IIIIIIHI rj; :: 'I'V-ii..-;^"!- III nil III mil illill llllli 'MM' nil ... llllllll iiiiiiiiiiiiiiiii ..J If 1 fU In' 1 1 1 i ft lii li'l 'ifl HI III llii 1 Ill f 111 1',' III! 1 !l 4 ill 1 III 1 1 1 1 ll«L III 111 III III! I m m M \m 1 H i m i\i\\ ^m-: m^M'"' S^' JVo DE No. ^ >N COMPLIMENTS AMERICAN BOOK CO, A. P. OUNN, Oen'l Ag't, 204 PINE STREET, SAN FRANCISCO. Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/americancompreheOObailrich • • •• • • • • • , «• • • •• . • • • ••• AMERICAN COMPREHENSIVE ARITHMETIC BY M. A. BAILEY, A.M. PBontssoR or Matiikmatics in tub Kansas Stats Normal School, at Emporia, Kansas NEW YORK :• CINCINNATI :• CHICAGO AMKRICAN BOOK COMPANY ••••••• I ■ / ,.. -.. ' ^ OOPTSIOBT, 18W, BT AMERICAN BOOK COMPANY. BA1UIY*8 kM.. OOMPE. ABITH. EDUCATloVl neP^» PREFACE In this work, the divisions of arithmetic are presented in their natural order. The fundamental operations upon integers, common fractions, decimals, denominate numbers, and numbers expressed by letters, are followed by their applications to business and to various employments. The introduction of the chapter on literal quantities is somewhat of an innovation, but is in accord with the views of leading teachers, and is based upon sound principles. The elementary parts of algebra should be studied before the advanced parts of arithmetic, because they are more easily com- prehended, and because they afford valuable assistanct in difficult problems. For the same reasons, the solution of many classes of problems should be taught by algebraic processes, before they are taught by the intricate methods required by analysis. The fact, too, that most pupils must leave school quite early, is a strong reason for introducing into arithmetic the elements of algebra and the conclusions of geometry. From the beginning, the pupil is taught to develop observa- tion and thought power. Instead of being required to memorize a large number of rules and directions, he is encouraged to fix well in mind the result to be obtained, to consider carefully the 54f;»34 PRKFACE means at his command, and to employ those means according to his best judgment. This book is intended to complete the course in arithmetic required by district and city schools. M. A. BAILEY. CONTENTS Integers rA«v Notation and Numeration 7 Addition 16 Subtraction 28 Multiplication 35 Division 44 Operations Combined 60 Factoring 68 Common Fractions Notation and Numeration 83 The Operations 84 Analysis 98 Decimal Fractions Notation and Numeration 103 The Operations 107 Special Cases 114 Denominate Numbers — Enolish System Notation and Numeration 119 The Operations 130 Dknominate Numbers — Metric System Notation and Numeration 186 The Operations 142 Literal Quantities Notation and Numeration 145 Addition 148 Subtraction 150 Multiplication 152 Division 154 Factoring 150 Fractions 157 Simple Equations 159 Two Unknown Quantities 167 6 6 CONTENTS PAOB pROPORTioir 170 Solution op Problems 180 Tercentage Simple Cases 101 Profit ami Loss 106 Commission 108 Taxes 200 Trade Discount 201 Insurance 202 Stoclts 206 Interest Simple 217 Annual 233 Compound 234 Involution and Evolution Involution 238 Evolution 239 Mensuration No Dimension 240 One Dimension 249 Two Dimensions 251 Tliree Dimensions 260 Similarity 268 Proofs 272 Occupations With the Lumber Dealer 279 With the Carpet Dealer 285 With the Paper Hanger 286 With the Mason 287 With the Farmer 289 Miscellaneous Longitude and Time 292 Problems in Physics 294 Accounts 295 Definitions and Indkx 301 Answers 311 AMERICAN COMPREHENSIVE AMTHMETIO: NOTATION AND NUMERATION TERMS A number answers the question, How Illustration many ? How many apples ? Numeration is the process of reading 6. numbers. 6, a number, or integer. Notation is the process of writing num- Read, Six. bers. Written, 6. NUMERATION In reading an integer, there are three steps : The number is pointed off from right to left into periods of three figures eoA^h. The periods are named from right to left to learn the name of the left-hand period. The periods are read and named, but the period, 000, is not read and units' period is not named. Pointing off into periods of three figures each Beginning aZ the right, point off into periods : 1. 3600028.371. 4. 378002000028706954. 2. 23600028371. 5. 4070080009000056207. 3. 423600028371. 6. 53008700960057008301. 7. How many figures may there be in the left-liand period ? a How many figures must there be in each of the other periods ? Ex. 1. 3,600,028,871. Ex. 2. 2.3,000,028,371. Ex- 3. 423,600,028,371. 7 8 NOTATION AND NUMERATION m '-i. o ' ' ^V*"' Or « Naming the periods tt. 1,023,438,019,000,320,078,736. 6. 21,023,438,019,000,320,678,735. c. 321,023,438,019,000,320,678,735. from right to left, the periods are : unitSy thousands, millions^ billions, trillions, quadrillions, quintillions, sextillions . . . . ; from left to right: .... sextillions, quintillions, quadrillions, trillions, billions, millions, thousands, units. Higher periods after sextillions are : septillionSj octillions, nonillions, decillions, undecillions, duodecillions, .... 9. Memorize the names of the periods from units to sextil- lions ; from sextillions to units. 10. In a, beginning at the right, point to and name each period ; then beginning at the left, point to and name each period. 11. In the same way, point to and name the periods in b and c. 12. In the same way, point to and name the periods in exam- ples 1 to 6. Ex. 10. Units, thousands, millions, billions, trillions, quadrillions, quin- tillions, sextillions ; sextillions, quintillions, quadrillions, trillions, billions, millions, thousands, units. Reading the periods Each period is made up of three orders, units, tens, and hundreds. III. 368 equals 3 hundreds, 6 tens, 8 units. In units' order, except when 1 is in tens' order, 1, 2, 3, 4, 5, 6, 7, 8, 9 are read one, two, three, four. Jive, six, seven, eight, nine. In tens' order, 2, 3, 4, 6, 6, 7, 8, 9 are read twenty, thirty, forty, fifty, sixty, severity, eighty, ninety. 1 in tens' order is read with the units' fic^ure. Thus: 10, 11, 12, 13, 14, 15, 16, 17, 18, 19; ten, eleven, twelve, thirteen, four- teen, fifteen, sixteen, seventeen, eighteen, nineteen. In hundreds' order, 1, 2, 3, 4, 5, 6, 7, 8, 9 are read one hundred, two hun- dred, three hundred, four hundred, five hundred, six hundred, seven hundred, eight hundred, nine hundred. in any order is never read. INTEGERS 9 Reading the periods d. 1,703,100,306,016,429. e. 25,603,007,075,018,208. / 436,000,056,000,001,200. 13. In d, read units' period ; thousands' period ; millions' period. 14. In cf, read billions' period ; trillions' period ; quadrillions'. 15. In e, read quadrillions' period; trillions' period; billions* period; etc. 16. In /, read each period in succession, beginning at the left. Ex. 13. Units' period, 429, four hundred twenty-nine. 4 in hundreds' order is read four hundred; 2, in tens' order, twenty; 9, m units' order, nine. Thousands' period, 016, sixteen. is not read ; 1 in tens' order is read with 6, the figure in units' order, sixteen. Millions' period, 306, three hundred six. 3 in hundreds' order is read three hundred; is not read ; 6, in units' order, six. The process as a whole 17. Read 37000016328. Ans. 37 billion, 16 thousand, 328. Pointing off into periods of three fig- ures each, 37,000,016,328 ; numerating to learn the name of the highest period, units, thousands, millions, billions ; reading and naming the periods, 87 billion, 16 thousand, 328. NoTR. — The plural form of the period's name should be used in numerating, bat the singular form, in reading. The word " and " should not be used. Read: 20305070; 3005007001. 10000001; 2000000002. 88000050; 9120000878. 6180700623074078. 58097060078039007. 136840050000200003. 24. 3000045; 3000000452. 31. 2780010204009086981. la 3678; 33678; 235678. 25. 19. 2000; 20000; 200000. 26. 20. 8008; 80008; 800008. 27. 21. 7017; 70017; 70(K)17. 2a 22. 6010; 70107; 701070. 29. 23. 3124567; 1234567890. 30. 10 NOTATION AND NUMERATION NOTATION —ARABIC In writing an integer there are two steps : The left-hand pe)iod is tvritten with one, two, or three figures. The other periods are tJien written in succession with three fig- ures each. Writing the left-hand period The Arabic notation employs two devices : The number in each order is represent- ed by one of the symbols: 0, 1, 2, 3, 4, 6, Three hundred eight, 6, 7, 8, 9. Z hundred ten S, The name of the order is represented by ong relative position. Write as a left-hand period : 32. Seven ; seventy ; seven hundred ; nine hundred sixty-eight. 3a Twenty; sixteen; thirty-six; forty-five; twelve; five. 34. Ninety; four hundred; five hundred sixty; one hundred one ; nineteen ; seventy-five. 35. Nine hundred ninety-nine; two; sixteen; eighty; forty- five; thirteen. Ex. 32. 7 ; 70 ; 700 ; Writing the other periods The same devices are employed as before, but each period must contain three figures. See examples 7 and 8. Write as a full period : 36. Zero; eight; eighteen; ninety; fifty-six; forty. 37. Thirty ; one ; six ; seventy-five ; one hundred ; three. 3a Five hundred six ; nine ; seventeen ; sixty-eight ; four ; seven; seventy; twenty. 39. Sixteen; thirty-six; forty-five; five; fourteen; ninety-nine. Ex.36. 000; 008; 018; INTEGERS 11 The process as a whole 40. Write thirty-six trillion, one hundred seventy-six million, six. Am. 36,000,176,000,006. We think 36 trillion, and write 36, ; we think no billion, and write 000, (the work now appears 36,000,) ; we think one hundred seventy-six million, and write 176, (36,000,176,) ; we think no thousand, and write 000, (36,000,176,000,) ; we think six, and write 006. (36,000,176,000,006.) Write: 41. Two billion, seventeen thousand, one hundred twenty-six. 42. 307 quadrillion, 20 billion, four hundred seventy-seven. 43. 300 sextillion, 4 trillion, 30 million, 98 thousand, sixty. 44. 65 quadrillion, 700 billion, 99 million, 999 thousand, 999. 45. Nineteen million, 75 thousand, seven hundred twenty-four. 46. Five million, 7 hundred 20 thousand, 6 hundred thirty. 47. 30 quintillion, 300 trillion, 475 million, 4 thousand, 16. 4a 555 sextillion, 505 million, five hundred thousand, 500. 49. 826 quadrillion, 469 billion, 8 million, 95 thousand, 18. Numeration table 50. Memorize the following table both by orders and by periods. By Periods 1000 units = 1 thousand. 1000 thou. = 1 million. 1000 mil. = 1 billion. 1000 bil. = 1 trillion. 1000 tril. = 1 quadrillion. 1000 quad. = 1 quintillion. Ill III III III III III III Ui 368, 496. 743. 978. 432. 389. 986, 725. MX- qnin- qoad- tril- bil- mil- thoa- a- tillionii tillioni, rillloni, lioni, Uout lionf, landi, nita. Bv Orders 10 units = 1 ten. 10 tens = 1 hundred. 10 bund. = 1 thousand. 10 thou. = 1 ten thousand. 10 t. thou. = 1 hundred thousand. 10 h. thoa , = 1 million. . . .... 12 NOTATION AND NUMERATION NUMBERS NAMED — TO A THOUSAND In naming numbers to a thousand, it is thought best to consider individuals as collected into groups of terij ten of these groups as collected into a higher group, and ten of these higher groups as collected into a still higher group. The individuals are called wmYs; the groups, orders. Thus: Ten units are considered a group of ten; ten teiw, a group of a hundred; ten hundreds, a group of a thousand. The individuals forming a group of ten are given independent names, and these names are repeated with the names of the orders to form larger numbers. Thus : 0?ip, two, three, four. Jive, six, seven, eight, nine, one ten, one ten one, one ten two, .... one ten eight, one ten nine, two ten, two ten one, two ten two, .... two ten eight, two ten nine, nine ten, nine ten one, nine ten two, .... nine ten eight, nine ten nine, one hundred, one hundred one, .... one hundred nine ten nine, nine hundred, nine hundred one, .... nine hundred nine ten nine. Note. — One ten, one ten one, one ten two, .... two ten, two ten one have been abbreviated to ten, eleven, twelve twenty, twenty-one, .... From the above, it will be seen that numbers to a thousand may be expressed by only ten different symbols, if the number in each order is expressed by a figure, and if the names of orders are expressed by relative position. Thus : 90, 91, 92, 93, 94, 96, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 197, 198, 199, 900, 901, 902, 903, 904, 905, 906, 907, 908, 997, 998, 999. The Arabic notation affords these devices, and therefore har- monizes with the plan of naming numbers. The oversight of these devices by the Romans and most other nations of antiquity, nnrtr\-n . + o fr^r. +-k. inte(;krs 13 numbers named — above a thousand When the number of individuals is larger than a thousand, it is thought best to consider the individuals as collected into groups of a thousand, a thousand of these groups as collected into a higher group, a thousand of these higher gi'oups as col- lected into a still higher group, and so on. The individuals are caMed units ; the groups, pmocfe. Thus: A thousand units are considered a group of a thousand ; a thousand thou- sandSf a group of a million; a thousand millions, a group of a billion; and so on. The individuals forming a group of a thousand are named as on the preceding page, and these names are repeated with the names of the periods to form larger numbers. Thus : 1 thousand, 1 thou. 1, 1 thou. 999, 999 thou. 999, 1 million, 1 mil. 1, .... 1 mil. 999 thou. 999, .... 999 mil. 999 thou. 999, From the above, it will be seen that numbers larger than a thousand may be expressed by only ten different symbols, if the number in each period except the left-hand is written with three figures, and if the names of periods are expressed by relative position. Thus : 1,000; 1,001; 1,002; 1,003; 1,999; 999,999; 1,000,000 ; 1,000,001 ; 1,000,002 ; . . . . 1,999,999 ; 9l)JM>99,999 ; Note. — The people of the United States and of France follow the foregoing plan indefinitely ; the English, only to a million. When the number of individuals is larger than a million, the English and most European nations think it best to consider the individuals as collected into groups of a milUony a million of these groups as collected into a higher group, a million of these higher groups as collected into a still higher group, and so on. The individuals are called units; the groups, periods. Thus : A million units are considered a group of a million ; a million millions, a group of a billion ; a million billions, a group of a trillion ; and so on. 14 NOTATION AND NUMERATION NOTATION — UNITED STATES MONEY Since 10 mills make 1 cent, 10 cents make 1 dime, and 10 dimes make 1 dollar, the devices already explained may be employed in writing United States money if it is agreed that the orders from the left shall be dollars, dimes, cents, mills. Thus : U. S. Five dollars no dimes eight cents three mills U. S. 5 dollars dimes 8 cents 3 mills U. S. 6083 This plan is modified, however, by writing the symbol, $, before, and a period, after, the number of dollars. Thus : 15.083. Write: 51. 8 dollars 19 cents 2 mills. 56. 600 dollars four cents. 52. 16 dollars 16 cents 4 mills. 57. 151 dollars ninety cents. 53. 18 dollars 4 cents 6 mills. 5a 225 dollars eight mills. 54. 425 dollars sixty-two cents. 59. 15 dollars 15 cents 5 mills. 55. 521 dollars 38 cents. 60. 27 dollars eleven cents. Ex. 51. $8,192. 19 cents equals 1 dime 9 cents. The word "dime" is rarely used in writing or reading. Bead : 61. ^1.846; $14.18; $95.95. 66. $43,065; $18.03; $15.92. 62. $91.12; $84.98; $41.05. 67. $91,164; $37.33; $41.25. 63. $82.07; $46.04; $37.07. 68. $126,003; $5.05; $23.50. 64. $86.06; $39.09; $27.64. 69. $8971.04; $3.07; $94.01. 65. $51.22; $49.87; $91.14. 70. $6128.97; $9.21; $79.49. Ex. 61. 1 dollar 84 cents 6 mills. The natural reading would be, 1 dollar 8 dimes 4 cents 6 mills, but it is customary to read dimes and cents together as cents. INTEGERS 16 NOTATION — ROMAN The Roman Notation uses seven capital letters, viz. : I, 1 ; V, 6; X, 10; L, 50; C, 100; D, 500; M, 1000. Since it does not use an independent symbol for each of the first ten numbers, it is out of harmony with the plan of naming them, and is rai-ely used except for ornamental purposes. Independent names are given to one, two, three, four, five, six, seven, eight, nine, ten; the Roman notation uses independent symbols for one^ Jive, ten, fifty, one hundred, five hundred, and one thoitsand. The Roman notation employs these devices : Repeating a letter repeats its value. Wheti a letter is placed after one of greater value, its value is to be added to that of the greater. When a letter is placed before one of greater value, its value is to be sub- tracted from that of the greater. When a letter is placed between two letters, each of greater value, its value is to be subtracted from the sum of the other two. A bar placed over a letter multiplies its value by 1000. Ill, 3 ; CC, 200 ; XXX, 30. XI, 11 ; LXXX, IX, 9 ; CD, 400. XIV, 14 ; DXL, 640. X, 10,000 ; CMIII, 900,008. Express by the Arabic : 71. IT, XX, CCC, MMM. 72. LI, CV, DC, LXX. 7a IV, XL, XC, CD, CM. 74. XIX, CXL, DXC, MCD. 75 D, M, CDCCCXXXIX. Express by the Roman : 76. 1,2,3,4,5,6,7. 77. 8, 9, 10, 11, 12, 13. 7a 28, 83, 95, 101, 190. 79. 256,379,400,568. aa 1492, 1520, 1876, 1896. ADDITION TERMS The whole is equal to the sum of all its parts. Thus : 9 cents = 4 cents -f 5 cents. If the whole is wanting, this becomes what = 4 cents + 5 cents ? It means " what is the sum of 4 cents and 5 cents ? " Addition is the process of uniting two or 4) more numbers into one. The numbers to be f I ^'"^ united are addends; the result, the sum, or 5, sum. amount. 6 + 4 = 10, The sign of addition is +. rea^^ The sign of equality is =. 6 plus 4 equals 10. COUNTING BY ONES Addition may be performed by counting by ones. 1. In this way, find the sum of 4, 3, and 2. Counting 4, then 3, then 2, and making a mark at each count, we have ////, ///, //; counting these together, we have one, two, three, four, five, six, seven, eight, nine. 2. Counting by ones, find the sum of 5 and 6. 3. Counting by ones, find the sum of 8, 9, and 7. 4. Would you care to find the sum of 968 and 754 in this way ? Why not? 16 INTEGERS 17 COMMON METHOD There are forty-five combinations of the first nine numbers taken two and two. They should be memorized as wholes. 5. Memorize the follomng : 1 2 23 34 345 466 1, ;?; 1, 3; 2, 1, 4; 2, 1, 5; 3, 2, 1, 6; 3, 2, 1, 7 4667 6678 66789 4, 3, 2, 1, 8; 4, 3, 2, 1, P; 6, 4, 3, 2, 1, 10 6789 6789 789 6, 4, 3, 2, ii; 6, 6, 4, 3, 7j?; 6, 6, 4, i5 78 9 89 89 9 9 7, 6, 6, i^; 7, 6, i5; 8, 7, 75; 8, i7; 9, 18, Thus : 4, 8, 2t i« are different ways of expressing 8. Call the sums : 83927948697179696 6. 6, 2, 3, 2, 8, 7, 4, 1, 4, 2, 7, 1, 6, 1, 6, 4, 4. 97868196939482178 7. 8, 9, 7, 9, 8, 4, 9, 6, 2, 9, 6, 9, 3, 8, 6, 4, 9. 63828346567964762 a 6, 3, 4, 6, 3, 1, 2, 2, 3, 3, 6, 6, 1, 3, 2, 2, 4. 19621674896271799 9. 6, 8, 6, 6, 7, 8, 1, 1, 6, 4, 7, 1, 6, 2, 7, 6, 3. 82938986761459769 la 7, 9, 3, 7, 6, 9, 2, 1, 3, 9, 8, 7, 1, 6, 6, 8, 8. 29836948693172698 11. 3, 2, 4, 6, 6, 7, 6, 1, 6, 3, 6, 3, 6, 7, 6, 4, 4. 67489617879483979 12. 7, 1, 6, 6, 4, 6, 9, 6, 8, 7, 6, 8, 1, 8, 7, 8, 2. Ex. 6. 18, 6, 12, 4, 16. AMBR. ARirn. — 2 18 ADDITION Steps in adding Is the sum of the units more tJuxn 9 : 13. When 6 is added to 24, 23, 27, 25, 28, 36, 41, 32? 14. When 5 is added to 12, 35, 38, 27, 43, 54, 19, 21? Ex. 13. Yes, no, yes, yes, .... Declare the tens of the sum : 15. When 36 is increased by 9, 2, 8, 4, 6, 3, 1, 5, 7. 16. When 24 is increased by 8, 7, 2, 9, 1, 3, 6, 5, 4. 17. When 68 is increased by 5, 9, 2, 8, 4, 3, 7, 1, 6. la When 79 is increased by 8, 3, 1, 4, 9, 5, 6, 7, 2. 19. When 27 is increased by 5, 9, 2, 8, 4, 3, 6, 1, 7. 2a When 35 is increased by 3, 1, 7, 2, 6, 4, 9, 6, 8. Ex. 15. Forty, thirty, forty, forty, .... Declare the units of the sum : 21. When 9 is added to 15, 29, 37, 46, 53, 22, 41, 68. 22. When 8 is added to 94, 38, 75, 26, 52, 63, 89, 71. 2a When 7 is added to 35, 48, 84, 76, 92, 53, 69, 21. 24. When 6 is added to 20y 38, 47, 56, 63, 72, 51, 49. 25. When 5 is added to 17, 23, 31, 45, 56, 62, 34, 49. 2e When 4 is added to 14, 58, 75, 66, 82, 91, 27, 19. 27. When 3 is added to 18, 16, 27, 39, 45, 64, 23, 12. Ex. 21. 4, 8, 6, 5, 2, 2a Declare the sums in examples 13 to 27 inclusive. Ex. 13. 30, 29, 33, 31, 29. Count from 1 to 100 by 9's ; by 8's ; by 7's ; by 6's ; by 5's ; by 4's ; by 3's ; by 2's. Ex. 29. By 9's ; 1, 10, 19, 28, 37, INTEGERS 19 A single order 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 30. 8, (5, 9, 3, 4, 7, 8, 4, 9, 5, 3, 8, 7, 6, 8. 31. 6, 4, 8, 2, 6, 6, 4, 6, 9, 8, 8, 6, 4, 9, 8. 32. 6, 6, 9, 9, 8, 4, 7, 6, 8, 8, 7, 9, 4, 8, 3. 33. 6, 7, 8, 9, 4, 6, 8, 9, 8, 7, 6, 5, 4, 3, 2. 34. 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 8, 8, 7. 35. 3, 6, 9, 6, 3, 2, 4, 8, 4, 2, 5, 7, 9, 8, 6. 36. 6, 7, 9, 8, 8, 4, 7, 6, 9, 8, 3, 4, 7, 7, 9. 37. 9, 4, 3, 2, 8, 7, 6, 3, 9, 4, 8, 6, 3, 3, 6. 3a 6, 8, 3, 9, 7, 4, 2, 5, 8, 7, 7, 4, 7, 6, 8. 39. 8, 8, 4, 4, 6, 6, 5, 8, 9, 7, 6, 3, 7, 4, 8. 40. 8, 9, 7, 6, 4, 3, 9, 8, 2, 6, 3, 8, 9, 4, 3. 41. 6, 6, 4, 3, 3, 4, 8, 7, 6, 9, 4, 6, 8, 7, 6. 42. 6, 6, 6, 3, 7, 2, 8, 1, 9, 4, 7, 9, 3, 8, 7. 43. 2, 2, 8, 6, 5, 4, 7, 9, 8, 6, 3, 6, 5, 7, 4. 44. 8, 8, 6, 5, 3, 2, 9, 4, 7, 6, 3, 8, 7, 6, 3. 45. 8, 6, 5, 3, 9, 8, 6, 6, 4, 8, 8, 5, 7, 2, 3. 46u 2, 9, 1, 3, 8, 2, 5, 3, 7, 6, 4, 7, 5, 5, 6. 47. 2, 3, 4, 6, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2. 4a 4, 6, 6, 4, 7, 8, 6, 7, 6, 3, 2, 9, 7, 6, 6. 49L 1, 1, 3, 5, 6, 2, 3, 4, 7, 1, 1, 8, 2, 5, 4. 50. 7, 6, 3, 2, 9, 8, 8, 4, 3, 8, 7, 6, 3, 1, 9. Ex. 30. 8,14, 23, 26, 30, 37, ... . Do not say ''8 and 6 are 14, and U are 23." Ex. 31. By this method, we add one figure at a time, but when the sum of two adjacent addends is 10, it is customary to add tl»e two at once ; we say 10, 20, 26, 35, 41, 20 ADDITION More than one order 66. Add and explain : 368, 496, 709. 368 The sura of the units is 23 units, or 2 tens iQQ ts and 3 units ; we write 3 in units' column and _, Q 17 carry 2 to tens' coluran. ^* The sum of the tens is 17 tens, or 1 hundred 1573 *"^ ^ tens; we write 7 in tens' column and carry 1 to hundreds' column ; etc. To provcy add in the opposite direction. See p. 79. Note. —For convenience in proving, the sum of each column should be written by itself as soun as it is obtained. 67. Add and explain : 468, 963, 735, 843, 987, 638, 475, 327. 6a Add and explain : 745, 908, 722, 496, 783, 954, 738, 873. 69. 70. 71. 72. 73. 74. 777 398 666 999 995 672 787 846 694 888 881 547 898 946 784 765 654 857 798 775 996 877 343 351 879 238 886 895 637 893 878 876 259 788 448 752 788 439 448 399 549 695 988 777 679 757 870 452 899 989 876 499 626 537 789 646 947 787 765 794 787 988 589 696 693 686 877 568 487 789 247 964 889 859 444 789 692 415 787 363 668 947 631 213 978 496 843 787 786 429 475 132 246 987 325 737 13i^ 547 745 579 417 217 824 724 993 813 629 425 INTEGERS 21 COUNTING BY COMBINATIONS 10-19 The combinations of three or more digits whose sums are 10 to 19 should be learned so that the results can be recognized and called at a glance. 8 S Thus: 8 should be recognized 5 ass, 4 2 i5; 8 as 9, 15; etc. Declare the sums rapidly : 8 1 8 8 8 7 6 6 4 4 1 9 7 8 7 3 3 4 5 4 7 1 75. 2 _9 2 2 3 1 3 9 6 8 8 7 8 4 6 7 8 6 4 9 7 9 3 5 4 3 4 2 3 6 8 3 5 76. 6 3 3 2 1 2 5 9 1 4 4 6 6 8 9 7 9 6 7 7 6 9 3 5 4 3 3 1 3 4 5 3 3 77. 2 4 1 1 2 2 3 4 4 1 2 4 5 5 6 6 7 7 4 6 4 6 6 4 5 2 1 3 2 6 1 4 3 6 2 4 5 3 7 3 2 2 1 5 7a 3 2 5 1 4 2 2 7 2 2 2 5 6 6 6 4 2 5 4 1 2 3 2 1 3 1 4 7 3 1 8 4 6 1 1 4 1 1 1 1 1 1 1 4 79. 3 2 2 4 2 1 3 4 6 3 4 7 6 6 9 3 1 3 4 1 4 2 1 3 1 1 3 8 4 6 8 4 6 1 1 1 4 2 2 2 8 4 2 4 80. 4 2 7 5 2 7 5 1 4 4 4 Ex. 75. 19, 17, 18, 17, . . . . Ex. 78. 19. 13 ,19. 13, . . . . 22 ADDITION Steps in adding Is the sum of the units more than 9 : 81. When 15 is added to 72, 68, 47, 31, 45, 23, 36, 39? 82. When 18 is added to 94, 47, 39, 28, 55, 63, 92, 61? Ex. 81. No, yes, yes, no, yes, .... Declare the tens of the sum : 8a When 46 is increased by 13, 16, 17, 12, 18, 14, 19. 84. When 38 is increased by 15, 19, 17, 12, 14, 16, 13. 85. When 92 is increased by 17, 19, 18, 14, 16, 12, 15. 8a When 29 is increased by 11, 15, 18, l6, 19, 17, 12. 87. When 37 is increased by 17, 14, 16, 19, 13, 14, 11. 8a When 55 is increased by 12, 19, 17, 16, 13, 14, 18. Ex. 83. Fifty, sixty, sixty, fifty, sixty, sixty, .... Declare the units of the sum : 89. When 17 is added to 71, 85, 96, 84, 23, 72, 99, 87. 90. When 16 is added to 15, 29, 37, 46, 53, 22, 41, 74. 91. When 13 is added to 17, 19, 26, 28, 35, 41, 63, 94. 92. When 14 is added to 68, 94, 85, 96, 47, 53, 72, 91. 93. When 18 is added to 29, 37, 46, 55, 83, 92, 24, 38. 94. When 12 is added to 24, 69, 78, 37, 43, 82, 91, 65. Ex. 89. 8, 2, 3, 1, 95. Declare the sums in examples 81 to 94, inclusive. Ex. 81. 87, 83, 62, 46, 9a Count from 1 to 200 by 19's; by 18's; by 17's; by 16's ; by 15's ; by 14's ; by 13's ; by 12's ; by ll's. Ex. 96. By 19's. 1, 20, 39, 58, 77, 96, ... . INTEGERS 23 Addends less than 20 121. 122. 123 124. 125. 126. 127. 12a 129. 13a 131. 97. 13, 14, 15, 16, 17, 18, 12, 14, 11, 19, 18. 9a 12, 16, 13, 18, 11, 17, 15, 16, 13, 18, 11. 99. 15, 17, 12, 13, 18, 11, 16, 15, 12, 14, 17. 100. 19, 13, 14, 17, 14, 19, 13, 17, 15, 12, 14. 101. 16, 11, 15, 14, 17, 12, 14, 13, 17, 11, 13. 102. 17, 14, 18, 19, 15, 13, 17, 19, 16, 15, 12. 103. 18, 19, 17, 12, 13, 14, 19, 18, 14, 17, 19. 104. 19, 16, 14, 13, . 12, 18, 15, 16, 17, 19, 16. 105. 11, 19, 15, 13, 12, 16, 17, 13, 19, 18, 16. loa 16, 18, 14, 12, 10, 11, 13, 15, 17, 19, 16. 107. 19, 15, 13, 12, 11, 18, 16, 14, 13, 18, 17. lOa 16, 16, 15, 14, 19, 17, 16, 17, 18, 14, 13. 109. 13, 16, 19, 19, 17, 17, 15, 15, 11, 11, 17. 110. 18, 18, 16, 16, 13, 19, 17, 11, 13, 14, 14. 111. 17, 17, 16, 14, 15, 13, 19, 17, 15, 11, 12. 112. 15, 14, 12, 10, 11, 12, 13, 14, 17, 19, 18. ua 16, 12, 16, 16, 15, 14, 18, 19, 17, 16, 15. U4. 11, 13, 15, 17, 19, 17, 15, 13, 11, 12, 14. 115. 16, 17, 19, 16, 11, 13, 16, 14, 13, 17, 16. ua 17, 17, 16, 16, 14, 12, 11, 19, 16, 15, 13. 117. 14, 13, 19, 16, 17, 12, 16, 16, 13, 19, 18. lia 12, 11, 18, 16, 14, 13, 17, 19, 16, 16, 17. 119. 19, 18, 17, 16, 15, 11, 16, 16, 12, 11, 16. 12a 17, 16, 14, 13, 12, 16, 16, 11, 17, 19, 16. Ex. 97. 13, 27, 42, 68, 76, 93, 105, 24 ADDITION Addends greater than 19 132. By combinations 10 to 19, add: 24, 72, 98, 69, 34, 73, 28, 43, 52, 89, 73, 58. 8 8 2 We see s as 11, and say 11 ; we see 9 as 14, 4 8 8 and say 25 ; » as 15, and say 40 ; 9 as 17, and say 57 ; 2 as 6, and say 63. We see the 6 to carry and 5 as 18, and say 18 ; 6 4 8 6 7 we see 8 as 17, and say 35 ; 2 as 18, and say 53 ; 8 9 as 18, and say 71. Note. — The explanation may be given as on p. 20 ; the sum of the units is 63 units, or 6 tens and 3 units ; etc. 138. 938 544 629 887 368 299 886 743 592 447 918 925 759 967 721 i98] (69 S4] 6S 73 ^28] [4S] 71 62 [89] \7S^ 68] ,00 713 c^ @ ( 625 198 419 344 "W ■ 137 333 582 548 991 727 377 949 ^m 194 454 4tr 643 798 249 386 818 666 726 977 638 293 247 586 393 269 475 628 849 137 966 294 888 376 789 649 555 228 494 347 325 ^i3ei 597 487 676 345 799 288 376 467 156 165 246 254 334 343 628 627 ,^37. 896 288 144 727 419 589 476 665 647 925 194 886 827 889 929 588 484 INTEGERS 25 Practice for any method 140. 141. 142. 143. 144. 63954 611043 545982 662347 845954 87445 626915 606819 257938 358346 ]:;•_".> I 537454 146608 689477 214324 :>1G87 442014 889478 893298 843567 91532 780894 395777 478469 413578 72927 122993 865092 328947 495219 51470 726915 323459 886539 254183 82044 484471 478949 445328 145482 94514 705606 338948 768894 562531 12592 560247 667848 325537 214644 88416 827922 769949 892468 523671 97654 439706 876329 729848 954751 92214 364399 448869 849889 552169 82405 671361 732964 946677 773713 96185 225163 985263 833849 631526 :>5391 954670 868897 486695 652847 69039 258520 402986 329875 156879 71623 184109 399478 684498 235659 89555 194020 456984 725578 921982 66526 184059 737842 398765 567399 68507 680637 698325 438942 936783 21661 457351 447983 668937 356773 92717 721295 892567 865432 692683 •S329 665260 443846 598428 578756 31786 539891 329847 697646 578325 97394 382579 666744 393948 565731 86929 426869 889847 872896 757785 31993 389495 653894 498695 567320 51395 708040 655;«3 213151 315625 r:i:.::. 943511 773311 431512 567921 1 Ic... 443225 773221 772202 521673 26 ADDITION PROBLEMS First fonn of analysis 145. If there are 96 apples in one pile and 88 apples in another, how many are there in both ? OR gg In both, there are the sum of 96 apples and 88 apples, or 184 apples. To Moritt vohai each term represents is of service. 146. A man paid $ 45 for a horse, and $ 95 more for a carriage than for the horse. Hfew much did he pay for both ? $ M5i horse 95 increase "^^^ carriage cost the sum of $45 and $95, r—r' or $ 140 ; both cost the sum of $ 45 and $ 140, 140y carriage or $186. 185y both i^ A drawing is often of service. 147. One village is 37 miles west and another 48 miles east t\ of Denvor. Mow far apart are the villages? 4 D B Since it is 37 miles from A to D, and 48 ^j A to D miles from D to B, from A to B it is the sum, 4Sy-^ to B or 86 miles. 85, A to B 148 A T>inn i^nid 350 for a horse, $125 for a carriage, $12 for a hai ' 5 for a saddle. How much did he pay in all ? 149. The village A is 36 miles east of Chicago, and B is 43 miles east of A. How far is it from Chicago to B ? 150. A grocer sells 18 pounds of coffee, 16 pounds more of tea than of ( d 96 pounds more of sugar than of tea. How many pou..- ;1 (loes he sell ? 151. A has 1 *» has $ 29 more than A ; C has as much as A and B ar : and D has as much as A, B, and C together. H' all possess? INTEGERS 27 152. How many strokes are made in a day by a clock which strikes the hours ? 153. Mr. Brown owes three bills, one for $1987, another for $ 1849, and a third for $ 2789. How much does he owe in all ? 154. A man bought 4 horses, paying for them $ 3985, $ 5025, $2789, and $6898; he sold them for $2399 above cost. How much did he receive for them ? 155. A man paid $ 348G for a lot, $ 2878 for a house, $ 1695 for furniture, and $387 for repairs; he disposed of his entire property for $285 above cost. How much did he receive? 156. A nurseryman sold 862 pear trees, 965 more apple trees than pear trees, 688 peach trees, 466 more plum trees than apple trees, and 568 ornamental trees. Find the entire number of trees sold. 157. The village A is six hundred sixty-five miles east of Chicago, and B is eight hundred eighty miles west of Chicago. What is the distance from A to B ? 15a The Duke of Wellington's army at Waterloo consisted of 26,661 English infantry, 8735 English cavalry, 6877 English artillery, and 33,413 allies. How large was his army ? 159. North America has an area of 9,349,741 square miles ; and South America, 6,887,794. What is the area of the American continent ? 160. Mt. Etna is 10,874 feet above the level of the sea; Mt. lUanc is 4870 feet higher than Mt. Etna; Mt. Everest is 13,258 feet higher than Mt. Blanc. What is the height of Mt. Everest ? 161. A train travels 719 miles one day, 698 miles a day for the next three days, 692 miles a day for two days, and 718 miles the seventh day. How many miles do6 it travel during the week? 162.A- farmer raised on one fjitib teen mastered, the results in the following examples may be called rapidly. In multiplication, we have two factors to name their product; in division, the product and one factor to name the other. 6. Oiven the product and one factory name the other : 9 14 7 15 10 6 16 8 11 17 7 ?, j?7; ?, ?, iS*; ?, ?, ?, ^0; ?, ?, 5^; ?, 5^; ?, ^^; ?, 5^; 18 12 9 6 19 13 20 10 8 21 14 7 ?, ?,?,?, ^6; ?,A9j ?,5P; ?, ?,?,^(?; ?, ?, ?, ^^ 22 11 15 9 23 24 16 12 8 7 25 10 ?, ?,^; ?,?,^; ?,4^; ?, ?, ?, ?, ^; ?,-^5; ?, ?, 5(?; 17 26 13 27 18 9 11 28 14 8 19 ?,5i; ?, ?,5je; ?, ?,?,5^; ?,55; ?, ?,?,^e; ?,57; 29 30 20 15 12 10 31 21 9 32 16 8 ?,^; ?, ?, ?, ?, ?,^^; ?,^^; ?,?,^^; ?, ?, ?,64; 13 33 22 11 34 17 23 35 14 10 ?,^5; ?, ?, ?,5^; ?, ?,6, p.48. i^ofll9U9}i; ▲ MBR. ARITH. — 4 60 DIVISION Times contained A whole may contain a part an exact number of times. A whole contains its half two times ; its third, three times ; its fourth, four times ; and so on. 12 cents contains 6 cents 2 times ; 4 cents, 3 times ; and so on. Thus : 8 = 4 + 4 ; or 8 = i of 8 + i of 8. 6 = 2 + 2 + 2 ; or 6 = J of 6 + } of 6 + i o' ^• Dividing a number of cents by 4 cents is finding how many times the number contains 4 cents ; dividing a number of eggs by 5 eggs is finding how many times the number contains 5 eggs ; and so on. Thus : 8 cenU -i- 4 cents = 2, because 8 cento = 4 cento x 2. 8 cento contains 4 cento 2 times, because 8 cento = 4 cento x 2. How many times does : 47. 26 days contain 13 days ? 24 days contain 12 days ? 4a 20 hours contain 2 hours ? 16 hours contain 4 hours ? 49. 18 gallons contain 2 gallons ? 96 gallons contain 6 gallons ? 5a 25 quarts contain 5 quarts ? 60 quarts contain 4 quarts ? Ex. 47. 2 times. 26 days = 13 days x 2. How many times is : 51. 2 pints contained in 18 pints ? 3 pints in 12 pints ? 52. 4 pecks contained in 20 pecks ? 5 pecks in 30 pecks ? 53. 3 pounds contained in 15 pounds ? 6 pounds in 72 pounds ? 54. 5 ounces contained in 30 ounces ? 7 ounces in 42 ounces ? Ex. 51. 9 times. 18 pinto = 2 pinto x 9. Note. — The examples on pp. 47 and 48 should be reviewed as follows: Ex. 7, p. 47. 144 contains 12, 12 times ; . . . . Ex. 26, p. 48. 119 contains 12, 9 times with 11 remaining. INTEGERS 51 SHORT DIVISION 55. Divide 8609 by 12 and explain. In full. 86 Imndred8-f-12=7 hundreds and 2 hun- 12 ) 8609 dreds remaining ; we write 7 in hundreds' cohinin. 717I. 2 hundreds and tens = 20 tens ; 20 tens -=- 12 It =1 ten and 8 tens remaining ; we write 1 in tens' column. 8 tens and 9 units = 80 units ; 89 units -r- 12 = 7 units and 5 units remain- ing ; we write 7 in units' column and 12 under 6, with a line between, to show that 6 is still to be divided by 12. Abbreviated. 7, 1, 7, ^j. Note. — It is a great loss of time to say, " 86 divided by 12 is 7 with 2 remain- ing." While these words are being said, the process is delayed. To provey multiply the divisor by the quotient and add the remainder. The result should be the dividend. See p. 79. Divide and prove, explainitig in fuU : 5e 57. 58. 59. 60. 2)864 3)981 9)648 8)1238 12)9754 61. 62. 6a 64. 65. 9)369 11)858 7)504 6)3255 4)2367 Divide and prove, abbreviating : 66. 67. 6a 69. 70. 9)495 6)558 8)696 7)3728 5)4163 71. 72. 73. 74. 75. 12)864 9)718 7)623 8)3001 12)7000 76. 77. 7a 79. 80. 11)979 6)425 4)916 3)7070 2)5007 Bx. 66. Ana. 66. Say 6, 6. 52 ] DIVISION For mental and written work For mental practice, name each figure of the quotient without writing it, and without stating the full answer after it is obtained. 81. 12)567024 82. 9)3063205 83. 11)30670508 84. 9)1023456789 85. 11)781605 86. 7)3643036 87. 9)20345607 88 8)3321456648 89. 9)612036 90. 5)3245321 91. 8)72867408 92. 7)1111111111 9a 8)921608 94. 6)3215750 95. 7)30000005 96. 6)7340962416 97. 7)333333 98. 3)2222667 99. 6)55555554 loa 5)9765432120 Id. 6)100308 102. 6)5307943 loa 5)98765043 104. 4)9998887772 105. 5)666665 106. 4)5202415 107. 4)12345672 loa 3)9012506370 109. 4)999904 110. 9)1396735 111. 3)12345672 112. 2)9012506370 113. 3)444444 114. 7)5803214 115. 2)56708914 116. 12)9876954312 117. 2)973514 iia 8)5905309 119. 12)63178908 120. 11)5443322344 Ex. 81. For mental work, say 4, 7, 2, 5, 2. INTEGERS 53 LONG DIVISION 121. Divide 8609 by 12 and explain. 717-^ 86 hundreds -^12 = 7 hundreds and 2 hundreds i&\o^nq~ remaining ; we write 7 in hundreds' column. 8 A ^ hundreds and tens = 20 tens ; 20 tens - 12 — '^— = 1 ten and 8 tens remaining ; we write 1 in tens' ^0 column. 1^ 8 tens and 9 units = 89 units ; 89 units -r- 12 S9 =7 units and 6 units remaining ; we write 7 in Sj^ units' column and 12 under 5 with a line between, ~e to show that 6 is still to be divided by 12. 122. Divide 34056 by 17, prove, and explain. ^ i7 34 thousands -=- 17 = 2 thousands and thou- 17) S4O66 sands remaining ; we write 2 in thousands' S4 column. hundreds -4- 17 = hundreds ; we write in hundreds' column. 6 tens -f- 17 = tens and 5 tens remaining ; we write in tens' column. Proof. 5 tens and 6 units = 56 units ; 56 units -i- 17=3 2003— units and 6 units remaining ; we write 3 in units' jrj^^ column and 17 under 5 with a line between, to z — show that 5 is still to be divided by 17. 14021 Proof. Multiplying the divisor by the quotient 200 3 mid adding the remainder, we obtain the dividend. 34056 Dividej provcj explain : 7649 -t- 21; 4179 + 22. 12a 30000 -f- 16; 6875 + 16. 0066 61 6 124. 3174 + 23; 8754 + 24. 129. 85038 + 17; 6283 + 17. 125. 2800 + 13; 6200 + 13. 130. 72004 + 18; 5796 + 18. 126. 9963 + 14; 3528+14. 131. 38794 + 19; 4986 + 19. 127. 8072 + 15; 6123 + 15. 132. 79498 + 25; 3899 + 23. 54 DIVISION 133. Divide 896 by 112. 8 To find the quotient figure when the first jyo\QQ^ figure of the dividend is larger than the first HSS)<5ifo figure of the divisor, we use only the first figure 896 of each term. Approximately, 896 -»- 112 = 8 -j- 1, or 8. 8 x 112 = 896. 134. Divide 2992 by 374. 8 To find the quotient figure when the first figure of the dividend is smaller than the first 874)2992 figure of the divisor, we use only the first two 2992 figures of the dividend and the first figure of the divisor. Approximately, 2992 + 374 = 29 h- 3, or 9. 9 x 374 = 3366. Since 9 is too large, we try 8. 135. Divide 3808 by 476. 8 To determine whether the quotient figure is 476)3808 too lai^e, it is rarely necessary to use more S808 ^^""1 ^6 fi^^^ ^^o figures of the divisor. Approximately, 3808 ^ 476 = 38 -j- 4, or 9. 9 x 47 = 423. Since 9 is too large, we try 8. Perfonn the indicated operation : 136. 730-^365. 141. 8300^426. 146. 16748 -- 4187. 137. 980-245. 142. 8502-5-321. 147. 10380 -- 2076. 13a 845-169. 143. 7321-375. 14a 47472-5934. 139. 984-5- 123. 144. 6325-523. 149. 35315 -r- 7063. 140. 981-5-109. 145. 8053^437. 150. 61263-5-6807. INTEGKKS 66 151. Divide 78264648 by 98076. 98076)7SiS6464S(798 686532 961144 882684 784608 784608 78-5-9 = 8, 8x98 = 784. Since 8 is too large, we try 7. Since the product of 98076 by 7 must con- tain six figures, we place 2, the right-hand figure of the product, under the sixth figure of the dividend. Note. — It is sometimes more convenient to place the quotient at the right. 152. Divide 7384 by 100. 100 )7384 7^lk If the divisor is 10, we cut off one figure from the right of the dividend ; if 100, two figures ; and so on. 153. Divide 3218738 by 92000. 92.000^)3218.738^(34^ 276 458 368 90 If the divisor ends with ciphers, we cut off the ciphers, and the same number of figures from the right of the divisor. We divide the parts left and prefix the remainder to the part cut off, to find the true remainder. NoTB. — A cross may be placed after units, and a point before the last figure cut off. 154. Prove the answer of the last example. » ; cost 5 apples = 6 times cost 1 apple. In some problems, relations must be ascertained from a knowl- edge of how objects are constructed, or from a knowledge of the sciences. 3a State the principles of construction, the required term, and the relation. How many minute spaces has the hour hand of a watch passed since 5 o'clock when the minute hand is at 3 ? Principles : there are 60 minute spaces on a dial ; the minute liand passes 60 spaces while the hour hand passes 5 spaces ; etc. (liven term : minute hand has passed 15 minute spaces since 5. Required term: number minute spaces passed by hour hand since 5. Relation : number spaces passed by hour hand = ^^ number spaces passed by minute band. In some problems, the relations must be ascertained from a knowledge of business usage. The gain equals the selling price minus tfie cost; the loss equals the cost minus the selling price. A person (principal) may give pay (commission) to another (agent) for buying or selling articles for him. Then : The entire cost equals the buying j^rice plus the commission; the proceeds equal the selling price minus the commission. ZL State the relation. If an orange sells for 6^ at a gain of 2 ^, how much is the cost ? Relation : cost = selling price - gain. 32. State the relation. If an agent buys an article for $50 and charges a commission of $1^ how much does it cost the principal ? 3a State the relation. If an agent sells an article for $50 and charges a commission of $ 1, how much does the principal receive ? 64 OPERATIONS COMBINED Write the relations; do not solve the problems. 34. In jumping, A beats B by 2 feet. If A jumps 10 feet, how far does B jump ? 35. In jumping, A beats B by 2 feet. If B jumps 10 feet, how far does A jump? 36. If 3 men can do a piece of work in 6 days, in how many days can 1 man do it ? 37. If 1 man can do a piece of work in 18 days, in how many days can 3 men do it ? 3a With his present force, a contractor can do a piece of work in 18 days. By what number must he multiply his force to finish the contract in 2 days ? 39. If it takes a man 10 minutes to saw a log into 3 pieces, how long will it take him to saw it into 4 pieces ? 4a When the hands of a watch are opposite to each other, how many minute spaces are there between them ? 41. Three boys bought a top for 10^ ; the first gave 2^, and the second, 4^. How much did the third give ? 42. A and B travel in the same direction, A at the rate of 5 miles per hour, and B at the rate of 7 miles per hour. How many miles does B gain in 9 hours ? 43. Conditions as in Ex. 42, if A has a start of 4 hours, in how many hours will B overtake him ? 44. If they start from the same place and at the same time and travel in opposite directions, in how many hours will they be 36 miles apart ? 45. By selling a watch for $90, a man would gain $20; at what price must he sell it to gain $25? 