PROBLEMS. 374. — 1. What is the cost of .6725 of a hundred-weight of butter, at 40 cents per pound ? 2. What decimal of a hundred-weight of butter, at 40 cents per pound, can be bought for $26.90 ? 3. What Avill 1 hogshead 4 gallons 1 quart of wine cost, at $5 per gallon ? 4. How much wine, at $5 per gallon, can be bought for $336.25 ? 5. The distance between two places on the same parallel of latitude is 17° 30' ; how far apart are they, a degree in that latitude being 54 miles ? Ans. 945 miles. 6. A boy has 1 pk. 6 qt. ^ pt. of chestnuts ; what part of a bushel has he ? 7. How many ounces of gold weigh as much as 4 pounds of lead? Ans. b^. 8. What decimal of a ton of nails, at 5 cents a pound, can be bought for $2.40 ? 9. A grocer has 8316 eggs to pack in 11 boxes ; how many dozen must he pack in each box? Aiis. 63. COMPOUXD NUMBERS. . 157 10. What will 23 A. 120 sq. rd. of land cost, at $.50 per square rod ? Ans. $1900. 11. How many sheets of paper are 12 reams 5 quires 18 sheets? 12. What number of silver spoons, each weighing 1 oz. 9 pwt., can be made from 2 pounds of silver ? TEST QUESTIONS. 375. — 1. What is a Number? A denominate number? A simple denominate number? A compound denominate number? 2. What is Reduction ? How do reductions descending and ascending differ ? Wliich is performed by multiplication ? Which by division ? 3. What is Cancellation ? Upon what principle does -cancellation depend? (Art. 129 — 3.) How is a fraction reduced to its lowest terms? Upon what principle does the process depend? (Art. 162.) 4. In what two ways may a fraction be multiplied by an integer? How may a number be multiplied by a fraction ? (Arts. 189-195.) 5. In what two ways may a fraction be divided by an integer? In what two ways may a number be divided by a fraction ? (Arts. 203-200.) SECTION XXXVI. ABBITIOJf OF COMPOU^'D LUMBERS. 376.— Ex. 1. What is the sum of 2 T. 15 cwt. 25 lb. ; 3 T. cwt. 64 lb. ; and 7 cwt. 16 lb. ? Solution. — Since only units of like kind 2 T. 15 Clot. 25 lb. can be added (Art. 46 — 1), write the numbers so 3 64- *'^^*' ""Its of the same denomination shall stand ,y 1^ in the same column. Begin at the right, and add the numbers of b i. o civt. o lb. each denomination in the order of tlie de- nominations. The sum of the pounds is 105 lb., or 1 cwt. 5 lb. Write the 5 lb. as the pounds of the sum, and add the 1 cwt. with the sum of hundred-weights. The sum of the hundred-weights is 23 cwt., or 1 T. 3 cwt. Write the 3 cwt. as the hundred-weights of the sum, and add the 1 T. with the column of tons. The sum of the tons is 6 T., which we write as the tons of the sum. Therefore, 6 T. 3 cwt. 5 lb. is the sum required. 14 158 COMPOUND yVMBEBS. 2. What is the sum of 810 yd. 1 ft. 10 in. ; 617 yd. 2 ft. 11 in. ; 85 yd. 2 ft. 8 in. ; 679 yd. 5 in. ; and 6 yd. 3 in. ? 3. What is the sum of 3 lb. 9 oz. 18 pwt. 11 gr. ; 1 lb. 4 oz. 19 pwt. 20 gr.; and 1 oz. pwt. 23 gr.? 377. Rule for Addition of Compound Numbers.— TFV'^^e the numbers so that units of the same denomination shall stand in the same column. Begin ivith the lowest denomination, and add the numbers of each denomination separately . If the sum is less than one of the next higher denomina- tion, ivrite it as a part of the required result. If the sum is equal to or exceeds one or more units of the next higher denomination, write the excess _ if any, as a part of the required result, and add the number of units of the higher denomination with the numbers of that denomination. FR'OBZEMS. 1. What is the sum of 13 lb. 6 oz. 11 pwt. 9 gr. ; 1 lb. 4 oz. 13 pwt. 20 gr. ; and 1 oz. pwt. 13 gr. ? 2. Find the sum of 4 gal. 1 qt. 1 pt. 1 gi. ; 4 gal. qt. 1 pt. 3 gi. ; 5 gal. 3 qt. pt. 2 gi. ; and 10 gal. 2 qt. 1 pt. 4 gi. 3. Find the sum of 71° 9' 59.5"; 20^ 24' 18.4"; and 19° 30' 34". 4. Find the sura of 46 bu. 2 pk. 6 qt. 1 pt. ; 43 bu. 2 pk. 2 qt. 1 pt. ; 86 bu. 1 pk. 3 qt. ; 68 bu. 3 pk. 1 qt. 1 pt. ; 76 bu. 2 pk. 3 qt. ; and 69 bu. 2 pk. 1 qt. 1 pt. Ans. 391 bu. 2 pk. 2 qt. 5. Find the sum of 30 A. 120 sq. rd. ; 42 A. 60 sq. rd. ; 80 A. 20 sq. rd. ; and 150 sq. rd. .4ns. 154 A. 30 sq. rd. 6. What is the sum of 100 rd. 5 yd. 2 ft. ; 150 rd. yd. 2 ft. ; and 105 rd. 3 yd. 1ft.? 100 rd. 5 yd. 2jt. 150 2 Solution.— Since 1 yd. = 1 ft. 105 3 1 6 in., we may substitute this vahie ' 'i for the \ yd., and thus obtain, as 1 mi. 36 rd. 3z yd. 3 ft. .,„ expression of the result, 1 mi. Or, 3G rd. 4 yd. ft. G in. 1 vii. 36 rd. 4 yd. ft. 6 in. COMPOUND NUMBERS. 169 7. What is the siim of 2 A. 120 sq. rd. 10 sq. yd. ; 3 A. sq. rd. 12 sq. yd. ; and 140 sq. rd. 20 sq. yd. ? Ans. 6 A. 101 sq. rd. llf sq. yd. Or, 6 A. 101 sq. rd. 11 sq. yd. 6 sq. ft. 108 sq. in. . 8. What is the sum of 66 y. 99 d. 8 h. 50 rain. ; 9 y. 1 d. 2 h. 57 min. ; 6 y. 70 d. 1 h. ; and 5 h. 50 min. ? 9. What is the sum of 13 cu. yd. 8 cu. ft. 1030 cu. in.; 20 cu. yd. 11 cu. ft. 903 cu. in.; and 107 cu. yd. 11 cu. ft. 1240 cu. in. ? Aiu. 141 cu. yd. 4 cu. ft. 1445 cu. in. 10. Find the sum of | of a mile and |^ of a rod. Solution.- Since | mi. = 177 rd. 4 yd. Oft. 10 in. and i'>-d-= 4 2 6J I mi. + I rd. ^178 rd. 3^ yd. ft. 3~ in. Or, 178 rd. 3 yd. 1 ft. 9- in. 11. What is the sum of 5.141 tons and .3218 of a ton, ex- pressed as a compound number ? 12. What is the sum of .005 of a common year and f of a week ? Ans. 6 d. h. 36 min. 13. What is the sum of 3 gal. 2 qt. pt. 1.4gi. ; | of a gal- lon ; and .875 of a hogshead ? ) 14. I have in one range of wood 13 cd. 3 cd. ft. ; in a second, 21 cd. 48 cu. ft. ; and in a third, 42 cd. 4 cd. ft. 8 cu. ft. How much have I in all ? Ans. 77 cd. 2^ cd. ft. - 15. A ship sailing from Boston, in latitude 42° 20' north, to Cape Horn, 55° 58' 15" south, passes through how many de- grees of latitude ? >' ' 16. Washington is 77° 2' 48" of longitude west of Greenwich, and the extreme west point of Alaska is 91° 14' 12" west of Washington. What is the longitude of that point, reckoned from Greenwich? Ans. 168° 17' W. 17. What is the sum of the following measurements : 2 yd. 2 ft. 7 in. ;. 71 yd. ; 3 yd. 1 ft. 11 in. ; li rd. 5 yd. 1 ft. 6 in. ; and 2 rd. 16 ft. 6 in. ? Ans. 8 rd. yd. ft. 9 in. 160 COMPOUND NUMBERS. SECTION XXXVII. &JJBTEACTIOX OF COMPOIWD XUMBERS. 378.— Ex. 1. From 17 bu. 2 pk. 6 qt. take 8 bu. 3 pk. 4qt. * f '^ 7 (D 1 r> Solution. — Since only units of the same 1/ bu. ^pk. 6 gf. Yn\A can be subtracted the one from the other o o 4- (Art. 57 — 1), write the subtrahend under the 8 bu 3 pk 2 at minuend, so that units of the same kind shall stand in the same column. Begin at the right, and subtract the units of each denomination of the subtrahend from those of the same kind in the minuend. 4 qt. from 6 qt. leave 2 qt., which is the difference of the quarts. Since 3 pk. cannot be taken from 2 pk., take 1 bu. from tlie 17 bu., leaving 16 bu., and add it, reduced to pecks, to the 2 pk., thus obtaining 6 pk. ; then, 3 pk. from 6 pk. leave 3 pk., which is the difference of the pecks. 8 bu. from 16 bu. leave 8 bu., which is the difference of the bushels. Therefore, 8 bu. 3 pk. 2 qt. is the difference required. 2. From 35 lb. 14 oz. take 19 lb. 15 oz. Ans. 15 lb. 11 oz. 379. Rule for Subtraction of Compound Numbers.— Write the siihtjxdiend under the minuend , so that units of the same denomination shall stand in' the same column. Begin with the loivest denomination, and suhtract the numher of units of each denomination of the sub- trahend from the numher of units of the same de- nomination of the minuend, if possihle, and write the difference beneath as apart of the required differ- ence. If the number of any denomination of the subtra- hejid is greater than that above it, increase the upper number by adding to it as rjvany units as are one of the next higher denomination, and, subtract; then, regarding the number of units of the next higher denomination of the minuend as one less, proceed as before. coif FOUND NUMBERS. 161 PR OBZ, EMS. 1. From 12 cwt. 85 lb. 11 oz. take 7 cwt. 58 lb. 6 oz. 2. From 1 lihd. 34 gal. 2 qt. 1 pt. 8 gi. take 45 gal. 3 qt. Ipt. 2 gi. Ans. 51 gal. 3 qt. 1 gi. 3. From 14 yd. 2 qr. take 9 yd. 3 qr. 4. From 78° 55' 0" take 71° 4' 20". ^jis. 7° 50' 40". 5. From 116 cd. 4 cd. ft. 6 cu. ft. 1620 cu. in. take 105 cd. 5 cd. ft. 7 cu. ft. 1511 cu. in. 6. Subtract 7 lb. 7 oz. 10 pwt. 23 gr. from 21 lb. 4 oz. 14 pwt. 13 gr. Ans. 13 lb. 9 oz. 3 pwt. 14 gr. 7. Subtract 5 mi. 215 rd. 5 yd. from 8 mi. 216 rd. 3 yd. Atis. 3 mi. 3|- yd., or 3 mi. 3 yd. 1 ft. 6 in. 8. Take .0038 of a year from f of a week. Solution. -Since | wk. = ^ d. 19 h. 12 min. sec. and .0038 y. = 1 9 17 16.8 ~iuh. — .0038y. = Id. 9h. 54 min. 43.2 sec. 5 '' 9. Take |^ of a square yard from 1 rd. 21 sq. ft. 56 sq. in. 10. Take | of a great gross from 9.125 gross. 11. From 1^ of a bushel take 3 pk. qt. 1 pt. 12. A man has travelled 4 mi. 64 rd. How much farther must he go to have travelled 6 miles ? . 13. From a cask containing 36 gal. 1 qt. of molasses, 19 gal. 2 qt. 1 pt. have been drawn. How much remains in the cask ? DIFFEEENCE OF DATES. 380. — In computation of the Difference of Dates, centuries are numbered from the beginning of the Christian era, the months from the beginning of the year, and the days from the beginning of the month. Thus, May 23, 1871, is the 23d day of the 5th month of the 71st year of the 19th century. In estimating the difference between dates, the entire calendar montlis are found, and the remaining days counted. Any number of days less than 30, in business transactions, are usually regarded as the same number of thirtieths of a month. 14 » It32 COMPOUND NUMBERS. 381. — Ex. 1. A man left home on a journey, July 17, 1867, and returned November 12, 1869. How long was he absent '( Yroiry Solution. 1869 y. 11 mo. 12 d. July 17, 1867, to July 17, 1869 = 2 y. 1867 7 17 " July 17, 1869, to Oct. 17, 1869 = 3 mo. " Oct. 17, 1869, to Nov. 12, 1869 = 26 d. 2 ^J. 3 mo. 26 d. fience, the entire difference is 2 y. 3 mo. 26 d. 2. A man was born May 16, 1819; how old was he Septem- ber 23, 1862 ? Ans. 43 y. 4 mo. 7 d. 3. If a note dated February 25,. 1868, Avas paid July 11, 1869, how long did it remain unpaid ? Ans. 1 y. 4 mo. 16 d. 4. The late civil war, which continued 4 y. 1 mo. 14 da., ter- minated May 26, 1865. When did it begin ? Ans. April 12, 1861. 5. The American Revolution began April 19, 1775, and ter- minated January 20, 1783. How many years did it continue? SECTION XXXVIII. MULTIPLICATIOJr OF COMFOVXB JfUMBERS. 382— Ex. 1. Multiply 6 gal. 3 qt. 1 pt. by 5. 6 gal. 3 qt. 1 pt. Solution. — Write the multiplier under the /- lowest denomination of the multiplicand, and, beginning at the right, multiply the number 34- gal. 1 qt. 1 pt. of each denomination in the order of the de- nominations. Five times 1 pt. are 5 pt., or 2 qt. 1 pt. Write the 1 pt. as the number of that denomination in the product, and reserve the 2 qt. to be added to the product of the quarts. Five times 3 qt. are 15 qt. ; 15 qt. and 2 qt. are 17 qt., or 4 gal. 1 qt. Write the 1 qt. as the number of that denomination in the product, and reserve the 4 gal. to be added to the product of the gallons. Five times 6 gal. arc 30 gal. ; 30 gal. and 4 gal. are 34 gal., which write as the gallons of the product. The entire product is 34 gal. 1 qt. 1 pt. 2. Multiiily 12 bu. 3 pk. 1 qt. by 7. 3. Multiply 7 yd. 3| qr. by 8. Ans. 63 yd. 2 qr. COMPOUND NUMBERS. 163 383. Rule for Multiplication of Compounii Numbers.— TJ^V-j^e the multiplier luider the lowest cleuoDii nation of the multiplicand. Begin udth the lowest denomination, and multiply the juvmber of each denomination in its order . If the product is less than one of the next higher denomina- tion, write it as a part of the required product. If tlie product is equal to or exceeds one oi^ more units of the next higher denomination, ivrite the ex- cess, if any, as a part of the required product, and add the number of units of the next higher denomi- nation to the product of that denomination. PROBLE 3rS. 1. Multiply 16° 58' 26f" by 9. Ans. 152= 46' 2". 2. Multiply 15 lb. 5 oz. 13 pwt. by 11. 3. Multiply 2 gal. 1 qt. 1 pt. 2 gi. by 19. Ans. 46 gal. 1 qt. pt. 2 gi. 4. Multiply 1 T. 17 cwt. 92 lb. by 28. 5. One ship is in 5° 15' 45" north latitude, and another is 5 times as far north. Wha.t is the latitude of the latter ? Am. 26° 18' 45" north. 6. If a team can draw in one load 1 cd. 1|- cd. ft. of wood, how much can it draw in 14 loads ? 7. I bought 4 packages of medicine, each containing 3 lb. 4s6ol9 16gr. What is the weight of the whole ? Ans. 131b. 7^ 23 19 4gr. 8. A farm consists of 9 fields, each containing 12 A. 72 sq. rd. What is the extent of the farm ? 9. If a steamer sail 211 mi. 192 rd. a day, hoAv far will it sail in 15 days? 10. How much time in 100 years, each 365 d. 5 h. 48 min. 49.7 sec. long ? Ans. 36524 d. 5 h. 22 min 50 sec. 11. A lot of laud is divided into 6 house-lots, eac i of which contains 1 A. 4 sq. rd. 120 sq.ft. How much land is there in all the lots? A7is. 6 A. 26 sq. rd. 175|- sq. ft. 164 COMPOUSD NUMBERS. SECTION XXXIX. BIYISIOX OF COMPOUJS'D J\^UMBERS. 384.— Ex. 1. Divide 34 gal. 1 qt. 1 pt. by 5. 5)34 gal. 1 qt. 1 Jit. Solution.— Write the divisor at the left of ~T> J Q i 1 ^^® dividend, and beginning at the left, divide ga . q . jj . ^^q number of each denomination in its order. One fifth of 34 gal. is 6 gal., with a remainder of 4 gal. Write the 6 gal. as the gallons of the quotient; the 4 gal. = 16 qt., which added to the 1 qt. = 17 qt. One fifth of 17 qt. is 3 qt., with a remainder of 2 qt. Write the 3 qt, a-s the quarts of the quotient ; the 2 qt. are 4 pt., which, added to the 1 pt., = 5 pt. One fifth of 5 pt. is 1 pt., which write as a part of the quotient. The entire quotient is 6 gal. 3 qt. 1 pt. 2. Divide 89 bu. 1 pk. 7 qt. by 7. A7is. 12 bu. 3 pk. 1 qt. 3. Divide 39 lb. 7 oz. 8 pwt. 9 gr. by 6. A71S. 6 lb. 7 oz. 4 pvrt. 17|^gr. 385. Rule for Division of Compound Uumhers. — Be£iji7i7'7?d with the highest denomination, divide the mvnvber of each denomination in its order, and write the several quotients as thepaHs of the same denominations of the required quotient. If there are paHial remainders, reduce each to the next lower denomination, and add the same to the number of that denomination before dividing it. When divisor and dividend are both compound numbers, they must be reduced, to simple denominate numbers of the same denomination before dividing. V no Til, KMS . 1 . DivMe 67 yd. 2 qr. liy 8. Am. 8 yd. If qr. 2. Diviie 23 cu. yd. cu. ft. 12 cu. in. by 4. Ans. 5 cu. yd. 20 cu. ft. 435 cu. in. 3. Divide 53 T. 1 cwt. 76 lb. by 28. COMPOUND NUMBERS. 165 4. In 9 equal lots of land, taken together, there are 112 A. 8 sq. rd. What is the extent of each lot ? Ans. 12 A. 72 sq. rd. 5. A quantity of tea, consisting of 19 equal parcels, con- tains 3 cwt. 32 lb. 8 oz. What is the weight of a single parcel ? Ans. 17 lb. 8 oz. 6. Divide 30° 2' by 2° 30' 10". S0° 2' = 108120" Solution. — When, as in this prob- lem, the divisor and dividend are simi- 2 oO 10 =^9010 lar compound numbers, reduce both to mo-io I' an in" -i a the lowest denomination mentioned in lUai^U ■ JUIU 1^ either, and divide as in simple numbers. 7. How many kegs, each containing 6 gal. 3 qt. 1 pt., can be filled from a cask containing 34 gal. 1 qt. 1 pt. ? 8. If 8 yd. If qr. are required for a suit of clothes, how many suits can be made from 67 yd. 2 qr. ? 9. How many loads, of IT. 17 cwt. 92 lb. each, are there in 53 T. 1 cwt. 76 \b. of hay? Ans. 28. LONGITUDE AND TIME. 386. — The earth turns on its axis from west to east once in 24 hours. This causes the sun to appear to pass around the earth from east to west in the same time. The sun appears sooner to places east of any given point on the earth than to those west of it. Hence, of any two given places, the one farthest east has the later time, and the one farthest west the earlier time. Since the cirdlimference of any circle is 360°, the sun ap- pears to pass over ^ of 360° of the earth's circumference, or 15° of longitude, in 1 hour ; -^ of 15°, or 15', in 1 minute °. and g^o of 15', or 15", in 1 second. Hence the following COMPARISON OF LONGITUDE AND TIME. 15° of Longitude correspond to 1 hour in time. 15' of Longitude " 1 ininute in time. 15" of Longitude " 1 second in time. 168 COMPOUND NVMBERS. 387. — Ex. 1. The difference in longitude between Washing- ton and San Francisco is 45° 23' 27" ; what is the difference in time ? 15 ) 45° 23' 27" Solution.— Since 15° of difference ^ , ^ . crxD-i in longitude correspond to 1 hour dif- o h. Imxn. oo-- sec. r. . ^. ,., ,.„, . -, 5 terence in time, lo dirierence m longi- tude correspond to 1 minute difference in time, and \r>" difference in longitude correspond to 1 second difference in time, 45° 23' 27^'' of dif- ference in longitude must correspond to jJ- as many hours, minutes and seconds respectively, or to 3 h. 1 min. 33f sec. • 2. The difference in time between Washington and San Francisco is 3 h. 1 min. 33f sec. ; what is the difference in longitude ? 3 h. 1 min. 33^ sec. Solution. — Since 1 second difference ^ in time corresponds to 15'^ difference ^ in longitude, 1 minute difference in 45° 23' 27" time to 15' of difference in longitude, and 1 hour difference in time to 15° difference in longitude, 3 h. 1 min. 33f sec. difference in time must cor- respond to 15 times as many seconds, minutes and degrees of longitude respectively, or to 45° 23' 27". 388. Rules for Longitude and Time.— i. Divide the dijfer- ence of longitude hy lo, and the ninnher of degrees, minutes and seconds of the quotient, respectively , will he the hours, minutes and seconds of the difference of time. 2. Multiply the difference of time by 15, and the number of the seconds, minutes and hours of the product, respectively, will he the seconds, minutes and degrees of longitude. PltOBLJE^rS. 1. When it is 12 o'clock M. at that part of Alaska which is 87° 14' 30" west of Boston, what is tlie time in Boston ? Ans. 48 min. 58 sec. past 5 o'clock p.m. COMPOUND NUMBERS. 167 2. The time at Philadelphia is 5 h. min. 40 sec. earlier than that of Greenwich ; what is the longitude of Philadelphia ? Ans. 15° 10' W. 3. The difference in time between Portland and Chicago is 1 h. 9 min. 25\^^ sec. ; what is the difference in longitude ? A71S. 17° 21' 26". 4. When it is midnight at Canton, 113° 15' east, what time is it at New Orleans, 90° 7' west? Ans. 26 min. 32 sec. after 10 o'clock a. m. TEST QUESTIONS. 389. — 1. In what respects is the Addition of compound numbers like addition of simple numbers? Why cannot dissimilar denominate num- bers be added ? 2. What is the difference between Subtraction of simple and com- pound numbers ? How may the subtraction be proved ? 3. From what are the centuries numbered? Months? Days? What is the process of finding the diflerence between dates ? 4. In what respects is Multiplication of compound numbers like multiplication of simple numbers? Why cannot a compound number be multiplied by a denominate number? (Art. 71 — 1.) 5. When in Division of compound numbers the divisor is an abstract number, what kind of a number will the quotient be? When the divisor and dividend are similar compound numbers, what must be done before dividing ? What kind of a number will the quotient be ? 6. Why does the sun appear to pass around the earth from east to west ? Of two places on the earth having different longitude, which has the earlier time ? 7. What part of the earth's circumference does the sun appear to pass over in 1 hour? To what in time do 15° of longitude correspond? 15^ of longitude? 15'^ of longitude? What are the rules for longitude and time? 8. What is the Rule for addition of simple numbers? (Art. 49.) The rule for addition of compound numbers? For subtraction of simple liumbers? (Art, 60.) For subtraction of compound numbers? 9. What is the rule for multiplication of simple numbers? (Art. 75.) For the multiplication of compound numbers? For the division of simple numbers? (Art. 91.) For the division of compound numbers? 168 ANALYSIS. SECTION XL. ANALYSIS BY ALIQUOT PARTS. 390. — Ex. 1. How many hundred- weight is one half of a ton ? One fourth of a ton ? One fifth of a ton ? One eighth of a ton ? One tenth of a ton ? 2. What part of an acre is 80 square rods? Is 40 square rods ? 32 square rods ? 20 square rods ? 3. What part of a year is 6 months? Is 4 months? 3 months? 2 months? What part of a month is 15 days? Is 10 days? 3 days? 4. What will one half a ton of hay cost, at $22 j^er ton ? One fourth of a ton, at $24 per ton ? 5. How much will 2 A. 80 sq. rd. of land cost, at $30 per acre ? 5 A. 40 sq. rd., at $20 per acre ? 6. How much will it cost a man for 2 mo. 3 d. board, at $20 per month ? DEFINITION". Analysis by Aliquot Parts, or Practice, is a concise method of computation by employing aliquot parts. (Art. 279.) WniTTEN EXERCISES. 391. — 1. What is the rent of a store for 1 year 5 months 10 days, at $300 a year ? Solution. The rent for 1 y. is $300.00 4 mo. "j of $300= 100.00 Imo. " ^ of 100= 25.00 lOd. " I of 25= 8.33J " ly. 5 mo. 10 d. " $433.33 j 2. At $80 a ton, wliat will 5 T. 15 cwt. 50 lb. of iron cost? Ans. $462. AXALYNZS. 169 3. At S.96 per gallon, what will 3 gal. 2 qt. 1 pt. of molasses cost? 4. How much will 9 months 24 days of labor amount to, at $600 a year ? Ans. S490. 5. What is the cost of constructing 20 mi. 120 rd. of road, at 84000 per mile? Ans. $81500. 6. At $6 a hundred-weight, what must be paid for 137;"» pounds of fish? ^?js. $82.50. 