148i When the multiplier is a composite number, and none of its factors exceed 12. Ex. 1. What will 35 loads of coal weigh, if each load weighs 2T. lcwt. 2qr. 61b.? Operhtion. "\Ve find the num T. cwt. qr. lb. Der 35 equal to the 2 12 6 = weight of 1 load. product of 7 and 5; 7 we therefore multiply ;—A TT;—o—• 1.1 T i J the weight of 1 load 14 10 3 17= we.ght of 7 loads. fcy ?, ^ then ^ 'product by 5 ; and the 72 14 2 10 = weight of 35 loads, last product is the answer. Hence, when the multiplier is a composite number, Multiply by its factors in succession. Examples. 2. Bought 90 hogsheads of sugar, each weighing 12cwt. 2qr. 11 1b.; what was the weight of the whole? 3. What cost 18 sheep at 5s. 9Jd. apiece? 4. What cost 21 yards of cloth at 9s. lid. per yard? 5. What cost 22 hats at lis. 6d. each? 6. If 1 share in a certain stock be valued at 13£. 8s. 9£d., what is the value of 96 shares? 7. If 1 spoon weighs 3oz. 5pwt. 15gr., what is the weight of 120 spoons? 8. If a man travel 24m. 7fur. 4rd. in 1 day, how far will he go in 1 month? 9. If the earth revolve 0° 15' per minute, how far does it revolve per hour? 10. Multiply 39A. 3R. 17p. 30yd. 8ft. lOOin. by 32. 11. If a man be 2d. 5h. 17m. 19sec. in walking 1 degree, how long would it take him to walk round the earth, allowing 365£ days to a year? 149i When the multiplier is not a composite number, and exceeds 12; or when a composite number one of whose factors exceeds 12. Ex. 1. What is the value of 453 tons of iron at 18£. 17s. lid. a ton? 10 tons = 188 19 2, X 5 = 944 15 10 = Value of 50 tons 10 100 tons =1889 11 8,X4=7558 6 8 = Value of 400 ton» Ans. 8559 16 3 = Value of 453 tons. Since 453 is not a composite number, we cannot resolve it into factors; but we may separate it into parts, and find the value of each part separately: Thus, 453 = 400 50 -f- 3. In the operation, we first multiply by 10, and obtain the value of 10 tons, and this product we multiply by 10, and obtain the value of 100 tons. Then, to find the value of 400 tons, we multiply the last product by 4; and to find the value of 50 tons, we multiply the value of 10 tons by 5; and to find the value of 3 tons, we multiply the value of 1 ton by 3. Adding the several products, we obtain 8559£. 16s. 3d. for the answer. Hence, Having resolved the multiplier into any convenient parts, as of units, tens, Sfc., multiply by these several parts, and add together the products thus obtained for the required result. Examples. 2. Multiply 2hhd. 19gal. Oqt. lpt. by 39. 3. Multiply 3bu. lpk. 4qt. lpt. lgi. by 53. 4. Multiply 16ch. 7bu. 2pk. Oqt. Opt. by 17. 5. What will 57 gallons of wine cost at 8s. 3£d. per gallon? 6. Bought 29 lots of wild land, each containing 117A. SR. 27p.; what were the contents of the whole? 7. Bought 89 pieces of cloth, each containing 37yd. 3qr. 2na. 2in.; what was the whole quantity? 8. Bought 59 casks of wine, each containing 47gal. 3qt. lpt.; what was the whole quantity? 9. If a man travel 17m. 3fur. 13rd. 14ft. in one day, how far will he travel in a year? 10. If a man drink 3gal. lqt. lpt. of beer in a week, how much will he drink in 52 weeks? 11. There are 17 sticks of timber, each containing 37ft. 978in.; what is the whole quantity? 12. There are 17 piles of wood, each containing 7 cords 98 cubic feet; what is the whole quantity? DIVISION OF COMPOUND NUMBERS. 150. Division of Compound Numbers is the process of dividing compound numbers into any proposed number of equal parts. Ex. 1. Divide 139£. 13s. lid. 2far. equally between 5 persons. Ans. 27£. 