rrtJ U .0^^ >- IvlBRARV OF THK University of California. (-.TKX OK Receive J ^^l^ ■ iSgy . Accession No. 6 7v^^ • Class No. CJIT ! I Evening Normal School ! mm Twm li This Book must be returned to the Principal at the end of the term. *tii i nim i fnm i Digitized by tine Internet Arciiive in 2007 with funding from IVIicrosoft Corporation http://www.arcliive.org/details/commonschoolaritOOfrenrich PUBLISHERS^ NOTICE. FRENCH'S ARITHMETICS. This Series consists of Five Books, viz, : I. FIRST LESSONS IN NUMBERS. II. ELEMENTARY ARITHMETIC, in. MENTAL ARITHMETIC. IV. COMMON SCHOOL ARITHMETIC. V. ACADEMIC ARITHMETIC. (lu Preparation.) Tlie Publishers present this Series of Text-Books to American Teach- ers, fully believing that they contain many new and valuable features that will especially commend them to the practical wants of the age. The plan for the Series, and for each book embraced in it, was fully matured before any one of the Series was completed ; and as it is based upon true philosophical principles, there is a harmony, a fitness, and a real progressiveness in the books, that are not found in any other Series of Arithmetics published. Entered, according to Act of Congress, in the year 1869, by HARPER & BROTHERS, In the Clerk's Office of the District Court of the United States for the Southern District of New York. ■'^-^^^^^ ^ ==:^ Business men generally agree in the statement that, after leaving school. they were obliged to learn, and in many cases to devise for themselves, methods of computation adapted to their use in business. The universality of this experience led the Author of this Series of Arithmetics to a careful and protracted investigation into the philosophy of the development of the mathematical powers of the mind, and to a critical examination into the present methods of teaching Arithmetic. From these investigations he has become convinced that, in order to make practical arithmeticians^ a radical change in the plan of text-boolis upon the subject is necessary. The acknowledged requisites of a good method of instruction are these: 1st. It must be adapted to the nature of both subject and learner ; 2i. It must be an uninterrupted progress from the easy to the difficult ; 3i. It must use facts known to the learner, in imparting to him a knowl- edge of the unknown subject ; 4ih. It must regard the natural order of development of the subject, and present it in that order ; hth. It must arrange the substance of the facts presented under each general division, into brief summaries, recapitulations, or general principles ; Qth. It must thoroughly reach the understanding of the learner. This book fully recognizes these requisites, in every Chapter, Section, and Case. The attention of teachers and parents is particularly invited to the following distinctive features of the work : Order of Subjects. — A philosophical arrangement of subjects has been carefully observed. The four classes of numbers, Integers, Decimals, Com- pound Numbers, and Fractions, are presented in the order here stated. But Factors and Multiples precede Fractions, as a knowledge of the former subject is essential to a clear understanding of the latter. The Chapter '1 Converse Operations presents, in immediate connection, those operations J?, numbers which are the converse of each other. The succeeding Chapters >i'esent the subjects of Percentage, Ratio and Proportion, Evolution, Pro- gressions, Mensuration, and Miscellaneous Problems. Principles and Rules. — Each new process or method of computation is introduced inductively, and the Principles which follow are evident sequents of the Inductions. The learner is required to make practical ^applications of these Principles in solving a number of Problems, thus giving him a thorough comprehension of the principles upon which the practical calculations of after life are based. Rules are then presented, v.'hich arc uniformly based upon the Principles previously established. IV PKEFACE. IllustTations.— The pictures are designed to aid the pupil in acquir- ing a clear understanding of the subjects and principles which they illustrate, and also to cultivate a taste for the beautiful in art. The Italic Figures, which were cut expressly for this Scries, also add much to the beauty and attractiveness of the book. Problems.— Much labor has been bestowed upon the preparation of the Problems, to make them the vehicle of practical knowledge. Nearly every occupation, trade, profession, and art has its own peculiar business terms, and its peculiar articles of commercial exchange, with their appropriate market values. The use of these business terms, and transactions in these articles of exchange, make up the every-day matters of the business w^orld. Thus, the merchant, the manufacturer, the grocer, the druggist, the phy- sician, the lawyer, the printer, the bookseller, etc., etc., has each his own statistics of business ; and from these have been prepared problems that convey to the pujsil a great amount of knowledge of the principles, customs, and details of business. Such problems are all the more interesting from their practical utility, while they are none the less illustrative of the Prin- ciples of Numbers. The Problems are all prepared from materials collected for this book, and present statements and business terms in a correct business way. In this way the arithmetic of the school-room may be made to meet, in a considerable degree,, the practical wants of after life. Abstract problems possess little interest for pupils, and hence few of them are found in this book : but the number of practical problems drawn from the every- day transactions of business, greatly exceeds that in any similar work. Useless Matter. — Reduction of State Currencies is nearly obsolete, Alligation Alternate is merely a curiosity of numbers, English Money is of little value to American youth, and many denominations in Compound Numbers often heard of in the school-room are unknown in business. Those who look for these matters in this book will look in vain. New and Distinctive Features will be found in the Notation, Multiplication, and Division of Integers and Decimals ; Compound Num- bers ; Factors and Multiples ; Division of Fractions ; Converse Operations ; Table of Legal Eates of Interest, from official sources; Classification of all computations in Percentage under Five Ceneral Cases ; Rules for Interest and Partial Payments ; Average of Accounts ; Proportion; Evolution; Pro- gressions; Review and Miscellaneous Problems; and the deductions of Principles from Inductions, and the basing of Rules upon Principles. In the preparation of this work for American Schools, the Author has had constantly in mind the present condition and the "future requirements of American youth. The work differs, both in general plan and details, from other works upon the same subject. It is confidently believed that the adoption and introduction into schools of the distinctive features of the book will eliect a change in making good^ pra-^Aical AritJiimticiam, A "Word with. Teacliers, — A hint, a snggestion^^r an item of information not found in the body of a text-book, -will often awaken thonglit, and start a train of inqui- ries in a class, that will greatly increase their interest in the study. Connected with the subject Arithmetic, and not found in text-books, are many items of interest and im- portance to pupils, to which their attention should, at the proper time, be directed. This Manual is intended to give you brief hints and suggestions, that will enable you the better to give instruction to your pupils in this important branch of study. Page 9. Arithmetic, as a Science, logically investigates and philosophically classifies and arranges the principles and rules of the subject ; as an Art, it applies the principles and rules for computations, to the practical affairs of lire. 10. As the Roman Notation is not presented in this book, it may be well to spend time enough in giving oral instruction upon the subject, to make pupils familiar with the following facts and their application : I. The Roman Notation employs seven letters to represent numbers. 11. Each letter has a fixed value when used alone. Thus, I. =1, V. =5, X. =10, L. =50, C.=100, D.=500, and M. = 1,000. III. Repeating a letter repeats its value. Thus, II.=2, XXX.=30, CC.=200. IV. Whejb any letter stands after another expressing greater value, the number ex- pressed is the sum of the values of the two letters. Thus, VI.=6, XV,=15, CXVI.=116. V. When any letter stands before another expressing greater value, the number ex- pressed is the difference of the values of the ttco letters. Thus, IV. =4, IX. =9, XC.=90, CM. =900. VI. When a letter stands betioeeniwn others, loth of greater value, its value is taken from the sum of their values. Thus, XIX.=19, CIV.=104, MXL. =1,010. 13. Beginners require abundant practice both in writing and reading numbers. Give them numerous exercises on each new period of figures. 15. Explain clearly that the simple value is the number of ones or units expressed by the figure, and the local value is the value given to these units by the place; i. e., one is the value of the figure determined by its form, the other by the place it occupies. Bull pupils may be aided in learning to write and read numbers, by allowing them at first to write a Skeleton of JVotaiion, consisting of periods of ciphers; thus, 000,000,000,000; and under this to write the figures of any given number, in their proper places. They should also learn the Family Name of each period, — as ones, thou- sands, millions; and the nam.es of ike places in each family, — as ones, tens, hundreds. 19, Give original problems under each Section and Case in this Chapter, before as- signing the problems from the book. This may be done in various ways ;— the following has been found a good one : At the close of a recitation, or other convenient time, put problems on the blackboard, and let the class copy them upon their slates. At the com- mencement of the next recitation, call for the results (and solutions also, if you choose), to all the problems given out at the previous lesson, 25. More mistakes occur in addition than in any other process. Thorough drills in adding columns whose sums reach 100, will greatly lessen this defect. These 26 problems (41—67) afford this kind of drill ; they sliould not be skipped. VI , MANUAL. 30. Those teachers who prefer to use the "borrowing 10" method of subtraction, will find an explanation of that method in the Elementary Arithmetic of this Series. 32. An explanation oi Left-hand Subtraction sometimes assists pupils in acquiring a clear understanding of the process commonly called " borrowing ten." The following course will enable you to make the explanation : \st. Solve the problems in Case I., page 28, commencing at the left hand to subtract 2(7. Take a subtrahend whoso right-hand figure only exceeds the corresponding figure i.i the minuend (as 582 — 347), commence at the left to subtract, calling the tens figure of the minuend 1 less, and adding the 1, as a ten, to the ones of the minuend. M. Proceed in a simila»manner with problems in which other orders of units of the subtrahend exceed the like orders in the minuend, as shown in the following; Explanation.— Since 7 is more than 3, Ave have 4—1 = 3. Since 2 is more solution. than 0, we have 12—7—5. Since 4 is not more than 6, we have 10—2=8. 53062 Since 8 is more than 2, we have 5— 4=1. And 12— 8=4. 17248 With practice, pupils will acquire great facility in left-hand subtraction. 35814= 34. The market grades or qualities of some kinds of goods are indicated by certain marks upon the cask or package. Thus, in sugars, we have A or " straight A," 4ior '• diamond A," ® or " circle A," B, and C ; etc. V 38. In explanations of solutions, use the true multiplicand for the multiplicand. 52. A clear understanding of this Principle will aid the pupil in the induction to division of decimals. 57. These problems should be solved by both long and short division. 59. In long division, require pupils to memorize the catch-words, divide, multiplij, subtract, bring down. And indelibly impress upon their memories this fact ; For every figure of the dividend used after the first quotient figure is obtained, there must be a figure in the quotient. 61. Referring to 93, Note 1, teach the pupil how to write a quotient containing a fraction; as 142|. 69. Instruct pupils, when explaining solutions, to tell what is given, and what is re- quired. Thus, " In problem 4 are given a number and all its 2>arts but one. To find this part, I subtract the sum of the given parts from, the number."' In problem'9 exjjlain the use of the '* or repeating marks. 70. Problem 16. " As per annexed schedule " is a {30). common business reference to a bill or memorandum on Bountij, $949 the same paper. Call the attention of pupils to the signi- ^ ''^°- ®' ^i^Z. ^Vii fication of the commercial terms found in the problems. -tn n n yyZ 2'*! 71. -P^'oWem 30. Cultivate neatness, order, and method 7 " " 18— 126 in blackboard work. Thus, the solution of this problem 36 mo. $1512^ might be placed upon the blackboard as here shown. $1512-*-36=$42 perm,o. Explain the meaning of Average, and how it is found. 72. The Areas in this Table are from standard authorities. A little ingenuity will enable you to greatly increase the number of problems of the kind found on this page. 76. The only place in which and is properly used in reading numbers, is in a mixed mimber, after the integer. Thus, 5.7, 5 and 7 tenths; '^J'-, 4 and 9 sixteenths. 78. Pupils should carefully compare Art. 129, 130, with Art. 25, 26, 27. 80. Familiarize pupils with the reading of such numbers as 1400.010 ; 1000.410 ; .1400. 81. ) Instruct pupils to place the decimal point in the result, before adding 57 83. ) or subtracting the ones. Require them to subtract without writing 5.91 decimal ciphers in the minuend over decimal figures in the subtrahend, 94. Encourage pupils to solve problems in different ways. Thus, " In how many ways can this problem (23) be solved ?" " Which way do you prefer, and why?" MANUAL. Vii 97. First Reference. — The coins are shown in the cut, page 95, in perspective, and of course only the longest diameter is correct in measurement. Second Reference.— Gold and silver coins are alloyed, to make them hard enough for nse as a circulating medium, without depreciating perceptibly in value by wear. Third liefet^ence.—FuTpils must make all their computations in decimals ; and express parts of a cent, in final results, in fractions, when they are halves, fourths, or eighths. 113. " Fixed Standards" are weights and measures established by the General Gov- ernment, or recognized and sanctioned by custom. 115. Abbreviations of denominations should always be written after the numbers, and be followed by periods. The signs $ and £ are written before the nnmbers. 118. By English Statute Law, a heaped bushel is 18^- inches in diameter, 8 inches deep, and heaped in a true cone to the height of 6 inches. Tliis cone, 18| inches in diameter and 6 inches high, is 1 peck. 125. ) The denominations of square and cubic measures are used only in computa- 126. ) tions, the measurements being taken in linear units. 131. Each lot of a Government Section of land is divided into 2 Forties. Hence, 1 section (640 A.)=4 quarter-sections (160 A. each)=8 80-acre lots (or 80's)=16 40's. 135. 1 solar year is 5 h. 4S min. 48 sec. longer than a common year. 4 " years are 23 " 15 " 12 " " 4 " years. 100 " " " 24 da. 5 « 20 " " 100 " " 400 " " " 96 " 21 " 20 " " 400 " " Hence, if 97 days be added to every 400 years, the calendar will be only 2 h. 40 min. ahead of true time. These 9T days are distributed among 97 leap-years. (See 234.) 137. Require pupils to point out, on all diagrams, the lines defined. 138. A geogi-aphic mile=Jjj of 69.16 Eng. mi.=l'.15T*j mi.=l mL 48.85 rd. 142. The commonly recognized units of the other Tables are, for Canada Money, Dollar; Sterling Money, Pound; Wood Measure, Cord; Surveyors' Linear Measure, Foot and Chain ; Surveyors* Square Measure, Square Chain and Acre ; Time, Day and Year ; Circles, Degree, Right Angle, and Circumference. 144. As none of the tables or denominations of the Metric System have come into actual use, a presentation of theTables is all that the present state of the subject demands. 150. ) Problems like 17 are easily solved by left-hand subtraction. 152. > Commencing at the left, we have 17 mi. — 14 mi, =3 mi. 3 rd. (4 rd.-l rd.)— rd.=3 rd. 5i yd. + 2 yd. ^7 mi. 4 rd. 2 yd. 1 ft. r-7iyd.,and7iyd.— 4yd.=3iyd.=3yd.lft.6in. 1 ft, ^ *^ ^ ^ 6 in. + 1 ft. = 2 ft. 6 in., and 2 ft in.-2 ft. =0 ft. 6 in. ^ "**• 3rd.3yd.O ft. 6 in. 167. If the multiplier or divisor is less than 1, the first two principles will be re- versed. 178. The following method of finding the least common multiple is preferred by many. Take, for example, Problem 13, page 178. We first arrange the numbers from least to great- J?, i>^, 'i^, J^, 35, 45, 60, 72 5 est, and cancel or drop such as are factors of any 7, 9, 12, 72 of the others. We next divide through by any prime 7, 3, 4, 24 number that is a factor of any two, or by any number Y, Y, 4, 8 that is a factor of all the given numbers; and divide ^r ^ j^ ^j these results and the undivided numbers in the same >yy^2'>i.5')(.3y.3y.4:=2520 manner ; and so on, until the final quotients are prime to one another. The divisors and final quotients are the factors of the least common multiple. 183. Show wherein these General Principles are similar to the General Principles of Division, Art. 278. 187. The fractional unit J is 6 times as great as the fractional unit ^V That is, The less the denominator of a fraction, Vie greater is its fractional unit. VUl MANUAL. 193 Mixed numbers are readily subtracted by left-hand subtraction. 7^ = 7s*s- For example, take Ex. 2, page 193. Since l^ is more than ^^, we have 5|=3|2 0-3=3, and -^r-tS^BS- S^i 206. Give several original problems like Problems 9, 10. Require pupils to write out a full explanation of a solution, and therefrom deduce the Principle stated in this Note. 211. Decimal figures which continually repeat, are called a Repetend. Its value is expressed by a fraction with the repeating figures for the numerator, and as many 9's for a denominator. Thus, i = .66G.. .. = | = ? ; |=.428571. . . . = ||fi|i = ?. 221. C. is the abbreviation for the Latin Centum, signifying one hundred; and M. for the French Mille, signifying one thousand. 223. A logical explanation of any solution requiring more than one process or com- putation, or of the reasons upon which any principle or process is based, is an Analysis. This section applies particularly to the solution of problems which involve more than one process or computation. 2.34. Pupils should face south, and hold their books erect before thera, while study- ing or explaining this astronomical cut. 252. The tax on any sum from $1 to $10,000 can be taken from a Table that gives only the tax on $1, $2, $3.. to $10 inclusive, if the table is carried to six decimal plices. Thus, if the tax on $1 is $.023145, on $10 it is $.23145; on $100, $2.3145; on $10 10, $23,145; on $10,000, $231.45; and so of $2, $20, $200, $2,000, etc. 262. Teachers in Vt., N. II., or Conn, should require pupils to solve the problems on page 262, both by the U. S. Court Rule, and the Rule for their own State. 291. Explain that the third term is divided by the first, to find the value of a unit ; and the result is multiplied by the second term, to find the value of the number of units ; the same as in Analysis, Art, 370- Also, require pupils to solve the problems by analysis, after solving them by proportion. 293. Pupils will learn to state problems very rapidly, if they are taught to first write the terms in two lines, as they occur, writing the second set of conditions under cor- responding terms of the first. For example, in Problem G, page 294, the pupil writes 10 h. 1365 bar. 13 da. tt , ., -.Q-u ? " sa '» ^^^ ^^^^ ^^^^ ''"^y *° arrange the couplets. (P. 202, prob. 24). (P. 292, prob. 24). In solving problems, some I jSfi. fi 2 ^/^-i * - x y teachers write the work in one '^ ^P / J's J^^' 17 17 ^Sf X-^ ^'^ '^ of the forms here shown, in ~^i SXo 34lr^ preference to the forms given 68~ft — 68^ fi on pages 290, 293. "^ ' "^ " 300. Other roots are indicated by placing over the radical sign the figure denoting the required root; as -\/} -\/, ^, etc. 310. Since 2^ x 2» = 2*, the V of a number = V of V- And since 2' X 2> y, 2» = (2»)» =2», and 2» X 2» = (2^)= = 2«, the V of a number = V of V of V? or V of V> or V of V* 313. j In Progressions, if any three of the five things are given, the other two may 317. ) be found. The rules here given cover the ordinary applications of the subject. 321. Have a figure drawn to illustrate each definition and problem in this Chapter. 325. i These Principles (V., p. 325, III., p. 327) may be made Dlain to the pupil by 32 7. J practical appU cations. \ y h A Uhii is a single thing, or one, of any kind. y 2. A JVmnber is a unit, or a collection of units. Note. — Any number is cither concrete or abstract. V 3. A Conc7*ei:e JV*umber is a number applied to some object ; as, four men, ten apx^les, fifty days. y 4t An Abstract JVumber is a number not applied to any object ; as, four, ten, fifty. V 5. An Jnteger is a number tbo units of wbicb arc whole or undivided. Note. — Integers arc also called ^ylLole Numbers. y 6, TTniiy is the abstract unit 1. 7. A.rWl7neHe is the science of numbers, and the art of computation. (See Manual, page 5.) 8. A SotuHo7l is a process of computation used to ob- tain a required result. 9. A "I^roblein is a question requiring a solution. ' 10. An ^xptanaiioJi is a statement of the reasons for the manner of solving a problem. 11. A "Prlncijjle is a general truth upon which a proc- ess of computation is founded. 12, An Example is a problem used to illustrate a prin- ciple, or to explain a method of computation. V 13. An Analysis is a statement of the different steps in a solution. 14. A ^ule is a brief direction for performing any com- putation. ,, Note. — These general definitions apply to all classes of numbers. 10 INTEGERS. SECTION IL JVOTATIOjY AJVD jYZTMB^ATIOJSr, l^ 15. JVoiaiiori in arithmetic is WBtKSSit- expressing numbers bjr ton charac^ero, oallod figm p oo . These figures are 0123Jf56789 called. nmcgJit, one, two, three, four, five, six, seven, eight, nine. The figure 0, also called Cipher, denotes nothing, or the absence of number ; and the other figures represent the first nine integers, and are sometimes called Digits. y 16. JVume7"aH07i is tho art of reading numbers ^«« pr e ss e d -by^^g'^geS. (See Manual, page 5.) To express numbers greater than 9, two or more of the ten figures must be combined. 17. In writing numbers, every ten ones taken together are called a ten. Ten is written 10 2 tens are called twenty., written 20 5 tens " fifty, " 50 8 tens " eighty, " 80 9 tens " ninety, " 90 When two figures are written together to express a num- ber, the left-hand figure expresses tens, and the right-hand figure ones. Thus, Sixteen consists of 1 ten and 6 ones, written 16 Thirty-five " 3 tens " 5 ones, " 35 Seventy-two " 7 tens " 2 ones, " 72 Ninety " 9 tens " ones, " 90 18. Every 10 tens taken together are called a hundred. One hundred is written 100 Two hundred " 200 Seven hundred " 700 r. NOTATION AND NUMERATION. 11 When three figures are written together to express a number, the left-hand figure expresses hundreds, and the other two figures express tens and ones. Thus, Four hundred twenty-seven consists of 4 hundreds 2 tens and 7 ones, and is written Jf27 2 hundreds 5 tens and 6 ones, or two hundred fifty- six, is written 256 7 hundreds 1 ten and 8 ones, or seven hundred eighteen, is written 718 5 hundreds 3 tens and 9 ones, or five hundred thirty- nine, is written 539 4 hundreds 6 tens and ones, or four hundred sixty, is written 4^0 1 hundred tens and 5 ones, or one hundred five, is written ^05 EXEJtCISES. 1. Write in words, 10, 30, 70, 23, 99, 16, 11, 12. 2. Write in words, 100, 400, 700, 350, 280, 190. 3. Write in words, 596, 281, 694, 375, 333, 899. 4. Write 108, 904, 301, 707, 510, 811, 600, 150. Express by figures the following numbers : 5. Fifty, ninety, forty-one, sixtj^-six. 6. Fourteen, one hundred, four hundred, six hundred. 7. Two hundred sixty, five hundred ninety. 8. Seven hundred ten, three hundred twenty-six. 9. Five hundred eighty-one, six hundred fifteen. 10. Two hundred four, five hundred three. 11. Seven hundred six, eight hundred one. 12. Six hundred fifty, seven hundred tw^elve. 13. Five hundred sixty-three, two hundred ninety. 14. One hundred nineteen, nine hundred ninety-nine. 19. In writing numbers, every 10 hundreds taken to- gether are called a thousand, every 10 thousands taten together are called a ten-thousand, and every 10 ten- thousands are called a hundred-thousand. When a figui'e stands at the left of hundreds in a num- ber, it express thousands ; when at the left of thousands, it 12 INTEGERS. expresses ten-thousands ; and when at the left of ten thousands, it expresses hundred-thousands. One thousand is written Five thousand " Nine thousand " Ten thousand " 2 ten-thousands, or twenty thousand, " Sixty thousand " One hundred thousand " Three hundred thousand " Eight hundred thousand " 20. Every three figures in any number, counting from the right, are called a Period. Periods of figures are separated from each other by commas. The first, or right-hand period, consists \ of ones, tens, and hundreds ; and the second period of ones, tens, and hundreds of thousands. Nine thousand one hundred is written Six thousand two hundred fifty " Two thousand seven hundred forty-two " Ten thousand four hundred " Fifty-six thousand six hundred seventy " Nineteen thousand one hundred thirty-six " Forty thousand seven hundred nine " Eighty-two thousand six " Three hundred seventy-six thousand Five hundred ten thousand Six hundred thousand five hundred Three hundred fifty-two thousand seven hundred eighty-two EXEItCISES. 15. Read 5,000; 4,200; 1,3G0; G,384; 3,569; 8,113. 16. Read 9,011; 5,608; 3,008; 1,040; 4,076. 17. Read 30,000; 57,000; 42,300; 05,850; 83,294. 18. Read 15,203; 47,056; 50,912; 90,052; 89,005. 19. Read 80,000; 25,030; 60,200; 40,475; 30,800; 55,703. 1,000 5,000 9,000 10,000 20,000 60,000 100,000 300,000 800,000 752,194 5 8 9,100 6,250 2,71^2 10,400 56,670 19,136 1^0,709 82,006 376,000 510,000 600,500 352,782 NOTATION AND NUMERATION. 13 Write the following numbers : 20. Two thousand ; seven thousand live liundred. 21. Four thousand one hundred sixty. 22. Nine thousand six hundred fifty-three. 23. Three thousand eight hundred eleven. 24. Seven thousand forty- one. 35. One thousand one ; two thousand fifty. 36. Five thousand four hundred nine. 37. Sixteen thousand five hundred. 28. Eighty-one thousand two hundred seventy. 29. Eleven thousand nine hundred eighty-five. 30. Read 275,000 ; 100,000 ; 860,000 ; 493,600 ; 815,350. Write the following numbers : 81. Two hundred thousand. 32. Six hundred fifty thousand eight hundred. 33. One hundred nine thousand seven hundred twenty-six. 34. One hundred five thousand eighty. See Manual. 21. The third period of figures con- | „• | | sists of ones, tens, and hundreds of mill- ° •- o " **^^s- ^ 45 9,20 8,103 In any full period the riglit-hand I : • -s • : •! • • figure is ones, the middle figure tens, % i i I s I I = i and the left-hand figure hundreds. Thus, in any number consisting of three full periods, there are ones, ones of thousands, and ones of millions ; tens, tens of thousands, and tens of millions ; and hundreds, hundreds of thousands, and hundreds of millions. One million two hundred thirty-one thou- sand three hundred sixty-four is written l,S31,36Jf. Twenty-five million " 25,000,000 Nine hundred million " 900,000,000 Four hundred six million " ^06,000,000 EXERCISES. 35. Read 4,000,000; 80,000,000; 73,000,000; 9,721,312. 36. Read 18,271,100; 300,000,000 ; 253,729,594 ; 604,000,000. Write the following numbers : 37. Nine million ; fourteen million. 14 INTEGERS. 38. Four hundred fifty-two million. 39. Nine hundred one million. 40. Three hundred million two hundred sixty-five thousand. 41. Five hundred nine million six hundred twelve thousand nine hundred eighty-five. 22. The first period is caU'ed the period of ones or units, the second the period of thousands, and the third the period of millions. The fourth period is that of billions, the fifth that of trillions, and the sixth that of quadrillions. 493,307,508,210,064,119 I : : I : : I : •' I : : I : : "i : : c S S e ?' S c " S "2 ■» S "So!" "2 =■ " 5c2 5^2 Sc^ Cc** Bell o2« .S-2§ .5^3 J^§ J^§ jIg Jl§ EXER CIS JE S. 42. Read 4,359,006,110; 19,000,000,000; 40,060,139,194. 43. Read 5,236,481,279; 10,500,600,000; 92,675,244,000. 44. Read 3,000,000,000,000,000 ; 396,728,136,294. 45. Read 17,252,005,030; 18,000,039; 410,000,060,000. Write the following numbers : 46. Five billion two hundred million twenty-two thousand eight. 47. Forty-five billion one hundred fifteen million one hundred sixty-four thousand eighty-nine. 48. Fifty-two trillion. 49. One hundred nine quadrillion. 50. Nine billion three hundred six thousand. 51. Four hundred' seventy-eight quadrillion two hundred thirty- four trillion eight billion five hundred sixteen million seven hun- dred thousand five hundred eight. 52. Six hundred nineteen million thirty. 23. The ones, tens, hundreds, thou- ^ c c c c c ^ ^ c sands, etc., of any number are called 111 III ||| units of different orders; ones being ^ ^ ^ s si a ss.2 simple units, or units of the first order ; 593,298,756 NOTATION AND NUMERATION. 15 tens, units of tlie second order ; hundreds, units of the third order, and so on. 24 • Every figure has an absolute or simple value, and a local value. Its simple value is the number of ones it ex- presses when taken alone. Its local value is the order of units it expresses in a number. Thus, 8 when taken alone expresses 8 things, 8 ones, or 8 simple units ; but when taken with other figures it expresses different units, accord- ing to its place. In 80, it expresses 8 tens ; in 800, 8 hun- dreds ; in 8,000, 8 thousands, and so on. (Sce Manual, page 5.) 25. A unit of any order is ten times as great in value as a unit of the next lower order. Thus, a ten is 10 times a one ; a hundred 10 times a ten ; a thousand 10 times a hundred, and so on, as shown in the following NOTATION AND NUMERATION TABLE. 10 ones are 1 ten, 10 tens " 1 hundred, 10 hundreds " 1 thousand, 10 thousands " 1 ten-thousand, 10 ten-thousands " 1 hundred-thousand, 10 hundred-thousands " 1 milUon, and so on. Iten is 10 ones, 1 hundred " 10 tens, 1 thousand " 10 hundreds, 1 ten-thousand " 10 thousands, 1 hundred-thousand " 10 ten-thousands, 1 million " 10 hundred-thousands, and so on. 26. ^rinc/ples of JVotation. I. The values of the different places in a number increase from right to left in a tenfold ratio, n. The place ivhich any figure occupies in a number deter- mines the value expressed by it in that number. III. The highest period of any number must stand ai the left, and tJie succeeding periods in their order. 16 INTEGERS. rV. Every full period must consist of three figures, — hun- dredSy tens, and ones ; the place of any unit not named in the given number being filled by a cipher. Y. Tlie three places of any period not named in a number must be filled by three ciphers. 27. 'Principles of A''iejne7^aiio7i, I. Every integer consisting of more than three figures should be separated into periods. II. Each period of an integer is read separately, as hundreds, tens, and ones ; the name of the period being pronounced after the ones. III. In reading any number, the names of places and periods filled with ciphers are omitted. EXEMCISBS. 53. Read 80; 290; 763; 409; 7,000; 2,009; 5,080. 54. Read 9,393; 6,500; 50,000; 83,400; 14,008; 10,086. 55. Read 512,094; 809,123; 559,026; 300,006.; 110,090. Write the following numbers : 56. Eighty ; three hundred ; nine hundred ten, 57. Fifty-five ; seven hundred sixteen ; four hundred one. 58. Eight thousand ; fifty thousand ; ninety-two thousand. 59. Six hundred twelve thousand one hundred sixty-five. 60. Fifteen thousand seventeen. 01. Four hundred thousand fifty-six. 62. Sixty million ; seven hundred million. 63. One hundred eighty-two million three hundred fifty-five thousand four hundred eighty-eight. 64. Two hundred nine million eighteen thousand nine hundred ten. 65. Read 320,000,296 ; 200,165,000 ; 693,100,083 ; 501,080,276. 60. Read 433,279,187,695 ; 309,400,060,009. 67. Read 393,000,000,000,0.00,000 ; 117,371,545,903. Write the following numbers : 68. Sixteen trillion three hundred ninety-six billion. 69. Two hundred forty-seven billion fifty-six thousand. 70. Seventy-one trillion two hundred forty-one. 71. Two hundred sixty seven quintillion. ADDITION. 17 SECTION III. INDUCXIOI^ J^^Ty DEE^INITIONS. 28» 1. MxVRiA had 3 peaches, and George gave her 4 more. How many peaches had she then ? 2. Frank has 5 large rabbits and 6 small ones. How many rabbits has he ? 3. How many apples are 6 apples, 4 apples, and 7 apples ? 4. How many birds are 5 birds, 7 birds, 3 birds, and birds ? 5. Ella has 5 roses, Mary has 8, Olive has 4, Alice has 7, Louise has 9, and Flora has 6. How many roses have all the girls ? V 29. jlddiliO?i is 'i b a prooooo ef uniting two or more numbers to form one number. 30. The Amotl7Zt or Su7?i is the result obtained by- Addition. 31. The ^ai'ts are the numbers which are united to form the sum. 32. The Sig7i of Additlo?iy made thus +, is called Plus ; and when written between numbers, it signifies that they are to be added. 33. The St(/?i of J^quality, made thus =, when writ- ton between numbers or sets of numbers, signifies that they are equal to each other. Thus, 4 + 5 = 9; 16 ==3 + 7 + 6. Note.— A number with the sign $ before it expresses dollars. 6. What is the sum of 5 cents, 9 cents, and 8 cents ? 7. Add 9, and 5, and 3, and 4, and 7. 8. Add 6 books, 8 books, 5 books, 4 books, and 9 books. 9. 12 days + 3 days + 7 days + 1 day = how many days ? 10. What is the amount of 5 pens, 11 pens, 8 pens, and 2 pens? 11. 15 pictures + 7 pictures + 3 pictures + 8 pictures + 9 pic- tures = how many pictures ? 12. The parts are 12, 7, 4, 1, 5, and 8. What is the sum ? 18 INTEGERS. 34( ADDITION TABLE. nJO 123456789 "jo 000000000 013 3 456789 k(01 33456 789 »^|555555555 5 5 6 7 8 9 10 11 13 13 14 ,(0 123456789 ^U 1 1 1 1 1 1 1 1 1 123456789 10 /jjO 123456789 "16 666666666 6 7 8 9 10 11 13 13 14 15 OJ0123456789 '^\2 222222223 23456 789 10 11 lyjO 133456789 '17777777777 7 8 9 10 11 13 13 14 15 16 qjO 123456789 •^13 333333333 3 4 5 6 7 8 9 10 11 12 QJO 123456789 0]8 888888888 8 9 10 11 13 13 14 15 16 17 /LJO 123456789 ■^14 444444444 4 5 6 7 8 9 10 11 13 13 q(01234 5 6789 ^1 9999999999 9 10 11 13 13 14 15 16 17 18 C^SE I. The sum of all the figures of any place not more than 9. 35. We can add apples to apples, dollars io dollars, pens to pens, or hours to hours ; but we can not add apples to dollars, nor pens to hours. For 4 apples + 9 dollars = neither 13 apples nor 13 dollars. Again, we can add ones to ones, tens to tens, or hun- dreds to hundreds ; but we can not add ones to hundreds, nor tens to thousands. For 4 tens + 9 thousands == neither 13 tens nor 13 thousands. 36. Example. What is the sum of 4,216, 3,152, and 1,321 ? Explanation. — Since we must add ones to ones, tens to tens, hundreds to hundreds, etc., ^°^^"^^; it is most convenient to write the parts with like orders of units in the same column. We then add each column separately, writing the sum directly under the column added. The sum of the ones, 1 -f 2 + 6, is 9 ; the sum Paris. 8G8 Sum. of the tens. ADDITION. 19 2 + 5 + 1, is 8 ; tlio sum of the hundreds, 3 + 1 + 2, is G ; and the sum of the thousands, 1 + 3 + 4, is 8. The result, 8,689, is the sum required, csee Manual.) In this manner solve and explain the following FM OBJuE3rS. Find the sum in each of the first five problems : (1) (2) (3) (4) (5) 54 71 556 3,615 215,124 35 26 ^ 2,371 583,643 6. A man gave $22 for a coat, and $25 for an overcoat. How much did he pay for both ? $jf,7. 7. A gentleman paid $125 for a horse, and $163 for a carriage. How much did both cost him ? $287. 8. A farmer has 14 cows, 11 oxen, and 23 yoimg cattle. How many head of cattle has he ? J^S. 9. One day a miller sold 321 barrels of flour, the next day 143 barrels, and the third day 235 barrels. How much flour did he sell in the three days ? 699 larrels. 10. What is the sum of $5,418, $51, and $430 ? 11. Add 13,300 miles, 2,051 miles, 1,435 miles, and 3,013 miles. 13. la the Congressional Library at Washington there are 50,700 volumes, and in the library of the Smithsonian Institute 35,000 volumes. How many volumes in both libraries ? 75, 700. 13. Three men together purchase a vessel, A paying $11,735, B, $10,050, and C, $8,130. What is the cost of the vessel ? $29,895. 14. One day a produce dealer bought from three men 730 bush- els, 145 bushels, and 1,134 bushels of oats. How many oats did he buy ? 1^989 tmshels. 15. An army containing 41,430 men received two reinforcements, the first of 13,325 men, and the second of 34,334 men. How many men were then in the army ? 78,889. 16. A dealer in real estate sold three city lots for $1,220 each, another lot for $3,135, and a farm for $12,310. For how much did they sell? $17,995. 2.J INTEGERS. Cj?4.SiG II. The sum of all the figures of any place more than 9. 37. 5 + 8 + 3 = 16, and 16 = 1 ten and 6 ones. 5 hundreds + 8 hundreds + 3 hundreds = 16 hundreds, and 16 hundreds = 1 thousand and 6 hundreds. 7 tens + 9 tens + 8 tens = 24 tens, and 24 tens = 2 hun- dreds and 4 tens. Hence When the sum of the units of any order exceeds 0, the tens of this sum are units of the next higher order. 38, Ex. "What is the sum of 3,475, 2,694, and 1,383? ExPL:iNATioN. ■ — We write the parts as in Case I., draw two horizontal lines underneath, ^^^^^^ solution. as shown in the Fkst Sohition, and then add. 3Jf75 The sum of the ones, 3 + 4 + 5, is 12, or 2 ^^^| ones and 1 ten. We write the 2 ones below — -^ the lower line as the ones of the sum, and the 1 ten in tens' place, between the two lines, to 7532 bo added with the column of the tens. The sum of the tens, 1 + 6 + 9 + 7, is 23, or 8 tens and 2 hundreds. We write the 3 tens as the tens of the sum, and the 2 hundreds in hundreds' place between the two lines. The sum of the hundreds, 2 + 3 + 6 + 4, is 15, or 5 hun- dreds and 1 thousand. We write the 5 himdreds as the hundreds of the sum, and the 1 thousand in thousands' place between the lines. The sum of the thousands, 1 + 1 + 2 + 3, is 7, and this we write as the thousands of the sum. The result, 7,532, is the sum required. Explanation. — In the Second Solution we write the parts as before, draw one horizontal second soi.utiox. Iin3 underneath, and then add. The sum of 3Ji.75 the ones is 12, or 2 ones and 1 ton. We write i&ni the 2 ones below the line for the ones of the sum, and add the 1 ten with the column of 7o3^ tens. The sum of all the tens is 23, or 3 tans ADDITION 21 and 2 limidreds. We write the 3 tens as the tens of the sum, and add the 2 hundreds with the column of hun- dreds. The sum of all the hundreds is 15, or 5 hundreds and 1 thousand. We write the 5 hundreds as the hundreds of the sum, and add the 1 thousand with the column of thousands. The sum of all the thousands is 7, and this wo write as the thousands of the sum. The result, 7,532, is the sum required. JPJt OB LJ^31S. 17. The numbers on these packages of express freight show their weight in pounds. What is the weight of the marked packages that have been un- loaded ? 971 pounds. 18. How many pounds do the cask, barrel, and half-barrel weigh ? 1, 133. 19. What is the weight of all the marked boxes on the express wagon? Jj., 110 pounds. 20. How much does all the marked freight on the wagon weigh ? SI. What is the total weight of all the marked packages shoM^n in the x^icture ? 5,373 pounds. 22. In the Old Testament arc 39 "books, and in the New Testa- ment 27 books. How many books are in the Bible ? 60. 23. A cabinet-maker paid $125 for black-walnut lumber, and $90 for mahorran-.^ How much did the lumbcT cost him ? $215. 22 INTEGERS. 24. A merchant taiJor bought three pieces of broadcloth, the first containing 27 yards, the second 45 yards, and the third 84 yards. How many yards did he buy ? 25. One day a man traveled 241 miles by railroad, 57 by steam- boat, and 14 by stage. How far did he travel ? 312 miles. 26. Henry is 16 years old, his father is 29 years older than he, and his grandfather 32 years older than his father. How^ old is his grandfather ? r^^ years. 27. From the creation of the world to the Christian era were 4,004 years. How many years from the creation to the end of the present year ? 28. A fruit dealer bought 56 barrels of russet apples, 74 barrels of pippins, 69 barrels of spitzenbergs, and 83 barrels of greenings. How many apples did he buy ? ^82 darrcls. 39. The facts deduced in Arts. 35 and 38 maybe stated as Principles of Addiiioti, I I. Only abstract numbers or like conerete numbers can be \ added. y I II. Only like orders of units in different numbers can be ; added. i m. WJieyi the sum of the units of any order exceeds 0, the I tens of this sum are units of the next higher order. 40. Upon these principles is based the (Rule for Addiiioji of Tnteffei's, I. Add like orders of units, and icriie the ones of the sum in the result. H. Add the tens of the sum of any order loiih the next higher order. in. Write the whole sum of the highest order of units given. Note. — Since the tens of the sum of any column must be added -with the next left-hand column, it is iu general more convenient to commence at the rirrht to add. ADDITION. 23 rjz or, Tj E3IS. 29. A nursery-man sold during the year 3,729 apple-trees, 1,415 pear trees, 974 peacli trees, 567 plum-trees, 918 cherry-trees, and 1,584 ornamental trees. How many trees did he sell ? 9,187. 30. One year a farmer raised 649 bushels of oats, 422 bushels of corn, 178 bushels of wheat, and 96 bushels of barley. How much grain did he raise ? l^SJ^B bushels. 31. A merchant pays his book-keeper $1,250 a year, two clerks $825 each, and a boy $175. How much do their salaries amount to? $3,075. 32. An Erie canal-boat has on board 273 barrels of flour for Utica, 385 barrels for Albany, and 465 barrels for New York. Hov/ much flour has she on board ? 1,123 barrels. 33. A grocer, in purchasing his first stock, paid $466 for sugai-s, $387 for syrups, $196 for teas, and $1,760 for other goods. What was the cost of his stock ? ^ $2,809. 34. At an auction a woman bid off one carpet for $24, anothei- for $36, some oil-cloth for $7, and some window-shades and fixtures for $12. What was the amount of her bill ? ^7^ 35. At the New York Cattle Market the number of beeves on sale last Tuesday was as follows : number remaining over from Monday, 396; received by Erie Railroad, 1,516; by Hudson River Railroad, 1,044 ; by Harlem Railroad, 1,185 ; by Camden & Amboy Railroad, 296 ; by Hudson River boats, 210 ; by New Jersey Cen- tral Railroad, 329 ; and on foot, 311. How many beeves were on sale? 5,287. 36. A man built a house, which cost, for brick and stone, $375 ; for lumber, $540 ; for other materials, $224 ; for excavation, $72 ; for mason work, $284 ; for carpenter work, $580 ; and for painting, glazing, and paper hanging, $225. How much did the house cost him? ■ $2,300. 37. At one time the rolling stock of the New York Central Rail- road was 211 locomotives, 196 first-class passenger cars, 41 second- class and emigrant cars, 61 baggage, mail, and express cars, 2,760 freight cars, and 350 gravel cars. What was the whole number of cars? 3,619. 24 I X T E G E R S. 88. I paid $325 for a lot, |1,42G for building a house upon it„ $589 for building a barn and carriage house, $74 for fences, anc' $48 for grading the lot. For liow much must I sell the property to gain $338 ? $2,800. 39. On Monday morning a merchant had $1,767 in the bank. That day he deposited $94; on Tuesday, $118; on Wednesday, $78; on Thursday, $141 ; on Friday, $52 ; and on Saturday, $279. How much had he on deposit at the end of the week ? $2,524. 40. Find the sum of 15 million 9 thousand 17, 9 million 503, 675 thousand 899, and 245 million 320 thousand 8. 270 million 11 tlioiisand J^32. 41. In each of the two following sets of numbers, find the sunn of all the numbers above e. X 21,365_^ 42. From a to d. 2,194,756_ 43. From I to e. 40,373,254 18,890" —a -h 54,363" — /! 27,5411^ 53,027 —e 34,198 44,254" -f 87,079 -0 -h 73,250~ — i 19,000 48,408 -k 91,510 -I 60,009 38,482~^ ~-7i 19,504 —0 05,587 28,385 -P -<1 -r y 78,126 44. Above/. 90,000,383 45. From a to g. 0,275,851 —b — c —d 46. From d to i. 12,593 47. From h to m. 6,005 ^ 48. Below «. 373,58^ 49. Eclow h. 218 ^ 0. From /to 7.7. 1,694,583 51. From g to o. 657,679_ -I -J 53. From e to I. 73,418 ~ 52. From,; to i7. 500,000,290 54. From a to ;. 1,547 55. Above i. 4,293,500_^ 56. Below?. 400,000 '^ 57. From/ to 72. 44^ 58. From Ic to r. ' 68,974_^ 59. From d to p. 13,987,457 ^ 60. What is thc-Gum of the two answers of Problem 41 ? 61. The sum of the two answers of Problem 42 ? 62. The sum of the four answers of Problems 43 and 44 ? 63. Add the answers of Problems 45, 46, 47, and 43, 64. Add the answers of Problems 49, 50, 51. 65. Add the answers of Problems 52, 53, 54. 66. Add the answers of Problems 55, 56, 57, and 58. ADDITION. ^ 1^ 67. What is the sum of the two answers of Problem S^^^^^^ifl^-S^^ See Manual. 68. How many rods of fence will it take to inclose a field that is ^8 rods long on each side, and 29 rods wide on each end ? i j 6 7 8 9 10 11 13 13 14 15 ^J6 666666666 0133456789 2|l 3 4 3 3 1 3 5 6 7 8 9 10 11 3 3 3 3 3 3 3 3 4 5 6 7 8 9 « j 7 8 9 10 11 13 13 14 15 16 (|7 777777777 0133456789 3]l 4 5 3 3 6 7 8 9 10 11 13 3 3 3 3 3 3 3 ft ( 8 9 10 11 13 13 14 15 16 17 o|8 888888888 1 3 3 4 5 6 7 8 9 0133456789 Mt 5 6 4 4 1 3 7 8 9 10 11 13 13 4 4 4 4 4 4 4 3 4 5 6 7 8 9 Q ( 9 10 11 13 13 14 15 16 17 18 y]9 999999999 0133456789 c-a.se I. No figure of the Subtrahend greater than the corresponding figure of the Minuend. 48. We can subtract apples from apples, dollars from dollars, pens from pens, or hours from hours ; but we can not subtract apples from dollars, nor pens from hours. For 13 apples — 4 dollars = neither 9 apples nor 9 dollars. Again, we can subtract ones from ones, tens from tens, or hundreds from hundreds ; but we can not subtract ones from hundreds, nor tens from thousands. For 9 thousands — 4 tens = neither 5 tens nor 5 thousands. 28 INTEGERS. 49. Ex. From 5,267 subtract 2,215. solution. Explanation. — Since we must subtract 526 7 Minuend. ones from ones, tens from tens, hundreds 22 15 Subtrahend. from hundreds, etc., it is most convenient SO 52 Remainder. to write the figures of the subtrahend under the figures of like orders in the minuend. We then subtract each figure of the subtrahend from the figure above it in the minuend, writing the result directly below in the remainder. 5 ones from 7 ones leave 2 ones ; 1 ten from 6 tens leave 5 tens; 2 hundreds from 2 hundreds leave hundreds ; and 2 thousands from 5 thousands leave 3 thousands. The result, 3,052, is the difference or remain- der required. In the same manner solve and explain the following PB OBZEMS. (1) (3) (3) (4) (6) From 85 459 4978 13379 feet $2384 Subtract 43 348 3264 3148 feet. 1073 6. From a chest of tea, which contained 76 pounds, a grocer sold 43 pounds. How much tea remained in the chest ? S3 pounds. 7. From a flock of 396 sheep a drover bought 194. How many sheep were left in the flock ? 202. 8. A man bought a house and lot for $2,375, paying $1,225 down. How much did he then owe on the place ? $1, 150. 9. A contractor received $7,875 for building a railroad bridge, and it cost him $5,450 to build it. How much was his profit ? 10. In a city school there are 849 pupils, of whom 437 are girls. How many are boys ? ^12. 11. A and B together bought a steamboat for $78,385, and A furnished $45,385 of the purchase-money. How much did B fur- nish ? $33,000. 13. From 9 million 548 thousand 276, subtract 5 million 84 thousand 153. lierminder, 4,514,123. SUBTRACTION. 29 C^^SJE II. Any figure of the Subtrahend greater than the correspond- ing figure of the Minuend. 50i If the minuend is 5, and the subtrahend is 2, the difference is 3. 5 2 5 + 4 = 9 2 + 4 = 6 5 + 7 = 12 2 + 7=9 3 3 3 If 4 be added to both minuend and subtrahend, the dif- ference is 3, as before. Again, if 7 be added to both min- uend and subtrahend, the difference is still 3. Hence, The difference or remainder is not affected by adding the same number to both minuend and subtrahend. 51. Ex. 1. From 40,658, subtract 21,385. Explanation. — "Writing the numbers as in Case sonjuoy. I., we commence at the right to subtract. 