^^i^i^spv^ :£' 1^^^^ wia:wmi^rm£' ':mmfm.M. I (Sjmull Utifemitg Jtatg 4.J.00SX: 8817 arV1797l''°™" """"™"y Library 'llllMffiiiiiiiiiii!iifii!i7 ^"^ practice. / ounj ^924 031 285 749 The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031285749 N ARITHMETIC IN THEORY AND PRACTICE .s- ARITHMETIC THEORY AND PRACTICE. BY THE LATE BROOK-SMITH, M.A. LL.B. ST John's college, Cambridge; barrister at law; ONE OF THE MASTERS OF CHELTENHAM COLLEGE. ILonlion : MACMILLAN AND CO. AND NEW YORK. 1889 [T/ie J^ig/il of Translation is resefved.'\ First Edition i860. New Editions 1872, 1873, 1877, 1879, 1880, 1881 {eUdrotyped), ■ Reprinted 1882, 1883, 1884, i886, 1887, 1889. PREFACE TO THE SIXTH EDITION. In the following pages I have endeavoured to reason out in a clear and accurate manner the leading propositions of the science of Arithmetic, and to illustrate and apply those propositions in practice. Every writer on Arithmetic at the present day feels the necessity of explaining the principles upon which the rules of the subject are based, but every writer does not as yet feel the necessity of making these explanations strict and complete ; or, failing that, of distinctly pointing out their defective cha- racter. Difficulties are still avoided or slurred over, and in- complete proofs without one word of remark or warning are used as though they were full and satisfactory. This surely ought not to be. If the science of Arithmetic is to be made an effective instrument in developing and strengthening the mental powers, it ought to be worked out rationally and con- , clusively. In the practical part of the subject, I have advanced some- what beyond the majority of preceding writers ; particularly in Division, in Greatest Common Measure, in Cube Root, in the Chapters on Decimal Money and the Metric System, and more especially in the application of Decimals to Per- centages and cognate subjects. So long as the mania for neat answers continues to exist, so long will Decimals fail to take their legitimate place in the class-room, and be relegated to the office and the counting-house. The Chapter on Weights and Measures and the Metric System is longer than usual, but not I hope uninteresting. VI PREFACE. The tendency of the present day is to make use of the Metric System in international transactions and scientific pursuits; but to the retail dealer and his customers our present system presents so many advantages, that I almost doubt the possi- bility of uprooting it. They first divide into halves, into quarters, and into eighths, then into thirds and into sixths, but very rarely into fifths or tenths. The probability therefore is that, in this country, the two systems will exist side by side, the scientific and the practical; just as we have two systems of logarithms, and two ways of measuring angles. In the earlier part of the work I have used the Method of Reduction to the Unit, but I am far from advising an exclusive adherence to that Method; when the student has gained a clear and firm grasp of ratio, it would be unwise of him to neglect the powerful instrument that has come into his pos- session. I have to express my many and great obligations to the College Lectures on Arithmetic of Professor Kelland, and of the late Professor De Morgan: — I have also consulted with advantage the works of Loomis, Lionnet, Serret, and Bertrand. Any student wishing for a more detailed explanation of the earlier portion of Arithmetic, will meet with it in the admirable treatise of Messrs Sonnenschein and Nesbitt. Of the ex- amples about one half are original; the other half are selected from University and College Examination papers, and from papers given in the various competitive Examinations. I shall be thankful to receive and glad to acknowledge any suggestions or corrections that may be offered to me on any part of the work. Cheltenham College, June, 1 88 1. CONTENTS. CHAPTER I. Definitions, Art. i. Names of Numbers, 7. Notation and Nume- ration, 15. CHAPTER II. THE FUNDAMENTAL OPERATIONS. Addition, 21. Subtraction, 27. Multiplication, 34. Division, 44. CHAPTER III. Explanation of signs, 54. Propositions in Addition and Subtraction, 56 ; in Multiplication and Division, 60. Involution, 65. CHAPTER IV. Greatest Common Measure, 67. Least Common Multiple, 77. Criteria of Divisibility, 84. Proof of Multiplication by casting out the nines, 91. Primes and Prime Factors, 92. CHAPTER V. FRACTIONS. Definitions, 98. Notation and Numeration, 10 r. Reduction of Mixed Numbers to Fractions, 107. Fundamental Proposition, 109. Multiplication and Division by a whole number, 1 10. Reduction of a Fraction to its lowest terms, 113. Least Common Denominator, 117. Addition of Fractions, 119. Subtraction, 121. Propositions in Addition and Sub- traction, 124. Multiplication, 126. Division, 132. Greatest Common Measure and Least Common Multiple of Fractions, 135. Complex Fractions, 136. CHAPTER VI. DECIMALS. Definitions, 140. Numeration, 142. Conversion of a decimal to a vulgar fraction, 145. Addition and Subtraction of Decimals, 146. Multiplication, 148, Division, 150. Contracted Addition and Subtraction, 151. Con- tracted Multiplication, 152. Contracted Division, 153. Reduction of Vulgar Fractions to Decimals, 154. Repeating Decimals, 158. Addition and Subtraction of Repeating Decimals, 160. Multiplication and Division, l6r. VlU CONTENTS. CHAPTER VII. EVOLUTION. Definitions, 163. Propositions in Square Root, 164, i6g. Pointing, 168. To extract the Square Root of an Integer or a Decimal, 170; Con- tracted Process, 173; of Fractions and Mixed Numbers, 175. To extract the Cube Root of an Integer or a Decimal, 178; Contracted Process, 181 ; of Fractions and Mixed Numbers, 183. Extraction of the Fourth, Sixth and Ninth Root, 184. CHAPTER VIII. RATIO AND PROPORTION. Definitions, 185. Compounding Ratios, 190. Proportion, 191. Main Proposition, 192. Alternately and Inversely, 193. Compounding, 194. Mean Proportional, 194. CHAPTER IX. Introduction, 195-197. Unit of Time and Table, 198; of Length, 200; of Surface, 203; of Volume, 205; of Weight, 206; of Capacity, 208; of Money, 209. Proposed Decimal Coinage, 212. Money, &c. of the United States, 214. The Metric System, 216. CHAPTER X. REDUCTION AND THE COMPOUND RULES. Definitions, 227. Reduction Descending, 229; Ascending, 230. Com- pound Addition, 234. Compound Subtraction, 238. Compound Mul- tiplication, 241. Compound Division, 248. Reduction to the Unit, 255, i; Averages, 25S, ii; Revolution of Wheels, &c., 255, iii-x. Reduction and the Compound Rules; Fractions, 256. Reduction to the Unit; Time and Work ; Hands of a Clock, &c., 266. Reduction and the Compound Rules — Decimals, 267. CHAPTER XI. Decimal Money, 272. Some Appjjications of Decimal Money, 277. The Metric System, 278. CHAPTER XII. Definition of Practice, 279. Principles on which Practice depends, 281 ; Simple Practice, 283; Compound Practice, 284. Directly and Inversely Proportional, 285-288. Rule of Three, 289. Alternation of Terms in a Rule of Three statement, 291. Double Rule of Three, 292. Chain Rule, 298. Proportional Parts, 301. Partnership, 304. Alligation, 305. CHAPTER XIII. Per-Centages, 310. Commission, Brokerage, Insurance, 319. Profit and Loss, 324. Interest, 327. Simple Interest, 330. Present Worth and \Discount, 335. Discounting Bills, 340. Compound Interest, 342. Present Worth and Discount at Compound Interest, 344. Stocks, 345. Ex- changes, 353. CHAPTER XIV. Square and Cubic Measure, 358. Duodecimals, 362. Examination Papers, p. 369. Answers, p. 393. ARITHMETIC. CHAPTER I. DEFINITIONS. NAMES OF NUMBERS. NOTATION AND NUMERATION. I. Whatever is susceptible of increase or diminution is called a -magnitude. 1. A magnitude may be continuous, that is, whole and undivided, as the length of a field, the water of a reservoir; or it may be made up of separate and distinct parts, as a heap of pebbles, a flock of sheep. 3. When a magnitude is continuous, we take some well-defined magnitude of that kind which we call its unit, and by repeating this unit a sufficient number of times, we make up the given magnitude. If the magnitude be made up of distinct objects, we take an object of that kind as our unit, and observe how many such units must be taken to make up the given magnitude. 4. When a magnitude is represented as made up of repetitions of its unit, it is called a quantity, and the result of the comparison of the given magnitude with its unit is called a number: thus the length of a field, a heap of pebbles are magnitudes; twenty yards, a hundred pebbles are quantities ; twenty and a hundred are numbers. In Mathematics we are only concerned with those magnitudes that can be referred to a unit and expressed as quantities. 5. Numbers are often spoken of as concrete or abstract according as the nature of the unit in any particular case is or is not men- tioned ; thus twenty yards, a hundred pebbles, are called concrete numbers, but twenty, a hundred, are called abstract. g B-S. A. ' I 2 NAMES OF NUMBERS. § s- An abstract number is therefore a number in its proper sense, conveying the idea of times or repetitions ; and a concrete number is simply a quantity. 6. Arithmetic is the science of numbers : — it explains their nomenclature and notation, it investigates their properties, and points out methods of making calculations by means of them. NAMES OR NUMBERS. 7. To a unit standing by itself we give the name one unit : to one unit and one unit taken together, we give the name two units ; to two units and one unit three units : to three units and one unit four units : and so we may proceed to an endless extent. Or, speaking only of the numbers employed, without reference to any particular unit, we may say — one and one taken together is called two : two and one, three: three and one, /ours and so on. But if to each of the successive numbers thus formed we were to give an in- dependent name, our range of numbers would of necessity be very limited. We will therefore shew how by a few independent names, and by a systematic combination of them, we can express all the numbers we require. 8. The names of the first numbers in order are — one, two, three, four, five, six, seven, eight, nine. These nine numbers are called simple numbers, and units of the frst order. Their names are per- fectly arbitrary. 9. To nine and one we give the name ten. Ten forms a single unit of the second order, and we repeat or count by ten, as we before counted by one ; thus one-ten, two-ten, three-ten, four-ten, nine-ten; or treating ten as a simple number (9), and remembering that "ty " or tig is equivalent to ten, we say ten, twenty, thirty, forty, ninety. The names of the numbers between ten and twenty are formed irregularly; the first is eleven, supposed to come from the Gothic ainhf {ain one, and hf ten), the next is twelve from the Gothic tvalif{tva two, and hf ten), and the others explain themselves; § 9- NAMES OF NUMBERS. 3 thus we have eleven, twelve, thirteen, fourteen, fifteen, nineteen. The names of the nine numbers between twenty and thirty, thirty and forty, are formed by placing the names of the first nine numbers (8) in order after twenty, thirty, 10. Ten units of the second order form a single unit of the third order, which is called a hundred: and we count by hundreds as we counted by simple units, thus one hundred, two hundred, three hundred, nine hundred. The names of the numbers between one hundred and two hun- dred, two hundred and three hundred, are formed by placing the names of the first ninety-nine numbers in order after one hun- dred, two hundred, 11. Again ten units of the third order form a single unit of the fourth order, which is called a thousand: also ten thousands form a single unit of ih&Jifth order, and ten ten-thousands or a hundred thousand (10) form a single unit of the sixth order : but instead of calling these numbers by independent names, we consider a thou- sand as a second principal unit, and count by units, tens, and hundreds of thousands. The names of the numbers between one thousand and two thou- sand, two thousand and three thousand, are formed by placing the first nine hundred and ninety-nine numbers in order after one thousand, two thousand, 12. Again ten hundreds thousands or a thousand thousands (11) form a single unit of the seventh order, which is called a million ; and we consider a million as a third principal unit, and count by units, tens, hundreds, thousands, ten-thousands, and hundred-thou- sands of millions. The names of the numbers between one million and two ipillions, two millions and three milliojis, are formed by placing in order all the numbers from one to nine hundred and ninety-nine thousand nine hundred and ninety-nine after one million, two millions, 13. Lastly ten hundred thousand millions, or a million millions (12) is called a billion, a million billions z. trillion, a laillion trillions I — 2 4 NAMES OF NUMBERS. % 13. a quadrillion, and so on ; and we count by units, tens, hundreds, thousands, ten-thousands, and hundred-thousands of billions, tril- lions, quadrillions,... precisely as we do in millions. The names billions, trillions, were proposed by Locke {Essay con- cerning Human Understanding, Book ii, i6) in place of millions of millions, millions of millions of millions, and are very convenient for scientific purposes, but the wants of ordinary life seldom require that we should proceed beyond millions. In France and some parts of the United States of America, a thousand millions is called a billion, a thousand billions a trillion, a thousand trillions a quadrillion, and so on ; hence our billion is their trillion, our trillion their quintillion, &c. 14. It appears then that (i) We employ practically no more than thirteen independent words : — one, two, three, four, five, six, seven, eight, nine, ten, hun- dred, thousand, million. (2) Ten units of any order always make one unit of the next higher order. (3) Every number is made up of units of successive orders, the number of units in any order never exceeding nine. Ex. I. Five hundred and sixty-nine, may be expressed as Five hundred six tens and nine units. Ex. 2. Seven millions six hundred and four thousand and thirty- two, may be expressed as Seven millions six hundred-thousands no ten-thousand four thousand no hundred three tens and two units. Exercise i. Express the following numbers as made up of units of each successive order from the highest to the lowest : — 1. Sixty-seven thousand and forty-eight; forty thousand and forty. 2. Nine hundred and six thousand five hundred and four. 3. Forty-five millions seven hundred and thirty-four thousand six hun- dred and ninety-one. 4. Twenty-seven thousand and five millions eighty-seven thousand three hundred and three. 5. What is the number next less than a thousand? a million? 6. Find the regular expression equivalent to Eighteen hundred and seventy-seven; £/i?w«j hundred and one ; 7ra««^-^/4r^e hundred thousands. § ij. NOTATION AND NUMERATION. 5 NOTATION AND NUMERATION. 15. Notation is the art of representing the names of numbers by means of a few written characters. The system we use is called the Arabic, because it was introduced into Europe by the Arabs in the twelfth century, but it is undoubtedly of Hindu origin. The converse process of reading the names of numbers when expressed in written characters is called Numeration. 16. To express the name of any number in figures. (i) Represent the first nine numbers by the following nine characters, c^e.& figures or digits : I, 2, 3, 4, s, 6, 7, 8, 9, and represent the absence of units by a tenth figure o, called nought or cipher. (2) Express the given number as made up of successive orders, from the highest to the lowest, taking care that no order shall be missing (14, 3) ; and as the number of units in each order is either one of the first nine numbers, or is nought, replace the number of units in each case by its corresponding figure. (3) Write now the figures of each order in succession in horizon- tal line beginning with the highest ; and adopt the convention that Figures occupying the first, second, third. . .place from the tight shall represent units of the first, second, third... orders and the given number will be completely represented in figures. Ex. I. Write in figures, Five-hundred and sixty-nine. This number is, — 5 hundred 6 tens and 9 units (14, 3) ; and is therefore written in figures thus 569 Ex. 2. Write in figures, Seven millions six hundred and four thousand and thirty-two. This number is — 7 millions 6 hundred thousand o ten-thousand 4 thousand o hundred 3 tens and 2 units (14, 3) ; and is therefore written in figures thus 7604032. 17. If the given number contains thousands, we shall find it convenient— /^z>j/ to write down in figures the number expressing the thousands, .a«^ ^<7 its right the rest of the number :— and if the 6 NOTATION AND NUMERATION. § 17- rest does not contain three figures, we must prefix the requisite number of ciphers. Ex. 3. Write in figures Twenty-five thousand three hundred and forty-six. The number of thousands is expressed by 25, and the rest by 346 ; therefore the given number is expressed by 25,346. Ex. 4. Write in figures Eighty-thousand and forty-six. The number of thousands is 80, and the rest is 46 or 046 ; there- fore the given number is 80,046. And if the given number contains higher orders than hundred thousands :— Write down in figures the number of units of the highest principal order, to its right the number of the next principal order,.. .then the number of millions, and lastly the rest of the number: — and if any order except the first does not contain six figures, prefix the requisite number of ciphers. Ex. 5. Express in figures Forty-six bilhons three hundred thou- sand and sixty-nine millions four thousand and forty. The number of billions is expressed by 46, of millions by 300069, and the rest by 4040 or by 004040 ; therefore the number is ex- pressed by 46,300069,004040. 18. To the general rule in Notation there corresponds the following rule in Numeration : — Proceeding from left to right, write down the name of each figure, annexing the order of its units. Ex. I. 736 represents Seven hundreds three tens and six units : or more simply, Seven hundred and thirty-six. Ex. 2. 910 represents Nine hundreds one ten and no units, or simply, Nine hundred and ten. But if the number does not contain more than six figures we shall find it convenient : — To mark off the last three figures to the right then to write down the period to the left as a number by itself annexing thousands, and then the other period. Ex. 3. 763054 is marked off thus :— 763,054, and represents Seven hundred and sixty-three thousands and fifty-four. § 17- NOTATION AND NUMERATION. 7 And if the number contains more than six figures : — Mark off the last six figures to the right, then the next six, and so on until not more than six remain : proceeding now from left to right, write down each period as a number by itself, annexing the order of units it represents. Ex. 4. 34567008093402 is marked off thus : — 34,567008,093402, and is read Thirty-four billions five hundred and sixty-seven thou- sand and eight millions, ninety- three thousand four hundred and two. 19. In the system of Notation we have explained, a figure has two distinct valuesr— one absolute or intrinsic, depending on its shape, the other local, depending on the place it occupies in a given number : thus in 65345251 the figure 5 in one place represents five tens, in another yfz/i? thousands and in zxiQ'Ca.tx five millions. 20. The number of units in any order which is taken to form a unit of the next higher order is called the base of the system. Hence in our system the base is ten, and it is therefore called the decimal system. Had the base been twelve, it would have been called the duodecimal. We may also remark that when the base is ten we require ten figures, nine significant and the cipher ; and in like manner if the base were twelve, we should require twelve figures. Exercise 2. Express in figures : — 1 . Seven hundred and seven thousand and seventy. ■i. Twelve millions twelve thousand and twelve. 3. Six hundred and forty millions sixty-four thousand six hundred, 4. Eight hundred and seven billions eighty thousand and eight millions six hundred thousand and fifty. Express in words the following numbers : 5. 90960; 70047; 600304; 780983; 6008029; 50706004. 6. 407017170; 370094586061; 304007.36020095347. 7. Express in words the following number, both after the English and the French method,— 696536483318640035073641037. 8. Write down the greatest and least numbers of three figures. How many numbers are represented by three figures? CHAPTER II. THE FOUR FUNDAMENTAL OPERATIONS. ADDITION. 21. Addition is the operation by which we find a single number that is equal to two or more given numbers put together. This single number is called the sum of the given numbers. 22. Case I. To find the sum of two simple numbers. For example find the sum of 7 and 3. 3 is the sum of 2 and i (8), and therefore the sum of i, i and i :— and adding each of these ones in succession to 7, we have 7 and i is 8, 8 and i is 9, 9. and i is 10; that is, the sum of 7 and 3 is 10. Hence, to add 3 "to 7 we have simply to count forwards 3 steps from 7 thus — 8, 9, 10 : and the last number gives the sum. And in the same way may be found the sum of every two simple numbers. 23. We shall afterwards see that finding the sum of any two or more numbers depends on finding the sum of every two simple numbers : it is therefore necessary that all such sums should be thoroughly learnt by heart. 24. Case II. To find the sum of diUy two or more numbers. (i) Suppose each of the numbers decomposed into its simple units, tens, hundreds, . . . ; and find the sum of all the simple units, the sum of all the tens, the sum of all the hundreds,. . . : the sum of these partial sums will give the sum required, for it will contain all the parts which make up all the given numbers and no more. For convenience in effecting these partial sums write the numbers under one another, so that units of the same order may be in the same vertical column, and draw a line below, In ^^i this way the sum of 231, 423 and 13J is found to be 9 units, 13c 8 tens and 7 hundreds, or 789. 789 § H- ADDITION. (2) If the sum of the units in any order exceeds 9, we avail ourselves of the following principle usually termed carrying : The tens of any order in a partial sum may be carried as units to the next higher order j for ten units of any order are equivalent to one unit of the next higher order (14). (3) Lastly instead of finding all the partial sums, and then carry- ing the tens of any order as units to the next higher order, we may add the number carried from any column to ihs first number in the next column, instead of to the sum ofHhat column; and as the carryings are made from any column to the next one on the left, we ought to begin the work at the first column on the right. We have then the following general Rule : (1) Write the numbers under one another, so that units 0/ the same order may be in the same vertical column, and draw a line underneath. (2) Begin at the first column on the right, and find the sum. of the numbers in that colum-n : set down below the column the units' figure of the sum and carry the tens to the first figure of the next column. (3) Having carried thus, find the sum of the second column : set down and carry as before. (4) Proceed in this way through all the columns, and below the last write down its sutn at full length. 25. A Proof is a second operation which serves as a test of the correctness of the first. The Proofs of Addition depend on this principle — The sum of several numbers is not affected by the order in which they are add- ed together. For the sum of the units in the three following groups, iiiim, mil, III, will be the same in whatever order we add them together : thus the sums of 7, S and 3 ; of 5, 3 and 7 ; of 3, 5 and 7 are the same. 26. Proof. As we usually begin at the bottom of each column and add upwards, in the proof begin at the top of each column and add downwards ; — if at each step the results coincide, we may pre- sume that the work is correct. lO ADDITION. § 26. Ex. Find the sum of 82093, 9386, 51764, 475 and 123897. 82093 9386 51764 475 123897 267615 Beginning at the units' column and repeating the result only of each step, say 12, 16, 22, 25 : set down s, and carry 2 to the first figure of the second column; then say 11, 18, 24, 32, 41 : set down i and carry 4; 12, 16, 23, 26, set down 6 and carrj' 2 : and so proceed through all the columns. For the Proof, begin at the top of each column and add downwards: thus for the units' column say, 9, 13, 18, 25; for the second column 11, 19, 2g, 32, 41, &c. And as at each step these results coincide with the former ones we presume that the work is correct. Exercise 3. I. Find the sum of 4578, 34796, 2385, 97, 50405 and 56789. ■i. Add together 125, 309486, 7098, 67, iiooo, 2106 and 10785. 3. Find the sum of six numbers each equal to 7903856. 4. Add together the sum of five numbers each equal to 4597, and the sum of four numbers each equal to 89796. 5. Add together, Eighty millions sixty-seven thousand and eighteen; nine millions seven hundred and six thousand five hundred and nine ; eight hundred and one millions nine hundred and seventy thousand seven hun- dred and sixty ; seven millions and seventy-seven ; sixty millions six hundred, and six thousand and sixty-six ; and five hundred and fifty-five thousand and fifty. 6. Find the sum of 69798856 and the six following numbers. 7. In 1871 the population of England and Wales was 22704108, of Scotland 3358613, of Ireland S402759, of Islands in the British Seas 144430, and of the Army and Navy, &c. 207198: find the total population of the United Kingdom at that date. 8. Find the net revenue of the United Kingdom for the year 1871-2 as derived from the following sources : — Customs ^^20243044, Excise ;£'23386o64, Stamps ;£'9739548, Land Tax and House Duty ;^2352i8i, Income Tax ;f 9328103, Post Office ;^494i5io. Telegraph Service ;^75 16 u, Crown Lands ;^4468oi, and Miscellaneous ;^4o6o3i5. § ^1. SUBTRACTION. 1 1 SUBTRACTION. 27. Subtraction is the operation by which we find what num- ber is left when a smaller number is taken from a greater. The greater number is called the Minuend, the smaller one the Subtrahend, and the number left the Remainder. 28. The number left is the difference between the two given numbers ; it is also the excess of the greater number over the less ; it is also the number which must be added to the less number to make it equal to the greater. 29. Subtraction is the inverse of Addition : — in Addition two numbers are given to find their sum ; in Subtraction the sum of two numbers is given and one of the numbers to find the other. 30. Case I. When the Subtrahend is a simple number, and the Minuend less than that number increased by 10. For example subtract 3 from 7. Now 3 is the sum of i, i and i (22) : and subtracting each of these ones in succession from 7 we have — I from 7 leaves 6 (8), 1 from 6 leaves 5, and i from 5 leaves 4 ; therefore 3 from 7 leaves 4. Hence, to subtract 3 from 7 we have simply to count backwards 3 steps from 7, thus — 6, 5, 4 ; and the last number gives the Remainder. Or, we may ask to what number must we add 3 to get 7 : the answer is 4 ; hence if from 7 we take away 3 the remainder is 4. 31. Case II. When Minuend and Subtrahend are 3.i\y numbers. (i) Suppose the numbers to be decomposed into their simple units, tens, hundreds..., and subtract the units, tens, hundreds... of the Subtrahend, from the corresponding orders of the Minuend ; the sum of these partial remainders will give the remainder required, for it gives what is left when all the parts of the Subtrahend have been taken away from the Minuend. For convenience place the Subtrahend under the Minuend so that the units of the same order may be in the same vertical column, and draw a line below. ^^g In this way the remainder in subtracting 235 from 978 is 235 found to be 3 units 4 tens 7 hundreds, or 743. 743 12 SUBTRACTION. §31. (2) If the units of any order in the Subtrahend exceed those of the Minuend we avail ourselves of the following principle, usually termed borrowing ; The Minuend and Subtrahend may be increased by the same number without altering their difference : hence we may increase the number of units in any order of the Minuend by 10, if we increase that of the next higher order in the Subtrahend by I. (3) Lastly if we proceed from right to left, the partial remainders at each step will hsjinally obtained, for the borrowings are always made from one, order to the next higher order ; and having found all the partial remainders we have found the final remainder. We have then the following Rule : (i) Write the Subtrahend under the Minuend, so that units of the same order may be under one another, and draw a line below. (2) Beginning at the units' figure, subtract each figure of the Sub- trahend from the one above it in the Minuend, and place the remainder im.mediately below j and if in any case the figure of the Subtrahend be greater than the one above it, add 10 to the latter figure and then subtract, taking care to add I to the next figure of the Subtrahend. 32. We may use either of the following proofs in Subtraction : — (i) Add the Remainder to the Subtrahend: — we ought to obtain the Minuend (28). (2) Subtract the Remainder from the Minuend: — we ought to obtain the Subtrahend. Example. Subtract 28549 from 54627. , ,,. , Beginning at Jhe units' figures, proceed thus:— 54627 Mmuend <■ „ , „ , , a ' , 9 "0™ 17 leaves 8 : set down 8; ^^549 Subtrahend ^ ^nd i is 5, 5 from 12 leaves 7: 7; 26078 Remainder 5 and i is 6, 6 from 6 leaves o: o; 8 from 14 leaves 6 : 6 • 2 and I is 3, 3 from 5 leaves 2 : vi. But the student should accustom himself to repeat the following numbers only : — 9, 8: 5, 7: 6,0: 8, 6: i,/i. §32- SUBTRACTION. 1 3 where the first number is the figure or increased figure of the Subtrahend, and the second the corresponding figure of the Remainder. Proofs. Adding the Remainder to the Subtrahend each step gives the corresponding figure in the Minuend. Or, subtracting the Remainder from the Minuend we get at each step the figure in the Subtrahend. Remark. In France it is usual to borrow from the next order in the Minuend; and the preceding example would be gone through thus.- — 9 from i7 is 8; 4 from 11 is 7; 5 from 5 is o; 8 from 14 is 6; 1 from 4 is ^. In this way the borrowing is delayed one step later than in our method ; and this is a disadvantage. 32*. Since the Remainder added to the Subtrahend gives the Minuend, we may at each step find what figure added to the figure of Subtrahend will give the figure of the Minuend. The preceding example would be gone through thus, where the dark figure is the one to be set down on the Remainder : — 9 and 8 is 17; 5 and 7 is 12; 6 and is 6; 8 and 6 is 14; 3 and 2 is J. This method is especially convenient when we are re- jTrnm c ^/i R quired to take from a number the sum of several numbers. — -7— In the following example we proceed thus : — Take J Q'?4 13, IS and 3 is 18; 7, 10, 17 and 7 is 24; ( 869 lo, 19, 25 and 8 is 33; 3 and 2 is 5. Remd. 2873 33. The complement of a number is its defect from 10 units of the number's highest order: thus the complement of 57364 is its defect from 10,0000. But 1 00000 is 99999 and i : hence we may subtract each of the figures of 57364 from 9, except the units' figure and that we must subtract from 10 : therefore its complement is written off from left to right at sight thus, 42636. Exercise 4. 1. Subtract 38967 from 123456. 2. Find the difference between 78954 and 836523. 3. What number must be added to 7965499 to give 541850036? 4. From Twenty-three millions thirty thousand two hundred and thirty subtract Eight millions eight hundred and seven thousand and sixty-five: 5. From the sum of 75301 and 6456 subtract the sum of 3087 and 56299. 14 MULTIPLICATION. §33. 6. From the difference between 3285 and 456 subtract the difference between 19011 and 17455. 7. What number must be added to the sum of 750 and 3287 to make the result equal to the sum of 505, 650, 19 and 9003? 8. Write off the complement of 43 ; 574; 9999; 542056; 859064. MULTIPLICATION. 34. Multiplication is the operation by which we find the sum of a given number repeated as many times as there are ones in another given number. The number to be repeated is called the Multiplicand, the other the Multiplier, and the sum found the Product. The Multiplicand and Multiplier are called the Factors of the Product. 35 . If we interchange Multiplicand and Multiplier, the Product remains the same. For write down l in a horizontal line 7 times, and repeat this line 5 times. The sum of each horizontal line is 7, and there are 5 such lines, therefore the sum of all the ones is 7 multiplied by 5. Again the sum of each vertical line is 5 and there are 7 such lines, therefore the sum of all the ones is 5 multiplied by 7 : that is 7 multiplied by 5 is the same as 5 multiplied by 7. 36. If the Multiplier be a product of two nutnbers, we may multiply by each of those numbers in succession. Let the Multiplier be 15, the product of 5 and 3 : then to multiply ()^ by IS, we may multiply 67 by 5 and the result by 3. For write 67 in a horizontal line 5 times, and repeat this line 3 times, then 67 will be repeated 5 times 3 or 1 5 67 67 67 67 67 times, and therefore the sum of all the numbers 67 67 67 67 67 written down is 67 multiplied by 15. But the ^^ ^7 67 67 67 sum of each horizontal line is 67 multiplied by 5, and there are 3 such lines, therefore the sum of all the numbers written down is this product multiplied by 3 ;— that is to multiply 67 by 15 we may multiply 67 by 5 and the result by 3. I I I I I I I I I I I I I I I §37- MVL TIPLICA TION. IS 37. Case I. When Multiplicand and Multiplier are simple mtmbers. For example multiply 7 by 3. Here we have to find the sum of 7 repeated 3 times, or the sum of 7, 7 and 7 ; but this sum is 21 ; that is 7 multiplied by 3 is 21. And in the same way must be found the product of every two simple numbers. 38. We shall afterwards see that the product of any two numbers depends on the product of two simple numbers ; it is therefore necessary that the product of every two such numbers should be thoroughly learnt by heart. Every such product will be found in the following Table, called the MultipUcation Table. I 2 3 4 5 6 7 8 9 10 II 12 2 4 6 8 10 12 14 16 18 20 22 24 3 6 9 12 IS 18 21 24 27 30 33 36 4 8 12 i6 20 24 28 32 36 40 44 48 S 10 IS 20 2S 30 35 40 45 50 55 60 6 12 18 24 30 36 42 48 54 60 66 72 7 14 21 28 3S 42 49 56 63 70 77 84 8 16 24 32 40 48 56 64 72 80 88 96 9 18 27 36 45 54 63 72 81 90 99 108 10 20 30 40 SO 60 70 80 90 100 no 120 II 22 33 44 SS 66 77 88 99 no 121 132 12 24 36 48 60 72 84 96 108 120 132 144 The first horizontal line gives the numbers i, 2, 3,... 12, and each line is derived from the preceding by adding i, 2, 3,. ..12 in order ; hence the second line gives the respective products of these numbers and 2 ; the third line the respective products of these numbers and 3 ; and so on. Hence to find the product of 7 and 3 we look for the vertical column with 7 at the top and in the third line we l6 MULTIPLICATION. §38. find 21, or (35) we look for the vertical column with 3 at the top and in the seventh line we find 21, the^product required. For the products of simple numbers the Table is carried to 9 times 9 only ; but as the products of 10 and 1 1 are easily learnt and the products of 12 are much used in business transactions, the Table is usually carried to 12 times 12 ; and the pupil should gradually extend it to 12 times 20. 39. Case II. When the Multiplicand is any number and the Multiplier a number not greater than 12. (i) We have to find the sum of the Multiplicand repeated as many times as there are ones in the Multiplier (34) : that is we have to find the sum of the simple units, tens, hundreds,... of the Multiplicand repeated as many times as there are ones in the Multiplier (24) ; or in other words we have to multiply the simple units, tens, hundreds,... of the Multiplicand by the Mul- tiplier. For convenience write the Multiplier under the units' figure of the Multiplicand and draw a line under- 3122 neath. In this way, the product of 3122 by 3 is found to — jl be 9366. 9366 (2) If any partial product exceeds 9, we carry as in Addition ; that is we set down the units' figure of such product and carry the tens' figure to the next partial product (24, 2). (3) We begin at the units' figure of the Muhiplicand and proceed from right to left for the reason given in Addition (24, 3). We have then the following Rule : — (i) Write the Multiplier under the units' figure of the Mul- tiplicand and draw a line underneath. (2) Begin at the units' figure of the Multiplicand, and multiply each figure in succession by the Multiplier, setting down and carrying precisely as in Addition. Ex. I. Multiply 3468 by 7. 3468 3468 X 7 24276 24276 §39- MULTIPLICATION. 1/ Here say 7 times 8 is 56 : set down 6 and carry 5 ; 7 times 6 is 42 and £ carried is 47 : set down 7 and carry 4 ; 7 times 4 is 28 and 4 is 32 : set down 2 and carry 3 ; 7 times 3 is 21 and 3 is 24: set down 24. But in effecting this operation we ought only to say 56; 42, 47; '8, 32; 21, 24. 40. Case III. When the Multiplier is a simple number fol- lowed by one or more noughts. (i) Take any number 3468 and multiply it by 10 in the ordinary way : we see that the product is formed 3468 by simply placing a nought to the right of the given ,„ number. In like manner, we multiply a number by 100, or by 10 times 10, by placing 2 noughts to its right ; and by locx) by placing 3 noughts to its right ; and so on. (2) Suppose now that the Multiplier is \oo, — then since 400 is the product of 4 and 100, we may multiply the Multiplicand by 4 and the result by 100 (36) ; that is we may multiply by 4 and then place 2 noughts to the right of the result. We have then this Rule : — Multiply the Multiplicand by the simple numier, and to the right of the result place as many noughts as there are noughts to the right of the Multiplier. Ex. 2. Multiply 5867 by 70 ; and by 400. S867 5867 70 400 410690 2346800 41. Case IV. When Multiplicand and Multiplier are any numbers. Multiply 5867 by 2479. The product is the sum of 5867 repeated 2479 times ; and 2479 i^ ^'^ sum of 9, 70, 400 and 2000 ; if there- fore we repeat 5867 9 times, then 70 times, then 400 times, and then 2000 times, we shall have repeated it in all 2479 times; that is the product of 5867 by 2479 is the sum of the products B.-S, A. 2 iS MULTIPLICATION. § 4t. of 5867 multiplied by 9, by 70, by 400, and by 2000. The first of these partial products is found by Case II., and the others by Case III. For convenience, write the Multiplier under the Multiplicand and draw a line below; and under this line place as they arise the products by 9, by 70, by 400, and by 2000 ; and arrange that in Multiplicand, Multiplier, and the partial products, units of the same order may be under one another, and therefore the arrange- ment will hold for the final product : thus — S867 5867 (B) 2479 2479 52803 = product by 9 52803 410690= 70 41069 2346800 = 400 23468 1 1734000 = 2000 I1734 14544293 = 2479 14544293 And if instead of multiplying by 70, 400 and 2000 we were to multiply by 7, 4 and 2, and to place the units' figure of the respective products under the figure we were multiplying by, we should obtain the same result, and our work would stand as at (b). We have then the following general Rule : (i) Write the Multiplier under the Multiplicand, so that units of the same order may be imder one another, and draw a line underneath. (2) Begin at the units'' figure of the Multiplier and multiply by each of its figures in order, -writing down each partial product so that its first figure shall be under the figure of the Multiplier that produces it. (3) Add together these partial pro- ducts; their sum is the product required. 42. Case V. When Multiplicand and Multiplier are both ter- minated by noughts. Multiply 347000 by 2100. We multiply by 2100 by multiplying by 21 and placing 2 noughts to the right of the result (40), and we multiply 347000 by 21— which we will suppose to be done in one operation by Case II.— by first setting down 3 noughts and then multiplying 347 by 21 ;— that is we multiply 347000 by §42- MULTIPLICATION. I9 2100 by multiplying 347 by 21 and then placing 3 noughts and then 2 noughts to the right of the result, that is, as many noughts as there are in Multiplicand and Multiplier together. Hence we have this Rule : — Suppose the noughts at the right of Multiplicand and Multiplier suppressed, find the product of the resulting numbers, and to the right of this product place as many noughts as were -supposed to be suppressed in Multiplicand and Multiplier together. Ex. 3. Multiply 365700 by 70 ; and by dTooo. 365700 70 365700 67000 25599000 2SS99 21942 24501900000 43. We may adopt the following Proofs in Multiplication : Proof I. Interchange Multiplicand and Multiplier : the product ought to be unaltered (35). Proof II. By casting out the nines. We cast the nines out of a number thus : add together all its figures, omitting every 9, and if the sum be greater than 9, replace it by the sum of its figures, and if the new sum be greater, replace it by the sum of its figures, and so proceed till we have a sum less than 9. Cast the nines out of Multiplicand and Multiplier. Multiply the results, and cast the nines out of their product, noting the new results now cast the nines out of the Product, and if the result coincide with the one' previously noted, we presume that the work is correct. This process is explained at Art. 91. Ex. 4. Multiply 5867 by 2479, annexing the proofs. 5867... 8 V S / 2479 2479 - ■■4 \ / _S867 52803. ..5 \ / I73S3 41069 MuM 8^(4 Mult'. 14874. 23468 / \ 19832 1 1 734 / X 12395 14544293.. .5 6^."^. - 14544293 2—2 ^0 mvltiplication: § 43. Beginning at the left hand we cast the nines out of the Multiplicand thus — 13, rp, 26 ; replace 26 by the sum of 1 and 6, or 8 ; — out of the Multiplier thus — 6, 13 ; replace 13 by the sum of i and 3, or 4 : — now multiply 8 by 4 giving 32, which replace by the sum of 3 and 2, or £ ; and note this result. Again we cast the nines out of the Product thus — 5, 10, 14, 18, 20, 23; replace 23 by the sum of 2 and 3, or 5; and as this result coincides with the previous one, we presume the work is correct. Ex. 5. Multiply 43896 by 357, and by 735 : making in each case only two partial multiplications. 43896 43896 357 735 307272 307272 1536360 153 6360 15670872 32263560 Here we first multiply by 7 and then the result by 5 : that is, we first multiply by 7 and then by 35 (36). In multiplying by 7 we set down the first figure of its product under 7 (41, 2) ; and in multiplying by 35 we set down the first figure under the 5 of 35. Exercise 5. 1. Multiply 6508794 by 8, by 7, by 9, by ii and by 12. ■'■■ Multiply 56380477 by 18, by 35, by 48, by 72 and by 132— in each case by two successive multiphcations, and also by the usual method. 3. Multiply 67836479 by 356, by 4378 and by 78539. 4. Multiply 70870096 by 404, by 3009 and by 900807. 5. Multiply 57483000 by 40, by 900, by 430 and by 4670000. 6. What is the difference between 23456 multiplied by 996, and the remainder in subtracting 4 times 23456 from 23456000? 7. If the area of England and Wales is 58325 square miles, and there is an average population of 317 persons to every mile, find the total population. 8. If a locomotive travelled from London to Rugby at a uniform speed of 1223 yards a minute, it could perform the distance in 121 minutes and have 380 yards to spare— find the distance between the two places in yards. 9. Multiply 324567 by 486, and by 936, and by 13212, making in each case only two partial multiplications. Multiply 765389 by 64164, and by 189279, and by 83256, making in each case only three partial multiplications. 10. There are 5 piles of shot in an arsenal : the first consists of 8436 68-pounders, the second of 1 1440 i8-pounders, the third of 24395 g-pounders, the fourth of 2567 95-pounders, and the fifth of 7580 36-pounders : find the weight in pounds of the five piles. §44. DIVISION. 21 DIVISION. 44. Division is the operation by which we find how many times one given number contains another given number. The first of these numbers is called the Dividend, the second the Divisor, and the number telling how many times the Quotient. 45. Since the Quotient tells how many times the Dividend contains the Divisor, it follows that the Dividend is the product of Divisor and Quotient. But since the terms of this product are interchangeable (35), the Divisor may be taken to represent a number of times, and then the Quotient will represent the number to be taken each time. Hence we may further say : — Division is the operation by which we break up a given number into as many equal parts as there are ones in another given number, and thus find one of these parts. It is under this view of Division that we have adopted the terms Dividend and Divisor. 46. The Quotient can always be found by successive ,, ^Dl 1 ■ la 1 subtractions of the Divisor ; for to divide 26 by 8 means g that we are to find how many times 26 contains 8, and 18 the operation at the side shews that 26 contains 8, 3 _i times with a remainder 2. It also shews that 26 divided 'g into 8 equal parts gives 3 in each part, leaving 2 un- — operated upon, that is with remainder 2. Here 26 is called the Dividend, 8 the Divisor, and 3 the Quotient. 47. When there is no remainder the division is said to be exact, and the Dividend is the product of Divisor and Quotient ; but when there is a remainder the Dividend is the product of Divisor and Quotient increased by this remainder. 48. Case I. When the Divisor is a simple number and the Dividend less than 10 times that number. Divide 35 by 7. By successive subtractions of 7, putting i for each subtraction in the Quotient (46), we find that 35 contains 7 S times exactly ; that is, the Quotient is 5. Or, from the Table we know that 7 taken 5 times is 35 ; that is, the Quotient is 5. 22 DIVISION. § 48- Divide 38 by 7. Now 35 contains 7 5 times, and 38 exceeds 35 by 3, where 3 is less than 7 : hence 38 contains 7 S times with a remainder 3 : that is the Quotient is 5 with remainder 3. 49. Case 11. When the Dividend is any number and (i) the Divisor is not greater than \^. Short Division. (2) the Divisor is any number. Long Division. For example divide 368549 by 678. Instead of successive sub- tractions, putting I for each subtraction in the Quotient (46), we will assume that at any step we may subtract any number of times the Divisor, if we put this number of times in the Quotient. For convenience draw curved lines on either side of the Dividend, placing the Divisor on the left, and reserving the right for the Quotient ; thus : — 678 ) 3685,49 ( 500 678 ) 3685,49 ( 543 339000 40 3390 2954,9 3 2954 2712 2712 (b) 2429 2429 2034 2034 395 395 From the left of the Dividend cut off a number not less than 678 but less than 678 multiplied by 10, or than 6780, that is cut off 3685 : this is our first partial dividend and represents hundreds. It will be found that 3685 contains 678 5 times but not 6 times : hence the Dividend contains 678 500 times but not 600 times (40) : put 500 in the Quotient and subtract 678 multiplied by 500, that is 339000, leaving a remainder 29549. Form a second partial dividend as the first was formed : it will be 2954, representing tens. It will be found that 2954 contains 678 4 times but not 5 times : hence this new dividend contains 678 40 times but not 50 times : put 40 in the Quotient and subtract 678 multiplied by 40, that is 27120, leaving a remainder 2429. Lastly our third partial dividend is 2429 itself, and it contains 678 3 times, but not 4 times : put 3 in the Quotient and subtract 678 multiplied by 3, that is 2034, leaving a final remainder 395. Hence the Quotient is 543 with a remainder 395. §49- DIVISION. 23 Now with regard to this operation we may observe : — (i) The partial dividends represent succeeding orders of units : and each partial dividend is formed from the partial remainder by placing to its right, or as it is usually expressed bringing down, the next figure of the Dividend. (2) In determining the figures of the Quotient we are only con- cerned with \h& partial dividends, and therefore we need only write down at each step the figures necessary to their formation. (3) Since the partial dividends represent every order of units from the highest down to simple units, the figures of the Quotient will represent the same, and no figure of the Quotient can exceed 9 ; we may therefore lose sight altogether of the order of units in each partial dividend, and take them for what they are intrinsically, writing down the figures of the Quotient as they arise one after another. Our work will then stand as at (b). From these considerations we have the following general Rule : — (i) On either side of the Dividend draw curved lines : place the Divisor on the left, and the figures of the Quotient as they arise on the right. (2) Frotn the left of the Dividend cut off a number not less than the Divisor but less than 10 times the Divisor, giving the first partial Dividend. Find how often the Divisor is contained in this dividends P^l the figure in the Quotient, multiply the Divisor by it, and subtract the product from the partial dividend. (3) To the remainder bring down the next figure of the Divi- dend forming the second partial dividend: proceed just as before, and continue the process till all the figures of the Dividend have been brought down, and the final remainder obtained. Short Division. When the Divisor is not greater than 12, we shall find no difficulty in forming the partial dividends without writing down the preceding products and remainders : and in this case, it will be convenient to write each figure of the Quotient under the units' figure of the partial dividend that gives rise to it. In every other respect we proceed as in Long Division. 24 DIVISION. § 50- SO. The Proof usually adopted in Division is this -.—To the pro- duct of the Divisor and Quotient add the Remainder; if the result coincides -with the Dividend (47) lue presume that the work is correct. Ex. I. Divide 612469 by 7. 7 ) 612469 7 ) 612469 87495-4 87495... 4 7 612469 From the left of the Dividend cut off a number not less than 7 but less than 70 : that is cut off 61, our first partial dividend. Now 7 is contained in 61 8 times and 5 over; put the 8 under the i in 61, and to the remainder 5 bring down the next figure of the Dividend 2, giving 52, the second partial dividend. But 7 is contained in 52 7 times and 3 over: put 7 in the Quotient, and to the remainder 3 bring down the next figure 4 giving 34, the third partial dividend : and so proceed. The above operation is usually performed in saying ; 7 in 61, 8 and 5 over; in 52, 7 and 3 over; in 34, 4 and 6 over; in 66, 9 and 3 over; in 39, 5 and 4 over; but the pupil should accustpm himself to bring down the figures of the Divi- dend to each remainder without previously mentioning the remainder, saying only: 61. 8; 52, 7; 34, 4; 66, 9; 39, 5 and 4 over. For the Proof we multiply the Quotient 87495 by 7 and add the remainder 4 ; and as the result coincides with the Dividend we may suppose that the work is torrect. Ex. 2. Divide 104051456 by 12. 12 ) 104051456 Here we say: — 8670954.. .8 104, 8; 80, 6; 85, 7; II, o; \^ "4i 9; 65. S; 56, 4 and 8 over. 10405 1456 Ex. 3. Divide 39875365 by 8654. 8654 ) 39875365 ( 4607 8654 34616 4607 52593 60578 Proof. 51924 519240 34616 6387 39875365 § so. DIVISION. 25 Cut off from the left of the Dividend a number not less than 8654 but less than 86540; that is cut off 39875, the first partial dividend. It con- tains the Divisor 4 times; put 4 in the Quotient, multiply 8654 by 4, placing the product under 39875, and subtract, leaving 5259. To the remainder 5259 bring down the next figure of the Dividend 3, giving 52593, the second partial dividend. It contains the Divisor 6 times, put 6 in the Quotient, multiply 8654 ^Y 6> placing the product under 52593, and subtract, leaving 669. To 669 bring down the next figure 6, giving 6696, the third partial dividend. It contains the Divisor o times ; put o in the Quotient, and the remainder is now 6696. To 6696 bring down the last figure 5, and proceed as before. Proof. Multiplying the Divisor 8654 ^1 1^^ Quotient 4607 and adding the remainder 6387, we obtain the Dividend. 51. We may dispense with writing down the products under each partial dividend, by carrying on the several steps of the multiplication and subtraction at the same time. To render each step of the subtraction possible, we may increase any order in the dividend by the requisite number of tens, if we increase the next order in the product by the same number of units (31, 2). And instead of subtracting the less from the greater at each step, we shall find it much more convenient to add to the less the number which gives the greater : thus instead of saying 28 from 35 leaves 7 we shall say 28 and 7 is 35, where the dark figure 7 is to be set down as soon as it has been pronounced. Therefore in the following example we say — 8654)68015,637(7859 74376 5 1443 81737 3851 28 and 7 is 35, where we have borrowed 30: 35 and 3 carried is 38; 38 and 3 is 41 : 42 and 4 46; 46 and 4 is 50: 56 and 5 61; 61 and 7 is 68. For the next step we say— 32 and 4 is 36; 40, 43, and 4 is 47; 48, 52 and 1 is 63 ; 64, 6^ and 6 is 74. 26 DIVISION. § 52- 52. Case III. When the Divisor is the product of two or more factors. If we multiply a number by 3 and its product by 5 we multiply the number by.lS ; hence, conversely, if we divide a number by 5 and its quotient by 3 we divide the number by 15. Divide 4586 by 105; the factors of 105 s gg being 3, 5 and 7. Dividing in succession by y^T^ 2 3,5 and 7 the final quotient is 43, and the 7)305. ..3. ..9. ..11 partial remainders are 2, 3 and 4: we must 43. ..4.. .60.. .71 now find the final remainder. For the first remainder to disappear 4586 must be diminished by 2 ; for the second remainder also to disappear 1528 must be diminished by 3, and therefore 4586 by 3 times 3 or 9 ; hence for both remainders to disappear 4586 must be diminished by 9 and by 2, or by II. For the last remainder to disappear 305 must be di- minished by 4, and therefore 1528 by 5 times 4 or 20, and therefore 4586 by 3 times 20 or 60; hence for all the remainders to disappear and the quotient 43 to be exact, 4586 must be diminished by 60 and by 11, or by 71. All which shews us, that when 4586 is divided by 3 the remainder is 2, when divided by 15 the remainder is 11, and when divided by 105 the remainder is 71. Hence we deduce the following Rule : — (i) The Quotient is found by dividing in succession by each of the factors of the Divisor: (2) The final remainder at each step is found by multiplying its particular remainder by all the Divisors preceding its own, and adding the preceding final remainder Ex. 4. Divide 87654321 by 378, whose factors are 6, 7 and 9. Dividing in succession by 6, 7 and 9 the f,\a r particular remainders are 3, 4 and 6. The ' — ~k^ — final remainder at the first step is 3; the ' '__4_9_5S"-3 final remainder at the second step is found sr ^ ' ' thus: — multiply its remainder 4 by the pre- '" " ceding divisor 6 giving •24, to which add the preceding remainder 3 giving 27. The final remainder at the third step is found thus: — multiply its remainder 6 by the preceding divisors 7 and 6 giving 352, to which add the preceding final remainder 27 giving 279. § S3- DIVISION. 27 53. Case IV. When ,the Divisor is terminated by one or more ciphers. (i) To divide a number by 10, 100, 1000,... Let the dividend be 345678. Now 345678 is the sum of 345670 and 8 ; but 345670 is the product of 34567 and 10 (40) ; therefore the quotient of 345670 by 10 is 34567 and therefore the quotient of 345678 by 10 is 34567 with a remainder 8. But if in 345678 we cut off one figure to the right, thus 34567,8 ; — the number on the left 34567 will give the quotient, and the number on the right 8, the remainder. In like manner to divide 345678 by 100 we cut off 2 figures to the right, thus 3456,78 : the quotient is 3456 and the remainder 78. And to divide by 1000 we cut off 3 figures, thus 345,678 : ,the quotient is 345 and the remainder 678. (2) Divide now by 700 : 700 is the product of 100 and 7, hence we first divide by 100, and then by 7 : that is we , ^ first cut off two figures to the right giving the — -— ' quotient 3456 and a remainder 78, and then we divide 3456 by 7 giving a quotient 493 and a remainder 5 : but this remainder 5 we multiply by the first divisor giving 500, and to it add the first remainder 78 giving a final remainder 578 (52), or in other words to the particular remainder 5 bring down the figures cut off 78, giving 578. Hence we have the following Rule : — Cut off all the ciphers 071 the right of the Divisor and as many fiptires from the right of the Dividend :— for the quotient^ divide the remaining figures of the Dividend by the remaining figures of the Divisor, and for the final remainder bring dbwn to the particular remainder the figures cut off from the Dividend. Ex. 5. Divide 3687594 by 80 and by 53000. 8,0 ) 368759,4 S3)00° ) 3687,594 (69 46094 74 318 507 477 30594 28 DIVISION. § 53- In the first example in dividing by 8 the remainder is 7, to which we bring down the figure cut off 4, giving 74 for the final remainder. In the second example the remainder in dividing by 53 is 30, to which we bring down the figures cut off 594, giving 30594 for the final remainder. Exercise 6. I. Divide 793046584 by 8, by 12, by 7, by 9, and by 11. 3. Divide 93538476? ^1 37. ^7 43. ^y 348. and by 4836. 3. Divide 237876093 by 5605, by 9089, by 40857 and 57085. 4. Find the quotient by short division of 76538959 by 28, and by 64, and by 72, and by 96. 5. Divide 3790603808 by 132, and by 196 (whose factors are 4, 7 and 7), and by 378 (whose factors are 6, 7 and 9). 6. Divide 78534826 by 60, by 800, by 12000, and by 3200. 7. Divide 3854269734 by 310, by 5900, by 587000, and by 90900. 8. Find the quotient of Fifty-seven miUions six thousand and seventy by six thousand and sixty-seven. 9. The product of two numbers is 17037006 and one of them is 4858, what is the other? 10. What number multiplied by 2951 will give 20376655? II. If the divisor be 3857, the quotient 489, and the remainder 1305, what is the dividend? 12. What number must we subtract from 57385, so that it can be exactly divided by 387 ; and what number must we add? 13. How many times in succession can 3589 be subtracted from 241462? What will the final remainder be? 14. How many times in succession must 1739 be added to 83487, to make the final sum 200000? 15. If the distance of the Sun from the Earth be 95000000 miles, and light travels from one to the other in 495 seconds, what is the velocity per second? 16. Write in figures, Twelve thousand twelve hundred and twelve. 17. Write down 576987, and under it write the eighth succeeding num- ber, and under this latter the next eighth succeeding number, and so proceed till nine numbers have been written down ; find their sum. 18. Take a set of numbers each of which is greater than the one pre- ceding, for example 67. 259. 3°0S. 463^20. 574008, 7654321; subtract each number from the one following, and add the remainders and the first number together; the sum ought to be the last number. CHAPTER III. PROPOSITIONS IN THE FUNDAMENTAL OPERATIONS. 54. The sign of Addition is + plus : thus 7+3 means that to 7 we are to add 3, and 7 + 3 + 2 means that to 7 we are to add 3 and to the sum we are to add 2 : although the result will be the same in whatever order we take these numbers together (25). The sign of Subtraction is — minus: thus T — }, means that from 7 we are to subtract 3, and 7-3 — 2 means that from 7 we are to subtract 3 and from the remainder we are to subtract 2. The sign of Multiplication is x or . multiplied by or into : thus 7 X 3 or 7 . 3 means that 7 is to be multiplied by 3, and 7x3x2 means that 7 is to be multiplied by 3 and the product is to be multipUed by 2 ; although the result will be the same in whatever order these numbers are multiplied together (61). Also 7 . 3 . 2 is called the continued product of 7, 3 and 2 ; and each number is called a. factor of the product. The sign of Division is + divided by or simply by: thus 24+3 means that 24 is to be divided by 3, and 24+3 + 2 means that 24 is to be divided by 3 and the quotient is to be divided by 2. Also 24+3x2 means that 24 is to be divided by 3 and then that the quotient is to be multiplied by 2. To these four signs of operation we shall afterwards add twp others ; those of Involution and Evolution. 55. The sign of Equality is = equals or is equal to : thus 7 + 3 = 10 means that the sum of 7 and 3 is equal to 10. When two or more numbers connected by signs of operation have a line called a vinculum drawn over them, or are enclosed in a bracket, the whole is to be treated as a single number : thus 7 + (5 — 3) means that to 7 we are to add the difference of 5 and 3 ; and (i8 + 7) + s means that \h&sum of 18 and 7 is to be divided by 5. It therefore follows from the last Art. that 7 + 3+2= (7 + 3) + 2; 7x3X2 = (7X3)x2 : and 24+3 x2=(24+3) x2. 30 Propositions in the § 65- The signs of Addition and Subtraction, ( + ) and (-), were first intro- duced by Michael Stifel, in ». work published by him at Nuremberg m 1544. The sign of Multiplication ( x ) was introduced by Wilham Oughtred in his C/flt^«iI/artm«ftV«, published in 1631. The sign of Division { -^ ) was invented by Dr John Pell, an English mathematician of the 17th century. The sign of Equality (=) was introduced by Robert Recorde m his Whet- stone of Wit, a work on Algebra, published in 1557. The Vinculum was first used by Vieta, the Italian mathematician, and the Bracket by Albert Girarde, a Dutch writer on Algebra. ADDITION AND SUBTRACTION. 