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Theft, mutilation, and underlining of books are reasons for disciplinary action and may result in dismissal from the University. To renew call Telephone Center, 333-8400 UNIVERSITY OF ILLINOIS LIBRARY AT URBANA-CHAMPAIGN L161—O-1096 ne y nA ; nal el ae, : H J A , iy a Ne : t WAG yh is aon i Digitized by the Internet Archive in 2022 with funding from University of Illinois Urbana-Champaign httos://archive.org/details/northamericanariO0emer_0O o} d ) rn if a lie Prat cane, AS s) , F EMERSON’S THIRD PART. THE NORTH AMERICAN ARITHMETIC. ‘PART THIRD, FOR ADVANCED SCHOLARS. BY FREDERICK EMERSON, LATE PRINCIPAL IN THE DEPARTMENT OF ARITHMETIC, BOYLSTON SCHOOL, BOSTON. | BOSTON: RUSSELL, ODIORNE, & METCALF. NEW YORK, Collins & Hannay. PHILADELPHIA, Hogan & Thomp- son. BALTIMORE, David Cushing. H ARTFORD, Pdi Huntington. WINDSOR, Ide & Goddard. HALLOWELL, Glazier, Masters, & Co. CINCINNATI, C, P. Barnes. RALEIGH, Turner & Hughes. 183.45" & Entered, according to act of Congress, in the year 1834, by FrepErick KareRsox, in the Clerk’s office of the District Court of the District of Massachusetts. STEREOTYPED BY THOMAS G. WELLS AND CO. BOSTON, a — Say eMATIOS Llbieyhs PREFACE. Tue work now presented, is the last of a series of books, under the general title of ‘Tue Norta American ArirH- mETIC, and severally denominated Part First, Part Second, and Part Third. 3 Part First isa small book, designed for the use of child- ren between five and eight years of age, and suited to the _ convenience of class-teaching in primary schools. | Parr Seconp consists of a course of oral and written ex- ercises united, embracing sufficient theory and practice of arithmetic for all the purposes of common business. Part Tuirp comprises a brief view of the elementary principles of arithmetic, and a full development of its higher operations. Although it is especially prepared to succeed the use of Part Second, it may be conveniently taken up by scholars, whose acquirements in arithmetic are considerably - less than the exercises in Part Second are calculated to ai- ford. While preparing this book, I have kept in prominent » view, two classes of scholars; viz.— those who are to prose- a cute a full course of mathematical studies, and those who are to embark in commerce. In attempting to place arith- _ metic, as a science, before the scholar in that light, which shall prepare him for the proper requirements of college, 1 have found it convenient to draw a large portion of the ex- amples for illustration and practice, from mercantile trans- actions; and thus pure and mercantile arithmetic are united. No attention has been spared, to render the mercantile information here presented, correct and adequate. Being _ convinced, that many of the statements relative to commerce, which appear in books of arithmetic, have been transmitted down from ancient publications, and are now erroneous, I have drawn new data from the counting-room, the insurance office, the custom-house, and the laws of the present times. The article on Foreign Exchange is comparatively exten- sive, and I hope it will be found to justify the confidence of merchants, Its statements correspond to those ofthe British ‘Unwersal Cambist,’ conformably with our value of foreign. coins, as fixed by Act of Congress, in 1834. BO3134 4 PREFACE. Although a knowledge of arithmetic may, in general, be | well appreciated as a valuable acquisition, yet the effect produced on intellectual character, by the exercises neces- sary for acquiring that knowledge, is not always duly con- sidered. In these exercises, the mental effort required in discovering the true relations of the data, tends to strengthen the power of comprehension, arfd leads to a habit of investi- gating; the certainty of the processes, and the indisputable correctness of the results, give clearness and activity of thought; and, in the systematic arrangement necessary to be observed in performing solutions, the mind is disciplined to order, and accustomed to that connected view of things, so indispensable to the formation of a sound judgment. These advantages, however, depend on the ‘manner in which the science is taught; and they are gained, or lost, in proportion as the teaching is rational, or superficial. Arithmetic, more than any other branch of learning, has - suffered from the influence of circumstances. Being the vade-mecum of the shop-keeper, it has too often been viewed as the peculiar accomplishment of the accountant, and neglected by the classical student. The popular sup- position, that a compendious treatise can be more easily mastered than a copious one, has led to the use of text- books, which are deficient, both in elucidation and exer- cises. But these evils seem now to be dissipating. —The elements of arithmetic have become a subject of primary instruction; and teachers of higher schools, who have adopt- ed an elevated course of study, are no longer satisfied with books of indifferent character. It has been my belief, that a treatise on arithmetic might _be sé constructed, that the learner should find no means of proceeding in the exercises, without mastering the subject in his own mind, as he advances; and, that he should still be . enabled to proceed through the entire course, without requir- ing any instruction from his tutor. Induced by this belief, I commenced preparing The North American Arithmetic about five years since; and the only apology I shall offer, for not earlier presenting its several Parts to the public, is the unwillingness that they should pass from my hands, while I could see opportunity for their improvement. Boston, October, 1834. FE’. EMERSON. A KEY to this work (for teachers only) is published separately. ws 1. ARITHMETIC. ARTICLE I. DEFINITIONS OF QUANTITY, NUMBERS, AND ARITHMETIC. QUANTITY is that property of any thing which may be increased or diminished—it is magnitude or multi- tude. It is magnitude when presented in a mass or con- tinuity ; as, a quantity of water, a quantity of cloth. It is multitude when presented in the assemblage of several things ; as, a quantity of pens, a quantity of hats. The idea of quantity is not, however, confined to visible ob- jects ; it has reference to avery thing that is susceptible of being more or less. NUMBERS are the expressions of quantity. Their names are, One, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, &c. In quantities of multitude, One expresses a Unit; that is, an entire, single thing; as one pen, one hat. Then each succeeding number ex- presses one unit more than the next preceding. In quantities of magnitude, a certain known quantity is first assumed as a measure, and considered the unit; as one gallon, one yard. Then each succeeding number ex- presses a quantity equal to as many times the unit, as the number indicates. » Hence, the value of any number de- pends upon the nice of-its unity. When the unit is applied to any particular thing, it is called a concrete unit ; and pe consisting of concrete « 6 ARITHMETIC. | II. units are called concrete numbers: for example, one dollar, two dollars. But when no particular thing is indicated by the unit, it is an abstract unit; and hence arise abstract numbers: for example, one and one make two. jens Without the use of numbers, we cannot know precise- ly how much- any quantity is, nor make any exact com- parison of quantities. And it is by comparison only, that we value all quantities; since an object, viewed by itself, camot be considered either great or small, much or hit- tle; it can be so only in its relation to some other object, that is smaller or greater. | ARITHMETIC treats of numbers: it demonstrates their various properties and relations; and hence it 1s called the Science of numbers. It also teaches the methods of computing by numbers; and hence it is call- ed the Art of numbering. iH, NOTATION AND NUMERATION. Notation is the writing of numbers im numerical char- acters, and NumERATION is the reading of them. The method of denoting numbers first practised, was undoubtedly that of representing each unit by a separate mark. Various abbreviations of this method succeeded; such as the use of a single character to represent five, another to represent ten, &c.; but no method was found perfectly convenient, until the Arabic FIGURES or DIGITS, and DECIMAL system now in use, were adopted. These figures are, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; denoting respec- — - tively, nothing, one unit, two units, three units, &c. To denote numbers higher than 9, recourse is had to a law that assigns superior values to figures, according to the order in which they are placed. viz. Any figure placed to the left of another figure, expresses ten times the quantity that it would express if it occupied the place of the latter. Hence arise a succession of higher orders of units. As an illustration of the above law, observe the dif- ferent quantities which are expressed by the figure 1. — : > watt - ‘ I. NOTATION AND NUMERATION. 7 » When standing alone, or to the right of other figures, 1 represents 1 unit of the first degree or order; when stand- ing in the second place towards the left, thus, 10, it represents 1 ten, which is 1 unit of the second degree ; when standing in the third place, thus, 100, it represents 1 hundred, which is 1 unit of the third degree: and so on. The zero or cipher (0) expresses nothing of itself, yore employed only to occupy a place. jb ay The units of the second degree, that is, the tens, are denoted and named in succession, 10 ten, 20 twenty, 30 thirty, 40 forty, 50 fifty, 60 sixty, 70 seventy, 80 eighty, 90 ninety. The units of the third degree, that is, the hundreds, are denoted and named, 100 one hundred, 200 _ two hundred, 300 three hundred, and so on to 900 nine hundred. The numbers Benen 10 and 20 are denoted and named, 11 eleven, 12 twelve, 13 thirteen, 14 four- teen, 15 fifteen, 16 sixteen, 17 seventeen, 18 eighteen, /19 nineteen. Numbers between all other tens are de- noted in like manner, but their names are compounded of the names of their respective units; thus, 21 twenty-one, 22 twenty-two, 23 twenty-three, &e.: ; ol thirty-one, 32 thirty-two, &c. &c. This nomenclature, although not very imperfect, might be rendered more consistent, by substituting regular compound names for those now ap- plied to the numbers between 10 and 20. This alter- ation would give the names, 11 ten-one, 12 ten-two, 13 ten-three, &e. As the first three places of figures are appropriated to simple units, tens, and hundreds, so every succeeding three places are appropriated to the units, tens, and hundreds of succeeding higher denominations._ For illustration, see the following table. a S a a wn p 2 Ss = u re 2 oa ua =e ae ae ee ae oO S) ° ° oS 2 aig ae = Fal n a se © iS) as, = et = oe & ra q 5 we ee tr MEE ee et Caco « BRE Ae pe = ae ~ @ RES Siete: oS 70 ite Can # @ eH ® BR SB aa (FRING Gea Ek ON Coe DR GO EK ee ON 450725 206 194 007 185 039 000 164 396 205 013008741 By continuing to adopt a new name for every three degrees of units, the above table may be extended indef- - . 8g _ > ARITHMETIC. IL initely. Formerly, the denominations higher than thousand; were eachmade to embrace six degrees of units; taking i inf thousands, tens of thousands, and hundreds of thousands b The mode of applying a name to every three degrees’ however, is now universal on the continent of Europe b and is becoming so in England and America. e The learner may denote in figures, the following num} bers, which are written in words. | Example 1. Four hundred seventy-eight million, tw¢ hundred forty-one thousand, and one hundred. 2. Seven million, six hundred ninety-two thousand _ and eighty-nine. 3. Nineteen million, twenty thousand, and five. 4. Eight hundred billion. | 5. One billion, six hundred forty-four thousand, five hundred and thirteen. 6. One trillion, five hundred thirty-four billion, thred million, eighteen thousand, and four. _ ; 7. Two hundred Pian: sixteen thousand and one. 8. Eleven billion, one million, and sixty. 9. Five trillion, eight billion, four million, nine thou sand, and seven. | 10. One hundred trillion, twenty billion, thr ee hun) dred million, two thousand, and four. 7 ae 11. Thirty-one trillion, five hundred, and sixty. : 12. Six quadrillion, two hundred ‘and ‘fourteen trillion! 13. Two hundred forty-nine quadrillion, seventy-fivi thousand, and twenty-two. 14. Forty-six quintillion, one quadrillion, nmeteen bil ion, seven hundred and eight. | 15. Nine hundred sextillion, three hundred twenty, five trillion, two thousand, and fourteen. ) | INDICATIVE CHARACTERS OR SIGNS. The sign-+ (plus) between numbers, indicates a they are to be added together; thus, 3-++ 2 is 5. The sign— (minus) mdicates, that the number placer after it, is to be subtracted from the number placed be fore it; thus, 5—2 is 3. Lf. ADDITION. S _ The sign X (into) indicates that one number is to be multiplied into another; thus, 4 x 2 is 12. _ Whe sign + (by) indicates that the number on the left hand is to be divided by the number on the right hand; ee 12-3 is 4oc _ The sign== (equal to) indicates that the number before it, is equal to the number after it; for example, 4+ 2— 6. @ 2-4. 5X3—15. 15+3=5. iil. ADDITION. ADDITION is the operation by which two or more num- bers are united in one number, called their sum. It is the first and most simple operation in arithmetic, effecting the first and most simple combination of quantities. The primary mode of forming numbers, by joining one unit to another, and, this sum to another, and. so on, ex- hibits the principle of addition. When numbers, which are to be added, consist of units of several degrees, such as tens, hundreds, &c., it is found. convenient to add together the units of each degree by themselves; and since ten units of any degree make one unit of the next higher degree, the number of tens in the sum of each degree of units is carried to the next higher degree, and added thereto. , é' RULE FOR ADDITION. Write the numbers, units un- der wnits, tens under tens, &c. Add each column sep- wately, beginning with-the column of wnits. When the sum of any column is not more than 9, write it under ‘he column: when the sum is more than 9, write only the units’ figure under the column, and carry the tens to the wext column. Finally, write down the whole sum of the jejt hand column. es | 1. What is the sum of 370+ 90264-+ 1470+ 40060? _ 2. What is the sum of 4000-++ 570-+99-+ 54-273 + 59073-+-4000-+-61998 +752? | 10 ARITHMETIC. Iv 3. What is the sum of 243-+- 5021 +-7628-+- 927+ 6. +5823-+742-+796 + 5009 -+ 325 +7426 +31186 4 987-+-6954 +2748? | 4. What is the sum of two thousand and seven, forty four million five hundred and sixty-one, one hundred mu) lion, six billion twenty-eight thousand and eleven ? YZ SUBTRACTION. | SuBTRACTION is the operation by which one numbe is taken from another. The number from which another is to be taken | called the minuend, and the number to be taken is calle the subtrahend. The number resulting from the ope: ation shows the remainder of the minuend, after th subtrahend has been taken out; it also shows the diffe ence between the minuend and subtrahend, or the exce, of the former above the latter. The subtrahend and ré mainder may be considered the two parts into which tk minuend is separated by the operation; and in this viev subtraction is the opposite of addition, in as much addition unites several quantities in one sum, and subtra tion separates a quantity into two parts. | Subtraction is performed by taking the units of eac degree in the subtrahend, from those of correspondir| degree in the minuend, and severally denoting the ri mainders. When the units of any degree in the subir hend exceed those of the same’ degree in the minuen we mentally join one unit of the next higher degree the deficient place in the subtrahend, and consider tl units of the higher degree to be one less than they a’ dencted: this process is the reverse of carrying , addition. One other method may be adopted in th, case; viz. Increase both the minuend and subtrahen by mentally adding ten to the deficient place in t) former, and, one to the next higher degree of units | the latter. This method is justified by the self-evide) |W: MULTIPLICATION. il truth, that, if two unequal quantities be equally increased, > their difference is not thereby altered. RULE FOR SUBTRACTION. Write the smaller number | under the greater, placing units under wnits, §c. Be- | gin with the units, and subtract each figure in the lower number from the figure over it. When a figure in the upper number is smaller than the figure under it, consid- er the upper figure tv be 10 more than it is, and the neat upper figure on the left hand, to be 1 less than it is. proor. Add together the remainder and the smaller number: their sum will be equal to the greater number, if the work be right. _ 4. What is the difference between 70240 and 69418? ' 2. How much is the excess of the number 482724 ’ above the number 194750? ' 3. Suppose 479021 to be a minuend, and’ 38456 ' the subtrahend; how much is the remainder? ' 4. 905106392 — 904623724 —>? ‘5. Subtract fifty-one thousand from one hundred bil- * lion, cighteen thousand, five hundred and one. i j et ening FE eh oe EE a SES ONT cee Le | MULTIPLICATION. }) ¢ t MULTIPLICATION is the operation by which a number is produced, equal to as many times one given number, as there are units in another given number. It is an | abridged method of finding the sum of several equal | quantities, by repeating one of those quantities. | The number to be multiplied or repeated is called the | multiplicand; it may be viewed as one of several equal | quantities, whose sum is to be produced by the operation. {The number to multiply by 1s called the multiplier; it indicates how many such quantities as the multiplicand ' are to be united, or, how many times the multiplicand is ' to be repeated. Thenumber resulting from the operation | 3s called the product. 12 ARITHMETIC. Vv The multiplicand and multiplier, considered as con: curring to form the product, are called factors of the product. Either factor may be used as the multiplier of the other; that is, the multiplicand and multiplier may change places, and the product will be still the same, For example, 4x 3=12. 3xX4=—12. When a product arises from more than two factors, . the numbers may be denoted thus, 6 x 3 x 5=90; but. in forming the product, a distinct operation is necessary te bring in each factor, after the two first. The numbers, 6, 3, 5, would, therefore, be multiplied into each other tins, 6X 3—=I18: 18 x 5=90. Factors may be arranged in any succession whatever, since the mere order in which they are brought imto the, operation cannot affect their final product. For exam- ple, 5X3 xX4=60. 4X38xX5=60. 3X5X4=60, The products of small numbers may be committed to) memory; but when the product of factors consisting of several figures is required, it is necessary to multiply’ each figure in the multiplicand by each figure in the multiplier, and denote the several products in such ordei, that they shall represent their respective values. When simple units are employed as the multiplier, the product _ of each figure in the multiplicand is of the same degree! as the figure multiplied; that is, units multiplying units give units, units multiplying tens give tens, units multi- plying hundreds give hundreds, &c. When tens are employed as the multiplier, the product of each figure in the multiplicand is one degree higher than the figure multiplied; that is, tens multiplying units give tens, tens| multiplying tens give hundreds, tens multiplying hundreds give thousands, &c. When hundreds are employed as the multiplier, the product of each figure in the multi- plicand is two degrees higher than the figure multiplied; and so on. RULE FOR MULTIPLICATION. Write the multiplier under the multiplicand, placing units under units, §e. | When there is but one figure in the multiplier, begin with the units, multiply each figure in the multiplicand separately, and place each product under the figure im co | ¥, MULTIPLICATION. 13 the multiplicand from which it arose; observing to carry _ the tens to the left as in addition. — - When there is more than one figure in the multiplier, multiply by each figure separately, and write its product ina separate line, placing the right hand figure of each line under the figure by which you multiply; and finally, add together the several products. The sum will be ° the whole product. __ Abbreviations of the above rule may frequently be adopted, as follows. When there are ciphers standing between other fig- ures, in the multiplier, they may be disregarded. When ciphers stand on the right of either factor, or both, they may be disregarded till the multiplication is performed, and then annexed to the product. When either factor is 10, 100, 1000, &c., merely place the ciphers in this factor on the right hand of the other factor, and it becomes the product. When the multiplier is a number that can be produc- ed by multiplying two smaller numbers together, multi- ply the multiplicand first by one of the smaller numbers, and the product thence arising by the other. 1. Suppose 479265 to be a multiplicand, and 9236 the multiplier; how much is the product ? 2. Suppose 26537 to be one factor, and 873643 another; how much is their product ? 3. Suppose the numbers 725, 38046 and 91, to be factors; how much is the product? 4. What is the product of 62392 x 4003 ° 5. What is the product of 248000 < 9400? 6. What is the product of 24 x 300 X13 10002? 7. Multiply one hundred five million, by one thousand. For the purpose of determining whether any error has happened in the process of multiplication, the following ' method of trial, which depends on the peculiar property of the number 9, and which is called casting owt the nines, may be practised. Add together the figures of the product, herizontally, 14 ARITHMETIC. ) Viv rejecting or dropping the number 9 as often as the sum amounts to that ‘number, and proceeding with the excess,| and finally denote the last excess. Perform the same operation upon each of the factors ; then multiply together the excesses of the factors, and cast out the nines from! their product. If the excess of this smaller product be equal to the excess of the larger product first found, the work may be supposed to be right. It is, however, to’ be observed, that, although this test furnishes satisfactory evidence of the correctness of an operation, it is not an infallible proof; for, if a product chance to contain an error of just 9 units of any degree, the excess of its, horizontal sum is not thereby altered. : In order to perceive why the excess above nines found in the horizontal sum of a product, must be equal to the excess found in the product of the excesses of the fac- tors, observe that, by the law of notation, a figure is increased nine times its value by its removal one place to the left; and hence, however far a figure is removed from the place of units, when its nines are excluded, its remainder can be only itself. Therefore, any number, and the horizontal sum of its figures, must have equal remainders when their nines are excluded. This being understood, observe that, since factors composed of. entire nines will give a product consisting of entire nines, it follows, that any excess above nines im a product, must arise from an excess above nines in the factors. Therefore, the product of the excesses of the factors, must contain the same excess that is contained in the product of the whole factors. : Was DIVISION. Divisron is the operation by which we find how many times one number is contained in another. Itis the con- verse of multiplication; the product and one factor being given, and the other factor resultmg from the operation. : iY VI. DIVISION. 15 The number which corresponds to the product in multiplication, is the number to be divided, and is called the dividend. ‘The given factor is the number to divide by, and is called the divisor. The factor to be found, that is, the number which shows how many times the dividend contains the divisor, is called the quotient. As multiplication has been shown to proceed from addition, so division may be shown to proceed from subtraction.. If we repeatedly subtract the divisor from the dividend till the latter is exhausted, the number of subtractions performed will answer to the number of units in the quotent. For example, if the dividend be 24, and the divisor 6, the quotient may be found by sub- traction thus, 24—6—18, 18—6—12, 12—6=6, 6—6=0. Llere 6 is subtracted four times from 24, and there is nothing remains; therefore,4 is the number of times that 6 is contained in 24. In division, this oper- ation is denoted thus, 24+6==4; or thus, 74 —=4. Division not only investigates the number of times the - dividend contains the divisor, but it also serves to divide the dividend into as:many equal parts as the divisor con- tains units; the quotient being one of ihese paris. ‘Tis effect of the operation may be understood by consider- ing, that, since the divisor and quoticat are factors of the dividend, they must each indicate how many of the other the dividend contains. It may be observed, that all the preceding operations begin at the place of simple units; division, however, must begin at the highest degree of units; for, the number of times that the divisor is contained in the higher units of the dividend must be taken out first, in order that any remainder, or excess above an exact number of times, may be carried down to the lower degrees of units, and divided therewith. When the divisor is not contained an exact number of times in the dividend, there will be a remainder at the end of the operation. This remainder, being a part of the dividend, is to be divided; but its quotient will be smaller than a unit, since a quantity in the dividend just equal to the divisor, gives only a unit in the quotient. or 16 ARITHMETIC, VI. Quantities smaller than a unit, that is, parts of a unit, are’ called Fractions. Such quantities are commonly ex-' pressed by two numbers, placed one above the other. with a line between them, thus $. The lower number, called the denominator, shows how many equal parts the, unit is divided into; and the upper number, called the numerator, shows how many of the equal parts are em-. braced in the fraction. When the unit is divided into’ two equal parts, the parts are called halves; when divided. into three equal parts, the parts are called thirds; when divided into four equal parts, the parts are called fourths; and so on; the number of the denominator giving the name. For example, if the unit be divided into five equal | parts, one of the parts is denoted thus, 4, and called one-. fifth; two of the parts, thus, 2, and called two-fifths; and so on. In this method, the unit may be divided into any number of equal parts, and any number of such parts may be denoted. sig With this elementary view of fractions, it may be per- ceived, that when there is a remainder of 1 unit, it is to. be divided into as many equal parts as there are units in| the divisor, and one of these parts is to be annexed to the quotient. ‘This is performed by merely writing the 1 | as a numerator, and the divisor as the denominator, on’ the right of the quotient. If the remainder be 2 units, there will be 2 such parts of a unit as the divisor indicates | to be annexed to the quotient, and, therefore, the nume- rator will be 2. If the remainder be 3 units, the numera- tor will be 3; andsoon. Hence, whatever the remainder | may be, it becomes, in the quotient, the numerator of a | fraction, the divisor being the denominator. | RULE FOR DIVISION. When the divisor does not ex-— ceed 9, draw a line under the dividend, find how many | times the divisor is contained in the left hand figure, or two left hand figures of the dividend, and write the figure — expressing the number of times underneath: if there be a remainder over, conceive it to be prefixed to the next fig- ure of the dividend, and divide the neat figure as before Thus proceed through the dividend. When the divisor ts more than 9, find how many times ‘ | VE, DIVISION. 17 it ts contained in the fewest figures that will contain it, on the left of the dividend, write the figure expressing the number of times to the right of the diwidend, for the first quotient figure ; multiply the divisor by this figure, and subtract the product from the figures of the dividend considered. Place the next figure of the dividend on the right of the remainder, and divide this number as before. Thus proceed through the dividend. [If there be a final remainder, place it as a numerator, and the divisor as a denominator, on the right of the quotient. PROOF. Multiply the whole numbers of the divisor and quotient together, and to the product add the numerator of any fraction in the quotient: the sum will be equal to the dividend, tf the work be right. ; Abbreviations of the above rule may frequently be adopted, as follows: When there are ciphers on the right hand of a divisor, cut them off, and omit them in the operation; also cut off and omit the same number of figures from the right hand of the dividend. Finally, place the figures cut off from the dividend, on the right of the remainder.- When the divisor is 10, 100, 1000, &c.5 cut off as many figures from the right hand of the dividend as there are ciphers in the divisor; the other figures of the dividend will be the quotient, and the figures cut off will be the remainder. When factors of the divisor are known, diwide the dividend by one of these factors, and the quotient thence arising by the other: the last quotient will be the true one. To find the true remainder, multiply the last re- mainder by the first divisor, and to the product add the first remainder. 1. Divide 4062900311 by 9, and prove the operation. 2. How many times is 502 contaimed in 742607107 3. Suppose 52076348 to be a dividend, and 8649 the divisor; what is the quotient? 4. If 26537009535 be divided into 27856 equal parts, what will be one of those parts? 5. Divide 16500269842 by 86000 ; abbreviating. Q* F a is ARITHMETIC. VIL 6. Divide 8065743924 by 10000; by abbreviation. 7. Divide 290516 by 63; using factors of the divisor. 8. 142375800392 + 5274 —=what number? | Vil. PROPERTIES OF NUMBERS. Before proceeding to examine the properties of num-, bers, a few arithmetical terms, which we shall here collect and define, should be perfectly understood. As. an exercise in this article, the learner may give, upon his slate, an example of each term defined, and each prope erty described. | A UNIT, or UNITY, is any thing considered individual- | ly, without regard to the parts of which it is composed. An INTEGER is either a unit or an assemblage of units; } and a FRACTION is any part or parts of a unit. One number is said to MEASURE another, when it divides it without leaving any remainder. : A number which divides two or more numbers with- out a remainder, is called thelr cOoMMON MEASURE. When a number can be measured by another, the for- mer is called the MULTIPLE of the latter. If a number can be measured by two or more numbers, it is called their coMMON MULTIPLE. ' A COMPOSITE NUMBER is that which can be measured | by some number greater than unity. The auiquotT parts of a number, are the parts by’ which it is measured, or into which it can be divided. | An EVEN NUMBER is that which can be measured, or’ exactly divided by 2. An ODD NUMBER is that which cannot be measured by 2; it differs from an even number by 1. | A PRIME NUMBER is that which can only be measur- ed by unity, that is, by 1 ‘| One number is PRIME TO ANOTHER, when unity is | only number by which both can be measured. VIL PROPERTIES OF NUMBERS. 19 A SQUARE NUMBER is the product of two equal fac- tors; or, the product of a number multiplied by itself. The sQuARE ROOT is the number, which, being mul- tiplied by itself, produces the square number. A cuBE is the product of three equal factors; or, the product of a number twice multiplied by itself. The cube rRooT is the number, which, being twice multiplied by itself, produces the cube. Property 1. The sum, or the difference of any two even numbers, is an even number. Prop. 2. The sum, or difference, of two odd num- bers is even; but the sum of three odd numbers is odd. _ Prop. 3. The sum of an even number of odd num- bers is even; but the sum of an odd number of odd num- bers is odd. Prop. 4. Thesum, or the difference of an even num- ber and an odd number, is odd. - Prop. 5. The product of an even, and an odd num- ber, or of two even numbers, is even. Prop. 6. An odd number cannot be divided by an even number, without a remainder. Prop. 7. A square number, or a cube number, aris- ing from an even root, is even. : Prop. 8. The product of any two odd numbers is an odd number. Prop. 9. The product of any number of odd num- bers 1s odd: hence the square, and the cube of an odd number are odd. Prop. 10. If an odd number measure an even num- ber, it will also measure the half of it. Prop. 11. Ifa square number be either multiplied or divided by a square, the product or quotient is a square. Prop. 12. If a square number be either multiplied or divided by a number that is not a square, the product or quotient is not a square. Prop. 13. The difference between an integral cube and its root, is always divisible by 6. Prop. 14. The product arising from two different prime numbers cannot be a square. 2) ARITHMETIC. VIL Prop. 15. ‘The product of no two different numbers, prime to each other, can make a square, unless each of those numbers be a square. Prop. 16. Every prime number above 2, is either ] greater or 1 less than some multiple of 4. | Prop. 17. Every prime number above 3, is either | greater or 1 less than some multiple of 6. ‘ | Prop. 18. The number of prime numbers is unlimit ed. ‘The first ten are, 1, 2, 3, 5, 7, 11, 13, 17, 19, 23 The learner may find the succeeding ten. sib emma | PROBLEMS. | | A PROBLEM is a proposition or a question requiring something to be done; either to mvestigate some truth 01 property, or to perform some operation. | he following Problems and Rules are founded in thé) correspondence of the four principal operations of arid metic; viz. Addition, Subtraction, Multiplication, anc Division. —_- ii PROBLEM I. ‘The sum of two numbers, dag one ol the numbers being given, to find the other. RULE. Sub: tract the given number from the given sum; the remain der will be the number required. 1. Suppose 37486 to be the sum of two numbers, one of which is 8602; what is the other? 2. 33000 news-papers are sold im London, daily: of these, 17500 are morning papers, the rest, evening: how many of the latter? PROBLEM Il. ‘The difference between two numbers, and the greater-number being given, to find the smaller, RULE. Subtract the difference from the greater number; the ene will be the number required. . | If 1406 be the difference between two numbers, and the greater number be 4879, what is the smaller? | VIIL. PROBLEMS, 21 _ 4, The area of North and South America is 18000000 square miles: that of North America is 11000000: what is that of South America? PROBLEM II. The difference between two numbers, 4nd the smaller number being given, to find the greater. RULE. Add the smaller number and the difference ‘ogether; the sum will be the number required. 5. Suppose 86974 to be the difference between two aumbers, and the smaller number to be 7064; what is he greater number? 6. The British House of Lords consists of 427 mem- vers; the number in the House of Commons is 131 great- st. How many are there in the House of Commons? PROBLEM Iv- The sum and difference of two num- vers being given, to find the numbers. RULE. Sub- ract the difference from the sum, and divide the re nainder by 2; the quotient will be the smaller number. Then add the given difference to the smaller number, vid this sum will be the greater number. ~%. What are the two numbers whose sum is 1094, and whose difference is 154? 3 _§. The United States Congress, consisting of a Sen- ite and House of Representatives, has 288 members. The House has 192 members more than the Senate. dow many in each branch? _ PROBLEM V. The product of two factors, and one of the factors being given, to find the other. RULE. Divide the product by the given factor, and the quotient vill be the required factor. 9. 1246038849 is the product of some two numbers, me of which is 269181: what is the other? 10. Suppose a session of Congress which continues (80 days, to cost 504000 dollars; what is the expense yer day, to the United States? PROBLEM Vi. The dividend and quotient being given | 22 ARITHMETIC. Vit to find the divisor. RULE. Divide the dividend by th given quotient, and the quotient thence arising will | the number sought. | 11. Suppose 101442075 to be a dividend, and 402 the quotient; what is the divisor ? 4 12. 17155 pounds of beef having been equally divide among a number of soldiers, each one found that hj share was 47 pounds. What was the number of soldiers PROBLEM Vil. .'The divisor and quotient being giver| to find the dividend. RULE. Multiply the divisor an) quotient together ; the product will be the required aq idend. 13. If 800027 be a divisor, and 97563 the quotied what number is the dividend ? 14. A quantity of beef was divided equally amoal 2742 soldiers, and each soldier received for his shar 152 pounds. What quantity was divided ? | PROBLEM VIII. The product ‘of three factors, an two of those factors bemg given, to find the third factoy RULE. Find the product of the two given factors, an, by this number divide the given product; the quotes will be the factor required. 15. Suppose the product of three factors to be 1344 one of these factors being 12, and another 8; what is th/ third factor ? 16. How many days will 9720 pounds of hay last 1| horses; allowimg each horse to eat 45 pounds a day ? PROBLEM IX. Two numbers being given, to find thet greatest common measure; that iS, the greatest numbe which will divide them both without aremainder. RULE Diwide the greater number by the smaller, and this di’ visor by the remainder, and thus continue dividing th’ last divisor by the last remainder, till nothing remains The divisor last used will be the number required. When the greatest common measure of more than tw numbers is required, first, find the greatest common mea sure of any two of the numbers, then find the my Ce VIII. PROBLEMS. 238 common measure of the number found and another of the riven numbers, and thus proceed, till all the gwen num- vers are brought in. / 17. What is the greatest common measure of 918, 1998, and 522? 918)1998(2 54) 522(9 | 1836 — 486 | 162)918(5 36)54(1 | 810 la | 108) 162(1 18)36(2 108 36 | 34)108(2 ) 108 Ans. 18. The truth of the rule in this problem will be discovered oy retracing the first of the above operations, as follows. Since 54 [the last divisor] measures 108, it also measures LOS -+-54, or 162. Again, since 54 measures 108 and 162, it also measures 5 X 162+-108, or 918. In the same manner it will be found to measure 2 X 918 + 162, yx 1998. Therefore, 54 measures both 918 and 1998. [tis also the greatest common measure; for, suppose there de a greater— then, since the greater measures 918 and 1998, it also measures the remainder, 162; and since it iigasures 162 and 918, it also measures the remainder (08; in the same manner it will be found to measure ‘the ‘emainder, 54; that is, the greater measures the less, which is absurd. 18. What is the greatest common measure of the num- ers, 323 and 425? 19. What is the greatest common measure of 2314 ind 4626? _ 20. What is the greatest common measure of 1092, 1428, 1197 and 805? ' 21. Suppose a hall to be 154 feet long, and 55 wide; what is the length of the longest pole, that will exactly measure both the length and width of the hall ? | 22. A owns 720 rods of land, B owns 336 rods, and C 1736 rods. They agree to divide their land into equal house lots, fixing on the greatest number of rods for a lot, f ' 24 ARITHMETIC. Vit that will allow each owner to lay out all his land. Ho many rods must there be in a lot? Pl PROBLEM X. ‘Two or more numbers being given, t) find their least common multiple; that is, the least num ber that will contain each of the given numbers a whol number of times. RULE. Divide two or more of th given numbers by any prime number that will measur) them, repeat the operation wpon the quotients and und) vided numbers, and thus continue, till they become prim, to each other. Multiply the several divisors, the la, quotients, and undivided numbers together; the prod@ will be the least common multiple. | If, among the numbers to be divided, any number is measure of another, the measuring number may he | jected; that is, dropped from the operation. It is obvious, that one number is the multiple of anothal when the former contains all the factors of the latte). The factors of 6 are 3 and 2, and the factors of 9 are and 3. Now 54 contains all these factors, (3 x 2 3) 3== 54), and 54 is a common multiple of 6 and 9, but | is not their least common multiple—it is 3 times as gre: as the least, owing to the existence of the factor, 3,1 both 6 and 9. Hence we observe, that a common facto of two or more numbers must enter but once into th multiplication, to give the least common multiple. Th above rule effects the necessary exclusion. 23. What is the least common multiple of 12, 25, at) and 45. 3) 12 25 30 45 We find, after dividin\ by4eles 210N15 . | twice, that 4 and 2 ap! ——____ | pear; and, . by» droppigy 4 5 2 38 |the 2 because it measure, the 4, we avoid angie 3X5X4xX5 xk 3=900 division. ins. 900. | 24. What is the least common multiple of 6, 10, 6 and 20? 25. What is the least common multiple of 25, 35, 60) and 72? | ; | x: COMPOUND NUMBERS. 25 and 245? 27. What is the least common multiple of 18, 82, 94, 788, and 356? 