46. If $ 4.20 is paid for 3 days' work, how much will be paid for 10 days' work ? INTEGERS 65 The process as a whole The relation will often suggest a better method than the set forms of analysis. Seepages S6, 55, 4^^ 57. 47. If 3 apples cost 6^, how much will 12 apples cost ? Skt Form. —Since 3 apples cost 6j^, 1 apple will cost ^ of 6/* or 2j* ; since 1 apple costs 2 f, 12 apples will cost 12 times 2 f, or 24^. Better Form. —The cost of 12 apples is 4 times the cost of 3 apples, or 4 times 6 f, or 24 f. Good judgment should be used in selecting the forms. 4a What is the profit on buying 6 cows at $26 each and sell- ing them at $ 28 each ? GrooD Judgment. — The profit on 1 cow is the difference between $28 and $26, or $2 ; the profit on 6 cows is 6 times $2, or ^ 12. Poor Judgment. — Since 1 cow costs $26, 6 cows will cost 6 times $26, or 8 156 ; since 1 cow sells for $28, 6 cows will sell for 6 times $28, or $ 168 ; the profit is the difference, or $ 12. In vfritten work, norite what each term represents. 49. Through an agent, I sell 8 horses at $58 each, commission $ 1 per head, and buy with the proceeds cows at $ 26 each, com- mission $ 1 per head. How much should the agent remit ? f 58, sell. p. 1 horse f 26, cost 1 cow i, commission 1, commission 67 f proceeds 1 horse 27, entire cost 1 cow S 27) 456 {16 f^ 456, entire proceeds 24, amount to remit Be sure to prove every answer. 5a Prove in Ex. 49, that the agent should remit $ 24. $27, cost 1 cow J6 Since the entire cost of tlie cows 1^, entire cost cores P^"^ ^^« '^"^^""^ remitted Js the pro- ^ . ^ .^^ , ceeds from the sale of the horses, the _24y amount remitted ^„3^^^ j^ ^^^ 456, amount to invest in cows AMER. ARITH. — 5 66 OPERATIONS COMBINED 51. In an election, Mr. Jones received 3689 votes, but was de- feated by 216 votes. How many votes did his opponent receive ? 52. In an election, Mr. Brown received 3905 votes and defeated his opponent by 216 votes. How many votes did his opponent receive ? 53. I bought a horse for $ 15,284, paying $ 2684 cash and the balance in monthly payments of $1575 each. How many monthly payments did I make? 54. A farm house is worth $ 2450 ; the farm is worth 12 times as much, less $ 600 ; and the stock is worth twice as much as the house. How much are the house, stock, and farm worth ? 55. In still water, a crew can row 10 miles per hour j the cur- rent runs 2 miles per hour. How many miles can they row down stream in 1 hour ? State the relation. 56. How many miles can they row up stream in 1 hour ? State the relation. 57. In how many hours can they row 48 miles down stream and return ? State the relations. 5a A crew can row down stream 12 miles per hour and up stream 8 miles per hour. What is the rate of the current ? State the relations. 59. In how many hours could the crew in example 58 row 30 miles in still water ? State the relations. 60. How much will 5 barrels of potatoes cost if 13 barrels of apples cost $ 39, and 6 barrels of apples cost as much as 9 barrels of potatoes ? State the relations. 61. If 5 horses eat 14 bushels of oats in 2 Aveeks, how long would it take them at the same rate to eat 56 bushels ? 62. E owes a debt of $ 365. How many sheep must he sell at $ 15, commission $ 1 each, to discharge the debt ? How much money will he have left ? 63. A man owed ^2896; he paid $499 at one time, and all but $ 375 a second time. How much did he pay the second time f INTKGKRS 67 64. If 37 horses cost $ 1295, how much will 48 horses cost ? State the relations. 65. How much will 126 barrels of beans cost if 9 barrels cost $ 22 ? State the relation. 66 If an orchard is sold for $ 375 at a loss of $ 28, what was the cost ? State the relation. 67. If a stock of goods costing $4376 is sold at a gain of $ 1094, what is the selling price ? State the relation. 6a B sells a house through an agent and receives $ 1975. If the agent's commission is $ 97, what is the selling price ? 69. If a hound runs 78 rods while a hare runs 64 rods, how far will the hound run while the hare runs 1856 rods? State the relations. 70. Suppose a body falls 16 feet the first second, 48 feet the next, 80 feet the next, and so on, constantly increasing, how far will it fall in 5 seconds ? State the relations. 71. Three men can do a piece of work in 5 days. In what time can 1 man and 8 boys do it, if 1 man does the work of 2 boys ? State the relations. 72. Conditions as in Ex. 71, how many boys would be required to do the work in one day ? State the relations. 73. A has 7 loaves of bread ; B, 5 ; C, none. The three eat all of the bread, each the same amount. C pays to A and B 12 ^. How much should each receive ? 74. A liveryman makes an annual profit of $125 from each horse ; his income each year is $ 2125 ; his horses cost $ 87 per head. How much did he pay for the horses? State the rela- tions. FACTORING TERMS 8 = 4x2. If both multiplicand and multiplier are wanting, this becomes 8 = what X what? It means, "what are the numbers whose product is 8 ?" Factoring is the process of finding num- bers whose product is given. The numbers required sae factors or measures; the prod- uct, a multij)le. Every integer is the product of itself and one. If a number has no set of integral factors besides itself and one, it is a prime number. If it has another set besides itself and one, it is a composite number. Numbers are prime to each other when their greatest common factor is one. Numbers are severally prime when each is prime to each of the others. What are the factors of 30? Am. S and 75, S and iO, 5 and 6^ or SO and J; SO = e X 15, SxlO, 6 x6t or SO X J. 7, a prime number; 30, a composite number. 4, 8, 9, are prime to each other. 4, 9, 25, are severally prime. FROM THE COMBINATIONS If the combinations in multiplication are known, the factors of all numbers less than 100 may be called rapidly. See pp. 36, 4^. State sets of two factors for : 1. 99, 98, 96, 95, 94, 93, 92, 91, 90, 88, 87, 86, 85, 84, 82, 81, 80, 78, 77, 76, 75, 74, 72, 70, 69, 68, 66, 65, 64, 63, 62, 60, 58, 57, 56, 55, 54, 52, 61, 50, 49, 48, 46, 45, 44, 42, 40. Ex. 1. Ans. 96 = 2 X 48, 3 X 32, 4 X 24, 6 X 16, 8 x 12 ; . . . . INTEGERS 69 BY INSPECTION Whether one of the factors of a iiuiiiImt is 2, 3, 4, 5, 8, 9, 11, or a product of two or more factors severally prime, may be found by the following principles. See p. 80. A number is divisible by 2, when the number denoted by its last digit is di- visible by 2, or is 0. A number is divisible by 5, when the number denoted by its last digit is di- visible by 5, or is 0. A number is divisible by 4, when the number denoted by its last two digits is divisible by 4. A number is divisible by 8, when the number denoted by its last three digits is divisible by 8. A number is divisible by 3, when the sum of its digits is divisible by 3. A number is divisible by 9, when the sum of its digits is divisible by 9. A number is divisible by 11, when the difference between the sum of its digits in the odd places and the sum of its digits in the even places, is divisible by 11, or is 0. A number is divisible by the product of any number of its factors which are severally ])rinie to each other. Illustrations 27725 is divisible by 2, because 6 is divisible by 2. 28725 is divisible by 5, because 5 is divisible by 5. 7.1112 is divisible by 4, because 12 is divisible by 4. 21816 is divisible by 8, because 816 is divisible by 8. 27810 is divisible by 3, because 18, the sum of its digits, is divisible by 3. 27810 is divisible by 9, because 18, the sum of its digits, is divisible by 9. 1639 is divisible by 11, because 11, the difference between 16, the sum of its digits in the odd places, and 4, the sum of ita digits iu the even places, is divisible by 11. 27720 is divisible by 7. 8, 9, and Ls therefore divisible by 7 X * X 5. 70 FACTORING Of (he. numbers 2^ S, J^ 5, 8y P, //, use those tvhich are severalbj prhnej and form comhinalions : 2. Of 2 and one other, a Of 3 and one other. 4. Of 4 and one other. 5. Of 5 and one other. 6. Of 8 and one other. Ex. 2. 2 X 3, 2 X 6, 2 X 0, 2 X 11. StaJte why 27720 is divUible : 12. By 2; 3; 4; 6. la By 7; 8; 9; 11. 14. By 2 X 3; 2x5; 2x9. 15. By 2x11; 3x4; 3x5. Ex. 13. By 7, by trial. By inspection tell why . 2a 2448 is divisible by 72. 21. 6930 is divisible by 55. 22. 4788 is divisible by 63. 23. 8184 is divisible by 88. Ex. 20. 2448 is divisible by 8 and 9, fore by 8 X 9, or 72. Find all the factors of: 28. 12540, less than 100. 29. 27720, less than 100. Ex. 28. 2, 3, 4, 5, 11 ; 6, 10, 22, 12, Note. — It is best to test for 2, 3, 4, 5, factors which are severally prime. 7. Of 9 and one other. a Of 2, 3, and one other. 9. Of 2, 5, and one other. 10. Of 5, 8, and one other. U. Of 5, 8, 9, and one other. Ex. a 2 X 3 X 5, 2 X 3 X 11. la By 3x8; 3x11; 4x5. 17. By 4x9; 4x11; 6x9. la By 5x8x9; 8x9 xll. 19. By 3 X 7 ; 5x8x9x11. Ex. la 3 and 8 are severally prime. 24. 20934 is divisible by 18. 25. 14630 is divisible by 77. 2a 30144 is divisible by 24. 27. 98000 is divisible by 35. which are severally prime, and there- 30. 17622, less than 100. 31. 26585, less than 100. 16, 33, 20, 44, 55 ; 30, 66 7, 8, 9, 11 ; and then to combine those INTEGERS 71 Finding prime numbers Whether an integer is a prime number is found by trial. In the trial, it is necessary actually to divide only by 7 and by prime numbers greater than 11; divisibility by 2, 3, 5, 11, and by all composite numbers may be tested by inspection. 32. Is 397 a prime number ? Ans. Yes. 397 is not divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, IS, 14, 16, 16, 77, 18, /i?, nor 20. 397 -=- 20 = 19 +. It is not divisible by a number larger than 20, for the quotient would then be a number that has already been tried. NoTK. —In this example, to test divisibility by 7, 13, 17, and 19, the division must be performed. 33. Make a list of all the prime numbers to 100. How many are there ? 34. Is 431 a prime number ? 323? 131? 523? 601? 319? 35. Find the prime number next greater than 401. Expressing by factors 36. Express 2772 by factors. 37. Express by prime factors. 1 2)2772 2772 = 12 x 11 x 21. 11^2 '91 ^" prime factors, ^-^ 2772 = 2x2x3x11x3x7 '^^ = 22 X 32 X 11 X 7. 2772 = 12 X 11 x21. NoTK. — A number written over a factor shows how many times the factor is used. Express by factors : Express by prime factors : 3a 72; 630; 2808. 42. 120; 220; 5616. 39. 56; 836; 7425. 43. 108; 144; 1445. 4a 48; 840; 1232. 44. 156; 160; 7392. 41. 45; 945; 5929. 45. 135; 288; 3025. Note.— To express by prime factors, it is best to re^ 6. 2x 3x 5 4x6 2 X 12x30 (2 X 3 X 5) X (4 X 6) = 2x(3x4)x(5x6), or 2 X 12 X 30. 51. Divide 2x12x30 by 4x6. 3 5 Jg X j2 X g0 i4x(5 2x3x5 12 -=- 4 = 3 ; 30 -T- 6 = 5 ; the result is 2 X 3 X 6. NoTB. — This method is culled cancellation. See pp. 90, 92. INTEGERS 73 Written work Add : Subtract : 52. 89 X 15, 16 X 15, 13 x 15. 54. 75 x 15 from 89 x 15. 53. 7x12, 8 X 12, 9 X 12. 55. 16 x 17 from 73 x 17. Multiply : Divide : 56. 2 X 3 X 4 by 6; by 8. 5a 12 x 15 x 21 by 3; by 6. 57. 8 X 9 X 7 by 3; by 7. 59. 21 x 35 x 42 by 7; by 3. Expi'ess the quotients by factors : ea (18x9xl6)-i-(2x3x3). 65. (48x96x98)-?- (16x24x49). 61. (16 X 18 X 12)--(4x9x2). 66. (36 x 24 x 90) h- (18x18x12). 62. (72x24x36)H-(18x3x3). 67. (18 x 12 x 30)-5-(5 x 6 x 3). 63. (75xl8x21)-^(35x9x5). 6a (94 x 78 x 65) -- (47 x 13 x 39). 64. (56x49x63)-h(14x7x9). 69. (57 x 91 x 77) -h (19x13x11). Mental work Divide, declaring the results by factors : 7a 85 X 6 X 7, by 17. 7a 17 x 19 x 18, by 34. 71. 95 X 8 X 3, by 24. 79. 23 x 12 x 15, by 69. 72. 72 X 8 X 7, by 56. 80. 26 x 27 x 28, by 63. 73. 90 X 3 X 7, by 54. at 96 x 35 x 17, by 48. 74. 75 X 4 X 8, by 50. 82. 80 x 9 x 11, by 88. 75. 18 x 8 X 4, by 36. 83. 62 x 8 x 63, by 93. 76. 16 X 7 X 8, by 64. 84. 22 x 3 x 33, by 66. 77. 19 X 9 X 4, by 38. 85. 18 x 4 x 62, by 93. Ex.73. Ans. 6x7. The factors of 64 are 18 and 3; 90+18 = 6; 3 + .3 = 1. NoTK. — For practical illustrations of Exs. 70 to 85, see Notes, pp. 88, 89. 74 FACTORING GREATEST COMMON DIVISOR The greatest common divisor of numbers less than 100 may be known from the combinations. See pp. 36 and 40. Findf from the combinationsy the O. C. D. of: 86. 40, 96. 89. 30, 48. 92. 48, 80. 87. 24, 66. 9a 64, 72. 9a 46, 94. 8a 35, 91. 91. 66, 60. 94. 49, 77. Kx. 86. 8. 40 + 8 = 5 ; 96 -^ 8 = 12 ; 5 and 12 are prime to each other. Smaller numbers may be found which have the same G. C. D. I. Hie G. C. D. of two numbers is the O. C. D of the smaller, and of the remainder found by dividing the greater by the smaller. 95. In Exs. 86 to 94, find smaller numbers with same G. C. D. Ex. 86. Ans. 40, 16. Since 06 = 40 + 40 + 16, the G. C. D. of 40 and 96 is the G. C. D. of 40, and 40 + 40 + 16, or of 40 and 16. 96. Making use of the first expedient, solve examples 86 to 94 Ex. 86. Ans. 8. The G. C. D. of 40 and 96 is the G. C. D. of 40 and 16, or 8. n. One of several numbers may be divided by a fa/dor prime to any other, without affecting the G. C. D. 97. In Exs. 86 to 94, find smaller numbers with same G. C. D. Ex. 86. Ans. 8, 96. Since 40 contains 6, which is prime to 96, 5 may be canceled from 40 without affecting the G. C. I). 98 Making use of the second expedient, solve examples 86 to 94. Ex. 86. Ans. 8. The G. C. D. of 40 and 96 is the G. C. D. of 8 and 96, or 8. Find the G. C. D of: 99. 36, 40, 48, 72. 102. 55, 77, 88, 99. 100. 49, 5(S, 70, 77. 103. 45, 75, 90, 60. 101. 60, 84, 96, 72. 104. 42, 56, 70, 77. INTEGERS 76 III. The G. C. D. of tv3o or more numbers is the product of all the common factors that may be used as successive divisors until the quotients are prime to each other. Find the O. C. D of: 105. 288, 432, 720. 109. 945, 1260, 2625. 106. 495, 660, 990. 110. 1680, 4200, 5040. 107. 475, 855, 760. 111. 1875, 3750, 5000. loa 1728, 1296, 1872. 112. 2850, 3800, 4750. Ex. 105. By expedient III. 12 I 288 432 720 Since 12 and 12 are component -^ — -^. ^ ^ factors of the G. C. I)., and the quo- 2 3 5 or 144, is the G. C. D. Note. — In finding common factors by inspection, it is best to try 12, 11, 9, 8, 7, 5, 4, 3, 2, in order. If no common factor is seen by inspection, smaller numbers should be found which have the same G. C. D. ^ee p. 7 4. Find the G. C. D. of: 113. 153, 374. 117. 143, 1765, 2912. 114. 625, 1728. lia 495, 1452, 9317. 115. 1144, 1365. 119. 1152, 1728, 3375. 116. 1177, 2675. 120. 875, 448, 567. Ex. 113. By expedient I. 153)374(2 ^0^ The G. C. D. of 163 and 374 is the 68)153(2 G. C. D. of 1 63 and 08, or the G. C. D. 136 of 68 and 17, or 17. 17 Ex. 113. By expedient IL 153 374 The G. C. D. of 163 and 374 is the 51 G.C. D. of 51 and 374, or the G. C. D. 27 of 17 and 374, or 17. 76 FACTORING LEAST COMMON MULTIPLE Multiples of small numbers are easily found from the combina- tions. See pp. 36, 4^. 1. Hie least common multiple of two or mot'e numbers is the product of all their prime factors, each taken the greatest number of times it is found in any one of them. 121. Write all the multiples of 2, to 36 ; all the multiples of 3, to 36. 122. Make a list of all the multiples of 2, less than 36, that exactly contain 3; of all the multiples of 3, less than 36, that exactly contain 2. 12a What is the least common multiple of 2 and 3 ? Is it exactly contained in each of the common multiples ? Why ? 124. Answer examples 121 and 122, with reference to 4 and 6, instead of 2 and 3. 125. What is the least common multiple of 4 and 6? Is it exactly contained in each of the common multiples ? Why ? 126. Why is 2 X 3 the least common multiple of 2 and 3, while 4 X 6 is not the least common multiple of 4 and 6 ? Ans. 2x3 is the L. C. M. of 2 and 3 because 2 and 3 are prime to each other ; 4 X G is not the L. C. M. of 4 and 6 because 4 and 6 are not prime. Find, from the combinations, the L. C. M. of: 127. 10, 18. 131. 16, 20. 135. 21, 14. 12a 14, 12. 132. 12, 16. 136. 12, 18. 129. 10, 12. 133. 24, 16. 137. 24, 32. 130. 10, 14. 134. 25, 20. 13a 27, 18. 139. Taking the product of prime factors, solve Exs. 127 to 138. Ex. 136. The L. C. M. must contain 12, or 2 x 2 x 3, 12 — Q ^ S \f « *"^ ^^'^ retain these factors. The L. C. M. must contain 18, or 2 x 3 x 3. We already 18 •=^ 2 X 3 y. S l^ave 2 and one of the 3's ; we retain the 8x2 X3x 3=36, L. C. M. other. Then 2 x 2 x 3 x 3 is the L. C. M. INTEGERS 77 The product of the prime factors may be found more easily by the following principles : II. To find the L. C. M of two numbers^ divide one of them by their O. C. D and midtiply the quotient by the other. III. To find the L. C. M. of more than two nu7nber.% find the L. CM of two of them, then of the result and a third, and so on. IV. If one of the numbers exactly contains another, the smaJUer may be neglected. 140. In Exs. 127 to 138, find the L.C.M. by principle II. Ex. 136. 12, 18 As before, the L. C. M. must contain 12 y. S =i S6 L C M 12, or 2 x 2 x 3. Instead of retaining ' ' * ' these factors, we retain 12 itself. The The G.C. D. of 12 and 18 is 6 ; L. C. M. must contain 18, or 2 x 3 x 3. 18 -4- 6 is 3 ; 12 X 3 is 30, L. C. M. In 12, we already have 2 once and 3 once, or 6, the G. C. D. of 12 and 18, and we simply retain the other 3, or 18 -f- G. C. D. NoTK. — By retaining 12, instead of 2 x 2 X 3, we are saved the effort of first separating 12 into its factors, and later of multiplying them together. Find mentally the L. C, M. of: 141. 60, 72. 145. 16, 20. 149. 21, 45. 142. 25, 30. 146. 20, 45. 150. 16, 30. 143. 24, 27. 147. 24, 36. 151. 24, 30. 144. 49, 63. 14a 42, 56. 152. 60, 75. Ex. 141. 00 X 6, or 360. The G. C. D. of 60 and 72 is 12 ; 72 h- 12 = 6 ; 60 X 6 = 360, L. C. M. By these principles, find mentally the L. C. M. of: 153. 2, 3, 4, 5, 6, 7. 155. 10, 12, 15, 6, 5, 4. 154. 5, 8, 9, 6, 12, 10. 156. 18, 8, 9, 36, 24, Vz. Ex. 153. Ans. 420. We neglect 2 and 3 because they are contained in A The L. C. M. of 4 and 6 is 20 ; of 20 and 0, 60 ; of 60 and 7, 420. 78 FACTORING Another method of finding the L. C. M. is to divide by prime divisors that are common to any two of the numbers. 157. Find the L. C. M. of 63, 108, 28, 42. Since 3 is contained in 63 and 108, 3 is used once. Since 3 is contained in 21 and 36, 3 is used a second time. Since 2 is con- tained in 12 and 28, 2 is used once. Since 2 is contained in 6 and 14, 2 is used a seanul time. Since 7 is contained in 7 and 7, 7 is i S 1 1 used once. Since 3 is among the quotients, 3^X2^X7= 756, L. CM. 3 is used a third time. If no two numbers appear to have a common factor, it is neces- sary to find their G. C. D. 15a Find the L.C.M. of 692, 703, 171. s 63 108 28 ^ s 21 36 28 U 2 7 12 28 U 2 7 6 u 7 7 7 3 7 7 37 19 703 171 16 19 171 Th« G. C. D. of 592 and 703 is -7^ Z r 37. The G. C. D. of 19 and 171 is 1^ ^ ^ 19. 37xl9xl6x 9=101,232, L. C. M. 159. Solve Ex. 157 again. Divide by 4; then select such divisors as you please. Why is the answer, 1512, incorrect? Ans. Because the greatest number of times 2 is used in any number is twice^ viz., in 108 and 28. Dividing by 4, took 2 twice out of 108 and 28, but left 2 once in 42, making three times that it appears by this process. Note. — It is ansafe to divide by other than prime numbers. Find the L.C.M. of: 160. 30, 32, 45, 48, 64, 75. 16a 169, 221, 204. 161. 125, 245, 147, 225. 164. 266, 285, 209. 162. 36, 72, 128, 126, 243. 165. 708, 1062, 1475. INTEGERS 79 Casting out 9*8 Addition, subtraction, multiplication, and division may be proved by casting out U's. In addition^ the sum of the excesses of 9*8 in tlie several addends^ slioidd equal the excess ofO^s in the sum. 5078 8 ^Ve add the digits of the first addend, 5, 11, 18, A93A s ^^ ' ^**^" ^^^ digits of the sum, 26, and write 8. ■* ■* This is the excess of the 9's, for 5678 h- 9 gives 8 ^^^ _^ for a remainder. We proceed in the same way 1A£39 1 ^'^** ^^'^ °^^^^ addends : 4, 13, 16, 20 ; 2 : 3, 9, 11, "^^ 18 ; 9 ; (a sum of 9, or any multiple of 9, we count as 0). We proceed in the same way with the excesses, 8, 10 ; 1. We proceed in the same way with the sum, 1, 6, 7, 10, 19 ; 10 ; 1. The sum of the excesses of 9's in the several addends is 1 ; the excess of 9*8 in the sum is I ; the answer is probably correct. In subtraction, the excess of 9*s in the minuend, should equal the excess of 9^s in the subtraJiend and the remainder. '^^^^ ^ Minuend, 7, 13, 21, 24; 6. S997 J^ Subtrahend, 3, 12, 21, 28 ; 10; 1 : remainder, ~J^ 7 ^' ^' ^^' 23; b: their sum, 6. In multiplication, the excess of9*s in the product of the excesses of the factors, should equal the excess in the answer. ^^^ ^ Multiplicand, 9, 16, 23 ; 6 : multiplier, 4, 10, 468 18 ; 9 ; : their product, 0. 463,0^4 ^^«^"<^^' ^' ®' ^2, 14, 18 ; 9 ; 0. In division, the excess of 9'8 in the product of the excesses of the quotient and divisor, plus the excess in the remainder, should equal the excess in the dividend. 488)472,870968'^ Quotient, 9, 15, 23; 5: divisor, 4, 12, 20; 2: their product, 10 ; 1 : the remainder, 4, 12, 19 ; , ^ 10 ; 1 ; their sum, 2. — - Dividend, 4, 11, 13, 21, 28, 29 ; 11 ; 2. iO » 80 FACTORING RELATIONS 166. When 13,825, or 13,820 -f 5, is divided by 2, why is the remainder the same as when its last digit is divided by 2 ? Give the rule for the divisibility of a number by 2. The first part, 13,820, or 1382 tens, is divisible by 2 because ten is divisible by 2. Since the first part is divisible by 2, the divisibility of the number depends upon the last digit. 167. When 13,825, or 13,800 + 25, is divided by 4, why is the remainder the same as when the number denoted by its last two digits is divided by 4? Give the rule. 16a When 13,825, or 13,000 + 825, is divided by 8, why is the remainder the same as when the number denoted by its last three digits is divided by 8 ? Give the rule. 169. What is the remainder when 1 with any number of ciphers is divided by 9 ? 170. What is the remainder when 2, 3, 4, 5, 6, 7, or 8, times 1 with any number of ciphers, is divided by 9 ? 171. When 13,825, or 10,000 + 3000 + 800 -f- 20 + 5, is divided by 9, why is the remainder the same as when the sum of its digits is divided by 9 ? When 10,000 is divided by 9 the remainder is 1 ; when 3000 is divided by 9 the remainder is 3 ; etc. See Ex. 170. 172. If a number divided by 9 gives the same remainder as the sum of its digits divided by 9, what is the rule for the divisi- bility of a number by 9 ? 173. Show that 1 with any odd number of ciphers lacks 1 of being a multiple of 11. 174. Show that 1 with any even number of ciphers exceeds by 1 a multiple of 11. 175. How much does 2, 3, 4, 5, 6, 7, 8, 9, times 1 with any odd number of ciphers, lack of being a multiple of 11 ? 176. How much does 2, 3, 4, 5, 6, 7, 8, 9, times 1 with any even number of ciphers, exceed a multiple of 11 ? INTEGERS 81 177. When 75,316, or (70,000 4-300 +6) + (5000 +10), is divided by 11, why is the remainder tlie same as when the sura of the digits in the odd places minus the sum of the digits in the even places, is divided by 11 ? 70,000+300 + exceeds a multiple of 11 by 7 + 3 + 6 (Ex. 176); 5000+10 lacks 6 + 1 of being a multiple of 11 (Ex. 175); 75,316 exceeds a multiple of llby (7 + 3 + 0)-(5 + l). 17a If a number divided by 11, gives the same remainder as the difference between the sum of its digits in the odd places and the sum of its digits in the even places, divided by 11, what "is the rule for the divisibility of a number by 11 ? 179. A contractor is to build houses 24, 36, 48, and 60 feet long, and 16, 32, 32, and 48 feet wide. What length of clapboard can be used most conveniently for the sides? for the ends? State the relations. Relation : number of feet in length of clapboard for the sides = the largest number that is exactly contained in 24, 36, 48, and 00, or their G.C.D. . . . 180. A lady wishes to buy a piece of cloth which she can cut without waste into an exact number of pieces either 3, 4, or 5 yards long, as she may decide later. What is the smallest num- ber of yards the piece can contain ? 181. A real estate agent wishes to divide 3 pieces of land 325, 675, and 950 feet wide, into town lots of equal width. What is the largest possible width for each lot ? 182. Ropes 48, 52, and 56 feet long are to be cut into the long- est possible equal lengths. How long must each piece be ? 183. A, B, and C start together around a circular track. A goes once around in 6 minutes ; B, in 8 minutes ; C, in 9 minutes. AVhat is the least number of minutes before they will be together again at the starting point ? State terms and relations. (riven terms : number of minutes passed when A is at the starting point ill 6, 12, 18, 24, 30, 36, 42, 48, 54, 00, 66, 72, ... ; B, 8, 16, 24, 32, 40, 48, 66, 64, 72, . . . ; C, 9, 18, 27, 36, 45, 54, 0.3, 72. . . . Relation : number of minutes before they are again together = least num- ber that will exactly contain 6, 8, and 9, or their L. C. M. ▲ MBR. AHITH. — 6 82 FACTORING 184. After how many minutes will A and B first be together at the starting point ? A and C ? B and C ? las. D, E, and F start together around* a circular track 5280 feet in length. D rides 1760 feet per minute ; E, 1320 ; and F, 1056. How many times must D ride around the track before they are all together again at the starting point? State the relations. Relations: number of minutes D goes once around = 5280 -=-1760; number minutes B = 5280 ^ 1320 ; number minutes C = 5280 h- 1056. The minute's when they are first together = the L. C. M. of the minutes each makes the circuit ; number times A goes around = L. C. M. -;- number min- utes A makes the circuit. 186. By counting eggs, 4, 6, or 10 at a time, a farmer had none left over in each case. "What is Che least number he could have had ? State the relations. 187. By counting eggs 4, 6, or 10 at a time, a farmer had 3 left over in each case. What is the least number he could have had ? State the relation. 18a By counting eggs 4, 6, or 10 at a time, a farmer had 3 eggs left over in each case; counting 11 at a time, he had none left. Wliat is the least number he could have had ? State the relations. Relations : ppssible numbers = 3 -f common multiples of 4, 6, and 10 ; the least number = the least of these results divisible by 11. SohUion : the L. C. M. of 4, 6, and 10 is 60 ; 63, 123, 183, 243, 303, 363, . . . are numbers representing in order 3 + common multiples of 4, 6, and 10 ; 303 is the least of these which contains 11. COMMON FRACTIONS FIRST CONCEPTION AN EXPRESSION OF DIVISION Division may be expressed by writing the dividend above, and the divisor below, a horizontal line. Such an expression is a common froA'tion; the dividend is the nu- merator; the divisor, the denomi- nator. The numerator, or the denomina- tor, or both, may contain fractions ; such an expression is a complex fraction. We sometimes speak of a frac- tion of a fraction; a compound fraction. An integer plus a fraction is a mixed number. The plus sign is usually omitted. State the terms, and the meaning of the fractions : } ; 5« i Illdstrations ^, common fraction, 5 4y numerator. 5y denominator. Read, 4-^5. S '^ 6 8 complex "s fractions. 9 a compound fraction. 5 -f- ;, or 5 t a mix4>d 5' number. I ; 3 is the numerator ; 4, the denominator ; it means 3 -r- 4. I is the numerator ; |, the de- nominator ; it means f -^ }. NoTR. —The pupil should read the first part of p. 45, the whole of p. 49, and the explanation of Ex. 55 ou p. 51. 88 84 SECOND CONCEPTION A unit may be divided into two or more equal i)arts, and one or more of these parts may be considered. The number showing into how many parts the unit is divided, is written below a hor- izontal line, and is the denomi- nator. The number showing how many parts are considered, is written above the line, and is the numerator. The whole expression is a CO m mon frart ion. According to this conception, is I a fraction? No. It is called an improper fraction, i.e., not properly a fraction. According to this conception, IS 1^ a fraction? No. t called a complex fraction. EQUAL PARTS OF A UNIT Illustrations It is T — r AB is divided into 8 equal parts ; AC contains 5 of them ; AC = 6 eighths of AB ; expressed, AC = \ of AB. g -, common fraction. o 8t denominator. 5, numerator. It means tliat a unit is divided into 8 equal parts, and that 5 of tliese parts are considered. \, read, one half; f, read, three quarters, or three fourths. It is impossible to divide a unit into 5 equal parts and then con- sider 8 of them. It is impossible to divide a unit into I equal parts. Define by ea^h conception : 1. A common fraction. 2. The numerator. 3. The denominator. 4. A complex fraction. 5. A compound fraction. 6. A mixed number. Note. — Sometimes deductions are made from the first conception ; and some- times, from the second. For illustTations of the first, see pp. 87 and llO; of the second, pp. lOl^ 102. COMMON FRACTIONS 85 CHANGE OF FORM To lower terms Dividing both numerator and denominator by the same num- ber does not change the value of afrojction. By this principle, we reduce fractions to their simplest forms. Which fraction is the more readily comprehended, f f or J ? Why? 7. Reduce ^fj to lowest terms. 216 18 3 a Reduce f^fj to lowest B I ' I ' I ' I AB = i, orf, of AC; ••.* = ?. I is obtained from ^ by dividing both terms by 2. I, because the terms are smaller. By inspection, we find common factors. Dividing both terms by 12, m = H ; by 6, H = f . r-ry in When no ( 5ob7 19 hy inspprtior 6789 28 here it is 293 common factor is found , we find the G. C. D. ; Reduce to lowest terms : 9- li; A- 1* Mi «• 17. a «• laA; M- i*-+li«- la H «• 11- H; H- "• Mi W 19. a H- "• tti ii 16- Hi H- 20. a H Reduce to lotoest terms : ^ ttHi /A^- 24. im; m 27. tmi; «M. »• AV\,i H«- aa- Wi! Hi- 2a TTtt» AWr* ^- Ht*» llli* *^ TtTTJ Ht' 29. AWriHi Itt- 86 CHANGE OF FORM To higher terms Multiplying both numerator and denominator by the same number does not change the value of a fraction. • By this principle, we prepare I ' I ' I "n^ AB = i, or I, of AC ; I is obtained from J by multiply- fractions for addition and sub- ing both terms by 2. traction. aa Change } to 12th8. S 12 Change: 31. jtol2th8. 32. |to40ths. 3a ^to35th8. 34. {to48ths. To make the denominator 12, we must multiply 3 by 4 ; multiplying both terms by 4, | = /j. 35. ■jS^to24th8. 36. T^to39ths. 37. -^toTTths. 3a '^to28ths. 39. |^tol35l8ts. 40. i^to6825th8. 41. ^to2856ths. 42. ^to2277ths. 43. Reduce J, f , f to equivalent fractions having L. C. D. g Q 2Q The L. C. D. is 12 ; 12 - 3 = 4 ; — -, — , — • we multiply both terms of f by 4 ; 12 12 12 12 ^ 4 = 3 ; we multiply both terms of J by 3 ; 12 -=- 6 = 2 ; we multiply both terms of \ by 2. Reduce to equivalent fractions having L. C. D : 44. I, I, |. 49. I, H, A- 54. f, if, IH 55- T> TT? TTS"' 56- l> «. Mi 57- 1, ^, m 5®- 'Si wky T9Z *5. i, \, \. so. i, %, if. 46. \, 1, A- 51. f , i, A- *'• f. A. A- 52. i, A. H- 4a \, ^, A- S3, f A, A- €OMMON FRACTIONS 87 To whole or mixed numbers A fraction is an ex})ression of divi- sion. It means that the numerator is to be divided by tlie denominator. By this principle, we change to a form that can be more readily com- prehended. 59. Reduce fj to a whole or mixed number. 13 3'-' IS Or, since there are 3 thirds in 1, in 8 thirds there are as many Vs, as 3 is contained times in 8, or 2|. f^=45~13. Performing the operation, we obtain o/y. Reduce to a wliole or mixed number : ea V; i^. 63. tl; fj. ^ «; M- 6*- tt; ff- 62. H; H- 65. ^; ff. 66. i-til. 67. if-fp. 6a ^^^. 69. i^i. 70. IfJfL. 71. ^ifF. Whole or mixed numbers to common fractions A mixed number is an integer plus a fraction. If the integer is reduced to an equivalent fraction having the denominator of the fraction, the two paHs may be united. By this principle, we prepare frac- tions for multiplication and division. 2f = 2 + f = ? + ! Or, since there are 3 thirds in 1, in 2 there are 2 times 3 thirds, or 6 thirds; with 2 thirds, 8 thirds. 72. Reduce 4j\ to an improper fraction. Reduce to an improper fraction : 7a 5J; 4f 75. 5H; 6W. 74. 7i;8i. 76. 3H;^A- 4A = HH-A = lf. 77. 35t|; U^' 7a 87H; 86H. 88 ADDITION ADDITION Before fractions can he addedj they must be reduced to equivalent fractions having a com- ~ = A mon denominator, . Tfie least common denominator should alusxys 4~lg be found. 79. Find the sum of } and |. UEOUN , I = I'j ; I = t'i ; the sum of 5 ^ the 12th8 is 17 twelfths, or 1 5 g unit and 5 twelfths ; we write ^ ^2 io fractions' column, and carry 1 to units* column. If NoTB. — When the fractions are ready for addition, the work appears as at the left; when finished, as at the right. In practice, it should not be begun in one place and finished in another. 80. Find the sum of 48 J|, 32f^, 76f4. ^rs '^* ?^^ 160 ^^ = ^tH ; H = m ; H = m- The sum of the tf^^j iw ^^^^ 286ths is 619 286ths, or 2 units and 49 285ths ; we 76— tiS write 49 285ths in fractions' column, and carry 2 to ^^ units' column. See p. 20. OOMPUTID 9 s 8 S u 9 4 n It ..^U9 619 Note. — It is convenient to express the L. C. D. by its factors. Thus : the L. C. D. is 1>5 X 3 ; <)5 is contained in this 3 times ; 57, or 19 X 3, is contained 5 times ; 19 is contained 5 x 3, or 15 times. See p. 73. Find the value of: ^ l+i; A + i + l- 84. 98i + 25f + 5^. 82. i-f f; i^ir + f + l. 85. 65f + 91f + 8f. 83. -1 + ^; H + i + f 86. 27J + 18| + 7f COMMON FRACTIONS 89 SUBTRACTION Before fractions can be subtrax^ted, they must be reduced to equivalent fractions having a com- mon denominator. 5^_9 4 IB g _ 8 The least common denominator should always s~J2 be found. 87. From } subtract }. OOMPLBTCD 9 8 Ig 1 IS » g leaves ^ ; we write ^ in frac- ^ tions' column. 7i Note. — When the fractions are ready for subtraction, the work appears as at the left; when finished, as at the right. In practice, it should not be begun in one place and finished in another. 8a From 300^ subtract 200j^. ^nn^ '""a A = t!t ; 1^1 = 4»(jV ; 21 204ths from 8 204th8 we *^ Hi cannot take ; we add 204 204ths to 8 204ths making pnnL ♦, 212 204ths, and 1 to units making 1 unit; 21 '-'H ^^ ** ^ 204th8 from 212 204ths leaves 191 204tlis ; we writ6 qqIQI 191 1^^ 204tfts in fractions' column ; 201 units from 300 iou W4 units leaves 99 units. See p. SO. NoTK. — It is convenient to express the L.C.D. by its factors. Thus: the L. C. D. is .'il X 4 ; 61 is contained fn this 4 times ; 68, or 17 X 4, is contained 3 times. iS«e p. 73. Find the value of: » i- I 91. J- } i-i;H-f 92- ^-n-' 68,V-27ii. i-i< H-i 93- H-n-' 75A-29H- |-»i ji-f 94. 4}-2J; 67^-30?!. 90 MULTIPLICATION MULTIPLICATION Common method Multiply the numerators for a new numerator and the denominators for a new denominator, canceling wJien pos- sible. Mixed Jiumbers should be reduced to improper froAUions, c } of AB = AC ; §of }of AB = 2 AC = AD J of AB = AD. . f of }of AB = i()f AB, or § of I = i. 95. Multiply J by 4. 96. Multiply f} by 5J by 2\. Find the value of: 97. Jx6; 8xt; 9xi. 9a }Xt\; 7x|; 9Xt3j. 99. Jxl6; 18 X I; 72 x J. 100. 84 XtV; |x96; f x 48. Divide both 4 and 8 by 4 ; i.e., cancel 4 from both 4 and 8. See p. 72. 6J = Y ; ^=\^ cancel 7 from 21 and 49 ; 7 from 7 and 7 ; 4 from 36 and 4 ; 3 from 3 and 3. 102. fof^fe; }of J; y^oflG. 103. iof25; f of 24; |of 16. 104. I of 18; ^x91; | x 65. 105. «of21;y9yOf 44; ^7^ of 65. 101. 96xy\; |Jx84; ^X64. loa |of20; ^of 91; t\ of 55. Find the value of: 107. H X if X t^ X A- lOa 5^ X 6} X 6} X 8|. 109. if X H X H X H- no. f X f J X 1 Jj X II X «. Ill- Ht X tt X H X « X H- 112. tt X ^V X tf X 12f 113. 15| X 4f X 8rV X -h' 114. 55i X If X Hi X if 115. f of If of A of H of if. 116. l^x5Jx Axff xl^. COMMON FRACTIONS 91 9 S_ 98th» 8 ■ i~6«'A« Dividing the numerators and the denominators gives tlie same result. 9-4-3 ^3 8 + 4 2 DIVISION European method Divide the numerators for a new nw- merutor and the denominators for a new denominatorf changing to equivalent frac- tions with their least common denomi- nator, if necessary. Mixed numbers should be reduced to improper fractions. 117. Divide } by |^ mentally. When J is reduced to 12th8, tlie numerator be- ^ _^ 5 __ 9_^ comes 9 ; when ^ is reduced to 12ths, the numer- j^ ' 6 10 a^or becomes 10 ; 9 h- 10 = ^q. The quotient of the denominators, 12 -f- 12, is 1. NoTB. — When two fractions have the same denominator, since the quotient of the denominators is always 1, it is necessary to think of the numerators only. lia Divide 9J by lOf mentally. 9'-^ tO- = 75 When reduced to Sths, the numerator "« ■ '"4 86' becomes 73 ; of 10|, 86 ; the answer is J J. Find the value mentatty: 119- i + f 130. 16 -Hf 141. 5*^3}. 12a f-t-J. 131. 18 -f 142. 2i^li. 121. i + }. 132. 27 -hf. 143. ^^^• 122. f-!-f. 133. 63-*-}. 144. n-^H- 123. }-Hj. 134. 20-s-f 145. 7} + 8}. la*- i + l- 135. 25 -J- f 146. 6| + 8}. 125. t + f 136. 49 -hf 147. 9f -h 6f 126. J + f 137. 84-*-}. 14a ^^^' 127. 1 + }. 13a 48 -hf 149. 8|^6i. 12a i + f 139. 54 H- }. 15a 91-*- 3}. xaft | + |. 14a 77 + }. 151. 7^^2i, 92 DIVISION Common method Invert the divisor and proceed Inverting the divisor and proceed- as in multiplication. 152. Divide } by 4. 8 4 32 153. Divide 4 by J. ing as in multiplication gives the same result. } X f = |. 15*. Divide J by }. 8 8 64 155. Divide 16J by 3}. 10 S Divide, inverting the divisor: 156. if by 18. 159. 36byT>p 157. Hbyie. leo. 25 by f 15a iV^y21. 161. 48 by i J. Divide, inverting the divisor : 165. 14tf by41. 17a 113 by 13^. 166. 17|4byl3. 167. 18}fby33. 16a 16||by79. 169. 19ifby56. 171. 594 by 18^. 172. 190 by 26Jj. 173. 850 by 38^. 174. 221 by 583V 1€2. ttbyf 163. A by H- 16*. HbyA- 175. 42fby36yV 176. 57f by 12^ 177. If2| by ^, 17a 2^byfH 179. 2^ by ^. COMMON FRACTIONS 93 SPECIAL METHODS laa Multiply 475 by 361. 475 gQ* gjjo We first multiply 476 by J ; 2 times 476 = I 950 ; 950 ^ 3 = 3l«j. It is well to write 950 316- to the right as we multiply, and to divide by 3 aosn without writing the 3. "^^^^ We then multiply 476 by 36 in the usual way. 1425 17416'- laL Multiply 638f by 27i. 638l ^*'^ J7| SI 4^51 U We first multiply J by | ; then, 638 by § ; 4466 1276 ^J\ then, } by 27 ; then, 638 by 27. 17672j-^ S 182. Divide 65J by 4. / \ j^? 6 tens -f- 4 = 1 ten and 1 ten remaining ; we I £. write 1 in tens' column. IS— 1 ten and 5 units =16 units ; 16 units -f- '' 4 = 3 units and 3 units remaining, we write 3 in units' column. 3 unita and 2 thirds = Y ; V •<- ^ = H- ^^^ P- ^i- 183. Divide 275 by 3J. _f \ ^j- Multiplying both dividend and divisor by the o-)4io 'game number does not affect the quotient. 11)825 ^^ multiply both dividend and divisor by ^j' something that will make the divisor an in- teger. Here, we multiply both terms by 3. 94 i m THE FOUR OPEllATIONS Add: THE FOUR OPERATIONS 184. 185. 186. 187. 23A 14A 12A m 7A 13A ISA 12H Subtract: 18a 189. 190. 191. 263Hi 7345», 673H 325A 266H 192. 193. 194. 195. 145}J 122H 164^1 142Hi 585H 425^ 424tt 222H Multiply: 196. 197. 19a 199. 324J 48 425 75f 625f 28| 738| 2311 200. 201. 202. 2oa 687f 27i 584J 48| 938f 69^ 865| 36f Divide : 204. 205. 206. 207. 4)723^ 3)9345i 4)6785| 7)92861 2oa 209. 210. 211. 4i}935 62^)3863^ 4281)3285 2661)3586 COMMON Fractions 95 COMPLEX FRACTIONS To simplify complex fractions, it is often best to multiply both terms by the least common denominator of the several fractions. 212. Simplify 1^. Tkis = l±^, = f 2ia Simplify ^. 214. Simplify i^. I 5 g 215. Simplify j*~*. ThaA = S. 87 216. Simplify 1^1^-1 2%w = f xjx|xjx «x? = l. The L. C. D. of the several frac- tions is 12; 12xi=6; 12x^ = 4; 12 X f = 9 ; 12 X i = 2. The numerator of the complex frac- tion is f ; the L. C. D. of f , J, and | is 15. Before applying this principle, compound terms must be simplified. |x| = i. When the L. CD. is lai^e, it is often better to perform the indicated operations. If a fraction is used an odd num- ber of times as a divisor, we invert ; if an even number of times, we do not invert. NoTB. — In Ex. 214, the L.C.D. of the simple fractions is 12. If we had mnl- tiplied both i and ] by 12, the numerator would have been multiplied by 12 x 12, or 144. See p. 72. 96 COMPLEX FRACTIONS It should not be forgotten that complex fractions may always be solved by performing the indicated operations. 217. Solve examples 212, 213, 216 by performing the indicated operations. 2ia Solve again example 216, as on page 95, and explain why J is inverted ; J, not inverted ; J, inverted ; f , not inverted. I is used once as a divisor (it is in the denominator of tlie complex frac- tion), invert; |, twice (preceded by ♦-,-,' and in the denominator of the complex fraction), do not invert; J, once (the second complex fraction is preceded by •-*-'), invert; ^, tteice (tiie whole complex fraction is preceded by '-{-,* and ^ is in the denominator), do nut invert. Using expedients of page 95, simplify : 219. 1. V i * 22a V i. 223. ± x3|. 224. ^ xlf. h- ■A *- -i 99fl n -1 3i -2| 229. 54 x8i ^ ^tt 222 li; 5i. 225. gxf 230. -r# — ^X- 231. 232. 6|of3;^-2f COMMON FRACTIONS 97 MISCELLANEOUS 233. Analyze aud read | by the tirst conception ; by the second. 234. Give the meaning of | by the first conception; by the • coikI. 235. Why is J not a fraction by the second conception ? Why is it a fraction by the first conception ? 236. Reduce -j^y to lowest terms ; state the principle, and the object of the reduction. 237. Reduce J|4 to lowest terms, and explain how we proceed when no common factors can be found by inspection. 23a Change f to 24ths ; state the principle. Why do we re- duce fractions to higher terms ? 239. Change |J to a whole or a mixed number ; explain. What is the object of this reduction ? 240. Reduce 6| to an improper fraction ; explain in two ways. 241. Why are whole and mixed numbers reduced to fractions ? 242. Add and explain : 22 j, 16J ; subtract and explain ; multi- ply and explain ; divide and explain ; find the product of the sum and the difference. 243. Multiply 632| by 24} ; (a) by the common method ; (6) by the special method. Which do you prefer ? 