SECTION III.. Interest, Etc.

INTEREST, ETC.

245. In the questions contained in this Section, almost the only process which will be employed is the taking a fractional part of a sum of money, which has been done before in several cases. Suppose it required to take 7 parts out of 40 from £16, that is, to divide £16 into 40 equal parts, and take 7 of them. Each of these parts is

16 16 16 × 7 £----, and 7 of them make ---- × 7, or ------ pounds (116). 40 40 40

The process may be written as below:

£16 7 ----- 40)112(£2 . 16s. 80 -- 32 20 --- 640 40 --- 240 240 --- 0

Suppose it required to take 13 parts out of a hundred from £56. 13. 7½.

56 . 13 . 7½ 13 ---------------- 100) 736 . 17 . 1½ ( £7 . 7 . 4 ¼ ¹/₄₁ 700 --- 36 × 20 + 17 = 737 700 --- 37 × 12 + 1 = 445 400 --- 45 × 4 × 2 = 182 100 --- 82

Let it be required to take 2½ parts out of a hundred from £3 12s. The result, by the same rule is

£3 12s. × 2½ 5 -------------------, or 123 £3 12s. × ---; 100 200

so that taking 2½ out of a hundred is the same as taking 5 parts out of 200.

EXERCISES.

Take 7⅓ parts out of 53 from £1 10s.

129 Answer, 4s. 1---d. 159

Take 5 parts out of 100 from £107 13s. 4¾d.

Answer, £5. 7. 8 and ³/₂₀ of a farthing.

£56 3s. 2d. is equally divided among 32 persons. How much does the share of 23 of them exceed that of the rest?

Answer, £24. 11. 4½ ½.

246. It is usual, in mercantile business, to mention the fraction which one sum is of another, by saying how many parts out of a hundred must be taken from the second in order to make the first. Thus, instead of saying that £16 12s. is the half of £33 4s., it is said that the first is 50 per cent of the second. Thus, £5 is 2½ per cent of £200; because, if £200 be divided into 100 parts, 2½ of those parts are £5. Also, £13 is 150 per cent of £8. 13. 4, since the first is the second and half the second. Suppose it asked, How much per cent is 23 parts out of 56 of any sum? The question amounts to this: If he who has £56 gets £100 for them, how much will he who has 23 receive? This, by 238, is 23 × ¹⁰⁰/₅₆ or ²³⁰⁰/₅₆ or 41¹/₁₄. Hence, 23 out of 56 is 41¹/₁₄ per cent.

Similarly 16 parts out of 18 is 16 × ¹⁰⁰/₁₈, or 88⁸/₉ per cent, and 2 parts out of 5 is 2 × ¹⁰⁰/₅, or 40 per cent.

From which the method of reducing other fractions to the rate per cent is evident.

Suppose it asked, How much per cent is £6. 12. 2 of £12. 3? Since the first contains 1586d., and the second 2916d., the first is 1586 out of 2916 parts of the second; that is, by the last rule, it is ¹⁵⁸⁶⁰⁰/₂₉₁₆, or 54¹¹³⁶/₂₉₁₆, or £54. 7. 9½ per cent, very nearly. The more expeditious way of doing this is to reduce the shillings, &c. to decimals of a pound. Three decimal places will give the rate per cent to the nearest shilling, which is near enough for all practical purposes. For instance, in the last example, which is to find how much £6·608 is of £12·15, 6·608 × 100 is 660·8, which divided by 12·15 gives £54·38, or £54. 7. Greater correctness may be had, if necessary, as in the Appendix.

EXERCISES.

How much per cent is 198¼ out of 233 parts?--Ans. £85. 1. 8¾.

Goods which are bought for £193. 12, are sold for £216. 13. 4; how much per cent has been gained by them?

Answer, A little less than £11. 18. 6.

A sells goods for B to the amount of £230. 12, and is allowed a commission of 3 per cent; what does that amount to?

Answer, £6 . 18. 4¼ ⁷/₂₅.

Commission is what is allowed by one merchant to another for buying or selling goods for him, and is usually a per-centage on the whole sum employed. Brokerage is an allowance similar to commission, under a different name, principally used in the buying and selling of stock in the funds.

Insurance is a per-centage paid to those who engage to make good to the payers any loss they may sustain by accidents from fire, or storms, according to the agreement, up to a certain amount which is named, and is a per-centage upon this amount. Tare, tret, and cloff, are allowances made in selling goods by wholesale, for the weight of the boxes or barrels which contain them, waste, &c.; and are usually either the price of a certain number of pounds of the goods for each box or barrel, or a certain allowance on each cwt.

