University of California • Berkeley The Theodore P. Hill Collection 0/ Early American Mathematics Books THE PEOGEESSIVE RACTICAL ARITHMETIC, CONTAINING THE TUEOUY OF NUMBERS, IX CONNECTION WITH CONCISE ANALYTIC AND SYNTHETIC METHODS OF SOLUTION, AND DESIGNED AS A COMPLETE TEXT BOOK ON THIS SCIENCE, FOR COMMOxN SCHOOLS AND ACADEMIES. BY DANIEL W. FISH, A.M., AUTUOE or THK TABLE-BOOK, PBIMAKT AND ISTELLEOTCTAL AEITHMETICB, AND RUDIMENTS. NEW YORK : IVISON, PHINNEY, BLAKEMAN & CO., CHICAGO : S. C. GRIGGS & CO. 1 864. RO B I ISr S ON'S The most Complete, most Peactical, and most Scientific Series of Mathematical Text-Books ever issued in this country. I. Robinson's Progressive Table Book, II. Kobinson's Progressive Primary Arithmetic, - - - - III. Kobinson's Progressive Intellectual Arithmetic, - - - IV. Robinson's Rudiments of "Written Arithmetic, ... V. Robinson's Progressive Practical Arithmetic, ... VI. Robinson's Key to Practical Arithmetic, VII. Robinson's Progressive Higher Arithmetic, - - - - VIII. Robinson's Key to Higher Arithmetic, IX. Robinson's New Elementary Algebra, X. Robinson's Key to Elementary Algebra, XI. Robinson's University Algebra, XII. Robinson's Key to University Algebra, XIII. Robinson's New University Algebra, XIV. Robinson's Key to New University Algebra, ... - XV. Robinson's New Geometry and Trigonometry, - XVI. Robinson's Surveying and Navigation, XVII. Robinson's Analyt. Geometry and Conic Sections, - XVIII. Robinson's Differen. and Int. Calculus, (in preparation,). XIX. Robinson's Elementary Astronomy, XX. Robinson's University Astronomy, XXI. Robinson's Mathematical Operations, XXII. Robinson's Key to Geometry and Trigonometry, Conic Sections and Analytical Geometry, Entered, accordiii;,' to Act of Congress, in the year IS58, by IIOKATIO N. IIOBINSON, LL.D., and again in tlie year 1SG3, by - DANIEL W. FISH, A.M., In the Clerk's OflQco of tlie District Court of tlio United States, for the Northern District of Now York. PREFACE. Progress and improvement characterize almost every art and science ; and within the last few years the science of Aritlimetic has received many important additions and improvements, which have appeared from time to time successively in the different treatises pub- lished upon this subject. In the preparation of this work it has been the author's aim to com- bine, and to present in one harmonious whole, all these modern im- provements, as well as to introduce some new methods and practical operations not found m other works of the same grade ; in short, to present the subject of Arithmetic to the pupil more as a science than an art; to teach him methods of thought, and how to reaso7i, rather than xchat to do; to give unity, system, and practical utility to the science and art of computation. The author believes that both teacher and pupil should have the privilege, as well as the benefit, of performing at least a part of the thinkincj and the labor necessary to the study of Arithmetic ; hence the present work has not been encumbered with the multiplicity of •'notes," "suggestions," and superfluous operations so common to most Practical Aritlmictics of the present day, and which prevent the cultivation of that self-reliance, that clearness of thought, and that vigor of intellect, which always characterize the truly educated mind. The author claims for this treatise improvement upon, if not superi- ority over, others of the kind in the following particulars, viz. : In the mechanical and typographical strjle of the icork ; the open and attractive page ; the progressive and scientific arrangement of the subjects ;' clearness and conciseness of definitio7is ; fullness and accu- racy in the i}eio and improved methods of operations and analyses; brevity and perspicuity of rules ; and in the very large number of Ciii) iV PREFACE. examples prepared and arranged tcith special reference to their prac- tical utility, and their adaptation to the real business of active life. The answers to a part of the examples have been omitted, that the learner may acqtiire the discipline resulting from verifj-ing the opera- tions. Particiilar attention is in\'ited to improvements in the subjects of Common Divisors, Multiples, Fractions, Percentage, Interest, Propor- tion, Analj'sis, Alligation, and the Roots, as it is believed these articles contain some practical features not common to other authors upon these subjects. It is not claimed that this is a perfect "work, for perfection is impossi- ble ; but no effort has been spared to present a clear, scientific, com- prehensive, and complete system, sufficiently full for the business man and the scholar ; not encumbered with unnecessary theories, and yet combining and systematizing real improvements of a practical and useful natujre. How nearly this end has been attained the intelligent and experienced teacher and educator must determine. CONTENTS. SIMPLE NUMBERS. PAGr Definitions, T Roman Notation, 8 Table of Pwoman Notation, 9 Arabic Notation, 10 N umeiation Tabic, 15 Laws and Rules for Notation and Numeration, 17 Addition, 20 Subtraction, 28 Multiplication, S5 Contractions, 42 PAGE Di vision, 4T Contractions, 68 Applications of preceding Rules, .... 60 General Principles of Division, Gi Exact Divisors, C6 Piiuie Numbers, CT Factoring Numbers, 67 Cancellation, 69 Greatest Common Divisor, 73 Multiples 79 Classification of Numbers, 8-i COMMON FRACTIONS. Definitions, &c., 86 Gener.al Principles of Fractions, 89 Reduction of Fractions, 83 Addition of Fractions, 96 Subtraction of Fractions, 93 Multiplication of Fractions, 101 Division of Fractions, 106 Promiscuous Examples, 112 DECIMALS, Decimal Notation and Numeration,. 116 Reduction of Decimals, 121 Addition of Decimals, 134 Subtraction of Decimals, 126 Multiplication of Decimals, 127 Division of Decimals, 123 DECIMAL CtJRRENCT. Notation and Numeration o. Decimal Currency, 131 Reduction of Decimal Currency, 132 Addition of Decimal Currency, 134 Subtraction of Decimal Currency,. . . 185 Multiplication of Decimal Curren- cy 136 Division of Decimal Currency, 137 Additional Applications, 139 "When the Price is an Aliquot Part of a Dollar 139 To find the Cost of a Quantity, 140 To find the Price of One, 141 To find the Quantity, 141 Articles sold by the 100 or 1000, 143 Articles sold by the Ton, 143 Bills, 144 Promiscuous Examples, 148 Tl CONTENTS. COMPOUND NUMBERS. PAGE Eeduction, 150 DeflnitioDS. «fec., 150 English Money, 151 Troy Weight, 153 Apothecaries' Weight, 154 Avoirdupois Weight, 155 Long Measure, 15S Surveyors' Long Measure, KiO Square Measure, ICl Surveyors' Square Measure, 164 Cubic Measure, 1C5 Liquid Measure, 1G7 Dry Measure, 1 03 Time,..- 170 Circular Measure, 172 PAOB Counting; Paper; Books, Ac, 173 lieduction of Denominate Fractions, 175 Addition of Compound Numbers,. .. 183 Addition of Denominate Fractions, . 185 Subtraction, 186 To find the Difference in Dates, 1S8 Table, 1S9 Subtraction of Denominate Frac- tions, 190 Multiplication of Compound Num- bers, 191 Division,... 193 Longitude and Time, 195 Duodecimals, 193 Promiscuous Examples, 202 PERCENTAGE. Definitions, &c., 205 Com mission and Brokerage, 210 Stocks, 214 Profit and Loss, 217 Insurance, 223 Taxes, 224 Custom House Business, 227 Simple Interest, 230 Partial Payments or Indorsement.',.. -37 Problems in Interest, 243 Com])ound Interest, 24G Discount, 249 Banking, 252 Exchange, , 256 Equation of Payments, 201 RATIO AND PROPORTION, Patio, 209 Proportion, 272 Simple Proportion, 273 Compound Proportion, 279 Partnership, 2S1 An.-ilysis, 2SS Alligation Medial, 297 Alligation Alternate, 293 Involufion, Sfl.T Evolution, 304 Square Root, 305 Cube Root, 812 Arithmetical Progres.sion, 313 Geometrical Progression, 821 Promiscuous E.xamples, 824 Mensuration ...., .. 832 PHACTICAL ARITHMETIC. DEFINITIONS. 1. Quantity is any thing that can be increased, diminished, or mca'^ured. S. Mathematics is the science of quantity. 3. A Unit is one, or a single thing. 4. A Number is a unit, or a collection of units. , 5. An Integer is a whole number. 6. The Unit of a Number is one of the same kind or name as the number. Thus, the unit of 23 is 1 ; of 23 dollars, 1 dollar ; of 23 feet, 1 foot. 7. Like -Numbers have the same kind of unit. Thus, 74, 16, and 250 ; 7 dollars and 62 dollars ; 19 pounds, 320 pounds, and 86 pounds; 4 feet 6 inches, and 17 feet 9 inches. 8. An Abstract Number is a number used without refer- ence to any particular tliiug or quantity. Thus, 17; 365; 8540. 9. A Concrete Number is a number used with reference to some particular thing or quantity. Thus, 17 dollars; 365 days ; 8540 men. XoTES. 1. The unit of an abstract number is 1, and is called Trmfij. 2. Concrete numbers are, by some, called Denominate Numbers- Denomination means the name o"f the unit of a concrete number. 10. Arithmetic is the Science of numbers, and the Art of computation. 11. A Sign is a character indicating an operation to be performed. 12. A Rule is a prescribed method of performing an op- eration. Define quantity. Mathematics. A imit. A number. An integer. The unit of a number. Like numbers. An abstract number. A concrete number. The unit of an abstract number. Denominate numbers. Arithmetic. A sign, or sjTnbol. A rule. 8 SIMPLE NUMBERS. NOTATION AND NUMERATION. IS. Notation is a method of z^n'^jHy or expressing numbers by charactei's ; and, 1 4: . Numeration is a method of reading numbers expressed by characters. 15. Two systems of notation are in general use — the Roman and the Arabic. Note. The Roman Notation is supposed to have been first used by the Romans ; hence its name. The Arabic Notation was intro- duced into Europe by the Arabs, by -whom it was supposed to have been invented. Eut investigations have shown that it was adopted by them about 600 years ago, and that it has been in use among the Hin- doos more than 2000 years. From" this latter fact it is sometimes called the Indian Notation. THE ROMAN NOTATIOX 16. Employs seven capital letters to express numbers, thus : Letters, I V X L C D M one' five one hundred, buudred, thoasand. AAoli^r^c ,. i re ona' five one values, one, five, ten, fifty, 17, The Roman notation is founded upon five principles, as follows : 1st. Repeating a letter repeats its value. Thus, II repre- sents two, XX twenty, CCC three hundred, 2d. If a letter of any value be plaeed after one of greater value, its value is to be united to that of the greater. Thus, XI represents eleven, LX sixty, DC six hundred. 3d. If a letter of any value be placed before one of greater value, its value is to be taken from that of the greater. Thus, IX represents nine, XL forty, CD four hundred. Define notation. Numeration. ^Miat systems of notation are now in general use ? From what are their names derived ? What are used to express numbers in the Roman notation r "What is the value of each ? What is the first principle of combination ? Second ? Third ? NOTATION AND NUJIERATION. 9 4th. If a letter of any value be placed hetiveen two letters, each of greater value, its value is to be taken from tlie united value of the other two. Thus, XIV represents fourteen, XXIX twenty-nine, XCIV ninety-four. oth. A bar or' dash placed over a letter increases its value one thousand fold. Thus, V signifies five, and V five thou- sand ; L fifty, and L fifty thousand. TABLE OF KOMAN NOTATION. I is One. II Two. III Three. IV Four. V Five. VI Six. VII Seven. VIII Eight. IX X^ine. X Ten. XI Eleven. XII Twelve. XIII Thirteen. XIV Fourteen. XV Fifteen. XVI Sixteen. XVII Seventeen ^VIII Eigliteen. XIX X'ineteen. XX is Twenty. XXI " Twenty-one. XXX " Thirty. XL " Forty. L " Fifty. LX " Sixty. LXX " Seventy. LXXX " Ei<>:hty. xc " Ninety. c " One hundred. cc " Two hundred. D " Five hundred. DC " Six hundred. M " One thousand. [dred MC " One thousand one hun- MM " Two thousand. x " Ten thousand. C " One hundred thousand M " One million. NoTE. The system of Roman notation p\n-poses of niimcrical calculation ; it is principally confined to the s not well adapted to the numbering of chapters and sections of book Express the fjllowing numbers by letters : 1. Eleven. 2. Fifteen. piiblic docmnents, &c. Ans. AnS' XL Fourth ? Fifth ? Repeat the table. ^Miat is the value of I-VTI ? CLXXIII? XCVin? CDXXXII? XCIX? DCXIX? V-MDCCXLIX? MDXXVCDLXXXIX ? To what uses is the Roman notation now principally confined ? 1* 10 SIMPLE NUMBERS. 3. Twenty-five. 4. Thirty-nine. 5. Forty-eight. 6. Seventy-seven. 7. One hundred fifty-nine. 8. Five hundred ninety-four. 9. One thousand five hundred thirty-eight. 10. One thousand nine hundred ten. 11. Express the present year. THE ARABIC NOTATION IS. Employs ten chai-acters or figures to express numbers. Thus, Figures, 012345G789 Names and ') "^^"S^* °^^' *^^'°j three, four, five, Bix, Beven, eight, nine, values, 5 ^i,,°'„^ 19. The first character is called naught, because it has no value of its own. The other nine characters are called signif- icant figures, because each has a value of its own. 20. The significant figures are also called Digits, a word de- rived from the Latin terra digitus, which signifies ^;(^er. 21. The naught or cipher is also called nothing, and zero. The ten Arabic characters are the Alphabet of Aritlimetic, and by combining tliem according to certain principles, all numbers can be expressed. We will now examine the most important of these princii)les.* 92. Each of tlie nine digits has a value of its own ; hence any number not greater than 9 can be expressed by one figure. * Fractional and Jecimtil not.-ition, and the notation of compound numbers, will bo discussed .1 their appropriate places. "\Miat are used to cxpro'ss nunibtTs in the Arabic notation ? '^^'ll,^t is the value of each ? What general name is given to the signiticant figures ? AVhy ? Numbers less than ten, how c.xiivessed ? NOTATION AND NUMERATION. H S3. As we have no single character to represent ten, we express it by writing the unit, 1, at the left of the cipher, 0, thus, 10. In the same manner we represent 2 tens, 3 tens, 4 tens, 5 tens, G tens, 7 tens, 8 tons, 9 tens, or or or or or or or or twenty, thirty, forty, fifty, sixty, seventy, ciglity, ninety, 20; 30; 40; 50; GO; 70; 80; 90. S4:. When a number is expressed by two figures, the right hand figure is called units, and the left hand figure tens. We express the numbers between 10 and 20 by Avriting the 1 in the place of tens, with each of the digits respectively in the place of units. Thus, eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, nineteen. 11, 12, 13, 14, 15, 16, 17, 18, 19. In like manner we express the numbers between 20 and 30, between 30 and 40, and between any two successive tens. Thus, 21, 22, 23, 24, 25, 26, 27, 28, 29, 34, 47, 56, 72, 93. The greatest number that can be expressed by two figures is 99. 25. We express one hundred by writing the unit, 1, at the left hand of two ciphers, or the number 10 at the left hand of one cipher; thus, 100. .In like manner we write two hun- dred, three hundred, &c., to nine hundred. Thus, one t^\•o three four five six seven eight nine hundred, hundred, hundred, liun 3C)7 15G7 Analysis. An-anging tlie numbers as before, we add the first column and find the sum to be 17 units; writing the 7 units under the column of units,- we add the 1 ten to the column of tens, and find the sum to be 10 tens ; writing the 6 tens under the column tens, we add the 1 hundred to the column of hundreds, and find the sum to be 15 hundreds ; as this is the last column, we write down its amount, 15 ; and wc have the lohole amount, 15G7, as before. Notes. 1. T'nits of the same order are written in tlie same rolunm ; and when the sum in any column is 10 or more than 10, it procUiucs our or more unitx of a hi;^hfr order, whitli must be addtd to the next column. This process is sometimes called " carryinn; the tens." 2. In adding, learn to pronounce the partial results without namiii;? the numbers separately ; thus, instead of saying 7 and 5 are 12, and 6 are 17, simply pronounce the results, 7, 12, 17, &c. Give explanation. ing the tens ? Second explanation. "SMiat is meant by carry- ADDITION. 23 4L'^. From the preceding examples and illustrations we deduce the following Rule. I. 11 rite the numbers to he added so tliat all the units of the same order shall stand in the same column ; that is, units under units, tens under tens, 8fc. II. Commencing at units, add each column separateli/, and write tl/e sum underneath, if it be less than ten. III. If' the sum of an?/ column be ten or more than ten, write the unit figuret only, and add the ten or tens to the next column. IV. Write the entire sum of the last column. Proof. 1st. Begin witli the right hand or unit column, and add the figures in each column in an opposite direction from th:it in which they were first added ; if the two results agree, the work is supposed to be right. Or, 2d. Separate the numbers added into two sets, by a hori- zontal line ; find the sum of each set separately ; add these sums, and if tlie amount be the same as that first obtained, the work is presumed to be correct. Note. By the methods of proof here given, the numbers are rmited in new combinations, which render it ahnost impossible for two pre- cisely simiUir mistakes to occur. The first method is the one commonly used in business. EXAMPLES FOK PRACTICE. (2.) (3.) (4.) (5.) (6.) mill's. jncliL-3. tons. ft-et. buslii'ls. 24 321 427 1342 3420 48 479 321 730G 7021 96 1G5 903 5254 327 82 327 278 1929 8G29 22531 97 250 1292 108G5 Rule, first step ? Second ? Third ? Fourth ? Proof, first method ? Second ? Upon what principle are these methods of proof founded ? 24 SIMPLE KUMBERS. (7.) (8.) (9.) (10.) hours. years. gallons. rods. 347 7104 3462 47637 506 3762 863 3418 218 9325 479 703 312 4316 84 26471 424 2739 57 84 11. 42 + 64 -|- 98 -f 70 + 37 == how many ? Jns. 311. 12. 312 -|- 425 + 107 -|- 391 + 76 = Low many ? A71S. 1311. 13. 1476 -j- 375 + 891 -f 66 4-80.= how many? Ans. 2888. 14. 37042 -f 1379 -|- 809 + 127 + 40 rr: how many ? Ans. 39397. 15. "What is the sura of one thousand six hundred fifty-six, eight hundred nine, three hundred ten, and ninety-four ? Ans. 2869. 16. Add forty-two thousand two hundred twenty, ten thou- sand one hundred five, four thousand seventy-five, and five hundred seven. Ans. 56907. 17. Add two hundred ten thousand four hundred, one hun- dred thousand five liundred ten, ninety thousand six hundred eleven, forty-two hundred twenty-five, and eiglit hundred ten. Ans. 406556. 18. What is the sum of the following nimihers : seventy- five, one thousand ninety-five, six thousand four hundred thir- ty-five, two hundred sixty-seven thousand, one thousand four hundred fifty-five, twenty-seven million eighteen, two hundred seventy million twenty-seven thousand? Ans. 297303078. 19. A man on a journey traveled the first day 37 miles, the second 33 miles, the third 40 miles, and the fourth 35 miles ; how far did he travel in the four days? , 20. A wine merchant has in one cask 75 gallons, in another 65, in a third 57, in a fourth 83, in a fifth 74, and in a sixth 67 ; how many gallons has he in all? Ans. 421. ADDITION. 25 21. An estate is to be shared equally by four heirs, and the portion to each heir is to be 3754 dollars ; what is the amount of the estate? ^ws. 15016 dollars. 22. IIow many men in an army consisting of 52714 in- fimtry, 5110 cavalry, 6250 dragoons, 3927 light-horse, 928 artillery, 250 sappers, and 40(! miners ? 23. A merchant deposited 56 dollars in a bank on Monday, 74 on Tuesday, 120 on Wednesday, 96 on Thursday, 170 on Friday, and 50 on Saturday ; how much did he deposit during the week? 24. A merchant bought at public sale 746 yards of broad- cloth, 650 yards of muslin, 2100 yards of flannel, and 250 yards of silk ; how many yards in all ? 25. Five persons deposited money in the same bank ; the first, 5897 dollars; the second, 12980 dollars; the third, 65973 dollars ; the fourth, 37345 dollars ; and the fifih as much as the first and second together ; how many dollars did they all deposit ? Ans. 141072 dollars. 26. A man willed his estate to his wife, two sons, and four daughters ; to his daughters he gave 2630 dollars apiece, to his sons, each 4647 dollars, and to his wife 3595 dollars; how much was his estate ? Ans. 23409 dollars. (27.) (28.) (29.) (30.) (31.) 476 908 126 443 180 390 371 324 298 976 915 569 503 876 209 207 245 891 569 314 841 703 736 137 563 632 421 517 910 842 234 127 143 347 175 143 354 274 256 224 536 781 531 324 135 245 436 275 463 253 RP. 26 SIMPLE NUMBEKS. 32. A man commenced farming at the west, and raised, tlie first year, 724 bushels of corn ; the second year, 'S-idS bushels ; the third year, 'JS72 bushels; the iburth year, 9964: bushels; the lifth year, 11078 bushels; how many bushels did he raise in the five years ? Ans. o5136 bushels. 33. A has oG48 dollars, B has 70oo dollars, C has 429 dollars more than A and B together, and D has as many dol- Lirs as all the rest ; how many dollars has D ? How many have all? Ans. All have 43590 dollars. 34. A man bought three houses and lots lor 15780 dollars, and sold them so as to gain. 695 dollars on each lot ; for how much did he sell them? Ans. 17865 dollars. 35. At the battle of Waterloo, which took place June 18th, 1815, the estimated loss of the French was 40000 men ; of the . Prussians, 38000 ; of the Belgians, 8000 ; of the Hano- verians, 3500 ; and of the English, 12000 ; what was the entire loss of life in this battle ? 36. The expenditures for educational purposes in New England for the year 1850 were as follows : Maine, 380623 dollars; New Hampshire, 221146 dollars; Vermont, 246604 dollars ; Massachusetts, 1424873 dollars ; Rhode Island, 136729 dollars; and Connecticut, 430826 dollars; what was the total expenditure ? Ans. 2840801 dollars. 37. The eastern continent contains 31000000 square miles; the western continent, 13750000; Australia, Green- ];md, and other isl;uids, 5250000 ; what is the entire area of the land surface of the glolie ? 38. The population of New York, in 1850, was 515547; Boston, 136881; riu'L-idelphia, 310015; Chicago, 29963; St. Louis, 77860 ; New Orleans, 116375; Avhat was the en- tiro j)opulation of these cities ? Ans. 1216671. 39. Tlu; ])opulation of the globe is es(iin;ited ;i< follows: North America, 39257819; South Amerie:.. 18373188; Eu- rope, 265368216: Asia, 630671661; Africa, 61688779; Oceanic:!, 231II(IR2; whnt is (he total jippulation of the globe according (o this estimate? Ans. 1038803715. ADDITION. 27 40. The railroad distance from Nbav York to Albany is 144 miles ; fVoin Albany to Jkitialo, 2'JS; from BuHalo to Cleveland, 183 ; from Cleveland to Toledo, 109 ; from Toledo to Spring- field, oGo ; and from Springfield to St. Louis, 95 miles; wliat is the distance from Kew York to St. Louis? 41. A man owns farms valued at 5G800 dollars; city lots valued at 867G0 dollars; a house worth 12500 dollars, and other property to the amount of 6785 dollars; what is the entire value of his property ? Jns. 162845 dollars. (42.) (43.J (44.) (45.) 15038 2G881 41919 93803 7404 12173 19577 41371 S4971 39665 74736 110525 30359 33249 66768 102936 G293 6318 12673 17087 2875 4318 7193 13251 IGGGO 34705 51365 112110 G4934 80597 155497 220619 8O0O1 95299 183134 225255 7444 8624 16845 68940 570G8 53806 111139 176974 17255 18647 35902 86590 32543 41609 82182 149162 40022 35077 75153 109355 5 GO 63 46880 132936 283910 338 GO 41842 82939 112511 17548 26876 44424 72908 28944 36642 65586 157672 16147 29997 52839 86160 38556 44305 83211 119557 234882 262083 522294 839398 39058 39744" 78861 117787 152526 169220 353428 471842 179122 198568 386214 571778 7626 8735 17005 41735 1218099 1395860 28 SIMPLE NUMBERS. SUBTRACTION. MENTAL EXERCISES. 48. 1. A farmer, having l-i cows, sold G of them; how many had he left? Analysis. He had as many left as 14 cows less 6 cows, wWch arc 8 coMS. Therefore, he had 8 cows left. 2. Stephen, having 9 marbles, lost 4 of them ; how many had he left? 3. If a man earn 10 dollars a week, and spend 6 dollars for provision?, how many dollars has he left ? 4. A merchant, having IG barrels of flour, sells 9 of them; how many has he left ? 5. Charles had 18 cents, and gave 10 of them for a book; how many had he left ? G. James is 17 years old, and his sister Julia, is 5 years younger; how old is Julia? 7. A grocer, having 20 boxes of lemons, sold 1 1 boxes ; how many boxes had he left ? 8. From a cistern containing 25 barrels of water, 15 bai-- rels leaked out ; how many barrels remained ? 9. Paid IG dollars for a coat, and 7 dollars for a vest; how much more did tlie coat cost than the vest ? 10. How many arc 18 less 5 ? 17 less 8 ? 12 less 7 ? 11. How many are 20 less 14 ? 18 less 12? 19 less 11 ? 12. How many are 11 less 3? IG less 11? 19 less 8? 20 less 9 ? 22 less 20 ? 44. Subtraction is the process of determining the difler- CTice, between two numbers of the same unit value. 4»"5. The Minuend is the number to be subtracted from. 46. Tiie Subtrahend is the number to be subtracted. T-^ — . Define siibtrartion. ^liimcnd. Subtrahend. SUBTRACTION. 29 4'^'* The Diiferenee or Eemaiuder is the result obtained b}' the process ot'subti'action. Note. The iniiuiend and subtrahend must be lilic numbers ; thus, 5 dollars from 9 dollars Ivave 4 dollars ;o apples from 9 apples leave 4 apples ; but it would be absurd to say o apples fiom 9 dollars, or 5 dollars from 9 apples. 4:8. The sign, — , is called minns, which signifies less. When placed between two numbers, it denotes that the one after it is to be taken from the one before it. Thus, 8 — 6 = 2 is read 8 minus G equals 2, and denotes that G, the siihtrahend, taken from 8, the minuend, equals 2, the remainder. CASE I. 49. When no figure in the subtrahend is greater than the corresponding figure in the minuend. 1. From 574 take 323. OPERATION. Analysis. We write the less num- -Ti ber under the p-reater, with units under o-,.^ units, tens under tens, ccc, and draw Subtrahena, 6:6 ^ j.^_^^ Underneath. Then, beginning at Eemaindcr, 251 the right hand, Ave subtract separately each figure of the subtrahend from the figure above it in the minuend. Thus, 3 from 4 leaves 1, which is the difference of the units ; 2 from 7 leaves .j, the difference of the tens ; 3 from 5 leaves 2, the difference of the hundreds. Hence, we have for the whole difference, 2 hundreds 5 tens and 1 unit, or 251. EXAMPLES FOR rRACTICE. (2.) (3.) (4.) ■ (5.) 5nnuend, 876 676 3C7 925 SuUralH-nd, 334 415 152 213 542 2G1 215 712 Remainder, Case lis what? Give explanation. 80 SIMPLE NUMBERS. (6.) (7.) (8.) /(9.) From 876 732 987 498 Uake 523 522 782 178 Remainders. 10. From 3276 take 2143. 1133. 11. From 7634 take 3132. 45U2. 12. From 41763 take 11521. 30242. 13. From 18346 take 5215. 13131. 14. From 397631 take 175321. 222310. 15. Subtract 47321 from 69524. 22203. 16. Subtract 16330 from 48673. 32343. 17. Subtract 291352 from 895752. 604400. 18. Subtract 84321 from 397502. 313241. 19. A farmer paid 645 dollars for a span of horses and a carriage, and sold them for 522 dollars; ho\v much did he lose ? 20. A man bought a mill for 3724 dollars, and sold it for 4856 dollars; how much did he gain .'* Ans. 1132 dollars. 21. A drover bought 1566 sheep, and sold 435 of them; how many had he left? Ans. 1131 sheep. 22. A piece of land was sold for 2945 dollars, which was 832 dollars more than it cost ; what did it cost ? 23. A gentleman willed to his son 15768 dollars, and to his dniigliter 4537 dollars ; how much more did he will to the son than to the daughter? Ans. 11231 dollars. 24. A merchant sold goods to the amount of 6742 dollars, and by so doing gained 2540 dollars ; what did the goods cost him ? 25. If I borrow 15 175 dollars of a person, and pay him 4050 dollars, how much do I still owe him ? 26. In 1850 the white population of the United States was 19,553,068, and the slave population 3,201,313; how much was the difference? 27. Tiie population of Great Britain in 1851 was 20,936,468, and of England alone, 16,921,888 ; what was the difference? SUBTRACTION. 31 \ CASE II. 50. When any figure in the subtrahend is greater than the corresponding figure in the minuend. 1. From 84G take 351). OPERATION. Analysis. In this example we (7) (13) (IC) cannot take 9 units from 6 units. Minuend, 8 4 6 From the 4 tens we take 1 ten, which Subtrahend, 3 5 9 equals 10 units, and add to the G ~~, [^^i ^ units, makiiif^ 16 units ; 9 units from llemaiuder, 4 O / , ,. . , ,. . , • , 16 units leave / units, which we write in the remainder in units' place. As we have taken 1 ten from the 4 tens, 3 tens only are left. We cannot take o tens from 3 tens ; so from the 8 hundreds we take 1 hundred, which equals 10 tens, and add to the 3 tens, making 13 tens ; 5 tens from 13 tens leave 8 tens, which we write in the remainder in tens' place. As we have taken 1 hundred from the 8 hundreds, 7 hundreds only are left ; 3 hundreds from 7 hundreds leave 4 hundreds, which we write in the remainder in hundreds' place, and we have the whole remainder, 487. , Note. The numbers written over the minuend are used simply to explain more clearly the method of subtracting ; in practice the pro- cess should be performed mentally, and these numbers omitted. The following method is more in accordance with prac- tice. OPERATION. Analysis. Since we cannot take 9 units from 6 1^:2 units, we add 10 units to 6 units, making 16 units; Jl 3 9 units from 16 units leave 7 units. But as we have 846 added 10 units, or 1 ten, to the minuend, we shall 359 have a remainder 1 ten too large, to avoid which, we .Qj add 1 ten to the 5 tens in the subtrahend, making 6 tens. We can not take 6 tens from 4 tens ; so we add 10 tens to 4, making 14 tens; 6 tens from 14 tens leave 8 tens. Now, having added 10 tens, or 1 hundred, to the minuend, we shall have a remainder 1 hundred too large, unless we add 1 hundred to the 3 hundreds in the subtrahend, making 4 hun- dreds ; 4 hundreds from 8 hundreds leave 4 hundreds, and we have for the total remainder, 487, the same as before. Case II is what ? Give explanation. Second explanation. 32 SIMPLE NUMBERS. Note. Tlie process of adding 10 to the minuend is sometimes called borrou'ing 10, and that of adding 1 to the next figure of the subtrahend, carrying one. 51. From the preceding examples and illustrations we have the following general Rule. I. Write the less number under the greater, placing units of the same order in the same column. II. Begin at the right hand, and take each figure of the sub- trahend from the figure above it, and write the result under- neath. III. If any figure in the subtrahend be greater than the cor- responding figure above it, add 10 /o that upper figure be- fore subtracting, and then add 1 to the next left hand figure of the subtrahend. Proof. Add the remainder to the subtrahend, and if their sum be equal to the minuend, the work is supposed to be right. EXAMPLES FOK PRACTICE. (2.) (3.) (4.) (5.) Minuend, 873 7432 1969 814a Subtrahend, 538 C711 1408 4377 Remainder, 335 (6.) (7.) (8.) (9.) gallons. bushels. miles. days. From 3176 9076 7320 5097 Take 2907 4507 3871 3809 (10.) (11.) (12.) (13.) dull.irs. rods. acivs. feet. From 7G377 67777 900076 767340 Take 457G1 46699 899934 5039 "What do Ave mean by borrowing 10 ? By carrying ? Rule, first step ? Second ? Third ? Proof ? SUBTRACTIOX. 83 16. 17. 18. 19. 20. 21. 22. ^2;). 2L 25. 26. Ans. 4111107. Ans. 2479679. Ans . 935993. Ans . 608889. Ans. 3968579. Ans. 50000001. Ans. 3819851. Ans. , 800924. A^ns. 7023024. 14. 479 — 382 =z how many ? J?^s. 97. 15. 6593 — 1807 = how many? Ans. 4786. 17380 — 3417 = how many ? Ans. 13903. 80014 — 43190 =z liow many ? Ans. 3GS24. 282731 — 90756 = how many? A71S. 191975. Flora 234100 take 9970. . From 345673 take 124799. From 4367676 take 256569. From 3467310 take 987631. From 911000 take 5007. From 197()0:)0 take 1361111. From 290017 take 108045. ' Take 3077097 from 7045676. 27. Take 9999999 from 60000000. 28. Take 220202 from 4040053. 29. Take 2199077 from 3000001. 30. Take 377776 from 8000800. 31. Take 501300347 from 1030810040. 32. Subtract nineteen thousand nineteen from twenty thou- sand ten. Ans. 991. 33. From one miUion nine thousand six take twenty thou- sand four hundred. Ans. 988606. 34. What is the difference between two million seven thousand eighteen, and one hundred five thousand seven- teen ? EXAMPLES COMDINING AUDITION AND SUKTRACTIOX. t§Q. 1. A merchant gave his note for 5200 dollars. He paid at one time 2500 dollars, and at another 175 dollars; what remained due ? Aiis.. 2525 dollars. 2. A traveler who was 1300 miles from home, traveled homeward 235 miles in one week, in the next 275 miles, in the next 325 miles, and in the next 280 miles ; how far had he still to jro before he would reach home? Ans. 185 miles. 3. A man deposited in bank 8752 dollars ; he drew out at one time 4234 dollars, at another 1700 dollars, at another 9G^ 2* 84 SIMPLE NUMBERS. dollars, and at another 49 dollars ; how much had he remain- ing in bank ? Ans. 1807 dollars. 4. A man bought a farm for 47 C") dollars, and paid 750 dollars for fencing and other improvements ; he then sold it for 384 dollars less than it cost him ; how much did he receive for it? Ans. 5131 dollars. 5. A forwarding merchant had in his warehouse 7520 bar- rels of flour; he shipped at one time 1224 barrels, at another time 1500 barrels, and at another time 1805 barrels; how many barrels remained? G. A had 450 sheep, B had 175 more than A, and C had as many as A and B together minus 114 ; how many sheep had C ? Ans. 9G1 sheep. 7. A farmer raised 1575 bushels of wheat, and 900 bushels of corn. He sold 807 bushels of wheat, and 391 bushels of corn to A, and the remainder to B ; how much of each did he sell to B ? Ans. 7G8 bushels of wheat, and 509 of corn. 8. A man traveled G784 miles ; 2324 miles by railroad, 1570 miles in a stage coach, 450 miles on horseback, 175 miles on foot, and the remainder by steamboat ; how many miles did he travel by steamboat ? Ans. 22 G5 miles. 9. Three persons bought a hotel valued at 35G80 dollars. The first agreed to pay 7375 dollars:, the second agreed to pay twice as much, and the third the remainder ; how much was the third to pay ? Ans. 13555 dollars. 10. Borrowed of my neighbor at one time 750 dollars, at another time 379 dollars, and at another 450 dollars. Having paid him 1000 dollars, how much do I still owe him? Ans. 579 dollars. 11. A man wbrth G709 d. The Multiplier is the number which shows how many times the multiplicand is to be taken. q37. The Product is the result obtained by the process of multiplication. 58. The Factors are the multiplicand and multiplier. Notes. 1. Factors are producers, and the multiplicand and mul- tiplier are called factors because they produce the product. 2. ^Multiplication is a short method of performing addition when the numbers to be added are equal. «59. The &ign,'X, plficed between two numbers, denotes that they are to be multiplied together ; thus 9X6:= 54, is read 9 times G equals 54. MTTLTIPLICATION TABLE. 1X1=1 2X 1= 2 3X 1= 3 4X 1= 4 IX 2= 2 2X2=4 3X -2= 6 4X2=8 IX 3=3 2 X 3= 6 3X3=9 4X 3 = 12 IX 4=4 2X 4= 8 3X 4 = 12 4 X 4 = 16 1 X o= 5 2X 5 = 10 3X 5 = 15 4 X 5 = 20 1X0=6 2X 0=12 3X 6 = 18 4X 6 = 24 *! X 7=7 2X 7 = 14 3X 7 = 21 4X 7 = 28 1 X 8= 8 2X 8 = 16 3 X 8 = 21 4X8 = 32 1X9=9 2X 9 = 18 3 X 9 = 27 4 X 9 = 3() 1 X 10=10 2 X 10 = 20 3 X 10 = 30 4 X 10 = 40 1 X 11 = 11 2 X 11 = 22 3 X 1 1 = 33 4 X 1 1 = l-l 1 X 12 = 12 2 X 12 = 24 3X 12 = 36 4 X 12 = 48 Define multiplication. Multiplicand. ^Multiplier. Prodtict. Fac- tors. Multiplication is a short niftb.od of what ? What is tiie sign of multiplication ? MULTIPLICATION. 87 5X 1= 5 GX 1= 6 7X 1= 7 8X 1= 8 oX 2r=10 GX 2 = 12 7X 2=14 8X 2=1G 5X 3 = 1,; GX 3 = 18 7X 3 = 21 8 X 3 = 24 5 X 4 = 20 OX 4 = 24 7X 4 = 28 SX 4 = 32 5 X 5 = 25 GX 5 = 30 7 X 5 = 35 8 X 5 = 40 X C = 30 G X G = 36 7X G = 42 8X G = JS r>X 7 = 3J GX 7 = 42 7X 7 = 49 8 X 7 = 56 oX 8 = 40 X 8 = 48 7 X 8 = 5G 8 X 8 = 64 oX = 4.5 6 X = 54 7 X = G3 8X 9 = 72 5 X 10 = 50 6 X 10 = GO 7 X 10 = 70 8 X 10 = 80 5 X 11 = 55 G X 11 = G6 7X 11 = 77 8 X 11=: 88 5 X 12 = 60 GX 12 = 72 7 X 12 = 84 8 X 12 = 96 OX 1 = 9 10 X 1= 10 IIX 1= 11 12 X 1= 12 OX 2 = 18 10 X 2= 20 11 X 2= 22 12 X 2= 24 OX 3 = 27 10 X 3= 30 11 X 3= 33 12 X 3= 36 OX 4 = 36 10 X 4= 40 11 X 4= 44 12 X 4= 48 9X 5 = 45 10 X 5= 50 11 X o — 00 12 X 5= 60 9X 6 = 54 10 X 6= 60 11 X 6= 06 12 X 6= 72 9X 7 = 63 10 X 7= 70 11 X 7= 77 12 X 7= 84 OX 8 = 72 10 X 8= 80 11 X 8= 88 12 X 8= 96 9X = 81 10 X 9= 90 11 X 0= 99 12 X 9=108 9X 10 = 90 10X10 = 100 11 xio = iio 12X10=120 9X11 = 99 10X11 = 110 11 Xll = 121 12X11 = 132 9X 12 = 108 10 X12 = 120 11 X12 = 132 12X12 = 144 CASE I. 60. When tliG multiplier consists of one figure. 1. Multiply 374 by G. Analysis. In this example it is» required to take 374 six times. If we take the units of each order 6 times, Ave shall take the entire number 6 times. Therefore, -writing the multi- plier under the unit figure of the mul- tiplicand, we 2)roceed as follows : 6 times 4 units are 24 units ; 6 times 7 tens are 42 tens ; 6 times 3 hundreds are 18 hundreds ; and adding these partial products, we obtain the entire product, 2244. OPEKATION. Multiplicancl, 374 Multiplier, 6 units, 24 tens, 42 hundreds, 18 Product, 2244 Case I is what ? Give explanation. S8 SIMPLE NUMBERS. The operation in this exainjile may be performed in another way, wliicli is the one in common use. OPERATION. Analysis. AVriting the numbers as before, we o-i begin at the right hand or unit figure, and say: 6 /» times 4 units are 24 units, wliicli is 2 tens and 4 units ; write the 4 units in the product in units' 2244 place, and reserve the 2 tens to add to the next prod- uct ; 6 times 7 tens are 42 tens, and the two tens re- served in the last product added, are 44 tens, which is 4 liaadicds and 4 tens ; write the 4 tens in the product in tens' place, and reserve the 4 hundreds to add to the next product ; 6 times 3 hundreds are 18 hundreds, and 4 hundreds added are 22 hundreds, which, l^eing written in the ])roduct in the places of hundreds and thousands, gives, for the entire product, 2241. 61. The unit value of a number is not changed by re- peating the number. As the multiplier always expresses times, the product must have the same unit value as the mul- tiplicand. But, since the product of any two numbers will be the same, whichever factor is taken as a multiplier, either factor may be taken for the multiplier or multiplicand. Note. In multiplying, learn to pronounce the partial results, as in addition, without naming the numbers separately ; thus, in the last example, instead of saying 6 times 4 are 24, 6 times 7 are 42 and 2 to carry are 41, 6 times 3 arc 18 and 4 to carry are 22, pronounce only the results, 21, 44, 22, performing the operations mentally. This will greatly facilitate the process of multii)lymg. EXAMPLES FOR rUACTICE. (2.) (3.) (4.) MultipIicaiKl, 7-324 G812 34G51 •MultipliiT, 4 21)200 40872 5 rioiiiicf, 17325} ('>■) (G.) (7.) (8.) 821oG 92714 28093 -iG217 7 8 9 Second explanation. Repeating a number has what effect on the unit value ? The product must be of the same kind as what ? I MULTIPLICATION. 89 9. Multi[)ly 327 W, hy o. 10. Multiply 840371 by 7. 11. Multii)ly 137G29 by 8. 12. Multiply 93702 by 3. 13. Multiply 543272 by 4. 14. Multiply 703164 by 9. Ans. 1G3730. Ans. 5882597. A/is. 1101032. Aus. 28128G. Alts. 21730S8. Ajis. G32847G. 15. AVliut will be the cost of 344 cords of wood at 4 dol- lars a cord ? Ans. 137G, 16. How much will an army of 785G men receive in one week, if each man receive 6 dollars? Ans. 4713G dollars. 17. In one day are 8G400 seconds; how many seconds in 7 days ? Ans. 604800 seconds. 18. What will 7 640 bushels of wheat cost, at 9 shillings a bushel ? A}is. G87G0 shillings. 19. At 5 dollars an acre, what Avill 2487 acres of land cost? Ans. 12435 dollars. 20. In one mile are 5280 feet; how many feet in 8 miles ? Ans. 42240 feet. CASE II. 6sS. "When the multiplier consists of two or more fio-ures. 1. Muhiply 746 by 23. OPERATION. Multiplicand, 746 Multiplier, 23 2238 1492 Product, 17158 23 o 5 times the mul- ^ ^ tiplicaiui. mS times the mul- ' ( tiplicand. times the mul- tiplicand. Analysis. AVrit- ing the multiplicand and multiplier as in Case I, we first mul- ti])!)' each figure in the multiplicand by the unit figure of the mul- tiplier, precisely as in Case I. AVe then multijily by the 2 tens. 2 tens times 6 units, or 6 times 2 tens, are 12 tens, equal to 1 hundred, and 2 tens ; we place the 2 tens under the tons figure in the product already obtained, and add the 1 hundred to the next hundreds produced. 2 tens times 4 tens are 8 hundreds, and the 1 hundred of the last product added are 9 hundreds ; we write the 9 in hundreds' place in the product. 2 tens Case II is Avhat ? Give explanation. 40 SIMPLE NUMBERS. times 7 hundreds are 14 thousands, equal to 1 ten thousand and 4 thousands, wliich we -write in their appropriate places in the ])roduct. Then adding the two products, we have the entire product, 17158. Notes. 1. When the multiplier contains two or more figures, the several results obtained by multiijlying by each figure are called 7;a;//((/ products. •_'. AN'hen there are ciphers hctwccn the significant figures of the niuUipUer, pass over them, and multiply by the significant figures only, S3. From the preceding examples and illiistralions we deduce the ibllowing general Rule. I. Write (he multiplier ^mder the 7nulti2jlicand,pl(icinff units of the same order under each otlier. II. Multiply the multiplicand by each fgure of the multi- plier successively, beginning with the unit fgure, and torite the first figure of each partial product under the fgure of the mul- tiplier used, meriting down and carrying as in addition. III. If there are partial products, add ihem, and their sum icill he the product required. 6 J:. PnooF. 1. IMultiidy the multiplier by (he multipli- cand, and if the product is the same as the first result, the Avork is correct. Or, 2. Multiply the multiplicand by the multiplier diminished ]>y 1, and to the product add the multiplicand ; if the sum he the same as the product by the whole of the multiplier, the work is correct. EXAMPLES FOR rUACTXCE. (3.) (4.) 8721 17605 47 ' '204 (2.) Multiply 4732 liy 36 28392 61047 70420 1410G 34884 35210 Ans. 170352 409887 359M--0 What are partial products ? When there are ciphers in the multi- plier, how proceed ? llule, first step ? Second ? Third r Proof, first method ? Second ? MULTIPLTCATION. 41 (6.) (7.) 81092 379G7 194 42G 8. IIow many yards of linen in 759 pieces, eacli piece con- taining 25 yards ? Ans. 18975 y;, if there be 362 pages in each book ? Ans. 1257950 pages. 13. In a garrison of 4507 men, each man receives annually 208 dollars ; how much do they all receive ? 14. Multiply 7198 by 216. Ans. 1554768. 15. Multiply 31416 by 175. Ans. 5497800. 16. Multiply 7071 by 556. Ans. 3931476- 17. Multiply 75649 by 579. Ans. 43800771. 18. Multiply 15607 by 3094. Ans. 48288058. 19. Multiply 79094451 by 76095. Ans. 6018692248845. 20. Multiply five hundred forty thousand six hundred nine^ by seventeen hundred fifty. Ans. 946065750. 21. Multiply four million twenty-five thousand three hun- dred ten, by seventy-five thousand forty-six. Ans. 302083414260. 22. Multiply eight hundred seventy-seven million five hun- dred ten thousand eight hundred sixty-four, by five hundred forty-five thousand three hundred fifty-seven. A}is. 478556692258448. 23. If one mile of railroad require 116 tons of iron, Avorth 65 dollars a ton, what will be the cost of sufllcient iron to construct a road 128 miles in length? Ans. 965120 dollars. 42 SIMPLE NUMBERS. CONTRACTIONS. CASK I. G5. When the multiplier is a composite number. A Composite Uumber is one tli;it may be j^ioduced by multiplying together two or more numbers; thus, 18 is a com- posite number, since G X 3 = 18 ; or, 9 X 2 := 18 ; or, 3 X 3X 2 — 18. 06. The Component Factors of a number are the sev- eral numbers which, multiplied together, produce the given number; thus, the component liictors of 20 are 10 and 2, (10 X 2 = 20 ;) or, 4 and o, (4 X 5 = 20 ,) or, 2 and 2 and 5, (2 X 2 X 5 z= 20 ) XoiE. The pupil must not confound the fnr/o)s with the parts of a muiibtr. 'Ihus, the Jhc/ois of whuh 12 is toniposed, are 4 and 3, (^4 X 3=12;) wlnle the par/s of which 12 is composed are 8 and 4, (8 -f- 4 = 12,) or 10 fuid 2, (10 + 2 = 12.) The factors are jnulliplicd, A\hde the pa) ts are added, to produce the number. 1. "What will 32 horses cost, at 174 dollars apiece.'' MuUiplicanJ, 1st factor, OPKRATiox Analysis. The fac- 174 cost of 1 horse. tors of 32 are 4 and A 8. If we multiply the cost of 1 horse by 4, G9G cost of 4 horses. -^e obtain the cost of 4 2d factor, 8 horses ; and by nudti- c -/.o X r-o-Ti plvinp: the cost of 4 rroiiuit, 0008 cost or 62 horses. ,' • " , „ , . horses by 8, we obtani the cost of 8 times 4 horses, or 32 horses, the numLer bought. G'7. Hence we have the following Rile. I. Separate the composite number into two or more fiictors. II. Mtdtiplij the multiplicand by one of these factors, and What arc contractions? Case I is ^\hat? Define a composite number. Component factors. AVhat caution is given ? Give ex- phmation. Kule, first step ? Second? MULTIPLICATION. 43 that product ly anotJier, and so on viitll all the factors have lean used succcssivelij ; the last product will be the product re- quired. Note. The product of any number of factors will lie the ^aiiic in Avhatcvcr order they aic niuhiphcd. Tliu^;, 4 X >^ X 5 = 60, and 5 X 4 X 3 = G0. EXAMTLKS FOR TKACTICE. 2. Multiply 3472 by 48 r= G X 8. Ans. 1CGG5G. 3. Multiply 147G1 by 64 = 8 X 8. 4. Multiply 87034 by 81 = 3 X 3 X 0. Ans. 7049754. 5. Multiply 4732G by 120 =: G X 5 X 4. G. Multiply G0315 by 9G. Ans. 5700240. 7. Multii)ly 291042 by 125. Ans. 3G380250. 8. If a vessel sail 43G miles in 1 day, bow lar Avill slie .-ail in 5G (lays ? Ans. 2441G miles. 9. IIow much will 72 acres of land cost, at 124 dollars an acre ? Ans. 8928 dollars. 10. There are 5280 feet in a mile; how many feet in 84 miles? Ans. 443520 feet. 11. "What will 120 yoke of cattle cost, at 125 dollars a yok e? CASE 11. 68. When the multiplier is 10, 100, 1000, &c.- If we annex a cipher to the multiplicand, each figure is re- moAed one place toward the left, and consequently the value of the whole number is increased tenfold, (Jl^.) If two ciphers are annexed, each figure is removed two places toward the left, and the value of the number is increased one hundred fold ; and every additional cipher increases the value teni'old. d.^. Hence the following Rule. Annex as many cipliers to the multiplicand as there are cipliers in the multiplier ; the number so formed will be the product required. Case II is what ? Give explanation. Rule ? 44 SIMPLE NUMBERS. EXAMPLES FOR PRACTICE. 1. Multiply 347 by 10. Ans. 3470. 2. Multiply 4731 by 100. Ans. 473100. 3. Multiply 13071 by 1000. 4. Multiply 89017 by 10000. 5. If 1 acre of land cost 36 dollar?, what -will 10 acres cost? Ans. 3 GO dollars. G. If 1 bushel of corn cost 65 cents, what will 1000 bushels cost ? Ans. G5000 cents. CASE III. 70, When there are ciphers at the right hand of one or both of the factors. 1. Multiply 1200 by GO. OPERATIOX. Analysis. Both multiplicand and Multiplicand, 1200 multiplier may be resolved into their jj , .. /.Q component factors ; 1200 into 12 and u ip.er, ^,^^^ ^^^^^ ^^ .^^^^ ^ ^^^^^ ^^ j^, ^^^^^^ Product, 72000 several factors be multiplied together they will produce the same product as the given numbers, (6Y.) Thus, 12 X 6 ==72, and 72 X 100 =r 7200, and 7200 X 10 ^ 72JOO, which is the same result as in the operation. Hence the followii; i^g Rule. MuttijiJy the' signijicant Jigures of the muItipUcnnd hj tliose of the multiplier, and to tlie product annex as many ciphers as there are ciphers on the right of both factors. ]Mult By iply EXAJIl'LES (2.) ' 4720 340000 1888 141G . 1G04800000 POU PRACTICE « . I034(K)00 10,)000 5170 1034 108570000U000 Case III is what ? Give explanation. Rule. ( MULTIPLICATION. 46 4. Multiply 70340 by 800400. Ans. 563001 3G000. 5. Multiply 3400900 by 207000. Ans. 70398G300000. 6. Multiply 634003000 by 40020. Ans. 25372800060000. 7. Multiply 10203070 by 50302000. A71S. 513234827140000. 8. Multiply 30090800 by 600080. Ans. 18056887264000. 9. Multi|)ly eighty million seven thousand six hundred, by eight million seven hundred sixty. Ans. 640121605776000. 10. Multiply fifty million ten thousand seventy, by sixty- four thousand. ■ Ans. 3200044480000. 11. Multiply ten million three hundred fifty thousand one hundred, by eighty thousand nine hundred. Ans. 837323090000. 12. There are 296 members of Congress, and each one re- ceives a salary of 3000 dollars a year ; how much do they all receive ? EXAMPLES COMBINING ADDITION, SUBTRACTION, AND MULTIPLICATION. 1. Bought 45 cords of wood at 4 dollars a cord, and 9 loads of hay at 13 dollars a load; what was the cost of the wood and hay ? Ans. 297 dollars. 2. A merchant bought 6 hogsheads of sugar at 31 dollars a hogshead, and sold it for 39 dollars a hogshead ; how much did he gain? 3. Bought 288 barrels of flour for 1875 dollars, and sold the same for 9 dollars a barrel ; how much was the gain ? Ans. 717 dollars. 4. If a young man receive 500 dollars a year salary and pay 240 dollars for board, 125 dollars for clothing, 75 dollars for books, and 50 dollars for other expenses, how much will he have left at the end of the year ? Ans. 10 dollars. 5. A farmer sold 184 bushels of wheat at 2 dollars a bushel, for which he received 67 yards of cloth at 4 dollars a yard, and the balance in groceries ; how much did his gro- ceries cost him ? 46 SIMPLE NUMBERS. 6. A sold a farm of 320 acres at 36 dollars an acre ; B sold one of 244 acres at 48 dollars an acre ; uliicli received the greater sum, and how much? Ans. B, 102 dollars. 7. Two persons start from the same point and travel in opposite directions, one at the rate of 35 miles a day, and ihe other 29 miles a day ; bow far apart will they be in 16 days? Ans. 1024 miles. 8. A merchant tailor bought 14 bales of cloth, each bale containing 26 pieces, and each piece 43 yards; how many yards of cloth did he buy? Ans. 15652 yards. 9. If a man have an income of 3700 dollars a year, and his daily expenses be 4 dollars ; what will he save in a year, or 365 days ? Ans. 2240 dollars. 10. A man sold three houses ; for the first he received 2475 dollars, for the second 840 dollars less than he received for the first, and for the third as much as for the other two; how much did he receive for the three ? Ans. 8220 dollars. 11. A man sets out to travel from Albany to Buffalo, a distance of 336 miles, and walks 28 miles a day for 10 days ; how far is he from BuflTalo? 12. Mr. C bought 14 cows at 23 dollars each, 7 horses at 96 dollars each, 34 oxen at 57 dollars each, and 300 sheep at 2 dollars each ; he sold the whole for 3842 dollars ; how much did he gain? Ans. 310 dollars. 13. A drover bought 164 head of cattle at 36 dollars a head, and 850 sheep at 3 dollars a head ; how much did he pay for all ? 14. A banker has an income of 14760 dollars a year; he pays 1575 dollars for house rent, and four times as much for fitmily expenses ; how much does he save annually ? Ans. 6885 dollars. 15. A flour merchant bought 936 barrels of flour at 9 dol- lars a barrel; he sold 480 ban-els at 10 dollars a barrel, and the remainder at 8 dollars a barrel ; how much did he gain or lose ? Ans. Gained 24 dollars. DIVISION. 47 DIVISION. MENTAL EXERCISES. 71 . 1. IIow many hats, at 4 dollars apiece, can be bought for 20 dollars ? Analysis. Since 4 dollars will buy one hat, 20 dollars will buy as many hats as 4 is contained times in 20, which is o times. There- fore, 5 hats, at 4 dollars apiece, can be bouglit for 20 dollars. 2. A man gave 16 dollars for 8 barrels of apples; what was the cost of each barrel ? 3. If 1 cord of wood cost 3 dollars, how many cords can be bought for 15 dollars ? 4. At 6 shillings a bushel, how many bushels of corn can be bought for 24 shillings ? 5. When flour is 6 dollars a barrel, how many barrels can be bought for 30 dollai-s ? 6. If a man can dig 7 rods of ditch in a day, how many davs will it take him to dig 28 I'ods ? 7. If an oi-chard contain 56 trees, and 7 trees in a row, how many rows are there ? 8. Bought 6 barrels of flour for 42 dollars ; what was the cost of 1 barrel ? 9. If a farmer divide 21 bushels of potatoes equally among 7 laborers, how many bushels will each receive ? 10. How many oranges can be bought for 27 cents, at 3 cents each ? 11. A farmer paid 35 dollars for sheep, at 5 dollars apiece ; how many did he buy ? 12. IIow many times 4 in 28 ? in IG ? in 36 ? 13. How many times 8 in 40 ? in 56 ? in 64 ? 14. How many times 9 in 36? in 63? in 81 ? 15. IIow many times 7 in 49 ? in 70 ? in 84 ? 48 SIMPLE NUMBERS. 7?2. Division is the process of finding how many times one number is contained in another. 73. The Dividend is the number to be divided. 74. The Divisor is the number to divide by. 7^1, The Quotient is the result obtained by the process of division, and shows how many times the divisor is contained in the dividend. Notes. 1. "When the dividend does not contain the divisor an exact number of times, the part of the di\idend left is called the remainder, and it must be less than the divisor. 2. As the remainder is always a part of the diA-idend, it is always of the same name or liind. 3. "When there is no remainder the division is said to be complete. 76, The sign, -4-, placed between two numbers, denotes division, and shows that the number on the left is to be divided by the number on the right. Thus, 20 -7- 4 :=: 5, is read, 20 divided by 4 is equal to 5. Division is also indicated by writing the dividend ahovcy and . . 12 the divisor heloxo a short horizontal line ; thus, — = 4, shows that 12 divided by 3 equals 4. CASE I. 77. When the divisor consists of one figure. 1. How many times is 4 contained in 848? OPERATION. Analysis. After writing the divisor on the left of the dividend, with a Hne , N. o i q' between them, we begin at the left hand ^ and say : 4 is contained in 8 hundreds, Qnotient, 212 2 hundreds times, and write 2 in hun- dreds' place in the quotient ; then 4 is contained in 4 tens 1 ten times, and write the 1 in tens' place in the quotient ; then 4 is contained in 8 units 2 units times ; and writing the 2 in units' place in the quotient, we have the entire quotient, 212. Define division. Dividend. Divisor. Quotient. Kcmnindcr. "What is complete division ? "Wliat is the sign of division. Caso I is what? Give first exi)lntiation. DIVISION. 