244. Divide } by f ; (a) by the European method ; (b) by the common method. Which do you prefer for mental work ? 245. Divide 157J by 6 ; (a) by the common method ; (6) by the special method. Which do you prefer ? 246. Simplify t-Hl ; (a) by performing the indicated opera- tions ; (6) by the expedient on page 95. Which do you prefer ? 247. Simplify | — v| -h ^^ ; (a) by performing the indicated oi)erations ; (b) by the expedient. Which do you prefer ? AMBS. ARITH. — 7 98 ANALYSIS ANALYSIS The pupil will derive great benefit from writing out the rela- tions. State the relations and analyze: 24a If 1 yard of cloth costs 12^, how much will 5^ yards cost? Belation : cost 5J yards = 6^ times cost 1 yard. Analysis: since 1 yard costs 12^, 5^ yanis will cost 6} times 12^, or mf. Note. — Some prefer two relation* : cost 5 yards = 6 times cost 1 yard ; cost I yard = icost 1 yard. Analysis : the cost of 5 yards is 5 times 12 cents, or 60 cents ; the cost of I yard is I of 12 cents, or 6 cents ; the cost of 5k yards is the sum, or 66 cents. 249. If I of a yard of cloth costs 12^, how much will 1 yard cost? Belation : cost 1 yard = j cost } yard. Analysis: since | of a yard costs 12/», 1 yard will cost | of 12^, or 16^. Note. — Some prefer two relations: cost of i of a yard = i cost of J yard; cost of I = 4 times cost of J. Analysis: since i of a yard costs 12 cents, i of a yard will cost k of 12 cents, or 4 cents ; |, or 1 yard, will cost 4 times 4 cents, or 16 cents, 250. 15 is f of what number ? Belation : the required number is J of the given number. Analysis : since J of a number is 16, the number is f of 15, or 20. Note. — Some prefer two relations: k of the number = J of } the number; J, or the number, = 4 X i the number. Analysis : since J of a number is 15, i of the number is i of 15, or 5 ; }, or the number, is 4 times 5, or 20. 251. If 1 bushel of potatoes costs 36 ^, how much will 2f bush- els cost ? 252. If 2^ bushels of apples cost 90 ^, how much will 1^ bushels cost? 253. 20 is f of what number? 30 is f of what number? 45 is f of what number ? 254. A watch was sold for $ 12. The selling price was f of the cost. What was the cost ? COMMON FRACTIONS ;;1J^ ; 255. A man lost ^\ of liis money and afterwards found | of what he had lost. What part of the original amount had he then ? 256. A man bequeathed ^ of his estate to his wife, J of the remainder to his daughter, and the part remaining to his son. What part of the estate did the son receive ? 257. In a certain school, f of the scholars belong to the fourth class, J to the third, | to the second, and the remainder to the first class. What part of the school belongs to the first class ? 25a Of a farm of 160 acres, f is used for grazing, f for corn, ^ for wheat, and the rest for oats. How many acres of oats are there ? 259. If IJ yards of cloth are required for each coat, how many yards will be required for 17 coats ? 260. How many coats may be made from 33| yards of cloth, if 1 J yards are required for each coat ? 261. Of a farm of 160 acres, 48 acres are in wheat, 36 acres in oats ; the remainder is pasture. What part of the whole is used for pasture ? 262. If my whole farm is planted with corn, wheat, and oats, what part of the whole is in corn, if I have twice as many acres of corn as of wheat, and three times as many acres of oats as of corn ? 263. A grocer sells 18 bunches of radishes at 4|^ a bunch, 12 quarts of peas at 2}^ a quart, and takes his pay in eggs at 13^^ a dozen. How many dozen eggs does he receive ? 264. Henry is 18 years old, or } as old as Albert. How old is Albert? 265. A farmer sold a quantity of rye for $ 24, which was only J of its value. How much did he lose ? 266. If j of a ship is worth $ 12,000, how much is i of it worth ? 267. Jane spent 30 cents, or jf of her money, for a book ; with the remainder she bought apples at 2^ apiece. How many apples did she buy ? IOC ANALYSIS 26a If 1 yard of cloth costs $ 5J, how much will 12J yards cost? 269. If 4 yards of cloth cost 99^, how much will } of a yard cost ? 27a A lady gave \ of all her money for a dress, and \ of it for a, shawl. What part remained ? 271. A farmer, having lost 24 sheep, had only J of his floik remaining. How many sheep had he at first ? 27X If 6 apples cost 7^, how many apples can be bought for 14^? 273. B owned | of a ship, and sold } of his share. What part of the whole ship did he still own ? 274. ) of 6 is } of what number ? J of 10 is j of what number ? 275. If 12 apples cost 2^ ^, how many apples can be bought for 26^? 276w A coat cost $ 20, and f of the cost of the coat is ^ of the price of the suit. What is the price of the suit ? 277. A farmer has J of his cattle in one field, J in a second, and the remainder, or 15 head of cattle, in a third. How many cattle are in the herd ? 27a If I of a yard of silk costs } of a dollar, how much will I of a yard cost ? 279. A is f as old as B ; B is f as old as C ; C is 60 years old. How old is A ? 280. If the difference between } and j of my age is 5 years, how old am I ? 2aL If 7 men can do a piece of work in 4^ days, how long will it take 6 men to do the same work ? 282. If I give 18| bushels of potatoes at 60 cents a bushel for cloth at 22J cents a yard, how many yards of cloth do I receive ? 283. If a man can do a piece of work in 9 days by working 7J hours i^er day, iu how many days of 8J^ hours each, can he do the same work? COMMON FRACTIONS 101 PROOFS I. Multiplying the numerator multiplies a fraction. Multiplying the numerator multiplies the number of equal parts that are Uiken, without affecting the size of the parts, and thus multiplies the frac- tion. I I. Multiplying the denominator divides a fraction. Multiplying the denominator multiplies the number of equal parts into which the unit is divided, thereby dividing the size of each part, without affecting the number of parts taken, and thus divides the fraction. III. Dividing the numerator divides a fro/ction. Dividing the numerator divides the number of equal parts that are taken, without affecting the size of the parts, and thus divides the fraction. IV. Dividing the denominator multiplies a fraction. Dividing the denominator divides the number of equal parts into which the unit is divided, thereby multiplying the size of each part, without affect- ing the number of parts taken, and thus multiplies the fraction. V. Multiplying both numerator and denominator by the same number does not change the value of a fraction. Since multiplying the numerator multiplies the fraction, and multiplying the denominator divides the fraction, multiplying both terms by the same number, first multiplies and then divides the fraction by the same number, and does not, therefore, change the value of the fraction. VI. Dividing both numerator and denominator by the same iium- ber does not change the value of a fraction. Since dividing the numerator divides the fraction, and dividing the denomi- nator multiplies the fraction, dividing both terms by the same number, first divides and then multiplies the fraction by the same number, and does not, therefore, change the value of the fraction. Note. —The pupil should distinguish carefully the difference between proof $ and illustrations. See pp. 86, 86, 90, 91. 102 PROOFS VII. To multiply fractionSf multiply the numerators for a new numerator and the denominators for a new denominator. To prove that - x ^ = ^JSA, *^ S 5 SxS ? X 4 - ^_ ^_j. {MuiUplying th» numtra 3 3 * tor tnuttiplie»tht fraction.) Since 4 is 6 x ^ inuliiplying by 4 is multiplying by a number 5 times too large ; tlie product in 6 times too large, and must be divided by 5. 2x4 ^ 5 _ 2x4 {MHltiptying tks denomi- 3 3x6 nator dttidet tke/raettom.) . 2 4^2x4 "3 6 3x6* VI IT. To divide frcuiionSf divide the numerators for a new numerator and the denominators for a new denominator. Toprovethat *+? = ^j!l?. ^ 9 3 9+3 8 ft _ 8 •t-2 (Dinidtng Vu numerator g"*" ~9 divid0$ tk4 fraction.) Since 2 is 3 times f, dividing by 2 is dividing by a number 3 times too large ; the quotient is 3 times too small, and must be multiplied by 3. 8 -i-2 J. 3 _ 8 -s-2 (Diridinff the denominn- 9 9 T- 3 tor muUiplies ths fraction.) . 8 ^ 2_ 8-^-2 IX. To divide fractionSy invert the divisor and proceed as in multiplication, Toprovethat f + ? = ^x^- 9 3 9 2 8 , o _ 8 (Multiplying the detiomi- Q "^ ^ ~Q 9* naior divides Ote fraction.) Since 2 is 3 times |, dividing by 2 is dividing by a number 3 times too large ; the quotient is 3 times too small, and must be multiplied by 3. 8 w *> _ 8x3 (Multiplying the numera- 9x2 "9x2* tor multiplies Vie fraction.) , 8^2_8^3 •9 3-92 DECIMALS TERMS AND RELATIONS A common fraction may have for its denominator 10, 100, 1000, a decimal fraction. A decimal fraction may be re- solved into a series of decimal frac- tions whose denominators are 10, 100, 1000, .... and whose numera- tors are less than ten. Since .... 10 thousandths make 1 hundredth, 10 hundredths make 1 tenth, . this series of fractions may be expressed by the devices of the Arabic notation. See p. 10. By the Arabic notation, the names of the orders are expressed by rela- tive position. Thus, 378 means 3 hundreds 7 tens 8 units. If 378 is to mean 3 tentfis 7 hundredtJis 8 thousandthsy an additional device is necessary. This device consists in placing before tenths a period, called the decimal point. A decimal fraction whose denomi- nator is expressed by a decimal point, is a decimal. 103 Illlstrations ^,/\y, a decimal fraction. Since 378 = 300-1-70-1-8, jhh — T^^yi iViJ — !*<»» Write T^5, using the de- vices of the Arabic notation. S7S _S_,_7_ S 1000 10 100 1000 S tenths 7 hundredths 8 thou- sandths, decimal 378, .378 .378, a decimal. 104 TERMS AND RELATIONS Write the steps to show : 1. ThatVV(F = .37 3. That T^HHhr = -0379 2. That tI J^=.a37 4. That ,Tft^ = -0037 Ex. 2. 37 = 000 + 30 + 7 ; jUn = A%% + dh + lo'ou = A + lU + TO^ffj = tenMa 3 hundredths 7 thousandths = decimal 037 = .037. Write the steps to show : 5. That .93 = T^. 7. That .0973 = ,^Jt^. 6. That .093 = T^^f^. a That .0097 = y^ftVr. Ex. 6. .003 = tetUhs hundredths 3 thousandths = i^« + ^U + r/sv = jlh + tAb = r!l5- 9. State the numerator of each decimal in examples 1 to 8, and observe that it is the decimal regarded as an integer. Ans. Ex. 4. In .00:37, or xo'oVo* ^^^ numerator is 37, or it is the decimal .0037 regarded as an integer. 10. State the denominator of each decimal in examples 1 to 8, and observe that it may be ascertained by numerating from the decimal point to the right-hand figure. Ans. Ex. 4. In .0037, or roVoof ^^® denominator is 10,000. Numerating .00^37 ; tenths, hundredths, thousandths, ten-thousandths. 11. Show that the decimal orders are: tenths, hundredths, thousandths, ten-thousandths, hundred-thousandths, millionths,... Ans. Y^ = ^ ; yjj = jl^jf ; y^j,^ = ji^^gn ; ti^H^ = Tiri^uff J • • • 12. Name the decimal orders to quadrillionths ; give the table of decimal orders to quadrillionths. See p. 11. 13. Show that the decimal periods are : thousandths, millionths, billionths, trillionths, quadrillionths, .... Seep. 11. Ans, y^rtnT ~ TooWW » lOOoOOO — 1000000000 i • • • • DECIMALS 105 NUMERATION Read a decimal as a common froL'tion; first its numeratOTy then its denominator in the ordinal form. Read the numerator^ the denominator, the fraction : 14. .5; .06; .007 17. .0002006; .00000213 15. .16; .027; .0039 la .003080654 16. .368; .0407; .00362 19. .0123005678937286 Ex. 18. 'Hie numerator is 3,080,654 ; the denomiuator, 1 billion ; the frac- tion, 3,080,654 billion ths. NoTK. — In Ex. 19, we may numerate by periods; thousandths, millionths, billlonths, trillionths, quadrillionths, ten-quadrillionths. A mixed decimal is made up of an integer and a decimal. • Analyze and read : 20. 175.68; 400.006; .406 23. 23.0708; 35682.3 21. 3000.003; 20.586; 1.7 24. 7583.64; 700.0075 22. 1.0003; 62.832546 25. 328.00062; 10.0001 Ex. 20. 175 is the integer ; .68, the decimal ; read, 175 and 68 hundredths. NoTK. — In reading mixed decimals, the word "and "should be used at the decimal point, but nowhere else. See p. 9, Note. Without this understanding. /our hundred and six thoitsandths might be written either 400.006, or .406. A complex decimal is made up of a decimal and a common fraction. A common fraction can not occupy a decimal order, but is of the same denomination as the order which it follows. Read: 26. .061; -^H 2a .OOi; .OJ; .^ 27. .0008i; .OOOJ 29. 20.086J; 5.0123| Ex. 26. Ans. 6^ hundredths; 1} Ex. 28. Am. .| is an improper thousandths. expression. 106 WUIATION ROTATION Write the numercUor as in common fractions, and pface the deci- mal point so as to make the name of tfie last order the name of tlie denominator. aa Write 638 millionths. ,000638 We write the numerator as in common frac- tious, 638 ; looking at 8 we thinlc millionths because this is tlie name of the denominator ; at 3, hundred-thousandths ; at 0, ten-thousandths; supplying (it now appears 0038), we tiiinlc tlitm- sandllis; supplying; 0(00038), we think hundredths; supplying 0(000038), we think tenths ; writing the decimal point, we have .000038. Write decimally : 31. Five hundred millionths. 32. Seventeen, and 7 tenths. 3a (>38 hundred-millionths. 34. Twenty-three trillionths. 35. 85 thousand 2 billionths. 36. 7007 ten-thousandths. 37. 3G0 hundred-thousandths. 3a Six hundred four thousandths. 39. 9431, and 906532 ten-milliontli 4a Two thirds ten-thousandths. 4L 2 billion, and two billionths. 42. 83 thousand, and 5 thousandtlis. 4a 99 thousand 999 ten-millionths. 44. 1 million 6 thousand 4 billionths. Read: 45. 75.0G432 4e .000000} 47. 35.02767 4a 999.9999 49. 9.0999999 50. 200.0016 51. 1400.014567 52. 62.00634010 53. .09022635 54. 304.00672 55. .98637208 56. .83000008 57. 9200.0929 5a 9000.0009372 DECIMALS 107 ADDITION Write units of the same order in the same column. 59. Add 36.03, 7.864 36.03 7.864. ^^^ explanation, see p. SO. The sum of the 43.894 tboiuiandths is 4 thousandths : etc. SUBTRACTION Write units of the same order in the same column. 60. From 38.03 subtract 25.128 38.03 S5.128 ^^^ explanation, see p. SO. 8 thousandths from '. thousandths we cannot take : etc. 1^.90£ MULTIPLICATION Multiply as in integers, and point off as many decimal places in the product as there are decimal places in both multiplicand and multiplier. 61. Multiply .473 by .23, and explain. .473 To multiply fractions, we multiply the numera- 09 tore for a new numerator, and the denominators ! — for a new denominator. See p. 90. 1419 The numeratore are 47.3 and 2.3 ; we multiply giQ them as in integere. The denominatore are KXK) and 100 ; their product is 100,000 ; we point off .10879 3 + 2, or 5 decimal places. 62. Multiply 378.6954 by 1000. ^ Moving the decimal point one place to the ripht 1000 multiplies by 10 ; two places, by 100 ; etc. See 378695.4 P' 40, Ex. SI. 108 DIVISION DIVISION Divide as in integers, and point off as 7naiiy decimal places in the quotient as those in tJie dividend exceed those in the divisor. 63L Divide .0414 by .23 and explain. £S).04J4(.1S To divide fractions, we divide the numera- •^ tors for a new numerator and the denominators jgi for a new denominator. See p. 91. 2g, I'lte numerators are 414 and 23; we divide ^ them as in integers. The denominators are 10,000 and 100 ; their quotient is 100 ; we point off 4-2, or 2 decimal places. GC Divide 426 by .08. Before dividing, it is necessary to annex .0 8) 4x6.00 ciphers to tlie dividend until the number of 5S25 decimal places equals or exceeds those in the divisor. Since tliere are two places in the divisor and none in the dividend, we annex two ciphers to the dividend. 65. Divide 8% by 432 true to two decimal places. 432)896.00(2.07^^ ^ . ^ , . , 86A ® annex two ciphers to make the places in — — - — the dividend exceed those in the divisor by 3024 ^1 should be reduced to lowest terms. 176 NoTF. — If the exact qnotient is not required, it is customary to write ' + ' instead uf the common fraction. 66. Divide 268 by 10,000. 10000 ) 268 Moving the decimal point one place to the Q^0g left divides by 10 ; two places, by 100 ; etc. See p. 55, Ex. 152. 67. Divide .06 by 6000. 6.000x).000it06 If the divisor ends with ciphers, we move QQQQl the decimal point in both dividend and divisor. See p. 55, Ex. 15S. DECIMALS 109 THE FOUR OPERATIONS Add. 6a $988,058, $75,896, $75, $68,299, $86.43, $45,256. 69. $77,948, $89.23, $57,637, $88,009, $6,783, $.086, $55. 70. $ 235.06, $ 578.085, $ 735.88, $ 967.08201, $ 3.6872, $ 20.98, $66, $8.7963, $48.23415, $ 9996.6, $ 8763. Subtract : 71. 35.1789 from 103.45 72. .463268 from 13.603 73. 21.9875 from 86.013 74. 178.369 from 10,000 Multiply : 79. 3.876 by .3816 80. .4081 by 1001 81. .6375 by .5082 82. 700.87 by .642 Divide : 87. .00072 by .016 8a 63.904 by 4.86 89. .97230 by .313 90. 27013 by 176 91. 17.181 by 567 92. 5.0000 by 789 93. 1.4511 by .214 75. .166 from 55.03 7a 8.954 from 9.862 77. .88875 from 100.02 7a 88.86 from 856.9 83. 23.478 by 3300 84. 2.0087 by 2500 85. 6.0961 by 1600 sa 365.08 by 68.092 94. .0005 by 5000 95. 5000 by .0005 96. .1265 by 7.91 97. .9614 by .907 9a .9396 by 67.12 99. 6800 by .0034 100. .0006 by 3400 XoTR. — The pupil should turn to p. 54 and solve examples 141 to 145, carry ing out each quotient two ur three decimal places. 110 COMMON FRACTIONS TO DECIMALS COMMON FRACTIONS TO DECIMALS 101. Reduce | to a decimal. 8)6.000 I means 6 -»- 8 ; performing the indicated oper- ,625 atioD, we obtain .625. 102. Reduce ^ to a decimal. 7)S.OOOOOO Oar laai remainder is 2, and the next dividend 001:71 li ^ ^» ^*"^ ^^ ^*^ *^ fi"'- Hence, the quotiem .^iS0il4^ will repeal the figures 2, 8, 6, 7, 1, 4, and will not ooiue out exact. It is, therefore, impossible to reduce f to a decimal exactly. I. Result exact 7)2.000 If the exact result is required, it is customary ^ ^B to carry out the division a few places, and to write -^^7 the remainder as in I. II. Result approximate. 7)2.000 If an approximation is desired, In place of the JSSS-k- remainder, we write * + * as in IL III. Result as a repetend. 7)2.000000 M we desire to show the figures which repeat, 9>JiK7lL ^'^ carry out the division until the quotient begins JSifOfi4 ^ repeat, and place dots over the first and last figures of the repetend, as in III. The result is a circulating decimal; it is read: decimal, repetend, 2, 8, 5, 7, 1, 4, repetend. loa What fractions cannot be reduced to decimals exactly ? An3. Those fractions which have prime factors other than 2 or 5 in their denominators. The factors of 10 are 2 and 6. Hence, if a fraction has any prime factor other than 2 or 6 in its denominator, the reduction will not be exact. 104. Can y|y be reduced to a decimal exactly ? Why ? 105. If ^^ is reduced to a decimal exactly, how many decimal places will there be in the result? Ans.B, ,i,=ixjxixjxixixjxi = (.6)». DECIMALS 111 DECIMALS TO COMMON FRACTIONS 106. Reduce .G25 to a common fraction. .625 = AVff = g- ^® write the decimal as a common fraction, and reduce to lowest terms. 107. Reduce .83J to a common fraction. §^ = — = 5. We first reduce 100 300 6 to a simple fraction. .831 = §^ = ?^ = ?. We first reduce the complex fraction, ^ Reduce to common fractions in lowest terms : loa .032; .0016 115. .12^; .33J 109. .075; .0128 11& .87^; .37f ua .025; .1126 117. .83J; .62^ 111. .005; .3125 lia .125f; .0375 112. .004; .1728 119. .0875; .OOOJ 113. .225; .4375 120. .086}; .216J 114. .024; .1626 121. .1875; .080J Reduce to decimals exactly : 122. i, J 125. i, J 12a .75i 12a i, i 126. ^, ^ 129. .871 124. h i 127. iVV» tW 130. .OO^V 131. Can jj be reduced to a decimal exactly ? Why ? 132. Reduce jj to a decimal, writing the divisor under the remainder after the third decimal place of the quotient. 133. Reduce fi to a decimal, writing * -f ' after the fourth deci- mal j)lace of the quotient. • 134. Reduce H to a circulating decimal, writing dots over the first and last figures of the repetend. 112 COMPLEX DFXIMALS COMPLEX DECIBIALS If a complex decimal is to be subjected to any operation, it is best first to reduce to a simple decimal. 135. Prepare 36.00}, 7.1} for the operations. S6.00i = 36.0025; TJ^ = 7.1S75 When a complex decimal cannot be reduced to a simple deci- mal, it is generally sufficient to carry out to three or four decimal places and to substitute * -f ^ for the common fraction. 136. Prepare 16.2|, 48.32} for the operations. 16.2\^ 16,2S3^", 48.S2\ = 48,328-^- If absolute accuracy is required, the common fraction cannot be neglected. 137. Add exactly 16.21, 48.32f 16,2\ ^16£S\ 18,32- =s 48,32- Since ^ is tenths and | is hundredths^ it is neces- • ^ sary to reduce | to himdredths. 64.561 In division, it is frequently possible to simplify by multiplying both dividend and divisor by some number. 13a Divide 60 by 2.8^. 2.8\)60( We multiply both terms by 7, and proceed as 2,0^ 0)42.0^ in Ex. 67. 21 139. Divide .365f by 3.8}. 3.8 g). 365 ^{ We multiply both terms by 6, and proceed as 23.2)2.195( ^^ DECIMALS 118 CIRCULATING DECIMALS If a circulating decimal is to be subjected to any operation, it is best first to reduce to a common fraction. To reduce a circulating decimal to a common fraction, for the numerator, write the repetend, and for the denominator as many 9^s cw there are figures in the repetend. 140. Reduce .15 and .207 to common fractions and illustrate the rule. 99 33 999 111 .15 X 100 = 15.1516 .207 x 1000 = 207.207 .is X 1= .1515 .207 X 1= .207 .16 X 99=15 .207 X 999 = 207 .16 = U .201 r = H5 141. Reduce .0315 to a common fraction. ,0315 : = Mf^ = 'OSh lOU ss 100 104 3300 26 825 By the rule, . 0316 = . 03U- 142. Reduce 1.5 to a common fraction. 1.5 = .15 X 10 1.6x100 = i.6x 1 = 161.515 1.515 . . . . 2Q^l^y^20 = — 1.6x99=150 99 99 1.6 = W = UJ Find the value of: 14a .3 + .1^ + 1.6 147. 6 4- .6 + .6 144. 23.36-16.4 14a 9.16 -.t3 145. .25 X 1.6 149. 5.'00'3 X 6 146. .625 + 2.7 isa 15.324-1-81 AMER. ARITH. — 8 114 SPECIAL CASES SPECIAL CASES The decimal and the per cent equivalents of a few common 1 rivctions are used so frequently that they should be memorized. 151. Memorize Uie decimal and the per cent equivalents : i, .50, 50%. J, .40, 40% f .37i, 37i%. i, .33i, 33i%. f, .60, 60%. |, .62J, 62i%. f .66|, 66}%. f .80, 80%. J, .STJ, 87J%. J, .25, 25%. i, .16}, 16|%. f .Hi, 11J%. }, .75, 75%. }, .83}, 83J%. ^, .08}, 8}%. I, .20, 20%. }, .12}, 12}%. ^, .06}, 6J%. Thoa: i, 60 hundredths, W per cent. Note. — When the denominator of a decimal is 100, it may be expressed by the sign %, read, per cent. State rapidly the decimal equivalents: IM. h h h i- 155. }, }, }, ^. 15a i, }, ^, f IM- i t. »» h 156. }, }, J, }. 159. f , }, ,V, }. !»*• i t» A» A- 157. }, f , }, tV. 1«X f , J, f , f Ex. 152. .20, .33J, .25, .50. State rapidly the per cent equivalents: 161. h h h h 16* tV. h h f . 1«7. f, h h i- 162. I, ^, }, i. 165. }, f, }, tIj. 16a ^j, }, f, f. 1«3- i. h A» f 166. i I, }, }. 169. h iV. h h Ex, 161. 76%, 83}o/o, 87i%, 20%. Staie rapidly the fractional equivalents: 170. 87}%, 661%, 33}%. 173. 37}%, 40%, 8}%. 171. 16}%, 60%, 25%. 174. 62}%, 16|%, 20%. 172. 62}%, 20%, 12}%. 175. 66|%, 50%, 11}%. Ex. 170. I, I, I DECIMALS 115 In multiplication and division, it is easier to use the common fractions than to use the decimal or the per cent equivalents. Find mentally the value of: 176. 24 X i, 24 X .16i, 24 x 16|%. 179. 84 x 83^%. 177. 36 X i, 36 X .66|, 36 x 66J%. 180. 76 x 75%. 17a 64 X I, 64 X .37i, 64 x 37^%. 181. 81 x lli%. Find mentally the value of: 182. 72^1, 72H-.37i, 72-5-37i%. 185. 49 -^87^%. 183. 60 -f- 4, 60-^.83J, 60-f-83i%. 186. 65 -5-83i%. 184. 84 H- J, 84-5-.66J, 48-h66{%. 187. 64 What is the cost of: 188. 48 articles at 6J^? 8^^? 12J^? 16|^? 33^^? 66|^? 189. 72 articles at 87i^? 75^? 83^^? 37^^? 12^^? 25^? 190. 36 articles at $1.33J? $1.25? $1.75? $1.66|? $2.16J? Ex. 190. $48; $1.33^ = 9H; 36 X $1^ = $48. Hoio many articles can be bought for : 191. $24@37i^? 75^? 66J^? 40^? 12^^? 80^? 33^^? 192. $30@83i^? 62^^? 25^? 37J^? 60^? 66§^? 50^? 193. $40 @ $1.25? $1.33i? $1.66|? $2.50? $3.33J? Ex. 193. 32 ; $ 1.26 = $ | ; $40 -f- $ f = 32. 194. What is the cost of cloth per yard, when $ 1 is the cost of 2 yards? 3 yards? 4 yards? 5 yards? 6 yards? 8 yards? D yards? 10 yards? 12 yards? 16 yards? 195. How many yards of cloth can be bought for $1 at 50^ a yard? 33^^? 25^? 20^? 16}^? 12^^? llj^? 8^^? 6J^? 10^? 9^^? 116 SPECIAL CASES In multiplication and division, it is easier to use the common fractions than to use the equal (aliquot) parts of 100. State the rule for: 196. Multiplying by 20. 201. Dividing by 50. 197. Dividing by 20. 202. Multiplying by SSJ. 19a Multiplying by 25. 203. Dividing by 33^. 199. Dividing by 25. 204. Multiplying by 12^. 20a Multiplying by 50. 205. Dividing by 12^. Ex. 198. Multiply by 100 and divide by 4 ; since 25 = i^ft. Ex. 199. Multiply by 4 and divide by 100 ; since 25 = ija. Find the value of: 206. 3683x25. 210. 4686x50. 214. 9678 x 16|. 207. 3876-5-25. 211. 4686 -i- 50. 215. 9678-f-16|. 20a 4773 x33J. 212. 7265x20. 216. 6272 x 12J. 209. 4873 ^33f 213. 7265-5-20. 217. 6272 -5- 12f When the price of articles is given by the 100 or 1000, it is easier to reduce the number of articles to hundreds or thousands. 21a How much will 13625 bricks cost at $4.75 per M (1000) ? Suggestion.— There are 13.625 M. at $4.75 per M. 219. How much will 1875 pounds of beef cost at $ 7.25 per C (100)? 220. How much will 1975 pounds of hay cost at $ 6.75 per ton (2000 pounds) ? 221. How much will 2146 feet of lumber cost at $ 22.50 per M ? 222. How much will 875 pounds of binder twine cost at $ 6.25 per C ? DECIMALS 117 ANALYSIS In answers to problems involving U. S. money, the mills are usually dropped, but 5 or more are reckoned as an additional cent. 223. How much will 13 J yards of cloth cost at $ 1.25 a yard ? 13 J yards ? Ans. Iie.sec^ 16.6625); |16.88($ 16.876). Before multiplying, it is often wise to determine the lowest order required in the product, and to neglect unnecessary orders in the multiplicand. 224. In $ 1.23456789 x 52, what decimal orders may be neg- lected in the multiplicand ? Find the product true to cents. The product must be true to cents, or to two decimal places. It is well to find one more decimal order than is required, because the figure in the last order will not always be exact. The highest order in the multiplier is tens; tens X ten-thousandths = thousandths ; all to the right of ten-thousandths may be neglected. 225. In $20.123456789x3200, what decimal places may be neglected in the multiplicand ? Find the product true to cents ; to mills. 226. Find the value of $(1.06)^*' x 200, true to cents. We will find 3 decimal places in the answer. The highest order in the multiplier is hundreds; hundreds x hundred-thousandths = thousandths; all to the right of hundred-thousandths may be neglected ; i.e., we must retain 6 decimal places. (1.06)10= 1.06 X 1.06 X 1.00 x . . . . where 1.00 is taken 10 times as a factor. 1.06 x 1.00, or (1.06)«= 1.1230; 1.1230 x 1.1230, or (1.06)* = 1.20247; 1.26247x1.26247, or (1.00)" = 1.60.383; 1.69383x1.12:16, or (l.OO)* x(1.06)«, or (1.06)W, = 1.79083, $1.79083 x 200 = <>368.17. The pupil should perform these multiplications. 227. How many decimal places are there in the exact value of (1.06)'«? 22a If $1 amounts to $(1.06)* in 5 years, how much will $234.63 amount to in the same time? 118 ANALYSIS 229. Find the cost of 5| cords of wood at $ 3.25 a cord ; of 7 J tons of hay at $4.75 a ton ; of 53} bushels of corn at 23^ a bushel. 230. The distance around a wheel is 3.1416 times its diameter. How many times will a wheel 4 feet in diameter turn around in going a mile (5280 feet) ? 231. A wheel 3 feet in diameter turns 286 times in going a certain distance. What is the distance ? 232. If a man's income is $2000 a year (365 days), and his average expenses are $3.68 a day, in how many days will he save $75? Prove the answer. 233. What is the average expense per day of a newspaper that costs $ 1.50 per year ? 234. At an average expense of 5^ per day, how much does a man spend for tobacco in 20 years ? 235. There are 2150.42 cubic inches in a bushel and 231 cubic inches in a gallon. How many bushels are there in 86.378 gal- lons? 236. When pork sells in the market for $4.25 per hundred pounds, how many pounds can be bought for $ 340 ? 237. From a granary containing 287 bushels of corn, .36 of the whole was sold at 23^ per bushel. How much was received from the sale ? 23a A man worked 221 days at the rate of $ 1.15 per day, and took his pay in wheat at 62^^ per bushel. How many bushels did he receive ? 239. A farmer raised 6768 bushels of corn from 160 acres of land, at an average cost of $ 6.25 per acre. How much will he gain if he sells the whole at 18 ^ per bushel ? 240. On a certain day the sales at a lumber yard were : 5280 feet rough lumber at $14.50 per M; 3060 feet dressed lumber at $17.25 per M; 4327 feet yellow pine flooring at $42.50 per M; 224 pounds lime at 6\^ per pound. To what did the sales amount ? DENOMINATE NUMBERS ENGLISH SYSTEM Substances may be measured in various ways to determine how much they possess of certain attributes, such as length, surface, volume, capacity, weight, value, .... A denominate number answers the question, How much ? See p. 7. Thus : How much length? How much surface? How much volume? How much capacity ? How much weight ? How much value ? The answers ; 4 feet, 06 square feet, 64 cubic feet, 50 quarts, 2 tons, $50, are denominate numbers. In naming units of measure, it is thought best to consider several smaller units as forming a unit of larger measure ; several of this larger unit as forming a unit of still larger measure ; . . . . Thus : 12 inches make 1 foot; 3 feet make 1 yard; 5\ yards make 1 rod. Each of these denominations is a unit of measure. It is difficult to conceive of a system of measures more unscien- tific and more confusing than the English. Not only do the same denominations occur in different tables ^vith different values, but there are four different tables for weight, two for capacity, and a multitude of miscellaneous units. The same number of each unit does not make one of the next higher; the names of the smaller and the larger units in the same table have nothing in common ; the reduction from one denominatioo to another in the same or different tables, involves much labor. Compare with metric systemy p. 136. 119 120 ENGLISH SYSTEM LENGTH The length of a line is the number of linear units which it contains. Thus: the line, A, contains the linear unit, ^ j^ If , 3 times ; its length is 3 units. If 3f is 1 • ' ' ' • ' inch, ^ is 8 inches ; if Jf is 1 foot, ^ is 3 feet ; . . . . The units of length are : incfi, in. ; foot, ft ; yard, yd. ; rod, rd. ; miley mi. ; chain, ch. ; link, li. The surveyors* chain is 4 rd., or 66 ft, or 792 in., long, divided into 100 links of 7.92 in. each ; the engineers* chain is 100 ft long, divided into 100 li. of 1 ft each. ScBTBYOKS^ Long Measdrx 7.92 in. = 1 li. 26 li. = 1 rd. 80 ch. = 1 mi 1. What instruments for measuring length have you seen? What does a carpenter use ? a tailor ? a dry-goods clerk ? 2. Draw a line 12 units long. If each unit is an inch, what unit is the whole line ? Draw a line 5 J units long. If each unit is a yard, what unit is the whole line ? a Draw a square 6 units long and separate into 36 equal squares. If each linear unit is a mile, what is the length in rods of one side of each small square ? of the large square ? 4. In example 3, separate one of the smaller squares (a section) into 4 equal squares. How long is a quarter-section ? 5. Draw a line an inch long ; separate into 2 equal parts ; each half into 2 equal parts ; each quarter into 2 ABCDSF ' ' i/ ^^^^^^ P^^^s. What is the length of ABl of ^C? oiAM"^ oiAE? oi AF? Long Measdrjb 12 in. = 1 ft 3 ft. =1 yd. 16> ft. 1 ^ "^ 320 rd. =lmi. NOTE.- 3 mi.; . . — Other units are : hi M DENOMINATE NUMBERS 121 SURFACE The area of a surface is the number of square units which it contains. Thus: the surface, A^ contains the square unit, 3f, 6 times ; its area is 6 square units. If one side of itf is 1 in., ^ is 6 square inches ; if one side of Jf is 1 ft., .4 is 6 square feet ; . . . . The tables are formed by squaring the units of long measure : 12 in. = 1 ft., 144 sq. in. = 1 sq. ft. ; 3 ft. = 1 yd., 9 sq. ft. = 1 sq. yd.; 5 J yd., or 16^ ft. = 1 rd. ; 30^ sq. yd., or 272J sq. ft. = 1 sq. rd. Other units are : acre, A. ; township, Tp. Square Measure Sdrteyors* Square Measure 144 sq. in. = 1 sq. ft 625 sq. li. = 1 sq. rd. 9 sq. ft. = .1 sq. yd. 16 sq. rd. = 1 sq. ch. 30i sq. yd. 1 _ . , 10 sq. ch. = 1 A. 272^ sq. ft. j ~ ^ 640 A. =1 sq. mi. 160 sq. rd. = 1 A. 36 sq. mi. = 1 Tp. Note. — The surveyors' chain was made 4 rd. long, because a tenth of an acre is 16 sq. rd., or a square whose side is 4 rd. 6. Separate a square 3 units long into 9 equal squares. If each linear unit is a foot, what unit is the large square ? 7. Draw a square 5^ units long, making a dot at the end of each unit. Connect the dots by lines parallel to the sides. How many whole squares are there ? How many half squares are there ? What part of one of the 25 equal squares is the small square in one corner ? If each linear unit is a yard, what is the area of the large square ? a Draw a rectangle 4 units long and 2 units wide ; separate into strips, and separate one of the strips into squares. What is the relation between the number of squares in each strip and the number of linear units in the length ? between the number of strips and the number of linear units in the width ? What is the area of the rectangle ? .f=m) 122 ENGLISH SYSTEM VOLUME The \roluine of a solid is the nninber of cubic units which it contains. Thu« : the solid, A, contains the cubic unit, Jf, 3 times ; its volume is 3 cubic units. If one side of Jf is 1 in., ^ is 3 cubic inches ; if 1 ft., ^ is 3 cubic feet ; . . . . The tables are formed by cubing the units in long measure : 12 in. = 1 ft. ; 1728 cu. in. = 1 cu. ft ; . . . . A pile of wood 8 ft long, 4 ft. wide, and 4 ft high, is called a cord, cd. Cubic Measurb Wood Mbasubb 1728 cu. in. = 1 cu. ft 16 cu. ft = 1 cd. ft 27 cu. ft, = 1 cu. yd. 8 cd. ft = I cd. Note. — A perch of ston© or masonry is 164 ft. long, 1) ft. wide, and 1 ft. high, aoJ contains *iu cu. ft. 9. Draw a cube whose length is 3 linear units, and separate into small cubes. If each linear unit is a foot, how long is one side of the cube? What is the area of each face? What is the entire surface of the cube? If each linear unit is 1 yd., answer the above questions. la In example 9, if each linear unit is a foot, what unit is the large cube ? each small cube ? What is the volume of the large cube in cu. ft ? in cu. yd. ? What is the volume of each small cube in cu. ft ? How many times does the large cube contain the small cube ? U. Draw a prism 4 units long, 2 units wide, and 2 units high ; ////yA separate into layers, and separate one of the layers into cubes. What is the relation between the num- ber of cubes in each layer, and the product of the i linear units in the width and the height ? between the number of layers and the number of linear units in the length ? "WTiat is the volume of the prism ? DENOMINATE NUMBERS 128 CAPACITY The capacity of a vessel is the number of volume units which it contains. c Thus : C will hold the measure, M, 20 times ; its capacity is 20 volume units. U M is & pint, the capacity of C is 20 ^ pints ; if a quart, C is 20 quarts ; if a gallon, C is 20 gal- ^ Ions; .... There are two sets of measures, one for liquids, and one for dry commodities. The units are : gillj gi. ; pintj pt. ; quart, qt. ; gallon, gal. ; barrel, bbl. ; bushel, bu. Liquid Measukb Dry Measurb 4 gi. = 1 pt. 2 pt. = 1 qt. 2 pt. = 1 qt. 8 qt = 1 pk. 4 qt. = 1 gal. 4 pk.= 1 bu. 12. Name an article that is sold by the gal.; by the barrel. Name an article that is sold by the dry qt. ; by the peck ; by the bushel. 13. Draw a rectangle, 11 J in. long and 5f in. wide; draw the fisL^ BEFC. Cut out the rectangle id EFZ), roll ^Z> f bji to BC and paste BEFC; a cylinder holding a * liquid quart will be formed. fc ^ 14. In the same way prepare a cylinder holding a dry quart, making the dimensions of the rectangle 12J in. and 5J in. Notice that the dry quart is larger than the liquid quart. 15. Since a cube is a volume unit, capacity may also be ex- pressed by cubic measure. A bu. is 2150.42 cu. in. ; a gal. is 231 cu. in. How many cu. in. in a dry qt. ? in a liquid qt. ? 16. Capacity may be determined by weighing, and in many states the number of pounds of the different grains to be reckoned a bu. is fixed by law. How many bu. of corn are there in 224 lb., if a bu. = 66 lb. ? 124 ENGLISH SYSTEM WEIGHT For weight, there are four sets of measures: troy, for the precious metals; apothecaries', for dry medicines; apothecaries' fluid, for liquid medicines; avoirdupois, for the coarser articles of merchandise. The troy units are : grain, gr. ; pennyrceighty pwt. ; ounce, oz. ; pound, lb. The apothecaries': grain, gr. ; scruple, 3; dram, 3; ounce, 5 ; pound, lb; minim, ^, -, pint, O. ; gaUon, Cong. The avoirdupois : ounce, oz. ; pound, lb. ; hundredweight, cwt. ; ton, T. Tbot Atoirbcpois 24 gr. =lpwt 16 OS. = 1 lb. 20 pwt = 1 o£. 100 lb. = 1 cwt 12 oz. = 1 lb. 20 cwt = 1 T. Apothbcaribs* Apothecaries* Fluid 20gr. = 13 60nv,=lf3 33 =13 8f3 =lf5 8 3 =15 16f5=10. 12 5 = 1 lb 8 O. = 1 Cong. Note. — The English, or long ton of 2240 lb., is used at the U. S. Custom House. 17. The grain is the same for all measures ; 5760 gr. = 1 troy or apothecaries' pound ; 7000 g^. = 1 avoirdupois pound. How many gr. make a troy or apothecaries' oz. ? an avoirdupois oz. ? la 60 drops make a teaspoonful; 8 teaspoonfuls make an ounce. 30 drops are what part of a teaspoonful ? How many doses of 10 drops to a dose, are there in an ounce of medicine ? 19. A pint of water weighs an avoirdupois pound (nearly). Why do 16 oz. instead of 12, make a pint apothecaries'? 20. In writing prescriptions, physicians employ the small letters of the Roman notation, placing the symbols first, and writing j for final i. Read: 5ij 3iij 3ij gr. xij. DENOMINATE NUMBERS 126 TIME AND ARC A day, da., is the time of the revolution of the earth upon its axis; a month, mo., the time of the revolution of the moon around the earth; a year, yr., the time of the revolution of the earth around the sun. The other units are : second, sec. ; minute, min. ; hour, hr. ; tceek, wk. ; century, cen. Circular measure is used in measuring arcs and angles. The units are: second, "; minute, '; degree, °', sign, S; circumference, C. The sign is rarely used. TiMB Circular Measure 60 sec. = 1 min. 60 min. = 1 hr. 24 hr. = 1 da. 7 da. =1 wk. 62 wk. 1 da. = 1 yr. 12 mo. = 1 yr. 366 da. =1 yr. 366 da. = 1 leap year. 100 yr. = 1 cen. Note. — 30° = 1 sign ; 12 signs = 1 circumference ; 90^ = 1 right angle. 21. How many degrees are there in a right angle ? in the cir- cumference of a circle ? What is the angular space about a point ? 22. Name the months and give the number of days in each. January is the first month ; name and give the ordinal numeral for each month. 2a A year lacks 674 sec. of being 365^ days. The fourth of a day is disregarded for 3 years and a whole day is added to February every year that is divisible by 4. How much of an error does this plan make in 400 yr. ? 24. How is this error of 3 da. in 400 yr. corrected ? Is 1900 a leap year? is 2000? Ans. TJte addition of 1 day to centennial years is omitted unless the centennial year is divisible by 400. 126 ENGLISH SYSTEM VALUE The units in U. S. and Canada money are: mill, m. (Latin, mille, 1000) ; cetUf t (Latin, centum, 100) ; dime, d. (Latin, decern, 10); dollar, $; eagle^ £. The units in English money are : farthing, far. or qr. ; penny, y\ur2A, j)enc€, d. ; ahilling, s.; pound, £ ($4.8665). The French units are : centime, decime, fi-anc (19.3 cents). The German units are : pfennig, mark (23.85 cents). Ukitbd Statbs Mohbt 10 f =ld. 10 d. =«1 #10 =1E. English Momst 4 far. = 1 d. 12 d. = 1 s. 208. =£1 FSBNCH MOVBT 10 centimes = 1 decline 10 decimee = 1 franc Gbbman MomtT 100 pfennigs = 1 mark MoTB. — The English use also croum (5 shillings) ; florin (2 shillings) ; sover- eign {£!). The pupil should thoroughly memorize the following equiva- lents: 12 units =1 dozen 12 dozen = 1 gross 12 gross = 1 great gross 7000 gr. = 1 lb. avoir. 00 ni =1 teaspoonf ul 48 lb. = 1 bu. barley 56 lb. =1 bu. com 32 lb. = 1 bu. oats 56 lb. =1 bu. rye 60 lb. = 1 bu. wheat water 60 lb. =1 bu. potatoes 24 sheets = 1 quire 20 quires = 1 ream 480 sheets = 1 ream 231 cu. in. = 1 gal. 2150.4 cu. in. = Ibu. 4 bu. (approx.) = 5 cu. ft. 7| gal. (approx. )= 1 cu. ft 24 hours = 360° 5760 gr. = 1 lb. troy 5760 gr. = 1 lb. apoth. 62| lb. = weight 1 cu. f u DENOMINATE NUMBERS 127 EXERCISES U. S. Money How many : 25. m. make 3 1 ? 26. d. make $ 1 ? 27. ^ make $ 1 ? 2a ^ make IE.? 29. d. make IE.? 30. nickels make $ 1 ? 31. quarters make $ 1 ? 32. m. make Id.? 33. m. make IE.? 34. m. make 1 ^ ? 35. ^ make Id.? 36. ^ make IE.? 37. halves make $ 1 ? 38. ^ make 1 quarter ? Foreign Monet Hoic many : 39. far. make £1? 40. s. make £1? 41. s. make 1 crown ? 42. s. make 1 florin ? 43. far. make Is.? 44. centimes make 1 franc ? 45. cents make 1 franc ? 46. d. make Is.? 47. d. make 1 crown ;? 4a d. make £1? 49. far. make Id.? 50. dollars make £1? 51. pfennigs make 1 mark ? 52. cents make 1 mark ? Troy and Avoirdupois Weights How many : 53. gr. make 1 lb. (troy) ? 54. gr. make 1 oz. (troy) ? 55. pwt. make 1 lb. ? 56. pwt. make 1 oz. ? 57. oz. make 1 lb. (troy) ? 5a gr. make 1 lb. (avoir.)? 59. gr. make 1 oz. (avoir.)? 60. gr. make 1 lb (apoth.) ? 61. oz. make 1 lb (apoth.) ? 62. cwt. make IT.? 63. lb. make IT.? 64. gr. make 1 T. ? 