8. If 25 oranges are worth 50 lemons, and 6 lemons are worth 24%, how much are 15 oranges worth? 9. If 10 men, by working 10 hr. per da., can build a boat in 8 da., how many da. would it take 12 men to do the work, if they wrought only 8 hr. per da. ? 10. If a 5 loaf weighs 10 oz. when flour is $8 a bbl., how much should an 8 loaf weigh, when flour is $5 a bbl.? Written Exercises+ EXAMPLE.—If 16 men in 20 da. 10 hr. long can build 5 mi. of railroad, in how many da. 12 hr. long can 64 men build 30 miles of railroad? 64 × 12 × 5 : 16 × 10 × 30 :: 20 : (25) EXPLANATION. -We write 20 da. for 3d term, because our answer, or 4th term, is required in da. Since 64 men will perform the work in less time than 16 men, our 4th term must be less than the 3d, and we therefore write 64 men for our 1st term and 16 men for the 2d. Since it would take a shorter time to do a piece of work by working 12 hr. per day than by working 10 hr. per day, our 4th term must be less than the 3d, and we therefore write 12 hr. for the 1st term and 10 hr. for the 2d. Since it would take more time to build 30 mi. of road than 5, the 4th must be greater than the 3d term, and we therefore write 5 mi. for the 1st and 30 mi. for the 2d term. We then reduce the compound ratio to a simple one (269), and find the 4th term by Rule (284), or by (281, RULE I.) 285. RULE.-Write as the third term that number which is of the same kind as the required answer. Then, of the other terms, form a compound ratio, being careful to write the terms in pairs, and to state each pair of terms, as if the value of the 4th term depended on that pair alone. After which, find the fourth term in accordance with the Principle of Proportion (281). This and all similar examples may be solved by Analysis. Thus, SOLUTION.—If 16 men can build 5 mi. (or any other number of miles) of railroad in 20 da., 64 men will build the same amount in the same part of 20 da. that 16 is of 64, that is of 20 da., or 5 da.; and if they can build 5 mi. in 5 da., they can build 30 miles in § of 30 da. = 30 da. And if they can build it in 30 da., by working 10 hr. per da., they can, by working 12 hr. per day, build it in 19, or § of 30 da., or 25 dá., Ans. This and similar examples may also be solved by Cause and Effect. Thus, We have now to find the number of days in the Sec. cause. And since this is a proportion, the product of the extremes equals the product of the means, and we therefore can readily fill the blank () by Rule I, or II. (281). PROBLEMS. 1. If 6 men build a wall 3 rd. long in 5 da., how many men would build a wall 15 rd. long in 10 da. ? Ans. 2. 1f 24 cows eat 36 bu. corn meal in 12 da., how many bu. will last 20 cows 60 da. ? Ans. 150 bu. 3. If $75 gain $12 in 2 yr., what sum will gain $150 in 18 mo. ? Ans. $1250. Ans. $7680. 4. If 16 men earn $640 in 26 da., what will 48 men earn in 104 da. ? 5. If 3 men in 18 da. earn $162, how many dollars will 9 men earn in 12 da., with like wages? Ans. $324. 6. If 10 persons spend $600 in 8 mo., how much would 8 persons spend in 12 mo., at the same rate? Ans. $720. 7. If 7 persons spend $273 in 2 mo., in how many mo. would 5 persons spend $650? Ans. 6 mo. 8. If $250 gain $15 in 1 yr., in how many months will $750 gain $37.50? Ans. 10 mo. 9. If $300 gain $9 in 3 mo., what will gain $50 in 10 mo. Ans. $500. 10. If 8 men dig a ditch 6 rd. long in 10 da., how many men would dig 27 rd. long in 12 da.? Ans. 30 men. 11. If 8 men dig a ditch 6 rd. long in 10 length of time would 30 men dig 27 rd. long? 12. If 8 men dig a ditch 6 rd. long in 10 would 30 men dig in 12 da. ? da., in what Ans. 12 da. da., how far Ans. 27 rd. 13. If 12 horses eat 18 bu. of oats in 6 da., how many bu. would last 20 horses 30 da.? Ans. 150 bu. Ans. 40 da. 14. If 6 horses eat 4 bu. oats in 3 da., how many days will 100 bu. last 10 horses? 15. If 4 horses eat 12 bu. of oats in 6 da., how many horses will require 75 bu. in 30 da. ? Ans. 5 horses. 16. If 11 men mow 45 A. of grass in 6 da. of 10 hr. each, how many men will be required to mow 81 A. in 12 da. of 11 hr. each ? Ans. 9 men. 17. If 11 men mow 45 A. in 6 da. of 10 hr. each, how many hr. per da. must 9 men mow, to finish 81 A. in 12 da. ? Ans. 18. If 11 men mow 45 A. in 6 da. of 10 hr. each, how many A. will 9 men mow in 12 da. of 11 hr. each? Ans. 81 A. 19. If 108 men build a fort in 24 da. of 12 hr. each, in how many da. would 84 men build it, working 101 hr. per da.? Ans. 37 da. Distributive Proportion. 286. Distributive Proportion is, as its name implies, proportion applied to the distribution of a quantity into parts which have a given ratio. It is sometimes called Partitive Proportion. +Oral Exercises EXAMPLE.-What parts of 20 are as 3 to 1. SOLUTION. Since 20 is to be divided into parts which are as 3 to 1, it must be divided into 3 + 1, or 4 equal parts, and 3 and 1 of these parts taken. 20 ÷ 4 = 5, or 1 part; three of these parts is 15, and one of them is 5. Therefore, the parts of 20 which are as 3 to 1, are 15 and 5. found 9. Anna and Mary, aged respectively, 5 and 7 yr., $1.20, and divided it in the ratio of their ages. How much did each receive? 10. Jane, Ann, and Eliza contributed to the purchase of a present of an $18 clock to their teacher, in the ratio of 1, 2, and 3. How much did each contribute? 11. E and F spent $29, F spending 61 times as much as E. How much did each spend ? 12. Frank and James have 75; Frank has as much as James. How much has each ? 13. Fifty has two parts, one of which is the other. What are the parts? +Written Exercises+ EXAMPLE.-Resolve 75 into two parts, of which the first shall be to the second as 7 to 8. 7: SOLUTION. 7 8, ratio of parts. 7+ 8 = 15, no. of parts. 75 ÷ 15 = 5, one of the parts. 5 x 735, 1st Part, = } 5 x 840, 2d Part, Ans. EXPLANATION. - From the conditions of the Example, 75 is to be divided in the ratio of 7 to 8, or in such a manner that the sum of the same equimultiple of 7 and of 8 shall be equal to 75. And since 75 is to equal an equimultiple of 7 and 8 at the same time, it must be an equimultiple of their sum, 15. Now 75 contains 15, 5 times; therefore, one part of 75 must contain 7, 5 times, and the other part must contain 8, 5 times. Hence the parts are 5 x 7 = 35, and 5 x 8 = 40. SOLUTION BY PROPORTION. 7 + 8 = 15, + 8 = 15, 15: 7:: 75: (35) 1st Part. 287. RULE.-Divide the given number by the sum of the terms of the ratio, and multiply the quotient by each term. Or, As the sum of the terms of the ratio is to one of them, so is the number to be resolved, to its corresponding part. What parts PROBLEMS. 1. Of 48 are as 1, 2, and 3? 2. Of 108 are as 2, 3, and 4? Ans. 8, 16, and 24. Ans. 24, 36, and 48. |