4$ 2. How many times is 4 contained in 2884 ? OPERATION. Analysis. As we cannot divide 2 thousands by 4V^884 ^' ^^'^ ^'^^^^ ^^^ ^ thousands and the 8 hundreds to- ^^ gether, and say, 4 is contained in 28 hundreds 7 hun- 721 dreds times, which we write in hundreds' place in the quotient ; then 4 is contained in 8 tens 2 tens times, wliich we write in tens' place in the quotient ; and 4 is con- tained in 4 units 1 unit time, which we write in units' place in tlie quotient, and we have the entire quotient, 721. 3. How many times is 6 contained in 1824 ? OPERATION. Analysis. Beginning as in the last example, we 6) 1824 say, 6 is contained in 18 hundreds 3 hundi-eds times, which we write in hundreds' place in the quotient ; ^^^ then 6 is contained in 2 tens no times, and we write a cipher in tens' place in the quotient; and taking the 2 tens and 4 units together, 6 is contained in 24 units 4 units times, which we write in units' place in the quotient, and we have 304 for the entire quotient. 4. How many times is 4 contained in 943 ? OPERATION. ■• Analysis. Here 4 is contained in 9 1 \n_) 3 hundreds 2 hundreds times, and 1 hundred over, which, united to the 4 tens, makes 235 ... 3 Rem. 14 tens ; 4 in 14 tens, 3 tens times and 2 tens over, which, united to the 3 units, make 23 units ; 4 in 23 units 5 units times and 3 units over. The 3 which is left after performing the division, should be divided by 4 ; but the method of doing it cannot be explained until we reach Fractions ; so we merely indicate the division by placing the divisor under the dividend, thus, |. The entire quotient is written 235|, wliich may be read, two hundred thirty-five and three divided by four, or, two hundred thirty-five and a remainder of three. From the foregoing examples and illustrations, we deduce the following *o Rule. I. Write the divisor at the left of the dividend, with a line between them. • , ^*— — ^y Second. Third. Rule, first step? RP 8 50 SIMPLE NUMBERS. II. Beginning at the left hand, divide each figure of the dividend by the divisor, and write the result under the divi- dend. HI. If there be a remainder after dividing any figure, re- gard it as prefixed to the figure of the next lower order in tlie dividend, and divide as before. IV. Should any figure or part of the dividend be less than the divisor, write a cipher in the quotient, and prefix the num- ber to the figure of the next lower order in the dividend, and divide as before. Y. Jf titer e be a remainder after dividing the last figure, place it over the divisor at the rigid hand of the quotient. Proof. Multiply the divisor and quotient together, and to the product add the remainder, if any; if the result be equal to the dividend, the work is correct. Notes. 1. This method of proof depends on the fact that division is the reverse of multiplication. The dividend answers to the product, the divisor to one oi the factors, and the quotient to the other, 2. In multiplication the two factors are giA-en, to find the product . in di\T.sion, the product and one of the factors are given to find the other factor. EXAMPLES FOR PRACTICE. 1. Divide 7824 by 6. OPERATION. Divisor. 6)7824 Dividend. 1304 Quotient. PROOF. 1304 Quotient. 6 Divisor. 7824 Dividend. (2.) (3.) 4)05432 5)89135 6)1789^ 7)4708935 (G.) 8)1462376 9)7468542 Second step? Third? Fourth? Fifth? Proof? ITow does divis- ion differ from multiplication ? DIVISION. 51 8. Divide 3102455 by 5. 9. Divide 1762891 by 4. 10. Divide 546215747 by 11. 11. Divide 30179G24by-12. 12. Divide 9254671 by 9. Quotients. Quotients. G20491. 440722a. 49655977. 25 149 68 jC 1028296^." Rem. 13. Divide 7341568 by 7. 14. Divide 3179632 by 5. 15. Divide 19038716 by 8. 16. Divide 84201763 by 9. 17. Divide 2947691 by 12. 18. Divide 42084796 by 6. Sums of quotients and remainders, 20680083. 28. 19. Divide 47645 dollars equally among 5 men; how mucli -will each receive ? Ans.^ 9529 dollars. 20. In one week are 7 days; how many weeks in 17675 days? Ans. 2525 weeks. 21. How many barrels of flour, at 6 dollars a barrel, can be bought for 6756 dollars? Ans. 1126 barrels. 22. Twelve things make a dozen ; how many dozen in 46216464? Ans. 3851372 dozen. 23. How many barrels of flour can be made from 347560 bushels of wheat, if it take 5 bushels to make one barrel ? Ans. 69512 barrels. 24. If there be 3240622 acres of land in 11 townships, how many acres in each tOAvnship ? 25. A gentleman left his estate, worth 38470 dollars, to be shared equally by his wife and 4 children ; how much did each receive ? Ans. 7694 dollars. CASE II. 78. When the divisor consists of two or more figures. Note. To illustrate more clearly the method of operation, we will first take an example usually performed by Short Division. Case n is what ? 52 SIMPLE NUMBERS. 1. How many times is 8 contained in 2528 ? OPERATION. Analysis. As 8 is not contained in 2 thou- 8 "i 25'^8 ( 316 sands, we take 2 and 5 as one number, and con- ni sider how many times 8 is contained in this partial dividend, 25 hundreds, and find that it 12 is contained '3 hundreds times, and a remainder, 8 To find tliis remainder, we muhiply the divisor, "TT 8, by the quotient figure, 3 hundreds, and sub- tract the product, 24 hundreds, from the par- **'" tial dividend, 25 hundreds, and there remains 1 hundred. To tliis remainder we bring down the 2 tens of the dividend, and consider the 12 tens a second partial dividend. Then, 8 is contained in 12 tens 1 ten time and a remain- der ; 8 multiplied by 1 ten produces 8 tens, which, subtracted from 12 tens, leave 4 tens. To this remainder we bring down the 8 units, and consider the 48 units the third partial dividend. Tlien,8 is con- tained in 48 units 6 units times. Multiplying and subtracting as before, we find that nothing remains, and we have for the entii'e quotient, 316. 2. How many times is 23 contained in 4807 ? OPERATION. Analysis. AVe first find how Divisor. Divid'd. Quotient. many times 23 is contained in 48, 23 ) 4807 ( 209 the first partial dividend, and place 46 the result in the quotient on the ~T~~I right of the dividend. "We then niultiijly the divisor, 23, bv the 907 • " * quotient figure, 2, and subtract the product, 46, from the part of th^ dividend used, and to the remainder bring down the next figure of the dividend, which is 0, making 20, for the second partial dividend. Then, since 23 is contained in 20 no times, we place a cl|)hcr in the quotient, and bring down the next figure of the dividend, making a third partial dividend, 207 ; 23 is contained in 207, 9 times ; multi- plying and sul)tracting as before, notliing remains, and we have for the entire quotient, 209. Notes. 1. AMicn the process of dividintj is performed mentally, and the results only are written, as in Case I, the operation is termed Short Division, 2. When the -whole process of division is written, the operation is termed Loncf Division, Give first explanation. Second. "WTiat is long division ? "SMiat is short division ■ AMien is each used ? DIVISION. 53 3. Short Division is generally used ■vvhcn the divisor is a number that -will allow the process of dividing to be performed mentally. From the preceding illustrations we derive the following general Rule. I. Write the divisor at the left of the dividend, as in short division. II. Divide the least number of the left hand fgures in the dividend that icill contain the divisor one or more times, and place the quotient at the right of the dividend, with a line be- tween them. III. Multiply the divisor by this quotient fgure, suhtract the product from the partial dividend used, and to the remain- der bring down the next figure of the dividend. IV. Divide as before, until all the figures of the dividend have been brought down and divided. V. If any partial dividend will not contain the divisor, place a cipher in the quotient, and bring down the next figure of the dividend, and divide as before. VI. If there be a remainder after dividing all the figures of the dividend, it must be written in the quotient, with the divi- sor underneath. XoTES. 1. If any remainder be equal to, or greater than the divisor, the quotient figure is too small, and must be increased. 2. If the product of the di^^sor by the quotient figure be greater than the partial dividend, the quotient figure is too large, and must be diminished. 7f?. Proof. 1. The same as in short division. Or, 2. Subtract the remainder, if any, from the dividend, and divide the difference by the quotient ; if the result be the same as the given divisor, the work is correct. 80. The operations in long division consist of five prin- cipal steps, viz. : — 1st. "Write down the numbers. Rule, first step ? Second ? Third ? Fourth ? Fifth ? Sixth ? Fkst dii'ection ? Second ? Proof ? Recapitulate the steps in their order. 54: SIMPLE NUMBERS. 2d. Find Iioav many times. 3d. Multiply. 4tli. Subtract. 5 ill. liriiig down another figure. EXAMPLES FOR PRACTICE. 3. Find how many times 36 is contained in 11798. OPERATIOX. PROOF BY MULTIPLICATION. Divisor. Diviileuil. 36) 11798 108 99 72 ( 327 Quotient. 327 36 1962.. 981 Quotient. Divisor. 278 252 26 Eemaintler. 11772 26 Remainder. 11798 Dividend. 4. Find how many times 82 is contained in 89634. OPERATION. 82 ) 80634 ( 1093 82 PROOF BY DIVISION. 89634 Dividend. 8 Remainder. 763 738 254 246 8 Quotient. 1093 ) 89626 ( 82 Divisor. 8744 2186 2186 5. Find how many limes 154 is contained in 32740. Ans. 958. Aiis. 7198. Ans. 31416. Ans. 7071. A71S. 83209. 6. Divide 32572 by 34. 7. Divide 1554768 by 216. 8. Divide 5497800 by 175. 9. Divide 393M76 by 550. 10. Divide 10983588 by 132. DIVISION, 55 11. Divide 12. Divide 13. Divide 14. Divide 1.1 Divide 16. Divide 17. Divide 18. Divide 19. Divide 20. Divide 21. Divide 73484248 by 19. 8121918 by 21. 10.557312 by 16. 93S40 by 63. 352417 by 29. 51«46734 by 102. 1457924G51 by 1204. 729386 by 731. 4843167 by 3605. 49816657 by 9i01. 75867308 by 10115. Ans. 3867592. Ans. 386758. Ans. 659832. Divide 28101418481 by 1107. Divide 65358547823 by 2789. Divide 102030405060 by 123456. Divide 48659910 by 54001. Divide 233183S961 by 6739549. Hem. 33. Rem. 9. Rem. 32. Rem. 1051. Rem. 579. Rem. 1652. Rem. 6884. Rem. 4808. Quotients. Rem. 25385201. 974. 23434402. 645. 826451. 70404. 901. 5009. 346. 7. 22. 23. 24. 25. 26. 27. A railroad cost one million eight hundred fifty thousand four hundred dolhirs, and was divided into eighteen thousand five hundred and four shares ; what was the value of eaeh share? A.ns. 100 dollars. 28. If a tax of sBventy-two million three hundred twenty thousand sixty dollars be equally assessed on ten thousand seven hundred thirty-five towns, what amount of tax must each town pay ? Ans. 673 e^^jyV'V dollars. 29. In 1850 there were in the United States 213 college libraries, containing 942321 volumes ; what would be the average number of volumes to each library ? Ans. 442 4^f^ vols. 30. The number of post offices in the United States in 1853 was 22320, and the entire revenue of the post office department was 5937120 dollars ; what was the average revenue of each otfice .'' Ans. 266 dollai'S. 56 SIMPLE NUMBERS. CONTRACTIONS. CASE I. 81. When the divisor is a composite number. 1. If 3270 dollars be divided equally among 30 men, how many dollars •will each receive ? OPERATION. Analysis. If 3270 dollars be divided 5)3270 equally amoilg 30 men, each man will receive — ~ as many dollars as 30 is contained times in 6)6o4 3270 dollars. 30 may be resolved into the 109 Ans. factors 5 and 6 ; and we may suppose the 30 men divided into 5 groups of G men each ; dividing the 3270 doUars by 5, the number of groups, we have 654, the number of dollars to be given to each group ; and dividing the 604 dollars by 6, the number of men in each group, we have 109, the number of dollars that each man will receive. Hence, Rule. Divide the dividend by one %f the factors, and the quotient thus obtained hj another, and so on if there be more than tu'o factors, until every factor has hccn made a divisor. Tlie last quotient will be the quotient required. EXAMPLES FOR PRACTICE. 2. Divide 3r.90 by 15 = 3 X 5. Ans. 246. 3. Divide 3528 by 24 = 4 X 6. Ans. 147. 4. Divide 7280 by 35 = 5X7. Ans. 208. 5. Divide 6228 by 36 = 6 X 6. Ans. 173. 6. Divide 33642 by 27 r= 3 X 9. Ans. 1246. 7. Divide 153160 by 56 = 7 X 8. Ans. 2735. 8. Divide 15625 by 125 = 5 x 5 X 5. Ans. 125. 82. To find the true remainder. 1. Divide 1143 by 64, using the foctors 2, 8, and 4, and fiud the true remainder. What are contractions ? Case I is what ? Give explanation. Rule. DIVISION. 67 OPERATION. Analysis. Divid- 2)1143 ing 1143 by 2, ^ve ^TTZ~ , have a quotient of 8)0/1 1 rem. ,- 1 , . , / 5(1, and a remainder 4)71 3 X -^ = G " of 1 undivided, which, 17--3X8X 2 = 48 being a part of the given dividend, must 5.3 true rem. also be a part of the true remainder. The 571 being a quotient arising from dividing by 2, its units are 2 times as great in value as the units of the given dividend, 1143. Dividing the 571 by 8, we have a quotient of 71, and a remainder of 3 undivided. As this 3 is a part of the 571, it must be mnltipHed by 2 to change it to the same kind of units as the 1. This makes a true remainder of 6 arising from dividing by 8. Dividing the 71 1)y 4, we have a quotient of 17, and a remainder of 3 undivided. Tliis 3 is a part of the 71, the iniits of which are 8 times as great in vahie as those of the 571, and the units of the 571 are 2 times as great in value as those of the given dividend, 1143 ; therefore, to change this last remainder, 3, to units of the same value as the dividend, we multiply it by 8 and 2, and obtain a true remainder of 48 arising from dividing by 4. Adding the three partial remainders, we obtain 55, the true remainder. Hence, KuLE. I. Multipli/ each loartial remainder^ except the first, hj all the preceding divisors. II. Add the several products tvith the first remainder, and the sum will be the true remainder. EXAMPLES FOR PRACTICE. Hem. 2. Divide 34712 by 42 = 6X7. 20. 3. Divide 40137G by G4 = 8 X 8. 32. 4. Divide 139074 by 72 =r 3 X 4 X 6. 42. f). Divide 9078126 by 90 = 3 X 5 X 6. 6. 6. Divide 18730627 by 120 = 4 X 5 X 6. 67. 7. Divide 73G0479 by 96 = 2 X 6 X 8. 63. 8. Divide 24726300 by 70 = 2 X 5 -X 7. 60. 9. Divide 5610207 by 84 = 7 X 2 X G. 15. Explain the process of finding the true remainder when dividing by the factors of a composite number. 58 SIMPLE NUMBERS. CASE 11. S3. When tho divisor is 10, 100, 1000, &c. 1. Divide 374 acres of land equally among 10 men; how many acres will each have ? OPERATION. Analysis. Since -we have shown, lICloTI-l ^^^^ ^° remove a figure one place toward the left by annexing a cipher Quotient. 37 4 Rem. increases its value tenfold, or multi- or, 37 J*^)- acres. plies it by 10, (6§,) so, on the con- trary, by cutting off or taking away the right hand figure of a numher, each of the other figures is removed one place toward the right, and, consequently, the value of each is diminished tenfold, or divided by 10, (32.) For similar reasons, if we cut off two figures, w-e divide by 100, if three, we divide by 1000, and so on. Hence the Rule. From iJte rifjht hand of the dividend cut off as many figures as there are ciphers in the divisor. Under the figures so cut off, place the divisor, and the ichole will form the quotient. EXAMPLES FOR PRACTICE. 2. Divide 47 GO by 10. 3. Divide 3G2078 by 100. 4. Divide 130G321 by 1000. 5. Divide 9700347 by 10000. G. Divide 20371G0310 by 100000. CASE III. 84. When there arc ciphers on the right hand of the divisor. 1. Divide 437GG1 by 800. OPERATION. Analysis. In this example Ave 8l00Vi37Gl01 resolve 800 into the factors 8 and ' ^— ■ . 100, and divide first by 100, by cut- 547 61 Rem. tii^nr ofi' two riglit hand figures of the Caro II is what? Give explanation. Rule. Case III is what? Give cxplaiiation* DIVISION. 69 dividend, (§3,) and \\c have a quotient of 437G, and a remainder of 61. AVe next divide by 8, and obtain 547 for a quotient; and the entire quotient is ijil^^^i^. 2. Divide 347 IG by 900. OPERATION. Analysis. Dividing 9100)34711 G as in the last example, we 77 , have a quotient of 38, and 38 Quotient. 0, 2d rem. ^^^.^ remainders, 16 and 5 X 100 -j- 16 = 51 G, true rem. 5, Multiplying 5, the o8|^|, Ans. last remainder, by 100, the preceding divisor, and adding IG, the first remainder, (§2,) we have 516 for the true re- mainder. But this remainder consists of the last remainder, 5, pre- fLxcd to the figures 16, cut off from the dividend. Hence, 8«5, When there is a remainder after dividing by the sig- nificant figures, it must be prefixed to the figures cut ofT from the dividend to give the true remainder ; if there be no other remainder, the figures cut off from the dividend will be the true remainder. EXAMPLES FOn PKACTICE. Quotients. Rem. 3. Divide 347 IG by 900. 38 516 4. Divide 1047G34 by 2400. 436 1234 5. Divide 47321046 by 45000. 1051 26046 6. Divide 2037903176 by 140000. 63176 7. Divide 976031425 by 92000. 3425 8. Divide 80013T7G321 by 700000. 376321 9. Divide 19070367428 by 4160000. 4584 927428 10. Divide 379025644319 by 554000000. 89G44319 11. The circumference of the earth at the equator is 24898 miles. How muny hours would a train of cars require to travel that distance, going at the rate of 50 miles an hour ? Ans. 497 48. 12. The sum of 350000 dollars is paid to an army of 14000 men ; what does each man receive? Ans. 25 dollars. How is the true rcraamdcr found ? 60 SIMPLE XUMBERS. EXAMPLES IX THE PRECEDING RULES. 1. George AYashington was born in 1732, and lived 67 years ; in what year did he die ? Ans. in 1799. 2. How many dollars a day must a man ppend, to use an income of 1095 dollars a year ? Ans. 3 dollars. 3. If I give 141 dollars for a piece of cloth containing 47 yards, for how much must I sell it in order to gain one dollar a yard ? Ans. 188 dollars. 4. A speculator who owned 500 acres, 17 acres, 98 acres, and 121 acres of land, sold 325 acres ; how many acres had he left? Ans. 411 acres. 5. A dealer sold a cargo of salt for 2300 dollars, and gained 625 dollars ; what did the cargo cost him ? Ans. 1675 dollars. 6. If a man earn 60 dollars a month, and spend 45 dol- lars in the same time, how long will it take him to save 900 dollars from his earnings ? 7. If 9 persons use a barrel of flour in 87 days, how many days will a barrel last 1 person at the same rate ? Ans. 783 days. 8. The first of three numbers is 4, the second is 8 times the first, and the third is 9 times the second ; what is their sum? Ans. 324. 9. If 2, 2, and 7 are three factors of 364, what is the other factor? ^ Ans. 13. 10. A man has 3 farms ; the first contains 78 acres, the second 104 acres, and the third as many acres as both the others ; how many acres in the 3 farms ? 11. If the expenses of a boy at school are 90 dollars for board, 30 dollars for clothes, 12 dollars for tuition, 5 dollars for books, and 7 dollars for pocket money, what would be the expenses of 27 boys at the same rate ? Ans. 3888 dollars. 12. Four chililren inherited 2250 dollars each; but one dying, the remaining three inherited the whole ; what was the share of each ? Ans. 3000 dollars. PROMISCUOrS EXAMPLES. 61 13. Two men travel in opposite directions, one at the rate of 35 miles a clay, and the other at the rate of 40 miles a day ; how far a[)art are they at the end of 6 days? 14. Two men travel in the same direction, one at the rate of 35 miles a day, and the other at the rate of 40 miles a day ; how far apart are they at the end of 6 days ? 15. A man was 45 years old, and he had been married 19 years; how old was he when married? Ans. 26 years. IG. Upon how many acres of ground can the entire popu- lation of the globe stand, supposing that 25000 persons can stand upon one acre, and that the population is 1000000000? Ans. 40000 acres. 17. Add 384, 1562, 25, and 946; subtract 2723 from the sum ; divide the remainder by 97 ; and multiply the quotient by 142 ; what is the result? Ans. 284. 18. How many steps of 3 feet each would a man take in •walking a mile, or 5280 feet? Ans. 1760 steps. 19. A man purchased a house for 2375 dollars, and ex- pended 340 dollars in repairs ; he then sold it for railroad stock worth 867 dollars, and 235 acres of western land val- ued at 8 dollars an acre ; how much did he gain by the trade ? Ans. 32 dollars. 20. The salary of a clergyman is 800 dollars a year, and his yearly expenses are 450 dollars; if he be worth 1350 dol- lars now, in how many years will he be worth 4500 dollars ? Ans. 9 years. 21. How many bushels of oats at 40 cents a bushel, must be given for 1600 bushels of wheat at 75 cents a bushel ? Ans. 3000 bushels. 22. Bought 325 loads of wheat, each load containing 50 bushels, at 2 dollars a bushel ; what did the wheat cost ? 23. If you deposit 225 cents each week in a savings bank, and take out 75 cents a week, how many cents will you have left at the end of the year ? Ans. 7800 cents. 24. The product of two numbers is 31383450, and one of the numbers is 4050 ; what is the other number? 62 SliirLE KUMBEKS. 25. The Illinois Central Railroad is 700 miles long, and cost 31647000 dollars; what did it cost per mile? Atis. 45210 dollars. 2C. Wliat number is that, which being divided by 7, the quotient multiplied by 3, the product divided by 5, and this quotient increased by 40, the sum will be 100 ? Ans. 700. 27. How many cows at 27 dollars apiece, must be given for 54 tons of bay at 17 dollars a ton ? 28. A mechanic receives 56 dollars for 26 days' work, and spends 2 dollars u day for the whole time ; how many dollars has he left ? Ans. 4 dollars. 29. If 7 men can build a house in 98 days, how long would it take one man to build it ? Ans. 686 daj's. oO. The number of school houses in the State of New Yoik, in 1855, was 11,137 ; suppose their cash value to have been 5,301,212 dollars, v/hat would be the average value? Ans. 476 dollars. 31. A cistern wtose capacity is 840 gallons has two pipes ; through one pipe 60 gallons run into it in an hour, and through the other 39 gallons run out in the same time ; in how many hours Avill the cistern be fdled ? Ans. 40 hours. 32. The average beat of the pulse of a man at middle age is about 4500 times in an hour ; how many times does it beat in 24 hours? Ans. 108000 times. 33. How many years from the discovery of America, in 1492, to the year 1900? 34. According to the census, Maine has 31766 square miles; New Hampshire, 9280; Vermont, 10212; Massachu- setts, 7800; Rhode Island, 1306; Connecticut, 4674; and New York, 47(ii)0; how many more square miles has all New England than New York? 35. What is the remainder after dividing 62530000 by 87900? v(»s. 33100. 36. A pound of cotton has been .-pun into a lliread 8 miles in Icugtli ; allowing 235 pounds for waste, how nnniy ]>(>nnds w ill it take to spin a thread to reach round the earth, suppos- ing- the distance to be 25000 miles ? Ans. 3360 pounds. I i ri:uMiscrous kxamplks. 63 37. John has 854G dollars, which is 342 dollars less than 4 times as much as Charles has ; how many dollars has Charles? Ans. 2222 dollars, 38. The quotient of one number divided by another is 37, the divisor 24,0, and the remainder 230 ; what is the divi- dend .? Ans. 0295, oil "What number multiplied by 72084 will i>r!)duce 51'jn048? Ji>s. 72. 40. There are two numbers, the greater of which is 73 times 109, and their diliereiice is 17 times 28; wliat is the less number? Ans. 7481, 41. The sum of two numbers is 3G0, and the less is 114 ; what is the product of the two numbers ? Ans. 28044. 42. What number added to 2473248 makes 25G87o4? Ans. Oo.lOG. 43. A farmer sold 35 bushels of wheat at 2 dollars a bush- el, and 18 cords of wood at 3 dollars a cord; he received 9 yards of cloth at 4 dollars a yard, and the balance in money ; how many dollars did he receive ? Ans. 88 dollars. 44. A farmer receives 684 dollars a year for produce from his ftirm, and his expenses are 375 dollars a year ; how many dollars will he save in five years ? 45. The salt manufacturer at Syracuse pays 58 cents for wood to boil one barrel of salt, 10 cents for boiling, 5 cents to the state for the brine, 28 cents for the packing barixd, and 3 cents for packing and weighing, and receives 125 cents from the purchaser ; how many cents does he make on a barrel ? Ajts. 21 cents. 4G. A company of 15 persons purchase a township of western land for 28G000 dollars, of which sum one man pays GOOO dollars, and the others the remainder, in equal amounts ; how much does each of the others pay ? Ans. 20000 dollars. 47. If 25G be multiplied by 25, the product diminished by G25, and the remainder divided by 35, what will be the quo- tient ? A/(S. 1G5. 48. Two men start from different places, distant 189 miles, and travel toward each other ; one goes 4 miles, and the other 5 miles an hour ; in how manv hours will thev meet ? 64 SIMPLE NUMBERS. GENERAL PRINCIPLES OF DIVISION. §6, The quotient in Division depends upon the relative values of the dividend and divisor. Hence any change in the value of either dividend or divisor must produce a change in the value of the quotient. But some changes may be produced upon both dividend and divisor, at the same time, that uill not affect the quotient. The laws which govern these changes are called General Principles of Division, which we will now examine. I. 54-^9=:G. If we multiply the dividend by 3, we have 54 X 3-^9=102^9 = 18, and 18 equals the quotient, 6, multiplied by 3. Hence, Multiplying the dividend by any number, multiplies the quotient by the same number. IL Using the same example, 54 -^ 9 zr: 6. If we divide the dividend by 3 we have i■3^-^9 = 18-^9 =2, and 2 = the quotient, 6, divided by 3. Hence, Dividing the dividend by any number^ divides the quotient by the same number. in. If we multiply the divisor by 3, we have 54 _|_ 9 X 3 = 54 -^ 27 = 2, and 2 = the quotient, G, divided by 3. Hence, Multiplying the divisor by any number, divides the quotient by the same number. IV. If we divide the divisor by 3, we have 54-^-^ = 54 4-3 = 18, Upon what docs the value of the quotient depend ? AMiat is the first general principle of division ? Second ? Thii-d ? Fourth i GENERAL PRINCIPLES OF DIVISION. 65 and 18 = the quotient, G, multiplied by 3, Hence, Dividing the divisor by any number, midtiplics the quotient by tlie same number. V. If we multiply both dividend and divisor by 3, we have 54 X 3 -^ 9 X 3 z= 1G2 H- 27 = 6. Hence, MuUiplyincj both dividend and divisor by the same num- ber, does not alter the value of the quotient. VI. If we divide botli dividend and divisor by 3, we have 54-^ a =18 — 3=: G. Hence, Dividing both dividend and divisor by the same num- ber, does not alter the value of the quotient. 87. These six examples illustrate all the different changes we ever have occasion to make upon the dividend and divisor in practical arithmetic. The principles upon which these changes are based may be stated as follows : Prin. I. Multiplying the dividend midtiplies the quotient ; and dividing tlie dividend divides the quotient. (8®. I and II.) Prin. II. Multiplying the divisor divides the quotient ; and \ dividing the divisor multiplies the quotient. (8G. Ill and IV.) Pkix. III. Multiplying or dividing both dividend and divisor by the same number, does not alter the quotient. {^& , ^ V and VI.) 88. These three principles may be embraced in one ■/ GENERAL LAW, A change in the dividend produces a like change in the quotient ; but a change in the divisor produces an opposite change in tlie quotient. Note. If a number be multiplied and the product divided by the same number, the quotient will be equal to the number multiplied. Thus, 15 X 4 = CO, and 60 -e- 4 = 15. Fifth ? Sixth ? Into how many general principles can these be con- densed r AVhat is the tirst ? Second ? Thii-d ? In what general law are these embraced ? 66 PROPERTIES OP NUMBERS. EXACT DIVISORS. 80c An Exact Divisor of a number is one lliat gives a whole number lor a quotient. As it is frequently desirable to know if a number has an exaet divisor, we will present a few directions that will be of assistance, particularly in finding exact divisors of large numbers. XoTE. A number whose unit figure is 0, 2, 4, 6, or 8 is called an Even Number. And a nvunber whose luiit tigure is 1, 3, 5, 7, or 9, is called an Udd Number. 2 is an exact divisor of all even numbers. 4 is an exact divisor when it will exactly divide the tens and units of a number. Thus, 4 is an exact divisor of 2G8, 756, 1284. is an exact divisor of every number whose unit figure is or 5. Thus, 5 is an exact divisor of 20, 955, and 2840. 8 is an exact divisor when it will exactly divide the hun- dreds, tens, and units of a number. Thus, 8 is an exact divisor of 1728, 5280, and 2135G0. U is an exact divisor when it will exactly divide the sum of (lie digits of a number. Thus, in 2486790, the sum of the digits 2 + 4 + 8 + 6 + 7+94-0 = 36, and 36-^9 = 4. 10 is an exact divisor when occupies units' place. 100 when 00 occupy the places of units and tens. 1000 when 000 occupy the places of units, tens, and hun- dreds, &c. A composite number is an exact divisor of any number, when all its factors are exact divisors of the same number. Thus, 2, 2, and 3 are exact divisors of 12 ; and so also are 4 (=2X2) and 6 (= 2 X 3). An even number is not an exact divisor of an odd luimber. If an odd number is an exact divisor of an even number, "What is an exact divisor ? "What is an even number? An odd num- ber ? When is 2 an exact divisor ? 4? -5? 9? 10? 1.00? 1000? When is a eomiiosite number an exmt divisor ? An even number is not an exact divisor of what ? An odd number is an exact divisor of what ? PACTORTXtt NUMBERS. 67 twice that odd number is also an exact divisor of tho even number. Thus, 7 is au exact divisor of 42 ; so also is 7 X 2, or U. PRIME NUMBERS. 9©, A Prime Ifumber is one tliat can not be resolved or s(-:[)arated into two or more integral factors. For reference, and to aid in determining the prime factors of composite numbers, we give the following : — 1 59 139 2 Gl 149 3 G7 151 5 71 157 7 73 103 11 79 167 13 83 173 17 89 179 19 97 181 23 101 19] 29 103 193 31 107 197 37 109 199 41 113 211 43 127 223 47 131 227 53 137 229 PRIM E NU MBER 3 FRO M 1 1 ro 1000. 233 337 439 557 653 769 883 239 347 443 563 659 773 887 241 349 449 569 661 787 907 251 353 457 571 673 797 911 257 359 461 577 677 809 919 203 367 4G3 587 683 811 929 269 373 4G7 593 691 821 937 271 379 479 599 701 823 941 277 383 487 601 709 827 947 281 389 491 607 719 829 953 283 397 499 613 727 839 967 293 401 503 617 733 853 971 307 409. 509 619 739 857 977 311 419 521 631 743 859 983 313 421 523 641 751 863 991 317 431 541 643 757 877 997 331 433 547 647 761 881 FACTOllINO NUilBERS. CAsr; I. 91. To resolve any composlto numljor into its prime factors. WTiat is a prim 3 number ? Li f.istoriiig numbers, Case I is A\'hat ? 2 2772 2 138G 3 G93 3 231 7 77 11 11 1 68 PROPERTIES OF NUMBERS. 1. "What are the prime factors of 2772 ? OPEPvATiON. Analysis. We divide the given number by 2, the least prime factor, and the resuh by 2 ; this gives an odd number lor a quotient, divisible by the prime factor, o, and the quotient resulting from this division is also divisible by 3. The mxt quotiont, 77, we divide by its least prime factor, 7, and we obtain the quotient 11 ; this be- ing a prime number, the division can not be car- ried further. The divisors and last quotient, % 2, 3, 3, 7, and 11 are all the prime factors of the given number, 2772. Hence tlie Rule. Divide the given number hi/ ani/ prime factor ; di- vide the quotient in the same manner, and so continue the division until the quotient is a prime number. The several divisors and the last quotient will be the prime factors required. Proof. The product of all the prime factors will be tha given number. EXAMPLES FOR PRACTICE. 2. What ai-e the prime factors of 1 1 40 ? Ans. 2, 2, 3, 5, 19, 3. Wliat are the prime factors of 29925 ? 4. 'NMiat are tlie prime factors of 2431 ? 5. Find the prime factors of 12G73. G. Find the prime factors of 2310. 7. Find the prime factors of 2205. 8. What are the prime factors of 13981 ? CASE II. 9°3o To resolve a number into all the difibrent sets of factors possible. 1. In 3o how many sets of factors, and what arc they? Giro explanation, llulo. Trooi. Caoc II i^ v/kit f 36 =<^ CANCELLATION. 69 OPERATION. Analysis. ^VritIng the 36 at 2 V 18 ^^^ ^^^^ °^ ^^^^ ^^^^ ^^' ^^® arrange o y 1 9 all the difl'erent sets of factors into , (." -which it can be resolved under each other, as shown in the opera- " ^ " tion, and we find that 36 can be 2X2X9 resolved into 8 sets of factors. 2X3X6 3X3X4 2X2X3X3 EXAMPLES FOR PRACTICE. 2. How many sets of factors in the nunaber 24 ? What are they ? A)is. 6 sets. 3. In 125 how many sets of factors ? What are they ? Ans. 2 sets. 4. In 40 how many sets of factors, and wliat are they ? Ans. 6 sets. 0. In 72 how many sets of factors, and what are they ? Ans. 15 sets. CANCELLATION. 93. Cancellation is the process of rejecting equal fl^ctors from numbers sustaining to each other the relation of dividend and divisor. It has been shown (77) that the dividend is equal to the product of the divisor multiplied by the quotient. Hence, if the dividend can be resolved into two factors, one of which is the divisor, the other factor will be the quotient. 1. Divide 63 by 7. OPERATION. Analysis. We see in Divisor, "^p X 9 Dividend. this example that 63 is composed of the factors 7 9 Quotient. ^nd 9, and that the factor 7 is equal to the divisor. Therefore we reject the factor 7, and the remaining factor, 9, is the quotient. Give explanation. What is cancellation ? Upon what principle is it based f Give first explanation. 70 PROPERTIES OF NUMBERS, "Whenever the dividend and divisor are each composite numbers, the factors common to botli may first be rejected without altering tlie final result. (8T5 Prin. III.) 2. What is the quotient of 24 times 56 divided by 7 times 48? 24 X 56 7 X 48 OPERATION. ;? X X S Analysis. "We first in- =. 4 ^ns. dicate the op- eration to be performed by ■writing the numbers -which constitute the dividend above a line, and those Avhich constitute the divisor below it. Instead of multiplying 24 by 56, in the dividend, we resolve 24 into the factors 4 and G, and 5G into the factors 7 and 8 ; and 48 in the divisor into the f\ic- tors 6 and 8. "We next cancel the factors 6, 7, and 8, which are common to the dividend and divisor, and we have left the factor 4 in the dividend, which is the quotient. Note. "NMion all the factors or numbers in the dividend are can- celed, 1 should be retained. ©jj. If any two numbers, one in the dividend and one in the divisor, contain a common factor, we may reject that factor. 3. In 54 times 77, how many times 63 ? Analysis. In this example we see that 9 will divide 54 and 63 ; so we reject 9 as a fitctor of 54, and retain the factor 6, and also as a fi^ctor of 6.3, and retain the factor 7. Again, 7 will divide 7 in the divisor, and 77 in the dividend. Dividing both numbers by 7, 1 will be retained in the divisor, and 11 in the dividend. Finally, the l)roduct of 6 X 1 1 =: 66, the quotient. OrERATION. 6 11 $i X 711 00 4. Divide 25 X 16 X 12 by 10 X 4 X 6 X 7. OPERATION. Analysis. In 5 4^ this, as in the jire- ^^ X l'0 X ^^ 5x4 -Z<1 = 2f. ceding example, we reject all the fac- 10X^X0X7 7 tors that are com- ^ mon to both divi- » dend and divisor. Give second explanation. CANCELLATION. 7i and \ro have remaining the factor 7 in the divisor, and the factors 5 and 4 hi the dividend. Completing the work, we have ^j^ =: 2&, Aiis. From the preceding examples and illustrations we derive the following Rule. I. Write the numbers composing the dividend above a horizontal line, and the numbers composing the divisor below it. II. Cancel all the factors common to both dividend and divisor. III. Divide the product of the remaining factors of the div- idend by the product of the remaining factors of tlie divisor, and ilie result will be the quotient. Notes. 1. Kejccting a factor from any niunber is dividuig the number by that factor. 2. When a factor is canceled, the unit, 1, is supposed to take its pla(?e. I 3. One fiictor in the dividend will cancel only one equal factor in the li divisor. I| 4. If all the factors or numbers of the divisor are canceled, the !j product of the remaining factors of the dividend Avill be the quotient. II 5. By manj' it is thought more convenient to write the factors of \ the dividend on the right of a vertical line, and the factors of the divisor on the left. EXAMPLES FOR PR'ACTICE. 1. What is the quotient of 16 X 5 X 4 divided by 20 X 8 ? FIRST OPERATION. SECOND OPERATION. i0' 4 o 2, Ans. 2. Divide the product of 120 X 44 X 6 X 7 by 72 X 33 X 1 4. 'j Hule, ftrst step ? Second ? Third ? AMiat is the effect of rejecting a factor ? What is the quotient when iill the factors in the divisor are canceled ? 72 mOPERTIES OF NUMBERS. FIRST OPERATION. / ^^X^0X44X(>X^_IOX2 3^ ¥ = 6|, Am. SECOND OPERATION. 10 > 3 xn U 11 20 2 6|, -<4ws. « 3. Divide the product of 33 X 35 X 28 by 11 X 15 X 14. Ans. 1 4. 4. AVhat is the quotient of 21 X H X 26 divided by 14 X 13? Ans. 33. 5. Divide the product of the numbers 48, 72, 28, and 5, byl the product of the numbers 84, 15, 7, and 6, and give the! resuh. Ans. 9|. 6. Divide 140 X 39 X 13 X 7 by 30 X 7 X 2G X 21. Ans. 4 J. 7. What is the quotient of6GX9Xl8X5 divided byj 22 X 6 X 40? Ans. 10| 8. Divide the product of 200 X 3G X 30 X 21 by 270 X] 40 X 15 X 14. Ans. 2. 9. Multiply 240 by 56, and divide the product by GO mul-l tiplied by 28. Ans. 8. 10. The product of the numbers 18, 6, 4, and 42 is to be divided by the product of the numbers 4, 9, 3, 7, and 6 ; ■what] is the result ? Ans. 4. 11. IIow many tons of hay, at 12 dollars a Ion, must bol given for 30 cords of wood, at 4 dollars a cord ? Ans. 10 tons.! I GREATEST COMMON DIVISOR. 78 12. How many firkins of buttei', each containing 56 pounds, at 13 cents a poiniLl, must be given for 4 barrels of sugar, each containing 182 pounds, at 6 cents a pound ? Ans. 6 firkins. 13. A tailor bought 5 pieces of cloth, each piece containing 24 yards, at 3 dollars a yard. How many suits of clothes, at 18 dollars a suit, must be made from the cloth to pay for it? Ans. 20 suits. 14. How many days' work, at 75 cents a day, will pay for 115 bushels of corn, at 50 cents a bushel? Ans. 76§ days. GREATEST COMMON DIVISOR. 06. A Common Divisor of two or more numbers is a number that will exactly divide each of them. 97. The Greatest Common Divisor of two or more num- bers is the greatest number that will exactly divide each of them. Numbers pi'ime to each other are such as have no common divisor. Note. A common divisor is sometimes called a Common Measure } and the greatest common divisor, the Greatest Common Measure. CASE I. 98. "When the numbers are readily factored. 1. What is the greatest common divisor of G and 10 ? Ans. 2. OPERATION. Analysis. We readily find by inspection G . . 1 that 2 will divide both tlie given numbers ; "~ ~ hence 2 is a common divisor; and since the '_l quotients 3 and 5 have no common factor, but are prime to each other, the common divisor, 2, must be the greatest common divisor. 2. What is the greatest common divisor of 42, 63, and 105 ? "\Miat is a common divisor ? The greatest common divisor ? A common measure ? The greatest common measui-e ? "WTiat is Case I ? Give analysis, KP. 4 3 42 . . OS . . 105 7 14. .21 . . 35 2 o . 5 74 PROPERTIES OF NUMBERS. OPERATION. Analysis. AVe observe that 3 vill exactly divide each of the given numbers, and that 7 will exactly divide each of the resulting quo- tients. Hence, each of the given numbers can be exactly divided by 3 •^ ^ ' -*■' -^'*** times 7; and these numbers must be component factors of the greatest common divisor. Now, if there were any other component factor of the greatest common divisor, the quotients, 2, 3, 5, would be exactly divisible by it. But these quotients are prime to each other. Hence 3 and 7 are all the component factors of the greatest common dinsor sought. 3. "What is the greatest common divisor of 28, 140, and 280? Analysis. "We first divide by 4 ; then the quotients by 7. The re- suking quotients, 1, 5, and 10, are prime to each other. Hence 4 and 7 are all the component factors of the greatest common divisor. 4 X 7 ziz 2^, Ans. From these examples and analyses we derive the following Rule. I. Write the numhers in a line, ivith a vertical line at the left, and divide hy any factor common to all the numhers. II. Divide the quotients in like manner, and continue the \ division till a set of quotients is obtained that have no common- factor. III. Multiply cdl the divisors together, and the product will he the greatest common divisor sought. EXAMPLES FOR PRACTICE. 1. What is the greatest common divisor of 12, 36, 60, 72 ? A71S. 12. 2. "What is the greatest common divisor of 18, 24, 30, 36, 42? A71S. 6. Rule, first step f Second ? Third ? ^Pif^ '' on 28 .. :ration. 140.. 280 7 . . 35 . . 70 1 .. 5 . . 10 GREATEST COMMON DIVISOR. 75 3. What is the greatest common divisor of 72, 120, 240, 384? ^ns. 24. 4. What is the greatest common divisor of 3G, 126, 72, 21 G? Ans. 18. 5. AVhat is the greatest common divisor of 42 and 112 ? Ans. 14. 6. "Wliat is the greatest common divisor of 32, 80, and ^ 25G? Ans. 16. 7. What is the greatest common divisor of 210, 280, 350, 630, and 840? -Ans. 70. 8. What is the greatest common divisor of 300, 525, 225, and 375 ? Ans. 75. 9. What is the greatest common divisor of 252, 630, 1134, and 138G? Ans. 126. ^^. 10. What is the greatest common divisor of 96 and 544 ? ^^^k Ans. 32. ^B^ 11. What is the greatest common divisor of 468 and 1184? Ans. 4. 12. What is the greatest common divisor of 200, 625, and 150? Atis. 25. CASE II. 99. When the numbers can not be readily factored. As the analysis of the method under this case depends upon ] three properties of numbers whicli have not been introduced, we present them in this place. I. An exact divisor divides any number of times its dividend. II. A common divisor of two numbers is an exact divisor of their sum. III. A common divisor of tvrc numbers is an exact divisor of their difference. "SMiat is Case IT ? V>Tiat is the first principle upon which it is founded ? Second ? Third ? 76 PROPERTIES OF NUMBERS. 1. "What is the greatest common divisor of 84 and 203 ? OPERATION. 203 168 35 28 7, Ans. m Analysis. We draw two vertical lines, and place the larger number on the right, and tlie smaller number on the left, one line lower down. ^\'e then divide 203, the larger number, by 84, the smaller, and write 2, the quo- tient, between the verticals, the prod- uct, 168, opposite, under the greater number, and the remainder, 35, below. We next divide 84 by this remainder, writing the quotient, 2, between the verticals, the product, 70, on the left, and the new remainder, 14, below the 70. We again divide the last divisor, 3.5, by 14, and obtain 2 for a quotient, 28 for a product, and 7 for a remainder, all of which we write in the same order as in the former steps. Finally, dividing the last divisor, 14, by the last remainder, 7, and we have no remainder. 7, the last divisor, is the reatest common divisor of the given numbers. 84 2 70 2 14 2 14 2 In order to show that the last divisor in such a process is the greatest common divisor, we will first trace the worlv. in the reverse order, as indicated by the arrow line below. OPERATION. 84 70 14 14 7 divides the 14, as proved by the last division ; it will also divide two times 14, or 28, (I.) Xow, as 7 divides both itself and 28, it Avill divide 35, their sum, (11.) It will also divide 2 times 35, or 70, (I ;) and since it is a common divisor of 70 and 14, it must divide their sum, 84, which is one of the given numbers, (IT.) It will also divide 2 times 84, or 108, (I;) and since it is a common divisor of 168 and 35, it must divide their sum, 203, the larger number, (II.) Hence 7 is a common diviso)- of the given numbers. Again, tracing the work in the direct order, as indicated below, we 203 168 35 28 Give analysis. GREATEST COMMON DIVISOR. 77 know that the greatest common divisor, whatever it he, must divide 2 times 84, or 168, (1.) Then since it will divide both 168 ^' y 203 84 70 14 •^ 1G8 35 28 7 and 203, it must divide their difference, 3j, (III.) It will also divide 2 times 3j, or 70, (I ;) and as it will divide both 70 and 84, it must divide their difference, 14, (III.) It will also divide 2 times 14 or 28, (I ;) and as it will divide both 28 and 35, it must divide their difference, 7, (III;) hence, it cannot he greater than 7. Thus we have shown, 1st. That 7 is a common divisor of the given numbers. 2d. That their greatest common divisor, whatever it be, cannot be greater than 7. Hence it must be 7. Fi'om this example and analysis, we derive the following Rule. I. Draw two verticals, and write the tivo numbers, one on each side, the greater number one line above the less. ir. Divide the greater number hy the less, iv'riting the quo- tient between the verticals, the product under the dividend, and the remainder beloio. III. Divide the less number by the remainder, the last divisor hy the last remainder, and so on, till nothing remains. The last divisor will be the greatest common divisor sought. IV. If more than two numbers be given, first find the greatest co7nmon divisor of tivo of them, and then of this divisor and one of the remaining mimhers, aitd so on to the last ; the last common divisor found will be the greatest common divisor of all the given numbers. Notes. 1. "WTien more than two numbers are given, it is better to begin with the least two. 2. If at any point in the operation a prime number occur as a re- mainder, it must be a common divisor. or the given numbers have no common divisor - Rule, first step ? Second ? Thu'd ? Fourth ? 'NMiat relation have numbers when their difference is a prime number ? 78 PROPERTIES or NUMBERS. EXAMPLES FOR PRACTICE. 1. What ic the greatest commoi: divisor of 251 and 5512? OPERATION. Ans. 221 2 4 208 1 1 13 G 5512 442 1092 884 208 13 78 7r> 2. Find the greatest common divisor of 154 and 210. " Ans. 14. 3. What is the greatest common divisor of 31G and 664? Ans. 4. 4. What is the greatest common divisor of 679 and 1869? Ans. 7. 5. AVhat is the greatest common divisor of 917 and 1495? Ans. 1. G. What is the greatest common divisor of 1313 and 4108? Ans. 13. 7. What is the greatest common divisor of 1649 and 5423 ? Ans. 17. The following examples may be solved by either of the fore- going methods. 8. John has 35 pennies, and Charles 50 : how shall they arrange them in parcels, so that each boy shall have the same number in each parcel ? Ans. 5 in each parcel. 9. A speculator ha^ 3 fields, the first containing 18, the second 24, and the third 40 acres, which he wishes to divide into the largest possible lots having the same nnmber of acres in each ; how many acres in each lot ? Ans. 2 acres. MULTIPLES. 79 10. A farmer had 231 bushels of Avheat, and 273 bushels of oats, which he wished to put into the least number of bins containing the same luunber of bushels, without mixing the two kinds ; what number of bushels must each bin hold ? Ans. 21. 11. A village street is 332 rods loner; A owns 124 rods front, B 116 rods, and C 92 rods; they agree to divide their land into equal lots of the largest size that will allow each one to form an exact number of lots ; what will be the width of the lots ? Ans. 4 rods. 12. The Erie Railroad has 3 switches, or side tracks, of the following lengths: 3013, 2231, and 2047 feet; what is the length of the longest rail that will exactly lay the ti'ack on each switch ? Ans. 23 feet. 13. A forwarding merchant has 2722 bushels of wheat, 1822 bushels of corn, and 1226 bushels of beans, which he "wishes to forward, in the fewest bags of equal size that will exactly hold either kind of grain ; how many bags will it take ? Ans. 2S tj^j,^ ^j^e factors of 2X 3 X 13 X 11 X 7 X 5 = 30030, J«s.' '^''^-^ ^ ^^ We here have all the prime factors of 78, and also all the factors of 66, except the factor 11. Annexing 11 to the series of factors, 2X3X13X11, and we have all the prime factors of 78 and 66, and also all the factors cf 42 except the factor 7. Annexing 7 to the scries of factors, 2X3X 13X 11 X 7, and we have all the prime factors of 78, 66, and 42, and also all the How can a common multiple of two or more numbers be found ? Pirst principle derived from prmie numbers? becoud ? Third? Give analysis. LEAST COMMON MULTIPLE. Si factors of 30 except the factor 5. Annexing 5 to the series of Victors, 2 X 3 X 13 X 11 X "^ X 5, and we have all the prime factors of each of the given numbers ; and hence the product of the series of factors is a common multiple of the given numbers, (II.) And as no factor of this series can be omitted -without omitting a factor of one of the given numbers, the product of the series is the least common multiple of the given numbers, (HI.) From this example and analysis avc deduce the following Rule. I. Resolve the given numbers into their prime factors. II. Take all the prime factors of the lar-gest number, and I such prime factors of the other ninnbers as are not found in the largest number, and their product will be the least common multiple. Note. When a prime factor is repeated in any of the given numbers, it must be used as many times, as a factor of the multiple, as the greatest number of times it appears in any of the given numbers. EXAMPLES FOR PRACTICE. 2. Find the least common multiple of 7, 35, and 98. Ans. 490. 3. Find the least common multiple of 24, 42, and 17. Ans. 2856. 4. What is the least common multiple of 4, 9, 6, 8 ? Ans. 72. 5. What is the least common multiple of 8, 15, 77, 385? Ans. 9240. 6. What is the least common multiple of 10, 45, 75, 90 ? I Ans. 450. 7. What is the least common multiple of 12, 15, 18, 35 ? A»s. 1260. Eule, first step ? Second ? "What caution is given > 4* 82 PROPERTIES OP NUMBERS. SECOND METHOD. and 12 ? What is the least common multiple of 4, 6, 9, Analysis. We first -write the given numbers in a series, ■with a vertical line at the left. Since 2 is a factor of some of the given numbers, it must be a factor of the least common multiple sought. Dividing as many of th^ numbers as are divisible by 2, we write the 2 OPERATION. 4.. G ..9.. 12 2 2 ..3. .9.. 6 3 3 . .9.. 3 3 3 2 X 2 X 3 X 3 = 36, Ans. quotients and the undivided number, 9, in a line underneath. We now perceive that some of the numbers in the second line contain the factor 2 ; hence the least common multiple must contain another 2, and we again divide by 2, omitting to write down any quotient when it is 1. We next divide by 3 for a like reason, and still again by 3. By this process we have transferred all the factors of each of the numbers to the left of the vertical ; and their product, 36, must be the least common multiple sought, (104, III.) 2. What is the least common multiple of 10, 12, 15, and 75 ? Analysis. ^Ve read- ily see that 2 and 5 are among the factors of the given mnnbers, and must be factors of the least common multiple ; hence we divide every number 2,5 operation. 10.. 12.. 15.. 75 2,3 6.. 3.. 15 5 5 300, Ans. 2X5X2X3X5 that is divisible by either of these factors or by their product ; thus, we divide 10 by both 2 and 5; 12 by 2 ; 15 by 5; and 1o by 5. We next divide the second line in like manner by 2 and 3 ; and aftenvards the third line by 5. By this process wo collect the factors of the given numbers into groups ; and the product of the factors at the left of the vertical is the least common multiple souglit. 3. What is the least common multiple of G, 15, 35, 42, and 70 ? Give explanation. 2, 5 LEAST COMMON MULTIPLE. 83 OPERATION. Analysis. In this opcr- 1 5 . . 42 . . 70 ation we omit the 6 and 3(5, "■" because they are exactly con- o . . z . . iu tained in some of tlie otlier 3X7X2X0 = 210, Ans. given numbers; thus, 6 is contained in 42, and 'So in 70 ; and ■whatever Avill contain 42 and 70 must contain 6 and 'do. Hence Me have only to find the least common multiple of the re- maining numbers, 1 j, 42, and 70. From these examples we derive the following Rule. I. Write the numbers in a line, omitting any of the smaller numbers that are factors of the larger, and draio a vertical line at the left. II. Divide bij any frime factor, or factors, tlicd may he con- tained in one or more of tlie given immbers, and write the quotients and undivided numbers in a line underneath, omitting the Vs. III. In like manner divide the quotients and undivided num- bers, and continue the process fill all tlie factors of the given numbers have been transferred to the left of tlie vertical. Then multiply these factors together, and their product ivill be the least common multiple required. EXAMPLES FOR rRACTICE. 4. What is the least common multiple of 12, 15, 42, and 60? Ans. 420. 5. What is the least common multiple of 21, 35, and 42 ? Ans. 210. 6. "What is the lea^^t common multiple of 25, 60, 100, and 125? Ans. 1500. 7. What is the least common multiple of IG, 40, 9G, and 105? Ans. 3360. 8. What is the least common multiple of 4, 16, 20, 48, GO, H and 72? Ans. 720. 9. What is the least common multiple of 84, 100, 224, and 300? Ans. 16800. Rule, first step ? Second ? Third ? 84 PROPERTIES OF NUMBERS. 10. What is the least common multiple of 270, 189, 297, 243? A71S. 187110. 1 1. What is the least common multiple of 1, 2, 3, 4, 5, 6, 7, 8,9? Ans. 2520. 12. What is the smallest sum of money for Avhich I could purchase an exact number of books, at 5 dollars, or 3 dollars, or 4 dollars, or 6 dollars each ? A?7S. 60 dollars. 13. A farmer has 3 teams; the first can draw 12 barrels of flour, the second 15 barrels, and the third 18 barrels ; what is the smallest number of barrels that will make full loads for any of the teams? Ans. 180. 14. What is the smallest sum of money with Avhich I can purchase cows at $30 each, oxen at $55 each, or horses at $105 each? A7is. $2310. 15. A can shear 41 sheep in a day, B 63, and C 54; what is the number of sheep in the smallest flock that would furnish exact days' labor for each of them shearing alone ? Ans. 15498. 16. A servant being ordered to lay out equal sums in the purchase of chickens, ducks, and turkeys, and to expend as little money as possible, agreed to forfeit 5 cents for every fowl purchased more than was necessary to obey orders. In the market he found cliickens at 12 cents, ducks at 30 cents, and turkeys at two prices, 75 cents and 90 cents, of which he im- prudently took the cheaper; how much did he thereby for- feit ? Ans. 80 cents. CLASSIFICATION OF NUMBERS. Numbers may be classified as follows : BOO. I. As Uven and Odd. 107. II. As Prime and Composite. What is the first classification of riTimbcrs ? What is an even num- ber ? An odd number ? JBecond classification ? A prime number ? A composite number ? CLASSIFICATION OP NUMBERS. 