A stockbroker buys £1700 stock, brokerage being at £⅛ per cent; what does he receive?--Answer, £2. 2. 6.

A ship whose value is £15,423 is insured at 19⅔ per cent; what does the insurance amount to?--Answer, £3033. 3. 9½ ²/₅.

247. In reckoning how much a bankrupt is able to pay his creditors, as also to how much a tax or rate amounts, it is usual to find how many shillings in the pound is paid. Thus, if a person who owes £100 can only pay £50, he is said to pay 10s. in the pound. The rule is easily derived from the same reasoning as in 246. For example, £50 out of £82 is

50 50×20 £---- out of £1, or ----- shillings, 82 82

or 12s. 2½ ¹⁵/₄₁ in the pound.

248. INTEREST is money paid for the use of other money, and is always a per-centage upon the sum lent. It may be paid either yearly, half-yearly, or quarterly; but when it is said that £100 is lent at 4 per cent, it must be understood to mean 4 per cent per annum; that is, that 4 pounds are paid every year for the use of £100.

The sum lent is called the principal, and the interest upon it is of two kinds. If the borrower pay the interest as soon as, from the agreement, it becomes due, it is evident that he has to pay the same sum every year; and that the whole of the interest which he has to pay in any number of years is one year’s interest multiplied by the number of years. But if he do not pay the interest at once, but keeps it in his hands until he returns the principal, he will then have more of his creditor’s money in his hands every year, and if it were so agreed will have to pay interest upon each year’s interest for the time during which he keeps it after it becomes due. In the first case, the interest is called simple, and in the second compound. The interest and principal together are called the amount.

249. What is the simple interest of £1049. 16. 6 for 6 years and one-third, at 4½ per cent? This interest must be 6⅓ times the interest of the same sum for one year, which (245) is found by multiplying the sum by 4½, and dividing by 100. The process is as follows:

(230) (a) |£1049 . 16 . 6 +-------------- a × 4 | 4199 . 6 . 0 a × ½ | 524 . 18 . 3 +--------------

(82) 100) 47,24 . 4 . 3(£47 . 4 . 10¹¹/₁₀₀

20 ---- (228) 4,84 12 ------ 10,11

(b) £47 . 4 . 10¹¹/₁₀₀ Int. for one yr. +------------------ b × 6 | 283 . 9 . 0⁶⁶/₁₀₀ b × ⅓ | 15 . 14 . 11³⁷/₁₀₀ +--------------------- £299 . 4 . 0³/₁₀₀ Int. for 6⅓ yrs.

Here the 4s. from the dividend is taken in.

Here the 3d. from the dividend is taken in.

EXERCISES.

What is the interest of £105. 6. 2 for 19 years and 7 weeks at 3 per cent?

Answer, £60. 9, very nearly.

What is the difference between the interest of £50. 19 for 7 years at 3 per cent, and for 8 years at 2½ per cent? Answer, 10s. (2½)d.

What is the interest of £157. 17. 6 for one year at 5 per cent?

Answer, £7. 17. 10½.

Shew that the interest of any sum for 9 years at 4 per cent is the same as that of the same sum for 4 years at 9 per cent.

250. In order to find the interest of any sum at compound interest, it is necessary to find the amount of the principal and interest at the end of every year; because in this case (248) it is the amount of both principal and interest at the end of the first year, upon which interest accumulates during the second year. Suppose, for example, it is required to find the interest, for 3 years, on £100, at 5 per cent, compound interest. The following is the process:

£100 First principal. 5 First year’s interest. --- 105 Amount at the end of the first year. (249) 5 . 5 Interest for the second year on £105. -------- 110 . 5 Amount at the end of two years. 5 . 10 . 3 Interest due for the third year. ------------ 115 . 15 . 3 Amount at the end of three years. 100 . 0 . 0 First principal. ------------ 15 . 15 . 3 Interest gained in the three years.

When the number of years is great, and the sum considerable, this process is very troublesome; on which account tables are constructed to shew the amount of one pound, for different numbers of years, at different rates of interest. To make use of these tables in the present example, look into the column headed “5 per cent;” and opposite to the number 3, in the column headed “Number of years,” is found 1·157625; meaning that £1 will become £1·157625 in 3 years. Now, £100 must become 100 times as great; and 1·157625 × 100 is 115·7625 (141); but (221) £·7625 is 15s. 3d.; therefore the whole amount of £100 is £115. 15. 3, as before.