65. lb. make 1 long T. ? 66. oz. make 1 lb. (avoir.)? 128 ENGLISH SYSTEM Apothecaries* Wbioht How many : 67. m, make 1 f 3 ? ea f3 make 1 f5? 69. f 3 make 10.? 70. f 5 make 1 Cong.? 71- f 3 make 1 Cong. ? 72. drops make 1 f 3 ? 7a drops make 1 teaspoonful ? 74. gr. make 1 !b ? 75. gr. make 15? 76. 3 make 1 !b ? 77. 3 make 1 lb ? 7a 3 make 15? 79. drops make 1 f 5 ? aa tcaspoonfulsmakelf5? How many : 8L in. make 1 yd. ? 82. in. make 1 rd. ? aa ft. make 1 rd. ? 84. ft. make 1 mi. ? 85. rd. make 1 mi. ? 8a in. make 1 hand ? 87. ft. make 1 fathom ? Loxo Mbasckb 88 li. make 1 eh. ? 89. li. make 1 rd. ? 90. ch. make 1 rd. ? 91. ch. make 1 mi. ? 92. in. make 1 li. ? 9a rd. make 1 furlong ? 94. mi. make 1 league ? How many '.* Squabb Measure 95. sq. in. make 1 sq. ft.? 102. 96. sq. in. make 1 sq. yd.? 103. 97. sq. ft. make 1 sq. rd.? 104. 9a sq. rd. make 1 A. ? 105. 99. sq. ft make 1 sq. yd.? loa 100. A. make 1 section 9 107. 101. A. make 1 quarter section ? 108. sq. ch. make 1 A. ? sq. li. make 1 sq. ch. ? sq. li. make 1 sq. rd. ? A. make 1 sq. mi. ? sq. mi. make 1 Tp. ? A. make 1 half section ? sections make 1 Tp. DENOMINATE NUMBERS 129 Cubic and Wood Measures How many : 109l cu. in. make 1 cu. ft, ? no. cu. in. make 1 cu. yd. ? 111, cu. ft. make 1 cu. yd. ? 112. cu. in. make 1 cd. ft. ? ua cu. in. make 3 cu. ft. ? 114. cu. ft. make 1 j^erch ? 115. cu. in. make 1 perch ? 116. cu. ft. make 1 cd. ? 117. cu. ft. make 1 cd. ft. ? lia cu. ft. make 3 cu. yd. ? 119. cu. ft. make 2 cu. yd. ? 12a cu. ft. make 3 cd. ? 121. cd. ft. make 4 cd. ? 122. cu. ft. make 4 cd. ? Liquid and Dry Measures How many : 123. pt. make 1 qt. (dry) ? 124L qt. make 1 pk. ? 125. pt. make 1 bu. ? 126. pt. make 1 pk. ? 127. pk. make 1 bu. ? 12a lb. in 1 bu. corn ? 129. lb. in 1 bu. oats ? 130. pt. make 1 qt. (liquid) ? 131. qt. make 1 gal. ? 132. qt. make 1 bbl. of 31| gal. ? 133. pt. make 1 bbl. of 45 gal. ? 134. qt. make 1 bbl. of 40 gal. ? 135. lb. in 1 bu. wheat ? 136. lb. in 1 bu. potatoes ? Miscellaneous How many : 137. da. make 1 leap year ? 13a degrees make 1 C. ? 139. cu. in. make 1 gal. ? 140. cu. in. make 1 bu. ? 141. degrees make IS.? 142. da. in year 19()0 ? 14a da. in year 1004 ? AMEK. ARITH. — 9 144. units make 1 dozen ? 145. units make 1 score ? 146. sheets make 1 quire ? 147. sheets make 1 ream ? 14a units make 1 gross ? 149. da. in year 2000? 150. da. in year 2004 ? 130 ENGLISH SYSTEM TO LOWER DENOMINATIONS 151. Reduce 2 T. 4 cwt. 3 lb. 6 oz. to oz. T. cwt. lb. oz. 2 4 3 6 JO 44 cwt. 100 4403 lb. 16 1 T. ia 20 cwt. ; 2 T., 2 times 20 cwt., or, with 4 cwt, 44 cwt. 1 cwt. is 100 lb. ; 44 cwt, 44 times 100 lb., or, with 3 lb., 4403 lb., etc. Ans. 70,454 oz. 70454 OZ. 152. Reduce .875 bu. to integers of lower denominations. .875 bu. . 1 bu. is 4 pk.; .875 bu., .875 times 4 pk., or ± 3.5 pk. 3.500 pk. 1 pk. is 8 qt ; .5 pk., .5 times 8 qt , or 4 qt. 3 Ans. 3 pk. 4 qt 4.0 qt, 15a Reduce | lb apothecaries' to integers of lower denomi- nations. |lb = |x 7^, or e|s. ^S=|x«,or5i3. S in)isl2 5;^n)i8f times 12 5 , or 6| 5 . 15 is 83; §5 is f times 8 3, or 5^- 1 3 is 3 3; i 3 is i times 3 3, or 1 3. Ans. 055313. Bedtice : 154. 3 bu. 2 pk. 1 pt. to pt. 156. 5iij, 3 vij to drops. Seep. 124. 155. £8 3s. 5d. 2 far. to far. 157. 1 lb 8 5 7 3 2 3 19 gr. to gr. Reduce to integers of lower denominations : 15a .628 T. 162. .125 yr. 166. £ .1416. 170. .7854 lb. 159. .345 mi. 163. .385 A. 167. .3285 bu. 171. .2825 C. 160. \\ A. 161. |hr. 164. TT cu. yd. 16a I mi. 165. I lb. troy. 169. ^^ T. 172. I gal. 173. I Tp. DENOMINATE NUMBERS 131 TO HIGHER DENOMINATIONS 174. Reduce 2736 oz. avoirdupois to integers of higher de- nominations. 16 )2736 oz. 10 oz. are 1 lb. ; 2736 oz., as many lb. as 16 100 )171 lb. is contained times in 2730, or 171 lb. ; etc. 1 cwt. 71 lb. ^ns. 1 cwt. 71 lb. 175. Reduce 4 cwt. 3 lb. 8 oz. to the fraction of a T. Soz. = ^,or'-lb. ^' 16' ' 4 cwt. 3 lb. 8 oz. = 6456 oz. 9I Jh '^ ^inn r.^ '^ ^n* 1 '^' = 32000 oz. 3i lb. = -^ m or — cwt. ^^^ ^^^.^^ ^ ^^^,^ ^, ^ R07 R07 ~ ^^^ ^' 176. Reduce 7 5 3 3 2 3 to the decimal of a ft). s)2'Sy R\ ^ 66fi4- X 3 3 are 1 3 ; 2 3, as many 3 as 3 is con- ^ ^ i tained times in 2, or, with 3 3 , 3.666+ 3 ; etc. 12)7451-^1 ^n,..621+lb. .6^i+lb Reduce to integers of higher denominations: 177. 8269 gi. 181. 32844 s. 178. .'^S46 gr. troy. 182. 46381 in. 179. 8637 gr. ai)othecaries'. 183. 47351 sq. in. 180. 6732 in. surveyors'. 184. 48394 cu. in. liedme to common fractions of the highest denomination: 185. 8 5 7 3 1 3 10 gr. 18a 5 ch. 60 li. 3.96 in. 186. 1 A. 9 sq. ch. 2000 sq. li. 189. 2 cd. 7 cd. ft. 12 cu. ft. 187. 1 lb. 9 oz. 2 pwt. 12 gr. 190. 2 T. 15 cwt. 60 lb. 4 oz. Reduce to decimals of the highest denomination: 191. 3 gill. 2 qt. 1 pt. 194. 7.16 T. 6 c\vt. 5 oz. 192. 5 bu. 3 ])k. 7 pt 195. 40° 20' 15". 193. J^6 8s. 6d. 196. 5 yr. 4 mo. 3 wk. 5 da. 132 ENGLISH SYSTEM FROM TABLE TO TABLE 197. Reduce 288 lb. avoirdupois to lb. troy. 288 X 7000 5760 = 360 7000 gr. = 1 lb. avoirdupois. 6760 gr. = 1 lb. troy. 19a Reduce 87 lb 6 S apothecaries' to lb. avoirdupois. 87.5 X 5760 7000 = 72 7000 gr. = 1 lb. avoirdupois. 6760 gr. = 1 lb. apothecaries*. 199. Reduce 10 bbl. to bu., approximately. J0x^x^xi=SS.6 2 15 5 31) gal. = 1 bbl. ; 7J gal. = 1 cu. ft. ; 6 cu. ft. = 4 bu. 200. Reduce 10 bbl. to bu., true to 1 decimal place. m .. ^S ^ 231 _ »» p 31J gal. = 1 bbl.; 231 cu. in. = 1 gal.; 1 21504 2150.4 cu. in. = 1 bu. 20L Reduce 11° 13' 47" to time. uiin. sec. ii° = 44 360° = 24 hr. 13' = 52 1° = ^ or T^s hr., or 4 min. ^7" = 3* 1' = A Of 1*5 ™i°' ^^ ^ ^^' u U 55j-^ 202. Reduce 6 hr. 3 min. to arc. 6 hr. = 90° 3 7iun. = 45' 90° 45' Reduce and explain : 203. 630 lb. avoir, to apoth. 204. 6 lb. 9 oz. troy to avoir. 205. 30 cu. ft. to bu. approx. 206. 10 bu. to bbl. approx. 1" = «V or T»5 sec. 24 hr. =360°. 1 hr. = W or 15°- 1 min. = l^ or i°, or 16'. 207. 10 bu. to bbl., to 1 dec. place 208. 5 bu. to gal. approximately. 209. 5° 5' 5" to time ; 6° 6' to time 210. 5 hr. 5 min. 5 sec. to arc. DENOMINATE NUMBERS 183 4 4 2 7 3 5 1 8 8 ^\ 1 S 1 6 ADDITION 211. Add : 4 rd. 4 yd. 2 ft. 7 in., 3 rd. 6 yd. 1 ft. 8 in. nl. v«l. ft. in. The sum of the in. is 15 in., or 1 ft. 3 in. ; we write 3 in in. column and carry 1 to ft. column. ■* write 1 in ft. column and carry 1 to yd. column. Since i yd. = 1 ft. 6 in., Se.e p. 20. to avoid fractions in the answer, we may proceed as above. 212. To Dec. 5, 1883, add 165 da. Dec. 17 Oy Mar. 80, r«^ i9n A.r^ I a I^^c. 5 + 166 = Dec. 170. Since there are Jan. 139, Apr. 49, g^ ^^ ^^ ^^^ ^^^ 17^ ^ j^^ 139 ; ... . Feb. 108, May 19. 213. To Apr. 30, add 3 mo. 3 da. Jtdy 30, July 33, Aug. 2. 214. Add: 5 T. 4 cwt. 16 lb. 5 oz., 16 T. 17 cwt. 13 oz. 75 lb., 3 T. 7 cwt. 12 oz. 215. Add : 5 rd. 3 yd. 2 ft. 3 in., 13 rd. 6 yd. 2 ft. 9 in., 7 rd. 2 yd. 5 ft. 216. Add: 3 lb. 3 oz. 7 pwt. 22 gr., 16 lb. 11 oz. 19 pwt. 3 gr., 7 lb. 14 gr. 217. Add: 20 yr. 5 mo. 6 da. 4 hr., 30 yr. 8 mo. 7 da. 16 hr., 9 mo. 7 da. 7 hr. 2ia Add : ^ mi. and J rd. ; first, reduce each fraction to in- tegers of lower denominations. 219. Add J mi. and | rd. ; first, reduce each fraction to the deci- mal of a rod. 220. Add f lb. and f pwt; first, reduce each fraction to grains. 221. To July 23, add 93 da. To Aug. 5, add 100 da. 222. To July 23, add 3 mo. 3 da. To Mar. 20, add 200 da. Apr. .30+3 mo. = July 30. July 30 + 3 da. = July 33, or Aug. 2. 134 ENGLISH SYSTEM STJBTRACTION 22a Subtract 20 gal. 2 qt 1 pt from 30 gal. 9Q Q Q ^ P^ ^^^^ ^ P^ ^® cannot take ; we add 2 pt to pt., making 2 pt, and 1 qt. to 2 qt, ^0^1 making 3 qt 1 pt from 2 pt leaves 1 pt. ; we g 2 1 write 1 in pt column. See p. SO. 224. Find the exact number of days from July 20, 1897, to Nov. 19, 1897. Jly July Sly Oct, */ Aun 1Q V/M» '" ^^^^ ^^^ are 11 da. left ; in Aug., 31 ; 31, Aug. jy, jsov. jj^ g^ ^ ^^ SOj Sejit. 122y Am. NoTB. — This method is used in Equation of Payments. 500 p. 231. 225. Find the time from July 20, 1897, to Nov. 19, 1897. 1897 11 19 1897 7 20 ^**^* '^' ^®^^' ^ ^^^ 1897th yr., the 11 mo., . and the 19th da. S 29 NoTK. — This is the method commonly used in Partial Pa3rment8. See p. 223. 226. Subtract 5 lb. 6 oz. 3 pwl 9 gr. from 8 lb. 4 gr. ; 4 mi. 100 rd. 4 yd. from 5 mi. 227. Subtract 6 rd. 4 yd. 6 in. from 16 rd. 1 ft 1 in. ; 3 A. 142 sq. rd. from 5 A. 22a From ^ da., subtract 4J hr. Give the answer (a) as a common fraction of a dayj (b) as a decimal of a day; (c) in hr., min., and sec. 229. Find the exact number of days from Sept. 3, 1899, to Apr. 4, 1900. 23a Find the time from Sept. 3, 1899, to Apr. 4, 1900, as practiced in partial payments. 231. How many days' difference by the methods in examples 229 and 230 ? DENOMINATE NUMBERS 135 MULTIPLICATION AND DIVISION 232. Multiply 2 T. 3 cwt. 46 lb. by 8. T. cwt. lb. IS S 46 8 times 46 lb. are 368 lb., or 3 cwt. and » 68 lb. ; we write 68^ in lb. column, and carry 3 cwt. See p. S8. 17 7 68 23a Divide 17 T. 7 cwt. ^ lb. by 8. 17 T -T- 8 = 2 T. and 1 T. remaining ; we T- cwt. lb. ^rite 2 in T. column. 8 ) 17 7 68 1 T. and 7 cwt. = 27 cwt. ; 27 cwt. ^ 2 S 4.6 8 = 3 cwt. and 3 cwt. remaining ; etc. Seep. 51. 234. Divide 19 rd. 6 yd. 10 in. by 2 rd. 2 yd. 2 ft. 2 in. 19 rd. 5 yd. 10 in. =3952 in. a ^ m ,^^^. ,o/« We reduce both dividend and divisor 2Td.2 yd. 2 ft. 2 in. = m m. ^ ^^^ ^^^^ denomination. 494)3952 {8 Multiply : 235. 10° 36' 48" by 8. 23a 6 rd. 4 yd. 2 ft. 3 in. by 6. 236. 2 bu. 3 pk. 4 qt. by 9. 239. 4 da. 8 hr. 25 min. 40 sec. by 7. 237. 3 gal. 1 qt. 1 pt. by 5. 240. 4 sq. yd. 2 sq. ft. 56 sq. in. by 4. Divide : 241. 3 da. 6 hr. 2 min. by 4. 245. 6 yr. 2 mo. 2 da. by 12. 24i 3 mi. 6 ft. 10 in. by 6. 246. 20 bbl. 3 gal. by 3 qt. 1 pt. 243. 15 T. 16 cwt. 4 lb. by 9. 247. 3 rd. 6 yd. 2 ft. by 6 yd. 1 ft 244. 2 bu. 3 pk. by 5 pk. 1 qt. 24a £6 63. 6d. by 3d. 3 far. 136 METRIC SYSTEM METRIC SYSTEM — LAWS The principal unit of each table is derived ihmh the meter. The other units are formed by prefixing to the principal unit the Latin sub-multiples: w///i, 1000; centij 100; decif 10, and the Greek multiples: deca^ 10; hectOy 100; kilo, 1000; myriVi, 10,000. i'lie abbreviation of each sub-multiple is the small form of its first letter ; of each multiple, the capital form. SUB-MCLTIPLSt AHD MULTIPLM 10 milli uniU = 1 centi unit 10 deca unUs = 1 hecto unit 10 centi units = 1 deci unit 10 hecto units = 1 kilo unit 10 deci units = 1 unit 10 kilo units = 1 myria unit 10 unitfl = 1 deca unit 249. Substitute meter for unit in the table above, and write the new table thus formed ; compare with long measure, p. 137. 25a In the same way, substitute are, and compare with the table of land measure, p. 138. 251. Substitute Mere, liter, gram, and compare with tables for wood measure, capacity, weight, pp. 139, 140, 141. 252. Give the sub-multiples for 100, 10, 1000. State the mean- ing of deci, milli, centi. Read m, d, c. 25a Give the multiples for 100, 10, 10000, 1000. State the meaning of deca, kilo, hecto, myria. Read K, M, D, H. 254. How many milli make 1 unit ? How many centi ? How many deci ? 255. How many units make 1 deca? 1 hecto? 1 kilo? 1 myria ? 256. How many m make 1 M ? How many c make 1 H ? Ans. 10,000,000 m = 1 M ; m = 1000 ; M, 10,000. Since one is a sub- multiple and the other a multiple, we multiply 1000 by 10,000. 257. How many D make 1 M ? How many m make Id? Ans. 1000 I) = 1 M ; D = 10; M, 10,000. Since both are multiples, we divide 10,000 by 10. DENOMINATE NUMBERS 137 LENGTH The principal iiiiit of long measure is the meters ra. It is designed to be 1 ten-millionth of the distance from the equator to the i)ole; its equivalent is 39.37 in. The other units are formed by prefixing to the principal unit the sub-multiples and multiples, as explained on p. 136. Approximately, 1 m = 1.1 yd.; 1 Km = f mi. Long Measure 10 mm = 1 cm 10 Dm = 1 Hm 10 cm =1 dm 10 Hm = 1 Km 10 dm = 1 m 10 Km = 1 Mm 10 m = 1 Dm 25a State the principal unit of long measure ; its abbreviation ; how obtained; its English equivalent; approximate equivalents for 1 m, 1 Km. 259. Explain how the table of long measure is made up from the table of sub-multiples and multiples. See Ex. 249. 260. To illustrate long measure, prepare a strip of paper 39| in. long ; divide it into 10 equal parts (dm) ; divide each dm into 10 equal parts (cm) ; divide the first cm into 10 equal parts (mm). 261. To illustrate long measure, place 5 nickels (5-cent pieces) in a row; the row will be nearly 1 decimeter long, since the diameter of each nickel is nearly 2 centimeters. Place them in a pile ; the pile will be nearly 1 centimeter high, since each nickel is nearly 2 millimeters thick. 262. How tall are you ? How long is your arm ? What is the length of your forefinger? What is tlie thickness of your tlmmb nail ? What is the width of your thumb nail ? 263. Prove that 1 Km = f mi. (nearly). Equivalents ; 1 ni = 39.37 in., 1 mi. = 6280 ft. 264. What is the distance around the earth in meters? in inches ? in miles ? 138 METRIC SYSTEM SURFACE Square measure is formed by squaring long measure. Thus: 10 mm = 1 cm ; 100 sq mm = 1 sq cm ; . . . . The principal unit of land measure is the are, a ; it is a square, 10 m by 10 m. Approximately, 1 are = ^ of an acre. SouARB Mjcasusb Laitd Mbasuri 100 sq mm = lBqcm 10ma=lca 100 sq cm = 1 sq dm 10 ca = 1 da 100sqdm=lsqm 10da=la 100 sq m = 1 sq Dm 10 a = 1 Da 100 sq Dm = 1 sq Urn 10 Da = 1 Ha 100 sq Hm = 1 sq Km 10 Ha = 1 Ka 100 sq Km = 1 sq Mm 10 Ka = 1 Ma NoTB. —The final vowel of each sab-multiple and multiple is dropped before art : millare, centare ; not milliare, centiare, .... The table of land measure is often given : 100ca = la; 100 a = l Ha. State the principal unit of land measure ; its abbreviation; how obtained ; its approximate equivalent 266. Read: 50.23 sq mm; 238.6 sq Mm; 560 a; 378.6 Ha 50.23 sq Dm. 267. How many sq mm make 1 sq Km ? How many sq Dm make 1 sq Mm ? 26a Is the multiple in land measure 10 or 100 ? In the abbre- viated form, why does it appear to be 100 ? 269. How could you illustrate land measure on the school grounds ? 270. How much land would you like for a flower garden? What is the area of this floor ? How much land is needed for a good farm ? 271. Prove that 1 are is nearly ^^ of an acre. Equivalents: 1 a = 10 m X 10 m ; 1 m = 39.37 in. 272. An emigrant exchanges his farm of 4 hectares, at 12 francs an are, for land in America at $ 5 an acre. How many acres does he secure ? DENOMINATE NUMBERS 139 VOLUME Cubic measure is formed by cubing long measure. Thus: 10 mm = 1 cm ; 1000 cu mm = 1 cu cm ; .... The principal unit of wood measure is the sterCf s ; it is a cube, Imxlmxlm. Approximately, 1 stere = J of a cord. Cubic Measure Wood Measure 1000 cu mm = 1 cu cm 10 ms = 1 cs 1000 cu cm = 1 cu dm 10 cs = 1 ds 1000 cu dm = 1 cu m 10 ds = 1 s 1000 cu m = 1 cu Dm 10 s = 1 Ds 1000 cu Dm = 1 cu Hm 10 Ds = 1 Hs 1000 cu Hm = 1 cu Km 10 Hs = 1 Ks 1000 cu Km = 1 cu Mm 10 Ks = 1 Ms Note. —The table of wood measure is often given, 10 ds = 1 s, the other units being omitted. 273. Explain how the table of cubic measure is made up from the table of long measure. 274. Explain how the table of wood measure is made up from the table of sub-multiples and multiples. See Ex. 251. 275. State the principal unit of wood measure, its abbreviation ; how obtained ; its approximate equivalent. 276. How many cu dm make 1 cu Hm ? cu Dm make 1 cu Mm ? 277. How could you illustrate a stere on the school grounds? 27a What is the contents of this room in cu m ? How much earth will make a gootl load for two horses? Approximately, how many steres are there in a cord of wood ? 279. Prove that 1 stere is nearly J of a cord. Equivalents: 1 stere = lmxlmxlm; lm = 39.37 in. 280. Which is the cheaper, to buy wood at $ 3 a cord or at $ 1 a stere ? 281. What is the cost of excavating 40 cu Dm of earth at 12^ a cu m? 140 METRIC SYSTEM CAPACITY The principal unit of capacity is the liters 1 ; it is a cube, 1 dm by 1 dm by 1 dm ; its equivalent is .908 qt. dry, or 1.05 qt liquid. Approximately, 1 1 = 1 qt Table of Capacity 10 ml = 1 el 10 Dl = 1 HI 10 cl =1 dl 10 HI = 1 Kl lOdl = 1 I lOKl^ 1 Ml 10 1 = 1 Dl 282. State the principal unit of capacity; its abbreviation ; how obtained ; its exact equivalent ; its approximate equivalent. 283. How much milk do you need for a cup of coffee ? How much milk per day would you use for a family of six? How much will a teacup hold ? How much will a tablespoon hold ? 284. How many liters in 23.645 cu m ? How many cu m in 385,623 1 ? 285. Prove that 1 liter is .908 qt dry. Equivalents : 1 liter = 1 dm X 1 dm X 1 dm ; 1 m = 39.37 in ; 1 bu. = 2150.4 cu. in. 286. Prove that 1 liter is 1.05 qt liquid. Equivalents : 1 liter = Idmxldmxldm; lm = 39.37 in. ; 1 gal. = 231 cu. in. 287. Counting a liter as a quart, what is the capacity in cu m of a bin that will hold 256 bu. ? 28a If the bin in Ex. 287 is filled with wheat, what is the value of the wheat at 22^ a Dl ? 289. To illustrate measures of capacity, draw a rectangle, as ABCDy whose length is 29.16 cm, and breadth, 14.78 cm. Draw the flap BEFC. Cut out AEFDy roll over AD to EC f! ^ and paste BEFC; a cylinder holding a liter will be formed. Compare with the liquid and CF dry quarts. See Exs. I4 and 15. DENOMINATE NUMBERS 141 WEIGHT The principal unit of weight is the gram, g ; it is the weight of a cii cm of pure water at its maximum density ; its exact equivsr lent is 15.432 gr. Approximately, 1 Kg. = 2^ lb. avoirdupois. Table of Weight 10 mg = 1 eg 10 Dg = 1 Hg 10 eg = 1 dg 10 Hg = 1 Kg 10 dg = 1 g 10 Kg = 1 Mg 10 g = 1 Dg Note. — There are two additiunal uuits : 10 myriagrams = 1 quintal, Q ; 10 quiutals = 1 tonntau, T. 290. State the principal unit of weight ; its abbreviation ; how obtained ; its exact equivalent ; the approximate equivalent of 1 Kg in avoirdupois pounds. 291. How many dg in 3 Hg ? How many eg in 63 Dg ? How many mg in 46 Kg ? 292. How many grams in 1273.14 gr. ? How many Hg? How many grains in 9 Dg 8 g ? 293. To illustrate weights, procure a piece of tin foil that weighs as much as a nickel (5-cent piece); cut the foil into five equal parts. Each part will weigh a gram, since a nickel weighs 5 grams. Procure a stone that weighs 2 lb. 3 oz. ; it will weigh 1 Kg. 294. What is your weight ? How much beefsteak would you buy for breakfast for three ? What is the weight of a nickel ? How heavy a letter of the first class will go for 2^ ? What does an average horse weigh ? 295. Prove that 1 cu m of water weighs nearly 1 long ton. 296. Prove that 1 Kg is nearly 2^ lb. av. ; nearly 2| lb. troy. 297. Prove that the weight of 1 gram is 15.432 gr. Relations : 1 g = weight of 1 cu cm of water; 1 m = 39.37 in. j 1 cu. ft. water weighs 62J lb. 142 METRIC SYSTEM APPROXIMATIONS — MENTAL EXERCISES The approximate equivalents given in the several tables should be memorized, and should be used except when exact results are required. Reduce approximately: 29a 5 bu. to.l. 299. 10 g to gr. 300. 30 gr. to g. 301. 10 bu. to 1. 302. 12 1 to gal. 303. 22 lb. to Kg. Reduce approximately : 310. 16 Km to mi. 311. IG Dl to bu. 312. 75 1 to qt 3ia 32 1 to bu. 314. 40 1 to gal. 315. 10 mi. to Km. Reduce approximately : 322. 9 Kg to lb. 323. 8 acres to a. 324. 60 m to ft 325. 4 m to in. 326. 3 m to yd. 304. 60 gal to 1. 305. H4 1 to gal. 306. 64 1 to bu. 307. 15 gal. to 1. 30a 256 1 to bu. 309. 9 cords to 8. 316. 42 1 to gal. 317. 60 qt. to 1. 3ia 25 1 to cu. in. 319. 10 1 to cu. in. 320. 400 a to acres. 321. 72 s to cords. 327. 1 T. of water to cu. ft. 32a 6 tons of water to cu m. 329. 10 tons of water to cu m. 330. 10 cu m of water to tons. 331. 7500 sq. miles to sq Km. DENOMINATE NUMBERS 143 EXACT RESULTS — WRITTEN EXERCISES The exact equivalents of the principal units should be memo- rized; the exact equivalents of the other units should be com- puted from the principal units. 332. Keduce 24 Km to mi. 24 X 1000 X S9.37 ^ ^,f^ 12 X S280 24 Km =14.9+ mi. 333. Reduce 5 Ha to sq. dm. SxlOOx 100 X 100 =5,000,000 6 Ha=^ 5y000fi00 sq dm. The principal unit of long meas- ure is the meter, 39.37 in. ; I Km = 1000 m ; 1 mi. = 6280 ft. The principal unit of land meas- ure is the are, 100 sq m ; 1 Ha = 100 a ; 1 sq m = 100 sq dm. 334. Reduce 6 cu Dm to cubic yd. 5 X 1000 X 39.37^ _ f,f.oQ '7 I '^^® principal unit of cubic meas- — — — — —00^i^./+ ^Jg ig I Q^ m^ (39.37)» cu. in.; 1728 cu. in. = 1 cu. ft. ; 27 cu. ft. 5 cuJhn = 6639.7+ cu. yd. = 1 cu. yd. Reduce 365 cu m to Dl. 365 X 1000 10 = 36,500. 365cum = 36,500 Dl. The principal unit of capacity is the liter, 1 cu dm. 1 cu m = 1000 cudm. What is the capacity, in cu. ft., of a cistern that will hold 5 tonneaux of water? Find the exact answer to two decimal places ; the approximate answer. 337. 6.4 Dl of potatoes to the are is equivalent to how many bushels to the acre ? Find the exact answer to two decimal places ; the approximate answer. 33a If a bushel of oats weighs 32 lb., how many Kg will a HI of oats weigh ? Find the exact answer to two decimal places ; the approximate answer. 144 DENOMINATE NUMBERS MISCELLANEOUS 339. Change 20 mi. 70 rd. 6 ft to feet; change 106,760 ft. to integers of higher denominations. 340. Express } rd. in yards and feet; express 3 yd. 2 ft. as a fraction of a rod. 341. Reduce .3975 of a mi. to integers of lower denominations ; reduce 127 rd. 3 ft 3.6 in. to the decimal of a mile. 342. Reduce 72 lb. avoirdupois to lb. troy; reduce 87.5 lb. apothecaries' to lb. avoirdupois. 34a Reduce 157.5 gal. to bu. approximately; reduce 16.8 bu. to gal. approximately. 344. Reduce 157.5 gal. to bu., exact to 1 decimal place ; reduce 16.9 bu. to gal. exactly. 345. Reduce 38** 50' 30" of arc to time ; reduce 2 hr. 35 min. 22 sec. of time to arc. 34e Find the sum of 18 gal. 3 qt, and 6 gal. 3 qt 1 pt ; find the difference between 25 gal. 2 qt 1 pt, and 18 gal. 3 qt 347. To Mar. 3, add 182 da. ; from Sept. 1, subtract 182.da. 34a To Mar. 3, add 6 mo. 2 da, ; from Sept. 5, subtract 6 mo. 2 da. 349. Find the time from the discovery of America, Oct 21, 1492, to the Declaration of Independence, July 4, 1776. 350. Find the exact number of days from the fourth of July to Christmas. 351. How many cu cm are there in 9 1 ? 1 in 9000 cu cm ? 352. Give the weight in Kg of 5 cu m of water ; give the num- ber of cu m in 5000 Kg of water. 35a How much does a grocer gain by buying 3 bu. of chest- nuts at $ 3 a bu. dry measure, and selling for 5p a half pint liquid measure ? 354. How much does an apothecary gain by buying 50 lb. of medicine at 20^ a lb. avoirdupois weight, and retailing it at 10^ an ounce apothecaries' weight ? LITERAL QUANTITIES NOTATION AND NUMERATION A number may be considered un- der conditions that are directly opposed. One of the conditions is regarded as positive, and is represented by *-!-'; the opposite is regarded as nega- tive, and is represented by * — / The sign * + ' has, therefore, an arbitrary signification ; the sign * — ' denotes the opposite of ' + ' in the same position. A number may be represented by ft letter, and this letter may be sub- jected to the various operations. The same quantity, base, may be used several times as an addend. The expression is abbreviated by writing the base and, before it, a num- ber, coefficient^ denoting how many times the base is used as an addend. If both base and coefficient are num- bers, the sign ' x * is necessary. AMKR. ARITH. — 10 146 Illustrations 6° may be regarded as 6*- above zero, or 6° below zero ; 6 mi. may be regarded as 6 mi. north, or 6 mi. south ; 6 may be regarded as 6 to be added, or 6 to be subtracted. In + 6°, ' + ♦ has the arbi- trary signification above zero ; in — 6°, ' — ' means the oppo- site, or below zero. In 8 + 6, ' + ' has the arbitrary significa- tion add; in 8— 6, '-' means the opposite, or subtract. Take a number, z ; multiply by 4, 4a:; add 8, 4x + 8; di- vide by 2, 2 X + 4 ; subtract 2 times the number, 4. a -f- a + a + a = 4 a ; 2 + 2-^2-1-2 = 4x2. In the former, a is the bast and 4 the coefficient; in the latter, 2 is the base and 4 the coefficient. 146 NOTATTf>N AM> NUMERATION The same quantity, busej may be used several times as a factor. The «a, 2a-36. 39. 2a5«-2aj« + 3a?, -4«» + 3aB* + a?, Sir" -a^-3x, -3iB*. 4a 4a«6-3a6« + 6c«, 2a«6 + 4a6«-36c«, - 3 a«6 - ae>« + 6c». 150 SUBTRACTION SUBTRACTION III. To suhtractj change the si^n of the subtrahend and proceed as in addition. Let us take '^l ~^ "^ +^ + 3 -3 4-3 -3 To prove the remainders — 1 +1 — 5 +6 In these examples, let us assume that + means to the right, and — to the left. From + 2 subtract + .3. -f 2 means 2 to the right, one, two. If we were to add + 3, we would count 3 to the right, but since sub- 1 traction is the opposite of addition, we must count 3 to the opposite of the right, or 3 to the left, one, two, three ; counting, we have 1 to the left, or — 1. From — 2 subtract — 3. — 2 means 2 to the left, one, r^ 1 two. If we were to add —3, we would count 3 to the left, but since subtraction is the opposite of addition, we must count 3 to the opposite of the left, or 3 to the right, one, two, three ; count- ing, we have 1 to the right, or + 1. In like manner, the other results may be proved to be — 5 and + 5. By diagram^ subtract the lotcer number from the upper. 2 6 3-4-8+3 41. 3 -2 8 -6 -3 -5 IV. To subtract quantities having a common factory subtract the factors not common^ and retain the common factor. Let us take ^^ 6a To prove the remainder 2 a 8 a — 6 a means 8 x a — G x a, or that a is taken 8 times as an addend, and then 6 times less as an addend, or 8 — 6, or 2, times as an addend, or 2 a. Hence, the rule. Note. — The pupil should compare this with the same principle on p. 72. LITERAL QUANTITIES 151 42. From - 12 subtract - 19. 43. From 16 subtract 20. -12 16 -19 _20 T^ -4 44. From a-2b-3c + 5d subtract — a + Sb-^2c — 2d. (I — -^ b — 3c -{- od ^ means + 1 a ; —a means —la; changing .a-\-'Sb-\-2c — 2d the sign of the subtrahend, mentally, and adding, ]a-5b-5c + 7d we get 2 a, etc. Subtract : 17 -72 + 75 -72 + 81 -86 45. 18 -80 -85 + 83 -71 + 73 -50 -35 -56 + 35 -62 -35 46. -80 + 40 -48 + 60 + 35 + 62 Subtract : -4a -9a + 6a -8a + 7a -10a 47. H-5a -8a H-lGcZ^ — 7a + 6a -Urn' + 3a + 11 a^ - 7a -5c» + 18e -167l« 4a -f4c* -h 4cP -17e -16m» - 9a^ + 14n» Subtract : 49. 4a-66 + 10c — 12d from 26-2a + 6c + 2d. 5a 2a:* + 5a?y + a» + 2y» from 4«' — 6a^ + 2a2! + 2y". 51. 4a;'-2ajV-f 3a^ + 23/» from 3 «» + 3 a?»y - 4 ajy" - 3 3/». 52. 3a^2/ + 9aV + 6ary»-8 from 4ajV + 8«y - 6ajy»- 2. 53. -aV-3a*6c-2a6«c+6c» from aV+2a*6c+4a6«c+36c«. 152 MULTIPLICATION MULTIPLICATION V. The product of like signs is -f ; t?ie product of unlike signs is —. Let us take +3x-f2, -3x-2, +Sx-2, -S x + 2. To prove the products -|- 6, -h 6, — 6, — 6. + 3 X + 2 means that + 3 is taken 2 times as an addend, or + 3 + 3, or +6. — 3 X — 2 means that — 3 is taken 2 times as a subtrahend, or — ( — 3) -(- 3), or + 3 + 3, or +6. 4- 3 X - 2 means that + 3 is taken 2 times as a subtrahend, or — ( + 3) -(+3), or -3-3, or -6. — 3 X + 2 means that — 3 is taken 2 times as an addend, or ( - 3) + (-3), or -3-3, or -6. Hence, the rule. Declare the products : 54. +6x + 6; -5x-6j -6x + 6; 4-6x-6. 55. -6x-7; -6x + 7; -f6x-7; -f-6x4-7. VI. To multiply when the bases are the same, write the common hasey and over it, the exponent of the multiplicand plus the exponent of the multiplier. Let us take a* x a' To prove the product a* .8 — a X a a X a X a — /,« Analyzing the product, we see that the base, a, is the common base ; that the exponent, 5, is the exponent of the multiplicand plus the exponent of the multiplier. Declare the products : 56. a*xa^\ ai^xx'] f xf; b^xb-, - a^ x - a^; - a-^ x + a. 57. a^xa-j a-xa^; x*xa^', y^ X y, +q^x-x-, -{• x^ x - x"- LITERAL QUANTITIES 163 SB, Multiply —8a by -3a*. The product of like signs is -f . — o * To multiply when the bases are the same, — 3 a* write the common base, and over it, the expo- + 24" a* ^^^^ ^^ ^^® multiplicand plus the exponent of the multiplier. 59. Multiply a» - 2 ab + b^ by a-\-b. a*-2ab + 6» a 4- 6 Beginning at the left, we multiply by a; a» — 2 a*6 + ab* a x a* = a'; a x - 2a6 =- 2a26 ; ax&2=a62. „ ^ , , , , Then we multiply by 6 ; 6 x a* = a'^ft ; 6 X — a»6-2a6'-f &' 2ab = -2ab^; bxb^ = b». o?- a*b- ab^+b" Multiply : eo. a6 X a6 ; a'6 x ab* ; a^b x a*b^ ; a^6' x a*6'. 61. -3ax-6a; -36x26; a*x-2a*. 62. -3a^zx2a:*; 2a5x3a^; -SxyxSx'y^. 63. 2a6x3a6; -2a«6x3a6*; 2a26«x3a6. 64. —6xyzx2; — 2x — 6«y2; — 2a^x3 xyhi. Multiply 65. a+b a-6 a-6 a*-2a6-h6' 65. e& 67. 6a a + 6 a — h a-\-b a'4-2a6+6» 69. 3a + 46 3a-46 70.* a^ + a^ ^-xy^ a*-l^ a« + 6« 72. ^ + xy + i^ T^-xy+y" 73. 74. 75. 76. (a + 6)* (o-f6)* (a 4- 6)* (a-\-b^cf 164 DIVISION DIVISION VII. The quotient of like signs is -k- ; the quotient of unlike signs is — . -t-6 -6 -f6 -6 Let us take ^3' ^3* ZTg 4.3* To prove the quotients -f 2, +2, — 2, — 2. + 8 x + 2 = + 6 -8x+2=-6 -8 x-2= + 6 +8x-2=-6 Since the product divided by either factor is equal to the other factor, ±-?--4.2- Zl?-J.9. ±_?~_2. ~^-_9 Hence, the rule. Declare the quotients : 77. +6-f-+2; -6-!--2; -|-6-t--2; -6-f--h2. 7a -f8-^-2; -8-J- + 2; -8-i--2; +8-S- + 2. VIII. To divide when the bases are the same, write the common base, and over it, the exponent of the dividend minus the exponent of the divisor. Let us take c^-i-a^ To prove the quotient a* cfi _ a Xffxaxaxg a* a X a X a = a X a = a^ Analyzing the quotient, we see that the base, a, is the common base ; that the exponent, 2, is the exponent of the dividend minus the exponent of the divisor. Declare the quotients : 79. a* ^ a* ; a^ -i-a^; a^-i-a-, a* -r- a* ; — a^ -j- — a*. 80. a^^x; x^-5-a^; ic^-^ic*; a^-^x*-, +a^-^ — a;*. LITERAL QUANTITIES 166 ai. Divide 9 c* by — Sc*. 9c*-i 3c^ = — 3 c*. The quotient of unlike signs is — . To divide when the bases are the same, write the common base, and over it, the exponent of the dividend minus the exponent of the divisor. 82. Divide a* + b*-2 a'b'' by h^-\-a?-2 ab. a* - 2 aft + 62)a* - 2 a'^b'^ + b\a^ + 2 a6 + 6« a*-2a»6 + a^b^ + 2a'fe + 2a^b Ha^b-i+b* 4 a^b^ + 2 ab* 2 ab* + b* 2 ab* 4- 6* Before dividing, it is necessary to arrange the terms according to the descending powers of some letter. a* is contained in a*, a* ; a^ x a"^ is a* ; a^ x — 2ab is — 2 a'6 ; a* x 6'^ is a^&'^ ; subtracting the terms in the order of the descending powers of a, + 2 a«6 - 3 a%^ + b* ; etc. NoTB. — a<-fa'6 + a2i2 -)-«58 -1-54 ig arranged according to the descending powers of a. The expoueuts decrease in order from the largest. Divide : 83. — a' -J- — a ; -6a«6-j-4-a6; — 4 oft' -!- aft. 84. -a^ + x»; _6ic«-j.-3a;«; -12x2y-!-2a^. 85. _a^.^_a;5; -4a5--2a6; 6 arV H- 2 a;. 86. ar*H--a^; 6a^^H--2a6«; 8ir*y2^-2xy. 87. ar'-j--ar»; 4a«6*^a»6»; 6a^-hy'. Divide : 88. .r^-{-2xij-{-f by x-\-y] a? -2xy + f hy x — y. 89. a^-y* by a;-y; ar» + ^ by a; -f- y. 90. a^ + y* by x-f y; 25a5«-20a^+4y* by 6a;— 2y. 91. a^ + aY + y* by 7?-xy-\-%^\ aj»-y* by x-y. 92. 6ar*-5a^-6y* by 2aj-3y; x^-y" by a:*-y». 166 FACTORING FINDING FACTORS Factor: 9a 3a«6 + 3a6»; 6aj»- 12aj« + 24»-6. 94. 12oV-12a»6*; 14 aj^ - 28 a^ 4- 7 icy + 14. 95. 24a>&»~36aV; 12«y-10ajy* + 8a!y ~4ajV- 96. 20a*6*-30a»; 10 aj^ - 20 aV + 10 ay> - 5 y*. 97. 12a»6« + 9a*6*; 18a?*y + 30aV- 24aV-f- 12a^. Ex. 93. 8 ah{a + 6). 8 a6 is contained in 8 a>6, a times ; in 3 ah\ f> timea GREATEST COMMON DIVISOR Find the O, C, D, : 9a Ixy, Uf, 21y». 103. 15a*6, 20a«6», 40(i^6*. 99. 12«*, 6a^, 18 a^. lOi. 6a^6*, 18a«6», 2a<6. 100. 9a^, 12 a^, 18 a?*. 105. 5a*6, 10a»6*, a«6. 101. 4y», 16 y*, 22 y». 106. 9a*6«, 18a»6», 21a«6». 102. 12 a:*, 10 a^, 18. 107. 16a*6, 18 a^ 12a»6». Ex. 9a. Ana. 1 y. Use principle III, p. 76. LEAST COMMON MULTIPLE Find the L. C. M. : loa 2a% Sa*b\ 4ab\ 113. 4ar»?/, 6a^, 3y*. 109. oa^b^y 6a% Sab*. 114. 2xy, oa^y, 3x. ua Sa% 6a*6^ 9a*6*. 115. a;», 2y«, 3y. 111. 8a»6, 10a*6«, 12a*6^ ua ar'^', 6/, 7 a;*. 112. ea% 5a»6*, 15a*6«. 117. 8a^, 12ar^y*, 16 a^. Ex. 108. 12 a*b*. Use principle I, p. 76. FRACTIONS— LITERAL QUANTITIES 157 REDUCTION 16 a*b* lia Reduce ~ — - to lowest terms. 24 a6' 16a*6* 2 a' Dividing both terms by 8a6*, we 24a6»^36' obtain |^. See p. 85. 119. Reduce -^, -— r, to fractions having the same de- 36 9 0*^ 15 ao nominator. 30a«6 20 a 66 Th® L.C.D. is i5ab^. 3 6 is con- T? — T^' T? — r»» t:^ — r5 tained in the L. C. D., 16 a6: 15 ab x 2a 46 a6* 45 ab^ 4o a6« ^ g^ ^,^ ^^^ ^ ^^ ADDITION AND SUBTRACTION 120. Simplify - + -• c a a c a* 4- c* ^® reduce to equivalent fractions - + - = • having the L. C. D., and unite the ^ " ^ numerators. See pp. 88, 89. m. Simplify l^-2£^. 2(4a;-5)-3(2a;-3) j^ removing -3(2 a: -3) from the 6 parenthesis, we say, — 3 times 2 x = 8«— 10— 6a:-|-9 2x—l -Qx; -3 times -3 = + 9, because ™ g = g • 2 X — 3 is to be multiplied by — 3. MULTIPLICATION AND DIVISION 122. Simplify — —^ x — • We multiply the numerators for a Jo 6 3 a new numerator and the denominators 6o* 66 ___ 2a for a new denominator, canceling when 26 6« 3 a "" 6^* possible. See p. 90. 123. Simplify ^ ^^ 8 X ^^ ^'^* = —. We invert the divisor, and proceed 9 «y 16 2 aa iQ multiplication. /See p. 52. 158 FRACTIONS EXERCISES Reduce to lowest terms: ^^ 6aVj^. Sx^ 75aW^. 96 ar».v» * 12a«6' 15a!V ' 126a^6V' 144 ary ^^y . 8ai» 32afec . 72 xV * 24aY* 12aV * 128a«W 2l6xy' Reduce to fractions having the L, C. D, : -^ 3x 2x 7x 5« ,^ 3x-2 6a;-4 "^ T' T' y T* ^ -T-' -Tr- ias. ?, 4 4?.. m.^--^^ «'-^ a a^>' a** 6« 8aj ' 12x ^*nd f^€ raiM« of: 132. 5^ + ^£^. 134. ^£:=I-2^z:i. 6 9 3 2 ^^3^ 5x — 2y 7j:-3y 5x — 3y Sx—7y S ^ 12 ' 8 12 * /^'nd fAc va7M€ o/; ^25^ 32/. -16 y" 35a^ 137. 12^x^2_ 55 15 y* 32 aV 44 a» ^^ 45 6^ ■ 356» 139. 2-«\. 125 if' 236' LITERAL QUANTITIES 169 EQUATIONS IX. To transpose a term from one member of an equalion to the other, change its sign. Illl'stratiox. 8 — 6 = 2; transposing — 6 to the right-hand member, 8 = 6 + 2. X. Either of two factors is equal to their product divided by the other. Illustration. 6 x 8 = 48 ; whence 6 = Y, or 8 = y . XI. Multiplying or dividing both members of an equation by the same number, cannot affect the equality. Illustration. 12 = 8 + 4 ; multiplying both members by 2, 24=16+8 ; dividing both members by 2, 6 = 4 + 2. XII. Raising both members of an equation to the same power, or depressing both members to the same root, cannot affect the equality. Illustration, jc* = 4 ; squaring both members, x* = 16 ; extracting the square root of both members, x = 2. Find the value of x: 140. 6x-12 = 4x+-4. 147. 6aj* = 46. 141. 6 = 6x + 15-8a;. 14a 6Va5 = 26. 142. 9a;+-10-18 = 7aj. 149. 6a^ = 48. 143. 7a;-10=:3a;+-6. 150. 6^ = 12. 144. 33a;+-8-f-7=-61. 151. 3aJ* = 48. 145. -17-8 + 30 = -2aj. 152. 3\/x=6. 146. lox-4-+7x=20a?. 153. 2ar' = 64. Ex. 140. Transposing to the left-hand side all terms which contain x, 6x-4z = 12 + 4; uniting, 2 x = 16 ; x = 8. Ex. 147. 6 X* = 46 ; x* = 9 ; extracting square root, x = 3. Ex. 150. 6v^ = 12 ; v^ = 2 ; cubing, x = 8. Note. — To depress a term to the cube root, is t4» fitui one of the three equal factors whose product is that term. Thus, the cube root of 8, written v^, is 2, because 2x2x2-8. 160 PROBLEMS PROBLEMS The solution of problems by the literal method differs from the analysis method (see p. 62) in that the relations are stated directly, the required terms are represented by letters, and these letters are subjected to addition, subtraction, multiplication, and division. In explaining a problem : After every statement give a reason, unless the reason adds nothing to clearness. After the equation is ft>rmed, do not explain the solution, hut declare the result. In the proof f state the first relation and show how it is met; state the second relation and show how it is met; and so on. 154. The sum of two numbers is 32, and one of them is 3 times the other. What are the numbers ? Relations : 32 = the sum of two numbers ; the larger = 3 times the smaller. First Solution Let X = the smaUer, 8 Let a; = the smaller ; then 3 x must equal the 3 a; = the larger, 24 larger, because the larger is 3 times the smaller. 4 a; = 32 4x = 32, because their sum is 32. Whence jp _ o X = 8, the smaller ; and 3 x = 24, the larger. Second Solution Let X = the larger, 24 32 — a; = the smaller, 8 Let x = the larger ; then 32 - x must equal a; = 3 (32 — x) *^® smaller, because the sum of the two is 32. __Qn o ^ — 3(32 — x), because the larger is 3 times the ^ — ^^ — ^^ smaller. Whence, x = 24, the larger ; and 4 X = 96 32 - X = 8, the smaller, jc = 24 Proof. — The first relation is, the sum of the numbers is 32 ; 24 + 8 = 32. The second PROOF relation is, the larger is 3 times the smaller; 1.24 + 8 = 32 24 = 3x8. 2. 24 = 3 X 8 Note. — There is no rule as to which of the required terms shall be represented by a:, nor as to which of the relations shall be used first. LITERAL QUANTITIES 161 155. There are 54 children in a schoolroom, and twice as many- boys as girls. How many boys are there ? 15a There are 135 books on three shelves ; on the second shelf there are twice as many as on the first, and on the third, three times as many as on the second. How many books are there on each shelf ? 157. A man's property is worth $3600. His barn is worth twice as much as his house, and his land is worth as much as his house and barn together. What is the value of each ? 15a A farmer has 208 animals, consisting of horses, sheep, and cows. What is the number of each, provided that the number of sheep equals three times the number of cows, and the number of cows equals three times the number of horses ? 159. A is 5 times as old as B ; and C, 3 times as old as B ; the sum of their ages is 81 years. What is the age of each ? 16a The sum of $ 264 was raised by 4 persons, A, B, C, and D ; B contributed twice as much as A ; C, 4 times as much as A and B together ; and I), one half as much as B and C together. How much did each contribute ? 161. A certain fish is 6 ft. 6 in. in length ; its tail is twice as long as its head, and its body is as long as its tail and head together. What is the length of its body ? 162. Divide 96 into 3 parts, such that the first part shall be three times the second, and the third twice the sum of the other two. 163. A certain number is expressed by three digits whose sum is 12. The digit in hundreds' place is twice the digit in units' place, and the digit in tens' place is three times the sum of the other two. ^Vhat is the number ? 164. A boy has 30 pieces of money, nickels, dimes, and quar- ters; the number of quarters is six times the number of dimes; the number of nickels is a third more than the number of quar- ters. How much money has he ? ▲MKH. ARITH. — 11 162 PROBLEMS 165. Had the cost of a horse been three times as much and $70 more, it would have been 9 445. What was the cost? 166. If 63 is subtracted from a number, three times the re- mainder will he twice the sum of the original number and 16. What is the number ? Relation : 3 thnes cost + $70 - Relation : 3 x (No.-63) =2 x (No. $445. + 16). Let a; = cost, 125 Let a; = the No., 221 3 X 4- 70 = supposed cost X — 63 = the rein. 445 = supposed cost 3 (a; — 63) = three times rem. .-. 3a? + 70 = 445 2 (a; -f- 16) = twice sum 3a: = 445-70 .-. 3(a;-63) = 2(a;4-16) 3a; = 375 3a; -189 = 2a; + 32 x=125 a; = 221 Proof Proof 1. 125 X 3 4- 70 = 446 1. 3(221 - 63) = 2(221 + 16) 167. One number is three times another ; if I take the smaller from 24 and the greater from 46, the remainders are equal. What are the numbers ? 16a If to twice a certain number I add 18, 1 obtain 120. Find the number. 169. Anna is four years younger than Mary; if three times Anna's age is taken from five times Mary's, the remainder will be 62 years. What is the age of each ? 170. Find a number whose excess over 50 is equal to twice what it lacks of being 113. 171. Divide 60 into two parts such that one part may exceed the other by 24. 172. The joint ages of a father and son are 70 years ; if the age of the son were doubled, the result would be five years more than his father's age. What is the age of each ? LITERAL QUANTITIES 163 173. The difference between two numbers is 28; and if four times the less is added to the greater, the sura is 43. What are the numbers ? 174. The sum of two numbers is 34 ; the larger increased by 66, is 9 times the other. Find the numbers. 175. Divide 180 into two such parts that one of them dimin- ished by 35, shall be equal to the other diminished by 15. 176. After 30 gallons had been drawn out of one of two equal casks and 82 gallons out of the other, there remained five times as much in one cask as in the other. What was the capacity of each cask if both were full at first ? 177. Divide 41 into three such parts that the second shall be 4 more than the first, and the third 3 less than the second. 17a A man borrowed as much money as he had, and then spent ^ 4 ; he then borrowed as much as he had left, and spent $ 3 ; again he borrowed as much as he had left, and spent $ 2 ; he then had nothing left. How much had he at first ? 179. An estate valued at $ 4800 was divided in such a manner that the wife's share plus $400, was equal to three times the share of the children. What was the wife's share ? 180. If $300 is subtracted from B's income, five times the remainder will be three times, the sum of $ 3100 and the original income. What is his income ? 181. A merchant began business with a certain capital; the first year he doubled it ; the second year he gained a sum equal to the original capital plus $ 100 ; the third year he lost as much as he had gained the first year, and then had $ 3100. What was his original capital ? 182. A lady bought two pieces of cloth; the longer lacked 9 yards of being three times the length of the shorter. She paid $2 per yard for the longer, and $3 for the shorter, and the shorter piece cost as much as the longer. How many yards were there in each piece? 164 EQUATIONS WITH FRACTIONS EQUATIONS WITH FRACTIONS 18a Simplify ^ + ^ = -^-1-6. 