5 ones Ji.0658 from 8 ones leave 3 ones, which we write as the ^ 1^8o ones of the remainder. Since we can not subtract 19273 8 (tens) from 5 (tens), and since the difference will not be affected by adding the same number to both minuend and subtrahend (50), we add 10 (tens) to the 5 of the minu- end, and 1 (hundred = 10 tens) to the 3 of the subtrahend. Vfe then subtract 8 from 15, and 4 from 6, writing the results 7 (tens) and 2 (hundreds), as the tens and hundreds of the remainder. Since we can not subtract 1 (thousand) from (thousand), we add 10 (thousands) to the of the minuend, and 1 (ten-thousand = 10 thousands) to the 2 of the subtrahend. We then subtract 1 from 10, and 3 from 4, and write the results, 9 (thousands) and 1 (ten-thousand), as the thousands and ten-thousands of the remainder. The result, 19,273, is the remainder required. 30 INTEGERS. Ex. 2. From 923 subtract 48. Explanation. — Since we can not subtract 8 (ones) soltttion. from 3 (ones), we add 10 (ones) to the 3 of tlie ^^^ minuend, and 1 ten ( = 10 ones) to the 4 of the "^ subtrahend. Then 8 from 13 leaves 5. Since we ^^^ can not subtract 5 tens (4 + 1) from 2 tens, we add 10 (tens) to the 2 of the minuend, and 1 (hundred = 10 tens) to the subtrahend. Then, 5 from 12 leaves 7, and 1 from 9 leaves 8. The result, 875, is the remainder re- quired. Ex. 3. From 1,000 subtract 257. Explanation. — In solving this example, we first boltjtion. add 10 (ones) to the minuend, and 1 (ten) to the 1000 subtrahend, and subtracting 7 from 10, we obtain _l^Z 3 ones. "We next add 10 (tens) to the minuend, 743 and 1 (hundred) to the subtrahend, and subtract- ing 6 from 10, we obtain 4 tens. We then add 10 (hun- dreds) to the minuend, and 1 (thousand) to the subtrahend, and subtracting 3 from 10 and 1 from 1, we obtain 7 hun- dreds and thousands. The result, 743, is the remainder required. See Manual. In the same manner solve and explain the following PB OBLEMS. (13) (14) (15) (16) From 93 416 1483 men 433150 miles Subtract 28 198 _C45 men. 145316 miles. 17. A man having $725 on deposit, draws out |268. How much has he left in the bank ? $Jf57. 18. A merchant sold a bill of goods for $173, and his profits were $37. How much did the goods cost him ? $136. 19. A man's income is $1,675, and his expenses are $948. How much does he save ? $727. 20. 1,378 tons — 985 tons = how many tons ? 393, 21. 1,045 bushels — 66 bushels = how many bushels ? 979, SUBTRACTION. 31 22. How much more does the heaviest bale of cotton weigh than the large box near the ship ? 382 pounds. 23. How much more do the 3 cotton bales weigh than the heav- iest marked package ? 24. "What is the difference between the weight of the 2 barrels and the bale of cotton which the 3 men are handling ? 25. Which weighs the most, the ])ale on which one of the men is standing, or the 4 lightest marked packages, and how much ? 26. All the marked boxes are to be shipped on board the vessel lying at the wharf, and all the other marked freight has been landed from her. Will she take in more or less freight than she has discharged, and how much ? BJfS pounds less. 27. A Boston provision dealer having 1,296 barrels of pork, ships 748 barrels to Liverpool. How many barrels has he then on hand ? 28. In an election for Member of Congress, A received 12,031 votes, and B, 10,032 votes. Which candidate was elected, and by what majority ? A,h/ a majority of 1,999. 29. A yoke of oxen, before being fattened, weighed 2,586 pounds, and after being fattened, 3,174 pounds. How much had they gained ? 588 pounds. 30. An ox weighed 1,326 pounds on foot, and 996 pounds when slaughtered. What was the difference between the live and dressed weights 830 pounds. 32 INTEGERS. 52. The faots deduced in Arts. 48, 50, may be stated as l^rinciptes of Subtraction, I. Only abstract numbers or like concrete numbers can be subtracted the one from the other. II. Only like orders of units can be subtracted the one from the other. m. The difference or remainder is not affected by adding the same number to both minuend and subtrahend. 53. Upon these principles is based the ^ute for Subtraction of Integers* I. Subtract units from units of like orders, writing each dif- ference for the same order of units in the result. n. Wheii any order of units in the subtrahend is greater in value than the corresponding order in the minuend, add 10 units of the same order to the minuend, and 1 unit of the next higher order to the subtrahend. Notes. — ^1. When no figure of the subtrahend exceeds in yalue the cor- responding figure of the minuend, we may commence at the right or at the left to subtract. 2. "When one or more figures of tlie subtrahend exceed in value the cor- responding figures of the minuend, it is in general more convenient to commence at the right to subtract. See Manual. PJt OJiJuEMS. (31) (32) (33) (34) From 3250 10000 25600 gallons 342651 reams Subtract 89 24 8008 gallons. 142652 reams. 35. A man's property is valued at $75,000, of which $48,766 is in real estate. How much is his personal property worth ? 86. A grain buyer in Milwaukee receives an order for 12,500 bushels of " No. 1 " wheat, and has only 7,645 bushels in store. How much must he purchase to fill the order ? J!j,,855 htisheh. 37. A religious society, after raising $17,675 by subscrii^tion, contracted for the erection of a church for $22,500. How much remains yet to be raised ? $Jt,825. SUBTRACTION. 33 38. Christopher Columbus was bom a.d. 1437, discovered Amer- ica A.D. 1492, and died a.d. 1506. How old was he when he dis- covered America ? How old when he died ? 65 years ; 69 years. 39. 26,957,239 bushels of salt were used in the United States in 1860, and of this amount 14,094,227 bushels were imported. How many bushels were of home manufacture ? 12,863,002. 40. The distance from Albany to Buffalo is 297 miles, and from Albany to Rochester, 229 miles. How far is it from Rochester to Buffalo ? 68 miks. 41. At a saw-mill 100,000 feet of pine lumber were sawed in one month, and 47,250 feet of it were sold. How much remained at the mill? 52,750 feet. 42. One year the receipts of the Tliird Avenue Railroad of New York city were $564,839, and the expenses were $307,188. How much were the profits ? $257, 651. Lake Superior has an elevation of 023 feet above tide ; Lake Huron, of 591 feet ; Lake Erie, of 565 feet ; Lake Ontario, of 232 feet ; Great Salt Lake, of 4,200 feet ; and Lake Titicaca, of 12,785 fcet. 43. How much higher is Lake Superior than Lake Erie ? 44. How much higher is Lake Superior than Lake Huron ? 45. Great Salt Lake is how much higher than Lake Ontario ? 40. How much fall is there in Niagara River ? 47. Lake Superior is how much higher than Lake Ontario ? 48. How much higher is Lake Titicaca than Lake Erie ? 49. How much higher is Lake Huron than Lake Ontario ? 50. How much fall is there between Lake Huron and Lake Erie ? 51. Lake Titicaca is how much higher than Great Salt Lake ? 52. From a farm of 417 acres, the owner sold 132 acres to one man, and 96 acres to another. How much land had he left ? 53. A bank teller received a salary last year of $1,250. His per- sonal expenses were $753, and he bought a village lot for $213, and paid out $149 for improvements upon it. How much money had he at the end of the year ? $135. 54. A drover having 319 head of cattle, sold 98 head to one butcher and 127 head to another. How many cattle had he left ? 2* 3^ INTEGERS. 55. An estate worth $35,474 is encumbered to tlie amount of $17,625. How much is it worth above the incumbrance ? 56. A farmer having 113 sheep, sold 57 of them, and afterward bought 83 more. How many had he then ? 139. 57. A man at his death left an estate worth $48,765. He owed $13,596, and bequeathed $12,750 to his widow, $5,875 to charitable institutions, and the balance to his only son. How much did his son receive ? $16^544' 58. A regiment was mustered into the service with 976 men, and afterward received 274 recruits. During service its losses were 38 killed in battle, 94 wounded, 54 taken prisoners, 69 discharged for sickness, 13 died from sickness, and 47 deserted. Of how many men did the regiment then consist ? 935. 59. A lumber dealer sold a quantity of plank for $746, making a profit of $148. How much did the lumber cost him ? $598. 60. A hardware merchant, who owes a grocer $113 on account, sells him a cook stove for $32, a parlor stove for $28, and some pipe for $7, and pays the balance in cash. How much money does the grocer receive ? 61. A clergyman had his life insured for $3,500. At the time of his death $376 of his salary was unpaid ; he owned a house and lot worth $3,275, but upon it there was a mortgage for $1,390 ; and his other debts amounted to $294. How much did he leave his family? $5,467. 62. A block of stores, valued at $37,675, and goods worth $69,325, were destroyed by fire. The buildings were insured for $31,875, and the goods for $49,290. What was the loss on the buildings ? 63. What was the loss on the goods ? 64. How much did the loss on the goods exceed the loss on the buildings? $U,235. 65. June 1, a grocer bought 1,754 pounds of ^ sugar, 1,249 pounds ©, 2,154 pounds B, 1,864 pounds C, 2,752 pounds W. I., and 1,954 pounds N. O. August 1, he had on hand 967 pounds 0, 856 pounds ®, 1,182 pounds B, 1,692 pounds C, 2,158 pounds W. I., and 369 pounds IST. O. How much sugar of each brand had he sold in the month ? See Manual. MULTIPLICATION. 35 SECTION V. MUJO TITZIC:>± TIOJV. INDXJCTION^ ^I^r) DEB^INITIONS. 54* 1. James found 4 hens' nests in the bam, and in each nest were 5 eggs. How many eggs did he find ? 2. If a cooper can make 7 barrels in a day, how many barrels can lie make in 5 days ? 3. How many blades in 9 4-bladed knives ? 4. A lady bought 7 spools of thread, at 7 cents a spool. How much did it cost her ? 5. If 9 pounds of flour will last a family one week, how many pounds will last them 5 weeks ? 6. How many dollars can a man earn in 6 days, if he earns $3 a day? 55. J^fulHpticaHon is a short process of finding the sum of as many times one of two numbers as there are ones in the other. 56. The ^rodtici is the result obtained by Multiplica- tion. 57. The jF^act07*S are the numbers used to obtain the product. 58. The J)€ictHplicand is that factor which is to be taken any certain number of times. 59. The Muttiptier is that factor which shows how many times the multiplicand is to be taken. 60. Continued Mu2tip2icaHo7i is the process of find- ing the product of more than two factors. 61. The Sig7i of MulHpticaHo7Z^im.diQ thus x, when placed between two numbers, signifies that they are to be multiplied together. It is read "times," or "multiplied by." Thus, 5 x 8 is read "5 times 8," or "5 multiplied by 8." 36 INTEGERS. 7. 4 X 6 slates are how many slates ? 8. What is the product of 6 x 7 oranges ? 9. What is the product of 7 x 9 ? 10. The factors are 5 and 8. What is the product ? 11. The multiplicand is 9, and the multiplier 8. What is the product ? 13. What is the product of 3 x 3 x 7 ? 13. 4 X 2 X 6 pen-holders = how many pen-holders ? 62. MULTIPLICATION TABLE. 1 23456789 11111111 23456 7 89 oiO 123456789 '^|6 666666666 6 12 18 24 30 36 42 48 54 nil 2 23456789 32222222 4 6 8 10 12 14 16 18 «(0123456 789 *]7 777777777 7 14 21 28 35 42 49 56 63 qjO 1 '^(3 3 3 23456789 33333333 6 9 12 15 18 21 24 27 q(0123456789 0]8 888888888 8 16 24 32 40 48 56 64 72 4 23456789 444444 44 8 12 16 20 24 28 32 86 q(01234 5 6789 ^\9 999999999 9 18 27 36 45 54 63 72 81 KJO 123456789 ^15555.5 55555 5 10 15 20 25 30 35 40 45 in^ 0123456789 •^"ho 10 10 10 10 10 10 10 10 10 10 20 SO 40 50 60 70 SO 90 CA.SE I. The Multiplier One Figure. 673 + 673 + 673 + 673, or 4 63* Ex. How many are times 673 ? Explanation. — ^In the Solution by Addition we find the sum of 4 673's to be 2,692. But since 3 ones + 3 ones + 3 ones + 3 ones = 4 times 3 ones, and 7 tens -f- 7 tens -f 7 tens H- 7 tens = 4 times 7 tens, and 6 hundreds -}- 6 hundreds + 6 hundreds 4- 6 hundreds = 4 times 6 hun- 80LTTTI0N BY AUDITION. 673 673 673 673 2692 MULTIPLICATION. 37 dreds, we write the 673 but once, and write a 4 under its right-hand figure, to show how many times it is to be taken, as shown in the following Solution by Multipli- cation : ^n this Solution we multiply each ^^ ^['^^Jl^^,,,,^, figure of the multipHcand by the multi- ^ 7 ^ Multiplicand. plier. Thus, 4 times 3 ones are 12 Jf. ituiupUer. ones, or 2 ones and 1 ten. We write 26 92 Product. the 2 ones as the ones of the product, and reserve the 1 ten in the mind to be added to the prod- uct of the tens. 4 times 7 tens are 28 tens, and 28 tens + 1 ten = 29 tens, or 9 tens and 2 hundreds. We write the 9 tens as the tens of the product, and reserve the 2 hun- dreds in the mind to be added to the product of the hun- dreds. 4 times 6 hundreds are 24 hundreds, and 24 hun- dreds + 2 hundreds = 26 hundreds, or 6 hundreds and 2 thousands. We write the 6 and 2 as the hundreds and thousands of the product. The result, 2,692, is the sum or product required. 64. From this explanation we learn that The muUiplicand is that factor which icould he used in solving a problem or example by Addition. 65. We may add abstract numbers or like concrete num- bers. (See 40.) Hence, The muUiplicand may be either an absti^act or a concrete number. 66. The multiplier is used simply to show how many limes the multiplicand is taken. Hence, The multiplier is alivays an abstract number. 67. In Addition the sum is of the same kind as the parts added. Hence, The product is always of the same kind as the multiplicand. 38 INTEGERS. FROBJOEMS. (1) (2) (3) (4) (5) Multiply 43 133 491 6243 13562 pounds, by A ^1 -1 1 ? 6. How much "will 3 cows cost, at $52 apiece ? 7. How much will a mechanic earn in 6 months, if he earns $35 a month ? $210. 8. A grocer bought 7 barrels of sugar, each containing 245 pounds. How much sugar did he buy ? 1, 715 pounds. 9. How much will 9 tons of hay cost, at $14 a ton ? $126. 10. How far will a railroad train run in 7 hours, at the rate of 39 miles an hour ? 273 miles. 11. A canal-boat captain bought 4 horses, paying $156 apiece for them. How much did they cost him ? $62Ji,. 12. A manufacturer pays his hands $2,356 a month. How much do their wages amount to in 6 months ? $H, 136. 13. In a certain army corps there were 8 regiments of 966 men each. What was the number of men in the corps ? 7,728. 14. At $5 a week, how much will a year's board cost, there being 52 weeks in a year ? $260. 15. If the price of thrashing wheat is 4 cents a bushel, how much must be paid for thrashing 17,944 bushels ? 71,776 cents. 16. A miller sells 74 barrels of flour, at $9 a barrel. How much does the sale amount to ? $666. 68. In explaining the solution of Problem 16, we would say, " 74 barrels sell for 74 times as much as 1 barrel, or 74 times $9." Hence, $9, being a concrete number, is the true multiplicand. But since 74 times 9 is the same as 9 times 74, in solving the problem we may take 74 for the multipli- cand, and use 9 as the multiplier. That is, I. In the solution of problems, either factor may he used as the multiplicand. See Manual. n. In the explanation of the solution of problems containing concrete number's, the concrete number is the true multiplicand. MULTIPLICATION. 39 17. How many windows in the front of all the stories of this building except the first, there being 20 windows to each story? 18. How many iron pillars or columns in the front of the three upper stories, there being 21 columns to each story ? 19. How many windows in 3 street cars, there being 18 windows in each car ? 20. If 24 passengers ride in a car at one trip, and the fare is 6 cents, how much fare will the conductor collect ? 21. A woman who keeps a fruit stand sold 178 oranges one day, at 4 cents apiece. How much did she receive for them ? 22. How much did it cost to paint the two signs on this build- ing, at $3 a letter ? 23. How much would it cost all the persons in the foreground of this picture to go to Harlem and back on a horse-car, the fare being 7 cents each way ? 210 cents. 40 INTEGERS. 24. The Third Avenue Railroad, in New York City, is 7 miles long. How many miles does a car run in making 11 round trips ? 25. A wood dealer sold 6,591 cords of wood, at $7 a cord. How much were his receipts ? $^6, 137. 26. In 1 mile there are 1,760 yards, or 5,280 feet. How many yards in 8 miles ? 14,080. 27. How many feet in 8 miles ? 42,246. 28. Ten brick-layers finish the walls of a building in 9 days, lay- ing 9,475 bricks each day. How many bricks are in the walls ? 29. At $9 apiece, how much will it cost for the transportation of 33,875 soldiers from Washington to Chicago ? $304,875. C^SE II. The Multiplier any number of tens, hundreds, thousands, and so on. 69, Ex. Multiply 3,176 by 10. Explanation. — ^We write the numbers as shown bolution. in the Solution, and multiply each figure of the 3176 multiplicand by the multiplier, as in Case I. ^^ Comparing the multiplicand and product, we 3176 find that the figures of ihe product are the same as those of the multiplicand, with a cipher on the right. Hence, 70. Annexing a cipher to any number multiplies it hy 10. 71. Annexing a second cipher multiplies by 10 again ; that is, annexing two ciphers to any number multiplies it by 10 times 10, or 100. 72. Annexing three ciphers to a number multiplies it by 10 times 100, or 1,000. 73. Annexing four ciphers multiplies by 10,000 ; annex- ing Jive ciphers, by 100, 000 ; and annexing six ciphers, by 1,000,000, I MULTIPLICATION. 41 74. Ex. Multiply 291 by 60. Explanation. — 60, or 6 tens, = 6 times 10 or soltttion. 10 times 6. Hence, 60 times 291 is 10 times as 291 much as 6 times 291. We may therefore multi- ^^ ply 291 by 6, and to the product thus obtained 17 Jf6 annex a cipher. 75. To multiply hy 600, we multiply hy 6, and annex two ciphers to the product ; to multiply hy 6,000, we multiply hy 6, and annex three ciphers. 76. We multiply hy any number of tens, hundreds^ thousands, or units of higher orders in the same manner. Pit O BJjEMS. 30. How many oats in 10 wagon loads of 65 bushels each ? 31. A barrel of flour contains 196 pounds. How many pounds in 100 barrels ? 19,600. 82. The United States Government purchased 10,000 rifles, at $24 each. How much did they cost ? $U0, 000. 33. Multiply 23,947 by 1,000. Product, 23,947,000. 34. Multiply 175,941 by 100,000. Product, 17,594,100,000. 35. A pork packer sells to a provision dealer 125 barrels of pork, at $20 a barrel. What is the amount of his bill ? $2,500. 36. At $50 an acre, how much will a farm of 176 acres cost ? 37. A gentleman bought a city lot, having a front of 22 feet, at $90 a foot. How much did it cost him ? $1,980. 38. A drover sold 500 head of cattle, at $67 a head. How much did he receive ? $33, 500. 39. A manufacturer sells 594 sewing-machines, at $60 each. How much does he receive for them ? $35, 640. 40. In a certain county, 237 drafted men purchased their exemp- tion by paying $300 each. How much did their exemption cost ? 41. The President's Cabinet consists of 7 members, who receive a salary of $8,000 each. "What do their united salaries amount to ? 42. 400 X 495 = how many? 198,000, 43. What is the product of 32,721 x 50,000 ? 1,636,050,000. 42 INTEGERS, C-A.sk III The Multiplier more than One Figure. 77. Ex. Multiply 3,528 by 472. Multiplying S528 7056 Explanation.— Since 472 consists of 2 ones, 7 tens, and 4 hun- dreds, or 2, 70, and 400 ; and since we can not multiply 3,528 by the whole 472 at once, we multiply it first by 2, then by 70, and then by 400, and afterward add the re- sults, or Partial Products. The re- sult thus obtained, 1,665,216, is the sum of 2 X 3,528, 70 x 3,528, and 400 X 3,528, or 472 x 3,528. In the First Solution, each of the four steps stands by itself ; in the Second Solution they are placed to- gether. In the Second Solution, the ciphers on the right of the second and third partial products serve merely to fill the places of ones and tens ; and since the sum of any number of O's is 0, they may be omitted without affect- ing the total product, as shown in the Third Solution. In this Solution the second partial product is found by multiplying by 7, instead of 70 ; and the third partial product by multiplying by 4, instead of 400. But we must always FIEST SOLTTTION. Multiplying hy TO. 8528 7^ 2^6960 Multiplying lyma. 3528 J^OO lJt.11200 Adding Partial Products. 7056 246960 1^11200 1665216 SECOND SOLUTION. 3528 Ji.72 Partial Products, 7056' 246960 1411200 1665216 TIIIED SOLXmON. 3528 472 Partial Products. 7056 24696 1 4112 1665216 Partial Products. MULTIPLICATION. 43 Write the first figure of each partial product directly under the figure of the multiplier used to obtain it. Note.— In the explanation of Solutions, the value of each partial product should be named. This is done by reading it as though the ciphers were written. PR OBZJSMS. (44) (45) (46) (47) (48) 34 73 281 2976 127493 23 45 _54 ^ 647 49. A prairie farmer planted 176 acres of com, wliich yielded 73 bushels to the acre. How many bushels in the crop ? 12,8J^8. 50. At a certain recruiting station, in 1862, 43 men enlisted each day for 36 days. How many men enlisted? l,5JiS. 51. An overland emigrant train traveled 23 miles a day for 17 days. How far did they travel ? S91 miles. 52. The skipper of a fishing smack received, as his share of the season's catch, 273 barrels of mackerel, which he sold at $11 a bar- rel. How much were his receipts ? $3, 003. 53. In a pump manufactory, 125 of the workmen receive $39 a month each. How much do their wages for a month amount to ? 54. How much do their wages amount to in 1 year, or 12 months? $58,500. 55. A livery-stable keeper who has 23 horses, finds that it costs him $83 a year for the keeping of each horse. What is the cost of keeping all of them ? $1,909. 56. A steamer sailed from New York for Liverpool with 376 first-class passengers. How much did their fares amount to, at $135 apiece? $50,760. 57. A railroad company contracted for 43 locomotives, at $18,725 apiece. What was the amount of the contract ? $805, 175. 58. In a certain paper-mill 184 reams of paper are made daily. How many reams are made in a year of 313 working days ? 59. Multiply 1,372 by 861. Product, 1,181,292. 60. What is the product of 4,293 multiplied by 2,726 ? 11,702,718. 61. 417,293 X 581 = how many? 242,U7,233. 44 , INTEGERS. One or more ciphers bet-ween other figures of the Multiplier. 78. Ex. Multiply 2,566 by 3,007. Explanation. — In the Second ^'^'^"^ solution, second solxttion. Solution we have multiplied by 2566 2566 7 ones and 3 thousands (see ^^^^ JOOJ 76), omitting to multiply byO 17962 17962 X ^A-u ^ ^ X n 0000 7698 tens and hundreds, because OOOO times 2,566 is 0, as shown in '^^gg 7 7 159 62 the First Solution. Hence we ^ ^^^^ may always Omit to multiply by ciphers that stand between other figures in the multiplier. PROBLEMS. (62) (63) (64) (65) Multiply 426 1728 4765 29873 by 203 _506 _807 5008 66. At $105 each for horses, how much will it cost to moitnt a cavalry regiment of 1,043 men ? $109,515. 67. In one season a manufacturer sold 307 reapers, at $135 apiece. How much did he receive for them ? $41,445- 68. A furrier bought 108 buffalo-robes, at $17 apiece? How much did they cost him ? $1,836. 69. A canal 203 miles long was kept in repair one season at an expense of $383 a mile. What was the expense for the whole canal? $77,749. 70. At a cotton manufactory 1,396 yards of cloth are made each day. How many yards are made in 307 days ? 4^8j572. 71. In a certain cotton factory are 203 looms, which turn out 69 yards of cloth per day, each. How many yards of goods are made daily ? 72. At the same rate, what is the total product of the factory in a year of 308 working days ? 4j^Hj ^^6 yards. M ULTIPLIC ATION, 45 One or more ciphers on the right of either or both factora 79. Ex. 1. Multiply 89 by 1,600. To multiply by 1,600, we first multiply by 16, and then to the product annex two ciphers, as in Case II. (See 75.) Ex. 2. Multiply 54,000 by 37. , Explanation. — "We first multiply 54 by 37, and obtain 1998. Since the 54 is thousands, this product must be thousands (see 67) ; and we therefore write three ciphers in units' pe- riod ; that is, annex three ciphers to the prod- uct of 37 X 54. SOLUTION. 89 1600 89 lJi.2Ji.00 BOLUTIOIf. 5Jf.000 87 378 162 1998000 60LCTI0N. 73000 2600 Jf.38 lJf6 189800000 Ex. 3. Multiply 73,000 by 2,600. Explanation. — We first multiply 73 by 28, obtaining 1,898. But since the 73 is thousands, we annex three ciphers to this product for those at the right of the mul- tiplicand. The result thus obtained is 26 X 73,000 ; and to make it 2,600 x 73,000, we annex two ciphers more. Hence, when there are ciphers on the right of one or both factors. To the product of the other figures annex as many ciphers as there are ciphers on the right of both factors. PROBLEMS. 73. In one barrel of beef there are 200 pounds. How many pounds in 37 barrels ? 7jJi,00. 74. A carriage maker sold 30 top carriages, at $285 apiece. How much did they come to ? $8,550. 75. A Missouri farmer had 130 acres of wheat, ajtd the yield was 27 bushels to the acre. How mu^h wheat did hef raise ? ^.^^ TH-^ '"''^ - 46 INTEGERS. 76. In printing an edition of 5,000 copies of a book, 16 sheets of paper were used for each book. How much paper vf as used ? 77. A Georgia planter had 94 acres of cotton that yielded 460 pounds to the acre. How much cotton did he raise ? 78. At an ax factory 280 axes were made each day for 156 days. How many axes were made ? Jf3, 680. 79. What will be the cost of 25,000 army blankets, at $5 apiece ? 80. In one ream of paper there are 480 sheets. How many sheets in 260 reams ? 124,800. 81. A Pennsylvania oil well flowed 110 days, at the rate of 340 barrels of oil a day. How much oil did it yield ? 37,400 larrels. 82. How much will be the cost of building a line of telegraph 680 miles long, at $1,250 a mile ? $850,000. 83. What is the product of 760,000 and 53,000 ? 40,280,000,000. 84. The multiplicand is 694,000, and the multiplier 56,700. What is the product ? 39,349,800,000. 80. The facts deduced in Arts. 64 to 70 may be stated as Principles of Mtcltiplication, In the Explanation of Solutions : I. The true multiplicand is that factor which mould be used in solving the problem by Addition. II. The multiplicand may be either an abstract or a concrete number. m. The multiplier must always be an abstract number. IV. Tlie product is always of the same kind as the true mul- tiplicand. In tbe Solution of Problems : V. In the solution of problemSy either factor may be used as the multiplicand. VI. Annexing a cipher to any number multiplies it by 10. VII. The sum of all the partial products arising from multi- plying one of two numbers by the ones, tens, hundreds, etc., of the other, is the product of the two given numbers. MULTIPLICATION. 47 81. Upon these principles is based the ^ute for MuUipUcation of Integers. I. The multiplier one figure. Commence with the ones, and multiply successively each figure of the multiplicand by the multiplier. In the product write the ones of each result, and add the tens to the next result. II. The multiplier more than one figure. Multiply by each figure of the multiplier except ciphers, place the right-hand figure of each product under that figure of the multiplier used to obtain it, and add the partial products. m. Ciphers on the right of either or both factors. To the product of all the other figures annex as many ciphers as there are ciphers on the right of both factors. FR OB IjEMS. 85. Every mile of a road 4 rods wide contains 8 acres. How- many acres are there in 168 miles of such a road ? 1^344' 86. An army train was 9 hours in passing a given point, 84 wagons passing each hour. How many wagons were in the train ? 87. How many poles will be required for a telegraph 328 miles long, if 16 poles are required for one mile ? 5,^48. 88. A phonographic reporter, in taking down a speech, wrote 68 words a minute for 137 minutes. How many words were in the speech? 9,316. 89. How much will it cost to build a railroad 134 miles long, at $65,475 a mile ? $8, 773, 650. 90. Last season a cheese factory used the milk of 470 cows, and made 290 pounds of cheese to the cow. How much cheese was made ? 136, 300 pounds. 91. In one square mile there are 640 acres, and in the State of Iowa there are 50,914 square miles. How many acres in the State? 32,584,960. 92. The New American Cyclopsedia contains 13,805 pages, of 3,036 ems each. How many ems in the work ? 41,911,980. 93. The factors are three hundred ninety-seven thousand five hundred, and nine thousand eight hundred. What is the product ? Three Mllion eight hundred ninety-five million five hundred thousand. 48 INTEGERS. 94. How many miles will a railroad conductor travel in a year, if he goes over a road 108 miles long once every day ? 39, ^20. 95. It is estimated that Mississippi River deposits 3,703,758,400 cubic feet of solid matter in the Gulf of Mexico evei*y year. How many cubic feet have been deposited in 503 years ? 1,858,784,716,800. 96. How many yards of sheeting are there in 91 bales, each bale containing 33 pieces of 45 yards each ? 94j 185. 97. At $13 a month, how much will the pay of the privates of a certain regiment, 896 in number, amount to in 13 months ? $139,776. 98. A shoe dealer bought 37 cases of French calf boots, each case containing 13 pairs, at |5 a pair. What was the amount of the purchase? $1,620. 99. In a woolen factory there are 48 looms. How many yards of cloth will be made in the factory in 308 days, if 37 yards are woven upon each loom daily? 269,568. 100. Just before an expected battle, 04 rounds of cartridges were given to each of 70,000 men. How many rounds were distributed to all of them ? 4, 480, 000. Po-wers. 82. A ^ower is the product of two or more equal fac- tors ; as 49, which is the product of 7 x 7. 83. A Square is the product of two equal factors ; as 25, which equals 5x5. 84. A Cube is the product of three equal factors ; as 64, which equals 4x4x4. 85. The JF'ourtli ^ower\^ the product of four equal fac- tors ; the ^iftJi ^ower the product of five equal factors ; the Sixt?i ^07Per the product of six equal factors, and so on. Notes.— 1. The process of finding the square of a number is called squar- ing it ; and the process of finding its cube is called cubing it. 3. The process of finding any power of a number is called raising it to that power. MULTIPLICATION. 49 86. An Index is that one of two numbers which denotes the power to which the other number is to be raised. It is written at the right, and a httle above the other number. Thus, in the expression 8^ 3 is an index, and it denotes that 8 is to be cubed. So, also, IV indicates the square of 21, and 59* indicates the fourth power of 59. 87. Ex. Kaise 15 to the fourth power. solution. Explanation. — We first find the ^ ^ square of 15, by multiplying it by _ itself. We then find the cube by 7^ multiplying the square by 15. We -^^ then multiply the cube by 15, and the 22 5 Square. result is the fourth power ; because -^^ it is the product of 15 x 15 x 15 x 15. 1125 In squaring a number there are ^^^ 2 equal factors and 1 multipHca- 3 3 7 5 Cube. tion ; in cubing it, 3 equal factors ^ ^ and 2 multiplications ; and in raising 16 8 7 5 it to the fourth power, 4 equal fac- 3 375 tors and 3 multiplications. 5 6 2 5 FourVipower. In finding any power of a number ^ the number of multiplica- tions is one less than the number of factors, PB, OBLJEMS, 101. What is the square of 9 ? SI. 102. What is the cube of 5 ? 125. 103. Square 423. Cube 47. 178,929; 103,823. 104. Raise 12 to the fourth power. 20,736. 105. What is the cube of 52 ? The cube of 901 ? 106. What is the square of 2,016 ? 4^064,256. 107. Raise 218 to the sixth power. 107,334,407,093,824. 108. Raise 63 to the seventh power. 109. Raise 14 to the eighth power. 110. Raise the following numbers to the powers denoted by their . 18S 11^ 3 50 INTEGERS. SECTION VI. ^ITISIOJV, iNDXJCTionsr ^nd definitions. 88. 1. Emma exchanged a 50-cent fractional-currency note for 5-cent pieces. How many 5-cent pieces did she receive ? 3. How many oranges can I buy for 28 cents, if I pay 4 cents apiece for them ? 3. A silversmith sold 30 teaspoons, in sets of 6 spoons each. How many sets of spoons did he sell ? 4. A farmer put 15 bushels of oats into bags, putting 3 bushels in each bag. How many bags did he use ? 5. One day Henry saw a pic-nic party of 36 persons passing by, in 4 carriages. How many persons were there for each carriage ? 6. Two boys received 20 cents for carrying a lady's trunk to the depot, and they shared the money equally between them. How many cents had each boy ? 7. In my garden are 21 fruit-trees, standing in 3 equal rows. How many trees in 1 row ? 8. A grocer paid $54 for 9 barrels of cranberries. What was the price per barrel ? In solving each of the first four problems, we find how many times one of two numbers is contained in the other ; and in solving each of the other problems, we separate one of two numbers into as many equal parts as there are ones in the other. 89. !Dirtsio?z is the process of finding how many times one of two numbers is contained in the other ; or of finding one of the equal parts into which a number may be divided. 90. The Quotient is the result obtained by Division. 91. The dividend is the number to be divided. 92. The divisor is the number by which the dividend is to be divided. DIVISION. 51 NoTES.—l. A Partial Dividend is that part of the dividend used to obtain one figure of the quotient, when the whole dividend is too large to obtain the entire quotient at one operation. 2. Division is Exact when all the dividend is divided and the quotient is a whole number. 3. A Remainder is that part of the dividend left undivided, either when the division is only partiaUy completed, or when exact division is impos- sible. 93. The Sign ofDivision, made thus -^, when placed between two numbers, signifies that the number before it is to be divided by the number after it. It is read, "divided by." Thus, 175 -~ 25 is read " 175 divided by 25." Notes. — 1. Division is also expressed by writing the dividend above, and the divisor below a horizontal line. Thus, ^l§- is read, " 175 divided by 25." 2. In writing numbers for solution, the divisor may be written either at the right of the dividend, thus, 175 1,25, or at the left of it, thus, 25 J 175. 9. How many times are 6 cents contained in 54 cents ? 10. What is the quotient of 40 divided by 8 ? 11. 56 -T- 7 = how many ? -^ = how many ? 12. The dividend is 24, and the divisor is 4. What is the quotient ? 13. If as many of 17 apples be divided among 5 children as will give them whole apples, how many whole apples will each child have ; and how many apples will be the remainder ? 14. Divide 42 figs among 6 girls, and tell me the dividend, the divisor, and the quotient. Why is there no remainder ? 94. DIVISION TABLE. 0123456789 [1 0123456789 6 12 18 24 30 36 43 48 .54 [ 6 0123456789 2 4 6 8 10 12 14 16 18 (^ 2 0123456789 7 14 21 28 35 42 49 56 63 1 7 0123456789 3 6 9 12 15 18 21 24 27 1 3 0123456789 8 16 24 32 40 48 56 64 72 [ 8 0i23456789 4 8 12 16 20 24 28 32 36 [4 0123456789 9 18 27 36 45 54 63 72 81 [ 9 0123456789 5 10 15 20 25 30 35 40 45 [ 5 0123456789 10 20 30 40 50 60 70 80 90 [IQ 0123456789 52 INTEGERS. 95. 2 is contained in 6, 3 times ; in 6 tens or 60, 3 tens or 30 times; in 6 hundreds or 600, 3 hundreds or 300 times. 25 is contained in 75, 3 times ; in 75 tens or 750, 3 tens or 30 times ; in 75 hundreds or 7,500, 3 hundreds or 300 times. In other words, if we divide ones, the quotient must be ones ; if we divide tens, the quotient must be tens ; if we divide hundreds, the quotient must be hundreds ; if we divide thousands, the quotient must be thousands, and so on. Hence, in division of integers. Any quotient figure must he of the same name or order of units as the right-hand figure of the partial dividend used to obtain it. see ManuaL C^SE I. The Divisor One Figure. FIEST METHOD. 96. Ex. 1. Divide 936 by 3. SOLUTION, 936 3 Divisor. 312 Quotient. 3 Explanation. — We place the divisor at the right of the dividend, separating them by Dividend. a hne, and draw a hne under the divisor, to separate it from the quotient. Then commencing at the left hand, ^ we divide each figure of the — dividend by the divisor, thus : 3 is contained in 9, 3 times ; and as the 9 is hun- dreds, the 3 is hundreds (95), and we write it as the first figure of the quotient. We have now used the 9 hundreds of the dividend, and subtracting it from the dividend, we have no hundreds left. The next part of the dividend to be used is the 3 tens, which we bring down for a partial dividend. 3 is contained in 3, 1 time ; and as the 3 is tens, the 1 is a ten (95), and we write it as the second figure DIVISION. 53 of the quotient. We have now used the 3 tens of the divi- dend, and subtracting it from the dividend, we have no tens left. The next part of the dividend to be used is the 6 ones, which we bring down for another partial dividend. 3 is contained in 6, 2 times ; and as the 6 is ones, the 2 is ones (95), and we write it as the third figure of the quo- tient. We have now used all the figures of the dividend, and the result, 312, is the quotient required. Ex. 2. Divide 17,668 by 7. Explanation. — 7 is not contained in 1 any number of times ; we must there- fore take 17 for the first partial dividend. Since 2 times 7 are 14 and 3 times 7 are 21, and 17 is more than 14, but less than 21, 7 is contained in 17, 2 times. As 17 is thousands, the 2 is thousands (95), and we write it as the first or thousands' figure of the quotient. We have now used 2 (thousands) times 7, or 14 (thous- ands) of the dividend ; and to find how many thousands remain undivided, we subtract the 14 (thousands) from the 17 (thousands), and obtain a remainder of 3 (thousands). We next bring down the 6 (hundreds) of the dividend, and uniting it with the 3 (thousands), we have 36 (hundreds) for a second partial dividend. Since 5 times 7 are 35, and 6 times 7 are 42, and 36 is more than 5x7 and less than 6 X 7, 7 is contained in 36, 5 times. As 36 is hundreds, the 5 is hundreds (95), and we write it as the second or hundreds' figure of the quotient. We have now used 5 (hundreds) times 7, or 35 (hundreds) of the partial divi- dend ; and to find how many hundreds remain undivided, we subtract the 35 (hundreds) from the 36 (hundreds), and obtain a remainder of 1 (hundred). We next bring down the 6 (tens) of the dividend, and uniting it with the 1 (hun- dred), we have 16 (tens) for another partial dividend. SOLUTION 17668 7 U 252A 86 85 16 U 28 28 54 INTEGERS. Since 16 is more than 2 times 7 and less than 3 times 7, 7 is contained in 16, 2 times. As 16 is tens, the 2 is tens (95), and we write it as the third or tens' figure of the quo- tient. We have now used 2 (tens) times 7, or 14 (tens) of the last partial dividend ; and to find how many tens re- main undivided, we subtract the 14 (tens) from the 16 (tens), and obtain a remainder of 2 (tens). We now bring- down the 8 (ones) of the dividend, and uniting it with the 2 (tens), we have 28 for a final partial dividend. 7 is con- tained in 28, 4 times ; and as 28 is ones, the 4 is ones (95), and we write it as the last or ones' figure of the quotient We have now used 4 times 7, or 28 ; and subtracting this from the last partial dividend, we have no remainder. All the figures of the dividend have been used ; and the result, 2,524, is the quotient required. 97. The quotient in Division is sometimes an abstract, and sometimes a concrete number. It is therefore neces- sary, before proceeding to the solution of problems, that we determine when the quotient is abstract, and when con- crete. To do this, we will take the following examples : (1) 15 cents [ 3 cents 5 (2) - 1513 5 (3) 15 cents ^3 5 cents (4) 15 13 cents Impossible Writing in the places of numbers, words indicating the kinds of numbers used, we have : (1) (2) Concrete [ Cmicrete Abstract [ Abstract Abstract Abstract (3) (4) Cmicrete yAbsPract Abstract [ Concrete Concrete Impossible DIVISION. 55 98. These illustrations fully establish the following facts : I. The quotient will he an abstract number, when the divi- dend and divisor are both abstract or both concrete numbers. (Ex. 1, 2.) II. The quotient will be a concrete number, ivhen the dividend is a concrete, and the divisor an abstract number. (Ex. 3.) III. Either the divisor or the quotient must always be an abstract number. (Ex. 1, 2, 3.) IV. An abstract number can not be divided by a concrete number. (Ex. 4.) PBOBZEMS, (1) (3) (3) (4) 648 1 2_ 8484 [$4 373 tons t;^ 7393 [6 5. A stage company paid $396 for 3 horses. How much did they cost apiece ? $132. 6. How many suits of clothes can be made from 1,348 yards of broadcloth, allowing 4 yards for each suit ? 312. 7. At a mortgage sale, 3 city lots were sold for $1,596. How much was that for one lot ? 8. A railroad company bought 1,456 cords of wood, which they transported on platform cars, each carrying 8 cords. How many car loads were there ? 182. 9. An Ohio farmer raised a crop of 1,965 bushels of wheat, which he exchanged with a miller for flour, receiving 1 barrel of flour for every 5 bushels of wheat. How much flour did he receive for his wheat crop ? 393 barrels. 10. If 6 masons lay 15,894 bricks in a day, how many bricks can 1 mason lay? 2,649. 11. If the yearly expenses of a family of 7 persons are $3,065, what are the expenses of 1 person ? 13. An army-wagon train, 9 miles long, contains 1,944 wagons. How many wagons is that to the mile ? 216. 56 INTEGERS. SECOND METHOD. 99. Ex. Divide 25,216 by 8. Explanation. — We write the dividend and divisor as in the bolutiox. First Method, but below the ^i^i<^ena.^5 2 1 6 [Sm^isor. dividend we draw a horizontal 315 2 Quotient. Hne, under which to write the quotient. 8 is contained in 25 (thousands), 3 (thousands) times. This quotient figure we write directly below the last figure (5) of the part of the dividend used to obtain it. We multiply the divisor (8) by the 3 (thousands), and sub- tract the product from the 25, performing both computa- tions mentally. We now mentally unite the remainder 1 (thousand) with the next figure of the dividend, 2 (hun- dreds), and divide the result, 12 (hundreds), by the divisor. 8 is contained in 12 (hundreds) 1 (hundred) time. We write the 1 as the second figure of the quotient, and then multiply 8 by it, and subtract the product from the partial dividend, 12 (hundreds), performing the computations men- tally, as before. We next mentally unite the remainder, 4 (hundreds), with the 1 (ten) of the dividend, and divide the result, 41 (tens), by the divisor. 8 is contained in 41 (tens) 5 (tens) times. We write the 5 as the third figure of the quotient, and then multiply 8 by it, and subtract the prod- uct from the partial dividend, 41 (tens), performing the computations mentally, as before. We mentally unite the remainder, 1 (ten), with the 6 (ones) of the dividend, and divide the result (16) by the divisor. 8 is contained in 16 (ones) 2 (ones) times, and we write 2 as the fourth figure of the quotient. We have now used all the figures of the dividend ; and the result, 3,152, is the quotient required. In the Second Method, the same computations are ]3er- f ormed as in the First Method ; but the results of the sub- tractions and multiplications are not written, and hence fewer figures are used. DIVISION 57 100. I^ong division is that method of dividing, in which all the products and partial dividends are written. 101. Short 1)ivision is that method of dividing, in which only the dividend, divisor, and quotient are written. Seo Manual. VJt O BIjJEMS. 13. A city corporation paid $9,376 for 3 steam fire-engines. What was the cost of each ? $J^, 688. 14. A man whose wages were $3 a day, earned $891 in a year. How many days did he work ? 297. 15. A steamboat which was owned by 4 men in equal shares, was sold for $39,724. How much was each man's share of the re- ceipts? $9,931. 16. A farmer has 1,458 bush- els of wheat, which he intends to carry to market in 2-bushel bags. How many bagfuls will he have? 729. 17. If 3 horses eat 2,025 pounds of hay in a month, how much will 1 horse eat ? 18. If I feed a horse 7 half- bushel measures of oats in a week, how many weeks will 483 half-bushels last him ? 19. How long will 3 horses be in eating the same quantity of oats ? 23 weeks. 20. In a barrel in the gran- ary are 96 quarts of corn, from which Clara feeds her ducks and chickens. How many quarts of corn are there for each oiie of the fowls ? 21. How many weeks will the corn last, allowing 6 quarts a week for the poultry ? 16. 22. A man leased a farm for $865, at the rate of $5 an acre. How many acres were in the farm ? 173. 3* 58 INTEGERS. 23. A forwarder shiiDped 24,744 bushels of grain in 6 equal car- goes. How many bushels were in each cargo ? ^ 12 J^. 24. A builder paid $5,145 for bricks, at $7 a thousand. How many thousand did he buy ? 735. 25. It cost $6,504 to build a plank-road 8 miles long. What was the cost per mile ? $813. 26. How long will it take a ship to make a Toyage of 889 miles, if she sails 7 miles an hour ? 127 hours. 27. A boatman carried 8,532 barrels of flour from Oswego to New York in 9 down trips. How many barrels did he take each down trip ? 28. A party of 8 men spent $1,072 on a journey to California, and they shared the expense equally. How much did each man pay? CASE II. The Divisor more than One Figure. 102. Ex. Divide 13,091 by 63. Explanation. — When the divisor con- solution. sists of more than one figure, the division inr is commonly most easily performed by Lonof Division, or the First Method ex- ^-^^ plained in Case I., as shown in the Solu- 53 [2J,.7 tion. ^71 103. It is sometimes difficult to tell, _ - — without trial, how many times the divisor is contained in a partial dividend. For example, divide 25,474 by 47. first trial. We can not readily tell how many times 25 Jf7 Ji. 47 is contained in 254, but we will sup- ^^^ pose that it is contained 4 times. Writ- 6 6 ing 4 as the first quotient figure, we mul- tiply and subtract, and obtain a remainder of 66. Since this remainder, which is a part of the partial dividend, is greater than the divisor, 47 is contained in 254 more than DIVISION. 59 4 times. We next try 5 as the quotient second tkial. figtire ; and the remainder, 19, is less than 25 Jf7 Jf 55 235 47. Hence, 47 is contained in 254, 5 times. We will now suppose that 47 is contained -^ ^ ^ in 197, the next partial dividend, 5 times. Writing 5 as the second figure in the quo- tient, we multiply, and obtain a product of 235. Since this product is more than 197, 47 is not contained in 197 as many as 5 times. Hence, I. When any remainder is greater than the dwisoVy the quo- tient figure is too small; and n. When any product is greater than the partial dividend, the quotient figm^e is too great. seo Manual. PHOBJOEMS. 29. A drover paid $313 for 13 beeves. How much did they cost him per head ? $24. 30. A man on a journey traveled 608 miles in 19 days. At what rate per day did he travel ? 33 miles. 31. If 21 acres of land produce 945 bushels of barley, what is the yield per acre ? 4^ iusheU. 32. A farmer paid $6,804 for a farm of 108 acres. What was the price per acre ? 33. A turnpike-road 46 miles long was kept in repair a year at ■an expense of $1,242. What was the cost per mile ? $37. 34. In a certain school the aggregate or total attendance for a term of 65 days was 7,410. What was the average daily atten- dance ? 114. 35. How much must a man earn in each of the 313 working days of a year, to earn $1,252 in a year ? 36. A dairy-man packed 10,304 pounds of butter in 56-pound tubs. How many tubs did he fill ? I84.. 37. Last season it cost a milkman $576 to winter 32 cows. What was the cost per cow ? $18, 38. If 1 sheet of paper is required for 24 pages of a book, how many sheets will be required for a book of 432 pages ? 60 INTEGERS, 39. How many cars, each carrying 49 passengers, will be required to carry 392 passengers ? 40. The entire cost of constructing a railroad 84 miles long was $4,065,264. What was the cost per mile? $48,396. 41. A person whose property is worth $9,375, spends yearly $625 more than his income. In how many years will he spend all his property ? 15. 42. In how many hours will an express train run the entire length of a railroad 544 miles long, running 32 miles an hour ? 43. The stock of a bank, consisting of 875 shares, is worth $117,250. What is the value of each share ? $134. 4:4:. A grain buyer shipped 401,400 bushels of wheat from Mil- waukee to Buffalo in 24 equal cargoes. How many bushels were there in each cargo ? 16,725. 104. Ex. Divide 48,507 by 69. Explanation. — In the First Solution we see that the second remainder, 20, is the same as the second partial dividend. In the Second Solution we omit to mul- tiply by the quotient figure, 0, and form the third partial dividend by annexing 7, the next figure of the dividend, to the 20, the second partial dividend. Hence, Whenever the^ partial dividend is less than the divisor, ive place in the quotient, and hring down the next figure of the dividend for a new partial dividend. PROBIjEMS, 45. How much does a man earn each month, $1,260 a year ? 