7. What is the cost of 360 yards of camlet, at $.621 per yard ? Solution. At $.50 a yard, the coat of P>GO yd. is | of $360-^$180.00 " .12^ " " 380yd." j of 180^ 45.00 "$.62^ " " 360 yd.'' $225.00 $.62^ = i of $1 + i of ^ of $1. At i of a dollar a yard the cost of 360 yards is \ of $360, or $180. At \ of | of $1 a yard the cost is J of $180, or $45. Hence, at $.62^ a yard, the cost is $180 + $45, or $225. 8. What is the cost of 180 bushels of corn, at $1.12| per bushel ? 9. What is the cost of 460 yards of cloth, at $5.75 per yard ? 10. What is the cost of 172 bushels of rye, at $.87^ per bushel ? Solution. At $1.00 per hu., the cost of 172 bit. is $172. 00 « .12- " " 172hu.''~of $172= 21.50 "$ .87J " " 172 bu." $150.50 Since $.87^ is \ of a dollar less than a dollar, at $.87i^ per bu. the cost of 172 bu. will be \ of $172 less than $172, or $150.50. 11. What is the cost of 1671 poimds of tea, at $.83^ per pound ? Ans. $1392.50. 12. What is the cost of 4 T. 16 cwt. 20 lb. of hay, at $25 per ton? tItis. $120.25. 15 170 RECTANGULAR MEASUREMENTS. SECTION XLI. BECTAJfGULAR MEASUKEMEJ^TS. Cj^SE I. Surfaces. 392. — Ex. 1. How many square inches are there in a rec- tangular surface which is 15 inches long and 1 inch wide ? 2. At 3 cents per square foot, how much must be paid for a rectangular board which is 25 feet long and 1 foot broad ? 3. How many square rods are there in a walk which is 50 rods long and 1 rod wide ? DEFINITIONS. 393. The Dimensions of a rectangular surface are the length and breadth, or width, of that surface. 394. The Unit of Measure for surfaces is always a square whose dimensions are known ; as 1 square inch, 1 square foot, etc. 395. The Area of a surface is the number of times the sur- face contains a given unit of measure. Thus, the rectangle in the margin will be seen to contain 12 square inches, if it be supposed to be 4 inches long and 3 inches wide. For, upon each inch of length there may be conceived to be 1 square inch, making a row of 4 square inches, and as tliere will be as many such rows as there are inches in the width, or 3 rows, the area of the rectangle must be 3 times 4 square inches, or 12 square inches. 306. Principle. — The area of a rectangle is equal to the number of square vnifs denotcil by the product of the number of linear xinits in the Irnr/th, multiplied by the mimber of the same linear units in the -width, the square units having the same name as the linear units. RECTAJS^GVLAR MEASUREMENTS. 171 WJtITTX:N EXERCISES. 307. — Ex. 1. How many square feet of surface has a rec- tangular table whose length is 7 feet 5 inches and width 5 feet 4 inches ? 89X64 = 5696 No. of sq. in. Solution.— 7 ft. 5 in. == 89 5696 sq. in. - 39 sq. ft. 80 sq. in. !""^^''' ^"'^ ^ ^^- ^ '''■ = ^^ The product of 89 by 64, or 5696, must denote the number of square inches of surface ; and 5696 square inches are equal to 39 sq. ft. 80 sq. in., which is the surface required. 2. In a floor 16 feet long and 11 feet wide, are how many square feet? Ans. 176. 3. The area of a rectangular floor is 176 square feet, and its length is 16 feet. What is its width ? ]^'^0 H- XO = H ft. Solution.— Since the product of the num- ber of linear units in the length by the number in the width is equal to the number of square units in the area, the number of linear units in the required dimension must equal the quotient of the number of the square units of the area divided by the number of linear units in the given dimension, or 11 feet. 4. The area of a board is 45 square feet, and its width 1^ feet. What is its length ? 398. Rulesfor Measurements of Rectangular Surfaces.— i, Mul- tiply the length by the width, and the product will denote the area. 2. Divide the area by either of the dimensions, and the quotient will denote the other dimension. PROBZESrS. 1. How many acres are there in a rectangular field whose length is 80 rd. and width 20 rd. ? Ans. 10. 2. The area of a field is 4608 sq. rd., and its Avidth is 16 rd. ; what is its length ? 3. How many square yards of carpeting will cover a room 13| ft. square ? Ans. 20|. 172 RECTANGULAR MEASUREMENTS. 4. A i^ath is 18 ft. 8 in. long, and 5 ft. 3 in. wide. What is its area in square feet ? Solution.— 18 ft. 8 in. = 18| ft. = \^ ft., and 5 ft. 3 in. = 5^ ft. = ^ ft. V X V ^^ 9^- ilence, the area required is 98 sq. ft. 5. If it take 32^ sq. yd. of carpeting to cover a floor whose width is 14 ft., what is the length of the floor ? CA.se II. Solids. 399. — Ex. 1. How many cubic feet are there in a rectangu- lar beam whose length is 20 ft., width 1 ft. and thickness 1 ft. ? 2. What is the value of a stick of timber 12 ft. long, 1 ft. wide and 1 ft. thick, at 10 cents per cubic foot? DEFINITIONS. 400. The Dimen.sions of a rectangular solid, or volume (Art. 319), are the length, breadth or width, and thickness, depth or height. 401. The Unit of Measiii'e for solids, or volumes, is always some cube (Art. 317) whose dimensions are known ; as, 1 cubic inch, 1 cubic foot, etc, 402. The Cubic Contents, or capacity, of a solid, or volume, is the number of times the solid or volume contains a given unit of measure. Thus, tlie rectangular solid, or volume, in the margin will be seen to contain 72 cubic inches, if it be supposed to be 6 inches long, 3 inches wide and 4 inches thick. For, upon each of the 18 square inches of the lower fiice there m;iy bo conceived to be 1 cubic inch, making a layer of 18 cubic inches ; and as there will be as many such layers as there are inches of thickness, or 4, the contents of the volume must be 4 times 18 cubic inches, or 72 cubic iuche». yyy .^ 7^7 Z17 RECTANGULAR MEASUREMENTS. 173 403. Principle. — The cubic contents of a solid, or volume, are equal to the number of cubic units denoted by the product of the number of the same linear units in the length, width and thickness, the cubic units having the same name as the linear units. WRITTEN EXERCISES. 40-1:. — Ex. 1. How many cubic feet are there in a rectangular block of marble which is 8 ft. long, 3 ft. 6 in. wide and 2 ft. 3 in. thick ? 8 3^ Solution.— 3 ft. 6 in. = 3^ ft., and 2 ft. 3 in. = 2\ ft. 28 The product of the number of units in the length, width and thickness is 63, which must be the number of 2— cubic feet required. 63 m.ft. 2. What are the cubic contents of a body 20 ft. long, 6 ft. Avide and 4 ft. thick ? 3. A block, containing 15625 cubic inches, is 2 feet 1 inch wide and 2 feet 1 inch thick. What is its length ? Aiis. 2 ft. 1 in. 4. A rectangular body whose cubic contents are 480 cu. ft. is 20 ft. long and 6 ft. wide. What is its thickness ? SO X 6" = 120 Solution. — Since the number of cubic feet of contents must be the product of tlie number of 120)4-80 units in the length, width and thickness, the quo- / tient of 480 divided by 120, the product of the units in the two given dimensions, must be the number of feet of thickness required. 480 -^ 120 = 4. Hence, 4 feet must be the thickness required. 405. Rules for Measurement of RecianguSar Solids.— i. Multi- ply the length, width and thickness together, and the product will denote the cubic contents. B. Divide the cubic contents by the product of any two of the dimensions, and the qiootient will denote the other diinension. 15 « 174 REVIEW PROBLEMS. probz,i:ms. 1. How many cubic feet is the capacity of a bin whose in- side measures 12 feet long, 6^ feet wide and b\ feet deep? Alls. 400. 2. What are the contents of a cube wiiose edge measures 5.5 feet? Ans. 166.375 cu. ft. 3. How many cord feet are there in a load of wood 8 feet long, 3| feet wide and 5 feet high ? 4. How much wood of the usual length is there in a range 163 feet long and 4 feet high ? 5. If a load of wood is 8 feet long and 3 feet wide, how high must it be to contain a cord ? Ans. 5 ft. 4 in. SECTION XLII. BEVIEW PROBLEMS. mentaij exercises. 406. — Ex. 1. How much will 2 pecks of berries cost, at 12^ cents per quart ? 2. How much is gained by selling a gross of buttons, which cost 75 cents, for 8^ cents per dozen ? 3. If the forenoon and afternoon sessions of a school are each 3 hours, what part of the two sessions are two recesses of 20 minutes each ? 4. When it is 9 o'clock in the morning at Philadelphia, what time is it at a point 15° 45' west of Philadelphia ? 5. How much must be paid for a board 22 feet long and 1 foot 6 inches wide, at the rate of $30 per thousand square feet ? 6. If you should leave home and travel till your watch is 35 minutes fast, how far in longitude would you have travelled, and in what direction ? 7. Two boys were cni])loyed to measure the length of a ditch ; one reported it to be 1 rd. 16 ft. 11 in., and tlic other 1 rd. 5 yd. Ift. 11 in. Tlic true length was 2 rd. 5 in. How much did cacii of the mcasurenienls differ from the true length? REVIEW ITxOBLEMlS. 175 WRITTEN EXERCISES. 407. — Ex. 1. How many hogsheads are 2217 quarts? 2. How mauy quarts are 8 hhd. 50 gal. 1 qt. ? 3. In 1000000 seconds are how many weeks ? 4. In 1 wk. 4 d. 46 min. 40 sec. are how many seconds ? 5. How much wood in a range 40 feet long, 1\ feet high and 4 feet wide ? Ans. 9f cords. 6. What must be the height of a range of wood which is 40 feet long and 4 feet wide, to contain 9 cd. 3 cd. ft.'? 7. What decimal part of 3 gallons is 3 pints ? 8. What is the sum of f of a foot, f of a yard and -J of a mile ? Am. 280 rd. yd. 2 ft. 1\ in. 9. What is the value of .131 of 5 hours? 10. What decimal part of 5 hours is 40 minutes? 11. How many years after the battle of Lexington, A})ril 19, 1775, was that of New Orleans, January 8, 1815 ? 12. How many days will there be from October 10, 1871, to March 17, 1872? 13. If the 16th of May is Sunday, what day of the week is the 20th of the next October ? Solution. — The difference in the given dates is 157 days, or 22 weeks 3 days. 3 days after Sunday, the ending of the 22 weeks, must be Wednesday, the day of the week required. 14. What part of a day is 6 h. 3 min. 4 sec. ? 15. If a clock tick 172800 times a day, how many times will it tick in 6 h. 3 min. 4 sec. ? 16. When 2 bu. 1 j)k. of clover-seed is sold for $5.94, what is the price of a peck ? 17. How much molasses in 43 casks, each holding 97 gal. 1 pt. 2 gi. ? Am. 4179 gal. qt. pt. 2 gi. 18. How long must a rectangular lot of land be, whose width is 16 rods, to contain 2 acres ? Ans. 20 rods. 19. How much is f of 45 T. 15 cwt. 25 lb. ? 20. How many yards of cambric, which is f of a yard wide, will line 6f yards of cloth, which is 1^ yards wide ? 21. What will it cost to carpet a room 20 feet wide and 18 feet long with carpeting 4 feet wide, costing $2.33|- per yard ? 176 PERCENTAGE. SECTION XLIII. FERCEJ^'TAGE. 408.— Ex. 1. How much is -^ of 100 yards? yfj? 2. How much is ^ of 100 ? ^-^ ? y-^^ ? ilnr ? 3. What part of 100 yards is 1 yard? Is 11 yards? 4. How many hundredths of a hundred-weight are 3 pounds? 5. How many hundredths of anything is \ of it? -^ of it? 6. What part of 100 hundredths is 25 hundredths? DEFINITIOETS. 409. Any Per cent, of a number is so many hundredths of that number. The term j)e.r cent, is a contraction of the Latin per centum, and means by the hundred. Thus, 6 per cent, of 25 is .06 of 25 ; and 5 per cent, of 4 is .05 of 4. 410. The Sign % is generally used by business-men, instead of the words per cent. Thus, 8fo means 8 per cent. 411. Any per cent, less than 100 per cent, may be expressed by a decimal or common fraction ; any per cent, equal to or greater than 100 per cent, may be expressed by an integer, a mixed number or an improper fraction ; and a fractional part of 1 per cent, may be expressed as a common fraction at the right of the figure in the hundredths order. Thus, 1 per cent, may be written, ll 6 per cent. " 7 per cent. " 7-j^ per cent. " 12^ per cent. " 100 per cent. " i^5 per cent. " l^o'^.l.l^o. " —^ n, n. .01, or 100 6%, .06, 11 6 100 7%, .07, i< 7 100 n,3 H^- 07— •^' 10' « 'To 100 1^1%, ■i^i « ^4 100 100%, 1.00, " 100 inn PER CENT A GE. 177 WRITTEN EXERCISES. 412. Write and read — 1. 4%. 2. 5%. 3. 16%. 2\ per cent. 16f per cent. 217 per cent. 4. ouyc 5. 17%. 6. 106% 10. Write decimally 3% ; 11% ; 14% ; 93%. 11. Write as a common fraction 5% ; 17% ; 31%. 12. Express as a number of hundredths \; |- ; f . 13. Express in per cent. | ; | ; 2^ ; 6^ ; 4^. GENERAL CASES OF PERCENTAGE. 413. — Ex. 1. To take 5% of a number is to take how many hundredths of that number ? 2. What is 5% of 100 bushels? Of 200 yards? 3. What is 1% of $500? 7% of 8500? 4. What per cent, of 100 bushels is 5 bushels? Is 6 bushels ? Is 25 bushels ? 5. What per cent, of $500 is $5 ? $35 ? 6. Five bushels are 5 % of what number of bushels ? 7. Thirty-five dollars are 7 % of what number of dollars ? 8. Twelve dollars are 2% of Avhat number of dollars? 9. Of what number are 15 yards 6% ? 10. Of what number are 27 gallons 9% ? 11. To what will $400 amount if increased by 5% of itself? 12. How much is $200 diminished by 25% of itself? 13. A certain number increased by 5% of itself is 420; what is that number ? 14. A certain number diminished by 25% of itself is 150 ; what is that number ? DEFINITIONS, 414. Percentage is the process of computing by hundredths. 415. The Base is the number or quantity of which the hun- dredths are computed. 416. The Rate Per Cent., or Rate, is the number denoting the number of hundredths of the base which are taken. 178 PERCENTAGE. 417. The Percentage is the result of finding a number of hundredths of the base. The percentage is also sometimes called the per cent. 418. The Amount is the base with the percentage added to it. 419. The Difference is the base with the percentage sub- tracted from it. CA.SE I. Base and Rate giren, to find the Percentage, Amount or Difference. 420.— Ex. 1. What is 20% of 455 yards, and what is the amount and difierence ? Solution. — Since A55 vd ^^^ ^^ ^"7 "i^^ber Qf^ is .20 of that number, 20% of 455 yards Or, 01.00yd. j^^gj be .20 of 455 yards, or 91 yards. 455 yd. X ^ = y- yd. =^91 yd. Or, since '20% of any number is i of Amount = 4^5 yd. + 91 yd. = 54-6. that number, 20% of -r..r- , - <- 1 n-1 1 r> n t i ^^^ yards must be \ Dv^erence = 400 yd. - 91 yd. = 064 yd. ^^ 455 y^^^,,^ ^^ g^ yards. The amount is 455 yards plus the percentage, or 546 yards ; and the difference is 456 yards minus the percentage, or 364 yards. 2. How much is 15% of $460? 3. How much is 2% of 6550 pounds, and what is the amount ? 421. Principles. — 1. The percentage is equal to the product of the base by the rate. 2. The amount is equal to the sum of the base and percentage. 8. The difference is equal to the base less the percentage. 422. Rules for Finding the Percentage, Amount or Difference.— 1. Multiply the base by the rate, and the produet will be the percentage. 2. Add the percentage to the base, and the sum will be the amount; or subtract the percentage from the base, and the result will be the difference. PERCENTAGE. 179 I'll OB LEMS. Find the percentage of — 1. 516 bushels at 5%. 2. $360 at 15%. Ans. $54. 3. 455 gallons at 3%. 4. 93 acres at 7%. Ans. 6.51 A. 5. 812 men at 25%. How much is — 6. 7% of 210 pounds? 7. 1% of 703 yards? 8. 33^% of 942 sheep? 9. 125% of 1215 tons? 10. 350% of $1600? 11. If a man, whose income is $1250 a yeai", spends 80% of it for his living, how many dollars does he spend ? Ans. $1000. C^SE II. Base and Percentage given, to find the Rate. 423.— Ex. 1. What per cent, of 455 yd. is 91 yd.? 91 _ 455' Or, 91 .20 = 20% Solution.— 91 yd. are /^V of 455 yd., or .20, or 20% of 455 yd. Or, 91 yd. are ^V^ of 455 yd. -^-^-^ = .20 ; hence, 91 yd. equal 1^. of 100% =20% -20 of 455 yd., or 20%. 2. What per cent, of $390 is $11.70? Ans. 3. 3. What per cent, of 3240 T. is 21.6 T. ? Ans. | of 1. 42-i:. Principle. — The rate is equal to the quotient of the per- centage divided by the base. 425. Rules for Finding the Rate Per Cent.— i. Divide the per- centage by the base. Or, 2. Tahe such a part of 100 per cent, as the percent- age is of the base. PMOBI^JEMS. What per cent, of — 1. 516 bu. is 25.80 bu. ? 2. 455 gal. is 13.65 gal.? 3. 93 A. is 6.51 A. ? 4. 1.4.701b. is 14.701b.? 5. $1600 is $4000? Ans. 250. 10. $6300 is $173.25? 6. 460ft. is 368 ft.? ^ns. 80. 7. 51 4 i rd. is 84 rd. ? 8. 942 sheep is 314 sheep ? 9. 703yd. is 5.261 yd.? 180 PERCENTAGE. 11. What per cent, of 2 A. 72 sq. rd. is 144 sq. rd.? Solution. 2 A. 72 sq. rd. = S92 sq. rd. 12. A pound Troy is what per cent, of a pound Avoirdupois ? 13. A merchant had 1 T. 15 lb. 10 oz. of sugar, and sold 10 cwt. 46 lb. 4 oz. What per cent, did he sell ? Ans. 51|f . CASE III. Rate and Percentag-e Given, to find tlie Base. 426.— Ex. 1. 91 yd. are 20% of how many yards? _9i _ . _. _ Solution. — 20% of any number is .20 of that ' 'SO "^ number. Since 91 yd. are 20 % of some number, Or 1 % of tliat number must be i^^ of 91 yd. ; and Q-] \/ FT -^ lepr 100%, or the number required, must be 100 times -^-^ of 91 yd., or V'j*' of 91 yd., which is the same as the quotient obtained by dividing 91 yd. by .20. Or, Since 100% is 5 times 20%, and 91 yd. are 20% of some number, that number must be 5 times 91 yd. ; or, since 20% is | of a number, and 91 yd. are 20% of some number, that number must be the quotient obtained by dividing 91 yd. by \. 2. Of what number are 71 mi. 16|% ? 3. Of what number are 203 men 50% ? Ans. 406 men. 427. Principle. — The hose is equal to the quotient of the per- centage divided by the rate. 428. Rules for Finding the Base of the Percentage.— i. Divide the percentage hy the rate. Or, 2. Tahe as Tnany times the percentage as 100 per cent, is times the rate. PJlOIiLJSMS. Of what are — 1. 13.65 gal., 3% ? 4. 5.27J yd., f % ? 2. 6.51 A., 7% ? 5. 85600, 350% ? 3. 822.61, 1% ? A71S. $2584. 6. 1724 ft., 400%? Ans. 431 ft. PERCENTAGE. 181 7. If the percentage be 23.8 lb. and the rate 4%, what must be the base ? 8. If the rent of a store at $1020 is 12% of its valuation, what is its valuation ? Ans. $8500. 9. A farmer sold 85 A. 100 sq. rd. of land, which was just 25% of his farm. What was the extent of his farm ? C^SE TV. Amount or Difference and Rate Given, to Find the Base. 429. — Ex. 1. A certain number increased by 8% of itself is 189 ; what is that number? 1.08)189.00(175 108 - Solution. — A number increased by 8% of it- 810 self is 108% of itself, or 1.08 times itself. 756" If 189 is 1.08 times a number, that number r irx must be y.^-j of 189, or 175. 540 2. A certain number diminished by 25% of itself is 327. What is the number ? .75)327.00(4.36 300 Solution. — A number diminished by 25% of 270 itself is 75% of itself, or .75 of itself. 225 If 327 is .75 of a number, that number must be -J^ .^\ of 327, or 436. . 450 3. If the amount is 124.20, and the rate 15%, what is the base? Am. 108. 4. If the difference is 278.30, and the rate 45%, what is the base? 430. Principle. — The base is equal to the quotient of the amount divided by 1 plus the rate, or to the quotient of the differ- ence divided by 1 minus the rate. 431. livi\e.— Divide the amount by 1 plus the rate, or divide the difference by 1 minus the rate. 16 182 PERCENTAGE. PROBLEMS. 1. An infantry regiment, after losing 1\% of its men, had 740 left. How many had it at first ? Ans. 800. 2. By running 15% faster than usual, a locomotive ran 552 miles in a day. What was the usual daily speed ? Ans. 480 miles. 3. A farmer purchased a farm for a certain sum, expended for tools and stock, 11% of the price of the farm, and found that the whole cost was $7215. What was the cost of the farm alone? 4. Osgood raised 800 bushels of corn, which was 25% more than ^ of what Benton raised. How many bushels did Ben- ton raise? Ans. 1280. SECTION XLIV. PROFIT AKB LOSS. 432. — Ex. 1. I sold a barrel of flour, which cost $8, at an advance of 25 % , or for \ more than the cost. How much did I gain? 2. John bought a hat for $5, and sold it at a loss oi1^%, or for "I" less than the cost. How much did he lose ? 3. If I sell a horse, which cost me $120, at a profit of 10%, how much do I get for him ? 4. I had 40 sheep, but have sold 4 of them. What per cent, of the 40 have I sold ? 5. What per cent, shall. I gain by selling for §10, flour which cost me $8 ? 6. What per cent, does a merchant lose by selling goods at ■| of their cost ? 7. I sold flour at $10 and gained 25 % . What did it cost me ? DEFINITIONS. 433. Profit and Loss are terms used to denote the gain or loss in businci^s transactions. The gain or loss may be regarded as a certain per cent, of the cost. Hence, PBOFIT AND LOSS. 183 434:. The Base of computation of profit or loss is the cost ; and, by the principles of percentage, we have the — 435. Rules for Profit or Loss.— _Z. Multiply the cost by the rate, and the product will he the profit or loss. 2. Divide the profit or loss by the cost, and the quo- tient will be the rate. 3. Multiply the cost by 1 plus the rate of profit, or by 1 minus the rate of loss, and the product will be the selling price. Jj.. Divide the profit or loss by the rate; or, divide the selling price by 1 plus the rate of profit, or by 1 minus the rate of loss, and the quotient ivill be the cost. I'JiOBLJEMS. Ex. 1. I bought a house for 83500, and sold it at a gain of 12|-%. How much was the profit? 2. Goods which cost 85400, were sold at 9% below cost. How much was the loss? Ans. $486. 3. A farmer had 460 sheep, which cost him $3 each, but he lost 5% of them. How much was the loss? Ans. $69. 4. A merchant bought 112 barrels of flour, at 87 a barrel, and sold it so as to gain 15%. How much was the profit? 