56. To a number we may add the sum of two others by adding thetti in succession. Thus 29 + (7 + S) = 29 + 7 + 5. For to 29 we are to add 7 increased by 5 ; if therefore we first add 7 we must increase the result by 5, thus giving 29 + 7 + 5. In the same way we may shew that From a number we may subtract the sum of two numbers by subtracting them in succession; thus : — 29-(7 + 5) = 29-7-S- 57. To a number we mxiy add the difference of two others by adding the first and subtracting the second. Thus 29 + (7-5) = 29 + 7-5. For to 29 we are to add 7 diminished by 5 ; if therefore we first add 7 we must diminish the result by 5, thus giving 29 + 7 — 5. In the same way we may shew that From a num.ber we may subtract the difference of two others by subtracting the first and adding the second; thus : — 29-(7-5)=29-7 + 5- 58. Again, let the number to be added or subtracted be an expression made up of additions and subtractions, as 7 — 5 + 3 : then looking upon 7 — 5 as a single number, we have from the preceding cases 29 + (7-5 + 3)'=29+(7-5) + 3 = 29 + 7-5 + 3 and 29-(7-5 + 3) = 29-(7-5)-3 = 29-7 + 5-3. §59- FUNDAMENTAL OPERATIONS. %X 59. If an addition and a subtraction, or vice versd, have to be performed in succession, we may invert their order, provided the resulting expression be possible. For 9 = 9-3 + 3 therefore 9 + 5=9-3 + 3 + 5-(9 - 3) + 3+ S (SS) = (9-3) + 5 + 3, (25) therefore 9 + 5-3=9-3 + 5 + 3-3 = 9-3 + 5- 60. Hence it is easily shewn that (i) Additions and subtractions may be performed in any order. (2) An expression m.ade up of additions and subtractions tnay be made equal to the difference of two sums; thus : — 9-8 + 7-6+ 5 -4=(9+7 + 5) -(8 + 6 + 4). MULTIPLICATION AND DIVISION. 61. The product of two or more numbers will remain the same, however we may change the order of its factors. (i) Let there be two factors ; we may change their order (35), thus 7. 5 = 5.7- (2) Let there be three factors ; we may change the order of the last two, thus 8.7.5=8.5.7. For write „ „ „ „ o- ,. IV ■ J 8888888 o in a horizontal line 7 times, and repeat 8 8 8 8 8 8 8 this line 5 times. Now the sum of the 8 8 8 8 8 8 8 numbers in each line is 8 . 7, and there are 0000000 ,,. ,. . 0000000 5 such lines, therefore the total sum is 8.7.5. Again the sum of the numbers in each vertical line is 8 . 5, and there are 7 such lines, therefore the total sum is 8.5.7: that is 8 . 7 . 5 = 8 . 5 . 7. (3) Let there be any number of factors as in 4 . 2 . 7 . 5 . 3 ; any one of these factors, 5 for example, may occupy the first place. For 4. 2. 7. 5 = (4. 2). 7.5 = (4. 2). 5. 7 (61,2) = 4.2.5.7, therefore 4.2.7.5.3=4.2.5.7.3; that is in the given product, 5 has changed ^places with the preced- ing factor 7 giving 4.2.5.7.3; in like manner it may now change 32 FUNDAMENTAL OPERATIONS. § 6l. places with the preceding factor 2 giving 4.5.2.7.3; and lastly it may change places with the first factor 4 giving 5.4.2.7.3. In the same way any other factor may occupy the second place, any remaining factor the third place, &c. ; thus the factors may occupy any places we please. 62. If we multiply separately each of the parts which compose one number by each of the parts which compose another number, the sum. of these partial products will give the product of the two numbers. (i) Let the Multiplicand be 11 or 4 + 5 + 2, and the Multiplier 3 ; then their product is the sum of 3 numbers each equal 4 + c + 2 to 4+5+2 (34); and from the arrangement at the side, 4+5 + 2 this sum is the sum of 4 taken 3 times, of 5 taken 3 4 + 5+2 times, and of 2 taken 3 times, that is of 4 . 3, of 5 . 3 and of 2 . 3 ; or (4 + 5 + 2)x3=4.3 + 5.3 + 2.3. (2) Let the Multiplicand be 11 and the Multiplier 13 or 6 + 7, then iix(6 + 7) = (6 + 7)xii (35) = 6.11+7.11 (62, i) = 11.6+11.7. (3S) (3) Let the Multiplicand be 4 + 5 + 2 and the Multiplier 6 + 7, then (4 + 5+2)x(6 + 7) = (4+s+2)x6 + (4 + 5 + 2)x7 (62,2) =4.6 + 5.6 + 2.6 + 4.7+5.7 + 2.7. (62, i) Hence the truth of the proposition. Remark. The theory of Multiplication is involved in the three cases of this proposition, as may be seen by referring to Arts. 39, 40 and 41. 63. (1) In like manner if the Multiplicand be the difference of two numbers as g-2, and the Multiplier another number as 3, we may shew that (9-2)x3=9x3-2x3, and interchanging Multiplicand and Multiplier we have further 3 X (9-2)^9. 3 -2. 3, or =3-9-3.2, §63. INVOLUTION. 33 Cor. Hence to multiply a number by 99, 999,. •• that is by 100— I, 1000— I,... we may multiply by 100, 1000,... and subtract the number: to multiply by 98, 998,... we may multiply by 100, ic»o,... and subtract twice the number. (2) Again, if the Multiplicand be an expression made up of additions and subtractions as 9-7 + 5 — 2, and the Multiplier any number as 3, then looking on 9 - 7 + 5 as a single number we have (9-7 + S-2)x3 = (9-7 + s)x3-2.3 (63,1) =(9-7)x3+S-3-2.3 (62,1) = 9-3-7-3 + S-3-2.3- (63.1) 64. If Divisor and Dividend be multiplied by the same number, the Quotient will remain the same, but the Remainder will also be multiplied by thai number. For example, if the Quotient of 38 by 7 is 5 with Remainder 3, the Quotient of 38 . 4 by 7 . 4 will also be 5, but with Remainder 3 . 4. Now 38 = 7-5 + 3, (47) and multiplying each member of this equality by 4, we get 38.4=7.5.4+3.4 (62,1) =(7 -4) -5 +3 -4, (60) which shews that the Quotient of 38 . 4 by 7 . 4 is 5 with Re- mainder 3 . 4. INVOLUTION. 65. Involution is the operation by which we find any power of a given number. The second, third, ionxih.,... power of a number is the product of 2, 3, 4,... factors each equal to that number: thus the second power of 4 is 4 . 4, the third power of 5 is 5 . 5 . 5. The second and third powers of a number are usually called its square and its cube. The square, cube, fourth,... power of a number is represented by placing a small 2, 3, 4,... to the right of the number, a little above the Kne : thus the square of 6 is represented by 6^. Hence the first power of a number will be the number itself, or 5' is the same as 5. These small figures denoting the powers of a number are called its indices or exponents. B.-S. A. 3 34 INVOLUTION. § 66. 66. The product of two or more powers of the same number is a power of that number whose index is the sum of the indices of the factors. Since 7^ is the product oif 2 factors each equal to 7, and 7' the product of 3 factors each equal to 7, we have ^2^^3 = 7x7x7x7 y.'j = ']^'^'^. Again 72 x 7^ x 7* = 72+^ x 7* = 7^+3+4^ and so if there are more factors. Cor. To raise a power to another power we multiply the first index by the second. For (7<)3=7*x7''x7'' = 7«+*+^ = 7<'". Ex. I . Find the value of (1536- 1392) -(29 + 7). This expression means that we are to divide the difference of 1536 and 1392 by the sum of 29 and 7 ; but the difference of 1536 and 1392 is 144, and the sum of 29 and 7 is 36, therefore (1536- i392)H-(29 + 7) = 144-^-36-4. Ex. 2. Find the value of (iS36-487)-i392-=-29-H7x5. This expression consists of 3 terms: the first is (1536-487), the second 1392-^-29, and the third 7.5; it therefore means that from the difference of 1536 and 487, we are to subtract the quotient of 1392 by 29, and to the result we are to add 'Cae. product of 7 and 5 ; therefore (1536 - 487) - i392H-29-f 7 . 5= 1049 -484-35 = 1036. Ex. 3. Find the value of (194 + 65) X 7 -H (352-220)-=- 1 1 -952-=-(9i -35). This expression means that to (194-1-65) x 7 we are to add (352 - 220)4-11 and from the sum subtract 962-^(91-35); hence given expression = 259 x 7 + 132-7-11 - 952-^56 = 1813 + 12- 17 = 1808. Ex. 7. FUNDAMENTAL OPERATIONS. 35 Exercise 7. Find the value of 1. 271048 + 37506 + 6709 + 209 + 740087. 2. 67351-3985-4537 + 80359-57036. 3- (7805 + 3907) + (7805 -3907); (7805 + 39°7)- (7805- 39°7)- 4- (7805 + 39°7)>< (7805 -39°7); 7805 + (3907 -1996) -854- 5- 7805 + (3907 - 1996) X 854 ; 324-19x17; (324-i9)xi7. 6. 756x3-25x16+3^; 1536-^8 + 9.125-100. 7. 5^-^4-1392-^(29-5); i536-i392-=-29+i6. 8. 588o+(i67-i32)x6; 5880-^(35-^7) + 17 . 12. 9. (67893-8637)-r-823 + 7546x (2356-945) -9870. 170. 10. 59256-^72x91-^(130-117); i9.20.2i-=-(3.5.7). 11. Express the following statement by signs :— Multiply the difference between 325 and 293 by the quotient of 306 by 17, and to the product add the sum of 1000 and 99. Find the value of the expression. 12. Express by signs : — From 34856 subtract the product of 763 and 41, and to the remainder add the quotient of 1998 by the difference between 663 and 441; and then find the value of the expression. 13. The product of 175 and 53 is 9275 : how much must be added to it to give the product of 177 and 53; and to give the product of 175 and 59 (see Art. 62)? 14. The product of 3267 and 71 is 231957: how much must be sub- tracted from it to give the product of 3267 and 68 ; and to give the product of 3258 and 71 (see Art. 63)? 15. Multiply 387659 by 85672 in 3 lines. 16. Find the prime number expressed by 1251^+ 2920^. 17. The following results are true when a and b stand for any numbers, which will allow them to be arithmetically possible. Shew that they are true in each case, when 3=935 and * = 396 : — (i) (a + *) + (a-*) = 2-3- (2) \a + b)-(a-b) = '!..b. (3) (fl + *)x(a-i)=o»-i52. (4) {a^bf-{a-Vf=i^.a.b. (5) aH2.o.* + *2=(a + *)2. (6) a^-^.a.b->rb-' = {a-bY. (7) (a^b).[d?-a.b^b-^^a^^t^. (8) (a-b).{a^-\-a.b + l>')=d^-lr'. 3—2 CHAPTER IV. GREATEST COMMON MEASURE: LEAST COMMON MULTIPLE, &C. GREATEST COMMON MEASURE. 67. One number is a measure, or a submultiple, or an aliquot part of another, when it divides that other exactly (47) : thus 4 is a measure, or a submultiple, or an aliquot part of 20, but not of 21. One number is a multiple of another, when it can be divided by that other exactly : thus 20 is a multiple of 4, and of 5, but not of 6. The Greatest Commoti Measure (G.C.M.) of two or more numbers is the greatest number that divides each of them exactly. The Least Common Multiple (l.c.m.) of two or more numbers is the least number that can be divided by each of them exactly. 68. A Prime nutnber, or a Prime, is a number divisible only by. itself and by i : thus 2, 3, 5, 7, 11 are primes. Two numbers are prime to each other, when their only common measure is i : hence (i) Two primes must be prime to each other. (2) A firime num.ber must be prime to every number not divisible by it. 69. A measure of each of several numbers is a measure of their sum. Let 7 be a measure of each of the numbers, then each number is composed of parts each equal to 7, and therefore their sum is composed of parts each equal to 7; that is, 7 is a measure of their sum. Cor. a measure of any number is a measure of any multiple of it. 70. A measure of each of two numbers is a measure of their differences also of the difference between the first and any multiple of the second. §7o. GREATEST COMMON MEASURE. 37 Let 7 be a measure of each of them, then each is composed of parts each equal to 7, and therefore their difference is composed of parts each equal to 7 : that is 7 is a measure of their difference. Again, since 7 is a measure of the second number, it is a measure of any multiple of the second number : and is therefore a measure of the difference between the first number and any multiple of the second. 7 1 . To find the G. C. M. of any two numbet's. (i) The G.CM. of Dividend and Divisor is also the G.C.M. of Divisor and Remainder, and vice versa. The Remainder is the difference between the Dividend and a multiple of the Divisor, and therefore every common measure of Dividend and Divisor is a measure of the Remainder (70), and is therefore a common measure of Divisor and Remainder. Again, the Dividend is the sum of a multiple of the Divisor and the Remainder, therefore every common measure of Divisor and Remainder is a measure of the Dividend (69), and therefore a com- mon measure of Divisor and Dividend. Hence Dividend and Divisor, and Divisor and Remainder, have the same common measures ; and therefore the G. C. M. of Dividend and Divisor is the G.C.M. of Divisor and Remainder. (2) Find the G.C.M. of 304 and ,0,2 304)1072(3 Divide 1072 by 304 giving the Re- l6o)3o4( i mainder 160; then divide 304 by 160 j6o giving the Remainder 1 44 ; then divide 1 44 ) 1 60 ( i 160 by 144 giving the Remaindei: 16: i^ lastly divide 144 by 16 giving no Re- j^^ mainder. Now the G.C. M. of 304 and 1072 is also the G.C.M. of 160 and 304 ; and the G.C.M. of 160 and 304 is also theTJ.CM. of 144 and 160; and the G.CM. of 144 and 160 is also the G.C.M. of 16 and 144 ; but the G. C. M. of 16 and 144 is 16 ; therefore the G. c. M. of 304 and 1072 is 16. 38 GREATEST COMMON MEASURE. §7> We have then the following Rule : — Divide the greater number by the less, the less by the first re- mainder, the first remainder by the second remainder, and continue this process till we come to u remainder that exactly divides the preceding remainder: such last remainder is the G.C.M. Remark i. We shall find it convenient to place the numbers in a horizontal line, drawing a vertical line between them, and then carry on the divisions alternately on the right and left of this line. We may neglect the quotient figures. Remark 2. At any step of the process we may find the differ- ence between the greater number and the nearest multiple of the less, regardless whether that nearest multiple be less or greater than the greater number. Ex. I. Find the G.C.M. of 93883 and 166581. 93883 72698 166581 93883 93883 84740 166581 187766 21185 18286 72698 63SSS 9143 8697 2I185 18286 2676 9143 8697 446 446 2676 223 446 446 • rt r lu - 223 223. 72. Every common measure of two numbers is a measure of their G.C.M.. For being a measure of the two numbers it is a measure of the first remainder, and being a measure of the less number and the first remainder, it is a measure of the second remainder ; and in like manner it is a measure of each succeeding remainder : it is therefore a measure of the last remainder, that is of the G.C.M. of the two numbers. T^. To find the g. cm. of three or more numbers. Let 304, 1072, 40 and 36 be the given numbers : and let 16 be the common measure of 304 and 1072. Now every common measure §73- GREATEST COMMON MEASURE. 39 of the given numbers is a common measure of 304 and 1072, and therefore of i6 their g.C.m. (72), and therefore of 16, 40 and 36. Again, every measure of 16 is a common measure of 304 and 1072 (69 Cor.), and therefore every common measure of 16, 40 and 36 is a common measure of 304, 1072, 40 and 36. Hence 304, 1072, 40 and 36 have the same common measures as 16, 40 and 36 : and therefore the G. C. m. of the given numbers is the G.C.M. of 16, 40, 36. In like manner if the G.C.M. of 16 and 40 be 8, the g.cm. of 16, 40 and 36 is the G.C.M. of 8 and 36 : and therefore the G.C.M. of the given numbers is the G.C.M. of 8 and 36. And in this way step by step may be found the G.C.M. of as many numbers as we please. Hence we have the following Rule : — Find the G.C.M. of the first two numbers j then the G.C.M. of this G.C.M. and the third numbers then the G.C.M. of this last G.C.M. and the fourth number; and continue this process unto the last number: the last G.C.M. is the G.C.M. of the given numbers. 74. If two numbers be multiplied by a third, the G.C.M. of the products will be the g.C.m. of the given numbers multiplied by the third num.ber. Thus if the G.C.M. of 304 and 1072 be 16, the G.C.M. of 304. 7 and 1072. 7 will be 16.7. Referring to Art. 71 we see that the remainders in finding the G.C.M. of 304 and 1072 are successively 160, 144 and 16 : hence (64) the remainders in finding the G. cm. of 304 . 7 and 1072 . 7 will be 160 . 7, 144 . 7, and 16.7; that is if the G.CM. of 304 and 1072 is 16 the G.C. M. of 304 . 7 and 1072 . 7 is 16.7. Cor. I. Conversely, since the G.C.M. of the second set of numbers is the G.C.M. of the first set multiplied hy 7, the G.CM. of the first set is the G. c M. of the second set divided by 7. Cor. 2. If we divide two numbers by their G. C. M. the quotients will be prime to each other, for their G.CM. will be i. 40 GREATEST COMMON MEASURE. §74- Ex. 2. Find the G.C. M. of 285600 and 621600. 800 ) 285600 621600 3) 357 777 119 126 259 238 7 21 21 It is seen that the g.c.m. of 119 and 259 is 7, therefore the G.C.M. of 357 and 777 is 7 . 3, and the g.c.m. of the given numbers is 7 . 3 . 800, or 16800. 75. If a number divide a product of two factors and be prime to one of them, it must divide the other. For example, if 4 divide 7 . 24 and 4 is prime to 7, then 4 must divide 24. The G.C.M. of 4 and 7 is I, therefore the G.C.M. of 4. 24 and 7. 24 is 1 . 24 (74) or 24. Now 4 is a measure of 4. 24, and by hypothesis of 7 . 24, therefore it is a measure of their G.C.M. 24 (72) ; that is 4 divides 24. 76. The G.C.M. of two numbers will remain the same, if one of them be multiplied or divided by a number prime to the other. Thus if 5 be prime to 24, the G. c. M. of 24 and 16 is the same as - the G. c. M. of 24 and 5.16. Eveiy measure of 16 is a measure of 5 . 16 (69 Cor.) : therefore every common measure of 24 and 16 is a common measure of 24 and s . 16. Again, since 24 is prime to 5, every measure of 24 is prime to 5, therefore every measure of 24 which measures 5 • 16 must measure 16 (75), and therefore every common measure of 24 and 5 . 16 is a common measure of 24 and 16. Hence 24 and 16, and 24 and 5 . i6 have the same common measures, and therefore the same g.C.m. Ex. 3. Find the G.C.M. of 39835 and 162424 39835 7967 7196 257 ^03^ ^'""'^ '5835 by 5, for 5 is prime to 162424, and divide 23901 '^^*^''- ^y 8, for 8 is prime to 7967 ; we have therefore "3598 *° ^"^^ *^ G.cyL. of 7967 and 20303. After the first 1799 ^*^P ^^ '^^^^ '° '5n 32, 8 2 )3, 18, 5, 32, 8 9, S, 16 which shews that the L. c M. of the numbers in the first line is the L.C.M. of the numbers in the second line multiplied by 5. Again, of the numbers in the second line, 3 and 8 being factors of 18 and 32 may be suppressed ; and of the others 18 and 32 are divisible by 2, which is prime to 5 : hence the L.CM. of the numbers in the second line is the L.CM. of the numbers in the third hne multiphed by 2. Lastly, there is no number that will divide any two of the numbers 9, 5 and 16 ; hence their L. CM is their product 9 . 5 . 16 ; and therefore the L.CM. of the numbers in the second line is 9.5. 16.2 ; and of the given numbers 9.5 . 16.2. 5. We have then the following Rule :— Write down the given numbers in a horizontal line, suppressing those that are divisors of any of the others. Seek out two numbers at least, that have a prime divisor 2, 3, 5, 7, H--- ; divide them by 46 LEAST COMMON MULTIPLE. § 83. 3) 8, 9, 10, 12, 25, 27, IS 5) 8, 3, 10, 4> 25, 9> 5 this prime, placing the quotients and the undivided numbers in the line below. Proceed in the same way with the numbers in the second, and each succeeding line, till we come to a line where no two numbers have a common divisor. The product of the numbers in the last line and of the several divisors is the L. c. M. of the given numbers. Example. Find the l.c.m. of 15, 16, 18, 20, 24, 25, 27 and 30. 2 ) *5, 16, 18, 20, 24, 25, 27, 30 S ) 15, 16, 18, 20, 24, 25, 27, 30 3 ) 16, 18, 4, 24, 5, 27. 6 3 ) 16, 6, 8, s, 9 8, 2, s, 9 16, 2, 5, 3 .-. L.c.M. = 2. 3. 5.8.5. 9=10800. .-. L.C.M.=5. 3. 3. 16.5.3=10800 In the preceding Example we might begin by taking either 2 or 3 or 5 as a divisor ; but instead let us take at once 2.3.5 or 30 as the divisor : then the first number 16 can be divided by 2 giving 8 : the second number 18, by 2 . 3 or 6 giving 3 : the next number 20, by 2 . 5 or 10 giving 2 : and so on. The work will therefore stand thus 2 . 3 . 5 or 30 ) ^5, 16, 18, 20, 24, 25, 27, 30 8, 3, 2, A, 5) 9) .-. L. CM. = 30. 8. 5. 9 =10800. Exercise 9. Find the l.c.m. of I. 156 and 429; 1287 and 6281; 729 and 1681. ■L. 7469 and 102025 ; 8975 and 25489; 12432 and 36075. 3. 7247 and 9365; 2341 and 203667; 25664 and 781033. 4. 629, 851 and 253; 1003, 2301 and 4017. 5. 2691, 6435 and 8349; 2523, 5887, 203 and 8631. 6. the first 12 numbers; the first 1 2 odd numbers. 7. the even numbers from lo to 28 inclusive. 8. the odd numbers from 13 to 27 inclusive. 9. 12,16,18,28,32,40,42; 9.12,15,18,21,24,27,30. 10. 15,18,24,33,40,54,55; 32.63,25,36,42,49,84. Ex. 9. LEAST COMMON MULTIPLE. 47 II- 35, 62. 63, 77, 132, 117, 143; 18, 21, 11, 45, 99, 154, 168. 12. 2S< 14. 35. 12. 21. 105, 72, 112; 27, 91, 42, 39, 63, 156, 234. 13. 187, 121, 119, 154, 385, 165, 231, 340. 14- 24, 35, 52, 60, 91, 108, 126, 156, 315. 15. 27, 87, 189, 126, 145, 210, 203, 261, 385. 16. Seven bells are tolling, and they toll at intervals of 3, 5, 7, 8, 9, 10 and 12 seconds respectively. What interval will elapse between their once tolling together, and tolling together again? 17. Two cog-wheels containing 80 and 128 cogs respectively are work- ing together : after how many revolutions of the smaller wheel will two cogs which once touch, touch again? 18. Find the least number which divided by 6, by 8, and by 9, gives in every case the remainder 4. 19. Find the least number which divided by 675, 1050, and 4368, will leave the same remainder 32. 20. A heap of pebbles can be made up exactly into groups of 25 j but when made up into groups of 18, 27, and 32, there is always a, remainder of 1 1 ; find the least number of pebbles such heap can contain. 21. Three horses are running round a race-course of 5280 yards: the 6rst horse runs 440 yards a minute, the second 352 yards, and the third 264 yards : find the time between their once coming all together, and their coming all together again. CRITERIA OF DIVISIBILITY. 84. One number is divisible by another when it can be divided by that other exactly (47), that is when the former is a multiple of the latter (67). 85. The sum of several multiples of a given number is a multi- ple of that number. For each multiple may be considered as made up of parts each equal to the given number, therefore their sum will be made up of parts each equal to the given number, or will be a multiple of the given number. Cor. Any multiple of a multiple of a given number is a multiple of that number. 86. Any number is divisible by 2, 4, Z,.. .if the number expressed by the last one, two, three,.. .figures to the right be divisible by 2, 4, &,... 48 CRITERIA OF DIVISIBILITY. § 86. Any number 87654 is the sum of 87650 and 4 ; and the first part 87650 is divisible by 10 (53) and therefore by 2 (85 Cor.), for 10 is a multiple of 2: therefore the whole 87654 is divisible by 2, if 4 be divisible by 2. Again, 87654 is the sum of 87600 and 54; and the first part 87600 is divisible by 100 (53) and therefore by 4, for 100 is divisible by 4 ; therefore the whole 87654 is divisible by 4, if 54 be divisible by 4. In like manner 87654 is divisible by 8, if 654 be divisible by 8, &c. Cor. Hence those numbers only, whose units' figures are o, 2, 4, 6 and 8, are divisible by 2 : such numbers are called even numbers ; and numbers not divisible by 2, or numbers whose units' figures are i, 3, 5, 7 and g, are called (?^i^ numbers. 87. Any number is divisible by 5, 25, 125,... if the number ex- pressed by the last one, two, three,... figures to the right be divisible '^J'S, 25, 125,... Proceed exactly as in the last Article. Cor. Hence those numbers only, whose units' figures are o and 5, are divisible by 5. 88. 71^1? remainder in dividing any number by 9 is the same as iti dividing the sum of its figures by 9. Take i followed by any number of ciphers, as loooo ; now 10000 = 9999+ i=a multiple of 9+ i. Multiply these equals by any simple number 6, then 60000 = a multiple of 9 + 6. (69) Now 87654 is the sum of 86000, 7000, 600, 50 and 4 ; and 80000 = a multiple of 9 + 8, 7000 = a multiple of 9 + 7, 600 = a multiple of 9 + 6, 50 = a multiple of 9 + 5, 4= 4; therefore 87654 = a multiple of 9 + 8 + 7 + 6 + 5 + 4; (85) § 88. CRITERIA OF DIVISIBILITY. 49 and therefore the remainder in dividing 87654 by 9 is the sarne as in dividing the sum of its figures 8+7 + 6 + 5 + 4 by 9. Cor. I. Hence any number is divisible by 9, if the sum of its figures be divisible by 9. Cor. 2. Since 9 is a multiple of 3, any number 87654 = a multiple of 3 + (8 + 7 + 6 + 5 + 4) ; therefore any number is divisible by 3, if the sum of its figures be divisible by 3. 89. Any number is divisible by 11, if the difference between the sum of its figures in the odd and in the even places be o, or be divisible by 11. Take i followed by an even number of ciphers as loooo ; now 10000 = 9999+ i = a multiple of 11 + i, therefore 30000 = a multiple of 1 1 + 3. (62) Again, take I followed by an odd nixvc^zx of ciphers as 1000; now 1000 = 990+ 10 = 990+11 — I =a multiple of 11 — i ; therefore 3000 = a multiple of 1 1 - 3. (63) Any number 87654 is the sum of 80000, 7000, 600, 50 and 4 ; and 80000 = a multiple of 1 1 + 8, 7000 = a multiple of 11— 7, 600 = a multiple of 1 1 + 6, 50 = a multiple of 1 1 — 5, 4= 4; therefore 87654=a multiple of 11 +4-5 + 6-7 + 8 (85) = a multiple of 1 1 +(4 + 6 + 8) -(5 + 7); (59,2) and therefore 87654 is divisible by 1 1 if the difference between the sum of its figures in the odd and in the even places be o, or be a multiple of II. 90. Criteria of Divisibility when the Divisor is not greater than 12. Since 2 and 3 are prime to each other, any number divisible by 2 and by 3 is divisible by 2 . 3 or 6 ^J^) : and any number divisible by 3 and by 4 is divisible by 12. B.-S. A. 4 50 CRITERIA OF DIVISIBILITY. § 90. Therefore any number is divisible by 2, if its units' figure be even (86) ; 3, if the sum of its figures be divisible by 3 (88) ; 4, if the number formed by its last 2 figures to the right be divisible by 4 ; 5, if its units' figure be o or 5 (87) ; 6, if it be divisible by 2 and by 3 (90) ; 7, by trial ; 8, if the number formed by the last 3 figures to the right be divisible by 8 ; 9, if the sum of its figures be divisible by 9 ; 10, if its units' figure be o ; 1 1, if the difference between the sum of its figures in the odd and in the even places be o, or be divisible by 1 1 (89) ; 12, if it be divisible by 3 and by 4. 9 1 . Proof of Multiplication, by casting out the nines. ■- — Casting the nines out of the Multiplicand and Multiplier, suppose the remainders to be 5 and 3, so that Multiplicand = a multiple of 9 + 5, Multiplier =; a multiple of 9 + 3. Nowr we may multiply Multiplicand and Multiplier together by multiplying each part of the first number by each part of the second (62). The first part of this product will be the product of two multiples of 9, and therefore a multiple of 9 (85 Cor.), the second part will be 5 times a multiple of 9, the third part 3 times a muhiple of 9, and the fourth part 5 . 3. But the sum of the first three parts will be a multiple of 9 (85), hence Product = a multiple of 9 + 5 . 3, and therefore the remainder in dividing the Product by 9 is the same as in dividing 5 . 3 by 9 ; that is, in casting the nines out of the Product and out of 5 . 3 the results ought to be the same : hence the usual Rule (43). ' Remark. This Proof will not point out an error in the product (i) if a o be set down for 9, or vice versd ; or (2) if a figure has been set down as much too high, as another has been too low ; §91- PRIMES AND PRIME FACTORS. 5 I or (3) if a partial product has been set down in a wrong place ; or (4) if one or more noughts have been omitted in the product. PRIMES AND PRIME FACTORS. 92. We shall take it for granted that every number is either a prime or the product of two or more prime factors. In the latter case the number is sometimes called composite. To decompose a number into its prime factors is to find those prime numbers which when multiplied together produce the given number, as when 210 is put under the form 2 . 3 . 5 . 7, or 504 under the form 2' . 3^ . 7. 93. To ascertain what numbers are prime. (i) Every number whose units' figure is o, 2, 4, 6, or 8 is divisi- ble by 2 (86), and therefore every such number except 2 itself is not a prime. Every number whose units' figure is o or 5 is divisible by 5, and therefore every such number except 5 itself is not a prime. Hence the units' figure of every prime number, except 2 and 5, must be I, 3, 7 or 9. (2) If then the units' figure of the given number be i, 3, 7 or 9, try as divisors one after another the primes 3, 7, 11, 13,... until we either obtain an exact quotient, in which case the number is not prime, or until we obtain a quotient less than the divisor, in which case it is prime ; for if the given number were divisible by- a prime which gives a quotient less than the divisor, it must also be divisi- ble by this quotient ; and this quotient is either a prime, or the product of two or more primes, and therefore the number would be divisible by a prime which gives a quotient greater than the divisor ; but it is not : hence the number is a prime. Ex. I. Is 689 a prime number ? 689 is not divisible by 3 (88, Cor. 2), nor by 7 by trial, nor by 1 1 (89), but is divisible by 13 : therefore 689 is 7iot a prime. Ex. 2. Is 947 a prime ? 947 is not divisible by 3, 7, 11, 13, 17, 19, 23 or 29; and when divided by the next prime 31 (93), the quotient is less than 31 : therefore 947 is a prime. 4—2 52 PRIMES AND PRIME FACTORS. | 94. 94. To decompose a number into its prime factors. Let the number be 44856: it is divisible by 2 giving 2 )44856 the quotient 22428; 22428 is divisible by 2 giving the 2 )22428 quotient 11214; 11214 is divisible by 2 giving the quo- 2 )11214 tient 5607 ; and 5607 is not divisible by 2. 3 )S6o7 Again, 5607 is divisible by 3 (88, Cor. 2) giving the 3) '069 quotient 1869: 1869 is divisible by 3 giving the quotient —^ 623 ; but 623 is not divisible by 3. ■ Nor is 623 divisible by the next prime 5 (87) ; but it is divisible by 7 giving the quotient 89, and 89 is not divisible by 7. Lastly, 89 is not divisible by the next prime 1 1 ; moreover any greater prime than 11 would give a quotient less than itself: there- fore 89 must be prime (93). Hence 44856 = 2.2. 2.3 . 3. 7. 89 = 2'. 3^. 7. 89. And as the same method is applicable to every number, we have the following Rule : — Divide in succession, and in each case as often as possible, by each of the prhnes 2, 3, 5, 7, 11,... which can be used as divisors, until we come to a quotient which is prime (68)/ these divisors and the last quotient put under the form of- a product express the de' composition of the given number into its prim^ factors. Ex. 3. Resolve S^Szyjy into its prime factors. 3^=9 ) 8862777 ^y adding the figures together, we find that the 3^=9 ) 984753 number is divisible by 9 or 3^: in like manner the 7) 109417 quotientis divisible byg or 3^; butthequotient 109417 7 ) 15631 is not divisible by 3. We now try 7 as many times in 7 ) 2233 succession as possible, and we get the quotient 319 1 1 I 31Q which is not divisible by 7 ; we therefore try 1 1, giving ^ the quotient 29, and 29 is prime; therefore 8862777 = 3^3=. 7.7.7.11. 29=3^. 7S. II . 29. 95. If a number be prime to each of two others, it is prime to their product. Thus if 8 is prime to 7 and to 15, it is prime to 7 . 15. For if not, 8 and 7.15 have a common measure : let it be 2. Now since 8 is §95- PRIMES AND PRIME FACTORS. S3 prime to 7 and to 15, every measure of 8 is prime to 7 and to 15, and therefore 2 is prime to 7 and to 15. But again, 2 divides 7 . 15 and it is prime to 7, therefore it divides 15 (75): but it is prime to 15, which is impossible. Therefore 8 is prime to 7 . 15. Cor. I. If one number be prime to another, it is prime to any power of that other. Thus if 3 be prime to 7 it is prime to 7 . 7 or 72, and therefore to 7^. 7 or 7', and therefore to 7*, 7°,... Cor. 2. If one nutnber be prime to another, any power of the former is prime to any power, of the latter. Thus if 3 be prime to 7, it is prime to any power of 7 (Cor. i), that is, any power of 7 is prime to 3, and therefore to any power of 3 (Cor. i). 96. The G.C.M. of two or m,ore numbers may be found by de- composing them into their prime factors, and forming the product of the least powers of those factors which are common to all the given numbers. Suppose the given numbers when decomposed into their prime factors to be 2'.3^5^7; 2*. 33. 52; 2^32.5.7'; and2^3^5^7^ then their G.C.M. will be 2^ . 32 . 5. For take the number in which the lowest power of 2 occurs, namely 2' . 3* . 5^ . 7 ; this number is divisible by 2^, but since 2 is prime to 3, 5, and 7, it is prime to 3*- S'- 7 (95)> and therefore 2''.3*. 5^7 is not divisible by a higher power of 2 than 2^; and therefore a higher power of 2 than 2^ cannot occur as a factor in the G.CM. Similarly a higher power of 3 cannot occur as a factor in the G.C.M. than 3"; nor of S than 5 ; and 7 cannot occur at all, for 2*. 32. 5 is not divisible by 7. Nor can any other prime num- ber occur as a factor in the G.C.M. such as 11, 13,... for it would be prime to each of the given numbers (95). Therefore the G.C.M. ' required cannot be a number greater than 23. 3'. 5. But again, each of the given numbers contains all the factors of 2^*. 3^.5, 9r 2'. 32. s is a common measure of those numbers ; hence 2^. 3^. 5 is the G.C.M. required. . 54 PRIMES AND PRIME FACTORS. § 97. 97. The L.C.M. of two or more numbers may be found by de- composing them into their prime factors, and forming the product of the highest powers of all the factors that occur z« the given numbers. Suppose the given numbers, when decomposed into their prime factors, to be ■2>.t-f-T, 2^3'-S'; 2^3^5.7'; and 2=. 3'. 53.73, then their L.C.M. will be 2* . s'' . 5' . 7'. For take the number in which the highest power of 2 occurs, namely 1^ .f.f.^'^^, in order that the L.C.M. required may be divi- sible by this number, it must be divisible by 2', that is, by the highest power of 2 that occurs in the given numbers. In like manner, it must be divisible by the highest powers of all the other numbers that occur as factors in the given numbers, that is, by 3*, s', and 73 ; or the L.C.M. must be divisible by 2°, 3*, 58, and 7'- And since each of these numbers is prime to all the others, the L.C.M. must be divisible by 2= . 3* . 5' . 73 (^j^)^ and cannot therefore be less than iK 3*. f. 73 itself. But again, iK 3^ 53. 73, containing as it does all the factors of each of the given numbers, is a common multiple of them ; hence 2^ . s* . 53 . 78 jg the L. CM. required. Exercise 10. Ascertain which of the following numbers are prime, and, when not prime, give their least divisor : — (I) 289, 461, 667, 851. (.) 953, ,517, 17,9. (3) ^j„j_ jggg^ ^^^y_ How many prime numbers are there between (4) 330 and 350. (5) s56and58o. (6) 790 and 825? Decompose the following numbers into their prime factors :— (7) 462, 460, 320, 1188. (8) 1309, 945, loooo. {9) 1827, 1485, 226800. (lo) 55020, 16632, 402325. (") 93565, 53S99. 47089. (12) 88725, 98735, 190463. (13) 508079, 4149173. (14) 259811, 73896433. Find the G. c. M. and L. c. M. of the following numbers by decomposing them into their prime factors :— (15) 756 and 6435. (i6)- 756. 3SO and 9075. U7 735, 1575 and 2205. (18) 14553, 22869 and 53361. (19) 13 • 17 • 19. 17 • 19 • 21, and 19 . 21 . 13. CHAPTER V. FRACTIONS.. 98. A MAGNITUDE may contain its unit a number of times exactly, or a number of times with a part remaining over less than the unit, or the magnitude itself may be less than its unit. In the latter cases we divide or break up the unit into a certain number of equal parts, and taking one of these equal parts or sub-multiples of the unit as our sub-unit, we find how many times it must be repeated to make up the remaining part, or to make up the given magnitude. Thus if we divide the unit into 8 equal parts, the sub-unit will be one-eighth of the unit : and if this sub-unit has to be repeated 5 times, the magnitude is 5-eighths of the unit. 99. When a magnitude contains its unit a number of times exactly, the resulting number (4) is called an integer or whole number ; when a magnitude contains a sub-multiple of the unit, a number of times exactly, the resulting number is called z. fraction; thus 5 is an integer, 5-eighths is z. fraction. The sum of a whole number and a fraction is called a mixed number, as 7 and 5-eighths. 100. The number which points out into how many equal parts the unit has been divided is called Xhs denominator of the fraction ; and the number which points out how many times one of these parts has been repeated is called the numerator. The numerator and denominator are called the terms of a fraction. loi. Notation and Numeration of Fractions. We express a fraction in figures by writing the numerator above the denominator and drawing a bar between them; thus the fraction 5-eighths S6 FRACTIONS. § loi. which has 5 for its numerator and 8 for its denominator is written |. And the mixed number 7 and 5-eighths is written 7 + f, or rather 7I ; for the addition-sign is almost always omitted. Conversely, a fraction expressed in figures is read by first reading the numerator and then the denominator with the termination "ths"; thus 1% is read eight-thirteenths. The exceptions are that fractions with denominator 2 or 3 are read as so many halves or thirds, and with denominator 4 as so many quarters as well as fourths. A mixed number is read by connecting the integer and the fraction by "and" ; thus 7| is read seven a«rf five-eighths. 102. Not only do we measure magnitudes which are less than the unit by a sub-hnit, but sometimes those which are equal to or greater than the unit ; hence we may have such fractions as f, ^. Also if we suppose the unit to be divided into i part, so that the sub-unit is the same as the unit, we may have such fractions as f, \ : which differ only inform from the integers 5, 9. Again, in measuring a magnitude by a sub-unit, we may have a part remaining over less than the sub-unit, and this remaining part we may measure by a subordinate sub-unit: hence we may have such fractions as -^ , i^ M a ' 20 ' 4 ■ And, lastly, if we take our sub-unit so that a.fractidnal number of sub-units make one unit, we may have such fractions as Si' Si' 4H' 103. Again, A fracHon expresses the quotient of the numerator by the deitominator. For to take an eighth part of 5 is to take an eighth part of each of the units which make up 5, and is therefore one-eighth repeated 5 times, or is 5-eighths. Or we may proceed thus:-Since i unit is 8-eighths, therefore S units is 40-eighths, and therefore 5 divided _by 8 is 40-eighths §io3. FRACTIONS. 57 divided by 8, and is therefore 5-eighths ; that is Cor. I. Hence f is not only read ^-eighths, but also 5 by 8. Sometimes, the position only of the figures is regarded, and | is read S over 8 ; but this reading ought to be discouraged. Cor. 2. If we multiply a fraction by its denominator we obtain its numerator. Since f is the eighth part of 5, | repeated 8 times gives 5, or fx8 = S. 104. Fraction of a Fraction. If we take a fractional magnitude, and regarding it as a new unit, divide it into any number of equal parts and take one or more of these parts we shall get a fraction of a fraction : as f of f . loj. We distinguish fractions into the following kinds : (i) h proper fraction is one in which the numerator is less than the denominator, as f . (2) An improper fraction is one in which the numerator is not less than the denominator, as % ^. (3) A simple fraction is one in which numerator and denomi- nator are both whole numbers, as f, ^. (4) A complex fraction is one in which numerator or denomi- nator or both are not whole numbers, as 7' Sf 4f' Zi + n' (5) A compound fraction is a fraction of a fraction, as f of |. 106. The following definitions will also be required : — (i) The reciprocal of a fraction is the fraction formed by inter- changing its terms ; thus the reciprocal of f is J, of 6 or f is \. (2) Fractions whose denominators may be any number we please, whole or fractional, are called vulgar, that is, common or ordinary fractions; whereas fractions whose denominators are 10, 100, 1000,... are called <&«»?«/ fractions. 5 8 FRACTIONS. § 107. REDUCTION OF MIXED NUMBERS TO FRACTIONS. 107. To express a whole nuMber or a mixed number as a fraction. (i) Take the whole number 7. Since i is equal to 2-halves, or 3-thirds, or 4-fourths, . . . therefore 7 is equal to 7 times 2-halves, or 7 times 3-thirds, or 7 times 4-fourths, ... ; that is 7x2_7x3_ 7x4 ^7x5_ 2 3 4 5 •' hence, For the denominator take any integer, and for the numerator the product of the given number and the denominator. (2) Take the mixed number 7|. By the preceding case ^_7X5_35. ^" S "'s' therefore 7l=7 + ? = 35,3 but these fractions have the same sub-unit, namely i-fifth; and of such sub-units the first contains 35 and the second 3; therefore their sum contains 3S + 3 of them : that is 7a_35, 3_3S + 3 7-5-j+---^, hence, For the denominator take the denominator of the fraction, and for the numerator the product of the whole number and the denominator increased by the numerator. 108. Conversely, To express an improper fraction as a mixed number. Take the improper fraction ?§. Since 5-fifths is equal to i, 35-fifths is equal to 7, and 38-fifths is equal to 7 and 3-fifths : that is where the division of 38 by 5 gives the quotient 7 and the remainder 3 : — hence we have this Rule : Divide the numerator by the denoininator ; the quotient will be the integer of the mixed number, the remainder will be the ttume- rator of its fraction, and the denominator of the given fraction its denominator. § io8. REDUCTION OF MIXED NUMBERS. 59 Ex. I. Express 325 as a fraction, with denominator 999. ^ 999 999 ^ ^' ^ 999 ' Ex. 2. Express 59^ as an improper fraction. ^ 59x73 + 17 _ 4324 S9« = ^ 59 _n. 177 73 73 ' -^'^^ 43=4 Ex. 3. Express ^§^ as a mixed number. 274 ) 6901 ( 25 1421 therefore ^u = 2Si^. SI Exercise ii. I. What fraction do we form in dividing a unit into 17 equal parts, and taking 12 of them; into loooo equal parts, and taking looi? ^. Express in figures: One fifth; one quarter; nine halves; twenty- three thirds; twenty-five forty- ninths ; five thousand and ninety millionths. 3. Express in figures : Three, and a half; two, and a quarter; five, and five ninths ; seven, and three elevenths ; sixteen, and nine twenty-oneths ; three hundred, and ninety-one thousandths. 4. Express in words : |, |, \, \\, ^%\, 3I, gH, 27||. 5. Express as a fraction : 12 with denominator ri; 9 with den. 17; 56 with den. 37; 779 with den. 13; 87 with den. 97. Express the following mixed numbers as fractions : 6. "5. i3f. 12H, 33t^t, 46^^, 87^^^^, 156TS, 95Tf- 7. loi^f, 19U. 323ff. 49lt. 68BUI 9879Mfl- Express the following improper fractions as mixed numbers : g 132 143 £43 I^ 875 523 747 1000 19583 ■ II ' 13 ' 12 ' 17 ' 8 ' 23 ' 45 ' 23 • 144 ■ 3003 4515 76845 879641 830526 38542 71 17 ' 71 ' 999 ' 3125 ' 9891 ' 3769 10. Express the reciprocals of the following fractions as mixed numbers: 1. 15 17 2^ ll^ _99. '13 15' 49' 65' 5874' 2515' 4567' looooo' II. Find as mixed numbers : The seventh part of xooo; the seventeenth part of 2345 ; the eighty-ninth part of 3567. 9- 6o FRACTIONS. § 109. 109. If we midiiply the numerator and deno7ninator of a simple fraction by the same whole number, the value of the fraction is unaltered. For example, - = ~ — - . For f means that a unit has been a » . 3 5 divided into 8 equal parts, and that 5 of these parts have been taken ; if now we divide each of these 8 parts into 3 equal parts the unit will be divided into 8 times 3 equal parts, and the 5 parts previously taken will give 5 times 3 of these new parts : hence, to divide the unit into 8 equal parts and to take 5, is equivalent to dividing it into 8 . 3 equal parts and taking 5 . 3 of them : that is, 5 5-3 8 8.3- Cor. If we divide the numerator and denominator of a simple fraction by the same whole number, supposing both to be divisible by that number, the value of the fraction is unaltered. 1 10. To multiply a simple fraction by a whole num.ber, we may multiply the numerator by that number: or, if the denominator be divisible by that mtm.ber, we m.ay divide the denominator. (i) The parts composing the fractions | and ^-^ are each o 8 equal to one-eighth, and the number of parts taken in the second fraction is 3 times the number taken in the first, therefore the second fraction is 3 times the first, or 5 x 3 = ^-^ . 8 8 (2) By the preceding case, -^x3 = ^— ^; but ^^ = - (109), c c 4-3 4-3 4-34 therefore ^ x -5 = - . 4-3 4 111. To divide a simple fraction by a whole number, we may divide the numerator, if it be divisible, by that number j or we may multiply the denominator. (i) The parts composing the fractions ^~- and - are each o 8 equal to one-eighth, and the number of parts taken in the first §111. REDUCTION OP COMPOUND PR ACTIONS. 6 1 fraction is 3 times the number taken in the second, therefore the second fraction is the quotient of the first by 3, or ^-^ -^3 = 5. o 6 (2) But if the numerator be not divisible by 3, as in | , multiply ■> 5 ■ 3 numerator and denominator by 3, then § = -^- (109) ; and by the a 0.3 preceding case, 8::-^3 = g^; therefore g-^3=g^- 112. To express a compound fraction as a simple one- Q y J Take the compound fraction - of — : it means that regarding 11 9 12 — as a whole we are to divide it into 9 equal parts, and take 8 of 12 J J those parts ; but when we divide — into 9 equal parts, each part will be equal to — =-9 or to (iii, 2), and 8 of these parts will II - II . 8 , , , ^ give X 8 or (no); therefore " 12 . 9 12 . 9 ^ ' 8 , II II . 8 8. It ,, , -of — = = . (61) 9 12 12 . 9 9 . 12 Ti-1 3r8.ii 3,8.11 In like manner - of - of — = - ot 4 9 12 4 9. 12 ^ 3-8.ii . 4.9.12' hence we have this Rule : — Multiply all the numerators together for the numerator required and all the denominators for the denominator. Remark i. We must express mixed numbers as improper fractions before applying the Rule. Remark 2. If there are factors common to numerator and denominator, they may be cancelled, or struck out, before obtaining the final result. Thus, in -— — '- — . rejecting the factor 3 common .4.9-i2\ ^ 1.8. II / . . to numerator and denominator, we have : agam, rejecting I 2 II 4 • 3 • 12 the common factor 4 we have — ; lastly, rejecting the com- I . 3 • 12 62 FRACTIQNS. §112. mon factor 2 we have ----- or -= . The work usually stands 1.3.6 18 thus : a 3 ^8 ,11 3.8. II II ^ of ~ of — =5 = -5 . 4 9 12 4.9.*2 18 3 6 Remark 3. In canceUing, if a quotient be i, it is usually omitted : thus, over the cancelled factors 3 and 2 in the above numerator no number appears. Ex. I. Express — , -^ and — as fractions with denominator 4^2. ^ 12 16 27 ^^ Now c J 7 7-36 252 432 -^ 1 2 = 36 and -^ = -^— 2-, = -^ ; 12 12.36 432' Ex. 2. 432-=- 16=27 432-=-27=i6 9 9-27 243 _ 16 16.27 432' 8 8.16 128 27^27. 16 "432" 8x9-8-7^- 3Jx4 = fx4 = ^=,5i. 35^. _ 5 64-7 64- II 121 3= of - of " of 13 24 2A ofA = ?^ofi^ofHof??of A 12 7 13 24 II 12 2 4 _a§ X i3 X ii X 38 X 5 5 = i|. 3 Exercise 12. i. Find fractions equal to 2, — , and i^, having 84 for their denominator. 7 12 14 ° ^ 2. Transform — , — , and g?,intoequalfractions whoseden'shallbe 325. 3. Find fractions equal to i^ , i- , — , and ?^, having 756 for their com- 17 26 32 35 mon numerator. 4. Transform — - . ^f^ , and ^^ , into equivalent fractions whose de- 9' 367 6140 nominators shall be 13, 21 and 20 respectively. 6- Express 17I, 25^ and 13I as fractions, with denominator 18. 6. Express 41I, 23^, 7|| and 9/^ as fractions, with denominator 240. §112. LOWEST TERMS. 63 Find the value of: ,. i!x., 12 g — X12; 5ifx7; 9AX17; 6^x II. ,. ^,. ^9- = "7 143 n — ^-^13; -!— f-ii; 7t-=-i2; 120 ^ 200 ' " 9if-i5- Reduce the following compound fractions to simple ones : 8 fi5 g. - of — ; 9 32 iiofM; 28 51 55 28' 4 '^ ^5 10. -of6fofi^; 2»of-%f^; i of - of ^ of 4ii 9 15 45 91 9 ^5 '6 11. 2i of i|of 7f of ^; -^ of6|of i7iof lofofg,;^. 10 05 3^ 12. 'lof^ofgAofsJofJ^of^. REDUCTION OF A FRACTION TO ITS LOWEST TERMS. 113. A simple fraction is said to be in its lowest terms, when no fraction of equal value can be found whose terms are less than those of the given fraction. Thus -I is not in its lowest terms, for it is equal to | ; but | is in its lowest terms, for no equal fraction can be found whose numerator and denominator are respectively less than 3 and 4. 114. If the numerator and denominator of a fraction be prime to each other, the num.erator and denominator of any fraction of equal value will be equimultiples of the numerator and denomi- nator of the given fractio7i. Take any fraction | whose numerator and denominator are prime to each other, and suppose | is equal to the fraction - where a and b stand for two whole numbers ; then a and b are equimultiples of 4 and 5. If we multiply each of the equal fractions ^ and | by ^ the pro- ducts will be equal ; but |- multiplied by b gives a (103), and 64 FRACTIONS. § 114. I multiplied by b gives 2^ — (i 10) ; ° S therefore a = — — ; and since « is a whole number, 4 x ^ must be divisible by 5 : but 5 is prime to 4, therefore it must divide b (75). In dividing b by 5, let the quotient be denoted by the whole number c, therefore we have b=iy.c and therefore a = — =^>^c; thus a is the same multiple of 4 that /5 is of 5 ; or a and b are equi- multiples of 4 and 5. Cor. Hence all the fractions which are equal to a given frac- tion, whose numerator and denominator are prime to each other, are found by multiplying its numerator and denominator by the numbers 2, 3, 4, 115. For a fraction to be in its lowest terms, it is necessary and sufficient that numerator and denominator be prime to each other. (i) It is necessary, — for if they are not prime to each other, they have a common measure; divide them by this common measure, and we have a fraction equal to the proposed in lower terms. (2) It is sufficient, — for if they are prime to each other, every equal fraction must have its terms equimultiples of the terms of the proposed fraction ; and must therefore be in higher terms than that fraction. Cor. If we divide the numerator and denominator of any fraction by their G.C.M., the numerator and denominator of the new fraction will be prime -to each other, and therefore it will be in its lowest terms. Hence, to reduce a fraction to its lowest terms, we divide the numerator and denominator by their G. C. M. Example. Reduce f^Jf to its lowest terms. 703s 105 ) 2415 ( 23 105 ) 7035 ( 67 7H5 315 735 2415 ''3'° i°S 210 o o 210 •'■ fMs = ff fraction required. §ii6. LOWEST TERMS. 65 1 16. In practice^ however, we seldom need to find the G. C. M. of ntunerator and denominator : we see by inspection, or find by trial, some factor common to both j and, having expelled that factor, we proceed again in the same way, and continue the process until the terms are prime to each other. Thus in the preceding example we see that 5 is a factor of both terms (86), hence we have ^^^ : we now see that 3 is a factor of both (87), hence we have -jH : again we find that 7 is a factor of both, hence we have || : but 23 is a prime, and does not divide (fj, therefore 23 and 67 are prime to each other (68), and lowest terms. The work usually stands thus : 2416 _ 483 _ Ifll _ 23 7035 1407 469 07' Exercise 13. Reduce the following fractions to their lowest terms, without finding the G.c.M. of numerator and denominator (116) : 1 19 _ 208 _ 304 639 _ 660 _ 672 '■ 768' 864' 1072' ^' 637' 1155' 1056' 1584 . 4^ . 3^- 591? . ""395 ^' 5940' 7315' 4914' ' 9'63' 19635' 12540 9971S g 3°4^9 ■ 47 5^°° ^' 21945' 113960' ■ 88641' 639930' Reduce the following fractions to their lowest terms by finding the G. cm. of numerator and denominator : 1261 _, 8991 . 10307 . g 18577 . ''4^93 . 61013 '^' 1649 ' 10989 ' 94637 ■ ' ?ooo6 ' 48569 ' 63529 ■ 4223 7 95469 10 '^°'94 ■ 333567 ^' 75582 ' 359784 ■ ' 1973594 ' 436203 70609 256417 j^ 1832051. 496606401 ^ "■ 232001' 7010117' ■ 2592525' 1006110363' Express the following fractions as mixed numbers, reducing the fractional part to its lowest terms : • 69 371 6244 1619 1. 15157-. 75648 '3- 7: T7' 777- "^' 257' 161 ' 864 • B.-S. A. S 66 FRACTIONS. §117. LKAST COMMON DENOMINATOR. 117. To reduce fractions to others having a least common de- nominator is to find equal fractions having a common denominator, and that denominator the least that can be taken. Let the given fractions be in their lowest terms, then the terms of the equal fractions must be equimultiples of the terms of the given fractions (114), and therefore every common denominator of these equal fractions must be a common multiple of the given denominators. Again, take any common multiple of the given denominators ; find the factors by which we must multiply each denominator to produce this common multiple, multiply the terms of each fraction by their factor, and we have equal fractions with this common multiple for each denominator. Hence every common denominator of the given fractions is a common multiple of their denominators, and every common multiple of the denominators is a common denominator of them : therefore the L.c.M. of the denominators is the least common denominator of the given fractions. Therefore, to reduce fractions to others having the least common denominator, we have this Rule : — Find the L.C.M. of the given denominators, and take it for the least common denominator: divide it by the denominator of the first fraction, and multiply the terms of this fraction by the quotients and do the same with all the other given fractions. Ex. I. Find fractions equivalent to |, ii, A and ^ having the least common denominator. 2x3 or 6 ) 8, 12, ij, -ix .■. Least common denominator= 8x3x5x7 = 840, Now 840^-8 = 105, therefore 3 = 3iii£S ^ 3£S . 8 8X/05 840' 8404-12 = 70, ii^n_x_7o^77?. 12 12x70 840' §117. LEAST COMMON DENOMINATOR. 6/ o . - ,c .T. r 8 8x56 448 840-i- 15 = 56, therefore — = —. = 52l . IS 15x56 840' „ 1 7 1 7 X 40 680 840-i-2i=40, -i = -^ — 3L- = 21 21 X 40 840 Ex. 2. Arrange in order of magnitude the fractions | , — , — 17 8 12 15 and — . 21- Reduce these fractions to equivalent ones with the least common de- nominator ; these by Ex. t we find to be 315 770 448 ,, 680 840' 840' 840 840' and, as the parts composing each of these fractions are all equal, the one in which the greatest number of parts is taken will be greatest ; hence, arranged in order of magnitude, we have 315 448 680 770 _ 840' 840' 840 840' that is. I, A, £7 and ^ 8' 15' 21 12 3 ^ II are in order of magnitude; ^ being least and — greatest. . O 12 Exercise 14. Reduce to equivalent fractions with the least common denominator : 6' 9' 15' lo' „ 3 5 7 8 24. • 5' 7' 9' ^i' 35' 23 IZ 19 47. ^" 27' 18' 24' 54' 73 ii £3 35 if 7. £9 £7. ij i^S 25 *■ 60' jj' 24' 40' 100'. 36' 8' 54' 48 ' "32" 19 ri 8^ 25 81 27 ; 14 15 il 19 ^ ^ *• 21' 12' 15' 27' 3s' 40' 15' 16' i8' 20' .24* 27" Express the following numbers as fractions haying the least common denominator : o- 3t:t> tT' '^^' r;' 'atiri otff> 'Ottfi "itj 77^, 95if 40 2/ 45 5—2 6 7 8 II 7' 8' 9' 12- 3 9 II 17 4' 20 15 12 _3 57 459 8756 10 ■ ' 1000' loooo' I 00000 68 FRACTIONS. Ex. 14. 7- Arrange in order of magnitude : 7 10 i6 Ts' ^' 35 ' 18 3i 14. 25' 47' 19' L5 4 ' i\, ^ofgf 8. Which is greater, 19 19+8 19 or 19-8 9- Arrange in order of magnitude 8 9 8 + 9 _ 13 IS 13+15. 17' ^S' 17 + 25' 14' 16' ' 14+16' S 6 7 9 S+6+7+9 6' 7' 8' 10' 6 + 7 + 8+10 lO. Find the least and the greatest of the following numbers: 3 7 5 1 9_. ' 13' 16' 20' 19 IS 29 37. 18' 14' 27' 35' lOI 13 . 7if. 7A. 1 091 143 11. Find a fraction intermediate in value to I and - whose denominator 6 7 is 84: to - and — whose denominator is 720. * 5 72 12. Of =^ , and -^ which is intermediate in value? 87 341 428 2 5 7 1? ?i mon numerator. ic!. Reduce-, |, J, — , — to equal fractions with the least com- ^ 9' 6' 8 19 23 ^ ADDITION, SUBTRACTION, &C. OF FRACTIONS. 118. The definitions that have been given of Addition, Sub- traction, Multiplication and Division have been given with reference to whole numbers only : it will now be necessary to extend them, that they may be applicable and consistent whether we are operating on whole numbers' or on fractions. ADDITION. 119. ADDITION is the operation by which we find a single num- ber that is equal to two or more fractional numbers put together. This single number is called the sum of the given numbers. 120. To find the sum. of two or more given fractions. (i) Let the given fractions have the saine denominator : for example, find the sum of |, | and | . These fractions have all the § I20. ADDITION OF FRACTIONS. 69 same sub-unit, one-seventh; and of such sub-units the first fraction contains 4, the second 5, and the third 8; therefore their sum contains 4 -t- 5 + 8 of them ; or 4.5, 8 _ 4+5+8 . 1^ 1 1 ■ ■ 1 ' that is, — We add the numerators of the given fractions together for the numerator of the sum, and take their denominator for its denominator. (2) Let the given fractions have different denominators : for example, find the sum of |, |, |, |. Since the parts com- posing these fractions are all different, the first step will be to find fractions equal to them composed of the same parts, that is, which have the same denominator; and of all cqmmon denominators the least will b? the most convenient. The least common denominator is found to be 24, so that ?4-3+£ + I = i6.18,20,21 3 4 6 8 24 24 24 24 ^16+1_^?0+21 (,2p_ j) _75_26 24 8 We have then this Rule, — Reduce the given fractions to their least common denominators "■dd the numerators together for the numerator of their sum, and take the least cotntnon denominator for its denominator. Remark i. The sum should always be expressed in its lowest terms ; and, if an improper fraction, should be reduced to a mixed number. Remark 2. Compound fractions should be reduced to simple ones, before the application of the Rule. Remark 3. Numbers, whether -whole or fractional, may be added together in any order we please: for reduce all of them to a common denominator, then their sum is found by finding the sum of their numerators; but these, being all integers, may be added in any order we please (25), and therefore the numbers themselves may be added in any order we please. Hence, instead of reducing "JO FRACTIONS. § 120. mixed numbers to improper fractions, we may add separately the integral and the fractional parts of such mixed numbers ; thus Ex. I. Find the sum of ^, J-|, ff and H. ly 19 19 19 25 I 18 I 88 I 17_ 26+18+58+l7' 19 19 19'^19 19 ■ .118 19 =6^9- Ex. 2. Find the sum of s|, 8| , 13I of ^^ , t^ and l^f . r7 12 niof -5-=^of J!^=^=d*: 101-8-5- 6 3)8, 9. 5. M, 24 3, S, 8 .•. least common denominator = 3 x 3 x 5 x 8 = 360 hence, 5|H-8^i3iof,^-f^H-i7t-I = si+8|+4f + 8f^ + i7M =4^ + l + l + f + 5^2 + H = 4,2 + SIS + 280 I 28S , 160 , 216 360 360 S60 360 360 = 42+1288 ^ 360 ^ 45 ^ -'46 Exercise 15. Find the sum of • ' 1 ' 1- li 9- M. ^21 ,si in 18 '• :7' 17^ 17' 17' '^^' ^^' 7f ■ :??' ^^' -M' ^' '3l4f. 23456' 12' 18' 20' 15' 21* II' 7^' -g' fl' Ex. i;. SUBTRACTION OF FRACTIONS. 7 1 i£3 ^ 56s. 133 235 261. 3„fS I f^J 9_f . 3- 226' 339' 678' 182' 325' 351' 4°'8^i2 16 + 16°' *'■ 4- 4t. 87. 3rT. 8^, Ittt. 2t3^, 3w. 4^7, ,„. j^^. ,,^. J^^- 5. nsi>6^T'A, 10311, |of4iS°f; ^. ^^. ^^ ^J. 1^- Find the value of 6- 3A+7ii + 24of ?+6A; ;^-+ ,4;°f ^f + -^ + 8A of 4JH-30JI. B II 19 15 7. %f4l+7i + -^-of6J + *°f5i + |of9t. 8. ^ of ^ of 52A + ^- of - of 506J + ^ of I of 1864. 4 7 5 9 6 8 5. £ of 5 + 4 „f _L+3 of Ci+i!\ + l_ of C? + 4y ^7 7 5 10 5 \2 14; 70 \7 sy 23 I I ^ 10. Of the numbers if^-J-, - + -- , and -^ \--, which is greatest? 3 4 257 SUBTRACTION. 121. Subtraction is the operation by which we find what number is left when a smaller fractional number is taken from a greater. The number left is called the Remainder. 122. As in integers, the number left is the difference between the two given numbers, is the excess of the greater number over the less, and is the number which must be added to the less number to make it equal to the greater. 123. To subtract 07u fraction from another fraction. (i) Let the given fractions have the same denominator : for example, subtract ~-^ from ~- . Each of these fractions has the same sub-unit, one-seventeenth ; and of such sub-units the Minuend contains 9 and the Subtrahend 4, therefore the Remainder contains 9—4 of them; or i _ 4 9-4 . 17 17 17 ' that is,-^ We subtract the numerators of the given fractions for the numerator of the Remainder, and take their denominator for its denominator. 72 FRACTIONS. § 123. (2) Let the given fractions have different denominators: for example, subtract | from LI . Since the parts composing these fractions are different, the first step will be to find fractions equal to them composed of the same parts, that is, which have the same denominator; and of all common denominators the least will be the most convenient. The least common denominator is found to be 24, so that 11_3_.22__9_ 12 8 24 24 _ 2 2-9 24 = 13 24 ■ We have then this Rule, — Reduce the given fractions to others having the least common denominator, subtract their numerators for the numerator of the Remainder, and take the least common denom.inator for its denominator. Remark i. The Remainder should always be expressed in its lowest terms ; and, if an improper fraction, should be reduced to a mixed number. Remark 2. Compound fractions should be reduced to simple ones, before the a.pplication of the Rule. / ' ADDITION AND SUBTRACTION. 124. The propositions relating to the addition and subtraction of expressions made up of additions and subtractions enunciated in Arts. 54—57 are equally applicable to fractions as to whole numbers: thus (i) To a fraction we may add the sum of two others by adding them in succession (54). For8| + (3f+2|) = ^Y + (3^ + ^B) (x,7) _735 , 300+2'38 84 84 _ 73 5+1300+2381 / \ - -^ (120) _ 735+300+288 84 (54) § 124. ADDITION' AND SUBTRACTION. "Jl 84 84 84 ^ I 4 ■'T 6 (2) Again, From a fraction we may subtract the sum of two others, by subtracting them, in succession. TT^r- 5)3 /,4.-,S\_735 /300 , 238\_ 736 300+238 i-or ^-^-K^-i'^^i^)—^-\^ + -^)—^^ 75J— _ 7 3B-(3 0+238) _ 735-300-238 / c^\ ~ 84 ' 84 ^^^' 735-300 238 l\1\\ 84 84' ^ ' _735 300 , , 238 84 84 84 4 -'7 % (3) And in precisely the same way may the other propositions in Arts. 55, 56 be shewn to hold true of fractions. 125. We may also shew in the same way that (i) Additions and subtractions of fractions may be performed in any order. Hence instead of reducing mixed numbers to improper fractions, we may subtract separately the integral and the fractional parts of the mixed numbers ; thus 8f-3f = 8 + |-(3 + f) = 8 + |-3-| (124,2) = 5 + 3-*-. ^4 7 (2) An expression made up_ of additions and subtractions of fractions may be made equal to the difference of two sums; thus R3_,4 , 25_3 + 2_l=/83+-25 + 2\_C,4 + 8 + lV »J-37- + 2g-g + 3 2 \°4 6^3/ \-'7^8^2; Ex. I. 19 19 19 19' ,12 17_1_2 , ,_1I_12+-2. = 14 19'"l9~19 19 19^19 19 ?L2 _ ,11 = i:12 _.ll =4.12 + X-^AS 19 3ig 5ig 19 '^IB^ig ^18' 74 FRACTIONS. § 125. Ex. 2. Subtract 3j|^ from 8$ ; and 3^ from 8iV (1) 8i-3T^ = Sii-M = S^. (2) 8Jr-3l=Sf|-if=4M + l*=4M- Or we may adopt the following arrangement : — (I) 8^ ...44 (2) 8tL...3s 3i--3S 3I -44 SA 4tt In the second example as we cannot subtract ^ from ||- we borrow I or II, and then say ^ from || leave \\, and |i and || is || : or simply 44 from 55 11, and 35 46. Ex.3. 