46. In a factory 275 yards of cloth are made daily, days will be required to make 57,475 yards ? 47. The total cost of constructing a telegraph line 359 miles long was $360,795. What was the cost per mile ? $1,005. FIRST SOLUTION. Jf8507 Ji.8S 20 00 69 JOS 207 207 SECOND SOLUTION. U8507 U8S 69 70S 207 207 whose salary is $105, ily. How many DIVISION. 61 48. The yield of a Pennsylvania oil well for 1 month was 30,186 gallons, which was put into casks, each holding 43 gallons. How many casks were used ? 49. What is the quotient of 120,772,144 -v- 592 ? 20J^,007. 50. The dividend is 524,177,472, and the divisor 5,824. What is the quotient? 90,008. 51. Divide 1 billion 963 million 198 thousand 504 by 28 thou- sand 9. '^^ thousand 56. C^SE III. Remainders after Dividing last Partial Dividend. 105. Ex. Affcer dividing 1,315 acres of land into as many farms as possible of 92 acres each, how much land will be left? Explanation. — ^The divisor, 92, is con- solution. tained in the last partial dividend, 395, IS 15 \ 92^ 9 ^ 4 times, with a remainder of 27. As there ^ [ 1 If. is no fisfure of the dividend that has not <5 P 5 been used, we have no figure to write _i_i. with the 27 to form a new partial divi- ^ 7' Remainder. dend, and the work is completed, leaving a remainder of 27. Hence, after dividing 1,315 acres of land into 14 farms of 92 acres each, there are 27 acres re* maining undivided. See ManuaL FJtOBLJEMS. 52. How many payments of $350 each must I make, to pay for a house and lot worth $3,400 ? 9, and one payment of $250. 53. A farmer made 963 gallons of cider, which he put into casks holding 41 gallons each. How many full casks had he ? • 23, and 19 gallons over. 54. The dividend is 80,963, and the divisor 376. What is the remainder ? 123. 55. A forwarder ships 15,500 barrels of flour to New York, by canal. If 856 barrels make a boat load, how many full boat loads has he to ship ? 18, and 92 Ixirrels remainder. 62 INTEGERS. 56. If the directors of a railroad company appropriate $32,000 for the purchase of passenger cars, and the cars cost $1,875 each, how many cars can be bought with the appropriation ? 17, mid leave a surplus of $125. 67. What will be the remainder, after dividing 62,676 by 573 ? 58. How many bales of 396 pounds each can be made from 84,000 pounds of cotton ? Memainder^ JfS pounds. 59. 1,405,169 -=- 3,376 = how many ? Bemaijider^ 953. 60. The dividend is 5 million 92 thousand 209, and the divisor 10 thousand 23. "What is the remainder ? 61. Divide 56,432,782 by 27,541. Bemainder, 1,273. The Divisor any number of Tens, Hundreds, Thousands, and so on. 106. Ex. 1. Divide, 67,200 by 100. Explanation. — By the First Solution it ^^Rst solution. will be seen that, after removing as many figures from the right hand of the divi- dend as there are ciphers in the divisor, the remaining figures of the dividend are the same as the quotient. Therefore, In the Second Solution we have brought down, for the quotient, all the figures of the dividend except as many of its right-hand figures as there are ciphers in the divisor. Ex. 2. Divide 49,392 by 1,000. Explanation. — By the First Solution we see that, if we omit the three right-hand figures of the dividend, the other figures are the same as the quotient, and that the three right-hand figures thus omitted are the same as the remainder. There- fore, 67200 100 600 672 720 700 200 200 8KC0ND SOLUTION. 67200 JOO 672 FIK8T B0LT7TI0IT. 4.9392 40 00 9392 9000 392 1000 [49 DIVISION. 63 In the Second Solution we have brought second solxttion. down, for the quotient, all the figures of Jf9892 \10 the dividend except three, (^. e., as many ^P 392 of the right-hand figures as there are ciphers in the divisor), and for the remainder we have written the three right-hand figui'es. Hence, I. Removing the units' figure from a number, divides the number by 10. II. Removing the units and tens, or the two right-hand figures from a number, divides it by 100. III. Removing the three right-hand figures from a number, divides it by 1,000. IV. Removing the four right-hand figures, divides by 10,000 ; removing five figures, by 100,000; removing six figures, by 1,000,000; and so on. V. Ttie figures removed are the remainder, and the other figures are the quotient. 107. "When the divisor contains more than one ten, hun- dred, thousand, etc., the manner of obtaining the final remainder is more difficult. For illustrating the method, we will take the following examples : (1) (3) (3) (4) 75 1 15 75_y3_ 5 25 15 13 75 [5 I 15 [5 7500 [ 100 5 l5~ [ 3 75 1500 y 100 15 5 5 5 By carefully examining these four examples, we see in Ex. 1 that the quotient of 75 divided by 15 is 5 ; in Ex. 2, 3, that the quotient is not changed by dividing both divisor and dividend by 3 or 5 ; and in Ex. 4, that the divisor, 1,500, and the dividend, 7,500, may both be divided by 100, and the results used as divisor and dividend, without af- fecting the final quotient. That is. Both divisor and dividend may be divided by the same num- ber, without affecting the value of the final quotietit. FIU6T SOLUTION. SECOND SOLUTIOIf. SJf.\00 64 INTEGERS. 108. Ex. Divide 15,284 by 3,400. -, r o ^ i Explanation. — In the First Solution we 1S600 divide as tauglit in Case III., and obtain a iqrJl quotient of 4, and a remainder of 1,684. In tlie Second Solution we first divide- botli divisor and dividend by 100, (which i52\8Jf we do by cutting off, by a vertical line, 13 6 the two right-hand figures from each IGSA term), obtaining 34 for a new divisor, 152 for a new dividend, and a remainder of 84. Dividing 152 by 34, we obtain a final quotient of 4, and a second re- mainder of 16. Since this remainder was obtained by sub- tracting 136 hundreds from 152 hundreds (see Eirst Solu- tion), it must be 16 hundreds ; while the first remainder, 84, is the tens and ones of the dividend. We therefore unite these two remainders, by annexing the 84 to the 16 (hundreds), and we have 1,684 for the final remainder, the same as in the Eirst Solution. Hence, In dividing by any number of tens, hundreds, thousands, and so on, TJie final remainder consists of the figures which were cut off from the dividend, annexed to the figures of the last remainder. Pit OBLEMS. 62. A dealer sold 10 sewing-machines for $G50. How much were they apiece ? 63. A cental of grain is 100 pounds. How many centals in 47,300 pounds ? Jf73. 64. The capital or stock of a certain mining company is $235,000, and it is divided into $1,000 shares. How many shares of stock are there ? 65. At $3,000 each, how many steam-tugs can be bought for $6,000 ? 66. A builder paid $5,760 for boards, at $30 per thousand. How many thousand feet did he buy ? 192, DIVISION. 65 67. A carriage maker received $1,200 for light carriages, at $200 apiece. How many carriages did he sell ? 68. What is the quotient of 393,400,000 divided by 100,000 ? 69. A government agent paid $56,400 for horses for the army, at $150 apiece. How many did he buy ? 376. 70. A farmer having $83, wishes to jourchase yearling calves, at $10 a head. How many can he buy ? 8, and Jmm $3 left. 71. How many freight cars will be required to transport 58,293 barrels of flour, if 100 barrels make one car load ? 582 full cars, and 1 car carrying 93 larrels. 72. A company purchase a hotel in New Orleans for $165,675, and the payments are to be $25,000 annually. How many pay- ments must they make ? G of $25, 000 each, and 1 of $15, 675. 73. Divide 103,285 by 36,000. Memainder, 13,285. 74. What is the quotient of 17,630,000 divided by 24,000 ? 109. The facts deduced in Arts. 95, 98, 106, 107, may now be stated as General ^rmciples of 2)lrisio?i, I. A concrete number can he divided by either a conci^ete or an abstract number. II. An abstract number can be divided by an abstract number only. III. The quotient is an abstract number ^ when the divisor and dividend are both abstract or both concrete numbers. rV. The quotient is a concrete number, when the divisor is an , abstract, and the dividend a concrete number. V. Any quotient figure is of the same order of units as the last figure of the dividend used to obtain it. VI. 2he removal of the right-hand figure from a number divides that number by 10. VH. Both divisor and dividend may be divided by the same number, without affecting the value of the final quotient. Vm. The right-hand figure of any remainder is of the same order of units as the last figure of the dividend used. 66 INTEGERS. llOt Upon these principles is based the 2iule for 2)lvlsio7i of Integers, I. For Long Division. 1. Flace the divisor at the right of the dividend, separate them by a line, and draw a line under the divisor to separate it from the quotient. 2.- Find how many times the divisor is contained in the first partial dividend, and write the result for the first figure of the quotient. 3. Multiply the divisor by this quotient figure, suUract the product from the partial dividend used, and to the remainder annex the next figure of the dividend for a new partial divi- dend. 4. In the same manner, continue to divide, multiply, suh- tract, and hring down, until all the figures of the dividend have teen used, II. For Short Division. 1. Write the divisor and dividend as in Long Division, and draw a line under the dividend to separate it from the quotient, 2. Mnd hoio many times the divisor is contained in the first partial dividend, as in Long Division, and write the result under the last figure of the dividend used, for the first figure of the quotient, 3. Multiply, subtract, and form a neiu partial dividend, as in Long Division, performing the operations mentally. 4. Divide this partial dividend, and write the result as the second figure of the quotient. 5. Proceed in the same manner until all the figures of the dividend have been used, in. For dividing by any number of tens, hundreds, thou- sands, and so on. 1. Gut off the ciphers by a line, and also an equal number of figures from the right oftlie dividend. DIVISION. 67 2. Divide the remaining figures of tlie dividend ly the re- maining figures of the divisor. 3. For the true remainder, annex to the last remainder the figures cut off from the dividend. JPJt OBIjEMS. 75. An army contractor paid $276,560 for beef, at $16 a barrel. How much beef did he buy ? 17,285 barrels. 76. In a cotton-factory are 54 looms, which were bought at a total cost of $9,720. What was the^cost of each loom ? 77. A canal 97 miles long was constructed at a cost of $14,131,930. What was the cost per mile ? $14^j 690. 78. Peter having an ear of corn in which were 864 kernels, planted it in hills of 6 kernels each. How many hills did he plant ? 79. He planted the corn in 8 equal rows. How many hills were there in each row ? 18. 80. The fare of 219 passengers by steam-ship from New York to Havre, was $36,135. How much was the fare of each passenger ? 81. At $60 a head, how many cows can be bought for $1,650 ? 27, with $30 left. 82. A city builder received $780,000 for building brown-stone front houses, at an average price of $10,000 each. How many houses did he build ? 83. A manufacturer sold reapers at $130 each, and received $6,240. How many reapers did he sell ? 4^. 84. How many barrels, each holding 200 pounds, will be required for packing 47,875 pounds of pork? 239, with 75 pounds of porTo left. 85. A wholesale grocer bought 3,440 pounds of tea, in 80-pound chests. How many chests did he buy ? J^. 86. How many canal boats can a transportation company buy with $34,000, at $1,000 each ? 87. In the schools of a certain city 28,497 pupils are taught by 483 teachers. What is the average number of pupils to a teacher ? 68 INTEGERS. 88. A paper maker having 361,920 sheets of foolscap, put it up for market in quires of 34 sheets each. How many quires were there ? 89. He sold the paper by the ream of 30 quires. How many reams did he sell ? 75^., 90. The United States Government paid $103,600 for 740 army wagons. How much was that for each wagon ? $1^0. 91. What is the quotient of 447 billion 670 million 621 thou, sand 104 divided by 4 million 930 thousand 76 ? 90 tJiousaml 8O4. 92. Divide 660,886,723 by 982. Remainder, 723. 93. The dividend is 468,002,659, and the divisor 9,497. What is the remainder ? 9^,493. 94. A planter raised 86,301 pounds of cotton on 233 acres of land. How many pounds was the yield i)er acre ? 387. 95. The New York and New Haven Railroad track is 401,380 feet long. How many miles from the New York to the New Haven Railroad depot, there being 5,380 feet in a mile ? 76. 96. In a certain county $3,039,688 were paid in bounties to 3,394 volunteers. What bounty was paid to each soldier ? $852. 97. How long will 564,000 rations last an army brigade of 5,875 men ? 96 days. 98. A produce buyer purchased 417 bushels of wheat, 873 bush- els of oats, and 314 bushels of barley. The bins in his storehouse will hold 73 bushels each. How many bins will each kind of grain fill ? Wheat, 5 Una, and 57 bushels over. Oats, 12 "■ " 9 " " Barley, ^ " " 26 " " 99. If all of the grain was of one kind, how many bins would it fill ? , 20 hashels, remainder. 100. A man buys a farm of 113 acres, at $54 an acre. He pays $1,392 down, and agrees to pay the balance in 6 equal yearly pay- ments. How much of the debt must he pay each year ? $785. REVIEW PROBLEMS. 69 SECTION VII. ^:^riByr i>^osi.bms ij\r ijytbgb^rs, 1. The parts of a number are 73, 437, 856, and 32,519. What is the number ? See Manual. 2. The minuend is 59,408, and the subtrahend 14,642. What is the remainder ? 3. The sum of two numbers is 1 million 56 thousand, and one of the numbers is 304 thousand 9. What is the other number ? 4. A reward of $7,650 was shared among 4 detectives, the first of whom received $2,225, the second $1,750, and the third $1,875. How much did the fourth receive ? 5. The multiplicand is 185,046, and the multiplier 4,309. What is the product ? 6. What is that number, the factors of which are 384, 27, 90, and 10,000 ? 7. The dividend is 1,728,000, and the divisor 1,800. What is the quotient ? 8. If the dividend is 5,443,200, and the several successive divisors are 9, 14, and 600, what is the final quotient ? 9. A farmer's expenses and receipts one year were as follows : ^oi 7r/Cea-/. '^ ^a'/ ^ z r z o jv. 135. Ex. What is tlie sum of 56.125, 9.356, and 123.25 ? Explanation. — Since only like orders of solution. units in different numbers can be added (see 5 6.125 89, II.), we write the numbers with like orders y ^'i't'i of units — both decimal and inteoral — in the same columns. The decimal points then stand 1-88.731 in a column. We commence at the right, and add as in integers. Since the sum of thousandths is thou- sandths, the sum of hundredths is hundredths, and the sum of tenths is tenths, and there are thousandths, hundredths, and tenths in the given parts, there must also be thou- sandths, hundredths, and tenths in their sum. We therefore place the decimal point in the sum before the 7, and directly under the decimal points in the parts. 136. ^ule for oiddltlon of decimals, I. Write the numbers so that the decimal points shall stand in a column. II. Add in the same manner as i7i integers, and place the decimal point in the sum directly under the decimal points in the parts. see Manual. PM OBLEMS. (1) (2) (3) (4) (5) .331 .48 3.62 162.71 .0052 .746 .697 517.83 48.086 .02081 .984 .8 21.9 3915.3004 .016 .258 .5764 674.08 ■ .721 .0000375 6. A farmer brought three loads of wood to market, the first load containing .8 of a cord, the second .75, and the third .9375. How many cords did he bring in all ? 2.^875. 4* 82 DECIMALS. 7. A peddler traveled 6.75 miles one day, 4.6 miles the next, 7.384 miles the third, and 2.14 miles the fourth. How far did he travel in the four days ? 20.87 Jf, miles. 8. How many acres in four ^elds, there being 9.5 acres in the first, 11.4 acres in the second, 8.75 acres in the third, and 12.675 acres in the fourth ? Jt2.325. 9. A lady bought 16.25 yards of silk, 12.75 yards of alpaca, 6.5 yards of merino, 11.875 yards of delaine, and 23.5 yards of French calico. How many yards of goods did she purchase ? 70.875. 10. A farmer sold eight lots of hay, as follows : 7.637 tons, 3.5 tons, 17.396 tons, 5.824 tons, 12 tons, .95 of a ton, 8.0625 tons, and 6.4 tons. How much hay did he sell ? 61.7695 tons. 11. In six consecutive days a company of California miners obtained 3.5286 ounces, 1.4 ounces, 3.125 ounces, 7.0064 ounces, .65 of an ounce, and 2.72 ounces of gold. What was the whole amount for the six days ? 18.43 ounces. 12. Capt. Allen's farm consists of 7.4 acres of woodland, 16.275 acres of pasture, 23 acres of meadow, 6.025 acres of orchard and garden, and 72.3 acres of tilled land. How much land is there in his farm ? 125 acres. 13. At London the average fall of rain is, for MONTHS. INCHES. MONTHS. INCHES. MONTHS. INCHES. January, 1.483 May, 1.853 September, 2.193 February, .746 June, 1.83 October, 2.073 March, 1.44 July, 2.516 November, 2.4 April, 1.786 August, 1.453 December, 2.426 What is the average annual fall ? 22.199 inches. 14. A real-estate agent received for his services in selling seven farms, $137.25, $94.5, $216,375, $56.4, $113.7, $80,625, and $296.3. What were his total receipts ? $995.15. 15. A grocer bought six hogsheads of molasses, containing 117.5 gallons, 124 gallons, 129.3175 gallons, 104.75 gallons, 130.0625 gal- lons, and 131.5625 gallons. How much molasses did he buy ? 16. What is the sum of 967 thousandths, 54 hundred-thou- sandths, 953 and 5 tenths, 7 and 375 thousandths, 1000 and 1 ten- thousandth, 6 and 75 hundredths, 8 and 80,808 hundred-thou- sandthSj and 483 ? 2,460 and 40, 072 Imndred-thousandths. SUBTRACTION. 83 SECTION IV. STr:B T^AC TIOJV, 137. Ex. 1. From 11.278 subtract 4.825. Explanation. — Since only like orders of units solutiok. in different numbers can be subtracted the one 11.278 from the other (see 52, 11.), we write the num- Jf.8 2 5 bers with the units of the subtrahend, both deci- 6.4-63 mal and integral, under like orders of units of the minuend. The decimal point of the subtrahend then stands directly under that of the minuend. We subtract as in integers. Since the difference between thousandths and thousandths is thousandths, the difference between hun- dredths and hundredths is hundredths, and the difference between tenths and tenths is tenths, and there are thou- sandths, hundredths, and tenths in the given numbers, there must also be thousandths, hundredths, and tenths in their difference. We therefore place the decimal point in the difference before the 4, and directly under the decimal points in the given numbers. Ex. 2. From 52 subtract 9.785. Explanation. — Since decimal ciphers may be soltjtion. annexed to a number without changing its value 5 2 .0 (see 133, VI.), we annex ciphers to the minuend ^- '^ ^ ^ until it has as many decimal figures as the sub- Jf.2 .2 15 trahend, and then subtract and place the deci- mal point as in the Solution of Ex. 1. 138. ^ule for Subtraction of decimals, I. Write the numbers loitli the decimal point of the suUra- hend directly under that of the minuend. II. Subtract in the same manner as in integers, and place the decimal point in the remainder directly under the decimal point in the subtrahend, see Manual, / 84 DECIMALS. PROBLEMS, (1) (2) (3) (4) From .3758 386.25 57.4628 2940 Subtract il^ 50.76 82 41.93 .0492 5. A merchant sold 19.25 yards of sheeting from a piece which contained 43.75 yards. How many yards were left ? ^^.5. 6. A company that contracted to build a turnpike 17.5 miles long, have completed 9.875 miles. How much have they yet to build ? 'y^QoS miles, 7. One year a stock farmer put 43.5 tons of hay into his bams, and the following spring he had only 8.75 tons. How much had he fed to his stock ? 5^.75 tons. 8. A seedsman having 73.625 bushels of choice potatoes, bought enough more to increase his stock to 120 bushels. How many potatoes did he buy ? Jf6.375 dusMs. 9. The owner of a schooner sold .3125 of her to the captain. What part of the vessel did he still own ? .6875. 10. There are 192.8125 barrels of water in a cistern which will hold 320.5 barrels. How much more water will the cistern hold ? 11. A man bought a horse for $118,375, and afterward sold him for $130.25. What was his gain ? $11,875. 12. A load of hay with the wagon weighed 2 and 65 thou- sandths tons, and the wagon weighed 1 and 9 hundredths tons. What was the weight of the hay ? 975 thousandths of a ton. 13. A woman sold a house and lot, which cost her $2250.5, for $1900.75. How much did she lose on it ? $349.75. 14. A person traveled 1,200 miles in 4 weeks, going 276.5 miles the first week, 318.37 miles the second, and 294.2 miles the third. How far did he travel the last week? 310.93 miles. 15. A vessel of 400 tons burthen, bound up the lakes, ships at Buffalo 93.4 tons of railroad iron, 56.81 tons of salt, and 211.7 tons of general merchandise. How much does she lack of a full cargo ? 10. A man having three farms, containing, respectively, 296,5 acres, 145.75 acres, and 96 acres, sold to one man 72.5 acres, and to another 86 acres, and gave to each of his two sons 105.25 acres. How many acres had he left ? 169.25. MULTIPLICATION. 85 SECTION V. 139. 2 times 4 are 8, 3 times 3 are 9, 5 times 7 are 35 (or 35 ones) = 3 tens and 5 ones. In each of these illustra- tions the two factors are ones and the product is ones. The product of ones multiplied ly ones is alimys ones. 140. In multiplying 24.3 by 2, the 8 of the prod- uct is obtained by multiplying the 4 ones of the multipKcand by the 2 ones of the multiplier. Hence, the 8 is ones, and the decimal point must be placed at the right of it. In multiplying 4.17 by 2.1, the product of the 4 ones of the multi- plicand and the 2 ones of the multi- pher is the 8 of the second partial product. Hence, that figure is ones, and the 8 of the final product is also ones. In Ex. 3, the 8 of the third partial product is the product of the ones of the multiplicand and the ones of the multi- plier. Hence it is ones, and the 9 of the final product is also ones- In Ex. 4, the 8 in the sec- ond partial product is ones, and the 9 in the final product is also ones ; and in Ex. 5, the 7 of the third partial product is ones, and also the 8 of the final product. 141. In Ex. 1 there is one decimal figure in one of the factors, and one decimal figure in the product. In Ex. 2 there are three decimal figures in the factors and three in the product. In Ex. 3 and 4 there are four decimal figures Ex 2. J,. 17 2J 8.3 J^ 8,75 7 Ex. 1. Jf8.6 Ex.8. J^.26 2.13 127S U2G 8.52 9.07 3 8 Ex.4. Jf.316 32.U 1726 Jf. 8.6 3 2 1291^8 139.8381/. Ex. 5. 2,501/3 3.1/2 50086 100172 7.5 12 9 8.561/706 86 DECIMALS. is ten-tliou- sandths : 1 u \6\3 2\A\0 2 3 \ 6 9 1 2 L? 1 ^1 1 T 5 1 9\ 6 •9 1 1 1 ll III in tlie two factors and four in the product ; and in Ex. 5 there are six decimal figures in the factors and six in the product. From these examples we learn that The member of decimal figures in a product must equal tlie number of decimal figures in its factors. 1 42. By examining this Example we see Ex. that the 2 of the third partial product is 2.103 the product of ones multiplied by ones, 1.32 and therefore must be ones ; and that ' of ones and tenths is tenths ; of ones and hundredths, ) .^ i,^j,^iredths and of tenths and tenths ' of ones and thousandths, [ is thou- The and of tenths and hundredths ) sandths product "j of tenths and thousandths, and of hundredths and hun dredths of hundredths and thousandths is hun- dred-thousandths. Hence, Tlie number of tlie decimal place in which the product of any tivo decimal figures belongs, counting from ones, is equal to the sum of the numbers of the decimal places of the two figures multiplied. 113* Since 3 times ones is ones, we place the decimal point in Ex. 1 at the right of the in the product, as shown in (1). But the may be omitted from the multiplicand without changing its value, and the product will then be .759, as shov/n in (2). In multiplying 4 ten-thousandths by 2 thousandths, since the 4 is four places and the 2 is three places to the right of ones, the product, 8, must be seven places to the right of ones, and the other six places must be filled by ciphers. Hence, Ex.1. 0) 0.253 £ 0.759 (2) .253 5 .759 Ex.2. .OOOJf. .002 0000008 ^ m TIT' IVEi MULTIPLICATION. X^T^ *• T'?^?7 When there are not as many figures i7i the product as tKere are decimal figures in the factors, decimal ciphers must le prefixed to the other figures to supply the deficiency. 14li Upon the principles deduced in Art 141, 142, 143, is based the ^tele for MuUipUcation of decimals. I. Write the numters and multiply as in integers. II. Place the decimal point in the product, so that it shall contain as many decimal figures as hoth factors. I'M OBIjEMS. 1. How many gallons of oil are there in 16 barrels, each contain- ing 31.5 gallons ? 50^. 2. In a certain manufactory 3.7 tons of coal are used each day. How much will be used in 26 days ? *96.2 tons. 3. A farmer sows 1.75 bushels of wheat to the acre. How much seed will he require to sow 19 acres ? 33.25 lusJiels. (4) (5) (6) (7) 33.125 miles 14.6 27.31 24753 27 _3^ 4.5 3.16 8. Bought 137.5 acres of land, at $76.25 per acre. What did the whole cost ? $10484.375. 9. If 62.5 tons of iron be required for the track of one mile of railroad, how much iron will it take for 371.75 miles? 10. One pound English money is worth $4.84 United States money. What is the value in United States money of 16.87 pounds English money ? $81.6508. 11. What was the length of an army wagon- train that passed a given point in 4.08 hours, passing at the rate of 3.025 miles per hour ? 12.342 miles. 12. A mowing-machine company bought 137.5 tons of iron, at $32.75 a ton. To what did the purchase amount ? $4503.125. 13. Brass is .8 copper and .2 zinc. How much copper and how much zinc must be used to make .875 of a ton of brass ? Copper, .7 of a ton; zinc, .175 of a ton. 88 DECIMALS. 14. A man owning .8 of a mill, sold .3125 of his share. What part of the mill did he sell ? .25 of it. 15. If the length of a military step is 2.25 feet, how far will a soldier march in taking 1,762 steps ? 3964.5 feet. IG. One week a butcher bought 354 lambs, at $4.7 per head. How much did they cost him ? 17. If the average rate of speed of a railroad freight train, in- cluding stops, is 11.88 miles per hour, how far will it run in .85 of an hour ? 18. Limestone loses .35 of its weight when weighed in water. If a piece of limestone weighs 17.137 ounces in air, how much less will it weigh in water ? 5.99795 ounces. 19. A cubic inch of silver weighs 6.0G13 ounces, and gold weighs 1.83865 times as much as silver. What is the weight of a cubic inch of gold ? ll.lU6092Jt5 ounces. (20) (21) (22) (23) 72.65 .92 .000873 .00096 .6 ^ .26 .01298 24. What is the product of .0625 and .48 ? .03. 25. What is the weight of 25.75 feet of copper pipe, if one foot Weighs .375 of a pound ? 8.65625 ^pounds. 26. If one man can mow 1.875 acres in a day, how many acres can 13 men mow in 7.5 days ? 182.8125. 27. I made 325 gallons of cider. How much had I left, after selling 9 barrels, each containing 31.5 gallons ? Ji^l.5 gallons. 28. On invoicing his stock, a merchant finds that he has 7 pieces of cotton goods, of 43.75 yards each, 4 j)ieces of 46.5 yards each, 3 pieces of 39.5 yards each, one piece of 24.375 yards, and one of 19.675 yards. How many yards has he in all ? 65Jf.8. 29. A farmer sowed three fields to rye. The first, of 13.5 acres, yielded 23 bushels per acre ; the second, of 9 acres, yielded 30.25 bushels per acre ; and the third, of 11.75 acres, yielded 24.44 bush- els per acre. What was the total yield ? 869.92 Imshels. 30. How many bushels of oats must a livery stable keeper buy to last 11 horses 19 weeks, if he feeds to each horse 2.625 bushels a week ? 548.625. DIVISION. 89 SECTION VI. ^Dirisiojy. 145. We have learned, in Art. 95, that in the division of integers any quotient figure must be of the same order of units as the right-hand figure of that part of the dividend used to obtain it. If we divide 6 tenths, or .6, by 3, (i) (2) the quotient is 2 tenths, or .2. If :^ I ^ ^ [ 5 we divide 6 hundredths, or .06, by ,2 ,02 3, the quotient is 2 hundredths, or .02. The quotient of 6 thou- sandths, or .006, divided by 3, is ^^\ ^^^ 2 thousandths, or .002, and the 'l^V^ .0006 [3 quotient of .0006, divided by 3, is -002 ,0 002 .0002. In other words, When tenths are divided by an integer, the quotient is tenths ; " hundredths " " " " hundredths; " thousandths " " " " thousandths; " ten-thousandths" " " " ten-thousandths; and so on. When the divisor is an integer y any quotient figure will be of the same order of units, integral or decimal, as the right-hand figure of the partial dividend used to obtain it. 146. Ex. Divide 16.285 by 5. Explanation. — We write the terms, and solution. commence at the left to divide, as in in- IG .28 5 {5 tegers. Since the first partial dividend, 16, 3.257 is ones, the first quotient figure, 3, must be ones, and the next quotient figure will be tenths. We there- fore place the decimal point after the 3 ones, before writing any of the other figures of the quotient. The decimal point must always be placed in the quotient, be- fore writing the tenths' figure. 90 DECIMALS. 147. Ex. 1. Divide .0056 by 4 Explanation. — Since 4 is contained in solution. tenths times, and in hundredths times, -0056 { ^ we write ciphers in the places of tenths and ,00 IJf. hundredths after the decimal point in the quotient. Ex. 2. Divide .1272 by 8. Explanation. — Since 8 is contained in 1 solution. tenth times, we write a cipher in place of 'J-^'7 ^ {8 tenths after the decimal point in the quotient. ,015 9 Hence, When the first decimal figure or figures of the dividend will not contain the divisor, a decimal cipher or ciphers must be written in the quotient. 148. Ex. 1. Divide 12.6 by 24. Explanation. — 12.6 -4- 24 = .5, with a re- solution. mainder of 6 tenths. 6 tenths = 60 hun- ^^-^ _^ dredths (see 129), and 60 hundredths -^ 24 = — [.525 2 hundredths, with a remainder of 12 hun- ^0 dredths. 12 hundredths = 120 thousandths ^^ (see 129), and 120 thousandths -^ 24 = 5 -?^^ thousandths. Hence, -^ ^ When there is a remainder after using all the figures of the dividend, the division may ie continued, each new partial divi- dend being formed by annexing a decimal cipher. Ex. 2. Divide 31.5 by 8. Explanation. — ^In this Solution we form each solution. partial dividend after the second, by mentally ^ ^-^ I ^ annexing a decimal cipher to the partial re- 3.9375 mainder. JPH OBTjEMS. 1. A father divided 280.5 acres of land equally among 3 sons. How much land did each receive ? 93.5 acres. 3. In how many weeks will a man whose wages are $9 a week, earn $157.5 ? 17.5. DIVISION. 91 3. If a ditcher digs 8 rods of ditch in one day, how long will it take him to dig 118 rods ? H.75 days. 4. A farmer made 45 barrels of cider from 292.5 bushels of apples. How many apples did it take for a barrel of cider ? (5) (6) (7) (8) 209.58 1 6 $7209 [ $8 341.5 [ 77 .1537 1 29 9. If 18 silver spoons weigh 33.75 ounces, what is the weight of 1 spoon ? 1,875 ounces. 10. If one sheet of paper makes 48 pages of a book, how many sheets will it take for a book of 348 pages ? 7.25. 11. A tailor cut 6 coats from 13.75 yards of broadcloth. How much cloth did he put into a coat ? 149. 2 is contained in 6, 3 times ; 2 tens in 6 tens, 3 times ; 2 Imndreds in 6 hundreds, 3 times ; and so on. So, also, 2 tenths, or .2, is contained in 6 tenths, or .6, 3 times ; 2 hundredths, or .02, is contained in 6 hundredths, or .06, 3 times ; 2 thousandths, or .002, is contained in 6 thousandths, or .006, 3 times ; and so on. That is. When the divisor and dividend are of the same order of units, either integral or decimal^ the quotient is ones. This truth is shown in the following examples : (Ex. 1) (Ex. 2) (Ex. 8) (Ex. 4) (Ex. 5) (Ex. 6) 6001200 60{20 6 12 .6 [.2 .06^{.02 .006 [ .002 3 3 8 3 3 3 150. Ex. 1. Divide 36.45 by .15. solution. Explanation. — Since the right-hand figure of both divisor and dividend is of the same order of units (hundredths), the right-hand figure of the quotient must be ones, and consequently the entire quotient is an in- ^5 teger. ^tz^ S6.Jf5 .15 SO 2Jf3 6J, 60 92 DECIMALS. Ex. 2. Divide 2.68 by .25. Explanation. — The right- partial solution. hand figure of both divi- ^-^^ I -^^ sor and dividend being hun- I -^ ^ dredths, the 10 of the quo- ^ ^ tient is an integer, as shown in the Partial Solution. But since there is a remainder, we place a decimal point after the 10, and continue the division by annexing decimal ciphers to the partial remainders. (See 148). Ex. 3. Divide 15.695 by 7.3. Explanation. — The quotient of 15.6 (the first three figures of the dividend) divided by 7.3, is an integer, because the right-hand figure of each term is tenths. We therefore place an inverted caret (V) after the 6 of the dividend, to show what figures are used to obtain the integral part of the quotient. Placing the decimal point after the quotient figure, 2, we complete the division and obtain the entire quotient, 2.15. The number of decimal figures in the quotient will equal the number of decimal figures left in the dividend^ after taking from it as many decimal figures as there are decimal figures in the divisor. Note.— When decimal ciphers are annexed to form partial dividends, they must be counted as decimal figures of the dividend. 151. Ex. Divide 42 by .56. Explanation.— The right-hand figure of the soltttion. divisor is hundredths ; and as the right-hand |^-^^ j "^ figure of the dividend must also be hundredths !_. [ 7 5 to obtain ones for the quotient (see 149), we ^^0 annex two decimal ciphers to the dividend be- Z. fore dividing. 15.6' 9 5 7.3 lJf-6 2.15 109 78 365 865 DIVISION. 93 The dividend must contain at least as many decimal figures as the divisor. 152. Upon the principles deduced in Art. 145, 147, 148, 150, is based the ^iite for division of decimals. I. When the divisor is an integer. 1. Jf necessary, annex decimal ciphers to the dividend, till the figures of the dividend will contain the divisor. 2. Divide as in integers. 3. Place the decimal point in the quotient so that it shall con- tain as many decimal figures as the dividend. II. When the divisor is a decimal or a mixed number. 1. Place an inverted caret after the figure of the dividend that is of the same order of units as the right-hand figure of the divisor. 2. Divide as in integers, and place the decimal point so that the quotient shall contain as many decimal figures as there are decimal figures at the right of the inverted caret in the dividend. PM OBLEMS. 12. How many dress patterns, of 11.5 yards each, are there in a piece containing 46 yards of delaine ? 4- 13. If one length of 6-inch stove pipe can be made from 3.14 pounds of Russia iron, how many lengths can be made from 72.23 pounds ? 23. (14) (15) (16) (17) 75.6 [.9 21.25 [.4 99^24.75 3.985 [159.4 18. How many casks, each holding 41.315 gallons, will be re- quired to hold 11278.995 gallons of alcohol ? 273. 19. A merchant exchanged 35.0625 yards of cloth for wood, at the rate of 4.125 yards for 1 cord. How much wood did he re- ceive ? 8.5 cords. 20. A miller received $3,009 for ship-stuflfs, at $21.25 per ton. How many tons did he sell ? m.6. 94 DECIMALS. 21. How long will it take to manufacture 1321.65 barrels of flour, at the rate of 53.4 barrels per day ? 2Jf.75 days. 22. At Catskill, N.Y., on the 26th of July, 1860, the extraordi- nary fall of 18 inches of rain occurred in 7.5 hours. What was the average fall per hour ? S.J/, inches. 23. If 7 men cradle 116.55 acres of grain in 4.5 days, how many acres does 1 man cradle in 1 day ? See Manual. 3.7. 24. The winter term of a country school continued 13 weeks of 5.5 days each, and the aggregate attendance for the whole term was 3074.5 days. What was the average daily attendance ? Ji3. 153t Ex. How many goblets, each weighing 7.5 ounces, can a manufacturer make from 176 ounces of silver ? Explanation. — Since he will not make the solution. decimal part of a goblet, the result in this -^'^^ J^ problem will be an integer; and the solution is . , [ 2 3 complete when the ones of the quotient are ^^0 obtained. Since the 260 is tenths, the 35 is ?H tenths, and the true remainder is 3.5. Hence, ^'^ The right-hand figu7'e of any remainder will ahvays ie of the same order of units, integral or decimal^ as the last figure of the dividend used to oUadn it. PMOBLEMS. 25. Into how many building lots, each containing .375 of an acre, can 5 acres of land be divided ? 13^ with .125 of an acre left. 26. An oil refiner has on hand 22,240 gallons of oil. How many casks can he fill, if he puts 42.5 gallons in each cask ? 523 f and have 12.5 gallons left. 27. A forwarder has 2,150 tons of freight to ship by canal. If 110.5 tons make one boat-load, how many boat-loads has he ? 50.5 tons more than 19 'boat-loads. 28. If a teamster draws 1.125 cords of wood at a load, how many loads will 41.75 cords make ? 37 ^ and .125 of a cord more. 29. How many potash kettles, each weighing 362.5 pounds, can be made from 20,500 pounds of iron ? Remainder^ 200 pounds. UNITED STATES MONEY. 95 154. United States Money, or JF'ederat J^foneyy con- sists of dollars, cents, and mills. 10 mills are 1 cent. 100 cents are 1 dollar. 1 dollar is 100 cents. 1 cent is 10 mills. The unit of United States Money is the dollar. 155. Since 100 cents are 1 dollar, 1 cent is 1 hundredth of a dollar ; and since 10 mills are 1 cent, 1 mill is 1 tenth of a cent, or 1 thousandth of a dollar. Hence, Cents may always he written as hundredths, and mills as thousandths of a dollar. 1 cent is written $.01. 47 cents are written $.Ji7. 1 mill is written $.001. 50 cents 4 mills are written $.504. 25 dollars 5 cents 8 mills are written $25,058. 100 dollars 5 mills are written $100,005. 96 DECIMALS. Note. — The Table of United States Money, as established by Act of Con- gress, August 8, 1786, is as follows :' 10 mills are 1 cent ; 10 cents " 1 dime ; 10 dimes " 1 dollar ; 10 dollars " 1 eagle. But dimes are always read as tens of cents, and eagles as tens of dollars. Thus, 7 eagles 2 dollars 4 dimes 5 cents is $73.45, and is read " 72 dollars 45 cents," or " 72 and 45 hundredths dollars." JEXEMCISES, 1. Read $.06, $.44, $.80, $3.15, $70.40, $9.08. 2. Read $.005, $.456, $.047, $.192, $.601, $.309. 3. Read $19,476, $500,104, $1,008, $297,027. 4. Write 20 cents ; 5 cents ; 93 cents. 5. Write 10 dollars 50 cents ; 150 dollars 88 cents. 6. Write 4 mills ; 26 cents 9 mills ; 5 cents 3 mills. 7. Write 5 dollars 17 cents 5 mills. 8. Write 200 dollars 4 cents 8 mills. 9. Write 30 dollars 6 mills. 156. Decimal parts of a dollar less than mills or tTiou- sandths are read as decimals of a mill. $.0005 is 5 tenths of a mill. $.00025 is 25 hundredths of a mill. $.0064 is 6 and 4 tenths mills. $.3765 is 37 cents 6 and 5 tenths mills. $45.40375 is 45 dollars 40 cents 3 and 75 hundredths mills. EXEJt CISES. 10. Read $.0004, $.0056, $.00075, $.3715. 11. Read $.47675, $93.7564, $300.85354. 12. Write 5 tenths of a mill ; 75 hundredths of a mill. 13. Write 5 and 8 tenths mills ; 4 cents 2 and 9 tenths mills. 14. Write 56 cents 4 and 72 hundredths mills. 15. Write 8 dollars 10 cents 1 and 38 hundredths mills. 157. A Coin is a piece of metal on which certain char- acters are stamped, by authority of the General Govern- ment, making it legally current as money. UNITED STATES MONEY. 97 United States coins arc made of gold, silver, nickel, and copper, as shown in the COIN TABLE. Gold, NAMES OF COINS. VALUES. METALS. NAMES OF COINS. VALUES. 50-dollar piece, $50.00 [Dollar, $1.00 Double eagle, 30.00 Half-dollar, .50 Eagle, 10.00 Silver, - Quarter-dollar, .25 Half-eagle, 5.00 Dime, .10 3-dollar piece, 3.00 Half-dime, .05 Quarter-eagle 3.50 , 3-cent piece, .03 Dollar, 1.00 5-cent piece, .05 2-cent piece. .03 Nickel, . 3-cent piece. .03 Cent, .01 . Cent, .01 Notes.— 1. Gold coins of the values of $.50 and $.25 were coined by pri- vate assayers in California, the former in the years 1853-53, and the latter in 1854. 3. Half-cent copper coins have not been coined since the first issue of the nickel cent in the year 1857. 3. 3-cent pieces of copper and nickel -were first issued in the year 1865, and 5-cent pieces of the same metals in 1866. 4. The 50-dollar piece shown in the cut, page 95, is about .8 as large across as the real coin ; the other coins shown in the cut are full size. See Manual. 158. AI^OJ is a baser metal mixed with a finer ; as silver with gold, or copper with silver. 159. The United States gold and silver coins consist of 9 parts or .9 by weight of pnre metal, and 1 part or .1 of alloy ; the alloy of gold coins being equal parts by weight of silver and copper, and that of silver coins pure copper. Nickel and copper coins are not alloyed. see Manual. 160. In final results of computations, and in business transactions $.005 are written $.00^, and read one half cent. $.0035 (I $.00i one fourth " $.0075 (( $.00f. three fourths " $.00135 u $.00i. one eighth " $.00375 u $.00|, three eighths " $.00635 u $.00|, five eighths " $.00875 u $.001, seven eighths " See Manual. 98 DECIMALS, COMIFXJT^TIONS OJB^ xj. s, m:o:ney. 161. Since the dollar is the unit of United States Money, (see 154), and cents, mills, and parts of a miU are decimals of a dollar, it follows that United States Money is added, sultracted, multiplied, and divided in the same manner as other decimals. SOLUTION. 162. Ex. 1. What is the sum of $275,10, $27 5.10 $18.37i, $.883, and $31 ? 18.3 7 5 Explanation. — "We write the numbers with .8 8 3 3 1 dollars under dollars, cents under cents, and mills under mills ; and then add the parts, and place the decimal point in the sum, as in Addition of Decimals. Ex. 2. From $52.75 subtract $10.96|. Explanation. — "We write dollars under doUars, and cents under cents ; and then subtract, and place the decimal point in the remainder, as in Subtraction of Decimals. Ex. 3. Multiply $45,625 by 5.6, Explanation. — "We write the multiplier under the multipHcand ; and then multiply, and place the decimal point in the product, as in Multiplication of Decimals, Omitting two decimal ciphers from the right of the product, (see 133, VI), the required product is $255.50. $325,358 SOLUTION. $5 2.7 5 10.9675 $Ji.l.7825 SOLUTION. $1^5.625 5.6 273750 228125 $255.5000 $lJfJt.5. 82 25 8. $17,625 Ex. 4. Divide $1445.25 by 82. Explanation. — "We write the divisor at the right of the divi- dend; and then divide, and place the decimal point in the quo- tient, as in Division of Deci- mals. 625 5_7Jt_ 512 Jf.92 205 16J_ UIO It-IQ UNITED STATES MONEY. 99 Ex. 5. How many times are solution. $5.06i contained in $567 ? $567.000 ^. $5.0625 Explanation. — We divide as in 5 0625 | ^ ^ ^ Division of Decimals. Since both 60750 dividend and divisor are con- 50625 Crete numbers (dollars), the quo- 101250 tient must be an abstract num- 101250 ber. (See 97). 163i In business transactions, when the mills in any fined result are 5 or more, they are regarded as 1 cent ; and when less than 5, they are rejected. 164i The commercial character @ signifies at, or by the yard, pound, gallon, bushel, or other unit named in the problem. Thus, "2 dozen eggs, ® $.28," signifies ''2 dozen eggs, at $.28 a dozen." Pit OBLEMS. 1. Martha paid $.87|- for a grammar, $.25 for a slate, $.75 for a reader, and $.12^ for a writing-book. What was the amount of her purchases ? $2, 2. A farmer killed an ox, and sold the four quarters for $9,935, $9.62^, $8.11, and $8 ; the hide for $6.89 ; and the tallow for $8.92. How much did the ox bring him ? $51.48. 3. One year a gentleman's income tax was $34.26 ; his state tax was $42.11; village tax, $18.04 ; school tax, $7.65 ; road tax, $.62 J- ; and military tax, $1. What amount of taxes did he pay that year ? $103.68^. 4. A owes to B, $374 ; to C, $47.50 ; to D, $193.1875 ; to E, $21.81 ; to F, $6.75 ; to G, $3,125 ; and to H, $11.0625. What is the amount of his indebtedness ? $657,435, What is the amount of each of the following bills ? 7. Traveling Expenses. Railroad fare, $18,625 Steamboat " 7.25 Carriage hire, 5.00 Hotel bills, . 31.875 Other expenses, 17. 6 7 5. For Furniture. 6. For Hardware. 1 Set of Chairs, $7.50 1 Plow, . . $6.75 1 Table, . . . 4.75 1 Spade, . 1.125 1 Eocking-chair, 3.25 1 Hammer, . .5625 1 Bedstead, . . 7.25 1 Pitchfork, .875 1 Mirror, . . 1.375 Nails, . . 1.375 100 DECIMALS. 8. A man having $300.82 in bank, draws out $00.29. What ia then his balance in bank ? $2Jf0.53, (9) (10) (11) (13) From $593,025 $1132.053 $251 $1574 take 490.54 90 8.375 .856 13. A man gave me his note for $75, and he has since paid all but $24.50 of it. How much has he paid on the note ? $50.50. 14. A dress-maker earned $34 in a month, and her expenses were $26.67. How much did she save ? $7.33. 15. A grain buyer purchased a lot of wheat for $1078.25, and the following week sold it for $1219.125. How much did he clear on the wheat ? $140.87^-. 10. I paid $2841.375 for an interest in an iron-foundery, and afterward sold it for $3129.16, How much was my gain ? (li) (18) (19) (20) Multiply $21.25 $2.4375 $0.80 $4,025 by 46 .215 12.5 11.25 21. How much will 24 bushels of turnips come to, at $.375 a bushel ? $9. 22. A lady bought 32 yards of carpeting, @, $1,121^. How much did it cost her ? $36. 23. How much must I pay for 23 rolls of paper-hangings, © $.375 ? $8.6^. 24. A farmer sold 390 pounds of wool, @ $.075. How much did it coiue to ? $267.30. 25. A mechanic worked 4.9 days, for $1,875 per day. How much did he earn ? $9.18^. 20. If the interest on $1 for 1 year is $.07, what is the interest on $24.75 ? $1.73^. 27. A drover bought a flock of 123 sheep, ® $2.5025. What was the cost of the flock ? $315.18}. 28. A milk-man's sales average 219 quarts a day, at $.05 a quart. What are his daily receipts ? 29. How much do his sales amount to in a year, or 305 days ? $3996.75. UNITED STATES MONEY. 101 30. I paid $19.9375 for iron castings, at ^.0625 a pound. How many pounds of castings did I buy ? 319. (31) (32) (33) (34) $5.25 [14 $362.25 [23 $180.40 [$.1025 $63 [56 85. A shipment of 1,583 bushels of corn was sold for What was the price per bushel ? ^.56f . 36. A tanner paid $156.82|- for 25.5 cords of hemlock bark. How much was that a cord ? $6.15. 37. Divide $6 into 150 equal parts. 38. How many bushels of potatoes can be bought for $57.3125, at $.875 per bushel? 65.5. 39. A builder contracted to put up a brick dwelling for $3,725. Tlie building materials cost him $1641.0625, and he paid out for labor $1296.50. Did he make or lose money on the contract, and how much ? Be made $787.43§. 40. A man exchanged a horse worth $187.50, and a watch worth $64,875, for a span of horses worth $310, paying the balance in money. How much money did he pay ? $57,625. 41. A farmer carried some pork to market, which he sold for $57.62|-, and some poultrj'-, which brought him $4.18f. He paid out $13.50 for a coat, $4.48 for some groceries, and $29.74 for a bill of hardware. How much money had he left ? $14.09^. 42. A young man bought a farm of 84 acres, at $75 an acre, and made a cash payment on it of $1,750. How much did he run in debt ? $%550. 43. A man sold a quarter of beef, which weighed 156 pounds, at $.08^- per pound, and expended the money for nails, at $.05^- per pound. How many nails did he purchase ? 2S4 pounds. 44. The salary of the President of the United States is $25,000. How much is that per day, allowing 365 days to the year, and deducting the Sundays ? 45. I bought a lot of teas for $376.75, and paid $31.18f for transportation on them. For how much must I sell them to make $103.12^ ? 46. A druggist bought 7 barrels of turpentine, each containing 31.5 gallons, at $1.37|- per gallon. Wliat did the whole cost ? 102 DECIMALS, What is the sum of each of the followiaj (47) BOOT AND SIIOK TKADE. ONE day's bales. 1 Pair Calf Boots, . . .$6.50 1 '' Stoga, . . . . 4.75 1 " Coarse, , . . . 2.25 1 " Ladies' Gaiters, . 2.75 1 " Misses' " . . 1.75 1 " " Slippers, . 1.25 1 " Children's Shoes, . .5625 1 " Gents' Slippers, . 2.25 1 " Ladies' Rubbers, . .875 1 ^' Boys' Boots, . . 2.1875 1 " Misses' Rubbers, . .75 Renai rinsf" . 3.83 abstracts of business ' (48) DRY GOODS TKADK. ONE week's sales. Monday, . . Tuesday, . . Wednesday, Thursda}'', . Friday, . . . Saturday,. . Cash, . Credit, Cash, . Credit, Cash, . Credit, \ Gash, . ( Credit, ( Cash, . 1 Credit, { Cash, . ( Credit, .$39.24 . 23.19 . 61.73 . 12.48 . 71.04 . 56.31 . 58.98 . 00 . 49.06 . 87.50 . 65.81 .129.17 49. A house agent rents 7 tenements at $1.12j a week, 5 at §1.25, and 11 at $1.50. What do the rents amount to in a year, or 52 weeks ? $1592.50. 50. A merchant bought 3 barrels of sugar, containing, respec- tively, 236, 249, and 261 pounds, at $.09|^ per pound. What was the amount of the bill ? $70.87. 51. If 5 tons of coal are equal to 9 cords of wood for fuel, and a family bums 31.5 cords of wood in a year, how much will they save by changing from wood to coal, when wood is worth $4.25 per cord, and coal $6.80 per ton ? $14.87^- a year. 52. The rates of telegraphing from New York to Washington are 50 cents for the first 25 words, and 5 cents for each additional word. At these rates, what will be the cost of sending a telegram of 117 words ? $5.10. 