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BEV18ED AinO EBTLABOEIX' THE NOVA SCOTIA ARITHMETIC FREPARED AND DESIGNED FOR SCHOOLS AND ACADEMIES, FULLT KXPLAINIKO THE PRINCIPLES OF THE SaEKCB. w W. E. MULLHOLLAND, Teadher of lf»th«n«tlci, &«., in the ProTineitl Vormal Sehcoli Tmro, H. S. VBI8 BOOK IS ▲ OOKPLITB STSTEK OF ARITHMBno. XT COXTAIKI ILL THS N1CSB8ABT EULBS BELATINO TO TBB BECXKAL OVBBBVGT, WITH BUXEBOUS BXAXPLBS. Anihoriied by the Cooneil of Pnblle InatractioB fbrNoTftBooUA, 4 A. A W. MACKINLAY, PUBLISHERS, HALIFAX, N. a 1864. PBOVINCB OF NOTA SCOTIA. Bt IT RBXcmciiBD, That on thin fifteenth day of June, A. D. 1863, A. 4 W. iSAOmi- tAT, of the city of HallCuc, In the said Province, have deposited In thia office the title of a book, the copyright whereof they claim In the words tbllowlng : "The Nova Scotia Arithmetic ; prepared under the direction of the Superintendent of Education : fully explaining the principles of the science. OesiRned fur Schools and Academies. This book Is a complete system of Arithmetic It contains all the necessary rules relating to the Decimal Currency, with numerous examples. Ualitax, N. S. A A W. Ilacklnlay. 1863." In confomlty to duptcr om bondnd and nineteen of the Bevlsed statutes. CHARLES TUPPER. Frovinciat Secretary. B7 WJI. H. KEATING. I>^puty Seerttarif. §-' -* PREPACB. !> h 'r There are only two reasons that warrant, in our estU mation, the publication of another Arithmetic. There is, first, its own intrinsic merit, which may arise either from the exposition of some new principle or principles, or from some new method of arrangement or illustration. Then, there is its suitableness to local circumstances, it may be, to some peculiarity in the educational or financial condition of those for whom it is mainly intended. The title of our book sufficiently indicates the grounds of its appearance, — as to which of the above causes it owes its existence. For a number of years, the educational authorities in the Province have been making an efibrt to introduce Arithmetic into the common schools at a much earlier period than heretofore. If some children are signalized for their observational powers, even from their youngest years, there are others equally so for their calculating or arithmetical powers ; and hence it has been argued, and we think with great cogency, that it is as incumbent on the Teacher to use every legitimate means for the culture of the latter as it is for that of the former. But to teach arithmetic to the young when they enter Bchool, in adap- tation to, and for the development of, their intellectual viU nUCFACE. l< endowment, requires special treatment, — even the teach- ing of their understanding tlirough the medium of their perceptive faculties ; in other words, arithmetic must bo taught concrctel}'-, not abstractl3\ Hence the cause of the division of the following treatise into two parts — the first being intended for beginners of five or six j^cars of age, and the second for those who, having gone through that course, have reached their eighth or ninth year. Again, the value of mental arithmetic is now universally admitted, and is being introduced into all schools of note from the primary to the more advanced. It is of great practical utility to persons engaged in business, but it is of still greater benefit for disciplining the mind, exer- cising and unfolding at once the powers of attention, memory, reasoning and abstraction. It ought to precede the use of the slate, accompany it at every stage, and also succeed it. It holds a similar place to Arithmetic on tho slate that Mental Composition does to that on paper. Hence, to save the expense of a separate treatise on Mental Arithmetic, special exercises in this department have been given throughout the work. Bat the most important local circumstance that seems to demand a new Arithmetic, is the recent introduction of the decimal currency into our Province. When the Bill (bat legalized this currency had passed the Legislature, it was considered advisable in the Normal and Model Schools to work the arithmetical exercises both ways ; first, according to the pounds, shillings, and pence system, and then the decimal, and thereby still to make use of the old aritlimetical books. This mode of introducing ^tke4€!imid.citfTency into our schools is making but slow nUFACE. is tcach- ' their jst bo ise of — the lars of irough ersally )f note r great lit it is , excr- cntion, precede ge, and letic on paper. :iso on rtment .1 seems Ition of be Bill [tnre, it Model ways; system, luse of fduclng it slow progress, and thus has rendered it escpcdient to try some other means more likely to effectuate the object ; and this has been the issuing of the following work. These arc the principal causes that have led to tho publication of the Nova Scotia Arithmetic, As to tho plan pursued, or the way in which the work is done, it does not become us to speak. We may, however, say a word or two on a few of its more prominent features. It will be seen that an unusually large space has been devoted to the discussion of what is generally termed the fundamental rules, viz. : Addition, Subtraction, Multipli- cation and Division. Under Multiplication, a considera^ rablo number of contracted rules has been given, which, it is hoped, will be found not only a pleasing feature of the work, but well adapted for mental culture. Mental exercises follow the exposition of each rule. These will not only make the pupil familiar with the prin- ciple involved, but enable the teacher readily to see whether the pupil has thoroughly understood its appli- cation. Decimals, to a limited extent, have ii^en introduced im mediately after the fundamental rules. There is no reason wh}' a child may not work decimals up to a certain stage, as well as integral numbers. And the sooner ho becomes acquainted with these the better, now that our currency is based on this form of notation. The more important Tables of Weights and Measures, &c., are presented in a collected form, at the end of Part First. In Part Second, they arc treated separately, and ai considerable length, in their origin, history and applica- tion ; and this on the principle that the pupil ought to be \n S* nucFACC. thoroughly drilled in one thing till he i.s completely master of it, before he proceeds to another. To ensure this thoroughness to a greater degree, miscellaneous exercises have been appended to every portion of the worlc. It will be observed that Pi'aetice is taken up before Proportion, as it is more applicable to ordinary mercan- tile transactions. In the discussion of Proportion, we have endeavored to show that a knowledge of Addition, Multiplication and Reduction is all that is necessary to produce efficiency in this department of arithmetic, and to point out, step by 6tep, that Interest, Discount, Commission, Fellowship, <&c., are mere applications of this rule. It has been our aim so to explain the rationale of eveiy rule that we have oftentimes left ourselves comparatively little space for examples. The intelligent Teacher can easily supply this deficiency. H i PREFACE TO THE SECOND EDITION. -•♦•- In issuiDg a new edition of the Nova Scotia Arithmetio the author and the publishers desire to tender their sin- cere thanks to the Press and the Teachers, generally, throughout the Province, for the favorable opinion they have formed of the book ; and to express the hope that this edition will be found yet more worthy of their com- mendation. The only exception, we believe, taken to this Arithme- tic, has been thefeioTiess of the examples. This objection is now, we trust, fully met. Nearly 1000 new examples, arranged in consecutive order and rising, step by step, from the easy to the more difficult, are now appended. Several errors, both in the figures and in the letter- press, unavoidably marred the first edition. These have all been carefully corrected. Every example that is given in the book has been worked and proved ; and we cherish the hope that few or no mistakes will now be found to exist. Notwithstanding the work has been much enlarged, and the quality of the paper and the ink greatly improved the publishers have decided to issue this edition at the same price as the former, hoping thereby to place it within the reach of all. It 91 1 HINTS FOR THE TEACHING OP ARITHMETIC. BY THE SUFEBUrrENDEHT OF EDUOATION. I. Arithmetic, like Penmanship, can only be effectively taught in accordance with what is usually designated the Training System ; that is, the pupils must do the exercises themselves. II. In the first part of the following Arithmetic^ the pupils are supposed to perform the exercises, mentally, with the aid of visible objects ; and, afterwards, with the aid of objects unseen, but with which they are perfectly familiar. These exercises should be engaged in simultaneously, by all the children from 5 or 6 to 8 or 9 years of age, not longer tlian 5 or 8 minutes at a time, though oftentimes repeated throughout the day. To do full justice to thic: initiatory department, the teacher should possess a Black- board, an Arithmeticon, or Ball frame, a box of bricks, a good collection of the current coin of the country, of the weights and measures, &c. III. In the various stages of slate arithmetic, the pupils must be thoroughly classified, if the emulative prin- ciple of tlieir nature, tlirough the sympathy of numbers, is to be operated upon and taken advantage of. In a miscellaneous school the arithmeticians may be con- IV HINTS ON TEACHING ARITHMETIC. veniently divided into four groups, embracing:-!. The youngest children who are not yet using slates. 2. Those working the simple and compound iundamental rules, viz. : Addition, Subtraction, Multiplication and Division. 3. Those in Vulgar Fractions, R-actice, Proportion, Inter- est, &c. 4. Those in Decimals and the higher rules. Each of these groups jaay be subdivided, coiTcsponding to the exact attainments of the pupils, though one person could easily overlook the subdivisions of any one gi-oup. rV. When any pupils evince general superior excel- lence in accuracy and expertness, they should at once be transferred to the higher groups, and others from the lower invited to take their place. This furnishes a much more powerful stimulant than the taking of places, and is preferable in a moral point of view. V. An hour should be devoted to this branch every day ; and the first half hour of every alternate day all the classes in group second should be collected into the gallery, or raised up benches, at one time, and thoroughly drilled in one or other of the simple rules ; at another time, those in the compound rules ; and, again, those in the more advanced rules — Practice, Proportion, Fractions, &c. These exercises should be conducted, at one time, with the mind alone, and at another, with the slate. These frequent revisals are of paramount impor- tance. In fact, it is the want of being thoroughly grounded in the common rules, that accounts for so few persons being good arithmeticians; and so it is in other branches of education. VI. In no other branch can Monitors, or Pupil teachers, be employed with equal benefit. The pupils, working s^ ■It.- HINTS ON TEACUINO ARITUMETIC. 1 exercises from the text-books or black1x>ard, can be arranged in the raised benches, and superintended by the Monitors, while the master retains the class-room or the platfoim for special instruction in the principles of the science. Vn. Before beginning the lesson, the teacher should see that the slates are all clean ; and, in order to secure this, each slate should have a small piece of sponge attached to it. At the commencement of the lesson, a Monitor, appointed for the purpose, should pass along each bench, carrying a small vessel filled with clean water, into which each pupil dips his sponge. On no account should the pupils be allowed to drop saliva on their slates, or to rub them with their sleeve or any other part of their dress. When the slates are thus prepared, all the pupils should be required to show their slate pencils, and, at a given signal, the work should commence. These prelimi- nary exercises contribute very materially to form and cherish habits of cleanliness and neatness. ti n lers. ing PAUT I. 1 A few special explanations to the teacher in this depart- ment. Art. 1. Children, at a very early period, obtain some idea of number,,there being comparatively few of five years of age that cannot count ten or even twenty. As to the source or origin of such an idea, we do not here say anything. One thing, however, is certain, that the first conception of number in the mind of the young, is uni- formly associated with the object or objects. They have evidently, at first, no idea of number in the abstract, but in the concrete. They understand perfectly the mean- ing of 5 apples or 2 marbles, whereas they know nothing of 5 or 2 in themselves. Tiiis is sufficient to furnish a Key to the teacher in his first lesson to young children, and points out very forcibly the absurdity of teaching the symbols first, and, thereafter, the things S3'^mbolized, This is reversing nature's metliod, and hence we need not wonder that the acquisition of multiplication and reduc- tion tables has always been regarded as a work of such extreme difficulty, as to require some six months for its performance. If it be true that the early notions of children, in regard to number, are associated with objects, then it folio wq that we ought to introduce them to the knowledge of abstract numbers through the medium of objects which appeal to their senses. For this purpose, every common school, and especially ever}?- infant and pri- mary school, should be ftirnished with a ball-frame, the decimal currency coins, as well as the sterling money, and a full set of weights and measures. The blackboard may be used with great effect, both along with, and after, the ball-firame, making strokes or dots, representative ot units. u I 12 PIRST LESSON IN ARITHMETIC — COUNTINti. The second stage in this initiatory process is the num- bering of objects with which the children are perfectly familiar^ but which are not present. The third stage is the enumeration of units, tens, hundreds. Then the children will be prepared to take up the abstract numbers, not as unmeaning symbols, but as the signs of things — of realities. And if this course be pursued, and the children, as they advance, rendered thoroughly familiar with every principle of the science — so familiar that they are able to frame rules for themselves — Arithmetic would become not only one of the best cultivators of the human mind, but one of the most interesting and delightful studies. FIRST LESSON IN ARITHMETIC. COUNTING Art. 2. Here every effort should be made to see that the 3^oung are carried on from the known to the unknown. Every child knows something, less or more, connected with the subject 3^ou wish to introduce to his notice, and this should be made the basis of instruction, and of all future progress. There is, for example, scarcely a child of five or six years old that cannot count ten, and counting should thus plainly form the first lesson in Arithmetic. Take a box of pencils, and, lifting the one after the other, make the scholars of the most initiatory class count ; then take the frame and do the same with the balls. At the commencement, the lessons given to the very j^oung should be short and frequent. After they have been well exercised, by a succession of lessons, in plain counting, yoxa should proceed a step farther, and bring out the idea, that 2 are just two ones, and 3 three ones of the same sort only. To show this more clearly, take two different objects and count Ihem alternately; such as slates and books, lead and slate pencils, boys and girls ; and when you reach ten, ask if there are really ten slates, or ten books. This will lead them to think, and to qee that though there are ten things or objects, there are only five slates and five pencils ; and that though there ON SYMBOLS. 13 are ten children or scholars, there are only five boys and five girls ; and that the great lesson taught by all this, is, that in adding one unit to another, every succeeding one must be of the same nature. Another lesson on tae same subject may consist in taking the scholars on fiom ten to twenty. Put the ten balls of the first line a little asunder ; then take the second ten, one by one, and, as you count eleven, show that it is made up of ten and one ; twelve, of ten and two ; thirteen, of ten and three, and so on to twenty. And if you go farther, point out that a new name is given at the end of every ten. So much as to counting right on, as high as 50 or 100. Another form of the same exercise is counting back- wards, first from ten, then from twenty, &c. Try and vary every lesson, so as to impress the minds of the young with the idea that 20 is just twenty ones ; and 100 a hundred ones ; and the longer these simple points are dwelt upon, you are only weaving into their minds, more thoroughly, that which lies at the basis of the science of numbers. ON SYMBOLS. Art. 3. Here give a short oral lesson on signs. Take a signboard, such as the children may have often seen. It may be the signboard over a shoe-maker's door, with a boot or shoe painted upon it. From this, picture out in words the difference between the sign and the thing sig- nified. Another illustration may be taken from letters being the signs of sounds, and words, of ideas. And so it is in figures. AV'e have here also certain signs or sym- bols expressive of a certain number of units. It would be a ver}' cumbersome, if not an impracticable operation, to write down three ones or six ones, in the shape of dots or strokes, if we wished to convey the idea of three or six ; and, therefore, to simplify matters, symbols are employed. Here the teacher may wi'ite on the blackboard a series of ones, and place their corresponding figure underneath : I II III Mil mil mill mini iiiiiiii iiiiiim 12346 6 7 8 9 llllllllll 10 u FUNDAMENTAL RULES. ADDITION. These characters, nine of "which are significant, and the cipher 0, are sometimes called digits, from the \s'ord that signifies fingers. Here the teacher must explain everything connected with the different designations given to these numbers, such as integer, even, odd, simple, compound, abstract, prime, composite, &c. He should also show the Boman form of notation, as well as the one usually called the Ajabic. In part Second, these subjects are luily considered. FUNDAMENTAL EULES. f^t: ADDITION. i Akt. 4. Here, by an oral lesson, picture out the idea, that the quantity of any thing can either be made larger or smaller, added to, or taken from, increased or dimin- ished. For example, ask the scholars of this class to hold up their left hand, and then put to them the question : How many fingers have you upon it? Five, they will, with one voice, reply. Then tell them to hold up their right hand, and ask : How many fingers or digits have you upon it ? The same answer. Tell them to put all the fingers of both hands together, and How many have you now ? Ten^ will be the immediate reply. Now what is this ? It is adding. You have five fingers on one hand and five on the other, and by putting them both together, you have — Ten; and this is Called the sum, and the act of bringing them together into one, is called — Addition. Poes any of j'ou know the symbol used to represent addition ? No reply. Would you like to know it? Yes. It is this-j- ; and the name given to it is plus. Can you point out the sign or symbol which expresses the sum of any two or more numbers? JS^o. "Well, then, I wUl tell you. It is =, i|||| + Mill = llllllllll ; that is, 5 5 10 five plus five has for its sum ten, or is equal to ten, and hence it is sometimes called the sign of equality. Her« FCMDA^LNTAL ItL'LXS. — ADDlllOM. 15 give a gi'cat number of exercises, such as the following : (1.) II + 111 = mil; (2.) II + III! = mill; (3.) 2 3 5 2 4k a lll + im = imm,andsoon. 3 4 7 When the children are perfectly familiar with these symbols, by having a great number of examples presented to them, tlie next step is to accustom them to add. For this purpose, very simple questions must at first be taken. Indeed, it would be desirable to begin with the constant addition of one. Thus ;-l ball and 1 ball are 2 balls, 2 balls and 1 ball are 3 balls, 3 balls and 1 ball are 4 balls, and so on. Vary the exercise by substitution of every object in the room, in plate of balls : such as books, slates, pencils, panes of glass, seats, &c. ; any thing to keep up the interest, and prevent the attention of the children from flagging. Then take the ball-frame and add 2. 2 and 2 balls are 4, and 2 balls are G balls, &c. Go on to 10. When the children are well acquainted with this form, take one more systematic : a regular addition table. 2 and 2 are 3 - 4 — 5 — and 4 5 6 7 2 3 4 5 3 and are 5 6 7 8 2 3 4 5 4 and are 6 7 8 9 2 3 4 5 5 and are 7 8 9 10 and as a and so on again to 10. Alter this, various interesting amusing questions may be put. Take the following sample: (1.) William has 6 marbles in one pocket, 5 in another; how many has he altogether? (2.) There are 7 girls in a class ; if 3 more be added, how many will there then be? (3.) Henry got 9 pence, then 5 pence, and a penn}' ; how much money had he? (4.) A man took 3 horses to be shod ; one had to get all liis feet shod, another had to get 3, and the last had only to get 1 ; how many shoes would the blacksmith require to fhmish? (5.) How many feet have a horse, a cow, a sheep, a goose and a hen ? Another sort of exercise may be resorted to on the ball-frame ; and that is the breaking of a simple number down to its constituent parts. Thus laying off on one line 5 balls, and on another 3, the child- ren should be asked, how many should be laid off, on a IG rUNDAilE^^TAL RULE9. — SUBTRACTION, M third line, to make up tlie same number as the two lines contain. Then the}' might be asked what other numbers would make up 8. All to be proved on the frame, that the childi'en may learn to take nothing upon trust. SUBTRACTION. Art. 5. Here explain the term by a short oral lesson. Reversing the former, ask the children to hold up both hands. How many fingers have you on both ? Ten. Take one hand down, and how many fingers do you hold up ? Five. AVell, then, do you now see that you can do some- thing else besides adding to any quantity ? Yes, we can take from. Do an}' of you know what that is called ? No, It is called subtraction, and the symbol that represents it is just one line, — which is called minus. ||||| — || = 5 2 III J llllll — nil = ll« Give a large number of exerci- 3 6 4 2 ses on this plan, and afterwards with the ball-frame in the same systematic way as in Addition. Thus lay off 10 balls, and remove 1, and ask the children, 1 ball from 10 balls will leave how many balls ? balls, for 9 balls and 1 ball will give 10 balls. And so on, till the whole is gone over in the same way. The subtraction of the numbers successively from 10, as 9 from 10, 8 from 10, &c., and similar exercises, should follow. Vary the exercises, as much as possible, to keep up the interest. Take the fol- lowing as a sample: (1.) John has eight pence in his pocket, and he gave James 5 pence for a knife ; how much has John now? (2.) A window has 12 panes of glass ; 5 of them were broken ; how many were entire ? (8.) A dozen birds sat on a tree ; 7 flew away ; how many remained ? (4.) In two weeks a tradesman lost 4 days' work through ill health ; what number of days did he work? (5.) A boy in a hay-field worked ten hours a day; how many in the 24 did he not work? &c. After some practice in these and similar exercises, double sub- traction niay be taken up. Thus, John had 6 marbles, and he gave 2 to James and 3 to Robert ; how many had he left? A butcher had 12 sheep ; he killed 3 of them and lost 4 ; how many had he left? It may be well now to FUNDAMLMaL 1:LLLS. — MULTIPLICATION. 17 join Addition and Subtraction, in evcrj' variety of way. Thus, John .hud C marbles ; he won ft'om Thomas 5 and from William 2 ; and afterwards lost 8 first game, and 5 second ; how many had he then ? Go on in this strain day after day, and week after week, the teacher keeping a note-book, and jotting down some new form ever}' night, so as to go on steadily and progressively. i MULTIPLICATION. , as fol- his how sof ;ire? how >st 4 did rs a ,fter jub- j>les, had land to Art. C. For explanation of the term, give a short oral lesson. Ask the children how many rov.s of panes there are in one of tbo windows of the school-room ? They all cry, foxir. How many panes are there in each row? Three. And how do 3'ou find how many there are alto- gether? The children, looking at the window, begin at once and count one and one are two, and one is three, and so on till they make 12. The teacher again asks, could you not shorten this ? At once they say. Three mid three are six^ and three are nine, and three are twelve. Is there not a shorter way still ? Yes. And they say 4 times 3 are 12. And what is the lesson we are taught by all this? That whenever a number exactl}'' contains a certain num- ber several times, it can be done in a short or abridged form. 12 contains 3 exactly 4 times. 12 is, in this case, said to be a multiple of 3 ; that is, it contains 3 four times, and the act of doing it is called Multiplication. Multiplication is, therefore, nothing but a short or an abridged or a contracted way of doing addition. The sign of this is X • To familiarize children with the fundamental idea of Multiplication, the teacher should write down lines on the blackboard, as follows : and cause the children to repeat the results. II or two ones are 2, or 2 times 1 are 2 III or three ones are 3, or 3 times 1 are 8 II II or four ones are 4, or 4 times 1 are 4 IIIJI or five ones are 5, or 5 times 1 are 5 mill or six ones are 6, or 6 times 1 are 6 and so on, until the first line la familiar. Then the same 18 FUNDAMKNTAL UULES. -^ MULTIPLICATIOSI. ¥\ exercise may be done with the ball-lVamo, or with objects : 4 boys have each a slate, how many have the}' all? The second Hne of the Multiplication Table should be gone over in the same manner, cither with the ball-frame or strokes on the blackboard, thus : II, 2 ones are 2, or 2 times 1 are 2 II II, 2 twos are 4, or two times 2 are 4 II II II, 3 twos are 6, or 8 times 2 are G II II II II, 4 twos are 8, or 4 times 2 are 8 II II II II II, ^ twos are 10, or 5 times 2 are 10, dbc. III, 3 ones are 3, or 3 times 1 are 3 III III, 2 threes are G, or 2 times 3 arc G ill III III, 3 threes are 9, or 3 times 3 are 9 III ill iii III, 4 threes are 12, or 4 times 3 are 12 iii iii iii lii in, ^ threes are 15, or 5 times 3 are 15 and so on, till the whole of the Multiplication Table is accurately learned by means of visible objects, either on the blackboard or ball-frame, or with slate-pencils. It is not meant that the whole table should be taught at once. Numerous examples of a simple kind should be given at each step. Take the following as a specimen : 1. How man}^ hands have 4 or 5 or 6 boj^s? 2. How many feet have 2 or 3 or 4 sheep? 3. How many legs have 6 or 7 or 8 cows? 4. How many half-pennies are there in 4 pence? 5. How many units are there in 3 tens? 6. Bought 3 eggs at 2 pence each ; how much should I pay? 7. What should you pay the milk-man for 4 pints of milk, at 2 pence a pint? 8. Find the cost of 4 loaves, at 5 pence each. 9. If a man travels 3 miles in 1 hour, how far will he travel from 6 o'clock in morning till noon ? 10. If one large ball is worth 5 little ones, how many little ones will 4 large ones be worth? To resolve numbers into their factors, the simplest plan is to arrange the numbers in various forms. Thus, for example, to decompose 12, arrange the numbers as follows : FUNDAMENTAL RULES. DIVISION. 19 l8t ■{ mm mm That is, 6 X 2 = 2 X 6 = 12 (llll 2ncl.{ III! That is, 4 X 3 Mill = 3 X 4 = 12 The children should be made to arrange counters, mar- bles, bricks, or any such things in rectangles, like the fore- going, and from these to deduce their factors. Such exercises will be both instructive and amusing. DIVISION. Art. 7. A short oral lesson, to show why this term is used for another rule. In the first reading class, how many girls are there ? Six, Well, I wish to give them an equal share of 12 apples ; how many have I to give each ? Tkco, And why? Because twice 6 are 12, and by giving 2 to each, I divide equally the apples. This is done a great deal quicker than by subtracting 2 from 12, leaving 10, and 2 from 10, leaving 8, &c. ; and from its showing how often one number is contained in another, it is called Division, and has for its sign -7-. Division, then, is just a short or an abridged way of working Subtraction, in the same way as Multiplication is a short way of working Addition. "When the mind is thus prepared for understanding the nature of the work, the Multiplication Table should be referred to, and used as follows : II, 2 ones are 2, the half of 2 is 1 III, 3 ones are 3, the third of 3 is 1 II II, 2 twos are 4, the half of 4 is 2 III 111 III, 3 threes are 9, the third of 9 is 3 llll llll mi im, 4 fours are 16, the fourth of 16 is 4 and so on with the whole table. Then give general exer- cises, such as the following : 1. How many apples at Id. each, can I buy for 4d. ? 2. How many at 2d. each for 6d., for 8d., for lOd. ? 20 FUNDAMENTAL RULES. — ilLVCTIONAL NUBIBKM. 8. Four whips cost 8d. ; what is the price of 1 ? 4. Three loaves cost 9d. ; what is tlic price of 1 ? 5. Divide a shilling among 3 boys ; how much has each? 6. Tlirow 18 marbles equally into 3 holes ; how many will there be in each hole ? 7. There are 36 boys in a class, seated equally in 4 forms ; how many will be on each form ? 8. A horse galloped 40 miles in 5 hours ; what rate is that per hour ? Note. At the end of every new lesson, give a few exercises on the preceding one. When Subtraction is going on, combine it with Addition ; when Multiplication, with Addition and Subtraction ; and when Division, with Addition, Subtraction and Multiplication. Here there should be given a large number of miscella- neous exercises on the fundamental rules. FRACTIONAL NUMBERS. Art. 8. Give an oral lesson, to show the real meaning of the word Fractional, Children, do you see this apple ? Yes, Is it a whole apple or part of one ? It is a whole one. Could you give me another name for. .whole? Yes, entire. Any other? No answer. Well, I will tell you. An Integer has just the same meaning. Well, this whole or integer apple I am going to cut into two equal parts. Done, and holding up the two parts. What do you call each of these ? The half. Suppose I divide these halves again equally, how would the apple be divided ? Into four parts. And can I go on dividing these as I will ? Yes. And these parts, when applied to number, are called ? 27b answer. Well, then, I will tell you. They are called Fractions, which just means broken. A fraction, then, is just a part of an integer or whole number. Come, now, and I will show you the symbols employed to represent these ft'actions or broken numbers. There is a half i, a third ^, a fourth i^ a fifth {, two thirds }, four fifths (. rCNDAMENTAL liULKS. — FKACTIONAI NUMBERS. 21 Having thus flxcd in tlie mind of the young the idea of a fk*actional number, it is very easy to go on, and, by dividing an apple into a great number of parts, to give more enlarged vic\ys on the same subject. Again, tulco an orange. If I wi?ih to divile this orange equally among three boys, what must I do? How much must each get? Here is half an orange, and I divide it between two boys ; How much of the half docs each get ? How much of the whole? Very likely this question would not be answered. Aa easy demonstration of it would consist in dividing a whole orange into two, then each half into two equal por- tions, and the children would at once see that one half of half an orange was exactly the same thing as one fourth of the whole orange. The pupils having been thus rendered familiar with the elementary ideas of fractions, they should then be led on to the addition, subtraction and multiplication of fVac- tional numbers, the teacher still availing himself of tangi- ble, ocular demonstrations, or visible objects; and he need be at no loss for these. Here follow a few exercises in the fundamental rules. 1. Two halves and one half are how many halves? 2. Three fourths and four fourths, how many fourths? 8. Four fifths and six fifths, how many fifths ? 4. A boy has 8 half apples, and gives 1 of these halves to his neighbor; how many halves has he left? How many whole apples ? 6. If I were to give one boy 6 fourths of an apple, and another boy 2 fourths, which would have more ? how much more ? how many apples more ? Note. All this is to be done with the apple, by sub- mittinsT it to the children's observation. 6. Take 4 tenths fVom 6 tenths; 8 tenths A:om 10 tenths ; 2 hundredths ftom 3 hundredths. 7. From 5 thirds take 2 thirds. 8. Let 6 boys hold up each 10 fingers. Now, if we take two tenths of these 50 tenths away, how many tenths will be left? 22 APPLICATION OP NUMBER TO MONET. 9. Six boys got half an apple each, how many halves had they all ? How many wholes ? 10. In three apples how many halves? 11. How many thirds are 4 times 2 thirds? 12. How many tenths do 4 times 3 tenths make? The teacher can lay the whole school under contribu- tion in these exercises. APPLICATION OF NUMBER TO MONET. # t Art. 9. What is this (the teacher holding up a cent in his hand) ? A cent, James, suppose I were to give you this cent, what would you do with it ? J would buy a pear. Well, then, j^ou go to Mrs. Thompson's, and you ask her to give you a pear for a cent ; she gives you one, and you put down the cent on the counter. But, suppose you found the pear, the moment j^ou took it into your hand, was rotten, what would you do ? / would say to Mrs. Thompson, this pear is rotten ; give me another pear that is worth a cent, or, of the same value. When, then, you buy the pear, you give what you con- sider is of equal value, and what do j'ou call the cent ? Money — a piece of money. Any other name ? A coin. The cent, then, is of equal value, and you exchange it '—for the pear. Could you get the pear for any thing else ? Yes : if I were to give Mrs. Thompson a ball, which would be of a little more value than the pear. Would Ehe like this as well? J^o, Why? Because it is not nearly so convenient; and, very likely, she might not get any per- son to purchase the ball. Do you know any other kind of money? Yes, a dime. How many pears could you get for a dime ? Ten, And why ? Because the dime is worth ten cents. What coin is this ? A quarter dollar. And how many pears could j-ou buy for it ? Twenty-five, because there are twenty-five cents in a quarter dollar. What is this ? Malf a dollar, or a Florin. And how many pears could you pun^hase for it? and so on, till you go over all the decimal currency, holding up every' coin before the childreQ, im^ allowing them to touch it/if they will. : © j u; y APPLICATION OF KUMBE« TO MEASURES. 23 Take a farthing, a penny, a shilling and a pound, and go over them in the same way. Now the children are prepared to make the application in accounts, like the following : 1. John gave 5 cents to James, and 10 cents to Andrew, and had four remaining ; how many cents had he at tirst ? 2. How many dimes are there in half a dollar? in a whole dollar ? in two dollars ? 3. If John has two pennies, how many farthings would vou require to give him for them ? How many half pence ? 4. John has 4 pennies and Jane has 8 ; how many have they between them ? 5. How many pence in one shilling? in two? in three ? 6. How many shillings in one pound? in two? in three ? 7. John goes to the grocers and buys 3 lbs. of sugar at 6d. a pound ; how much should he bring home out of 2s.? 8. James bought a pair of shoes for a dollar and a half ; a jacket for 2 dollars, and a cap for 75 cents ; how -^uch did he pay for all ? These may suffice as a specimen. The teacher can mul- tiply these to any amount, and the more they are within the range of common life, the more clearly will the chil- dren see that Arithmetic has to do with every day trans- actions. APPLICATION OF NUMBER TO MEASURES, WEIGHTS, &a Art. 10. The teacher, having beside him a few inch, foot and yard measures, will proceed to the first lesson, by giving to his pupils the names, if they do n't happen to know them, by comparing their lengths, laying them along- side each other, then drawing them on the blackboard, and requiring the pupils to do the same on their slates or blackboard. Having ascertained by actual measurement that there are 12 inches in a foot, and 3 feet in a yard, the various objects in the school may be taken and tested, the pupils being required to state what they believt to be 24 APPLICATION OP NUMBER TO \VEIOHT8, ETC. ' ( ' i the length of each object in rotation, and then to prove it by meastiring it themselves. Similar exercises to the fol- lowing may be given : 1. In 5 yds. how many feet? inches? 2. John is 5 feet 3 inches, and William is 4 feet 9 inches ; what is the difference ? 3. How many yards of cloth, at 2s. per yard, can be purchased for $ 3 ? 4. I bought 6 yards of cloth for 12 dollars ; what was that for each yard ? The application of number to square measure and measure of capacity, may be taught exactly in the same way. The application of Number to Weights may now be considered. The teacher, being provided with a pair of scales, and the more common of the Avoirdupois, Troy and Apotheca- rie's Weights, will proceed to explain every thing con- nected with the scales, and then give the names of the dif- ferent weights. Then he will take the ounce weight, and putting it into one of the scales, he will proceed to put the dram weights into the opposite scale, and the pupils will then see how many drams are in an ounce, which the teacher will mark on the blackboard, and the pupils on their slates. Again, putting the pound weight into one scale, and the ounce weights into the other, it will be seen how many ounces are in a pound. And so onward, through all the weights. Exercises can easily be given, corresponding with those on the measures. DECIMAL NOTATION. Art. 11. The pupils are now in a state of prepared- ness for the use of the slate. They ought to be kept in the foregoing preliminary exercises for two years at least, so that on the supposition that they entered school at six years of age, they are now about eight. The time they have spent at this initiatory work h^ not only had the efifeot of weaving into their DECIMAL NOTATION. 25 mental frame-work a practical acquaintance with the ele* ments of the science, but will prove of incalculable ser^ vice in their whole subsequent career as arithmeticians < and even irrespective of any practical benefit, the devel*^ opment of mind, in adaptation to its germinative condi* tion, which the great majority thus exercised have experi- enced, is far more than a compensation for all the time they have spent, and all the toil they have undergone. The first thing to be done in slate arithmetic, is to give the pupils clear and correct notions of the decimal system of notation. As this system is entirely a conventional arrangement, the teacher cannot reason it out with the pupils as he would a theorem in Geometry. He must explain and illustrate, by many familiar examples, the principle on which this convention is based, showing that it is neither more nor less than the localizing of figures, by giving every figure placed on the left hand side of another, ten times its former value. Thus I place the figure 6 on the blackboard. It stands there in its own free, indepen- dent condition, as the representative of 6 units of any thing, and is therefore called an absolute figure. • I place another 6 before it, or on its left hand side, 66 ; and accor- ding to the conventional arrangement, it is no longer a representative of 6 units, but of 6 tens ; and as it gets this property from the position it occupies, it is technically called its local value. And suppose I were to prefix another 6 to the two already on the blackboard, its value would be ten times greater than the former 6, that is, it would no longer represent tens, but hundreds, or ten tens. And so onwards. This arrangement, so beautiful and so simple, was no doubt suggested to its inventor hy the ten digits or fingers of the human hand. But be its direct origin what it may, ' to us it is perfectly manifest that the principle underlying the whole is that of Classification. So multitudinous and varied are the objects of nature, and so limited is our capacity, that to facilitate the memory of these objects, even in reference to the matter of number, some arrange- ment or classification was indispensable, in which classifi- cation we have the finest material or food provided for the exerciso of our rational faculties. I 86 DECIMAL NOTATION. This principle, we hold, lies at the very foundation of the decimal system of notation ; at all events, we know of no more effectual way of rendering it plain and palpable to the understanding of the young, than by resorting to it. Suppose that I had a very large box, say, of lead pen- cils, and that I wish to divide them ; what am I to do ? I place them before the children, and proceed to reduce them to something like a methodical arrangement, by tying them up in bundles of ten, until I have gone over them all, when I have, say, eight remaining. I place these eight aside, and write the figure 8 on the blackboard, telling the children that that figure represents the eight pencils I hiive set aside, or 8 ones. I now tie every ten of these bundles together, and find, say, five remaining. Place these bundles of ten to the left of the eight ones, and write on the blackboard the figure 5 to the left of the 8, telling the children that that figure represents the five bundles of ten. Take the last made bundles, which contain each ten tens, or one hundred ones, and tie them also together by tens. Suppose I have two of these, with tlirce of the bundles of one hundred remaining. I place aside to the left of the five bundles of ten, these three of one hundred, and write on the black- board the figure 3 to the left of the 5, to represent three bundles of one hundred. Lastly, I place to the left of the three bundles of one hundred, the remaining two bundles, which contain each ten hundreds or one thousand, and •write the figure 2 to the left of the 3, to represent the 2 thousands. I have now on the blackboard 2 3 5 8, which represents respectively two bundles of one thousand each, three bundles of one hundred, five bundles of ten, and eight ones. Or we may take another way of representing the prin- ciple of decimal notation, by the drawing of lines on the blackboard. I wish, for example, to write in figures fifteen. I draw fifteen lines on the boaril, placing live in one place and ten in another, because fifteen is thus compouiul. I put down 5, and place 1 as representative of ten before it, and I call this 15. And so on, up to 20, when the same pro- cess should be repeated. Two children might be requested DECIMAL NOTATION. 27 to hold up each his ten fingers, and one three fingers. "We have thus 2 tens and 3 ones ; but 2 tens and 3 ones make 23. In this way the representation of numbers from 1 to 99 can be easily illustrated. The same process must be again gone over, and the children taught that to put a figure two places to the left, increases its value one hundred times. For some time, however, it will be desirable not to pro- ceed beyond thousands. Exercises in tens and units ; hundreds, tens and units ; thousands, hundreds, tens and units, should be given, so as to render the children perfectly at home in the ^decimal system of notation. Example 1. Add together 26 and 32. Write on the blackboard tens. units. 2 6 8 2 5 8 Here we have 6 units and 2 units, which make 8 units, and 2 tens and 3 tens, which make 5 tens ; and so we have 5 tens and 8 units, which make 58. And so on, to hundreds and thousands. Ex. 2. Add 5 shillings and 4 pence, and 4 shillings and 3 pence. Shillings. Pence. 5 4 4 3 9 7 Here we have 4 pence and 3 pence, which make 7 pence, and 5 shillings and 4 shillings, which make 9 shillings. fe 28 EZERCISSI. ..f EXERCISES. 1. A boy spent in the grocer's store 78. 6d., and in the baker's 8s. 4d. ; how much did he spend altogether ? 2. John paid 3s. 8d. for a jacket, and 3s. 2d. for a vest ; what did he pay for jacket and vest together ? 3. James sold 6 lb. 8 oz. of sugar to one boy, and 8 lb. 3 oz. to another ; how much did he sell altogether? 4. The length of this room is 30 ft. 6 in., and that of the class-room 15 ft. 4 in. ; what is the length of the two together? The exercises in Subtraction, Multiplication and Divi- sion, should be precisely of the same nature as the forego- ing, and gone over in the same careful and accurate manner. The more practical the questions, the more will they tend to show that Arithmetic is no abstract, barren science, but one of the utmost utility, and necessary for the most common transactions of life. To give variety and consecutiveness to these exercises, the teacher ought to keep a memorandum-book, and jot down every night the general character of those that are to be given out the following diay. ▲BITHMETICAL TABLES. 8» «H ^^ g, 2 ►» « a ►• a 2 S »0Q ?• "3 a 0:= a & fl p^ b:3 3 o-g 01 .a u » ij PL^S^UU P-ri v^ p>4 »i^ --^ ■ KHi.^* t- H "^ 1 1 1 i 1 I O « d ^ >** i 1 1 • V 1 1 ■»* ' ' III! writ ^^ 1 1 1 1 I S OB J3 m 1 r^w 59 g' &§:& bCbcbC a 6 a 2«.S.S.S 222 t:§=f== ^tJ"S^ CO a>^.aid rt CS 3J (i^a^cnooco ^Ci^Pm ^SS-'S ^c^eo 53 ^ ^ ^ 8 I— s,sS!§!g|S 1-1 I3S ^ S S K!88S[S|g,H • S j8 ^ S s ess Q C »H c^ r-t i-( ri 3 o> 00 't- 1-1 |N eo in iC CO l (N CO ^8|^ % ■* |S Oil- g M IS!^ ^!s'gl8 H •* OO s i-l ©< (M ?< CO CO ^ ,-r irp fiQ SO «o A 00 ■* t- 'o !eo ec « M ;« CO eo 1 1 1 N "* « 00 © 1-4 a r-t rH 00 I-( |(M '■* M |CNl |C^ IH N eo ■* to « b- 00 o> rl r-t rH a o ••1 a fe as o IH o > 2 o< o 5 9 « C D O ^^ to M a O.SZ OS r es a oeooocM N O 1 •S3 » P '^ « 1 I 3 « s o o h O Oi Ph s 00 H •s •-I CD M< o •s a a V Put 00 -a o rH I— 3 .a bo ci"— •« >- fc d « a 2^ 2H o a OS a rH rH rH rH ~ J) a I I I rH^ I u a a _ a o s-F S 30 ARITHMETICAL TABLES. |: i ^ » » OB 2 ♦-"C-z Oj^^u-d m2 = = = tc cyan C5 a, aa o'-J cZ cj ^ r^ „ ^ pH PH rH IH ,-t rH r-< pH rH ,-t ^ ^ «S ^ I 3 • I I CO ^- o a o o See ta go V (;^ v •♦"foOOSt'-ou- 9t CS3' 3308^3335 «.2^'2' 0'Ci';5 CU( M O":!: (» =Q 35 X d, o « « O 4>»^^ « 3 3 . «-i 2 k" 3 9) a • • o « H a V a 4j 4, a, ^ ^ u S e4 cS P 3 3 QQaOCQ 1-1 65 I W i •S-a oa go 2 0) o u 5"! ^"2 o3 '>^>i .2 « •23 fl 3 3 ^'J r-tr-4 D V . cc M ' •< a S S ' OB ' fl ■ o 00 .q H lit! »-l'»T-t< s i^ •a a •d § a B 3 3 00 I li H P> Q 8^ I- ■* to PAET II. S> ^ u P Si 2 I- o «t « g si CD tD kO o Article 1. Arithmetic may be considered either as a science, or as an art. As a science, it teaclies the prop- erties of numbers ; as an art, it enables us to apply this knowledge to practical purposes ; the former may be called theoretical, the latter practical arithmetic. 2. The grand elementary principle or idea in this science, is unity or a unit ; and the only way of impressing it upon the mind is by presenting to the senses a single object, as one apple or one pear. 3. There are three signs by which this idea of one is expressed and communicated : the word one, the Roman character I, and the figure 1. 4. If one be added to one, the idea thus arising is different from the idea of one, and is complex. This new idea has also three signs, viz. : Two, II, and 2. 6. If we begin with the idea of the number one, and then add it to one, making two ; and then add it to two, making three, and so on, it is plain we shall form a series of numbers, each of which will be greater by one than that which precedes it. Now, one, or unity, is the basis of this series of numbers, and each number may be expressed in the three waj^s before mentioned. 6. Number, then, is the name by which we signify how many objects or things are considered. 7. Numbers are considered either as abstract or con- crete. Abstract numbers are those \^ich have no reference to any particular kind of unit ; thus, five, as an abstract number, signifies five units only, without any regard to particular objects. Concrete numbers are those which have reference to some particular kind of unit ; thus, when we speak of five 31 82 AKITHMETIC. I |l horses, seven cows, the numbers, five, seven, are Said to be concrete numbefs, having reference to the particular units, one horse, one cow, respectively. 8. All numbers in common arithmetic are expressed by means of the figure 0, commonly called zero or a cypher, which has no value in itself, and nine significant figures: 1, 2, 3, 4, 5, G, 7, 8, 9, which denote respect- ively the numbers one, two, three, four, five, six, seven, eight, and nine. These ten figures are sometimes called Digits ; but this name is often improperly limited to the nine significant figures above mentioned, which are then called nine digits. This admirable method of writing all numbers by the use of the ten digits, is of great antiquity. It is some- times called the Arabic method, on the supposition that it was invented by the Arabians. This, however, was not the case, as we have the highest authority for believing that the knowledge of this method was communicated to the Arabians by the Hindoos, and as it cannot be tr&,ced to any remoter period, the latter people are entitled to all the honor of its invention. But if the Arabians cannot claim its invention, they have the honor of introducing it into Europe. When they invaded Spain, about the begin- ning of the eighth century, it was in common use among them, and it is probable that a knowledge oi it was soon afterwards communicated to the inhabitants of Spain, and gradually to those of the other European countries. 9. When any of these figures stands by itself, it expresses its simple or intrinsic value ; thus 9 expresses nine abstract units, or nine particular things ; but when it is followed by another figure, it then expresses ten times its simple value. Thus 94 expresses ten times nine units, together w^ith four units more ; when it is followed by two figures, it then expresses one hundred times its simple value ; thus, 943 expresses one hundred times nine imits, together with ten times four units, together with three units more ; and so on, by a ten fold increase, for each additional figure that follows it. The value which thus belongs to a figure, in conse- quence of its position or place, is called its Local Value. AUITIIMKTIO. S3 aid to Acular rcsscd I or a lificaiit Dspcct- scvcn, called to the L-e then by the 3 some- 1 that it w'as not elieving ;atod to e triiced ed to all 3 cannot Lucing it le begin- among as soon lain, and Is. Itself, it q^resses when it kn times ^e units, by two simple le units, [h three for each conse- LOCAL A system in which ten is the commoi ratio, i ^ called Decimal. Our system is, therefore, a decimal one. li the common ratio were sixty, it would be a sexagesimal system; if twelve, a duodecimal. Both of these wciv; formerly in use, and, as will be seen from the tables, still, to some extent, retained. 10. It appears, tlien, tliat in common Arithmetic, we proceed towards the left from luiits to tens of units ; from tens of units to tens of tens of units, or hunch-eds of units ; from hundreds of units to tens of hundreds of units, or thousands ; from tliousands of units to tens of thousands of units ; from tens of thousands of units to tens of tens of tliousands of units, that is, to hundreds of thousands of units ; hence to tens of hundreds of thou- sands of units, or millions of units, and so on to billions, trillions, quadrillions, &c. Thus 10 represents one ten of units, together with no unit ; or, as it is briefly read, ten. 11 represents one ten of units, together with one unit, and is briefly read, eleven. Similarly, 12, 13, 14, 15, 16, 17, 18, 19, respectively represent one ten of units, together with two, three, four, five units, &c. The next ten numbers are expressed by 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, which respectively represent two tens of units, together with no, one, two, three, four, five, six, seven, eight, nine units ; they are briefly read twenty, twenty-one, &c. The next ten numbers are expressed by 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, which are respectively read thirty, thirty-one, thirty-two, &c. ; we thus arrive at 40 (forty), 50 (fifty), 60 (sixty), 70 (seventy), 80 (eighty), 90 (ninety) — 99 being the largest number which can be expressed by two figures, since it represents nine tens of units, together with nine units. The next number to this is 100, which represents ten tens of units, or one hundred of units, together with no tens of units, together with no units ; or is briefly read, one hundred. By pursuing the same system in higher numbers, the figure occupjang the fourth place from the right hand will represent so many tens of hundreds of units, or thousands of units ; the figure in the fifth place will represent so many tens of thousands of units ; and so on 2* 34 NI'MKIIATION. .ii 6473 represents five thousands of units, together with four hundreds of units, together with seven tens of units, together with three units, or, as it is briefly read, five thousand four hundred and scvcntv-thrcc. NOTATION AND NUMERATION. 11. Notation is the art of expressing any number by figures which is already jrivon in words. Numeration is the converse of Notation, })cing the art of expressing any number in words which is already given in figures. Numeration Table. M •8 < CO 43 987 654 321 •a ^ The terms hundred, thousand and million, are primitive words, and bear no analogy to the numbers which they de- note. The terms billion, trillion, quadnllion, &c., are formed from million and the Latin numerals, his, tres, quatuor, &c. Thus, prefixing his to million, by a slight contraction, for the sake of euphony, it becomes billion, &c. with inits, , Ave >cr by ION is igany NUMERATION. 85 { 3 53 S 9 HP 3 2 1 rimitive [hey de- pc, are |s, tres, slight 12. To express in words the numbers denoted by lines of figures » RuLK. (1.) Commencing: at the rifjfht hnnrl side, divirlo the given fiij^ures into periods of tlu'ee flgurcH cacli, till not more tluin tiu'ee rcinuin. (2.) Then, coniiuenciiig at the left liiind, annex to tlie value expressed by the lii^urea of each i jriod, except that of the units, the name of the period, uccording to the numeration table. Thus, the expression, 37053907, becomes, by division into periods, 37,053,907, and is read thirtij-seven millions, fifty-three thousand, nine hundred and seven, the term units or ones, at the last, being omitted, as was explained in Art. 10. Note. — Before the pupil be allowed to put the : ')ove rule into prao- tioe, he should bo well exercised on the priuoiples laid down in Art. 10. Thus, how many tens of units are there in 46, now many tens of tens of units, or hundreds, together with how many tens of units or tens, to- gether with how many units or ones are there in 467; and so on to thousands, &o. EXERCISES IN NUMERATION. "Write down in words, or name the numbers signified by the following exercises : Ex. 1. 24 Ex.4. 1000 Ex. 7. 1284 2. 124 6. 1200 8. 6789 8. 465 6. 1230 9. 36847 Ex. 10. 10 076 Ex. 11. 876543219 13. To express numbers by figures. Rule. Make a sufficient number of dots or cyphers, and divide them into periods of three each. Then, com- , mencing at the left, place, in their proper jjosition, beneath the dots or cyphers, the significant figures necessary for expressing the proposed number. If any places remain unoccupied, let them be filled with C3'phers. Thus, the method of expressing the number two hundred and five millions, twenty thousand, seven hundred and nine, will be found in the following manner ; 000, 000. 000 2 5 2 7 9 and hence, by filling the unoccupied places with cjrphers, we get 205,020,709. »., I !^ 36 NOTATION. EXERCISES IN NOTATION. Express the following in figures : 1. Fifty-four. 2. One thousand and twenty-four. 3. Sixtj^-four millions, three hundred and ten thou- Bands, four hundred and six. • 14. Roman System of Notation. Our ordinary Numeral Characters have not been always, nor every where, used to express numbers. The letters of the alphabet, as being well known, naturally presented themselves for the purpose, and accordingly were very generally adopted. For example, by the Hebrews, Greeks, Eomans, &c., each of course using their own Alphabet. The pupil should be acquainted with the Roman Nota- tion, on account of its being used to denote chapters, sections, and other divisions of books and discourses. It will be found, by referring to the following table, that the Romans used ver}^ few characters — fewer, indeed, than we do, although our system is still more simple and effective, from our applying the principle of " position," unknown to them. They expressed all numbers by the following symbols, or combinations of them : I, V, X, L, C, D, M. The rules for Roman Notation are as follows : By a repetition of a letter, the value denoted by the letter is represented as repeated: as 11 represents two; XX represents twent3\ The annexing a letter of a lower value to one of a higher, denotes the sum of both, or their joint value : as VI denote 6, XII denote 12. The prefixing a letter of a lower value to one of a higher, denotes their difference : as IV denote 4 ; IX denote 9. Every 3 annexed to I3, increases the value ten times. Thus, Io3 denote 5000. Every C a:;d 3 to the left and right of CT3, increases the value ten times. Thus, CCIqq denote 10,000 ; CCCI033 denote 100,000. NOTATION. 87 thou- ilways, letters Bsented re very Greeks, labet. 1 Nota- bapters, ses. g table, , indeed, pie and )sltion," iymbols, by the ^ts two; 16 of a lilue: as le of a 4; IX times, icreases 110,000 ; A line over any number increases its value a thousand fold. Thus, X denotes 10,000, L denotes 50,000. In constructing their sj-stem, they evidentl}^ had a qui- nary in view, that is, one in which five would be the com- mon ratio ; for we find that they changed their character, not only at ten, ten times ten, &c., but also at five times five, &c. As far as notation was concerned, what they adopted was neither a decimal nor a quinary system, nor even a combination of both. They appear to have supposed two primary groups or periods, one of five, the other of ten, and to have formed all the other groups or periods from these, by using ten as a common ratio of each resulting series. They anticipated a change of character, one unit before it would naturally occur. In this point of view, four is one unit before five ; forty, one unit before fifty — ten being now the units under consideration ; four hundred, one unit before five hundred — hundreds having become the units contemplated. TABLE OF ROMAN NOTATION. Characters XIX XX XXX, &c. Anticipated > ^r change. 3 Change L LX, &c. Anticipated > y^ change 5 Change C CC, &c. Anticipated ) qj\ change. > Change. D or Iq Anticipated > q^t Change. M V X Characters. Nos . Expres'd. I 1 II 2 ni 3 ^^^^l^"^^ 4 Change. V 5 VI 6 VII 7 VIII 8 Anticipated ) tv change. 5 9 Change. X 10 XI 11 XII 12 XIII 13 XIV 14 XV 15 XVI 16 XVII 17 XVIII 18 Nos. Expres'd 19 20 30, &0. 40 50 60, &o 90 100 200, &0. 400 500 900 1,000 5,000 10,000 It may be doubted whether any other ratio of increase would, on the whole, be more convenient, than that of the 38 ARITHMETICAL SIGNS. decimal system. If the ratio were less, it would require more places of figures to express larger numbers ; if the ratio were larger, it would not indeed require so many llgures, but the operations would manifestly be more diffi- cult than at present, on account of the numbers in each order being larger. Besides, the decimal system is suffi- ciently comprehensive to express with all desirable facilitj^, every conceivable number, the largest as well as the smallest ; and yet so simple that a child may understand and apply it. In a word, it is in every way adapted to the practical operations of business, as well as the most abstruse math- ematical investigations. In whatever light, therefore, it is viewed, the decimal notation must be regarded as one of the most striking monuments of human ingenuity, and its beneficial influence on the progress of science and the arts, on commerce and civilization, must win for its unknown author the everlasting admiration and gratitude of mankind. ARITHMETICAL SIGNS. 15. The sign = (equal) indicates that the numbers between which it is placed, are equal to one another. Thus, 12 pence = one shilling. The sign -f- (P^^s), placed between two numbers, indi- cates that they are to be added together. Thus, 5 + 7 = 12. The sign — (minus), placed between two numbers, indicates that the latter is to be take7i from the former. Thus, 12 — 7 = 5. The sign X (multiplied by), denotes that the number which precedes it is to be multiplied by that which follows it. Thus, 4 X 6 = 24. The sign -7- (divided by), signifies that the number which precedes it is to be divided by that which follows it. Thus, 24 -^ 6 = 4. But division is more generally indi- cated by placing the di\ddend above a line, and the divisor 24 under it. Thus, — = 4. H|| ARITHMETICAL SIGNS. ^ The signs • . . - -,. expression 2 ; *3 • *. n . o ^^c^:*'^ proportion. Thus fh^ to 9. •^••6:9, denotes that 2 is to 3, as is q The siofn ' ' / . within them « ^ ^ ' '"""""^ ™«bers the Vinci ,Tr "i^^lT^ ^' all n„„,bers under »i^:l«'^tX^-l-„tside the viL. oL^^^^ and conversely. ""^ ^"y' «"««' affect 3 i„ the same wty that tleC:b:r i: toTetulliS' f. f^ »«'nber of timej by *. placed above tt 'ro^f Tt"l}f' ''« ^-P-'e-ed Vt^ •' "^" "i' the second, Mrd «L /' *'i® expression, V, Sign of Cm6« root ; as ^S' 97 « '-'"'^ 27 = 3, or (27) J =3. ^ >/ S^n Of the ^,,^^, ,„„^ ^ ^ (1296)* = 6. ^' ^^ beSe!'