5. How much will be my loss on 3 casks of molasses, of 63 gallons each, which cost me 80 cents a gallon, if I am obliged to sell it at 10% below cost? 6. Williams bought coal at $7.50 per ton, and sold it at $7.05 per ton. What was the loss per cent. ? . Ans. 6. 7. A field produced the value of 867 one year, and 816.75 more the next year. What was the gain per cent. ? 8. If I sell 1^ of a house at f of the cost of the whole house, do I gain or lose, and at what rate? Ans. Gain 20%. 9. By selling coal at 88.05 per ton, I gained 15%. What v/as the cost per ton ? Atis. 87. 10. A watch not proving as good as I expected, I was con- tent to sell it at a loss of $6, which was 7^% of the cost. What was the cost? Ans. $80. 184 PROFIT AND LOSS. 11. By selling an article at 87|^% of its cost, I lost $125. What was the cost ? 12. I sold molasses at 115% of its cost, and thereby gained 9 cents on a gallon. What was the cost per gallon ? ' 13. Henry bought cloth at $6.50 per 5^ard ; at what price must it be marked to allow of 12% profit ? 14. At what price must I sell a house which cost me $5400, to gain 20% ? Ans. $6480. 15. John paid $225 for a horse, and $15 for his keeping. For how much must he be sold above the first cost to allow of 5|% profit ? Ans. $28.20. 16. I bought 60 yd. of cloth for $240 ; Avhat must be my selling price per yard to make 12^% ? 17. I sold a house for $6976, and thereby gained 9%. What was the cost ? Ans. $6400. 18. By selling tea at 76 cts. per pound, I sufier a loss of 20%. What was the cost ? TEST QUESTIONS. 436. — 1. What is any Per Cent, of a number? Of what is the term a contraction ? What sign is generally used for the words Per Cent. ? 2. How may any per cent, less than 100 per cent, be expressed ? A per cent, equal to or greater than 100 per cent. ? 3. What is Percentage? The base of percentage ? The rate? The amount? The difference? 4. To what is the percentage equal ? How is the percentage found when the base and rate are given ? How is the amount found ? How is the difference found ? 5. To what is the Rate equal ? How is the rate found when tlie base and percentage are given? 6. To what is tlie Base equal? How is the base found when the rate and percentage are given ? Plow is the base found wlien the amount or difference and rate are given ? 7. Wliat are Profit and liOss? What is the base of computation of profit and loss? To what is the profit or loss equal ? The rate? The cost ? 8. To what is tiie Sei.ling Price equal? How is the cost found when the selling price and the rule uf pruiit ur loss are known ? COMMISSION. 185 SECTION XLV. COMMISSIOJf. 437.— Ex. 1. How much should I receive for selling goods to the amount of ^500, if allowed 2^ % ? 2. If I am allowed 2|% for making purchases, how much per ton should I receive for purchasing coal at $8 per ton ? 3. To how much will a collector be entitled for collecting 500, at 2% ? 4. How much will remain of a collection of $300, after de- ducting the collector's fees at the rate of 1^ % ? DEFINITIONS. 438. An A^cnt, Comuiission Merchant or Broker is a person who, by authority, buys or sells goods or property, or collects money for another, 439. A Consignee is a person to whom goods are sent for sale, and a Consignor is the person sending the goods. 440. Commission is the percentage allowed an agent or com- mission merchant as pay for transacting business. 441. The Base of commission is the sum expended or received. 442. The Amount is the sum expended or received plus the commission. 443. The Net Proceeds of a sale or collection are the sum left after the commission and other charges, if any, are deducted. 444. Rules for Commission.— i. Multiply the base by the rate, and the product is the coTmnission. 2. Subtract from the base the cojmnission, and the other charges, if any, and the result will be the proceeds. 3. Divide the coinmission by the base, and the quo- tient is the rate. Jf. Divide the commission by the rate, or divide the aTYiount by 1 plus the rate, aivd the quotient is the hojse. 16* 186 COMMISSION. PROnLEMS. 1. An agent has sold goods for me to the amount of $5000, at 2^% commission. What is his commission ? Atis. $125. 2. A commission merchant sells 4520 bu. of wheat, at $2.50 per bushel. How much is his commission at 3% ? 3. If I collect as an agent $390, and am entitled to 5% commission, how much must I pay over ? Ans. $370.50. 4. An agent bought 80 barrels of beef, at $22 per barrel, and paid $16 for insurance and $9 for cartage. His commis- sion v\-as 2^%. What was the amount of his bill to his employer? Ans. $1824.60. 5. A commission merchant having sold some goods, paid $4168.80 to his employer, and retained as his commission $151.20. What was the rate of his commission ? 6. My agent collected $390, and, retaining his commission, paid over $370.50. At what rate was his commission ? 7. A commission merchant bought some goods, paid for cartage $21.50, and charged for storage $31, and for commis- sion $112.50. His entire bill was $5165. What was the rate of commission? Ans. 2^%. 8. I paid a commission merchant $12.56 for selling goods, at the rate of 4%. What amount did he sell? Ans. $314. 9. A treasurer's commission for collecting taxes in one year, at 1|%, is $413.10. What was the amount collected? 10. A commission merchant in Chicago received $1665.62|- with which to purchase flour at $6.50 per barrel, after deduct- ing his commission of 2|^%. How much was his commission, and how many barrels did he purchase? Ans. Commission, $40,621; barrels, 250. 11. An agent received $5922 with which to purchase goods, after deducting his commission of 5%. How much was his commission, and what was the sum to be expended? Ans. Commission, $282 ; sum, $5640. 12. I remit to an agent $360.70, with which to purchase goods; deduct his commission of 5%, and pay $3.70 for insurance. What sum can he expend for the goods ? INSURANCE. 187 SECTION XLVI. IJfSUBAJ^CE. 445. Insurance is indemnity secured for loss. 446.- Fire IiisiuMnce is indemnity secured for loss of property by fire or lightning. 447. Marine Insurance is indemnity secured for loss of prop- erty by casualties of navigation. 448. Health, or Accident, Insurance is indemnity secured for loss by sickness or accident. 449. Life Insurance is indemnity secured for loss by death. 450. The Policy is the contract between the insurer and the insured. 451. The Premium is the sum paid for insurance. 452. In Property Insurance the premium is computed at a certain rate per cent, on the value insured. 453. In Life Insurance the premium is computed at a certain sum or rate per |100 or $1000 insured. 454. Rules for Insurance.— i. Multiply the value insured by the rate, and the product is the premium. 2. Divide the premium by the value insured, and, the quotient is the rate. 3. Divide the value insured by 1 minus the rate, and the quotient luill be the ainount to be insured to cover the value insuj^ed, and pj-emimn of insurance. PliOBT.EMS. 1. What premium must be paid for insurance of $3000 on a house, at 2^% ? Ans. $75. 2. What is the expense of insuring f of a mill valued at $8400, at 5^^, the policy being $1 ? Ans. $316. 8, Hall paid $91, including the policy at $1, for insuring 188 INSURANCE. $2500 on his house and $2000 on his saw-mill. What was the rate of insurance ? Ans. 2 % . 4. If $75 is paid for insuring $3000 on a house, what is the rate of premium ? 5. What sum must be insured on $5600, at 3%, to cover property and premium in case of loss? Ans. $5773.19 +. 6. For what sum must property, valued at $4000, be insured, at 5%, to cover f of the property, the premium and the policy at $2? 7. A man 40 years old has obtained a life policy for $6000, at the rate of $29.60 per $1000. What is the annual premium ? 8. A man 31 years old took out a life policy for $7500 for the benefit of his family, on the plan of semi-annual pay- ments, at the rate of $23.78 per $1000. He died at the age of 35. How much did the amount due his family exceed the payments he had made ? Ans. $6073.20. 9. If $3160 must be insured on a house to cover | of its value, the premium at 5% and the policy at $2, what is the value of the house ? TEST QUESTIONS. 455.— 1. What is a Cojimission Merchant or Broker ? A con- signee? A consignor? 2. What is Commission ? The base of commission ? The amount ? What are the net proceeds ? 3. To what is the Commission equal ? How is it found when the base and rate are given ? 4. To what are the Net Proceeds equal ? How are they found when the base and commission are given ? 5. To what is the Kate equal ? How is the rate found when the com- mission and base are given ? 6. To what is the Base equal ? How is it found when the commission ■and rate are given ? When the amount and rate are given ? 7. What is Insurance? Fire insurance? Marine insurance? Health and accident insurance ? Life insurance ? 8. What is the Policy ? The premium ? How is the premium com- puted in property insurance ? In life insurance ? BEVIi:W PROBLEMS. 189 SECTION XLVII. REVIEW PROBLEMS. MJENTAZ EXEKCISES. 456. — Ex. 1. At what price must tea which cost 75 cts. be sold, to make 16f% profit? 2. I have 60 cts. ; how much of it must I spend to have 88% of it left? 3. If I had 35 sheep and sold 14, what per cent, of them did I sell ? 4. I bought a cart for $75, and sold it for $90. What per cent, did I gain ? 6. In a certain school, 5 of every 8 pupils are girls. What per cent, are boys ? 6. 69^ of a number is what per cent, of 30% of the number? 7. A merchant sold tea at a loss of IG cts. a pound, which was 25% of the cost. What was the cost ? 8. Henry is 11 yr. old and John is 125% as old. How many years is the difference in their ages ? 9. If 37|-% is lost by selling goods for $90, what was their cost? 10. 75 is 25% more than what number? 42 is 20% more than what number ? 11. What amount of money must be forwarded to a com- mission merchant, to cover a purchase of $180 and his com- mission of 10%? 12. A horse was sold for $90, at which price 12 J % was gained. What per cent, would have been gained by selling him for $100 ? 13. I bought a wagon for $72, which Avas 20% more than its value. I sold it at 5% less than its value. How much did Hose? 14. Higgins sold a cow for $30, and by the transaction lost 16-|%. He sold another cow at an advance of 16% for just enough to cover, by the profits, the loss upon the first cow. What did he get for the last cow ? 190 REVIEW PROBLEMS: WniTTEN EXERCISES. 457^ — Ex. 1. If a cubic foot of pine timber weighs, when green, 44 lb. 12 oz., and when seasoned, 30 lb. 11 oz., what per cent, does it lose in seasoning ? A^is. 31y^^. 2. How large a sale must a merchant make, at a profit of 15%, to clear $3750? Ans. $25000. 3. What must be the selling price and profit of coal whose first cost is $6, freight 10% and rate of gain 20% ? A}is. Selling price, $7.92 ; profit, $1.32. 4. If a merchant closed out his goods at a loss of 10%, how much did he lose on calico that cost 12^ cts. per yard, and on sugar that cost 15 cts. per pound? Ans. On calico, 1^ cts. per yard. On sugar, 1| cts. per pound. 5. My horse cost | as much as my carriage ; what per cent, of the cost of the one was the cost of the other ? Ans. The horse, 60% of the carriage. The carriage, 166|%> of the horse. 6. I sold one half of a lot of goods which cost me $456, at a loss of 25%, and the other half at a profit of $69.54. What was the gain per -cent, on the whole transaction? Am. 2|. 7. I bought a farm for a certain sum, and after expending 5|% of the cost for repairs and improvements, and paying a tax of li%, I sold it for $6420, which just made up what I had paid out. What was the original cost ? ^ns. $6000. 8. By selling cloth at $6 per yard, I gain 25%?. What per cent, shall I gain by selling it at $5.28 ? Solution. — If the gain at $6 per yard 1.25)$6.0000($4-80 is 25%, $6 must be goo 125%, or 1.25 times 1000 1000 the cost, and the cost must be yVu of $6, or $4.80. By selliiif;; that .28-$4.80^i.48; -§^ = .10^10% which cost $4.80 at ^^^ $5.28, the gain is $.48, which is 10% of $4.80, RE VIE W PR OBLEMS. 191 9. A man drew from a bank $264, which was 8^% of his deposit. What was his deposit ? 10. I sold a watch for $69, and lost 20 per cent. What per cent, should I have gained if I had sold it for $93.50 ? 11. Do I make or lose in selling an article marked 25% above cost, if I deduct 20% ? Soi>TTTiON.— The marked price is 25% above cost, or 125% of cost 20%, or \ of 125% of cost, is 25% of cost. 125% —25% of cost is 100% of cost, or the cost. Hence, I neither make nor lose. 12. I sold goods marked 40% above cost, at a deduction of 35%. What per cent, did I lose? Ans. 9. 13. A merchant sold two bills of goods of $75 each ; on the one he made 20%, and on the other he lost 20%. What was his gain or loss ? Ans. Loss, $6.25. 14. Find the net proceeds of the following account of sales, rendered by Barnard & Smith, commission merchants, to Henry Law, consignor : Sales of Wheat for Acct. of Henry Law, Elgm, 1871. To WHOM Sold. Description. Price. Jan. 4 Albert Ward. 101 hi. No. 1 Spring. $1.50 $151 50 ti 8 Snyder & Co. 552 " Mixed Spring. 1.40 772 80 Feb. 3 0. Sinith & Co. 810 " Amber State. 1.45 449 50 " 10 H. A. Stein. 500 " White. 1.60 800 00 " 17 Thomas Prince. 75 " Mixed Spring. 1.40 105 00. Mar. 2 H. Dunster. 124 " Amber State, 1.45 179 80 CHAEGES. $2458 60 Freight and Drc Insurance on $2( $51.00 80.00 11.00 122.98 214 100 @ lj% Commission, 5% , on $2458.60 Net proceeds 98 Barnard & Smith. Chicago, March 4, 1871. 192 SIMPLE INTEREST. SECTION XLVIII. SIMPLE lA^TEREST. 458. — Ex.'l. When the allowance for the use of money is 6 per cent., how many hundredths of the money is the allowance? 2. When the allowance for the use of money is 7 per cent., how many hundredths of the money is the allowance ? 3. When the allowance for the use of money is 5 per cent, per year, what is the allowance for the use of $1 for 1 year ? For 2 years ? For 3^ years ? 4. When the allowance for the use of money is 6 per cent, per year, how many hundredths of the money is it for 12 months ? For 2 months ? 5. At the rate of 6 per cent, for the use of money per year, to how much will $300 amount in 2 years ? 6. When the allowance for the use of $200 per year is $30, W'hat is the yearly rate ? 7. If I should lend $500 at a yearly rate of 4 per cent, for its use, to how much would it amount in 2\ years ? DEFINITIONS. 459. Interest is an allowance for the use of money. 460. The Principal is the sum for the use of which interest is paid. 461. The Amount is the sum of the principal and interest. 462. The Rate of Interest is the rate per cent, of the prin- cipal allowed for its use one year. 463. A Legal Rate of Interest is any rate allowed by law, and Usury is interest reckoned at a higher rate than the law allows. When in a contract between parties, no rate of interest is named, in most of the States and on debts due the United States, the legal rate is 6% ; in some of tlio States it is 7^c, in several ^'/c, and in others 10%. In this book, when no particular rate is named or implied, Qfo is understood. SIMPLE INTEREST. 193 464. In the Computation of interest it is customary to reckon 30 days a month, and 12 months, or 360 days, a year ; and in finding the difterence between dates, to Take the number of entire calendar ino)iths, and the actual number of days left. Thus, from January 20 to August 4 is 6 mo. 15 da., or 6^ mo. Months are often conveniently expressed as twelfths of a year, and 3 days as a tenth of a month. Thus, 6 mo. 15 da. may be expressed as G.5 mo., or as j^ of a year. 465. In the process of reckoning interest, partial results, if necessary, may be carried to four orders of decimals. But in answers it is sufficiently exact to reject mills if less than 5, and if 5 or more than 5, to call them 1 cent, 466. Simple lutei'est is interest on the principal alone. CASE I. Principal, Rate and Time Given, to Find the Interest or Amount. GENERAL METHOD. 467. — Ex. 1. What is the interest and what is the amount of $576 for 2 y. 6 mo., at 6% ? $576 .06 ^3 J- 56 Solution. — 2 y. 6 mo. are 2| years. ■ q1 The interest of $576 for 1 year at 6% is .00 of $576, 2 or $34.56. $69.12 Since the interest for 1 year is $34.56, for 2^ years it 1 7 28 "^"•''* ^® ^^ ^^^^^ $34.56, or $86.40. '- The principal added to the interest, or the amount, ia $86.40 $662.40. 576.00 362.40 2. What is the interest of $760 for 4 y. 8 mo. ? 3. What is the interest of $662.50 for 3 y. 4 mo.? 17 194 SIMPLE INTEREST. 4. What is the interest of §480.50 for 2 y. 7 mo. 12 da., at 8% ? $480.50 .08 $38 AAOO Solution. — 2y. 7 mo. 12 da. are 31.4 mo., 31.4 y. The interest of $480.50 for 1 year, at 8^, is 15376 .08 of $480.50, or $38.44. 3844 Since the interest for 1 year is $38.44, for i 1 notp 31.4 mo., or ^\--^ of a year, it must be ^2* of , ^ — $38.44, or $100.58. 12 )$1207.016 $100.58 5. What is the interest (Tf $78.50 for 3 y. 3 mo., at 7% ? 6. What is the amount of $110.25 for 1 y. 8 mo., at 6^^ ? Ans. $121.28. 468. Principle. — The interest is equal to the product of the principal, rate and time. 469. Rules for Interest by the General Method.— i. Multiply the principal hy the rate, -and that product by the time expressed as years. Or, 2. Multiply the principal hy the rate, and that product by the time expressed as months, and divide the result by 12. S. Add the interest and principal, and the result will he the amount. mOIiLJEMS. What is the interest of — 1. 85631 for 1 year, at 6% ? Ans. $337.86. 2. $860 for 2 years, at 7% ? Am. $120.40, 3. $325for 5years, at8%? 4. $1450 for 2 years 6 months, at 6% ? Ans. $217.50. 5. $111.42 for 4 years 2 months, at 5% ? 6. $19000 for 2 y. 4 mo., at 7^% ? Ans. $3236.33^ 7. $6600 for 3 y. 6 mo. 20 da., at (5% ? Ans. $1408. SIMPLE INTEREST. 195 8. What is the interest of $750 from Jan. 9, 1869, to Nov. 9, 1870 ? Ans. $82.50. 9. What is the amount of $3350 for 5 y. 9 mo., at 8% ? 10. What is the amount of $1242 from July 3, 1868, to Jan. 18,1870? ^ns. $1356.89. SIX PER CENT. METHOD. 470. The Interest on any principal, at 6%, For 12 months, or 1 year, is .06 of the principal. " 2 months, " 1 month, j month, " .001 j^ month,'' .000^ ^ month, " .000^ " 6 days, " 3 days, 1 day. Hence, the following 7 year, " .01 ^, year, " .005 30 471. Principles. — 1. The interest at 6 per cent, for any num- ber of months is equal to one half as many hundredths of the principal as there are inonths ; and 2. The interest at 6 per cent, for any number of days is equal to one sixth as many thousandths of the principal as there are days. WBITTEN EXEMCTSES. 472.— Ex. 1. What is the interest of $576 for 1 y. 7 mo., at fit/.? nol SoXiTTTiON. — 1 y. 7 mo. are 19 months. '2 Since the interest at 6 % is one half as many hundredths 5 184- ^^ *^^® principal as there are months in the time, it must be Qgg one half of 19 hundredths, or .09|, of $576, which is $54.72. $54.72 2. What is the interest of $950 for 2 y. 8 mo., at 6% ? 3. What is the interest of $420 for 3 y. 4 mo., at 7% ? 196 SIMPLE INTEREST. 4 What is the interest of $455 for 2 y. 6 mo. 12 da., at 7% ? $455 152 Solution. — 2 y. 6 mo. 12 da. are 30.4 mo. Since the interest at Q'^'o is one half as many hundredths of the principal as there are montlis in the time, 2275 it must be one half of 30.4 hundredths, or .152, 455 of $455, which is $69,160. r )^PQ 1RC) "^ ^ interest is \ more than 6 fo interest ; hence, OJ^OJ.IOU $69,160 plus 1 of $69,160, or $80,687, is the in- terest required. 11.5266 + $80,687 5. What is the interest of $940 for 33 da. at 6% ? '005- Solution. — Since the interest at 6% is one sixth as ~ many thousandths of tlie principal as there are days in Jf-i UU fijg time, it must be one sixth of 33 thousandths, or .005|, 470 of $940, which is $5.17. $5.17 6. What is the interest of $756 for 8 mo., at 6% ? An8. $30.24. 7. What is the interest of $631.20 for 11 mo., at 8% ? 473, Rules for Interest by the Six Per Cent. Method.— i. Mul- tiply the principal by one half as many hundredths as there are months, or by one sixth as many thou- sandths as there are days in the time, and the result will be the interest at 6 per cent. ^. For interest at any other rate than 6 per cent., increase or diminish the interest at 6 per cent, by such a paH of itself as will mahe the required interest. PROIiLEMS. Wliat is the interest of — 1. $38.60 for 6 mo. 24 da., at 5% ? 2. $1090 for 14 da., at 6% ? Arts. $2.54. 3. $400.50 for 7 mo. 6 da., at 7% ? 4. $5000 for 63 da., at 9% ? An^. $78.75. SIMPLE INTEREST. 197 5. $342 for 93 da., at 7% ? 6. S1200 for 1 mo. 21 da., at 6% ? Ans. $10.20. 7. $1560 for 1 y. 8 mo., at 7% ? 8. $1920 for 2 y. 3 mo., at 5% ? Ans. $216. 9. $500 from January 15 to December 2, at 72% ? 10. What is the amount of $1345 from April 9, 1870, to September 5, 1871, at 7% ? $1477.59. 11. What is the amount of $3000 from June 11, 1870, to Aug-ust 17, 1871, at 8% ? Ans. $3284. SPECIAL METHODS FOR DAYS. 474-. Tlie Interest of any principal at 6% for 2 months, or 60 days, is one hundredth of the principal. Hence the fol- lovring — 475. Principle. — The interest of any jyrindpal at Q'fo for any number of days is as many hundredths of the jmncipal as 60 is contained times in the number of days. WMITTEW JEXERCISJES. 476.