CASE II. When the multiplier or quantity exceeds 12, and is the product of two factors in the Multiplication Table; that is, of two numbers which being multiplied together, amount to the same as the multiplier. Multiply the sum by one of the two numbers, and then muitiply the product by the other. When the quantity, or multiplier, is such a number that no two numbers in the Multiplication Table will produce it. Multiply the sum by two numbers whose product will amount to nearly the same as the multiplier; then multiply the sum by the number which will make the product of the two numbers equal to the multiplier, and add its product to the sum produced by the two numbers. 451 5 0 15 0 10 466 5 10 Here note, I multiply by 10, then by 6, because 10 times 6 make 60; then I multiply the same sum by 2, that I multiplied, first, by 10, and add its product to the other product, which makes the amount of the answer. When the multiplier is greater than the product of any two numbers in the Multiplication Table. Multiply the sum by 10, and that product by 10, which is equal to multiplying by 100; then multiply the product by the number of hundreds in the multiplier, and if the sum be even hundreds, the product will be the answer. If there be odd numbers over even hundreds, as 70, 80, or 87, &c., multiply the amount or product of the first multiplication by 10, by the number of tens over 100; thus, if there be 70 over, multiply by 7. If, in additon to tens, there are smaller numbers, as 7, 8, 5, &c., the sum must be multiplied by such number; and the amount of all the multiplications being then added together, their sum will be the answer. In the foregoing example, I first multiply by 10, three times, which gives the amount of the sum multiplied by 1000; then by 4, which gives the amount of 00`.The sum is yet to be multiplied by 300. To do this, I take the product of the sum multiplied by 100, viz. 1117. 13s. 4d. and multiply it by 3, which gives the product of the sum by 300. The sum of these is the answer. 1. What do 84 pounds of sugar cost at 9d. per pound? Ans. £3. 3s. 2. What do 18 yards of cloth cost at 19s per yerd. 3. Sold 7 tons of iron at £32 much is the amount? Ans. £17. 2s. 10s. per ton; how Ans. £227. 10s. 4. What is the weight of 4 weighing 7 cwt. 3 qrs. 19 lb? 5. What is the weight of 6 Jing 3 cwt. 2 qrs. 9 lbs? hogsheads of sugar, each Ans. 31 cwt. 2 qrs. 20 tbs. chests of tea, each weighAns. 21 cwt. 1 qr. 26 lbs. 6. What is the value of 79 bushels of wheat, at 11s. 54d. per bushel? Ans. £45 6s. 101. 7. What is the value of 94 barrels of cider, at 12s. per barrel? 2d. Ans. £57 3s. 8d. 8. What is the value of 114 yards of cloth, at 15s. 3rd. per yard? Ans. £87 5s. 7id. 9. What is the value of 12 cwt. of sugar, at £3 7s. 4d. per cwt.? 10. What is the worth of 63 gallons per gallon? Ans. £40 8s. of oil at 2s. 3d. Ans. £7 1s. 9d. 11. What is the amount of 120 days wages at 5s. 9d. per day? Ans. £34 10s.| 12. What is the worth of 144 reams of paper at 13s. 4d. per ream? Ans. £96. 13. What will 1 cwt. of sugar cost, at 10‡d. per lb?*. Ans. £5 0s. 4d. 14. If I have 9 fields, each containing 12 acres, 2 roods. and 25 poles; how many acres have I in the whole? Ans. 113A. 3R. 25P. 15. What will 1 ton of lead cost, at 1 pence per pound? Ans. £37 6s. 8d. 16. If a man can travel 25 ms. 3 fur. 20 ps. 4 yds. in 1 day, how far can he travel in 6 days? Ans. 152ms. 5fur. 4ps. 2yds. Q. 1. What does Compound Multiplication teach? 2. In what is it particularly useful? 3. Which is made the multiplier-the price, or the quantity? 4. How do you proceed when the multiplier does not 5. How do you proceed when the multiplier exceeds 12? *It must be recollected that lcwt. is 112lbs. Compound Division teaches the manner of dividing numbers of different denominations. CASE I. When the divisor does not exceed 12. Begin at the highest denomination, and after dividing that, if any thing remain, reduce it to the next lower denomination, adding it to that denomination in the sum, and proceed in this manner until the whole is divided. If the number of either denomination should be too small to contain the divisor, reduce it to the next lower denomination, and add it thereto, as directed in case of a remainder. The denominations in the quotient must be kept separate. PROOF. Multiply the quotient by the divisor, and the product, if right, will be equal to the dividend. In doing the 6th sum, which is divided by 12, I find the divisor is contained once in 21; and setting down 1 I find 9 pounds remaining; which, reduced to shillings, and added to the 16 shillings in the sum, make 196 shillings. The divisor being contained 16 times in 196, with 4 remaining, I set down 16, and reducing the 4 shillings to pence, and adding them to the 11 pence, in the sum, the amount is 59 pence. The divisor is con |