- -^-"es therefore. The sign ... .^.^^ By-h^ir^r t:i?4r:r™^^ «» -^th these advantage, they will sTrv^^„^'°^' «? they ought, agrea? ^n be no doubt that The einr^*»«.'»barrass hta.' C characters, and not by words T-f?" "?'' 11. tends to processes which are to beX-ZV T^"^ *«« of the they are indicated the tett^'*™«f both. ^t or rep' form but Vhen the J have as lication." umber is " Involu- s to the added ; not the ind and Jd under Qd COM- or more ning but 'ts which r, called ore con- denomi- lays, or 'undamen- tion, and however, ier each s under draw a SIATPLE ADDITIOX. Find the sum of f)i« « i ^^ full sum c^r tu ^" ^'^^ same wav anf] J; -J .^ ^^^^ »um ot the extreme infv u y ' ^"^^ ^'^ite down iha. «um, so marked down wUl ^^i^""^ •'°'««n- The "„tir« ^oceeding by 5469 743 27 6239 . carry on tl.t 1 ? ""."' "»<^«'- ^e eolmnn ^ ' '"^ ^^'e- Now, the sum of i f 42 SIMPLE ADDITION. \ 18. Tlie preceding example might have been worked thus, putting down at lull length the. local value of tfie figures : Thus, 54G9 = 5000 + 400 + 60 + 9 + 743 = 700 + 40 + 3 + 27 = 20+7 Now, adding the columns, we get the sum — = 5000 + 1100 + 120 + 19 = 5000 + 1000 + 100 + 100 + 20 + 10 + 9 (Since 1100 = 1000 + 100, 120 = 100 + 20, 19 =10-1-9) = 6000 + 200 + 30 + 9 (Collecting the thousands together, the hundreds to- gether, and so on), = 6239. Note.— To the farmer and mechanic, the accountant as well as the mathematician, accuracy is indispensable; and as this can only be acquired by long experience, each sum should be proved, either by a direct method, by analysis, of the foregoing rule (we prefer the latter, as by it, these ^-ules are rendered practical by their application), and no pupil should be satisfied with the correctness of the result, till so tested. 19, First Method of Proof. Begin at the top, and add the columns downward, in the same way as they were before added upward, and if the sums agree, the work is presumed to be right. Note.— The reason of this proof is, that, by adding downward, tho order of the figures is inverted; and therefore, any error made in the first addition, would probably be detested in the second. This method of proof is genemlly used in business. Second Method of Proof. Separate the given numbers into two or more divisions. Find the sums of these divi- sions, severally, and add these partial sums together. If the last result be equal to that found by the commo method, the work is right. Note.— This method of proof depends on the axiom, that the whole of a quantity is equal to the sum of all its parts. Third Method of Proof. Commencing at the left hand, add the several columns without carrying^ and set down the ftill sum of each column, with the units in their proper ^;ii f SIMPLE ADDITION. 43 place, and the tens below the figure immediately to the left. Add together these lines thus resulting, and if the last result agi'ee with that obtained, by the common method, it may be concluded that botli are right. Thus, in the annexed example, the sum of the left hand column is 25, which is set down in full : the sum of the next column is 30 ; the cypher is set in its proper place, and 3 under the 5 ; and so with the rest. The sum of the two lines thus obtained, is equal to the sum found by the ordinary method. This method of addition might be used instead of the common method ; and, as it requires nothing to be carried, it may be employed with advantage when the calculator is liable to inter- ruption. Fourth Method of Proof. Cast the 9s out of each of the given numbers separately, and place each excess at the right of the number. Then cast the 9s out of the sum of these excesses ; also cast the 9s out of the amount ; and if these two excesses are equal, the work may be supposed to be right Thus, as in the example : 5946 excess of 9s 9738 2697 9868 5946 9738 2697 9868 28249 25029 322 28249 u (( (( (£ (( = 6 = = 6 = 4 28249 excess of 9s = 7 = 7 Note. — ^This mode of proof is based on a peculiar property of the number nine, whioh will be further illustrated under Multiplication. Mental Exercises in Addition. 1. Add 17, 8, 11, 14, 15, 10, 6, 5, 9; how many? 2. There were 4 boxes of oranges. The first contained 527, the second 265, the third 69, and the fourth 72 ; how many oranges were there altogether ? Note. — It is not necessary to multiply examples here, as the teacbe< can supply them to suit the capacity of his pupils. u SiMPLE ADDITION. rt 18. Exercises in Slate Arithmetic. • I 1. 13678632 46795320 4679532 19848015 8309169 93310668 2. 6584334156 1290887091 7876a2123 356796432 45699543 9065409345 8. Add together 7384, 326, 6780, and 57. Ans. 14547. 4. Add together 89, 4500, 423, 2024, 5408, 60546, and 9401. Ans. 82391. 5. 986759 + 4976346 + 29483 + 898647 + 3984753 + 6489778 + 57893 + 2468144 + 576989 + 498653. Ans. 20967445. 6. £ 7654 + £ 50121 + £ 100 + £ 76767 + £ 675. Ans. £135317. 7. $ 10600 + $ 7676 + $ 6760 + $ 90017. Ans. $115053. 8. A merchant owes to A £ 1500 ; to B £408 ; to C £ 1310 ; to D £ 50 ; and to F £ 1900 ; what is the amount of all his debts? Ans. £5168. 9. A merchant received the following sums : $ 200, $315, $317, $10, $172, $513, and $9: what is the sum of all? Ans. $1536. 10. A merchant bought 7 casks of merchandize. No. 1 weighed 310 lbs. ; No. 2, 420 lbs. ; No. 3, 388 lbs. ; No. 4, 335 lbs. ; No. 5, 400 lbs. ; No. 6, 412 lbs. ; No. 7, 429 lbs. ; what is the weight of the whole ? Ans. 2694 lbs. 11. A farmer sold 9 loads of hay. The first weighed 2065 lbs. ; second, 1896 lbs. ; third, 1467 lbs. ; fourth, 2000 lbs. ; fifth, 1867 lbs. ; sixth, 2891 lbs. ; seventh, 1872 lbs. , eighth, 784 lbs., and the ninth, 1740 lbs. ; what is ihe total weight? Ans. 16582 bis. SUinUACTlUN. 45 12. Add together the following numbers Fifteen thousand, seven hundred and ninety-six ; four hundred and nine ; two hundred and thirtj'-four thousand and fifty ; four millions, three thousand and seventy-six ; four thousand and thirty-six ; ten thousand, nine hundred and one. Ans. 42682G8 Note.— The pupil should not be allowed to leave Addition until he can, •with ffreat rapidity, continually add any of the nine digits to a given quantity. Thus, beginning with 9, to add 6, he should say: 9, 15, 21, 27, 33, ^c, without hesitation, or further mention of the numbers. For example, he should not be allowed to proceed thus : 9 and 6 are 15; 15 and (3 are 21, &c.; nor even, 9 and 6 are 15; and 6 are 21, &c. He should be able, ultimately, to add the following, or any other : 5638 in this manner : 2, 8, 16 (the sum of the first col- 4756 umn, of which one is carried, and 6 to be set 9342 down) ; 5, 10—13; 4, 11—17; 10, 14—19. 19736 SUBTRACTION. 20. Subtraction is the method of finding what number remains when a smaller number is taken from a greater number. * The number found by subtracting the smaller of two numbers from the greater, is called the Remainder. There are two kinds of Subtraction, Simple and Com- pound, which difffer from each other in precisely the same way in which Simple and Compound Addition differ from each other. The sign — (minus) placed between two numbers, sig- nifies that the second number is to be taken from the first number. (Art. 15.) SIMPLE SUBTRACTION. 21. Rule. Place the less number under the greater, so that units come under units, tens under tens, hundreds under hundreds, and so on ; then draw a straight line underneath. 46 SUBTRACTION. Ir : I Take, if possible, the number of units in each figure of the lower line from the number of units in each figure of the upper line, which stands immediately over it, and put the remainder below tlie line just drawn, units under units, tens under tens, hundreds under hundreds, and so on ; but if the units in any figure of the lower line exceed the number of units in the figure above it, add te.i to the upper figure, and then take the number of units \n the lower figure from the number in the upper figure thus increased ; put down the remainder as before, and then carry one to the next figure of the lower line. The entire difference or remainder, so marked down, will be the difference or remainder of the given numbers. 22. In order that the difference of two numbers may remain the same, if we increase one of the numbers we must increase the other by the same quantity : for exam- ple, five apples minus two apples equal three apples, or 6 — 2 = 3, and increasing each of the numbers by 1, 6 — 3 = 3. In like manner, 5 tens — 3 tens = 2 tens, and increas- ing each number by 1 ten : 6 tens — 4 tens = 2 tens, the same as before. This axiom will enable us to explain the rule of Sub- traction. «* 23. Subtract 356 from 634. Hundreds. Tens. Units. Here, as wc Cannot take 6 units from X f ^ 4 units, we borrow one of the tens from 3 5 6 2 7 8~ tens to take from 2 tens, (why not from 3 tens ? because we took or borrowed one to put along with the 4 units) as this cannot be done, we take or borrow a hundred or 10 tens from the 6 hundreds, and then we have 5 tens from 12 tens (for 10 tens -f- 2 tens =12 tens) and 7 tens remain ; lastly, we have 3 hundreds from 5 hundreds, (why not 6 hundreds ? because we took or borrowed one of them to put along with the two tens as above) and 2 hundreds remain. In the course of the foregoing demonstration we had 5 tens to take from 12 tens, now the result will not be altered if we increase each of the numbers by 1 ten, that the 3 tens, and then 6! units from 14 units, and 8 remain : we have now 5 SUBTRACTION. 47 is, if wc say, 6 tens from 13 tens ; in like manner, in the place of saying 3 hundreds from 5 hundreds, we may say, without altering the result, 4 hundreds from 6 hundreds : thus establishing the common rule of Subtraction, when we "borrow" 1 from a figure in tlie top line, we must "carry " 1 to the next figure in the bottom line. 24. As a farther illustration of the above, let us take the following examole. Subtract 4938 from 5123 accord- ing to the rule 6123 Here we cannot take 8 units from 3 units, we 4938 therefore add 10 units to the 3 units, which are thus increased to 13 units ; and taking 8 units from 185 13 units we have 5 units left ; we therefore place 5 under the column of units ; but having adde*' 1 ten units to the upper line or number, we must add the same num ber of units (1 ten units) to the lower number, so that the difference between the numbers may remain the same ; and adding 1 ten units to the 3 ten units in the lower number, we obtain 4 tens or 40 instead of 3 tens or 30. Again, we cannot take 4 tens from 2 tens ; we therefore add 10 tens or 1 hundred to the 2 tens, which thus becomes 12 tens, or 120 ; and then taking 4 tens or 40 ii'om 12 tens or 120, we have 8 tens or 80 remaining ; we therefore place 8 under the column of tens : but having added 1 hundred to the upper number, we must add 1 hundred to the lower number, for the reason given in Art. 21 ; and adding 1 hundred to 9 hundreds in the lower number, we obtain 10 hundreds or 1000 instead of 900. i Again, we cannot take 10 hundreds or 1000 from 1 hundred and we therefore add 10 hundreds or 1 thousand to the 1 hundred, which thus becomes 11 hundreds or 1100 ; and taking 10 hundreds or 1000 from 11 hundreds or 1100, we have 1 hundred or 100 left; we therefore place 1 under the column of hundreds ; but having added 10 hundreds or 1 thousand to the upper number, we must add 1 thousand to the lower number, for the reason given above ; and adding 1 thousand to the 4 thousands in the lower number, we obtain 5 thousands or 5000 ; and 5 thousands or 5000 taken from 5000 leaves ; therefore the whole difference or remainder is 185. /"^ 48 ftUBTllACTIUN. ill D! ' 24. The above Example might have been worked thus, putting down at Aill length the local values of the figures : 6123 = 6000 +100-1-20+3 = 4000 + 1000 +100 + 10 + 10 + 3 (collecting the first 10 with the 100, and the second 10 with the 3.) = 4000 + 1000 + 110+18 and 4938= 4000 -f- 900 -[- 30 + 8 100 + 80 + 5 = 185 The truth of all results in Subtraction may be proved by adding the less number to the difference or remainder ; if the sum equals the larger number, the result obtained by subtraction may be presumed to be correct. Second Method. Cast the 93 out of the larger number, and place the excess to the right. Next, cast the 9s out of the less number and place the exc ss to the right of this number also, then take the latter excess, if possible, from the former excess, and note the difference : if the latter excess cannot be taken from the former, add 9 to it, and then take the difference and set it down as before ; lastly, cast the 9s out of remainder, and if this excess be equal to the difference of the above excesses the result obtained by subtraction may be considered right. Thus, 5123 excess of 9s =2 adding 9 = 11 4938 " 9s =6 and 6 * 186 excess of 98 = 5 = 5 Mental Exercises in Subtraction. 1". A chest of oranges contained 703 ; but 285 were bad. How many good ones were there ? Ans. 418. k 2. A boy had a 103 marbles, and lost 86 of them, how mmiy had he left? Ans, 17. I 8. A man bought a horse for $186 and sold him again for $226. a. How much did he gain by the bargain. ^ Ms, $39. \. MULTIPLICATION. 49 4. A farmer procured 2896 poles for fencing, he took 1689 of them to go round a pasture. How many has ho left? Ana 1207. 5. A farmer raised 896 bushels of wheat, and sold 675 bushels of it ; how many did he reserve for his own use ? An8, 221. Exercises in Addition and Subtraction. 1. 16 -f 17 + 19 + 10 + 7 — 8 + 11 — U + 16 — 20. Ans. 54. 2. A man owing $767, paid at one time $190, at anotlier time $131, at another time $155 ; how much did he then owe? Ans, 291 dollar^. Exercises for the Slate. | 1. From 45079 take 32048. Ans, IZOZir 2. From 1896481035 take 736792632. Ans. 1159688403; •. 3. From 4673842150 take 2186367025, Ans. 2487475125. 4. 153425178—53845248. Ans. 99579930. 5. Required the difference between three, and three huiidred thousand. Ana, 299997. 6. Mont Blanc, the highest mountain in Europe, is 15,680 feet high ; and the height of Chimborazo, the highest in America, is 21,427 feet ; how much is the latter high'^r than the former? Ans. 5747 feet. 5 MULTIPLICATION. 25. Multiplication is a short method of finding the sum of any given number repeated as often as there are units in another given number (Art. 16.) thus : when 3 is multiplied by 4, the number produced by the multiplica- tion is the sum of 3 repeated four times,, which sum is equal to 3 + 3 + 3 + 3 or 12. The number to be repeated, or added to itself, is called the Multiplicand. The number which shows how often the multiplicand is to be repeated or added to itself, is called the Multiplier. 50 M SIMPLE MULTIPLICATION. The number found by the multiplication is called the Product, The multiplicand and the multiplier are sometimes call- ed ''Factors' because they are factors or makers of the product. 26. The process of Multiplication is founded upon the axiom that any quantity taken a certain number of times is the same as the parts of that quantity taken the same number of times ; for example, 3 times 27 = 3 times 20 + 3 times 7. The truth of this axiom will be rendered more apparent by arranging counters, as in the following figure. • •• • • • • • • • ••• • • • •• • • = 4 times 5 = 4 times 3 + 4 times 2. In the same manner any other case may be illustrated. 27. Multiplication is of two kinds, Simple and Com- pound. It is termed Simple Multiplication, when the Multiplicand is either an abstract number, or a concrete number of one denomination. It is termed Compound Multiplication when the mul- tiplicand contains numbers of more than one denomination but all of the same 'kind. , I 28. One of the factors, namely the multiplier, must necessarily be an ' abstract number^' since it would be ab- surd to speak of 6 shillings multiplied by 4 shillings. We can multiply 6 shillings by 4, i. e., we can find how many shillings there are in four times six shillings ; but there is no meaning in 6 shillings multiplied by 4 shillings. SIMPLE MULTIPLICATION. 29. Rule. — ^Place the multiplier under the multiplicand in such a way that units maj^ be under units, tens under tens, and so on. Mnlti]>ly each figure of the multiplicand, beginning with tlie units, b}'' the figure in the units' place of the multiplier (by means of 1 lie table given for multipli- cation in the first part) ; set down and carry as in addi- tion. Then multiply each figure of the multiplicand, be- ginning with the units, by the figure in the tens* place of the mi^ti|)lier, placing the first figure so obtained under g r5» B f«gayT i «»< i i «ag»ll8W^'iM*M«* W |« «T| i I iW< w*^ SIMPLE MULTIPLICATION. n led the es call- of the poll the f times le same 3 times pparent !S 2. rated, d CoM- len the oncrete lie mul- lination , must be ab- fllings. id how but lllings. Ilicand I under fcand, place lltipli- addi- l, be- ice of ider 5;Jo78 6888G 220G2 the tens of the line above, the next figure under the hun- dreds, and so on. Proceed in tlie same way with each succeeding figure of the multiplier. Then add up all the results thus obtained, by the rule of Simple Addition. Note— If the multiplier does not exceed 12, the multiplication can and should be effected in one line. Ex. Multiply 7654 by o97. Proceeding by the rule given above we obtain, 7G54 Here we place the 7 units under the 4 397 units, the 9 tens under the 5 tens, and so on. When 7054 is to be multiplied by 7, we first tako 4 seven times which by the table gives 28, i. e. 8 units and 2 tens ; we therefore put the 8 in the units' place and carry on the two tens : again 5 tens taken 7 times gives 35 3038038 tens, to which add 2 tens, and we obtain 37 tens, or 7 tens and three hundreds ; we put down the 7 in the tens' place, and carry on the 3 hundreds : again, 6 hundreds are to be taken 7 times which gives 42 hundreds, to which add 3 hundreds and we obtain 45 hundreds, or 4 thousands and 5 hundreds ; we put down the 5 in the hundreds' place, and carrj^ on the 4 thousands : again,' 7 thousands taken 7 times gives 49 thousands, to which we add the 4 thousands, thus obtaining 53 thousands which we write down. Next, when we multiply 7654 by the 9, w6 in fact mul- tiply it b}' 90 ; and 4 units taken 90 times give 360 units, or 3 huudreds, 6 tens and units : therefore, omitting the cypher, we place the 6 under the tens place, and carry on the 3 to the next figure, and proceed with the operation as in the line above . When we multiply 7654 by 3, we in fact multiply by 300 ; and 4 multiplied by 300 gives 1200, or 1 thousand 2 hundreds tens, and units : therefore omitting the cyphers, we place the first figure 2 under the hundreds* place, and proceed as before. Then adding up the three lines of fiaurcs as in simple addition, we obtain the pro- duct of 7054 by 397. 30. If the multiplier, or multiplicand, or both, end with cyphers, we may omit them in the working ; taking I 52 SIMPLE MVLTIPIICATICK. care to affix to the product as man}" cyphers as we have omitted from the end of the multiplier or multiplicand,* or both. Thus, if 263 be multiplied by G200, and 570 be multi- plied by 3200, we have 263. 670 ^ 6200 3200 526 1678 V 1630600 114 171 1824000 r i . The reason is clear ; for, m the first case, when we mul- tiply by 2, in fact we multiply by 200 ; and 3 multiplied by 200 gives 600. In the second case, the 7 multiplied by the 2 is the same as 70 multiplied by 200 ; and 70 multi- plied by 200 gives 14000. I 31. If the Multiplier contains any cypher in any other place, then, in multiplying by the different figures of the multiplier, we may pass over the cypher, taking care, how- ever, when we multiply by the next figure, to place the first figure arising, from that multiplication, under the third figure of the line above, instead of the second figure. The reason of this is clear ; for, if we were multiplying by 206, when we multiply by the 6, we take the multipli- cand 6 times ; when we multiply by the 2, we really take the multiplicand, not 20 times, but 200 times. 32. When two numbers are to be multiplied together, it is a matter of indifference, so far as the product is con- cerned, which of them be taken as the multiplicand or multiplier ; in other words, the product of the first, multi- plied by the second, will be the same as the product of the second multiplied by the first. Thus, 2X4 = 2 + 2 + 2-f2 = 8 4X2 = 4-1-4 =8 Therefore the results are the same, that is, 2 X 4 = 4X2. 33. We have hitherto confined our attention to products formed by the multiplication of two factors only. Pro- ducts may arise from the multiplication of three or more SIMPLE MULTIPLICATION. 53 ire have and,* or multi- v^e mul- iltiplied )lied by ) multi- ly other 3 of the 'c, how- lace the ler the figure, iplying lultipli- lly take jether, lis con- md or multi- ofthe 4 = ^ducts Pro- more factors. This is termed Continued Multiplicfition. Thus, 2X3X4, denotes the continued multiplication of the factors 2, 3, and 4 ; and means that 2 is first to be multi- plied by 3, and the product thus obtained to be then mul- tiplied by 4. The result of such a process would be 24, which is, therefore, the continued product of 2, 3, and 4. We may thus express it : 2 X 3 X 4 = 24. 34. Methods of Proof. First, Make the multiplicand the multiplier, and the multiplier the multiplicand ; proceed as before, and if the results are the same, the work may be considered right, i Second, Begin at the left hand of the multiplicand, and add together its successive digits towards the right, till the sum obtained equals or exceeds the number nine. If it equals it, drop the nine, and begin to add again at this point, and proceed till you obtain a sum equal to, or greater than, nine. If it exceeds nine, drop the nine as before, and carr}^ the excess to the next figure, and then •J "iinue the addition as before. Proceed in this way till ; J.. : ave added all the figures in the multiplicand, and re- jected all the nines contained in it, and write the final excess at the right hand of the multiplicand. Proceed in the same manner with the multiplier, and write the final excess under that of the multiplicand. Multiply these excesses together, and place the excess of nines in their product at the right. Then proceed to find the excess of nines in the product obtained by the original operation; and if the work is correct, the excess thus found will be equal to the excess contained in the product of the above excesses of the multiplicand and the multiplier. Example. Multiplicand, 12345 = 6 excess Multiplier, 2231 = 8 excess 12345 37035 24690 24690 27541695 = 48 = 3 •Proof. 3 54 STMPLE MULTIPLICATION. Note. — This method of proof, though perhaps sufficiently sure for common purposes, is not always a test of the correctness of an opera- tion. If two or more figures in the work slioulJ be transposed, or the value of one figure be just as much too great ns another is too small, or if a nine be set down in place of a cypher, or the contrary, the excess of nines will be the same, and still the work may not be correct. Such a balance of errors will not, however, be likely to occur unless designedly effected. Tliird. Commencing at the units of the multiplicand, add together the digits in the odd places, rejecting 11 as often as it occurs (the same as the 9 in the former), and reserve the result. Proceed in the same manner with the digits in the even places, and from the former result, increased if necessary by 11, take this result, and place the excess opposite the multiplicand. In a similar way, find the excesses in the multiplier and product ; then mul- tiply the excesses of the two factors together, and find, in the same manner, the excess of their product. If this excess, and the excess of the product of the two factors be the same, the work is generally correct. If they differ, it must be wrong. Ex. Multiplicand, 84963 = 9 — 10 = 10 Multiplier, 44085 = 9 — 1 = 8 Product, 3745593855 = 1 — 9 = 3, 80 excess = 3. In the above example (the work of which, for brevity, is omitted), we say 3 + 9 are 12, which exceeds 11 by 1 ; 1 -j- 8 are 9 ; then 6 + 4 are 10, which being taken from 20, = (9 + 11), the excess is 10. Then in the multi- plier, the sum of 5 + + 4 are 9, from which 1, the excess of 8 + 4 above 11, being taken, the excess is 8. In the product, 8 + 5 are 13 ; 2, the excess, and 9 are 11, which is rejected ; 5 and 7 are 12, the excess of which is 1 ; then taking the even numbers 5 and 3 and 5 are 13 ; the excess 2 + 4 + 3 are 9 ; and 9 from 12 (= 11 + 1), and the excess is 3. The product of the first two excesses is 80, and the excess of lis is 3, the same as the excess of the product of the factors. Fourth. Divide the multiplicand by 11, and set the excess or remainder to the right. Do the same with the multiplier ; then divide the product of these excesses by SIMPLE MULTIPLICATION. ^ 11, and set clown the remainder. Also divide the product of the factors by 11, and if the excess of lis be the same as tlie excess of 1 Is in the product from the above excesses, the work is right. Thus, in the foregoing example, the excess of the multiplicand is 10, and the excess in the multiplier 8, which being multiplied by 10, gives 80, the excess of which is 3. The excess of the product is also 3, which proves the work to be right. Kv/TE. — This method is not applicable in this place, as the pupil is not supposed to be able to do Division. The reasons for these last three methods of proof ia-re given in Art. 93, props. 15 and 18. Mental Exercises in Multiplication.^ 1. A ship sails 11 miles in an hour ; how far will she sail in 7 hours ? in 4 hours ? in 8 hours ? 2. At 3 shillings a pound, what will 12 lbs. of tea cost? 6 lbs.? 8 lbs.? 3. If the interest of 1 dollar is 6 cents for 1 year, what is the interest on 8 dollars for the same time? of 10 dollars ? 4. If a box hold 36 apples, how many will 6 boxes of the same size hold? 8 boxes? 10 boxes? 5. If a man earn 4 dollars in one -v^eek, how many will he earn in 8 weeks ? IMental Exercises in Multiplication, Addition, and Subtraction combined. 1. 3 fives, and 6 fives, and 7 fives, are how many fives? are how many ? 2. John has 5 pens ; Robert has 5 times as many, and Charles 6 times as many as John ; how many times 5. penS) and how many pens do they all possess ? n^ 56 SIMPLE MULTIPLICATION. I . 3. Lucy has 10 cents ; her father gave her 6 more, her brother 5, and she worked a purse which she sold for 37 cents ; how many cents had she remaining after buying 3 books at 8 cents each ? 4. 6 + 8 + 7 multiplied by 7, + how many are 160? Exercises for the Slate. 1. 1816 X 10 2. 40376 X 40 1. 18160 2. 1615040 3. 40376 X 400 4. 81967 X 13 ANSWERS. 3. 16150400 4. 1065571 7. 14S X 53 8. 958 X 34 9. 31416 X 175 10. 15607 X 3094 11. 1368752 X 72 12. 1267887621 X 468 .13. 14638887425 X 3672 5. 7854 X 16 6. 4079 X 12000 5 125664 6. 48948000 ANSWESS. 7844 32572 5497800 48288058 98550144 693371406628 63753994624600 14. Find the continued product of 12 17, and 19. Ans. 3876. 15. Find the continued product of 3781, 3782, and 8783. Ans. 54095923986. • 16. How many yards of linen are there in 759 pieces, each containing 25 yds. ? Ans. 18975. 17. It is found by microscopic observations, that in each square inch of the human skin there are about 1000 pores ; and the surface of the body of a middle-sized man contains about 2304 inches, or 16 square feet. Required, the number of pores in the surface of such a body, 999 being supposed to be contained in each inch ? Ans. 2301696. CONTRACTIONS IN MULTIPLICATION. 35. The general rule is adequate to the solution of all examples that occur in Multiplication. In many instances, re, her for 37 ring 3 5 160? :X 16 12000 25664 948000 La. 28 |4600 9. 876. 2, and 986. )ieces, 175. Ihat in 1000 id man [uired, r, 999 t96. SIMPLE MULTIFLICATIOK. 57 |f all inces, however, by the exercise of judgment in applying the pre- ceding principles, the operation may be very much abndged. 36. The contractions under this rule, if perfectly familiar both in their extent and application, will enable the pupil to abridge very materially nearly one-half of the usual business processes in multiplication. The abridged processes can be applied with a very great saving of time and labor, in computing interest and other usual counting-house calculations. The contractions, which are limited in their application, if they have but little practical value, will ser\'e as excel- lent exercises for mental training. iToTK. — ^The following contractions may be studied by the pupils as soon as they are prepared to understand them. The principle of each contraction should be fuUy shown to the pupils. 37. Any number which may be produced by multiply- ing two or more numbers together, is called a Composite number. Thus, 4, 15, 132 are composite numbers ; for 4 = 2 X 2 ; 15 = 5 X 3 ; 132 = 4 X 3 X 11. Note. — The factors which, bein^ multiplied together, produce a com- posite number, are sometimes called component parts of the number. The process of finding the factors of which a given number is composed, is called resolving the number into factors. The factors into which a number may be resolved, must not be con- founded with the parts into which it may be separated. The former has reference to Multiplication, the latter to Addition. Thus 56 may be re> solved into two factors, 8 and 7; it may be separated into two or more parts, 50 or 5 tens and 6, or 25 and 31. 38. To Multiply by a Composite Number. Resolve the multiplier into two or more factors, and proceed as in Con- tinued Multiplication, (Art. 33.) 39. To Multiply by 5. Add a cypher, or rather con- ceive it to be added to the multiplicand, and take half the result. This is evidently the same as multiplying by 10, and tak ng half the product. 40. To Multiply by 15. Conceive a cypher to be added to the multiplicand, and to the result add half of itself «8 »8 SIMPLE MULTIPLICATION. , 41. To Multiply by 25. Conceive two cj-phers to be added to, the multiplicand, and take one fourth of the re- sult. In like manner, because 125 is one eighth of 1000. To multiply by 125, conceive three cyphers to be added to the multiplicand, and take one eighth of the result. 42. To Multiply by 75. Conceive two cyphers to be added to the multiplicand,, and from the result take one fourth of itself. 43. To Multiply by 175. Divide 700 times the multi- plicand by 4. 44. To Multiply by 9, 99, 999, or any other number of 9s. Annex as many ciphers to the multiplicand as there are 9s in the multiplier ; from the result subtract the given multiplicand, and the remainder will be the answer required. 44. To Multiply by 11. Set down the unit figure of the multiplicand for the first figure of the product, then add the digit in the ten's place in the multiplicand to the units of the same for the second figure of the product ; proceed in the same manner with the others. Thus — 186743 X 11 We first set down the 3, then 4 + 3 = 11 7, in the place of tens ; 7 + 4 = 11, • which gives 1 in the place of hundreds, 2054173 and one to carry ; 6 + 7 + 1 = 14, setting down the 4 in the place of thousands, and carrying 1 to the next place ; and so on, we obtain the correct re- sult. This is simply multiplying by 10 first, and adding once the multiplicand at the same time. 45. To Multiply by any Multiple of 11, By the figure denoting how many times 1 1 are contained in the multi- plier, multiply first the unit's figure of the multiplicand, then the sum of the units and tens, then the sum of the tens and hundreds, &c., and lastly the left hand figure. Thus, 176 X 55 = (5 X 11) 6 5X6 6 + 6 9680 30, then (7 + 6) 5 + 3 = 68, and (1 + 7) 46, and finally 5X1 + 4 = 9. i SIMPLE MULTIPLICATION. 59 to be le re- 1000. led to to be e one multi- .her of i there ct the inswer ure of b, then to the oduct ; 3 = = 11, idreds, = U, figure Imulti- llcand, )f the 7) 46. To find the Product of any Two Numbers between 10 and 20. Arrange in decimal order, first the right hand fignre of the product of the units, then the right hand figure of the sum of the units, then the product of the tens. Note.— In writing figures in the decimal order, whenever the product or sum contains more than one figure, the left haud figure must be added to the figure in the next place on the left. Ex. 16 X 17 = 272. Proceeding by the rule given above, the right hand figure of the product of the unit? (6 X 7) is 2 ; the right hand figure of the sum of the units (6 + 7) is 3, plus the tens of the product of the units (4) is 7 ; the product of the tens (1 X 1) plus 1, i& 2 = 272. The reason of tne rule may be illustrated by writing at fUU length the local value of the figures Thus, . 16 r= 10 + 6 and 17 = 10 + 7 We have first the product of the units, then the 7 units by the 10, and the 6 units by the 10, which is the same as (7 -|- 6) X 10, or 13 in the ten's place ; and lastly we have the product of the 1 ten by 1 ten = 100, or 1 in the place of hundreds. 42 13 1 272 Note. — All the following rules may be illustrated in nearly the same manner. Exercises. 1. 2. 3. 4. 5. 17. IC X 12 18 X 12 19 X 16 15 X 14 18 X 13 6. 7. 8. 9. 0. 14 X 19 15 X 13 17 X 14 19 X 13 13 X 12 11. 12. 13. 14. 15. 19 X 19 19 X 18 19 X 17 19 X 15 17 X 15 5. 18 X 13 0. 13 X 12 15. 17 X 15 17. In one foot tnere is 12 inches; how many inches are there in 13 feet ? in 16 feet? 18. In one pound, Troy, there are 12 ounces ; how many are there in 15 lbs. ? in 16 lbs. ? I I i 60 SIMPLE MULTIPLICATIOK. Examples. 21 X 21 31 X 21 41 X 21 61 X 21 5. 61 X 21 6. 71 X 21 7. 81 X 21 8. 91 X 31 9. 10. 11. 12. 31 X 81 41 X 81 51 X 31 61 X 31 19. In one pound, Avoirdupois, there are 16 ounces ; how many are there in 1 7 pounds ? how many in 1 3 pounds ? 20. How many rods in a piece of land that is 18 rods long, and 17 rods wide? 47. To find the Product of any Two Numbers^ of two figures each, when the unit* a figure in each is 1. rrange in decimal order the unit figure (1), the sum of the tens, and the product of the tens. Ex. 61 X 71 = 4331. The unit figure is 1 ; the sum of the tens (6 + 7) = 13 ; the product of the tens (6 X 7), plus the right hand figure of the sum o^ the tens (1), is 43. 1. 2. 8. 4. 13. In one guinea there are 21 shillings ; how many are there in 61 guineas? 14. If I pay 31 cents for one pound of butter, how many cents will I have to pay for 51 pounds? Note. '-By the use of the extended Multiplication Table, we can use the above rule when there are three figures, aa 121 x 171, or we may make use of the former rule to multiply 12 by 17. The same may be said of all the others. 48. To find the Product of Two Numbers, when the sum of the units is ten, and the preceding figure or figures are alike in each. Multiply the preceding figure or figures of one of the numbers by the preceding figure or figures of the other, increased by one, and prefix the product to the product of the units. When the product of the units is less than ten (10), a cypher must be written in the ten's place. Ex. 21 X 29 = 609. The preceding figure of one number (2) by the same figure increased by one (2 + 1), is = 6 ; the product of the units (9 X 1), is_9 ; written with a cypher prefixed, 09. how 1. 2. 3. 4. 62 X 68 63 X 67 84 X 86 91 X 99 BlUPLK MULTIPLICATION. Exercises. 61 141 X 149 161 X 169 95 X 96 ^S4 X 186 195 X 195 194 X 196 174 X 176 33 X 37 minute, how ''«*^^Ji9ufes-ti%^',tV< ?"" ^""^^-o containing, ti''^'irtf^eteneisW^'l''ll/l""'f^<*re alike, SZ the unit's figure of one^ f ^. * ^""^^""^ °^ the tens, Jl the units Thus? 24 xl I aS'Jf "J? '""^ ^'^^^ot of the units is less thanten riof , !" ^^^^ **« P^duct in the place of tens. ^ "•*• * ''^P^'er must be written the^nit^flg^te ST - M- ^h*^"*" ^'^ *« *«»« ^ 16 d1u« ^ 22 X 82 f- 23 X 83 3. 47 X 67 4- 49 X 69 Exercises. 89 X 99 99 X 19 ^6 X 56 76 X 36 -^- ^SXS8] 10. 33 X 73 --.>.., ^. 7g 11. 163 X 43 3. ^^l^atwill46bus.oatscosat36 ; ''' ^ '^^ i4. 93 sheep at 13 dollars elet? """*' ^'^ '^^^^^ sum of the units is 10 aJ^ /l <• J"^^ ^^^wiJcr* t^;;^^;, ,,,, ce^n^me unit, Jlgl::iVf;Z\^^^^ less 1, to the difference bltwepn f h! ^ *^^ ^^^«'' number the larger number and 100 ^^ '^"^"^ ^^ the units of Ex. 126 X 134 =: ifiQia^ rw,. 62 SI.'fPLE MULTIPLICATION. Note. — The product of a number multiplied into itself is called the square of that number. Exercises. 1. 21 X 39 5. 43 X 57 9. 64 X 66 2. 81 X 49 6. 51 X 69 10. 55 X 65 8. 27 X 33 7. 52 X 48 11. 101 X 119 4. 36 X 44 8. 53 X 67 12. 149 X 151 13. How many square feet are there in a lot of land that is 146 feet long, and 134 feet wide. 61. To find the Proditct of Two Numbers containing two figures each, when the figures in the place of units are alike. Arrange in decimal order, first the right hand figure of the product of the units ; then the right hand figure of the product of the sum of the tens by the unit's figure of one of the numbers ; then the product of the tens. Ex. 27 X 47 = 1269. The product of the units (7 X 7) is 49 ; the right hand figure is 9. The sum of the tens (4 + 2), multiplied by the units (7), gives 42 ; plus 4 (from the 49) are 46 ; the right hand figure is 6. The product of the tens (4 X 2) is 8 ; plus the 4 (from the 46) are 12 = 1269. Exercises. • 1. 2. 3. 4. 22 X 32 22 X 72 23 X 33 24 X 44 6. 26 X 46 6. 26 X 56 7. 28 X 48 8. 28 X 88 9. 10. 11. 12. 29 X 59 29 X 69 29 X 89 29 X 99 13. In one day there are 24 hours ; how many are there in 84 days ? 14. A person bought 65 turkeys at 65 cents each; what did they cost him ? 52. To find the Product of Two Numbers containing two figures each when the figures in the place of tens are alike. Arrange in decimal order the right hand figure of the SIMPLE MULTtPLICATXON. 63 59 ,69 89 199 are ich; hng \are the product of the units, the right hand figure of the pro- duct of tlie sum of tlio units by one of tlie tens, and then tlie product of tlic tons. E\-. 46 X 47 = 2102. The product of the units (7 X 6) is 42 ; the riglit hand ligure is 2. The sura of the units (0 -f- 7), multiplied by one of the tens (4), gives 62 ; plus 4 (from the 42) are 56 ; the right hand figure is 6. The product of the tens (4 X 4) is 16 ; plus 6 (from tho 66) are 21 = 2162. 1. 37 X 87 2. 88 X 38 3. 43 X 47 Exercises. 4. 88 X 88 5. 99 X 91) 6. 87 X 83 7. 72 X 78 8. 97 X 97 3. 106 X 107 53. To find the Product of any two Mixed ^ timbers whose fractional parts are halves. I'refix the ^ loduct o( the whole numbers, i>lus one half their sum, tr the fracti. i i. Thus, ^X^= 12i. When the sura of the whole number is not an even number, add one half of tne next lower number to the product, and prefix the sum to J. Thus, 2^ X SJ = SJ. Exercises. 8-^ X 12^ H X 11^ H X m 9} X r^'- 9. 10. 11. 12. m X m m X Hi- 12^ X 12^ 11^ X 12J 1. 6^ X 8^ 5. 2. ^ X lOj 6. 3. 7^ X 8-]. 7. 4. 7x X 111 8- 13. If a man w^alk 8| miles in one hour, how many miles can he walk in 15^ hours? 14. "What will 16^ lbs. of laisins cost at 12^ cents a pound ? 64. To find the Product of Two Mixed Numbers, when the whole numbers are alike, and the sum of the fraction is 1. Multipl}'' one whole number by the other, increased by 1, and prefix the product to the product of the fractions. Thus, 6f X 6f = 42 A . i ■■ ! i : DIVISION. • Exercises. H X H 6^X 5f 4. 7i X 7i •'>. 7t^ X 7^ 6. 8-^ X 8J 7. 8AX 8H 8. 10-ft. X lOA 9. 10-1-^ X 10t\r 64 1. 2. 3. 10. Coal from Pictou is sold for 8f dollars a chaldron ; what is paid for S^ chaldrons ? 55. To find the Product of any Two Mixed Numbers, when the difference of the whole numbers is 1, and the sum of the fractional parts is 1 . Prefix the square of the larger number, less 1, to the difference of the square of the frac- tion of the larger number and 1. Thus, 5^ X 6f = 35f^. Exercises. 1. 2. 3. 2^ X H 6| X H 7i X Si 4. 7^XS^ 5. 8^ X 9f 6. 13^ X 14f 7. 15^ X 14^ 8. 17^ X m 9. 19i X 20f 66. To find the Product of any Two Numbers ending in 5. Prefix the product of the figures preceding the 5 in each number, plus ^ their sum to 25. Thus, 185 x 45 =: 8325 When the sum of the preceding figures is an odd num- ber, add to the product ^ of the next smaller nmmber^ and prefix the mm to 75. Thus, 55 X 25 = 1375 1. 25 X 25 2. 85 X 45 3. 85 X 45 Examples. 4. 75 X 85 5. 95 X 85 6. 105 X 125 7. b. 9. 145 X 175 165 X 135 125 X 145 DIVISION. 57. Division is the method of finding how often one number, called the Divisor, is contained in another num- ber, called the Divipend. The result is called the Quo- tient. The number which is sometimes left, after dividing, is called the KEUAiia)£R, and is always of the same name as the dividend. DIVISION. 65 1 : m :io-^r :iOt\ Idron ; mhers, he sum larger le frac- = 35H. X20^ iding in he 5 in 45 = [d num- er, and X175 X135 XU5 ken one num- Quo- [viding, name • • • • • • Division is the opposite of Multiplication, and as the latter is an extension of Addition, so, in like manner, Division may be regarded as an extension of Subtraction. 58. If any number be divided into two or more groups of units, then tlie collected units will contain the divisor as often as it is contained in the parts. Thus, 20 = 12 + 8 ; therefore ¥=¥+!• To illustrate this axiom, let 20 counters or dots be arranged, as in the annexed figure : here we first observe that 4 can be taken out of 20 5 times. In the group to the left there are 12 counters, out of which 4 can be taken 3 times, and in the group of the right there are 8 counters, out of which 4 can be taken 2 times. That is, 4 will be contained in 20 the same number of times that it is con- tained in 12, together with the number of times it is con- tained in 8. It is on this principle that we perform opera- tions of Division. 59. Division is of two kinds — Simple and Compound. It is called Simple Division, when the dividend and divi- sor are, both of them, either abstract numbers, or concrete numbers, of one and the same denomination. It is called Compound Division, when the dividend, or when both divisor and dividend, contain numbers of differ- ent denominations, but of one and the same kind. 60. In Division, if the dividend be a concrete number, the divisor may be either a concrete number or an abstract number, and the quotient will be an abstract number or a concrete number, according as the divisor is concrete or abstract. For instance, 5 shillings taken 6 times, give 30 shillings ; therefore, 30 shillings divided by 5 shillings, give the abstract number 6 as quotient ; and 30 shillings divided by 6, give the concrete number 5 shillings as quotient. I 66 SIMPLE DIVISION. — SHOUT PmSION. SMPLE DIVISION. 61. Simple Division is generally divided into two kinds — Siiort Division and Long Division. AVhen the process of dividing is carried on in the mind, and the quotient only is set down, the operation is called Short Division. When the result of each step m the operation is written down, the process is called Long Division. SHORT DIVISION. 62. Rule. Place the divisor and dividend thus : Divisor, ) Dividend, Quotient. By the Multiplication Table, find how often the divisor is contained in the first figure, or, if necessary, in the num- ber expressed by the first two, or the first three figures of the dividend, and write down the figure denoting the num- ber of times directly under the figure or figures divided. Find the product of this figure and the divisor, and take it from the number expressed by the figure or figures of the dividend formerly used. If there is a remainder after dividing anj" figure, prefix it mentally to the next figure of the dividend, and divide as before ; and if the divisor is not contained in any figure of the dividend, place a cypher in the quotient, and prefix this figure to tlie next one of the dividend, as if it were a remainder. If there be a re- mainder at the conclusion, write it, with the divisor under it, at th** end of the quotient. In oraer to render the division complete, it is obvious that the whole of the dividend must be divided. But when there is a remainder after dividing the last figure of the dividend, it must of necessity be smaller than the divisor, and cannot be divided by it. We therefore represent the division by placing the remainder over the divisor, and annex it to the quotient. The reason of this is clear ; for, were it required to divide 7 apples among 3 boys, each boy would get 2 whole apples, and there would be 1 remaining. Now, each 11 o kinds e mind, s called written s: i B divisor the num- giires of the niim- divided. d take it rures of er after igure of ivisor is cypher tt one of I be a re- )r under lobvious it when of the Idivisor, ]ent the )r, and lired to get 2 , each SHORT DIVISION. 67 7)1738 248^ boy is as much entitled to a share of this 1 apple, namely, ^, as he is to the 2 whole apples. We therefore write the quotient 2-]-. Ex. Let it be required to divide 1738 dollars among 7 persons. Putting, tiviH, the divisor and dividend as di- rected, we say 7 is cuntained in 17 (or, for brevity, 7 in 17) twice and 3 over, and we write 2 under the 7 in the dividend ; then the remainder 3 and the next figure 3 annexed to it, expresses 33 — 7 in 33 four times and 5 over ; writing 4 under the 3, aud to the remainder 5, annexing, mentally, the 8, we have 58 : lastly, 7 in 58, 8 times and 2 over ; wo set 8 under the 8 in the dividend, and after it, the remainder 2, with the divisor 7 under it, to complete the quotient. Hence, the share of each would be 248 dollars, with a seventh part of 2 dollars that remain at the end of the operation. The expression f , meaning, as we have seen, a seventh part of 2 dollars, is less than one dollar ; and being, there- fore, as it were, a broken quantity in comparison of a wJiole dollar, it is called a Fraction of a dollar ; while the dollar, in reference to the fraction, is called the Integer. As 2 dollars are twice as great as 1 dollar, a seventh part of 2 dollars is evidently 2 times or twice as great as a seventh part of 1 dollar. It is plain, therefore, that the expression f is the same in value as two sevenths of 1 dollar ; and hence it is read two sevenths of a dollar. In like manner, ^ is 7'ead five eighths ; ^ one sixth, &c. It may be shown, also, in a similar manner, that, if a shilling be the integer, the fraction f means either two sevenths of one shilling, or one seventh of two shillings ; while, in reference to a ton, f means equally two sevenths of one ton, or one seventh of two tons. In any fraction, expressed in the same manner that has been now explained, the upper number is called its Numer- ator, and the lower its Denominator. Thus, in |, 2 is the numerator, and 3 the denominator. Note.— The theory and management of fractions form an important branch of Arithmetic, which will' be discussed at due length, in a more advanced part of the work. With the few viers and explanations given above, as well as those given in Art. 8, Part I., the pupil shouM be made thoroughly acquainted. tf \^ I 1 1 68 LONG DIVISION. Mental Exercises in Division. 1. If one 3''ard of cloth can be bought for 5 dollars, how many can be bought for 50 dollars ? how many for 35 dollars ? 2. How many eggs can be bought for 63 cents, at the rate of a dozen for 9 cents ? 3. How long will it take a horse to travel 76 miles, if he travels 8 miles in each hour ? 4. If one man can do a piece of work in 56 days, how many men will be required to do the same in 7 days ? 6. X have 47 pounds of tea in a chest; how many packages of 8 pounds can I put up, and how many pounds will remain in the chest? If I make 6 packages out of the whole, how many pounds will be in each package ? Exercises for the Slate. 1. 2. 3. 470850 -1- 3 1829765 -r 4 4265983 ■- 5 4 3782047 -r- 6 5. 7165537 ; 7 6. 27459332 : 8 7. 8. 9. 74593822 -r- 9 53248675 -^ 11 49276189 -r 12 1. 2. 3. 156950 457441J 853196f ANSWERS. 4. 630341^ 5. 1023648f 6. 34324161 LONG DIVISION. 7. 8. 9. 8288202 f 4840788^7^ 4106265t^ 63. Rule. Place the divisor and dividend thus : Divisor, ) Dividend, ( Quotient. Take off from the left hand of the dividend the least num- ber of figures which make a number not less than the divi- sor ; then find (by the Multiplication Table), how often the first figure on the left hand side of the divisor is contained in the first figure, or the first two figures on the left hand side of the dividend, and place the figure which denotes this number of times in the quotient ; multiply the divisor by this figure, and bring down the product, and subtract ^t from the number which was taken off at the left of the LONG DIVISION. 69 dollars, ly for 35 s, at the miles, if ays, how lys? ►w many \f pounds )ut of the ? 322 575 .89 — 9 11 12 1202 I 788tZj. 265tV is: ist num- the divi- kfben the mtainei 1ft hand [denotes divisor liubtract of the dividend ; then bring down the next figure of the dividend, and place it to the right of the remainder, and proceed as before. If the divisor be greater than this remainder, affix a cypher to the quotient, and bring down the next figure from the dividend to the right of the remainder, and proceed as before. Carry on this operation till all the figures of the dividend have been thus brought down, and the quotient, if there be no remainder, will be thus deter- mined, or if there be a remainder, the quotient and the remainder will be thus determined. NoTB— If any product be greater than the number which stands above it, the last figure in the quotient must be changed for one of smaller value; but if any remainder be greater than the divisor, or equal to it, the last figure of the quotient must be changed for one of greater value. Ex. Divide 2338268 by 6758. Proceeding by the Rule given above, we obtain— 6758 ) 2338268 ( 346, Quotient. 20274- • 31086 27032 40548 40548 The reason for the Rule will appear from the following considerations. The divisor represents six thousand, seven hundred and fifty-eight ; the &:st five figures on the left hand side of the divider d represent two millions, three hundred and thirty- eight thousand, and two hundred. Now, the divisor is contained in this 300 times ; and 6758 X 300 = 2027400, or, omitting the two cyphers at the end for convenience in working, we properly place 4 under 2 in the line above ; we subtract the product thus found, and we obtain a remainder of 3108, which repre- sents three hundred and ten thousand, and eight hundred. Bring down the six, by the Rule, (it is proper, for prevent- ing mistakes, to put a dot below each figure of the given 70 LONG DIVISION. > i y K II hi dividend,, when it is brought dovm) ; this 6 denotes 6 tens or 60, but the cypher is omitted for the reason above stated ; the number now represents three hundred and ten thousand, eight hundred and sixty ; 6758 is contained 40 times in this, and 6768 X 40 = 27032. We omit the cypher at the end as before, and subtract the 27032 from the 31086 ; and, after subtraction, the remainder is 4054, whicli represents forty thousand, five hundred and forty. Bring down the 8, by the Rule, and the number now rep- resents forty thousand, five hundred and forty-eight ; 6758 is contained 6 times exactly in this number. Therefore, 346 is the quotient of 2338268 by 6758. 64. The above example, worked without omitting the cyphers, would have stood thus : 6758)2338268(800 + 40-1-6. 2027400 3108C8 270320 40548 40548 , ■ Hence It appears that the divisor is subtracte 1 from the dividend 300 times, and then 40 times from what remains, and then 6 times from what remains, and there being now no remainder, 6758 is contained exactly 346 times in 2338268 (Art. 58). Note.— It will help us to understand how greatly division abbreviates subtraction, if we consider how long a process would be required to disj- cover, by actually subtracting it, how often 6758 ia contained in 2338268. 65. If the divisor terminate loith cyphers^ the process can be abridged by the following method. Rule. Cut off the cyphers from the divisor, and as many figures from the riglit hand of the dividend as there are cj'phers so cut off at the right hand of the divisor ; then proceed with the remaining figures according to the Rule (Art. 63) ; and to the last remainder annex the figures cut off from the dividend for the total remainder es 6 tens >n above I and ten ained 40 omit the 032 from • is 4054, md forty, now rep- ;ht; 6768 3758. itting the from the t remains, eing now times in abbreviates lired to dia- in 2338268. le process and as as there divisor ; ig to the le figures LONG DIVISIOH. ttis^^ -^ ^^ Ex. Divide 537523 by 3400. Proceeding by Rule — 34,00)5375,23(158 34- • ' 197 170 275 272 Therefore 3400 is contained in 537523, 158 times, with re- mainder 323. The reason of the Rule will appear from the following considerations. 637523 is 5375 hundreds and 23, of which 537500 con- tains 3400 158 times, with a remainder 300 over ; and as 23 does not contain 3400 at all, the quotient will evidently be 158, with remainder 300 + 23, or 323. Note. — ^The same rale applies when the divisor and dividend both ter- minate with cyphers. Methods op Proof. First, Find the product of the divisor and quotient, and add to it the remainder ; if the sum be equal to the dividend, the work is correct. Second. First cast the 9s out of the divisor and quo- tient, and multiply the excesses together ; to the product add the excess of 9s in the remainder, if any after divi- sion ; cast the 9s out of this sum, and set down the excess ; finally cast the 9s out of the dividend, and if the excess is the same as that obtained from the divisor and quotient^ the work may be considered right. Tliird. Commencing at the units of the divisor, add together the digits in the odd places, rejecting 1 1 as often as possible, and reserve the result ; proceed in the same maimer with the di^t«s in the even places, and from the r tl' i ;i H Kl 72 DIVISION BT COMPOSITE NUMBERS. former result, increased, if necessary, by 11, take this result, and place the excess below the divisor. In a simi- lar way find the excesses of the dividend, quotient and remainder ; then multiply the excess of the divisor by the excess of the quotient and add to the product the excess of the remainder, and if the excess of this sum be equal to the excess of the dividend, the work is right ; if they differ, the work is wrong. EXERaSES. 764235 -f- 51 657442 -i- 81 945054 -^ 43 3387612 -^ 121 51846734 -7- 102 980263711 -i- 809 1700649160000 -r 759 9302688 H- 14356 9. 75843639426 -r- 8593 10. 1111111111111 — 854 11. 1000000000000000 -T- 111 12. 1000000000000000 -T- 81 13. 267817938473 -4- 8760 1 2. 3. 4. 5. 6. 7. 8. AKIWEBfl. 14985 17982 21978 27996^* 608301AV 1211698^^ 2240644479f^ 648 8826211119} 1301 066874 J^i 9009009009009i|x 12345679012345H 30572824^VV^ DIVISION BY COMPOSITE NUMBERS. 66. A number which cannot be separated into factors, which are respectively greater than unity, is called a Prime Number. Thus, 3, 5, 7, 11, 13, are prime numbers. A number which can be separated into factors respect- ively greater than unity, or which, in other words, is pro- duced by multiplying together two or more numbers respectivel}^ greater than unity, is called a CoMPOsrrE Number. Thus, 4, which = 2X2; 6, which =2X3; 8, which = 2X2X2, are composite numbers, because they are composed or consist of the product of two or more numbers, each of which is greater than unity. Numbers whicb haT« no oonunon factor greater thaa i DlVIfllOir BT C0MP08ITX lOTMBEES. 73 take this n a simi- bient and or by the le excess be equal ; if they fXBS* 4985 7982 1978 996t'W 8301AV 1698A% 4479*41 648 6211IIJ* 6874tii 9009itx 2845^1 2824JVW factors, a Prime respect- is pro- numbers jMPOsrrE 2X3; because ' two or it thfm I unity, are said to be prime to one another. Thus, the numbers 3, 5, 8, 11, are prime to each other. Note. — The learner must be careful not to confound numbers Trhicb are prime to each other with prime numbers: for numbers that are prime to each other may themselves be composite numbers. Thus, 4 and 9 are prime to each other, while they are composite numbers. 67. "When the divisor is a composite number, and made up of two or more factors, neither of which exceeds 12, the dividend may be divided by one of the factors in the way of Short Division, and then the result (or quotient) by the other factor ; and so on, till all the factors are employed. The last quotient will be the answer. 68. To find the trtie remainder. If the divisor is resolved into but two factors, multiply the last remainder by the first divisor, and to the product add the first remainder, if any, and the result will be the true remainder. When more than two factors are employed, multiply each remainder by all the preceding divisors, to the sum of their products add the first remainder, and the result will be the true remainder. Or multiply the last remainder by the preceding divisor and add in the preceding remainder ; then multiply this sum by the next preceding divisor, and to it add the next preceding remainder, and so on, till all the remainders ave been added or taken in. The last result will be the true remainder. Ex. Divide 507 by 64. f2 64 8 4 507 253 — 1 rem. =a 1 31 — 5 rem. Now 6x2 =a 10 7 — 3 rem. And 3 X 8 X 2 = 48 59 true rem. i r I I 74 ■**r-:i>^ GEXfEBAL nUMCIVLES IK DIYlfllOM. Second method : r 2 507 ei-i 8 263 — 1 31 — 6 ::} 29 X 2 + 1 = 59 3 X 8 + 5 = 29 The reason for the above Bales is manifest from the following considerations : 81 is 4 times 7, together with 8. and 253 is 8 times 31, together with 5. and 507 is 2 times 253, together with 1. or is 2 times (8 times 31 + 5), together with 1. or is 16 times (4 times 7+3) together with 10 + 1. or is 64 times 7 + 48 + 10 + 1. or is 64 times 7 + 59. Exercises. 2. 3. 4. 5. 795456 41763481 317682 -^ 18 25760 234765 -^ 192 •84 -^625 56 AlfSTTEBS. 9469|^ 66821|if 17649 460 1222JJI GENERAL PRINCIPLES IN DIVISION. 69. From the nature of Division, it is evident that the value of the quotient depends both on the divisor and the dividend, 70. If a given divisor is contained in a given dividend a certain number of times, the same divisor will obviously be contained, In double that dividend, twice as many times. In tJu'ee times that dividend, thnce as many tin^os, &c. Hence, If the divisor remains the same, multiplying the divi- demd by any number, is in effect multiplying the quotient by that number. i OEKKIUX PRXNCIPLEt OF DIVtSlOST. w = 59 om the + 1. EBS. n RI41 182 at the 1(1 the idend ously ?, &c. divu nt by Thus, 6 i8 contained in 12, 2 times ; in 2 times 12, or 24, 6 is contained 4 times (i. c. twice 2 times) ; in 3 times 12, or 86, G is contuiucd 6 times (i. c. thrice 2 time»), &c. 71. Again, if a given divisor is contained in a gi\'cn dividend a certain niuubcr of times, the same divisor is contained, In half that dividend, half as many times. In a third that dividend, a third as many times, &c. Hence, If the divisor retnaiiis the same^ dividiny the dividend by fifty number, is in effect dividing the quotietit by that nuvi- ber. Thus, 8 is contained in 48, 6 times : in 48 ■—- 2, or 24 (half of 48), 8 is contained 3 times (i. e. half of 6 times) ; in 48 -^3, or 16 (a third of 48), 8 is contained 2 times (i. e. a third of 6 times), &c. 72. If a given divisor is contained in a given dividend a certain number of times, then, in the same dividend, Twice that divisor, only half as many times ; Three times that divisor, a third as many times, &c. Hence, If the dividend remains the same, nmltiply^nr/ the divisor by any number, is in effect dividiny the quotient by that number. Thus, 4 is contained in 24, 6 times ; 2 times 4, or 8, is contained in 24, 3 times (i. e. half of 6 times) ; 3 times 4, or 12, is contained in 24, 2 times (i. e. a third of 6 times), &c. 73. If a given divisor is contained in a given dividend a certain number of times, then in the same dividend. Half that divisor is contained twice as many times ; A third of that divisor, three times as many times, &c. Hence, If the dividend remains the same, dividing the divisor by any number, is in effect mvltivlying the ovotient by that number. Thus, 6 is contained in 86, 6 times ; 6 -i- 2, or 3 (half of 6), is contained in 36, 12 times (i. e. twice 6 times) ; 6 -T- 3, or 2 (a third of 6). is contained in 36, 18 times (I e. thrice 6 times), &c. I ii i It 76 QtzrnuL mHcirtts or mnsioH. 74. From the preceding articles, it is evident that any given divisor is contained in any given dividend Just as many times as ttoice that divisor is contained in ttoice that dividend ; three times that divisor in three times that divi- dend, &c. . . Conversely, any given divisor is contained in any given dividend just as many times as half that divisor is con- tained in half that dividend ; a third of that divisor in a third of that dividend^ &c. Uence, 75. If the divisor and dividend are both multiplied, or both divided by the same number^ the quotient will not be altered. Thus, 6 is contained in 12, 2 times ; 2 times 6 is contained in 2 times 12, 2 times ; S times 6 is contained in 8 times 12, 2 times, &c. Again, 12 is contained in 48, 4 times ; 12 -7- 2 is contained in 48 -f- 2, 4 times ; 12 -r 3 is contained in 48 -r- 3, 4 times, &c. 76. If the sum of two or more numbers is divided by any number, the quotient will be equal to the sum of the quotients which will arise Arom dividing the given numbers separately. Thus, the sum of 12 + 18 = 30 ; and 30 -r 6 = 5. Now, 12-^6 = 2; and 18 ^ 6 = 3 ; but the sum of 3 + 2 = 5. Again, the sum of 32 -|- 24 + 40 = 96 ; and 96 -r- 8 = 12. Now, 32 -h 8 = 4 ; 24 -^ 8 = 3 ; 40 -• 8 = 5 ; but 4 + 3+5 = 12. 