8|-3i| + 2lof^of4|-(si|-2|) 4 5 ST = 8^-35-5+-?^ of -S'- of ^-i:ll+2^ * 7 4 •'14 7 •'12^ 8 = 4+S+6+Z_/13 , 11\ ^ 4^7^8 \l4^12y -4-1- 126+144+147 156+1S4 168 1,68 = 4+417_3i0^.+ 107, ^ 168 168 ^168 = 4101 ^168 • Exercise 16. Find the difference between '■ Tk ^^^ Ti' * ^"^ 7^ ^* ^"d I' 3A and -. ^. 9tV and 2/^; 37if and 19I?; '^ and - ; i^ and ^r. 15 12 39 26 3- p and ^-| ; 8| and 5^ ; 3^ and 2^. In the following examples write down the remainders ^ 12H 3^ Jii sS 3II Ex. i6. MULTIPLICATION OF FRACTIONS. 75 Find the value of S- 3 -2 8 24 3 5 +-! _ ii _ 69 12 '•7 18 • 6. 3,5,4 5. — T r — : — -p f 20 12 15 6 3*+2f- -6f+iA; I — 4 12 I -87 + I I "6i- 7- i°^f-^°^3f.| ofsl; 15_£4. 16 15 14 II 12 ' 8. e-i)^G-i)' (i^ *i)-(f-i)^ II 12 4> ' 5y l-i 9. 6Jof 2A-(6i-2T\); ^^°f 7^°^ 3S-i^ + ^. 2 S 7 1 1 10. Of the fractions "^ , — , -^, — find how much the sum of the 16' 24 36 54 greatest and least exceeds the difference of the other two. 11. Find the least fraction which added to the sum of 5, -^ and — o 10 12 will make the result an integer. 12. From the sum of 25! and 16^^ take the difference between 18A and sH- 13. Subtract 2 of — of 6^ from | of 5S; and - of — of 1089 from - of — of 41404. 7 13 14. Add together — , if, A and — and subtract the result from 4J. 15. By how much does sf + gf - S| fall short of 8^, of 2^^ ? 3 5 16. Find the sum of the greatest and least of the numbers 5, ■— , i , — , the sum of the other two, and tlje difference of these sums. MULTIPLICATION. 126. Multiplication is the operation by which we do to one given number called the Multiplicand, what we do to unity to obtain another given number called the Multiplier. The result is called their Product. Since S is i repeated 5 times, to multiply a number by 5 is to repeat that number five times : hence the above definition includes the multiplication of integers. 76 FRACTIONS. § 127. 127. To multiply any nuniber by a fraction. For example, multiply f by | . Now | is obtained by dividing I into 7 equal parts and taking 5 of those parts ; hence, to multiply a number by | (126), we must divide the number into 7 equal parts and take 5 of those parts ; that is, we must first divide by 7, and then multiply by 5. But | divided by 7 gives -^ (i 1 1), and this result multiplied by 5 gives |^ (l lo) ; that is 3 „ 5_3X5 . 4 7 4x7 ' hence we have this Rule, — Multiply the numerators for the nume- rator of the Product, and the denominators for its denominator. Remark 1. If the Multiplicand or Multiplier be an integer, we may consider it as a fraction with denominator i (loi). Remark 2. If Multiplicand or Multiplier, or both, be mixed numbers, we may reduce them to improper fractions, and apply the preceding Rule, thus ^3 X ^-^=5^ X B2 = 35 X52 ^8 •'is 8 15 8X16 " Remark 3. Before obtaining the final result, we may cancel any factor common to numerator and denominator, thus ^8 •'is g Xg S •'6' 2 3 128. Comparingtheprocesg^of the last Art. with that of Art. 112, it is seen that to take a fractional part of any number is equivalent to multiplying that number by the fraction : thus, ^ of ^ is equi- 74 valent to 2 x | . 4 7 Again, if we multiply | by 5, and then divide by 7, we shall ob- tain the same result as in first dividing by 7 and then multiplying by 5 ; and, as the same remark holds good whatever fraction we take for Multiplier, we may say that To multiply by a fraction, we may multiply by the numerator and then divide by the denominator, or we may divide by the de- 3 , 4' ^7 _3X5 4xr ' X 8 15 _ 3X5 „ ~4X7 ^ _ 3X5XS 8 15 § 128. MULTIPLICATION OF FRACTIONS. TJ nominator and then multiply by the numerator; the order of the operations being indifferent to the result. 129. To find the continued product of three or more fractions. For example, find the continued product of | , | and -~ . Now (127) therefore |x| ■4x7x16 ' hence we have this Rule, — Multiply all the numerators together for the numerator of the continued Product, and all the denominators for its denominator: cancelling all the factors common to nume- rator and denominator before obtaining the final result. 130. The Product of two or more fractions remains the same, however we change the order of the factors. For the product has for its numerator the product of the nume- rators of the factors, and for its denominator the product of the denominators, and the order of the factors in this new numerator and new denominator may be changed in any way we please (61). Cor. This proposition carries with it the following : To M,ultiply by a number which is the product of two or more fractions we may multiply by each of those fractions in succession; thus !'<(?>=l^) = |'« f x ^ = f x J - or |xIxquotient = f x|, (130) that is quotient = f x|; the same result as before ; that is, to divide a number by | we multiply the number by | ; but | is the reciprocal of | ; hence to divide a number by a fraction we have this Rule :— Multiply the number by the reciprocal of the Divisor. B-S. A. 6 8^ FRACTIONS. % 134- Remark i. If the Dividend be a whole number, or if Dividend or Divisor or both be mixed numbers, we reduce them to improper fractions, and then apply the Rule. Remark 2. Before obtaining the final result, we may cancel any factors common to numerator and denominator ; thus 3 14 ■ ^7 14 ■ 7 He «-0' 16' 2 8 Cor. To divide by ~ is to multiply by |; hence, to divide a number by a fraction, — We multiply by the denoininator and then- divide by the numerator, or ive divide by the numerator and then multiply by the denominator (128). Ex. I. Divide 3| by 1 1| ; and ^\ by 3^^ of 2 j^ • 3 9 CtI •53_^.I7_15^35_^ V iS'^il^TS-? ^ ' -^4 ■ 18 4 ■ 18 «■ ai 14 14' (-A 7l_^-3 3_ of 7i 6_3 . ^-z- „f 2-2r_6 3 . 27_B'irv *■— t1 '^^i T% ■ 3i4°*2io- 8 "=-F#°'f;S-8"~T-¥''?r~'6- 2 2 2 3 Ex. 2. Divide 4164^ by 11, and by 132. II ) 4164- Dividing by 11 the integral remainder is 6 and T ^5" the full remainder 6| or ^f-: but -^^— i-ii is f . §- Again, dividing by 12 the full remainder is 6| 3 iff or ^#, and ^#-1. is If. Ex. 3. io|x3f-4f-(2| + 3|K4f + 7|^(3f + 4|) _3J yi^ — ^-l'?+L5^-^^ -1- Bl_:_/'1_5 , 24\ 3 4'6 ^3'^4''-5"^8-''^4"^e' «• 6 7 5 ^ «• z'* 12 2'2' ar I'T-j' 3 2 2 3 _25_35", B 3- -?4 fi _ 200-35+2 _ 18 6 24 24' § 134^ DiVISIOlSt OP FRACTIONS. §3 Exercise i8. I. Find the Complete quotient in dividing 567 by 13; 8768 by 45; 84SH ty 12; 6739^1 by 37 and by 73. Divide - X43byH; 3,ty3j; H fey g , ,,, ,, ,^», , ^8^,i^o_ 3- 7Aby5l; %f:^by3f; 2jbyfof4j; iof4|by7A. 3 14 5 5 4- 3Mby2lof3^; 3i of ^A by s| of 8J ; , 3i of 4 by 2,V of ^. . 3^X3fbygx9, ^-%-l-^\^\--h^ 6. 3H°f^^^-*-A-4iof5; l-lfH-^by^-^.|2. ^ ^ + J + g+l + x^^y^i+^ + ^' .A of. I by. I -.J. 8. . + I+'+i. + i- + ±byi+i + i + i + i. 3 8 12 15 18 -' 2345 Find the value of 9. 3^451-^; (i+M)-4l; (fx^^3i)-(ix^+4o). 10. 6| + W+3A)-6f; 6|-+i| + 3A-i-6f; ^x i|x i2i-=-6f 13. Multiply the sum of - , if and % by the difference of -- and — , and ^ ^ ' 2' ' 6 ^ 15 20' divide the product by — of i^4- 10 14. Of the fractions — of i\, — of 3*. —r of /L-f-r, divide the sum of 21 24 •' 28 ' the greatest and least by the intermediate one. 15. Add - of 5 to 5 of 2j and multiply the result by T- of |j -h (- + -)• 16. Divide the sum of 2J, 3I and 5I by the sum of 4I and 8|, and to the quotient add the difference of loA and 5^. c 8 1 7. Divide the sum of % and - by 7i and subtract the quotient from 3J cf6H. ■ ^5 18. To the sum of i\ and 3 J add the difference between 4! and 5I, and multiply the result by the quotient of 7^ by 6%. 6 — 2 84 FRACTIONS. % 135. GREATEST COMMON MEASURE, &C. 135. We have said (67) that The G. C. M. of two or more numbers is the greatest number that divides each of them exactly ; and The L.c.M. of two or more numbers is the least number that can be divided by each of them exactly ; and these definitions will be applicable when the given numbers are fractions, provided that we understand by exactly, that the complete quotients must be integers. Take any fraction in its lowest terms, as | , and suppose it to be multiplied by another fraction so that the product is an integer ; then, since 9 is prime to 8, the numerator of this second fraction must be divisible by 9, and must therefore be 9 itself or a multiple of 9, and the denominator must be i or a measure of 8 including 8 itself; thus 8v9. 8„9. 8v9. 8„9 9 1' 9 2' a^4 ' 98 all give integral results. If therefore | is to be divided by a fraction and the quotient is to be an integer, its denominator must be 9 or a multiple of 9, and its numerator i or a measure of 8. Let us now take any number of fractions in their lowest terms, then any fraction by which we can divide each of them so that all the quotients shall be integers, must be one whose denominator is a common multiple of their denominators and whose numerator is a common measure of their numerators ; and of all such fractions the greatest is the one that has the least denominator and greatest numerator: hence The G.CM. of two or more fractions is a fraction whose denomi- nator is the L.C.M. of their denoininators, and whose nume- rator is the G. G. M. of their numerators. And in like manner, The L. c. M. of two or more fractions is a fraction whose numerator is the L.C.M. of their numerators, and whose denominator is the G.CM. of their denominators. § 135- COMPLEX FRACTIONS. 85 Ex. Find the G. C. m. and L. C. M, of ^, 3^ and ^. These fractions are equal to ^, ^ and Jf, and L.c.M. of their denominators is 75 and G. CM. of their numerators is 4 ; .-. their g.C.m.=^. Again, L.c.M. of their numerators is 48, and G.C.M. of their denominators is 5 ; .-. their L. CM. =-^-9f. Exercise 19. Find the g.C.m. and l.CM. of (I) f, 1, f, f. (2) S, i, f. %■ (3) f, *, f. *■ (4) i^, 2A, 3^. (5) \h 3if, lA, 3l- (6) iH, SM, HA, 6Jg. (7) iM. 2^r 3M, m- (8) i^, 3A, 2M, m, si COMPLEX FRACTIONS. 136. As yet we have treated of simple fractions only, the main propositions of which we have explained and illustrated. We shall now treat of complex fractions, and in doing, so we shall make use of the following general definition, which has been already shewn to hold for simple fractions (103) : A Fraction expresses the quotient of the numerator by the denominator. With this definition we experience no difficulty in considering fractions whose terms are themselves fractions, or mixed numbers, or the sum or difference of two fractions or mixed numbers, or any arithmetical expression whatever we have as yet become acquainted with. 137. A complex fraction is read by inserting the word by, for divided by, between the readings of numerator and denominator, thus ^ is read i\ by 1% 2| 86 FRACTIONS. § 137. In the sunt of a whole number and a fraction, when the fraction is complex as well as simple (106), the sign is sometimes omitted, as in S~tt which means 5 + ^; and in z. product when one of the factors is enclosed in a bracket the sign is often omitted, as in |(§-f),whichmeans|x(|-|). Complex fractions can always be reduced to simple ones ; f(Def.) = |x|(I34) = ^^ 41 + 2^ 5 + 4-1 _ _ pi +-2 816 18 139. Complex fractions are subject to the same rules as simple fractions ; thus, (i) If we multiply or divide numerator aitd denominator of a complex fraction by any nu7nber, the value of the fraction is unaltered (109). For take an-y complex fraction and reduce its numerator and denominator to simple fractions, and reduce the multiplier also to a simple fraction : suppose the numerator to be | , the denominator \ , and the multiplier | ; then, since y x | = i, we have 2 43^4-3^4'*7'^63^7^4^6 '^^^°l 6 _ 2X6 X " ' 3x7 4x6 2x6 2 V 8 ■ 3x7 3 7 . " 4x6 4 x^' 6X7 5 7 § 139- COMPLEX FRACTIONS, 8/ and putting now for numerator, denominator, and multiplier their original values, the truth of the proposition is established; (2) Again, — To multiply a complex fraction by any number, we may multiply the numerator or divide the denominator by that mimber (no). For, making the same, reductions and suppositions as in (i), we have i.vS_2„5„6_2„6„6_ 3 7 . t'^7-3'^4'*7~3'*7^4~ 4 ' S i> 2 2. 2 _ = 2 J. 6x6 ^ _£_ _ "3 _ 3 . 3 4X7 4xT 4vl 4^6' 5xa 5 6 6 ■ 7 and putting for numerator, denominator, and multiplier their original values,' the truth of the proposition is established. In precisely the same way we may shew (3) , To divide a complex fraction by any number we may divide the numerator or multiply the denominator by that number (in). (4) If we multiply a complex fraction by its denominator we obtain its numerator (103 Cor. ii). (5) If we multiply a complex fraction by its reciprocal the product is unity. (6) To multiply two or more complex fractions together we multiply their numerators for the numerator of the product, and their denominators for its denominator (127). (7) To divide one complex fraction by another we multiply the first by the reciprocal of the second {\ii^. Ex. I. Simplify %--^i^i^- have 1 Multiplying numerator and denominator of first fraction by 8, we 5l 45 _ 3 3l 30 2' 88 FRACTIONS. § 139. (2) In the second fraction, the l.cm. of the denominators of the fractional parts is 12, and multiplying numerator and denominator by 12, we have 8|--4f_ ^ 106-56 ^ _5o ^ 25 3f + 7A 45 + 89 134 67' Ex.2. 5 + _--=s + ^ = 5+_--.= s+-J_^ = S^^. 3 + 3•^ — 3 + -^ •^ ,^i -^ 3S + I •* 36 Ex. 3. 7 III III 643 -+- + - - + - + - _ + i+J- 234 2 3 4 _ 12 12 12 I I 1222 126 00 70 + + _ + _ + _ + _H_ + _^_ 2i 3i 4i 5 7 9 31S 315 31S 13 105 ~ 286 *a 286 ^ 4 22 \4 38 3/ 30s _ 607+1050—152 228 228 305 30s Ex. s. From 17^ subtract the difference of 9J and 4f , and divide the remainder by the product off and 9?. Quotient required = ^-TAf^^lii) ^ i7A-9l+4f ,, ) ^ |X9? ix^=. K^^'^i) ,2 + 5S_63^6o 84_,.3a,^ 325 ^* 325 63 "I- 3 o 03 e6e 63 32s 2* 32s ex. 20. complex fractions. 89 Exercise 20. Reduce to simple fractions 3f iS 2I j^ |of|_ f+l. _5|_. ^I°f3i '■ 6' Si' 3f' 13A' 5i^ ' 8A' 1I+7A' 'i + 3l' 4+_3. I2?__I_ 2ofi + 4of- 1 + iof^-- 7 " 119 ^39 ■ 3 5 5 7 ■ 3 5 9 ^i 4 ,^' I 120' 4 ,6 I , 2 ' ,2^,5 S -Sof— 1 + — X -of of- ^■^-^z~~k 7 II isg "9 5 7 3 5 3 7 9 2| • 8^' |of4f ■ '7*' iSJ-r-f ' 14 6J iif 4 4 r 4 i + ^x- + -of- 3|x3f-2ix2i . i + 6^x(i+6|) . 4ix4ix4i-i 1x4-1 ' 3l-2i ' i + 5l'<{i+Sl)' 4l>«4i-i 3 3 3f+i^l-3l . 7T-3i ■ 5i+if . nf-ioj; . lof+nj ^ 7 11 „6 !!„fS' 6 + 4I "61-21' iif+ioj' iof-9i 2 j^" 7^-4^ of ^ 7 " 7 3 V'i 7/ , ^ I .- I ,1 6. ^ : — of , ; 2 J X -=— , 21^ 4 ,1 ,1 ' + — '+^j 3*+-J '-^- 5 + 2 - 9 4 ^"^3 7^ 4 ^i 5» + 48 ,. 5l-4i . ''8f-22A jg_l5i^fj ' 3f+H 3l-2|' I4H-8U' 228 II 21 77 5 If IJ JTS Ij ,+£ + £+' -i5_0f^0f3j+23Vofl=ff0f; _w ._^ n 2, of, 3 "'■4 5 6 209" y -- J== • -»o --» -,g Find the value of 90 FRACTIONS. Ex. 20. ■°- 8 + 2ixi*''8o' 3* 92 4^' 3i -, .5 ixii 8o' 3i 9 2 4f' 3i j,^5 9 ^ 5 of £|-£| of i-9 + 3 „f 6i. 2 „f ^ „f ^i+i-i + i). 16 6J 20 7 3I 3 Vs 3i-i 2/ "• (3A-^4?)of(iofH-7i)of^; ^^+(^^^^ ,o\)^{^\-r ^^ . 13. — —— . + 2t^ 3 °' 2q' '+ I ' 7 ^ e • ,fof7|-^i 3 3 + _L^ 8-^ 6— i-^ 5 5 + i ,_1 ._! M. ^i+-V' 3* + —-^; 6R^. _, 5 + 5 7l + ^^^ --JL. 3 V ^; ^^j_ '5- '7 .X ■> ax.2iof|-^^— —Li. '"^ 012— Of- 4 (-h)- I I I , I . , „ ^ , ,24 1 6. Find what fraction the sum of — , — ? , — and — is of 2t of i J of — 24 56 21 12 ' 3 17. Of the fractions — , -^ and -4^ , express the difference of the first 33 loi 194 two as a fraction of the difference of the last two. 18. To the continued product of. 6J, 7|- and 81, add ^, and divide 4r the sum by 9I of lo-g of i2|. 19. From result by 2JI 19. From — f take the sum of - of 21 and | of ~ , and divide the 3i 3 67 o 7rir 3x-L 20. Subtract - of = — ^ — 5 + - of " + from loi times the 3 iof33i ^ ^^± 7„f . iG 9 " sum of — and - of -^ of -^ . 10 2 15 20 CHAPTER VI. DECIMALS. 140. In the ordinary system of notation (16) a figure immediately to the right of another represents units of an order ten times less than that other: thus, if a certain figure in a number represents hundreds, the next figure to the right will represent tens, and the next simple units ; and if by a natural extension of this system we agree to carry our orders beyond simple units, the next figure will represent tenths, the next hundredths, the next thousandths, &c. But then the order of some one figure in a number must be pointed out, from which we can derive the orders of all the others ; and it has been agreed that the figure to whose right a point (•), called decimal point, is placed shall be the units' figure, and to dis- tinguish it from the sign of Multiplication, it is placed towards the top of the figure. Thus, if we wish to represent 25 units, 3 tenths, 4 hundredths, 7 thousandths, 8 ten-thousandths, we write 25-3478. If any of the decimal orders are wanting we supply their places by ciphers (16, 2) ; thus 25 units, 4 hundredths and 8 ten-thou- sandths is written 25 -0408. Lastly, if there be no units, we may suppose a cipher to occupy the units' place ; thus 4 hundredths, 8 ten-thousandths is written 00408 or simply •0408. 141. A number thus expressed, composed of units and decimal orders of unity, or of decimal orders of unity only, is called a decimal number, or simply a decimal. The part to the left of the point is called the integral, and to the right the decimal part of the given number. 92 DECIMALS. § 142. 142. Numeration of Decimals. Take the decimal 25'3478. This number represents 25 and 3 tenths, 4 hundredths, 7 thousandths, and 8 ten-thousandths : but I of any order is equal to 10 of the next lower order (140), therefore 3 tenths and 4 hundredths is 34 hundredths, 34 hundredths and 7 thousandths is 347 thousandths, 347 thousandths and 8 ten-thousandths is 3478 ten-thousandths, and therefore 25-3478 is 25 and 3478 ten-thousandths : In 3'I4I592 the last decimal figure 2 represents millionths, therefore as before the number is 3 and 141592 millionths : In like manner "00036 is 36 hundred-thousandths : that is ; — IVe read off the decimal part as an integer annexing that decimal order of unity which the last figure represents. Remark. In practice, however, we do not annex the decimal order, but saying {decimal) point read off the figures of the decimal separately in order: thus 25 '3478 is read 25, point 3, 4, 7, 8; 3-141592 is read 3, point i, 4, i, 5, 9, 2 ;— "00036 is read point o, o, o, 3, 6. Sometimes instead of point, we say decimal or decimal point J but this would require that in every different scale we should use a different word ; thus in the scale of 12 we should have to say duodecimal or duodecimal point, and in the scale of 20 vigesimal or vigesimal point. ' 143. The value of a decimal is not changed by writing ciphers to the right of the last figure. Thus -307 is equal to -30700 : — ^for the ciphers written on, do not alter the position of the other figures relatively to the decimal point, and therefore do not alter their value ; and of themselves they have no value. Cor. An integer may be expressed as a decimal by writing ciphers in the decimal part, thus 307 is equal to 307-000. 144. To multiply a decimal by 10, 100, 1000,... we remove the decimal point I, 2, ■i,... places to the right; to divide a decimal by 10, 100, 1000,... we remove the decimal point i, 2, 'i,... places to the left. i 144- DECIMALS. 93 For in removing the decimal point one place to the right, the value of each of the figures composing the number is increased ten- fold, or the number (62) is increased ten-fold, that is, the number is multiplied by 10 : — and in the same way the rest of the pro- position may be proved. Ex. I. 25-04089 X 100 = 2504-089; 25'04o89H-ioo=-2504o89. Ex.2. 25-04 X 10000=250400; 25-04-;- 1 0000 =-002 504. 145. To convert a decimal to a decimal fraction and vice versA. (i) Take the decimal 25-3478: — this number represents 25 and 3 tenths, 4 hundredths, 7 thousandths and 8 ten-thousandths; therefore 2 1;-^478 = 2 c -1- -^- -t- -^ a — ''— h ^ — JJ+/" ^^lo 100 1000^10000 _j- , 347 8 =-}.c 3478 ■'10000 10000' or — 263478 10000 ' vchere the numerator is the given number with the decimal point taken away, and the denominator represents the decimal order of the last decimal figure and is therefore i followed by as many ciphers as there are decimal figures : hence we have this Rule : — Write down the given nurnber suppressing the decim.al point for the numerator, and for the denominator write i followed by as many ciphers as there are figures in the decimal part. To express the given decimal as a mixed number, apply the Rule to the decimal part only. (2) Conversely. To convert a decimal fraction to a decimal. — • Write down the numerator and cut off from, its right by the decimal point as m.any figures as there are ciphers in the deno- minator. If the number of figures be less than the number of ciphers, prefix in the numerator the requisite number of ciphers. Ex. 5. Express 206-0875 as a decimal fraction, and then as a vulgar fraction. -jrifi-nR-Ti: — 2060875 —82435 _ 16487 20b OS75 - ■ jo^oo j^ „- - -^0 - , or = 206 /^^ = 206-^ = 206^ • 10000 400 80 94 DECIMALS. % 145. Ex. 6. •000-^6= — ^5 — = — ? — . ■> lOOOOO 25000 Ex.7. «I8934 = 67-8934; _7a_ = _0'L3. = -o73. ' 10000 ' ='-'^' 1000 1000 '-' Ex. 8. 26 .„YS 6 MJ7^ = 26-00875. 100000 lOoooo '■' Exercise 21. Express as decimals I. Three and seven tenths. Five and forty- three hundredths. ■i. Four tenths seven hundredths and six thousandths. 3. Eight tenths five thousandths and three millionths. 4. Nine hundredths. Seven ten-thousandths. Five millionths. 5. Twenty-one and four tenths and four hundred-thousandths. 6. 65 and 8 hundredths 9 hundred-thousandths and 7 ten-millionths. Read off the following decimals, annexing the decimal order of the last decimal figure (142), 7- S'37; 'oo^S; "56789; -002405. 8. 9-87654321; 35-00000456765. 9. Multiply 8-003056 by 100, by loooo, and by loooooop. 10. Multiply -01728 by 10, and by 1000; -005236 by 100,000,000. II. Divide 73-56 by 10, and by 1000; 3-7165 by 100, and by loooo. 12. Divide 57324 by looooooo; -i by 100; -ooi by 1000. Express as decimal fractions, and then as vulgar fractions in their lowest terms, 13- 4"375; "8125; -37875; 23-04096. 14. -00068755 5-0096875; -222464. Express as mixed numbers with the fractional parts in their lowest terms, 15. 13-0675; 9-2221875; 23-006875. 16. 89-0131072; 12-08056640625. Express the following decimal fractions as decimals: ,7. i^; 3i5_^. 7^5. _i_. 48725 . 37 100' 1000' 10 ' 1000' loooo' 100000' jg_ 3°°5°7 . _78539_ . 20304005 looooooo ' 1 00000000 * 1 00000 19. 325 millionths; 4 ten-thousandths ; 79 hundred-millionths. i 146. ADDITIOM AND SUBTRACTION. 95 ADDITION AND SUBTRACTION OF DECIMALS. 146. The Rules for the Addition and Subtraction of Decimals proceed on precisely the same principles as the rules for the Addition and Subtraction of whole numbers, and may be enunciated in the same words. But in Decimals, we conveniently provide that units of the same order may be in the same vertical column (24, i) by placing the decimal points of the given numbers under one another ; and then the decimal point of the sum or difference will be under the other decimal points. Remark. In the Minuend we may if necessary suppose ciphers to be written to the right of the last decimal figure (143). Ex. I. Find the sum of 3"I4I5926 ; 271828 ; -434294 and 144. Ex. 2. From 35'oo6 subtract '067835. Ex. 3. Find the complement (32) of •4771213. (i) 3-I4IS926 (2) 35-006000 (3) r 271828 -067835 -4771213 -434294 34'938i65 -5228787 144- 150-2941666 In Ex. 1 we have supposed the vacant decimal places in the second, third, and fourth numbers to be occupied by ciphers; in Ex. 2 we have placed ciphers in the Minuend, but in Ex. 3 we have supposed them to be placed there. 147. The Propositions relating to expressions made up of ad- ditions and subtractions, which have been shewn to hold for integers (56 — 59) and for fractions (124, 125)^ also hold for decimals. For the decimals in any such expressions may be replaced by their equivalent decimal fractions, these may be subjected to the required operations, and then reconverted to the original decimals. Exercise 22. Add together 1. 47-6054; 6752; -0543; 75-572 and -987654. 2. -i; -00095; 84-0563; 7-3 and 325-65432. 3- 673"4S; -008742; 0-064063; 47-83504 and 961. 4- 37'045! 6'3; -0098; 8-6943; 617-241 and 'oi. 9^ MaiTlPUCATlOM OP DECIMALS. Ex. i4. 5. Subtract 8'23456 from 37* 12; "987604 from 7 'oiis. 6. Subtract '99999 from 9 ; 36*0005 1 from 45. Find the value of 7- 3"584 + 387'6 + 5-894003 + -00397 + S'BSg. 8. 8939 + 8-939 + 89-39 ■<" '8939 + -0008939 + 893-9. 9- 36-73-5"894; 5'5--o73S9; -ooi--ooooi. 10. 5-0009 - -089898 ; 0-4763 - '387387. II- 7-654327 - -3793086 + 9-06996- -00999 + -345. 12. 16-945 - 2-994387 - -06735 - -0007 + -953 + o'8. 13. Find the sum of 47 tenths, 345 hundredths, 17 thousandths and 4256 millionths. 14. Express as decimals 347 ten-thousandths and 347 millionths, and subtract the latter from the former. 15. Find the complement of -7781513; -9542425; '000356; 97-654321; and 998-899. 16. Whether is 3-1415926535 more accurately represented by 3-1415926 o"^ 3'i4i5927i ^nd why? MULTIPLICATION OF DECIMALS. 148. Take any two decimals and replace them by their equi- valent decimal fractions ; their numerators are the given numbers with their decimal points suppressed and their denominators are respectively i followed by as many ciphers as there are decimal places in the given numbers (141). The product of these decimal fractions is a decimal fraction whose numerator is the product of their numerators and whose denominator is I followed by as many ciphers as there are ciphers in the two denominators together (42) : and this result becomes a decimal by cutting oflf from the right of the numerator as many decimal places as there are ciphers in the denominator, that is, as there are decimal places in the two given numbers (145, 2). Hence we have this Rule : Multiply the given numbers as if they were integers, and cut off frojn the product as many decimal places as there are in the two given numbers together. N.B. If the number of figures in the product is less than the number of figures to be cut off, first prefix the reqitisite number of ciphers (145, 2). § 148. DIVISION OF DECIMALS. 97 Ex. Multiply 25-347 by 2-69 ; and 75 by "oocxjS. (I) 25-347 (2) 75 2"69 'cxjooS 228123 ■oo6oo='cx36. (143) 152082 50694 68-18343 In Ex. i , the number of decimal places in Multiplicand and Multiplier is 3 and 2 respectively, therefore the number in the product is 3 + 2 or 5. In Ex. -i, multiplying by 8 we get 600, but as there must be J decimal places in the Product we first prefix two ciphers, and the product becomes •00600 or 'ooS. 149. The Propositions relating to Multiplication which have been shewn to hold for integers (60 — 63), and for fractions (130), also hold for decimals. See Art. 147. DIVISION OF DECIMALS. 150. In multiplying two decimals we proceed as if they were integers, and cut off in the product as many decimal places as there are in the two numbers (148): hence, conversely, having given the product of two decimals, and one of these decimals, we find the other by dividing as if they were integers, and cutting off in the quotient as many decimal places as the number in the product exceeds the number in the divisor ; or, using the terms of Division, we have this Rule : Divide, as if Dividend and Divisor were integers, and cut off in the Quotient as many decimal places as the number in the Dividend exceeds the number in the Divisor. N.B.— If necessary, write ciphers to the right of the Dividend, so that the number of its decimal places shall be equal to the number in the Divisor, and be careful to bring down all these ciphers in the division. Also, if the number of figures in the Quotient be less than the number of places to be cut off, we must first prefix the requisite number of ciphers. Cor. If the decimal points be equally removed in Dividend and Divisor, the Decimal point in the Quotient will be unaltered ; that is, the quotient will be unaltered. We can therefore always B.-S. A. 7 98 DIVISION OP DECIMALS. § 150 arrange that the Divisor shall be a whole number, and then division of decimals proceeds on the same principles as division of whole numbers ; and in short division of decimals we shall find it con- venient to adopt this method. Remark. The above Rule with its corollary, like the Rule in Multiplication, may be established by replacing Dividend and Divisor by their equivalent decimal fractions. Ex. I. Divide '003456 by fi, and 2678508 by '072. (i) 1 1 ) -03456 (2) ( 8 ) 26785-08 -0031418 72 j 9 ) 3348-135 372-015 In (i) we remove the decimal point i place to the right in Dividend and Divisor (150, Cor.), and in (2) 3 places to the right, so that the Divisors become 11 and 72 respectively; and then we proceed precisely as in integers. Ex; 2. Divide 3-1415926535 by 987-543. 987-543 ) 3'i4i5926535 (-003181221 1789636 8020935 1205913 2183705 2086190 1 1 1 1040 123497 Here there are 10 places of decimals in the Dividend, and we have brought down 1 ciphers giving 12, and there are 3 in the Divisor, therefore there must be 1 2 - 3 or 9 in the Quotient, and therefore we must prefix two ciphers (00) before writing the decimal point. Exercise 23. Multiply 35-603 by 27-61; 421-619 by '547; 1-0759 by 3-16. -01385 by 61-37; 13-676 by -00048; -0204 by 40-2. 3. -346875 by -119808; -015625 by -0064; -0701 by 700-01. 4. i-23by-ooii; -007853 by -00476 ; -57689 by i'32. 6- 4-037^1 by -01207; 29000 by -oi; 56*875 by '0144. Ex. 43. DIVISION OF DECIMALS. 99 6. 325 tenths by 547 millionths; 128 thousandths by 78115 ten- millionths. Find the value of 7- (37'i-i9"o8)x'703; 37"i - i9'o8x-703; (•o6)3 + (-o45)' + -ooo25. 8. "4 X -CSX -006 X '0007 X 800000; •845X -0017 X 7-4X-09X loooo. Divide 9. -iby-oi; 'oiooi by •001 ; 927 by '06; 99 by "0009 . 10. I422'3 by "oii; "90804 by 1*2 ; 4-068 by •0018. 11. -0007672 by -00056; -08748 by 10-8 ; 418-25 by -128. 12. 8-79462 by -084; -376809 by -132; 3-14159 by 14-4. 13. -000144 by -on; 1-0665 by -00135; 34S'6by3-78. 14. 8886-66 by -00037; 14S 81 7 by -0563; 1114-869146005 by -385. 15. 7006-652 by 12-34; -2219904 by -3854; 1065-855558 by 7695-708. 16. -0003738028 by -0476; -0064096 by 2 -003 ; -014904 by 3^^. 17. -213419596 by -0100103; {6-25)= by (-025)'; -001 by "looi. 18. Express 2 and 22 hundredths, and 74 ten-thousandths, as decimals, and find the quotient of the first by the second. Find, to the number of places of decimals indicated, the value of 19- 765439-^ 3S9'2 1 to 5 places; •S-^76'9i342 to 6 places. 20. -046-^-00762089 to 4 places; -32165H--0035216 to 3 places. 21. 4-oo654-r-329-265 to 7 places; 3I4i59-26-t--oo8597 to 4 places. 22. -01385x61-37^2-77; 3833336+(8-99 X 20-8). 23. -399 X -007 ■=■ -000019 ; (2-05)'' X2-24-i- -0041. 24. 15-8402 + 3-689-^672-4 to 6places; 206-59-^1872 x -001 to 5 places. -0075 x2-1 4-255 X -0064 _ Sl , 21-25 25. Simplify: .^^^^ , -00032 ' 74 •04687'6 ' 26. Add together 1-465, -0095, 37-15, 28-457 and 16-1685, and divide the sum by -0296. 27. Find the sum, difference, product and two quotients of 30-33 and •0337 ; and find the sum of all the results. 28. Simplify ^-^ x ^-^ , and divide the result by -001 25. ^ -152 2-95 29. Express as mixed numbers 999f| x 2-3 and 10000^^7-5 x -5909. 30; Express 4578 thousandths and 397 millionths as decimals, and find the quotient of the first by the second to 5 places of decimals. 7-2 100 CONTRACTED OPERATIONS §iSi- CONTRACTED ADDITION AND SUBTRACTION. 151. Sometimes there are a great number of figures in the decimal parts of the numbers to be added together, and yet we only wish their sum to be accurate within a certain Umit, as, for instance, a thousandth. But as thousandths occupy the third place of decimals, we require that the sum should be correct to the third place, and if we neglect all the remaining figures in it, the error will not amount to one thousandth. We therefore proceed thus : write down each number as far as 3 places of decimals, and then I or 2 places more to be sure that in making the addition we are carrying the correct figure to the third place, and then proceed in the usual way. The same remarks are applicable to Subtraction. Ex. (i). Find the sum of 3-14159265, 27789789, 54"5678678 and 543777777 correct to 2 places of decimals. (2) Find within one ten thousandth the difference between 52'34563456 and 7'6666666. (0 3'i4 2778 54-56 54377 15 97 78 n (2) 52-3456J34 7-6666|6 44-6789 629-27 CONTRACTED MULTIPLICATION. 152. In multiplying two numbers we may begin with the left- hand figure of the Multiplier, provided we place each succeeding partial product one place farther to the right than the preceding one : and adopting this method we shall find it convenient to write the figures of the Multiplier in the reverse order. Thus, if it be required to multiply 32-52678 by 957-34, the work will stand as at (A) : 32-52678 4 3759 32-52678 43759 2927410 2 (A)' 162633 90 22768 746 975 8034 130 107 1 2 31 139-18 75652 (B) 2927410 162633 22768 975 13° 31139-18 2 746 8034 107 1 2 75652 (c) 32-52678 4 3759 2927410 162634 22768 976 130 31139-18 § 152. IN DECIMALS. lOI Now if we wish to retain (for example) only two places of decimals in the product, we must dispense with, as far as possible, all the work to the right of the vertical line at (a). But the figures in the column to the left of this line o, 3, 8, 5, o are respectively the units' figures of the products of 9 in the Multiplier and 7 in the Multipli- cand (with 7 carried), of 5 in the Multiplier and 6 in the Multiph- cand (with 3 carried), &c. : hence, if we remove all the figures of the Muhiplier one place farther to the left, these products will be formed by the figures of the Multipher and the figures of the Multiplicand immediately above them (as at B) : an arrangement which secures that the units' figure of the Multiplier shall be under the second place of decimals in the Multiplicand. But, again, in dispensing with all the work to the right of this vertical line, we lose the figure carried in finding the sum of the first column on the left ; to compensate for this loss, instead of carrying to the first figure of each partial product the preceding tens' figure, we carry the nearest ten: thus, for any number from 5 to 14 we carry i, from 15 to 24 we carry 2, from 25 to 34 we carry 3, &c. Thus, in multiplying by 5 (at c) we say 5 times 6 is 30 and 4 carried is 34 ; where the 4 carried is from 5 times 7, or 35 : and in multiplying by 3 we say 3 times 5 is 15 and i carried (from 6) is 16. We have then the following Rule : — Count off froTn the decimal part in the Multiplicand as many figures as we are required to retain decimal places in the Product j under the last of these figures place the units' figure of the Multi- plier, writing its other figures in a reverse order: and if any figure of the Multiplier has not a figure above it in the Multipli- cand place a cipher there. Begin the multiplication with the right-hand figure of the Multi- plier and multiply in succession by each of the others, setting down iti the product from the figure above the one we are multiplying by, but carrying to it the nearest ten from its product with the pre- ceding figure. Place the first figure of all these partial products in the same vertical line; add, and cut off from the result the required number of decimal places. I02 CONTRACTED OPERATIONS § 151. Remark. The last figure in the product may not be quite cor- rect, but to ensure its accuracy we must carry the process one place farther than is required to be retained. Ex. Multiply -43429448 by •6931472 so as to retain 7 places of deciinals ; and S947'i83 t)y '093187 so as to retain 4 places of decimals. ■43429448 S947'i839 27413960 781 3900 2605766 5352465 390865 178415 13029 5947 434 4758 174 416 3° 554'2ooi •3010299 CONTRACTED DIVISION. 153. The following Rule for Contracted Division requires no explanation beyond what has been given in Contracted Multiplica- tion, or will be afforded in the Examples given below. Determine first of all — by inspection, or by an equal removal of the decimal points in the Dividend and Divisor (150, Cor.), or by taking one step in the ordinary way — the highest order of units in the Quotient, and then the number of figures in the Quotient : from the left of the Divisor cut off one more than this number of figures, and strike out the rest. Proceed one step with this new divisor, but in multiplying its first figure by the quotient figure, carry the nearest ten from- the preceding figure. Instead of bringing down a figure to the remainder, strike off another figure from the Divisor, and proceed as before. If the number of figures in the Divisor be less than the number of figures to be cut off, proceed in the ordinary way until the number of figures still to be found is one less than the number of figures in the Divisor, and then apply the Rule. Ex. I. Divide 496"94325 by •17614352 so as to retain integers only in the quotient. § "SB- IN DECIMALS. IO3 17>6j>4352 ) 49694325 ( 2821 Now 496 divided by -17 or 4969 di- 14465 vided by i-y gives tliousands, therefore 374 there must be 4 figures in the Quotient : 22 retain 5 figures in tlie Divisor and strike 4 out the rest. Ex. 2. Divide •549532676 by 9"3i2i67 so as to retain 5 places of decimals in the Quotient. The highest order of units in the Quotient is manifestly hundredths; 93J.2,l6f ) 549532,676 (-05901 therefore there must be 4 significant ^3924 figures in the Quotient, and the first ^^ must be o : hence we cut off" 5 figures to the left of the Divisor and strike out the rest. 22 Ex. 3. Divide 578564327 by -8345 so as to retain 5 places of decimals in the Quotient. 8>3>4.5 ) 578564327 ( 6-93306 The first figure of the Quotient will >j>i-»j / 77864 represent simple units, therefore there 27593 must be 6 figures in all : but as there are 2558 only 4 figures in the Divisor, we must 54 take 3 steps in the usual way and then 4- apply the Rule. Exercise 24. I. Find within a hundredth, the sum of 27-035035, 3-7676, "2596596 and -00345; and the difference of 315-857142 and 47-950375. ■i. Find within a thousandth, the sum of -0795, 617-34833, -08391 and 25-808080; and the difference of 3-183546 and -93681. 3. Find correct to five places of decimals the sum of -385385, 19-777777, •05 and 6-7897897; and the difference of 23-34534534 and 7-8888888. Multiply 375-76843 by 3-14159, retaining 4 places of decimals. 5. Multiply 65-00763 by -9876, retaining 5 places of decimals. 6. Multiply 583-26784 by '00985, retaining 2 places of decimals. 7. Multiply 678-3089 by 45-657, retaining integers only. 8. Multiply -86858896 by 1-0986123, retaining 5 places of decimals. 9. Multiply -0008127 by 483-2716, retaining 6 places of decimals. 10. Divide 3789-436 by 265-5984 so as to retain 1 places of decimals. ri. Divide 742-876315 by 4967-358 so as to retain 4 places of decimals. 12. Divide 59-3264 by -09352 so as to retain [ place of decimals. 4' 104 REDUCTION OF Ex. 24. 13. Divide i7'3S9267 by •6574 so as to retain 6 places of decimals. 14. Divide 2 by I5'3i486s so as to retain 5 places of decimals. 15. Divide io-936954 by '3547808034 so as to retain 3 places of decimals. 16. Find, in each case within a millionth, the reciprocals of 2'302585i and 3-14159^6535. Find, to the number of places of decimals indicated, the value of 17. i'05o6252, i'oso625', i "060625*; in each case to 4 places. 18. i'0375' and 987-625 x 1-0375* to 5 places of decimals. 19. 1-0425* and 357-6-T-I-0425'' to 4 places of decimals. 3^575 — \__ 3^515 j J 5_^[ ; in each case to 5 places. -045 1-0458' -045 ( 1-0458)' ='^ REDUCTION OF FRACTIONS TO DECIMALS. REPEATING DECIMALS. 154. To reduce a vulgar fraction to its equivalent decimal frac- tion, we find how often the denominator is contained in some power of 10, and we multiply numerator and denominator by the quotient : thus, if we take the fraction -^ , the multiplying factor is — 4^, and therefore 16 100... 500... 5 ^ "" 16 T6~ 16 100... 100... ' But we have still to find what power of 10 must be taken, or, which is the same thihg, how many ciphers must be annexed to the numerator 5 ; and this is found by actual division, thus : 16 ) 50000 3125 where we see that the division terminates on bringing down 4 ciphers; therefore 5 3125 \=— — - =-3125 : 16 loooo ^ •' ' hence, to reduce a vulgar fraction to a decimal we have this Rule : — IVriU down the numerator, annexing ciphers, divide by the de- nominatoj; and cut off as many decimal places in the quotient as we have brought down ciphers. § IS4- VULGAR FRACTIONS TO DECIMALS. 10$ Cor. Again, a fraction represents the quotient of the numerator by the denominator (103), and by division of decimals we can express this quotient as a decimal. A comparison of the two pro- cesses will shew that they are in reality identical. 155. But it will often happen that however many ciphers we may bring down the division will riot terminate, and consequently that the given fraction cannot be expressed exactly as a decimal. Take any fraction in its lowest terms, as ^ , then the numerator of the equivalent decimal fraction is 500... 5 X 100... or ; 12 12 and, as 12 is prime to 5 (68), this result can only be an integer if icK)... is divisible by 12 (75), that is, if some power of 10 be divisi- ble by the given denominator. But 10 is the product of 2 and 5, therefore the second power of 10 is the product of two 2s and two 5s, the third power of three 2s and three js,... hence, if the prime factors of the denominator be 2 and 5 only, however often they may be repeated, thpre is always a power of 10 that can be divided by that denominator. But if the denominator contain any other prime factor than 2 or 5, this factor will be prime to 10, and therefore to every power of 10 (95), and consequently however far we carry on the division it will never terminate. We conclude then that If the denominator of the given vulgar fraction {itt its lowest terms) be composed of the prime factors 2 and 5 only, the fraction can be expressed as an exact or terminating decimal; otherwise, it can not. 156. But though a given vulgar fraction may not be capable of being expressed as a decimal exactly, yet it can be expressed to any degree of accuracy we please. Take the fraction ^^ ; II ) 30000000 •2727272 and we see that -^ is greater than -1 but less than -3 ; is greater than -27 but less than -28 ; is greater than 272 but less than -273 ;... that is, if the decimals •2; -27; -272; -2727; -27272;... Io6 REDUCTION OF §156. -. __ — . ^ be taken to represent ^, the errors are respectively less than 1 J- 10' io'd' 1000' 10,000' loo.ooo'"' and thus by taking a proper number of figures in the decimal part, we can represent ^ to any required degree of accuracy. 157. In non-terminating decimals the figures of the quotient must recur. Take the fraction ^ : to convert it to a decimal we annex ciphers to 4 and divide by 7. Since the division does not terminate, we cannot have the remainder o, and the only other remainders that can arise are i, 2, 3, 4, 5 and 6, and consequently after 6 steps at most we must come to the proposed numerator or to a remainder that has occurred before, and therefore from that point we must have a recurrence of the remainders, and therefore of the quotient figures. 158. (i) When, starting from a certain point in the decimal part of a number, the figures repeat themselves indefinitely and in the same order, the number is said to be a repeating, recurring or circulating decimal. (2) When the recurrence takes place from the first figure of the decimal part, the number is a pure repeating decimal ; otherwise it is a mixed repeating decimal: thus ■272727... is a pure, and 25'34s67567... is a /«z>^ ^' I' ■°'t6S75' ''^3- 6- i7TTr. 25-1. 6|4, 13-iftr, and 20x/At- 7. Express as decimals and find the difference of l^andil; ^-^^6.1^. 625 16 32 15625 Find a decimal which shall not differ from 8. -5- by a ten-thousandth ; i^^f by a thousandth. o. - -' by a millionth ; — by a hundred-millionth. y- 17 -^ 14 17x31 10. Find the difference between ffi and 3T4159265 to 6 places of decimals. 11. Which of the following fractions can be expressed as finite decimals? 19 ^ ^ -21 167 5" ^31 64' 192' 405' 560' 625' 875' 288" 12. Write down those numbers between i and 20, of which if any one be the denominator of a fraction in its lowest terms, that fraction can be converted into a terminating decimal. Express the following vulgar fractions as repeating decimals : 13 IS '7. '9- % 5 44 ■ 3' 9 3*" 450 103 180 ■ 14. 343 21 li 401 375' 11' ^^' 352' . i8A\, _5_ 214 396' 37 233 185' 16. 809 ,,9 506 62s 296' ^^"' 505' 576' , 997 ■ 1375' ^^' '£ lOIO 18. 80 200 425 541 41' 271' 328' 1084' , S 97 '• 7' 13 4fT, gj. 7^- 20. 7^5 135 1000 313 52 ' 143' 407 " 416' 354^ 1025' 661 63 728' 259' 1000 100 1 22. 7624 20000 16 17 4329' 5291 ' 17' 19 no REPEATING DECIMALS. Ex. 25. Convert the following repeating decimals to vulgar fractions in their lowest terms : n- '4 "S^. ■?*. '^59. 8-961. ■24. -tii?, 3-?8o4g, 5-5S356- 25. -STH^S, -620268, •48339- •26. "OS. ^'oSt. ■=0495. 3'545- 27. •01136, -00729, 42-20837. 28. -5925, 8-43117, -254629. 29. 37-40185, -754347. -OSOISS- 30- 3-86436i§, 5-789306. 31. 2-619047, 18-7621951. 32- :oo8497i35, 13-94430769. ADDITION AND SUBTRACTION OF REPEATING DECIMALS. 160. The sum of two or more repeating decimals may be found to a limited degree of accuracy by the method pointed out in Art. 145. If we wish to find the sum exactly, we must bear in mind that after a decimal has begun to repeat, the repetition may be sup- posed to begin at any subsequent figure (158, S), and that a period of 2 figures is the same as one of 4, 6, 8,... figures, thus -27 is the same as '2727, -272727..., that a period of 3 figures is the sg.me as one of 6, 9, 12... figures, &c. : hence, if several decimals have to be added, and one period consists of 2 figures, another of 3 figures, and another of 4 figures, we may consider each of them to consist of a period of 12 figures, where 12 is the L.C.M. of 2, 3, and 4; and as the whole body of figures within these 12 places will constantly be repeated, their sum will be constantly repeated. We have then this Rule : — Extend each decimal as far as the farthest non-repeating figure in any of them : find the L. C. M. of the number of figures in each period, and extend each repeating decimal so many places further, and one or two places more to make sure we are carrying the cor- rect figure to the last place of the second extension : add in the usual way, then in the sum the first extension will give the non-repeating part, and the second the repeating part. If in this Rule we write subtract for add and remainder for sum, we shall have the Rule for the Subtraction of Repeating Decimals. § i6o. REPEATING DECIMALS. Ill Ex. I. Find the sum of 31416, 8'2si42857, -034, 23"257635 and 5 "45627 ; (i) correct within one ten-thousandth, (2) quite correctly. (I) 3-1416 8-2514 ■0344 23-2576 5-4562I72 28 44 35 (2) 3'i4i6 8-2514 ■0344 23-2576 5-4562 285714 444444 356356 727272 28 44 35 72 40-1413 40-1413 813787 Ex. 2. Find the difference between 27-035471 and 5-98765 (i) correct to one millionth, and (2) absolutely correct. (l) 27-035471171 5-987657165 21-047814 (2) 27-03541717171171 5-9876I576576I57 21-0478 140595 MULTIPLICATION AND DIVISION OF REPEATING DECIMALS. 161. If the Multiplicand be a repeating decimal, and the Multiplier an integer or finite decimal, the Product will be a re- peating decimal of the same kind as the Multiplicand, but will sometimes admit of simplification. To form the Product we pro- ceed in the usual way, carrying out the work one or two places beyond the period, to make sure that the figures carried are always correct. 37'8345^« l_ 264-84216 (A) 37-823636 or 37 '82 A II II 416-0599 = 416-06 37"83459 537 26484216 I 1350378 19917297 37-6285714285 7 263-3999999 =263-4. 37-83459 537 (B) 26484 113503 I 99 I 729 216 783 729 21317-17729 21 78 72 In the last Ex. preceding, as at (a), we see that the periods in the partial dividends are 216, 378 and 297 ; if therefore we draw or suppose to be drawn vertical lines under the period of the Multiplicand and extend the above periods we obtain the product required, as at (b). 112 REPEATING DECIMALS. § i6i. Remark. If the Multiplier had been 537 or 5-37 or ■S37,." the only difference would have been in placing the decimal point in the Product. 162. The Product or Quotient of two repeating decimals may be found — (i) to a limited degree of accuracy by contracted Multiplication or Division of decimals (152, 153); and (2) exactly, by converting them into fractions, performing the required operation, and reducing the resulting fraction to a decimal. Ex. I. Multiply 275436 by 8-347, (i) retaining 5 places of decimals, and (2) exactly. 27-54363|3 ■> ^ g.-,^_ 275436- 2754 ^ 8347-834 m 777438 27 5436 xB 347- ^^^ X — — 220 34QOQO 8263091 30298 683 I 101745 ^2?a682 ^ f5*3 192805 9909 900 19280 900 100 '193 ^ 20693534 jg 90000 ^ =229-92815. 229-92815 Ex. 2. Divide 66-02637 by 248-722, (i) correct to 5 places of decimals and (2) exactly. 66-02637-248-722 ^ 6602037-6602 ^ 248722-2487 99900 ■ 990 6595435 990 24,8 J2>2 ) 6602037 ( -26543 = -^^ X 3l65is X I 13526 ^ 3241 1090 6 " 2035 95 3241 21 -^^-J* I22IO : -2654381. Ex.26. REPEATING DECIMALS. 113 Exercise 26. Find the sum of 1. 6-3, U6-43 and 375-8^43 ; 7843, 12-4718 and -0631. 2. 13-076, 19-445 and 31-263; 36-2514285? and 63-74857144. 3. 27-64235, 9-264263?, 5-4945, 1-498 and -603306. 4. 3-80?, 6-76435, 8-5486, -0037, -6571428 and 87-^8$. Find the difference between 5. -328 and -318; 25-47 and 16-8578; 753-6716 and 19.004459. 6. 17-573 and 14-5?; 6-7345$ and 3-0726; ■?I4285 and -001136. 7. Find the complement of "04563 ; "O^S^; 25-6440376. Multiply 8. -37644 by 91 -37644 by ii;_ -37^4^ by 37; 2-3857144 by 56. 9. ■4322443(8 by 88; 3-54468 by -144; '63542? by r3-2. 10. 27-38443 by 26-7; -78539811373457; 9"3856787 by 7-659. 11. Divide 3-457954 by 8; 37'63sMby7; 8-9854 by 12. 12. Divide 235-47 by 24x20; 539-63436 by 112. Find the value of 13. 4-8x44; 1-18 X -538461; 43-491x6-24; 6-36 X -571428. H- 7"657'44X4'^; ■4?X4-96; 7-63x8-83; 19-74 x 2(/45. 15. 22-gi7x-566; 5598-9443 X 8-247; 44-20645x1-5823707. 16. -5-^-45; 3-8-J-2-73; i-g6-v--583; 3-6-^ -047; 60-45 -j-7-38. 17. 43■49l-^6•23; •37-7--148; -01236^-051; 28-583-^37-14. 18. 6-891^15-45; 9"S3-^3'2o83; 411-3519-^19-5881; 7 7 -6 704 7 -="9 -486; 14-476196-^2-15^6. 19. -4 of -0006 of ; 1-83 of -954 of -428571 of 2-25. it ■°°^4 2'S+ 1-25-2-125 1+5-4x6-4 .. r i ■ ^ c c '°- i^ +2-3-4-^5 ' iT^fei'- 5-a$ofii-i6^4-59^of36. 2-6 of 2-8 3 4I of 4-036 _ '03 - -63 ^'" 6-4 of -857144 3-75ofi-7' -123 22. Divide 9-614 by -0000019 and -j by -oooj, and multiply the sum of the quotients by "0005. Miscellaneous.' 23. Express - (6i-H2|-3) as a decimal, .ind -7 + — of -825 + 4-13 as a vulgar fraction. 24. Reduce. ( — o{ 2-4S of -02 )-hiooo to a decimal. \25 i°o / B.-S. A. 8 114 REPEATING DECIMALS. Ex. i6. 25. Which is the greatest and which is the least of the expressions (i) ^ + 5; (.) 1-414^1; (3) ^ + 7 + f? 3 4 » B / 76. Find the sura, difference, product and two quotients of lo-oi and •0091, and find the sum of the results. 37. Find within a thousandth the value of 3239"i6 x i'0475*. 28. Arrange in order of magnitude : (i) j-H; (2) 3+—^; (3) 3-14159^6. 7 + 76 29. Find the G.c.M. of I353'6 and 23i'48; of 29-75 and ii3"9; of 36-795 and 57-98; and of 376-1034 and 1081. Make the number of decimal places in the two numbers the same (143); find thdr G. c. M. as if they were integers, and cut off that number of decimal places in the result. See Art, 135. 30. Find to 6 places of decimals the value of Express the first term as a decimal, and derive each term from the preceding by dividing by 5, placing the results under one another. U\ ^ y.\-L 3|3-4i 3-4-5 J_ ) ^^' 10= ( 10=' ^1.2 10^ 1-2.3 ■ io«r Express each term in the bracket separately as a decimal. (4) '+7^ + ; ' ■ ' • ' 1.2 1.2.3 1-2.3.4 1.2.3.4.5 The second term is derived from the first by dividing by 2 ; the third from the second by dividing by 3, &'c. ..iiiiiiiii (5 - + -. -5+- . ^ + -.-= + -. -„- + ... 5 3 5' 5 5' 7 5' 9 5' First express as decimals -, — 5 , — j , . . . then write under one another 5 5 5 I I I I 1 c6' . . . and add. i 3 i'" 5 5" ,^. ,liiiiiii ) 4 (5 3 5' 5 6° 7 S' ' 239 First express as decimals and add the 1st, $rd, c,th,... terms within the bracket, then the 2nd, 4th, 6th,... terms; find the difference, and multiply by 16 ; and from the result subtract — expressed as a decimal. 239 CHAPTER VII. EVOLUTION. 163. Evolution is the operation by which we find any root of a given number. The square root, cube root, fourth root, fifth root,... of a given number is the number whose square, cube, fourth power, fifth power,... is equal to the given number. Thus the square root of 25 is S, because 5^ is 25 ; the fourth root of 256 is 4, because 4* is 256. The root of a number is denoted by writing ^ (really r) before the number, and placing against it a small figure denoting which root is to be taken ; thus the square root of 25 is denoted by 1^25, or simply V 25; the cube root of 64 by 4/64 ; the fifth root of 128 by 4/128. SQUARE ROOT. 164. (i) In multiplying a number by itself we see that its square and the square of its units' figure have the same units' figure ; thus 537^ has the same units' figure as 7^. Now the squares of the simple numbers I, 2, 3, 4, 5, 6, 7, 8, 9 are respectively I, 4, 9, 16, 25, 36, 49, 64, 81, and if a number ends with o, its~ square also ends with o ; hence the squares of all numbers, integral or decimal, must end with either, o, I, 4, 5, 6, or 9 ; and therefore it follows that A number ending with 2, 3, 7 or 8 cannot be the square of any number, integral or decimal. (2) If a number ends with i, 2, 3,... ciphers, its square must end with 2, 4, 6,... ciphers (41) : also, if a number does not end with a cipher, its square does not end with a cipher ; therefore A whole number ending with an odd number of ciphers cannot be the square of a whole number. _^ 8—2 1 16 SQUARE ROOT. § 164. (3) If a number has i, 2, 3,... places of decimals, the last figure being significant, its square must have 2, 4, 6... places of deci- mals (148) ; therefore it follows that A decimal whose last figure is significant, and which has an odd number of decimal places, cannot be the square of any number integral or decimal. 165. When the square root of a whole number is not a whole number, neither is it a fraction whose numerator and denominator are whole numbers. For, if possible, let ^53 be 7f or — , where — is in its lowest terms ; then by definition 67 67 672 ^ , , ^ -^ X -^ or -^ must be equal to 53; but 67 and 9 are prime to each other, therefore 67" and 9* are prime to each other (95), and therefore 67'' is not divisible by 9^ or 672 -hr is not a whole number. 92 166. A number which can neither be expressed (measured) by a whole number, nor by a fraction whose numerator and denominator are whole numbers, is called an incommensurable or. irrational number; thus ij2, VS, \/53 are incommensurable numbers. Again, numbers whose square roots can be expressed exactly either by a whole number or by a fraction are called perfect squares: thus 36 whose square root is 6, and - whose square root 2 4 ^ is -, are perfect squares ; whereas 53, and - are not perfect squares. 167. The squares of '. 2. 3. 4, 5. 6, 7, 8, 9, 10, are respectively I, 4, 9. 16, 25, 36, 49, 64, 81, 100, which shews that the square of a number of i figure consists of either l figure or 2 figures. If we affix a cipher (o) to each of the above numbers we must affix two ciphers (00) to their squares ; hence the square of a number of 2 figures consists of either 3 or 4 figures. §i67. SQUARE ROOT. 117 If we affix 2 ciphers (oo) to each of the above numbers, we must affix 4 ciphers (oooo) to their squares ; hence the square of a number of 3 figures consists of either 5 or 6 figures. And, in like manner, every additional figure in the number makes two additional figures in the square ; hence The square of any number consists of twice as many figures, or twice as many less i, as there are in the given number. ' 168. Hence conversely, if the square of a number be given, we may suppose its first two figures on the right to correspond with the first figure to the right in the number ; the next two figures in the square with the second figure in the number, and so on ; and the last two or last one figure in the square to correspond with the last figure in the number. Thus we may mark off the squares 529 and 459684 in this way, ,5 ,29 45 S6 J84, which shews that the numbers of which these are the squares con- sist of 2 and 3 figures respectively. The parts into which these numbers are marked off are cailsA. periods j thus in 529 the periods are 5 and 29 ; and in 459684 they are 45, 96 and 84. It is more usual, however, to place a dot over the units' figure, and every alternate figure in 'the square; hence there will be as many dots as there are periods, and each period will consist of the figure over which the dot is placed, and the figure to its left, if there be one : for example, in 97344 we point thus, 97344 ; so that the first period is 9, the second 73, and the third 44. 169. The extraction of the square root of a number depends on the following proposition : If from the square of a number we subtract the square of one part of it, the remainder is a product of two factors : one factor is twice that part increased by the other part, and the other factor is the other part. Take 57, which is made up of the parts 50 and 7 ; now 572=(5o + 7)x(5o + 7) = 50x50+50.7 + 7.50+7.7 (62) = 502+2.50.7 + 7.7 (35) = 5o2+(2.5o+7)x7; (62) therefore 57' - 50' = (2 . 50 + 7) v 7. Il8 SQUARE ROOT. §169- Cor. When the part whose square is taken away is much greater than the other part, it follows from the above result that if we divide the remainder by twice the part whose square is taken away, the quotient will give a near approximation to the part left — and of course in excess. 170. To extract the square root of a given number. Take 71289. Placing dots over 71289 ( 200 . ' , , .. 40000 60 the units' and every alternate figure, , , c^-5- — .U 2-n^ C .^,. 400 + 60 = 460)31289 7 we see thatTnere are 3 figures m the 5q 27600 root, and that the first period is 7: 520 + 7=527) 3689 and, since 7 lies between 4 and g, 3689 the squares of 2 and 3, the root must lie between 200 and 300. Put down 200 as one part of the required root, and subtract its square from the given number, leaving 31289. Divide the remainder 31289 by 2.200 or 400 ; the quotient is greater than 70 but less than 80 ; therefore the remain^ing part of the root is less than 80 (169). Let us try 70 ; but (400 + 70) x 70 or 470.70 or 31900 is greater than 31289; therefore 70 is too great. Let us try 60; now (400 + 60) x 60 = 27600, which is less than 31289; therefore the remaining part of the root lies between 60 and 70, and the required root lies between 260 and 270. Subtract 460x60 or 27600 from 31289, leaving 3689: but we have now subtracted from the given number 200^ + (2 . 200 + 60) X 60 or 260^, (i 69) we may therefore consider the root as consisting of two parts, one is 260, and the other still to be found is less than 10. Divide the remainder 3689 by 2 . 260 or 520 : the quotient is greater than 7 but less than 8 ; therefore the other part is less than 8. Let us try 7 : now (520 + 7) x 7 = 3689, and this subtracted from the remainder leaves o; therefore 7 is the part required exactly, and therefore the root required is 267. 171. In the preceding example the given number was ^.perfect square, but the method is equally applicable if the number be not a perfect square; except that in the latter case the root can be found to a limited degree of accuracy only. §171. SQUARE ROOT. 119 Find the square root of 5713. S?i3-o6o6 ( 70 49°° 5 140 + 5 = 145 ) 813 '5 725 'oS i5o+'S = i5o'5 ) 88-00 -004 151 + ■o8=i5i'o8 ) 127600 1 2 -0864 151 ■16+ '004= 1 51 "164 ) •663600 ■604656 ■058944 Pointing the units' and each alternate figure, we find there are two periods, and that the first is 57; and since 57 lies between 49 and 64, the root lies between 70 and 80; therefore 70 gi%'es the root correct within 10. Put down 70 as a part of the root, and subtract its square, leaving 813. Divide 813 by 2 . 70 or 140: the quotient is greater than 5 but less than 6: therefore the remaining part of the root is less than 6 {169). Let us try 5: now (140 + 5) x 5 = 725, which is less than 813; therefore the root lies between 75 and 76, and therefore 75 gives the root correctly within I. Subtract 725 and there remains 88; but we have now subtracted 7o» + (2. 7o + 5)x5or75«, (169) hence we may consider the square root of 5713 to consist of two parts, one of which is 75, and the other a part to be found, but which is less than I. Dividing the remainder 88 by 2.75 or 150: the quotient is greater than '5 but less than '6; therefore the remaining part of the root is less than -6. Let us try -5 : now (150+ ^5) x -5 is 75^25, which is less than 88; therefore the root lies between 75-5 and 75^6, and therefore 75-5 gives the root correctly within T. Proceeding in precisely the same way, the next step shews that 7S"S8 gives the root correctly within '01, and the next that 75^584 gives it correctly within "ooi ; and continuing this process we can obtain the root to any degree of accuracy we please. Let us now in these examples strike off all unnecessary ciphers, bring down the periods only when we actually require them, sup- press the decimal point in all places except the root, and take the remainders and divisors for what they really are, and not for what they represent, and our work will stand thus : 120 SQUARE ROOT. §171. 