— Ex. 1. What is the interest of $240 for 93 da., at 6% ? Solution. — TJie interest at Int. of $240 for 60 da. = $2.40 ^% for 60 days is one hun- Q6) drcdtli of the principal, or ~^— $2.40. /^U Since for any number of 2160 days it is as many hundredths fiO) '^'P'P'^ ^0 ■ ^^ *^^® principal as 60 is con- — — tained times in the number of $3.72 days, for 93 da. it must be ^V of 93 times $2.40, or $3.72. Or, Solution. — Since the in- $24-0 = Principal. terest for 60 days is $2.40, for $2.40 = Int. for 60 da. ^0 days, or \ of 60 days, it 2 20= "' 30 da. """"^ ^^ ^ °^ ^'^■^^' °'' ^^•"^' .12= " 3 da. and for 3 days, or y^y of 30 days, it must be ^V of $1.20, or $.12. >y2 = " Q^ da. '^^Q sum of these results is $3.72, which is the interest required. 17 « 198 SIMPLE INTEREST. 2. What is the interest of $120.60 for 11 da., at 6% ? 3. What is the interest of $500 for 123 da., at 5% ? Ans. $8.54. 477. Rule for Interest by Special Method for Days.— i, i2e- TYhove the decimal point in the principal tivo orders to the left, for the interest at 6 per cent, for 60 days. For the interest for any other number of days, multiply the interest for 60 days by the number of days, and divide by 60 ; or, talce any convenient multiples or aliquot parts of the interest for 60 days. PROBLEMS. What is the interest of — 1. $318.20 for 36 da., at 6% ? Ans. $1.91. 2. $415 for 19 da., at 6% ? 3. $31.25 for 16 days, at 6% ? Aiis. $.08. 4. $1120 for 153 days, at 7% ? 5. $6000 for 8 days, at 12% ? Ans. $16.00. 6. $311.50 for 35 days, at 6% ? Ans. $1.82. 7. $65.20 for 130 days, at 7% ? 8. Find the amount of $17000 for 75 days, at 8%. CASE II. Principal, Interest and Time Given, to Find the Rate. 478.— Ex. 1. At what rate will $576 earn $86.40 in 2 years 6 months ? w 3 >r -r- -T- at 1^ for 2^ years is 5!l4.40. $86.40 -^$14.40 = 6 ''^i"<=e the principal at 1% in 2\ years earns $14.40, to earn $86.40 it nuist be at as many per cent, as $86.40 is times $14.40, or at 6%. 2. At what rate will $1450 earn $217.50 in 2 years 6 months ? 479. Rule for Finding the Rate of Interest. -i)^z;z^e the given interest by the interest of the principal for the given time at 1 per cent. SIMPLE INTEREST. 199 PltOJiLJEMS. At what rate will — 1. $1760 earn $246.40 in 2 years? Am. 7%. 2. $110.25 gain $11.02i in 1 year 8 months? Am. 6%. 3. $6600 gain $1412.40 in 3 years 6 months 20 days? 4. $19000 gain $3236.33^ in 2 years 4 months? Ans. 7j\%. 5. At what rate will $100 double itself in 16 years 8 months? Am. 6%. 6. At what rate will any principal double itself in 12|^ years? 7. If you should pay, January 22, 1871, $735, as principal and interest for $700 borrowed July 22, 1870, what would be the rate? Am. 10%. C.A.SE III. Pi-incipal, Rate, and Interest, or Amoimt Given, to Find the Time. 480.— Ex. 1. In what time will $576 gain $86.40, interest being at 6% ? $576 X.06 = $34.56 Solution.— The interest of $576 for 1 year, at 6%, is $34.56. $86.4-0 -^ $34-56 = 2- Hence, it must require as many years to gain $86.40 interest as $86.40 is times $34.56, or '2\ years, which are 2 years 6 months. 2. In what time will $400 amount to $435, at 7 % interest ? $435 — $400 = $35 Solution. — The amount $435 less the (Ji(Qo principal $400 is $35, which is the interest. X.(J7 — ^^8 rpj^g interest of $400 for 1 year, at 7%, is ^$28=lj $28. * Hence, it must require as many years to gain $35 interest, or for $400 to amount to $435, as $35 is times $28, or 1;[ years, wliich are 1 year 3 months. 3. In Vthat time will $650 gain $55.25, interest being at 6% ? 481. Rule for Finding the Time that a Sum has been at Interest. — Divide the given interest by the iiiterest of the princi- pal for one year at the given rate. 200 SIMPLE INTEREST. mo BL JEMS. In what time will — 1. $7000 gain $588, at 7% interest ? A^is. 1 y. 2 mo. 12 d. 2. S6000 gain $355, at 5% interest? 3. $100 double itself, at 7% interest? Ans. 14 y. 3f mo. 4. $1250 amount to $1253.12^ at 6% interest? Atis. 15 days. 5. Any principal double itself, at 8 % interest ? 6. $1828.60 amount to $2331.33f, at 10% interest? Ans. 2 y. 9 mo. Time, Rate, and Interest, or Amount (Jiren, to Find the Principal. 482. — Ex. 1. What principal will earn $86.40 in 2 y. 6 mo., at 6% ? Soi^rTiON.— The interest of $1 for 2 ■ ^1 X Ofi X S- = ^ 15 years 6 months, at 6^^, is $.15. ^ ' Hence, it must require as many dollars $86.40 -^$.15 = 576 to earn $86.40 as $86.40 is times $.15, or $576. 2. What principal will amount to v435 in 1 year 3 months, at 7% ? .Solution. — The amount of $1 for 1 $1 + $.08- = $1.08 J year 3 months, at 7f,, is $1.08}. Hence, it mu.st require as many dollars $435 -^ $1. 08 J = 400 to amount to $435 as $435 is times $1 .08 1, or $400. 3. What principal will gain $45.24^ in 3 years, at 6% ? Ans. $251.80. 483. Rule for Finding the Principal that has been at Interest— Divide the given interest or aniojvnt by the interest or amount of $1 for the given time and rate. riton T.r.MS. What principal will — 1. Gain $18.75f in 90 days, at 6% ? Ans. $1250.50. 2. Gain $246.40 in 2 years, at 7% ? SIMPLE INTEREST. 201 3. Gain $352.50 in 1 year 2 months 3 days, at 5% ? Ans. $6000. 4. Gain $3236. 33| in 2 years 4 months, at 7y\% ? 5. Amount to $355.60 in 1 year 7 months 10 days, at 8% : Ans. $315. 6. Amount to $200 in 14 years 3f months, at 7% ? Ans. $100. TEST QUESTIONS. 484. — 1. What is Interest? What is the principal ? The amount? The rate ? 2. What is a Legal Rate of interest? What is usury? What is the rate in most of the States, when no rate is named ? What are the rates in other States ? On debts due the United States ? 3. In the Computation of interest, how is it customary to reckon time? In finding the ditierenee between dates? How far in the pro- cesses of reckoning interest may partial results be carried? What is sufhcient when there are mills in the answer ? 4. What is Simple Interest? To what is simple interest equal? What is the rule for the general method of computing interest ? 5. To what is interest at Qfo for any number of months equal? For any number of days? What is the rule for the 6^ method of com- puting interest? 6. What part of the principal is the interest at 6^ for 2 months? How do you most readily find the interest at 6^^ for 60 days? For any other number of days ? 7. When the principal, interest and time are given, how do you find the rate? When the principal, rate and interest, or amount, are given, how do you find the time? When the time rate and interest are given, how do you find the principal ? 8. What is Percentage? To what is the percentage of a number equal? To what is the rate equal ? To what is the base equal ? 9. To what is Profit and Loss equal? What are the rules for profit or loss ? 10. What is the base of Commission? The amount? The net pro- ceeds ? What are the rules for commission ? 11. How do property and life insurance differ? What are the rules for insurance ? 202 FAHTIAL PAYHENTS. SECTION XLIX. PARTIAL PAYMENTS. 485. A Note, or a Promissory Note, is a written promise to pay a certain sum of money for value received. 486. The Maker of a note is the party who signs it. 487. The Payee of a note is the party to whom, or to whose order, it is to be paid. 488. An Indorser of a note is a party who writes his name upon the back of a note or other obligation, to transfer it or to guarantee its payment. 489. The Face of a note is the sum named in it. The number of dollars should be written in wards. 490. A Time Note is one made payable at a specified time. When no time for its payment is specified, the note is due on demand. 491. Three days, called Days of Grace, are usually allowed for the payment of such a note. Thus, a note payable 30 days after date is really due on the last day of grace, or 33 days after date. 492. A Negotiable Note is one so made that it can be sold or transferred. FOEM OF A NEGOTIABLE NOTE. fQOOj^Q' g^_ Louis, January 4, 187L For value received, I promise to pay to the order of 50 Andrew Hale Eiglit Hundred Sixty ^ Dollars, on demand, ivith interest. Daniel Wright. Andrew Hale can transfer this note by sim])ly writing his name on its back, or he may transfer it to John Jones by writing upon its back, Pay to the order of John Jones. Andrew Hale. PARTIAL PAYMENTS. 203 493. Partial Payments are part payments of notes or other obligations bearing interest, 494. Indorsements of partial payments are statements of the payments on the back of the instrument. 495. The Supreme Court of the United States, and most of the States, adopt for partial payments a rule based upon the following : 496. Principles. — 1. Payments must he applied first to the discharge of interest due, and the balance, if any, toward the discharge of the principal. 2. Interest must not be added to the principal to draw interest. 3. Interest must accrue only on unpaid principal. WJtITTEN EXEMCISES. 497.— Ex. 1. A note for $2000 was given July 1, 1870. Upon it was paid, as by indorsement, January 1, 1871, $260; July 1, 1871, $50; and May 1, 1872, $96. Required the balance due November 25, 1872. Solution. Principal, $2000.00 Int. to Jan. 1, 1871, 6 mo., *" 60.00 Amount, $2060.00 First payment, Jan. 1, 1871, 260.00 New principal, $1800.00 Int. to May 1, 1872, 16 mo., I44.OO Amount, $1944.00 Second payment, July 1, 1871, $50 Third payment. May 1, 1872, 96 I46.OO New principal, $1798.00 Int. to Nov. 25, 1872, 6.8 mo. 61. IS Amount or balance due, $1859. IS 204 PARTIAL PAYMENTS. 498. Rule of the Supreme Court of the United States for Partial Payments.— i^^j/it^ the cunount of the principal to the time when the payment, or the sum of the payments, equals or exceeds the interest due, and subtract the payment or the sum of the payments. Regard the remainder as a neiv principal, and proceed as before. ' ritOBLJSMS. 1. Find the amount due December 29, 1871, at 7%, upon a note for $960, dated Albany, N. Y., March 11, 1869; and on which there has been paid, November 1, 1870, $63.52 ; and April 17, 1871, $70.60. 2. Find the balance due April 1, 1872, on the following note: $500. Harrisbitrg, May 16, 1869. For value received, on demand, I promise to pay Clia^rles Berger, or bearer. Five Hundred Dollars, with interest, ivithout defalcation. John Hofland Indorsements: Jfov. 22, 1869, received Forty-five Dol- 50 lars; May 28, 1870, received Seventy j^ Dollars. Ans. $460.47. 3. Find the balance due July 25, 1872, at 6%, on the fol- lowing note : $6000. Providence, January 1, 1870. On demand, for value received, I promise to pay the bearer Six Thousand Dollars, with interest. James D. Mowry. Indorsements: Jvly 1, 1870, received One Tlwusand Dollars; May 1. 1872, received Three Thousand Dol- lars. PARTIAL PAYMENTS. 205 4. Find the balance due August 1, 1870, at 8%, on the fol- lowing note : $4000. Richmond, June 5, 18G9. / proiyiise to pay, to the order of Andrew L. Brown, Four TliousaJid Dollars, with interest. Value re- ceived. H. M. Reeves. Indorsements: August 17, 1S69, received One Thou- sand Bollars; January 29, 1870, received One Hun- dred Dollars. ^^s^ $3198.91. 499. Business men often settle notes and interest accounts, running not more than a year, by the following, called — 500. The Merchants' Rule for Partial Payments.—i^^Mt^ the amount of the principal at the time of settlement. Find the amount of each payment from the time itivas made until settlement; and from the amount of the principal subtract the amounts of the pay- ments. In mercantile accounts, settlements are made, according as the custom may be, either at the end of the civil year or at the end of some under- stood number of months. PltOTiT.EBIS. 1. A note for $1500, on demand with interest, dated Jan- uary 1, 1870, had paid on it. May 1, 1870, $800, and May 16, 1870, $300. How much was due at the end of the first six months, interest being at 7% ? Ans. $440.54 2. Find the balance due at the end of the year on a note for $650, given April 9, 1870, on which has been paid, June 9, $150, and November 30, $200. Ans. $322.33. 3. Find the balance due August 5, 1871, on a note for $1275, given September 30, 1870, on which has been paid, December 5, 1870, $55 ; January 9, 1871, $760 ; June 3, 1871, 8400. 18 206 PRESENT WORTH AND DISCOUNT. SECTION L. FEi:SEJVT WORTH A^D DISCOUNT. 501. Discount is a sum deducted from a price or debt. Its computation may have reference to time, or not, according to the kind of deduction understood. COMMEECIAL DISCOUNT. 502. Commercial Discount is a per cent, deducted from a price or from the face of a bill, without reference to time. 503. The Net Price of an article is the selling price less the discount. 504. The Cash Talue, or Net Proceeds, of a bill is its face less the discount. 505. The Base of commercial discount is the selling price, or the face of a bill. 506. The Rate is the rate per cent, of deduction. 507. The Discount is the percentage of the deduction. 508. Rules for Commercial Discount— MuUiplj/ the selling price, or the face of the hill, hy the rate per cent, of deduction, and the result will be the coimnercial discount. Subtract frorn the selling price, or from tlie face of the bill, the coimnercial discount, and tlie result will be the net price, cash value, or net proceeds. PliOBLEMS. 1. What is the net cash price of flour invoiced at $12.40 per barrel, on 30 days' time, or 5% oflT for cash ? A)).nn ^^^^' ^' '^^^ ^^^^^ of credit is months. 800 X 6' = 4-800 is the same as the credit of 2 times $1500 )$6000 ^^^^' °^ ^1200, for 1 month. ^ ^ The credit of $800 for G months is -¥• the same as the credit of 6 times -r -, , , ^T -I $800, or $4800, for 1 month. Jwiuary 1 + 4 mo. = Mai) 1 „ ,, ,.^ » ^, •^ ^ Hence, the credit or the entire indebtedness, or $1500, is the same as that of $1200 + $4800, or $6000, for 1 month, which is equal to the credit of $1500 for as many months as $1500 is contained times in $6000, or for 4 mouths. January 1 + 4 months equals May 1. We may also consider the debtor as entitled to the use, or interest, of each of the debts for its term of credit. Hence, the following Interest of $100 for mo. =$00. 00 Solution. - Eeckon- " 600 " Smo.= 6.00 i"g ^^^^ interest at 6%, onn u ^,„. _ SAOO t'^e aggregate of interest ^'^^ ^ ^^^•— ^^-^^ for the terms of credit is Debts, $1500 Total Int., $30.00 $6 + $24, or $30. 23G . AVERAGE OF PAYMENTS. • The interest of $1500, or the sum of the debts, at the same rate for 1 month is $7.50. Wilder should therefore have the use of the $1500 as many months from January 1 as $7.50 is contained times in $30, which is 4 ; and 4 months from January 1 is May 1. Any rate of interest might have been used in the computation, and the result would have been the same. 2. Three debts are due me — one of $120 in 5 months, an- otlier of $125 in 4 months, and a third of $500 in 8 months. What is the average time of their payments ? 581. l{\x\Q%.— Multiply each of the debts by its term of credit, and divide the sum of the products by the sum of the debts ; the quotient will be the average term of credit. Or, Find the iivterest of each debt for its term of credit, and divide the sum of their interests by the interest of the sum of the debts for one month o?^ one day ; the quotient will be the average teimi of credit. The date of the debts, plus the average term of credit, will be the average time. In finding the average term of credit when any of the debts have cent«i, it is customary to neglect them if less than 50 ; and if 50 or more to regard them as $1 . In a result, if there be a fraction of a day, reject it when less than \ ; and when otherwise, call it 1 day. TItOJiljEMS. 1. A merchant owes $60 due in 72 days, $85 due in 128 days, $70 due in 176 days, and $105 due in 320 days'. Re- quired tlie average time at which the Avholc will be due. 2. January 1, Alfred Day bought bills of goods payable as follows: $70 at date, $110 on March 2, $80 on May 5, $120 on July 20, $48 on September 27, and $50 on October 7. Recjuired the average time of payment. An&. May 22. 3. July 5, Johnson Paterson bought bills of goods payable as follows: $500 on August 5, $600 on Septend)er 5, and $1000 on September 20. Required the average tiuie of payment. AVERAGE OF I'A I'M EATS. 237 C^SE II. Terms of Credit Beginning: .at Different Times. 582. — Ex. 1. I bought goods of James Hunt & Co. as fol- lows : March 1, a bill of $500, ou 4 months ; March 22, a bill of 6200, on 2 months, and April 29, a bill of 8680, on 5 mouths. What is the average time of payment of the whole ? Solution. • 3Iarch22+2mo.=May22, 000X0 =$00000 March 1+4," =JuJy 1, 500X^0 = 30000 April 39 +5 " = Sept. 39, 680X130= 88^00 $1380 )$ 108400 (7 8^^ 9660 11800 May 33 + 79 days = Aug. 9. llOJfi 760 Interest of $300 for days = $00.00 " 500 " 40 days = 3.33 « 680 " 130 days = 14.73 Or, Sum of bills, $1380 Total interest, $18.06 Interest of $1380 for 1 day = $.33 $18.06 -^ $.33 = 78§. May 33 + 79 da. =Aug. 9. The several bills are due July 1, May 22 and September 29, respectively. Selecting the earliest day of maturity as the day from which to reckon, $200 has no term of credit, the $500 has a credit of 40 days, and the $680 has a credit of 130 days, from May 22. The average term of credits, by either form of solution, is 79 days, nearly. Hence, May 22 -f- 79 days, which is August 9, is the average time required. The first form of solution is called the Product Method, and the second the Interest Method. Accountants generally prefer the latter. The date of the first debt's maturity was selected to reckon from for con- venience. Had the latest date of maturity been selected, the average time would have been counted back from that date. 238 AVERAGE OF PAYMENTS. 2. Kobert Hendricks gave me, June 4, a note for $315.63 on 4 months ; June 15, a note for $535.47 on 2 months ; 'and July 3, a note for $300 on 3 months. Regarding these notes without grace, should he wish to take them up by giving one note for their amount, when should it be payable ? Ans. September 10. 583. ^w\b.—Find the date at which each debt becomes due. Select the earliest date (it which any of the debts matures, and rechoning fj^om it, as in the previous case, find the average term of credit; and the selected date, plus the average term of credit, will be the average time. In working by the interest method, it may be most convenient to take for the selected date the ^rst day of the month in which the first credit begins. When the terms of credit are all equal, we may simply find the aver- age date of the debts, and add the common term credit, for the average time. jphoblems 1. When should a note to settle the following account be made payable ? Lewis Manly, To Stone, Dexter & Co, Dr. 1871. May 13 To Merchandise @ .^ mo., as per bill <$soo 15 <( u " " @ 2 mo., « 800 00 June 15 Cash 99 83 $899 98 Ans. August 17. 2. Purchased of Jonas Mungor, on a credit of 90 days, January 6, a bill of $G00, and February 15,. a bill of $200. Required the average date of purchase and the average time of payment. Ans. January 16 ; April 16. AVERAGE OF PAYMENTS. 239 3. What is the average time of the following bills, allow- ing to each term of credit 3 days' grace ? — Sold, April 3, a bill of $500 on 3 months ; April 4, a bill of $200 on 2 months ; April 4, a bill of |200 for cash ; and April 10, a bill of $500 on 3 months. Ans. June 21. CASE III. Debit and Credit Accomit. 584. — Ex. 1, From what time should a note draw interest for the balance of the following account, allowing 3 days' time to the i^ms on time ? Dr. James Blake in account with George Hill. Cr. 1870. 1870. May 8 To Mdse., Cash $ 40 00 May 10 By Cash $ 30 00 " 17 " " 30day.fc in gold, and is payable quarterly. 4|'s of 1886 are bonds which are payable after 1886. The interest on them is at the rate of 4J % in gold, and is payable quarterly. 4's of 1901 are bonds which are payable after 1901. The interest on them is at the rate of 4% in gold, and is payable quarterly. 606. Bonds issued by cities, counties, States and corpora- tions are usually named according to the rate of interest they bear. Thus, Virginia 6's are bonds bearing interest at 6fo, issued by the State of Virginia. WBITTEN EXERCISES. 607. — Ex. 1. What is the cost, including brokerage, of 400 shares of railroad stock, at 95% ? S0I>UTI0N. (95% + j%) of $100 - $95.25, cost of 1 share. $95.25 X 400 - $38100, cost of 4OO shares. 2. What is the market value of 50 shares of National Bank stock, at 115? STOCKS AND INVESTMENTS. 245 3. How much, including brokerage, must be paid for $1000 Maine 6's at 101 ? Solution. (101^0+ jfc) of$l = $1.0lj, amount paid for $1. $1.01JX1000=$1012.50, " « $1000. 4. When gold is quoted at 11 2^, what is the value in cur- rency of $5000 in gold ? Ans. $5625. 5. I bought Government securities of the par value of $3000 at lOOf , and sold them at 109^. How much did I gain ? Am. $255. 6. When gold is worth 112^, what is the value in gold of $5625 in currency ? Solution. At 112^$! of exchange by the num- ber denoting the face of the bill. 2. To find the face of a bill that can be bought for a given sum, divide the given sum by the ivumber denoting the cost of $1 of e.vchange. EXCHANGE. 249 PROBLEMS. 1. What will be the cost of a sight draft on New York for $5850, at 1% premium ? Am. $5879.25. 2. When exchange is at ^% premium, what is the face of a sight draft that costs $5879.25 ? 3. How much must be paid in Pottsville for a draft of $750 on Pittsburg, exchange being at l|-% discount? Ans. $738.75. 4. What is the face of a draft that can be purchased for $4301, when exchange is at 2^% discount? Ans. $4400. 