18a;-|-6 = — 10«-|-90 Multiplying both members of 1ft I in QA a an equation by the same number l»xH-lUa;-yu-b ^^^ ^^^ ^^^^^ ^^^ equality. 28 x = 84 We multiply both members by ^ = 3 theL.C.D., 16. ^xl6=18a;; ? X 16 = 6 ; - ^-= X 16 = - lOx; 6 X 16 = 90. We then simplify as in the 6 3 preceding case. 3-x x-5 13 184. Simplify 3 (3 — a;) — 2 (x — 5) = 39 we multiply both members by 9-3x-2x+10 = 39 theL.C.D.,6. 3i:*x6=3(3-x), x = -4 =39. F«Md x; 185.^- 3a:-4_-^ 5 190. 5a;+l 3a;-2 lOar+1 2 3 6 5aj-4 1 6 3 191. 3a;-2 4a; + 5 25a; 3*4 12 "'•T- 3a;4-2 1 4 4 192. a; — 1 a; + l a; — 9 5 6 10 ,«« 4a; 18a ^- 3x--f 2 33 ^4 4 193. a; + l a;-l 2a;-17 6 5 5 189. i?- 5 3.T-2 29 ^4 4 194. a;-2 a;-3 5a;-h3 3 4 12 LITERAL QUANTITIES 165 PROBLEMS 195. Three times the num- ber of hours before nooi; is equal to f of the number since mid- night. What is the time ? Relations : 3 times hr. before noon = \ hr. since midnight ; hr. since midnight + hr. before noon = 12. Let X = no. hr. since mid., 10 12 — a; = 710. hr. before noon, 2 3(12-a:) = ^ 5 15(12 -«)= 3a; 180-15x = 3a; 180 = 18a; a; = 10 196. Divide 56 into two such parts that { of the less, di- minished by \ of the greater, may equal 12. Belations : the larger number + the smaller number = 56; f of the less — J of the greater = 12. Let X = the smaller, 24 56 — a; = the larger, 32 5 a; 56 — x 6 4 = 12 10a;- 3(56 -«)= 144 10a;-168H-3a;=144 13 a; = 312 a; = 24 Phoof 1. 3x2 = f xlO 2. 10 -f 2 = 12 Proof 1. 32 + 24 = 56 2. f of 24- i of 32 = 12. 197. In a school of 495 pupils there are J as many boys as girls. How many girls are there in the school ? ISa John and James together have $ 98 ; if James has J as much as John, how mucli has each ? 199. Mr. Drake, who owns f of a tract of land, has 14 acres less than Mr. Brown, who owns -j\ of it. How many acres does the whole tract contain ? 200. In a certain orchard there. are 40 more apple trees than peach trees, and ^f of the whole number are peach trees. How many peach trees are there ? 201. Jane's age is } of Ann's, and the sum of their ages equals 21 years. What is the age of each ? 202. Find a number whose f part exceeds its -j^ part by 3. 166 PR()BI>EMS 203. A man has f as many horses as cows, and the cows are 15 more in number than the horses. How many horses has he ? 204. The width of a room is j\ of its length ; if the width had been 4 feet more and the length 2 feet less, the room would have been square. Find its dimensions. 205. I bought a number of apples at the rate of 3 for 1 cent ; sold one third of them at 2 for a cent, and the remainder at 5 for 3 cents, gaining 7 cents. How many did I buy ? 206. A man sold a horse for ^ of its cost, plus $ 75, and thereby gained $ 15. How much did the horse cost ? 207. Find three consecutive numbers such that if they are divided by 7, 10, and 17 respectively, the sum of the quotients will be 15. 20& A man bought a horse and carriage for $ 280 ; if ^ the price of the horse is subtracted from -J the price of the carriage, the remainder will be the same as if 1.9 times the price of the carriage is subtracted from 2 times the price of the horse. What was the price of each ? 209. Divide $ 115 between two men so that f of what the first receives shall be equal to J of what the second receives. 210. B's expenses are f of A's, plus f 30; and A's expenses plus ^ of B's, amount to $ 795. How much does each spend ? 211. A father divided $ 1.43 among his three sons so that the first had f as much as the second ; and the third, f as much as the first and the second together. How much did each son re- ceive ? 212. A man paid $806 for four horses; for the second he gave ^ more than for the first; for the third, \ more than for the second; and for the fourth as much as for the first and the third together. What was the cost of the fourth horse? 213. A farm of 263 acres was divided among four heirs, so that A had H as much as B ; C, as much as A and B ; and D, \ as much as A and C. What was the share of each ? LITERAL QUANTITIES 167 TWO UNKNOWN QUANTITIES If we have two equations with two unknown quantities, the first step is to get one equation with one unknown quantity. This step may be taken by addition and subtraction, by substitution, or by comparison. Bt Addition and Subtraction 214. [2x + 3y=^ 4y=10 18 ^ (^ , find X and y. (2) 6a;-8y = 20 6a;-i-9y = 54 17y = 34 y = 2 (3) (5) 3aj-8 = 10 3a;=18 a; = 6 PROOF 1. 18 - 8 = 10 2. 12 + 6 = 18 3x-43/ = 10 2ajH-3y=18 (1) (2) 3a; = 10 + 4y ?*i5±M+3y=18 (3) o 2(10 + 4y) 4-9.7 = 54 20 4-8?/ + 9^ = 54 17y = 34 y = 2 x = 6 We may eliminate x by multiplying (1) by 2, and (2) by 3, and subtracting. Multiply- ing (1) by 2, we obtain (3) ; multiplying (2) by 3, we obtain (4). Subtracting (3) from (4), we obtain (6); whence, y = 2. Substituting this value of y in (1), we ob- tain 3 X — 8 = 10 ; whence, x = 6. By Substitution We may eliminate x by finding the value of X in terms of y in (1), and substituting tliis value in (2). From (1), 3x = 10 + 4y, and x = — i — ^; substituting this value of x in 3 (2), we obtain (3). Clearing (3) of fi-actions, and proceeding as usual, we obtain y = 2. Substituting and proceeding as before, we obtain x = G. 168 TWO UNKNOWN QUANTITIES Bt Compa&isom 3x-4y=10 (1) 2x + 3y = 18 (2) 3 18 « = Sy (3) 10-h4y ^ l8-3y 3 2 2(10 H-4y) = 3(18 -3y) 20H-8y = 54-9y 17y = 34 y = 2 x = 6 We may eliminate x by finding the value of X in terms of y in (1) and (2), and plac- ing these values equal to each other. From (,),. = 10±iJ!;from(2).x=li^; placing these values equal to each other, we obtain (3). Clearing of fractions and proceeding as osuali we find that y = 2, and x = 6. Find the values of x and y : 2a?-h3y = 8, aJ + y = 3. 215. 216. 217. 2ia 219. 220. r3a?-y = 3, I4aj + y = ll. 2x + 3y = 10, a; + 2y = 7. ic-y = 4, 2a; + y = 14. |15x + 8y = l, ll0a;-7y = -24. 2x 3y_. ■3""T-^' 2 4 221. 222. 223. 224. 225. 226. r3a; + y = 13, |ar-3y = -9. \x + y=^7, U- y = l. 3ic 2x + 2y = 8, -y = 3. [5x Sx -2/ = 4, + y = 4. Ix 2y -3y = 21, -5a; = -16. (5x 3 + 'f = 94, Ix [ 2 5y_7 3 Note. —In examples 220 and 226, the first step is to clear of fractions. LITERAL QUANTITIES 169 227. A certain number expressed by two digits, is equal to 4 times the sum of those digits ; if 27 is added to the number, the digits will be reversed. Find the number. Relations : units' digit + 10 times tens' dij?it = number ; 4 (units' digit + tens' digit) = number ; number + 27 = tens' digit + 10 times units' digit. Let x = units* digits 6 y = tenf^ digit, 3 a? 4- 10 1/ = number Let x = units' digit and let y = tens' 4 a; 4- 4 y = nwwifter <^*6*^; ^^^° x + 10y = the number, be- _i_in —A I A n\ cause a number is equal to its units' .-. X + luy — 4a; + 4y (i; ^j^j^. ^ j^ ^.^^^^ ^^ ^j^^, ^-^^^^ + . . . . a; + 10?/ + 27 = y4-10a; (2) 4x + 4y = the number, because the X — Q number is equal to 4 times the sum of _ „ its digits ; x-\-\0y = 4z + 4y, because y — *^ things equal to the same thing are equal Proof to each other ; etc 1. 4(3 + 6)= 36 2. 36 + 27 = 63 22a The sum of two numbers is 5 times their difference ; twice the greater, increased by 4 times the less, is 56. Find the numbers. 229. A farmer received $ 8.75 for 6 bu. of potatoes and 5 bu. of apples. What was the price per bushel of each, if, at the same rate, 7 bu. of potatoes and 3 bu. of apples were sold for $ 8.65 ? 230. A man, having $ 2.50 to divide among a certain number of boys and girls, found that if he gave each of them 10^, he would be 30^ out of pocket ; so he gave each of the boys 8^ and each of the girls 9^, and had 10^ left. How many were there of each ? 231. A certain fraction becomes ^Ij when 3 is added to each of its terms, but becomes J when 3 is subtracted from each of its terms. Find the fraction. 232. If A gives B $ 10, B has twice as much as A has left ; if B gives A $30, A will have twice as much as B has left How much has each ? PROPORTION SIMPLE PROPORTION - TERMS Division may be expressed by writing the dividend before and the divisor ajXer a colon. Such an expression is a ratio. See p. JfS, The dividend is the antecedent; the divisor, the consequent. The colon is read * is to.* Two fractions may be equal; in the same way, two ratios may be equal, a proportion. The sign of equality is often abbre- viated by writing the extremities of the sign * =,' thus making * : : ' read * as.' The first and last terms of a propor- tion are extremes; the second and third, means. The means of a proportion may be equal ; then each is a 7nean proportional between the extremes. iLLUSTRATIOltV 3 : 4, ratio. Meaning, 3-4-4. 3, antecedent. 4, consequent. Z is to 4. I = f. 3:4 = 6:8, a proportion. 3 : 4 : : 6 : 8, 3 is to i as 6 is to 8. 3 and 8 are extremes f 4 and 6, means. 4:6 = 6:9. 6, a mean proportional between 4 and 9. Note. — All problems in proportion may be solved by analysis. The pupil is advised to master simple proportion, but to use analysis for the solution of prob- lems which fall under the other divisions of proportion. 170 PROPORTION 171 I. In a proportion, the product of the extremes is equal to the product of the means. « « J « ... i^t antecedent Sd antecedent . . Proof. By definition, — = - r— ; clearing of frac- J8t consequent M consequent' tions, 1st antecedent x Sd consequent = 1st consequent x Sd antecedent. Illustration. 3 : 4 : : 9 : 12, 3 x 12 = 4 x 9. II. If three teiinis of a proportion are giveuy the other term may he found. Proof. Since the product of the extremes = the product of the means, either extreme = the product of the means dividetl by the other extreme ; either mean = the product of the extremes divided by the other mean. Illustration. If3:4=9:x, x= i^ ; if 3 : 4 = x : 12, x = ^^Ll?. 3 ' '4 III. Tlie mean proportional between two quantities is the square root of their product. Proof. To find the mean proportional between two numbers, as 4 and 9, we may form a proportion whose extremes are 4 and 9, and whose means are x and x. Thus, 4 : x = x : 9 ; whence x"^ = 4 x 9, and x = Vi x 9. 1. Write a fraction whose value is ^ ; a ratio whose value is J. 2. Write a proportion each of whose ratios is equal to | ; each of whose ratios is equal to j. 3. Write an equation which is an equality of two fractions; write the same equation as a proportion. 4. Is 4 : 8 : : 5 : 6 a proportion ? Why not ? W^hat is the test of a proportion ? 5. Find the mean proportional between 9 and 16; 3 and 12; 4 and 25. S. Define ratio; antecedent; consequent; proportion; means; extremes. 172 SIMPLE PROPORTION PROBLEMS The method of solving a problem by proportion differs from the analysis method, in that each relation is expressed as a proportion whose third term has the same denomination as the answer, and whose fourth term is the required term. See p. 62. 7. If 2 apples cost 8^, how much will 3 apples cost ? Relation : 2 apples : 3 apples = cost 2 apples : cost 3 apples. £:3 = 8:z Since the answer is to be cents, we make Sf g X S ^® third term, and write the ratio 8f :x. * ~ ]j~ ~ ^^ Will 3 apples cost more or less than 2 Cost 3 aooleM — if # apples ? More ; then we make the second ~ * term of the other ratio greater than the first, and we have 5 apples: 3 apples = 8f:x; whence x = 12^, the cost of 3 apples. a If 3 apples cost 12^, how many apples can be bought for 8^? Relation : 12 ^ : 8^ = apples for 12 f : apples for 8^. li : 8 = S : X Since the answer is to be apples, we make 8 X S ^ apples the third term, and write the ratio * - jg = * 8 apples : x. No atmles = f ^^^^ ^ ^ ^"^ ™^'® ®' ^^^ apples than 12 ^ ? Less ; then we make the second term of the other ratio less than the first, and we have Iff: 8f = S apples :x; whence X = 2, the number of apples. 9. If 3 men can do a piece of work in 12 days, how many days will 2 men require ? Relation : 2 men : 3 men = days for 3 men : days for 2 men. g: S = IS :x Since the answer is to be days^ we make S X 12 ^2 days the third term, and write the ratio * = -^— =^^ ISdays-.x. No davs = 18 ^'^^ ^ ^^'^ require more or less days than 3 men ? More ; then we make the second term of the other ratio greater than the first, and we have 2 men : S men — IS days : x ; whence x = 18, the number of days. Note. — Ask about the term in the question, in the denomination of the answer. PROPORTION 173 10. How much will 54 bu. of potatoes cost if 16 bu. cost $ 5.60 ? 11. How many eggs can be bought for 70^ at the rate of 2 lor o^? 12. If 16 men can dig a trench in 24 days, how many men will be required to dig it in 32 days ? 13. If 16 men can dig a trench in 24 days, how many days will 12 men require ? 14. If $150 gains $12 in 8 months, in what time will it gain $17? 15. If a man travels 32 miles in 4 hours, how many miles can he travel in 5 hours at the same rate ? 16. A flagstaff 82 ft. high casts a shadow 62 ft. long. Under the same conditions, what must be the height of a steeple which casts a shadow 93 ft. long ? 17. If it costs $12 to carpet a room with carpet 4 ft. wide, how much will it cost if the carpet is 3 ft. wide, provided there is no waste and the cost per linear yard is the same ? See Ex. 31. la How many bushels of wheat can be bought for $102.12, if 24 bu. can be bought for $ 13.32 ? 19. If a man gains $ 1500 from his business in 1 yr. 6 mo., how much will he gain in 3 yr. 9 mo. at the same rate ? 20. A garrison of 600 men has provisions for 80 days; how many men must leave to make the provisions hold out 20 days longer ? 21. A's rate of working is to B's as 3:5. How long will it take B to do what A does in 48 days ? 22. If 40 men can build a wall in 6 days, at the same rate how long will it take 16 men to do | the same work ? 2a If 3 men can do a piece of work in 51 days, how many men must be added to the number to do the work in 17 days ? 24. James can do 2\ times as much work in a given time as John. How long will it take James to do what John does in 38 hours? 174 COMPOUND PROPORTION COMPOUND PROPORTION A compound fraction is a fraction of a fraction ; in the same way, a comjjound ratio is a ratio of a ratio. To find the value of a compound fraction, we multiply the numerators for a new numerator and the denominators for a new denominator ; to find the value of a compound ratio, we multiply the antecedents for a new antecedent and the consequents for a new consequent A compound proportion is a proportion having one or both of its ratios compound. 25. Find the value of the compound ratio, *" [ » o^ « * [ > o^ ^S0:2l)- 0:12 3: 2 4:7 26. In the compound proportion, 9 : 6 14:18 =2 : X, find the value of x. J 7 xgx ^^x^ — • The product of the extremes is equal to "" 4x 9 \ 14 "" the product of the means. 27. If 2 men in 14 da. of 10 hr. each earn $ 280, how many hr. per da. must 3 men work to earn $ 120 in 5 da. ? S: g ] Since the answer is to be hr.per 280: 120 \ = 10:z da.y we make 10 hr. the third term, 5: 14 -' and write the ratio 10 hr. : x. — S X 280 X 5 -^'^^'P^^- hr. per da. than 2 men? Less; then we make the second term of the other ratio less than the first, and we have S men : 2 men. Will $120 require more or less hr. per da. than $280 ? Less ; then we make the second term of the other ratio less than the first, and we have $280:^120. Will 5 days require more or less hr. per da. than 14 days ? More ; then we make the second term of the other ratio greater than the first, and we have 5 days : 14 days. NoTK 1. — We compare each set of ratios with the second ratio separately. We ask about the term in the question, in the denomination of the answer. Note 2. — Elxamples in Compound Proportion are found on p. 177. PROPORTION 175 ANALYSIS METHOD It is recommended that Analysis be substituted for Simple Proportion. 2a If 2 apples cost 8^, how much will 3 apples cost? See Ex. 7. Relation : 3 apples will cost f as much as 2 apples. -x8 = IX Since 3 apples will cost § as much as 2 apples, 3 apples will cost | of 8^, or 12^. Coat 3 apples = Igf. 29. If 3 apples cost 12^, how many apples can be bought for «^? See Ex. 8. Relation : Sf will buy ^^ as many apples as 12^. Jg^ Since 8 f wiU buy ^ as many apples as 12 ^, Sf will buy ^j of 3 apples, or 2 apples. No. apples = f . 30. If 3 men can do a piece of work in 12 days, how many days will 2 men require ? See Ex. 9. Relation : 2 men will require | as many days as 3 men. Ix IS = 18 Since 2 men will require | as many days as £ 3 men, 2 men will require f of 12 days, or 18 No. days =18. ^^y^' 31. If it costs $ 12 to carpet a room with carpet 4 ft. wide, how much will it cost if the carpet is 3 ft. wide, provided there is no waste and the cost per linear yard is the same ? See Ex. 17. . Necessary knotcledge : carpete are sold by the linear yard ; the less the width of the carpet, the greater the length. Relation : for the room, carpet 3 ft. wide will cost | as much as carpet 4 ft. wide. i.x lt= 16 Since, for the room, carpet 3 ft. wide will S cost \ as much as carpet 4 ft. wide, the 3 ft Cost = $ 16. carpet will cost J of 9 12, or $ 10. NoTK. — The pupil should solve the examples on p. 173 by this method. 176 ANALYSIS METHOD It is recommended that Analysis be substituted for Compound Proportion. 32. If 2 men in 14 da. of 10 hr. each earn $ 280, how many hr. per day must 3 men work to earn $ 120 in 5 days ? ^x — x^xiO = « ^^^^^^^- 3 ™^° ^^^^ require | as many 5 280 5 hr. per day as 2 men ; to earn i 120 will require Ans Shr.perda. U8a8ma»y '^'••/>«rd«ya8toeam ^280; 5days * will require V as many hr. per day as 14 days. Since the first set require 10 hr. per day, the second will require | x ^f^ X V >< ^^ ^^- P^*" ^*^y> ^^ ^ hr. per day. NoTB. — In such problems, the relation is more plainly seen, if we ask aboat the term in the question^ in the denomination of the answer. Thus: will 3 men require more or less hr. per day than 2 men ? Less, i as many ; will 6 120 require more or less hr, per day than S28D? Less, US as many .... 3a If 5 compositors, in 16 days, of 8 hours each, can compose 20 sheets of 24 pages in each sheet, 50 lines in a page, and 40 letters in a line, in how many days, of 4 hours each, will 10 com- positors compose a volume to be printed in the same type, containing 40 sheets, 16 pages in a sheet, 60 lines in a page, and 50 letters in a line ? f V A V -^ V ^« V ^<^ V ^^ V r/: - t# Belations: 4 hr. per day will 4 10 20 24 50 40 require \ as many days as 8 hr. . ^^ _ per day : 10 compositors will re- Ans. Sz days. . t -, c ' quire /^ as many days as 5 com- positors ; 40 sheets will require f g as many days as 20 sheets ; 16 pages will require J| as many days as 24 pages ; 60 lines will require f g as many days as 60 lines ; 60 letters will require |§ as many days as 40 letters. Since the first task requires 16 days, the second will require | x /j x |g X if >< f M IB X 16 days, or 32 days. 34. If 9 men can perform a certain labor in 17 days, how long will it take 3 men to do twice as much ? 35. If a man working 6 hours a day, 6 days in a week, and 42 weeks in a year, earns $ 1323, how much mil he earn if he works 8 hours a day, 5 days in a week, and 50 weeks in a year ? PROPORTIOX 177 36. If 25 men can build a wall 200 ft. long in 40 days, how many days will 4 men require to build a similar wall 600 ft. long? 37. If 12 iron bars 4 ft. long, 3 in. wide, and 2J in. thick, weigh 1050 lb., what is the weight of 26 iron bars 6 ft. long, 4 in. wide, and 2 in. thick ? 3a If 36 men can dig a trench 150 ft. long, 6 ft. wide, and 4 It 6 in. deep, in 24 days, how many days will 32 men require for a trench 210 ft. long, 5 ft. wide, and 4 ft. deep ? 39. If 5 men in 6 days of 8 hours each can mow 16 acres, how many days of 6 hours each must 10 men work to mow 24 acres ? 40. What is the weight of a block of stone 10 ft. 6 in. long, 6 ft. 8 in. wide, 5 ft. 3 in. thick, if a block of the same stone 8 ft. long, 3 ft. 6 in. wide, and 3 ft. thick, weighs 12,600 lb. ? 41. If 20 men build 240 rd. of fence in 24 da. of 9^ hr., how many hours a day will 16 men have to work to build 300 rd. of fence in 50 da. ? 42. If a bin 16 ft. long, 6 ft. wide, and 4 ft. high, holds 308 bu. of grain, what must be the height of a bin 24 ft. long, 5 ft. 4 in. wide, to hold 462 bu. ? 43. If there is no waste in either case, how much will it cost to cari)et a room with carpet 3 ft. wide at 90^ per yard, if it costs $ 25 to carpet the same room with carpet 4 ft. wide at $ 1.25 per yard ? 44. In the reprint of a book consisting of 810 pages, 50 lines, instead of 40, are contained in a page, and 72 letters, instead of 60, in a line. Of how many pages will the new edition consist? 45. If 15 men cut 480 steres of wood in 10 days of 8 hours each, how many boys will it take to cut 1152 steres, only J as hard, in 16 days of 6 hours each, provided that, while working, a boy can do only | as much as a man, and that \ of the boys are idle at a time, throughout the work ? AMKR. ARITU. — 12 178 ANALYSIS METHOD It is recommended that Analysis be substituted for Partitive and for Conjoined Proportion. 46. Divide 360 into three parts proportional to 3, 4, and 6. Ji 0/ 360 = 90, first Relations : first given part = ^, of the jt sum of the given parts ; second given part ^ of 360 = ISO, secotid = ^ of the sum of the given parts ; third J. given part = A of the sum of the given ^ of 360 = 250, third parts. Therefore, first required part = ^^ o^ the ^^^^^ gum of the required parts, or ^^ of 360, or 1. ^ + ISO + 160 = 360 90 ; second required part = i^ of the sum 2. 90 : ISO = 3:4 of the required parts, or ^ of 360, or 120 ; 3. 90 : 150 = 3:6 etc. 4. ISO: 150 = 4 '6 47. A, B, and C go into partnership. A puts in $960; B, $510; and C, $1440. If they gain $ 727.50, how much should each receive? 960 ^'« = ^^ X 7S7.60 = S40 nelations : A's investment = ^%^^ of the whole investment ; B's = /^^ of the whole ; B's = ^ X 7S7.60 = 127.60 C's = ^ fg of the whole. ^^^^ Therefore, A's gain = /^ of the whole C'« = M^ X 7S7.50 = 360 gain, or /^ of « 727.60, orij 240 ; 48. If 10 barrels of apples are worth 7 cords of wood, and 14 3ords of wood are worth 5 tons of hay, how many barrels of apples are worth 50 tons of hay ? 5(? X ^ X — = SOO Belations : 1 ton of hay is worth i^ cords 5 7 of wood ; 1 cord of wood is worth J^ bbl. of Ans. SOO bbl. apples. ^PPjf- ^ u i. ,o w, Therefore, 1 ton is worth i^ x V hhl., and 50 tons are worth 60 x ^5* x V- bbl., or 200 bbl. 49. Divide $510 into three parts which shall be to each other as 2, 3, and 5. PROPORTION 170 50. An insolvent debtor fails for .f 3780, and is able to pay only $1550. If A's claim is $378, how much will he receive ? 51. Divide $ 873 among A, B, and C, so that for every $ 2 that A receives, B shall receive $4, and C, $3. 52. A and B buy goods to the amount of $600, of which A pays $250, and B, $350. If they lose $150, what will be the loss of each? 5a A bankrupt owes A $ 600 ; B, $ 800 ; C, $ 1000 ; I), $ 1200 ; but his property is worth only $ 1440. How much should each of his creditors receive ? 54. Divide a man's estate of $29,000 so that his wife shall receive $7 for every $5 received by each of his two sons, and every $ 4 received by each of his three daughters. 55. A, B, and C engage in business, A putting in $ 982 ; B, $365; and C, $843. If their profit in one year is $1460, what is each one's share ? 56. A, B, and C plant 1200 acres of corn, A planting 2 times as many acres as B ; and B, 3 times as many acres as C. They sell the entire crop, amounting to 45 bu. to the acre, at 22^ per bu. What is each man's share of the profit ? 57. If 10 lb. of cheese are equal in value to 7 lb. of butter, and 14 lb. of butter to 5 bu. of corn, and 12 bu. of corn to 8 bu. of rye, how many pounds of cheese are equal in value to 4 bu. of rye ? 5a If 15 bu. of wheat are wortli 18 bu. of rye, and 5 bu. of rye are worth 8 bu. of corn, and 9 bu. of corn ai'e worth 12 bu. of oats, and 16 bu. of oats are worth 20 lb. of coffee, how many pounds of coffee should be exchanged for 20 bu. of wheat ? SOLUTION OF PROBLEMS METHODS OF PROCEDURE There are three methods of solving problems : analysiSj the lit- eral metfiodf and proportion. In each, the relations between the given terms and the required terms are expressed by equations. By analysis, the required term must form one member of an equation, and the given terms the other; no operation is performed upon the required term. By the literal method, no attempt is made to place the required term by itself, but the equation is stated naturally, and the operations are performed upon the terms without discrimination. By proportion, the equation is stated as an equality of two ratios, the antecedent of the second ratio being of the same denomination as the answer, and the consequent, the required term. 180 Illcsthations At 4f each, how many apples can be purchased for 8^ ? Analysis Belation : number of apples = the number of times cost of 1 apple is contained times in cost of all. Solution: since 1 apple costs 4f, as many apples can be bought for Bf, as 4* is contained times in 8f , or 2 apples. Literal Method Belation : cost of all = cost of 1 apple X no. of apples. Solution : let x=no. of apples ; 4x = cost of all ; 4 x = 8 ; x = 2, no. of apples. Proportion Belation : cost of 1 ; cost of all = 1 apple : all apples. Sxj 4 Solution .-4:8 2, no. apples. 1 : x: x=- SOLUTION OF PROBLEMS 181 PROBLEMS OF PURSUIT Analysis proceeds indirectly, introducing a special method for each special case; the literal method proceeds directly, expressing the required term by a letter, which is used as a number. Both methods should be mastered ; the former will give power for an indirect, and the latter for a direct, attack upon a problem. 1. At what time between 3 and 4 o'clock are the hands of a watch opposite to each other ? Necessary knowledge: at 3 o'clock the min. hand is at 12, and the hr. hand at 3; when the hands are opposite to each other, they are 30 min. spaces apart ; while the min. hand advances 60 spaces, the hr. hand advances 5 spaces. Analysis Relation : the min. hand will ad- vance as many min. spaces as the number of spaces it gains in 1 min. is con- tained times in the 1^ number of spaces to be fC gained. The min. hand has advanced from ^ to D, or 46 spaces + B to C; because from ^ to B is 16 spaces, and from C to D is 30 spaces. The hr. hand has advanced from B to C; therefore, the min. hand has gained 46 spaces. Since the min. hand advances 60 spaces while the hr. hand advances 6 spaces, the min. hand gains 65 Proof spaces in 60 min., or y, or \\ of a space in 1 min. Since it gains \\ot 1. \9^ = 12 x 4^*^. aspace in 1 min., it will take as many 2. 40 j^ = 46 -H 4 j^. min. to gain 46 spaces, as }| is con- tained times in 46, or 49 ^^ min. 2. How many minute spaces does the minute hand of a watch gain on the hour hand in 1 minute ? Literal Method Relations : spaces passed by min. hand = 12 times spaces passed by hr. hand ; spaces passed by min. hand = spaces passed by hr. hand + spaces gained by rain. hand. Let X = sp. pas. hr. A., 4j^ 12x = »p. pas. min. A., 49^ X + 45 = sp. pas. min. h. .-. 12x = a; + 45 12x-x = 46 llx = 45 x = 4^ 182 PROBLEMS OF PURSUIT 3. If the three hands of a watch all turn on the same point, how many minute spaces does the second hand gain on the hour hand in 1 minute ? 4. When the hands of a watch are first 20 min. spaces apart between 5 and 6 o^clock, how many spaces has the min. hand gained on the hr. hand since 5 ? Draw a diagram. 5. When the hands are at right angles between 2 and 3 o'clock, how many spaces has the minute hand gained since 2? Draw a diagram. 6. At what time between 5 and 6 o'clock are the hands of a watch first 20 minute spaces apart ? 7. At what time between 2 and 3 o'clock are the hands of a watch at right angles ? a Between 4 and 5 o'clock, when the hour hand is as much after 4 as the minute hand is before 10, how many minute spaces have the hour and minute hands together passed since 4 o'clock ? How many spaces do they together pass in 1 minute ? 9. In Ex. 8, what is the time ? Solve both with and without the use of x. 10. At what time between 8 and 9 o'clock are the hands of a watch together ? 11. A and B start from the same point and travel in the same direction. If A travels 6 miles an hour, and B 4 miles an hour, how far apart are they after 6 hours ? 12. In how many hours will A overtake B, if the latter has 5 hr. the start ? la If they travel in opposite directions, how far apart are they at the end of 6 hours ? 14. If they are 60 miles apart and travel toward each other, how far will A travel before they meet ? 15. Two men, A and B, 26 miles apart, set out toward each other, B 30 minutes after A; A travels 3 mi. an hr., and B 4 mi. an hr. How far will each have traveled when they meet ? SOLUTION OF PROBLEMS 188 16. A fox has 60 of its leaps the start of a hound. While the fox makes 5 leaps the hound makes 4 ; 3 leaps of the fox cover the same distance as 2 leaps of the hound. How many leaps must the hound make to catch the fox ? Analysis Literal Method litlation: the bound must make as Relations : diHi&nce hound mns many leaps as the distance (in fox = length of 1 leap x no. of leaps ; leaps) he gains in 1 leap is contained distance hound runs = distance times in the distance (in fox leaps) to fox runs + start of fox. be gained. Solution: the distance the bound Let 5 x = no. ?p. /ox, 300 goes in 1 leap = the length of I fox then 4x = no. Ip. hound, 240 leaps, because 2 leaps of the hound cover the same distance as 3 leaps of ^©^ 3 a = length 1 h. Ip. the fox. The disUnce the fox goes dur- ^^en 2 a = length 1 /. Ip. ing the same period is | fox leaps, be- cause the fox makes 6 leaps while the 12 ax = distance h. runs hound makes 4. Therefore, in 1 leap, iq ax + 120 a = distance h. runs the hound gams (§ — J) fox leaps, or ^ of a fox leap. .•• 12 ox = 10 ox + 120 o It would take the hound as many „ ^^ .o^^ 1 J. ■ on. t A 1 • 2 ax = 120 a leaps to gam 60 fox leaps, as J is con- tained times in 60, or 240 leaps. x = 60 17. A fox pursued by a hound makes 3 leaps while the hound makes 2 ; but the latter in 3 leaps goes as far as the former in 7. Find the length of 1 hound leap in terms of fox leaps. la Find the distance in terms of fox leaps that the hound gains in one leap. 19. If the fox has 60 of her own leaps the start, how many times will the hound leap before he catches the fox ? 20. If the fox has 60 of the hound leaps the start, how many times will the fox leap before she is overtaken ? 21. A hare is pursued by a hound. The hare makes 5 leaps while the hound makes 3 leaps ; 2 leaps of the Itound cover the same distance as 5 leaps of the hare. If the hare has 50 of her leaps the start, in how many leaps will the hound overtake her ? 184 BUYING AND SELLING BUYING AND SELLING 22. By selling eggs at 6^ each, I shall lose 24^; by selling at 10^ each, I shall gain 24^. How many eggs have I ? Analysis Literal Method Belation: I have as many eggs as Belations: costof all = 1st sell, the gain on 1 egg is contained times in price of all + 24<^ ; cost of all = the entire gain. 2d sell, price of all - 24 f. Solution : since the difference in sell- Let x = no. eggsy 12 ing price on 1 egg Is 4^, the difference 6x + 24 = cost all in gain on 1 egg must be 4^. ' lOx - 24 = cost all Since I lose 24* in one case, and gain .-. 6x + 24 = 10 x — 24 24f in the other, the difference in gain 48 = 4 x on all the eggs is 48^. x = 12 Since the difference in gain on 1 egg is 4ft I must have as many eggs to gain Proof 48^, as 4 f IB contained times in 48f, or 1. 12 x 6 + 24 = 96. 12 eggs. 2. 12 X 10 - 24 = 96. 2a If I gain 2^ apiece by selling eggs at 72^ a dozen, how much apiece do I gain by selling them at 60^ a dozen ? 24. If I gain 2^ apiece by selling eggs at 72^ a dozen, how much apiece do I lose by selling them at 24^ a dozen ? 25. If I sell eggs at 96^ a dozen, I gain on all 48^ more than if I sell them at 72^ a dozen. How many eggs have I ? 26. If I sell eggs at 24^ a dozen, I lose 30^ on all; if I sell them at 60^ a dozen, I gain 15^ on all. How many eggs have I ? 27. If I sell eggs at 12^ a dozen, I lose 3^ apiece. How much a dozen must I charge to gain 3^ apiece ? 2a I sell 8 eggs for a certain price. Had I sold 2 more for the same money, the price of each egg would have been dimin- ished 1^. For how much did I sell each egg? 29. B bought apples at 2 for a cent and the same number at 3 for a cent; he sold them all at 5 for 2^, and thereby lost 2/. How many did he buy ? SOLUTION OF PROBLEMS 185 LABOR PROBLEMS aa A agreed to Vork 24 days for $ 2 a day and his board, and to pay 50^ a day for board when idle ; at the end of the time he received $ 38. How many days W^as he idle ? Literal Method Relation : amount received for la* bor — amount paid for board = 9 38. 20 = 5» 2 = 4 Analysis Relation: he was idle as many days as the amount lost for each idle day is contained times in the entire ^^^ ^ = **^' ^^' *^'^' * amount lost. then 24 - x = no. da. work, 20 Solution: on each idle day he 4S - 2 x = amt. for labor lost e 2 that he might have earned, ^=amt.for board and 60 /> for board, or $2.60 in all. 2 * If he had worked the whole time, 48 — 2x— - = 38 he would have received $ 48 ; he lost o« a — 7ft through idleness, $ 48 - $ 38, or « 10. ^ 4 x - x - 70 Since he lost $2.60 for 1 idle day, he must have been idle as many days as $2.60 is contained timee in Pboof $10,or4day8. j 20 x $2 - 4 x $.60 = $38. 31. A agrees to work for 40 days at $ 1.50 a day and his board, and to pay 50^ a day for board when idle. How much does he lose each idle day ? 32. If he receives $ 20 at the end of the time, how many days was he idle? 33. A agrees to work 30 days at $ 3 a day, and to forfeit $ 1 a day for every day he is idle. How much does he lose each idle day ? 34. If he receives $ 60 at the end of the time, how many days did he work ? 35. A agrees to work 30 da. at $ 3 a day and his board, and to pay $1 a day for his board when idle; at the end of the time he receives $ 40. How many days was he idle ? 186 INVOLVING A PART INVOLVING A PART 36. If A can do a piece of work in 6 days, and B in 3 days, in how many days can they do the work together ? Relation : both can do the work in as many days, as the part they can do in 1 day is contained times in the whole. Solution : in 1 day, A can do ^ of it ; B, | of it ; both, the sum of ^ and |, or y'5 of it. It will take them as many days to do the whole as ^ is con- tained times in } I, or 1 1 days. 37. A pastures 5 cows, and B, 4 cows. If the whole expense is $ 18, how much should each pay ? Relation : A's expense will be 5 times the cost of pasturing 1 cow ; B's expense, 4 times the cost of pasturing 1 cow. Since the expense for cows is $18, the expense for 1 cow is (of 918, or $2. A's expense is 5 x 82, or $10; B's, 4 x $2, or $8. 3a A crew row down stream 8 mi. an hr. and up stream 6 mi. an hr. How far down stream can they row and return in 7 hr. ? Analysis Literal Method Relation: they can go as many Relation: no. hours down + no. miles, as the number of hours re- hours up = 7. quired to go and return 1 mile are Let x = no. miles, 24 contained times in 7 hr. jc Solution : to row 1 mi. down g = ^^' ^^' ^^^^ stream requires ^ hr. ; up stream, ^ hr. ; both ways, ^ + J. or j'^ hr. r = no. hr. up They can go and return as many mi. in 7 hr. as ^f is contained times in •*• I "*■ I ~ "^ 7, or 24 mi. 8 6 3a; + 4x = 168 x = 24 39. A can do a piece of work in 4 days, and B in 5 days. VThat part of the work can A do in 1 day ? 40. What part can both do in 1 day ? How much more can A do in 1 day than B ? How many days will it take them both to do it? 41. C and D together can do a piece of work in 6 days; C alone^ in 8 days. How many days will it take D alone ? SOLUTION OF PROBLEMS 187 42. A, B, and C can do a piece of work in 20 days; A and B, in 40 days ; A and C, in 30 days. In how many days can each alone do it ? 43. Two pipes can fill a reservoir in 8 days ; with the help of a third pipe, they can fill it in 3 days. How many days will it take the third alone to fill it? 44. A and B can do a piece of work in 6 hr. After A has worked alone for 3 hr., B commences and, working alone, finishes the work in 10^ hr. In how many hours can A do the work alone ? 45. A can ride on a bicycle 12 miles an hour, and return on the cars 30 miles an hour. What part of an hour does it take him to ride 1 mile on his bicycle ? 46. How many miles can he ride on his bicycle and return by the cars, in 7 hours ? 47. If a steamer sails 9 mi. an hr. down stream, and 5 mi. an hr. up stream, how far can it sail down stream and return in 28 hr. ? 48. A puts 8 cows into a pasture for 6 months ; B, 10 calves for 8 months ; B pays $ 16. How much should A pay, if 4 cows eat as much as 5 calves ? 49. A and B rent 32 A. of land for $ 63. A agrees to take 12 A. of timber, and B, 20 A. of meadow land. How much should each pay, if 3 A. of timber rent for the same as 4 A. of meadow ? 50. Henry has 8 marbles, James 10, and Walter none; they divide the marbles equally among them, and Walter pays 6^ to Henry and James. How much should each receive ? 51. A, B, and C enter into partnership. A puts in $ 6000 for 4 mo.; B, $8000 for 3 mo.; and C, $4000 for 6 mo. If their profits amount to $ 5040, what is each man's share ? 52. A, B, C, and D plant 5464 acres of corn. A puts in 1} times as many acres as B ; C, 2f times as many ; and D, 5^ times as many. If they market their crop for $ 43,712, what is each man's share of the profits ? Letx = : number ¥= : tncreote 6x_ 8 : no. increased 6x_ 3 " 16 6z = 45 X = 9 188 A PART MODIFIED A PA£T MODIFIED Sa What number increased by } of itself becomes 15? LlTBRAL MODIFIKD LITERAL MkTHOD Relation : required uoiuber + f Relation : required number + } of itself = 15. of itself = 15. Solution : a number increased by § of itself becomes | of itself. Since I times the number is 15, the number is 15 -4- ^, or 9. NoTB. I X No. » 15; No. « 15 •!> |, because either of two factors is equal to their product divided by the other. Analysis: since } of the number is 15, \ of the number is ^ of 15, or 8 ; }, or the number, is 3 times 3, or 9. See p. 98, Ex. £50, Note. Proof. 9 + J of 9 = 16 54. By selling a watch for $40, a man lost ^. What was the cost? Business usage : the gain or loss is always some part of the cost. Relation : cost — i of the cost = selling price. Solution : the cost diminished by | of itself becomes f of itself. Since | times the cost is $40, the cost is $40 -^ j, or $60. 55. An agent sold an article for $ 100 on a commission of ^. What were the proceeds ? Business usage : when an agent buys, his commission is some part of the buying price ; when an agent sells, his commission is some part of the sell- ing price. Relation : proceeds = selling price - commission. Solution : since an agent sold on commission at y^,, his commission was ^ of $ 100, or $ 10 ; the proceeds were $ 100 - $ 10, or $ 90. 56. An agent bought an article for $ 100 on a commission of ■j^. What was the entire cost ? Relation : entire cost = buying price + commission. Solution : since an agent bought on commission at j^j,* ^Js commission was ^ of $ 100, or $ 10 ; the entire cost was $ 100 + § 10, or $ 1 10. SOLUTION OF PROBLEMS 189 57. What number increased by 5 times itself becomes 30 ? by 4 times itself, 20? sa What number increased by J of itself becomes 20 ? by ^ of itself, 30 ? 59. What number diminished by J of itself becomes 30 ? by | of itself, 20? 60. A horse cost $ 60 and was sold at a gain of \. What was the selling price ? 61. A horse cost $ 60 and was sold at a loss of J. What was the selling price ? 62. By selling a horse for $ 40, 1 lose \. How much did he cost ? 63. By selling a horse for $ 36, I gain ^. How much did he cost? 64. By selling a horse for $ 40, I gain \. By how much must I increase my price to gain ^? 65. If my gain was j, or $40, what was the selling price ? the cost ? 66. If my loss was ^, or $20, what was the cost? the selling price ? 67. If I buy at $3 and sell at $4, what part do I gain? ea If I sell J of an article for what the whole cost, what part of the whole do I gain ? 69. An article was sold at a gain of -fj^ ; if it had cost $ 120 more, the same selling price would have entailed a loss of -^. Find the cost. 7a A man sold an article for 40^, and thereby gained J as much as if he had sold it for 60^. What was the cost? 71. An agent sold flour for $200 at a commission of ^. What was the commission ? the proceeds ? 72. An agent bought flour for $200 at a commission of -^. What was the commission ? the entire cost to his employer? 7a An agent sold flour at a commission of ^, and received as commission, $25. What were the proceeds ? pp:rcentage TERMS AND RELATIONS In many cases, it has become custom- ary to use the term per cent, written %, in place of hundredths. The number on which the per cent is computed is the base; the product is the percentage; the base -f the percent- age, the amount; the base — the per- centage, the difference. % means hundredths. See p, 114* A per cent expression may be changed to a fraction or an integer, and conversely. Illustration .06 of 200 = 12 Or, 6% of 200 = 12 200, base 12, percentage 212, amount 188, difference ejo means .06 5. .16 to per cent 6. .33J to per cent. 7. ^ to per cent, a 4 to per cent. Reduce : 1. 16% to a decimal. 2. 33 1 % to a decimal. a 33 J % to a fraction. 4. 400% to an integer. .33^ = \. 9. State rapidly the per cent equivalents : f , |, f , f , |^, J, J, J, TVif^iiii*»TV»f Seep.m. 10. State rapidly the fractional equivalents: 62J%, 87^%, 37.1%, 33.^%, 16|%, 2^%. 40%, 75%, 60%, 20%, 80%, 8^%, fo, 12^-%, 50%. See p. IIJ,. 190 Ex. a 33}% Ex. 4. 400% Ex. 7. \ Ex. a 4 .331=331% m = 4oo% PERCENTAGE 191 CASE I — DIRECT U. What is 6% of 60 ? 12. What is 16 J% of 486 ? 50 486 .06 .16^ 3.00 "TT 6% of 60 is .06 of 50, or 3. 