85 108. III. As Integral ax\(\. Fractional. An Integral Number, or Integer, expresses whole things. Thus, 281 ; 78 boys; 1000 books. A Fractional Number, or Fraction, expresses equal parts of a thing. Thus, half a doUar; three-fourths of an hour; seven-eighths of a mile. 100. IV. As Abstract and Concrete. ISO. V. As Simple and Compound. A Simple Number is either an abstract number, or a concrete number of but one denomination. Thus, 48, 926; 48 dollars, 926 miles. A Compound Number is a concrete number whose value is expressed in two or more different denominations. Thus, 32 dollars 15 cents ; 15 days 4 hours 25 minutes ; 7 miles 82 rods 9 feet 6 inches. 113. VI. As Like and Unlike. Like Numbers are numbers of the same unit value. If simple numbers, they must be all abstract, as 6, 62, 487 ; or all of one and the same denomination, as 5 apples, 62 ap- ples, 487 apples ; and, if compound numbers, they must be used to express the same kind of quantity, as time, distance, &c. Tiius, 4 weeks 3 days 16 hours; 1 week 6 days 9 hours ; 5 miles 40 rods ; 2 miles 100 rods. TTnlike Numbers are numbers of ditferent unit values. Thus, 75, 140 dollars, and 28 miles; 4 hours 30 minutes, and 5 bushels 1 peck. What is the third classification ? "WTiat is an integral number ? A fractional number ? "What is the fourth classification ? An abstract number ? A concrete number ? What is the fifth classification ? A simple number ? A compound number ? Sixth classification ? ^Miat are like numbers ? Unlike numbers ? 86 FRACTIONS. FRACTIONS. DEFINITIONS, NOTATION, AND NmiERATION. It^, If a unit be divided into 2 equal parts, one of the parts is called 07ie half. If a unit be divided into 3 equal parts, one of the parts ii called 07ie t/drd, two of the parts two thirds. If a unit be divided into 4 equal parts, one of the parts is called oite fourth, two of the parts two fourths, three of the parts three fourths. If a unit be divided into 5 equal parts, one of the parts is called one fifth, two of the parts two fifths, three of the parts tJ tree fifths, &c. The parts are expressed by figures ; thus, One half is written J. One fifth is written i One third " Two fifths (; f Two thirds " o S One seventh T24iT' T"boCr » S'OOTJ' U4 7b* 1Q 17. 1 . 91^. 380 ^h. -'■'-'• T(J¥ ' TrrTTTT ' 'STS'^T ' 4'&ai42g"' ISl. Fractions are distinguished as Proper and Improper. A Proper Fraction is one wliose numerator is less than its denominator; its value is less than the unit, 1. Thus, t^, yV' T^U' §f ^^^ proper fractions. An Improper Fraction is one whose numerator equals or exceeds its denominator ; its value is never less than the unit, 1. Thus, \, §, -'4^, -3g^-, -f-g, -'^^5°- are improper fractions. ISS. A Mixed Number is a number expressed by an in- teger and a fraction; thus, 4J^, 17^|, Dy^j arc mixed numbers. IS!t. Since fractions indicate division, all changes in the terms of a fraction will afiect the value of that fraction according to the laws of division ; and we have only to modify the lan- guage of the General Principles of Division (97) liy substi- tuting the words numerator, denominator, ^n^ fraction, or value What is a proper fraction ? An Improper fraction ? A mixed number ? "V^Tiat flo fractions indicate ? REDUCTION. 89 of the fraction, for the words dividend, divisor, and quotient, respectively, and we shall have the following GENERAL PRINCIPLES OF FRACTIONS. 124. Prin. I. 3Iidtiplying the numerator midtiplies the fraction, and dividing the numerator divides the fraction. Prin. II. Multiplying the denominator divides the fraction, and dividing the denominator midtiplies the fraction. Prin. III. Multiplying or dividing both terms of the frac- tion by the same number does not alter the value of the fraction. These three principles may be embraced in one GENERAL LAW. \^5. A change in the numerator produces a like change in the value of the fraction ; hut a change in the denomina- tor produces an opposite change in the value of the fraction. REDUCTION. case I. „ 126. To reduce fractions to their lowest terms. \ A fraction is in its lowest terms when its numerator and de- nominator are prime to each other ; that is, when both terms ii have no common divisor. 1. Reduce the fraction ^| to its lowest terms. first operation. Analysis. Dividin?^ both I 4 8 — 24 — 12 — 4 j„, terms of a fraction by the same BO 30 15 5' ^"A- , , 1 number does not alter the value \ of the fraction or quotient, (124, III ;) hence, we divide both terms of ||, by 2, both terms of the result, |A, by 2, and both terms of this result by 3. As the terms of | are prime to each other, the lowest terms of || are |. We have, in effect, canceled all the fac- tors common to the numerator and denominator. First general principle ? Second ? Third ? General law ? 'NMiat is meant by reduction of fractions ? Case I is what ? AVTiat is meant by lowest terms f Give analysis. 90 FRACTIONS. SECOND OPERATION. In this operation we have divided 29 \ 48 __ 4 jlfis^ both terms of the fraction by their greatest common divisor, (97,) and thus performed the reduction at a single division. Hence the Rule. Cancel or reject all factors common to both numera- tor and denominator. Or, Divide both terms by their greatest common divisor. EXAMPLES FOR PnACTICE. 2. Reduce ^^f to its lowest terms. Ans. -i. 3. Reduce §|f to its lowest terras. Ans. |. 4. Reduce HJ- to its lowest terms. Ans. |-^. 5. Reduce f || to its lowest terms. 6. Reduce ^'rVVs to its lowest terms. 7. Reduce f^'^j to its lowest terms. 8. Reduce -^^^^ to its lowest terms. 9. Reduce -fg^f °- to its lowest terras. Ans. ^|. 10. Reduce If |§- to its lowest terras. Ans. ^%. 11. Reduce -i^i^js to its lowest terms. Ans. ^l^. 12. Express in its simplest form the quotient of 441 divided by 4G2. Ans. f^. 13. Express in its simplest form the quotient of 189 di- T^:r- vided by 273. Ans. 14. P^xpress in its simplest form the quotient of 1344 di- vided by 1536. Ans. f. CASE II. 127. To reduce an improper fraction to a whole or mixed number. 1. Reduce ^-5*- to «i whole or mixed number. OPERATION. Analysis. Since ■.V*- = 324— 15 = 2IA = 21-?, Ans. ^^ fifteenths equal 1,324 fifteenths are equal to as many times 1 as \o is contained times in .'521, which is Slj'j times. Or, since the numerator is a dividend and the denom- Rulc. Case 11 is what ? Give explanation. REDUCTION. 91 inator a divisor, (1 !§,) ■\ve reduce the fraction to an equivalent whole or mixed number, by dividing the numerator, 324, by the denominator, 15. Hence the EuLE. Divide the numerator hy the denominator. Notes, 1. "When the denominator is an exact divisor of the numer- ator, the result ■will be a whole number. 2. In all answers contaLiimg fractions, reduce the fractions to their lowest terms. EXAMPLES FOR PRACTICE. 2. In Jy^- of a Aveek, how many weeks ? Ans. ] f^. 3. In -1-^^ of a bushel, how many bushels ? Ans. 23|. 4. In A |i of a dollar, how many dollars? 5. In ^-^^- of a pound, how many pounds ? Ans. 54t5-. 6. Reduce -f J^ to a mixed number. 7. Reduce -^^g^- to a whole number. 8. Change ^If- to a mixed number. Ans. 18§. 9. Change ^ff- to a mixed number. 10. Change -S-V^V"^ to a mixed number. Ans. 1053||-. 11. Change ^^Igj^^ to a whole number. Ans. 7032. CASE III. 128. To reduce a whole number to a fraction hav- ing a given denominator. 1. Reduce 46 yards to fourths. OPEEATiox. Analysis. Since in 1 yard there are 4 fourths, 4g in 46 yards there are 46 times 4 fourths, wliich are 4 184 fourths z= l|^. In practice we multiply 46, the number of yards, by 4, the s;iven denominator, 4 > -^'25' and taking the product, 184, for the numerator of a fraction, and the given denominator, 4, for the de- nominator, we have ^^. Hence we have the Rule. Mnltiflij the. %ohole number by the given denominator ; take the product for a numerator, under which write the given denominator. Rule. Case III is what ? Give explanation, llule. li 92 FRACTIONS. XoTE. A whole number is reduced to a fractional form by writing 1 under it for a denominator ; thus, 9 ::= #. EXAMPLES FOR PRACTICE. 2. Reduce 25 bushels to eighths of a bushel. Ans. -§a. 3. Reduce G3 gallons to fourths of a gallon. Ans. -2^2. 4. Reduce 140 pounds to sixteenths of a pound. 5. In 56 dollars, how many tenths of a dollar? Ans. ^^ty. 6. Reduce 94 to a fraction whose denominator is 9. 7. Reduce 180 to seventy-fifths. 8. Change 42 to the form of a fraction. Ans, ^-f-. 9. Change 247 to the form of a fraction. 10. Change 347 to a fraction whose denominator shall be 14. Ans. iff 8. CASE IV. 129. To reduce a mixed number to an improper fraction. 1, In 5| dollars, how many eighths of a dollar ? OPEKATION. 5 3 Analysis. Since in 1 dollar there are 8 eighths, o in o dollars there are 5 times 8 eighths, or 40 — eighths, and 40 eighths -(- 3 eighths zr: 43 eighths, ■*g*) Ans. or ^-, From this operation we derive the following Rule. MulUphj the whole nnniher hy the denominator of the fraction ; to the product add the nionerator, and under the sum write the denominator. EXAMPLES FOR PRACTICE. 2. In 4i dollars, how many half dollars? Ans. f. 3. Ir. 71 f weeks, how many sevenths of a Aveck ? 4. In 341^ acres, how many fourths ? 5. Change 12/j years to twelfths. G. Change SG/^- to an improper fi-action. 7. Reduce 2\ ,-Jn to an improper fraction. 8. Reduce 225 i| to an improper fraction. Ans. X.^£J.. Ans. -W- Ans. -^HF- Ans. Hi^' Case IV is what ? Give explanation. Rule. REDUCTION. 93 9. In 96/j% ^low many one hundred twentieths ? 10. In 1297g\, how many eighty-fourths ? Ans. -LQ|9^. 11. What improper fraction will express 400§^- ? CASE V. ISO. To reduce a fraction to a given denominator. As fractions may be reduced to lotver terms by division, they may also be reduced to higher terms by multiplication j and all higher terms must be multiples of the lowest terms. (iOS.) 1. Reduce f to a fraction whose denominator is 20. OPERATION. Analysis. We first divide 20, the 20 -^ 4 zz: 5 required denominator, by 4, the denomi- nator of the given fraction, to ascertain _ ^ '^ ;^ J.5 ^;j5. if it be a multiple of this term, 4. The 4 >< 5 division shows that it is a multiple, and that 5 is the factor which must be em- ployed to produce this multiple of 4, We therefore multiply both terms of | by 5, (124,) and obtain if, the desired result Hence the Rule. Divide the required denominator by the denominator of the given fraction, and midtiply both terms of the fraction by the quotient. EXAMPLES FOR PKACTICE. 2. Reduce | to a fraction whose denominator is 15. Ans. -jfij. 3. Reduce f to a fraction whose denominator is 35. 4. Reduce |f to a fraction whose denominator is 51. Ans. §f. 5. Reduce f g to a fraction whose denominator is 150. 6. Reduce ^§g to a fraction whose denominator is 3488. Ans. m%' 7. Reduce ^|^ to a fraction whose denominator is 1000. Case Y is what ? How are fractions reduced to higher terms ? AiVTiat are all higher terms ? Give analysis. Rule, 94 FRACTIONS. CASE YI. ISA. To reduce two or more fractions to a com- mon denominator. A Common Denominator is a denominator common to two or more fractions. 1. Reduce ^ and f to a common denominator. OPERATION. Analysis. AVe multiply the terms of the 3 \/ 5 first fraction by the denominator of the second, — z=z 43^ and the terms of the second fraction by the 4X5 denominator of the fii-st, (1*24.) This must re- ^ sy A ^'^^'^ ^^'^^ fraction to the same denominator, — — 8 for each new denominator Avill be the product 5X4 of the given denominators. Hence the Rule. Multiply the terms of each fraction hy the denomina- tors of all the other fractions. Note. ^lixcd numbers must iirst be reduced to improper fractions. EXAMPLES FOR PRACTICE. 2. Reduce §, 4, «ind f to a common denominator. //«c 16 12 18 JinS. ^^, 245 21. 3. Reduce ^ and | to a common denominator. A„f. 2 7 2 8 4. Reduce |, -/j, and | to a common denominator. A„o 28 8 210 300 ■^nS. -560, J605 360' 5. Reduce f , |, §, and ^ to a, common denominator. G. Reduce j^g, ^, and f to a common denominator. Jiic 2*^ H4 9 6, jins. 5-3 J, -4 5 J, i^-7- 7. Reduce f, 2^, |, and ^ to a common denominator. 8. Reduce 1§, j%, and 4 to a common denominator. /f „ o 150 2 4 3 2 Cnse YI ic vlmt ? "NVJmt is a rommnn denominator ? Give ana]3'sis. REDUCTION. 95 CASE VII. IS'3. To reduce fractions to the least common de- nominator. The Least Common Denominator of two or more fractions is the least denominator to wliich they can all be reduced, and it must be the least common multiple of the lowest denom- inators. 1. Reduce ^, |, and -f'2 to the least common denominator. OPERATION. Analysis. We first find 2,3 9 •> "1 " Q g 22 the least common multiple An of the given denominators, _JL! which is 24. This must be 2 X 3 X 2 X 2 = 24 the least common denom- inator to which the frac- Ans. tions can be reduced, (III.) We then multiply the terras of each fraction by such a number as will reduce the fraction to the denominator, 24. E.e- ducing each fraction to this denominator, by Case V, we have the answer. Since the common denominator is already determined, it is only necessary to multiply the numerators by the multipliers. Hence the following [ Rule. I. Find the least common multiple of the given de- nominators, for the least common^ denominator. II. Divide this common denominator hy each of the given denominators, and multiph/ each numerator hy the correspond- ing quotient. The products toill he the new numerators. examples for practice. 2. Reduce j^-g, f-^, %l, and /^ to^their least common de- nominator. Ans. -'--A Mr i-il t2.^ 3. Reduce i, f, ^^g, /j to their least common denominator. J^, C )fi«! 192 63 32 -^'"' 3 3 5' 3 3 5' 3J6' 3aff' _"^^^mt is Case VIT? "What must be the least common denominator ? Give analysis. Rule, first stop. Second. 96 FRACTIONS. 4. Reduce f, ^i, f , and 6 to their least common denorai- 5. Reduce 5^, 2^, and 1| to their least common denomina- tor. Ans. -\S-VS-V- 6. Reduce j\, f, f, and \ to their least common denomi- rmtnr >lw big* 7. Reduce f, |^, f , 2|, and j^j to their least common de- nominator. Ans. ii§, T^gV, yVff' 1-6 8' iVs- 8. Change %, {^, 3|, 9, and |- to equivalent fractions hav- ing the least common denominator. 9. Change f^, If, I, }|,and 6 to equivalent fractions hav- ing the least common denominator. 10. Change 2/5, f J, 4, 1§, ^^, and f to equivalent frac- tions having the least common denominator. 11. Reduce f, §, ^, and /^ ^0 a common denominator. 12. Reduce |, |-, 2£, and ^ to a common denominator. 13. Reduce ||, -/j, §, and 3^ to equivalent fractions hav- ing a common denominator. Ans. ||, '^^, §g, ^^. 14. Change ^j, f , and |- to equivalent fractions having a common denominator. Ans. f^Viy toViT' tWit- 15. Change ^\, 7^, f^, and 5 to equivalent fractions hav- ing a common denominator. • Ans. ||, i^^P-, |g, 3^^-. 16. Change 277, 6^, /jy, 7, f, and 1^ to equivalent fractions having a common denominator. ADDITION. 133. 1. What is the sum of ^, f , f , and | ? OPERATION. Analysis. Since the i + i + i + I- = -'# = 2, ^«s. g"'en fractions^ have a common denominator, 8, their sum may be found by adding their numerators, 1, 3, 5, and 7, and placinfj; the sum, 16, over the common denominator. We thus obtain -^ =r: 2, the required sum. 2. Add Z^, ^(j, tV, ^s, and j%. Ans. 2^. 3. Add ^, 1^, -^.j, -i^, W, and H- ^^«- 2/7- Give first explanation. ADDITION. 97 4. What is the sum of /^, /g, /^, ^f, ^f, and fi? 5. What is the sum of jV^, j%%, -rVa, t¥(J' ^nd jQ? ? 6. What is the sum of ^^% ^-^%, J-|a ^ |j., and f f f ? i34. 1. What is the sum off and | ? OPERATION. Analysis. In a J- 2 — 2 7 110—. 2 7 + la =z^l An. 1. From /jj take f^. OPERATION. Analysis. Since the given ^Tj — Jj z=z '^ij = -?, Ans. fractions have a common denom- inator, 10, we find the (lifl'erence by subtracting 3, the less numerator, from 7, the greater, and write How arc mixed numbers added ? Give note. SUBTRACTION. 99 the remainder, 4, over the common denominator, 10. "We thus obtain -^\ := |, the required difference. 2. From f take |. * Aiis. 3. From {4 take }i. Ans. 4. From f ^ take ^^y. Ans. 5. From fl take ^|. -^ks. 6. From -/tj^q take yVj. -4??s. 7. From iff take ^Jg. ^ns. 1 4' J 3 TO"' 1 J* 6 27* 106. 1. From # take |-. OPEKATION. Analysis. 8 5. — ^2 30 — 3 2^,3 — _2 — J /f„c As in -whole numbers, we can subtract like numbers only, or those having the same unit value, so, we can subtract fractions only when they have a common de- nominator. As f and | have not a common denominator, we first reduce them to a common denominator, and then subtract the less numerator, 30, from the greater, 32, and write the difference, 2, over the common denominator, 36. We thus obtain j^g iz:: Jg, the required difference. Hence the following Rule. I. Wl/cn necesscwy, reduce the fractions to a commoyi denominator, II. Subtract the numerator of the suhtrahend from the numerator of the minuend, and place the difference over the common denominator. EXAMPLES FOR PRACTICE. _5 Tff* _9 ¥5' 2. From 4 take f . Ans. 3. From J-f take f. Ans. 4. Subtract /^ from |. Ans. ■^^^. 5. Subtract -j*-^ from -f^^. Ans. 4^. 6. Subtract ^a from j^ag. Ans. -jW. 7. Subtract f^j^/v from fg-. A71S 637 iij^g"* 8. What is the difference between 9^ and 2|- ? Give explanations Rule, first step. ■ Second. 100 FRACTIONS. OPERATION. Analysis. We first reduce the frac- 91 rr 9 4j tional parts, ^ and -|, to a common denom- 23 _- 2-9- inator, 12. Since we cannot take ^\ from — ~ ^\, we add 1 = if to ^\, wliich makes \^, G/j Ans. and j\ from \^ leaves ^. We now add 1 to tiie 2 in the subtrahend, (50,) and say> 3 from 9 leaves 6. We thus obtain 6^^, the difference required. Hence, to subtract mixed numbers, we may reduce the fractional parts to a common denominator, and then subtract the fractional and integral parts separately. Or, We may reduce the mixed numbers to improper fractions, and subtract the less from the greater by the usual method. 9. From 81 take 3|. 10. From 25 1 take 9yV 11. From 4| take i|. 12. Subtract 1} from 6. 13. Subtract 120^9^ from 4501. 14. Subtract j%\ from 3-j-V 15. Find the difference between 49 and 75;^. 16. Find the difference between 227f and 196§. 17. From a cask of wine containing 314- gallons, 17|- gal- lons were drawn ; how many gallons remained? Ans. 13|. 18. A farmer, having 450-^^ acres of land, sold 3043 acres; how many acres had he left ? Ans. 145^^. 19. If flour be bought for G^ dollars per barrel, and sold for 72 dollars, what will be the gain per barrel ? 20. From the sura of f and 3J- take the difference of 4J and 5|. Ans. 3||. 21. A man, having 25 1 dollars, paid 6^ dollars for coal, 2^ dollars for dry goods, and f of a dollar for a pound of tea; how much had he left? Ans. $1G|5. 22. What number added to 2| will make 7^ ? Ans. 4^^. 23. What fraction added to j-4- will make ^^ ? Aiis. -^s. In how many ways may mixed numbers be subtracted ? What ore • they ? Ans. HI Ans. 16tV Ans. 330i|. Ans. 33¥5- MULTIPLICATION. 101 24. A gentleman, having 2000 dollars to divide among his three sons, gave to the first 91 2| dollars, to the second 545^ doHars, and to the third the remainder ; how much did the tliird receive ? Ans. $i>-i2^2- 25. Bought a quantity of coal for loG^^^ dollars, and of limiher for 3502 dollars. I sold the coal for 184^ dollars, and the lumber for 41G2 dollars. How much was my whole gain? Ans. $114jig. ItlULTIPLICATIOX. CASE I. 137. To multiply a fraction by an integer. 1. If 1 yard of cloth cost f of a dollar, how much will 5 yards cost ? oPERATiOiV. Analysis. Since 1 yard cost 3 \/ 5 — - JL5 — : 33 J^ns. ^ fourths of a dollar, 5 yards will cost 5 times 3 fourths of a dollar, or 15 fourths, equal to 3| dollars. A fraction is multiplied by multiplying its numerator, (124.) 2. If 1 gallon of molasses cost -^^ of a dollar, how much •will 5 gallons cost ? OPERATION. Analysis. Since 5, the 7 v5 — n — 13 An f C'onf; lined in ] 2 ? Ans. 31. Rule, first step. Second. "Wliat other method is mentioned ? DIVISION. Hi 9. How many times is -/g contiilned in ^^ ? Ans. 2^. 10. How many times is /g^ contained in ^§ ? 11. How many limes is ^ of |^ contained in f of 2J^? 12. Wliat is the quotient of -^^y of 4, divided by f of 3| ? 13. What is the quotient of ^ of ^ of oG divided by 1 ^^ times •? ? ^/is. o-S. 1-4. Wliat is the value of ^^ ? 4? ■^8 OPERATION. oJ._ J __7 . 3o_;3^ ^ Tliis example 7" — — Ti : ^=^ X — ^^^ ■&> Ans. is only another ^ij 8 " /^ lOp iorm for ex- 5 • r • pressmg divis- ion of fractions ; it is sometimes called a complex fraction, and the process of performing the division is called reducing a com jdcx frac- tion to a simple one. We simjily reduce the upper number or dividend to an improper fraction, and the lower number, or divisor, to an improper fraction, and then divide as before. p.? 15. Vrhat is the value of — ? Ans. fa. lU IG. What is the value of '-? Ans. 20. 4 T 17. ■ What is the value of ? Ar.s. ^V-- ^ 2 of 3 18. What is the value of ^^ -? Ans. 1. JL 2 of 5 19. What is the value of ~ ^ ? Ans. i. fof4i 20. If a horse cat f of a bushel of oats in a day, in how many days will ho eat 5i bushels? Ans. 14. 21. If a man spend If dollars per month fur tol)acco, in what time will he spend lOf dollars? Ans. Gj months. "SMiat is a complex fraction ? 112 FEACTIONS. 22. How many times will 4| gallons of campliene fill a vessel tliat holds i of | of 1 gallon ? Ans. l(>i. 23. If 14 acres of meadow land produce 322 ton.- of liav bow many tons will acres produce ? Ans. 11 §. 24. If 2 yards of silk cost $3^, how much less than SI 7 will 9 yards cost ? Ans. S2|. 25. If f of a yard of cloth cost -f^y of a dollar, how much will 1 yard cost ? 26. A man, having $10, gave f of his money for clover seed at S3i- a bushel; how much did he buy? Aus. 2 bush. 27. How many tons of hay can be purchased for SllOy'^i at $0f per ton? Ans. 12 /q. PROJIISCUOUS EXAMPLES. 1. Reduce ^, |, f, and ^ to equivalent fractions whose de- nominators shall be 24. Ans. if, |§, ^j, ^V 2. Change 4 to an equivalent fraction having 01 for its denominator Ans. |f. 3. Find the least common denominator of £, 1§, ^ of |, 2, i of I of 1 J,. 4. Add 4^, J, I of 1^, 3, and J-J-. 5. Find the difference between f of G/j and | of 4,\. 6. The less of two numbers is 4750*, and their difference is 123f ; what is the greater number? Ans. 4885j'5. 7. What is the difference between the continued products of 3, I, i 42, and 3^, §,4, f? Ans. Sh}. 4 2J- 8. Reduce the fractions — and -^ to their simplest form. JL 11 '■ 9. What number multiplied by f will produce 1825| ? Ans. 3043 i. 10. A farmer had {■ of his sheep in one pasture, ^ in an- other, and the remainder, which were 77, in a third jiasture; how many sheep had he ? Ans. 140. 11. What will 7^ cords of wood cost at l of 9i dollars per cord? Ans. $24^ J. PROMISCUOUS EXAMPLES. 113 12. At i of a dollar per bushel, how many bushels of apples can be bought for 5 1 dollars ? 13. Paid $1837| for 7350^- bushels of oats ; how much was that per bushel ? Ans. ^ of a dollar. 11, If 235 J- acres of land cost $4725 1, how much will G28 acres cost.? Ans. $12G01. 15. A man, owning | of an iron foundery, sold J of his share for $51.0 J; what was the value of the foundery ? Ans. $1055^. 16. 14^ less ^"^^^" is f of ^ of what number? I'^^ty Ans. 27. 17. A merchant bought 4f cords of wood at $3^ per cord, and paid for it in cloth at f of a dollar per yard ; how many yards were required to pay for the wood ? 18. How many yards of cloth, f of a yard wide, will line 201 yards, 1^ yards wide ? Ans. 34^. 1 9. If the dividend be |, and the quotient tj*^, what is the divisor ? 20. If the sum of two fractions be |, and one of them be ^»(j, what is the other ? Ans. /^y, 21. If the smaller of two fractions be ||, and their differ- ence /tj, what is the greater ? ■ Aiis. |f. 22. If 3 1 pounds of sugar cost 33 cents, how much must be paid for 65^ pounds ? 23. If 324 bushels of barley can be had for 2591 bushels of corn, how much barley can be had for 2000 bushels of corn ? Ans. 2500 bushels. 24. A certain sum of money is to be divided among 5 per- sons ; A is to have |, B |, C yV' ^ ?o» ^^(^ E the remainder, which is 20 dollars ; what is the whole sum to be divided ? Ans. $50. 25. What number, diminished by the difference between | and f of itself, leaves a remainder of 34 ? Ans. 40, 26. If I of a farm be valued at $1728, what is the value of the whole ? 114 FRACTIONS. 27. Bought 320 sheep at $2|- per head ; afterward bought 43.3 at §1^ per head ; then sold f ofthe whole number at Sl| per head, and the remamder at ^2^ ; did 1 gahi or lose, and how inueh ? Ans. Lo.^t S44J-. 28. h' o be added to both terms of the fraction -|, will its value be increased or diminithed? Ans. Increased y|j. 2'j. If be added to both terms of the fraction f , Avill its value be increased or diminished? Ans. Diminished ^\. 30. How many times can a bottle holding |^ of § of a gal- lon, be filled from a demijohn containing f of If gallons ? Ans. 7^. 31. Bought I of 71 cords of wood for I of $32 ; how much did 1 cord cost ? 32. Purchased 728 pounds of candles at 1 6| cents a jiound ; had they been purchased for 3| cents less a pound, how many pounds could have been purchased lor the same money ? Ans. 953^1- 33. AYluit number, divided by If, will give a quotient of 91? Ans. 123|. 34. The pi-odiict of two numbers is 6, and one of them is 184G; what is the other? Ans. gfj- 35. A stone mason worked II5 days, and after paying his board and other expenses with f of his earnings, he had S20 left ; how much did he receive a day? 3G. If f of 4 tons of coal cost $5^, what will f of 2 tons cost ? Ans. $5. 37. In an orchard £ of the trees arc apple trees, yV pt'acli trees, and the remainder are pear trees, which are 20 more than -^ of the wliolr ; liow many trees in the orchard? Ans. 800. 38. A man gave G| pounds of butter, at 12 cents a pound, for ^ of a gallon of oil ; how much was the oil worth a gal- lon ? Ans. 100 cents. 39. A gentleman, having 27 H acres of land, sold ^ of it, and gave g of it to liis son; what was the value of the re- mainder, at $o7^ per acre ? Ans. ^4u77gV PROMISCUOUS EXAMPLES. 115 40. A horse and wagon cost $270; the hor.sc cost 1]- times as much as the wagon ; what was the cost of the wagon ? 41. AVhat number tukeii from 2^- times 12|- will lea^e 20,^ > Ans. Ux. 42. A merchant bouglit a cargo of flour for $2173^, and sold'it for f- 2_ of tlie cost, thereby losing I of a dollar i)er bar- rel ; how many barrels did he purchase ? Ans. 126. 4o. A and B can do a piece of work in 14 days ; A can do I as much as L ; in how many days can each do it ? Ans. A, 32 1 days ; B, 24i days. 44. How many yards of cloth f of a yard wide,are equal to 12 yards f of a yard Avide ? Ans. 1 1 J. 45. A, B, and C can do a piece of work in 5 days ; B and C can do it in 8 days ; in Avhat time can A do it ? 4G. A man put his money into 4 packages; in the fir.-t ho put |, in the second i, in the third -|, and in the fourth the re- mainder, which was $24 more than -jL of the whole ; how much money had he? ' Ans. $720. 47. If $71 will buy 3^ cords of wood, how many cords can be bought for $10i ? Ans. Uh 48. How many times is ^ of | of 27 contained in | of ^ of 422? 41). A boy lost J- of his kite string, and then added 30 feet, when it was just \ of its original length; what was the length at first? Ans. 100 feet. 50. Bought f of a box of candles, and having used \ of them, sold the remainder for ^\ of a dollar; how much would a box cost at the same rate ? Ans. $5|-|. 51. A post stands ^ in the mud, J- in the water, and 21 feet above the water; what is its lenn^th ? 52. A father left his eldest son -^ of his estate, hrs youngest son \ of the remainder, and his daughter the remainder, who received $17235 less than the youngest son; what wa-^ the value of the estate ? Ans. $21114iJ. 116 DECIMALS. DECIIilAL rr.ACTIOXS. I'^IS. Decimal Fractions are fractions ■which have for their denominator 10, 100, 1000, or 1 with any number of ciphers annexed. XoTF.s. 1. The -word decimal is derived from the Latin decern, ■which signifies ten. 2. Decimal fractions are commonlj' called decimals. 3. Since ^V =^ iVo' lio = Tofo' ^^•' ^'^ denominators of decimal fractions increase and decrease in a tenfold ratio, the same as simple numbers. DECIMAL NOTATION AND NUMERATION. 144:. Common Fractions are the common divisions of a unit into any number of equal parts, as into halves, fiftlis, twenty-fourths, &c. Decimal Fractions are the decimal divisions of a unit, thus: A unit is divided into ten equal parts, called tenths ; each of these tenths is divided into ten other equal parts called hun- dredths ; each of these hundredths into ten other equal parts, called thousandths ; and so on. Since the denominators of decimal fractions increase and decrease by the scale of 10, the same as simple numbers, in writing decimals the denomina- tors may be omitted. In simple numbers, the unit, 1, is the starting point of notation and numeration; and so also is it iiv decimals. We extend the scale of notation to the left of units' place in wrltiii;^' integers, and to llie right of units' place in writing (Irciiiials. Thus, the first place at the left of units is tens, and (lie first ))lace at the right of units is tenths ; the second place at the left is hundreds, and the second ))laoe at the right is hundredths ; the third place at the left is thousands, and the third place at the right is thousandths ; and so on. What are dociTnal frnctirms? TTnw do they difTor from common fractions r Ilow are they written ? NOTATION AND NUMERATION. 117 Tlie Decimal Point is a period ( . ), which must always be placed before or at the left hand of the decimal. Thus, •j^j is expressed .G 5 4 « u r I Tacr -^i Note. The decimal point is also called the Separafrix. This is a correct name for it only when it stands between the integral and deci- mal i^arts of the same number. .5 is 5 tenths, which = Jg. of o units ; .05 is 5 hundredths, " = y'^y of 5 tenths ; .005 is thousandths, " = ^-L. of 5 hundredths. And universally, the value of a figure in any decimal place is yij the value of the same figure in the next left hand place. The relation of decimals and integers to each other is clear- ly shown by the following NUMERATION TABLE. -3 J3 £ a -S t: 2 « .2 t« . 5 5 § I ^ I a ^ ,Z J^ S ^^ ~ -^ 5=~ Z ^ 475 3.G2418G95 — V — Integers. Decimals. By examining this table we see that Tenths are expressed by one figure. Hundredths " " " two figures. Thousandths " " " three " Ten thousandths " " " four " And any order of decimals by one figure less than the corre- sponding order of integers. 145, Since the denominator of tenths is 10, of hun- ^\Tiat is the decimal point ? AMiat is it sometimes called ? "WTiat is the value of a figure in any decimal place ? Its DECDIALS. dredths 100, of thousands 1000, and so on, a decimal may be expressed by writing tlie numerator only ; but in tiiis ca2i 57 columns; that is, units under units, tenths under p-.^ tenths, hundredths under hundredths, &c. This ' brings the decimal points directly inider each • other. Commencing at the right hand, we add 045.9524: each column separately, and cany as in -whole numbers, and in the result we place a decimal point between units and tenths, or directly under the decimal point in the numbers added. From this example we derive the following Rule. I. Write tlie numbers so that the dccimql points shall stand directly under each other. II. Add as in whole nwnbers, and place the decimal point, in the result, directly under the points in the numbers added. / EXAMPLES FOR PKACTICE. 2. Add .199 3. Add 2.7509 .25 .054 4.015 0.75 27.38203 375.01 2.5 415.05703 Sum, 3.8599 Amount, 4. Add 1152.01, 14.11018, 152348.21, 9.000083. Ans. 153523.330203. 5. Add 37.03, 0.521, .9, 1000, 4000.0004. Ans. 5038.4514. 6. What is the sum of twenty-six, and twenty-six Imn- dredths ; seven tentlis ; six, and eighty-three tliou.^andths ; four, and four thousandths? Ans. 37.047. Explain the operation of addition of decimals. Give rule, first step. Second. ADDITION. 125 7. "What is the sum of thirty-six, and fifteen thousandths ; three Jiundred, and six hundred five ten-tliousandths ; five, and three miUionths ; sixty, and eighty-seven ten-millionths? Ans. 401.0755117, 8. "\7hat is the sum of fifty-four, and thirty-four hun- dredtlis ; one, and nine ten-lhousandtlis ; three, and two liun- Vlred seven millionths ; twenty-three tliousandths ; eight, and nine tenths; four, and due hundred thirty-five tliousandths? Ans. 71.399107. 9. ITow many yards in three pieces of cloth, tlie first piece containing 18.375 yards, the second piece 41.G25 yards, and the third [)iece 35.5 yards? 10. A's farm contains 01.843 acres, B's contains 143.75 acres, C's 218.4375 acres, and D's 21.9 acres; how many acres in the four farms ? 11. My farm consists of 7 fields, containing 12 J acres, 18f acres, 9 acres, 24| acres, 4|| acres, 8f^y acres, and 15^^ acres respectively ; how many acres in my farm ? Note. Reduce the common fractions to decimals before adding. Ans. 93.6375. 12. A grocer has 2J- barrels of A sugar, 53. barrels of B sugar, 3 1 barrels of C sugar, 3.0 G42 barrels of crushed sugar, and 8.925 barrels of pulverized sugar ; how many bar- rels of sugar has he ? Ans. 23.8642. 13. A tailor made 3 suits of clothes; for the first suit he used 2^ }ards of broadcloth, 3j\ yards of cassiraere, and f yards of satin ; for the second suit 2.25 yards of broadcloth, 2.875 yards of cassimere, and 1 yard of satin ; and for the third suit 5-^\ yards of broadcloth, and 1^ yards of satin. ILnv many yards of each kind of goods did he use? How many yards of all ? Ans. to last, 18.375. 126 DECIMALS. SUBTRACTIOX. From 91.73 take 2.18. Analysis. In each of these three examples, wo write tlie subtrahend under the minu- end, placing units under units, tenths under tenths, &;c. Commencing at the right hand, we subtract as in whole numbers, and in the remainders we place the decimal points directly under those in the numbers above. In the second example, the number of decimal places in the minuend is greater than the number in the subtra- hend, and in the tliird exam- ple the number is less. In both cases, we reduce bouh minuend and subtrahend to the same number of decimal places, by annexing ciphers; or we suppose the ciphers to be annexed, before performing the subtraction. Hence the RuLK. I. Write the numbers so tliat the decimal points shall staiTil directly vnder each other. 11. Subtract as in tcliole numbers, and place the decimal point in the result directly tinder the points in the given numbers. A. Find the {lifFerence between 714 and .910. Ans. 71.3.084. f). How much jireater is 2 than .298? Ans. 1.702. G. From 21.004 fake 75 hundredths. 7. From 10.0302 take 2 ten-thousandths. Ans. 10.03. 153. 1. OPER.\TION. 91.73 2.18 Ans. 89.55 2. From 2.9185 take 1.42. OPERATION. 2.9185 1.42 « Ans. 1.4985 3. From 124.65 take 95.58746 OPEEATION. 124.05 95.58740 Alts. 29.0G254 8. From 900 take .009. Ans. 899.991. 9. From two tii9. The dollar is the itnit of United States money; dimes, cents, and mills are fractions of a dollar, and are sepa- rated from the dollar by the decimal point ; thus, two dollars one dime two cents five mills, are written $2,125. By examining the table, we see that the dime is a tenth part of the unit, or dollar; the cent a tenth part of' the dime or a hundredth part of the dollar; and the villi a tenth part of the cent, a hundredth part of the dime, or a thousandth part of the dollar. Hence the denominations of decimal currency increase and decrease the same as decimal fractions, and are expressed according to the same decimal system of notation ; and they may be added, subtracted, multiplied, and divided in the same manner as decimals. Dimes are not read as dimes, but the two places of dimes and cents are appropriated to cents ; thus, 1 dollar 3 dimes 2 cents, or $1.32, are read one dollar thirty-two cents ; hence, When the number of cents is less than 10, we write a cipher before it in the j)lace of dimes. Note. The half cent is frequently written as 5 mills ; thus, 24 J cents, ■written $.245. 100. Business men frequently write cents as common fractions of a dollar ; thus, three dollars thirteen cents are Avritten $3^^^, and read, three and thirteen hundredths dollars. In business transactions, when the final result of a computation contains 5 mills or more, they are called one cent, and when less than 5, they are rejected. EXAMPLES FOR PRACTICE. 1. Write four dollars five cents. Ans. $-4.05. 2. Write two dollars nine cents. 3. Write ten dollars ton cents. 4. Write eight dollars seven mills. Ans. $8,007. ^Miat i>; the unit of TT. S. rnrroncv ? AMint is the Eronornl law of inrreaso and drcroa^c ? In pvaotiro, how many decimal places are civen to cents? In hnsiness transactions, how are cents frequently written ? "WTiat is done if the mills exceed 5 ? If less than 5 ? REDUCTION. 133 5. "Write sixty-four cents. Ans. $0.G4. G. AVrite three cents two mills. 7. AVrite one hundred dollars one cent one mill. 8. Read $7.93 ; $8.02 ; $G.542. 9. KeadSo.272; $100,025; $17,005. 10. Kead $16,205; $215,081; $1000.011; $4,002. REDUCTION. 116 1 . By examining the table of Decimal Currency, we see thai 10 mills make one cent, and 100 cents, or 1000 mills, make one dollar ; hence, 2o change dollars to cents, multiply ly lOQ ; 'that is, annex 'tivo ciphers, 7o change dollars to mills, annex three ciphers. To change cents to mills, annex one cipher. EXAMPLES FOR PRACTICE. 1. Change $792 to cents. Ans. 79200 cents. 2. Change $36 to cents. 3. Reduce $5248 to cents. 4. In G.25 dollars how many cents? Ans. G25 cents. Note. To chanso dollars and cents to cents, or dollars, cents, and mills to mills, remove the decimal point and the sign, $. 5. Change $63,045 to mills. Ans. 63045 mills. 6. Change 16 cents to mills. 7. Reduce S3.008 to mills. 8. In 89 cents how many mills ? ■ 16^. Conversely, To change cents to dollars, divide hy 100 ; that is, point off two figures from the right. To change mills to dollars, point off three fgures. To change mills to cents, point off' one figure. How are dollars changed to cents ? to mills ? How are cents changed to mills? How are cents changed to dollars ? MiUs to dollars ? to cents? 131 DECIMAL CURRENCY. teXA-MPLES FOR PRACTICE. 1. Change 875 cents to dollars. Ans. $8.75. 2. Change 1504 cents to dollars. 3. In 13875 cents how many dollars? 4. In 16525 mills how many dollars? 5. Reduce 524 mills to cents. 6. Reduce 6524 mills to dollars. ADDITION. 16!S. 1- A man bought a cow for 21 dollars 50 cents, a horse for 1 25 dollars 37J- cents, a harness for 4G dollars 75 cents, and a carriage for 2 1 dollars ; how much did he pay for all ? OPERATION. % 21 50 Analysis. Writing dollars under dol- ,^.' __ lars, cents under cents, &c., so that the ~ '[.- decimal points shall stand under each '* other, we add and point off as in addition 2! 0-00 of decimals. Hence the following Ans. S403.G25 Rule. I. Write dollars under dollars, cents under cents, S)-c. II. Add as in simple numbers, and place the point in the amount as in addition of decimals. examples for practice. 2. What is the sum of 50 dollars 7 cents, 1000 dollars 75 cents, 60 dollars 3 mills, 18 cents 4 mills, 1 dollar 1 cent, and 25 dollars 45 cents 8 mills ? Ans. $1137.475. 3. Add 304 dollars 54 cents 1 mill, 486 dollars 6 cents, 93 dollars 9 mills, 1742 dollars 80 cents, 3 dollars 27 cents G mills. Ans. $2089.686. 1. Add 92 cents, 10 cents 1 mills, 35 cents 7 mills, 18 cents G mills M cents 4 mills, 12J- cents, and 99 cents. Ans. $3,126. Explain the process of addition of decimal currency. Pvule, first step. Second. SUBTRACTION. I'J.j 5. A farmer receives 89 dollars 74 cents for wheat, 13 dol- lars 3 cents lor corn, G dollars 374 cents for potatoes, and 11) dollars G2i cents for oats ; Avliat does he receive for the Avhole ? Jns. $128.77. G. A lady bonght a dress for 9 dollars 17 cents, trimmings for 874 cents, a pa[)er of i)ins for (j^ cents, some tai)e ibr 4 cents, some thread for 8 cents, and a comb i'or 11 cents; what did she pay for all ? • Ans. $] 0.3375. 7. Paid for building a house S2175.75, for painting the same $240.37;i, tor furniture $605.40, for carpets $140.12i^; what was the cost of the house and furnishing ? 8. Bought a ton of coal for S6.08, a barrel of sugar for $26.G25, a box of tea for §1G, and a barrel of flour for $7.40 ; what was I he cost of all? 9. A merchant bought goods to the amonnt of $7425.50 ; he paid for duties on the same $253.9 G, and for freight $170.09 ; what was the entire cost of the good.^ ? 10. I bought a hat for $3.62^, a pair of shoes for $1£. an umbrella for $1|, a pair of gloves for $.G2J, and a cane for $.874 ; what was the cost of all my purchases ? Ans. $8.25. SUBTRACTIOX. IG-l. 1. A man, having $327.50, paid out $1SG.75 for \ horse ; Iww much had he left ? OPERATiox. Analysis. Writing the less number im- $327.50 ^^^' ^^^ greater, dollars under dollars, cents 18G 75 under cents, Src, we subtract and ])oiiit off in the result as in subtraction of decimals. Ans. $140.75 Hence the following IvULE, I. W}-ife the siihtrahend under the minuend, dollars under dollars, cents zmder cents. S)-c. 11. Suhtract as in simple numbers, and place the point in the remainder, as in siibtraction of decimals. Explain the process of subtraction. Give rule, first step. Second. 136 DECIMAL CURRENCY. EXAMPLES FOR PRACTICE. 2. From $365 dollars 5 mills take 2G7 dollars 1 cent 8 mills. A?is. $97,987. 3. From 50 dollars take 50 cents. Ahs. $49.50. 4. From 100 dollars take 1 mill. Ans. $99,999. 5. From 1000 dollars take 3 cents 7 mills. 6. A man bought a farm for $1575.24, and sold it for $1834.16; what did he gain ? Ans. $258.92. 7. Sold a horse for 145 dollars 27 cents, which is 37 dol- lars 69 cents more than he cost me ; what did he cost me ? 8. A merchant bought flour for $5.62^ a barrel, and sold it for $6.84 a barrel ; how much did he gain on a barrel ? 9. A gentleman, having $14725, gave $3560 f(jr a store, and $7015.871 for goods; how much money had he left ? 10. A lady bought a silk dress for $13|, a bonnet for $5^, a pair of gaiters for $1 1, and a fan for $| ; she paid to the shop- keeper a twenty dollar bill and a five dollar bill ; how much change should he return to her? Ans. $3.75. Note. Heduce the fractions of a dollar to cents and mills. 11. A gentleman bought a pair of horses for $480, a har- ness for $80.50, and a carriage for $200 less than he paid for both horses and harness; what was the cost of the carriage? Ans. $360.50. IMULTIPLICATION. J 05. 1. If a barrel of flour cost $6,375, what will 85 barrels cost ? OrERATION. $6,375 Analysis. We multiply as in simple 85 numbers, always regarding the midtiplicr _ as an ahsfrncf number, and point oil' liom the right hand of the result, as in multijjli- .51000 cation of decimals. Hence the following Ans. $541,875 Give analysis for multiplication in decimal currency. DIVISION. 137 Rule. Multiply as in sim'ple numhers, and place the point in the product, as in multiplication of decimals. EXAMPLES FOR PRACTICE. 2. If a cord of wood be worth $4,275, wliat will 300 cords be worth? ^"S- $1282.50. 3. What will 175 barrels of apples cost, at $2.45 per bar- rel ? Ans. $428.75. 4. What will 800 barrels of salt cost, at $1.28 per barrel? 5. A o-rocer bought 372 pounds of cheese at $.15 a pound, 434 pounds of coffee at $.12i a pound, and 16 bushels of pota- toes at $.33 a bushel ; what did the whole cost ? 6. A boy, being sent to purchase groceries, bought 3 pounds of tea at 56 cents a pound, 15 pounds of rice at 7 cents a pound, 27 pounds of sugar at 8 cents a pound ; he gave the grocer 5 dollars ; how ranch change ought he to receive ? 7. A farmer sold 125 bushels of oats at $.374- a bushel, and received in payment 75 pounds of sugar at $.09 a pound, 12 pounds of tea at $.60 a pound, and the remainder in cash; how much cash did he receive ? Ans. $32.92.j-. 8. A man bought 150 acres of land for $3975 ; he after- ward sold 80 acres of it at $32.50 an acre, and the i-emainder at $34.25 an acre ; how much did he gain by the transaction ? Ans. $1022.50. DIVISION. 166. 1. If 125 barrels of flour cost $850, how much will 1 barrel cost ? OPERATION. Analysis. We divide as in 125 ) $850.00 ( $6.80, Ans. f™P'e numbers, and as there rr-n is a remainder after dividing the dollars, we reduce the div- 1000 idend to cents, by annexinj? two 1000 ciphers, and continue the di- vision. Hence the following Rule. Give rule for division m decimal currency. 138 DECIMAL CURRENCY. Rule. Divide as in simple nwnbers, and place the point in t/ie quotient^ as in division of decimals. Notes. 1 . In bu>iuess transactions it is never necessary to carry the division further than to mills in tlie quotient. 2. If the dividend will not contain the divisor an exact nun.ljer of times, ciphers may be annexed, and the division continued as in divis- ion of decunals. In this case it is always safe to reduce the dividend to mdls, or to 3 more decimal places than the divisor contains, be- fore commencing the division. EXAMPLES FOR PRACTICE. 2. If 33 gallons of oil cost $41.25, what is the cost per gal- lon ? Ans. $1.25. 3. If 27 yards of broadcloth cost $94.50, what will 1 yard cost? 4. If G4 gallons of wine cost $136, what will 1 gallon cost? Ans. $2,125. 5. At 12 cents apiece, how many pine-apples can be bouglit for $1.32? Am. 11. G. If 1 ])0und of tea cost 54 cents, how many pounds can be bought for $405 ? 7. If a man earn $180 in a year, how much does he earn a month ? 8. If 100 acres of land cost $2847.50, what will 1 acre cost? Ans. $28,475. 9. What cost 1 pound of beef, if 894 pounds cost $80.40? Ans. $.09. 10. A farmer sells 120 bushels of wheat at $1,124- a bushel, for wliich he receives 27 barrels of flour; what does the flour cost him a barrel ? 11. A man bought 4 yards of clotli at $3.20 a yard, and 37 pounds of sugar at $.08 a pound ; he paid $G.80 in cash, and ihe remainder in butter at $.1G a pound ; how many pounds of Iiiittcr did it take? Ans. 56 ]iouiids. 12. A man l)ouglit an eqtial number of calves and sheep, paving $166.75 for them ; for the calves he ]>aid $4.50 ;i licad, and for the sheep $2.75 a head ; how many did he buy of cacli kind? Ans. 23. APPLICATIONS. 139 13. If 154 pounds of sugar cost $18.48, what will 1 pound cost ? 14. A merchant bought 14 boxes of tea for $500; it being damaged he was obliged to lose $100.75 on the cost of it; how much did he receive a box ? Ans. $32.37 i-. Additional Applications. CASE I. S67. To find the cost of any number or quantit}', When the price of a unit is an aliquot part of one dollar. 168. An Aliquot Part of a number is such a part as will exactly divide that number ; thus, 3, 5, and 7^ are aliquot parts of 15. Note. An aliquot part may be a "whole or a niLxcd number, while a factor niu^t be a whole number. \ ' ALIQUOT PARTS OF ONE DOLLAR. 50 cents = ^ of 1 dollar. | 12^ cents irr ^ of 1 dollar. 33^ cents = ^ of 1 dollar. 25 cents := ]- of 1 dollar. 4 20 cents = i of 1 dollar. IGg cents = ^ of 1 dollar. 10 cents z=z jify of 1 dollar. 81^ cents =: jL- of 1 dollar. 0^ cents z=z -j'g of 1 dollar. 5 cents rr ^'o of 1 dollar. 1. What will be the cost of 3784 yards of flannel, at 25 cents a yard ? OPERATION. Analysis. If the price were $1 a yard, 4 ') 378 t ^^^ ^^^'^ Mould be as many dollars as there arc yards. But since the price is -I- of a dollar a Ans. $J40 yard, the whole cost will be \ as many dollars as there are yards ; or, 1 of 3784 := 3784 -^ 4 =: $946. Hence the Rule. Take such a fractional fart of the given mauler as (he price is part of one dollar. EXAMPLES FOR PRA'CTICE. 2. What cost 903 bushels of oats, at 33^ cents per bushel? Ans. $321. Case I is Avhat ? AMiat is an aliquot part of a dollar ? Give ex- planation. Rule, 140 DECIMAL CURRENCY. 3. AVhat cost 478 yards of delaine, at 50 cents per yard ? 4. AVliat cost 42GG yards of" sheeting, at 8^ cents a yard? Ans. $355. oO. 5. What cost 1250 bushels of apples, at 12^ cents per bushel? Ans. $156.25. 6. What cost 3126 spools of thread, at 6| cents per spool? Ans. $195,375. 7. At IGg cents per dozen, what cost 1935 dozen of eggs? Ans. 322.50. 8. What cost 56480 yards of calico, at 12^ per yard ? 9. At 20 cents each what will be the cost of 1275 salt barrels? -^ns. $255. CASE II. 169. The price of one and the quantity being given, to find the cost. 1. How much will 9 barrels of flour cost, at $6.25 per barrel ? OPERATION. Analysis. Since one barrel cost $6.25,9 $6.25 barrels will cost 9 times $6.25, and $6.25 X q 9 := $56.25. Hence j Ans. $56.25 Rule. 31idtipli/ the jnice of one by the quantity. EXAMPLES FOR PRACTICE. 2. If a pound of beef cost 9 cents, what will 8G4 pounds cost? Ans. $77. 7(). 3. What cost 87 acres of govornmont land, at $1.25 per acre ? 4. What cost 400 barrels of salt, at $1.45 per barrel ? Ans. $580. 5 Wliat cost TG chests of tea, each chest containing 52 pounds, at 44 cents per pound? Case II is ■\vh;it ? Give explanation. Rule. APPLICATIONS. 141 CASE III. 170 . The cost and the quantity being given, to find the price of one. 1. If 30 bushels of corn cost $20.70, what will 1 bushel cost ? OPERATION. Analysis. If 30 bushels cost $20.70, 1 310 ) S'^IO 70 bushel -will cost ^^ of $20.70; and $20.70-^ <—^- 30;= $.69. Hence, $.69 Rule. Divide the cost by the quantity. EXAMPLES FOR PRACTICE. 2. If 25 acres of land cost $175, what will 1 acre cost? 3. If 48 yards of broadcloth cost $200, what will 1 yard cost? Ans. $4.1 6§. 4. If 96 tons of hay cost $1200, what will 1 ton cost? 5. \'i 10 Unabridged Dictionaries cost $56.25, what will 1 cost ? Ans. $5.62J.. 6. Bought 18 pounds of tea for $11.70; what was the price per pound ? Ans. $.65. 7. If 53 pounds of butter cost $10.07, what will 1 pound cost ? 8. A merchant bought 800 barrels of salt for $1016 ; what did it cost him per barrel ? 9. If 343 sheep cost $874.65, what will 1 sheep cost ? Ans. '$'2.00. 10. If board for a family be $684.37^ for 1 year, how much is it per day? Ans. $1.87^. CASE IV. 171. The price of one and the cost of a quantity being given, to find the quantity. 1. At $6 a barrel for flour, how many barrels can be bought for $840 ? Case III is what ? Give explanation. Rule. Case IV is what f 142 DECIMAL CURRENCY. OPERATION. Analysis. Since .*6 Avill buy 1 barrel Q \ g4() of flour, $840 Avill buy J as many barrels as there are dollars, or as manv barrels as Ans. 140 barrels. $6 is contained times in 8840 ; 840 -^ 6 rr 140 barrels. Hence, Rule. Divide the cost of the quantity by the price of one. EXAMPLES FOU PRACTICE. 2. How many dozen of eggs can be bought for $.5.55, if one dozen cost $.15 ? Ans. 37 dozen. 3. At $12 a ton, how many tons of hay can be bought for $216? Ans. 18 tons. 4. How many bushels of wheat can be bought for $2178.75, if 1 bu.'occed to the highest denomination required. The last quotient, ivith the several remainders annexed in a reversed order, will be the answer. Note. Eoduction descending and reduction ascending mutually prove each oilier. EXAMPLES FOU PIIACTICE. 1. Ill 14194 farthings how many pounds? 2. In 14 £ 15 s. 8 d. 2 far. how many farthings ? 3. In 15359 farthings how many pounds? 4. In 46 Rov. 12 s. 2 d. how many pence ? 6. In 11186 pence how many sovereigns? COMrOUNl) NUMBERS. 168 "WT^IGIITS. fiS4. V/eight is a, measure of the quantity of matter a boily contains, determined according to some fixed standard. Tliree scales of weight are used in the United States nnd Great Britain, namely, Troy, Apothecaries', and Avoir- dupois. I. Tkoy Weight. I8»>. Troy Weight is used in weighing gold, silver, and jewels ; in philosophical experiments, &c. T\BLE. 24 grains (gr.) make 1 pennyweight,, .pwt. or dwt. 20 jjL'iinyweights " 1 ounce, oz. 12 ounces " 1 pound, lb. UNIT EQUIVALENTS. invt. pr. 1 = 24 lb. 1 = 20 = 480 1 = 12 = 240 = 5760 Scale — ascending, 21, 20, 12; descending, 12, 20, 24. EXAMPLES FOR TRACTICE. 2. How many pounds in 85894 grains ? OPEEATION. 24 ) 85894 gr. 1. How many grains in 10 oz. 18 pwt. 22 gr.? 141b. operatic:^. 141b. 10 oz. 18 pwt. 12 22 gr. 178 oz. 20 3578 pwt. 24 14334 715G 20 ) 3578 pwt. 4- 22 gr. 12) 178 oz. 4- 18 pwt. 141b. + 10oz. Ans. 141b. 10 oz. 18 pwt. 22 gr. 85894 gr., Ans. 3. In 5 11). 7 oz. 12 pwt. 9 gr., how many grains? 4. In 32457 grains how many pounds ? Define weight. Troy weight. Rejoeat the table. Give the scale, 7* 154 EEDUCTION. 5. Reduce 41 7 GO grains to pounds. Ans. 7 lb. 3 oz. G. A miner had 14 lb. 10 oz. 18 pwt. of gold dust; how much was it worth at $.75 a pwt. ? Ans. $2683.50. 7. How many spoons, each weighing 2 oz. 15 pwt., can be made from 5 lb. G oz. of silver ? Ans. 24. 8. A goldsmith manufactured 1 lb. 1 pwt. IG grs. of gold into rings, each weigliing 4 pwt. 20 gr. ; he sold the rings for $1.25 apiece ; how much did he receive for them ? Ans. $62.50. II. Apothecaries' Weight. 186. Apothecaries' WeigM is used by apothecaries and physicians in compounding medicines ; but medicines are bought and sold by avoirdupois weight. TABLE. 20 grains (gr.) make 1 scruple, sc. or 9- 3 scruples 8 drams 12 ounces 1 dram, dr. or 3 ■ 1 ounce, oz. or § . 1 pound, lb. or tb. lb. UKIT EaUIVALENTS. SC. gr. dr. 1 = 20 1 = 3 — 60 1 = 8 = 24 = 480 1 := 12 — 96 = 288 z= 5760 Scale — ascending, 20, 3, 8, 12; descending, 12, 8, 3, 20. EXAMPLES FOR PRACTICE. 1. IIow many gr. in 12 lb 8§ 33 1 9 15 gr.? OPEIIATION. 12 ft 85 3 3 19 I5gr. 12 152 § 8 1219 5 3 3658 3 20 gr 2. How many lb in 73175 OPERATION. 2|0 ) 7317|5 gr. 3) 3G58 9-f 15gr. 8 ) 1219 5 + 1 9 12) 152g-{-3 3 12tb-f-8S Ans. 121b 8S 3 3 19 I5gr. 73175 gr., Ans. Define apothecaries' weight. Repeat tlie table. Give the scale. COMPOUND NUMBERS. 155 3. In IG lb. 11 oz. 7 dr. 2 sc. 19 gr., how many grains? 4. Reduce 47 ft 6 § 4 3 to scruples. A)is. 13092 sc. 5. How many pounds of medicine would a physician use in one year, or 365 days, if he averaged daily 5 prescriptions of 20 grains each ? Ans. G ft. 4 § 1 9. III. Avoirdupois Weight. 1S7. Avoirdupois Weight is used for all the ordinary pur- poses of weighing. TABLE. 16 drams (dr.) make 1 ounce, oz. 16 ounces " 1 pound, lb. 100 lb. " 1 hundred weight, . cwt. 20 cwt., or 2000 lbs., " 1 ton, T. UNIT ZaUIVALENTS. OZ. dr. lb. 1 = 16 cwt. 1 =: 16 =r 2j6 T. 1 rr: 100 = 1600 — 25600 1 =: 20 = 2000 = 32000 = 512000 Scale— ascending, 16, 16, 100, 20 ; descending, 20, 100, 16, 16. Note. The Imig or grross ton, hundred weight, and quarter were formerly in common vise ; hut they are now seldom used except in estimating English goods at the U. S. custom-houses, and in freighting and wholesaling coal from the Pennsylvania muxes. LONG TON TABLE. 28 lb. make 1 quarter, marked qr. 4 qr. = 112 lb. " 1 hundred weight, " cwt. 20 cwt. = 2240 lb. " 1 ton, " T. Scale — ascending, 28, 4, 20; descending, 20, 4, 28. The following denominations are also in use. • 56 pounds make 1 firkin of butter. 100 " " 1 quintal of dried- salt fish. 100 " " 1 cask of raisins. 196 " " 1 barrel of flour. 200 " " 1 " " beef, pork, or fish. 280 " " 1 " " salt at the N. Y. State salt works. 