77. If a given divisor is contained a certain number of times in a given dividend, then in the same dividend. If the divisor be increased by oTie half ome thirds three fourths^ or any other fractional part of itself, the quotient will be the same fractional part of itself too small. If the divisor be diminished by any fractional part of itself, the quotient will be the same fractional part of itself too great. it.. owfciuL nmncirLzs or Dxvisioif/ 77 78. If tho divisor remain the same, and the dividend be increased otis half^ the quotient will be too great by one third of itself; and if increased one thirds it will be too great by one fourth of itself, &c. If the dividend be diminished by one third of itself, the quotient will be too small by one "half of itself; and if diminished by one fourth^ too small by one third of itself, <&c. 79. It is evident, from the foregoing principles, what- ever operation is performed on the divisor, in order to find the true value of the quotient, the same must be performed on the dividend im of -r-8 )ut 4 )erof \ three Itieut MULTIPLICATION AND DIVISION BY FRACTIONAL NUMBERS. 80. In Arithmetical operations, it is often necessary to multiply or divide by numbers containing fractions ; and, though the method of doing this will be l\illy given in the Articles on Multiplication and Division of Fractions, it may be proper here to point out the modes of proceeding in the simplest cases. Every such operation requires both multiplication and division ; and hence it may be introdu- ced with propriety here. Ex 1234 X 29^ Here we multiply 1234 by 29 m 29 j- the usual manner ; but, before add- ing, we divide 1234 by 2, and, wri- 11106 ting the quotient, 617, under the 2468 partial products, adding the three 617 lines together, we find the required product is 36,403. 36403 The answer would also be found by doubling 29J, which gives 69 ; and then by multiplying 1234 by 59, and taking half the product. ft of Itself 'I ' il |! I I 78 CONTRACTIONS IN DIVISXON. Ex. 2. If the circumference of a carriage-wbcel be 14§ feet, how often will it turn round in ^oing a mile, the mile containing 5280 feet? 14f) 5280(3G0 In this example, the denominator is 3 ; 3 3 we treble both divisor and dividend (Art. — 78), thus obtaining i4 and 15840 ; and 44 15840 then by dividing the latter by the former, 132 • • we get 360, the number of times required. 264 264 Exercises. Answers. Exercises. Answers. il.' 13746 X 2^ 84365 2. 477121 X H 715681^ 8. 4275 X 4^ 18168| 4. 41785 ^ 2^ 5. 24579 4- 121 6. 24248 — 2f ' 16714 1920J| 9984^7 CONTRACTIONS IN DIVISION. 81. To Divide by 5. Double the dividend, aud cut off the last figure of the result, the half of which figure will be the remainder. This is the same as doubling the divi- dend, and dividing by 10, (Art. 78). 82. To Divide by 15, 35, 45, or 55. Double the divi- dend, divide the result by 30, 70, 90, or 110, respectively, and for the true remainder take half the remainder thus found. 83. To Divide by 25. Multiply the di\idend by 4, cut off two figures from the result, and take one fourth of the number expressed by them for the remainder. 84^ To Divide by 125. Multiply the dividend by 8, cut off three figures from the result, and take one eighth of the number expressed by them for a remainder. 85. To Divide by 75. Multiply the dividend by 4, di- vide the product by 300, and for the true remainder take one fourth of the remainder thus found. CANCELLATION. 79 be 14§ LC mile • is 3 ; i (Art. I ; and brmer, quired. SWERS. 6714 9984^7 cut off re will e divi- e divi- ively, thus 4, cut lof the :i 86. To Divide by 175, 225, and 275. Multiply the dividend by 4, divide the result in the first case by 700, in the second by 900, and in the third by 1100, and take one fourth of the remainder, in each case, for the true re- mainder. Exercises. ANSWERS. 1. 317684 -j- 5 63536^ 2. 1623841617 -^ 25 649536641^ 3. 813764 -r- 125 6510tW 4. 12347681 : .75 164635^1 5. 247632 : 175 1415t^^ 6. 487695 : 225 2167^1^ 7. 8941364 ~ 275 325142^5^ 8. 12345678 -r- 125 98765^ CANCELLATION. 87. We have seen that Division is finding a quotient, which, multiplied into the divisor, will produce the divi- dend (Art. 57). If, therefore, the dividend is resolved into two such factors that one of them is the divisor, the other factor will, of course, be the quotient. Suppose, for example, 42 is to be divided by 6. Now, the factors of 42 are 6 and 7, the first of which being the divisor, the other must be the quotient. Therefore, cancelling a factor of any number^ divides the numher by that factor. Hence, 88. When the dividend is the product of two factors, one of which is the same as the divisor. Cancel the factor common to the dividend and divisor; the other factor of the dividend will be the answer. ■by 8, jighth 4, di- take li h* '•li 80 CONSTRUCTION OF QUESTIONS. Ex. Divide the product of 34 into 28 by 34.' By cancellation. 3^)3^X28 28, Answer. Cancelling the factor 34, which is common to both the divisor and dividend, we have 28 for the quotient, the same as before. Common method. 34 28 272 68 84)952(28 68- 272 272 89. The method of contracting arithmetical operations, by rejecting equal factors, is called Cancellation. 90. ¥^hen the divisor and dividend have common factors. Cancel the factors common to both; then divide the product of those remaining iii the dividend by the product of those rc- maining in the divisor, Ex. 1260 ~ 210 ; 1260 = 15 X 7 X 12, 210 = 5 X 3 X *" X 2 3 and ^ X 3 X 5? X t)l^ X 5T X 12 Here we say, first, that 5 is contained in 15, 3 times ; again, 3 will cancel 3, 7 will cancel 7, and 2 into 12, 6 times. Note— The farther devdopmtnt and application of the principles of this most useful Rule, may be seen in the Reduction of Compound Frac- tions to Simple ones; in Multiplication and Division of Fractions; in Simple and Compound Proportion. ON THE CONSTRUCTION OF QUESTIONS. 91. When a class of beginners is being exercised in any or all of the fundamental Rules, it is often a great labor for the teacher to look over the work of each pupil ; in such cases, the following elegant and really useful method of constructing questions is well deserving the CONSTRUCTION OF QUESTIONS. 81 5 X lies of Frac- is; in Id in rreat pil; leful the attention of the teacher. It will be observed that the questions may be formed as quickly as the figures can be written, and that the law of the figures, testing the accu- racy of the answer, may be seen at a glance. Ex. 167,832 Addition. Here, in each row, the cor- 476,523 responding figures on the left hand, add- 18,981 ed to those on the right, produce nines ; 67,932 and the same law is observed in the answer. The only precaution is simply 731,268 that the sum of the last column should not exceed nine. Ex.1 78649,21350 Subtraction, ITere the rows are 64327,35672 formed in the same way as in Addition, and the figures in the 14321,85678 answer follow the same law. Ex. 2. 7467,1421 Instead of the figures in the ques- 3648,5240 tion forming nines, any other num- ber may be selected. In the above 3818,6181 example, 8 is the number taken; but here, also, the figures in the answer follow the law of nines. Multiplication. Ex. 1. 256,9,743 X 34 = 87371262. Here the figures to the left and right of the nine, are formed in the same way as in addition, and the figure? in the product follow the same law. When there are three figures in the multiplier, there must be two central nines, and so on. The only precaution, is simply that the product by the last figure of the multiplier, when added to the other par- tial products, does not produce a number greater than 9, as was remarked in Addition, which may be easily obviat- ed b}^ taking low numbers for the highest digit's place in both multiplicand and multiplier. Ex. 2. 136,8,752 X 18 = 2463,7536. Here there is a central 8, the other figures make up eights, the addition of the figureg in the multiplier make up 9, and the answer is 82 PROPERTIES OP NUMBERS. the law of Nines. When the addition of the figures in the multiplier makes nine or nines, the figures in the mul- tiplicand may make up any particular number whatever. Division. Ex. 1. 51 ) 764,235 ( 14985. Here the divi- sor is written at pleasure ; the first three figures, in the dividend, are found by multiplying the divisor by 15, and taking 1 from the unit's figure. The remaining figures make up nines, as in Addition ; the quotient figures follow the law of Nines. Ex. 2. 62; 31)557442(17982. Here, to modify the form, we first write 62, and multiply by 9 (any other num- ber will do), and thus form the dividend as in the last example ; then take the half of 62, which gives 31, for the divisor. The figures in the quotient follow the law of Nines, as in the last example. Ex. 3. 86 ; 43 ) 945,054 ( 21978. Ex. 4. 484 ; 242 ) 33876612 ( 139986. Ex.5. 484:121)33876612(279972. PROPERTIES OF NUMBERS. 92. The progress, as well as the pleasure, of the stu- dent in Arithmetic, depends very much upon the accuracy of his knowledge of the terms which are employed in mathematical reasoning. Particular pains should there- fore be taken to understand their true import. Def. An integer signifies a wliole number. 2. Whole numbers or integers are divided into prime and composite numbers. 3. A Composite number, we have seen, (Art. 66), is one which may be produced by multiplying two or more num- bers together. 4. A Prime number is one which cannot be produced by multiplying any two or more numbers together ; or which cannot be exactly divided by any whole number except a unit and itse^. PROPERTIES OP NUMBERS. 88 5. An Even number is one which can be divided by 2, without a remainder, as 4, 6, 10, 12. 6. An Odd number is one which cannot be divided by 2 without a remainder, as 3, 7. Note. — All even numbers, except 2, are composite numbers; an odd number is sometimes a composite ^ and sometimes a prime number. 7. On» number is a measure of another, when the for- mer is contained in tlie latter any number of lataios without a remainder. Thus, 3 is a measure of 15, &d. 8. One number is a multiple of another "When the for- mer can be divided by the latter without" a remainder. Thus, 6 is a multiple of 3 ; 20 is a multiple of 5, &c. Note. — A multiple is therefore a composite number, and the number thud contained in it, is always one of its factors. 9. The Aliquot parts of a number are the parts by which it can be measured or divided without a remainder. Thus, 5 and 7 are the aliquot parts of 35. 10. The Reciprocal of a number is the que nt arising from dividing a unit by that number^ Thus, the recipro- cal of 2 is ^ ; of 3 is J, &c. 11. A perfect number is one which is equal to the sum of all its aliquot parts. Thus, 6 = 1 -[- 2 + 3, the sum of its aliquot parts, and is therefore a perfect number. All perfect numbers terminate with 6 or 28. 93. By the term properties of numbers, is meant those qualities or elements which are inherent and inseparable from them. Some of the more prominent are the following : Prop. 1. The sum of any two or more even numbers is even number. 2. The difference of any two even numbers is an even number. 3. The sum or difference of two odd numbers is even; but the sum of three odd numbers is odd. 4. The sum of any even number of odd numbers is even, but the sum of any odd number of odd uumbera ig odd. i 84 PROPERTIES OP NUMBERS. if if 5. The sum, or difference, of an even and an odd num- ber is an odd number. 6. The product of an even and an odd number, or of two even numbers, is even. 7. If an even number be divisible by an odd number, the quotient is an even number. 8. The product of any number of factors is even^ if any one of them be even. 9. An odd number cannpt be divided by an even number without a remainder. 10. The product of any two or more odd numbers is an odd number. 11. If an odd number divides an even number, it will also divide the half of it. 1 2 If an even number be divisible by an odd number, it ^vui also be divisible by doicble that number. 13,. Any number that measures two others, must like- msc ociaasure their sum, their difference, and their product, i:'t. A number that measures another, must also measure it 3 mzdtipie or its prodvxit by any whole number. .15. Any number expressed by the decimal notation, divided by 9, will leave the same remainder as the sum of its figures or digits divided by 9. Thus, take any number, 6357 : now, separating it into its several parts, it becomes 6000 = 6 X 1000 = 6 X (999 + 1) = 6 X 999 + 6. In like manner, 300 = 3 X 99 + 3, and 50 = 5 X 9 + 5. Hence, 6357 = 6 X 999 + 6 + 3X 99 + 3 + 5x9 + 5 + 7, or6X 999 +3X 99 +5X9 + 6 + 3 + 5 + 7 ; and 6357 -7- 9 = (6 X 999 + 3 X 99 + 5 X 9 + 6 + 3 + 5 + 7) -^ 9. But 6 X 999 + 3 X 99 + 5 X 9 is evidently divisible by 9 ; therefore, if 6357 be di- vided by 9, it will leav3 the same remainder as 6 + 3 + 5 + 7-^9. The samt will be found true of any number whatever. Note.— This property of the number 9 aflFords an ingenious method of pro V lag the fundamental rules. The same property belongs to the number 3 ; but it belongs to no other figure. The preceding is not a »«c««Miry but an incidtntal property of th« li'j: PROPERTIES or NUMBERS. 85 It arises from the law of increase in the decimal notation. If the rattix of the system were 8, it would belong to 7; if the radix were 12, it would belons; to 11 : and, universally, it belons^s to the number that is one less than the radix of the system of notation. 16. If the number 9 is multiplied by any single figure or digit, the sum of the figures composing the product will make 9. 17. If we take any two numbers whatever, then one of them or their sum, or their difference, is divisible by 3. Thus, take 11 and 17 : though neither of the numbers them- selves, nor their sum, is divisible by 3, yet their difference is, for it is 6. 18. Any number divided by 11, will leave the same re" mainder as the sum of its alternate digits in the even places, reckoning from the right, taken fVom the sum of its alternate digits in the odd places, increased by 11, if necessary 19. Every prime number, except 2, if increased or di- minished by 1, is divisible by 4. 20. Every prime number, except 2 and 3, if increased or diminished by 1, is divisible by 6. 21. Every prime number, except 2 and 5, is contained, without a remainder, in the number expressed in the com- mon notation, by as manj'^ 9s as there are units, less one, in the prime number itself. Thus, 3 is a measure of 99 ; 7 of 999,999. 22. Ever}^ prime number, except 2, 3 and 5, is a meas- ure of the number expressed in common, by as many Is as there are units, less one, in the prime number. Thus, 7 is a measure of 111,111 ; and 13 of 111,111,111,111. 23. All prime numbers, except 2 and 5, must terminate with 1, 3, 7 or 9 ; all the other numbers are composite. H% To find the prime numbers in any series of num- bers. Write in their proper order all the odd numbers con- tained in the series. Then, reckoning ftom 3, place a point over every third number in the series; reckoning 11 86 OBEATEST COMMON MEASURES. from 5, place a point over every fifth number ; reckoning from 7, place a point over every seventh number and so on. The numbers remaining witliout points, together with the number 2, are the prime numbers. Take tlie series of numbers up to 40. Thus : 1, 2, 3, 5, . . . . 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, • • 39 ; then adding the number 2, the prime numbers are 1, 2, 3, 5, 7, 11, 13, &c. GRE4TEST COMMON MEASURE. 95. A Common Divisor of two or more numbers, is a number wliicli will divide each of them without a remain- der. Thus, 2 is a common divisor of 6, 8, 12, 16, 18, &c. 96. The Greatest Common Divisor, or, as it is called, the greatest common measure of two or more numbers, is the greatest number which will divide them without a re- mainder Thus, 6 is the greatest common measure of 36, 12 18 and 24. ^ Note. — ^We may here remark, that the measure of two or more quan- tities can sometimes be found by inspection. The following facts may assist the learner in finding the common measure: 1. Any number ending in 0, or an even number, may be divided by 3. 2. Any number ending in 5 or , may be divided by 5. 3. Any number ending in 0, may be divided by 10. 4. When the two right hand figures are divisible by 4, the whole num- ber may be divided by 4. 5. If the three right hand figures of any number are divisible by 8, the whole is divisible by 8. To find the Greatest Common Measure of Two Numbers 97. Rule. Divide the greater number by the less ; if there be a remainder, divide the first divisor by it. If there be still a remainder, divide the second divisor by this remainder, and so on, always dividing the last prece- ding divisor by the last remainder, till nothing remains. The last divisor will be the greatest common measure required. OREATKST COMMON MEA8URS8. •T Ex. Required, the greatest common measure of 475 and 689. Proceedinjjj by the rule given above— 475)589(1 475 114)475(4 456 19)114(6 114 ibers Iss; if It. If )r by )rece- lains. lasure Therefore, 19 is the greatest common measure of 475 and 589. Reason for the above process, - Any number which measures 589 and 475 also measures their difference, or 589 — 475, or 114 (Art. 93, prop. 13) ; also measures any multiple of 114, and therefore 4 X 114 or 456, (Art. 93, prop. 14). And any number which measures 456 and 475, also measures their difference, or 475 — 456, or 19 ; and no number greater than 19 can measure the original numbers 589 and 475 ; for it has just been shown that any number which measures them must also measure 19, Again, 19 itself will measure 589 and 475, For 19 measures 114 (since 114 = 6 X 19), Therefore, 19 measures 4 X 114, or 456, Art, 93, pro. 14). Therefore, 19 measures 456 + 19, or 475, (Art. 93, pro. 13), Therefore, 19 measures 475 + 114, or 589 ; Therefore, since 19 measures them both, and no number greater than 19 can measure them both, 19 is their greatest common measure. Exercises. Find the greatest common measure of 1. 16 and 72, Ans. 8. 6. 532 and 1274, Ans. 14. 2. 55 and 121, 11. 7. 2145 and 3471, 39. 3. 272 and 425, 17. 8. 4872 and 81, 3. 4. 128 and 324, 4. 9. 126 and 162, 18. 5. 825 and 960, 15. 10. 176 and 1000, 8. 88 OmSATEST C0MU02( MEASURES. To find the Greatest Common Measure of Three or more Numbers, 98. Rule. Find the greatcs common measuro of the first two numbers ; then the greatest common measure of the common measure so found and the thu'd number ; then that of the common measure last found and the fourth number, and eo on. The last measure so found will be the greatest common measure required. "Ex, Find the greatest common measure of 16, 24 and 18. Proceeding by the Rule given above — 16)24(1 16 8)16(2 16 Therefore 8 is the greatest common measure of 16 and 24. Now find the greatest common measure of 8 and 18. 8)18(2 16 2)8(4 8 Therefore, 2 is the greatest common measure required. Reason of the above process. It appears, from Art. 93, prop. 13, that every number which measures 16 and 24, measures 8 also ; Therefore, every number which measures 16, 24 and 18, measures 8 and 18 ; Therefore, the greatest common measure of 16, 24 and 18, is the greatest common measure of 8 and 18. But 2 is the greatest common measure of 8 and 18 ; Therefore, 2 is the greatest common measure of 16, 24 and 18. LXAST COMMON ITULTIFLS^ EXEBCISES. Find the greatest common measure of 1. 14, 18 and 24. 2. 13, 52, 416 and 78. 3. 805, 1311 and 1978. 4. 504, 5292 and 1520. LEAST COMMON MULTIPLE. M Ans. 2; Ans. 13. Ans. 23. t Ans. 4. 99. One number is said to be a multiple of another, ^•hen til ' former can be divided by the latter without a re- mainder (Art. 92, def. 8). Hence, 100. A Common Multiple • ^ two or more numbers, is a number which can be divide by eac/i of them without a remainder. Thus, 144 is a common multiple of 8, 9, 18 and 24. 101. The Continued Product of two or more given numbers will always form a common multiple of those numbers. The same number may have an unlimited num-< ber of common multiples ; for, multiplying their continued product by any number, will form a new common multiple (Art. 93, prop. 14). 102. The Least Common Multiple of two or more num- bers, is the least number which will contain each of the given numbers an exact number of times without a re- mainder. Thus, 12 is the least common multiple of 4 and 6, for it is the least number which can be exactly divided by them. Note. — ^The least common multiple of two or more numbers, is evi- dently composed of all the prime factors of each of the given numbers repeated once and only once. For, if it did not contain all the prime factors of any one of the given numbers, it could nut be divided by that number. On the other hand, if any prime factor is employed more times than it is repeated as a factor in someone of the g?ven li umbers, then it would not be the ieatt common multiple. > IMAGE EVALUATrON TEST TARGET (MT-3) 1.0 I.I 11.25 l^|2£ 125 lis _^^ 12.2 *J 13.6 i^K US 122 |A0 US u 2.C U ||6 Photographic .Sciences Corporation 23 WIST MAIN STREET WEBSTER, N.Y. MS80 (716) •72-4503 '^ ? ^ LBAST COMMON MULTIPLE. 1 03 . To find the Least Common Multiple of Ttoo Numbers, KuLE. Divide their product by their greatest common measure. Or, divide one of tlieiii by their greatest com- mon measure, and multiply the quotient by the other. The result of eitlier method will be the least common mul- tiple of the numbers. Ex. Find the least common multiple of 18 and 30. Proceeding by the Rule given above, the greatest com- mon measure of 18 and 30 (Art. 98), is 6, and 18 X 30 = 640, and 540 -r. 6 = 90. Therefore, 90 is the least common multiple of 18 and 80. Reason of the above process. 18 = 3 X 6, and 30 = 5 X 6. Since o and 5 are prime factors, it is clear that 6 is the greatest common measure of 18 and 30 ; therefore, the least common multiple must contain 3, 6 and 5 as factors. Now, every multiple of 18 must contain 3 and 6 as fac- tors ; and every multii)le of 30 must contain 5 and 6 as factors (Art. 93, prop. 14). Therefore, every number, which is a multiple of 18 and 30, must contain 3, 5 and 6 as factors ; and the least num- ber which contains them is 3 X 5 X 6, or 90. Now, 90 = (3 X 6) X (5 X 6), divided by 6 ; = 18 X 30, divided by 6 ; = 18 X 30, divided by the greatest common measure of 18 and 30. 104. Hence, it appears that the least common multiple of two numbers which are prime to each other, or have no common measure but unity, is their product. 105. To find the Least Common Mnltiple of TJiree or More Numbers, Rule. Find the least common multiple of the first two numbers ; then the least common multiple of that multi* pie, and the third number, and so on. The last common multiple so found will be the least common multiple re* quired. c t 1 8 1 m 11 i ff LEAST COMMON MULTIPLE. ft ree or Ex. Find the least common multiple of 9, 18 and 24. Proceeding by the Rule given above, Since 9 is the greatest common measure of 18 and 9, their least common multiple is clearly 18. Now, find the least common multiple of 18 and 24. The greatest common measure of 18 and 24 is 6 (Art. 98). Therefore, the least common multiple of 18 and 24 is equal to (18 X 24) -^ 6, or 432 -r- 6 = 72. Therefore, 72 is the least common multiple required. Reason for the above process. Every multiple of 9 and 18 is a multiple of their least common multiple, 18 ; therefore, every multiple of 9, 18 and 24, is a multiple of 18 and 24 ; and, therefore, the least common multiple of 9, 18 and 24 is the least com- mon multiple. 106. When the Least Common Multiple of several Num- bers is required^ the most convenient practical method is that given by the followijiy Rule. Rule. . Arrange the numbers in a line from left to right, with a comma placed between every two. Divide by the smallest number which will divide any two or more of them without a remainder, and set the quotients and the undivided numbers in a line below. Divide this line, and set down the results as before ; thus continue the operation, till there are no two numbers which can be divided by any number greater than 1 . The continued product of the divisors into the numbers in the last line will be the least common multiple required. Note. — The least divisor of every number is a priixte number. For every whole number is either prime or composite; hence, dividing by Xh9 smallest number which will divide two or more of the given numbers, is dividing them by a prime number. The result will evidently be the same if, instead of dividing by the smallest number, we divide the given numbers by any prime number that will divide two or moreof tliem without a remainder. The preceding operation, it will be seen, resolves the given numbers into their prime factors, then multiplies all the different factors together, taking each &ctor as many times in the product, as are equal to tho greatest number of times it is found in mther of the giyen numbers. .4 i LEAST COMMON MULTIPLE. Ex. Find the least common multiple of 6, 8 and 12. OPERATION. 6 = 2X3 By resolving the given nam- 8 = 2X2X2 bers into their prime factors. 2 = 2X2X3 it will be seen that 2 is found and 2X2X2X3 = 24 once in 6 ; tmce in 12 ; and three times in 8. It must therefore be taken three times in the product. Again, 3 is a factor of 6 and 12, consequently must be taken only once in the product. Thus, 2X2X2X3 = 24, which is, therefore, the least common multiple. £x. Find the least common multiple of 12, 18 and 36. Proceeding by the Rule given. First operation. Second operation. 9 ) 12, 18, 36 2)12, 18, 36 Third operation. 12 ) 12, 18, 86 2) 6, 9, 18 2)12, 2, 4 3)1, 18, 3 3) 3, 9, 9 2) 6, 1, 2 1, 6, 1 3) 1, 3, 3 3, 1, 1 and 12 X 3 X 6 = 216 Now, 9X2X2X3 = 108. 1, 1, 1 2X2X3X3 = 36. Ans. Explanation. In the first operation, we divide by the smallest numbers which will divide any two or more of the given numbers without a remainder, and the product of the divisors, &c., is 36, which is the answer required. In the second and third operations, we divide by num- bers that will divide two or more of the given numbers without a remainder, and in both cases obtain erroneous answers. NoTB— 1. It will be seen ftom the second and third operations giren above, that ** dividing by any number which will divide two or more of the given numbers without a remainder/* according to the rule given by some authors, does not always give the ha%t common multiple. 3. The reason for dividing by the tmallest number, is because the divi- ■or may otherwise be a composite number, and have a factor common to ■ome one of the quotients or undivided numbers in the last line: eonse- qnently, th« oontinatd prodaot of them would be too lai^e for the least LEAST COMMON MULTIPLE. 93 /' oommon multiple. Thus, in the second operation, the divisor 9 is a composite number, containing the factor 3 comniuii to the 3 in the quo- tient; consequently, tlie product is three timet to.t large. In the third operation, the divisor 1*2 is a composite number, >iiid contains the factor 6 common to the (i in the quotient} therefore, thu product is six timet too large, 107. The process of finding the least common multi- ple may otlen be shortened by canceling in any line any number which is exactly contained in any other number in i the same line. Ex. Find the least common multiple of 4, 6, 10, 8, 12 and 15. 2) ^, 0, 10, 8, 12, 15 2) ^, 4, 6, 15 2, 3, 15 Since 4 and 6 will exactly divide 8 and 12, we cancel them. Again, since 5 in the second line will exactly divide 15 in the same line, we therefore cancel it, as also 3 in the third line, and proceed with the remaining num- bers as before. Thus, 2X2X2X15 = 120. Ans. Exercises. Find the least common multiple of the following num- b le value of bole num- >ning from Eice on the md place, )lace from en times. iminishes one phfece farther from the units' place. Thus, .4 = ^, but .04 r= 1^0 ; for the denominator to a decimal fraction is 1, with as many cyphers annexed to it as there arc figures in the numerator, (Art. 110). Annexing cyphers to decimals does not alter their value ; for each significant figure continues to occupy the same place as before. Tiius, .5 = ^i^j ; so .50 = ^q, or -ff^, by dividing the numerator and denominator by 10. It will be perceived that the efitect of annexing and pre- fixing cyphers to decimals is exactly opposite to that which is produced by affixing and prefixing cyphers to integers. 114. To read Decimal Fractions, Beginning at the left hand, read the figures as if they were whole numbers, and to the last one add the name of its order. Thua, .7 is read 7 tenths. .36 " " 36 hundredths. .475 " " 475 thousandths. .6342 '' '' 6342 ten-thousandth. Note.— Sometimes we pronounce the word decimal when we come to the separatHz, or decimal point, and read the figures as If they were whole numbers, or simply repeat them one after another. Thus, 125.427 is read one hundred and twenty -be, dtcimal four hundred and twenty- seven; one hundred and twenty-i'i ve, decimal four, two, seven; or point four, two, seven. Exercises in Notation and Numeration op Decimals. Write the fractional parts of the following numbers in decimals. (1.) 25^7^ (4.) 4x^ (7.) (2.) 30f^ (5.) 6xMir (8.) (3.) 72 1«^ (6.) 7,S^ (9.) 10. Write 9 tenths, 25 hundredths, 45 thousandths.] 11. Write 71 thousandths, 7 millionths. J 98 ADDITION IN DECIMAL FKACTIONS. ■ I 115. Decimals are addedj subtracted^ multiplied and divided, in the same manner as whole numbers. Addition of Decimal FnAcrioNS. 110. RiTLR. Place the numbers under each other, units under units, tens under tens. &c. ; one-tenths under one- tcntlia, onc-hundrcdths under one-hundredths, &c. Add OS in whole numbers, and place the decimal point in the ,sum under the decimal point above. j Ex. Add together 27.5037, .042, 432 and 2.1. Proceeding by the Rule given above, 27.5037 Or, filling up the vacant .042 places by Art. 13. 432. . 27.5037 2.1 .0420 432.0000 461.6457 2.1000 .t I 461.6457 Note. — ^The same method of explanation holds for the Ainaamentai rules of decimals, as also the proof, wbidi has been ^ven at length in explaining: the Rules for Simple Addition, Simple Subtraction, and the other fundamental rules in whole numbers. Mental Exercises in Addition of Decimals. 1. What is the sum of .1 and .05? 2. What is the sum of .6 and .45 ? 3. What is the sum of 4.25 dollars and 92 cents? • ^ Note.— It is unnecessary to multiply questions for mental exercise, \inder any of these rules, as any teacher can supply an unlimited quan- tity to suit the capacity of his pupils. Exercises for the Slate. 1.) .234 + 14.3812 + .01 + 32.47 + .00075. Ans. 47 09595 (2.) 232.15 + 3.225 + 21 + .0001 + 34.005 '+ .001304. Ans. 290.38U04. SUBTRACTION OF DECIMALS 99 ,X) .08 + 1C5 + 1.327 + .0003 + 27C0.1 + 0. Ans. 293:).:)()73. (4.) 725.1201 + 31.0007C + .04 + 50.9 + 113.71 3. Ans. yo3.773HG. (5.) G7.8125 + 27.105 + 17.5 + .000875 + 1>55. Ans. 807.117875. (C.) l.«3 + 5.C74 + .3125 + 18.3 + KM) + 38.C2 + 4.3957 + .5. Ans. 1G9.0322./ SunTIlACTION OF DECIMALS. 117. RrLF.. Place the less number under the greater^ units under units, tens under tens, &c. ; one-tenths under one-tenths, &c. Suppose cyphers to be supplied, if neces- sary, in the upper line, to tlie right of the decimal ; then proceed as in Simple Subtraction of Whole Numbers, and place the decimal point under the decimal point above./ Ex. Subtract 5.473 from 6.23. Proceeding by the Rule given above,'' 6.23 6.473 .767 Exercises in Subtraction.^ (1.) (2.) (3.) (4.) 213.5 — 1.8125. 603 — .6584003. .582 — .09647. 3.468—1.2591. Ans. 211.6875. Ans. 602.3415997. Ans. .48553. , Ans. 2.2089. Ans. 23.8933. » (5.) 34.528 — 10.6347. Multiplication op Deciblals. 118. Rule. Multiply, as in whole numbers, and point off as many figures from the right of the product for deci- mals, as there are decimal places in both the multiplicand and multiplier. If tiie product does not contain so many figures as there are decimals in both factors, supply the deficiency J^yjpre^ fieing cyphers. • '"^ ■4,. # t 100 MXTLimiCATION OF DICIMALB. Ex. 1. Multiply 5.84 by .21. Proceeding by the Rule given above, 5.34 Now, the number of decimal places in the .21 multiplicand -\- the number of those in the multiplier = 2 -{- 2 = 4. 534 1068 11214 Therefore, the product = 1.1214. £x. 2. 5.84 X .0021. Here we must have 6 places of decimals in 5.34 .0021 534 1068 the product ; but there are only 5 figures, and therefore we must prefix one zero, or cypher, and place a point before it, thus, .011214. 11214 Beaaon for the above procees. The reason for pointing off as many decimal places in the product as there are decimals in both factors, may bo illustrated thus : Suppose it is required to multiply .25 by .5. Supplying the denominators, .25 = ^^j, and .5 = y\j (Art. 110) ; now, multipljring the denominators together, and also the numerator, we obtain -^ X A == iVlftri or .125 (Art. 112) ; ^hat is, the product of .25 X •5 contains just as many decimals as the factors themselves. In like manner, it may be shown that the product of two or more decimiil num- bers must contain as man^^ decimal figures as there are places of decimals in the given factors. So, multiplying by any number of tens, it is only necessaiy to remove the point one place toward the right for each ten, EXEBCISBS W MULTIPUGATIOK OF DECIMALS, Answers. Answers. 1. 62.38 X 7. 436.66. 4. .1 X .1 X .001. .00001, 2. 3.81 X 41.7 ' 158.877. 5. 417 X .417. 173.889. S. .31 X .32. .0992. 6. .417 X .417. 1 .173889. OOKTRAmOXS IX MLLTiri.ICATIOX OV PKCIMALS. lOl is in the le in the ;imalB in ires, and ' cypher, 2U. plaoesin may bo ipplying 110); also the 1. 112) ; s many , it may al num- lere are tiplying ove the Answers. .00001. 73.889. ,73889. Answers. 7. 7i0.iC X .00002'>. l.TDHU. 8. 7.G X .071 X 2.1 X 2J). 32.Hr,ini. Answers. 9. i.o:> X I'Ort X 1.0") l.i:i7C25. 10. 7.1U X <'»3.1. 472.G19 Contractions in Mri.TirMCATiox of Dfjimals. 119. Whim the luinibcr of (Icfimal placcH in the multi- ])licr and multiplieund is hirgo, the luirubGr of ilceinialH in the product must also ha iar^c. But decimals below the Hfth or sixth place express so small parts of a unit, that when obtained they are commonly rejected. It is therefore desirable to avoid tlie unnecessary labor of obtaining those which are not to be used. Ex. It is required to multiply 1.35G9 by .3G742, and retain only five places of decimals in tlie product. 1.85691 It is evident, from the nature of dec- •36742 imal notation, that if -we begin to mul- i tiply by the left hand figure of the mul- 4 07073 tiplier first, instead of the right hand, bl41 46 and advance the partial product of each 949837 ilguro in the multiplier one place to the 54 2704 right instead of the left, the operation 2i71382 "Nvill correspond with the descending scale, and at the same time will give the true product. But since only five places of decimals are required, those on the right of the perpendicular are useless. Our present object is to show how the answer can be obtained without .4985558722 them. 1.35G91 .3G74 2 4070 7 814 1 95 54 3 .49855 Beginning at the right hand, as oeforo, we first multiply the fourth figure, or ten thou- sands' place of the multiplicand, by the tenths* or left hand figure of the multiplier (for 4 decimals in the multiplicand and 1 decimal place in the multiplier will give 5 places in the product, (Art. 118,) and place the first figure of the partial product under the figure multiplied. In obtaining the second partial product (i. e. multiplying by 6), it is plain we s i> I ! 102 CONTRACTIONS IN MULTIPLICATION OF DECIMALS. omit the rigid haDcl figure of the multiplicand, for if multiplied, its product Will fall to the right of the per- pendicular line, and therefore will not be used But if we multiply 9 into G, the product will be 54 ; consequent!}', there would be 5 to carr}^ to the next product. We there- fore carry 5 to 3G, which makes 41. Again, in the third partial product (i. e. in multiplying by 7), we may omit the two right hand figures of the multiplicand, for their pro- duct will fall to the right of the perpendicular line. But by recurring to the rejected figures, it will be seen that the product of 7 into 6 is 42, and 6 to carry makes 48 ; we therefore add 5 to the product of 7 into 5, because 48 is nearer 50 than 40 ; consequently, it is nearest the truth to carry 5 than 4. In the fourth partial product, we may omit the three right hand figures, and in the fifth or last, the four right hand figures. Again, it may be seen from the last operation, that if we had placed the figure occupying the tenths* place in the multiplier under tlie fourth figure of the multiplicand, and written the other figures fn reversed order, then multiplied as before, the result would have been the same. Tims, 1.35691 we place 3 under 9, and multiply, and set 2 4763 down the right hand figure of the partial pro- duct, 7, and carry 2, &c. Again, we multiply 4 0707 6 into 6, and to this product add the carrying 8141 figure arising from 6 into 9, viz., 6, and place 950 the right hand figure of this product under the 54 7, &c. ; proceeding in the same way with the 3 other figures, only setting down the product of each figure of the multiplier into the one .49855 directly above it for the right hand figiu'e of the partial product I 1 : r ' Note. — In finding what is to be carried for the rejected figures, it is generally sufiicient to go one figure back, but in doubtful cases it may be well to go farther, but, even then, the last figure cannot be depended on. It ia therefore better to work for one figure more than it is neces- sary to have true, and to reject it at the conclusion. And in lengthened computations, such as many of those in compound interest and annui- ties, it may be right to work for two or three additional figures. When the work is carried to one or two places more than required, it is not necessary to go back farther than ouo figure to obtain the carrying figure. I, for if the per- But if qucntl}^, "^e there- lio tliiril omit tlio eir pro- e. But leii that 1 48 ; we 36 48 is truth to ve may or last, at if we in the nd, and iltiplied Thus, ind set ial pro- lultiply irrying d place der the ith the duct of \e one Hire of es, it 13 J it may pended p neces- jthened annuU When t is not [figure. < CONTRACTIONS IN MULTIPLICATION OF DECIMALS. 108 120. 2b Multiply Decimals and retain only a given number of Decimal Figures in the Product, Rule. Count off, after the decimal point in the multi- 'iilicand (annexing cyphers if necessary), as many figures of decimals as requisite to have in the product. Below the last of these, write the unit figure of the multiplier, and write its other figures in reversed order. Then multi- ply by each figure of the multiplier, thus inverted, neglect- ing all the figures of the multiplicand to the right of that figure, except to find what is to be carried ; and let all the partial products be so arranged that their right hand figures may stand in the same column. Lastly, from the sum of these partial products, cut off the assigned number of decimal places. In carrying ft'om the rejected figures, we should take what is nearest the truth, whether it be too great or too small. Ex. 1. Multiply 7.24651 by 16.3476, so that there may be only four places of decimals in the product. 7.2465 1 6743.6 1 72465 1 43479 6 2173 9 5 289 8 6 60 7 2 434 118.463 0[3 Here, the unit figure of the multiplier is written under the fifth figure of the multiplicand, because we carry it one place farther than necessary in the pro- duct ; 1, the figure which precedes it (6) is put ajier it ; 3, the figure w}3ich follows it, is written or set b^ore it, &c. We then proceed as in the former examples, rejecting the first column, and pointing off four places of decimals : the number required. Ex. 2. Multiply .681472 by .01286, so as to retain only 4 places of decimals. ,6814 7*2 In this example, since the multiplier con- 6 8210.0 tains no integer, a cypher is placed below the fifth place (one more than required in the product to insure accuracy) of the mul- tiplicand; and then the multiplier being written in reversed order, the work proceeds as in the last example. 68 1 13 6 54 3 .0087[4 I f 104 DIVISION OF DECIMALS. Exercises To be worked by the contracted method, and proved by the former method, 1. 1.23467 X 4.896 to 5 places. 2. 4.8367 X 12.63 to 3 places. 3. 17.674 X 3.298 to 4 places. 4. 1.65 X 1.65 to 6 places. 6. .0006 X 3.48 to 4 places. 6. 186.784 X .2986 to 4 places. 7. .00678 X .46743 to 6 places. 8. .863541 X .10983 to 5 places. 9. 6.74321 X .0006 to 8 places. 10. 4867.632 X 123.45 to 2 places. 11. 98.98 X 98.98 X 98.1 to 4 places. 12. .167 X 167.1 X 16.7848 to 6 places. NoTB. — The same contractions may be used in Decimals as were used in Multiplication of Integers, always bearing in mind to count off from the product the same number of decimals as there are decimal places both in the multiplier and multiplicand. ' Division of Decimals. 121. First, When the number of Decimal Places in the Dividend exceeds the number of Decimal Places in the Divisor. Rule. Divide as in whole numbers, and mark off in the quotient a number of decimal places equal to the excess of the number of decimal places in the dividend over the number of decimal places in the divisor ; if there are not figures sufficient, prefix cyphers as in Multiplication. Ex. 1. Divide 1.1214 by 5.34. Proceeding by the Rule given above, 5.34) 1.1214 (.21 1068- 534 534 Now, the number of decimal places in the dividend — the number of decimal places in the divisor = 4 — 2 = 2. Therefore, the quotient = .21. DIVISION OF DEGIVALS. 105 d by the rere used off from al places aces m s in the r in the cess of er the ire not 1' Ind — = 2. Ex. 2. Divide ,011214 by 5.34. 6.34). 011214 (21 1068- 534 534 Now, the number of decimal places in the dividend — the number of decimal places in the divisor =6 — 2 = 4; therefore, we prefix two cyphers, and the quotient = .0021. Heason for the above process. We have seen in Multiplication of Decimals that the product has as many decimal figures as the multiplier and multiplicand (Art. 118). Now, since the dividend is equal to the product of the divisor and quotient, it follows that the dividend must contain as many decimal places as the divisor and qtiotient together; consequently, the quotient will contain a number of decimal places equal to the num- ber in the dividend less those in the divisor. 122. Secondly, When the number of Decimal Places in the Dividend is less than the number of Decimal Places in the Divisor. Rule. Affix cyphers to the dividend until the number of decimal places in the dividend equals the number of decimal places in the divisor ; the quotient up to this point of the division will be a whole number ; if there be a re- mainder, and the division be carried on farther, the figures in the quotient after this point will be decimals. Ex. 1. Divide 1121.4 by .534. .534)1121.400(2100 1068 • 534 534 "We have 3 places in the divisor and only one place in the dividend ; we therefore afl^ two cyphers to the div** dend. 6» 106 DIVISION OF DECIMALS. Ex. 2. Divide 172.9 by .142 to 3 places of decimals, rroceeding by the Rule given on foregoing page, .142 ) 172.900,000 ( 1217.605 142 30 9 28 4 2 50 142 1080 994 86 85 2 800 710 ) 1 90 123. Instead of the foregoing Rules, the following may be used with advantage, particularly with beginners. Rule. If the divisor and dividend do not contain the same number of decimal places, supply the deficiency by annexing cyphers. Then, rejecting separating points, divide as in whole numbers, and the quotient will be a whole number. If there be a remainder, after all the figures of the dividend have been used, cyphers may be annexed, till nothing remains, or till as many figures are found as may be judged necessary. The part of the quo- tient thus obtained, will be a decimal. If, after rejecting the decimal points, the divisor be gi'eater than the dividend, the quotient will contain no whole number. Ex. Divide 1346.5 by 43.68. Proceeding by the Rule given above, 4368 ) 134650 ( 30.826465, &c. DIVISION OP DECIMALS. 107 Here, by annexing a cypher to the dividend, and reject- ing the point, we have 4368 for the divisor, and 134650 for the dividend. Hence, dividing in the common way, we find 30 for th^ integral part, and aimexing cyphers to the remainders, and continuing the operation, we get .826465^ &c. Tlie answer, therefore, is 30.826465. Reason for the above Rule and process. The value of 1346.5 is not changed by the annexing of a cypher, (Art. 113) ; and the removal of the points merely multiplies each of the given numbers by 100. (See Art 75.) It is evident, therefore, that the value of the quotient will not be affected ; since, while the dividend is multiplied by 100, the divisor is increased in the same ratio. The reason of remoying the points, is to make the divi- sor and dividend whole numbers, and thus render the ope- ration, as much as possible, the same as in simple division. Ex. 2. Divide .1342 by 67.1. Here, by annexing three cyphers to the divisor, and re- jecting the decimal points, we get for the divisor 071000, and for the dividend 1342. Then, the divisor being greater than the dividend, the quotient will contain no integral part ; and the annexing of a cypher to the dividend gives one cypher for the quotient; the annexing of a second cypher to the dividend gives another cypher ; but the annexing of a third gives 2. Hence, the quotient is .002. The work is left for the learner to perform. Exercises. 1 1. 10.836 : 5.16. Ans. 2.1. 2. 34.96818 -T- .381. Ans. 91.78. 3. .025075 -r- 1.003. Ans. .025. 4. .02916 -r .0012. Ans. 24.3. 5. .00081 -1- 27. Ans. .00003. 6. 1.77089 -r- 4.735! Ans. 374. 7. 1 -^ .1 -r .01 -r .0001. Ans. 10,100, 10000. 124. When the divisor consists of many figures, the work will be shortened, if, instead of annexing a cypher to each remainder, a figure be cut off fiori^ the divisor. In 108 MISCELLANEOUS EXERCISES. this case, each product is to be increased by carrying from the product of the figure last cut off, and of the figure last placed in the quotient. Ex. 1. Divide 2.3748 by 1.4736, so that the quotient may contain three places of decimals. 14736)23748(1.611 In this example, the numbers • • • 14736 being prepared according to the former rule, and the first figure of 9012 the quotient being found, instead 8842 of adding a cypher to the remain- der, 9012, we omit the last figure 170 of the divisor, to denote which a 147 point is placed below it. Then 6, being but in the quotient, we mul- 23 tiply 6, the figure cutoff, by it, and 15 without setting anything down, we — carry 4, because the product 36 is 8 nearer 40 than 30. After that, 3 is cut off in like manner, and then 7. The quotient is found to be 1.611, or more nearly 1.612. because the remainder 8 is rather more than the half of 14, Miscellaneous Exercises on the foregoing Rules. 1. Find the sum and diflference of 9090909 and 90909. Ans. 9181818; 9000000. 2. A person, whose age is 73, was 37 years old at the birth of his eldest son ; what is the age of his son ? Ans. 36. 3. Find the value of the following expression : 15 X 87153 --(73474 — 67152)-7- 4 + 40734 X 2. Ans. 356954^ 4. The annual deaths in a town being 1 in 45, and in the country 1 in 50 ; in how many years will the number of deaths out of 18675 i^ersons lining in the town, and 79250 persons living in the country, amount together to 10,000? Ans. 5 years. MISCELLANEOUS EXKRCISES. 109 5. Find the value of 494871 — 94853 + (45079 — 3177) — (54312 — 3987) ■— (1763 + 231) + 379 X 879. Alls. 147802420. 6. What number divided by 528 will give 36 for the quotient, and leave 44 as a remainder? Ans. 19052. 7. The Iliad contains 15683 lines, and the -.Enead con- tains 9892 lines ; how many days will it take a boy to read through both of them, at the rate of eighty-five lines a day ? Ans. 300 days, and 75 lines rem. 8. At a game of cricket, A, B and C together score 108 runs ; B and C together score 90 runs, and A and C to- gether score 51 runs ; find the number of runs scored by each of them. Ans. A, 18 ; B, 57 ; C, 33. 9. The remainder of a division is 97, the quotient 665, and the divisor 91 more than the sum of both. What is the dividend? Ans. 567342. 10. The quotient arising from the division of 9281 by a certain number is 17, and the remainder is 373. Find the divisor. Ans. 524. 11. What number multiplied by 86 will give the same product as 163 by 430? Ans. 815. 12. Find the greatest number which can divide each of the two numbers, 849 and 1132; also, the least number Which can be divided by each of them. Ans. 283 ; 3396. 13. From 46 hundredths take 46 thousandths. Ans. 0.414. 14. In one rod thgi*e are 16.5 feet ; how many are there in 41.3 rods? Ans. 681.45 feet. 15. How many suits of clothes will 29.6 yards of cloth make, allowing 3.7 yards to a suit? Ans. 8 suits. 16. How many bales of cotton are there in 56343.75 pounds, allowing 375 pounds to a bale? Ans. 150.25 bales, t, 110 RBDUCTION, ETC. REDUCTION, ADDITION, SUBTRACTION, &c., OF COM- POUND OR DENOMINATE NUMBERS. 125. Our operations hitherto have been carried on with regard only to abstract numbers, or concrete numbers of one denomination. It is evident that if concrete numbers were all of one denomination ; if, for instance, shillings were the only units of money, yai'ds of length, years of time, and so on, such numbers would be subject to the common rules for abstract numbers. Again, if the con- crete numbers were different denominations, and these de- nominations differed from each other by 10, or multiples of 10, then all operations with such concrete numbers could be carried on by the rules which have been given for Deci- mals. But, generally, with concrete numbers, such a rela- tion does not hold between the different denominations, and therefore it is necessary to commit to memory tables which connect the different units of money together, the different units of time together, the different units of lengths together, and so on. We shall now put down some of the most useful of these tables, with a brief remark on each, and show the manner of applying them to the Reduction, Addition, Subtraction, Multiplication and Division of each denomination sepa- rately. TABLE OF MONEY. English or Sterling Money. 2 Farthings make h Half-penny. 2 Half-pence t( 1 Penny. 12 Pence «( 1 Shilling. 20 Shillings (( 1 Pound. Pounds, shillings, pence and farthing were formerly de- moted by £, S, D, and q, respectively, these letters being ihe fii'st letters of the Latin words libra, solidus, denarius, and quadrans, the Latin names of certain Roman coins or sums of money. £, S, D, are still the abbreviated forms COIN'S. Ill for pounds, shillings, and pence, respectively ; but J an- nexed to pence, denotes 1 farthing; J denotes a half- penny ; J denotes three farthings— showing that one far- thing, two farthings, and three farthings, are respectively ]th, Jths or ^, and jths of the concrete unit one penny. The mark /. which is often placed between shillings and pence is a corruption of the long /. The following coins are at present in common use in England : COPPER COINS. A Farthing, the coin of least value. A Half-penny = 2 Farthings. A Penny = 4 Farthings. GOLD COINS. A Half-Sovcreign = 10 Shillings. A Sovereign = 20 Shillings. SILVER COINS. Three-penny piece = 3 pence. Four-penny piece = 4 pence. A Shilling =12 pence. A Florin = 2 Shillings. A Half-crown = 2 Shillings and 6 pence. A Crown = 5 Shillings. Money, as expressed by means of these denominations, is commonly called Sterling Money ^ in order to distinguish it from Stock, &c., which is only nominal. Though all commercial transactions are conducted by means of the money enumerated in the preceding table, there are coins or denominations frequently met with, and some of them more particularly in old documents, of which the most important, and their value in current money, is here annexed : £. S. D. £. S. D. A Groat or Fourpenny, A Noble, 0— 6—8 0— 0—4 An Angel, O— 10— A Tester, 0— 0—6 AMarkorMerk, 0—13 — 4 A Seven-Shilling A Carolus, 1— 3-0 Piece, 0— 7—0 A Jacobus, 1 — 5 — A Guinea, 1 — 1 — A Moidore, 1— 7—0 A Half-Guinea, 0—10—6 A Six-and-Thirty, 1-16-0 The office at which coin is made and stamped, so as to pass or become current for legal money, is called the Mint. 112 KEDUCTION. The standard of gold coin in England is 22 parts of pure gold and 2 parts copper^ melted together. From a pound, Troy, of standard gold, there arc coined at the Mint 46Jg sovereigns, or £4G. U. 6., so that the weight of each is exactly 5dwt. 3^Jj^ grs., or nearly 123.- 274 grs. ; and the mint price of standard gold is £ 3. 17. 10 J per ounce, (12 oz. Troy, = 1 pound Troy). The standard of silver coin is 37 parts of pure silver, and 3 parts of copper. From a pound, Troy, of standard silver are coined 66 shillings. Therefore, the mint price of silver is 5s. 6d. per ounce, standard. In the copper coinage, 24 pence are coined from a pound, Avoirdupois, of copper. Therefore, 1 penny should weigh 2^th of a pound, Avoirdupois, The copper coinage is not, according to the present law, a legal tender for more than 1 2d. ; nor is the silver coinage for more than 40s. — ^the gold coinage being the standard of Britain. REDUCTION. 126. Reduction is the method of expressing numbers of a superior denomination in units of a lower denomina- tion, and conversely. Thus, £ 1 is the same value as 240 pence, and £21 as 5040 d., and conversely ; and the pro- cess by which we ascertain this to be so, is tei ned JReduC" tion. First, To eatress a number of a higher denomination in UT^its of a lower denomination. Rule. Multiply the number of the highest denomina- tion in the proposed quantity by the number of units of the next lower denomination contained in one unit of the highest, and to the product add the number of that lower denomination, if there be any in the proposed quantity ; repeat this process for each succeeding denomination, till the required one is anived at. nF.BUCTIOH US per ounce, £z. How many penee are there in £ 28, 15s. Od.? Proceeding by tlio Rule given on preceding page> £23, 158. Od. 20 460 + 15, or 475s. 12 ' 5700d., or £23. 15s. = 5700d. Reason for the process. There are 20 shillings in £ 1. Therefore, there are (23 X 20) shillings, or 460s. in £23, and so there are 460s, + l^-? or 475s. in £23, Ids. Again, since there are 12 pence in 1 shilling, therefore there are (475 X 12) d., or 5700d. in 475s.: i. e. in £23, 15s. In practice, we do not affix the shillings to the product by means of the sign +> but merely add them as we pro- ceed with the work. Note.— It has been stated (Art. 28) that one of the factors of any pro- duct must be an abstract number; and as neither £, 23 nor 20s. are such, we should have reasoned thus : as there are SOs. in £ 1 , therefore there are 23 times 20 shillings in £ 23 ; i. e. we should have multiplied 208. by 23, instead of 23 by 20— for £ 23 x 20 equals £ 460— but for convenience we multiply £ 23 by 20, and call the product shillings (Art. 33), and so with the pence, &c. Secondly. To express a number of inferior denomina^ Hon in units of a higher denomination. Rule. Divide the given number by the number of units which connect that denomination with the next higher, and the remainder, if any, will be the number of surplus units of the lower denomination. Carry on this process till you arrive at the denomination required. lU REDUCTION OK STERLING MONEY. Vfl Ex. How many pounds and shillings are there in 6700 pence ? rrocecding by the Rule given above, 12)5700 2,0) 47,5 £23, 15s. Od. In dividing 475 by 20, we cut of! the and 5, by Art. 65. Reason for the above process. Since 12 pence = 1 shilling, therefore, in any given number of pence, for every 12 pence there is 1 shilling, so that in 5700d. or (475 X 12) d. there are 475 shillings. Again, since 20s = £ 1, therefore, in any given number of shillings, for every 20 shillings there is £ 1 . Hence, in 475s. or (20 X 23 + 15) s., there are £23, and 158. over. Note.— Since each of the above Eules is the converse of eaoh'^other, the accuracy of any result obtained by either of them may be tested by working the result back again by the other Rule. REDUCTION OP STERLING MONEY. Mental Exercises. The pupil should be made well acquainted with the pence and shilling table 1. Change 102 pence to their value in shillings. 2. In 5s. 8d. how many pence are there ? 8. What is the difference, in farthings, between 15 shil- lings and ^ of a guinea? 4. What is the difference, in pence, between ^ of a sov- ereign and a Mark ? Exercises for the Slate. K Reduce (verifying each result) 1. £57 to pence ; and 613 guineas to farthings. Ans. 13680d.; 617904q. I COMPOUND ADDITIOV. 115 2. £ 15, 12 i^ to pence ; and 5000 ^incas to pcnc(». Ana. a744(l. ; 1260000d. 5. 89. 4 jd. to half pence ; and £ 1 , Os. OJd. to farthings. Ans. 201 half ponce; 975q. 4. 7S8 half-crowns to fiirthiiiGjs ; 22} guineas to six- peft<*es. Ans. 88o(;0q. ; U 15 sixpences. 6. IIow many half-crowns, how many sixpences, and how many fom-pences are there in 25 pounds? Ans. 200 half-crowns, 1000 sixpences, 1500 fourpcnces. 6. In 351 seven-shilling pieces, how many half-guineas are there, and how many moidores ? Ans. 234 half-guineas ; 91 moidores COMPOUND ADDITION. 127. Compound Addition is the method of collecting several numbers of the same kind, but containing diffcrout denominations of that kind, into one sum. BuLE. Arrange the numbers so that those of the same denomination may be under each other, in the same col- imin, and draw a line below them. Add the numbers of the lowest denomination together, and find by reduction how many units of the next higher denomination are contained in the sum. Set down the remainder, if anj', under the column just added, and carry the quotient to the next column ; proceed thus with all the columns. Ex. Add together £2, 4s. Tj^d., £3, 5s. lO^d., £15, 15s. Od., and £33, 12s. ll^d. Proceeding by the RiUe given above, £ B. d. .247^, i 3 5 10^ \ '. 15 15 . , ^ i ,33 12 11^ 54 18"""5i (•' 116 ADDITION OF STERLING MONEY. Beason for the foregoing process. The sum of 2 farthings, 1 farthing and 2 farthings = 5 farthings, = 1 penny and 1 farthing. We therefore put down |-, that is, 1 farthing, and carry 1 penny to the col- umn of pence. Then, (1+ 11 + 10 + 7) pence = 29d. = (12 X 2 + 5) d. or 2 shillings and 5 pence. We therefore put down 5d., and carry on the 2 to the column of shillings. Then, (2 + 12 + 15 + 5 + 4) s. = 38s. = (20 X 1 -|- 18)s. = £1 and 18s. We therefore put down 188., and carry on the 1 pound to the column of pounds. Then, (1 -I- 33 4- 15 -I- 3 -I- 2) pounds = £54. '^ ^Therefore, the result is £54, 18. 5 J. ADDITION OF STERLING MONEY. . .- Mental Exercises. 1. A gentlemen paid £ 8 for a cow, £ 22, 10. for a horse, and £5, 6. 3. for a saddle ; how much did he pay for all? 2. John had 10 pence, Hugh gave him 11 pence and James gave him 1 shilling and lOj^. ; how much had he then? 3. A farmer has three cows worth £ 21, a calf worth £ 2, and three sheep worth £ 1, 18. 9. ; how much are they all worth? Exercises for the Slate. 1. Find the sum of £28, 14. Gf., £27, 18. 4^., £79, 12. 6., £19, 18. lOf and £85, 14. 3J. Ans. £241, 18. 7i. 2. Find the sum of £678, 10. 2., £325, 6. 5., £487, 18. 9., £507, 0. 11. and £779, 10. 8. Ans. £2778,6. 11. 3. Find the sum of £306217, 13. 9J., £5o, 0. 9., £450812, 15. 2^., £9837. 1. 5^. and £2939, 3. 11 J. Ans. £769861, 15. 2^. iH# COMPOUND SUBTRACTION. 117 igs = 5 fore put the col- h5)d. ►wn 5d., (20 X 1 yn 188., Then, a horse, for all? $nce and had he )rth £ 2, they all .., £ 79, \, 7i. £487, ;. 11. 0. 9., I*- \ 2i. 4.* Find the sum of £485, 12. 7f., £49, 16. 8^., £ 186, 18. llf, £787, 10. 8|., £289, 9. 9i., £843, 11. 4f, £374, 16. 7., £285, 4, 9 J and £599, 19. 8. Ans. £ 3852, 15. 10. 5.' Add together £18, 14. 8J., £12, 13. 9f , £21, 12. 10., £32, 9. 10^., £63, 13. 9^., £16, 4. 8^.. £35, 14. 9^., £17, 16. 7^., £23, 15. 9^., £35, 17. 2f, £8,19.8., £ 12, 10. OJ. and £ 13, 8. 8^. Ans. £ 313, 12. 6 J. COMPOUND SUBTRACTION. 128. Compound Subtraction is the method of finding the difference between two numbers of the same kind, but containing different denominations of that kind. BuLE. Place the less number below the greater, so that the numbers of the same denomination may be under each other in the same column, and draw a line below them. Begin at the right hand, and subtract if possible each number of the lower line from that which stands above it, and set the remainder underneath. But when any number in the lower line is greater than the number above it, add to the upper one as many units of the same denomination as make one unit of the next higher denomination ; sub- tract as before, and carry one to the number of the next higher denomination in the lower line. Proceed thus throughout the columns. Ex. Subtract £ 88, 18. 8^ from £ 146, 19. 6j. Proceeding by the Rule given above. £ s. d. 146 19 H 88 18 ^ 68 9f lieason for the above process. Since j^d. is greater than a ^d., we add 4 farthings or 1 penny, thus raising it to 5 farthings ; and when 2 farmings are subtracted from 5 farthings, we have 3 farthings left. / «i i ' ! 118 SUBTRACTION OF STERLINQ MONET. ;? :|m We therefore place down Jd., and in order to increase the lower number equally with the upper number, we add 1 penny to 8 pence. Now, 9 pence cannot be taken from 6 pence. "We there- fore add 12 pence or 1 shilling to 6 pence, thus raising the latter to 18d. ; we take the 9 pence from 18 pence, and put down the remainder 9d. ; then adding Is to 18s., the latter becomes 19s. ; 19s. taken from 19s. leave no remainder. We then subtract £ 88 from £ 146, as though they were abstract numbers. It is manifest that in this process, whenever we add to the upper line, we also add a number of the same value to the lower line, so that the final differ- ence is not altered (Art. 22). SUBTRACTION OF STERLING MONEY. Mental Exercises. 