Sufficient tables for all common purposes are contained in the article on Interest in the Penny Cyclopædia; and ample ones in the Treatise on Annuities and Reversions, in the Library of Useful Knowledge.

251. Suppose that a sum of money has lain at simple interest 4 years, at 5 per cent, and has, with its interest, amounted to £350; it is required to find what the sum was at first. Whatever the sum was, if we suppose it divided into 100 parts, 5 of those parts were added every year for 4 years, as interest; that is, 20 of those parts have been added to the first sum to make £350. If, therefore, £350 be divided into 120 parts, 100 of those parts are the principal which we want to find, and 20 parts are interest upon it; that is, the principal is £(350 × 100)/150, or £291. 13. 4.

252. Suppose that A was engaged to pay B £350 at the end of four years from this time, and that it is agreed between them that the debt shall be paid immediately; suppose, also, that money can be employed at 5 per cent, simple interest; it is plain that A ought not to pay the whole sum, £350, because, if he did, he would lose 4 years’ interest of the money, and B would gain it. It is fair, therefore, that he should only pay to B as much as will, with interest, amount in four years to £350, that is (251), £291. 13. 4. Therefore, £58. 6. 8 must be struck off the debt in consideration of its being paid before the time. This is called DISCOUNT; and £291. 13. 4 is called the present value of £350 due four years hence, discount being at 5 per cent. The rule for finding the present value of a sum of money (251) is: Multiply the sum by 100, and divide the product by 100 increased by the product of the rate per cent and number of years. If the time that the debt has yet to run be expressed in years and months, or months only, the months must be reduced to the equivalent fraction of a year.

This rule is obsolete in business. When a bill, for instance, of £100 having a year to run, is discounted (as people now say) at 5 per cent, this means that 5 per cent of £100, or £5, is struck off.

EXERCISES.

What is the discount on a bill of £138. 14. 4, due 2 years hence, discount being at 4½ per cent?

Answer, £11. 9. 1.

What is the present value of £1031. 17, due 6 months hence, interest being at 3 per cent?

Answer, £1016. 12.

253. If we multiply by a + b, or by a-b, when we should multiply by a, the result is wrong by the fraction

b b --- + b, or ---------, a a - b

of itself: being too great in the first case, and too small in the second. Again, if we divide by a + b, where we should have divided by a, the result is too small by the fraction b/a of itself; while, if we divide by a-b instead of a, the result is too great by the same fraction of itself. Thus, if we divide by 20 instead of 17, the result is ³/₁₇ of itself too small; and if we divide by 360 instead of 365, the result is too great by ⁵/₃₆₅, or ¹/₇₃ of itself.

If, then, we wish to find the interest of a sum of money for a portion of a year, and have not the assistance of tables, it will be found convenient to suppose the year to contain only 360 days, in which case its 73d part (the 72d part will generally do) must be subtracted from the result, to make the alteration of 360 into 365. The number 360 has so large a number of divisors, that the rule of Practice (230) may always be readily applied. Thus, it is required to find the portion which belongs to 274 days, the yearly interest being £18. 9. 10, or 18·491.

274 18·491 ------ 180 is ½ of 360 9·246 --- 94 90 is ½ of 180 4·623 -- 4 is ¹/₉₀ of 360 ·205 ------ 9)14·074 ------ 8)1·564 ----- ·196 13·878 = £13 . 17 . 7 Answer.

But if the nearest farthing be wanted, the best way is to take 2-tenths of the number of days as a multiplier, and 73 as a divisor; since m ÷ 365 is 2m ÷ 730, or (²/₁₀)m ÷ 73. Thus, in the preceding instance, we multiply by 54·8 and divide by 73; and 54·8 × 18·491 = 1013·3068, which divided by 73 gives 13·881, very nearly agreeing with the former, and giving £13. 17. 7½, which is certainly within a farthing of the truth.

254. Suppose it required to divide £100 among three persons in such a way that their shares may be as 6, 5, and 9; that is, so that for every £6 which the first has, the second may have £5, and the third £9. It is plain that if we divide the £100 into 6 + 5 + 9, or 20 parts, the first must have 6 of those parts, the second 5, and the third 9. Therefore (245) their shares are respectively,

100 × 6 100 × 5 100 × 9 £-------, £------- and £-------, or £30, £25, and £45. 20 20 20

EXERCISES.

Divide £394. 12 among four persons, so that their shares may be as 1, 6, 7, and 18.--Answer, £12. 6. 7½; £73. 19. 9; £86. 6. 4½; £221. 19. 3.

