either of the products (except the last) be more than 9, set down its right hand figure only, and add its left hand figure or figures to the next product. The whole of the last product must be set down. PROOF. Multiply by double the multiplier, and the product will be double the former product. EXAMPLES. Multiplicand 2 4 3 2 7 4 2005 2 40092 Multiplier 3 4 1 2 8 2 2 8 0 2 0 8 4 8 1 1 0 4 2 60 6 7 0 7 0 5 3 4 6 2 1 4 4 0 9 5 6 0 4 2 0009 2 1 1 9 When the multiplier exceeds 12, work by RULE II. Multiply by each figure of the multiplier separately, first by the one at the right hand then by the next, and so on, placing their respective products one under another, with the right hand figure of each product directly under that figure of the multiplier by which it is produced. And these products together; and their amount will be the product required. Note. When cyphers occur at the right hand of either or both of the factors, omit them in the operation, and annex them to the product. Note. When the multiplier is the exact product of any two factors in the multiplication table, the operation may be performed thus: multiply by one of the factors, and then multiply the number produced by the other factor. 1. Richard has 125 nuts, and George has 6 times that number. How many has George? Ans. 750. . There are 20 boxes of raisins with 14 pounds in each box. How many pounds are there in all? Ans. 280. 3. The price of one orange is 9 cents: how many cents will five oranges come to, at the same price? 4. There are 12 pence in one shilling. pence are there in 40 shillings? Ans. 45. How many Ans. 480. ADDITION AND MULTIPLICATION. 1. Multiply 25 by 10, and 36 by 14, and 124 by 45. Add the several products, and tell their amount. Ans. 6334. 2. There are ten bags of coffee weighing each 120 pounds; and 12 bags weighing each 135 pounds. What is the weight of the whole? Ans. 2820 pounds. 3. A merchant bought five pieces of linen containing 25 yards each, and 2 pieces containing 24 yards each, and 1 piece containing 26 yards. How many yards were there in the whole? Ans. 199. SUBTRACTION AND MULTIPLICATION. 1. Multiply 342 by 22, and from the product subFacit 7124. tract 400. 2. There are 15 bags of coffee, each of which weighs 112 pounds. The bags which contain the coffee weigh 22 pounds. How much would the coffee weigh without the bags? Ans. 1658 pounds. 3. There are 12 chests of tea, each of which weighs 96 pounds. The chests which contain the tea weigh cach 20 pounds. What would the tea weigh without the chests? Ans. 912 pounds. DIVISION. By division we ascertain how often one number is contained in another. The number to be divided is called the dividend. The number of times the dividend contains the divisor is called the quotient. If, on dividing a number, there be any overplus, it is called the remainder. The dividual is a partial dividend, or so many of the dividend figures as are taken to be divided at one time, and which produce one quotient figure. When the divisor does not exceed 12, work by RULE I. See how often the divisor is contained in the first left hand figure or figures of the dividend.* If it be contained an exact number of times, set down that number; and then see how often it is contained in the next figure. But if it be contained any number of times with a remainder, set down the number of times, and conceive the remainder to be prefixed to the next figure; then see how often the divisor is contained in these, and proceed as before till the whole is divided. PROOF. Multiply the quotient by the divisor, and to their product add the remainder, (if any,) and the result will be equal to the dividend. Dividend EXAMPLES. Divisor 3)963 5)2960 12)112813 12)970811280 *The multiplication table shows how often any number, not exceeding 12, is contained in any other number not exceeding 144; as that 4 is contained in 12 three times, because 3 times 4 are 12; 10 is contained in 115.eleven times with 5 over; because 11 times 10 are 110, which, with 5, make 115. When the divisor exceeds 12, work y RULE II. or LONG DIVISION. Take for the first dividual as few of the left hand figures of the dividend as will contain the divisor, try how often they will contain it, and set the number of times on the right of the dividend-multiply the divisor by this number-subtract its product from the dividual, and to the remainder affix the next figure of the dividend, to form a second dividual or if this be not sufficient, set a cypher on the right of the dividend, and affix the next figure, and so on, till a sufficient number of figures are affixed-try how often the divisor is contained in this second dividual, and proceed as before. Continue this process till all the dividend figures are employed as above directed; or till the number they form, when affixed to a remainder, is not large enough to contain the divisor. When the work is done, the figures on the right of the dividend form the quotient. |