71289 ( 267 57i3-o6o6 ( 75-584 4 49 46)312 145)813 276 725 527 ) 3689 1 505 ) 8800 3689 7525 1 5108 ) 127500 1 20864 151 164 ) 663600 604656 58944 172. We deduce then the following rule : (i) Place a dot over the units' figure, and over every alternate figure to its left, and to its right : if there be no units' figure, suppose a cipher placed there, and remember that each period consists of the figure over which the dot is placed and the figure to its left. (2) Find the number whose square is immediately below the number in the first period : place that number in the root, and sub- tract its square from the first peiiod. (3) To the remainder bring down the next period, giving the first dividends double the figure in the root, and see how often it is contained in the dividend when its last figure is omitted j set down the quotient as the second figure in the root and annex it to the trial divisor we have just used, giving the first divisor. Multiply this divisor by the second figure of the root, and if the product be not greater than the first dividend subtract it from the dividend, but if the product be greater, use a lower number for the root figure until it becomes less; subtract, and we thus get the second remainder. (4) To this remainder bring down the next period: to the first divisor add its last figure, and see how often it is contained in the second dividend when its last figure is omitted j and proceed pre- cisely as before. (5) Continue the process till all the periods have been brought down : if there be no final remainder the given number is a perfect square, and its root is found j if there be a final remainder we have §172- SQUARE ROOT. 121 obtained the root 7iot exactly but within a unit of its last decimal order, and the remainder shews by how much the square of the root obtained differs from the given number. Remark. If at any step the quotient figure is o, set down o in the root, annex it to the trial divisor, bring down the next period and proceed as before. Ex. Find the square root of '08042896, and of 34'85o. (i) 6-08042896 ( -2836 (2) 34-8506 ( 5-90338 25 48)404 109)985 384 98_i 563 ) 2028 1 1803 ) 40000 1689 35409 5666 ) 33996 1 18063 ) 459100 33996 354189 1180668 ) 104911C0 9445344 1045756 In Ex. ( i) we put a cipher in the units' place, over which we placed the first point, and then over each alternate figure to the right; it is more usual simply to suppose the cipher to be put in the units' place. In Ex. (2) we have found the square root, accurate within one hundred- thousandth, or -ooooi ; and the remainder shews that the square of the root obtained differs from the given number by -0001045756. 173. When the number of figures to be found in the root is large, the work may be considerably contracted, as the following considerations and example will shew. When a certain number of figures have been found in the root, the divisor will contain at least the same number of figures, and the remainder either the same number or one less ; and as the subsequent steps of the operation can alter only in a very slight degree the figures already obtained in the divisor and remainder, it follows that we may obtain by contracted division as many additional figures within one as we have already got : hence, when one more than half the required number of figures in the root has been obtained, we may cut off the last figure to the right in the divisor and proceed as in contracted division. 122 SQUARE ROOT. §173- Example. Find to 12 places of decimals the square root ot 3-141592653589. Here we must have 13 figures in the root, therefore we find 7 in the usual way, and 6 by contraction. 374,1 5>92.65.3S39 ( 17724S3850905 3">i4>iS.92.65,3589 ( 1772453 I I 27 ) 214 27 ) 214 189 189 347 ) 2515 347)2515 2429 2429 3542) 8692 3542 ) 8692 7084 7084 35444) 160865 35444 ) 160865 141776 141776 354485 ) 1908935 35448s ) 190893s 1772425 177242s 3544903) 13651089 3544903 ) 13651089 10634709 10634709 35449068)3016380 00 3>5449P,6 ) 3016380 ( 850905 2835925 44 2835925 354490765) 180454 5600 180455 177245 3825 17750000 17724s 35449077009 ) 3209 3210 3190 4 I 69308 I 3190 3544907701805 ) 18 760569190000 20 17 724538509025 18 936030680975 2 174. In the preceding examples we have, for the sake of clear- ness, written down each product at full length, and then performed the subsequent subtraction ; it will, however, both save time and be conducive to accuracy, if we combine the two operations (51). Thus, taking the last example, the work will appear as follows : 3\i4.i5-92>65,35,89 ( 1772453 27 214 347 2515 3542 8692 35444 160865 35448 5 1908935 3544903 13651089 3 >S .4 ,4v9 ,0 >6 3016380 (850905 18045s 3210 2 o § 175- SQUARE ROOT. 123 FRACTIONS AND MIXED NUMBERS. 175. The square of 3. fraction is found by squaring its numr and denr : hence conversely the square root of a fraction is found by extracting the square root of its numr and denr. (i) If the denr of the given fraction, or of the fractional part of the mixed number, be a perfect square, we apply the Rule directly, whether the numr be a perfect square or not, thus : Ex. I. /25 _ V25 _ 5 V 64 V64 8' (2) But if the denr of the given fraction or of the fractional part of the mixed number be not a perfect square, we reduce the fraction or the mixed number either (i) to an equivalent fraction whose denr is a perfect square, and extract the square root of numr and denr, or (2) to a decimal, and proceed in the ordinary way ; thus : /I _ . /l-^-i3 = /IE:4 ^ nAo4 ^ '°-i98°39- ^ .784464... Vi3~V 13x13 V 169 V169 13 or= ^-615384=784464... or = ^2572 = 5-072205... Exercise 27. Extract the square root of I- 54756; 804609; 822649; ,12809241; 97574884; 21224449. 2. 93-7024; -128881; -02819041;, -00822649; 3659-0401; -49112064. 3- 5923303369; 3226694416; 7578747136; 5777216064; 6407522209. 4. 236-144689; 285970-396644; 400-54818769; -00501361708761. 5. 41605-800625; 8260628544; 3601 1 7-609604; 93870306991561. 6. 120888-68379025; 5783663477-05072081; 787026841863680889. 7. AVi; 72002^, 32II; 56411^; 30831^1; i8-7; 3-361 ; 4738-027. 124 FRACTIONS AND MIXED NUMBERS. Ex. 27. •00125 8. 24-2064, 3i24-8i and 2-42064 X 312-481; -004x15-625; . „ -■ 9. Find within one ten-thousandth the square root of the following numbers, writing down in each case the difference between the square of the root obtained and the given nurnber (172, 5) : 20; -i; 3-14159265; -012; i75'250564; 12-566360. 10. Find to 7 places of decimals the square root of •0068: ^; — ; iiA; 27TV3; ^-^; Vo^-V'ooi. 5' 13' '' T<^ y ■^ 159-658625 94 X 3129-96 X 23^365375ooo 1 y^ 58-3^/ 22-391272393 9^< /3I88-28 -974102607 •6 / 4-8925 96-6 / 3I93-I72S 6 / 4-8950 97-2 / 3 > 98-0675 •6/ •685699 97-8 3i98-753'99 •OS 97-85 •OS 97-90 •OS 97-95 •007 97^967 Point the units' and every third figure: the first period is 34, which lies between 27 and 64, the cubes of 3 and 4; hence the cube root of the given number lies between 30 and 40, and therefore 30 gives the cube root within 10. Put 30 in col. i : multiply by 30, giving 900, which put in col. ii ; -multiply by 30, giving 27000, which put in col. iii and subtract, giving the remainder 7829. Complete cols, ii and i thus — add 30 to col. i, giving 60; multiply this by 30, giving 1800, which add to col. ii, giving 2700; 128 CUBE ROOT. § 179. and now add 30 to eol. i, giving 90 : — therefore the cols, stand thus, 90, 2700, 7829. Use the trial divisor 2700; the quotient is greater than 1 and less than 3: therefore the other part of the root must be less than 3. Let us try 1 ; the remainder will be 2061 : therefore the root lies between 32 and 33, and therefore 32 gives the root within i. Complete cols, ii and i, and the cols, will stand thus — 96, 3072, 2061. Use the trial divisor 3072 : the quotient is greater than •6 but less than ■7; therefore the other part of the root must be less than •7. Let us try -6: the remainder will be i83'024; therefore the root will lie between 32 '6 and 327, and therefore 32-6 will give the root within ■!. In like manner the next step shews that 32'65 gives the root within "oi : and the next that 32"657 gives the root within '001 : and continuing this process we can obtain the root to any degree of accuracy we please. Let us now, in these examples, strike off all unnecessary ciphers, bring down the periods only when we actually require them, suppress the decimal point in all places except the root, and take the remainders and divisors for what they really are, and not for what they represent, and our work will stand thus : I. II. III. I. II. III. 4 16 948i§8id ( 456 3 9 34.829-006 ( 32-657 4 32 64 3 18 27 8 48 30818 6 27 7829 5768 4 625 27125 3 184 I^S 64^5 3693816 92 2884 2061000 5 650 3693816 2 1S8 1877976 130 6075 94 3072 183024 s 8136 2 5796 15965862s 1356 615636 ^6 312996 23365375 6 5832 22391272393 972 318828 974102607 6 48925 978s 31931725 5 48950 979° 31980675 s 685699 97957 3 198753 199 180. We deduce then the following Rule : — (i) Arrange for three columns, and head them respectively i, ii, and iii ; in col. in put the number whose cube root is to be extracted, and to its right provide a place for the root. §i8o. CUBE ROOT. I29 (2) Place a dot over the units' figure, and over every third figure to its left, and to its right if the number be a decimal: if there be no units' figure suppose a cipher placed there j and re- member that each period consists of the figure over which the dot is placed and the two figures to its left, if there are so many. (3) Find the number whose cube is immediately below the first period; place it in the root; place it also in col. i ; multiply col. i by the root figure, and place the product in col. ii ; multiply col. ii by the root figure and subtract the product from the first period in col. iii. (4) To find the next figure in the root proceed thus : — complete the columns, that is, bring down the next period in col. iii : add the root-figure to col. i; multiply the sum by the root-figure and add the_ product to col. ii ; add the root-figure to col. i : see how often the trial divisor {col. ii) is contained in col. iii when the last two figures are omittedj place the quotient in the root; bring it down to col. i ; multiply col. i by the root-figure ; put the product under col. ii, two places to the right, and add; tnultiply the sum by the root-figure, and place the product under col. iii and subtract, if the product be not too great : — if the product be too great, we must try the next lower figure in the root, or the next again, till we get a product sm.all enough, and then subtract. (5) Find the next and each succeeding figure in precisely the same way, and continue the process till all the periods have been brought down : if there be no final remainder, the given number is a perfect cube, and its cube root is found: if there be a final remainder, we have obtained the cube root within a certain degree of exactness only, determined by a unit of the last order in the root, and the remainder shews by how much the cube of the root obtained differs from the given number. Remark i. If at any .step the root-figure is o, bring down o to col. i, and 00 to col. ii, and proceed as before. Remark 2. There can seldom be any doubt about the root- figure except at the second step ; and here the numbers to be operated on are so small, that the process may be gone through mentally before the root-figure is actually put down. B.-s; A. -9 130 CUBE ROOT. § i8i. i8i. When one more than a third of the figures required in the root have been obtained by the ordinary method, the rest may be found by contraction. For when a certain number of figures have been obtained in the root, there will be generally twice that number in the trial divisor and as many or as many less one in the remainder; and from this remainder as a dividend we can find as many root-figures as it contains figures, less one or less two at most. Instead therefore of bringing down new periods, we confine our work within the limits indicated by the remainder, and neglect those portions of col. ii and col. i which do not affect the figures we are seeking. The way in which this is done will be understood from the following example. Ex. Find the cube root of 78539816339745 to 12 places of decimals. I. n. III. 9 81 ■78539* '63397456 ( -92263 9 162 729 18 243 56398 9 544 49688 272 24844 6710163 2 548 5089448 ?,74 25392 1620715397 2 55H 1531147176 2762 2544724 89568221450 2 5.128 76609659447 2764 2550252 / 12958562003 { 507432201 2 16.5996 / 12768760950 27666 255191196 / 189801053 / 178763634 6 166032 27672 255357228 / 11037419 6 830349 / 1 02 1 5066 276783 25536553149 / 822353 3 830358/ 766130 276786 ^^-^2553738350,7 56223 3 ^--' 13839 51075 ,27^6789 2553752190 5148 13839 5107 25537660,2,9 41 2 26 25537662 2 15 25.53,7,6,6,4 § i8i. CUBE ROOT. 13 1 The total number of figures in the root is to be 12, and since the integer next greater than i- of 12 is 5, we shall find the first 5 figures in the usual way, and the rest by contraction. Arrange the 3 columns, and put the given number in col. iii; place a dot over the units' figure, and over every third figure to its right. The first period is 785, and therefore the first root-figure is g. Put 9 in the root, prefixing the decimal point; put 9 also in col. i, multiply this 9 by 9, giving 81, which put in col. ii; multiply this 81 by 9, giving 729, which put under the first period in col. iii, and subtract, giving 56. To find the second figure in the root, complete the columns : that is bring down the next period in col. iii, giving 56398; add 9 to col. i, giving 18; multiply this 18 by 9, giving 162, which add to col. ii, giving 243; add 9 to col. i, giving 27. See how often the trial divisor 243 is contained in 563, the quotient is 1 : put 2 in the root, bring it down to col. i, giving 272 ; multiply this 272 by 2, and put the product 544 under col. ii, two places to the right, and add, giving 24844; multiply this by 2, and place the product 49688 under col. iii, and subtract, giving 6710. In the same way find the third and each succeeding figure. Suppose that we have now found five figures in the root, and completed col. i and ii, as if to find the 6th figure: — then the columns will be 276789 25537383507 12958562003 as they appear just below the broken line. As we do not intend to bring down another period in col. iii, we must cut off r figure in col. ii, and 2 figures in col. i, so that in multiplying up from one col. to the next, units of the same order may be under one another. Cut off therefore 7 in col. ii, and 89 in col. i. Use col. ii as a trial divisor, the quotient is 5 ; put 5 in the root, multiply col. i by 5, and add the product to col. ii; multiply col. ii by 5, and subtract the product from col. iii. To find the next figure, complete the columns, which now means add the same number as before to col. ii. Cut off 9 from col. ii, and 67 from col. i. Use col. ii as a trial divisor, the quotient is o ; put o in the root, and the columns will remain unaltered. Again, cut off another figure in col. ii, namely 2, and the rest of the figures in col. i. Use col. ii as a trial divisor, the quotient is 7 : and though there is no figure in col. i to set dawn from, yet from 7 times 2 plus 4 or from 18 we carry 2 ; add 2 to col. ii, multiply the sum by 2, and subtract the product from col. iii. Again, add 2 to col. ii, completing it; cut off its last figure 4, and then the rest of the work is effected by contracted division only. In this way the cube root of the given number is found to be •92263507432201 ; which is correct to the last place of decimals. 9—2 132 CUBE ROOT. § 182. 182. In the preceding examples we have, for the sake of clearness, written down the products at full length, and then performed the subsequent addition or subtraction ; it will how- ever both save time and be conducive to accuracy, if we combine the two operations (51). Thus, taking the last example, the work will appear as follows : 9 8 I •78639S163397456 ( -92263 I S 243 56398 272 24844 6710163 274 25392 16207 r5397 2762 2544724 89568221450 2764 2550252 12958562003(507432201 27666 255191196 189801053 27672 255357228 11037419 276783 26S36553149 822353 276786 2553738350,7 56223 ,2767,89 2553752190 5148 25537660,2,9 41 25537662 15 2 5v5,3>7 ,6,6,4 FRACTIONS AND MIXED NUMBERS. 183. The cube of a fraction is found by cubing its numerator and denominator ; hence, conversely, the cube root of a fraction is found by extracting the cube root of its numr and denr. (i) If the denominator of the given fraction, or of the fractional part of the given mixed number, be a perfect cube, we apply the Rule directly, whether the numr be a perfect cube or not; thus Ex. I Y^7 _ 4/27 _ 3 V 64 4/64 4' Ex. 2. 64 4/64 " EX.4. ^^^^m-~:J^-^-'^~-^-^^f^-^■e,^e^^.. I 183. FRACTIONS AND MIXED NUMBERS. 133 (2) But if the denominator of the given fraction or of the fractional part of the givfen mixed number be not a perfect cube, we reduce the fi-action or the mixed number either to (1) an equivalent fraction whose denominator is a perfect cube, and extract the cube root of numerator and denominator : or to (2) a decimal, and proceed in the ordinary way : thus T7^ , V5- 7 5 X 49 _. 4/245 _ 6-2573248- _.„ „„„ ^-•s- VrvT^-i/Si- — 'i — -^939035... or = 3/7i4285 = -893903S... Ex. 6. m=im = ^^^= '4^^^- = 2-o5783S.... or = ^8714285 = 2-0578352... FOURTH ROOT, SIXTH. ROOT, AND NINTH ROOT. 184. (l) Since 5*= (5^)2 (66), we may find the fourth power of a number by squaring the number, and then squaring its square : hence, conversely, we may extract the fourth root of a number by extracting its square root, and then' the square root of its square root. (2) In like manner, since 5"= (5^)' and =(5^)^66), we may extract the sixth root of a number by extracting its cube root, and then the square root of its cube root ; or by extracting its square root, and then the cube root of its square root. (3) And since S*=(5*)^={(5^)^}^ we may extract the eighth root of a number by extracting its square root, then the square root of its square root, and lastly the square root of that square root. (4) Lastly, since 5' = (53)', we may extract the ninth root of a number by extracting its cube root and then the cube root of its cube root. Exercise 28. Extract the cube root of 1. 912673; 12812904; 233744896; -0070778880; -000026730899. 2. -024137569; 176464-0815290; -081690010219; 355496768704. 3. -0567 1 1623688; 2-222447625; 1987677 1 7056; 8452-26465300. 134 FOURTH, SIXTH, AND NINTH ROOTS. Ex. 28. 4. 702121283072; -097781036543; 967-068262369; Si3S37536'S"o- 5. 93162981941037; 7-986807258669; 179301192-791869. 6- 97135540969725504; 696536477676927488079289747. ?• tItT' ^: '^^^s 4561^; 46A\; 57^1; 3845-49^- '33' 49'3 8. Find to 4 figures before contraction the cube root of 1034; 5-912; -009968; 5; — ; 1\. 9. Find to 5 figures before contraction the cube root of 3-467; 2-71828182846; |; -67; -078759; -325142. Extract the cube root of 'o. -78539; 18^; -002; and -024, each to 14 places of decimals. 11. 1-187; '003; 397"9.'>3; '013; and 81-812703, each to 20 places. 12. 2 + v'3; i-sjy, 7 + v'7, each to 13 places of decimals. 13. Extract the fourth root of 2x1309379856; 2-71828182846; 4351^1; -0008217; 62; 203^. 14. Extract the sixth root of 260184053769595201; S368|4if; 75-347; !■ 7 15. Find the eighth root of 57! | ; , -003532 ; ^ . 7 16. Extract the ninth root of g •689869781056; 3000; — ; 7 + V7; 82#|. 17. Find the value of '/-lli^; U ^^-^ ^3\^±3I-^lVli . V33-75 V -03375 'y8-«e/ooi 18. Find the square root of 3, and then shew that the cube root of 26 - 1 5 .^3 is equal to 2 - V3. 19. Divide the sum of the cube roots of 377149-515625 and 12771m by the square root of 5017^!. 20. In A.D. 1696, De Lagny, a Member of the Academy of Sciences, and afterwards a Fellow of the Royal Society, said that the most skilful computer could not in less than a month find within a unit the cube root of 696536483318640035073641037 ; find the cube root (r) within a unit, and (2) to 15 places of decimals. CHAPTER VIII. RATIO AND PROPORTION. 185. Ratio is the relation which two quantities of the same kind bear to each other, with respect to the number of times that the first contains the other ; thus, the ratio of 11 yards to 4 yards is determined by the number of times that 11 yards contains 4 yards, and therefore by the fraction 3J . The ratio of 11 yards to 4 yards is written thus 1 1 yards : 4 yards, and is read 1 1 yards to 4 yards. The two quantities must be of the same kind that the division may be possible, and the quotient will always be a number ; thus, the ratio of 11 yards to 4 yards, of 1 1 hours to 4 hours, of 11 pence to 4 pence, are all determined by the number — ; that is 1 1 yards : 4 yards =11 hours : 4 hours =11 pence : 4 pence = 11:4. Hence in treating of ratios we usually consider the tei-ms to be numbers : for at any time we can pass from quantities of the same kind to the numbers which measure them, anrl vice versa, whenever we find it necessary to do so. 186. The first term of a ratio is called the Antecedent and the second the Consequent. One ratio is said to be the inverse of another when the Antecedent and Consequent of the one are respectively the Consequent and Antecedent of the other : — thus the inverse ratio of 11 : 4 is the ratio of 4 : 11. 187. One ratio is greater or less than another according as the fraction determining the former is greater or less than the fraction determining the latter. Thus the ratio of 3 to 4 is greater than that of 5 to 7, because f is greater than f. 188. As the value of a fraction is not altered by multiplying or dividing numerator and denominator by the same number, the value 136 PROPORTION. §188. of a ratio is not altered by multiplying or dividing both its terms by the same number : thus the ratio of 2 to 3 is the same as that of 4 to 6, of 12 to 18, of 20 to 30, &c. 189. The ratios that each of a set of numbers have to each other is expressed in the form of a continued ratio : thus the ratios that each of the numbers 3, 5, 7, 9 have to one another is written 3:5:7:9. The values of these ratios will not be altered if we multiply or divide all the terms by the same number. 190. We compound raxios by multiplying the antecedents for a new antecedent, and the consequents for a new consequent : thus the ratio compounded of the ratios 2 : 3, 5 . 7 and 14 : 15 is 2x5x14 : 3x7x15. PROPORTION. 191. Four quantities are in proportion when the first has the same ratio to the second which the third has to the fourth. A pro- portion, therefore, is the equality of two ratios ; thus, since 2 : 3 and 10 : 15 are equal ratios, we have the proportion 2 : 3=10 : 15, which is read 2 to 3 equals 10 to 15. This proportion is also written, as in Geometry, thus 2 : 3 :: 10 : 15, and is then read 2 is to 3 as 10 is to 15. Of this proportion 2 and 15 are called the extremes, and 3 and 10 the means: 2 : 3 tht first ratio, and 10 : 15 the secoitd ratio. 192. In every proportion the product of the extremes is equal to the product of the means. Take the proportion a : b = c : d, where a, b, c, d are numbers, either integral or fractional ; then a _ c b~ d' ^, , ax.d bit.c , . therefore - — = - (139) b y.d b'/.d ^ -'^' §192. PROPORTION. 137 and since these fractions have a common denominator their numerators must be equal, therefore a x d= by.c. Cor. Hence, d= and a = —3- , that is, — Either extreme is equal to the product of the means divided by the other extreme. And in like manner, — Either mean is the product of the extremes divided iy the other mean. Remark. In any proportion we may replace the numbers of each ratio by quantities of which they are the measure, thus if 2 : 3=10 : 15, we may have 2 men : 3men = ;^io : ;^I5, but we cannot apply the terms of this proportion to the results just obtained, for of course we cannot multiply a quantity by a quantity ; but if the terms of one ratio be quantities and of the other numbers, we can apply them : thus if 2 : 3=^10 : ^is; 1 /- ;£lox3 then ;£isx 2=^10x3; and ^15 = — ^ — . 193. If four numbers are in proportion, they are in proportion when taken inversely. For example, \i a : b = c : d, then inversely b : a = d : c. Since a : b = c : d, therefore bxc=axd, (192) b'xc axd and therefore = , axe axe b d or J "" ", ' a c or b : a = d : c. Remark. This proposition holds whether the terms of both ratios are numbers, or the terms of one ratio be numbers and of the other quantities, or the terms of both ratios be quantities : thus if 2 men ; 3 men = ^io ; £'iS> then 3 men : 2 men = ^iS : £10. 138 PROPORTION. §193'- ^93*' If four numbers are in proportion, they are in proportion •when taken alternately. For example, \i a : b^c : d, then alternately a: c^b : d. Since T = ji o a therefore axd=b><.c, (192) , , . axd by-c and therefore -,= „ a b or ~ = -^> c d or a : c = b : d. Remark. The result of this Article is. applicable when the four terms of a given proportion are either all numbers or all quantities of the same kind : but not when the terms of each ratio are quantities of different kinds : for example, although 2 men : 3 men=;£io : ;£iS> we cannot have 2 men : .£10 = 3 men : £1^,, for each of the two ratios is impossible. But since .£2 : .£3 = ^10 .-^15, there is now no objection to our alternating the second and third terms, and therefore we have ;£2 :.£io = ^3 :/i5. 194. If we compound the corresponding ratios of two or more proportions the resulting ratios will form a proportion. For example, if a -.b =p : q, c : d=q : r, e ■.f=r -.s, then ax-cxe : bx. di=-, b q' d r / J ^, J. a c e p q r therefore •;x^x-=-^x5-x-- b d f q r s' axcxe px qxr bxdxf q xrxs' axcxe : bxdxf=pxqxr : qxrxs. (129) §194- PROPORTION. 139 COE. If in corresponding ratios the consequent of one be the antecedent of the next, the common terms may be made to dis- appear in the compound proportion : thus in the last proportion we may write axcxe : dx dx/=^ : s. (i88) 194*. When, in a proportion, the means are equal, each of them is called a fnean proportional between the two extremes ; thus, in the proportion 2 : 4=4 : 8, 4 is a mean proportional between 2 and 8. And because 4x4 = 2 X 8 or 4'' = 2X 8; therefore 4=^2 x 8; (192) that is, — A mean proportional between two numbers is the square root of their product. Ex. I. Find a 4th proportional to 2f, 3I, and 4^. Now 2f : Sf = 4r • 4th proportional required ; therefore 4th proportional = ^^iL+l = 15 x | x — = ^ = 6|. ^^ 4054 Ex. 2. The ratio of /i to j5 is 2 : 3 ; and of 5 to C is 5 : 6 : find the ratio of A to C. Now _ = _and^ = g; therefore -x- = -x-g, or- = -; or A : C=5 : 9. Ex. 3. The ratio of ^ to ^ is 2:3; of -S to C is 5 : 6 ; and of C to Z> is 7 : 8 : find the continued ratio oi A, B,C and D. A : B = 2 : 1= 70:105, multiplying each term by 5 x 7; B : C=6 : 6=105 : 126; C ■.D=1 : 8=126 : 144; therefore .^ : 5 : C : Z> = 70 : 105 : 126 : 144. Ex. 4. A cask of 72 gallons consists of 1 1 parts brandy and I part water : how much water must be added that it may consist of 9 parts brandy and i water ? 72 gallons is made up of 12 parts, therefore i part is 6 gallons and II parts is 66 gallons. But, while the brandy remains the same 66 gallons, 14° PROPORTION. Ex. ■zg. the water is to be increased so that the brandy is to the water as 9 : i ; but 9 : 1=66 : •j\, therefore \\ gallons of water must be added. Exercise 29. I. Simplify the ratios: 21 ! 35; 441 : 2401; i : f ; Si : 6|; 3-16 : rW; t : 4 : i : J; I : f : I : I- vs. Express in its simplest form the ratio compounded of the ratios : 2:3. 3:4. 4:5; 7 : 15. 9 : 16' ^4 = 36! if : 2f; ^-h •• 7t- 3. Of the following ratios which is greatest? £:7or7:9; 8: 15 or 15: 29; 2I : 6f or 4! : iif. 4. Do the ratios ^ : ifr ^"d 6J : i^^ constitute a proportion? 5. Do the numbers 4^, tV> 13 ^"d ^tI form a proportion? If not, find what the fourth number must be, so that a proportion may be formed. 6. What is the first term of the proportion of which the other three terms in order are 8, 9 and 12? 7. If one mean of a proportion is 5!^, and the two extremes are 1% and 7^, what is the other mean? 8. Find a fourth proportional to 6j, 4fandi3j; "0004, i'4and*02; •!, "oi and 'ooi; ■0625, 3-85 and i2'5; 8352, 3-69 and 30'57, within -0001. 9. Find a mean proportional to 3f and lof; "i and '001 ; 387-908 and •0187 to 4 places of decimals. 10. Compare the rates of two locomotives, one of which travels 397 J miles in ii| hours, and another which travels 262^*3 miles in S| hours. II. If, when A makes a profit oi £1, B makes £■>,•, and when B makes a profit of £^, C makes ;^6 ; and when C makes a profit of £6, D makes £1 : compare the profits of ^, B, C and D. 12. One grocer to 19 lbs. of coffee adds 5 lbs. of chicory, and another to 27 lbs. of coffee adds 7 lbs. of chicory: compare the amount of coffee in the two mixtures. 13. A mixture is composed of 9 parts brandy and i water; 4 gallons of water are added, and the mixture contains 6 times as much brandy as water: how many gallons of brandy does it contain? 14. A barrel consists of 3 parts ale to 2 parts stout; how much of the mixture must be drawn off' and replaced by stout, that the new mixture may be half and half? 15. A greyhound pursues a hare and takes 3 leaps for every 4 leaps of the hare, but 2 leaps of the hound are equal to 3 of the hare : compare the rates of hound and hare. CHAPTER IX. CONCRETE NUMBERS. TABLES OF TIME, LENGTH, &C. 195. If one quantity contains another of the same kind an exact number of times, the first is said to be a multiple of the second, and the second a submultiple or aliquot part of the first. 196. Of quantities of the same kind we take an arbitrary but well-defined quantity of that kind as our unit, and finding how many times or parts of a time it is contained in each of them, we express them either as whole numbers or as fractions. But in this way very large quantities will be expressed by very high numbers, and very small quantities by fractions, which give by inspection little idea of their relative values ; to obviate this inconvenience, we take such multiples and submultiples of the unit as will enable us to avoid very high numbers in one case, and troublesome fractions in the other. Thus of length, we take a yard as our unit, but to measure long lengths we use the mile, a high multiple of the yard, and to measure short lengths we use the inch, a small submultiple. But in England, not only do these multiples and submultiples proceed, in any particular case, without uniformity, but the relation observed between them in any one case, is no guide in any other ; thus the relation between the multiples and submultiples of the unit of length, is not that which obtains between those of weight, or oi time, or oi capacity, &c. 197. The principal quantities with which we shall be concerned are those of length, surface, volume, weight, capacity, time, and money. And as in England the unit of time is the only one derived from a fixed quantity in nature, we shall begin with the measures of time. 142 TIME. % 198. TIME. 198. The unit of time is the day, or, strictly speaking, the mean solar day. A solar day is the interval between two successive noons ; but as these intervals are of unequal length, we take the mean or average of all the solar days in the year, and to this mean solar day we give in civil reckonings the name of DAY. The sub- multiples of the day are the hour, the minute and the secondj and its multiples are the week, the month, and \h&year: their relations are set forth in the following table : 60 seconds (s.) . . . are i minute (m.). 60 minutes .... i hour (h.). 24 hours I DAY. 7 days i week. 4 weeks I month. 26s days or ^66. days . . 1 year. 199. The solar year consists of 365'242242 mean solar days, or very nearly 365J days ; hence to make the civil year correspond with the solar, we take three consecutive years of 365 days, and a fourth, called leap-year, of 366 days, those being leap-years of which the number is divisible by four; and this is called the Julian correction, after Julius Gsesar. But in this way we insert 100 days in 400 years, which is too much, for "242242 x 400 is 96'8968 or 97 days nearly ; to riiake the necessary correction, those years whose centuries are not divisible by four are not leap-years; thus 1700, 1800, 1900 are «(?/ leap-years, but 2000 is a leap-year; and this is called the Gregorian correction, after Pope Gregory XIII. The civil year is divided into 12 calendar months, of which February contains 28, or in leap-year, 29 days : April, June, September, November, 30 days; and each of the rest 31 days. LENGTH. 200. The unit of length is the yard. In the Office of the Warden of the Standards at Westminster there is a solid bar of bronze 38 inches long and i inch square ; near to each end a small cylindrical hole is sunk in which is inserted a gold plug ; and the § 200. LBNGTIt. 143 distance between the centres of the gold plugs when the bronze is at a temperature of 62° F. is the imperial standard yard (41 and 42 Vict. c. 49, s. 10). 201. By the Act, 5 Geo. IV. c. 74, the imperial standard yard was declared to be the length of the pendulum, vibrating seconds of mean time in the latitude of London in vacuum at the level of the sea, in the proportion of 36 to 39'i393. In this way, it was enacted that a new standard yard should be constructed, if at any time it should be required ; but when the old standard was rendered useless by the burning down of the Houses of Parliament, and its restoration rendered necessary, so much doubt was thrown by men of science as to how far the standard could be accurately restored by the above method, that the present standard was con- structed from a comparison of copies that had been carefully made of the old standard (18 and 19 Vict. c. 72). 202. The multiples and submultiples of the yard, and their relation to each other, are set forth in the following table : — xi inches (yci.) . . . ax& i foot (it.), ifiet . ... I YARD (yd.). Siyanis 1 rod ox pole. 4 poles or 22 yards . . i chain. \a chains \ furlong. % furlongs or 1760 yards . i mile. The following special measures are also used : — 4 inches are I hand (used in measuring the height of horses) ; 6 fee/ axe \ fathom j and 100 links are i chain or 22 yards. Linen and woollen drapers divide the yard into quarters, and each quarter again into quarters called sixteenths or nails ; thus 2J inches are i sixteenth or nail, and 4 nails or 9 inches are i quarter of a yard. Sometimes 5 quarters of a yard is called an ell. SURFACE. 203. The unit of surface is a square yard; that is a square each of whose sides is a yard in length. Take such a square, and divide each of two adjacent sides into 3 equal parts, and through the points of division draw straight 144 SURFACE. § 203. lines parallel to the sides; we have then 3x3 or 9 squares, each of whose sides is a foot in length ; that is I square yard is equal to 9 square feet. In like manner i square foot is equal to 12 x 12 or 144 square inches, &c. The following is the Table of Surface : 144 square inches . . are I sq.foot. 9 sq.feet .... 3o|- sq. yards or 272J sq.feet I sq.yard. I sq. rod or sq. pole, or I 'berch.'\ , ■ ^ I used in mea- / I rood. J- • 1 J I surmg land. 40 perches .... 4 roods .... I acre. 640 acres or 1760^ sq. yards i jjc. »zz7i?. 204. Since 22 yards are i chain, 22 x 22 or 484 sq. yards are I sq. chain, and 10 sq. chains or 4840 sq. yards are i acre. Also since 100 links are i chain — 100^ or 10,000 sq. links . . are i sq. chain. and 10 sq. chains or 100,000 sq. links i acre. VOLUME OF SOLIDITY. 205. The unit of volume is a cubic yard ; that is a cube, each of whose sides is a yard in length. Take such a cube, and divide each of three adjacent sides into 3 equal parts, and through each of the points of division draw planes parallel to the sides, then we have 3x3x3 or 27 cubes, each of whose sides is 1 foot in length : that is i cubic yard is equal to 27 cubic feet. In like manner i cubic foot is equal to 12^ or 1728 cubic inches, &c. Hence we have the following Table of Volume : — § 205. WEIGHT. 145 1 728 cubic inches . are i cu. foot. 27 cu.feet . . . . 1 cu. yard. I'jix? cu. yards ... i cu. mile. WEIGHT. 206. The unit of weight is the Pound; formerly called the Pound Avoirdupois to distinguish it from the Pound Troy; but as the Pound Troy has now no legal existence, the epithet Avoirdupois is unnecessary. In the ofSce of the Warden of the Standards at Westminster there is deposited a cylinder of platinum marked P. S. 1844, i lb., and this weight is the imperial standard pound, and is the only unit or standard measure of weight from which all other weights, and all measures having reference to weight, are derived (41 & 42 Vict, c.49, s. 13). One sixteenth part of the Pound is an ounce, and one seven- thousandth part of the Pound is a grain j therefore 7000 grains is 16 ounces and 437 J grains is an ounce. Also 480 grains is an ounce troy (41 & 42 Vict. c. 49, s. 14). 207. The multiples and submultiples of the Pound, and their relation to each other, are set forth in the following Table : — 16 drams or ^'i']\grs. are i ounce (oz.). i^o grains . 1 ouiice troy {oz.tr.). 16 ounces or yooagrs. i pound (lb. ; libra). I /^pounds . . I stone. Impounds . . . I j'Kflr/^r (of a; hundred- weight). 4 quarters or 1 12 lbs. i hundred-iueigkt (cwt. = C, wt.). 20 hundred-weight . i ton. 208. All articles sold by weight are sold by imperial standard weight ; except that (1) Gold and silver, and articles made thereof, also platinum, diamonds, and other precious metals or stones, may be sold by the ounce troy or by any decimal parts of such ounce ; and (2) Drugs, when sold by retail, may be sold by a peculiar subdivision of the troy ounce, called Apothecaries' Weight (41 & 42 Vict. c. 49, s. 20) . B.-S. A. 10 146 CAPACITY. § •209. 209. As many questions in Arithmetical papers have reference to Troy weight, although with the exception of the Troy ounce it has ceased to exist, it is necessary for us to give the Table : — 24 grains . . . are i penny-weight (dwt. = d, wt.). 20 dwts. or 480 grains . i ounce troy (oz. tr.). 12 ounces tr. or ^"jto grs. . i pound troy (Ih. tr.). APOTHECARIES' WEIGHT. 20 grains . . . . are i scruple (3). :i scruples or 60 grs. . . \ drachm {1). 8 drachms or 480 gj-s. . 1 ounce troy C^). CAPACITY. 210. The unit of capacity is the GALLON : and is the only stand- ard measure of capacity from which all other measures of capacity, as well for liquids as for dry goods, are derived. The Gallon is defined to contain 10 lbs. weight of distilled water (41 & 42 Vict, c. 49, s. 15): — and as I cubic inch of distilled water weighs 252'458 grains, it follows that 10 lbs. or 70,000 grains of water, will fill 7o,ooo■^252■458 or 277-274 cubic inches. The GALLON its multiples and submultiples are used to measure liquids, and some dry goods, as grain, fruit, &c., according to the following Table : — 4 gills . . . are i pint. 1 pints . . I quart {or qnaxter 0/ a gallon). 4 quarts . . i GALLON. 2 gallons . . . I peck. | ^^^ A pecks or S gallons i bushel. > 0,7, r goods only. 8 bushels . . 1 quarter {of a ton). I 211. Apothecaries'' Fluid Measure is founded on the fact that A pint of pure water, Weighs a pound and a quarter^ or 20 ounces. The following is the Table : — 60 minims (minima) . are \ fluid drachm (fl. dr.). i fluid drachms . i fluid ounce i^. oz.). ■20 fluid ounces . i pint (O ; Octarius). % pints .... \ QhlASXi {Q; Congius). § 212. MONEY. 147 MONEY. 212. The unit of money is THE POUND or sovereign. The sovereign is made of standard gold, which is composed of 1 1 parts pure gold and i part alloy : 480 oz. troy of this metal are coined into 1869 sovereigns, so that a sovereign contains I23"274... grains of standard gold. The other gold coin is the half-sovereign. The SHILLING is made of standard silver, which is composed of 37 parts pure silver and 3 parts alloy ; 12 oz. troy of this metal is coined into 66 shillings, so that a shilling contains 87,^ grains of standard silver. The other silver coins are — the crown or five- shilling piece, the half-crown, the florin or two-shilling piece, the sixpence, the four-pence and the three-pence, which are respec- tively the half, the third and the quarter of a shilling. Though the four-pence is still in circulation, it has ceased to be coined for some time. The PENNY is a bronze or copper coin made of metal which is composed of 95 parts copper, 4 tin, and i zinc ; I lb. of this metal is coined into 48 pence, so that a penny weighs \ oz. The other copper coins are the half-penny, and the farthing or quarter- penny. The following is the Table of Money :— ■2 farthings . . are I half-penny, ^farthings or 2 half-pence i penny. \i pence .... l shilling. 20 shillings . . . . i POUND or sovereign. 213. Pounds, shillings and pence are represented by £,, s, and d; the initials of the Latin words Libra, solidus, and denarius. Far- things are written as fractions of a penny, and any lower sum is expressed as a fraction of a farthing, just as if the farthing were written as an integer ; thus 2 shiUings and 7 pence 3 farthings is written 2s. y^d. ; and 2 shillings and 7 pence 3 farthings and ^ of a farthing is written 2s. 7f ^^. The guinea is no longer in circulation ; but we sometimes give the name to 21 shillings or^i. is. 214. A sovereign is of equal value with the metal composing it ; for a person may take gold to the Mint, and on having it reduced 10 — 2 148 PROPOSED DECIMAL COINAGE. § 714. to standard gold (209) may have delivered to him an equal weight of gold coin, free of charge ; whereas a shilling passes for more than its intrinsic value, and a penny for many times its intrinsic value. For this reason gold coin is a legal tender to any amount ; but silver coin only up to 40 shillings, and copper coin only up to 12 pence. PROPOSED DECIMAL COINAGE. 215. The following decimal system of coinage has been recom- mended by a Committee of the House of Commons, and has re- ceived the general assent of those arithmeticians and men of science who wish to see the introduction of a decimal system into England ; there is little doubt, therefore, that should a change take place, this system will be adopted and sanctioned by Parliament. In it, the pound or sovereign as at present defined will still be our unit of money : its submultiples will follow the decimal system and take the names oi florin, cent and milj so that 10 mils {m.)=i cent (c). 10 cents — 1 florin {/.). 10 florins = I POUND or sovereign {£). Hence £2$. 7/ 8^. <)m. may be written ;^25789 (140), and maybe operated on as easily as the decimal 25789. The gold coins will remain as at present, the sovereign and the half-sovereign. The silver coins will be the double florin, the florin, half-florin, 2 cents iA^d.) and i cent (2|^.). The copper coins will be the 5 mils {l\d.), 2 mils ff oi\d.) and I mil (1^ of \d.) : so that the present half-penny and farthing may remain in circulation with their values depreciated 4 per cent. 216. The great drawback to the proposed system is that no sum of money of the present coinage, except a multiple of 6d., can be paid exactly in the new coinage, for \d. is if\m. ; hence a fare, a toll, or a stamp of id., must be paid for either by i^n. or t,m., that is either by \m. too little or by %m. too much; or id. either by %m. or 9»2., that is either by \m. too little or \m. too much, &c.; and therefore in the transition from the present to the proposed system there must be considerable inconvenience. § 217- THE METRIC SYSTEM. I49 UNITED STATES. 217. The weights and measures of the United States differ from those of England in the following particulars only : (i) Their cwt is really 100 lbs., and their ton 2000 lbs. (2) Their tmit of liquid measure is our late wine gallon, and is about | of our standard gallon. (3) Their unit of dry measure is our late Winchester bushel, and is about f| of our standard bushel. (4) Their money is alto- gether different from ours. 218. Money, The unit of money is the DOLLAR ; and its sub- multiples follow the decimal system according to this Table: — 10 mills {m.) = I cent (ct.). 10 cents = I dime (d.). 10 dimes = i DOLLAR ($). 10 dollars = i eagle (E). Dimes are usually expressed as tens of cents, and mills are either neglected or expressed as fractions of a cent, so that accounts are kept in dollars and cents only: thus, 36-645 dollars is not read 3E. 61. ()d. 4cts. sm. but 36I. 64icts. The gold coins are composed of 9 parts pure gold and i part alloy; and are the eagle, weighing 258 grains, the half-eagle, &c. The silver coins are composed of 9 parts pure silver and i part copper ; and are the DOLLAR, weighing 4\i\ grains, the half- dollar, &c. The copper or bronze coins are the cent and the half-cent. The dollar is worth 4J. 2d. nearly, and therefore the cent \d. nearly. THE METRIC SYSTEM. 219. Formerly there existed in France the same want of uni- formity in forming the multiples and submultiples of the units of weights and measures as exists at the present time in England ; but soon after the Revolution of 1789 a Commission was nominated consisting of Borda, Condorcet, Monge, Lagrange and Laplace, for the puipose of preparing a new system of weights and measures. The system they recommended was established by the Legislature in 1801, and now prevails in almost its entirety, throughout the whole of France, under the name of the metric system. Since then ISO LENGTH, SURFACE, VOLUME. §219. it has been introduced into Belgium, Holland and Switzerland ; and to a greater or less extent into Italy, Germany, the United States and England. In the formation of the multiples and submultiples the decimal system is followed exclusively ; the Greek prefixes to any unit denoting multiples and the Latin prefixes denoting submultiples : thus, the metre being taken as the unit of length, its multiples are and its submultiples are deci-metre, denoting -^ metre centi-metre „ ^ „ milli-metre „ -^ „ deca-metre, denoting 10 metres hecto-metre „ 100 „ kilo-metre „ 1000 „ myria-metre „ 10,000 „ The multiples of the myriametre and the submultiples of the millimetre have received no special name. 220. Length. The unit of length is the metre : — it is also the fundamental unit, because from it every other unit of weight or measure is derived ; and hence the name metric system. A metre is defined to be the ten-millionth part of the distance from the pole to the equator measured along the surface of the ocean ; but is, in reality, the length of a rod of platinum, deposited in the Archives of the State, which is called the standard metre, and gives the legal length at the temperature of melting ice. 221. Surface. The unit of surface is the square metre. From Art. 195 we see that — \casq. m.iHimetres = \ sq. centimetre. 100 sq. centimetres = i sq. decimetre. 100 sq. decimetres — i sq. metre. 100 sq. metres = i sq. decametre, Sa'c; hence 735'SS°69 sq. metres is 7 sq. decametres, 35 sq. metres, 55 sq. decimetres, 6 sq. centimetres, and 90 sq. millimetres. In measuring land the square decametre is called the are, and the only multiple and submultiple are the hectare (or square hectometre) and centiare (or square metre) : thus an estate may contain 67834*09 ares, or 678 hectares, 34 ares and 9 centiares. The surface of a country is expressed in square kilometres. §222. LENGTH, SURFACE, VOLUME. IS I 232. Volume. The unit of volume is the cubic metre. From Art. 197 we see that 1000 cu. millimetres— \ cu. centimetre. 1000 cu. centimetres = i cu. decimetre. 1000 cu. decimetres = I cu. metre. 1000 cu. metres =1 cu. decametre, &-•£■.,■ but the multiples of the cubic metre are seldom used : hence 34S6'03425g63 cu. metres is 3456 cu. metres, 34 cu. decimetres, 259 cu. centimetres and 630 millimetres. In measuring wood the cu. metre takes the name of stire. 223. Capacity. The unit of capacity is the litre; it is the capacity of a cubic decimetre, and therefore the capacity of the kilolitre is that of a cubic m^tre. 224. Weight. The unit of weight is the gramme: it is the , weight in vacuo of a cubic centimetre of distilled water at 4" C. that is at its greatest density. Hence the kilogramme is the weight of a cubic decimetre of such water, that, is of a litre; and 1000 kilogrammes, called a millier or tonneau de mer, is the weight of a cubic metre of such water, that is of a kilolitre. But in reality a kilogramme is the weight of a cylinder of plati- num whose height is equal to its diameter, deposited in the Archives of the State, and called the legal standard. 225. Money. The unit of money is the franc: it is a coin weighing 5 grammes, and is composed of 9 parts pure silver and I part copper. The submultiples of the franc are the decime and the centime, which are respectively the tenth and the hundredth of the franc : so that 10 centimes = i decime. 10 decimes or 100 centimes =1 franc. Accounts however are kept in francs and centimes only, so that 76"85 francs is read 76 francs 85 centimes. The gold coins are composed of 9 parts pure gold and i part alloy, and are the 20-franc piece, the lo-franc piece and 5-franc piece. The legal value of gold is 15 J times that of silver, and as 152 MEASURES OF LENGTH. §225- the franc weighs 5 grammes, the weight of the 20-franc piece = 100 grammes-r-i5|^=:6'4Si6i grammes. The silver coins are composed of 9 parts pure silver and i part copper ; and are those of 5 francs, 2 francs, i franc, 50 centimes and 20 centimes. The copper or bronze coi;ns are composed of 95 parts copper, 4 tin and i zinc ; and are those of 10, 5, 2 and i centime. 226. We will now give the principal weights and measures of the metric system with their equivalents in the English system, and from them all the others may be easily derived. MEASURES OF LENGTH. Myriametre = 10,000 metres = 6'2i382 miles. Kilometre = 1,000... =io93'633o6 yards. Metre = i ... =i ''°^^^^ - 39'37°79 inches. Hence a kilometre is about f mile and a metre about ^ of a yard. LAND MEASURE. Hectare = i sq. hectometre = 10,000 sq. metres= 2'47ri4 acres. Are = I sq. decametre = 100 ... =ii9'6o332 sq. yards. Centiare - = i ... = ^'19603 Hence a hectare is about i\ acres and an are about 4 perches. MEASURES OF CAPACITY. Kilolitre =1000 litres =1 cu. metre =220 "09668 gallons. Litre = i ,, = i cu. decimetre = j , .'"' ' I 176077 pmts. Hence a litfe is about \ quart or if pints. - WEIGHTS. Millier or ) _ Brammes. ^ 1 1 cu. metre of) = 19-68412 cwts. Tonneau de mer) ' ' (distilled water) =2204'62i2 lbs. Av. Quintal = ioo,ooo=t:V ... =220_' 46212 Kilogramme = 1000=1 cu. decim. =! 'iS432"3487 grams, Gramme = 1 = 1 cu. centim. = I5'43235 ... Hence a tonneau de mer is a little less than a ton, and a kilogi-amme about 2^ lb. Av. CHAPTER X. REDUCTION AND THE COMPOUND RULES. — INTEGERS. REDUCTION. 227. The unit of any quantity, its multiples and submultiples, are called its denominations: thus, of length, the mile, furlong, yard, foot, &c. are its various denominations. The highest multiple and the lowest submultiple are called respectively the highest and lowest denomination of the quantity : thus, of length, the highest denomination is the mile, and the lowest the inch. When a quantity is expressed in one denomination only it is called a simple quantity — as 7 yards ; £,^; 36 gallons. When a quantity is expressed in several denominations it is called a compound Q^s.n'iity — as £t. 9J. 2%d. ; 1 5 days 1 7 h. 25 m. 37 s. 228. Reduction is the process by which we express (i) a simple or a compound quantity in terms of its lower denominations, or (2) a simple quantity in terms of its higher denominations. 229. I. To express a quantity in terms of its lower denomina- tions. Descending Reduction. For example, express £/\,']. 14. 6f in farthings. £. s. d. £. s. d. 47 . 14 . 6| 47^ 14 • 6i _^ 954 , X 12 954 11454, X 4 12 - - "454 4 45819 farthings. 45819 farthings. Since £1 is ^oj-., £4"] is 47 x 20j-. or 940J., and ;^47. 14^. is 940 + 14 or 954J.; that is, we multiply 47 by 20 and add 14. Again since is. is I'ld., 954J. is 954xi2(/. or 11448^^., and 954J. 6d. is 11448 + 6 or 11454;/.; that is, we multiply 954 by 12 and add 6. In like manner to reduce to farthings, we multiply 11454 by 4 and add 3, giving 45819/".; hence ;^47. 14. 61=45819/- 1 54 REDUCTION. § 329. And as a process similar to the above may be pursued in all cases of descending Reduction, we have the following Rule : Multiply the number in the highest denomination by the number which tells how many of the next lower denomination makes one of the highest, and add in to the product the given number of that lower denomination {if any), and continue the process till we come to the denomination required. 230. II. To express a simple quantity in terms of its higher denominations. Ascending Reduction. For example, reduce 45819 farthings to pounds. 4 ) 45819 12 ) ii454i 20 ) 95,4. 6i £A7- 14- 6i Since 4/. is id. there areas many pence in 45819/1 as 45819 contains 4; but 45819 contains 4, 1 1454 times and 3 over; therefore 45819/! is ii454|(/. Again, since i2ii. is is. there are as many shillings in ii454(/. as 11454 contains 12 : but 11454 contains 12, 954 times and 6 over; therefore 11454^. is 954J. 6d. and 45819/". is 954r. 6^d. Lastly, since los. is £1 there are as many pounds in 954J. as 954 con- tains 20 : but 954 contains 20, 47 times and 14 over (53) ; therefore 954J. is £^1- 14-f- and 458x9/. is £^']. i^s. 6^d. And as a process similar to the above may be pursued in all cases of ascending Reduction, we have the following Rule : Divide the number in the given denomination by the num.ber which tells how many of this denomination tnake- one of the next higher denomination, setting down the remainder as of the same denomination as its dividend: and continue this process till we come to the denomination required. 231. Proof. Reduction descending and ascending are inverse processes ; if therefore we perform one process on a given quantity, and on the result the other process, we ought to get the original quantity. Thus, if by the descending process we find that £i,T. 14. 6| is 45819/;, we ought by the ascending process to find that 45819/ is £i,']. 14. 6|. §231- REDUCTION. 155 Hence every example of Reduction with its Proof is an example of both processes, and gives at the same time a guarantee for the coiTectness of the work. Ex. I. Reduce 365 days 5 h. 48 m. 51 s. to seconds. da. h. m. s. !^. 365 S 48 SI P^oof- 6p ) 3155693 ,1 1095, X 8] 24 6p ) 52594,8 .. .51s. 8765, X 60 (8 )_876s...48 m. 525948,' X 60 ^'^ \ 3 ") I09 S--5 h. 315 56931 sec. 365 d. 5 h. 48 m. 51 s. Here we first multiply the number of days by 24, or by 3 and by 8 in suc- cession, adding in 5 to the second product; thus giving the number of hours, Ex. 2. Reduce 993629 ounces to tons, &c. oz. tons. cwts. qr. lbs. oz. 4)993629 Proof. ^7 14 I 25 13 16 4 ) 248407 ... I 954 > x4 / 4) 62101. .. 13 oz. 2217, x7 I 7) 15525...1 15519, x4 4 ) 2217 .. .25 lbs. 62101, x4) ^^ 2p ) 55,4 ... I qr. 248404, x4i 27 tons 14 cwt. I qr. 25 lbs. 1 3 oz. 993629 oz. 232. There is a difficulty attending the application of the above Rule in reducing yards to poles and square yards to square poles, which we shall now explain. To reduce yards to poles we have to divide by i\ ; but since 5^ yds. is 1 1 half-yds. we multiply the yards by 2, and divide by 1 1, the remainder being half-yds.; and note that i half-yd. is ijft. or I ft. 6 in. To reduce square yards to square poles we have to divide by 30J ; but since 30^ sq. yds. is 121 qr.-sq. yds, we multiply the sq. yds. by 4 and divide by 121, the remainder being qr.-sq. yds.; and note that i qr.-sq. yd. is 2J sq. ft. or 2 sq. ft. 36 in. 156 REDUCTION. §2^2. Ex. 3. Reduce 60665 1 inches to miles. 12 ) 60665 1 3 ) 5° 554 ■ • ■ 3 in. The remainder in dividing 16851...1 ft. by II isphalf-yds., or44yds. 2 or 4 yds. i ft. 6 in., and this II ) 3370 2 added to the previous re- 4P ) 3°6, 3--9 hf-yds. or 4 yds. i ft. 6 in. mainder, i ft. 3 in., gives a 8 ) 76...2 3P. I ft. 3 in. finalremainder4yds.2ft.9in. 9 m. 4 f. 23 p. 4 yds. 2 ft. 9 in. Ex. 4. Reduce 37S43473 sq. inches to acres. sq. in. j 12 ) 37543473 I 12 ) 3128622 . ..9 9 ) 260718 . ..81 in. 28968. ..6 ft. 4 ( II ) 11S872 121) -^ ' " ) r°S33--9 4P ) 95,7 ---75 qr- yds. or 18 sq. yds. 6 ft. 108 in. , 4 ) 23 ...37 p. 6 ft. 81 in. " ' 5 A. 3 r- 37 P- 19 yards 4 ft. 45 in. The remainder in dividing by 121 is 75 qr.-yds., or i8| sq. yds. or 18 sq. yds. 6 ft. 108 in.; and to this we add the previous remainder, 6 ft. 81 in., giving the final remainder 19 sq. yds. 4 ft. 45 in. 233. There are some cases in Reduction where we cannot pass directly step by step from the given denominations to the one proposed. We must in such cases pass through an intermediate denomination common to both, and it will be advisable to keep such common denomination as high as possible. Ex. 5. Reduce ;^25. i6j. 6\d. to francs of 9jrf. each. ^. J.. (/_ Here the common denomination is 25 . 16 . 64 half-pence : but ;^25. 16. b\ is 12397 5i6j. half-pence, and i franc is 19 half-pence; JO ) i2397t 652 fr. ^^ therefore reduce 12397 half-pence to pg francs by dividing by 19, and the result 47 is 652 francs with 9 half-pence or /[\d. 9 = 42"- over. §233- REDUCTION. 157 Ex. 6. Express i cwt. i qr. 25 lbs. in Troy weight. cwts, qrs. lbs. J_ I 25 _5, X 28 165, X 7000 (8 ) 1 15500 grs. 48o< 3 ) 144375 \ip ) 4812^5 2406 oz. tr. 120 grs. As Troy weight now consists simply of ounces troy of 480 grains, the common denomination is grains; we therefore reduce i cwt. i qr. 25 lbs. to grains, and then express the re- sult in Troy ounces by Reduction ascending. Exercise 30. Reduce each of the following compound quantities to its lowest expressed denomination, giving the proof in each case : — I- ;^343- 13-5; ;^"7- i6- Si: :^47- 19- "S; £.^19- 18. o\. •z. 76da. iph. 43min. 57 s. ; 43 weeks 5 h. 49 m. 57 s. 3. 7 miles 6 f. 32 p. 4 yds. ; 25 fur. 39 p. 3 yds. 2 ft. 8 in. 4. 25miles6fur. i7p. 4yds. 3in.; 25 miles 459yds. 31 in. 5. 1 7 acres 25 p. 25 yds. ; 3 roods 17 p. 2 r yds. 8 ft. 6. 8 acres 2 r. 34 p. 3 ft. 87 in. ; 53 acres 21 p. 8 ft. 125 in. 625 cu. yds. 19 ft. 1609 in.; 89 tons 17 cwt. 27 lbs. 15 oz. 165 oz. tr. 280 grs. 2767 cu. yds. 24 ft. ^^<) in. 18 cwt. 73 lbs. 9 dr. 19 lbs. 80Z. I4dwts. 17 grs. 7- 8. 9- 10. 8 tons 8 cwts. 98 lbs. 3045 grs.; S C. 7 O. 1 7 fl. oz. 5 fl. dr. 45 m. 11. 95 gall. I pt. 2 gills ; 54 qrs. 7 bsh. 6 gall. 12. 75 yds. 3 qrs. 3nl. 2 in.; 425tons igcwt. loolbs. 150Z. 200grs. Reduce each of the following simple quantities to its highest denomina- tion, giving the proof in each case : — 13- 3456791/.; 987653 half-pence; 1 300 1 3 farthings 14. 1,000,019 farthings; 1,000,000 min. ; 4568657 sec. 15- 3055709 sec. ; 25432245 sec.; 268543 in. 16. 1847638 ft.; 57383 yds. ; 3136749 in. '7- 7865432 sq. in.; 12565257 sq- in-; 657345 sq- ft- 18. 895487 sq. yds.; 10,000,000 sq. in. ; 25607809 sq. in. 19. 986877 cu. in. ; 2099520 cu. in.; 87654 lbs.; 20. 378539 o'^-; 1693539 drams ; 20,000,000 grs. 1S8 COMPOUND ADDITION. Ex.30. 21. 425095 grs. of gold ; 882743 minims; 1879 pints. 22. 24357 gil's; 4357 gallons of wheat ; 97324 pints (dry). 23. Find the number of ounces troy in Three hundred millions three thousand eight hundred and forty grains of gold. 24. Find the number of miles, furlongs, &c. in Fifty millions five thou- sand six hundred and ninety inches. 25. Reduce 247 guineas 13J. 9>\d. to half-pence; 54J guineas to six- pences. 26. Reduce 534S crowns to threepences; 23567 florins to half-crowns. 27. Reduce ;^25. 18. 9 to thalers of y. each; ;^443. i6. 8 to dollars of 4^-. ^d. each; £i},'j. 13. 6^ to francs oi^\d. each. 28. Find the number of square yards in a square mile; and then find how many perches there are in a square mile. 29. Reduce 2897 inches oi cloth to yards, &c. Express 988 feet in cloth measure; and 34 ells 4 qrs. 3 nls. in yards, &c. 30. Express 2 cwts. 3 qrs. 17 lbs., S cwts. 18 lbs. 14 oz. in Tr. ounces. 31. A cash-box contains 89 sovereigns, 35 half-sovereigns, 19 half- crowns, 25 florins, 31 shillings and 15 sixpences: find the sum of all these coins in pence. 32. What will a penny a minute amount to in a year of 365 days 6hrs.? 33. A child can paper i pin in i second ; how many pins can it paper in a working week of 52 J hours? 34. How many seconds are there from 6 A.M. Jan. r to 6 A.M. Feb. i ; from S A.M. 3 Mar. to 9 p.m. 22 May; from 10 A.M. 15 June to 7 A.M. 8 Dec? COMPOUND ADDITION. 234. Compound Addition is the operation by which we find a single quantity which is equal to two or more qtiantities of the same kind put together. This single quantity is called the sum of the given quantities. 235. We proceed upon precisely the same principles in the Compound that we do in the Simple Rules, and the only difiference in the corresponding operations is this — that in the Simple Rules 10 units of one order always make i unit of the next higher order, whereas in the Compound Rules the Relation between two sue- §^3S- COMPOUND ADDITION. 1 59 cessive orders is never uniform, but has to be sought from the ' Table of the quantity under consideration. Making then the necessary alteration for this difference, the following is the Rule for Compound Addition : — (i) Write down quantities under one another, so that units of the same denomination may be in the same vertical column, and draw a line underneath. (2) Begin at the first column on the right, and find the sum. of the numbers in that column; reduce the sum to the next higher denomination: set down the remainder under the column and carry the quotient to the first figure of the next column. (3) Having carried thus, find the sum of the second column : reduce, set dowti and carry as in the first column, and proceed in this way through all the cobimns: — Proof. The proof given in Simple Addition (25) is equally applicable in Compound Addition. Ex. I. Find the sum of ;^I2S. 14. gf ; £^6. 9. 3J ; ^^267. 18. 7^; £A9- 13- 6; £iS7- I9- "f and £^7. 12. 5^. £■ s. d. 12? . ■ 14 . 9. 36. , 9 . 3| 7? 267 . 18 . 49 13 • 6 157 • 19 • lif 87 . 12 . Si 72s . 8 . 71 Begin at the first or farthings column, and say 5, 6, 8, 11; now iif. is 2j(/.; set down ^d. and carry 2 to the first figure of the second column. For the second column say — 7, 18, 24, 31, 34, 43; now 43 and 47^. is zs. lid. ; set down n and carry 3. 7 times 13 is 91 and 3 is 94, and g^s. is £4. 14J. : set down 14, and carry 4. And now proceed as in Simple Multiplication. 243. Case II. When the Multiplier is the product of two or more numbers each not greater than 12, multiply by each of those numbers in succession, and the last result will be the Product required (36). Ex. 2. Multiply 2 tons 8 cwts. 3 qrs. 13 lbs. by 96. Since96 is 8.12 or 12. Swemay first multiply by 8 and the result by 1 2 ; or we may first multiply by 12 and the result by 8; thus I. J. d. 325 ■ 13 • 6J 1 §243- COMPOUND MULTIPLICATION. 1 6/ tons cwts. qrs. lbs. tons cwts. qrs. lbs. 2 8 3 13 8 2 8 3 13 12 19 10 3 20 12 29 6 I 16 8 234 n o 16 234 u o 16 244. Case III. When the Multiplier exceeds or falls short of a product within the table by a number not greater than 12, multiply by such product (Case II.) and then by this tmmber and add or subtract for the required Product (62, 63}. Ex. 3. Multiply ;^ 1 7. 