56 " " 1 bushel " " " " " " " 32 " " 1 " " oats. 48 « " 1 '< « barley. 56 " " 1 " *' corn or rye. 60 " " 1 « " wheat. Define avoirdupois weight. Repeat the table. Give the scale. The long ton table. What other denominations are in use J What is the value of each ? 156 REDUCTION. EXAMPLES FOR PRACTICE. 1. In 25 T. 15 cwt. 70 lb. how many pounds ? OPERATION. 25 T. 15 cwt. 70 lb. 20 515 cwt. 100 2. In 51570 pounds how many tons ? OPERATION. 100) 515701b. 210)5115 cwt. + 70 lb. 25T.-[-15cwt. Ans. 25 T. 15 cwt. 70 lb. 515701b., Ans. 3. Keduee 3 T. 14 cwt. 74 lb. 12 oz. 15 dr. to drams. 4. Reduce 1913551 drams to tons. 5. A tobacconist bought 3 T. 15 cwt. 20 lb. of tobacco, at 22 cents a pound; liow much did it cost him ? Ans. $1G54.40. 6. How much will 115 pounds of hay cost, at $10 per ton ? 7. A grocer bought 10 barrels of sugar, each weighing 2 cwt. 17 1b., at 6 cents a pound; 5 barrels, each weighing 3 cwt. G lb., at 7^ cents a pound ; he sold the whole at an average price of 8 cents a pound ; how much was his whole gain? J^ns. $51.05. 8. Paid $360 for 2 tons of cheese, and retailed it for 12^ cents a pound ; how much was my whole gain ? Ajis. $140. 9. If a person buy 10 T. 6 cwt. 3 qr. 14 lb. of English iron, by the long ton weight, at 6 cents a pound, and sell the same at $130 per short ton, how much will he gain ? Ans. $115.85. 10. A farmer sold 2 loads of corn, weighing 2352 lbs. each, at $.90 per bu. ; what did he receive ? A7is. $75.60. 1 1. How many pounds in 300 barrels of flour ? 12. A grocer bought 3 barrels of salt at $1.25 per barrel, and retailed it at £ of a cent per pound ? what did he gain ? Ans. $2.55, STANDARD OF WEIGnT. 188. In the year 1834 the U. S. government adopted a uniform standard of weights and measures, for the use of the custom houses, and the otherbranches of business connected with the general government. Most of the States which have adopt- ed any standards have taken those of the genci'al government. COMPOUND NUMBERS. 157 IS??. Tlie United States standard unit of weight is tlie Troy pound of the mint, wliich is tlie ?ame as the imperial standard pound of Great Britain, and is determined as fol- lows : A cuhic inch of distilled water in a vacuum, weighed by brass weights, also in a vacuum, at a temperature of 62° Fahrenheit's thermometer, is equal to 252.458 grains, of v/hich the standard Troy pound contains 57G0. I!I4>. I'he U. S. Avoirdupois pound is determined from tlie standard Troy pound, and contains 7000 Troy grains. Hence, the Troy pound is ^l%% = -fft ^^ ^^ avoirdupois pound. But the Troy ounce contains -ff- == 480 grains, and the avoirdupois ounce ^f §^ = 437.5 grains ; and an ounce Troy is 480 — 437.5 =1 42.5 grains greater than an ounce avoirdu- po'S. The pound, ounce, and grain, Apothecaries' weight, are the same as the like denominations in Troy weight, the only diflerence in the two tables being in the divisions of the ounce. 11^1. COMPAKATIVE TABLE OF WEIGHTS. Troy. Apothecaries'. Avoirdupois. 1 pound z= 5760 grains, z=z 5760 grains, rr: 7000 grains. 1 ounce rr: 480 " r= 480 " z= 437.5 " 175 i^ounds, zir 175 pounds, z=. 144 pounds. EXAMPLES FOB PRACTICE. 1. An apothecary bought 5 lb. 10 oz. of rhubarb, by avoirdupois weight, at 50 cents an ounce, and retailed it at 12 cents a di'am apothecaries' weight ; how much did he gain ? Ans. $33.75. 2. Change 424 drams apothecaries' weight to Troy weight. Ans. 4 lb. 5 oz. 3. Change 20 lb. 8 oz. 12 pwt. Troy weight to avoirdu- pois weight. Ans. H^j-b ^'^* 4. Bought by avoirdupois weight 20 lb. of opium, at 40 cents an ounce, and sold the same by Troy Aveight at 50 cents an ounce; how much was gained or lost? Ans. $17.83J. ^^^lat is the TJ. S. standard of \veio;ht ? How obtained ? How is the avoirdupois pound determined ? How is the apothecaries' pound determined ? What are the values of the denominations of Troy, avoii'- dupois, and apothecaries' weight f 158 REDUCTION. MEASURES OF EXTENSION. fi3*2. Extension has three dimensions — length, breadth, and thickness. A Line has only one dimension — length. A Surface or Area has two dimensions — length and breadth. A Solid or Body has three dimensions — length, breadth, and thickness. I. Long Measure. 193. Long Measure, also called Linear Measure, is used iu measuring lines or distances. TABLE. 12 inches (in.) make 1 foot, ft. 3 feet " 1 yard, yd. 5^ yd., or 16^ ft., " 1 rod rd. 40 rods " 1 furlong, fur. 8 furlongs, or 320 rd., " 1 statute mile,, .mi. UNIT EQUIVALENTS. ft. in. yd. 1 =: 12 rd. "l = 3 = 36 f„r. 1 = 5i rr 16^ = 198 „,i 1 zr: 40 r= 220 = 6G0 =: 7920 1 =z 8 =: 320 = 1760 zr: 5280 = 63360 Scale — ascending, 12, 3, 5|, 40, 8 ; descending, 8, 40, o^, 3, 12. The following denominations are also in use: — 3 barleycorns make 1 i^eh, ? "'^^ ^^f^7™^V'V" "'^'"""° •' ' } the length of the foot. 4 inches « 1 hand ^ "^^'^ ^" measuring the height of ' } horses directly over the fore feet. 6 feet " 1 fathom, used in measuring dc])ths at sea. , 1 1 j; <,f^f„f„ .v,;i„ u 1 !-• •! S nsed in measuring dis- 1.15 statute mues " 1 geographic mile, < , ^ ° '^ o I ' ^ tances at sea. 3 geographic " " 1 league. 60 " " " ^ 1 d o- • . 5 "^ latitude on a meridian or of 69.16 statute " " ^ ^ oegree j ]^^„^^^^q p,^ ^j^g equator. u 60 degrees " the circumference of the earth. How mnnv flimnnsioiis has extension ? Define a line. Surface or Rvea. A solid or body. Define lonj]; measure. \Miat arc the denom- inations ? The value of eaoh. "What other denominations arc used ? COMPOUND NUMBERS. 159 Notes. 1. For the purpose of measuring cloth and other goods sold by the yard, the yard is divided into halves, fourths, eighths, and fcLx.- teenths. The old table of cloth measure is practically obsolete. 2. The geographic mile is ^^g- of-j^g or -jieg-g of the distance round the center of the earth. It is a small fraction more than 1.15 statute miles. 3. The length of a degree of latitude varies, being 68.72 miles at the equator, 68.9 to 69.05 miles in middle latitudes, and 69.30 to 69.31 miles in the polar regions. The mean or average length is as stated in the table. A degree of longitude is greatest at the equator, where it is 69.16 miles, and it gradually decreases toward the poles, where it is 0. EXAMPLES FOR PRACTICE. 1. In 2 mi. 4 fur. 32 rd. 2 yd. how many inches ? OPERATION. 2 mi. 4 fur. 32 rd. 2 yd. 8 20 fur. 40 832 rd. 5J- 416 4162 4578 yd. 3 13734 ft. 12 2. In 164808 inches Low many miles ? OPERATION. 12) 164808 in. 3) 13734 ft. 54 ] 4578 yd. 2" J' 2 11 )9156 4| 0)83|2 rd. + 1- yd. = 2 yd. 8 ) 20 fur. -f 32 rd. 2 mi. -|- 4 fur. Ans. 2 mi. 4 fur. 32 rd. 2 yd. 164808 in., J«s. 3. The diameter of the earth being 7912 miles, how many inches is it? Ans. 501304320 inches. 4. In 168474 feet how many miles ? 5. In 31 mi. 7 fur. 10 rd. 3 yd., how many feet ? 6. If the greatest depth of the Atlantic telegraphic cable from Newfoundland to Ireland be 2500 fathoms, how many miles is it ? Ans. 2 mi. 6 fur. 29 rd. 1^^ ft. 160 . REDUCTION. 7. If this cable be 2200 miles in length, and co.-t 10 cents a foot, what -was its whole cost? Ans. SllGlGOO. 8. A pond of water measures 4 fathoms 3 feet 8 inclie? in depth ; how many inches deep is it ? Ans. 232. 9. How many times will the driving wheels of a locomo- tive turn round in going from Albany to Boston, a distance of 200 miles, supposing the wheels to be 18 ft. 4 inches in cir- cumference? Ans. 57C00 times. 10. If a vessel sail 120 leagues in a day, how many stat- ute miles does she sail ? Ans. 414. 11. How many inches high is a horse that measures 14^ hands? Ans. 58. surveyors' long measure. 134. A Gunter's Chain, used by land surveyors, is 4 rods or 6G feet long, and consists of 100 links. TABLE. 7.92 inches ' (in.) make 1 link, 1. 25 links " 1 rod, rd. 4 rods, or 66 feet, " 1 chain . .ch. : 80 chains " 1 mile, . . mi. UNIT EQUIVALEXTS. 1. iP. H. 1 = 7.92 ,,, 1 r= 25 = 198 . i"— 4 = 100 z= 792 1*= 80 = 320 = 8000 =r 63360 Scale— ascending, 7.92, 25, 4, 80 ; descending, 80, 4, 25, 7.92. Note. Rods are seldom used in chain measure, distances being taken in chains and links. , EXAMPLES FOR PRACTICE. W^ 1. In 3 mi. 51 ch. 73 1. how many links? » 2. Reduce 29173 I to miles. 3. A certain field, enclosed by a board fence, is 17 ch. 311. long, and 12 ch. 87 1. wide ; how many feet long is the fence whicli encloses it? ^ns. 3983.7G ft. Repeat the table of surveyors' long measure. Give the scale. COMPOUND NUMBERS. 1.61 II. Square Measure. S9*9. A Square is a figure having four equal sides, and four equal angles or corners. 1 square foot is a figure having four sides of 1 ft. or 12 in. each, as shown in the diagram. Its contents are 12 X 12 1= 144 square inches. Hence The contents or area of a square, or of any other figure having a uniform length and a uniform breadth, is found by multiplying the length by the breadth. Thus, a square- foot is 12 in. long and 12 in. wide, and the con- tents are 12 X 12 = 144 square inches. A board 20 in. long and 10 in. wide, is a rectangle, containing 20 X 10 = 2U0 square inches. 1J>6 . Square Measure is used in computing areas or sur- faces ; as of land, boards, painting, plastering, paving, &c. TABLE. 1 1 n = = 1 ft 12 iu. :1 ft. 144 square inches (sq. iu.) make I square foot, marked sq. ft. 9 square feet " 1 square yard, " sq. yd. 30-^ square yards " 1 square rod, " sq. rd. 40 square rods . " 1 rood, " E,. 4 roods " 1 acre, " A. 640 acres " 1 square mile, " sq. mi. UNIT EQUIVALENTS R. 1: 4: A. 80. mi. 1 ;^ 1 = 640 = 2o60 = 102400 = 3007600 Bq.rd. 40 r= 160 = pq. yd. 1 : 30^-: 1210: 4S40: sq. ft. 1 9 2721 : 10S90 : 43r)00 : : 27878400 : sq. in. 144 1296 39204 1568160 6272640 4014489600 Scale — ascending, 144, 9, 301 40, 4, 640; descending, 640, 4, 40, 301 9, 144. Define a square. How is the area of a square or any rectanprular fiijnre found ? For what is square measure used ? Repeat the table. Give the scale. 162 REDUCTION. Artificers estimate their work as follows : By the square foot : glazing and stone-cutting. By the square yard : painting, plastering, paving, ceiling, and paper-hanging. By the square of 100 feet: flooring, partitioning, roofing, slating, and tiling. Brick-laying is estimated by the thousand bricks; also by the square yard, and the square of 100 feet. Notes. 1. In estimating the painting of moldings, cornices, &c., the measuring-line is carried into all the moldings and cornices. 2. In estimatmg brick-laying by the square yard or the square oi 100 feet, the work is understood to be 11 bricks, or 12 inches, thick. EXAJIPLES FOR PRACTICE. 1. In 10 A. 1 R. 25 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in. now many square inches ? OPERATIOX. 10 A. 1 R. 25 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in. 4 41 R. 40 1665 sq. rd. 301 416^ 49966 503821 sq. yd. 9 4534441 sq.ft. ■! 144 ^ 36 = I sq. ft. £ 1813912 with 136 sq. in. '^ 1813776 453444 65296108 sq. in., Ans. 2. In 65296108 sq. in. how many acres ? IIow do artisans estimate work ? COMPOUND NUMBERS. 163 OPERATION. 144 ) 6o29Gl()8 sg. in. 9 ) 4 j344o sq. ft. -f 28 sq. in. 30f I 50382 sq. yd. + 7 sq. ft. 4 4 121 ) 201 r>28 fourths sq. yd. 4. 0)166|5 sq. rd. -\-\-- = 15| sq. yd. 4)4111. -[-25 sq.rd. 10 A. -(- 1 R. Ans. 10 A. 1 11. 25 sq. rd. 15f sq. yd. 7 sq. ft. 28 sq. in. f 10 A. 1 II. 25 sq. rd. 15 sq. yd, 7 sq. ft. 28 sq. in. Or j C sq. ft. 108 sq. in. Or 10 A. 1 II. 25 sq. rd. IG sq. yd. x sq. ft. 136 sq. i in. Analysis. Dividing by the numbers in the ascending scale, and arranging the remainders according to their order in a line below, we find the square yards a mi.xed number, 15|. Ijut ^ cf a sq. yd. = I of 9 sq. ft. =: 6| sq. ft. ; and | of a sq. ft. =i | of 144 sq. in. rr 108 sq. in. Therefore | sq. yd. =:r 6 sq. ft. 108 sq. in. ; and adding 108 sq. in. to 28 sq. in. we have 136 sq. in., and G sq. ft. to 7 sq.ft. we have 13 sq. ft. = 1 sq. yd. 4 sq. ft., and writing the 4 sq. ft. in the result, and adding 1 sq. yd. to 15 sq. yd. we have for the reduced result, 10 A. 1 11. 25 sq. rd. 16 sq. yd. 4 sq. ft. 136 sq. in. ?,. Reduce 87 A. 2 R. 38 sq. rd. 7 sq. yd. 1 sq. ft. 100 sq. in. to square inches. ^ns. 55035oOG8 sq. in. 4. Reduce 550355068 square inches to acres. 5. A field 100 rods long and 30 rods wide contains how many acres ? Ans. 18 A. o R. G. How many rods of fence will enclose a faim a mile square? -4?«s. 1280 rods. 7. How much additional fence will divide it into four equal square fields ? ^"S- 640 rods. 8. TTow many acres of land in Boston, at $1 a square foot, will $100000 purchase? Ans. 2 A. 1 R. 7 sq. rd. 9 sq. yd. 3^ sq. ft. 9. How many yards of carpetinjr, 1 yd. wide, will be required to carpet a room 18^ ft. long and 16 ft. wide ? Ans. 32| yd. 164: REDUCTION. 10. "What would be the cost of plastering a room 18 ft. long, 1G4- ft. wide, and 9 ft. high, at 22 cts. a sq. yd.? Ans. $22.44. 11. What will be the expense of slating a roof 40 feet long and each of the two sides 20 feet wide, at $10 per square? ^«s- Si GO. surveyors' square measure. 1397. This measure is used by surveyors in computing the area or contents of land. TABLE. 625 square links (sq. 1.) make 1 pole, P. 1(5 poles " 1 square chain,, .sq. ch. 10 square chains " 1 acre V. 6-iO acres " 1 square mile,. ..sq. ini. 36 square miles (6 miles square) " 1 township,. .. . ...Tp. LNIT r.aUIVALEXTS. r. Fq. 1. sq. ch. 1 ^^^^ "-'^ A 1 = 16 = 1000 gn mi l'=: 10 — 160 = 100000 Tp 1 = 610 =r 6400 = 102400 = 64000000 1 "i= 36 = 23040 — 231)400 = 368G400 = 2304000000 Scale — ascending, 625, 16, 10, 640, 36; descending, 36, 640, 10, 16, 623. Notes. 1. A sqiiare mile of land is also called a section. 2. Canal and railroad engineers commonly use an engineers' chain, ■which consists of 100 links, each 1 foot long. 3. The contents of land are commonly estimated in square miles, acres, and hundredths ; the douommation, rood, is fast going mto dis- use. EXAMPLES FOR PRACTICE. 1. How many poles in a township of land ? 2. Reduce 3686400 P. to sq. mi. 3. In 94 A. 7 sq. ch. 12 P. 118 sq. 1. how many square links ? 4. Wimt will be the cost of a farm containing 4")r)0000 square links, at $<50 per acre ? Ans. $22750. Repeat the table of surveyors' square measure. Give the scale. COMrOUXD XUMBEKS. 165 III. Cubic Measure. ^^=^z—zi y^ C' — ^ -i^d. t9S. A Cube is a polkl, or body, having six equal square sides, or faces. If eacli side of a cube be 1 yard, or 3 feet, 1 foot in thickness of tliis cube Avill contain 3X3X1 v=. 9 cubic feet, and the whole cube will contain 3 X 3 X 3 =r 27 cubic feet. A solid, or body, may have the three dimensions all alike or all different. A body 4 ft. long, 3 ft. wide, and 2 ft. tliick contains 4 X 3 X 2 =: 24 cubic or solid feet. Hence we see that llic cubic or solid contents of a body are found by midtipJy- iny the length, breadth, and thickness together. lf5S>. Cubic Measure, also called Solid Measure, is used in estimating the contents of solids, or bodies ; as timber, wood, stone, &c. TABLE. 1728 cubic inches (cu. in.) make 1 cubic foot,. . " 1 cubic yard,. " 1 cord foot, . . . 27 cubic feet 16 cubic feet 8 cord feet, or 128 cubic feet !et, or ? ;eet, I 24|- cubic feet ..cu. ft. .cu. yd. .cd.*^ft. 1 cord of wood,. . . .Cd. , S perch of stone ? -r> i 1 < ^ > Pch. ; or masonry, ^ Scale — ascending, 1728, 27. Tae other numbers are not in a regular scale, but are merely so many times 1 foot. The unit equivalents, being fractional, are consequently omitted. NoTr,=;. 1. A cubic yard of earth is called a load. 2. Railroad and transijortation companies estimate li^lit freight hj the space it occupies in cubic feet, and hea^^ freight by weit^ht. 3. A pile of wood 8 feet lonp:, 4 feet wide, and 4 feet high, contain'5 1 cord ; and a cord foot is 1 foot in len2;th of such a pile. 4. A perch of stone or of masonry is 16.^ feet long, 1.^ feet ■wide, and 1 foot hiofh. Pefine a cube. How are the contents of a cube or reetan^nlar solid found? For what i?i cubic measure used? Repeat the table. Give the scale. How is railroad freight estimated ? "What is under- titood by a cord foot ? By a perch of stone or masonry i 166 REDUCTION. 5. Joiners, bricklayers, and masons make no allo-vvancc for -windows, doors, &e. liricklayers and masons, in estimating their -work by cubic measure, make no allowance for the corners of the Avails of houses, Cellars, &c., but estimate their woi^k by the girt, that i.-, the entii-e length of the wall on the out.-ide. 6. Engineers, in making estimates for excavations and embankments, take the dimensions with a line or measure divided into fiet and deci- mals of a foot. The estimates are made in feet and decimals, and the results are reduced to cubic yards. EXAMPLES FOB PRACTICE. , 1. Inl25cu.ft. 840cu.in.ho\vmanycu. in. ? ^hs. 216840. 2. Reduce 5224 cubic feet to cords. Ajis. 40||. 3. In a solid, 3 ft. 2 in. long, 2 ft. 2 in. -wide, and 1 ft. 8 in. thick, how many cubic inches? A?is. 19760. 4. How many small cubes, 1 inch on each edge, can be sawed from a cube 6 feet on each edge, allowing no Avaste for sawing? Ans. 373248. 5. In a pile of wood 60 feet long, 20 feet wide, and 15 feet liigh, how many cords ? Ans. 140|, 6. IIow many cubic feet in a load of wood 10 feet long, 3^ feet Avide, and 3 J- i'eet high ? Ans. 113 J cu. ft. 7. If a load of wood be 12 feet long and 3 feet wide, hoAv high must it be to make a cord? Ans. 3| ft. high. 8. The gray limestone of Central New York weighs 175 pounds a cubic foot. What is the weight of one solid yard ? Alls. 2 T. 7 cwt. 25 lb. n. A cellar wall, 32 ft. by 24 ft., is 6 ft. high and U ft. thick. IIow much did it cost at $1.25 a perch? Atis. $50,909+ 10. IIow much did it cost to dig the same cellar, at 15 cents a cubic yard ? Aiis. $2o.60. 1 1 . l\ry sleeping room is 10 ft. long, 9 ft. wide, and 8 ft. high. Tfl brcnthe 10 on. ft. of air in one minute, in how long a time will I breathe as much air as the room contains ? Ajis. 72 min. 12. In a sciiool room 30 ft. long, 20 ft. wide, and 10 ft. high, with 50 persons breathing each 10 cu. ft. of air in one minute, in how long a time will they breathe as much as the room contains? Ans. 12 min. IIow are excavations and embankments measured ? COMPOUND NUMBERS. 167 MEASURES OF CAPACITY. I. Liquid Measure. SOO. liquid Measure, also called Wine Measure, is u-^ed in measuring liquids ; as liquors, molasses, water, &c. TABLE. 4 gills (gi.) make 1 pint, pt. 2 pints " 1 q^i'iit, qt. 4 quarts " 1 gallo" g^l. 3U gallons " 1 barrel, bbl. 2 barrels, or 63 gal. " 1 hogshead,, .hhd. XTNIT EQUIVALENTS. pt. . pi. qt. 1=4 ,,1. 1 = 2 =r 8 bW. 1 = 4 = 8 = 32 ],hd. 1 = 3U =: 126 = 252 = 1008 1 r= 2 =: 63 =r 252 zrr 504 =r 2016 Scale — ascending, 4, 2, 4, 31^, 2; descending, 2, 311 4, 2,4. The following denominations are also in use : o 36 gallons make 1 barrel of beer. 54 " or li barrels " 1 hogshead " " 42 " " 1 tierce. 2 hogsheads, or 120 gallons, " 1 pipe or butt. 2 pipes or 4 hogsheads, " 1 tun. Notes. 1. The denominations, barrel and hogshead, are used in es- timating the capacity of cisterns, reservoirs, vats, &c. 2. The tierce, hogshead, pipe, butt, and tim are the names of casks, and do not express any fixed or definite measures. They are usually ganged, and have their capacities in gallons marked on them. 3. Ale or beer measure, formerly used in measuring beer, ale, and milk, is almost entirely discarded. What is Uquid measure ? Repeat the table. Give the scale. "WTiat other denominations are sometimes used ? How are the capacities of cisterns, reservoirs, &c., reckoned ? Of large casks ? 163 REDUCTION. EXAMPLES FOR PRACTICE. 1. In 2 hhd. 1 bar. 30 gal. 2 qt. 1 jjt. 3 gi. how many gills ? OPERATION. 2 hhd. 1 bar. 30 gal. 2 qt. J_ [Ipt. 3gi. obbl. 31J- 185 1871 gal. 4 752 qt. 2 1505 pt. 4 2. In G023 gi. how many hhds. ? OPERATION. 4 ) 6023 gi. 2 ) 1505 pt. + 3 gi. 4 )752 qt. + 1 pt. 311 188 gal. 2 J 2 63 )376 [gal. 2)_5bbl.-f--V-gal.:=30^ 2 hhd. -f 1 bar. Jns. 2 hhd. 1 bar. 30i gal. 1 pt. 3 gi. But ^ gal. =r 2 qt., making the A?is. 2 hhd. 1 bar. 30 gal. 2 qt. 1 pt. 3 gi. 6023 gi., A)is. 3. Reduce 3 hogsheads to gills. 4. Reduce 6048 gills to hogsheads. 5. In 13 hhd. 15 gal. 1 qt. how many pints? 6. In 6674 pints how many hogsheads ? 7. What will be the cost of a hogshead of wine, at 6 cents a gill? Ans. $120.96. 8. A grocer bought 1 barrels of cider, at $2 a barrel ; after converting it into Ainegar, he retailed it all at 5 cents a quart ; how much was his whole gain ? Ans. $43. 9. At 6 cents a pint, how much molasses can be bought for $3.84? Ans. 8 gal 10. How many demijohn?, that will contain 2 gal. 2 qt. 1 pt. each, can be filled from a hogshead of wine ? Ans. 24. ir. Dry INIeasure. 20 Bo Dry Measure is used in measuring articles not liquid, as grain, fruit, salt, roots, ashes, &;c. "NMiat is dry measure ? COMPOUND NUMBERS. 169 TABLE. 2 pints (pt.) make 1 quart, qt. 8 quarts " 1 peck, pk. 4 pecks " 1 bushel, . bu. or bush. UNIT EQUIVALENTS, qt. pt. pk. 1=2 bu. 1 = 8 = 16 1 r= 4 =z 32 zr: 64 Sc.VLE — ascending, 2, 8, 4 ; descending, 4, 8, 2. Note. In England, 8 bu. of 70 lbs. each are called a quarter, used in mcasuiing grain. The weight of the English quarter is 1 of a long ton. EXAMPLES FOR PRACTICE. 1. In 49 bu. 3 pk. 7 qt. 1 pt. how many pints? 2. In 3199 pt. how many bu.shels ? 3. Ileduce 1 bu. 1 pk. 1 qt. 1 pt. to pints. 4. Reduce 83 pints to busheLs. 5. An innkeeper bought a load of 50 bushels of oats at 65 cents a bushel, and retailed them at 25 cents a peck ; how much did he make on the load ? Ans. $17.50. STANDARD OF EXTENSION. S©?5. The U. S. standard unit of measures of extension^ whether linear, superficial, or solid, is the yard of 3 feet, or 36 inches, and is the same as the imperial standard yard of Great Britain. It is determined as follows : The rod of a pendulum vibrating seconds of mean time, in the latitude of London, in a vacuum, at the level of the sea, is divided into 391393 equal parts, and 360000 of these parts are 36 inches, or 1 standard yard. Hence, such a pendulum rod is 39.1393 inches long, and the standard yard is ff^jf^^ of the length of the pendulum rod. 203. The U. S. standard unit of liquid measure is the old English wine gallon, of 231 cubic inches, which is equal to 8.83888 pounds avoirdupois of distilled water at its maximum density, that is, at the temperature of 39.83° Fahrenheit, the barometer at 30 inches. Repeat the table. "What is a quarter ? What is the U. S. standard unit of measurement of extension ? How is it determined f What is the U. S. standard unit of liquid measure ? R.P. 8 170 REDUCTION. ^O^t. TJie U. S. standard unit of dry measured the Brit- ish ^ViIlche.ster bushel, wliich is 184- inelies in diameter and 8 inches deep, and contains 2150.42 cubic inches, equal to 77.6274 pounds avoirdupois of distilled water, at its maximum density. A gallon, dry measure, contains 2G8.8 cubic inches. Note. 1. The wine and drj' measures of the same denomination are of different capacities. The exact and the relative size of each may be readily seen by the following 3|0»5. COMPARATIVE TABLE OP MEASURES OF CAPACITY. Cn. in. in Cu. in. in Cu. in. in Cu. in. in cue i;all(jn. oue quart. one pint. one gill. Wine measm-e, 231 57| 28| TgV Dry measiu-e, (\ pk.,) 26S| Q>~i\ 33| 8f " 2. The beer gallon of 282 inches is retained in use only by custom A bushel is commonly estimated at 21504 cubic inches. EXAMPLES FOR PHACTICE. 1. A fruit dealer bought a bushel of strawberries, dry measure, and sold them by wine measure ; how many quarts did he gain ? Ans. o^f quarts. 2. A grocer bought 40 quarts of milk by beer measure, and sold it by wine measure ; how many quarts did he gain ? Aris. 8f f quarts. 3. A busheh or 32 quarts, dry measure, contains how many more cubic inches than 32 quarts wine measure ? Ans. 302f cu. in. Tlme. S06. Time is used in measuring periods of duration, as years, days, minutes, &c. T.\r.i,E. 60 seconds (sec.) make 1 minute, min. 60 minutes " 1 hour h. 21 hours " 1 day, da. 7 days " 1 week, wk. 365 days " 1 common year,. . .yr. 366 days " 1 leap year, yr. 12 calendar months " 1 year, yr. 100 years " 1 century C. "What is the U, S. standard unit of dry measure ? How is it ob- tained ? "What is the relative sizo of the wine and the dry p:nllon ? Wliat is the .size of a beer gallon ? What is time ? Ilcpcat the tabic. COMrOUND NUMBERS. 171 UNIT EQUIVALENTS. mm. sec. 1,. 1 = 60 d.. 1 = 60 := 3600 wk. 1 = 24 = 1440 = SG'IOO 1 = 7 = 168 =: 10080 — 604800 y mo. ^ 365 z= 87GO z= 525600 = 31536000 1 — 12 — ) 366 — 8784 — 527040 — 31622400 Scale — ascending, 60, 60, 24, 7 ; descending, 7, 24, 60, 60. The calendar year Is divided as follows : — Names. January, February, ' March, April, . May, ' June, ! July, ' August, r September, ? October, I November, December, No. of nio. Sp;iPon, 1 Winter, 2 it 3 Spring, 4 (. 5 a 6 Summer, 7 (t 8 « 9 Autumn, 10 « 11 It 12 Winter, AbbreTiations. Jan. Feb. Mar. Apr. Jun. Sept. Oct. Nov. Dec. No. of days. 31 28 or 29 31 30 31 30 31 31 30 31 30 31 365 or 366 Notes. 1. The exact length of a solar year is 365 da. 5 h. 48 min. 46 Bee. ; hut for convenience it is reckoned II min. 14 sec. more than this, or 3G5 da. 6 h. = 30.5^ da. This \ day in 4 years makes one day, which, every fourth, bissextile, or leap year, is added to the shortest month, givins; it 29 days. The leap y^ars are exactly divisible by 4, as 1856, 1860, 1864. The number of days in each calendar month may be caisily remembered by committing the followmg lines : — " Thirty flays hath Si'iitfinher, April, Junp. ai)il November; AJl the rt'st liavp tliirty-one, Pave February, which alono Hath twenty-eiglit ; and one day more AVe add to it one year in four." 2. In most business transactions 30 days are called 1 month. EXAjrPLES FOR PRACTICE. 1. Reduce 365 da. 5 h. 48 min. 46 sec. to seconds. 2. Reduce 315.36926 seconds to days. Give the scale. "What is the len<;tli of each of the calendar months ? What is the exact length of a solar year ? Explain the use of bissextile •r leap year. What is the length of a month in busiiiess transactions? 172 KEDUCTION. 3. In 5 wk. 1 da. 1 li. 1 min. 1 sec. how many seconds ? 4. In 31140G1 seconds how many weeks? 5. How many times does a clock pendulum, 3 ft. 3 in. long, beating seconds, vibrate in one day? Ans. 86400. 6. If a man take 1 step a yard long in a second, in how long a time will he walk 10 miles ? Ans. 4 h. 53 min. 20 sec. 7. In a lunar month of 29 da. 12 h. 44 min. 3 sec. how many seconds? Ans. 2551443. 8. How much time will a person gain in 40 years, by rising 45 minutes earlier eveiy day ? Ans. 456 da. 13 h. 30 min. Circular Measure. S®7. Circular Measure, or Circular Motion, is used prin- cipally in surveying, navigation, astronomy, and geography, for reckoning latitude and longitude, determining locations of places and vessels, and computing difference of time. Every circle, great or small, is divisible into the same num- ber of equal parts, as quarters, called quadrants, twelfths, called signs, 360lhs, called degrees, &c. 'Consequently the parts of different circles, although having the same names, are of different lenjrths. c TABLE. 60 seconds (") make 1 minute, . . . '. 60 minutes " 1 decree, . . . ^. 30 doi^rees " 1 Kia:n S. 12 signs, or 3G0°, " 1 circle, C. UNIT EQUIVALENTS. / /' 1 = 60 P. 1 r= 60 r= . 3600 r. 1 m 30 — i.soo := insnoo 1 = 12 = 360 — 21600 =: 1296000 Scale — ascending, 60, 60, 30, 12; descending, 12, 30, 60, 60. Notes. 1. Minutes of the earth's circumference are called geo- graphic or nautical mile?:. 2. The denomination, sir/us, is confined exclusively to Astronomy. Define circular measure. How are circles rlividod ? llcpcat the table. Give the scale. "What is a geographic mile ? "What i* S i| sign ? COMPOUND NUMBERS. • ITS 3. Degrees are not strictly divisions of a circle, but of tlie space about a point in any pbne. 4. 90° make a quadrant, or right angle, and G0° a sextant, or -I of a cii'cle. EXAMPLES FOll PRACTICE. 1. Reduce 10 S. 10° 10' 10" to seconds. 2. Reduce lUGGlO'' to signs. 3. How many degrees in 11 -100 geograpliic or nautical r.:iles? Ans. 190°. 4. If 1 degree of the earth's circumference is 69^ statute miles, how many statute miles in 11-100 geograpliic miles, or 190 degrees? A7is. 13148. 5. How many minutes, or nautical miles, in the circum- ference of the earth? Ans. 21600' or mi. G. A ship during 4 days' storm at sea changed her longitude 397 geographical miles; how many degrees and minutes did she cluuige ? Ans. 6° 37'. 2®8. In Counting. 12 units or things. . . .make. ... 1 dozen. 12 dozen " 1 gross. 12 gross " 1 great gross. 20 units " 1 score. S®!>. Paper. 24 sheets make 1 quire. 20 quires " 1 ream. 2 reams " 1 bundle. 5 bundles " 1 bale. SiO. Books. The terms folio, quarto, octavo, duodecimo, &c., indicate the number of leaves into Avhich a sheet of paper is folded. A sheet folded in 2 leaves is called a folio. A sheet folded in 4 leaves " a quarto, or 4to. A sheet folded in S leaves " an octavo, or 8vo. A sheet folded in 12 leaves " a 12mo. A sheet folded in 16 leaves " a IHmo. A sheet folded in IS leaves " an ISmo. A sheet folded in 24 leaves " a 24010. A sheet folded in 32 leaves " a 32mo. A\Tiat is a deo;ree ? Repeat the table for counting. For reckoning paper. For indicating the size of books. f, 4 REDUCTION. EXAMPLES FOR PRACTICE. 1. If in Birmingliam, England, 150 million Gillott pens are manufactured annually, liow many great gross will tliey make ? Ans. SG8U5 great gross 6 gross 8 dozen. 2. In 100000 sheets of paper, how many bales? Aiis. 20 bales 4 bundles 6 quires 16 sheets. 3. What is the age of a man 4 score and 10 years old? 4. How many printed pages, 2 pages to each leaf, will there be in an octavo book, having 8 fully printed sheets ? A71S. 128 pages. 5. How large a book will ten 32mo. sheets make, if every page be printed? Ans. G40 pages. PROMISCUOUS EXAMPLES IX REDUCTION. 1. How many suits of clothes, each containing 6 yd. 3f qr., can be cut from 333 yards of cloth ? Ans. 48. 2. A man bought a gold chain, weighing 1 oz. 15 pwt., at seven dimes a pennyweight; what did it cost? Ans. $24. cO. 3. A physician, having 2 tb 3§ 5 5 19 10 gr. of medicine, dealf it out in prescriptions averaging 15 grains each; how many prescriptions did it make ? Ans. 886. 4. A man bought 1 T. 11 cwt. 12 lbs. of hay, at IJ- cents a pound ; wliat did it cost.'' Ans. Ci^38.90. 5. What will be the cost of a load of oats weighing 1456 pounds, at 37^ cents per bushel ? AicS. $17.0625. 6. If one bushel of wheat will make 45 pounds of flour, how many barrels will lOOO bushels make ? Ans. 229 bbl. 116 lb. 7. A load of wheat weighing 2430 pounds is worth how much, at $1.20 a bushel? Ans. $48.60. 8. Paid $12.50 for a barrel of beef; how much was that per pound? Ans. 6^ cents. 0. If a silver dollar measure one inch in diameter, how many . To reduce a denominate decimal to integers of lower denominations, 1. Reduce .78125 of a pound Troy to integers of loiver de- nominations. Give explanation. Rule. Case V is what ? 180 REDUCTION. Analysis. We first multiply by 12 to reduce the given number from pounds to ounces, and the result is 9 ounces and the decimal .375 of an oz. We then multiply this decimal by 20 to reduce it to pennyweights, and get 7 put. and .5 of a pwt. This last decimal Me multiply by 24, to reduce it to grains, and the result is 12 gr. Hence the answer is 9 oz. 7 pwt. 12 gr. Rule. I. Multiply the given decimal by that mimber in the scale which loill reduce it to the next lower denomination^ and' point off as in multiplication of decimals. 11. Proceed with the decimal part of the product in the some manner until reduced to the required denominations. The in' tegers at the left will he the answer required. OPERATION. .78125 lb. 12 9.37500 oz. 20 7.50000 pwt. 24 12.0000 gr. 9 oz. 7 pwt. 12 gr., Ans. 2. What is 3. What is 4. Reduce inations. 5. Reduce 6. What is 7. What is 8. What is 9. What is 10. W^hatis EXAMPLES FOR PRA.CTICE. the value of .217°? Ans. 13' 1.2''. the value of .G59 of a week ? Ans. 4 da. 14 h. 42 min. 43.2 sec. .578125 of a bushel to integers of lower denom- Ans. 2 pk. 2 qt. 1 pt- .125 bbl. to integers of lower denominations. Ans. 3 gal. 3 qt. 1 pt. 2 gi. the value of .628125 £? the value of .22 of a hogshead of molasses ?. Ans. 13 gal. 3 qts. 3.52 gi. the value of .G7 of a league ? Ans. 2 mi. 3 rd. 1 yd. 3-^ in. the value of .42857 of a month ? Ans. 1 2 da. 20 h. 34 min. IH^ sec. the value of .78875 of a long ton ? Ans. 15 cwt. 3 qr. 2 lb. 12.8 oz. Give explanation. Rule. DENOMINATE FRACTIONS. 181 11. Wliat is the value of 5.88125 acres ? Arts. 5 A. 3 R. 21 P. 12. Reduce .0055 T. to pounds. Ans. 11 lb. 13. Reduce .034375 of a bundle of paper to its value in lower denominations. Ans. 1 quire 9 sheets. CASE VI. SI6. T6 reduce a compound number to a decimal of a higher denomination. 1. Reduce 3 pk. 2 qt. to the decimal of a bushel. OPERATION, Analysis. Since 8 quarts make 8 2.00 qt. 1 peck, and 4 pecks 1 bushel, there T~7~ will be \ as many pecks as quarts J_lll!_ ^^" (1§3), and ^ as many bushels as .8125 bu., Ans. pecks. Or we may reduce 3 pk. 2 qt. to Or, 3 pk. 2 qt. — 2b qt. ^^^ fraction of a bushel (as in 2 1 1 ), 1 bu. = 3- qt. and we have |4 o^ ^ bushel, M'hicli, 1^ = .8125 bu., Ans. reduced to a decimal, equals .8125. Hence the Rule. Divide the lowest denomination given by that num- ber in the scale which icill reduce it to the next higher, and ttn- nex the quotient as a decimal to that higher. Proceed in the same manner until the whole is reduced to the denomination required. Or, ' Reduce the given number to a fraction of the required dc' nomination, and reduce this fraction to a decimal. EXAMPLES FOR PRACTICE. 2. Reduce 3 qt. 1 pt. 1 gi, to the decimal of a gallon. Ans. .90G25ga1. 3. Reduce 10 oz. 13 pwt. 9 gr. to the decimal of a pound Troy. Ans. .88906251b. 4. Reduce 1.2 pints to the decimal of a hogshead. Ans. .00238 + hhd. 5. What part of a bushel is 3 pk. 1.12 qt. ? Ans. .785 bu. Case VI is what ? Give explanations. Rule. IS2 ADDITION. 6. What part of an acre is 3 R. 12.56 P. ? 7. Reduce 17 jd. 1 ft. 6 in. to the decimal of a mile. Ans. .00994318 -f mi. 8. Reduce .32 of a pint to the decimal of a bushel. Ans. .005 bu. 9. Reduce A} feet to the decimal of a fathom. Ans. \8125 fothom. 10. Reduce 150 sheets of paper to the decimal of a ream. Ans. .3125 Rm. 11. Reduce 47.04 lb. of flour to the decimal of a barrel. 12. Reduce .33 of a foot to the decimal of a mile. 13. Reduce 5 h. 3G min. 57y^jy sec. to the decimal of a day. ADDITION. SST. 1. A miner sold at one time 10 lb. 4 oz. IG pwt. 8 gr. of gold ; at another time, 2 lb. 9 oz. 3 pwt. ; at another, 11 oz. 20 gr. ; and at another, 25 lb. 16 pwt. 23 gr. ; how much did he sell in all ? AtsULYSIS. Arranging the num- bers in columns, placing units of the sayne denomination under each otli- er, wc first add the vniits in the right hand column, or lowest de- nomination, and find the amount to be 51 grains, which is equal to 2 pwt. 3 gr. AYe write the 3 gr. under the column of grains, and add tlie 2 pwt. to the column of pwt. Wc find tlie amount of the second col- unni to be 37 pwt., which is equal to 1 oz. 17 ]nvt. "Writing the 17 ])wt. under the column of pwt., we add the 1 oz. to the next column. Adding this column in the same manner as the preceding ones, wc find the amount to be 25 oz., equal to 2 lb. 1 oz. Placing the 1 oz. niiilcr the column of oz.. we add the 2 lb. to the column of lb. Addintr the last column, we find the amount to be 3911). Hence the following OPEUATION. IV.. oz. pwt. gr. 10 4 16 8 2 9 3 11 20 25 16 23 39 1 17 3 "What is addition of compoimd nvmibers ? Give explanation. COMrOUND NUMBERS. 183 TtuLE. I. Write (he nuvibers so tJiat tJiosc of the same unit VGiKe will stand in the same column. II. Beginning at the right iiaiul, add each denomination as in simple numbers, carrying to each succeeding denomination one for as many units as it takes of the denomination added, to make one of t'ue next higher denomination. EXAMPLES FOR PRACTICE :. (-'•) ( :-) £. s. d. lb. §• 5- 9- gr- 43 13 8 12 8 7 2 15 51 6 4 10 4 1 10 67 11 3 15 00 2 1 19 76 18 10 11 6 12 244 10 1 13 4 4 2 00 (4.) (5.) T. CAVt. lb. oz. cir. bu. pk. qt. pt. 4 7 IS 4 10 1 3 7 1 15 98 15 5 3 2 2 3 9 10 G 15 1 G 1 1 15 4 17 45 5 2 4 1 9 12 42 11 2 G. What is tlie sum of 4 mi. 3 fur. 30 rd. 2 jd. 1 ft. 10 in,, 5 mi. Gfur. 18 rd. 1yd. 2 ft. Gin., 10 mi. 4 fur. 25 rd. 2 yd. 2 ft. 11 in., and G fur. 28 rd. 4 yd. 2 ft. 1 in. ? ^ 7. Find the sum of 197 sq. yd. 4 sq.ft. 104J- sq. in., 122 sq. yd. 2 sq. ft. 27^ sq. in., 5 sq. yd. 8 sq. ft. 2f sq. in., and 237 sq. yd. 7 sq.ft. 128|sq. in.? Ans. 563 sq. yd. 4 sq. ft. 118.825 sq. in. Note. When common fractions occur, they shonlrT be reduced to a common denominator, to decimals, or to integers of a lower denomi- nation, and added according to the usual method. Give the Rule. 18i 1 ADDITION. (S.) A. E. P. sq. yd. i ;q. ft. s q. in. 26 3 28 25 8 1 25 19 2 38 30 7 1 50 456 2 20 16 6 98 503 1 8 i2(i: ) 5 85 W = 72 503 1 8 13 1 13 (0.) (10.) mi. fur. rd. yd. ft. in. hhd . gal. qt. pt. 1 7 30 4 2 11 27 65 3 2 3 4 00 2 1 10 112 60 2 3 10 7 25 1 2 11 50 421 29 00 2 1 16 3 16 ~3i~ 1 8 3 ^ 14 39 1 2 (11.) (12.) bu. pk. qt. pt. 3T- da. h. min. sec 23 3 7 1 25 300 19 54 35 34 2 1 21 40 12 40 24 42 3 5 3 112 14 15 17 51 1 4 1 6 10 11 45 59 23 3 3 4 1 1 1 1 1 11 57 109 11 37 16 13. If a printer one day use 4 bundles 1 ream 15 quires 20 sheets of paper, the. next day 3 bundles 1 ream 10 quires 10 sheets, and the next 2 bundles 13 sheets, how much does he use in the three days ? Aiis. 2 bales 1 ream 6 quires 19 sheets. 11. A tailor used, in one year, 2 gross 5 doz. 10 buttons, another year 3 gross 7 doz. 9, and another year 4 gross 6 doz. 1 1 ; how many did he use in the three years ? Ans. 10 gross 8 doz. 6. COMPOUND NUMBERS. 185 15, A ship, leaving New York, sailed east the first day o° Ao' 50" ; the second day, 4° 50' 10'' ; the third, 2° 10' 55 ' ; tlie fourth, 2*^ oO " ; how lar was she then east from the phice of starting? Ans. 12° 47' 34''. IG. A man, in digging a cellar, removed 127 cu. yd. 20 cu. ft. of earth ; in digging a drain, G cu. yd. 25 cu. it. ; and in digging a cistern, 17 cu. yds. 18 cu. ft. ; what was the amount of earth removed, and what the cost at 16 cents a cu. yd.? Ans. 152-]- cu. yds. ; $24.37^. 17. A farmer received 80 cents a bushel for 4 loads of corn, weighing as follows : 2564, 2713, 3000, and 3109 lbs. ; how much did he receive for the whole? Ans. $162.657-j- 18. A druggist sold for medicine, in three years, at an aver- age price of 9 cents a gill, the following amounts of brandy, viz.: 1 bbh4gah 1 pt. ; 30 gal. 2 qt. 1 gi. ; 2 bbh 15 gal.; how much did he receive for the whole ? Ans. §415.17. 5218. To add denominate fractions. 1. Add 1^ of a mile to J- of a furlong. OPERATION. Analysis. We find the f mi. = 6 fur. 26 rd. 11 ft. "^^^^^ of each fraction in in- 1 fLu._^— 13 rd 5-i ft tegers of less denominations -;;; — - (213), and then add their Ans. i lur. 00 values as in comjjound num- Or, i fu r. -^ 8 = jV nii- ^""^ ( '^ * ^ )• ^V mi. + -I mi. = li mi. = 7 fur. . ^^''/^"^ . ^"^^ 'f""'? ^^% ■^ * given tractions to fractions of the same denomination (212), then add them, and find the value of their sum in lower denominations (213). 2. Add I of a rod to f of a foot. Ans. 13 ft. U in. 3. What is the sum of |^ of a mile, f of a furlong, and f of a rod ? Ans. 7 fur. 27 rd. 8 ft. 3 in. 4. What is the sum of f of a pound and |^ of a shilling ? Ans. 13 s. 10 d. 2§ qr. 5. What is the sum of § of a ton and f of 1 cwt. ? Ans. 12 cwt. 42 1b. 13f oz. Give explanation of the process of addmg denominate fractions. 1S6 SUBTRACTION. 6. "What is the sum of | of a day added to ^ an hour ? Alls, i) h. 30 mill. 7. What is the sum of ^ of a week, f of a day, and ^ of an hour ? Ans. 1 da. 22 h. 15 min. 8. Add f of a hhd. to f of a gal. 9. What is the sum of f of a cwt., 8f lb., and Sy^j oz. by long ton table ? Ans. 73 lb. 1 oz. 3J-J- dr. 10. What is the sum of | of a mile, f of a yard, and ^ of a foot ? 11. Sold 4 village lots; the first contained J- of ^ of an acre ; the second, G0|^ rods ; the third, f of an acre ; and the fourth, § of f of an acre ; how much land in the four lots ? Ans. 3 E. 2G P. 120^-^^ sq. ft. 12. A farmer sold three loads of hay ; the first weighed 1^ T., the second, 1 ^3^. T., and the third, 18| cwt.; what was the aggregate weight of the three loads ? Ans. 3 T. 5 cwt. 91 lb. lOS oz. SUBTRACTION. *I19. 1. If a druggist buy 25 gal. 2 qt. 1 pt. 1 gi. of wine, and sell 18 gal. 3qt. 1 pt. 2 gi., how much has he left? OPERATION. Analysis. Writing the subtrahend gal. qt. pt. gi. under the minuend, placing units of the 2'3 2 1 1 same denomination under each other, 18 3 2 ve begin at the right hand, or lowest ^ ~^ o 7^ ^ denomination ; since we cannot take 2 gi. from 1 gi., we add 1 pt. or 4 gi. to 1 gi., making 5 gl. ; and taking 2 gi. from 5 gi., we write the remain- der, 3 gi., underneath the cokimn of gills. Having added 1 pt. or 4 gi. to the minuend, we now add 1 ])t. to the pt. in the subtra- liond, making 1 pt. ; and 1 pt. from 1 pt. leaves pt.,Mhich wc write in the remainder. Next, as we cannot take 3 qt. from 2 qt., wt' add 1 gal. or 4 qt. to 2 qt., making 6qt., and taking 3 qt. from (p., we write the remainder, 3 qt., under the denomination of quarts. Add- ing 1 gal. to 18 gal., we subtract 19 gal. from 2o gal., as in simple "WTiat is subtraction of compound nimibcrs ? Give explanation. COMPOUND NUMBERS. 187 numbers, and write the remainder, 6 gal., under the column of gal- lons. Hence the foUoMing Rule. I. Write t/ie subtrahend under tlie nunucnd, so that units of the same denomination shall stand under each otJier. II. Beginniiuj at the right hand, subtract each doioiniiiation separately, as in simple numbers. III. If the number of any denomination in the subtrahend exceed that of the same denomination in the minuend, add to the number in the minuend as many units as make one of tlis next higher denomination, and then subtract; in this case add 1 to the next higher denomination of the subtrahend before subtracting. Proceed in the same manner icith each denomi- nation. EXAMPLES FOR PRACTICE. lb. oz. •) pwt. gT- (3.) A. R. P. From 18 6 10 14 25 2 1G.9 Take 10 5 4 6 19 3 25.14 Rem. 8 1 6 8 5 2 31.76 (i-) (5.) T. cwt. lb. 5^- da. h. min. sec. 14 11 69f 38 187 16 45 50 10 12 98| 17 190 20 50 40 20 361 19 bo 10 6. A Boston merchant bought English goods to the amount of 4327 £ 13 s. 7^d., and he paid 1374 £ 10 s. llf d. ; how much did he then owe ? 7. From 300 miles take 198 mi. 7 fur. 25 rd. 2 yd. 1ft. Ans. 101 mi. 14 rd. 2 yd. 2 ft. 8 in. 10 in 8. What is the difference in the longitude of two places, one 75= 20' 30" west, and the other 71° 19' 35 ' west ?' Ans. 4° 55". 9. From 10 tb 7 § 4 3 1 9 15 jjr. take 3ft)8§ 2323 18 or. Ans. 6 1b 11 § 1 3 1 9 17gr. Give the Rule. 188 SUBTRACTION. 10. The apparent periodic revolution of the sv.n is made in 36.3 da. 6 h. 9 min. 9 sec., and that of the moon in 29 da. 12 h. 44 inin. 3 sec. ; what is the difference ? Ans. 33.5 da. 1.5 h, 2.5 min. 6 sec. 11. A man, having a hogshead of wine, drank, on an aver- age, for five years, including two leap years, one gill of wine a day ; how much remained ? Ans. 5 gal. 3 qt. 1 pt. 1 gi. 12. A section of land containing 6-iO acres is owned by four men ; the first owns 196 A. 2 R. 16^ P. ; the second, 200 A. li R. ; the third, 177 A. 36 P. ; how much does the fourth own ? Ans. 65 A. 3 R. 7.75 P. 13. From a pile of Avood containing 75^ Cd. was sold at one time 16 Cd. 5 cd. ft.; at another, 24 Cd. 6 cd. ft. 12 cu. ft. ; at another, 27 Cd. 112 cu. ft. ; how much remained in the pile ? Ans. 6 Cd. 3 cd. ft. 4 cu. ft. 14. If from a hogshead of molasses 10 gal. 1 qt. 1 pt. be drawn at one time, 15 gal. 1 pt. at another, and 14 gal. 3 qt. at another, how much will remain ? 220. To find the difference in dates. 1. What length of time elapsed from the discovery of America by Columbu?, (jlct. 14, 1492, to the Declaration of Independence, July 4, 1776? FIRST OPERATION. ANALYSIS. "\Ve place the earlier date yr. mo. da. mider the later, writing first on the left ^ ^ ' ■* the number of the year from the Chris- 1492 10 14 tian era, next the number of the month, 2S3 8 50 counting January as the first month, and next the number of the day from the first day of the month. Instead of the number of the year, month, and day, some use the number of years, months, and days that SECOND OPERATION. !"'^'^ elapsed since the Christian era, thus : ,,.. „in_ fjj,/ instead of saying July is the 7th month, UJ,'') 6 3 ^^^ ^^V ^ months and .'5 days have ■lACj] n -I o elajisod, and instead of sayini;: October is tlie lOth montli, we say 9 months and 283 8 20 13 days have elapsed. How is the difference of dates found ? COMPOUND NUMBERS. 189 Both methods Mill obtain the same result ; the former is generally used. Notes. 1. When hours are to be obtahicd, -we reckon from 12 at night, and if minutes and seconds, we write them still at the right of hours. 2. In finding the time between two dates, or in computing interest, 12 months are considered a year, and 30 days a month. When the exact number of days is required for any period not exceeding one ordinary year, it may be readily found by the following TABLE, Slioxcing the miynber of days from any day of one month to the same day of any otlier mo7ith within one year. rilOM ANY TO THE SAME DAY OF THE NEXT. DAY OF Jan. 365 Feb. 31 Mar. 59 Apr. 90 May. 120 151 July 181 Aii^r. 212 .«.pt. 243 Oct. 273 Nov. 304 Dec. January . . . 334 February . . 334 36.3 28 59 89 120 150 181 212 242 273 303 March .... 306 337 365 31 61 92 122 153 184 214 245 275 April 275 306 334 365 SO 61 91 122 153 183 214 244 Mav 2-15 276 304 335 365 31 61 92 123 153 184 214 June 214 245 273 304 334 365 30 61 92 122 153 183 July 184 215 243 274 304 335 36o 31 62 92 123 153 August . . . 153 184 212 243 273 3C4 334 365 31 61 92 122 September . 122 153 181 212 242 273 303 334 365 30 61 91 October. . . . 92 123 151 182 212 243 273 304 335 365 31 61 November . 61 92 120 151 181 212 242 273 304 334 365 30 December. . 31 62 90 121 151 182 212 243 274 304 335 365 If the days of the different months are not the same, the number of days of difference should be added vihen the earlier day belongs to the moni\\ from which we reckon, and subtracted when it belongs to the month to which we find the time. If the 29th of February is to be included in the time computed, one day must be added to the result. EXAMPLES FOR PRACTICE. 2. George Washington was born Feb. 22, 1732, and died Dec. 14 1799 ; wdiat was his age ? Ans. 67 yr. 9 mo. 22 da. How can the nmnber of days, if less than a year, be obtained ? 190 SUBTRACTION. 3. How much time has elapsed since the declaration of independence of the United States ? 4. How many years, months, and days from your birthday to this date ; or wiiat is your age ? 5. How long from the battle of Bunker Hill, June 17, 1775, to the battle of Waterloo, June 18, 1815 ? Ans. 40 yr. 1 da. 6. What length of time will elapse from 20 miimtes past 2 o'clock, P. M., June 24, 1856, to 10 minutes before 9 o'clock, A. M., January 3, 1861 ? Ans. 4 yr. 6 mo. 8 da. 18 h. 30 min. 7. How many days from any day of April to the same day of August ? of December? of February? 8. How many days from the 6th of November to the 15tli of April? Ans. 160 days. 9. How many days from the 20th of August to the 15th of the following June ? Ans. 299 days. 2S3. To subtract denominate fractions. 1. From f of an oz. take |^ of a pwt. OPERATION, Analysis. We per- |oz. z= 7 pwt. 12 gr. form the same reduc- z pwt. =r 21 o-r. tions as in addition of denominate fractions, 6 pwt. 15 gr., Ans. (218 ), and then sub- tract the less value from Or, f oz. X 20 r= Y pwt. the greater. ¥ — i = ¥- pwt. = 6 pwt. 15 gr. to' 2. What is the difiercnee between ^ rod and f of a foot ? Ans. 7ft. 6 in. 3. From |- £ take f of |- of a shilling. 4. From f of a league lake -/j of a mile. Ans. 1 mi. 2 fur. 16 rd. 5. From S-p^ cwt. take 1 qr. 2^ lb. A)is. 8 cwt. 2 qr. 14 lb. 5 oz. lOj^dr. 6. From -^ of a week take ^ of a day. A71S. 1 da. 4 h. 48 min. Give explanation of the process of subtracting denominate fractions. COMPOUND NUMBERS. . 191 7. Two persons, A and B, start from two places 120 miles apart, and travel toward each other; after A travels f , and 1> #, of the distance, how far are tliey apart ? Ans. 41. mi. 7 fur. 9 rd. 8 ft. 7^ in. 8. From a cask of brandy containing 9G gallons, -^ leaked out, and f of the remainder was sold ; how much still remained in the cask ? Ans. 25 gal. 2 qt. 3^ gi. MULTIPLICATION. 223. 1. A farmer has 8 fields, each containing 4 A. 2 R. 27 P. ; how much land in all ? OPERATION. Analysis. In 8 fields are 8 times as much A. It. r. land as in 1 field. We write the muki])lier 4 2 27 under the lowest denomination of the mul- 8 tiplicand, and proceed thus; 8 times 27 P- ~ — ' --" are 216 P., equal to 5 11. 16 P.; and we '^' write the 16 P. under the number multiplied. Then 8 times 2 R. are 16 11., and 5 11. added make 21 R., equal to 4 A. 1 R. ; and we Avrite the 1 II. under the number multiplied. Again, 8 times 4 A are 32 A., and 4 A. added make 36 A., which we write under the same denomination in the multipUcaud, and the work is done. Hence, Rule. I. Write the multiplier under the loioest denomina- tion of the midtiplicand. 11. Multiphj as in simple members, and carry as in addi' iion of compound numbers. EXAMPLES FOR PRACTICE. (3.) mi. fur. rd. ft. 9 4 20 13 6 (2-) bu. rk. qt. pt. 4 2 5 1 2 9 13 57 3 4 12 Multiplication of compovmd numbers, how performed ? Rule. 192 MULTIPLICATION. (4.) £. s. d. 5 18 4 (5.) lb. oz. pwt. gr. 3 4 22 4 7 (6.) T. cwt. lb. 14 16 48 oz. 12 13° 10' 35" 11 9 8. In 6 barrels of grain, each containing 2 bu. 3 pk. 5 qt., how many bushels? ^«5- 17 bu. 1 pk. 6 qt. 9. If a druggist deal out 3 lb 4 S 1 5 2 9 16 gr. of med- icine a day, how much will he deal out in 6 days ? 10. If a man travel 29 mi. 3 fur. 30 rd. loft, in 1 day, how far will he travel in 8 days ? 11. If a woodchopper can cut 3 Cd. 48 cu. ft. of wood in 1 day, how many cords can he cut in 12 days? Ans. 404- Cd. 12. What is the weight of 48 loads of hay, each weighing 1 T. 3 cwt. 50 lb. ? T, 1 cwt. 3 OPERATION, lb. 50 6 7 1 00 weight of 6 loads. 8 56 8 00 weight of -IS loads. Analysis. "When the multi- plier is large, and a composite number, we may multiply by one of the factors, and that product by the other. Multiplying the weight of 1 load by 6, we obtain the weight of 6 loads, and the weight of 6 loads multiplied by 8, gives the weight of 48 loads. 13. If 1 acre of land produce 45 bu. 3 pk. 6 qt. 1 pt. of corn, how much will 64 acres produce? Ans. 2941 bu. 14. How much will 120 yards of cloth cost, at 1 £ 9 s. 8^ d. per yard ? 15. If $80 will buy 4 A. 3 R. 26 P. 20 ?q. yd. 3 sq. ft. of land, how much will $4800 buy? Ans. 295 A. 10 pq.yd. 16. If a load of coal by the long ton weigh 1 T. 6 cwt. 2 qr. 26 lb. 10 oz., what will be the weight of 73 loads ? Ans. 97 T. 11 cwt. 3 qr. 1 1 lb. 10 oz. COMPOUND NUMBERS. 193 17. The sun, on an average, clianges his longitude 59' 8.33" per day; how much will be the change in 3 Go days? 18. If 1 pt. 3 gi. of wine fill 1 bottle, how much will be re- quired to fill a great gross of bottles of the same capacity .? DIVISION. 223. 1. If 4 acres of land produce 102 bu. 3 pk. 2 qt. of wheat, how much will 1 acre produce ? OPERATiox. Analysis. One acre will produce ^ pt, bu. ijU. qt. pts. as much as 4 acres. "Writing tlie divi- "^ ) ^^'^ ^ ^ sor ou tlie left of the dividend, we divide 2^ 2 6 1 102 bu. by 4, and we obtain a quotient of 2 p_ Explain the process of dividing compound nimibera. 194 DIVISION, Rule. L Divide the highest denomination as in simple members, and each succeeding denomination in the same man- ner, if there be no remainder. II. Jf there be a remainder after dividing any denomina- tion, reduce it to the next lotver denomination, adding in the given number of that denomination, if any, and divide as be- fore. III. The several partial quotients will be the quotient re- quired. Notes. 1. "Wlien the divisor is large and is a composite number, ■we may shorten the work by dividing by the factors. 2. \Vlien the divisor and dividend are both compound nirmbers, they must both be reduced to the same denomination before dividing, and then the i3rocess is the same as in simple numbers. £. 5)25 (3.) s. 8 EXAMPLES FOR PRACTICE. d. 4 4)3 1 (5.) da. h. 5 22 8 mm. 00 T. 7) 45 (4.) cwt. 15 lb. 25 6 10 (6.) 75 ))25° 42' 40" 6 17 30 2 34 16 7. Bought 6 large silver spoons, which weighed 1 1 oz. 3 pwt.; ■what was the weight of each spoon ? 8. A man traveled by railroad 1000 miles in one day; what was the average rate per hour ? Ans. 41 mi. 5 fur. 13 rd. 5 ft. 6 in. 0. If a family use 10 bbl. of flour in a year, what is the average amount each day? Ans. 51b. 5 oz. 14-5T}dr. 10. The aggregate weight of 123 hogsheads of sugar is 57 T. 19 cwt. 42 1b. 14 oz.; what is the average weight per hojishead? Ans. 9 cwt. 42 lb. 10 oz. 11. IIow many times are 5 £ 10 s. 10 d. contained in 537 £ 10 s. 10 d.? Ans. 97. ■ ■■ . - ■ - -^ — ' — ■ Give the rule. "SMien the divisor is a compo'site number, how may we proceed ? ^Y^len the divisor and dividend arc both compound numbers, how proceed ? COMPOUND NUMBERS. 195 12. A cellar 50 ft. lonpj, 30 ft. wide, and 6 ft. deep was ex- cavated by 5 men in G davs ; how many cubic yards did each man excavate daily? Ans. 11 cu. yd. 3 cii.ft. 13. If a town !') miles square be divided equally into 150 farms, what will be the size of each farm ? Ans. lOG A. 2 R. 26 P. 20 sq. yd. 1 sq.ft. 72 sq. in. 14. How many times are 4 bu. 3 pk. 2 qt. contained in 336 bu. 3 pk. 4qt.? Ans. 70. 15. A merchant tailor bought 4 pieces of cloth, each con- taining 60 yd. 2.25 qr. ; after selling ^ of the whole, he made up the remainder into suits containing 9 yd. 2 qr. each; how many suits did he make ? Ans. 17. LONGITUDE AND TBIE. S'21. Every circle is supposed to be divided into 360 equal parts, called degrees. Since the sun appears to pass from east to west round the earth, or through 360°, once in every 24 hours, it will pass through 2^y of 360^, or 15" of the distance, in 1 hour ; and 1° of distance in J^ of 1 hour, or 4 minutes; and 1' of distance in •^'g- of 4 minutes, or 4 seconds. TABLE OF LOXGITUDE AND TIME. 3G0° of longitude z=z 24 hours, or 1 day of time. 5° " l( = 1 hour >( X ic „ u z=i 4 minutes i( (( 1' " ti r= 4 seconds « u CASE I. 3S«>. To find the tlifforencc of time between two places, when their longitudes are given. 1. Tlie longitude of Boston is 71' 3', and of Chicago 87^ 30' ; what is the difference of time between these two places ? Explain how distance is measured by time. Repeat the table of longitude and time. Case I is what ? OPEKATION. 