By examining these proportions, it is seen that the ratio of the third term to the fourth, depends on two ratios 1st, the ratio of 2 men to 6 men; 2d, the ratio of 5 da. to 10 da. The first ratio is 3, and the second 2; their product, 3×2=6, is the ratio of the third term to the required term. Hence, The ratio, which the 3d term has to the 4th, is compounded of the ratios of 2 to 6 and of 5 to 10: write the two simple proportions Thus, $ 2 men : men :: 20 fourth term. Or, 2X5: 6X10 :: 20: fourth term. 20X6X10 1200; 2X5=10; and 1200+10 $120. Ans. *2. If a man travel 24 mi. in 2 da., by walking 4 hr. a day; at the same rate, how far will he travel in 10 da., walking 8 hr. a day? Ans. 240 mi. Rule for Compound Proportion.-1. Write for the 3d term, that number which is of the denomination required in the answer. 2. Take any two numbers of the same kind, and arrange them as in Simple Proportion. Art. 203, Rule 2, page 201. 3. Arrange any other two numbers of the same kind, in like manner, till all are used. 4. Multiply the third term by the continued product of the second terms; divide the result by the continued product of the first terms; the quotient will be the fourth term, or answer. NOTES.-1. If the terms in any couplet are of different denominations, reduce them to the same. If the third term consists of more than one denomination, reduce it to the lowest named. 2. The examples may be solved by two or more statements in Simple Proportion, or by Analysis, Art. 263. Also, by Cause and Effect, Art. 204, see statement, page 208. KEVIEW.-205. What does Compound Proportion contain? To what is it applied? In stating a question, what number is put for the third term, Rule? How are the other numbers arranged? 205. How is the fourth term found? NOTE 1. If the terms in any couplet are of different denominations, what is required? What if the third term contains more than one denomination? 3. If 6 men, in 10 days, build a wall 20 ft. long, 3ft. high, and 2ft. thick in how many days could 15 men build a wall 80 ft. long, 4ft. high, and 3 ft. thick? 3 The terms of the 2d cause and 1st effect form the divisors; thos of the 1st cause and 2d effect, the dividends. The x shows the place of the required term. EXPLANATION. When the terms forming the divisor and the dividend, contain one or more common factors, shorten the operation by Cancellation. In such cases, arrange the divisors on the left of a vertical line, and the multipliers (dividends), on the right. OPERATION BY CANCELING. 4. If 16 men build 18 rods of fence in 12 days, how many men can build 72rd. in 8 da. ? Ans. 96 men. 5. If 6 men spend $150 in 8mon., how much will 15 men spend in 20 mon. ? Ans. $937.50 6. I travel 217 mi. in 7 days of 6 hr. each, how far can I travel in 9 da. of 11 hr. each? Ans. 511 mi. 7. If $100 gain $6 in 12 months, what sum will $75 gain Ans. $3.375 in 9 mon.? 8. If 100 lb. be carried 20 mi. for 20 cts., how far may 10100lb. be carried for $60.60? Ans. 60 mi. 9. To carry 12 cwt. 3 qr. 400 mi., costs $57.12: what will it cost to carry 10 tuns 75 mi. ? 10. If 18 men, in 15 da., build a wall high, 4ft. thick; in what time could 20 87 rd. long, 8ft. high, and 5ft. thick ? Ans. $168. 40 rd. long, 5 ft. men build a wall Ans. 5828 da. 11. If 180 men, in 6 da. of 10 hr. each, dig a trench 200 yd. long, 3 yd. wide, 2 yd. deep; in how many days can 100 men, working 8 hr. a day, dig a trench 180 yd. long, 4 yd. wide, and 3 yd. deep? Ans. 24.3 da. For additional problems, see Ray's Test Examples. XV. ALIQUOTS, OR PRACTICE. ART. 206. One number is an aliquot part of another, when it will exactly divide it (Art. 110). Thus, 5 cents, 10 cts., 20 cts., &c., are aliquot parts of $1. TABLE OF ALIQUOT PARTS OF A DOLLAR. ART. 207. To find the cost of articles, when the price is an aliquot part, or aliquot parts of a dollar. 1. What cost 24 yards of muslin, at 25 cts. a yd.? SOLUTION.-If the price was $1 a yard, the cost would be $24; and, since 25 cents is of a dollar, the cost at 25 cts. a yard will be the cost at $1; and of $24-$6. Ans. 1 2. What cost 16 yards of calico, at 37 cts. a yard? SOLUTION.-If the price was $1 a yard, the cost would be $16. At 25cts. a yard, the cost would be the cost at $1; Again; since 12 cts. is of 25 cts., the cost at 12 will be the cost at 25 cts.; of $4-$2. of $16-$4. cts. a yard, But, the cost at 371cts. a yard, is equal to the sum of the costs at 25cts. and at 124 cts.; $4+$2=$6. Ans. REVIEW.-206. When is one number an aliquot part? Give examples. 3d Bk 14 *3. What cost 24 yd. silk at 62 cts. a yd.? Ans. $15. Rule for Case I.-Take such aliquot parts of the cost at $1, as may be necessary to find the cost at the given price 4. What cost 48 yd. of linen, at 683 cts. per yd.? OPERATION. $48 cost of 48 yards, at $1. Ans. $43.25 Ans. $33= What will be the cost 5. Of 173 bu. corn, at 25 cts. a bu.? 6. 7. 8. 45 lb. cheese, at 314 cts. a lb.? Ans. $14.061 54 yd. calico, at 433 cts. a yd.? Ans. $23.621 32 bu. rye, at 933 cts. a bu.? 9. What cost 20 yd. cloth, at $3.12 3 The cost at $1 a OPERATION. $20 yard is multiplied by the number of dollars (3); and for the 12 cts., an 12 cts. 60 = = Ans. $30.00 per yd.? cost of 20 yd. at $1 cost at $3.00 2.50 cost at .12 Ans. $62.50-cost at $3.12 aliquot part, () of the cost at $1 is taken; the results are added. ART. 208. CASE II.-To find the cost of a quantity consisting of several denominations. 1. What will be the cost of 4 A. 1 R. 20 P. of land, at $16.40 per A.? OPERATION. Rule for Case II.-Multiply the price by the number of the denomination at which the price is rated, and find the cost of the lower denominations, by taking aliquot parts. Add the different costs; the sum will be the cost of the whole. 15. 16. 3 lb. 13 oz. 15 dr. spice, at $2.56 a lb. 17 A. 3 R. 39 P., at $3.20 per A. $9.91 $57.58 For additional problems, see Ray's Test Examples. REVIEW.-207. What is Case 1? What the Rule for Case 17 208. What is Case 2? What the Rule for Case 2? |