98. Allowing 63 gallons to fill a hogshead, 42a tierce, ‘and 32 a barrel, what is the smallest quantity of molasses, that can be first shipped in some number of full hogs- | heads, then discharged and reshipped in some number of ‘full tierces, and again Biaghasead and reshipped in some number of full barrels ? / ~29. Acertain flour dealer, who purchased his flour from a mill on the opposite side of a river, owned four _ boats, one of which would carry 8 barrels of flour, another 9, another 15, and another 16. What is the smallest “number of barrels he could purchase, that would make - some number of full freights for either of the boats ? i - 26. What is the least common multiple of 105, 140, : ' | IX. y Pd 7 ! COMPOUND NUMBERS. Compounp Numsers are those‘which are employed _to express quantities that consist of several denominations; -each denomination being denoted separately. Under this Bead are classed, all the subdivisions of measures; of ‘Yength, surface, solidity, weights, money, time, &c. The following tables of denominations of compound _numbers, show how many units of each lower denomina- _tion are equal to a unit of the next higher, and, exhibit each lower denomination as a fraction of the next higher. MONEY, WEIGHTS, AND MEASURES. ENGLISH MONEY. The denominations of English Money are, the pound, _£, the shilling, s., the penny, d., and the farthing, qr. 4 farthings Bes Fi = 1d. | Lar. (ie eee ge Le a a =1s. : eee Dy. OW Pgh. 05 eet tae es | 3 quarters ....... a= 1 Fle.) Vor.. oS eer ae | S.qQuarters s:.1 44> —I1E. e.| | I qr. lof 1E. ef 5 quarters ....... a=) Fre. 1 qr’ Soe oF ee ‘ DRY MEASURE. The denominations of Dry Measure are, the bushel, bu., the peck, pk., the gallon, gal., the quart, gi., and By pint, pt. PT i a re ==Iqt. | 1 pt. ee | 4 quarts ...... pate Leal Varts :... 222 of Leal Siquartsa he. 0. .ae = 1p I qt WD WOCKA eh ss 35s = Ibu. |} Ipk. ....—=4o0f lbu ' COMPOUND NUMBERS. a7 ‘ WINE MEASURE. The denominations of Wine Measure are, the ton, 7., the pipe, p., the puncheon, pun., the hogshead, hhd., the © tierce, tier., the barrel, b/., the gallon, gal., the quart, gt. the pint, pt. and the gill, gi. Met cills . .:.... sail pte tibeiwi. com cohhot lmmepints ....... == 1 qt. Ppt a Sof qh wet quarts... ... saul galviit hgts syed oofkieal | B13 gallons ..... — 1bl. lgal... = 4 of 1bl. me gallons). . ...- = Itier. | 1gal... = 4, of I tier. | 63 gallons ...... == [hhd. | lgal... = 4 of 1hhd | 84 gallons ...... — pun. | Igal... = 4 of 1 pun. mo gallons .:..... —Ip. 1 Bal Seer oleh ty PIPES: Hore se LPuet hp asete tof PT: BEER MEASURE. The denominations of Beer Measure are, the butt,,bé:, the hogshead, hhd., the barrel, bl.,the kilderkin, kil., the firkin, fir., the gallon, gal., the quart, gt., and the pint, pt. REOtGe o 2c ee == iat. | lpt. ... = 4 of 1lqt. BMAENS © ci wie 83 == i galon) Pgteri.: == 4 ofiligal: Gallons ....... == fir. | Igal....-= 4 of I fir. Seorkins ........ == [Tkil. | Ifir. ».. = 4 of 1kil. Qkilderkins ..... = Ibl. Blea. Be 4 of 1bl. 3 kilderkins ..... =="Thhds Pelkih 2.9 4) of Lbhd. 2hogsheads ..... —I1bt. | Lhhd: .. — 3 of 1bt. “NOTE. In the United States, the Dry gallon contains 2684 cubic inches, the Wine gallon 231 cubic inches, and the Beer gallon 182 cubic inches. By an Actof the British government, however, the distinction between the Dry, Wine, and Beer gallon was abolished in Great Britain, in 1826, and an Imperial Gallon was established, as well for liquids as for dry substances. The Imperial gallon must contain ‘10 pounds, Avoirdupois weight, of distilled water, weighed in air, at the temperature of 62° , of Fahrenheit’s thermometer, the barometer standing at 30 inches.” This quantity of water will be found to measure 2772/4 cubic inches. ‘The same Act estab- lishes the pound Troy at 5760 grains, and the pound _ Avoirdupois at 7000 grains. 28 ‘ARITHMETIC. IX. LONG MEASURE. The denominations of Long Measure are, the mile, Mis ~ the furlong, fur., the rod or ‘pole, r., the yard, yd., the foot, ft., and the inch, in. 12‘inchessa vy 3. == 1] fee JL mis os eee of ft SBOE gs pW ni dacs == lyd. || 1ft. = 3 of Lyd RaROOS 414-3 5h = Ir. lyd.... = +4, of In@ 40 SOEs yb cette ts eo Adu fdr. oS. = qo of 1 fur 8 furlongs... -.. — 1m. |) 1 for. = + of lm SQUARE MEASURE. The superficial contents of any figure having four sides and four equal angles, is found in squares, by multiplying | together the length and breadth of the figure. 4j The denominations of Square Measure are, the mile, | mm., the acre, A., the rood, R., the rod, r., the yard, yd. i the foot, ft., and the inch, in. | 144 inches....... 21 ft. y Lin. .. == qhy of I ft. CUBIC MEASURE. The cubical contents of any thing which has 6 sides— its opposite sides being equal—is found in cubes, by) multiplying together, the length, breadth and depth. The deaaininations of Cubic Measure are, the yard, yd., the foot, ft., and the inch, in. 1728 inches. cit vues) Bi ea ee = ray Of I ft. ya fe Tc nee = 1yd. | 1ft. .. = 3 oflyd. 40 feet of round timber, or 50 feet of hewn timber make a ton. 16 cubic feet make a foot of wood, and 8) feet of wood make a cord. TIME. The denominations of Time are, the year, Y., the day, | d., the hour, h., the minute, m., aod the second, 5 fi = 7 9 feeb shims bh =< Lyd. ft. ..o== dof hyd 304 yards>...... = ir, lyd == 737 Of Ir | 40:70dS -¢.5eSde. —I1R. | 1r......= gy of URE A POOKS is ark x Ji A. 4 LR: == of iol 640 acres yee oe ew Aba AAS = g}y of Im. | — i | 1 f } 60 seconds ....., 1m. / 1s. rire ae ce | 60 minutes ...... — Ih. | lm = 7, of th DAT OUTE IS fais. « dtd = 1d: | Lhe. = 5, of ld BOD MEY Voie. > 8 == 1 Y, lia legs’ ik = zt, of LY 1X. COMPOUND NUMBERS. 29 The earth revolves round the sun once in 365 days, . 5 hours, 48 minutes, 48 seconds: this period is therefore — a Solar year. In order to keep pace with the solar year, in our reckoning, we make every fourth year to contain 366 days, and call it Leap year. Still greater accuracy requires, however, that the Leap day be dispensed with 3 times, in every 400 years. Whenever the number which denotes the year can be measured by 4, the year is Leap year— the centurial years excepted. The year is also divided into 12 months—See Almanac. THE CIRCLE. The divisions of the circle, C., are, the sign, S., the _ degree, (°), the minute, (’), the second, (”). ee This table is applied to the Zodiac; and by it are com- puted, planetary motions, latitude, longitude, &c. mOseconds « .. «..’. — j! | Lik’ shes he == gy of 1° Se mUteS) <. 6.» = ry: } Wnts tie == gy of 1° 30 degrees ....... — i i2 . == 3 of 1S. 12 signs ee ‘aati | is. == zy Of IC. GEOGRAPHICAL MEASURE. - The circumference of the globe—like every other cir- cle—is divided in 360 equal parts, called degrees. Hach degree is divided mto 60 equal parts called miles, or minutes. Three miles are called a league. On the equator, 69+ statute miles are equal to 60 geo- graphical miles, or | degree, nearly: and, on the meridian, at a mean, 69); statute miles are equal to a degree. REDUCTION OF COMPOUND NUMBERS. Repvucrion is the operation of changing any quantity from its number in one denomination, to its number in another denomination. } RULE FOR REDUCTION. When a greater denomina- tion is to be reduced to a smaller, multiply the greater denomination, by that number which is required of the smaller, to make a wnit of the greater; adding to the _ product, so many of the smaller denomination as are ex- pressed in the given quantity. Perform a like operation on this product, and on each succeeding product. * . 30 ARITHMETIC. 1X. When @ smaller denomination is to be reduced to a greater, divide the smaller denomination by that number which is required of the smaller, to make a unit of the next greater: the quotient will be of the greater denomi-_ nation, and the remainder will be of the same denomina- tion with the dividend. Performa like operation on this: | quotient, and on each succeeding quotient. © : . Reduce £351 13s. Od. 1 qr. to its value in farthings. | . How many pounds, &c. are there in 6169 pence > | In 591b. 13dwt. 5gr. Troy, how many grains? ~~ Change 20571005 drams to its value in tons, &c. In 231ib 33 05 0D 5gr. how many grains ’ . How many English ells are there in 352 nails ? . Reduce 7 bushels and 6 quarts to pints. . How many hhds. are there in 9576 pints of wine? . 9. How many pints im 1 bl. 1 fir. 1 pt. of beer ? 7 ~ 10. How many miles, &c. are there in 26431 rods? 11. In 3 square miles, how many square rods ? yy 12. In 1259712 cubic inches, how many cubic yards ? } 13. Reduce 1 solar year, 7d. and 10h. to seconds. cag tree eee a ADDITION OF COMPOUND NUMBERS. The operation of adding compound numbers, differs from that of adding simple numbers, only, with respect | to the irregular system of units, which determines the | principles of carrying from one denomimation to another. | RULE. Write the numbers so that each denomination | shall stand in a separate column. Add the numbers of | the lowest denomination together, and divide their sum | by that number which is required of this denomination to | make @ unit of the next higher: write the remainder un- | der the column added, and carry the quotient to the next | column. Thus proceed through the denomination. 14. What is the sum of £9 8s. 4d., £250 8s. 5d. 3qr., £9 7s. 4d., £20 16s. 4d., and 3s. 6d. 2qr. 7 | 15. Add together 100z..14dwt. 16gr., 5lb. 9oz. | 6dwt. 22¢r., 4lb. Loz. 18 awt. Qgr., and 11 dwt., Troy | 16. Add together 15T. 19cwt. 3qr. 21b. 702., 25 | 13cwt. 2qr. 20lb. 150z., and 3qr. 26]b. x. COMPOUND NUMBERS. ot _ 17. How much is 18 yd. 3qr. 3na., Loyd. 2qr. 3na., %yd. lqr. 2na., and 57yd. 3qr. 2na. of cloth? 18. Add together 25bu. 3pk. 7qt., 100bu. 2pk. 4qt., ’115bu. 2pk. 2qt. lpt., and 57bu. 3pk. of corn. 19. Add together 4p. 125gal. 3qt., 75gal. Qqt. 1 pt., 35p. 92gal., and 39 gal. 3qt. 1 pt. of wine. , _ 20. How many acres are 13A. 3R. 38r., 87A.2R. 33r., 26A. 2R., 41 A. 2R. 28r., and 36r.? ‘ | 21. How much hewn timber is 9T. 19ft. 1725in., 150T. 39ft. 1695in., and 500T. 31 ft. 915in.? SUBTRACTION OF COMPOUND NUMBERS. RULE. Wriie the several denominations of the smaller juantity under the same denominations of the greater juantity: then, begin with the lowest denomination, and verform subtraction on each denomination separately. Whenever a number expressing a denomination in the spper line is smaller than the number under it, increase he upper number by as many as make a unit of the next ugher denomination, and consider the number of the vext higher denomination in the upper line, to be 1 less han tt stands. 22. Subtract ilb. 100z. 16dwt. from 3lb., Troy. 23. From 6T. 3cwt. take 7cewt. 2qr.15lb., Avoir. 24. From 2th 73 take 73 63 2D 5er., Apoth. wt. 25. Subtract 3qr. 3na. from 5yd. 2qr. Lna. of cloth. 26. Subtract 8bu. 1 pk. 6qt. 1 pt. from 50bu. of corn. 27. From 3hhd. 25gal. take 41 gal. 2qt. of wine. 28. From 6bl. 1kil. take 1 fir. 6 gal. Sqt. of beer. 29. Subtract 3yd. 10in. from 5yd. 2 ft. oi Todas 30. Subtract 57 A. 2R. 31r. fromim., Square mea . 31. Subtract 2Y. 90d. 4h. 55m. from 4Y., Time. » a MULTIPLICATION OF COMPOUND NUMBERS. RULE. Begin with the lowest denomination, and mul- ‘ply each denomination separately; divide each product by the number which is required of its own denomination | } ea 32 ARITHMETIC. io make a unit of the next higher; write the remainde under the denomination multiplied, and carry the quotiel to the product of the next higher denomination. | | | 32. Multiply £215 19s. 6d. by 72 or its factors. 33. Multiply 2lb. 50z. 7dwt. 10gr., Troy, by 56. 34. What is 16 times 18cwt. 3qt. 15]b., 1402.? 35. What is 81 times 36bu. 3pk. 6qt. 1 pt., Dry me: 36. Multiply 4p. 105gal. 3qt. of wme by 60. 37. Multiply 2m. 7fur. 35r., Long mea., by 63. | 38. Multiply 4m. 320A. 1R. 9r., Square mea., by 15 39. Multiply 2Y. 250d. 14h. 30m., Time, by 96. | | % x DIVISION OF COMPOUND NUMBERS. | RULE. Divide each denomination separately, begin ning with the highest. Whenever a remainder occurs reduce it to the next lower denomination, add it to th number expressed in the lower denomination, and divid’ it therewith. | 40. Divide £251 15s. 7d. 2qr. into 46 equal parts.) 41. Divide 15lb. Soz. 7dwt. 5gr., Troy, by 13. 42. Divide 12T. 27 lb. 150z., Avoirdupois, by 5. 43. Divide 136 K.e. 3qr. 3na. of cloth by 31. | 44. Divide 1621 bu. 2pk. of corn into 50 equal parts’ 45. Divide 1 pipe of wine equally among 9 owners. | 46. Divide a Leap year into 100 equal paris. | | FEDERAL MONEY. | f Ve | The denominations of Federal Money are, the eagle the dollar, the dime, the cent, and the mill. 16 mill’ make 1 cent, 10 cents 1 dime, 10 dimes 1 dollar, and 1! dollars 1 eagle. Dollars, ¢, and Cents, ets. are the onl denominations commonly mentioned in business— eagle) being counted as tens of dollars, dimes being counted a tens of cents, and mills not being denoted. j 100 cents ........ == $1 || Lcent... =), of $i] The cents in any number of dollars are expressed by the same figures which express the dollars, with twe ! IX. COMPOUND NUMBERS. 33 siphers annexed; $15==1500 cents. The dollars in any number of cents are distinguished by cutting off two fig- wes from the right for cents; 325 cts. — $3.25. Operations on numbers expressing Federal money, are yerformed as on simple numbers; care must however be aken, in addition and subtraction, to place dollars under lollars, and cents under cents; these denominations being veparated by a point. 47. What is the sum of $34.21, $7064.04, 36cts., $10004.85, $96, $900.10, $14, $1.99, and #76529 ° "48. Subtract $4926 from $ 12262.37. 49. Subtract $297.18 from $ 100000. 50. Suppose $295.48 to be a multiplicand, and 25 the nultiplier; what is the product? _ In multiplication, only one of the factors can be Federal oney, and the product will be of the same denomination js this factor. If, therefore, there be cents in either fac- jor, two figures must be pointed off for cents, from the ight of the product. _ 51. What is the product of 96 cts. multiplied by 43 ? _ §2. What is the value of 1304 pounds of coffee at 9 vents per pound ? _ 53. How many times $7 are there in $29.46? , In division, when both the dividend and divisor are “ederal money, they must both be of the same denomi- ation. If therefore, one of the numbers contain cents, nd the other dollars only, the latter number must have wo ciphers annexed to it. 54. How many barrels of Lines at $4. 36 per barrel, van be purchased. for $4370? | 55. Divide $4279.50 into 746 equal parts. | 56. If 407 pounds of Hyson tea cost $395, what is fe cost of 1 pound? _ 57. How many times are 95cts. contained in $56? _ 58. A merchant sold 1248 yards of cloth, at such price s to gain 1 cent on every nail. How much did he gain > [ 59. What is the gain on a hogshead of molasses, sold t an advance of 3 cents per gallon? 60. A jeweller sold a silver pitcher 3lb. 80z. 16dwt., jt 7 cents a pennyweight. What did it amount to ? 34 ARITHMETIC. Ix 61. What is the freight of 60480 pounds of cotton fron Charleston to Liverpool, at $4 per ton ? y MISCELLANEOUS EXAMPLES. 62. How many bottles, holding 1 pint and 2 gills eack are required for bottling 4 barrels of cider? 63. How much will 46 bushels of oats cost, at 4 pene 2 farthings for every two quarts ? | 64. A brewer sold 96 hogsheads of beer for £38) 16s. What was the price of 1 pint at the same rate ? 65. A certain tippler spent 12 cents a day for arder| spirit, during 39 successive weeks, and then died, the vi tim of his folly. What did the spirit all cost ? 66. Bought five loads of wood; the first containing — cord 32 cubic feet, the second 1 ee 64 cubic feet, th, third 112 cubic fect, the fourth 1 cord 28 cubic feet, ani the fifth 1 cord 20 cubic feet. How many cords wer there in the whole ? 67. Bought goods to the amount of £25 13s. 10d) 2qr.; and afterwards sold goods to the same man, amount. ing to £30 10s. 4d. 2qr. What is the balance of mone) in my favor? \ 68. A farmer sold five lots of land, at $9 an acre; thi first lot contaming 30A. 2R. 2r., the second 41A. 3B 8r., the third 14A.1R. 10r., the fourth 25 A. 36r. z ant the fifth 54A. 6r. What did the whole amount to? — 69. How many cubic inches in a brick 8 inches long) 4 inches wide, and 2 inches thick ? 70. How many cubic inches in the cube of 2 inches meni in the cube of 3 inches?..... in the cube of ? inches?..... in the cube of 5 inches? 71. If the cube of 4 inches be taken from the cube ol L foot, how:many cubic inches will remain ? 72. If the cube of 4 inches be taken from the cubs of 2 feet, how many cubic inches will remain ? . 73. A young man, on commencing business, was worth £643 10s.; the first-year he cleared £54 11s. 7d. 2 qr. the second year, £87 Os. 10d. Iqr.; but the third year he lost £196 7s. 11d.3qr. How much was he then worth: oa ee | | q LX. COMPOUND NUMBERS. 35 _ 74. A gentleman had a hogshead of wine in his cellar, rom which there leaked out 17gal. 3qt. lpt. How nuch then remained ? 75. Aman started ona journey of 20 miles 6 fur. 29r., nd stopped to rest at a house, 4m. 4 fur. 20r. from the lace of starting. How far had he still to go? _ 76. Ina pile of wood, 96 feet long, 5 feet high, and 4 eet wide, how many cords? | 77. How much would 13 hogsheads of sugar cost, at } cents per pounss allowing each hogshead to contain, sewt. 3qr. 24]b.? | 78. Acent weighs 8 peaniweignts 16 grains. What 3 the weight of 100 cents? 79. How many yards of cloth are there in 19 pieces; ach piece containing 27 yd. 3qr. 2na.? 80. If a man sell 2b]. 1 kil. 1 fir. 6 gal. Qqt. 1 pt. of eer in #6 week, how many barrels would he sell m 6 weeks ° 81. If 1 pint and 3 gills of wine will fill a bottle, how wich will fill a gross, or 12 dozen bottles ? i 82. A father left an estate worth £5719 17s., to be ivided equally among 11 children. How much was each ‘ne ’s share ? 83. Sixteen men own 24 tierces of molasses, in equal lie. What is one man’s share ? 84. A company of 23 men bought 1850 acres 10 rods ‘f wild land, and divided it equally among them. How such land bad each mah ? 85. What must be the length of a lot of land, that is & ods wide, in order that the lot shall contain 1 acre? | Observe in the above question, that 1 acre contains 60 square rods; and, that this number of square rods is ae product of the two factors that denote the width and lie of the lot. See Prosiem V, page 21. _ 86. What must be the depth of a house lot, that mea- “ures 72 feet on the front, to contain 9432 square feet ? _ 87. What must be the length of a stick of hewn timber, nat is 10 inches wide and 1 ft. 3in. deep, in order that he stick shall contain 1 ton? | Observe in this question, that the number of cubic i k le 36 ARITHMETIC. x inches in a ton, is the product of the three factors whic] denote, in inches, the width and depth and length of th stick. See ProgLemM VIII, page 22. 88. What must be the length of a pile of wood that j 4 feet wide and 3 feet high, in order that the pile sha contain 1 cord, that is, 128 cubic feet? 89. Suppose a pile of wood to be 11 feet long and‘ feet wide; how high must it be, to contain 2 cords 4 fee of wood and 10 cubic feet ° | x | FRACTIONS. A FRACTION signifies one or more of the equal part mto which a unit, or some quantity considered as an in ieger, or whole, is divided. A fraction is expressed by two numbers or termi written one above the other, thus, ¢. The lower tem —called the denominator — denotes the number of equi parts into which the integer is divided; and the uppe' term—called the numerator—indicates what numbe| of those equal parts the fraction expresses. We may not only consider a fraction as a certain num ber of parts of a unit, but, may also view it as a x | of a certain number of units. Thus, 3 may either b/ considered as 2-thirds of 1, or, 1-third of 2; for 1-thir of 2 is the same quantity as 2-thirds of 1. Hence, if th numerator of a fraction be viewed as an integer, an divided into as many equal parts as the denominator 4 dicates, the fraction may be regarded as expressing of! of these parts. Thus, if 4 be divided into 5 equal party the fraction ¢ expresses one of these parts. Fractions ‘generally have their origin from the divisia of a number by another which does ‘not measure it; th excess. of the dividend, above what can be measured b the divisor, being the numerator, and the divisor bem the denominator, as shown in Art. VI. If the numerator of a fraction be made equal to th } | , ! ’ x | FRACTIONS. 37 } denominator, the fraction becomes equal to unity; thus $=1. If the numerator be greater than the denominator, the fraction is equal to as many units as the denominator is contained times in the numerator ; for example P= 3. ‘Hence, a fraction may be viewed as an unexecuted divi- sion; the divisor being written under the dividend. It follows, also, that since any number divided by 1 gives the same number in the quotient, any number may be ‘expressed as a fraction by making 1 its denominator. For example, 17 may be expressed thus, 4’. ' The following propositions concerning fractions, should be distinctly noticed. PROPOSITION I. As many times as the numerator is made greater, so many times the fraction is made greater; and, as many times as the numerator is made smaller, so many times the fraction is made smaller. Hence, a jrac- tion is multiplied by multiplying the numerator, and divided by dividing the numerator. _ PROPOSITION Il. 4s many times as the denominator is made greater, so many times the fraction is made smaller ; and as many times as the denominator is made smaller, so many times the fraction is made greater. Hence, « fraction is divided by multiplying the denominator, and multiplied by dividing the denominator. PROPOSITION II. Whenthe numerator and denomina- tor are both multiplied, or both divided by the same num- ber, the quantity expressed by the fraction is not thereby changed. | A PROPER FRACTION is a fraction whose numerator us less than its denominator; as 35. |; An IMPROPER FRACTION isa fraction whose numerater equals, or exceeds its denominator; as 4, }. _Anumber consisting of an integer with a fraction an- mexed, as 142, is called a MIXED NUMBER. | A coMPOUND FRACTION is a fraction of a fraction; as of 3. 4 of 45 of 2. A COMPLEX FRACTION is that which has a fraction either in its numerator, or in its denominator, or in both . at of them; thus, 6 Ox a 7 ) 4 * i | 38 ARITHMETIC. Dy | REDUCTION OF FRACTIONS. .| | REDUCTION OF FRACTIONS consists in changing them) from one form to another, without altering their value. | CASEI. To reduce a fraction to its lowest terms; that is, to change the denominator and numerator to the smallest numbers that will express the same quantity. RULE. Divide both terms of the fraction by thew greatest common measure, and the two quotients will be the lowest terms of the fr action. See Pros. IX, page 22. | _. When the greatest common measure is readily per: ceived, the fraction may be reduced mentally. For in- stance, tne greatest common measure of the terms of the fraction 75, 1s 4, and the only notation necessary in the reduction, is, 7 =. | Dividing the terms of a fraction by a common measure} which is not the least, will reduce it in some degree, and when thus reduced, it may be reduced still lower by another division, and so on, till no Bela be will measure both the terms. For example, to reduce 33, divide by 2, and the result is 34; again, divide by 3, i and the result is) 3. Here the fraction is known to be in its lowest terms,| because the terms are prime to each other. ' 1. Reduce 84 to its lowest terms, by repeatedly dividing the Ae by any common measure. 2. Reduce +55 to its lowest terms, by dividing the, terms by their greatest common measure. 3. Reduce each of the following fractions to its lowell 32 270 384 156 720 3108 terms. Zoo: 306° isd’ 336° 736° 3559" a CASE Il. To reduce a whole number to an imprope) fraction. : | RULE. Multiply the whole number by the proposes denominator, and the product will be the numerator. When the quantity to be reduced is a miaed number: the numerator of the fraction in the mixed number mus) be added to the product of the whole number, and thei’ sum will be the numerator of the improper fraction. &. FRACTIONS. 39 4. Reduce 16 to a fraction whose denominator is 9. 16 In 1 unit there are 9-ninths; 9 Seema se 9 times as arr Many ninths as there are units in 144 Ans. +44 any vee ite Reduce 75 to a fraction whose denominator is 13. Reduce 3 to a fraction whose denominator is 342. How many fifteenths are there in 74? How many eighths of a dollar in $647? Reduce 36% to an improper fraction. ) 364 In this example, we add the i 3-sevenths to the sevenths pro-. — ,., | duced by the multiplication of 36. 256 Ans. 27° | by 7, and thus obtain 238 10. Reduce 254% to an improper fraction. 11. Reduce 61552, to an improper fraction. _ 12. How many sixteenths of a dollar in $5412? eT CASE Il. To reduce an improper fraction to a whole ‘number, or a mixed number. RULE. Divide the numerator by the denominator, and the quotient will be the whole, or mized number. 13. Reduce 38? to a whole, or mixed number. 8)362 Since $ are equal to 1 unit, oe there are as many units in 282 as there are times 8 in 362. 14. Reduce #393 to a whole, or mixed number. 15. How many units are there in #°31° ? 16. How many dollars in 29? of a dollar? CASE Iv. To reduce a compound fraction to a sim- ple, or single fraction. RULE. « Multiply all the numerators together for a new numerator, and all the denominators for a new denomi- | nator: then reduce the new fraction to its lowest terms. | When any numerator is equal to any denominator, the operation may be abbreviated by rejecting both. | If part of the compound fraction be an integer, or a mixed number, it must first be reduced to an improper | fraction. : . oD 40 ARITHMETIC. X! . Reduce ¢ of 2 of # of 6 to a sunple fraction. Here the common term, 3, 1s | Baba =33=8 | omitted in the multiplication. | 18. Reduce 2 of 4 toa Simple fraction. ) 9. Reduce + of + i) f O of 32 ta a simple fraction. | 20. Reduce 44 f Soir of 2-7; to a simple fraction. f 21. Reduce 50 of 5 toa simple fraction. | €ASE V. To reduce a fraction from one denomination | to another. i RULE. Multiply the proposed denominator by the numerator of the given fraction, and divide the product by the denominator of the given fraction; the quotient | “ee be the numerator of the proposed denominator. | . Reduce 2 to afraction whose denominator shall be | a an in other words change 5-sixths to omaaes | 14 g is equal to % of 77, and 2 is’ 5 5 times as mine a therefordl| 6)70 , | find 5 times 14-fourteenths and — fins. 11¢ | take ¢ of this product for the | 11% 14 | required fourteenths. j 23. How many fifths are there in 2? 1 24. 3% is equal to how many twenty-fourths ? | 25. Hahn $ to a fraction whose denominator is 4. | 26. How many twelfths of 1 shilling in + of 1s.? | =i CASE VI. To reduce the lower denominations of a compound number to the fraction of a higher denomination. } RULE. Reduce the given quantity to ; the lowest denomi- nation mentioned, and this number will be the numerator: | ihen reduce a unit of the higher denomination to the same | denomination with the numerator, and this number wild be the denominator. | 27. Reduce 7oz. 18dwt. 13gr. to the fraction of a | ound. | We find, that 7o0z. 18 dwt. 13gr. when reduced to grains, gives 3805 for the numerator; and 1 pound when reduced to grains, gives 5760 for the denominator. Therefore, $422 — +%4 is the fraction required. 28. Reduce 4s. 9d. Sqr. to the fraction of £1. ie FRACTIONS. 41 29. Reduce 34 inches to the fraction of a yard. 30. What fraction of a hogshead is 9 gal. 22 pt.? 31. Reduce 5cwt. 8lb. 40z. to the fraction of a ton. CASE Vil. ‘To reduce the fraction of a higher denomi- nation to its value in whole numbers of lower denomination. RULE. JWultiply the numerator by that number of the next lower denomination which is required. to make a wnit of the higher, and divide the product by the denominator; the quotient will be awhole number of the lower denomi- nation, and the remainder will be the numerator of a frac- tion. Proceed with this fraction as before, and so on. It will be readily perceived, that the fraction of a higher denomination is reduced to the fraction of a lower, by multiplying the numerator by the number of units of the lower, required to make a unit of the higher. ‘Thus, 3 of a bushel is 4 times as many fifths of a peck; that is, of apeck. Again, 7 of a peck is 8 times 12-fifths, that is, %° of a quart; and again, %° of a quart is 2 times 96-fifths, that is, 422 of a pint. If the denominator be multiplied, instead of the numerator, the effect is the re- verse, and the fraction is reduced to a higher denomination. Thus, 2 of a pint, (the 5 being multiplied by 2,) becomes 7a of a quart; 3%; of a quart, (the 10 being multiplied by 8,) becomes = of a peck; and ,%; of a peck, (the 80 being multiplied by 4,) becomes 335 of a bushel. 32. Reduce 43 of a gallon to its value in quarts, &c. 11 We find by multiplication, that A +3 Of a gallon is 4% of a quart; i 2)44 and, by division, that 73 of a ph quart is 3qt. and 4% of a quart. 3. 8 We then find, that 58, of aqt. is 2 46 of a pint; and, that 4§ of a pt. 12)16 is 1 pt. and +4 of a pt. And thus, by finding the units of one de- : ; nomination at a time, we finally —— obtain the whole answer, which, 12)16 denoted as a compound number, 144—14 |is 3qt. I pt. 13g). 3. Reduce 2 of £1 to its value in shillings &c. 4% ¥ e ~~ Bc 42>, ARITHMETIC. | 34. Reduce 4 + of a yard, to its value in feet, &c. | 35. In 75 of Tow. how many quarters, pounds, &c.? 36. Reduce 33 of a bushel to pecks, quarts, and pints. | CASE VII. To reduce fractions toa common denomi-( nator; thatis, to change two or more fractions which have | different denominators, to equivalent fractions, that shall have the same denominator. | RULE ist. Multiply each numerator into all the denom- | inators except itsown, for a new numerator. Then mul- tiply all the denominators together for a new denomination | and place it under each new numerator. | RULE 2nd. Find the least common multiple of the given | denominators for the common denominator; then divide | the common denominator by each given denominator and multiply the quotient by its given numerator; the several products will be the several new numerators.- (See PROBLEM X, page 23.) The ist. of the above rules is convenient when m | terms of the fractions are small numbers, but the 2nd. is; otherwise to be preferred, as it always gives a denomina- tor which is the least possible. Other methods of finding a common denominator will occur to the student, after further practice. ‘4 If any of the fractions to be reduced to a common de- nominator be compound, they must first be simplified. — 37. Reduce 3, 44, =4 and 42 to a common denomi- nator. Tn this example, the least common denominator is found | to be 840. Then the several numerators of the common denominator are found as follows. | 840+ 8—105, and 105 5==525. Ans. 2—228) 340 + 12—= 70, and 7011770. =o 840+ 14=—= 60, and 60 9=540. 38 840+15= 56, ane 56 X 13= 728. ee | 38. Reduce 3 7 +s and 3 to a common denominator. 39. Reduce 2 oe 4354 and 5to a common denominator. | 40. Reduce 24 3 eT and 42 2 to. a common denominator. 41. Reduce 4 and 2 of + te a common denominator. t 4 | x. FRACTIONS. 48 CASE IX. To reduce a complex fraction to a simple raction. RULE. Jf the numerator or denominator, or both, be vhole or mixed numbers, reduce them to improper frac- tons: multiply the denominator of the lower fraction into he numerator of the upper, for a new numerator; and nultiply the denominator of the upper fraction into the ) aie of the lower, for a new denominator. 42. Reduce 2. to a simple fraction. 7 | 9 ‘Th mation, os OS Ans. 88 € operation =5 1533 = 93 Ans 2 48. geet a each of the following complex fractions. 10 53, 32, 21, 65, 28, Ale > ets g coins | ° ADDITION OF FRACTIONS. Fractions are added by mere-y adding their numera- ors, but they must be of the same integers; we cannot nmediately add together 3 of a yard and 2 of an inch, sr the same reasons that we cannot immediately add gether 5 yards and 3 inches. They must, also, be of ie same denomination; we cannot immediately add to- ether fourths and fifths. RULE. Reduce compound fractions, (if there be any), » simple fractions, and reduce all to a common denomi- ator; then add together the numerators, and place their i um over the common denominator. If the result be an nproper fraction, reduce it to a whole or mixed number. 1 44. Add together, 33, 3, 8$ and 4. | By operations not here de- I ( 360 noted, we find the common de- ey “280 nominator to be 360; and also A 3 935 find the several new numerators. , 82 216 The sum of the fractions is 3¢5 , 2 wi a [= — 2271, which, added to the Q271 9919271 | whole numbers; gives the total 360 380: Re SEO} (sum, 13444 44 ARITHMETIC. xX 45. Add together, 97, 1244, 74, 2 and 213. 46. What is the sum of -+2-+4$+2+4? 47. What is the sum of 19,3,-+7 0 48. What is the sum of 4 of 4 49. Find the sum of $ of a sh In this example, first reduce the 3 0 and the fraction of a penny. 50. Find the sum of -?; of a gallon and Z ofa gill. | 51. What is the sum of 52 days and 52,3, minutes ? | 52. Whatis the sum of 4 of a cwt., 82Ib. and 3%, 02.| | @ of a penny ! a shilling to pence, —_— : SUBTRACTION OF FRACTIONS. As in addition of fractions we find the sum of the} numerators, so in subtraction of fractions we find th difference of their numerators. RULE. Jf either quantity be a compound fraction, ré duce it to a simple fraction, and if the two fractions hav’ different denominators, reduce them to a common denomh ‘nator. Subtract the numerator of the subtrahend fror the numerator of the minuend, and place the remainde: over the common denominator. co ug When the minuend is a mized number, and the frac tion im the subtrahend is greater than that in the minu end, subtract the numerator of the subtrahend from th denominator, and to the difference add the numerator 0 the minuend; and consider the integer of the minuen to be 1 less than it stands. It is not always obvious, which of two fractions ex) presses the greater quantity. In such case, the fraction are denoted with a character between them, thus, 53 DT and the greater is discovered by reducing them to acom| mon denominator. , a | 53. What is the difference between 247 and 262? 72 Here the fraction in the su 3 ee trahend: is the greater, and w 263 27 ne | 947 56 are obliged to convert a unit int : ie seventy-seconds to obtain a quan 143 43 by i ae Te tity from which to subtract 3$. 54. What is the difference between ;2; from 18? xX, FRACTIONS. 45 55. Perform subtraction on 33 wii. 56. What will remain if 5145 be taken from 842? 57. Subtract $ of 7 from 36-7. 58. What is the difference between 44), and 10? ' 59. What will remain if 2 of Z be taken from a unit? | 60. What is the difference between #7; and 13? 61. 43— 5 of ¢ of 2 is equal to what quantity? MULTIPLICATION OF FRACTIONS. _ The following rules for multiplication of fractions, are vased on the Propositions 1, and 11, stated in page 37. , CASEI. To multiply a fraction by a whole number. | RULE. LHither multiply the numerator, or divide the enominator by the whole number. | CASE TI. To multiply a whole number by a fraction. ) RULE. JSMultiply the whole number by the numerator, nd divide the product by the denominator. (CASE II. ‘To multiply a fraction by a fraction. | RULE. Multiply numerator by numerator, and denom- ator by denominator, for a new fraction. ' When both factors are mixed numbers, it is generally ‘lore convenient to reduce them to improper fractions ‘nd then proceed according to the rule under Case itr. | The effect of multiplying any quantity by a proper ‘action is, to give in the product, such a part of the uantity multiplied as the fraction indicates. Thus the ‘roduct must be less than the multiplicand. _ This effect tthe operation will appear consistent with the principle of uultiplication, when it is considered, that multiplying any umber by 1, gives only the same number in the pro- uct; and, therefore, multiplying by less than 1, must give product less than the number multiplied. , 62. Multiply23by 9. 23x9 — 25X92 — 25 — 82 | 63. Multiply 49 by &. (See rule under Case 11.) | 64. Multiply 35 by 2. (See rule under Case 111.) 65. Multiply 6% by 344. (Remark under Rule m1.) 66. What is the product of 34 by 15? AG ARITHMETIC. | | 67. What is the product of 9241 by = fo! | 68. What is the product of ¥ z, by 23? | 69. What is the product of i 5 by 12%? RY 70. Which is the most, 3X 65, or, 65 xo? a 71. What is the product of 2941: 3 by 25? f In this example, it will be most convenient to find tl product of the whole numbers without regard to the fra tion first; then find the product of the fraction in a sep) rate operation, and, finally, add the two products togethe! 72. What is the product of 361 by 3443? 73. How many square inches of paper “in a sheet th is 142 inches long, and 113 mches wide? DIVISION OF FRACTIONS. | The rules for division of fractions, like those for a | plication, are ceey on Propositions 1, and 11. | | CASEI. To Taiie a fraction by a whole number. RULE. Hither divide the numerator, or multiply ti denominator, by the whole number. CASE II. To divide a whole number by a fraction. | 1 RULE. Multiply the whole number by the denomim) tor, and divide the product by the numerator. | CASE 111. To divide a fraction by a fraction. RULE. Invert the divisor, and then proceed as in mu tiplying a fraction by a fraction. | Observe, that the operation of this last rule is, to mu} tiply the denominator of the dividend by the numeraty, of the divisor for a new denominator, and the numerat| of the dividend by the denominator of the divisor | new numerator. Compound fractions are to be reduced to simple one, and mixed numbers to improper fractions, before th adoption of either of the above rules. | 74. Divide 7 by 8. 7+S—=asieg=—ay. Ans. & 75. Divide 14 by +. (See rule under Case 11.) © 76. Divide 2; by 2 ‘(See rule under Case III.) 77. Divide the compound fraction 3 2 of 2, by 6. j XK. FRACTIONS. AT 78. Divide 325 by the mixed number 53. 79. What is the quotient of $4 divided by 13? — 80. What is the quotient of 57 divided by ;5? 81. What is the quotient of 4 divided by 4? 82. Divide § of 5 by % of $ of 2. | 83. What is the quotient of 912 divided by 15? | 84. What is the quotient of 2062 divided by 944? . 85. How many times is 24 contained in 319? - 86. How many times is 194 contained in 994? 87. How many times ¢ of an inch in =; of a yard? First, reduce the 3°, of a yd. to the fraction of an inch. 88. How many times 2 of a gill in 3 barrels ? 89. Suppose a wheel to be 11,7; feet in circumference; 1ow many times will it roll round in going 393 rods ? MISCELLANEOUS: EXAMPLES. In the following examples, all fractions which appear athe answers, must be reduced to their value in whole tumbers of lower denominations, whenever there is op- iortunity for such reduction. 90. What distance will a carrun in 93 hours, allowing ts velocity to be 232 miles an hour? 91. Suppose a car wheel to be 8 feet 7 inches in cir- uumference, how many times will it turn round in running 165 miles ° 92. If 32 cwt. of sugar be taken from a hogshead con- ainmg 14cwt. 1qt. 64lb., how much will remain in the iogshead ? - 93. What is the sum of 16Zcwt., 7ewt. 3qr. 84ib., IT. 194cwt., 2cwt. liqr., and 7 of a ton? 94. A farmer owning 1322 acres of land, sold 464A. 3R.12r. How much land had he remaining ? _ 95. What is the value of 367 acres of land, at $472 yer acre? | 96. What is the value of 154 barrels of flour, at $4.625 per barrel ? 97. What is the value of a- load of wood, containing i feet, [§ of a cord,] at $5.25 per cord? Or, what is 3 £ $5.25? Or, $5.25xXS—? ) 48 ARITHMETIC. 4 3544 rods on every side? (See page 28.) 99. What quantity of land in a lot, which is 654 ro long and 47} rods wide ? | 100. What quantity of wood is there in a pile, 14- feet long, 342; feet wide, and 6,3; feet high: ? i 101. Supiobs a lot of land to be 61rods wide, ho long must it be, to contain 1 acre? (See Pros. v, pa ) 98. How much land is there in a square lot, | . i 21. Consider that 1 acre contains 160 rods.) 102. What quantity of loaf sugar must be sold at 19 ; cents per pound, that the price shall amount to $524 | 103. What cubical quantity of earth must be ae in digging a pit, 135 feet deep, 123 feet long, and § feet wide ? a) 104. What quantity of hewn timber is there in a sti that is 124 feet long, 24 feet deep, and 12 foos wide ?) 105. Suppose a stick of timber to be 14, 5 foot dee) and 8 inches wide; what must be the leneth oF the stic in order that its quantity shall be 1 ton of hewn tim) | (See Pros. viii, page 22. Consider a ton as the pr’ duct of three factors. ) 106. Suppose wood to be piled ona base 18 feet lo and 73 feet wide, what must be the height of the pile, contain 91 cords? | 107. What quantity of molasses in 4 casks, seal? severally, 55;gal., 31ljgal., 277; gal., and 584%, gal.? 108. What i is the cost of 486 bushels of corn, at 69) cents per bushel ? ne 109. Suppose 63. gallons to have leaked from a hog wine, at 874 cents per gallon? | 110. How many bottles, each holding 13 pint, are rm 111. Suppose 44 gallons of cider to have evaporate from a barrel; what number of bottles, each holding 1p 112. What is the value of 1425 tons of coal, at 7 dollars per ton? the rate of 2 of a dollar per bushel? _ Xf?) a é eh. ! head of wine, what 4 is the value of the remainder of a quired for bottling 3 barrels of cider? | 3sel., will be required to bottle the remainder ? 113. What is the value of % of abushel of wheat, i) : Ks eae, | | x. FRACTIONS. 49 ie 114. If 1 hogshead [63 gal.| of molasses cost $262, what is the cost of 1 gallon ? / 115. What is the cost of 7hhd. 63 gal. molasses, at 1, cents per gallon? _ 116. What is the cost of 25 yards 34 2 quarters of rib- bon, at 194 cents per yard? mys If 54 + cords of wood cost $265 , what is the cost lof 1 cord? 118. What is the value of 162 tons of hay, at 114 dollars per ton? 119. What is the value of 1lb. 60z. 12 dwt. of silver, te 20: cents per pennyweight ? | 120. If 167 yards of broad-cloth cost $86.24, what is the cost of 1 yard? » 121. At 5s. 34d. per yard, what is the cost of 783 yards of cambric, in pounds, shillings, and pence? | 122. If 49232 yards oftcloth cost £68 4s. 10d., what jis the cost of 1 yard? / 123. If 183-yards of cotton cost 12s. 9d., what is the ‘cost of 1 yard? 124. What is the value of 57683]b. of coffee at 102 ;pence per pound ? | 125. At what price per pound must I sell 4323 pounds of coffee, in order to receive £27 3s. for the w hole: P | 126. If £448,5 be equally divided among 76 ai what will each man receive ? 