1. A man owing £ 19, 10. 0., paid all but 15 shiilings ; how much did he pay ? 2. A merchant sold a piece of cloth for £10, 5. 0., which was £ 1, 2. 6. more than he paid for it ; how much did it cost him ? 3. A man bought 1 barrel of flour at £ 1, 17. 6., a barrel of com meal at £ 1, 2. 6. ; he sold both together for £2, 18. 9. ; how much did he lose by the bargain? 4. A man having £ 20, gave £ 1 for a hat, £ 3, 15. for a coat, and £1, 5. for a pair of boots ; how much had he left? Exercises for the Slate. 1. From £ 19, 3. 10. take £ 8, 15. 3 J. Ans. £ 10, 8. ej. 2. From £ 575, 15. 1 J. take £ 124, 13. 4. Ans. £451, 1. 9^. 3. From £ 192, 11. 4f take £88, 16. 9^. Ans. £ 103, 14. 7. 4. What sum added to £ 947, 19. 7f . will make £ lOQO ? :. Ans. £52,0. 4 J. .1 If Jl ) f/ COMPOUND BfULTIPUOATION 119 irease the we add 1 We there- aising the }, and put the latter jmainder. they were process, a number nal differ- shillings: LO, 5. 0., ow much 17. 6., a together t^ain? 16. for a had he 8. ej. 1.9^ 14. 7. £1000? 5. A fticnished house is worth £ 4759, 10. 9^. ; unfhr- nished, it is worth j£ 1494, 11. 92. By how much does the^ value of the furniture exceed the value of the liousc ? Ans. £1770, 7. IJ. COMPOUND MULTIPLICATION. 129. Compound Multiplication is the method of find- ing the amount of any proposed compound number, that is, of any number composed of different denominations, but all of the same kind, when it is repeated a given num- ber of times. Rule. Place the multiplier under the lowest denomina- tion of the multiplicand ; multiply the number of the low- est denomination by the multiplier, and find the number of units of the next denomination contained in this first product. If there be a remainder, place it down, adding on the number of units just found to the second product ; for this second product, multiply the number of the next denomination in the multiplicand by the multiplier, and after carrying on to it the above-mentioned number of units, proceed with the result as with the first product. Carry this operation through with all the different denomi- nations of the multiplicand. Ex. Multiply £ 56, 4. 6^. by 5. Proceeding by the Rule given above £ s. d. 56 4 f 281 2 8^ Reason for the above process. Jd. multiplied by 5 is the same as ( ^ -j" i+ i- + i + i) d. = 5 balf pence = 2^d. We therefore put down Jd., and carry on 2d. to the denomination of pence. 6d. nultiplled by 5 = 80d. ; therefore, (6X5+2) = 82d. ==i (2X1? + 8) d. = 2b. + 8d. We therefore put I I- 1 1 I ■ liM ,.:',li^ 120 MULTIPLICATION OF STEBLIMO MOMET. down 8d., and canyon 2s. to the denomination of shillings. 4s. multiplied by 5 = 20s. ; therefore, (4 X 5 + 2) s. = 22s. = (20 + 2) s. = £ 1 + 2s. We therefore put down 2s., and carry on £ 1 to the denomination of pounds. Now, by simple multiplication, £56 X 5 = £280; therefore, £ (1 + 56 X 5) = £ (1 + 280) = £ 281. Therefore, the total amount is £ 281, 2. 8^. 130. When the multiplier exceeds 12, it will be thj easiest method to split the multiplier into factors or into factors and parts. Thus, 15 = 3X5; 17 = 3X5-t. 2 ; 23 = 5 X 4 + 8, and so on. Ex. Multiply £ 55, 12. 9J by 23 £ 55 t. 14 d. H 4 222 11 1, value of £ 55, 12. 9|, multiplied by 4. ft i 1112 15 5, value of £222, 11. 1., multiplied by 5, or of £ 55, 12. 9^., multiplied by (4 X 5 or 20). 166 18 8 J, value of £ 55, 12. 9 J., multiplied by 3. 1279 13 8J, value of £55, 12. 9^., multiplied by (20 4- 3), or 23. Note 1.— When the multiplioand contains farthings, if one of the &o« tors of the multiplier be even, it will often be advantageous to use it first, as the farthings may disappear. Note 2.~Should the multiplier consist of many factors, it will be fbnnd in that case convenient to reduce the multiplicand to the lowest denomination contained in it, then multiply this result by the multiplier, and then to reduce the result back again. MULTIPLICATION OF STERLING MONEY. Mental Exercises. 1. What will 17 sheep come to, at 15 shillings each? 2. What wiU 125 pairs of boots comie to, at 258. a pair? n li t OOXFOOIID DIYISION. 121 1. Xl, 10.8 X 90 2. £2, 19.6 X 121 3. £1,12.5 X 76 4. £2, 15. 2i X 196 6. £3, 6. 5^ X 3178 6. £ 2, 6. 9 J X 938 8. What cost 86 yds. of cloth, at £ 2, 8. 9. a yard ? 4. A and B engage to saw 60 cords of wood, at 2 f shil- lings a cord ; A saws 3 cords for every 2 cords sawed by B. How much will A receive more than B? Exercises for the Slate. Ans. £ 138, 0. 0. Ans. £359, 19. 6. Ans. £121, 11. 3. Ans. £540, 16. 9. Ans. £ 10556, 18. 4^. Ans. £2194, 10. 7. 7. What cost 112 pounds of indigo, at lis. 4J^d. per pound? Ans. £63, 14. 0. 8. What is the amount of duty on 149 lbs. of West India coffee, at 7Jd. per pound? Ans. £4, 16. 2f. COMPOUND DIVISION. 131. CoMPOXJND Division is the method of finding a compound number, that is, a number composed of several denominations, but all of the same kind, into as many equal parts as the divisor contains units ; and also of find- ing how often one compound number is contained in another of the same kind. When the divisor is an abstract number. Rule. Place the numbers as in Simple Division ; then find how often the divisor is contained in the highest de- nomination of the dividend ; put this number down in the quotient. Multiply as in Simple Division, and subtract. If there be a remainder, reduce that remainder to the next inferior denomination, adding to it the number of that denomination in the dividend, and repeat the division. Carry on this process through the whole dividend. I ) i' i' i r I i; 123 COHPOUKD DIVISION. Ex. Divide £ 199, 6. 8. by 130. Proceeding by the Bule given on preceding page. 130)199, 6. 8!(£1. 130 69 20 180 ) 1386 ( 10s. 130 86 . 12 * 130) 1040 (8d. r 1040 Therefore, the answer is £ 1, 10. 8. Beason for the above process. We first subtract £ 1, taken 130 times, froi:i £ 199, 6. 8.» and there remains £ 69, 6. 8. Now, £69, 6. 8. rr-. 1386s. 8d. ; from this amount we subtract 10s., taken 130 times, and there remains 86s. 8d. Again, 86s. 8d. = 1040d. ; from this amount we sub- tract'8d., taken 130 times, and nothing remains. Therefore, £ 1, 10. 8. is contained 130 times in £ 199 6. 8. Note. — ^When the divisor is not greater than 12, the division can be easily performed in one line. Thus, for example, divide £ 8, i8. 6. by 12. 12)8 s. 18 6 Since we cannot divir*e 8 by 12, we re- duce £ 8 to shillings, imd adding in the term 18s., we have 178s. to divide by 12. We obtain 14s., with remainder 10s. ; and 14 lOi Bince 10s. = 120d., therefore, adding in the term 6d., we * -ii DITISION OF STEKL12fO MONET. 1^ have to divide 126d. by 12. We obtain lOd., with remain- der 6d. ; and since 6d. = 24q., we divide 24q. by 12, and thus obtain 2q., or j<:l. DIVISION OF STERLING MONEY. Mental Exercises. 1. How many lbs. of sugar, at 9d. per lb., may be bouglitfor 117d.? 2. How much coffee, at 8d. per lb., may be bought for 4s. 8d. ? 3. How much wheat, at 8s. per bushel, may be bought for £2, 16.0.? 4. In 243 farthings, how many pence? how many shil- lings ? liow many pounds ? 5. A goldsmith sold a tankard for jC 10, 8. 0., at the rate of 5s. 4d. per ounce. How much did it weigh ? 6. In £84 how many shillings? In these shillings, liow many guineas ? 7. Paid £ 7, 8, 6, for fire-wood, at 9s. per cord ; how many cords did I get ? Exercises for the Slate. 1. £409, 6. 2. -^ 8. 2. £386, 16. 5J. -f- 11, 8. £12, 18. 4^. -7-89. 4. £ 130264, 9. 6. -7- 9416, 5. £1746-7-2737. Ans. £51, 3. 3 J. Ans. £35, 3. 3 J. Ans. £ 0, 6. 7j. Ans. £13, 16. 8|. Ans. £0, 12. 9^3^. 132. When the divisor Is a composite number, it may sometimes be found convenient to break up the divisor into factors (Art. 67). Thus, Ex. 1, Divide £37, 14. 0. by 24. 24 = 4 X 6. X B. d. 4)37 14 24 { 8 6 1 U 5 m 124 nvxnon ov smLnto moitbt. In i Ex. 2. Divide £ 181, 2. ^ by 43, and also by the fko- tors 6 and 8, and show that the results coincide. £ 8. d. 48)181 2 8j^(je2. 96 85 20 48 ) 702 ( 14s. 48* 222 192 80 12 Now, dividing by the factors 6 and 8, we get £ s. d. 6)131 2 8^ 48 " U) 21 17 li+4 2 14 7J + 5 Therefore, the answer is £2, 14. 7^., 84 rem. 1 34 48 ) 368 ( 7d 886 82 4 48 ) 130 ( 2q., or «^d. 96 84 farthings, or 8^d. Therefore, the answer is £2, 14. 7j^., 34 farthings re- mainhig; or £2, 14. 7jd. jj. ; or £2, 14. 7id. JJq. Note.— The true remainder in the second operation is Ibund by Art. 68. the fto- >TB 6 and } 34 BTBRLIKQ MOKET. 125 Division op Steblinq Monet. — CojittTmed. In the following examples, diN^de by the numbers them- selves, and tlien by any factors composing them, and show th the results arc the same. 1. £440, 16. 9j. -r- 15. Ans. £29, 7. 9^. |iq. 2. £ 123, 13. 0}. -T- 99. Ans. £ 1, 4. 11 J. -ftq. 3. £ 678, 19. 9J. -f- 32. Ans. £ 21, 4. 4|. f|q. 4. £ 113, 14. 9. -T- 96. Ans. £ 1, 3. 8|., 9d. rem. 6. £ 100, 0. 0. -r- 121. Ans. £ 0, 16. 6^., 11 J. rem. 133. If the Divisor be 10, 100, 1000, &c., the operation of Division is usually performed by pointing off* as deci- mals, one, two, three, &c., figures accordingly, at the right hand of the dividend. Ex 1. Divide £5362, 10. 0. by 100. Long Method. Usual Method. £ 8. d. 100)5362 10 0(£53. i 500- £ •. d. 53,62 10 20 362 300 12,50 12 -i ' , 62 i 20 6,00 100)1250(123. igs re- 100- • Art. 68. I 250 li 200 1 50 1 ^^ 1 100) 600 (6d , 1 600 Therefore, the quotient is £53, 12. 6. 126 STERLING MONIT. Reasi/fi for the preceding process, je5362, 10s. ~ 100 = £\W + W^aS. = £53.62 + ^s. + x\)V. = £53 +<>24.^fl^-l- 10)g. ^ £53, 12. +<» $0.14^ $ (a) Ml (a) .11 (a> .lOJ- (a> .13 (a) .09J $ 208.838 (8.) Pictou. Aug. 17th, 1862. Messrs. Henry & Brothers, To J. L. Hoffman & Co. Dr. I k 15260 lbs. Pork, 7265 lbs. Cheese, 11521 bush. Oats, 1560 bbls. Flour, a) $0.05j- $ (a) 0.08| (a) 0.50 (S> 6.12J WEIOHT3 AND SlEASUItES. 135 Credit By 1150 lbs Soda, $0.06] S ** 8256 lbs. Sugar, 0.07 " 6450 gal. Molasses, 0.37J " Cash to balance account, What is the amount of Cash requisite to balance the account? Ans. $13703.78 Note.— The pupils should be required to reduce the ezampies given under Sterling to Nova Scotia currency, and perform the sameoperationr as required in the several questions, and verify the results. 139. It has been remarked (Art. 138, Note) that the currency of New Brunswick and Canada is taken at the same value as in the above-mentioned places ; but Nova Scotia currency does not pass for the same amount in those Provinces as it does here. The following table will there- fore be found useful to those having small transactions with the other British North American Provinces. N. Scotia. N. Brunswick. $5.00 = $4.86$ $0.25 ss $0.25, or 24|c. Table. Canada. N. F. Land. P. E. Mand. 4.86$ =£1,4.0. =£1,10.0. 0.25, or 24i. = £0, 1. 2. = £o, 1. 6. NoTB. — When the amount is large, it comes under the rule of Exchange, nrhich will be fully treated of in a more advanced portion of the work. To reduce small amounts Nova Scotia Currency To New Brunswick Currency, deduct ^ ; deduct ^; (( To Canada To P. E. Island " To New Foundland " add \ ; deduct ■^, TABLE OP WEIGHTS AND MEASURES. 140. Weights and Measures were invented 869 B. C. ; fixed to a standard in England, A D. 1257 ; regulated, 1492 ; equalized, 1826. 186 MEA&URE OF LENGTH. Agreeable to the Act of Unijormityj which took effect let January, 1826, The term Measure may be distinguished into seven kinds, viz. : Length, Surface, Volume, Capacity, Specific Grav^ ity. Space, Time and Motion, The several denominations of these measures have ref- erence to certain Standards, which are entirely arbitrary, and consequently vary among different nations. In Eng- land and her Colonies, the standard of Length is a Yard, Surface is a Sjare Yard, ^^J^y of an acre^ Solidit}'- is a Cubic Yard, Capacity is a Gallon, Weight is a Pound, The standards of angular measure and of time are the same in all European and most other countries. The Imperial standard yard, and the Imperial standard pound, Troy, of 1 758 and 60, in the custody of the Clerk of the House of Commons, having been destroyed by the fire at the House of Parliament in 1834, Restored Stan- dards of Weights and Measures have been legalized by 18 and 19 Vict., Cap. 72. MEASURE OF LENGTH. i Table of Lineal Measure. 141. It is recorded that the various denominations of length were constructed from a corn of barley, 3 of which, taken from middle of the ear, and well dried, made an inch. Other terms were taken from the human body, such as the Digit or finger's breadth. Palm, Hand, Cubit or length of the arm from the elbow to the wrist ; Fathom, from the extremity of one hand to that of the other, the ai*ms oppositely extended. It is stated that Henry I., in 1101, commanded that the ulna or ancient ell, which answers to the modern yard, should be made the length of his arm ; and that the other measures of length were hence derived, whether Lineal^ Superficial or Solid, MEAJSritt or LENUTB. 18T H The Hestored Standard of Lincul Measure, whose length is called a yard, is a solid square Bar, thirty-eight inches long, and one inch square, in transverse section, the Bar being of Bronze or Gun Metal, at the temperature of 62** of Fahrenheit's Thermometer, marked Copper, 16 oz.; Tin, 2} oz. ; Zinc, 1 oz. ; and near to each end a Cylindri' cal Hole is sunk to the depth of half an inch — the distanco between the centres of the two holes being 3 Feet or 36 Inches, or one Imperial Standard Yard, The standard of Square and Cubic measures will there- fore depend entirely upon it. At present, we have no means of ascertaining why this particular length was originally fixed upon ; but as it is most essential that it should always remain the same, it will be found convenient to refer it to something else, which we have no reason to suppose ever undergoes any change. Now, the length of a Pendulum, vibrating seconds, or forming 86,400 oscilations in the interval between the sun's leaving the meridian of a place and returning to it again, is always the same at n fixed place and under the same cir- cumstances ; and if this length be divided into 891,892 equal parts, the yard is defined to be equivalent to 360,000 of these parts ; also, conversely, since a yard is equal to 36 inches, it follows that the length of the seconds' pendu- lum, expressed in inches, is 39.1392. The Pendidum referred to, ^ one vibrating seconds at Greenwich or in London, at the level of the sea, in a vacuum or non-resisting medium ; and if the standard yard be at any time lost or destroyed, it is easy to have recourse to experiment for its recovery. The standard yard being the general unit of lineal measure, it follows that all lengths less than a yard will be expressed by fractions ; and it is on this account that a lineal inch, or ten thousands of the expressed portions of the pendulum, is conveniently adopted as a unit of lineal measure, when applied to small magnitudes. Hence, also, by the same means, the standard superficial and solid measures will be accui*ately ascertained and kept correct. 188 MEASURE OF LEKGTB. In this measure, which is used to measure distances, lengths, breadths, heights, depths, and the like, of places or things, 12 Inches (in length) make 3 Feet 6 Feet 6} Yards 40 Poles 8 Furlongs 3 Miles 69^ Miles iak( i 1 Foot, wiitten 1 ft. u 1 Yard, " 1 yd. (( 1 Fathom " 1 fath. u 1 Rod, Pole or Perch, "1 po. (( 1 Furlong, '* 1 ftir. (( 1 Mile, " 1 m. »( 1 League, " 1 lea. (( 1 Degree, " 1 deg.orl** An inch is the smallest lin^gLineasure to which a name is given ; but subdivisions arerised for many purposes. Among mechanics, the Inch is commonly divided into eighths and sixteenths. By the officers of the revenue, and by scientific persons, it is divided into tenths, hundredthSj &c. The inch, three-fourths inch, half-inch and quarter- inch, divided into twelfths, are used by architects. The following measurements may be added, as useful in certain cases : 4 Inches make 1 Hand (used in measuring horses). 22 Yards make 1 Chain, ) yy^ , . ^^„„„„i„„ i„„ j 100 Links make 1 Chain; { ^«^^ ^^ measuring land. A Palm = 3 inches, a Span := 9 inches, a Cubit = 18 in. A Pace = 5 feet, 1 Geographical Mile =: ^th of a degree. CLOTH MEASURE. 142. This measure, which is a species of Long Meas- ure, is used for all kinds of cloth, muslin, ribbon, &c. The yard, in Cloth Measure, is the same as in Long Measure, but differs in its divisions and subdivisions. i Inches make 1 Nail. 4 Nails " 1 Quarter, 1 qr. 4 Quarters " 1 Yard, 1 yd. 6 Quarters *' 1 English Ell. 6 Quarters " 1 French Ell. 8 Quarters " 1 Flemish EU. itances, ' places b. ith. 0. ir. I. a. Bg. or 1 ^ a name irposes. ed into lue, and dredthSy [juarter- LsefUl in s). hd. 18 in. degree. Meas« sC. Long OLOTH MEASURE. Mental Exerciser. 189 1 In 8 ft. Tin. how many inches are there? 2 In 2 fath. 1 ft. 9 in. how many inches are there ? 8. How many feet of line will reach the bottom in 9? fathoms of water, the deck of the vessel being 15 ft. above the water ? 4. A school-room is 2 rods 15 feet in length, and 2j- rods in breadth ; how many inches does the length exceed the breadth ? 5. James lives ^ of a mile from the school-house ; if he takes 2 steps every yard, how many steps will he take in going to school ? 6. What is the use of Long. Measure? What is the Standard, and how obtained ? 7. If it requires 2^ yds. of cloth to make a coat, how many nails of cloth will there be in 16 coats ? Exercises for the Slate Beduce, verifying the result in each case, the following : 1. dm. 7 fur. 8po. to yards ; and 573 miles to inches. Ans. 6864 yds.; 36305280 in. 2. 1364428 in. to leagues; and 74m. 3 fur. 4yds. to inches. Ans. 7 lea. 4 fur. 10 po. 5 yds. 2 ft. 4 in. and 4712544 in. 3. 7 fur. 200 yas. to chains; and 6 cubits, 1 span to feet. Ans. 79 ch. 2 yds.; 9 ft. 9 in. 4. 1000000 inches to miles. Ans. 15 m. 6 fur. 10 po. 2 yds. 2 ft. 4 in. 5. A person engages to walk 16 times, without stop- ping, between two places whose distance is 1 m. 3 fur. 14 po. 4 yds 2 ft. 3 in. ; what will be the whole distance he will have to walk? Ans. 22 m. 5 ft. 37 po. 4 yds. 1 ft. 6 in. 140 MEASURE or SURFACE. MEASURE OF SURFACE Table ok Square Measure. 143. The Imperial square yard contains 9 imperial square feet, and tlie Imj)erial square foot 144 square inches ; the circular foot (that is, a circle whose diameter is 1 foot} contains 113,097 square iiiches ; and the square foot contains 183,346 circular inches (t^atis, inches whose diameters are each 1 inch). This measure is used to measure all kinds of s iperficies, such as land, paving, flooring, in fact everything in which length and breadth are to be taken into account. A square is a figure which has four equal sides, each perpendicular to the adjacent ones. A square inch is a square, each of whose sides is an inch in length ; a square yard is a square, each of whose sides is a yard in length, &c. The table of Square Measure is formed from that of Long Measure, by multiplying each lineal dimension by itself. Thus, A Square Foot is = 12 X 12 = 144 square inches. 144 Square Inches mt>ke 1 Square Foot, 1 sq. ft. or 1 ft. 1 Square Yard, 1 sq. yd. or 1 yd. i Square Pole, 1 sq. po, or 1 po. 1 Square Rood, 1 ro. 1 Acre, 1 ac. 1 Square Mile. 1 Yard of Land. 1 Hide of Land. 1 Barony. 25000 Square Links = 1 Bood. 100000 Square Links z=. 1 Acre. 10 Squai'e Chains ^^ 1 Acre. 9 Square Foot 30^^ Square Yards 40 Square Poles 4 Roods 640 Acres 30 Acres 100 Acres 40 Hides NoTB 1.— The chain, referred to iotlie above and preoedinft table, ia illed GunUr't Chain; it is 4 perches in length, and is diyided into 100 equal parts, called linka^ Four perches being equiyalent to 792 called GunUr's Chain; it is 4 perches in length, and is diyided into lual parts, called links. ¥< inches, it follows, from dividing by 100, that the length of a link is 7 93-100 inches. Sunrerors ooDinuia Vkv r\!s.t«« •«<) l.'.rlrc. but exhibit the I9|ult in MEASURE OF SURFACE. 141 NoTR 3.~In tntny parts of Enffland, Ireland and Seotlaod, the old measure is still retained. The following table will therefore be found useful in some cases : 6| Yards 6 Yards 7 Yards 8 Yards 1 Perch, Cunningham measure. 1 Perch, Woodland or Burleigh measure. 1 Perch, Irish measure. I Perch, Forest u The roads, in Ireland, are measured by the Irish perch. In some parts of Ireland, land is rented, bought and sold by the Irish acre, or, as it is sometimes called, Irish plajitation measure; in some, by Cunningham measure^ and in others, by English statute measure. In Irish measure^ 64 acres ire equivalent to 49 acres in Forest measure ; G25 to 784 in Cunningham measure ; 86 to 49 in Woodland or Burleigh measure; 121 to 196 in Eng^ Ush statute measure; and 1369 to 1764 Scotch acres. These numbers may be found by multiplying each of the numbers expressing the length of the perches, in the dif- ferent kinds of measure, by itself. The Scotch chain of 74 feet, or 24 ells of 37 inches •achr is partially used yet. Ex. 1. Reduce 13ac. 8ro. 14 po. to square yards. Proceeding by the Bule given (Art. 126), ao. la 4- ro. 3 po. 14 55 40 . 66420 558^ 66978^ M 142 MEA8CRB OF SURFACE. Ex. 2. Reduce 1896784 sq. yds. to acres. Proceeding by Rule given (Art. 126). Square yards. 1896784 4 80^ reduced to fourths ,= 121 1 11) 11)7587136 689739"7 4,0) 6270 ,3"6 j - quarters ISi yds. 4) 1567"23per. 391ac. 3ro. 28per. ISJyds. Ex. 8. How many acres are there in a field which is 11 chains, 78 links Ion/, and 4 chains, 72 links broad? 12 ch. 78 links = 1278 links. 4 ch. 72 links = 472 links. 2556 8946 6112 6,03216 4 0,12864 40 6,14560 BOi 436800 8640 4,40440 Ans. 6ac. Oro. 5po. 4tyd8. MEASURE OF BUnPACE. Reason for the preceding process. H3 •When links are multiplied by links, the product is called square links. Thercforp, 1278 X 472 = ()U321G squaro links, and as there are 100000 square links in an aero, wo divide the tbrnier hy the latter for acres ; that is, we cut off five figures toward the right hand. Again, a Hood being tlie next unit of measure, the re- mainder will contfdn 4 times as many roods as acres. Wo therefore multiply by 4, and divide as before. In the same manner, we multiply by 80J for yards, Ac. Mental Exercises. How many square inches are there in 2 sq. ft. 12 sq. Change } of a sq. ft. to sq. in., and add 6 sq. in. to 1. m.? 2. it. 8. What will it cost to dig a garden 900 feet long and 80 feet wide, (S> 8 cents per sq. yard? 4. What cost 240 sq. rods of land, at $1000 an acre? 5. What is the use of Square Measure? By what measure would you ascertain the distance from your home to the school-room? Would you use the same kind to measure your slate ? 6. How wide must a piece of land be that is 16 rods long, to make an acre ? 7. Find the numbers for the foregoing table. Exercises for the Slate. Verify each oi the following results : 1. Reduce 35 ac. 2ro. to poles. Ans. 5680 po. 2. Reduce 3 ro. 37 po. 26 yds. to inches. Ans. 6188724. 8. Reduce 15 ac. 3 ro. to links. Ans. 1575000. 4. Reduce 93827 perches to acres. Ans. 586 ac. 1 ro. 27 po. 144 ARTIFICERS MEAsiURE. I I' L 5. Find the sum of 25 ac. 2 ro. 16 po. ; 80 ac. 2 ro. 25 po. ; 26 ac. 2 ro. 35 po. ; 63 ac. 1 ro. 31 po. ; and 34 ac. 2 ro. 29 po. Ans. 181 ac ro. 16 po« G. A man having 56 ac. 2ro. 34 po., sold 48 ac. 3ro. 38 per. ; how much had he remaining ? Ans. 7 ac. 2 ro. 36 po. 7. Multiply 380 ac. 3 ro. 32 po. by 106. Ans. 40380 ac. 2 ro. 32 po. 8. A person, at his death, bequeathed his farm, con- taining 1867 ac. 3 ro. 14 po. to his three sons, in equal shares ; how much does each get? Ans. 622 ac. 2 ro. 18 po. ABTIFICERS' MEASURE. 144. Flooring, roofing, &c., are measured by the jsquare of 100 feet. Plastering, by the square yard. Plasterers, generally, deduct ^of all the openings from the total content, for the true content. Bricklayers* work, by the pole of 16 J feet, the square of which is 272|^ feet, though this is partly a cubic measure, as the brickwork is reckoned to be 14 inches, or 1^ brick thick. Masons calculate by the rood of 36 sq. yds. This is pai'tly a cubic measure, also, as the work is reckoned to be 22 inches thick. Ex. 1. Required, the number of square feet in a roof which is 46 feet, 10 inches long, and 21 feet, 3 inches of j^after. 46 ft. 10 in. = 562 inches. 21 ft. 3 in. = 255 inches. i 144) 143310 sq. inches. 995 sq. ft. 80 inches. ▲STIFIC£BS* MEASURE. 145^ Ex. 2. Find the expense of paving a floor, whose length is 33 ft. 2 in., and breadth 18 feet, fa> 6s. per square yard. 33 ft. 2 in. = 398 inches. 18 ft. = 216 144 ^12)85968 112) 7164 9) 597 66 J^ sq. yds., fa> 6s. per sq. yd £3, 6. 4. = cost, (a) Is. per yd. 6 £ 19, 18. 0. = cost, to) 6s. per yd. It may be well to remark that each of the foregoing ex- amples might have been worked in the fractional form, or by reducing the lower denominations into decimals of the highest involved. There is, however, a method of working examples in square and cubic measure, without reducing the different denominations to the same denomination. This method is styled Cross Multiplication or Duo- decimals. Duodecimals are calculations by feet, inches and parts, which decrease and increase by twelves. Hence, the}'' take their name. Table. 12 Thirds make 1 Second, Marked 1" 1 12 Seconds " 1 Prime, " 1' 12 Primes or Inches " 1 Foot, " ft. The inches or primes, seconds, &c., being distinguished by the signs ', ", '", "", &c., the place of any product may be known by adding the signs of the factors together. Thus, 9 feet X 5" = 45" = 3', 9" ; 4' X 5" = 20'" = 1", 8'". \' 146 ▲RnFICESS^ ICEASUBi;. To Multiply Feet and Inches, or Feet, Inches and Parts Duodecimally. Rule. Place feet below feet, inches below inches, &c. First multiply the whole multiplicand b}' the feet in the multiplier, as m compound multiplication ; then by the inches in the same manner, placing the product, however, one place farther to the right ; then the seconds in the same manner, &c. The sum of these products will be the product of the two quantities. Ex. 1. Multiply 4ft. Gin. oy 2ft. Sin. Proceeding by the Rule given above, 4f!^. 6 in. 2 ft. 3 in. 9 1 0' 1' 6'' I 10 ft. 1' 6", or 10 ft 18 in. NoTB.—In the above example of Gross Multiplicatioit, we see tn mixed decimal and dtiodecimat scale of notation is employed, the figt."* i of the feet being expressed and miritipUed'in the ordinary way ; whereas, in other places the number 12 is always used instead of 10. Cross Mul- Uplicatioa is not, therefore, properly termed Duodecimal Multiplication or Duodecimals, because, although the different denominations are con- nected with each ottker by the number 12, still the different digits of those denominations are connected with each other by the number 10. After multiplying by the feet, it will often be found more convenient and expeditious to perform the rest of the operation by means of aliquot parts (Art. 92, de£, 9). Ex. What is the superficial content of a floor whose length is 40 feet 6 inches, and breadth 28 feet 9 inches ? Ft. ' Ft. ' 40 6 40 6 ^8 9 14 0' = 6'X28 320 80 6' = Joflft.= 20 3' 3' = J of 6' = 10 1' 6" 1164 4' 6' or 1164 sq. ft. 54 sq. in. 28 9 :14 ft. 1134 0' 30 4' 6' 1164 4' 6" or 1164 ft. 54 sq. in. MCAStmi: OF flOLlDXTT. Mental Exercises. lil 1. How many square inches in a board 15 inches wide and 11 feet long? How many i^q. ft. ? 2. What will it cosb to plaster a room 9 feet long, 8 feet wide, and 9 feet high, at 15 cents per yard? 3. How many feet are there in a board 3 feet 2 inches long, and 23 inches wide ? Exercises for the Slate. 1. What is the superficial content of a plank 16 feet 8 inches long, and 15 inches broad ? Ans, 20 ft. 10', or 120 in. 2. Find the content of a board 9 ft. 8 in. long, and 4 ft. 61 in. broad. Ans. 43 ft. 10', 10", or 43 ft. 130 sq. inches. 3. How many square yaras of paving in a court 88 ft. 7 in., by 20 ft. 9 in. ? Ans. 88 yds, 8 ft. 7', 3" 4. How many yards of flooring in a house of two stories, 40 ft. 9 in. long, by 32 ft. in. within, deducting from both flats the well-hole of stair, 10 ft. by 6 ft. 6 in. ? Vns. 130 yds. 4 ft. 5. How many square feet of board ^ill be in a log 28 feet 3 inches long, 16^ inches deep, cutting 10 boards? Ans. 388 feet, 63 inches. MEASURE OF SOLIDITY. Table of Solid or Cubic Measure, 145. The Imperial cubic {or solid) yard contains 27 im- perial cubic feet, and the imperial cubic foot contains 1728 cubic inches. The cylindric foot (that is, u cylinder 1 foot long and 1 foot in diameter) contains 1357.17 cubic inches. The conical foot (that is, a cone 1 foot in height, and 1 foot at the base) contains 452.39 cubic inches. A Cube is a solid body, and contains length, breadth, md thickness, having six equal sides. 148 VtikSURE OF flOLIDlTT. "I I A Cube Number is produced by multiplying a number twice into itself. Hence, the following table is formed fVom the table of lineal measure by multiplying each lineal dimension by itself twice. Cubic Measure is used in measuring solid bodies, or things which have length, breadth, and thickness: such as timber, stone, boxes of goods, the capacity of rooms, ships, &c. ^1728 Cubic Inches 27 Cubic Feet 40 Cubic Feet of Rough, or ) 50 Cubic Feet of Hewn Timber ] 42 Cubic Feet of Timber 128 Cubic Feet 5 Cubic Feet = 1 Cubic Foot. = 1 Cubic Yard, = 1 Ton. = 1 Shipping Ton. = 1 Cord of Wood, = 1 Barrel bulk. NoTB 1.— By a ton of round timber, is meant such a quantity of tim- ber in its rough or natural state, as when hewn, will make 40 eubio feet, and is suppoMd to be equal in weight to 50 feet of hewn timber, NoTB 2.<— The truth of the foregoing Tables of Square and Cubio Measures, should be illustrated to the pupils by means of small blocks, of one inch each, and by diagrams. Ex. What is the solid content of a block which is 10 feet 3 inches long, 2 feet 4 inches wide, and 3 feet 5 inches deep? 10 ft. 3 in. = 123 Inches, 2ft. 4in. = 28 ] • 984 246 8444 Sq. Inches: 8 ft. 5in,~= 41 Inches. , < 3444 13776 \ 141204 Cubic Inches. = 81 cub. ft. 1236 cub. inche8.\ . A MEASUBC OF SOLIDITT. By DuodecimalB* 149 Ft. 10 2 8 4 0" 0"' inches. 4' = i = 4' = i = l' = i = Ft. 10 2 3 4 20 8 6' 5' 20 3 6' 5 23 11' 3 0" 5' 23 3 11' 5' 71 9 9' 11' 0" 7" 71 7 1 9' 11' 8" 11' 11" 81 8' b.ft 7" 0'" . 1236 cub. or 81 cu 81 8' 7" Mental Exercises. 1. How many solid inches does a block 2 inches long, 2 inches wide, and 1 inch thick, contain ? 2 inches thick ? 4 inches thick ? 2. How many cubic inches does a block that is 1 foot long, 1 inch thick, and 1 inch wide, contain ? How many if 2 inches wide ? 4 inches wide ? 3. What is the cost, per foot, of wood, at $1.60 per cord, and what will a pile of such wood, 5 ft. 6 in. long, 4 ft. 2 in. high, and 4 fett deep, cost ? ' Exercises for the Slate. 1. ^ How many cubic inches in 767 cubic feet? Ans. 1325876. 2. _ Reduce 157248 cubic inches to yards ? Ans. 3 cub. yds. 10 cub. ft. 3. Find the sum of 3 c. yds. 23 c. ft. 171 c. in. ; 17c.' yds. 17 c. ft. 31c. in. ; 28c.3'ds. 26 c. ft. 1000 c. in., and 34 c. yds. 23c. ft. 1101 c. in. * Ans. 85 c. yds. 9 c. ft. 575 c. in. 4. Required, the content of a log 28 ft. 8 in. long, 17, in. wide, and 24 J deep.;. . . Ans. 82 ft. 10 in. 11", 8'". i1 i 150 MEASCBE OF CAPACITY. 6. Find the content of a tank 12 ft. 6 in. by 11 ft. 4 in. at top, by 10 ft. 8 in. at bottom, and 6 ft. deep. Ans. 764 ft. 72 in. Note. — When the article to be measured tapers in either length, breadth, or depth, take half the sum of that at the two ends for a mean. MEASJRE OF CAPACITY. Table of Liquid and Dry Measure. 146. The Imperial Gallon is the standard unit of the measure of capacity, and is defined to be 277.274 cubic inches, the lineal inch being that above mentioned. The gallon, and its multiples and parts, are used to measure both liquids, as wat3r, spirits, &c. ; and dry goods , as malt, corn, &c., and the system is therefore called the Imperial Liquid and Dry Measure* 4 Gills make 2 Pints 4 Quarts 2 Gallons 4 Pecks 8 Bushels 86 Bushels (( 1 1 1 1 1 1 1 Pint, marked 1 pt. Quart, Gallon, Peck, Bushel, Quarter, Chaldron, 1 1 1 1 1 1 Cubic Inches 84.6592. 69,3185. 277.274. 654.548. bus. 2218.192. qar. 17745.586. ch. qt. gal. pk. HEAPED MEASURE. 147. Potatoes, Tiu*nips, Fruit, Lime, Coals, and a few ; other articles, are bought and sold by heaped measure. 4 Pecks 8 Bushels 12 Tubs 1 Peck, 1 Bushel, 1 Sack or Tub, 1 Chaldron, Cubic Inches 708.87148. 2815.4871. 8446.45776. 101357.49309. The diameter of the exterior brim of the bushel is to be 19^ inches, and the height of the heap at least 6 inches. The content of the heap is therefore 597.29518 cubic HEAPED MEASURE. 151 inches, which, added to 2218.192, the content of the bushel, gives 2816.4871 cubic inches for the content of the heaped bushel, and the contents of the other measures are in pro- portion. The outside diameter of the measures less than a bushel, are as follows : Half-bushel = 15^ Inches. Peck = 12i Inches. Gallon = 9J Inches. Half-gallon = 7J Inches. Mektal Exercises. 1. At 43 cents a peck, what cost 14 bus. 8 pks. of wheat? 2. At 3 cents per quart, what will 5 bus. 3 pks. 2 qts.* of salt come to ? 3. A man sold 63 gal. molasses ^ $0.40 per gal., and received his pay in corn (a> $0.84 per bus. How many] bushels did he receive ? 4. In 48 pecks how many pints ? 5. By what measure would you buy potatoes? oats? oil? boards? ribbon? cloth? Exercises for the Slate. / 1. In 186040 pks. how man}'- chaldrons? Ans. 1291 ch. 34 bus. 2. In 365843 gills how many gallons ? Ans. 11432 gal. 2pt. 3gi. 3. Add together 39 gal. 3 qt. 1 pt. ; 48 gal. 2 qt. 1 pt. ; 66 gal. 1 pt. ; 74 gal. 3 qt. ; 84 gal. G qt. 1 pt. Ans. 306 gal. qt. 4. Find the diflTerence between 23chal. 5 bus. 2pk. and 14 chal. 6 bus. 3 pk. Ans. 8 chal. 34 bus. 3 pk. 5. Multiply 57 gal. 3qt. separatel}^ by 10 and 257. Ans. 577 gal. 2 qt ; ^14841 gal. 3 qt. f 152 FLUID MEASURE — MEASURES OP WEIGHT. APOTHECARIES' FLUID MEASURE. 60 Minims (m) = 1 Fluid Dram, Marked fZ* 8 Drams = 1 " Ounce, " f^. ' 16 Ounces = 1 " Pint, " O. 8 Pints = 1 " Gallon, " gall. Note. — In some places a pint equals 20 ounces. A minim may be reckoned 2 drops, a dram about a tea-spoonful, and 1 ounce about 3 table-spoonfuls. MEASURES OF WEIGHT. Table op Troy Weight. 148. The origin of all weight in England was derl'^ed from a grain of wheat. Vide Statute of 51 Henry lil. ; 31 Edward I., and 12 Henry VII., which enacted, that 32 of them gathered from the middle of the ear, and well dried, were to make 1 pennyweight ; 20 pennyweights 1 ounce ; and 12 ounces 1 pound. It was subsequentl}'' thought better to divide the penny- weight into 24 equal parts, called grains. William the Conqueror introduced into England what is called Troy-weight, from Troyes^ a town in the province of Champagne, in France, now in the department of Aube, where a large and celebrated fair was held. It seems to have been brought hither from Egypt. It has also been derived from Troy-^iovant^ the monkish name for London. This weight was formerly used for weighing articles of every kind. It is now only employed in weighing gold, silver, diamonds, and other articles of a costly nature ; also in determining specific gravity. The different units are grains (grs.), pennjn^'eights (dwt.), ounces (oz.), and pounds (lbs.), and are connected thus: 24 Grains miake 20 Pennyweights " 12 Ounces " 1 Pennyweight, Idwt. 1 Ounce, 1 oz. 1 Pound, lib. APOTHECARIES' WEIGHT. 153 Note 1.— It was called Apennyioeight from its being the weight of the silver penny then in circulation. The term ounce comes from the Latin word uncia, which signifies a twelfth part. In the abbreviation, dwt., for pennyweight, d is from the Latin word denarinitt a penny; wt, the first and last letters of the English word weight. Oz, is from the Spanish onza, an ounce. Note 2.— Diamonds, and other precious stones, arc weighed by ** Carats," each carat weighing 3 17-101, or nearly 3 1-6 grams, Troy. The term carat, applied to gold, has a relative meaning only ; any quan- tity of pure jsjold , or of gold alloyed with any other metal, being su|»^os- ed to be divided into 24 equal parts (carats) : if the gold be pure, it is said to be 24 carats fine; if 22 parts be pure gold, and 2 parts alloy, it is said to be 22 carats fine. Standard gold is 22 carats fine; jewellers* gold is 18 carats fine. Thos we generally perceive *M8 " on the cases of gold watches. This indicates that they are ** 18 carats fine," the lowest degree of purity which is marked; but many articles are manufactured as low as 9 carats fine. Table of AroTHECAftiEs' Weight. 149. Apothecaries' weight only differs from Troy weight in the subdivisions of the pound, which is the same in both. This table is only used in mixing medicines, as they buy and sell by Avoirdupois weight. The different units are grains (grs.), scruples (f)), drams (3)j ounce (J), pound (lb.), and these are donnected thus: 20 Grains make 1 Scruple, 1 sc. or 19. 3 Scruples " 1 Dram, 1 dr. or 1 3. 8 Drams " 1 Ounce, 1 oz. or 1 ?. 12 Ounces " 1 Pound, 1 lb. or 1 &. Mektal Exercises in Tkoy and Apothecaries Weight. 1. How many grains ai*e there in 1 dwt. ? 2. In 3 dwt. 6 gr. how many grains ar*^ there ? 8. In 3 scruples how many grains ave there? 4. How often is 86 drams contained in f lb. ? 6. 8di*. 19 20 gr. are how many times 22 grains? 6. J of a pound + 2oz. ai*e equal to how many dwt. ? 7. "When gold is selling for 16J dollars an ounce, what is the value of a i)ouud? 7* 154 AVOIRDUPOIS WEIGHT. ^!. I Exercises for the Slate. 1. Reduce 53G lb. to scrnplcs. Ans. 15486880. 2. Reduce 171b. 23 2 q to grains. Ans. 98920 gr. 3. Reduce 29 lb. 7 oz. 3 d>vt. to grains. Ans. 1 70472 gr. 4. Reduce 6237 drams (apoth.) to lbs. Ans. 641b. 11 oz. 5 dr. 6. A gentleman's silver plate weighs as follows, viz. : salvers, 121b. 3oz. 20 gr. ; tankards, 91b. 6oz. ISdwt. 17 gr. ; spoons, 211b. 14dwt. 6gr. ; knives and forks, 161b. 8 oz. 15 dwt. 21 gr. ; and salts, 2 lb. 6 dwt. Required, the total weight of his plate. Ans. 61 lb. 2oz. 10 dwt. 16 gr. 6. A person having a piece of pure gold weighing 13 oz. 5 dwt. 9 gr., offered to exchange it for a piece of plate weighing 12 oz. 3 dwt. 3gr., provided he would get 84 cents per dwt. for the difference in weight. How much would he receive ? Ans. $18.69. AVOIRDUPOIS WEIGHT. 150. %'hen William the Conqueror introduced Troy weight, the English were dissatisfied with it, because it did not weigh so much as the pound then in use. Hence arose the term Auoir-du-poids, which was a medium between the French and the ancient English weight. Avoirdupois Weight was first made a legal tender in the reign of Henry VII., and its particular use was to weigh provisions and coarse heavy articles. Henry fixed the stone at 14 lbs., which has since been confirmed by Act of Parliament. The Imperial Pound, Avoirdupois, which is the stan- dard unit by means of which all heavy goods of large masses is weighed, is defined to he the weight of one-tenth part of an imperial gallon, or of 27.7274 cubic inches of dis- tilled water, ascertained at a time when the barometer stands at 30 ins., and the height of Fahrenheifs thermometer is 62** ; and this standard may consequently be verified or recovered at any time, when it may he necessary to appeal to experiment. NEW SYSTEM OF WEIGHT. 155 880. Ogr. 2gr. 5 dr. }, viz. : Iwt. 17 , 161b. ed, the 6gr. ing 13 f plate get 84 7 much 8.69. Troy it did |e arose en the ider in ras to fixed ^ed by stan- large •tenth )f dis- ^meter ^meter ied or ippeal If the weight of a cubic inch of distilled water be divid- ed into 605 equal parts, and each of such parts be defined to be a hal/'f/rauu it follows that 27.7274 cubic inches con- tain very nearly 7000 such grains ; and it is hence declared by Act of Parliament that 7000 grains exactly, shall here- after be considered as a pound, avoirdupois ; and that 10 grains shall be equivalent to 1 scruple; and 3 scruples to 1 dram. But these latter denominations are seldom neces- sary, unless gi'cat nicety is required. This weight receives its name from avoirs^ the ancient name of goods and chattels, and poids, signifying weight, in the ordinary language of the country at the time of the Normans, The Restored Imperial Standard Pound, Avoirdupois^ is constructed of platinum, the form being that of a cylin" der, nearly 1.35 inch in height, and 1.15 inch in diameter, marked P. S. 1844, 1 lb. The different units are drams (drs.), ounces (oz.), pounds (lb.), tons (ton), and they are connected thus: 16 Drams 16 Ounces 28 Pounds 4 Quarters 20 Hundredweight" 1 Ton, Grains, Troy, make 1 Ounce, loz. = 437.50 " 1 Pound, 1 lb. = 7000 " 1 Quarter, Iqr. = 196000 " 1 Hundi-edweight, 1 cwt.= 784000 1 ton =15680000 In general, 1 Stone (1st.) = 14 lbs. avoirdupois, but for butchers' meat or fish, 1 stone = 8lbs. In Ireland, in the sale of some articles, the stone is =r 16 lbs. avoirdupois. NEW SYSTEM OF WEIGHT. 151. By an Act of the Provincial Parliament, passed A. D. 1859, the hundredweight is to contain 100 lbs. avoir- dupois, instead of 112 lbs., and the ton 2000 lbs. instead ©f 2400 lbs., as formerl}-. f •'\\\.{ 156 NEW SYSTEM OF WEIGHT. The different units are the same as in the old system. Thus: Troy ffrains. make 1 Ounce, loz. = 437.5 *' 1 round, lib. = 7000 ** 1 Quarter, Iqr. = 176000 " 1 Hundredweight, lcwt.= 700000 1 ton =14000000 16 Drams 16 Ounces 25 Pounds 4 Quarters 20 Hundredweight " 1 Ton, Note.— The old system of weight is called long, and the new system short weight. Ex. 1. Reduce 23 cwt. 3 qr. 14 lb. long weight, to pounds. Proceeding by the Rule given (Art. 126), cwt. qr. lb. 23 3 14 4 95 28 774 190 2674 pounds. SECOND HETEOD. owt. qr. lb. 23 3 14 2300 276 84 14 23 X 100 > 23 X 12 ( 3 qr. X 28, 14 lbs. = 112 X 23. 2674 pounds. I MEW 8T8TEM OF WEIGHT. IW Ex. 2. Reduce 18967432 dr. to hundredweight, short weight. Proceeding by Rule given (Art. 126), 4 ) 180G7432 16 dr. •i 4) 4741858,0) 4) 1185464,2 j 4 ) 296366,0 \ 740,9 1,*JJ 8 dr. 80E In dividing by 100, wc merely cut oflf the two right hand figures. Therefore, che unswor is 740 cwt, 91 lb. 8oz. «dr., or 740 cwt. 3qr. 161b. ^oz. 8 dr. Mentai Exercises. 1. How many oz. are theie in 16 dr.? hotv many in 80dr.? 2. In 3 oz. 3 dr. how many arams are there ? 3. In 2 ton 12 cwt. 3 qr. how many qr. ? 4. At 9 cents a pound, ^ hat cost 3 cwt. 2 qr. 16 lb. short weight, of sugar? 5. In 25 lb. how many grains ? Exercises for the Slate. 1. Reduce 6cT';t Iqr. 18 lb. long weight, to drams. Ans. 183308. 2. In 30 ton 18 cwt. 2qr. 201b. 12 oz. 15 dr. short weight, how rrxny drams? Ans. 15,838,927. 3. A man bought 15 loads of hay, each weighing 1 ton 270^ lb. ; what was the weight of the whole? Ans. 17 ton 551b. 4. In 25 packages, each containing 20336oz., how many pounds ? and what will the whole cost at 22i|- cents per lb.? Ans. $285.9 ?v> 158 SPACE. — ^TIIklB ^ MEASURE OF SPACE. Angular Measure or Division of the Circle. 152. 1 Second is written Isec, or 1". 60 Seconds make 1 Minute, Imin. or 1'. 60 Minutes " 1 Degree, Ideg. or 1^. 90 Degrees " 1 Right Angle, Irt. ang. or 90^. The circumference of every circle is considered to be divided into 360 equal parts, each of which is often called a degree, as it subtends an angle of 1^ at the centre of the circle. - MEASURE OF TIME. * Table of Time. 153. 1 Second is written thus : 1". 60 Seconds make 1 Minute, 1'. 60 Minutes 24 Hours 7 Days 28 Days 28, 29, 30 or 31 Days 12 Calendar Months 365 Days 366 Days u t( u 1 Hour, Ihr. 1 Day, Iday. 1 Week, Iwk. 1 Lunar Month. 1 Calendar Month. 1 Year. 1 Common Year. 1 Leap Year. The number of days in each month is easily remem- bered by means of the following lines : Thirty days hath September, April, June, and November ; February hath twenty-eight alone, And all the rest have thirty-one ; But leap-year, coming once in four, February then has one day more. A day, or rather a mean solar day, which is divided into 24 equal portions^ called mean solar hours, is the standard MEASURE OF TIME. 159 unit for the measurement of time, and is the mean or average time which elapses between two successive tran- sits of the sun across the meridian of any place. The time between the sun's leaving a certain point in the Ecliptic^ and its return to that point, consists of 365.242218 mean solar days, or 365 days, 5 hours, 48 minutes, 47^ seconds, very nearly, and is called a solar year. Therefore, the civil or common year, which contains 365 days, is about |-th of a day less than the solar year ; and this error would of course, in time, be very considera- ble, and cause great confusion. Julius Csesar, in order to correct this error, enacted that every 4th year should consist of 366 days. This was called Leap or Bissextile year. In that year February had 29 days, the extra day being called the Intercalary day. But the Solar year contains 365.242218 daj^s, and the Julian year contains 365.25, or 365] days. Now, 365.25 — 365.242218 == .007782. Therefore, in one year, taken according to the Julian calculation, the sun would have returned to the same place in the ecliptic .007782 of a day before the end of the Julian year. Therefore, in 400 years the sun would have to come to the same place in the ecliptic .007782 X 400, or 3.1128 days before the end of the Julian year; and, in 1257 years, would have come to the same place .007782 X 1257, or 9.7819, or about 10 days before the end of the Julian year. Accordingly, the vernal equinox which, in the year 325, at the Council of Nice, fell on the 2 1st March, in the year 1582 (that is, 1257 years later), happened on the 11th of March. Therefore, Pope Gregory caused 10 days to be omitted in that year, making the 15th of October immedi- ately succeed the 4th, so that, in the next year, the vernal equinox again fell on the 21st of March ; and, to prevent the recurrence of the error, ordered that, for the future, in every 400 years, 3 of the leap years should be omitted, viz. : those which complete a century, the numbers express- ing which century, are not divisible by 4. Thus, 1600 and 2000 are leap years, because 16 and 20 are exactly divisi- ble by 4 ; but 1700, 1800, and 1900 are not leap years, because 17, 18, and 19 are not exactly divisible by 4, 160 MEASURfi OF TIMS. This Gregorian Style, which is called the New StyUy was adopted in England on the 2d of September, 1752, when the error amounted to 1 1 da3^s. The Julian calculation is called the Old Style. Thus, Old Christmas takes place 12 days after New Christmas. In Russia, they still calculate according to the Old Style; but in the other countries of Europe, the New Style is used. Sir Harris Nicolas, in his Chronology, gives the dates at which the New Style was adopted in different countries. Of course it was almost immediately adopted by most of the Roman Catholic Courts of Europe. Mental Exercises. 1. How many seconds are there in 3 min. 48 sec. ? 2. If a person works fths of a day, at 12^ cents an hour, how much should he receive ? 8. In 37 days there are how many weeks? 4. If the sun travels round the earth in 24 hours, through how many degrees will he pass in 1 hour ? in 4 minutes ? 5. The longitude of Boston is 71*^ West, and that of Chicago 87^ West ; what is the difference of time ? When it is 6 o'clock in the latter olace, what time is it at the former ^ Exercises for the Slate. 1. One year being equivalent to Z&o\ days, find how many seconds there are in 27 years 245 da3^s. Ans. 873223200. 2. From 9 o'clock, P. M., Aug. 5, 1852, to 6 o'clock, A.M., March 3, 1853, how many hours are there? and how liiany seconds ? Ans. 5025 hr. ; 18090000 sec. 3. How many days are there from the 12th of Aug. till the 24th of next April, in a common year? Ans. 255. 4. If the 8th of August be on Monday, on what day ^of the week will the 1st of November be ? Ans. TuMiday. •it- TARlOtS SUBJECTi. lei 6. If a leap year commeDce on Wednesday, what day of September will be the first Monday of that month ? Ans. The 7th. A Catalogue op Useful Things. A Standard Gallon contains 101b. avoirdupois, distilled water. A Barrel of Beer, 36 gal. A Quintal of Fish, 1121b. Do. inlJ. States, 1001b. A Firkin of Butter, 56 lb. A Decker of Gloves, 1 doz. pr. A Bushel of Oats, 341b. A Gallon of Flour, 7 lb. A Barrel of Flour, 196 lb. There are several measures mentioned in commerce, as Tierce, Hogshead, Puncheon, Pipe, Butt, and Tun ; but these may be considered rather as the names of the casks in which commodities are imported, than as expressing any definite number of gallons. It is the practice to gauge all such vessels, and to charge them according to their actual contents. Commercial Numbers. 12 Articles 13 12 Dozen 20 Articles 5 Score 6 Score 80 Deals 90 Words in Common makel Dozen. '* 1 Long Dozen. ** 1 Gross. 1 Score. 1 Common Hundred. 1 Qt. Hundred. 1 Quarter. 4 Quarters make 1 Hundred. 24 Sheets Paper** 1 Quire. 20 Quires ** 2 lieams ** 10 Reams ** 5 Doz. Skins of Parcliment ** Ream. Bundle. Bale. 1 Roll Chancery, 8 Words in Exchequer, or 72 in ) . vctWa Law, > '*'"*'• Sizes of Books. The terms Folio, Quarto, Octavo, &c., applied to books, denote the number of leaves into which a sheet of paper is folded. A sheet folded into 2 leaves forms a Folio. A sheet '* " 4 leaves a Quarto, or 4 to. A sheet " " 8 leaves a Octavo, or 8vo. A sheet " " 12 leaves a Duodecimo, or 12mo. A sheet ** " ?8 leaves an 18 mo. A sheet ^^ " 36 leaves a 36 mo. let WEiaHTS AND MEASXTRES OF 0. 8. CURRENGT. Sizes of Drawing-Pafer. Wove Antique, 4ft . 4in . X 2ft . 7 Double Elephant, 8 4 X2 2 Atlas, 2 9 X2 2 Columbies, 2 10 X 1 11 Elephant, 2 4 X 1 11 Imperial, 2 5 X 1 9 Super-Royal, 2 3 X 1 7 Royal, 2 X 1 7 Medium, 1 10 X 1 6 Demy, 1 8 X 1 3 iii:.' lili li;' marked ct. " d. " doll, or $. " E. , WEIGHTS AND MEASURES OP THE UNITED STATES CURRENCY. 154. Federal Money is the currency of the United States, The denoruinations are Eagles^ Dollars, Dimes, Cents, and Mills, 10 Mills (m.) make 1 Cent, 10 Cents " 1 Dime, 10 Dimes '* 1 Dollar, 10 Dollars " 1 Eagle, The National Coins of the United States are of three kinds, viz. : Gold, Silver, and Copper or Bronze. The gold coins are the eagle, the double^eagle, half^eagle, quarter'eagle, and gold dollar. The eagle contains 258 grains of standard gold ; the half-eagle and quarter-eagle like proportions, The silver coins are the dollar, half-dollar, quartevm dollar, the dime, half-dime, and three-cent-jdece. The dollar contains 412^ grains of standard silver ; the others, like proportion. The copper and bronze coins are the cent and halj^cent. The present standard for both gold and silver coin is 900 parts of pure metal and 100 parts of alloy. The alloy of gold coin is composed of silver and copper, the silver not to exceed the copper in weight. The alloy of sliver coin Is pure copper. FRENCH MONET, WEIGHTS, AND MEASURES. 1G3 Measure op Length, etc. The standard measure of Length, Surface, and Solidity are the same as those used in the British dominions. r Liquid Measure. The standard Unit of Liquid Measure, adopted by the United States, is the Wine Gallon, of 231 cubic inches (old wine gallon of England), which is equal to 68372.175 grains of distilled water, at the maximum density, weighed in air at 30 inches barometer, or 8.338 lbs. avoirdupois, very nearly. Dry Measure. The Standard Unit of Dry Measure, adopted by the United States, is the Winchester Bushel, which is equal to 77.627413 pounds, avoirdupois, of distilled water, at the maximum density, and contains 2150.4 cubic inches, of 7.75557 imperial gallons. 1 bushel, imperial, is equal to 1.032 bushel, U. S., nearly. Note.— The standard bushel of the State of New York is the imperial bushel of 80 pounds. Measures op Weight. The standard measures of all weights in the United States are the same as those of England. FRENCH MONEY, WEIGHTS, AND MEASURES. 155. The new System of Money, Weights, and Meas- ures of France, was formed according to the decimal nota- tion. French Money. The Franc is the unit money of the new system of French cun'ency . It is a silver coin, consisting of / £1, 18. 6^. each? Ans. £26497, 7. 11. 8. A person bought 1763 yds. of cloth, ^ $1.10 per jard, and sold at 6s. lid. per yd. What was his profit? Ans. £124, 17. 7. 4. Divide £3, 13. 9. between two persons, so that one shall receive half as much as the other. Beduce the result to dollars and cents. Ans. $9.83 J, and $4.91f. 5. A servant*s wages are £ 10, 8. a year. How much ought he to receive in 7 weeks (supposing the year 52 weeks)? Ans. $5.60. 6. A factor bought 56 pieces of stuff for £ 1569, 17. 4. at 4s. lOd. a yard ; how many yards were there in each piece? Ans. 116 yds. 7. A merchant bought 450 yards of cassimere at 8s. 6d. per yd. ; while lying in his wareroom 14 yds. were ren- dered useless ; he sold the remainder at $2.06f per yard. What did he gain? Ans. $186,063. QVBSttOMS AKD EXAIIPLKS. 167 1 6. Two boys run a race of 1 mile ; one of them gains 5 feet in every 110 yds. How far will the other be left behind at the end of the race? Ans. 26Syd8. 9. A merchant buys 10 gallons of spirit, at 12s. a gal- lon ; 15 gallons, at $ 2.90 a gallon ; and 18 gallons, nt 15s. 9d. a gallon. What will be the price of a gallon of the mixture, so that he may gain $ 9.10 on his outlay? Ans. $:ll0. 10. An apothecary bought 1001b. opium (Troy wt.), at 168. 3d. a pound, and sold it at 19s. Od. per poun 1, avoir- dupois ; did he gain or lose? Ans. Lost £3, I. G^. 11. A gentleman sent a tankard to his silversmith, which weighed 100 oz. 16 dwt., and ordered him to make it into spoons, each weighing 2oz. 16 dwt. ; how many spoons did he receive? Ans. 36 spoons. 12. Divide £ 17, 3. 5. by £ 14, 3. 6. to ^places of deci- mals. Can these sums be multiplied together ? Ans. 1.2113. 13. A certain number of men, twice as many women, and three times as many boys, earned in 5 days £ 7, 15. ; each man earned 30 cents, each woman 10 pence, and each boy 13 J cents a day. How many were there of each? Ans. 6 men, 12 wo. 18 boys. 14. "What is the expense of plastering the walls of a room, the perimeter of which is 58 ft. 4 in., and the walls 11 ft. high, deducting three doors, each 7 ft. 10 in. by 3 ft. 3m., two window-openings each lOft. 6 in. by 6ft., and a fire-place 6 ft. by 5 ft. 1 in., ^15 cents per yard ; likewise a cornice, at 5 cents per foot, allowing a foot in length at each corner? Ans. $9.92^f. Iti. An fipothecary bono:ht 126 ,?rnllo-n=! of esserr .^^of lomoii, L $;».07} \)vv (iinit, vr.ul -i>. i ■ - ;'' ounce, tlui■ 4. i = 1. 6. £4. 8. 5^. r Note.— If there are ehillings in the price, find the amount for the ahillings first, and that for the pence afterward, and add theee two amounts for the whole. BULBS FOB HElfTAL ABITHMBTIO. 169 (2.) To find the price of any number of dozens. RuLB. Regard the pence in the price as shillings, and multiply by the number of dozens. Ex. Find the price of 72 (6doz.). at 8Jd. 8. d. 8Jd. regarded as shillings, = 89 6 je2 12 6 ] (3.) Wien tJie number of articles does not consist of exact dozens. ' Rule. Find the price of the nearest number of dozens, by the last Rule, and increase or diminish it, as the case may require, by the price of the other articles. Ex. Find the price of G9 articles, to) 8Jd. 8}d. regarded as shillings, = 69 = 6 dozen, wanting 3. Multiplying by a. d. 8 9 6 £2 12 6 Deducting the price of 3, 2 2^ £2 10 32 (4.) To find the price of any number composed of «ta> teens. Rule. Reduce the price to farthings, multiply by the number of sixteens, and divide the product by 3, for shil- lings. If there is a remainder, after dividing by 3, reckon each one over as 4 pence. Ex. Find the price of 16 articles, at 42d. each. Also, 64 articles, at 6Jd. 4id. = 19 ejd. = 27 1 64 = 4 X 16 4 3)19 6s. 4d. 8 3)108 36s. or£l. 16. 0.' f 4 170 BULES FOK MENTAL AKITIIMETIC. n..!' i I* Reason for the preceding process, in iirticli'H, at ^(1., = 4(1. Thercrore. imiltiphing the farthings in the price by the nnniber of sixtcc'iis in the whole number, the product will be so many four-penny- pieces ; hence, dividing by 3, will give ebiilings. (5.) To find the price of 48. Rule. Reduce the price to farthings, and reckon the product as shillings. Ex. Find the price of 48 articles, at 5Jd. 5Jd = 28 farthings ; Reckoned 23 shillings, = £ 1, 3. 0. Reason for the above Rule. 48 farthings being equivalent to 1 shilling, therefore each farthing in the price will be equal to 1 shilling. (6.) To find the price of 96. Rule. Reduce the price to farthings, double the unit figure for shillings, and consider the other as pounds. £x. Find the price of 96 articles, at 5}d. 5}d. reduced to farthings, = 23 2 £. 2, 6. 0. 'I Tlie reason for the above may be easily deduced from the preceding Rule, (7.) To find the price o/ 112 articles. Rule. Find the price of 16 by the former rule, and multiply by 7. Ex. What cost 1121b. (7 X 16) of butter, at 9Jd. 161b. ^9Jd. = 13 7 . ^ £4,11. 0. / •• I" RULES FOR MENTAL ARITHMETIC. 171 (8.) ft) flnd the prirr of 100 articles. Rule. Reduce the price to farthings, and take those farthings as pence, and their double as shillings. Ex. Find the price of 1001b. of nails, at S^d. per lb. 8^d. = 18 farthings, X 2 = 2G ; as shillings, = £ 1, C 0. and 13 " X 1 = 13 ; as pence, = 1. 1. £1,7. 1. Reason for the above process. 100 articles, at |-d. each, will amount to 2 shillings and 1 penny. Therefore, twice as many shillings, and once as many pence, as there are farthings in the price, will give the true result. If there are shillings in the price, reckon each as £ 5. (9.) To find the price of 120 articles^ at any given number of pence, (2.) At any given number of pence and faHhings, Rule. (1.) Regard the pence in the given price as pounds, and divide by 2. (2.) Regard the farthings in the price as pounds, and divide by 8. Ex. 1. What will 1201b. tea cost, at Is. 6d. per lb. Is. 6d. = 18, which, divided by 2, = £ 9, 0. 0. Ex. 2. Find the price of 1201b. soda, at 4Jd. 4jKi. = 18 farthings, divided by 8, = £ 2, 5. 0. (10.) Find the price of any number of articles^ at any given price, in cents or dollars and cents. Rule. Multiply the number of articles by the cents or dollars and cents in the price, and point off as in decimals. Note.— As dollars and cents are in decimal notation, several contrac- tions can be made in the operation, to which the attention of the learner should be directed by the teacher. f; I ■i 172 RULlfl FOS araHTAL ARTTBltSTZC. To Calculats Inteoest Mentallt. 157. When money or property is hired or let put, a cei-tain fitipulated sum, or an equivalent, has to be paid by the borrower, for the loan of that property ; this, in the case of money, is called Interest. Interest, therefore, is the sum of money paid for the loan or use of some other sum of money, lent for a certain time, at a fixed rate ; generally, at so much for each JClOO or $100, for one year. The money lent is called The Principal. The interest of jC 100 or dollars, for a year, is called the Mate per Cent. The principal, -|- the interest, is called the Amount. Computations in Interest and Discount may sometimes be performed mentally with considerable facility, especially by means of contrivances which have been fallen upon to shorten the work in particular cases. The following are some of the more useful of these. (1.) To find the interest of any number of pounds^ for a given number of months y (d) 5 per cent. Rule. Take the pounds as pence, and multiply by the number of months. Ex. Find the interest of £ 186, 10. for 5 months, at 5 per cent. Proceeding by the above Rule, £ 186, 10., taken as pence, = 186^ pence : == £ 0, 15. 6;^. Interest for 1 month. 5 months. £3, 17.8^. * !l M RULES FOR MENTAL ARITHMETIC. Reason for the jyreeeding process. 173 The iiitewst of £ 100, at 5 per cent., for one 5'ear, is £5, and the interest of £ 1 would be xi^yth of this, or 1 shil- ling, for one year, and therefore 1 penny for 1 month. Hence, £ 186, 10., for one month, would be 186^ pence, or 15s. 6^d., and for 5 months, 5 times as much, or £3,17. 8^. (2.) To Jind the interest of any number of pounds^ SfC.^for any given number of months^ at 6 per cent, per annum. Rule. Multiply the principal by the months ; increase the unit figure by a fifth of itself, to find the pence of the answer, and take the others as expressing shillings. For OS. add l^. For 10s. add |d. For 15s. add fd. To 16s. 8d. and above, add Id.^ Ex. Required, the interest of £ 16, 3. 4. for 7 months, ' at 6 per cent. Proceeding by the above Rule, £ s. d. 16 3 4 7 11,3 3 4 = £0, 11. 8f. So Ss. 4d. = 4 of £ 1 . Therefore, ^ of | or ^^ |. £0, 11. 3|. Reason for the above process. It is evident from the preceding rule, that the interest of £1, for 1 month, at 6 per cent., will be l|d., and as 200 times l-|^d. = £ 1, the product of the principal and months, divided by 200, will give pounds. Wo therefore cut off the unit figure, which is the same as dividing by 10, and taking the rest as shillings instead of pounds, divides by 20. (20 X 10 = 200). The figure cut off is plainly tenths of shillings, and increasing of it by a fifth of itself, gives twelfths of a shilling or pence. ,.| f^ t f. a I ?l 174 RULES FOR MENTAL ARITHMETIC . (3.) To find the interest of any number of dollars and cents, for a given number of mouths^ at 6 "per cent, 'per annum. Rule. Multiply the principal by half the number Oi months, and consider the product as cents. Ex. Required, the interest of $126.15, for 8 months, at ,6 per cent. Proceeding by the Rule given above, $126.15 X 4 = 504.60 cents, or $5.04,^. (4.) To find the interest of any number of pounds, ^c, for a given number of days, at 5 per cent, per annum. Rule. Multiply the principal by one third the num- ber of daj^s, or multiply a third of one of them by the other ; a tenth of the result is the answer in pence, nearly. To correct it, reject a penny for every 6 shillings, or 3^d. for every pound contained in it. Should this correction be considerable, add a penny for every 6 shillings in it to the answer. Ex. Required, the interest of £141, 10. for 27 days, at 5 per cent. Proceeding by the Rule given above, £141, 10. 