8. 5 J by 139. Since 139 is 132+7, or 144-5, we ™^y multiply by 132 and then by 7, and add, or we may multiply by 144 and then by 5, and subtract. I. S. d. I- s. d. 17 • 8 . 5i 17 . , 8 . Si 12 209 , . I , 12 ■ 3 209 . I . 3 II 132 2508 . IS ■ 12 product by 2299 . 13 • 9 product by 144 121 . 19 . of 7 87 . 2 . 2i S 2421 . 12 . 9f 139 2421 . 12'. 9l 139 245. Case IV. When the Multiplier is a number beyond the range of the table. For example suppose the Multiplier to be 2479. Let us multiply the given quantity by 10 3 times in succession ; the first result will give the product by 10, the second by 100, and the third by 1000; now let us multiply the given quantity by 9, the first result by 7, the second by 4, and the third by 2 ; in this way we have found the products of the given quantity by 9, by 70, by 400, and by 2000, and the sum of these partial products will give the Product by 2479 (41). We have then the following Rule : — Multiply by 10 as many times in succession as there are figures in the Multiplier less i ; then multiply the Multiplicand by the units^ figure of the Multiplier, the first product by the ten^ figure, the second product by the hundreds' figure. The sum of these partial products will give the Product required. 1 68 COMPOUND MULTIPLICATION. §245- Ex. 4. Multiply £1. 15. 7f by 2479. £. s. d. £. s. d. 3 . 15 . 7}x9= 34 . o . 9f product by 9 10 37 • 16 . Si><7= 264.15. 2^ 70 10 378 .4.7 X4= 1512 . 18 . 4 400 lo 3782 . 5 . 10 x2= 7564 . II ■ 8 2000 9376 . 6 . oj 2479 When the Multiplier is a large number, as in this Example, and we are told to proceed by Compound Multiplication, the following is the simplest method. £■ s. d. 4 )_7437 3 • 15 • 7f 1859 ■•• K 2479 17353 ■ 9376 . 6 . oj 12 ) 19212 1601 ... od. 2479_ 20 ) 38786 1939 ... f«. 7437 ;£'9376 We first multiply 3/. by 2479 giving 7437/ or 1859^. and \d., set down \d, and carry 1859^/. to ^'^ pence: then multiply 'jd, by 2479 ^"d add in 1859, giving 19212;/. or 1601^. and od., set down od. and carry 1601 to the shillings: then multiply 15J. by 2479 and add in 1601 giving 38786^., Or ;^r939. 6.!., set down 6j. and carry 1939 to the ;^'s : lastly multiply £■!, by 2479, an\d. over. 7p^ ( 1 Again, since 26/. divided by 678 is ^/ £356 or Hts/., complete Quotient =;^5. 8. 8^^'^. 26/.=6i^. 252. When the Divisor is the product of two or more factors, divide by each of them in succession (52), andfnd the remainder as in Simple Division. Ex. 2. Divide ^3458. I7- 95 by 72 :— (0 giving the Quotient and Remainder, and (2) the complete Quotient. £. s. d. The final remdr. is 4x9 + 2 or 38/. . . Quotient is ;^48.o.9J, and pji/.over. The second remdr. from the farthings is 4t or =/ ; and V-h8 is ^4 or if, 48 • o . 9m . . complete Quotient =;£'48. o. 9i-||. -, 9 ) 3458 . 17 ■ 9i '- 8)384. 6 , . 54-2/ 48. .. f9) 3458 . , s. 17 ■ . 9i-38/: d. ■ 9i 'I 8)384. 6 . ■ &U §■252. COMPOUND DIVISION. 173 Ex. 3. Divide 53 miles 3 fur. 23 po. 4 yds. 2 ft. by 35. miles fur. po. yds. ft. in. yds. ft. 7 ) 53 3 23 4 2 16 1 3 2 o 8... 4 in. 20 .32111. 5)7 14 832 2. ..32 in. 18 2 2 . In dividing by 5 the remainder from the poles is 3 poles, or i6i yards, or 16 yds. I ft. 6 in., which added to 2 yds. o ft. 8 in. in the Dividend gives 18 yds. 2 ft. 2 in., to be further divided by 5. The second partial remainder is 4 in., therefore the full remainder is 4 x 7 + 4, or 32 in. or 1 ft. 8 in. The complete Quotient = i mile 4 fur. 8 po. 3 yds. 2 ft. 2 f| in. 253. When the Divisor is 10, 100, 1000,... cut off \, 2, ■},... figures to the right in each succeeding Dividend; the figures to the left •will at each step give the Quotient and the figures to the right the remainder (53). Ex. 4. Divide ;£8S432. 19. iijby 1000. £■ s. d. £. s. d. 1000 ) 85,432 . IQ ■ ■ Hi 85.432 . 19 • iii 20 8,659 8,659 7'S'9 ^ „ 12 3,678 678 339 7.919 1000 500 4 i4(/. over, and the 3.678 The Quotient is ^^85. 8. 7f, and 678/. or 14J. complete Quotient is ^^85. 8. 7jTff4- Ex. 5. Divide ;^8S432. 19. 11^ by 900. 900 ;^94. 18. 6 and gs. ii^d. over; and the complete Quotient £gi. 18. 6f ff§. Ex. 6. Divide £227. 9. 4if by 42. Ex. 7. Divide 58 gallons 2 qts. if pts. by 5. £. s. d. £. s. d. 9 ) 85432 • 19 . 114 9 ) 85432 . 19 . iii 100 ) 94,92 . II . , 1J...1/ 100 ) 94,92 . II . li^ 18,51 18,51 6,13 6,13 ,53 ,53l 53i-_478_239 100 goo 450 final remdr. is 5; 1x9+1 or 478/ or gs. 1 1 J. .•. the Quotient is £. 6 ) 227 d. 4i^ gallons 5)58 pt. li 7 )37 • 18 • 2iH '-5.8. 3iA Itt 174 COMPOUND DIVISION. § 253- In Ex. 6 the first remainder from the farthings is sf or V, and V-^6 is W; the second remainder is i|-| or fl' ^n<5 tt"^? is ^.. In Ex. 7 the remainder from the pints is 4I or Vi ^.nd "-i-5 is ^. i^ei^ If the value, weight, length,... of any number of units be given, we can by this case of Compound Division find the value, weight, length,... of one unit of the same kind ; for example : — If the area of a field containing 32 equal allotments be ig acres 3 r. 20 p., the area of one allotment is found in dividing 19 acres 3r. 2op. by 32. Exercise 34. Divide 1. ;^286. 3. 2by;^i. II. ij; ;^727- i- 7i by ;^i5- 9- 4i- 2. ;^I44. 13. iii by gj. ii%d.\ £i3l6. 11. loj by loj. 2ld. 3. i'9961. 7. 6J by ;^3S. 16. 7i; ^^601803. 15. Sf by £16. 13. 44. 4. How many francs of gj * quired is — *' ;^i97. II. 6+2j(/. or;^i97. II. 8i. Ex. 2. If ;£i97. II. 6 be given for 23 pieces of cloth, find to the nearest penny the price given for each piece. From the last Ex. it appears that ^8. 11. 9 a piece would give it^d. too little, and £fi. 1 1 . 10 would give 8rf. too much : hence to the nearest penny the price would be;£^8. 11. 10. VI. Barter and Exchange. Ex. I. How many pounds of tea at 3J. 2j(/. a lb. must a grocer give in exchange for 35 yards of cloth at \2s. 5^. a yard? We must first find what 35 yards at lis. t^d. a yard amount to; and the number of times 3^. ^\d. is contained in this amount will give the number of pounds of tea. s. d. s. d. 3 • 24 12 . 5 12 12 38 149 2 2 77 298 35 1490 894 77- 7) 10430 II ) 1490 135 and 35 half-pence over. Therefore he will give 135 lbs. and is. f)\d. 12—2 1§0 SOM£. APPLICATIONS OP REDUCTION AND § ^Sg. Ex. 2. How many francs of ()\d. each will be given in exchange for 475 thalers at 2s. i \\d. each ? d. d. 9i 35i ■2 2 T9 71 The value of 475 thalers is 33725 half- 475 pence, and i franc is ighalf-pence; there- 475 fore there will be as many francs as 33725 ^^^' — - half-pence contains 1 9 half-pence : or the 19 ) 337^5 ( 1775 number of francs is 1775. 142 95 o VII. Mixtures — Alligation. Ex. I. A tea-merchant mixes 25 lbs. of tea at \s. ()d. a lb., 40 lbs. at 2s. c,d., and 27 lbs. at y. 2d. : at what rate per lb. must he sell the mixture? s. d. s. d. 25 lbs. at X . 9 = 43 . 9 40 lbs. at 2 . 5 = 96 . 8 , . , , , , jN i]35_ at 3 . 2 = 8? . 6 "■ ^PPfi^rs from the work that he 02 ) 221; . II ( 2 . si ™'^^5 92 l^s-j ^""i 'li^t i's cost is 184 225i. lid., and that the cost of i lb. is 41 2j.5j(/., with a remainder of 80/^ or 20^.; , _^ hence by selling the whole at 2S. 5ji/. 503 a lb. he would lose 2od., but if he sells so as only not to lose he must sell at 92 460 ^'^ 2J. ^\d. a lb., and then he would gain — 12/ or ^d. 92 80/ Ex. 2. At what rate per lb. must he sell the mixture, so as to gain £2. 6. 3 upon the transaction ? J. d. 225 . II 46 . 3 J. d. "^^^ '^^ '^o^t 1^™ 225J. Hi/, and his gain is 92 ) 272 . 2~( 2 . Ii4 to be 46J. 3(/., therefore he must sell it for the 88 sum of 225^. \\d. and 46J. 3^^., or for 272J. id. ; 1058 ( II and dividing this sum by 92 we get is. n\d., / 1 the selling price of i lb, o §255- THE COMPOUND RULES. INTEGERS. l8l Ex. 3. How many lbs. of tea-dust (which cost him nothing) must be put in the mixture, to enable him to sell the tea at ^s. ^d. per lb. and gain at the same time 8s. 6d. on the transaction ? s. d. 225 . II 8 . 6 Including his gain of 8j. 6d. he must receive for 234 . 5 the mixture 234J. ^d. ; and dividing this sum by 12 2J-. tjd., it appears that he must sell 97 lbs. ; that is, to 29 ) 2813 ( 97 the 92 lbs. of tea he must add 5 lbs. of dust ; that is No. of lbs. of tea-dust =97 -92 = 5. 203 VIII. Weekly and Daily Expenditure. A man has a yearly income of ;^486. 1 5 and he sets aside ;^63 for charity, insurance and other purposes ; what is the greatest sum he can spend per week, without getting into debt .■" The sum he may spend in one year or 52 weelcs is ;^423. 15; and dividing this sum by 52, we see that he may spend ^^8. 2. iif £■ s. d. 48b , . 15 . 6.S ■ . i 4 ) 4^3 • IS ■ ° every weelc, and have \d. over at the end [13 ) 105 . 18 . 9 of the year. If he spends ;^8. 3 per week, 8. 2 .iii...4/ he will nm into debt. IX. Division OF Money. Ex. Divide ^16. 5. 6 among A, B and C so that A may have £\. 2. 6 more than B, and B i6j. 9^. more than C. /■. S. d. £. s. d. . 16 . 9 to be given to B 16 . 5 • 6 . 16 J} • • • •^- 2 . rb . I . 2 3) 13 • 9 • 6 2 , , 16 . 4 ■ Q • 10 C's share. .-. 5 . 6 . 7 B's . . . and . q • I A's . . . Here B has 16s. gd. more than C, and A has £1. 2. 6+i6s..gd. more than C: if we talce away these sums, to be subsequently given to B and A respectively, their shares will be equal to that of C :— dividing then the remainder £1^. 9. 6 by 3 we get C's share. 1 82 SOME APPLICATIONS OF REDUCTION AND §255- X. Men, Women and Boys. Ex. Divide ^15. 12. 6 among 7 men, 9 women, and 11 boys, so that each man may receive three times as much as a boy, and each woman twice as much as a boy. The 7 men will receive as much as 7 x 3 or 2 1 boys, and the 9 women as much as 9 X 2 or 18 boys ; therefore 7 men 9 women and 1 1 boys will receive as much as 21 + 18+ 11 or 50 boys; but they receive ;£'i5. 12. 6; therefore i boy will receive ^£^15. 12. fi-i-jO or 6s. ^d.; and a woman will receive (>s. %d. x 2 or 1 2s. 6d. ; and a man 6^. ^d. x 3 or i &s. gd. 7 men =21 boys 9 women =18. 1 1 boys =11. . 50 £■ s. 12 2 d. . 6 . 6 6 12 . 3 boy's share. . 6 woman's . 18 . , 9 man's . . Exercise 35. 1. Find the average of the following scores at cricket :— 29, 67, 35, 18, o. 43. 19' 137. SI. 38. o and 91. 2. Reduce 5792685 inches to miles, &c., and prove the result. 3. If 9 yards of cloth cost £•;. o. of, what will 32 yards ccst? 4. Divide ;^i89. 5. 7i among 3 men, so that one of them may have 15 guineas more than either of the other two. 5. A dealer buys a chest of tea weighing 185 lbs. at the rate of 4 lbs. for p. 6d. ; — he finds 6 lbs. of it to be worthless, but sells the remainder at an advance of i^d. a lb. on the cost price: find his gain. 6. 1869 sovereigns weigh 480 oz. Troy; find the weight of 100 17 84 sovereigns in standard weight, the lowest denominations being lbs., oz. and grs. 7. Find the least sum of money that must be subtracted from £663. 14. 8 to make the remainder divisible by 37. 8. A person has an income of ;^67o. 13. 8, and for the first 7 months he spends on an average ;^s8. 16. 9I a month: how much must he spend during each of the remaining 6 months, so as not to run into debt? Ex. 35- THE COMPOUND RULES. INTEGERS. 1 83 9. How many lbs. of tea at 3J. ^\ll. a lb. must be given in exchange for 46 yards of silk at 8j. o\d. a yard? 10. A grocer mixes 40 lbs. of tea at is. i,\d. a lb., 48 lbs. at is. ?>\d, a lb., and 64 lbs. at jj. •2f ; what will he gain or lose by selling them at ;^i. 10. 6 a-piece? 17. A merchant bought thirty-five pieces of cloth measuring on an average 29 yards each at 3J. \o\d. a yard, and sold them at 5^. id. a yard : what profit did he make? 18. A purse and the money it contains are worth £,\. 18. 6, and the money is 10 times the value of the purse: how much does the purse contain? ig. A grocer mixes 19 lbs. of tea at is. io%d. a lb., 26 lbs. at is. i\d. a lb. and 27 lbs. at is. (>\d. a lb. ; at how much per lb. must he sell the mixture so as to gain ^^2. 3. 4 on his outlay? 20. Divide £,10. 1. 6 into two sums of money, one of which contains as many half-crowns as the other contains shillings. III. 21. The daily receipts of a grocer for the week are as follows: — Monday, £^. 15. 3J; Tuesday, £5. 13. oj; Wednesday, £■}. 16. loj; Thursday (being a hoUday), nothing; Friday, £3. 18. 11, and Saturday, £ii. 19- 7i; fi°fl. 18. 8 is paid for the fare of t7 first class passengers at i\d. a mile, of 26 second class at i%d. a mile, and of 40 third class at id. a mile. VII. 6r. If a person spend ;,f 135. 8. 8| in 23 weeks, how much will he spend in a year? 62. A has ;^ioo. II. 4i, and B has 64392 farthings: if A receives 34567 farthings, and B receives ;^58. 16. 7I, which vrill then have the most, and by how much ? 63. Reduce 294322493 sq. in. to acres, &c. and prove the result. 64. Divide ;^ii9. 16. 3 among 36 persons in such a way that 17 of them may each receive i %s. gd. more than each of the rest. 65. Reduce 35 tons 19 cwts. 99 lbs. 12 oz. 135 grs. to grains. First, reduce 12 oz. i$Sffrs. to grains. 66. The construction of a mile of a certain railway costs a million pounds : at this rate what is the cost of construction, computed to the nearest penny, of every inch of its length ? Ex.35- COMPOUND RULES. INTEGERS. 1 8/ 67. A gi-ocer buys 4 cwt. of sugar at 6d. per lb., and 8 cwt. at ^\d. per lb. He sells 6 cwt. at i\d. per lb., at what rate per lb. must he sell the remainder so as neither to gain nor lose ? 68. A merchant laid out £(i^. 6s. in spirits which he bought at 1 2 J. \od. a gallon; he retailed it at i6j. 6d. & gallon, making a profit of £,\\. \\s.\ how many gallons must he have lost by leakage? 6g. ;£'75 is paid in wages at the end of the week to a certain number of men, twice as many women, and three times as many children ; each man earns 4J. ^d. a day, each woman ^s. gd., and each child 2s. ^d.; how many children are there ? 70. There are five presses striking oiF at the same rate florins, shillings, sixpences, fourpences and threepences, and the value of the money coined in a day of 8 hours [is ;^267i. 2. 8 : how many coins does a press strike off in one hour? 71. If a person has a yearly income of £sS°> ^^^ ^^ spends at the rate of ;^78. 10. 8J in 85 days, how much will he be able to lay by at the end of the year? , 72. Find the circumference of the wheel of a locomotive which makes on an average 4 revolutions in a second, and which performs a journey of 76 miles in i hour 36 min. 73. Find the greatest weight that is contained exactly in 3 tons 5 lbs. and in 20 tons 3 cwts. 2 qrs. 74. If the value of the United States dollar be 4s. 2d., how many dollars must be given for ;^Soo? Find also the least number of pounds that contains an exact number of dollars. 75. Divide ;^I5. 6^. among 12 men, 17 women and 26 children in such n way that a man shall receive 3 times as much as a child, and a woman twice as much as a child : what does a woman receive? 76. A gives B 112 gallons of brandy at 32J. 6d. a gallon, and receives in return ^40. 12. 6 and 780 yards of cloth: what is the price of the cloth per yard? 77. What is the least sum of money that can be paid in francs of lod. each, in half-crowns, in thalers of 2s. iid. each, and in dollars of 4J. 2d. each? 78. Employ short division in dividing 195477 by 7920; write down the final remainder, and compare the process by which 195477 in. may be reduced to furlongs, yds., ft. and in. I°o REDUCTION AND THE Ex. 35. 79. Divide ;^ii5. 2. 6 among 20 women and 25 men, so that each woman may receive 15^. more than each man: how much will each woman receive? 80. How long would a column of men 2145 fe^' ii length take to march through a street 3 furlongs long at the rate of 75 paces a minute, each pace being 2 j feet ? 81. If 25 ounces of gold is worth fy'j. 6. loj, what will be the worth of 15 bars each weighing 5 lbs. 3 oz. ? 82. By the payment of 2S. id. in London a banker will give credit at Calcutta for i rupee : how many rupees may be received in Calcutta for the payment of ^^5025. 6. 8 in London? 83. In how many days of 8 hours each will a person be able to count 1,000,000 sovereigns at the rate of 80 per minute? How many will remain to be counted on the morning of the 26th day? 84. A merchant buys 84 gallons of whiskey at i6s. gd. a gallon, and sells it at i6s. dd. a gallon, making a profit of 10 guineas: how many gallons of water did he add to the whiskey? 85. On the reduction of the income-tax from 91/. in the pound to j^d., a person saves ;^29. 15. 10; find his gross income. 86. A man bought 150 apples at 2 a penny, and 150 more at 3 a penny, and mixed them and sold the whole at 5 for 2d. ; how much did he lose, and where did the loss occur? 87. The total stock of gold coin and bullion in the Bank of England on a certain day being of the value of ;!f 165481 26, and the weight of it 354160 lbs., determine the value of an ounce of gold. 88. The weekly wages at a mill amount to ;^i86. 4J. In the mill a certain number of women are employed at 2S, lod. a day, five times as many men at 5J. 6d. a day, and 6 times as many boys at 2S. ^d. n. day :. how many men are employed ? 89. I lb. Troy of standard silver is coined into 66 shillings : what would be the error in weight if 264 crowns were considered as each weighing i oz, Av.? what in value? 90. A body of military one furlong in length is about to pass through a defile 3 miles 44 yds. long at quick step, which is 108 steps of 2 ft. 8 in. each per minute ; what time must elapse before the last man clears the defile? ^256. COMPOUND RULES. FRACTIONS. I §9 REDUCTION AND THE COMPOUND RULES— FRACTIONS. REDUCTION. 256. While " times" denotes the multiplication of a quantity by an integer, " of" denotes its multiplication by a fraction, and either "times" or "of" its multiplication by a mixed number : — thus we say 3 times J gallons, ^ofj gallons, and either 4f times 7 gallons or 4f of 7 gallons, but each expression denotes the multiplication of 7 gallons by 3, by f , and by ^f respectively. 3 times 7 gallons is then 7 gallons x 3, and it is often written 3x7 gallons ; but since the Multiplier must be a number, we must either read the expression as 3 times 7 gallons, or we must suppose Multiplicand and Multiplier to be interchanged. Also 3x7 gallons is often used for (3 x 7) gallons, but unless we especially wish to shew that the operation has been performed although desirous of exhibiting the factors, no harm will arise from neglecting the bracket. 257. In Reduction we have to consider the two following cases : — (i) To reduce a fraction of one denomination to a lower de- nomination; and conversely, (ii) To reduce a quantity of one denomination to 2^ fraction of a higher denomination. I. II . 7 For example, reduce ;^ ^ to shillings, and -^ s. to pence. _ii II J. /ii \ 11x20 and -^.y. = -^-of I2^.= (ix 12") d. = '^~^d. 16 16 \ib J 16 hence to reduce a fraction of a pound to shillings we multiply the fraction by 20, and to reduce a fraction of a shilling to pence we multiply the frac- tion by 12. And as a like method is applicable to fractions of other denominations we have this Rule — Multiply the fraction of the given denomination by the number •which tells how many of the lower denomination make one of the given detiotnination. 190 REDUCTION AND TBE % 257- Ex. I. — of a day = ^ hours = — hrs. = 7 J^ hrs. 27 27 9 8x24x60 . 1280 . ,„ . and = mm. = mm. = 420 J mm. 27 3 8 X 24 X 60 X 60 and = sec. = 25000 sec. 27 II. 257*. Reduce 5jrf. to the fraction of a shilling, and 3x5 J. to the fraction of a pound. Now Si'^. = Siof id. = sloi ^s. = Uixl-\s. = ^^s. 12 \ 12/ 12 and 3iV- = 3A of is.^i^-, oi£.^= (3^ x ~);^. = $ £: 20 \ 20/ 20 hence to reduce any number of pence to the fraction of a shilling we divide the number by 12, and to reduce any number of shillings to the fraction ol a pound we divide by 20. And as a like method is applicable to other denominations we have this Rule — Divide the number of the given denomination by the number which tells how many of that denomination make one of the higher denomination. Ex. 2.- i8Jgrs.= '-^dwts.=-^dwts. = ^dwts.; 24 96 32 18? 75 c and = ^~ oz. = -^^-5 — oz. = -^-„ oz. : 24 X 20 ' 96 X 20 ' 128 24x20x12 96x20x12 1536 and = ^M lb. = __15 lb. = -5. lb. 258. Sometimes in reducing a fraction of one denomination to a fraction of another denomination we have to employ both the descending and the ascending process : for example, Reduce — of a guinea to the fraction of ;^i : here the denomination common to guineas and pounds is shillings : we §258- COMPOUND RULES. PRACTtONS. I9I therefore reduce the given quantity to shillings, and the result to the fraction of a pound : thus '-3 of a guinea = iii^' .. = 32 ,. =^ ^9_ ^^39 . 14 14 2 2 X 20 40 259. The preceding cases enable us (i) To reduce a fraction of one denomination to a compound quantity of lower denomination : and (ii) To reduce a compound quantity to a fraction of a higher denomination. Ex. I. Find the value of — of a lb. Troy. 45 37 lb. Troy = 3liiiloz. = i4? oz. =9lf oz. 45 46 15 ^ £3 oz. = iSiii? dwts. = ^ dwts. = 1 7^ dwts. 15 IS 3 I , , I X 24 - dwt. — grs. = 8 grs. .-.^^Ib. Troy = 9oz. 17 dwts. 8 e:rs. 45 260. If the denominator of the fraction and the several multipliers by which we pass from the given denomination to the lower denominations have no common factor, we shall find it easier to proceed by Compound Division than by Reduction : for example, Find the value of /- . 7 Since £- = — (103) we have simply to divide ;^5 by 7, thus 7 7 14 • 3if or, without any arrangement, we may write off at once ;^^ = i4-f- %l¥- Q In like manner, - cwt. — - 3 qrs. 15 lbs. 8 oz. 14I drs. 192 REDUCTION AND THE § 261- 261. And if they have a common factor, we shall often find it convenient to combine the processes of Reduction and Division: for example, Find the value of — of a pole. ^ pole = ^ X i- yds. = ?# yds. 44 44 2 ^ = 4 yds. I ft. lojin. where the result ^ yds. is got by Reduction, and the final result by o Division. II. 262. Ex. 2. Reduce 5 cwts. 3 qrs. 24 lbs. to the fraction of a ton. lu '24 6 24lbs.=-qr. = -qr. . . 3 qrs. 24 lbs. = 3y qrs. = — cwt. =-5 cwt., 4 20 and 5 cwts. 3 qrs. 24 lbs. = Sjf cwts. = "-^ ton= -p- ton. 20 5"0 Ex. 3. Express £yj. 16. 6f in pounds only. 12 48 16 and .-. ^i'sr- 16. 6i=;^37Tl- 263. Sometimes instead of proceeding step by step through all the intervening denominations from the lowest to the one required, as in the two preceding Examples, we reduce the given quantity to its lowest denomination and reach the result in a single step. Ex. 2*. Reduce 5 cwts. 3 qrs. 24 lbs. to the fraction of a ton. cwts. qrs. lbs. . ■. 5 cwts. 3 qrs. 24 lbs. =668 lbs. __5 • 3 • 24 ^ 668 ^^^ 23, X4I 28x4x20 92, X 7J 167 668" ="^0 Ex. 36. COMPOUND RULES. FRACTIONS. 193 Exercise 36. Reduce : . \a\d. to the fraction of a shilling ; i jf J. to the fraction of a pound. ^. f of a farthing to the fraction of is. ; \d. to the fraction of £,\. 3- ^SSt'^- to shillings, and to the fraction of a pound. 4. \% dwt. to the fraction of i lb. Troy ; f | min. to the fraction of a week. 5. lifl oz. to the fraction of a cwt. ; 1 8454^4 seconds to hours. 6. io67Tf7- cu. inches to the fraction of a cu. yard. 7. 4I feel to the fraction of a furlong; |f sq. yds. to that of an acre. 8. 49^4^ hours to the fraction of a year of 365I days. Reduce 9. IJj. to farthings ; ;^f^ to shillings ; ;^^| to pence. lo- £ttt!s to the fraction of a penny ; rl^J. to the fraction of a farthing ; ts'it of a guinea to the fraction of a farthing. 1 1 . -t-Jtf owt. to the fraction of i lb. ; ^ ton to qrs. ; i% oz. to grs. 12. 14 lb. Troy to dwts. ; ^W gallon to the fraction of a pint. 13. -nrV^ of a mile to poles; 3^^^ miles to yards. 14. f§^ of an acre to sq. yards; SyfJ acres to perches. Reduce 15. ^^\ lb. to the fraction of i lb. Troy; Jf nail to that of a foot. 16. tf of a yard to the fraction of an ell ; f § link to the fraction of a ft. 17. 124 oz. Av. to the fr. of i lb. Tr.; 35 lbs. Bf oz. Tr. to that of a cwt. Reduce 18. 16^. 9fi/., 6s. iii(/., I4J-. ^i^d. to the fractions ol £1. 19- ;£'4.9:. 2j, ;^3- 19- 8i and ;^5. 16. nil to pounds only. 20. 4 oz. 15 dwts. 15 grs. to the fraction of a pound Troy; and then to the fraction of a lb. 21. IS cwts. 3 qrs. 17 lbs. 8 oz. to cwts. and to the fraction of a ton. 22. s bushels 3 pks. i gall, to the fraction of a quarter. 13. 3 qrs. 27 lbs. 9 oz. 124 dr. to the fraction of a cwt. 24. 3 furlongs 29 po. 4 yds. i ft. 9 in. to the fraction of a mile. B.-S. A. 13 194 REDUCTION AND THE Ex. 36. Reduce 45. 8 perches 27 yds. 6t% ft. to perches, and to the fraction of an acre. 26. 72 days 6 hrs. 56 m. 15 s. to the fraction of a year of 365^ days. Find the value in lower integral denominations of ^7- \l^-, ^V^d., £.Vi, £^; n%s., £%\^, £^j, ;^5flM- 28. I lb. Troy; fib.; i|oz. Troy; 2^J lbs. Troy; H cwt. 29. ifll of a ton ; ^ of an acre ; jf I furlongs ; -^^ of a mile. 30. \i of a day ; tVs of an acre ; 2^^ of ^ of a cwt. ; 3-1*^ ells. o* 3 •* . j2 ^i. -?; of a year of 365 days; ^^ of a ton ; — — of a week. ^ 12^ •' ^ ^ ■' 20 "4 32- III .-- + - + - . of — "^-^ of a gallon ; of - of a perch. Q + 3 'i 34 44 4 Find the value of 34. |- cwt. +81 lbs.+ 3i^ oz. ; f of a ton + | of a cwt. +| of a lb. 35- SA miles- 7/ir fuf. + 35|| po.; ^V sq- mile + ^T acre+|- of a rood. 36. f of a week + f of a day + f of an hour + 1 of a minute. MULTIPLICATION, DIVISION, &C. 264. (i) To multiply a quantity by a fraction — multiply by the numerator and divide the product by the denominator {128). (li) To multiply a quantity by a mixed number — (i) multiply separately by the integer and by the fraction and add (131 Rk.) ; or (2) reduce the mixed number to a fraction and proceed by {\). (iii To divide a quantity by a fraction — multiply by the denomi- nator and divide the product by the tmmerator (134 Cor.). §264. COMPOUND RULES. FRACTIONS. I9S (iv) To divide a quantity by a mixed number — reduce the mixed number to a fraction, and proceed by (iii). (v) To take a fractional part of a quantity — multiply the quan- tity by the fraction (128) ; that is, proceed by (i) or (ii). Ex. I. Multiply ^25. 14. 9 J by ^. Ex. 2. Divide 15 tons 13 cwts. 2 qrs. 20 lbs. by f. tons. cwts. qrs. lbs. 15 . 13 . 2 . 20 7 £. 25- 14- ■ d. •9f 7 11 ) 180 . 3 ■ ■ 8i 16 . 7 ' ■7iA 3 ) 109 ■ IS . 3 ■ o 36 . II . 3 ■ i8f Ex. 3. Multiply ;£4S. 12. 6f by 12^. Ex. 4. Divide 345 lbs. 9 oz. i6dwts. 20grs. by I3f. £■ s. d. 96 45 . 12 • 6| I3f =y 9 lbs. Qz. dwts. grs. 5 )4io . 13 ■ of 345 . 9 • 16 . 20 5 )82. 2.7if ^ 7.^ 16 . 8 . 61^' product by ^ '2 ) 2420 . 8 . 17 . 20 547 . 10 . 9 12 8 ) 201 ■ 8 ■ 14 . igf 563 • 19 • 3ift 12^ 25 . 2 . II . 20^ Ex. 5. Find the difference between | of ^35. 14. 7I and 1% of -£5- 14. 9f- 35 • H • 8 ) 250 . 2 . d. 7i 7 4i £,. s. d. 5 • 14 • 9l 5 6 ) 28 . 14 . of 31 • 5 • 16. 5 • 3i ■ 4 . IS . 8^ J product by |. II . 9 • 7I 2 14 . 19 . iiff Difference reqd. 16 . 5 . 3ii (vi) When the quantity is simple or can be easily reduced to a simple quantity, the arrangement above is usually dispensed with ; we cancel, multiply, and proceed by reduction (259) or by compound division (260). 13—2 196 REDUCTION AND THE § 264. Ex. 6. ^ of 2 guineas=- of 42^.=^ — —s. — yis.=£i. \os. Ex. 7. 3J of half-a-crown=— x -s. =~s. = gs. 44<^. (260) Ex. 8. -- of £4. los. = — X goj-. = ^s. = 347^- =;^i. 14- Sif- Ex. 9. Find the value of | of a guinea +f of a crown 4 1 of 7s. 6^. — f of 2(^. - of a guinea=- of 21J. = — J. =0 .15. 9 4 4 4 = of a crown =1 of 5J. =-^s.= i . loj 2 of 7.r. 6d. =^ of i^j. = 2j. = 4.6 5 5 2 2 * 4 2 ■• 1.2.0 Value reqd. Ex. 10. %f^is + 3f of;^i + iof ^of ^of/i+-of3ofi.r. 7 3 7 5 3 7 =£l- 11 fs. =£7- 17- S^f (260) Ex. 1 1. From -^ of a mile take - of 3^ poles. ij y Qg fur. po. yds. ft. in. -%mile=:~x8fur.=— fur. = 3 .4.2.1.4. lo Its Q ^of34poles = |x^po.=^po.= 2 . 3 . 2. 9f 3 ■ I • 3t- I • 6f I • 6 (236) . ■. Difrerence=3 .1.4.0.0^ Ex. 37. COMPOUND RULES. FRACTIONS. I97 Exercise 37. Find the value of ' • I. ;^3-i6-8|xf; ;^6.i8. yffxA- ■L. £-n. 12. 54 X iii't; £■^^1- 19- 8|ix 30J. 3. 7 lbs. 9 oz. IS dwts. 'i.ixx gfs. X 4I ; 2 qrs. 3 lbs. i if oz. x n|^. 4. 19 hrs. 43 m. 561^ s. X 12^'^ ; 10 ac. 3 r. 37 p. 15I yds. x lof. 5. £^Z. 8. 4iH-|; ;^i8. 16. 7if-3l- 6. ;f34. 16. 94-91; ;f8. 18. sll-if. 7. ;^3976. 18. 7f-H23U; ;^879. 16. 7|iK8H. 8. 4lbs. 90Z. i4dwts. I5|^grs.-i-|; 8 days 15 h. 48m. 57^4 s.-^ 4^. 9. 13 cwts. 3qrs. iSlbs. 15I oz.-r-3f ; 6m. 7 fur. 15 p. oA yds. -J- 3^. 10. ^ of jfs. 19. 3J; H of ;^3- i4- lOj^ II. 3f of ;^8. 13. 8i|; 29J of ;£-i2S. 17. 6|H. 12. If of 789lbs. 120Z. I4||drs.; J|of657gal.2qts. ipt-s^tgiUs. 13. £\o. 12. 4|x iiV-;^S- 18. 61^10,^. 14- 7lT0f;£'7- iS-3i-iAof.!f9- 5- iof-3Aofj^3. 12. iij. 15. 5 yds. 2 ft. 5|m. X7f-9yds. 2 ft. 7lim. - 15 yds. i ft. 9|in.H-3|. 16. Take A of £.\- lo- 9 from H of ;^6. 6. 9. 1 7. Find the difference between f of I j'. 6d. and I of is. 6d. 18. By how much is if of ;£'2. 7. iif greater than 2| of ;f r. 3. 2^ ? 19. Find the difference between a thirty-fifth part of 95 guineas and a twenty-third part of ;^65. 18. 8. 20. Find the value of 500 times the difference between an eighty-fourth part of 2^ cwts. and a thirtieth part of i cwt. o qr. 3 lbs. Find the value of 21. I of ^ of i6j. 6 3 and I cwt ^3^7,6>^5 3 /"■^ 7 6 or I cwt a • (134) (3) If 6f cwts. of sugar cost £z7- 2. 6, orif^ £37.2.6, xv ^ £37- 2. 6 £y7. 2. 6 then I cwt ^^^ — or^^'^^Zq — • S "5 Hence whether the number of cwts. be integral or fractional or mixed, the value of i cwt. may always be got by dividing the given sum by this number. And generally If the value, weight, length oi any number of things be given, the value, weight, length of i of them maybe always found by division. Ex. If 6j lbs. of silver be worth £i!^ 9. 4|, what is the value of 3 lbs. 9 oz. 12 dwts. ? £ii,. 9. i,\=£'n. ci%s. = £'i\ -^ = ;^24i|. at 3 lbs. 9 oz. 1 2 dwts. = 3 lbs. 9f oz. = 3 — lbs. = 3I lbs. 204 REDUCTION AND THE § 266. Since 6f lbs. is worth ;^24^|, lib ^1^. (266,3). and 3f lbs gj ; orslbs-ooz. i2dwts ^^ X -S X A or ^511 or £11. 15. 6. II. Time and Work, by Reduction to the Unit. (i) If A can do i piece of work in 6 days, A can do J^ i day. (2) If A can do i piece of work in f day, A can do J \ day. and A can do f i day, or A can do ^ i day. '5 (3) If .<4 can do i piece of work in 6f days or in &A days, A can do 3^ i day "s" or A can do ?g i day. Hence the number of pieces of work may always be found by dividing i by the number of days, whether this number be in- tegral, or fractional, or mixed. Conversely, and in like manner, the number of days may always be found by dividing, i by the number of pieces of work, whether this number be integral, or fractional, or mixed. Ex. I. A can do a piece of work in lof days, B in 9I days, and C in 5]^ days ; in how many days can A, B and C working to- gether do the piece of work ? §266. COMPOUND RULES. FRACTIONS. 205 ^ in I day can do — j of the piece of work, B can do -r and C — r- ; I of 9t 5tt . •. A, B, C working together can in i day do — = + -r H ^ I of the piece of work, 32 48 64 or -^— 192 and .'. A, B and C working together can do the piece of work in -y^ days, or i-?^ days, or if ^ days. Ex. 2. A and B can do a piece of work in 8 days, A and C in lof days, and B and C in 9I days : in how many days can A alone do it ? Now A and 5 in i day can do - ai the piece of work, o and A and C — - . I of .". A and 5 in i day and A and C in i day can d.o\-=-\ ) ,of it, \8 io|y or ^ in 2 days and B and C in one day can do I 5 H 5 ) of it, \o lof/ but B and C in i day can do ^ of it ; 94 .•. A alone in 2 days can do ( - H ; j I of it, \8 lof 9f/ i + ^-4oriIofiU 8 32 48 . VI. A person has a number of oranges to dispose of: he sells half of what he has and one more to A, half of the remainder and one more to B, half the remainder and one more to C, and half the remainder and one more \.o D: by which time he has disposed of all he had. How many had he at first .' When he had given half\i\% oranges to D he had one left; therefore he had 2 x i or 2 before D came, and therefore he had 3 before he had given the i orange to C: but this is the number he had left when he had given half his oranges to C, therefore he had 2x3 or 6 before C came, and therefore he had 7 before he had given the r orange to B, and therefore (as before) he had 2 x 7 or 14 before B came; therefore he had 15 before he had given the i orange to A, and therefore he had 2 x 15 or 30 before A came : that is, he had thirty oranges at first. Exercise 39. 1. Reduce HM of £,^ t° '^^ fraction of a thaler, when 6| thalers are equal to los. 2. If I of an estate be worth ;^220, what is the value of y'y of the same? 3. A alone can do a piece of work in lo days, B alone can do it in 14 days: how many days would the two together be in doing it? 4. At what time between 2 and 3 o'clock are the hands of a clock exactly together? 5. If ii of a number exceeds \ of half the number by 40 Jj what must the number be? 6. Find the least sum of money that contains an exact number of thalers of 2J. iii 4°6S/. to the dec. of ;iiri. 6. 47"733 lbs. to the dec. of a ton; 10 drs. to the dec. of i lb. 7" S27'3994 yds., to the dec. of a mile; 37*9872 sec. to the dec. of adigr. 8. 420"8i38 sq. yds. to the dec. of an acre; i oz. to lie dec. of i cwt. 9. "625 of ;^i to the dec. of a guinea, and oflialf-a-guinea. JO. 3-589 po. to the dec. of a chain; "4326 yd. to the dec. of an ell. 11. I oz. Av. to the dec. of i oz. Troy; "54375 lb. Troy to ounces Av. 12. "67375 sq- chain to sq. yards; i perch to the dec. of i sq. furlong. 13. Find the decimal of a leap year which differs from a week by less than the ten-thousandth of a leap year. 14. Reduce sid., Id., 8i. iiy.,£i. 14. 10^ to the decimal of u. 15. Express loj. ijS4i6 where we divide 65 by 12 ; dd. = £-oil; \o\d. =.£ ■043^75 where we divide 90 by 12 ; and 1 11-^. = .£'0489583 IIS by 12. §272. DECIMAL MONEY. 22/ (3) If now the number of shillings be even, we write down their decimal of ;^i, and then write down the decimal of the pence, &c. ; thus, 4J. 3j^.=;£-2i354i6, i8j. io^^.=;^-94375. But if the number of shillings be odd, the 5 in the second place of their decimal must be added to the first figure of the decimal of the pence, &c. ; thus, 7J. 3K = ;iC'363S4i6, 19^- loj^- = ;i^'99375- In the same way we can write at sight 3J. lofrf. =^-1947916, ^5. 18. I if =;£s '9489583; iSJ. o|rf.=^7S3i2S, £%. 17. ioi=;£8-89375 ; and so for every other sum of money. 273. Conversely, — To find the value of any decimal of £\. For the shillings double the figure in the first place of decimals, and if the figure in the second place be 5 or greater add i ; thus, ^■6345 = I2J. ... ^78456= ISJ. ... Remove these figures and in the second example we have ;^ -03456; but /•■n.icfi /•34'S^ /• 34-S6xM _ /•3 4-56x(i-jg) A"J4io i- yooo -^loooxfl"* 960 = 34-56 X (I -gij)/; or =34'S6x(i-T^iy)/; if therefore from 34-56 we subtract ^-^^y of 34-56 we shall have the result in farthings ; that is, we must multiply 34-56 by 4, setting down 2 places to the right, and subtract ; thus 34-56 1-3824 33-1776 therefore ;£ -78456 =15^. 8ii776. In like manner we find the values of ^3-948125 and of .£-65916 ; thus^' 15—2 228 DECIMAL MONEY. % lU- 48-125 9"i6 . I '92500 "3666 46'2 8'8 .-. ;£3-948i25=;^3- 18. iiJ-2. .-. ^-65916= 13J. 2£-8; and similarly for every other decimal oi £\. 274. To reduce any sum of money to the decimal of £1, not ^oing beyond 3 places. Now 3i^. = ;£'oi3^ (272, 2) where the numerator of the fraction is the number of farthings in the given sum ; hence up to 3(/. this fraction is less than \, above 2>d- and up to ^d. it is greater than \ and less than ij, above <)d. and up to \^d. it is greater than \\ and less than 2 ; therefore, that the third place of decimals may be as nearly correct as possible, we must up to 3(/.. neglect the fraction altogether, and the decimal will be given by the number of far- things ; above 2>d. and up to f)d. we must add i ; above 9^. and up to \id. we must add 2 ; thus, 2j^.=;^'oo9 and 6j. ■2,\d.=£-y)<), 7id.= £032 ... 1 5 J. 7i^.=^782, iiid. = £-o48 ... 17s. iiid.=£-8<)S. 275. Conversely, — To find the value of any decimal of £1 of three places, to the nearest farthing. Remove the figures giving the shillings and suppose the remain- ing decimal to be ;^'034. Now ^-034 = 34 X (I -^)/ = 34/-M/, (273) hence if the number formed by the figures in the second and third places of the remaining decimal be not greater than 12 the fraction to be subtracted is less than \, and therefore this number will give the number of farthings ; but from 13 to 37 the fraction is greater than \ and less than i\, therefore we must subtract i ; and from 38 upwards we must subtract 2. Thus, £-%o7 = i6j. \ld., £-i,y2. = 8j. 7i^., Is -399 = ;^5 • 7- 1 1 f . ^7790 = ^7- I S- 9i §276- DECIMAL MONEY. 229 276. In the proposed decimal coinage, a sovereign (£1) is taken as the unit (212), and its submultiples the florin, cent, and mil are respectively the tenth, hundredth, and thousandth part of the sovereign ; so that in a decimal of ;^l the figure in the first place of decimals represents florins, in the second cents and in the third mils. Hence to convert ordinary money into decimal money is simply to express ordinary money as a decimal of £\ : and to con- vert decimal money into ordinary money is simply to express a decimal of £1 in ordinary money : and these processes both accurate and approximate have been pointed out in the preceding Articles (272 — 275). Ex. I. Reduce ;^i5. Tc. 8m. to cents; £g. 8/. 6c. to mils; 25684 mils to pounds, &c. ; and 5 f. 6 m. to the decimal of a florin. To reduce pounds to cents we multiply by 100 (229); to reduce pounds to mils we multiply by 1000 ; to reduce mils to pounds we divide by 1000 (230) ; and to reduce cents to florins or the decimal of a florin we divide by 10 (267, ii) : hence (i) £i5-7':-Sm. = £ii-o78 = i5or8c. (144) = 1507c, 8m. (2) £g. 8/. 6c. =£9-86 =g86om. (3) 25684OT.=;^25-684 =;^25. 6/ 8c. 4m. (144) (4) sc 6m. = s-6c. =-56/ Ex.2. Find the sum of ;£i2. 3/ 5 »«., £23. 6m., , ;£i8. 9/, £2$. 46 c. and £37. 7/ 8 c. 9 '«• £■ £i'i.i/.im. =12-305 ;^23. 6m. =23-006 ;^i8. 9/. =i8-9 ;^25. 46c. =25-46 £3'!.'!f.8c.gm. =iri^9 ii7-46o=;^ii7. 4/ 6c. 230 DECIMAL MONEY. §276- Ex. 3. Subtract ^123. 7/ Zc. from £<)9,T. 6c. sm.; and divide the difference by S"643. £. ^987. 6c. 5»«. = 987-065 jfi23. 7/. 8ir. =123-78 6"643)863-28s ( 152-9833 29898 16835 5549° 4703 189 20 3 . . Quotient required=;$'i52-9833.. = ;^iS2. 9/ 8c. 3-3.. /». Ex. 4. Express £2. 15- 7^ and ^4. 18. lof accurately in the proposed decimal coinage ; that is in £./. c. m. At sight £i. 15. 74=;if3'78i25 (272) =£z- 7/ 8<:. i-25»«. (276) At sight ^4. 18. ioi=;^4-94479id =£'i- 9f- V- 479i^'«- Ex. 5. Express £2. 7/. Sc. 9 m. and ;^i. 9/ 3 c. 6'4S83 m. accu- rately in ordinary money : that is, in £. s. d. £2. If. 8c. 9OT. =;f2-789 39. (273) = £2. 15. 9i-44- 37'44 £1. 9/. ic. 6-4683»^.=;^r9364583 ^^'«^? ''4583 =;fi. i8.8i, jTr-- Ex. 6. Express ;^6. 8. 2^ and ;^7. 13. lof to the nearest mil'm £./. c. m. At sight £6. 8. 2i=;^6-409 (274) =£6. 4/ o^- 9»'- (276) Atsight;^7. 13. ioi=;^7-69S =;£■;. 6f. gc. c,m. Ex. 7. Express ;£'3. 4/ 5^:. 6»«., ;£8. 8/. ic. 8m., £g. g/. gc. 6m. to the nearest farthing in £. s. d. Now £i. 4/ 6^. 6/«. =;£3-466=;^3 • 9 • iJ (275) ;f8. 8/ 2f. 8;«. =;^8'828 = ^8 . 16 . 6i ;£'9- 9/ 9^- Swz. =;£'9-996=;f9 ■ '9 • " Ex.43- DECIMAL MONEY. 23 1 Exercise 43. I. Reduce £^%. if. "jc. to mils; and 876345 mils to £./. c. m. ■i. Reduce £25. 2$m. to mils; arid 3809 mils to florins. 3. Reduce ;^8. 4/. ^\c. to mils; and 16^/. to mils. 4. Find the sum of £121. 9/. v. $m., £8g. 4/ 71S- '6. 9j, and his assets realize £19^7. 13. 4j : how much will his estate pay in the pound, and how much will a creditor receive for a debt of .^359- IS- 3i? 43;7>5,-8,3,8,54 ) 1957-66875 ( -4473814 359-765625 20733333 4'83744 322998 1439062 i66go 143906 3563 25183 63 1079 20 287 I 4 ;^i6o-952i .-. the estate pays 8^. iijt/. in the pound, and the creditor receives ;^i6o. 19. oj. 234 DECIMAL MONEY. Ex.44. Exercise 44. I. Find the value of 4785 articles at \2s. C^d. each. ■i 876^1 £2. 17. Sieach. 3 SoSpxi £4- 19- 5i each. 4 594'?^ £3- II- iijeach. 5 356'9375 bushels at 53J. loti. a quarter. 6 3479 things at Si-, sfof. a dozen. 7 7857 £^- 9- i4 a score. 8 26953 oranges at 4^. 7j. 14. 6J, what will be the price of i cwt.? 27, If a bar of gold weighing 7 lbs. 5 oz. 12 dwts. i5grs. be worth ^330. 15. 5 J, what is the price per oz.? 28. A, B, and C advance respectively ;£'i9i. 12. 7f, ;^6i. 14. 8, and ;£'i22. I. 9J in a mining adventure which yields ;^Sii. 12. 6J. How much will this give for every £\ advanced, and how much of it vifill each of them receive? MISCELLANEOUS. 19. Find as a decimal the average of 2i|, 73^, o, 3 "065, 82, 17^, 6i- and 9^. 30. In how many years will the error amount to a day in considering the year to consist of 365^ days instead of 365'2422i8 days? 31. The gallon contains 277-274 cu. inches, and a gallon of water weighs 10 lbs., find the weight of a cu. foot of water. If mercury be 1 3-568 times heavier than water, find in ounces the weight of a cu. inch of mercury, 32. How many parcels of gold dust each weighing 17-36 grains can be made up out of i lb. 2 oz. i dwt. 3 grs.; and how much will remain over? 33. The average year of the Gregorian calendar is greater than the true year by 24-3648 seconds. Find, as a decimal of a day, the length of the true year. The Gregorian calendar intercalates 97 days in i^QQ years. 34. The circumference of every circle is 3-14159 times its diameter. The diameter of a carriage wheel is 4 ft. SJin.; find its circumference and how many revolutions the wheel will make in going 2 miles 7 f. 15 po. 35. From a rod 2-078 inches long, portions are cut off each equal to •0037 of an inch long; find how many such portions can be cut off, and what length will remain over. 36. The weight of a cubic foot of sea- water is 2 qrs. 8 lbs. 8foz.; a cask floating therein displaces gcu. feet 937icu. inches of water; find the weight of the cask, or, which is the same thing, of the water displaced. 236 METRIC SYSTEM. § 278. THE METRIC SYSTEM. 278. The explanation of the Metric SystemNgiven in Arts. 216 — 222 should be carefully read before commencing the fol- lowing examples. Ex. I. Express 31415" '92 in succeeding denominations from the highest to the lowest. Since 10 units of one denomination make i unit of the next higher denomination, and the units' figure represents grammes, the tens' figure will represent dekagrammes, the hundreds' figure hectogrammes, ; hence 314158' -92 = 3 myriag. i kilog. 4hectog. i dekag. 5 gr. Qdecig. 2 centig. or =31 kilog. 415 gr. 92 centigr. or = 31 kilog. 415-92 gr. Ex. 2. Express 9 hectares 25 ares 8 centiares as a decimal of a sq. kilometre. A hectare is a sq. hectometre, and 100 sq. hectometres make i sq. kilometre (218) : hence 9 hectares 25 ares 8 centiares=9'25o8 sq. hectom. (218) = •092508 sq. kilom. Ex. 3. How many hectares are there in a field which contains I3acres 3r. I7p.? 40) 17 4 ) 3'4^S 2-47114) i3"8s62S ( S"6o723 150055 1787 57 8 I hence we see that 13 acres 3 r. 17 p. = 13-85625 acres, and from the Table i hectare=2-47ii4 acres; j-i , r?"8562e .-. no. of hectares = -^^ — - — ^ = «-6o72^. 2-47114 "^ ' •' §278. METRIC SYSTEM. 237 Ex. 4. If a gallon of water weigh 10 lbs., find its volume in cu. centimetres. 2'.2p,4>f',2 ) lo-ooooo ( 4-53593 118152 7921 1307 205- 7 o From the Table (226) i kilog. = 2'20462 lbs. . . 10 lbs. =4-53593 kilog. =4535'93 grammes, = 4535-93 cu. centimetres of water, {221) that is, a gallon contains 4535-93 cu. centimetres. Ex. 5. When silk is sold at \<)f. 25 c. the metre, find the cor- responding price per yard in shillings and pence : supposing ;^i to be equal to 25/i 20 c. 25 20 2520 72 and I metre = 39- 37 1 inches = ' yard; ... 39:^1 yard jg gold for £^ ; 36 72 J r 55x36 or I yard £ 72 X 39-371 r 27'5 or -X . 39'37i £ 4848 39'/3,7/i ) 27-5000 ( -69848. .2M 38774 46-55 3340 I go 33 2 -. price per yard is 13s. 114-55. 238 SOME APPLICATIONS OF Ex. 45. Exercise 45. Express 1. A length of 345678'»"09 in kilometres, and succeeding denominations. 2. A capacity of 207856'" "508 in kilolitres, . . . 3. A surface of fifiySpseo"' in sq. kilometres, . . . . 4. A volume of 357"° -08267 in cu. metres, g. A length of 345 kilometres 7 hectometres 6 metres 8 decimetres and 9 millimetres in metres and kilometres. 6. A weight of 3567 kilogrammes 8 hectogrammes 9 grammes and 5 decigrammes in tonneaux de mer, and in decigrammes. 7. An area of 708 hectares 9 ares and 5 centiares in hectares, and in centiares ; also in sq. kilometres. 8. A volume of 457 cu. metres 24 cu. decimetres and 60 cu. centimetres in cu. metres. 9. A weight of 67 tonneaux de mer 8 quintaux 5 kilogrammes and 8 grammes in cu. metres, &c. of water. 10. Find the sum of 9 hectares 35 ares 8 centiares, 15 hectares 8 ares and 63 centiares, 23 hectares 85 centiares, and 17 hectares 80 ares 90 cen- tiares. 11. From 4 dekagrammes 83 decigrammes subtract 29 grammes 687 milligrammes. 12. Find as a decimal of a kilogramme the value of ■5678 millier-l- 9-3257 kilog. - 349-82 gr. 13. Find the value of 34-j of 87 cu. metres 62 cu. decimetres and 300 cu. centimetres. 14. What will be the price of 47 hectares 5 ares 65 centiares of land at 89 francs 76 centimes the are? 15. If 7'" -89 weigh 107 kilog. 288-22 gr., find the weight of i litre. 16. If 8 metres 8 decimetres of cloth cost 253 francs 88 centimes, find the cost of 1 7 metres 64 centimetres. 17. The weight of a volume of mercury is 13-598 times that of an equal volume of water : find the weight of 567 -859 cu. centimetres of mercury. (221.) In the following examples we shall use these relations : I metre= i -09363 yds. = 39-37079 in. I sq. meire=i'ig6os sq.yds.; i cu. metre =i'^o8oi cu. yds. I hectare or sq. hectometre=v\'ji\i, acres. Ex. 45. THE PRECEDING RULES. 239 I litre— vj(>o'j'] pinls — ' 11010 gallons. I kilogramme =i"iOi,6i. lbs. Av.= I6432"3487 grains. I gramme= i cu. centimetre of distilled water. 18. Express a yard in terms, of the metre, and an inch in terms of the centimetre. 19. Find the length of a tunnel 2 miles 63 chains 18 yards long in kilo- metres and metres. 20. Snowdon is 3571 feet above the level of the sea: express this height in metres, &c. 21. The length of the seconds pendulum in the latitude of London is 39' 1 393 inches : reduce this length to the decimal of a metre. 22. The mean diameter of the earth is 79I2'409 miles: express this length in kilometres. 23. Find in miles, chains and yards the length of a railway 96 kilom. 325 metres long. 24. Mont Cenis tunnel is 12234 metres long: express its length in miles, chains and yards. 25. Mont Blanc is 48io'88 metres high: express this height in feet. 26. The standard height of the barometer at Paris is 76 centimetres: find this height in inches. 27. The metre is the ten-millionth part of the distance from the pole to the equator, measured on the surface of the ocean : find the earth's circum- ference in miles. 28. Express an acre in terms of the are; and a sq. mile in kilometres. 29. The area of England is 50387 sq. miles: find the area in sq. kilom. 30. Find in hectares, &c. the area of an estate which measures 387 acres 3 r. 24 p. 31. The area of France is 53027894 hectares: express this area in sq. miles. 32. How many acres, &c. are there in a field containing 7 hectares 25 ares 8 centiares? 33. In making a railway cutting 2558 cu. yards 25 cu. feet of earth have been removed: find this quantity in cu. metres, &c. 34. The volume of a room is 870 cu. metres 936 cu. decimetres : express this volume in cu. yards, feet and inches. 35. How many gallons, &c. are there in a cask containing 29 veltes — velte being equal to 7 '4 505 litres? 240 APPLICATIONS OP PRECEDING RULES. Ex. 45. 36. How many hectolitres are there in 57 gallons 3J pints? 37. Express the lb. Troy in grammes; the lb. Av. as a decimal of a kilogramme ; and the cwt. as a decimal of a millier. 38. A sea-service mortar weighs 4 tons 19 cwts. 2 qrs. 19 lbs.: find its weight in tonneaux and kilogrammes. 39. By how many kilogrammes is a French gun weighing 263 myria- grammes heavier than an English gun weighing 49 cwts. 3 qrs. 24 lbs.? 40. When cloth is sold at 15J. ij\d. a yard, what is the corresponding price in francs and centimes per metre, \i £1 be worth 25 fr. 25 c? 41. When wheat is sold at 32 francs 50 centimes the hectolitre, what is the corresponding price in English money per bushel, if 25 fr. 10 c. equal £1 ? 42. When 325 sq. yards of building land is sold for ^'^ip, find the corresponding price in francs per sq. metre, when £\ is equal 25 fr. 30 c. 43. The rent of a farm of 45 hectares 75 centiares is 3695 francs : find the rent per acre in £. s. d., when 25 fr. is worth ;^i. 44. The pressure of the atmosphere is 14J lbs. to the sq. inch : find the pressure in kilogrammes to the sq. centimetre. 45. A shot projected with a charge of -figs kilog. of powder had a range of 734 metres; express these quantities in grains and yards. 46. A lo-inch gun threw a solid shot of 134 lbs. si oz. a distance of 4875 yards; express the bore, weight and range in centimetres, kilo- grammes and metres respectively. 47. In the year 1870, 14945 kilolitres 615 litres of brandy were im- ported into the United Kingdom, and paid a duty of ioj-. id. a gallon : find the amount of duty paid. 48. A litre of alcohol, i.e. a cu. decimetre, weighs 792 kilog. : find the weight of a cu. foot in lbs. 49. When the French Post-office allows 10 grammes, the Enghsh allows \ oz.; by how many grains is the latter weight within the former? 50. A cask of brandy containing 4*732 hectolitres is bought for 2835 fr. and duty is paid thereon at the rate of 10s. s!■ 1 2 J. itd. is the difference between ^4 and 7j. kd.; if now we give £i, for each of a given number of articles and receive back 7J. (>d. for each, we shall give £z- 12 J. (>d. for each of them. (3) If one price be an aliquot-part of another price, the value of a given number of things at the first price is this aliquot-part of their value at the second price. Thus 5 J. is J of ;£i ; the value of any number of things at is. is f of their value at ;^i. What has been said of value in this Art. applies to weight, length, area, &c. 282. Practice is Simple or Compound, according as the quantity whose value, weight, &c. is to be found is simple or compound. 283. SIMPLE PRACTICE. i. Ex. I. Findthevalueof345 oz.ofgoldat;^3. 17J. lo^i^peroz. .^• 345 3_ £■ s. d. 1035 = value of 345 oz. at 3 . o . operoz. 172 . 10 . o = 10 . o 86. 5.0= 5.0 43 -^-Ss 2.6 4-6.3= 3 10^. s.6d. id. lid. hoi£i i of los. I of SJ. ^ of 2J. 6d. i of id. ^ • 3 ■ 4 = 14 1343 . 6. ioJ= 3 . 17 . loj The value of 345 oz. at £1 per oz. is;f345, ^nd. ? If the tradesman paid in full, or ;^i in the £,1, the creditor would receive j£'545. \']s. 6d., hence if he paid los. in the £\, the creditor would receive J of ;^545. 17^. 6d., &c. lOS. \s. id. Zd. £. J. d. 545 17 6 \ 272 18 9 4 34 2 4*i i 8 10 m i 6 16 SH £■ 545-875 322 8 . m I Of. i IJ. A- td. i ^d. i id. i 27' '9375 27-2937 13-6468 6-8234 1-7058 322 -4072 =;^322. 8. ij. Ex. 7. What does a tax of 7^/. in the ;£l amount to on an income £. s. 1285 ■ IS 10 is. ^ M. i- id. I 64 • - s - 32 . 2 5 ■ 7 -^ lOf If* (283, 3) £i7 o^i 284. COMPOUND PRACTICE. Ex. I. Find the value of a bar of gold weighing 5 lbs. 10 oz. 12 dwts. 6|grs. at £3. 17s. iid. per oz. £. J. d. 5 lbs. 10. oz. = 70 oz. 3 • 17 • II 10 38 19 2 7 272 14 2 10 dwts. J of I oz. I 18 iij 2 dwts. \ of 10 dwts. 7 9i 6grs. 1 of 2 dwts. iiiS fgr. 4of6grs. ii«^ ;^275 oJH Ex. 2. What is the cost of 17 tons 12 cwts. 3 qrs. 18 lbs. of goods at £6. IS J. 9^. per cwt. ? 246 PRACTICE. §284. 17 tons 12 cwts. = 352 cwts. i and here we shall first find the cost of 352 cwts. by Simple Practice, and not by Compound Multiplication as in the last Example. 1 5 J. ''I 2 qrs. 4. I qr. A 14 lbs. 4 lbs. J of I cwL ^ of 2 qrs. \oi\ qr. 1 of I qr. 2389 3 I . . s. d. rS • 9 35^ 1112 264 13 • 4 . 7 • loj 13 . Hi 2389 . 16 . \^^...^ 4 ■ io4f...io ;^2395 7iA In the two last Examples it would have been easier to have expressed the price as a decimal of ;£i, and then to have proceeded as before ; thus ^■ 35^2 3-89583 I3'S750 70 339'37S 2727083 2036-25 dwts. ^ of I oz. I '9479 2 qrs. ■J- of I cwt. 3"3937 2 dwts. \ of 10 dwts. •389s iqr. ^ of 2 qrs. I -6968 6grs. i of 2 dwts. ■0487 14 lbs. \ of I qr. ■8484 igr. i^ of 6 grs. ■0061 4 lbs. f of I qr. ■2424 275-1005 = ;^275. 2. O 239S'38i3 =£'ii95- 7- lh Ex.3. If lib. standard is lib. 2 oz. 11 dwts. 16 grs. Troy, what is the weight Troy, of i cwt. 2 qrs. 25 lbs. 10 oz. 6 drs. ? I cwt. 2 qrs. 25 lbs. = 193 lbs. st. lb. Tr. oz. I . 2 dwts. . II . grs. 16 14 weight of I lb. st. 12 lbs. 80Z. 2 oz. 4 drs. 2 drs. \ of I lb. St. J of 8 oz. 1^ of 2 oz. ^ of 4 drs. 116 116 I o II o o 16 7 I 16 20 II 4 2 . 13' 6ri 235 i9tV . 96 lbs. . 96 lbs. . lib. 80Z. 2 OZ. 4 drs. 2 drs. . I93lbs. 10 OZ. 6 drs. Ex.46. • PRACTICE. 247 Exercise 46. Find the value of 1. S087 articles at 3J. 4^'., at %s. ^d. and at y. io\d. each. 2. 389s things at ^s. t^d., at gj. id. and at 5J. 7-J(/. each. 3- 25983 lbs. at M., at 8^1/. and at ^\d. a lb. 4. 365 days at sfi/., at 5^. 3(/. and at £i. 4. 4^ a day. 5. 3119 things at ^4. 7. 7 and at ;^5. 18. 5 each. 6. 1493 things at lis. i\d., at i6j. i,d. and at;^io. 9. 9 each. 7. 2750 things at 4J. sji/. each; 653 things at %s. 9|(/each. 8. 862 things at lu. 5^;/. each; 276 things at 14^'. \6^d. each. 9- 567384 lbs. of cotton at 1%d. and at V^d. a lb. 10. 2157 things at £1,. 7. 4J each; 14765 things at £\. 17. 8J each. 11. 313 things at £i,. 6. 9! each; 4321 things at £4.. 17. 3! each. i2- 3655 things at £1. 19. 5J and at £g. 16. 10^ each. 13. 4678 things at i2j. o^d. and at;^2. 12. 2f each. 14. 6^"] things at gs. l%d. and at £g. 18. loj each. 15- 497 things at £i. 16. 8, at;^3. 17. 6 and at £i. 8. 4 each (283, 2). 16. 969 things at lo^d., at igs. ii^d. and at £^. 19. 8i each. I?- 3S46 things at ;^i. 6. 75, at ;!f3. 18. lof and at;^5. 15. 7I each, la 4562J things at;^3. 15. gj each; 356I tons at £1. 13. sj a ton. i9' 73941 things at ;^i2. 8. 8J each; 3764! things at;^2. 14. 'ji each. 20. 2611 yards at £2. 17. 7^ for a dozen yards. 21. 34897 things at 15J. 7|,. 16. 8 and cost required are of another kind. Also if the number of yards be increased 2, 3, . . . times, the cost will be increased 2, 3... times; that is the length and the cost are in direct proportion. (287.) But the corresponding values to 23 yards and 37 yards are £2>- 16. 8 and cost required respectively ; therefore the ratio of 23 yards to 37 yards is the same as the ratio of £3. 16. 8 to cost re- quired (285), and we have this proportion or (as it is usually called) statement — 23 yards : 37 yards =;^3. 16. 8 : cost required, where it is arranged that the cost required is the fourth term, and therefore the given cost is the third ; and since the proportion is direct, the first and second terms of the first ratio correspond to the first and second terms of the second ratio. Replace the quantities in the first ratio by the numbers which measure them, and we have 23 : 37 = £3- 16. 8 : cost required, therefore cost required =^^-^ — '■ . (192) 23 Hence to find the fourth term, or cost required, we multiply the third term by the number in the second term, and divide by the number in the first term. 252 RULE OF THREE. §289. Again— If I cwt. 3 qrs. 16 lbs. be worth £11. 3. 9, what is the value of 5 cwts. i qr. 20 lbs.? Here if the weight be increased the value wiU be increased, and therefore the proportion is direct (288) ; hence, as before, we have this statement : — I cwt. 3 qrs. 16 lbs. : 5 cwts. i qr. 20 lbs. =>f 23. 3. 9 : value required. 4 4 7 21 28 28 212 608 But in order to replace the quantities in the first ratio by their corresponding numbers, we must bring them to a common deno- mination ; bring them to lbs., and the statement becomes 212 lbs. : 608 lbs. =^23. 3. 9 : value required, therefore 212 : 608 = ^23.3.9 '• value required ; (185) and therefore value reqd^* ^ ^' "" . (192) Lastly — If 12 men can do a piece of work in 15 days, in how many days will 20 men do the same work ? Here if the number of men be increased 2, 3, ... times, the number of days will be diminished 2, 3, ... times, therefore the men and the days are in inverse proportion. But 12 men correspond to 15 days and 20 men days requiredj therefore 15 days is to days required in the inverse ratio of 12 men to 20 men, that is, in the ratio of 20 men to 12 men : hence we have the statement 20 men : 12 men =15 days : days required, where, as before, the quantity required is the fourth term ; and since the proportion is inverse, the second 3.nA first terms of the first ratio correspond to theyfw/ and second terms of the second ratio. 1 290. From these considerations we have deduced the following Rule:— Take out the two given quantities of the same kind and corre- §■290- RULE OF THREE. 253 sponding to them the other given quantity and the quantity required. If the proportion be direct^ make the ratio of the first two equal to the ratio of the second two : if inverse, make the inverse ratio of the first two equal to the ratio of the second two. If necessary, express the first and second terms in the same denomination. Multiply the third term by the number in the second term, and divide the product by the number in the first term; the result will be the ^^ quantity required." Remark. The first ratio is not altered in value, if we multiply or divide both its terms by the same number (188) ; nor is the answer altered if we multiply or divide the first and third terms by the same number. Ex. I. If 29 tons of coal cost £2^. 17. 2, what will 356 tons cost? If the weight be increased 1, 3, ... times, the cost will be increased 1, 3, ... times, therefore the cost is directly proportional to the weight. Now 29 tons correspond to ^25. 17. -i,, and 356 tons cost required ; therefore we have 29 tons : 356 tons=;^25. 17. 2 : cost required. 20 517 12 6206 356 37236 31030 186:8 12) 29: ) 2209336 ( 76184^. 179 20 ) 6348. 8 53 243 ;^3i7- 8. 8. 116 o ,.-. Cost required =;^3 1 7. 8. 8. When we have reduced the third term to pence, we may consider the statement to stand thus : — 29 : 356=62061/. : cost required (in pence), we therefore multiply the third term by 356, and then divide by 29, giving 76i84(/. which is equal to ;,f3i7. 8. 8. 2S4 RULE OF THREE. §290. Ex. 2. If a bar of gold weighing 8 oz. lodwts. 15 grs. be worth £V>- 3- 9) what will a bar weighing 3 lbs. 8 oz. 15 dwts. 3 grs. be worth ? Since as in Ex. i the value is directly proportional to the weight, we have oz. dwts. grs. oz. dwts. grs £. s. d. 8 . 10 , ■ 15 : 44 . IS •3 = 33 • 3 • 9 : ^^lue required. 20 20 20 170 89s 663 24 24 12 695 3583 340 1790 1593 4°9S 341 8*9 238? ^1593 9* 341 6372 13 4779 13 ) S43213 12 ) 41 7854 A 20 ) 348>2 . I ;^I74 . 2 . \\-^ value reqd. Here we reduce the ist and 2nd terms to a common denomination (grs.), and for convenience we reduce the 3rd term to pence, so that we have 4095 : 2 1483 = 79651^. •- value required (in pence). Divide now the ist and 3rd terms by 5; then divide the ist and 2nd terms by 9, and then again by 7 ; and we shall have 13 • 341 = i593'^* ' value required (in pence). Proceeding now in the usual way, we find the value required to be '^T-n^ik^d. ox £,\^^. 2. iJA- Ex. 3. If I7cwts. 3 qrs. 16 lbs. of barley cost £i. 18. 9, what weight may be bought for £1. 16. 3? The weight is directly proportional to the money, therefore, stating as in Ex. I, we have cwts. qrs. lbs. £%. 18. 9 : £1. i6. 3 = 17 • 3 ■ 16 ■: weight required. 20 20 S ^78 12 2*45 429 ■43 45 cwts. 5.2. I4x% weight reqd. 56 89 , 12 , I . , 24 9 675 fii)8o5 ; . . 20 *35 '^^ 13 ) 73 ■ , ■ 22t-\ § 290. RULE OF TBREE. 255 » Here we reduce the ist and 2nd terms to the common denomination, pence ; and then dividing each of them, first by j and then by 3, we get 143 : 45 = 17 cwts. 3 qrs. 16 lbs. : weight required, and without reducing the. 3rd term, we multiply by 45 or 5 x 9, and then divide by 143 or 11 x 13, giving the weight required. Ex. 4. If 4|oz. of gold be worth ^19. 12. 6, what is the value of 3 lbs. lif oz.? Here the value is directly proportional to the weight ; .■. 4J0Z. : 47I0Z. =^19! : value required ; II .-. Value required=;£--|? x 2? x ■^-£-J- 3 = ;^I9I . 17 . 9H- Ex. 5. If for a given sum of money 4 tons 3 cwts. 48 lbs. can be carried 8f miles, how far can 5 tons i qr. 14 lbs. be carried for the same sum? 4 tons 3 cwts. 48 lbs. = 83,;^ cwts. = 83^ cwts. 5 tons I qr. 14 lbs. = iooy*A cwts. = loof cwts. Now if the weight be increased 2, 3, ...fold, the distance must be de- creased 2, 3, ...fold: therefore the weight is inversely proportional to the distance. But 83f cwts. correspond to 8| miles and I oof cwts distance required; therefore the inverse ratio of the two weights is equal to the ratio of their corresponding distances, thus loofcwts. : 837- cwts. =8f miles : distance required ; » 8 J- . ■ J f? 584 8 64 ., .". distance required=-^-^ x =-J x - — = —s miles 9 f 8e3 9 fi = 7^ miles. Ex. 6. If 7 men can mow a field in 18} hours : in how many hours can 15 men mow the same field? If the number of men be increased, the number of hours in which they can mow the field will be proportionally decreased, hence the inverse ratio of 7 men to 15 men equals the ratio of i8| hours to hours required ; that is, 2S6 RULE OF THREE. § 290. 15 men : 7 men=i8J hours : hours required; .-. hours required =— x — = — 4 15 4 = 8J hours. Ex. 7. If a capital of ^3250 realize a profit of ;rfi46. 5^., what profit will a capital of ;£ioo realize at the same rate.? Though the three given quantities are all money, yet in stating the question we must distinguish between them: for £^1^0 and ;^ioo repre- sent capitallsxA out, while £\\6. i,s. and "profit required" represent ^ro;f^ arising therefrom : and as the profit is directly proportional to the capital laid out, we have ;^3'2S° : £t-°°— £^\^- 5 ' profit required. .•. Profit required=/'i46J x' 3250 291. When the quantities in the second and third terms are of the same kind, there is no objection to our alternating these terms (193*) : thus in the preceding Example we have -£3250 ; ;£ioo =£idfi\ : profit required, therefore ;^32So : ;^i46i=^ioo : profit required, which expresses that the ratio of the first capital to its profit is equal to the ratio of the second capital to its profit. Such a mode of statement as this latter one is found very con- venient in those classes of questions where 100, or some other fixed number, is taken as a standard. Ex. 8. A public company whose capital was ;^5oooo is wound up, and its assets are found to be £\'iiZ'j. \os. : how much will be paid in the pound ? The dividend on ;^5oooo is ;^i2i8y5, at the same rate what is the dividend on £1 ? hence we have (291) £ioooa : £i'i'i^Ti=£>^ ■ dividend on ;i£'i ; .