Divide £20 among 6 persons, so that the share of each may be as much as those of all who come before put together.--Answer, The first two have 12s. 6d.; the third £1. 5; the fourth £2. 10; the fifth £5; and the sixth £10.

255. When two or more persons employ their money together, and gain or lose a certain sum, it is evidently not fair that the gain or loss should be equally divided among them all, unless each contributed the same sum. Suppose, for example, A contributes twice as much as B, and they gain £15, A ought to gain twice as much as B; that is, if the whole gain be divided into 3 parts, A ought to have two of them and B one, or A should gain £10 and B £5. Suppose that A, B, and C engage in an adventure, in which A embarks £250, B £130, and C £45. They gain £1000. How much of it ought each to have? Each one ought to gain as much for £1 as the others. Now, since there are 250 + 130 + 45, or 425 pounds embarked, which gain £1000, for each pound there is a gain of £¹⁰⁰⁰/₄₂₄. Therefore A should gain 1000 × ²⁵⁰/₄₂₅ pounds, B should gain 1000 × ¹³⁰/₄₂₅ pounds, and C 1000 × ⁴⁵/₄₂₅ pounds. On these principles, by the process in (245), the following questions may be answered.

A ship is to be insured, in which A has ventured £1928, and B £4963. The expense of insurance is £474. 10. 2. How much ought each to pay of it?

Answer, A must pay £132. 15. (2½).

A loss of £149 is to be made good by three persons, A, B, and C. Had there been a gain, A would have gained 4 times as much as B, and C as much as A and B together. How much of the loss must each bear?

Answer, A pays £59. 12, B £14. 18, and C £74. 10.

256. It may happen that several individuals employ several sums of money together for different times. In such a case, unless there be a special agreement to the contrary, it is right that the more time a sum is employed, the more profit should be made upon it. If, for example, A and B employ the same sum for the same purpose, but A’s money is employed twice as long as B’s, A ought to gain twice as much as B. The principle is, that one pound employed for one month, or one year, ought to give the same return to each. Suppose, for example, that A employs £3 for 6 months, B £4 for 7 months, and C £12 for 2 months, and the gain is £100; how much ought each to have of it? Now, since A employs £3 for six months, he must gain 6 times as much as if he employed it one month only; that is, as much as if he employed £6 × 3, or £18, for one month; also, B gains as much as if he had employed £4 × 7 for one month; and C as if he had employed £12 × 2 for one month. If, then, we divide £100 into 6 × 3 + 4 × 7 + 12 × 2, or 70 parts, A must have 6 × 3, or 18, B must have 4 × 7, or 28, and C 12 × 2, or 24 of those parts. The shares of the three are, therefore,

6 × 3 × 100 4 × 7 × 100 £----------------------, £----------------------, 6 × 3 + 4 × 7 + 12 × 2 6 × 3 + 4 × 7 + 12 × 2

12 × 2 × 100 and £----------------------. 6 × 3 + 4 × 7 + 12 × 2

EXERCISES.

A, B, and C embark in an undertaking; A placing £3. 6 for 2 years, B £100 for 1 year, and C £12 for 1½ years. They gain £4276. 7 How much must each receive of the gain?

Answer, A £226. 10. 4; B £3432. 1. 3; C £617. 15. 5.

A, B, and C rent a house together for 2 years, at £150 per annum. A remains in it the whole time, B 16 months, and C 4½ months, during the occupancy of B. How much must each pay of the rent?

Answer, A should pay £190. 12. 6; B £90. 12. 6; C £18. 15.

This question does not at first appear to fall under the rule. A little thought will serve to shew that what probably will be the first idea of the proper method of solution is erroneous.

257. These are the principal rules employed in the application of arithmetic to commerce. There are others, which, as no one who understands the principles here laid down can fail to see, are virtually contained in those which have been given. Such is what is commonly called the Rule of Exchange, for such questions as the following: If 20 shillings be worth 25½ francs, in France, what is £160 worth? This may evidently be done by the Rule of Three. The rules here given are those which are most useful in common life; and the student who understands them need not fear that any ordinary question will be above his reach. But no student must imagine that from this or any other book of arithmetic he will learn precisely the modes of operation which are best adapted to the wants of the particular kind of business in which his future life may be passed. There is no such thing as a set of rules which are at once most convenient for a butcher and a banker’s clerk, a grocer and an actuary, a farmer and a bill-broker; but a person with a good knowledge of the principles laid down in this work, will be able to examine and meet his own future wants, or, at worst, to catch with readiness the manner in which those who have gone before him have done so for themselves.