127. If ofa yard ~ eee eost $3, what is the price of 1 yard? on Go fe 198. If 743 barrels i stow cost $214, what is the ‘cost of 1 barrel of the apples ? 129. If 44 gallons of molasses cost $23, what is the ‘cost of 1 quart: P 130. If 14 hogshead of wine cost $2504 what is the icost of 1 quart : ° | -131. Bought 5 yards of sine at $24 per yard; 154 yards of ribbon, at 124 cents per yard; 17 pairs of gloves, at 68% cents per pair; ‘and 164 1 yards of lace, at $34 per yard. What is the whole cost ? | 132. Bought 64 pounds of tea, at 875 cents per pound; 153 pounds of su oe 4 113 cents per pound; 13% pounds = 50 ARITHMETIC. x of coffee at 125 cents per pound; and 163 gallons ol molasses, at 4 of a dollar per gallon. What is the whok cost ? 133. Bought 93 barrels of cider, at $24 per barrel 8 barrels of apples, at $12 per barrel; 16 boxes of raisins. at $2.624 per box; 234 pounds of almonds, at 142 cent: per pound. What is the whole cost ? _ 134. Bought 3584 bushels of wheat, at $ ofa dolla] per bushel; 420 bushels of rye, at 964 cents per bushel 1464 bushels of corn, at 2 of a dollar per bushel; ane} 6511 bushels of oats, at 232 cents per bushel. What is the whole cost ? 135. A purchased of B, 752 tons of iron at $9. 614 perton. What quantity of coffee, at 123 cents per pound, must A sell B, to cancel the price of the iron? 136. © purchased of D, 1397 hogsheads of molasses, at 153 cents per gallon; and D, at the same time, pur. chased: of C, 896) tons of iron, at $94 per ton. How much was the balance— and to raheniy was it due? 137. What is the sum of 4 13> qa %, 34,45, 42, 7 $9 50 99 479 tp eodigy of et | 138. Suppose +; of 3% of 74 to be a minuend, and of 2-of t of # a subtrahend; what is the remainder? 139. What is the product of 4 of 2 of ¢ of 100, multi plied by 4 of 2 of Z of 20f 752 - 140, What is the quotient of 2 of 2 of 32, divided by! + of 42 of 2 of 32 of 2? 141. Suppose the sum % two fractions to be 2, and’ one of the fractions to be 353; what is the other? (See| PROBLEM I, page 20.) 142. Suppose the greater of two fractions to be 44,) and their difference to be 4 43 ; what is the smaller fraction? (See Pros. 11, page 20.) ( 143. Suppose the smaller of two fractions to be 34, and their difference to be 3; what is the greater fraction | (See Pros. 111, page 21.) 144. What are the two fractions, whose sum is 42, ari whose difference.1 is f7? (See Pros. iv, page 21.) | 145. If .°5 be the product of two factors, one of which’ is 77, What is the other? (See Pros. v, page 21.) SS —— en haa = XI. DECIMALS. 51 » 146. Suppaseny % to be a dividend, and <5 a quotient; what is the divisor? (See Pros. vi, page 21.) 147. What must be that div iead: whose divisor is 22 and whose quotient is 3? (See Pros. vii, page 22. 3) , 148. Suppose the product of three factors to be 32, one of those factors being 2, and another 7; ; what is the tae third factor? (See Pons VIII, page 22. iP 149. A merchant owning 3% of a ‘ship, sold 2 of what he owned. What part of the whole ship did he sell ? } 150. A merchant owning 32 of a ship, sold $ of what he owned. What kee of the ship did he still own? | 151. If I buy = of 3 of a ship, and sell 2 of what I ‘bought, what part ‘of the ship shall I have left? | The kind of fractions, which have been treated in this article, are called Vulgar fractions, or Common frac- “tons, in distinction from another kind, called Decimal fractions, or simply Decimals. * dF DECIMAL FRACTIONS. A DECIMAL FRACTION is a fraction whose denominator jis 10, or 100, or 1000, &c. The denominator of a decimal fraction is never written: the numerator is written with a point prefixed to it, and the denominator is understood ito be a unit, with as many ciphers Rpu ered as the le ator has places of figures. Thus, .5 is 7%, .26 is #5, 907 is #95 1000 ° When a whole number and decimal fraction are written together, the decimal point is placed between them. Thus, 68.2 is 6875, 4.87 is 4fr. In the notation of whole es any figure, wherever 3 may stand, expresses a quantity 7/5 as great as it would express if it were written one place further to the left: and so it isin the notation of decimal fractions—the same system is continued below the place of units. The first place to the right of units is the place of tenths; the second, 52 ARITHMETIC. XI of hundredths; the third, of thousandths; the fourth, : ten-thousandths; and so on. Ciphers placed on the right hand of decimal figures, és not alter the value of the decimal; because, the figures still remain unchanged in their distariee from the unit’s! place. For imstance, .5,°.50, and .500 are all of equal value,— they are each equal to 5. But every cipher that is placed on the left of a decimal, renders its value ten times smaller, by removing the ficures one place further ats the unit’s place. Thus, if we prefix one cipher to 5, it becomes .05 [725]; if we prefix two ciphers, it becomes .005 [550]; and so on. ‘ To READ DECIMAL FRACTIONS—Enumerate and read the figures as they would be read if they were whole num= bers, and conclude by pronouncing the name of the lowest denominalton. ‘| 1. Read the several numbers in the following columns. | | .99 .2008 4.008 24.09 | .064 -00006 §.37002 630.1174 } 0003 ~=—s-«.03795~=——S~St«é«C 9999 6.972479 | 5237 .180009_-. 5.0001 28.797 2. Write in decimals the following mixed numbers. Sehey 21560 «© 8800 Sisco 24790 326 750 Siot00 © 47 ro0000 38 7000 T9000 «= 97 zdd0 6 x5000 CSaps00— sW9TaSGa CO too | OS tan 10000000 | ADDITION OF DECIMALS. 3. Add the following numbers into one sum. 1051. 7 + 70.602 -++ 4.06 + 807.2659. 151.7 — In arranging decimals for addition, | 70.602 | weplace tenths under tenths, hun- 4.06 dredths under hundredths, &c. We _ 807.2659 then begin with the lowest denomi-| "1033.697 9 nation, and proceed to add the col- say yumnns as in whole numbers. { i ; ) XI. DECIMALS. 53 4. What is the sum of 256.94-+-9121.7-+- 8.3065? 5. Add together .6517-+ 19.2-+ 2.8009 + 51.0007 + 00009 + 22.206 +-4.732. In Federal Money, the dollar is the unit; that is, dol- lars are whole numbers; dimes are tenths, cents are hundredths, and mills are thousandths. ‘2 6. Add together $18.25, $4.09, $2.40, $231.075, ,$64.207, $50.258, $10. 09 and ets. \ ae Write the following sums of money in the form of decimals, and add them together. $1 and Icent, 37 cents, $25 and 7 chimes, 65 cents, $15, 9 dimes, 8 nulls; 4 cents and 3 mills, -§ of a mill, 7 and 8 cents, 3° of a mill, 36; cents, 10 eagles and 25 dollars, and 7 cents. } SUBTRACTION OF DECIMALS. _ 8. Subtract 4.16482 from 19.375. 19.375 After placing tenths under tenths, 4.16482 &c., we subtract as in whole num- 15.21018 bers. ‘The blank places over the 2 ee + and (S are: viewed | 16. What part of .65 is .408° _. 17, What part of $2 is $7? (Ans. 3.) 18. What part of $2 is $7.49? | 19. What part of 90 cents is $1.35? 20. What part of $4.375 is $28? 7 ARITHMETIC XI 21. What part’of 5.8 is 31.42? 22. What part of .253 is .97? 23. What part of $ is 75? | The expression oF a4 sa oh Is, at first, a complex fraction, of which ; is the numerator, and 4 the ieatcal nator. The expression may be simplified by reducin these fractions to a common denominator, and taking th new numerators for the terms of the relation. See a to reduce a complex fraction to a simple one, page 43. 24. What part of 12 is sin 25. What part of 3 si 26. What part of 7 Is $ 27. What part of 6} is ter P 28. What part of 18 is 23 > 29. What part of 24 feet is 102 inches? 30. What part of 145 days is 23.1, 5 hours ? 31. What part of 24 “gallons i is 3 quarts 23 gills? @ 32. What part of 54 rods is 3 rods 22 ft.? 4 33. What part of 4 1s 3? : 34. What part of $is $? 35. What part of 3 7. ions 36. What part of $3 * 96: 37. What part of 1g is 3? | 38. What part of 3 ‘shillings i is 5s. 7d.? - i 39. What part of £1 14s. is £5 2s. 73d.? : 40. What part of 784 days is 125 days 178 hours? 41. What part of 22 tons is 4 tons 64 pounds? i 42. If 35 horses eat 12278 pounds of hay in a wee what will 17 horses eat, in the same time ? 4 The most obvious view of the solution of this question | is this —If 35 horses eat 12278 pounds, 1 horse will eat | zs Of 12278 pounds, which is 35033 pounds; and 17 }| horses will eat 17 times 35033 pounds, which is 623 pounds. A more concise view, however, may be taken, | as follows. 17 horses are 42 of 35 horses, and the will eat 32 of the 12278 pounds of hay. Therefore, we shall obtain slic answer by multiplying 12278 pounds by the fraction 3. 12278X 346230 Ans. | : eT. RELATIONS OF NUMBERS. ror 43. If a car run 552 miles upon a rail-road, in 24 hours, how far will it run in 13 hours ? 44. If a car run 3 miles [960 rods] in 38 minutes [480 seconds], in what time will it run 300 rods? 45. If a hogshead of wine [63 gallons] cost $98.50, _ what will 45 gallons cost, at the same rate? 46. If the annual expense of supporting a fort manned _ with 600 soldiers be $ 182571, what is the exneng of a fort manned with 424 soldiers ? 47. If I can buy 325 barrels of flour for $1425, pave _ many barrels can I buy for $521? 48. Ifa ferry boat cross the river 18 times in 5 ea in how many hours will it cross 4 times ? 49. If 9barrels of flour cost $ 32, what will 28 bl. cost? In this example, the relation in which 28 barrels stand to 9 barrels is expressed by an improper fraction; 28 barrels being *° of 9 barrels. Therefore the answer is obtained by multiplying $32 by 32; that is, by multiply- ing $32 by 28, and dividing the product by 9. 50. If it take 300 yards of cloth to make the uniform clothes for 52 soldiers, how many yards are required to clothe 784 soldiers ? 51. If 12 horses eat 20 bushels of oats in a week, how many bushels will 45 horses eat in the same time ? 52. Ifa post 5 feet high cast a shadow 3 feet, on level -ground, what is the height of a steeple, which, at the same time, casts a shadow 176 feet? 53. If $40 will pay for 144 yards of cloth, how many yards can be bought for $75 ? 54. If 95 bushels of corn cost $68.25, what will 320 bushels cost, at the same rate ? 55. Suppose a ship’s expenses in Liverpool to be £131 13s. 10d. for 22 days; what-would be her ex- penses in the same port for 35 days? 56. If 144 bushels of corn will grow upon 38 acres 1 rood 15 rods of land, how much land is necessary to produce 500 bushels ? 57. Bought 269 yards of cloth, at the rate of $ 100 for _ 30 yards. What did it amount to ? 7% ae » 18 ARITHMETIC. XII, } 58. Bought 24yd. Sqr. Ina. of cloth, at the rate of | $12.30 for 4yd. Igr. 2na. What did it amount to? 9) Since it is necessary, in this example, to consider 24 | yd. 3qr. Ina. as a fractional part of 4 yd. I qr. 2na., they) first step in the operation is, to reduce both quantities of | cloth to nails. , 59. If 13gal. 2 at. 1pt. of wine cost poe 16, what bi sk gal. 3 qt. 1 pt. cost, at the same rate? 4q If 26 barrels of flour cost £ 28 14s. 6d. how many oe will £35 10s. 4d. pay for? q 61. If 6gal. Qqt. 1 pt. of wine will fill 31 bottles, how) many bottles are required for 11 gal. 3 qt.? 4 62. If 144 gross of buttons cost £22 19s., how many gross can be bought for £12 5s. 54d.? 63. If 2hhd. 19 gal. 2qt. of wine cost £93 1s. a what will 25hhd. 36 gal. cost? 64. If 15 yards of cloth cost $39.45, how many rcs can be bought for $21? (See remark ciel example 12.) 65. At the rate of # 94 for 78 days’ work, in how many days can a labourer earn $'72.375 ? 66. At the rate of ¢ 240 for 9.5 acres of land, what is the value of 7.25 acres? 67. Atthe rate of $182.50 for 8 acres of land, wall is the value of 12.7 acres? | 68. At the rate of 75 cents for 92 of a bushel of corn, what is the, value of .648 of a bushel ? | 69. In how many minutes will a locomotive car rum 49.9 miles ; il pedine: it to run at the rate of 2.5 miles in 5.75 minutes ? 7 70. If 43.64 pounds of satis be worth $9.075, whale is the value of 108.9 pounds? oe} 71. If 14 dollars will pay for the carriage of a ton 75.6 e | miles, what distance can a. ton be carried for 16 dollars | 75 cents, at the same rate ? 72. If 4 of a yard of cloth cost $7, Be is the cost | of ; of a yard? (Recur to example 23.) 73. If a rail- road car run 260 miles in 12 hours, wha distance will it run in 102 hours? (See example 24.) } fe CHI. RELATIONS OF NUMBERS. 79 | 74. If a man earn ol. 15 in 2 of a day, how much can e earn in $ of a day? (See example 25.) m5. Suppose 3 of an acre of land to be worth 54 dol- us; what is > ot an acre worth ? “To solve this question, by the eee to which the cholar has been led, he will consider ¢ as a denominator ad 4 as the numerator of a complex fraction, expressing ‘hat part of 54 dollars 3 of an acre is worth; and, after bdacin this complex fraction to a simple one, will mul: ply the simple fraction into 54 dollars, for the answer. low the effect of the process is the same as that of mul- plying the 54 by 5, and dividing the product by 4; and us last method is to be preferred, because it is shorter. “hus, 54x96, and rina Deen 76. If of aship cost $15000, what does 4 L of tier cost? ay, Iii i of a lot of new piletnd be worth 300 ialiinss what 45 of the lot worth? 78. If a horse trot 1840 rods in 49 of an hour, how rany rods does he trot in 3% of an hour? 79. If 964 yards of cloth cost $642, what will 28 ards cost, at the same rate ° 80. If 154 yards of cloth cost $75, what will 142% ) ost, at the same rate ? 81. If 97 barrels of flour be consumed by a company 118 days, how long will 253 barrels last ? 82. Ifa mill grind 18;% bushels of corn in | hour and 2 minutes, in what time val it grind 253 bushels ? 83. Ifa ship sail 924 miles in 83 hours, j in how many ours does it sail 65,°, miles ° - 84. If a barrel of flour will support 12 men for 25 ays, how long will it support 8 men? Since the flour will support 12 men 24 days, it would apport 1 man 12 times 25 days, or 300 days; and since would support 1 man 300 days, it will support 8 men of 300 days, or 874 days. Thus, to obtain the answer, e multiply 25 days by 12, and divide the product by 8. . little attention to the conditions of. this question, and le process of the operation, will enable the learner to erceive, at once, that the answer is 1 of 25 days. 86 ARITHMETIC. XID) 85. If a quantity of beef will support 436 men 73 days how long will it support 240 men? 86. Ifa barrel of beer will last 10 men 16 days, hoy long will it last 23 men? The beer would last 1 aaa 10 times 16 daysye or 6 ee and it will last 23 men 35 of 60 days, or 244 day =—=2,/, days. The question” is, however, more conve niently viewed thus,—Since the beer will last 10 me 16 days, it will last 23 men? a3 Of 16 days; and, hence 16 is to be multiplied by 38 87. Suppose a certain quantity of hay will feed 8 sheep 71 days; how long will it feed 230 sheep ¢ 88. If 256 men can make a certain piece of road i 240 days, in what time will 190 men make it? 89. If 9 yards of silk, that is 3 cieeist wide, will lin a cloak, how many yards, that is 5 quarters wide, wi line the same cloak ? | 90. If 110 yards of paper, that is 52 inches wide, wi cover the walls of a room, how many yards; that is 2 ae wide, will cover the same walls ? Suppose a man can perform a piece of work | 5 ' 45 ay s, by working 7 hours a day, in what time will h_ perform it, if they work 10 hours a day ? 92. Suppose a company of men can perform a piec of work in 155 days, by working 12 hours a day, in whe time will they perform it, by working 5 hours a day? 93. How many days will it take 119 horses to eat th hay that 44 horses would eat in 60 days ? 94. The hind wheels of a coach, which are 180 inche in circumference, will turn round 4825 times in runnin acertain distance, how many times will the forward wheel turn round, they being 145 inches in circumference ? 95. If a ship, by sailing 9 miles an hour, will effect — passage to Europe in 55 days, in how many days woul she effect the passage by ples 13 miles an hour? 96. Ifa vessel, by sailing 10} miles an hour, will mak a passage from Bangor to New Orleans in 11 days, 1. how many days would she make the passage by sain 12! miles an hour ? 97. Suppose A rides 62 miles an hour, and perform XIII. RELATIONS OF NUMBERS. St a certain journey in 14°; days; in what time will B; who rides only 4,’; miles an hoor; perform the same journey? 98. If 6 persons expend $300 in 8 months, how much will serve 15 persons for 20 months? Since 6 persons expend $300 300 750 in 8 months, 15 persons would, 15 20 in the same time, expend 'P of a $300, which is 4750. Then, 6)4500 — 8)15000 since 15 persons would expend 750 1875 $750 in 8 months, they would, in 20 months, expend * of $750, which is $1875. The adjoined operation corresponds to this solution. | 99. If the wages of 6 men for 14 days be $84, what will be the wages of 9 men for 11 days? - 100. If 3 pounds of yarn make 9 yards: of cloth, 5 quarters wide, how many pounds would be required to make a piece of cloth 45 yd. long and 4qr. wide? - 101. If aclass of 25 girls perform 1750 examples in arithmetic, in 15 hours, how many examples of equal length may a class of 30 girls perform, in 18 hours ? | 102. If the use of $100 for 90 days, be worth $1.50, what is the use of $73 for 85 days worth? 103. If the use of $ 100 for 30 days be worth 75 cents, what is the use of $1240 for 57 days worth ? 104. If a man euvel 217 miles in 7 days, travelling 6 hours a day, how many miles will he travel in 9 days, if he travel 11 hours a day? | When he travels 6 hours a day, he divéiees 217 miles i 7 days, and were he to proceed thus for 9 days, he would advance 7 of 217 miles, or 279 miles. Since, by travelling 6 hours a day he would, in 9 days, advance 279 ml., ~ by travelling 11 hours a day , he would advance Ksof 279 ml., which is 5112 ml., or 5115 ml. 105. If a man perform a aounien of 1250 miles in 15 lays, by travelling 14 hours a day, how many days will it take him, to per phen a journey of 1000 miles, by travel- ling 13 hours a day? 106. If 10 cows eat 7} tons of hay ii in 14 weeks, how imany cows will eat 224 tons in 28 weeks ? | S2 ARI'FHMETIC. XID eB) 107. If 6 men will mow 35 acres of grass in 7 days by working 10 hours a day, how many men will be re quired to mow 48 acres in 5 days, when they work 1 hours a day? | 108. If 14 men can cut 87 cords of wood in 3 days | when the days are 14 hours long, how many men will ct 175 cords, when the days are 11 hours long ? 109. If 16 men can build 18 rods of wall in 12 days. how many men must be employed to build 72 rods of th’ same kind of wall in 8 days? | 110. If 25 persons consume 600 bushels of corn i 2 years, how much will 139 persons consume in 7 years” Since 25 persons consume 600 bushels in 2 years 139 persons would, in the same time, consume 4% 600 bushels, which is 3336 bushels. Then, since 13) persons would consume 3336 bushels in 2 years, the’ will, im 7 years, consume 4% of 3336 bushels, which 1) 11676 bushels. . 111. If 154 bushels of oats will serve 14 horses fet 14 days, how long will 406 bushels serve 7 horses? | 112. If 25 men can earn $6250 in 2 years, how lon) will it take 5 men to earn $11250? | 113. If 9 men can mow 36 acres of Soa ls in 4 days | how many acres will 19 men mow in 11 days ? i. 114. Ifa family of 9 persons spend $450 in 5 months” how much would be sufficient to maintain the family months, after 5 more persons were added ? | _ 115. Ifa stream of water running into a pond of 19) acres, will raise the pond 10 inches in 12 hours, hov much would a pond of 50 acres be raised by the sam stream, in 10 hours? 116. If the wages of 4 men, for 3 days, be $11.04 how many men may be hired 16 days for $103.04? 117. If 3 men receive £8 18s. for working 194 day: what must 20 men receive for working 1004 days: ; | 118. If 1112 bottles are sufficient to receive 5: casks of Mads. how many bottles are sufficient to receive 1¢ casks of wine? 119. If 725 bottles hold 4 barrels of wine, how many notiles are required to hold 3 tierces of wine ? a | } XM. RELATIONS OF NUMBERS. $3 120. If 240 men, in 5 days, of 11 hours each, can dig a trench 230 yards long, 3yards wide, and 2 yards deep, in how many days, of 9 hours each, will 24 men dig a trench 420 yards long, 5 yards wide, and 3 yards deep ? Since 248 men, in 5 days, of 11 hours each, can dig a trench 230 yards long, 3 yards wide, and 2 yards deep, 24 men, working in days of the same length, would dig a trench of the same dimensions in 248 of 5 days, which is §13$—514 days; and, working in days of 9, instead of 11 hours each, the trench would occupy them '¥} of 514 days, which is 633, days. Again, since the trench to be dug by 24 men is 420, instead of 2350 yards long, this length, (the width and depth remaining unchanged) would occupy them $39—33 of 633 days, which is 115%. days. Again, since the trench to be dug by 24 men is 5, instead of 3 yards wide, this width (the depth remain- ing unchanged) would occupy them 3 of 115%, days, which is 1922338 days. Lastly, since the trench to be dug by 24 men is 3, instead of 2 yards deep, it will occupy them 3 of 192434 days, which is 288355— 2883, days, the answer. 121. If12 men can build a brick wall 25 feet long, 7 feet high, and 4 feet thick, in 18 days, in how many days will 20 men build a brick wall 150 feet long, 8 feet high, and 5 feet thick ? 122. If 15 men can dig a trench 75 feet long, 8it. wide, and 6ft. deep, in 12 days, how many men mus? be employed to dig a trench 300ft. long, 12 ft. wide, and 9ft. deep, in 10 days? 123. If the carriage of 44 barrels of flour, 108 miles be worth $215, what is the carriage of 36 barrels, 162 miles worth ? . : 124. If 175 bushels of corn, when corn is worth 60 cents a bushel, be given for the carriage of 100 barrels of flour, 58 miles, how many bushels of corn, when corn is worth 75 cents a bushel, must be given for the carriage of 90 barrels of flour, 200 miles ? 125. If 12 ounces of wool make 25 yards of cloth, that is 6 quarters wide, how many pounds of wool would make 150 yards of cloth, 4 quarters wide ? 84 ARITHMETIC. XII MISCELLANEOUS EXAMPIES. 126. A owned 3 of a ship, which he sold for $ 365¢ and B owns -%; of her, which he wishes to sell at the sam rate. What must be B’s price? Since the price of 3% of the ship is $3650, the pric of the whole ship must be % of § 3650, which is $ 36906 and +5 of $36900 is $11070, which must be B’s price 127. If 3650 be 3 of some number, what is 33; of th same number ? y 128. A merchant has bought 4 of a company’s stock for $92000. What would be the price of <4; of the stock at the same rate ? | 129. A merchant owning 3/5 of a ship, sold 2; of whe he owned for $1841. What is the value of the whol ship, according to this sale ? | 130. 1841 is ~% of 4% of what number ? 131. After a certain tract of land had been equall divided among 16 owners, one of them sold 2 of his shar at $5, an acre, and received $444. How much land wa’ there in the whole tract ? 132. If 4% of a yard of cloth be worth 2 of a dollar what is the value of -2; of a yard? | a Since 7% of a yard is worth = of a dollar, a yardi worth 1 of 2 of a dollar, which is 22 of a dollar; ant ¢z of a yard is worth {3 of £2 of a dollar, which is 7 of a dollar, or 61328 — ¢ 143 — 1.5594. | _ 183. If § of a yard of lace be worth 12 of a dollar what is +2, of a yard worth? | 134. If 4 of a barrel of flour cost 4 dollars, what 1: the cost of 63 barrels, at the same rate? —- | 135. If 13% bushels of corn cost 7 dollars, what i: . the price of 93 bushels, at the same rate ? ¢ _ 136. If 422 pounds of indigo be worth $87.625, wha’ is the value of 1923 pounds ? ¥ , 137. A garrison of 900 men have provision for 4 months. How many men must leave the garrison, that the provision may last the remainder 9 months ? : 138. Ifa loaf of bread weighing 32 ounces be sold for eight cents, when flour is worth $6.50 per. barrel, what Y ; XIII. MISCELLANEOUS EXAMPLES. 85 ought the eight-cent loaf to weigh, when flour is worth only $5 a barrel ? ' 1389. Acompany of 75 soldiers are to be clothed; each _ suit is to contain 33 yards of cloth, 6 quarters wide, and to be lined with flannel 3 of a yard wide. How many _ yards of flannel will be required ? _ 140. If a garrison of 1500 men consume 750 barrels of flour in 9 months, how many barrels. will 2150 men consume in 15 months? _ 141. How many tiles 8 inches square, will cover a hearth 16 feet long, and 12 feet wide? » 142. If the expense of carrying 17cwt. 3qr. 14lb. 85 miles be $23.84, what will be the expense of carrying S3cwt. 2qr. 150 miles, at the same rate? 143. Two men bought a barrel of flour; one paid 3} dollars, and the other paid 37 dollars. What part of the flour should each of them have ? _ 144. If the corn contained in 8 bags, holding 2 bushels ‘3 pecks each, be worth $14.25, what is the value of the corn contained in 7 bags, each holding 2bu. 3pk. 7 qt? 145. A ship of war sailed with 650 men, and provision for a cruise of 15 months. At the end of 3 monihs she captured an enemy’s vessel, and put 75 men on board of her. Five months after, she captured and sunk another vessel, and took on board ihe crew, consistmg of 350 men. How long did the provision last, from the com- mencement of the cruise ? 146. A built 156 rods of wall in a certain time, and B in the same time built 13 rods to every 12 that A built. ‘They were paid $1.25 per rod. How much did B re- -Ceive more than A? _ 147. A father bequeathed $6000 as follows; viz. 2 to his wife, 4 to his.son, + to his daughter, and the remain- der to his servant. How much did each receive? 148. If 42 of a pound of sugar be worth 2 of a shilling, what is the value of ¢ of a cwt.? 149. If 754% gallons of waaay one hour, run into a cistern, which will hold 64 hogsheads, and by a pipe 244 gallons an hour run out, in how many hours, minutes and seconds will the cistern be flled ? 8 \’ bel ® ? “ 86 ARITHMETIC. XIV.) XIV. PERCENTAGE. | Under this head may be classed, those computations! which investigate the value of a given number of hun-) dredths of any quantity. The number of hundredths to be taken or considered in any number, is called the per cent. The term, per cent., is an abbreviation of pet) eentum, which signifies by the hundred. Any per cent. is conveniently expressed bya decimal Thus, 1 per cent. of any number is .01 of that aeiitiks | 8 per cent. is .08; 25 per cent. is .25; &c. 1. A merchant, who has 426 dollars deposited in tH bank, wishes to draw out 5 per cent. of his deposite.: How many dollars must he draw ? Since 5 per cent. of any quantity is zjy of 1426 that quantity, the ee to be solved in this 05 example is— What is +2, of 1426dollars? Or, $71.30! decimally— What is .05 of 1426 dollars? The answer is conveniently found by multiplying 1426 by .05._ The whole number in the product expresses dollars, ane the decimal expresses cents. 2. A trader, who went to the city with 321 dollars, to purchase goods, laid out 9 per cent. of his money for. coffee. _ How many dollars did he pay for coffee ? | 3. What is 1 per cent. of 100 dollars ?* | . What is 1 per cent. of 834 dollars ? OO NED ov What is 3 per cent. What is 3 per cent. What is 6 per cent. What is 6 per cent. What is 7 per cent. of 100 dollars ?. of 42 dollars ? of 100 dollars ? of 99 dollars ¢ of 100 dollars ? 10. What is 7 per cent. of 1000 dollars ? 11. What is 8 per cent. of 26 dollars ? 12. What is 9 per cent. of 354 dollars ? 13. What is 10 per cent. of 2244 dollars ? _ XIV. PERCENTAGE. 87 14. What is 16 per cent. of 13 dollars ? 15. What is 37 per cent. of 211 dollars? 16. What is 99 per cent. of 100 dollars ¢ 17. What is 100 per cent. of 48 dollars ° 18. A trader laid out 1214 dollars as follows. He paid 24 per cent. of the money for broadcloths; 38 per cent. for linens; 8 per cent. for calicoes; and the remain- der for cottons. How many dollars did he pay for each _kind of goods? When the rate per cent. is a vulgar fraction, or a mix- ed number, the fraction may be changed to a decimal. Observe, that, 1 per cent. when expressed decimally, is .01; therefore a fraction of 1 per cent. when reduced to a decimal, becomes so many tenths, hundredths, &c. of a pundredth. For example, as 4 of 1 unit is .25 of a unit, so of 1-hundredth is .25 of a “hundredth, and is denoted thus, .0025. 19. What is 34 per cent. of 243 dollars? 3 per cent. —=.03 243 5 per cent. —.005 -035 .035 1215 729 8.505 Ans. $8.504 20. What is 44 per cent. of 2746 dollars ? 21. What is 74 per cent. of 41 dollars ? 22. What is 123 per cent. of 358 dollars ° 23. What is 2 per cent. of 100 dollars ° 24, What is 2 per cent. of 61 dollars? 25. What is 1 per cent. of 9487 dollars ? 26. If 84 per cent. be taken noe 36 dollars, how many dollars will there be remaining ? 27. A merchant who had 400 barrels of flour, shipped 421 per cent. of it, and sold the remainder. How many barrels did he sell ? 28. A trader bought 800 pounds of coffee; and, in getting it to his store, 24 per cent. of it was wasted. How many pounds did he lose?» What did the remainder amount to, at 13 cents a pound? 88 ARITHMETIC. Bs XIV) 29. Two men had 120 dollars each. One of them paid out 14 per cent. of his money, and the other 17%) per cent. How many dollars did one pay more than the | other ? | 30. Find 74 per cent. of $344. | When there is a fraction in the 344+ 100—3. a4 } rate per cent. which cannot be 74 exactly expressed by a decimal— z : 2408 | as in this example— we first find 1142) 1 per cent. of the given sum, by 5 dividing it by 100; that is, by cut- $25.22) ting off two decimal figures, and then multiply this quo- tient by the mixed number expressing the ie per cent. 31. What is 44 per cent. of 624 dollars ? | 32. What is 62 per cent. of 38 dollars? d 33. What is 34 per cent. of 2310 dollars? i 34. What is 94 per cent. of 17 dollars? 35. What is 83 per cent. of 152 dollars? ) 36. Find the difference between 52 per cent. of 41 | dollars, and 44 per cent. of 39 dollars. 37. What is 7 per cent. of $24.32? Here we have cents { decimals] in the num- 24.32 ber on which the percentage is to be taken. 07 | We however multiply as usual in decimal ae multiplication; ; and the first two decimal fig- 1.7024 ures in the product express cents, the third figure ex- presses mills, and the fourth expresses tenths of a mill. 38. What is 14 per cent. of $641.94 ? 39. What is 44 per cent. of $37.26? 40. What is 1i2 percent. of $150.75 ? 41. What is 124 per cent. of $25.32? 42. If a horse and gig cost 400 dollars, and the gig cost 32 per cent. of the sum, what did the horse cost? — | 43. Find the difference between 131 per cent. of . $18.09, and 7 per cent. of $41. 44. Find the difference between 9 per cent. of $ 16, | and 84 per cent. of $17.30. oo) — —— SS exIV. PERCENTAGE. 89 45. A young man, who had 94 dollars deposited in the Savings Bank, drew out 25 dollars. What per cent. of his deposite did he draw out ? We perceive, that the sum 94)95.0(.96 54 —96 24 he drew out, was 23 of the sum inte Salih ik} he had deposited: and, since PaaS the rate per cent. of any sum ree is a certain number of hun- Sree paid ? dredths of that sum, the ques- 56 tion to be solved is— How many hundredths is 25- ninety-fourths >— To solve this question, we change 25 Ss, q ) £ 4 to a decimal; restricting the decimal to hundredths; that is, carrying the quotient no further than two places. Any remainder which might allow the quotient to be carried further, may, in cases like this, be expressed in a vulgar fraction. Ans. 2644 per cent. 46. A man, who was owing a debt of 240 dollars, has paid 32 dollars of it. What per cent. of the debt has he 47. A merchant gave his note for 235 dollars, and soon after paid 110 dollars of the sum. What per cent. did he pay; and what per cent. still remamed due ? 48. If the cloth for a coat cost 12 dollars, and the making 7 dollars, what per cent. of the whole expense is the making ° 49. What per cent. of 100 dollars is 6 dollars ? 50. What per cent. of $28.50 is $1.10? 51. What per cent. of $94.12 is $4.42? 52. What per cent. of $57.08 is 32 cents ° 53. What per cent. of $10.10 is 7 cents ? 54. What per cent. of $48.11 is 99 cents ? 55. What per cent. of $75 is $4.18 ? To find the value of a rate per cent. on any sum of ‘English money,— First, change the lower denominations of money in the sum, to a decimal of the highest denomi- nation; and then proceed to multiply by the rate, as if the sum were dollars and cents. The whole number in the product will be of the same denomination of money . with the whole number in the multiplicand; and the 8 * ‘ om ' 90 ARITHMETIC. XIval decimal in the product must be changed to the lower denominations. 4 56. An English gentleman took passage from Liver- pool to Boston, in the ship Dover, having £672 12s. 4d. He paid 5 per ane of his funds for his passage. How much did he pay ? a 12) 4. .6308 20|12.333-- 20 672.616 _ 12.6160 ; 05 12 33.63080 7.3920 4 ~ 3.680Ans. £33 19s. 7d. qr. + | 57. What is 8 per cent. of £47 18s. 7d.? 58. What is 3 percent. of £9 14s. 3qr.? 59. What is 16 per cent. of £22 16s.? 60. What is 25 per cent, of 19s. 8d. 2qr.? 61. What is 6 per cent. of £2584? 62. What is 50 per cent. of 18s. 10d. 2qr.? 63. What is 44 per cent. of £214 15s. 10d.? COMMISSION. 7 ComMISssION is the compensation made to factors and brokers for their services in buying or selling. It is reckoned at so much per cent. on the money employed in the transaction. 64, What is the commission on £500 at 2} per cent.? 65. Suppose I allow my correspondent a commission of 2 per cent., what is his demand on the disbursement — of £369? 66. If I allow my factor a commission of 3 per cent. for disbursing £748 11s. 8d. on my account, what does his commission amount to ? 67. How much does a broker receive on a sale of «stocks amounting to 52648 dollars, allowing his commis- sion to be 4 of 1 per cent.? XIV. STOCKS. 91 68. What is the amount of commission on 395 dollars 75 cents, at 33 per cent.? 69. A commission merchant sold goods to the amount of 6910 dollars and 80 cents, upon which he charged a commission of 24 per cent. How much money had he to pay over to his employer ? 70. Sold 94 tons, 17cwt. 3qr. of iron, at 96 dollars a ton, at a commission of 24 per cent. on the sale. What did my Bominianott amount to? How much had I to pay over? STOCKS. STOCK is a property, consisting in shares of some es- tablishment, designed to yield an income. It includes government securites, shares in incorporated banks, in- surance offices, factories, canals, rail-roads, &c. The nominal value, or par value of a share, is what it originally cost; and the real value, at any time, is the sum for which it will sell. When it will sell for more than it originally cost, it is said to be above par, and the excess is stated at so much per cent. advance. When its real value is less than the original cost, it is said to be below par, and is sold at a discount. 71. Sold 10 shares in the Commonwealth Insurance Company, at 5 per cent. advance, the par value of ashare being 100 dollars. How much did I receive ? 72. Bought 15 sharesin the Boston Bank, at 3 of 1 per cent. advance, the par value being 50 dollars a share. How much did I give for them? 73. Sold 64 shares in the State Bank, at 14 per cent. advance, the par value being 60 dollars a share. How much did I receive for them 74. Sold 9000 dollars United States 5 per cent. stock, t an advance of 74 per cent. What was the amount of the sale ? _ 75. Sold 18 shares in an insurance office, at 12 per ent. discount, the par value being 100 dollars a share. . How much did they come to? | ) 76 Bought 16 shares in the Massachusetts Bank, at 1, per cent. advance, the par value bemg 250 dollars. i share. What was the amount of the purchase? | it 92 ARITHMETIC. “ 77. Bought 54 shares in the New York City Bank, a. 74 per cent. advance, the par value being 100 dollars ; / snare. How much did they cost me? | 78. I directed a broker to purchase 25 shares of rail road stock, at a discount of 13 per cent. the par valu being 100 per share. Allowing the broker’s commissioi to be ~ per cent., what will the whole cost me? 79. What will 16 shares in the Philadelphia Bank cost the par value being $100 per share, the price being 3; per cent. above par, and the broker charging a commis sion of $ per cent.? INSURANCE. INSURANCE is security given, to restore the value o/ ships, houses, goods, &c., which may be lost by the perils of the sea, or by fire, &c. The security is | in consideration of a premium paid by the owner of the property insured. The premium is always a certain per cent. on the value of the property insured, and is paid at the time the msurj ance is effected. | The written instrument, which is the evidence of the contract of indemnity, is called a policy, 80. What jis the amount of premium for insuring) 19416 dollars at 25 per cent.? 81. I effected an insurance of 3460 dollars on my dwelling house for one year at 2 of 1 per cent. What did the | premium amount to ? | 82. If you obtain an insurance on your stock of zodid valued at 7325 dollars, at} of 1 per cent. what will the premium amount to ? it 83. If you should take out a policy of 3168 dollars, on your store and goods, at a premium of 41 cents on a hundred dollars, what would be the amount of premium / 84. An insurance of 18000 dollars was effected on | | | 1 xv. INTEREST. 93 Ibe ship Sturdy, on her last voyage from Boston to Cal- ‘utta, at a premium of 3 per cent. out and home. What ied the premium amount to ? } 85. An insurance of 3500 dollars on stock in a cotton actory was effected at 31 per cent. for one bab What was the amount of premium? ' 86. A gentleman procured an insurance for one year mm his house valued in the policy at 8756 dollars, and on is furniture valued at 2139 dollars, at a premium of 39 fats on a hundred dollars. How much did the pre- nium amount to? | XV. INTEREST. _INTEREsT is a premium paid for the use of money. ' Tt is computed by percentage; a certain per cent. on he money being paid for its use, for a stated time. ) The money on which interest is paid, is called the rincipal. The per cent. paid, is called the Rate. The incipal and interest added together, are called the Imount. When a rate per cent. is stated without the mention of ny term of time, the time is understood to be 1 year. The rate of interest is regulated by state laws, and is ot uniform in all the states. We shall, however, first reat of 6 per cent. per annum, as this is the rate most ommonly paid. . | As interest is always expressed by some rate per cent., he most convenient way of computing it is, to find the ‘eeimal expression of the rate for the time, and multiply fe principal by this decimal: the product is the interest. hus, if the rate for 1 year be 6 per cent. or .06, for 2 fears itis 12 per cent. or .12, for 3 years it 1s 18 per ent. or .18, and soon. The interest of 24 dollars for } years, at 6 per cent. a year, is found thus, 24.18 = $4.32, the interest sought. | | 94 ; - ARITHMETIC. x Rate PER CENT. FoR Montus. The decimal ey pression of the rate for months, when the rate is 6 pt cent. a year, is easily obtained; for, if the rate fore 1 months be 6 per cents or .06, for 1 month it is 7 of per cent. which is 4 per cent. or .005; for 2 months is 1 per cent. or .01; for 3 months it is 15 per cent. ¢ .015; for 4 months it is 2 per cent. or 025 for 5 mont it is Qi per cent. or .025; for a year and L month it | 64 per cent. or .065; for a year and 2 months it is 7 pe oan or .07; for a year and 11 months it is 114 per cen’ “115. i ve If the rate of interest be 6 per cent. for a year what is the rate for 1 month?...... for 6 months ?..... for 7 months ?....... for 8 months ?...... for 9 months? © 2. At 6 per cent. a year, what is the rate for a yea and 1 month?...... a year and 3 months?...... a yea and 4 months ?...... a year and 10 months ? "i RATE PER CENT. FoR Days. Observe, that the rat for 2 specie which is 60 days, 1 is 1 per pained or .OL and for 75 of 60 days, which is 6 days, it is 7’ of 01 which is .001. Now since the rate for 6 days is 1-thou sandth, the rate for any number of days is as many thou sandths as there are times 6 days. Therefore, to fin the rate for days, at 6 per cent. per annum, adopt thi following RULE. Denote the days as so many thou -sandths, and dwide the expression by 6: the quotien ay be the rate. | If the rate of interest be 6 Pe cent. for a year, wha is eh rate for 1 day?...... for 2 days?...... for 3 days , note for 4 days?...... for 5 days?...... for 6 days? ...1) for 7 days?...... for 9 days?...... for 24 days?...... foi 26 days? 4, At 6 per cent. a year, what is the rate for 2 months and 12 days?...... 3 months and 10 days?...... for 6 months and 18 days?...... for 10 months and 29 days? 5. What is the interest, and what the amount of 546 dollars 72 cents, for 4 years 7 months 19 days, at 6 Ber cent. a year? » “4 ,} “XY. INTEREST. 95 4 To find the rate for 4 years, we multi- 546.72 ply the rate for 1 year by 4; thus, .06 X 27816 -4=.24. To find the rate Fer 7 months, 398032 we multiply the rate for 1 month by 7; 54672 thus, .005 X 7=.035. To find the rate 437376 jor 19 days, we denote 19 as eh alt dk 389704 and divide the expression by 6; thus, 499344 | 019+6—.00316-+. Now the sum ———-— of these rates, .24-++-.035-+ .00316—= 1 : 27816, is the rate for the whole time; “*"'‘“ and by this sum we multiply the principal. 698.7956352 ‘The interest found, is $152.07,5+3; 7° “which, added to the principal, gives the amount, $698 —79,5-+-. The rate for 19 days is not exact, as the deci- nal does not terminate; it is, however, sufficiently near exactness. 6. What is the interest of 148 dollars 92 cents, for 3 years, at 6 per cent. per annum ? 7. What is the interest of 57 dollars 10 cents, for 5 years, at 6 per cent. a year? -. 