-. Dividend on £i=£^^^i~lJ' =£-243'js 50000 §291- RULE OF THREE. 2$? Ex. 9. A bankrupt's debts amount to £a357- 7- 8 and his assets to ;£929. 16. 5^ : how much will be received on a debt of ^325. 14. 7i ? Here the whole debt of £4357. 7. 8 is discharged by a payment of ;f929. 16. 5J: and the question is by what payment will £^2$- 14. 7i be discharged at the same rate. Hence ;^435y383 = ;^929'823 = 32573i : Payment required. ^3 ^8929 ■293157900 6514030 2931579 260585 6515 977 43,5,7\3>8„3 ) 302872-176 ( 69-5078 4.142919 221274 3405 355 7 .-. Payment required =;f69. 10. if. Ex. 10. A person after paying an income-tax of yd. in the £1 has a net income of ;£i247. 10. 5, what is his gross income ? For every igs. ^d. of net income he has a gross income oi £1, hence 19J. t^d. : £1247. 10. 6 = ;^! : gross income required T2 20 233 2495° 12 233 ) 299405 ( 1285 664 1980 1 165 o .". Gross income=;^i285. Ex. II. Two clocks are exactly together at 12 o'clock noon, on a certain day : one of them gains 7 sec. and the other loses 6 sec. in 12 hours. After what interval will one have gained half an hour on the other? and what o'clock will each then shew ? The one clock gains on the other at the rate of 13 sec. in 12 hours, or 26 sec. in i day; and the question is, after what interval will it have gained 30 min. or 30 x 60 sec. on the other. Now E.-S. A. 17 2S8 RULE OF THREE. § 291- 26 sec. : 30 X 60 sec. = i day : interval required ; . . Interval required = 5_^^ — days=^ — days. =6(j days 5 h. 32^ m. and therefore the true time at the end of the interval is 5 h. 32^^ m. p.m. But in -°,^ days the first clock has gained ^^x 14 sec. or ifi^min., and the second has lost -%°- x 12 sec. or 13JJ min. .■. the first clock will shew 5 h. 48y^^ m. p.m. and the second „ „ ,, 5 h. iSx'Vni. P.M. Exercise 47. i . If 3 qrs. 7 lbs. of tobacco cost ;f 1 7. 13. 6, what is the value of 5 cwts. I qr. 23 lbs.? 2. If 17 cwts. 2 qrs. 14 lbs. can be obtained for ;^8. 13. 3J, what weight can be obtained for ;^2i. 10. 2 J? 3. If 19 men can finish a work in 437 days, how long will it take 23 men? 4. The clothing of a regiment of 735 men costs ;f 1398. 1 5. 6, what will the clothing of a regiment of 903 men cost at the same rate? 5. A person in 87 days spends ^38. 19. 4J, in how many days will he spend ;^i63. 9. 9J at the same rate? 6. The interest on ^271. 18. 4 for 77 days is £%. 3. 8J : find the in- terest on the same sum for 245 days. 7. The interest on ;^2 32. 11. 7j for 10 months is ;^io. 2. 8J: what sum will yield the same interest in 17 months? 8. How many ducats of 4J. i\%d. each are equal in value to 55926 rix- doUars of 4^. \o\d. each? 9. A draper having sold 147 yards of cloth at the rate of £1. 9. 3f for if yardj found that he had gained £\^. 10. 9. What did the dloth cost him? 10. What (do the taxes on a house rented at ;^327. 12. 6 come to, when the taxes on a house rented at 35 guineas are £^. 8. 7J? 11. If 35 ells of velvet cost ;^i3. 19. 8J, what is the price of 63 yards? 12. When 19 yds. 2 qrs. 3 nls. of cloth cost £'i. 12. i\, how many ells can be obtained for ;^io. o. of? Ex. 47. RULE OF THREE. 259 13. When the carriage of 3 cwts. 2 qrs. 14 lbs. for 37 miles is iSj. i\d., what weight can be carried the same distance for £ii,. 16. 3? 14. If I ton i6cwts. 3 qrs. 20 lbs. cost £^. 15. 8, how much will 3 tons \i\ cwts. cost at the same rate? 15. A garrison of 638 men has provisions for 124 days; how long will the provisions last if the garrison be reinforced by 418 men? 16. A gang of reapers can reap 84 ac, 3 r. 14 p. in 13! hours: in how many hours can they reap 401 ac. 8 p.? 17. If a sequin be worth ^s. i^\d., and a carlino £ii. 12. 3J, how many sequins are equivalent to 450 carlini? 18. A grocer bought 2 tons 3 cwts. 3 qrs. of sugar for ;^i20, andjaaid jof, for expenses; how much must he charge per cwt. to have a clear profit oi£fi\. ss.f 19. If the 4^?. loaf weighs 2 lbs. 3 oz. when wheat is at js. i J find to the nearest farthing at how much in the £1 the rate must be laid to realise this sum. How much must be paid by an estate valued at ;^453. 12. 8? 42. If £$7.. 18. 9 be given for 40 yards 2ft. loin. of brickwork, how many yards, &c. can be built for j^ioi. igj.? 43. The weight of a 32-pounder gun being 3 tons 4 cwts., and that of an i8-pounder 2 tons 2 cwts., how many iSpounders will be equal in weight to 189 32-pounders? 44. When (he price of copper is y. ii^d. a lb. and the price of tin I J. I^d. a lb., how much copper must be given in exchange for 5 cwts. 3 qrs. 16 lbs. of tin? 45. A cup weighing 11 oz. 18 dwts. 8grs. is worth £i. 18. 9: what is the value of a goblet weighing 3 lbs. 8 oz. 19J dwts. at the same rate? 46. A floor can be covered with 32 J yards of drugget 7 quarters wide : how many yards of Brussels carpet 26 in. wide will cover the same room? Ex. 47. RULE OF THREE. 26 1 47. If 4 men working 15 hours, 3 men working 12 hours, and 8 men working 3 hours earn ;^3. 5J., what will a man's wages for 6 days come to if he works 1 1 hours a day ? 48. A regiment of a thousand men are to have new coats : each coat is to contain i\ yards of cloth of ij yard wide, and to be lined with shalloon | yard wide : how many yards of shalloon will be required ? 49. A clock which was i"i minute fast at a quarter to 11 p.m. on Nov. 28, was exactly right at 11 '30 P.M. the following day. How many minutes was it slow at a quarter to 2 P.M. on Dec. 7? 50. Supposing the number of sheep in the country to be 13000000, what would be the value of the wool in a year, estimated at ;^8. 15J. per cwt., if 15 sheep yield 29 lbs. of wool? 51. If 7 women earn as much as 4 men, and 48 men assisted by 14 women earn ;^42. ^s., what number of women assisting 20 men will earn ;^27. 4. 6 in the same time? 52. If 7 oxen or 1 1 horses can eat the grass of a field in 37 days, in how many days will 5 oxen and 8 horses eat it? 53. If 3 lbs. 7f oz. of cotton can be spun into a thread 264 miles loio yards long, what weight of this thread would be sufficient to reach round the globe, a distance of 25000 miles? 54. A room, 31 ft. 6 in. long and 23 ft. 10 in. broad, is to be covered with carpet f yard wide: what length of carpet will be required? 5«. When 54 yards 2 J feet of carpet are required to cover a floor 23 ft. 6 in. long and 15 ft. 9 in. broad, what must be the width of the carpet? 56. If 12 oxen and 35 sheep eat 6 tons 7 cwts. of hay in 4 days, hoy( much will it cost per week to feed 4 oxen and 6 sheep ; the price of hay being ;^3. 15^. per ton, and 2 oxen eating as much as 5 sheep? 57. A tug 90 feet long, steaming at the rate of 8 miles an hour, is towing a ship 480 feet long by a halser of 60 fathoms : a yacht whose length is 280 feet is sailing in the same direction at the rate of 10 miles an hour. What time will elapse between the yacht first coming up with the ship and finally clearing the tug? 58. A bankrupt's estate was calculated to give a dividend of I2J. 8i/. in the £\\ but bythe unexpected recovery of a debt o{ £8'j. 18. 9 the dividend was raised to 13J. lod. in the £1. Find the amount of the bankrupt's liabilities. 59. Two clocks, of which one gains 4 m. 17 s. and the other loses 3 m. 13 s. in 24 hours, were both within 2^min. of the true time, the former fast and the latter slow, at noon on Monday last: they now differ from one another by half an hour ; find the day of the week and the hour of the day. 262 RULE OF THREE. Ex. 47. 60. In running a 3 mile race on a course ^ of a mile round, A overlaps B at the middle of the seventh round. By what distance will A win at the same rate of running ? 6i. When the price of oats is 30if. a quarter, it costs 17^. 6 = i6 cannon : cannon required; 270 : 420 2.3 I . J ^ 4X 9x420 x^ .•. cannon required= 10 x J — ^ — j ^ -^.x 7x270x1 = 36. Ex. 4. A garrison of 4500 men is supplied with provisions for 1 5 weeks at the rate of 13 oz. per day per man : how many men must leave that the same provisions may supply those that remain 27 weeks at 10 oz. per day per man I Here we must find how many men the provisions will supply on the second hypothesis. men. weeks. oz. 4500 16 n men required 27 10 If the number of weeks be increased, the number of men must be decreased — proportion inverse. If the number of ounces per day be increased, the number of men must be decreased — proportion inverse : — Kence f = 4.R00 men : men required ; 10 : 13) . , 15 X 13 .-. men required = 4500 x — = 3250; .•. no. of men that must leave=45oo — 3250 = 1250. 268 DOUBLE RULE OF THREE. § 297. Ex. 5. A wall 1690 feet long has to be built in 30 days; and it is found that 7 men in 14 days have completed only 490 feet ; how many additional men must be employed that the wall may be finished in the required time ? The question really is — how many men must be employed to complete the remaining 1300 feet in the remaining 16 days? Now the length of the wall is directly proportional to the number of men ; and the number of days is inversely proportional to the number of men : hence 400 : lioo) . , , \ = 1 men : men required ; . , 1200 X 14 .•. men required = 7 x -^ = 15 ; ^ 490x16 .■. no. of additional men= 15 - 7 = 8. Exercise 48. I. If 67 tons carried 87 miles cost £^\. 5. 9, what will 73 tons carried 93 miles cost? If 61 tons carried 81 miles cost ;^20. 11. 9, how far can 77 tons be carried for;^3i. i. 6? If 37 tons carried 57 miles cost ^^8. 15. 9, what weight can be carried 83 miles for £11. 15. 9? 1. If 939 men consume 364 quarters of wheat in 7 months, how many will consume 1404 quarters in 13J months? 3. If 15 men take 17 days to mow 300 acres of grass, how long will 27 men take to mow 167 acres? 4. If ;iS'i37S put out at simple interest for 3 years produce ;^i44. 7. 6, in what time will the interest on ;^2420 amount to ;^i90. n. 6 at the same rate? 5. If a tradesman with a capital of £iooa gains £,^0 in 7 months, how long will he be in gaining £^o. ^s. with a capital of ^315? 6. When wheat is at 15J. a bushel 8 men can be fed for 12 days at a certain cost. For how many days can 6 men be fed for the same cost when wheat is \is. a bushel? If 15 men are fed for 7 days at a certain cost when wheat is i 2j. a bushel, what must be the price when 10 men are fed for 8 days at the same cost? 7. If a penny loaf weigh 6 oz. when wheat is i,s. 6d. a bushel, what should be the weight of the shilling loaf when wheat is 8j. ^d. a bushel? If the penny loaf weigh 6 oz. when wheat is 5^. 6d. a bushel, what should be the price of a loaf weighing 4ilbs. when wheat is at 8j. j,d. a bushel? Ex. 48. DOUBLE RULE OF THREE. 269 8. If 36 men can do a piece of work in 10 days working 10 hours a day, in how many days could 40 men complete the same work, working 9 hours a day? If 27 men can do a piece of work in 14 days working 10 hours a day, how many hours a day must 12 men work to do the same in 45 days? 9. If a quantity of provisions would serve a besieged garrison of 2000 men for 15 weeks at the rate of 18 oz. a day for each man, how many ounces must each man receive that the garrison increased to 2500 may be able to hold out 3 weeks longer? If a quantity of provisions will serve a besieged garrison of 1 500 men for 12 weeks at the rate of 2002. a day for each man, how many men would the same provisions maintain for 20 weeks at the rate of 8 oz. a day for each man? 10. If 20 men could perform a piece of work in 12 days, find the number of men who would perform another work three times as great in one-fifth of the time. 11. If 9 men can reap 15 ac. i r. 28 p. in 5 days of loj hours each, how many men will reap 401 ac. 8 p. in 7 days of iij hours each? 12. If the carriage of 6cwts. 3 qrs. for 124 miles cost £,},. 4. 8, what weight would be carried 93 miles for £1. o. 7i? If the carriage of £ cwts. i qr. 12 lbs. for 39 miles be £1. 8. 6, what must be paid for the carriage of 7 cwts. 16 lbs. for 48 J miles? 13. How many men would it employ for 5 J days to cultivate a field of 2f acres, if each man completed 77 sq. yards in 9 hours, and the day con- sisted of 10 hours? 14. If 6 iron bars 4 ft. long 3 in. broad and 2 in. thick weigh 288 lbs., find the weight of 15 bars each (s\ ft. long 4 in. broad and 3 in. thick. 15. If 5 men can reap a rectangular field whose length is 800 feet and breadth 700 feet in 3J days of 14 hours each, in how many days of 12 hours each can 7 men reap a field 1800 feet long and 960 feet broad? If 5 men in 6 days can reap a field 1200 feet long and 800 feet broad working 6 hours a day, what is the breadth of a field whose length is 1280 feet which 6 men can reap in 5 days working 8 hours a day? 16. If s horses require as much corn as 8 ponies, and 15 quarters of corn last 12 ponies for 64 days, how long may 25 horses be kept for ;^4r. 5^., when corn is 2 2 J. a quarter? If 21 horses and 217 sheep can be kept 10 days for £,i,6. 8. 4, what sum will keep 9 horses and 60 sheep for 27 days, supposing that 3 horses eat as much as 50 sheep? 270 DOUBLE RULE OF THREE. Ex.48. If 6 oxen or 1 3 sheep eat 26 cwts. 3 qrs. 20 lbs. of hay in 29 days, how much ought to be paid for the supply of hay to J7 oxen and 40 sheep during the month of March, hay being worth 4'53 guineas per ton? 17. If 10 cannon which fire 3 rounds in 5 minutes kill 270 men in an hour and a. half, how many cannon which fire 5 rounds in 6 minutes will kill 500 men in one hour at the same rate? 18. If 25 labourers can dig a ditch 220 yards long 3 ft. 4in. wide and 2 ft. 6 in. deep in 32 days when the day is 9 hours long, how many labourers would be able to dig a ditch half a mile long 2 ft. 4 in. deep and 3 ft. 6 in. wide in 36 days wlien the day is 8 hours long? 19. Two gangs of 6 and 9 men are set to reap two fields of 35 and 45 acres respectively. The first gang works 7 hours in the day, and the latter 8 hours. If the first gang complete their work in i2 days, in how many days will the second complete theirs? 20. If a beam which is 10 in. wide 8 in. deep and 5 ft. 6 in. long weigh S cwts. I qr., find the length of another beam the end of which is a sq. foot and which weighs a ton. 21. If the wages of 25 men amount to £16. .13. 4 in 16 days, how many men must work 24 days to receive .^103. lOJ., the daily wages of the latter being one-half those of the former? 22. If when copper is ;^7. 14. 4J per cwt. I can get 3 cwts. 2qrs.i4lbs. of brass for .1^27. o. 3I, how much brass shall I get for ;^-iS3. 17. 6 when copper is at ;if9tV per cwt.? 23. If 17 cwts. I qr. can be carried 840 English miles for £\i. vs. 6, how many Irish miles may 2 tons 6cwts. be carried for £11. los.: the Irish mile being i"if of an English mile? 24. If 5 labourers or 3 navvies can excavate 28 cu. yards of earth in ro hours, in how many hours can 6 labourers and 8 navvies excavate 120 cu. yards, the soil in the latter case being one-fourth more difficult to work than in the former? 25. A person is able to perform a journey of 142-2 miles in 4^ days when the day is io'i64hours long : how many days will he be in travelling 505-6 miles when the days are 8-4 hours long? 26. If the 4(/. loaf weigh i lb. 9^- oz. when wheat is at 93-, 3/. a bushel, how much bread can be got for 5^-. l\d, when wheat is at ^f>s. a quarter? If the 6d. loaf weigh 4-35 lbs. when wheat is at 575J'. a bushel, what ought to be paid for 49-3 lbs. of bread when wheat is at y^s. a bushel? Ex. 48. DOUBLE RULE OF THREE. 27 1 27. If 48 pioneers in J days of 12 J hours long can dig a trench 13975 yards long 4'5 yards wide and 2'5 yards deep, how many hours per day must go pioneers work during 42 days in order to dig a trench i636"6875 yards long 4'875 yards wide and 3'2 yards deep? 28. An oz. of standard gold is worth ;^3. 17. loj and contains -grd of pure gold ; find the value of (neglecting tlie alloy) (i) 10 lbs. of jewellers' gold containing •583 pure gold : (2) a Mohur weighing 7 dwts. 23grs. containing '993 pure gold: (3) a Star Pagoda weighing idwta. 4jgrs. containing 792 pure gold. 29. A contractor engaged to finish a work of 3150 cu. yards in 50 days, and employed at once 60 men upon it, Ijut at the end of 35 days he finds only 1 800 cu. yards completed : how many more men must lie put on to complete the work in the given time ? 30. An engineer engages to complete a tunnel 3J miles long in 2 years 10 months ; for a year and a half he employs 1 200 men, and then finds he has completed only three-eighths of his work : how many additional men must he employ to complete it in the required time? 31. If 7 women earn as much as 4 men, and 48 men assisted by 14 women earn 121 guineas in 17 days, what number .of women assisting 20 men vdll earn ;^2i. 3. 6 in.one^liird of the time? 32. If 16 men in 18 days can perform a piece of work which occupies ,36 boys for 24 days, how many men must help 21 boys to perform the same piece of work in 16 days? 33. Two cogged wheels, of which one has iS cogs and the other 28, work in each other. If the first turn 16 times in i\ seconds, how often will the other turn in 21 seconds? 34. Two sets of men perform the same amount of work- Each man in the first set is stronger than each one in the second in the ratio of 7 to 6: the first set works 6 days a week for 10 weeks, and the second set 5 days a week for 7 weeks. If there are 9 men in the first set, how many are tliere in the second? ■35. If 6 men can reap 15 acres in 3 days of 14 hours •each, and 10 boys can reap rof acres in 5 days of 9 houra «ach, find the ratio of the work of a maji to the work of a boy. Find also how many acres 4 men and 7 boys can reap in 7 days, working 12 hours a day. 272 DOUBLE RULE OF THREE. Ex. 48. 36. If 5 pumps each having a length of stroke of 3 ft. working 15 hours a day for 5 days empty the water out of a mine, how many pumps with a length of stroke of ijft. working 10 hours a day for 12 days will be required to empty the same mine, the stroke of the former set of pumps being performed 4 times as fast as those of the latter? 37. An engine of 40 horse-power can raise 53 tons 11 cwts. i qr. 20 lbs. through lift, in one minute; how long will an engine of 30 horse-power take to fill a reservoir containing 900000 cu. feet, the water being raised through 50 feet, and a cu. foot of water weighing 1000 oz.? 38. In a boat-race theC. crew rowed 39 strokes per minute and the O. crew 41, but ig strokes of the former were equivalent to 20 of the latter. The C. crew rowed over the course of 4 miles in 25 minutes; find the number of feet and the number of seconds by which the race was won. 39. When wheat was at 75^. a quarter the 4 lb. loaf was sold for ^\d., but when wheat rose Jj. a quarter the price of the 6 lb. loaf was raised to IS. Suppose the cost of converting wheat into bread be at the rate of 7.S. 4(/. per cwt., how much would the bakers lose or gain on every £1 . 14. 6, find the value of SS° Madras gold rupees each weighing 7 dwts. 12 grs. of which 916 parts out of 1000 are fine. £,»,(>. 14. 6=;^46|f, .-. 40 lbs. staTidard=;^i869: hence ;^s reqd=550 Madras rupees, 2 = 15 dwts. Madras standard, 20 X 12 = 1 lb 1000 = 916 lbs. fine, 11 = 12 lbs.' English standard, 40=;^i869; 5.50x15x916x12x1869 ^ ^ * ^ 2x20x12x1000x11x40 = ;^802. 10. oil. 18—2 276 CHAIN RULE. Ex. 49- Exercise 49. I. If 16 darics made 17 guineas, 19 guineas made 24 pistoles, 31 pis- toles made 38 sequins; how many sequins were there in 1581 darics? I. 121 Irish acres are equal to 196 Imperial acres, and 126 Imperial acres are equal to 100 Scotch acres; how many Scotch acres are equal to 385 Irish acres? 3. If -LI of A count for 13 of B, 6 oi B for 18 of C, and 13 of C for ■2 oi £>\ how many of A count for 100 of Z)? 4. If 12 oxen be worth 29 sheep, 15 sheep worth 25 hogs, 17 hogs worth 3 loads of wheat, and 8 loads of wheat worth 13 loads of barley; how many loads of barley must be given for 340 oxen? 5. Required the relation between the metre and the Rhineland foot, supposing the metre equal to 39'37t inches, and that 37 English feet are equal to 36 Rhineland feet. 6. Suppose 10 lbs. of London equal ii lbs. of Rome, and 26 marks of Spain equal 16 lbs. of London, what is the ratio between the Roman pound and the Spanish mark? 7. I lb. of standard silver, of which 222 parts out of 240 are pure, is coined into 66 shillings : what weight of pure silver is there in 20 shillings? 8. If 40 lbs. of standard gold, of which 11 parts out of 12 are fine, be coined into 1869 sovereigns; how many grains of pure gold are there in i sovereign ? 9. If 100 francs are equal in value to 27 thalers, and 33 thalers are equal to £$, how many francs and centimes are equal to £1 ? 10. How many pounds of tea at 2s. n\d. a lb. must be given in ex- change for 2 tons 3 cwts. of sugar at £2. 7. 6 per cwt.? II. The stadium contained 600 Greek feet; and the Roman mile con- tained 1000 paces of 5 Roman feet each : 24 Greek feet were equal to 25 Roman feet, and 12 Roman miles to 11 English miles: find the length of the stadium in yards. 12. A sovereign standard gold weighs 5'i36 dwts., a shilling standard silver weighs -jV of a pound Troy : what weight of standard silver is equal in value to 4 oz. of standard gold? 13. If 22 oz. of fine gold make 24 oz. of standard gold, and the Mint gives 1869 sovereigns for 40 lbs. of standard gold, how many sovereigns ought to be given for 67 lbs. 10 oz. of fine gold? 14. The Cologne mark of fine silver weighs 3609 grains : find its value when 37 oz. of fine silver make 40 oz. of English standard silver, and i oz. of standard silver is worth f,s. i Jrf. Ex. 49. CHAIN RULE. 277 15. From a Cologne mark of fine silver weighing 3609 grains are coined 14 thalers; find the value of a thaler, when an ounce of English standard silver, of which 57 parts out of 40 are fine, is worth 5J. i\d. 16. If 47J South German florins, which are -^ fine, weigh goo grammes, and a kilogramme be equal to 15432 grains; what is the value of this florin in English money when standard silver ^ fine is worth 5s. i|(/. ? 17. If I lb. of standard silver, of which 37 parts out of 40 are fine, be worth 66s., find the value of an Arcott Rupee weighing 7 dwts. 9 grs., of which 941 parts out of 1000 are fine. 18. If I lb. of standard gold, of which 11 parts out of n are fine gold, be worth £'i6. 14. 6, find the value of 595 gold rupees of Bombay, each weighing 7 dwts. loj grains, of which 187 parts are fine gold and 13 alloy. 19. The length of the minute-hand of the clock of the Palace at West- minster is 1 1 feet : what distance will the end of it travel through in a year of 365J days, if 7 times the circumference of a circle be 22 times its diameter? 20. The distance from Paris to Lyons by railway is 506 kilometres and the first-class fare is 56 fr. 80 c. ; find the rate per mile in pence, &c. sup- posing 5 francs to be equal to 4J. and 8 kilometres to be equal to 5 miles. 21. The distance by rail from Turin to Venice is 435 kilometres and the first-class fare is 51 lire 45 centimes; find at the same rate, in English money, the fare from London to Edinburgh a distance of 401 miles; reckoning 5 lire equal to 4^. and 8 kilometres to 5 miles. 22. The distance by railway from London to Bristol is 118 miles, and the fare 20J'. rod. : find at the same rate the fare in florins and cents from Innsbruck to Verona a distance of 3i's German miles, when 117 English miles equal 25 German and 20s. equal 12 fl. 30 cent. 23. When cloth is sold at i^s. g^d. a yard, what is the corresponding price in francs and centimes per metre; a, metre being equal to 39 "3 71 inches and £i to 25 fr. 25 c. ? 24. When wheat is sold at 22 fr. 50 c. the hectolitre; find the price per bushel in English money, supposing the hectolitre equal to 22 gallons and 25 fr. 10 c. equal to/'r. 25. The rent of a farm of 45 hectares 75 centiares is 3695 francs; find in English money the rent of 87 acres 3 r. 25 p. at the same rate; supposing 100 hectares equal to 247 acres, and 25 francs equal to £1. 278 PROPORTIONAL PARTS. §301- PROPORTIONAL PARTS. PARTNERSHIP. 301. To divide a given quantity into proportional PARTS is to divide it into parts which shall have the same ratio to each other that certain given numbers have. For example — Divide .£735 among A, B, C, and £>, so that their shares may be proportional to the numbers 3, 5, 7, and 9. Now 3 + 5 + 7 + 9=24: if therefore we divide the given quantity into 24 equal parts, and give 3 of these equal parts to A, 5 to B, 7 to C, and 9 to 2?, we shall have given away the whole quantity, and the shares of A, B, C, and D wiU be respectively 24 ''3. 24 x5, 2^ x7, and -^xg, and will therefore be as 3 : 5 : 7:9 (188). 302. If the given numbers are fractions we may follow the same method; but it will be more convenient to find integral numbers proportional to the given numbers. For example, Divide £73$ among A, B and C, so that their shares may be proportional to i^, 2f and 3I. Nowii : 2f : 3f=3 . ». £5 23 4 = 18 : 32 : 45. (188) But 18 + 32 + 45 = 95; therefore ^'sshare=^5^i8, B's=^x32, and Cs=^^5 ^45. 303. We have then the following Rule : — Divide the given quantity by the sum of the given numbers expressing the ratios of the parts; multiply the quotient by each of these numbers, and the products will give the parts required. Ex. I. Divide 837 into three parts proportional to the numbers 5, 9 and 13. §303- PROPORTIONAL PARTS. 279 5 + 9 + 13 = 27; .. istpart = ^x5 = 3ix 5 = 165; 837 2nd part = -^^x 9 = 31 x 9 = 279; 27 3rdpart = ^x 13 = 31 X 13 = 403.. 27 Ex. 2. Divide ^Jfp.. 10 among three persons so that their III shares may be to each other in the ratio of —r , -^ , — 5 . '5 ^s 3f — 1- J- = 2 3 ±_ 4 ■ 2| ■ 3i 3 ■ 8 ■ IS = 80 : 45 : 32. But 80 + 45 + 32 = 157; . . ist person's share = " — x 80=/- x 80= /■200; 157 ^2 2nd .... = =jf|x4S=;^ii2. 10; and 3rd . . . . = =;^-X32=;,^8o. 2 Ex. 3. Divide £n2,. 17. 6 among A, B, and C so that B'% share may be half as much again as ^'s, and C'% share one-third as much again as both ^'s and ^'s. Half as much again of a quantity is the quantity and half the quantity, or - of the quantity. In lilce, one-third as much again of a quantity is - of that quantity. .•. ^'s share = - of /4's share, 2 and .•. A\ share and ^'s share =- of ^'s share, 2 and C's share = - of - of ^'s share = — of /4's share, 32 3 .-.^'s share : ^'s : C"s=i : ? : — « 3 = 6 : 9 : 20. 28o PROPORTIONAL PARTS. § 303- But 6 + 9 + 20=35; .-. ^'sshare = ^^x6=;^23x6=i6j. (,d.->i(> = £^. 19. o, 35 40 ^'s = =i6j. 6 and of this capital A contributes ;^360 ; . •. ^'s share of the profits= ^^ '*'" '"" ^ x 350 ^ lOio =£n- 4- loJA- Similarly ^'s share = ^^^5^^°' ^ x 4^0 = &c. Ex. 8. A and B enter into partnership with a capital of ;£sooo, of which A contributes ;£33oo, and B the remainder. At the end of 3 months they admit C with a capital of ^1500, the next month they admit D with a capital of ^1950. Their profits at the end of the year amount to ^1729. 13. 9: — how much will each of the partners receive ? A has had ;f3300 in the partnership for 12 months, which is to be con- sidered as equal to .ii'3300 X 12 for i month. In like manner -5's capital is to be considered as equal to ^1700 x 12 for i month; C's as equal to ;^iSOo x 9 for 1 month; and D'% as equal to ;rf 1950 x 8 for i month. These equivalent capitals are for the same time, and therefore the division of profits will be in proportion to these several capitals only. The work may be arranged thus : — £. £■ 3300 X 12 = 39600 ^'s equivalent capital. 1700 X 12 = 20400^'$ ijoox 9 = 13500 C's 1950 X 8=i66ooZ''s . . . . . 89100 .-. A's share = -^''t'^' '^' ^ x 39600 =;g768. 15. o; 89100 .g's share = '^'^^^" '^' ^ x 20400 = &c. 89100 Ex. 9. A and B engage in trade, their capitals being in the ratio of 7 : II. At the end of 3 months A withdraws \ of his capital, and a month afterwards B adds twice as much as A has withdrawn. How should a profit of ;^337. 7. 6 be divided at the end of the year ? § 304- PARTNERSHIP. 283 If we represent yi's capital by 7, ^'s will be represented by 11 : the unit being a matter of indifference. At the end of 3 months A withdraws t,\, and therefore has remaining 4I for 9 months ; and at the end of 4 months B adds 2 X 2 J or 4f, and therefore has invested i sf for the next 8 months. Proceeding as in the last Example : — 7x3 = 4|X9= 42 [• = 63 /i's equivalent capital, 44 ) lS|x8 = "SiJ ^'^^=^^^^=.6,M^.. ., A', ,hare=^M7lllg >, 63=^ 337-375 X 189 232^ 697 =;^9i- 9- 7fi-988 —£9^- 9- 8 very nearly; and .".^'s share =£'^'i5- 17- 10 . . . Exercise 50. 1. Divide 28561 into parts proportional to i, i, 4, 7; and also into parts proportional to -^j J, -J, J. .i. Divide 15223 into parts which have the same ratio to one another as 6 7 9 fi s 'S' 'Bt TT7) lifi TT- 3. A, B and C contribute respectively to an undertaking ;^96, ;^I75 and ^^284, and they gain ;^I29. \os.: how shall this gain be divided between them? 4. Copper coins are composed of 95 parts copper 4 tin and i zinc. What weight of each metal would be required to coin copper coins of the value of ;£'iooo? ^pennies weigh i oz. Av. 5. In England gunpowder is compounded of 75 parts nitre 10 sulphur and IS charcoal; in France of 77 parts nitre 9 sulphur and 14 charcoal: if a ton of each gunpowder be mixed, what weight of each ingredient will there be in the mixture? 6. Sugar is composed of 49 '856 parts oxygen, 43-265 carbon and 6-879 hydrogen : how many lbs. of each of these materials is there in 11 cwts. 2 qrs. 12 lbs. of sugar? 7. A certain kind of brass is compounded of 325 parts copper 165 zinc 8 lead and 2 tin ; what weight of each metal is there in 43 kilog. 850 gr. of this brass? 284 PROPORTIONAL PARTS. Ex. 50. 8. Divide j^'sgj. 8. 6 into three parts which shall be to one another as 9. A person in his will directed that \ of his property should be given to A, ^ to ^, J to C, and \ \.o D: shew that this disposition cannot be carried out. If his property amount to ^471. 12. 6, dispose of it so that their shares may have to one another the relation he intended. 10. Divide a sovereign between A, B, and C, so that B may have one- third as much again as A, and C one-fourth as much again as B: — and again, so that B may have one-fourth as much again as A, and C may have two-thirds of what A and B have together. 11. The sum of £11%. 3. 6i is to be divided between A, B, C and D in such a way that for every £i given to A, B is to receive £i, C £8, and D £g. What sum did each receive? 12. A, B and C had each a cask of whiskey containing respectively 36, 54 and 78 gallons. They blended their whiskeys and then refilled their casks from the mixture : how much of the whiskeys of A and B axe con- tained in C's cask? 13. Four merchants A, B, C and D trading with a capital of ^23800, find after a certain time their respective shares increased by £26. 11. 8, ;^37- 4- 4> ;^53. 3- 4 and £6^. i6s. How much did they respectively subscribe to the original capital ? 14. A and B hold a pasture in common for which they pay ;^i3. los., A puts in 23 horses for 27 days, and ^21 horses for 39 days ; how much ought each to pay of the rent ? 15. B and J become partners, & bringing ;^5oo and J £},oa. At the end of 4 months B doubles his capital and a new partner R is introduced who brings ;^35o; and at the end of 6 months J trebles his capital. The year's profits amount to ;^7So: how ought it to be divided between them? * 16. A starts business with a capital of £1^0 on the morning of 19th March, and on the i6th July admits a partner B with a capital of £180. The profits at the end of the year amount to ;^94. 9>s. : what is each person's share of them? 17. A and B enter into partnership with capitals as 4 : 5. At the end of 3 months they withdraw respectively J and f of their capitals. When the year closes they find their profit to be ;^436. 9. 6 : how must it be divided between them? 18. A debt of £34.. 15. 10 is paid in crowns, shillings, and pence, whose numbers are as 4 : 7 : 10. Find the number of each coin. Ex. 50. PARTNERSHIP. 285 19. A and B rent a field for 21 guineas. A puts in 10 horses for i\ months, 30 oxen for 2 months and 100 sheep for 3J months; B 40 horses for 7.\ months, 50 oxen for ij months and 115 sheep for 3 months. If the food consumed in the same time by a horse, an ox and a sheep be in the ratio 3:2:1, what portion of the rent must each pay? 20. The sum of £,l(>i is to be divided between 10 men 32 women and 48 children : if each man's share is to be equal to the share of 2 women, and the 32 women are to have twice as much as the 48 children, how much will each woman receive? 21. The sum of ;^i77 is to be divided among 15 men 20 women and 30 children in such a manner that a man and a child together may receive as much as 2 women, and all the women together may receive £(>o. What will a man and a child receive? 22. Divide 23 cwt. i qr. 9 lbs. into two parts which shall be to each other as iij cu. feet is to 3^ cu. yards. 183 23. The sum of three fractions is — -\ and 22 times the first, 23 limes 242 the second and 24 times the third give equal products. Find the fractions. 24. Divide 32 gall. 3 qts. i\ pt. into four measures, so that the first shall be to the second as 9 to 14, the second to the third as 21 to 25, and the third to the fourth as 20 to 33. 25. I lb. of tea, of coffee, and of sugar together cost gj. SJaT. : find the price of each having given that 7 lbs. of tea cost as much as 16 lbs. of coflFee, and 3 lbs. of coffee as much as 11 lbs. of sugar. 26. The total amount paid in wages to 6 men 8 youths and 1 1 women is ^^23. 9J.' How much does each person receive, when for every half- crown earned by a man a woman earns is. gj " .. - y- ,, 33\\ 3 .. ^-f- 9'1- ^^ 27 /I J. .. xs. id. " ^^ 37/ ) I „ 3-f- i<^- 3oN 3 lbs. at 2S. 6d. 33 N ^ 6 „ 2J. gd. iiy ' 4 .. 3'- i^- 40 y ' I ., 3^- 4'^- 288 ALLIGATION. §308- 308. We deduce then the following Rule : — Write the given prices under one another in order, and to the left write the mean price. Link all the prices, so that one under and one above the mean price shall always go together: and put against each price the difference between the price with which it is linked and the mean price j — these differences, or any equimulti- ples o/them, will give the quantities required. Remark. It is from this process of linking that the name Alligation is derived. 309. Sometimes it is required to make the mixture, so that there shall be a given quantity of one kind : — Thus in !Ex. 3 suppose everything else the same, but that we must take 20 lbs. of the tea at 2J. dd. Now the first solution shews that a mixture can be made at the required price, by taking the following quantities in order, or any"" equimultiples of them — 3 lbs., 6 lbs., 4 lbs., I lb. multiplying then each of them by *j*, we get 20 lbs., 40 lbs., 26flbs. 6|lbs. Sometimes, again, the quantity to be contained in the whole mix- ture is given : thus in Ex. 3, suppose everything else the same, but that the whole mixture must weigh 20 lbs. Now the first solution gives a whole mixture of 14 lbs. : if therefore we multiply each of the quantities composing it by f£ or ^, their sum will be 20 lbs. Exercise 51. I. A grocer mixes 47 lbs. of tea at ^s. i|(f. a lb., 25 lbs. at is. i,d. a lb., and 20 lbs. at is. io\d. a lb., what is the price of i lb. of the mixture? If he had also .added 8 lbs. of sloe-leaves at o,\d. a lb., what thea would be the price ? ■!,. A goldsmith melts together 27 oz. of standard gold (309), 19 oz. of gold 15 carats fine, 20 oz. iSj carats fine, and 3I oz. of copper: what is the fineness of the mixture ? Pure gold is 24 carats fine; hence standard gold is 22 carats fine. 3. A farmer buys wheat at 39^. per quarter and some of a superior quality at 6s. per bushel : in what proportion must he mix the two so as to sell the mixture at 46^. per quarter ? Ex. SI. ALLIGATION. 289 4. A person wishes to melt equal quantities of gold 943 and 827 milliemes fine, with alloy, so as to get a gold 467 milliemes fine ; what quantities of each must he take? 943 milliemes fine means that 943 parts out of 1000 are fine. 5. A person buys some tea at (>s. per lb. and some at \s. per lb. In what proportion must he mix them so that by selling his tea at ^s. 30?. per lb., one-sixth of his receipts may be clear profit? 6. Mix spirit at %s., wine at 7J. and cider at \s. a gallon with water, so that the mixture may be worth gf. a gallon. 7. It is required to mix teas at 2j. loJoT., 2j. 6d., and is. ^d. a lb., with sloe-leaves at ^d. a lb., so that the mixture being sold at is. lod. a lb., one- fourth of the receipts may be clear profit. 8. How much gold at £^. 5J-. per oz., silver at i,s. an oz. and copper considered as of no value comparatively, may be melted together that the compound may be worth ;^2. 15J. per oz. ? 9. We wish to mix wines at 35 c. and at 55 c. the litre to form a mix- ture at 42 c. the litre : how much of the second wine must we take for 182 litres of the first? How much must we take of each kind to have a mixture of 640 litres ? 10. A grocer having four sorts of tea at js. gd., 2s., 2s. 6d. and is. gd. a lb. would have a mixture of 81 lbs. at 2S. ^d. a lb. What quantity must he take of each sort? 11. A person bought 80 lbs. of sugar of two different sorts for ;^i. 5. 4. The better sort cost ^d. per lb., and the worse ■i,\d. per lb. Find how many lbs. there were of each sort. 12. A greengrocer sells potatoes at is. Sd., is. iid. and 3^. ^d. a bushel : what quantities of each kind must he sell that the average price obtained shall be y. a bushel? I3> What quantities of coffee at is. -j^d. a lb. and of chicory at 5|i/. alb. must a person take to make a mixture of 33 lbs. worth is. i\d. a lb. ? 14. The specific gravity of lead is ii'^n, of cork "24 and of fir '45: how much cork must be added to 60 lbs. of lead that the united mass may weigh as much as an equal bulk of fir ? 15. A silversmith gave ;^48. 10. 10 for 16 lbs. 8 oz. of silver, giving 5J. i^d. an oz. for one part and 4^. ^\d. for the rest : how many oz. of each kind did he buy? 16. The price of gold is £■}. 17. loj per oz.; a composition of gold and silver weighing 18 lbs. is worth £6iT. "js., but if the proportions of gold and silver were intefchanged it would be worth only £iSg. is. Find the proportion of gold and silver in the composition, and the price of silver per oz. B.-S. A. 1 9 CHAPTER XIII. PER-CENTAGE. 310. Per centum or per cent, means for a hundred: thus, S per cent, means 5 for a hundred. 311. Public companies, bankers, merchants, &c., regulate their transactions, and calculate their profits and losses with reference to 100 as a standard. In this way they can readily accommodate their dealings to the varying circumstances connected with them, and at once compare the results of their several undertakings. Thus, suppose a company with a capital of ;£ 50000 makes a profit of ;^4i25, and another with a capital of ;^325oo makes a profit of .^2637. 10s., it is not easy to see which of the two is the more prosperous; but reducing them to a common standard of £100, the profit of the first is ^8 J and of the second £7^. 312. Tables recording the increase or decrease of population, the number of persons engaged in trade, agriculture, under educa- tion, &c. are for the same reason constructed with reference to 100. Suppose the population of a town has increased during the last 10 years from 34575 to 37341 ; the increase is 2766 on 34575, or 8 on 100, or 8 per cent. 313. Again, when a quantity is made up of several parts, it is usual to consider the whole quantity as composed of 100 units (of weight, measure, value, &c.), and then each of the parts will con- tain a certain number of these units ; that is, a certain per cent, of the given quantity. Thus, 15 per cent, of gunpowder is charcoal; that is, if a mass of gunpowder contain 100 units of weight, 15 of these units, when the mass is decomposed, will be charcoal. § 314- PER-CENTAGE. 29I 314. The 100 we refer to may be £\oo, 100 persons, 100 oz., &c. and the number giving the per cent, is so many units of the same kind ; but since ;^ioo : ;^5 = icx) persons : 5 persons = 100 oz. : 5 oz,, = 100 : 5, we usually consider the 100 as an abstract number, and there- fore the number giving the per cent, will also be an abstract number ; but if necessary we can at once pass to any corresponding concrete numbers. 315. Every question in Per-Centage can be readily solved by Rule of Three, and the rules given below can be immediately de- duced from the Rule of Three statement. 316. Case I. To find the value of a given per cent, of a given quantity. 5 per cent, of a quantity means S parts for a hundred of the quantity, and is therefore equal to ;|- of the quantity. In like ^^^ 7 12 manner 7 per cent, or 12 per cent, of a quantity is —or— of the quantity. Hence we have this Rule (128), — Multiply the quantity by the number expressing the rate per cent, and divide by 100. Ex. I. A rent collector receives ;^I365. 17. 6, and he retains 3^ per cent, for his trouble: how much does his per-centage amount to ? £>■ 136s '875 3 I- s. d. 136s • 17 . 6 3 4097 . 12 . 6 682 . 18 • 9 47,80 . 11 • 3 20 16^1 12 1,35 4 1,40 4097-625 68^ "9375 47-805625 ■. Per-Centage = ;^47. 16. ii|. 19—2 292 PER-CENTAGE. § 3>5- Ex. 2. How many persons are engaged in agriculture, when they constitute 17 per cent, of a population of 6457312 .? No. reqd = — x 6457312 = 1097743-04= 1097743. 100 317. Case II. To find how much per cent, one quantity is of another quantity. For example, — How much per cent, is 19 parts out of 45? that is, how many parts must be taken out of 100, when 19 are taken out of 45 ? hence 45 • 19=100 : per cent, reqd; .•. per cent. reqd=— x 100. Again, — What per cent, will a profit of .£4375 give on a capital of ^50000? that is, what will be the profit on 100 when ^4375 is the profit on ;^5oooo ? hence ;£'5oooo : ;^437S=ioo : per cent, reqd ; .-. per cent. reqd = — ^^ x 100. 50000 We may therefore adopt this Rule, — Find what fraction the first quantity is of the second, and multiply by 100. Ex. 3. How much per cent, is 9^. in the pound ? * Since £1 = 2401/. ; .'. per cent. reqd = -^x 100= 3J. Ex. 4. The population of a town has increased during the last 10 years from 38851 to 44565 : find the increase per cent. Since 44565-38851 = 5714, there is an increase of 5714 persons on 38851 persons ; , S714 38851)571400(14-707 . . per cent. reqd=-^5 — x 100 182890 SisoSi 27486 = i47o7- ^?I 318. Case III. The per-centage on a quantity being given and the rate per cent., to find the quantity. § 3i8. PER-CENTAGE. .293 Suppose the rate per cent, to be 4f ; that is, suppose the per- centage on 100 to be 4j, then 4f : 100 = given per-centage : quantity reqd ; J 100 .•. quantity reqd = — j x given per-centage. Hence this Rule, — Divide 100 6y the rate per cent., and multiply the quotient by the given per-centage. Ex. 5. The profits of a Company during the past year amount to ;f2457. II. 3, which will just allow of a dividend of 3f per cent. ; find the amount of the Company's capital. Since the dividend is equal to 3f per cent, of the capital, .•. Company's capital =^£'2 45 y-jj x — 5- 16 15 =;f6653S. Exercise 52. I. How much is i2f per cent, on ;^668. 6. 8? and how much is \\ per cent, on the result? 1. Find 5 per cent, on £,611. 13. 9; and from the result deduce 4! per cent, on the same sum. 3. The population of York in i85i was 40433, and it increased 8'32 per cent, between 1861 and 1871; find the population in 1871. 4. The population of the City of London decreased 33T1 per cent, between 1861 and 1871 : in 1861 it was 113387; find what it was in 1871. 5. The population of Huddersfield increased ioi'43 per cent, between 1861 and 1871; in 1871 it was 70253, find what it was in 1861. 6. The population of a town increased 35 per cent, between 1851 and 1861, and 19 per cent, between 1861 and 1871; the population in 1871 was 93177, find the population in 1851. 7. A bought goods to the value of £,},^i. 15J. and sold them to 5 at a gain of IS per cent, on his outlay, and B sold them to C at a loss of 15 per cent, on his outlay; how much did C give for them? 8. A sells goods to ^ at a gain of 22^ per cent., and B sells the same goods to C at a gain of 7 J per cent.: C gave ;^263. 7. 6 for the goods, how much did A give for them? 294 PER-CENTAGE. Ex. 51. 9. How much per cent, is i part out of 8; 7 parts out of 24^; 37 of 75; 43iof 165; 869! of 4655? 10. What rate per cent, is equivalent to 6d. in the^^'i ; 3J. ioJi^i. 12. 6, on which he re- ceives a commission of jf per cent. , while his office and other expenses amount to 22 J per cent, of his commission. How much clear profit does he make, and how much does he remit to his principal? 9. A broker at the public sales buys 5 chests of indigo weighing 18 cwts. 3 qrs. 22 lbs. nett, at JJ. lod. a lb. ; find the brokerage at \ per cent. 10. A person at the age of 45 insures his life in each of two offices for £,'i?>io% the premiums being at the rate of ;^3. 19. 10 and £,'^. 14. 7 P^'' cent, respectively. Find his annual payment. 11. At what rate per cent, is disco.unt allowed when a tradesman de- ducts £^. o. 9J from a bill of ;^89. 15. 10? 12. A tradesman insures his warehouse and goods for ;^r25oo ; f of this sum being for his warehouse and f for his goods. The warehouse is insured at I2.r. 6ii. and the goods at 24J. 6i/. per cent.: find the amount of his premium. 13. For what sum must a merchant insure a cargo worth £iii'j. 8. 6 at 3f per cent., so that in case of loss both cargo and premium may be covered? 14. A ship worth ;£^I5325 is to be insured, so that its value and all the expenses connected with its insurance may be covered. The premium is 2J guineas per cent., policy duty 4^-. per cent., and brokerage J per cent. ; what is the amount of the whole expense paid on insurance? 1 5. What sum must be paid on the insurance of a cargo of the value of £m7- lo- 6 so that in case of loss the cargo and all expenses of insurance may be recovered ? The premium is at the rate of 4^ guineas per cent. , policy 4^. per cent., and agent's commission J per cent. m. PROFIT AND LOSS. 324- Under the head of Profit and Loss, we estimate a profit or a loss not absolutely, but in relation to the cost price. If one article costs is. and is sold for 6j., and another article costs \os. and is sold for i is., the absolute gain is in both cases the same, but relatively to the cost price the first gain is double of the second. 325. Men of business adopt 100 as a standard cost price, and reduce the gain or loss on a particular cost price to the correspond- 300 PROFIT AND LOSS. § 325. ing gain or loss on loo; that is, to a gain or loss of so much per cent. This may be effected by the statement (291). Cost price : gain or loss thereon = 100 : gain or loss per cent. (A.) Again, when the cost price is represented by 100, the selling price is represented by loo+gain per cent., or 100— loss per cent., according as a gain or loss has been made, and we have the statement „ ^ . ,,. . ( 100 + gain per cent., or ,_. Cost price : sellmg price =100 : { f (B). '^ ° '^ ( 100— loss per cent. ^ ' Lastly, if an article be sold at two different prices, these selling prices will be to one another as their representative selling prices ; for from (B) we have (193) 1st selling price : cost price = istrepve selling price : 100, and cost price : 2nd selling price =100 : 2nd repve selling price; and, compounding these statements (194), we have 1st selling price : 2nd selling price (C). = 1st repve selling price : 2nd repve selling price. 326. But although questions in Profit and Loss can always thus be solved by Rule of Three, yet it is often useful to remember that since a gain of 15 per cent, means a gain of 15 on 100, where 100 represents the cost price, it is a "gain of — - of the cost price. And in like manner a loss of 9 per cent, means a loss of -^ of 100 cost price. Ex. I. Tea is bought at 3J. dd. a lb. and sold at y. \d^d. What is the gain per cent. ? The gain on 3^. 6d. is ^\d., hence (A) 3J. dd. i ^\d. = 100 : gain per cent. ; .•. gain per cent. = — x 100 = ^ x 100 42 84 = iof. § 326. PROFIT AND LOSS. 30I Ex. 2. If cloth be bought at lbs. M. a yard, and sold at a loss of 12J per cent., what price did it fetch? A loss of nj per cent, means that if the cost price be represented by 100 the selling price will be represented by 100 - 12 J or by 87^, therefore (B) 100 : 874= i6j. 8a?. : selling price; .•. selling price = -^ x 16^. %d.=!-y. i6j. id. 100 — i\s. "jd. Or we may proceed thus : — s. d. 16 . 8 = cost price, 12 J per cent. =1^ | 1 . i= loss, 14 . 7 = selling price. Ex. 3. If a cwt. of sugar cost £^. 6. 8, at what price per lb. ought it to be retailed, to gain 1 5 per cent. ? From (B) 100 : iiS=;^2. 6. 8 : selling price per cwt. ; .•. selling price per cwt,= — -x. £-=£— x - 100 3 20 3 ■■ i6r i6t 4 .•. selling price per lb. = d, Ex. 4. By selling a watch for ^34. los. there is a loss of 8 per cent. ; what will be the loss or gain per cent, by selling it for £38? The two selling prices are £34.. tos. and ;^38, and the first representative selling price is 100 - 8 or 92, therefore (C) £34. 10 : ;^38=92 : 2nd repve selling price ; a8 76 + 304 .-. 2nd repve selhng price= ^ x 92=^x92==^—!: 343 °3 3 = 101^; 2 . •. Gain per cent.= t^. 302 PROFIT AND LOSS. § 3*6- Ex. 5- A clockmaker by selling a clock for ^\. I2. 6 loses ^\ per cent. ; at what price should he have sold it to gain 6^ per cent. ? The representative selling prices are too-7j and 100 + 6J, or 92 J and 106J, hence (C) 92^ : \o(i\=£i,. 12. 6 : selling price reqd ; 8S .-. selling price reqd = ^ x Al=;^^ x ^=£--^ =£5tts *e Ex. 6. If 5^ per cent, be gained by selling butter at £i. 5. 6 per cwt., how much per cent, will be gained by selling it at I J. 302i ii 600 512-8333 4 1 4 i i 1 2051-3333 256-4166 64-1041 2371-8540 3 6 mo. I mo. 16 da. 6 da. 7115-5620 1185-9270 197-6545 98-8272 39-5.309 86-37 Soi6=;^86. 7- 6 , Interest =;^86. 7. 6^^' 6200 310 SIMPLE INTEREST. §331. Ex. 3. What is the simple interest oi £■^^0. 17. 6 at 4J per cent, for 17 weeks ? See Remark (4). £. 320-875 4_ 1283-500 \ I i6q"4375 I443-9375 17 400 ) 24S.46"9375 13 ) 6i'3673 4'7205=;^4. 14. 4|. Ex. 4. Find the simple interest of £\2o. 13. 8 at 4J per cent, from 19th Mar. 1868 to 8th Sept. 1870. See Remarks (5) and (6). Mar. 12 April 30 The day on which the money is lent and the day on June 30 which it is paid make only one day: hence we reckon July 31 only 12 days in Mar. but 8 in Sept., giving 173 days: Aug. 31 therefore the whole time is 2 years 173 days, or 903 Se pt- 8 days. ' years_73o 420-6833 903 9 3786-1497 9°3 113584491 3418893-1791. ..Rk. (6) 340753473 1139631 73,000 ) 3418,893-1791 ( 46-8341 113963 498 11396 608 ^6.83883 '1^ 468 Deduct xrJwth 'i?i ;^46-834i .-. interest =;£'46. 16. 8 J. Exercise 55. Find the simple interest and amount of >■• £m°- 12. 6 for 4 years at 3 per cent. ■i. £g6. 1$. i ... 8 years at 4 per cent. 3- AJ1057. 7. 6 ... 9 years at 5 per cent. 4- ;^450. lo. o ... 7 years at 3 J per cent. 5- £T5- ^- S ... I year at 2| per cent. 6. £33- 13- 4 7- ;^SS- i6- 8 8. £654. i. 6 9- £lh- 8- 9 lO. ^1725. 18. 6 II. £416. 18. 6 12. ;f385- 9- 10 13- ^543- 17- 6 14. ;^345o. 12. 7 '5- A5- 15- 6 Ex.55. SIMPLE INTEREST. 31I ibr 15 years at 4|- per cent. .. I year at 3J per cent. ... 4 J years at 4^ per cent. . . . f year at 3 J per cent. ... 3^ years at 4J per cent. . . . 4f years at 3^ per cent. ... \ year at 5J per cent. . ! . 2 J years %\ per cent. ... 8 J years at 4^ per cent. ... if years at 2| per cent 16. £'j(>4. 18. 9 at 4f per cent, for i year 219 days. 17. £'iyT- 10. 2j at 3I per cent, for 3 years 73 days. 18. £i'}- 16. 9 at ;^3. 12. 6 per cent, for 4 years 7 months. 19. 1000 guineas at 4I per cent, for i year 5 months. 20. ;^46o. 3. 6 at 4! per cent, for 3 years 8| months. 2i- £hi°- '4- 8 at 4! per cent, for 2 years 9 months 25 days. 22. ;^526o. 10. 2 at 2j per cent, for 6 years 5 months 21 days. 23. 5004 guineas at i\ per cent, for i year 7 months 18 days. 24. .^^386. 14. 4I at 5I per cent, for 5 months 16 days. 25. £i'(>6. 13. 4 at 3^ per cent, for 19 weeks. 26. ^^987. 15. 8i at £\. 13. 4 per cent, for 37 weeks. 27. ;^228. II. si at 4i per cent, for i year 23 weeks. 28. £78";. o. 74 at 4J per cent, for 92 days. 29. ;fiio8. 13. 9 at 5 J per cent, for 191 days. 30. ;^i84. 3. 9 at 5J per cent, from July 17th to Dec. 5th. 31- ;^853. o. 10 at 3J per cent, from June i8th to Sept. T5th. 32. .!^3o57. 14. 7 at 4J per cent, from April 6th to Oct. 27tli. 33. ;^I253. 8. 5 at 3f per cent, from Jan. i6th to Mar. 23rd (leap year). 34. £473. 3. 6 at 3^ per cent, from April 14th to July 6th. 35- £^6i- 'S- II at 5i per cent, from 9 Nov. 1867, to 3 Mar. 1868. 36. ;if 1327. 3. 8 at 54 per cent, from Oct. i8th, 1869, to May 27th, 1871. 332. Case II. Having given the interest or amount, rate j)er cent., and time, to find the principal. 1st. Let the interest be given. Find the interest of ^100 at the given rate per cent, for the given time ; then, since the interest is directly proportional to the principal producing it, we say this interest : given interest =;^ 100 : principal reqd. 312 SIMPLE INTEREST. §332. 2nd. Let the amount be given. Find the amount of ;£loo at the given rate per cent, for the given time ; then, since the amount is directly proportional to the principal (330), we say this amount : given amount = ;£ 100 : principal reqd. Ex. I. What sum of money must be lent, that its interest may come to ^29. 6. 8 at 2^ per cent, in 2 years 7 months ? £. s. d. 2 • 10 , . o=int. on ;^ioo for i year. 2 .s • . 6 mo. * I . 5 . I mo. k. '0 . 4 . 2 2=int. on ;f 100 for 2 years 7 months; •'• £6' 9. 2 : ^^29. 6. 8.=;^ioo : principal reqd; .-. principal reqd=jfioo X ^=£^^4. 3. loiU- Ex. 2. What principal will amount to £45^. 16. 8 in 2 years 8 months at 4| per cent. ? Int. of ;^ioo for 2yrs. 8mo.=;^4f x 2|=j^i2|; .•. amount =£ji2%; and .-. ;^ii2f : ;^4S2-|=;^ioo : principal reqd; .-. principal reqd=^^ x ;fioo=;f ^*=;^40i. 18. sJ A- Ex. 3. What sum must have been lent to have amounted to ;£465. 13. 10 in 127 days at 4} per cent.? £■ 950 127 9501s 100 X double the rate p. c. The interest of 6650 ;f 100 is r'65273, and therefore we have '^4°° iof65273 : 465'69ifi=;^ioo : sum lent. 120650 40216 4021 402 165289 Deduct... 16 101-65273 ) 46569-166 (48s-i20=;£'438. 2. 5. 5908 074 825 437 12 215 2050 17 §333- SIMPLE INTEREST. 3l3 333. Case III. Having given the principal, time, interest or amount, to find the rate per cent. Find the interest on the given principal for the given time at I per cent. ; then, since the rate per cent, is directly proportional to the interest, say this interest : given interest =1 : rate per cent, reqd ; that is,— rate per cent, required is found by dividing the given interest by the interest at i per cent. Ex. I. At what rate per cent, will ^lifl. 10s. amount to ;^i63. 13. I li in 4j years ? Given Int. =;^i63. 13. irj-;^i42. io=j£'2i. 3. ni. Int. on;^i42. 10 at i per cent, for 4J years=;^6. i. ij; .-. Rate per cent. = ^"" ^" "* = 34- ^ £6. I. i4 Ex. 2. At what rate per cent, will the interest on £34S- iS^- become ^192. 17. 6 in 8} years? £■ 34575 9 311175 i I 8643 30-2532 ) 192-8750 (6-37535 .-.Rate per cent. = 6-37535 113558 =6f very nearly. 2 2798 1621 108 17 2 Ex. 3. At what rate per cent, will a sum of money double itself in 12^ years? In I2i years the interest is equal to the principal, .-. interest on ;£'ioo for I2i years =;£'ioo ; but interest on ;£'ioo at i per cent, for 12J years = ;^i 24, .•. rate per cent. = -> — -= =8. 334. Case IV. Having given the principal, rate per cent., and interest or amount, to find the time. 314 SIMPLE INTEREST. §334- Find the interest on the given principal for one year ; then, since the time is directly proportional to the interest, say one year's int. : given int. = i : no. of years reqd ; that is, — the number of years is {aw\Aby dividing the given interest by the interest for i year. If it be manifest that the time is less than a year, then, as before, — the number oi days is found by dividing the given interest by the interest for i day. Ex. I. In what time will £i,i^ amount to ^635. 7. 6 at 5J per cent.? Given interest =;^63 5. 7. 6-;^42S=:2io. 7. 6. Int. on ;^425 for i year= '''^^ ^ ^^ ; 100 nT r ;if2Iof X 100 .■.No.ofyears=^^^-^=9. Ex. 2. In what time will a sum of money treble Itself at 8 per cent.? The time will be the same whatever sum of money be taken as the principal : suppose, then, the principal to be ;^ioo, .•. given interest = 2 x principal =;^20o, and interest on ;^ioo for i year=;f8; ■M c £'^°° .-. No. of years = ^==2^= 2 J. Ex. 3. In how many days will the interest on .£498. 16. 8 amount to ^10. 9. 3J at 6| per cent. ? 498-833 (331, Rk. 6) 12 598599S 124708 61IO7 2036 "9 2037 20 '4 ■°83>7>i ) lOH^S ( 125 2092 418 o .'. No. days is 125. Ex. 56. SIMPLE INTEREST. 3 1 S Exercise 56. What principal will amount to I. ;^762. 16. 4! in 5 years at 34 per cent.? 2' ;^235. 13. 4 in if years at 3J per cent.? 3- ;^843. 13. 6 in 15J years at 2| per cent.? What principal will produce 4- £,i^- 12. 6 interest in 2 J years at 3^^ per cent.? 5- ;^23. 18. 10 interest in i\ years at /^\ per cent.? 6- ;f 13- IS- 9i interest in f year at 38 per cent.? What principal will amount to 7- ;f 1357- 14- 3 in 2 years 7 months at 4J per cent.? i^- ;^7^S- 12. 6 in 2 years 9 months 18 days at t\ per cent.? 9. Find the principal whose interest amounts to £,t,1. 16. 8 in i year 9 months 24 days at 3! per cent.? 10. Find to the nearest penny the sum that must be invested at 3! per cent, for 21 years to amount to ;^iooo. What principal will amount to II- £,%^^- 13- 4 at 3^^ per cent, in 240 days? 12. ;^S8. 4. 6 at 5j per cent, from June i6th to Nov. 7th? 13- ;^73- 5- o a-t 4I per cent, from April 22nd to July asth? 14- jf5^53- 8. 9 at 44 per cent, from Feb. i8th to Sept. i6th? At what rate per cent, will 15- ;^i70. 6. 3 amount to £,1^0. 15J. in 3 years? i6- ;^33- 6- 8 amount to ;£^38. 4. 2 in 4J years? 17- ;^i36- 17- 6 amount to ;^i64. 5. o in 6J years? At what rate per cent, will the interest on 18. £1%. 15- o amount \^o£,^. o. 5J in 4^ years? 19. 500 guineas amount to ;^i03. 9. i,\ in 3 years 7 months? 20. ;^239. 16. 6 amount to ;^22. 13. o in 2 years 10 months? 21- £lia- 12. 6 amount to ;^I58. 14. 6J in 5 years 7 months 20 days? 22. The interest of a sum of money at the end of 6J years is three- eighths of the sum itself; what rate per cent, was charged? 23. What must be the rate per cent, that the interest at the end of 16 years 8 months may be equal to seven-eighths of the sum lent? 24- £,T- 19- 6 was charged for the loan of £,1^},. 10. o for 87 days; what was the rate per cent. ? 3l6 SIMPLE INTERKST. Ex. 56. 25. What is the rate per cent, when the interest on ;^i85. 14J. for 125 days amounts to ;^3. o. 5 ? 26. At what rate per cent, will the interest on ;^r368. 15J. become ;f 14. 4. 7J from July 5th to Nov. 20th? V}. In how many years will ;^340. 12, 6 amount to £'i%\- \os. at 4 p.c? 28. In how many years will the interest on £,},i. lis. amount to £,A- °- si ^' 2 J per cent.? 29. In how many years will ;^i45i. 6. 6 amount to £1661. 4. 2J at 4J per cent. ? 30. In how many years and months will the interest on 1000 guineas amount tO;^i03. 9. 4^ at 2j per cent.? 31. In how many years will ;i£'286o. 16. 9f amount to ;^3S29. 11. 3 at 5 J per cent. ? 32. In how many years, months and days will ;£'22Si. 17. 6 amount to ^^2728. 1. of at 3J per cent.? 33. In how many years will a sum of money amount to half as much again as itself at 7^ per cent.? 34. In how many years will a sum of money double itself at 6J p. c. ? 35. In how many days will the interest on ;^243. 6. 8 amount to £2. o. 5 at 3^ per cent.? 36. In how many days will £556. 17. 6 amount to £$6^. 18. 9 at 4f per cent.? 37. In how many days will the interest on £^11. 15J. amount to £3. 19. 9 at 4J per cent.? 38. On Jan. ist, 1870, a person borrowed ;£'4835 at 3| per cent., pro- mising to return it as soon as it amounted to ;£^60oo: on what day did the loan expire? 39. A sum of money amounts in 10 years at 3^ per cent, simple interest to £i°^- 15' 'i; ™ ho'w many years will it amount to ^703. 16. 6|? 40. A person lent another a sum of money for 72 days at 3 per cent, per annum. A' fie end of that time he received ;^293. 12. oj; what was the suni lent ? 41^ The sum of £32j is borrowed at the beginning of the year at a cer- tain rate of interest, and after 9 months ;£'400 more is borrowed at double the previous rate. At the end of the year the interest on both loans is ;^I3. 3. 6. What is the rate of interest at which the first sum was borrowed? §33Sr PRESENT WORTH AND DISCOUNT. ll^ 42. What sum of money laid out at 4 per cent, will give \d. interest a day? and what sum at 3 J per cent, will give a guinea interest per day? 43. The simple interest on ;^58i2. loj-. for "3 of a year is £'&^. 3. 9 ; find the interest on £,7,, | ^ ^ 2624 xif 4 525 8 = 4f- 320 PRESENT WORTH AND DISCOUNT. Ex. 57. Exercise 57. Find the present worth of I. ^9^6. 10. odue 2 years hence at 4^^ per cent. ■■=• ;^6o5. 10. 6 ... 3 years £,\. if,5. ... 3. ;^io79. 2. 5 ... lyr. 6mo 5 4- ;f 132. S'f- ... ijyears 4,\ 5. ;^748. II. 8 ... 3years 5 6. ^1250. lOJ. ... 2i years 34 7. 25 guineas ... 18 months ;^3. 12. 6... 8- .^245-i3-4 -■ 34 years 44 9. ;^46. 16. 8 ... 9 months 3I 10. ;£'843. 12. 6 ... I54years i\ 11. ;£'i243. 2. 6 ... 3 yrs. 5 mo £l-'i.(>- ■■■ 12. ;fii44. 8. I ... 4 yrs. 90 days 2J Find the discount of 13. ;^4i20. 8. 7 due 9 months hence at 4 per cent. 14. ;^i2382. 4J. ... 4 months 3i 15- jf55447 ••• 24 years 44 16. ;^52o. 17.6 ... 34 years 44 17. ;^46i. 15. io4 ... 3 months 74 i8- ;^7S3-". 6 ... 5 months 3! 19- .^450 ...2 yrs. 9 mo 44 20. 1000 guineas ... i yr. 115 days 34 21. jS^^Ji. 12. 6 ... 3 yrs. 9 mo. 18 da. ... 6J 22. ;^376. 10. 6 ... 60 days 6j 23- ;^8765. 18. 9 ...272 days 4I 24. £fi()i. \os. ... 55 days 4! 25- ;^3245- iS'f- ... 136 days 5i 26. What is the present value of ;^i due i year hence at i per cent.? 27. If the present worth of ;£'328. 13. 5 due 3 months hence be £i2$. 8. 4, what rate per cent, is allowed? 28. If the discount of .jf 13735 at 3J per cent, be ;^335, how long was the sum paid before it was due? 29. On what sum of money due at the end of i year and 4 months does the discount at 4! per cent, amount to £48. gs.} Ex. 57. PRESENT WORTH AND DISCOUNT. 321 30. A tradfisman on being paid ready money deducts igj. sJif. from an account of £,10, 5. 61 due at the end of 12 months; what rate of interest does he allow ? 