16|% of 486 is J of 486, or 81. Note. — A per cent expression must be reduced to a decimal or to a common fraction, before it can be used as a multiplier. 13. A farmer having 360 sheep, lost 5%. How many had he left? S60 = no. sheep Qg He lost .05 of 360 sheep, or 18 sheep ; he had left the difference, or 18.00 = no. lost 342 sheep. S42 = wo. left 14. What is 16% of 5250 ? 18% of 3825 ? 66f % of 3483 ? 15. A man wills 17% of his property to his son, and 18% to his daughter. If his property is worth $ 8200, how much will each receive? 16. A has $625; B has 86% as much as A; C, 78% as much as B. How much has C ? 17. A lady buys 55 yards of muslin at 8^ a yard, and 27 yards of cloth at 65^ a yard. If she pays 66% of the bill, b<>\v unieh does she still owe ? la A man's salary is $75 a month; if he spends 65% ot his salary each month, how much will he have at the end of 6 months ? 19. A gentleman gave $ 11.20 to his children ; his sons received 65% of the money, and his four daughters the remainder. How much did each daughter receive ? 20. A man has a library of 1600 volumes ; 14% are biography ; 62%, history, and 83^% of the remainder, fiction. How many volumes of fiction are there in his library ? 192 INDIRECT CASES CASE II — INDIRECT 21. ber? 3 is 6% of what num- .06 The equation is readily formed by substituting ' = * for 'is*; 'x' for *of»' and *No.' for 'what number.* Since either factor is equal to the product divided by the other factor, No. = 3 ^ .00, or 60. Note. — A per cent expressioD must b« reduced to a decimal or to a common fraction, before it can be used as a divisor. 23. A farmer lost 18 sheep, or 5% of his flock. How many had he at first? 22. Of what number is 81, 16i%? 161% ^o.=81 No. = 81-t-- = 486 The sign of multiplication may be omitted. Since either factor is equal to the product divided by the other factor, No. = 81 -i- i, or 486. 5% F= no. lost 18 = no. lost .-. 5%F=^18 360 Of what number: 24. Is 12, 50%? 25. Is 368, 23%? 26. Is 57, 15%? 27. Is 522, 18%? Proof 360 = no. at first .06 18 = no. lost 2a Is 10.5, 12i%? 29. Is 40.5, 7 J %? 3a Is 4578, 84%? 31. Is 7735, 85%? 32. What is a man's income if 31 J % is ^600 ? 33. A general lost 16 J % of his army; 315 killed, 110 prisoners, and 70 deserters. How many men had he at first ? 34. A man paid $750 for a house, which was 24% of what he paid for 160 A. of land. What was the cost of the land per acre ? 35. Mr. Turner earns $85 a month; his salary for the year is 68% of his brother's salary for 8 months. How much does his brother receive a month ? »1 = n x4SG R- 81 = i = ie\f. The \ must be reduced to % be- cause the 1 answer is to be expressed in 0/,. PERCENTAGE 193 CASE III — INDIRECT 36. 3 is what % of 60 ? 37. 81 is what % of 486 ? 3 = Ex 50 The equation is readily formed by substituting * = ' for ' is ' ; • K,' for %, and 'x' for 'of.' 38. A farmer having 360 sheep, lost 18. What % did lie lose ? 18= Rx 360 ^ - y lielation : the number lost is some R = ^^ = — z=5% % of the whole number. 360 20 ' What % : 39. Of 45 is 15? 42. Of 1311 is 437? 40. Of 72 is 18 ? 43. Of 2288 is 286 ? 41. Of 78 is 63? 44. Of 2700 is 300? 45. A man worth $250,000 lost $27,500. What % remained? 46. Mr. Brown works 6 days each week for 44 weeks of the year. What per cent of the time is he idle, counting 300 working days to the year ? 47. The population of a certain town is 57,500 ; of these, 3450 are Irish; 22,540, Spanish ; and the remainder, English. What per cent of the population is English ? 4a A man having $ 1275, spent $210 for a carriage, $112 for a horse, and $86 for a harness. What per cent of his money remained ? 49. A rectangular field 75 feet long, is J^ as wide. The breadth is what per cent of the length ? 50. The product of two numbers is 11,250; the first is 126. What per cent of the first is the second? AMER. ARITH. — 13 194 INDIRECT CASES CASE IV — INDIRECT 51. What number increased by 6% of itself, becomes 53 ? 100% X=theno. 6% N=the increcue Proof 60 = the no. ,06 3 = the increase fo N=th€ no. incr. 53 = the no. incr, 106%N=53 1.06 63 = the no. incr. 52. What number diminished by 16| % of itself, becomes 405 ? 100% N= the no. 16\% N= the decrease 8S\% N= the no. deer. 405= the no. deer. ,\ 83l%N=405 N =405+1=^486 Pboof 486 = the no. .83^ 81 = the decrease 406 = the no. deer. 53. After losing 5% of his sheep, a farmer had 342 left. How many had he at first ? 100%F=no. at first 5% F=no. lost 95%F=no. left 342 = no. left .-. 95%F=S42 F=^ = 360 .95 WhaJt number increased by : 54. 66f % of itself becomes 30 ? 55. 6% of itself becomes 212 ? 56. 25% of itself becomes 60 ? Proof 360 = no. at first .05 18 = no. lost 342 = no. left What number diminished by : 57. 25% of itself becomes 30 ? 58. 6% of itself becomes 188 ? 59. 10% of itself becomes 81? PERCENTAGE 195 MISCELLANEOUS PROBLEMS 6a After 16% of a heap of grain was taken away, there remained 252 bushels. How many bushels were there at first? d. In 1890, the population of a town was 17,280, which was 35% more than in 1880. What was the population in 1880 ? 62. My crop of corn this year is 12^ % greater than last year, and I have raised, during the two years, 3825 bushels. What was my last year's crop ? 63. A young man having received a fortune, deposited 60% of it in bank ; he afterward drew 30% of his deposit, and then had $ 7560 in bank. What was his entire fortune ? 64. Mr. Black's property is valued at $ 15,000, and 85% of his property is 2% more than his brother's property. What is the value of his brother's property ? 65. At a forced sale, a bankrupt sold his farm for $ 6642, which was 18% less than its real value. What was the value of the farm ? 66. Into a vessel containing pure vinegar, there were poured 12^ gallons of water, which was 16|% of the mixture. What was the quantity of pure vinegar ? 67. A man paid $ 16.20 for the use of land which cost $ 360. What % did the owner realize on his investment ? 68. From a cask containing 24 gal. 3 qt., 17 gal. 3f qt. leaked out What % leaked out ? 69. A fruit grower, having sent 2200 baskets of peaches to Philadelphia, found that 11% of them had decayed. If he sold the balance at 76^ a basket, what sum did he receive for his peaches ? 70. On Jan. 1, a man weighed 160 lb. In January he lost 2^% in weight, and in February gained 2j%. What % of his weight Jan. 1 was his weight on the first day of March ? 71. How many gallons of water must be mixed with 70^ gallons of wine, so that the mixture may contain 6% of water ? 196 PROFIT AND LOSS PROFIT AND LOSS In buying or selling, the gain or loss is If I buy for $ 100 and always some % of the cost. »«" ^^r « 110, the gain is ^ ' "^ «10, or 10% of the cost. This principle is established by business usage. 72. A boat was bought for $ 9136.50, and sold at a loss of 3%. What was the selling price ? f 9136.50 = cost Note. — Since our smallest coin is Q^ one cent^ we give the answer to the nearest cent. The loss is $274.10. f 274-095 = lo88 Count 5 or more mills as 1 ^ ; neglect 9136.50 less than 5 mUls. $886240 = selling price 73. The profit on the sale of a horse was $39.20, or 14%. What was the cost? 14% C:^ gain Proof f 39.20 = gain 1280 =cost .-. 14% C= 39.20 ^ n_S9.20_^gQ 1120 ^"TTT" 280 Cost = $280. $ 39.20 = gain 74. A house was sold for $320.32 at a gain of 12%. What was the cost ? 100% C = cost 12% C = gain Proof 112% C = selling pnce ' ^ 286 = cost 320.32 = selling price .'. 112% C = 320.32 J^/^'^ = ga^° = 2J2 ♦ 320.32 = selling price Cost = 286. PERCENTAGE 197 75. I sold a plow for $ 16.32, and thereby lost 4%. What was the cost ? 76. My sales exceeded the cost by $ 240 ; the sales were $ 560. What was the gain % ? 77. I sold silks for $270.00, thereby losing 10%. What sell- ing price would have made the gain $ 26.50 ? 7a A merchant bought 540 yards of muslin at 7 f, and sold it at a reduction of 2^%. What was the entire loss ? 79. I bought a quantity of wheat at 75^ a bushel. At what price must it be sold to gain 16%? 80. I bought 75 horses at $ 5(j each, expenses, $S. If I lost 8 of them, at what average price must I sell them to gain 15%? 81. By selling for $2000, I gained 14% of the selling price. Had I gained 14% of the cost, how much would 1 have received? 82. How shall I mark shoes that cost $2.50, so that I may deduct 20% from the marked price and still make 10%? 83. I sold a carriage at 20% gain, and with the money bought a horse which I sold for $168.30, thereby losing 15% of the selling price of the carriage. What was the cost of the carriage? 84. I began business with $24,000; gained 16% the first year and added it to the capital. What was my capital at the begin- ning of the second year ? 85. 4% of my capital was invested in 18 watches at $180 a dozen; 7% of my capital was invested in clocks at $2.10 each. How many clocks had I ? 86. A merchant marked goods at 16 J % above cost, and sold the goods at this marked price for $ 350, What was the cost ? What % would he have gained by selling the goods for $375? 87. A man gained 33^% on the sale of a horse, 20% on the sale of a carriage, and 50% on the sale of a buggy. If the selling price of each was $ 150, what was his entire gain ? 88 A man lost 10% of his money, then gained 10% of what he had left, and then had $396. How much had he at first ? 198 COM^IISSION COMMISSION A person may be employed A farmer takes to a grocer 10 bar- to buy or sell for another. The ^*» ^^ apples to be sold at 1 2 a barrel, employer is the principal, the »«^»ng ^ Pay 1^% commission for 1 , ,, ^ XL • selling them. It is just that the employed, the agent; the price ^^^^ ^^^^^^ ^^^p ,,,^ ^^ ^^^ ^^^^ paid for the service, the com- a manufacturer sends his agent mission ; the amount returned % 1060 to buy wool, after deducting to the principal, the nef proceeds. » commiasion of 6%. If the agent „ ,, takes 6% of $1050, he will take 6% Business Usaob * u / u * ., , f of what he pays for the wool, and If an agent aelUy his commis- also 6% of what he keeps. It is not sion is some per cent of the sales, just that he should receive 5 % of If an agent buys, his commit- ^*»a^ *»« keeps, because he performs sion is s(me per cent of tiie pnr- ''^ ^^^^ ^°' ^^' ^"^ ^^ ^ ^"^^•'^^ ^ , 6 % of what he pays for the wool. 89. Find the amount of sales, when an agent receives $ 40 at a commission of 20%. 20% S = 40 jg __ ^^ _ £QQ Since the agent sells, his commis- .20 sion is 20 % of sales. Sales = f 200. 90. Find the amount of sales, when the principal receives $ 40 ; agent's commission 20%. 100% S = sales 20% S = com. Proof 80% S = proceeds f 50 = sales f 40 = proceeds j^ foS = 40 10 = com. ,Q j60 ^^~sd^^^ ♦40 = proceeds Sales=f50. PERCENTAGE 199 91. At 3%, what is an agent's commission on $ 1150 sales ? 92. At 3%, what is an agent's commission on $ 1150 purchase ? 93. Find the commission when an agent receives $259.60 to be invested in goods, after deducting his commission of 10%. 94. How many pounds of sugar at 8^ a pound can an agent buy for $ 15.99, after deducting his commission of 2^% ? 95. Find the rate per cent of commission when $ 2.40 is paid for a sale of $ 160. 96. Find the amount of sales when a commission of 2\fo pays the agent $ 6.48. 97. Find the net proceeds from the sale of 36 barrels of sugar at $4, commission 6J%. 98. Find the amount of the purchase when an agent receives i 351 to invest in sugar after deducting his commission of 8^. 99. My principal instructed me to invest $ 1220 in wool, and sent me a draft for $ 1220 plus 2% commission. What was the amount of the draft ? 100. My principal instructed me to invest $1244.40 in wool after deducting my commission of 2%. What was my commis- sion? 101. After selling wheat, an agent deducts $56 commission, and sends his principal $ 2744. What rate of commission does he receive ? >02. An agent bought 4000 bushels of oats ; his commission at 2Xfo was $ 28 ; charges for freight and storage, $ 5'2. How much per bushel did the oats cost the principal ? 103. I sold on a commission of 4^%, 320 barrels of sugar at $ 18, and 60 barrels of oil, 44 gallons to the barrel, at 15^ a gallon. How much did I remit to my employer ? 10*. An agent sells 1200 barrels of apples at $4.50 a barrel, and charges 2^% commission. After deducting his commission of 8% for buying, he invests the net proceeds in flour at $5 a barrel. What is his entire commission ? 200 TAXES TAXES In some states, every male citizen is annually taxed a small amount, jwll tax, without regard to his property ; property owners pay an additional tax. In these states, the entire tax is dimin- ished by the sum of the poll tax, and the remainder is divided by the assessed value of all the taxable property, to find the tax on each dollar. Tfie collector usually receives some % of the amount collected. 105. A tax is $16,020; the taxable property $784,760; the number of polls at $1.25 is 260. A*s property is $ 7800; what is his tax ? f 1.25 f 7800 = As property eeo .02. fS25 = poll tax $156.00 = As property tax _16fi20 = whole tax ^ ^^ ^ ^,^ ^^^ ^^^ f 15f695 = property tax 784750 )15695.00(. 02 f i57.^5 = A's whole tax 15695.00 106. The tax on A*s property at 4 mills on a dollar is $ 615.80. What is the assessed value of his property ? 107. The school tax of a certain town is $ 4782. If the rate of taxation is 3 mills on a dollar, what is the amount of taxable property ? loa A town is to be taxed $ 13651.48 ; the taxable property is $865,432; the number of polls at $1 is 670. What is B's tax, whose property is assessed at $9720? 109. The whole amount of taxable property in a certain town is $ 386,722 ; there are also 1560 polls at $ 1.50 each. If the rate of taxation is 5 mills on a dollar, what is the tax ? 110. A tax of $ 14,846 is to be assessed on a certain village ; the property is valued at $ 1,060,000, and there are 2123 polls at $ 2. What is the assessment on a dollar ? What is Mr. Doan's tax, whose property is assessed at $ 11,600 ? PERCENTAGE 201 TRADE DISCOUNT Wholesale dealers and manufacturers issue price lists of their goods. As the market varies, they change the rate of discount instead of changing the fixed price. They frequently offer other discounts and an additional discount for cash. 111. A man sold a bill of goods, list price $ 100, on 4 mo. at .'), 10, and 6% off for cash. What was the net price for cash ? ^100 = list £ — ^^^ ^*^* Meaning : If the goods are paid for at the ^95 = after 1st end of 4 mo., 5% will be deducted from the list g r/i _ Qnfi fiio price, and 10% from the remainder. If cash is ! — ~ * paid at the time of purchase, an additional $85.50 = after 2nd discount of 6% will be deducted from the last 5.13 = cash dis. remainder. $80.37= net price 112. A man sold a bill of goods, list price $ 100, on 4 mo. at 10, 0, and 6% off for cash. What was the net price for cash ? 113. If the purchaser decides not to pay cash, is one of the above offers better for him than the other ? 114. A merchant offered a bill of goods, list price $100, on 3 mo. at 8, and 4% off for cash. What was the cash price ? 115. A merchant offered a bill of goods, list price $100, on 3 mo. at 4%, and 8% off for cash. What was the cash price ? 116. If the purchaser had decided not to pay cash, which of the offers would have been to his advantage ; that in Ex. 114, or that in Ex. 115 ? 117. A bill of goods, list price $100, was offered at 45% dis- count; or at 30% and 15% off. Which offer was better for the purchaser ? lia Which is better for the purchaser, 55% discount, or two successive discounts of 50% and 10%? 202 INSURANCE INSURAJNCE An agreement to pay for loss or damage is insurance. The indemnity may be against loss on i^vo^rty, property insurance; by fire,^re insurance; or by accidents of navigation, marine insur- ance. The indemnity may also be against personal injury, or loss of life, life insurance. The contract or agreement between the insurance company and the person protected is the policy; the sum paid for insurance, the pretnium. The premium is some per cent of the amx)unt insured. 119. A store worth $ 20,000 was insured for ^ of its value at 1\%' What was the premium ? $16000 =amt. insured Q2 1 Relation : the premium is some % of the amount insured. $ 200.00 = premium 120. Find the premium to be paid for insuring a person's life for $ 5000, at an age for which the rate is 2J%. 121. For what sum should a cargo worth $ 60,000 be insured at 4%, so that, in the event of loss, the owner may receive both the value of the cargo and the premium ? 122. A man paid $125 for insuring at |%. What was the amount insured? 123. I paid $ 53, including cost of policy, $ 1.75, for insuring my property to the amount of $ 8200. What was the rate ? 124. What is the amount of the annual premium at $ 35.20 per 3 1000, on a life policy of $ 6500 ? 125. I paid $40.50, at 1^%, for insuring my house for | its value. What is the value of the house ? 126. A man 35 years old, takes out a $6000 life insurance policy payable when he reaches the age of b5. If he pays an annual premium of $ 63.10 per $ 1000, how much money will he have paid in if he lives till the policy falls due ? PERCENTAGE 203 DIFFICULT PROBLEMS Care should be used in deciding what the *% is of.' 127. I mix 20% of rosin with 8 lb. of tallow. How many pounds of rosin in the mixture ? 100% .yf = mixture S0% M = rosin 80% M= tallow The 20% is of the mixture and not of the S lb. = tallow rosin ; another form of statement would be •'• 80%M-8 *'20% of a mixture is rosin." .80 Ans. S lb. rosin. Generally, it is best to represent what the *% is of by 100% of that thing. 12a A quantity of sugar was sold at 10 per cent gain. If it had cost $ 120 more, the same selling price would have entailed a loss of 10 per cent. Find the cost of the sugar. 100% C = cost 100 % C = cost 100% C+ 120 = sup. cost 10% C = gain 10%C-\- IS = sup. loss 1lO%C = sell p, 90% C+ 108= sell. p. .'. 110%C=90%C+108 110% C-90% C=108 20% C= 108 €='-^ = 640 .20 Cost = f 540. Sometimes it is more convenient to represent what the '% is of ' by a number. 129. I mix 20% of rosin and 80% of tallow. AVhat per cent of the weight of the tallow, is the weight of the rosin ? Let 6 lb. = wt, mixture Since rosin is some % of tallow^ jl^L. l = Rx4 J lb. = t€t. rotin R = lz^25H. 4 lb. = u>t. tallow 4 204 DIFFICULT PROBLEMS In some problems, there is % of two different bases, and neither base is given. 130. How many per cent above cost must a man mark his goods, in order to take off 10 per cent and still make a profit of 12^%? 100 %C = cost 200% M= marked p. li^-% C = gain 10% M = deduction IJil^C^sellp. 90%M = seilp. .-. 90%M=llg{%C Mark, p,i»t5% above cost. In some problems, two equations with two unknown quantities axe necessary. 131. My agent sold my flour at 4% commission. Increasing the proceeds by $168, I bought wheat, paying 2% commission; wheat declining 3%, ray loss, including commissions, was $30. AVhat was the selling price of the flour ? 100 % S = selUng flour 100 % C = cost teheat 4%S = com. S%C = com. 96% S = proceeds 102 % C = invst. wheat 96% S -\- 168 = invst. wheat 5%C = dec. and com. 4%S^5%C=S0 (1) 96% S- 102 % C = - 168 (2) Since the loss, including com- missions, was §.30, we obtain (1). 96%S-\- 120%C— 720 (3) Since the investment in wheat 96% S- 102%C = - 168 (2) was $168 more than the proceeds 222^ C — 888 r4'i of the flour, we obtain (2). C = 400 We solve these equations as on COH + Com. Z UOS. P, '«' ■• •'\^, (>) ^''^ (•"'^ ■' (^) '"■" Sell. p. Flour = %iSO. (3) gives (4) ; . . . . PERCENTAGE 205 132. What per cent would a dishonest dealer gain by using a false weight of 15 oz. instead of a pound ? 133. What per cent does a customer lose, if his grocer uses a false weight of 15 oz. instead of a pound ? 134. The lead ore from a certain mine yields 40% of metal, and of the metal, |% is silver. How many ounces of silver will be obtained from a ton of ore ? 135. A brewery is worth 4% less tlian a tannery, and the tan- nery 16% more than a boat; the owner of the boat has traded it for 75% of the brewery, thus losing $103. How much is the tannery worth ? 136. 8 lb. of a certain article loses 3 oz. in weight by drying. What per cent of the original weight is water? 137. A man sold an article at 20% gain ; had it cost f 300 more, he would have lost 20%. What was the cost ? 13a How must I mark goods so that I may deduct 10% from the marked price and still make 17% ? 139. A merchant marked damaged goods at a certain per cent below cost, but sold them for cost at an advance of 11 J % on the marked price. What was the marked price ? 140. An agent sold my corn, and, after reserving his commis- sion, invested all the proceeds in corn at the same price; his com- mission, buying and selling, was 3% each, and his whole charge $ 12. For how much was the corn sold ? 141. A dealer sold wheat, losing 4% ; keeping $ 18 of the pro- ceeds, he gave the remainder to an agent to buy corn, 8% com- mission; his loss together with the commission was $32. How much did the agent pay for the corn? STOCKS AND BONDS TERMS AND RELATIONS For the prosecution of a business enterprise, individuals sometimes form a stock company. Officers are elected, and a charter (authority to do business as an individual) is ob- tained from the state. The engraved part of a certificate states the number of parts (shares) into which the property (stock) is divided, the face value (par value) of each part, etc. The written part states the number of shares bought, the name of the purchaser, etc. 206 From a study of the above certificate, the pupil should an- swer the following questions : Where is the Altar mine situated ? Who was the president of the Altar Mining Co.? the secretary ? Into how many shares is the property divided ? What is the par value of 1 share ? What is the capital stock ? Who is the owner of this certificate ? How can this certificate be transferred? STOCKS AND BONDS 207 Usually, shares do not sell for their par value, but for less John Fluker bought the stock when i;it a discount), or for more (at »^ ^^ *^7^ '««"«'^' ^^«2 a share, or ^ . ^ ^' ,. \ at 2% of the par value, or at 98% a premium), according to the discount. prospects of the company. If the business is success- ful, the profits (dividends) are The profits of the company at the T • 1 J J- i. xi end of the first year were ft 660,000, divided according to the num- , ,. ., , ,«„/ I \ a ° and a dividend of 6 % was declared. ber of shares. Brokers buy and sell stock After the dividend, James Lyman for their employers, charging bought the 50 shares of John Fluker, something (brokerage) both for through a broker, at $110 a share, buying and selling. P^^'^g * % ^'^^^^^^^e. 1. What is the number of this certificate of stock ? 2. To whom was the certificate issued ? For how many shares ? a ^Vhat was the par value of each share? How do you know ? 4. What was the market value of each share at the time of issue ? How do you know ? 5. If the earnings were $ 660,000, what dividend might have been paid on each share ? 6. How many dollars were paid as dividend on each share ? 7. What dividend, in all, did John Fluker receive? a What was the par value of each share after the dividend ? Does the par value ever change ? 9. What was the market value of each share after the divi- dend ? Does the market value change ? la What was John Fluker's entire gain from the stock ? 11. What % brokerage did James Lyman pay? What part of a dollar on each share, if the brokerage is some % of the par value? 12. How much brokerage in all did James Lyman pay ? What was the entire cost of the stock to James Lyman ? 208 PROBLKMS PROBLEMS The par value of one share is f lOOy unless otherwise stated. This understanding saves confusion, and makes it unnecessary to call attention to the par value. The cost, the selling price, the dividend, the brokerage, each is some % of the imr value. These tenns must be some- per cent either of the market value or of the par value. The par value never changes, the market value is constantly changing ; hence the former is selected. All examples in stocks may be analyzed by using $1, 1%, or 1 share as the unit. The last is the simplest, because it eliminates % as much as possible. The student should be able to declare in dollars the cost of one share, the selling price of otie sliare, the dividend on one sliare, and the brokerage of one share, however the same is stated. la Cost. What is the cost of 1 share of 6% stock at 60 ? Ans. 9 60. The cost is 60 % of $ 100, or $ 60. 14. Selling price. What is the selling price of 1 share of 8% stock at 70? Ans. $ 70. The selling price is 70 % of $ 100, or $ 70. 15. Dividend. What is the dividend on 1 share of 6% stock ? Ans. $ 6. The dividend is 6 % of $ 100, or $ 6. 16. Brokerage. What is the brokerage on 1 share of 6% stock ati%? Ans. $ |. The brokerage is | % of $ 100, or $ |. Note. — On the New York Stock Exchange, brokerage higher than 4% is not allowed. How many shares in : 17. $600 6% stock? 19. $1200 2% stock? la $800 3% stock? 20. $1500 8% stock? Ex. 17. 6 shares. The par value of one share is $ 100 ; ^600 is the par value of as many shares as $ 100 is contained times in $600, or 6 shares. STOCKS AND BONDS 209 Brokerage \ffoyfind the cost of: 21, $4()0stock at75. 2a $ 800 stock at 80. 22. $320 stock at 60. 24. $040 stock at 90. Ex. 21. 1300.50. The entire cost of 1 share is $ 75^ ; the cost of 4 shares is 4 times 975|, or $300.50. Brokerage \%,find the net proceeds of: 25. $400 stock sold at lb\. 27. $ 800 stock sold at 80. 26w $ 600 stock sold at 90^. 2a $ 240 stock sold at 60. Ex. 25. $300. The net proceeds of 1 share is $75^ - $ J, or $75 ; of 4 shares, 4 times §75, or $300. Find the dividend on : 29. $ 400 3% 's bought at 80. 31. $ 1200 4%'s sold at 75. 30. $ 300 6% 's bought at 00. 32. $ 1500 5%'s sold at 90. Ex. 29. $ 12. The dividend on 1 share is $3; on 4 shares, 4 times $3, or $ 12. How many shares may be bought for : 3a $900 at 89i, brok. \%? 35. $1200 at 59J, brok. J%? 34. $600 at 74f, brok. \%7 36. $1600 at 79|, brok. \%? Ex. 33. 10 shares. The cost of 1 share is $90; $900 will buy as many shares as $90 is contained times in $900, or 10 shares. How much stock gives an income of: 37. $200; stock 5% ? 39. $600; stock 3% ? 3a $400; stock 4% ? 40. $800; stock 8% ? Ex. 37. $4000. The income on 1 share is $5 ; ii will take as many shares to yield $200 as $5 is contained times in $200, or 40 shares. 40 shares = $4000 stock. What % mil I realize on my investment when : 41. 5%'s are bought at 40 ? 43. 3% 's are bought at 60 ? 42. 8%'8 are bought at 60 ? 44. 4%'s are bought at 80? Ex. 41. 12^ %. One share costs $40 and gains $5 ; the gain % is $5+$40, orl2J%. ▲MKR. ARITH. — 14 210 PROBLEMS 45. Find the cost of $6000 S% stock at 80, brokerage J%. Given terms: no. shares, 60 ; first cost of 1 share, $80; brokerage on 1 share, $ J. §80 = cost 2 sh. - = brok. I sh. Since the first cost of 1 share Is 8 $80, and the brokerage $ ^, the en- I _ . tire cost is the sum, or $80 J. The 80^ - entire c. 1 sh. ^^^ ^^ ^ ^^^^^ ^ ^ ^^^ ^ ^^^ ^^ ^ $4006.25. $4006.i5 = entire c. all. 46. Find the proceeds of $5000 3% stock at 80, brokerage J%. Given terms : first sell. p. 1 share, $ 80 ; brokerage 1 share, $ ^. §80 = sell. p. 1 sh. i _ brok. 1 sh. Since the selling price of 1 share 8 is $80, and the brokerage $|, the ^ proceeds will be the difference, or 79-^= proceeds. ^^g^ ^Pl^^ proceeds of 60 shares 60 will be 60 times $ 79|, or $ 3993.76. $3995.75 = entire proc. 47. How much 3% stock (50) can be bought at 98 for $ 2456.25, brokerage J% ? Given terms : first cost 1 share $ 49 ; brokerage on 1 share, $ |. f 49. = cost 1 share. ^125 — brokerage. Since the first cost of 1 share is $49, and the brokerage $ J, the en- $ 49.185 = entire cost 1 share. ;. ' . .„ .,„ ^ ^^«^ oW a<. 49.2e5)24f6.S5{50 ^ ^^^^^ ^^^ ^ ^^^^j^^^ ^^ z4o6 z5 4a What is the dividend on $ 2500 3% stock (50) ? Given terms: dividend on 1 share, $ 1.60 ; no. of shares, 60. ft 1.50 Since the dividend on 1 share is 50 ■ $1.60, on 60 shares it is 60 times 1 75 $1.60, or $76. STOCKS AND BONDS 211 49. How much 5% stock at 80, must be bought to get an annual income of $240? Given «cn»« .• dividend on 1 share, $5; entire income, $240; cost of 1 share, 9 80. Since the gain on 1 share is $5, it will take 5)g40 as many shares to gain $240, as $5 is con- ^ tained times in $ 240, or 48 shares, or $ 4800 stock. 50. Find the rate per cent of dividend when $ 6200 stock yields $310. Given terms : no. shares, 62 ; entire income, $ 310. Since 02 shares yield an income of $310, 6SM10{5 1 share will yield gJj of $ 310, or $ 5. SIO Since the dividend on 1 share is $5, the stock is 6% stock. 51. If 3% stock is at 80, what rate % will a person receive on his investment ? Given terms: investment in 1 share, $80 ; dividend on 1 share, $3. Re- lation : income = B x investment. ' S = R X 80 g ^ Since the income is some % of the invest- ■R = — = ^'i % ment, the rate is $ 3 -r- $ 80, or ^ %. 52. What must be the price of a 3% stock to equal a 4% stock at 80? Given terms : income on 1 share tirst stock, $ 3 ; income on 1 share second stock, $4; investment in 1 share second stock, $80. Relations: income on 2d = ^ x invest, in 2d ; rate on first investment = rate on second investment. 4 = Rx 80 y,_ 4 _ A(v Invst. = — zz60 80 ^ .05 5% Invst = 3 ^^** ^ '^^^^ ^% stock, $60 53. What must be the price of stock when $8300 stock is bought for $4150? Given terms no. shares, 83; cost of all, $4150. 212 PROHLEMS 54. What is the cost of 4000 3^% government bonds ($ 1000) at 01 1, brokerage J % ? 55. I buy 200 shares railroad stock at 102J, and sell them at 104 J, brokerage J% in each transaction. How much do I gain? 56. My broker buys for me 300 New York state bonds at 72^, brokerage J%. How much do the bonds cost me? 57. What are the proceeds from the sale of 45 shares of Ohio state bonds at 93, brokerage |%. 5a Find the proceeds of $ 1000 New York city 4J's at 101 J, brokerage J%. Note. — N. Y. city 41*8 means New York city bonds yielding 4i% dividend. 59. What are the proceeds from the sale of 80 shares of Ohio state bonds at 90, brokerage J % ? 6a If Altar mining stock sells for 83J, what are the proceeds from 150 shares, brokerage i% ? 61. How many shares of stock at 62 J, brokerage J%, can be bought for $ 9337.50 ? 62. A broker invested $24,062.50 in Union Pacific bonds at 4% discount, brokerage J%. How many shares of stock will his principal receive ? 63. A pays a debt of $ 2045 with bank stock at 102^. How much stock w^ill the creditor receive ? 64. The Altar Mining Company declares a dividend of 6J%. How much will a stockholder receive who holds 50 shares ? 65. What income will be realized from investing $58,080 in 4J% stock at 90 J, allowing \% for brokerage? 66. A speculator invested $2400 in mining stock which pays him a dividend of 4J%. If he agrees to take his dividend in coal at 6^ a bu., how many bushels should he receive ? 67. Stock bought at 93 J^ and yielding 5%, bears $225 annual income. How many shares are there? 6a A man receives $882 as a 7% dividend on his stock. How many shares does he hold ? STOCKS AND BONDS 213 69. How much must be invested in U. S. 4's at 121 J, brokerage J%, that I may pay a debt of $900 from one year's income ? ' 70. If I liave to pay an assessment of $G10 on 305 shares of stock (50), what is the rate of assessment ? 71. I paid $ 1G,G05 for stock at 92, brokerage \%. If it yields $3000 the first year, what is the rate of dividend ? 72. The earnings of a mining company, whose capital is $120,000, amount in one year to $24,000; their expenses are $ 9000. What rate of dividend can they declare ? 73. What per cent on my investment will I realize by buying 6% stock at 115? 74. If 1 invest $ 18,000 in U. S. 4's at 102, my annual income is what per cent of my investment ? 75. Philadelphia G's are bought at 81}. What is the rate of investment? 76. How much must a broker pay for Virginia G's to realize S% on his investment ? 77. What must be the price of 3J% stock to equal 4}% stock at 90? 7a What must be the market value of 60 shares 4% stock to equal G% stock at 75? 79. A man's rate of dividend is 4%. What must be the price of stock if he realizes 3% on his investment? 80. What income is derived from $500 G% stock (25) ? 81. A sells GO shares of stock at 80, and invests the proceeds in sto(^k at 75. If the latter yields 4% annually, what is his income? 82. I own 500 shares mining stock bought at 102. If the stock yields 3% annually, what is my income in dollars? What |)er cent of my investment is my income ? 83. What is the better investment for $ 1500, 6% stock at 75, or 4% stock at 60? What is the difference, in dollars, in the annual income? 214 PROBLEMS 84. A stockholder owns 3% railroad stock worth 90, and 2J% mining stock worth 70. If his income from each is $600 a year, in which stock has he the larger investment ? 85. How much stock, brokerage J%, must be sold at 137 J to buy with the proceeds $3600 of stock at 68 J, brokerage \%? 86. A man receives an annual income of $270 from 4J% stock, and $ 160 from 2% stock. If he sells the former at 85, and half the latter at 95, brokerage i% in each transaction, how much does he receive ? 87. If 4% stock is at 60, what rate per cent will a person re- ceive on his investment ? 8a A man invested a certain sum of money in 5% stock at 80, and twice as much in 4% stock. If his income from the former is $ 300, and from the latter 1 J times as much, what was the price of a share in the latter investment ? 89. B's income on 80 shares of stock is $ 320. If this yields him 5% on his investment, what is his entire stock worth? 90. A man owns 70 shares 2J% stock and 45 shares 4% stock. Which is worth more in market, and by how much, if the rate realized on each investment is 5% ? 91. \Yhat amount must be invested in 3% stock at 90 to yield an annual income of $ 225 ? 92. By selling 60 shares of 4% stock at 90, and investing the proceeds in other stock at 75, my income is increased by $ 84. What is the entire income from the latter investment ? 93. A man owns $8000 5% insurance stock. He exchanges this for mining stock at 80, which increases his income $150. What rate per cent dividend does the mining stock yield ? 94. My income on 70 shares of stock is $ 245. What rate of dividend does the stock pay ? 95. I sell $9000 5% stock at 80, and invest the proceeds in S% stock at 60. I increase or decrease my income, and by how much? STOCKS AND BONDS 216 DIFFICULT PROBLEMS In some problems, it is best to represent a required term by x, 96. Suppose 10% state stock is 20% better in market than 4% railroad stock. If A's income is $500 from each, how much money has he paid for each, the whole investment bringing Given terms: income on 1 share state stock, $10; income on 1 share rail- road stock, $4 ; whole income on state stock, $500 ; whole income on K. U. stock, $600. lielations: cost 1 share state stock = 120% cost 1 share R. R. stock. G,§j % of investment = $ 1000. Let X = cost 1 share R. R. stock, $ 90 125 X = entire cost R. R. stock, $ 11250 1.2 X = cost 1 share state stock, $ 108 60 X = entire ajst state stock, $ 5400 185 X = entire investment 6A%of 185x = 1000 a: = 90 97. I received a 10% stock dividend, and then had 102 shares (1^50 each) and $15 of another share. How many shares had I before the dividend ? 9a Thomas Reed bought 6% mining stock at 114|, and 4% furnace stock at 112 J, brokerage J%. The latter cost him $430 more than the former, but yielded the same income. How much did each cost him ? 99. I bought stock at 10% discount, which rose to 5% premium, and sold for cash. Paying a debt of $33, I invested the balance in stock at 2% premium, which, at par, left me $11 less than at first. How much money had I at first ? 100. W. T. Baird invested a certain sum of money in Phila- delphia G's at 115J, and three times as much in Union Pacific 7's at 89 J, brokeraj^e ]^% in each case. How much was invested in each kind of stock, if the annual income was $ 9920 in all ? INTEREST TERMS AND RELATIONS Money paid for the use of money is interest; the money loaned is the /)r//a/; the sum of the principal and the interest is the amount. 1. If $ 100 is loaned for 3 yr. at 6%, and no payment is made until the end of the third year, how much interest is then due ? There are three conceptions : That the principal alone bears interest, simple interest. * That the principal, and the interest on the principal at the end of each year, bear interest, annual interest. That the principal, the inter- est on the principal at the end of each year, and all other in- terest at the end of each year, bear interest, compound interest. Unless otherwise stated, simple interest is always un- derstood. Illi:stration If $6 is paid fur the use of 9 100, $ 100 is the principal; $0, the inter- est; 9 100, the amount. Simple Interest 1 yr. 2yr. 3 yr. Total. f6 f6 f6 $18.00 The interest on $100 is $0 each year. Annual Interest 1 yr. 2 yr. 3 yr. Total. $6 f6 f6 .36 .36 _ .36 f6 $6.36 $6.72 $19.08 The interest on $100 is $6 each year. The interest on the first $ 6 is 36^ the 2d year, and 30 f the 3d year. The interest on the second §6 is 36^ the 3d year. Compound Interest 1 yr. 2 yr. 3 yr. Total. $6 $6 $6 .36 .36 .36 .0216 $6 $6.36 $6.7416 $19.1016 In addition to the annual interest, the first 36^ gains 2. 16 ^ 216 INTEREST SIMPLE INTEREST First Conception : The principal cdoue bears interest, 2. What is the interest of $1 217 for 1 yr. at G% ? 3. What is the interest of $1 for 1 mo. at 6%? 4. What is the interest of $ 1 for 1 da. at 6^ ? The interest of $ 1 for 1 yr. at 6% is .06 of a 1, or 0^ Since the interest of $1 for 12 mo. isO)?, for 1 ino., it is j^i of 0^, or o m. Since the interest of f 1 for 30 da. is 5 m., for 1 da., it is ^^ of 5 m., or ^ of a mill. To BE memorized: The interest of $ 1 for 1 year at 6fo *'* ^^; for 1 monthy ^ofa cent; for 1 day, \of a mill. 5. What is the interest of $ 1 for 5 yr. 5 mo. 16 da. 2kt6%? f.30 .025 The interest of § 1 for 5 yr. at 6 % is .30 ; for QQ2* 5 mo., $.025; for 10 days, $.002|; for the ^ whole time, f .327f. .327\ 6. What is the amount of $ 3G0 for 5 yr. 5 mo. 16 da. at 6% ? ^360 .327\ 117.96, interest 360. Assume $1. The interest of $1 for 6 yr 6 mo. 10 da. at 6% is $.327| ; for $300, 300 times «.327i{, or $117.96. The amount is $477.96. 477.96, amount Find the interest of f 1 at 6% 7. For 4 yr. 2 mo. 16 da. 8. For 5 yr. 7 mo. 9 da. 9. For 3 yr. 1 mo. 17 da. 10. For 7 yr. 7 mo. 7 da. 11. For 2 yr. 6 mo. 12 da. Find the amt. of f 125 at 6% : 12. For 2 yr. 2 mo. 2 da. 13. For 6 yr. 6 mo. 6 da. 14. For 5 yr. mo. 19 da. 15. For 5 yr. 7 mo. 27 da. 16. For 4 yr. 1 mo. 20 da. 218 SIMPLE INTEREST If the rate is not 6%, the interest may be found at 6 per cent, and the result modified for the required rate. 17. What is the amount of $ 3G0 for 3 yr. 3 mo. 5 da. at 5% ? tS60 '195'1 Assume $ 1. The interest of $ 1 for 3 yr. @ 6% is $.18 ; for 3 mo., 70.oO @ 6%. ^.015; for 6 da., $.000^ ; for the ii.75 @ 1%. whole time, $.196J. fro'rr f7i,Krrt 'I^e interest of f 360 is 360 times 58,75 @ 5%. ^^^^^^ ^^ ^^^^_ ^^ gy^ ^ ,^^^ ^^ ^^0. $68.76. fjfl8.75y amount, la How would you modify the interest @ 6% to find the interest at 8%? at 10%? at 7%? at 9%? Am. At 8 %, I would add a third of the interest, because 8 is 6 plus | of 6 ; at 10 %, I would divide the interest by 6 and multiply by 10. When the time does not exceed 4 months, a modification of the 6% method is in general use. Moving the decimal point 2 places to the left in the principal, gives the interest for 60 dat/s at 6%. Proof : the interest of $ 1 for 60 da. @ 6% is 1^, or yiff o^ <> 1- 19. What is the interest of $ 167.80 for 3 mo. 3 da. @ 10%? 91.678 60 .889 80 The interest for 60 da. @ 6% is .0839 S % 1 .678 ; for 30 da. , \ of that, or $ .839 ; 2.6009 @ 6% ^^^ ^ ^^^ t'o of that, or $.0839. 4.334 @10% Find the interest of $438.96 for : What is the interest of: 20. 3 yr. 10 mo. 17 da. @ 6%. 25. $ 295.40 for 36 da. @ 5% ? 21. 4yr. mo. 25 da. @ 9%. 26. $360.20 for 93 da. @ 7% ? 22. 2 yr. 6 mo. 17 da. @ 10%. 27. % 235.90 for 93 da. @ 9% ? 23. 6yr. 1 mo. 18 da. @ 7%. 28. $ 840.50 for 33 da. @ 8 % ? 24. 4 yr. 3 mo. 12 da. @ 4J%. 29. $ 501.02 for 63 da. @ 6% ? INTEREST 219 Accurate interest, that is, interest found by counting 365, in- stead of 360, days to the year is sometimes computed. 30. Find the accurate interest of $ 625 for 80 days at 6%. ^^ X .06 X 626 = 8.219. Since the interest for 366 days is .06 S65 ' of the principal, the interest for 80 days Accurate Int. = ^8.219. « sVs x .06 x $ 625, or $ 8.219. 31. Find the interest of $ 625 for 80 days, counting 360 days to the year, subtract ^ of the result, and observe that accurate interest has been computed. 32. What is the accurate interest of $ 100 from Jan. 1, 1897, to Jan. 1, 1898, at 6% ? the interest as commonly found ? Ana. $6 in each case. The two methods agree for a whole number of years ; for a fraction of a year accurate interest is less. 33. Show that accurate interest for any number of days les& than a year, may be found by deducting j^^ of the interest found by counting 360 days to a year. Let R = the int. for 1 yr. R = int. 1 da., 365 da. to ayr. mt. . . ^i. • * * * j v ^gj ' ^ That IS, the mterest found by n counting 360 da. to the year mul- = int. 1 da.f 860 da. to a yr. tiplied by \\, or diminished by ^^0 ^^ becomes the accurate interest. ^ x^ or-?-x^ = -^ 360 365' 360 73 365 Find the accurate interest of : 3C $200 from Dec. 13, 1896, to May 1, 1897, @ 6%. 35. 1440 from Jan. 20, 1896, to Apr. 5, 1897, @ 8%. 36. f 450 from June 16, 1895, to Nov. 8, 18%, @ 10%. 37. $300 from Dec. 1, 1806. to May 10, 1898, @ 4%. 220 SIMPLE INTEREST INDIRECT CASES A» a basis, always assume 1 of the denomination required. 3a Wliat principal will gain $ 76.80 in 3 yr. 2 mo. 12 da. @ 8%? J8 ^01 Aasome f 1. $1 in 3 yr. 2 mo. S^n )78 f^nOtSiOa 002 12 da. @ 8% will gain $ .266 ; it will -^ iuit as $ .266 is contained times in 9 76.80, 'SM or $300. 