18s. 9d. 2far. operation. Having divided 139£. by 5, we find £. a. d. far. the quotient to be 27£., and 4£. re5)139 13 11 2 maining. We place the quotient 27£. ———— —- under the 139£., and the remainder 2 7 1 8 9 2 4£. reduced to shillings = 80s.; 80s. -J- the 13s. in the dividend = 93s.; 93s. -f- 5 = 18s. and a remainder of 3s. We write the quotient 18s. under the shillings in the dividend; and the remainder 3s. reduced to pence = 36d.; 36d. -f- lid. in the dividend = 47d.; 4 7d. + 5 = 9d. and a remainder of 2d. We write the quotient 9d. under the pence in the dividend; and the remainder 2d. reduced to farthings = 8far., -f- the 2far. in the dividend = l0far.; l0far. -f- 5 = 2far. The quotient 2far. we write under the farthings in the dividend; and thus find the answer to be 27£. 18s. 9d. 2far. Rule. — Divide as in division of simple numbers, each denomination in its order, beginning with the highest. If there be a remainder, reduce it to the next lower denomination, adding in the number already contained in the dividend of this denomination, if any, and divide as before. Proof. — The same as in simple numbers. Note. — When the divisor and dividend are both compound numbers, they must be reduced to the same denomination, and the division then is that of simple numbers. Examples. Notb. — The answers to the following examples are found in the corresponding numbers of examples in Multiplication of Compound Numbers. 2. 3. £. s d. T. cwt. qr. lb. oz. dr. 8)19 10 4 9)5 5 7 19 11115 7 4. 5. lb. oz. pwt. gr. lb. oz. pwt. gr. 5)39 10 13 3 8)261 11 0 0 6. 7. lb § 3 9 gr- deg- m- fur. rd. ft. In. 11) 427 10 0 2 14 12)858 44 4 6 7 0 8. 9. m. fur. ch. p. 1. A. R. p. yd. ft. in. 12) 215 7 9 3 1 11)181 3 11 6 4 41 10. Divide 54yd. 2qr. 3na. equally among 5 persons. 11. Divide 123'un 3hhd. 36gal. 3qt. by 7. 12. Divide 209hhd. 55gal. 3qt. Opt. lgi. by 7. 13. What is the value of 118bu. lpk. 5qt. H- 6? 14. What is the value of llOy. 343d. 3h. 41m. 12s. -h 8? 15. Divide 149deg. 9m. 5fur. 13rd. 3yd. 1ft. by 9. 16. A man divides his farm of 214A. 3R. 12p. equally among his 9 sons; how much does each receive? 17. If one man perform a certain piece of labor in 3d. lGh. 54m., how long would it take 12 men to perform the same work? 18. A farmer has 29 bushels of oats which he wishes to put in 8 sacks; how much must each sack contain? 151. When the divisor is a composite number, and none of its factors exceed 12. Ex. 1. If 35 loads of coal weigh 72T. 14cwt. 2qr. 101b., what will 1 load weigh? Operation. We find the fac T. cwt. qr. lb. tors of 35 to be 5 5)72 14 2 10 = weight of 35 loads, and 7. We there 7 ) 14 10 3 1 7 = weight of 7 loads. for? . t , th° '° weight of 3o loads 2 12 6 = weight of 1 load. by 5, and obtain the weight of 7 loads; and the weight of 7 loads we divide by 7, and thus find the weight of 1 load. Hence, when the divisor is a composite number, Divide by its factors in succession. 2. If 90 hogsheads of sugar weigh 56T. 14cwt. 3qr. 151b., what is the weight of 1 hogshead? 3. What will be the price of 1 sheep, if 18 cost 5£. 4s. 3d.? 4. If 21 yards of cloth cost 10£. 8s. 3d., what is the price of 1 yard? 5. What is the value of 1 hat, when 22 cost 12£. 13s. Od.? 6. When 96 shares of a certain stock are valued at 1290£. 4s. Od., what would be the cost of 1 share? 7. If 120 spoons weigh 321b. 9oz. 15pwt., what does 1 weigh? 8. If a man in 1 month travels 746m. 5fui\, how far does he go in 1 day? 9. If the earth revolves 15° on its axis in 1 hour, how far does it revolve in 1 minute? 10. Divide 1275A. 2R. 16p. 22yd. 8ft. 32in. equally among 32 men. 11. If a man walk round the earth in 2y. 68d. 19h. 54m., how long would it take him to walk 1 degree, allowing 365| days to a year? 152. When the divisor is not a composite number, and exceeds 12, or when a composite number one of whose factors exceeds 12, the whole operation can be written, as in the following example. |