53. A man bought 3 80-acre lots, and 1 40-acre lot, of Govern- ment land, at $1.25 an acre. He sold one half the land at 3 times, and the balance at 4 times, its original cost. For how much did he sell the land ? $1,225. 54. A commission-merchant in Dubuque shipped 17 tons of prairie fowls to Philadelphia, where they were sold at $.145 per pound. How much did the shipment amount to, a ton being 2,000 pounds? $4,930. MEASUREMENTS. 103 SECTION VIII. sir^BACBS ^JV2> sozi^s, IJEiniN ITIONS. 165. A Zfine is length or distance. 168. A Siraig?it Jjine is the shortest distance between two points. A- 167. A 'Pe7''pe7idicu2ar is a line which stands upon another without inclining to either side. Thus, the hne GB is perpendicular to the hne AB. 168. An Angle is the difference of direc- tion of two lines that meet in a point ; as, the opening between the lines AB and BG. 160. A ^zg/a Angle is one ^ formed by two lines perpendic- ular to each other. Thus, the angles ABG and DEF are right angles. 170. A Surface y or StiperJlcieSy is a figure that has length and breadth. 171. A liecta7igle is a four- sided figure having only right an- gles. Thus, the surfaces ABGD and EFGH are rectangles. 172. A Square is a figure bound- ed by four equal sides, and having four right angles. 173. A Square IncJi is a square 1 inch long and 1 inch wide. 174. A Square F'oot is a square 1 foot long and 1 foot wide ; a Square ^od is a square 1 rod long and 1 rod wide ; and a Squa?^e J)file is a square 1 mile long and 1 mile wide. B E 1 Square Inch^ 1 Inch long 104 DECIMALS. t y ji w 175. Area is tlie extent of any limited surface. Thus, if a figure extends over a surface of 15 square inches, its area is 15 square inches. 176. A Solid or E D Q P ^ody is a figure that has length, breadth, and thickness. 177. A :Eecia7igti- tar Solid is a body that has six sides or surfaces, each of which is a rectangle. Thus, the solids AJWDEF and MNOPQR are rectangular solids. 178. A Cube is a body bounded by six equal square sides or surfaces. 179. A Czibtc Inch is any body or portion of space 1 inch long, 1 inch v/ide, and 1 inch thick. 180. A Cubic I^oot has six equal surfaces each 1 foot square ; and a Cubic Yard has six equal sur- faces each 1 yard square. Notes.— 1. Length, width, and thickness are called Dimendom. 2. A line has one dimension, length; a surface has two dimensions, length and width ; and a body has three dimensions, length, width, and thickness. 18L Capacity is the extent of any body or any portion of space having length, v^idth, and thickness. Thus, if a body or a portion of space occupies 15 cubic feet, its capac- ity is 15 cubic feet. Note. — Areas and capacities are also called Contents. 182. j^xte7ision is either length, area, or capacity. MEASUREMENTS 105' Measzeres of Extension are of the three kinds named in the three following definitions ; 183. Z/inear JKeasure is the measure of lines. 184. Superficial J^feasure^ or Square J)feasu7^ey is the measure of surface. 185. Solid Measure y or Cubic Measure ^ is the measure of capacity. C-A.SEJ I. Measurement of Surface. 186. Ex. How many square inches in one side of a piece of paper 7 inches long and 5 inches wide? Explanation. — If on the paper you draw lines just 1 inch apart, both lengthwise and crosswise, the surface of the paper will be divided into squares each 1 inch long and 1 inch wide. Since in each of the 5 rows there are 7 square inches, there are in all 5 times 7 square inches, or 35 square inches. We see from the figure that there are 5 rows of 7 square inches each, or 7 rows of 5 square inches each. (See 80, V.) Hence, The number of units in the area of any right-angled surface is equal to the number of units in the product of its two dimeiz- sions. 187. Ex. The contents of a certain field are 1,152 square rods, and the field is 36 rods long. How many rods wide is it? Explanation. — Since the field is 3G rods long, there are 38 square rods in a strip 1 rod wide. And since there are 1,152 square rods in the field, there must be as many strips, each 1 rod wide, as the number of times 3(1 SOLUTION. 1153 [30 72 72 106 DECIMALS. square rods are contained in 1,152 square rods, which is 32. Since each strip is 1 rod wide, the lot must be 32 rods "v\dde. Hence, Either dimension of a right-angled surface is equal to the quotient obtained by dimding the area by the other dimension. 188. Upon the principles deduced in Arts. 186, 187, is based the ^ute for Jifeasitremeni of ^ectajigiilaQ" Surfaces. L To find the area. Multiply the length by the breadth, II. To find either dimension. Divide the area by the other dimension, Pit Oli IjEMS. 1. If upon a blackboard 19 feet long and 5 feet wide, I draw lines 1 foot apart, both lengthwise and crosswise of the board, into how many strips, lengthwise, wi.ll I divide the surface of the board ? How many square feet will there be in 1 strip ? How many square feet on the surface of the board ? 95. 2. How many square rods in a garden 8 rods long and 6 rods wide? 3. My slate is 12 inches long and 8.5 inches wide. How many square inches on one side ? 102. 4. How many yards of carpeting will it take to carpet a room 11.5 yards long and 5.5 yards wide ? 63.25. Note.— Numbers expressing width and length are frequently written with the word " by," or the sign of Multiplication between them. Thus, 7 by 9 inches, or 7 X 9 inches, means 7 inches wide and 9 inches long. 5. Hov/ many square inches in a pane of 9 x 14 window-glass ? 6. My hall is 2.75 by 8.5 yards. How much must I pay for oil- cloth to cover the floor, at $1.12^ per yard ? 7. How many square feet in a city lot 75 x 125.3 feet ? 8. How many square rods in a plat of ground 5.5 rods square ? 9. The area of a door-way 31 inches wide is 2,604 square inches. Wliat is its height ? SJi, inches. MEASUREMENTS. 107 10. A farm in the form of a rectangle is 80 rods wide, and con- tains 19,200 square rods. What is its length ? ^J^O rods. 11. The ceiling of a room is 18 feet long, and its contents are 288 square feet. How wide is it ? 16 feet. 12. A carpenter put 3,375 feet of inch boards into the floor of a church 45 feet wide. What was the length of the church ? 75 feet. 13. I have 16,875 strawberry plants, in 75 equal rows. Hovr many plants in each row ? 225. C^SE II. Measurement of Capacity. 189. Ex. How many cubic feet in a block of stone 5 feet long, 4 feet wide, and 3 feet thick ? Explanation. — If 5 blocks, eacli containing 1 cubic foot, be placed side by side, they will form a row 5 feet long, 1 foot wide, and 1 foot thick. ] If 4 such rows be placed side by side, they will form a layer 5 feet long, 4 feet wide, and 1 foot thick. If 3 such layers be placed one exactly upon the other, they will form a pile 5 feet long, 4 feet wide, and 3 feet thick. Hence, 4 times 5 cubic feet = 20 cubic feet, the number in 1 layer ; and 3 times 20 cubic feet = 60 cubic feet, the number in the pile or in the block. There are as many cubic feet in one row of these blocks as the pile is feet SOLtlTIOJf. 5 cubic feet. J. 2 cubic feet. 6 cubic feet. 108 D E C I M AL S. long, as many rows of blocks in one layer as the pile is feet wide, and as many lay- ers in the pile as the j)ile is feet high. The number of units in the capacity of any right-angled body or portion of ^ace is equal to the ^lumber of units in the product of its three dimen- sions. 190. Ex. A paper-box maker made a package of 432 boxes, putting 8 boxes in each row, and 6 rows in each layer or tier. How many boxes high was the package ? Explanation. — Since in 1 row there were 8 boxes, in the 6 rows of 1 layer or tier there were 6 6 x 8 = JfS times 8 boxes, or 48 boxes. Since in the whole package there were 432 boxes, and in 1 layer 48 boxes, there were in the pile as many layers as the number of times 48 is contained in 432, which is 9. As there were 9 layers, each one box high, the package was 9 boxes high. Hence, Any one of the three dimensions of a right-angled body or portion of space is equal to the quotient obtained by dividing the capacity by the product of the other two dimensions. 191. Upon the principles in Arts. 189, 190, is based the ^ule for Jifeasurement of (Rectangular Solids » I. To find the capacity. Multiply the length, width, and thickness together. n. To find any one of the three dimensions. Divide the capacity by the product of the other two dimen- sions. SOLUTION. J^32 Jf.32 9 ME ASUEEMENTS. 109 14. A pile of bricks consists of 7 layers, and each layer contains 8 rows of 9 bricks each. How many bricks in the pile ? 50^. 15. How many blocks, each measuring 1 cubic inch, can you put into a box 7x6x4 inches inside ? IG. An embankment 12 feet high and 4 feet thick contains 6,000 cubic feet. How long is it ? 125 feet. 17. A pile of wood 8 feet long, 4 feet wide, and 4 feet high, con- tains 1 cord. How many cubic feet in a cord ? 18. How many cubic feet in a stick of timber 85 feet long, 3 feet wide, and 1.5 feet thick ? 105. 19. The contents of a pile of vv^ood 4 feet wide and 5 feet high are 1,280 cubic feet. "What is its length ? GJi.feet. 20. How many cubic yards of earth will be removed in digging a cellar 27 feet long, 24 feet wide, and 7 feet deep, there being 27 cubic feet in a cubic yard ? 1G8. 21. What is the capacity of a space 22.5 feet long, 6.4 feet wide, and 3.25 feet deejj ? J!i,68 cvMcfeet. 22. A music dealer found that a packing box that would hold a melodeon which was 18.5 inches wide and 28 25 inches high, must have a capacity of 21323.1 cubic inches. Allowing the lum- ber to be 1 inch thick, what were the outside dimensions of the box ? 20.5 ly 30.25 hj ^2.8 inches. 23. In a granary is a bin 7 x 8 x 2.25 feet. What is its capacity ? 24. The walls of a stone building 45 feet long and 24 feet wide, are 36 feet high and 1 foot thick. How many cubic feet of ma- sonry in the walls, no allowance being made for openings ? Jf,82Jf. 25. How many cubic inches in a cubic foot 1 26. How many cubic yards in a cubic mow of hay which meas- ures 1 rod, or 5.5 yards, in each of its three dimensions ? 166.375. 27. Two of the dimensions of a stone column which contains 196.8 cubic feet, are 12.3 feet and 4 feet. What is the shape of one end of the columu ? It w square. 110 DECIMALS. SECTION IX. 1. A contractor employed 37 laborers 56 days, 13 laborers 84 days, 13 laborers 43 days, and 17 laborers 21 days. What was the total amount of their wages, at $.87i per day ? $3532.37 f 2. A grocer buys 4 barrels of kerosene, each containing 31.5 gallons, for $55.12^, and he wishes to sell it at a profit of $.18|- per gallon. At what price per gallon must he sell it ? 3. At $7.87| per bushel, how many bushels of grass seed can be bought for $66.93f ? ^-^. 4. A fish dealer has 45 barrels of mackerel, which he wishes to repack into kits holding 12.5 and 25 pounds each, and to use an equal number of each size. How many kits must he use ? 480. 5. If 67.5 bushels of oats are required to feed one horse through the winter, how many horses can be wintered on 950 bushels ? 14j with 5 JnisheU left. 6. If in 1 hour 1,354 gallons of water run into, and 1010.8 gal- lons run out of, a reservoir which will hold 23,381 gallons, and the reservoir now contains 12999.2 gallons of water, in how many hours will it be full? 30.25. 7. The terms of a weekly newspaper are, to single subscribers, $1.50 ; clubs of three, $3.75 ; clubs of five, $5.00 ; clubs of ten, $8.75. The paper has 694 single subscribers, 63 clubs of three, 47 clubs of five ; 34 clubs of ten, and a free exchange list of 50 cojjies. What is the total circulation, and what are the receipts from sub- scriptions? Circulation, 1,508 ; receipts, $1809.75. 8. A ton of iron ore from Iron Mountain yields .50 of a ton of pure iron. How much iron will 736.72 tons of ore yield ? 9. A drover bought 69 beeves at $28.75 a head, and sold 42 of them at $36.5 a head, and the rest at $37.75 a head. How much did he gain by the transaction ? $568.5. 10. A and B start from the same place at the same time, and travel in opposite directions, A traveling at the rate of 23.16 miles per day, and B at the rate of 19.21 miles per day. How far wall they be apart at the end of 17.4 days ? 737.238 miles. REVIEW PROBLEMS. Ill 11. In a scliool-room are 8 rows of double desks, and 7 desks in a row. How many pupils can be seated at the desks in the school- room ? 112. 13. A copper-plate engraver bought a plate of copper 16.35 by 25.3 inches, @ $.03 jDer square inch. How much did it cost him ? 13. I bought a boat load of wood for $183, and by retailing it at $4.50 a cord, I gained $70. How many cords of wood were there in the load ? 56. 14. I borrowed some money, and paid $41.79 for the use of it, paying $.105 for the use of each dollar borrowed. How much money did I borrow ? $398. 15. A government township of land is 6 miles square. How many square miles does it contain ? IG. My garden, which is square, is inclosed by a tight board fence 8.5 feet high, and the fence contains 2,805 feet of lumber. What is the length of one side of my garden ? 82.5 feet. 17. One year a farmer's account of grain sold was as follows : SJ'/f^u^/fe^ e>a/j @ / . (f^S m '^ 'tfui^j^ '' /^/c^ 18. How many square feet in the four walls of a room 18 feet long, 13.25 feet wide, and 9.5 feet high ? 19. A publisher's expenses in publishing an edition of 3,000 copies of a certain book are, for stereotyping, $515 ; paper, $365 ; binding, $370 ; engraving, $80. What must be the retail price of the book, that the author may receive a copyright of 8 cents per copy, the publisher's profit be 30.5 cents, and the retail bookseller's profit 30 cents per copy ? $1.25. 30. Find the amount of six thousand one hundred nineteen mill- ionths, four hundred eight and twenty-six thousandths, two million twenty thousand two hundred and seven hundred three ten-thou- sandths, and thirty thousand sixty-five hundred-millionths. Two million twenty thousand six hundred eight and ten million two hundred seventy-one thousand nine hundred sixty-Jive hundred-millionths. 112 DECIMALS. 21. A steamship made a voyage in eiglit days, sailing the follow- ing distances : 217 miles, 198.85 miles, 246.7856 miles, 208 miles, 227.6987 miles, 200.045 miles, 241 miles, and 205.08675 miles. What was the length of the voyage ? 174446G05 miles. 22. A farmer bought a yoke of oxen, which he fattened, slaugh- tered, and sold. The following is taken from his account of the entire transaction : Disbursemerds. ^a(c//oi /^o/e oa;en '^/y/6>. 00 Sodi( o/ /a'^/encna >//{& ,jame //.SO- - ^/S% SO Receipts. -^ ^{'nc/auui/eiJ/ /S^S, /2^ //S", anc/ 7^6'0 Aounc^^ @,^. /f //Aie '' /S//, /i'^ /SJ, -^ /SP '^ ^' ./Si- S '/( 46 45 Barley, . . . 48 Buckwheat, 40 45 40 50 52 52 46 42 42 52 50 50 48 42 48 46 42 42 Clover seed, 60 60 60 60 60 60 60 64 60 60 60 60 60 Indian corn. 52 56 56 52 56 56 56 56 56 56 56 52 54 56 58 56 56 56 56 56 56 Oats, . . . . 32 28 32 32 35 100 to 3bu. 32 30 30 32 32 35 30 30 30 32 32 34 32 32 36 32 Rye, 54 56 54 56 56 56 32 56 56 56 56 56 56 56 56 56 56 56 56 Timothy s'd 45 45 45 45 45 44 46 Wheat,. . . 60 56 60 60 60 60 60 60 60 60 60 60 60 60,60 60 60 60|60^ 60i60 Table XII.— Troy Weight. 233. The table of Troy weight consists *of the denomina- tions pounds, ounces, pennyweights, and grains. These denominations are used in weighing the precious metals and jewels, and in philosophical experiments. 24 gr. (grains) are 1 pwt. (pennyweight.) 20 pwt. " 1 oz. 12 oz. " lib. 1 lb. is 12 oz. 1 oz. " 20 pwt. 1 pwt. " 24 gr. Scale.— Ascending, 24, 20, 12; descending, 12, 20, 24. Note. — Physicians in prescribing, and apothecaries in mixing, medicines that are dry, divide the Troy pound according to the following table of apothecaries' weight. 20 gr. (grains) are 1 sc. or ^ (scruple.) 3 3 " 1 dr. or 3 (dram.) 8 3 "1 oz. or 3 (ounce.) 12 I " 1 lb (pound.) Dry medicines are sold by avoirdupois weight. 134 COMPOUND NUMBERS. EXEMCISES. 28. Read 5 T. 16 cwt. ; 33 T. 1 cwt. 54 lb. 7 oz. ; 2 T. 375.25 lb. 29. Read 4 lb. 7 oz. 10 pwt. 20 gr. 30. Write fifty tons two hundred seven and five tenths pounds. 31. Write 6 pounds 4 ounces 19 pennyweights 12 grains. JPJB OBIjEMS. 95. 7 T. 15 cwt. 45 lb. 9 oz. are how many ounces ? SJiS,729. 96. Change 1,999 oz. to hundred-weight. 1 cwt. 2^ lb. 15 oz. 97. How many grains are 1 lb. 9 oz. Troy weight ? 10, 080. 98. Reduce 5,190 grains to ounces. 10 oz. 16 pwt. 6 gr. 99. One day, 9 T. 56 lb. of Oswego corn-starch were packed in pound papers. How many papers were put up ? 100. A jeweler made 456 finger rings, each containing 4.25 pwt. of gold. How much gold did he use ? 101. Reduce .6 gr. to the decimal of an ounce. 102. 11 oz. 11 pwt. 11 gr. are how many grains? 103. One year the Lebanon Shakers put up 2 T. 16 cwt. 95.75 lb. of garden seeds in papers, each containing .25 of a pound. How many papers of seeds of this weight did they put up ? 104. .00021 T. is what decimal of a pound ? .J^2 lb.=6.72 oz. 105. One spring a Vermont farmer made 161,268 oz. of maple sugar. How many tons did he make ? 5 T. 79.25 lb. 106. Reduce .003125 lb. Troy to the decimal of a grain. 107. What is the length of a roll of gold wire that weighs 2 lb. 9 oz., if it weighs 1 gr. to the foot ? 3 mi. 108. How many pounds of bluing will a manufacturer use in putting up 845,000 1-ounce boxes ? 52812.5 lb. = 26 T. 812 lb. 8 oz. 109. A wholesale dealer bought 2 T. 8 cwt. of carpet tacks in 8-oz. papers. How many papers of tacks did he buy ? 9,600. 110. .375 lb. = how many pennyweights? 111. 1,250 flat-irons, weighing 5 lb. each, weigh how many tons ? 3.125 T. :=3 T. 250 lb. NOTATION AND REDUCTIONS. 135 Table XIII,— Time. 234. Time is a lim- ited portion of duration. The table of time con- sists of the denomina- tions centuries, years, months, weeks, days, hours, minutes, and sec- onds. These denomina- tions are used in ex- pressing portions of time or duration of different lengths. The day and the year are the natural divisions of time, the other denominations, except centuries, being parts of one or the other of these. 60 sec. (seconds) are 1 min. (minute.) 60 min. are 1 h. (hour.) 1 century is 100 yr. 24 h. " 1 da. (day.) I'-p-year 'f^-y^f.' 7 da. " 1 wk. (week.) 53 wk. 1 da., ? a 1 common yr. or 365 da. \ (year.) lco..on,r. 'f^-y^t: 53 wk. 3 da. ) „ . , „^ ^^ or 366 da. J 1 leap-yr. Ida. " 24 h. Ih. " 60 min. 100 yr. " 1 century. 1 min. " 60 sec. Scale.— Ascending, 60, 60, 34, 365 or 366, 100 ; descending, 100, 366 or 365, 34, 60, 60.— The 7 is ( Dmitted from the scale. The length of a solar year is 365 da. 5 h. 48 min. 48 sec. The following years are leap-years, of 366 days each : I. Every centennial year that is exactly divisible by JfiO ; as 400, 800, 1600 ; 2000, 2400, 2800. And, n. Every year not a centennial year that is divisible by Jf. ; as 1804, 1808, 1812 ; 1876, 1880, 1892. For explanation of the Calendar, Bee Manual. 136 COMPOUND NUMBERS, The year is divided into 12 calendar months, and these are divided into 4 seasons. SEASONS. JIONTUS. ACBKEVIATIOXS. DATS. Winter. I 1st I 2d mo. January, Jan. 31 a February, Feb. 28 or 29 ( 3d ] 4th ^ 5th a March, Mar. 31 Spring. u April, Apr. 30 a May, 31 ( 6th u June, 30 Summer. ] 7th u July, 31 ' 8th u August, Aug. 31 C 9th a September, Sept. 30 Autumn. jlOth u October, Oct. 31 'nth a Kovember, Nov. 30 Winter. 12th u December, Dec. 31 Notes.— 1. February has 28 days in a common year, and 29 in a leap-year. 2. lu most business transactions, 30 days are regarded as a montli. EXEH CISES. 32. Read 5 yr. 4 mo. 15 da. ; 3 "wk. 5 da. 10 h. 45 min. 30 sec. 33. Eead 12 yr. 134 da. 17.35 h. 34. Write 4 years 9 months 3 days 30 minutes 15 seconds. PJt OB JO J3MS. 112. How many minutes in the three spring months of a common year? 132,480. 113. IIow many hours from Independence day at noon, to Christmas day at noon ? 4j ^'^^' 114. Reduce 573,596 min. to higher denominations. 56 tch 6 da. 7 7i. 5G min. 115. How many seconds in a solar year? 116. How long a time will it take a clock that ticks once every second, to tick one million times ? 11 da. 13 h. 46 min. 40 sec. 117. My watch ticks 4 times each second. How many times wiU it tick in a leap-year ? 118. 3 wk. 6 da. 23 h. 30 min. 45 sec. are how many seconds ? NOTATION AND REDUCTIONS. 137 Table XIV,— Circular and Angular Pleasure, 235. A Circle is a round, plane surface. See Manual. 236. A Circumference is the bounding line of a circle. At 237. An Arc is any part of the circumference of a circle. 238. A diameter is the dis- tance across a circle through its center, di" 239. A ^adiics is the distance from the center to the circumference of a circle. Note. — The radius of a circle is always equal to one half its diameter. 240. If the surface about a point in a plane be divided into 360 equal parts or spaces, by lines drawn from the point, 360 equal angles will be formed, and any one of these angles will be a degree. And since an angle is the differ- ence of dkection of two lines, or the opening between two lines that meet in a point (see 168), it follows that A degree is one of the 360 equal angles which will just fill the space about a common point in a plane. 241. The lines which form the sides of these angles may be of any length ; and if about their common point of meeting as a center, a circumference be drawn, cutting all these lines, it will be divided into 360 equal parts, and one of these parts will be the measure of a degree, or of the angle at the center of the circle. Hence, The Meastire of an cingte at the center of a circle is that part of the circumference included between the sides of the angle. 242. If the circumference of any circle be divided into 360 equal parts, each of these parts is also called a degree. Note.— Since circles may be great or small, the degrees in their circum- ferences will he correspondingly great or small. An angle of 1 degree is constant ; while the measure of the angle, or 1 degree in a circumference, varies with every change in the dimensions of the circle. 138 COMPOUND NUMBERS. 243. The table of cir- cular and angular meas- ure consists of the denominations circles, degrees, minutes, and seconds. These denomi- nations are used By Surveyors^ in de- termining the directions or bearings of land boundaries and other lines ; By Navigators, in de- termining the position of vessels at sea ; and By Geographers and Astronomers, in determining latitude, longitude, and the position and motion of the heavenly bodies ; and in computing difference of time. 60" (seconds) are 1' (minute.) 1 C. is 360° 60' " 1° (degree.) 1° " 60' 360° " IC. (circumference.) 1' " 60" Scale. — Ascending, 60, 60, 360 ; descending, 360, 00, 60. Notes.— 1. A right angle or a quadrant is an angle of 90°, and is always included between two lines perpendicular to each other ; its measure is one fourth of a circumference. Hence, we say 4 right angles or quadrants are 1 circumference. 2. Navigators call one sixth of a circumference a sextant. Hence, in nav- igation, 60° are 1 sextant ; and 6 sextants are 1 circumference. 3. Astronomers divide the zodiac, or the snn's apparent path in the heav- ens, into 13 equal parts, of 30° each (for the 12 months of the year), which they call signs. Hence, in astronomical calculations, 30° are 1 S. (sign), and 12 S. are 1 great circle of the heavens. 4. A minute of the circumference of the earth is 1 geographic mile, which is 1.15 English miles, or 1 mi. 48 rd. See Manual. EXEJRCISES, 36. Read 10° 40' 35"; 8 S. 25° ; 72° C 23.75". 37. Write 19 degrees 53 minutes 42 seconds. 38. Write one hundred five degrees twenty-six geographic miles. NOTATION AND REDUCTIONS. 139 PR OBLEMS. 119. Reduce 47° 13' to seconds. 169,980". 120. How many degrees in 59,300''? 16° 28' 20". 131. When a planet has moved 1,426,444" in the heavens, has it described more or less than one complete revolution in its orbit ? 36° Uf' 4" more than 1 revolution. 122. When the sun has seemed to pass over 8 S. 62.25° of the zodiac, how many seconds has he seemed to move ? 123. How many seconds are .015° ? 5^". 124. 3 quadrants 57° 58' are how many minutes ? 125. 67,875" = how many degrees ? 18° 51' 15". 126. Reduce 3' to the decimal of a de^rree. .05°. Table XV. Counting, 244. The table of counting consists of the denominations ones, dozens, gross, and great gross. These denomi- nations are used in counting several class- es of articles for mar- ket purposes. 12 ones or things are 1 doz. 12 doz. " 1 gro. 1 great gro. is 12 gro. 1 gro. "12 doz. 1 doz. " 12 thinsrs. (dozen.) (gross.) 12 gro. " 1 grt. gro. (great gro.) Scale. — Ascending and descending, uniformly 12. Notes.— 1. Six things of a kind arc often called a set ; as a set of chairs, spoons, forks, plates, etc. 2. Twenty things of a kind are sometimes called a score; as a score of times, three score of years, etc. 140 COMPOUND NUMBERS. Table XVI^—JPaper, 2i5» The table of paper consists of the denommations bales, bundles, reams, quires, and sheets. These denomi- nations are used in the paper trade. 24 sheets are 1 quire. 30 quires " 1 rm. (ream.) 2 rm. " 1 bundle. 5 bundles " 1 bale. 1 bale is 5 bundles. 1 bundle " 2 rm. 1 rm. " 20 quires. 1 quire " 24 sheets. Scale. — Ascending, 24, 20, 2, 5 ; descending, 5, 2, 20, 24. Note.— Paper is bought at wholesale by the bale, bundle, and ream ; and at retail by the ream, quire, and sheet. Table XVII —Copying, 246. Lawyers' clerks and copyists in public offices are often paid by the folio for making copies of legal papers, records, and documents. 72 words are 1 folio, or sheet of common law. 90 " " 1 " " " " chancery. Table XVIII.—BooJcs. 247. This table consists of the terms used to indicate the number of leaves of a book made from one sheet of paper. When a sheet is The book is folded into called 2 leaves, a folio, 4 " a quarto or 4to, 8 " -an octavo or 8vo, 12 " a duodecimo or 12mo, 16 " a 16mo, 18 " an 18mo, 24 " a 24mo, 32 " a 32mo, 64 " a 64mo, Note.— A sheet of medium size print paper is 23 X 36, 24 X 37.5, or 25 X 88 inches. These are the sizes commonly used for printed books. And 1 sheet of paper • makes 4 pp. (pages.) 8 " 16 " 24 " 32 " 36 " 48 " 64 " 128 " NOTATION AND REDUCTIONS. 141 i:XEIt CISES. 39. Read 5 grt. gro. 11 gro. 4 doz. ; 6 gro. 5 doz. 8 steel pens. 40. Write five rm. fifteen quires eleven sheets. 41. Write 19 gross 7 dozen ; 8 great gross 7.75 dozen. JPJB OBLEMS. 127. How many gross are 3,156 buttons ? H gro. 11 doz. 8. 128. Hovs^ many crayons are there in 25 boxes, each containing 1 gross? 8,600. . 1 29. How many gross of screws will a joiner use in the 26 work- ing-days of a month, if he uses 56 screws j^er day ? 10 gro. 1 doz. J^. 130. On inventorying his goods, a hardware merchant finds that he has 7 gro. 8.5 doz. wardrobe hooks on hand. What number of hooks has he ? 1, 110. 131. One day a paper dealer sold 6 bales 1 bundle 4 rm. of S" manilla wrapping pajDer. How many sheets of paper did he sell ? 3 / V^ 132. 7,260 sheets of foolscap x)aper are how many reams ? 15 rm. 2.5 quires. 133. How many sheets of print paper in a 12mo book of 456 pp. ? 134. One month a lady copied 648.5 common-law folios for a lawyer, at $.10 per folio. How much did she receive ? 135. A stationer has grt. gro. 11 doz. lead pencils. How many pencils has he ? 136. One winter a wood turner manufactured 56,870 clothes- pins, which he packed in boxes of 1 great gross each. How many boxes had he ? 10 gro. 11 doz. 2 pins more than 32 boxes. 137. A printed case in the Supreme Court (or Chancery) con- tained 456,120 words. How much was the printer's bill, at $.12|- per folio? $633.50. "^ 138. How many reams of print paper will be required to supply 3,250 subscribers with a weekly newspaper one year ? 352 rm. 1 quire 16 sheets. 139. A village grocer shipped 5,160 eggs to the city, in 5 barrels. How many eggs did he pack in a barrel ? 86 doz. p^^ \ \«' 142 COMPOUND NUMBERS. 140. A bookseller's stock of steel pens consists of 7 packages of 1 doz. boxes each, 9 boxes of a broken package, and 5 doz. 8 pens of an opened box. How many pens lias he, each full box contain- ing 1 gross? 13,460. 141. A publisher issued an edition of 5,000 copies of an 18mo book of 316 pp. How much paper did he use, allowing 1 quire to each ream for waste ? 65 rm. 15 quires 19 sheets. Table XIX. Government Standard Units of Measures and Weights, 248i The standard units of the money, measures, and weights now in use, as adopted by the United States Gov- ernment in the year 1834, and from which the other denom- inations in the respective tables are determined, are — TABLES. rNITS. VALUES. United States Money, Dollar, .900 silver, .100 alloy. Lines, Surfaces, and Solids, Yard, 3 feet, or 36 inches. Liquid Measure, Gallon, 231 cubic inches. Dry Measure, Bushel, 2150.43 cubic inches. TroyAVcight, Pound, 5,760 grains. Avoirdupois Weight, Pound, 7,000 Troy grains. See ManuaL Notes.— 1. The yard in use at the Custom-Houses is divided decimally into tenths and hundredths. 2. For ordinary purposes, 2150.4 cu. in. are called a bushel. 3. In the actual standard weights used by the 1 oz. = 480 gr. General Government, the Troy ounce is divided .1 " - 48 ||_ decimally into tenths, hundredths, thousandths, .01 || = 4.8 ^^ and ten-thousandths. The values of these divis- .001 ^'^' = .48 ^^ ions are shown in the margin of this note. .0001" - .048 Table XX.— Comparative Values. 249. I. Of Measures of Capacity. DENOMINATIONS. LIQUID MEASURE. DRY MEASURE. 1 gal., 231 cu. in., 268.8 cu. in. (.5 pk.) 1 qt, 57.75 " " 67.2 " " ll)t., 28.875" " 33.6 " " NOTATION AND REDUCTIONS. 143 n. Of Weights. DKN0MINATI0N8. TROY -WEIGHT. AVOIEDUPOIS WEIGHT. lib., 5,760 gr., 7,000 gr. 1 oz., 480 " 437.5 " Notes.— 1. Multiplying the number of cubic inches 231 268.8 in a liquid gallon by 7, and the number in a dry gallon 7 6 by 6, we find that 7 liquid gallons contain 4,3 cubic _^g_^/^ 1612.8 inches more than 6 dry gallons. Hence, in ordinary computations, it is sufficiently accurate to estimate 7 liquid gal. = 6 dry gal. 3. Multiplying the number of grains in a pound Troy by 175, and the number in a pound avoirdupois by 144, we have 5,760 X 175 = 7,000 x 144 Hence, 175 pounds Troy = 144 pounds avoirdupois. PB^OBZEMS, 143. How many more cubic inches in 568.5 gallons dry measure, than in the same number of gallons liquid measure ? 21489.3. 143. If you dip 33 quarts of water into a tub that will hold 33 quarts of wheat, how much will the tub lack of being full ? The above problem can be solved in at least five different ways. 144. What is the difference in the weight of 43.375 pounds of lead and 43.375 pounds of silver ? 145. 100 pounds avoirdupois are how many pounds Troy ? 146. A brewer has a vat that will hold 5,000 gallons of beer. How many bushels of barley will it hold ? 147. A grocer buys 3 bu. of chestnuts, at $5.00 a bushel, wooden measure, and retails them at $.30 a quart, tin measure. How much does he gain ? $7.40. NoTK 3. — Among many retailers, dry measure is called Wooden Measure, and liquid measure. Tin Measure. 148. An express agent in "Washington collected charges at the Mint, for transporting 563 pounds, commercial weight, of silver from California. How many Mint pounds of silver were trans- ported ? 682 lb. 11 oz. 16 pwt. 16 gr. 144 COMPOUND NUMBERS. 250. The Metric System of Weights and Measures, In the year 1866, the Congress of the United States passed a bill authorizing the use of a new system of weights and measures. In this system the principal de- nomination is the Metre, from which all the other denom- inations in all the tables are derived. Hence, this system is called the Metric System. The principal denomination for the Measure of Surface is the Are ; for the Measure of Capacity, the Litre ; and for Weight, the Gram. see Manual. The lower denominations in each table are tenths, hun- dredths, or thousandths of these ; and their names are formed by prefixing deci, centi, or milli to the name of the principal denomination. The higher denominations are 10, 100, 1,000, or 10,000 times the principal denomination of any table ; and theii* names are formed by prefixing deka, hecto, kilo, or myria to the name of that principal denomination. TABLE OF DENOMINATIONS AND THEIR RELATIVE VALUES. PREFIXES FOR LOWER DENOMINATIONS. Milli (mill-ee) .001 of Centi (sent-ee) .01 of Bed (des-ee) .1 of NAMES OF PRINCIPAL DENOMINATIONS. Metre (mee-ter) Are (are) Litre (li-ter) Gram PREFIXES FOR UIGUER DENOMINATIONS. Deka (dek-a) 10 Hecto (hec-to) 100 Kilo (kill-o) 1,000 Myria (inir-e-a) 10,000 MEASURES or LENGTH. 10 millimetres are 1 centimetre. 10 centimetres " 10 decimetres " 10 metres " 10 dekametres " 10 hectometres " 10 kilometres " 1 decimetre. 1 metre. 1 dekametre. 1 hectometre. 1 kilometre. 1 myriametre. 1 millimetre 1 centimetre 1 decimetre I METRE 1 dekametre 1 hectometre 1 kilometre 1 myriametre .001 metre. .01 metre. .1 metre. 39.37 i7ichcs. 10 metres. 100 metres. 1,000 metres. 10,000 metres. off Tl^ NOTATION AND EE D U C T 10 N S„C*9 . IW ^ifl MEASURES OF SUEFACE. 1 centare is .01 are. I ARC u i 1^^ su. 1 ph. Ij, qt. 2. How much will 25 doz. pocket-knives cost, at £3 2 s. 6 d. a dozen? &78 2s.6d. (3) (4) (5) 22 cd. 7 cd. ft. 12 cu. ft. 2° 43' 19" 5 mi. 37 ch. 56 1. 9 m_ 14^ 6. What is the weight of 6 sets of silver forks, each fork weigh- ing 1 oz. 15 pwt. 12 gr. ? 5 Tb. 3 oz. 18 pwt. 7. A stone-mason contracts to build the cellar walls for 13 dwell- ings. If it takes 7 cd. 5 cd. ft. 4 cu. ft. for each cellar, how much stone will it take for all of them ? 99 cd. 4 cd. ft. ^ cu. ft. 8. If in digging the cellars, 76 cu. yd. 15 cu. ft. of earth be taken from each, how much earth will be taken from all of them ? 995 cu. yd. 6 cu. ft. 9. A butcher slaughtered 18 sheep, and their average weight was 35 lb. 15 oz. What was their total weight ? 10. A train of 63 coal cars was loaded at a coal mine in Penn- sylvania, 3 T. 5 cwt. 2 qr. of coal being put upon each car. How much coal did the train carry ? 11. Multiply 27 mi. 218 rd. 4 yd. 2 ft. 8 in. by 145. 4,014 mi. 58 rd. 4 yd. 2 ft. 8 in. 12. If 5 men can make 38 rd. 5 yd. of post-and-rail-fence in a day, how much fence can they build in 30 days ? 3 mi. 207 rd. 1 yd. 1ft. 6 in. 13. A piece of land near a city was divided into 38 lots, each containing 50 sq. rd. 24 sq. yd. Horw much land was there in tho piece ? 12 A. 10 sq. rd. ^.5 sq. yd. 156 COMPOUND NUMBERS. 14. If a steam-boat runs a mile iii 4 min. 30 sec, in how long a time will it make a trip of 295 miles ? 221i.l min. SO sec. 15. If a housekeeper uses on an average 1 gal. 3 qt. 3 gi. of molasses in a month, how much will she use in a year ? 21 gal. S qt. 16. A farmer drew 45 loads of hay to market, and each load weighed 1 T. 375 lb. How much hay did he draw ? 17. If school 13 in session 5 h. 25 min. each day, how long is it in session during a term of 17 weeks of 5 school days each ? 18. If the rate of speed of a railroad train is 25 mi. 315 rd. an hour, how far will it run in 24 hours ? 19. How much wine in eight casks, if each contains 28 gal. 2 qt. 1.5 pt. ? 20. Charles is 7 yr. 251 da. old, and his grandfather is 9 times as old as he. How old is his grandfather ? SECTION VI. CJ^SE I. 258* The Divisor au Abstract Number. Ex. Divide 46 mi. 126 rd. 1.5 yd. by 9. Explanation. — ^We write solution. the dividend and divisor as -/^g mz. 126 rd. 1.5 yd. [9 in integers, and commenc- 5 mi. ^P rd. 3 yd. Oft. 8 in. ing at the left, we divide tlie units of each denomination of the dividend by the divisor, in order, from the highest to the lowest. The dividend being a concrete number, the quotient must be a concrete number (see 109, IV.) ; the quotient arising from dividing the units of any denomination must be of the same denomination (see 109, V.) ; and any partial remain- der must be of the same denomination — or order of units — as the partial dividend used (see 109, VIII.). One ninth DIVISION. 157 of 46 mi. is 5 mi. witli a remainder of 1 mi. We write the 5 mi. as the miles of the quotient, reduce the 1 mi. remain- der to rods, and to it add the 126 rd., making 446 rd. One ninth of 446 rd. is 49 rd. with a remainder of 5 rd. We write the 49 rd. as the rods of the quotient, reduce the 5 rd. remainder to yards, and to it add the 1.5 yd., making 29 yd. One ninth of 29 yd. is 3 yd. with a remainder of 2 yd. We write the 3 yd. as the yards of the quotient, and reducing the 2 yd. remainder to feet, we have 6 ft. One ninth of 6 ft. is no whole feet ; we therefore write ft. in the quotient, and reducing the 6 feet to inches, we have 72 in. One ninth of 72 in. is 8 in,, which we write as the inches of the quotient. The result, 5 mi. 49 rd. 3 yd. 8 in., is the quotient required. In integers and decimals, the units of each order are divided separately, and the units in any partial remainder are caUed 10 times as many units of the next lower order. In compound numbers the units of each denomination are divided separately, and the units in any partial remain- der are changed to units of the next lower denomination by reduction. That is, I. In division of integers and decimals, any partial remain- der is tens of the order of units next lower than the partial dividend used ; And, II. In division of compound numbers, each unit of any partial remainder is as many times 1 of the next lower denom- ination, as there are ones of the lower denomination in a unit or 1 of the partial dividend used. PMOBLEMS. 1. A silver-ware manufacturer used 5 lb. 6 oz. 12 pwt. of silver in making 9 goblets. How much silver did he use for each ? 2. In settling an estate, a farm of 184 A. 46.25 sq. rd. was divided equally among 15 heirs. How much land did each heir receive ? 12 A. 4S.75 sq. rd. 158 COMPOUND NUMBERS. 3. A farmer cut 11 T. 17 cwt. of hay from 6 acres of meadow. What was the yield per acre ? 1 T. 1,950 IK (4) (5) 56 gal. 2 qt. 1 pt. 1 gi. [ 7 46 mi. 230 rd. 4.5 yd. [ 9 6. If 2,864 cu. yd. 24 cu. ft. of stone are used in making 586 rd. of Macadamized road, how much stone will be used in making 1 rd. ? Jt cu. yd. 21i, cu. ft. C^SE II. 259. The Divisor a Concrete Number. $4 are contained in $12, 3 times ; $.04 in $.12, 3 times ; 4 lb. in 12 lb., 3 times ; 4 oz. in 12 oz., 3 times ; et-c. That is. We can divide dollars by dollars, cents by cents, pounds by pounds, ounces by ounces, etc. But we can not divide dollars by cents, nor pounds by ounces. For $12 -^ $.04 = neither $3 nor $.03 ; so, also, 12 lb. -f- 4 oz.= neither 3 lb. nor 3 oz. Hence, Only concrete numbers of the same denomination can be divided, the one by the other. Ex. 1. How many times are 4 lb. 9 oz. contained in 27 lb. 6 oz. ? Explanation. — Since both divi- solution. dend and divisor are compound ^^ ^^« ^ oz. = 4^8 oz. numbers, we reduce them to simple ^ ^^' ^ ^^' — '^^ ^^' 73 oz, nomination, (ounces), and divide as j^S8 in integers. Ex. 2. Divide 86 yd. 1 ft. by 8 ft. 4 in. Explanation. — Since the lowest „^ , ^^^l'^^^^^ ^^^ . , .,...,., . 86 yd. 1ft. =3.108 m. denomination m either term is g % t ^^^ _ ^ joo in inches, we first reduce both terms to inches, and then divide as in 3,108 in. | 100 in, integers and decimals. 31.08 DIVISION. - 159 Ex. 3. 105 wk. are how many solution. times 3 wk. 4 da. ? 105 lok. = 735 da. Explanation. — After dividing ^ ^^^- -^ ^^- — ^^ ^^• all the units of the dividend, we annex a decimal cipher to the remainder, and continue the division as in decimals. 735 da. I 25 da, 50 -ZTT" 235 225 PROBLEMS, 100 7. I have a measure that is 5 ft. 8 -^^^ in. long. How many times the length of my measure is a pole 10 yd. 7 ft. 4 in. long ? 8. If a man feeds his horse 1 pk. 6 qt. of oats a day, how long will 5 bu. 1 pk. last him ? 12 days. 9. A joiner used 2 gro. 7 doz. 3 screws in hanging and trim- ming the doors of a house, using 1 doz. 5 screws to each door. How many doors were in the house ? 22. 10. At how many loads can a teamster draw 93 cd. 1 cd. ft. 8 cu. ft. of wood, drawing 1 cd. 2 cd. ft. 8 cu. ft. at a load ? 11. How many demijohns, each containing 3 gal. 2 qt. 1 pt,, can be filled from 97 gal. 3 qt. 1 pt. of wine ? 27. 260. Upon the principles deduced in 258, 259, is based the mule for division of Compound JV^umders. I. When the divisor is an abstract number. 1. Divide the units of each denomination separately, and ivrite the several results for the same denominations of the quotient. 2. Reduce each partial remainder to the next lower denomi- nation, and add to it the units of that denomination, for the next partial dividend. 11. When the divisor is a compound or a concrete number. Reduce both dividend and divisor to the lowest denomination contained in either, and divide as in integers and decimals. 160 COMPOUND NUMBERS. PJt OBIjEMS. 13. Divide 71 mi. 237 rd. 3 yd. 1 ft. 6 in. by 9. 7 mi. 310 rd. ^ yd. 2 ft, 13. A wood-chopper cut 68 cd. 3 cd. ft. of wood in 26 days. How much did he cut per day ? 2 cd. 3.5 cd. ft. 14. An importer paid £663 15 s. for 50 gold watches. What did they cost apiece ? £13 5 s. 6 d. . 15. A farmer raised 488 bu. 1 pk, of barley from 14 acres of land. What was the yield per acre ? 34.875 lu.j or 3^ hi. 3 pk. 4 qt. 16. If a vessel sails 250 mi. in 2 da. 4 li. 5 min., what is the average rate of speed ? 1 mi. in 12 min. 30 sec. 17. If a paver in 24 days can put down 76 sq. rd. 15 sq. yd. of pavement, how much can he put down in one day ? 3 sq. rd. 5 sq. yd. 6 sq. ft. 18. If a train of 27 cars carry 57 T. 1 cwt. 2 qr. 24 lb. of ii'on ore, what is the average per car ? 19. A family consumes 7 lb. 11 oz. of meat each week; how long will it take them to consume 192 lb. 3 oz. ? 25 weeks. 20. In 256,728 cu. in. how many gallons liquid measure ? How many quarts dry measure ? 21. If 6 men in 12 days mow 86 A. 64 sq. rd. of grass, how much will 1 man mow in 1 day ? 1 A. 32 sq. rd. 22. If a railroad train runs 144 mi. 291 rd. in 5.5 h., how far does it run per hour ? 26 mi. Ill rd. 1ft. 6 in. 23. How many silver spoons, each weighing 1 oz. 9 pwt., can be made from 12 lb. 18 pwt. of silver ? 24. How long must a field be to contain 14 A., if it is 35 rd. wide? 25. How many bars of railroad iron, each 18 ft. long, will be required for a railroad 521 miles long ? 26. How many rolls of wall-paper, 20 in. wide and 9 yd. long, will be required to paper the walls of a room 14 by 16 ft. and 9 ft. high, no allowance being made for openings in the walls ? 12. REVIEW PROBLEMS. 161 SECTION VII. I. How many yards of carpeting will be required to cover a floor 24 ft. long and 18 ft. wide ? J^S. 3. A note dated June 17, 1865, was made payable^ January 10, 1866. How long had it to run ? f '^' £% * 3. How many bushels of wheat can a farmer store in a molasses hogshead which will hold 138 gallons ? (See 248, Note 3.) IS lu. SpTc, 4. A manufacturer of patent medicine puts 1 3 23 10 gr. of cal- omel into each bottle of medicine. How much calomel does he use for 50 dozen bottles ? llrb.51 4z. 5. If I start at latitude 15° 35' 40" north, and travel due north 2,159 geographic miles, in what latitude will I be ? 61° 34' 40" north. 6. What is the capacity in bushels of a bin 8 ft. long, 7 ft. wide, and ft. deep ? 270. 7. A gardener raised 31 bu. 1 pk. 5 qt. of marrowfat peas for seed, and put them up in papers, each holding .25 pt. How many pa,pers did he put up ? 0. How many pieces of ribbon, each 1 eighth of a yard long, can be cut from 5 pieces, each containing 35 yd. 3 qr. ? 9. An apothecary pays $3.50 per pound avoirdupois for 6 pounds of rhubarb. If he sells it in prescriptions, at the rate of $.30 an ounce, how much does he gain ? 10. What is the weight of 1,250 barrels of flour? 1S2.5 T. = m T. 10 cwt. II. How many pump logs, each 12 ft. long, will it require to bring water to my house from a spring 1.375 miles distant ? 13. Plow many bunches of lath will be required for the walls of a room 18 ft. long, 15 ft. wide, and 13 ft. liigh ? 13. In digging a ditch 130 rd. long and 3 ft. wide, 1,330 cu. yd. of earth were removed. How deep is the ditch ? 6 ft. 14. How many more minutes in the summer months than in the winter months of a common year ? 162 COMPOUND NUMBERS. 15. How many bunches, each containing 500 shingles, -will be required to cover a roof, each side of which is 75 ft. long and 23.5 ft. wide ? J^O.5. 16. How many bricks lying flatwise will be required for a walk 25 rd. 4 ft. long and 5 ft. wide ? 9371.25. 17. What is the difference between a figure which contains .5 of a sq. ft. and one which is .5 of a foot square ? 18. How many times will a wagon wheel 12 ft. 6 in. in circum- ference turn round in going 11 mi. 28 rd. ? 19. How many more farthings in £19 4.5 d. than in £17 19 s. 8.875 fai-. ? 20. A merchant bought 15 pieces of merino, each piece contain- ing 42 yd., at 2 s. per yd. What was the amount of his bill ? 21. A stationer paid 7 s. a gross for 1,080 pen-holders. How much did they cost him ? &2 12 s. Gd. 22. I own a tract of Western land 215 rd. long and 140 rd. wide. How many acres in the tract ? 188.125. 23. When hay sells at $1.37|- per hundred, what is the price per ton? 24. How many cords of stone will be required to lay a wall 218 ft. long, 3 ft. high, and 1.5 ft. thick ? 7 cd. 5 cd.ft. 5 cu.ft. 25. A pile of wood 183 x 6 x 4 ft. contains how many cords? 26. .02875 bu. is what decimal of a quart ? .92 gt. 27. How many bricks will be required for the two side walls of a building 50 ft. long, 20 ft. high, and 1 ft. thick ? 28. What is the capacity of a cistern 14 ft. long, 11 ft. wide, and 7.5 ft. deep ? 137 hU. 9 gal. 29. A pile of wood which contains 12.25 cd., is 56 ft. long and 8 ft. high. What is its width ? 30. A man owns 3.5 A. of land. If he lays it out into village lots, each 5 by 8 rd., how many lots will he have ? 31. What is the entire weight of 15 bar. of pork, 12 bar. of flour, 6 casks of nails, and 8 bar. of K Y. salt ? ^ T. 192 11. 32. How many quart cupfuls, tin measure, in 2 bu. 3 pk. of chestnuts. (See 249, Note 3.) 33. How much will 18 barrels of pork cost, at $.12|^ per pound? REVIEW PROBLEMS. 163 34. A milkman furnishes 420 qt. of milk daily to tlie customers on his route. How many barrels of miilk does he furnish in 2 weeks ? 35. A man bargained for 3 qt. of blackberries daily during the blackberry season, which lasted 21 days. How many berries did he buy ? 36. How many square yards in the four walls of a room which is 40 ft. long, 30 ft. wide, and 20 ft. high ? 37. How many square yards in the ceiling of the same room ? 38. A coal dealer bought 225 tons of coal, at $4.75 per gross ton, and retailed it for $6.50 per net ton. What was his gain ? $569.25, 39. How much land is there in a field that is 11 x 13 ch. ? 40. Reduce 1,000,000 cu. in. to higher denominations. 41. Reduce 1,000,000 sq. in. to higher denominations. 25 sq. rd. 14.75 sq. yd. 5 sq. ft. 64 sq. in., or 25 " 15 " 3 " 28 " 42. A farmer raised 5 acres of potatoes, which yielded 175 bush- els to the acre. He sold them to a grocer for $2.75 a barrel. How much did he receive for them ? 43. How many days were there in the last century ? 44. How many seconds in the circumference of a cart wheel ? 45. A printer used 3 rm. 2 quires 12 sheets of paper for quarter- sheet posters. How many posters did he print ? 6,000. 46. 10,000 silver dollars, of 412.5 gr. each, weigh how many pounds ? 716 lb. 1 oz. 15 pwt. 47. A farmer sold 25 T. 625 lb. of hay for $8.37|- per load of 1,125 lb. How much did he receive ? 48. If you were to count 80 1-dollar bills every minute, for 10 hours a day, how long would you be in counting $1,000,000 ? 49. A fruit-dealer bought 3 bu. 3 pk. (=3.75 bu.) of cranberries, at $6.75 per bushel, and retailed them at $.25 per quart, tin meas- ure. What was his profit ? $9.68%* 50. A produce dealer buys 5,600 bu. of oats in St. Louis, Mo., <^ $.37|^, and sells them in New Haven, Conn., ® $.75. How much are his profits, freights being $.31|- per bu., payable at St. Louis ? $l,JiO0. CHAPTER 4, FACTORS AND MULTIPLES SECTION I. ^BF'IJ^I TIOJV'S, 261. An SJxact divisor of a number is any factor of that number. Thus, 2, 3, 4, and 6 are exact divisors of 12. 262. An ^ven JViimber is one that is exactly divis- ible by 2 ; as 2, 4, 20, 36, 758. 263. An Odd JVumber is one that is not exactly divis- ible by 2 ; as 1, 3, 7, 29, 245. 264. A Composite JVzwiber is one that can be sepa- rated into factors. Thus, 18 is a composite integer, and its factors are 2 and 9, or 3 and 6, or 2, 3, and 3. 265. A !Przme JVtmiber is one that can not be sepa- rated into integral factors ; as, 3, 5, 7, 29, 257. Note. — ^When the factors of a number are prime numbers, tliey are called Prime Factors. Thus, 4 and 6, or 3 and 8, or 2 and 12, are factors of 24; but the prime factors of 24 are 2, 2, 2, and 3. 266. A Common !Divisor of two or more numbers is any factor found in each of them. Thus, 4 is a common divisor of 24, 36, and 48. NoTB. — Two or more numbers are prime to eacli other when they have no common factor. The number 1 is not regarded as a factor. 267o The Greatest Common divisor of two or more numbers is the greatest factor found in each of them. Thus, 12 is the greatest common divisor of 24, 36, and 48. 