5. What must be paid in Vicksburg for a draft of $600 on St. Louis, at 60 days, exchange being at 101, and interest at 9%? Solution. Bank discount of $1 for 63 days, at 9% = $.01575. $1 - $.01575 = $.98435, cost of $1 at par. $.98425 + $.01 = $.99425, cost of $1 at 101. $.99425 X 600 = $596.55, cost of the draft. Or, Bank discount of $600 for 63 days, at 9% = $9.45. $600-$9.45=-$590.55, cost of draft at par. $600 X .<9i = $6, premium of draft at 101. $590.55 + $6 = $596.55, cost of the draft. 6. What will be the face of a draft payable 60 days after sight, that can be bought for $596.55, exchange being 1 % pre- mium, and interest 9 % ? 7. How much must be paid for a draft of $750, payable 10 days after sight, exchange being at |% discount, and interest iit 6% ? Ans. $744,621. 8. I owe a note, in New York, of $3000, with interest for 1 year at 7%. What must be the face of a sight draft, exchange at 101^, which I can remit, and thus exactly discharge the note and interest ? J.m. $3250.12^. 250 EXCHANGE. FOREIGN EXCHANGE. 619. Forei^ Bills are drafts drawn in one country and pay- able in another. 620. The Par of Exchange between two countries is the vahie of the currency of one. country estimated in the currency of the other. 621. Bills of Exchange between the United States and foreign countries for convenience are generally drawn and negotiated on London or Paris. Foreign bills are drawn in sets of three, of the same tenor and date, and called, respectively, First, Second and Third of Exchange. They are forwarded diflerently, to prevent delay by accident. When one is paid, the others are cancelled. BILLS ON ENGLAND. 622. Bills on England, or London, are drawn in English, or sterling money, the denominations of which are shown in the following — TABLE. 4- farthings (qr.or far.) are 1 penny . . . d. 12 pence " 1 shilling. . s. 2 shillings " 1 florin . . . fl. 10 florins, o?' 20 shillings, " 1 pound . . . £. 623. The Value of a pound sterling, which is represented by the English gold coin called the Sovereign, previous to the change in the United States coinage in 1834, was $4f, or $4.44A In the present gold coinage of the United States a sovereign of standard weight is equal to $4.8634. Allowing for the wear of coins, we have what the Government has established as the Custom-house or legal value of the pound, which is $4.84. 624. In the Computation of Exchange the old value of $4^ is usually considered the Base. Hence, when English excliange is quoted at 109, it is at the Custom- house value, and when at 109|, it is at about the intrinsic value, or true par. EXCHANGE. 251 WRITTEN EXERCISES. 625.— Ex. 1. What is the cost in New York of a bill of exchange on London of £400 10 s. 6 d., at 109f, including brokerage at :^% ? Solution. — £400 10 s. £400 10 s. 6 d. = £400.525 ed., decimally expressed, is *//, V 7 M ^ £400.525. £1, at 109|, and ^^^^X400.525 = $1958.12+ ^.^kerage at \7o, will cost $40X1.10. Hence, £400.525 will cost 400.525 times ^^"V^" ' o'^ $1958.12 +. 2. What is the face of a bill on London which, at 109f , and brokerage at \%, can be bought for $1958.12? Since the given number is expressed by 5 orders of figures, it may be separated into three periods, and its square root will consist ot 3 figures, and will express hundreds, tens and units. Eepresenting the hundreds, or tens of tens, by t, and the units of the tens by u, the square of the root thus represented will equal (- + 2tu + u^. This root, then, will be found in 552 units of hundreds. The greatest number contained in the left-hand period that is a square of tens is 4 hundreds, whose root, 2 tens, we write for the tens of tens of the root. Subtracting 4 hundreds and its equal f from the expression above, 152 hundreds remain, and 2ti(, + u\ which is equal to (2< + u)u. Since 2t of the factor 2t + u is large compared with the other term, we use it as a trial divisor in finding the value of u. Now, 2i = 2 X 2 tens = 4 tens, and dividing the 15 tens of the remainder by the 4 tens, we have 3, which we write for the units of the tens of the root. Adding to 2t or 40, 3, the equal of u, we have 43, and multiplying this by 3, the equal of w, we obtain 129. Subtracting 129 from the dividend, and its equal {2t -{-u)u from the expression above it, and bringing down the remaining period of the power, we have as a remainder 2325. We now consider 55225 as the square of 23 tens and some number/)f units of units. Eepresenting the 23 tens of the root, which have been found, by t, and the \inits of the root to be found, by n, the square of tlie root is represented by <* + 2lu + u'^. The equal of t"^, or 230^, has been sub- tracted ; hence, the remainder, 2325, is equal to 2tu + u"; or (2/ + u)u. We find the value of 2t to be 46 tens, or 460, and making use of it as a trial divisor for finding the value of u, we have for that value 5, which we write for the units of the root. Adding to 460, the equal of 2/. 5. the equal of w, and multiplying this by 5, the equal of u, we obtain 2325. Subtracting 2325 from the divi- E VOL VTION. 269 dend, and its equal {2t-{-u)u from the expression above it, we find no remainder. Hence, tlie square root of 55225 is 235. If the root had consisted of four or more figures, the explanation of the process of solution would have been similar. Thus, if the root had consisted of four figures, expressing thousands, hundreds, tens and units, we should represent tlie thousands or tens of hundreds by /, and the units of hundreds by u, and having found their values, should regard the part of the root found, as tens of tens, and represent it by t, and the units of tens by u, and so on. If the given number had been a decimal fraction, its periods would have been pointed off to the right, beginning with units; for the square of .1 is .01, the square of .09 is .0081, etc. 3. Find the square root of 4096. Ans. 64. 4. Fiud the square root of 64516. Ans. 254. 673. Rule for Square ^oo\.— Point off the given munher into periods of two orders each, beginning with the units, and proceeding toward the left a.nd right. Find the greatest square in the highest period, con- sidered as units, and place its root at the right for the first figure of the required' root. Subtract this square from the highest period, and to the remainder bring down the next period for a dividend- Tahe for a trial divisor twice the root already found, considered as tens ; divide the dividend, omitting its right-hand order, by the trial divisor, and ivrite the quotient for the second figure of the required root. To the trial divisor add the part of the root found by it, multiply the result by that part of the root, and subtract the product froiyv the dividend. Continue the process, if there are other periods, as before. The trial divisor being less than the true divisor, the probable root found by it may prove too large ; if so, diminish by 1 or more, and re- new the process. When occurs in the root, instead of indicating the multiplication by and subtracting, it is simpler to annex a cipher to the trial divisor, and to the dividend bring down another period. 28* 270 EVOLUTION. If there be a remainder after all the periods have been used, periods of decimals may be formed by annexing ciphers, and the work continued. PROBLEMS. What is the square root — 1. Of 9216 ? Ans. 96. 2. Of 4096 ? 3. Of .0676 ? Am. .26. 4. Of 717409 ? Am. 847. 5. Of 46656? Ans. 216. 6. Of 7569 ? 7. Of 62504836 ? Ans. 7906. 8. Of 21.16 ? Ans. 4.6. 9. Extract the square I'oot of 5 to three decimal orders, or to within less than yuVo- -^^s. 2.236. 10. Extract the square root of .5 to four decimal orders, or to within less than ^q^qq . Ans. .7071. 11. A general has 11664 men. How many must he place in rank and file to form them into a square ? Ans. 108. 12. A square lot contains 18225 square feet. What is the length of its equal sides ? Ans. 135 feet. 13. What is the value of i/.00062o ? Ans. .025. 14. A circular garden contains 6561 squar.e feet. What is the length of one side of a square containing the same number of square feet ? 15. A man wishes to lay out a farm in a square containing exactly 140 acres 100 square rods. What must be the length of each side? Ans. 150 rods. 674. Since the square of a fraction is found by squaring the numerator and denominator, the square root of a fraction is found hy taking the sqiuire root of the numerator and denom- inator. 16. What is the square root of ||^ ? ^ 224 16 fie 4 A^o feoLVT.ox.-^-^-; ^~ = l,Ans. 17. Find the square root of ^^^. Ans. f^. 18. Find the square root of aVA- -^^^^^ t^- 19. Find the square root of lljf. Aiu. y, or 3f. 20. What is the vahie of i 30]:? Ans. y, or 5^. EVOl^UTION. 271 675. When the numerator and denominator of a fraction are not both squares, to find the approximate square root, First reduce the fraction to a decimal, and then tahe the root. Or, Multiply the momerator by the denominator, and divide the root of the product hy the denominator. 21. What is the square root of f, to within less than y^? Solution.-^ = .4.285 + I V .4-285 = .65 +. Or, V5_ \SX7 _ III.— 4:58 __ ^^ 7~\'n ~\49 ~ 7 --^^ + . 22. Find the square root of ^, to within less than y oVo- Ans. .807. 23. What is the square root of ff |-, to within less than yoVo ? 24. What is the square root of Gf, to within less than yowo ^ GENERAL METHOD FOR CUBE ROOT. 676. A General Method of finding the cube root of num- bers may be deduced from raising numbers of different orders of units to the third power, and noticing the various relations existing between the roots and their cubes. 677. The cube of 1 unit of units, or 1, is 1 ; of 1 unit of tens, or 10, is 1000; of 1 unit of hundreds, or 100, is 1000000; of 1 unit of thousands, or 1000, is 1000000000; and so on. Hence, The cube of units must be found in the orders of units, tens and hundreds, or in the units' period ; of tens, in the orders thousands, ten-thousands and hundred-thousands, or in the thousands' period ; of hundreds, in the millions' period ; of thousands, in the billions' period, and so on. Or, Numbers with 1 figure have in their cubes 1, 2 ov 3 figures. 2 figures " " 4,5 or 6 " 3 figures " " 7, 8 or 9 " 4 figures " " 10, 11 or 12 " and so on. 272 EVOLUTION. 678, Take any number composed of tens and units, as 47, and cubing it as in actual multiplication, notice of what parts the cube is composed, and the orders in which these parts are found. Thu5, ^7 = 4 ^^^^ "^ ^ wwifs ^ tens + 7 units Jj, tens X 7 units + 7 ^' =10800 ) 84-763 = Sf^u-V8tu^-^v? = Stu = SX6tensX7= 1260 {Sf + Stu + m^jm 3f^ + 3tu + w2 = 12109 12109 X 7 = 84763 = J Sf^ +Stu +m')m Since the given number is expressed by 6 orders of figures, it may be separated into 2 periods, and its cube root must consist of 2 figures, ex- pressing tens and units. Eepresenting the tens by t, and the units by n, 300763 is represented by t^ + sC-u + Stv? + V?. In reversing the process of invohition, we must first find l^, and as this must be a number of thousands, we find the greatest number contained in the left-hand period that is a cube of tens. Tliis number is 216 thou- sands, Avhose root is 6 tens, which we write for the tens of the required root. Subtracting 216 thousands from the cube, and its equal, t^, from the expression above, 84763 remains, equal to Zt'^u + 3te^ + u^, or, since u is a common factor of these terms, equal to (3<^ + Ztu + m^)m. Of the two factors of the latter expression, one terra, 3<^ of the factor within the parentheses can be found ; and since this, compared with the other terms, is large, we find and make use of its equal as a trial divisor for finding the equal of the factor u. Now, 3<^ = 3 times 6 tens square, or 108 hundreds ; and dividing the 847 hundreds of the remainder by the 108 hundreds, we have 7, which we write for the units of the required root. Adding to the 108 hundreds, 126 tens, the equal of Ztu, and 49, the equal of w^, we have 12109 ; and multiplying this by 7, the equal of u, we obtain 84763, which we write equal to (3<^ -f 3lume ? How are the cubic contents of a rectangular volume found? (Art. 405.) 11. What is a Prism ? A cylinder? How t>r^ the cubic contents of a prism or of a cylinder found ? 12. What is a Pyramid ? A cone ? The slant height of a pyramid or a cone ? A frustum of a pyramid or cone ? How are the cubic con- teuts of a pyramid or a cone found ? The cubic contents of a frustum of a pyramid or a cone? 13. What is a Sphere ? To what is the surface of a sphere equal ? To what are. the cubic contents equal ? 14. What are Similar Figures ? What relations have like dimen- sions of similar figures? The areas of similar figures? The cubic* contents of similar figures? 15. How does a curved line differ from a straight line? An acute angle from an obtuse angle ? What is the perimeter of a triangle ? Of a square ? Of a circle ? 16. What kind of a prism is a rectangular solid ? What kind of a pyramid is one having a four-sided base? How does a cone difl'er from a pyramid ? How does a sphere differ from a cylinder? 2J 290 MEASUREMENTS OF BOARDS AND TIMBER. SECTION LXVIII. MEASUREMEJfTS OF BOARDS AJ{I) TIMBER. 728. A Board Foot is 1 foot long, 1 foot broad and 1 inch thick; hence, 12 hoard feet are 1 cubic foot. 729. Lumber, or sawed timber, is estimated in board feet. ^^■^"^^B^^^ MM^rfy^^r 730, Eoiiml Timber is usually estimated in cubic feet. 731. Squared or Hewn Timber is estimated either in board feet or cubic feet. 732. The Mean Girt of a tapering log is the circumference, clear of bark, one-third the distance from the larger to the smaller end of the log. 733. The Mean Breadth and Thickness of tapering squared timber is the breadth and thickness of the timber measured at the middle of its leno-th. CASK I. Dimensions of Lumber, or Squared Timber, giren, to And the Contents, 734. — Ex. 1. AVhat is the number of board feet in a board 24 feet long, 18 inches wide and 1 inch thick ? Solution.— 18 inches - ] \ feet. 24 X if X 1 = ^^2^^ == ^^- ^^^^ number of board feet required. MEASUREMENTS OF BOARDS AND TIMBER. 291 2. How many cubic feet in a piece of hewn timber 45 feet long, whose breadth and thickness are 15 and 14 inches ? Solution. — 15 and 14 inches = {| and if feet. 45 X B X 11 = 45 X 15X14 65.625 cubic feet. 144 8. What are the contents in board feet of a joist IG feet long, 4 inches thick, and tapering in breadth from 6 inches to 4 inches? Ans. 26|. The mean of the tapering breadth is 5 inches, which is used in the computation. 735. Rule for Finding the Contents of Lumber and Squared Tmber.—Multiplj/ the length in feet by the hreadth and thichness in inches, and divide by 12 for board feet, or by IJfJi- for cubic feet. If the hunber or squared timber tapers, the mean breadth and thick- ness must be used in the computation. A common rule for estimating tapering squared timber is — Add together the areas of the two ends in square inches, and multiply half the suin by the length in feet, and divide by 12 for board feet, or by IJfJf for cubic feet. pitoBLJi:3i:s. 1. What are the contents of a plank whose length is 20 feet, breadth 16 inches and thickness 3 inches? Ans. 80 board feet. 2. A beam is 30|- feet long, 22 inches broad and 15 inches thick. What are its contents ? Ans. 843^ board feet. 3. The length of a piece of timber is 9.8 feet, and its mean breadth and thickness 2.6 and 1.5 feet. What are the cubic contents ? Ans. 38.22 cubic feet. 4. Bought 20 joists, each 18 feet long, 5 inches wide and 3 inches thick, at S30 a thousand feet, board measure. What did they cost me? , ^ns. 813.50. 5. The breadth and thickness of one end of a stick of timber, whose length is 17 feet 3 inches, are 36 and 20 inches, 292 MEASUREMENTS OF BOARDS AND TIMBER. and of tlie otlier end 18 and 10 inches. What are its cubic contents by the rule, allowing the mean breadth and thickness to be 27 and 15 inches ; and what the true contents, measured as a frustum of a pyramid ? Ans. By the rule, 48,5156 + cubic feet ; true con- tents, 50.3125 cubic feet. C^SE II. Lengdi and Mean Girt of Round Timber g-iren, to find tlie Contents. 736. — Ex. 1. What are the cubic contents of a piece of round timber Avhose mean girt is 100 inches and length 18 feet? Solution.— 1^ of mean girt = \ of 100 = 25; square of \ of mean cfirt = 252 = 625 ; 625X18 = 11250; and 11250 -^- 144 = 78.125, the nnmlier of cubic feet ref|nired. 2. The mean girt of a log is 88 inches, and the length of the log is 40 feet. What are the cubic contents ? Avs. 134.44 cubic feet. 737. Rule for Finding the Cubic Contents of Round Timber.— Multiply the square of one fourth of the mean girt in inches by the length iit feet, and divide by 1^^. Note. — The rule gives about one fifth less than the exact quantity, one fifth heing til lowed for crooks and waste in working. The exact cubic contents may be found very nearly by multiplying the square of ane Jiflh of tlie mean girt in feel by twice the length in feet. PROJil. EMS. 1. The length of a log is 32 feet 6 inches, and its mean girt, after allowing for the bark, is 60 inches. What are the contents by the rule, and what by the note under the rule ? Ans. By the rule, 50.78125 cubic feet; by the note, 65 cubic feet. 2. What is the value of a pine log 30 feet long, and whose mean girt is 10 feet, at $20 per ton of 40 cubic feet? Ans. S93.75. 3. The circunifcrcncc of a piece of round timber is 6 feet 8 inches, and its length 24 feet. What are its contents by the rule, and what as a cylinder? (Art. 724 — 1.) MEASUREMENTS OF STONE, AND BRICK- WORK. 293 SECTION LXIX. MEASVREMEKTU OF STOKE, AKD BRICK- WORK. 738. Stone Masonry is usually estimated by the cubic foot or by the perch. 739. Brick-Layiiiisr is generally estimated by the thousand bricks. 740. A Perch of stone-Avork is 16^- feet long, 1 foot deep and 1^ feet thick, and is equivalent to 24| cubic feet. 741. Bricks are of various dimensions. Philadelphia or Baltimore front bricks are 8|, 4| and 2| inches ; North River bricks, 8, ?>\ and 2|" inches; Maine bricks, 7|, 3| and 2f ; and Milwaukee bricks, 8|, 4| and 2|- inches. C^SE I. Dimensions of Stone-ivork j^iren, to And the Number of Perclies. 742. — Ex. 1. How many perches of stone-work in a M'all 66 feet long, 4 feet high and 3 feet wide ? Solution. 66y.4'Xo=792, mimher of cubic feet. 792 -^ 2^.75 = 32, number of j^erches. 743. Rule for Finding the Number of Perciies of Stone-work.— Find tlie contents in cubic feet, and divide hy 2Jj..7o. PJtOBI.EMS. 1. What are the contents in perches of a stone wall whose dimensions are 24 feet 3 inches, 10 feet 9 inches and 2 feet? Ans. 21.065 + perches. 2. What will it cost, at 83.25 per perch, for the stone- and mason-work of a cellar 8 feet deep, under a house whose length and width are 411 ^nd 33 feet, the wall of the cellar to be 1\ feet thick, and no allowance to be made for corners or openings? Ans. $234. 25 » 294 MEASUREMEl^TS OF STONE, AND BRICK -WORK. ca-sk; II. Dimensions of Bricks and Thickness of Mortar of Brick-ivork giyen, to find the ^"nmber of Bricks. 744. — Ex. 1. The width of a wall is 10^ inches, laid of Maine bricks, in courses of mortar \ of an inch thick. How many bricks has it in a cubic foot ? Solution. 7.5 + (.25 X2)^2-= 7.75, length of brick and joint. 2.S75 +(.25X2)^2 = 2. 625, width of brick and joint. 7.75 X 2. 625 = 20.34375, area of face. 10.5^3 = 3.5 ; 20.34375 X 3.5 = 71.2 + cubic inches. 1728 - 71.2 = 24.269 -^, the number of bricks. 745. Rule for Finding tiie Number of Bricliien the staves are but slightly curved. 753. The Ullage, or wantage, of a cask is the (quantity it lacks of being full. 754. Rules for Gauging.— i. Multiply the product of the square of the mean diameter and the length or depth, of the cash, expressed in inches, hy .003 Jf, and the result jvill be its capacity in gallons. GA UGING. 297 2. Multiply the square of one third of the sum of the head, ineaiv aiul hung diaiueters , expressed in inches, by the height of the liquid in inches, and that product by .003 Jj., and the result will he the contents of ayv ullage cash. PROBLEMS. 1 . How many gallons is the capacity of a cask whose length is 40 inches and mean diameter 25 inches? Ans. 85 gallons. 2. Required the quantity of vinegar in a cask whose bung and head diameters are 37 and 28 inches, and the height of the liquid 10 inches. Ans. 37.026 gallons. 3. How much Avill a cask of molasses cost whose mean diameter is 30 inches and length 36 inches, at 'S.55 per gallon ? Ans. $60,083%. TEST QUESTIONS. 755. — 1. What are the dimensions of a Board Foot? How many board feet are one cubic foot ? 2. What kind of timber is Lumber? In what is squared or hewn timber estimated ? What are the rules for finding the contents of lumber and squared timber ? 3. In what is RocND Timber estimated ? What is the mean girt of a tapering log? What is the mean breadth and thickness of tapering squared timber ? What is the rule for finding the cubic contents of round timber? 4. By what is vStoxe Masoxry estimated ? What is a perch of stone- work? What is the rule for finding the number of perches of stone-work ? 5. IIow is Bricklaying usually estimated ? What is the rule for finding the number of bricks in brick-work? 6. How is Grain usually estimated? How much does the standard bushel contain? What is the rule for finding the quantity, in bushels, of grain in a bin or wagon ? For finding the quantity of grain when heaped upon a floor? 7. How is hay bought and sold ? About how many cubic feet of clover make a ton ? About how many cubic feet of meadow-hay make a ton ? What is the rule for finding the quantity of hay in tons ? 8. What is Gauging ? What are the rules for gauging ? 298 METRIC SYSTEM. SECTION LXXII. METRIC SYSTEM. 