APPENDIX TO THE FIFTH EDITION OF

DE MORGAN’S ELEMENTS OF ARITHMETIC.

I. ON THE MODE OF COMPUTING.

The rules in the preceding work are given in the usual form, and the examples are worked in the usual manner. But if the student really wish to become a ready computer, he should strictly follow the methods laid down in this Appendix; and he may depend upon it that he will thereby save himself trouble in the end, as well as acquire habits of quick and accurate calculation.

I. In numeration learn to connect each primary decimal number, 10, 100, 1000, &c. not with the place in which the unit falls, but with the number of ciphers following. Call ten a one-cipher number, a hundred a two-cipher number, a million a six-cipher number, and so on. If five figures be cut off from a number, those that are left are hundred-thousands; for 100,000 is a five-cipher number. Learn to connect tens, hundreds, thousands, tens of thousands, hundreds of thousands, millions, &c. with 1, 2, 3, 4, 5, 6, &c. in the mind. What is a seventeen-cipher number? For every 6 in seventeen say million, for the remaining 5 say hundred-thousand: the answer is a hundred thousand millions of millions. If twelve places be cut off from the right of a number, what does the remaining number stand for?--Answer, As many millions of millions as there are units in it when standing by itself.

II. After learning to count forwards and backwards with rapidity, as in 1, 2, 3, 4, &c. or 30, 29, 28, 27, &c., learn to count forwards or backwards by twos, threes, &c. up to nines at least, beginning from any number. Thus, beginning from four and proceeding by sevens, we have 4, 11, 18, 25, 32, &c., along which series you must learn to go as easily as along the series 1, 2, 3, 4, &c.; that is, as quick as you can pronounce the words. The act of addition must be made in the mind without assistance: you must not permit yourself to say, 4 and 7 are 11, 11 and 7 are 18, &c.; but only 4, 11, 18, &c. And it would be desirable, though not so necessary, that you should go back as readily as forward; by sevens for instance, from sixty, as in 60, 53, 46, 39, &c.

III. Seeing a number and another both of one figure, learn to catch instantly the number you must add to the smaller to get the greater. Seeing 3 and 8, learn by practice to think of 5 without the necessity of saying 3 from 8 and there remains 5. And if the second number be the less, as 8 and 3, learn also by practice how to pass up from 8 to the next number which ends with 3 (or 13), and to catch the necessary augmentation, five, without the necessity of formally undertaking in words to subtract 8 from 13. Take rows of numbers, such as

4 2 6 0 5 0 1 8 6 4

and practise this rule upon every figure and the next, not permitting yourself in this simple case ever to name the higher one. Thus, say 4 and 8 (4 first, 2 second, 4 from the next number that ends with 2, or 12, leaves 8), 2 and 4, 6 and 4, 0 and 5, 5 and 5, 0 and 1, 1 and 7, 8 and 8, 6 and 8.

IV. Study the same exercise as the last one with two figures and one. Thus, seeing 27 and 6, pass from 27 up to the next number that ends with 6 (or 36), catch the 9 through which you have to pass, and allow yourself to repeat as much as “27 and 9 are 36.” Thus, the row of figures 17729638109 will give the following practice: 17 and 0 are 17; 77 and 5 are 82; 72 and 7 are 79; 29 and 7 are 36; 96 and 7 are 103; 63 and 5 are 68; 38 and 3 are 41; 81 and 9 are 90; 10 and 9 are 19.

V. In a number of two figures, practise writing down the units at the moment that you are keeping the attention fixed upon the tens. In the preceding exercise, for instance, write down the results, repeating the tens with emphasis at the instant of writing down the units.

VI. Learn the multiplication table so well as to name the product the instant the factors are seen; that is, until 8 and 7, or 7 and 8, suggest 56 at once, without the necessity of saying “7 times 8 are 56.” Thus looking along a row of numbers, as 39706548, learn to name the products of every successive pair of digits as fast as you can repeat them, namely, 27, 63, 0, 0, 30, 20, 32.

VII. Having thoroughly mastered the last exercise, learn further, on seeing three numbers, to augment the product of the first and second by the third without any repetition of words. Practise until 3, 8, 4, for instance, suggest 3 times 8 and 4, or 28, without the necessity of saying “3 times 8 are 24, and 4 is 28.” Thus, 179236408 will suggest the following practice, 16, 65, 21, 12, 22, 24, 8.