87° 30' 71 3' 16° 27' 4 19 g LONGITUDE AND TDIE. Analysis. By subtraction of compound numbers v.e first find the difference of longitude be- tween the two places, which is 16° 27'. Since 1° of longitude makes a difference of 4 minutes 1 , I '. TZ" A of time, and 1' of longitude a 1 h. o rain. 48 sec, Ans. ,.^. ' , , , p • dmerence or 4 seconds oi tunc, we multiply 16° 27', the difference in longitude, by 4, and we obtain the difference of time in minutes and seconds, wliich, reduced to higher denominations, gives 1 h. 5 min. 48 sec, the difference in time. Hence the Rule. 3Tultiply the difference of longitude in degrees and minutes by 4, and tJie product will be the difference of time in minutes and seconds, wliich may he reduced to hours. Note. If one place be in cast, and the other in west longitude, the difference cf longitude is found by adding them, and if the sum be greater than 180°, it must be subtracted fiom 360°. EXAMPLES FOR PRACTICE. 2. New York is 74' 1' and Cincinnati 84^ 24' west longi- tude ; what is the difference of time ? Ans. 41 min. 32 sec. 3. The Cape of Good Hope is 18^ 28' cast, and the Sand- wich Islands 155' west longitude; what is the difference of time ? Ans. 1 1 h. 33 min. 52 sec. 4. Washington is 77° 1' west, and St. Petersburg 30' 19' east longitude ; what is their difference of time ? Ans. 7 h. 9 rain. 20 sec. 5. If Pekin is 118' east, and San Fiancisco 122' west longitude, what is their diflerence of time ? G. If a message be sent by telegraph without any loss of time, at 12 M. from London, 0° 0' longitude, to Washington, 77° V Avest, what is the time of its receipt at Washington? Note. Since the sun ajipcars to move from on«t to west, when it is exactly 12 o'clock at one place, it will be pant 12 o'clock at all places east, and before 12 at all places west. Hence, knowina; the difference of time between two places, and the exact time at one of them, the exact time at the other will be foimd by addinfj their difference to the given time, if it be east, and by siiblracfing if it be west. Ans. 6 h. 51 min. b(j sec, A. M. Give explanation. Kule. COMPOUND NUMBERS. 197 7. A steamer arrives at Halifax, G3° 36' west, at 4 o'clock, P. M. ; the fact is telegraphed to St. Louis, 90^ 15' west, without loss of time; what is the time of its receipt at St. Louis ? Ans. 2 h. 13 mill. 24 sec, P. M. 8. If, at a presidential election, the voting- begin at sunrise and end at sunset, how much sooner will the polls open and close at Eastport, Me., G7^ west, than at Astoria, Oregon, 124° west ? Ans. 3 h. 48 min. 'J. When it was 1 o'clock, A. M., on the first day of Jan- uary, 18o9, at Bangor, Me., 68° 47' west, what was the time at the city of Mexico, 99° 5' west? Ans. Dec. 31, 1858, 58 min. 48 sec. past 10, P. M. CASE II. 2*J equal to -^^ of 100 per cent. which is 4 per cent. Hence the Rule. Divide the percentage hj the base, and the quotient will be the rate per cent, expressed decimally. Or, Take such a part of 100 as the percentage is part of the base. EXAMPLES FOR rUACTICE. 2. What per cent, of $ 150 is $00 ? Ans. 20. Ans. 12.i. 3. What per cent, of $1400 is $175? 4. What per cent, of $750 is $1 G5 ? 5. Wiiat per cent, of $2-10 is $13.20? Ans. 5^-. r.. Wliat per cent, of $2 is 15 cents ? 7. What per cent, of G bushels 1 peck is 4 bushels 2 pecks G quarts ? Ans. 75 per cent. , 8. What per cent, of 15 pounds is 5 pounds 10 ounces avoirdupois weight ? Ans. 373- per cent. 9. What per cent, of 250 head of cattle is 40 head ? Cose II is what ? Give explanation. Rule. 1 PERCENTAGE. 209 10. From a hogshead of sugar containing 7G0 pounds, 100 pounds were sold at one time, and 90 pounds at another; what per cent, of the whole was sold ? 11. A man, having GOO acres of land, sold ^ of it at one time, and ^ of the remainder at another time; what per cent, remaaied unsold ? Ans. 50 per cent. CASE III. SS7, To find a number when a certain per cent, of it is given. 1. A man lost $5, which was 4 per cent, of all the money he had ; how much had he at first ? OPERATION. Analysis. We are here required to $5 -f- .04: = $125. find the base, of which $5 is the per- Or, centage. Now, percentage equals base -^ X 100 zr: Si 25 multiphed by the rate per cent. ; con- * versely, base equals percentage divided by rate per cent. Or, $5 is 4 per cent, of all he had ; | of $5, or -I, equals 1 per cent, of all he had, and 100 times | equals 100 per cent., or all he had. Hence the Rule. Divide the percentage ly the rate per cent,, ex- pressed decimalhj, and the quotient loill he the lase, or number required. Or, Take as many times 100 as the percentage is times the rate per cent, EXAMPLES FOR PRACTICE. 2. 1 6 is 8 per cent, of what number ? Ans. 200. 3. 42 is 7 per cent, of what number ? 4. 75 is 12^ per cent, of what number ? Ans. 600. 5. 33 is 2f per cent, of what number? Ans. 1200. 6. $281.25 is 374 per cent, of what sum of money ? Ans. $750. 7. A farmer sold 50 sheep, which was 20 per cent, of his whole flock ; how many sheep had he at first ? Case ni is what ? Give explanation. Rule. 210 PERCENTAGE. 8. I loaned a man a certain sum of money ; at one time he paid me $59.75, which ^as 12^^ per cent, of the whole buni loaned to him ; how much did I loan him ? 9. A mercliant invested $975 in dry goods, wliieli was 15 per cent, of his entire capital ; what was the amount of his capital? -^ns. $G5U0. 10. If a man, owning 40 per cent, of an iron fpundery, sell 25 per cent, of his share for $12-40.50, what is the value of the whole foundery ? ^'is. $12405. 11. A merchant pays $75 a month for clerk hire, which is 25 per cent, of his entire pro.Hts ; how^ much are his profits for one year, after paying his clerk hire .'* Ans. $2700. 12. A produce buyer, having a quantity of corn, bought 2000 bushels more, and he found that this purchase was 40 per cent, of his whole stock ; how much had he before he. bou!?ht this last lot ? Ans. 3OU0 bushels. COMMISSION AND BROKERAGE. flS8, An Agent, Factor, or Broker, is a person who trans- acts business for another, or buys and sells money, stocks, notes, &c. !S30. Commission is the percentage, or compensation allowed an agent, factor, or commission merchant, for buying and selling goods or pi'oduce, collecting money, and transact- ing other business. S'^IO. Brokerage is the fee, or allowance paid to a broker or dealer in money, stocks, or bills of exchange, for making exchanges of money, buying and selling stocks, negotiating bills of exchange, or transacting otlier like business. Note. The ratos of commi'ision and brokcra2;Garc not recrnlatod hy law, but are usually reckoned at a certain per cent. ui)on the money employed m the transaction. Define an agent, factor, or broker. "\Miat Is meant by commission ? Brokerage ? * COMMISSIOx\ AND BROKERAGE. 211 CASE I, 241. To find the commission or brokerage on any smn of money. 1. A commission merchant sells butter and cheese to the amount of §loJ:0 ; what is his commission at o per cent. ? OPEKATION. Analysis. 1540 X -05 =1 $;77, Alts. Since the com- Or, t5 J = 5-V , and g-'j X 1540 = $77. mh^lon on $1 is 5 cents or .Oo of a dollar, on 81-340 it is $1540 X .05:= $77. Or, since 5 per cent ^^ loT^^-sV o^ the sum received, the commission is ^L of $1540 r= $77. Hence the RuLK. Multiply the given sum hj the rate per cent, ex- pressed decimally, and the result ivill he Uie commission or bro- kerage. Or, 2'akc such a part of the given sum as the number expressing thz per cent, is part o/lOO. EXAMPLES FOn PUACTICE. 2. A commission merchant sells goods to the amount of $6756 ; what is his commission at 2 per cent. ? Ans. $iy5.12. 3. What commission must be paid for collecting $17380, at 3J- per cent. ? Am. $608.30. 4. An agent in Chicago purchased 4700 bushels of wheat, at 75 cents a bushel ; what Avas his commission at 14 per cent, on the jiurchase money? 5. A broker in New York exchanged $25875 on the Suf- folk Bank, Boston, at ^ per cent. ; how much brokerage did he receive? Ans. $64.6875. 6. An auctioneer sold at auction a house for $3284, and the furniture for $2170.50 ; what did his fees amount to at 2^ per cent. ? 7. A broker negotiates a bill of excliange of S2890 for ^ per cent, commission ; how much is his brokerage ? Ans. $23.12. Case I is what ? Give explanation. Rule. 212 PERCENTAGE. 8. An agent buys for a manufacturing company 2G750 pounds of wool, at 32 cents a pound, and receives a commis- sion of 2f per cent. ; what amount does he receive? Arts. $235.40. 0. If I sell 400 bales of cotton, each weighing 570 {)ounds, at 9 cents a pound, and receive a commission of 2} per cent., how much do I make by the transaction ? Ans. $401.70. 10. A commission merchant in New Orleans sells 450 bar- rels of flour at $7.G0 a barrel; 38 firkins of butter, each con- taining 56 pounds, at 25 cents a pound ; and 105 cheeses, each weighing 48 pounds, at 9 cents a pound ; how much is his commission for selling, at 5i per cent.? Atis. $242,308. 11. A lawyer collected a note of §950, and charged GJ- per cent, commission ; what was his fee, and what the^ sum to be remitted? Ans. Fee, $G1.75 ; remitted, $888.25. 12. An insurance agent's fees are G per cent, on all sums received for the com])any, and 4 per cent, additional on all sums remaining, at the end of the year, after tlie losses- are paid ; he receives, during the year, $30456.50, and ])ays losses to the amount of $19814.15; how much commission does he receive during the year ? Ans. $2253.084. CASE II. 24:3. To find the commission or brokerage, when it is to bo deducted from the given sum, and the bal- ance invested. 1. A merchant sends his agent $1260 with which to buy merchandise, after deducting his commission of 5 per cent. ; ■what is the sum invested, and liow much is the commission ? OPEnATION. $1260 -4- 1.05 =r $1200, invested. $1260 — $1200 = $60, commission. Or, i?,?y -j- T^.T = U ; ^^260 H- U = $1200, invested; And $1260 — $1200 = $60, ron>nus.inn. Case U is what ? Give o.xiilanation. Rule. COMMISSION AND BROKERAGE. 213 Analysis. Since the commission is 5 per cent., the agent must receive $1.05 for every $1 he expends ; he can invest as many- dollars as $1.0J is contained times in $1260, which is $1200; and the difl'erence between the given sum and the sum invested is his commission. Or, the money expended is |gj] of itself, the commission is ^|-q- of this sum, and the commission added to the sum expended is ii| of the whole sum. Since $1260 is 1§| z=z fi,$126U ^U = '^l^OO, the sum expended; and $1260 — $1200=; $60 the commission. Hence the Kt'LE. I. Divide the given amount by 1 increased ly the rate per cent, of commission, and the quotient is tlie sum invested. II. Subtract the investment from the given amount, and the remainder is the commission, EXAMPLES FOR PKACTICE. 2. A man sends $3246.20 to his agent in Boston, request- ing him to lay it out in shoes, after deducting his commissioa of 2 per cent; how much is his commission ? Ans. $G3.65. 3. What amount of stock can be bought for $9G82, and al- low 3 per cent, brokerage ? Ans. $9400. 4. A flour merchant sent $10246.50 to his agent at Chica- go, to invest in flour, after deducting his commission of 3^ per cent. ; how many barrels of flour could he buy at $5..50 per barrel? Ans. 1800 barrels. 5. An" agent receives a remittance of $4908, with which to purchase grain, at a commission of 4^- per cent. ; what will be the amount of the purchase ? 6. Remitted $603.75 to my agent in New York, for the pur- chase of merchandise, agent's commission being 5 per cent. ; what amount of broadcloth at $5 per yard should I receive? Ans. 115 yds. 7. A commission merchant receives $9376.158, with or- ders to purchase grain ; his co.Timission is 3 per cent., and he charges 1^- per cent, additional for guaranteeing its delivery at a specified time ; how much will he pay out, and what are his fees ? Ans. Fees, $403,758. 214 PERCENTAGE. 8. A real estate broker, whose stated commission is 1^ per cent., receives $13842.07, to be used in the purchase of city lots ; how much does he invest, and what is his commis- sion ? Alls. $1.") 604 invested; $238.07 commission. 9. A broker received $10050, to be invested in stocks after deducting ^ per cent, for brokerage ; Avhat amount of stock did he purchase ? STOCKS. 9^3. A Corporation is a body authorized by a general law, or by a special charter, to transact business as a single indivi(hiai. iM^^^, A Charter is the legal act of incorporation, and de- fines the powers and obligations of the incorporated body. ^^4:e3. A Firm is the name under which an unincorporated company transacts business. S46. Capital or Stock is the property or labor of an indi- vidual, corporation, company, or firm ; it receives differen'.; names, as Bank Stock, Railroad Stock, Government Stock, &c. 24:7. A Share is one of the equal parts into which the stock is divided. S^!^. Stockholders are the owners cf the shares. S49. The Nominal or Par Value of stock is its first cos*, or original valuation. Note. The original value of a share varies in different companies. A share of bank, insurance, railroad, or like stock is usually |ilOO. Sr5^. Stock is At Par when it sells for its first cost, or original valuation ; imi , Above Par, at a premium, or in advance, wlien it foils ibr more tlian its original cost; and ^vi*2. Below Par, or at a discount, wlicn it sells for less than its original cost. Define a corporation. A charter. A firm. Capital or stock. Shares. Stockholders. I'ar value. At par. Above par. Ik'low par. STOCKS. 215 2*TS. The Market or Real Value of stock is what it will bring per share in money. •254. A Dividend is a sum paid to stockholders from the profits of t lie business of the company. S#5«5. An Assessment is a sum i-equired of stockholders to meet the losses or expenses of the business of the comiiaiiy. '^9i&. Premium or advance, and discount on stock, divi- dends, and assessments, are computed at a certain per cent. upon the original value of the shares of the stock. CASE I. 2tl7. To find the value of stock when at an ad- vance, or at a discount. - 1. What will $3240 of bank stock cost, at 8 per cent, ad- vance ? OPEKATION. Analysis. Since $1 -f- .08 = $1.08 $1 of the stock at $3240 X $1.08 — $3409.20, Ans. P^r value will cost $1 plus the premi- um, or $1.08, $3240 of the same stock will cost 3240 X $1.08 = 83499.20. If the stock were 8 per cent, below par, $1 minus the discount, or $1.00 — $.08 =: $.92, M-ould show what $1 of the stock would cost. Hence the Rule. M}dtipJy the par value of the stock hy the nianhcr indicating the price of $1 of the same stock, and the product ivill he the real value. Note. In all examples relating to stocks, $100 is considered the par value of a share of stock, unless otherwise stated. KXAMPLES FOR PRACTICE. 2. If the stock of an insurance company sell at 5 per cent. below par, what will $1200 of the stock cost? Ans. $1140. 3. What is the market value of 35 shares of New York Central Railroad stock, at 15 per cent, below par ? iSIarkot value. A dividend. An assessment. Case I is what ? Give explanation. Rule. 216 . PERCENTAGE. 4. What must be paid for 48 shares of Panama RaHroad •«?tock, at a premium of 5^ per cent., if the par value be $150 per share ? Atis. $7596. 5. What costs $53G4 stock in the Minnesota copper mines, at 9 per cent, above par ? 6. A man purchased $G275 stock in the Pennsylvania Coal Company, and sold the same at a discount of 12 per cent.; what was his loss ? Ans. $753. 7. Wliat must be paid for 125 shares of United States stock, at 4|- per cent, premium, the par value being SlOOO per share? Ans. $130937.50. 8. Bought 42 shares of Illinois Central Railroad stock, at 14 per cent, discount, and sold the same at an advance of 12^ percent.; how much did I gain .'* Ans, $1113. 9. What is the market value of 175 shares of stock in the Suffolk Bank, at |- per cent, advance ? Ans. $17031.25. 10. Bought 75 shares of stock in the Bank of New Orleans, of $50 each, at 3 per cent, discount, and sold it at 2|- per cent, advance; what was my gain ? Ans. $190,875. 11. B exchanged 28 shares of bank stock, of $50 each, v/orth 7 per cent, premium, for 25 shares of railroad stock, of $100 each, at 12^ per cent, discount, and paid the difference in cash; how much cash did he pay ? Ans. $G89.50. CASE II. S58. To find how miicli stock may be purchased for a given sura. 1. How many shares of bank stock, at 3 per cent, advance, may be bought for $5150? OPEKATION. Analysis. Since the stock $5150 -^ 1.03 r= $5000 = is at 3 per cent, advanco, $1 50 shares Ans. of stock at par will cost $1.03; and if we divide $5150, the whole sum to he expended, by $1.03, the cost of $1 of stock, the quotient must be the amount of stock purchased. Hence the Case II is what ? Give explanation. PROFIT AND LOSS. 217 EuLE. Divide the given stun hy the cost of $1 of stock, and the quotient will be the nominal amowit of stock purchased. 2. How many shares of railroad stock, at 5 per cent, ad- vance, can be purchased for $6300 ? Ans. 60 shares. 3. I invested $6187.50, in Ocean Telegraph stock, at 10 per cent, discount ; how much stock did I purchase ? Ans. $6875. 4. I sent my agent $53500 to be invested in Illinois Cen- tral Railroad stock, which sold at 7 per cent, advance ; what amount did he purchase ? Ans. $50000. 5. Sold 50 shares of stock in a Pittsburg ferry company, at 8 per cent, discount, and received $1150; what is the par value of 1 share ? Ans. $25. PROFIT ANB LOSS. SSO. Profit and Loss are commercial terms, used to ex- press the gain or loss in business transactions, which is usually reckoned at a certain per cent, on the prime oiv first cost of articles. CASE I. 26®. To find the amount of profit or loss, "when the cost and the gain or loss per cent, are given. 1. A man bought a horse for $135, and afterward sold him for 20 per cent, more than he gave ; how much did he gain ? OPEKATION. Analysis. Since $1 $135 X .20 r= $27, Alls. g^ins 20 cents, or 20 per dr. 20 i.<2:iq^vi «;97 *^"'^t., $135 will gain $135 '1"" ° ^ X-20 = $27. Or, smce 20 per cent, equals ^VV = h the whole gain will be ^ of the cost. Hence the following Rule. Ilidtiply the cost hy the rate per cent, expressed decimally. Or, Take such part of the cost as the rate per cent, is part o/lOO. Rule. "S\Tiat is meant by profit and loss ? Case I is what ? Give explanation. Rule. E.P 10 218 ' PERCENTAGE. EXAMPLES FOR PRACTICE. 2. A grocer bought a hogshead of sugar for $84.80, and sold it at 12^ per cent, j^rofit ; what was his gain ? 3. A miller bought 500 bushels of wheat at $1.15 a bushel, and he sold the flour at 16| per cent, advance on the cost of the wheat; what was his gain? Ans. $1)5.83 A. 4. Bought 76 cords of wood at $3. 62 J- a cord, and sold it so as to gain 26 per cent. ; what did I make ? 5. A hatter bought 40 hats at $1.75 apiece, and sold them at a loss of 14f per cent. ; what was his whole loss ? 6. A grocer bought 3 barrels of sugar, each containing 230 pounds, at 8^ cents a pound, and sold it at 18jSj- per cent, profit ; what was his whole gain, and what the selling price per pound ? Ans. Whole gain, $10.35 ; price per pound, 9f cents. 7. A sloop, freighted with 3840 bushels of corn, encoun- tered a storm, when it was found necessary to throw 37^ per cent, of her cargo overboard ; what was the loss, at 62^- cents a bushel ? Ans. $900 loss. 8. A genfleman bought a store and contents for $4720 ; he sold the same for 12^ per cent, less than he gave, and then lost 15 per cent, of the selling price in bad debts; Avhat was his entire loss? ^??s. $1209.50. 9. A man commenced business with $3000 capital ; the first year he gained 2 2 J- per cent., which he added to his capi- tal ; the second year he gained 30 per cent, on the whole sum, which gain he also put into his business ; the third year he lost 16f per cent, of his entire capital ; how much did he make in the 3 years ? A7is. $981.25. CASE II. SGI. To find the gain or loss per cent., when the cost and selling price are given. 1. Bought wool at 32 cents a pound, and sold it for 40 cents a pound ; what per cent, was gained ? Case n is what ? Give explanation. Rule. PROFIT AND LOSS. 219 OPERATION. 40 — 32 = 8 ; 8 -H 32 = ^8^ = .25, Ans. Or, 40 — 32 z= 8 ; 8 -^ 32 = ^^ = ^; "i X 100 = 25 per cent. Analysis. Since the gain on 32 cents is 40 — 32 = 8 cents, the whole gain is -r^^ = \ of the i^urchase money ; and \ reduced to a decimal is 2.3 hundredths, equal to 2o jier cent. Or, if the gain were equal to the purchase money, it would be 100 jjer cent. ; but since the gain is -.f^ r= 1 of the purchase money, it will be \ of 100 per cent., equal to 2j per cent. Hence the following Rule. 3Iake the difference hetween the purchase and selling prices the numerator, and the purchase price the denominator ; reduce to a decimal, and the result will be the per cent. Or, Take such a part of 100 as the gain or loss is part of the purchase price. EXAJrrLES FOR PRACTICE. 2. A man bought a pair of horses for $275, and sold them for $330 ; what per cent, did be gain ? Ans. 20 per cent. 3. If a merchant buy cloth at $.60 a yard, and sell it for $.75 a yard, what does he gain per cent. ? 4. A speculator bought 108 barrels of flour at $4,624 a barrel, and sold it so as to gain $114,881; Avhat per cent, profit did he make? ^„^^. 93 per cent. 5. Bought sugar at 8 cents a pound, and sold it for 9^ cents a pound ; what per cent, was gained ? 6. A drover bought 150 head of cattle for $42 per head, and sold them for $5400 ; what Avas his loss per cent. ? Ans. 141 per cent. 7. If I sell for $15 what cost me $25, what do I lose per °^'''^- ^ A71S. 40 per cent. 8. Bought paper at $2 per ream, and sold it at 25 cents a quire; what was the gain per cent. ? Ans. 150 per cent. 0. If I sell ^ of an article for f of its cost, wdiat is gained P^^ cent. ? Ans. 50 per cent. 10. If 4 of an article be sold for what ^ of it cost, what is ' tbe loss per cent. ? Ans. 37^ per cent. 220 PERCENTAGE. 11. If I sell 3 pecks of clover-seed for what one bushel cost me, what per cent, do I gain ? A7is. 33^ per cent. 12. A, having a debt against B, agreed to take $.87^- on the dollar ; what per cent, did A lose ? 13. A grocer bought 7 cwt. 20 lb. of sugar, at 7 cents a pound, and sold 3 cwt. 42 lb. at 8 cents, and the remainder at 8^ cents ; what was his gain per cent. ? Ans. IS^^g per cent. 14. Bought 2 hogsheads of wine, at $1.25 a gallon, and sold the same at $1.G0 ; what was the whole gain, and what the gain per cent. ? Aiis. Gain 2S per cent. 15. A grain dealer bought corn at $.55 a bushel and sold it at $.66, and wheat for $1.10, and sold it for $1.37^; upon which did he make the greater per cent. ? Ans. 5 per cent., upon the wheat. CASE III. S6*3. To find the selling price, when the cost and the gain or loss per cent, are given. 1. Bought a horse for $136 ; for how much must he be sold to gain 25 per cent. ? OPERATION. Analysis. Since $1 of cost $1 -f- .25 =z $1.25. sells for $1.25, $136 of cost will $1.25 X 136 z=z $170, Ans. sell for 136 times 81.2.5, which n ioo _1_ _?.i. — a->f. — 5 equals -Si 70, the selhng price. '^i, 1 o'i -f 1^0 — 1 ''^~^' Or, since the cost is \^, and $136 X i = $170, A71S. ^Yic gain ^2J_, the selling price Mill be |3lrr:i of the cost, or I of $136 = $170. If the horse had been sold at a loss of 25 per cent., then $1 of cost would have sold for $1 minus .25, or $.75, &c. Hence, Rule. 3Iultiphj $1 increased hy the gain or diminished hij the loss per cent, hy the number denoting the cost. Or, Take such a part of the cost as is equal to \%% increased or diminished hy the gain or loss per cent. Case in is what ? Give explanation. Rule. PROFIT AND LOSS. 221 EXAMPLES FOR rRACTICE. 2. If 124- liundred weight of sugar cost $140, how must it be sold per pound to gain 25 per cent. ? Ans. 14 cents. 3. Bought a hogshead of molasses for 30 cents a gallon, and paid IGf per cent, on the prime cost, for freight and cart- age ; how much must it sell for, per gallon, lo gain 334- per cent, on the whole cost? Jns. $.4G§. 4. For what price must I sell coifee that cost 10^ cents a pound, to gain 17^ per cent.? 5. If I am compelled to sell damaged goods at a loss of 15 per cent., how should I mark goods that cost me $.02^ ? $1.20? $3.87^? Ans. $.53^; $1.02; $3.29|." 6. A man, wishing to raise some money, offers his house and lot, which cost him $3240, for 18 per cent, less than cost; "A^'hat is the price ? 7. C bought a farm of 120 acres, at $28 an acre, paid $480 for fencing, and then sold it for 12^ per cent, advance on the Avhole cost; what was his whole gain, and wdiat did he receive an acre ? Ans. $480 gain ; $36 an acre. 8. Bought a cask of brandy, containing 52 gallons, at $2. GO per gallon ; if 7 gallons leak out, how must the remain- der be sold per gallon, to gain 37^ per cent, on the cost of the whole? Ans. $4.13^. 9. A merchant bought 15 pieces of broadcloth, each piece containing 23^ yards, for $840, and sold it so as to gain 18|- per cent, ; how much did ho receive a yard ? CASE IV. ^63. To find the cost, when the selhng price and the gain or loss per cent, are given. 1. A merchant sold cloth for $4.80 a yard, and by so doing made 33-J j^er cent. ; how much did it cost ? OPERATION. 1 + .331 = 1.33^ ; $4.80 ^ 1.33i = $3.G0, Ans. Or, $4.80 = f of the cost ; $4.80 -^f = $3.60. Case IV is what ? 222 PERCENTAGE. Analysis. Since the gain is 33^ per cent, of the cost, $1 of the cost, increased by !33^ per cent., will be what $1 of cost sold for : therefore there will be as many dollars of cost, as l.oo^- is con- tained times in $4.80, or $3.60. Or, since he gained 33^ per cent. = 1 of the cost, $4.80 is f of the cost ; $4.80-:- |- = $3.60, Note. If the rate per cent, be loss, we subtract it from 1, instead of adding it. Hence the foUowuig Rule. Divide the selling price hy 1 increased hij the gain or diminished hy the loss per cent., expressed decimally, or in the form of a common fraction, and the quotient will he the cost. EXAMPLES FOR PRACTICE. 2. By selling sugar at 8 cents a pound, a merchant lost 20 per cent. ; what did the sugar cost him ? Ans. 10 cents. 3. Sold flour for §6.12^ per barrel, and lost 12J- per cent. ; what was the cost ? Ans. $7.00. 4. A grocer, by selling tea at $.96 a pound, gains 28 per cent. ; how much did it cost him ? Ans. '^.1 b. 5. Sold a quantity of flour for $1881, which was ISf per cent, more than it cost ; how much did it cost ? 6. Sold 25 barrels of apples for $G9.75, and made 24 per cent. ; how much did they cost per barrel ? 7. Sold 9i cwt. of sugar at $8^ per cwt., and thereby lost 12 per cent. ; how much was the whole cost? 8. Having used a carriage six months, I sold it for $9G, which was 20 per cent, below cost ; what would I have received had I sold it for lo per cent, above cost? Ans. $138. 9. B sells a pair of horses to C, and gains 12^ per cent. ; C sells them to D for $570, and by so doing gains 18J per cent.; how much did the horses cost B? Ans. $42G.66j. 10. A grocer sold 4 barrels of sugar for $24 each ; on 2 barrels he gained 20 per cent., and on the other 2 he lost 20 per cent. ; did he gain or lose on the whole ? Ans. Lost $4. 11. A person sold out his interest in business for $4900, which was 40 per cent, more than 3 times as much as he began with; how much did he begin with ? Ans. $1166.66|. Give explanation. Rule. INSURANCE. 223 INSURANCE. QSl. Insurance on property is security guaranteed by one pcirty to another, for a stipulated sum, against the loss of that property by fire, navigation, or any other casualty. 1365. The Insurer or ITnclerwriter is the party taking the risk. S66. Tiie Insured is the party protected. ISIIT. The Policy is the written contract between the parties. S^§. The Premium is the sum paid by the insured to the insurer, and is estimated at a certain rate per cent, of the amount insured, which rate varies according to the degree of hazard, or class of risk. Note. As a security against fraiid, most msiirance companies take risks at not more than two tlni'ds tlie full value of the property insured. S6^. To find the premium when the rate of insur- ance and the amount insured are given. 1. What must I pay rinnually for insuring my house to the amount of $.3250, at 1^ per cent, premium ? opERATiox. Analysis. We $3250 X -01^ or .0125 = $40.G25. multiply the amount Or 11 npr pt — ^&^ — J-- insured, $3250, by V-^li ^4 PCI »-t- iTSTS — so ' , til S3250 X ,V = SJ0.C2i. tltXii^. 'JZ $40,625, is the premium. Or, the rate, 11 per cent., is ^r= J^ of the amount insured, and -gL of $3250 is $40,621. Hence the Rule. 3faUiply the amount insured hj the rate per cent., and the product will be the premium. Or, Take such a part of the amount insured as the rate is part of 100. Define insurance. Insurer, or underwriter. Policy. Premium. To what amount can property usually be msured ? Give analysis of example 1. Rule. 224 PERCENTAGE. EXAMPLES FOR PRACTICE. 2. TVhat is the premium on a policy for $750, at 4 per cent. ? Ans. $30. 3. "Wliat premium must be paid for $4572.80 insurance, at 2J- percent.? Ans. $114.32. 4. A house and furniture, valued at $5700, are insured at If per cent. ; what is the premium ? Ans. $99.75. 5. A vessel and cargo, valued at $28400, are insured at S^ per cent. ; what is the premium ? Ans. $994. 6. A woolen factory and contents, valued at $55800, are insured at 2f per cent. ; if destroyed by fire, what would be the actual loss of the company ? Ans. $54237.60. 7. What must be paid to insure a steamboat and cargo from Pittsburg to New Orleans, valued at $47500, at f of 1 percent.? Ans. $356.25. 8. A gentleman has a house, insured for $8000, and the fur- niture for $4000, at 2 1 per cent,; what premium must he pay? Ans. $285. 9. A cargo of 4000 bushels of wheat, worth $1.20 a bushel, is insured at f of 1^ per cent, on % of its value ; if the cargo be lost, how much will the owner of the wheat lose ? Avs. $1036. 10. What will it cost to insure a factory valued at $21000, at |- per cent. ; and the machinery valued at $15400, at | per cent.? A71S. $264.25. TAXES. t 270, A Tax is a sum of money assessed on the person or property of an individual, for public purposes. 371. When a tax is assessed on property, it is apportioned at a certain per cent, on the estimated value. When assessed on the person, it is apportioned equally among the male citizens liable to assessment, and is called a poll tax. Each person so assessed is called a poll. "S\niat is a tax ? How is a ta.\ on property apportioned ? On the person, how ? TAXES. 225 *I73. Property is of two kinds — real estate, and personal property. ^?»?. Eeal Estate consists of immovable property, such a- lands, houses, itc. ^74. Eersonal Property consists of movable property, such as money, notes, furniture, cattle, tools, &c. ^7*5. An Inventory is a written list of articles of proper- ty, with their value. SI^H. Before taxes are assessed, a complete inventory of all the taxable property upon which the tax is to be levied must be made. If the assessment include a poll tax, then a complete list of taxable polls must also be made out. I. A tax of $3165 is to be assessed on a certain town; the valuation of the taxable property, as shown by the as- sessment roll, is $000,000, and there are 220 polls to be as- sessed 75 cents each ; what will be the tax on a doUai-, and how much will be A's tax, whose property is valued at $3750, and who pays for 3 polls? OPERATION. $.75 X 220 izr $165, amount assessed on the polls. $31(35 — $165 = $3000, amount to be assessed on the property. $3000 -:- $600,000 = .005, tax on $1. $3750 X -005 z=. $18.75, A's tax on property. $.75 X 3 =: $2.25, A's tax on 3 polls. $18.75 + $2.25 z= $21, amount of A's tax. Hence the following Rule. I. Find the amount of poll tax, if any, and subtract this sum from the whole amount of tax to be assessed. II. Divide the sum to be raised on property, hy the xohole amount of taxable property, and the quotient will be the per cent., or the tax on one dollar. III. Midtiply each man's taxable property by the per cent., or the tax on $1, and to the product add Jus poll tax, if any ; the result will be the whole amount of his tax. A\Tiat is real estate ? Personal property ? An inventory ? Explain the process of levying a state or other tax. Rule. 10* 22G PERCENTAGE. Note. Having fomicl the tax on ^'1, or the per cent., -which in the preceding example -vve find to be 5 mills, or 1 per cent., the operation of assessing taxes may be greatly facilitated by finding the tax on i^2, $3, Sec, to ,$10, and 'then on $20, $30, &c., to $100, and arranguig the nuiiibcrs as in the following TABLE. Trop. Tax. Prop. Tax. Prop. Tax. Prop. Tax. $1 gives $.005 $10 $.05 $100 $ .50 $1000 $5.00 2 " .01 20 .10 200 1.00 2000 10. 3 " .015 30 .15 300 1.50 3000 15. 4 " .02 40 .20 400 2.00 4000 20. 5 " .025 50 .25 500 2.50 5000 25. 6 " .03 60 .30 600 3.00 6000 30. 7 " .035 70 .35 700 3.50 7000 35. 8 " .04 80 .40 800 4.00 8000 40. 9 " .045 90 .45 900 4.50 9000 45. (( (( 11 (( « i( (( (i (( (( « i( $15.00 4.00 .20 .025 1.50 EXAMPLES FOR PRACTICE. 2. According to the conditions of the last example, liow much would be a person's tax whose property Avas assessed at $3845, and who paid for 2 polls ? Finding the amount from the table, The tax on $3000 is . . . 800 " 40 " 5 " 2 polls " Total tax is $20,725 3. How mucli would be Ws tax, who was assessed for 1 poll, and on property valued at $5390 ? A}is. $27.70. 4. A tax of $9190.50 is to be assessed on a certain village ; the property is valued at $1400000, and there are 2981 poll:^, to be taxed 50 cents each ; what is the assessment on a dollar ? what is C's tax, his property being assessed at $12450, and ho paying for 2 polls ? Ans. $.005^ on $1 ; $09.47^, C's tax. 5. "What is the tax of a non-resident, having property in the same village valued at $5375 ? Ans. $29.5025. Explain tho table and its use. CUSTOM HOUSE BUSINESS. 227 G. A mining corporation, consisting of 30 persons, are taxed $4342.75; their property is assessed for $188000, and " each poll is assessed G2J- cents ; what per cent, is their tax, and hovv much must he pay whose share is assessed for $2500, and who pays for 1 poll ? Ans. 2^^ per cent.; $58,125. 7. In a certain county, containing 25482 taxable inhab- itants, a tax of $103294.60 is assessed for town, county, and state purposes ; a part of this sum is raised by a tax of 30 cents on each poll ; the entire valuation of property on the as- sessment roll is $38260000 ; what per cent, is the tax, and how much will a person's tax be who pays for 3 polls, and whose property is valued at $9470 ? Ans. to last, $24,575. 8. The number of polls in a certain school district is 225, and the taxable property $1246093.75 ; it is proposed to build a union school house at an expense of $10000 ; if the poll tax be $1.25 a poll, and the cost of collecting be 2^- per cent., what will be the tax on a dollar, and how much will be E's tax, who pays for 1 poll, and has property to the amount of $11500 ? Ans. $.008, tax on $1 ; $93.25, E's tax. 9. In a certain district the school was supported by a rate- bill ; the teacher's wages amounted to $200, the fuel and other expenses to $75.57 ; the public money received was $98, and the whole number of -days' attendance was 3946 ; A sent 2 pupils 118 days each; how much was his rate-bill ? Ans. $10.62 CUSTOM HOUSE BUSINESS. *|'J"7'. Duties, or Customs, are taxes levied on imported goods, for the support of government and the protection of home industry. S78. A Custom House is an office established by govern- ment for the transaction of business relating to duties. ^7I>. A Port of Entry is a seaport town having a cus- tom house. DeSue duties. A custom house. 228 PERCENTAGE. 980. Tonnage is a tax levied upon a vessel, independent of its cargo, for the privilege of coming into a port of entry. 281. Revenue is the income to government from duties and tonnage. Duties are of two kinds — ad valorem and specific. 982. Ad Valorem Duty is a sum computed on the cost of goods in the country from which they were imported. 983. Specific Duty is a sum computed on the weight or measure of goods, without regard to their cost. 284. An Invoice is a bill of goods imported, showing the quantity and price of each kind. 285. By the New Tariflf Act, approved March 2, 1857, all duties taken at the U. S. custom houses, are ad valorem. In collecting customs, it is the design of government to tax only so much of the merchandise as will be available to the importer in the market. The goods are weighed, measured, gauged, or imported, in order to ascertain the actual quantity and value received in port; and an allowance is made in every case of waste, loss, or damage. 286. Tare is an allowance of the weight of the package or covering that contains the goods. It is ascertained by actually weighing one or more of the empty boxes, casks, or coverings. In common articles of importation, it is sometimes computed at a certain per cent, previously ascertained by frequent trials. 287'. Leakage is an allowance on liquors imported in casks or barrels. 288. Breakage is an allowance en liquors imported in bottles. Note. Actual leakage or breakage is allowed, tlicre being no fixed or legal rato. 289. Gross Weight or Value ia the weight or value of the goods before any allowance has been made. 290. Net Weight or Value is the weight or value after all allowances have been deducted. Define Tonnage. Revenue. Ad valorem duty. Spiecifio duty. An invoice^ Tare. Leakage. Breakage. Gross weight or value. Net ■weight or value. I CUSTOM HOUSE BUSINESS. 229 Note. Draft is an allowance for the waste of certain ai'ticles, and is made only for statistical purposes ; it does not atfect the amount of duty. The rates of this allowance are as follows : On 112 1b , lib. Above 112 lb. and not exceeding 224 lb., 2 lb. " 224 lb. " " " 336 lb., 3 lb. " 336 1b. " " " 1120 lb., 4 1b. " 1120 1b. " " " 2016 1b., 7 lb. " 2016 1b 9 1b. 1. What is the duty, at 24 per cent., on 50 gross of London ale, invoiced at $1.20 per dozen, 2J per cent, being allowed for breakage ? OPERATION'. Analysis. TVe S1.20 X 12 X 50 = ST20, gross value, first find the cost of $720 X .025 = 818, leakage. the ale, at the in- §720 — 818 = $702, net value. voice price,which is $702 X .24 = $168.48, duty. ^^^O. From this sum we deduct the allowance for breakage, $18, and compute the duty on the re- mainder. Kence the following ^ KuLE.. Deduct allowances, if necessary, and compute the duty, at tlie given rate, on the net value. Note. — In the following examples, the legal rate of duty will be given, according to the Tariff of 1851. EXAMPLES FOR PRACTICE. 2. What is the duty at 19 per cent, on 224 yards of plaid silk, invoiced at $.95 per yard ? Ans. $40.43 + . 8. What is the duty at 24 per cent, on 50 barrels of sperm oil, each containing originally 31 J gallons, invoiced at $.54 per gallon, allowing 2 per cent, for leakage? Ans. $200.03 + . 4. What is the duty at 15 per cent, on 175 bags of Java coffee, each containing 115 lbs., valued at 15 cents per pound? Ans. 1452.81+ 5. John Jones imported from Havana 25 hhds. of W. I. molasses, which was invoiced at 36 cents per gallon ; allowing 5 per cent, for leakage, what was the duty at 24 per cent. ? Ans. $135,399 + . Define draft. Give analysis. Rule. 230 SIMPLE INTEREST. SIMPLE INTEREST. S91« Interest is a sum paid for the use of money. *^i>S. Principal is the sum for the use of which interest is ])aid. !!l9el. Eate per cent, per annum is the sum per cent, paid for t!ie use of $100 annually. Note. The rate per cent, is commonly expressed decimally, as him- dredths (331). !S!>4. Amoiint is the sum of the principal and interest. ^9<5. Simple Interest is the sum paid for the use of the principal only, during the whole time of the loan or credit. S®©. Legal Interest is the rate per cent, established by law. It varies in different States, as follows : Alabama, 8 per cent. Arkansas, 6 Connecticut, G Delaware, 6 Dist. of Columbia, ... 6 Florida, 8 Georgia, 7 Illinois, 6 Indiana, 6 Iowa, 7 Kentucky, 6 Louisiana 5 Maine, (3 jNIaryland, 6 Massachusetts, 6 Michigan, 7 (( (( li l( 1( (( (( (( (( l< l( (1 (i (t « Mississippi, 8 per cent. Missouri, G " " New Hampshire, . . . .G " " New Jersey, 6 " " New York, 7 " " North Carolina, 6 " " Ohio, G " " Pennsylvania, G " " lihode Island, G " " South Carolina, 7 Tennessee, G Texas, 8 United States (debts), 6 Vermont, 6 Virginia, 6 Wisconsin, 7 Note, ^^^len the rate per cent, is not spccilicd, in accoimts, notes, mortgages, contracts, &c., the legal rate is always understood. 29t, Usury is illegal interest, or a greater jier cent, tha.i the legal rate. CASE I. 998. To find the interest on any sum, at any rate per cent., fur years and months. Define interest. Principal. Rate per cent, per annum. Amount. What is simple interest ? Legal interest ? Usury ? Case I ? PERCENTAGE. 231 In i^Grcentage, anj' per cent, of any given number is so many hundredtlis of that number ; but in interest, any rate per cent, is confined to 1 year, and the per cent, to be obtained of any given number is greater tlian the rate per cent per annum if the time be 7nore tlian 1 year, and less than the rale per cent, per annum if the time be less tlian 1 year. Thus, tlie interest on any sum, at any rate per cent., for 3 yeai-s G montlis, is 3i times the interest on the same sum for 1 year; and tlie interest for 3 months is ^ of the interest for 1 year. 1. Wliat is the interest on S75.19 for 3 years G months, at G per cent. ? OPERATIOX. $7o.l9 .06 Analysis. The interest on $75.19, for 1 yr., ^ at 6 per cent., is .06 of the principal, or $4.5114, ' and the interest for 3 yr. 6 mo. is 3 A ^r 3^ times ^2 the interest for 1 jr., or $4.5114 X 3-^, which is 2-557 $15,789 -j-, the Ans. Hence, the following 135342 $15.7899, Ans. Rule. I. Multiply the principal hy the rate per cent, and the product loill be the interest for 1 year. II. Multiply this product hy the time in years and fractions of a year, and the result will be the required interest. EXAMPLES FOR PPvACTICE. 2. What is the interest of $150 for 3 years, at 4 per cent. ? Ans. $18. 3. What is the interest of $328 for 2 years, at 7 per cent. ? 4. What is the interest of $125 for 1 year G months, at 6 percent.? Ans. $11.25. 5. What is the interest of $200 for 3 years 10 months, at 7 per cent ? Ans. $53. G6 -f- . 6. What is the interest of $76.50 for 2 years 2 months, at 5 per cent. ? Ans. . $8,287 -f . Explain the difference between percentage and interest. Give analysis. Rule. 232 SIMPLE INTEREST. 7. What is the interest of $1276.25 for 11 months, at 7 percent.? Ans. $81.89 + . 8. What is the interest of $2569.75 for 4 years 6 months, at 6 per cent. ? 9. What is the interest of $1500.60 for 2 years 4 months, at 6^ per cent. ? Ans. $218.8375. 10. What is the amount of $26.84 for 2 years 6 month?, at 5 percent.? Ans. $30,195. 11. What is the amount of $450 for 5 years, at 7 per cent. ? 12. What is the interest of $4562.09 for 3 years 3 months, at 3 per cent. ? A7is. $444.80 + . 13. What is the amount of $3050 for 4 years 8 months, at 5^ per cent. ? Ans. $3797.25 + . 14. What is the interest of $5000 for 9 months, at 8 per cent.? Ans. $300. 15. If a person borrow $375 at 7 per cent., liow mucli will be due the lender at the end of 2 yr. 6 mo. ? 16. What is the interest paid on a loan of $1374.74, at 6 per cent., made January 1, 1856, and called in January 1, 18G0? Ans. $329,937 + . 17. If a note of $605.70 given May 20, 1858, on interest at 8 percent., be taken up May 20, 1861, what amount will then be due if no interest has been paid? Ans. $751,068. CASE II. 299. To find the interest on any sum, for any time, at any rate per cent. The analysis of our rule is based upon the following Obvioics Relations between Time mid Interest. I. The interest on any sum, for 1 year, at 1 per cent., is .01 of that sum, and is equal to the principal with the scparatrix removed two places to the left. II. A month being j'j of a year, yV of the interest on any sum for 1 year is the interest for 1 month. "What is Case II ? Give the first relation between time and interest. Second. PERCENTAGE. 233 III. The interest on any sum for 3 days is t^^j =z -^\j = .1 of the interest for 1 month, and any number of days may readily be reduced to tetiihs of a month by dividing by 3. IV. The interest on any sum, for 1 month, multiplied by any "■iven time expressed in months and tenths of a month, will produce the required interest. I. What is the interest on S724.G8 for 2 yr. 5 mo. 19 da., at 7 per cent. ? OPERATION. Analysis. We remove 2yr. 5mo. 19da. = 29.G^mo. the separatrix in the given in \ (S'Toi/'Q principal two places to the ^ leit, and we have §/.24(j8, $.6039 the interest on the given sum 29.64- for 1 year at 1 per cent. -— — (300 I.). Dividing this by "^^^'^ 12, we have $.6039, the inter- 36234: est for 1 month, at 1 per cent. 54351 (II.) Multiplying this 12078 quotient by '29.6^, the time ' ~ expressed in months and deci- $17.895o7 jj^j^jg Qf ^ j^Qj^^i^^ (jjj_ jy_^) we have $17.89557, the in- $125.26899, Ans. terest on the given sum for the given time, at 1 per cent. (rV.). And multiplying this product by 7 (7 times 1 per cent.), we have $125,208 -|-, the interest on the given principal, for the given time, at the given rate per cent. Hence, Rule. I. Remove the separatrix in the given principal tivo places to the left ; the result will he the interest for 1 year, at 1 per cent. II. Divide this interest hy 12 ; the result ivill be the interest for 1 months at 1 per cent. III. Multiply this interest hy the given time expressed in months and tenths of a month ; the result will he the interest for the given time, at 1 per cent. IV. Multiply this interest hy the given rate ; the product will be the interest required. Give the thh-d. Fomth. Give analysis. Rule. 234 SIMPLE INTEREST. Contractions. After removing the' separatrix in the principal two places to the left, the result may be regarded either as the in- terest on the given principal for 12 months at 1 per cent., or for 1 month at 12 per cent. If we regard it as for 1 mouth at 12 per cent., and if the given rate be an aliquot part of 12 per cent., the interest on the given principal for 1 month may readily be found by taking such an aliquot part of the interest for 1 month as the given rate is part of 12 per cent. Thus, To find the interest for 1 month at G per cent., remove the sep- aratrix two places to the left, and divide by 2. To find it at 3 per cent., proceed as before, and divide by 4 ; at 4 per cent., divide by 3 ; at 2 per cent., divide by 6, &c. SIX PER CENT. METHOD. 300. By referring to S9@ it will be seen that the legal rate of interest in 21 States is 6 per cent. This is a sufficient reason for introducing the following brief method into this work : Analysis. At 6 per cent- per annum the interest on $1 For 12 months is $.06. " 2 months {j%-=i- of 12 mo.) " .01. " 1 month, or 30 days( yL of 12 mo.) " .00J=$.005 {^^ of $-0G)- " 6 days (I of 30 days)" " .001. " 1 " (>-of Gda.==gVof30days) " .000^. Hence we conclude that, 1st. The interest on $1 is S.005 per mouth, or ^.01 for every 2 months ; 2d. The interest on $1 is ^.OOOi per day, or 8.C01 for every G days. From these principles we deduce the Rule. I. To find the rate : — Call every year $.0G, every 2 months $.01, every 6 days $.001, and any less number o/ days sixtJis of 1 mill. II. To find the interest : — Jltdttply the 2)rincqKd hy the rate. Notes. — 1. To find the interest at any other rate per cent, b^- this method, first find it at G per cent., and tlien increase or diminish the result by as many tnnes itself as the given rate is grcator or less than 6 per cent. Thus, for 7 per cent, add i^, for 4 per cent, subtract ^, &c. What contractions are given? Giveanalj'sisof the C per cent, method. Rule. Its application to any other rate ])er cent. SIMPLE INTEREST. 235 2. The interest of $10 for 6 clays, or of $1 for GO ilays, is $.01. Therefore, if the principal be less thari $10 ami the tiino less than (i days, or the principal less than $1 and tlie time less than GO days, the interest will be less tlian $.01, and may be disregardeu. 3. Since the interest of $1 for GO days is $.01, the interest of $1 for any number of days is as many cents as 60 is contained times in the number of days. Therefore, if any principal be multiplied by tlie num- ber of days in any given number of moutlis and days, and tlie proiluct divided by GO, the result will be the interest in cents. That is, Jfii'ii- j^ly the principal by' the number of days, divide the product bi/ GO, and punit ojf two decimal places in the quotient. The result will be the interest in ihe same daiuwination as th& principal. EXAMPLES FOR PRACTICE. 2. What is the interest of $100 for 7 years 7 months, at 6 per cent. ? Ans. $45.50. 3. What is the amount of $47.50 for 4 years 1 month, at 9 per cent. ? j47is. $G4.95G -j- . 4. What is the amount of $2000 for 3 months, at 7 per cent.? ^ ' Ans. $2035. 5. AVhat is the interest of $250 for 1 year 10 months and 15 days, at G per cent. ? A7is. $28.12^. G, What is the interest of $30.75 for 2 years 4 months and 12 days, at 7 per cent. ? Ans. $G.088 + . 7. What is the amount of $84 for 5 years 5 months and 9 day.«!, at 5 per cent. ? 8. What is the interest of $51.10 for 10 months and 3 days, at 4 per cent. ? 9. What is the interest of $175.40 for 15 morrths and 8 days, at 10 per cent. ? Ans. $22.31 + . 10. What is the amount of $1500 for G months and 24 days, at 7^ per cent. ? Ans. $1563.75. 11. What is the amount of $84.25 for 1 year 5 months and 10 days, at 6|- per cent. ? 12. What is the interest of $25 for 3 years 6 months and 20 days, at G per cent. ? Ans. $5.33^. 13. What is the interest of $112.50 for 3 months and 1 4ay, at 9^ per cent. ? Ans. $2.70 -}-• What contractions are given ? 236 PERCENTAGE. 14. What is the interest of $408 for 20 days, at G per cent.? Ans. $1.36. 15. "What is the interest of $500 for 22 days, at 7 per cent. ? IG. What is the amount of $4500 for 10 days, at 10 per cent.? Ans. $4512.50. 17. What is the amount of $1000 for 1 month 5 days, at Cf])ercent. ? A>is. $100G.56i. 18. Find the interest of $973.G8 for 7 months 9 days, at 4-1 per cent. 19. If I borrow $275 at 7 per cent., how much will I owe at the end of 4 months 25 days ? 20. A person bought a piece of property for $2870, and agreed to pay for it in 1 year and G months, with GJ- per cent, interest; what amount did he pay ? A)7s. $3149.825. 21. In settling with a merchant, I gave my note for $97.75, due in 11 months, at 5 per cent.; what must be paid when the note falls due ? A)is. $102.23+. 22. How much is the interest on a note of $384.50 in 2 years 8 months and 4 days, at 8 per cent. ? 23. What is the interest of $97.86 from May 17, 1850, to December 19, 1857, at 7 per cent. ? Ans. $51.98 -f-. 24. Find the interest of $35.G1, from Nov. 11, 1857, to Dec. 15, 1859, at G per cent. Ans. $4,474. 25. Required the interest of $50 from Sept. 4, 1818, to Jan. 1, 1860, at 3^ per cent. 26. Required the amount of $387.20, from Jan. 1 to Oct. 20, 1859, at 7 per cent. Ans. $408,957 -f . 27. A man, owning a furnace, sold it for $6000 ; the terms were, $2000 in cash on delivery, $3000 in 9 months, and llie remainder in 1 year 6 months, witli 7 per cent, interest; what was the whole amount paid ? Ans. $G2G2.50. 28. Wm. Gallup bought bills of dry goods of Geo. Bliss & Co., of New York, as follows, viz.: Jan. 10, 1858, $350; April 15, 1858, $150 ; and Sept. 20, 1858, $550.50 ; he bought on time, paying legal interest ; what Avas the whole amount of his indebtedness Jan. 1, 1859? Ans. $1092.66 +. PARTIAL PAYMENTS. 237 PARTIAL PAYMENTS OR INDORSEMENTS. S© 3 . A Partial Payment is payment in part of a note, bond, or other obligation ; when the amount of a payment is written on the back of tlie obligation, it becomes a receipt, and is called an Indorsement. $2000. Springfield, Mass., Jan. 4, 1857. 1. For value received I promise to pay James Parish, or order, two thousand dollars, one year after date, with interest. George Jones. On this note were indorsed the following payments : Feb. 19, 1858, $400 June 29, 1859, $1000 Nov. 14, 1859, $520 What remained due Dec. 24, 1860? OPERATION. Principal on interest from Jan. 4, 1857, $2000 Interest to Feb. 19, 1858, 1 yr. 1 mo. 15 da., 135 " Amount, $2135 Payment Feb. 19, 1858, 400 Remainder for a new principal, $1735 Interest from Feb. 19, 1858, to June 29, 1859, 1 yr. 4 mo. 10 da., 141.69 Amount, $LS7G.69 Payment June 29, 1859, 1000. Eemainder for a new principal, $876.69 Interest from June 29, 1859, to Nov. 14, 1859, 4 mo. 15 da., 19.725 Amount, $896,415 Payment Nov. 14, 1859, 520. Remainder for a new principal, $376,415 Interest from Nov. 14, 1859, to Dec. 24, 1860, 1 yr. 1 mo. 10 da., 25.09 Remains due Dec. 24, 1860 $401.505 -f- What is meant by partial payment ? By an indorsement ? 233 PERCENTAGE. ^-^^^•^Q- New York, May 1, 1855. 2. For value received, we jointly and severally promise to pay Mason & Bro., or order, four hundred seventy-five dol- lars fifty cents, nine months after date, v.'ith interest. Jones, Smith & Co. The following indorsements were made on this note : Dec. 25, 1855, received, $50 July 10, 185G, " 15.75 Sept. 1, 1857, " 25.50 June 14, 1858, " 104 How much was due April 15, 1859 ? OPEKATION. Principal on interest from ^lay 1, 1855, $475.50 Interest to Dec. 25, 1855, 7 mo. 24 da., 21.63 Amount, $497.13 Pajinent Dec. 25, 1855, 50. Ijlemainder for a new principal, $447.13 Interest from Dec. 25, 1855, to June 14, 1858, 2 yr. 5 mo. 19 da., 77.29 Amount, $524,42 PajTnent July 10, 1856, less than interest^ then due, > .$15.75 Pa}Tnent Sept. 1, 1857, ) 25.50 Their sum less than interest then due, . . . $41.25 Paj-ment June 14, 1858, 104. Their sum exceeds the interest then due, $145.25 Remainder for a new principal, $379.17 Interest from June 14, 1858, to April 15, 1859, 10 mo. 1 da., 22.19 Balance due April 15, 1859, $401.36 -|- These examples liave been wrouglit according to llic method prescribed by the Supreme Court of the U. S., and are suf' ficient to illustrate the following PARTIAL PAYMENTS. 289 United States Rule. I. Find the amount of the given principal to the time of the frst payment, and if this payment exceed the interest then due subtract it from the amount obtained, and treat the remainder as a neiv principal. II. But if the interest be greater than any payment, cast the interest on the same principal to a time when the sum of the payments shall equal or exceed the interest due ; subtracting the sum of the payments from, the amount of the principal, the re- mainder will form a new principcd, on ivkich interest is to be computed as before. ^^^^•'^^- San Francisco, June 20, 1858. 3. Three years after date we promise to pay Ross & "Wade, or order, five hundred fourteen and -^^^ dollars, for value received, with 10 per cert, interest. Wilder & Bro. On this note Avere indorse 1 the folloAving payments :' Nov. 12, 1858, $105.50 ; March 20, 18G0, S200 ; July 10, 1860, $75.60. How much remains due on the note at the time of its maturity? Ans. $242.12 -f. ^^^^"^- Charleston, May 7, 1859. 