31. A offers for an estate ;^378oo, and B offers ;^'4S400 to be paid at the end of 4 years. Which is now the better offer and by how much, allowing J per cent, interest? 32. Find the difference between the discount on .£'196. 4. i,\ due 6 months hence at 8 per cent., and the interest on the same sum for the same time at the same rate. 33. If the discount on £'j'i. g. 9 due 8 months hence be ;^3. o. 4^, at what rate per cent, is the discount calculated? 34. The discount on a sum of money due 3J years hence at sf per cent. is;^i6. 14. 9; find the sum. 35. How many years hence is ;^589. 6. 3 due, when its present value at 3i per cent, is 500 guineas? 36. Find the difference between the amount of ^£'494. 10 for 2 years, and the present worth of the same sum due at the end of 2 years, at 3! per cent. 37. Find the difference between the interest and discount oi£i 114. 1 1. 8, the time being if years, and the rate 4 per cent, per annum. 38. Find the discount on ;^i7o. 18. 5 due 52 days hence at ^yi. per cent, per day. 39. A farmer buys 75 sheep for ;^I20 payable at the end of a twelve- month, and the same day sell's them at 35J. a head ready money ; what did he gain by the transaction, reckoning interest at 5 per cent, per annum? 40. Find the difference between the interest on ^246. 13. 4 for 2f years at 5 J per cent., and the discount on ^i.'&i. 19. 6 due i% years hence at the same rate. Explain the result (336, 2). 41. If a person's salary be paid at the beginning instead of at the end of the month, what part of the month's salary ought to be abated, reckon- ing 4J per cent, per annum? 42. A tradesman marks his goods with two prices, one for ready money and the other for credit of 6 months: what ratio should the two prices bear to each other, allowing interest at 7J per cent, per annum? If the credit price of an article be ^33. 4^'., what is the cash price? 43. The discount on ^i.'Ji for a certain length of time is £'i.i; what is the discount on the same sum (i) for twice that length of time, and (2) for half that length of time? 44. The interest on £1^1. los. for a certain time is .^34. 7. 6 ; find the discount on the same sum for the same time. B-S. A. 21 322 PRESENT WORTB AND DISCOUNT. Ex. 57. 45. If the discount on a sum of money due at the end of 8 months at 6f per cent, be ;£'43. 15. gj; find the present worth of the sum. 46. The interest on a certain sum of money for 1 years is £'}\- 16. jj, and the discount for the same time is £(>},. ifs. Find the rate per cent, per annum and the sum (337, 3). 47. The difference between the interest and the discount on a certain sum of money at 4f per cent, for 2J years is j£2. 12. 7S; find the discount on the sum (337), and the sum itself. 48. A man bought a horse for 30 guineas and sold him immediately for ;^38. 10s. payable at the end of 6 months. If the use of the money be reckoned at 6 J per cent, per annum, what is now his gain per cent.? 49. I purchase a piece of land for ;^3500 and sell it the same day for 4000 guineas, to be paid in two equal instalments at the end of 3 and 6 months respectively: how much do I make by my bargain, the use of money being worth 6 per cent.? 50. A person's salary of 1000 guineas is paid in four quarterly payijients at the end of each quarter : what sum at the beginning of the year is equi- valent to these quarterly payments, reckoning interest at S per cent.? 51. What sum must be paid now in order that a person may receive ;^2So at the end of every year for the next three years, the rate of interest being 3^ per cent.? 52. ^125 is due at the end of 3 months and ;£'90 at the end of 7 months; what sum at the present time is equivalent to both these sums, calculating interest at 4^ per cent.? In what time will the result amount to ;^I25 +;f 90 at the same rate of interest? DISCOUNTING BILLS. 340. A bill of exchange is a written instrument in which one person orders another to pay to him, or to some other person, a sum of money at a specified time. Thus : ;£'5oo. London, 1st January, 1881. Two months after date pay. C. .D. or order Five hundred pounds, value received. £ *^ To E. F., §" A. B. o Park Street, Liverpool. -< Here E. F. engages to pay to C. D. or his order £^00 at the end of two months from ist Jan. 1881. §340. DISCOUNTING BILLS. 323 A promissory note, or note of hand, is a written instrument in which one person promises to pay another a sum of money at a specified time. Thus : £^0Q. London, ist fanuary, 1881. Three months after date, I promise to pay C. D. or order Six hundred pounds, value received. A. B. Here A. B. engages to pay to C. D. or to his order ^600 at the end of 3 months from ist Jan. 1881. 341. A bill of exchange or a promissory note always runs 3 days beyond the time specified, and these three days are called days of grace. Thus a bill drawn on 1st Jan. at 2 months is nominally due on 1st March, but really on 4th March. Moreover, calendar months are always reckoned, so that a bill at 3 months, whether drawn on 30th or 31st Jan., is nominally due on the 30th April, and really on 3rd May. If now the holder of a bill wishes to realize it, he presents it to a banker or bill-discounter, and if the banker or bill-discounter be satisfied of the credit of the parties to the bill, he discounts it ; that is, he pays the sum specified on the bill, deducting discount for the time it has still to run. But with bankers and bill-discounters, dis- count is the interest oi'CeA sum specified, whereas, properly speaking, it is the interest of the present worth of that sum (336, 2). And as the present worth of a sum due at a future time is less than the sum itself, the true discount is less than the banker's or mercantile discount ; and therefore the banker obtains a small advantage. Ex. A bill for ;^343. ij. 6 is drawn on 15th April at 4 months, and discounted on 8th May at 4j per cent. ; how much did the holder receive? (See Art. 331, Remarks S & 6.) May 23 xhe bill is really due on 18 Aug., and therefore when it June 30 ^,^g discounted it had to run from 8 May to 18 Aug., or 102 AuE 18 days: the discount, therefore, on this bill is the interest of ^£■343. 15. 6 at 4j per cent, for 103 days. 102 21 — 2 324 DISCOUNTING BILLS. § 341- &. 343"775 9 3093-975 i I 171-887 3265-862 102 6531 724 326586 2 73000) 333 1 17-924 I I 1039 1 1 104 mo 4-5637o=^4- II- 3J discount, and 339-2ii3=;^339. 4. 2J paid to holder of bill. Exercise 58. I. A bill is drawn for ;^33. igJ. on July 17th at 2 months, and dis- counted Aug. nth at 3f per cent.; how much did the holder receive? ■i. Find the discount on a bill for ;^843. 12. 6 drawn Dec. i8th at 6 months and discounted Jan. 26th at 5J per cent. 3. What does a biU-discounter give as the present worth of a bill for £562. 2. 6 drawn Sept. 4th at 5 months and discounted the same day at 6J per cent.? How much is the result less than the irue present worth? 4. A bill is drawn for;^32i. 4. 3 on Dec. 31st at 2 months and dis- counted Jan. 14th at 4I per cent. : how much is charged for discount? By how much does the discount charged exceed the irue discount? 5. What deduction does a banker make in discounting a bill for £l'}i6. 6. 9 drawn Oct. loth at 9 months and discounted March 15th at 6J per cent.? Find also the true discount. 6. How much does a banker give as the present worth of a bill for £S'^SS- 8. 6 drawn Nov. 6th at 10 months and discounted by him on Feb. 2ist at 4 J per cent.? Find also the irue present worth. 7. Find the discount charged in discounting a bill for ;^s8. 4. 3 drawn April 9th at 7 months and discounted June 19th at 5J per cent. Find also the true discount. 8. Find the present worth of a bill for ^£'657. i. 6 drawn Sept. 24th at 4 months and discounted Dec. 12th at 6 J per cent. How much is this less than the true present worth? Ex. s8. _ COMPOUND INTEREST. 3^S 9. A bill is drawn for ^^1687. gj. on March 31st at 3 months and dis- counted same day at 5 J per cent. ; how much is received, and what is the true present worth? 10. On the 31st Oct. a bill is drawn at 6 months for ;£^309. 15. 4 and discounted Jan. 27th at 7 per cent.; what was charged for discount, and how much does this charge exceed the true discount? COMPOUND INTEREST. 342. In Compound Interest, the interest of each period is added to its principal, and the amount forms a new principal for the next period. The period is always understood to be a year, unless the con- trary is stated ; but it may be half a year, or a quarter, or a month,... Case I. Having given the principal, rate per cent., and time {number of periods), to find the interest or amount at compound interest. From the definition of compound interest we proceed thus : — Find the amount of the given principal for one period at simple interest; this amount is the principal for the second period. Find the amount of this second principal in the same manner; and con- tinue the process till the amount for the last period has been found. This last amount is the amount required; and if we subtract from it the given principal we obtain the interest. Remark I. If there are more than 2 periods, employ decimals; and as we only wish the result to be correct up to the third place (275), we need not retain more than 4 places, or, if there are many periods, 5. Remark 2. We may multiply by 3, and divide by 100 at the same time, if we set down the figures of the product two places farther to the right ; thus 857463 2-5724 326 COMPOUND INTEREST. % 342. where we neglect the figures beyond the fourth place, but before we set down in the fourth place, we add the nearest ten (152) from the preceding figure. Ex. I. Find the compound interest of £mo. 16. 9 for 3 years at 4j per cent. £. 4So'8375 18-0335 1 I'liyo J I St year's interest. 469 '9980 amount in i year. 187999 1 i'i749 J 489'9728 amount in 1 years. i9'5989 1 I "2249 J 5io'7966 amount in 3 years. 4S°'8375 principal. S9'9S9i=;^S9- i9- ^i interest. Ex. 2. Find the amount at compound interest of £iT. 3. 6 for \\ years at 4} per cent, per annum, payable half-yearly. Here there are 3 periods of half-a-year each, and the rate per cent, per period is \ of 4f or 2|. £■ 87-ifso "■7435 ■2179 ■1089 interest for ist period. 89'24S3 amount in 1 period. 1-7849 ■ •2231 •1115 91-3648 amount in 2 periods. 1-8273 ■2284 ■1142 93'S347 amount in 3 periods or \\ years. .-. amount reqd=;^93. 10. 8J. Ex. 3. Find the compound interest of ^45. 12. 6 for 3 J years at 3^ per cent, per annum, payable yearly. Having found the amount for 3 periods of i year each, we have still to find the interest of the remaining \ year ; this is done either §342- COMPOUND INTEREST. 327 by considering the \ year as a new period and finding its interest at if per cent., or by finding the interest of the next full period and taking \ of it : — we take the first method. 11 \ £. 45-6250 1-3688 ■2281 47-2219 I "4 160 48-8746 1-4662 •^ 443 50-5851 •5058 •2525 •1264 amount in 3 years, int. for last \ year. amount in 3^ years, principal, interest reqd. 51-4702 45-6250 5-8452 .-. interest reqd=;^5. 16. lof. Ex. 4. Find the amount at compound interest of £j$. 14. 9 for 2 years 7 months, 21 days at 4iper cent. Having found the amount for 2 years, we shall find the interest for the remaining 7 months 21 days, by finding the interest for the 3rd year, and taking aliquot parts ; thus — £■ 75'7375 3"029S •1893 78-9563 3- 1 58-2 •J973 82-3118 amount in a yrs. 2-2445 int. for 7 mo. 21 da. 84-5563 amount reqd. 6 mo, I mo. 15 da. 6 da. 82-3118 prin. for 3rd yr. 3-2924 •2057 3-4981 int. for 3rd yr. 1-7490 •2915 •1457 •0583 2-2445 .•. amount reqd =;^84. 11. i\. Ex. 5. Find the difference between the simple and compound interest of .£549. 17 • 6 at 5 J per cent, in 3 years. 328 PRESENT WORTH AND DISCOUNT. §342- £■ 549-8750 27'4938 ''•7493 580-1181 29-0059 2-gooD 612-0246 30-6012 3-0601 I =30' 243 1 simple interest for i year. 907293 3 years. 645-6859 549'875 95'8i09 compound interest. 90-7293 simple 5-0816 difference=;f5. i. 7J. 343. Case II. Having given the interest or amount, rate ^er cent., and time, to find the jirincipal. 1st. Let the interest be given. Find the interest of ;^i at the given rate for the given time; then, since the interest is directly proportional to the principal, we say its interest : given interest =;^i : principal reqd ; therefore the principal is found by dividitig the given interest by the interest of £\ at the given rate for the given time. 2nd. Let the amount be given. In like manner, the principal is found by dividing the given amount by the amount of £1 at the given rate for the given time. PRESENT WORTH AND DISCOUNT. 344. We find the present worth and discount at compound interest thus : — Find what £\ amounts to at the given rate for the given time; the present worth of this sum (amount) due at the end of the given time is ;^i ; and since the present worth is directly proportional to the sum due, we say this sum : sum due ='^1 : present worth reqd ; therefore the present worth is found by dividing the sum due by the §344- PRESENT WORTH AND DISCOUNT. 329 amount of £1 at the given rate for the given time. The discount is found by subtracting the present worth from the sum due. Ex. I. What sum of money must be put out at compound interest for 3 years at 4 per cent, for the interest to come to £12. ISJ.? foo .„4 1-04 416 i-"o8i6 !• 124864 •12,4,8,6,4 ) 327500 ( 262-285 7-77720 28536 3563 1066 68 6 . . principal reqd = ;£^262. 5. 84. ■124864 int. on ;£i. Ex. 2. Find the discount on £y2,\. 16. 9 due 4 years hence at 3j per cent, compound interest. T-OOO 30 S r oi£^ '>4.7.S>2.'2 ) 324'837S ( •283-0774 present w 95 3331 35313 I-03600 310S 517s 887 84 4 1-071225 32137 5356 £yn . 16 . 9 sum due. 1-108718 33261 5543 283 . I . 6^ present worth. 41 . IS . 2^ discount. I- 147522 amount Exercise 59. Find the amount at compound interest of ';^225. 10. o for 1 years at 3 per cent. ^•7853- 16. 8 £^i- 18. 9 ;^756. 3- 4 £(>il- 5- o ;^2554. 12. 9 S 7 3J 24 6i 33° PRESENT WORTH AND DISCOUNT. Ex. 59. Find the compound interest of 7- £.'i'^i- 4- o for 3 years at 4 per cent. 8. 19 guineas ... 5 8 9. fy^. 10.6 ... 4 4I lo- .^87. 13. 8i ... 4 il II. ;,fl627. 15. 6 ... 3 61 12- ;£'i57-i4-8 ... 6 3^ 13. ;^i86. 14. 9 for 2^ years at 6 per cent, payable half-yearly. 14. £64,6.1^. i\ ... if. 8 quarterly. 15. £8io.2.6i ... 3 4I half-yearly. 16. £■2350. i. Q ... 2 3^ quarterly. 17- £l'2i^- 3- 4 for 2^ years at 3^ per cent, per annum. 18. ;^9i7. 5. 9 for 2f years at 4I per cent, per annum. 19- £'ii9- 18. 4 for 4 years 5 months at sf per cent. 2°- .)o2352. 14. 8 for 2 years 10 mo. 15 da. at 6J per cent. 21. Find the difference between the simple and compound interest of ^£■1750. lOJ. for 3 years at 5 J per cent. 22. A and £ each lend ;£'787. 15^. for S years at 7^ per cent., the former at simple and the latter at compound interest ; find the difference between the amounts they will receive at the end of the given time. ■23. What is the difference between the simple interest of 1000 guineas for 4 years at 3J per cent., and the compound interest of the same sum for the same time at 3 J per cent.? 24. Find the difference between the simple and compound interest of ;^3333- 6. 8 for 3J years at 34 per cent. 25. The difference between the simple and compound interest of a cer- tain sum of money for 3 years at 4J per cent, is ;^8. 13. 7j: find the sum. 26. A person at the beginning of each year lays aside ;^28o, and employs the money at 3J per cent, compound interest : how much will he be worth at the end of 5 years? 27. The population of a city is 765240 and its annual increase is at the rate of 27 per cent. : what will be the number of its inhabitants at the end of 5 years? 28. The population of England and Wales in April, 1871, was 22704108, and the annual increase was i'24 per cent. ; what would be the population, estimated at this rate, in April, 1875? Ex.59- COMPOUND INTEREST. 331 ag. What is the difference between the simple and compound interest of ;^52 7. 17. 6 for 2 years 9 mo. 25 days at 4£ per cent.? 30. A banker borrows money at 3^ per cent, per annum, and pays the interest at the end of the year ; he lends it out at 5 per cent, per annum payable quarterly, and receives the interest at the end of the year ; by this means he gains £iao a year: how much money does he borrow? 31. What sum will amount to ^^405. 3. \\ in 4 years at 5 per cent. compound interest ? 32. What sum of money must be paid now in order to receive ,^360. 10s. two years hence, allowing 3J per cent, compound interest? 33. A offers ;!£'8oooo for an estate; B offers ;^9500o to be paid at the end of 4 years. Which is now the better offer, and by how much, allowing 4i per cent, compound interest? 34. What principal put out at compound interest for 3 years at \\ per cent, will ainount to ;^647. 15J.? 35. Find the discount on £11^0 due 4 years hence at 4J per cent, com- pound interest. 36. What is the present worth of a legacy of ;^23So to be paid to a person on his coming of age, and who is now 18, reckoning 3I per cent, compound interest? 37. What sum of money will in ij years amount to £1^^. Ss. at 5^ per cent, compound interest, payable quarterly? 38. Find the discount on ;^245o. 18. 9 due 3J years hence at 3! per cent, compound interest. 39. ;f loooo is due at the end of 4 years : find the difference between its present worth calculated at 54 per cent, simple, and 5J per cent, compound interest. 40. What sum of money ought to be paid now in order to receive £36^ at the end of each year for the next 3 years, allowing compound interest at the rate of 4^ per cent.? STOCKS. 345. When the English Government wishes to raise a sum of money which cannot be met by the annual revenue, it usually sells Annuities of £2, in sufficient quantities to realize the sum required. These Annuities are payable half-yearly, and can be 332 STOCKS. § 345. transferred, wholly or in part, from one person to another : the Government reserving to itself the right of redeeming them at ;^ioo each whenever it pleases, but allowing no corresponding right of redemption to the annuitant. Strictly speaking therefore, the Government does not borrow money, nor raise a loan: what it does, is to incur the liability of paying regularly every half-year the amount of these Annuities. The price such an Annuity will realize depends on a great variety of circumstances ; it has ranged between £6,']'% and ^107, and at the present time. May 1881, is about £,\02. 346. When Government first issued Annuities, they were for limited periods, and a duty or tax was set apart as a fund for their complete discharge, and thus the Funds meant the duties or taxes set apart for the payment and redemption of Government Annuities. But when Government began to issue Annuities for unlimited periods, charging their payment upon the annual revenue, and setting aside no fund for their redemption, the Funds came to mean the sum of money which Government would require to redeem these Annuities at ;!f 100 each, although no fund what- ever had been set apart for their redemption, and to invest money in the Funds, to purchase any portion of that sum. 347. The capital of public companies, as of the Bank of Eng- land, the East India Company, &c., is called Stock, and the division of profits at the end of each half-year is called the dividend. In imitation of this language, the capital required to redeem the various Government Annuities is called the Stocks, and the half-yearly payment of the Annuities, although it is in- variable, is called a dividend. Hence every Annuity represents £\oo Stock, or £100 in the Funds : and so for any number of Annuities and portions of an Annuity. 348. When the price of ;^ioo Stock, or of an Annuity, is ex- actly £\oo,'\\. is said to be at far, if below ;£'ioo at a discount, and if above ^100 at 2i premium.. § 349- STOCKS. 333 349. The most important of the Stocks are the following : (i) Consolidated Annuities or Consols, so called from the consolidation of the stock of various Annuities into a joint 3 per cent, stock. They amount to about ;^395,ooo,ooo stock. (2) Reduced Annuities, so called because they have been re- duced to a ^3 Annuity from a higher Annuity : they amount to about ;£ 105,000,000 stock. (3) New three per cent. Annuities, which have originated from the conversion of a higher Annuity to a 3 per cent. Annuity : they amount to about ;^i96,ooo,ooo stock. Besides these, there is a small amount of new ■z\ per cent. Annuities, and still smaller amounts of new 3 J per cent., and new 5 per cent. Annuities. 350. Consols are paid half-yearly on 5 Jan. and 5 July : Re- duced Annuities and New Three per cent. Annuities on 5 April and 5 Oct. : the price of the two latter ought therefore to be always the same, but ought to differ from that of Consols either in excess or defect by J of £1, or £\ : but owing to the comparative scarcity of Consols on the Stock Exchange, their price is ad- ventitiously higher than that of New or Reduced. 351. Consols and other Stocks are usually quoted on the Ex- change between two prices, for example, 3 per Cent. Consols loij to loif, Consolidated Bank %\ ... sf prem., London and North- Western Railway 159! ... 160, and these figures express that buyers are bidding at the first price, and that sellers are offering at the second price : and the sale is effected according to the temper of the market at one of the two prices or at an intermediate one. The purchase and sale of Government Stock is made through a broker, who charges at the rate of £^ or 2s. 6d. per cent, upon the amount of Stock he buys or sells : hence, if the broker buys Consols at gaf the actual buyer will give 92f+J or 92I, and if the broker sells at 92J the actual seller will receive 334 STOCKS. § 351- 92J— ^ or 92|. The broker's charge on transactions in Foreign, Railway and other Stock is not uniform : but is usually a per- centage on the proceeds. Remark. When in any question we wish the broker's charge to be allowed, it will be followed by the letter B. 352. All questions in Stocks may be solved by Reduction to the Unit, or by Rule of Three ; as may be seen from the following Examples. Ex. I. When the 3 per cent. Consols are offered at gof, how much stock can be bought with ^825 ? B. Here gof + l or pof will purchase £100 stock, at the same rate how much stock will £81^ purchase? therefore we have £9°i '• ;£^825=;if 100 stock : stock reqd; .-. stock reqd=;^ioox82Sx-4-=;^909. 1. 9!^. Ex. 2. How much must be given for £iyso stock in the 3i per cents, when the price is 96J ? B. Here ;^ioo stock can be bought for 96^+4 or 96!, at the same rate what can ^^1750 stock be bought for? .-. ;^ioo stock : ;ifi75o stock=;^96f : cost reqd; .-. cost reqd=;^?']' x -^-=-=;^i686. 11. 3. o 100 Ex. 3. How much will be received from the sale of £2/^^o. 10 Stock, when the quotation is £')6\ for sale ? B. Here jf 100 stock realizes 96J - i or £g6^, at the same rate what will ;if 2450. TO stock realize? .■. £100 stock : ^£'2450. 10 stock=;^96| : money realized. Ex. 4. What amount of India Five per Cent. Stock at 1 1 1 1 must be sold to realize £1776. 15^.? Here ^100 stock realizes ;^iii|-, at the same rate how much stock will be required to realize ^1776. 15? •■•A^^iiii : ;^i776f = j£'ioo stock : stock reqd. § 352. STOCKS. 335 Ex. 5. What rate per cent, is obtained on money, invested in the Two-and-a-half per Cents, at 74f ? B. Here 74t+i' or £,1a,\ produces £i\ per annum, at the same rate what will j^ioo produce? .•. £'li,\ : ;^ioo=;^'2j : rate per cent. reqd. Ex. 6. At what price would a person have to purchase Three- and-a-half per Cents, to get 4 per cent, for his money ? B. Here, by investing £100, he wishes to get £\ per annum, at the same rate how much must he invest to get 3^? ■ '■ £^ •■ £3i=£i°° • £Slls price reqd of 3 J per cent, stock. But £8'!^ includes the broker's charge of £^, therefore the quotation must be £8^^. Ex. 7. What income is derived from £zn^- 12. 6 Stock in the New Three-and-a-half per Cents.? £ 3775'62S 3 11326875 1887812 , r 1 .'. Income=^i32. 2. iij. 132-14687 Ex. 8. A person invests £i^\S in India Four-per-Cent. Stock at looj; find the amount of his half-yearly dividend. Here ;£^ioof produced a yearly Dividend of £a„ or a half-yearly Dividend of £2, at the same rate what Dividend will ;^iS45 produce? .■. ;^ioo| : ;£^I545=;£^2 : half-yearly Dividend reqd. Ex. 9. How much money must a person invest in the Three per Cents, at 92J to obtain an annual income of ;^i87. 10? B. To obtain an income of £3, he must invest £g2i + £i or ;^92f, at the same rate how much must he invest to obtain ;^i87. 10? ••• £3 ■■ £iS'!i=£g2i : money reqd. Ex. 10. A person invests ;£85o in Consols when they are at 89^, and sells out when they are at 93! : what is his gain .' B. Here an Annuity which . costs £8g^ is sold for £g3i ; therefore on £8g% tliere is a gain of £3^; at the same rate what will be the gain on ;^860? •■• £89^ •• ;^85o=;^3l = gain reqd. 33^ STOCKS. § 35^' Ex. u. A person buys Railway Stock at Sgf, and sells out at 103J, and clears ;£385 : how much money did he invest? Here what cost him ;^89f he sells for £io'^\, and therefore, on in- vesting £i,c,l, he gains ;^i3i ; ■'• i^lA '■ £'^i—£,^9i '• money reqd to be invested. Ex. 12. Whether is it better to invest in the Three per Cents, at 92I, or in the Four per Cents, at io8f ? B. Here £,^}, invested in the 3 per cents, produces yearly ^■^ : how much would ;^93 invested in the 4 per cents, at loSJ produce? The answer is got from the following statement : 108J : 93=;£^4 : what ;^93 produces in 4 per cents.; . •. what ;f93 produces in 4 per cents. =£,\ x 93 x -A.^'=£'i^' .-. the 4 per cents, is the better investment. Ex. 13. Find the difference per cent, in income between invest- ing in the Three per Cents, at 92^, and in the Four per Cents, at io8|? B. .^93 invested in the 3 per cents, produces yearly £3 ; ■••^' ^i-^^- /• 4 /• 8 Similarly ^ I . . . . 4 per cents *'io8i ""^ "^ai?' .•. difference in income on £i=£ £ — = £ — -, 217 31 217 and difference in income per cent. =£ — - =£0. 9. ^i^. Ex. 14. A person transfers .£3622. 10 Stock from the Three per Cents, at 92J to the Four per Cents, at loof : find how much of the latter Stock he will hold, and the alteration in his income. The quantity of Stock held is inversely proportional to the price ; . . ;^ioo| : £<)^\= £}fi'i'i\ ■■ Stock in 4 per cents. ; /-724S 369 8 , . •. Stock in 4 per cents. =£^ x 2-2 x 8^=^3321- Now income from 3 per cents. =^36224 x ~ =^'°^- '3- 6, and . . . .4percents.= ;£'332ix^=;^i32. 16. 9JI; .-. alteration in income =;^24. 3. 3i|. § 35^- STOCKS. 337 Ex. 15. The Three per Cent. Consols are paid on 5 Jan.; what rate per cent, is obtained in buying Consols on 23 April at93|? From 5 Jan. to 23 April is 108 days: and Int. on^S'ioo for 108 days at 3 per cent, is 17^. i^d. Now ;^93. ij. o Price of Consols including broker's charge; and o. 17. 9 Growing Dividend ; 92. 17. 3 Net Price. Hence ;f 92. 17. 3 :;^ioo = ;^3 : Rate per cent, or Rate per cent.=;£'3. 4. 7i. Exercise 60. How much Stock can be purchased for I. 5^2317. 4. 9 in the Three per Cents, at 95I? B. I. ^(^12^1^6. 13. 8 in India Stock at 252? 3. ;£'r756. 9. 6 in Itahan Five per Cents, at 65I? 4. ^^645 1. 3. 6 in Bank Stock at 2 17 J? S- £63°- '7- ^ ™ ^- Railway Stock at 27J below par? 6. £sio. 13. 4 in M. Railway Stock at 8J above par? How much money can be obtained from the sale of 7- £9'1^3- 6- 8 ConsoHdated Annuities at 97? E. 8. ;^7925. 8. 4 Reduced Annuities at 98!? B. 9- £156. 18. 9 New South "Wales Stock at io6|? 10. £i2io. 10. 6 Stock in the Dutch Four per Cents, at 64^? How much money must be given for the purchase of II. ;£'673. 6. 8 Stock in the Russian Five per Cents, at 90J? 12. ;^5S5o Railway Stock at 97|? 13. ;^3257. 15. 6 Consols at 91I? B. 14. ;f loooo Bank Stock at 2ii|? 15. What amount of Consols must be sold to realise. ;^352£. 2. 9 when the sale price is gif? B. 16. What amount of Railway Stock must be sold, when the quotation is I2| above par, to realize £i6iz- 10. 6? E,-S. A. . 22 338 STOCKS. Ex. 60. What interest per cent, per annum is obtained from investing money in 17. Three per Cent. Consols at 91I? B. 18. Dominion of Canada Five per Cent. Stock at 105I? 19. L. Railway Stock at 117I, whose dividends are at the rate of 5 J per cent, per annum? ■20. M. Railway Stock at 88|, whose half-yearly dividends are at the rate of i\ per cent. ? At what price must a person purchase 21. Three per Cent. Consols to obtain 3 j per cent, for his money? 11. Italian Five per Cents, to get 8J per cent, per annum on the money he invests ? 23. Two-and-a-half per Cent. Stock to obtain 3f per cent, on any in- vestment he may make in them? 24. Bank Stock paying half-yearly dividends at 4J per cent, to get £>i- 6- 3 per cent, per annum for his money? 25. A person holds ;^38s2. 12. 6 Stock in the L. and Y. Railway: find his half-yearly dividend calculated at the rate of 7I per cent, per annum. 26. What annual income does a person derive from;£'i7835. 18. 9 Stock in the Three per Cents, and ^^3247. 8. 11 Stock in the New Three-and-a- Half per Cents.? What income per annum will a person obtain from investing 27. ^^3350 in the Three-and-a-Half per Cents, at 97f ? B. 28. £'^'764$. 17. 6 in the Three per Cent. Consols at 91I? B. 29. ;f3923. lis. in the Argentine Six per Cents, at 64J? 30. £iiJi. los. in G. Railway Stock at 89J, whose dividends are at the rate of 4J per cent, per annum ? 31. £3614. 19. 6 in L. and Y. Railway Stock at isgj, whose dividends are at the rate of 7I per cent, per annum ? 32. 78765 fr. 85 c. in Three per Cent. French Rentes at 63 fr. 35 c. ? How much money must a person lay out in 33. Three per Cent. Consols at 90^ to secure an annual income of ^146. loj.? B. 34. Three-and-a-Half per Cent. Reduced Annuities at gSf to obtain an income of 200 guineas per annum ? B. 35. L. Railway Stock at 115^, paying half-yearly dividends at the rate of 2f per cent., to obtain an income of ;^i55, 13. 4 a year? Ex. 60. STOCKS. 339 36. Egyptian Seven per Cents, at ySf to get an annual income of ;f728.8.9? 37. Bank Stock at 250!, paying annual dividends at the rate of 10 per cent., to derive an income of ;^54S. 18. 9 a year? 38. If I lay out ^1359. 15^. in the purchase of Consols at 92I and after- wards sell at 94i, what profit shall I make ? B. 39. A person expended £26^^. los. in the purchase of New Three-and- a-Half per Cents, at 97!, and after a time sold out at 96^ : find his loss. B. 40. A person laid out ;^2SSo •" 3- Three-and-a-Half per Cent. Stock at 91, and after receiving the half-year's dividend he sold out at 9o|: how much did he gain? 41. A person held ;^3S59. los. Stock in the L. Railway, having bought in at 1 1 74: he received the half-year's dividend at the rate of 5 J per cent, per annum, and then sold out at iigf : how much money has he made? 42. A person bought Dutch Four per Cent. Stock at 654, and sold it when the price had risen to 6gi, thereby clearing ;^I2S. ri. 6: how much money did he lay out? 43. A person bought M. Railway Stock at 88f, and after receiving the half-yeaar's dividend at the rate of 4^ per cent, per annum sold out at 93|- and made a profit of ^^142. los. : how much Stock did he buy? 44. If a person invest £126^4. in South Australian Stock at 99!, at what price must he sell to gain^f 1581. igf.? 45. If a person invest ;£^3S35o in the Three per Cent. Consols at 92^, at what price must he sell out after receiving the dividend to make a profit of 46. Whether is it better to invest in the Three per Cents, at 89J, or in the Three-and-a-Half per Cents, at 95? B. 47. Whether is it better to invest in the Two-and-a-Half per Cents, at 65!, or in the Four-and-a-Half per Cents, at 103!? 48. Compare the two investments : Three-and-a-Quarter per Cent. Stock at 92I and Three per Cent. Stock at 87I. 49. Whether would it be better to invest in L. Railway Stock at 117^ whose dividends are sJ pe"^ <=ent. per annum, or in M. Railway Stock at 89^ whose dividends are 4^ per cent, per annum? and compare the two in- vestments. 50. Which is the better investment per cent, per annum, and by how much: Bank Stock at 22i| paying dividends at the rate of 8 per cent., or Three per Cent. Consols at 93 J? 23 — 2 34° STOCKS. Ex. 60. 51. A person holds ^3545 Consolidated Annuities : if he sells out at 92!: and invests the proceeds in the Two-and-a-Half per Cents, at 75I, how much of the latter Stock will he hold? 52. A person held ;^682o. 17. 6 Stock ia the Three-and-a-Half per Cents. ; he sold out at 92I and transferred the proceeds to Railway Stock at 117J : how much Railway Stock does he hold? 53. A person invested ;£^S330 in the Three per Cents, at 91, and when they, had risen if per cent, he sold out and invested the money in the Stock of the Dominion of Canada at 1024 : how much Canadian Stock does he hold? 54. A person invested ^^4950 in the Four per Cents, at 99^, and when they had risen to par he sold out and invested the money in Consols at 6J discount : what amount of Consols does he hold? 55. A person laid out;^749. 5^. in the purchase of Five per Cent. Stock at par, and after receiving the half-yearly dividend he sells out at 4 premium and invests the proceeds in C. Railway shares at 87^: how much Railway Stock does he hold ? 56. If a person transfer ;^300o Stock in the Three per Cents, at 89! to the Three-and-a-Half per Cents, at 98J, find what amount of the latter Stock he will hold, and the alteration in his income. 57. A person transfers ;if 1708. 7. 6 Stock in the Four per Cents, at 1024 to the Three per Cents, at 88|^ : find the alteration in his income. 58. A person expended ;^3565 in the purchase of Two-and-a-Half per Cents, at 61% and when they had fallen to 58 he sold out and invested the money in the Four per Cents, at 96J : find his gain or loss in income. 59. A person derived an income oi £126. 19. 2 from Stock in the Four per Cents.: this Stock he sold out at par and invested the proceeds in M. Railway Stock at i48f, whose annual dividends are at the rate of 7 per cent. : find the alteration in his income. 60. A person has an annual income of;^i9i. %s. from Stock in the Three per Cents. : if he were to sell out at 92 J and invest the money in Five per Cent. Stock of New Zealand at 105, how much of the latter Stock would he hold, and what would be the alteration in his income? 61. What rate per cent, is obtained in buying Three per Cent. Reduced Annuities on 17 Feb. at 92 J? B. The Dividends are paid on 5 April and 5 Oct. 62. A man invests ;i^8o63 in the Three per Cents, at 94J : what will be his clear income after an income-tax of \od. in the pound has been de- ducted? B. Ex. 60. STOCKS. 341 63. What sum must a man invest in the Three per Cents, at pij, in order to have a clear income of ;^230 after paying an income-tax of \od. in the pound? 64. What must be the price of the Three per Cents, so that by investing £,}fl^io, a man may have a clear income of;£'i030. ioj. after an income-tax of \\d. in the pound has been deducted? 65. When the Tvifo-and-a-Half per Cents, are at 83J, what ought to be the price of the Three-and-a-Half per Cents, to give the same rate of interest? 66. A gentleman in Australia has been receiving 11 per cent, on bis capital in the colony ; he brings his capital home, invests it in the Three per Cents, at 944, and his income in England is ;f 2400 a year : what was his income in Australia? 67. The income derived by a legatee from money invested in his behalf in the Three per Cents, at 93J is ^^68. 3. 6. What was the amount of the legacy? 68. A person holds ;^4675 Stock in the Five per Cents. : what sum must he lay out in the purchase of Four-and-a-Half per Cents, at 102J so that his income from both sources may be ;£'843. \os. ? 69. A person invests £i'i^T. los. in the Three per Cents, at 83, and when the funds have risen to 84 he transfers three-fifths of his capital to the Four per Cents, at 96 : find the alteration in his income. 70. A person sells out of the Three-and-a-Half per Cents, at 92J and realizeS;£^i8s5o. If he invests two-fifths of the produce in the Four per Cents, at ^d, and the remainder in the Three per Cents, at 90, find the alteration in his income. 71. A man invests £i'2g'!' tos. in the Three per Cents, at 95J. He sells out one-third when the funds have fallen to 94, ;£'i6oo Stock when they have risen to 96^, and the reifiainder at par. What sum does he gain? If he invests the proceeds in the French Three per Cents, at 67*50, what would be the difference in his income? 72. A and B are two railway companies that pay respectively 4^ per cent, and i| per cent, per annum on their ;^ioo shares. When the price of a share in A is 10 1 J and in B 32J, in which company is it more advantageous to invest? and what is the difference of income that would arise from the investment of ;^ 174 15 in one rather than in the other? 73. Which is the better investment, — Three per Cent. Stock at- 87I, or shares at ;^233 each, on each of which a dividend oi £•]. 13. 4 is paid annually? How much more money must be invested in one rather than in the other to produce an annual income of ;^46o? 342 STOCKS. Ex. 60. 74. A person possesses 5^3200 Three per Cents, which he sells at 99I : he invests the proceeds in railway shares at ^56 a share, whfch shares pay 5 per cent, interest on £,^i, the amount paid on each share. How much is his income altered by the transaction? 75. What amount of Stock must be sold out of the Three per Cents, at 87 J to pay the present worth of £ifni. 17. 6 due 10 months hence at 3J per cent.? 76. If the Three per Cents, be at 92!, and the Four per Cents, at 123 J, in which should one invest? and how much is one investing in each when the difference in income is half a crown? 77. If the French Three per Cents, be at 60 when the English are at 95, the exchange between the two countries being 25 francs to the pound, how much French Stock in francs can be bought by selling ;^6ooo Stock out of the English funds? 78. A person derived an income of;^434. 15J. from the Three per Cent. Consols : he sold out at 92 J and invested the proceeds in the Four-and-a- Half per Cent. French Rentes at 9870, the rate of exchange being 25 fr. 60 c. for;^i : what income does he derive from the Rentes in francs? 79. A person having to pay ;^io85 at the end of 2 years invested a certain sum of money in the Three per Cent. Consols, allowing the dividends to accumulate until the payment of the debt, and also an equal sum the next year ; supposing the investments to be made and the debt to be paid when Consols are at 73, what must be the sum invested on each occasion that there may be just sufficient to pay the debt at the proper time? 80. If a person invest in the Three per Cents, so as to receive 3 per cent, clear on his investment when there is an income-tax of gd. in the pound, what percentage clear does he receive (i) when the income-tax is reduced to 51/. in the pound, and {2) when it is raised to is. in the pound ? 81. In the Three per Cents, what fraction of a given amount of Stock is paid for annual interest, (i) without any deduction, and {2) after a deduction of gd. in the pound for income-tax ? What is the amount of Stock for which ;^ii6. 10s. is paid as annual interest after gd. in the^i has been so deducted? 82. A proprietor of Three per Cent. Consols receives his half-yearly dividend and lays it out in the purchase of more Consols at 90. His next half-year's dividend is;^457. los. : how much does this dividend exceed the former? Ex. 60. EXCHANGE. 343 83. A person after paying an income-tax of \pf. for ;^i : Bills on St. Petersburg at 3 months at 28 d. to V]\d. for I rouble : Bills on Vienna at 3 months at izfl. to 13^?. \okr. iox £1 : and Bills on Lisbon at 90 days at 52^^. to 52 rf. for I milreis. In the exchange on Paris ^i is called ^i^ fixed price, and the varying number of francs and centimes the variable price ; on Lisbon i milreis is ih&fixed price, and the number of pence the variable price. MONEY TABLE. France, Belgium, 1 Switzerland J I franc = 100 centimes Italy Spain .... Greece .... I lira I peseta I drachme = 100 centesimi = 1 00 centesimos = loo lepta ■ =0 9|ne German Empire. Austria . . . I marc I florin = 100 pfennige = 100 kreuzers = . = 1 Hi . Ill . Russia . . . Denmark . . . Holland . . . 1 rouble I rigsdaler I florin = 100 copeks = 96 skillings = 20 stivers = 3 ■ = 2 . i4 ■ 2i . 8 . Portugal . . . United States . I milreis . I dollar = 1000 reis = 100 cents = 4 • = 4 • 6 . India .... I rupee = 16 annas = 1 . loi . Francs, lire, pesetas, and drachmai are declared by the French Monetary Convention to be of equal value, and to be interchangeable with one another in those countries where these coins are in circulation. § 357- EXCHANGE. 347 Ex. I. Find the par of exchange between the U. S. gold eagle, weighing 258 grains ^ fine, and the sovereign of which 1869 weigh 40 lbs. of gold ^ fine. £,% reqd= i eagle, I =258 grs. % standard, 10=9 grs. fine, n = i2 British standard, 5760 = 1 lb. 40=jfi869; , , 258 XQX I2X i860 _ - .-. £fi reqd=-5 ^ 2 2=2-054838, ■^ ^ 10x11x5760x40 ^^ ^ i.e. I eagle=;£^2'o54838... .•. I dollar gold=;£'''2oS4838... = 4'io9676j. = 4J. i'3i6i^. =4^. i^V^. nearly, , ^ looooooo^ „., „ "'^"^^'=15i^8i8*='^'^^^S...«. Remark. Formerly the rate of exchange of New York on London was a variable number of dollars for £^i. los., this fixed price being adopted because it was equal to 100 dollars at 4J-. ()d. each ; but now the rate of exchange is a variable number of dollars and cents for £\. The far oi exchange is, as we have just found, 4'866s $ for £1. Ex. 3. Find the relation between the sovereign and the Napoleon, as determined from the intrinsic value of the two coins : — (i) 40 lbs. British standard gold, W fine, is coined into 1869 sovereigns ;, (2) i kilog. French standard gold, ^o" ^"6, is coined into 155 Napoleons ; (3) i kilog. is equal to 15432 grains. Napoleons reqd =;£■!, 1869=40 lbs. British standard, 12 = 11 lbs. fine, 1 = 5760 grains '5432 = 1 kilog 9=10 kilog. French standard, 1 = 155 Napoleons; T1.T 1 J 40X11X5760X10X155 ,.„fi„„fi .-. Napoleons reqd= 2— -J ^ — — = i-2&iio&, '^ ^ 1869 X 12 X 15432 X 9 i.e. ;^i = i*26iio6 Napoleons=25'222i2 francs = 25/. 22 f. 348 EXCHANGE. § 357- Ex. 3. What is the value of a sovereign in France, or at the French mint ? A person taking a kilogramme of standard gold, -^ fine, to the French Mint will receive 155 Napoleons, or 3100 francs; but then he must pay a mintage of 6 f. 70 c, so that in reality he only receives 3093 i. 30 c. : hence we have Francs reqd= i sovereign, 1869 = 40 lbs. British standard gold, 12 = 11 lbs. fine gold, 1=5760 grains 15432 = 1 kilog 9 = 10 kilog. French standard, I = 3093"3 francs; , , 40 X II X 5760 X 10 X 3093-3 ,„ .-. francs read =2 -■ — i± ^^-^^^ = 'ii-if,'i, 1869x12x15432x9 i.e. £i = iif. i-jc. Ex. 4. The price of standard gold is £1. 17. loj per oz., and in Paris it is at 4j per mille premium : find the rate of exchange in Paris; and if the short exchange be 25-60, find how much per cent., gold is dearer in London than in Paris-. By short exchange is meant exchange on bills at sight, or at short sight, which is usually 3 days. Since £'^. 17. io§ per oz. is equal to ;^i869 per 4a lbs., we have from the last example : ;^i in Paris = 2 5" 1 70 f. 4J per mille prem. = • 1 20 ; .•. rate of exchange = 25*290. But short exchange = 2 5'6o ; .-. diff.= -31. Hence ;!fi in London will purchase 25-60 f., but in Paris only 25*29 f., therefore it is -3 1 f. dearer in London than in Paris : but 25-60 : ioo=-3i : 1-21, or gold is nearly \\ per cent, dearer in London than in Paris. § 357- EXCHANGE. 349 Ex. 5. Exchange £(>j6. 17. 6 for francs, when the rate of exchange is 2^/. 17I c. for £1. francs. 25"i7S 677 176225 176225 151050 , 6d. is ■ 1 7043 '475 3'i46875 17040-328125 = 17040/. 33 c. Ex. 6. Find the arbitrated rate of exchange between London and Paris when the course of exchange between London and Amsterdam is 12 florins 3J stivers for £1, and between Amster- dam and Paris 209J francs for 100 florins, i stiver = 5 cents. Francs reqd=;f i, ;^i = ia'i62S florins, 100=209! francs; , I2'l625 X 2ooi . .•. francs reqa= ^-^ ^ = 25*45 irancs, TOO or London gives 25/ 45f. for ;if i. Ex. 7. A New York merchant remits 27940 florins to Amsterdam l)y way of London and Paris, at a time when the exchange of New York on London is 4'88J$ for £1, of London on Paris is 2Sfr. 40 c. iot £1, and of Paris on Amsterdam is 212 francs for 100 florins; ^ per cent, brokerage being paid in London and in Paris. Dollars reqd = 27940 florins, 100=212 fr., ioo = iooj fr. with brokerage, 2540=iooi^^;^s ....,: i=4-i „ „ , 27040x212x801x801x4885 , Dollars reqd=-^^^=^ ~ -„ — 5 ^ 100x100x2540x8x8x1000 = 11420-317 . = 11420$ 32 c, 35° EXCHANGE. Ex. 6i. Exercise 6i. 1. How many francs will be given in Paris for ;^688. 14. 8 when the course of exchange is 25 f. 42 J c. for £\ ? 2. How many dollars must be given for a letter of credit on London for ;^2346. loj., when the exchange is 489 c. for £1 ? 3. Exchange ;^598. 16. 9 into money of the German Empire, the course of exchange being 20 m. 54 pf. for £\ . 4. Exchange £i'2'jo. 13. 9 for florins and kreuzers of Vienna, the rate of exchange being 12 fl. 85 kr. for £1. 5. Reduce ^^1857. 14. 3 to rupees, &c. at the rate of is. iifo'. for i rupee. 6. How many florins and cents, at Amsterdam must be given for a bill on London for ;^74S. 3. 6, at the rate of 12 fl. 7J c. for ^i ? 7. What sum of money in London must be given for a bill of is643'25 $ on New York, when the rate of exchange is reckoned at 4-85 $ for £1 ? 8. Reduce 37847 lire 60 c. to Hamburgh currency, the rate of exchange being 124! lire for 100 marcs. 9. A traveller goes to Paris with ^^57. los., which he exchanges for French money at the rate of 25 f. 35 c. for .^i. During his stay in France he spends 830 f. 50 c., and on leaving exchanges the remainder of his French money for English at the rate of 25 f. 20 c. for £1. What sum will he receive ? 10. A person on leaving England exchanged his money for French money at the rate of 25 francs for £1 ; and on arriving at Vienna receives 1 35 (P^PS'^) florins for 15 20-franc pieces: what was his loss In English money, supposing a florin to be worth is. S^d. ? and what was his loss in French money? 11. Some years ago to pay 18 kreuzers I gave a thaler, and received back 22 kreuzers 10 silber-groschen and half a gulden : — i thaler was 30 silber-groschen, and i gulden was 60 kreuzers : how many gulden were worth 4 thalers ? 12. When;^i isequivalent to 25f. 35 c. to 20m. 64pf. and to i3fl. S '''•i what is the value of 36980 marcs in English, in French, and in Austrian money? 13. What is the arbitrated rate of exchange between Hamburgh and Paris in francs per 100 marcs, when the course of exchange between London and Paris is 25"4S francs for £1, and between London and Hamburgh 2048 pfennige for £1 ? Ex. 6i. EXCHANGE. 35^ 14. When the exchange between London and Paris is 25 fr. 60c. for £,i and between Lisbon and Paris 55 7 J centimes for i milreis, what is the value of the milreis in pence? 15. The rate of exchange between London and Petersburg is ■^x^d. for one rouble, between Vienna and Petersburg is 95!^ florins for 60 roubles, and between Paris and Vienna is 93^ florins for 200 francs : find the arbitrated rate between London and Paris in francs for ;^i. 16. Find the arbitrated rate of exchange between Vienna and London in florins and cents for ;^i, when the exchange between Paris and Vienna is 222J francs for 100 florins, between Paris and Berlin 124J francs for 100 marcs, and between Berlin and London 20 marcs 50 pf. for £u 17. When the exchange between London and Lisbon is tJ^\d. for I milreis, Lisbon and Paris 552 francs for 100 milreis, Paris and Hamburgh I24i francs for 100 reichmarcs, Hamburgh and Amsterdam 166 marcs 40 pf. for ICO florins ; what is the corresponding exchange between Amsterdam and London in florins and stivers for £1 ? 18. A merchant in London owed another in Petersburg 2460 roubles 50 copeks, which he remitted through Paris when the exchange between London and Paris was 25 fr. 35 c. for £,1, and between Paris and Petersburg 339 centimes for i rouble. Shortly afterwards the exchange between London and Paris was 25 fr. 62 J c. for ^^i, and between Paris and Peters- burg 337 c. for one rouble. How much would he have gained by the delay? 19. Find the value oi £1 in marcs and pfennige of North Germany, having given that i kilogramme of fine gold is coined into 139^ 20-marc pieces, that i lb. of standard gold is coined into 46^ sovereigns, that standard gold is ^ fine, and that i kilogramme is 15432 grains. 20. Calculate the par of exchange between the dollar and the shilling when British standard silver is valued at 5^. o\d. per oz., having given that I dollar weighs 41 2 J grains, and is i\ fine : and I lb. Troy standard silver, |^ fine, is coined into 6(1 shillings. 21. When British standard silver is valued at 5J. \%d. an ounce, find how many francs are equal to los., having g^ven that I lb. Troy standard silver, fj fine, is coined into 66 shillings, I kilog. French st. silver, ^ fine, is coined into 200 francs : and I kilog. is equal to 15432 grains. 352 EXCHANGE. Ex. 6i. 2 2. Having given the same elements .as in the last question, find what must be the price of British standard silver, an ounce, in order that the par of exchange as determined from the silver coinage of the two countries shall be the same as that determined from the gold coinage, — that is, that ^\ shall be equal to 25 f. 22 c. (See Ex. i, p. 347.) 23. How many ounces of bar gold 21 J carats fine will be required to be sent to France to pay a debt of 27654 f. 40c. ? (See Ex. 3, p. 348.) 24. The exchange of London on Amsterdam at 3 months is 12 fl. 135 stiv. : what is the short price, allowing 5 per cent, per annum ? 25. The short exchange of London on Paris is 25 f. 32jc.: find the ex- change at 3 months, reckoning 4J per cent, per annum. 26. The exchange of London on Paris at 3 months is 25 f. 70 c, and of Paris on London is 25 f. 250.: find the difference between the rates of the short exchanges, reckoning 3J per cent, per annum. 27. At how much per mille premium is gold quoted at Paris when the exchange there is 25-27 francs per £\ ? (See Ex. 3, p. 348.) 28. When British standard gold is quoted at Paris at 3 per mille pre- mium and is -^ per cent, dearer there than in London, what is the short exchange on London ? 29. Gold is quoted at Paris at 30 per mille premium and the short ex- change on London is 26 "50 francs for £-1. : find how much per cent, gold is higher in London than in Paris. (See Ex. 4, p. 348.) 30. A merchant in Hamburgh delivers goods to a merchant in London at 5 m. 16 pf. per pfund, a pfiind being equal to i '068 lbs. The London merchant remits payment at 2052 pf. per £\, and sells the goods at ;^27 for 100 lbs. What is his gain or loss per cent.? 31. A Lyons merchant could sell silk at home at 7 f. 10 c. per metre, gaining thereby 6J per cent. : but at Vienna he could sell it at 10 fl. 25 kr. for 3 metres net, and gain thereby 8J per cent. What rate of exchange is hereby established between Austria and France in florins for 200 francs ? 32. The sum of ;^iooo is laid out in London in bills on Vienna at iifl. 35 c. for^^i. The bills are sold in Hamburgh at 54 fl. 90c. for 100 marcs, less i month's discount at 4 per cent, per annum, commission on sales \ per cent., and brokerage for sales and returns \ per cent. The returns are made in bills on Madrid at 3 m. 92 pf. for i peso, and are sold in London at i,^d. for i peso, less brokerage for purchase and sales \ per cent. Find the profit on the original outlay. CHAPTER XIV. SQUARE AND CUBIC MEASURE. DUODECIMALS. 358. A Parallelogram is a quadrilateral figure whose oppo- site sides are parallel ; they are also equal. A Rectangle is a parallelogram which has all its angles right angles. A Square is a rectangle which has all its sides equal. 359. A Parallelopiped is a solid figure bounded by six quadrilateral figures of which every opposite two are parallel. A Rectangular Parallelopiped or Rectangular Solid is a parallelopiped bounded by six rectangles ; as a brick. A Cube is a rectangular solid bounded by six squares : as a die. 360. To find the area of a rectangle. Let ABCD be a rectangle of ^ which the length AB is 7 feet and the breadth AD is 5 feet. Divide AB into 7 parts each equal to I foot, and AD into 5 parts each equal to i foot ; through the points ^ of division draw straight lines parallel to AB and AD ; then the rectangle will be divided into 7x5 squares, each side of which is i foot in length: that is, the area of the rectangle is 7 x 5 square feet (203). B.-S. a. 23 B / n c 354 SQUARE AND CUBIC MEASURE. §360- Again, suppose the number of feet in the length and in the breadth to be mixed numbers or fractions : for example, let the length AB be 7^ feet and the ^ ^ k breadth AD be 5f feet. Pro- duce AB to K so that AK is D five times AB, and AD to M so that AM is 3 times AD ; and draw straight lines pa- rallel to AK and AM, as shewn in the figure. Now AJC is 7f x 5 feet, or 39 feet, and AMis 5f x 3 feet, or 17 feet ; therefore the area of the rectangle is 39 x 17 square feet. But the rectangle AKLM is divided into 5x3 rectangles, each equal to the rectangle ^.5CZ?; therefore the area of the rectangle ABCD is — of the area of the ^ IS rectangle AKLM, or is equal to ^Q X 17 "^O 1 7 — sq. feet, or — x — sq. feet, or 7f x s| sq. feet. Hence, whether the number of feet in the length and breadth be integral or fractional, their product is the number of square feet in the area. In like manner, if the numbers giving the length and breadth be given in inches, or in yards, or in miles, their product will give the area in square inches, or in square yards, or in square miles. We have then this Rule, — Express the length and breadth in units of the same denomina- tions the ■product will give the area in square units of that de- nomination. Cor. If we divide the number of square units in the area by the num,ber of units in either side, we shall find the number of units in the other side. 361. To find the volume of a rectangular parallelopiped, or rectangular solid. § 36i- SQUARE AND CUBIC MEASURE. 355 Let the figure represent a rectangular solid, whose length AB is 7 feet, breadth BC 5 feet, and height AD 6 feet. Divide AB into 7 parts, BC into 5 parts, and AD into 6 parts, each equal to i foot ; and through the points of division draw planes parallel to the sides ; then the figure will be divided into 7x5x6 cubes, each of the sides of which is i foot ; that is, the volume of the rectangular solid is 7 x 5 x 6 cubic feet. In the same way as in Art. 360, it may be shewn that if the number of feet in the length, and breadth, and height be mixed numbers or fractions, still their product will give the number of cubic feet in the volume. Also, if the numbers giving the three dimensions be given in inches, or in yards, or in miles, their pro- duct will give the volume in cubic inches, or in cubic yards, or in cubic miles. We have then the following Rule, — Express the length, breadth and height in units of the same de- nomination: their product will give the volume in cubic units of that denomination. Cor. If we divide the number of cubic units in the "Volume by the product of the number of units in any two of its dimensions, we shall find the number of units in the third dimension ; and if we divide the num.ber of cubic units in the volume by the number of units in any one dimension, we shall find the product of the nu7nber of units in the other two dim.ensions, or the number of square units in that face of the solid whose sides are those two dimensions. Ex. I. Find the area of a rectangular floor 23 ft. 8 in. long and IS ft. 10 in. wide. 23—2 356 SQUARE AND CUBIC MEASURE. §361. Length = 2S4in. Area=23Jx 15J sq. ft. Breadth = 190 m- _ 7 1 9 5 „„ r, 25560 3 284 _6745 J f 12 ) 53960 sq. in. ^"^iT^I '■ I 12) 4496 - 8 _ 1 3 iv \ : ■ =374tI sq.ft. 9) 374.. 104 in. 41 sq. yds. 5 ft. 104 in. =4' sq. yds. 5 ft. 104 in. Ex. 2. Find the area of the walls of a rect. room 23 ft. 8 in. long, 15 ft. 10 in. wide, and 11 ft. 11 in. high. The four walls form together a rectangle whose length is the circuit oi the room, and width the height of the room. Circuit = 2x(23ft. 8in. + i5ft. loin.) = 2 X 39 ft. 6 in. = 79 ft.; .-. Area of walls = 79 x i if^ sq. ft. = ii^ sq.ft. 12 ^ = 941 sq. ft. 60 in. = i04sq. yds. 5 ft. 60 in. 104 sq. yds. 5 ft. 60 in. Ex. 3. From a rectangular plot of land 55 yards 2ft. gin. wide, what length must be cut off to make a garden of one acre .' 23 56yds. 2ft. 9in. = 55yyds. = 55iJyds. and I acre = 4840 sq. yds.; .'. Length of garden ='t-l_ yds. = ^-^^^ 55H-' 671 5280 = 86 yds. I ft. 4/1 in. Length = Width = 284 in. 190 in. 474 2 Circuit = 948 in. Height= 143 m. 2844 in. 3792 948 ( 12) ^M X2- 135564 sq. in. ) 11297 9 ) 941 .. 60 in. §36i. SQUARE AND CUBIC MEASURE. 357 Ex. 4. Find the cost of carpeting a room 23 ft. 8 in. long and 15 ft. 10 in. broad, with carpet 27 in. wide, at ^s. 2,d. a yard. Area of floor =23! x 15! sq. feet; ,-. length of carpet = ^^^"^'^^ ft. _ 231X1 si 284 in. 190 in. 25560 284 yds. .•. cost of carpet = 2iX3 a3lxi5^X5i _ q S 2iX3 27 71 Q15 21 4 I 36493 162 = ;^I4. II. SJ^. 9 ) 53960 sq. in. in area. 3 ) 5995' 5 1 2 ) 1998-5185 in. in length. 3 ) 166-543^ ft 55'5i44 yds 5^. = i i• = 12' = 144" = 1728'" = = ; I cubic foot / therefore in linear measure the inch is the same as the prime, in square measure as the second, and in cubic measure as the third. §362. DUODECIMALS. 359 We can therefore readily pass from quantities expressed in duo- decimals to those expressed in feet and inches, and conversely ; thus, Ex. (i). 8 ft. 2' 3" = 8 ft. 2' j^ = 8 ft. 2jin. Ex. (2). I5sq.ft. 10' ii"8"' = i5sq. ft. 131"^= 15 sq.ft. I3if in. Ex. (3). 24 cu. ft. 6' 7" 8'" 9'' 4' = 24 cu. ft. 956"'ijf tf 7" 8" = 24 cu. ft. 956^ in. 956"' Conversely, Ex. (4). S yds. 2 ft. 5i in. = 17 ft. s'f = 17 ft. 5' 9". Ex. (5). 19 sq. ft. 1 18| in. = 19 sq. ft. 1 18''| = 19 sq. ft. 9' 10" 8'". Ex. (6). 3Scu.ft. i267|in. 12)1.67- = 35 cu. ft. i267."'f 12 )105 -7'" = 3Scu.ft. 8'9"/"6"8'. 8.9"-7"' 363. Let the figure ABCD represent a square foot. Divide AD into 12 equal parts, then AG B I each part as AF is i' ; through the points of divi- sion draw straight lines parallel to AB, dividing the sq. foot into 12 equal rect- angles, and therefore each of them is a superficial prime. But each of these rectangles is I ft. long and l' broad; therefore a rectangle i ft. long and i' broad is i' in area : hence a rectangle 7 ft. long and l' broad is 7' in area, and a rectangle 7 ft. long and S' broad is 7 x 5 or 35' in area ; and this result is usually expressed by saying that Feet into primes give primes. Again, divide AF into 12 equal parts, each part will be i" ; through the points of division draw straight lines parallel to AB, dividing the rectangle ABEF, which is a superficial prime, into 12 equal rectangles ; therefore each of these rectangles is a superficial 360 DUODECIMALS. § S^S- second. But each of them is i ft. long and i" broad ; therefore a rectangle i ft. long and i" broad is i" in area ; and, as before, a rectangle 7 ft. long and s" broad is 7 x 5 or 35" in area ; that is. Feet into seconds give seconds. In like manner, Feet into thirds give thirds, Feet into fourths give fourths, &c. Again, divide AB into 12 equal parts, and through the points of division draw straight lines parallel to AF, dividing the rectangle ABFE into 12 equal rectangles ; therefore each of them is ^^ of a superficial prime, or is a superficial second. But each of them, as AGHF, is 1' long and i' broad ; therefore a rectangle i' long and i' broad is i" in area; and, as before, a rectangle 7' long and 5' broad is 35" in area : hence Primes into primes give seconds. And if we divide .<4i^into 12 equal parts, each will be i"; and if through the points of division we draw straight lines parallel to A G, we may shew that Primes into seconds give thirds; and, as before. Primes into thirds give fourths, 8i.c. Also, Seconds into seconds give fourths. Seconds into thirds give fifths, &c. ; that is, the number of the dashes expressive of the denomination of the product is the sum of the dashes of its factors. 364. By pursuing a similar method we may shew that Feet into suplfeet give solid feet. Feet into supl primes primes, Feet into supl seconds seconds, &c. Primes into supl primes seconds. Primes into supl seconds thirds, &c. Seconds into supl seconds fourths, &c. ; that is, the number of dashes expressive of the denomination of the product is the sum of the dashes of its factors. Hence in duo- decimals the rule given for multiplying a length by a breadth is equally applicable in multiplying a surface by a length. §36S- DUODECIMALS. 361 H GC 365. Let us now find the area of a rectangle 9 ft. 4' 10" long and 7 ft. broad. Let ABCD be this rect., and let AE be 9 ft., EF\i& 4 and FB be 10", and .<4Z? be 7 ft. Now the whole rect. ABCD is the sum of the 3 rects. AH, EG, and FC. But the rect. i^Cis 7 ft. by 10", .-. its area = 70" or = 5' 10", rect.£'G is 7 ft. by 4', .•. its area = 28' or = 2 sq. ft. 4', and rect.^.^ is 7 ft. by 9 ft., .•. its area =63 sq. ft., .•. area of the whole rect. =65 sq.ft. 9' 10". Instead of finding the area of each rect. separately and adding the results, we shall find it more convenient to reduce the first result at once, and to carry to the second, as in Compound Multiphcation : thus ft. 9 . 4 • 10 _7 sq. ft. 65 . 9 . 10 where we say 7 ft. x 10", or simply 7x10=70 and 70" = 5' 10"— set down 10" and carry 5'. 7x4 is 28, and S (carried) is 33 : 33' = 2 sq.ft. 9' — set down 9' and carry 2. Lastly, 7 ^ 9 = ^3, and 2 (carried) is 65— set down 65 sq. ft. Again,— find the area of a rectangle 9 ft. 4' 10" long, and 8' broad. As in the last example,, the whole rectangle may be made up of 3 rects. whose respective lengths are 9 ft., 4' and 10", and whose breadth in each case is 8': making therefore the same arrangement, we proceed thus : ft. 9 . 4 . 10 sq. ft. 6 . 3 • 2 . 8 saying 8' x 10" = 80'" (the area of the first rect.) : but 80'" = 6" 8"'— set down 8'" and carry 6". 8x4=32, and 6 (carried) is 38; but 38", &c. 362 DUODECIMALS. §365 Lastly. Find the area of a rect. 9 ft. 4 10" long and 7 ft. 8' broad. Let yi^ilfTV represent this rect. where ^ ^ AD is 7 ft. and DN is 8' : draw DC parallel to AB. The whole rect. is made up of the two rects. A C and DM. But the rect. /i C is 9 ft. 4' 10" by 7 ft., and its area has been found to be 65 sq. ft. 9 10", and the rect. DM is 9 ft. 4 10" ^ '^ by 8', and its area has been found to be 6 sq. ft. 3' 2" 8'" ; adding these results, the area of the whole rect. is 72 sq. ft. i' o" 8"'. Or, combining the work of the last two Examples, the whole will stand thus : — ft. 9 • 4 = area of rect. AC. . DM. sq.ft. 72 . I . 