0. 9 1273, 10. 0. Dividing by 10 = 127f^, or 127^d. = Rejecting a penny for every 6 shillings, 10s. 11, H. £0, 10. 5i. (5.) To find the interest of any number of pounds. 4*c., for a given number of days, at 6 per cent, per annum. Rule. Divide the product of the principal and days by 100 ; take one third of the quotient for shillings, and one sixth of the remainder for farthings, and from the sum KULES FOR MENTAL AUITIIMi:TIC. 175 thus obtained, reject a penny for every six shillings con- tained in it. The remainder will be the interest required, nearly. If the interest is large, correct it as in 5 per cent. Ex. Find the interest of £163, 5. 3. for 25 days, at 6 per cent. Proceeding by the Rule given above, £163, 5. 3. 5 816, 6. 3. 5 /I A O ' 11 Q Dividing by 100, V = £0, 13. 4. V for farthings, 0. 3^. £0, 13. 7^) Rejecting Id. for 6s., 2^. £0. 13. 5. Interest. (6,) To fiiid the interest of any number of dollars and cents, for any given 7iumher of dags, at 6 per cent, per annum. Rule. Multiply the principal by one sixth of the num- ber of days, and consider the product as mills. Ex. Required, the interest of $146.16, for 18 days, at 6 per cent. Proceeding by the above Rule, $146.16 X 3, = 438.46 mills, or $0.44 cents, nearly. Note 1.— When the interest is large, or even amounts to dollars, de- duct li cents from it, for the true interest. Note 2. — The renson for the deduction :n the preceding rules, is in consequence of reckonine the commercial year as 12 months, of 30 days each, or 360 instead of 365 days, which gives the interest one seventy- second part of itself more than it should be. This will be better under- stood when the learner hag worked the Buub of Proportion. (See Note, Rule 3, Art. 221). BI r I! 176 FRACTIONS. FRACTIONS. Vulgar Fractions. 158. If a qnantit}', considered as a whole, be divided, into any number of equal parts, a Fraction, in reference to that quantity, signifies one or more such parts ; and the quantity divided is called, in respect to the fraction, the Integer or Unit. A fraction is expressed by two numbers, or Terms,' called the numerator and the denominator. The Denominator is written below the numerator,' and expresses the number of parts into which the integer is divided ; and the Numerator expresses the number of such parts denoted by the fraction. Thus, |, which is YQVidi four-fifths^ is a fraction, and sig-' nifies that a unit, a day, for instance, is divided into 5 equal parts, and that four of these parts are taken. ;■ If the numerator of a fraction be taken as an integer ,^ and be divided into as many equal parts as there are units in the denominator, the fraction may also be regarded as expressing one of these parts. Thus, for instance, if 4 days be divided into 5 equal parts, the fraction | expresses one of these parts, and is therefore the same as one-fifth of 4 days. Henco, f of a pound means either three times one-fourth of a pound, or one-fourth of three pounds, . A Proper Fraction is that whose numerator is less than its denominator. An Improper Fraction is that whose numerator is not less than its denominator. Thus, f and ^ are proper fractions ; and |, ^ and V are improper fractions. A number consisting of a whole number, with a fraction annexed, as 6f , is termed a Mixed Number. FRACTIONS. 177 A Simple Fraction expresses one or more of the equal parts into which a unit is divided, without reference to any other fraction. A Compound Fraction expresses one or more of the equal parts into which another fraction, or a mixed number, is divided. Thus, I is a simple fraction ; but J of ^ of -|4, and J of ^ of If, are compound fractions. Compound fractions haA^e the word of interposed between the simple fractions of which they are composed. A Complex Fraction is that which has a fraction either in its numerator or denominator, or in each of them. 51 7 54- ■ Thus, -7=r , ^ and ~, are complex fractions. "When fractions are considered in the abstract, without reference to any particular integer, we say briefly four- sevenths, instead of four-sevenths of one, the former ex- pression being understood to be equivalent to the latter. "We thus see that the two definitions of a fraction, given in the text, though they suggest to the mind ideas which are apparently different, are in effect the same. Agreea- ble to either, the lower number is very properly called the denominator, as it gives name (such as fifths, sixths, &c.,) to the parts into which the integer is divided ; and the upper number is called the numerator, as it numbers the parts expressed by the fraction, or is the number of the integers, which are divided by the denominator. According to these views, every fraction vould have its numerator less than its denominator, and would conse- quently be less than its unit, as its name imports. Such a view, however, is too limited to answer the gen- eral purposes of calculation ; and it is often necessary to consider, as fractions, quantities which equal or exceed the integer, or which indicate the quotients of numbers divided not Only b}' the greater, but also by equal or by smaller numbers, without the actual performance of the division. Quantities of the latter kind are therefore called improper fractions; and, for the sake of distinction, others, in refer- 8* 11 \ ' }■ 1, 1 'i' 1 ' 178 FRACTIONS. ,»i ill [O'^/f, ^y^;;|^ ence to them, are called proper fractions. It is evident, from either view of the nature of a fraction, that it is less or greater than a unit, accordingly as its numerator is less or greater than it ? denominator, and that it is equal to a unit, if its numerator and denominator be equal. In addition to the preceding, it may be proper to state, that fractions are of the same nature as the subordinate quotients in compound division. Thus, if 27cwt. be divided by 20, the quotient is 1 ton 7cwt., or Ij/o-ton, each of these expressing precisely the same quantity, since 1 cwt. is one twentieth, and, conse- quently, seven cwt. seven twentieths of a ton. From these principles, and those laid down in Art. 75, wc arrive at the following important conclusions : 159. If the terms of any fractmn he both multiplied^ or 'both divided, by the same number^ the resulting fraction is eguivale7it to the original o?i£. Thus, if the numerator and denominator of the fraction f be multiplied by 3, the fraction resulting will be ^, which is of the same value as f . Reason of the process. In the fraction f , the unit is divided into 7 equal parts, and 2 of these parts are taken ; in the fraction ^, the unit is divided into 21 equal parts, and 6 of these parts are taken. Now, there are 3 times as many parts taken in the sec- ond fraction as there arc in the first ; but 3 parts of the second fraction are only equal to 1 part in the first frac- tion. Therefore, the G parts taken in the second fraction equals he 2 parts taken in the first fraction : therefore, 160. Hence, it follows ,hat a whole number may be converted into a fraction witlx any donominator, b}^ multi- plying the numbei by the required denominator for the numerator of the fraction, nnd placing the required de- FRACTIONS. 179 nominator underneath ; for 6 = f , and to convert it into a fraction, with the denominator 5, or any other number, we have ^ — T — 1X6 — 6' » 161. Again ^ to mvltiply the numerator of a fraction by any numbers is the same in effect as dividing the denomi- nator by it, and conversely. For, if the numerator of the fraction •§ be multiplied by 4, the resulting fraction is ^^ ; and if the denominator be divided by 4, the resulting fraction is f . Now, the fraction ^^ signifies that unity is divided into 8 equal parts, and that 24 such parts are taken ; these are equivalent to 3 units. Also, f signifies that unity is divi- ded into 2 equal parts, and that 6 such parts are taken ; these are equivalent to 3 units. Hence, ^^ and J are equal. Again, if we divide the numerator of the fraction f by 2, the resulting fraction is f ; and if we multiply the de- nominator by 2, the resulting fraction is ^^, Now, % signifies that unity is divided into 8 equal parts, and that 3 of such parts are taken ; and j^ signifies that the unit is divided into 16 equal parts, and that 6 of such parts are taken. But each part in f is equal to 2 parts in •f^ ; therefore, f is of the same value as '^--^ or y^. The same operations can be performed on fractional as on integral quantities. , Hence, the doctrine of Fractions comprises the addition^ subtraction, multiplication and division of fractions. Before entering on these operations, it is proper to show how such quantities may be modified, without changing their A^alae, so as to fit them for the operations to be per- formed on them, and for the uses to which they may be applied. This will Gonsititute Reduction of Fractions, \ <\ f* 1 i I f 180 FRACTIONS. !•!!). 1^; tm Rbbttctxon op Vulgar Fractions. 162. When the numerator and denominator of a frac- tion are not prime to each other, they have a common fac- tor greater than unity (Art. 92, def. 4). If we divide each of them by this, there results a fraction equal to the former, but of which the terms, that is, the numerator and denominator, are less, or lower than those of the original fraction; and it may be considered the same in lower terms. When the numerator and denominator of a fraction are Prime to each other, that is, have no common factor greater than unity, it is clear that its terms cannot be made lower by division of this kind, and on this account the fraction is said to be in its Lowest Terms. 163. To Reduce a Fraction to its lowest terjns. Rule. Divide the numerator and denominator by their greatest commori. measure. Ex. Reduce f Jf | to its lowest terms. The greatest common measure of 6'.t65 and 7335 is 15 (Art. 97), and 3 ) 6465 ( 3 ) 7335 15 I 15 I { 5 ) 2155 { 5 ) 2445 431 489 Therefore, the fraction, in its lowest terms, = J-|^. Reason for the above process. If the numerator and denominator of a fraction be divi- ded by the same number, the value of the fraction is not changed (Art. 159) ; and the greatest number which will divide the numerator and denominator is their greatest common measure. 1 1 FRACTIONS. Mektal Exercises. 181 Reduce each of the following fractions to its lowest terms. 1. 1 5. U 9. fi 13. U 2. A 6. a 10. n 14. » 3. ii 7. u 11. it 15. U 4. if 8. is 12. if 16. u Exercises for the Slate. 1. Ui Ans. *f 4. tHu Ans. TUt 2. &2^ iC ^ 5. mn (( -m, 8. ^m7- i( ^wv 6. mm (C A 163. To Reduce Fractions to equivalent ones, with a common denominator. Rule. Find the least common multiple of the denomi- nators ; this will be the common denominator. Then divide the common multiple so found by the denominator of each fraction, and multiply each quotient so found into the numerator of the fraction which belongs to it for the. new numerator of that fraction. Ex. Reduce -^^ ^^, ^^, ^l into equivalent fractions, with a common denominator. Proceeding by the Rule given above, We find the least common multiple to be 528 (Art. 105). Therefore, the fractions become respectively, J^Xi-i = m (since W = 44), -A- X H = If i (since W = 33), HXU = fit (since W = 22), U X if = m (since V¥ = 16), Or, the fractions with a common denominator, are uh m^ uh fif. Reason for the above process. The least common multiple of the denominators pf the given fractions will evidently contain the denominator of l\ \ i 182 FRACTIONS. any one of the fractions an exact number of times. If both tlic numerator and dvi'nomiaator of that iraction be multiplied by that number, the value of the fraction will not be altered (Art. 159) ; and the donominntor will then be equal to the least common multiple of all the denomi- nators. If this be done with all the fractions, they will evidently be, in like manner, reduced to others of the same value, and having the least common multiple of all the de- nominators for the denominator '"^f each fraction. Note. — If the denominators have no common measure, we must then multiply each numerator into all the denomi- nators, except its own, for a new numerator for each frac- tion, and all the denominators together for the common denominator. 1. 2. 3. 4. 5. Exercises for the Slate. Reduce the following sets of questions to others having common denominators. Exercises. Answeb!^. f, f , I and T^, U^ f ^ M. U- h U and ^, ee, Uh M. h -A-. U, ih and ^, W, m^ m^ ^^^ ^^' U. hh H^ T^^ and i, ^^$1 iUh mt, ^SS-cr. UU- A? tIttj TTyo Ans. 1514|0-8- 6. »J§|* " 59i§g- 7. «^|r " 96,Vg- 165 To Reduce a Mixed Number to an Improjjer Fraction. Rule. Multiply the whole number by the denominator of the fractional part, and to the product add tlie numera- tor ; the sum will be the required numerator, below which write the given denominator. Ex. Reduce 7-^^ +o an improper fraction. Proceeding by the Rule given above, 11 2JL±-i = i{. Ans. Mental Exercises. 1. In 16 J how many halves ? 2. A lady purchased ^%^ of b. yard of silk for a dress, what number of yards did she buy ? I J ■I IMAGE EVALUATION TEST TARGET (MT-3) ./^..^^ K<^ e ^^<^ 4^ 4^ ^ 1.0 2.2 I.I lU lit 140 70 1.25 WMI* U 11.6 Photographic Sciences Corporation 23 WtST MAIN STRiiT VVIBSnR,N.Y. USM (7I6)«73-4S03 164 FRACnOKS. 3. Reduce 6, 7, 8, 9, 12 and 14 to improper ft'actions, the teacher naming the denominator of the fraction to which he wishes the number reduced. 4. If an acre of land produces 275bus. of corn, how many thirds of a bushel does it produce ? What will it come to at 14 cents each third? % Exercises for the Slate. 1. S^Vf Ans. ^11 5. mm Ans. ^iSfi 2. mu " ¥it¥ 6. mm " VAV 8. iQim U 7JJP 7. 427TTVr " Hf§l* 4. 106iif U 9^J|5 8. lOOf^^ " 7^Ji3 166. 7b Reduce a number^ or a fraction^ of any de- nomindtion^ to a fraction of another denomination. Rule. "Reduce the given number, or fraction, and also the number or fraction to the fraction of which it is to be reduced, to their respective equivalent values in terms of some one and the same denomination ; then the fraction of which the former is made the numerator, and the latter the denominator, will be the fraction required." £x. 1. Reduce 4s. 6d. to the fraction of £ 1. Proceeding by the above Rule, 4s. 6d. = 54 pence £ 1 =240 pence. Therefore, fraction iiaquired = ^j^, or ^. ■ Reason for the above process, £1, or unity, is here divided into 240 equal parts ; and 54 of such parts being taken, the part of unity, or £1, which they make up, is represented by ^(r, or ^. I actions, ction to rn, how t will it any de- 71. ion, and hich it i3 } in terms fraction he latter 'I bs ; and or £ly rnACTioxs. 185 Ex. 2. Reduce | of Icwt. to the fraction of 2 7 lbs, I of Icwt. = 112 times f of 1 lb. ~ 8 = Alii (Art. 87.) 27 lbs. = 27 lbs. Therefore, *^^^ ^ = ^^ X ^V = J? Ex. 3. Reduce £v^ to the fraction of a farthing. ^ of £1 = (-jL X 20 X 12 X 4) farthings, = W. = '^2 farthings, and 1 farthing = 1 farthing, therefore the fraction required. Exercises for the Slate. 1. Reduce 2. Reduce 3. Reduce 4. Reduce 5. Reduce r2s. 6d. to the fraction of l£. Ans. |, £ 18, 7. 6. to the fraction of 2 £ " V^. 6s. 7Jd. to the fraction of 7s. 9d. " |^. Is. 2d. to the fraction of 27s. " y|j. 2 ac. Iro. to the fraction of 9ac. 2ro. Ans. 7^. 6. Reduce 6ft. 3Jin. to the fraction of 13ft. ST^^in. Ans. J^Jf . 167. To find the value of a Fraction in the denomi" nations contained in the integer. Rule. Divide the numerator, considered as so many times the integer, by the denominator. Ex. 1. Required, the value of £f. ' Proceeding by the Rule given above, The numerator considered as so many times the unit, = £5, which, divided by 6, = 6)£5, 0. 0. £0, 16. 8. 4 i t 1S6 FRACTIONS. Ex. 2. Required, the value of J of £5, 13. 9. Here, the integer is £5, 13. 9.; then £5, 13. 9. the numerator expressing tliree times 3 this quantity, we multiply by 3, and di- vide by the denominator 8 ; tlie quotient, ^) 17, 1. 3. £2, 2. 7J, is the required valus. TZ Z IT* Exercises. 1. Required, the value of £ 1^. Ans. £0, 3. 9. 2. Required, the value of -f^ of £148, 5. Ans. £80, 17. 3^. 8. Required, the value of -^ of 48ac. Iro. 13 po. Ans. 4ac. Iro. 23 po. 4. Required, the value of 6JJbus. Ans. 6 bus. 3pk. l^^gal. Note. — The method of reducing Compound to Simple Fraotions will be given in Multiplication of Fractions, and that of reducing Complex Fractions to Simple ones» in Division of Fraotions. Addition of Vulgar Fractions. 168. Rule. Reduce the given fractions to others hav- ing a common denominator, if they be not such already. When they are in this state, add all the numerators, and below their sum write the common denominator ; the result will be the sum of the fractions, which, if it be an Improper fraction, must be reduced to a whole or mixed number. If some of the given quantities be mixed numbers, the fractional parts are to be added by the preceding [)art of the Rule ; and the whole numbers, with any intejral part that may be obtained by summing of the fractious, are to be added by simple addition. FK ACTIONS. 187 I. i, 13. 9. 3 r, 1. 3. }, 2. TJ, :o, 3. 9. 17. 3^3^. 18 po. •o. 23 po. . lyVgal- 'raotions will ling Complex )thers hav- already. :ators, and the result it be an or mixed ibers, the ^ig [)art of ^jrcU part LS, are to Ex. 1. Add J, I and -j^. Proceeding by tlic preceding Rule, First finding the least common multiple of the denomi- nators — 48 (Art. 105). Therefore, the fractions become * =36) J = 40 > 48, Common Denominator. t7^ = 21J Ex. 2. Add together 14J, 12f and ^. In this case, by reducing the given fractions to equiva- lent ones having a common denominator, we have 14| = 15 14f = 15 ) 121 =32Uo, C. D. ^= 6J 27^^ M = Hh Answer. Mental Exercises. What is the sum of 4f and 5f ? What is the sum of 91 and 8f ? What is the sum of 10| and 8fV. A farmer sold 7| tons of hay to one man, and 3^g- tons to another man ; how much did he sell altogether ? 5. A merchant bought ^ of a ship, and afterwards -^ more ; what part of the ship had he ? 1. 2. 3. 4. Exercises for the Slate. Find the sum of the following : 1. -f^, A and fa. 2. 4, T^TT, ^ and ^p. 3- h -^2^ t and ^^, 4. i, 6| and 4x. 5. lOOf , 64^ and 420f . 6. 26H, 174 J and 8 J. 7. 387i, 285 J, 394i and 1481 1. Ans. l-r\jV Ans. 2^. Ans. mi, Ans. 7^Y. Ans. 585|. Ans. 444f. Ans. 2548|^. • I 188 FRACTIONS. i u I, ll m li; i Subtraction of Vulgar Fractions. 1G9. Rule. Set the less quantity below the greater, and prepare them both as in Addition of Fractions. Then, if possible, subtract the lower numerator from the upper ; below the remainder write the common denominator ; and if there be whole numbers, find their diiference, as in Simple Subtraction. But if the lower numerator exceed the upper, subtract it from the common denominator ; to the remainder add the upper numerator ; write the common denominator be- neath the sum, and carry one to the whole number in the lower line. r Ex. 1. From 4 J take 2 J. Proceeding by the above Rule, of Z] 3 ( 8, Common Denominator. n s Ex.2. From 6^ take 3 J. Proceeding by the above Rale, 3i^ = 27 } 36. C. D. 2Ji li. Answer. Reason for the above process. In the second example, after reducing -rV ^°d J to frac- tions, with a common denominator, viz. : -f^ and gj, we find that we cannot take f J from -f^. We therefore add 1 t<> A (®^ 51 to ih)y ^^^ now take f J from J|, which leaves a remainder of ^^. Again, having added 1 to the upper number, wc must add 1 to the lower number, so that the difference between the two numbers may not be altered ; and adding 1 to 3, we obtain 4, which, taken from 6, leaves 2. Therefore, the difference or remainder is 2^. f It FRACTIONS. 189 Note.— It is more conTenient, in practice, to take the lower numerator from the denominator, as the Rule directs. The reason is the same. greater, . Then, upper ; 3r; and }, as in subtract der add ator be- r in the to frao- SJ, we fe add 1 leaves re must )etween 1 to3, Ebre, the Mental Exercises. 1. 2. 3. 4. What is the difference between |f and f f ? What is the difference between 2^ and 1 J ? What is the difference between } + ^ and J- + J ? Ellen bought 14J yards of merino for a dress, and 2tV J'ards of colored cambric for lining ; how many yards were there in both ? how many more in the dress than in the lining? Exercises for the Slate. Find the difference between— 1. T^jj and T^g. Ans. 2^ 2. 4 J and j^. Ans. ^ 3. i\ and Z^. Ans. -^2 4. 50t^ and 472^. Ans. 3^^ 5. IS-jS^andO^y^. Ans. 3|f J 6. ^ and ^, Ans. 1 J 7. loy^andljift. " 13fi| 8. 46^ and 15 ^. " 8l|i Multiplication of Vulgar Fractions. 170. EuLE. If any of the quantities be mixed num- bers, reduce them to improper fractions. Find the product of the numerators for the numerator of the required result, and the product of the denomina- tors for its denominator. Compound Fractions are reduced to Simple ones in the same manner. Ex. Multiply J by -ft . Proceeding by the above Rule, Reduced, = f^. Reason for the above Rule, If i be multiplied by 6, the result is V. But this result must be 17 times too large, since, instead of multiplying II' ir i 190 FRACTIONS. by 6, wo have only to multiply by -^^ which is 17 titnes smaller than G, or, in other words, is one-seventeenth part of G. Consequently, the product above, viz. : *q?, must be divided by 17, and V -h 17, = T^^ffe, or U- It has been shown that a fVaction is reduced to its lowest terms by dividing ita numerator and denominator by their greatest common measure, or, in other words, by the product of those factors which are common to both. Ilence, in all cases of multiplication of fractions, it will be well to split up the numerators and denominators as much as possible into the factors which compose them ; and then, after putting the several fractions under the form of one fraction, the sign of X? placed between each of the factors in the numerator and denominator, to cancel those factors which are common to both, before carrying into effect the final multiplication. Ex. 1. Multiply I by |. Product =: 5 >^ 4 cancelling the 5 in each, Ex. 2. Multiply J, J, |, -f^ and ii together. Product — 3* 2 ' 5 « 8 ' 11 jrroaucii— ^^a^exnx 12 Splitting up, r= 3«2x5x2« 4 xn ^Jr O r' 4x3s2x3xiix2>6 Answer. Cancelling rV- Note.— Instead of splitting up the factors, it is often more convenient to divide by some number wliich is common to one of the factors in each, (numerator and denominator), placing the quotient above the factor divided, in the numerator, and below the factor divided in the denomina- tor. Thus, 3 ^ 5 $ 11 i^a^fi^TT^f^* Cancelling 11 and 3 in each. 8 6 Next dividing 8 and 4 by 4. Again, dividing 2 and 6 by 2, and 2 and 12 by 2, we have 5 in the numerator and 3 and 6 in the denominator, which gives ~g = ^, 17 times onth part , must be ed to its lominator 3r words, 1 to both, it will be I as mucti em ; and e form of icli of the icel those ying into Iconyenietit }rs in each, the factor Idenomina- iach. nd 6 by Ir and 3 rRACTIONS. 191 Mental Exercises. 1. What is the product of 9 multiplied by | ? by J ? by 2. What is the product of 12 X 8? 12 X |? 12 X V 12 X V*? 3. A man who had found a purse containing 13 dollars, paid -fV of the money for advertising it ; how much did he pay for advertising it ? 4. At 20 dollars a ton, what is the cost of J of a ton of hay ? what cost | of a ton ? 5. An inch is -j^ of J of a yard ; what fraction of a yard is 8 inches ? 6. A grocer sells sperm oil for 1^^ dollars a gallon j what is the cost of f of a gallon ? Exercises for the Slate. 1- M X ff . Ans. ^^. 2. -f^ X ^!. Ans. 5^/^. 3- 7^ X i of J. Ans. %. 4. 12 by I of 5. Ans. 40. 5. |§of3f by l^of Hof |. Ans. 1. 6. ^of ^ by 5f of3. Ans. 5|. 7. m X m\ X m x f^ X l\kh Ans. T^j. 8. ^ X § X J X t X |. Ans. ^. 9. 5^2^ X 3i of -rf^ of 34 by yj^ of ^% of 1^ of 19. Ans. 2. Division op Vulgar Fractions. 171. Rule. Reduce mixed or whole numbers to improper fractions, and compound to simple ones, if any such be given. Then invert the divisor, i. e. take its numerator as a denominator, and its denominator as a nu- merator, and proceed as in multiplication. A Complex Fraction is reduced to a Simple one by di- viding its numerator by the denominator. Ex. 1. Divide -f^ by -j^. Proceeding by the Rule given above, 2 18 • 12 ~" 1$ ^ 7 ■" 21 u 192 FRACTIONS. r if Reason for the preceding Rule, If tW l^e divided by 7, the result is -j—^ or yj^, (Art. 161). This result is 12 times too small, or, in other words, is only one-twelfth part of the required quotient, since, instead of dividing by 7, we have to divide by -j^, which is only one-twelfth part of 7 ; and the quotient of ^^ divi- ded by -^21 nnist therefore be 12 times greater than if the divisor were 7. Hence, the above result, yf^, must be multiplied by 12, in order to give the true quotient. Therefore, the quotient = t^^, X 12 = i^, or J^. Ex. 2. Divide 4f by 6 J, re- ducing the fractions. H = V- . V -^ V. Ex. 3. Required the value of the complex fraction, H Here 5* = V. and 6f = V- Therefore, V -r "^j = VXA. 172. To Divide a Fraction by a tvhole number. Rule. Divide the numerator by the whole number when it can be done without a remainder ; but when this cannot be done, multiply the denominator by the whole number, and reduce, if necessary. Ex. "What is the quotient of ^ divided by 5. First Method. Second Method (Art. 161). ^^ -^ 5 = jfe. Ans. H-^^ = ^ = A' Mental Exercises. 1. What is the quotient of \^ -— 6 ? 2. What is the quotient of *^ >ri- ^ ? 3. If 10 yards of Venetian stair-carpeting cost 9^ dol- lars, what eoiBt 1 yard? wh^^t is the cost of 6 yards? w FRICTIONS. 198 4. If 6 yards of 3-ply carpeting cost 7| dollars, what is the cost of 1 yard? what is the cost of 21 yards? 5. In what two ways can we multiply a fraction by a whole number ? In what two ways can we divide :i fraction by a whole number ? 6. A lady purchased 7 yards of linen for G^ dollars; what would 8 yards cost at the same rate ? 1. 2. 8. 4. 6. 6. Exercises for the Slate. Divide if by f. . Divide |i by 3 J. Divide ^| by -fl. Ans. Divide 2^ by 4^. Ans. Divide 3£ of 3^ of i by 75. Ans. Divide 3^ of 5| of 3| by 9J of ^^ of 7^. Ans. 1^. Ans. ^l: A- ^ •iV. Ans. 3/7f. 7. A merchant wishes to lay out 657^ dollars for wheat, which is worth 1 ^ of a dollar a bushel ; how many can he buy ? Ans. 584| bus. 8. Paid 575f dollars for 96 J yards of cloth ; what was the cost per yard? Ans. $5ff^. Reduction of Decimal Fractions. 178. Cet'tain Vulgar Fractions can be expressed accu- rately as Decimals, Rule. Make the numerator of the given fraction the dividend ; place a dot or decimal point after it, and affix cyphers for decimals. Divide by the denominator, as in Division of Decimals, and the quotient will be the deci- mal, or the whole number and decimals required. £x. 1. Convert f into a decimal. 8)5.000 .625 h 194 FRACTIONS. There arc three places of decimals in the dividend, and none in the divisor. Therefore, there are three places in the quotient. In rcdncinj? any such fraction us t^^j or T^jJ^y to a decimal, wc may proceed in the Hjune way hh if we were ro(hicing J, taking care, however, in the result, to move the decimal point one place further to the left for each cypher cut off. Thus, I = .6, ^VTT- .0001'7'. 9. 24^^^. 24.0084'97133'. 10. 17^1^. 17.018'57142'. 6. mh 176. To convert Pure Circulating Decimals into their equivalent Vulgar Fractions, Rule. Make the period or repetend the numerator of the fraction, and for the denominator put down as many nines as there are figures in the period or repetend. Exs. Reduce the following pure circulating decimals, .3', .2'7', .8'57142', to their respective equivalent vulgar fractions. Proceeding by the Rule given above, .3' = t = h .27' = f J = T^r. .8'57142' = fliilf , 55 237 6 ^ 1587 3 TlTlT 7 * 16873' = f. The truth of tliese results will appear from the following considerations. Let the circulating decimal, .3333 .. .., be represented by the symbol x; then x = .3333 Therefore, 10 times a; = 10 times .3333 = 3.333 .... (Art. 118). FRACTIONS. 197 563636 .. 539689 .. .12666 .. 623623 .. :ed num- rSWBRS. 78977'2'. ,2'85714'. .00017'. t4'97133'. 8'57142'. into their lerator of as many ecimals, It vulgar Now, 10 times ar, diminished by 1 times x, will leave 9 times aj, and 3.333.... — .333.... = 3.333... t illowing isented in 3 or 9 times a? = 3 Therefore, 1 time x, that is, x or .333... := J =: ^. , Next, let the circulating decimal, .2727 , be repre- sented by X, Then x = .2727 Here, since there are two figures in each period, wo multiply by 100, and we have 100 times x = 100 times .2727.... = 27.2727 (Art. 118). Therefore, 100 times x, diminished by 1 time a?, will be equal to 27.27... — - .2727... or 99 times x = 27. Therefore, x or .2727.... = U = tt- In lilce manner, 999999 times x =i 857142, Note.— The object in each case is to multiply the recurring decimal by such a power of 10 as will bring out the period a whole number. Otherwise. To investigate this problem otherwise, let us recur to the origin of circulating decimals, or the manner of obtain- ing them. For example, ^= .1111, &c., or .1'. There- fore, tlie true value of .1111, &c., or .r, must be ^ ftom which it arose. For the same reason, .2222, &c., or .2', must be twice as much or f (Art. 160 and 170) ; .3333.... or .3' = f ; .4' = | ; .5' = f , &c. Again, ^ = .010101...., or .01' ; consequently, .0101...., or .01' = ^ ; .0202 , or .0'2' = ^ ; .0'3' = ^ ; .0'7' = Z^, &c. So, also, ^^7 = .001001001..., or .O'Ol'. Therefore, .001001...., or .001' = ^^ ; .0'02' = ^f y, &c. In like manner, f = .1'42857' (Art. 174) ; and .1'42857' 198 FRACTIONS. = ^yf f H » ^or, multiplying the numerator and denomina- tor of I by U2857, wc have UUU (Art. 159). So, f is twice as much as ^ ; ^ three times as much as | ; ^ four times as much, &c. • 177. To convert Mixed Circulating Decimals into their equivalent Vulgar Fractions. Rule. Subtract the figures which do not circulate from the figures taken to the end of the first period, as if both were whole numbers ; make the result the numerator, and write down as many nines as there are figures in the circu- lating part, followed by 'as many zeros as there are figures in the non-circulating part, for the denominator. Exs. Reduce the following mixed circulating decimals, .14', .0138', .24'18', to their respective equivalent vulgar fractions. Proceeding by the above Rule, .0138' = "'°~"' = ^y^ = »V, in its lowest terms. .24'i8' = ^-;-^ = im = im- The truth of these results will appear from the following considerations : Taking the last as an example, and separating the mixed decimal into its terminate and periodical parts, we have .24'18' = .2 + .04'18'. Now .2 = ^^j (Art. 110) ; and .04'18' = ^^ ; for, the pure period 4'18' = ||§, (Art. 176) ; and since the mixed period, .04'18', begins in hundredths* place, its value is evidently only ^ as much ; but Jit -T- 10, = ^8^ (Art. 172). Therefore, .24'18' = ^ _(_ ^y(y. Now, to perform the addition, we must reduce FRACTIONS. 199 them to a common denominator, when they become (Art. 163), ^^ + sWiT, (and since 999 = 1000 — 1), 2 ^ (1000— 1) J ^18 2 ^ 1000 — 2 I 4 1 g "^ 9 9 90 ~r ^9^17 9 9 90 "T FFFIT* • 2000 — 2 j 4j 8_ 2000-1-418 — 2 9990 I ^9^0 9990 ' 2^ ^8 - 2 24 16 XS.3.S. 9 9 9 9'FFO t^Q 5 • 178. Quantities composed of whole numbers, and either pure or mixed circulating decimals, may be reduced to their equivalent vulgar fractions in the same manner, taking care to use cyphers in the denominator for the non- circulating decimals only, Ex. 1. Reduce 17.6' to the form of a vulgar fraction. Proceeding by the preceding Rule, 17.6' = ^^^-^^ = If 9 = 5^3. Ex. 2. Find the vulgar fraction equivalent to the quan-' tities 17.63', and 1'7.6'. 17.63' = ll^^=-ll-i = W + ^%^' First Method. 1'7.6' = 17.61'76' = '^^'eVao '^^ = 'UW = 'iW' Second Method. Affixing a cypher for each whole number, and forming the denominator as before directed, we have 17.6' =. *^f^^. Exercises. For exercises to Articles 176, 177 and 178, verify the result of those in Article 175. % 200 FRACTIONS. 179. ^ 7b Reduce a Number or Fraction of any De^ nomindkor to the Decimal of another Denomination, Rule. Reduce the given number or fraction to a frac- tion of the proposed denomination (Art. 166), and then reduce this fraction to its equivalent decimal. Ex. 1. Reduce f of £ 1 to the decimal of 1 guinea. |of £lz=2jL20g _gg^ 1 guinea =: 21s. Therefore, the fraction required = ^. 21 r 71 8.0. (3|l.l'42857' .3'80952', Decimal required. Ex. 2. Reduce 13s. 6Jd. to the decimal of £1. 13s. e^d. = 6j.9d. £1 = 9fO- Therefore, the fraction ^J^ ilF = IM = .6760416'. We may work such an example as the above more expe- ditiously, by first reducing Jd. to the decimal of a penny, which decimal will be .25, and then reducing 6.25d. to the decimal of a shilling, by dividing by 12, which decimal will be .52083', and then reducing 13.52083's. to the deci- mal of £1, by dividing by 20, which process gives .6760416' as the required decimal of £1. The mode of operation may be shown thus*. 4 1.00 2,0 6.25 13.52083' .6760416'. I. a frac- 1 then i€L. expe- >enny, Ito the jcimal deci- , gives FRACTIONS. 201 £x. 3. Reduce 3 roods 11 po. to the decimal of an acre. 4,0 11.000 3.275 .81875, Decimal required. Exercises. Reduce 1. 6s. 4d. to the decimal of £1. Ans. .316'. 2. £3, 11. 9 J. to the decimal of £2, 10. Ans. 1.43625. 8. 2oz. 13 dwt. to the decimal of 1 lb. Ans.. 22083'. 4. 27^ gals, to the decimal of 1^ qts. Ans. 82.5. 5. 3 reams to the decimal of 19 sheets. Ans. 75.789. 6. 52- yds. to the decimal of 2 Fr. ells. Ans. 1.916'. 180, 2b Reduce a Decimal of any Denomination to its proper value. Rule. Multiply the decimal by the number of units connecting the next lower denomination with the given one, and point off for decimals as many figures in the pro- duct, beginning from the right hand, as there are figures in the given decimal. The figures on the left of the deci- mal point will represent the whole numbers in the next denomination. Proceed in the same way with the decimal part for that denomination, and so on. Ex. Find the value of .5375 of £ 1. Proceeding by the Rule given above, £.5375 20 10.7500 12 9.0000 Therefore, the value of .5375 of £ 1 = 10s. 9d. 9* ■II 13 202 FllACTIONS. W I With respect to the reason of the process, it is only ne- cessary to observe, that it is exactly the same as finding the vakie of £ tWAj by Art. 167, the pointing off of the decimals serving the purpose of dividing by the denomi- .ziator. i* Exercises. Required, the value of the 1. .0675 cwt. .0625 of a guinea. .875 of a league. 1.605 of £3, 2. 6. 1.005 of 15 guineas. A' foot, long measure. .063' of 100 guineas. 2. 3. 4. 5. 6. 7. 8. .2383' of a degree. following decimals : Ans. 7^^ lbs. Ans. Is. 3Jd. Ans. 2 m. 1100 yds. Ans. £5, 0. 3}. Ans. £15, 16. ej. .6q. Ans. 5^ inches. Ans. £6, 13. 0. Ans. 14'. 18". Addition and Subtraction of Circulating Decimals. 181. In arithmetical operations, where circulating dec- imals are concerned, and the result is only required to be true to a small number of decimal places, it will be suffi- cient to carry on the circulating part to two or three deci- mal places more than the number required, taking care that the last figure retained be increased by 1, if the suc- ceeding figure be 5, or greater than 5 ; but when it is required to find the true sum, difference, &c., the periods must be rendered similar and coterminous, or reduced to their equivalent vulgar fractions before being operated upon. 182. To make any number of Penodical Decimals similar and coterminous. Rule. Extend each of the circles as many places be- yond the longest finite part as is denoted by the least com- mon multiple of the number of places in the given circles. Ex. Make .31'6', .4'287' and .514273' similar and co- terminous. only ne- s finding ff of the denomi- li lbs. i. 3Jd. )Oyds. 0.3}. }. .6q. uches. 13. 0. V, 18". SCIMALS. ;ing dec- ed to be be suffl- ee deci- ing care the suc- m it is I periods iced to Iterated icimals jes be- st com- jircles. id co- FRACTIONS. 203 Here the longest finite part in these decimals is 514 in the last, and the least common multiple of 2, 4 and 3, the number of places in the circles, is 12. Therefore, extend each of the circles to 12 figures beyond the 4, when the same figures will again fall under one another, if repeated. Thus, .316' .4'287' .514273' = .31'6'i 161616161616 = .4'28 7'42874287428 = .514 2'73'273273273 1 7 2 183. When you add or subtract the right hand column, include the carriage that would arise by extending the circles still farther ; or, in other words, to find the carry- ing figure, go towards the left hand as many places as the periods have been extended, to make them similar and coterminous. Thus, if they are extended 6 places go back 6 ; if 10, go back 10, &c. Exercises. 1. Add 31.61'4', 3.416'23', 80.57'631' and 6.3'. Ans. 121.940'02861547416'. 2. Add 8.1576'4', .5'1387625', 8.13'042' and .4. Ans. 17.2019'4576208'. 3. From 21.63' take 3.8761. Ans. 17.7571. . . 4. From .8102 take .5376. Ans, .2726433904651. Multiplication of Circulating Decimals. 184. WJien the Multiplicand only is interminate. Rule. Place the multiplier as usual below the multipli- cand, and when you multiply the right hand figure, include the carriage that would arise by extending the repeating figures, and before adding up the partial products, extend the circles or repeating figures as in Addition. it i 1 I if i •1 204 FRACTIONS. Exercises. 1. Multiply 1.38 by 2.76. 2. Multiply 15.396 by 7.89. 8. Multiply 20.4387 by 6.73. Ans. 3.83. Ans. 121.48209. Ans. 137.552746. 185. When the Multiplier is interminate also. Rule. Reduce the multiplier to a vulgar fraction, and multiply by the numerator and divide by the denominator. Or, reduce both factors to vulgar fractions, and multiply. When there is an integer, the decimal portion only may be reduced to a vulgar fraction : then multiply, as in Art. 80. Exercises. 1. Multiply 4.108 by .387. Ans. 1.593750841. 2. Multiply 681.738 by 2.363'. Ans. 1611.17588. 8. Multiply 27.1824 by 3.285714. Ans. 89.3138710. Division op Circulating Decimals. 186. When the Dividend only is interminate. Rule. Divide as in finite decimals, but annex the repeaters or circulating figures instead of cyphers. 187. WJien the Divisor is interminate. Rule. Reduce the divisor to a vulgar fraction, then multiply by the denominator, and divide by the numerator. Exercises. , Verify the results of those in Artieles 184 and 185. 3.83. :8209. >2746. ion, and minator. aultiply. nly may } in Art. )0841. 7588. 8710. lex the I, then jrator. practice. 205 Miscellaneous Exercises on Articles 158-187. 1. A clerk spent 26f rloUars for a coat, 9} dollars for pants, 6J dollars for a vest, 54 dollars for a hat, and C^ dollars for a jiuir of boots ; how much did the suit cost him? Ans. $54|. 2. What number added to J of (^ + ^ — t\ + ^) makes 3^^ ? Ans. 25^. 3. If I pay away ^ of my money, then ^ of what re- mains, and then ^ of what still remains, what fraction of the whole will be left ? Ans. ^. 4. Find the sum of the greatest and least of these fractions : f , -j^, | and ^ ; the sum of the other two ; and the difference of these sums. Ans. ^|^. 5. A man has f of an estate ; he gives bis son one-half of his share. What portion of the estate has he then left ? Ans. A- 6. Multiply 3^ by 3^, and divide ?2i by Hi and 3 4 find the difference between the sum and diflference of these results. Ans. IJ. 7. Add together j^, ^, J and } ; subtract the sum from 2 ; multiply the result by § of f| of 8, and find what frac- tion this is of 99. Ans. jf^^. 8. Work each of the foregoing decimally, and thus verify the results. PRACTICE. 188. Practice is a compendious mode of finding the value of any number of articles by means of Aliquot parts, when the price of an unit of any denomination is given. Practice may be separated into two cases, Simple and Compound, 1 r. i 1) M llH 206 PRACTICE. I. — Simple Practice. In this case, the given number is expressed in the same denomination as the unit whose value is given ; as, for instance, 21 oz. at Is. 3d. per oz. ; or 4G0 ai'ticles at £1, 2. G. each. (1.) To find the price of a commodity^ when the price of each article as well as the quantity^ is of one denom^ {nation. Rule. Multiply the price of each article by the number of articles. Note. — All the exercises under this article should be worked by dollars and cents also, and thus verify the results. Ex. Required, the price of 368 cwt. (a) £2 per cwt. Here, the price of 368 368 cwt. ^ £2 per cwt. cwt., at £1 per cwt., is , evidently £368, and the £368 = price, (a) £1 per cwt. price of the same at £ 2 per 2 cwt., must obviously be twice as much. ^ 736, = price at £2 per cwt. Exercises. 1. 2. 344 fa) £2. Ans. £683. 421 (a) £3, " £1268. 3. 161 fa) £5. Ans. £805. 4. 121 rS) £2. " £242. (2.) When the i^rice is an aliquot part of a higher denomination. Rule. Take a like part of the number of articles, and the result will be the price in the higher denomination. Ex. 1. What is the value of 127 lbs. of tea, at 2s. 6d. per lb. 1271bs. (a) 2s. 6d. per lb. £ 127 = price, at £ 1 each. 2s. 6d. = 4 of £ 1. £ 15, 17. 6. = price at 28. 6d. each. le same as, for at £ 1 } he imcG denom- number )er cwt. I. £805. £242. higher ;s, and [on. 2s. 6d. ich. PRACTICK. 207 In this example, since 127 lbs., at X 1 each, would cost £ 127, it is evident that, at 2s. Gd., the same number wou'd cost one-eij^hth as much, 2a. Od. boiny; that part of Xl. We therefore divide £ 127 by 8, and the quotient, £ 15, 17. 6., is the price required. Ex. 2. Find the price of 1C3 articles, at 3d. each. Here, 103, at 1 shilling, would cost 1G3 shillings; and the price of the same, at 3(1., < /.« /^ o i i 11 -A 4.\ u r *.u 1C3 ^ 3d. each, would evidently be one-fourth of that, or £2, 0. 9. i63 = price, at Is. each. 3d. = J of Is. = 40, 9. = price at 3d. each. ould be rify the 1 £2,0. Exercises. 9. cwt. \ 1. 346 at 10s. Ans. £173,0. 0. 2. 327 at Gs. 8d. £109,0.0. 3Wt. 3. 271 at 3s. 4d. £ 45, 3. 4. 4. 141 at Gd. £ 3,10. 0. per cwt. 6. 146 at 2d. £ 1, 4. 4. 6. 189 at 4d. £ 8, 3. 0. (3.) To fijid the price of any number of articles^ at 2 shillings each. Rule. Double the last figure of the numVer for shil- lings, and take the number expressed by the preceding figures as pounds. Ex. What is the cost of 587 yards of muslin, at 2s. per yard ? 587 yds. (a) 2 shillings. 2 £58, 14. The Reason of this Rule is, that 2 shillings are one- tenth of a pound, and the work is merely a contracted form of dividing by ten. Thus, in the above example, by dividing by 10, we would have £58, with the fraction £-^(j, or, b}^ doubling both the terms (Art. 159), £JJ, or 14 shillings. I 208 riiACTICE. (4.) If the rate be an even number of ahillinga, KcLE. Multiply by half llio mimbcr of shlllinf^s, and in multiplying, double the last tigure for shillings ; the rust will bo pounds. Ex. Required, the price of 180 lbs. of indigo, at 8 shil- lings per lb. 18Glb8. at8s. 4 £ 74, b. 0. The operation of this rule is evidently an extension of the last. In the annexed example, the price is -^ or -^ of a pound ; and henco wo multiply by 4, and the doubling of the last figure of the product for shillings is equivalent to a division by 10. (5.) When the number of ahillinga ia odd. Rule. Find, by the last rule, the amount at one shil- ling less than the given rate, and for that shilling, take one-twentieth of the price at one pound, or add the aliquot part that it is of the even number of shillings in tho price. Ex. What is the cost of 787 pounds of tea, at 7 shil- lings per lb. ? 787 3 £236, 2. 0. = price, at 6s. per lb. Is. = £-Jij 39, 7. 0. =: price, at Is. per lb. £275, 9. 0. 7s. In this example, the price is first found at 6s. per pound, by multipljdng by 8, and doubling the last figure of tho product for shillings; then, for the remaining shilling, a twentieth part of £787, the price at 20s. per lb., is taken, and the sum of both results is £275, 9. 0., the price at 7 shillings. rs, and ^s ; the , 8 shil- ision of >r -iV of oubling livalent ae shil- g, take aliquot in tho 7 shil- bound, lof the ling, a ^aken, \q at 7 rU ACTICK. 209 (Q.)Wh€n the nutnber of articles is even and the price odd, huff the nnmhr man '*'' taken ^ muhiply l)}/ the whole number of shillingM and double the last figure for shillings, Ex. Required, the coit of 12G articles, at 11 shiliingi each ? 2)126 The preceding rules should be worked mentally as well as on the slate. They might have been given under the Article on Mental Calculation, but have been inserted here, instead, to render the present Article complete in itself. (7.) To find the price of any^ number of articles when the price is of more than one denomination. Rule. Multiply the number of articles by the integer of the price, and take parts of the integer for the lower denominations. The sum of the results thus obtained, will be the price required. 63 11 shillings. 1 ■ 1 £69,6.0. i| Exercises. i i 1. 397 at 2s. Ans . £39, 14. 0. p 2. 418 at 3s. u £62, 14. 0. p 3. 763 at 6s. ^<, £228, 18. 0. 4. 937 at 9s. (; £421, 13. 0. 5. 675 at 13. u £438, 15. 0. 6. 319 at 16. (( £255, 4. 0. ^ 7. 868 at 17s. i( £737, 16. 0. 8. 480 at 18s. (( £432. 0. 0. ' i 210 TRACTICE. Ex. 1. Find the value of 484 things, at £ 2, 16. 8. eack 484, at £2, 16. 8. £484, at £2, = price at £1 each. 10s. Od. = ^of£l 6s. 8d. = ^of£l £968, at £ 2, = price at £ 2 each. 242, 0. 0. ==. price, at 10s. each. 161, 6. 8. =: price, at 6s. 8d. each. £1371, 6. 8. = price, at £2, 16. 8. each. Ex. 2. Wliat cost 643 yards of linen, at 3s. 9d. per yd.? 643 yds. at 3s. 9d. per yd. £643, 0. 0. = cost of 643 yd. at £l per yd. 2s. 6d. = ^of £1 I 80, 7. 6. = costof 643yd.at2s.6d. " i ls.3d.= ^of 2s.6d.| 40, 3. 9. = cost of 643 yd. at ls.3d. '* . £120, 11. 3. = cost of 643 yd. at 3s.9d. " * SECOND METHOD. 643 yds. at 3s. 9d. per yard. 643s. o O value, at Is. per yard. 6d. = ^ of Is. 3d. = ^ of 6d. 1929s. Od. = value, at 3s. per yard. ^ 321s. 6d. = A^alue, at Os. 6d. per yard. s| 160s. 9d. = value, at Os. 3d. per yard. | 2^0)241,1. 3d. 3s. 9d. £ 120, 11.3. Ans. as before. Ex. 3. Find cost of 1263 yards ribbon, at 6-2-d. per yd. 1263 3^ds. (a) 6f^d. per yd. 1263s. Od. price, at Is. Od. per j^d. 6d. = ^ of Is. M. =z ,-V of 6d. Jd. = } of J-d. 631s. 6d. = price, at Os. 6d. per yd. 52s. 7^d. z= price, at Os. O^d. per yd. 26s. 3Jd. = price, at Os. O^d. per yd. 2^0 ) 71vOs. 5i-d. = price, at Os. 6Jd. per yd. £35, 10. 6^. Answer. PRACTICE. 211 i. eacLu , each. . each. . each, er yd. ? per yd. .6d. " i .3d. '' , .9d. u yard. ^ yard. J yard. ; ►er yd. r yd. jryd. jr yd. fer yd. SECOND METHOD. 1263 yds. at 6|d. per yard. 6d. id. ^ of Is. } of 6d. 631s. 6d. z= cost, at 6d. per j^ard. 78s. ll^d. = cost, at fd. per yard. 2,0 ) 71 ^Os. b^d. z=z cost, at 6|d. per yard. £35, 10. of Ans. Note. — The student must use his own judgment in selecting the most convenient * aliquot * parts, tfking care that the sum of those taken make up the given price of the unit. In taking aliquot parts, it sometimes shortens the work to take the same part twice, as the result may thus be copied without working for it again. Thus, 18s. 6d. may be divided into 10s., 4s., 4s. and Cd. Some- times the price at a smaller rate may be found, and from it the price of a greater may be obtained by multiplication. Thus, 16s. 4d. maybe divided into 2s., 14s. and 4d.; the price at 14s. being 7 times the price at 2s. Exercises. Find the value of 1. 454 things, (a) 2s. 9d. each. Ans. £62, 8. 6.] 2. 80 things, (a) 4s. 4id. each. Ans. £ 17, 8. 4. 3. 898 things, fa) 18s. 7fd. each. Ans. £837, 3. 11J-. 4. 4681 things, rd) 82-d. each. Ans. £ 170, 13. 2f .5. 7382 things, ^ £3, 15. 4^ each. Ans. £27820, 18. 3. 6. 43.35 things, (a) 8s. ll|d. Ans. £ 19, 7. 5| + ^^q. i 7. 147.625 things, at 19s. 7id. Ans. £ 144, 14. Of. + ^q. The price may often be determined very easilj^, by find- ing the amount at a rate higher than the given rate, and deducting from the amount the price at the difference be- tween the given rate and the assumed one. Ex. What cost 189 tons, at 17s. 6d. per ton? 189, at 17s. 6d. 189, 0. 0. =: price, at £ 1 per ton. 2s. Cd. = ^ of £ 1. 23, 12. 6. = price, at £ 0, 2. 6. per ton. £ 1 65, 7.6. = price, at 0^ 1 7., 6. per ton. i 212 1. 2. 3. 4. rUACTICE. Exercises. 358 articles, at 13s. 4d. 699 articles, at Os. lO^d. 276 articles, at £2, 15. 0. 721 articles, at £0,19. 0. Ans. £ 238. 13. 4. Ans. £26. 4. 1^. Alls. £ 756, 5. 0. Ans. £684, 19. 0. II. — Compound Practice. In this case, the given number is not wholly expressed in the same denomination as the npit, whose value is given ; as, for instance, 79g lbs. at 2s. 9d. per lb., 126 cwt. 3 qrs. 14 lbs. at £ 2, 9. 6. per cwt. (8.) When the quantity is not expressed by a whole num" ber of one denxmiination. Rule. Compute the price of the integral part by some ;of the methods already given. Then find the price of the fractional parts, or lower denominations, from the giyen ' rate, by means of aliquot parts, or otherwise. The sum of all will be the whole price required. Or, (9.) Find the price of the entire given quantity at £ 1 ' or $ 1 for each unit of the integral part, valuing the sub- ordinate parts at the same rate. Then the operation will proceed in the same manner already explained, without .farther work for the subordinate parts. Ex. Find the cost of 167^ cwt. at £2, 5. 6 per cwt. 167^ cwt. at £2. 5. 6. per cwt. 167, 0.0. = price of 167, at £1 per cwt. 2 5s = I of£l. 6d. = TfJj of 5s. J = ^ of 1 cwt. (.-. ^of £2, 5. 6.) i=^oft. ^ = i-off. 334, 0. 0. 41,15. 0. 4, 3. 6. price of 1 67, at £ 2 per cwt. price of 167, at 5s. per cwt. price of 167, at 6d. per cwt. 1, 2. 9. = price of | cwt. at £2, 5. 6. price of I cwt. at £2, 5. 6. priceofjcwtat £2, 5. 6. 0,11. 4^. 0. 5. 8|. £381,18. 3f . J. 4. 5. 0. }. 0. ressed given; 3qrs. e num' Y some of the I giyen le sum at £1 e sub- >n will ithout rt. jr cwt. kr cwt. [r cwt. \v cwt. 5. 6. 5.6. 5.6. PRACTICE. SBCOlfD METHOD. 5d. = 1 of £1 6s. = ^ of 5s. 167^^1 cwt. at 2 1= J of 1 cwt. 334, 0. 0. 1= J of I " 41,15.0. i=^off " 4. 3. 6. 213 £2, 5. 6. per cwt. 1, 2. 9. 0,11.4^. 0, 5. ^. £1,19. 9f. £379,18. 6. price 167 cwt. at £2, 5. 6. 1,19. 9}. price | cwt. at 2, 5. 6. £381,18. 3}. price 167|^ cwt. £2, 5.6. u THIBD METHOD. 167| cwt. at £2, 5. 6. per cwt. 167, 17. 6. = value 167|^ cwt. at £1 per cwt* 2 5s.= Jof£l 6d.=TJ<,bf5s. 335, 15. 0. (( (( (( £2 (( 41, 19. 4*. ^^ (; (( (( 5s. (( 4, 3. 18. Hi. 3J. — (( (( (C (( 6d. (( 381, £2, 5. 6. (( BY DOLLA R8 . 4ND CENTS. 167J cwt. at $9.10 per cwt. 167.875 cwt. at $9.10 per cwt. 9.10 1678750 1510875 $1527.66250 = £381,18. 3J. y 214 PRACTICE. SECOND METHOD. I,:!)' I 167J cwt. at $9.10 per cwt. 167^ 63.70 546.0 910. I = J of 1 cwt. at 2 — 4.55 = ^ of I cwt. at 2.275 I = Jeff cwt. at 1.1375 $1527.6625=£381, 18. 3J. The examples under this and the following rules in this article should also be worked by Art. 133. Exercises. 1. 165J at £2, 5. 6. 2. 7538jat£0,2.4. 3. 164^ at $9.10. 4. 239 lbs. 12 oz. at $2.16§ 5. 257^J at £2, 9. 4. Ans. £377, 7. 3}. Ans. £879,10. 5. Ans. $1500.,^6i. Ans. $519.45|. Ans. £636,3.10J^. (10.) In calculating the price of Jmndred, quarters, and pounds, the price, at £1 per cwt. will be found by multi- l^lying the pounds by 2|, and considering the product as pence ; and by multiplying the quarters by 5, and consid- ering the product as shillings. Ex. Required the price of 319 cwt. 3 qr. 16 lbs. at £2, 12. 6. per cwt. 319 cwt. 3. 16. at £2, 12. 6. per cwt. 5. n 319, 17. lOf = price at £1 per cwt. 2 10&.>siof £1 2s. £d. v^ J of lOs. £639, 15. 8^ = price at £2 per cwt. 159, 18. 11| = " 10s. " 39, 19. %{l=L " 2s. 6d. " £839, 14. 4^. = " £2, 12. 6. " in this ^3f. 0.5. Ml. .45^. . 10^. •s, and multi- ict as onsid- at£2, wt. • cwt. cwt. u (i PRACTICK. Reason for ihe preceding proceu. 215 It is evident that as £1 per cwt. each quarter would ct>8t| 5 shillings, and each pound the twenty-eighth ef this, or; 2f pence, Therefore, multiplying the pounds by 2f , we say, one-' seventh of 16 is 2f, and twice 16 are 32, and 2f are 34f pence, or 2s. lOfd., which is the value of the pounds at £l per cwt. We then set down lOf d. and carry 2 to 5 times 3, or 15 shillings, the price of 3 quarters ; and we thus find the price of 8 qr. 16 lbs., at £1 per cwt., to be 17s. lO-f pence ; to which £319, the price of 319 cwt. at the same rate is prefixed. The rest of the operation pro-t' ceeds in the usual manner. The above might have been calculated by Rule 7. Thus, 319 cwt. 3 qr. 16 lbs. at £2, 12. 6. £839, 14. 4J. 7^ $ame bjf Dollars and CenU, 319 cwt. 3 qr. 16 lbs at $10.50 per. cwt. In some cases it will be found more convenient to mul- tiply the price by the quantity. Thus, $10.50 319. 3. 16. ' 94.50 105.0 3150. 2 qr. — ^ of 1 cwt. . 5.25 1 qr. = ^of 2qr. 2.625 14 lb. = i of 1 qr. 1.3125 2 lb. = f of 14 lb. .1875 • $3358i6750 s £839, U. 4^. m £638, 0.0. = price 319 cwt. at £2 per cwt; 10s.=iof £2 159, 10. 0. — " " at £0,10.0. '* 2s.6d— ^oflOs 39, 17. 6. = " " at 0. 2. 6. ". 2 qr,— j^^ 1 cwt. 1, 6.3.== " 2qr.at£2,12.6." | 1 qr.=:^ 2 qr. 0,13.1^.= " 1 qr. " " «} 14 lb.=^ 1 qr. 0, 6. 6}.= " 14 lb. " " " 21b.=fl41b. 0, 0.11J.= " 2 1b. " " " 2X0 PKACTICE. *' imm* It will be found in nearly all the examples in practice, that dividing by the aliquot parts &c., gives origin to vul- gar fractions, which, of necessity, must be added to find the true resuts. This may be obviated by proceeding on the same principle, but converting the fractions into deci- mals, which will be found to be sufficient to carry to two places each. Thus the work may be as follows, 819 cwt. 8 qrs. 16 lbs. at £2, 12. 6. per cwt. 5 2f 10s.= j of £X 2s. 6d.=i 10s. £319, 17. 10.29 2 639, 15. 8.58 159, 18. 11.14 89, 19. 8.78 £839, 14. 4.50 or £839, 14. 4}. Here, in taking | of 16, we have 2 to carry, and 2 re- maining, then conceiving a cipher annexed to this remain- der, and dividing 20 by 7, we set down the quotient 2, and conceiving a cypher annexed to the remainder 6, we have contained most nearly ^ times in 60. We then proceed as before, and find the price at £1 per cwt. £319, 17. 10.29 nearly. After this the work proceeds as before, only that in each line the pence and the decimal are multiplied and divided as if they were a single whole number, the point being re- tained. Thus in finding the price at 2s. 6d., after ha^ng found £39, 19, we have 2s. lid., or 35d. remaining; we then divide 35.14 by 4, as if it were all one number, and find for the quotient 878 or with the point, 8.78. Note. — ^In working by this method it is evident that each penny is supposed to be divided into 100 equal parts. Therefore 25d.=i, .50d= J , .75d.= J. In valuing the d:n mal found in the answer, Ihe pupil should consider to which of these it is nearest, and vaikie it accordin^y. practice, In to viil- d to find eding on nto deci- ■y to two BWt. bd 2 re- remain- i2, and ^e have proceed . 10^9 n each iivided ingre- iaving g ; Wfe r, and t that parts. iffhich — nUCTICB. EXEBCISES. Cwt qn. lbs. 1. 134. 1. 21. at £0,18. 4. 2. 812. 3. 7. at £6, 12. 8. a. 786. 2. 8. at $3.75 4. 179. 3. 25. at $14.25 217 Ans. £123, 4. 8^. Ans. £5391, 13. ij. Ans. $2949.64^. Ans. $2564.6l|. (11.) In computing the price of tons, hundreds , and quar* tersy at £1 per ton^ take the tons and hundreds as pounds and shillings, and multiply the quarters by 3 for pence. Ex. Find the price of 163 tons 2 cwt. 3 qr. at £!^, 16. 3. per ton 163 tons 2 cwt. 3 qr. at £2, 16. 3. per ton. 3 £163, 2. 9. = price at £1 per ton. 10s. z= J of £1. 5s = j- of 10s. Is. 3d. = j^ of 5s. 326, 5. 6. = 81, 11. 4.50 = 40,15. 8.25= «' 10, 3. 11.06 = •« £458, 16. 5}v81 £2 10, 1. 8, £2, 16.3, By DoUars and Cents. 163 tons 2 cwt. 3 drs. at $11.25 per tan $11.25 163 tons 2 cwt. 3 qr. 33.75 675.0 2 cwt. =: -A of 1 ton 2 qr9. = J of 2 6wt. 1 qr. = ^ of 2 qr. 1125. 1.125 .28125 .140625 1835.296875=; £458. 16. 52 J 10 \.\ H m 218 PRACTICE. Exercises. * 1. 175 tona 18cwt. 1 qr. at £38, 13.0. perton £6799. 0.4J 2. 219 tons 16 cwt. 3 qrs. at $45.50 «* $10002.60^^ 3. 2 tons 15 cwt. 2 qrs. at $7. 84^ " $21.77 nearly 4. 163 tons 2 cwt. 1 qr. at £2, 19. 6. " £485, 5. 2^ (12.) In computing the price of acres, roods and perches at £1 per acre, multiply the perches by 1 j- for pence, and the roods by 5 for shillings. • The reeuon for this rule and the following depend mi the ,same principle as Bide 10. • £x. Required the price of 127 acres 3 ro. 14 po. at £2, 11. $. per acre. 127 acres 3ro. 14 po. at £2, 11. 6. per acre. 5 1^ £127, 16. 9. = value at £1 per acre. 2 10s.: ls.= 6d.= :Jof£l. ,\r 10s. : j^of Is. £255, 13. 6. = yalue at £2 per acre 63, 18. 4.50 = " " IQ. " 6, 7. 10.05= " " 1. " 3, 3. 11.02= « '^ 6." £2,11.6. cr £329, 3. 7i^57= By Dollars and Cents, 127 acres 3 ro. 14 po. at $10.30 per acre. $10.30 127 1 7210 2060 1030 1 $1308.10=price of 127 ac. at 10.30 perac. 2 ro.r=j- of 1 ac. 5.15=price of 2 ro. at 10.30 per ac. lro.=Jof2ro. 2.575= " I ro. 10 po.=i 1 ro. .64375= *' 10 po. 4po,=^x\, 1 ro.. .2575= « 4 po. (( it ((. $1316.72625= " 127 ac. 3 ro. 14 po. or £329, d 7^. 99. 0. 4 J 02.60t4 7 nearly 5, 6. 2i perches ice, and ' ^dti the . at £2, ir acre. r acre. Br acre e." 6. acre. rac. ac. practice. Exercises. 219 Ao. ro. po. 1. 23. 3. 6. at £2, 12. 6. Ana. £62, S.GJJ^ 2. 225. 1.19. at £0,13. 2^ per ro. Ana. £595, 6. 11 2+2^ q. 3. 45. 2. 35. at£0, 16. 6. Ans. £37, 14. 4^. 4. 311. 2. 26. at $4.55 Ans. $1418.06| ri3.) In computation in Troy weight at £1, per o«., take the ounces as pounds, the penny-weights as shillings, and half the grains as pence. Ex. Required the price of 10 oz. 3 dwt. 14 grs. Troy weight at 17s. 6d. per ounce. 16 oz. 3 dwt. 14 grs. at £0, 17. 6. per oz. i £16, 3. 7, =:price at 2«. 6d.= J £1 de. 2, 0. 6 J. =price at £1, per oz 2. 6d. " (( 17.6. £14, 3. H. = By Dollars and Cents, 16 oz. 3 dwt. 14 grs, at $3.50 per oz. $3.50 16 $56.00 z^price of 16 oz. at $3.50 per oz. 2dwt.=TV loz. .35 = " 2dwt. " ldwt.=^2dwt. .175= « Idwt. 12grs.=:^ 1 dwt. .0875= " 12grs. 2gr8.=::^ 12 grs. .014583'=" 2grg. u $56.627083'= " =£14, 3. If. Exercises. 16 oz. 3 dwt. 14 grs. 1. 15 oz. 6 dwt. 17 grs. at £0, 5. 10. Ans. £4, 9. 5^ 2. 93 oz. 7 dwt. 15 grs. at £0, 10. 4. Ans. £48, 4. 11^ 3. 263 oz. 16 dwt. 9 grs. at £0. 11. 3. Ans. £148. 7. 11 J 4. 3 lbs, 11 oz. 12 dwt. at $1.16^ per oz. Ans. $55.45^ ! I) 220 PRACTKI. (14.) In computing the price of yards ^ quarteray and j»| nails at £\ per yard^ tako cnch quarter at 5 shillings, and each nail at Is. 8d. £z. Find the cost of 188 yds. 8 qrs. 2 nls. at 15. 6. per yd. 183 yds. 8 qrs. 2nls. at jCO, 15. 6d per yd. Is. 3di lOsrr^of £1 6s = i of 10s. 6d. = tV of 58. CI88. 17. 6.= 91, 18. 9.= 45, 19. 4.5( 4, 11. 11. 2S price at £1 per yard, 10. 4( 6d. 15,6. £142, 10 0iyJ6 = By Dollars and Cents, 188 yds. 8 qrs. 2 nls. at $3,10 ptr yd. 8.1 183 549 $567.80=oost of 183 yds. at $340 per yd. 2 qrs.= J of 1 yd. 1.55= " 2qr8. " lqr.=Jof2qrs. 0.775= " 1 qr. «* 2 nl5=J of 1 qr. .3875= " 2 nls. « $570.0125= £142, 10. OJ. 183 yds. Sqrs. 2n. Miscellaneous Exercise? ok the Foregoing Rules. Ans. £80, 12. 6. Ans. £9, 15. 0. Ans. £17, 8. 4. Ans. $249.70. 1. 645 things at £0, 2. 6. each. 2. 52 things at £0, 3. 9. each. 3. 80 things at £0, 4. 4^ each. 4. 454 things at $0.55 each. 5. 51143 tMngs at $19.55^ each. Ans. $100005d.74^V* 6. 4013{ things at £2. 16. 6^ each. Ans. £11342, 13. 5^ PRACTICK. 221 rSf and igs, and per yd. per yd. ryard. d.« wyd. eryd. ES. 6. 0. 4. 0. - 7. Find the value of 5 cwt. 2 qr. 14 lis. at £2 5. C. per cwt. Ans. 12, 15. 1J|. 8. Find the value of 7 cwt. 1 qr. lo^^ lbs. at 2£, 0. 7. per cwt. Ans. £14, 19, lO^/i^. 9. Find the value of 9 yds. 2 ft. 10 in. at Si. 12] per 3'ard. Ans. $11.18}. 10. Find the value of 317 gal. 8 pts. at $2.10 per gal. Ans. $666.48}. 11. What is the cost of d$ qr. 6 bus. 3 pk. at £1, 18. lOi per quarter? Ans. £75, 10. 0;J.+t7^. First Principles. 189. Examples which are usually classed under partic- ular Rulef, such as the Rule of Proportion, &c., can nev- ertheless be readily solved independently by means of the foregoing principles which are therefore called First Prin- ciples. To work a question by first principles is to employ only the elementary rules. Multiplication, Division, Reduction, &c., and thus show that no new rule is necessary to solve questions in Proportion, <&c., whatever expediencies such rules may in some cases possess. The following examples which are worked out, are in- tended to exemplify various methods of reasoning ; the number of such examples must in this place be very limit- ed, and therefore the student is strongly recommended to apply to all questions which are hereafter given under par- ticular Rules, an independent method of solution, as well as the one denoted by the Rule to which they are resiicc- tively affixed. Ex. 1. If 6 men can do a piece of work in 10 days, how many men would do the same in 15 days? 6 men can perform the whole work in 10 days, .*. 60 men can do it in one day. And 60 men performing it in 1 day, it will take ^ as many to do it iu 15 days. .•. f^ r= 4, the number required. I I mi 222 PRACTICE. . Ex. 2. How many yards of cloth .at Tis. Od. per yard, should be given for ID lbs of tea, 3s. 8d. per lb? Change into the equivalent question : IIow many yards at 66d. should be given for 19 yards at 44d. ? 44d. for each yard In 19 yards amounts to Id. for each yard in 19X44, or to 66d. for each yard in ^^* ytls. = ^^ = 12J yards. Ex. 8. Find the interest of £182, 10. for 20 days at 6 per cent. £100 will yield for interest £6 in 865 days, .-. £1 will give £j^ in 1 day and in 20 days, 20 times as much. ^ '•■ -^oV = ^-nfe: =lnterest of £1 for 20 doys, and £182. 10. will give 182^ times as much, . ...^iiLL2^±_2 = jeo, 12. 