4. For value received, I promise to pay George Babcock three thousand dollars, on ^demand, with 7 per cent, interest. John May. On this note were indorsed the following payments : — ' Sept. 10, 1859, received 625 -Jan. 1, 1860, " 500 Oct. 25, 1860, " 75 April 4, 1861, " 1500 How much was due Feb. 20, 1862 ? Ans. $1344.35 -f . Give the ITnited States Court rule for computing interest Adhere partial pajTuents have been made. 2-iO PERCENTAGE. ^QI'^titV New Orleans, Aug. 3, 1850. 5. One year after date I promise to pay George Bailey, or order, nine hundred twelve -^jj dollars, with 5 per cent, in- terest, for value received. James Powell. The note was not paid when due, but was settled Sept. 15, 1853, one payment of §250 having been made Jan. 1, 1852, and another of $316.75, May 4, 1853. How much was due at the time of settlement ? Ans. $4G7.53 -|- . ^l^^-5^^- Cincinnati, April 2, 1860. 6. Four months after date I promise to pay J. Ernst & Co. one hundred eighty-four dollars fifty-six cents, for value received. S. Anderson. The note was settled Aug. 26, 1862, one payment of §50 having been made May 6, 1861. How much was due, legal interest being 6 per cent. ? Atis. §154.188 -\- . Note. A note is on interest after it becomes due, if it contain no mention of interest. 7. Mr. B. gave a mortgage on liis farm for §6000, dated Oct. 1, 1851, to be paid in 6 years, with 8 per cent, interest. Three months from date he paid $500 ; Sept. 10, 1852, $1126 ; March 31, 1854, $2000 ; and Aug. 10, 1854, $876.50. How much was due at the expiration of the time ? Ans. $3284.84 -f- . S03. The United States rule "for partial payments has been adopted by nearly all the States of the Union ; the only prominent exceptions are Connecticut, Vermont, and New Hampshire. Connecticut Rule. I. Payments made one year or more from the time the in- terest commenced, or from another payment, and payments less than the interest due, are treated according to the United States ride. Gire Connecticut rule for partial pajments. PARTIAL PAYJIENTS. 241 II. Payments exceeding the Interest due, and made within one year from the time interest commenced, or from a former payment, shall draw interest for the balance of the year, jjro- vided the interval does not extend beyond the settlement, and the amount must be subtracted from the amount of the principal for one year ; the remainder will be the new principal. III. If the year extend beyond the settlement, then find the amount of the payment to the day of settlement, and subtract it from the amount of the principal to that day ; the remainder 10 ill be the sum due. ^^^'^- Woodstock, Ct., Jan. 1, 1858. 1. For value received, I promise to pay Henry Bowen, or order, four hundred sixty dollars, on demand, with interest. James Marshall. On this note are indorsed the following payments : April 16, 1858, $148 ; March 11, 1860, $75 ; Sept. 21, 1860, $56. How much was due Dec. 11, 1860? Ans. $238.15-}-. 30«l. A note containing a promise to pay interest an- nucdly is not considered in law a contract for any thing more than simple interest on the principal. For partial payments on such notes, the following is the Vermont Rule. I. Find the amount of the principal from the time interest commenced to the time of settlement. II. Find the amount of each payment from the time it was made to the time of settlement. III. Subtract the sum of the amounts of the payments from the amount of the principal, and the remainder will be the sum due. S600. Rutland, April 11, 1856. 1. For value received, I promise to pay Amos Cotting, or order, six hundred dollars on demand, with interest annually. John Brown. Give the Connecticut rule for partial payments. The Vermont rule. E.P 11 242 PERCENTAGE. On this note were indorsed the following payments : Aug. 10, 1856, $156; Feb. 12, 1857, $200; June 1, 1858, $185. What was due Jan. 1, 1859 ? Ans. $105.50-j-. S^4. In New Hampshire interest is allowed on the an- nual interest if not paid when due, in the nature of damages for its detention ; and if payments are made before one year's interest has occurred, interest must be allowed on such pay- ments for the balance of the year. Hence the followinrf New Hampshire Rule. I. Find the amount of the principal for one year, and de^ duct from it the amount of each payment of that year, from the time it was made up to the end of the year ; the remainder will he a new principal, with which proceed as before. II. If the settlement occur less than a year from the last an- nual term of interest, malce the last term of interest a jiart of a year, accordingly. ^^^^- Keene, N. H., Aug. 4, 1858. 1. For value received, I promise to pay George Cooper, or order, five hundred seventy-five dollars, on demand, with in- terest annually. David Greekman. On this note were indorsed the following payments : Nov. 4, 1858, $64; Dec. 13, 1859, $48; March 16, 1860, S248 ; Sept. 28, 1860, $60. What was due on the note Nov. 4, ISfiO? Ans. $215.3^. 3^«5. When no payment whatever is made, upon a note promising annual interest, till the day of settlement, in New Hampshire the following is the Court Rule. Compute separately the interest on the principal from the time the note is given to the time of settlement, and the interest on each year's interest from the time it should be paid to the time of settlement. The sum of the interests thus obtained, added to the principal, will he the sum due. The New Ilampshire rule. The New Hampshire court ride. PAKTIAL PAYMENTS. 243 ?:^ ^ Keexe, N. H., Feb. 2, 1855. 1. Three years after date, I promise to pay James Clark, or order, five hundred dollars, for value received, with interest annually till paid. ■ John S. Briggs. What is due on the above note, Aug. 2, 1859 ? A71S. $G49.40. Problems in Interest. SI06. In examples of interest there are five parts involved, the Principal, the Eate, the Time, the Interest, and thJ Amount. CASE I. 11©?. The time, rate per cent., and interest being given, to find the principal. 1. What principal in 2 years, at 6 per cent., will gain $31.80 interest ? OPERATION. Analysis. Since $1, in $.12, interest of $1 in 2years atGpercent. 2 years, at 6 per Cent., AviU $31.80 -^ .12 = $265, Ans. S"i'^ 8.12 interest, the prin- cipal thatAvill gain $31.80, at tlie same rate and time, must be as many dollars as $.12 is con- tained times in $31.80; dividing, we obtain $265, the required principal. Hence, Rule. Divide the given interest hy the interest of $1 for the given time and rate, and the quotient will he the -principal. EXAMPLES FOR PRACTICE. 2. What principal, at G per cent., will gain $28. 12^- in 6 years 3 months. ? Ans. $75. 3. What sum, put at interest for 4 months 18 days, at 4 per cent., will gain $9.20 ? Ans. $600. 4. What sum of money, invested at 7 per cent., will pay me an annual income of $1260 ? Ans. $18000. 5. What sum must be invested in real estate, yielding 10 per cent, profit in rents, to produce an income of $3370 "^ Ans. $33700. How many parts are considered in examples in interest f What are they ? What is Case I ? Give analysis. Rule. 214 PERCENTAGE. CASE II. 308. The time, rate per cent., and amount being given, to find the prineipal. 1. AVliat principal in 2 years G months, at 7 per cent., will amount to $88,125? OPEEATiON. Analysis. $1,175 Amt. of $1 in 2 years 6 months, at 7 per cent. Since $1, ill $88,125 H- 1.175 == $75, Ans. ^ ^'^,'''\ t ' montlis, at 7 per cent., will amount to 81.17.J, the principal that will amount to $88,125, at the same rate and time, must be as many dollars as $1,175 is contained times in $88,125 ; diviihng, we obtain $75, the reqmred principal. Hence the Rule. Divide the given amount hj the amount of $1 _/br the given time and rate, and the quotient will he the jjrincipal required. Wk* EXAMPLES FOR TRACTICE. ' 2. "What principal, at G per cent., will amount to $655.20 .,_ in 8 months ? ^ Ans. $630. ^| 3. What principal, at 5 per cent., will amount to $106,855 jj in 5 years 5 months and 9 days ? Ans. $81. 4. "What sum, put at interest, at 5J- per cent., for 8 years 5 months, will amount to $1897.545 ? Ans. $1297.09+. 5. What sura, at 7 per cent., will amount to $221,075 in 3 , years 4 months ? Ans. $179.25. G. What is the interest of that sum, for 11 years 8 days, at 10^- per cent., which will at the given rate and time amount to $857.54? Ans. $460.04. CASE III. P JJ01>. The principal, time, and interest being given, to find *Jie rate per cent. ,|| 1. I lent $450 for 3 years, and received for interest $G7.50; what was the rate per cent. ? Give case 11. Analysis. Hulc. Case III. PROBLEMS IN INTEREST. 245 OPERATION. Analysis. Since at $ 4.50 1 per cent. $450, in 3 3 years, -will gain $18. oO >^ , o - rt . interest, the rate per S 1 3. OO, int. of $150 for 3 years at 1 percent. ^ . i • t .i ' cent, at MJiicn the same $67.50 -^ 13.50 = 5 i.ercent., Ans. principal, in the same time, will gain $67.50, must be equal to the number of times $13.50 is contained in $67.50; di\ iJing, we obtain 5, the required rate per cent. Hence the Rule. Divide the given interest hy the interest on the prin- cipal for the given time at 1 per cent., and the quotient will he the rate ptc^' cent, required. EXAMPLES FOR PRACTICE. 2. If I pay S45 interest for the use of $500 3 years, •what is the rate per cent. ? Ans. 3. 3. The interest of $180 for 1 year 2 months 6 days is $12.78 ; what is the rate per cent. ? Ans. 6. 4. A man invests $2000 in bank stock, and receives a semi-annual dividend of $75 ; what is the rate jier cent. ? 5. At what per cent, must $1000 be loaned for 3 years 3 months and 29 days, to gain $183.18 ? Ans. 5^, 6. A man builds a block of stores at a cost of $21640, and receives for them an annual rent of $2596.80 ; what per cent, does he receive on the investment? Ans. 12. CASE IV. SIO. Principal, interest, and rate per cent. Leing given, to find the time. 1. In what time will $360 gain $86.40 intei'cst, at 6 per cent. ? orERATiON. Analysis. Since in $ 360 1 year $360, at 6 per .06 eent., will gain $21.60, the number of years in which the same princi- pal, at the same rate, will gain $86.40, will be $21.60 Iuterostof$3C0mlyearat6percent. ,^|,ichthe same princi- $86.40 H- 21,60 = 4 years, ylws. pal, at the same rate, Analysis. Rule. Case IV. Analysis. 2-i6 PERCENTAGE. as many as $21.60 Is contained times in $86.40 ; dividing, we ob- tLiin 4 years, the required time. Hence the Rule. Divide the given interest hy the interest on the prin- cipal for 1 year, and the quotient will he the time required in years and decimals. Note. The decimal part of the quotient, if any, may be reduced to months and days (by 209). EXAMPLES FOR PRACTICE. 2. The interest of $325 at G per cent, is $58.50 ; \vhat is the time ? j Ans. 3 years. 3. B loaned $1G00 at G jier cent, until it amounted to $2000 ; what was the time ? Ans. 4 years 2 months. 4. How long must §204 be on interest at 7 per cent., to g amount to $217.09 ? Ans. 11 months. 5. Enc-a'Tinsr in business, I borrowed $750 of a friend at 6 COO ' per cent., and kept it until it amounted to $942 ; how long did I retain it ? Ans. 4 years 3 months G days. G. IIow long will it take $200 to double itself at 6 per cent, simple interest ? Ans. 1 G years 8 months. 7. In what time Avill $G75 double itself at 5 per cent. ? Note. The time in years in which any sum wiU double itself may be foimd by dividuig 100 by the rate per cent. co:\iPorxD interest. 31 \l\. Compound Interest is interest on both principal and interest, when the interest is not paid when due. Note. The simple interest may be added to the prmcipal annually, semi-annually, or quarterly, as the parties may agree ; but the taldng of compound interest is not logal. 1. Wliat is the compound interest of $200, for 3 years, at G ])tr cent.? Rule. In what time wUl any sum double itself at interest ? "^Miat 'fi compound interest ? lii COMPOUND INTEREST. 24.7 OPERATION. $200 Principal for 1st year. $200 X -06 = 12 Interest for 1st year. $212 Principal for 2d year. $212 X -06 — 12.72 Interest for 2d year, $224.72 Principal for 3d year. $224.72 X -06 = 13.483 Interest for 3d year. $238,203 Amount for 3 years. 200.000 Given principal. $38,203 Compound interest. Rule. I. Find the amount of the given princij^al at the given rate for one year, and make it the -principal for the second year. II. Find the amount of this new principal, and make it the principal for the third year, and so continue to do for the given number of years. ^ III. Subtract the given principal from the last amount, and the remainder will be the compound interest. Notes. 1. A^Tien the interest is payable semi-annually or quar- terly, find the amount of the given principal for the first interval, and make it the principal for the second interval, proceeding in all respects as •svhen the interest is payable yearly. 2. When the time contams years, months, and days, find the amount for the years, upon which compute the interest for the months and days, and add it to the last amount, before subtracting. EXAMPLES FOR TKACTICE. 2. What is the compound interest of $500 for 2 years at 7 per cent. ? Ans. $72.45. 3. What is the amount of $312 for 3 years, at 6 per cent, compound interest ? Ans. $371.59-|-. 4. What is the compound interest of $250 for 2 years, payable semi-annually, at 6 per cent.? Ans. $31.37-f-. 5. What will $450 amount to in 1 year, at 7 per cent, com- pound interest, payable quarterly ? Ans. $482.33. 6. What is the compound interest of $236 for 4 years 7 months and G days, at G per cent. ? Ans. $72.6G-j-. Explain operation. Give rule. 248 PERCENTAGE. 7. "What is the amount of $700 for 3 years 9 montlis and 24 days, at 7 per cent, compound interest ? Ans. $906.55-|-. A more expeditious method of computing compound interest than the preceding, is by means of the following TABLE, Showing the amount 0/"$!, or £1, at 3, 4, 5, 6, and 7 per cent., compound interest, for any number of years, from 1 to 20. Yrs. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 3 per cent. 4 per cent. 5 per cent. 1.030,000 1.040,000 1.050,000 1.060,900 1.081,600 1.102,500 1.092,727 1.124,864 1.157,625 1.125,509 1.109,859 1.215,506 1.159,274 1.210,653 1.194,052 1.229,874 1.206,770 1.304,773 1.343,916 1.384,234 1.425,761 1.468,534 1.512.590 1.557,967 1.265,319 1.315,932 1.368,569 1.276,282 1.340,096 1.407,100 1.477,455 1.423,312 1.551,328 1.480,244 1.628,895 1.604,706 1.652,848 1.702,433 1.753,506 1.539,454 1.710,339 1.601,032jl.795,856 1.665,074 1.885,649 1.731,67611.979,932 1.800,044 ;2.078,928 6 per cent. 1.060,000 1.123,600 1.191,016 1.262,477 1.338,226 7 per cent. 1.418,519 1.503,630 1.593,848 1.689,479 1.790,848 1.872,981 2.182,875 1.947,900 2.292,018 2.025,817|2.406,619 2.106,849 2.526,950 1.898,299' 2.012,196' 2.132,928 2.260,904 2.396,558 1.806,111,2.191,123 2.653,298 2.540,352 2.692,773! 2.854,339| 3.025,600 3.207,135 1.07,000 1.14,490 1.22,504 1.31,079 1.40,255 1.50,073 1.60,578 1.71,818 1.83,845 1.96,715 2.10,485 2.25,219 2.40,984 2.57,853 2.75,903 2.95,216 !3.15,881 3.37,293 3.61,652 3.86,968 8. AVhat is the amount of $800 for 6 years, at 7 per cent. orERATION. From the table $1.50073 Amount of $1 for the time. 800 Principal. $1200.58400, Ans. 9. What is the compound interest of $120 for 15 years, at 5 per cent.? Ans. $129.47+. Of what use is the table in computing compound interest ? DISCOUNT. 249 10. "What is the amount of $.10 fur 20 years, at 7 per cent.? Aks. $.38 GOG. DISCOUNT. SI^. Discount is an abatement or allowance made for the payment of a debt before it is due. •Ilel. The Present "Worth of a debt, pa^-able at a future time without interest, is such a sum as, being put at legal in- terest, will amount to the given debt when it becomes due. 1. A owes I> $321, payable in 1 year; what is the pres- ent worth of the debt, the use of money being Avorth 7 per cent. ? OPERATiox. Analysis. The Am't of $1, 1.07 ) $321 ($300, Present value, amount of $1 for 321 1 year is $1.07; _ therefore the prcs- $321 Giveu sum or debt. ent worth of every 300 Present worth. $1.07 of the given $21 Discount. fl'^^^t is $1; and the present Avortli of $321 will be as many dollars as $1.07 Is contained times in $321. $321 -^ 1.07 = $300, Ans. Hence the following KuLE. I. Divide the given sum or debt hy the amount of $1 for tlte given rate and time, and the quotient will be the pres- ent worth of the debt. II. Subtract the present worth from the given svm or debt, and the remainder will be the discount. Note. The terms present worth, discount, and debt, are equivalent to principal, interest, and amount. Hence, Avhen the time, rate per cent., and amount are given, the principal may be found by (30§) ; and the interest by subtracting the principal from the amount. EXAMPLES FOR PRACTICE. 2. What is the present worth of $180, payable in 3 years 4 months, discounting at 6 per cent. ? Ans. $150. Define discount. Present worth. Give analysis. Rule, 11* 250 PERCENTAGE. 3. What is the present worth of a note for $1315.389, due ill 2 years 6 months, at 7 per cent.? Ans. $1119.48. 4. What is the present worth of a note for S8GG.038, due in 3 years 6 months and G days, when money is worth 8 per cent. ? What the discount ? Ans. $190.15-}-, discount. 5. What is the present worth of a debt for §1005, on which $475 is to be paid in 10 months, and the remainder in 1 year 3 months, the rate of interest being G per cent. ? KoTE. "When payments are to be made at different times without interest, find the present worth of each payment separately, and take their sum. Ans. $94*.40-[-. G. I hold a note against C for $529,925, due Sept. 1, 1859 ; what must I discount for the payment of it to-day, Feb. 7, 1859, money being worth G per cent. ? Ans. $17,425. 7. A man was offered $3675 in cash for his house, or $4235 in 3 years, without interest; he accepted the latter offej- ; how much did he lose, money being worth 7 per cent. ? Ans. $175. 8. A man, having a span of horses for sale, offered them for $480 cash in hand, or a note of $550 due in 1 year 8 months, without interest ; the buyer accepted the latter offer ; did the seller gain or lose thereby, and how much, interest be- ing G per cent. ? Ans. Seller gained $20. 9. What must be discounted for the present payment of a debt of $2637.72, of which $517.50 is to be paid in 6 months, $793.75 in 10 months, and the remainder in 1 year G months, the use of money being worth 7 per cent. ? Ans. $187.29 -\-. 10. What is the difference between the interest and discount of $130, due 10 months hence, at 10 per cent. ? Ans. $.83^. PROMISCUOUS EXAMPLES IX PERCENTAGE. 1. A merchant bought sugar in New York at 6^ cents per pound ; the wastage by transportation and retailing was 5 per cent., and the interest on the first cost to the time of sale was 2 per cent. ; how much must he ask per pound to gain 25 per cent.? Ami. R'-f- cents. PROMISCUOUS EXAMPLES. 251 2. A person purchased 2 lots of land for $200 each, and sold one at 40 per cent, more than cost, and the other at 20 per cent, less ; how much did he gain ? • A?is. $40. 3. Sold goods to the amount of $425, on G months' credit, which was $25 more than the goods cost; what was the true profit, money being worth G per cent. ? Aiis. $12.62 -f-. 4. Bought cotton cloth at 13 cents a yard, on 8 months' credit, and sold it the same day at 12 cents cash; how much did I gain or lose per cent., money being worth 6 per cent. ? Ans. Lost 4 per cent. 5. A farmer sold a pair of horses for $150 each; on one he gained 25 per cent., on the other he lost 25 per cent. ; did he gain or lose on both, and how much ? Ans. Lost $20. 6. A man invested f of all he was worth in the coal trade, and at the end of 2 years 8 months sold out his entire interest for $3100, which was a yearly gain of 9 per cent, on the money invested ; how much was he worth when he commenced trade? Ans. $3750. 7. In how many years Avill a man, paying interest at 7 per cent, on a debt for land, pay the face of the debt in interest ? Ans. 14f years. 8. Two persons engaged in trade ; A furnished f of tlie capital, and B f ; and at the end of 3 years 4 months they found they had made a clear profit of $5000, which was 12^- per cent, per annum on the money invested ; how much cap- ital did each furnish ? Ans. A, $7500 ; B, $4500. 9. Bought $500 wwth of dry goods, and $800 worth of groceries ; on the dry goods I lost 20 per cent., but on the groceries I gained 15 per cent. ; did I gain or lose on the whole investment, and hov/ much ? Ans. Gained $20. 10. What amount of accounts must an attorney collect, in order to pay over $1100, and retain 8^ per cent, for collect- ing ? Ans. $1200. 11. A merchant sold goods to the amount of $667, to be paid in 8 months ; the same goods cost him $600 one year previous to the sale of them ; money being wortli 6 per cent., what was his true gain ? Ans. $5,346 -\-. 252 PERCENTAGE. 12. A nurseryman sold trees at $18 per lumdred, and cleared ^ of his receipts ; what per cent, profit did he make ? Ans. 50 per cent. 13. If I of an article be sold for what f of it cost, what is the gain per cent. ? Atis. 40 g. 14. A lumber merchant sells a lot of lumber, which he has had on hand G months, on 10 months' credit, at an advance of 30 per cent, on the first cost ; if he is paying 5 per cent, inter- est on capital, what are his profits per cent. ? A71S. 213.. 15. A person, owning | of a piece of property, sold 20 per cent, of his share ; what part did he then own ? Ans. i. IG. A speculator, having money in the bank, drew GO per cent, of it, and expended 30 per cent, of 50 per cent, of this for 728 bushels of wheat, at $1.12i per bushel ; how much was left in the bank ? Ans. $3G40. 17. I wish to line the carpet of a room, tliat is G yards long and 5 yards wide, with duck | yard wide ; how many yards of lining must I purchase, if it will shrink 4 per cent, in length, and 5 per cent, in width ? Ans. 43|f . 18. A's money is 28 per cent, more than B's ; how many per cent, is B's less than A's ? Ans. 21^. 19. A capitalist invested f of his money in railroad stock, which depreciated 5 per cent, in value ; the remaining f lie in- vested in bank stock, which, at the end of 1 year, had gained $1200, which was 12 per cent, of the investment ; what was the whole amount of his capital, and what was his entire loss or gain ? Ans. $25000, capital ; $450, gain. 20. C's money is to D's as 2 to 3 ; if 4- of C's money be put at interest for 3 years 9 months, at 10 per cent., it will amount to $1933.25 ; how much money has each ? Ans. C, $2812 ; D, $4218. BANKING . S14:. A Bailk is a corporation ciuirtered by law for the purpose of receiving and loaning money, and furnisliing a paper circulation. "Wlmt is a bank ? BANKING. 253 315. A Promissory Note is a written or printed engage- ment to pay a certain sum, either on demand or at a specified time. 38 60 Bank Notes, or Bank Bills, are the notes made and issued by banlis to circuhxte as money. They are joayable in specie at the banks. SB 17. Tlie Face of a note is the sum made payable by the note. ?II8. Days of Grace are the three days usually allowed by law for the payment of a note after the expiration of the time specified in the note. Si9. The Matui'ity of a note is the expiration of the days of grace ; a note is due at maturity. S5®. Notes may contain a promise of interest, which will be reckoned from the date of the note, unless some other time be specified. The transaction of borrowing money at banks is conducted in accordance with the following custom : the borrower pre- sents a note, either made or indorsed by himself, payable at a specified time, and receives for it a sum equal to the face, less the interest for the time the note has to run. The amount thus withheld by the bank is in consideration of advancing money on the note prior to its maturity. 3^21. Bank Discount is an allowance made to a bank for the payment of a note before it becomes due. JI2ilo The Proceeds of a note is the sum received for it when discounted, and is equal to the face of the note less the discount. CASE I. S?I3. Given the face of a note to find the proceeds. The law of custom at banks makes the discount of a note Define a promissory note. Banit notes. Tlie face of a note. Days of grace. The maturity of a note. Explain the process of discounting a note at a bank. Defijie bank discoixnt. The proceeds of a note. What is Case I ? 254: PERCENTAGE. equcil to tlie simple interest at the legal rate for the time spe- cified in the note. Hence the Role. I. Compute the interest on the face of the note for three days more than the specified time ; the result will he the discount. II. Subtract the discount from the face of the note, and the remainder will he the proceeds. EXAMPLES FOR PRACTICE. 1. What is the discount, and what the proceeds, of a note for $450, at 60 days, discounted at a bank at 6 per cent. ? Ans. Discount, $4,725 ; proceeds, $445,275. 2. What are the proceeds of a note for $368, at 90 days, discounted at the Bank of New York ? Ans. $361,345 -f-. 3. What shall I receive on my note for $475.50, at 60 days, if discounted at the Crescent City Bank, New Orleans ? Aus. $471.33-1-. 4. What are the proceeds of a note for $10000, at 90 days, discounted at the Philadelphia Bank ? Ans. $984.5. 5. Paid, in cash, $240 for a lot of merchandise. Sold it the same day, receiving a note for $250 at 60 days, which I 20t discounted at the Hartford Bank. What did I make by this speculation ? Ans. ^t.oi^. 6. A note for $360.76, drawn at 90 days, is discounted at the Vermont Bank. Find the proceeds. Ans. $355.168 4-. 7. Wishing to borrow $530 of a Avestern bank which is discounting paper at 8 per cent., I give my note for $536.75, payable in 60 days. How much do I need to make up the required amount ? Ans. $.7645. Notes. 1. To indicate the matiiritj' of a note or draft, a vortical line ( I ) is used, with the day at which the note is iiominally due ou the left, and the date of maturity on the right ; thus, Jan. ' | m • 2. ^\^■lcn a note is on interest, payable at a future specified time, the amount is the face of the note, or th.e sum made payable, and must be made the basis of discount. Give rule. BANKING. 255 Find the maturity, term of discount, and proceeds of the following notes : — =i^'>'"^• Boston, Jan. 4, 1859. ^s. Three months after date, I promise to pay to the order of John Brown & Co. five hundred dollars, at the Suffolk Bank, value received. James Barker. Discounted March 2. (Due, April*! . Ans. \ Terra of discount, 3G da. Proceeds, §497. ^"•^^'- St. Louis, June 12, 1859. 9. Six months after date, I promise to pay Thomas Lee, or order, seven hundred fifty dollars, with interest, value re- ceived. , Byron Quinby. Discounted at a broker's, Nov. 15, at 10 per cent. Due, Dec. 121 1 5' Ans. { Term of discount, 30 da. Proceeds, $7GG.434+. CASE II. 3S4. Given the proceeds of a note, to find the face. 1. I wish to borrow $1:00 at a bank. For what sum must I draw my note, payable in 60 days, so that when discounted at 6 per cent. I shall receive the desired amount ? OPERATION. Analysis. $400 is the SLOOOO proceeds of a certain note, .0105 r= disc, on $1 for 63 da. the face of which Ave are required to find. We first $ .9895 =: proceeds of $1. obtain the proceeds of ,'^1 $400 -i- .9895 r= $404,244 = by the last case, and then face of the required note. divide the given proceeds, $400, by this sum ; for, as many times as the proceeds of $1 is con- tained in the given proceeds, so many dollars must be the face of the required note. Hence the Give Case II. Analysis. 256 PERCENTAGE. KuLE. Diride the jiroceeds hy the proceeds of $1 for the time and rate mentioned, and the quotient toill he the face of ilic note. EXAMPLES FOll FRACTICE. 2. What is the face of a note at GO days, which yields $680 when discounted at a New Haven bank ? Ans. $687,215+. 3. What is the face of a note at 90 days, of which the pro- ceeds are $1000 when discounted at a Louisiana bank ? Ans. $1013.085 -[-. 4. Wishing to borrow $500 at a bank, for what sura must my note be drawn, at 30 days, to obtain the required amount, discount being at 7 per cent. ? Ans. $503.22 -{-• 5. James Hopkins buys merchandise of me in New York, at cash price, to the amount of $1256. Not having money, he gives his note in payment, drawn at 6 months. What must be the foce of the note ? Ans. $1302.341 +. EXCIL.\NGE. S^«>. Exchange is a method of remitting money from one place to another, or of making payments by written orders. 3'26. A Bill of Exchange is a written request or order upon one person to pay a certain sum to another person, or to his order, at a specified time. 327. A Sight Draft or Bill is one requiring payment to be made "at sight," which means, at the time of its presenta- tion to the person ordered to pay. In other bills, the time specified is usually a certain number of days "after sight." There are always three parties, and usually four, to a trans- action in exchange : II2S. The Drawer or Maker is the person who signs the order or bill. Give the rule. Define exchange. A bill of exchange. A sight draft. The diawer. EXCHANGE. 257 3^9. The Drawee is the person to whom the order is addressed. SJI®. The Payee is the person to whom the money is or- dered to be paid. 5511 . The Buyer or Remitter is the person who purchases the bin. lie may be himself tlie payee, or the bill may be drawn in favor of any other person. 5512. The Indorsement of a bill is the writing upon its back, by which the payee relinquishes his title, and transfers the payment to another. The payee may indorse in blank by writing his name only, which makes the bill payable to the hearer, and consequently transferable like a bank note ; or he may accompany his signature by a special order to pay to another person, who in his turn may transfer the title in like manner. Indorsers become separately responsible for the amount of the bill, in case the drawee fails to make payment. A bill made payable to the hearer is transferable without in- dorsement. 330. The Acceptance of a bill is the promise which the drawee makes when the bill is presented to him to pay it at maturity ; this obligation is usually acknowledged by writing the word " Accepted," with his signature, across the face of the bill. Note. Three days of ^ace are usually allowed for the pa;(Tnent of a l)ill of exchange after the time specified has expii-ed. But in New York State no grace is allowed on sight drafts. From these definitions, the use of a bill of exchansfe in mon- etary transactions is readily perceived. If a man wishes to make a remittance to a creditor, agent, or any other person residing at a distance, instead of transporting specie, which is attended wi-rii expense and risk, or sending bank notes, which are liable to be uncurrent at a distance from the banks that issued them, he remits a bill of exchange, purchased at a bank or elsewhere, and made payable to the proper person in or The drawee. The payee. The buyer. An indorsement. An acceptance. What of grace on bills of exchange ? 25S PERCENTAGE. near the place where he resides. Thus a man by paying Boston funds in Bctton, may put New York funds into tlie hands of his New York agent. SS4. Tlie Course of Exchange is the variation of the cost of sight hills from their par value, as affected by the rela- tive conditions of trade and commercial credit at the two places between which exchange is made. It may be either at a pre- mium or discount, and is rated at a certain per cent, on the face of tlie bill. Bills payable a specified time after siglit are subject to discount, like notes of hand, for the term of credit given. Hence their value in the money market is affected by both the course of exchange and the discount for time. 8S*>. Foreign Exchange relates to remittances made be- tween different countries. S3€. Domestic or Inland Exchange relates to remit- tances made between diflerent places in the same country. An inland bill of exchange is commonly called a Draft. In this work we shall treat only of Inland Exchange. CASE I. SSr. To find the cost of a draft. $500. Syracuse, May 7, 1859. 1. At sight, pay to James Clark, or order, five hundred dollars, value received, and charge the same to our account. To M. Smith & Co. Messrs. Brown «&; Foster, \ Baltimore. ^ What is the cost of tlie above draft, the rate of exchange being 1^ per cent, premium? operation. Analysis. Since cx- $500 X 1 .0 1 5 r= $507.50, Ans. ^^'"^"^ ^« "^ H Fj; c^'»t premium, each dollar ot the draft Avill cost $1,015 ; and to find the whole cost of the draft, How is exchange conducted? Explain coiirse of cxchanc:e. For- cis;n exchange. Inland exchange. Define a draft. "What is Case I ? Give analysis. EXCHANGE. 259 we multiply its face, $.300, by 1,015, and obtain $507.50, the re- quired Ans. $480. Boston, June 12, 1859. 2. Thirty days after sight, pay to John Otis, or bearer, four hundred eiglity dollars, value received, and charge the same to account of Amos Tkenchard. To John Stiles & Co., ) New York. ) What is the cost of the above draft, exchange being at a premium of 3 per cent. ? OPERATION. Analysis. Since $1.0000 time is allowed, the .OOoa :r= discount for 33 days. di'aft must suflar dis- count in the sale. The $ .9045 =r proceeds of SI. .03 ^z rate of exchange. discount of $1, at the legal rate in Boston, for $1.0245 = cost of $1 of the draft. the specified time, al- $480 X 1.0245 = $491.76, Ans. lowing grace, is $.0055, which, sulitracted from $1, gives $.9945, the cost of $1 of the draft, provided sight ex- change were at par ; but sight exchange being at preiuiiim, we add the rate, .03, to .9945, and obtain $1.0245, the actual cost of $1. Then, multiplying $480 by 1.0245, we obtain $491.70, the Ans. From these examples we derive the following Rule. I. For sight drafts. — Ilultipli/ the face of the draft hfl 1 2}lus the rate xvhen exchange is at a premium, and hj 1 minus the rate lohen exchange is at a discount. II. For drafts payable after sight. — Find the proceeds of $1 at hank discount for the specified time, at the legal rate where the draft is purchased ; then add the rate of exchange tvhen at a premium, or subtract it when at a discount, and multiphj the face of the draft hy this result. EXAMPLES FOR PRACTICE. 3. A mercha,nt in Cincinnati wishes to remit $1000 by Give analysis, Rule I; II. 260 PERCENTAGE. draft to his agent in New York ; what will the bill cost, ex- change being at 3 per cent, premium? Ans. $1030. 4. What will be the cost in Rochester of a draft on Albany for $400, payable at sight, exchange being at f per cent, pre- mium ? 'Ans. $403. 5. A merchant in St. Louis orders goods from Kew York, to the amount of $530, which amount he remits by draft, ex- tliange being at 2| per cent, premium. If he pays $20 for transportation, what will the goods cost him in St. Louis ? Ans. $564,575. G. What will be the cost, in Detroit, of a draft on Boston for $800, payable GO days after sight, exchange being at a pre- mium of 2 per cent.? Ans. $806.20. 7. A man in Philadelphia purchased a draft on Chicago for $420, payable 30 days after sight ; what did it cost him, the rate of exchange being 1 J- per cent, discount? Ans. $411.39. 8. A merchant in Portland receives from his agent 320 barrels of floui', purchased in Chicago at $10 per barrel ; in I)ayraent for which he remits a draft on Chicago, at 2|- per cent, discount. Tlie transportation of his flour cost $312. What must he sell it for per barrel to gain $400 ? Ans. $12. CASE 11. 338. To find tlio face of a draft which a given sum will purchase. 1. A man in Indiana paid $3G9.72 for a' draft on Boston, drawn at 30 days ; what Avas the face of the draft, exchange being at 3^ per cent, premium? OPERATION. Analysis. We find, $3G9.72 — 1.027 = $360, Ajis. ^'y <^^^^ ^' t^"* ^ '^^""^^ ■ ' for $1 will cost $1,027; hence the draft that will cost $309.72 must be for as many dollars as 1.027 is contained times in $360.72 ; dividinp:, we obtain $300, the Ans. From this example and analysis we derive the following "NMiat is Case II ? Give analysis. EQUATION OF PAYMENTS. 261 Rule. Divide the given cost by the cost of a draft for $1, at the given rate of exchange ; the quotient ivill he the face of the required draft. EXAMPLES FOR PRACTICE. 2. What draft may be purchased for $243. GO, exchange being at 1^ per cent, premium? Ans. $2-iO. 3. What draft may be purchased for $79.20, exchange be- ing at 1 per cent, discount ? ^ Ans. $80. 4. An agent in Pittsburg holding $282.66, due his em- ployer in New Haven, is directed to malvc tlie remittance by draft, drawn at 60 days. What will be the face of the draft, exchange being at 2 per cent, premium ? Ans. S280. 5. An emigrant from Bangor takes $240 in bank bills to St. Paul, Min., and there pays 4- per cent, brokerage in ex- change for current money. What would he have saved by purchasing in Bangor a draft on St. Paul, drawn at 30 days, exchange being at 1^ per cent, discount ? Ayis. $5.599 -[-. 6. A Philadelphia manufacturer is informed by his agent in Buffalo that $3600 is due him on the sale of some property. He instructs the agent to remit by a draft payable in GO days after sight, exchange being at f per cent, premium. The agent, by mistake, remits a sight draft, which, when received in Phila- delphia, is accepted, and paid after the expiration of the three days of grace. If the manufacturer immediately puts this money at interest at the legal I'ate, will he gain or lose by the blunder of his agent ? Ans. He will lose $8.24-j-. EQUATION OF PAYMENTS. «l«IO. Equation of Payments is the process of finding the mean or equitable time of payment of several sums, due at different times without interest. S-l®. The Term of Credit is the time to elapse before a debt becomes due. Rule. Define epilation, of payments. Term of credit. 262 EQUATION OF PAYMENTS. ?I41. The Average Term of Credit is the time to elapse before several debts, due at different times, may all be paid at once, witliout loss to debtor or creditor. S49. The Equated Time is the date at which the several debts may be canceled by one payment. CASE I. 343, Wlien all the terms of crcclLt begin at the game date. 1. On the first day of January I find that I owe Mr. Smith 8 dollars, to be paid in 5 months, 10 dollars to be paid in 2 months, and 12 dollars to be paid in 10 months; at what time may I pay the whole amount ? OPERATIOX. $ 8 X 5 zr: 40 10 X 2 == 20 22 X 10 = 120 30 180 -^ 30 =: 6 mo., average time of credit. Jan. 1. -f- 6 mo. = July 1, equated time of payment. Analysis. The whole amount to be paid, as seen above, is 830 : and we are to find how long it shall be withheld, or what term of credit it shall have, as an equivalent for the various terms of credit on the different items. Now, the value of credit on any sum is meas- ured by the product of the money and time. And we say, the credit on $8 for 5 mo.r=the credit on $40 for 1 mo., because 8 X ^ = 40 X 1. la the same manner, Ave have, the credit on $10 for 2 mo.= the credit on $20 for 1 mo. ; and the credit on $12 for lOmo.m: the credit on $120 for 1 mo. Hence, by addition, the value of the several terms of credit on their respective sums equals a credit of 1 month on $180; and this equals a credit of 6 months on $30, be- CLiuse 30 X 6 r= 180 X 1. Rule. I. Multiply each payment by its term of credit, and divide fho sum of the products hj the sum of the payments ; the quotient will he tha average term of credit, Average term of credit. Equated time. GiVe Case I. Analvsis. Kulc. AVERAGING CREDITS. 263 II. Add the average term of credit to the date at which all the credits begin, and the result will he the equated time of payment. Notes. 1. The periods of time used as multipliers must all be of the same denomination, and the quotient will be of the game denomi- nation as the terms of credit ; if these bo months, and there 1)0 a re- mainder after the division, contiiuie the division to days by reduction, always taking the nearest unit in the last result. 2. The several rules in equation of payments are based upon the principle of bank discount ; for they imply that the discount of a sum paid before it is due equals the interest of the same amount paid after it is due. EXAMPLES FOR PRACTICE. 2. On the 25th of September a trader bought merchandise, as follows : $700 on 20 days' credit ; $400 on 30 days' credit ; $700 on 40 days' credit: what was the average term of credit, and what the equated time of payment ? , ( Average credit, 30 days. Ans. ■{ ^ ' ■^ ^ «K (. Equated time of payment, Oct. 2o. 3. On July 1 a merchant gave notes, as follows : the first for $250, due in 4 months ; the second for $750, due in 2 months; the third for $500, due in 7 months: at what time may they all be paid in one sum ? Ans. Nov. 1. 4. A farmer bought a cow, and agreed to pay $1 on Mon- day, $2 on Tuesday, $3 on Wednesday, and so on for a week ; desirous afterward to avoid the Sunday payment, he offered to pay the whole at one time : on what day of the week would this payment come ? Ans. Friday. 5. Jan. 1, I find myself indebted to John Kennedy in sums as follows : $G50 due in 4 months ; $725 due in 8 months ; and $500 due in 12 months : at what date may I settle by giving my note on interest for the whole amount? Ans. Aug. 21. CASE ir. S44. When tlio terms of credit begin at different dates, and tlie account has only one side. H'lii'S. An Account is the statement or record of mercantile transactions in business form. Give Case II. Define an accotint. 264: EQUATION OF PAYMENTS. S-16. The Items of an account may be sums due at the date of the transaction, or on credit for a specified time. An account may have both a debit and a credit side, the former marked Dr., the latter Cr. Suppose A and B have dealings in which there is an interchange of money or prop- erty ; A keeps the account, heading it with B's name ; the Dr. side of the account shows what B has received from A ; the Cr. side shows what he has parted with to A. SJiT. The Balance of account is the difference of the two sides, and may be in favor of either party. If, in the transactions, one party has received nothing from the other, tlie balance is simply the whole amount, and the account has but one side. Bills of" purchase are of this class. Note. Book accounts bear interest after the expiration of the term of credit, and notes after thoy become due, 348. To Average an Account is to find the mean or equitable time of payment of the balance. 349. A Focal Date is a date to which all the others are compared in averaging an account. 1. When does the amount of the following bill become due, by averaging ? J. C. Smith, 1859. To C. E. BoEDEN, Dr. June 1. To Cash, $450 " 12. " Mdse. on 4 mos., 500 Aug. IG. " Mdse., 250 FIRST OPERATION. SECOND OPERATION. Due. da. Items. Prod. Jime 1 Oct. 12 i Aug. 16 133 76 450 500 250 1200 66500 19000 85500 85500-1-1200 = 71 da. . ^11 da. after June 1, ^^*- \ or Aug. 11. Due. da. Items. Prod. June 1 Oct. 12 Aug. 16 133 .0 57 450 500 250 59850 14250 74100 1200 74100 -I- 1200 r= 62 da. . 5 62 da. before Oct. 12, ^"*-^ or Aug. 11. Define items. Balance. To average an account. A focal date. i AVERAGING ACCOUNTS. 265 Analysis. By reference to the example, it will be seen that the items are due June 1, Oct. 12, and Aug. 16, as shown in the two operations. In the first operation we use the earliest date, June 1, as a focal date, and find the difference in days between this date and each of the others, regard being had to the number of days in cal- endar months. From June 1 to Oct. 12 is 133 da. ; from June 1 to Aug. 16 is 7G da. Hence the first item has no credit from June 1, the second item has 133 days' credit from June 1, and the third item has 76 days' credit from June l,*as appears in the column marked da. After this we proceed precisely as in Case I, and find the average credit, 71 da., and the equated time, Aug. 11. In the second operation, the latest date, Oct. 12, is taken for a focal date ; the work is explained thus : Suppose the account to be settled Oct. 12. At that time the first item has been due 133 days, and must therefore draw interest for this time. But interest on $450 for 133 days = the interest on $598o0 for 1 da. The second item draws no interest, because it falls due Oct. 12. The third item must draw interest 57 days. But interest on $250 for 57 days = the interest on $14250 for 1 day. Taking the sum of the products, we find the whole amount of interest due on the account, at Oct. 12, equals the interest on $74100 for 1 day ; and this, by division, is found to be equal to the interest on $1200 for 62 days, which time is the average term of interest. Hence the account would be settled Oct. 12, by paying $1200 with interest on the same for 62 days. This shows that 1200 has been due 62 days ; that is, it falls due Aug. 11, without interest. Hence the following EuLE. I. Find the time at lohich each item becomes due, by adding to the date of each transaction the term of credit, if any be specified, and write these dates in a column, II. Assume either the earliest or the latest date for a faced date, and find the difference in days between the focal date and each of the other dates, and write the results in a second column. III. Write the items of the account in a third column, and multiply each sum by the corresponding number of days in the preceding column, writing the products in a final column. IV. Divide the sum of the products by the sum of the items. The quotient will be the average term of credit when the — — ■ • — ^ — . _-. — . — ) Give analysis. Rule. RP. 12 266 EQUATION OF PAYMENTS. , earliest date is the focal date, or the average term of interest when the latest date is the focal date ; in either case always reclcon from the focal date toicard the other dates, to find the equated tune of payment. examples for practice. 2. John Brown, 1859c, To James Greigg, Dr. Jan. 1. To 50 yds. Broadcloth, (d) $3.00, ... $150 " IG. " 2000 " Calico, " .10,... 200 Feb. 4. " 75 " Carpeting, « 1.33^,,. 100 March 3. « 400 " Oil Cloth, " .40, ... IGO If James Greigg wishes to settle the above bill by giving his note, from what date shall the note di-aw interest ? Ans. Jan. 27. 3. Adram Russel, 1859 To Wtnkoop & Brc, Dr. March lo To Cash, .., c r c o r » ..,.«.. -^ ... $300 April 4. " Mdse., .cooo.o,„ooococcc.-. 240 June 18. " " on 2 mo., ooooooc-oc c ^ 100 What is the equated time of payment of the above account ? Ans. May 20. 4. John Otis, 1858. To James Ladd, Dr. June 1. To 500 bu. Wheat, Co) $1.20, . . , - . . $G00 1.50 „ 300 1.30, u . . ■ . n 8o2 1.00,.. .„,,.. 7G0 l.OU, p • o c • • i 0\} When is the whole amount of the above bill due, per average? Ans. June 18. 5. My oxponditnres in building a house, in tlie year 1856, were as follows : Jan. 1 G, $530.78 ; Feb. 20, $ 125.30 ; JMarch 4, $259.25 ; April 24, $78G.3G. If at the last date I agree to (( 12. (( 200 a n li Ci 15. (( 040 li u ii (( 25. tk 700 ii u ii (( 30. t( 500 u ii ii AVERAGING ACCOUNTS. 267 sell the house for exactly what it cost, with reference to interest on the money expended, and take the purchaser's note for the amount, what shall be the iace of the note, and what its date ? ^^^ ( Face, $2007.75. "*° \ Date, March 8, 185 G. G. Thomas Whiting, 185'J, To IsKAEL Palmer, Dr. Jan. lo To GO bbls. Flour, O $7.00, . $420 " 28. " 90 bn. Wheat, " 1.50, ... . 135 Mar 15. " 300 bbls. Flour, " 6.00, . . . . 1800 If credit of 3 months be given to each item, when will tho above account become due ? Ans. May 30. CASE III. |gi?j©o Wlicn the terms of credit begin at different times, and the account has both a debt and a credit side. 1.. Averanre the followina; account. Dr. David Wake. Cr. 1858. 1 June 1 t( 16 Oct. 20 To Mdse « Draft, 3 mo. . " Cash, 1858. 1 400 00 1 July 4 800 00 Aug. 20 250 00 Sept. 20 By Mdse. " Cash. 200 150 500 00 00 00 Focal date. Dr. oper.vtion. Cr. Due da. 141 31 Item.s. 400 800 250 1450 850 600 Prod. Duo da. Items Prod. June 1 Sept. 19 Oct. 20 56400 24800 July 4 Aug. 20 Sept. 20 108 61 30 200 150 500 850 21600 9150 15000 45750 Bal mces. 81200 45750 35450 35450 -i- 600 =z 59 da., average term of interest. Oct. 20 — 59 da. Aug. 22, balance due. ■V\rhat is Case III ? Explain operation. 268 EQUATION OF PAYMENTS. AxALYSls. In the above operation ue have written the dates, showing when the items become due on either side of the ac- count, adding 3 days' grace to the time allowed to the draft. The latest date, Oct. 20, is assumed as the focal date for both sides, and the two columns marked da. show the dili'ereuce in days between each date and the focal date. The products are obtained as in the last case, and a balance is struck between tiie items charged and the products. These balances, being on the L)r. side, show that David Ware, on the day of the focal date, Oct. 20, owes i-GOO with interest on $35450 for 1 day. By division, this interest is found to be equal to the interest on .$600 for 59 days. The balance, $600, therefore, lias been due 59 daj's. Reckoning back from Oct. 12, we find the date when the balance fell due, Aug. 22. ]Ience the following Rule. I. Find the time when each item of the account is due ; and write the dates, in two columns, on the sides of the account to which they respectively belong. II. Use either the earliest or the latest of these dates as the focal date fur both sides, and find the products as in the last case. III. Divide the bcdance of the products by the balance of the account; the quotient tvill be the interval of time, ivhich must be reckoned from the focal date toward the other dates when both balances are on the same side of the account, but from the other dates when the balances are on opposite sides of the account. 2. "What is the balance of the following account, and when is it due? John Wilson. Dr. Cr. 1859. 1859. Jan. Feb. it 1 4 To Mdse. . " Cash . . 448 364 232 00 00 00 Jan.i 20 Feb.' 16 " l2o By Am't bvo't forward " 1 Carriage " Cash 560 264 900 00 00 00 ^^^^ ^ Balance, $6.= ' \ Due March J $680. 13. 3. If the followinj]; account be settled by giving a note, what shall be the face of the note, and what its date? Give analysis. Rule. RATIO. 2Gy Isaac Foster. Dr Cr. 18,5? • 1858. Jan 1 To Mdoc. on o ino. 11.3 80 May 11 Ry Cash 11 OO i< ii; ii (. i» ,3 u 37 48 July 112 t* it lo 00 June 57. The Antecedent is flie first term. SslS. The Consequent is the second term. 2I»1I>. No comparison of two numbers can be fully ex- plained but by instituting another comparison ; thus, the com- NoTE. It is thniiKht best to omit the questions at the bottom of tlie pa^es. in the reiiiainiii;^ part of this work, leaving the teacher to use such as may be deemed ap- propnute. ^ 270 RATIO. parison or relation of 4 to 8 cannot be fully expressed by 2, nor of 8 to 4 by h If the question were asked, what relation 4 bears to 8, or 8 to 4, in respect to magnitude, the answer 2, or J-, would not be complete nor correct. But if we make i/ni/i/ the standard of comparison, and use it as one of the terms in illustrating the relation of the two numbers, and say that the ratio or relation of 4 to 8 is the same as 1 to 2, or the ratio of 8 to 4 is the same as 1 to J-, unify in both cases being the standard of comparison, then the whole meaning is conveyed. SIC5©. A Direct Eatio arises from dividing the consequent by the antecedent. 261. An Inverse or Eeciprocal Ratio is obtained by di- viding the antecedent by the consequent. Thus, the direct ratio of 5 to 15 is -^-^- =: 3 ; and the inverse ratio of 5 to 15 is ~^- = J-. 3(|S|. A Simple E,atio consists of a single couplet ; as 3:12. S63. A Compound Ratio is the product of two or more simple ratios. Thus, the compound ratio formed from the simple ratios of 3 : G and 8:2isfXf = 3X8:GX2 = JL2 — 1 3^4. In comparing numbers with each other, they must be of tlie same lind, and of the same denomination. S6«$. The ratio of two fractions is obtained by dividing the second by the first ; or by reducing them to a common de- nominator, when they arc to each other as their numerators. Thus, the ratio of fV = "I is f "i" tV = t§ = 2, which is the same as the ratio of the numerator 3 to the numerator G of the equivalent fractions f'jj and -f^. Since the antecedent is a divisor and the consequent a divi- dend, any change in cither or both terms will be governed by the general principles of ilivision, (^7.) AV<' liave only to substitute the terms antecedent, consequent, and i-atio, for divi- sor, dividend, and quotient, and these principles become RATIO. 271 GENERAL PllINCIPLES OF RATIO. Prin. I. MnUijyJy'mg the consequent muUiplies the ratio; dividimj the consequent divides the ratio. Piiix. II. Multiplying the antecedent divides the ratio ; di- viding the antecedent multiplies the ratio. PiiiN. III. Multiphjing or dividing both antecedent and con- sequent by the same number does not alter the ratio. These three principles may be embraced in one GENERAL LAW. A change in the consequent produces a like change in the ratio ; but a change in the antecedent produces an opposite change in the ratio. JlCm. Since the ratio of two numbers is equal to the con- sequent divided by the antecedent, it follows, that 1. The antecedent is equal to the consequent divided by the ratio ; and that, 2. The consequent is equal to the antecedent multiplied by tl\e ratio. EXAMPLES FOR PRACTICE. 1. What part of 9 is 3? I := i ; or, 9 : 3 as 1 : J, that is, 9 has the same ratio to 3 that 1 bi's to 1^. 2.. What part of 20 is 5 ? Ans. |. 3. What part of 36 is 4? Ans. ^. 4. What part of 7 is 49 ? Ans. 7 times. 5. What is the ratio of IG to 88 ? Ans. 5i G. What is the ratio of G to 8^? Ans. \l. 7. What is the ratio of 64- to 78? Ans. 12. 8. What is the ratio of 16 to 66 ? Ans. 4 >. 9. What is the ratio of f to f ? Ans. |. 10. What is the ratio of I to jV? ^"^^- f- 11. What is the ratio of 3,1 to 1G|? Ans. 5. 12. What is the ratio of 3 gal. to 2 qt. 1 pt. ? Ans. /j. 272 PROPOETION. 13. What is the ratio of 6.3 s to 8 s. 6 d. 7 Ans. Iff 14. What is the ratio of 5.6 to .56? Ans. y'^. 15. Wliat is the ratio of 19 lbs. 5 oz. 8 pwts. to 25 lbs. 11 oz. 4 pwts. ? Ans. 1^. 16. What is the inverse ratio of 12 to 16? Ans. f. 17. What is the inverse ratio of f to |? Ans. ■^^. 18. What is the inverse ratio of 5f to 17^? Ans. \. 19. If the consequent be 16 and the ratio 2f, what is the antecedent? Ans. 7. 20. If the antecedent be 14.5 and the ratio 3, what is the consequent? Ans. 43.5. 21. If the consequent be J and the ratio f, what is the an- tecedent? Ans. 1|-. 22. If the antecedent be % and the ratio ^, vrhat is the consequent? Ans. yV PROPORTION. S67. Proportion is an equality of ratios. Thus, the ratios 6 : 4 and 12 : 8, each being equal to f, form a proportion. 