VIII. Now, carry the last still further, as follows: Seeing four figures, as 2, 7, 6, 9, catch up the product of the first and second, increased by the third, as in the last, without a helping word; name the result, and add the next figure, name the whole result, laying emphasis upon the tens. Thus, 2, 7, 6, 9, must immediately suggest “20 and 9 are 29.” The row of figures 773698974 will give the instances 52 and 6 are 58; 27 and 9 are 36; 27 and 8 are 35; 62 and 9 are 71; 81 and 7 are 88; 79 and 4 are 83.

IX. Having four numbers, as 2, 4, 7, 9, vary the last exercise as follows: Catch the product of the first and second, increased by the third; but instead of adding the fourth, go up to the next number that ends with the fourth, as in exercise IV. Thus, 2, 4, 7, 9, are to suggest “15 and 4 are 19.” And the row of figures 1723968929 will afford the instances 9 and 4 are 13; 17 and 2 are 19; 15 and 1 are 16; 33 and 5 are 38; 62 and 7 are 69; 57 and 5 are 62; 74 and 5 are 79.

X. Learn to find rapidly the number of times a digit is contained in given units and tens, with the remainder. Thus, seeing 8 and 53, arrive at and repeat “6 and 5 over.” Common short division is the best practice. Thus, in dividing 236410792 by 7,

7)236410792 --------- 33772970, remainder 2.

All that is repeated should be 3 and 2; 3 and 5; 7 and 5; 7 and 2; 2 and 6; 9 and 4; 7 and 0; 0 and 2.

In performing the several rules, proceed as follows:

ADDITION. Not one word more than repeating the numbers written in the following process: the accented figure is the one to be written down; the doubly accented figure is carried (and don’t say “carry 3,” but do it).

47963 1598 26316 54792 819 6686 ------ 138174

6, 15, 17, 23, 31, 3″ 4′; 11, 12, 21, 22, 31, 3″7′; 9, 17, 24, 27, 32, 4″1′; 10, 14, 20, 21, 2″8′; 7, 9, 1′3′.

In verifying additions, instead of the usual way of omitting one line, adding without it, and then adding the line omitted, verify each column by adding it both upwards and downwards.

SUBTRACTION. The following process is enough. The carriages, being always of one, need not be mentioned.

From 79436258190 Take 58645962738 ----------- 20790295452

8 and 2′, 4 and 5′, 7 and 4′, 3 and 5′, 6 and 9′, 10 and 2′, 6 and 0′, 4 and 9′, 7 and 7′, 9 and 0′, 5 and 2′. It is useless to stop and say, 8 and 2 make 10; for as soon as the 2 is obtained, there is no occasion to remember what it came from.

MULTIPLICATION. The following, put into words, is all that need be repeated in the multiplying part; the addition is then done as usual. The unaccented figures are carried.

670383 9876 ------- 4022298 18′, 49′, 22′, 2′, 42′, 4′0′, 4692681 21′, 58′, 26′, 2′, 49′, 4′6′, 5363064 24′, 66′, 30′, 3′, 56′, 5′3′, 6033447 27′, 74′, 34′, 3′, 63′, 6′0′. ---------- 6620702508

Verify each line of the multiplication and the final result by casting out the nines. (Appendix II. p. 166.)

It would be almost as easy, for a person who has well practised the 8th exercise, to add each line to the one before in the process, thus:

670383 9876 ------- 4022298 50949108 587255508 6620702508

8; 21 and 9 are 30′; 59 and 2 are 61′; 27 and 2 are 29; 2 and 2 are 4′; 49 and 0 are 49′; 46 and 4 are 5′0′.

On the right is all the process of forming the second line, which completes the multiplication by 76, as the third line completes that by 876, and the fourth line that by 9876.

DIVISION. Make each multiplication and the following subtraction in one step, by help of the process in the 9th exercise, as follows:

27693)441972809662(15959730 165042 265778 165410 269459 202226 83756 6772

The number of words by which 26577 is obtained from 165402 (the multiplier being 5) is as follows: 15 and 7′ are 2″2; 47 and 7′ are 5″4; 35 and 5′ are 4″0; 39 and 6′ are 4″5; 14 and 2′ are 16.

The processes for extracting the square root, and for the solution of equations (Appendix XI.), should be abbreviated in the same manner as the division.

The teacher will find further remarks on this subject in the Companion to the Almanac for 1844, and in the Supplement to the Penny Cyclopædia, article Computation.