8. What is the interest of 93 dollars 50 cents, for 4 rears, at 6 per cent. a year? 9. What is the interest of 608 dollars 62 cents, for a year and 9 months, at 6 per cent. a year? | 10. What will 713 dollars 33 cents amount to, in 2 rears and 10 months, at 6 per cent. per annum ? 11. What will 1256 dollars 81 cents amount to, in 8 nonths, at the rate of 6 per cent. a year? 12. What is the interest of 100 dollars, for 1 year 1i nonths and 24 days, at 6 per cent. a year? 13. To what sum will 37 dollars 50 cents amount, in | year 7 months and 21 days, at 6 per cent. per annum ? 14. What is the interest of 314 dollars 36 cents, for 1 year 1 month and 6 days, at 6 per cent. a year? 15. What is the interest of 37 dollars a cents, for 11 months and 15 days, at 6 per cent. a year f 16. What is the interest of 512 dollars 38 cents, for 7 months and 10 days, at 6 per cent. a year? 17. To what sum will 691 dollars 28 cents amount, in 1 year and 1 month, at 6 per cent. a year? 96 ARITHMETIC, J 18. What is the amount of 194 dollars 69 denies for year 5 months and 6 days, at 6 per cent. a year? 19. What will 32 dollars 47 cents amount to, in| months and 25 days, at 6 per cent. a year? | 20. What is the interest of 217 dollars 19 cents, fa year and 17 days, at 6 per cent. a year? 21. What is the amount of 143 dollars 37 cents, fie year 9 months and 4 days, at 6 per cent. per annum ?- 22. To what sum will 203 dollars 9 cents amount,” 2 years and 18 days, at 6 per cent. per annum ? 3 23. To what sum will 18 dollars 63 cents amount, 1 year 10 months and 19 days, at 6 per cent. ayear 2” 24. What is the interest of 600 dollars, for 7 mont! and 22 days, at 6 per cent. a year ? 25. What is the interest of 817 dollars 44 cents, f 11 months and 12 days, . at 6 per cent. a year? i 26. What is the interest of 155 dollars, for 1 year sa and 10 days, at 6 per cent. a year? To what sum will 109 dollars 12 cents amount, | 5 cab and § days, at 6 per cent. a year ? ‘a 28. What is the amount of 25 dollars 92 cents, | _ year 4 months and 7 days, at 6 per cent. a year? 29. To what sum will 65 dollars 48 cents amount, 1 year 1 month and 18 days, at 6 per cent. a year? — 30. What is the interest of 110 dollars 25 cents, fi 10 months and 4 days, at 6 per cent. a year? » 31. What is the mterest of 2814 dollars jn cents, fi 6 months and 3 days, at 6 per cent. a year? é 32. What is the amount of 84 dollars 33 cents, for months and 26 days, at 6 per cent. per annum? 33. What is the interest of 345 dollars 68 cents, fi 7 months and 13 days, at 6 per cent. a year? : 34. To what sum will 13 dollars 98 cents amount, , 2 years 4 months and 7 days, at 6 per cent. a year? — 35. What is the interest of 802 dollars 27 cents, for month and 5 days, at the rate of 6 per cent. a year! — 36. What is the interest of 1309 dollars, ast 2 mont] and 3 days, at the rate of 6 per cent. a year? F 37. To what sum will 23 dollars 8 cents amount, in! years 6 months and 22 days, at 6 per cent. a year? a 7 i XV. INTEREST. 97 _* 38. What is the interest of 2538 dollars 17 cents, for _$months and 28 days, at the rate of 6 per cent. a year? ' 39. What is the amount of 1800 dollars 34 cents, for 1 year and 2 days, at 6 per cent. a year? 40. What is the interest of 199 dollars 15 cents, for 1 | year and 23 days, at 6 per cent. a year? _ 41. To what sum will 49 dollars 5 cents amount, in | year 2 months and 3 days, at 6 per cent. a year? _ 42. What is the interest of 201 dollars 50 cents, for 7 | years, at 6 per cent? | 43. What is the interest of 3010 dollars 75 cents, for 3 months and 1 day, at the rate of 6 per cent. a year? ' 44. To what sum will 41 dollars 6 cents amount, in 1 _ year 5 months and 14 days, at 6 per cent. a year? '* 45. What is the amount of 50 dollars and 11 cents, for 1 year and 21 days, at 6 per cent. a year ? 46. What is the interest of 1100 dollars for a year and 15 days, at 6 per cent. a year? 47. What is the interest of 9 dollars 89 cents, for 1 year and 27 days, at 6 per cent. a year? 48. What is the interest of 80 dollars, for 1 year 5 __months and 12 days, at 6 per cent a year? _ 49. What is the*interest of 90 dollars, for 1 year 2 _ months and 6 days, at 6 per cent. a year? 50. Towhat sum will 55 dollars amount, in 3 years and 9 days, at 6 per cent. a year? 51. What is the amount of 4119 dollars 20 cents, for 1 year and 5 days, at 6 per cent. a year? | . To compute interest by pays, when the rate 1s 6 per cent. per annum. RULE. Multiply the principal by the number of days, and divide the product by6. The quo- tient is the interest in mills, when the principal consists of dollars only; but when there are cents in the prinet- pal, cut off two figures from the right of the quotient, and . the remaining figures will express the mills. This rule—like the rule for finding the per cent. for days— is based upon the supposition of 360 days to the year; and, since the year contains 365 days, the rule gives 7}; part more than a true six per cent. terest. 9 yg ARITHMETIC. XWe | 52. What is the interest of 86 dollars, for 20 days, 4 6 per cent. a year? 53. What is the amount of 108 dollars, for 25 days, at 6 per cent. a year? 54. What is the interest of 204 dollars, for 40 days, at 6 per cent. a year? 55. What is the interest of 1000 dollars, for 29 dayal at 6 per cent. a year? | 56. What is the amount of 98 dollars 60 cents, for 35) days, at 6 per cent. a year? 57. What is the interest of 250 dollars, for 18 aayall at 6 per cent. a year? 58. What is the interest of 61 dollars 25 cents, for 28) days, at 6 per cent. a year? 59. What is the amount of 215 dollars 78 cents, for 50° days, at 6 per cent. a year? | 60. What is ae interest of 71 dollars, for 41 days, at 6 per cent. a year? 61. What is the interest of 3333 dollars, for 10 days, | at 6 per cent. a year? 62. What is the amount of 37 dollars 58 cents, for 16 days, at 6 per ‘cent. a year? 63. What is the interest of 91 doHars 80 cents, for 57. | days, at 6 per cent. a year? 64. What is the interest of 4109 dollars, for 18 dayss | { at 6 per cent. ayear? 65. What is the amount of 5214 dollars, for 50 days, at 6 per cent. a year? | 66. What is the difference between the interest of | $1000 for 1 year, computed by the year, and the interest | on the same sum for the same time, computed by days; both at 6 per cent.? | It will be observed, that, in all the preceding examples, | the rate of interest has been 6 per cent. per annum. . The method of computing interest at any other rate per cent. is the same, and equally simple, when the time con- sists of years only; but when there are months and days inv the time, and the rate per cent. perannum is other than 6, - it will frequently be convenient to find the interest for a i ov. INTEREST. 99 year first; and then for the months, to take the aliquot parts of a year; and for the days, the aliquot parts of a month; as in the following examples. 67. What is the interest of 934 dollars 34 cents, for 3 years and 5 months, at 7 per cent. per annum? 934.34 .07 months is } of a Y. 3) 65.4038 interest for 1 year 3 196.2114 interest for 3 years. “Lmonth is fof 4 ms. 4) 21.8012 interest for 4ms. 5.4503 interest for 1m. $§ 223.4629 interest for 3Y.5ms. 68. What is the interest of 371 dollars 42 cents, for I year 9 months and 19 days, at 75 per cent. a year ? 371.52 .O75 185760 «. 260064 6 months is 4 of a year. 2) 27.86400 for 1 year. 3 months is} of 6ms. 2) 13.93200 for 6 months. 15 days is ¢ of 3ms. 6) 6.96600 for 3 months. 3 days is tof 15 days. 5) 1.16100 for 15 days. IdayisZof3days 3) .23220 for 3 days. .07740 for 1 day. $50.53268 for the whole time 69. What is the interest of 412 dollars 17 cents, for 1 year 7 months and 10 days, at 7 per cent. a year? 70. What is the interest of 15748 dollars, for a year, at 44 per cent.? 71. What is the interest of 125 dollars 50 cents, for 2 years at 7 per cent. a year? , 72. What is the interest of 969 dollars, for 4 years, at 8 per cent. a year? 73. What is the interest of 655 dollars 30 cents, for a year, at 7 per cent.? | 100 aR ee XV. f 74. Whats is the interest of 404 cli 39 cents, for a ; ge at 54 per cent.? | aa i) ‘what sum will 1060 dollars 90 cents amount, ia a a at 7 per cent.? 76. What is the interest of 1650 dollars, for a eu at 30 per cent. | 77. What will 1428 dollars amount to, in a year and 5 | months, at 5 per cent. a year? 78. What is the interest of 2194 dollars 50 cents, for | a year and 10 months, at 7 per cent. a year? 79. What is the interest of 20750 dollars 42 cents, | for 1 year 2 months and 20 days, at 44 percent. a year f 80. What is the interest of 1109 dollars 44 cents, for | 11 months, at 55 per cent. a year? 81. What is the interest of 717 dollars 19 cents, foe 5 months and 6 days, at the rate of 7 per cent. a year? 82. What is the interest of 2119 dollars 78 cents, | for 3 months and 24 days, at 44 per cent. a year? | 83. To what sum will 107 dollars 29 cents amount, in | 7 months and 5 days, at the rate of 7 per cent. a year? | 84. To what sum will 5128 dollars 60 cents amount, | in 3 months and 26 days, at 545 per cent. a year? } 85. What is the mterest of 8244 dollars, for 1 month '. and 20 days, at the rate of 8 per cent. per annum? | 86. What is the interest of 1062 dollars 80 cents, for 2 months, at the rate of 9 per cent. per annum : P 87. What is the interest of 4008 dollars 90 cents, for 9 months, at the rate of 75 percent. a year? i 88. What is the interest of 12416 dollars 25 cents, for | as pare at the rate of 4 per cent. a year? . To what sum will 103 dollars 70 cents amount, | in : year. 2 months and 13 days, at 7 per cent. a year? 90. ‘Lo what sum will 86 dollars 21 cents amount, in | 1 year 1 month and 27 days, at 7 per cent. a year? 91. What is the interest of 502 dollars 9 cents, for 1_ year 3 months and 7 days, at 7 per cent. a year? 92. What is the interest of 319 dollars 27 cents, for 2 years 7 months and 11 days, at 7 per cent. a year? 93. What is the amount of 753 dollars 50 cents, for } year 9 months and 21 days, at 30 per cent. a year? | @ “ei XV. INTEREST. 101 94. To what sum will 207 dollars 8 cents amount, ir 1 year 4 months and 5 days, at 7 per cent. a year? ~95. What is the interest of 99 dollars 10 cents, for 2 years 1 month and 23 days, at 7 per cent. per annum ? To calculate interest on English money, first reduce the shillmgs, pence and farthings, to the decimal of a _ pound; the operation will then be as simple as the opera- tion on Federal money. 96. What is the interest of £17 10s. 6d. for 2 years _ 6 months, at 4 per cent. a year? £ : £ Sa £ £ 17 10 617.525. Then, 17.525 *.10=1.7525. . £ SS ds. ari i 1.7625=1 15 0 2345 Ans. 97. What is the interest of £42 18s. 9d., for 1 year 7 months and 15 days, at 5 per cent. per annum? _ 98. What is the interest of £23 8s. 9d., for 6 years, at 7 per cent. a year? 99. To what sum will £140 12s. 34d. amount, in 1 year 4 months and 12 days, at 6 per cent. a year ? 100. To what sum will £463 19s. 6d. amount, in 2 years and 8 months, at 6 per cent. per annum ? 101. What is the interest of £104 16s. 105d., for 11 months and 27 days, at the rate of 7 per cent. a year ? 102. What is the interest of £90 5s. 3d., for 1 year 1 month and 9 days, at 7 per cent. per annum ° 103. What is the interest of £512 7s. 4d., for I year 2 months and 21 days, at 5 per cent. a year? 104. To what sum will £210 10s. 6d. amount, in 1 year 3 months and 18 days, at 7 per cent. a year? 105. What is the interest of £2148 13s. 3d., for 5 months and 17 days, at the rate of 54 per cent. a year? 106. What is the interest of £750 4s. 6d., for 2 years 3 months and 20 days, at 7 per cent. ayear ? 107. To what sum will £70 10s. amount, in 3 years 2 months and 10 days, at 74 per cent. a year? 108. What is the interest of £803 5s. 7d., for 10 months and 14 days, at the rate of 5 per cent. per Q* annum ? 102 ARITHMETIC. XV. 109. To what sum will £13 13s. 6d. amount, in L year 11 months and 19 days, at 5 per cent. per annum. PARTIAL PAYMENTS. Tn computing interest on notes, bonds, é&c. whereon partial payments have been made, it is customary, when settlement is made in a year, or in less than a year from. the commencement of interest, to find the amount of the whole principal to the time of settlement, and also the | amount of each payment, and deduct the amount of all the payments from the amount of the: principal. | The learner may compute the interest on the follow notes; considering the rate to be 6 per cent. per annum, | when no other rate is stated. (PLOT Mae Boston, January 14th. 1833. 7 For value received, I promise Samuel Burbank Jr. to | pay him or order the sum of one hundred and forty-one _ dollars and eight cents, in three months, with interest | afterward. Horace Chase. © | On the back of this note were the following endorsements. May | Ist. 1833, received seventy-five dollars. September 14th. 1833, re- ) ceived forty-five dollars. The balance of the note was paid January | 14th. 1834. How much was the balance ? *| First payment, $75. | 2nd. payt. $45. | Principal, $ 141.08 ) Interest,8m.14d., 3.17/Int., 4m. 90 | Int. 9m. 6.34 | Amount, $ 78.17 Amount, 45, 90 Amount, 447.42 it 8.17 124.07 | Amount of payments, $ rie Balance, $23.35 | | | (111.) ' - New York, May 25th. 1833. For value received, I promise Joseph Day to pay him) or order the sum of three hundred and oné dollars and | forty-seven cents, on demand, with interest. j Attest. John Smith. Samuel Frink. On the back of this note, the following endorsements were “made, July Ist. 1833, received sixty-seven dollars and fifty cents. Janu- | ary 4th. 1834, ‘received forty-eight dollars. April ‘11th. 1834, re- by ‘ived thirty-nine dollars. "The balance of this note was paid June ist. 1884. Required the balance. Xv. INTEREST. 103 (112.) ‘Philadelphia, June 26th. 1833. For value received, I promise Charles 8. Johnson to pay him or order ninety-three dollars and twenty-eight cents, on demand, with interest. James Orne. Attest. Levi Dow. _ On this note there were two endorsements, viz. Nov. 5th. 1833, received forty-three dollars and seventy-five cents. Feb. 22d. 1834, ‘received thirty-seven dollars. What was due, May 26th. 1834, when the balance was paid. (113.) Baltimore, March 4th. 1832. For value received, I promise Hay & Atkins to pay them or order the sum of four hundred and three dollars and fifty-six cents, in nme months, with interest afterward. Homer Chase. The following endorsements were made on the back of this note. Jan. ist. 1833, received one hundred and eighty-four dollars. August 18th. 1833, received one hundred dollars. ‘This note was taken up Dec. Ist. 1833. What was the balance then due upon it? | (114.) Hartford, July 11th. 1831. For value received we promise Joseph Seaver to pay him or- order the sum of two hundred and seventeen dollars and fifty cents, in four months, with interest after that time. Whiting & Davis: On this note there were three endorsements: viz. Nov. 16th, 1831, received ninety-three dollars. Feb. 12th. 1832, received fifty dollars. August 2d. 1832, received sixty-seven dollars and seventy-five cents. This note was taken up Oct. 4th. 1832. How much was then due upon it? (115.) Burlington, October Ist. 1832. For value received, we promise Hannum, Osgood, & Co. to pay them or order the sum of seven hundred and fourteen dollars, in three months, with interest afterward. Mason & Gould. The following payments were endorsed on the note. January | Ist. 1833, received three hundred and sixty-four dollars. May Ist. _ 1833, received one hundred and twenty-five dollars and fifty cents. August Ist. 1833, reccived eighty-six dollars. Nov. Ist. 1833, re- | ceived a hundred and ten dollars. ‘The balance due on this note | was paid Jan. Ist. 1834. How much was it? | | 104 ARITHMETIC. XV. If settlement is not made, till more than a year has_ elapsed after the commencement of interest, the preced-_ ing mode of computing interest, when partial payments have been made, ought not to be adopted; and indeed de is not in strict conformity with law. i The United States Court, and the Courts of the several States, in which decisions have been made and reported, with the exception of Connecticut and Vermont, and a slight variation in New Jersey, have established a gen-_ eral rule for the computation of interest, when partial payments have been made. This rule is well expressed in the New York Chancery Reports, in a case decided by — chancellor Kent, and here given in the Chancellor’s own) words, as follows. 4 ‘¢ The rule for casting interest, when partial payments have been made, is to apply the payment, in the first placa ; " ; to the discharge of the vnterest then due. If the payment exceeds the interest, the surplus goes towards discharging the principal, and the subsequent interest rs to be com-| puted on the balance of principal remaining due. bs | payment be less than the interest, the surplus of interest) must not be taken to augment the principal; but interest continues on the former principal until the period when the payments, taken together, exceed the interest due, and then the surpius is to be applied towards discharging) the principal; and interest is to be computed on the bale ance, as aforesaid.’ i The interest on the following notes, must be computed | by the above legal rule. | | iE i (116.) Washington, March 4th. 1832. For value received, I promise Nehemiah Adams to pay him or order the sum of one thousand two hundred dollars, on demand, with interest. Charles Train. — Attest. William Dorr. The following endorsements were made on this note. June 10th. 1832, received one hundred and sixty-nine dollars and twenty | cents. Oct. 22d. 1832, received twenty dollars. March 30th. 1833, received twenty-eight dollars. Nov. 5th. 1833, received six hun- dred and eighteen dollars and five cents. What was the balance | due, on taking up this note, March 5th. 1834? . ( | | | | | 3.4%. INTEREST. 105 7 Principal, - $ 1200. _ Interest from Mar. 4, to June 10, (3 m. 6 d.), - = 19.20 First Amount, - - 1219.20 . First payment - - : - - - - 169.20. Balance, forming a new principal, . - - - - 1050,00 Interest from June 10, to Oct. 22,(4m.12d.), $23.10 — | Second payment, - - - - . 20. 3.10 | Leaving interest unpaid, - - -~ - Interest from Oct. 22, to Mar. 30, (5m. 8 d.), 27.65 Third payment, Sie ae ak ta 28.00 ‘Leaving interest unpaid, = - - - - 2.75 Interest from Mar. 30, to Nov. 5,(7m.6d.), 37.80 40.55 | Second Amount, - - 1090.55 Fourth payment, - - - - - - - 618,05 Balance, forming a new principal, - - - - 472.50 Taterest from Nov. 5, to Mar. 5, (4 m.), - - - 9.45 Balance due on taking up the note, - - - - $481.95 BElv)) Richmond, Jan. 5th. 1833. For value received, I promise Joseph Tufts to pay him or order one hundred and forty-three dollars and fifty cents, on demand, with interest. John Hanes. Two payments were endorsed upon this note: viz. April 13th. 1833, received forty-five dollars and eighty-four cents. Dec. 22d. 1833, received fifty-four dollars and fifteen cents. The balance of this note was paid March 28th. 1834. How much was it? (118.) Raleigh, July Ist. 1832. For value received, I promise Charles Goodrich to pay him or order the sum of six hundred and twenty-five dollars and fifty cents, in three months, with interest after- ward. John Frink. Three payments were endorsed upon this note: viz. January Ist. 1833, received two hundred dollars. Nov. Ist. 1833, received twenty dollars. Jan. ist. 1834, received three hundred dollars. , The balance was paid May Ist. 1834. How much was it? (119.) Charleston, Dec. 22d 1830. _ For value received, I promised George Winship to pay him or order ninety-seven dollars and eighty cents, on demand, with interest, Thomas White. , | | rs ' I I 106 ARITHMETIC. | XV. The endorsements made on this note were the following. Oci. 12th. 1831, received twelve dollars eighty-five cents. July 20th 1832, received twelve dollars and seventeen cents. Feb. 26th 1833, received fourteen dollars and ninety-five cents. August 26th 1833, received thirty-six dollars and ten cents. Requir ed the | ance, which was paid Jan. 31st. 1834. } | | (120.) Augusta, January Ist. 1331. _ For value received, I promise Israel Capen to pay him or order eighty- -four dollars and forty cents, on de_ mand, with interest. Edward Ruggles. | On the back of this note were the following endorsements. Oct 9th. 1831, received nineteen dollars and thirty-two cents. Juh 15th. 1832, received twenty dollars. April 9th. 1833, receiver twenty-one dollars and eighty-one cents. Oct. 9th. 1833, receive( twenty-two dollars and fifteen cents. The balance of this not. was paid Feb. 19th. 1834. How much was it? ) t (121) New Orleans, Feb. 22d. 1830. For value received, I promise Maynard and Noye: to pay them or order the sum of nine hundred dollars, i ir three months, with interest till paid. Isaac Jettison. Attest. William Proctor. The following payments were endorsed upon the note. May 22d. 1830, received twenty-five dollars. Sept. 22d. 1830, received; fifteen dollarse May 22d. 1831, received thirty-five dollars. May 22d. 1832, received one hundr ed and forty-five dollars and twelve cents. Dec. 4th. 1832, received one hundred and twenty-five dollars and sixty cents. "May 22d. 1833, received two hundred and’ nineteen dollars and sixty cents. Dec. 31st. 1833, received two! hundred and sixty-eight dollars and twenty-five cents. The. Si ist Ae this note was paid Feb. 24th. 1834. What was the alance: (122.) Cincinnati, Dec. Ist. 1830. For value received, I promise Horatio Davis to pay him or order the sum of one thousand dollars, on demand, with interest tll paid. Edward.Lang. | Five partial payments were endorsed on this note: viz. Feb.: Ist. 1832, received seventy-five dollars. June Ist. 1832, received’ twenty dollars. August Ist. 1833, received twenty dollars. October Ist. 1833, received seven hundred and fifty dollars. Feb. 1st. 1834, received one hundr ed dollars. ‘The balance of this note was paid - J une Ist. 1834, How much was it? ~XV.~ INTEREST. 167 (123.) Louisville, April 4th. 1832. _ For value received, I promise Samuel H. Wheeler to ‘pay him or order the sum of three hundred and ninety- six dollars, on demand, with interest, at the rate of 7 per cent. a year, till paid. George Guelph. Partial payments were made on this note, as follows: Sept. 14th. 1882, received twelve dollars. May 4th. 1833, received eighteen dollars. Oct. 24th 1833, received forty-nine dollars twelve cents. The balance was paid May 30th. 1834. What was the balance ? (124.) Nashville, Sept. 7th. 1831. For value received, I promise Darius Pond to pay him or order the sum of four hundred and eighty-six dollars and ninety cents, on demand, with interest at the irate of 7 per cent. a year. Martin Smith. The following partial payments were endorsed on this note. March 22d. 1832, received one hundred and twenty-five dollars. Nov. 29th. 1832, received one hundred and fifty dollars. May ‘A8th. 1833, received one hundred and twenty dollars. ‘The bal- ance was paid April 19th. 1834. Required the balance. (125.) Albany, August 13th. 1830. For value received, I promise Theodore Leonard to pay him or order the sum of two hundred and ninety- eight dollars and nineteen cents, on demand, with interest at the rate of 7 per cent. a year. Stephen Kirkland. Attest. W. Stevenson. The following endorsements were made on this note. April 6th. 1831, received fifty-four dollars. Dec. 17th. 1831, received forty-two dollars. June, 21st. 1832 received sixty-one dollars. Feb. 26th. 1833, received thirty-seven dollars and eighty cents. July 8th. 1833, received seventy-five dollars. The balance was paid May 12th. 1834. How much was the balance ? COMPOUND INTEREST. Compound interest is that which is paid not only for the use of the principal, but also, for the use of the inter- est after it becomes due. _ When the interest is payable annually, find the interest for the first year, and add it to the principal, and this amount is the principal for the second year. Find the 108 interest on this second principal, and add as before; thi! amount is the principal for the third year: and so o7 through the whole number of years. When the interes| is payable half-yearly, or quarterly, find’ the interest fo! half a year, or a quarter of a year, and add it to thi principal, aud thus proceed through the whole time} Subtract the first principal from the last amount, and thi remainder is the compound interest. | 126. What is the compound interest of a thousam dollars for 3 years, at 6 per cent. per annum ? ¢ 1000. 60. 1060. 63.60 1123.60 ae ee 416 ———— 1191.016 1000. $191.016 127. What is the compound interest of 740 dolla fo 6 years, at 6 per cent. per annum ? 128. What is the compound interest of 500 dollar for 4 years, at 7 per cent. per annum? | 129. To what sum will 450 dollars amount, in 5 years at 5 per cent. per annum, compound interest ? | 130. What is the compound interest of £760 10: for 4 years, at 4 per cent. per annum? 131. A gave B a note for 300 dollars, with interest ; ARITHMETIC. XV principal. interest for the first year. amount, principal for the second year. | interest for the second year. second amount, principal for third year interest for the < _ird year. third amount. | first principal deducted. Answer. — 6 per cent. a year, payable semiannually. How mue i did it amount to in 2 years, at compound interest ? 132. At compound interest, what will 600 dollay " amount to in 14 year, at the rate of 6 per cent. a yea interest payable quarterly ? a PROBLEMS IN INTEREST. a ' In reviewing the subject of simple terest, we percely four several problems, which arise from its conditions and which we shall now distinctly notice. | XV. INTEREST. 109° PROBLEM I. The principal, time, and rate per cent. given, to find the interest. RULE. Multiply together the decimal expressing the rate per annum, the lime in years and the decimal 1 of a year, and the principal: the product will be the interest. This problem has already been exemplified in the pre- ceding pages of this article. PROBLEM II. The principal, time, and amount given, to find the rate per cent. per annum. RULE. Subtract the principal from the amount, and the remainder will be the interest for the given time. 4 4 Divide this interest by the given time expressed in years or the decimal of a year, and the quotient will be the ‘interest for one year. Divide the interest for one year by ‘the given principal, and the quotient will be the rate per cent. per annum. 133. At what rate per cent. per annum must 172 dol- lars 40 cents be put on interest, in order to amount to 332 dollars 74 cents, in 5 years : ? 134. Lent 51 dollars 25 cents, and in 1 year and 4 months it amounted to 55 dollars 35 cents. What wa the rate per cent. per annum? 135. Borrowed 340 dollars for 9 eertnes and at the expiration of the time it amounted to 355 dollars 30 cents. | What was the rate of interest per annum? | | 136. At what rate per cent. per annum must 874 cents ‘be put on interest, in order to amount to 98 cents, in2 years ? PROBLEM Ill. The principal, rate per cent., and amount given, to find the time. RULE. Subtract the principal from the amount, and . the remainder will be the interest. Divide the interest oy the principal, and the quotient will be the interest of 1 dollar. Divide the interest of 1 dollar by the rate, and the quotient will be the time. 137. In what time will 89 dollars 25 cents amount to 92 dollars 82 cents, at the rate of 6 per cent. a year? 138. In what time will 171 dollars 40-cents, amount to 231 dollars 39 oe at 7 per cent. a year? 0 110 ARITHMETIC. XME! 139. Borrowed 163 dollars 50 cents at 6 per cent. a year; at the time of payment it amounted to 176 dollars 58 cents. How long did I keep the money ? 140. In what time will 4810 dollars 25 cents, amount to 5002 dollars 66 cents, at 6 per cent. a year? 141. Lent 114 dollars at an interest of 7 per cent. a) year; on its return it amounted to 127 dollars 30 cents. — How long was it out? 142. In what time will $100, or, any other sum of k money double, at the rate of 6 per cent. per annum, ! simple interest ? . as Soe PROBLEM IV. The amount, time, and rate per cent. given, to find the principal. RULE. . Divide the amount by the amount of 1 dollar i for the time, and the quotient will be the principal. | This problem-forms the subject of the next article, | under the head of Discount. XVI. | DISCOUNT Discount is an allowance made for the payment of | money before it is due. | The present worth of a debt, payable at a future period without interest, is that sum of money, which, being put on interest, arid amount to the debt, at the period when the debt is payable. It is obvious, that, when money is worth 6 per cent. per annum, the present worth of $1.06, payable in a_ year, is $1. Hence, the present worth of any debt, pay> able in a year, is as many dollars as there are times $1.06 _ in the debt. And hence we deduce the following. | RULE. Divide the debt by the amount of 1 dollar for i the time, and the quotient is the present worth. Subtract | the present worth from the debt, and the remainder will be the discount. wen mM. Oo XVI. DISCOUNT. | 111 1. What is the present worth of 450 dollars, payable in 6 months, when money is worth 6 per cent. per annum ? 2. What is the present worth of 535 dollars, payable in 15 months, when money is worth 6 per cent. per annum? 3. When money is let for 6 per cent. per annum, what is the present worth of a note for 1530 dollars, payable in 18 months? 4. Sold goods to the amount of 1500 dollars, to be paid one half in 9 months, and the other half in 18 months: | what is the present worth of the goods, allowing interest to be 5 per cent. per annum? 5. What is the present value of a note for 2576 dol- lars and 83 cents, payable in 9 months, when interest is 6 per cent. per annum? 6. When interest is 6 per cent. a year, what is the difference between the discount on 1285 dollars for a year and 8 months, and the interest of the same sum for the same time ? 7. Purchased goods amounting to 6568 dollars 50 cents on a credit of 8 months: allowing money to be worth 4 per cent. a year, how much cash down will pay the bill ? 8. A man, having a horse for sale, was offered for it 225 dollars, cash in hand, or 230 dollars payable in 9 months: he chose the latter, although money was worth 7 per cent. a year. How much did he lose by his ignorance ? ~ 9. Bought a quantity of goods for 1831 dollars 53 cents cash, and the same day sold them for 1985 dollars 48 cents on a credit of 6 months, when money was 5 per cent. a year. How much did I gain upon the goods ? 10. What is the discount on 198 dollars 60 cents, for 9 months, when interest is 5 per cent. a year? ~ 11. What is the discount on 241 dollars 81 cents, for 7 months, when interest is 44 per cent. a year ? 12. What is the present worth of 741 dollars 65 cents, payable in 48 days; interest being 6 per cent.? 112 ARITHMETIC. XVII. XVII. BANKING. A BANK is an institution which trafficks in money. It is owned in shares, by a company of individuals, called _ stockholders; and its operations are conducted by a Presi- dent and board of Directors. It has a deposite of specie, and issues notes or bills, which are used for a circulating medium, as money. ‘These bills are mostly obtained from the bank in loans, on which interest is paid; and the amount of bills issued being greater than the amount of specie kept in deposite, a profit accrues to the bank. The interest on money hired from a bank, is paid at the time when the money is taken out—the hirer receiving as much less than the sum he promises to pay, as would be equal to the interest of what he promisesdo pay, from the time of hirmg the money until the time it is to be paid. From this circumstance, the interest on money hired from a bank is called discount, and the promissory note received at the bank is said to be discounted. A note, to be discounted at a bank, is usually made payable to some person, who endorses it, and who there- by binds himself to pay the debt, in case the signer of the note should fail to do so. Any person, therefore, who holds the note of another, payable at a future time, may endorse it, and obtain the money for it at a bank, — by paying the bank discount; provided the credit of the parties is undoubted. It is customary in banks, to compute the discount on every note for 3 days more than the time stated in the note; and the debtor is not required to make payment until 3 days after the stated term of time has elapsed. These-3 days are called days of grace. 1. What is the bank discount on 775 dollars for 30. days, and grace, when interest is 6 per cent. a year? 2. What is the bank discount on 900 dollars for 90. days, and grace, at the rate of 6 per cent. a-year? XVII. BANKING. 1138 3. How much is received on a note for 2540 dollars’ 80 cents, payable in 4 months, discounted at a bank, when interest is 45 per cent. a year? 4. A note for 452 dollars, payable in 7 months, is discounted at a bank, when interest is 6 per cent. per annum. What sum is received on it? 5. A note for 3000 dollars, payable in 70 days, is dis- counted at a bank, when interest is 6 per cent. a year. What sum is received on it? : - 6. Amerchant bought 1625 barrels of flour for 5 dollars a barrel cash, and on the same day sold it for 5 dollars 60 cents a barrel, on a credit of 8 months, took a note for the amount, and got it discounted at a bank, when money was 6 per cent. a year. How much did he gain on the flour ? 7. A man got his note for ¢ 1000, payable in 3 months, discounted at a bank, at the rate of 6 per cent., and im- mediately put the money he received for his note on interest for 1 year, at 6 per cent. He kept the money from the bank 1 year, by renewing his note every 3 months, and paying in the required bank discount at each renewal. At the end of the year he received the amount of the money he had put on interest, and paid his note at the bank. How much did he lose by this exchange ? In the above example, interest on the several discounts paid into the bank forms part of the loss. 8. A money broker subscribed for 20 shares in a new bank; at $100 a share. When the bank commenced operation he paid in 50 per cent. of the price of his stock, and in 6 months after, he paid in the remainder. In 12 months from the time the bank commenced, there was a dividend of 34 per cent. on the stock among the stock- holders; aud the same dividend accrued every 6 months thereafter. At the end of 3 years the broker sold his stock at 7 per cent. advance. Now, allowing that this broker hired his money, and paid 6 per cent. annually, how much did he make by the speculation ? In this example, the broker must charge annual interest on the interest he pays, and must give credit for annual interest on his share of the dividends. | 10% f 114 ARITHMETIC. XVI. XVII. | EQUATION OF PAYMENTS. 1 EQuATION OF PAYMENTS consists in finding a mean time for the payment at once of several debts, payable at different times, so that no loss of interest shall be sustained” by either party. For instance, if A owes B one dollar, payable in 1 oe | months, another dollar payable in 3 months, and a third | dollar payable in 4 months, at what time may the three sums be paid at once, without injustice to either of them It is evident, that the interest of 1 dollar for 2 months, is— the same as the interest of 2 dollars for 1 month; and, the interest of 1 dollar for 3 months, is the same as the inter- est of 3 dollars for 1 month; and the interest of 1 dollar _ for 4 months, is the same as the interest of 4 dollars for 1 month: 2 dollars, 3 dollars, and 4 dollars, added together, make 9 dollars for 1 month; but the three sums to be~ paid, when added together, make only 3 dollars, which sum being only a third part of 9 dollars, the term of | credit must be three times as long, or 3 months, which is the equated time. This result is obtained by multiply-— ing the sum, payable in 2 months, by 2; that payable in 3 months, by 3; and that payable in 4 months, by 4; and then adding the several products together, and dividing the sum of them by the sum of the debts. RULE. JWMultiply each debt by the time, in which tt is payable, and divide the sum of the products by the sum of the debts: the quotient will be the equated time. ; If I owe you 50 dollars payable in 4 months, | ie payable im 6 months, and 100 dollars payable in’ 7 months, in what time may the three sums be paid at once, without loss to either of us ? 2: A owes B 200 dollars, a0 dollars of which is to be paid 1 in 3 months, 60 dollars in 5 months, and the remain- dor in 10 months. At what time may the whole be paid at once, without injustice to either party ? tr / | XVII. EQUATION OF PAYMENTS. 115 _ 3. Bought goods to the amount of 1552 dollars, pay- able at four different times, as follows; 225 dollars and 75 cents in 4 months, 250 dollars and 25 cents in 6 months, 425 dollars and 50 cents in 8 months, 650 dol- | lars 50 cents in 10 months; but afterward agreed with my creditor to pay him all at once, at the equated time. ‘What was the time? _ 4. If lowe you three sums of money payable at differ- ent times, viz. 50 pounds in six months, 60 pounds in 7 months, and 80 pounds in 10 months, what is the equated ‘time for paying the whole at once? 5. Bought goods to the amount of 1000 dollars, 200 ‘dollars of which was to be paid down, 400 dollars in 5 months, and the remainder in 15 months; but it was afterward agreed, that the whole be paid at once. In what | time ought the payment to be made ? ; 6. A merchant has due to him a certain sum of money, to be paid as follows; ¢ in 2 months, 4 in 3 months, and the rest in 6 months. What is the equated time for pay- ing the whole ° | 7. Sold goods amounting to 1296 dollars, of which 346 dollars was to be paid in 24 months, 323 dollars in 6 months, and the balance in 10 months; but the purchaser afterward agreed to make but one payment of the whole. What term of credit ought he to have? 8. Bought goods to the amount of 640 dollars 80 cents, payable } down, + in 4 months, 4 in 8 months, and the balance in a year; but afterward made an agree- ment to pay the whole at one time. In what time ought Ito pay for the goods ° 9. A merchant has due to him $3800 to be paid in 60 days, $500 to be paid in 120 days, and $750 to be paid in 120 days. What is the equated time for these dues ? 10. A owes B $1200, to be paid in 8 months; but A offers to pay $400 in 4 months, on condition that the remainder shall continue unpaid an adequate term of time. In what time ought the remainder to be paid ? 11. If adebt of $1000 be payable at the end of 7 months, and the debtor agree to pay $300 at present, ba is the proper time for paying the rest? | 116 ARITHMETIC. XIX XIX. PROFIT AND LOSS. The ascertaining what is gained or lost n buying anq selling, and the adjusting of the price of goods so as tc gain or lose a certain sum, or a certain per cent., comé under the head of Profit and Loss. _ 1. Bought a piece of broadcloth containing 28 ya for 112 dollars, and sold it at 5 dollars 25 cents a yard) How much, and what per cent. was my profit? (See Arr. xiv, Example 45.) | ie Bought 3 pieces of broadcloth, containing 28 yards, each, at 5 dollars 25 cents a yard. At what price pet yard must I sell it, to gain 20 per cent.? 3. Bought cloth at 4 dollars 60 cents a yard, which, not proving so good as I expected; I sold at 3 dollars 91 conts a yard. What per cent. did I lose? 4. Bought 1250 barrels of flour for 6250 dollars. Al what price per barrel must I sell it, to make a profit of 124 per cent.? 5. Bought 30 hogsheads of molasses, at 20 dollars ¢ hogshead, in Havana; paid duties 20 dollars 66 cents; freight 40 dollars 78 cents; porterage 6 dollars 5 cents; insurance 30 dollars 84 cents. What per cent. shall 1 gain by selling at 26 dollars per hogshead ? 6. Bought wheat at 75 cents a bushel; at what price per bushel must I sell it, to gain 20 per cent.? 7. A merchant received from Lisbon 180 casks al au raisins, containing 80$1b. each, which cost him 2 dollars 18 cents a cask. At what price per cwt. must he sell) them, to gain 25 per cent. ? 4 8. If I sell sugar at 8 dollars per cwt., and thereby, lose 12 per cent., what per cent. do I gain or lose, by selling the same at 9 dollars per ewt.? 9. If I purchase 6 pipes of wine for 816 dollars, aul sell it at 59 dollars 50 cents a hhd. do I gain or lose, and what per cent.? | | | ‘Xx. PARTNERSHIP. Lie 10. If you purchase 5cwt. Igr. 12Ib. of rice, at 2 dollars 80 cents per cwt., at what price per pound must you sell it, to make 6 dollars on the whole ? i. If I purchase 13cwt. of coffee at 124 cents per pound, at what price per lb. must I sell it, “to a 80 dollars 8 cents on the whole? 12. A miller sold a quantity of corn at 1 dollar a bushel, and gained 20 per cent.; soon after, he sold of the same, to the amount of $37.50, and gained 50 per cent. How many bushels were there in the last paren, and at what did he sell it per bushel ? XX. PARTNERSHIP. Partnersulip is the union of two or more individuals \ ° ° mtrade. ‘The company thus associated is called a firm: wind the amount of property, which each partner puts ato the firm, is called his stock in trade. The profit or ‘oss is shared among the partners, when the stock of each s employed an equal length of time, in proportion to ach partner’s stock in trade; but, when the stock of he several partners is employed in the firm unequal ‘erms of time, in proportion to each one’s stock and the ime it was employed. 