0. Ex. 4. After taking from my purse ^ of my money, I find that § of what is there left amounts to 7s. 6d. ; what money have I in my purse at first ? Let Unity or 1 , denote the sum in the purse at first. After taking away i, £ remains. Now by the question J of J of. unity, or § of J of the sum in the purse at first = 7s. 6d. or ^ of the sum in the purse at first = 7s. Gd. .*. sum in the purse at first = 15s. Ex. 5. If a person, travelling 13J hours a day, per- form a journey in 27f daj'-s, in what length of time will he perform the same if he travel lOf hours a day ? If he travel 13 J hours a day, he does the journey in 27f days. If he travel 1 hour a day, he does the journey in (27fXl3J.) If he travel lOf hours a day, he does the journey in <£!l_^_lit:]which worked out gives 86f §ii days. lOf, RATIO. m r yard, r yards '3 at 6 iey, I what first, stion first >er- he 27f in in Ex. 6. Gunpowder is composed of nitre 15 parts, charcoal 3 pai'ts, and sulphur 2 parts ; And liow much of each is required for 16 cwt. The whole number of parts = (15-(-3+2, = 20 of ev- ery 20 parts. ,]^ or > is nitre, ^ften denot- ed by placing a colon between them. Thus the ratio of 7 to 13 is denoted by 7 : 13. As we have shown that the ratio of one number to another may be expressed by the fraction in which the former is the nu- merator and the latter the denominator, we see that 7:13 The two numbers which form a ratio are called its terms; the first number, or the number compared, being called the first term, or The Antecedent, and the second number or that with which the former is compared, the second term, or The Consequent, of the ratio. 194. If the two numbers to be compared together be concrete, they must be of the same kind. We cannot compare together 7 days and 13 miles in respect of magnitude ; but we can compare 7 days and 13 days ; and it is clear that 7 days will have the same rela- tion to 13 days in respect of magnitude, which the num- ber 7 has to the number 13, so thht the ratio of 7 days to 13 days, will be the same as the ratio of the abstract num- ber 7 to the abstract number 13, and may be expressed by the fraction -fr^. If however the concrete, though of the same kind, be not in the same denomination of that kind, it will be convenient to reduce them to one and the same denomination, in order to find the ratio. Thus, if one of the numbers be 7 day:? and the other 13 hours, the ratio of the former to the latter will not be that of 7 to 13, but that of 7 days to 13 hours, that is 168 hours to 13 hours, Which will clearly be the same as that of the abstract num- RATIO. 225 t ber 168 to the abstract number 13, aiicl so will be express- ed not by -/tj, but by ^r^. We see, then, that 7 days : 13 hours is the same as 1G8 : 13, and that each is = ^-f^. Thus it is plain that when the numbers are concrete, wo may reduce them to one and the same denomination, and then in considering their ratio, treat them as abstract uum])ers. 195. A Direct ratio is that wliich arises from the divid- ing the antecedent by the consequent. (Art. 192.) 196. An inverse or reciprocal ratio, is tlie ratio of the reciprocals of two numbers (Art. 92, Def. 10). Thus the direct ratio of the numbers 9 and 3, is 9 to 3, or § : the reciprocal ratio is ^ : ^ or ^ : ^- = | (Art. 171), that is, the consequent 3, is divided by the antecedent 9. An inverse or reciprocal ratio is expressed by inverting the fraction which expresses the direct ratio; or when the notation is by points, by inverting the order of the terms. Thus 8 is 4 invertedly, as 4 to 8. 197. A simple ratio is a ratio which has hut one ante- cedent and one consequent, and may be either direct or in- verse ; as 9 : 3, or I : -J. 198. A compound ratio is the ratio of the Products of the corresponding terms of two or more simple ratios. Thus, The simple ratio of 9 ; 3 is 3 And simple ratio of 8 : 4 is 2 The ratio compounded of these is, 72 : 12 = 6. From the principles of fractions already established, we may therefore deduce the following truths respecting rar- tios. 199. To multiply or divide both the antecedent and con- sequent of a raiio by the same number does not alter the ratio: for, multiplying or dividing both the numerator aiid denominator of a fraction by the same number does not alter its value. (Art. 159). 10* U 226 PROPORTION. Thus the ratio of 12 : 4 is 3 The ratio of 12 X 2 : 4 X 2 is 3 And the ratio of 12 -f- 2 : 4 -r 2 is 3 200. If to or from the terms of a ratio^ two other num" hers having the same ratio be added or subtracted, the sumf or remainders will also have the same ratio. Thus the ratiq of 12 : 3 is the same as that of 20 : 5. And the ratio of the sum of the antecedents 12 -f- 20 to the sum of the conse* quents 3 + 5, is the same. That is, 20 + 12 or 3 + 5, So also the ratio of the difference of the antecedents^ to the difference of the consequents, is the same. That is 20 — 12 : 5 — 3 is the same as 12 : 3. for 20 — 12 5-3 f = 4. PROPORTION. 201. Proportion is the equality of two ratios ; so that when the ratio of one number to a second is equal to the ratio of a third number to a fourth, proportion is said to exist among the numbers, and the numbers are called pro- portionals. Thus, the ratio of 8 to 9 is equal to that of 24 to 27, for the former ratio is f , and the latter ^^, which is also equal to f . The ratios being equal, proportion ex- ists among the numbers 8, 9, 24, 27 ; and those numbers are proportionals. Note. — The expression equality of ratios is not strictly accurate. The Katio of 3 to 9, viz. , 1-3, is not properly said to be equal to that of 5 to 15, viz., 1.3', the two are identical. It would be more accurate to say Proportion is the equivalence of two expressions denoting the same ratio. 202. When proportion exists among four numbers, that is, when the ratio of the first to the second is equal to that of the third to the fourth, this j^roportion or equality is often denoted by writing down the two ratios in the man-j ner mentioned in (Art. 193) in one line, and placing a ; ' jiat iat is m-i a PROPORTION. 227 double colon {^i i) between them. Thus, the existence of proportion among the numbers 6, 8. 24, 32, is indicated as follows: 6:8: : 24 : 32, wliich is commonly read, ''Six are to eight as twenty-four is to thirty-two," or, " as six to eight, so is twenty-four to thirtj^-two." * 203. In order to form a proportion four numbers are required. It may indeed happen that the second and thu'd are the same, in whicli particular case it might be said that only three numbers are required. Thus, 9 : 6 : : G : : 4 ; but even in such a case it is better to consider the second and third as distinct numbers, and to regard the proportion as consisting of foiu: numbers, of which indeed two are equal. The four numbers required to form a proportion are called its Terms 204. The ^rs^ and last terms are called th^ extremes; the other two, the means. 205. Direct proportion is an equality between two direct ratios. Thus, 12 : 4 : : 9 : 3 is a direct proportion. 206. Inverse or reciprocal proportion is an equality be- tween a direct and a reciprocal ratio. Thus, 8:4: : ^ : | ; or 8 is to 4 reciprocally, as 3 is to 6. 207. If four numbers are proportional, the product of the extremes is equal to the product of the means. 208. It has been stated that proportion is the equality of two ratios, and we have explained that the two numbers constituting a ratio must either be both abstract, or (if concrete) both of the same kind. In a proportion, if one of the ratios be formed by two abstract numbers, the other maj^ arise from two concrete numbers. For it has been explained (Art. 194) that if a ratio consist of two concrete numbers, we may reduce them both to the same denomina- tion, and then treat the resulting numbers as abstract, the ratio of those abstract numbers being the same as that of the two concrete numbers from which they have arisen. For the same reason, one of two ratios constituting a pro- portion may be formed from concrete numbers of one kind, while the other is formed from concrete numbers of a dif- i^ •' 228 PROrORTlON. ferent kind ; for 7 days : 13 days : : 7 miles : 13 miles, each ratio being in fact that of 7 to 13. Indeed it appears b^ (Art. 194) that the ratio of two concrete numbers may always be expressed by a ratio of two abstract numbers. If both or either of the ratios in. a proportion be formed from concrete numbers, we may thus replace each such ratio by one arising from abstract numbers, and in this way every term of the proportion will become an abstract number ; so that, notwithstanding the remark in (Art. 28), any one of the terms may then be multiplied or divided by the other. 209. If only three of the numbers m a proportion be given, we can by means of them find the fourth, and the method or Rule by which it may be found is one of very great importance in Arithmetic. Almost all questions which arise in the common concerns of life, so far as they require calculation by numbers, might be brought within the scope of the Rule of Proportion, which enables us to find the fourth term in a proportion, and which, on account of its great use and extensive application, is often called the Golden Rule. 210. The Rule of Proportion, then, is a method by which we are enabled, from three numbers which are given, to find a fourth which shall bear the same ratio to the third as the second to the first, that is, shall be the same multi- ple, part or parts of the third, as the second is of the first ; or, in other words, it is a Rule by which, when three terms of a proportion are given, we can find or determine the fourth Proportion m Arithmetic is usually divided into Simple and Compound. Simple Proportion. 211. Simple proportion is an equality of two simple ratios. It may be either direct or inverse (Arts. 197, 205, 206.) Rule. Leaving out of consideration supei-fluous quan- tities, find, ften be Ler the >*^'. PROPORTION. 2flt Ex. If 2 cwt. 3 qr. 16 lbs. of sugar cost £6, how much can be purchased for £16. 11. C.? £. £. 8. d. owt. qi'S. lbs. As 6 : 16, 11. 6 : : 2,3, 16. 4 11,2, 8 4 10s. = ,j of 1£ Is. = tV of 10s. 6d. = J of Is. 46, 1, 4, 1, 1, 22, 0, 0, 16.20 0,0, 8.10 6)47, 8. 22.30 7, 3. 27.05— or, 7 cwt. 3 qr. 27a»(y lbs. Note. The student should prove the following exer- cises by making the answer found one of the terms of a new question, as well as by other methods. Exercises. 1. If 27 yards cost £11, how much will £33 buy? Ans. 81 yds. 2. If 156 yards cost $700, how many will i:ir\.iiase 39 yards? Ans. $175. 3. If $204 pay for 10 cwt. 2 qr. 14 lbs. of sugar, what would 4 cwt. 1 qr. 14 lbs. cost at the same rate ? Ans. $84. 4. How many yards at 8s. 6d. per yard, must be given for 66 yds. at 5s. 9d. per yard? Ans. 44 yd. 2 qr. 2^. 6 If the rent of 5 acres be £4, 13s. 4d., what would be the rent of 75a. 2r. 10 p. at the same rate ? Ans. £70, 10s. 6d. 6. A person travelling at the rate of 20 miles a day, performs a journey in 18 days ; at what rate must he travel per day that he may return in 16 days? Ans. 22 J. f ; i 292 PROPOKTION. 7. Find the value of 23 yds. 1ft. of cloth, supposing 4 yds. 81 in. of the same qualit}'' to cost $15. Ans. $72. 8. If a property worth £185, 10s. pay a tax of $21.64 ^, what ought a property worth 1000 guineas sterling, to pay? ; Ans. $153.12^, 9. If 26 yds. of cloth cost 48s., what must it be soM at per foot, to gain 4s. on the purchase ? Ans. 8d. 10. If the prime cost of 247 cwt. 3 qrs. 14 lbs. of sugar amourt to £457, 148. 4^d. and the charges, freight, duty, &c., to £6, 3s. 9d. what does it stand per cwt.? Ans. $7.48|. 11. If 3f oz. avoir, cost $1.40, what will 30J lbs. cost? Ans. $191.70. 12. How many yards of drugget an ell wide will cover 40 yds. of carpet { yd. wide ? Ans. 24 yds. 13. A servant enters on a situation at 12 o'clock at noon on Jan, 1, 1863, at a yfsarly salary of 35 guineas sterling, he leaves at noon on the 27th of May following ; how xnany dollars should he receive for his services ? Ans. $73.50. 14. A person, after paying 7d. in the £ for poor*s-rate on his income, has £1632, 18s. lOd. remaining ; what had he at first? Ans. £1682. 15i The circumference of a circle is to its diameter as 8, 1416 : 1 ; find (in feet and inches) the circumference of a circle whose diameter is 22.5 feet. Ans. 70 ft. 8.232 in. 16. A watch is 10 minutes too fast at 12 o'clock (noon) on Monday and it gains 3' 10'' a day ; what will be the time by the watch at a quarter past 10 o'clock A. M. on the following Saturday ? Ans. IQh. 40' 36/^". 17. How many yards of carpet | yd. wide will cover a room whose width is 16 feet, and length 27.5 feet? Ans. 65^ yds. 18. A regiment of 1000 men are to have new coats ; each coat is to contain 2 J yds of cloth 1 J j^d. wide ; and « it is to be lined with shalloon of J j'^d. wide ; how many yards of shalloon will be required? Ans. 4166}. PBOPORn05. 2dd posing 4 I. $72. $21.64 i, •ling, to 3.I22J. 3 sold at s. 8d. of sugar it, duty, 7.48f. 3S. cost ? )1.70. 11 cover tyds. ;lock at guineas lowing ; 3.50. rate on had he L682. Leter as ence of • 2 in. noon) e time [)n the -7 II r F • over a rds. 3oats ; and <9 many 6f. 1 19. If two numbers are as 8 to 12, and the less is 820, what is the greater? Ans. 480. 20. If an ounce of gold be worth £4.189583, what is the value of .36822916 lbs.? Ans. £18, 10. 3. nearly. 21. If 1000 men have provisions for 85 days, and if, after 17 daj^s 150 of the men go away \ find how long the remaining provisions will serve the number left. Ans. 80 days. 22. If 4 horses and 6 cows together find sufficient gi*ass on a certain field ; and 7 cows eat as much as 9 horses ; what must be the size of a field relatively to the former, which will support 18 horses and 9 cows? Ans. 207 : 82. 23. What must be the breadth of a piece of ground whose length is 40^ 3'^ds, in order tliat it may be twice as great as another piece of ground whose length is 14| yds., and whose breadth is 13^^ yds. ? Ans. 9j> yards. Compound Proportion, 214. There are many questions which are of the same nature with those belonging to simple proportion, but which, if worked out by means of that Eule as before given, would require two or more distinct applications of it. Every such question, in fkct, may be considered to contain two or more distinct questions belonging to the Rule of simple proportion, and when each of these questions have been worked out by means of the Rule, the answer ob- tained for the last of them will be the answer to the original question. 215. The following example may serve to illustrate the preceding remarks. If the carriage- of 15 cwt. for 17 miles cost $17, what would the caiTiago of 21 cwt. for 16 miles cost at the same rate ? This question, though of a like nature with those which engaged our attention under the last Rule, is nevertheless of a more complicated desciiptlon ; for we observe that Instead of three quantities here are five, every one of which must necessaiily have a beaiing on the answer, so that none of them can be superfluous. J i i^'i m 284 PROPORTIoy. If, however, llie question be divided into two distinct questions, each of these, Avlien the superfluous terms are rejected, will bo found to compriso only three given terms of a proportion, from wliicli three terms the fourth is to be ascertained : bo that in this way we obtain the correct answer by applying the Rule of proportion twice over. . The first question maybe this : If the carriage of i ^ cwt. for 17 miles cost $17, vvhat would the carriage of 21 owt. cost? In th^'s question 1 7 miles would have no eff'ect upon the answer, because any other number of miles, or the words, a certain distance, might have been used instead of 17 miles. This number may therefore be neglected as superfluous. Solving the question by the Rule of simple proportion, we find the answer will be $23.80. The second question may be this : If the carriage of 21 cwt. for 17 miles cost $23.80, what will the carriage of 21 cwt. for 16 miles cost? In this question, for reasons similar to those given above, the 21 will be a superfluous quantity. Applying the Rule of simple proportion to the question we find the answer to be $22.40. From the connection of these two questions with that originally proposed we observe that the answer thus ob- tained through the means of two distinct applications of the Rule of simple proportion, must be the answer to the original question. 216. Compound Proportion is a shorter of more com- pendious method of working such questions as would re- quire two or more applications of the Rule of Simple Pro- portion, 217. For the sake of convenience, we may divide each question into two parts, the supposition and demand; the former being the part which expresses the conditions of the question, and the latter the part which mentions the thing demanded or sought. In the above question, the words, " if the carriage of \b cwt., for 17 miles cost $17, form the supposition ; and the words, what would the car' riage of 21 cwt.^ for 16 miles cost at the same rate? form the demand. Adopting this distinction, the following rule i distinct Tins are 3n terms is to be correct ver. f X ^ cwt. 21 V wt. )ct upon , or tlie stead of >cted as ' simple ?e of 21 :e of 21 3 given pplying find the itii that lus ob' ions of to the e com* uld re- ie Pro- le each the iUfi of is the , the $17, \e car- form rule PRoronxioN. 235 will be found apiDlicable for workinr; out questions 'u\ Com- pound Proportion. » 218. Rule. — Take from the supposition that quantity which corresponds with the quantity sought iu the de- mand ; and write it down as the third term. Then take one of tlie other quantities in the supposition and the cor- responding quantity in the demand, and consider them with reference to the third term only (regarding each oth- er quantity in the supposition and its corresponding quan- tity in tlie demand as being equal to eacli otlier) ; wiieii the two quantities are so considered, if from the nature of the case, tlie fourth term would be greater than the third, then, as in the Kule of Simple Proportion, put the larger of the two quantities in the second term, and the smaller in the first term ; but if less, put tlie smaller in tac second term, and the larger in the first term. Again, take another of the quantities given in the sup- position, and the corresponding quantity in the demand ; and retaining the same third term, proceed in the same way to make one of those quantities a first term and the other a second term. If there be other quantities m the supposition and de- mand, proceed m like manner with them. In each of these statings reduce the first and second terms to the same denomination. Let the common third term be also reduced to a single denomination if it be not already in that state. The terms may then be treated as abstract numbers. Multiply all the first terms together for a final first term, and all the second terms for a final second term, and re- tain the former third term. In this final stating multiply the second and third terms together and divide by the first. The quotient will be the answer to the question in the de- nomination to which the third term was reduced. 219. As a contraction in the use of this rule, divide an antecedent, and either the last term or any consequent, by any number that will divide them without remainders, and employ the quotients instead of those terms ; or if an an- tecedent and any consequent, or an antecedent and the last term be the same, reject them. (Art. 87— .90.) .11. 4 t • I \f t • I 236 PROPORTION. 7 horses \ '$ 20 days, \ \ ^1 850 J ) 7 clays, > 8112. ) 1- Ex. If 7 horses bo kept 20 clays for $56, how many will be kept 7 days for $112? The 7 horses in the supposition correspond to the re- quired quantity (number of horses) in the demand. Make this the third term. Then taking 20 days in the supposition, and the 7 days in the demand, and considering them with reference to our third term, wc observe that if the number of days bo diminished, the number of horses which can be kept in them for a given sum of money will be increased, and thus a fourth term will be greater than the third ; wc therefore place the 7 days in the first term, and the 20 days in the second. Again, taking the $56 in the supposition, and the $112 in the demand, and considering them with reference to the third tenn, we ob- serve that if the sum be increased the number of horses which can be kept by it in a given tii. .e will be increased ; so that here also a fourth term will be greater than the third ; we therefore place the $56 in the first term and the $112 in the second term. We thus obtain the following statement : — As 7 days : 20 days \ ,.n UQ^gg $56 : $112 ) * ^9^^^^t 56 X 7 20 X 112 :: 7 = 392 : 2360 : : 7 = 2240 X 7 -r- 392 = 40 horses. By Art. 219. 5 As :r : ^ ) . - s 8 X 5 = 40 horses, as before. Here, the 7 in first term cancels the 7 in the third term. Again dividing 56 in the first term and 20 in the second term by 4, we set down the quotients 14 and 5. For sim-. ilar reasons we omit 14 and write 8 instead of 112. We then multiply 5 and 8 together and find the answer as be- fore. w manj the rc- I in the . Then and the sidering d term, lays bo lich can ' money greater he first [ing the nd, and , we ob- ' horses reased ; lan the and the Uowing rses, )efore, term, econd r simo We as be- I'liOPOllTION. 237 The student should work each of the following ques- tions by simple statements, and thus verify the results. EXUICISES. 1. If 7 men can reap 6 acres in 12 hours, how many men will reap 15 acres iii 14 hours? Ans. 15 men. 2. If a family of 8 persons expend £200 in 9 months ; how much will serve a family of 18 persons for 12 mos.? Ans. £G00. 8. If 120 bushels of oats serve 14 horses 56 days ; how many days will 94 bushels serve 6 horses ? Ans. 102^f days. 4. If 180 men, in 6 days, of 10 hours each, can dig a trench 200 yards long, 3 wide, and 2 deep ; in how many days of 8 hours long, will 100 men dig a trench of 360 yards long, 4 wide, and 3 deep ? Ans. 48f days. 5. If the rent of a farm of 17 ac. 3 ro. 2 po. be £39, 4. 7., what would be the rent of another farm, containing 26 ac. 2 ro. 23 po., if 6 acres of the former bo worth 7 acres of the latter? Ans. $201.75. 6. If 1500 copies of a book of 11 sheets require 66 reams of paper, how much paper will be required for 5000 copies of a book of 25 sheets, of the same size as the former? Ans. 500 reams. 7. A pit 24 feet deep, 14 sq. feet, horizontal section, cost $12 to dig out ; how deep will a pit be of horizontal section, 7 ft. by 9 ft., which cost $18? Ans. 8 ft. 8. If 1 man and 2 women do a piece of work in 10 days, find in how long a time 2 men and 1 woman will do a piece of work 4 times as great, the rates of working of a man and woman being as 3 to 2. Ans. 35 days. «i ;M I I 2SS IKTEREST. INTEREST. 220. Some information regarding Interest has been al- ready given in (Art. 157.) It is repeated here, however in substance, to render the present article complete in itself. Interest is the sum of money paid for the loan or use of some other sum of money, lent for a certain time at a fixed rate ; generally at so much for each £100 or $100 fbr one year. The money lent is called the Principal. The interest of £100 or $100 for a year is called Th» Rate Per Cekt. The Principal -(- the interest is called The Amount. Interest is divided into Simple and Compound. When interest is reckoned only on the original principal, it is called Simple Interest. When the interest at the end of the first period, instead of being paid by the borrower, is retained by him and add- ed on as principal to the former principal, interest being calculated on the new principal for the next period, and this again, instead of being paid, is retained and added on to the last principal for a new principal, and so on ; it is called Compound Interest. Note. — ^The rate of interest has varied much at different periods, and in different countries, but it has been generally observed to diminish as' commerce extends. In Italy, about the be^^ning of the thirteenth century, it varied trom 20 to 30 per cent, per annum; and in the Neth- erlands, it was fixed by Charles V. in 15G0, at 12 per cent. By an act of the 37th year of Henry VIII., interest in England vas not to exceed . 10 percent. By the 21st of James I., it was reduced to 8 per cent. Soon after the Restoration, it was reduced still farther, to 6. per cent.; and in the 12th of Anne, to 5 per cent., the present rate. The legal rate of interest in Nova Scotia is at present 6 per cent. Simple Interest. 221. In interest five quantities are concerned, the prin- cipal, the rate, the time, the interest, and the amount, and any three of these except the principal, interest and amoaut being given the rest may be found. Henoe com- ^- INTEEEST. 339 } been al- however 3plete ID m or use time at a $100 fbr lied Thb HINT. When >al, it is instead nd add- t being tod, and ^ added > on ; it iods, and miiiish as hirteenth the Neth- (y an act to exceed per cent, ler cent. ; rhe legal le prm- (it, and st and B com- putations in interest admit of several problems all of which depend upon either Simple or Compound Propor- tion or both ; for the interest of any sum for any time is directly proportional to the principal sum, and also to the time of continuance, and conversely. (1.) To find tJie interest of a given sum for a year, at a given rate per cent, per annum. Rule. — Multiply the principal by the rate, and divide the product by 100. (Art. 133.) or, As 100 : rate : : prin- cipal : Its interest for one year. Ex. 1. What is the interest of £168, 16. 3. for 1 year at 6 per cent. ? £.8. D. Dollars. 168, 16.3. 676.25 6 6. 10;12, 17. 6. 20 40;51.50. $40.51 j. 6;90 4 3^60 . JfilO, 2. 6f /. • . . i. . I »■■■ . • The reason of the above operation is quite evident, as it is nothing more than this : As the principal £100 or $100 is to its interest £6, or $6, so is tlie principal £168, 16. 3. or $675.25 to its interest ; and it is evident that as often as the one principal contains its interest, so often will the other contain its interest : that is, by the nature of pro^- portion, the interest will be proportional to the principal. 240 INTEREST. Ex. 2. What is the interest of £127, 13. 4^. for 1 year at 5( per cent. ! i = J of 1 unit i = *off £. pi. D. 127, 13. Ai 638, 6.10J 63,16. Si 31,18. 4 15, 1». 2 Dollars. dlO.675 t = i of 1 f = *oft i = Joff 2553.875 255.337 127.668 63.884 7;5a, 1. Of. 20 10;01 £7. 10. 0. 30;00.264 $30.00. Non.— The vemiiiiden after pence may be nijeaCed ae they do not efieot the result The same may be obsenred with regard to the deci- mal form. • For mental exercises to this and snCceedittg Rales see (Art. 157;. Exercises. Find the interest of the following sums, Tot 1 year at the given rates per cent, per annum. 1. £376, 12. 8. at 4 per cent. 2. £774, 13. 3. at 5 per cent. 3. £637, 11. 0. at 5f per cent. 4. $2687.90 at 4| per cent. 5. $69.40 at 3 j^ per cent. 6. $189.63j^ at 6f per cent. Ans. £15, 1. 3}. Ans. £38, 14. 7j. Ans. £37, 3. 9J. Ans. $114.23), Ans. $2.42||. Ans. $12.80. (2.) To find the interest of a given principal for years and months. Rule. — Find the interest for 1 year by the last Rule, multiply the number of years and take parts of a yeai^ for the months. Or nrrKRBST. ui or 1 year an. .675 5{ .375 .337 .668 .884 iii .00. ly do not » thedeoU lales see year at .3}. .7J. . 9f. 23|. 2ii. 2.80. r years Rule, yeatf Multiply the principal by the rate ; then by the nmnber of years, taking aliquot parts for the months ; and, last of all, dividing by 100. Ez. Required the interest of £99, 2. 4^. for 2 years* 9 months, at 4 per cent, per annum. fXBST XKTHOD. £99, 2. 4^. 4 8^96, 9. 6 20 19.29 12 8;54 £3, 19. 3.54 = Ins. for 1 year 2 at 4 per cent. 6 i >»•*;- =: j^ of 1 jrear 3 mos. = j- of 6 mos. 7. 18. 7.08 = Int. for 2 years, 1. 19. 7.77 =r Int. for 6 montiis, 0, 19. 9.88 = Int. for 3 months. £10, 18. 0.73 = Int. for 2 y. 9 mo. 8BC0ND MSTHOD. £99, 2. 4], 4 396, 9. 6. 2 792, 19. 0. 6 = ^ 198, 4.9. 8 = J 99, 2.4^. 10;90, 6. 1^. 20 18;06 12 i c *. > i i i i « M • « £10, 18. 0.73. 9i% ijrrEREST. Or thus 499, 2. 4}. for 2 years and 9 months at 4 per cent, is equiyfdent to £99. 2. 4^. for 1 year at 11 per cent. There- fore ' • ~ £99. 2. 4J. 11 10;90, 6. 1^. 20 18;06 12 0;73 £10, 18. 0.73 as before. The rectson for the above methods is evident from (Art. 818). Exercises. Find the interests of the following principals, for the given times, and at the given rates per cent., per annum. 1. £3o5, 15. 0. for 4 y., at 4 poi- cent. Ans. £56, 18. 4}. -^. £^1.9, 0. 6. for 5f y., at 3|per cent. Ans. £68, 15. 9 J. 3. £712, 6. 0. for 8 mo., at 7^ per cent. Ans. £35, 12. 3^. 4. £25, 0. 0. for 1 y. 9 mo., at 5 per cent. Ans. £2, 3. 9. 5. $738.00 for 1 y. 2 mo., at 7 per cent. Ans. $60.27. 6. $894.00 for 1 y. 8 mo., at 6 per cent. Ans. $89.40. 7. $65256 for 4 mo., at 7 per cent. Ans. $1522.64. (3.) To find the interest of a given princijMl for year s^ months and days. 5 Rule. — ^Find the interest for the years and months by : the last Rule, then take such a fractional part of one month's interest, as is denoted by the given number of days. 1 Note 1. — In calculating interest, a month, whether it ' contains 30 or 31 days, or ev^n but 28 or 29, as in the ' case of February, is assumed to be one-twelfth of a year. Again 30 days are commonly considered a month ; con- sequently the interest for 1 day, or any number of days imder 90, is so many thirtietlis of a month's interest. t t I nfTEREST. •^^43 cent, is There- 8forc. m (Art. for the annum. 18. 4 j. 15. 9|. 12. 3j. ,2, 3. 9. 60.27. 189.40. '22.64. yearSf iths by lof one daj's. thcr it in the year. li con- daj^s This practice seems to have been originally adopted on account of its convenience. Though not strictly accurate, it is sanctioned by general usage. Allowing 30 days to a month, and 12 months to a year, a year would contain only 360 days, which in point of fact -jj^, or ^ less than an ordinary year. Hence to find the interest for any number of days with en^ tire accuracy, by means of aliquot parts, we must take so man> 365ths of 1 year's interest, as is denoted by the given number of days ; or find the interest for the days as above, from this subtract -^ of itself and the remainder will be the exact interest. (See Note 2, Art. 157, Rule 6.) Note 2, — ^If the interest has to be calculated from one given day to another, as for instance from the 30th of January to the 7th of February, the 30th of January must be left out in the calculation, and the 7th of Febru- ary must be taken into account, for the borrower will not have had the use of the money for one day till the 31st of January. Ex. Required the interest of $148 for 8 mo. 12d., at 6 per cent, per annum. $148 6perct. 6 mo. = i of 1 y, 2 mo. = J of 6 mo. 10 d. = i of 2 mo. 2 d = i of 10 d. 888 "444.00 148.00 24.666 4.933 621~599 6 mo. = 2 mo. = lOd. = 2d. = $6.21 jV £37 6 : i of 1 y. :iof6mo. :|of2mo. I of 10 d. 222 111 37, 0. 0. 6, 3. 4. 1, 4. 8. 1;55, 8. 0. 20 lli08 12 0,96 I £1, 11. 1. nearly. Hi ' iirrEREST. Find the Interest of the following sums for the given iime, and at the given rates per cent., per annum. 1. £127, 18, 4. for 6 mo. 10 d., at 4 per cent. Ans. £2, 13. 10|. 2. £362, 11. 0. for 9 mo. 20 d., at 4} par cent. Ans. 13, 17. 5^. 3. $287.50 for 1 y. 3 mo. 9 d., at 6 per cent. Ans. $22.00 nearly. 4. $18d for 11 mo. 20 d., at 5| per cent. Ans. $9.65. 5. $1763.16^ for 8 mo. 15 d., at 4^ per cent. Ans. $51.51^. (4.) To find the interest accurately of any sum for any time^ and at' any rate per cent,, per annum, Rtt]&. — Atf £100 or $100 is to the rate per cent, and as 1 year in the same name as the given time is to the given time, 80 is the principal to the interest on that principal for the given tiiiie. INTEREST. 245 le given 3. lOi 7. 5J. Barly. ^9.65. for any . and as le given rincipal Ex. Required tlie interest of £456, 10. 0. for 31 days at 6 per cent. Asl00:5J ) ..45610 865 : 31 j • • *^ ^"' 31 7300 :31 456 1368 10s. = J of £1 /. J of 81 as £. *». 15.10 Interest. 7800)14151. 10(£1, 18. 9i. 7300 6851 20 137030 7300 64030 58400 ~680 12 67560 65700 1860 4 7440 7300 140 In the above example, if we had adhered to the Rule given for Compound Proportion, we should have placed the principal in the second term instead of the rate, but the present mode which, for brevity, is that generally adopted in calculating interest, as it saves the trouble of reducing the £100 to the lowest denomination mentioned in the principal. 246 INTERFST. If the principal consists of Dollars and Cents the Rule in Compound Proportion will be equally short. Exercises. Find the interest of Answers. 1. £690, 10. 6. for So days, at 4 per cent. £6, 8. 7^. 2. £573, 8. 3. for 73 days, at 3^ per cent. i*4, 0. 3^. 3. £684, 7. 6. for 56 days, at 3^ per cent. £3, 13. 6. 4. 81719.46J for 86 days, at 4 per cent. ^16.20^. 5. $3311.50 for 292 days, at 2} per cent. $66.23. Calculate the amount o/i£643, 16. 6. 6. From Feb. 20 till July 18th, at 4^ per cent, Ans. £655, 11. 5]. 7. From May 26th till Nov. 26th, at 3^ per cent. Ans. £655, 3. 8J. 8. From June 3d till May 3d, at 6 per cent. Ans. £673, 5. 7^. 9. Required the interest of $63 from March 17th, 1840, till January 26th, 1842, at 6 per cent, per annum. Ans. $7.04^. 10. What is the interest of $213.33» from June 14 1841, till Sept. 22, lb43, at 4^ percent.? Ans. $21.82^^. 11. What is the interest of £52, 10. for 1 year and 2 months, at 6 per cent., per annum. Ans. £3, 13. 6. (5.) To find the interest of a given sum for any number of days. Rule. — Multiply the principal by twice the rate, and the product by the da^'s, and divide the result by 73;000. The division by 73000 may be performed by the follow- ing Rule : Below the dividend write one-third of itself, one-tenth of that third, and one-tenth of that tenth, re- jecting shillings and remainders ; then add the four lines together, divide the sum by 100.000 (or cut off five fig- lu'es toward the right), and reject a farthing for every £10 or a mill for every $10 in the result. INTEREST. 247 he Rulo wers. 0. Sf 13. 6. [6.20i. 36.23. 1.5]. 3. 8i. 5. 7^. 1, 1840, me 14 51.82ii. r and 2 3.6. \mber of Lte, and (73;00O. I follow- itself, ith, re- ir lines Ive fig- ry £10 The tenths will be obtained by setting the figures one place to the right hand, and rejecting the last of them, or the remainders may be treated as in dollars and cents. Ex. Find the interest of £372, 10. 10. for 309 days, at 4j- per cent, per annum. £372, 10. 10. 9 £3352, 17. 6. or £3353, 0. 0. nearly. 309 30177 10059 1036077 i 345359 Vn 34536 A 3453 14.19425 20 3;88500 12 10;62000 4 2;48000 Correction £14, 3. lOJ. £14, 3. lOJ. Ans. The reaso7t for the above process will be evident from the operation by Compound Proportion. u 948 INTEREST. If, instead of £100 and the rate per cent., their dou- bles be employed (Art. 199). Thus we have in this exer- cise As £200 : £9. ) 365 days : 309 days, £372, 10. 10. i 73000 24333' 2433' 243' 309 X 9 X 872, 10. 10. 9 100010 or 100000 £3352. 17. 6. or £3353, 0. 0. 309 1036077 345359 34536 3453 14; 19425 or £14, 3. lOJ. Now as 73000 becomes the division and the continued product of the time, double the rate and principal, viz^ £1036077 becomes the dividend, it is evident, from (Art. 79) that if we increase the dividend by ^ of itself we must increase the divisor also by ^ of itself ; again when we increase the dividend by -^ of this third and also -^ of that tenth, we must of necessity increase the divisor in the same proportion. We have now 100010 for the divisor and 1419425 for the dividend, rejecting the 10 and dividing by 100,000 or cutting off 5 figures, &c., we obtain £14, 3. 10^. for the quotient. Now we divide by 100,000 instead of 100010, therefore the quotient will be too great by the same frac- tional part of itself as the 10 is of 100010 (Art. 77). Hence the correction. (6.) To find the interest of a given principal for any number of days at 4 per cent, per annum. BuLE. — ^Multiply the principal by the days : to the pro- duct add one-tenth of itself: from the sum take four times INTEREST. 249 tinned vizn, (Art. f we when sor in any the same product wanting the last three figures : divide what remains by 10,000 (or cut off 4 figures) ; tlie quo- tient will be the answer nearly. When the interest is large reject a farthing for each JCIO or 1 mill for each $10 contained in it. For other rates than 4 per cent, increase or diminish the product of tlie principal and days, by the method of ali- quot parts, and then proceed by the rule. Kx. Required the interest of $35942.80 for 12 days, at 4 per cent., per annum. $35942.80 12 431313.60 43131.36 474444.96 1724.25 47;2720.7l $47,272 4 Correction. Here the product of tlie prin- cipal and days is 431313.69 and tlie tenth of this (found by set- ting each figure one place neai*- er the right hand side) being added to it tliesum is 474444.96. After this we multiply 43131 by 4 and increase the product by 1 (carried for 4 times 3, the first of the figures cut off). The re- sult 1724.25, is then subtracted, and the remainder 472720.71, divided by 10,000 in the way $47,268. pointed out in the last example. The quotient is $47,272 from which subtract 1 mill for each $10, viz., 4, the re- mainder $47.26J is the interest required. The reasons for the above process are similar to those given for Rule 5. It may also be reasoned otherwise. The exercises under Rule 4 may be used for Rules 5 and 6. The following rules serve for the resolution of the re- maining cases of interest. As these cases are of minor importance the rules are given without illustration. They are easily proved by the principles of proportion. (7.) To find what principal^ in a given time, would pi'O^ duce a given interest, at a given rate per cent, per annum. r - # 250 INTEREST. Rule. As the rate : JCIOO or j^ ) fu^ : * The given time : 1 y. ( ''^^^ i*^^- : the principal. i Exor. 1. Wliat sum will produce for interest £56, 14. in 2J- years at 4^ per cent. Ans. £560, 0. 0. 2. What principal at 5 per cent, per annum, will brin^ a yearly income of $1365? Ans. 627300.09. (8.) To find the time in which^ at a given rate per cent. per annum, a given principal would produce a given in" terest. Rule. As the principal : £100 or $100 ) . . i vnor • time reod The rate : the interest ] ' ^ ' ^ Exer. 1. In what time will £560 amount to £616, 14* at 4^ per cent, per annum ? Ans. 2 J years. 2. How long must $8000 be lent at simple interest, at ^} per cent, per annum, to amount to $9120? Ans. 4 years. (9.) To find at what rate per cent, a given principal would tfain a given interest in a giveii time. Rule. As the given time : 1 year ) the principal : £100 or $ J ' * ^ interest : rate. Exer. 1. At what rate will £157, 15. 4. amount to £295, 16. 3. in 25 years at simple interest? Ans. 3J per cent. 2. If $1 amount to $1.13 cents 7 J mills in 3^ years, at simple interest, at what rate per cent, per annum, must it have been lent ? Ans. 4.2307 per cent. (10.) To find what principal, in a given time would in^ urease to a given amount, at a given rate per cent, per annum 4 Rule. — To the product of the time and rate, add the product of £100 or $100 and 1 year, in the same name as INTEREST. 251 'H the given time. Then as the sum is to the above men- tioned product of £100 or $100 and 1 year, so is the amount to the principal. Exer. 1. What sum of money will amount to £256, 10. in 4 years, at 3V per cent., simple interest? Ans. £226, 0. 0. 2. Wiiat sum must be lent, at simple interest, at 4 per cent., per annum, tliat tlie amount, at the end of 2 years 10 months, may be $2511.70? Ans. $2256.01 J. M Xnum* the le as COMMISSION, INSURANCE, BROKERAGE, &c. 222. Commission is the per cent, or sum charcfed by agents for tiieir services in buying and scD'ag goods, or transacting otiier business. Note.— An accent who buys and sells goods for anotbe?- ia called a Commission Merchant^ a Factor ^ or Correspondent. Brokerage is of the same nature as (ion mission, but has relation to money transactions, rather than dealings in goods or merchandise. Insurance is a contract, by which one party on being paid a certain sum or Premium by another party on proper- ty which is subject to risk, undertakes, in case of loss, to make good to the owner the value of that property. Note.— The written instrument or contract is called the Policy. When duty is charged on the Policy it is 'calculated on even hundreds. Thus ^'650 would be $.'700. (1.) To Compute the Commission.^ Brokerage, Insurance, or any other allowance on a alven sum, at a given rate per cent. Rule. — Multiply the sum by the rate per cent., and di- vide the product by 100 ; or as £100 or dollars are to the rate per ^jeut., so k the given sum to the required allow- ance. 252 INTEREST. Ex. Requii'cd the premium of insurance on £512, 9. 4. at £6, 16. 6. per cent. £512, 9. 4. $2049.866' 6 $6.82^ 12s. 0. = tV of £6 4s. 0. r= ^ of 12s. 08. 6. = ^ of 4s. j£3074, 16. 0. 112299.199 307, 9. 7f 60ct.=J I 1024.933' 102, 9. 10^. 20ct.=il 409.973' 12, 16. 2J. 12Jct.=:i| 256 233' 139j90.339 Ans. $139,90]. 100) 3497, 11. 8 J. Ans. £34, 19. 6^. (2.) To find how rrmch must be insured on property worth a given sum^ so that^ in case of loss, both the value of the property and the premium may be repaid. Rule. — Subtract the rate from £100 or Dollars. As the remainder is to £100 or dollars, so is the value of the prop* erty to the sum to be insured. Ex. How much must be insured at 8J per cent, on goods worth £600, that in case of loss not only the value of the goods, but also the premium of insurance, may be paid ? Here, as £100 — 8J, or £91^ : £100 : : £600 : : £655. 14.9. The accuracy of this operation is proved by finding the premium on £655.14.9 at 8^ per cent. This is found to be £55.14.9. Hence, in case of the property being lost, the owner will receive, not only £600, the value of the goods, but also £55.14.9 the premium ; and he will there- fore sustain no loss whatever. The reason will appear manifest from considering that, in receiving £100, which has been insured at 8 J per cent, the owner would receive but £91^ in lieu of the goods, 8 j- having been paid for the insurance. Exercises. 1. What is the commission on £942, 16. 3. at 4j^ per cent.? Ans. £42, 8. 6^. INTEREST. 253 2, 9. 4. 66' $6.82i L99 233' < of the As the le prop- ent. on lG value may be £655. finding s found ng lost, of the I there- ig that, cent, goods, Hi per 1. 6i. 2. Required the brokerage on £946, 18. 10. at 5s. 6d. per cent. Ans. £2, 12. 1. 3. Find the commission on $2278.95 at 7^ per cent. Ans. $170.92^. 4. What sum must be insured at £1, 10. per cent, on goods worth £1200, so that in case of loss, both the value of the goods and the premium may be repaid ? Ans. £1218, 6. 6., nearly. 5. At $2.27^ per cent., what will be the cost of insuring goods worth $6240, so that in case of loss, the owner may be entitled to the value of the goods, and the premium? Ans. $145.26^. Application op Interest. 223. In the application of interest to business transac- tions the following particulars deserve attention. (1.) A promissory liole is a writing which contains a promise of the payment of mone> or other property to another, at or before a time specified in consideration of value received by the promiser or maker of the note. The words " value received " were at one time supposed to be an essential part of a bill or note, but as a valuable consideration is always presumed until the contrary is proved, they are not at all necessary, though usually inserted. ^2.) The person who signs a note is called the maker, drawer J or giver of the note. The person to whom a note is made payable, is called the payee; the person who has legal possession of a note is called the holder of it. (3.) A note which is made payable "to order" "or bearer," is said to be negotiable ; that is, the holder may 8ell or transfer it to whom he pleases, and it can be col- lected by any one who has lawful possession of it. Notes without these words are not negotiable. (See Nos. 1 and 2.) (4.) If the holder of a negotiable note which is made payable to order wishes to sell or transfer it, the law re- quires him to endorse it or write his name on the back of it. The person to whom it is transferred, or the holder of !|l 254 INTEREST. I ii Hi it, is then empowered to collect it of the drawer ; if the drawer is unable or refuses to pay it, then the endorser is responsible for its payment. (See No. 1.) (5.) When a note is made payable to the bearer ^ the holder can sell or transfer it without endorsing it, or in- curring the liability for its payment. Bank notes are of this description. (See No 2.) (6.) When a note is made payable to any particular per- son without the words order or bearer, it is not negotiable ; for it cannot be collected or sued^except in the name of the person to whom it is made payable. (See No. 3.) (7.) A note should always specify the time at which it is to be paid ; but if no time is mentioned, the presumption is that it is intended to be paid on demand, and the giver must pay it when demanded. (8.) According to custom, a note or draft is not pre- sented for collection until three days after the time speci- fied for its payment. These three days are called days of grace. Interest is therefore reckoned for three days more than the time specified in the note. When the last day of grace comes on Sunday, or a national holiday, it is cus- tomary to pay a note on the day previous. (9.) If a note is not paid at maturity or the time specified, it is necessary for the holder to notify the endorser of the fact in a legal manner, as soon as circumstances will admit ; otherwise the responsibility of the endorser ceases. (10.) Notes do not draw interest unless they contain the words " with interest." But if a note is not paid when it becomes due, it tlien draws legal interest till paid, though no n:ention is made of interest. (11.) Notes which contain the words ^'ivith interest,'* though the rate is not mentioned, are entitled to th^ legal rate established by the Kingdom, State or Province in which the note is made. In writing notes, therefore, it is unnecessary to specify the rate unless by agreement it is to be less than the legal rate. (12.) When two or more persons jointly and severally give their note, it may be collected of either of them. (See No. 4.) If the :ser is r, the or in- are of ir per- tiable ; of the hich it uption 3 giver 3t pre- speci- days of s more day of is ciis- ecified, of the les will [ceases, ain the ^vhen it though \terest" I9 legal Ince in specify ie legal rerally I. (See Ks'TEREST. 255 (13.) The sum for which a note is given, is called the principal, or face of the note, and should always be written out in words. (JV'o. 1.) $450. Halifax, April 16th, 1863. Sixty days after date, I promise to pay George Kinman, or order, Four Hundred and Fifty Dollars, with interest, value received. JOHN JOHNSON. (No. 2.) $630. Boston, May 23d, 1863. Thirty da3''s after date, for value received, T promise to pay John Holmes, or bearer, Six Hundred and Thirty Dollars with interest. JAMES GOODYEAR. (No. 3.) $850. Windsor, March 16th, 1863. Four months after date, I promise to pay Horace Wil- liams, Eight Hundred and Fifty Dollars, with interest, value received. SANDFORD ATWATER. (No, 4.) 1000. Truro, April 14th, 1863. For value received, we jointly and severally promise to pay to the order of John K. Blair, One Thousand Dollars, in one year from date, with interest. ROBERT L. ARCHIBALD, WILLIAM SIMMONDS. ! h mfi 256 I I I IM Discount. 224, Discount is the abatement or deckuition made for the payment of money before it is due. For example, if I owe a man $100, payaljle in one year without interest, the present worth of the note is less than $100 ; for if $100 were put at interest for 1 year at 6 per cent., it would amount to $106 ; at 7 per cent., to $107, £ As im : 5 12 mo. : 7 mo. I ::im '' ' 1 12 : 35 : ; 1 12)35 £2, 18. 4. interest of £100 for 7 mos. .-. £100 + £2, 18. 4. = £102, 18. 4. = Amount of £100 for the given time. £ B. D. £ £ Again, As 102, 18. 4. : 100 : : 463 20 20 2058 12 24700 2000 12 24000 46 Present worth. 24700)1 1112000(£449, 17. G}, The reason for this rule will be evident from the consid- eration, that £100 or dollars is the present worth of its amount regarded as a debt ; and, consequently, the analo- gy given above will become simply this : as the amount of. £100, considered as a debt is to £100, the present work of that debt, so is any other debt to its present worth. It is obvious also, that, for the first two terms of the analogy, we might use the amount of a7iy sum whatever^ and that sum itself ; but it is general^ more simple and easy to employ £100 or dollars and its amount. \\\U Pi! irs, foi* eioo or md the remain- [nonths, aos. Amount th. consid- of its analo* lount of rork of of the iatever^ )le and STOCKS. 259 By comparing the above result with that found by the common method we see that the error amounts to 7s. 7^. Where the principal is given in pounds, &c., it will be found more convenient to add the interest of £100 for the given time to jC 100 in the form of a fraction. Thus As 100 12 :5) : 7) •• 100 12 35 = £1^ interest of £100. Again, As lOOf^ 100 : 463 : £449, 17. 6^. Both of these are, in reality, the same as Rule (10) in interest. Stocks. 226. If the 3 per cent. Consols be quoted in the money-market at 96|^, the meaning of this is, that for £96, 12. 6. of money a person can purchase £100 stock, for which he will receive an acknowledgment which will enti- tle him to half-yearly dividends from Government at the rate of 3 per cent., per annum on the stock held by him. Similarly, if the shares in any trading company, which were originally fixed at a given amount, saj^, $100 each be advertised in the share market at 86, the meaning of this is, that for $86 of money one share can be obtained, and the holder of such share will receive dividends at the end of each half-year upon the $100 share according to the state of the finances of the company. Stock may therefore be defined to be the capital of moneyed institutions, asincorporatedBanks, Railroad, In- surance Companies, and Manufactories, &c. ; or to be the money borrowed by our or any Government at so much per cent., to defray the expenses of the nation. The amount of debt owing by the Government is called the National Debt, or the Funds. The Funds represent tlie credit of the country, which is bound to pay whatever debts are contracted by its Government. The Govern- ment, however, reserves to itself the option of pa>'ing off the principal at any future time whatever ; i)ledging itself ^it. M a. 4 1^60 STOCKS. nevertheless, to pay interest on it reguUrly at fixed peri- ods, in the meantime. If money would alwa3^s bring the same amount of in- terest, the average price of £100 or $100 stock would be always the same (viz., «£ 100, or $100 the price first given for it) — we say average price, because even then the price would evidently be somewhat less immediately ajter the pa5^ment of a dividend, than it would be immediately be- fore it : but not only does this cause affect the price of Stocks, but the continual fluctuations in the A'alue of money, arising from commercial or political changes or ex- pectations abroad and at home, are constantly disturbing it, even two or three times in the same day according to the news which reach us. The price of stock will rise and fall according as it seems most likely that money would fetch elsewhere a higher or a lower rate of interest ; i. e. would be more scarce, and in demand, as in prospect of war, or of active speculation, or be lying upon hand and plentiful, as when trade is looking dull, and there are no means of employing capital. Thus if at a time A wished to sell his stpck, money was elsewhere making 5 per cent., it is plain that no one would give him $100 for the right to receive only 4, but since $80 of common money would now bring $4 interest, he would be able to sell his $100 stock for $80, and the 4 per cents, would be said to be selling at 80. When the price of £100 or $100 stock is £100 or $100 in money, the stock is said to be at par. When the price is more than the original price it is said to be at a premium, and when less at a discount. All examples in stocks depend on the principles of proportion, those of most frequent occurrence will novo be explained, Ex. 1. Required, the sum which will purchase £1500 in the 8 per cents, at 82. In this case £100 stock is worth £82 in money. •.• As £100 : £1500 : : £82 money : Money required whence required sum of money = £1230. STOCKS. 261 I peri- of in- ulcl be , given 3 price er the 3ly he- :ice of ilue of ; or ex- ;urbing ling to ise and would t ; i. e. jpect of md and are no ,ey was would Ince $80 would cents. )r $100 is said tlSOO in lired Ex. 2. "What amount of stock in the Z^ per cents, at 90 will $4050 purchase? In this case $90 money will purchaise $100 stock. ,*. As $90 : $4050 : : $100 stock : requd. amt. of stock whence required amount of stock = $4500. Ex. 3. If I buy £1520 3 per cent, consols at OSJ, and pay i per cent, for brokerage, what does it cost me ? Every £100 stock costs me (£93^ + £i) or £933. •*. As £100 stock : £1520 stock : : 93^ : requ. s. of money whence required sum of money = £1419, 6s. Ex. 4, What sura shall I receive for £1920, 13. 4. in the 3^ per cents, at 98|, brokerage being £^ per cent? £100 stock realises (£98| — i) = £98J. •'. As £100 stock : £1920| stock : : £98J : required sum whence required sum = £1896, 13. 2. Ex. 5. If I invest $7927.50 in the 3 per cents, at 94|, what annual income shall I receive from the Investment? For every $94 § I get $100 stock, and the interest on $100 stock is $3, therefore for every $94^ of money I get $3 interest. ••. As 94 J : 100 : : $3 : required income whence required income = $252. NoTK 1. — If it be required to find the income arising from a certain quantity of stock, it is merely a question of simple interest. Note 2. — It maj' be noticed in the above examples, that when the question was simply to find the amount of stock, or money realized by sale of stock, the 3, 4, or other rates per cent, never entered into the itaiement; and when the question was simply to find the income arising from any sum invested in the funds, then the £100 or $100 never entered into the statement. Ex. 6. Which is the best investment, $4000 in the 3 per cents, at 89]^, or the 3^ per cents, at 98 j-. In the first case every $89 J- gives $3 interest. .*. every $1 of money gives $g|T or $yfy interest. In the second case every $98J gives $3J interest; .*. ex^vy $1 of money gives $_i or %^l^ interest. I 1 f III 2C2 STOCKS. and comparing the fractions jf 9 and ^Jri 11^2^ and gV!Mr? the second fraction is greater tlian the first, and therefore the secjoild investment is the best. Ex. 7. How mucli stock can be purchased by the trans- fer of £2000 stock from the 8 per cents, at 90 to tlie 3^ per cents, at 96, and wliat change will be effected in income by it? It is evident thai: we can purchase less for a certain sum at 96 than if it were only 90, so that we must state as follows ; As 96 : 90 : : £2000 stock : required amount of stock whence required amount of stock = £1875. Income in first case = £60 Income in second case = £65, 12. 6. Therefore income increased by £ 5, 12. 6. Note. ' The Iflst question might have been worked thus: first sell out the stock 8^1^ 90) and then invest the proceeds in 3i per cents, at 96. Ex. 8. A person purchiises $4000 3 per cent, consols at 97^, and sells out again when they have sunk to 83 J ; how much does he lose by the transaction ? He loses on every $100 stock ($97| — 83^), or Sl3f. • •.• his total loss = (S13| X 40) = $ 545.00. Exercises. 1. The 4 per cents, being at 82i, what must be given for £1000 stock, and what sum Avould be gained by selling out again at 86^ ? Ans. £821, 5. ; £41, 5. 0. 2. If I lay out 812,000 in the 3 per cents, when they are at 84§, what income should I thence derive? Ans. $426.66$. 3. What is the cost of £850 Bank stock at 90|^, ^ per cent, being paid for Ijrokerage ; and what sum would be lost by selling out at 89^? Ans. £771, 7, 6. ; £10, 12. 6. 4. A person transfers $4000 stock from the 4 per cents, at 90, to the 3 per cents, at 72 ; find the alteration in his Income. Ans. $10.00. lerefore e trans- i ^ per income 1 sum at follows ; stock 'St sellout it 96. consols to 83 J; $13f. '0. )e given selling I, 5. 0. len they I6.66J. \h i per rould be 12.6. ier cents. In in bis 10.00. pnorrr and loss. 2d8 IJ 5. A person having £4236 sterling invested in the 3 per cents, sells out at 7o and reinvests the proceeds in tliis country at G \yev cent, per annum, allowing exchange to be at par, what is the ditference of his income. Ans. An increase of $317.70 Nova Scotia currency. Profit and Loss. 227. All questions in Arithmetic which relate to gain ®r loss in mercantile transactions, fall under the head of Profit and Loss. Examples in prqfit and loss are worked by the principles of proportio7i. The method of working the first four examples is so ob- vious as not to require a formal rule. Ex. If 112 lbs. of rice be bought for £1, 15. 0. and sold at 4jd. per lb., what is the gain ? 112 at 4^ per lb, = £2, 2.0, Prime cost = £1, 15. 0. Gain £0, 7. 0. Ans. EXSROISES. 1. If a piece of linen, containing 25^ yds., cost £3, 8. 8., what is gained by selling it at 3s. 9^., per yard ? Ans. jei, 8. 5^^. 2. If a pipe ofwlne, containing 138 gals., cost $455, what is gained by selling it at $3,70 per gallon? Ans. $55.60. 8. If 113 pounds of tea be sold at 2s. 9d. per lb., what is the gain, the first cost being $48.02j ? Ans. $14,124- 4. If a roll of tobacco, containing 25lbs., be bought at 8s. 6d. per lb. ; what is gained by selling it iit 3Jd. per oz., a pound and a half being lost by weighing it out in small quantities, and by drying? Ans. £0, 14. 4. I II I 264 PROPTP IND loss. i J 228. From the prime cost and the selling price^ to find the gain or loss per cent. Rule. As the prime cost is to the gain or loss on that cost, so are £100 or $100 to the gain or loss per cent. Ex. If tea be bought at 2s. 8d. per lb„ and sold at 3s. IJd. per lb., what is gained per cent? Here, 3s. IJd. — 2s. 3d. = lO^d. gain on 2s. 3. Then, As 2s. 3d. : lOj- : : £100 regarded as a first cost : £38, 17. 9^, the gain on 100 pounds By Dollars and Cents, 8s. IJd. = 62J cents = selling price, 2s. 3 d. = 45 cents = prime cost, 17^ = gain on 45 cents. As 45 cents : 17 j cents : : $100 : $38.88}. '.* the gain is 38f per cent. i\ I |r! I'll"''- 229. To find how a commodity must be sold to gain or lose a certain rate per cent. Rule. As £100 or dollars are to the gain or loss per cent., so is the prime cost to the gain or loss on that cost ; and from this and the prime cost, the selling price will be found by addition or subtraction. Ex. How must nutmegs, which cost $1.20 per pound, be sold to gain 16 per cent. ? As $100 (regarded as the first cost) : $16 (gain on $100) : I $1.20 (first cost of 1 lb.) : $0,192, the gain per lb., which being added to $1.20, the amount is $1.39^. Note. It should be particularly remarked that by the gain or loss per cent, is to be understood the su?n that would be gained or lost at the given prices, not on a hundred pounds or a hundred dollars' worth sold, but n )er lb., t hy the vould be pounds pounds there be PROPrr AND LOSS. KXERCItCS. 265 1. If paper which cost 19s. 2d. per ream, be sold for 17s. lid. per ream, how much is lost per cent.? Ans. X6, 10,6j^, 2. How must linen which cost 3s. l^d. per yard, be sold to gain 16 per cent.? Ans. $0.72J^. 3. If Broadcloth cost $2.35 per yard, how much la gained per cent, by selling one part at $2.80 per yard, and how much by selling another at $2.90? Ans. 19{^7- and 23^f per cent. 4. Isinglass, which cost 8s. 6d. per lb., is sold at a loss of 1 1 per cent. : at what rate is it sold, and how much is lost on the sale of 17 cwt. 1 q. long wt. at the same rate ? Ans. 7s. €J§d. ; £90, 6. 5^, 5. Bought 2048 yards of cambric at 3s. 2^d. per yard, and sold the whole for $1443.95. Required, the whole gain and the gain per cent. Ans. $129.81$ ; $39.51 j. 6. If soap cost £3, 5. per cent., how must it be sold to gain 10 per cent, and how to gain 20 and 30 per cent. ? Ans. £3, 11. 6. ; £3, 18. ; £4, 4. 6. 230. From the gain or loss per cent, and the selling price^ to find the first cost. Rule. As £100 or $100 together with the gain per cent., or diminished by the loss per cent, are to £100 or $100, so is the selling price to the prime cost. Ex. 1. What was the first cost of flax seed, which beinf^ sold at $14.10 per hogshead, the seller gained 13 percent? As $100 + 613 : $100 : : $14.10 : $12.47f In this example it is evident, that what cost $100 is sold for $113, and the analogy used above is no more than this : as the selling price 8113 is to its first cost #100, so is the selling price $14.10 to the corresponding first cost $12.47f . Ex. 2. If 15 per cent, be gained by selling sugar at £2, 10. per cwt., how much is gained by selling it at £2, 5. 6. per cent. ? As £2, 10. : £2, 5. 6. : : £115 : £ld4, 13. and £104, 13 — £100 = £4, 13. Ans. s if I ! ;! ■! 266 RECIPROCAL ntOPORTIOX. This question might hare been worked by finding the first cost, and thence the selling price, as in the preceding cxauples. Exercises. 1. If 11 per cent, be lost by selling 128 yards of broad- cloth for £98, 18. 8., what was the first cost per yard? Ans. £0, 17. 4^. 2. If a book be sold for $19.95, and 17 per cent, be gained, what was the first cost? Ans. $17.05. 3. "What was the first cost of tar, which being sold at £0, 17. 4. per barrel, the merchant loses 9 per cent., and how much is lost on the sale of 63 barrels ? Ans. 19s. 0^. and £5, 8. 