0.8= ABMN. We have then the following Rule : — Write the breadth under the length so that units of the same denomination may be under one another. Multiply the length by the number of feet in the breadth, as in Compound Multiplications then multiply by the number of primes in the breadth, setting down one place farther to the right; then by the seconds, setting down one place farther again to the right, &'c.j add these partial products, and their stem will give the area of the rectangle in sq. feet, primes, &^c. Ex. I. Find the area of a rectangular floor whose length is 23 ft. 7' 9" and breadth 15 ft. 10' 7", 7 8 6; 9 • 10 6 3 ■ 2 8'" ft. / " IS 7 10 9 7 6'" 6 . 354 19 I 8 8 I 3 5 9 .S" sq. ft. 375 6 6 . . 3 12 ) 236 19 . 13 ■ 9 or 41 sq. yds. 6 ft. 78:^ in. §365- DUODECIMALS. 363 Ex. 2. Find the area of a parallelogram whose base is S yds. 2 ft. 8J in. and altitude 3 yds. 2 ft. 5 in. sq. ft. 202 ft. / II 17 8 . 9 II • S 19s . . 3 7 4 ■ 7 9" ix> . 9 =22 sq. yds. 4ft. 58! in. Ex. 3. Find the volume of a water-tank whose length is 20 ft. 8 in., breadth 11 ft. 10 in., and depth 8 ft. 5f in. 20 8 ir 10 .8" 227 17 4 2 244 8 6 5 8 9 1956 lOI IS 5 10 3 4 9 5 4'" . 2073 7 6 4 3 )^30- 3 76 .. 21 . Volume = 2073 cu. ft. 7' 6" 4" = 76cu. yds. 21 ft. 1084111. Exercise 62. 1. Find the area of a rectangle 4 ft. 11 in. long and 2 ft. 4 in. broad. 2. Find the area of a floor 19 ft. 4in. long and 16 ft. 8 in. wide. 3. How many sq. feet and inches are there in a sheet of glass 3 ft. 9 in. in length and 2 ft. 7 J in. in width? 4. How many sq. feet and inches remain out of 313 sq. feet of matting, after covering a floor 16 ft. 9 in. long by 12 ft. 11 in. broad ? J. How many sq. yds., &c. of carpet will coyer a floor whose length is 22 ft. 8|in. and breadth 16 ft. 7jin.? 6. Find the area of a square whose side is 7 ft. 10 in. 7. How many acres, &c. are there in a rectangular field 3265 links in length and 384 hnks in breadth ? 3^4 SQUARE AND CUBIC MEASURE. Ex. 62. 8. How many acres, &c. of land will be required to form a street J 10 yards long and 37 ft. 7 in. wide ? How many sq. yards of flagging will be required to form a pavement 5 ft. 9 in. wide down one side of the street ? 9. A rectangular garden is t chains 9 links wide and 300 yards long : find its area as a decimal of an acre. 10. Find the area of a rectangular court-yard 13 yards 2 ft. 7 in. long and 23 ft. 10 in. broad. 11. How many sq. yards of ground are covered by a plank I7"64 ft. long and 7^ inches wide ? 12. A garden roller is 3 ft. 7J in. wide and 5 ft. io| in. in circum- ference : how many sq. feet and inches of ground does it pass over in making one complete revolution? 13. How many sq. yards of brickwork are there in the face of a wall surrounding a circular reservoir, the perimeter of the wall being 333I yards and its height 6 ft. loj in. ? 14. A school-room is 35ft. Sin. long, 23ft. 4 in. wide and isifeet high : what is the area of its walls ? 15. A piece of canvas of uniform viddth is 7 ft. 3! in. long, and it covers 2 sq. yards 103 j in. : what is its vndth ? 16. The floor of a room is isf feet wide, and its area is 23 sq. yards: find its length. 17. The area of a rectangle is 1532 sq. ft. 117 in., and one side is 81 ft. 9in. : find the other side. 18. The surface of a rectangle 8 inches wide is the fifth part of a sq. yard : what is its length? ig. The area of a sq. garden is i ac. i p. 29 yds. 6f ft.: find the length of its side in yards. 20. A sheet of glass is 3 ft. 9 in. long and 2 ft. 7^ in. wide : by how much must the width be narrowed to leave a surface of one square yard? 21. A piece of cloth 2 yards 3 qrs. 2jnl. in length covers 21 sq. ft. lojin.: find its width. 22. A square space containing 140 sq. yds. 36 in. is to be lengthened by 4 ft. 3 in. in one of its dimensions and shortened by 3 ft. 4 in, in the other: what will then be its area? 23. A rectangular piece of ground is 60 yards long and contains half an acre. It consists of a walk 8 feet wide surrounding a grass-plot. Find the area of the plot. 24. How many boards 18 ft. 6 in. long and 7 in. wide will be required to floor a room 10 yards i ft. 9 in. long and 8 yards 6 in. wide ? Ex. 62. DUODECIMALS. 365 25. How many tiles 7 in. square will be required for the floor of a kitchen 19ft. 3 in. long by 13ft. gin. wide? 26. The length of a room is double its width, and the area of the floor is 136 sq. yds. i ft. 18 in. : find its length. 27. The breadth of a rectangle is one-third of its length, and its area is 63s sq. yards 5 ft. 48 in. ; find its length in feet. 28. The area of a rectangular field whose length is 3 times its breadth is 6ac. 960 yds.: find its perimeter. 29. A rectangular field is 1050 links longer in length than in breadth, and its perimeter is 3400 links : find its area in acres, &c. 30. How many cubic feet are contained in a beam 20 ft. 4 in. long, I ft. 5 in. broad, and 10 in. thick? 31. What is the cubical content of a cistern 6 ft. Sin. long, 5ft. 10 in. high, and 3 ft. 5 in. wide? 32. How many cubic feet and inches are there in a block of marble, each of its three dimensions being 4 ft. 9 in. ? 33. Find the solid content and also the surface of a cube whose edge is 4J feet. 34. How many cubic feet of air are contained in a room 40 ft. loj in. long, 25 ft. Sin. broad, and 16 ft. 9 in. high? 35.. How many cubical packages each having 4^ inches in an edge will fill a rectangular box whose length, breadth, and depth are respectively 4ft. 4 in., 3 fl. 3 in., and 2 ft. 2 in.? 36. Water is flowing into a cistern whose base measures 4840 sq. inches : how many cubic feet will have been supplied when the depth of water is 3 J feet ? 37. Find the length of an edge and the area of a face of a cube of which the solid content is 29 cu. feet. 541 in. 38. The depth of water in a cistern whose base contains 1344 sq. inches is 2 ft. 10 in. Find the depth of the same quantity of water in another cistern whose base contains iq88 sq. inches. 39. A rectangular cistern whose length is I3|ft. and breadth 6 ft. contains 294J cu. feet of water : what is the depth of the water ? and what is its weight in tons when a cubic inch of water weighs 252 '5 grains ? 40. A cubic foot of gold is extended by hammering so as to cover an area of 6 acres. Find within one ten-millionth of an inch the thickness of the gold as a decimal of an inch. 366 SQUARE AMD CUBIC MEASURE. Ex. 62. 41. Express in yards, feet, and inches (i) 23 ft. 11' 10" 8'". (2) 7isq. ft. 9'8"4"'6". (3) 83 cu. ft. 6' s" 10'" 8" 8'. (41) 143 ft. 6' 9" g'". (5) 197 sq. ft. o' i" 3'" 4" S'- (6) 1 763 cu. ft. 8' 10" 1 1'" o" 1" i''. (7) 844 sq. ft. s' 6" d" 8i' 9^. (8) 2341 cu. ft. 5' 6" 8'" 8" 9" 10". Find, by Duodecimals, the area of a rectangle whose sides are 42. 17 ft. 4 in. and 9 ft. 11 in. 43. 25 ft. 6 in. 7 parts (or seconds) and 11 ft. gin. 44. 7 yds. I ft. 6f in. and 13 ft. 5 J in. 45. 31 ft. 4 in. 6" and 17 ft. loin. 8". 46. 15 yds. 2 ft. 4^ in. and g yds. 2 ft. 4! in. 47. 451 ft. 6iin. and 71 ft. 3iin. 48. 207 ft. 4iin. and 95 ft. 7-j:Vi"- 4g. 17 yds. 2 ft. gin. 7 pts. and 11 ft. g in. lopts. 50. 31 yds. 6-1 in. and 5yds. ift. i\'-a\. 51. 19 yds. 2 ft. 6| in. and 7 yds. i ft 3! in. 52. 15 yds. 2 ft. 4 in. 10 pts. and 7 yds. i ft. 11 in. 7 pts. 53. Find by Duodecimals the area of a square whose side is 2 yds. I ft. 3|in.; 17 ft. 4 in. 6|pts.; 123 ft. 6Jin. Find by Duodecimals the volume of a rectangular solid whose dimen- sions are 54. 13ft. 8' 11", 14ft. gin., and 15ft. \' 1". 55. 18 ft. 7 in. 4 pts., 17 ft. 3 in. 9 pts., and 1 1 ft. 1 1 pts. 56. Find by Duodecimals fhe volume of a cube whose edge is lift. 6in. 5pts.; 5yds. 2ft. 7in. 10 pts.; 3yds. i ft. 74in. 57. Find the cost of a marble slab whose length is 5 ft. 7 in. and breadth i ft. 10 in. at 8j. ^d. a sq. foot. 58. What will be the expense of painting a wall 22 ft. 6 in. long by 10 ft. 8 in. high at is. \o\d. a sq. yard? 5g. Find the cost of carpeting a floor 23 ft. 4 in. long and 16 ft. 9 in. wide with carpet at 6s. yi. a sq. yard. 60. Find the side of a square court-yard the expense of paving which at 3^. ^d. per sq. yard is £■^9,. 10. g. 61. An upholsterer covers a floor 21 ft. 8 in, by 16 ft. 6 in. with carpet 27 inches wide : find the cost of the carpet at 3.r. \\d, per yard in length., Ex. 62. DUODECIMALS. 3^7 62. Find the cost of painting the four walls of a room whose length, breadth, and height are respectively 24 ft. 3 in., 16 ft. 8 in., and i \\ ft., at 3^. \^d. per sq. yard. 63- How much paper, f of a yard wide, will be sufficient to paper a room 22 ft. S in- long, 12 ft. i in. broad, and 11 ft. 3 in. high? and how much will it cost at i^\d. a yard ? 64. A room is 20 ft. 6 in. long by ij ft. 6 in. wide, and is 16 ft. high. It has two doors, each 8 ft. high by 3 ft. 9 in. wide, and three windows, one 5 ft. by 7 ft., the other two each 5 ft. by 4 ft. How much paper a yard wide will be required to paper it ? 65. A box with a lid is to be made of inch-and-a-half plank ; the ex- ternal dimensions to be 3 ft. 6 in., 2 ft. 6 in., and i ft. 9 in. How many square feet of plank Will be used in the construction ? 66. The floor of a hall is 260 feet long and 93 feet wide. Find the cost of suppljring it with carpet at 8j. 3(/. per square yard, and oil-cloth at 3J. gi/. ; the oil-cloth to be laid along the sides and ends a yard wide, and the carpet to extend 6 inches over the oil-cloth everywhere. 67. A rectangular piece of land is 1345 Jinks in length and 860 in breadth : how many square feet would be occupied on paper, by a plan of the land drawn from a scale of an inch and a half to the chain ? 68. How many pounds of gunpowder, of which each cubic foot weighs 932 oz., will fill a box whose Jieight is 2 ft. s in. , breadth i ft. 7 in., and length 5 ft. 9 in.? 69. If 67 gallonfe of water be -equal to i of cubic feet, how many gallons can be conteiaed in a seistern S ft. 10 in. long, 3 ft. 7 in. wide, and 2 ft. 8 in. deep? 70. Find the cost of digging a cellar 1 8 ft. 4 in. long, 1 2 ft. 9 in. broad, and 14ft. sin. deep, at M. per cubic yard. (By Duodecimals.) 71. A pond whose area is 4 acres, is frozen over with ice uniformly 6 inches thick. If a cubic foot of ice weigh 896 oz., find the whole weight of the ice in tons. 72. An iron bar is i| of an inch broad, | of an inch thick, and \\ feet long. Find its weight at 4I ounces per cubic inch. 73. If s6 cu. ft. 912 in. be the content of an open cistern, 6 ft. 2 in. long, and 3 ft. 4 in. wide, what will be the cost of lining the inside of it with lead at i&r. i^. per square yard? 74. What would be the cost of paving a hall, 50 yards long and 50 feet broad, with marble slabs i foot long and 9 inches broad, the price of the slabs being £i, per dozen? 368 SQUARE AND CUBIC MEASURE. Ex. 62. 75- A canal, 10 miles long, has an average width of 7 yards and is 5 f'SGt deep. How soon would the excavation of it be completed by 100 men, each removing, on an average, 15 cubic yards per day? 76. To what uniform depth must a piece of ground, 414 yards long and 57 yards wide, be excavated, that the earth taken out may form an embank- ment of 25530 cubic yards, supposing the earth to be increased one-ninth in volume by removal ? 77. A side of cne square is 9-49 inches ; and if this square were made 6"24 inches longer in one of its dimensions, and 6*24 inches shorter in its other dimension, the area of the rectangle thus formed would be equal to that oi another 7,Q^3x^. Find a side of the latter square. 78. The depth of a box is equal to a third of its length and the width is a third as much again as the depth. The content is 32 cubic feet : find the dimensions. 79. A postage stamp is f| of an inch long and j of an inch broad : how many postage stamps will be required to paper a room 14 ft. 9 in. long, 9 ft. 3 in. wide, and 10 ft. 6 in. high, the room containing two win- dows each s^ ft. by 4 ft., and three doors each 6 ft. by 3 ft. ? 80. A room whose height is 11 feet and length twice its breadth, takes 143 yards of paper, 2 feet wide, for its four walls: how much carpet will it require ? 81. The carpeting of a room, twice as long as broad, at 5J. per square yard, cost £fi. 2. 6 ; and the painting of the walls at 9. 15. 5, what is the price of a pound? 3. Find the number of square feet in 3 acres i rood 1 perches. 4. If 12 horses are fed for 17 days at the cost of £,\i.. is., how many days can 4 horses be fed for ;^ii. 14.S., the price of food and the rate of consumption being the same in both cases? 5. Find the cost of 6^-f lbs. at £1. 16. 10 the pound. 6. From the sum of ^ and f of ij subtract the sum of ^ and -3 of 34. 7. Express 89 gallons i quart i pint as the decimal of 572 gallons. 8. Find the number of grains in -35 of 3 lbs. i oz. Troy. 9. What is the simple interest on ^£'429. 3. 4 in 4 years at 4^ per cent, per annum? 10. Extract the square root of 3312400. II. I^or Direct Commissions. May, 1870. I. Find the number of pounds Troy in three hundred millions, three thousands, eight hundred and forty grains. ■!.. If 257 pounds of tea cost £-3,^. 16. oj, what is the price of a pound? 3. If a peck of corn weighs 7 lbs. 5 oz. Avoirdupois, what is the weight at that rate of 73 quarters i bushel? 4. A rectangular cistern 9 feet long, 5 feet 4 inches wide, and 2 feet 3 inches deep, is filled with liquid which weighs 2,520 pounds. How deep must a rectangular cistern be which will hold 3,850 pounds of the same liquid, its length being 8 feet, and its width 5 feet 6 inches? B.-S. A. 24 370 EXAMINATION PAPERS. 5. Divide /^ of ^ by 2^ of ij. How many square yards are there in the fraction of an acre which the result represents? 6. Multiply 28-8 by -0595, and divide the product by 9520. 7. Find, as a fraction in its lowest terms, the difference between j^^r and -g\-g, and then express that difference by a recurring decimal. 8. Find the value of '00625 of a sovereign. 9. What is the principal sum to be placed at simple interest at the rate of 4 J per cent, per annum, that in 16 months it may amount to £l9- ". 6? 10. Extract the square root of 452929. III. Control Department. April, 1872. I. In 164,723 pints how many quarters, bushels, pecks, &c.? ■i. If 15 men can reap 20 acres of corn in 6 days working 14 hours a day, how many men must be employed to assist 10 other men to reap 6 acres in if days of 8 hours a day? 3. Find (by Practice) the dividend on £,'j^i. 13. 6 at 14J. ^d. in the pound. 4. Find the amount of ;^8900 in three years at ^\ per cent, compound interest (neglecting fractions of a penny). 5. Add together if, 3^, i/y, and 4j. 6. Subtract 5^^ from 6|. 7. Multiply 4^, ^, i^, and i. 8. Divide sJ by i^. 9. Add together 84'69i2, -001567, 1-0056 and 549-2. 10. Subtract 23-69428 from 50-012. II. Multiply 40-061 by '0054. 12. Divide '055757592 by -009207. 13. Reduce 33 yards to the decimal of a mile. 14. Extract the cube root of 9555119848. 15. Reduce -267187^ to a vulgar fraction; and if the unit he£6, reduce the fraction to shillings, pence and decimals of a penny. 16. When the 3 per cents, are at 87J, and shares paying 5 per cent, are at 130J, which is the more profitable investment? and what sum does a person invest, when the difference of the incomes resulting from the two investments is ;^56i? OXFORD LOCAL EXAMINATIONS. 3/1 IV. UNIVERSITY OF OXFORD. LOCAL EXAMINATIONS. Senior Candidates. May, 1870. I . Write out Troy weight. How many grains are there in a, pound Avoirdupois? How many pounds avoirdupois are equivalent to one hundred and nine thousand three hundred and seventy-five pounds Troy? ■i. Divide two hundred and eight pounds two shillings and sixpence farthing by twenty-three; and multiply one mile seven furlongs thirty-nine poles by forty-one. 3. Obtain, by Practice, the cost of three hundred and fifty-five things at one pound sixteen shillings and eight pence each; and calculate a person's wages for five months three weeks and six days at one pound seven shillings and five pence per month. 4. Simplify f^^, and ^ of ^^ of iy^. Also add the results. 5. Add together f of ;^i. 6. 6j, -^ of a guinea, ^ of a sovereign, •437s of a shilling, and '1375 of half a sovereign. 6. Divide -045 by -0015, 4-5 by 150, and -45 by -15. 7. What part of a sovereign is three pence three farthings? and what decimal fraction of a pole is an inch? 8. Extract the square root of 893830609, and raise i'o5 to the fourth power. 9. Compare the simple and compound interest on £^1.. ioj-., at the end of four years, reckoning money at 5 per cent, per annum. 10. If a farthing be the interest on a shilling for a calendar month, what is the rate per cent, per annum? II. If twenty-seven hundred-weight twenty-one pounds cost three hundred and seventy-nine pounds two shillings and three pence three farthings, what will be the cost of three hundred-weight three quarters fifteen pounds? 12. If three persons are boarded four weeks for seven pounds, how many can be boarded thirteen weeks five days for one hundred and twelve pounds? V. Senior Candidates. May, 1871. I, How many persons can be paid i\s. 6d. each out of a sum of ;^I70. 10. 2? And, if the balance be also distributed equally among them, how much more will each receive? 24 — 2 372 EXAMINATION PAPERS. ■i. Write out the tables of Solid Measure and Measure of Capacity. If a gallon contains 277-274 cubic inches, how many cubic yards are there in 100 bushels? 3. Find the cost of 7 lbs. 2 oz. S dwts. 4 grs. at 3J. 7 Jaf. per dwt. 4. To 34 of Sff add - of (6J - ij) ; and from their sum subtract -^ . 3 T 5. Divide 2-375 by 250, by 2-25, and by -00005. 6. Reduce ^^ and -^^ to decimals; and -2361 to a vulgar fraction. 7. Find the value of 2-5 of i ton 6cwt. + 3-i25 of 2qrs. 16 lbs. + 3-75 of 448 oz. 8. Extract the square roots of 22099401 and 210J. 9. How many hours a day must 24 men work to accomplish as much in 5 days as 25 men could do in 4 days if they worked 6 hours a day? 10. What will a. debt of ;£^425o amount to, if it is left standing for 2 J years at J per cent, per annum compound interest? 11. If a school-room is 25 ft. long and 20ft. vdde, how many children will it accommodate, allowing for each of them 8 superficial feet at the least? And if the room is 10 ft. 4 in. high, what cubical space is there for each child? 12. Find the price of Three per Cent. Stock, when an investment of ;^434. 12. 6 produces an income of ;^i4. ^s. VI. UNIVERSITY OF CAMBRIDGE. LOCAL EXAMINATIONS. Senior Students. Dec. 1869. 1 . Divide 69 miles 7 f. 39 po. 2 ft. by 492. ■L. Add together \, |, |, 4, |, |; and subtract the result from ioot^- Reduce to their simplest forms _^; iofixlofl-fUiof2o); 3J^. I 3. Add together S36'42i. S3642T. S"3642i; and subtract the result from 1 00000. Provethat^=^3^'. -2ll 9321 CAMBRIDGE LOCAL EXAMINATIONS. 373 4. State your rule for the division of decimals ; divide 1 by "2, '002 by ■02, 2*2 by 2'i. , 5. Find the value of J of 17^. 8(/. + 2'62^ of IJ. -j off of 5 J. 41/. + '263 of 25^.; and reduce the result to the decimal of £«). 6. Extract the square root of 35'672 and of '4 to 4 places of decimals. 7. What would be the cost of paving a hall 50 yards long by 50 feet broad with marble slabs i foot long and 9 inches broad, the price of the slabs being ^'5 per dozen ? 8. Find the difference between the simple interest and the discount on ;^ioo for 5 years at J per cent. If the present worth of £,11^ due two years hence be ^£'200, what is the present worth of ;£^iooo due six years hence at the same rate? g. Of the boys in a school one third are over 15 years of age, one third between 10 and 15. A legacy of ;^ioo can be exactly divided amongst them by giving loj. to each boy over 15, 6j. '&d. to each between 10 and 15, and y. /^d. to each of the rest. How many boys are there in the school ? 10. If the price of candles 8J inches long be 9\. 10. 4 for 11 months at (>\ per cent, per annum ? 9. If the weight of a cubic foot of water is 62'3^ pounds avoirdupois, what is the error in calculating the weight of 1000 cubic feet on the supposition that a cubic fathom weighs 6 tons? 10. A publican mixes 4 gallons of gin which is worth 15^. a gallon, with 4 gallons of water and a gallon of base spirit worth \os. ; what will he gain per cent, on his outlay by selling the mixture at 2^-. \od. per bottle of six to the gallon ? 11. A man having ^^550 in cash, invests it in the Three-and-a-half per Cents, when they are at 88J ; afterwards when they are at 93 he sells out, and invests his money in a mortgage which brings him in 5J per cent. What difference does the transaction make in his income ? IX. INDIA FOREST DEPARTMENT. Jan, 1870. 1. Shew that the ,^2 lies between ^ and ^. ■i. Extract the cube root of 669"92i875 cubic feet, and reduce the result to inches. 3. What is the present worth of ;^i,842. IJJ., payable a quarter of a year hence, at 5 per cent. ? 4. What length of paper 22J inches wide would be required to paper the walls of a room 18 ft. 9 in. long, 13ft. 3 in. broad, and 14 ft. 6in. high? 5. Express 4 acres 2 roods 16 perches as the decimal of a square mile. 6. Multiply by duodecimals 7 ft. 3 in. 5 pts. by 5 ft. 7 in. 4 pts. and the product by 4 ft. 2 in. What does the product become when expressed in cubic feet and inches? 7. A man purchases ^^700 Stock in the Three per Cent. Consols at 94i, and also invests ;^s85 in the purchase of Russian Five per Cent. Stock at 974. How much Stock has he standing in his name ? If he sells out of the Three per Cents, at 95 and out of the Five per Cents, at 96J, does he gain or lose by the transaction, and how much ? 8. Divide 9I-S63 by 87-56. 376 EXAMINATION PAPERS. 9. A tradesman's stock in trade is valued on January ist, 1868, at ;£'8,ooo, he has also £-if,o in cash and owes ^1,870; during the year his personal expenses, ;^3oo, are paid out of the proceeds of his business, and on January rst, 1869, his stock is valued at £i,^io, he has £i'iO in cash and owes ;^i,5io. What is the whole profit on the year's transactions after deducting 5 per cent, interest on the capital with which he began the year ? 10. Two clocks point to 2 o'clock at the same instant on the afternoon of Christmas day; one loses 8 seconds, and the other gains 9 seconds in 24 hours : when will one be half an hour before the other, and what time will each clock then shew ? X. FOR ADMISSION INTO THE R. M. ACADEMY, WOOLWICH. July, 1870. I. If building-ground be bought for 15J. 90'. a square yard, what will be the cost of half an acre of such ground ? The purchaser of the half-acre builds a house upon it and lays out the ground at a further cost of £1,0^1,. i,s., what rent per annum must he obtain so as to realize 9 per cent, on his whole outlay ? ■1.. If 20 English navvies, each earning 3j. 6d. a. day, can do the same piece of work in 15 days that it takes 28 foreign workmen, each earning three francs a day, to complete in 20 days ; taking the value of the franc at \od., determine which class of workmen it is most profitable to employ. If a piece of work done by navvies cost .^3,000, what would be the cost of the same work done by foreign workmen ? ,97 (291 3. Reduce "^ I^I . Divide ^2-4 by -00625; and without using the common rule for the extraction of the square root, prove that i-8j is the square root of 4. What is meant by the "course of exchange" between two countries? A merchant in New York wishes to remit to London 5,110 dollars, a dollar being equal to 4,s. (sd. English, for what sum in English money must he draw his bill when bills on London are at a premium of a\ per cent. ? ROYAL MILITARY ACADEMY. 2i77 XI. Royal Military Academy. Dec. 1870. 1. Divide £,%6. 9. 7 by 82. 2. Find the difference in yards and fractional parts of a yard between 10 chains 5 links and i furlong 2 rods. 3. An account after a discount of aj per cent, is taken off becomes £,\(>. 14. 9. What was the sum due before the discount was subtracted? 4. If a litre is "22 gallons, find to the nearest penny in English money the value of a pint of liquid which is worth 10 francs the litre, 1200 francs being equivalent to £,\^. 5. Express •loi of 1 lb. 5 oz. as a decimal of i cWt. 6. A person borrows £,\oo, and at the end of each year pays £,^i to reduce the principal and to pay interest at 4 per cent, on the sum which has been standing against him through that year. How much will remain of the debt at the end of 3 years? XII. Royal Military Academy. July, 1871. I. Find the number of inches in the length i mile i furlong 3 poles, ■i. Find -the value of 17 quarters i bushel of com at the price of 4j. ^\d. the bushel. 3. What is the least weight which can be expressed either by a number of Troy pennyweights or by a number of Avoirdupois ounces ? Give the answer in Troy weight. 4. A vessel holds 2J pints. How many times can it be filled from a cask of 56 gallons, and will there be any remainder ? i;. Find the value of of £1. o. li. 6. A school of boys and girls contains 453 children, and the boys are ■525252... of the girls. How many boys are there ? 7. If a metric system of area were adopted wherein i acre i rood 3 perches is represented by 5- 12, express the unit of measurement in square yards and decimal parts of a square yard. 8. At what rate per cent, is the deduction made when igj. io\d. is taken from an account of ;^39. 15J. in consideration of immediate pay- ment? 9. What is the compound interest to the nearest penny on ;^83. 14. 7 in 7 years at the rate of 3 per cent, per annum? How much does this compound interest exceed the simple interest in the same time on this principal at the same rate per cent.? 37^ EXAMINATION PAPERS. XIH. EXAMINATION FOR ADMISSION INTO THE INDIA CIVIL ENGINEERING COLLEGE. June, 1871. 1. Convert ^ and -jtI^ into decimal fractions: divide the second result by the first, and convert the quotient into a vulgar fraction in its lowest terms. 2. Find the length of the side of a square which is equal in area to the rectangle, the sides of which are 5x3 yards i foot 11 inches, and 1,628 yards 11 inches. 3. Find the length of the edge of a cube which contains 450 feet 1,088 inches. 4. The external length, breadth, and height of a rectangular wooden closed box are 18 inches, 10 inches, and 5 inches respectively, and the thickness of the wood is half an inch. When the box is empty it weighs 15 lbs., and when filled with sand 100 lbs. Compare the weight of equal bulk of wood and sand. 5. A cask weighing 1 cwt. \n. lbs. 4 oz. floats in a square cistern of water, whose side is 2 ft. 6 in.; on the removal of the cask, find how much the water will sink in the cistern, supposing a cubic foot of water to weigh 63 lbs. 6. The interest on a certain sum of money for two years is £,'J\. 16. 7J, and the discount on the same sum for the same time is ^63. 17J., simple interest being reckoned in both cases. Find the rate per cent, per annum, and the sum. XIV. UNIVERSITY OF LONDON. MATRICULATION EXAMINATION. Jan. 1870. 1. Find the value of | - A + A. and divide \\ by the result. Divide '0075 by 25*6, and state the principle upon which you fix the position of the decimal point in the quotient. 2. Reduce nine inches and nine tenths to the decimal of a mile; and find the value of '0625 of i ton 1 cwt. 3 qrs. 12 lbs. 3. A sells goods to B for ;^iis. 19. 2, and gains 10 per cent, on the price he originally paid for them. B sells the same goods again, and loses 10 per cent, on the price at which he bought them. At what price did A buy the goods, and at what price did B sell them ? MATRICULATION ; LONDON. 379 4. What annual income will be produced by ;^i3,ooo invested in a. Tliree-and-half per Cent. Stock at 91 ? and by the same sum invested in a Four per Cent. Stock at 96 ? 5. Extract the square root of 10,074,538,384; and find the value of — p to four places of decimals. \/2+ I Prove that no square number can end with one of the digits 2, 3, 7, 8. XV. Matriculation Examination, yan. 1871. 5 4 8 21 6 12 1. Simplify /' , 236 and divide 'oooiss by 8-75. 2. State and prove the Rule for finding the Least Common Multiple of two given numbers. Define a Prime Number. Express 364, 2520 and 5445 as products of powers of prime numbers. 3. Find the value of '01625 of ;£204. 3. 4; and reduce 8 1b. 5 oz. 14 drs. to the decimal of a quarter. 4. What fraction when multiplied by itself produces - — -S- ? What is the length of each side of a square court which contains 43785'5625 sq. feet? 5. Supposing a gallon to contain 2 77 J cubic inches, find approximately the number of gallons of water which would cover a square mile of ground to the depth of an inch. XVI. Matriculation Examination. June, 1871. 1 . If the circumference of a coach- wheel measures 1 7 ft. '}\ in. , how often will it turn round in travelling a distance of 8 miles 264 feet? 2. At what rate of simple interest will ;^325 amount to ;£'379. 3. 4 in 5 years? What rate of interest for money can be obtained from Three per Cent. Stock, when ;^ioo of Stock can be bought for ;^85 in cash? 3. Express ^||, and ^ - 7e\ + J| - -^ in their lowest terms. 380 EXAMINATION PAPERS. 4. Find the value in decimals of 3+— 7 ^ + 76 and the quotient of the recurring decimal •2323... divided by the recurring decimal '28752875... 5. Extract the square root of 324'oooo5625 ; and find the value of V5 + V3 V5-V3 . V6-V3 V5+V3' in both cases to 6 places of decimals. 6. If a man can do a piece of work in 77 hours which a boy wants 121 hours for, in how many hours minutes and seconds can they do it conjointly? XVII. UNIVERSITY OF OXFORD. RESPONSIONS. Hilary Term, 1871. 1. Find the present value of £,\i-i(> due four years hence at 5 per cent, simple interest. 2. What will be the cost of carpeting a room 25 feet long and 13 ft. 6 in. broad with carpet costing los. a square yard? 3. What is the rent of a field containing 112 acres 2 roods 29 J poles at £^. 12. 10 an acre? 4. Divide I368'2394 by 2400'2i and by "00240021; and add together •14, '01 16, and '345. 5. Simplify (i) I* of -^ 4r 01 -^ ^ 64 . *__i6o_ iiof9A 2|ofi-5 VS 5 7 15/ V ^345/ (2) 6. Find the value of -5^of ;^i + - of ;^I40. 10. 6 +- of a guinea. 7. Find the Least Common Multiple of 5, 7, 16, 28, 48, 76: and the Greatest Common Measure of 1288, 1736, 104. 8. If 120 men can build a house 60 feet high in 15 days, how many will it take to build a house 55 feet high in 10 days? 9. Find the difference between the simple and the compound interest on ;f955 at 6 per cent., for 4 years. OXFORD RESPONSIONS. 38 1 10. A train 88 yards long overtook a person walking along the line at the rate of 4 miles an hour and passed him completely in 10 seconds: it afterwards overtook another person and passed him in 9 seconds. At what rate per hour was this second person walking? XVIII. Responsions, Trinity Term, 1871. I. Find the value of 50,000 boxes of matches at 3J. i^d. per gross (12 dozen). 1. The floor of a, room 19 feet 6 inches long by 13 feet wide is covered by a carpet each strip of which is 3 feet 3 inches in width. The strips of carpet cost 4^. 9^. per yard. What was the cost of the whole carpet? 3. What vulgar fraction is equivalent to the sum of i'45 and '175 divided by 2*5? Simplify 4 3 5. Find the value oi £1. 6. 8-7- ij off, and of f of I5i. + f of 4J. (id.-%oi ^. lid. 6. Multiply '035 by "00795, ^'^^ divide 2'5 by "36. 7. If a tithe-owner receives ;^I04. 15. i for a tithe rent-charge of the nominal amount of ;^ioo, what will he receive for a rent-charge of ;^ii3. 16. 6? 8. Find the Greatest Common Measure of 7056 and 7092, and the Least Common Multiple of 8, 12, and 20. 9. The Guernsey pound contains 18 ounces Avoirdupois, and the Guernsey shilling contains 13 English pence. If a Guernsey pound of butter costs is. 6d., Guernsey money, what will be the price in English money of 2 J lbs. Avoirdupois? 10. The interest on ^^250 for 73 days amounts to £ii. 15J. Find the rate of interest per cent, per annum. 11. If the price of £100 Three per Cent. Stock is 93J sterling, what would be the income obtained from investing ;^2ooo sterling in that Stock? ^^-i°^i I 4+^f-f 2331 3^2 EXAMINATION PAPERS. XIX. UNIVERSITY OF CAMBRIDGE. PREVIOUS EXAMINATION. March, 1869. I. In subtraction how do you evade the difificulty of taking a greater digit from a less? Illustrate your answer by taking 791 from 943. ■i. What is a measure? — a common measure? — the Greatest Common Measure? Find the g.c.m. of 13 x 17 x 19, 17x19x21, 19x21x13. 3. Divide 24763 by 56 by short division. Explain how you determine the remainder. Justify your method. 4. Add together 23-076, I9'445, 3I'263; and multiply 3"620I5 by i'oo236. 5. Add I of I to I of 2^, and multiply the result by f of |-^J+|. 6. What fracti e-' ■*)*:!• ""•;±:i I 2- 5 and to the difference of the first and third of these fractions add the second. B-S. A. 25 386 EXAMINATION PAPERS. 3. Multiply 76-371 by 8-54. Divide '37848 by '456 ; 3-7848 by -0456, and 3784-8 by -00456. 4. Divide i-pSot by 7-456. 5. What fraction of 1 qrs. 10 lbs. 7 oz. 9 drs. is i qr. 7 oz. 13 drs.? What part is;^i. 2 fl. 2 c. 5 m. of ;^6. 2. 6? 6. Find the expense of turfing a plot of ground, which is 40 yds. long, and 100 feet wide, with turfs each a. yard in length and i foot in width j the turfs, when laid, costing 6j. 91/. per hundred. 7. A square block of stone, 2 feet in thickness, is, in cubic content, 5 cub. ft. 24 in. : what is the length of its edge ? 8. Find by practice the value (i) of 5362 things at £\. 5. 3J, (ii) of 3 lbs. 4 oz. 7 dwts. at £^ 2. 6 per lb. 9. A gives away in charity - of his income, and pays — of it in rates and taxes — with these deductions he has\;^473. 13. i left, what is his gross income ? 10. A force of police 1921 strong is to be distributed among 4 towns in proportion to the number of inhabitants in each : the population being 4150, 12450, 24900, and 29050 respectively: determine the number of men sent to each. 11. Find the discount on ^388. 17. 9 due 18 months hence, interest being reckoned at 4 per cent. 12. Find the alteration in income occasioned by shifting .^3200 Stock from Three per Cents, at 86|, to Four per Cent. Stock at 114I: the brokerage being 5 per cent, on each transaction. XxIV. Previous Examination, Dec, 1870. 1. SuBTRAct two thousand two hundred and two millions two thousand and two from eight thousand six hundred and sixty-six millions sixty-six thousand and ' sixty-six. Divide the difference by sixty-four, and express the quotient in words. CAMBRIDGE PREVIOUS EXAMINATION. 387 2. Simplify: 3. Divide i'95i7 by 673000 and 64000 by 'oooS. Reduce 4^]^ and 5^^^ to decimals and '9375 and '4926 t" Vulgaf fractions. 4. Find the value of •0013671875 of a mile, and reduce 10 lbs. 5 oz. to the decimal of a ton. 5. Sound travels at the rate of 1140 feet a second. If a shot be fired from a ship moving at the rate of 10 miles an hour, how fat will the ship have moved before the report is heard at a place 14J miles otf? 6. Divide £1^ into an equal number of half-sovereigns, crowns, half- crowns, shillings, sixpences and fourpences. 7. Extract the squaire root of 32239684, of 17^ and of •121 to three places of decimals. 8. What is the rent of 10 acres 3 roods 26 poles at £^. 8. 103 per acre ? If 3 cwts. 3 qrs. 21 lbs. 12$ oz. cost £\. 8. 9, what is the price per cwt.? 9. If the cost of papering a room 8J yards long and 6^ yards wide with paper 2 feet wide at ^d. per yard be £^. 19. 8, find the height of the room. 10. Find the present value of ;^8o8. i. 4 due 3 years and 9 months hence at 4 per cent, per annum simple interest. 11. By selling out ;^45oo in the India Five per Cent. Stock at H2j and investing the proceeds in Egyptian Seven per Cent. Stock a person finds his income increased by ;^i68. 15J. What is the price of the latter Stock? 12. If 3 per cent, more be gained by selling a horse for ;^83. 5^. than by selling him for ;^8r, what must his original price have been? 13. At what distance Irom London will a train which leaves London for Rugby at 2-45 p.m. and goes at the rate of 41 miles an hour meet a train which leaves Rugby for London at i'45 p.m. and goes at the rate of 25 miles an hour, the distance between Rugby and London being 80 miles ? 25—2 388 EXAMINATION PAPERS. XXV. Previous Examination. Dec. 1870. I. Multiply three hundred and forty-three millions four thousand nine hundred and seven by two hundred and sixteen millions three thousand six Iiundred and six. ■i. Divide 830728 by 231 by short division and explain the process. 3. Simplify ^of6Uof24ii-4Hx3§l-3H ^ ^^^^ ^'' 8U X 6^+4^1 - 1\% X siK i4li "" *^^' *'' '^^''yrs'oo- 4. Subtract 3-05 from g'ois and divide I2'03i by 5300 and io'24 by •0128. Add together '35 of ioj., '54189 of i2j. ^. and -isS of i2j-. dd., and reduce the result to the decimal of ;^i. 5. A clock which was i'4 minutes fast at a quarter to 11 p.m. on Dec. 2 was 8 minutes slow at 9 a.m. on Dec. 7, When was it exactly right? 6. Extract the square root of 546121, of 65fx and of*i69 to three places of decimals. 7. After paying income-tax at the rate of 4;/. in the pound, a man has ;^49i. 13. 4 remaining. What is his income? 8. The sum which will pay ^'s wages for 61J days, will pay 5's wages for 8if days. For how many days will it pay the wages of A and B together? 9. Find the present value of £i^'i. i. 4 due 3 years and 4 months ^ hence at 4^ per cent, per annum simple interest. 10. How many pounds of tobacco at 5J. }, * » 11' 155' 217' 259' 333 387 and simplify 5. Find the area of a room 17 ft. 4 in. long, and 14 ft. 3 in. broad, and the cost of carpet i ft. ir| in. broad, at 5^. ^\d. per yard. 6. Find by Practice the cost of (i) 4321 articles at £t. ii. 8; (ii) 1978 at £,1- 9. 2; and (iii) 214 acres 3 rds. 29 poles at;^i25. 7. 6 per acre. 7. Divide (i) 3-003 by 148-28 ; (ii) '003003 by ■014828; and (iii) 300-3 by I -4828. Find the value of -42857! of £.%. 4. 3i + 'i875 of ;^5. 11. 8 + %f ■8t of 4J. 9| fin2; 4288179204. 49110419796; 144872064531; 63723226584. 10. 1515868. Ex. 6. 1. 99130823; 66087215. ..4; 113292369...1; 88116287...1; 72095144. 2. 25280669. ..14; 21753134. ..5; 2687887. ..91; 193421...811. 3. 42439.!. 5498; 26171...7874; 5822. ..6639; 4167.. .2898. 4- 2733534.. .7; 1195921...15; 1063041...7; 797280.. .79. 5. 28716695. ..68; 19339815...68; 10028052. ..152. 6. 1308913...46; 98168. ..426; 6544.. .6826; 24542. ..426. 7. 12433128...54; 653266.. .334; 6566.. .27734; 4240r... 18834. 8. 9396.. .538. 9. 3507. 10. 6905. II. 1887378. 12. 109; 278. 13. 67:999. 14. 67. 15. 191919 miles. ..95 miles, 16. 13212. 17- 5193171- 394 ANSWERS. Ex. 7. I- IOS5559- 2. 82152. 3. 15610; 7814. 4. 45653376; 8862. 5. 1639799; i: 5185- 6. 1877; 1217. 7. 73; 1504. 8. 1008; 1380. 9. 8969578. 10. 5761; 76.' II. (325-293)x(3o6-r-i7) + (iooo + 99); 1675. 12. 34856-763x41 + 1998-^(663-441); 3582. 13. 106; 1050. 14. 9801; 639. 15. 33211521848. 16. 10091401. 17. Each side is equal to (i) 1870; (2) 792; (3) 717409; (4) 1481040; (5) 1771561; (6) 290521; (7) 87949951 1 ; (8) 755301239. Ex. 8. I- 17; ^7; 55- 2. 3; I ; 707. 3. 37; 999; I. 4. 365; 571. 5. 127; 2476099. 6. 11; 21. 7. 59. 8. 69. 9. 3431. 10. 21. II. 84. 12. No, G.c.M. is 27. Yes. 1%. Yes; Yes, 14. 63 gallons. 15. 40 grains. 16. 35. 17. 25. 18. No. Ex. 9. 1716; 734877; 1225449. 2. 9896425; 637225; 4040400. 3. 67868155; 203667; 20044430912. 4. 159137; 4029051. 5. 1628055; 7258671. 6. 27720; 334639305- 7- 720720. 456326325. 9. 10080; 7560. 10. 11880; 352800. 11. 180180; 27720. 12. 25200; 9828. 13. 863940. 14. 98280. 15. 602910. 16. 2520 sec. = 42m. 17. 8. 18. 76. 19. 982832. 20. 875. 21. 60 min. Ex. 10. 17 : pr.; 23 : 23. 2. pr.; 37 : 7. 3- 41; pr-; S3- 4- 4- 5- S- 6. 5. 7. 2.3.7.11; 2^5.23; 2«.5; 2^.33.11. 3''-7-29; 3-S'"; 2'.3'.5''-7- 2^3-S-7-i3i; 23.33.7.11; 52.7.112.19. 3^5.7.11; 7.13.19.31; 72.31". 3-5'-7-i3''i 5-7'-i3-3i; 7^-i3''-23- 112.13.17.19; 72.172.293. 172.29.31; 132.178.89. 3''; 22.33.5.7.11.13 = 540540. i; 22.33.52.7.112=2286900. 3-5-7 = io5; 3'''5''-7' = iio25. 32.7.11 = 693; 33.72.112=160083. 19. 19; 13.17.19.21 = 88179. Ex. II. 5- V¥-; H^; ^«F; ^V-^; ^^• 6. iF; Y-; Hi-, V/; -W; ^liW; HV ANSWERS. 395 »T 1X2-3. ■ ISfliL. 814 8. 4 8 31.. ft 8452 5 . 9^696098 8. 12; ii; ii\^; 11; 109I; 22U; 16H; 43H; ISSIM- 9. I76if ; 63tf ; 76IM; i&i\m; SslH^; 1022m*- 10. ^f; 3A; 3if ; SSj^; ^Slf; 46if ; 884m. 11. 142^; i37if; 401V Ex. 12. • e . 11 .?S - 175.208. 245 ^ IBO _l6fi 7 5 C 160 *• TIi 84> ¥T* 2. -g^-g", ^"^^ , TTB' 3* ITTS"* TTBTT* TTT^» t^ 6. ifiF; V^; ^V-; ^^. 7- 5f ; i^; 8A; 37l; i6i; 7o|. 8- A^; iVsj tItT) tSw; fr! Tf- 9' tit! oT' ItIj 3tb^- 10. ii|;M;iH- II- 12; 243ii|. 12. 68|. Ex. 13. 2- H. h A- 3- A; H; I- 4- H. *- 5- h I- 6. Mjt'A- 7- \h A. tVi- 8. H; m^; *:• 9- M; U^- 10. -Uh; if- "• ts't; t^WiV- "• 11 13- 71; i7i; ssi- 14- 63; 94f; 87I- Ex. 14. - 7S 10 48 81 . 482 441 448 462 I- w» wi "inr* ■ffTTj Tin* Tirrj sv^> Tin- ^- iff. 144. i^> iff. lil; t*. w- **■ ^• 3- Ml. Mf. Hi. ^; lAWiftr. T^WJu. tI4«^, tIW^- 4- »f. I«.«W. w. iU; M4, -¥u\°-. IM. %¥. ^• s- n«. «i«. ffU. IfU. V^"-. fUf ; l?«, m*. IfU. f4!«- 6- VA^ f«4*. W#. fiff. ^UF; Wt?. ¥//. ^'AS Wi?. ^^• 7- il. A. ii; if. il. il; f °f9f. 3*. J^- -8- ^. i«- » 8+9 8 . 13 I3+IS 15. 6 8 5+6+7+9 f 9 9- TI. iyq:^' IT. TT. ,^+,6> XT. 7. 7". 6+y+8+io' *• ^' lo- A, A; ft. If; ^s^. 7A- "• H; n*- "- i^i- 4 2Q_ 4 20 42 420 420 '3- ^irtr> -ztX' TST. ^TTT. rsir- Ex. IS. I. 3; i6U; 33llf- '- 35i; 3M; 3IIII- 3- 2;2//^;3|i. 4- ^5; lo^fl; ^xVi^- 5- Sooo; ^M^. 6. i8||; 73U4- 7- ^sii- 8. 492I. 9. i. 10. | + J. 396 ANSWERS. Ex. 1 6. I- A; sA; 7«; ^i- -^^ 6A; i74f; ^; tV s- il*; sH; 'AV 4- ifiH; 5ll; lAV; isA'f; 'MI; ^itt- s- i; U- 6. o; o; i^. 7. ifl; ^. 8. ^; U; 'H; 9- loH; i|*- 10. iS. ir. tVit- 12. 29jA- 13- 3fl; 34- 14- i4li- IS- r^fffU- '6. ^J^. Ex. 17. I- Th, \\, TT. 17, 52, 6if 2. 45iiJf, upsJK-. 90. Sf- 3. 26, I267f. 8m- 4- imJ. ItAt- 5- 37f. 8- 6. 2f, 6oi=A, 3- 7- lil. 498i*- 8. 117, 441^, i947i, 37389^-. 168726TV 9. 66^, 33583, 14. 10. 22Tfj, 3|^, 14II. II. 26^, 18IIJ. 12. 32399in^' 998999i§U. 492|oi- i3- "^1°°- 14- I- IS- I09H- 16. iStnfir- Ex. 18. 43A. i94lf. 7<>i^. i82^A, 92IH- 2. 156, 9,»i, itS^. 4i SH- A 321.1 cR 66 4- 7"> TST' 4t- 5- °> irj- I*. 4172U. ^'A, A\- 10. Iim, 9i4K. li A. li- 12- iH. H- 13- A- 14- ilif- ilf- 16. smi- 17- 22^- 18. 8JU- IS Ex. 19. 1- ttt; 24- 2. -sV; 146I. 3- tU; 6°- 4- U; i8f 5- -bIt; 2361- 6- ^Vj; i8oi4- 7- -jMit; 7oH- 8. /A; 4458A- Ex. 20. I. I; 3i; H; IB; A; A\; US; i^- 2. i; i; ^A; H- 3- 3; A^; A%; A- 4- 7; sH; i«m4; 4lf- - ^933. 1126. 26 < O^- JL- -2-5^- J^ »7 T- T S- 3Tinrir' tttt> it's- "• 23-5-, tj, -s-s-j, -^. 7. i, i. 8. bA- 9- 4; 7It- 10- AW. if. AV- n- i; i- I2--II; «- 13- if; 2A\; 4^- 14- 3 AW. 4414111 ; 14II- 15. 6Jt; 2. 16. ^. 17- W- 18. fU4|- 19- 14?- 20. 38i. ANSWERS. 397 Ex. 21. T 1 5JB. - 13' gOg. - 7 g. O 3 . 11 . 1 0031 . 34^6 15- 13AV; QT^^iVir; 23t«t; Sp/sVA; i^^fVA- Ex. 22. I. 6876-219354. i. 417-11157. 3. 1582-357845. 4. 669-3001. 5. 28-88544; 6-024696. 6. 8-00001; 8-99949. 7. 405'970973- 8. 9932'i237939- 9. 30-836; 55-92641; -00099. 10- 4-911002; -088913. n. 16-6799884. 12. iS"635563. 13. 6-171256. 14- •034353- IS- -2218487; -0457575; -999644; 2-345679; i-ioi. 16. 3-i4r59^7- EX. 23. I. 982-99883; 230-625593; 3-399844. ■L. -8499745; -00656448; -82008. 3. -0415584; -0001; 49-07070:. 4- "001353; -00003738028; -7614948. 5. -0487291247; 290; -819. 6. -0177775; -ooi. 7. 12-66806; 23-68676; -OOI. 8. -0672; 9-56709. 9. 10; 10-01; 154s; iioooo. 10. 129300; -7567; 2260. II. 1-37; -0081; 3267-578125. 12. 104-69785..., 2-8546136..., -2181659... 13. -012; 790; 91-428571... 14. 24018000; 2590; 2895-764013. 15. 567-8; -576; -1385. 16. -007853; -0032; -0046. 17. 21-32; 2500000; -009990009990... 18. 300. 19. -00213...; -0065008... 20. 6-0360...; 91-336... 21. -0121681...; 36542894-0328... 22. -30685; 2050. 23. 147; 2296. 24. 15-845686; -0001 1... 25- -9; ss'i; 326-4- 26. 2812-5. ^7- 961-683232I. 28. 6400. 29. 2299-95; 5909-00011875. 30. 1x531-48614... Ex. 24. 1. 31-06; 267-90. 2. 643-3x9; 2-246. 3. 27-00295; I5-4S64S- 4. 1x80-5x03. 5. 64-20153. 6. 574- 7- 30969- 8. -95423- 9- -392754- 10. 14-26. IX. -1495- ". 634-3. 13- 26-451577. 14. -13059. 15- 30-799- 16- -434294; •318309. 17. 1-1037; I-I596; 1-3447- '8. 1-15866; 1144-32158. X-23X2; 290-5024. 20. 7238-88888; -703x8; 2148-58183. 19, 398 ANSWERS. Ex. 25. ^ I- •87s; -yi^; 'Si^s; -03375; 2-84375; -3216; -7578125. 2. -32768; -68359375; -0135546875; -222464; 1-0009765625; 1-687841796875. 3. -00081, -0722; -35912125; 57-666015625; -112. 4- S-8392S- 5- 2-183125. 6. 83-03355. 7. -0001; -008686. 8. -7142...; 1-923... 9. 89-235294...; 1-88872323... 10. o. II. \\, HJ. 12. 2, 4, 5, 8, 10, 16. 13. 1-6; 4-§; 3-614583; -464; -572. 14. -9146; 1-96; 9-369; i-139204.5. 15. 18-1893; -0126; 5-783; 1-2594. 16. 2-733to§; 5-287!; i-o6i9§; 1-0850694. 17. -72569; 3-3567; -20432; -98712. 18. 1-9512I; -73806; 1-29573176; •499077.4. 19. -714285; 7-461538; 4-803571428; -I26984; 7-1893. 20. 13-94230769; -944055; 2-457002; -75240384615. 21. 3*4556097; -907967034; -404633; -991006. 22. 1-761145; 3-780003; -9411764705882354; -§94736842105263157. 23- I; K; A; ift-; 8|f. 24. tVt; sH; 5i4?- 25- ^; Ml; ifi- ^6. ^V; ^tttt; tMt; sA- 27- A; ^^; 42^««- 28. \\; 8^^; ^. 29. 37lU; M4M ; Tf?fT- 30. 3H^; sil^- 31. 2M; iSfll- " 32. Tffij; i3fl- Ex. 26. 1. 408-6420311; 13-259435526. 2. 73-5249422!; 100. 3. 44-5012779837. 4- 107-77164749359. 5. -0006; 8-6iC)§; 734-668367003. 6. 2-9967; 3-661923286; -713149350^- 7- -95436; -9416; 74-3579649- 8. 3-3878; 4-1406; 13-927; 13-36. 9- 38-0375; -51014636; 8-3876426. ANSWERS. 399 10. 73I-I64S99; 27i5-i2i369J; 7i-88o3i7§23s. 11. •4322443t§; 5-37654897?; •74878703. 12. '490578703; 4-8181639610389. 13. i-lSg; -63; 27011^^=270-3300633967; 3-63. 14. 36-6; 1-36; 67-45; 580-96. IS- 12114^=12-8479236812570145903; 46178-683621; ^^-^^^ = 69-95100308641975. 8.4.9.9 16. 2-2; 1-42066; 3-47; 134-4; 8-18. 17. 6-945126; 2-5247; -24; -77. 18. '4459; 2-9714285; 21; 8-18746; 6-7647619. 19. -16; 1-6875. 20. -8863; 4^V; 37i?/ti- 21. 4-2168^3; -0445706. 22. 2530-6. 23. 3-083; ffj. 24. -0002938. 25. (i) is greatest and (2) least. 26. 1120-11200069. 27- 3887-7S5-" 28. m, 3-141S926 and 3 + — i^. 29. -36; -85; 1-115; ■0094- 30. (i) -249999... (2) -166666... (3) -00097061. (4) 1-718281... (5) -202732... (6) 3-141592... Ex. 27. 1- 234; 897; 907; 3579; 9878; 4607. 2. 9-68; -359; -1679; -0907; 60-49; -7008. 3- 76963; 56804; 87056; 76008; 80047. 4- 15-367; 534-762; 20-0137; -0708069. 5. 203-975; 90888; 600-098; 9688669. 6. 347-6905; 76050-4009; 887145333. 7- H; 268^; 5f; ni^; ss^r, 4-3; 1-83; 68-83. 8- 4*92; 55-9; 27-5028; -25; -083. 9. 4-4721..., -00032159; -3162..., -00001756; 1-7724..., -00019089; •1095..., -00000975; 13-2382..., -00062476; 3-5449--. -00004399. 10. -0824621...; -7745966...; -9607689...; 4-6612522...; 6-2164104...; 20 -493901 5...; -0967000. 11. -01705544487...; 3-12889756943...; 9-89444288483...; -97014250014-. •; -92195444573—; -26134268690...; 81236400...; •37I7679I43I-- 400 ANSWERS. 12. 8-6802649729I4I3...; 173205080756887...; •84515425472851... •3- 7'587869io639328i46...; -00594306318324145...; ■26590801173915521... 14. -264575131106459059050...; 1-252996408614166778849...; 7-100704190430692769065 . . . Ex. 28. 1. 97; 234; 6-i6; -192; -0299. 2. -289; 56-09; -4339; 7084. 3. -3842; 1-305; 5836; 20-37. 4- 8888; -4607; 9-889; 800-8. 5- 45333; i'9989; S^S'Sg- 6- 459684; 886437163. », 12. 36. t5. 177. 2-^- aS-* IK-^. 8. 10-112072614...; 1-8081931033...; -2152134170...; 1-7099759466...; •85058074056...; I-98I3073I75... 9- '•5I3507627I833...; i •395612425086...; •7539474411293...; -41351152199011...; -42864727229153...; -68763455217716... 10. -92263187770011...; 2-64875143904066...; -12599210498948...; •28844991406148... 11. 1-298024613276667544089...; -14422495703074083823.. ; 7'3554728o856778274984...; -23513346877200757489...; 4-341 1 7 1206625299230788... 12. 1-5511335180712...; -6446898273744...; 2-12868804063096... 13. 214-4014925321183...; 1-2840254166877...; 4-569463863518,..; •169312334656...; 2-806066263296...; 3-778994240959... 14. 799; 4-18446843506...; 2-0551537959234...; -86830146903555... 15. 1-6597013896947...; -4937448592186...; -9588131805648... 16. -95958894439...; 2-4341470225193...; -9474839202664...; I -2863841294624. . .; I -6335444707528. . . 17- T^=-53; 5-3i V/ = 8-578947368... , 18. Each expression = -267949i9243ii2... 19. 1-35. 20. 886437165-3932815252769293 .. Ex. 29. 1. 3 : 55 9 149; 7 : 8; 2 : 3:209 : 510; 12 : 6 : 4 : 3; 16 : 18 : 20 : 21. 2. i ; 5; 9 : 50; 100 : 483. 3- 7 : 9; 8 : is; 4I : "i- No. 5- Yes. 6. 6. 7. s^^V- 9^; 70; -0001; 770; -0135... 9. 6; -oi; 2-6933... 91:81. II. 16:24:30:35. 12. 323:324. 4. 10. 13. 72. 14- i- 15- 9 ANSWERS. 401 Ex. 30. I. %ii,^\d.\ 122721/; 46079/^; 422353 (iaT.). ■L. 6637437J.; 26027397J. 3. 13600 yds.; 205862 in. 4. 1635033111.; 1600SSS in. 5. 83061J sq. yds.; 374951^ sq. ft. 6. 54650895 sq. in. ; 333274481 sq. in. 7. 29194441 cu. in.; 129139623 cu. in. 8. 3220671 oz.; 534793 dr. 9. 79480 grs.; 113633 grs. 10. 132401045 grs.; 459705 m. 11. 3046 gills; 3518 gall. "• 2735! in.; 66796027621 grs. 13. ;fi440. 6. 7; ;f2057. 12. 2i; ;fi35. 8. 7J. 14. ;f 1041. 13. 8i; I year (of 365 days) 47 wks. 10 h. 40 m.; 7 weeks 3 d. 21 h. 4 m. 17 s. 15. 5 weeks 8 h. 48 m. 29 s. ; 42 weeks 8 h. 30 m. 45 s.; 4 miles I f. 36 p. I yd. i ft. 7 in. 16. 349 miles 7 f. 18 p. I ft.; 32 miles 4 f. 33 p. i yd. i ft. 6 in.; 49 miles 2 f . 31 p. 3 yd. i ft. 11 in. 17. I acre i r. 19yds. Sin. ; 2 acres 13yds. 7ft. 121 in.; 15 acres 14 p. 14 yds. 7 ft. 72 in. i8. 185 acres 2 p. 26 yds. 4ft. 72 in.; i acre 2 r. 15 p. 2yds. 2 ft. 100 in.; 4 acres 13 p. 5 yds. 7 ft. 109 in. 19. 21 cu. yds. 4 ft. 189 in.; 45 cu. yds.; 39 tons 2 cwt. 2 qrs. 141b. 20. 10 tons 1 1 cwt. 26 lbs. 1 1 oz. ; a tons 19 cwt. 7 lbs. 6 oz. 3 dr. ; I ton 5 cwt. 2 qrs. i lb. 2 oz. 125 grs. 21. 73 lbs, 90Z. 12 dwts. 7grs. ; or 885 Tr. oz. 295 grs. (Art. 209); II C. 3 O. 19 fl. oz. 23 m.; 234 gall. 3 qt. i pt. 22. 761 gall. 1 pt. I gill; 13 loads 3 qrs. 2 pk. i gall.; 38 loads 2 pk. I gall. 2 qt. 23. 625008 Tr. oz. 24. 789 miles 1 f. 33 P- S yd. i ft- 4 in- 25. i248i7(K); 2289 sixpences. 26. 106900 threepences;, 18853 half-crowns and i%d. over. 27. 172 thalers and is. ^d. over ; 2130 dollars and is. Sd. over; 3478 fr. and i4rf. over. 28. 3097600 sq. y4s. ; 102400 p. 29. 80 yds. I qr. 3 nl. i J in. ; 263 ells 2 qr. 1 pi. | in. ; 43 yds. 2 ft. f in. 30. 394 lb. 1 1 oz. II dwt. 16 grs. ; 703 lb. 5 oz. 18 dwt. 13 grs. 3*. 271921^. 32. £"9^- lo^- 33- 189000. 34. 2678400J.; 6958800J.; 15195600^. B.-S. A. 26 402 ANSWERS. Ex. 31. I. ^23089.9.8!. •.. ;^ioo87. 4. 8J. 3. ;^i45S68. 6. 7. 4. ;^49o8. I. 4l- 5- £h9i99- i7- 9i- 6. 55lb. 4i3grs. 7. 44 cwt. 1 qr. 1 3 oz. 8. 107 tons 8 cwt. 3 qr. 2 lb. 12 oz. 9. 51 lb. 1 1 oz. 4 dwt. 3 grs. 10. 8c.60.7fl.oz. II. 187 gall. 2 qt. I gill. 12. 17 yds. iqr. 2nl. ijin. 13. 1 1 ells 2 qr. 3 nl. 14. 21 Ids. 3 qr. I pk. i gall. 15. 65 days ,23 h. 26 m. 32 s. 16. 651110. I da. 15 h. 8 m. 17. 40 years 262 da. 54 m. 17 s. 18. 1 fur. II p. 3 yds. 2 ft, 10 in. 19. 2 miles i f. 15 p. 5 yds. 20. 29 miles 4 f. 6 p. 2 yds. I ft. 6in. 21. 2roods6p. rjyds. 8ft. i4iin. 22. 2 ac. I r. 28 p. 15 yds. I ft. 135 in. 23. 169 cu. yds. 6 ft. 763 in. 24. 4 tons. 25. 35lbs. 15 oz. ii6Jgrs. 26. 79 miles 4 f. 15 p. 27. i24ac. 7 p. 29yds. 4ft. 72 in. 28. ;^79. 11. 4i^\. 29. £ii. II. iin. 30. ;^92. 7. iJUl. 31. 19 cwt. I qr. 1 7 lbs. ^■i■^o^. 32. 2 fur. 8po. 5 yds. i^in. II Ex. 32. I- £m'i- 14- 94; ^446. 14- 9i; pfip- i^- iJ- ■'■ £ai^. is- 7l; £i^- i6-9f; £'^- 19- 9i- 3. j^26. 18. 8f. 4. ^361.2.24. 5. ;f37. 19. IlJ. 6. ;^6. 14. 7J. 7. 61b. II oz. 12 dr. 8. 6cwt. 3qr. 251b. 130Z. 9. i2tons i5cwt. I qr. 231b. looz. 10. 41b. 8oz. i6dwt. I7grs. 3 c. 40. I4fl. oz. 3fl. dr. 35 m. 12. 30 gall. 2qt. i pt. 3 gills. 13. s yds. I qr. I nl. J in. 14. 2 ells 2 qr. 3 nl. ij in. 15. 3 Ids. 3qr. stsK 2pk. i gall. 16. 4 days 1911. 53 m. 53 s. 17. 3 mo. I wk. s d. 18 h. 51 m. 18. 3 yirs. 220 d. 18 h. 51 m. 48 s. 19. 4 po. 2 yds. 2 ft. 4 in. 20. 2 fur. 28 p. 4 yds. i in. 21. 2imiles 4f. 3ip. 3yds. i ft. 6in. 22. iiper. 22yds. 6ft. i7in. 23. I rood 32 p. 20yds. 8 ft. 139 in. 24. 51 cu. yds. 11 ft. 964 in. 25. 4 po. 2 yds. I ft. 3-in.; 22 acres 2 r. 26 p. 25yds. 2 ft. 36 in. 26. 6 cwt. 76 lbs.. II bz. 1 3 dr. 27. 4 lbs. Av. 10 oz. 2154 grs. 28. I lb. 6 oz. 6dwts. 16 grs.; 13 oz. Av. 3i2jgrs. ■^9- ^i3-4.8i|f;;f8.-7. 6i^.. 30. £i- i1.B,\\l^; 3 cwts. iq|| lbs. 31. s fur. 10 p. 2 yds. 9^ in. 32. looz. i6dwts. i8ii|grs. 33- £1^1- 10. 34- 34- ;^ii8. 7. 6i. ANSWERS. 403 Ex. 33. 1. ;^iS9. 8. ro; /4i-i5-9; ;^874. 9. 94- 2. ;^34i. 18. ij; ^^2622. 9. iij; £(iOi,(>. la. ih 3. ;^4i52. 1. loj; ;f62879;3. 6; ;^43o84. 6. 10. 4. ;f98o49. 9. 9; ;£'2i7o. 6. 2|; £1^x10. ■!..o. 5. ;f2249. i^- '^i; ^^2320. 18. oj; ;f82r. i. 5. 6. ;^26o. 18. iii; ;^96i4- 5- I'i; ;^I3I44- 7- i.oJ- 7. ^^4921. 19. 6i;. ;^68866. 4. 4|. 8. innn- is- 8; ;^S3904. s- 14- 9. ;^44IS4- 18. 55 ;^l206l2. 15. 9J. 10. ^8729. I. 6; ;^225382. 12. i\. "• ;^49S93°8- 18. o; ;^56o5iogo. 3. o. 12. 1311b. idwt. i6grs.; 3591b. 40Z. 13. 1006 gall. 3 qt. I pt.; 6600 gall. 2 qt. i pt. 14. 59qrs.; 7i7qrs. 6bu^. 15. 158 weeks 6 d. I h. 8 m. 24 s. ; 1394 weeks 3 d. 7 h. 20 m. 24 s. 16. ii5oiniles 7 f. 6ch. ; 1241 ac. i r. 5 p. 17. 185 cu. yds. 23ft. 216 in.; 1672 cu. yds. 19ft. 216 in. 18. i72lbs. loz. 7ogrs.; i72olbs. 11 oz. 262igrs. 19. 107 tons 10 cwt. 12 lbs. ; 4485 tons 18 cwt. 12 lbs. 20. 5 miles IS p. 2 yds. 2 ft. 6in. ; 60 miles 4 f. 26 p. i yd. 21. I446miles if. 7p. 3yds. 10 in.; 2i43miles sf. 7 p. 3yds. 2 ft. 4 in. 22. 4ac. 5p. 4yds. 5ft. 36 in.; 6ac. 2r. 35 p. 7yds. 5ft. 108 in. 23. 302 ac. 3 r. 39 p. 16 yds. 3 ft. 130 in.; ■2840 ac, 2 r. 15 p. 21 yds. 7 ft. 2 1 in. 24. 26 yds. 3nl. iin.; 220 yds. 3qf. ml. lin. 26- ;^i38-9-5iT; ;f44i. lo- "H- 26- £1^1- 4- oil; ;^386- H- "II- 27. ;^82. 6. nil; ;f 167- I. "ixt- 28. ;^43S- 6- 8^^; ;^395- «>• o^M- 29. 282lbs. 90Z. 4f87S. S7- 1930; IIS2CU. in. 58. Receives ;,£'i3. 13. 9. 59. s8s- 6°- S8 miles. 61. ;^3o6. 4. I. 62. A, by ;^io. 13. 4i. 63. 46 ac. 3 r. 27 p. 13 yds, 8 ft. 6s in. 64. ;^2. 17. 8J; ;^3. 16. sJ. 6S- s6439438Sgrs- 66. ;^is- ,'5- 8. 67. 4K- 68. 10. 69. 4S. 70- 1636. 71. ;^i2. IS. ij. 7«- Syds. 2ft. sin. 73. 2 cwt. I qr. 1 7 lbs. 74- 24°°; ;^S- 7S- 6s. ^id. 76. 3J. 7i(/. 77- i^4- 7- 6- 78. S397 remdr.; 24fur. 27 p. i yd. i ft. 3 in. 79- £'^- ^9- ^.^ 80. 2omin. 81. ;^3679. II. loi. 82.' 48243, and. s £6- iS-f- 5. £1. gs. ly.; OS. if -8594; OS. 31-24. 6. ^^1640; ;^ii7- 10- 2V725- 7. £2i,. 10. ioi^oo8; ;^5i6i. 5- 84-78. 8. ;^23. 10. 10; £i. IS- 5- 9- ;^i9- ■=- 9f ; ^^3- i5- if-8- 10. 12 lbs. 20Z. 1-152 drs.; 177 lbs. 20Z. i dwt. 13-61 grs. 11. 2 tons 17 cwt. I qr. 27 lbs. 7 oz. 4 drs. 41 ANSWERS, 12. ipo. I yd. aft 6in. ; 25 days 21 h. j'oosS... m. 13. £iS. 19. 11; IS. 8^-808. 14. 2 hrs. 5 m. 4 s. ; i cwt. 2 qrs. 6 lbs. 6 oz. 15. 4ac. i2p. ; 52 miles 2f. 25 p. 3 yds. 16. I bz. 17 dwt. 22 grs. 17. 6j. 8Jf 18. ;^ir. 7. 51-691. 19. ;^9. 4. 8i. 20. t,s.\id: 21. lOJ. 2i-93S09. 22. 19J. ii(/. 23. £i^. 4. 4I. 24. 8j. id. 25. ;f2. 7.. I. 26. I ton 1 7 cwt. 2 qrs. 4 lbs. 27. 1 sq. yd. 6 ft. 32-4in.; ;^2. 10. 9. 28. Id. 84. 29. ;^i. li. 30. ;^i5. II. 2^875- Ex. 42. I. -i; -0625; -4729. 2. -615873; -4878. 3- "875; •3675; i4'49; 'otH^Ss. 4- •8293. 5- 59"5i- 6. •10885416. 7- '9S- 8. -01636. 9. -36. 10, 0625. 11. -0219238095. 12. -6305. 13. -13027. 14. -0203125. 15. -171496. 16. 3-28. 17. -0072562358... 18. 10; ■2-5. i-d; 2-083. 19. •2,36075. 20. -45. 21. -38257. 22. -36. 23. -003. 24. -68oi. Ex. 43. I. 45170W.; ;£'876. 3/4C. SOT. 2. 25025 »«.; 38-09/ 3. 8435 OT.; 2625 ?«. 4. ;^965. 8/ 6 Ex. 45. I. 345 kilom. 6 hectom. &c. 2. 207 kilol. 8 hectol. &c. 3. 66s296. i6. o^ 48. 94-44... lbs. 49- 8-49 gr. so. p. 4^37o7<}3- 2. 6. 48. /2057- H- »i- 49. ^595.6. iifiAr. SO. ;^i5. 18. 8|. 51- ;^6. 8. 7|. 5«- ;f55- 6- loi- 53. 2377 ac. I r. I sip 54- Sgcu.ft. i202^uV 5 in. 55. I43qrs. 7bu. 3||pk. 56. 530 lbs. Htioz. 57- 18? miles 3 f. HsHlyds. Ex. 47. I. ;f 118. 13.6. 2. 2 tons 3 cwt. 3 qrs. 3. 361 days. 4. ^1718. 9. loii. S- 365- 6. ;£'io. 2. SJ. 7. 4'i36. 16.3. 8. 54756- 9. ;^io6. II. 6. 10. ;^67. 6. 8^. 11. ;^2o. 2, 9. 12. 43ells3qr.ini. 13. s tons I cwt. 2 qrs. 14. ;^5- 7- gill- 15- 74t4 days. 16. 65. 17. 5389. 18. £^. 4. o. 19. lib. 15J0Z. 20- ^^257. loj. 21. ;^5. 22. 29^. «3- £'iS- M- ;^23. 13. iij; i6yds. 25. 39 hours. 26. 4682^V 27. ;fi. 4. 6. 28. 113 miles 592 yds. 29- j^i40o; 3tV- 3°- 10 hrs. 40 m. 36:^5. 31. ;f6l. I2J. 32. ;^2II. 19. 3. 33. ;^l8. 3. ii\. 34. 26olbs. 80Z. lydwts. 35. ;^i. II. loi. 36. ;f3. 7. 10. 37. ihr. lom. I3|f s. 38. 2400. 39. ^.lod. 40. 3p.m. Dec. 3. 41. 2s.%id.; £61. 1%. oi^%. 42. 78yds 2ft. iHn. 43. 288. 44. 2 cwt. I qr. 20,^ lbs. 45. £22.6.6^^. 46- 78|- 47- ;^i-iS-9- 48- 4i66f. 49, 9 min. 50- £^9^it>^^- '3- 4- 5i- ^8. 52. 25!. S3. 2cwt. 3qrs. 20lbs. Stroz. 54. 11 1 yds. Sin. 55. 27 in. 56. ;fio. 5. jjA- 57- 6min. S24s. 58- ^1507. ioj-. . 59. 8 p.m. Thursday. 60. y«,mile. 61. 17 qrs. 34 pks. (in 365 days). 62. 15. 63. i3Hm...; i,3ff m. 64. ;^6oi2. 6j.; 15^. 3i<^. 65. 104 days. 66. 10 days; I2fdays. 67. 275625. 68. i4 miles. 69. 3h.2Sm.p.m. 70- 2hrs. 32ra. 19JS. 414 ANSWERS. Ex. 48. i- ;£'28. 5. 9; 97 miles; 63 tons. 2. 1878. 3. s days 6 h. 10 m. 40 s. 4. 2^ years. £• 5 months. 6. 20 ; iSJ. 9 ;£'i22. \os; £gi. 17. 6, ;^73. los. 10. 5J., 6i. 8(/., 8j. 4(/.; 5J. 4 £'^1- 19^-. ^6- lo-f- 29. 56, 50, 32, 16, 46. Ex. 51. I. 2^.4111; 2J. 2|H- ■•'• iScarats. 3. 2:7. 4. 467,467, 836 of alloy. 5. 3:13. 6. 4, 5.3. 2; or 5, 4, 2, 3; org, 9, 5, 5. 7. 3,3,3,2. 8. 7, 2, z. 9. 98; 416, 224. 10. 22jlb., 91b., 18 lb., 3ijlb.; or gib., 224 lb., 3iilb., i81b. 11. 16 lbs., 64 lbs. 12. 3,3,5. 13. 21 lbs. and 12 lbs. 1-4. iton 7cwt. 2qrs. 26f lbs. 15. 1150Z., 850Z. 16. 20 ; 7; 5.f. ijfl 20 25 Ex. 52 >■■ £l'i- 9- 3; i8j. iJU- 2- £i^- I- 8i; ;^29. 10. 7iU. 3- 43797- 4- 75844- 5- 34877- 6. 58000. 7- ;^337- 19- 4l#- 8. ;^02jd. 16. 5p. c; ;^i 36762. lOJ. IV. I. 90,000. 2. ;,f9. o. iif ; 8imUes6f. 39p. 3. ;f650. 16. 8; ;f8. 3. 6J. 4- 41; II; ■• 5- £^- 6. 30; -03; 3. 7- -ri; "ooS- 8. 29897; 1-21550625. 9. j;:4.6.f-; ;^4- ". 8|t^|^. 10. 25P-C. II. ;^54- 3- 2i. ". 14. V. J. ii)6; Id. 2. 4C11. yds. 20ft. 635-2 in. 3. ^^312. 13. 85. 4. 18. 5. -0095; 1-05; 47500. 6. -00375; -17*5714*; H- 7. 3 tons 7 cwts. 3 qrs. 22 lbs. 8. 4701; 14!. 9. 5- 10. ;^48o2. 15. 3J. II. 62; 83icu.ft. 12. 914. VI. 1. I fur. Spo. 2 yds. 2 ft. 8||in. 2. 95liJ; i^; -sV; fh- 3. 45816-11479. 4- 10; -i; 1-647619. 5. iiJ.-ii^/.; -11125. 6* 5-9726...; -6. 7- ;if4i.66- 13- 4' 8. ;f5; ;^787. 8. oJt^V- 9. 300. 10. First kind ; iM5 or i'355oi---P-^- VII. 2. 761. ..16. 3- ;^78i328.4-4i; i6tonsi7cwts. 2qrs. 4. 2MU; 3f^; 'i?- 5- ;^40ii.6-6|. 6. -82008; -002; 200; 2; 20000. 7. •4047"-; 2 fur. 14 po. 2^fyds. 8. 2tons icwt. 2qrs. HUfflbs. 9. 2 men. 10. 15*. 11. ;^I4520. 12. 3-60555...; 7 sq- yds- 7ft- 6™- VIII. I. ;fi7-5-iiSi- ^- '^V 3- 7-377^91^- 4. .3-0688259...; ^«tV , S- 23-17x102... 6. ^155- IS- o|^|. 7. ihr. 57|m. 8. ;^3S- "- 6iAV- 9- icwt.2iilbs. 10. ii8f p. c. II. £^- 12- 3lU^- IX. 2. io5in. 3- ;^i820. 4- 164yds. 2ft. ii^ in. 5- -0071875. 6. 1 70 cu. ft. 3' 9" 2'" 9" 4" = 1 70 cu. ft. S42I; in. 7. Loses ;£-4. 8. 1-0490734- 9- £i°6- 10. At iih. -ion«i- A-M- on 10 April, 1871; 11^26^^™.; 10 h. 5617- •"• 422 ANSWERS. X. I. /1905. ISJ.; ^560. 2. English navvies ; ;^4000. 3- 15; ;^384; 3'6i»=¥5'-7 &c. 4. ;^io50. XI. I. ^s.i\d. 1. 9Ayds. 3. ;^i7-3-4- 4- 4.f- 8.5.7; 3^5.11=. 3- ;£'3. 6. 4i; -298828125. 4. fB; 209ift. 5. 14479673-94... XVI. 1. 2415. -i. i\; 3tV = 3'S294-- 3- Vs; tH-- 4. -3183098...; -808. 5. 18-0000015...; 7'745966-.. 6. 47 hrs. 3 m. 20 s. XVII. I. ;^i530. 2. ;^i8. 15^. 3. ;^297. 13. 58i|. 4. -57004987...; 570049-87...; -47691. S- *; tAV 6- ;^3o- 14- 8if. 7- 3'92o; »• 8. 165. 9. ;f2i. 9. 3i... 10. 2 miles. XVIII. I. ;^65. 2. I. 2. ;f6. 3. 6. 3. U. 4- m;l?- 5- A- '3-4; 3-f-7i«'- 6. -00027825; 6-94. 7- .^i 19- 4- 8i Vinni- 8. 36 : 120. 9. 3J. 7iJ. 10. 37^ p. c. II. .^64. 6. loiiff. XIX. I. Art. 31, 2; 152. 2. Art. 67; 19. 3. 442. ..11; Art. 52. 4. 7.V5249422t; 3'6286935S4. 5. l\. 6. I; ^17. +. •.=. 8. I2J. 4f f. 9. ;f4307- II- loi : £tS- 10. 4^. 10. iqr. I3j^lbs. u. gj. J^d.i^. 12. 2j^ or 2-50015... p. c. 13. Sp.c. 14. 3jhrs. 15. 5J-. io(/. ANSWERS. 423 XX. XXI. XXII. 4- 6. 9- II. I. 4- 7- 9- I. 4- 7- 8. 9- II. Art. 15. 2. Art. 67; 2003. 982'99883; 6'oi; 60100; "oSoi. ilffH; f- 7- ;^64. ;fi377- "-6; £:Z^- 3- •Jt't^- 4i P- c. 3- 3- 5. 2 7o' 3.^00633967- 8. 70 yds. 10. j^27s. 436985; &c. ;^34- 6- 4li. 60320000; 20500. 12. 20 yards from it. 2. 1 864000; 1848000. 3 5. ;^67. 12. oJt^^. 6. 2 cwts. 26 lbs. iiyj; 20. 8. 3366-204361; 137-65. ;^i8. 19. 11; u. 8f |. 10. ;,f56oo. II. 7ft. 12. 23p.c. 270728,547466. 2. ^; J. 7371 i 2089. 5- 35 and 25. 121-11458...; I4-S57i4i; Art. 157. 6-550477?; i8j. \\ -4048. ;^ioo. o. oJrlTi; £'^^- "• 3i - ism or 1 3- 76s -P-c. 117; 23400. 3; 7i- 10. ;^675; ;£'7o5. 16. 74|- 12. 3 : 2. XXIII. 1. 3- 6. 8. 10. XXIV. I. 3- 4- 6. XXV. I. 3- 6- 7- II. XXVI. I. 3- 4' 6 li; li; I; ^• 4- 5- T> 56725- 652-20834; -83; 83; 830000. ;^I3. lOJ. 7- I ft. 7 in. ;C6775- 2- H; ;^i3- i7- 4ii 9- .^611. 3. 4. 113; 339: 678; 791- "• i^"- °- 3- '^- o. 101,001,001. *. 66; 18. -0000029; 80,000,000; 4"o5536; 5-07964; fj; i||. 2 yds. I ft. 2|in.; -00460379464.28571. 5- 322yds. 2ft. 30 of each. 7- 5678; 4^! ■34785-- ;^26. 13. 6; £l. .. 6. 9- 4^13. 10. ;^702. 13- 4- 90. 12. i^75- '3- 34i miles- 74090,296787,694642. 2. 3596-52; Art. 62. ^. ij. 4. 1-965; -00227; 800; ly. i\\; -66125. 3 o'cl. p.m. Dec. 3- «• 739; 8i; •411096.- ;^5oo. 8. 35 days. 9- ;^502. 13- 4- "o. 4. ;^4725. 12. 39yds- 13- 9™- 1035yds.; 9^-3 fur. 3522178, Art. 41; 3998936616. 2. 1000 ft. 37; i4days7h. iim. 17 s.; 2674 days o h. 9 m. 59 s. A%; i- 5- 27sq.yds.4ft-; ;^'°-i6-8. ^32767. II. 8; ;^6840. 11. 8; ;C26947. o. lii- 424 ANSWERS. XXVII. I. 3- 5- 7- lO. XXVIII. I. 3- S- 6. •0202i2i2ii...; ^1134. 18. il. 774- 13- 70047.. .1145. £^Ol.6.6h £181 100.1. < •18984375 ; I5J-. lid. 102723... or ;,f 102. 14. 6. •2025...; 202-52225. ..;^2. II. if; .•6"4683. 9. ;^72. 9. 4. 10. 9J. 7i^. II. /lo. ;^ioi66. 13. 4 and ;^6ooo Stock. 2. 876576 hrs.; ;f9i3. 2J-. 4. 1287; 9009. 20 p. c. II. £6.y. 9. 4 months. 12. ;^2000. 2532837. ..285803562. 2. 4410075; 942. 223'358...; 20-0574... oz. 4. lof ; f. 42-75; -04275. 4275000; 2ft. Sin. 321 sq. yds. 3J ft.; ^72. 6. 3. 7. ;£'i627. 12. i ; 4p.c. 47Timr or 47-68... days. 9. 3 cwts. 2 qrs. 23 lbs. ;^i92. ir. i^s. iild.; \is. lod.; i^s.^d. 12. ^7.11.3. V CAMBRIDGE: l-KINTED BY C. J. CLAY, M.A. AND SONS, AT THE UNIVERSITY I'KESS.