1. A, B, and C entered into partnership, and the stock of each was employed in the firm one year. imperfect powers of another degree. For example, 4_ is a perfect power of the second degree, and its square root, which is 2, is rational; but an imperfect power of | the third degree, and its cube root, which is 1.587 +, is a surd: 8 is an imperfect power of the second degree, and its square root, which is 2.828-++, is a surd; but a perfect power of the third degree, and its cube root, which is 2, is rational: 16 is a perfect power of the se- cond and fourth degrees, and its square root, which is 4, and its biquadrate root, which is 2, are both rational; but an imperfect power of the third degree, and its cube root, which is 2.519-+, is a surd. These irrational numbers or surds occur, whenever we endeavor to find a root of any number, which is not a perfect corresponding power; and, although they can- not be expressed by numbers either whole or fractional, they are nevertheless magnitudes, of which we may form an accurate idea. Jor however concealed the square root of 2, for example, may appear, we know, that it must be a number, which, when multiplied by itself, will exactly produce 2. This property is sufficient to give us an idea of the number, and we can approximate continually by the aid of decimals. XXVIII. EVOLUTION. 159 A radical sign, written thus 4/, and read square root, is used to express the square root of any number, before which it is placed. The same sign with the index of the root written over it, is used to express the other roots: thus 4/ cube root: 4/ biquadrate root: ¥/ fifth root: &c. We will give the following radical expressions; 4/9 = 3 A/ 8 = 2; 4/81=8; 4/32—2; these expressions are read thus; the square root of 9 is equal to 3; the cube root of 8 is equal to 2; the fourth root of 81 is equal to 3; the fifth root of 32 is equal to 2. Hence it is evident that 46/9 XK 4/9=9; 4/8 X V/8 X 4/88; Kc. The explanation, which we have given of irrational numbers or surds, will readily enable us to apply to them the known methods of calculation. We know that the square root of 3 multiplied by itself must produce 38, which may be rales expressed, / 3 XK 4/ 3=3; also a/ BX BX V/8=3; VEXVI=HH V5 X 553 ran aor ar? SSX A/S XK A/S XK 4/5 = 5; Instead of the radical sign, a fractional exponent is also used to express the roots of numbers. The numerator indicates the power of the number, and the denominator 1 the root. ‘Thus, 4° expresses the square root of 41, or 4; 4°, the cube root of 4; 43, the biquadrate root of 4; 1 2 4°, the fifth root of 4; 8°, the cube root of the second power of 8; and since the second power of 8 is 64, and the cube root of 64 is 4, the expression 8? is equal to 4. The expression 48, is read thus, the square root of the third power of 4 is equal to 8. The expression 9° is equivalent to 4/9: and 8 is equivalent to 4/ 8°: also 4} is equivalent to 4/47: the expression 4} is also equal to 4, because 7 is equal to 5. A fines or vinculum, drawn over several numbers, sig- nifies that the numbers under it are to be considered jointly: thus, 4/25-+-11 is equal to 6, because 25-+-11 is 160. ARITHMETIC. XXIX. 36, and the square root of 36 is 6; but 4/25 --11 is equal | to 16, because the square root of 25 is 5, and 5--11 ig 16. The expression 4/27—6 -- 43 is equivalent to a/ 64 And 4/100—73=8. Also 20—/9+-7+1=17. Likewise 4/90 —9 —4 + ./53—45 + 6= 13. X XIX. EXTRACTION OF THE SQUARE ROOT. -The product of a number multiplied by itself, is call- ed a square; and for this reason, the number, considered _ | im relation to such a product, is called a sQUARE ROOT. For example, when we multiply 12 by 12, the product, | 144, is a square, and 12 is the square root of 144. If the root contains a decimal, the square will also_ contain a decimal of double the number of places; for example, 2.25 is the square of 1.5; and vice versa, if the square contains a'decimal, the square root will contain a decimal of half the number of places; for instance, 1.5 is the square root of 2.25. In the upper line of the following table are arranged several square roots, and in the lower line, their squares. | Square roots. |1]2/8/4[5]6]718] 9110111112 ? | Squares. | 1/4] 9 1625.36.49 64/s1]100 121/144) When the square of a mixed number is required, it may be reduced to an improper fraction, then squared, and reduced back to.a mixed number. The squares of the numbers from 3 to 5, increasing by 4, are as follows. | Square roots. | 3| 33 | 34, hie [4 41 | 41 | 43 | | Squares. | 9 10-2, 1221141116] 18,1,| 204! 22%! J ———___ From this table we infer, that if a square contains a fraction, its square root also contains one; and vice versa. XXIXx. SQUARE ROOT. 161 It is not possible to extract the square root of any _ number, which is not a perfect square; we can approxi Se mate the square root of such numbers, however, by the aid of decimals. When a root is composed of iwo or more factors, we may multiply the squares of the several factors together, and the product will be the square of the whole root; and conversely, if a square be composed of two or more factors, each of which is a square, we have only to mul- uply together the roots of those factors, to obtain the complete root of the whole square. For example, 2304 = 4X 16x 36; the square roots of the factors are 2, 4, and 6; and 2X4X6==48; and 48 is the square root of 2304, because 48 X 48=—= 2304. A square number cannot have more places of figures than double the places of the root; and, at least, but one less than double the places of the root. Take, for in- stance, a number consisting of any number of places, that shall be the greatest possible, of those places, as 99, the square of which is 9801, double the places of the root. Again, take a number consisting of any number of places, _ but Jet it be the least possible, of those places, as 10, the square of which is 100, one less than double the places of the root. As the places of figures in the square cannot be more than double the number of places in the root, whenever we would extract the square root of any number, we point it off into periods of two figures each, by placing a ‘dot over the place of units, another over hundreds, &c. Thus 1936. The places in the root can never be more or less in number, than the number of periods thus pointed off. When the number of places in the given sum is an odd number, the left hand period will contain only one figure, as 169; but the root will nevertheless consist of as many places as there are periods; for 13 is the square root of 169. . The terms, square and square root, are derived from geometry, which teaches us that the area of a square is found by multiplying one of its sides by itself. 62. ARITHMETIC. XXIX. The word arma signifies the quantity of space contained _ in any geometrical figure. te the length of aside of the an- | : i "i nexed square to be four feet, it i } | i i is evident that the figure contains ' | 4 times 4small squares, each of | | which is 1 foot in length and 1 TAA foot in breadth; and since a foot | | in length anda foot in breadth ™ aa . constitute a square foot, the whole square contains 4_ times 4, or 16 square feet. If, instead of 4 feet, thal length of a side were 4 yards, the whole square would” contain 16 square yards; &c. Hence it is evident that — the area, which is 16, is found by multiplying a side of — the square by itself. v A PARALLELOGRAM is an oblong figure, having two of its sides equal and parallel to each other, but not of the same length with the other two, which are also equal and parallel to each other. We find the area, “or contents of a parallelogram by multiply- ing the length by the breadth: If we sup- pose the annexed right angled parallelogram to be 8 feet long.and 2 feet wide, it is manifest that it contains 2 times 8, or 16 square feet; if the length were 8 yards and the breadth 2 yards, it would contain 16 square yards; if 8 miles long and 2 miles wide, 16 square miles; &c. .We see that the area of this parallelogram is the same i with that of the preceding square; there- fore the square root of the area of a paral- lelogram gives the side of a square equal in area to the parallelogram. : It is further to be observed, that the square root of the — area of any geometrical figure whatever, is the side of a_ square, equal in area to the figure. ; . | XXIX. SQUARE ROOT. 163 _ When the area of a square is given, the process of finding one of its sides, which is the root, 1s called the extraction of the square root, the principles of which we will now proceed to explain. | _ We have already learned, that a square number is a product resulting from two equal factors. For example, 2025 is a square number resulting from the multiplication of 45 by 45. To investigate the constituent parts of this groduct, we will separate the root into two terms, thus, 40--5, and multiply it by itself.im this form. We begin with multiplying 40-5 by 5, and set down the products separately, which are 200 4-25; we then multiply 40 +- 5 xy 40, and set down the products separately, which are 1600-+-200; the whole product, therefore, is 1600 -- 200 + 200 + 25 = 2025; thus we see, that the whole wroduct or square contains the square of the first term, 0 x 401600; twice the product of the two terms, 0X 5 X 2400; and the square of the last term, 1X 525. _ Now the extraction of the square root is the reverse if squaring or raising to the second power; therefore, he operation of extracting the square root of 2025, which ve know is the square of 45, must be performed in the nyerted order of raising 45 to the second power. We will now extract the square root of 2025, and ‘xplain the process, step by step. | | | 2025 (45 16 a Divisor. 40X2=80 425 dividend Divisor, increased by last fig. 85 425 product of 85 by 5. Explanation of the precess. We began by separating he given number into periods of two figures each, putting i dot over the place of units, and another over hundreds, fe thereby ascertained that the root would contain two alaces of figures. We then found that the greatest square in the left hand period was 16, and placed its root, which is 4, in the quotient, and subtracted the square from the left hand period, and to the remainder osrought down the next period for a dividend. % ' figure in the root must be such, that twice the product o ' dend would have been divided by 10. Thus 8 is cor) 164 ARITHMETIC. XXI¥ Then, knowing the figure in the root to be im the plac| af tens, and therefore equal to 40, and that the secon the first and second terms, together with the square ¢| the second, would complete the square, we took twic| the root already found, viz. 40 X 2== 80, for one of th factors, and using it for a divisor, found the second figut| in the root by dividing the dividend by this factor. Lastly, after finding that the second figure in the ro¢ was 5, we added it to the divisor, making 80-+ 588 and multiplied the sum by 5, the last figure m theroo| and thus obtained twice the product of the two term,| and the square of the last term; because, 80 is twice th first term of the root, and being multiplied by 5, whic’ is the last term, the result is twice the product of th| two terms; and 5 being multiplied by 5, the product | the square of the last term. | It will be cbserved, that 4, the first figure in the roo being in the place of tens, was called 40, and double for a divisor; but, if we had merely doubled the roc without any regard to its place, making the divisor ¢ and had cut off the right hand figure of the dividend an divided what was left, the result would have been th same; because, in this operation, both divisor and div) tained 5 times in 42. The figure obtained for the roo! in this abridged method, would be placed at the right han| of the divisor, instead of being added thereto; thus, 88| making the completed divisor the same as before. Thi course, being the most concise, will be adopted in th! rule for extracting the square root, which we shall here after give. ) Suppose 169 square rods of land are to be laid out 1) a square, and the length of one of its sides is required. | We know that the length of a side must be such | aumber of rods, as, when multiplied by itself, will pro} duce 169; therefore we must extract the square root 0! the given number of square rods, and that root will b the answer. /169=13 Ans. XIX. SQUARE ROOT. 165 , We here illustrate this 4 yo hay ast example geometrical- [7] , by a square figure A B _D, each side of which ¢}——-—- s 13 rods long. This jguare is divided into 169 mall squares, each of vhich is a square rod. . The whole figure is also wubdivided into four fig- res, two of which, ef g aan hf iB, are squares; adthe other two, Ahfe ? Svinnk,.: jad C if g, are oblongs. , The square ef g Dis 10 rods on aside, and, therefore, | ‘ontains 100 square rods. The oblongs are each 10 rods mg and 3 rods wide, and consequently each contains 30 quare rods. The other square, B hfi, is 3 rods ona jide, and contains 9 square rods. , To illustrate the process of extracting the square root, ye shall take the side A B, which is divided into two jatts, the first of which, A h, is 10 rods long; the other, ,B, 3 rods long. Ah being equal to ef, the square {A h, the first part, gives the area of the square D ef g; _B veing equal to h f, the area of the oblong A h fe, is ouna by multiplying the two parts, A h and h B, togeth- 5 tne area of the other oblongifg C, is the same; there- re, the area of the two oblongs ts twice the product {the two parts, Ah andh B.. The square of the last art, h B, gives the area of the squaeh Bif. , We have therefore the square of the first part A h, 0X 10—100 rods; twice the product of the two parts, hand h B, 10x 3X 2=60 rods; and the square of r last parth B, 3X 3=—9 rods. These being added »gether make 169 rods, the square of the whole figure — ABC D. : This illustration of a square corresponds exactly with jat of the first example, and of course the extraction of 2 square root must proceed on the principles there xhibited. t 166 ARITHMETIC. XXT From the illustrations of the two precedmg example we give the following rule for the extraction of the squa root. | é _ RULE. First—Point off the given number into perio of two figures each, by putting a dot over the place ¢ units, and another over every second figure to the le and also to the right, when there are decimals. — Secondly—Find the greatest square in the left has period, and write its root in the quotient. Subtract t square of this root from the left hand period, and to t remainder bring down the next period for a diwidend. Thirdly—Double the root already found, for a diviso Ascertain how many times the divisor is contatned ‘| the dividend, excepting the right hand figure, and pla the result in the root, and also at the right hand of t divisor. Multiply the divisor, thus increased, by t last fizure in the root, and subtract the product from ti dividend, and to the remainder bring down the ne period for a new dividend. | _ Fourthly—Double the root already found for a me divisor, and continue to operate as before, until all ti pertods are brought down. — | i It will sometimes happen, that, by dividing the divider as directed in the rule, the figure, obtained for the roo) will be too great. When this happens, take a less figur| and go through the operation again. When the places-in the decimal are not an even nun ber, they must be made so, by continuing the decime if it can be continued; if it cannot, by annexing a ciphe! that the periods may be full. r| If there be a remainder after all the periods are use a period of decimal ciphers may be added; or, if th! _ given number end ina decimal, the two figures that wou! arise from a continuation of the decimal. The operatic) may be thus continued to any degree of exactness. | If any dividend shall be found too small to contain th divisor, put a cipher in the root, and bring down the ne}| period to the right hand of the dividend for a new div’ dend, and proceed in the work. ¢ XXIX SQUARE ROOT. 167, _ When the square root of a mixed number is required, it will sometimes be necessary to reduce it to an improper fraction, or the vulgar fraction to a decimal, before ex- tracting the root. . _ If either the numerator or denominator of a vulgar frac- tion be not a square number, the fraction must be reduced to a decimal, and the approximate root extracted. i ) 1. Extract the square root of 4579600. 4579600(2140 Ans. ~~ 4 ( we siete so | Ist. divisor 4 57 first dividend. ' 4] 41 | Qd. divisor 42 1696 second dividend. oe 4.24 1696 | 7 ae 00 Tee. ‘ _ 2. What is the square root of 110 33! a ’ 110.24(10.499-++ Ans. I ie ime, sist. divisor, 2 10 first dividend. . Qd. divisor, 20 1024 second dividend. = 204 = 816 3d. divisor, 208 20824 third dividend. 2089 ~—«18801 ‘ Ath. divisor, 2098 202324 fourth dividend. 20989 188901 - 13423 remainder. Reducing 24 to a decimal, we found it to be infinite, inthe recurrence of 24 continually; therefore, in con- tinuing the extraction of the root, instead of adding eriods of decimal ciphers, we added the period 24 each time. The extraction of the root might have been con- tinued indefinitely; but having obtained five places of figures in the root, we stopped, and marked off the three last places of the root for decimals; because we made use of three periods of decimals in the question. . What is the square root of 2704? . Extract the square root of 361. a| . What is the square root of 3025! i ‘) . What is the square root of 121? | . Extract the square root of 289. . Extract the square root of 400. . What is the square root of 848241? ‘| . Extract the square root of 3356224. © « | . What is the square root of 824464 ? . Find the square root of 49084036. . What is the square root of 688900? . Find the square root of 82864609. . Find the square root of 3684975616. . What is the square root of 44890000? . What is the square root of 165649? . Find the square root of 90484249636. . Find the square root of 26494625227849. . Find the square root of 262400.0625. 3 . What is the square root of 841806.25 ? 4 . What is the square root of 39:037504? ~7 || . Find the square root of 213.715161. . Find the square root of .66650896. i . What is the square root of 13340752? | . What is the square root of 152; ? . Extract the square root of 31834. . What is the square root of 51,4, ? . Extract the square root of 2%. . What is the square root of 55634? . Extract the square-root of 109674. . Find the square root of 4120900. . Extract the square root of 5. , . Extract the square root of 8. . Extract the square root of 84. . Extract the square root of 99. . Extract the square root of 101. . Extract the square root of 120. | . Extract the square root of 124. a . Extract the square root of 143.. . Extract the square. root of 1.5. a ARITHMETIC. XXD What is the square root of 4761? xix. | SQUARE ROOT. 169 43. Extract the square root of .00032754. 44. Extract-the square root of 2.3. 45. Extract the square root of 2. 46. Extract the square root of 3. 47. Exxtract the square root of 2. | 48. Extract the syuare root of 32. 49. Extract the square root of 1132. 50. Extract the square root of 2673. The square root of the product of any two numbers is ‘a mean proportional between those numbers. Thus, 41s a mean proportional between 2 and 8; be- cause 2:44:86. But when four numbers are propor tionals, the product of the extremes is equal to the pro- ‘duct of the means;..that is, the product of the two given numbers is equal to the square of the mean proportional. 51. Find a mean proportional between 4 and 256. 52. Find a mean proportional between 4 and 196. 53. Find a mean proportional between 2 and 12.5. . 54. Find a mean proportional between 9.8 and 5. 55. Find a mean proportional between 25 and 121. 56. Finda mean proportional between 130 625 and 10. _ 57. Find a mean proportional between 52 and 545. * 58. Find a mean proportional between 2 and 34. 59. Find a mean proportional between 12 and 147. 60. Find a mean proportional between { and 4. 61. Find a mean proportional between .5 and 98. _ 62. Find a mean proportional between 40623;%, and 828. 63. Find a mean proportional between .25 and 1. 64. Find a mean proportional between .1 and S10. 65. Find a mean proportional between .04 and .36. 66. Find a mean proportional between .09 and .49. 67. Find a mean proportional between .2 and .018. When the square root of the product of the two givea numbers cannot be extracted without a remainder, the mean proportional is a’suRD, and may be approximated — by the aid of decimals. 68. Find a mean proportional between 6 and 12. 15 _ table purpose; each man gave as many cents, as there - ~ will it be ? 170 - ARITHMETIC. XXIX. 69. Find a mean proportional between 25 and 14. 70. Find a mean proportional between 64 and 21. 71.. Find a mean proportional between 46 and 55. 72. Find a mean proportional between 5 and 81. 73. Find a mean proportional between 77 and 19. 74. A number of men spent 1 pound 7 shillings in | company, which was just as many pence for each man, | as there were men in the company. How many were there ° | 75. A company of men made a contribution for a chari- were men in the company. The sum collected was 31 | dollars 36 cents. How many men did the company | consist of ? x 76. If you would plant 729 trees in a square, how | many rows must you have, and how many trees in a row ? | 77. A certain regiment consists of 625 men. How many must be placed in rank and file, to form the regi- ment into a square? 78. It is required to Jay out 40 acres of land in a square. Of what length must a side of the square be? — 79. It isrequired to lay out 20 acres of land in the form of a right angled parallelogram, which shall be twice | as long as it is wide. What will be its length and breadth? (See page 162.) 80. It is required to lay out 30 acres of land in the form of a right angled parallelogram, the length of which shall be three times the width. How long and how wide A TRIANGLE is a figure having three sides and three angles. When one of the angles is such as would form one corner of a square, the figure is calleda right-angled triangle, and the following propositions belong to it. PROPOSITION ist. The square of the hypotennedin is equal to the sum of the squares of the other two sides. | | ; | MXIX, SQUARE ROOT. © 171 | PROPOSITION 2d. The square root of the sum of the squares of the base and perpendicular is equal to the hypotenuse. _ PROPOSITION 3d. The square root of the difference of the squares of the hypotenuse and base is equal to the | perpendicular. _ PROPOSITION 4th. The square root of the difference of the squares of the hypotenuse and perpendicular is equal to the base. _ By observing the above. propositions, when any two _ sides of a right-angled triangle are given, we may always _ find the remaining side. For example, suppose the base of the preceding figure to be 4 yards in length, and the perpendicular to be 3 yards in height; then the square of the base is 16 yards, and the square of the perpendicu- lar 9 yards, and the sum of their squares is 25 yards. The square root of 25 yards is 5 yards, which is the length of the hypotenuse. 81. A certain castle, which is 45 feet high, is sur- rounded by a ditch, 60 feet broad. What must be the Jengih of a ladder, to reach from the outside of the ditch to the top of the castle ° 82. A ladder 40 feet long, resting on the ground at the distance of 24 feet from the bottom ofa straight tree, and leaning against the tree, just reaches to the first limb. What is the iength of the tree’s trunk ? 83. Two brothers left their father’s house, and went, one, 64 miles due west, the other, 48 miles due north, and purchased farms, on which they now live. How far from each other do they reside ” 84. James and George, flying a kite, were desirous of knowing how high it was. After some consideration, they perceived, that their knowledge of the square root, and of the properties of a right angled triangle, would enable them to ascertain the height. James held the line close to the ground, and George ran forward till he came directly under the kite; then measuring the distance from James to George, they found it to be 312 feet; and pulling in the kite, they found the length of line out, to be 520 feet. How high was the kite ? ail XXIX. 85. A ladder, 40 feet long, was so placed in a street, | as to reach a window 33 feet from the ground on one side, _ and when turned to the other side without changing the place of its foot, reached a window 21 feet high. The breadth of the street is required. ; 86. The distance between the lower ends of two equal_ rafters, in the different sides of a roof, is 32 feet, and the — height of the ridge above the foot of the rafters is 12. feet. Find the length of a rafter. 172 ARITHMETIC. A straight line, drawn [ through the centre of a square, Pa or through the centre of a right- angled parallelogram, from one angle to its opposite, is called’ a DIAGONAL; and this diagonal is the hypotenuse of both the right-angled triangles into w?.1ch the square or parallelogram is thus divided. i om NI 87. A certain lot of Jand, lying in a square, contains | 100 acres: at what distance from each other are the. opposite corners ? ; . 88. There is a square field contaming 10 acres: what is the distance of the centre from either corner ? = bounded by one curve line, called the circumference, every part of which is equally distant from the centre. = A straight line through the B= centre of a circle is called a = - diameter, and a straight line from the centre of a circle to SS SSS { ————————————————— = : the circumference is called a <= | radius. | pe eg | j ¥ [ A cIRCLE is a plane surface 3 ; The areas of all circles. are to one another, as the squares of their like dimensions. That is, the area of a greater circle is to the area of a less circle, as the square __ | ‘ XXIX. ~ SQUARE ROOT. 173 , of the diameter of the greater to the square of the diame- ter of the less. Or thus, the area of the greater is to the | area of the less, as the square of the circumference of _ the greater to the square of the circumference of the less. ) Therefore, to find a circle, which shall contain 2, 3, . 4, &c. times more or less space than a given circle, we have the followng— RULE. Square one of the dimensions of the given | eircle, and, if the required circle be greater, multiply the square by the given ratio, then the square root of the | product will be the like dimension of the requ red circle; but, if the required circle be less than the given one, | dwide the square by the given ratio; then the square root of the quotient will be the similar dimension of the circle required. 89. The diameter of a given circle is 11 inches: what is the diameter of a circle-containing 9 times as much space? 90. Find the diameter of a circle, which shall, contain one fourth of the area of a circle of 42 feet diameter. 91. What must be the circumference of a circular pond, to contain 4 times as much surface, as a pond, of 43 mile in circumference ? ) 92. Find the circumference of a pond which shall con- tain ; part as much surface, as a pond of 134 miles circumference. 93. Find the diameter of a circle, which shall be 36 times as much in area, as a circle of 184 rods diameter. The diameter of a circle is to the circumference in the ratio of 1-to 3.14159265, nearly: therefore, if we know _ the one, we can find the other. Thus, the circumference of a circle, the diameter of which is 8, is 3.14159265 X 8—25.1327412; the diameter of a circle, the circum- ference of which is 15.70796325, is 15.70796325 + 3. 14159265=5. To find the area of a cirlce, multiply the cireumference by the radius, and divide the product by 2. . 94. How many feet in length is the side of a square, equal ih area to a circle of peel diameter ? . 15 | * eo 174 ARITHMETIC. XXX » 95. Find the side of a square eaual in area to a circle | of 20 rods in diameter. a cl 96. Find the diameter of a pond, that shall contain 4 | as much surface, as a pond of 6.986 miles circumference. | 97. Find the length and breadth of a right-angled | parallelogram, which shall be 4 times as long as it is wide, | and equal in area to a circle of 43.9822971 rods circum- } ference. | 98. Find the circumference of a pond, which shall | contain as much surface, as 9 ponds of ¢ of a mile diam-_ | eter each. | XXX. EXTRACTION OF THE CUBE ROOT, A CUBE isrepresented by a solid block—like either of me those annexed—with six plane | surfaces; having its length, breadth, and height all equal. Consequently, the solid con- tents of a cube are found by ; multiplymg one of its sides Aa twice into- itself. For this Oe reason, the third power of any_ jij number 1s called a cube. al _ Therefore, if we multiply the square of a number by its om root, we obtain a product, which is called a cube, or a 7 cubic number. For instance, 4 multiplied by 4 produces 16, which is the square of 4, as shown on one of the : sides of this larger block; and 16 multiplied by 4 pro- | duces ef, which is the cube of 4, as shown by the whole of the larger block. 4 ‘Thus the cube of any quantity is produced by multiply- A ing the quantity by itself, and again multiplymg the pro- — - duct by the original quantity. When the quantity to be — Dice XXX. CUBE ROOT. 175 * cubed is a mixed number, it may be reduced to an im- proper fraction, and the fraction cubed, and then reduced back to a mixed number. _ As we can, in the manner explained, find the cube of a given number, so also, when a number is proposed, we ‘may reciprocally find a number, which being cubed will produce the given number. In this case, the number sought is called, in relation to the given number, the ‘cusE Root. ‘Therefore, the cube root of a given num- ber is the number, whose cube is equal to the given number. For instance, the cube root of 125 is 5; the cube root of 216 is 6; the cube root of } is 4; the cube root of 32 is 14. ; A cube cannot have more places of figures than triple the places of the root, and, at least, but two less ‘than triple the places of the root. - Take, for instance, a number consisting of any number of places, that shall be the greatest possible in those places, as 99, the cube of which is 970299; here the places are triple. Again, take a number, that shall be the least possible in those places, as 10, the cube of which is 1000; here the places are two less than triple. | . It is manifest from what has been said, that a cubic nuinber is a product resulting from three equal factors. For example, 3375 is a cubic number arising from 15 15X15. To investigate the constituent parts of this cubic number, we will separate the root, from which !t was produced, into two parts, and instead of 15, write 10-5, and raise it to the third power in this form. 10+ 5 | 10 -+ 5 Product of 10-+ 5 by 5, - - - - 50-+ 25 Product of 10-++-5 by 10, - - 100+ 50 The square, - - ae - 100+100 ++ 25 eth Gt aa ae Prod. of 100-100-++25 by5, - - 500+500--125 Prod. of 100-+100-+25 by 10, 1000+1000+-250° he third power, - - 1000-+-1500-+'750-}- 125 'e This product contains the cube of the first term, three ae | } ee ‘ - 3 % | wt a 176 ARITHMETIC. XXX) times the square of the first term multiplied by the sec- ond term, three times the first term multiplied by the square of the second term, and the cube of the second term: thus, 10 10x 101000; 10X10X3X 5=1500; 103 X 25 =750; 9X5 Xa = 125, Now, if the cube be given, viz. 1000-+ 1500-+-750- 125, and we are required to find its root, we readily perceive by the first term 1000, what must be the first term of the root, since the babe root of 1000 is 10; if, therefore, we Sables the cube of 10, which is 1000, from the given cube, we shall have for a remainder, 10X10X3X5 =1500, 10X3X25 =750, and 5X5X5 ‘==125; and from this remainder we must obtain the second term in the root. As we already know that the second term is 5, we have only to discover how it may be derived from the above remainder. Now that remainder may be expressed by two factors; thus, (10X10X3-—- 10X3X5-+5xX5) 5: therefore, if we divide by three) times the square of the first term of the root, plus threé times the first term multiplied by the second term, plus the square of the second term, ‘the quotient will be the; second term of the root, which is 5.. But, as the second term of the root is supposed to te unknown, the divisor also is unknown; nevertheless we have the first term of the divisor, viz. three ‘times the square of the root already found; and by means of this; we can find the next term of the root, and then complete’ the divisor, before we perform the caviar After find- ing the second term of the root, it will be necessary, im order to complete the divisor, to add thrice the product | of the two terms of the root, and the square of the sec- ond term, to three times the square of the first term eS ' viously found. The preceding analysis explains the fotlevsas tule a the extraction of the cube root. RULE. First—Point off the given number into periods) | of three figures each, beginning al the wait’s place, and pointing lo the left in integers, and to the right in deci- mals; making full periods of decimals by supplying the. deficiency, when any exists. : ws “ 4,5. 1 ,) «+ 1 ve (XXX. CUBE ROOT. 177 | @dly—Find the root of the left hand period, place it ‘im the quotient, and subtract its cube from the given ‘number. The remainder is a new dividend. | 3dly— Square the root already found and multiply its square by 3, for a divisor. ; ! Athly—Find how many times the divisor is contained in the dividend, and place the result in the quotient. | 5thly—mn order to complete the divisor, multiply the root previously found, by the number last put in the root, triple the product, and add the result to the divisor; also square the number last put in the root, and add tts square ‘0 the divisor. Lastly—JMultiply the divisor thus completed, by the WiAber last put in the root, and subtract the product from the dividend. The remainder will be a new dividend. Thus proceed, till the whole root ts extracted. We will extract the cube root of 34965783, denoting sach step of the operation, from first to last, by a refer-- ance to that part of the rule, under which it falls. | BADGRTeS ddly. Cube of 300, subtracted, - - 27000000(300 4 New dividend, - - - - - 7965783 3dly. 3003003 {adivisor| 270000 : {thly. Divisorinnewdividend, - - - - = (20 ithly. Triple prod.of 30020, 18000 | Square of last number, 400 Divisor completed, - 288400 Lastly. 288400 X 20, subt’ed, - - | 5768000 (3 New dividend, - - - - - 2197783 3dly. 3203203 [adivisor] 307200 tthly. Divisorinnewdividend, - - sive ithly. Triple prod. of 320X7, 6720 Square of last number, 49 Divisor completed, - 313969 ia 313969 X 7, subtracted, | 2197783 “ee © «© # ® 300 -+-20-+-7 —327 Ans. 178 ARITHMETIC. xXxy In completing every divisor, we have three produei to add together; viz. three times the square of the ro¢ already found; three times the product resulting from th multiplication of the reot already found, by the numbé Jast put in the root; and the square of the last number.*} If the ciphers be removed from the right hand of eae| of these products, the remaining figures in each succeed ing product will stand one place to the right of eac| preceding product; therefore, the work will be consider ably abridged by adopting the following— i RULE. First—Point off the given number into period} of three figures each, as before directed. ig 2dly—Find the root of the left hand period, place 4| in the quotient without regard to local value, and sub| tract its cube from that period; and to the remainder bring| down the next period for a dividend. a | 3dly—Square the root already found, without any re} gard to its local value, and multiply its square by 3, for a divisor. r] 4thly—Find how many times the divisor is containes in the diwidend, omitting the two right hand figures, ane place the result in the quotient. Sthy— To complete the divisor, multiply the root pre viously found, by the figure last placed in the quotient. without regarding local value, triple the product, ana write u under the divisor, one place to the right; square the figure last put in the quotient, and write its square under the preceding product, one place to the right. Add these three together, and their sum is the divisor comp lela | | | % Lastly—.Multiply the divisor thus completed, by the figure last placed in the root, and subtract the produgt from the dividend; and to the remainder bring down the next period for a new dividend. i Thus proceed, till the whole root is extracted. + Observe, that, when the divisor is not contained in the dividend, as sought in the fourth part of the rule, a cipher must be put in the root, and the next period brought down for anew dividend. Shi) e a sa nx Pe > ‘ iy ; | ix_xx. CUBE ROOT. 179 Observe, also, that when the figure obtained for the ‘root by dividing, as directed in the fourth part of the rule, jis found, on completing the divisor, to be too large, a ssmaller figure must be substituted in its place, and the divisor completed anew. _ There are always as many decimals in the root, as jperiods of decimals in the power. £ We will extract the cube root of 6589031 1319, in the abridged form; referring, as before, to the particular part of the rule, under which each step of the operation pro ceeds. First, 65890311319(4039 Qdly. Cubeof4,subt’d - - - (64 } Dividend, - - - - - 1890 Sdly. 4x4~x8[divr] - - 48 Athly. 48 was not con- ad tamed in 18. - - - | 0000 Lastly. New dividend, - - - | 1890311 Sdly. 40x40x3, - - 4800 4thly. 4800 in 18903, leg 3 times. dthly. Triple product i GE4A0 Sars Ob Square of 8, - - - 9 ie Divisor comp’d, - 483609 Lastly. 4836093, and subtracted, - - - - -| 1450827 New dividend, - - - 439484319 Sdly. 4034033, - .487227 4thly. 487227 in 4394 843, 9 times. Sthly. Triple product eae of 403x9, - 10881 Square of 9 - 81 Divisorcomp’d, 48831591 astly. 48331591 x 9, , andsubtracted, - - - 439484319 - Ans. 4089 180 ARITHMETIC. without reference to the parts of the rule. 178263.433152(56.28 Ans. | 125 : 75° | 53263 90 36 ~ 8436) 50616 0408 | 2647433 300 ; Ro Bucs, 944164) 1888328 947532 759105152 13488 64 94888144} 759105152 oie ie Wm 6 8 (ee ie: . Extract the cube root of 614125. i 2. What is the cube root of 191102976 ? 3. What is the cube root of 18399.744 ? 4. Find the cube root of 253895799552. 5. What is the cube root of 1740992427 ?- 6. Extract the cube root of 35655654571. 7. Find the cube root of 27243729729. 8. What is the cube root of 912673000000 ? 9. What is the cube root of 67518581248 ? XxX) We will now extract the cube root of 178263.43315: m the avridged form, as in the preceding example; | . Find the cube root of 729170113230343. . Extract the cube root of 643.853447875 Find the cube root of .000000148877. . What is the cube root of 123? . Extract the cube root of 517. . Extract the cube root of 900. . Extract the cube root of 74 . What is the cube root of $ : r . What is the cube root of 2 37° . What is the cube root of 725% ? . Extract the cube root of 26. | i | % . ¢ aoe ae —s = — nein — LS STI EA ~_ cs = om = \RXX. CUBE ROOT. isl | i To find two MEAN PROPORTIONALS between two given mumbers, divide the greater by the less, and extract the cube root of the quotient: then multiply the sube root by | the least of the given numbers, and the product will be the least of the mean proportionals; and the least mean proportional multiplied by the same root, will give the greatest mean proportional. 21. What are-the two mean proportionals between 6 and 750? _. 22. What are the two mean proportionals between 56 and 12096? To find the side of a cube equal in solidity to any given solid, extract the cube root of the solid contents of the given body, and it wiil be the required side. 23. There is a stone, of cubic form, containing 21952 solid feet. What is the length of one of its sides? 24. The solid contents of a globe are 15625 cubic inches: required the side of a cube of equal solidity. 25. Required the side of a cubical pile of wood, equal to a pile 28 feet long, 18 ft. broad, and 4 ft. high. . All solid bodies are to each other, as the cubes of their ‘diameters, or similar sides. 26. If a ball 6 inches in diameter weighs 32 pounds. what is the diameter of another ball of the same metal, weighing 4 pounds ? 27. If a ball of 4 inches diameter weighs 9 pounds, what is the diameter of a ball weighing 72 pounds ? 28. What must the side of a cubic pile of wood mea- sure, to contain 4 part as much as another cubic pile, _ which measures 10 feet ona side? 29. If 8 cubic piles of wood, each measuring 8 feet on a side, were all put into. one cubic pile, what would be the dimensions of one of its sides ? 30. The solid contents of a globe 21 inches in diame- ter are 4849.0596 solid inches; what is the diameter of a globe, whose solid contents are 11494.0672 inches ? 31. What are the inside dimensions of a cubical bin, that will hold 85 bushels of gram? (See note, page 27.) 32. What must be the inside dimensions of a cubical 16 ——— 182 ARITHMETIC. XXXII. bin, to hold 450 bushels of potatoes, 2815.489 cubic inches, (heaped measure), making a bushel ? 33. What must be the inside measure of a cubical | cistern, to hold 10 hogsheads of water ? 34. What must be the inside measure of a cubical cistern, that will hold 20 hogsheads of water ? | | 35. What are the inside dimensions of a cubical cis- tern, that holds 40 hogsheads of water? 36. Suppose a chest, whose length is 4 feet 7 inches, - breadth 2 feet 3 inches, and depth 1 foot 9 inches: what is the side of a cube of equal capacity ? 37. Suppose I would make a cubical bin of sufficient capacity to contain 108 bushels; what must be the dimen- sions of the sides ? : » XXXI. ROOTS OF ALL POWERS. The roots of many of the higher powers may be ex- tracted by repeated extractions of the square root, or cube root, or both, as the given power may require. Whenever the index of the given power can be resolved into factors, these factors denote the roots, which, being successively extracted, will give the required root. Thus, the index of the fourth power is 4, the factors” of which are 2X2; therefore, extract the square root of the fourth power, and then the square root of that square root will be the fourth root. The sixth root is the cube root of the square root, or the square root of the cube root; because 3 X2—6. The eighth root is the square root of the square root of the square root; be- cause 2X2 2=—8. The ninth root is the cube root of the cube root; because 3X 3—9. The tenth root is the fifth root of the square root; because 2K 5=10. The twelfth root is the cube root of the square root of the square root; because 2X 2X*3—12. The twenty- seventh root is the cube root of the cube root of the cube TOOL; BDECAUSE SX ho 0 faa | XXXII ROOTS OF ALL POWERS. 