0. 4. If by selling muslin at $0.62^ per yard 2 per cent, be lost, bow must it be sold to gain 25 per cent. ? Ans. $0.80 nearly. DIVISION INTO PROPORTIONAL PARTS AND RECIPROCAL PROPORTION. 231. To divide a given number i^ito parts which shall be jn'oportional to certain other given nuinlers, ■] Rule. State thus : As the sum of the given parts : any one of them : : the entire quantity to be divided : the cor- responding part of it. When all the parts except one, have been determined, that one may be found by adding the rest together and taking the sum from the number to be divided. It is bet- ter however, to find them all by proportion, as the operi^ tion cap then be tested by adding them together. R^aPROCAL PROPORTION. 2C7 ng the seeding r broad- ard ? 7. 4i. icnt. be L7.05. sold at at., and , 8. 0. er cent. Lcarly. shall be [rts : any the cor- jrmined, bher and Lt ia bet- |e opern- Ex. 1. Divide a purse containing IGs. 8d. among 3 boys according to their ages : A is 10, B is 12, and C is 14. Required, their sliares? A = 10 years of age. B =12 C =14 11 ♦J As 36 : 10 : : 16s. 8d. As 36 : 12 : : I63. 8d. As 36 : 14 : : 16s. 8d. 43. 7J. =: A's share. 5s. 6J. = B^s 6s. 5|-. = C's 16s. 8d. proof. Ex. 2, Divide $1000 among 4 persons A, B, C, and D, in the proportions of J, ^, I, and J, i=:i30 I = j9 y CO L. C. M. i = 12 As 77 : 30 : : $1000 : $389.61 A's share. The other shares may be found in the same mannner. 232. To divide a given quantity into parts which shall be in inverse or reciprocal proportion to the give7i numbers. Rule. Invert the terms on which the inverse proportion depends, proceed with these fractions as in the former rule. When other conditions in direct proportion are given, multiply each of these by the inverted term, and proceed with these fractions as before. Ex, A gentleman left $4000 to his tlu*ee children to be divided in inverse proportion to their ages, so that the younger may receive the greater share ; their ages are 10, 12, and 15 years, what must be the share of each child? iV = 6) tV = 5 V 60 L. C. M. tV = 4J As 15 : 6 : : $4000 : $1600 = share of youngest. As 15 ; 5 : : $4000 : $1333.33^= „ second. As 15 : 4 : : $4000 : $1066.66f = „ eldest. $4000.00 proof. , nn n i-^i '1 ' I,. t 268 RECIPROCAL PROPORTION'. i I Exercises. 1. Divide 398 into three parts, which shall be to one another as the numbers 5, 7, and 11. Ans. 86^1, 121^3^, and IOO^-Stj. 2. Divide 80 Miles into 4 parts, proportional to the numbers 10, 9, 8, and 7. Ans. 23y9y, 21-j3y, 18|t, and 1C^^%■. 3. How much copper and how much tin will be requir- ed to make a cannon weighing 16c. Iq. 20 lbs. ; gun-metal being composed of 100 parts of copper and 11 of tin? Ans. 14c. 3q. 5-j7fylbs, ; Ic. 2q. Uj\\. 4. 76 parts of nitre, 14 of charcoal, and 10 of sulphur, compose gun-powder ; how much of these ingredients will form 4650 lbs. of powder. Ans. 35341bs., 6511bs., 4651bs. 5. Divide 1065 into parts which shall be to each other in the ratio of 3, 5, 7 ; and also into parts which shall be in the ratio of ^, J, ^. Ans. 213, 355, 497 ; 525, 315, 225. 6. The standard, gold coin of England and her colonies is 'jaade of gold 22 carats fine, and a pound Troy of this metal yields 46f § sovereigns ; what weight of pure gold is there in 100 sovereigns ? Ans. lib. lloz. 10 dwt. 20^^^ grs. 7. Divide $1200 among three persons, so that the first shall have twice as much as the second, and the third twice as much as the other two together. Ans. $266.66^, $133.33^, $800. 8. British silver coin consists of 37 parts of pure silver and three of copper ; how much of each does the florin con- tain, each pound, Troy weight, being coined into 66 shil- lings? Ans. 8 dwt. 9y\ qrs. and 16-j*j- grs. 9. A common consisting of 200 acres, is to be divided among three adjoining proprietors in direct proportion to the extent of their estates, but in inverse proportion to their value ; A possesses 200 acres, valued at £60 per acre ^ B 300 acres, valued at £80 per acre, and C 600 acres at £40 per acre. Required, their shares of the common? Ans. A 30ac. Oro, SOf^po., B 33ac. 3ro. 33f]po., C. 135ac. 3ro. ISJ^po. 1l|!u„ FELLOWSHIP OR TARTNERSHir. 269 to one 90/3- I to tlie } requir- m-iuetal in? 4._3_8_ *l 1 1* sulphur, jnts will 651bs. cli other shall be I, 225. colonies 1 of this 5 gold is % grs. the first rd twice $800. •e silver >rin cou- lee shil- irgrs. 1 divided Irtion to *tion to ►er acrev Lcres at m? [po., C. ! FELLOWSHIP OR PARTNERSHIP. 232. Fellowship or Partxership is a method by which tlic respective gains or losses of partners in any mercantile transactions are determined. Fellowship is divided into Simple and Compound Fel- lowship : in the former, the sums of money put in by the several partners continue in business for the same time ; in the latter, for ditferent periods of time. Simple Fellowship. 233. 'Wlie,n the shares are in proportion to the stock or claims. Rule. — State thus : — As the whole stock : the whole gain or loss : : the stock of any partner : his gain or loss. This rule is merely a particular application of the one given in Art. 10, and therefore requires no separate illus- tration. Note. — The estate of a Bankrupt may be divided among his creditors by the same method. 234. When interest is alleged upon the stocks and the part' ners have either equal or propoi'tional shares. Rule. — Find the interest of the several stocks, subtract the amount from the gain, and divide the remainder equally among the partners, or according to their claims, as in the former rule. Exercises. 1 . A's and B's stocks are $1500 and $1 700 respectively Required the share of each in a gain of $960. Ans. A's share, $450 ; B's, $510. 2. A's stock £1750, B's £1250 ; whole gain £565, 12. 0. Required each person's share. A's, £329, 18. 8. ; B's, £235, 13. 4. 3. C's stock, £475, 15. 8., D's, £346, 12. 4., E's, £396, 17. 6. ; whole gain, £279. 4. 10. Ans. C's gain, £108, 19. ^., D's, £79, 7. 7|., and E's, £90, 17. 10^. *.•, J I I If ;f If . \ ¥ 'li ;: 11 i- i it 270 COUPOUKD FOLLOWSHIP. 4. Three merchants enter into partnership ; A advancecl $2720, B $2320, and C $1200 ; their net gain at the end of the year is $2080. How much will each partner draw, allowing 5 per cent, for stock, and dividing the remainder equally among the partners ? Ans. A, $725.33J. ; B, $^705.33^-. ; C, $G49.33-i. 5. Two merchants join stocks : W advances £2400, and M £3000 ; the interest on each man's stock is to be at 4 per cent. ; and W, for superintending the warehouse, is to have three shares, while M, acting as traveller, is to have five. Required each man's share of the first year's gain, which was £850? Ans. W, £333, 15. ; M, £516, 5. 6. R and I are in partnership ; R possesses $4000 of the stock and I $4800, for which they receive 4 per cent. ; the}'- admit C, their foreman into the firm, and, having no stock, he is only allowed one-fourth share of the profits, the remainder being equally divided between R and I. Re- quired each man's share of $5800,00, the first year's gain ! Ans. R, $2203,00: I, $2235.00; C, $1302.00. COMPOUND FELLOWSHIP. 235. Rule. — Reduce u;! the times into the same denom- ination, and multiply each man's stock by the time of its continuance, and then state thus : As the sum of all the products : each particular product : : the whole quantity to be divided : the corresponding share. Ex. A and B enter into partnership; A puts in $2400 for 13 months, and B $3200 for 10 months. Required the share of each in a gain of $2600. A's stock, $2400 X 13 = 31200. B's $3200 X 10 = 32000. 63200. Sum. Ag 63200 : 31200 : : $2600 : $1283.54f , A's share. Ag 63200 : 32000 : : 2G00 : $1316.45f , B's share. $2600.00. Proof. ALLEGATION. 271 ivanced e end of r draw, nainder 1.33-1. too, and be at 4 sc, is to to have p's gain, 16,5. HOOO of sr cent. ; ,ving no profits, II. Re- :'s gain ! G2.00. denom- le of its oduct : ; g share. $2400 lired the The reason for the above process is evident from the con- sideration, that a stock of $2400 for 13 months, would be equivalent to 13 times $2400 for 1 month : and one of $3200 for 10 months, to 10 times $3200 for 1 month. Hence if these increased stocks be employed, it is evident, that since the times are then to be regarded as equal, the work will proceed in the same manner as in simple fellow- 3hip. Exercises. 1. A's stock $1120 for 5 months, B's, $1066.665 for 6 months ; whole gain, $1326.50. Ans. A's gain $619.03^ ; B's, $707.46f . 2. A's stock £1 70 for 8 months, B's, £280 for 6 months ; whole gain, £250. Ans. A's gain £111, IG. 10|- ; B's, £138, 3. IJ. 3. A and B enter into partnership ; A contributes £3000 for 9 months, and B £2400 for 6 months ; they gain £1150. Find each man's share of the gain. Ans. A's share £750, and B's £400. 4. There were at a feast 20 men, 30 women, and 15 s«^rvants ; for every 10 shillings that a man paid a woman paid 6 shillings, and a servant 2 shillings ; the bill amounted to £41. How much did each man, woman, and servant pay? Ans. Each man £1 ; each woman 12s. ; and each servant 4s. ALLIGATION. re. re. 236, The process of finding the mean rate or price of a mixture compounded of several ingredients, or the quan- tity of ingredients of different lates necessary to make a mixture of a required rate, or quality, is called Alligation. Alligation is usually divided into two kinds. Medial and Alternate, ii ! •hi I I mU m i ' I, / ^ n s 272 ALLEGATION. Alligation Medial. 237. Alligation Medial is the process of finding the mean price of two or more ingredients, or articles, of difier- ent values. From the quantity and rate of several ingredients to he mixed ^ to find the mean rate. Rule. — Multiply each quantity by its rate, and divide the sum of the products by the sum of the quantities. Ex. A farmer mixed 20 bushels of wheat at 5s. per bushel, and 50 bushels at 3s. per bushel, with 10 bushels at 2s. per busliel. What is a bushel of this mixture worth ? iiO X 5 = 100 50 X 3 = 150 10 X 2 = 20 80 80)270 3 s. 4^d. Ans. Exercises. 1. A grocer mixes 10 lbs. of tea at 4s., 12 lbs. at 4s. 9d., 18 lbs. at 5s. 2d., and 20 lbs. at 5s. 6d. What is a pound of this mixture worth ? Ans. 5s. 2. A goldsmith melted together 8 oz. of gold, 16 carats fine, 10 oz. of 18 carats fine, 18 oz. of 20 carats fine, and 4 oz. of allo}'^. Of what fineness was the mass? Ans. 16-j-\r carats. 3. A vinter mingles 15 gallons of canary, at $1.60 per gallon, with 20 gallons at $1.50 per gallon, 10 gallons of sherry, at $2.1.) per galloi , and 24 gallons of white wine, at $0.80 per gallon. What is the worth of a gallon of this mixture? Ans. $ 1. 3 7:|^ nearly. Alligation Alternate. 238. Alligation Alternate is the process of finding what quantity of anj^ given number of ingredients, whose prices are given will form a mixture of a given mean pnce. ing the r difier- ts to he divide les. 5s. per bushels worth? at 4s. at is a 5s. 16 carats ine, and irats. .60 per llloiis of pe wine, of this jarly. I finding whose |n2>n*ce. ALLEGATION. 273 Alligation Alternate embraces three varieties of examples, which are sometimes called, Alternate, Partial, and Total. 1. To find the quantity of each ingredient, when its price and that of the required mixture are given. Rule. — ^Let the rates of the ingredients, all in the same denomination, be written in a line ; and let the mean rate in the same denomination be written above them. Take one rate which is greater, and one which is less, than the mean rate, and write the difference between each of them and the mean rate below the other. Proceed thus with the rates two by two, if there be more than two, till one or more differences stand below each. Then, if only one difference stand below any rate, it will be the quantity required at that rate ; but if there be more than one, their sum will be the required quantity. Note. — The connecting or linking of the rates with crooked or curved lines, in the use of this rule, is attended with no advantage, and has an awkward appearance. Should that method be preferred , however, it can present no difficulty, as each rate less than the mean rate is to be con- nected with one greater, and each greater with one less, and the ditter- ences are to be set below the rate to which the line directs. Ex. A man mixed four kinds of oil, worth 8s., 9s., lis., and 12s. per gal. ; the mixture was worth 10s. per gal. Required, the quantity of each. 10 8 9 II 12 Here the mean rate, 10 shillings, is set above the other rates. Then the difference between 12 and the mean is set below 9, and the difference between 9 and the mean is set below 12. Again, the difference between 11 and the mean is set below 8, and the diiference between 8 and the mean is set below 1 1 . Hence we find, that for 1 gallon at 8s. we must take 2 at 9s., 2 at lis., and 1 at 12s. AYith respect to the reason of the operation, it is obvious from the previous article, that 1 gallon at 8s., and 2 gal- lons at lis. each, would make a mixture worth 10s. per gallon ; and likewise that a mixture of 1 gallon a^ ^"^m < 1 ''\i. m hi f! 271 ALLIGATIUN. I :il: Li i r i ' 1 and 2 at 9s., would be worth the same per gallon ; and ifc is evident, that both mixtures, taken together, must mako a mixture of the same value still, per gallon : and in th^ same way ever}' operation in this rule may be explained. NoTK.— It h rwa-^ToBl that other answers may be obtained by connect- ing the prices in a difierent manner. It is also manifest, if we multiply or (lif ido the answers already obtained by any number, the results wiV fulfil the conditions of the question ; consequently the number ot answers Is unlimited. (2.) When the quantity of one of the ingredients and the mean price are given. Rule. — lind the difference between the price of each ingredient and the -mean price, as before. Then, As the difference of that ingredient whose quantity ia given : the rest of the differences severally : : the quantity given : the quantity required of each ingredient. Ex. How many pounds of sugar at 10, and 15 cents a pound, must be mixed with 20 lbs. at 9 cents, so that the mixture may be worth 12 cents a pound? 12 9 10 15 3 3 2 3 3 5 As 3 : 3 : : 20 : 20 lbs. at 10 cents a pound. As 3 : 5 : : 20 : 33^ lbs. at 15 cents a pound. It appears, therefore, that 20 lbs. at 10 cents, 33^ lbs. at 15 cents, and 20 lbs. at 9 cents, will compose a quantity worth 12 cents per pound, at an average. This question belongs to what is usually called Alligation partial. (3.) WJien the quantity to he mixed and the mean price of the required mixture are given. Rule. — Find the difference between the price of each ingredient and the mean price of the required mixture, as before. Then, ; and iti st inako X in th0 ainecl. r connect- I multiply ;sult8 wiV )t answers 5 and the of each antity ia quantity 5 cents a that the 1^ lbs. at quantity lUgation in price lof each Iture, as ALLIGATION. 273 As the sum of the differences : each particular difference : : the whole quantity to be mixed : required of each ingredient. Ex. A grocer has raisins worth 8, 10, and 16 cents a pound : how many of each kind may be taken to form a mixture of 112 lbs. worth 12 cents a pound? 12 8 10 16 A 4 2 4 4 6 or, dividing by 2, 2 2 3 As 7 : 2 : . 112 : 32 lbs. at 8 cents a pound. As 7 : 3 : : 112 : 48 lbs. at 16 cents a pound. It appears, therefore, that 32 lbs. at 8 cents, the same quantity at 10 cents, and 48 lbs. at 16 cents, will form a compound of 112 lbs. worth 12 cents per lb. This question belongs to what is usually called Alligation total. Note.— To prove questions in this ruUy find the value of all the in- gredients at their given prices; if this be equal to the value of the whole mixture at the given price, the work is right. Exercises.* 1. A goldsmith has gold of 18, 20, 22, and 24 carats fine : how much of each may be taken to form a mixture 21 carats line ? Ans. 3 of 18 car., 1 of 20, 1 of 22, and 3 of 24 car. fine. 2. How much wool at 20, 30, and 54 cents a pound must be mixed with 95 lbs. at 50 cents, to form a mixture worth 40 cents a pound ? Ans. 133 lbs. at 20 cents, 95 lbs. at 30 cents, and 190 lbs. at 54 cents. 3. How much water must be added to a cask of spirits containing 84 gallons, worth 13s. 6d. per gallon, to reduce the value to lis. 4^d. per gallon? Ans. 15^^^ gal. 4. A grocer having four sorts of tea, at 50 cents, 60 cents, 80 cents, and 90 cents per lb., wishes to formamix- i ig V V "r 1 1; f-: a:' 'I I * »■ 276 CONJOINED PROPORTION. ture of 87 lbs., worth 70 cents per lb. Whr.t quantity must there be of each sort ? Ans. lA} lbs. at 50 cents, 29 lbs. nt 00 cents, 29 lbs. at SO cents, and 14^ at 9u cents. CONJOINED PROPOKTION. 239. "When each antecedent of a compound ratio is equal in value to its consequent, the proportion is called Conjoined Proportion. Conjoined Proportion is often called the cliain rule. It is chiefly used in comparing coins, weights and measures of two countries, through the medium of those of other countries, and in the higher operations of exchange. The odd term is called the demand. Rule. — Taking the tenns in pairs, place the first term on the left hand for the antecedent, and its equal on the right for the consequent, and so on. Then if the answer is to be of the same kind as the first term, place the odd term under the antecedents ; but if not, place it under tlie consequents. Cancel the factors common to both sides, and if the odd terra falls under the consequents, divide the product of the factors remaining on the right by the product of those on the left, and tne quotient will be the answer ; but if the odd term falls under the antecedents, divide the product of tho factors remaining on the left by the product of those on the right, and the quotient will be the answer. Note. — In arranging the terms, it should be observed that the first antecedent and the last conseqibent will always be of the same kind. All the quantities of the same kind must be reduced to the same denom- ination, if they be not bo already. I ISiM m ,y must cents, I cents. atio is 3 called ule. It leasures )f otlier 3. The term on he right is to be under sqiients. the odd It of the lose on It if the )duct of If those the first find. I denom- CONJOINED PROPORTION. 277 Ex 1. If 20 lbs. Nova Scotia make 12 lbs. in Spain; and 15 lbs. in Spain 20 lbs. in Denmark; 40 lbs. in Den- mark 60 lbs in Russia: how many pounds in Russia are equal to 100 lbs. Nova Scotia? rroceediiii? by tht> Rule given above. HO lb. N. S. = 3, n lb. Spain. 10 U). Spain = *;20 lb. Den. 10 lb. Den. =z 4, ^0 lb. Russ. 10, UO 10 X 4 X 3 = 120 lbs. Ex. 2. If $18 U. S. are worth 8 ducats at Frankfort; 1 2 ducats at Frankfort 9 pistolep ' Geneva ; and 50 pistoles at Geneva, 24 rupees at Boni,^>ay how many rupees at Bombay are equal to $100 U States? Proceeding by the Rule given above. %U ^,4 00 U, 2 100, 2 2 X 2 X 4 = 16 rupees. Note. — Questions in this rule may be proved by reversing the opera- tion, taking the consequents for the anteoedetits, and the answer for the odd term, or by as many simple statements as the question requires. Exercises. 1. If 10 yds. at New York make 9 yds. at Athens ; and 90 yds. at Athens, 112 yds. at Canton ; how many yds. at Canton are equal to 50 yds. at New York? Aus. 56 yds. at Canton. 2. If 50 yds. of cloth in Boston are worth 45 bbls. of flour in Philadelphia ; and 90 bbls. of flour in Philadelphia 127 bales of cotton in New Orleans ; how many bales of cotton at New Orleans are worth 100 yds. of cloth in Boston? Ans. 127 bales N. O. 3. If 140 braces at Venice are equal to 156 bonnets at Leghorn ; and 7 bonnets at Leghorn equal to 4 yds. of cloth at Halifax ; how many braces at Venice are equal to 16 5'ds. of cloth at Halifax? Ans. 25^. If ) i' i si' 'i I ]U IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I I^|2j8 |25 140 |2.C III '-2^ III '-^ u^ < 6" ► Photografiiic Sciences Corporation ^ \ ^ \\ *.^ <«> ^.>. «^ 33 WIST MAIN STRUT WnSTIR.N.Y. I4S80 (716) •72-4503 4^ \ 278 EXCUANOE. EXCHANGE. 240. Exchdnge, in commerce, signifies the receiving or paying of money in one place, for an equal sum in another, by draft or bill of exchange. A Bill of Exchange is a written order, addressed to a person, directing him to pay, at a specified time, a certain sum of money to another person, or to his order. The person who signs the bill is called the drawer or maker; the person in whose favor it is dra^^n, the buyer or remitter; the person on whom it is drawn, tlie drawee; and after he has accepted it, the axepter; the person to whom the money is directed to be paid, the payee; and the person who has legal possession of it, the holder. On the reception of a Bill of Exchange, it should be immediately presented to the drawee for his acceptance. 241. The acceptance of a bill, or draft, is a promise to pay it at maturity or the specified time. The common method of accepting a bill, is for the drawee to write his name under the word accepted^ across the bill, either on its face or back. The drawee is not responsible for its pay- ment, until he has accepted it. If the payee wishes to sell or transfer a bill of exchange, it is necessary for him to endorse it, or write his name .on the back of it. If the endorser directs the bill to be paid to a particular person, it is called a Special endorsement and the person named, is called the endorser. If the endorser simply writes his name on the back of the bill, the endorsement Is said to be blank. When the endorsement is blanks or when a bill is drawn payable to the bearer^ it may be trans- ferred from one to another at pleasure, and the drawee is bound to pay it to the holder at maturity. If the drawee, or accepter of a bill, fail to pay it, the endorsers are re- sponsible for it. K ring or nother, ;cl to a certain awer or e buyer drawee; rson to and tlie lould be ince. )mise to lommon rite his X on its its pay- :cliange, Lame .on irticular person simply k'sement lank^ or )e trans- :awee is I drawee, are ro- rXCUANGE. 279 242. When acceptance or payme?it of a bill is reftised,' the liolder should duly notit* the endorsers and dr. wer of the fact by a legal protest, otherwise they will not be respon- sible for its payment. A protest is a formal declaration in writing, made by a civil olficer, termed a Notary Public^ at the request of the holder of a bill, for its 7ion-accepta7ice, or non-payment, Tlie time specified for the paj^ment of a bill is a matter of agreement between the parties at the time it is negotiated. Some are payable at sight, others in a certain number of days or months after sight, or after date. 243. Bills of exchange are usually diAided into inland and foreign bills. When the drawer and drawee both leside in the same country they are termed iidand bills or drafts; when they reside in different countries, foreign hills. In negotiating foreign bills, it is customary to draw three of the same date and amount^ which are called the Firsts Second and Third of Exchange; and collectively, a Set of Exchange, These are sometimes sent by different conveyances, and when the first that arrives, is accepted, or paid, the others become void. The object of this arrangement is to avoid delays, which might arise from accidents, miscarriage, &c. FoRsi OF A. Foreign Bill op Exchange. No. Halifax, 4th May, 1868. Exchange for £1000, 0. 0. Stg. Sixty days after sight of this First Exchange, second and third of the same tenor and date unpaid, pay to the order of Charles Rlngland & Co., One Thousand Pounds Sterling, value received, which place to the account of JOHN L. NEWCOMB. To Messrs. John Felthau & Co., Bankers, London. 1 ir I 'J I T- ^ 2«0 ^ H I!' 1-1 it EXCIIANrjE. FoKM ON AN Inland Bill or Draft. SlOO Halifax, Ma^^ 4th, 18C3. Thirty daj's after date, pay to the order of Messrs. Newman & Co., One Hundred Dollars, value received, and charge the same to MACEY & BROWN. To Messrs. Blair & Brown, ) Windsor. ) 244. The term par of exchange^ denotes the standard by which the comparative worth of the money of different countries is estimated. It is either intrinsic or commercial. The intrinsic par is the real value of tlie monej' of dif- ferent countries, determined by the vmght and purity of their coin. The commercial par is n nominal value, fixed by law or commercial usage, by which the worth of the money of different countries is estimated. The intrinsic par remains the same, so long asthes^flTi^- ard coiiis of each country are of the same metals^ and of the same weight and purity; but in case the standard coins are of different metals, the. intrinsic par must varj"^, as the com- parative values of the metals vary. The commercial par is conventional, and may at any time be changed by law or custom. • 245. By the term, course of exchange^ is meant the cur^ rent price which is paid in one place for bills of a given amount drawn on another place. The course of exchange is seldom stationary or at par. It varies according to the circumstances of trade. When the balance of trade is against a country, that is, when the exports are less than the imports, bills on the foreign country will be above par, for the reason that there will be a greater demand for them to pay the balance due abroad. On the other hand, when the balance of trade is in favor of a country, foreign bills mil be below par, for the I'eason that few will be required. KXCIIAN(JE. 2^51 18G3. Messrs. etl, and dard by iifferent imercial. y of dif- \urity of y law or loney of le stand' id of the ioins are ihe com- at any Ithe cur- Si. given at par. When jrhentho [country , greater I On the ror of a )n that It should be remarked that tlie course of exchange can never exceed very nnich the intrinsic par value; for it is plain that coin or bullion^ instead of bills, will be remitted, whenever the course of exchange is such that the expense of insuring and transporting it from the debtor to the creditor country, is less than the premium for bills, and the exchange will soon sink to par 246. All the calculations in exchange may be performed by the rule of proportion : and the operations may often be abbreviated bj' the method of aliquot parts. General Rule. — ^Place, as the second term in the analo- gy, that sum whose value is to be found in the money of another countrj' ; make that term of the rate which is of the same kind with the second term, the first term of the analogjs and the remaining term of the rate, the third term : then work the analogy in the usual way. English and American Exchange. 247. Formerly £90 sterling were worth £100 British North American Currency, this is called the old par value; now, however, £90 are worth more, and as the rates of ex- change on Great Britain are reckoned, in Nova Scotia, New Brunswick, Canada, and the U. States, at a certain per cent, on the old commercial par instead of the new, the rate of exchange must reach a nominal premium before it is at par, according to the new standard. This nominal premium is just the difference per cent, between the old and new par value. In Nova Scotia, the new par value is 12^ per cent, greater than the old. Hence, when bills of exchange on England are selling at 12^ per cent, premium, they are said to be at par. In New Brunswick and Canada, it is 9J per cent, more than the old par value ; consequently when exchange is 9^ per cent, premium, it is said to be at par in those places. In the U. States, according to the wd jwr, the value of a pound sterling is $4.44|, as fixed by act of Congress in 1799. According to the new par it is $4.84|. Hence, b I id L !l i 1 Ill I ' 282 EXCHANGE. per cent, premium, added to $4.44 J, will give the new par value of £l sterling. 248. To find the value of any sum stg. in Nova Scotia, New Brunswick, Canada, and U. States currency. Rule. — State tlms :■ — As 90 : 100 -f- the rate of premium :: $4, the result being multiplied by the number of pounds sterling will give the equivalent number of dollars. Or, To $4.44 J. add the premium, at the given rate, the sum being multiplied by the number of pounds stprling, will give the equivalent number of dollars. Note. — Exchange whioh is varionsly quoted in the newspapers as llSi, 113, 170, &c., means 124, 13, 70, &c., per cent, premium on the old par value. Ex. A merchant in Halifax negotiated a bill on Liver- pool, G. B., for £1000 stg., at 13^ per cent, premium: what did he pay for it ? riBST MSTBOD. As 90: 113^:: $i 2 2 l$0 227 45 And ^ X £1000 = $5044.44J N. S. Currency. eVOOMP METHOD. $4,444' .60 ^.044' 1000 $5044.44^ U. S. Currency, $4,444' 13^ per cent. 5777? 2222' .59999' or .60 in- EXCHANGE. 283 new par ;a Scotiat remium:: f pounds s. Or, , the sum ling, will )ersaB U2i, the old par on Liver- premium : Currency. [r cent. .60 We might hare used 4| and 113^ instead of $4,444' and 13^ per cent, thus : — 113 J = 113.5 per cent. 4|_ 454.0 |of 1 =iof4= 50.444' 5.04 444 X 1000 = $5044.44|. THIRD METHOD. £1000 X $4,441 = $4444.444' at old par value. Then $4444.444' X .135= 599.999 the premium. $5014.444' 249. To find the value in sterling moTiey of any sum : currency. Rule. — State thus: — As 100 + the rate per cent, pre- mium: 90:: sum: currency, (which, if in dollars, must be divided by 4) : the equivalent in pounds and decimal of a pound sterling. Or, Multiply the amount currency by 100 and divide the pro- duct of 4 J and the rate per cent, of exchange, the quotient will be the equivalent in pounds and decimal of a pound sterling. Ex. A merchant paid $1650 for a bill of exchange on London, at 13 J per cent, premium : for what amount ster- ling was the bill drawn? FIBST METHOD. As 113J : 90 : : $1650 : 1308.37 and 1308.37 -4- 4 = £327.092 stg. = £327, 1. 10. stg. ; SKCOND UETHOD. $1650 100 ■} i ; Mi i li V 1 f 1 i] 113J X 4t =504J) 165000 (£327, 1, 1&. stg. N 284 ESCHANOE. i^ k ,i I I I I Exchange between England and Newfoundland. 250. In Newfoundland accounts are kept in pounds, shillings, and i)cnce. Rates of exchange are reckoned at a certain rate per cent, on the sterling. Trade between Newfoundland and England is in a state of equilibrium when exchange is 120 per cent., or 20 per cent, premium, which, if added to the sterling, will give the equivalent number .of pounds, &c., Newfoundland currency. Ex. 1. What will a bUl cost on London, for £3486, 10. at 21^ per cent, premium? £3486, 10. 0. 20 = i of 100 697, 6. 0. 1 = aV of 20 34, 17. 3^. i = iof 1 17, 8.7?. £4236, 1. Hi. Currency. Ex. 2. When exchange on England is at 21^ per cent, premium, how much sterling can bo obtained for £4236, 1. 11 J. currency? As 121^: 100:: £4286, 1. 11». £4236, 1. llf , or £4236, 2. 0. nearly 200 2s. = T»o 847200 20 243)847220(£3486, 10. 0. SterUng. Exercises. 1. A merchant in Paris draws a bill of 1500 francs upon a merchant in London for goods supplied. What sterling money will the latter have to pay, exchange being 24.25 francs for £1 sterling? Ans. £61, 17. Iff. 2. AVhat is the course of exchange between London and Lisbon when 594 milrees, 480 rees, are received for £158, 16. 9. ? (1 mikee = 1000 reers,) Ans. 64.124d. or 5s. 4id. nearly. EZCHAN«K. 285 LAND. pounds, rate per [and and ^e is 120 id to the ids, &c., 5486, 10. 3. A merchant negotiated a bill on Glasgow for £12C7, 10. at 12 J per cent, premium. What did he pay for it? Ans. $6337.50. 4. Merchant in Liverpool, G. B., draws a bill of £1020, 0. 0. upon a merchant in Halifax for goods supplied. What amount of currency must the latter pay, exchange being at 15 per cent? Ans. $5213.33^. 6. A merchant in London draws a bill of $1867 upon Boston. What amount sterling will he have to pay, ex- change between Boston and London being 150 per cent. ? Ans. £ 280, 1, 0. i« per cent. £4236, 1. sterling. [ncs upon sterling ig 24.25 idon and )r £158, Abbitration of Exchange. 261. When the course of exchange between the first place and the second, the second and third, the third and fourth, &c., of any number of places, are given ; the method of finding the course of exchange between the first place and the last, corresponding to these courses, or of valuing any sum of the money of the first place in that of the last, through the medium of the others, is called Arbitration OF EXCHANQE. As the actual course of exchange between the first place and the last, is almost always, from various circumstances, different from the arbitrated ourse, this method is of use in enabling a merchant, in one place, to discover whether he should draw and remit directly between his own place and another, or circuitously, through other places. When there is but one intervening country, the operation is termed Simple Arbitration, when more thafi one, it is termed Compound Arbitration. Problems in Arbitration of exchange may be solved by conjoined proportion, or by one or more analogies in simple proportion. ^ J learly. 286 BX0BAM«1. I Ex. 1. What is the value of $1 N. S. Currency in New Brunswick, exchange between England and Nova Scotia at 18 j- per cent, and between England and New Brunswick 9^ per cent ? Proceeding by the Rule given for conjoined proportion. 112* =90 n" = 109A _ Jl 226 219 iii = $0.97^ N. B. Currency. Hence the rate of exchange between Nova Scotia and New Brunswick is 2% per cent, discount. Ex. 2. What is the value of a franc in N. S. Currency, exchange between England and Nova Scotia at iSj- per cent., and between England and France at 24 francs, 87 centimes, per pound sterling ? $113J N. S. = £22, 10. or 690 sterling. £l stg. = 24 francs, 87 centimes. 1 franc. Reducing the quantities of the same kind to the tame denomination. , 113^ = (»^ 11 2 = 2487 20 200 4540 27367 = $0.16^ Exercises. 1. A trader in London owes a debt of 608 pistoles to one in Cadiz : is it more advantageous to him to remit di- rectly to Cadiz, or circuitously through Fran'^e. The ex- change being £1 = 25. 4 francs, 19 francs = 1 Spanish pistole, 4 Spanish pistoles = £3. Ans. Through France. 2. If the exchange of New York on London is 8 per cent, prem., and that of Amsterdam on London is 12 florins for £1, what is the arbitrated course of exchange between New York and Amsterdam ; that is, how many florins are eqoal to $1 U. S. Ans. $1 s 2^ florine. syinNew , Scotia at QBwick 9^ roportion. •ency. A and New Currency, at X^ per francs, 87 the same )i8toles to remit di- Tlie ex- II Spanish iFrance. is 8 per 12 florins between lorins are Iflarixui. INVULLTIOK. INVOLUTION. 1187 252. A power of any number is the product obtained by the continual multiplication of that number, taken a certain number of times as factor. A number, in relation to an}' power of it, is called the root of that power. .*, When the proposed number is used twice as factor, the product is called the Second Power, or the Square, of that number ; when three times, the Third Power, or Cube ; when four tima, the Fourth Power, &c. Powers are often denoted by writing after the proposed number, a little higher, the number which shows how often the proposed number is repeated as factor. This number is called the Index, or the Exponent, of the Power, Thus 5 X 5, or 25, is the second power, or square, of 5, and may be "\vritten 5^, where 2 is the index ; while 7X7 X 7 X 7, or 2401, is the fourth power of 7, and may bo written 7* , where 4 is the index, &c. Also, 5 is the second or square root of 25, and 7 is the fourth root of 2401. The method of finding any assigned power of a given number, or as it is also expressed, the method of raising a number to any proposed ])ower, is called Involution, 253. To Jind any assigned power of a given number ; to raise a given number to any proposed power. Rule. — ^Find the continual product of the given number repeated as a factor, as often as there are units in the index of the proposed power. The process may often be abbreviated bj' multiplying together powers already found. In this case, the index of the power thus found is equal to the sum of the indices of the powers multiplied together. "When the given number is either wholly or partly a decimal, the operation may often be much abbreviated, by the rule for contracting the multiplication of decimals given in Art. 119. M 2C8 INV )Lt'TlON. if^ '^ :ii I Ex. 1 . Required the fifth power of 23. Multiply I II Here, l)y multiplying 23 by itself, wo find 529 for the second power of 23. By multiplying this by 23, we get 121G7forthe third power. By pro- ceeding in like moaner, we find the fourth power to bo 279841, and the fifth to be 6436343. Multiply { ^11 Multiply j l"^ Multiply I ™J Ans. 6436343 = 1st power, 2d " 8d " 4th " 5th " The answer might also have been found by multiplying the second power, 529, by itself, and the product of 279841, ■which is the fourth power, by 23. The same result would also be obtained by multiplying the third power by the second. Ex. 2. Required the fifth power of J. The fifth power of 3 is 243, while that of 8 is 32768. The answer, there- fore, is ir5}J^. The reason of this is evident from the method of multiplying fractions. Ex. 3. AVhat is the third power of 1 J ? This, by reduc- tion to an improper fraction, becomes J, and by involving the numerator and denominator each to the third power, we find the answer ^ , or 1 f ^. Each of the last examples might have been worked by reducing the fractions to decimals, and then working by the general rule. Ex. 4 . Required the sixth power of 1 . 1 2 true to 5 places of decimals. By raising this to the third power, in the way already shown, we find 1 .404928, and this being multiplied by itself as in the margin, we find for the sixth power 1.973822. 1.404928 = Sd power. 829404.1 1404928 56J971 5620 1264 28 11 1.973822 = 6th " = 1st power. = 2d = Sd i4 tl = 4th »* = 5th *» ' multiplying ct of 2 79841, result would »y the second. e fifth power iswer, there- int from the lis, by reduc- )y involving hird power, worked by working ])y |e to 5 places Sd power. 6th it BTOLUTION — SQUAltl ROOT. KXKRCIRES. 289 Involve the following 7iu7nb€r3 to the powers denoted by their respective indices. 1. 2. 8. 4. 2880* 185» 9* 86* Ans. 8294400 •• 2460376 " 887420489 •* 4704270176 5. (»)• G. (2})» 7. (8?)* 8. 1.06»> Ans. ViS 1 " 4.638089 j EVOLUTION. 264. EyoLxrrxoN is the method of finding, or, as it Is usually termed, extracting, an assigned root of a given number. The Index of a root is n fraction whose denomination denotes the order of the root, and whose numerator is unity. The root of a number is also expressed by prefixing to the number the sign Vi '^i^h the number above it, which denotes the order of the root. In case of the square, or second root, however, the number 2 is omitted. Thus the fourth root of 16 is denoted by 16^ or a^IO ; and means a number whose fourth power is 16. Note. — ^The sign ^, called the radical sign, is the letter r, the initial of the Latin word radiXy a root, changed in form by rapidity in writing it, and by its appropriation to A particular use. SQUARE ROOT. 265. The square root of a given number is a number, which, when multiplied by itself, will produce the given number. 266. The number of figures in the integral part of the Square Root of any whole number may readily be known from the following oonaideratione : ' s 290 S<^>LAi:£; KOOT. The square root of 1 is 1 100 13 10 10000 is 100 1000000 is 1000 &c, is &c. Hence it follows that tho square root of any number between 1 and 100 must lie between 1 and 10, that is, will have one figure in its inte^al part ; of any number between 100 and 10000, must lie between 10 and 100, that is, will have two figures in its integral part ; of any number between 10000 and 1000000, must lie between 100 and 1000 ; that is, must have three figures in its integral part ; and soon. AVhore- fore, if a point be placed over the units* figure of the number, and thence over every second figure to the left of that place, the points will show the number of figures in the integral part of the root. Thus the square root of 99 consists, so far as it is integral, of one figure ; that of 198 of two figures; that of 176432 of three Agurea ; and so on. 257. To extract the square root of a given number, EuLE. — ^Place a point or dot over the units* place of the given number, and thence over every second figure to the lefb of that place, thus dividing the whole into several periods. Find the greatest number whose square is contained in the first period at the lefb ; this is the first figure of the root, which place in the form of a quotient to the right of the given number. Subtract its square from the first period , and to the remainder bring down the second period. Divide the number thus formed, omitting the last figure, by twice the part of the root already obtained, and annex the result to the root and also to the divisor. Then multiply the di- visor, as it now stands, by the part of the root last obtained, and subtract the product from the number formed, as above mentioned, by the first remainder and the second period. If there be more periods to be brought down, the operation must be repeated ; and if any remain, proceed in the same manner to find decimals, adding, to find each figure, two cyphers, or if the given number end in an interminate decimal, the two figures that would next arise from it« continuation. r between [ have one 1 100 and have t'vvo ecn 10000 %t is, must Whcre- ire of the the left of figures in root of 99 mt of 198 and so on. mber, ,ce of the [ure to the o several stained in ire of the right of ^st period, . Divide by twice the result !y the di- pbtained, as above ^1 period. )pcration Lhc same jure, two ■rminate [from it« SQUARE ROOT. ?0l NoTR. — It there he not whole numbers, or intefci**! P»»'t !a the giren number, we must, in pointing, begin with the leoond figure from that which would be the units' place, if there were a whole AuxabM, aad nark suceessiTely oter every seeond figui'e to the right. £x. 1. Find the square root of 529. 520(23 4 129 229 ^2X2 = 4;> 4; After pointing, according to Rule, we take the first period, or 5, and find the greatest number whose square is con- tained in it. Since the square of 2 is 4, and that of 3 is 9, it is clear that 2 is the greatest number whose square is co'itaine the pro- iroved ia lecl for A ist found Hon, not 60 \thtj h% CUBE ROOT. 293 complete powers ; otherwise, find their product, extract its square root, and divide the result by the denominator. The root may also be found by reducing the fraction to a decimal, and taking the root of the result. EX£BCISES. Find the square roots of 1. 2601 Ans 51 10. f Ans. .61237243 2. 5329 (( 73 11. 13t 3.63318042 3. 784 (( 28 12. 1728 41.5692193 4. 566.44 u 23.8 13. IS 1.1726039 5. 7.3441 (( 2.71 14. lA 1.01857743 G. .81796 (( .9044 15. 33 5.7445626 7. 1169.64 (( 84.2 16. Hf 1.16' or 1^ 8. .07 (( .26457 17. 6| 2.4784789 9. .006 (( . 0774597 18. 794i 28.1815542 CUBE ROOT. 259. The Cube Root of a given number is a number which, when multiplied into itself, and the result again multiplied by it, will produce the given number. Thus 6 is the cube root of 21G ; for 6 X 6 X6 is = 216. 260. To extract the cube root of a given number. Rule. — Place in succession, and at moderate intervals, two cyphers and the given number, as the commencements of three columns. Beginning at the unit figure of the given number, point off as many periods as possible, of three figures each. For the first figure of the root, take the root of the greatest cube contained in the left-hand period. Place the figure in the first column ; and, having added it to what stands above it, multiply the sum by the same figure, writing the product in the second column. Add, in like manner, in the second column, and multiply the sum by the same figure ; set the product in the third column , and subtract it from what stands above it. Perform a pro* i ! i J(^ 294 circE rooT. CGSs exactly similar in llie first and second columns ; and» after that, add the llgnre found for the root, to what stands in the first column. Annex one cyi)hor in the first column, and two in the second ; and in the third the next period of the given immber ; or if there be no fijjures remaining, annex three cyphers. To find the next figure of the root, divide the number in the third column, by the one in the second. Place this figure in the first cohimn, and proceed in the same manner. Then annex cyphers, &c., and con- tinue the process till nothing remains, or till the root is carried out as far as may be considered necessary. Care must be taken to insert tiie decimal point in the root, when the figures in the integral part of the given number have been all employed. Ex. Extract the cube root of 926859375. 926859375 (^975 729 197859 183673 9 9 81 81 162 18 9 24300 1939 270 7 26239 1988 277 7 2822700 14575 284 7 2837275 2910 5 14186375 14186375 2915 Here, the greatest cube contained in the first period, 962, is 729 : the root of which is 9, the first figure of the required root. This is placed under the first cypher ; and, going throvgk the fona adding these, we get 9, the product ins ; nnd, I at stands t column, l)eriod of Mnaiuing, the root, >ne in the I proceed and con- le root is • int in the he given 975 period, of the and, )r9duot CUBE ROOT. 295 of which by 9 is set in the second column. Then, by ad- dition, we have 81, the product of which by 9 is 729. This is set under the first period, 92G, and subtracted from it ; and to the remainder, 197, the second period, 859, is an- nexed. Then commencing at the first column, we add 9 ; and multiply the sum, 18, by 9, we set the product, 162, in the second column, and adding it to the number above it, we get 243. "We next add 9 in the first column ; and an- nexing one cypher in that column, and two in the second, we finish all that is preparatory to the finding of the second figure of the root. To find that figure, we divide the number in the third column by the one in the second. The quotient would appear to be 8 ; this, however, would be found on trial to be too large, and we therefore take 7, whicli answers. We add this to the first column, and multiply the sum, 277, by 7, setting the product in the second column. Then, by adding, we get 26239, the product of which by 7 is put in the third column. By taking this from the number above it, we find for remainder 14186, to which the third period, 375, is annexed. We then add 7 in the first column, and multiplying the sum by 7, we obtain 1988, which is added to the number in the second column. The operation pre- paratory to the finding of the third figure of the root, is completed by adding 7 in the first column, and annexing one cj'^pher in it, and two in the second. To find the third figure, we divide, as before, the number in the third column by the one in the second. We thus obtain 5, which is added to the first column, and the sum, 2915, being multiplied by 5, and the product being added to the number in the second column, the sum, 2837275, is multiplied by 5 ; and the product being exactly equal to the number in the third column, there is no remainder, and the work terminates, the root being 975. 261. To extract the cube root of a vulgar fraction. Rule. — If, when the fraction is in its lowest terms, the numerator and denominator be exact cubes, extract their roots for the numerator and denominator of the answer. If only the denominator be an exact cube, the answer will 1^ fi h 296 COMPOUND INTEREST. be obtained by finding the cube Toot of the numerator, and dividing it by that of the denominator. If the denomina- tor be not a cube number, multiply the numerator by the square of the denominator ; take the cube root of the pro- duct, and divide it by the denominator. In every case, the vulgar fraction may be reduced to a decimal, the root of which taken by the former rule, will be the required root. The cube root of a mixed number is generally best found, by reducing the fractional part to a decimal, by annexing this decimal to the integral part, and then extracting the root by the general rule. 1. 2. 3. 4. 5. Exercises. Find the cube roots of the following numbers. Ans. 4.9/3190 8.025S57 123 517 900 " 9.654894 123456789 "497.983b59 12346678 "231.120418 6. 7. 8. 9. 10. 1234567 Ans. 107.276572 44.6 " 3.546823 ^ " .6436595897 " .9614997135 376 " 7.217652 COMPOUND INTEREST. 262. The method that naturally presents itself for find- ing the amount of a sum at compound interest, is to find its amount at simple interest at the end of the first year ; then to take this amount as a new principal, and find its amount in like manner, which would be the amount at compound interest at the end of the second year, and the principal for the third year : the amount of which must be found in like manner. Continuing the process, we should thus find the amount at the end of the proposed time. This will be illustrated in the following example. Ex. 1 . Required the amount of £2500 at the end of 4 years at 6 per cent, per annum, compound interest. COMPOUND INTEREST. 297 ator, and enomina- or by the the pro- uced to a rule, will sat found, [innexing sting the irs. 7.276572 3.546823 36595897 U997135 7.217652 for find- 3 to fiud it year; find its ount at and the nust be should This id of 4 Here, the amount for one year is £2650 ; the amount of which for one 3* car also is £2809, the amount at compound interest for 2 years. The amount of this again, for 1 year, or the amount of the given sum at the end of the third year, is £2977, 10. 9^. ; the amount of which for another year is £3156, 3. 10., the amount of £2500 for four years. When the time is short, this method may be practised without much trouble : but when it is long the labor becomes very great. In this case the following method should bo employed. 263. To find the amount^ or the interest^ of any sum^ at compound interest for a given time, and at a given rate. Rule. — Find the amount of £1 or $1 for the time of the first payment ; raise this amount to such a power as is de- noted by the number of payments, and multiply this power by the principal for the amount, from which subtract the principal for the interest. Ex. 1. What is the amount of £162, 10. for 6 years, payable half-yearly, at 5 per cent, per annum, compound interest ? As £100 : £5 I . . ^^ . ^^^^^ Interest for 6 months. » And £1 + £.025 = £1.025, amount of £1 for do. £1.02512 =£1.3448888. £1.3448888 X£l6?, 10. 0. = £218.54443. £218, 10. 10. amt. J £l62, 10. for 6 years, &c. The above might have been more easily found thus : The amount of £100 for 6 months, at 5 per cent., is £102, 10., or £102.5 ; and consequently that of £1 for the same time is £1.025. Reason for the above process will appear from the follow^ ing considerations. Since the amount of £1 for a year, or any given time, will evidently be a hundredth i)art of the amount of £100 for the same time : and as £1 is to its amount for a year, or any given time, so is any other principal to its amount I 298 COMPOUND INTEREST. for the same time. Hence, to take a particular instance, the amount of £1 for a year at 5 per cent, will be £1.05 ; and by the nature of compound interest, this will be the principal for the second year. Then, as the principal, £1 : £1.05, its amount:: the principal, £1.05 : £1.052, the amount at the end of the second .year, and the principal for the third. Again, as £1 : £1.05, its amount : : the prin- cipal, £1.052 : £1.053, the amount at the end of the third year, and the principal for the fourth. It will thus appear, that the amount of £1 for any number of years, will be equal to £1.05 raised to the power denoted by the number of years. The amount of £1 being thus determined, it is plain, that the amount of any other principal will be had by multiplying the amount of £1 by that principal, since the amount will evidently be i>roportional to the principal. Ex. 2. Required the amount of $780 for 12 years at 5 per cent., per annum, compound interest. The amount of $1 for a year is the hundredth part of the amount of $100, or $1.05, the twelfth power of which is $1.796856. This being multiplied by $780, the result is $1400.76-/^-, the amount. Exercises. « Find the amounts of the following sums, at the given rates per cent., per annum : 1 . £251 , 16. 6. for 9 yrs, at 5 per ct. &c. Ans. £390, 13. 3 2. £212, 0.0. "15 u 4 n u " £381,16.0 3. $500.00 *^ 15 U Q U U " $1198.28 4. $960.00 , "10 (( fj (( (( " $1888.46^ 5. £151,12.1^^12 " 4f " " " £264, 11.8i 6. £1000, 0. 0. " 22 U Q tt U " £3603, 10. 8 » 264. To jind the principal^ which, at a given rate, and in a given time, would amount to a given smn. Rule. — Divide the given sum by the amount of £1 or $1 for the given time. Ex. "What sum must be lent out at compound interest, at 5 per cent, per annum, so as to amount to £3000, 6. 8. Qstance, (£1.05; be the pal, £1 : )52, the >rincipal :he prin- he third I appear, , will be number led, it is I be had il, since rincipal. ^ars at 5 irtof the which is iresult is e given ,13.3 1, 16. ;.28 1.46^ 11. 8^ 10. 8» te, and or$l terest, I, 6. 8. MISCELLANEOUS EXERCISES. 299 at the end of 21 years? Here the amount of £1 for 21 years being 2.786962, we have for answer £3000.3'-r *2. 785962 = £1076.9469 = £1076, 18. llj. For exercises to this rule^ prove those of the former. MISCELLANEOUS EXEKCISES. i. The sum of two numbers is 980, and their difference 62 : what are the numbers? Ans. 459 and 521. 2. "What number multiplied by 28^, will produce 145 ? Ans. S^W* 8. If an army of 24000 men have 520000 lbs. of bread, how long will it last them, allowing each man l^lb. per day? Ans. 20 days. 4. For what sum must a note be drawn, payable in 4 months, the proceeds of which shall be $1800, discounted at a bank at 7 per cent. ? Ans. $1843.003. 5. A merchant bought 500 yds. of cloth for $1800 : how must he retail it by the yard to gain 25 per cent. ? Ans. $4.50. 6. A man bought 640 lbs. of beef for £1250, and sold it at a loss of 12 per cent. : how much did he get a barrel? Ans. £1, 14. 4^. 7. How many dollars, each weighing 412^ grs., can be made from 16 lbs. 5 oz. of silver? Ans. $229-^. 8. How many revolutions will the hind wheel of a car- riage 5 ft. 6 in. in circumference, make in 2 miles 4 fur- longs? Ans. 2400. 9. What cost 15f lbs. of cheese, at $8| per hundred pounds? Ans. $1,828. 10. How many yards of carpeting f yd. wide will it take to cover a floor 18 ft. long and 15 ft. wide? Ans. 40 yds. ^ 800 MI8CELLAXBOU8 EXERCISES. N: 11. If ft man travelling 14 hours per day, perform half his journey in 9 days, how long will it take liim to go the other half tnivcUing 10 hours a day? Ans. 12f days. 12. .If 150 apples cost Us. 4^d.^ how many of them must be sold at the rate of 8 for C^d., and how many at the rate of 3 for 24d., that the gain on the whole may be 10 per cent. ? " Ans. 90 at 3 for 2id., and 60 at 8 for 6^d. 13. A merchant engages a clerk at the rate of £20 for the first year, £25 for the second, £30 for the third, &c., thus augmenting his salary by £5 each year. How long must the clerk retain his situation so as to receive on the whole as much as he would have received had his salary been fixed at £52, 10. per annum? Ans. 14 years. 14. A son asked his father's age, the father replied : " Your age is twelve years ; to which if five-eighths of both our ages be added, the sum will be equal to mine." What, was the father's age? Ans. 62 years. 15. If one person lies in bed 9 hours per day, and another G hours, how much time will the one gain over the other in 20 3'ears? Ans. 2 y. 182^ d. 16. A man and his wife can drink a barrel of beer in 30 days, and the man alone can drink it in 40 days : how long will it last the wife? Ans. 120 days. 17. A teacher being asked how many scholars he had, replied, J study Arithmetic, | study Latin, -^ study Alge- bra, 2^ study Geometry, and 24 study French. How many scholars had he ? Ans. 120. 18. If A could reap a field in 13 days, and B in 16 days, in what time would both together reap it ? Ans. In 7^ days. 19. Suppose 17 gallons of spirits, at 10s. 6d. per gallon, to be mixed with 7 gallons at a different price. What was the price of the latter per gallon, if 20 per cent, be gained by selling the mixture at 1 3s. per gallon ? Ans. lis. 7^d. 19. If gold be beaten out so thin that an oz. avoirdu- pois will cover 20 square yards, how many leaves of this thickness will make an inch thick, the weight of a cubic fiM; of gold being 10 cwt. 95 lbs. ? Ans. 291600 leaves. MISCELLANEOUS EXERCISES. 301 brm half to go the days. lem must t the rate )e 10 per •r G^d. ' £20 for ird, &c., iow long ^e on the s salary years. replied : ghths of o mine." years. lay, and over the 82id. eer in 30 low long days; he had, [y Alge- wmany 120. 6 days, hays. [gallon, lat was [gained 17^d. 'oirdu- )f this cubic ives. * 20. A man owes a dol)t, to be paid in four equal instal- ments at the end of 4, 0, 12, and 20 months respectively ; and ho finds, that discount being allowed, according to the true method, at 5 per cent., per annum, $3000 paid at present will discharge the whole debt. How much did ho owe? Ans. e3136eiHk- 21. The liabilities of a bankrupt are 863240, and his assets $12648 ; what per cent, can he pay? Ans. 20 per cent. 22. Four men, A, B, C, and D, spent £255, and agreed that A should pay J ; B J : C J ; and D J. How much must each pay? Ans. A £51 ; B £34 ; C £68 ; D £102. 23. A man wished to tie his horse by a rope so that he could feed on just an acre of ground. How long must the rope be? Ans. 7.13645 r. 24. How many bushels will a cubical bin contain whose side is 9 feet? Ans. 585.80357 bush. 25. In how many years will the error of the Julian cal- endar involve the loss of a day? Ans. 128^ yrs. 26. A contractor sends in a tender of £5000 for a cer- tain work ; a second sends in a tender of £4850, but stipu- lates to be paid £500 every three months ; find the differ- ence of the tenders, supposing the work in both cases to be finished in two 3 ears, and money to be worth 4 per cent, simple interest. Ans. Diff. =r £10. 27. Exchange between London and Paris is 25.5 francs per pound sterling; between Paris and Amsterdam is 117 ft'ancs for 55 florins ; between Amsterdam and Hamburg is 11 florins for 33 marks; what is the exchange between London and Hamburg ? Ars. 14^ marks. 28. AVhat must A bequeath to B so that B may receive £1000, after a legacy duty of 10 per cent, has been de- ducted? Ans. £1111, 2. 2f. 29. A railway train travels 27 miles per hour, including stoppages, and 30 miles per hour when it does not stop ; in what distance will it lose 20' by stopping ? Abs. 90 miles. i Jp \y 803 MISCKLLAMX0U8 XZERCI8£8. 80. The expenses of constructing a railway is £2,000,* 000, of which |th part was borrowed on mortgage at 5 per cent, and the remaining {Ihs was held in shares ; what must be the average weekly receipts so as to pay the share- holders 6 per cent., the expenses of working the railroad being 45 per cent, of the gross receipts? Ans. JC4020, 19. C}. 81. IIow much sugar at 6d. and 8d. per lb. roust bo mixed with 12 lbs. at 7d., 16 lbs. at 94d., and 20 lbs. at lOd., that the mixture may sell at 8^d.? Ans. 8 lbs. of each. 82. A cistern can be fllled by a pipe in 20 minutes, and emptied by another in 25 minutes ; in what time may the cistern be filled, when empty, by opening both pipes at the same time ? Ans. I hour 40 min. 83. A market-woman bought 120 apples at 2 a penny, and 120 more at 3 a penny, but not liking her bargain, she offered them to her neighbor at 5 for 2 pence, being what they cost her, she said. Did she gain or lose by this trans- action, and how much ? Ans. Lost 4 pence. 84. Sold cloth at 15s. 4d. per yard by which I lost 8 per cent., whereas I ought to have gained 15 per cent. ; how much was it sold at per yard below value ? Ans. 3s. lOd. 35. In what time will £2500 double itself at 4 per cent, simple interest ? Ans. 25 years. 36. Of 138,918 persons, 30.66 per cent, can read and write ; 58.89 per cent, can do neither ; and the rest can only read; find the numbers in each class. Ans. 42592.2588 can read and write; 81808.8102 can do neither; 14516.931 can read only. 37. Find the value of 57 kilogr. 8 decagr. 4 grains of any article which cost £17, 11. 4|- per kilogramme. Ex- press the result in dollars and cents. Ans. $4011.86]. 38. Find the squares of 1039681 and 328776 ; and divide the greater result by the less, to the first significant figure in the decimal pUces. Ans. 10.000000000009. MISCILLANtOrS F.XIItCISIf. SOd :2,ooo,- At 6 per ; what i sliurc- 'ailrood '. 6J. lust bo lbs. at mch. es, and lay the 3 at the min. penny, lin, she ; what 3 trans- ince. lost 8 cent. ; HOd. cent, ars. d and \t can )2can 39. IIow much in the 3 per cents, at 96 must bo sold out to pay a bill of £1054, 9 months before it becomes due, true discount being allowed at 4} per cent, per annum? Ans. £1CC6, 13. 4. 40. How much ought the price cf the three per cent. consols to sink bolow par, in order that a broker may bo enabled to obtain four per cent, on money. Ans. 25 per cent. 41. The cost of carpeting a room twice as long tm it is broad at 5s. per square yanl umonnted to £6, 2. 6. ; and the painting of the walls at 9d. per square yard amounted to £2, 12. G. Find the height of the room. Ans. 10 ft. ns of Ex- 61. and ficant 09. 304 IVPPLCUENTART IZIROXSrS. ADDITION. (PAGES 40 — 45.) Es. 1. Find the sum of 4T38685,237869513,148t9434- 3978, 865, 4647,250,68539582,78602045,370489000,70555- 91231,276,9123456789,5000. Ans. 165733641864. 2. Add together 629405,7629,31000401,263012,13005- 12,390217,13268. Ans. 33604444. 3. What is the sum of 6457,29301,82406,7589,63489, 101364,46745? Ans. 337351. L Addtogether432678902,8l0046734,2167005,327861, 298000428. Ans 1038220930 6. Add together 493742,56710607,23461,400072,681- 1004,8999003,26501. Ans. 73464390. 6. Add the following amounts, $2425, $3282, $2793, $2354, $4262, $9158, $2653, $3424, $1266, $8742, $2126, $5387. Ans. $47872. 