756c The Metric System is a system of weights and meas ures based upon a unit called a meter. 757. The Meter is one ten-millionth part of the distance from the equator to either pole, measured on the earth's surface at the level of the sea. 758. The Jfames of derived metric de- nominations are formed by prefixing to the name of the primary unit of a meas- ure — Milli (mill'e), a thousandth; Centi (sent'e), a hundredth; Deci (des'e), a tenth ; Deka (dek'a),' ten ; Hecto (hek'to), one hundred ; Kilo rkil'o), a thousand ; Myria (mir'ea), ten thousand. This system, first adopted by France, has been extensively adopted by other countries, and is much used in the sciences and the arts. It was legalized in 1866 by Congress to be used in the United States, and is already employed by the Coast Survey, and to some extent by the Mint and the General Post-OfEce. The illustration adjoining shows the length of 10 centimeters, or a tenth of a meter, compared with 4 inches, or a third of afoot. The nickel 5-cent pieces are each two hun- dredths of a meter in diameter; hence, 50 of them placed side by side in a straight line will measure 1 meter. The simplicity and utility of the system, recognized now by all civil- ized nations, is likely to lead to its general adoption, and to the great advantage of home and foreign trade. In the tables the units most used are denoted by CAPITALS or by plain Roman type. METRIC SYSTEM. 299 LINP]AR MEASURES. 759. The Meter is the primary unit of lengths. TABLE. 10 millimeters [mm.) are 1 centimeter [cm.) = .S9S7 in. 10 centimeters ' 1 decimeter = S.937 " 10 decimeters ' 1 METER (m.) = 89.37 " 10 meters ' 1 dekameter = S9S.7 10 dekameters ' 1 hectometer = S28 ft- 1 " 10 hectometers ' 1 KILOMETER {km.) = .62137 mi 10 kilometers ' 1 viyriaincter = 6.2137 " The Meter is used in ordinary measurements; the Centimeter, or Millimeter, in reckoning very small distances ; and the Kilometer, for roads or great distances. A Ceniimeter is about | of an inch ; a Meter is about 3 feet 3 inches and I of an inch ; a Kilometer is about 200 rods, or f of a mile. SURFACE MEASURES. 760. The Square Meter is the primary unit of ordinary sur- faces ; and, 761. The Are (air), a square each of whose sides is ten meters, is the unit of land measures. TABLE. 100 sq. millimeters (sq. mm.) are 1 sq. centimeter {sq. cm.) = .155 sq. in. 100 sq. centimeters " 1 sq. decimeter = 15.5 sq. in. f 1550 sq. in., or 100 sq. decimeters " 1 sq. meter [sq. m.) = < ^ ^^^ ^ , Also, 100 centiares, or sq. meters, are 1 are (ar.) = 119.6 sq. yd. 100 ares " 1 hectare {ha.) = 2.Jfll acres. A Square Meter, or 1 Centiare, is about 10| square feet, or 1^ square yards, and a Hectare is about 1\ acres. CUBIC MEASURES. 762. The Cubic Meter, or Stere (stair), is the primary unit of a volume. TABLE. 1000 cu. millimeters {cu. mm.) are 1 cu. centimeter {cu. cm.) = .061 cu. in. 1000 cu. centimeters '• 1 cu. decimetet = 61.022 cu. in. 1000 cu. decimeters " 1 cu. meter (cm. m.) = 35.314 cu. ft. 300 METRIC SYSTEM. The Stere (stair) is the name given to the cubic meter in measuring wood and timber. A tenth of a stere is a Decislere, and ten steres are a Dekastere. A Cubic Meter, or Stere, is about 1 \ cubic yards, or about 2i cord feet. LIQUID AND DRY MEASURES. 763. The Liter (leeter) is the primary unit of measures of capacity, and is a cube, each of whose edges is a tenth of a meter in length ; and, 764. The Hectoliter is the unit in measuring large quanti- ties of grain, fruits roots and liquids. TABLE. 10 miUili Icrs f m/ . ) a?-e i e e n t i i i f c r ( c^. ) = .888 fl. oz. 10 centiliter.^ " 1 deciliter -= .845 liq. gill. 10 deciliters " 1 liter (l.) = 1.0567 liq. qt. 10 liters " 1 dekaliter = 2.6417 gal. 10 dekaliters " i iikctof.itrr (W.) = 2bu. 8.35pk. 10 hectoliters " 1 kihlitcr = 28 hu. ij pk. A Centiliter is about ^ of a fluid ounce; a Liter is about l^j liquid quarts, or -f^ of a dry quart ; a Hectoliter is about 2| bushels ; and a Kiloliter is 1 cubic meter or stere. WEIGHTS. 765. The Gram is the primary unit of weights, and is the weight in a vacuum of a cubic centimeter of distilled water, at the tem- perature of 39.2 degrees Fahrenheit. A CnBic Ckntimetf.r. TABLE. 10 milligrams img.) arc 1 centigra^n 10 centigrams 10 decigrains 10 grams 10 dekagrams 10 hectograms 10 kilograms 10 myriagrams 10 quintals 1 decigram " 1 ORAM (g.) " 1 dekagram " 1 hectogram " 1 KILOGRAM (k.) " 1 myriagram " 1 quintal .1543 gr. T. UJfS " 15.432 " .3527 av. oz. 3.5274 " 2.2046 av. lb. 22.046 220.46 " 2 TO N N K A u ( i. ) = 2204.6 METRIC SYSTEM. 301 The Grnm is used in weighinsj gold, jewels, letters and small quantities of things. Tlie Kilogram, or, for hrevity. Kilo, is used by grocers ; and the Tonneuit (tonno), or Metric 'Ton, is used in hnding the weight of very lieavy articles. A Gram is about 15J grains Troy; the Kilo, about 2^ pounds avoirdu- pois ; and the Metric Ton, about 2205 pounds. A Kilo is the weight of a liter of water at its greatest density, and the Metric Ton of a cubic meter of water. 766. Metric Numbers are written with the decimal poiut (.) at the right of the figures denoting the unit. Thus, 15 meters 3 centimeters are written 15.03 m. 767. When metric numbers are expressed by figures, the part of the expression at the left of the decimal point is read as the number of the unit, and the part at the right, if any, as a number of the lowest denomination indicated, or as a decimal part of the unit. Thus, 46.525 m. is read 46 metres and 525 millimeters, or 46 and 525 thousandths meters. 768. In writing and reading metric numbers, according as the scale is 10, 100 or 1000, each denomination should be allowed one, two or three orders of figures. WRITTEN EXERCISES. 769. — Ex. 1. Express by figures two kilometers one hundred sixty-nine meters seventy-five centimeters as meters. Ans. 2169.75 m. 2. Express nine hundred sixteen millimeters as a decimal of a meter. Ans. .916 m. 3. Express four hundred fifty kilometers three hundred twelve meters as kilometers. Ans. 450.312 km. 4. Express thirty-eight hectares three ares ninety-four centi- ares as hectares. Ans. 38.0394 ha. 5. Express twenty-five square meters seventy-one sc^uare centimeters as square meters. Ans. 25.0071 sq. m. 6. Express five cubic meters one thousand seventy-six cubic centimeters as cubic meters. Ans. 5.001076 cti. m. 20 302 METRIC SYSTEM. 7. Express four hundred twenty-two kilos thirty-five grams as kilos. Ans 422.035 k. 8. Express one hundred one tonneaux nine hundred nine kilos as tonneaux. Ans. 101.909 t. 9. Express fifty-five liters five centiliters as liters. Ans. 55.05 i. 10. Express one thousand thirty-seven hectoliters twenty- five liters as hectoliters. > Ans. 1037.25 hi. Write and read — 11. 2169.75 m. 12. 195.007 km. 13. 31.9 cm. 14. 8.0394 ha. 15. 104.147 cu. cm. 16. 106.07 St. 17. 51.001001 cu. m. 18. 31.15 hi. 19. 67.3051. 20. 6.005 gr. 21. 316.08 k. 22. 163.455 t. COMPUTATIONS. 770. The Computations in metric numbers are similar to those in United States money. From the nature of the scales, the numbers are operated with in like manner as are simple integers and decimals. Thus, 16.55 m., or 16 meters 55 centimeters, may be changed to centi- meters by removing the decimal point two orders to the left, in the same manner as 16.55 dollars may be clianged to cents; and 1634 millimeters may be changed to meters by pointing off three orders from the right, in the same manner as 1634 mills may be changed to dollars. 771. Units of the common system may be readily changed to units of the metric system l)y aid of the following TABLE. 1 inch = S.54 centimeters. 1 cu. inch = 16.89 cu. centimeters. 1 foot = S0./f8 centimeters. 1 cu. foot = 28820 cu. centimeters 1 yard = .9144. meter. 1 cu. yard = .7646 cu. meter. 1 rod ^ 5.029 meters. 1 cord = 8.635 steres. 1 mile = 1.6098 kilometers. 1 fl. ounce = 2.958 centiliters. 1 sq. inch = 6.4528 sq. centimeters. 1 gallon = 8.786 liters. 1 sq. foot = 929 sq. centiJiieters. 1 bushel = .3524 hectoliter. 1 sq. yard = .8861 sq. meter. 1 grain Troy = 64.S milligrams. 1 sq. rod = 25.29 centiares. 1 pound Troy = .873 kilo. 1 acre = 40.47 ares. 1 pound av. = .4586 kilo. 1 sq. mile = 269 hectares. 1 ton = .907 tonneau. METRIC SYSTEM. 303 I'RO BI^EMS. 1. Reduce 14937 meters to kilometers. Ans. 14.937 k. 2. Reduce 160000 square meters to hectares. Ans. 16 ha. 3. What decimal of a tonneau are 83000 grams ? 4. How many centiliters are 56.55 hectoliters ? 5. Reduce 20 miles 40 rods to kilometers. 1.6093 km. X20^ 32.186 km. SoLUTioN.-Since l .005029" x4o= . soiie yn. ;":,;; r^'ui: 32.38716 " 1.6093 km., or 32.186 km. Since 1 rod is 5.029 m., or .00)029 km., 40 rods are 40 times .005029 km., or .20116 km. 32.186 km. + .20110 km. are 32.38716 km., the result required. 6. How many miles are 32.3871 kilometers ? 7. The length of the tunnel through Mt. Cenis is about 12.22 kilometers ; what is its length in miles? 8. The French post-oflBce allows 7.5 grams for a single post- age — the United States, \ of an ounce avoirdupois. How^ many grains Troy does the latter exceed the former? Ana. 103.01 9. How^ many cubic centimetres are 31.631 cubic meters? An%. 31631000. 10. How many hectares in a rectangular farm whose length is 1500 meters and width 800 meters ; and what is its value at $80 per acre? Ans. 120 ha. ; value, $23721.60. 11. A square mile is how many hectares? 12. If a .'^ack of flour of 150 kilos be sold at 54.17 francs, what "would be the corresponding price of a cental in United States money, allowing 5.14 francs to a dollar ? Ans. $3.18 + . 13. A bin is 3.75 meters long, 2.50 wide and 1.80 deep. How many hectoliters will it contain ? 14. What must be the height of a range of wood nhich is 25 meters long, 1.12 meters wide, to contain 35 steres? Ans. 1.25 meters. 15. When wine is at 2 francs a liter, what is it a gallon in United States money, the value of a fx-anc being $.18y^^? Ans. $1.40+. 304 SERIES OR PROGRESSION. SECTION LXXIII. SERIES OP, PROGRESSION. 772. A Series, or Progression, is a succession of numbers in which each succeeding number is formed from the preceding one by adding or subtracting the same quantity, or by the multiplication by a constant factor. The Terms of a series are its numbers ; the first and last terms are its Extremes, and the other terms its 3feans. 774. A series or progression is ascending when the terms in- crease regularly from the first, and descending when the terms decrease regularly from the first. ARITHMETICAL PROGRESSION. 775. An Arithmetical Prog'ression is a series whose terms in- crease or decrease by a common difierence. Thus, 3, 5, 7, 9, 11, and 16, 14, 12, 10, are arithmetical progressions in which the common difference is 2. 776. The first term, the last term, the common difference, the number of the terms, and the sum of the terms, are the elements of a series. The relations of these are such that when three of them are known the others may be determined. WRITTEN EXERCISISS. 777. — Ex. 1. The first term of an ascending arithmetical series is 5, and the common difierence 2. What is the 4th term? Solution. 1st term = 5 Sd " =5-¥2 =5 + 2X1 = 1st term + com. dif. X 1= 7 Sd " =5 + 2 + 2 = 5 + 2X2= Ut term + com. diff. X2= 9 4/h " =5 + 2 + 2 + 2 = 5 + 2X3= 1st term t com. dif. \S= 11 2. The first term of a descending arithmetical series is 11, and the common difi"orGnce 2. Wliat is the 4th term ? SEEIES OB PJiOGIii:SSION. 305 Solution. 1st term = 11 U " -^ 11-3 =11 — {2X1) = 1st term — com. diff. X 1 -^ 9 Sd '-. =11 — 2 — 2 =11 — {2X 2) = 1st ter}ii — com. diff. X 2 -= 7 4lh " = 11 — 2 — 2 — 2= 11 — [2X8) = 1st term — com. dif. ,< ,i = 5 3. What is the sum of the arithmetical series 3, 7, 11, 15, 19? Solution. 3, 7, 11. lo, 19, w the arithmetical series. 19, 15, 11, 7, 3, is the series inverted. 22 + 22 + 22 + 22 ^ 22 = 110, the sum of twice the scries. 11 + 11 + 11 + 11 + 11 — 55, the sum of the series. Or, since the sum of the extremes, or of any two terms equally distant from them, is the average of the several term.s of the series, ^ .^ ^ X 5 = 55, is tlie sum of the series. 778. Rules for Arithmetical Progression.— i. Multiply the coimnon dif}x'rencc by the niunber of terms less one; add the product to the smaller extreme, and the sum will he the greater ; or, subtract the product from the greater extreme, and the remainder will be the smaller. 2. Multiply half the sum of the extremes by the number of terms, and the result will be the surn of the series. rROBLEMS. 1. A man being asked the age of his eldest child, replied that his youngest child was 2 years old, the number of his children was 6, and the common difference in their ages was 3 years. What was the age of the eldest child ? Ans. 17 years. 2. A man travelled 10 days, increasing the distance gone over H niiles each day compared with the day preceding, so that he travelled the tenth day 24 miles. How far did he travel the first day? Ajis. 10^ miles. 3. What will $400, at 7% simple interest, amount to in 44 years? A^is. $1632. 26* 306 SEBIES OR PBOGJRESSIOir. 4. There are a number of rows of corn, the first of which contains 3 hills, the second 7, the third 11, and so on to the last, which has 43 hills. How many hills are there in all the rows? Ans. 253. 5. If a stone fall through 16.1 feet in the first second, 48.3 feet in the second second, 80.5 feet in the third second, and so on, how deep will be the shaft of a mine where a stone takes 7 seconds to reach the bottom ? 6. If you should begin witli a capital of $3500, and decrease every year $60, what would remain of your capital at the end of 10 years ? GEOMETRICAL PROGRESSION. "JIO. A Geometrical Progression is a series whose terms in- crease or decrease by a constant factor. 780. The Rate, or Ratio, of a series is the constant factor. Thus, 3, 6, 12, 24, 48, is a geometrical series whose rate is 2 ; and 75, 15, 3, is a geometrical series whose rate is ^. 781. An Infinite Scries is a descending series of an infinite number of terms. Of such a series the last term must be smaller than any assignable quantity ; hence, it may be considered 0. 782. As in an arithmetical series, the relation of the ele- ments of a geometrical series are such that when any three of them are known the others may be determined. WRITTEN EXEJtCISES. 783. — Ex. 1. The first term of a geometrical series is 5, and the common rate 2. What is the fourth term of the series ? SOLUTIOX. 1st term = g £d " =5X2 =5X2^ =- 1st term X rate = 10 Sd " =5X2X2 = 5X2^== 1st term X square of the rate =-- 20 4th '• =5X2X2X2 = 5X2^ = 1st term X cube of the rate = 40 2. What is the sum of a geometrical series whose first term is 5, last term 135, and rate 3? SEIiIi:S OB FEOGBESSIO^^. 307 Solution. 15 + 45 + 135 + 405 = 3 times the tium of the series. 5 + 15 + 4^ + 135 = once the .siun 0/ the series. — 5 + 405 = twice the sum of the scries. Hence, ^,r^ ~ ^^'^ "^ ^^^'^ *""*■ °/ '^''^ series. Had the series been descending, the first term 135, and rate \, by inverting the series, making the first term the last, and the fate 3, tlie sokition would then be the same as now given. 3. What is the sum of au infinite series whose first term is 4, and rate \ ? Solution. The series extended =-4, 1, r' t,' • ■ • 0. U 16 The series inverted = . . . —> —> 1, 4> 16 U whose rate is 4. Then, 4 times the sum of the series ^ ^ ...+-+ - +i + ^ + -Z5 Once the sum of the scries ^ ^ • • ' ~^ Jg~^ 7 +1 + 4 3 times the sum of the series = 16 Hence, the sum of the series = -- = 5— . 784. Rules for Geometrical Progression.— i. Multiply the first tervh by the rate raised to a power whose expo- nent is one less than the nmnher of terms, and the product ivill he the required term. 2. Multiply the last term hy the rate; subtract the first term from, the product, and divide the differ- ence by the rate less one, and the result ivill be the sum of the series. If the series is descending, use the series inverted, maldng the first term the last, and the rate greater than one; and, If the series is infinite, multiply the larger term, by the rate of the series inverted, and divide the quotient by the rate less one. 308 LIFE INTERESTS AND REVERSIONS. PROJiT^EMS. 1. A person, travelling, goes 5 miles the first da)', 10 miles the second, 20 miles the third, and so on. If he travel 7 days, how far will he go the last day ? Ans. 320 miles. 2. What is the amount of 6100 for 9 years, at 0% compound interest? Ans. $155.13. Here Si 00 is the first term, 10, or 1 more than the number of years, is the number of terms, and 1.05 is the rate. Rr^'inired the last term. 8. What is the sum of the series 2, 1, \, \, etc., to infinity? Am. 4. 4. If a ball be put in motion by a force which would move it 10 rods the first minute, 8 rods the second, 6.4 the third, and so on in the ratio of .8, how far would it move? 5. If a farmer should sow 5 grains of wheat, and its produce every year for 9 years, how many bushels would there be in the last harvest, supposing that each harvest amounts to 10 times the quantity sov/ed, and that 8000 grains make 1 pint ? Ans. 9765 bu. 2^ pk. SECTION LXXIV. LIFE IJ^TE RESTS AJs^B BEYERSIOKS. 785. An Annuity is a sum of money to be paid annually, or at regular intervals of time. 786. A Life Interest is an annuity to continue for life or lives. Thus, a pension for life and a widow's dower, or life estate, are each a life interest. 787. A RcTcrsioiiary Intei'cst is an interest which does not commence until alUr a certain period, or until after a certain event. 788. The United States Treasury Dci)artment, in the com- putation of life interests, uses the following tables, kno\vu ass the Carlisle Tables : LIFE INTERESTS AND REVERSIONS. 309 CARLISLE TABLES, Of the Expectancy of Life, and of the Present Value of Life Annuities. M^. B5 Present Value of An- nuity of $1 for the years in 2d column, interest at 6 per cent. Present Value of $1 to be received at the end of the years in 2d col- umn, interest at 6 per cent. Age. c ^ 1 B. X Present Value of An- nuity of $1 for the years in 2d column, interest at 6 per cent. Present Value of $1 to bo received at the end of the years in 2d col- umn, interest at 6 per cent. 38.72 14.9202 .104788 41 26.97 13.2043 .207741 1 44.68 15.4325 .074065 42 26.34 13.0737 .215580 2 47.55 15.6225 .062645 43 25.71 12.9395 .223635 3 49.82 15.7521 .054874 44 25.09 12.8032 .231812 4 50.76 15.8008 .051953 45 24.46 12.6576 .240548 5 51.25 15.8252 .050490 46 2.3.82 12.5059 .249646 6 51.17 15.8213 .050722 47 23.17 12.3454 .259278 7 50.80 ' 15.8029 .051830 48 22.51 12.1751 .269494 8 50.24 15.7742 .053551 49 21.81 11.9889 .280669 9 49.57 15.7385 .055689 50 21.11 11.7946 .292324 10 48.82 15.6972 .058168 51 20.39 11.5846 .304922 11 48.04 15.6523 .060860 52 19.68 11.3701 .317792 12 47.27 15.6055 .063670 53 18.97 11.1481 .331108 13 46.51 15.5573 .066559 54 18.28 10.9201 .344791 14 45.75 15.5072 .069566 55 17.58 10.6805 .359172 15 45.00 15.4558 .072650 56 16.89 10.4364 .373815 16 44.27 15.4028 .075832 57 16.21 10.1839 .388767 17 43.57 15.3501 .078996 58 15.55 9.9287 .404275 18 42.87 15.2956 .082267 59 14.92 9.6788 .419268 19 42.17 15.2384 .085695 60 14.34 9.4368 .433789 20 41.46 15.1778 .089331 61 13.82 9.2154 .447078 21 40.75 15.1151 .093095 62 13.. 31 8.9900 .460612 22 40.04 15.0500 .097002 63 12.81 8.7636 .474183 23 39.31 14.9972 .101248 64 12.30 8.5245 .488530 24 38.59 14.9068 .105592 65 11.79 8.2793 .503231 25 37.86 14.8307 .110157 66 11.27 8.0211 .518737 26 37.14 14.7521 .114876 67 10.75 7.7552 .534690 27 36.41 14.6685 .119892 68 10.23 7.4813 .551125 28 35.69 14.5829 .125024 69 9.70 7.1926 .568446 29 35.00 14.4982 .130105 70 9.18 6.9022 .585867 30 34.34 14.4123 .135258 71 8.65 6.5945 .604328 31 33.68 14.3240 .140560 72 8.16 6.3045 .621730 32 33.03 14.2343 .145938 73 7.72 6.0341 .637953 33 32.36 14.1366 .151789 74 7.. 33 5.7894 .652634 34 31.68 14.0344 .157932 75 7.01 5.5887 .664681 35 31.00 13.9291 .164255 76 6.69 5.3762 .677427 36 30.32 13.8174 .170957 77 6.40 5.18.33 .6S8999 37 29.64 13.7021 .180796 78 6.12 4.9971 .700193 38 28.96 13.5833 .185000 79 5.80 4.7763 .713420 39 28.28 13.4579 .192530 80 5.51 4.5719 .725687 40 27.61 i 13.3289 .200208 81 5.21 4.3604 .738376 SIO LIFE INTERESTS AND BEVERSIONS. PROHrUMS. 1. What is the present value of a "widow's dower whose yearly rent is $520, and whose age is 49 years ? Solution. Expectancy of life at Jf9 years of age = 21. 81 years. Present value of >$1 annuity for 21.81 years = $11.9889. Present value of $520 annuity for 21.81 years = $11.9889 X 520 = $6234-23. 2. What is the ready-money value of a legacy of $1000, to be received after 35 years ? Solution. Present value of $1, to be received after 35 years = $.130105. Present value of $1000, to be received after 35 years = $.130105 X 1000 = $130.10\ 3. A person 62 years old has a yearly pension of 896 ; what is its present value ? Ana $863.04. 4. A widow aged 51 has set off, as her dower, property whose appraised value is $3800. What is the present value of the reversionary interest ? Ans. $1158.70. 5. Smith, who is 70 years old, has a life annuity of S700 per annum. What is its present value? 6. AMiat should be the present value of a legacy of $4000, to be received after 10 years 9 months ? TEST QUESTIONS. 789. — 1. What is the Metric System? A meter? How are the names of derived metric denominations formed ? 