183. The following is a GENERAL RULE for extracting the roots of all powers. RULE. First—Prepare the givennumber for extraction, by pointing off from the unit’s place, as the required root directs; that is, for the fourth root, into periods of four figures; for the fifth root into periods of five figures, §c. 2dly.—Find the first figure of the root by trial, and _ subtract its power from the left hand period. 3dly.— To the remainder bring down the first figure in the next period for a dividend. 4thly.—Involve the root to the next inferior power to that which is given, and multiply it by the number denot- ing the given power, for a divisor. 5thly.—Find how many times the divisor ts contained in the dividend, and the quotient will be another figure of the root. | 6thly.— Involve the whole root to the given power, and subtract it from the two left hand periods of the given number. Lastly.— Bring down the first jigure of the next period to the remainder, for a new dividend, to which find a new divisor, as before. Thus proceed, till the whole root ts extracted. Observe, that when a figure obtained for the root by dividing, is found by involving, to be too great, a less figure must be taken, and the involution performed again. We will extract the fifth root of 36936242722357. 36936242722357(517 Ans. 55—. 3125 | 54 & 5 = 3125 first divisor. 5686 first dividend. Rit S45085251 Ben <5 ==33826005, second divisor. 943371762 2d. dividend. YW ie 369362427 22357: . What is the fifth root of 5584059449 . Find the fifth root of 2196527536224. . Extract the fifth root of 16850581551 ° . Find the seventh root of 2423162679857794647 - - m C9 to = 184 ARITHMETIC. XXXIV) XXXII. EQUIDIFFERENT SERIES. Aseries of numbers composed of any number of terms, which uniformly increase or decrease by the same num- | ber, 1s called aw EQUIDIFFERENT SERIES. ‘This series has, very commonly, but without any propriety, been | called Arithmetical Progression. | When the numbers increase, they form an ascending — series; but when they decrease, a descending series. | Thus, the natural numbers, 1, 2, 3, 4, 5, 6,7, 8,9, form | an ascending series, because they continually increase by _ 1; but 9, 8, 7, 6, &c. form a descending series, because | they continually decrease by 1. | ‘he numbers, which form the series, are called the terms of the series. The first and last terms in the series _ are called the extremes; and the other terms, the means. Lhe number, by which the terms of the series are continually increased or diminished, is called the common difference. Therefore, when the first term and comu:ion difference are given, the series may be continued to any — length. For imstance, let 1 be the first term in an equi- different series, and 3 the common difference, and we shall have the following increasing series; 1, 4, 7, 10,) 13, &c., in which each succeeding term is’ found by adding the common difference to the preceding term. THEOREM I. When four numbers form an equidiffer- ent serves, the sum of the two extremes is equal to the sum of the two means. Thus, 1, 3, 5, 7, is an equidifferent series, and 1+7==3-+-5. Also in the series 11, 8, 5, 2. To Be, THEOREM Il. Jn any equidifferent series, the sum of the two extremes is equal to the sum of any two means, that are equally distant from the extremes; and equal to double the middle term, when there is an uneven number of terms. Take, for example, the equidifferent series, 2, 4, 6, 8, 10, 12, 14;2+14—=4+412; and2+14= 6+10; also 2+14=—8-+18. XXXII. EQUIDIFFERENT SERIES. 185 Since, from the nature of an equidifferent series, the second term is just as much greater or less than the first, as the last but one is less or greater than the last, it is evident, that when these two means are added together, the excess of the one will make good the deficiency of the other, and their sum will be the same with that of _ the two extremes. In the same manner it appears, that the sum of any other two means equally distant from the extremes, must be equal to the sum of the extremes. THEOREM Ill. The difference between the extreme terms of an equidifferent series is equal to the common difference multiplied by the number of terms less 1. _ Thus, of the six terms, 2, 5, 8, 11, 14, 17, the common _ difference is 3, and the number of terms less 1, is 5; then _ the difference of the extremes is 17—2, and the common difference multiplied by the number of terms less 1, is i Xx 5; and 17—2=35. The difference between the first and last terms, is the increase or diminution of the first by all the additions or subtractions, till it becomes equal to the last term: and, as the number of these equal additions or subtractions is one less than the number of terms, it is evident that this common difference being multiplied by the number of terms less 1, must give the difference of the extremes. THEOREM Iv. The sum of all the terms of any equi- different series is equal to the sum of the two extremes 4 ft multiplied by the number of terms and divided by 2; or, which amounts to the same, the sum of all the terms is equal to the sum of the extremes multiplied by half the thenum- ber of terms. For example, the sum of the following geties, 2, 4, 6, 8, 10, 12, 14, 16, is 2116 X4=-72. This is made evident by writing under the given series the same series inverted, and adding the corresponding terms together as follows. The given series, 2, 4, 6,°°,10, 12 14,16. Same series inverted, 16,14,12,10, 8, 6, 4, 2. Sums of the series, 13,15,18, (8. tsoioe15,18 This series of equal terms, (18), is evidently equal tc twice the sum of the given series; but the sum of these 16% | IgG ARITHMETIC. XXXII | LY equal terms is 18 X 8== 144; and since this sum is twice: | as great as that of the given series, the sum of the given series must be 72. | Any three of the five following things being given, the | other two may be readily found. The first term. The last term. The number of terms. | The common difference. The sum of all the terms. PROBLEM. I. The extremes and number of terms be- ing given, to find the sum of all the terms. } RULE. Multiply the sum of the extremes by the num-_ ber of the terms, and half the product will be the sum | of all the terms. See Theorem 4th. | 1. The first term in an equidifferent series, 1s 3, the last term 19, and the number of terms is 9. What is the | sum of the whole series ? 2. How many strokes does a common clock strike in 12 hours ? ) | 3. A hundred cents were placed in a right line, a yard | apart, and the first a yard from a basket. What distance | did the boy travel, who, starting from the basket, picked them up singly, and returned with them one by one to | the basket? q 4. If a number of dollars were laid in a straight line | for the space of a mile, a yard distant from each other, _ and the first a yard from a chest, what distance would | the man travel, who, starting from the chest, should pick | them up singl~, returning with them one by one to the | chest ? , 3! PROBLEM Il. The extremes and number of terms — given, to find the common difference. | RULE. Subtract the less extreme from the greater, and | divide the remainder by the number of terms less 1, and the quotient will be the common difference. | It has been shown under Theorem 3d. thatthe differ- t : 4 » XXXIT = EQUIDIFFERENT SERIES. 187 | ence of the extremes is found by multiplying the common difference by the number of the terms less 1; conse- quently, the common difference is found by dividing the Biference of the extremes by the number of the terms. less 1. 5. A man had 10 sons, whose ages differed alike; the youngest was 2 years old, and the eldest 29. What ,was the difference of their ages? - | 6. The extremes in an equidifferent series are 3 and $7, and the number of terms 43. Required the common difference. __ 7. A man is to travel from Boston to a certain place in 9 days, and to go but 5 miles the first day, and tc ‘Increase his journey every day alike, so that the last day’s journey may be 37 miles. Required the daily increase, and also the number of miles travelled. _ PROBLEM IN. The extremes and common difference given, to find the number of terms. RULE. Divide the difference of the extremes by the common difference, and add 1 to the quotient; the sum will be the number of terms. ‘ The difference of the extremes divided by the number of the terms less 1, gives the common difference; con- ‘sequently, the same divided by the common difference must give the number of terms less 1: hence, this quotient augmented by 1, must give the number of terms. 8. The extremes in an equidifferent series are 3 and 39, and the common difference is 2: what is the number -of terms? _ 9. A man going a journey, travelled 7 miles the first day, and increased his journey every day by 4 miles, and the last day’s journey was 51 miles. How many days did he travel, and how far ? *10. A mancommenced a journey with great animation, and travelled 55 miles the first day; but on the second day he began to be weary, and travelled only 51 miles, ‘and thus continued to lose 4 miles a day, till his las: day’s journey was only 15 miles. How many days did he travel ? ‘| | ‘| | | 188 ARITHMETIC. XXXII. PROBLEM IV. 'To find an equidifferent mean between two given terms. RULE. Add the two given terms together, and half their swin will be the equidifferent mean required. | 11. Find an equidifferent mean between 3 and 15. 12. What is the equidifferent mean between 7 and 53? 13. Find an equidifferent mean between 5 and 18. i PROBLEM V. ‘To find two equidifferent means between | the given extremes. } | RULE. Dwwide the difference of the extremes by 3, and. the quotient will be the common difference, which, being continually added to the less extreme, or subtracted, from | the greater, gives the two required means. | ij 14, Find two equidifferent means between 4 and 13. ° 15. Find two equidifferent means between 5 and 22. | 16. Find two equidifferent means between 4 and 53. | PROBLEM VI. ‘To find any numbers of equidifferent means between the given extremes. | RULE. Divide the difference of the extremes by the re- quired number of means plus 1, and the quotient will be the common difference, which being continually added to | the less extreme, or subtracted from the greater, will give | | the mean terms required. 17. Find five equidifferent means between 4 and 28. 18. Find six equidifferent means between 6 and 55. _ | 19. Find 3 equidifferent means between 34 and 142. 29. Find one equidifferent mean between 56 and 100. XXXII. ' 7 CONTINUAL PROPORTIONALS. The numbers of a series in which the successive terms: increase by a common multiplier, or decrease by a com- mon divisor, are CONTINUAL PROPORTIONALS. : This series of numbers has been commonly called a Geometrical Progression; but, perceiving no appropriate | | ; XXXHT. CONTINUAL PROPORTIONALS. 189 meaning in this term, we choose to call the series, what ‘itis in truth, a series of Continual Proportionals. __ The common multiplier, or common divisor, by which: the successive terms are increased or deminished, is ealled the ratte of the series, or the common ratio. Ue 2 3 4) 5 6 ’ | Thus, 3, 6, 12, 24, 48, 96 is a series of continual pro- portionals, in which each successive term is produced by multiplying the preceding term by 2, which is the com- lion ratio. The numbers 1, 2, 3, 4, &c. standing above the series, mark the place, which each term holds — in the series. Also, 729, 243, 81, 27,9, 3, 1, is a series of continual proportionals, in which each successive term is found by dividing the preceding term by 3, the common ratio. In an increasing series, the ratio is the quotient, which results from the division of the consequent by the ante- cedent; but in a decreasing series, the ratio is the quo- tient resulting from the division of the antecedent by the consequent. In every series of continual proportionals, any four successive terms constitute a proportion. Thus, in the first of the above series, 3: 6=12: 24, and 6: 1224: 48, also, 12: 24—48:96. In the second series, Oey eto 3): 27, 2435 81 Rte Oy oli: QI 9 3, 27: 9==3:1. Therefore, when there are only four erms, the product of the extremes is equal to the pro- duct of the means. Furthermore, in any series of continual proportionals, he product of the extremes is equal to the product of my two terms equally distant from them; and equal to he second power of the middle term, when there is an meven number of terms. For instance, take the con- inual proportionals 2, 4, 8, 16, 32, 64, 128; then 2x '28—=4 X 64; also 2X 128=8 X 32; and 2K 128=— 16 X 16. When the first term and the ratio are given, a series pf continual proportionals may be extended to any num ver of terms by continually multiplying by the ratio ir mM increasing series, or dividing in a decreasing series 190 ARITHMETIC. XXXII For example, the first term being 2, and the ratio 3, if we make it an increasing series by continually multiplying by the ratio, we obtain the following series, 2, 6, 18, 54, 162, 486, which may be extended to any number of terms; but, if we make it a decreasing series by continu! ally dividing by the ratio, we obtain the following series, 2, 3) 3) He) Siro a¥z) Which may also be extended to any number of terms. In the series 2, 6, 18, 54, 162, 486, we obtain the| second term by multiplying the first term by the ratio; the third term by multiplying the second term by the ratio; the fourth term by multiplying the third term by the ratio; the fifth term by multiplying the fourth term by the first term by the fifth power of the ratio. The fifth power of 3 is 243, which being multiplied into the first term, the result is 486, the same as in the series. 4 Hence we see, that any term in any increasing series of continual proportionals may be found by multiply iti the first term by that power of the ratio, which is denoted. by the number of terms preceding the required one.: For instance, the ninth term in an increasing series is found by multiplying the first term by the eighth power | of the ratio; thus let 2 be the first term, and 3 the com>: mon ratio; then 235 gives the ninth term, which is| 13122. . a If the series be a decreasing one, any term in it may be found py dividing the first term by that power of the ratio; which is denoted by the number of terms preceding. the required one. For instance, the seventh term in a decreasing series is found by dividing the first term by the sixth power of the ratio; thus, let 24576 be the first. term in a decreasing series, and 4 the common ratio;| then 24576 + 4° gives the seventh term, which is 6. a We will now state several problems, which occur iP Sa hig proportionals, and give the rules for performing: them. | | —— aus : XXXIIl. CONTINUAL PROPORTIONALS. 191 PROBLEM I. ‘The first term and the ratio being given, to find any other proposed term. RULE. Raise the ratio to a power, whose index is equal to the number of terms preceding the required term: then, uf it be an increasing series, multiply the first term by this power of the ratio; but, if it be a decreasing series, divide the first term by it: the result will be the required term. | 1. Required the eighth term in an increasing series whose first term is 6, and ratio 2. * 2. Required the ninth term in a decreasing series, the ‘first term of which is 131072, and the ratio 4. _ 8. What is the seventh term in an increasing series, ithe first term being 3, and the ratio 1.5? __ 4, What is the sixth term in an increasing series, whose first term is z¢357, and ratio 7 ? _ 5. What is the tenth term in a decreasing series, the first term being 387420489, and the ratio 9? | i | | } One of the principal questions, which occurs in a series of continual proportionals, is to find the sum of the series. We shall, therefore, illustrate the method. Let there be given the following series, consisting of seven terms, whose common ratio is 3; viz. 2, 6, 18, 54, 162,486, 1458. Let each term in this series be multiplied by the ratio 3; and let each product be removed one place to the right of the terms in the given series. The given series, 2,6, 18, 54, 162, 486, 1458 multiplied by ratio. 6, 18, 54, 162, 486, 1458, 4374. Now the last term in the second series is produced by multiplying the last term in the given series by the ratio; and it is evident that if the given series be subtracted from the second series, the remainder will be the last erm in the second series diminished only by the first ‘erm in the given series, and this remainder will be twice the sum of the given series; consequently, if we divide t by 2, the quotient will be the sum of the given series out 2 is the ratio less 1. Hence _ PROBLEM II. The extremes and the ratio being given, ‘o find the sum of the series. e 192 ARITHMETIC. XXXI0 RULE. Multiply the greater extreme by the ratio, fron the product subtract the less extreme, und divide the re mainder by the ratio less 1, and the quotient will be th sum of the series. 6. The first term in a series of continual proportional) is 1, the last term is 65611, ana the ratio is 3. What 1) the sum of the series ? wae 65611 < 3= 196833 | 1 ratio 3—1=2)196832 98416 Ans. 7. The extremes of a series of continual proportional are 3 and 12288, and the ratio is 4. What is the sun of the series ° 8. The first term in a series of continual proportional in 12500, the last term is 4, and the ratio 5. What i) Hy sum ef the series? | . The first term in a series of continual proportional Me the last term 1792, and the ratio is 2. What is the sum of the series? 10. The extremes ima series of continual proportional are 5 and 37.96875, and the ratio 1s 1.5. _ What is the sum of the series ? f 11. The first term ina series of contmual proportional: is 100, the last term .O1, and the ratio 2.5. Requing the sum of the series. | 7 a | | ‘| | | PROBLEM I. ‘The first term, the ratio, and the num: ber of terms given, to find the sum of the series. ! RULE. Find the last term by problem 1, and the su of the series by problem 2. | 12. The first term in an increasing series of continua} proportionals i is 6, the ratio 4, and the number of terms 8. What is-the sum of the series ? | 13. The first jenn in an increasing series of f contiaill proportionals is 1, the ratio 4, and the number of terms 13. Required the sum of the series. . 14. The first term in a decreasing series of contin proportionals i is 1, the ratio 3, and the number of term 12. What is the sum of the series? “ -——_— XXXII. CONTINUAL PROPORTIONALS. 193 15. A man offers to sell his horse by the nails in his shoes, which are 32 in number. He demands one mill . for the first nail, 2 for the second, 4 for the third, and so on, demanding for each nail twice the price of the pre-’ / ceding. It is required to find what would be the price of the horse. 16. An ignorant fop wanted to purchase an elegant house, and a facetious man told him he had one, which he would sell him on these moderate terms; viz. that he , should give him one cent for the first door, 2 for the —S _ second, 4 for the third, and so on, doubling the price for | every door, there being 36. It is a bargain, cried the simpleton, and here is a half-eagle to bind it. What was the price of the house? PROBLEM IV. The extremes and the number of terms being given, to find the ratio. RULE. Divide the greater extreme by the less, and the quotient will be that power of the ratio, which is denoted by the number of terms less 1; consequently, the corres- ponding root of this quotient will be the ratio. This problem is the reverse of problem 1, and the reasoning which precedes that problem, ae netnay eluci- _ dates the rule in this. The first term im a series of proportionals is 192, the Jast term 3, and the number of terms 7. What is the ratio? 192-3—64; the number of terms less 1, is 6; there- | fore the sixth root of 64, which is 2, 1s the ratio. 17. Ina series of continual proportionals, the first /term is 7, the last 45927, and the number of terms, 9. _ What is the ratio? | | 18. The first term in a series of continual proportion- als is 26244, the last term 4, and the number of terms 5. Required the ratio. 19. The first term ina series of continual proportionals is 1, the last term 1029427, and the number of terms 8. , What is the ratio? | ; | 20. The first term ina series of continual proportion- als is 78125, the last term z45, and the number of terms 11. Required the ratio. 17 194 ARITHMETIC. XXXL! PROBLEM V. ‘To find any number of mean proportion-_ als between two given numbers. ; RULE. The two given numbers are the extremes of a series consisting of two more terms than there are means required; hence the ratio will be found by problem 4.) Then the product of the ratio and the less extreme will be one of the means; the product of this mean and the ratio will be another mean; and so on, till all the required means are found. ' When only one mean is required, it is the square root} of the product of the extremes. . | 21. Find 3 mean proportionals between 5 and 1280. | Here the series is to consist of five terms, and the) extremes are 5 and 1280; hence the ratio is found by the’ fourth problem to be 4; and by the repeated multiplica=| tion of the least term by the ratio, the means are found to be 20, 80, and 320. . 22. Find four mean proportionals between 4 and 2401. 23. Find five mean proportionals between the num- bers, 279936 and 6. 24. Find a mean proportional between 1 and 2809. COMPOUND INTEREST BY SERIES. It has been shown in ArT. xv, page 107, that cora- pound interest is that which arises from adding the interest to the principal at the end of each year, and taking the | amount for a new principal. Now, the several amounts! for the several years form a series of continual propor- tionals; and, to find the amount for any number of years, we may adopt the following— RULE. Find the last term of an increasing series of continual proportionals, whose first term is the principal, whose ratio is the amount of 1 dollar for 1 year, and whose’ number of terms is the number of years plus1. The last term is the required amount. See Problem Ist. In the examples under this rule, no more than six deci- _ mal places need be included. ] 25. What is the amount of $100, at 6 per cent. | compound interest, for 4 years ? | f ) pi, XXXII CONTINUAL PROPORTIONALS. 195 J ) 26. What is the amount of ‘75, at 5 per cent. com- _ pound interest, for 9 years? 27. What is the amount of §294, at 4 per cent. com- pound interest, for 7 years? 28. Find the amount of $18.25, at 7 per cent. “compound interest, for 12 years. 29. Find the amount of $751.30, at 5 per cent. compound interest, for 8 years. 30. Find the arhount of $4798, at 6 per cent. com- pound interest, for 12 years. 31. What is the amount of $5.14, at 7 per cent. compound interest, for 16 years? What is the interest ? 32. What is the compound interest of $1000 for 20 years, at 6 per cent.? What is the amount? + COMPOUND DISCOUNT. Discount corresponding to simple interest has already been treated, in Art. xvi; but discount corresponding to compound interest is now to be computed. On the supposition that money can be let out at com- pound interest, the present worth of a debt, payable at a future period without interest, is that principal, which, at compound interest, would give an amount equal to the debt, at the period when the debt is payable. RULE. Find the last term of a decreasing series of continual proportionals, whose first term is the debt, whose ratio is the amount of 1 dollar for 1 year, and whose number of terins is the number of years plus 1. The last term is the present worth. See Problem Ist. 33. What principal, at 10 per cent. compound interest, will amount, in 4 years, to $8.7846? 34. What is the present worth of $68.40, payable 11 years hence; allowmg discount according to 5 per cent. compound interest ? 35. What is the present worth of $350, ls in 5 years; allowing discount at the rate of 6 per cent. com- pound interest ! > 36. What is the present worth of $3525, ae in 3 years; discount being allowed as in the last example ? 196 ARITHMETIC. XXXIV 37. How much must be advanced to discharge a debt of $700, due in 8 years; discounting at the rate of 5 per. cent. compound interest ? 38. What. is the present worth of $1000, due in 20 years; discounting at the rate of 6 per cent. compound interest? How much is the discount ? | XXXIV. ANNUITIES. An ANNUITY is a fixed sum of money payable periodi- cally, for a certain length of time, or during the life of | some person, or for ever. } Although the term annuity, in its proper sense, applies only to annual payments, yet payments which are made semiannually, quarterly, monthly, &c., are also called annuities. Pensions, salaries, and rents, come under the head of annuities. Annuities may, however, be purchased by the present payment of a sum of money. The party selling annuities, is usually an incorporated trust company, insti- tuted and regulated upon principles similar to those of an insurance company. ‘The company has an office, called an annutty office, where all its business is transacted. The present worth of an annuity which is to continue for ever, is that sum of money, which would yield an in- terest equal to the annuity. But the present worth of an annuity which is to terminate, is a sum, which, being put on compound interest, would, at the termination of the annuity, amount to just as much as the payments of the annuity would amount to, provided they should severally be put on compound interest, as they became due. . The sum to be paid for the purchase of a life annuity — which is the same as its present worth— depends not only upon the rate of interest, but, also upon the probable con- tinuance of the life or lives on which the annuity is grant- ed. In order to bring data of this kind into numbers, the bills of mortality in different places have been examined, ee | XXXIV. ANNUITIES. 197 and from them, tables have been constructed, which show how many persons, upon an average, out of a certain number born, are left alive at the end of each year; and from these tables others have been constructed, showing the expected continuance of human life, at every age, according to probabilities. We shall not, however, treat the subject of life annuities in this work, and would refer : readers, who wish to become thoroughly acquainted with its theory, to the writings of. Simpson, De Moivre, Bai- ley, Price, and Milne. : PROBLEM I. ‘To find the amount of an annuity, which has been forborn for a given time. Before presenting the rule, let us inquire what would be the amount of an annuity of $100, forborn 4 years, allowing 5 per cent. compound interest? The last year’s payment will, obviously,.be $100 without interest; the last but one will be the amount of $100 for 1 year; the last but two will be the amount of $100 for 2 years; and so on: and the sum of the amounts will be the answer. Now the last payment with the amounts for the several years, form a series of continual proportionals. We, therefore, adopt the following— RULE. Find the sum of an increasing series of con- -tinual proportionals, whose first term is the annuity, whose ratio is the amount of 1 dollar for 1 year, and whose number of terms is the number of years. This sum is the amount. See Art. xxxit1, Problems 1st and 2nd. 1. What is the amount of an annuity of $200, which has been forborn 14 years; allowing 6 per cent. interest ? 2. What is the amount of an annuity of $50, which has been forborn 20 years; interest being 5 per cent.? 3. What is the amount of an annual rent of $150, for born 7 years; allowing interest at 5 per cent.? 4. If an annual rent of $1054 be in arrears 4 years - what is the amount, allowing 10 per cent. interest ? 5. Suppose a person, who has a salary of $600 a year, payable quarterly, to allow it to remain unpaid for 3 years, how much would be due him; allowing quarterly com- pound interest at 6 per cent. per annum? Vie 198 ARITHMETIC, XXXIV.” 6. What is due on a pension of $150 a year, payable half-yearly, but forborn 2 years; allowing half-yearly com- pound interest, at 44 per cent. per annum? 7. What is due on a pension of $300 a year, payable quarterly, but forborn 24 years; allowing quarterly com- pound interest, at.5 per cent. per annum. PROBLEM II. To find the present worth of an annuity which is to terminate in a given number of years. Before giving the rule, let us inquire, what is the pres- ent worth of an annuity of $100, to continue 4 years, allowing 5 per cent. interest? The present worth is, obviously, a sum, which, at compound interest, would produce an amount equal to the amount of the annuity. Now we can find the amount of any sum at compound interest, by multiplying the sum by the amount of 1 dollar for a year, as many times as there are years. Hence, to find a sum which will produce a given amount in a given time, we must reverse the process, and divide by the amount of 1 dollar for the time. Applying this principle — | to the example in question, we find by the preceding rule, that the amount of the annuity is $431. Then, dividing this amount by the amount of 1 dollar for 4 years, we find the present worth to be $354.593-+ RULE. Find the amount of the annuity as if it were in arrears for the whole time, and divide this amount by the — amount of 1 dollar at compound interest for the same time; the quotient will be the present worth. 8. What is the present worth of an annuity of $500, to continue 10 years; interest being 6 per cent.? 9. What is’ the present worth of an annuity of $80, to continue 22 years; interest being 5 per cent.? The operations in this rule being tedious, we introduce, upon the next page, a table, showing the present worth of $1 annuity, at 4, 5, 6, and 7 per cent., for every number of years, from 1 to 30. To find the present worth of an annuity by the use of this table, multiply the present worth of 1 dollar for the number of years, by the annuity. i ) ‘XXXIV. Y’rs.|4 per cent. 7 per cent. — | SO Ss | Ea _———_ 10 12.1656 12.6593 13.1339 13.5903 14.0291 14.4511 14.8568 15.2469 15.6220 "15.9827 16.3295 16.6630 16.9837 17.2920 ANNUITIES. 5 per cent. |6 per cent. .9523 9433 1.8594 1.8333 2.723 2.6730 3.5459 3.4651 4.3294 4.2123 5.0756 4.9173 5.7863 5.5823 6.4632 6.2097 fall © 6.8016 T7217 7.3600 8.3064 7.8868 8.3632 8.3838 9.3935 8.8526 9.8986 9.2949 10.3796 9.7122 10.8377 | 10.1058 11.2740 | 10.4772 11.6595 | 10.8276 L2:5883" |) hin set 12.4622 | 11.4699 12.8211 | 11.7640 13.1630 | 12.0415 13.4885 | 12.8033 13.7986 | 12.5503 _ 14.0939 | 12.7838 14.3751 | 13.0031 14.6430 | 13.2105 14.8981 | 13.4061 15.1410 | 18.5907 15.3724 | 13.7648 10. What is the present worth of an pari of $21.54, for 7 years; interest being 6 per cent.? 11. What is the present worth of an annuity of 4 936, 'e° 20 years, at 5 per cent.? 12. What is the present worth of an annuity of $258. for 17 years, at 4 per cent.? i 200 ARITHMETIC. XXXIV) 13. Find the present worth.of an annuity of $'796.50,| to continue 28 years; interest being 7 per cent.? | 14. A young man purchases a farm for $924; and) agrees to pay for it in the course of 7 years, paying 4 part of the price at the end of eachyear. Allowing inter- est to be 6 per cent., how much cash in advance will pay the debt’. ee | 15. Allowing interest to be 5 per cent., which will be} in my favor, to pay $15 a year for 10 years, or, to pay! $ 160 in advance ?—by how much? When an annuity does not commence until a given! time has elapsed, or some particular event has taken place, it is called a REVERSION. | _ PROBLEM Ht. ‘To find the present worth of an annuity) in reversion. RULE. Find, (by Problem 2nd.), the present value of | the annuity from the present time till the end of the period! of its continuance: find, also, its value for the time be- fore it is to commence: the difference of these two results’ will be the present worth. | | 16. What is the present worth of an annuity of $200, | to be continued 7 years, but not to commence till 2 years | hence; interest being 6 per cent.? . ~ FI 17. Find the present worth of a reversion of $152 a| year, to commence in 6 years, and to continue 18 years; | interest being 4 per cent. | As. What is the present worth of a reversion of $75 | a year, to commence in 5 years, and to continue 24) years; interest being 6 per cent.? - 19. What must be paid for the purchase of a reversion of $450 a year, to commence in 5 years, and to continue | 13 years; interest being 5 per cent.? | 20. Find the present worth of a reversion of $942.30 | a year, to commence in 2 years, and to continue 11 | years; interest being 7 per cent. . | 21. A father leaves to his son, a rent of $310 per’ annum, for 8 years, and, the reversion of the same rent — to his daughter for 14 years thereafter. What is the | present worth of the legacy of each, at 6 per cent.? © | We ‘ : * i XXXV. ALLIGATION. 201 | 22. What is the present worth of a reversion of $100 ayear, to commence in 4 years, and to continue for ever; interest being 6 per cent.? _ This annuity continuing for ever, will, when it com- mences, be worth that sum of money which would yield $100 a year, at 6 per cent. interest. Therefore, after finding the principal, whose interest is $100 per annum, deduct from it a compound discount for 4 years; the re- mainder will be the present worth. / | 23. What is the present worth of a reversion of $824 a year to commence in 7 eas and to continue for ever; mterest being 5 per cent.? _ 24. What is the present worth of a reversion of $530 ayear, to commence in 22 years, and to continue for ever; interest 7 per cent.? _ 25. How much must be paid, at present, for a share ma fund, which, after the lapse of 20 years, will yield in income of $400 a year; interest 6 per cent.? 26. How much must be paid, at present, for the title ‘oan annuity of $1000, to commence in 40 years; inter- ast being 5 per cent.? ; XXXV. ALLIGATION. ALLIGATION relates to finding the mean value of a mix- ure composed of several ingredients of different values, ind is considered under two heads, viz. Alligation Medial, md Alligation Alternate. ALLIGATION MEDIAL. We rank under the head of Alligation medial, those juestions, in which the several ingredients and don re- ‘pective values. are given, and the mean value of the -ompound is required. For example, a wine merchant bought several kinds of vine, as follows; 160 gallons at 40 cents per gallon; 74 q | 202 ARITHMETIC. XXXYj gallons at 60 cents per gallon; 225 gallons at 48 cenit per gallon; 40 gallons at 85 cents per gallon; and mixer| them together. It is required to find the cost of a gallor of the mixture. i | Now, if we find the whole cost of the several kind: of wine, and divide it by the whole number of gallons, i| is evident, that the quotient will be the cost of a single gallon of the mixture. | 160 gallons, at 40 cents per gal., cost § 64.00 75 gallons, at 60 cents per gal., cost $ 45.00 225 gallons, at 48 cents per gal., cost $108.00 | 40 gallons, at 85 cents per gal., cost $ 35.00 500 the whole number of gallons, cost $252.00 $ 252.00 + 500 =.504, or 5 cents and 4 mills. - Therefore, to find the mean value of a compound, composed of several ingredients, of different values, we give the following | 5 RULE. Find the value of each ingredient, add these values together, and divide their sum by the sum of the ingredients. The quotient is the mean value. 7 4 1. A farmer mixed together 5 bushels of rye worth 70 cents a bushel, and 10 bushels of corn worth 60 cents a bushel, and 5 bushels of wheat worth $1.10 a bushel, What is a bushel of the mixture worth ? ka 2. A grocer mixed together 38]b. of tea at 50 cents a pound, 15; 1b. at 80 cents, 124]b. at 60 cents, 831b. at 96 cents, 77]b. at 32 cents, and sold the mixture at a) profit of 20 per cent. At what price per pound did he. sell it ? . i 3. A goldsmth melts together 11 ounces of gold 23. carats ane, 8 ounces 211 carats fine, 6 ounces of pure gold, and 2 ounces of alloy. How many carats fine is, the mixture? - 4 We remark, that a carat is a 24th part. Thus, 23 carats fine, means 33 of pure metal. Pure gold is 34. ; Alloy is considered of no value. >| 4. On a certain day, the mercury in the thermometer . was Observed to stand, 2 hours at 60 degrees, 3 hours at | XXXV. ALLIGATION ALTERNATE. 203 '§2°, 4 hours at 64°, 3 hours at 67°, l hour at 72°, and i" es at 75°. _ What was the mean temperature for that - ay: a 5. A dealer bought 245 gallons of sirop at 34 cents a gallon, and 243 gallons at 38 cents a gallon, and mixed oth quantities “and 14 gallons of water together, and sold ‘the mixture at a profit of 50 per cent. At what price per gallon did he sell it ? 6nA goldsmith melts together 3 ounces of gold 18 earats fine, 2 ounces 21 carats fine, and 1 ounce of pure la. What is the fineness of the compound ? ALLIGATION ALTERNATE. Under the head of Alligation Alternate are included those questions, in which the respective rates of the dif- ferent ingredients are given, to compose a mixture of a fixed rate. It is the reverse of Alligation Medial, and may be proved by it. . If we would find what quantities of two ingredients, different in value, would be required to make a com- pound of a fixed value, it is evident, that, when the value of the required compound exceeds that of one ingredient just as much as it falls short of the value of the other, we must take equal quantities of the ingredients to make the compound; because there is just as much lost on the one, as is gained on the other. If the value of the compound exceeds that of one in- gredient twice as much as it falls short of the value of the other, we must take ofthe ingredients in the ratio of $ to +, or 1 to 2. For instance, if we would mix wines, at 4 dollars and 1 dollar a gallon, in such proportion that the mixture should be worth 2 dollars a gallon, we must take 1 gallon at 4 dollars to 2 gallons at 1 dollar; because there is just as much lost on 1 gallon at 4 dollars, as is gained on 2 gallons at-1 dollar. If we would mix wines, at 6 dollars and 2 dollars a gallon, in such proportion as would make the mixture worth 3 dollars a gallon, we should take of the two kinds | 204 ARITHMETIC. XXXV_ { in the ratio of } to 4, or 1 to 3; for, in this instance, there is as much lost on 1 gallon at 6 dollars, as is gainec on 3 gallons at 2 dollars. a | We see by the preceding ratios, that the nearer the value of the mixture is to that of one of the ingredients,| the greater must be the relative quantity of this ingredi-| ent, in forming the compound; and the farther the value of the mixture is from that of one of the ingredients, the’ less must be the -relative quantity of this ingredient in| making the compound. 4 Hence, if we make the difference between the rate of each ingredient and that of the compound, the denomina- tor of a fraction having | for its numerator, these fracuons| express the ratio of the ingredients required to make the compound; and, when these fractions are reduced to ¢ common denominator, the numerators express the requir ed ratio of the ingredients. | If, for example, it be required to mix gold of 12 caratg| fine with gold of 22 carats fine, in such proportion thai| the mixture may be 18 carats fine, we can ascertain the| ‘proportion of each kind in the following manner. The difference between 18 and 12 is 6; making 6 the de- nominator of a fraction with 1 for its numerator, we have the fraction 3; taking the difference between 18 and 22, we in like manner obtain the fraction 4; therefore, | | fractions, ¢ and 4, express the required proportion of each sort of gold. These fractions, when reduced to a common denominator, are 34 and 34, and the numerators| express the required proportion of each sort. Therefore, we must take 4 grains of 12 carats fine, and 6 grains of 22 carats fine; or, in that ratio. | If, for a second example, we would make a mixture {8 carats fine from gold of 15 carats and 20 carats fine, we should, in the same manner, obtain the fractions, 2 and 2, to express the required proportion of the two sorts of gold; consequently, in this instance, we should take 2 grains of 15 carats fine, and 3 grains of 20 carats fine. Therefore, since the fineness of the compound is the same in both the preceding examples, if we would make a compound 18 carats fine, from the four kinds of gold XXXV. ALLIGATION ALTERNATE. 205 mentioned in the two examples, we should take 4 grains of 12 carats, 6 grains of 22 carats, 2 grains of 15 carats, and 3 grains of 20 carats fine. : _ Now, these results may be readily obtamed by writing _ the rates of the given simples one under another, in reg- ular order, beginning either with the least or greatest, and alligating one of a less with one of a greater rate than that , of the compound, and writing the difference between | the rate of each simple and the rate of the compound, _ against the rate of the simple with which it is alligated. _ Thus, |12—— 4 grains 12 carats fine { 15 2) CC 15 CC C6 | Sag] 3 «20 t ee a eA De, We may connect the rates of the simples differently, and obtain equally correct, but different results. Thus, |12—, 2 7 ‘ 19/281 § 23 3 It must be observed, that the two simples linked to- _ gether, must always be one of a less, and the other of a greater rate, than the rate of the compound. : By connecting a less rate with a greater, and placing the differences between them and the mixture rate alter- nately, the gain on the one is precisely balanced by the loss on the other. This being true of every two, it is true of all the simples in the question, whatever may be their number. It is obvious, that a question in Alligation Medial admits : of a great variety of answers, all agreeing with the requi- _ sition of the question; for we may variously alligate the _ values of the ingredients, and thus obtain various results, all of which will be correct; and we may add all these _ together, and the results will be correct answers. We may _-also multiply, or divide the quantities found; for, if two quantities of two simples make a balance of loss and gain in relation to the value. of the compound, so must also the double or treble, the half or third part, or any other ratio of the quantities. 