3GS. Proportion is indicated in two ways. 1st. By a double colon placed between the two ratios ; thus, 2 : 5 : : 4 : 10. 2d. By the sign of equality placed between the two ratios ; thu.^s, 2:5 = 4:10. 3@0. Since each ratio consists of two terms, every pro- portion must consist of at least four terms. 370. The Extremes are the first and fourth terms. S73. The Means are the second and third terms. 3?s5. Three numbers may be in proportion when the first is to the second as the second is to the third. Thus, the num- bers 3, 9, and 27 are in proportion since 3 : 9 : : 9 : 27, the ratio of each couplet being 3. In such a proportion the second term is said to be a 7nean proportional between the other two. It7ti. In every proportion the product of the extremes is equal to the product of the means. Thus, in the proportion 3 : 5 : : 6 : 10 we have 3X10=5X6. SIMPLE PROPORTION. 273 «>> > 77^, Four numbers that are iDroportional in the direct order are pi'oportional by inversion, and also by alternation, or by inverting tlie means. Thus, tlie proportion 2 : 3 : : 6 : 9, by inversion becomes 3 : 2 : : 9 : G, and by alternation 2 : G : : t^73. From the preceding principles and illustrations, it foliows that, any three terms of" a jjroportion being given, the fourth mav readily be found by the followinfr lluLE. I. Divide the product of the extremes bi/ one of the means, and the quotient ivill be the other mean. Or, 1 1. Divide the product of the means by one of the extremes, and tlie quotient tcill be the other extreme. EXAMPLES FOR PRACTICE. Find the term not given in each of the following proportions. 1. 48 : 20 : : ( ) : 50. Ans. 120. 2. 42 : 70 : : 3 : ( ). Ans. 5. 3. ( ) : 30 : : 20 : 100. Ans. 6. 4. 1 : ( ) : : 7 : 84. A)is. 12. 5. 48 yd. :( ):: $67.25 : $201.75. Ans. 144 yd. 6. 3 lb. 12 oz. : ( ) : : $3.50 : $10.50. Ans. 11 lb. 4 oz. 7. ( ) : $38.25 : : 8 bu. 2 pk. : 76 bu. 2 pk. Ans. $4.25. 8. 4i- : 381 : : ( ) : 7G^. Ans. 8^. 9. ( ) : 12 :: a: If. Ans. 7. 10 -5- • ^ 'i • • ' • 2 /i)i ( ) = - 5 bar., Ans. Since the product of the extremes is equal to the product of the means (3^3), and the product of the means divided by one of the extremes will give the other; (blank) dollars wiU be equal to the product of 3 X 30 divided by 5, which is $18, Aas. 2. If 15 barrels of flour cost $90, how many barrels can be bought for $30 ? Analysis. In this ex- ample a cause is required, and the statement may be read thus : If 15 barrels cost $90, how many or (blank) barrels will cost $30 ? The product of the extremes, 30 X 15, di- vided by the given mean, 90, will give the required term, 5, as shown in the operation. Hence we deduce the following Rule. I. Arrange the terms in the statement so that the causes shall compose one coi/plef, and the effects the other, put- ting ( ) in the place of the required term. II. If the required term he an extreme, divide the -product of the means hy the given extreme ; if the required term he a mean, divide the product of the extremes hy the given mean. Notes. 1. If the terms of any couplet be of different denominations, they must be reduced to the same unit value. 2. If the odd term be a compound ninnbcr, it must be reduced to its lowest unit. 3. If the divisor and dividend contain one or more factors common to both, they should be canceled. If any of the tciTns of a proportion contain mixed numbers, they should first be changed to improper frac- tions, or the fractional part to a decimal. 4. "NVhen the vertical line is used, the divisor and the required terra are written on the left, and the terms of the dividend on the right. 276 PEOPORTION. «T9 We will now give another method of solving ques- tions in simple proportion, without making the statement, and which may be used, by those who prefer it, to the one ah-eady given. >Ye v/ill term it the Second Method. Eveiy question which properly belongs to simple propor- tion must contain four numbers, at least three of which must be given (IIT6). Of the three given numbers, one must always be of the same denomination as the required number. The remaining two will be like numbers, and bear the same relation to each other that the third does to the required num- ber ; in other words, the ratio of the third to the required number will be the same as the ratio of the other tv/c num^ bers. Regarding the third or odd term as the antecedent of the sec- ond couplet of a proportion, we find the consequent or required term by multiplying the antecedent by the ratio (SGd). By comparing the two like numbers, in any given question, \\'ith the third, we may readily determine whether the answer, or required term, will be greater or less than the third term ; if greater, then the ratio will be greater than 1, and the two like numbers may be arranged in the form of an improjier fraction as a multiplier ; if the answ^er, or required term, is to be less than the third term, then the ratio wnll be less than 1, and the two like numbers may be arranged in the form of a proper fraction, as a multiplier. 1. If 4 cords of wood cost $12, what wall 20 cords cost? OPERATION. AnATA'SIS. It will t^^ X 20 be readilv seen in tliis 12 X -^-j written = $60. examiilc,"that 4 cords ^ and 20 cords arc the like terms, and that $12 is the third term, and of the same denomination as the answer or required term. If 4 cords cost $12, will 20 cords cost more, or less, than 4 cords? evidently more : then the answer or required term will be greater SIMPLE PROPORTION. 277 than the third term, and the ratio greater than 1. The ratio of 4 cords to 20 cords is ^O-, or 5 ; hence the ratio of $12 to the answer must be o, and the answer will be ^^ or 5 times $12, which is $60. 2. If 12 yards of cloth cost $48, what will 4 yards cost ? OPERATION. Analysis. In this example we 4-8 X 1*2 = $16, Ans. see that 12 yards and 4 yards are the like terms and $48 the third term, and of the same denomination as the required answer. If 12 yards cost $48, will 4 yards cost more or less than 12 yards? less: then the ratio will be less than 1, and the multiplier a proper fraction. The ratio of 12 yards to 4 yards is -^% ; hence the ratio of $48 to the answer is ■^-^, and the answer will be\% times $48, which is $16. Hence the following Rule. I. Jiith the (wo given manhers, which are of the same name or kind, form a ratio greater or less than 1, accord- ing as the answer is to be greater or less than the third given number. II. MaUlpty the third number by this ratio, and the product loill be the required number or answer. Note. 1. !Mixed numbers should first be reduced to improper frac- tions, and the ratio of the fractions foimd according to (365). 2. Reductions and cancellation may be applied as in the first method. The following examples may be solved by either of the foregoiug methods. EXAMPLES FOB PRACTICE. 1. If 48 cords of wood cost $120, how much will 20 cords cost? Ans. $50. 2. If 6 bushels of corn cost $4.7^'>, how much will 75 bush- els cost ? Ans. $59.37^-. 3. If 8 yards of cloth cost $3J-, how many yards can be bought for $50 ? Ans. 114f yds. 4. If 12 horses consume 42 bushels of oats in 3 weeks, how many bushels will 20 horses consume in the same time ? 5. If 7 pounds of sugar cost 75 cents, how many pounds can be bought for $9 ? Ans. 84 lbs. 6. What will 11 lb. 4 oz. of tea cost, if 3 lb. 12 oz. cost $3.50? Ans. $10.50. 278 SIMPLE PROPORTION. 7. If a staff 3 ft. 8 in. long cast a shadow 1 ft. 6 in., ■\vliat is the heiglit of a steeple that casts a shadow 75 feet at the same time ? Ans. 183 ft. 4 in. 8. At $2.75 for 14 pounds of sugar, what will be the cost of 100 pounds? ■ A71S. $19.64f. 9. How many bushels of wheat can be bought for $5 LOG, if 12 bushels can be bought for $13.32 ? 10. What will be the cost of 28 J^ gallons of molasses, if 15 hogsheads cost $23G.25 ? Aus. §7.12i-. 11. If 7 barrels of flour are sufficient for a family 6 mouths, how many barrels will they require for 11 months? 12. At the rate of 9 yards for £5 12 s., how many yards of cloth can be bought for £44 16 s. ? Ans. 72 vds. 13. An insolvent debtor fails for $75 GO, of which he is able to pay only $3100 ; how much will A receive, whose claim is $75G ? Ans. $310. 14. If 2 pounds of sugar cost 25 cents, and 8 pounds of sugar are worth 5 pounds of coffee, what will 100 pounds of coffee cost ? Ans. $20. 15. If tlie moon move 13° 10' 35'' in 1 day, in what time will it perform one revolution ? IG. If 8^ bushels of corn cost $4.20, what will be the cost of 13^ bushels at the same rate ? Ans. $6.48. 17. If IJ yards of cotton cloth cost G]- pence, how many yards can be bought for £10 6 s. 8 d. ? Ans. 694|yds. 18. If 124- cwt. of iron cost $42^, how much will 4S^ cwt. cost? Ans. $163,50 4". 19. What quantity of tobacco can be bought for $317.23, if 8| lbs. cost $lf ? Ans. 15 cwt. 22.7-f- lbs. 20. If \r^l bushels of clover seed cost $156], how inm-h can be bought for $95.75 ? Ans. 9 bu. 2 pk. 2§ 0. The Partners are the individuals thus associated. ?5©1. Capital, or Stock, is the money or property invested in trade. S92. A Dividend is the profit to be divided. *^93. An Assessment is a tax to meet losses sustained. «s CASE I. 394. To find each partner's share of the profit or loss, when their capital is employed for equal periods of time. 1. A and B engage in trade; A furnishes $300, and B $400 of the capital; they gain $182; what is each one's share of the profit ? Analysis. Since the whole capital employed is $300 + $400 = $700, it is evident that A furnishes i-^^ =z I of the capita], and $182 X i = $78, A's share of the gain. ^ fw = t of the $182Xf = $104,B-s .. " .= 'T'-'^^- :^"f^'"^! ^ ^ ^ ' eacn man s share of the profit or loss will have the same ratio to the whole profit or loss that his share of the stock has to the whole stock, A will have ■2- of the entire profit, and B i, as shown in the operation. We may also regard the whole capital as the first cause, and each man's share of the capital as the second cause, the whole profit or loss as the frst effect, and each man's share of the profit or loss as the second effect, and solve by proportion thus ; OPERATION. . $300 $400 $700, whole stock. 300 Too = h A's share of the stock Too = h B-8 <( <( a PARTNERSHIP. let cause. 2d cause. 1st effi ct. 2d effect. $700 : $300 : : $182 : ( ) $700 : $400 : : $182 : ( ) /100 ^00' /^00 ^00^ ( ) A$2~^ ( ) l^^'^s 285 ( ) — ^lO** B's profit. ( ) — ^ ' 8, A'8 profit. Hence we have the folio win a; Rule. Multiply the whole profit or loss hy the ratio of the lohole capital to each mail's share of the capital. Or, 21ie whole capital is to each man's share of the capital as the whole profit or loss is to each mans share of the profit or loss. .2 Three men trade in company; A furnishes $8000, B $12000, and C 20000 of the capital; their gain is $1G80; Avhat is each man's share ? Ans. A's $33G ; B's $504 ; C's $840. 3. Three persons purchased a house for $2800, of Avhich A paid S1200, B $1000, and C $600 ; they rented it for $224 a year ; how much of the rent should each receive ? 4. A man failed in business for $20000, and his available means amounted to only $13654 ; how much will two of his creditors respectively receive, to one of whom he owes $3060, and to the other $1530 ? Ans. $2089.062 ; $1044.531. 5. Four men hired a coach for $13, to convey them to their respective homes, which were at distances from the place of starting as follows: A's 16 miles, B's 24 miles, C's 28 miles, and D's 36 miles; what ought each to pay? . , ^ A $2. C. $3.50. "^* 1 B $3. D $4.50. 6. A captain, mate, and 12 sailors took a prize of $2240, of which the captain took 14 shares, the mate 6 shares, and each sailor 1 share ; how much did each receive ? 7. A cargo of corn, valued at $3475.60, was entirely lost ; ^ of it belonged to A, ^ of it to B, and the remainder to C ; how much was the loss of each, there being an insurance of $2512? ^«s. $120.45, A's. $240.90, B's. $602.25, C's. 286 PARTNERSHIP. 8. Three persons engaged in the lumber trade ; two of the persons furnished the capital, and the third managed the busi- ness; they gained $2571. 2-i, of which C received $6 as often as D $4, and E had -i as much as the otlier two for taking care of the business ; how mueli was eacli one's share of the gain ? Ans. $1285.62, C's. $857.08, D's. $428.54, E's. 9. Four persons engage in the coal trade ; D puts in $3042 capital ; they gain $7500, of which A takes $2000, B $2800.75, and C $1685.25 ; how much capital did A, B, and C put in, and how much is D's share of the gain ? . ( A, $6000. C, $5055.75. ^"^- X B, $8402.25. D's gain, $1014. CASE II. 09«]». To find each partner's share of the profit or loss when their capital is employed for unequal periods of time. It is evident that the respective shares of profit and loss will depend upon two conditions, viz. : the amount of capital in- vested by each, and the time it is employed. 1. Two persons form a partnership; A puts in $450 for 7 months, and B $300 for 9 months; they lose $156 ; how much is each man's share of the loss ? oPERATiox. Analysis. The $150 X 7 = $3150, As oapit.-ii for 1 mo. "se of $4 JO Capital $300 X 9 = S2700, 15-3 « « « for 7 months is the same as the use of ' " 7 times $4,30, or apitai $3150 for 1 month; „ ■ and of $300 for 9 months is the same as the use of times $300, or $2700 for 1 month. The en- tire capital for 1 month is equivalent to $3150 -|- $2700 r= .$58 JO. If the loss, $150, l)e divided lietwcen the two partners, according to Case I, the results will be the loss of each as shown in the operation. $5850, en;iio « 31 5.0 5850 Y^j, A's share of the ent 2 7 00 _ 5if 50 T3' ^'^ " " $156 X YjT = $84, A's loss. $156 X T^;y=$72, D's " PARTNERSHIP. 287 Examples of this kind may also be solved by proportion as in Case I, the causes being compounded of capital and iijiie; thus, $5850 : $3150 : : $156 : ( ) $5850 : $2700 : : $15G : ( ) ( ) ( ) = $8i, A'a loss. ( ) = $12, B's loss. Ilcnce the following Rule. Multiply each man's capital hy the time it is em- piloyed ill trade, and add the jn'oducts. Then multijily iJ'C entire profit or loss by the ratio of the sum of the products to each product, and the results will be the respective shares of profit or loss of each piartner. Or, Multiply each man's capital by the time it is employed in trade, and regard each product as his capital, and the sum of the products as the entire capital, and solve by proportion, as in Case I. EXAJIPLES FOR rEACTICE. 2. Three persons traded togetlier; B put in $250 for 6 months, C $275 for 8 months, and D $450 for 4 months ; they gained $825 ; how much was each man's share of the, gain ? 3. Two merchants formed a partnership for 1 8 months. A at lir,--t put in $1000, and at the end of 8 months he put in $000 more ; B at first put in $1500, but at the end of 4 months he drew out $300 ; at the expiration of the time they found that they liad gained $1394.64; how much was each man's share of the gain ? Ans. A's $715.20 ; B's $679.44. 4. Three men took a field of grain to harvest and thresh for i of the crop ; A furnished 4 hands 5 days, B 3 lianda 6 days, and C 6 hands 4 days ; the whole crop amounted to 372 bushels ; how much was each one's share ? 5. "William Gallup began trade January 1, 1856, with a capital of $3000, and, succeeding in business, took in M. H. Decker as a partner on the first day of March following, with 283 ANALYSIS. a capital of $2000; four months after they admittedJ. New- man as third partner, Avho put in $1800 capital; they con- tinued their partner.-^hip until April 1, 1858, when they found tliat $4388.80 had been gained since Jan. 1, 1856; how much was each one's share ? ( §2106, Gallup's. Ans.-\ $1300, Decker's. ( $ 982.80, Newman's. 6o Two persons engaged in partnership with a capital of $5 GOO; A's capital was in trade 8 months, and his share of the profits was $560 ; B's capital was in 10 months, and his share of the profits was $800 ; what amount of capital had each in the firm ? Ans. A, $2613.33^; B, $298G.6Gf. 7. A, B, and C, engaged in trade with $1930 capital; A's money was in 3 months, B's 5, and C's 7 ; they gained $117, which was so divided that a of A's share was equal to ^- of B's and to ^ of C's ; how much did each put in, and what did each gain? ( A, $700 capital ; $26 gain. A)is. ]B,$GS0 " $39 " C, $600 " $52 « ANALYSIS. 30®. Analysis, in arithmetic, is the process of solving problems independently of set rules, by tracing the relations of the given numbers and the reasons of the separate steps of the operation according to the special conditions of each question. 397. In solving questions by analysis, we generally reason from the ffiven number to tinity, or 1, and then from unity, or 1, ^o the required number. 398. United States money is reckoned in dollars, dimes, cents,andmills(180),one dollar being uniformly valued in all tlie States at 100 cents ; but in most of the States money is sometimes still reckoned in pounds, shillings, and pence. Note. At tho time of the adoption of our decimal currency by Congress, in 1786, the colonial citrrenci/, or bills of credit, issued by the colonies, had depreciated in value, and this depreciation, being imequal in the different colonies, gave rise to tho different values of the State currencies ; and this variation continues wherever the denominations of bhillings and pence are in use. ANALYSIS. 289 S©9. In New England, Indiana, \ Illinois, Missouri, Virginia, Kentucky, > $1 m 6 s. = 72 d. Tennessee, Mississippi, Texas, ^ New York, Ohio, Micliigaii, $1 := 8 s. := 06 d. New Jersey, Pennsylvania, Dela- ) T.r ,1 t$l = 7s. 6d.=:00d Ware, Maryland, ) South Carolina, Georgia, \ |l=4s. 8d. = 56d. Canada, Nova Scotia, I $1 = 5 s. = 60 d. EXAMPLES FOR PRACTICE. 1. What will be the cost of 42 bushels of oats, at 3 shillings f-er bushel, New England currency ? OPERATION. 3 42X3= 126 s. 126 -^6 =$21 Or, — 121, Ans. Analysis. Since 1 bushel costs 3 shil- lings, 42 bushels will cost 42 times 3 s., or 42X3 = 126 s.; and as 6 s. make 1 dollar New England currency, there are as many dollars in 126 s. as 6 is contained times in 126, or $21. 2. What will 180 bushels of wheat cost at 9 s. 4d. per bushel, Pennsylvania currency ? .OPERATION. ! a'^02 Or, 00 113 $224 28 2 $224, Ans. Analysis. Multi- plying the number of bushels by the price, and dividing by the value of 1 dollar re^ duced to paxce, we we have $224. Or, when the pence in the given price is an aliquot part of a shilling, the price may be reduced to an improper fraction for a multiplier, thus : ,9 s. 4 d. rr 91 s. =: 2j? s., the multiplier. The value of the dollar being 7 s. G d. z=l 1\ s. :=: ^2*, we divide by ^, as in the operation. 3. What will be the cost of 3 hhd. of molasses, at 1 s. 3 d. per quart, Georgia currency ? R.p. 18 2r!:ii Pl> 9, 290 ANALYSIS. OPERATION. o Analysis. In this example we first 003 reduce 3 lilid. to quarts, by multiplying y by G3 and 4, and then multiply by the ^ price, either reduced to pence or to an ^'^ improper fraction, and divide by the 405 00 value of 1 dollar reduced to the same denomination as the price. $£02.50 4. Sold 9 firkins of butter, each containing 56 lb., at 1 s. G d. per pound, and received in payment carpeting at 6 s. 9 d. per yard ; bow many yards of carpeting would pay for the butter ? OPERATION. Analysis. The operation in this is similar to the preceding examples, except that we di- 5Q vide the cost of the butter by the price of a $i ,^2 ^mit of the article received in payment, reduced ^ to the same denominational unit as the price 112 yd. °^ ^ "^"^^^ °^ ^^^ article sold. The result will be the same in whatever currency. 5. "What will 3 casks of rice cost, each weidiinc 126 7 DO pounds, at 4 d. per pound, South Carolina cuiTency ? Ans. $27. 6. How many pounds of tea, at 7 s. per pound, must be given for 28 lb. of butter, at 1 s. 7 d. per pound ? Ans. 6^. 7. Bought 2 casks of Catawba Avine, each containing 72 gallons, for $648, and sold it at the rate of 10 s. 6 d. per quart, Ohio currency ; how much was my whole gain ? Ans. $108. 8. What will be the expense of keeping 2 horses 3 weeks if tlic expense of keeping 1 horse 1 day be 2 s. 6 d., Canada currency? Ans. $21. 9. How many days' work, at 6 s. 3 d. per day, must be given for 20 bushels of apples at 3 s. per bushel ? Ans. 9f . 10. Bought 160 11). of dried fruit, at Is. 6 d. a pound, in New York, and sold it for 2 s. a pound in Philadelphia ; how much was my wliole gain ? Ans. $12,663. 11. A merchant exchanged 434- yards of clotli, worth 10 s. G d. per yard, for other cloth worth 8 s. 3 d. per yard ; liow many yards did he receive ? Ans. 55 j^j. ANALYSIS. 291 12. What will be the cost of 300 bushels of wheat at 9 s. 4 d. per bushel, Michigan cun-encj ? Aiis. $350. 13. If f of f of a toa of coal cost $2|, how much Avill f of 6 tons cost ? OPERATION. $ j 1'2^ Analysis.. Since | of f = ^| of a ton costs 1^ I o^4 $2| = $132, 1 ton will cost 28 times ^^\ of $^^2, w .0 or $^'- X ft ; and -f of G tons z=i ^ tons, wiU ' ^ cost -3^ times f| of §-i/ — $16. t^lG, Ans. 14. If 8 men can build a wall 20 ft. Ions, 6 ft. high, and 4 ft. thick, in 12 days, working 10 hours a day, in how many da/s can 24 men build a wall 200 ft. long, 8 ft. high, and 6 ft. thick, Avorking 8 hours a day ? OPERATION. 1^ $ 10 J^00io $ -7 X — X — X X — X— = 100 da. I ^S $ 20 ^ Analysis. Since 8 men require 12 days of 10 hours each to build the wall, 1 man would require 8 times 12 days of 10 hours each, and 10 times (12 X 8) days of 1 hour each. To build a wall 1 ft. long would require ^ as much time as to build a wall 20 ft. long ; to build a wall 1 ft. high would require \ as much time as to build a wall 6 ft. high; to build a Avail 1 ft. thick, \ as much time as to build a wall 4 ft. tliick. Now, 24 men could build tliis Mall in ^^ as many days, by Avorking 1 hour a day, as 1 man could build it, and in \ as many days by AA'orking 8 hours a day, as by Avorking 1 hour a day ; but to build a wall 200 ft. long Avould require 200 times as many days as to build a Avail 1 ft. long ; to build a Avail 8 ft. high Avould require 8 times as many days as to build a Avail 1 ft. high ; ?nd to build a wall 6 ft. thick would requh-e 6 times as many days as to build a Avail 1 ft. thick. 15. If 2 pounds of tea are AA'orth 11 pounds of coffee, and 3 pounds of coffee are worth 5 pounds of sugar, and 18 pounds of sugar are AA'orth 21 pounds of rice, hoAv many pounds of rice can be purchased Avith 12 pounds of tea? 292 ANALYSIS. OPERATION. Analysis. Since 18 lb. of su- ^1'7 gar are equal in value to 21 lb. of 3^^ 5 rice, 1 lb. of sugar is equal to ^^ jg 2]^ . of 21 lb. of rice, or fi = | lb. of w j/^s rice, and o lb. of sugar 'are equal — — to 5 times |- lb. of rice, or ^^P- lb. ; 3 I 385 if 3 lb. of coffee are equal to 5 lb. Ans. T>8Mb. °^ ^^°f"' °^" ^ ^^' °^ "'^'^' ^ ^^' ^^ ^ ' coffee is equal to -J- of ^q°- lb. of rice, or f| lb., and 11 lb. of coffee are equal to 11 times ff lb. of rice, or -Y/ lb. ; if 2 lb. of tea are equal to 1 1 lb. of coffee, or -W^- lb. of rice, 1 lb of tea is equal to ^ of -\\5. lb. of rice, or -^g^i lb., and 12 lb. of tea are equal to 12 times -%\^ lb. of rice, or -SfA lb. = 128 A lb. 16. If 16 horses consume 128 bushels of oats in 50 days, how many bushels will 5 horses consume in 90 days? Ans. 72. 17. If $10i- will buy 4§ cords of wood, how many cords can be bought for $24| ? Ans. 11. 18. Gave 52 bai-rels of potatoes, each containing 3 bushels, v.'orth 33^- cents a bushel, for 65 yards of cloth; how much was the cloth worth per yard ? Ans. $.80. 19. If a staff 3 ft. long cast a shadow 5 ft. in length, Avhat is the height of an object that casts a shadow of 46f ft. at the same time of day ? Ans. 28 ft. 20. Three men hired a pasture for $63 ; A put in 8 sheep 74 months, B put in .12 sheep 4^ months, and C put in 15 sheep 6| months ; ho\t much must each pay ? 21. If 7 bushels of wheat are worth 10 bushels of rye, and 5 bushels of rye are Avorth 14 bushels of oats, and 6 busliels of oats are worth $3, how many bushels of wheat will S^30buy? ^»s. 15. 22. If $480 gain $84 in 30 months, what capital will gain $21 in 15 months? Ans. $240. 23. How many yards of carpeting § of a yard wide are equal to 28 yards ^^ of a yard wide? Ans. 3 14-. 24. If a footman travel 130 miles in 3 days, when the days are 14 hours long, in how many days of 7 hours each will he travel 390 miles? Ans. IS. ANALYSIS. 293 25. If 6 men can cut 45 cords of wood in 3 days, liow many cords can 8 men cut in 9 days ? Ans. 180. 26. B's age is IJ- times the age of A, and C's is 2y\j times the age of both, and the sum of their ages is 93 ; what is the age of each? A>2s. A's age, 12 yrs. 27. If A can do as much work in 3 days as B can do in 4i days, and B can do as much in 9 days as C in 12 days, ;>.ii or §16 more tlian he did on is- 09 . pay. Every barrel of second "" o"^ " '1 quality made a difference of $2 16-1-2= 8 bbl., 2d quality. j^^ ^j^g ^^^^ . j^g^^^ ^^^^.^ ^^.^^.g ■^0 — 8 = 12 bbl., 1st « as many barrels of second qual- ity as $2, the difference in the cost of one barrel, is contained times in $16, &c. 30. A boy bought a certain number of oranges at the rate of 3 for 4 cents, and as many more at the rate of 5 for 8 cents ; he sold them again at the rate of 3 for 8 cents, and gained on the whole 108 cents; how many oranges did he buy ? 294. ANALYSIS. oPERATiox. Analysis. For § 4" f ^^ if; 15 ^2=: ff, average cost. those he bought § — ff = If =r 1 J cts., gain on each. at the rate of 3 for 108 -^li = 90, number of oranges. ^ ^ents he paid | of a cent each, and for those he bought at the rate of 5 for 8 cents he paid | of a cent each; and -<-ff — -4| cents, what he paid for 1 of each kind, wliich divided by 2 gives f 2 cents, the average price of^all he bought. He sold them at the rate of 3 for 8 cents, or | cents each ; the dif- ference between the average cost and the price he sold them for, or f — if = If — H cents, is the gain on each ; and he bought as many oranges as the gain on one orange is contained times in the whole gain, &c. 31. A man bought 10 bushels of wheat and 25 bushels of corn for $30, and 12 bushels of wheat and 5 bushels of corn for $20 ; how much a bushel did he give for each ? Analysis. We may divide or multiply either of the expressions by such a number as will render one of the commodities purchased, ahke in both expressions. In this example we divide the first by 5 to make the numbers denoting the corn alike, (the same result would be produced by multiply- ing the second by 5,) and we have the cost of 2 bushels of wheat and o bushels of corn, equal to SO. Subtracting this from 12 bushels of wheat and 5 bushels of corn, which cost 820, we find the cost of 10 bushels of wheat to be $14 ; there- fore the cost of 1 bushel is J^ of $14, or $1.40. From any one of the expressions containing both wheat and corn, we readily find the cost of 1 bushel of corn to be 64 cents. 32. A, B, and C agree to build a barn for $270. A and B can do the work in 16 days, B and C in 13^ days, and A and C in lly days. In how many days can all do it working together ? In how many days can each do it alone ? "What part of the \y.\y ought each to receive ? OPEKATION. W. C. 1st lot, 10 25 $30 2d « 12 5 $20 1st • • 5 = 2 5 $6 10. • e • . $14 1 bu. W. = $1.40 1 bu. C. — $ .64 _£- 5 4 • 1 -' A- = ;20cla ., C alone. 80 ^0 ■ SO so — _9_ 6_ _3_ . 1-|- •A-= 261 d a., A " 80 80 80 80 6 9 7 . 2 ; 1-^ 2 40 da. ., B " 80 80 80 8 A X8|- = *>■ the part of the I whole C did. 3 X8f . 3 cc (( a A « 80 9' •80 X8| 1' u IC (1 B « ANALYSIS. 295 OPERATION. Analysis. Since JL := _5_.^ what A and B do in 1 day. A and B can do tlie 3_— - 6 " BandC " " Avoik in 16days,tlley ^T_ ^r -JL, " A and C « " Can do ^q- of it in 1 ■io + -s'V + 10 = 80' '"''^* ^' ^' '^'"^ ^ ^° ''^ '^''^y ' , ^ ^'^'■^ ^' "^ 2 dajs. 13|- or ^°- days, they .18 _^ 2 z= -T^g-, what A, B, and C do in 1 day. can do -^\ of it in 1 1 _|_ 9- :rr 8| days, time A, B, and C, will do the day ; A and C, in llf whole work together. _ or \°- days, they can do -/g of it in 1 day. Then A, B, and C, by working 2 days each, can do -^q-\- _3 I 7_ 18 of tiie 40T^80 80 "* *- ^ work, and by work- ing 1 day each they can do i of l|, or -^y $270 X I = $120, C"s f-hare. of the work ; and it $270Xf = $90. A's " will take them as $270 X 7 = $60, E-s '•• many days working together to do the Avliole work as -^^ is contained times in 1, or 8f days. Now, if we take what any two of them do in 1 day from what the three do in 1 day, the remainder will be what the third docs ; we thus find that A does -f^, B -^^, and C -^^. Next, if we denote the whole work by 1, and divide it by the part each does in 1 day, we have thQ number of days that it will take each to do it alone, viz. : A 26| days, B 40 days, and C 20 days. And each should receive such a part of $270 as would be ex- pressed by the i^art he does in 1 day, multiplied by the number of days he works, which will give to A $90, B $60, and C $120. 33. If G oranges and 7 lemons cost 33 cents, and 12 oranges and 10 lemons cost 54 cents, what is the price of 1 of each ? A)is. Oranges, 2 cents; lemons, 3 cents. 31. If an army of 1000 men have provisions for 20 days, at the rate of 18 oz, a day to each man, and they be reinforced by GOO men, upon what allowance per day must each man be put, that the same provisions may last 30 days ? Ans. 74- oz. 3'J. There are 54 bushels of grain in 2 bins ; and in one bin are G bushels less than ^ as much as there is in the other ; how many bushels in the larger bin ? Ans. 40. 296 ANALYSIS. 3G. The sum of two numbers is 20, and their difference is equal to ^ of the greater number; what is the greater number? Ans. 12. 37. If A can do as much work in 2 days as C in 3 days, and B as mucli in 5 days as C in 4 days; what time will B require to execute a piece of w^ork which A can do in 6 weeks? Ans. 11^ weeks. 38. How many 5^ards of cloth, f of a yard wide, will line oQ yards 1;^ yards wide ? Ans. GO. 39. How many sacks of coffee, each containing 104 lbs, at 10 d. per pound N. Y. currency, will pay for 80 yards of broadcloth at $3|- per yard ? Ans. 24. 40. A person, being asked the time of daj', replied, the time past noon is equal to -i of the time to midnight ; v/hat was the hour ? Ans. 2, P. M. 41. A market woman bought a number of peaches at the rate of 2 for 1 cent, and as many more at the rate of 3 for 1 cent, and sold them at the rate of 5 for 3 cents, gaining 55 cents ; how many peaches did she buy ? Ans. 300. 42. A can build a boat in 18 days, working 10 hours a day, and B can build it in 9 days, working 8 hours a day ; in liow many days can both together build it, working 6 hours a day ? 43. A man, after spending ^ of his money, and ^ of the remainder, had $10 left ; how much had he at first ? 44. If 30 men can perform a piece of work in 1 1 days, how many men can accomplish another piece of work, 4 times as large, in J- of the time ? Ans. 600. 45. If 1G|- lb. of coffee cost $3|, how much can be bought for $1.25? Ans. Glib. 46. A man engaged to write for 20 days, receiving $2.50 for every day he labored, and forfeiting $1 for every day he was idle ; at the end of the time he received $43 ; liow many days did he labor ? Ans. 18. 47. A, B, and C can perform apiece of work in 12 hours; A and B can do it in 16 hours, and A and C in 18 hours; what part of the work can B and C do in 9| hours ? Ans. f. ALLIGATION MEDIAL, e .- 297 ALLIGATION". ^€?>0. Alligation treats of mixing or compounding two or more ingredients of different values. It is of two kinds — Alli- gation Medial and Alligation Alternate. 401 . Alligation Medial is tlie process of finding the aver- age price or quality of a compound of several simple ingredi- ents whose prices or qualities are known. 1. A miller mixes 40 bushels of rye worth 80 cents a bushel, and 25 bushels of corn Avorth 70 cents a bushel, with 15 bushels of wheat worth §1.50 a bushel; what is the value of a bushel of the mixture ? OPF.RATIOX. AxALYSIS^ Since 40 bushels 80 X 40 = $32.00 of rye at 80 cents a bushel is 70 X 25 rr: 17.50 worth $32, and 2.J bushels of corn 1 50 y 15 = 22 50 ^^ "'^ cents a bushel is worth — $17.50, and 15 bushels of wheat 80 ) 72.00 at $1.50 a bushel is worth $22.50, fit; nn A„g therefore the entire mixture, con- sisting of 80 bushels, is worth $72, and one bushel is worth J^ of $72, or 72 -I- SO := $.90. Hence the following Rule. Divide the entire cost or value of the ingredients bg the sum of the simples. EXAMPLES FOR rRACTICE. 2. A wine merchant mixes 12 gallons of wine, at %l per gallon, with 5" gallons of brandy worth $1.50 per gallon, and 3 gallons of water of no value ; what is the worth of one gal- lon of tlie mixture ? Ans. $.975. 3. An innkeeper mixed 13 gallons of water with 52 gallons of brandy, which cost him §1.25 per gallon ; what is the value of 1 gallon of the mixture, and what his profit on the sale of the whole at C|- cents per gill ? Ans. %l a gallon ; $65 profit. 4. A grocer mixed 10 pounds of sugar at 8 cts. with 12 pounds at 9 cts. and IG pounds at 11 cts., and sold the mixture at 10 cents per pound; did he gain or lose by the sale, and hov/much? A71S. He gained 16 cts. 13* 293 ALLIGATION ALTERNATE. 5. A gi'ocer bought 7i dozen of eggs at 12 cents a dozen, 8 dozen at 10^ cents a dozen, 9 dozen at 11 cents a dozen, and 101 dozen at 10 cents a dozen. He sells them so as to make 50 per cent, on the cost j how much did he receive per dozen? Alls. 10} cents. 6. Bought 4 cheeses, each weighhig 50 pounds, at 13 cents a pound; 10, weighing 40 jiounds each, at 10 cents a pound; and 24, weighing 25 pounds each, at 7 cents a pound; I sold the whole at an average price of 9i cents a pound; how much Vt'as my whole gain ? Ans. $G. 4:®?5. Alligation Alternate is the process of finding the proportional quantities to be taken of several ingredients, v.'hose prices or qualities are known, to form a mixture of a required price or quality. CASE I. 4® 3. To find the proi^ortional quantity to be used of each ingredient, when the mean price or quality of the mixture is given. 1. What relative quantities of timothy seed worth $2 a busliel, and clover seed worth $7 a bushel, must be used to form a mixture worth $5 a bushel ? OPERATION. Analysis. Since on every in- 1 2 ) gradient used whose price or qnal- 3 j ^"^* ity is less than the mean rate there will be a gaii>, and on every in- gredient whoso price or quality is greater than the mean rate there will be a loss, and since the gains and losses must be exactly equal, the relative quantities used of each shoidd be such as repre- sent the unit of value. By selling one bushel of timothy seed worth $2, for $0, there is a gain of $3 ; and to gain $1 would require J of a buslicl, which we place opposite the 2. By selling one bushel of clover seed worth $7, for $.5, there is a loss of $2 ; and to lose $1 would require ^ of a bushel, wliicli we place opposite tlie 7. In every case, to find the unit of value we must divide $1 by tlie gain or loss per bushel or pound, &-c. Hence, if, every time we take !^ of a bushel of timothy seed, we take i of a bushel of clover seed, the gain and loss will be exactly equal, and we shall have \ and -^ for the proportional quantities. i ALLIGATION ALTERNATE. 299 OPERATION. 1 2 o 4 3 1 4 4 4 X 'J 1 1 7 1 2 2 I 10 1 o 3 If we wish to express the proportional numbers in integers, we may reduce these fractions to a common denominator, and use their numerators, since fractions having a common denominator are to each other as their numerators. ( 365) thus, \ and i are equal to I and |, and the proportional quantities are 2 bushels of timothy seed to 3 bushels of clover seed. 2. Wliat proportions of teas worth respectively 3, 4, 7 and 10 shillings a pound, must be taken to form a mixture worth 6 shillings a pound ? Analysis. To preserve the equalit}' of gains and losses, we must always compare two prices or simples, one greater and one less than the mean rate, and treat each pair or couplet as a separate example. In the given example we form two couplets, and may compare either 3 and 10, 4 and 7, or 3 and 7, 4 and 10. A7e find that ^ of a lb. at 3 s. must be taken to gain 1 shilling, and ^ of a lb. at 10 s. to lose 1 shilling ; also -| of a lb. at 4 s. to gain 1 shilling, and 1 lb. at 7 s. to lose 1 shilling. These proportional numbers, obtained by comparing the two couplets, are placed in columns 1 and 2. If, now, we reduce the num])ers in columns 1 and 2 to a common denominator, and use their numerators, we obtain the integral numbers in columns 3 and 4, which, being arranged in column o, give the proportional quantities to be taken of each.* It will be seen that in comparing the simples of any couplet, one of which is greater, and the other less than the mean rate, the pro- portional number finally obtained for either term is the difference between the mean rate and the other term. Thus, in comparing 3 and 10, the proportional number of the former is 4, which is the difference between 10 and the mean rate 6 ; and the proportional number of the latter is 3, which is the difference between 3 and the mean rate. Tlie same is true of every other couplet. Hence, when the simples and the mean rate are integers, the intermediate steps taken to obtain the final proportional numljers as in columns 1, 2, 3, and 4, may be omitted, and the same results readily found by taking the difference between each simple and the mean rate, and placing it opposite the one with which it is compared. * Prof. A. B. Oanfipld, of Oneida Conference Seminary, N. Y., used this method of Allisation, essentialij'. in the instruction of his classes as early as 1846, and he was duubtless the author of it. oOO ALLIGATION ALTERNATE. From the foregoing esamples and analyses we derive the following Rule. I. Write the several prices or qualities in a column, and the mean price or quality of the mixture at the left. II. Form couplets by comparing any price or quality less, vjith one that is greater than the mean rate, placing the part ichich inust he used to gain 1 of the mean rate opposite the less simple, and the part that must be used to lose 1 opposite the grecder simple, and do the same for each simple in every couplet. III. If the proportional numbers are fractioncd, they may be reduced to integers, and if tioo or more stand in the same hori- zontal line, they must he added ; the final results will he the pro- portional quantities required. Notes. 1. If the numbers in any couplet or column have a com- mon factor, it may bs rejected. 2. We may also multiply the numbers in any couplet or coliunn by any multiplier we choose, without affecting the equality of the gains and losses, and thus obtain an indefinite number of results, any one of which being taken will give a correct ihial result. EXAMPLES FOR PRACTICE. 3. A grocer has sugars worth 10 cents, 11 cents, and 14 cents per pound; in what proportions nmy he mix them to form a mixture worth 12 cents per pbund? Ans. 1 lb. at 10 cts., and 2 lbs. at 11 and 14cts. 4. What proportions of water at no value, and wine worth $1.20 a gallon, muit be used to form a mixture worth 90 cents a gallon ? Ans. 1 gal. of water to 3 gals, of wine. 5. A farmer had sheep worth $2, $2^, $3, and $4 per head; what number could he sell of each, and realize an average price of §2^ per head ? A^ns. 3 of the 1st kind, and 1 each of the 2d and 3d, and 5 of the 4th kind, 6. What relative quantities of alcohol 80, 84, 87, 94, and 96 per cent, strong must be used to form a mixture 90 per cent, strong ? Ans. G of the first two khids, fouj' of the 3d, 3 of the 4th, and 16 of the 5tli, OPERATION. 30 3^ 4 4 45 t\ 8 8 8i A ^\ 5 5 10 ALLIGATION ALTERNATE. 301 CASE II. ^®4. When tlic quantity of one of the simples is limited. 1. A miller has oats worth 30 cents, corn worth 45 cents, raid barley worth 84 cents per bushel ; he desires to form a mixture worth 60 cents per bushel, and which shall contain 40 bushels of corn ; how many bushels of oats and barley must he take ? Analysis. By 20 ) the same process 60 ^ 45 " yL 8 8 40 V Am. ^^ "^ Case I we rQ \ -find the propor- tional quantities cf each to be 4 bushels of oats, 8 of corn, and 10 of barley. But we wish to use 40 bushels of corn, which is 5 times the propor- tional number 8, and to preserve the equality of gain and loss M'e must take 5 times the proportional quantity of each of the other simples, or 5 X 4 =r 20 bushels of oats, and 5 X 10 =r 50 bushels of barley. Hence the following Rule. Find the proportional quantities as in Case I. Divide the given quantity by the proportional quantity of the same ingredient, and multipty each of the other proportional quantities hy the quotient thus obtained. EXAMPLES FOR PRACTICE. 2. A merchant has teas worth 40, 60, 75, and 90 cents per pound ; how many pounds of each must he use with 20 pounds of that Avorth 75 cents, to form a mixture at 80 cents. Ans. 20 lbs. each of the first three kinds, and 130 lbs. of the fourth. 3. A farmer bought 24 sheep at $2 a head ; how many must he buy at $3 and $5 a head, that he may sell the whole at an average price of $4 a head, without loss ? Ans. 24 at $3, and 72 at $5. 4. IIov/ much alcohol worth 60 cents a gallon, and how much v.'ater, must be mixed with 180 gallons of rum worth SI. 30 a gallon, that the mixture may be worth 90 cents a gallon ? Ans. 60 gallons each of alcohol and water. ^Q2 ALLIGATION ALTERNATE. 5. How many acres of land Avorth 35 dollars an acre must be added to a farm of 75 acres, -vvortli $50 an acre, that the average value may be $iO an acre ? Ajis. 150 acres. 6. A merchant mixed 80 pounds of sugar -worth G^ cents per pound with some worth 8^ cents and 10 cents per pound, so that the mixture was worth 71 cents per pound ; how much of each kind did he use ? CASE III. 4:©^, When the quantity of the whole compound is limited. 1. A grocer has sugars wortli G cents, 7 cents, 12 cents, and 13 cents per pound. He wishes to make a mixture of 120 pounds worth 10 cents a pound; how many pounds of each kind must he use ? OPERATION. Analysis. By Case I we find the propor- tional quantities of each to be 3 lbs. at 6 cts., 2 lbs. at 7 cts., 3 lbs at 12 cts., and 4 lbs. at 13 cts. By adding the propor- tional quantities, we find that the mixture would be but 12 lbs. while the required mixture is 120, or 10 times 12. If the whole mixture is to be 10 times as much as the sum of the proportional quantities, then the quantity of each simple used must be 10 times as much as its respective proportion- al, wliich would require 30 lbs. at 6 cts., 20 lbs. at 7 cts., 30 lbs. at 12 cts., and -JO llis. at 13 cts. Hence Ave deduce the foUowhig Rule. Find the proportional numbers as in Case I. Di- vide the given quantity hj the sum of the proportional qvan- tities, and multipuly each of the proportional quantities hy the quotient thus obtained. EXAMPLES FOR PRACTICE. 2. A farmer sold 170 sheep at an average price of 14 shillings a head ; for some he received 9 s., for some 12 s., for some 18 s., and for others 20?.; how many of each did lie sell ? Ans. GO at 9 s., 40 at 12 s., -20 at 18 s., and 50 at 20 s. \h 1 1 3 1 1 3 30 h 2 2 20 i 3 3 30 h 4 m 4 40 12 120 INVOLUTION. ' 3u3 3. A jeweler melted together gold 16, 18, 21, and 24 cariits line, so as to make a compound of 51 ounces 22 carats fine ; how much of each sort did he take ? Ans. 6 ounces each of the first three, and 33 ounces of the last. 4. A man bought 210 bushels of outs, corn, and wheat, and paid for the whole $178.50 ; for the oats he paid $^, for the corn $3, and for the wheat $1^ per bushel ; how many bush- els of each kind did he buy ? Ans. 78 bushels each of oats and corn, ainl 54 bushels of wheat. 5. A, B, and C are under a joint contract to furnish 6000 bushels of corn, at 48 cts. a bushel ; A's corn is worth 45 cts., B's 51 cts., and C's 54 cts. ; liovv many bushels must each put into tlie mixture that the contract may be fulfilled ? 6. One man and 3 boys received $84 for 56 days' labor; the man received $3 per day, and the boys SJ-, $f, and $lf re- spectively ; how many days did each labor ? Ans. Tlie man 16 days, and the boys 24, 4, and 12 days respectively. INVOLUTION. 406, A Power is the product arising from multiplying a number by itself, or repeating it several times as a factor ; thus, in 2 X 2 X 2 = 8, the product, 8, is a power of 2. 4®T. The Expongnt of a power is the number denoting how many times the factor is repeated to produce the poAver, and is written above and a little tc the right of the factor; thus, 2 X 2 X 2 is written 2-^, in which 3 is the exponent. Exponents likewise give names to the powers, as will be seen in the followimr illustrations : 3 = 3^ = 3, the first power of 3 ; 3X3 = 3- = 0, the second power of 3 3X3X3 — 33 =: 27, the third power of 3. 4:'?)8<. The Square of a number is its second power. 4:®S&. The Cube of a number is its third power. 410, Involution is the process of raising a number to a given power. 304 EVOLUTION. 411. A Perfect Power is a number that can be exactly produced by the involution of some number as a root ; thus, 25 and 32 are perfect powers, since 25 = 5 X 5, and 32 = 2 X 2X2X2X2. 1. What is the cube of 15? OPERATION. Analysis. We multiply 15 X 15 X 15 =: 3375. Ans. ^^ ^J' l'^' ^"^^ the product by 15, and obtain o375, which is the 3d power, or cube of 15, since 15 has-been taken 3 times as a factor. Hence, we have the following Rule. Multipli/ the number hj itself as many limes, less 1, as there are units in the exponent oj the required power. EXAMPLES FOR PRACTICE. 2. What is the square of 25 ? Ans. 625. 3. What is the square of 135 ? Ans. 18225. 4. Wliat is the cube of 72 ? Ans. 373248. 5. Yfhat is the 4th power of 24? Ans. 331776. G. Eaise 7.2 to the third power. Ans. 373,248. 7. Involve LOG to the 4th power. Ans. 1.2G247696. 8. Involve 12 to the 5th power. Ans. .0000248832. 9. Involve 1.0002 to the 2d power. Ans. 1.00040004. 10. What is the cube of | ? OPERATION. 2 2 2 2X2X2 2^ 8 5 5 5~5X5X5~ 53~125 It is evident from the above operation, that A common fraction may he raised to any poiocr, hy raising each of its terms, separately, to the required power. 11. What is the square of f ? Ans. -^^. 12. Wliat is the cube of -ij ? Ans. f 4-f f 13. Raise 2 4f to the 2d power. Ans. 61 2^-^ EVOLUTION. 4B2, A Soot is a factor repeated to produce a power; thus, in the expression 5X5X5=: 125, 5 is the root from whicli the power, 125, is produced. SQUARE ROOT, ?05 4:13. Evolution is the process of extracting the root of a number considered as a power, and is the reverse of Involution. 414. The Eadical Sign is tlie character, V) which, placed before a number, denotes that its root is to be extracted. 415. The Index of the root is the figure placed above the radical sign, to denote what root is to be taken. When no index is written, the index 2 is always understood. 416. A Surd is the indicated root of an imperfect power. 4tT. Roots are named from the coi-responding powers, as will be seen in the following illustrations : The square root of 9 is 3, wa-itten ■\/d z= 3. The cube root of 27 is 3, written ^27 = 3. The fourth root of 81 is 3, written ^^81 = 3. 418. Any number whatever may be considered a power •whose root is to be extracted ; but only the perfect powers can have exact roots. SQTJAHE EOOT. 419. The Square Eoot of a number is one of the two equal factors that produce the number ; thus the square root of 49 is 7, for 7 X 7 = 49. 422®. In extracting the square root, the first thing to be determined is the relative number of places in a given number and its square root. The law governing this relation is exhib- ited in the following examples : — Koots. Squares. Roots. Squares. 1 1 1 1 9 81 10 1,00 99 98,01 100 1,00,00 999 99,80,01 1000 1,00,00,00 From these examples we perceive 1st. That a root consisting of 1 place may have 1 or 2 places in the square. 2d. That in all cases the addition of 1 place to the root adds 2 places to the square. Hence, 306 EVOLUTION. If we point off a number into two-fg^ire periods, commen- cing at the rigid hand, the number of full periods and the left hand full or partial period will indicate the number of places in the square root ; the highest period answering to the highest figure of the root. 4:*2i, 1. What is the length of one side of (i square plat containing an area of 5417 sq. ft. ? orERATiox. Analysis. Since the given figin-e is 54,17 I 73.6 a square, its side will be the square root 49 of its area, which we will proceed to com- pute. Pointin": off the given number, the i.-±\j tji.t 2 periods show that there will be two in- 1 4o 42 J tcgral figures, tens and units, in the root. 14G.0 88.00 ^^^^ ^^"^ ^^ *^^ ^'°'^^ must be exti-acted 1 4 r G 87 or from the first or left hand period, 54 hun- dreds. The greatest square in 54 hun- 4 dreds is 49 hundreds, the square of 7 tens ; we therefore write 7 tens in the root, at the right of the given number. Since the entire root is to be the side of a square, let us form a Fii; I. square (Fig. I), the side of which is 70 feet long. The area of this square is 70 X 70 ziz 4900'sq. ft., which we subtract from the given number. This is done in the operation by subtracting the square number, 49, from the first period, 54, and to the remainder bringing down the sec- ond period, making the entire remainder 517. If we now enlarge our square (Fig. I) by the addition of 517 squai'e feet, in such a manner as to preserve the square form, its size will be that of the required square. To preserve the square form, the addition must be so made as to extend the square equally in two directions ; it will therefore be composed of 2 oblong figures at the sides, and a little square at the corner (Fig. II). Now, the width of this addition will be the additional length to the side of the square, and consequently the next figure in the root. To find width we divide square contents, or area, by length. But the length of one side of the liy;le square cannot be found till the width of the addition be determined, because it is equal to this widtli. We mIU tlierefore add the lengths of the 2 oblong figures, and the sum will be sufficiently near the whole length to be used as a trial divisor. SQUARE ROOT. 307 Fig. II. 7-. 3 70 o 70 Trial Divisor = 140 Complete Divisor = 143 Each of the oblong figures is equal in length to the side of the square first formed ; and their united length is 70 + 70 = 140 ft. (Fig. III). This num- ber is obtained in the operation by doubling the 7 and annexing 1 cipher, the result being written at the left of the dividend. Dividing 517, the area, by 140, the approximate length, we obtain 3, the probable width of the addi- tion, and second figure of the root. Since 3 is also the side of the little square, wo can now find the entire length of the addition, or the complete divisor, which is 70 -f 70 -|- 3 = 143 (Fig. III). '^' ■ This number is found in the oper- ation by adding 3 to the trial di- visor, and writing the result un- derneath. Multiplying the com- plete divisor, 143, by the trial quotient figure, 3, and subtracting tlie product from the dividend, Ave obtain another remainder of 88 square feet. With this remainder, for the same reason as before, we must proceed to make a new enlargement ; and mx- bring down two decimal cii)hers, because the next figure of the root, being tenths, its square will l^e hundredths. The trial divisor to obtain the width of this new enlargement, or the next figure in the root, will be, for the same reasgn as before, twice 73, the root already found, with one cipher annexed. But since the 7 has already been doubled in the operation, we have only to double the last figure of the complete divisor, 143, and annex a cipher, to obtain the new trial divisor, 146.0. Dividing, we obtain .6 for the trial figure of the root ; then proceeding as before, we obtain 146.6 for a complete divisor, 87.96 for a product ; and there is still a remainder of .04. Hence, the side of the given square plat is 73.6 feet, nearly. From this example and analysis we deduce the following Rule. I. Point off the given numher into periods of two ff/ures each, counting from laiifs place toward the left ajid riglt t. II. Find the greatest square numher in thelefl hand period, and ivrite its root for the^, first figure in the root ; subtract the square numher from the left hand period, and to the remainder ^ring down the next period for a dividend. i508 EVOLUTION". III. At (lie Icji of the dividend lori'e twice the jirst figure of the root, and annex one ciphe7',fo7' a trial divisor ; divide the dividend hy the trial divisor, and icrite the quotient for a trial figure in the root. IV. Add the trial figure of the root to the trial divisor for a complete divisor ; midtiply the complete divisor Ig the trial figure in the root, and subtract the product from the dividend, and to the remainder hring doivn the next period for a new dividend. V. To the last complete divisor add the last figure of the foot, and to the sum annex one cipher, for a new trial divisor, tvilh which proceed as before. Notes. 1. If at anytime the product be greater than the dividend, diminish the trial figure of the root, and correct the erroneous -work. 2. If a cipher occur in the root, annex another cipher to the trial di-visor, and another period to the dividend, and proceed as before. EXAMPLES FOR PRACTICE. 2. What is the square root of 406457.2516? OPERATIOX. 40,64,57.25,16 | 637.54, Ans. 36 Trial divisor, 120 464 Completo " 123 369 Trial " 1260 9557 Comiilete " 1267 8869 Trial " 1274.0 688.25 Complete " 1274.0 637.25 Trial « 1275.00 51.0016 Completo " 1275.04 51.0016 Notes. 3. The decimal points in the work may be omitted, rare being taken to point off in the root accordmg. to the number of deci- mal periods used. 4. The pupil will acquire greater facility, and secure greater accura- cy,, by keeping units of like order under each other, and each divisor opposite tlie corresponding dividend, by ,the use of the lines, as shown in the operation. 3. What is the square root of 576 ? Ans. 24. SQUARE ROOT. 4. What is the square root of 6561 ? 5. What is the square root of 444889 ? 6. What is the square root of 994009 ? 7. What is the square root of 29855296 ? 8. What is the square root of 3486784401 ? Ans. 59049. 9. What is the square root of 54819198225 ? Note. The cipher in the trial rlivisor may be omitted, and its place, after division, occupied by the trial root figui-e, thus forming in suc- cession only complete divisors. 309 An. •. 81. Ans. 667. Ans. 997. Ans. 5464. 10. What is the square root of 2 ? 2. 1.4142 +, Ans. 1 100 24 96 400 281 281 11900 2824 11296 V 60400 28282 56564 11. Extract the square roots of the following numbers; V3 = 1.7320508 + V5 =z 2.2360679 4- V6 irr 2.4494897 4- V7 = 2.6457513 -j- V8 = 2.82842714- VIO = 3.1622776 4- ' 12. What is the square root of .00008836 ? Ans. .0094. 13. What is the square root of .0043046721 ? Ans. .06561. Notes. 