2. What is the Primary Unit of lengths? Of ordinary surfaces? Of land measures? Of volumes? Of capacity ? Of large quantities of grains, fruits, etc.? 3. How are Metric Numbers written ? How many orders of figures are allowed to each metric denomination ? How is the part of a metric expression at the left of the decimal point read? At the right of the decimal point? PROBLEMS FOR ANALYSIS. 311 4. What is a Skries or a Progression ? What are the terms of a series? The extremes? The means? When is a series ascending? When descending? 5. What is an Akitiimetical Progression? What is the rule for finding either extreme of an arithmetical series ? For iinding the sum of the series ? 6. What is a Geometrical, Progression ? The rate or ratio of a geometrical series? What is an inlinite series? What is the rule for finding the last term of a geometrical scries ? For finding the sum of an ascending series ? The sum of a descending series ? The sum of an infinite series ? 7. What is an Annuity ? A life interest ? A reversionary interest ? What is used by the United States Treasury Department in the computa- tion of life interests? 8. How do the metric measures differ from the measures in common use? How do arithmetical and geometrical jirogressions differ? How does a reversionary interest differ from a life interest ? How does a root differ from a power ? The square root from the cube root ? Is the root of a proper fraction smaller or larger than its corresponding power? SECTION LXXV. PROBLEMS FOR AJ^ALYSIS. MENTAI. EXERCISES. 790. — 1. Two boys on counting their marbles found that one had 9 more than the other, and that together they had 49. How many had each of the boys ? 2. The greater of two numbers is 7f , and their difference is 3|. What are the numbers ? 3. A man being asked how many cows he had, answered that if he had 2 more, twice the number woukl be 26. How many cows had he ? 4. If John were 3 years younger, twice his age would be 18 years. How old is he ? 5. A man wishing to contribute money to an equal number of poor men and women, gave to each man 9 dimes and to each woman 3 dimes. If he gave them in all $120, how many men and women were there respectively? 312 FEOBLEMS FOR ANALYSIS. 6. If 9 be taken from the sum of two numbers of v/hich 7 is one, there Aviil be 13 left. What is the larger number? 7. A and B by -working together can do a piece of work in 2 days. B can do it alone in 5 days. How much of it can both together do in 1 day? What part of it can A alone do in 1 day ? In what time can A alone do it ? 8. Edward and Philip stai't from the same place and travel the same road. Edward starts 4 days before Philip, and ti'avels 20 miles a day. Philip follows, travelling 25 miles a day. In what time vvill Philip overtake Edward ? 9. A thief, having 50 steps the start of an officer, takes 4 «teps while the officer takes 3 ; and 2 steps of the officer are equal to 3 of the thief. How many steps can the thief take before the officer can catch him ? Solution. — 2 steps of the officer are equal to 3 of the thief; hence, 6 steps of tlie officer are equal to 9 steps of the thief While the officer takes 6 steps, the thief takes 8 ; hence, while the thief takes 8 steps, the officer gains upon him 1 step of the thief Then while the officer is gaining 50 of the steps of the thief, the thief can take 8 times 50 steps, or 400 steps. 10. An officer is in pursuit of a thief Avho has some miles the start. The thief goes 20 miles a day, and the officer 25. If it take the officer 8 days to overtake the thief, how many days had the latter the start ? 11. If 4 men can earn $16 in 2 days, how long will it take 6 men to earn $48 ? 12. If 6 men can do a piece of work in 8 days, what num- ber of men can do ^ of it in f of the time ? 13. Arthur is IG years old, and Albert is 4. In how many years will Arthur be only twice as old as Albert ? Solution. — 4 years ago Arthur was 12 years old; hence in 12 years from that time, or in 12 years less 4 from this time — that is, in 8 years — Arthur will be 2-1 years old, or twice as old as Albert. 14. Mary is 15 years old, and her mother 36. In what time will Mary be only \ as old as her mother ? 15. A father is 35 years old, and his son is 5. In what time will the son be \ as old as his father? PROBLEMS FOR ANALYSIS. 313 16. John gave each of his brothers 4 apples, giving all he had. If he had had 12 more apples he could have given each of his brothers 7 apples. How many brothers had he? Solution. — To give 7 apples to each brother he would have required 3 apples more for each; but he would have required 12 more for all. Hence, there were as many brothers as 3 apples are contained times in 12 apples, which are 4. 17. Jane wishes to purchase a certain number of oranges. If she pays 6 cents each, she will have 10 cents left ; but if she pays 8 cents each, it will take all her money. How many oranges does she want ? 18. James wishes to divide some peaches among his friends. If he gives each of them 3 ho will have 9 left ; but if he tries to give each of them 5, he will not have enough by 5. How many friends has he ? 19. Eggs are sold at the rate of 4 for 5 cents. At what rate were they bought if the profit is 25 per cent. ? 20. A has 5 times as much money as B has, and the sum. of the interest received by both for 2 years, at 7 per cent., is $70. What sum has each ? ' 21. If a wagon cost $80, wdiat would be the cost of a har- ness if -^ of the cost of the wagon were ^ of the cost of the harness ? 22. A has 4 times as much money as B has, and the sum of the interest received by both for 2 years, at 7 per cent., is 870. What sum has each? 23. A and B have $12, and i of A's money equals \ of B's. How many dollars has each ? Solution. — \ of A's money equals \ of B's ; hence, f , or the whole, of A's money must equal f of B's. If A's money equals f of B's, and B's must equal f of itself, $12 ran^t equal | plus f , or f , of B's. If f of B's money is $12, ^ is i of $12, or $2, and |, or A's money, is $4, and |, or B's money, is $8. 24. A pole 60 feet long broke into two parts, one of which was f of the other. What was the length of each part ? 25. Edward savs to Thomas, " ^- of my age lacks 2 years 27 314 PROBLEMS FOR ANALYSIS. of being | of yours, and the sum of our ages is 33 years." What is the age of each ? 26. The time between three and four o'clock is such that f of the minutes past three are equal to f of the minutes before four. What is the time ? 27. What is the time in the afternoon when the time past noon is equal to \ of the time to midnight ? WRITTEN EXERCISES. 791. — Ex. 1. Two travellers leave the same place at the same time. One goes 20 miles a day and the other 23|-. How far apart will they be at the end of 28 days, if they both travel in the same direction ? How far if they travel in oppo- site directions? Ans. 98 miles; 1218 miles. 2. A merchant, who commences business with a capital of $12000, gains at the rate of 69000 in 4 years, by trading in flour, and at the rate of S9000 in 6 years, by trading in grain. If his annual expenses are $4500, in what time will he have lost all? ^ns. 16 years. 3. I have a stick of squared timber 20 feet 6 inches long, 16 inches wide and 12 inches thick. If 3|- solid feet should be sawed off at one end, how long would the stick then be ? 7790 ,--, V?- ^^,o- * Solution. — -^■^ = -If = ^^-^ *■"■• = ^fi- H ^'- 1' 23 solid incites are 1 solid foot ; 20 jt. 6 in. — 2Jt. 7^ in. = 17 ft 10^ in. hence, the piece sawed off must contain 3J- times 1728 solid inches, or 6048 solid inches. The 6048 solid inches must be the ])roduct of the numbers, in inches, which denote the dimensions of the piece sawed off. The product of the width bv the thickness, or 16 X 12, is 192; hence, the quotient of 6048 divided by 192, which is 31.5, must denote the number of inches of the ])iece sawed off. The stick with 31.5 inches, or 2 ft. 7^ in., sawed off must be 20 ft. 6 in. — 2 ft. 11 in., or 17 ft. 101 in. long. 4. From a plank which is 16 feet 5 inches long I wish to cut off a strip containing a square yard. At what distance from the edge must the line be drawn ? Aiis. 6j-^ inches. PROBLEMS FOR ANALYSIS. 315 5. My bushel and half-peck measures, which are of a cylin- drical form, are respectively 18|^ and 'd\ inches in diameter. What must be the depth of each ? 6. I have a range, 56 feet long, of firewood, cut 4 feet long. When such wood is worth $6 per cord, how high must the range be piled to be worth $52.50 ? Ans. 5 feet. 7. A laborer agreed to work 12 weeks upon the conditions that he should receive $18 per week for every week he worked, and for every week he was idle he should pay $3.50 for his board. At the expiration of the time he received $151.50. How many weeks did he work ? $ 18 y. 12 =^$'2 16 Solution. — Had he labored $2i6-$i5i.5o=$64.5o ^ z^^:,:T,^.::t:t $18 + $3.50 = $21.50 ceived but $151.50, he lost by idle- $64.50-^$21.50 = 3 ness $216- $1 51.50 or $64.50. ^ Each week he was idle he lost his 12 weeks— J weeks = 9 weeks. ""^^ges and $3.50, amounting to $21.50; hence he was idle as many weeks as $21.50 is contained times in $64.50, or 3 weeks. Since he was idle 3 weeks, he worked 12 weeks less 3 weeks, or 9 weeks. 8. James received $1 a day for his work, and paid $.25 for every day he was idle. At the end of 18 days he received $6.75. How many days was he idle ? 9. A man agreed to carry 28 packages to a certain place on the conditions that for every one promptly delivered he should receive 30 cents, and for every one delayed he should forfeit 50 cents. He forfeited $2.80 more than he received. How many packages did he deliver promptly? Ans. 14. 10. How many cows can be kept on a farm of 48 acres if for every 5 cows there must be 2 acres of meadow, and for every 3 cows 2 acres of pasture-land ? 2 2 16 SoiiUTlON. — Since for every 5 cows there must be 5 ' J 25^ 2 acres of meadow, and for every 3 cows there must j„ be 2 acres of pasture, for 1 cow there must be f of an .48 ~^ jY = -40 acre of meadow and f of an acre of pasture, or j-f of an acre. Hence, as many cows can be kept on a farm of 48 acres as 48 contains times xf , or 45. 816 PROBLEMS FOR ANALYSIS. 11. A farmer has 150 acres. He cultivates 5 acres for every 3 horses he has, and allows 10 acres of pasture for every 4 horses. How many horses can he keep ? Ans. 36. 12. A certain field will furnish pasturage for 3 horses, or 4 cows, 56 days. For what time will it furnish pasturage for 1 horse and 1 cow grazing together? Ans. 96 days. 13. A bankrupt's stock was sold for $1660, at a loss of 17% on the cost price. Had it been sold in the course of trade, it would have realized a profit of 20%. How much was it sold below the trade price ? Ans. ^740. 14. A merchant sells tea to a trader at a profit of 60% ; but the trader becomes bankrupt and only pays 75 cents on a dollar. How much per cent, does the merchant gain or lose ? Ans. He gains 20%. 15. The head of a fish is 28 inches long ; the tail is as long as the head and | of the body ; and the body is as long as the head and tail. What is the length of the fish ? 28 in. + 28 in. = 56 in. = j the length , Solution. - Since ^ f 1 I the tail IS as long as '' ^' tlie head and ^ of the S6 in. X .^ = 22.^ in = the length of body, and the body is the fish. as long as both the head and tail, 28 inches plus 28 inches, or 56 inches, must equal \ the length of the body. Since the body is \ the length of the fish, 56 inches must equal \ the length of the fish, and 56 inches X 4, or 224 inches, must equal the length of the fish. 16. A sum of money was divided among 3 men. The first received $96, the second received half as much as the third, and the third received as much as the other two. How much did the second and the third receive ? 17. A, B and C, dividing a plantation consisting of 120 acres, agreed that B should have a third part more than A, and C a fourth part more than B. What number of acres will each have? Ans. A, 30 ; B, 40 ; C, 50. 18. If a steamboat, running uniformly at the rate of 12 miles per hour in still water, were to run 4 hours with a current of 4 miles per hour, then to return against that curi-ent, what PROBLEMS FOR ANALYSIS. 317 length of time from the time she started would she require to reach the place whence she started ? Ans. 12 hours. 19. A merchant purchased goods for $1200, and sold them at a loss of 12^%. He then purchased more goods with the proceeds, and sold them at a gain of 14%. Did he gain or lose by these transactions, and how much ? Ans. Lost $3, 20. I wish to plant 5292 trees equally distant in straight rows, and to make the length of the grove 3 times the width. How many of the shorter rows shall I have ? Solution. — Since the length is 3 times the width, one third of the trees are to form an exact square. The square root of one third of the number 5292 is 42, which must denote the number of trees in a side of the square. Since there must be 3 such squares, there must be 3 times 42, or 126, short rows. 21. A farmer has 2 10-acre lots ; one is a square, and the other is a parallelogram 4 times as long as it is wide. How many rods of fence will each lot require to exactly enclose it ? Ans. The square, 160 rods ; the parallelogram, 200 rods. 22. What is the mean proportional between 4 and 9 ? Ay^ 9 = 36 Solution. — Since the mean proportional between / q/^ _ _ /J the extremes of a proportion is one of the equal means of the proportion (Art. 547), the mean pro- portional between two numbers is equal to the square root of the product of those numbers. The product of 4 by 9 is 36, and the square root of 36 is 6, the mean proportional required. 23. What is the mean proportional between 7 and 252 ? Ans. 42. 24. A cheese, when put into one scale of an incorrect balance, was found to weigh 31^ pounds, but when put into the other it weighed only 20 pounds. What was its true weight ? Ans. 25 pounds. 25. I have corn of four different qualities, worth respectively 65, 72, 80 and 90 cents per bushel. In what proportions may 27* 318 PROBLEMS FOR ANALYSIS. these kinds be taken to form a quantity worth 75 cents per busliel ? Solution. At 65 c. to gain 1 c. take -— bu. ; -- hu, X 10^= Ibu. "72c. " Ic. " ^ hu.; j bu.XlS = 5bu. " 80c. to lose Ic. " - hu.; ^ hu.XlO = 2bu. 5 5 "90c. " Ic. " ~hu.; 4^bu.X15=lbu. 15 15 On 1 bushel, worth 65 cents, taken at 75 cents, there is a gain of 10 cents ; hence, to gain 1 cent we take y^j of a bushel. On 1 bushel at 80 cents there is a loss of 5 cents ; hence, to lose 1 cent, we take \ of a bushel. Therefore, we take j'^ of a bushel at 65 cents as often as we take 5 of a bushel at 80 cents ; or, multiplying these fractions by the least common multiple of their denominators, we find we may, also, take 1 bushel at 65 cents as often as we take 2 bushels at 80 cents. In like manner we take \ of a bushel at 72 cents as often as we take -^ of a bushel at 90 cents ; or we may take 5 bushels at 72 cents as often as we take 1 bushel at 90 cents. It is also evident that any number of times these proportions may be taken. Hence, the different qualities of corn may be taken in the proportions of tV, \, i and ^; 1, 5, 2 and 1 ; etc. 26. In what proportions may coffees, at 85, 40, 50 and 55 cents a pound, be mixed to produce a quantity worth 45 cents per pound ? Ans. In the proportions of 1 lb. at 35 c, 2 lb. at 40 c, 2 lb. at 50 c, and 1 lb. at 55 c. 27. How much gold, at 20, 21 and 23 carats fine, must be mixed with 12 ounces 20 carats fine, so that the mixture may be 22 carats fine? Solution. At 2 Oca. to gain lea. take -oz.; . . . j oz.X 24"= 12 oz. " 21 ca. " lea. " loz.; . . . 1 oz.X 24 = 24 oz. " 23 ca. to lose lca+ 1 ca. take loz+ 1 oz. ;2oz.X24'^4^ 02. 12 oz. ^joz. = 24 oz. PROBLEMS FOR ANALYSIS. 319 We find the proportions, without regard to any of the quantities being limited, to be \ oz. at 20 carats fine, 1 oz. at 21 carats, and 2 oz. at 23 carats. But of the 20 carats fine it is required to take 12 oz., or 24 times \ oz. ; hence the other proportions must also be taken 24 times as large. We must then take for the required mixture 12 oz. at 20 carats fine, 24 oz. at 21 carats, and 48 oz. at 23 carats. 28. A grocer mixed 20 pounds of sugar, worth 15 cents a pound, with others at 16, 18 and 22 cents. How many- pounds of each were taken to make a mixture worth 17 cents per pound ? Am. 20 lb. at 15 c, 41b. at 16 c, 4 lb. at 18 c., 8 lb. at 22 c. 29. A farmer has oats worth 46, 48, 51 and 54 cents a bushel. What quantity of each of these kinds must bo taken to make 8 bushels worth 50 cents a bushel ? Solution. At ^6 c. to gain 1 c. take - hu. 4 " 48c. '' Ic. " 61 c. to lose 1 c. " 64 c. " ie. 2 1 hu. - hu. -hu. X .i = i hu. 4 I hu. X4 = 2hu. Ihu.X 4 = 4 ^^'" - hu. X 4 = 1 hu. 2hu. S hu. -^ 2 hu. = 4- 8bu. We find the proportions, without regard to the total of the quantities being limited, to be \ bu. at 46 cents, h bu. at 48 cents, 1 bu. at 51 cents, and \ bu. at 54 cents. The sum of these quantities is 2 bushels ; but the total of the mixture must be 8 bushels, or 4 times as large; hence, each of the quantities found must be taken 4 times as large. We, therefore, must take for the required mixture, 1 bu. at 46 cents, 2 bu. at 48 cents, 1 bu. at 51 cents, and 1 bu. at 54 cents. 30. What quantities of sugars, worth 3.12, %.\\ and ^.08 per pound respectively, must be taken to form a mixture con- taining 35 pounds, at 9 cents per pound ? 320 PEOBLEMS FOR ANALYSIS. 31. A grocer requires a cliest of tea containing 75 pounds, worth 66 cents per pound. What quantities of several kinds, worth 42, 48, 72 and 78 cents per pound, must he mix to form it? Am. 9 lb. at 42 cents, 12 lb. at 48 cents, 36 lb. at 72 cents, and 18 lb. at 78 cents. Note.— The last seven problems are examples in what is called AUigatim Alternate, which is the process of finding the proportions of several articles of different values that may form a quantity of a given average value. 32. Express as a series .135135 +, in which the figures 135 continually repeat in the same order. Solution.— The decimal .135135 +, or .135 . . . , may be regarded as a geometrical progression, in which the rate is ^oV?. Marking the re- peating figures by placing a dot over the first and last of the set, and we have .135 = ^^-^-^ + xirfof ^t + , the series required. Note.— A decimal in which a figure or a set of figures is repeated in the same order indefinitely is called a Circulate, and the figure or set of figures repeated is called a Re- petend. 83. Express as a common fraction in its lowest terms the circulate .27. Solution. 100 times the circulate .27 = ^7.27 1 time " " .27 ... . 99 times " " =27. Hence, once the circulate, or -27 = —- = —, the fraction required. That is, a repetend is equal to a common fraction having for its de- nominator a-s many nines as there are figures in the repetend, and for the numerator the figures of the repetend. 34. Express as a mixed decimal .2259. Aiis. .2||-|. 35. Express as equivalent common fractions .7, .90 and .702. 36. Express as common fractions in their lowest terms 7.936, 8.936 and 32.715. A,n «&i "Rs T-sni 37. E.>ipress as conimou fractions in their lowest " terms .074, .8145 and ,138. ^^. ^_ ^,^^_ ^, GENERAL REVIEW. 321 SECTION LXXVI. GENERAL REVIEW. 792. — 1. Add tliirty-fivc milliou eight hundred forty thou- sand three hundred fifty ten-thousandths, four hundred sixty- three thousand nine hundred and eight-hundredths, and three hundred four thousands and three hundred four millionths. 2. What number divided by 417 will give the quotient 105 and the remainder 113? 3. If j^ of a ton of hay cost $18.50, how much will tAvo loads cost, one weighing |- of a ton and the other ^ of a ton ? Am. $27.75. 4. 20004 + (20.104 X 5.07) — (6.44 -- .0005) = what? 5. Which is the greater — a garden 40 rods square, or one containing 40 square rods, and how much ? 6. From 95 mi. subtract 57 mi. 192 rd. 4 yd. 3 ft. 18 in. 2 7. What part of 2| is f of | of f ? Ans. |. 8. Find the difierence between 7 thousand and 7 thou- sandths, and divide the remainder by 7 millionths. 9. What is the value in compound numbers of .3945 of a day? Ans. 9 h. 28 m. 4.8 sec. 10. How much greater is the quotient of f -^- f than the product of I X I? 11. A hall, 50 feet long and 30 feet wide, has around it a mop-board 9 inches high. The hall has one door 6 feet wide and 2 doors 3 feet wide. How many square feet are there in the surface of the mop-board ? 12. If a merchantman, sailing 9|- knots an hour, is chased by a gun-boat steaming lOf knots, how far ahead must the sailing vessel be to escape 3 knots ahead into a port from which she is 15|- knots at the commencement of the chase? 13. If by selling a horse at $80 I lose 12^% of the first cost, shall I gain or lose, and what per cent., by selling him at $90 ? 14. How many times has February 29th occurred since the year 1799? 322 GENERAL REVIEW. 15. A merchant bought f of a hogshead of molasses at f of a dollar a gallon. At what price per gallon must he sell it to make §4.20 ? Ans. $.70. 16. I bought apples at $5 per barrel, and lost one fourth of them. At what price must I sell the rest that I may gain 10% on the whole cost? Ans. ^1.Z^. 17. If 6 men in 8 hours thresh 30 bushels of wheat, in how many hours can 5 men thresh 50 bushels ? 18. A has a farm |- of a mile square, and B has one contain- ing y^g- of a square mile. How do the farms compare in size ? 19. An agent received 867.50 for collecting 84500. What was the rate of his commission ? 20. What is the least number which, being divided by 3, by 5, by 7, by 9 and by 10, leaves in each case a remainder of 2? 