18 } { 206 ARITHMETIC. XXXV. : We shall give the questions in Alligation Alternate: under four cases. 4 CASE I. The ratios of the several ingredients being given, to make a compound of a fixed rate. | RULE. First— Write the rates of the several ingredients | in a column under one another. i | 2dly— Connect with a continued line the rate of each, ingredient less than the rate of the compound, with one or more rates greater than the rate of the compound; and each of a greater rate than the rate of the compound with. one or more of a less rate. | 3dly— Write the difference between the rate of each ing | gredient and the rate of the compound, opposite the rate | of the ingredient with which it is connected. q Athly—If only one difference stand against any rate, ut will be the required quantity of the ingredient of that, rate; but, if there be several, their sum will be the quan-— tity required. | 7. A goldsmith has gold of 17, 18, and 22 carats fine, , and also pure gold. What proportion of each sort must he take, to compose a mixture 21 carats fine? ) 8. Having gold of 12, 16, 17, and 22 carats fine, what proportion of each kind must I take, to make a compound 18 carats fine ? 9. A merchant has spices at 30, 33, 67, and 86 cents a pound. How much of each sort must he take, to make & mixture worth 56 cents a pound ? 10. A wine merchant has Canary wine at 50 cents a gallon, Sherry at 76 cents, and Claret at 175 cents per gallon. How much of each sort must he take, to make a mixture worth 87 cents a gallon? | 11. A goldsmith wishes to mix gold of 16, 18, 19, and 23 carats fine, with pure gold, in such proportions that the composition may be 20 carats fine. What quan- tity of each must he take ? 12. It is required to mix different sorts of wine, at 56, 62, and 75 cents per gallon, with water, in such propor- tions that the mixture may be worth 60 cents a gallon. How much of each must be taken ? ‘XXXV. ALLIGATION ALTERNATE. 207 : | 13. How much corn at 52 cents a bushel, rye at 56 eents, wheat at 90 cents, and wheat at 1 dollar a bushel, must be mixed together, that the composition may be ‘worth 62 cents a bushel ? 14. A silversmith wishes to mix alloy with silver of ‘10, and 7 ounces fine, and pure silver, in such proportion ‘that the mass may be 9 ounces fine: 12ozs. fine being pure. How much of each must he take? CASE II. When one of the ingredients is limited to a ‘oertain quantity. RULE. Find the quantity of each ingredient, as in Case 1st. in the same manner, as though no such limitation ‘were made; then as the difference against that simple, whose quantity is given, is to each of the other differences, $0 is the given quantity of that simple to the quantity re- “quired of each of the other simples. 15. A trader has 90 pounds of tea worth 40 cents a : pound, which he would mix with some at 50 cents, some at 85 cents, and some at 90 cents. How much of each of the ciiier sorts must he mix with the 90 pounds, to make a mixture worth 60 cents a pound ? First solution. Second solution. 40—— as 40— 25 9|2°— 50—|-, 30 60135 | 4 60 85— | 20 9O0—— i | 90.4) mus 30 : 25—90:75 thus 25 : 3090 : 108 a0 > 10— 90°: 30 25. % 20=3 90" --72S 30 : 20=—90 : 60 25-1 0==90 + ..36 30]b. at 85 cents, and 72\b. at 85 cents, and 60 pounds at 90 cents. 36 pounds at 90 cents. 16. A farmer wishes to mix corn at 54 cents a bushel, rye at 61 cents a bushel, and wheat at 96 cents a bushel, with 3 bushels of wheat worth I dollar and 10 cents a bushel. How much of each of the other three must be mixed with the 3 bushels of wheat at 1 dollar and 10 cents a bushel, that the mixture may be worth 75 cents a bushel ? Ans. 75 lb. at 50 cents, Sah 108 lb. at 50 cents, 208 ARITHMETIC. | XXXV. 17. How much gold of 16, 20, and 24 carats fine, and how much alloy, must be mixed with 10 ounces of 18 carats fine, that the composition may be 22 carats fine ? 18. How much silver of 6.5 ounces fine, and of 10.5 ounces fine, and alloy, must be mixed with 17.1 ounces of pure silver, that the mass may be 9.5oz. fine? It must be observed, that pure silver is 12 ounces fine. CASE 111. When two or more of the ingredients are limited in quantity. . RULE. Find, as in Alligation Medial, what will be the | rate of a mixture made of the given quantities of the lim- | ited ingredients only; then consider this as the rate of a limited ingredient, whose quantity is the sum of the quan- tities of the limited ingredients, from which, and the rates of the unlimited ingredients, proceed to calculate the | several quantities required, as in Case 11. . ~19. T have 18 gallons of wine at 48 cents a gallon, 8 gallons at 52 cents, and 4 gallons at 85 cents, and would mix the whole with two other kinds of wine, one at a dollar and 26 cents, the other at 2 dollars and 12 cents a gallon. How much of the wine at a dollar and 26 cents, and of that at 2 dollars and 12 cents, must I mix with the other three, that the mixture may be worth a dollar a gallon ? 18 gal. at .48- come to $8.64 ’ ‘> ggal. at 62°“ 4.16 - 4gal. at .85 3 3.40 The 30 gal. come to 16.20, which is .54 a gallon. 54 cents a gallon being the mean value of the 30 gallons, contained in the three kinds that are limited, I must now inquire how much of each of the other two sorts of wine at 1 dollar 26 cents, and 2 dollars 12 cents, must be mixed with 30 gallons at 54 cents a gallon, to make a mixture worth one dollar a gallon. | 54—, 26-4119 138 - POD iat. |-vi mie a aie Pohang gal. gal. gal. — gal. ' Now as 188 : 46—30: 10 ) XXXV. ALLIGATION ALTERNATE. 209 Therefore, I must take 10 gallons each of the two _ sorts, which are worth 1 dollar 26 cents, and 2 dollars 12 cents a gallon. 20. How much gold of 14 and 16 carats fine must be _ mixed with 6 ounces of 19, and 120z. of 22 carats fine, that the composition may be 20 carats fine ° 21. A silversmith has silver of 6, 7, and 9 ounces fine, _ which he wishes to mix with 9 ounces of 10 ounces fine, and 9 ounces of pure silver, to make a mass, that shall be 8 ounces fine. How much of each of the three first must he take? 22. A lady purchases 7 yards of eelcor at 22 cents a _ yard, and 7 yards at 20 cents a yard, and wishes to know _ how many yards of two other kinds, one at 16 cents and the other at 17 cents a yard, she must purchase, to make the average price of the whole 18 cents a yard. Find the two quantities. CASE IV. When the whole compound is limited to a certain quantity. RULE. Find an answer, as in Case 1, by alligating then, as the sum of the quantities thus found, is to the given quantity, so is the quantity of each ingredient found by alligating, to the required quantity of it. 23. A goldsmith has gold of 15, 17, 20, and 22 carats fine; and would melt together of all these sorts so much, as to make a mass of 40 ounces 18 carats fine. How much of each sort is required ? 15—— E Or thus we. jd ee “aca - sae 3 22—— j A) Torres phon ; 10:40 == 4 216 10:40=—2: 8 10:40—2: 8 10 : 40=—4: 16 10:3.40=1;.4 10.: 40==3.2.12 10 : 40=3: 12. 10:40—1: 4 Ans. 1602. of 15; 802. of 17; 402. of 20; and 1202. | of 22 carats fine. 24. Having three sorts of raisins at 9, 12, and 18 cents is* ILO ARITHMETIC. XXXVI a pound, what quantity of each sort must I take, to fill a cask containing 210 pounds, that its contents may be | worth 14 cents a pound? 25. Of four different kinds of apples at 31, 37, 46, and | 74 cents a bushel, what quantity of each must be taken, to fill a bin containing 9 bushels, to make its contents worth 50 cents a bushel? XXXVI. PERMUTATIONS. PERMUTATION—which is also called variation—means the different ways in which the order or relative position of any given number of things may be changed. The only object to be regarded in Permutation, is the order in which the things are placed; for no two arrangements are to have all the quantities in the same relative position. For example, two things, a, and b, are capable of only two changes in their relative position, viz. ab, ba; and this number of changes is expressed by 1X2; ‘but three things, a, b, and c, are capable of six variation: viz. abc,acb, bac, bca,cab, cba, and this number of permutations is expressed by 123; and four things, a, b, c, and d, are capable of 24 variations, viz, abc d, - abdc,acbd;acdb, adbct, adc b; 6b aca, bes de, bead, becda, bdac, bdca; cabd;,cadp, c ba die b'da) cid. ab; c:d boas da 0 cy 2 eee ac, dbca, dcab, sidan: and this number of per- mutations 1s expressed by A Mehl ote In like manner, when there are 5 things, every four of them, leaving out the 5th, will have 24 variations; con- ———— sequently by taking in the 5th, there will Ve 5 times 24 _ variations. PROBLEM I. ‘To find the number of permutations that can be made of any given number of things, all different from each other. RULE. Multiply the terms of the A Ee series of ; | 4 ] } ! va ; 5 . | XXXVI. PERMUTATIONS. | 211 numbers, from 1 up to the given number of things, con- -_tinually together, and the product will be the answer. 1. How many changes can be made in the order of the six letters, abc def? 2. How many changes may be rung on seven bells ? 3. Five gentlemen agreed to board together, as long as they could seat themselves every day in a different position at the dinner table. How long did they board together ? 4. How many changes may be made in ‘the order of the words in the following verse? Proci tot tibi sunt, virgo, quot sidera coelo. _ 5. How many different sums of dollars can be expres- sed by the nine digits, without using any one of them more than once in the same sum ? 6. How many different arrangements may be made in seating a class of 20 scholars ? 7. A gentleman, who had a wife and eight daughters, one day said to his wife, that he intended to arrange the family in a different order every day at the dinner table, and that he would never give one of his daughters in marriage, till he had completed all the different arrange- ments of which the family was capable. How many years from that day must elapse, before either of his daughters can be married ? _ When several of the things are of one sort, and several of another, &c. the changes that can be made upon the whole is not so great, as when all the things are different. For instance, we have seen that the letters a b c admit of six variations; but, if two of the quantities be alike, as aa b,-the six variations are “~duced to three, aa b, baa, aba, which may be expressed by =X". We have also seen that the letters ab c d admit of 24 variations; but if we have aa b b, the 24 variations are reduced to six, im. aabwv,abba,abab, bbaa, baab, bab a, ‘and this number of variations may be expressed by (1xX2x3xK4 : Sele, ae. Beoxciscr. “Hence, we have, as follows,— | | * | 212 ARITHMETIC. XXXVI PROBLEM I. To find the number of changes that may | be made in the arrangement of a given number of things, whereof there are several things of one sort, several of another, &c. | RULE. Take the natural series of numbers from 1 up to the given number of things, as if they were all differ- ent, and find the product of the terms. Then take the natural series from 1 up to the number of similar things of one sort, and the same series up to” the number of similar things of a second sort, §c., and divide the first product by the joint product of all these series, and the quotient will be the answer. 8. Find how many changes can be made in the order of the letters a aa bbc. If the letters in this question were all different, they would admit of 1X2X3xX4x5xX6 =720 variations; but since a is found 3 times, we must divide that number of variations by | <2X3; and, since b occurs twice, we must again aivide by 1X2; therefore the number of variations will be ee = 60. ; 9. How: many changes can be made in the order of the lettersaaabbbbecdee? 10. How many variations may take place in the suc- cession of the following musical notes, fa, fa, fa, sol, sol, la, mi, fa? 11. How many whole numbers can you make out of — the number 1220055055, using all the figures each time? 12. How many variations can be made in the order of the figures in the number 97298279289 ? PROBLEM IH. Any number of different things being given, to find how many changes can be made out of them, by taking a given number of the things at a time. RULE. Take a series of numbers commencing with the given number of things and decreasing by 1, till the number of terms is equal to the number of things to be taken at a time, and the product of all the terms of this’ series will be the answer. | To illustrate the rule, we will take the four letters abed, and find the number of variations that can be “XXXVIL COMBINATIONS. 213 made upon them, by taking two at atime. In the first place, we will write the letter a on the left hand of each of the other letters, and the variations will be three, viz. ab, ac, ad; we will do the same with each of the other letters, thus, b a, bc, bd; ca,cb,cd; da,db,dc. Now we have all the changes that can be made upon the four letters, taking two at a time, and they are 4X3=12. __ We will also find, in the same manner, how many changes can be made on the same four letters, by taking three at a time; writing @ on the left, thus, a b c, a b d; achb,acd; adb,adc, we have 3X2=6 variations. Now, since each of the letters is to be written in the same manner on the left, we shall have four such classes of variations, and the whole number will be 4x*3X*2—=24 variations. __ 13. How many changes can be made upon the letters abcdef, by taking three at a time? 14. How many different whole numbers can-be ex- pressed by the nine digits, by using two at a time? 15. How many different whole numbers can be ex- pressed by the nine digits, by using four at a time? 16. How many different numbers can you express ‘with the nine digits and a cipher, by using five at a time? : XXXVII. COMBINATIONS. CoMBINATION consists in taking a less number of things out of a greater without any regard to the order in which they stand. This is sometimes called Election or Choice. No two combinations can have the same quantities; for imstance, the quantities, a and b, admit of only one com- bination, because a b and b a are composed of the same quantities; but, if a third quantity ¢ be added, we can imake three combinations of two quantities out of them, ‘because the third quantity e may be added to each of the two former, thus, a b, a c, b c; this number of combina- : 3 tions may be expressed by — If we adda fourth letter, Ae i | { 214 ARITHMETIC. XXXVI d, we can make six combinations of two letters out of the | four, since the new quantity d, may be combined with. each of the former ones; thus, ab, ac, bc, ad, bd, cd; and this number of combinations may be expressed | 4x3 by Tyas : If we would make a combination of four, it is evident. that only one such combination can be made out of the. letters a bc d; but if a fifth letter, e, be added, we can | make five such combinations; thus, abcd, abce,' abed,aecd,bede; and this number of combina-— 5X 4X 3X2 tions may be expressed by 337: PROBLEM 1. ‘To find the number of combinations from | any given number of things, all different from each other, taking a given number at a time. | RULE. Take a series of numbers, the first term of which | is equal to the number of things out of which the combi-_ nations are to be made, and decreasing by 1, till the num~ ber of terms is equal to the number of things to be taken | at a time, and find the product of all the terms. Then take the natural series 1, 2, 3, &c. up to the | number of things to be taken at a time, and find the pro duct of all the terms of this series. 4 Divide the former product by the latter, and the quo- trent will be the answer. a 1. How many combinations of 3 letters can be made out of the 6 letters abc def? 4 2. How many different yoke of oxen may be selected _ from twelve oxen? . 3. How many different span of horses can be selected from eighteen horses ? é | 4. A drover agreed with a farmer for a dozen sheep, — to be selected out of a flock of two dozen; but while he; was making the selection, the farmer told him, he might, take the whole flock, if he would give him a cent for. every different dozen that could be selected from it. To | this the drover readily agreed. How many dollars did | the whole flock come to, at that rate ? 5. A general, who had often been successful in war, Se — er ea XXXVII. COMBINATIONS. 215 was asked by his king, what reward he should confer upon him for his services. The general only desired a farthing for every file, of 10 men in a file, which he could make with a body of 100 men. How much did the general’s modest request amount to ? PROBLEM H. ‘To find the various combinations of a given number of things, which may be made out of an equal number of sets of different things, one from each set. RULE. Multiply the number of things in the several sets continually together, and the product will be the answer. ; A combination of this kind is called the composition of quantities. The rule may be illustrated thus. If there are only two sets, and we combine every quantity of one set with every quantity of the other set, we shall make all the compositions of two things in these two sets; and the number of compositions is evidently the product of the number of things in one’ set by the number of things in the other set. Again, if there are three sets, then the compositions of two in any two of the sets, being com- bined with every quantity of the third set, will make all the compositions of three in the three sets. That is, the compositions of two in any two of the sets, being multi- plied by the number of things in the third set, will give all the compositions of three in the three sets; and this 1esult is the joint product of all the numbers in the three sets. , 6. Suppose there are four companies, in each of which there are 9 men; in how many ways can 4 men be chosen, one out of each company ? 7. Suppose there are five parties, at one of which there are 6 young ladies, at another 8, at a third 5, at a fourth 7, at a fifth 10. How many choices are there, in selecting 5 young ladies, one from each party? _ 8. How many changes are there in throwing fou dice, each die having six sides ? : 9. A certain farmer has 5 barns, in one of which he has 15 cows, in another 11, in another 5, in another 9, 216 ARITHMETIC. XXXVI. and in another 7. How many different selections may be made, in choosing 5 cows, one from each barn ? 10. How many variations can be made in selecting a flock of a dozen sheep from 12 folds, one from every) fold, in each of which there are 10 sheep? 11. In a certain school there are seven classes, the first containing 12 boys, the second 7, the third 9, the fourth 10, the fifth 11, the sixth 8, and the seventh 13. How many variations can be made in selecting 7 boys, one from each class ? XXXVIII. EXCHANGE. | j ' : | | | | | | | | Scholars, who are to prosecute a course of classical studies, and those, who are not expected to engage in any extensive mercantile business, may omit the exercises in this article. ExcuancE is the act of paying or receiving the money of one country for its equivalent in the money of another country, by means of Bills of Exchange. 'This operation, therefore, comprehends both the reduction of moneys and the negotiation of bills; it determines the comparative value of the currencies of different nations, and shows how foreign debts are discharged, and remittances made from one country to another, without the risk, trouble, or expense of transporting specie or bullion. A Bill of Exchange is a written order for the payment of a certain sum of money, at an appointed time. It is | a mercantile contract, in which four persons are mostly concerned, as follows. First—The Drawer, who receives the value, and is.also called the maker and seller of the Bill. Second—The debtor in a distant place, upon whom the Bill is drawn, and who is called the Drawee. He also is called the Acceptor, after he accepts the Bil, which is an engagement to pay it when due. Third—The person who gives the value for the Bill, who is called the Buyer, Taker and Remitter. | XXXVILII. EXCHANGE. Q17 I Fourth—The person to whom the bill is ordered to be | paid, who is called the Payee, and who may, by endorse- | ment, pass it to any other person. ' Many mercantile payments are made in Bills of Ex- | change, which pass from hand to hand, until due, like any other circulating medium; and the person who at any ) time has a Bill in his possession, is called the holder. _ ~-To transfer a Bill payable to order, the payee should express his order of paying to another person, which 1s always done hy an endorsement on the back of the Bill. _ An endorsement may be blank. or special. A blank endorsement consists only of the endorser’s name, and the Bill then becomes transferable by simple delivery. A special endorsement orders the money to be paid to a particular person, who is called the endorsee, who must also endorse the Bill, if he negotiates it. A blank en- -dorsement may always be filled up with any person’s fame, so as to make it special. Any person may endorse -a Bill, and every endorser, as well as the acceptor, is a “security for the Bul, aud may be sucd for payment. In reckoning when a Bill, payable after date, becomes “due, the day on which it is dated, is not included. When the time is expressed in months, calendar months are un- derstood; and when a month is longer than the succeed- ag it js a rule not to go, m the computation, into a third ‘month. Thus, if a Bill be dated the 28th, 29th, 30th, ‘or 31st, of January, and payable one month after date, the term equally expires on the last day of February. ' An endorsement may take place at any time after the Bill is issued, even after the day of payment is elapsed. ' When the holder of a Bill dies, his executors may en- dorse it; but, by so doing, they become answerable to their endorsee personally, and not as executors. A Bill payable to bearer is transferred by simple deliv- ery, without any endorsement. _ Bills should be presented for acceptance, as well as | for payment, during the usual hours of business. The common way of accepting a Bill is for the drawee to write his name at the bottom or across the body of it, with the word, accepted. 19 | 218 ARITHMETIC. XXX VII When acceptance or payment has been refused, the Holder of the Bill should give regular and immediate notice to all the parties, to whom he intends to resort for payment; for if he do not, they will not be liable to pay. | With respect to the manner, in which notices of non- acceptance or non-payment are to be given, a difference. exists between Inland and Foreign Bills. | In the case of Foreign Bills, a Protest is indispensa-_ bly necessary: thus, a Public Notary appears with the Bill, and demands either acceptance or payment (as the case may be;) and on being refused, he draws up an in- strument, called a Protest, expressing that acceptance or payment (as the case may be) has been demanded and. refused, and that the holder of the Billintends to recover any damages which he may sustain in consequence. This instrument is admitted, in foreign countries, as a legal _ proof of the fact. The Protest on a Foreign Bill should be sent as soon as possible, to the drawer or negotiator; and if it be for non-payment, the Bill must be sent with the Protest. A Protest is not absolutely necessary to entitle the holder to recover the amount of an Inland Bill from the drawer or endorser: it is sufficient if he give notice, by letter or otherwise, that acceptance or payment (as the case may be) has been refused, and that he does not mean to give credit to the drawee. If the person, who is to accept, has absconded, or cannot be found at the place mentioned in the Bill, Pro- test is to be made, and notice given, in the same manner as if acceptance had been refused. It is customary, as a precaution against accident or miscarriage, to draw three copies of a Foreign Bill, and to send them by different conveyances. They are denomi- nated the First, Second, and Third of Exchange; and when any one of them is paid, the rest become void. When acceptance is refused, and the Bill is returned by Protest, an action may be commenced immediately against the Drawer, though the regular time of payment be not arrived. His debt, in such case, is considered as contracted the moment the Bill is drawn. XXXVIII. EX CHANGE. 219 FORM OF A BILL OF EXCHANGE. New York, Nov. 4, 1834. Excuance for £3000 sterling. At thirty days sight of this, my first of Exchange, (second and third of the same tenor and date not paid), pay to Robert N. Foster, or order, Three Thousand Pounds Sterling, with or without further advice from me. ' Enwin D. Harper. Messrs. Knox and Farnuam, Merchants, London. InLanD Excuance relates only to remitting Bills from one commercial place to another in the same country; by which means debts are discharged more conveniently than by cash remittances. Suppose, for example, A of New Orleans is creditor to B of Boston 1000 dollars, and C of New Orleans is debtor to D of Boston 1000 dollars; both these debts -may be discharged by means of one Bill. Thus, A draws for this sum on B, and sells his Bill to ©, who remits it to D, and the latter receives the amount, when due, from ‘B. Thus, by a transfer of claims, the New Orleans debtor pays the New Orleans creditor, and the Boston debtor the Boston creditor, and no money is sent from “one place to the other. This business is usually con- ducted through the medium of Banks, which are in the habit of buying both foreign and inland Bills of exchange, and transmitting them to the places on which they are drawn for acceptance. Inland Bills of Exchange are sometimes called Drafts; and the following short form of the instrument is adopted. : : | : : $ 56070, Boston, Dec. 8, 1834. Three months after date, pay to the order of Charles S. Hooper, Five hundred dollars 50 cents, value received, and charge the same to account of ' Hannum & Lorine. To SteruEN FroTHINGHAM, Merchant, Norfolk. 220 ARITHMETIC. XXX VITL | Some explanation of mercantile language used in rela-_ tion to Bills of Exchange seems necessary, that the | learner may have a clear idea of the questions, which will be given for practice. When a merchant in the United States draws on his | banker in London, his draft is styled ‘‘ Bill on London” or ‘‘ United States on London;’’ and if he sells his Bill at more than adollar for 54d. sterling, the exchange is — said to be above par; and if ‘he sells at less than a dollar | for 54d. sterling, below par: if a merchant in London draws on his banker in the United States, his draft is styled ‘‘ London on United States;’’ and if he sells his— Bilkat more than 54 d. sterling for the dollar, the exchange is said to be above par; and if he sells at less than 54d. sterling for the dollar, below par. / . If the merchant in London draws on his banker in Paris, it is ** London on Paris,’’ or ‘‘ London on France.” If the merchant in Charleston, S. C. draws on his banker in New York, it is ‘‘ Charleston on New York.”’ &c. | GREAT BRITAIN. {n Great Britain, accounts are kept in pounds, shillings, pence, and farthings, Sterling. The par value of the United States dollar is 4s. 6d. sterling; therefore, the dollar is equal to 7 of a pound sterling. Hence, any sum of sterling money, (the shil- lings and pence, if any, being expressed in a decimal,) may be reduced to Federal money, by multiplymg by 40 — and dividmg by 9: and, any sum in Federal money may be reduced to sterling money, by multiplying by 9 and dividing by 40. Or, if sterling money be increased by 4 of itself, the sum expresses the same value in the old currency of New England: and, conversely, if the old currency of New England, be decreased by i of itself, the result is the expression in sterling money. @ 1. United States on London. Reduce £784 14s. _ 103d. sterling to Federal money, at par. 2. London on United States. Reduce 3487 dollars 75 cents to sterling, at par. : | XXXVIIL. EXCHANGE. 221 8. United States on London. Reduce £2006 11s. sterlmg to Federal money; exchange at 4 per cent. below par. 4. London on United States. Reduce 4287 dollars 50 cents to sterling; exchange at 4 per cent. above par. 5. London on United States. Reduce 3646 dollars 50 cents to sterling; exchange at 2 per cent. below par. 6. United States on London: Reduce £4109 Ils. 10d. sterling to United States currency; exchange at 7 per cent. above par. 7. United States on London. Reduce £5129 15s. 6d. sterling to Federal money; exchange at 5 per cent. above par. The law assimilating the currency of Ireland to that af England, took effect in January 1826. ~ All invoices, contracts, &c. are considered there, in law, British cur- rency, unless otherwise expressed. 8. United States on Dublin (Ireland). Reduce £1834 2s. 104d. sterling to Federal money; exchange at 4 per cent. above par. FRANCE. Accounts were kept in France previous to 1795, ac- cording to the old system, in livres, sous, and deniers. 12 deniers= 1 sol or sou; 20 sous =1 livre; 6 livres ‘ = 1 ecuor crown, silver. By the new system, accounts are kept in francs, decimes, and centimes. 10 centimes = 1 decime. 10 decimes = 1 franc. The value of the franc is 182 cents in Federal money. 80 francs ==81 livres. 9. United Stateson France. Reduce 7232 francs 38 centimes to Federal money; exchange at 1 dollar for 5 francs 30 centimes. 10. France on United States. .Reduce 4093 dollars 80 cents to money of France; exchange at 5 francs 30 ~ centimes for the dollar. ioe pd). ARITHMETIC. XXXVIIL.- 11. France on United States. Reduce 1834 dollars 65 cents to French currency; exchange at 5 francs 40— centimes for a dollar. 12. United States on France. Reduce 20828 frances 67 centimes to Federal money; exchange at 1 dollar for — 5 francs 38 centimes. 13. United States on France. Reduce 12893 francs © 27 centimes to Federal money; exchange at 1 dollar for 5 francs 33 centimes. HAMBURGH. Accounts are kept here in marks, scnillings or sows, and pfenirgs, Lubs. 12 pfenirgs = 1 sol or schilling, Lubs; 16 schillings—=1 mark, Lubs; 3 marks 1 reichsthaler or rix dollar specie. Accounts are also kept, particularly in exchanges, in pounds, shillings, and pence, Flemish. 12 pence or grotes= 1 shilling. 20 shillings =1 pound, Flemish. The word Lubs originally meant money of Lubeck, which is the same with that of Hamburgh, and the term is intended to distinguish this money from the Flemish denominations, and also from the money of Denmark and ~ other neighboring places. The mark Lubs is worth 22 shillings Flemish, or 32 grotes; consequently the sol Lubs is 2 grotes Flemish, and the shilling Flemish 6 schillings Lubs. Banco or bank money, in which exchanges are reckon- ed, and curréncy, are the two principal kinds of money. Banco consists of the sums of money deposited by merchants and others i the bank, and inscribed in its — books; which sums are not commonly drawn out, but are transferred from one person to another in payment of a debt or contract. Current money, or currency, consists of the common ~ coins of the city, in which expenses are mostly paid. _ The bank money is more valuable than currency, and ~ bears a premium varying from 18 to 25 per cent. This ~ XX XVIII. EX CHANGE. 223 premium is called the agio. For instance, when the agio is 20 per cent. 100 marks banco are valued at 120 marks currency. The mark banco is valued in the United States at 334 cents. 14. United StatesonHamburgh. Reduce 1148 marks, 5 schillings, 4 pfenirgs banco to Federal money; exchange at 33 cents per mark banco. 15. Hamburgh on United States. Reduce 1245 dol- lars 75 cents to money of Hamburgh; exchange at 3 marks banco per dollar. 16. United StatesonHamburgh. Reduce 6194 marks 12schillings banco to Federal money; exchange at 34 cents per mark banco. 17. United Stateson Hamburgh. Reduce8246 marks 8 schillings banco to Federal money; exchange at 35 cents per mark banco. 18. Hamburgh on United States. Reduce 757 dollars 90 cents to money of Hamburgh; exchange at 1 mark banco for 33 cents. AMSTERDAM AND ANTWERP. In these places accounts were formerly kept in florins, Stivers, and penning; or in pounds, shillings, and pence, Flemish. 16 pennings= 1 stiver, 20 stivers ==1 florin or guilder. In Flemish, 12 grotes or pence, or 6 stivers— 1 shilling, 20 shillings, or 6 florins ==I pound; — 24 florins, or 50 stivers = I rix dollar. By the new system, adopted in 1815, accounts are kept throughout the kingdom of the Netherlands in florins or guilders, and cents. . . 100 cents 1 florin or guilder. The par value of the florin, in Federal currency, is 40 cents. 19. United States on Amsterdam. Reduce 13790 ‘florins 15 stivers to Federal money; exchange at 36 cents per florin. 224 ; ARITHMETIC. XXX VILL. | 20. United States on Antwerp. Reduce 6281 florins 88 cents to Federal money; exchange at 40 cents per florin. | 21. Amsterdam on United States. Reduce 2482 dol-_ lars 334 cents to Dutch money; exchange at 36 cents per florin. 22. Antwerp on United States. Reduce 3436 dollars 72 cents to Dutch money; exchange at 38 cents per florin. . 23. United States on Antwerp. Reduce 7294 florins 50 cents to Federal money; exchange at 42 cents per florin. : 24. United States on Amsterdam. Reduce 10148 florins to Federal money; exchange at 41 cents per florin. PORTUGAL. In Portugal, accounts are kept in milrees and rees; and also in old crusados. | 1000 rees—=1 milree. . 400 rees=1 old crusado or crusado of exchange. 480 rees—=1 new crusado. There are three sorts of money used in Portugal; viz. effective money, i. e€. specie; paper money, which is ata discount; and legal money, consisting of half specie and half paper. The value of the milree in Federal money is 1 dollar 24 cents. d 25. United States on Lisbon. Reduce 964 milrees 475 rees to Federal money; exchange at 1 dollar 24 cents per milree. 26. Lisbon on United States. Reduce 1274 dollars 66 cents to money of Portugal; exchange at 1 dollar 25 cents per milree. | 27. United States on Portugal. Reduce 1248 milrees” 645rees to Federal money; exchange at 1 dollar 26 cents per milree. a 28. United States on Portugal. Reduce 1846 milrees 500 rees to Federal money; exchange at 1 dollar 23 cents per milree. XXXVIII. EXCHANGE. 225 The exchanges of Brazil, in South America, are similar to those of Portugal; there is, however, a difference in the value of their moneys; that of Portugal is half specie and half paper, called legal money, and that of Brazil is effective. SPAIN. The most general mode of keeping accounts in Spain is in reals, of 34 maravedis; but there are nine different reals, each divided into 34 maravedis, but differing in value. Four of these reals are of general application, and five of local use. . The four principal moneys of Spain are the real vellon, the real of old plate, the real of new plate, and the real of Mexican plate; and in order to obtain a distinct view of them, it may be proper to make the real vellon the basis of all the rest. The real vellon is the twentieth part of the hard dollar, (peso duro), universally known by the name of the Span- ish dollar, which is the same in value with the dollar of the United States. The division of the real vellon is into quartos, ochavos, and maravedis. 2 maravedis—1 ochavo, 2 ochavos =1 quarto, 8 quartos, or 34 maravedis=1 real vellon; but maravedis are commonly used to express any fraction ofa real: thus, we say 1 real 33 maravedis. _ The real of old plate is better than the real vellon, in the proportion of 32 to 17. Thus 17 maravedis of old plate are equal to 32 maravedis vellon; the quartos and ochavos are in the same preportion. The real of old plate is not a coin; it is the most general money of ex- change. 8 of these reals make the piastre, which is also ealled the dollar of exchange. 102 of these reals are equal to the hard dollar. When plate only is mentioned, old plate is understood. '- The real of new plate is double the real vellon; there- fore 34 maravedis of new plate are equal to 68 of vellon; ) | | | | | | 226 ARITHMETIC. XXXVI quartos and ochavos in proportion. This real is a coin, but not a money of account in any general way; it is the tenth part of the hard dollar, and is estimated in the United States at 10 cents, and the real vellon at 5 cents. The real of Mexican plate is divided into halves and quarters, called medio and quartillo. It is the eighth part of the hard dollar, and is the chief money of account in Spanish America, where it is divided into sixteenths. The doubloon de plata, or doubloon of exchange is four times the value of the piastre, or dollar of exchange. The ducado de plata, or ducat of exchange is worth 11 reals 1 maravedi old plate, or 20 reals 2513 maravedis vellon. 29. United States on Spain. Reduce 3148 dollars (of exchange) 6 reals 32 maravedis plate. to Federal money, exchange at 67 cents per piastre ? 30. Spain on United States. Reduce 1821. dollars 60 cents to Spanish money; exchange at 68 cents per dollar of exchange. 31. United States on Spain. Reduce 1286 doHars (of exchange) 7 reals 17 maravedis plete to Federal money; exchange at 64 cents per dollar of exchange. 32. United States on Spain. Reduce 2136 doubloons of exchange to Federal money; exchange at 68 cents per dollar (of exchange). 33. United States on Buenos Ayres. Reduce 4680 rials of Mexican plate to Federal money; exchange at 12 cents per rial. | SWEDEN. In Sweden, accounts are kept in rix dollars specie, skillings, and rundstycken or ore. 12 rundstycken or ore=1 skilling; 48 skillings —=1 rix dollar specie. : The Swedish dollar agrees in value with the dollar of the United States. . 34. United States on Sweden. Reduce 3955 rix dol- Jars 24 skillings to Federal money; exchange at 1 dollar 2 cents fer rix dollar. . XXX VIII. EXCHANGE, 227 35. Sweden on the United States. Reduce 1344 dollars 87 cents Federal money to Swedish money; ex- change at 1 rix dollar for 1 dollar 2 cents. 36. United States on Sweden. Reduce 2481 rix dollars 36 skillings to Federal money; exchange at 1 dollar per rix dollar. 37. Sweden on the United States. Reduce 819 dol- lars 873 cents Federal money to Swedish currency; ex- change at 1 rix dollar per dollar. 38. United States on Sweden. Reduce 1234 rix dollars 12} skillings to Federal money; exchange at 98 cents per rix dollar. 39. United States on Sweden. Reduce 1126 rix ‘dollars 42 skillings to Federal money; exchange at 1 dol- lar 3 cents per rix dollar. RUSSIA. In Russia accounts are kept in rouoles and copecks. The rouble is also divided into 10 grieven. 10 copecks==1 grieve or grievener, : 10 grieven or 100 copecks, = 1 rouble. The silver rouble is estimated in the United States at 75 cents; but the commercial business of Russia is car- ied on, in a paper currency much inferior to that of specie. The variable agio of the paper, substituted for the silver rouble, makes exchanges with Russia extreme- ly fluctuating, as the paper rouble improves or declines in value. 40. United States on Russia. Reduce 4182 roubles 64 copecks to Federal money; exchange at 25 cents per rouble. 41. Russia on the United States. Reduce 2614 dol- lars 15 cents to Russian money; exchange at 1 rouble for 25 cents. 42. United States on Russia. Reduce 5416 roubles 50 copecks to Federal money; exchange at 28 cents per ‘rouble. 43. Russia on United States. Reduce $3148.56 to Russian currency; exchange at 1 rouble for 30 cents. * 228 ARITHMETIC. XXXVIII 44. United States on Russia. Reduce 8672 rouble: 75 copecks to Federal money; exchange at 32 cents pel rouble. PRUSSIA. : | In Prussia accounts are generally kept in thalers ol rix dollars, good gresehen, and pfenings. 12 pfenings =1 good grosehen, 24 good evosehen = 1 rix doth) The Prussian rix dollar is in value % of the dollar | the United States. 45. Uuited States on Beets, Reduce 4162 rix dok lars 18 good groschen to Federal money; exchange at 6€ cents per rix dollar. 46. Prussia on United States. Reduce 3148 dollars 32 cents to Prussian money; exchange at 1 rix dollar | 64 cents. 47. United States on Prussia. Reduce 1428 rix dok lars 14 good grosehen to Federal money; exchange at 67 och per rix dollar. | Prussia on United States. Reduce 2136 dollars Pederal money to Prussian money; exchange at 1 rix dol- lar for 48 cents. ) DENMARK. In 1813 a new monetary system was established in Denmark, in which system the rigsbank dollar is the money unit. ‘The denominations of money are the same as in the old, or Hamburgh system, but of only half the value. 12 pfenings =1 skilling, 16 skillmgs ==1 mark, 6 marks =1 rigsbank dollar. The Danish rigsbank dollar is equal to 50 cents in the United States. | 49. United States on Denmark. Reduce 3214 rigs bank dollars 4 marks 8 skillings to Federal money; ex change at 50 cents per rigsbank dollar. | : | | XXXVI. EXCHANGE. 229 | 50. Denmark on the United States. Reduce 2082 dollars 35 cents Federal money to Danish money; ex- “change at 2 rigsbank dollars per dollar. 51. United States on Denmark. Reduce 1968 rigs- bank dollars 5 marks 12 skillings to Federal money; exchange at 48 cents per rigsbank dollar. _. 52. Denmark on United States. Reduce 3007 dollars Federal money to Danish money; exchange at 2 rigsbank dollars 2 skillmgs per dollar. | | NAPLES. _ In Naples accounts are kept in ducati, carlini, and grani. The ducat is the money unit, and is divided into 10 carlins, each of 10 grams, and, by the public banks, into 5 tarins of 20 grains each, making ‘he ducat always 100 grains. 10 grani 1 carlino, 10 carlini=1 ducato. The value of the silver ducat, in Federal money, is 80 cents. . _ 53. United States on Naples. Reduce 4022 ducati 8 -carlini to Federal money; exchange at 80 cents per ducat. 54. Naples on United States. Reduce 1835 dollars 73 cents Federal money to Neapolitan money; exchange at 1 ducato for 78 cents. 3 55. United States on Naples. Reduce 3508 ducats 5 carlini to Federal money; at 82 cents per ducat. 56. Naples on the United States. Reduce 1817 dol lars 54 cents I’ederal money to Neapolitan money; ex change at 1 ducat for 76 cents. SICILY. In Sicily accounts are kept in oncie, tari, and grani. 