7. Required the amount of the following : $46519, $32- 271, $17436, $81587, $28333, $52745, $23052, $20158, $71- 232, $39467, $18643, $42027, $73235, $24103. Ans. $570808. 8. Required the amount of the following : £421536, £310101, £797019, £233680, £124402, £255353, £8520- 67, £618041, £100266, £971134, £536920, £703352, £420- 503, £312675. Ans. £6657039. 9. 607253+2320124-211849-1-3804364-578551+231. 849+ 145763 + 605037 -f- 7601554-357676-|-644844-|-276- 232+8033834-725918. Ans. 6460458. 10. Add together one thousand, four hundred and eighty-three ; seven hundred and ninety-six ; thirty-nine ; forty thousand, seven hundred and forty-four ; five thous- and, eight hundred and sixty ; fifty thousand and seven. Ans. 98929. .48T9434- [)0,t0655- i4:1864. 12,13005- 104444. 89,63489, m351. 5,327861, J20930. 0072,681- L64390. 2, $2793, 2, $2126, 1 147872. ;519,$32. 58, $71- 170808. 421536, £8520- 2, £420- 17039. il+231- t4-|-276- io458. red and ty-nine ; thous- )even. 8929. SUPPLSftlAKTART IZKROISM. 306 11. Add together the followiDg numbers : twenty-two millions, six hundred thousand, five hundred and three ; five hundred and sixty-three millions, seventy-six thousand and thirty-four ; one hundred and eleven millions, six hun dred and fifty thousand and fifty ; throe hundred and twen- ty-six millions, seven thousand, nine hundred and ninety on«i ; one billion, seven hundred and ten millions, ont thousand, seven hundred and ten ; one billion, three hua- cUeil thousand and five. Ans. 3733636293. SUBTRACTION. (PAOBS 40— 49.) Ix. 1. From 9876102 take 1060671. Am. 8825411 5. From 20030000 take 72534. Ana. 1995746S .''" g. From 84200591 take 8888888. Ans. 25SU70S ^:Ui^^, From 95246300 take 9438676. . Ans. 85807625 6. From 96531768 take 873625. Ans. 95658143 .QC 8. Prom 6764+3764 take 6500+2430. Ans. 1598. 7. From 6008+9270 take 5136 — 2352. Ans. 12494 B. From 9687— 3401 take 3021+1754. Ans. 1511. 9. A gentleman worth $163,250 bequeathed 81520C^ to each of his two sons, $16500 to his daughter, and to his wife as much as his three children, and the remainder to a hospital : how much did bis wife receive, and how much tk« hospital f Wife $46900. Hospital $69450. 10. A spe«ttlator gained $3560, and afterwards lost $2500 ; at another time he gained $6233, and then lost $d460 ; how much more did he gain than lose f Ans. $3893. ll. Sobtraot fire hundred and eighty^^four thou« w .fi I ll !' 806 SUPPLSMKNTABT XZERCISK8. Band and seyentj-flix, from fifteen millions, one hundred thousand and three. Ans. 14516927. 12. A man bought a house for MDCCCO XXXVII dollars, and sold it for DCXYIIII dollars less than he gave : how much did he sell it for ? Ans. $1318. -•-iij MULTIPLICATION. Ex. 1. Multiply 2, Multiply Ujhc.8. Multiply #1 f PAGES 49 — 6i. 61835720 by 1320. Ans. 81623150400. 67243 by 99999. Ans. 6724232767. 81890420 by 85672. Ans. 2732116062240. 2364793 by 8485672. Ans. 20066857745896. 1256702 by 999999. Ans. 1256700743298. 73084163 by 7584. Ans. 554270292192. 79548050 by 97280. Ans. 7738434304000. 8. Multiply eight hundred and seventy-seven mil* lions, five hundred and ten thousand, eight hundred and sixty-four, by five hundred and forty-five thousand, three hundred and fifty-seven. Ans. 478556692258448. 9. Find the continued product of 6565, 6786 and 9898. Ans. 440956790820. 10. A convoy of army bread, consisting of 896 wagons, and each wagon containing 18600 loaves, having been intercepted by the enemy, what is the number of loaves lost ? Ans. 16665600. 4. Multiply 5. Multiply 6. Multiply 7. Multiply DIVISION. 7-"; A (PAGES 64— 77.) Ex. 1. Divide 72091365 by 6201. Ans. 13861 iU[ 2. Divide 4637064283 by 57606. Ans. 80496HIH 8. Divide 353008972662 by 6406. Ans. 65299477. 8UPPLBMENTART EZERCI3JBS. 307 handred 16927. XXXVII than he ^1318. 3150400. 4232757. S062240. r7468&6. 3743298. [)292192. r304000. ven mil- ed and d, three 258448. 786 and 790820. of 89e' having iber of 65600. 199477. Ex. 4. Divide 67157148372 by 90009. Ana. 746115t?g8i 5. Divide 908070605040 by 654321. Ana. 1387805f||§Jf 6. Divide 65358547823 by 2789. Ans. 23434402 1^7*^ 7. Divide 3968901531620 by 687637943. Ans. 677immm DIVISION BY COMPOSITE NUMBERS. 1. Divide 41763481 by 625. Ans. 66821f ff. 2. Divide 4276318 by 450. Ans.9502§Sf. 8. Divide 618374121 by 125. Ans. 4946992f f^. 4. Divide 1267894a2 by 9072. Ans. 13975|^f f. 6. Divide 9060708093 by 24192. Ans. 374533|^VkV MULTIPLICATION AND DIVISION OP Ex. VBAOTIONAL NUMBERS. (PAGES 77 — 78.) 1. Multiply i27346 by 2^. 2. Multiply 477121 by H. 8. Multiply 567893 by 3^. 4. Multiply 5691036 by 7|. 6. Divide 25974 by 17^. .6. Divide 367849 by 12J. 7. Divide 2906709 by 121J. 8. Divide 4632987 by 133f . Ans. 68365. Ans. 715681^. Ans. 2224247^^. Ans. 44390080|. Ans. 1484^. Ans. 29825|f Ans. 24000^^. Ans. S4660f^f GREATEST COMMON MEASURE. (FAOEB 86—89.) : Find the greatest Common Measure of 1. 105 and 165. ; 2. 285 and 465. 3. 1879 and 2425. - 4. 108, 126 and 162. 6. 140,210, and 315. 6. 285714 and 999999. . . 7. 324, 612 and 1032. >c8. 16, 24, 48 and 74. ^.9. 837, 1134 and 1347. Ans. 15. Ans. 15. Ans. 1. Ans. 18. Ans. 35. Ans. 142857. Ans. 12. Ans. 2. Ans. 3. I i 808 SUPPtEMENTART EXERCISES. la 28, 84, 154 and 343. 11. 396, 5184 and 6914. 12. 56, 84, 140 and 168. Ana. 7. Ans. 2. Ans. 28. LEAST COMMON MULTIPLB. * (PAGES 89 — 93.) Find the least Common Multiple of the following num« bers. Ex. 1. 8, 16, 18 and 24. 2. 9, 15, 12, 6 and 6. 8. 5, 10, 8, 18 and 15. 4. 36, 25, 60, 72 and 36. 6. 27, 54, 81, 14 and 63. 6. 9, 12, 72, 36 and 144. 7. 8, 12, 20, 24 and ^6. 8. 63, 12, 84 and 7. 9. 72, 120, 180, 24 and 86. 10. 256, 612 and 1728. 11. 375, 860 and 3400. Ans. 144. Ans. 180. Ans, 360. Ans. 12600. Ans.. 1134. Ans.. 144. Ans.. 600. Ans. 262. Ans. 360. Ans. 13824. Ans. 51000. 12. 226, 266, 289, 1023 and 4096. Ans. 2017790775 .oxciifALS. .:: ADDITION OF DECIMALS. (PAaEB98— 99.) Sx. 1. Add together 467.3004; 28.78249; 1.29468; and 3.78241. Ans. 601.16998. 2. Add together 21.6434; 800.7 ; 29.461 ; 1.7606 and 3.46. Ans. 867.006. 3. Add together 46.001 ; 163.4234 ; 20.3046 ; 924.00369 and 62. 1346. Ans. 1214.86719. 4. Add together 293.0072 ; 89.00301 ; 29.84667 ; 924.00369 and 72.39602. Ans. 1408.26569, 5. Add together 1 . 721341 ; 8.620047; 61.720346 ; 2.684 and 62.304607. Ans. 127.06034, 6. 1.293062 + 3.00042 + 9.7003146 + 3.600426+ 7.0040031+8.7200489. Ans. 33.3182746. 7; 394 . 61+81.928+3624 .8103+640.203+6291.- 802+721,004+3920.304. Ans. 15674.1613. 8. Add together 26 hundredths, 8 tenths, 66 thous- andths, 16 hundredths 142 thousandths, and 8CPPLBMXMTART IXESCIflBS. 809 Ans. 7. Ans. 2. AuB. 28. lag num* 5. U4. 3. 180. Sp 360. 12600. . 1134. 8, 144. 8.. 600. 8. 252. 8. 360. 13824. 51000. 790775 1.29468 ; 1.16998. |l ; 1.7506 7.005. 20.3045 ; ,86719. 19.84667 ; [25559, .720345 ; 105034, 100426+ 12746. ;+6291.- 174.1613. »6 tbouB- ith8, and B9 hundredths. Ans. 1.807. 9. Add together 9 tenths, 92 hundredths, 162 thousandths, 489 thousandths, and 92 mil- lionths. Ans. 2.471092. 10. Add together 95 thousandths, 61 millionths, 6 tenths, 11 hundredths, and 265 hundred- thousandths. Ans. 0.807711. 11. Add together 96 hundred-thousandths, 92 mil- lionths, 25 hundredths, 45 thousandths, and seven tenths. Ans. 0.996052. IS. Add together five hundred and nine hun- dredths ; three hundred and seventy-five ; ■twenty thousand and eighty-four, and sev- enty-eight, hundred thousandths; eleven mil- libns, two thousand, and two hundred and nine millionths; eleven hundred-millionths ; one billion, and one billionth. Ans. 1011022464.090989111. SUBTRACTION OF DECIMALS. Page 99. Ex. 1. From .0516 tike .0094187. Ans. .0421813. 2. From 17.5 take 13.0046. Ans. 4.4954. a. From 456.0546 take 364,3123. Ans. 91.7423. 4. Take 57.704 from 713.00683. Ans. 655.30283. b. From 1460.39 take 32.756218. Ans. 1427.633782, 6. Take 9.163 from 81.6823401. Ans. 72.5193401. 7. What is the difference between 25 and .25 ? Ans. 24.75. 8. What is the difference between 3.29 and .999 ? Ans. 2.291. 9. What is the difference between 10 and .00000001 ? Ans 9.99999999. 10. From one thbusaifdth take one millionth. Ans. 0.000999. 11. From one billionth take one trillionth. Ans. 0.000000000999. 12. From 84S6 handred-millibnths take 426 ten- biUionths. -Ans. 0.0000843174. .«i. i • •• 810 lUPPLIMIMTlRT SZIROISSt. MULTIPLICATION OP DECIMALS. (PAGES 99 — 104.) To be proved, true to 6 places of decimals, by Rule Art. 120, page 103 — Ex. 1. Multiply 40 . 4869 by 1 .2904. Ans. 52 . 17977576. 2. Multiply 100 . 0008 by . 000306. Ans. 0.0306002448. 8. Multiply 75 . 85060 by 62 . 3906. Ans. 4701 . 169144860. 4. Multiply 31 , 50801 by 17 . 0352. Ans. 536.660075952 6. Multiply . 000713 by 2 . 80561. Ans. 0.00164889993. 6. Multiply 42 . 10062 by 8 . 821013. Ans. 160 . 86701632806. . 7. Multiply . 884567834 by . 00000008. Ans. . 00000006676542672. 8. Multiply 840003 . 1709 by 112 . 10371. Ans. 94167471 . 869654039. 9. Multiply 65 ten thousandths by 100,0. Ans. 6 . 5. . 10. Multiply 248 thousandths by 10,000. Ans. 2480. . 11. Multiply 2564 . 21035 by 4 . 300506. Ans. 11027 . 40199543710. 12. Multiply 446 . 3214032 by 10 000. Ans. 4468214 ; 032. DIVISION OF DECIMALS. (PAOEB 104—108.) Ex. 1. Divide 123 . 70536 by 54 . 25. Ans; 2 . 28024-. 2. Divide 4 . 00334 by 6 . 31. Ans. . 6344- -. 8. Divide 73 . 8243 by . 061. Ans. 1210. 2344+. 4. Divide . 00033 by . Oil. Ans. . 03. 6. Divide 684234 . 6 by 2682. Ans. 255 . 1210+. 6. Divide 46 . 634205 by 4807. 65. Ans. . 0097. 7. Divide 176432.76 by. 01257. Ans. 14036019.0930. 8. Divide 1.157625 by 1.05X1.06. Ans. 1.05 Rule • 7977576. 6002448. 9144360. 30075952 4889993. 1632806. 6542672. 9654039. .ns. 6 • 5. IS. 2480. >543710. tl4 ; 032. 2802+. 63444-. ^344-f. 0.03. 1210+. . 0097. . 0930. J. 1 . 06 SUPPLBMKNTART KZERCISCt. :iu 9. Divide 82. 86164 by 7.6X0.071. Ana. 60 . 9. 10. Dvide 8444 . 443752 by 6 . 84. Ans. 1234 . 5678. 11. Divide 6 . by 1024. Ans. . 0048828125. 12. Divide 134642 . 156 by 1622 , 2. Ans. 82 . 9997+. MISCELLANEOUS EXERCISES ON THE FOREGOINa RULES. Ex. 1. dif- The sum of two numbers is 980, and their fercnce C2 : what are the numbers ? An8.459 and 521. 2. The product of two numbers is 4410, and one is 63 : what is the other ? Ans. 70. 3. What number multiplied by 28^ will produce 145? Ans. Sv^oV 4. What number multiplied by 6^, will produce the product of 7^ multiplied by 5^? Ans. 6/3. 5. If an army of 24000 men have 520000 lbs. of bread, how long will it last them, allowing each man 1 ^ lbs. per day ? Ans. 20 days. 6. The age of the father and son together is GO years and if 18 be taken from the father's age and added to the son's their ages will be equal : what is the age of each ? Ans. Father 48, son 12. 7. From a certain sum 152 persons took $17 each, and there remained $13 : what was the sum ? Ans. $2597. 8. If you multiply a certain number by 7 you will augment it 1548 ; required the number. Ans. 258. 9. Two lines are 41*. 06328 and . 0438 of an inch long respectively. How many lines as long as the latter can be cut off from the former ? What will be the length of the remaining line. Ans. 937 lines. Ans. . 02268 in Jength of remaining line. REDUCTION. REDUCTION OF STERLING MONEY. (PAGE 114) Ex. I. In 4927 farthings how many pounds, shillings and pence ? Ans. £5. 2. 7} 2. In 189345 pence how many fibrins ? Ans. 7889} florins. y V i'!. 6 po. 59200. ches? 51562. ches? 77760. inch. 806}. cubic 12800. Ins ot icbes. |l896. feet, con- ' t'diuH 1220 feet, tlie second 45936 feet anl tiie i) J 320O feet, how luaity cordd are iu all ? Ai-:4. 394^. AKTIFICCRd. WORK AND DOODCCIUAL UDLTIPUCATlON. (PA0E3 1<(4~147.) Ex. 1. IIow many square feet are there in a piece of marble 15 feet, 7 iu. long, and 1 ft, 10 in, wide, Ans. 28 feet, 6'. 10", or 28 feet 82 inches. 2. How many cubic feet in a stick of timber 15 ft. 3 in. long, 2 ft. 4 in. wide, and 1 ft. 8 in. thick ? Ans. 59 ft. 8'. 8", or 69 ft. 528 in. 3. IIow many cubic feet in a block of granite 18 ft. 6 in. long, 4 ft. 2 in. wide, and 3 ft. 6 in. thick ? Ans. 268 feet, 6', 11", or 2G8 feet, 996 in. 4. now many square feet in a stock of 10 boards, 15 ft. 8 in. long, and 1 ft., 6 in. wide ? Ans. 235 feet. 5. How many square feet in a stock of 15 boards, 20 feet 3 inches long, and 2 feet 5 inches wide ? Ans. 734 feet, 0'. 9", or 734 feet, 9 inches. 6. How many squares of flooring in a house of 3 stories, 40 ft. 6 in. long, and 35 ft. 9 in. wide, allowing stair-way in all the flats, 9 ft. by 6 ft. 3 in. and places fori chimneys, 2 ft. 6 in. by 4 ft. Ans. 41 squares, 14 ft. 126 in. 7. Multiply 16 ft- 3', 4" by 6 ft.5', 8", 10"'. Ans. 105 ft. 5'. 4", 6"', 5"", 4""'. 8. Multiply 20 ft., 4', 8", 5"', by 7 ft., 6', 9'', 4"', Ans. 154 ft., 3', 1", 5'", 4'"', 6""', 8""". 9. How many cords of wood in a pile 50 ft., 6 in. long, 8 ft, 3 inches wide, and 7 ft. 4 in. high ? Ans. 23 c., Ill ft,, 36 in. 10. If a cistern is 30 ft., 10 in. long, 12 ft. 3 in. wide, and 10 feet 2 iu. deep, how many cubic feet, &c. will it contain? Ans. 3840 ft., 0', 6" or 3840 ft. 5 in. 11. What will it cost to plaster a room 20 ft. 6 in. lon^, 18 ft. wide, and 10 ft. high at 12 J cents per square yard ? Contents, Ans. 126§. Cost $15,819+. 12. The soil taken from a certain muund covered an acre of land to the depth of 2 feet ; how many cubic yards of earth were there? Ans. 3226 J. i •16 •OfPUMiyTART IZSBOItM. MEASURE OF CAPACITY. (PAOES 150— ISl.) Ex. 1. In 15 bushels, 1 peck, how many quarts ? Ans. 488, 2. In 763 bus. 3 pks. how many qts ? Ans. 24440. 8. In 56 quarters, 5 bushels, how many pints ? Ans. 28992 pints. REDUCTION. Ex. 4. Reduce 45672 quarts to bushels. &c. Ans. 1427 bushels, 1 peck. 6. Reduce 260200 pints to quarts. Ans. 130100. 6. In 3674 gals, how many cubic inches ? Ans. 1018704 • 676. 7. How many cubic feet in the hold of a ship, which contains 1000 bushels imperial ? Ans. 1283 ft. 1168 inches. 8. How many bushels of coals will a box hold vbich is 8 ft. long, 4 ft. wide, and 4 ft. 1^ inch deep ? Ans. 81 . 015 uearly. 9. In 186048 pecks how many chaldrons t Ans. 1291 oh. 84 bushs. 8 pecks. 10. Reduce 865848 gills to gallons. Ans. 11482 gals. 1 qt, 9 gilLi. MEASURE OP WEIGHT. TROY WHQHT, (PAGE 1S3) Ex. 1. Reduce 29 lbs., 7 oz., 3 dwts. to grains. Ans. 170472. 2. Reduce 87 lbs. U oz. to dwts. Ans. 9000. 3. Reduce 175 lbs., 4 oz., 5 dwts., 7 grs. to grs. Ans. 1010047. 4. Reduce 12256 grains to pounds, &c. Ans. 2 lb. 1 uz. 10 dwtd. IG grs. 6. In 42072 dwts. how many pounds, &c. Ans. 177 lbs. 9 oz. 12 dwts. 6. In 167040 grains, how many pounds ? Ans. 29 lbs. 7. Reduce 11 oz. 12 dwt. 12 grs. to grains.*^ Ans. 5580. lUPPLEMIlfTAnT IXERCTdEd. .017 grs. Lwt8. lbs. 1580. . } i 8. How many grains are there in 18 Ibp. 3 o/. jO dwt. ; and what will the cost umouut to ut 3.^ cvuia per grain? Anw. $;J0:)2 (U. 9. IIow many papers of 4^ grains each can bo made out of 1 lb., 6 oz., 15 dwt., 12 grs. ? Ana. IK!)!}. 10. In 700 lbs. troy of silver, how many poniKla avoirdupois? Ans. 57G lbs. avoir. 11. In 840 lbs., 6 oz., 10 dwt. ; how many pounds, &c., avoirdupois ? Ans. 691 lbs., 10 oz., 5 ^^-^ drms. 12. A merchant bought 1500 lbs. of lead, troy weight, and sold it by avoirdupois weight ; how many pounds did he lose ? Ans. 265f lbs. troy, or 218 i|| lbs. avoir. APOTHECARIES WEIGHT. (PAGE 153.) Ex. 1. In 130 pounds, how many scruples ? Ans. 37440. 2. In 6237 drams how many pounds ? Ans. 24 lbs. 6 oz., 13 drs. 8. Reduce 25463 scruples to ounces, &c., Ans. 1060 g, 7 3, 2 9. 4. Reduce 11 lbs., 10 5, 6 5, 1 B, to grains, Ans. 68540. 6. Reduce 17 lbs., 2 8, 29, to grains, Ans. 98920. 6. In 34678 grs. how many ounces ? Ans. 72 S, 1 3, 2 B, 18 grs. 7. What is the weight of 186 powders, each mado up of three ingredients, 1^, 2^ an 1^ grs. respectively ? Ans. IS, 73, 19,10 grains. AVOIRDUPOIS WEIGHT. (PAOEB 165—157.) Ex. 1. In 16256 ounces, how many hundred weights, old or long weight ? Ans. 9 cwt., qrs., 8 lbs. New. 10 cwt., 16 lbs. 2. Reduce 1876338 drams to hundred weights, &c., of both kinds. Ans., old, 65 cwt., 1 qr., 21 lbs. 7 oz. 2 drs. New, 73 cwt., 29 lbs., 7 oz., 2 drs. 3, Reduce 18967 lbs. to tons. Ans., old, 8 tons,9 cwt., 1 qrs*, 11 lbs. New, 9 tons, 9 cwt., 67 lbs. i*'^ 818 SUPPLEMENTARY EXERCISES. 4. Reduce 137 tons old weight to pounds, Ans. 306880. 6. Three loads of hay weighed 1896, 4327 and 1234 pounds respectively, how many tons are there in all ? Ans. old 3 tons, 6 cwt., 2 qrs., 9 lbs. Ans. new 3 tons, 14 cwt., 2 qrs., 7 lbs. 6. A person purchased 16 cwt., 3 qrs., 15 lbs., new weight, of pork at 5^ cents per pound, what did it cost him ? Ans. $92 . 95. 7. In 48 pounds avoirdupois, how many pounds troy ? Ans. 58^^ lbs. 8. A druggist bought 1260 lbs. of alum avoirdu- pois, and retailed it by troy weight ; how many pounds did he sell more than he bought ? Ans. 271 lbs. 3 oz. 9. Reduce 1867 lbs., 14 oz. 13 drs. avoir, to pounds, &Cf troy. Ans. 2270 lbs., oz., 11 dwt., 16^ grs., nearly. 10. In mill lbs. avoirdupois, how many pounds troy ? Alls. 135030 lbs., 420 grs. 11. Reduce 186321 drs. to pounds. Ans. 727 lbs., 13 oz., 1 dr. MEASURE OF TIME, &c. (PAGES UB - 160.) Es. 1. Reduce 25 days, 6 hours to minutes. Ans. 86360. 2. Reduce 365 days, 6 hours to seconds. Ans. 31557600. 3. In 847125 min,, how many weeks, &c. ? Ans. 84 wks. 6 ho., 45'. 4. Reduce 5623480 seconds to days, Ans. 65 days, 2 h., 4 m., 40 s. 5. How many seconds in a solar year ? Ans. 31556927^. 6. How many seconds in 30 years, allowing 365 days, 6 hours to a year ? Ans. 946728000. 7. How many years of Sabbaths are there in 70 years? Ans. 10 years. 8. In 110 degrees, 20 minutes, how many seconds ? Ans. 397200". ■ i06880. 17 and in all if , 9 lbs. , 7 lbs. 5 lbs., b did it 92 . 95. pounds 8i lbs. voirdu- pounds ). 3 oz. oir. to nearly, pounds 20 grs. ., 1 dr. 86360. ►7600. •., 45'. 40 s. mil, 1^365 ]8000. lin 70 ^ears. >nds ? 1200". 'i SUPPLE&fENTART EXERCISES. 319 9. In 11 signs, 45 degrees, how many seconds ? Ana. 135O0OO''. 10. Reduce 7654314 seconds to degrees, Ans. 2126", 11', 54". 11. In 121:678 seconds French, how many French grades ? How many degrees ? Ans., grades 12 ge.,46', 78". Ans. dogi-ees, 11°, 13', 15 . 672". 12. How many seconds are there between the first of August at 12 o'clock noon and 4 o'clock, P. M., on the 16th of November? Ans. 9259200". COMPOUND ADDITION. Ex.1. Add together £12. 13.9^, £156. 13. 4^, £12. 0.0, £18. 2. 6|, £146, £1276. 19. llf, £15.10, £12789. 16. 10^, and £0. 18. 4|. Ans. £14428. 14. llj, 2. Find the sum of £478. 9. 11, £147. 13. 0}, £111.- 11. 11^, £112. 11. Hi, £167. 19. 9^. £13. 13. 6|, £127.19.- 9i and £100. 10. 11. Ans. £1260. 10. 11. 3. Add £16. 8. 9. £8, 5. 6, £25. 6. 8 together. Ans. £50. 0. 11. 4. Find the sum of £68. 17. 10|, £10. 9. 6, £43-- 10. Hi and £65. 14. 8^. Ans. £188. 13. 0^. 5. Find the sum of $189. 62^, $196.78, $12.96^ $134.72^, $14.00^, $17.03^, $190.90 and $365.71^. Ans. $1121. 79|. 6. Find the sum of $126.33^, £14.3.9, $20.10, $180.15, £13.13.9^, , £127.18.7^ and $200.15. Ans. in $1149. 96^^ Ans. in £287.9.9^. 7. $78.62+$33.904-$120.16i+£19.19.10H-^14.. 13.6J-|-$122.82-f$0.05+£l4.6.7i. Ans. in $551. 56^. Ans. in £137.17.9t^. 8. Add together 126 miles, 3 fur., 14 po., 5 yds., 8 m., 4 f., 13 po., 126 m., 7 fur., 18 per., 3 yds., 163 m, 7 fur., 14 per., 1367 m., 3 fur., 10 per., 1 yd. and 176348 yards. Ans. 1888 m., 3 f., 13 po., 5 yds. 9. Add 126 m., 3 fur., 14 per., 16 ra., 3 f. 10 po., 14 m. 7 fur., 16 po . 5 yds., 126 m., 3 f., 7 per. 3 yds., 18 m. 2 f., 16 per., aud 183467 ft. together. Ans. 337 m. 1 fur. 23 po., 3y.. 2 ft. 10. Add together 5 leagues, 2 m-, 4 fur. 7 po., 4 y.; 18 lea,, 2 m., 3 fur., 21 per,, 5 yds. ; 85 lea., 6 fur,, 10 po,, 2 yds, Ans. 109 lea,, 2 m., 6 fur. i 320 SUPPLEMENTARY EXEBCISES. 11. Add together 19 po., 12 fl, 8 In. ; 64 po., 13 ft,, 8 in, ; 28 po., 10 It., 5 in, ; GO po., 9 ft., 11 in. Ans. 4 fur,, 13 po., 18 ft., 3 in, 12. Add tog-ether 19 yds,, 3 qrs., 3 na., ; 21 yds., 2 qrs,, 1 na, ; 42 yds., 1 qr., 2 na. ; 30 yds., 3 qrs., 2 na, Ans. 114 yds., 3 qrs. 13. Add together 65 yds., 3 qrs., 1 na. ; 81 yds., 2 rs,, 2 na. ; 100 yds., 3 qrs., 1 na. : 95 yds., 1 qr., 1 na I 6 yds. 3 na. ; 28 yds., 2 qrs. » » Ans. 387 yds., 1 qr 14. Add togethsr 12 ac, 3 ro., 30 po.; 16 ac. 29 po. ; 19 ac, 1 ro., 19 po. ; 9 ac, 2 ro., 3 po. ; 17 ac, 3 ro.; 10 ac, 21 po. ; 18 ac, 1 ro., 14 po. Ans. 114 ac, ro,, 36 po. 15. Find the sum of 13 ac, 3 ro, 27 po, ; 45 ac , 1 ro., 27 po, ; 63 ac, 2 ro,, 17 po. ; 26 ac, 2 ro.. 26 po,; 16 ac, 3 ro , 34 po. ; 21 ac , 8 po. ; 55 ac, 2 ro., 31 po. ; 37 ac, 2 ro,, 18 po. ; 44 ac 2ro., 20 po, ; 57 ac, ro., 19 po. ; 61 ac 3 ro,, 18 po. ; 39 ac, 2 ro. ; and 5 ac,, 1 ro., 30 po. Ans. 489 ac. 1 ro., 35 po. 16. Find the sum of 25 ac, 2 ro., 16 po. ; 30 ac,, 2 re, 25 po. ; 26 ac 2 ro., 35 po. ; ^3 ac. 1 ro., 31 po. *, 34 ac, 2 ro., 29 po. ; 5 ac, 2 ro., 15 po. ; 25} yds., 101 sq. in. ; 9 ac, 1 ro. 35 po., 12^ yds,, 87 in. ; 42 ac 3 ro., 24 po., 23| yds., 57 in. ; 12 ac, 2 ro. 5 po., 13| yds., 23 in. } It ac, ro., 24 po., 30 yds., 113 In. Ans. 268 ac, 3 ro., 2 po., 14 yds., 7 ft., 21 in. 17. One room in a house contains 15 sq. yds., 5 ft., 7 in. of plastering ; another 10 yds., 7 ft., 30 in. ; another 9 yds., 6 ft., 25 in. ; another 7 yds., 5 ft. 63 in. ; how much plastering is there in all of them ? Ans. 43 yds. 5 ft., 125 in. 18. Find the sum of 3 c. yds. 23 c, ft. 171 c, in. ; 71 c. yd., n c. ft. 31 c. in. ; 28 c y., 26 c. ft. 1000 c. in. and 34 c yd. 23 c ft., 1101 c in. Ans. 85 c. yds. 9 c. ft. 575 c in. 19. Add together 196 sq. ft. 11', IF, 2'", 27 ft., 6', 10" ; 136 ft., 4', 11''' and 1 ft.. 3', 10". Ans. 362 ft., 2'. 8", 1'", or 362 ft., 385 in. 20. Add together 127 gals., 3 qts., 1 pt. ; 127 gal., Iqt.; 44 gals., 3 qts., 1 pt. ; 93 gals., 1 qt. ; 73 gals. 3 qts*, 1 pt. and 126 gals., 3 qts., 1 pt. Ans. 594 gals. 21. A merchant bought a cask of oil, containing 73 gals., 3 qts. ; another 60 gals., 2 qts. ; another 40 gals., )0., 13 ft,, I ft,, 3 in, 21 yds., rs., 2 na. 3 qrs. II yds., 2 r., 1 na,; ^ds., 1 qr 16 ac. 29 ic, 3ro.; o,, 36 po. 45 ac, 1 .. 26 po,; , 31 po. ; iC, ro., 1 5 ac,, 1 )., 35 po. ; 30 ac,, , 31 po. •, ^ds., 101 -c. 3 ro., yds., 23 t., 21 in. Is., 5 ft., another . ; how 125 in. c, in. ; )0 c. in. ^5 c. in. |6', 10'' ; 1385 in. 17 gal., [gals. 3 |4 gals, baining " gals., SUPPLEMENTARY EXERCISES. 321 1 qt. ; another 65 gals., 2 qts. : how much oil did he buy ? Ans. 240 gals. 22. Add together 21 lbs., 7 oz., 12 dwt. 10 gvi^. ; 28 lbs. 5 oz., 8 dwt. 7 grs. ; 7 lbs., 6 dwt., 15 grs. ; 41 lbs., 6 oz., 20 grs. ; and 9 lbs., 7 grs. Ans. 107 lbs., 7 oz., 8 dwt., 11 grs. 23. What is the sura of 16 lbs., 6oz., 6 dwts. 19 grs. ; 100 lbs., 8 oz., 16 dwt. ; 97 lbs., 2 oz , 10 grs. ; 115 lbs., 9 grs ? Ans. 330 lbs., 2 oz., 3 dwt., 5 grs. 24. Find the sum of 7 qrs., 6 bush., 1 pk., 3 qts. ; 27 qrs., 6 bush., 6 qts. ; 34 qrs., 1 bush., 6 qts. ; 65 qrs*, bush., 3 qts. Ans. 135 qrs., 3 bush., 3 pks., 2 qts. 25. In 4 fluid pints, 14 f §, 6 f 3. 36 m. ; 19 pts. 11 f S, 1 3, 56 m. ; and 3 pts., 15 f §, 16 m. how many pints ? Ans. 28, 0. 9 f. §. 3. 48 m. 26. Add together 3 lbs., §., 7 3. ; ID. ; 13 lbs., 11 §, 7 3, 2 B, 19 grs. ; 14 lbs., lOg, 7 3, 2^, 18 grs., and 6i, 29, 17 grs. Ans. 32 lb., 6^, 14 grs. 27. Add together 3 drs., 2 scr., 19 grs. ; 2 drs., 2 Bcr., 11 grs. ; 7 drs., 17 grs. ; 6 drs., 1 scr., 9 grs. ; and 5 drs., 1 scr., 13 grs. Ans. 3 oz., 2 drs., 9 grs. 28. Add together 19 cwt., 3 qrs., 14 lbs. ; 16 cwt., 3 qrs., 11 lbs. ; 27 cwt., 3 qrs., 27 lbs. ; 18 cwt., 2 qrs., 11 lbs. ; and 196 cwt., 1 qr., 1 lb., old weight. Ans. Old, 279 cwt., 2 qrs., 8 lbs. New, 313 cwt., 12 lbs. 29. Find the aggregate of 126 cwt., 1 qr., 11 lbs. ; 813 cwt., 24 lbs. ; 18 cwt., 1 qr., 21 lbs. ; and 11 cwt, 2 qrs., 24 lbs. New weight. Ans. New, 469 cwt., 3 qrs., 5 lbs. Old, 419 cwt., 1 qr., 24 lbs. 30. Find the sum of 11 cwt., 2 qrs., 13 lbs. ; 12 cwt., 2 qrs., 8 lbs., 14 oz. ; 33 cwt., 2 qrs., 11 lbs., 14 oz. ; 9 cwt., 2 qrs., 13 oz. ; 126 cwt., 1 qr., 27 lbs., 13 oz. ; 17 lbs., 8 oz. and 1267 oz., old weight. Ans. Old, 194 cwt., 2 qrs. 19 lbs., 1 oz. New, 218 cwt., 3 lbs., 1 oz. 31. What is the sum of 10 wks., 5 d., 12 h. 40' , 21 wks.. 3 d.. 9 h.. 15'; 40 wks.. 4 d.. 17 h.. 32', and 42 wks.. Id.? Ans. 115 wks. 15 h., 27 m. 32. Add together 7 years, 28 wks., 3 sec. ; 26 y., 5 ^.-T 322 SUPPLEMENTARY EXERCISES. w., 5 d. ; 58 y., 6 d., 23 h., 59 sec. ; 43 w., 23 h., 60 m., 12 sec, and 124 y., 14 w., 19 h.37 sec. Ans. 21G years, 39 w., G d., 17 h., 51 m., 51 sec. 33. When B. was born, A.'s age was 2 y,, 3 mo,, 3 w., 2 d., 10 h., 11m.; when C. was born, B.'s age was 3 y., 2 mo., 3 w., 6 d., 11 h., 39 ni., 44 sec; when D. was boin, C's age was 11 y., 11 m, 3 w, 6 d., 21 li., 10 m., 49 sec. What was A.'s age when D.'s age was 14 y., 11 m., 3 w., 1 hour ? Ans. 32 y., 6 m., 2 w., 1 d., 20 h., 7 m., 38 sec. COMPOUND SUBTRACTION, Ex. 1. From £196.3.9i take £127,13.10. Ans. £68.9.11]^. 2. From £147.13,6f take £27.14.7^. Ans. £119.18.1H. $519,79^. 3. From £160^.6^ s. 3f d, take £100^, 8 s. Ans. £60.4.7|. $240.92-|J. 4. What sum added to £l89.l3.7f will make £1000.10? Ans. $3243,27t\. 5. From 69 miles, 3 f. 14 po. take 57 m., 2 f . 8 per. Ans. 12 . m. 1 f. 6 per. 6. From 36 lea. 1 m. 3 fur. 4 po. 3 y. take 26 lea. 2 m. 1 f. 2 yds. Ans. 9 lea., 2 m., 2 f.,4 po., 1 yd. 7. Subtract 29 yds., 2 qrs., 3 na., from 85 yds., 1 qr„ 2 na. Ans. 55 yds., 2 qr., 3 n. 8. Subtract 65 yds., 2 qr., 1 n. from 100 yds, Ans. 44 yds., 1 qr., 3 n. 9. A. gives to B. 16 ac, 3 ro,, 14 po, off a farm of 126 ac, 2 ro., 19 po. ; how much has he left ? Ans. 109 ac, 3 ro., 5 po. 10. A person has a farm of 93 ac, 2 ro., 10 po., 4 yd., how much miist he purchase to increase it to 100 ac, 3 ro ? Ans. 7 ac, 29 po., 26^ yds. 11. A person had a farm of 186 ac, 3 ro., 10 po„ and he sold off it, at different times, 20 ac, 2 ro. ; 21 ac,3 ro.; 14 po., and 14 ac, 2 ro., 39 po,, how much had he left ? Ans. 129 ac, 2 ro., 37 po. 12. A person having 167 cords, 93 ft, 16 in. of wood, sells 136 cords, 118^ feet; what quantity has he remaining? Ans. 30 cords, 102 ft., 890 inches. m., 12 51 sec. , 3 mo,, ige waa hen D. ., IGm., t y., U 83 sec. ^.9.11^. H9,79i. 0.92.|i. i make 3,27y2^, [. 8 per. *. 6 per. ; lea. 2 ., 1 yd. yds., 1 r., 3 n. (r., 3 n. farm of [ft? l> 6 po. 4 yd., 3ro? \k yds. ),, and 1,3 ro.; ft? 17 po. in. of las he iches. SUPPLEMENTARY EXERCtSGS. 823 < 13. From 196 cubic fe*^t, 144 inches, take 186 cubic feet, 1701 in. Ans. 9 ft., 171 cu. in. 14. From 163 sq. ft, 6'. 7" take 27 ft. 3'. and U", and reduce the reniainder to sq. in., Ans. 10(316 in. 15. From 36 chaldrons, 13 bush., 1 qt., take 27 chs.^ 9 bush., 3 gals., 3 qts. Ans. 9 ch., 3 bush., 4 gals, 2 qts. 16. From 85 bush., 2 pks., 4 qts., take 49 bush , 3 pks., 6 qts. Ans. 35 bush.,2 pks., 6 qts. 17. From 16 lbs., 11 oz., 19 dwt., take 11 lbs., 3 oz., 15 dwts., 23 grs. Ans. 5 lbs,, 8 oz,, 3 dvvts., 1 gr. 18. From 18 lb., 3 g, 7 3, 2 9 take 17 lbs., 11 g, 3 3, 2 B. 19 grs. Ans. 4 §. 3 3. 29. 1 gr. 19. From 18 tons, 16 cwt., 3 qrs., 11 lbs., take 14 tons, 13 cwt., 2 qrs., 27 lbs., old weight. Ans. Old wt., 4 tons., 3 cwt., 12 lbs. New wt., 4 tons, 13 cwt, 8 lbs. 20. From 14 lbs., 3 oz., 15 drs. take 12 lbs. 8 oz., 14 drs. Ans. 1 lb., 11 oz., 1 dr. 21. Subtract 13 lbs., 4 oz., 15 drs. from 126 lbs., 1 oz., 14 drs. Ans. 112 lbs., 12 oz,, 15 drs. 22. From 160 years, 11 mo., 2 wks., 5 d., 16 h., 80', 40", take 106 yrs., 8 mos., 3 wks., 6 d., 13 h., 45', 34" Ans. 54 y., 2 mo., 2 wk., 6 d., 2 h., 45'., 6". 23. The latitude of a certain place is 40°, 16', 10", and that of another 50°, 27', 35": required the difference. Ans. 10°, 11', 25". 24. The latitude of St Peter's, at Rome, is 41°, 53', 54" north, and that of St. Paul's, at London, is 51°, 30', 49", north. Find the difference. Ans. 9°, 36', 55". COMPOUND MULTIPLICATION. Ex. 1. What cost 7 acres of land, at £35. 6 s. 7 d, per acre? Ans. £247, 6.1. $989.21§. 2. What cost 18 barrels of flour, at £1, 6, 8^. per bbl. ? Ans. £24,*'0, 9. 3. Multiply £583. 0, 10 separately, by 13 and 16, and reduce the sum of both results to $. Ans. $67632.83 .\. 4. Multiply £2579, 0, OJ separately, by 147. 155, 474 and 2331. Ans. £379113, 9, 2^. Ans. £1222447, 9, 7i. £399745, 9, 8^. £6011656, 5, 8^. 5. Find the price of 1897^ lbs.of tea, at 75 cts, ^ 324 SUPPLEMENTARY EXERCISES. 8. 0. 10. and 29. 11. per pound, An8.S1423.12J. 6. What is the cost of 183 gals, oil at 62^ cts. per gallon? Ans. $114. 37^. 7. What does 127 J cords of wood cost, at $1.37 J ? Ans. $175.3U. ^43.16.6|. Multiply 175 miles, 7 fur. 18 po. by 84. Ans. 14778 m., 1 f., 32 po. Multiply 40 lea., 2 m., 5 fur. 15 po. by 50. Ans. 20441., 1 m., 4 fur. 30 po. Multiply 70 yds., 2 ft., 10 in. separately by 7 Ans. 2 fur., 10 po., 1 yd., 1 ft., 10 in. 1 m., 1 fur., 14 po., 1 fi., 2 in. Multiply 67 yds., 1 qr., 2 na., separately by 9 and 53. Ans. COGyds., 1 qr., 2 na. 3570 yds., 3 qrs., 2 na. 12. Multiply 310 ac, 3 ro., 3 po., by 81. Ans. 25172 ac, 1 ro., 3 po. 13. Multiply 380 ac, 3 ro., 32 po., separately by 12 and 106. Ans. 4571 ac, i ro., 24 po. and 40380 ac, 2 ro., 32 po. Multiply 146 sq. ft., 14 in. separately, by 12 Ans. 1753 ft., 24 in. 4090 ft. 104 in. Multiply 36 sq. yds., 3 ft., 16 in. separately, by Ans. 508 yds., 7 ft., 80 in. 1235 yds., 6 ft., 112 in. Multiply 126 cu. ft,, 133 in., by 14 and 56 sep- Ans. 1765 ft., 134 in. 261 yds., 13 ft., 536 in. A bushel contains 2218.192 cub. in., what is the contents of 196 bush. ? Ans 251 ft., 1037 . 632 in. 18. Multiply 14 ft., 3', 6" separately, by 14 and 35, Ans. 200 ft., 12 in. 500 ft., 30 in. 19. Multiply 126 ft., 3', 6" separately, by 182 and 400. Ans. 22985 ft., 12 in. 50516 ft., 8'. 20. If a ship sails 3^, 24', 10" per day, how far will *fihe sail in 60 days ? Ans. 204 =», 10', 0". 21. If 1 ac. of land produce 45 bush., 26 qts., how much will 100 ac. produce ? Ans. 4581 bush., 8 qts. 22. If one barrel of flour requires 4 bush., 3 pks., 5 qts., of wJieat, how much will 500 barrels require ? Ans. 2453 bush., 4 qts. 28, Multiply 16 lbs., 3 oz., 5 dwts., 9 grs. by 26. Ans. 423 lbs., 19 dwts., 18 grs. 24. Multiply 5 §, 73, 2a , 15 grs. separately, by 16 14. and 28. 15. 14 and 34. 16. arately. 17. SUPPLEMENTARY EXERCISES. 325 !3.12i. ts. per 4.37^. 1.37^? .16.6|. 32 po. 30 po. y by 7 10 in. , 2 in. y by 9 ., 2 na. , 3 po, ;ely by 24 po. ,32 po. , by 12 104 in. ?ly, by 80 in. 112 in. )6 sep- 36 in. is the 32 in. ind 35, 30 in. d400. ft., 8'. r will |0', 0''. , how 8 qts. pks., |4 qts. 126. grs. |byl6 and 48. Ans. 95 § . 6 3 . 2 9, and 287 5 . 4 5. 25. Multiply 3 tons. 27 lbs., 13 oz., old wt., by 11. Ans. Old, 33 tons, 2 cvvt., 2 qrs., 25 lbs., 15 oz. New 37 tons, 2 cwt., 1 qr., 15 oz. 26. Multiply 16 cvvt., 3 qrs., 24 lbs., new wt. by 76. Ans. New, 64tons,ll cwt., 24 lbs. Old, 57 tons, 12 cwt., 3 qrs., 16 lbs. 27. A person purchased 50 lambs at 15s., 6d., each ; 40 sheep at £1, 5, 6 each ; 30 cows at £12, 10, 6 for three, and 23 horses at £25 sterling each ; the expenses for get- ting them home amounts to 3 s. 6 d. per head. What is the whole cost ? Ans, $4463.60. COMPOUND DIVISION. Ex. 1. 2. 3. 4. 5. 6. , Divide £254, 7 s. 6 d. by 111. Ans. £2. 5. 10- Divide £160.4.8^ by 105. Ans. $6. 10. yV Divide £346, 9, 9^ by 155. Ans. £2. 4. 8^. Divide £111. 2. 6 by 180. Ans. £0. 12. 4, rem. §. Divide £2045. 16. 5^ by 4083. Ans. £0. 10. 0\, rem. ^ ^ Three persons purchased a ship at $111267.36. the first takes 2 shares, the second 5 shares, and the third 7 shares, how much do they severally pay ? Ans. $15895. 33f ; $39738.34^; $55633.68. Divide 75 ac, 3 ro., 39 po. by 26. Ans. 2 ac, 3 ro., 27 pr ., 19 yds., 7 ft., ly^ in. Divide 13 ac, 1 ro. bj i47. Ans. 14 po., 12 yds., 6 ft., 119^^ in. Divide 91 yds., 2 qrs.,lna. by903. Ans. If^fna. A tailor put 216 yds., 3 qrs. of cloth into 20 cloaks ; how much did each cloak contain ? Ans. 10 yds., 3 qrs., If. na. A person travelled 1000 miles ki 12 days: at what rate did he travel per day ? Ans. 83 m., 2 fur., 26 po., 11 ft. Divide 164 cords, 30 ft., by 17 in. Ans. 9 cords, 84 ft., 1016-fr in- Divide 410 cords, 10 ft., 21 in. by 61. Ans. 6 cords, 92 ft., 850^ in- A farmer raised 1000 bush., 3 pks. 16 qts. of wheat on 40 acres; how much was that per acre. Ans. 25 bush., 1| pts 7. 8. 9. 10. 11. 12. 13. 14. 326 SUPPLEMENTARY EXERCISES. 16. Divide 77 ac, 1 ro. 33 po., by 51. Ans. 1 ac, 2 ro., 3 po. 16. Divide 206 mo., 4 da. by 26. Ans. 7 !no., 3 wks., 6 days. 17. Divide 75 cwt., 2 qrs., 10 lbs. by 35. Ans. 2 cwt., 16 lbs. 18. Divide 312 lbs., 9 oz., 16 dwt., by 43. Ans. 7 11)8., 3 oz., 6 dwt. 19. Divide 17 lbs., 9 oz., 2 grs. by 7. Ans. 2 11)8., G oz., 8 dwt., 14 grs. 20. Divide 17 lbs., 5 oz., 2 drs., 1 ecr., 4 grs. by 12. Ans. 1 lb., 5 oz., 3 drs., 1 scr., 12 grs. 21. Divide 178 cwt., 3 qrs., 14 lbs. old weight, by 53. Ans. 3 cwt., 1 qr., 14 lbs. 22. Divide 365 days, 10 h., 40' by 15. Ans. 24 d., 8 h., 42', 40 ". MULTIPLICATION AND DIVISION OF COMPOUND l^UMBERS, BY NUMBERS CONTAINING FRACTIONS. Ex. 1. Multiply £163, 13 s., 6^ d.by 4^. An8.£804, 14, 10||. 2. Multiply $1896.13^ by 8|. Ans. 8l5379.74ff 3. Divide £1876, 14, 3^ by 4^^ Ans. £422, 18, 6 jf . 4. Divide 127 yds., 3 qrs., 2 nails by 14^. Ans. 8 yds., 3 qrs., 1^ nails. 6. Divide 196 cwt., 3 qrs., 14 lbs., new wt., by 14^. Ans. 13 cwt., 3 qrs., 18^^|. 6. Multiply 144 ft., 2', 11" by 16^. Ans. 2325 ft., 132| in. 7. Divide 178 cubic feet, 3', 4", 10'" by 14^. Ans. 12 feet, 1096^17. 8. Divide 1234 gals., 3 pts. by lUi^. * Ans. 11 gals. qts., O^fy. MISCELLANEOUS QUESTIONS ON ART'S 125—157. Ex. 1. How long will a person be in saving £150 ster- ling-, if he save 2s. 6d., N. S. currency per week ? Ans. 28 yrs., 44 wks. 2. How many dozen of tea-spoons, each spoon weighing 1 oz., 3 dwt., can be made out of 25 lbs., 10 oz., 10 dwt. of silver? Ans. 22 J doz. -i— . — . SCPPLEMEVTARY EXERCiaR;?. 827 0., 3 po. 5 days, , 16 lbs. , 6 dvvt. 14 grs. 1. by 12. 12 grs. ght, by , 14 lbs. 2', 40 ". ^OUND , 10||. 9.74f|. 8, 5H. nails. )y 14i. I2| in, -157. ster- wks. ipoon oz., doz. 3. If a pile of wood is 140 ft. long, and 3 ft., G in. wide, how high must it be to contain 18 curds ? Ans. 4 feet, 8.^51 in. 4. If 9 cows eat 21 tons, 7 cwt., 2 qrs., lli lbs., new weight, of bay in a year, how much will 2 cows eat in the same time ? Ans. 4 tons, 15 cwt., qrs., 2^ lbs. 5. If a gentleman's annual income be £1000, and his daily expenses £1, 17, 3^, how much does he save in 9 years ?. Ans. £2874, 16, 10^. 6. If a newspaper has a circulation of 1260 daily, and be sold at 21 d. each copy, what amount in dollars an cents will be realized by its sale for one year (313 days) ? Ans. $14789.25. 7. If I purchase 20 pieces of cloth, each piece 20 ells, for $2.50 per ell, what is the value of the whole in pounds ? Ans. £250. 8. How many yards of cloth may be bought for £21, 11, H, when 3^ yds. cost $10.85 ? Ans. 27 yds., 3 qrs., l^y nls. 9. The difference between two numbers is 12, and their quotient is 4 ; what are the numbers ? Ans. 4 and 16. 10. If 24 lbs of raisins cost 6 s. 6 d., what will 326^ lbs. cost at the same rate ? Ans. £4, 8, 4 ^y, 11. If 2^ oz., of silver cost $2.50, what is the price of 14 ingots, each weighing 7 lbs., 5 oz., 10 dwts. ? Ans. $1253.00. 12. How many gallons will a box hold, which is 3 feet 6 inches long, 2 ft, 6 in. wide, and 4 ft., 6 in. high ? Ans. 245 gals., 1 qt., 1 ipt„^^§^j, 13. How many feet of boards one inch thick, can be cut from a log which is 22 ft., 6 in. long, 141 in. deep on the one side, and 17^ in. on the other, allowing i inch waste for each cut ? Ans. Cutting boards 17^ wide, 393| feet. »' 14|do.387ft.,2',3". FRACTIONS. (( REDUCTION OF VULGAR FRACTIONS. (ART. 163, PAGE 180.) Reduce the following fractions to their lowest terms : Ex. 1. m, Ans. M. Ex. 4. §f|, Ans, J. 2. T^m, I. 5. Uh It «. 8. m, *• iS. 6. mh M i- '%:^ li 328 Ex. o 7.j.j?f, " 13 1 • Tuai. Ana, ISO 14. jii.i6 .. iooVr. 9. ■Keduce the fnU^r. • /»* H. iiaV looVr- on deno„.raS°-^ ^-'- to otLetltv..;, . j^ ''''^hM lilt twv, iWit. iVci!. im Reduce the foIIn'*-^'°'/"*- '*°« i«»0 *^^' ***' *^»' ben,. ^°"°"""fiffr'»«ion8towholeorm,- . Ex. I. u ,.. .. ""'eormuednum. mon denominator. ^' f ; l» ^ and /^, ^' t*> i^a^, t^ and ,tv 2. 3. 4. 5. 6. W 7. 8. 9. 1 9 6 10 v^ Ans. 23. 28. it II n , B«duce the foZ^^ 2 r' ''^' "' " '""^*^- X°7 'Td°"' *"' ""'''' '"■">^«'' to deno^initor sVube /l. '"' ^'' '^^ ^2 to fractions whose loff :: III- 8 296,1 ^"" a- 16U . ^^- 9- 4965 .. ^^^^- 129?* 4*- ff- 1880% : ,tl!ti- '3. Reduce 6, 32/4!»a';J ',^^/:f,^^i,,,,.4ag. E- 1. Reduce ^rTf", -• m"?:;;^*' ^^*' »" ^ ^f ^° . 2- Reducers 6 Vto the r''T'°"<"'-e2. Ans m 2. S. 4. 6. 6. " BrPFLDCIN'TAlIT IZEKUIBZI. 329 ins. 1^8, " AWr. " if. g" a corn- ed num* J. 2?. 28. )er8 to whose II to i 4. Reduoe S1.27i to the fhiotion of £1. Am. ^, 6. Reduoe 11 Ibi. 3 oz. troy to fraction of 3 lbs. 2 oz. RToirdupois. Ans, 3jJ^^ or2||J. 6. Reduce 1 gal. 3 pts. to fraction of 2 bu«. Am. .^, Art. 167. Page 186. Ex. 1. What is tiie value of i of £19. 10. 6 ? Ans. £2. 8. 9|. 2. What is the value of § of 1 ton? (old weight.) Ans. 18 owt. 1 qr. 9| lbs. 8. Find the value of ^ of 2 cords of wood. Ans. 1 cord, 102^ ft. 4. Find the value of j| of 3 ac. 2 ro. 14 po. Ans. 2 ac. 2 ro. 5^ po. 6. Required the value of 7^^ chal. Ans. 7 chal. 88 bus. 6. Find the value of ^j of %166.r>3i. Ans. 186.09J^. 7. What is the value of ^, of a yeai ' Ans. 21)8 days, 15 hours, 16 min. 21^ sec. 8. What is the value of ^ of a mile 7 Ans. 6 fbr. 21 po. 4 yd. 1 ft. 6 in. ADDITION OF VULGAR FRACTIONS, Am. 168. Paqb 186. Find the sum of the following. Ex. 1. 820» and 2. 686Jf , 427y, 1625f 8. 2U» '651, l|, and J.. 4. 826^^, 98^X5, 186JOj, 14i. and I Jj.W.andl*. 6. 189X.m.l46|, yj^.andJU 7. lU, fe 16J, 18|, and 196. 8. 614, 803^, and 4j|. 9. 418J. 6431, 618|. Am. 820-^ Am. 2689|V Ans. 6t| Ass. 675 Ans. 489 Ans. 85 2 Ans. 868 Ans. 1680 SUBTRACTION OF VULGAR FRACTIONS. Art. 169. Pagb 188. Find the difference between — 1. .^|. and |. Ans. 2. U and Ij. •• y, 8. if and |f ** 4. II and ^. *♦ 6. 69^-and ii9i ♦* 36i 6. 16Xand9f «• Q^X 7. £^ and 1 s. If Am. 28. h^ 8. f yd. and 14 nls. Ans. 1 qr. 8^ nls. 9. 1461 and 145Jl>. Am IJL. 10. 196^ and 192^^1^. 11. 147 and IZQJ^, Ans. lOlff 12. 1881 and 126||. Anik 66^. 830 8r?PLtair^ of if by ^ of | J of | of 6. 8. Divide " 4 " by _f_. 9. Divide--^- by J^ of ^y. 10. Divide -|- by J-. EEDUCTION OF DECIMAL FRACTIONS. Art. 178. Paqk 193, Reduce the following fraotions to decimals : 1 a. Ans. iH8|- :: "i? ii 2a 25. 243ff. Ans, 8. 4. 6. 6. P^rfV- m- K (< «( (I .675. .666. .204. .1001. .1844. .1216. 7. 8. 12. Am. 474ld I J Art. 170. Page 106. Redaoe tho following decimals to vulgar fractions : 1. .4. 2. .336. 8. .4598. 4. .53. 1. .18^. 2. .5925'. a. .0227. 4. .4746. Ans. J. •• mi «« fin 8tf- Arts. 177 and 178. Ans. /j. 5. .105^. 6. .IH'J7. 7. .t')()3. 8. .hoo. Ans. <« «t If •.' .1 1 itf 111 r- }. 41 if Art. 170. ft ff I'AQi:.-!! l'J8 AND 190. Sj. 3 0. 0. 14.06'.j4. 7. 15.0694, 8. 123.05491. Ans. i Ans. |. .72, 5808 9'dO Page 200, Ans. .226. •• 1.102882. 1. Beduce 4s. 6d. to the decimal of £1. 2. ** 14 02. 3 dwt. 19 gr. to the decimal of 1 lb. 3. *' 3 cwt. 2 qr. 11 lb. (old weight) to the decimal of 1 ton. Ans. .1798. 4. *' 3 qrs. 1 nl. to the decimal of 1 yd. 6. ** 3 fur. 14 per. 5 yd. 6 in. to the dec. of 1 mile. 6. " 16 hrs. 18 min. 14 sec. to the decimal of 2 days. 7. Express £4. 13. 6. in £.'& and decimal of ^. 8. 9. 10. 11. 12. «f II II II II .8126. .405902. .337928. 4.676. 5.41876. 44.8376. 19.876. 6 miles, 8 fur. 14 per. in the dec. of i. mile. 44 ac. 8 ro. 14 po. to the decimal of an acre. 19 gal. 8 qt. 1 pt. to the decimal of a gallon. 127 oz. 8 dr. 2 scr. 3 gr. to the decimal of a pound. Ans, 10.62204861. 10 feet, 10', 10" to the decunal of a foot " 10.9027. Aftt. 180. Paqb 201. Eequired the value of the following decimals : 1. .06789 of a ton (new weight). Ans. 1 cwt. 1 qr. 10 J^ lb. 2. .06789 of a ton (old weight). •• 1 cwt. 1 qr. 12^. 8. .329& of a mile. Ans. 2 fur. 25 po. 8 yd. 0||^.J ft. 4. .6783 of a league. Ans. 2 lea. fur. 11 po. IffJ^. yd. 6. .8763452 of a cord. Ans. 112 ft. 297jjf|| in. 6. .3298 of £3. 2. 6. 7. .489 of S4.26. 8. .826789 of 12 feet, 10 in. 9. .187543 of a pound, avoirdupois. 10. .187543 of 1 pound, troy. ** £1. 0. 7J|. «• ^2.08,1^. « 4ft-2Hf4B.i". «( 8 ^^ Mik ^ Aos. 2oz 5dwt.0J^j|.gr. 332 SUPPLEMSNTAltT BZEltCISSS. Abt. 181. Page 202. Make the following similar and coterminous : : (1.) 1.3fi780. Ans. 1.367§96789C7896789. 4.8y678. «« 4.89678067896789678. 16.87329876. " 16.87329876876876876. .18646. " .1864564564564564G. .000023. «* .00002323232323232. (2.) 486.7834. 298.6743. 273.1236. 1.4868. 101.89685. Ans. (( 4< z. 17. 2J, »93.206. 7. llf. 273. 67. 4. 0, 4*. J0.13.a. 18. 8. 0.16.9. 13. OJ, 12.6. 13.0. 6. 706. 18. 2^. 0. 2J. 14. 3|. 33. 28. 2.9f. .79J. 5.H. 8. 0|. SIMPLE PROPORTION. (ARTICLES 212—213. PAGE 228.) 1. If 367 cwt. 3 qrs. 14 lbs., new weight, of flour cost $1296.37, what cost 2 tons, 3 cwt., 3 qrs., old weight, at the same rate? Ans. $172.66J. 2. What cost 197 bush, of wheat, if 372 bushels of the same quality cost £139. 10. ? Ans. $295.50. 3. If 2^ bush, of beans cost $7.25, how many bushela can be bought for £18. 13. 6^ sterling, at par ? Ans. 32^ bush. 4. If 196 bush, potatoes cost £14. 14. 0, in P. E. Island, how many bushels can be purchased in the same place for $1896.75, N. S. currency? Ans. 7587 bushels. 6. If 160 bbls. of flour can be bought for $640, how many can be purchased for $2240.00. Ans. 560 bbls. 6. If a person walk 396 miles in 14 days of 12 hours each, in how many days of nine hours each, can he walk the same distance ? Ans. 18f days. ' 7. A hare -pursued by a dog, was 96 yds. before him at starting. The dog ran 7 yds. while the hare ran 5. How far did the dog run before overtaking the hare ? Ans. 336 yds. 8. If A can cut 3 cords of wood in 14i hours, and "A can cut 4 cords in 161^ hours, how many cords can they both cut in 2 days of 10 hours each ? Ans. 9^^j cords. 9. If a board be 11 inches wide, what must be its length to contain 72 sq. ft. AnS. 78^ . 10. At what time between 5 and 6 o'clock, are the hands of a watch together ? Ans. 27m., 16^ sec. past 5 o'clock. 11. If one man can do a piece of work in 14 days, and another person the same piece in 16 dys. what part of it can both do in 4 days ? Ans. ^$. 12. A town lot 375 feet, 6 in., by 76 ft, 6 in., cost $472.50 ; what will be the cost of a similar piece 278 feet, 9 inches by 151 feet. Ans. $700.84^. 13. A can do a piece of work in 5 days and B in 6 days, working 11 hours a day ; find in what time A and B can do it together, working 10 hours a day. Ans. 8 days. ' COMPOUND PROPORTION. (ARTICLE 214^-219. PAGE 233.) 1. If 360 sheep cost $980, what cost 127 oxen : 9 338 SUPPLEMENTARY EXERCISES. sheep costing as mnch as 1 ox ? Ads. $3111^. 2. What cost 196 lbs. tea, if 397 lbs. cost £22. 16. 8 sterling : 9 lbs. of the former being equal in value to 7^ of the latter ? Ans. |45.36, N. S. c»r. 3. If 196 men can reap o67 ac. Irish, in 18 days of 11 hours each, in how many days of 8 hours each, can 127 women reap 200 ac. English ; 6 men being able to reap as much as 7 women ? Ans.14 days, 7 h., 56 min. 4. If 126 men dig a trench 263 yds. long, 3 feet deep, and 2^ feet wide in 14 days of 9 hours each; what length of trench which is 11 feet wide, and 3^ feet deep can 15 men dig, working 10 days of 10 hours each ? Ans. 4 yds. 2 feet, 6) in. 5. If 136 bush, of oats serve 16 horses for 7 weeks, how many bushels will serve 144 horses for 76 days ? Ans. 1898ff. 6. If 3 men and 2 boys can reap 12 ac, 3 ro. in 14 days of 10 hours each, how many acres can 14 men and 5 boys reap in 10 days of 8 hours each : the working rates of the men and boys being as 3 to 5. Ans. 29 ac, 1 ro., 38 ^f po. 7. If 127i yds. of cloth 18 in. wide cost £228. 16. 8, what will 118^ yds, of yard wide cloth cost at the same rate ? Ans. £424. 9. 3i 8. If 15,000 copies of a book of 110 sheets require 166 reams of paper, how much paper will be required for dOOO copies of a book of 250 sheets of the same size as the former ? ' Ans. 125§f . 9. If a person is able to perform a journey of 146.812 miles in 4^ days when the day is 11.06 hours long; how many days will he be in travelling 1055. 6' miles when the days are 8.5 hours long ? Ans. 38.20441 days. 10. If 16 men in a tour of 6 months duration spend £900, how much will it cost 6 men and 3 boys for 2 rOionths ; each boy spending i as much as a man ? Ans. £133. ll.lOi. 11. If 365 oxen and 120 sheep eat 764 tons, new wt., of hay in 7 months, how many tons, old weight, will be required for 33 oxen and 325 sheep for 8 months at the same rate ; 15 sheep eating as much as 2 oxen ? Ans. 156 tons, 3 cwt., 3 qrs., 8 lbs. 12. If 126 men dig a trench 14 yds. long, 13 ft. wide 8UPPL1MINTART EZIRCISSS. 33<> and 12 ft. deep in 14 days of 8 hours each, in how many days of 10 hours each will 96 men dig a trench 11 yds. long, 10 ft. wide, and 9 ft. deep, supposing that the relative difficulties are as 3 to 6 ? Ans. HA't days. 13. If 163 men dig a trench 110 yds. long, 8 ft. wide, and 8 ft. deep in 6 days of 11 hours each, another trench is dug by half the number of men in 12 days of 7 hours each : how many gallons of water will the lattsr hold 7 Ans. 81409.796 gals. INTEREST. Rule 1. Abticles 220 and 221. Page 238. Find the interests of the following sums, for 1 year, at the given rates pw cent per annum : 1. £193 10 6 at 64- per cent. £111 11 11 at4J " " $1278.66^ at3| 2. 8. 4. 5. 6. 7. 8. «( tt $1126.72 at 4 £16 16 8 at 4 £18 17 61 at 8 $186,763 at 41 $1128.45 at 7| (t tt tt tt tt it tt tt tt tt Ans. £12 1 101 Ans. £4 12 o] Ans. $44.75] Ans. $45.0C Ans. £0 18 Ans. £1 9 4j Ans. $5.69t Ant. $80,888^ BuLE 2. Paob 240. Find the interests of the following sums of 0nd at the given rates per cent per annum : 1. £178 10 6 fbr 4 y. at 4h per cent 2. £396 10 11 &r 2i y. at 6 " 8. $196.30f| for 4 v., 7 mo. at 6 per cent 4. $1183.506 for 2 y.. 9 mo. at 7i " 6. £126 13 7i for 2 y., 1 mo. at 6 " 6. £1269 14 6 for 1 y. , 8 mo. at 7i " 7. $293i for 2 y. 7mo. at 41 " 8. $296.73 for 1 y.. 11 mo. at 5 9. £147 13 6 for 7 mo. at 4 10. $1289.55 for 4 mo. at 5|| tt tt tt msmej tot the given times, Ans. £31 4 8| *• £491144 $58,992. $244,097 £13 3 11 £153 8 6 $31,276 $28,486 £3 8 101 $28.64. tt tt tt tt tt tt tt tt B.VLK 8. Paqb 242. Find the interests of the following sums of money for the j^Ten times, and at the given rates per cent, per annum : 1. £168 13 6 for 7 mo., 6 days, at 54 per cent 2. £1695 14 9&r 2 y., 5 mo., 8 d., at 4 per cent 8. $128.46 for 1 y., 2 mo., 12 d. at Sh per cent 4. $1836.14i for 8 y., 120 d. at 4i per cent 6. £197 14 8i for 1 y., 160 d. at 5 per cent 6. $1367.92i^ for 130 d. at ^ per cent 7. £467 19 10^ for 4 mo., 11 d. at ih per cent 8. $1111.11<| for 8 mo., 15 d. at 2^ per cent Ans. £5 6 8. Ans. £1649 8| $5.1Sh $260.12 £14 6 6 $25,078. £7 18 8i $6,886 I- r 140 ■rPPLEHVNTART XXntCIIlS. \ : \ ■ Rum 4 A2fz> 6. Paoh 244-246. Find the interest of 1. $987.27 for 89 days at 6 per cent 2. £187 18 6 for 127 days at 6 per cent 8. £128 14 Qh for 189 days at 4i per cent 4. !^1234.66 for HI days at 21 per cent 6. £999 19 9<^ for 186 days at 4 per cent 6. $1483.055 for 86 days at 2h per cen^ Calculate the amount of £146 13 9^ 7. From 1st May, 1864, till 14th July, 1865, at 4^ per ot 8. From 22d July, 1863, till 15th Nov. , 1864, at 5 per ct 9. Required the interest on £1987 14 3 from 2d May till 18th October at 7 per cent 10. What is the interest of $1768.443 from 18thMarch, 1862, till 17th May, 1864, at 4| per cent? 11. Find the interest on #4987.72 from Ist Nov., 1848, till 2d January, 1864, at 3^ per cent. Rule 6. Page 248. Find the interest of the following principals, fbr the 4 per cent, per annum : 1. £488 13 6 for 89 days. ^1144.13i for 126 days #1001.15 for 1 year, 101 days £1635 15 6 for 2 years, 11 days. £487 13 2 for 2 years, 16 days. $489.19 fbr 1876 days. £156 14 for 128 days. BiTLB 7. Page 249. What sum will produce for intei'est £1 6 in 1 year at 5 per cent? What principal at 8^ per cent, will produce a year- ly income of £6 18 10 ? What principal lent out for 8 years, 8 ma, at 8i| per cent will produce for interest £48 2 67 What sum of money lent out at 4| per cent will produce for interest $111,804 2 y., 5 mo.? How many guineas must be lent for 117 days at 8^ 2. 8. 4. 6. 6. 7. 1. 8. 6. 6. per cent, so as to receive for interest £0 2 4yi^J^. What sum lent on the 16th Maieh, 1850, tiU 23d Jan., 1851, at &}■ per cent will produce for inte- rest $4.50|f 7. What sum must be lent from Mar. 14th till June 8th, at 6 per cent to produce for interest £3 9 10 ? BuLE 8. Page 250. 1. In yrhat time will $4000 amount to $4840 at ii per cent, per annum? 2. In what time will £2838 6 8 amount to £3216 16 8 «t 8 per cent? Ant. $14.44^ " £2 7 101 •« £2 17 71 «* $8.76 " £20 7 8 " $8.73i "£154 12 11 " £156 7 1 ** £64 8 5} " $182.04 " $2795.08i given times at Ana. £4 9111 " $16.80 " $51.18i " £182 16 11 " £39 17 4 " $100.68 M £2 2 8 «• £86 *« £198 6 8 £876 $978.97JJ ** lOgoinc «« « «• $168.00 " £247 «• 2 yean. *• 4h yeun. SUPPLEMENTARY EXERCISES. 841 I 1. 2. 3. 6. 6. (C <( 8. In wliat time will $1786.00 amount to f 207G.22i at 5 per cent. ? Alls. 3 yrs. 3 mos. How long must £243 10 be lent at simple int(in^«t, at 4% per cent., to amount to £271 9 0|J. ? Ans. 2 yi'S. 6 mos. If $706. 2G| be lent out on the 17th day of March, at 5i per cent. ; on what (lay was it paid wlien the amount was $723.40 ? Ans. Aug. 25th. £70 6 was lent on Juno Dtli, at 4 per cent, simple interest, and when paid amounted to £70 13 85 ; on what day Avas it paid ? Aus. Dec. Gth. Rule 9. Page 250. At what rate will £236 6 8 produce for mterest £17 14 6 in 2i years ? Ans. 3 per cent. If $3746.6' amount to $4629.47i in 4J years, at what rate was it lent ? Ans. 4| per cent. At what rate per cent., simple interest, will $160 double itself in 25 yrs ? Ans. 4 per cent. At what rate will $1900 amount to $2185 in 3 yrs ? If £225 amount to £256 10 in 4 years, at what rate was it lent ? £593 12 6 was lent 12th May, and the interest due 29th October was £11 1 2:J ; required the rate pr ct. Rule 10. Page 250. What sum will amount to $1488.00 at 6 per cent, in 4 years ? What sum will amount to $921.05 in 18 months at 6 per cent ? What sum must be lent at simple interest at 7 per ct. for 4 yrs and 5 mo., to amount to £571 8 0^ What sum will amount to £739 1 8^6^^. in 1 year, 11 mo., at 4-|- per cent. ? What sum of money lent out for 56 days at 3i per cent, will amount to £688 1 What sum of money lent out for 4i years at 2i per cent, will amount to £804 15 1-|- What sum lent out for 292 days at 2^ per cent, will amount to £844 8 7^ What sum lent June Sth, at 6i per cent., will amount to $506.35, Nov. 1st? What sum lent March 3d, at 5 per cent., will amount to $3653.55f October 28th. COMMISSION, INSURANCE, BROKERAGE, ETC. Art. 222. Page 251. Rules 1 and 2. What is the commission on $486.27 at 3i per cent. ? Ans. $17.02 What is the commission on £86 6 at 7& per cent. ? «« £6 15 10 Find the commission on £774 11 3 at 6 per cent. " £38 14 6| Find the commission on $1987.37i at 4i per ct. Ans. $81.98 nearly. What is the brokerage on £196 17 6 at 2s. 6d. per ct. Ans. £0 4 10^ What must be the sum insured at $5.75 per cent, on | goods worth $7754.50, so that, in case of loss, both the value of the goods and the premium may bo repaid ? " $8227.58^ 4. 6. 6. 1. 2. 8. 4. 5. 6. 1. 2. 8. 4. 6. 6. 7. 8. 9. €t (( 4( (( 5 per cent. Zh per cent. 4 per cent. $1200 $845.00 £486 9 4 « £68418 8 '« £684 7 6 ** £723 7 G « (( (( £827 17 6 $498,518 $3537.73* r.. 342 aUPPLEMENTART EXERCISES. 'fl I ■I i ij 7. At $9.10 per cent., what will it cost to insure goods worth $6240 BO that, ill case of loss, the owner may be entitled to the value of the goods and the premium ? $146.20^. 8. What must be the sum insured at 4i per cent, on goods worth £1010, so that, in case of loss, the worth of the goods and the premium may be recovered ? Ans. £2000. 9. At 7i per cent., what will be the cost of insuring goods or property worth 500 guineas, so that in cose of loss, the woi*th of the property and the premium of insurance may bo repaid ? Ans. £42 11 4JL. 10. Find the brokerage on $1121 .33^ at $0.82^ per cent. Ans. $9.25. 11. Find the expense of insuring $8200 on goods at $4.72:1 per cent., policy-duty $0.25, and allowing jl per cent, for commission. Ans. $175.20. 12. Find the cost of insuring goods worth $1063 at 8i per cent, policy duty $0.88} per cent., commission t^ per cent. Ans. $85.18|. 18. Calculate the expense of insuring $1920 at 4 per cent., and policy duty $0.25. Ans. $81.80 DISCOUNT. — Rule 1. Page 256. Find the present worth of the following bills at the given rates per cent, per annum. DRAWK. DISCOUNTED. 1. £388 2 6, Dec. 8, at 6 mos.. Mar. 25, at 6 per ct. 2. £649 13 4, Nov. 9, •« 8. $2274.55, Apr. 27, " 4. $66.00, Sept. 26, ** 5. £112 19 6, Aug. 12, «* 6. £467136i May 14, " 7. $874.33} Aug. 14, •« 8. $1790.50 June 23, " '9. $90.00 Mar. 31, " 10. $8582.50 Jan. 5, " 11. £146 7 11 Apr. 1, " 9 (( 7 (C 5 t< 7 (( 5 (( 4 (( 6 (( 7 (( 11 (( 8 (f Ap. 19, " 5i «* June 8, " 6 «» Nov. 80, " 6i «* Deo. 22, " si *« Aug. 14, « 6i *« Oct. 8, '« 4 " July 8, "51" May 8, " 6^ •* Oct. 8, " 4i «• «( 4( (( (( (( IC (( (< ANSWERS. £888 2 111. £688 8 2. $2218.46}. $64.98|. £111 12 el. £462 18 If). $867.14X. $1742.261. $87.13}. $8477.92i. £146 7 6. 12. 12 volumes of a work can be bought for a certain sum payable in 6 months : and 13 volumes of the same work can be purchased for the same sum, ready money ; what is the rate of discount ? Ans. 16| per cent. 18. For what sum must a note be drawn, at three months, so that the present worth, if discounted at the Bank at 6 per cent., may be $489.00? Ans. $496.59. 14. For what sum must a note be drawn for 2 months, which, when dis- counted at the Bank at 6 per cent., will liquidate a debt of $189,- 167? Ans. $191,186. BuLE 2. Page 258. Find the true present worth of the several amounts in the precedmg Bule: TRUE PRESENT WORTHS. 1. £883 4 21 2. £688 12 3. $2219.811 4. $65.00 6. £111 12 lOil 6. £462 14 1|. 7. $867.20i 8. $1748.63i^ 9. $87.22 10« £145 7 6i SUPPLFMENTART EXERCISES. 343 »rtli $6240 f the goods I46.20i >o Art. 230. Page 265. 1. If I sell coffee at 40 eta. per lb. and gain 17 -^^ per cent., what waa the prime cost 7 Ans. $0.84|. 2. By selling an article at $1.00, a person lost 5 per cent. ; what was the first cost, and what must he sell it at to gain 4^ per cent. ? Ans. Prime cost, $1.05^^ ; selling price, $1.10. • 8. A merchant sold a lot of cloths for $7205, which was 15 per cent, more than cost ; how much did they cost? Ans. $6817.391. 4. An importer sold a library for £769 10, which was 12i per cent advance on the cost ; what was that cost? Ans. £684 0. 5. A merchant paid 50 per cent, for importing goods which he sells for $800, thus gaining 25 per cent, on the whole cost ; what was the prime cost? Ans. $160. 6. Bought 120 gals, of spirits, into which I put 10 gals, of water, and sold the mixture at $1.80 per gal., realizing a profit of 80 per cent ; what was the prime cost 7 Ans. $1.60. ^ DIVISION INTO PROPORTIONAL PARTS AND RECIPROCAL PROPORTION. — Aets. 231-2. Page 266. 1. Divide $1896.85 among 3 men, according to their ages. A. is 40, B. is 50, and C. is 85. Required their shares. Ans. A.'s, $433.45^ ; B.'s, $541 .81f ; C.'s, $921.08f 2. Three men, in company, gain $1566. A.'s stock is $1600 ; B.'s, .$2400 ; and C.'s $3200. What is each man's share? ; Ans. A.'s, $348 ; B.'s, $522 ; C.'s, $696. ' 8. Divide $1896.82^ among three persons. A., B., and C. in the pro> portion of f i, ^. 1 Ans. A.'s share, $1083.6U ; B.'s, 541.80f ; C.'s, $270.90^. r 4. A person bequeathed £1890 10 6 to his three children to be divided in inverse proportion to their ages. Their ages are 18, 16, and loi yrs. What must be the share of each child? Ans. £546 18 l|f, £614 19 9fJ, £728 17 6|f . 6. Three persons make a joint stock. A. takes 8 snares ; B., 11 ; and C, 12. They lose $216. Required how much each man must advance. Ans. A., $55.74^ ; B., $76.64J} ; C, $83.61J^j. SlPPLriiENTARY EXERCISES. 345 ust they he ^•6020.85. gain 26 per £0 18 0. , I propose U8. |0.7'jJ. them ut ft ier bo as to £46 6. 0, sells one ist he raise le increased IS. 11^^. , what was $0.84J. ; what woa J, $1.10. 6 per cent. 317.391. 4 per cent ;684 0. he sells for the prime s. $160. (rater, and )er cent ; $1.60. ROCAL A. is 40, 21.08f 00 ; B.'s, $696. the pro> e divided loi yrs. 11 ; and advance. •«A- 6. Gunpowder being composed of 16 parts nitre, 3 parts charcoal, and 2 parts sulphur, find how much of each is required for 17' '2 U'?. Ans. Nitre, 1C44 lbs. ; charcoal, 208| lbs. ; sulpimr, 17U^ lbs. 7. Divide $1800 between A., B., and C, so that as oftou as A. get* $6 B. shall get $4, and us ot'toii »s B. gets ^l) C. shall get lii<'l. Aus. A.'s share, li^'.iOO ; B.'s, JniTlIU ; C.'s, !«!240. 8. If a cistern, when in good C(und, stg., he rate of £1 stg. |ir respec- ' 16 2 6 8. J407.0641. |06.8165'.. l)00000':>'j EVOLUTION. ~Abt. 256. Page 289. Find thr square root of the following : 1. 87. 2. 4761. 8. 7056. 4. 9801. 6. 152399025. 6. 10342656. Ans. 9.327+. 69. " 84. " 99. " 12345. " 3216. 7. f |. 9. 17|. 10. |. 11. 34967A. 12. 207||. Ans. ^ Ans. 4.1683+ 79066+ «• 186.9951+ *« 14.4116+ AsT. 260. Paqe 293. Required the cube root of the following numbers : 1. 91125. 2. 140608. 8. 671787. 4. 2515456. 5. 10218313. 6. 11643.176. Ans. t* «( (( (C (f 45. 52. 83. 136. 217. 22.6. 7. 20.570824. 8. .241804367. 1^: HI: Ans. 2.74. " 0.623. Ans. 8.5463+. " .643659+. COMPOUND INTEREST.— Arts. 262-264. Page 296. Find the amounts of the following sums, at the given rates per cent, per annum : for 10 years, at 7 per cent, at 6 1. $960 2. $1000 8. $1460 4. $5000 6. £198 10 6. £136 15 7. $10000 8. £1674 9. £1000 10. £198 16 8 Ans. «* 9 *' 12 ** 20 " 30 " 25 " 40 " 10 « 4 •* 11 (( «( it <( <( «( (( at 4 at 6 at 6 at 3 at 7 at 4 at 6 at 6 <( (< (( <( (C f( (( it it «{ (( (( (( (( (( «< «< $1888.464. $1551.328. $2337.606. $16086.676. £857 18 li. £286 6 6|. $149744. £2477 18 6|. £12161011. £375 12 li Art. 264. Page 298. 1. What sum of money, lent out at compound interest, at 6 per cent., will amount to $35948.37 in 15 years ? Ans. $16000.00. 2. A person wishes his son, who is three years old, to have $11782.80 when he comes to the age of 21 years ; how much must he deposit in the bank to amount to that sum, at 6 per cent compound interest 7 Ans. $4896.00. I I jH TABLE OF CONTENTS. HINTS FOR THE TEACHING OF ARITHMETIC 3 PREFACE , 7 FIRST LESSONS IN ARITHMETIC. Counting 13 On Symbols 13 PART I.— FUNDAMENTAL RULES. Addition 14 Subtraction, 16 Multiplication 17 Division 19 Fractional Numbers 20 Application of Numbers to Money 22 Application of Numbers to Measures, Weights, &c 23 Decimal Notation 24 Arithmetical Tables 29 PART II.— ARITHMETIC. Numeration Table 34 Exercises in Numeration 35 Exercises in Notation 30 Table of Rotnan Notation 37 Arithmetical Signs 3S Addition 40 Simple Addition 40 Subtraction 45 Simple Subtraction 45 Multiplication 49 Simple Multiplication 50 305 ilt TABLE or COXTKNTS. . 7 . 13 13 . 14 , 16 . 17 . 19 20 . 22 23 24 29 34 35 30 37 38 40 40 45 45 49 50 Contractions in Multiplication 56 Divisiorf. 64 Simple Division 66 Short Division 66 ivision by Composite Numbers 72 General Principles iu Division 74 Multiplication and Division by Fractional Numbers 77 Ciintractions in Divisions 78 Cancellation 79 On the Construction of Questions 80 Properties of N umbers ^ 82 Greatest Common Measure 86 Least common Multiple SU Decimals 94 Decimal Table 96 Additional of Decimal Fractions 98 Subtraction of Decimals 99 Division of Decimals 104 Keduction, Addition, Subtraction, &c. of Compound or De- nominate Numbers 110 Table of Money 110 Reduction 112 Reduction of Sterling Money 114 Compound Addition 115 Addition of Sterling Money 116 Compound Subtraction 117 Subtraction of Sterling Money 118 Compound Multiplication 119 Multiplication of Sterling Money 120 Compound Division 121 Division of Sterling Money 123 Currency of Nova Scotia. . . 130 Table of Decimal Currency 130 Measures of Lenath 130 Cloth Measure 138 Measure of Surface Table of Square Measure 140 Artificer's Measure 144 Measure of Solidity. Cubic Measure 147 .' tl^v" TAniT- or CONTENTg. \lm III I f I' iii.n / Measure of Capacity. Dry Measure ............... 150 Ileiped Measure 150 Measures of Weight. Troy Weight 1 52 Avoidupoia Weight l.')4 New Sjstcm of Weight ..155 Measure of Space 153 Measure of Time 158 Weights and Measures of tlie United States Currency IC'2 Measure of Length 1G3 Liquid Measure 1G3 Dry Measure 1 63 Measures of Weight 1G3 French Money , Weights and Measures 1G3 " Leneal Measure 164 ** Square Measure 104 •* Cubic Measure 1C5 " Liquid and Dry Measure 105 ** Weight 1G5 ** Circular Measure 1G5 Rules for Mental Arithmetic 108 To Calculate Interest Mentally 172 Fractions. —Vulgar Fractions 17t) Reduction of Vulgar Fractions 180 / Addition of vulgar Fractions 186 Subtraction of Vulgar Fractions 188 Multiplication of Vulgar Fractions 189 Division of Vulgar Fractions 191 Reduction of Decimal Fractions 193 Multiplication of Circulating Decimals 203 PRACTICE 205 Simple Practice 206 Compound Practice 202 RATIO 223 PROPORTION 226 Simple Proportion 228 Compound Proportion 233 .... ..150 ......IDO 152 1.^4 . ...155 153 158 1C2 163 1G3 163 1G3 163 164 164 165 165 . . ..165 ) • . • • • 100 168 172 176 180 .....188 188 189 191 193 203 205 206 202 223 • « . . • ii2xj ...228 ...233 TABLE cr coNTE^rrg. INTEREST. Simple Interest 238 Commissions, lutcrest, Brokerage, &o 251 Application of Interest 253 Discount 250 Stocks ; 259 Recipi'OCUDil Proportion • 2G() Fellowship and Partnership 269 Compound Fellowship 270 Alligation 172 Alligation Medial 272 ♦• Alternate 272 Comix)und Proportion 27Q Exchange 278 English and American Exchange 281 Arbitration of Exchange 285 Involution 287 Evolution 289 Square Roots '. 290 Cube Root 293 Compound Interest 290 Miscellaneous Exercises 299 9upplemontary Exerc'ses ....'*. 304