268. A Mtdtipte or SJxact !Dividend is a num- ber of which a given number is a factor. Thus, 27 is a multiple of 9. CHANGES OF DIVIDEND AND DIVISOR. 165 269. A Common J}fultipte of two or more numbers is a number of wliich each of the given numbers is a factor. Thus, 32 is a common multiple of 4, 8, and 16* 270. The Zeast Comm07i MulHpte of two or more numbers is the least number of which each of the given numbers is a factor. Thus, 30 is the least common multiple of 3, 5, 10, and 15. SECTION n. 271. The value of the quotient depends upon the values of both dividend and divisor* Hence, any change in either of these terms must produce a change in the quotient. C.A.SE I. Changes of Dividend. 272. The quotient of 30 -^ 5 is 6. n we multiply the div- idend by 2, and di- S0_ \5 6 2x30 [5 60_ \5 12=2x6 7x30 {5 210 \5 J^=z7x6 vide the product (60) by 5, the quotient is 12, or 2 times 6. Again, if we multiply the dividend by 7, and divide as before, the quotient is 42, or 7 times 6. Hence, Multiplying the dimdend multiplies the quotient. 3_0_ \3 10 30-^2 \ W {3 5=10- 30^5 [ 3 2=10-^5 273. The quo- tient of 30 -^ 3 is 10 ♦ If we divide the dividend by 2, and then divide the quotient (15) by 3, the result is 5, or 10 -f- 2* Again, if we divide the dividend by 5, and then divide the quotient (6) by 3, the result is 2, or 10 -~ 5. Hence, Dividing the dividend divides the quotient. 166 FACTORS AND MULTIPLES. C^SE II. Changes of Divisor. 120 \ 15 120 [2 X 15 120 [SO Jf=8-^2 m 120 J^xl5 2=8^4 274. The quotient of 120-f-15 is 8. If we multiply the di- visor by 2, and then divide 120 by the product (30), the quotient is 4, or 8 -^ 2. Again, if we multiply the divisor by 4, and divide as before, the quotient is 2, or 8 -h 4. Hence, 3Iultiplying the divisor divides the quotient. 90 lis 90_ \ 18-^2 10=2x5 90 1 18~6 90 [S 30=6x5 275. The quotient of 90-^18 is 5. If we divide the divisor by 2, and then divide 90 by the quotient (9), the result is 10, or 2 times 5, Again, if we divide the divisor by 6, and then divide the dividend (90) as before, the result is 30, or 6 times 5. Hence, Dividing the divisor multiplies the quotient. C^SE III. Like Changes of Dividend and Divisor. S2 {8 1 2x32 [ 2x8 64. [16 5x32 y 5x8 160 \^JfO 276. The quotient of 32 -h 8 is 4. If we multiply both dividend and divi- sor by 2, and divide the new dividend (64) by the new divisor (16) the quotient is 4, the same as before. Again, if we multiply both terms by 5, and then divide as before, the quotient is still 4. Hence, Multiplying both dividend and divisor by the same number does not change the quotient. 277. It has already been shown, in Art 107, that Dividing both dividend and divisor by the same number does not change the quotient. CHANGES OF DIVIDEND AND DIVISOR. 167 278. These three cases establish the General ^ri7iciples of Divisi07i, I. Tae quotient is multiplied by multiplying the dividend or dividing the divisor. n. The quotient is divided by dividing the dividend or mul- tiplying the divisor. III. The quotient is not changed by either multiplying or dividing both dividend and divisor by the same number. Sec Manual. Cancellation. 279. Ex. What is the quotient of 4 x 75 divided by 4 x 3 ? Explanation. — When division is exact, solution. the quotient consists of that factor of the U^'^^ l -^^'^ dividend not common to both dividend and 15 \S divisor. In this example 4 is a factor of both ^ dividend and divisor. And, since the quo- tient is not changed by dividing both dividend and divisor by the same number (278, m.), we divide 4 x 75 and 4 X 3 by 4, — or, which is the same thing, we omit the factor 4 from both terms, — and divide 75, the remaining factor of the dividend, by 3, the remaining factor of the divisor. Cancellation is the process of omitting or striking out equal factors from the dividend and divisor. 280. From the solution and explanation of the last ex- ample, we see that A factor is cancelled by dividing both dividend and divisor by that factor. Ex. 1. Divide 210 by 35. • solution. Explanation.— We can- ^^''^^' ^ l?£ ^'''''''^ eel the common factor, 5, -^ew Dividend, 4^ (^7 New Divisor. by dividing both dividend q Quotient and divisor by 5 ; and then divide the new dividend, 42, by the new divisor, 7. 168 FACTORS AND MULTIPLES. Ex. 2. Divide 21 x 64 by 56. Explanation. — Canceling tlie common fac- ^/°^^"°r^ tor, 7, we have 3 x 64 for a new dividend, " ^^ ^— and 8 for a new divisor. Then, canceling SxGJf. \^8 the common factor, 8, we have 3x8 for a 8x8 {^1 dividend and 1 for a divisor. The product £/ of 3 X 8, or 24, is the required quotient. Note. — From this solution and explanation wc learn that, When either dividend or divisor is canceled, a 1 helongs in its place. Ex. 3. Divide the product of 24, 80, 9, and 12.8, by the product of 3.2, 144, and 16. Explanation. — The fac- ^'^-st solution. tors of the divisor may r q 3 \ be written at the right ^^^ x ^0 x x XU [ U x XM x H of those of the dividend, 5x8x2 = 30 as shown in the First Solution; or under those second solution. of the dividend, as shown r q a in the Second Solution. ^^ x ^0 x x lt$ _ ^ o "When a factor is can- ~~r~ ~r: rr — —ox8x2=80 celed, we draw an ob- ^ lique line across it. ^ :PJtOBJLEMS, 1. The dividend is 714, and the divisor 42. What is the quotient ? 2. What is the quotient of 21 x 13 divided by 7 ? 3. How many times is 11 x 15 contained in 825 ? 4. Divide 28 x 7.2 by 16. 12.6. 5. The factors of a dividend are 8, 25, and .45 ; and of a divisor, 15, 2, and 1.2. What is the quotient ? 2.5. 6. How many tons of hay at $10 a ton, must be given in ex- change for 16 tons of coal at $5 a ton ? 7. In how many days can 45 men do as much work as 63 men can do in 35 days? -4^. 8. If a mechanic can earn $88 in 28 days, how much can he earn in 42 days ? PROPERTIES OF COMPOSITE NUMBERS. 169 9. I rent a store 18 months for $2,400. What is the rent per year? 10. If 15 acres of land produce 420 bushels of wheat, how much wheat will 35 acres produce, at the same rate ? 11. A ship's crew of 39 men have provisions enough to last them 76 days. If the crew is increased to 57 men, how long will the j^rovisions last them ? ^3 days. 12. In building a church, 9 bricklayers laid 407,880 bricks in 55 days. At the same rate, how many bricks can 11 bricklayers lay in 60 days ? J 3. If a telegram of 2,790 words can be transmitted in 45 min- utes, how many words can be telegraphed in 72 minutes ? 4j4^4- SECTION in. 1>1R0^jE^TIBS OJF" C0M^0SIT£J jyvm:sb^s, 281. The right-hand figure of any even number is 0, 2, 4, 6, or 8. Hence, Peopekt^y I. Any number is divisible by 2^ when its right- hand figure is 0, 2, Ji-, Gy or 8. (See 2^2.) 282. Tlie right-hand figure of the product of any even number of times 5 is ; thus 2 x 5=10, 6 x 5 = 30, 14x5 = 90. The right-hand figure of any odd number of times 5 is 5 ; thus, 3 X 5=15, 7 x 5=35 ; 19 x 5 = 95. Hence, Property II. Any number is divisible by o, ivkeii its right- hand figure is or 5. 283. Any number exj)ressed by more than one figin-e may be separated into two parts, one of which is a multiple of some power of 10, and the other is ones. Thus, 56 = 50 + 6, or 5 times 10' and 6 ones ; 256 = 200 + 56, or 2 times 10- and 56 ones ; 3256 = 3000 + 256, or 3 times 10' and 256 ones ; and so on. Hence, Property III. Any number is divisible by any power of 2 or 5, when the number expressed by as many of its right-hand figures as equal the index of the power is divisible by 2 or 5. 170 FACTO lis AND MULTIPLES. 284. The number 54,678, and the sum of the numbers 5, 4, 6, 7, and 8 (the digits of 54,678), are divisible by 3. The number 12,348, and the sum of the numbers 1, 2, 3, 4, and 8, are divisible by 9. Hence, Property IV. Any number is divisible by S or 9, luhen the sum of its digits is divisible by 3 or 9. 285. We can divide 84 by 3, and the result (28) by 7. We can also divide 84 by 21, the product of 3 times 7. Again, we can divide 756 by 4, the result (189) by 3, and this result (63) by 7. We can also divide 756 by all the successive divisors, 4, 3, and 7, and by their product, 84. Hence, Property V. Any number which is divisible by two or more factors successively, is also divisible by each of the factors, and by their product. Note. — A number may be divisible by two or more factors, and not be divisible by their product. Thus, 24 is divisible by 8 and by 12, but not by their product, 96. 286. If we divide any number by one of its prime fac- tors, and divide the result by another prime factor, and so on, until the quotient is 1, we shall use all the prime factors of the number for divisors. Thus, 72 -^ 2 = 36, 36 -=- 2 = 18, 18 -^ 2 = 9, 9 ^ 3 == 3, 3 -^ 3 = 1. The factors used as divisors are 2, 2, 2, 3, 3; and their product, 72, is divisible by the product of any number of these factors. Hence, Property VI. Any number is divisible by the product of any two or more of its prime factors. 287. The factors of 6 (2 and 3) are contained in 12 (=2 X 6), 18 (= 3 X 6), 30 (- 5 X 6), 54, 96, or any num- ber of times 6. Hence, Property VII. Any factor of a composite number is con- tained in any number of times that number. PRIME NUMBERS. 171 288. Since 35 is 5 times 7, and 21 is 3 times 7, 35 + 21, or 56, is 5 times 7 + 3 times 7, or 8 times 7 ; and 35 — 21, or 14, is 5 times 7 — 3 times 7, or 2 times 7. Hence, Property VIII. Any factor common to two numbers is also a factor of their sum, and of their difference. Note. — These properties apply more generally to uumbers in the deci- nml scale ; but to a limited extent to compound numbers also. SECTION IV. 289. All even numbers except 2, and all odd numbers end- ing in 5 except 5, are composite. (See 262, 282.) Hence, The right-hand figure of a prime number is 1, 3^ 7^ or 9. 290. Since 2 is a factor of 4 and of 6, it is a factor of 2 + 6 and of 4 + 6, and of 2 or 4 + any number of times 6 ; and, since 3 is a factor of 6, it is a factor of 3 + G, and of 3 + any number of times 6. (See 287, 288.) All the remainders that can be obtained in dividing num- bers by 6 are 1, 2, 3, 4, and 5 ; and we have just shown that, when the remainder is 2, 3, or 4, the number divided is composite. Hence, When any prime number is divided by 6, the remainder is 1 or 5. PH OBLEMS. 1. Which of the numbers 19, 45, 67, 91 are prime numbers? 2. Of the numbers 103, 126, 131, 217, which are prime numbers ? 3. Which of the numbers 111, 133, 14.7, 149, 219, 3.42, are com- posite numbers ? 4. Which of the seven numbers 293, 371, 385, 440, 524, 617, and 713 are prime, and which composite numbers ? 5. Determine which are prime, and which compos^^^Jf^'-tha numbers 911, 973, 103.3, 10.57, 3373, 3.407, 358.41^- r/'X:^^ 6. Find all the prime numbers less than lOO./f^ j^ 1= oar 172 FACTORS AND MULTIPLES. SECTION V. coMMOJV :d iris O'^s. C^SE I. Prime Factors or Divisors. 291. Ex. 1. \Vliat are the prime factors of 1,260? Explanation. — Since the right-hand figure of solution. 1,260 is 0, we divide by the prime number 2 ^^^^ l^ (see 281); and for the same reason, we divide 630 \2 the quotient (630) by 2. Since the right-hand g^^ i ^ figure of the second quotient (315) is 5, we next ~ . divide by the prime number 5 (see 282). — Since the sum of the digits of the thii'd quo- ^\p tient (63) is divisible by 3, we divide this 7 quotient by 3 (see 284) ; and for the same reason, we divide the fourth quotient (21) by 3. The last quotient (7) is a prime number. The product of the divisors 2, 2, 5, 3, 3, and the last quotient, 7, is 1,260 ; and hence they must be all the prime factors of that number. (See 57.) Ex. 2. What are all the factors or divisors solittiox. of 30? ^l^ Explanation. — ^We first find all the prime 15 \^5 factors of 30 to be 2, 5, and 3, and each of ~~g these is a divisor of 30 (see 285). Since 2, 3, and 5 are prime factors of 30 ; 6, or 2 times 3, 2x3= 6 10, or 2 times 5, and 15, or 3 times 5 are also 2x5=10 divisors of 30 (see 286). Hence, all the fac- <^x^=-^^ tors or divisors of 30 are 2, 3, 5, 6, 10, and 15. 292, Ultele for findi7ig 'Prime I^actors, I. Divide the number by any prime factor. II. Divide the quotient in the same manner ; and so on, till a quotient is obtained that is a prime number. The divisors and the last quotient will be the prime factors required. COMMON DIVISORS. 173 JPJt OBZJEMS. 1. Find tlie prime factors of 540. 2, 2, 5, 3, 3, S, 2. What are the prime factors of 1,650 and 1,755 ? 3. Separate 1,836 into its prime factors. 2, 2, 3, 3, 3, 17. 4. Separate 945 and 3,990 into their prime factors. 6. What are all the factors or divisors of 84 ? Tliere are ten. 6. What are the prime, and what the component factors of 164 ? There are 3 prime^ and 2 component factors. c^se: II. Common Factors or Divisors. 293. Ex. 1. Find a common divisor of soltttion. 15, 25, and 40. Sx^ Explanation. — We separate the given 10=5x2x2x2 numbers 15, 25, and 40, into their prime factors, and find that 5 is a divisor of each number. (See 268.) Ex. 2. Find all the common divi- souttion. sors of 54 and 72. H=2 xSxSxS Explanation. — We separate the 72=2x2x2x3x3 numbers 54 and 72 into their prime ^^^_^ factors, as in Ex. 1, and we find that gy^g—g 2, 3, and 3 are common factors, and 2x3x8=18 are therefore common divisors (see 266). Since 2, 3, and 3 are common factors, 6 or 2 times 3, 9 or 3 times 3, and 18 or 2 times 3 times 3 are also com- mon factors (see 286). Hence, 2, 3, 3, 6, 9, and 18 are all the common divisors of 54 and 72. FMOBLJEMS. 7. What number is a common divisor of 21 and 36 ? 3. 8. Find a common divisor of 4.5 and 10.5. 1.5. 9. What are the prime common factors of 20, 32, 50, and 18 ? 10. Find all the common divisors of 36, 42, and 90. 11. How many common divisors have 64, 112, 48, and 144 ? Four. 174 FACTORS AND MULTIPLES. C.A.SE III. Greatest Common Factor or Divisor. FIRST METHOD. 294. Ex. Wliat number is the greatest common divisor of 252, 210, and 168 ? Explanation. — We separate the solution. nnmbers 252, 210, and 168 into ^52^2x2x3x3x7 their prime factors, as in Case II. %lzlllxlxlx7 (see 293), and we find, that 2, 3, 2x3x7=Jf2 and 7 are the only prime factors common to all the given numbers. Since 2, 3, and 7 are all the common prime factors of the given numbers, their product, 42, is the greatest common factor of the given numbers, and hence is the greatest common divisor re- quired. (See 267.) SECOND METHOD. i. Ex. What is the greatest common divisor of 21 and 77? Explanation. — Since the common di- visor of two numbers can not be greater than the less number, we divide 77, the greater number, by 21, the less, and ob- tain a remainder of 14. If 14 is a divi- sor of 21, it is also a divisor of 77, H which equals 14 + 3 times 21 (see 287). 2 Dividing 21, the first divisor, by 14, the first remainder, we obtain a remainder of 7. a divisor of 14, it is a divisor of 21, which equals 7 + 14, and of 77, which equals 7 + 5 times 14 (see 287). Divid- ing 14, the last divisor, by 7, the last remainder, we find that 7 is a divisor of 14. Hence, 7 is a common divisor of 21 and 77. Having proved that 7 is a common divisor of 21 and 77, we must now prove that it is their greatest common divisor. SOLUTION. 77 21 ^l\7 21 u u ' 1 7 >. NOTN T, if 7 is COMMON DIVISORS. 175 Since any number that is a common divisor of 21 and 77, must be a divisor of 14 (see 288), a number greater than 14 can not be a common divisor of 21 and 77. Again, any number that is a common divisor of 14 and 21, must be a divisor of 7 (see 288). Hence, 7, the greatest common divisor of itself and 14, is the greatest common divisor of 21 and 77. PROBLEMS. 12. What is the greatest common divisor of 56 and 84 ? 28. 13. What is the greatest common divisor of .103 and .153 ? 51. 14. Find the greatest common divisor of 96, 130, and 168. 15. What is the length of the longest line that will exactly meas- ure two fences, one 96 rods and the other 76 rods long ? 296. The explanations and solutions given in 294, 295, are sufficient to estabHsh the following ^ule for fi7idi7iff a Greatest Com?7ion divisor, I. Separate the numbers into their prime factors. II. Multiply together all the factors that are common. Tlie product will he the greatest common divisor. Or, Divide the greater number by the less, the first divisor by the first remainder, the second divisor by the second remainder, and so on, until an exact divisor is obtained. This divisor will be the greatest common divisor. Notes. — ^1. By the Second Method, if more than two numbers are given, we must first find the greatest common divisor of two of them, then of their greatest common divisor and another of the numbers, and so on, till all the given numbers have been used. The last common divisor obtained will be the greatest common divisor of all the given numbers. 2. Only abstract numbers, or like concrete numbers of the same denom- ination, can have a common divisor. 3. The common divisor of two or more concrete numbers may be either an abstract or a concrete number. (See 98, I., II., III.) 176 FACTORS AND MULTIPLES. r It OB I, EMS. 16. What number is a common divisor of $25 and $60 ? 5, or $5. 17. Find a common divisor of 16 A. and 28 sq. rd. 18. "What is the greatest common divisor of 135 and 225 ? 19. How many common divisors have 1 pk. and 6 qt. ? One. 20. The sides of my garden are 168 ft, 280 ft., 182 ft., and 252 ft. What is the greatest length of boards that I can use in fencing it, without cutting any of them ? IJ^feet. 21. If 283.5 yd. Wamsutta, 567 yd. N. Y. Mills, and 445.5 yd. Lawrence Mills sheetings are in whole pieces of the greatest pos- sible equal length, how many yards are there in each piece ? J^OM. SECTION VI. COMMOjT M77ZTITZBS. C^SES I. Common Multiples or Dividends. 291. Ex. What number is a common multiple or divi- dend of 15 and 24? Explanation. — Since a common multiple solution. of 15 and 24 is a number of which both 15 ^^ x 2J^=360 and 24 are factors (see 269), and since any product must be a multiple of any set of factors which will produce it (see 57, 285), we multiply 15 and 24 together. The product, 360, is the common multiple required. VU OBIjEMS. 1. Find a common multiple of 3, 4, and 6. 2. Find a common multi^Dle of 5, 7, 32, and 10. 3. What number is a common multiple of 4.8, 9, and 5.25 ? 226.8 f or any integral nuniber of times 226.8. 4. What number is a common multiple of $15, $2, and $8.50 ? 5. Find a common multiple of 1 bu. 3 pk., 1 jok. 4 qt., and 5 qt. 1 pt. (= 5.5 qt.) 115 hu. 2 pic. COMMON MULTIPLES. 177 CASE II. Least Common Multiples or Dividends. 298. 24 is a common multiple of 4 and 6, because all tlie factors of 4 (2 and 2) and of 6 (2 and 3) are also primo factors of 24 (2, 2, 2 and 3). But 2, 2, and 3 are all tlio prime factors required to produce both 4 and 6 ; and a number that contains only the prime factors 2, 2, 3, will also contain 4 and 6 (see 285). Multiplying these three fac- tors together, we have 2 x 2 x 3 = 12 ; and since 12 con- tains all the prime factors of 4 and 6, and no other factors, it is their least common multiple. Ex. Find the least common multiple of 18, 24, and 30. Explanation. — Since the least common multiple of 18, 24, and solution. 30 must contain only the prime 2/^=3x3x2x3 factors of these numbers (see 30=2x3x5 270), we separate each of the numbers into its prime factors. PHme Factors required. Since 24 has the greatest num- 2x2x2x3x3x5=360 ber of prime factors, we next, for convenience, write all the factors of 24 (2x2x2x3) in a line. Then, comparing the factors of 18 v*^ith these factors, we find that we have all the factors but a 3 ; and we write a 3 with the prime factors required. Again, comparing the factors of 30 with the prime factors required, we find that we have all the factors but a 5 ; and we write a 5 with the prime factors required. We now have all the prime factors of the given numbers, and no others ; and multii)lying them together, we obtain 360, their least common multiple. PItOBJLEMS. 6. What is the least common multiple of 8, 12, and 14 ? 168. 7. "What is the least common multiple of $16 and $20 ? 8. Find the least common multiple of .6 and .8. .2Ji.. 9. What is the least number of which 75, 225, and 500 are factors? 178 FACTORS AND MULTIPLES. 299. (Hides for Finduig Multiples. I. For a Common Multiple. AMtiply the numbers together. The product, or any number of times the product, will he a common multiple. n. For the Least Common Multiple. 1. Separate the numbers into their prime factors. 2. 3Iultiply together all the prime factors of that number having the greatest number of prime factors, and those prime factors of the other - numbers not found in the factors of the number taken. The product will be the least common midtiple. Note. — Only abstract numbers, or like concrete numbers of the same denomination, can have a common multiple. See Manual. PH OB i: JEMS. 10. Find a common multiple of 36, 18, 24, and 12. 11. Find the least common multiple of the numbers given in Problem 10. 12. Find the least common multiple of 2, 8, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72. lU- 13. What number is the least common multiple of 9, 60, 45, 72, 15, 35, 18, 12 ? 14. Find the least common multiple of the nine digits. 2,520. 15. What is the least common multiple of 2 yd. 1 ft., and 2 ft. 8 in. ? 3 rd. 2 yd. 6 in. 16. What is the smallest sum of money for which a person can purchase, either oxen at $85 each, or cows at $35 each ? $595. 17. A can hoe a row of com in a certain field in 30 minutes, B can hoe a row in 20 minutes, and C in 35 minutes. What is the least number of rows that each can hoe, in order that all may finish together ? 18. In a factory are three wheels, which revolve once in 25, 30, and 50 seconds respectively. What is the least time in which all of them will make an exact number of revolutions ? 2 min. 30 sec. SECTION I. 300. The number one half may be obtained by dividing 1 into 2 equal parts ; one third, by dividing 1 into 3 equal parts ; one fourth, by dividing 1 into 4 equal parts ; one fifth, by dividing 1 into 5 equal parts ; and so on. Again, two tliirds may be obtained by dividing 2 into 3 equal parts, or by dividing 1 into three equal parts and taking two of those parts. Two fourths may be obtained by dividing 2 into 4 equal parts ; and three fourths, by dividing 3 into 4 equal parts. Two fifths, three fifths, and four fifths may be obtained by dividing 2, 3, and 4, respectively, into 5 equal parts. Halves, thirds, fourths, and fifths are written thus : 1 half, ^, 1 third, A, 3 thirds, f, 1 fourth, :|, 2 fourths, f, 3 fourths, f, 1 fifth, I, 2 fifths, f , 3 fifths, I, • 4 fifths, f . When 1 is divided into 6 equal parts, the parts are sixths / when into 7 equal parts, they are sevenths; when into 8 equal parts, eighths ; and when into 9 equal parts, ninths. Sixths, sevenths, eighths, and ninths are written thus : 1 sixth, ^, 2 sevenths, f, 1 eighth, ^, 2 ninths, f, 3 sixths, f, 3 sevenths, f, 3 eighths, f, 5 ninths, |, 5 sixths, I ; 6 sevenths, f ; 5 eighths, | ; 6 ninths, f . Numbers which express one of the equal parts of an integer ; as -J^, 4, i, 7^, 3V ; ov which express an equal part of two or more integers, or two or more equal parts of a one ; as, |, f , /y, -^Jq ; form a class of numbers called Frac- tions. Hence, ISO FK ACTIONS. 301. A J^raction is a number which expresses one or more of the equal parts into which a one or any other integer is divided. A fraction is expressed by two numbers, written one under the other, with a horizontal line between them. 302. The 2'erms of a fraction are the two numbers used to express it. Thus, the terms cf f are 5 and 7. 303. The ^e7iominat07* of a fraction is that term which expresses the number of equal parts into which gnd is divided ; ifc is written below the horizontal Hne. And 304. The Mi7neraior is that term which expresses the number of equal x^arts indicated by the fraction ; it is writ- ten above the line. Thus, in the fraction |, the 5 is the denominator, and expresses that a unit or 1 is divided into 5 equal parts, and 4 is the numerator, and expresses that 4 of the equal parts (fifths) are indicated by the fraction. 305. The Reciprocal of a number is the quotient of a one divided by that number. Thus, the reciprocal of 7 is 4, of $13 is %^., of 25 bu. is ^^ bu. 306. A ^racHo7ial Ujiit is one of the units of the numerator. Its value is expressed by the reciprocal of the denominator. Note. — A fractional unit may be either abstract or concrete. Thus, tlic fractional unit of f is \, of $| is %\, of f ft. is \ ft., of \ lb. is \ lb. 307. The value of a fraction depends upon the relative values of its numerator and denominator. I. Wlien the numerator and denominator are equal, the value of the fraction is 1 ; because as many fractional units are expressed as equal an integral unit or 1. Thus, ^, -]f, ||g. n. Wlien the numerator is less than the denominator, the value of the fraction is less than 1 ; because a less number of fractional units is expressed than equal an integral unit nr 1 Tlinc! 3 _7 15 158 m. When the numerator is greater than the denominator, NOTATION AND PRINCIPLES, 181 the value of the fraction is more than 1 ; because a greater number of fractional units is expressed than equal an integral unit or 1. Thus, -^, ^- ^"^ ^-'^■'- 808. A Proper I^r ac- tion is a fraction whose value is less than 1 ; as J, 1 0» 6' 1 G* 309. An Im^) roper ^i^aciion is a fraction whose value equals or ex- ceeds 1 ; as I, jg,i ii, ff. 310. A Mixed JViim- ber is a number expressed by an integer and a deci- mal, or an integer and a fraction ; as 3.7, 21.4, 9.35; 3i, llf , 1415. Note. — In reading a mixed num- ber, and belongs between the in- j— . teger and the fraetion or decimal. □— ' 311. Similar JF^rac- li07is are fractions that have a common fractional unit ; as 1l, % ; % %, %. 312. !2)is similar fractions are fractions that have different frac- tional units ; as f , 4 ; f , f , {7. 313. If we multiply the numerator of \ by 2, we obtain |. The fi-actional unit in \ and § is the same. If we multiply the numeratoi* of \ by 3, we obtain |, — a number of 3 times as many frac- tional units as i, each unit of both fractions being of the same value. That is, ^""^=1, |''- = f, |''^=|, etc. Hence, Multiplying the numerator multiplies the fraction. (See 272.) MPROPER FHRCTIONS 182 FRACTIONS. 314* If we divide the numerator of | by 2, we obtain |, a number of one half as many fractional units as f, each unit of both fractions being of the same value. So, also, if we divide the numerator of f by 3, we obtain f. That is, 2-5-2—1 2-^2—1 6^3—2 p+p HPTIfP B — 8> 4 — 4> "7 — "7» ^^^' -tience. Dividing the numerator divides the fraction. (See 273.) 315. If we multiply the denominator of | by 2, we ob- tain |, a fraction of the same number of fractional units as ^, each unit of the | being one half the value of a unit of the |. So, also, if we multiply the denominator of A by 2, we obtain {. That is, ^xo = |, ix2 = -h |x3 = r5J etc. Hence, Multiplying the denommator divides the fraction. (See 274.) 316. If we divide the denominator of | by 2, we obtain I, the number of fractional units in the | being the same as in the |, while the value of each unit is 2 times as great. So, also, if we divide the denominator of J by 3, we obtain i. That is, 1^2 =-|> 1^2 =i T o-^a = i etc. Hence, Dividing the denominator multiplies the fraction. (See 275.) 317. If we multiply both terms of \ by 2, we obtain §. The number of fractional units in § is 2 times as many as in \y but the value of each unit is only one half as much. In other words, the fraction \ is multipHed by 2 by multiply- ing its numerator by 2, and the result (f ) is divided by 2 by multiplying its denominator by 2. That is, ^x2=4> 1X2 — 2 3X4 — 12 pfp TTA-nr»o 4X2 — 8» "7x4 — 3s> etc. xience, Multiplying both terms of a fraction by the same number does not change its value. (See 276.) 318. If we divide both terms of | by 2, we obtain f . .The number of fractional units in | is one half as many as in |, but the value of each unit is 2 times as much. In other words, the fraction | is divided by 2 by dividing its numerator by 2, and the result (g) is multiplied by 2 by 4-^2 2 2^-2 1 H-^2 — 4' 4+2 — 2 J dividing its denominator by 2. That is, |ii — ^ 2+2 — 1 |, etc. Hence, EEDUCTIONS. 183 Dividing both terms of a fraction bij the same number does not change its value. (See 277.) 319. The deductions in the last six articles are the General Principles of I^ractions, I. A fraction is multiplied by multiplying its numerator or dividing its denominator. n. A fraction is divided by dividing its numerator or mul- tiplying its denominator. III. The value of a fraction is not changed by either mul- tiplying or dividing both terms by the same number, see Manual. 820. In integers, decimals, and compound numbers the successive orders of units increase and decrease by fixed scales. In fractions the scales (that is, the number of frac- tional units required to equal an integral unit) vary with every change of the denominator. This feature of fractions gives rise to the principal difference between computations in fractions and integers, decimals, and compound numbers. SECTION II. C^SE I. Fractions to Lo-west Terms. 321 • A fraction is in its Lowest Terms, when its numera- tor and denominator are prime to each other ; as |, f , |, 23 36' When the terms of a fraction are not prime to each other, they have some common factor. Ex. Reduce ^f to its lowest terms. Explanation. — Since the value of a frac- first solution. tion is n6t changed by dividing both terms ||— /^=| 184 FRACTIONS. by the same number (see 319, III.), we re- duce ^f to lower terms, by dividing its skcond solution. terms by the common factor 2, (If = 13) 5 ij=T and the result, -^%, we reduce to still lower terms, by dividing its terms by the common factor 3, (-»2— I) as shown in the Fu'st Solution. Since the terms 3 and 4 are prime to each other, | must be the lowest terms of the fraction 4f ; and coiisequently | is the result re- quired. Or, we can reduce ^f to its lowest terms at one operation, by dividing both terms by their greatest com- mon divisor, 6, as shown in the Second Solution. phoblems. 1. Reduce the fraction |^f to its lowest terms. s. 2. Reduce ^ and f i to their lowest terms. ^^ 4^ 3. Reduce ^, -j^^, and -^^ to their lowest terms. 4. In what lower terms can the value of ff be expressed ? In four different fractions. 5. Wliat are the lowest terms of ^ and f f ? 6. What are the lowest terms of the fractions ff , f^, -^j^ and CJLSE II. Fractions to Given Denominators. S22, Ex. Reduce | to a fraction having 42 for a denom- inator. Explanation. — Since the value of a fraction solution. is not changed by multiplying both terms by i? L^' the same number (see 319, TIL), we multiply 6 both terms of f by an integer that will give ^xg — sq 42 for a new denominator. We find this in- ^t * .^1_ ,, ^ teger by dividing 42, the required denomi- ' ''^~^^ nator, by 7, the denominator of •]. Then, multiplying both terms of f by 6, the integer thus found, we have |^, the fi'action required. s REDUCTIONS. 185 PJt OBLBMS. 7. Eeduce f to sixteentlis. ^|, 8. Reduce f to tenths, and to twenty-fifths. 9. Reduce |- to a fraction having 63 for a denominator. / '" 10. Reduce -| to 54ths, and ^V to 84ths. A|^ A|.. 11. Reduce f, f, and \ to sixtieths. 12. In ■jfg- are how many twenty-sixths, how many sixty-fifths, and how many ninety-firsts ? j? ±i>. z± C^SEJ III. Dissimilar Fractions to Similar Fractions. • 323. Ex. 1. Eeduce \ and | to similar fractions. Explanation. — Fourths can not be solution, reduced to fifths, nor fifths to fourths. |^^|~/j^ But since 4 times 5 = 20, we reduce \ §-^f=-^-§- to twentieths by multiplying iLs terms tt^„^^ -f a— 5 -i2 by 5 ; and since 5 times 4 = 20, we re- duce I to twentieths by multiplying its terms by 4. Ex. 2. Eeduce f , |, and f to similar fractions. Explanation, — Since the solution, product of the denomina- f^|^^— _'y)^ tors, 3 times 5 times 7, a^|^7— _s_a, = 105, we may reduce f^|^|=:--?-«j- these fractions to 105ths. t-t^^^p 2 a 0-^0 sa 90 inis we do by muitipiymg the terms of the first fraction, §, by 5 and 7 ; the terms of the second, |, by 3 and 7 ; and the terms of the third, f , by 3 and 5. That is, wo multiply the terms of each fraction by the denominators of the other fractions. Erom these examples it will be seen that The denominator of the similar fractions is a common mul- tiple of the denominators of all the given fractions. Notes. — 1. Fractions having like denominators are said to have a Com- mon De7iominator . 2. Reducing dissimilar to similar fractions is sometimes called reducing fractions to equivalent fractions having a common denominator. 186 FRACTIONS. PJtOBLEMS. 13. Reduce ^- and f to similar fractions. s a 14. Reduce f and f to similar fractions. 15. What similar fractions are equal to f and ^ ? IG. What similar fractions are equal to f , f , and | ? 17. Reduce ^ and -^j to similar fractions. or^^ is^ 18. Reduce f , f , f , and f to similar fractions. 19. Reduce -^^ f, f, and -^ to equivalent fractions having a common denominator. st^ko jl.ioo 44 ar gjinA 20. Reduce |, f , and ^ to equivalent fractions having a com- mon denominator. Dissimilar Fractions to Least Similar Fractions. 324. Since, in reducing dissimilar to similar fractions, the common denominator must be a common multiple of the denominators of all the given fractions (see 323), it follows that The common denominator of least similar fractions must be the least common multiple of the denominators of all the given fractions. 325. Ex. Keduce |, |, and ^ to least similar fractions. Explanation. — We first find the least common multiple of all the given denominators 6, 4, 9, to be 36 (see 299). Since 36 is the common denominator of the least similar fractions that are equal to the given fractions |, |, and ^, we reduce each of these fractions to 36ths by Case II. (See 322.) ] 6= Jf^ -.2x2 9= 3x3 3x 3x2x 2=36 36 [6 ^u 36 19 6 9 4 KS =M 4X9 =A AX-4 yxji. =U ce, •hh i=M. /^, U' KEDUCTIONS. 187 Notes.— 1. The fractional units of dissimilar fractions are unlike, but the fractional unit of their equivalent similar fractions is common. (See 311.) 2. The fractional unit of least similar fractions is the greatest fractional unit common to the given dissimilar fractions. See Manual. PJlOBI,E3lS, 21. Reduce -| and -| to least similar fractions, ^|-^ i.|;, 22. Reduce -^ and ^V to least similar fractions. _4_ r 23. What are the least similar fractions equal to | and f ? 24. What least similar fractions are equal to |, ^, y'^, and | ? 25. Reduce -^j, |, and f to least similar fractions. / 26. Reduce ^i, f, I, If, and ^ to least similar fractions. sfi an 55 .9S y.7 X5? :4 3^J J5> 7.3^J J3' 27. AYhat is the fractional unit of the least similar fractions to which f , f , yV? and y^^F ^an be reduced ? _ i^^-. Improper Fractions to Integers or Mixed Numbers. 326. Ex. 1. How many ones in ^^^- ? Explanation. — Since every 7 sev- solution. enths are 1, 21 sevenths are as many ^ sevenths. 1 7 sevenths. I's as the number of times 7 sevenths 3 are contained in 21 sevenths, which is 3 times. Ex. 2. Find the value of the improper fraction -^^-. Explanation. — Since every 5 fifths are soLUTioii. 1, 17 fifths are asmany I's as the number ^yy'i^'- \^5 fifths. of times 5 fifths are contained in 17 fifths. S§- The quotient figure is 3 ; and since the remainder is always like the dividend (see 109, Yin.), and the dividend is fifths, the remainder 2, is 2 fifths or f. Writing the | at the right of the quotient figure 3, we have 3|, the value required. Or, we may regard the numerator and denominator as dividend and divisor, and both concrete numbers (fifths). 188 FRACTIONS. Then, the quotient figure is 3, and the remainder, 2, is a less number to be divided by a greater, 5 ; and the result is | (see 301). Writing the | at the right of the 3, we have the abstract number 3?, as before. (See 109, III.) Note. — Any dividend may be written as tlie numerator, and the divisor as the denominator of a fraction. Fit OBJLEMS. 28. How many apples are ^ apples ? 29. In -^ miles are how many miles ? 30. Reduce the improper fraction -^ to a mixed number. .^|. 31. Reduce ■^- to an integer. Ih 32. How many yards are ^f yd. ? 33. Find the integer or mixed number equal to each of the im- proper fractions %^^, -Vs" ^la., V/, ^¥- ft-, ^fP cu. yd., ^^ lb. 34. 29 quarter-dollars are how many dollars ? 35. Reduce \^-, ^^ ^-^ ^-, and ^-^ to integers or mixed num- '43 bers. 6^- ), 51-, me. ca.se: "vi Integers or Mised Numbers to Improper Fractions. SOLUTION. FULL SOLUTION. 7 seventh^,. 327. Ex. 1. Reduce the integer 8 to fifths. Explanation. — Since 1 is 5 fifths, 8 are 8 times 5 fifths, or 40 fifths. Ex. 2. Reduce the mixed number 4| to an improper fraction. Explanation. — Since 1 is 7 sevenths, 4 are 4 times 7 sev- enths, or 28 sevenths ; and 28 sevenths + 3 sevenths are 31 sevenths. The reduction of 4| to sevenths is similar to the reduc- tion of a compound number of two denominations to the J 28 sevenths. 3 sevenths. 31 sevenths. _8 JfOfftJ^' Hence, 8 = -^/. COMMON SOLUTION. 28 + 3=31 Hence, #=V- REDUCTIONS. 189 lower denomination. Thus, the 4 ones corresponds to the higher denomination, and the 3 sevenths to the lower. In the second or Common Solution we reduce the 4 ones to sevenths and add the given 3 sevenths, in the same manner as we would reduce 4 wk. 3 da. to days. (See 223.) 1*JR OBI.E3IS. 36. Reduce 12 to sevenths, and 13 to ninths. ^^ &A^ ±-t7^, 37. In 5| are how many eighths ? .^?.. 38. Reduce 19f to an improper fraction. 39. Reduce b-^^ and 43^ to improper fractions. |7, ^M^- 40. What improper fractions are equal to 15fV and 17-| ? 41. Change 14||, 13/g^, and llf to improper fractions. 42. In 365^ days there are how many fourths of a day ? ■ia£,± da, 328. Brief directions for performing the processes in the preceding six Cases form the Steles for deductions of J^ractlons, I. Fractions to lowest terms. Cancel all the factors common to both tenns. n. Fractions to given denominators. Divide the given denominator by the denominator of the fraction, and multiply both terms of the /inaction by the quotient. m. Dissimilar to* similar fractions. Multiply both terms of each fraction by the denominators of all the other fractions. IV. Dissimilar to least similar fractions. 1. For the least common denominator, find the least common multiple of all the denominators. 2. For each new numerator, divide the least common multi- ple by the denominator of each fraction, and multiply the nume- rator by the quotient. V. Improper fractions to integers or mixed numbers. Divide the numerator by the denominator. 190 FKACTIONS. VI. Integers or mixed numbers to improper fractions. 1. Mulliply the integer by the denominator, and if there he a numerator, add it to the product. 2. Write this result and the given denominator for the terms of the required fraction. PJtOB ZEMS. 43. To what lower terms can -^^ be reduced ? i-j:^, / 5 2±^ ._7_^ 44. How many one hundred fifty-thirds are equal to eleven seven- teenths ? JL?_ ^53' 45. Reduce y^, ^, and ^ to equivalent fractions having a com- mon denominator. 46. What similar fractions arc equal to |-, I, and -^-^ ? 47. Reduce -^^^ |^f , -5^, and -^ to least similar fractions. 48. Reduce -^, ^f^, i^, ip-, and ^^ to integers or mixed numbers. 49. Reduce 59 to a fraction having 59 for a denominator. Re- duce it to 9ths. -H~i H~' 50. What least similar fractions are equal to \^ f , -^^, -^^ and ^\ ? 51. What is the greatest common fractional unit of H, |-|, -j'^, andfi? 7^^. 53. Find the lowest terms of ^W^, fii, and ^f^. 53. Change -^-^ to ninety-fifths, to one hundred seventy-firsts, and to two hundred ninths. ^|, //^, //g-. 54. Reduce the fractions |-|, f ^ y^? t> ^^^ H *^ eighty-fourths. 55. What are the lowest terms of -j^V?? fli» tHHti and ^^^ ? 56. Wbat similar fractions are equal to -^^ and -^ ? 57. Reduce |-, |, I'V, and ^V to similar fractions. 58. Reduce f , |-, f , |, f , f , and f to fractions having a common fractional unit. 59. What least similar fractions are equal to y^o, I, tV, and ^^ ? 60. Reduce lOOOyoW ^^^ ^^yV to improper fractions. 61. What improper fractions are equal to 67^^ and 133f ? 63. Reduce ^^^, -^^ ■^, and |^ to least similar fractions. . Their greatest common fractional unit is j-^^. ADDITION AND SUBTRACTION. 191 SECTION HI. 329. Since only like orders of units can be added one to another (see 39, II.), or subtracted one from another (see ^%, II.), and dissimilar fractions are of unhke orders of units, it follows that they must be reduced to similar frac- tions, (that is, to the same fractional unit), before they can be added or subtracted. C^SE I. All the Given Numbers Fractions. 330. Ex. 1. What is the sum of |, |, and \ ? Explanation. — The giv- fikst solutioit. en fractions being dissimi- i + § + i^=H + U+U = H=' Hi lar, we first reduce them ' _ SECOND SOLUTIOrr, to the similar fractions ;?. j_ .2 _|_ i — ^3+2^ +^0 _ _?_» — 74.5 |-§, §4, and Jg. Since the parts of these similar fractions are all of the same kind or denomination (sixtieths), and since the numerators express the numbers of the parts, we add the similar fractions by adding their numerators, 45 + 24 + 10 = 79 ; and since the fractional unit of the parts is g^^, we write the denominator, 60, under the 79, making l^. Then, reducing the Jg ^^ ^ mixed number, we have 1J§, the result required. Ex. 2. Subtract | from |. Explanation. — The given fractions ^^^^t soltttion. being dissimilar, we first reduce them '8~§-=io~jo—ii to the similar fractions |4 and M. 4 4 SECOND BOLTTTION. Since all the parts of these similar 7^—3^—2Jf-^sj.~±j. fractions are of the same kind or de- nommation (fortieths), we subtract 24 fortieths from 35 fortieths, and the difference, | J, is the result required. 192 FRACTIONS. In reducing dissimilar to similar fractions, the common denominator need be written but once, and the several numerators may be written above it, connected by the ap- propriate signs, as shown in the Second Solution of each of the two preceding examples. From these examples we learn that I. The numerators of similar fractions only can be added or subtracted; and H, The common denominator is icritten under the sum or difference. I*ROBLEMS.' 1. What is the sum of i and -| ? ±i 2. What is the sum ot 4 and 4 ? 4±, 3. What is the difference between -| and f ? /^ • 4. From -^^ subtract f . |2. 5. William gatliered |- bu. of butternuts one day, and f bu. the next. How many did he gather in the two days ? G. From | yd. of velvet a lady used -|- yd. How much velvet had she left ? 7. A Michigan farmer made ^ T. of maple sugar, and £)old -| T. How much sugar did he keep ? ^^ T. 8. The tide rose | ft. one hour, -^f ft. the next hour, and | ft. the third hour. How much did it rise in the three hours ? S-U. ft. C^SE II. Any of the Given Numbers Mixed Numbers. 331. Ex. 1. What is the sum of 5f, -^, 6^, and 11 ? Explanation. — We wi'ite the given numbers in columns, integers under integers, and frac- ^^°^"°%^ tions under fractions. Eeducing the frac- i~ ij tional parts to similar fractions, we have 5| qi — ^sj. = ^h i = ~4h and 6A ^ 6f i. Adding the i/ =. ii'''' fractions^ we have || or 1||. We write the 5S f ^ in the result, and add the 1 with the given ADDITION AND SUBTKACTION. 193 integers. 23, tlie sum of all the integers, written before the |A, gives 23|A, the required sum. Ex. 2. From 7^ subtract 3§. Explanation. — ^We write the subtrahend un- solutiox. der the minuend, and reduce the fractional ^j = 7';/^ parts to similar fractions. Since |§ can not be "^^ ~ "^sf subtracted from /g, and since the difference o§§- will not be affected by adding the same number to both minuend and subtrahend (see 52, III.), we add || (= 1) to the 3% of the minuend, ^nd 1 (= ||) to the 3 of the subtrahend. We then subtract |g from || {=m^), and 4 from 7, writing the results, 1 1 and 3, as the fractional and integral parts of the remainder. The result, 3§|, is the remainder required. See Manual. When any of the given numbers are mixed numbers, we may Regard the fraction?, as lower, and the integers as higher denominations, and add and subtract as in compound 7iumbers. PROBLEMS. 9. What is the sum of 4| and 3^ ? 8^'^. 10. From 6yV subtract 3f. 4j\. 11. A lady bought 15|- yd. of delaine, llf yd. of calico, and 4f yd. of merino. How many yards of dress goods did she buy ? 13. I bought 13f cd. of wood, and at the end of a year, had 1-jV cd. left. How much had I used ? iii-| al 13. A mechanic spent $9^'o ^^^ ^1^ week's wages, and had $3f left. What was the amount of his wages ? 14. A merchant sold a pair of fur gloves for $3^, upon which his profit was Iff. What was the first cost of the gloves ? 15. My farm consists of five fields that contain respectively 13^^ A., 15f A., 13^^ A., 11^ A., and 14| A. How many acres in my farm ? ^^?^V* 9 194 FRACTIONS- 332. Upon the principles deduced in 330, 331, is based the 'Utile for Addition and Subtraction of J^ractions, I. Reduce dissimilar to similar fractions. n. Add or subtract the numerators, and under the result write the common denominator. Notes. — 1. If the given fractions are reduced to least similar fractions, the numerators to be added or subtracted will be the smallest numbers possible. 2. In all final results reduce fractions to lowest terms, and improper frac- tions to integers or mixed numbers. JPM OB Z JEMS. 16. What is the sum of f and ^ ? as^ 17. From I subtract ^. iz, 18. George paid $|- for a pair of skates, and $^ for straps. What was the whole cost ? 19. From 7^^ subtract 6f . -<|.. 20. The parts are 4f , S^-, 3|, 4^, and l^^. What is their sum ? 21. Mary had $^, but she spent $f for a ribbon. How much money has she left ? ^jg-* 22. What is the sum of |, f , and f ? 23. From f subtract 1^. §f. 24. A lady purchased a shawl for $8f , and gave the merchant a 10-dollar bill. How much change should she receive ? 25. If a family bum f T. of coal in Dec, | T. in Jan., and ^f T. in Feb., how much do they burn in the three months ? 26. How much greater is ^ than -^^ ? ^y. 27. A merchant sold a lace collar for $||-, that had cost him |^. How much was his profit ? ^||. 28. A contractor having a contract to build 2d^ mi. of rail- road, has completed 14yV nai. How much has he yet to build ? , 29. From 32tV subtract ^. ^^iuT* ^^ 30. Add fl, i» 1^, -h^ and ^. 8um, 2^. MULTIPLICATION. 195 31. A stone-mason in building a wall, used f cd. of stone one day, -^^ cd. the second day, -^ cd. the third day, and y*j cd. the fourth day. How much stone did he use in the four days ? 32. A founder used ^ T. of iron in making f T. of castings. How much was the waste ? -^^ T, 33. If the less of two numbers is 7^, and the greater is 37^, what is the difference ? -^^f !• 34. Find the sum of 391f , 19^, 4tV, 57^, and -i^. 473i-§. 35. -^^ is how much greater than -^^ ? 36. What is the distance round a farm |- mi. long and ^^ mi. wide ? ^1^^ mi. 37. What is the sum of f? and ^ ? What is their difference ? Sum, 1/^2 ' differerwe, //g-. 38. The minuend is 11^, and the subtrahend 5^. What is the remainder ? 39. A cake of ice 1|4- ft- thick floats with -^^^^i. of its thickness above the water. What thickness of the ice is under water ? 1-?-^- ft ■^-i OS J'" 40. A farmer sold 13^ T. of his hay crop, put llf T. into his bam, and stacked 9^ T. How much hay did he raise ? SJ/.^^^ T, SECTION IV. MZrZ TITJDIC;^ TZOJV. C^SE I. One Factor a Fraction. 333. Ex. 1. Multiply j% by 5. Explanation. — ^In the First Solu- ^^^ soLtj-now. tion we have multiplied 8, the numer- 7J —TJ~^i^—^§ ator of the fraction, by 5, and in the „„„„„„ „ ^„ ' "^ ' SECOND SOLUTION. Second Solution we have divided _^^^ __|_^|: 15, the denominator, by 5 (see 319, 1.) The results in the two solutions are the same. 196 FRACTIONS. FIKST SOLUTION. tV /:,X5=ff = ^/^: SECOND SOLUTION. n 7 y _.7 — .9 5 - ■■^7%- Ex. 2. Multiply 7 by /j, or find j% of 7. Explanation. — To multiply a num- ber by 5 is to find 5 times the num- ber ; to multiply it by 1 is to find 1 time the number ; to multiply it by j\ is to find ^5 of it ; and to multi- ply it by ^4 is to find 5 times j\ or j\ of it. In the First Solution we divide 7 by 14, and obtain j\, which is j\ of 7 ; and we then multiply this result by 5, and obtain f f , or 2i Since 7 x -f^ = f 5 x 7 (see 80, V.), in the Second Solution we multiply 7 and y\ together, in the manner explained in Ex. 1. 1. Multiply ^^ by 8, or find 8 times ^. 2. How much is 1 3 times ^ ? 3. At $1 a yard, how much will ^ yards of alpaca cost ? 4. Multiply 18 by -^\, or find -^V of 18. Or which is j\ of 7.. ^A- S4i. 5. How much is A of 14 miles ? 6. What is the product of 19 and -^1 Of 31 and || ? Of and 34? J,-f,-, 19^, mh 7. How much will 12 gal. of kerosene cost, at $|f per gal. ? 8. A farmer bought f bu. of grass seed @ it cost him ? How much did Cj^SE II. Both Factors Fractions. 331. Ex. 1. Multiply | by |, or find I of §. Explanation. — | of any number is 3 times as much as | of it, and | of it is found by dividing it by 4. In the First Solution we multiply the denominator of | by 4 to find \ of I (see 319, n.). We then multiply the numerator of the result, 5^, by 3, to find 3 FIKST SOLUTION ■t SECOND SOLUTION. MULTIPLICATION. 