21. How much shall I gain by borrowing $3560 for 1 year 6 months 10 days, at 6%, and lending it at 7% ? 22. I obtained a discount at a bank at 7%, and left \ of the proceeds in the bank until the note was paid. At what rate did I get the money I used ? 23. If 14 men can perform a piece of work in 36 days, in how many days can they perform the same labor with the assistance of 7 more men ? 24. How much more time will it require a sum of money to double itself at 6 % interest than at 7 % ? 25. AVhat is the amount of $1450.40 from April 19, 1871, to August 3, 1872, at 6% ? 26. What is the difference between the simple and the com- pound interest on $5000 for ^ years, at 7 % ? Ans. $114.60. 27. A man bequeathed \ of his estate to his wife, ^ to a college and ^ to his eldest son; and these three legacies amounted to $18500. How much did each receive? Ans. Wife, $7500 ; college, $6000 ; son, $5000. 28. A merchant paid $4200 for cotton, and sold it at 10% advance, taking his })ay in prints, which he sold at a loss of 10%. Did he gain or lose, and how much? 29. What is the present worth of $770, due in 1 year 8 mouths, at G% ? Ans. $700. GENERAL REVIEW. 323 30. A and B traded in company. A put in $950, and B They gained $300. What was each partner's share of it? Ans. A's, S162f ; B's, $137|. 31. A person has a field measuring 3 acres 75 square rods, •which he wishes to exchange for a square one of inferior quality, but 3|- times as large. How many rods is the length of its side ? Ans. 44.0738. 32. Smith and Doland trade in company, Smith contributing $800 for 9 months, and Doland $600 for 8 months. They gain $450. What should each receive ? 33. What is the area of a triangle whose base is b\ yards, and whose altitude is 8^ yards ? Ans. 23f sq. yd. 34. Gold is soiling at 112. Find the interest in currency on 7 $1000 U. S. 5-20 bonds from June 1, 1870, to July 1, 1871. 35. A person being asked the time of day, replied that it was between 5 and 6 o'clock, and that the hour- and minute- hands of the watch Avere together. What was the time ? Ans. ^1-Yi minutes past 5 o'clock. 36. If the diameter of a 9-pound cannon-ball be 5 inches, what must be the diameter of a 28-pound ball? Ans. 7.3 in. 37. A merchant bought goods to the amount of $1400, on a credit of 6 months. At the end of 3 months he paid $600, and 1 month later $400. What extension ought he to have on the balance of the debt? Ans. 6|- months. 88. Wishing to find the distance between two trees, wliich cannot be directly measured on account of an intervening pond, I measure due west 50 rods from the foot of one of them ; tlien, turning north, measure 34 rods, when I find that I am just 20 rods west of the other tree. How far are the trees apart ? Ans. 45.34 + rods. 39. A gentleman selling a mortgage of $4410, for which ho received 5% interest, invested the proceeds in Government 8^ % bonds at 70. After receiving the interest for 5 years, on the bonds rising to 75, he sold out. What was his gain upon the Avhole transaction over what he would have received had he continued the mortgage ? Ans. $315. 324 GENERAL REVIEW. 40. After the outbreak of the Prusso-French war in 1870, the Prussian Government issued a 5% war loan at 88. The French 3 per cents, stood at 65|. State the ratio of the two rates of interest.^ Am. 6|| to 4^V 41. A joist is 7-|^ inches wide and 2^ thick, but I want one just twice as large, which shall be 3f inches thick. What must be the width ? Ans. 10 inches. 42. A and B can do a piece of work alone in 12 and 16 days respectively. They labor together on the work for 3 days, when A leaves it, but B continues, and after 2 days is joined by C, and they finish it together in 3 days. In what time would C do it alone ? 43. How many cubic yards of gravel will be required for a ■walk surrounding a rectangular lawn 200 yards long and 100 yards wide, the walk to be 3 yards wide, and the gravel 3 inches deep ? 44. A saves \ of his income; but B, who has the same income, spends twice as fast as A, and thereby contracts a debt ' of $120 annually. What is the income of each ? Ans. $360. 45. A stole a horse from B, and made off with him. Five days afterward, B gets intelligence of A, and follows him at the rate of 60 miles a day, by which he gains 20% upon A. How far must B ride to overtake A, and how many days ? Ans. 1500 miles ; 25 days. 46. A steamer, working with a given force, can run down the river at the rate of 12| miles per hour. Of this speed, f is due to the current. How long would the steamer require to go 15 miles up the stream ? 47o Subtract the square root of ^^4^ f^'om the cube root of the same. Ans. -j. 48. The plan of a town is 11 \ inches long and 14 inches broad, and the scale annexed to it is jui^t 2|- inches to 1100 >ards. What is the length of a mile upon this scale, and what will be the length and breadth of the pUxn if it bo enlarged to a scale of 6 inclics to a mile ? Ans. 4| inches ; 25 inches by 20 inches. APPENDIX. ROMAJ^' JfOTATIOM. 793. Roman Notation uses seven letters : I, V, X, L, C, D find M, which express, respectively, one, five, ten,fijtij, one hun- dred, five hundred and one thousand. All numbers can be expressed by these letters, used singly, or combined according to the following 794. Principles. — 1. When a letter is repeated, the number ivhich it expresses is repeated. Thus, 11 = 1 + 1 = 2; XXX = 10 + 10 + 10 = 30; CC = 100 + 100 = 200. 2. When a letter e.rpressing a certain number stands after one expressing a greater number, the sum of the numbers is denoted. Thus, VI = 5 + 1 = 6 ; XI = 10 + 1 = 11 ; LX = 50 + 10 = 60. 3. When a letter expressing a certain number stands before one expressing a greater number, the difference of the numbers is denoted. Thus, IV = 5 — 1 = 4 ; IX = 10 — 1 = 9 ; XL = 50 — 10 = 40. 4. JVlien a letter expressing a certain number stands between two letters expressing greater numbers, the least number is to be subtracted from the sum of the other tivo. Thus, XIV = (10 + 5) — 1 = 14 ; CXL = (100 + 50) — 10 = 140. 5. A bar, — , placed over a letter makes it denote thousands. Thus, V = 5000 ; D = 500,000 ; M = 1,000,000. EXERCISES. Write and read — 1. XIX. 3. LXXXV. 5. MDCCCLXXV. 2. CXVI. 4. DCLXX. 6. MLIXCXX. Express by letters — 7. Forty-eight. 9. One thousand six hundred eleven. 8. Two hundred five. 10. Eighteen hundred seventy-nine. 28 325 S26 • CONTRACTIONS. COJs'TRACTIOJs^S. To Add Two Coiumus at a Time. 795.— Ex. 1. Add, two columns at a time, 1235, 6714, 4566 and 4967. 1235 Solution.— 67 + 6 = 73, + 60 = 133, + 4 = 137, + 10 G71A "" ^^'^' + o = 152, + 30 = 182 ; write 82. 1 + 49 = 50, / nrr + 5 = 55, + 40 = 95, +7 = 1 2, + 60 = 162, + 2 = 40 ig4^ + 10 = 174; write 174. Ans. 17482. 4967 In practice; thus, 67, 73, 133, 137, 147, 152, 182; write 82. 17482 ^' ^^' ^•^' ^5, 102, 162, 164, 174; write 174. Ans. 17482. 796. Rule for Adding two Columns at a Time.— Jb the low- est niunher add the ones of the next ninnher above, then add the tens of that mnribev ; to the sinyv thus obtained add the ones of the next number above, then the tens of that number, and so on. JPJtOBrEMS. (1.) (2.) (3.) (4.) (5.) 36 4402 6645 47 4141 71 6307 5232 81 3226 58 1453 7070 92 1819 32 9205 3007 31 4234 43 1824 7084 35 1781 64 7132 2636 74 9603 50 1042 2273 60 2009 To Multiply by a Multiplier of Two Orders at Once. 797.— Ex. 1. Multiply 1246 by 32. 1246 Solution— 6 X 32 = 192; write 2. 4 X 32 = 128, 32 + 19 = 147 ; write 7. 2 X 32 = 64, + 14 = 78 ; write 8. 39872 1 X 32 = 32, + 7 = 39 ; write 39. Ans. 39872. 798. Rule for Multiplying by a Multiplier of Two Orders at Once.— Mnlii])ly each order of the multiplicand separately by the entire multiplier. Multiply — CONTRA CTIONS. 327 PltOBLEMS. 1. 7418 by 35 ; by 42. 2. 6320 by 15 ; by 53. 3. 91367 by 44; by 61. 4. 34205 by 67 ; by 88. To Multiply by an Aliquot Part of 10, 100, etc. 799.— Ex. 1. Multiply 3465 by 125. 8)34-65000 Solution. — Since 125 is one eighth of 1000, multiply ooTTTt" ^^' 1000, and take one eighth of the product. 4.33125 ^„g 433125. 800. Rule for Multiplying by an Aliquot Part of 10, 100, etc.— Multiply by 10, 100, etc., and of the product thus obtained take such a part as the given multiplier is of the multiplier used. mOBZEMS. Multiply — 1. 674 by 2| ; by 250. 2. 342 by 8^ ; by 33^. 3. 758byl2i; by 125. 4. 8910 by 16| ; by 333^. 5. 7648 by 25 ; by 250. 6. 68024 by 331; by 125. To Divide by an Aliquot Part of 10, 100, etc. 801._Ex. 1. Divide 433125 by 125. 433.125 Q Solution. — Since 1000 is 8 times 125, divide by 1000, and take 8 times the quotient. Ans. 3465. 3465.000 802. Rule for Dividing by an Aliquot Part of 10, 100, etc.— Divide by 10, 100, etc., as the problem may require, and multiply the quotient thus obtained by the num- ber which shows how lyuiny times the given divisor is contained in the divisor used. PItOBLEMS. Divide — • 1. 16850 by 25 ; by 125. 2. 5700 by 2^ ; by 84. 3. 25300 by 12| ; by 250. 4. 7360 by 16| ; by 33|. 5. 7600 by 250 ; by 333|. 6. 552642 by 50 ; by 125. 328 DUODECIMALS. DUODECIMALS. 803. A Duodecimal is a deuomiuate number in which a unit of any denomination is equivalent to twelve units of" the next lower denomination ; or, it may be regarded as a series of fractions whose denominators are successive powers of 12. Note. — Examples in DuoJecimals can generally be more readily performed by reducing the Duodecimals to Common or Decimal Fractions, but, since the special rules are used by gome mechanics in measuring surfaces and solids, it is thought best to give them here. In duodecimals the foot is taken as the unit ; twelfths of a foot are called primes ; twelfths of a prime, seconds ; twelfths of a second, thirds, etc. Primes are marked ^ ; seconds, '^ ; thirds, ^^^ ; fourths, ^^^^, etc., and the marks are called indices. For Lengths. 1ft. = 12' = 12 in. V = 12" = i " 1" = 12'" = — " 12 TABLES. For Surfaces. lft.= 12' =lUsq.in. 1' ■ = 12" = 12 " 1" = 12'" = i " 1'" = 12"" = L " For Volumes. 1ft. = 12' =1728 cu. in. 1' = 12" = lU " 1" = 13'" = 12 " 1'" - 12"" = 1 " ADDITION AND SUBTRACTION. 804. Duodecimals are added and subtracted in the same manner as compound numbers are. Thus, 23 ft. %'W' IV + 15 ft. 8^ 7'^ 8'^' = 39 ft. 6' 6'^ 1'" \ 12 ft. 3' 1" W" %"" — 8 ft. 8^ %" = 3 ft. 6^ 10^^ %'" W". MULTIPLICATION. 805. — Ex. 1. How many square feet in a board 12 feet 9 inches long and 2 feet 6 inches wide? 9' 12 ft. 2 6' 6 ft. 25 4 6' 6" 31 f 31ft '1 + , 10' 6" = - 31.875 sq.ft. Solution. — 9 in., or 9^,= t\ ft., and 6 in., or 6^ = ^% ft. 9^X6'=54'' = 4'6''. Write the &'', and add the A' to the next product. 12 ft. X 6' =-- 72'; 72' + ¥ = 76' = 6 ft. 4', which we write in the product. 9'X 2 = 18' = 1 ft. 6'. Write the 6', and add the 1 ft. to the 24 ft. ^- 1 ft. 3= 25 ft., whicli we next product. 12 ft. X 2 =^ 24 ft write in the proil? 332 EXAMIXATION PROBLEMS. 15. How many tons of hay at 32 dollars a ton, can be ex- changed for 44 tons of coal at 8 dollars per ton ? 16. Which of the numbers 84, 282 and 798 is divisible by the largest prime number ? 17. How much less is the greatest common divisor of 30 and 42 than their least common multiple ? 18. What is the smallest sum of money with which I can purchase colts at 25 dollars each, cows at 40 dollars each, oxen at 100 dollars each, or horses at 125 dollars each? (Articles 140— S96.) 19. How much does f + f exceed f — f? 20. The sum of two fractions is |-|, and one of them is f . What is the other ? 21. What number divided by 3|- will give ^? 22. If f of a yard of cloth costs |-^ of a dollar, what will be the cost of 1 yard ? 23. Show by examples the difference between a Compound Fraction and a Complex Fraction. 24. What is ^\ of a ship worth if \^ of it be worth 50000 dollars ? 25. A man spent f of -g- of his money one day, f of |- of it another day, and then had 3887 dollars remaining. How much money had he at first? 26. Show by examples that multiplying or dividing both terms of a fraction by the same number does not change the value of the fraction. 27. What common fraction is equal to the simi of .655, .33^ and .9375 ? 28. Express by figures six hundred thousand six, and six million sixty thousand six hundred six billionths. , 29. How much less than one million is one millionth ? 30. A 63-gallon cask is f full of wine ; if 27.625 gallons should leak out, the wine remaining will be what decimal part of a full cask ? 31. Show by an example how Integers and Decimals cor- respond in expression. EXAMINATION PROBLEMS. 333 32. What will 56.75 acres of land cost at $20.25 per acre ? 33. What is the value of ^^^ + II_ ^^^^ v .005 4 5.5 34. A grocer paid $58G. 50 for apples, giving S2.25 a barrel for 124 barrels, and $3.75 for the remainder. How many bar- rels did he buy ? 35. How many times is .029 exactly contained in .3786, and what will remain ? 36. A sum of money was divided among three boys ; the first received .375 of the whole ; the second, .6 of the whole ; and the third, ^2.12|-. What was the sum divided ? (Articles 297—407.) 37. How many inches are there in 1051 yards 2 feet 5 inches ? 38. Reduce 3186938 seconds to days. 39. What will 28 square rods 129 square feet of land cost, at 12 cents per square foot? 40. In walking from one town to another a man took 29700 steps of 2 feet 8 inches each. How many miles did he walk? 41. Show by an example that Reduction Descending and Reduction Ascending are reverse processes. 42. What is the sum and the difference of 75 yards 1 foot 9 inches and 46 yards 2 feet 11 inches? 43. On July 17, 1872, how old was a man Avho was born February 18, 1819? 44. How^ many cords are there in 3 ranges of wood, each being 12 feet long, 4 feet w'ide and 6 feet 4 inches high ? 45. What is the product of 16° 15' 16" multiplied by 11 ? 46. How much must be paid for 41 gal. 2 qt. If pt. of molasses, at 72 cents a gallon ? 47. What part of a cubic yard is a cube whose edge is one- half of a yard ? 48. What fraction of an ounce Troy is 15 pwt. 9y\ gr. ? 49. How much must be paid for one-seventieth of 336 bu. 3 pk. 4 qt. of corn, at SO cents a bushel ? 50. How many cubic feet in a rectangular beam 24 feet 6 inches long, 1 foot 9 inches wide and 1 foot 2\ inches thick ? 334 EXAMINATION PROBLEMS. 51. What decimal of a ton is f of an ounce? 52. How much hay, at $30 a ton, can be bought for 8131.25 ? 53. Express ^ oi a. day in hours, minutes and seconds. 54. From a piece of land 24 rods square I sold -^ of an acre to A, f of an acre to B, and .675 of an acre to C. How much of the land was left ? 55. A ship's chronometer, set at Philadelj)hia, longitude 75° 10' W., pointed to 3 h. 40 min. 24 sec. a.m. when the sun was on the meridian. In what longitude was the ship ? (Articles 408—531.) 56. If I sell land at S75 per acre, and thereby gain 25%, how much per acre did the land cost me ? 57. How much is 10% of 25% of 1680 bushels? 58. I bought a horse for $120, and sold him for 8160. What per cent, did I gain ? 59. My house is worth y% as much as my farm. What per cent, of the value of the farm is the value of the house ? 60. A received from B 85100, with which he bought flour at 85 per barrel, deducting his commission of 2% on the cost. How many barrels did he buy ? 61. What are the interest and the amount of 88500 for 2 years 7 months 21 days, at 7% ? 62. William bought cows at 880 each, and one-fifth of them died. At what price must he sell the rest, to gain 5% on the whole ? 63. How long must 8600 remain on interest at 6% to gain 81408? 64. What is the interest of 812750 for 5 years 5 months 18 days, at 6%? <6o. What must be the face of a note payable in 90 (lays, on which 85000 would be received from a bank, dis- counting at 5% ? 66. How much will the com])ound interest exceed the annual interest on 81000 for 3 years 6 months, at 6% ? 67. What is the difference between the true and the bank discount of 8500 for 3 months, at 8% ? EXAMINATION PROBLEMS. 335 68. What principal on interest at 7%, from April 9, 1871, to September 5, 1873, will amount to $1477.59 ? 69. The difference between the interest of $600 and that of $750, at b<'/f, for a certain time, is $18.75. "What is the time? 70. What sum, paid May 16, will settle a bill of $850.50 for goods bought April 19, on 60 days' credit, the rate of in- terest being 7 % ? 71. A note for $1740, dated June 15, 1869, with interest at 69^, was indorsed as follows : Jan. 1, 1870, received $100 ; July 15, 1870, received $112 ; June 1, 1871, received $200 ; and May 1, 1872, received $600. What was due Sept. 1, 1872? (Arlieies 535 -«51.) 72. Two numbers are 108 and 27 ; what is the ratio of the second to the first ? 78. Two numbers whose sum is 3410 are in the ratio of 5 to 6. What are the nund^ers ? 74. How many men can jierform a piece of work in 112 days which 12 men can perform in 84 days? 75. If 18 men can dig a trench 30 yards long in 24 days by working 8 hours a day, how many men can dig a trench 60 yards long in 64 days, working 6 hours a day ? 76. Show by examples the difference between Simple and Compound Proportion. 77. If A invests in a certain enterprise $600 for 4 months, B $300 for 7 months, and C $200 for 9 months, what part of the profits should each of them have ? 78. A, B and C are partners. A furnishes \ of the capital ; B, $500 ; and C, $400. At the end of the year the profits are $4200. What sum should each receive ? 79. I owe Jones two bills, one of $600 on a credit of 60 days, and the other of $800 on a credit of 30 days. These bills being without grace, in how many days should a note given for their amount be made payable ? 80. The balance of an account is $420, and is due, by aver- age, April 28. What was its cash value March 8, interest being at8%? 336 EXAMINATION PROBLEMS. 81. HoAV much must be invested in Government 4 per cents., at 93f, to realize a quarterly interest of S30? 82. What will be the cost of a draft of $12500, at 60 days, exchange being at 100|-, and interest at 7% ? 83. What rate is paid for money when l\^o is charged for exchange on a 30-day note discounted at 6% ? 84. I invested $1606 in 4J- per cents, at lOOi, and sold when they had fallen, losing $100, inclusive of the double brokerage of -1%. At what price did I sell? (Avtieles 652—792.) 85. What integral power of 5 is nearer than any other to 100000? 86. Show by an example the difference between the Square Root and the Cube Root of a number. 87. What is the difference, carried to 3 orders of decimals, between 1^3 and l/2 ? 88. What difference is there between the area of a floor 30 feet square and that of three others each 10 feet square? 89. Two ships, A and B, sailed from a certain port at the same time. A sailed north, 8 miles an hour, and B sailed east, 6 miles an hour. How far apart were they at the end of the third hour ? 90. What is the product of 7.3 multiplied by 1.92? 91. Show by examples the difference between Arithmetical and Geometrical Progression. 92. A room is 20 feet long, 16 wide and 10 feet high. What is the distance from an \ipper corner to the opposite lower corner ? 93. How many feet, board measure, in a rectangular beam 14 feet 9 inches long, 1 foot 8 inches wide and 1 foot 4 inches thick? 94. A garden is 100 feet long and 80 feet wide. What must be the dimensions of a similar garden to contain just one-half as many square feet ? ' u tl'-/' COWPERTHWAIT & CO:S EDUCATIONAL SERIES. HAGAE'S SERIES OF ARITHMETICS. RETAIL PRICE. I. Hagar's Primary Lessons in Numbers, . . $0.30 II. 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These books are all Fresh, Original, Thoroughly up to the Times, and especially Adapted to the Improved Methods of Instruction which now prevail in the best schools. They are in very general use in all parts of the United States, and we are proud of the fact, which has been so often stated to us, that They are Best Liked by the Best Teachers. Upon the Liberal Terms offered for First Introduction, to intro- duce these books will in most cases be MORE ECONOMICAL than to continue to use the old ones, which must, in many instances, soon be replaced by new books at full prices. Correspondence is solicited with refer- ence to the use of these publications in public or private schools. Copies for Examination or for First Introduction, in Exchange for other books in use, will be supplied at HALF RETAIL PRICES, on application to the Publishers, or to any of their Agents. COWPERTHWAIT & CO., Educational Publishers, 628 and 630 Chestnut Street, Philadelphia. B@^ New Illustrated Descriptive Catalogue sent free. CUWPERTHWAIT& Co:S EDUCATIONAL SERIES. RETJLII. I"«ICE LIST. GEOGRAPHIES. Mairen\ New Primary Geography, New Cummon School GVography, >'ew Physical Geography, Apgar's Geographical Drawing-Book, The Geographical Question-Book, . . Warren's Physical and Outline Charts, • Political ;ad Outline Charts'. . . Map-Drawing Paper, $0.75 1.S8 1.88 94 (Per 5«-. (Per Set,^ lu.UU .25 GPAMMAKS. Greene's Xew Iiitroduetiou to English Gramuiar, Xew English Giamiuar. Analysis of the English Language, ARITHMETICS. Hagar's Primary Lessons in Numbers, Elementary Arithmetic, Common School Arithnutic. l»i(;atiou Prohlems an«l Kev. .J6 1.05 .80 .30 • .50 l.uo l.uu READERS. Monroe's Fi.st RcaJer, Seo.nd Keader, .... Third Keader, Eoiirth Reader, . * . Fifth Reader, Sixth Keader, H1S":^0RIES. Goodrich's Chil.rs History of the United States Berard's School History of the Lnitcd States, '. MISCELLANEOUS Monroe's Manual of Physical and Vnca! 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