39. At what % will $ 300 gain $ 76.80 in 3 yr. 2 mo. 12 da. ? ,192 02 Assume 1%. $ 300 in 3 yr. 2 mo. S7.60 at 6% 'qoo ^2 da. at 1 % will gain $ 9.60 ; it will 9.60 ati% ^ ^^^^ ^ ""*"y % ^ ^'" $76.80, as J^^,-.^ ^x^ /O^ $9.60 is contained times in $76.80, 9.60)76.80(8 -^^^ or 8%. 76.80 *^ 40. In what time will $300 gain $76.80 at 8%? 300 3.2 yr. — ^^ Assume 1 yr. $.300 in 1 yr. @ 8% ^■* 2 J, rnn ^^^^ ^*" $ 24 ; it will take as many 24)76.80(3.2 "^ ' years to gain $70.80, as $24 is con- 72 80 tained times in $ 76.80, or 3 yr. 2 mo. 48 12 da. 12 da. 41. What principal will amount to $ 376.80 in 3 yr. 2 mo. 12 da, at8%? .18 Q2 Assume $1. $1 in 3 yr. 2 mo. 002 12 da. at 8% will amount to § 1.256 ; 1.256)376.800(300 '—3 it will t.ake as many dollars to amount 3768 'Vi') to $370.80, as $1,250 is contained -0^^ times in $376.80, or $300. .256 INTEREST 221 Each example shouhl be changed to a form already given. 42. At what % will $300 amount to $37G.80 in 3 yr. 2 mo. 12 da. ? This means, " At what % will $ 300 gain $ 76.80 in 3 yr. 2 rao, 12 da. ? " See Ex. 39. 4a In what time will $ 300 amount to $ 376.80 at 8% ? This means, " In what time will ^ 300 gain $ 76.80 at 8 % ? " /S^ee i^x. ^0. 44. At what % will a sum triple in 30 years ? This means, " At what % will $ 10 gain $ 20 in 30 yr.? " 45. In what time will a sum quadruple at 6%? This means, " In what time will $ 10 gain ^ 30 @ 6 % ? " 46. Is it proper to reason thus : " Since $ 1 amounts to $ 1.06, $ 5 will amount to 5 times $ 1.06, or $ 5.30 ? Yes. Because $5 will amount to 5 times as much as $ I. 47. Is it proper to reason thus: "If a principal amounts to 9 1.06 in 1 yr., in 2 yr. it will amount to 2 times $ 1.06, or f 2.12 ? " No. The amount is always once the principal plus the interest. 4a At what % will a sum double in 10 years ? 49. In what time will $ 260 gain $ 29.90 at 5%? 5a In what time will a sum triple at 10 per cent ? 51. In what time will $ 260 amount to $ 289.90 at 5%? 52. At what % will $ 260 gain $ 29.90 in 2 yr. 3 mo. 18 da. ? 53. At what % will $ 260 amount to $ 289.90 in 2 yr. 3 mo. 18 da. ? 54. Whatprincipal willgain$29.90in2yr.3mo. 18da.@5%? 55. What principal will amount to $ 289.90 in 2 yr. 3 mo. 18 da. 222 SIMPLE INTEREST MISCELLANEOUS Find: 56. Interest of $ 1025 for 3 mo. 6 da. @ 9%. 57. Amount of $ 2400 for 6 yr. 5 mo. 20 da. @ 6%. 5a Amount of $ 930 for 10 yr. 3 mo. 24 da. @ 4%. 59. Ainouut of $ 150 from Apr. 4 to Dec. 18 @ 8%. What principal : GO. Will produce $ 68.20 interest in 2 yr. 7 mo. @ 4%? 61. Will produce $ 1610 interest in 4 yr. 5 mo. 20 da. @ 6%? 62. Will amount to $ 742 in 6 yr. 18 da. @S%? 63. Will amount to $ 1065.75 in 10 mo. 15 da. @ 10«i^.? At what rate per cent : 64. Will $ 824 amount to $ 957.90 in 3 yr. 3 mo. ? 65. Will $ 235 produce % 84.60 interest in 6 yr. ? 66. Will $900 amount to $ 1200 in 10 years? 67. Will $ 840 produce $ 70.84 interest in 4 yr. 2 mo. 18 da. ? Find the time in which : 6a $ 286 Avill gain $ 70.07 @7%. 69. $ 800 will amount to $ 1040 @ 6%. 70. $ 340 will produce $ 13.60 interest @ 5%. 71. $760 will gain $285 interest® 10%. What will be the amount of: 72. $ 100 in 3 yr., if the amount in 1 yr. is $ 125 ? 73. $ 80 in 5 yr., if the amount in 2 y r. is $ 100 ? 74. $ 160 in 6 yr., if the amount in 2 yr. is $ 192 ? 75. $ 450 in 10 yr., if the amount in 3 yr. is $ 595 ? INTEREST 223 PARTIAL PAYMENTS — LONG NOTES A written promise to pay a sum of money on demand or at a stated time, is a note ; the sum promised, the jnin- cipal, or face of the note ; the time when a note is due, its mcUurity ; the person to whom the money is to be paid, the payee ; the person who signs the note, the maker or drawer. Whenever a payment, called an indorsement, is made, the date and amount of the payment are written upon the back of the note. June 4, 1893, Harold Blake bought a farm of Peter Ford for $1000, paying $475 cash and giving Note No. 1 for the balance. Mr. Blake agrees to pay the principal, with interest at 6% from date, on June 4, 1890. He may delay payment as long as Mr. Ford will con- sent. Thedratccr is Harold Blake ; the payee^ Peter Ford. Mr. Blake paid $114.20, Sep. 9, 1894; $8.29, May 16, 1896; $244.38, Aug. 6, 1896. f 625.00. Emporia, Kans., fun& f, 189 3. <5kv&& if&cLva^ after date J promise to pay to the order of ^eC&v c^<>^<^-.,^...^,ww^.wv^.^^.c^^Vi'=' AuTtdvtd tn^&ntif-fCve,.,..^^.,^....^^^Dollars, with interest a^ 6%, at ^/i& ^ivot o^atuynat Bank, of (^At^ttO'.^.s.^^ Value received. Ko. 1. Due fun& ^, /S^6. ^i ^ ^i ? :; ;| S =0 S On ».» "O d. S' ^ Co >J '^ 224 SIMPLK INTEREST The United States rule for partial payments is derived from the decision of the Supreme Court, for finding the amount due on a note when payments have been made. Unitkd States Rule Find the amount of the principal to the time of the first paj/mcnt; if the payment equals or exceeds the interest^ subtract the payment from the amount a}id treat the remainder as a new princijxtl. If the jxiyment is less than the interest, find the amount of the same princijxtl to the time when the sum of the payments shall equal or exceed the interest due, and subtract the sum of the payments from the amount. Proceed in the same manner with the remaining payments until the time of settlement, 76. What was due on note No. 1, Feb. 9, 1898 ? $525.00, P. S9.8U L »/p, ^94 564^81 114.20 450.61, P. 18.48 * I. 5/15, '95 469.09 SS.12 , I. s/c, '96 NoTK. — By arranging the dates as above, errors 503. 21 in subtracting will be discovered. Each date is 252.67 , Sum pay. subtracted from the next below. The difiference 249.54 P. between the first and last dates shoald equal the Sg 58 $272.12, Due 2/^?, '98 189S 6 4 1894 9 9 1 S 5 $114,20 1895 5 15 8 6 8,29 1896 8 6 1 2 21 244-98 1898 2 9 16 3 4 8 5 4 8 6 sura of the differences. * Since the interest exceeds the payment, we find the interest on $ 450.61 and not on $4G9.09, for 1 yr. 2 nao. 21 da. 77. On note No. 2, how much remained due Sept. 25, 1897 ? 7a On note No. 3, liow much remained due Oct. 9, 1897, counting interest at 6%? INTEREST 226 $750.00. Boston, Mass., June 17. 1892. On demand, for value received, I promise to pay John E. Wiley, or order, seven hundred fifty ^Y(r dollars, with interest at 6%. No. 2. J. J. Hill. Payments: Mar. 1, 1893, 075.50; June 11, 1803, $165; Sept. 15, 1803, $101; Jan. 21, 1804, 047.25; Mar. 12, 1895, $12.50; Dec. 6, 1805, 008; July 7, 1896, 169. $300.00. Wilmington, Del., Apr. 30, 1895. On demand, for value received, I promise to pay G. R. Bell, or order, three hundred -f^j^ dollars, with interest. No. 3. Russell Hibbs. Payments: June 27, 1896, 150 ; Dec. 9, 1890, 0160. $ 1500.00. Cincinnati, O., Jan. 10, 1895. On demand, for value received, I promise to pay Ernest Buck- man, or order, one thousand five hundred dollars, with interest **^%- ' L.H. Taylor. No. 4. Indorsements: Mar. 3, 1895, 0250; May 20, 1895, 0300; July 1, 1896, 0125; Nov. 15, 1895, 075. $625.00. Empobia, Kans., Dec. 15, 1895. On demand, for value received, I promise to pay James Tyner, or order, five hundred twenty-five dollars, with interest at 8%. No. 6. W. N. Simpson. IndorsemenU: Jan. 16, 1896, 076 ; Apr. 20. 1896, 060; June 16, 1896, 0140.60; Aug. 1, 1896, 080; Sept. 25, 1896, 0120. Notk. — For treatment of No. 4 and No. 5, see next page. AMBB. ABITH. — 15 226 SIMPLE INTEREST PARTIAL PAYMENTS - SHORT NOTES When notes, upon which payments have been made, are settled within a year of date, merchants commonly disregard the U. S. rule. Mbrcantile Rulb ' Find the amount of eacJi item to date of settlement. $ 1728.00. LiTTi.E Rock, Ark., Jan. 2, 1897. On demand, for value received, I promise to pay William Chamberlain, or order, one thousand seven hundred and twenty- eight ^5 dollars, with interest at 0%. No. 6. Henry Kixg. Indorsements: Mar. 1, 1897, ^300; May 16, 1897, $150; Sept. 1, 1897, #270; Dec. 11, 1897,9135. 79. How much was due Dec. 13, 1897, on Note No. 6 ? Principal, $1728.00 Interest to Dec. 13, 1897, 345 da., 99.36 $1827.36 First payment, $300.00 Int. to Dec. 13, 1807, 287 da!, . 14.35 Second payment, 150.00 Int. to Dec. 13, 1897, 211 da., . 5.28 Third payment, 270.00 Int. to Dec. 13, 1897, 103 da. , . 4.64 Fourth payment, 135.00 Int. to Dec. 13, 1897, 2 da., . . .05 879.32 Bal. due Dec. 13, 1897, $948.04 Owing $1728 Jan. 2, is equivalent to owing Dec. 13, $ 1728 + the interest of $ 1728 from Jan. 2 to Dec. 13, or $ 1827.36. Paying $300 Mar. 1, is equivalent to paying Dec. 13, $300 + the interest of $300 from Mar. 1 to Dec. 13, or $314.35. eo. On Note No. 4, how much remained due Dec. 12, 1895 ? 81. On Note No. 5, how much remained due Nov. 1, 1896 ? INTEREST 227 TRUE DISCOUNT The true present woi-th of a note, or of a sum of money due in the future, is that sum which, put at interest now, will amount to the given debt at the expiration of the time. 82. $ 275 is due me in 1 yr. 6 mo. What is the true present worth, interest at 6%? What is t]ie true discount ? The present worth is that sum which, put at interest to-day, will amount to $275 in 1.09)275.00(252.29 \ yr. 6 mo. at 6%. $275 face $1 in 1 yr. mo. at 6% will amount to ^^^^^ ^1 $109; it will take as many dollars to ^52.29 , true jyres. worth ^^^^^^ ^ ^ .^75 ^ 3 1 Oj, ia contained times $22.71, discount '^ * 275, or 8 252.29. ^ The true discount is $276 - $252.29, or $22.71. Note.— True discount is rarely found on notes or debts running less than four months. 83. Find the true present worth and true discount of Note No. 7, discounted at 10%, Mar. 1, 1896. ^ 300.00. Emporia, Kans., Jan. 2, 1896. Ninety days after date, value received, I promise to pay to the order of William Clarke, at the First National Bank, Three hun- dred and -^ Dollars, with interest at 10% after maturity. No. 7. W. C. Stevenson. Find the : 84. True present worth of $ 412 due in 6 mo., int. @ 6%. 85. True present worth of $ 324 due in 8 mo., int. @ 12%. 86. True present worth of $ 321 due in 1 yr. 9 mo., int. @ 7%. Find the : 87. True discount of $ 590 due in 2 yr. 8 mo., interest at 9%. 88. True discount of $ 33G due in 3 yr. 10 mo., interest at 12%. 89. True discount of $ 427 due in 5 yr. 11 mo., interest at 8%. 228 SIMPLE INTEREST BANK DISCOUNT In discounting notes, bankers use the following rule : To find the bank discount, compute the interest on the amount due at maturity for (three days more than*) the specified time. To find the proceeds, subtract the bank discount from the amount due at maturity. 90. Find the bank discount at 10% Jan. 2, 1896, on Note 7; iiud the bank proceeds ; find the date of maturity. fS.OO 60 Since the note does not bej^in to bear interest until U3 days after Jan. 2, tlie amt. due at maturity is the face of the note, or $300. The interest of $300 for 93 days at 10% is $7.75. 1.60 .15 f4.65, D.y 6% 7.75, D.y 10% f292,S5y Proceeds Apr. 4) <^«^€ ofmMurity 91. Write Note No. 7, substituting "with interest at 6%" in place of "with interest at 10% after maturity." Find the bank discount and proceeds at 10%, Feb. 2, 1896. fS.OO 60 1.50 .15 Since the note now begins to bear interest Jan. 2, the amount at maturity, or the face value of the note, is the amount of $300 for 93 days @ 6%, or $304.65. f4.65y Int. $804-65, amt. at maiurity $3.0465 60 .1015 f f 3.14s, D., 6% 5.25, D., 10% $299.40, Proceeds ♦ Tills phrase should be stricken out for the following states and territories in which the three days, called days of grace, have been abolished by statute : Alas,, Cal.. Colo., Conn., D. C. Fla., Ida.. 111., Me.. Md., Mass., Mont,, N. IL, N. J., N. Y., N. D., O., Ore., Pa., Utah, Vt., and Wis. Since the note was not discounted until Feb. 2, or 31 days after date, the specified time is 93 da.— 31 da., or 62 da. INTEREST 229 92. Write a bank note for 90 days, with interest after maturity, that will give bank proceeds at 10 %, of $292.25. .01 60 Assume 8 1. Since the bank proceeds of «1 at 10% for IW da. are 8. 074 J, it will -000^ * take as many dollars to yield $292.26, ni^t^ «% as e .974i is contained times in 8292.25, '^ ^ or 8 300. .025^ io% .974j)292M0{300 93. Find the date of maturity, and bank discount at 6 %, Jan. 10, 1896, on Note No. 8. f 400.00 * MiDDLETOWN, CoNN., Jan. 10, 1896. Sixty days after date, value received, I promise to pay to the order of the Middlesex County National Bank, four hundred and ^^ dollars, with interest at 6 % after maturity. No. 8. John S. Camp. 94. Find the bank discount at 6 %, Mar. 3, 1896, on Note No. 7. 95. Write Note No. 8, substituting " with interest at 5 % " in place of " with interest at 6 % after maturity," and find the bank discount at 6 %, Feb. 19. 96. Find the true discount at 6 %, of Note No. 8, Jan. 10, 1896. How much does the banker make by collecting bank, instead of true discount ? 97. Show that the difference between the bank discount and the true discount of Note No. 8, Jan. 10, 1896, is the interest on the true discount for 60 days. Note. — Do not count 3 da. of grace in either case. 9a What is the face of a note, which, when discounted for 2 mo. 24 da. at 5 %, will yield $224.00 bank proceeds? 99. Write a bank note for 3 mo., maker, John Brown ; payee, Henry Short; proceeds at 6 %, $ 732.92. 230 SIMPLE INTEREST SETTLEMENT OF ACCOUNTS — INTEREST METHOD In settling accounts between wholesale and retail dealers, oi where the amounts are large or of long standing, it is customary to take account of interest. 100. Thomas Stone bought merchandise of us as shown on the Dr. side of the following account, and made payments as on the Cr. side. Settlement was made Sep. 15, 189G; how much did Mr. Stone owe at that time, counting interest at 6% ? Dr. Thomas Stone Cr. 189G 1S96 Mar. 7 To Mdse 500 Mar.J) By Cash SOO Apr. 3 To Mdse SOD Apr. 1 By Cash 200 Aug. 2 To Mdse 700 July I By Cash 500 First debt f 500.00 First payment $S00.00 Int. to Sep. 16 IG.OO Int. to Sep. 15 9.50 Second debt SOO. Second payment 200.00 Int. to Sep. 16 8.g5 Int. to Sep. 15 5.57 Third debt 7pO. Third payment 500.00 lE t. to Sep. 16 5.1 S Int. to Sep. 15 6.33 §1529.38 Bal. due Sep. 15, 1896, $507.98. $1021.40 Owing $500 Mar. 7, is equivalent to owing Sep. 15, 8 500 + the inter- est of $500 from Mar. 7 to Sep. 15, or $516.00. Owing $300 Apr. 3, is equivalent to owing Sep. 15, $300 + the inter- est of $300 from Apr. 3 to Sep. 15, or $ 308.25 ; etc. The pupil should finish the ex- planation. Paying $300 Mar. 9, is equivalent to paying Sep. 15, $300 + the inter- est of $300 from Mar. 9 to Sep. 15, or $309.50. Paying $200 Apr. 1, is equivalent to paying Sep. 15, $200 + the inter- est of $ 200 from Apr. 1 to Sep. 15, or $ 205.57 ; etc. The pupil should finish the ex- planation. Note. — This method is identical with that used in settling short time notes, when payments are made within a year of date. 6'ee p. 226. interp:st 231 EQUATED TIME METHOD 103. Solve example 100 by the equated time method. Dr. 500 X 192 = 96000 SOO X 165 = 49500 700 y. 44= S0800 1600 176S00 1600 1000 600 Cr. SOO X 190 = 57000 goo X 167 = SS400 600 X 76= SSOOO 1000 128400 176S00 128400 47900 $500. 7.98 Int. f 47900 1 da. $607.98 Bal. due Sep. 16. Owing $ 600 Mar. 7, is equivalent to owing Sep. 16, $ 600 + the inter- est of $600 for 192 days, or the interest of $ 96,000 for 1 da. Owing $.300 Apr. 3, is equivalent to owing Sep. 16, $.300 + the inter- est of $300 for 106 days, or the interest of $49,600 for 1 da. Paying $ 300 Mar. 9, is equivalent to paying Sep. 18, $300 + the inter- est of $.300 for 190 days, or the interest of $57,000 for 1 da. Paying $200 Apr, 1, is equivalent to paying Sep. 16, $200 + the inter- est of $ 200 for 167 days, or the inter- est of $ 33,400 for 1 da. Sep. 16, he owes $ 1500 + the interest of $ 176,300 for 1 da., and has paid $ 1000 + the interest of $ 128,400 for 1 da. The bal. due is the difference, or $600 4- the interest of $47,900 for 1 da., or $507.98. 102. Find the equated time for the payment of the account in example 100. By the equated time^ is meant the time when the balance of the account i$ due. Sep. 15, Mr. Stone owed $ 500 -|- the interest of $ 47,900 for 1 da., or $ 500 + the interest of $500 for as many days as $600 is contained times in $47,1K)0, or for 96 da. The equated time, or the time when he owed just $500, wa8 96 da. before Sep. 16, or June 11. 232 SIMPLE INTEREST 103. Find the equated time for the payment of the Dr. side in example 100. The last debt was due Aug. 2. Reasoning as in the previous example, we find that Stone owed, Aug. 2, $1600 + interest of $110,300 for 1 da., or $1600 + interest of $ 1600 for 74 da. The equated timet or tlie time wlien he owed only $ 1600, was 74 days be- fore Aug. 2, or May 20. Aug. S. Dr, 500 X 14S= 74000 SOO X 121 = S6S00 700 y. 0- 1500 llOSOO 1500 )110300 74 Dr, James Whitman Cr. 1895 1895 Apr. 3 To Mdse. 600 Mayl By Cash 400 June 10 To Mdae. 500 July 15 By Cash 300 Sep, 15 To Mdse. 400 Aug. SO By Cash SOO 104. Settlement was made ori the above account Nov. 1, 1895. How much did Mr. Whitman owe at that time, counting interest at 6 % ? Solve by the interest method. Give the explanation in full. 105. Solve example 104 by the equated time method. Give the explanation in full. Which do you prefer; the interest method, or the equated time method? 106. In the above account, find the equated time for the pay- ment of the Dr. side ; that is, find the time when all of the Dr. side may be regarded as due. 107. In the above account, find the equated time for the pay- ment of the Cr. side ; that is, find the time when all of the Cr. side may be regarded as due. 108. In the above account, find the interest on the sum of the debts from the equated time to the date of settlement, Nov. 1 ; find the interest on the sum of the payments to the date of settle- ment ; find the difference. Note.— Ex. 108 illustrates another method of solving such problems. INTEREST 233 ANNUAL INTEREST Annual interest is rarely, if ever, computed in actual business. Hie principal bears interest, and the interest on the pnncipal at the end of each year hears interest. See p. 216. 109. What is the annual interest of $ 100 for 3 yr. 3 mo. 3 da. at 6%? The simple interest of $ 100 for 3 yr. 3 mo. 3 da. yr. MO. da. 96 2 3 S at 6% is a 19.56. The 9 6, interest on the principal at the end of 96 1 3 3 the first year, bears interest for 2 yr. 3 mo. 3 da. 96 3 3 the 9 6, interest on the principal at the end of the — second year, bears interest for 1 yr. 3 mo. 3 da. ; 96 3 9 9 tlie $ 0, interest on the principal at the end of the 9 19.55=int. onnnn. ^^"^^ y^^^^ bears interest for 3 mo. 3 da. The ^ - ,_ . . whole is equivalent to the interest of $6 for the J^3b — tnt. on int. ^^^^^ ^^ ^^j^^ intervals, or for 3 yr. J) mo. 9 da. 20.91 =annual int. '^^^^ interest of ^6 for 3 yr. 9 mo. 9 da. at 6% is 9 1.36 ; the annual interest, 9 20.91. What is the: 110. Annual interest of $ 420 for 3 yr. 2 mo. 5 da. @ 6% ? HI. Annual interest of $ 240 for 1 yr. 6 mo. 12 da. @ 9%? 112. Annual interest of $ 186 for 4 yr. 9 mo. 18 da. @ 6%? 113. Annual interest of $ 252 for 5 yr. 5 mo. 15 da. @ 8%? 114. Find the amount of a note for $ 1500, interest 6%, pay- able annually, given Sep. 3, 1893, and not paid until July 1, 1897. 115. Find the amount of a note for $ 2000, interest 10%, pay- able annually, given Jan. 16, 1895, and not paid until April 1, 1900. 116. Find the amount of a note for $ 600, interest 8%, payable annually, given Oct. 10, 1896, and not paid until May 16, 1902. 117. Find the amount of a note for $ 1200, interest 7%, pay- able annually, given May 4, 1897, and not paid until Feb. 19, 1899. 234 COMPOUND INTEREST COMPOUND INTEREST In all states, both law and business usage are opposed to the collection of compound interest. There are many problems, how- ever, especially in connection with insurance, annuities, and reserve funds, where this computation is necessary. Tlie principal bears interest^ the interest on the principal at the end of each year bears interest^ and all other interest at the end of each year bears interest. See p. ^16. lia What is the amount at compound interest of $ 1 for 3 yr. at 6%, During the first year f 1 is on interest ; it amounts to f 1.06. During the second year $ 1.06 is on interest ; since 9 1 amounts to $ 1.06, $ l.Ot) will amount to 1.06 times 8 1.06, or ^ (1.06)2. During the third year $ (1.06)- is on interest ; since $ 1 amounts to $ 1.06, I (l.OO)'J will amount to (1.06)^ times « 1.06, or$(1.06)«, or $1.191016. 119. What is the amount at compound interest of $ 1 for 4 yr. at G% ? Explain as in example 118. Amocnt of $ 1 AT Compound Interest Ykars r.% 6% 10% 3 (1.06)« = 1.167625 (1.06)» = 1.191016 (1.10)8 = 1.331000 4 6 120. Explain as in example 118 and fill out the blanks in the above diagram. 121. Prove that the amount of $1 for n years at r% is $(l+r)*. 122. What is the amount at compound interest of $ 1 for 2000 yr. at 6% ? Express the answer as suggested in Ex. 121. How many decimal places are there in the exact answer ? INTEREST 235 123. Using the table just prepared, find the compound interest of $ 129.36 for 3 yr. 3 rao. 3 da. @ 6%. $(106)^ or $1 in 3 yr. at 6% will amount to ij.'moje, anU.il, Syr. e(1.06)», or $1.191016; $ I2».3rt will 1X9.36 amount to 129.36 times $1.191016, or 154.07, amt. $129.36 ^^^'^l' , ,^ ^.r.r.„- 1 nir^n «.«# A 1 « «,« % 6% '% 8% 9% 10% 1 1.0500000 1.0600000 1.0700000 1.0800000 1.0900000 1.1000000 2 1.1025000 1.1236000 1.1449000 1.16(54000 1.1881000 1.2100000 3 1.1576250 1.1910100 1.2250430 1.2597120 1.2950290 1..33lO(KiO 4 1.2155063 1.2(524770 1.3107960 1.3604890 1.4115816 1.4041000 6 1.2762816 1.3382256 1.4025517 1.4693281 1.5386240 1.6105100 6 1.3400956 1.4185191 1.5007304 1.5868743 1.6771001 1.7715610 7 1.4071004 1.503(5:i03 1.6057815 1.7138243 1.8280391 1.9487171 8 1.4774554 1.6938481 1.7181862 1.8509302 1.9925626 2.1435888 9 1.5513282 1.6894790 1.8384592 1.9990046 2.1718933 2.3579477 10 1.6288946 1.7908477 1.9671514 2.1589250 2.3673(537 2.5937425 11 1.7103394 1.8982986 2.1048520 2.331(5390 2.58042(;4 2.8531167 12 1.7958563 2.0121965 2.252101G 2.5181701 2.812(5648 3.1384284 13 1.8856491 2.1329283 2.4098450 2.719(5237 3.0658046 3.4522712 14 1.9799316 2.2609040 2.5785342 2.9371936 3.3417270 3.7974983 16 2.0789282 2.3965582 2.7590315 3.1721691 3.6424825 4.1772482 16 2.1828746 2.5403517 2.95216,38 3.4259426 3.9703059 4.5949730 17 2.2920183 2.6927728 3.1588152 3.7000181 4.3276.334 5.0544703 18 2.4006192 2.8543392 3.3790323 3.99(50195 4.7171204 5.5599173 19 2.5269502 3.0255995 .3.6165275 4.3157011 5.141(;013 6.1159090 20 2.6532977 3.2071355 3.8696845 4.6609571 5.6044108 6.7275000 INTEREST 237 From the tablCj the amount for any time may he found. 135. What is the ainoiint of $ 100 for 90 years at 6% com- pound interest? S3.i^071355, amt. $ 1, 20 yr. «3.20 is on interest at the end oon'Ti^f-r of20yr. During the next 20 yr., iS.^UiliSOo ^ J ^^.jjj amount to $3.20 ; «3.20+ f 10.2857179, amt. ^i, 40 yr. will amount to 3.20+ times $3.20, 10.2857179 or $10.28. $10.28 18 on interest at the f 105.7959927 y ami. $1, 80 yr. end of 40 yr. During the next 1.7908477 ^0 y^» '^ ^ ^^'^'^ amount to $ 10.28; — ^— — — ^^^^— ^^ etc. 9 189.4645099, amt.^l, 90 yr. '^i^^ pupil should finish the f 18940. 45, amt. $100 explanation. If interest is payable at intervals of less than a year, the example should be reduced to an equivalent example, interest payable annually. 136. What is the amount at compound interest, of $ 100 for 3 yr. 2 mo. 4 da. at 8%, interest payable quarterly ? ^irr Smo Ada. ^^^^"^ quarterly is paying 4 of a year. The interest for a 12 yr. 8 mo. 16 da. P^"^ °"^ ^^^^^^ ®^ * ^^'^^ ^ ^% is the same as the interest for a whole year @ 2%; .*. the interest @ 8%, payable quarterly, is the same as the interest for 4 times 3 yr. 2 mo. 4 da., or for 12 yr. 8 mo. 16 da. @ 2 %, payable annually. The pupil should finish the explanation. 137. Show that the amount for 5 years at 12%, payable semi- annually, is the same as the amount for 2 times 5 years, at \ of 12% payable annually. 13a Find the amount at compound interest, of $ 100 for 'Jo yr. 6 mo. 7 da. at 4%. 139. Find the amount at compound interest, of $ 100 for 8 yr. 3 mo. 2 da. at 12%, interest payable 3 times a year. INVOLUTION AND EVOLUTION INVOLUTION— TERMS AND RELATIONS 8 = 2 X 2 X 2, or 8 = 2^ If the product is wanting, this becomes what = 2^? It means " what is the product when 2 is used 3 times as a factor?'' Seep. 146, Involution is the process of ^. o J- .V J i. t ^1 What 18 the product when 2 is hnding the product when the used 3 times as a factor ? same number is used several , u o o o o ^ns. o; z X ^ X ^ ^ o. times as a factor. The result is the power. ^ ^ '^' third pouter of 2. 1. Read: 5^ 5^ 5^ 5'. Ans. 6^, 5 square, or 5 to the second power ; 6', 6 cube, or 6 to the third power ; 6*, 6 to the fourth power ; 6*, 5 to the a;th power. 2. Write the squares of the integers from 13 to 25, and memorize the results. 3. Write the cubes of the integers from 1 to 10, and memorize the results. 4. Declare the results rapidly: 16^ 25^ 13^; 17^; 21"; 22^: 24*; 142; 152. 182. 192; 20*; 23*; 9«; 7^; 2»; 4»; 6»; 3^ 5^ 8^. To raise a factor to any power, write the base, and over it, the product of the exponent by the number denoting the required power. Illustration : (48)2 ^ 46 . (48)2 =: 4* x 4* = (4 x 4 x 4) x (4 x 4 x 4) = 46. 5. Find the value of: (2«/; (2^)3; (3*)^ (Sy. e. The square of 12 is 144 ; what is its 6th power ? Ans. 144^ 7. The cube of 12 is 1728 ; what is its 6th power ? Ans. 1728*. INVOLUTION AND EVOLUTION 239 EVOLUTION — TERMS AND RELATIONS 2» = 8. If the base is wanting, this becomes (what)' = 8 ?, what = VS ?, or what = S' ?. It means "what number must be taken 3 times as a factor to produce 8 ? " Evolution is the process of find- Tm. . • .i v u- t. ,. . , What IS the niimber which ing the number which must be must be taken 3 times as a factor taken several times as a factor to to produce 8 ? produce a eiven product. -'^'**' '^^ because ^ = 8. The result is the roo^ 2 is the tkirU root of S, a Read: V9; -5^; ^/Sl; -^81. Ans. VO, the square root of 9, or the second root of 9 ; v^, the cube root of 27, or the tliird root of 27 ; \/8T, the fourth root of 81 ; \/8T, the xth root of 81. 9. Read: 9*; 27^; 81^ 81*. Ans. These are read as in Ex. 8. Also, 9* may be read, 9 to the } power ; 27% 27 to the } power ; . . . . 10. Declare the results rapidly : V225; ViOO; V625 ; Vl69; V324; VTUG; ^/^56 ; V^SO; V3(U ; V441 ; V484 ; V529. U. Declare the results rapidly: "v/m; \/27; \/l26; ^3^; ■\/S; ot. In each case, the Icailing of the formula is followed. Compare with p. 241. 246 CUBE ROOT 59. Extract the cube root of l'879'()80'i)04. We separate the number into periods of three figures each, because the cube root of the first period will give the first figure of the root ; the cube root of the first two periods, the first two figures, etc. FIRST We extract the cube root of 1'879 to jbtain the first two figures of the root. The explanation is the same as already given. ITie pupil should give it in full. See p. S45. ^ ^ SECOND We extract the cube root of 1'879'080 to obtain the first three figures of the root. We may regard 1879'080 as 1879 thou- sands and 080 u nits. That is, we are again to extract the cube root of a number of two periods, the first being 1879; the second, 080. To obtain a, the first term of the root in the model, we must extract the cube root of a*. Hence, to obtain the first term of the root in this example, we must ex- tract the cube root of what corresponds to a», or of 1879. v'l879=12 ; a = 120, be- cause 1879 is not 1879 units, but 1879 thousands. The pupil should finish the explanation. ^,„„^ ^ THIRD We extract the cube root of 1 '870'080'904 to obtain the first four figures of the root. We may regard 1879080'904 as 1879080 thousands and 904 units. That is, we are again to extract the cube root of a number of two periods, the first period bein g 187908 ; the second, 904. v^l'879'080 = 123 ; a = 12.30, because 1879080 is not 1879080 units but 1879080 thousands. The pupil should finish the explanation. NoTB. — In practice only the tliird (a + 6)» = «» + 3a«6 + 3a6« -}- b^ = a» + (3aH3a6+62)6 FIRST V879'080'm{lt 1 SOO \S79 60 4 364 a = 10 b = g 728 151 SBCOITD V879'08O'9ok{liS 1 SOO 879 60 4 364 728 43200 151080 1080 9 44^89 1328tj7 a = 10 b = 2 a = 120 b = 3 18213 THIRD 1'879'080'B04{12S4 1 a = 10 b = 2 300 60 4 364 879 728 43200 1080 9 151080 44289 132867 4538700 14760 16 18213904 4553471 S 18213904 a = 120 b = 3 a = 1230 h = 4 form should be written. INVOLUTION AND EVOLUTION 247 Find the vcUue of: ea ^94818816 62. ^1067462648 61. ^177504328 63. /?/r^ Caution. — Be sure to point off, beginning V 006 oou ^j^j^ ^jjg decimal point. 6a Extract the cube root of }. «S __ t/ -, If ^^® denominator is not a perfect cube, \^ "" 'S/.666 666 -\- n is best to reduce the fraction to a decimal. FHnd the value of: 69. (H)^ 72. (J)* 75. (f)i 7a ^.001728 7a -s/vm 7a n. Irregular hexagon. Cizvle. Define : 23. A surface. 24. A polygon. 25. A triangle. 26. An equilateral triangle. 27. An isosceles triangle. 2a A scalene triangle. 29. A right-angled triangle. 30. An acute-angled triangle. 31. A regular polygon. 32. A regular pentagon. 33. A regular hexagon. 34. A regular heptagon. 35. An obtuse-angled triangle. 36. A circle. 37. Beginning with ^^ plane surface^^^ (see Note) define : paral- lelogram; rectangle; rhomboid; rhombus; square. 3a Beginning with quajdrilateraJy (see Note) define : parallelo- gram; rectangle; square; rhombus. Note. — A definition may bejjin with different terms, e.g.: A square is a plane surface bounded by two pairs of opposite sides, having: each pair parallel, having its angles all right angles, and having its sides all equal. Or, A square is a quadrilateral having each pair of its opposite sides parallel, hav- ing its angles all right angles, and having its sides all equal. Or, A square is a parallelogram having its angles all right angles, and having its sides all equal. Or, A square is a rectangle, having its sides all equal. That definition is the best which is tlie shortest, provided it begins with a term which is understood by the person for vchom the definition is given. MENSURATION 253 PARTS OF A POLYGON That part of a polygon on which it is supposed to rest is its base; the distance around a polygon, its perimeter; the perim- eter of a circle, its circiirnference. In a right-angled triangle, the side opposite the right angle is the hypotenuse; the other sides, legs. BC, Bufte. AB+BC-^AC, PeHmeUr. AB and BC, Leg». AC, l/ypotenune. The altitude of a triangle, parallelogram, or trapezoid, is a perpendicular to the base from d d the vertex opposite the base. AB is the altitude in each of these figures. Observe that the base must sometimes be extended. Since a triangle may be re- garded as resting on any one of its sides, it may have three bases with an altitude corre- sponding to each. The apothem of a regular poly- gon is the perpendicular from its center to any side. The diagonal of a polygon is a line which joins any two ver- tices not adjacent h^. Due d b X AB, Altitude. Bate, BC; AUitude, AD. Base. AB; Altitude, CH. Ba-, AC; Altitude, BE. (D ABy Apothem. ^ BD, DUtffonal. 254 TWO DIMENSIONS PARTS OF A CIRCLE A circle is a plane figure bounded by a curved line, every point of which is equally distant from a point within called the center. LiNKAR Parts We may consider the whole of the bounding line, the circumference; or a part of it, an arc. A line may cut the circumference in two points, a secant; or may touch it at one point, a tangent. A line may be drawn from the center to any point of the circumference, a radius. A line may connect any two points of the circumference, a chord. A chord may pass through the center, a diameter. M a Linear parts. ABEFGD, Circuwprene^. DG, Arc. HM, Secant. ST, Tangent. CB, Radiue. AB, Chord. DB, Diameter. Sdrfacb Parts We may consider the portion of a circle between a chord and its arc, a segment; or the portion between two radii and their arc, a sector. A sector may be half of the circle, a semicircle; a quarter, a quadrant ; a sixth, a sextant. Surfitce parta^ ABX, Segment. BCF, Sector. DOE, Semicircle. EGC, Quadrant. CFG, Sextant. Draw and define : 39. A circle. 4a A segment. 41. A sector. 42. A quadrant. 4a A sextant. 44. A semicircle. 45. A radius. 46. A chord. 47. A diameter. 4a A tangent. 49. A secant. 50. An arc. 51. The center. 52. A circum- ference. MENSURATION 255 COMPUTATIONS — LINEAR PARTS I. The circumference of a circle is equal to twice the radius times S.W6. II. The square of the hypotenuse of a right-angled triangle w equod to the sum of the squares of the other two sides. Note. — For illustration and explanation, the pupil should turn to p. 272. S3. The radius of a circle is 5 in. ; find its circumference. BelatioH : C = 2 x R x 3. I4I6 C= S X 5x 3.I4I6 = SI.4I6 Circumference = SI.4I6 in. We Bubstitata the valae given. 54. The circumference of a circle is 31.416 in. ; find its radius. Relation : C = £ x R x 3. 14I6 SI.4I6-2 X R X 3.14I6 ^^ SI.4I6 ^^ We sabstitate the ralae giren. 2 X S.I4I6 Radius = 6 in. The pnpll should always draw the fl^re. 55. The hypotenuse of a right-angled triangle is 10 in. ; its perpendicular 6 in. ; find its base. Relation : AB^ = AC^ + BC^ 100 = 36+ Bd^ 100-36= BCf^ = 64 BC = V64 = 8 C Base = 8 in. 56. A room is 20 ft. x 12 ft. x 9 ft. How far is it from a lower corner to the opposite upper corner ? d A^=144 + 4O0 = 5U ^/9 DC* = 81 + 644 = 625 DC = V62S = 25 Distance = 25 ft. A C, Diagonal ay off on BD and B'D' as for a pyramid. Construct on two sides, as EF and E'F*, regular polygons. Prepare flaps, cut and paste as before. To i>RAW THE Regular Hexagon. From E and F as centers, draw arcs of circles with a radius equal to EF. From 0, 'heir point of intersection, as a center, with a radius equal to EF, describe a cin^le. From F, lay off on tiie circumference, distances equal to EF\ con- ««ect the points thua found. E F 4-4 204 CONVEX SURFACES COMPUTATIONS — CONVEX SURFACES XI. The convex surface of a prism is the product of the perimeter nf its base by its altitude. XII. The convex surface of a cylinder is the product of the cir- cumference of its base by its altitude. XIII. Tlie convex surface of a pyramid is half the product of the perimeter of its base by its slant height. X I \\ The convex surface of a cone is half the product of the cir- cumference of its base by its slant height. XV. The convex surface of a frustum of a pyramid is half the product of the sum of the perimeters of its two bases by its slant height. XVI. 7%e convex surface of a frustum of a cone is half the product of the sum of the circumferences of its two bosses by its slant height. XVII. The surface of a sphere is four times the square of its radius times 3.1410- Note. — For illustrations, explanations, and proofs, the pupil should turn to p. 2<>1, and to pp. 275 and 276. 139. Find the convex surface of a cylinder, radius of the base 6 in., altitude 8 in. Belation: S= Cir. x Alt. S=gx Bx S.I4I6 X Alt. Conv. surf. = 301.5936 sq. in. = 2x6 X 3.1416 X8 = 301.5936 140. Find the convex surface of a square pyramid, one side of the base 4 in., slant height 12 in. Belation : S=i Per. x S. H. Conv. surf. — 96 sq. in. S = -gX 16 X 12 = 96 141. Find the convex surface of a frustum of a cone, radius of upper base 6 in., radius of lower base 8 in., slant height 12 in. Belation : 8 = ^{0 -h C')x S. R. S=l(Sx6xS.1416-\-2x8x3.1416)xl2 Conv. surf = 527.7888 sq. in. = 527.7888 MENSURATION 265 142. Find the radius of a sphere whose surface is 314.16 sq. ft. Relation: 5 = 4 x if» x S.I4I6 314-16 = 4 X if« X 3.14I6 jp - SI4.I6 _ 25 Badiua = 6 ft. 4 X S.I4I6 NoTB. — The relation is stated in its natural form, and the given terms are « ihstitated in the relation. Find the convex surface of: Find the surface of: 143. A sq. pyr. ; side of base, 146. A sphere, whose 6 in. ; S. H., 4 in. radius is 6 in. 144. A cone ; area of base 50.2656 147. A sphere, whose sq. ft. ; alt, 3 ft. diameter is 10 in. 145. Acyl.; radius of base, 6 in. ; 14a A cube, whose alt., 4 in. diagonal is 12 in. Find the : 149. Radius of a sphere whose surface is 201.0624 sq. in. 150. Circum. of a sphere whose surface is 804.2496 sq. ft 151. Edge of a cube whose surface is 864 sq. ft 152. Diagonal of a cube whose surface is 150 sq. ft. 153. Find the convex surface of a frustum of a square pyramid, one side of upper base 4 ra, of lower base 8 m, slant height 24 m. 154. Find the convex surface of a frustum of a cone, radius of upper base 6 ft, of lower base 11 ft., altitude 12 ft 155. What is the approximate area of the earth's surface, its diameter being nearly 8000 mi. ? 156. What is the convex surface of a rectangular prism, base 8 ft. X 6 ft, altitude 10 ft ? 157. Find the cost, at 1^ a sq. ft., of painting a church spire whose base is a pentagon, each side 6 ft, and whose slant height is 60 ft. 266 VOLUMES COMPUTATIONS -VOLUMES XVIII. The volume of a prism is equal to the product of the area of its base by its altitude. XIX. The volume of a cylinder is equai to the product of the area of its base by its altitude. XX. Tfie volume of a pyramid is equal to one third the product of the area of its base by its altitude. XXI. The volume of a cone is equal to one third the product of the area of Us base by its altitude. XXII. The volume of a frustum of a pyramid is equal to one third the product of the sum of the areas of its upper base^ lower basCj and mean proportional base, by its altitude. XXIII. Tfie volume of a frustum of a cone is equal to one third the product of the sum of the areas of its upper base, loiver ba>se, and mean proportional base, by its altitude. XXIV. The volume of a sphere is equal to four thirds times the cube of its ra/iins times S.IJ^IG. Note. — For illustrations, explanations, and proofs, the papil shoald turn to p. 122, p. 261, and pp. 276, 277, and 278. 158. Find the volume of a square prism, each side of the base 4 ft, altitude 12 ft. Relation : V= B x Alt. Volume = 19S cu. ft. V= 16x 12= 19S 159. Find the volume of a triangular pyramid, each side of the base 4 ft., altitude 12 ft. Helation : V = ^Bx Alt. Volume : 27. 71 cu. ft. V= iV6 X2x2x2x.l2 = 27.71 s MENSURATION 267 ISO. Find the volume of a frustum of a cone, radius 6f the upper base 6 ft., radius of the lower base 8 ft., altitude 12 ft. Relation : V=^iB+B'-\-B") A It. 36 X 3.1416= B 64 X 3.1410= W 4S X 3.1416 = B" 148 X 3. 1416 = sum areas 4 59S X 3.1416 = 1859. 827 g Volume = 1859.8272 cu. in. NoTB. — For meaning of mean proportional, see p. 171, ITT. Id. The volume of a sphere is 113.0976 cu. in. Find its surface. B" = y/36 X 3. 14I6 y. 64 x 3. 14I6 = 6x8x3. I4I6 = 4Sx 3.14I6 It is better to extract the sq. rt. of the factors before performing the multiplication. Relation: V = ^x R^ x 3.14I6. 113.0976 = ^ X iJ« X 3.14I6 339.2928 R* R 4x R^x 3.14I6 339.2928 4 X 3.14I6 3 27 Relation : 6' = 4 x i2« x 3.14I6. S = 4x 9x 3.14I6 Surface = 113.0976 sq. in. NoTB — The relation is stated in its natural form, and the given terms are substituted iu the relation. Find the volume of: 162. A tri. prism ; each side of base 4 in. ; alt. 10 in. 163. A cylinder; radius of base 6 in. ; alt. 10 in. Find the required jmrt : 166. The vol. cyl. 3141.6 cu. in.; alt. 10 in.; D of bane f 167. The vol. cone 188.406 cu m ; alt. 8 m; area of base? 168 The vol. cone 336 cu m; alt 12 m; Dofbasef 164. A tri. pyr. ; each side of base 6 in.; alt. 12 in. 165. A cone; diameter of base 12 in. ; alt. 15 in. 169. The vol. of a sphere 904.7808 cu. in. ; Df 170. The vol. of a sphere 33.5104 cu. in.; Sf 171. The vol. of a cube 110592 cu. in.; Sf SIMILARITY SIMILARITY Two figures may be alike in form, similar figures. In order that figures may be similar, two conditions must be fulfilled : For eveini angle of the one, there must be a corresponding equal angle in the other. Tlie aides about the equal angles must be proportional. Similar Triangles Similar Quadrilaterals .1^ Similar Prisms :^d Similar Cones MENSURATION 269 XXV. In similar Jigures, linear parts are to each ether as homologous linear jyarts. XXVI. In similar figures, surfaces are to ea^h other as the squares of homologous linear parts. XXVII. In similar figures, volumes are to each other as tJie cubes of homologous linear parts. 172. In the similar triangles, find the ratio of AB to ab. AB:ah::S:l Prin. xxv. 173. Find the ratio of area triangle ABC to triangle abc. ABC:abc::fi:fi Prin. xxvi. ::9 : 1 174. In the similar cones, find the ratio of their surfaces. ::16:l Prm. xxvi. 175. In the similar cones, find the ratio of their volumes. vol.A:vol.a::S^:£^ Prin. ixvlL .' .* 64 •' 1 17& If the volume of the large prism is 288 cu. in., what is the volume of the smaller ? vol l8t : vol. Sd : : BL^ : 6d* In similar figures, vol- £88 : vol. Sd : : €*: 3^ umes are to each other as £88 X S^ ^^^ cubes of homologous vol. 2d = — — — = S6 linear parts. 177. The convex surface of a sphere is 100 sq. ft. What is the convex surface of a sphere whose radius is twice as long ? LoNO Way Short Wat Let X = radius small sphere Since the radius of the ix = radius large sphere larger is twice the radius sur. small : sur. large : : x' : (2xy of the smaller, the surface 100 : sur. large : : x-* : 4 x* of the larger must be the sur larae - ^^ ^ ^'^^ - AM Square of 2, or 4 times the ^ ^ ^^ smaUer, or 400 sq. ft. 270 SIMILARITY 178. If a pipe 2 mches in diameter discharges 40 gal. per min- ute, how much will a pipe 3 inches in diameter discharge ? Section Ist .section Sd : : S* : ^, or 'n.e amount of water discharged 40: discharge M :: 4 ' 9 depends upon the area of a trans- discharge 2d = ^^7^ = 90 verse section ; this section is a circle. 