197 times ^, or |, of |. This result, if, = |, the -^^^^^^ sm^ution. result required. -i^^T—if In the First Solution we multiply the denominators 6 and 4 together for the denominator, 24, of the product ; and the numerators 5 and 3 together for the numerator, 15, of the product. The Second Solution shows the work in the usual form. Since the given numerators are factors of the numerator of the product, and the given denominators are factors of its denominator, we may cancel like factors from the nume- rators and denominators of the given fractions (see 328, I.). The product will then be in its lowest terms, as shown in the Third Solution. Ex. 2. Multiply 51 by 3|. solution. Explanation.— We first re- 5f x 3^=-^f x V-= W-=^-^ll duce the mixed numbers to improper fractions, and then multiply as in Ex. 1. Ex. 3. What is the product of | x 4| x 8 ? Explanation. — We reduce the mixed number 4^ to an solution. improper fraction, the integer | x^x ^=;^x|^x f =^^ 8 to the form of a fraction by writing 1 for its denominator, and then multiply as in Ex. 1. Notes. — ^1. The word of between fractions signifies multiplication. Thus, f of -,%=fXi% or -,%Xf ; f of ll=|Xll or llXf. 2. When a fraction is connected to any other number by of the expres- sion is commonly called a Compound Fraction ; as | of 4- of 2%, f of 13|, I off of 18. PROBLEMS. 9. Multiply f by ^, and f by if. /^, ^f . 10. How much is y^ of -f f of a mile ? ' f f mi. 11. Multiply 7f by 4|. 3^. 12. Multiply 4^ by Q-^-^, ^ by 3^-^, and 18^ by 9. 13. The factors are |f and ^. What is the product ? 14. What is the product of | of f of | ? I. 198 FEACTIONS. 335. From 333 and 334 we deduce the 2iule for Multiplication of ^Fractions, I. Reduce mixed numbers to improper fractions, and integers to the form of fractions. II. Multiply all the numerators together for the numerator, and all the denominators for the denominator, of the product. m OBJLBMS. 15. A fruit dealer put up 30 baskets of peaches, puttmg f of a bushel in each basket. How many peaches did he put in all the baskets ? 28^ lu. IG. If a man earns $78 in a month, how much will he earn in |- of a month ? S58^. 17. A man who owned f of a vessel, sold |- of his share. What part of the vessel did he sell ? -^^. 18. How much will |- of a yard of linen cost, at $|^ a yard ? 19. Multiply Hi by f . ^|-|. 20. How much is 8 times ^ ? 7^i-. 21. John's kite string is 118 yards long, and Frank's is ^ as long. What is the length of Frank's kite string ? 22. What is the product of 43 multiplied by -^V ? 16^^. 23. How many days' work can 54 men do in | of a day ? 24. Wliat is the product of f x -|- x f ; or what is the cube of f ? 25. How much will 4f bu. of sweet potatoes cost, at %1^ a bushel ? 26. How many square rods are there in a lot 15f rd. long and 12f rd. wide ? 198^. 27. What will be the cost of ff A. of land, at $156 an acre ? 28. What is the product of -j^ of f of ^ ? ^. 29. If it takes 1| bu. of wheat to seed 1 acre, how many bushels will it take to seed 17f acres ? 33^, 80. If in talking, a man speaks 75 words in a minute, how many words will he utter in ^^ of a minute ? 31. Raise ^ to the fourth power, f to the sixth power, f to the fifth power, and square ff . DIVISION. 199 32. f of ^ of f of a ream of paper is what part of a ream ? 33. A railroad train ran at the rate of 32 miles an hour, for 5-^ hours. How far did it run ? 116§^ Tni. 34. If it takes a man f of a day to mow an acre of grass, how long will it take him to mow -^^ of an acre ? 35. Cube 6^, and square 16f. ^Uwt^ ^^^«- 36. What is the product of f of f of | multiplied by f of f of | ? SECTION V. S)irisTOJ\r. FIEST SOLUTION. -3 — 2 — 7"* SECOND SOLUTION. •2i —7' TUIED SOLUTION. C^SE! I. The Divisor an Integer. 836. Ex. Divide f by 3. Explanation. — In the First Solution we divide the numerator, 6, of the divi- dend by the divisor, 3 (see 319, II.); and in the Second Solution we multiply the denominator, 7, of the dividend by the divisor. The result in each solution is f , the required quotient. Since to di- vide a number by 3 is to find I of it (see 334), in the Third Solution we find tiplication of fractions (see 335), and the result is f , the same as before. ritOBJOEMS. 1. Divide I by 5, and ^ by 12. /^^ /j. 3. What is the quotient of 15 divided by ^ (=^^ ? 3. If a family consume 5 bar. of flour in a year, in what time will they consume 1 bar. ? 4. A dealer in real estate sold |f of an acre of land in 6 equal building lots. How much land did each lot contain ? ^^ A. 5. If a carpenter can build 13^ rd. of picket fence in 3 days, how many rods can he build in 1 day ? ^f . 4 of # as in mul- 200 FRACTIONS. FIRST SOLTJTION. G3 18 C^SJS II. The Divisor a Fraction. FIRST METHOD. 337. Ex. 1. How many times is | contained in 7 ? Explanation. — Since the quo- tient is not changed by multiply- ing both dividend and divisor by the same number (see 276), wo multiply them both by 9, and thus obtain 63 for a new dividend and 8 for a new divisor. Then, 63 -^ 8 = -^g^- = 7|, the required quo- tient. In the First Solution the numbers are written as in divis- ion of integers and decimals ; but the common manner of writing the numbers is shown in the Second Solution. Ex. 2. Divide J by |. Explanation. — We first multiply both dividend and divisor by 4, the denominator of the divisor, and then divide the new dividend, -^g^, by the new divisor, 3, as in Case I. Hence, To divide hy a fraction consists of two operations, — a multiplication hy the denominator, and a division hy the nu- merator. "^ SECOND METHOD. 338. We have seen (975 ^^ ^^^ ^^^ ) ^^^ when the divi- sor is a concrete number, the dividend must also be a concrete number. We have also seen (303) that the denominator of a fraction gives denomination or name to the fractional units* We may therefore regard any numerator as one or more concrete units. Hence, When the divisor is a fraction, the dividend and divisor- should he reduced to similar fractions^ before dividing. SECOND SOLUTION. 7x9=63, and fx 9=8 63^8=-%^-=7^ Hence, 7 -^§=7-^-. FIRST SOLUTION. SECOND SOLUTION. ■h\=H DIVISION. 201 ■ 339. Ex. What is the quotient of J divided by | ? Explanation. — In the First Solution we reduce both divi- first solution. dend and divisor to similar frac- §-^§-= H -^ || ='H= l-fn tions (twenty-fourths), and then divide 21 twenty-fourths by 16 second solution. twenty-fourths, in the same man- ^-^§-=^ x §-=f-^= 1/^ ner as we divide 21 by 16. The result, 1/5, is the quotient required. If we change the places of the terms of the divisor, and multiply the dividend, J, by |, the fraction thus formed, we shall multiply the same numbers together as in the Eirst Solution. This is shown in the Second Solution. That is. To divide by a fraction,is the same as to change the places of the terms of the divisor, and multiply the dividend by the frac- tion thus formed. Note 1.— When the places of the terms of a fraction are changed, as 7-, J, the fraction is said to he inverted. PJt OBTjEMS. 6. Divide 6 by f . 10^, 7. What is the quotient of | divided by f ? l^-^. 8. How many times is 2^^ contained in 3|^ ? (^i^=^^ and 2|-=f .) 9. At $1 per cwt., how many hundred -weight of feed can be bought for $13 ? 10. How many quarts of chestnuts can be bought for $f , at %-^-^ per quart ? 11. If a man can plow ^ A. of fallow in a day, how long will it take him to plow 5|^| A. ? 13. What is the quotient of 1 divided by f ; or, what is the reciprocal of the fraction f ? |. Notes.— 3, From this problem we see that the reciprocal of a fraction is 1 divided by the fraction ; or, it is the fraction inverted. 3. Division of fractions is sometimes expressed, by writing the dividend above, and the divisor below a horizontal line. Thus, -Jj-itjstjI — 5T» Such expresssions are often called G&mplex Fractions. 9* 202 FRACTIONS. 310. The processes developed in 337, 338, 339, are all in- cluded in the following ^ute for Diyisio7i of Fractions, I. Reduce mixed numbers to improper fractions, and integers to the form of fractions. n. Multiply the dividend by the reciprocal of the divisor. PROBLEMS. 13. Divide ^ by 10, and || by 16. /^, /^. 14. How many yards of gingham © |^, can be bought for $4 ? 15. What is the quotient of 26 divided by H ? 30. 16. If 13 tea-spoons weigh -^ of a pound, how much does each spoon weigh ? 17. Divide I by {-|, and -^ by ^. ^4, 12. 18. At %l a pound, how much tea can be bought for $fi- ? ±f lb. 19. Divide 13f by 25, and 16|4- by 9. -<|, iff. 20. A locomotive ran 22i- miles in 35 minutes. What was the rate per minute ? /^ 'mi. 21. Divide 7| by f , and l^V by -^. 22. What is the cost of a pair of skates, if ^ of their cost is $^ ? 23. How many times is 17 contained in 234| ? 13 j-. 13 455. -• 24. -^j and ^ equal what numbers ? ^0^ lo. 25. If tV bu. of salt can be made from 48 gal. of salt water, how much salt can be made from 1 gal. ? 26. What is the quotient of f divided by 3f ? §. 27. If yV bu. of mortar cover 1 sq. yd. of wall, how many square yards will 5^ bu. cover ? 63. 28. Divide llf by 3f , and 16^ by 6^. 29. If 12^ lb. of rice cost $1^\, how much will 1 lb. cost ? Sf 30. How many gallons of oysters, at $lf a gallon, can be bought for $n^\ ? 31. If fl oz. of gold be obtained from 18 cwt. of gold quartz, what is the yield from 1 cwt. ? /t ^' REVIEW PKOBLEMS. 203 33. A lawyer's clerk wrote 36 pages in 6f hours. How much did he write in 1 hour ? 5f pages. 33. I bought 14f qt. of yinegar for $ff . What was the price per quart ? 34. At $5^ a bushel, how much clover seed can be bought for $f ? 35. If 8f qt. of strawberries can be bought for ff|, what is the price per quart ? Sj^o- 36. If 1 rod of fence require 74|- ft. of boards, how many rods will require 1811^?^ ft. ? 37. ?^5!! = what number ^ /^. f 01 8f ^^ 38. A plank 18| ft. long and ^ ft. thick, contains 2f cu. ft. What is its width ? ^ft. SECTION VI. OiBTIBW T'ROSI.BMS IJV IPHdoLC TIOJ\rS , 1. If a ship sails 1 mi. in -^^ h., how far will she sail in 14 h. ? 3. Add 3f t't, '^^h and 6f. 28-j^y. 3. $900 is y\ of what I paid for my house and lot. How much did they cost me ? $8,815. 4. What is the difference between \ and \ ? 5. A miller paid $3,156^^ for 1,540|- bu. of wheat. What was the price per bushel ? $1^^ 6. A regiment, when it was mustered out of service, consisted of 305 men, which was -^ of the original number. How many men belonged to the regiment at first ? 1,087. 7. Add^,3:V,and|. 8. A man having a lot containing f| A. of^fcmd, sold from it ■^ A. to one man, and ^ A. to another. How much land had he left? -^A 9. How long must I rent a house at $23|- a month, to cancel a debt of $433 ? 18 j- mo. 10. If 1^ rm. of letter paper cost $f , what is the price per ream ? 204 . FRA^CTIONS. 11. A jeweler melted together ^ oz. of gold, | oz. of silver, and ^ oz. of copper. How much did the mixture weigh ? 12. From 11-^ subtract lOyV fff. 13. -i\ of I of f of 15f = what number ? 5_^^. 14. A man bought a cow, paying $20i- down, which was -^ of the cost. How much did the cow cost ? ^22-^. 15. What is the sum of 3|-|, |, 2^, and If ? P|. IG. Divide -rV of 2| by -| of 8f . /_.. 17. A farmer has 4f mi. of rail fence on his farm, f^ mi. of stone fence, ^^^j^ mi. of board fence, and ^ mi. of picket fence. How many miles of fence has he on his farm ? 18. The greater of two fractions is f and the less is ||. "What is the difference ? 19. At $l/o- ^ hundred-weight, how much will it cost to trans- port 15 hundred-weight from Buffalo to Boston ? 20. The minuend is 1;^, and the subtrahend is |-|. What is the remainder ? of ^tt. 21. I sold a quantity of wool for |536f, which W9S 1^ times its cost. How much did it cost me ? $296§: 22. How much is -^V of /^ of \l of 3^ x 4f ? ^ ^.f. 23. Bell-metal is composed of f copper and ^ tin. How much of each of these metals is there in a church bell that weighs 1^ T. ? ■ Copper, -//^ T. ; Tin, //^ T. 24. Multiply T^ by ^ ; -^f by if ; ^ by ii ; and ^ by iff. 25. How much will 19^ bu. of apples cost, at the rate of %A^ for llf bu. ? ^7|. 26. How much will 35 men earn in 19|^ days, at $ly3_ a day ? 27. How many loads of sand at $f a load, will pay for 290| yards of plastering at %\ a yard ? QS. 28. How many yards of cloth \ yd. wide, will line 23^ yd., \\ yd. wide ? S3^. 29. A seamstress bought a sewing-machine for $56.50, paying $25 down. How much must she save from her earnings each month, to pay for it in 6 months ? , SECTION I. cojYyb^sb otb^atiojvs jjv- tub ^ibbb^- BJ\rT CZclSSBS OIP JVZrM2^B^S. 34 It Addition, subtraction, mnltiplication, and division are often called the Fundamental Rules of Arithmetic. 342. Addition is putting together, and subtraction is taking away, or taking apart ; multiplication is repeated addition, and division repeated subtraction of the same number. Hence, addition and multiplication are the reverse of subtraction and division. 343. Co7iverse Operations are those arithmetical pro- cesses which are the reverse of each other. C^SE I. Converse Operations in the Fundamental Rules. 344. The sum of the parts 73 and 48 is 121 ; 73 + 48^:121. This sum minus either part -j i21—JiB=73 i ^^^^^ *^® other part. Hence, Addition and subtraction are converse operations. 345. The product of the factors 57 and 26 is 1482 ; 57 x 26=1482. This product divided by either factor | ;^i^|lf ^^57 1 equals the other factor. Hence, Multiplication and division are converse operations. 346. From 344, 345, we learn that I. Either part is the difference between the sum and the other part. 206 CONVERSE OPERATIONS. II. TJie miniLend is the sum of the subtrahend and re- mainder. m. Either factor is the quotient of the product divided by the other factor. rV". ^e dividend is the product of the divisor and quotient. Note 1. — ^Addition may be proved by subtraction, and subtraction by ad- dition. So also multiplication may be proved by division, and division by multiplication. PBOBJLJEMS. 1. The sum of two parts is 319.5, and one of them is 96.875. What is the other ? 122.625. 2. The subtrahend is 27f , and the remainder 163^. What is the minuend ? -HH- 3. What number must I add to 4 rd. 7 ft., that the sum may be Imi.? 4. The sum of three parts is 298, and tvp-o of the parts are 47.5 and 5.95. What is the other part ? Note 2. — ^Any one of the parts is the difference between the sum and the Bum of the other parts. 2j^J^.55. 5. The sum of three parts is 43f , and two of them are 17f and f . What is the other part ? 6. The divisor is .25, and the quotient .344. What is the divi- dend? 7. The j)roduct of three numbers is 3402, and two of them are 9 and 27. What is the other number ? Note 3. — Any factor is the quotient of the product divided by the pro- duct of the other factors. 8. The product of three factors is 19 J, and two of them are If and 2|. What is the other ? 5±. 9. The sum of two numbers is 1,765, and their difference is 235. What is the greater number ? Note 4.— The sum of two numbers plus their difference equals two times the greater number. See Manual. 1000. 10. The sum of two numbers is 71|^, and their difference is 16^. What are the numbers ? ^^f , 27^. DIFFERENT CLASSES OF NUMBERS. 207 C^SE II. Multiplication and Division by Factors of Composite Numbers. 347. Ex, 1. Multiply 67 by 48. solution. Explanation.— Since 48 = 6 times 8, 48 -^^—^ ^ ^ times 67 = 6 times 8 times 67, which is ^ 3216. Ex. 2. Divide 3216 by 48. "^^f S216 Explanation. — Sin^e 48 is 6 times 8, 4^^ of any number is J of ^ of the number. We find J of i of 3216 by dividing first by 8 solxition. and then by 6. Hence, S216 ^8 ^ule for Multiplying or Dividi7ig by - — ^ - a Composite JVumber. 67 Multiply or divide successively by any set of factors of (he number. PJtOJiljJEMS, 11. Multiply 293 by 34. 13. How many square rods are there in a field 41.25 rd. long and 35' rd. wide ? • 144S.75. 13. How much will 4.5 bu. of wheat cost, at $1.93f a bushel ? (45=9 X. 5.) $8.71875. 14. Divide 3134 by 73. 15. A peat company sold 54 tons of peat for $203.50. What was the price per ton ? $3.75. 16. A farmer sowed 38 bu. 3 pk. of barley on 38 A. of land. How much did he sow to the acre ? llu. 1 pTc. 4 qt. 17. A man cleared 13f A. of woodland, cutting 49 cords of wood to the acre. How many cords did he cut*? 18. If 6.4 tons of porcelain clay cost $113, what is the cost of .81 of a ton ? $14.17^. 208 CONVERSE OPERATIONS. C^SE III. Multiplication and Division by Aliquot Parts. 348. An oHiquot ^ari of a number is any one of its exact divisors. Thus, 5 is \ of 10, 4 in. are \ ft., 6 h. are \ da., etc. The aliquot parts of any number may be found by divid- ing that number successively by 2, 3, 4, 5, 6, etc. 349. The Unit of a7i Miquot ^art is that number which is divided to obtain the part. 350* TABLE OF ALIQUOT PAETS. 100 ITon ift., 1 lb. Aliquot Parts of 1 10 or $1.00 1000 of 2000 lb. or Idoz. of 16 oz. lyd. 1 A. One half is * 5 50 500 1000 6 8 OZ. 1 ft. 6 in. 80 sq. rd. One third is \ 3* 334- 333i 666f 4 One fourth is \ 2* 35 250 500 3 4 oz. 9 in. 40 sq. rd. One fifth is ^ 2 20 200 400 32 " One sixth is \ If 16f 1661 333^- 2 One eighth is i n IH 125 250 2 oz. 4|-in. 20 sq. rd. One tenth is TO 1 10 100 200 16 " One twelfth is etc. ^ 8| 83^ 1 60LUTI0X. 93700 {6 156166§- 351. Ex. 1. Multiply 937 by 166|. Explanation. — Since 166 1 is ^ of 1000, 166 f times any number is J of 1000 times that number. We therefore multiply 937 by 1000, and divide the product, 937000, by 6. Ex. 2. What will 40 sq. rd. of land cost, at $275 per acre ? Explanation. — Since $275 is the price of 1 acre, 40 sq. rd. or | A. will cost \ of $275. We therefore divide $275 by 4. Ex. 3. Divide 2775 by 33i. Explanation. — Since 33 1 is | of 100, 33 J is contained in any number 3 times as many times as 100 is contained in that number. We therefore divide 2775 by 100, and multiply the quotient, 27.75, by 3. S275 [4 SOLUTION. 27.75 ^ 83.25 DIFFERENT CLASSES OF NUMBEKS. 209 SOLUTION, Ex 4. If 4 eggs cost $.11, what is the price per dozen ? $.11 Explanation. — 4 eggs are | of a dozen, ^ and 1 dozen eggs cost 3 times as much as $.S S \ dozen. We therefore multiply $.11, the price of 3 dozen, by 3. 352. These illustrations are sufficient to establish the following ^ules for Mtcliiplying a7id ^ivldinff by A.liquot ^arts, I. The multipher an aliquot part. 1. When the unit of the aliquot part is any power of 10 : — Multiply by the unit, and divide the product by the num- ber of aliquot parts in the unit. 2. When the unit of the ahquot part is 1 : — Divide by the number of aliquot parts in the unit. n. The divisor an aliquot part. 1. When the unit of the aliquot part is any power of 10 : — Divide by the unit, and multiply the quotient by the num- ber of aliquot parts in the unit. 2. When the unit of the aliquot part is 1 : — Multiply by the number of aliquot parts in the unit. PJtOB LJEMS. 19. Multiply 364 by li 21. How much will 12|^ bu. of millet cost, at $3.42 a bushel ? 23. What is the product of 333|- times 198 ? 25. How much will 83^ A. of land cost, at $92 an acre ? 27. Multiply 7.14 by l^f. 20. Divide 455 by If 22. If 12| bu. of millet cost $42.75, what is the price per bushel ? 24. What is the quotient of 66000 divided by 333^ ? 26. If 83^ A. of land cost $7666- .66 1. what is the price per acre ? 28. Divide 119 by 16f. 29. How many bushels of potatoes, at $.33^ a bushel, can be bought for $19.50 ? 58.5. 210 CONVERSE OPERATIONS. 30. At $.06|- a dozen, how much will 144 dozen clothes-pins cost? 31. At $.25 a yard, how much will 37.75 yards of shirting come 32. What is the cost of 376 bushels of com, at $1.12J per bushel ? 33. How much will 625 bushels of potatoes come to, at $.75 a bushel? ($l-$i=$f=$.75.) 34. How much will 250 lb. of iron cost, at $65 a ton ? $8J2-t. 35. If it costs $483 to build 66f rd. of Macadamized road, how much will it cost to build 83|- rd. ? $603.75. 36. A gardener raised 23 bu. of strawberries from a piece of land 8 rd. long and 4 rd. wide. "What was the yield per acre ? 115 du. SECTION II. Decimals to Fractions, and Fractions to Decimals. 353. All decimals may be written in two forms, the deci- mal and the fractional. Thus 7 tenths is .7 or /^ ; 59 thou- sandths is .059 or j j|o ^ ^ ten-thousandths is .0003 or TuSuTJj ®^^' -^^ the decimal form the denomination or unit is indicated by the position of the decimal point, and in the fractional form it is expressed by the denominator. Ex. 1. Express .075 in the fractional form. Explanation. — ^We write the number solution. without the decimal point, and express .075 = j^§iy = /g- its denomination or unit by the known denominator, 1000. BOLTITION. Ex. 2. Reduce .0081 to the .008^=j^f^ = g^f^^ = ^g^ fractional form. CONVERSE REDUCTIONS. 211 7 8 Explanation. — Since | expresses tlie quotient solution. of 7 divided by 8, we annex decimal ciphers to *- 7, and divide by 8, as in division of decimals. •'^^^ (See 152.) 354. From these explanations we deduce the following 'Rules fo7^ the Co?ire7^se deductions of decimals and J^ractions, I. A decimal to a fraction. Write the given number of decimal units, omit the decimal point, and express the denomination or fractional unit by a denominator. II. A fraction to a decimal. Annex a decimal cipher or ciphers to the numerator, and divide by the denominator. PMOBLEMS. 1. Reduce .375 to a fraction. 3. What fraction equals .16f ? 5. Eeduce 6.75 to a mixed fractional number. 7. .00004 of a mile = what frac- tional part of a mile ? 3. Reduce f to a decimal. 4. What decimal equals |^ ? 6. Reduce 6f to a mixed deci- mal number. 8. 2&l(io of ^ ii^il® = what deci- mal part of a mile ? 9. Reduce -^^ T. to the decimal of a ton. .01875 T. 10. What fractional part of a day = .2f da, ? t^ ^' 11. Reduce .06875 to the fractional form. 13. Reduce 74^^ ^^ ^ mixed decimal number. 7.075. 13. What fractional part of a cord equals .85 cd. ? 14. Reduce -^^ to a decimal. .SjJ- or .S8^j or .38J^^. Note. — Sometimes the decimal is interminable. In such cases a fraction may be written after the decimal figures ; thus, .3^, .73| ; or the quotient may be carried to any desired number of decimal places, and the sign -f- placed after it to show that the division is incomplete, or that there was a remainder after the last decimal figure of the quotient was obtained. Thus, | = .666+; f = .428571+. SeeManuaL 212 CONVERSE OPERATIONS. C^SE II. Denominate Decimals to Compound Numbers, and Com- pound Numbers to Denominate Decimals. 355. Ex. 1. Reduce .75 rd. to a compound number. Explanation. — ^We reduce the bolittiox. .75 rd. to yards by multiplying .75^ ret. by 5.5 ; the decimal part of this — '— result, .125 yd., to feet by multi- ^^'^^ plying by 3 ; and this result, .375 ft., to inches by multiplying '^- ^ ^ ^ 2/^' by 12, as in reduction of com- SI 5 ft pound numbers (see %%^). The * -, J ' 4 rd. and 4.5 in. taken together . form the required compound '^* number, 4 rd. 4.5 in. . H^^^^' '^^ ^^- = ^ V^' ^'^ ^*^- Ex. 2. Reduce 2 pk. 3 qt. 1 pt. to the decimal of a bushel. Explanation. — We write o«t,.^x^^ BOLTJTION. the denominations in order 1.0 pt. [2 in a column, with the lowest ^ ^ i g at the top. We reduce the ^ . 1 pt. to the decimal of a — ^— ^ quart by dividing by 2, as .609376 bu. in division of decimals, and Hence, SpTc.Sqt. lpt.=. 609375 hi. annex the result to the quarts, making 3.5 qt. We reduce the 3.5 qt. to the deci- mal of a peck by dividing by 8, and annex the result to the pecks, making 2.4375 pk. We then reduce this result to the decimal of a bushel by dividing by 4, as in reduction of compound numbers. (See 225-) Ex. 3. Reduce 4 yd. 4.5 in. to /^X^ri'^ the decimal of a rod. 037Sft~f3 Explanation. — Since there are ' ^ - Oft. in the compound number, J^.r25yd. H 8 5 we write a cipher in the place of 5.5 .75 rd. feet in the column, and then 27 o 275 proceed as in Ex. 2. CONVERSE REDUCTIONS. 213 356. From these explanations we deduce the following ^utes for t?ie Co7iverse deductions of denominate Decimals and Com^pound JVumbers, I. A denominate decimal to a compound number. 1. Multiply the decimal by the number which it takes of the next lower denomination to equal one of the given denomination. 2. Treat the decimal part of the product thus obtained in the same manner, and also the decimal part of each succeeding product, until there is no decimal in it, or until the lowest de- nomination is reached. 3. Write the integral parts of the several results and the final result in order, for the required compound number. II. A compound number to a denominate decimal. 1. Wi^ite the denominations of the compound number in a column, with the lowest at the top. 2. Divide the lowest denomination by the number which it takes of that denomination to equal one of the next higher, and annex the result to the given number of the next higher denomi- nation. 3. Treat the result thus obtained, and each succeeding result, in the same manner, until the whole has been reduced to the required denornination. mOJBZEMS. 15. In .8 lb. Troy there are how many ounces and pennyweights ? 17. Keduce .21675 of a ton to a compound number. 19. Reduce .26 of a bushel to a compound number, 21. How many days and hours in .75 of the year 1875 ? 22. What part of a diurnal revolution does the earth make in 15 h. 50 nun. 24 sec. ? .66. 16. 9 oz. 12 pwt. are what part of a pound Troy ? 18. Reduce 4 cwt. 33 lb. 8 oz. to the decimal of a ton. 20. Reduce 1 pk. .64 pt. to the decimal of a bushel. 214 CONVERSE OPERATIONS. 23. How much wheat must be sowed upon .85 of an acre, at the rate of 1 bushel to the acre ? S ph. 3 qt. ^pt. 24. What part of a rod = 3 yd. 2 ft. 3 in. ? .5. 25. Reduce 4 da. 4 h. 48 min. to the decimal of a week. 26. Reduce .45 ofa cord to a compound number. 8cd.ft.9.6cu.ft. CJ^SE III. Denominate Fractions to Compound Numbers, and Com- pound Numbers to Denominate Fractions. 357» Ex. 1. Reduce y\ sq. mi. to a compound number. Explanation. — BOLimoN. We reduce the t^t" «?• mi. x 6 40 = ^f f^ =232^8jA, j\ sq. mi. to acres, /r^- x ^^0 = ^ffn. =:116/^ sq. rd. by multiplying by tt «?• rd.x30^ = -fi- x ^^ = 1 1 sq. yd. 640 ; the fraction- Hence, /^ sq. mi. = 232 A. 116 sq. rd. 11 sq. yd. al part of this re- sult, /j A., to square rods, by multiplying by 160 ; and the fractional part of this result, j\ sq. rd., to square yards, by multiplying by 30| (= 30.25) ; as in reduction of compound numbers (see 225, I.). The 232 A., 116 sq. rd., and 11 sq. yd., taken together, form the required compound number. Ei. 2. Reduce 22 h. 13 min. 20 sec. to the fraction of a day. Explanation. — solution. We reduce the ^^ «^^- -^^0 = §4 = ^ min. 20 sec. to the ^^'^i^^- + i min. = 13^ min. = -^/- min. fraction of a min- '^- '^'^^^' ^ ^ ^ = jVk = f ^• ute, by dividing ^^ ^' + I ^^- =22§h.=^^ h. by 60, and annex H^h.~U = m = If da. or add the result Hence, 22 h. 13 min. 20 sec. = §fda. to the minutes, making 13 J min. We next reduce the 13 J min., = y^- min., to the fraction of an hour, by dividing by 60, and add the result to the hours, making 22 1 h. We then reduce this result, 22 1 h. = ^|^ h., to the fraction of a day, by dividing CONVERSE REDUCTIONS. 215 by 24, as in reduction of compound numbers (see 225 ^ 11.). The final result, -^f da., is the denominate fraction required. 358. From these examples we deduce the following ^utes for the Converse deductions of Denominate Infractions and Compound JVumbers, I. A denominate fraction to a compound number. 1. Multiply the fraction by the number which it takes of the next lower denomination to equal one of the given denomination. 2. Treat the fractional part of the product thus obtained in the same manner, and also the fractional part of each succeed- ing product, until there is no fraction in it, or until the lowest denomination is reached. 3. Write the integral parts of the several results and the final result in order, for the required compound number. n. A compound number to a denominate fraction. 1. Divide the lowest denomination by the number which it takes of that denomination to equal one of the next higher, ex- press the result in a fraction, and annex it to 'the given number of the next higher denomination. 2. Ti^eat the result thus obtained, and each succeeding result, in the same manner, until the whole has been reduced to the re- quired denomination. 27. In %r^ how many cents and mills ? 29. Reduce i^ of a ream to a compound number. 31. Reduce£|-| to a compound number. 33. Reduce f of a square mile to a compound number. PMOBIjEMS. 28. In 31 cents 2.5 mills how many dollars ? 30. Reduce 10 quires 16 sheets to the fraction of a ream. 82. Reduce 10 s. 7 d. 2 far. to the fraction of a pound. 34. Reduce 426 A. 106 sq. rd. 20 sq. yd. 1 sq. ft. 72 sq. in. to the fraction of a square mile. 216 CONVERSE OPERATIONS. 35. What part of a bushel is 3 pk. |- pt. ? ^^, 36. A tobacco grower had \\ of an acre of tobacco, which yielded at the rate of a ton to the acre. How much tobacco w^as in the crop ? 18 cwt. 3S-^ lb., or 1833 l lb. 37. If 11 silver forks weigh 1 pound of silver, how much will 1 set weigh ? 6 oz. 10 pwt. 21J>j gr. 38. What part of a hogshead is 60 gal. 2 gi. ? 39. What part of a bissextile year is 219 da. 14 h. 24 min. ? |-. 40. How many powders of 12 grains each will ^ ounce of qui- nine make ? SECTION III. thicb, qu^^jv ti tt, ajv2> cost, 359. In all transactions of purchase and sale, and of labor and wages, four elements are considered, viz.. Price, the Unit of Price, Quantity, and Cost. 360. ^rice is the sum paid or allowed for a unit, or a fixed number of units of the commodity ; as one, a dozen, a hundred. 361. The U7lit of ^?^ice is the number of units of the commodity upon which the price is based. 362. Quantity is the number of units or parts of a unit of the commodity. 363. Cost is the whole sum paid or allowed for the entire quantity. C^SE I. SOLUTION. Price and Quantity given, to find Cost. ^ ^ o> number of times 1 minus the rate is con- 5 2 5^0 tained in the difference. We therefore ^ ^^^ THE FIVE GENERAL CASES. 235 divide 64.4, the given difference, by .875, the difference of 1 at 12A^, and obtain 73.6, the required base. From these examples we learn that TJie base equals the quotient of the amount divided by 1 plus the rate^ or the quotient of the difference divided hy 1 minus the rate. l*It OB J. JEMS. 33. What number increased by '7% of itself is equal to 267.5 ? 34. A horse-dealer sold a span of matched horses for $1,155, which was 16 fo less than they cost him. What did they cost him ? 35. This year a clergyman's salary is $2,500, wbich is 25^ more than it was last year. What salary did he receive last year ? 36. The difference is 8,466, and the rate is 15^. What is the base? 9,960. 37. In Dec. a manufacturer made 4,865 yards of cassimere, which was 12|j^ more than he made in Nov. How much did he make in Nov.? 38. This year a man's house rent is $325, which is 18|.^ less than it was last year. What rent did he pay last year ? $490. 39. A house painter painted three houses, using 23|^ pounds of white lead for the first house, which was 20^^ less than he used for the second, and 17|^ more than for the third. How much white lead did he use for the second house ? How much for the third ? 29 lb. 6 oz. ; 20 lb. 389. Upon the principles deduced in 384-388 are based the ^ules for Co7nputations i7i ^erceiiiage, I. Base and rate given, to find percentage. Multiply the base by the rate. n. Base and percentage given, to find rate. Divide the percentage by the base. m. Rate and percentage given, to find base. Divide the percentage by the rate. 236 PERCENTAGE. IV. Base and rate given, to find either amount or dif- ference. Multiply the base by 1 plus the rate, for the amount ; and by 1 minus tJie rate, for the difference, V. Amount or difference and rate given, to find base. Divide the amount by 1 plus the rate ; and the difference by 1 minus the rate. Note. — Rules II. and III, are the converse of Rule I., and IV. and V. are the converse of each other. PM OJiI.E3IS, 40. If wheat yields 72^ of its weight in flour, how much flour can be made from 245 bushels of wheat ? 5^ IM. 41. After drawing 9 gallons from a cask of oil, the amount drawn was 40^ of the amount remaimng in the cask. How many- gallons were in the cask at first ? 3l4y. 42. I paid a tax of $61.40 on my farm, and with it a collector's fee of 5^. What was the whole amount paid 2 43. What % of 423 is 75^ ? 17^fc. 44. If Indian com contains 73^ of starch, how much starch is there in 1,192 pounds of com ? 45. If the ashes obtained from burning 2,275 j)ouncls of coal, weigh 68|^ pounds, what ^ of the coal remains in the ashes ? 3fc. 46. A merchant sold 51 yards from a roll of carpeting, and the amount sold was 37^^ of the whole number of yards in the roll. How many yards were in the roll ? 136. 47. A miller bought 2,175 bushels of wheat, 76;^ of which was winter wheat. How much of it was spring wheat ? 48. 63 is 64^ of what number ? 98/^75. 49. The number of children of school age in a certain'county is 11,275, and 3,157 children attend school. AVhat xjT^a?iO]srs iisj- oener^l taxes. 426. In the assessment of taxes, assessors must first find the rate, and then the tax. 427. The valuation of property is the base ; The rate % is the rate ; and The tax is the percentage. Hence, I. Valuation and tax giv- ) . j JSase and percentage giv- en, to find rate, ) \ en, to find rate. n. Valuation and rate ) . J Base and rate given, to given, to find tax, ) I find percentage. m. Tax and rate given, ) . j Percentage and rate given, to find valuation, ) i to find base. These three cases cover the ordinary computations in general taxes. PJ2 OS JO JEMS. 1. A school tax of $433.50 is levied in a district, and the property is assessed at $69,360. What is the rate ? .60, or $.006 j. on a dollar. 2. A tax of $95,935 is levied on a city, the assessed valuation of which is $7,674,800. What is the rate ? TAXES AND DUTIES. 249 3. If I am assessed at $1,250 on a house and lot, $300 on a vacant lot, and $3,000 personal property, how much will my tax be, the rate being $.0097 on a dollar ? $U-1H. 4. The assessed valuation of a village is $294,500, and a tax of $1,145 is to be laid. What must be the rate ? 5. A tax of $928.80 for building a bridge, is levied on a town, the assessed valuation of which is $967,500. What is the tax on property assessed at $1,250 ? $1.20. 6. A physician whose property was assessed at $2,750, paid a school tax of $23.37|-. What was the rate of taxation ? 7. One year, a man whose property was assessed at $1,350, paid .35^ village tax, .47^ school tax, 1.05^ county tax, and $1.00 poll tax. What was the amount of his taxes ? $26.^4^. 8. If the rate is $.001^ on a dollar, and the tax is $1178.85, what is the valuation ? 9. If a tax of $473.40 is levied on property assessed at $39,450, what is the assessed valuation of property that pays a tax of $29.70? .5-^.475. II. IN-TERNAIi REVENUE. 428. Internal Hevetiue is the income whicli Govern- ment receives from home business, products, and manu- factures. 429. Ificome Tax is a tax levied upon income. 430. A license I^ee is a tax levied for a license or per- mit to carry on any branch of business. 431. A Tax ozi Manufactures is a tax levied upon the value of home manufactures. COMIPXTTATIONS IInT INTEJE,N"AI:j rea^enxje. 432. Income taxes are computed at some legal rate upon the income minus the exemptions ; and Taxes on manufactures, at some legal rate upon the value of the manufactured goods. License fees are fixed sums established by law. 250 PERCENTAGE. 433. The assessed income {i. e. income minus exemptions), or the value of the manufactured goods, is the base ; The rate % is the rate ; and The tax is the percentage. Hence, Assessed income, or value of ^ r Base and rate giv- maniifactures, and rate given, toy is < en, to find percent- find tax, ) ( age. I'M OBLEMS. 10. A lawyer's income for the year 18G8 was $3,284, and his ex- emptions were $350 for house rent, and $1,000 for living expenses. How much income tax did he pay, the rate being 5^ ? $96.70. 11. A manufacturer's sales for the year amoimted to $58,750, upon which lie paid a government tax of .3^. What was the amount of the tax ? 13. A milliner pays a license of $10, and her assessed income is $835, on which the tax is 5^. How much revenue does she pay ? $51.25. III. CUSTOMS. 434 • Customs are duties paid to Government on im- ported goods and other property. Note. — The office at which customs or duties arc collected is a Custom- Jlouse ; and a seaport town in which a custom-house is situated is a Port of Entry. 435. An Ifivoice is a written account containing a list of merchandise sent to a purchaser, with prices and charges annexed. In custom-house transactions, certain deductions are made on some kinds of goods, before the duties are com- puted. These are tare, leakage, and breakage. 436. Tat^e is a deduction made from the weight of goods sold in chests, boxes, cases, casks, bags, or other envel- ope or covering, on account of the vreight of such covering. 437. ZfCakage is a deduction made from the quantity of liquors imported in casks. TAXES AND DUTIES. 251 438. breakage is a deduction made from the quantity of liquors imported in bottles. 439. G7'0SS }f eight is the entire weight of goods and case or covering. 440. JVet Weight is the gross weight minus the tare. 441; JVet Value is the value of goods at the original invoice price, after all deductions have been made. 442. Specific !Duty is duty on the number or quantity. 443. A.d Ydlorem ^uty is duty on the net value. Note.— A list of rates of duties established by Government is called a COMI^XJT^TIONS IN I3UTIES. 444. In ad valorem duties, The net value is the base ; The rate % of duty is the rate ; and The duty is the percentage. Hence, I. l^et value and rale ^f \ -^ S ^^^ ^^^ ^^^^ given, to duty given, to find duly, ) (find percentage. II. Specific duties are found by midtiplying the duly on one by the net number. PM OBI^EMS. 13. The gross weight of 175 boxes of raisins is 33^ lb. per box, and the tare is 25^. What is the total net weight ? 14. What are the duties, at $.25 per pound, on 150 chests of tea, invoiced at 62 lb. per chest ? 15. The duty on opium is 100^. What are the custom-house charges on 125 lb., invoiced at $5.37|- per lb. ? 16. A sugar refiner imports 72 hhd. W. I. sugar, weighing 475 lb. each, and 50 hhd. molasses containing 126 gal. each. What are the duties, sugar paying $.03 per lb., and molasses $.08 per gal., and the tare on the sugar being 12^^ ? $l,Ji01.75. 252 PERCENTAGE. JPBOBLJEMS IN T A X H: S A. N D DUTIES. 17. The valuation of the property of a certain county is $11,847,500, upon which a tax of $146,909 is levied. How much of this tax will be paid by the owner of a foundery which is as- sessed at $14,550 ? $18042. 18. A collector's fees for collecting a town tax were $197.73, and the whole tax was- $14,829. What rate fo did the collector receive ? mi- 19. In 1865, a 5^ tax was required upon the first $5,000 of a man's income, and 10^ upon all above $5,000, the exemptions being $600 for living expenses, $200 for house rent, and the amount paid for taxes. How much tax did a man pay, whose in- come was $17,675, and who had paid $453 for taxes? $1,392.20, 20. A real estate agent who charged the seller 2%. and the buyer 3;^, sold a house for $10,000. What was his commission ? 21. The duty on tobacco being $.35 per lb. ; and on segars $3 per lb. specific, and 50^ ad valorem ; what are the duties on 50 cases of tobacco invoiced at 65 lb. each, and 175,000 Havana segars, weighing 2,625 lb., and invoiced at $45 per M. ? $12,950. 22. Upon the property of a city assessed at $3,824,600, a tax of $72,667.40 is levied. Make a table, embracing the tax on $1 to $10, from which the tax list can be computed. See Manual. SECTION VIII. I jv T ^ :r b s t . 445. If a person hires a house or a farm, lie pays the owner for the use of it. If a person hires or borrows money, v/hen he pays the debt, he also pays an additional sum for the use of the money. And when a person pays a debt after it is due, he pays an additional sum for the credit ; i. e., for the use of the money after the debt is due. 440. I7iterest is the sum paid for the use of money. 447. ^ri7icipal is the sum for the use of which interest is paid. INTEllEST, 253 448. A?nount is the sum of principal and interest. 449. ^ate per Cent per Aiinum is the interest on $1 for 1 year. 450. Simple Interest is the sum paid for the use of the principal. 451. Compound Intei^est is the sum paid for the use of interest. 452. A Partial ^ay77ient is a payment of a part of an obligation due, or that is drawing interest. 453. legal ^ate is the rate of interest allowed by law. Note. — Any rate of interest greater than tlie legal rate is Usury. 454. TABLE OF LEGAL BATES OF mTEREST. WHEN NO BATE IS NAMED. KATES ALLOWED BY SPECIAL CONTRACT. H La. ■ Not i exceeding TN. Y., Mich., Wis., Sfo Fla. and La. H ]Mmn.,S.C., Geo., Utah, r Ohio, Iowa, Miss., Ark., ^ and Hudson Co., K J. lOfo •< Utah, and for borrowed tOfo Ala. and Tex. .( money in Mich, and 111. j Cal., Or., Kan., Neb., 1 W. T., Nev., and Col. 12fo j Wis., Tex., and on judg- ( ments in Minn. r All the other States, 15fo Neb. Qfo ) D. C, and bank inter- 20fo Kan. ( est in La. and Kan. Any rate per cent agreed upon. ( R. I, Minn., Cal., W. T., 1 Nev., and Col. c-a.se I. Computations of Simple Interest. 455. In all the previous Sections of this Chapter, rate % is a fixed sum without regard to time. But in interest, the entire rate per cent paid on $1, depends upon the time. Thus, if the rate is 6% per annum, the rate % on $1 for 1 year is .06 ; for 3 years it is 3 times .06, or .18 ; for 6 months or J year is ^ of .06, or .03, etc. Hence, The rate, or the percentage on $1, is the product of the rate per cent per annum and the time expressed in years. 254 PERCENTAGE. I. GENERAL METHOD. 456. Ex. What is the interest of $287.50 for 3 years, at Explanation. — Since the rate is 1% solution. per annum, the interest for 1 year $28 7.5 Principal is .07 times the principal, and the ^_^ ''^«^^- interest for 3 years is 3 times the $20,125 int. /or i yr. interest for 1 year. We therefore ^ multiply $287.50 by .07, and the $ 6 0, SI 5 int. for ^yr. product, $20,125, by 3. The final result, $60. 37^, is the required interest. Hence, Interest for years is the product of principal, rate, and time. FMOBIjEIIS. * 1. What is the interest of $515.50 for 1 year, at % ? $80.93. 2. What is the amount of $325 for 1 year, at 6^ ? (See 44§.) 8. What is the interest of $117.35 for 2 years, at 5^ ? $11.72^. 4. If I borrow $390 for 4 years, at 7^, what amount will be due at the expiration of the time ? $499.20. 5. What is the interest, and what the amount, of $1,068.50 for 1 year, at 8^ ? Amount, $1, 153.98. 457. Ex. What is the in- soltttion. terest of $654.75 for 1 jr. 5 $ 6 5 4.7 5 Principal. .0 6 Pate. Explanation. — Since 1 yr. $3 9.2 8 5 int. for i yr. 5 mo., or 17 mo., is \% yr., we ^ first find the interest for 1 yr., 2 7^995 as in (456) ; and then multiply 39285 this interest, $39,285, by |i, $6 67,8 45^12 the required time. That is, $55,65+ S^'.forwyr., when there are months in the < oriyr.^ mo. given time, we Multiply the interest for 1 year hy the number of months, and divide the product by 12. INTEREST. 255 Notes. — ^1. In computations, the partial results should be carried to four decimal places. 2. In final results, if the mills are 5 or more, it is customary to call them 1 cent, and if they are less than 5 to reject them. (Sce^ 163.) BJtOBLEMS. 6. What is the interest of $2,160 for 1 year 3 months, at 7^ ? 7. What is the interest of $39.25 for 3 yr. 8 mo., at 5^? 8. What is the amount of $1,278 for 11 mo., at 7^ ? $1,360.01. 9. rind the interest of $9,500 for 3 yr. 1 mo., at 4^. 10. How much interest, at 8^, must I pay, for the use of $2,575 from May 11, 1868, to Sept. 11, 1869 ? $27Jt.G7. 458. Ex. What is the solution. interest of $761.25 for 2 ^^ yr. 5 mo. 16 da. = 29.51,- mo, yr. 5 mo. 16 da., at 8^ ? $161.25 Principal. Explanation. — Since 30 .OSjiate. days are 1 month, every 3 $ 6 0.9 00 days are 1 tenth of a month ; 2 9.5 j- 16 days are J/ tenths or 20 3 5i tenths of a month ; and nVgY 2 yr. 5 mo. 16 da. are 29.5| 1218 29 5' ~ — mo. or -^^ yr. We there- $1798.58 { 1 2 12 <^ 1 ! Q Si fi^\ Int. for 2 yr. fore multiply the interest ^-/-^^.oo^^ ^mo.i&da. for 1 year by the number of months, and divide the product by 12, as in 457. PJIOBI^EMS. 11. What is the interest of $198.50 for 4 mo. 9 da., at 4^? 12. What is the interest of $10,796 for 2 yr. 1 mo. 24 da., at 7^ ? 13. Fmd the amount of $18,450 for 1 mo. 15 da., at ^'/o. 14. How much interest, at 6^, has accrued on a note for $94.75, that has been due 3 yr. 2 mo. 6 da. ? $18.10. 15. What is the amount of $978.18 from Sept. 24, 1867, to Oct. 25, 1869, at 8^? $1,U1.J^. 25G PERCENTAGE. II. SIX PER CENT METHOD. 459. The interest of $2 for 1 year, or 12 montlis, at Q%, or of $1 for the same time, at 12%, is $.12. Hence, I. The interest of any sum at 6% is the same as the interest of one half that sum at 12%. II. At 12%o per annum, the rate is 1%) per month. Ex. Find the interest, at Q%, solution. of $52.69 for 2 yr. 3 mo. 18 da. 2 yr. Smo.18 da. =27.6 mo. Explanation.— We first di- S S 2.6 9 \2 vide the principal, $52.69, by ^26.3^5 1 the phu. 2, to find the sum on which ^2 7 6 Rate at \% per mo. to compute interest at 12^ i 5 8 7 (I.). We then multiply this 18J^Jj.l5 result, $26,345, by the rate 52690 at 1%) per month, which is $7.2 71220 interest. .01 of the time expressed in months (H.). The final result, $7.27, is the required in- terest. ph OB lems. 16. How much interest, at 6^, will be due in 5 yr. on a loan of $5,790? $1,737. 17. What is the interest of $728.18 for 1 yr. 11 mo., at 6^? 18. If $2,765 be placed at interest at 6^, Mar. 14, 1869, what will be due Dec. 13, 1870 ? $3,054.86. 19. What is the interest of $20 for 12 yr., at 5^ ? (6^ — i of itself =5^.) $12. 20. Find the amount of $417.61 for 3 yr. 7 mo., at 8fc. (6^ + 1 of itself =8^.) ' $537.32. III. SEVEN PER CENT METHOD, FOR DAYS. 460. Computing interest on the basis of 30 days to a month, gives 360 days to a year. Hence, I. The product of any principal multiplied by any given number of days expressed as hundredths, is the interest at 360% per annum, or 1%) per day. 375 750 $7,8 7 5 \e $1,3125 \ 6 ,21875 INTEREST, 257 n. The product of any principal multiplied by any given number of days expressed as thousandths, is the interest at per annum, or .l%per day. III. The interest at 36% divided by 6 gives the interest at t IV. Interest at Gfo plus interest at 1% equals interest at 7%, Ex. What is the interest of $125 for 63 days, at 1% ? ExpL.iNATioN. — We first multiply solution. the principal, $125, by .063, and ob- $12 5 Pnn. tain $7,875, the interest at 36^ (n.). -^^^ We divide this result by 6, and ob- tain $1.31|, the interest at 6% (HI.). We then divide this result by 6, and obtain the interest at 1% ; and add- ing the last two results, we have .$1.53 1, the required interest (IV.). $ 1.53 12 5 int. PROBLEMS. 21. What is the interest of $735 for 27 days, at Ifo ? • $3.86, 23. What is the amount of $250 from Jan. 18 to March 30, 1868, at 7^ ? $253.50. 23. Dec. 24, 1868, I borrowed $25.50, and paid it June 1, 1869, with 7^ interest. What amount was due ? 24. What is the interest of $45.75 for 90 da., at 7^ ? $.80. 25. What is the amount of $1,250 for 63 da., at 7j^ ? $1,265.31, 461. ^ules for Computi7ig Interest, I. General Method. 1. For 1 year. Multiply the principal by the rate. 2. For 2 or more years. Multiply the interest for 1 year by the number of years. 3. For any other time. Multiply the interest for 1 year by the time expressed in months and tenths of a month, and divide the product by 12. 258 TERCENTAGE. II. Six Per Cent Method. Divide the principal by 2, and multiply the quotient by .01 of the time expressed in months. ni. Seven Per Cent Method, for Days. 1. Midtiply the principal by .001 of the number of days. 2. Divide the product by 6, and to the quotient add -^ of itself. Notes.— 1. 360 days + /^ of 360 days (5 days) = 365 days. Hence, if interest for days is required at 365 days to a year, subtract from itself ^\^ of the interest found by tlie "1% method. 2. To find the amount, we may first find the amount of $1 at the given rate for the given time, by any one of the above rules, and then multiply the principal by this amount. 3. The months and days may be reduced to the decimal of a year (sec 356, II.), and the interest for 1 year, at the given rate ^, may then be multi- plied by the time expressed in years and decimals of a year. r II o B Ij ems. 26. What interest must I pay for the use of $378.64 for 3 years, at 7^? 27. What* is the amount of $473 for 7 yr. 7 mo., at 5^ ? 28. What is the interest of $419.84 for 1 yr. 11 mo. 18 da., at 5^? $Jtl.28. 29. A debt of $1,560 was contracted May 23, 1868. How much was due June 18, 1869, interest at 6^ ? $1,660.10. 30. The balance due on a mortgage, Nov. 20, 1868, was $3,750. What was the amount due Aug. 20, 1869, interest 7^ ? 31. What is the amount of $75 for 8 yr., at 10^ ? $135. 32. A note for $1,116, bearing date Albany, N.Y., Oct. 9, 1866, was paid Oct. 9, 1869, with interest. What amount was paid ? 