5. The square root of a common fraction may be obtained by extracting the square roots of the numerator and denominator separately, provided the terms are perfect squares ; otherwise, the fraction may first be reduced to a decimal. 6. IMixed numbers may be reduced to the decimal form before ex- tracting the root ; or, if the denominator of the fraction is a perfect square, to an improper fraction. 14. Extract the square root of ^Wy » -^"^- ti- 15. Extract the square root of l^g'fl- Ans. ^. 16. Extract the square root of §. Ans. .816496 +. 17. Extract the square root of 17f. Ans. 4.1683 +• 310 EVOLUTION. APPLICATIONS. 4J2!S. An Angle is the opening between two lines tliat meet each otlier ; thus, the two lines, A B and A C, meeting, form an angle at A. ■4SS. A Triangle is a figure having three sides and three angles, as A, B, C. 4^4. A Eight- Angled Triangle is a tri- angle having one right angle, as at C. ^2^5, The Ease is the side on which it stands, as A, C. 4:S©, The Perpendicular is the side forming a right angle with the base, as B, C. 4:^7, The Hypotenuse is the side opposite the right angle, as A, B. 4:^8, Those examples given below, which relate to trian- gles and circles, may be solved by the use of the two following principles, which are demonstrated in geometry. 1st. Tlie square of tlie hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. 2d. The areas of two circles are to each other as the squares of their radii, diameters, or circumferences. 1. The two sides of a right-angled triangle are 3 and 1 feet ; what is the length of the hypotenuse ? Analysis. Squarine OPERATION. the two sides and add- 32 = 9, square of one side. i"?:. "e find the sum to 42 = 1 G, sc.uare of the other side- ^^ -^ ' '''''^ '^i"^^^ ^^"^ ^^"" — IS equal to the square of 25, square of hypotenuse. the hypotenuse, we ex- /25 5 A„g tract the square root, and ' obtain o feet, the hypot- enuse. Hence, To find tlie hypotenuse. Add the squares of the tioo sides, and extract the square root of the sum. To find cither of the shorter sides. Subtract the square of the given side from the square of the hi/poteniise, and extract the square root of the remainder. SQUARE ROOT. 311 EXAMPLES FOR TRACTICE. 2. If an army of 55225 men be drawn up in the form of a square, how many men will there be on a side ? Ans. 235. 3. A man has 200 yards of carpeting l^^ yards wide ; what is the lengtli of one side of the square room which this carpet will cover ? Ans. 45 feet. 4. How many rods of fence will be required to inclose 10 acres of land in the form of a square ? A/ts. 1 GO rods. 5. The top of a castle is 45 yards high, and the castle is sur- rounded by a ditch GO yards wide ; required the length of a rope that will reach from the outside of the ditcli to the top of the castle. Ans. 75 j-ards. G. Required the height of a May-pole, which being broken 39 feet from the top, the end struck the ground 15 feet from the foot. Ans. 75 feet. 7. A ladder 40 feet long is so placed in a street, that without being moved at the foot, it will reach a window on one side 33 feet, and on the other side 21 feet, from the ground ; ■wliat is the breadth of the street ? Ans. 56.64 -\- ft. 8. A ladder 52 feet long stands close against the side of a building ; how many feet must it be drawn out at the bottom, that the top may be lowered 4 feet ? Ans. 20 feet. 9. Tw'O men start from one corner of a park one mile square, and travel at the same rate. A goes by the walk ^ around the park, and B takes the diagonal path to the opposite corner, and turns to meet A at the side. How many rods from the corner will the meeting take place ? Ans. 93.7 -\- rods. 10. A room is 20 feet long, IG feet wide, and 12 feet high ; what is the distance from one of the lower corners to the op- posite upper corner ? Ans. 28.284271 -(-feet. 11. It requires 63.39 rods of fence to inclose a circular field of 2 acres ; what length will be required to inclose 3 acres in circular form ? Ans. 77.63 + rods. 12. The radius of a certain circle is 5 feet; what will be the radius of another circle containing twice the area of the first? A)is. 7.07106-1- feet. 312 ■ EVOLUTION. CUBE ROOT, 4:29. The Cube Root of a number is one of the three equal factors that produce the number. Thus, the cube root of 27 is 3, since 3 X 3 X 3 = 27. 4;3©. In extracting the cube root, the first thing to be determined is the relative number of places in a cube and its root. The law governing this relation is exhibited in the fol- lowing examples : — Roots. Cubes. Roots. Cubes. 1111 9 729 10 1,000 99 907,299 100 1,000,000 999 997,002,999 1000 1,000,000,000 From these examples, we perceive, 1st. That a root consisting of 1 place may have from 1 to 3 places in the cube. 2d. That in all cases the addition of 1 place to the root adds three places to the cube. Hence, If we pomt off a number into three-JTgi(.re periods, com- mencing at the right hand, the number of full periods and the left hand full or partial period ivill indicate the number of places in the cube root, the highest period answering to the highest figure of the root. 4:S1 . 1. What is the length of one side of a cubical block containing 413494 solid inches ? OPERATION — COMMENCED. AN.4.LYSIS. Since the block is a 413494 I 74 cube, its side -will be the cube root of 343 its soHd contents, which \\c will pro- ceed to compute. Pointing off the 14/UU /U4J4: given number, the two periods show that there will be two figures, tens and units, in the root. The tens of the root must be extracted from the first period, 413 thousands. The greatest cube in 413 thousands is 343 thousands, the cube of 7 tens ; we therefore write 7 tens in the root at the right of the given number. CUBE ROOT. 313 Since the entii-e root is to be the side of a cube, let us form a cubical block (Fig. I), the side of which is 70 inches in length. The contents of this cube are 70 X 70 X 70 = 343,000 solid inches, which Ave subtract from the given number. This is done in the operation by subtracting the cube number, 343, from the first period, 413, and to the re- ^ mainder bringing down the sec- iW cud period, making the entire PfjiiiiisSa^i^ca^^gsg^ I'emainder 70494. If we now enlarge our culncal block, (Fig. I), by the addition of 70494 solid inches, in sucli a manner as to preserve the cubical form, its size will be that of the required block. To preserve the cubical form, the addition must be made upon three adjacent sides or faces. The addition will therefore be composed of 3 flat blocks to cover the 3 faces, (Fig. II) ; 3 oblong blocks to fill the vacancies at the edges, (Fig. Ill) ; and 1 small cubical block to fill the vacancy at the cor- ner, (Fig. IV). Now, the thickness of this enlargement will be ihe additional length of the side of the cube, and, consequently, the second figure in the root. To find thickness, we may divide solid ^'-- ^^- contents by surface, or area. But the area of the 3 oblong blocks and little cube cannot be found till the thickness of the addition be determined, be- cause their common breadth is equal to this thickness. We will therefore find the area of the three flat blocks, which is sufla- ciently near the whole area to bo used as a trial divisor. As these are each equal in length and breadth to the side of the cube whose faces they cover, the M'hole area of the three is 70 X 70 X 3 =■ 14700 square inches. This number is obtained in the operation by annexing 2 ciphers to three times the square of 7 ; the result being written at the left hand of the dividend. Dividing, we obtain R.P. 14 314 EVOLUTION. Fis. Ill OPERATION — CONTINtTED. 413494 IT. LO 343 74 4, the probable thickness of the addition, and second figure of the root. With tliis assumed figiu-e, ■\ve will complete our divisor by adding the area of the 4 blocks, before undetermined. The 3 ob- long blocks are each 70 inches long ; and the Httle cube, being equal in each of its dimensions to the thickness of the addition, must be 4 inches long. Hence, their united length is 70 -|- 70 -j- 70 -|- 4 = 214. This number is obtained in the operation by multiplying the 7 by 3, and an- nexing the 4 to the product, the result being: written in column being I, on the next Hne below the trial divisor. Multiplying 214, the length, by 4, the common width, we obtain 856, the area of the four blocks, which added to 14700, the trial divisor, makes 1.3556, the complete divisor ; and multiplying this by 4, the second figure in the root, and subtract- ing the product from the divi- dend, M'e obtain a remainder o! 0270 solid inches. "Willi this re- mainder, for the same reason as before, we must proceed to make a new enlargement. But since we have already two figures in the root, answering to the two periods of the given number, the next figure of the root must be a decnnal ; and wc therefore annex to the remainder a period of three decimal ciphers, mak- ing 8270.000 for a new dividend. The trial divisor to obtain the thickness of tliis second cnlargc- 214 856 14700 1555G 70494 62224 8270.000 ment, or the next figure of the root, will be the area of three new flat blocks to cover the three" sides of the cube already formed ; and this CUBE ROOT. 315 surface, (Fig. IV,) is composed of 1 face of each of the flat blocks ah-cacly used, 2 faces of each of the oblong blocks, and 3 faces of tlie little cube. But we have in the complete divisor, 15556, 1 llice of each of the flat blocks, oblong blocks, and little cube ; and in the correction of the trial divisor, 856, 1 face of each of the oblong blocks and of tlie little cube ; and in the square of the last root figure, 16, a third face of the little cube. Hence, 16 -{r 856 -)- 15556 = 16428, the significant figures of the new trial divisor. This OrERATION — CONTINUED. ir. 413i94 I 74.5 343 214 856 14700 15556 70494 62224 222.5 111.25 1642800 16539.25 8270.000 8269.625 .oii) number is ob- tained in the operation by- adding the square of the last root fig- ure mentally, and coml^iu- ing units of like order, thus : 16, 6, and 6 are 28, and we write the unit figure in the new trial divisor ; then 2 to carry, and 5 and 5 are 12, Szc. We annex 2 ciphers to this trial divisor, as to the former, and dividing, obtain 5, the third figure in the root. To complete the second trial di- visor, after the manner of the first, the correction may be found bv annexing .5 to 3 times the former figures, 74, and midtiplying this number by .5. But as we have, in column I, 3 times 7, with 4 annexed, or 214, we need only multiply the last figure, 4, by 3, and annex .5, making 222.5, which multiplied by .5 gives 111.25, the correction required. Then we obtain the complete divisor, 16539.25, the product, 8269.625, and the remainder, .375, in the manner shown by the former steps. From this example and analysis we deduce the following Rule. I. Point off the giveii number into periods of three frjurcs each, counting from units' place toward the left and right. II. Find the greatest cuhe that does not exceed the left hand period, and write its root for the frst figure in the required root; sidjtract the cube from the left hand period, and to the remainder bring doivn the next period for a dividend. III. At the left of the dividend lorite three times the square of tlie first figure of the root, and annex two ciphers, for a trial divisor ; divide the dividend by the trial divisor, and write the quotient for a trial fgure in the root. 316 ETOLUTION, IV. Annex tlie trial figure to three times the former figure, and write the result in a column marked I, one line helow the trial divisor ; multiply this term by the trial figure, and write the product on the same line in a colwnn marked II ; add this term as a correction to the trial divisor, and the result icill be the complete divisor. V. Multiply the complete divisor by the trial figure, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. VL Add the square of the last figure of the root, the last term in column II, and the complete divisor together, and annex two ciphers, for a neic trial divisor; with which obtain an- other trial figure in the root. VII. Multiply the unit figure of the last term i?i column I by 3, and annex the trial figure of the root for the next term of column I ; midtiply this residt by the trial figure of the root for the next term of column 11 ; add this term to the trial divisor for a complete divisor, with which proceed as before. Notes. 1. If at any time the product be greater than the dividend, diminish the trial iigure of the root, and correct the erroneous -work. 2. If a cipher occur in the root, annex two more ciphers to the trial divisor, and another period to the dividend ; then proceed as be- fore -with column I, annexing both cipher and trial figure. EXAMPLES FOR PRACTICE. 1. What is the cube root of 79.112 ? OPERATION. 79.112 I 4.2928 + , ^ng. 64. 4800 15112 122 244 5044 10088 529200 5024000 12G9 11421 540621 4865589 55212300 ' 158411000 12872 25744 55238044 110476088 5526379200 47934912000 128768 1030144 5527409344 44219274752 3714637248 rem. CUBE ROOT. 317 2. What is the cube root of 84G04-j19 ? Ans. 439. 3. AVhut is the cube root of 2o57947G01 ? Ans. 1331. 4. What is the cube root of 109G3240788375 ? Ans. 22215. 5. What is the cube root of 27UG71777U3218989G ? Ans. 64G8G6. G. Yv^hat is the cube root of .091125 ? Ans. .45. 7. What is the cube root of .000529475129 ? Ans. .0809. 8. AViiat is the approximate cube root of .008649 ? Ans. .2052 + . Extract the cube roots of the followinj: numbers : — 72 = 1.259921 + ^S = 1.442249+ 4'4 — 1.587401 + -^5 = 1.709975+ 4'G — 1.817120+ ^7 = 1.912931+ APPLICATIOXS IN CUBE KOOT. 1. What is the length of one side of a cistern of cubical form, containing 1331 solid feet? Ans. 11 feet. 2. The pedestal of a certain monument is a square block of granite, containing 37324§ solid inches ; what is the length of one of its sides ? Ans. 6 feet. 3. A cubical box contains 474552 solid inches ; what is tiie area of one of its sides ? Ans. 42^ sq. ft. 4. How much paper will be required to make a cubical box which shall contain §|- of a solid foot? Ans. f of a yard, 5. A man wishes to make a bin to contain 125 bushels, of equal width and depth, and length double the width ; what must be its dimensions ? Ans. W^idth and depth, 51.223 + inches; length, 102.44G + inches. XoTE. Spheres are to each other as the cubes of theii- diameters or circumferences. G. There are two spheres whose solid contents are to each other as 27 to 343 ; what is the ratio of their diameters ? Analysis. Since spheres are to each other as the cubes of their diameters, the diameters will be to each other as the cube roots of the spheres ; and ^27 = 3, ^343 = 7 ; hence the diameters required are as 3 to 7. 318 ARITHMETICAL PROGRESSION. 7. The diameter of a sphere containing 1 solid foot is 14.9 inches ; Avhat is the diameter of a sphere containing 2 solid feet? Ans. 18.7 + inches. 8. If a cable 4in. in circumference, will support a sphere 2ft. in diameter, what is the diameter of that sphere which will be uipported by a cable 5in. in circumference? Ans. 2.32+ ft. ARITHMETICAL PROGRESSION. 4S2. An Arithmetical Progression, or Series, is a series of numbers increasing or decreasing by a common difference. Thus, 3, 5, 9, 11, &c., is an arithmetical progression with an ascending series, and 13, 10, 7, 4, «&c., is an arithmetical pro- gression with a descending series. ^JB3. The Terms of a series are the numbers of which it is composed. 4JI-1. The Extremes are the first and last terms. 4:»I«S. The Means are the intermediate terms. 4S@. The Common Difference is the difference between any two adjacent terras. 43^. There are Jive parts in an arithmetical series, any three of which being given, the other two may be found. The-y are as follows : the Jirst term, last te)-ni, common differ' ence, numljer of terms, and sum of all the terms. CASE I. 4:38. To find the last term when the first term, common diflbrcncc, and number of terms are given. Let 2 be the first term of an ascending series, and 3 the common difference ; then the series will be written, 2, 5, 8, 11, 14, or analyzed thus : 2, 2-f3, 2-|-3 + 3, 2-|-3-|-3-f-3, 2 -\- 3 -j- ■'! -J- '•> -j— 3. Here we see that, in an ascending series, we obtain the x"rnnd term I)y adding the common dilf(>rence once to the first tiiiii ; the tliird terra, by adding the conniion difierence twice to the first term ; and, in general, we obtain any term by ARITHMETICAL PROGRESSION. 319 adding the common difference as many times to the first term as there ai'e terms less one. Note. The analysis for a descending series would be similar. Hence, Rule. 31idtiply the common difference hy the number of terms less one, and add the product to the Jirst term, if the series be ascending, and subtract it if the series, be descending. EXAMPLES. 1. The first term of an ascending series is 4, the common difference 3, and* the number of terms 19 ; what is the last term ? A7is. 58. 2. What is the 13th term of a descending series Avhose first term is 75, and common difference 5 ? Ans. 15. 3c A boj bought 18 hens, paying 2 cents for the first, 5 cents for the second, and 8 cents for the third, in arithmetical progression ; what did he pay for the last hen ? 4. What is the 40th term of the series ^, -|, 1, 1^, &c. ? Ans. 10^-. 5. A man travels 9 days; the first day he goes 20 miles, the second 25 miles, increasing 5 miles each day; how far does he travel the last day of his journey ? Ans. GO miles. 6. What is the amount of $100, at 7 per cent., for 45 years ? $100 + $7 X 45 = $415, Ans. CASE II. 4Si5. To find the common diiference when the extremes and number of terms are given. Referring to the series, 2, 5, 8, 11, 14, analyzed in -^SS, we readily see that, by subtracting the first term from any term, we have left the common difference taken as many times as there are terms less one ; thus, by taking away 2 in the fifth term, 2 -|- 3 -[- 3 -}- 3 -j- 3, we have 3 taken 4 times. Hence, Rule. Divide the difference of the extremes by the number of terms less one. 320 ARITHMETICAL PROGRESSION. EXAMPLES. 1. The first term is 2, the last term is 17, and the number of terms is 6 ; what is the common difference ? Aiis. 3. 2. A man has seven children, whose ages are in arithmetical progression; the youngest is 2 years old, and the eldest 14; what is the cammon ditference of their ages ? A7is. 2 years. 3. The extremes of an arithmetical series are 1 and 50^, and the number of terms is 34 ; what is the common difference ? 4. An invalid commenced to walk for exercise, increasing the distance daily by a conuuon difference^ the first day he walked 3 miles, and the 14th day 9^ miles; how many miles did he walk each day ? Note. "When we have found the common difference we may add it once, twice, Sec, to the fii-st term, and we have tl\e series, and conse- quently the means. Ajis. 3, 3i, 4, A^, 5, 5i, &c. CASE III. 410. To find the number of terms when the ex- tremes and common difference are given. Examining the series, 2, 5, 8, 11, 14, analyzed in 4^38, we also see that after taking away the jrrst term irom any term, we have left the common difference taken as many times as the number of terms, less 1. Hence, Rule. Divide the difference of the extremes ly the common difference, and add 1 to the quotient. EXAMPLES. 1. The extremes are 7 and 43, and the common difference is 4 ; what is the number of terms ? Aiis. 10. 2. The first term is 2^, the last term is 40, and the common difference is 7^ ; what is the number of terms ? Ans. G. 3. A laborer agreed to build a fence on the following con- ditions : for the first rod he was to have G cents, with an increase of 4 cents on each successive rod ; the last rod came to 22G cents; how many rods did he build ? Ans. 5G i*ods. GEOMETRICAL PROGRESSION. 321 CASE IV. 441. To find the sum of all the terms when the extremes and number of terms ^.re given. To deduce a rule for finding the swji of all the terms, we will take the series 2, 5, 8, 11, 14, writing it under itself in an inverse order, and add each term ; thus, 9 _j_ 5 4- 3 + 1 1 -f 14 = 40, once the sura. 14-1-11+ 8 -]- 5 -f 2 = 40, " " « lG-f-lG-]-lG-[-lG-fl6==:80, twice the sum. Here we perceive that IG, the sum of the extremes, multi- plied by 5, the number of terms, equals 80, which is tivice the sum of the series. Dividing 80 by 2 gives 40, which is the sum required. Kence, Rule. MuMlphj the sum of tlie extremes hy the niimher gJ terms, and divide the 'product hy 2. EXA:\irLES. 1. The extremes are 5 and 32, and the number of terms 12 ; what is the sum of all the terras ? Ans. '2'22. 2. How many strokes does a common clock make in 12 hours ? Ans. 78 strokes. 3. What debt can be discharged in a year by weekly pay- ments in arithmetical progression, the first being $24, and the last $1224? J^ns. $32448. 4. Suppose 100 apples were placed in a line 2 yards apart, and a basket 2 yards from the first apple ; how far would a boy travel to gather them up singly, and return with each separately to the basket ? Ans. 20200 yards. GEOMETRICAL PROGRESSION. 44®. A GeometricpJ Progression is a series of numbere inci'easing or decreasing by a constant multiplier. When the multiplier is greater than a unit, the senes is 14* 322 GEOMETRICAL PEOGRESSION. ascending; thus, 2, 6, 18, 54, 1G2, is an ascending series, in which 3 is the rauUiplier. Wlaen the nudtiplier is less than a unit, the series is descend- ing; thus, 1G2, 54, 18, 6, 2, is a descending series, in which \ is the muhipHer. ^143. The Katio is the constant rauhiplier. 4l:'14. In every geometrical progression there are five parts to be considered, any tliree of wliich being given, the other hvo may be determined. Tliey are as follows: T\\c Jirst term, last term, ratio, number of terms, and the sum of all tJie terms. The first and last terms are the extremes, and the intex'me- dlate terms are the means. CASE I. 44i5. To find any term, the first term, the ratio, and number of terms being given. Tlie first term is supposed to exist independently of the ratio. Using the ratio once as a factor, we have the second term ; using it twice, or its second power, we have the third term ; using it three times, or its third power, we have the fourth term ; and, in general, the power of the ratio in any term is one less than the number of the term. The ascending series, 2, 6, 18, 54, may be analyzed thus: 2, 2 X ">, 2 X 3X3, 2X3X3X3. In this illustration we see that 1st term, 2, is independent of the ratio. 2d " 6 = 2X3 = the first term into the 1st power of ihfi ratio. 3d term, 18 = 2 X 3^ 1= the first term into the 2d power of the ratio. 4th term, 54 =: 2 X 3 ^ = the first term into the 3d power of the ratio. Hence Rule, ^ftdtipli/ the first term hj that power of the ratio denoted hy the number of terms less 1. GEOMETRICAL rROGllESSION. 323 EXAMPLES. 1. The first term of a geometrical series is 4, the ratio is 3 ; what is the 9th term ? Ans. 4 X S^ zrr 2G244. 2. The first term is 1024, the ratio J, and the number of terms 8; what is the hist term ? Ans. yg^. 3. A boy bought 9 oranges, agreeing to pay 1 mill for the first orange, 2 mills for the second, and so on ; what did the last orange cost him ? Ans. $.256. 4. The first term is 7, the ratio j-, and the number of terms 7 ; what is the last term ? Ans. iq^xjj' 5. What is the amount of $1 at compound interest for 5 years, at 7 per cent, per annum ? Ans. $1.40255 -[-• Note. In the above example the first term is $1, the ratio is $1.07, and the number of terms is 6. 6. A drover bought 7 oxen, agreeing to pay $3 for the first ox, $9 for the second, $27 for the third, and so on ; what did the last ox cost him ? Ans. $2187. CASE II. 4411. To find the sum of all the terms, the ex- tremes and ratio being given. If Ave take the series 2, 8, 32, 128, 512, in which the ratio is 4, multiply each term by the ratio, and add the terms thus multiplied, we shall have 8 + 32 + 128 + 512 + 2048 =. 2728 ={,^- [Irten'J"" But 2 + 8 + 32 + 128 + 512z:r 682 = j ^ telt!"" "' '''" Hence, by subtracting, we get 2048 — 2 — 204G = { o/lnthTtmaJ'^'^ Dividing by 3, the ratio less one, 204G -|- 3 = 682 =r | ji .""emc.^"'" " The subtraction is performed by taking the lower line or series from the upper. All the terms cancel except 2048 and 2. Taking their difference, which is 3 times the sum, and di- viding by 3, the ratio less one, we must have the sum of all the terms. Hence o2-i PROMISCUOUS EXAMPLES. Rule. JlluUipIt/ the greater extreme hy the ratio, subtract the less extreme from the product, and divide the remainder by the ratio less 1. Note. Let every decreasing series be inverted, and the first term called the last ; then the ratio will be greater than a luiit. If the series be infinite, the first term is a cipher. EXAMPLES. 1. The first terra is 2, the last term 512, and the ratio 3 ; ■u'liat is the sum of all the terms? Ans. 707. 2. The first term is 4, the last term is 2G2144, and the ratio is 4; what is the sum of the series? Ans. 349524. 3. The first term of a descending series is 1G2, the last term 2, and the ratio \ ; what is the sum ? Ans. 242. 4. "What is -the value of \, ^-^, y^, &c., to infinity ? Ans. \. I^^OTE. In the following examples v.'c first find the last term by the Rule under Case I. 5. What yearly debt can be discharged by monthly pay- ments, the first being $2, the second $6, and the third $18, and so on, in geometrical progression ? Ans. $531440. 6. If a grain of wheat produce 7 grains, and these be sown the second year, each yielding the same increase,Jiow mar.y bushels Avill be produced at this rate in 12 years, if 1000 grains make a pint? Ans. 252315 bu. 4i qt. 7. Six persons of the Morse family came to this country 200 years ago; suppose that their number has doubled every 20 years since, what would be their number now ? Note. The other cases in Progression will be found in tlio Higher Arithmetic. PROMISCUOUS EXAMPLES. \. One half tho sum of two numbers is 800, and one lialf the difference of the same numbers is 200 ; what are the numbers ? Am. 1000 and 600. 2. What number is that to which, if you add f of {^^ of itseLP, the sum will be 01 ? Ajis. 55. 3. What part of a day is 3 h. 21 min. 15 sec. ? Ans. ^-[^^. PROMISCUOUS EXAMPLES. S25 4. A commission merchant I'eceived 70 bags of wheat, each con- taining 3 bu. 3 pk. o qt. ; how many bushels did he receive ? 5. Four men, A, B, C, and D, are in possession of $1100; A has a certain sum, B has twice as much as A, C has $300, and D has $200 more than C ; how many dollars has A ? Ans. $100. 6. At a certain election, 3000 votes were cast for three candi- dates, A, B, and C ; B had 200 more votes than A, and C had 800 more than B ; how manv votes were cast for A ? Ans. GOO. 7. What part of 17^ IS 31? Ans. f|. 8. Tlie difference between 4 and -I of a number is 10 ; what is the number ? Ans. 560. 9. A merchant bought a hogshead of rum for $28.3<5 ; how much v.'ater must be added to reduce the first cost to 35 cents per gallon ? Ans. 18 gal. 10. A and B traded with equal simis of money ; A gained a sum equal to l of his stock ; B lost $200, and then he had ^ as much as A ; liow much was the original stock of each ? Ans. $500. 11. A farmer sold 17 bushels of barlej-, and 13 bushels of wheat, for $31.55 ; he received for the wheat 35 cents a bushel more than for the barley ; what was the price of each per bushel ? A71S. Barley, $.90; wheat, $1.25. 12. What is the interval of time between JIarch 20, 21 minutes past 3 o'clock, P. M., and April 11th, 5 minutes past 7 o'clock, A. M. ? Ans. 21 da. 15 h. 44 min. 13. What o'clock is it when the time from noon is ^^^ of the . time to midnight ? Ans. 5 o'clock 24 min. P. M. 14. What is the least number of gallons of witie that can be ship- ped in either hogsheads, tierces, or barrels, just filhng the vessels, without deficit or excess ? Ans. 126 gal. 15. A ferryman has four boats ; one v>-ill carry 8 barrels, another 9, another 15, and another 16 ; V)hat is the smallest number of bar- rels that will make full freight for any one, and all of the boats ? 16. A and B have the same income; A saves |- of his, but B, by spending $30 a year more than A, at the end of four years finds himself $40 in debt ; what is their income, and how much does each spend a year .f* C Income, $160. A71S. < A spends $140. ( B spends $170. 17. If a load of plaster weighing 1825 pounds cost $2.19, how much is that per ton of 2000 pounds ? Ans. $2.40. 18. If 21 yards of cloth If yards wide cost $3.37-|, what will be the cost of 36-1 yards l-l yards wide ? Ans. $52. 779. 19. I lend my neighbor $200 for 6 months ; how^ long ought he to lend me $1000 to balance the favor ? 20. Bought railroad stock to the amount of $2356.80, and found that the sum invested was 40 per cent of what I had left ; what sum had I at first ? Ans. $8248.80. 21. 20 per cent, of f of a number is what per cent, of # of it ? Arcs. 121 325 PROMISCUOUS EXAMPLES. 22. Divide a prize of $10200 among 60 privates, 6 subaltern oificers, 3 lieutenants, and a commander, giving to each subaltern double the share of a private, each lieutenant 3 times as much as the subaltern, and to the commander double that of a lieutenant ; how much is each man's share? Ans. Com. $1200; eacli man, $100. 23. A is 51 miles in advance of B, who is in pursuit oi him ; A travels 16 miles per hour, and B 19 ; in how many hours \\ ill B overtake A ? 24. How much wool, at 20, 30, and 54 cents per pound, must be mixed with 95 pounds at 50 cents, to make the whole mixture worth 40 cents per pound ? Am. 133 lb. at 20 ; 95 lb. at 30 ; 190 lb. at 54 cents. 25. If 240 bushels of wheat are purchased at the rate of 18 bushels for $22^, and sold at the rate of 22^ bushels for $33.75, M'hat is the profit on the whole ? Ans. $60. 26. 'My horse, wagon, and harness together are worth $169 ; the wagon is worth 4 times the harness, and the horse is worth double the wagon , what is the value of each ? C Horse, $104. Ans. < Wagon, $ 52. ( Harness, $ 13. 27. The shadow of a tree measures 42 feet ; a staff' 40 inches in length casts a shadow 18 inches at the same time ; what is the height of the tree ? Ans. 93^ ft. 28. If a piece of land 40 rods long and 4 rods wide make an acre, how wide must it be to contain the same if it be but 25 rods long P A}is. 6| rods. 29. A, B. and C are employed to do a piece of work for $26.45 ; A and B together are supposed to do -f of it, A and C -^^, and B and C ^f^, and paid proportionally ; how much must each receive ? 30. If 12 ounces of wool make 2^ yards of cloth that is 6 quar- ters wide, how many pounds of wool will it take for 150 yards of cloth 4 quarters wide ? 31. Six persons. A, B, C, D, E, and F, are to share among them $6300 : A is to have i of it, B^, C |, ]) is to have as much as A and C together, and the remainder is to be divided between E and F in the proportion of 3 to 5 ; how much does each one receive ? 32. AVhat is the amount of $200 for 8 years at 6 per cent, com- pound interest ? Ans. $318,769. 33. A garrison, consisting of 360 men, was provisioned for 6 months ; but at the end of 5 months they dismissed so many of the men that the remaining provision lasted 5 months longer ; how many men were sent away P 34. A certain principal, at compound interest for 5 years, at 6 per cent., •will amount to $015;). 113 ; in wiiat time will tli-o same j)rinciiial amount to the same sum, at 6 per cent, simple interest? Ans. 5 yr. 7 mo. lO.S-f-da. 35. Paid $148,352 for 0728 feet of jiine hunber ; how much waa that per tliousand ? I PROxAIISCUOUS EXAMPLES. 327 3G. Comparlns^ two luimbers, 4S0 was found to he their least common multiple, and 123 their greatest common divisor ; what is the product of the numbers compared ? Ans. 11,10S). 37. Eight workmen, hiboring 7 hours a day for 15 days, were able to execute -^ of a job ; in how many days can they complete the residue, by working 9 hours a day, if 4 workmen are added to their number? Ans. lo| days. 38. If a hall 36 feet long and 9 feet wide require 3li yards of carpeting 1 yard wide to cover the iioor, how many yards \\ yards wide will cover a floor GO feet long and 21 feet wide ? Alls. 144 yards. 39. A, B, and C traded in company ; A put in $1 as often as B put in $3, and B put in $2 as often as C put in $<3 ; B's money was in twice as long as ("s, and A's twice as long as B's ; they gained $d2.50 ; how much was each man's share of the gain ? C A's, $12. .4/t5. -{B's, $18. (C's, $22.50. 40. A and B found a watch worth $45, and agreed to divide the value of it in the ratio of |- to -| ; how much was each one's share ? , S $20, A's. ^ $2o, B's. 41. A man received $33.25 interest on a sum of money, loaned 5 years preA'ious, at 7 per cent. ; what was the sum lent ? Ans. $95. 42. The diameter of a ball weighing 32 pounds is 6 inches ; what is the diameter of a ball weighing 4 pounds ? A^is. 3 inches. 43. Divide $360 in the proportion of 2, 3, and 4. Ans. |80, $120, $160. 44. If by working 6f hours a day a man can accomplish a job in 121 days, how many days will be required if he work Si hours per day? Ans. 9^^^ days. 45. An open court contains 40 square yards ; how many stones, 9 inches square, will be required to pave it ? Ans. 640. 46. A drover paid $76 for calves and sheep, paying $3 for calves, and $2 for sheep ; he sold 1 of his calves and |- of his sheep for $23, and in so doing lost 8 per cent, on their cost ; how many of each did he purchase ? Ans. 12 calves ; 20 sheep. 47. If a cistern, 17-^ feet long, lO-J- ])road, and 13 deep, hold 548 barrels, how many barrels will that cistern hold that is 16 feet long, 7 broad, and 15 deep ? Ans. 384 bbls. 48. If 12 men, working 9 hours a day, for 15|- days, were able to execute f of a job, how many men may be withdrawn, and the resi- due be finished in 15 days more, if the laborers are employed only 7 hours a day ? Ans. 4 men. 49. A general formed his men into a square, that is, an equal number in rank and file, and found that he had 59 men over ; and increasing the number in both rank and file by 1 man, he wanted 84 men to complete the square ; how many men had he ? Ans. 5100. 823 PROMISCUOUS EXAEiriES. 50. Bought wheat at $1.50 per bushel, corn at $.75 per bushel, and barley at $.60 per bushel ; the wheat cost twice as much as the corn, and the corn twice as much as the barley ; of the sum paid, §243 and -|- of the whole was for wheat, and $153 and ^V of the whole Avas for the corn ; how many bushels of grain did I purchase ? Ans. 756. 51. Divide $630 among 3 persons, so that the second shall have ■| as much as the first, and the third -^ as much as the other two ; what is the share of each ? CI st, $240. , Ans.VlA, $180. ^3d, $210. 52. Bought a hogshead of molasses for $28, and 7 gallons leaked cut ; at what rate per gallon must the remainder be sold to gain 20 per cent. ? 53. 20 per cent, of |- of a number is how many per cent, of 2 times -|- of 11 times the number ? Ans. 1^. 54. B and C, trading together, find their stock to be worth $3o00, of which C owns $2100 ; they have gained 40 per cent, on their first capital ; what did each put in ? . 5 ^> $1000. ^^^^- \ C, $1500. 55. If the ridge of a building be 8 feet above the beams, and the building be 32 feet wide, what must be the length of rafters ? 56. If 12 workmen, in 12 days, working 12 hours a day, can make up 75 yards of cloth, :| of a yard wide, into articles of clothing : how many yards, 1 yard wide, can he made up into like articles, by 10 men, working 9 days, 8 hours each day ? Ans. 23-j'^g. 57- A grocer sells a farmer ICO pounds of sugar, at 12 cents a pound, and makes a profit of 9 per cent. ; tlie former sells him 100 pounds of beef, at 6 cents a ])ound, and makes a profit of 10 per cent. ; who gains the more by the trade, and how much ? Ans. The grocer gains $.415+ more. 58. In 1 yr. 4 mo. $311.50 amounted to $336.42, at simple interest ; what was the rate per cent. ? Ans. G. 59. Three persons engage to do a piece of work for $20 ; A and B estimate that they do 4- of it, A and C that they do |- of it, and B and C that thcv do | of it ; according to this estimate, what jiart of the $20 should each man receive ? Ans. A's, $1 If ; B's, $55 ; C's, $2f 60. Paid $375, at the rate of 2^ i)er cent., for insurance on a cotton factory and the machinery ; for what amount Avas the policy given ? 61. A merchant bought goods in Boston to the amount of $1000, and gave his note, dated Jan. 1, 1857, on interest after 3 months; six months after the note was given he paid $560, and 5 months after the first pavment he paid $406 ; Avhat was due Aug 23, 1859 ? Ans. $0G.63-(- 62. If I of A's money be equal to -| of B's, and ■§- of B's be equal to ^ of C's, and -?- of C's ])e equal to ^ of D'e, and D has $45 more tlmn C, howjtnuch has each ? 4 S A, 8378 ; C, $360 ; ^^"^- I B, $336 ; D, $405. PROMISCUOUS EXAMPLES. 329 63. A owed B $900, to be paid in 3 years ; but at the expiration of 9 mouths A agreed to pay $300 if B would wait long enough for the balance to compensate for the advance ; how long should B wait after the expiration of the 3 years ? Aiis. 13-|- mo. C4. A certain clerk receives $800 a year ; his expenses equal J^ of Mhat he saves ; how much of his salary does he save yearly P 65. A merchant sold cloth at $1 per yard, and made 10 per cent, profit ; what would have been iiis gain or loss had he sold it at $.87^ per ya.rd ? Aiis. Loss, 3|- per cent. 0J_9_ 66. What is the cube of ^~ Ans. U. 29 J^ ^^ 63 67. What is the cube root of — ^ A'ls. f . 68. A miller is required to grind 100 bushels of provender worth 50 cents a bushel, from oats worth 20 cents, corn worth 35 cents, rye worth 60 cents, and' wheat worth 70 cents per bushel ; how many bushels of each may he take ? 69. A man owes $6480 to his creditors ; his debts are in arith- metical progression, the least being $40, and the greatest $500; required the number of creditors and the common difference between the debts. a ^24 creditors. ■^^"'^- } $20 diilerence. 70. Two ships sail from the same port ; one goes due north 128 miles, and the other due east 72 miles ; how far are the ships from each other? Ans. 146.86 -|- 7iiiies. 71. If 10 pounds of cheese be equal in value to 7 pounds of butter, and 11 pounds of butter to 2 bushels of corn, and 14 bushels of corn to 8 bushels of rye, and 4 bushels of rye to 1 cord of wood ; how many pounds of cheese are equal in value to 10 cords of wood ? Ans. 550. 72. A and B traded until they gained 6 per cent, on their stock ; then I of A's gain was $18 ; if A's stock was to B's as f to ^, how much did each gain, and what was the original stock of each ? . 3 ^'s g'^i^i) '^'l^ ; stock, $750. "^"^•^B's " $37.50; " $825. 73. If 20 men, in 21 days, by working 10 hours a day, can dig a trench 30 ft. long, 15 ft. wide, and 12 ft. deep, when the ground is called 3 degrees of hardness, how many men, in 25 days, by v\ork- ing 8 hours a day, can dig another trench 45 ft. long, 16 ft. wide, and 18 ft. deep, when the ground is estimated at 5 degrees of hardness ? Ans. 84. 74. Wishing to know the height of a certain steeple, I measured the shadow of the same on a horizontal plane, 27-|- feet ; I then erected a 10 feet pole on the same plane, and it cast a shadow of 2|- feet ; what was the height of the steeple ? Ans. 1031 ft, 75. A can do a piece of work in 3 days, B can do 3 times as much in 8 days, and C 5 times as much in 12 days ; in what time can they all do the first piece of work ? Ans. ^ da. 830 PROMISCUOUS EXAMPLES, 76. A person sold tMo farms for $1890 eacli ; for one he roccived 25 per cent, more than its true value, and for the other 2o per cent, less than its true value ; did he gain or lose by the sale, and how much ? Alls. Lost $252. 77. Three men paid $100 for a pasture; A put in 9 horses, 1} 12 cows for twice the time, and C some sheep for 2-^ times as long as B's cows ; C paid one half the cost ; how many sheep had he, and how much did A and B each pay, i)rovided G cows eat as much as 4 horses, and 10 sheep as much as o cows ? ^ C had 25 sheep. Ans. ^ A paid $18. (B " $32. 78. A man purchased goods for $10500, to be paid in three equal installments, without interest ; the first in 3 months, the second in 4 months, the third in S months ; how much ready money will pay the debt, money being worth 7 per cent. ? Ans. $10203.94 -(-• 79. A farmer sold 50 fowls, consisting of geese and turkeys; for the geese he received $.75 apiece, and for 'the turkeys $1.25 apiece, and for the Avhole he received $52.50 ; how many were there of each ? Ajis. 20 geese, 30 turkeys. 80. There is an island 73 miles in circumference, and 3 footmen start together and travel around it in the same direction ; A goes 5 miles an hour, B 8, and C 10; in Avhat time will they all come together again if they travel 12 hours a day ? Ans. Gda. 1 h. 81. A, B and C are to share $100000 in the proportion of -^, l, and \, respectively; but C cl3-ing, it is required to divide the whole sum proportionally between the other two ; how much is each one's share? . 5 A's, $57142.854. ^"^•^B's, $42857. 14f 82. A, B, and C have 135 sheep; A's plus B's are to B's plus C's as 5 to 7, and C's minus B's to C's plus B's as 1 to 7 ; how many has each ? Ans. A, 30 ; B, 45 ; C, CO. 83. A man sold one hog, weighing 250 pounds, at 4 cents per pound ; a second, weighing 300 pounds, at 4^ cents ; and a third, weighing 309 ])ounds, at 5 cents ; what was the average price per jjound for the whole P Ans. 4-|j-| cents. 84. In a certain factory are employed men, women and boys ; the boys receive 3 cents an hour, the women 4, and the men 6; the boys work 8 liours a day, the women 9, and the men 12 ; the boys receive $5 as often as the Avomen $10, and for every $10 ))aid to tlie women, $24 are paid to the men ; how many men, women, and boys are there, the whole number being 59 ? Jns. 24 men, 20 women, 15 boys. 85. A fountain has 4 receiving pipes. A, B, C, and ]) ; A, B, and C will fill it in G hours, B, C, and ]) in 8 hours, C, D, and A in 10 hours, and D, A, and B in 12 hours; it has also 4 discharging pipes, W, X, Y, and Z ; AV, X, and Y will empty it in G hours, X, Y, and Z in 5 hours, Y, Z, and "W in 4 liours, and Z, "W. and X in 3 hours ; suppose the pipes all open, and the fountain full, in what time would it be emptied ? -^I;;*'. Gj'L h. PROMISCUOUS EXAMPLES. 331 86. How many building lots, each 75 feet by 125 feet, can be laid out on 1 A. 1 11. 6 P. 18^- sq. yd. ? Jns. 6. 87. A man bought a house, and agreed to pay for it $1 on the first day of January, $2 on the hrst day of February, $4 on the first day ]\lareh, and so on, in geometrical progression, through the year ; what was the cost of the house, and what the average time of payment ? . ^$1095. ■ ( Average time, Nov. 1. 88. A man sold a rectangular piece of ground, measuring 44 chains 32 links long by 3G chains wide ; how many acres did it contain ?^ Ans. 159 A. 2 11. 8.32 P. 80. What number is that which being increased by its half, its tliird, and 18 more, M-ill be doubled ? Ans. 108. 90. A merchant has 200 lb. of tea, worth $.62^ per pound, which he will sell at $.56 per pound, provided the purchaser will pay in coflee at 22 cents, which is Morth 25 cents per pound ; does the merchant gain or lose by the sale of the tea, and how much per cent. ? Ans. gained l-fj per cent. 01. A man owes a debt to be paid in 4 equal installments at 4, 9, 12, and 20 months, respectively; discount being allowed at 5 per cent., he finds that $750 ready money will pay the debt; how much did he owe .P Ans. $784.74-}-. 92. A and I> traded upon equal capitals ; A gained a sum equal to "I of his capital, and B a sum equal to -IL of his; B's gain was $500 less than A's ; what was the capital of each ? Ans. $4000. 93. I purchase goods in bills as follows: June 4, 1859, $240.75 ; Aug. 9, 1859, $137.25; Aug. 29, 1859, $05.04; Sept. 4, 1S59, $230.36; Nov. 12, 1859, $36. If the merchant agree to allow credit of 6 mo. on each bill, Avhen may I settle by paying the whole amount? " Ails. Feb. 1, 1860. 94. A young man inherited a fortune, i of which he spent in 3 months, and 4 of the remainder in 10 months, when he had only $2524 left ; how much had he at first ? Ans. $5889.33 -[-. 95. A man bought a piece of land for $3000, agreeing to pay 7 per cent, interest, and to pay princij^al and interest in 5 equal au- r^ual installments ; how much was the annual payment ? ^4)?.'?. $731.67 -[-. D6. I have three notes payable as follows : one for |200, due Jan. 1. 1859, another for $350, due Sept. 1, and another for $500, due April 1, 1860 ; what is the average of maturity ? A71S. Oct. 24, 1850. 97. A man held three notes, the first for $600, due July 7, 1859; the second for $530, due Oct. 4, 1859 ; and the third for $400, due Feb. 20, 1860 ; he made an equitable exchange of these Milh a s])eculator for two other notes, one of Mhich was for $730, due Nov. 15, 1859; what was the face of the other, and when due? . 5 Face, $800. ^"*- ) Due Aug. 29, 1859. 332 MENSURATION. ilEXSURATIOX OF LINES AND SUPERFICIES. 449'. In taking the measure of any line, surface, or solid, we are always o-overncd by some denomination, a unit of wliich is called the Unit of Measure. Thus, if any lineal measure be estimated in feet, the unit of measure is 1 foot ; if in inches, the unit is 1 inch. If any superficial measure be estimated in feet, the unit of measure is 1 square foot ; if in yards, the unit is 1 square yard. 44§. If any solid or cubic measure- be estimated in feet, the unit of measure is 1 cubic foot ; if in yards, the unit is 1 cubic yard. 419. The area of a figure is its superficial contents, or the surface included within any given lines, without regard to thickness. 430. An Oblique Angle is an angle greater or less than a right angle ; thus, ABC and C B D are oblique angles, CASE I. 451. To find the area of a square or a rectangle. 4.'52. A Square is a figure having four equal sides and four right angles. 453. A Kectaugle is a figure having four right angles-, and its opposite sides equal. Rule. Multiply the length ly the breadth, and the product icill be the square contents. EXAMPLES FOR PKACTICE. 1. IIov.- many square inches in a board 3 feet long and 20 inches wide? " ^ns._ 720. 2. A man bought a farm 198 rods long and loO rods wide, and af^reed to give $32 an acre ; how much did the farm cost him ? An^. 85940. 3. A certain rectangular piece of land measures 1000 links by 100 ; how many acres does it contain ? Ans. 1 A. CASE II. ^5S. To find the area of a rhombus or a rhomboid. 455. A SllOinbus is a figure having four equal sides and four oblique angles. 450. A Ehoniboid is a figure having its opposite sides equal and parallel, and its angles oblique. Note. The square, rectanp;lc, rhombus, and rhomlioid, havinp; their op- posite sides parallel, are called by the general name, parcdlehxjram. It is proved in geometry that any parallelogram is equal to a rec- tangle of the same length and width ; hence the MENSURATION. 333 Rule. Multiply the lengtJi by tJie shortest or perpendicular dis- tance between tioo opposite sides. EXAMPLES FOR rUACTICE. 1, A meadow in the form of a rhomboid is 20 chains long, and the shortest distance between its longest sides is 12 chains ; how many days of 10 hours each will it take a man to mow the grass on this meadow, at the rate of 1 square rod a minute ? Ans. 6 da. 4 h. 2. Tlic side of a plat in the form of a rhombus is 15 feet, and a perpendicular drawn from one oblique angle to the side opposite, will meet tliis side 9 feet from the adjacent angle ; what is the area of the plat ? Ans. 180 sq. ft. CASE III. 457. To find the area of a trapezoid. 45§. A Trapezoid is a figure having four sides, of wliich two are parallel. The mean length of a trapezoid is one half the sum of the. parallel sides ; hence the PtULE. Multiply one lialftlie sum of the parallel sides by the per- ITcndicular distance between them. EXAMPLES FOR PRACTICE. 1. "What are the square contents of a board 12 feet long, 16 inches wide at one end, and 9 at the other ? Ans. 12-^ sq. ft. 2. "^Vhat is the area of a board 8 feet long, 16 inches wide at each end, and 8 in the middle ? Ans. 8 sq. ft. 0. One side of a field is 40 chains long, the side parallel to it is 22 chains, and the perpendicular distance between these two sides is 2o chains ; how many acres in the field ? Ans. 77 A. 5 sq. ch. CASE IV. 459, To find the area of a triangle. 450. The Base of a triangle is the side on which it is supposed to stand. 461. The Altitude of a triangle is the perpendicular distance from the angle opposite the base to the base, or to the base produced or extended. 462. A Triangle is one half of a parallelogram of the same base and altitude ; hence the Rule. Mvltiply one half the base by the altitude, or one half the altitude by the base. Or/ Midtiply the base by the altitude, and divide the product by 2. EXAMPLES FOR PRACTICE. 1. How many square yards in a triangle whose base is 14S feet, and perpendicular 45 feet? Ans. 370yds. 334 MENSURATION. 2. The gable ends of a barn are each 28 feet wide, and tlie per- pendicular height of the ridge above the eaves is 7 feet ; how many feet of boards will be required to board up both gables P Alls.. IDG feet. CASE V. 46.3. To find the circumference or the diameter of a circle. 464. A Circle is a figure bounded by one miiform curved line. 46.5. The Circumference of a circle is the curved line bounding it. 466. The Diameter of a circle is a straight line passing through the center, and termina- ting in the circumference. It is proved in geometry that in every circle the ratio between the diameter and the circumference is 3.1416 -)-. Hence the Rule. I. To find the circumference. — Mult!pJ>j the diamefer by 3.1416. II. To find the diameter. — Multiply the circumference by .3183. EXAMPLES FOR PRACTICE. 1. What length of tire will it take to band a carriage wheel 5 feet in diameter ? Ans. 15 ft. 8.4 -|- in. 2. AVhat is the circumference of a circular lake 721 rods in diameter ? Ans. 7 mi. 25 rds. 1.54 -\- ft. 3. Vrhat is the diameter of a circle 33 yards in circumference ? Ans. 10.5 -\- yards. CASE VI. 467. To find flie area of a circle. From the principles of geometry is derived the following Rule. I. When both diameter and circumference are given ; — Mnltiplij the diameter by the circumference, and divide the inoduct &y4. _ II. AVTien the diameter is given; — Multiply the square of the diameter by .7854. III. When the circumference is given ; — Multiply the square of the circunij'erence by .07958. EXAMPLES FOR PRACTICE. 1. The diameter of a circle is 113, and the circumference 355 ; Avhat is the area ? Ans. 10028.75. 2. What is the diameter of a circular island containing 1 square mile of land ? Ans. 1 mi. 41 rd. 1.4 -F ft. 3. A man has a circular garden requiring 84 rods of fencing to inclose it; huw nuich laud in the garden ? Aiis. 3 A. 81.5-}- 1*. I MENSURATION. 335 JiIENSURATIOX OF SOLIDS. 468. A Solid or Body is a magnitude which has lengtli, breadth, and thickness. CASE I. 4G9. To find the cubic contents of a prism, cube, or cylinder. 470. A Prism, is a solid whn.se bases or ends are any similar, equal, and parallel plane figures, and whose sides are parallelograms. 4:71. A Cylinder is a body whose bases ^g^g or ends are equal and parallel circles, and ^ whose side is a uniform curved surface. 4'y2. The Altitude of a prism, cube, or cylinder, is the perpen- dicular distance between the two bases ; it is the leiujth of the body. To estimate the solid contents of any one of the bodies defined under this case Rule. Multiphj the area of the base hy the altitude. EXAMPLES FOR rUACTICE. 1. The side of a cubic block measures 8 inches ; hoAV many cubic inches does it contain ? Ans. 512. 2. The end of a prism 20 feet long is a right-angled triangle, the two shorter sides of wliich measure 9 and 12 inches ; what are the cubic contents of the prism ? Ans. 1\ cu. ft. 3. A stick of timber is 2-3 ft. 3 in. long, 1 ft. 8 in. wide, and 18 in. thick ; how much will it come to at 8 cents per cubic foot ? Ans. $5.05. 4. A cistern is 5\ feet in diameter, and 8 feet deep ; how many standard wine gallons will it contain ? ^4??.?. 1421.7984 gal. Notes. 1. The mean or average diameter of a barrel or cask may be found, by adding to the head diameter 2, or, if the staves be but little curv- ing, Yo °^ t'^^ difference between the head and bung diameters. The cask will then be reduced to a cylinder, and its contents found by the above rule. 2. The process of estimating the capacity of barrels or casks is called gaKging. 0. The head diameter of a cask is 22 inches, the bung diameter 28 inches, and the length 31 inches ; how many wine gallons will it contain? Ans. 71.2504. 6. The head diameter of a cask is 30 inches, the bung diameter 35 inches, and the length 40 inches ; what is its capacity ? 335 MENSURATION. CASE II. 473. To find the cubic contents of a pyramid or a cone. 474. A Pyramid is a solid whose base, is any plane figure, and whose sides are triangles terminat- / ing in a point at the top. ^75. x\ Cone is a solid whose base is a circle, and whose side is a curved surface terminating in a point at the tojj. Rule. Multiply the area of the base hy \ of the altitude. EXAMPLES FOR PRACTICE. 1. What are the solid contents of a pyramid 15 feet square at the base and 40 feet high ? Ans. 3000 cu. ft. 2. A pyramid has a triangular base, each side of Avhich is 30 inches, and the altitude of the pyramid is 4 feet ; what are the cubic contents ? Ans. 3.6 -f- cu. ft. 3. The base of a cone is 7 feet in diameter, and the altitude IG feet 9 inches ; what are the solid contents ? Ans. 214.87 -|-_cu. ft. 4. A heap of grain, in the form of a cone, is 4 feet high, and measures 15 feet round the base ; how many bushels does it con- tain? Ans. 19 bu. 5.9-j- qt. CASE III. 476. To find the surfoce or the solid contents of a sphere. 477. A Sphere or Glohe is a solid bound- ed by a single curved surface, which m every part is equally distant from a point within called Its center. Hence Rule. I. To find the surface ; —Multiply the square of the diameter hy 3.1416. II. To find the solid contents ; — Midtiply the cube of the diameter by .5236. EXAMPLES FOR PRACTICE. 1 . How many square inches on the surface of a globe 1 5 inches in diameter? yl».9. 706.S6. 2. The diameter of a sphere is 18 inches ; what is its solidity? 3. "What is the solidity of a ball that can just be put into a cylin- drical cup 5 inches in diameter and 5 inches deep ? Ans. 65.45 cu. in. ! m AMERICAN EDUCATIOML SERIES A coMrLETj :; r of mdatid ^ "Bxti^nt ibe i^txt-^aohs SCHOOLS, ACADEMIFS. '■ JTES, AND COLLEGES. ROBINSON'S SYSTEM i^ MATHEMATICS. PROGRESSIVE TABLE-BOOK. strated, - - - - PROGRESSIVE PRIMARY AR! i' ^ ^ETIC, - - - - PROGRESSIVE INTELLECTU L ARITHMETIC, - - RUDIMENTS OF WRITTEN >