179. If it costs $1200 to build a house 20 ft. by 30 ft., what will be the approximate cost of a house the same height, 40 ft. by 60 ft. ? ^ . . . _. -^ -^ -^ .^ ^r» Since the heights are the same, CoBt 1st : cost gd : : tOxSO .'40x60 ^,,^ ,,^^^^^ ^^ „^, ^j„,i,^^^ ,^,j^^ ^^^^^ :: 1:4 depends approximately upon tlie area Rs : whole : . ■ P :ifi .-. ' 1 . '4 B'8=l off 2 = 50f A's = fL50 CostSd = f4S00 of the base 18a A and B buy a grindstone 4 in. thick and 2 ft. in diameter. A uses the stone until the diameter of the part left is 1 ft. If the whole stone cost $ 2, how much ought each to pay ? The whole and B's part are not similar because they have the same thickness. The cost of each part de- pends upon the area of the base. All problems under Similarity may be solved in other ways ; in some cases, it is best to neglect the rules on p. 269. 181. Find the radius of a circle having an area equal to the sum of the areas of two circles whose radii are 3 in. and 4 in. Relation: area of a circle = R^ x S.14I6. Areal8t = 9x S.14I6 Area Sd =16 x S.I4I6 Note. — In many cases, it is wise to de- Sum = 25 X 3.I4I6 fer multiplying and dividing as lonj; as pas- Sum = R^x S.I4I6 sible. S«e p. 72. See, also, Ex. mo, p. 267. — - ^^ In this example, it would be unwise to .-. R^ X 3.1416 = 26X3. I4I6 multiply \) by 3.U16. R^ = 25 R = 5 in. MENSURATION 271 182. The volume of a cone wliose altitude is 8 in., is 50.2656 cu. in. What is the volume of a similar cone whose altitude is 12 in.? 18a The convex surface of a fiiistum of a pyramid whose altitude is 12 in. is 432 sq. in. What is the altitude of a similar frustum whose convex surface is 3888 sq. in. ? 184. The area of a circle is 10 sq. in. What is the area of a circle whose diameter is twice the diameter of the first ? 185. Two lead pipes are respectively 1 inch and 2 inches in diameter. The area of a horizontal section of the larger is how many times a similar section of the smaller ? 186. How many lead pipes 1 inch in diameter, will discharge as much water as one pipe 4 inches in diameter? 187. A cannon ball weighs 32 lb. What is the weight of a similar ball whose diameter is half the diameter of the first ? 18a What is the ratio of the surface of the two balls in Ex. 187 ? 189. A is 6 ft. tall; his bronze statue is 12 ft. tall. If the length of A's little finger is 2J in., what is the length of the little finger of the statue ? 190. If it costs $ 1 to paint a statue of A's size, how much will it cost to paint the statue in Ex. 189 ? 191. If a statue of A's size weighs 500 lb., how much will the statue in Ex. 189 weigh ? 192. If a bin 6 ft. deep holds 60 bu., what are the contents of a similar bin 12 ft. deep? 193. The volume of a sphere is 100 cu. ft. What is the volume of a sphere whose surface is 10 times as great ? 194. Four pipes, each 2 inches in diameter, empty into a tank. What must be the diameter of a single pipe to carry away all of the water ? 195. A and B bought a ball of twine, 8 inches in diameter, for $ I ; A wound from the outside until the diameter of the part that was left was 4 inches. How much should each pay ? 272 PROOFS AND ILLUSTRATIONS PROOFS AND ILLUSTRATIONS — LINEAR PARTS I. The circumference of a circle is equal to ttcice the radius times 3. 1 416. For proof, the pupil is referred to geometry. We may illustrate by measurement Take a circle, as the base of a pail ; with string and rule, measure its circumference ; measure its diameter ; divide the circumference by the diameter; the quotient will be 3.1410 approximately. II. The square of the ht/potenuse of a right-angled tnangle is equal to the sum of the squares of the other two sides. For proof, the pupil is referred to geometry. We may illustrate by measurement. Draw two lines at right angles ; from the right angle, lay off .3 in. on one line, 4 in. on the other, and connect the extremities ; by measurement, the hypotenuse will be 6 in. ; 5* = 3^ + 4^. AREAS III. The area of a rectangle is equal to the product of its base by its altitude. This rectangle may be regarded as separated into strips by lines parallel to the base. The number of squares in each layer is the same as the number of linear units in the base ; the number of squares in each strip multiplied by the number of strips, is the number of squares in the rectangle ; .-. the area of a rectangle is equal to the product of its base and its altitude. MENSURATION 273 IV. The area of a parallelogram is equal to tlie product of its base by its altitude. B C Let ABCD be a parallelogram, and BE its altitude. To prove that its area = AD x BE. Draw CF parallel to BE and prolong AD X 1800F = tnooc=:800R 10F= |0C= |OR 18° F= 1U0C= 80E When the thermometer reads 14° F, the temperature below fr«H»zlnff js 1S° F, 10° C, or S° R, i.e., 14° F = - U>° C, or -80R. 14° 15. The temperature of a room is 68° F. Find the temperature in C ; in R. 16. Lead melts at 335° C. Find its melting .point in F. 17. Silver melts at 800° R. Find its melting point in C. la Alcohol boils at 78° C. Find its boiling point in F and in R. 19. On a certain day the temperature fell from 86° F to 5° C. How many degrees R did it fall ? MISCELLANEOUS 295 ACCOUNTS Bookkeeping is a systematic method of recording business trans- actions. Many or few books may be used. Accounts may be kept with many or with few things. The underlying principle is simple: Everything tohich owes the proprietor is a debtor (Dr.); everything the proprietor owes is a creditor (Cr.). From this general rule, special rules may be derived : Persons. Dr. when he owes me, or I get out of his debt; Cr. when I owe him, or he gets out of my debt. Cash. Dr. all cash receipts; Cr. all cash disbursements. Expense. Dr. all outlays from ichich no direct return is expected. Proprietor. Dr. everything withdrawn; Cr. everything put into the business. In the following, only two books are used, — the daybook and the ledger; accounts are kept with only four things, — Cash, Expense, the Proprietor, and Persons. At the time of each trans- action, William Harris or his clerk entered it on the daybook, as on p. 29G; each day the transactions were posted from the day- book to the ledger, as on p. 297. 20. Enter the following in daybook and ledger. Jan. 1, 1897. William Harris began a grocery business with $2000 cash and 9 2000 uidse. I'aid rent for January by bank draft, No. 500, 9 60. Books and stationery, 9 lo. Cash sales, 9&. Jan. 2. Sold A. I). Ku.s.sel, on account : 300 lb. sugar @ O-J ; 25 lb. coffee @ 30# ; 18p cwt. flour @ $2.50. Cash sales, $07.60. Jan. 4. Bought of A. I). Hussel, on % : horse and wagon for use in the business. 9 175. Cash sales, $25. Jan. 5. Rec'd from A. D. Kussel, check on First Nat'l Bank, for $250, to apply on his %. With ; 2 lb. tea @ Qbf. Cash sales, $21.15. Jan. 19. Withdrew for private use, $25. Cash sales, $.34.50. Jan. 20. Rec'd from Henry Cook, cash in full of his %. Sold Jas. Otis, on % : 50 cwt. flour @ $ 3 ; 5 gal. kerosene @\bf\ 1 doz. cans apples @ 15 f. Cash sales, $30. Found net gain from Jan. 7 to Jan. 21, $100. MISCELLANEOUS 299 FORM OF BILL Emporia, Kans., Jan. 5, lvh @ ffi 5 e^. fUwv ® f 2.80 R&t/ci ^a^yK&nt, i 75 / 25 /^ 00 /8 00 Zihn, /'fawtQ,. FORM OF BANK DRAFT jYo. ^ 60. 2ri)t iFirgt National 13ank of Emporia. Emporia, Kans., fa/rv. f , 189 7. Pay to the order of lAh>v /ifaviui-. J50.00.^^.^^.^.^.^.^<^t^.. To FirH National Bank, Kansas City, Mo. Dollars. MN) Cashier. FORM OF BANK CHECK '..y\ ' Emporia, Kans., fam,. 5, 189 7 y STfjt iFirgt National 93ank. '^^ Pay to the order of KMtv /ifawU^^.^^^.^ s^.^^^'^.^^.^.^^otu^ humdv&cC fvfty ^Dollars. a. Jb. Ruf,^. f 250.00. 300 GOVERXMEXT DIVISIONS OF LAND GOVERNMENT DIVISIONS OF LAND In the western states, the laud has been divided by the United States government into squares 6 mi. each way, called townships; each township has been divided into squares 1 mi. each way, called sections; many sections have been divided into 4 equal squares, called quarter sections; and some quarter sections have been divided into 4 equal squares, called quarter quarter sections. ATOWN8HIP A SECTION 6 5 4 3 ? 1 7 8 9 10 11 12 IS 17 1« 15 14 13 19 W n at S3 M SO » » V » e 31 s S3 3i » 31 K.iSecUoa (SMA.) S.W.J (160 A.) of B-i-i .K-jsTT The sections forming a township are numbered from the N. E. comer as shown in the left-hand diagram. The subdivisions of a section are named as shown in the right-hand dia- gram. A division smaller than a quarter quarter section is called a lot. 29. Draw a north and south line to represent 1 mi. Upon this line construct a square, and divide it into 2 equal parts by an east and west line. Upon the south part, write its name. How many rods long is a half section ? How many rods wide ? How many acres does it contain ? 30. Divide the N. \ section into 2 equal squares by a north and south line. Upon the N. E. square, write its name. ^ What are the dimensions in rods of a quarter section ? How many acres does a quarter section contain ? 31. Divide the N. W. \ section into 2 equal parts by a north and south line. Upon the east part, write its name. What are the dimensions and the area of a half quarter section ? 32. Divide the west half of the N. W. quarter into 2 equal squares by an east and west line. Upon each part, write its name. MISCELLANEOUS 301 DEFINITIONS AND INDEX NoTB. Z, angle; ±, perpendicular; A, triangle; O, circle; ||, parallel. Acute angle Acute-angled triangle Addends Addition Agent Aliquot part Altitude I Amount , -I Analysis Angle Annual interest Antecedent Apothem Arabic notation Arc Are Area Arithmetic Assessment Average Bank discount Base J Bonds Bookkeeping Brokerage . . Cancellation — Chord 250 251 16 16 198 116 253 261 16 190 216 180 249 216 170 253 10 254 138 121 7 213 58 228 145 190 253 200 212 296 207 72 254 DBFINITION An Z less than 90°. A A having 3 acute A. Terms in Eiddition. I*roce8s of uniting numbers into one. One who transacts business for another. A number contained an integral number of times in a given number. A ± from the vertex opposite the base, to principal plus Sum of several numbers interest. A process of solving problems. The opening between two lines which meet. A conception of interest by which prin. and int. on prin. at end of each year, bear interest. First terra of a ratio. Perpendicular from the center to one side of a regular polygon. Method of expressing numbers by the char- acters, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Part of the circumference of a O. Unit of metric land measure. Number of square units in a surface. Science of numbers. A tax on stockholders to meet losses or expenses. The mean of several unequal quantities. Int. on amount due on a note at maturity. A quantity used several times as an addend or as a factor ; that on which % is com- putet! ; the part on which a figure rests. Written contracts under seal to pay cer- tain sums at specified times. Art of recording business tran-sactions. Commission charged by a broker. Dividing both dividend and divisor by the same number. A straight line connecting two points in a circtimference. 302 MISCELLANEOUS TKRM Cipher Circle Circulating decimal Circumference Coefficient Commission Common denominator Common factor Common fraction Common multiple . Complex decimal Complex fraction Composite number Compound fraction Compound interest Compound proportion — Compound ratio Cone Consequent Convex surface of a solid . . Creditor Cube I Cube root Curved line Curved surface Cylinder Debtor Decagon Decimal DBFINrriON 40 252 110 263 145 108 88 72 76 105 I 83 ! i 68 83 216 174 174 260 170 261 295 238 260 244 249 251 260 295 252 103 Naught; zero. A plane fi^re bounded by a curved line every point of which is equally distant from a [>oint within called the center. Decimal containing a repeteiid. The boundary line of a O. A number 8howin<2: how many times a base is used as an addend. An agenfs pay for buying or selling. A denominator common to two or more fractions. A factor common to two or more numbers. A fraction whose denominator is wriiteu below a horizontal line. A number which will exactly contain eacli of several numbers. A decimal ending in a common fraction, a^ A fraction having fractions in one or bot'.i of its terms. A number having another set of factors besides itself and 1. A fraction of a fraction. A conception of interest by which prin.. int. on prin. at end of each jr., and all other int., bear interest. A proportion having one or both of its ratios compound. A ratio of a ratio. A pyramid whose base is a O. Second term of a ratio. All its surface except the bases. One to whom something is owing. Product of three equal factors ; a solid having its three dimensions equal. One of the three equal factors of a number. A line which constantly changes its direc- tion. A surface such that no two points lie in same plane. A prism whose base is a O. One who owes a debt. A polygon of ten sides. A fraction written with the aid of a deci- mal point. DEFINITIONS 303 TKRM p. 103 108 DRFINITION Decimal fraction A fraction whose denominator is 10, 100, Decimal point 1000 A period written before tenths in a deci- Denominate number Denominator Diagonal 119 83 253 254 28 1!)0 21)1' 207 44 207 44 44 200 299 223 223 2G0 159 231 261 87 239 146 170 260 68 68 mal. A number which answers the question ' how much ? ' The part of a fraction which is a divisor ; the number showing into how many parts a unit is divided. A line in a polygon joining two vertices not adjacent. A line through the centor of a O joining two points of the circumference. Result obtained in subtraction ; base minus percentage. A deduction from the face value of a com- mercial paper ; the amount the market value of a stock is below the par value. A number to be divided ; earnings of stock. Process of finding the other of two num- Diameter Difference Discount I Dividend | Division Divisor bers when one of them and their product are given. A number by which to divide. Dodecahedron A solid bounded by twelve faces. Draft A written order directing one person to pay a sum of money to another. The person to whose order a draft or note is to be paid. The person who signs a draft or note. The intersection of two faces of a poly- gon. An expression of equality. Process of finding an average time at whit 1: several payments may be justly made. A A whose sides are equal. Fractions equal in value. Process of finding a root of a number. A number denoting how many times the base is used as a factor. First and last terms of a proportion. Its bounding planes. An exact divisor of a number. Drawee Drawer Edee Equation Equation of payments Equilateral triangle Equivalent fractions Evolution Exponent Extremes Faces of a solid Factor Factoring Process of finding numbers whose product is given. 304 MISCELLANEOUS F. DEFINITION 7 Symbols representing numbers ; diagrams representing forms. Fraction 83 An expression of division in which the dividend is written above and the divisor below a horizontal line ; one or more of the equal parts of a unit. Pmstuni 200 The i>ortion of a cone or a pyramid in- cluded between its base -and a plane || to the base. Grace, days of 228 An allowance of three days after a note is due for its payment. Gram 141 Metric unit of weight. Greatest common divisor. U The greatest number tlmt will exactly divide each of several numbers. Heptagon 252 A polygon of seven sides. Hexagon 252 A polygon of six sides. 250 A straight line II to the horizon. The side of a right-angled A opposite the right angle. 253 Icosahedron — 260 A solid bounded by twenty faces. Improper fraction 84 A fraction whose value is 1, or more than 1. Indorsement 223 The signature on the back of a note. Agreement to pay for loss or damage. A whole number. Insurance 20i Integer 7 Interest 21« Money paid for the ilsc of money. Boundary line between regions where the International date line 293 calendar day is different, coinciding ap- proximately with meridian of 180°. Involution 2:w Process of finding a power of a number. A polygon whose angles are unequal. Irregular polygon 252 Isosceles triangle 251 A A having two of its sides equal. Least common multiple . . 77 The least number that will exactly contain each of several numbers. Least common denomina- 76 The least common multiple of several de- tor nominators. Liabilities 297 249 140 Debts to be paid. That which has one dimension. Line Liter Metric unit of capacity. A quantity represented by letters. Division in which processes are written out. Distance E. or W. of the prime meridian, as measured on the equator. Literal quantity 145 Long division 58 Longitude 292 Market value 207 22.] Price of stock in the market. Maturity of a note Date at which a note legally becomes due. DEFINITIONS 305 TKBM F. DEFINITION Mean proportional Mffin^f 170 170 68 60 249 186 136 28 145 83 68 76 35 36 35 145 198 262 7 223 7 84 7 68 68 250 250 260 261 262 260 8 249 One of the means of a proportion in which the means are equal. Second and third terms of a proportion. Factors. Parts of equations connected by the sign =. Science of measurement. Metric unit of linear measure. Measures of numbers Members Mensuration Meter Metric system Minuend . . Minus sign Mixed number A decimal system of measurement, whose principal unit is the meter. Number in subtraction to be diminished. Sign which denotes subtraction, or the op- posite of ' +.' An integer plus a fraction. Number wh ch contains another an integral number of times. Number to be multiplied. . Process of finding the sum when the same number is used several times as an addend. Number by which to multiply. A quantity preceded by the minus sign. Sum left after every charge is paid. A polygon of nine sides. Art of writing numbers by symbols. Written promise to pay a sum of money at a stated time. Unit or collection of units. Multiple 1 Multiplicand Multiplication Multiplier Negative quantity Net proceeds Notation Note ... Number Numerator The part of a fraction which is the divi- dend ; the number showing how many parts are taken. Process of naming or reading numbers. Numbers having no common factor greater than 1. Numbers prime, each to each of the others. Line neither ± nor || to the horizon. An Z formed by two lines which meet so as to make the adjacent A unequal. An Z greater than a rt. Z. A A which has one obtuse Z. A polygon of eight sides. A solid bounded by eight faces. One of the three places in a period. Lines in the same plane which do not meet however far they may be produced. Numeration Numbers prime to each other Numbers severally prime. Oblique line Oblique angle Obtuse angle Obtuse-angled triangle. . . Octagon Octahedron Order Parallel lines AMBR. ARITH. — 20 306 MISCKLLANEOUS TKRM p. DRFIMITIOM Parallelogram 251 60 206 223 223 262 100 190 253 7 250 251 145 249 202 251 200 145 238 202 227 292 68 198 216 77 260 26 A quadrilateral having both pairs of ius opposite sides parallel. Curved marks inclosing one or more quan- tities considered as a whole. Face or nominal value. Parenthesis Par value Partial payments Payee Payments in installments of notes, etc. The person to whom a note is payable. A polygon of five sides. Hundredths. Computations with per cent. Distance around a polygon. In numeration, groups of 3 figures each; in evolution, groups of any number of figures. A line which meets another so as to make the adjacent A equal. A surface such that a straight line joining any two points in it, lies wholly within that surface. Sign which denotes addition or a positive quantity. That which has no dimension. Pentagon Per cent . . Percentage Perimeter Periods Perpendicular line Plane Plus sign Point Policy A contract between an insurance company and the person insured. A plane surface inclosed by straight lines. A tax levied at so much per head. A quantity preceded by the plus sign. A product obtained by using the same number several times as a factor. Sum paid for insurance ; excess of market value above par value. Present value of a debt due at a future Polveon Poll tax Positive quantity Power Premium Present worth Prime meridian . . . . time. The circle passing through the poles of the earth and through Greenwich, Eng. A number which has no integral factor besides itself and 1. One who employs an agent; a sum on which interest is reckoned. A fundamental truth. Prime number Principal | Principle Prism A solid having its two bases equal and || polygons, and its faces rectangles. An example in which the operations are not stated. Problem DEFINITIONS 307 TBKM p. DBFIMITION Proceeds Product Proof 228 36 79 170 200 254 251 44 254 170 251 85 252 28 48 110 251 251 250 251 15 239 251 254 254 264 254 254 206 Difference between face value of a note and the discount. See " net pro- ceeds." Result in multiplication. Operation checking accuracy of a calcula- tion. An equality of ratios. A solid having one base, and its faces A. Quadrant A sector equal to ^ of a O. Quadrilateral A plane figure bounded by four straight lines. Result in division. Rftdius A straight line drawn from the center of a O to the circumference. An expression of division in which tlie dividend is written before, and the divisor after, a colon. A parallelogram whose A are all rt. A. Process of changing the form of an ex- pression without chanj^ing its value. A polygon which is equiangular and equi- lateral. The result in subtraction ; the part left undivided in division. A figure or set of figures in a circulating decimal which repeat. A parallelogram whose A are not rt. A. A rhomboid whose sides are equal. An Z formed by two lines meeting so as to make the adjacent angles equal. A A containing 1 rt. Z. Method of expressing numbers by means of the characters. I, V, X, L, C, D, M. One of the equal factors of a number. A A which has no two of its sides equal. A straight line cutting the circumference of a O in two points. A portion of a circle between two radii and their included arc. A portion of a O between a chord and ite arc. A segment which is ) of a O. A sector which la J of a O. One of the ecjual parla into which a capital stock is divided. Ratio Rectangle Reduction Regular polygon Remainder | Repetend Rhomboid Rhombus Richt ancrle Right-angled triangle — Roman notation Root Scalene triangle Secant Sector Segment Semicircle SexUnt Share 308 MISCELLANEOUS F. DSriKITIOK Short division 61 268 216 170 261 260 160 294 260 238 251 260 240 293 139 206 28 28 16 251 254 200 60 260 294 201 159 Division in which the operations are per- formed mentally. Figures whose A are mutually equal and whose homologous sides are propor- tional. Interest on the principal alone. An equality of two simple ratios. Distance from vertex of a pyramid to base of one of its faces. That which has length, breadth, and thick- ness. Process by which the answer to a problem is obtained. The ratio of the weight of a substance to weight of same volume of water. A solid bounded by a curved surface, every point of which is equally distant from a point within, called the center. The product of two equal factors ; a rec- tangle whose sides are equal. A prism whose base is a square. One of the two equal factors of a number. Similar figures Simple interest Simple proportion Slant height Solid Solution Specific gravity Sphere Square Square prism Square root Standard time Railroad time. Stere Metric unit of wood measure. Stock Property owned by members of a charter company. Process of finding the other of two num- bers when one of them and their sum are given. A number to be subtracted. Subtraction Subtrahend Sum The result in addition Surface That which has two dimensions. Tangent A straight line which touches the circum- ference of a O at only one point. A sum assessed on a person or on property to meet public expenses. Expressions separated by the signs '-I-* or * — ' ; the numerator and denominator of a fraction ; nomenclature in the vari- ous operations. Solid bounded by four equal A. An instrument for measuring temperature. Discount from the list price of goods. Transferring a term from one member of an equation to the other. Tax Terms Tetrahedron Thermometer Trade discount Transposition DEFINITIONS 309 p. 251 261 251 260 227 84 167 260 260 60 261 DK7INITIOM TraDezium A quadrilateral having neither pair of its opposite sides ||. A quadrilateral having one pair only of its Trapezoid Triangle Triangular prism True discount opposite sides II . A plane figure bounded by three straight lines. A prism whose bases are i^. Principal minus true present worth. One, a single thing. Quantities whose values are sought. Union of three or more edges of a solid. A line ± to the horizon. Unit Unknown quantities Vertex Vertical line Vinculum A straight line placed over two or more quantities to show they are to be re- garded as a single term. The number of cubic units in its contents. Volume of a solid ANSWERS pp 25 to 41 pp. 41 to 41 Addition PP.2&-27 140. 2,0»),810 141. 1<5.2»R),«:J4 142. l«»,o<>:^,43«) 143. I\), 164. S21.()*H'. 155. S87:n 156. ()2:i8 trees 157. 1545 miles 158. 7:) ,<)}«) men 169. 10,2.S7,5:« square miles 160. 2«),002feet 161. 4915 miles 162. 2843 bushels Subtraction pp. 33-34 85. 12 acres 92. 8200 98. 87,590 .square miles 94. 1007 miles 96. 85056 96. 204 trees 97. 87536 98. 460 miles Multiplication p. 41 88. fi,4'J4,:<02 84. HH5,715 86. ii,7l9,:«.s Multiplication p. 41 86. 1,992,^8 87. 7,714,620 88. 7,180,9a5 89. 5,178,576 90. 8,300,03,62.') 101. 2,201,892 102. l,124,(Ui5 103. 2,070,7:« 104. 2<>,8.'i9,84 117. 478,70<).43<> 118. 742,9»«,9.')4 119. 41fi,(r>9,.^4 120. «)82.880,000 121. 431,257,7r»<> 122. 212,536,«r)4 123. 442,928,901 124. 2.'^,.■»4,^>.')4 126. ri04,849,8(»8 126. 620.919,a50 127. 621. aw.' "OS 128. 490,119.007 129. .'>,'>5,«»88,«W8 130. .•«5,580,:{94 181. 3i;7.W>8,400 132. 6V;.8.''.9.0»)8 133. l(M;,:vk').(v4s 134 4;<,48(),'.t90 pp. 41 to 54 Multiplication pp. 41-43 136. 482 138. 226,94<;,000 139. 1(>1»,'»*.W»,126 140. 8<),079,021 141. 2;i2..')78,200 142. l();i,<)19,026 147. S:K)24 148. 41,0<)2,000 stitches 149. S586 150. 226,176 letters 151. 91,289,79() miles 152. 70,215,270.400 cubic vards 153. S 16.978 164. S4«)81 155. 146,289 miles 156. 810,285 Division pp. 63-64 128. 364; 6:189; 21 124. 138:364; 18 126. 215; 5:476; 12 126. 711; 9:2.')2 127. 5:i8; 2:408:3 128. 1875:367; 3 129. 5002; 4:3<;9; 10 130. 4000; 4; 322 181. 2011 : 15: 262; 8 132. 3179; 23: 169; 12 136. 2 137. 4 138. 5 139. 140 141 142 143. 9 19; 206 26; 156 !!•; 19f> pp. 64 to 66 Division pp. 64-66 144. 12; 49 146. 18; 187 146. 4 147. 6 148. 8 149. 5 150. 9 166. 30mH 156. 175tXSll 167. 210il8» 168. 90]i;:i 169. 3089; 30 160. 4362; 1 161. 2931 162. .•082 163. 4107 164. 2573; 62 165. 3871 166. 38,413 167. 29,615; J 168. 24.542 169. 45,684 170. 41,075 171. 90,009 172. 80,808 173. l.'iJW 174. 2009 176. 806 176. 374 177. 423 178. :«33 179. 3949 180. 31 181. 200 182. 220 188. \ri; 103 184. 106 186. 237 186. 45 187. 468 188. 387 189. 267; 2076 190. 214 ; .T254 191. 25; T.m 192. 61; 176.3 193. 1.^; 2849 317 125; 1000 218: 2008 797 1«>36 194 196 196 811 312 AiNSWERS pp. 56 to 66 pp. 66 to 71 pp. 71 to 81 pp. 82 to 90 Division pp. 5(>-69 197. 476; 14134 198. 313; 12988 199. 127; 15333 200. 117; 899 205. 81300 206. 60 people 207. 64 bushels 208. 8136ilt 209. S26 210. 85500; 83000 211. 6 years 212. 113 days 213. 550 sacks 214. 12 miles 215. 5 farms, 100 acres 216. 82 217. 168 pp. 218. 3(>i months 219. 823 220. 65 times; 830 acres remain- ing 221. 226 222. 102 acres 223. 8152 224. 160 tons 225. 18 cents 226. (>4 hours 227. 15 feet Operations Com- bined pp. 61-66 13. 11 14. 35 15. 11 16. 3 17. 11 18. 4 19. 36 20. 8 21. 30 23. 58 51. 39a5 votes 52. 3(«9 votes 53. 8 payments 54. 836150 55. 12 miles Operations Com- bined pp. 66-67 8 miles 10 hours 2 miles 3 hours 810 8 weeks 27 sheep, 8 13 left 82022 81680 8 308 8403 67. 85470 68. 8 2072 2262 rods 400 feet 3 days 30 boys A.9f; D.3# 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 70. 71. 72. 73. 74. 81479 Factoring pp. 70-71 29. 2,2,2,5,3,3, 30. 31. 48 44 45 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 160 161 162 163 164 165 169 170, 34. 7, 11; 4, 6 9, 10, 12, 14, 15, 18, 20, 21, 22, 24, 28, 30, 33, 35, 36, 40, 42, 44, 45, 55, 56, 00, 63, 66, 70, 72, 77, 84, 88, 90, S« 2,3,3,11,89; 6,33.9,22,66, 99,18 5, 13 ; 65 1,2,3,5,7,11, 13, 17, 19, 23, 29, 31, 37, 41,! 43, 47, 53, 59, 179 61, 67, 71, 73, i 79, 83, 89, 97 ; 26 numbers '180 Yes, no, yes, j 181, yes, yes, no 1 182 175 176 Factoring pp. 71-81 409 2. 2, 2, 3, 5 ; 5, 11, 2, 2; 2. 2, 2, 2, 3, 3, 3, 13 2,2,3,3,3; 2. 2,2,2,3.3; 6, 17,17 2,2, 3, 13; 2, 2, 2, 2, 2,6; 2, 2.2,2,2,3,7, 11 , 3, 3, 3, 5; 3, 3, 2, 2, 2, 2, 2 ; 5, 5, 11, 11 165 95 144 106 840 626 950 17 1 13 107 13 11 9 7 14,400 55,125 217,728 34,476 43,890 53,100 1 2, 3, 4, 5, 6, 7, or 8, times 1 2, 3, 4, 5, 6, 7, 8, or 9, times 1 2, 3,4,5,6,7, 8, or 9, times 1 12 feet for ends, 16 feet for sides 60 yards 25 feet 4 feet Factoring p. 82 184. A and B, 24 min.; A and C, 18 min.; B and C, 72 min 185. 20 times 186. (iOcKfjs 187. 63 eggs Common Frac- tions pp. 85-90 21. iS; i 22. i; }i 23. f: t 24. il; }{ 25. ii; f^ 26. !; H! 27. 15; ii 28. ?lii; H 29. A; I 39. Ml 40. Itil 41. Mm 42. /AiV 54. ft'.,fA,m 55. m,i!l, i» 56. t;:. Ii}, lii 57. m\,^^,^^ 58. ilH,ifll, mi 66. 23/, 67. 64il 68. 55H 69. :mn 70. 152 AV 71. 80SI 77. W; W 78. ^?i*; *f|^ 84. 12838 85. l«v)SS 86. 53^ 92. II; 40}» 93. lii; 45/% 94. II; mn 107. zh 108. 2079 109. i 110. i 111. A 112. } ANSWERS 818 pp. 90 to 94 pp. 96 to 100 pp. 100 to 106 pp. 106 to 111 Common Frac- tions pp. 90-94 lis. 6 114. 5k 115. i 116. 2 165 166. 1,^ 167. m 168. A 169. fr 170. 8* 171. 32 172. 7J 173. 22 174. 31 175. lA 176. 411 177. 6i 178. 6A 179. 3i 184. 62 A% 185. r>l»|J5 186. 2()iii 187. 52.Vo 188. 81M 189. 73IH 190. 60!ii 191. 68i( 198. 23AV 193. 21111 194. 15<»;!1 195. 2()2ai 196. 15.V14 197. 32(M5 198. 178981 199. 171003A 200. 187<»/r 201. 2H.'>12!| 202. ♦ViiM.'iJ 203. Av.mn 204. 1801 905. 31151 206 umt 207 WiUkt 208. 2124 209. niH 210 211 l.iUI Common Common Decimals Fractions Fractions pp. lOG-lll pp. 1)6-100 p. 100 68. 9 thousand. 219. 1: i 270. A and 9 thou- 220. U; 14 271. 108 sheep sand 372 ten- 221. i; A 272. 12 apples i'* of whole millionths 222. 1; t. 273. 68. S VXiH.\m 223. f ship 69. S.374.(;!»3 224. 1 274. 10; 6 70. 1?2H2:<.4016() 225. 1 275. 120 apples 71. 68.2711 226. 1 276. «r,6 72. 13.1.351732 227. U 277. :«5 cattle 73. 64.0255 228. 5 278. SI 74. 9821.631 229. 9 279. A, 30 years 75. .'>4.864 230. 10 280. (K) years 76. .yas 231. 19i 281. 5 days 77. 9i).13126 232. 1 282. r)0 yards 78. 768.04 236. i 283. 8 days 79. 1.478<)94 287. 1 80. 408.5081 238. if 81. .32.'«»775 239 U 242! Sura, 39A; Decimals 82. 83. 449.95854 77477.4 dif.,5H; p. 106 84. .5021.75 prod., 3793; 85. {»7.'-.3.76 quo., IjVi; 31. .000500 86. 248.^9.02736 prod, of sum 32. 17.7 87. .015 and dif., 33 .000006.38 88. 13.1489-1- 233A«r 34. .000000000023 89. 3.10(53 + 243. 156584 35. .000085002 90. 153.482 -h 246. A 36. .7007 91. .03()3-H 247. !t 37. .oo;H)0 92. .00633 -f 251. 99^ 38. .604 93. 6.7808-}- 252. 54^ 39. 9431.0906532 94. .0(K)0()01 253. 24,25,81 40. .0000! 95. 10,000,()00 254. $16 41. 2.000,000,000- 96. .01599 -h 266. I .000,000,002 97. 1. 05997 -H 256. 4 42. 8:wu).oa') 9P. .01399 + 257. AV 43. .OJ^^nX);) 99. 2,000,000 258. 6 acres 44. .()()100<'»<)04 100. .000000176 + 259. 3U yards 55. 98 million (VM 108. tIi : tit 260. 18 coats thou.sand 2«»S 109. A: •!< 261. 41 hundred -mil - 110. A: A 868. 1 in com lionths 111. i4ii: A 863. 8 dozen 66. 8;J million 8 118. t4.; 411 864. 24 years hundred -mil- lis. A: A 866. $6 lionths 114. Ti.;4i 866. S 15000 87. 9 thousand 2 116. 4; 411 267. :i apples hundred, and 116. sm 268 <:ii 929 ten-thou- 117. 441; 1 269. $-V, sandths 118. AVb; A 314 ANSWERS pp. Ill to 118 pp. 180 to 181 pp. 181 to 184 pp. 184 to 189 Decimals pp. 111-118 119. /«: xA^ 120. iV»: hi 121. A: ^Afc 126. .1H75; .10 127. .117187:.; .OUli 128. .7525 129. .87.i75 130. .(J00625 131. No; 13 is a factor, other than 2 or 5 132. .H44>A 133. .H4t>H- 134. .846163 143. nn 144. m 145. m 146. MAP. 147. 7A 148. 8ilf8 149. :K>iIi 150. ,V 218. S«4.71876 219. » 135.94 220. $(>.67 221. « 48.29 222. $54.69 224. S 64.19 226. $ti4,395.06; S64,31>5.0t)l 227. 20 decimal places 228. 234.63 X «(1.06)S, or $ 313.91) 229. S 64.68 230. 420.16+times 231. 2695.4928 feet 232. 41.661 days 233. $.004 + 234. § 3(r).25 (5 leap years) 235. 9.278 +bu. 236 8000 pounds 237. $23.76 238. 40(1.64 bu. 239. $218.24 240. $327.24 Denominate Numbers pp. 130-131 154. 225 pt. 155. 7S46far. 156. 1860 drops 157. 10,079 gr. 158. 12cwt. 561b. 159. 110 rd. 2 yd. 7iin. ■ 160. 88 .sq. rd. 161. r.Oinin. 162. 1 mo. 15 da. (30 da. to the mo.) 163. 61 so. rd. 18 sq. yd. 1 sq. ft. 501 sq. in. 164. 11 cu. ft. 432 cu. in. 165. 6 oz. 13 pwt. 8gr. 166 2 s. 9d. 3.936 far. 167. lpk.2qt. 1.024 pt. 168. '2M rods 2 yards 1 foot 4 inches 169. 7 cwt. 77 lb. 12t oz. 170. 95 33 13 3.904 gr. 171. lOP 42' 172. 3qt. 1 pt. 173. 30 .sq. mi. 177. 8 bbl. 6 gal. 1 qt. 1 pt. 1 gi. 178. 80Z. 6gr. 179. in> 5 5 73 2 3 17 gr. 180. 8eh. 2rd. 181. £1642 4s 182. 234 rd. 1 yd. 1 ft. 1 in. ; 183. 1 sq. rod6sq. yards 2 sq. feet 83 sq. j inches Denominate Numbers pp. 131-134 184. 1 CO. vd. 1 cu. ft. 10 cu. in. 185. iti tti. 186. iiioo Tp. 187. lU^K 188. iV\AA>mi. 189. 2ii ed. 190. '211U T. 191. :u;2r.gal. 192. r).H59:i75 bu. 193. i:«;.425 194. 7.4<;01»526T. 195. .112 + C. 196. 5.41071 + yr. by 1st meth. ; 6.378 +yr. by 2d meth. 203. 7ty>i lb. 204 5.5r.f tb 205. 24 bu. 206. 21 J hbl. 207. 2.9rK5-f-bbl. 208. 46igal. S09. •20 min. 20J sec.; 24 min. 24 sec. 210. 76^ 16' 15" 214 2.1 T. 8 cwt. M2 lb. 14 oz. 215. 27 rd. 2 yd. 1 ft. 216. 27 lb. 3 OZ. 7 pwt. 15 gr. 217. 51 yr. 10 mo. 21 da.3hr. 218. 178 rd. 1 yd. 2 ft. 6 in. 219. 178.3.31 rd. 220. 4r>2<).6 gr. 221. Oct. 24; Nov. 13 222. Oct. 26; Oct. 6 226. 2 lb. 5 oz. 16 pwt. 19 gr. 227. 10 rd. 1 yd. 2 ft. 1 in.; 1 A. 18 sq. rd. 240. 241. Denominate Num- bers pp. 134-139 828. mda.; .5291+ da.; 12 hr. 41 min. Ml sec. 229. 213 da. 280. 7 mo. 1 da. 231. 2 da. 235. 84^ 64' 24" 236. 25 bn. 3 pk. 4 qt. 237. 16 gal. 3 qt. 1 pt. 238. 41 rd. 1 yd. 239. 30 da. 10 hr. 59 min. 40 sec. 17 8^. yd. to sq. in. 19 hr. 30 min. 30 sec. 242. KiO rd. 1 ft. 1! in. 243. 1 T. 15 cwt. 114 lb. 244. 2/r 245. 5 mo. 5 da. 4 hr. 1246. 723? 247. 3.1 248 404i 264. 40,000.000 m; 1,574.800,000 in.; 24,854.79+ mi. 267. 1,000,000.000,- 000 sq mm; 1 .000,000 sq Dm 268. 10 269. Lay off a square 10 m X 10 m 272. 185.28 A. 276. 1,000,000.000 cudm; 1,000,- 000,000 cu Dm 277. Lay off a cube 1 m X 1 m X 1 m ANSWERS 315 pp. 139 to 144 pp. 149 to 156 pp. 156 to 169 pp. 159 to 163 Denominate Num- bers pp. i;«>-144 280. Wood at $3 per cd. 281. ^4»00 284. ZitAHl; 385.(i23 cu m 8.192 cum $ 180.22 176.57+ cu. ft. ex. ; 180 cu. ft. approx. 887. 73.52+bu.ex.; 80 bu. approx. 838. 41.18+ Kg ex.; 45^, Kg approx. 1067(iO ft. ; 20 mi. TOrd.ri ft. 3 yd. 2 ft.; 1 rd. 127 rd. 1 yd. 3.6 in. ; .3H75 mi. 842. «7.r) lb. troy; 72 lb. avoir. l«5.8bu.; ir)7.5 16.9 bu. ; 157 JS gal. 2 hr. 35 min. 22 sec.; 38° 50* 30" 846. 25 gal. 2 at. 1 pt. ; 6 gal. 3 847. Sept. f; Mar. 3 848. Sept. 5; Mar. 3 849. 283 yr. 8 mo. 13 da. 860. 174 da. 861. g000cucm;91 862. 5000 Kg; 6cu 868. $13.34 864. 962.92 287 288. 886 839. 840. 841. 843. 844. 845. Literal Quantities pp. 14!^15«J 39. 3f«-3x-«+2 40. 3a^b-bci 51. -i« + 5x*y 52. x^y - arV - 12 2T/« + 6 58. 2u26^ + 5ai6c + 6 ab^c + 2 fee* 56. «»; x«; ... 57. «7; a*; ... 72. x* + xiy^-\- y* 75. a6 + 5a<6 + 10(/«62 + + 66 76. a8 + 6«-c« + 3 «62 + 3 a^b + 3ac2 — ."kiV — 6a6c+36c'^ -3 62c 79. a<; a*- ... 80. x; zo, or 1; ... 90. x*-x»y+x^i — xy* + y* ; 5x-2y 91. X2 + XV+I/2; x*+x^y + y* 92. 3x + 2y; x» + zV + af»y« + »/» 98. 3a6(a + 6); 6(z«-2z2 + 4z-l) 94. 12a262(6-a); 7(2x«i/-4xi/2 + xf/ + 2) 96. 12«262(2a6 -3); 2xw(6xy— .1 V +4-2X) 96. 10aS(2o&4 -3); 5i/(2x«-4x«v +2xy*-y*) Literal Quantities pp. 15(;-1."»9 97. 3a«62(4 +3«62); 6xi/(3x» +5x2y -Axy-i +'2yh 103. r)u26 104. 2a86 105. a26 106. 3«262 107. 2'i2 113. 12 xV 114. 30x«v2 115. 6x«y2 116. 42 xV 117. 48x«y6 a26. J_ 2 ' 6x 3 . 2x 8x6y' 3y« 3 . 2 5a6c' 3 1 . xji Aabc' 3 2o62 ^ a262' a262' 462 2o? a-ib^' q262 181. 21X + 9. 24x 16 X - 6 24 X 134. -I 186. 9^_5y 24 187. ^ 124. 125. 126 127. 129. ^? '2^, 139. 25y 3 a 5 6 140 x«8 141. x = 3 142 xr=4 148 x = 4 144 -2 Literal Quantities 15«^l(i3 145. x = -2J 146. x = 2 147. x = 3 148. x = 26 149. x = 2 150. x = 8 151. x = 2 152. x = 16 153. x = 2 155. 3() boys 156. 15 on 1st; 30 on2d;*i0on3d 157. $(;00, house; «1200, barn; $ 1800, land 158. 16 horses, 48 cows, 144 sheep 159. 45yr.,A'sage; B's.it; C'8,27 160. »12, A's share: S24, B's; $144, C's; »84, D's 161. 39 in. 162. 24, first part; 8, second ; 64, third 163. 21»1 164 $4 167. 11, the less; i 3;^, the greater 168. rn 169. 21 yr., Anna: 2.'» yr., Mary 170. iri 171. IH. the les-s: 42, the greater 172. 25 yr., son; 45. father 173. 3, the less: 31. the greater 174. 9, the less ; 25, the greater 176. 80, the less; 100, the greater 176. 96 gal. 816 ANSWERS pp. 163 t3-lti8 1(>8-I(il» p. 17D lems pp. 184-189 177. 12, first part; 221 x = 3, y=4 51. $ 114, A ; 24. 2 cts. 16, second; 222 3c = 4, y = 3 $:«8. B; 25. 2 doz. eggs 13, third 223 x = 2, i/«=l $21U,C 26. 15 eggs 178 S3 224 x = l.y = l 52. $«r2..'i0. A; 27. 84 ct«. per doz. 179. SSTiOO 225 X = 6, y = 7 $87.50, B 28. 5 cts. 180. «5400 226 x=42. y = W 68. $240, A; 29. 120 apples 181. $1500 228 8,12 $320, B; 31. $2 182. (}yd.;9yd. 229 $1, potatoes; $400, C; 32 20 da. 190. x = 6 55 cts., apples 9480, D 33 $4 191. z = 7 230 1<> girls, H* M. 817000, wife; 34 224 da. I2I da. 192. x=8 b(»y8 $6000, each 35. 193. a: = 8A 231 « son; $4000, 39. J 194. z=10 232 $50, A; $70, each daughter 40. A: A; 21 da. 197. 24 14. lU mo. 48. $10 203. 25 horses 15. 40 mi. Solution of Prob- 49. $28, A; $35, B 204. IG ft. wide, 22 16. 1*23 ft. lems 50. 2 cts., Henry; ft. long 205. 30 apples 17. $16 pp. 181-184 4 cts., James 18. 1.^ bu. 51. $ 1680 each 206. $120 19. ■S 3750 2. H min. space 52. $6720, A; 207. 49,50,51 20. 120 men 8. r^m min. sp. $3840, B; 208. $140, horse; 21. 2.SI da. 4. 5 min. sp. $10,752,0; $140, car- 22. Tida. 5. 25 sp. $22,400, D riage 23. <> men 6. 5fi min. past 5 57. 5; 4 209. $40, first; 24. IH hr. 7. 27A min. past 2 58. 15; 20 $75, second 34. 102 da. 8. 50; liSmin.sp. %: 40: 25 210. $600, A; 35. $1750 9. 46,^ min. past 4 $72 $:ftK), B 36. 750 da. 10. 43/i min. past 8 61. $45 311. 48 cts., first 37. :muo lb. 11. 12 mi. 62 $50 son; 56 cts., 38. 28 da. 12. 10 hr. 63. $27 second; 39 39. 6 da. 13. 60 mi. 64. $8 cts., third 40. 55.125 lb. 14. .% mi. 65. $60. cost; 212. $325 41. 7hr. 15. 12 mi.. A; 14 $ 100, selling p. 213. 50 A, A ; 65, B: 42. 4k ft. mi.,B 66. $200, cost; 115, C ; 33, D 43. $24 17. 1 hound leap $ 180, selling p. 215. x = l. y = 2 44 .>40 pp. = 5 fox leaps 67. J 216. 2 = 2, y = 3 45 24 Iwys 18 1 fox leap 68. 1 217. x=-l, v = 4 49. .■* 102, 1st part; 19. 72 times 69. $540 218. x = 6, y =2 $ln3, 2d part: 20. 252 times 70. 30 ct. 219. «=-!, v = 2 $2.5.5, 3d part 21. m leaps 71. $10, com.; 220. x = 12, y = 4 50. $1,t5 23. let. $190, proceeds ANSWERS 817 318 ANSWERS pp. 218 to 227 PP . 227 to 235 PP . 235 to 245 pp. 247 to 256 Interest Interest Interest Involution and pp. 218-227 1 pp. 227-235 pp. 235-2:>7 Evolution 24. $84.61 85. $.300 133. $733.99 pp. 247-248 25. S1.48 86. $2a'5.97 134. $ 1748.28 60. 460 26. S6.51 87. $114.19 138. $271.2.3 61. 562 27. «5.48 88. $ 1(».-..8G 139. $261.19 62. 1022 28. $6.16 89. $i;J7.18 63. 1013 29. $5.26 93. Due March 10; 67. .1539+ 34. $4.57 $4, bank dis. Involution and 68. .8735+ 35. $42.43 94. $1.60 Evolution 69. i 36. $02.88 95. $1.:J4 pp. 239-245 70. .12 87. $17.26 96. $3.96, true 71. .a5569+ 48. 10% dis. ; banker 14. 32 72. . 128. $381.18 64. 68 68. 5 in. 83. §2'.)7.44 pres. worth; $2.56 129. $19.94 66. 63 69. 24 in. 130. .5 18.57 66. 89 70. 28 in. true dis. 131. $9.45 67. 53 71. 40 ft. 84. $400 132. $.5^0.12 68. 55 72. 119 ft. ANS\ VEI IS 319 pp. 866 to 867 PP . 867 to 281 pp. 881 to 884 PP . 884 to 888 Mensuration Mensuration Occupations Occupations pp. •J5(>--JG7 pp. 2(J7-271 pp. 281-284 pp. 284-288 78. 374 ft. 167. WAim sq m 34. 3 sq. dm 188. 1:4 43. 24(» bd. ft. 80. 90. 1 rd. 189. Sin. 44. 22r) b sq. 55. 27J sq. ft. rolls ft.; 8:10 1. 8bd. ft. 56. Agrees nearly 90. $6.48 (strips 148. 48 sq. in. 8. 4 bd. ft. 57. 31i sq. ft. on ceiling run 144. 62.8;« sq. ft. 4 9 bd. ft. 68. Agrees nearly 61 bunches, by lengthwise) 145. l.'i0.71HHi sq. in. 5. 12 bd. ft. 59. 91. No ; strips 146. 452.3904 sq.ft. 147. 314.16 sq. in. 6. 15 b«l. ft. rule ; rafters, must match 8. H bd. ft. 17.77 ft. Ion- and be whole 148. 288 sq. in.. 10. 12 ImI. ft. 60. .•«»20ft. 06. 1(H.18 bbl.; entire surface 11. 20 bd. ft. 61. 320 bd. ft. 57.87 cu. yd. 149. 4 in. 12. 27 bd. ft. 62. [m bd. ft. 07. 5 masons 150. 50.26S6in. 14. :w4 bd. ft. 63. 12 sq. yd., bv 08. 1.^.7,520 bricks 151. 12 ft. 15. 40bd.ft. rule ; VM sq 99. S 748.44 ; 158. 8.6,fo0sq. 18. 240 boards 104. The number 164. 62.362 cu. in. 31. 11>2 ImI. ft. 78. 60 ft. of boartls of cu. ft. in 166. 565.^88 cu. in. 88. •Mi M. ft. 78. 115 ft. of walls is 4 166. 20 iu. 83. VA bd. ft. boards times as many 320 ANSWERS pp. 288 to 289 pp . 289 to 291 pp. 291 to 298 pp. 293 to 294 Occupations Occupations Occupations Miscellaneous pp. 288-289 pp. 289-291 p. 291 pp. 293-2iM 105. 42 cd. 117. 85+ bu. 181. 322 rd. 4. 6 hr. 45 min. 106. ItiiiJ? i>erch 118. 12Hbu. !1S2. 50.5(i+ rd. •M sec. 107. r»2i bbl.; 42 119. 4+ bins 133. 44.8 rd. 78^ 27' 35" W. cu. yd. 120. 2bu. 184. 80 trees; each * 3ftfg sec. past 108. Si« 122. 2 ft. 1 in. tree the cen- 2.30 A.M. 109. Ill bbl.; 4 cu. 123. 282.7+ bbl.; ter of a 2 rd. Ihr. .vd. count 7i gal. square West 110. 4fiO|bu. 1 cu. ft. 136. 15 ft. long 100 days 111. 36841 bu. 125. 24 T. 186. 40 bbl.; 126 hr. 7.7 112. 460tbu.; 126. 6Ami. 48U lb. 25,804.8 lb. 127. 7Arai. i 113. 5r»5? bu. ap- 128. 3240 bills; 37nb. proximately each hill the 2i 114. 7