33. What amount was due Aug. 5, 1869, on a note for $1,650, dated Philadelphia, Dec. 5, 1867 ? $1,815. 34. I bought a house and lot in Cleveland, for $4,750, paying $2,000 down, and giving a mortgage for the balance, due in 3 years. What was the amount of the mortgage when due ? $3,!245. 35. How much was due. May 3, 1869, on a note for $2,860. dated San Francisco, July 3, 1867 ? INTEREST. 259 36. What is the amount of $743.18 for 1 yr. 10 mo. 12 da., at 8^ ? 37. If I borrow $12,500 in New Haven, Conn., and loan it in New York, how much do I gain in 1 yr. 7 mo. ? $197.92. 38. If I loan $1,500, at 7^, Aug. 17, 1869, how much will be due June 30, 1871 ? $1,696.29. 39. What is the interest of $10 for 15 yr. 4 mo., at 6^ ? 40. A note for $293, dated Detroit, Apr. 26, 1867, was paid Jan. 26, 1869. What was the amount paid ? $328.89. 41. At 7^, what is the amount of $73.49 from Nov. 27, 1867, to Feb. 7,1870? $8J^.78. 42. A man bought a farm in Minnesota for $2,280, paying $1,000 down, and the balance in 10 months, with interest. How much was the last payment ? $1,354.67. 43. Find the amount of $856.75 for 2 years, at 5fo. $942.42-1. 44. How much interest will I have to pay on a loan of $7,650 for 20 days, at 7^ ? $29.75. 45. What is the amount of $25,390 for 7 months, at 10^ ? 40. Jan. 10, 1869, I borrowed $1,280 in Hartford, Conn., and paid it Aug. 7, 1869, with interest. How much did I pay ? 47. What is the interest of $1,310 for 1 yr. 1 mo., in Ya. ? 48. How much will be due June 19, 1871, on a note for $1,750, dated Boston, June 19, 1869, with interest ? $1,960. 49. Find the amount due Mar. 17, 1870, on a note for $217.85, dated St. Louis, Sept. 17, 1867, with interest. 50. A man who is paying $375 a year for house rent, borrows $5,000, at 6^, with which he buys the house. Does he gain or lose by the transaction ? JSb gains $75 per annum. CA.SE II. Compound Interest. 462. In computing compound interest, I. Hie amount of the principal for 1 year is (he principal for the second year, the amount of this principal for 1 year is the principal for the third year, and so on. n. The final amount minus the principal is the interest. 2G0 PERCENTAGE. Ex. What is the amount of $127.50 at compound interest for 2 jr., at 6% ? What is the interest ? Explanation. — Since the solution. amount is the product of the $12 7.50 Pnn. principal multiphed by 1 1-0 Q i + »*«^«- plus the rate (see 461, Note 11 6 50 2), we multiply the principal, 127 5 $127.50, by 1.06, and obtain $18 5.1 5\ p^I.-^Z/o^^vC $135.15, the amount for 1 1-0 ^ i+rate. year. We multiply this 8109 amount by 1.06, as before, 13 515 and obtain $143,259, the re- $ IJf 8.2 5 9 Amt.for 2 yr. quired amount for 2 years. 12 7.5 Prin. Then, subtracting the prin- $15,759 int. cipal, $127.50, from this amount, we have $15.76, the required interest. Note. — When interest is due semi-annually, quarterly, or monthly, the amount of the principal for the fixed period of time is the principal for the next period. FJIOBJLEMS. 51. What is the amount of $731.45 for 3 years, at 6^ compound interest ? $859.26. 53. What is the compound interest of $75.50 for 2^ years, at 6^, payable semi-annually ? $12.03. 53. The principal is $35.75, the time 4 years, and the rate 7^ compound interest. WTiat is the amount ? 54. How much will $535 amount to in 1^ years, at 7^ compound interest, payable semi-annually ? $582.08. 55. What is the compound interest of $437.50 for 1 yr. 3 mo., at Qfc, payable quarterly ? 56. At 34|^, interest payable monthly, what is the compound interest of $575 for 3-i- yr. ? 57. What is the difference between the simple and the compound interest of $5,435 for 4 years, at Qfo ? $121.94. INTEREST. 261 C^SE III. Partial Payments. 463. When partial payments are made upon notes, bonds, mortgages, or other obHgations bearing interest, the U. S. Courts have estabhshed the following principles : I. Paijments must he applied in the first place to the dis- charge of interest due, and the balance toward the discharge of the principal. II. Interest must not he added to the principal so as to draw interest. m. The principal must remain unaltered, when a payment is less than interest due. Ex. A note for S960, at Q% interest, was given Apr. 10, 1868. A payment of $225 was made Jan. 19, 1869, and another of $25, Nov. 3, 1869. What amount was due Jan. 5, 1870 ? Explanation. — We first find the amount of the principal from the date of the note to Jan. 19, 1869, the time of the first pay- ment (see 461, 11.), to be II, 04. 64. We next subtract the pay- ment, $225, from this amount (I.), and have a remainder of $779.- 64 for a new princi- pal. Since the pay- BOLTJTIOX. $9 60 Prin. 12 SJ^80 ,0 9 3 432 $ JfJf.6 Jf. Int. to Jan. 19, 1869. 9 60 Prin. $ 10 4.6 Jf Amt. due Jan. 19, 1869. 22 5 Payment, " " $T 19.6 4 New Prin. \2 $ 3 8 9.82 .115i 12994 194910 38982 38982 S 44.95 92 4 Int. to Jan. 5, 1870. 7 7 9.64 Prin. $824.59924 Amt. to Jan. 5, ISTO. 2 5 Payment, Nov. 8, 1869. $ 7 99.6 Ami. due Jan. 5, 1870. 262 PERCENTAGE. ment, $25, made Nov. 3, 1869, did not exceed the interest due (in.), we find the amount of $779.64 from Jan. 19, 1889, to Jan. 5, 1870. Then, subtracting from this amount, $824.59924, the payment of $25, made Nov. 3, 1869, we have $799.60, the required amount due Jan. 5, 1870. Hence, 461. (Rule for Computinff Interest 191 I*a7^tlal 'Paymejits, I, From the amount of the principal computed to the time when the payment or sum of the payments equals or exceeds the interest due, subtract the payment, or sum of the payments. n. The remainder is a new principal, with which proceed as before. JPM O B LE 31 S . 58. On Nov. 5, 1867, the face of a mortgage on a farm in Mich, was $2,875, and $1,000 was paid Aug. 23, 1868. What amount was due May 17, 1869 ? $2,140.51. 59. A mechanic bought a house and lot in Salem, Mass., for $1,750, paying $500 down. One year afterward he paid $387.50. How much then remained due ? $937.50. 00. Upon a note for $765, dated Buffalo, N. Y., Mar. 14, 1867, there was paid, Oct. 31, 1868, $50 ; and June 11, 1869, $285. The note was taken up, Sept. 25, 1869. How much was then due ? 61. May 7, 1867, a capitalist loaned $10,000, at ^fc. Dec. 28, 1867, $4,800 was paid ; and July 14, 1868, $3,750. What sum was due Jan. 3, 1869 ? $2,020.0^. VT., N. H., AND CONN. RULES. 465. In Vt. and N. H., a written stipulation to pay inter- est annually, allows the creditor simple interest on interests due on the principal and remaining unpaid after the end of each year. This allowance of simple interest for the use of interest due, is called Annual IntereM. see Manual. The differences between the U. S. Court Eule for com- puting interest in Partial Payments, and the rules in Vt., N. H., and Conn., are as follows : INTEREST. 263 466. The Vermont ^ule, I. Simple interest is allowed on all unpaid annual inter ests, from the time they become due to the time of final settlement. II. Simple interest is allowed on all payments, from the time they are made to the end of the year, or to the time of final settlement, 467. T/ie JVew Ilai7ipshire 'Eute, I. Interest is allowed on payments only when a payment, or the sum -of two or more payments, exceeds the interest due. n. In all other rejects the rule is the same as in Vermont. Notes. — 1. In computing annual interest in these States, the computa- tions must be made for intervals of 1 year, or to the time of final settle- ment, when that occurs within a year. 2. The interest in partial payments is computed by the U. S. Court Rule, unless annual interest is stipulated in the note or agreement. 468. 27ie Connecticut ^ute, I. Interest is allowed on payments to the end of a year, when they exceed the interest due, at the time they are made ; or to the time of settlement, when that occurs within a year. n. When more than a year passes from the date of any com- putation, without payments being made, interest is computed for the whole time, by the U. S. Court Rule. PJt OB TjEMS. 62-65. Find the amount due Oct. 1, 1869, on the following note, computing the interest by the U.S., the Vt., the N. H., and the Conn. Rule. ^/,tJTA.TIOI^S IN" BA-ISTKINGf-, 495. In computing bank discount, The face of the note to be discounted is the principal, or base ; The interest on the face of the note for the given time, at the given rate, is the bank discount, or percentage ; The proceeds of the note is the difi'erence ; and The product of the rate % per' annum and the time ex- pressed in years, is the rate. Hence, I. Face of note, rate and \ { ^^ , . time given, to find banki is \ Principal raie, and time discount, ) I g^ven, to find interest. II. Face of note, rate and ) • j Base and rate given, to time given, to find proceeds, ) \ find difference, III. Proceeds, rate and \ ( ^ . ^ lime given, to find fa^e of \ is \ Differenceand rate given, note, S Vofindhase. 272 PERCENTAGE. I'JtOBJjEMS. 1. "What is the bank discount upon a note for $3,500, due in 4 months, at 6;^ ? $51.25. 2. What are the proceeds of a note for $650, due in 90 days, if discounted at a N. Y. bank ? $638.25. 3. A note for $7,350, due Aug. 12, 18G8, was discounted at a bank in Charleston, S. C, May 20, 1868. What were the proceeds ? 4. What sum can be realized at a bank, upon a note for $11,500, due in 30 days, at Qfo discount ? 5. A bank loans $4,500 on a note payable in 4 months, discount- ing it at %fo. What is the face of the note ? $Jf, 62646. 6. I obtained $237, at a bank m New Orleans, on a note due in 8 months. What was the face of the note ? 7. A merchant wishes to borrow $2,000 at a bank. For what amount must he make his note, due in 60 days, if he gets it dis- counted at Qfo ? 8. What will be the proceeds of a note for $4,500, due in 6 months, discounted at a bank in San Francisco ? $4^271.25. 9. A manufacturer in St. Louis wishes to borrow $2,000 with which to pay his men. If he makes a bank-note, due in 30 days, what will be the face of the note ? $2,011.06. 10. A merchant buys a bill of goods in Milwaukee, amounting to $3,700, on 3 months' time, or 5^ off for cash. If he borrows the money at a Milwaukee bank, will he make or lose, and how much ? 11. I had a note for $460, dated Detroit, Jan. 9, 1869, and due in 8 months, with interest. March 19, 1869, I had it discounted at a bank, at 8^C How much did I realize from the note ? 12. For what sum must a bank-note, due in 5 months, be made, to have it produce $1,856 when discounted at the First National Bank of Boston ? $1, 904.57. 13. Find the face of a note, due in G months, on which I can borrow $875 at a Chicago bank. $921.86. 14. In payment of a debt, I took Chas. Marshall's note for $1,600, payable at the Sixth National Bank of Philadelphia, in 6 months, with interest. Four months afterward, I had it dis- counted at the First National Bank of Harrisburg. What were the proceeds ? EXCHANGE. 273 SECTION XII. B X C H A JV' G JE , 496. .Exchange is a commercial transaction, in which a party in one place pays money to a second party in another place, by means of an order upon a third party, and without the transmission of money. 497. A ^7^aft, or :Bill of Uxc?ia7ige, is a written or- der for money, drawn in one place and payable in another. Example.^ — A Chicago merchant, wishing to pay a debt in New York, buys at a bank in Chicago, a draft on a New York bank, payable to the order of the party in New York. The Chicago merchant sends this draft to his creditor in New York, and the latter indorses it, presents it to the New York bank, and receives the face of the draft in money. Notes.— 1. Any party may give a draft, or " draw " on another party, if the second party is debtor to the first. 2. A draft payable in the same country in which it is drawn, is an Inland Bill of Exchange ; and one drawn in one country, and payable in another, is a Foreign Bill of ExcJiange. 498. A SigJit ^7*aft is a draft payable " at sight," L c, when it is presented ; and 499. A 2tme ^raft is a draft payable at a future time named in it. Note. — Grace is allowed on time drafts, but not on sight drafts. 500. There are usually four parties to a transaction in exchange, viz.. The H) rawer or Maker of the draft ; The ^uyer or ^emiitery or the party who purchases the draft ; The ^7'aweey or the party on whom the draft is drawn ; The "Pa/yee, or the party to whose order the draft is made payable. Note.— The maker and remitter of a draft may be the same party, in which case there will be but three parties to the transaction. 1£* 274 PERCENTAGE. COMMON FORM OF DSAFT. STAMP. I c^« c^aj^ a/^Ai ^^/^ pay to tha order of ^ai/iei ^ Mlo^^eid ■■■■■ /'/eejf /func/uc/ Dollars, Value received, and charge to account of M^ ^oi/ ^/^. Note.— The words At SigU, in place of " Ten days after sight " would make the above a sight draft. COIMFXIT^TIONS IIS- EXCH^^STGE. 501. Drafts or bills of exchange are bought at par, at a premium, or at a discount. The face of the draft is the base ; The rate % of exchange is the rate ; The premium or discount is the percentage ; and The proceeds of the draft is the amount or difference. Hence, I. Face of draft and rate ) c % of exchange gimn, to fined is \ ^^"^^ ""'^ rate given, to premium or discount, ) ( find percentage. II. Proceeds and rate %\ i , -..^ r, 7 . i ^ j\ • \ Amount or difference of exchange muen, to jina> is -< , . f -, -, r jy 1 ^ f, \ I ana rate given, to find base. Notes. — ^1. The face of a draft plus the premium, or minus the discount, is the proceeds. 3. The subject of Foreign Exchange is not considered in this book. A full presentation of it will be found in the Academic Arithmetic of this Series. PU OBLEMS. 1. At 2>Ac^c^\: 200for3mo.= S GOOforlmo. of $400 has no 100 " 6 " = 2A00 " 1 " term of credit, and 2 00 " 9 " = 18 00 '^ 1 " only forms a part ~^ 12OO " ? " =$J/.800 " 1 « of the purchase money. The use $Ji.800 [ $1200 of $200 for 3 !{-'= Jf- mo., average term of credit. months is the j^iy i+^ rno. = Nov. 1. same as that of 3 times $200, or $600, for 1 month : the use of $400 for 6 months is the same as that of $2,400 for 1 month ; and tho use of $200 for 9 manths is the same as that of $1,800 for AVERAGE OF PAYMENTS. 277 1 month. The sum of the payments, due at different times, is $1,200, and of the equivalent sums for 1 month is $4,800. The use of $4,800 for 1 month is the same as the use of $1,200 for as many months as the number of times $1,200 are contained in $4,800. Dividing $4,800 by $1,200, we ob- tain 4—4 months, the average term of credit ; and July 1 + 4 months = Nov. 1, the time required. Hence, TJie product of the sum of all the payments multiplied by the average term of credit, equals the sum of the products of all the payments multiplied by their respective terms of credit. 511. On this principle is based the (Hiite for finding the a%'erage time ofpaymetit, 7i'hen t/ie ter77is of credit all begin at the same date, I. Multiply each payment by its term of credit, and add all the payments, and also all the products. II. For the average term of credit, divide the sum of the products by the sum of the paijments. III. For the average time of payment, add the term of credit to the given date. Tit OBL EMS, 1. March 8, I bought a building lot for $800, paying $200 down, and agreeing to pay $200 in 4 months, $200 in 8 months, and $200 in 12 months. Had I given my note for the payment of the whole amount at once, at what date should it have been made payable ? Sept. 8. 2. May 29, a merchant bought bills of goods as follows: $825 on a credit of 3 months, $675 on 4 months, $450 on 2 months, and $800 on 1 month. What is the average time for the payment of the whole amount ? Aug. I4. 3. On the first day of May, D hired a house at $300 per annum, agreeing to pay the rent quarterly. What would be the equated time for the payment of the whole ? Dec. 16. 4. To-day I owe $150 due in 30 days, $200 due in 60 days, and $250 due in 90 days. If I give my note for the whole amount, made payable at the average time, when vail the note be due?_^^ 278 PERCENTAGE. 5. What is the average time for the payment of 5 notes, all bearing date June 17 ; one for $300 due in 3 months, one for $500 due in 5 months, one for $150 due in 7 months, one for $350 due in 9 months, and one for $200 due in 1 year ? Cj^SE II. The terms of Credit beginning at different Dates. 512. Ex. March 21, 1 bought a horse for $175 on a credit of 4 months ; June 5, a harness for $55 on 3 months, and a top carriage for $225 on 4 months. What is the average time for the payment of the three debts ? Explanation. — solution. To find the dates ^^'^' ^i-\-J^rno. = July 2 1 1 . ■. , 1 June 6 + 3 " = Sept. 5 on which the sev- ,, ^|;^ ,, = Oct, 5 era] payments fall due, we add to $ 17 5 cash, July 21. each date the term 5 5 for J^6 da. = $ 253 for 1 da. 225 " 76 " = 17 100 " 1 " of credit. Since the first payment SJ^55 " 'i " = $ 196 30 '' 1 '' falls due July 21, ^ 19 630 \ $455 we take that date _^1^ I J,3=:J,3da.{ ""'^JTelr' as a focal date. lJf30 Comparing this 13_65_ focal date with 65 =/jV, less than | da., dropped. the dates on which July 21 + J^3 da. = Sept. 2. the several pay- ments mature, we have $175 due at the focal date, $55 in 46 dsijs, and $225 in 76 days. Proceeding as in Case I., we find the average term of credit to be 43 days after July 21, or Sept. 2, the average time required. Hence, When the terms of Credit begin on different dates, The earliest date on which any one of the payments matures, may be taken as the focal date ; and the time between this date and the date on which each of the other payments matures, may be taken as the term of credit of that payment. AVERAGE OF PAYMENTS. 279 513. On this principle is based the ^iile fo7' fiiidlng the average time ofpaj'77?e7it, when the tei^ms of credit begin at different dates. I. Find the time on which each payment becomes due, by adding the term of credit to the date of the transaction. II. For the focal date, take the earliest dale on which any one of the payments matures ; and for the term of credit of each payment, take the time between the focal date and the time on which the payment matures. m. For the average time of payment, find the average term of credit, and add it to the focal date. PR OBLEMS. 6. Find the average time for the payment of the following bills: Jan. 10, $415 on 3 months ; Feb. 25, 175 " 4 " Apr. 5, 350 " 3 " 35 days from Apr. 10, or May 15. 7. What is the equated time for the payment of three notes, one for $650, dated July 13, and due in 90 days ; one for $555, dated July 35, and due in 60 days ; and one for $445, dated Aug. 14, and due in 30 days ? Sept. 27. 8. George Adams bought provisions as follows : Mar. 14, 40 bar. beef, ® $17.50, at 3 months. May 1, 60 " pork, @ 34 " June 10, 150 " flour, @ 8 Terms cash. What is the average time for the payment of the whole ? 9. I make the following advances of money for a friend : May 19, $107; May 38, $35 ; June 37, $130; July 3, $70; Aug. 34, $80, and Sept. 11, $175. If I take his note for the whole amount, dated at the equated time, what will be the date of the note ? July 19. 10. B works for C 6 months from May 15, at $60 per month, his wages to be paid one half monthly, and the other half in 3 months. In lieu of receiving his pay according to contract, he takes C's note for the whole amount, bearing date from the average time, with interest. What is the date of the note ? Oct. 15. 280 PERCENTAGE. C^SK III. Accounts containing both Debits and Credits. 5!li Computations in compound average are based upon the following equitable principles : I. All payments made before the average term of credit ex- pires, should draw interest from the time they are made ; and II. All debits not paid till after the average term of credit expires, shoidd draw interest from the expiration of the aver- age term of -credit. 515. Ex. Find the average time for the payment of the balance of the following %. : Dr. KoBEKT Lansing. Cr. 1869 1869 i Jan. 4 To Mdse.@4 mo. 325 50 Feb. 15 By Cash, 200 00 u 16 " do. 4 " 37 50 Apr. 17 " do. 75 00 Mar. 1 " do. 4 " 162 50 u 26 " do. 4 " 250 00 EXPLANATION.- Ist, AVe find the average time of the debits to be June 12 ; and, 2d, The aver- age time of the credits to be Mar. 4 ; both by Case II. 3d, Since $275 was paid 100 days before it was due {i.e. be- fore the average time at Vv^hich the debits were SOLUTION. 1st.— Averaging the Debits, Jan. Jf. + Jfmo. = May J,.. " 16 -{-J^ " = " 16. Mar. i + 4 " = July 1. " 26 + Jf " = " 26. Focal date, May 4-. So 2 5.5 cash payment. May J^. 3 7.50 for 12 da. = $ 1^50 for 1 da. 162.50 250 $775.50 " $30625.0 23 265 ~~73600 69795 58 83 9Jf25 = 2075 = $30625 $7 75.5 39.4. — 39 da.i^^^'^''^^^.*^''"'' 3805^0 31020 ~70Jd May 4 + 39 da. 0/ credit. J- ^Q^ Average ti77ie /or June i-v j payment of d'ehita. AVERAGE OF PAYMENTS. 281 due), interest must be allowed on $775.50- $275 = $500.50, the balance, for 100 days (511, I.); or, whicli is the same thing, the term of cred- it for $500.50, the balance, must be extend- ed a sufficient time to average the use of the $275 paid 100 days before it was due. We find this addi- tional term of credit for the balance of the account, by mul- tiplying $275, the sum of the credits, by 100, the number of days before the maturity of the debits, and di- viding $27,500, the product, by J55500.50, the bal- ance of the %., as in 511. The 2(1.— Averaging the Credits. Focal date, Feb. 15. $200 cash payment, Feb. 15. T5^for Gl da. = S 45 7 5 for 1 da. $215 " ? " $4515 $4575 275 \$275 1825 1650 ~'rwo 1650 100 -I o a -iiy J S ■A'Berage term I 0.0 — 1/ aa.} of payments. Feb. 15 + 17 da. = Mar. 4\^l}'^'4mZs. Sd. — Averaging Debits and Credits. Focal date, June 12. From Mar. 4 to June 12 = 100 da. $775.50 - $275 - $500.50. $275 for 100 da. = $27500 for 1 da. 500.50'' ? " = 27500 " 1 " $27500.0 \ $500.5 25025 24750 20020 4730^^0 45045 2255 June 12 + 55 da. ■ ( Average time 54.9 = 55 da.} o/lal. o/c^c. Aug. ei^f/^fp'^y""'''* if \ofbal.ofal^ 282 PERCENTAGE. result, 55 days, added to June 12, the date of the maturity of the debits, gives Aug. 6, average time required. Had the date of the average term of the credits been later than that of the average term of the debits, i. e., after June 12, we should have dated back, or subtracted, the average time of the bal. of the %. from June 12. Hence, 516, ^if^e fo7^ Co77iputmg Compound Average, I. Find the average term of the debits and credits separately. n. For the average term of credit of the halance of the account^ tahe the average date of the larger of the two sides of the account for a focal date, multiply the smaller side hy the difference in time between its date and this focal date, and divide the product hy the balance of the account. HI. For the date of the average time, count foriuard from the last focal date, if the larger side of the account falls due later; and bachivard, if it falls due earlier, PJtOJiLE3IS. II. Balance the following %., and find when it is due : Dr. E. M. D ANIELS & Co. Cr. 1869 1869 June 14 To Mdse. ® 3 mo. 450 00 Sept. 3 By Cash, 400 00 Aug. 25 u u g a 175 00 Nov. 2 " do. 150 00 Oct. 11 u a Q u 425 00 u 30 " do. 225 00 Fel. 17, 1870. 12. What is the balance of the following %., and when is it due ? Dr. John G. Andeeson. Cr. 1869 1869 Oct. 9 To Mdse.® 3 mo. 300 00 Nov. 24 By Cash, 25 00 Nov. 18 " do. 3 " ; 329 00 Dec. 4 " do. 500 00 u 27 " do. 3 '' 142 00 a 30 " Note, 150 00 Dec. 19 " do. 3 " 256 00 $352; June 25, 1870. REVIEW PROBLEMS. 283 13. If a note, drawing interest, be given to balance tlie following %., for what sum will it be drawn, and What will be its date ? Dr. Ames & Potteb. Gr, 1869 1869 Mar. 17 ToMdse.(g^2mo. 325 00 July 25 By Cash, 125 00 Apr. 20 " do. 3 " 108 00 Aug. 17 " do. 300 00 July 18 " do. Cash, 264 00 Oct. 24 Draft on KY., 350 00 Aug. 11 " do. ®4mo. 50 00 j Sept. 35 •' do. 3 " 125 00 $97; May 25, 1868, SECTION XIV. it^riByr tiioszbms ij\r !Pjej^cbjvta.gb. 1. What is 13|^ of 837 bushels of wheat ? 1U.39 lu. 2. If I sell a sewing-machine for $50 that cost me $56, what fo do I lose ? lOff,. 3. How much will 100 shares of N. Y. C. R.R. stock cost me, at 154|^, brokerage Ifo ? 4. Of every 1,375 persons 25 years old, 1,265 will live to the age of 26. What fo of persons 25 years old die annually ? 8^. 5. A real estate agent receives $35 for selling a farm, his rates of commission being 1% from the buyer, and 1^^ from the seller. What is the price obtained for the farm? . $1,400. 6. A note for $36.50, dated June 27 of last year, was paid April 4 of this year, with interest at 10^^. What was the amount paid? 7. A commission-merchant receives $820 with v/hich to buy goods, after deducting his commission of 2-l-fo. How much does he expend for goods ? $800. 8. I borrow $4,000 on my note in Portland, Me., Feb. 21, and loan the money in Syracuse, N. Y., Feb. 23. If the money is paid to me, Nov. 12, and I pay my note Nov. 13, how much do I gain ? $26.78. 284 PERCENTAGE. 9. Three men own a mill. C's share, which is $2,550, is 60^ of B's, and B's share is 85^ of A's. What is the value of the mill ? $11,800. 10. A merchant paid a premium of $363.12| for a policy of in- surance covering $8,000 on his store, and $13,750 on his goods. What was the rate ? 1^%. 11. If a clergyman's salary is $1,500, and he pays $175 for house rent, what is his income tax ? 13. If the annual rate on a life at 50 years is $4,439 per $100, payable semi-annually, what will be each payment on a policy for $3,500 ? 13. What is the premium, at |^, for insuring a farm-house for $800, a barn for $750, and the hay and grain for $1,200 ? 14. One morning, five canal boats were weighed at the weigh- lock in Utica. The cargo of the first weighed 64 tons, and the cargoes of the others weighed, respectively, 85^, 67|^,^, 130^^, and 56^^ of that amount. What was the total weight of the five cargoes ? 27J^.Jf tons. 15. I buy a draft in Portsmouth, N. H., for $350, payable in Providence, R. I., 60 days after sight, exchange |^ premium. How much does the draft cost me ? $2JfS.62t. 16. In a village school of four departments, 33 pupils are in the first department, 44 in the second, 54 in the third, and 60 in the fourth. What ^ of all the pupils attend each department ? 17. A note for $360, drawing %fo interest, was dated April 10, 1868, and $335 was paid on it, Jan. 19, 1869. What amount was due Nov. 3, 1869 ? 18. A stock jobber bought 100 shares of Pacific R.R. stock, at 117. He sold 55 shares at 104|^, and exchanged the balance at 108 for Ocean Bank stock at 135. He afterward sold the bank-stock at 1441-. Did he gain or lose ? He lost $750.50. 19. A Va. planter took up a note for $843, Oct. 31, 1869, that was dated May 29, 1867. What interest had accrued ? $122.52. 30. A box of soap, marked 60 lb., loses l^fo by drying. What is its actual weight ? 31. If a man buys U. S. 5-30 bonds at 106|-, and sells the gold interest at 137|-, what fc in currency does his investment pay him ? KEVIEW PROBLEMS. 285 22. A fruit dealer bought quinces at $1.60 per bushel. After assorting them, he sold the best at 35^ profit, and the others at 15^ loss. What were his selling prices ? 23. In building a house which cost $1,480, 43^ of the cost was for labor, and the balance for materials. What was the cost of the materials ? $8JiS.60. 24. How shall a merchant mark carpeting that cost him $1.42 per yard, so that he can fall 8^ from the marked price, and still make 25^ ? At $1.68^} per yd. 25. A mechanic buys a city lot for $600, payments $250 cash, and the balance in 1 year without interest. In 6 months he pays the balance, less the discount, at Q^. How much does he pay ? 26. A field of 11 acres yielded 16.5 bushels of wheat to the acre ; the cost of seed and labor was $193.60, and the wheat brought $1.60 per bushel. What fo profit did the crop pay ? 50fc. 27. What is the face of a sight draft that costs $360, exchange being at 1^-% premium ? 28. An Oswego miller buys a draft for $2,500 on Chicago, at |^ discount. He remits the draft to a grain buyer in Chicago, with instructions to invest the proceeds, less his commission of 1^, in wheat. Is the miller's gain on the exchange more or less than the grain buyer's commission ? $6. 29. What is the face of a note due in 2 mo. 15 da., the pro- ceeds of which, discounted at the Firet National Bank of Burling- ton, Vt., are $370.12^ ? $375. 30. A drover paid $4,325 for cattle, and $1,498 for marketing them, and they sold for $6,375 at 60 days. What were his net cash profits, money being worth 8fc'i $Ji,68.12. 31. Memorandum :— Face of mortgage, April 23, 1866, $3,275. Indorsements,— Sept. 4, 1866, $845 ; Feb. 27, 1868, $150 ; Aug. 19, 1868, $75 ; Jan. 7, 1869, $1,250. What was due July 1, 1869, interest at 32. Find the time from which interest should be reckoned on the sum of the debts in the margin, if all of them are paid when the last one is due. Aug. 1. ? i $1,417.92 '. 400 due Mar. 3. 325 „ May 19. 1,000 »» n 25. 625 „ Sept. 4. 1,275 „ Nov. 12. 286 PERCENTAGE. 83. If you borrow $620 for 1 yr., at 8^ interest, and 5 mo. afler* -ward you pay $314, how much will you owe at the end of the year? 34. A's property is assessed at $6,750, and B's at $13,575, and A's tax is $52.65. How much is B's tax ? $105.89. 35. A liquor dealer imported 45 casks of wine, invoiced 36 gal. each, at $1.50 per gal. He paid $1,75 per cask for transportation, a specific duty of $1 per gal., and 25^ ad valorem duty. Deduct- ing the customary allowance of 2^ for leakage, what did the wine cost him ? $4.,69L70. 36. How many shares of bank-stock, at par, can a stock broker buy for $4,522.50, less his brokerage of^fo ? (37) (Dn-e ^eal a/'/ei e/a'/a, Q^/tiome^e >/(> /tay i^nt/ie^uji ^. ^^fnt^e^ Indorsements : June 18, 1869, $125 ; Oct. 25, 1869, $475. How much was due on settlement. Mar. 4, 1870 ? 38. When gold is worth 135, which is the better investment, U. S. 10-40's at 97i, or 5-20's at 107| ? 39. Find the average time for the payment of the following Statement of %. In % with % 113. Perkins ^ OTo., ^^ /&^ . 8. The antecedents of a compound ratio are 5, 73, and 30, and the consequents are 8, 3, and 5. Express the compound ratio. 10,800 : 120 = DO. SIMPLE PROPOIITION. 289 SECTION II. 526. ^roporHo7l is an equality of ratios. 527. Simple Proportion is an equality of two simple ratios. 528. The Sig7i of Proportion is the double colon (: :). It is written between the ratios, and is read " as," or " equals.'' Example. — The ratios 12 : 6 and 8 : 4 being equal, form the proportion 12 : G : : 8 : 4, which is read, " 12 is to 6 as 8 is to 4," or " the ratio of 12 to 6 equals the ratio of 8 to 4." Proportion may also be expressed by the sign of equality. Thus, 12 : 6 : : 8 : 4 may be written 12 : 6 == 8 : 4, or ^§- = f . 529. The Extremes of a proportion are the first and fourth terms ; and 530. The Means are the second and third terms. 531. We may write the proportion 12 : 6 : : 8 : 4 in the fractional form, and reduce the fractional expressions to similar fractions ; ^i^ : : | = 4| = = li ^^ W * = 1^4* ^^^ factors 12 and 4, of the first numerator, are the extremes of the proportion ; and the factors 6 and 8, of the second i^merator, are the means ; and the products of these two sets of factors are equal. Hence, *" 532. Principles of Proportion, I. The product of the extremes equals the product of the means. II. The first term is greater or less than the second, according as the third term is greater or less than the fourth. 533. Ex. 1. What is the fourth term of the proportion 21 : 6 : : 14 : — ? solutiok. Explanation. — 84, the product of the , $600. 10. Three men shipped a cargo of 1,500 barrels of flour to England, A furnishing 700 barrels, B 200 barrels, and C the balance. In a storm 195 barrels were thrown overboard. How should the loss be shared among the owners ? SECTION I. ^BFIJ^ITIOJVS ;dJV:D JVOTdTIOJST, 552$ A !^ool of a number is one of its equal factors. 553i The Square ^oot of a number is one of its two equal factors. , 554. The Ctcde !Eoot of a number is one of its tliree equal factors. Example. — 3 is a root of 9, of 27, and of 81, because 9 = 3 X 3 or 32, 27 =: 3 X 3 X 3 or 33, and 81 = 3 X 3 X 3 X 3 or 3^ 3 is the square root of 9, and the cube root of 27. Notes.— 1. One of the four equal factors of a number is its Fourth Boot, one of the five equal factors is its Fifth Boot, and so on. 2. A number whose square root can be obtained, is a Perfect Square; and one whose cube root can be obtained, is a Perfect Gvhe. 555* l7ivo2uHon is the process of finding any required power of a number. (See 82-87). 556. SJvottcHoji^ or ^xiracHo7i of ^oois, is the process of finding any required root of a number. Note.— Involution and evolution are converse operations. 557. JEJxtracHon of Square ^ooti^ the process of finding one of the two equal factors of a number; and 558. J^xiracHo7i of Citbe !Eoot is the process of finding one of the three equal factors of a number. 559. The Sig7Z of Square ^oot is V, called the Radical Sig7i; and 560. The Sig7i of Czibe !Soot is V. Thus, V/64 is read " The square root of 64 ; " and 1/125 is read " The cube root of 125." See Manual. Note.- A Surd is an indicated root which can not be obtained ; as ^5, y 7. EXTllACTION OF SQUARE ROOT. 301 SECTION II. 561. The least and the greatest number that can be ex- pressed by one figure are 1 and 9 ; by two figures, 10 and 99 ; by three figures, 100 and 999 ; and so on. The squares of these numbers are 1^= 1 10^= 100 100^= 10,000 9^=81 99^=9,801 999'=998,001 and so on. By examining these numbers and their squares, we see that The square of a number expressed by consists of One figure one or two figures ; Two figures three " four " Three " five " six " Four " seven " eight " and so on. That is, I. The square of any number consists of twice as many figures as the number, or one less, II. If a number be separated into periods of two figures each, beginning with ones, its square root will consist of as many figures as there are full and partial periods in the number. 562, If we write any digits, as 2 and 9, successively as ones, tens, hundreds, and so on, and square them, we shall have 2'= 4 20'= 400 200'= 40,000 9'=81 90'=8,100 900^=810,000 By examining these numbers, we see that The square of the ones is in the first period ; " " " tens " second " " « " hundreds " third « and so on. That is. The square of the left-hand figure of a root is wholly in the left-hand period of the number or power. 302 EVOLUTION. 563t If we square any numbers expressed by two figures, as 20 and 25, 60 and 63, 90 and 99, we shall have 20^=400 00^=3,600 90^^=8,100 35^=625 63^=3,969 99'^=9,801 By comparing these roots and their squares, we see that 4 is the greatest square in 6, the hundreds of 625 ; 36 " " " 39, " " 3,969; 81 " " " 98, " " 9,801. That is, The greatest square in the left-hand period of a number is the square of the left-hand figure of the root. 564. "We will now square the number 37, for the purpose of learning of what parts the square is composed. Ex. 37=30 + 7, and 3^=30 + 7 multipHed by 30 + 7. Explanation. — The square ^^^^^^^^ of the ones =49 ; the product 3 0-h 7 = 8 7 of the tens by the ones (7x30) 3 0+ 7 = 3 7 + the product of the ones by 2 10+A9 = 25 9 the tens (30 x 7), or two times 900+210 111 the product of the tens and 9 00+JL20+Jt.9 ~ 1369 ones =420 ; the square of the tens = 900 ; and the sum of these three partial products = 1,369. Hence, The square of a number consisting of two figures, is equal to the square of the tens, plus two times the product of the tens and the ones, plus the square of the ones. 565. Ex.1. What is the FIRST SOLUTION. square root of 1,369 ? 13-6 9 Explanation. — Separ- _l ating the number into Di'^idend. j^6 periods of two figures -^^ each, we find that the square root will consist of two figures. (561, II.) 3 7 Root 6 Trial divisor. 7_ 6 7 Compute divisor. EXTRACTION OF SQUARE ROOT. 303 6EC0ND SOLUTION. 13-6 9 9 U69 Jf69 37 67 Since 9 is the greatest square in 13, the first period, we write 3, its square root, for the first figure of the root (563). Taking 9, the greatest square, from the left-hand period, and annexing 69, the next period, to the remainder, we have 469. This num- ber is made up of two times the product of the tens and ones of the root, plus the square of the ones (564) ; i, e.f of 30 (=3 tens) x 2 x the ones of the root, +the square of the ones. Dividing 469 by the trial divisor, 60 (=2 times 3 tens, or 30), we obtain 7, which we write for the second figure, or ones, of the root. Since 60—2 times 3 tens, and 469=2 times 3 tens x the ones + the square of the ones, we add 7 to the trial divisor, 60, making 67. Then multiplying 67, the complete divisor, by 7, the last figure of the root, we obtain, 1st, 7 times 7 = the square of the ones ; and 2dj 7 times 60 = 2 times 3 tens x 7 ones = 2 times the product of the tens and the ones. The product, 469, is the same as the dividend, and 37 is the square root required. . In the Second Solution we have placed the quotient figure, 7, in the place of the in the trial divisor, thus com- pleting the divisor at once. BAT.TTTTDW. Ex. 2. What num- ber is the square root of 555,025? Explanation. — In extracting the square root of a number, only two periods of figures are considered at once. Therefore, in obtaining any figure of the root, after the first, we regard the figure or figures of the root already found as tens, and the figure sought as ones, and find each succeeding figure in the same manner as we find the second figure of a root consisting of two figures, as will be seen in the Solution. 1st dividend. 2d dividend. 55-5 0-25 49 650 576 7U25 71,25 7 If 5 Root. 1 If. If 1st divisor. Ilf85 2(Z divisor. 304 EVOLUTION. SOLUTION. 2 7,85 Ex. 3. Find the square root of ^olu 748.0225. 7^Jf8.0 2'^. Explanation. — Separating the num- ber into integral and decimal j)eriods, "f | ^ by counting left and right from ones, we proceed as in Ex. 1 and 2, putting ip^o a decimal point before the figure of the 5^8 5^65 root obtained from using the first deci- 2782 5 , . , ^ 27825 mal period. — Ex. 4. Extract the square root of g^^y. Explanation. — Since a frac- solution. tion is squared by squaring ^^ = ^ ^ ^57^ == ^^ each term separately (335), and since evolution is the converse of involution (556, Note), we extract the square root of each term separately. 566. Upon the principles deduced in 561-564, is based the !%tile for J^xtraciion of Square ^oot, I. To determine the number of figures in the root. Separate the numher into periods of tivo figures each, count- ing left and right from ones. n. For the first figure of the root. 1. Find the root of the greatest square in the left-hand period, for the first figure of the root. 2. Subtract this square from the first period; and to the remainder annex the next period, for the first dividend. m. For the second figure of the root. 1. Double the root already found, considered as tens, for the first trial divisor, by which divide the first dividend; and write the result for the second figure of the root, and also in the place of ones in the trial divisor, thus forming the complete 2. Multiply the complete divisor by the second figure of the EXTRACTION OF SQUARE ROOT. 305 root; subtract the product from the Jirst dividend; and to the remainder annex the next period for a new dividend. rV. For tlie succeeding figures of the root. Proceed with the second, and with each succeeding dividend, in the same manner as with the first, until all the periods are used. Notes.— 1. If any dividend is less than tlie divisor, annex a cipher to the root, and also to the divisor, and annex the next period to the dividend, for a new dividend. 2. If there is a remainder after all the periods have been used, i. e., in ex- tracting the square root of a surd, periods of decimal ciphers may he an- nexed, and the work extended to any required degree of exactness. 3. If the right-hand decimal period contains hut one figure, annex a decimal cipher. 4. To extract the square root of a mixed fractional number, first reduce it to a mixed decimal number, or to an improper fraction. PItOBl^:EM8. 1. Extract the square root of 5,476. 74. 2. Find the value of V75.69. 8.7. 3. What is the square root of .0389 ? 4. A square plat of ground contains 87,616 square feet. What is the length of one side ? 5. V881,731 = what number ? 6. Extract the square root of .455625. 7. What is the square root of 50,808,384 ? 7, 128. 8. The area of a square platform is 1,387^ sq. ft. What is the length of one side? ' 37.25 ft. 9. Fmd the value of V.000169. .013. 10. What is the square root of the fraction ff^ ? 11. V||i = what number? U- 12. The entire area of the six faces of a cubic block is 130f sq. in. What is one dimension of the block ? -4f in. 13. Find the square root of 914|^. P/^-. 14. What is the value of Vl5 ? 3.872 -f . 15. Extract the square root of 99. 9.9498 + . 16. Vil27.750734r=what number? 33.582. 306 EVOLUTION. SECTION III. JEXT^ACTiojv oj^ cu:bb ^oot. 567. If we cube an integral unit of each of the first four orders, we have 1«=1 10^=1,000 100^=1,000,000 1,000='=1,000,000,000 Since the cube of 1 is 1, and the cube of 10 is 1,000, the cube of any number between 1 and 10 must be a number between 1 and 1,000; Since the cube of 10 is 1,000, and the cube of 100 is 1,000,000, the cube of any number between 10 and 100 must be a number between 1,000, and 1,000,000; Since the cube of 100 is 1,000,000, and the cube of 1,000 is 1,000,000,000, the cube of any number between 100 and 1,000 must be a number between 1,000,000 and 1,000,000,000; and so on. That is. The cube of a number expressed by • consists of One figure one, two, or three figures ; Two figures four, five, or six " Three " seven, eight or nine " and so on. Hence I. The cube of any number consists of three times as many figures as the number^ or one or two less. II. If a number be separated into periods of three figures eachj beginning with ones, its cube root will consist of as many figures as there are full and partial periods in the number. 568 1 If we write any digits, as 2 and 9, successively as ones, tens, hundreds, and so on, and cube them, we have 2^= 8 20^= 8,000 200^= 8,000,000 9'=729 90^=729,000 900^=729,000,000 Examining these numbers, we see that The cube of the ones is in the first period ; " " tens " second " " " hundreds " third " and so on. Hence, EXTRACTION OF CUBE ROOT. 307 The cube of the left-hand Jigure of a root is ivholly in the left- hand period of the power. 569. If we cube any numbers expressed by two figures, as 20 and 25, 60 and 63, 90 and 99, we shall have 20^= 8,000 60^=316,000 90^=729,000 25^=15,625 63^=313,047 99=^=970,299 Comparing these roots and their cubes, we see that 8 13 the greatest cube in 15, the thousands of 15,625 ; 216 " " 313, " " " 313,047; 729 " " 970, " " " 970,299. Hence, The greatest cube in the left-hand period of a number is the cube of the left-hand Jigure of the root. 570. We will now cube the number 45, for the purpose of seeing of what parts the cube is composed. 45=40 + 5, and 45'=:40 + 5 multiphed by 40 + 5 multiplied by 40 + 5. FIEST SOLUTIOK. SECOND SOLUTION. 40 + 5 = 4^ (4-0 X 5) + 5' = . 225 40^ + (40 X S) = 180 lo"" + 2x{JfO X 5) + 5' = 2025 Jf.0 + 5 = Jf5 {40'' X 5)-\- 2x(40 X 5') + 5' = 10125 4 0^ ■\-2x{JfO'' X 5)+ {40 X 5') = 8100 40' + 3x{40'' X 5)-\- 3x(40 X 5') +5'= 91125 The several parts of the final product, reading from the left, are ls#. The cube of the tens, 6^,000 2d. Three times the square of the tens x the ones, 24,000 Sd. Three times the tens x the square of the ones, 3^000 Uli. The cube of the ones, 125 Thatis, 45^ = 5i,i^5 The cube of a number consisting of two figures, is equal to the cube of the tens, plus three times the square of the tens multiplied 308 EVOLUTION. hy the ones, plus three times the tens multiplied by the square of the ones, plus the cube of the ones. 571. Ex. 1. What is the cube root of 91,125? Explanation. — solution, 4 5 Root. Separating the 91-125 mimber into pe- ^ -^ riods of three fig- i>^idend. 2712 5 nres each, we find ^ ^ i^5 Jf.8 Trial divisor. 600 25 5 Jf.2 5 Complete divisor. that the cube root will consist of two figures (567, II.). Since 64 is the greatest cube in the left-hand period, 91, we write 4, its cube root, for the first figure of the root (569). Taking 64, the gTeatest cube, from the left-hand period^ and annexing 125, the next period, to the remainder, we have 27,125. This number is made up of 3 times the square of the tens x the ones, plus 3 times the tens x the square of the ones, plus the cube of the ones (570) ; i. e., of 3 x 40^* X the ones + 3 x 40 x the square of the ones + the cube of the ones. CaUing the first figure of the root tens, and multiplying its square by 3, we have 4,800 for a trial divisor. Dividing the dividend, 27,125, by the trial divisor, we obtain 5 for the second figure, or ones, of the root. Since 4,800=3 times the square of 4 tens, and 27,125 = 3 times the square of 4 tens x the ones, plus 3 times 4 tens X the square of the ones, plus the cube of the ones, we add to 4,800, the trial divisor, 600 ( = 3x4 tens or 40 X the ones), and also 25, the square of the ones, making 5,425, the complete divisor. Then, multiplying this com- plete divisor by 5, the second figure of the root, we obtain 27,125, which is made up of, 1st, 5 x 5 x 5, or the cube of the ones; M, 3 x 40 x 5 x 5, or 3 times the tens x the square of the ones; and Sd, 3 x 40 x 40 x 5, or 3 times the square of the tens x the c^nes. We have now used all of the given number, and 45 is the cube root required. EXTRACTION OF CUBE ROOT. 309 Ex. 2. Extract the cube root of 9,663,597. Explanation. — Since only two pe- ^°^ riods of figures are ^'^^^-^9 7 considered at once, in obtaining any fig- ^^^^^ ure of the root, after 213 1200 + 60 + 1 1261 132300 + 1890 + 9 134199 the firsWe regard |g|^^^ the ngure or figures —-^ of the root already found as tens, and the figure sought as ones; and proceed in the same manner as in obtaining the second figure. 572. Upon the principles in 567-570 is based the :%zg^e for JEJxtractlon of Cube "Eoot, I. To determine the number of figures in the root. Separate the number into periods of three figures each, counting left and right from ones. n. For the first figure of the root. 1. Find the root of the greatest cube in the left-hand period. 2. Subtract its cube from the period, and to the remainder annex the next period, for a dividend. m. For the second figure of the root. 1. Considering the 7'oot already found as tens, multiply its square by 3, for a tr^al divisor, by which divide the dividend, and write the residtfor the second figure of the root. 2. Add to the trial divisor 3 times the product of the tens and ones of the root already found, and also the square of the ones, for' a complete divisor. 3. Multiply the complete divisor by the last figure of the root ; subtract the product from the dividend ; and to the remainder annex the next period for a new dividend. rV". For each succeeding figure of the root. Consider that part of the root already found as tens, and proceed in the same manner as in finding the second figure. 310 EVOLUTION. Notes.— 1. If any dividend is less than the divisor, annex a cipher to the root; two ciphers to the trial divisor, for a new divisor; and the next period to the dividend, for a new dividend. 2. In extracting the cube root of a surd, periods of decimal ciphers may- be annexed, and the worlc extended to any required degree of exactness. 3. If a right-hand decimal period contains less than three figures, supply the deficiency by annexing a decimal cipher or ciphers. 4. If the given number is a fraction, take the cube root of the numerator and denominator separately ; and if it is a mixed fractional number, first reduce it to an improper fraction, or to a mixed decimal number. 5. Since the trial divisor is less than the true divisor, in obtaining the root figure we must malie allowance for this difierence. See Manual. PJJ O B JL JEJ M S. 1. What is the cube root of 103,823 ; and of 24,389 ? 47; SO, 3. V274.625 = what number ? 6.5. 3. .000729 is the cube of what number ? 4. What is the length of one side of a cubical block that contains 2 cu. ft. 1,457 cu. in. ? 1 ft. 5 in. 5. Find one of the three equal factors of 10,218,313. 217. 6. Vl31,09~6,512 = what number? 7. The length of a square stick of timber, which contains 13|^ cubic feet, is 32 times its width or thickness. What are its dimensions ? Note 6. — If the stick were cut, crosswise, into 33 equal parts, each part would be a cube. 2 4 ft. long, and 9 in. square. 8. In digging a cellar, the length of which was 4 times, and the width 6 times its depth, 192 cubic yards of earth were removed. What were the dimensions of the cellar ? 6, 2^, and 36 ft. 9. Extract the cube root of 187,149.248. 10. Of what number is 118,805,247,296 the cube ? 4,916. 11. In a granary is a bin that holds 270 bushels. Its length is 3 times, and its width If times its depth. What are its dimensions ? 12. Extract the cube root of -g^f^, and ^V- 13. VlGyV^T and V4^^ what numbers ? 2fj, and 1.6. 14. What must be the interior measurement of a side of each of two boxes, one of which will hold a bushel of grain, and the othev a gallon of oil ? 