of 120 hides, which would not be as many as we wanted by seven; we therefore multiply the price of one hide, 13s. 4d., by seven, and add it to the price of 120 hides, which we have found already, and the total is the sum required. £. s. d. 1 4 3 x 7 10 12 2 6 × 4 10 121 5 0 3 363 15 0 48 10 0 When the quantity is still larger, multiply the given price by a series of tens, to find the price of 10, 100, 1000. (13.) Thus, multiply £1. 4s. 3d. by 347. In this example the £1. 4s. 3d. is multiplied by 10, and the product, £12. 2s. 6d., we know to be the sum of £1. 4s. 3d. ten times told; again, the first product is multiplied by ten, and we know that the second product is the amount of £1. 4s. 3d. one hundred times told; the second product we multiply by three, and the third product is the amount three hundred times told; there then remains 47, or four tens and seven, we therefore multiply the £12. 2s. 6d., being the product of ten, by four, which produces £48. 10s. or the amount of the original sum forty times told; there then remains seven to multiply by; with this seven we multiply the first sum, £1. 4s. 3d.; the product, £8. 9s. 9d., we add to the other two products, and the total £420. 14s. 9d. is the amount of £1. 4s. 3d. three hundred and forty-seven times told. 8 9 9 420 14 9 (14.) What is the price of 126 stone of soap at 1s. 5d. per stone? Ans. £8. 18s. 6d. (15.) Bought 14 loads of hay at £3. 17s. 6d. per load; how much did they come to? (16.) Bought 1174 gallons of gin at lon; Ans. £54. 5s. 12s. 6d. per gal what is the amount of the whole? Ans. £73. 5s. 7 d. Note, when there is a fraction of, 4, or å, connected with the multiplier, take,, or 4, the given price, and add it in with the last product. (17.) What is the weight of 383 hhds. of sugar, the weight of each hhd. being 4 cwt. 3 qrs. 20 lb.? Ans. 1887 cwt. 2 qrs. 16 lb. (18.) Multiply 126 oz. 10 dwt. 20 gr. by 72. Ans. 9111 oz. (19.) Multiply 270 lb. 10 oz. 7 dr. 2 sc. 15 gr. by 7. Ans. 1896 lb. 4 oz. 7dr. 1 sc. 5 gr. (20.) Multiply 76 lea. 2 ml. 7 fur. 37 per. 5 yds. by 6. Ans. 461 lea. 2 ml. 7 fur. 27 pls. 2 yds. (21.) Multiply 40 yds. 3 qrs. 3 nls. 1 in. by 5. Ans. 204 yds. 3 qrs. 1 nl. in. (22.) Multiply 39 ld. 4 qrs. 7 bhl. 3 pk. 1 gal. 3 qt. Ans. 759 lds. 4 qrs. 7 bhl. 1 pk. 1 gal. 1 qt. by 19. Proof of Multiplication. Divide the product by the multiplier; the quotient will be the multiplicand, if the work is correct. COMPOUND DIVISION. Is exactly the reverse of Multiplication, and tends to undo what Multiplication has done; and to make the pupil more conversant with the proof in each of these E 5 rules, we will, for the sums given for practice in this rule, take the products of Multiplication for dividends, and the multipliers for divisors. The sums in this rule are to be worked in the same way as the sums with which you are now familiar in Simple Division; that is, when the divisor does not exceed twelve, the sum is within the rule of Short Division; when the divisor exceeds twelve, it is worked by Long Division. When by Short Division, you write the dividend as you may happen to have it, whether money, weight, or measure, in its several denominations; drawing a line underneath, you write the divisor on the left of the dividend, and proceed as in the annexed example. It is thus divided when the divisor is contained in the several denominations evenly, but this £. S. d 3) 3 12 9 1 4 3 will not always be the case; and let us now choose other terms that will not fit thus evenly, and examine the method then adopted. 6) 7 5 6 1 4 3 Divide £7.5s. 6d. by 6. In this case, having divided the pounds, we find one remaining, and one pound over not divided. Now to divide that pound you proceed thus: that pound contains twenty shillings, and the 5 shillings in the dividend make twenty-five; we then say six in twenty-five, four times six make twenty-four, which 4 we write under the shillings place; but as there were twenty-five shillings, there remains one shilling not yet divided, which when reduced to pence make twelve pence, added to sixpence in the dividend, make eighteen pence, and 6 into 18 is contained 3 times. Take another example, and let the dividend be rather £. S. d. 8) 461 18 6 57 14 larger. Divide £161 18s. 6d. by 8. You divide the pounds as in Simple Division and find there is a remainder of five, which five being pounds you must reduce to shillings by multiplying it by twenty and adding in the 18, which when done you will find is 118 shillings; now 8 is contained in one hundred and eighteen, 14 times, leaving a remainder of six, this six being shillings you reduce to pence by multiplying it by twelve, and adding in the 6 pence in the dividend, which makes 78; now 8 is contained in seventy-eight 9 times, leaving a remainder of six; this six being pence, and to reduce it lower would only express a part or parts of a penny, therefore, to avoid compound fractions, arithmeticians invariably take it for the numerator of their fraction, writing the divisor underneath it for a denominator, and the fraction thus produced they reduce to its lowest terms, thus, &÷2 = Now if you refer to the third example in Compound Multiplication, you will find the product of that sum taken for the dividend, and its multiplier for the divisor, and you will also perceive that the quotient of this sum is the multiplicand of the sum in Multiplication; which not only proves that both our sums are correct, but that the rules we apply for working them are correct also. Work for practice the following sums. Ex. (1.) Divide £73. 9s. 9d. by 2. Ans. £36. 14s. 10 d. Ans. £14. 16s. 9 d. Ans. £39. 15s. 6d. Ans. £63. 17s. 4 d. Ans. £120. 13s. 11⁄2d. Ans. £91. 14s. 24d. Ans. £14.11s. 103d. (9.) Divide £1577 2s. 11 d. by 11. Ans. £143 7s. 6d. (10.) Divide £1412 11s. 9d. by 12. Ans. £117 14s. 3 d. Thus far we have treated of division of compound numbers by divisors not exceeding twelve; it yet remains for us to speak of the method of proceeding when we have to divide by higher numbers. Some of these methods are the counterparts of those pursued under similar circumstances in Multiplication; another is to reduce the dividend to its lowest terms, and proceed as in Simple Division. Example. What is the value of 1lb. of tea, if 120lb. cost £27? In the example, as we cannot divide 120 into 27, we multiply it by 20, the shillings of a pound, and in the product, 540, we find 120 contained 4 times, with a remainder of 60, which we multiply by 12, the pence in a shilling, and divide the product 720, by 120) £27 20 540 (4s. 6d. 480 60 12 720 720 the divisor 120, which is contained 6 times, and this quotient 4s. 6d. is the answer, or price of one pound. Another method is by taking any two or three small numbers that, when multiplied by one another, will produce the divisor, and then proceed as in Short Division. Thus, 10 times 12 being 120, s. d. we divide £27, as in Short Division. £ 10) 27 0 0 12) 2 14 0 0 4 6 Third method; reduce £27 to its lowest denomination pence, and divide by 120, the quotient will be the answer in pence. |