20 grani= I taro, 30 tari ==1 oncia. Accounts are also kept in scudi, tari, and grani. 12 tari =1 scudo, or Sicilian crown, 5 scudi—2 oncie. 20 230 ARITHMETIC. XXXVI It must be observed, that the denominations of Siciliar’ money have but half the value of the same denomination; in Naples. The Sicilian oncia, for mstance, passes it Naples for only 15 tari, the Sicilian scudo for 6 tari, anc other denominations in the same proportion. : The scudo is equal to 96 cents, and the oncia to $2.4(| Federal money. | 57. United States on Sicily. Reduce 1214 oncie 20| tari 10 grani to Federal money; exchange at 8 cents per taro. | 58. Sicily on United States. Reduce 1457 dollars| 62 cents Federal money to Sicilian money; exchange at 1 taro for 8 cents. | : 59. United States on Sicily. Reduce 3010 scudi 9 tari 15 grani to Federal money; exchange at 96 cents per scudo. 60. Sicily on United States. Reduce 983 dollars 44 cents to Sicilian money; enchange at 1 scudo for 95 cents. ~ LEGHORN. In Leghorn accounts are kept in pezze, soldi, and denari di pezza. | 12 denari di pezza—1 soldo, | 20 soldi di pezza —1 pezza of 8 reals. The value of the pezza is 90 cents, Federal money. 61. United States on Leghorn. Reduce 2146 pezze’ 16 soldi 8 denari to Federal money; exchange at 92 cents per pezza of 8 reals. 62. Leghorn on United States. Reduce 1620 dollars: 45 cents to money of Leghorn; exchange at 1 pezza of 8 reals for 90 cents. | 63. United States on Leghorn. Reduce 3293 pezze 13 soldi 4 denari to Federal money; exchange at 93 cents per pezza of 8 reals. | 64. Leghorn on United States. Reduce 1214 dollars” 68 cents to money of Leghorn; exchange at 1 pezza of 8 reals for 91 cents. XXXVIII. EXCHANGE. 231 GENOA. In Genoa accounts are kept in lire, soldi, and denari di lira; or in pezze, soldi, and denari di pezza; all in money fuori banco, or current money. 12 denari—1 soldo, 20 soldi —1 lira; 12 denari—1 soldo, 20 soldi —1 pezza; The value of the pezza is to that of the lira as 4 to 23; that is, 4 pezze==-23 lire; therefore 4 soldi di pezza= 23 soldi di lira; 4 denari di pezza= 23 denari di lira. The par value of the lira is 154 cents, that of the pezza 89 cents, U. S. 65. United States on Genoa. Reduce 5254 lire 16 soldi 3 denari to Federal money; exchange at 16 cents per lira. 66. Genoa on the United States. Reduce 1532 dol- lars 30 cents to money of Genoa; exchange at 1 lira for 15 cents. 67. United States on Genoa. Reduce 8792 lira to Federal money; exchange at 164% per lira; | 68. Genoa on the United States. Reduce 2000 dol- lars to money of Genoa; exchange at J lira for 15 cents. VENICE. In Venice accounts were formerly kept, and exchanges computed in ducats, lire, soldi, and denari, moneta pic- cola. 12 denari = 1 soldo, 20 soldi = 1 lira piccola, 64 lire piccole= 1 ducat current, 8 lire piccole. =1 ducat effective. The par of the lira piccola, in Federal money, is 9% cents. Accounts are now kept, and exchanges computed in lire Italiane and centimes. 100 centesimi=— | lira Italiana. The common estimate of this money is, that 100 lire 232 ARITHMETIC. XXXVIIL.- piccole are equal to 514 lire Italiane. The lira Italiana is of the same value with the French franc. 69. United States on Venice. Reduce 14642 lire 4 soldi 8 denari piccoli to Federal money; exchange at 9 cents per lira piccola. = __ 70. Venice on the United States. Reduce 814 dollars 55 cents to Venetian money; exchange at 1! lira piccola for 8 cents. : We have given the two preceding examples in the old currency, for the sake of practice, although it has general- ly gone out of use. 71. United States on Venice. Reduce 6784 lire Italiane 40 centimes to Federal money; exchange at 184 cents per lira Italiana. | 72. Venice on the United States. Reduce’ 1817 dol- lars 82 cents to Venetian money; exchange at 1 lira Ita- liana for 18 cents. 7 73. United States on Venice. Reduce 5236 lire Italiane to Federal money; exchange at 183 cents per lira Italiana. TRIESTE. In Trieste accounts are kept and exchanges computed in florins and creutzers; or in rix dollars and creutzers. 4 pfenings = 1 creutzer, 60 creutzers 1 florm or gulden, 1 florin, or 90 creutzers=— 1 rix dollar of account. The rix dollar specie is equal to 2 florins. The par of the florin is 48 cents, in Federal money, which makes the dollar of account 72 cents, and the specie dollar 96 cents. 74. United States on Trieste. Reduce 2846 florins 25 creutzers to Federal money; exchange at 48 cents per florin. | 75. Trieste on United States. Reduce 1637 dollars 46 cents to money of Trieste; at 1 florin for 47 cents. 76. United States on Trieste. Reduce 2055 rix dol- lars, 25 creutzers to Federal money; exchange at 72 cents per rix dollar. ) XXXVIIL EX CHANGE. 233 77. Trieste on United States. Reduce 1738 dollars 88 cents to money of Trieste; exchange at 1 rix dollar for 70 cents. ROME. In Rome accounts are kept, by the old system, in scudi, paoli, and bajocchi; quattrini and mezzi quattrini are also sometimes reckoned. | 2 mezzi quattrini = 1 quattrino, 5 quattrini == 1 bajoccho; 10 bajocchi =1 paolo, 10 paoli, or 100 bajocchi =1 scudo, or Roman crown. The Roman crown is equal to the Federal dollar. The scudo di stampa d’oro, or gold crown, is equal to $1.53. | | - When the exchange between the United States and - Rome is at par, no reduction is required; for any number of scudi and bajocchi are equal to the same number of dollars and cents, and the reverse; for mstance, 125 scudi 75 bajocchi are equal to 125 dollars 75 cents. 78. Rome on the United States. Reduce 1871 dol- lars 19 cents to Roman money; exchange at 1 scudo 2 bajocchi per dollar. ‘79. United States on Rome. Reduce 2070 scudi 50 bajocchi to Federal money; exchange at 101 cents per _ scudo. | In 1809, the French moneys of account were introduc- edinto Rome. ‘The scudo was reckoned at 5 francs 35 centimes; the franc, therefore, was valued at 18 bajocchi 3.45 quattrini. MALTA. Acounts are kept in this island in scudi, tari, and grani, 20 grani= I taro, 12 tari ==1 scudo. The taro is likewise divided into 2 carlini, and a carline “into 60 piccioli. The pezza, or dollar of exchange, is equal to 24 scudi. 90* 1 934 ARITHMETIC. XXXVIII. The par value of the Maltese scudo is 40 cents in Federal money. The coins in circulation are chiefly Spanish dollars and doubloons, and Sicilian dollars and ounces. 'They are valued each at a certain rate, as follows, on which a variable agio is charged.. Spanish dollar 30 tari 10 grani. Spanish doubloon = 38 seudi 9 tari. Sicilian dollar 830 tari. Sicilian ounce ==6 scudi 38 tari. 80.-United States on Malta. Reduce 1108 Maltese scudi 9 tari to Federal money; exchange at 40 cents per scudo. 81. Malta on the United States. Reduce 874 dollars 76 cents to Maltese money; at 1 scudo for 38 cents. 82. United States-on Malta. Reduce 3964 Maltese scudi 6 tari to Federal money; at 41 cents per scudo. 83. Malta on the United States. Reduce 674 dollars 60 cents to Maltese money; at 1 scudo for 40 cents. SMYRNA. Accounts are kept here in piastres or gooroosh. The piastre, also called the Turkish dollar, is divided some- times into 12 temins, sometimes into 40 paras or medini; but the usual division is into aspers, the number of which varies. Thus, the English and Swedes divide the piastre into 80 aspers; the Dutch, French, and Venetians into 100 aspers; the Turks, Greeks, Persians, and Armenians into 120 aspers. An asper is a third part of a para. Bills o. exchange are often drawn on Smyrna in foreign coin, particularly in Spanish dollars, which are always to be had there; but, if drawn in a com not in current use, the exchange of the day is established to make the pay- ment. : The Turkish coins, owing to the frequent deterioration of them by the government, have been declining in their intrmsic worth for many years, and have no standard value. Foreign exchanges are conducted entirely ac- cording to the price of the day. : Fd XXXVIII. EXCHANGE. 235 84. United States on Smyrna. Reduce 53183 piastres of Turkey to Federal money; exchange at 20 cents per piastre. . 85. Smyrna on the United States. Reduce 912 dol- lars 27 cents to Turkish money; exchange at 1 piastre for 21 cents. 86. United States on Smyrna. Reduce 71614 Turk- ish piastres to Federal money; exchange at 22 cents per piastre. 87. Smyrna on United States. Reduce 1128 dollars 22 cents to money of Smyrna; exchange at 1 piastre for 204 cents. EAST INDIES. Before European colonies were established in the East Indies, particularly while the power of the Moguls pre- vailed in Hindostan, the monetary system was very simple. There was current throughout these vast dominions one principal coin of silver, denominated the sicca rupee. It was of a certain weight called the sicca. The siccu was ‘used also as a standard for weighing other articles. _. The British possessions in the. Kast Indies are divided ‘into three presidencies, viz. Bengal, Bombay, and Ma- dras. The monetary systems in these presidencies are different from each other. CALCUTTA IN BENGAL. Accounts are commonly kept here in current rupees, annas, and pice. 12 pice = I anna, 16 annas= 1 rupee, currency. The East India Company, however, keep their ac- counts in sicca rupees, similarly divided, which bear a batta or premium of 16 per cent. above current rupees. The current rupee of Calcutta is 447% cents, and the sicca rupee 51-7, cents, in Federal money. A Lac of rupees is 100000, and a Crore of rupees is 100 Lacs, or 10 millions of rupees. 236 ARITHMETIC. XXXVI 88. United States on Calcutta. Reduce 17438 rupees: 12 annas, currency of Calcutta, to Federal money; ex- change at 48 cents per rupee. 89. Calcutta on the United States. Reduce 6913 dollars 25 cents to money of Calcutta; exchange at 2 sicca rupees per dollar. et | 90. United States on Calcutta. Reduce 46173 cur- | rent rupees 9 annas to Federal money; exchange at 46. cents per rupee. 91. Calcuttaon United States. Reduce 28953 dollars | 63 cents to current money of Calcutta; exchange at 1 rupee for 44 cents. | 92. United States on Calcutta. Reduce a Lac of sicca rupees to Federal money; exchange at 53 cents per sicca rupee. BOMBAY. In the presidency of Bombay, accounts are kept in rupees, quarters, and reas. f 100 reas —=1 quarter, © | 4 quarters—1 rupee. : The current value of the Bombay rupee is equal to 50 cents in Federal money. | i 93. United States on Bombay. Reduce 10137 rupees ~ quarters 50 reas to Federal money; exchange at 50 cents per rupee. ' 94. Bombay on the United States. Reduce 6210- dollars 48 cents to money of Bombay; exchange at 48 cents per rupee. 95. United States on Bombay. Reduce 8413 rupees of Bombay to Federal money; at 49 cents per rupee. MADRAS. In the presidency of Madras, there are different mone- tary systems, which may be distinguished under the heads of the old system and the new. , According to the old system, accounts are kept in star _ pagodas, fanams, and cash. * XXXVIII. EXCHANGE. 237 80 cash ==1 fanam, 42 fanams = 1 pagoda. The current value of the star pagodais $1.80 Federal. By the new system, the silver rupee of Madras is made the standard coin, and money of account in this presidency. The current value of the silver rupee of Madras is 44,7 cents. It is divided into halves, quarters, eighths, and sixteenths. The sixteenth is the anna. 96. United Stateson Madras. Reduce 72183 rupees of Madras to Federal money; at 46 cents per rupee. 97. Madras on United States. Reduce 2684 dollars 85 cents to money of Madras; at 44 cents per rupee. 98. United States on Madras. Reduce 5367 rupees of Madras to Federal money; at 45 cents per rupee. CANTON IN CHINA. In China accounts are kept in tales, mace, candarines, and cash. 10 cash | ==1 candarine, 10 candarines= 1 mace, 10 mace ==} tale. The tale is reckoned at $1.48 in Federal money. 99. United States on Canton. Reduce 12144 tales 5 mace to Federal money; exchange at 1 dollar 48 cents per tale. - 100. Canton on the United States. Reduce 8754 dollars 89 cents to money of Canton; exchange at 1 tale per 146 cents. - 101. United States on Canton. Reduce 16235 tale to Federal money; exchange at 149 cents per tale. JAPAN. In the empire of Japan, which consists of severa: islands to the east of Asia, accounts are kept in tales. ‘mace, and candarines. 10 candarines= 1 mace, 10 mace = I tale. The Japanese tale is reckoned at 75 cts, Fed. money | : 238 ARITHMETIC. XXXVIIL. 102. United States on Japan. Reduce 3714 Japan- ese tales 8 mace 8 candarines to Federal money; exchange at -75 cents per tale. | 103. Japan on United States. Reduce 696 dollars 54 cents to money of Japan; exchange at | tale per 75 cents. 104. United States on Japan. Reduce 2468 tales 5 mace, money of Japan, to Federal money; exchange at 76 cents per tale. SUMATRA. This island is chiefly in possession of the natives; but the English have a small settlement at Bencoolen. At Bencoolen accounts are kept in dollars, soocoos, and satellers. 8 satellers—1 soocoo, 4 soocoos —1 dollar. This dollar is reckoned at $1.10 in Federal money, and is sometimes called a rial. 4 105. United States on Bencoolen. Reduce 1947 Bencoolen dollars 3 soocoos 4 satellers to Federal money; exchange at 110 cents per dollar of Bencoolen. 106. Bencoolen on United States. Reduce $2379.51 Federal money to money of Bencoolen; exchange at 1 dollar Bencoolen for 108 cents. ACHEEN. ( In the island of Sumatra). In Acheen accounts are kept in tales, pardows, mace, and copangs. 4 copangs 1 mace, 4 mace =1 pardow, 4 pardows = | tale. The mace is a small gold coin worth about 26 cents’ Federal money, which makes the tale $4.16. bs 107. United States on Acheen. Reduce 1432 tales, 3 pardows 2 mace to Federal money; exchange at 416 cents per tale. . 108. Acheen on United States. Reduce 3620 dollars — 964 cents to money of Acheen; at 1 tale for 412 cents. — oa ® " i XXX VIII. EXCHANGE. 239 JAVA. In Batavia, the capital of this island, the florin or guil- der of the Netherlands is the monetary unit; but instead of the decimal divisions, it is here sometimes divided into schillings, dubbels, stivers, and dotts. 5 doits = 1 stiver, . 2 stivers ==1 dubbel, 3 dubbels =1 schilling, 4 schilling=1 florin or guilder. The florin of Java, as the florin of the Netherlands, is equal to 40 cents Federal money. 109. United States on Batavia. Reduce 11841 florins 3 schillings 2 dubbels to Federal money; exchange at 40 cents per florin. 110. Batavia on the United States. Reduce $ 13746. 69 to money of Batavia; exchange at 1 guilder for 38 cents. 111. United States on Batavia. Reduce 42328 guil- ders 50 centimes to Federal money; exchange at 42 cents per guilder. | MANILLA. (In the island of Luzon). In Manilla, the capital of the Spanish East India pos- sessions, accounts are kept in Spanish dollars or pesos, reals, and maravedis. - 34 maravedis —1 real, 8 reals = 1 dollar. 112. United States on Manilla. Reduce 6341 dollars — 6 reals 17 maravedis to Federal money; exchange at 101 cents per Spanish dollar. 113. Manilla on United States. Reduce $5274.55. to money of Manilla; exchange at 1 Spanish dollar per dollar. COLOMBO. (In the island of Ceylon). In Colombo, accounts are kept in ri dollars, fanams and pice. 240 ARITHMETIC. | XXXVIIL| 4 pice =1 fanam, 12 fanams = 1 rix dollar. _... | The current value of this rix dollar is 40 cts. F. money, 114. United States on Colombo. Reduce 7328 rix dollars 9 fanams to Federal money; exchange at 40 cents per rix dollar. 115. Colombo on United States. Reduce $1426.71 Federal money to money of Colombo; exchange at 1 rix dollar per 38 cents. MAURITIUS, (Isle of France.) In Mauritius, accounts are kept in two different ways, viz. in dollars of 100 cents, which is the mode adopted in public or government accounts; and in dollars, livres, and sols, which method is mostly used by merchants. 20'sols =1 livre. 10 livres= 1 dollar. ‘These are called colonial livres, and are 10 cents each. 116. United States on Mauritius. Reduce 4132 dol- lars 7 livres 10 sols to Federal money; exchange at 1 dollar per dollar of Mauritius. 117. Mauritius on United States. Reduce $7547.47, Federal money, to money of Mauritius; exchange at 98 cents per dollar of Mauritius. ARBITRATION OF EXCHANGE. ARBITRATION OF EXCHANGE is a comparison of the courses of exchange between different countries, in order to ascertain the most advantageous course of drawing or remitting bills. It is distinguished into simple and com- pound arbitration. Simple Arbitration is a comparison between the ex- changes of two places through a third; that is, it is finding such a rate of exchange between two places, as shall be in proportion to the rates quoted between each of them and a third place. The exchange thus determined is called the arbitrated price. XXXVIIL. EXCHANGE. 241 ~ If, for example, the course of exchange between Lon- don and Paris is 24 francs for 1 pound sterling, and be- tween Paris and Amsterdam 54 pence Flemish for 3 francs, the arbitrated price between London and Amster- dam through Paris, is 36 shillings Flemish for 1 pound sterling; for, as 3fr. : 24fr.—54d. : 36s. Flem. Suppose the arbitrated price to be, as before stated, ~36s. Flemish for £1 sterling; and suppose the direct course between London and Amsterdam to .be 37s. Flemish; then London, by drawing directly on Amster- dam, must give 37s. Flemish for £1 sterling; whereas, by drawing through Paris, he will give only 36s. Flemish, for £1 sterling. « It is therefore the interest of London to draw indirectly on Amsterdam through Paris. On the contrary, if London remits directly to Amsiter- dam, London will receive 37s. Flemish for £1 sterling; but, by remitting through Paris, London will receive only 36s. Flemish. It is the interest of London, therefore, to remit directly to Amsterdam. 118. If the exchange of London with Genoa is 47d. sterling per pezza, and that of Amsterdam with Genoa 86 grotes Flemish per pezza, what is the proportional or -arbitrated exchange between London and Amsterdam through Genoa? that is, how many shillmgs and grotes Flemish are equal to £1 sterling ? Since 47d. sterling is equal to 1 pezza, and this pezza is equal to 86 grotes Flemish, the question may be stated thus; 47d. sterling : 240d. sterling =86 grotes Flemish: Ans. which is 36s. 77% grotes Fl. By the Chain Rule, (See Art. xxvr), the statement is as follows. : 1 pound sterling. ‘1 pound sterling = 240 pence. 47 pence = | pezza. 1 pezza = 86 grotes Fl: 12 grotes = 1 shilling FI. The product of the consequents being divided by the product of the antecedents, will give 36sh. 77 grotes ]. for the answer. 119. If the exchange on London with Hamburgh is 34 shillings 2 grotes Flemish banco for £1 sterling, and 21 242 ARITHMETIC. XXXVIIL. _ that of Amsterdam with Hamburgh 333 stivers per rix dollar of 2 marks, what is the arbitrated exchange be-— tween London and Amsterdam through Hamburgh? Since 2 marks are 64 grotes Flemish and 333 stivers are 663 grotes Flemish, the question may be stated thus, 64 grotes Fl.: 662 grotes Fl. = 34s. 2 grotes Fl.: Ans. By the Chain Rule, the statement is as follows, 1 pound sterling. 1 pound sterling — 34 s. grotes Flem. 8s. Flem. core (ON WAT K Se 2 marks == 33% stivers. _ 6 stivers = ion Flemish. 120. If the exchange of London on Leghorn is 514d. sterling per pezza, and that of Amsterdam on Leghorn 923 grotes Flemish per pezza, what is the proportional exchange between London and Amsterdam through Leg-— horn? : 121. If the exchange of London on Lisbon be 68d." sterling per milree, and that of Amsterdam on Lisbon 48 grotes Flemish per old crusado, what is the arbitrated ex-- change between London and Amsterdam through Lisbon ? 122. If the exchange of London on Madrid is 42d. sterlmg per dollar of plate, and that of Amsterdam on” Madrid 96 grotes Flemish per ducat of plate, what is the proportional exchange between London and Amsterdam through Madrid? | 123. If the exchange of London on Paris is 24 francs per £1 sterling, and that of the United States on Paris 185 cents per franc, what is the arbitrated or propor- tional exchange between London and the United States through Paris? 124. If the exchange of London on Amsterdam is 11 florins 16 stivers per £ sterling, and that of the United States on Amsterdam 38 cents per florin, what is the’ arbitrated exchange between the United States and Lon- — don through Amsterdam? 125. Ifthe exchange of the United States on Paris is” 18 cents per franc, and that of Amsterdam on Paris 547 grotes “semish for 3 francs, what is the proportional e,change between the United States and Amsterdam — through Paris? XXXVIII. EXCHANGE. . Q43 126. If the exchange of the United States on Lisbon is $1.24 per milree, and that of Paris on Lisbon 540 rees per ecu of 3 francs, what is the proportional ex- change between the United States and Paris through Lisbon? COMPOUND ARBITRATION. Compound arbitration is a comparison between the exchanges of more than thrce places, to find the arbitrat- ed price between the first place and the last, in order to ee on the most advantageous mode of negotiating ills. 127. Suppose the exchange between London and Amsterdam to be 35 shillings Flemish for £1 sterling; between Amsterdam and Lisbon, 42 pence Flemish per old crusado; and between Lisbon and Paris, 480 rees per ecu of 3 francs; wnat is the arbitrated price between London and Paris? First, 35s. Fl. : 42d. Fl. £1 sterling : A; which is 2s. sicrling. Secondly, 1 old crusado : 480 rees==2s. sterling : A; . which is 2s. 44d. sterling. Thirdly, 2 s.44d, sterling : £1 sterling =3 francs : A; which is 25 francs. Hence the arbitrated price is 25 francs for £1 sterling. But all such operations are best performed by the Chain Rule; thus, 1 pound sterling. 1 pound sterling 36 shillings Flemish. 33 shillings Fl. 1 old crusado. 1 old crusado 400 rees. 480 rees 3 francs. The product.of the consequents divided by that of the antecedents gives 25 francs per £ sterling, as before. 128. Suppose a merchant in London has a sum of money to receive in Cadiz, the exchange being at 38 d. sterling per dollar of plate; but, instead of drawing di- rectly on Cadiz, he draws on Amsterdam, ordering his agent there to draw on Paris, and Paris to draw on Ca- diz; the exchange between London and Amsterdam being Al 4 ' 44 ARITHMETIC. XXXVILI. at 35 shillmgs Flemish per pound sterlmg; between Am-— sterdam and Paris 534 grotes Flemish per ecu of 3_ francs; and between Paris and Cadiz 15 francs 50 cen- times per doubloon of plate. What is the arbitrated price between London and Cadiz ? 1 dollar of plate. 4 dollars of plate == 1 doubloon of plate. 1 doubloon of plate— 15% francs. + 3 francs = 531 grotes FI. 12 grotes FI. == 1 shillng FI. 35 shillings FI. == 240 pence sterling. The result is 3944d. sterling per dollar of plate. The circuitous operation is, therefore, the most advanta- Fi aaa SOS geous, as London gets 393d. nearly, instead of 38d. for é each dollar of plate. 129. London having a sum to receive in Lisbon, when — the exchange is at 64d. sterling per milree, draws on © Lisbon, but remits his bill to Hamburgh to be negotiated, and directs the returns to be made to him im bills on Leghorn; the exchange between Hamburgh and Lisbon ~ being 45 grotes Flemish per old crusado; between idam- burgh and Leghorn 85 grotes Flemish per pezza; and — between London and Leghorn 52d. sterling per pezza. What is the arbitrated price between London and Lis- bon? and what does London give per milree by the cir- — culitous exchange ? {30. A merchant in London has a sum to pay in Petersburg, and another to receive in Genoa; but there — being no regular exchange between these places, London draws on Hamburgh, and remits his bill to Petersburg, — directing Hamburgh to draw on Genoa; the exchange between London and Genoa being 463d. sterling per pezza; between Hamburgh and Genoa 81 grotes Flemish per pezza; and between Petersburg and Hamburgh 23 schillings Lubs per ruble. What is the exchange between London and Peter rsburg resulting from the operation? that is, how many pate sterling does London pay for the ruble ? 131. A merchant in the United States has funds in Paris, and owes a sum of money in Hamburgh; he draws | : | XXXVIII. FOREIGN COINS. 245 on London, remits his bill to Hamburgh, and directs London to draw on-Paris; the exchange between the United States and Paris being 18 cents per franc; be- tween London and Paris 24 francs 25 centimes per £ sterling; and between Hamburgh and London 13} marks banco per £ sterling. How many cents per mark banco does the American merchant pay by this course of ex- change ? 132. A merchant in the United States being indebted in London, remits bills on Paris to his correspondent in that city, and directs him to obtain bills of Paris on Lis- bon and remit them to his creditor in London; the ex- change between the United States and France being 18 cents per franc; between Paris and Lisbon 465 rees per ecu of 3 francs; and between London and Lisbon 63d. sterling per milree. In this course of exchange, how many pence sterling are paid with one dollar of the Unit- ed States ? In the preceding examples, no notice is taken of the expenses incident to exchange operations, such as com- mission, brokerage, interest, &c.; but in all transactions of business, it is necessary to make allowance for the difference of charges between direct and indirect ex- changes, in order to decide on the preference of the one to the other. | FOREIGN COINS. THE SILVER COINS of foreign countries, rendered current in the United States, by Act of Congress, are _as follows. Spanish dollars and parts thereof, at 100 cents the dollar. Dollars of Mexico, Peru, Chili, and Central America, of not less weight than 415 grains each, and those restamped in Brazil of the like weight, and of not less fineness than 10 oz. 15 dwt. pure silver in the Troy pound, all at 100 cents the dollar. The Five- franc pieces of France, weighing 384 grains each, and of not less fineness than 10 oz. 16 dwt. pure silver in the Troy pound, at 93 cents the piece. 21* 246 ARITHMETIC. XXXVII1. THE GOLD COINS of foreign countries, with their respective weights, and values, are stated in the following © TasLe. Those of the countries printed in Italics are rendered current, by Act of Congress. Weight. Value. Names of Countries and Coins. - dwt. gr. | $ cts m AUSTRIAN DOMINIONS. (oor) Ie > Souveremowye Ce Vise Gh Pale 3 14 3 37 7 Double Dudaty. 82h ae! Pe ee See @ Hungarian, Ducat, 2 53) 2296 BAVARIA Carolin, . Te TRA. 6 54] 495 7 Max d’or, or Maximilian, mes 4°40 S8Sr’s Ducat, zee WP ath, 2 53) 2275 BERNE. a Ducat, double in proportion oe 1-93 (1 1°98 Pistole. Set ees od 4 21 | 4 54 BRAZIL. Jahannes, half in proportion. . | 18 17 06 Doviraon, © ss Pees aa eS uae PUA. ks OEE Ga ee Moidore, half in | proportion, a ds a Crusade, .. Bes 164 63 5 BRUNSWICK. Pistole, double in atin ; 4 211} 4548 Diva ors eM as B $3) 2126 COLOGNE. Ducat ae Oi OPA Pe ice, 25352 Borg COLOMBIA. Doublbon! tgp eae ke 17° 6" (1S S35 DENMARK. Ducat, Current, a 2 1812 Dileat"Specie;s ois <0 2° 53-2 26M) Christian d’or, 4 4 02 1 EAST INDIA. Rupee, Bombay, 1818, TAL. [rooms Rupee, Madras, 1818, a) AD eee Pagoda, Star, 2 42) 1 7958 ENGLAND. Guinea, half in proportion, 5 81| 5 07 5 Sovereign, half in proportion, . 5 24| 4 846 XXXVIIL. Seven Shilling Piece, FRANCE. Louis, coined before 1786, Double Louis, before 1786, . Louis, coined since 1786, Double Louis, since 1786, Napoleon, or 20 francs, Double Napoleon or 40 francs, . Same as the new Louis Guinea, FRANKFORT ON THE MAIN. Ducat, Shee GENEVA. ¥ Pistole, old, Pistole, new, HAMBURG. Ducat, double in proportion, GENOA. Sequin, . HANOVER. - Double George d’or, ae in Ng Ducat, . Gold Florin: double! in pro a HOLLAND. Double Ryder, Ryder, , Poratss ss *: Ten Guilder Piece, 5 do. in n pro’n ‘n, MALTA, Double Louis, . Louis, Demi Louis, MEXICO. Doubloons, shares in pro’n, . MILAN. Sequin, . . Doppia or Pistole, Forty Livre Pieces, 1808, NAPLES. Six Ducat Piece, 1783, . Two do. or Sequin, 1762, Three do. or Oncetta, 1818, FOREIGN COINS. posed, eat PS) on on os Or “IO © ' Blo Alo Pwd Alo die — to BNO me OP oo 09 onal s% = 248 ARITHMETIC. XXXVIIL. NETHERLANDS. Gold Lion, or 14 Florin ia 5 731 5 04 6 Ten Florin Piece, 1820,. . 4 73|.4 019 PARMA. Quadruple Pistole, double in pro’n, 18.59. 16,632.85 Pistole or Droppia, 1787, : 414 ;,4194 Pistole or Droppia, 1796, bon 414 |} 4135 Maria Theresa, 1818, vis 4 33] 3 86 1 PIEDMONT. Pistole c’d since 1785, 3 10 pro’n, 5 20 | 5.41.0 Sequin, haif in proportion, 2) .5. | sae Carlino, c’dsince1785,4inpro’n, |29 6 |27 34 Piece of 20 Francs, or Marengo, 4 31) 3 564 POLAND. - Ducat Seek ss sa le ean PORTUGAL, | Dobraons Py. a ce) eh ol ee Dobra, toe 530.) oa ot & «Be Johannes, . .- eS 17 06 4 Moidore, half in proportion, aa 6.22 6.58 7 Piece of 16 Testoons, 1600 Rees, 2 6 |; 2.12.19 Old Crusado or 400 Rees, . . 15) bs pesig New Crusado or 480 Rees,. . 164 63 5 Milree, coined in 1755, . 193 73 PRUSSIA. | WoC Nei, oka koi ea ge 53 Ducat, 1787, 2 52 Frederick, dauble, 1769, mass 8 14 Frederick, double, 1800, ae: 8 14 Frederick, single, 1878, pL, Frederick, single, 1800, 4 7 ROME. Sequin, coined since 1760, : 2 44 Scudo of Republic, Se Wied) 8 RUSSIA. PUA Es Wanye a 2.26 Ducat, 1763;-4. : 2° 63 Gold Ruble, 1756, 01 Gold Ruble, 1799, 183 Gold Polten, 1777, 9 Imperial, 1801, ewe XXXVI. ’ Half Imperial, 1801, - Half Imperial, 1818, SARDINIA. Carlino, half in proportion, SAXONY. Ducat, 1784, Ducat, 1797, Augustus, 1754, Augustus, 1784, SICILY. Ounce, 1751, . Double Ounce, 1758,. SPAIN. Doubloons, 1772, double and sin- gle and shares in proportion, . Doubloon, Ay. dew Pistole, .. Coronilla, (Gold Dol. a or Vintern, 1801, SWEDEN, Ducat, SWITZERLAND. Pistole of Helvetic Republic, 1800, TREVES. Ducat, TURKEY. + Sequin fonducli, of Cons’ple, 1773, Sequin fonducli, of Cons’ Be 1789, Half Misseir, 1818, Sequin Fonducli, Yeermeeblekblek, TUSCANY. Zechino, or Sequin, Ruspone of kingdom of Etruria, VENICE. b] Zechino, or Sequin, shares in pro. WIRTEMBURG. Carolin, . Ducat, ZURICH. Ducat, double and half in pro’n, FOREIGN COINS. —t ret Ot CO Or Cr Ivo —" On Halo “Et On DB) Dla WD] AI co|co vl 249 G9 09 & WH © oo 09 LS) =P) moron iS) €9 0 > CQ & Or © 250 ARITHMETIC. XXXVIII. FOREIGN WEIGHTS AND MEASURES. The weights and measures of GREAT BRITAIN are the same as those of the United States, excepting the variations which are noted in the tables of “ Weights and Measures,’ page 27. The weights and measures of FRANCE being more nicely adjusted than those of any other country, will be here given the more fully on that account. It is, however, | to be observed, that these weights and measures are ac- cording to a new system, not yet in very common use. The fundamental standard adopted in France for the metrical system of weights and measures, is a quadrant of the meridian; that is to say, the distance from the. equator to the north pole. This quadrant is divided into ten millions of equal parts, and ones of these equal parts” is called the Metre, which is adopted as the unit o1 length, and from which by decimal multiplication and division all other measures are derived. In order to express the decimal proportions, the fol-' lowing vocabulary of names has been adopted. Fot multipliers, the word Deca prefixed, means - 10 times.” re dd eetoe ©" e 100 times. ee Chilo eo J 1000 times. eye © | hy _ 10000 times For divisors, the word Dect prefixed, expresses the 10th part. | ae enti. 8° eS — 100th part. “Milli « “ 1000th part. It may assist the memory to observe that the terms for multiplying are Greek, and those for dividing, Latin. - Thus, Deca-metre means 10 Metres. Deci-metre ‘* the 10th part of a Metre. Flecto-metre ‘* 100 Metres. cok Centi-metre ‘‘ the 100th part ofa Metre; &c. = XXXV11II. WEIGHTS AND MEASURES. 251 . Frencu Lona MEASURE. The Metre, which is the unit of long measure, is equal to 39.371 English inches. 10 milli-metres ==1 centi-metre, 10 centi-metres =1 deci-metre, 10 deci-metres =—=1 METRE 10 Metres. . ==1 deca-metre, 10 deca-metres 1 hecto-metre, 10 hecto-metres —1 chilo-metre, 10 chilo-metres —1 myria-metre. Frencu Square MEASURE. The Are, which is a square deca-metre (or 100 square Metres), is the unit of square or superficial measure, and is equal to 3.953 English square rods. 10 milliares . . ==1 centiare; 10 centiares . . ==1 deciare; 10 deciares . . ==1 ARE; 10 Ares . «© sz=l1 decare; 10 decares . . =I hectare; 10 hectares . . = 1 chilare; 10 chilares . . ==1 myriare. Frencu MEASURES OF Capacity. The Litre, which is the cube of a decimetre, is the unit of all liquid measures, and of all other measures of eapacity. The Litre is equal to 61.028 English cubic nches. 10 millilitres. . == 1 centilitre; 10 centilitres. . ==1 decilitre, 10 decilitres . . ==! LitRE; 10 Litres . . =1 decalitre; 10 decalitres. . 1 hectolitre; 10 hectolitres . ==1 chilolitre; 10 chilolitres. . ==1 myrialitre. Frencu Sotip MEASURE. The Stere, which is a cube of the metre, is the unit of solid measure, that is used for fire-wood, stone, &c. The Stere is equal to 35.31714 English cubic feet; it is the same as the chilolitre in measures of capacity. 252 “ARITHMETIC. XXXVIIL 10 decisteres. 1 Srerg; 10 Steres. . ==1 decastere. Frencu Werents. The Gramme, which 1s the weight of a cubic centi« metre of distilled water of the temperature of melting ice, is the unit of all weights. The Gramme is equal to 15.434 grains Troy. Grains Troy A milligramme is 1000th part of a gramme, = 0.0154 A centigramme is 100th part of a gramme,== 0.1543 A decigramme is 10th part of a gramme, = 1.5434 -A GRAMME ees 15.4340 A decagramme is 10 grammes, = 154.3406) A hectogramme is 100 grammes, = 1543.4000 Achilogramme is 1000 grammes, 15434.0000 | A myriagramme is 10000 grammes, == 154840.0000 ~ All the preceding French weighis and measures are de- duced from some decimal proportion of the metre. ‘Thus the chilogramme corresponds with the contents of a cubic vessel of pure water at the lowest temperature, the side of which vessel is the tenth part of the metre (the decimetre), and the gramme answers to the like contents ofa cubic vessel, the side of which is the hundredth part of the metre (the centimetre); for the contents of all cubic vessels are to each other in the triplicate ratio of 4 their sides. 100 lb. of HAMBURGH 106.8 jb. avoirdupois. The shipfund is 280 Ib. = 299 Ib. avoirdupois. 1 foot, Hamburgh = 11.289 inches, U. S. The Hamburgh ell is 2 feet ==22.578 inches, U.S. The Hamburgh mile = 4.684 miles, U.S. The fass of Hamburgh = 1.494 bushel of U.S. The last of grain is 60 fasses —89.64 bushelsof U.S. The ahm of Hamburgh = 38.25 gallons, U. S. 100 Ib. of AMSTERDAM 108.93 Ib. avoirdupois. 4 shipfunds is 1 ship-pound ==326.79 Ib. avoirdupois. — The Amsterdam last == 85.248 bushels, U.S. KXXVIII. WEIGHTS AND MEASURES. 253 The Aam (liquid) ==41 gallons, U. States. The Amsterdam foot = 11.147 inches, U. S. The ell of Amsterdam = 27.0797 inches, U.S. The ell of the Hague == 27.333 inches, U. S. The ell of Brabant == 27.585 inches, U. S. 100lb. of PORTUGAL = 101.19 lb. avoirdupois. An arroba is 32 lb. = 32.38 lb. avoirdupois. The moyo, a dry measure ==23.03 bushels, U. 8. ~The almude, a liquid measure 4.37 gallons, U. 5. The pe or foot, long measure 12.944 inches, U. 8. The palmo or standard span ==8.64 inches, U. 8. The vara is 5 palmos -~)==43.2 Mehes Uses. The Portuguese mile =21)26Cnilep Uie8. 100 lb. of SPAIN == 101.44 lb. avoirdupois. The arroba of wine == 4,245 gallons, U. 8S. The fanega, 75 of a cahiz 1.599 bushels, U. S. The Spanish standard foot = 11.128 inches, U. S. The vara, a cloth measure —= 33.384 inches, U.S. The legua or league 4.291 miles, U. S. 100|b. victualie, of SWEDEN = 93.76 |b. avoirdupois. The Swedish foot = 11.684 inches, U. S. The Swedish ell is 2 feet —= 23.368 inches, U. S. ~The Swedish mile — 6.64 miles, U. S. - The kann, (both dry and liquid) =1593 cubic in. U. 5S. 100 kanns 69.09 galls. wine,U.S. 100 kanns —-7.42 bushels, U. S. 100 lb. of RUSSIA == 90.26 lb. avoirdupois. 400 lb. make 1 berquit == 361.04 lb. avoirdupois. - A pood is 40 lb. Russian == 36.1054 Ib. avoir’s. A chetwert, a dry measure, ==5.952 bushels, U.S. The vedro, a liquid measure, ==3.246 gallons, Ue The Russian inch —=1 inch, U. S. _ The Russian foot —13.75 inches, U. S. - The arsheen, a cloth measure, = 28 inches, U. 5. The sashine or fathom = 7 feet, U. S. A werst or Russian mile == 3500 feet, U. S. 22 254 ARITHMETIC. XXXVIII 100Ib. of PRUSSIA = 103.11 Ib. avoirdupois The quintal is 110 lb. = 113.421 lb. avoir’s. The scheffel, a dry measure, ==1.5594 bushel, U. S. The eimer, a liquid measure, =—=18.14 gallons, U. S. : The Prussian foot == 12.356 inches,.\U.493 The Prussian ell = 26.256 inches, U. S. | The Prussian mile ==4..68 miles s1Us Site | 100 lb. DENMARK, = 110.28 lb. avoir’s. | The centner is 100 lb. ==110.28lb. avoir’s. | The shippond is 320 lb. = 352.896 lb. The bbl.or toende, a dry meas. 3.9472 bushels, U. S. The viertel, a liquid measure. =2.041 gallons, U. S. The Danish or Rhineland foot 12.356 inches, U. S. The Danish ell is 2 feet == 24.712 inches, U. S. The Danish mile = 4.684 miles, U. S. A cantaro grosso, NAPLES, =196.5 lb. avoirdupois. The cantaro piccolo = 106 lb. avoirdupois. The tomolo, a dry measure, ==1.451 bushels, U. S. The carro is 36 tomoli -==52.236 bushels, U.S. The barile, a liquid measure, ==11 gallons. U.S. The carro of wine is 24 barili 264 gallons. U. S. The palmo, long measure, —10.38 inches, U. S. The canna is 8 palmi = 83.04 onen U. S. 100 1b. or libras, SICILY, |= =70|b. avoirdupois. The cantaro grosso = 192.5 lb. avoirdupois. The cantaro sottile = 175 lb. avoirdupois. The salma grossa, adry measure, 9.77 bushels, U. S. The salma generale == '7385 bushels, U.S. The salma, a liquid measure, ==23.06 gallons, U. S. The palmo, a long measure, =9.5 inches, U. S. The canna is 8 palmi == 76 inches, U. S. 100 lb. of LEGHORN, —=75 lb. avoirdupois. The sacco, a dry measure, ==2-', bushels, U. S. The barile, a liquid measure, ==12 gallons, U. S. 155 braccia, cloth measure, © ==100 yards, U. S. The canna of 4 braccia == 93 inches, U. S. XXXVI. WEIGHTS AND MEASURES. 200 100 lb. peso grosso of GENOA, = 76.875 lb. avoir’s. 100 lb. peso sottile == 69.89 lb. avoir’s. The mina, a dry measure, == 3.426 bushels, U. S. ‘The mezzarola, liquid measure, = 39.22 gallons. U. S. ~The palmo, long measure, = 9.725 inches, U. S. The braccio is 24 palmi == 22.692 inches, U. S. 100 lb. peso grosso, VENICE, 105.18 lb avour’s. 100 Jb. peso sottile = 66.4 lb. avoir’s. The stajo, a dry measure, = 2.27 bushels, U. S. The moggio is 4 staja —=9.08 bushels, U. S. The bigoncia, liquid measure, 34.2375 galls. Uo. The anfora is 4 bigonzi. = 136.95 galls. U. S. The braccio for woollens, = 26.61 inches, U. S. The braccio for silks — 24.8 inches, U. 8. The Venetian foot — 13.68 inches, U. S. 100 lb. of TRIESTE, — 123.6 lb. avoirdupois. The stajo, dry measure, == 2.344 bushels, U. 8. The orna, or eimer, liquid = 14.94 gallons, U. S. The cil for woollens —926.6 inches, U.S. The ell for silks — 25.2 inches, U. S. The Austrian mile —=4.6 miles, U. S. 100 lb. or libras, ROME, — 74.77 lb. avoirdupois. The rubbio, dry measure, = 8.356 bushels, U. S. The barile, liquid measure, = 15.409 galls. U. S. The Roman foot — 11.72 inches, U. S. The mercantile canna — 78.34 inches, U. S. The Roman mile —7.4 furlongs, U. S. 100 Ib. or 100 rottoli, MALTA,=174.5 lb. avoirdupois. The salma, dry measure, —§.221 bushels, U. S. The foot of Malta —111 inches, U. 5. The canna is 8 palmi — 81.9 inches, U. S. The cantaro, kintal, SMYRNA,= 129.48 lb.avoirdupois. _ The oke or oka == 2.833 lb. avoirdupois. The killow, dry measure, — 1.456 bushels, U. 5. The pic, long measure, == 27 inches, U. S. 256 ARITHMETIC. XXXVI. A factory maund of BENGAL, =742 lb. avoirdupois. A bazar maund, = 82; lb. avoirdupois. The haut or cubit == 18 inches, U. S. The guz ==1 yard U: S. The coss or mile = 1.238 miles U. S. The maund of BOMBAY, == 28 lb. avoirdupois. The candy is 20 maunds = 560 lb. avoirdupois. A bag of tice weighs 6 maunds —168 lb. avoirdupois. The candy, dry measure, == 25 bushels, U. S. The haut or covid — == 18 inches, U. S. The maund of MADRAS, = 25 lb. avoirdupois. The candy is 20 maunds = 500 lb. avoirdupois. The baruay, a Malabar weight, —482.25 Ib. avoir’s. The garee, dry measure, = 140 bushels, U. S. The covid, long measure, == 18 inches, U. S. The pecul of CANTON, = 1333 lb. avoirdupois. The catty is100th part of a pecul, = 1.333 lh. avoirdupois. The covid or cobre, long meas. == 14.623 inches U. 8. The pecul of JAPAN, = 130 lb. avoirdupois The catti is 100th part of a pecul, =1.3 ib. avoirdupois. The ine or tattamy, long meas. —6.25 feat,“Ue'S: The bahar of BENCOOLEN, =560 lb. avoirdupois. The bamboo, liquid measure, -==1 gallon, U. S. The coyang is 800 bamboo: == 800 gallons, U. S. ‘The bahar of ACHEEN, = 423.425 lb. avoir’s. The maund of rice =75 lb. avoirdupois. The loxa of betel nuts = 10000 nuts. The loxa of nuts (when good) —168]b. avoirdupois. The pecul of BATAVIA, == 13572 lb. avoirdupois. 33 kannes, liquid measure, 13 gallons, U. S. The ell, long measure, = 27 inches, U. S. The candy of COLOMBO, -=—=8500lb. avoirdupois. XX XIX MENSURATION. 257 XXXIX. MENSURATION. MENSURATION is the art or practice of measuring, and has primary reference to the measurement of super- ficies and solids. Mensuration involves a knowledge of Geometry; and, as that science is not the object of this work, we shall confine our exercises under this head to,those measure- ments, which are most likely to be useful in the ordinary concerns of life. SUPERFICIES OR SURFACE. It has already been taught, that surfaces are measured in squares, and that the area of any square figure, or any parallelogram is found by multiplying together the length and breadth of the figure. For observations on the square and parallelogram, see page 162. Areaor A Ruomsus. «scales sages 233 j Hamburgh eR as ce CY DE” Malta” “mrs sehen sence 992 4 Amsterdam & Antwerp 223) Smyrna .--.---.+..s++-.-+s 234 Portugal ......-.-...-0- Od. 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