12.907+ in., and 6.135+ in. SECTION I. ■ 3) ^I^IJVI TZOJVS, 573. A Series is a succession of numbers increasing or decreasing, either by a common difference or by a common ratio ; as 3, 7, 11, 15 ; and 2, 6, 18, 54. Note.— The numbers that form a Series are the Terms. The first and last terms are thQ Extre7nes ; and the other terms are the Means. (See 539, 530.) 574. An Ascending Series is one in wHch tbe terms increase in regular order, from the first. 575. A !Descending Series is one in which the terms decrease in regular order, from the first. 576. An oirit?imeHcal Progression is a series whose terms increase or decrease by a common difference ; as 2, 7, 12, 17 ; and 24, 21, 18, 15. 577. A Geometrical Progression is a series whose terms increase or decrease by a common ratio ; as 2, 10, 50, 250 ; and 48, 24, 12, 6. SECTION II. ^ ^ITMM^ETICA. Z Z>Zl O G ZiBS SZOJV. 578. In the ascending arithmetical series 2, 5, 8, 11, 14, the common difference is 3, and the terms are formed as follows : 1st term, 2 ; 2d " 5 = 2 + 3, or 1st term + common difference » 3d " 8=2 + 3 + 3, " " + 2 times com. diff. ; 4th " 11 = 2 + 3 + 3 + 3, " " +3 " " 6th " 14=2+3 + 3 + 3 + 3, " '' +4 " " and the sum of the series is 2 + 5 + 8 + 11 + 14=40. 312 PROGRESSIONS. From this illustration we see, that in any arithmetical series there are five things to be considered : viz., the First Term, the Last Term, the Common Difference, the Number of Terms, and the Sum of the Series. Dividing 40, the sum of the series 2, 5, 8, 11, 14, by 5, the number of terms, we have 8, which is the average of all the terms, or the Average Term ; and adding 2 and 14, the extremes, we have 16, which is two times the average term. 579. From these illustrations we deduce the following Principles of A.rithmeUcal :Pro(/ressio7i, I. Any term in an ascending series is equal to the first term, plus the product of the common difference multiplied by the number of the term less 1. n. Tlie difference of the extremes is equal to the product of the common difference multiplied by the number of terms less 1. in. The sum of the extremes is equal to two times the average term of the series. JPJJ OBZE3IS. 1. The first term of T 'ii SOLUTION. an ascending antli- metical series is 6, 57 -1=56 'times com. diff. is added. the common differ- ^^ x S=168, sum of additions to 1st term. ence is 3, and the 6+ 168= 174, last term. (Seel.) number of terms is ^^^ 57. What is the 6 +{56 x 3)= m, last term. last term ? 2. The first term of , -J. .,, SOLUTION. a descending antn- metical series is 206, 21—1=20, times com. diff. is subtracted. the common differ- 20 xl0=200,sumof subtractions from 1st term. ence is 10, and the 206-200=6, last term. (See II.) number of terms is o>\ 21. What is the 206-{20xl0)=6, last term. last term ? ARITHMETICAL mOGRESSlON, 313 3. The first term is 5, tiie last term is 117, and the nmiiber oftermsislS. What is the common dif- ference ? 4. The extremes are 7 and 95, and the common diflfer- ence is 4. Find the number of terms in the series. 5. The extremes are 3 and 25, and the number of terms is 12. What is the sum of the series ? SOLUTION. 117—5=112, sum of additions to 1st term. 15 — 1= IJf, number of additions. 112-^14=S, common difference. (See II.) Or 117-5 _ ,.„ = 8. common difference. 15-1 SOLUTION. 95—7=88, sum of additions to 1st term. 88^4=22, number " " 22 + 1=23, number of terms. Or 95—7 [-1=23, member of terms. 4 SOLUTION. 3-^25=28,2 times the average term. (See III.) 28^2=14, tJie average term. 14x 12= 168, sum of the series. Or 3 -\-25 X 12=168, sum of the series. 580. Upon the principles and examples in 578, 579, are based the littles fo7' Computations In Arithmetical (Progression, I. To find either extreme. Multiply the common difference by the number of terms less 1 ; and add the product to the less extreme, or subtract it from the greater. n. To find the common difference. Divide the difference of the extremes by the number of terms less 1. m. To find the number of terms. Divide the difference of the extremes by the common differ- ence, and add 1 to the quotient. rV. To find the sum of the series. Multiply one half the sum of the extremes by the number of terms. see ManuaL 14 314 PROGRESSIONS. l^Ji OBZJ^MS, 6. The less extreme of an arithmetical series is 5, the common diflference is 7, and the number of terms is 13. What is the greater extreme ? 89. 7. Find the greater extreme of the progression of which 19 is the less extreme, 3 is the common difference, and 57 is the number of terms. 8. A boy 14 years old was apprenticed to a trade, and was to re- ceive $50 the first year, and an increase of $75 yearly, till he was of age. How much did he receive the last year ? $500. 9. The greater extreme of an arithmetical series is 215, the com- mon difference is 13, and the number of terms is 15. What is the less extreme ? 33. 10. A man who owns a plot of 18 building lots, asks $1,000 for the one nearest the city, and $20 less for each succeeding lot. What is his price for the lot farthest from the city ? 11. The extremes of a series of 60 terms are 13 and 249. What is the common difference ? ^. 12. If a laborer has $16 deposited in a savings-bank on Jan. 1, and $484 on Dec. 30 following, what are his average weekly deposits ? 13. The extremes are 4| and 67f, and the common difference is If. What is the number of terms ? ^6. 14. In how many years will the value of a piece of property be doubled, if it increases in value 16;^ the first year, and 7fo each suc- ceeding year ? 15. What is the sum of the natural series of numbers 1, 2, 3, 4, and so on to 1,000 inclusive ? 500,500. 16. What is the number of strokes made by a clock in 12 hours ? 17. The less extreme is |^, the common difference is |^, and the number of terms is 50. What is the greater extreme ? 18. If you deposit $25 in a savings-bank the first week of the year, and $5 each succeeding week, how much will you deposit in the year ? $280. 19. What is the 84th term of the series 90f , 90, 89^, etc. ? 30. Insert 32 arithmetical means between the extremes 13 and 244. 1st mmn, 20; S2d mean^ 237. GEOMETKICAL PROGRESSION. 315 21. If the water in a lake is 16,} feet deep 1 rod from a pier, and the bottom has a uniform slope of S^- feet to the rod, at wliat distance from the pier is the water 300 feet deep ? 82 rods. 32. Find the sum of 100 terms of the series 19^'^, 19|-, 20tV, etc. SECTION III. GJSOMBT'RIC;^!, 'P ^O G'B ^EJS S 10 J^, 581. In the ascending geometrical series % 6, 18, 54, 162, the ratio is 3, and the terms are formed as follows : 1st term, 2 ; 2d " 6=2x3, or 1st term x ratio, 2 X 3'; 3d " 18 = 2x3x3, " " X square of ratio, 2 X 3^; 4th " 54=2x3x3x3, " " x cube "" 2 X 3^; 5th " 162=2x3x3x3x3,'- " x 4th power, " and the sum of the series is 2 + 6 + 18 + 54 + 162=242 2 X 3*; From this illustration we see that, in any geometrical series there are five things to be considered : viz., the Fir^t Term, the La%t Term, the Ratio, the Number of Terms, and the Sum of the Series. If we take the above series, 2, 6, 18, 54, 162, multiply it by the ratio, 3, placing the terms of the products over the corresponding numbers of the series, and then subtract the series from the product (or 3 times the series), the terms consisting of like numbers will disappear, and we shall have 3 times the series, 6 18 54 162 486 Series, 2 6 18 54 162 (8 times— 1 time=)2 times the series=486 — 2=484 Since 484 is 2 times the series, 484 +-2 ==242 is 1 time the series, or the sum of the series. Note.— The pupil will notice that 2 is the first term of the scries, 4S6 b 8 times the last term, and the divisor, 3, is the ratio less 1. 316 PROGRESSIONS. 582. From these illustrations we deduce the following Principles of Geometrical Progress ion. I. The first term and the ratio are the only factors used in forming a series. II. In any term of a series, the first term is a factor once. m. In any term of an ascending series, the ratio is a factor as many times as the number of terms less 1. rV. The number of factors used informing any term, is equal to the number of the term. V. The product of the ratio and the greater extreme of a series, ^ninus the less extreme, is as many times the sum of the series, as is expressed by the ratio less 1. 1. The first term of an ascending geo- metrical series is 4, the ratio is 3, and the number of terms is 7. What is the last term? 2. The first term of a descending geo- metrical series is 96, the ratio is 2, and the number of terms is 6. What is the last term ? 3. The first term of a geometrical pro- gression is 7, the last term is 567, and the number of terms is 5. What is the ratio ? JPJiOBI^JE3IS, SOLXTTION. 7—1=6, times the ratio is a factor. (See in.) ^^=64, product of ratio used as a factor. 4 X 64=256, last term. (See I.) Or 4 X S''-^ =4 X 2^' ="256. last term. 60LUTI0X. 6—1=5, times tlie ratio is a dimsar. (See III.) 2^=32, product of ratio used as a divisor. 96^32=3, last term. Or ^6 -^2'-' =96 -^2" =3, last term. 567-7-7=81, product of ratio used as a factor. 5—1=4, times ratio is a factor. (See III.) V5J =3, ratio. See Manual, Reference 310. Or ^'1/567 W567 ^ ,. GEOMETRICAL PROGRESSION. 317 4. The extremes of solution. a geometrical pro- 1536-^6=256^ prod, of ratio used as a factor. gression are 6 and 256-^4=64, 64^4=16, 16-^4=4, 4-^4=1. 1 536 and the ratio -^ ('^^'^ times the ratio is a factor) + 1 (the time is 4 What is the ^^^^ ^^^^ extreme is a factor) =5^ number of number of terms ? ^^^^^«- (See IV.) SOLUTION'. 5. The first term ^^^ ^ 5=1875, 5 times the last term. IS 3, the last term is ^sjj^_s^ ^g^^, 4 times the series. (See V.) 375, and the ratio is ^s72-^4^ 468, sum of the series. 5. What IS the sum ' q^ of the series ? ?11^^ =468, sum of the series. 583t Upon the principles and examples in 581, 582, are based the (Rules for Computations in Geometrical "Progress ion, I. To find the greater extreme. Raise the ratio to a power 1 less than the number of terms, and multiply the result by the less extreme. n. To find the less extreme. . Baise the ratio to a power 1 less than the number of terms, and divide the greater extreme by this result. III. To find the ratio. Divide the greater extreme by the less, and extract that root of the quotient whose index is 1 less than the number of terms. TV. To find the number of terms. Divide the greater extreme by the less, and this and each suc- ceeding result by the ratio, till the quotient is 1. The number of divisions will be the number of terms. V. To find the sum of the series. Multiply the last term by the ratio, from the product subtract ih? first term, and divide the remainder by the ratio less 1. See ManuaL 318 PBOGKESSIONS. 6. The first term of an ascending geometrical series is 7, and the ratio is 3. What is the 6th term ? 1,701. 7. The less extreme is 13, the ratio is 4, and the number of terms is 7. What is the greater extreme ? 8. The 5th term of an ascending series is 5,625, and the ratio is 5. What is the iirst term ? p. 9. The greater extreme is 845,824, the ratio is 2, and the number of terms is 12. What is the less extreme ? 10. The extremes of a progression of 5 terms are 4 and 64. What is tlie ratio ? 2. 11. The four terms of a proportion are in geometrical progression, and the extremes are 8 and 2,744. What is the proportion ? Note 1.— First find the ratio. 8 : 56 : : S92 : 2744. 12. The extremes of a series are 3 and 234,375, and the ratio is 5. How many terms are there in the series ? 8. 13. The extremes are 1 and y^, and the ratio is -^. Find the number of terms. 14. The extremes of a series are 2 and 4,374, and the ratio is 3. What is the sum of the series ? . 6^560. 15. What is the sum of 9 terms of the series 2, 10, 50, etc. ? Note 2. — The greater extreme must first be found. 976 j 562. 16. What debt can be discharged in a year, by paying 1 cent the first month, 3 cents the second, 9 cents the third, and so on, in that ratio, for the 12 months? $2,657.20. 17. If it were possible for a person having only 1 cent, ip double his money every month for 4 years, how much money would he have ? $2,814,749,767,106.56. 18. What is the sum of the series 18 + 6 + 2 + f + f, and so on, to infinity ? Note 3.— When the number of terms in a descending geometrical series is infinite, the series is called an Infinite Series, and the last term is 0. 3-1 — "^ 19. Find the sum of the infinite series 100 + 25 + 61 + 1^? etc. 133^. 20. What is the sum of the series 1, |, -|-, \, . . . to ? INTEEEST BY P HOG RES SIGNS, 319 SECTION IV. IjYTBOiBS 2' :sr T^OG^BSSIOJSrS. 584. In Simple Interest, the amount of any sum is equal to the lorincipal, plus the product of the interest for 1 year multipHed by the time in years (see 461) ; the principal and the amount due at the close of the first year are both taken into account ; and the amounts due at the end of the several years form an arithmetical series. Hence, in computations of Shnple J?ite7'est by A.rlthnietlcal Progression, I. Tlie principal is the less extreme of an arithmetical series ; H. TJie interest for 1 year is the common difference ; in. TJie number of years plus 1 is the number of terms ; and lY. TJie amount is the greater extreme. 585. In Compound Interest, the amount of any sum^ for any year, is found by multiplying the amount due at the end of the preceding year by 1 plus the rate % (see 4G2); and the amounts due at the end of the several successive years form a geometrical series. Hence, in computations of Cot7?pound Interest by Geojnetrical Progression, I. l%e principal is the less extreme of a geometrical senes.; II. 1 plus the rate is the ratio ; III. The number of years plus 1 is the number of terms ; and IV. Tlie amount is the greater extreme. PROBLEMS. 1. What is the amount of $500 for 8 years, at 7^ ? at Qfo 'i 2. I have a lease of a building for 9 years, at $50 a year. If I allow Qfo interest on the rents from the time they are due, and pay the whole amount at the expiration of the lease, how much must I then pay ? Note 1. — As there is no principal at the commencement of the first year, there are only 9 terms. $558. 3. If I save $150 each year, and put it at interest at 10^, how much will my savings amount to m 10 yoars? $2,175. 320 PllOGRESSIONS. 4. If a soldier's pension of $100 per annum remains unpaid for 9 years, how much will then be due him, allowing Gfo simple interest ? Note 2.— Since the first year's pension lias been clue 9 years, and the last year's pension is now due, there are 10 terms in the series. $1,270. 5. What is the present worth of an annuity, or annual income, of $300 having 6 years to run, money being worth Sfo ? Note 3. — The amount of the annuity due at the end of the years, is the sum to be discounted, at ^%. 6. I hired a mill for 5 years, at an annual rent of $600, and paid the rent in advance, less the discount of 6^. How much did I pay ? 7. After three years, I shall come into possession of property that pays $400 annually. How much ready money can I borrow, by hy- pothecating 5 years of this income, and allowing 10^ for the loan ? Note 4. — The sum due in 5 years after the income commences, or 8 j^ears from the present time, is the amount to be discounted for 8 years, at 10^. 8. What is the present worth of an annuity 4 years in reversion (^. e. to commence in 4 years), and then having 7 years to run, money being worth 8^ ? 9. What is the present worth of a perpetual annuity of $1,500, to commence 7 years hence, if discounted at 6^ ? Note 5. — The annuity is the interest of a principal that will earn $1,500, at &%, 10. How much will $10,000 amount to in 5 years, at 6^ compound interest ? $13,382.26. 11. What is the compound interest of $425 for 5 years, at 7^^ ? 13. If you dci3osit$500 in a savings-bank that pays 5'^ on deposits, compound interest payable quarterly, how much will your money amount to in 3 years ? 13. What sum of money, at 6^ compound interest, will amount to $89.54 in 4 years ? $70.92. 14. The amount is $33,153.83, the time is 8 years, and the rate is 10^ compound interest. What is the principal ? $15,000. 15. Find the amount of an annuity of $185 for 4 years, at 7^ compound interest. $821.39. 16. A clerk deposited $75 in a savings-bank every 6 months, upon which he received Qfo interest compounded semi-annually. How much was standing to his credit at the end of 4 years ? 17. How much is an annuity of $1,300 per annum worth in 10 years, at 5^ compound interest ? $15,093.^7. SECTION I. 586. Mensieratlon embraces the processes of measuring and computing tlie length of lines, the area of surfaces, and the capacity of solids and spaces. (See Chap. 2, Sec. VIII.) 587. A Curve Z,i7ie continually changes its direction, no part of it being a straight line. 688. ^a7'aUel Zlnes run in the same direction, — at the same perpendicular distance apart. ^-^ — -^^ 589. An A.ciete ^7if/le is less than a right /^ \ angle; as ABG. 590. An Obtuse A.7igle is greater than a right angle ; as ABB. Note. — Acute and obtuse angles are Oblique Angles. 591. A "Plaiie Fiffm^e^ or a ^lane, is a level surface bounded by lines. 592. A ^otys^on is a plane bounded by straight lines. 593. A ^cffiUai'' Toty^on has all its sides, and also all its angles equal. 594. A 2rla77ffle is a polygon of three sides. 595. A mght-A.7iffled 2ria7iffle has one right angle ; as ABC. 596. A JIypothe7itise is the longest side of a right-angled triangle ; as ^ C. Note. — A triangle having three acute angles is an Acute- Angled Triangle ; one having one obtuse angle is an Obtuse-Angled Triangle ; one having all its sides equal is an Equilateral Triangle ; one having two sides equal is an Isosceles Triangle ; and one having all its sides unequal is a Scalene Triangle, 597. A Quadrilateral is a poly- ^''^ ^^''""'^^• gon of four sides. 598. A ^arallelosrram is a quadrilateral whose opposite sides are parallel, .ind consequently equal. 322 MENSURATION. Note.— If all the sides of an oblique-angled parallelogram are equal, it is a JRhombus, or Bhomb ; if only the opposite sides are equal, it is a Bhoirv- bold. 599. A Trapezoid is a quadrilateral having only two sides parallel. 600. A Trapezium is a quadrilateral having / ^~\ neither two of its sides parallel. / \ 601. A ^iaffonal is a straight line joining L A two oj)posite angles of a figure. Notes. — 1. A regular polygon of five sides is a Perv- tagon; one of six sides is a Hexagon; one of seven sides is a Heptagon ; one of eight sides is an Octagon ; one of nine sides is a Nbnagon ; and one of ten sides is a Decagon. 3. Any polygon of more than three sides may be divided, by diagonals all meeting at one angle, into as many triangles less 2 as the polygon has sides. 3. The total length of the sides of a polygon is its Perimeter ; and the length of the circumference of a circle is its Periphery. 602. The Sase of a figure is the side on which it is supposed to stand; 603. The Vertex is the point opposite, and furthest from the base ; and 604. The A.Uitude is the perpendicular height of \)aQi vertex above the base. 605. A l*rism is a solid whose bases or ends are equal, parallel polygons, and whose sides are parallelograms. 608. A Cylinder is a solid whose bases or ends are equal, par- allel circles. 607. A ^j/7'amid is a solid whose base is a polygon, and whose sides are triangles, terminating in a point or vertex. 608. A Cone is a solid whose base is a circle, and whose top is a point or vertex. MENSURATION OF LINES. 323 609. A Sp?ie7'e, or a Globe^ is a solid bouuded by one surface, whicli, in every jpart, is equally distant from a point within, called its center. Note.— One half of a sphere is a Hemisphere. 610. Similar Surfaces have their several angles equal each to each, and their sides about the equal angles proportional. 611. Simita?^ Solids are contained by the same number of similar surfaces, similarly situated. SECTION II. MJBJJV-SZra^ATIOJV OF ZIJVBS. 612. Some of the principles of mensu- ration can only be proved by a Geometri- cal analysis. Thus, this diagram illus- trates the first of the following Geomet- rical Principles, but the illustration is not an analysis of the principle. Geometrical ^rinci^les. I. The square of the hypoihenuse of a right- angled ti'iangle is equal to the sum of the squares of the other two sides. II. The diameter of a circle : the circumference : : 113 : 355. Note.— By II. we find that, if the diameter of a circle is 1, the circumfer- ence is 3.14159, nearly. For ordinary purposes it is sufficiently accurate to call the circumference of a circle, 3f times the diameter. PMOBJOEMS. 1. Robert lives 117 rods north, and the school-house is 156 rods east, from the corners. What is the distance across the fields from Robert's house to the school-house ? 195 rd. 2. What is the length of a hand-rail to a flight of 16 stairs, each 12 inches wide and 9 inches high? 20 ft. 3. My house is 24 ft. wide, the ridge is 9 ft. liigher than the side walls, and the eaves project 1 ft. 6 in. beyond the sides of the house. How wide is each side of the roof? 16 ft. 6 in. 4. A ladder 39 ft. long reaches to the top of a building, when its foot stands 15 ft. from the building. How high is the building ? 324 ENSURATION 5. From the top of a certain building, 36 feet liiffli, to the opposite Bide of the street, is 164 feet. How wide is the street ? 160 ft. C. Two streets, one 48 and the other 64 feet wide, cross at right angles. What is the distance between the diagonal corners ? 7. What is the side of a square whose diagonal is 50 feet? 8. Round a cylinder 5 ft. 10 in. high and 1 ft. in circumference, a string is wound spirally from bottom to top, passing 14 times round. How long is the string ? 9. The slant lieight of a cone is 21.8 inches, and the diameter of the base is 2.64 inches. How high is the cone ? 21.76 in. 10. A pole 45 ft. high is supported by three guys attached to the top, and reaching the ground at the distances of 60 ft., 108 ft., and 200 ft. from the foot of the pole. What are the lengths of the rods ? 75 ft.; 117 ft.; 205ft. 11. What is the circumference of a circle 8 feet in diameter? 13. What is the length of the tire on a carriage wheel 5 feet in diameter ? 13. What is the circumference of a lake 721 rods in diameter ? 14 What is the girth of an oak log which is 33 inches through? 15. Find the diameter of a circle which is 33 rods in circumference. 16. In a park is a fountain whose basin is 3ch. 201. in circumfer- ence. What is the diameter of the basin ? 17. The extreme end of the minute-hand of a town clock moves for- ward 19 inches in 12 minutes. How long is the minute-hand ? SECTION III. MBJVSU^A.TTOJ^ 01^ S 17 (R JFA C JS S, 613. By examining this diagram, we see Ut. The diagonal AG divides the parallel- ogram into two equal parts ; and conse- quently, the area of the triangle ABG is equal to one half the area of the parallelo- gram ABGD, or to \oiAB^ BG. (See 188). 2d. The areas of the triangles AFE and BFE are equal to one half the areas of the parallelograms A FED and BFEG^ respectively ; consequently, the MENSURATION OF SURFACES. 325 area of the whole triangle ABE is equal to one half the area of the parallelogram AJBCB, or to the area of the triangle ABO. Hence I. The area of a triangle is equal to one Jialf the areu of a parallelo- gram liaving the same tase and altitude ; or II. To one half tlie product of its lase multiplied ly its altitude. 614. If in the rhombus ABGD, the line DE be drawn perpendicular to the base, and the part AED be placed on the opposite side, the line AD on BG, the figure EFGD will be a square. By the same process, the rhomboid will be re- ^ ^ B F duced to a rectangle. Hence The area of any parallelogram is equal to the 'product of the hase multiplied hj the altitude. 615. If the right-angled triangles Aae and Bha be applied to the spaces Dde and Gce^ respec- tively, the figure abed will be a rectangle, equal in area to the trapezoid ABGB, because the A a b B side DC will be increased as much as the side J.-B is diminished ; AT} I fn AB + GD will equal al + cd; and ah or cd will equal ^ -. Hence 616. The area of a trapezoid is eqwil to the product of one half the sum of its parallel sides multiplied Tyy its altitude. 617. From what has been said in 613, 614, it is evident that I. The area of any polygon is equal to the sum of the areas of any set of ti'iangles into which it may he divided ; and II. The area of a regular polygon is equal to one half the product of the periphery multiplied lyy the altitude of one of its equal triangles. 618. Geo77iet7^lcat 'Prhiclples, I. The area of a circle is equal to the product of one half the circunrv- ference and one half the diameter ; or II. To one fourth the product of the circumference and diameter ; or ni. To the product of the square of the diameter multiplied lyy .755^. IV. The surface of a sphere or globe is equal to 4 times the area of a circle of the same diameter, V. The areas of similar figures a/re to each other as are the squares of any one of their similar dimensions. See Manunl ^6 MENSUPvATION. PItOB LEMS. 1. The base of a riglit-angled triangle is 12 inclies, and the perpen- dicular is 8 inches. What is the area ? JfS sq. in. 2. How many feet of boards will it take to cover the gable of a barn 33 feet wide, the ridge being 8 feet above the plates ? 3. The base of a triangle is 8 ft. 1 in., and its area is 2,861^ sq. in. What is its altitude ? J^ ft. 11 in. 4. The base of a rhomboid is 223 feet, and its altitude is 96 feet. What is its area? 5. Find the area of a trapezoid whose sides are 9 and 17 inches long, and 13 inches apart, 1 sq.ft. 25 sq. in. 6. How much lumber in an inch board 12 ft, long, 16 in. wide at each end, and 8 in. wide in the middle ? 12 sq. ft. 7. What is the area of a circle 20 feet in diameter ? 8. My horse is tied to a stake in the pasture, by a rope 11 ft. long. On how much land can he graze ? 380.1336 sq. ft. 9. The area of the bottom of a tin pan is 196 sq. in. What is its diameter? 15.79 in. 10. How many square inches of map surface on a 15-inch school globe? 706.S6. 11. The slant height of a pyramid is 11 inches, and the base is 4 inches square. How many square inches on the entire surface ? 12. The periphery of the base of a cone is 40 inches, and the slant height is 38 inches. How many square inches are there on the lateral or convex surface ? Note. — The cone may be regarded as a pyramid of an infinite number of sides, and the periphery of its base as the sum of the bases of all the tri- angles which form its convex surface. 5 sq.ft. 40 sq. in. 13. What is the surface of a prism 18 ft. long and 21 in. square ? 14. What is the surface of a round pillar 14 inches in diameter and 80 feet long? 15. How many feet of inch lumber in a box 6 ft. 6 in. long, 4 ft. 2 in. wide, and 3 ft. 2 in. deep, inside measurement ? 16. Two men start from the same place, at the same time. One of them travels south, at the rate of 3 miles an hour, and the other west at the rate of 4 miles an hour, for 7 hours. They then travel directly towards each other, at the rate of 8^ miles an hour, till they meet. How many hours do they travel, and how many square miles do they travel round ? MENSURATION OF CAPACITIES. 327 SECTION IV. MJSJSrSU^A.TIOJV' OJP" CA1>ACITIBS. 619. The capacity of a prism or a cylinder ^r 7\ 4, 5, or 6 feet in length, is 4, 5, or 6 times as >^B| , ji^^^ much as 1 foot in length of the same prism m| | ^ ^ ^^jj;^^ or cylinder. (See 191.) Hence, ^' m«...JlllLl i i n Tlie capacity of a prism or cylinder is equal /f!iii\^^===z:^.:,:^L=A to the product of the area of its tase multiplied P ^ ' ^^ " Sj^^ ly its length. \ ^^^^ B=— -^ ^ Note.— Lumber 1 incli thick or less is sold by surflicc measure. If more than 1 inch thick, it is computed at this thickness ; i. e., the product of the surface measure in square feet multiplied by the thickness in inches, is the number of feet of lumber of standard thickness. 620. Geo77ietricat 'Prhiciples, I. The solidity of a pyramid is -j- that of a prism^ and the solidity of a cone g- that of a cylinder^ having the same 'base and altitude. II. The solidity of a sphere is § that of a cylinder whose diameter and altitude are^ each^ equal to the diameter of the sphere. III. T7ie capacities of similar solids are to each other as are the cubes of any one of their similar dimensio7is. See Manual. PROBLEMS. 1. The ends of a prism 20 feet long are right-angled triangles, the two shorter sides of each of which measure 16 and 30 inches. Find the cubic contents of the prism. 3. How many feet of timber in a log 31 feet long and 17^ inches in diameter? 51.78 + . 8. I have a cylindrical cistern 6 feet in diameter and -8 feet deep. What is its capacity in hogsheads ? 26hhd. 54-05+ gal. 4. The area of the base of an octagonal pyramid is 78 sq. ft., and its altitude is 19 ft. 6 in. Wliat are its cubic contents ? 5. What are the cubic contents of a cone 7 ft. in diameter at the base, and 16 ft. 9 in. high ? 6. Find the solidity of a 13-inch school globe. 7. The capacity of a hollow globe of glass is 65.45 cubic inches. What is its diameter? 328 MENSURATION. 8. A leaden ball 1 inch in diameter weighs -^^ lb. How much, does a leaden ball 5 inches in diameter weigh ? 26-^^ j- lb. 9. A cast-iron ball 4 inches in diameter weighs 9 lb. What is the weight of a cast-iron ball 7 inches in diameter ? ^8^3, II), 10. A marble monument consists of a pedestal 18 inches square and 3 feet high, on which stands a pyramid 16 inches square and 7 feet high. What did it cost, at $16.25 per cubic foot ? $177.09. 11. A log chain and 3| quarts of water fill a cubical box whose inside edge measures 8 inches. How many cubic inches are in the chain ? 12. In a stick of timber 50 feet long, and 7 x 10 inches, there are how many feet, timber measure ? How many feet, board measure ? 13. Find the contents, in timber measure and in board measure, of a stick of timber 18 ft. long, 12 in. wide, 15 in. thick at one end, and 10 in. thick at the other. 14. Find the side of the largest square stick of timber that can bo cut from a log 2 feet in diameter. 17 indies^ nearly. 15. How much will the flooring for a two-story house 24 x 32 feet cost, at $40 per M., the flooring being 1| inches thick ? $76.80. 16. How much lumber in a stock of 9 boards 13 ft. long, 9 in. wide, and 1^ in. thick ? ^^^ft- 17. I wish to have built in my cellar, a cistern that shall hold 20 lihd., and to have it a cylinder 5 ft. 10 in. in diameter. How deep ^vill it be? 6 ft. 3.63 + in. 18. How much 2-inch plank must I bay for a 5-foot walk on the street sides of a corner lot 4x8 rods, the walk to be placed 2 ft. 6 in. from the fence ? And how much will it cost me, at $16 per M. ? 19. In a granary is a bin 12| ft. long, 8 ft. 7 in. wide, and 5.4 ft. deep. How many bushels of grain will it hold ? 20. What is the capacity, in hogsheads, of a rectangular cistern 12 ft. long, 8 ft. wide, and 6 ft. 4 in. deep ? 21. A laborer built a wall 5 rd. long, 5 ft. thick at the bottom, 2 ft. thick at the top, and 5 ft. high, in 2 days, building 2 ft. in height the first day. On which day did he lay the most wall ? 22. If a trqugli 5 feet long holds 12 pailfuls of water, how many pailf uls will a similar box hold that is 8 feet long. 23. If a pint of wine will fill 15 cone-shaped wine-glasses, how many times will a gallon of wine fill a similar glass of 1 } times the diameter at the top ? 35'1, CHAPTER 12. MISCELLANEOUS PROBLEMS. S^^Tff^ 1. The quotient is 436fH, and the divisor is 735. What is the dividend ? 2. A cubic foot of water weighs G2 lb. 8 oz. What is the pressure on 1 acre at the bottom of the sea, where the water is 1,000 fathoms deep? 3. Find the prime factors of .5313. 4. The expenses of an excursion party, consisting of 8 gentlemen and 9 ladies, were $3.40 a piece; which were paid by the gentlemen. How much did each pay ? 5. $.725 is what fraction of $1 ? G. How many feet in .735 of a mile ? 7. If the cost of manufacturing kip boots is $4.60 a pair, and they are sold at 25^ profit, what is the selling price ? 8. If 33 i- lb. of tea cost $29^-, how much will 12^^ lb. cost ? 9. How much lumber in a stock of 12 boards 14 ft. long and 10 in. wide? 10. A grocer bought 7 doz, brooms © $2.25, and retailed them at $".31^ apiece. How much did he gain on the lot ? 11. One week, 2,230 barrels of flour, which cost $9.25 per barrel, were received at the port of Cleveland, and it was sold at the rate of $3.15 per sack of 49 pounds. What was the gain ? 12. How many pickets each 3 inches wide, placed 3 inches apart, will it take for a fence round a lot 4 x 10 rods ? 13. What is the shortest distance that is an exact number of times a 1-ft. rule, a 2-ft. rule, a yard-stick, and a 10-ft. pole? 14. At $3.25 per C, how many broom handles can I buy for $26.52? 15. After 4,^ of a flock of sheep had been killed by dogs, and 68 had been sold to a butcher, y of the original flock were left. How many sheep were in the flock at first ? 16. If 21| bushels of oats are required to seed 9| acres, how many bushels will be required to seed a field of 17.2 acres ? 17. IIow many sheets of tin, each 14 x 22 in., will it take to cover a roof, each side of which is 80 ft. long and 18 ft, 4 in. wide ? 330 MISCELLANEOUS PROBLEMS. 18. A printer's price for "business cards is $3.75 for tlie first liundred, and $1.25 per liundred for any number after the first liundred. How much will 1,500 cards cost? 19. A man, dying, leaves an estate of $53,1G6, but it is incumbered to the amount of $17,496. His widow receives ^ of what remains after paying the incumbrances, and the balance is divided equally among 5 children. What is the widow's share ? How much does each child receive ? 20. A fruit dealer pays $4.25 per bushel for 3 bushels of chestnuts, and sells them at $.10 a pint, tin measure. What are his profits ? 21. What fraction equals .00096 ? 22. A grain buyer paid $1.40 per bushel for 2,380 bushels of wheat, and $.50 per bushel for transportation to New York, where he sold it at a loss of 20;^. How much did he lose ? 23. If 25 cows average 9 quarts of milk each, per day, through the year, and it is sold at an average of 7 cents per quart, and the expenses of keeping and labor are $78 per head, what are the annual profits ? 24. What will be the face of a sight draft on New York, that costs $664 in Louisville, Ky., exchange on New York being at 3|,^ premium ? 25. I invested one half of my capital in bank-stock, and the balance in R.R. stock. I gained 11^ on the bank-stock, and lost 7i% on the R.R. stock, and my net gain was $175. How much was my capital ? 26. If I buy 1,600 bushels of oats in Iowa, at a net cost of $.56 per bushel, transportation included, and sell them in New York @ $.54, how much do I gain ? 27. A grocer paid $22.75 for a barrel of mess pork, and retailed | of it at $.12^ per lb., and the balance @ $.14. What were his profits? 28. Portland, Me., is in latitude 43° 39' north, and L. Titicaca, on the same meridian, is in latitude 16° 42' south. How many miles, air line distance, from Portland to L. Titicaca ? 29. A note at 8 mo. for $750, with interest, was discounted by a Boston bank, 3 months after date. What were the proceeds ? 30. If three men can build a sidewalk 240 ft. long and 6 ft. wide, in 15 days, in how many days can 5 men build a walk 180 ft. long and 4 ft. wide ? 31. A mechanic contracted to work a year for $40 per month, his wages payable at the end of each month. Nothing was paid him till the close of the year, when he received the whole amount, with 8,^ intorest. How much did he receive ? MISCELLANEOUS PROBLEMS. 331 32. A tax of $3,156 is levied on a Union school -district, whose assessed valuation is $493,125. What is the rate ? 33. In the above-mentioned district A's property is assessed at $750, B's at $3,850, C's at $1,600, and D's at $14,500. How much is each one of them taxed ? 34. Memorandum : — Nov. 19, 1836, gave a note for $1,650, with 6^ interest. June 18, 1868, paid $125 ; Oct. 25, 1868, paid $475. March 4, 1869, took up the note. How much was the last payment ? 35. How many squares of flooring in a floor 44 x 75 ft. ? 36. In the right wing of an army there were 18,675 men, in the center 23,518, and in the left wing 11,498. The left wing was re-in- forced by 16,488 new troops, and the center by 3,486. The command- ing general then ordered 9,894 men from the left wing to the right, and 5,145 from the right wing to the center. How many troops were then in each division of the army ? 37. A pile of 4-foot wood 244 feet long and 5 feet high, was sold for $152.50. What was the price per cord ? 38. What is the commercial weight of a nugget of gold that weighs 3 oz. 3 pvrt. 19i gr. ? 39. A merchant sold broadcloth at 5^ less than the marked price, and yet made a profit of 25,^. At what % advance on cost were the goods marked ? 40. A miller pays $1.45 per bu. for 225 bu. of wheat. If 4.5 bu. make 1 bar. of flour, and he sells the ship-stufis for $54.75, at what price per bar. must he sell the flour, to realize a net profit of $104.50 by the transaction ? 41. A lOO-acre farm is a trapezoid in shape, the shorter of the two parallel sides being 64.7 rods, and the longer 135.3 rods. How far apart are these two sides ? 42. How many panes in each of three boxes of glass, marked, respec- tively, 8 X 10, 9 X 16, and 10 x 18 ? 43. For what amount must I draw my note, payable in 60 days, to obtain a discount of $250 from a Philadelphia bank ? 44. A teacher receives a salary of $1,050, and 6^ of liis expenses equals 20^^ of his savings. How much does he save yearly ? 45. If the transportation of 51.2 tons of freight costs $268.80, how much should be paid for the transportation of 32 J tons ? 46. I paid l{fc for an insurance of $1,075 on a building worth $l,500l If the building should burn, what would be my loss ? 332 MISCELLANEOUS PKOBLEMS. 47. A field, wliicli is 3^ times as long as it is wide, contains 22.4 acres. What arc its dimensions ? 48. A block of marble contains 54H cubic feet, and tlie leno-tli, breadth, and thickness are as 7, 4, and 1. What are the dimensions ? 49. I mi. + 71 rd. + ^ yd. = what distance ? 50. A note for $35G, dated Mar. 10, at 10 mo,, with interest at 7^-, was discounted at the American Exchange Bank of New York City, Aug. 35. What were the proceeds ? 51. A drover bought 135 head of cattle @ $23, and 147 head @ $19, and shipped them to New York, at a cost of $1,597. He sold 163 head @ $37, and the balance @ $31. Did he gain or lose, on the whole drove, and how much ? 52. The latitude of Chicago is 41° 54' N., and Mobile is 706A¥o miles S. of Chicago. In what latitude is Mobile ? 53. I sold a horse for ~g more than he cost me, receiving $216 for him. How much did he cost me ? 54. A cannon-ball 15 inches in diameter weighs 456 pounds. What is the diameter of a 260-pound shot ? 55. A mechanic having $852.75, paid f- of his money for a half-acre lot of land. How much would an acre cost, at the same rate ? 56. A sixty-day note for $237.40, dated Poughkeepsie, N. Y., June 21, was discounted at the Second National Bank of Troy, June 28. How much money was received ? 57. How far is it from one of the lower corners to the diagonal upper corner of a room 20 ft. long, 16 ft. wide, and 12 ft. high ? 58. When it is 9 o'clock, A. M., at Cincinnati, 84° 24' west, it is 9 o'clock 47 min. 12 sec, A.M., at Montpelier. What is the longitude of Montpelier? 59. A farmer exchanged 2 bu. of beans @ $1.31^, for sugars at $.10 and $.11 per lb., taking the same quantity of each kind. How many pounds of sugar did he receive ? 60. Wliat is the interest, in this State, on a mortgage for $490, for 1 yr. 5 mo. 24 da. ? 61. A andB were partners in business, with a capital of $1,250, of which A furnished $750. At the end of the first year A's share of the profits was $340.65, when B sold his interest to C for $637.50. At the end of the second year C's share of the profits was $247.80, when he bought A's interest for $870.74. How much did A and B each make ? And how much had C invested more than he had realized ? MISCELLANEOUS PROBLEMS. 333 62. From the product of the sum and difference of 3.G and 3.24, sub- tract the difference between the squares of 3.6 and 2.24. 63. A can build 50 rods of fence in 14 days, B can build it in 25 days, C in 8 days, and D in 20 days. In what time can they build it, if they all work together ? 64. The product is I, and the multiplier is J of f of *^. What is the multiplicand ? 65. One day a boy bought peaches at the rate of 3 for 4 cents, and sold them at the rate of 2 for 5 cents, and cleared $4.20. How many peaches did he buy and sell ? QQ. If each one of us breathes 30 cubic feet of air per hour, in how long a time will we breathe as much air as this school-room contains? 67. What is the equated time for the payment of $100 due in 6 mo., $120 due in 7 mo., and $160 due in 10 mo. ? 68. Pekin is in 118° E. longitiide, and San Francisco is in 122° W. longitude. When it is noon at Pekin, what is the hour at San Fran Cisco ? 69. The floor of a public hall 56 x 84 feet, is of boards 14 feet long and 6 inches wide, which are nailed with 10-penny nails, 8 to each board. Allowing 68 nails to a pound, how many pounds of nails are in the floor? 70. There are 22 1^ bricks to a cubic foot of brick wall. What part of the wall consists of mortar ? 71. How many bricks in the walls of a house 48 ft. long, 25 feet wide, and 18 ft. high, the walls being 1 foot thick, and allowing 2^^ for openings ? 72. If 2-^ yd. of cassimere @ $1| are worth as much as .7 of a ton of coal, how much is the coal worth per ton ? 73. When 8 eggs sell for $.25, what are they worth per dozen ? 74. What is the least common multiple of the fractions |, {-, f, and I ? Note. — Eeduce the fractions to least similar fractions. 75. What is the least common multiple of 222, 104, 68, 54, and 34 ? What is the greatest common divisor of the same numbers ? 76. The Oswego Starch Co. drew on a customer in Milwaukee for $1,275, at 60 days after sight. The bank charged ^% for collecting, and required 2 days for transmission each way. Exchange on Milwaukee being at !{% discount, what were the proceeds of the draft ? 334 MISCELLANEOUS PROBLEMS. 77. At $.36 per sq. yd. for plastering, and $.75 per roll for paper- hanging, liow much will it cost to plaster the walls and ceiling, and paper the walls of a room 18 x 16 x 9 ft., making allowance, in paper- ing, for 3 windows, each 3x6 ft., and 3 doors, each 3x7 ft., the wall- paper being 1 ft. 6 in. wide and 7 yd. in a roll ? 78. A 4-rod road extends along one end and one side of a farm which is 90.5 x 120 rd., the farm extending to the middle of the road. How much of the farm is in the road ? 79. How many days w^ill it take a ship to sail from St. John's, New- foundland, to Valentia Bay, Ireland, a distance of 1,950 miles, if she sails at the rate of 9.5 knots per hour? 80. What length of a board 9 inches wide will make a square foot ? 81. At $13.50 per M,, what is the value of a stock of 13 boards each 14 ft. long, 16 in. wide at one end, and tapering to a point ? 82. My agent in Toledo bought 5,000 barrels of apples @ $1.60, com- mission 2j%. I sent him a draft for the amount, which I purchased ^at \^/c discount. I paid $.30 a barrel to transport the apples to New York, and sold them @ $2.10. What were my profits ? 83. In a straight line between two buildings standing on opposite sides of a public square, is a post. The building, A, is 55 ft. high, and B 64 ft. From the foot of the post to the base of the building, B, is 76 ft. ; from the top of the post to the top of the same building is 95 ft. ; and from the top of the post to the top of the building, A, is 80 feet. What is the height of the post ? 84. What is the horizontal distance between the buildings? 85. What is the distance from the top of one building to the top of the other ? 86. The average diameter of the earth is 9,111 miles. How many square miles on its surface ? 87. Find the number of cubic miles in the earth. 88. If 6 masons build a pier 35 ft. long, 18 ft. high, and 4 ft. wide, in 15 days of 8 hours each, how many masons will be required to build a pier 48 ft. long, 21 ft. high, and 5 ft. wide, in 12 days of 10 hours each ? 89. Divide an estate of $7,500 among 3 children, 10, 12, and 15 years old, so that their respective shares, at Ifo interest, shall amount to the same sum when they are 21 years old. MISCELLANEOUS PROBLEMS. 335 90. A mortgage for $13,275, dated St. Louis, Mo., Oct. 10, 1865, bears the following indorsements : May 7, 1866, $1,350 ; Dec. 11, 18G6, $760; June 23, 1867, $500; Nov. 8, 1867, $850; July 20, 1868, $350. The mortgage was taken up Jan. 1, 1869. What amount was then paid? 91. What is the present value of a paid-up lease having 4 years to run, if the property will rent for $2,000 per annum, money being worth G% compound interest? 92. A carpenter, a mason, and a painter built a house, by contract, for $3,000. The carpenter worked 108 days, the mason 72 - days, and the painter 45 days, and the materials used cost $1,775. How much did each man receive for his labor? 93. Last year my expenses, which were 80^ of my last year's income, equalled 9G% of my expenses this year, and my income equalled 75^ of this year's income. Last year I saved $480. How much do I save this year ? 94. A broker bought 115 shares of Express stock at 79 1. He ex- changed 63 shares at 85 for U. S. 5-20's at 111, and the balance at i)ar for RR. stock at 78. He afterward sold the 5-20's at 116 L and the R.R. stock at 72. Did he gain or lose, and how much ? 95. Find the balance of the following account, and the equated time for its payment : Dr. Geo. H. Thomas. (7r. 1869 1869 Jan. 13 ToMdse.@4mo. 23 30 Feb. 35 By Cash, 25 00 Feb. 12 « (( A '( 42 83 Apr. 7 " " 75 00 Mar. 33 " « 6 " 169 33 May 33 u 30 00 Apr. 19 " " Cash, 73 19 July 7 " Note, 75 00 June 6 " «@30da. 48 53 (( 39 " Cash, 35 00 Aug. 16 u " 50 00 96. A widow who is left with a daughter 16, and a son 8 years old, is to have the income of property that pays an annual rent of $1,500 above taxes and repairs, till the daughter is 31 years old. The daugh- ter is then to have the income, till the son attains his majority, when the property is to be his. How much is each one's interest in the property worth to-day, money being worth Q% simple interest ? PACK Addition of Compoiind Numbers... 146 " " Decimals 81 " Fractions 191 " " Intesers 17 Anal5'sis ." 223 Aritlimetical Progression 811 Average of Payments 275 anking 2G9 Changes of Dividend and Divisor.. . 165 Commission 241 Common Divisors 172 '• Multiples 176 Compound N umbePvS 113 Compound Proportion 292 Converse Opeuations 205 Converse lleductions 210 Cube Eoot, Extraction of. 306 Decimals Definitions in Compound Numbers. " " Decimals " " Evolution " " Factors and Multiples " " Integers " " Mensuration " " Percentage " " Progressions Discount Division of Compound Numbers.. . . " " Decimals ^ *' " Fractions " " Integers Divisors, Common EVOLTTTION 800 Exchange 273 Extraction of Cube Hoot 806 " Square " 301 Factors and Multiples 164 Feactions 179 Geometrical Progression 315 Government Securities 267 Insurance 238 Integees 9 Interest 252 " by Progressions 319 Longitude and Time 224 Manual of Methods and Sugges- tions 5 Measurement of Eight- Angled Sur-. faces and Solids. paok Mensuratiox -{21 Mensuration of Capacities 827 " " Lines 323 " " Surfaces 8-24 Miscellaneous Problems 329 Multiples, Common 170 Multiplication of Compound Nos 154 " " Decimals 85 " " Fractions 195 " " Integers 35 Notation of Compound Numbers. . . 115 " " Decimals 74 " " Evolution 300 " " Fractions 179 " " Integers 10 " " Percentage 229 Partnership 294 Percentage 229 Percentage, the Five General Cases of 231 Prefiice 8 Price, Quantity, and Cost 216 Prime Numbers 171 Profit and Loss 243 Progression, Arithmetical 311 " Geometrical 315 Progressions 31 1 Properties of Composite Numbers.. 169 Proportion, Compound 292 " Simple 289 Eatio and Proportion 2S7 lleductions. Converse 210 " of Compound Numbers. 115 " " Fractions 183 Eeview Problems in Compound Numbers.. 161 " " Converse Operations 226 " " Decimals 110 " " Fractions 203 " " Integers 69 " " Percentage 283 " " Proportion 299 Simple Proportion 289 Square Eoot, Extraction of. 801 Stocks 245 Subtraction of Compound Numbers. 149 " " Decimals 83 " " Fracfions 191 " " Integers 26 .•,rTax«s and Duties 243 ^ .4Jntt&-d States Money 95 ■') 1 THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. j^iff^lfc ^EP 2 1993 • CBERKajsT / / / / LD21 / yS (7383 kl44 r / UNIVERSITY OF CALIFORNIA LIBRARY oVinn!^ r\ 2. Ek AiUlHMt,' the Natural Ore r: i- . is- f. Numbers; third, Ai- ;act ' ) cents. rithnietic, for the Slute; in which Mfvtl fids ui id upon rrincir'— »^'-i>n-»-»^fi >^- rT,.:n..iv • Tni. ,• iKihed upo Jllustratcd. 50 ceut md 1- hmStlo o'l r^'/iiooi Aric-hne-;- < w'Mi theii Frri-ticii; c Illnstraf .n-:. U.0' un.) Jrcaii^-:j on vne r^ei^iiC'. Oi .nu II. KAaF:!r'; ifm'^ books. '^•fw.UO^pcv dozcru >EP!E?, omTraein'^ ihe first si.v -^r • 3aeu niiTv^')f;r. nm zmmi^ m t^iui Aiiof Tke siAbriFoi. HS SCHOOL AND FAMILY SLA > Oardij, containing over Two Hni V • -;', ing, and Ariti^metic. S12 00 pc , •icu Tor school use posf '; T\nuUy Slate. It is