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If 6 men mow 30 acres of grass, how many acres will 4 men mow % It is evident the number of acres sought would be the same fractional part of 30 acres that 4 men is of 6 men; that is, the quantity required is # of 30 acres. If, now, we take into account the number of days, still supposing the number of hours in each day to remain the same in both cases, our question would become: If # of 30 acres can be mowed in 5 days, how much can be mowed in 9 days? The answer in this case is obviously # of # of 30 acres. Now, taking into account the number of hours each day, our question will become as follows: If # of + of 30 acres can be mowed in a certain time, when 8 hours are reckoned to each day, how much could be mowed when 10 hours are reckoned to each day? This leads to the following final result: * of # of # of 30 acres. * By cancelling, we reduce this last expression to 45 acres. From the above work we see that questions of Compound Proportion may be solved by the following

RULE.

Among the quantities given, there will be but one like the answer, which one we will call the odd quantity. The other quantities will appear in pairs or couplets. Form ratios out of each couplet in the same manner as in the Rule of Three ; then multiply all the ratios and the odd quantity together, and this will give the answer in the same denomination as the odd quantity.

NotE.-Before forming ratios from the couplets, they must be reduced to the same denominate value.

EXAMPLES.

1. If a person travel 300 miles in 17 days, traveling only 6 hours each day, how many miles could he have gone in 15 days, by traveling 10 hours each day? In this example, the answer is required in miles, therefore our odd term is 300 miles 2 The first couplet consists of days; and since in 15 days, other things being the same, he could not travel as far as in 17 days, we must divide 15 by 17, which gives # for the first ratio. The second couplet consists of hours; and since in 10 hours he could travel farther than in 6 hours, we must divide 10 by 6, which gives ** for the second ratio. Multiplying these two ratios and the odd term together, we get 300 miles x+}×*. Cancelling the 6 of the dencminator against 6, a factor of 300 (=310) of the numerator, we have 50 × 4 × 4 +44.1% miles, for the answer. 2. If a marble slab 10 feet long, 3 feet wide, and 3 inches thick, weigh 400 pounds, what will be the weight of another slab, of the same marble, whose length is 8 feet, width 4 feet, and thickness 5 inches 7 In this example, the answer is required to be given in pounds; therefore 400 pounds is the odd term. The first couplet consists of the lengths; and since 8 feet in length will give less weight than 10 feet, we must divide 8 by 10, which gives to for the first ratio. The second couplet consists of the widths; and since 4 feet wide will give more weight than 3 feet, we must di. vide 4 by 3, which gives # for the second ratio.

The third couplet consists of thicknesses; and since 5 inches thick will give more weight than 3 inches, we must divide 5 by 3, which gives # for the third ratio.

Multiplying the odd term and these ratios together, we get 400lbs. x+x+x+. Cancelling the 10 of the denominator against a part of the 400 of the odd term numerator, we get 40lbs. x+x+x+=*** =7114 pounds, for the anSWer.

3. 500 men, working 12 hours each day, have been employed 57 days to dig a canal of 1800 yards long 7 yards wide, and 3 yards deep; how many days must 860 men, working 10 hours each day be emlpoyed in digging another canal of 2900 yards long, 12 yards wide, and 5 yards deep, in a soil which is 3 times as difficult to excavate as

the first 7
In this example, the odd term is 57 days.

The different ratios will be as follows:
#}}=# ratio of the men.
+3= } ratio of the hours.
+++}=#4 ratio of lengths of the canals.

1,2– ratio of widths of the canals.
#= ratio of depths of the canals.
*— ratio of the difficulty in excavation.

Multiplying successively these ratios and the odd term,

we have
57 days x+}×{}×{}x + x+x+.
This becomes, after cancelling factors,
19 days x+, ×{x+x+x+x+=549 or days.

4. 15 men, working 10 hours each day, have employed 18 days to build 450 yards of stone fence; how many men, working 12 hours each day, for 8 days, will be requisite to build 480 yards of similar fence Ans. 30 men.

5. If it require 1200 yards of cloth #wide to clothe 500 men, how many yards which is #wide will it take to clothe 960 men? Ans. 3291} yards. 6. If 8 men will mow 36 acres of grass in 9 days, by working 9 hours each day, how many men will be required to mow 48 acres in 12 days, by working 12 hours each day? Ans. 6 men. 7. If 11 men can cut 49 cords of wood in 7 days, when they work 14 hours per day, how many men will it take to cut 140 cords in 28 days, by working 10 hours each day? Ans. 11 men. 8. If 12 ounces of wool make 24 yards of cloth, that is 6 quarters wide, how many pounds of wool will it take for 150 yards of cloth, 4 quarters wide? Ans. 30 pounds. 9. If the wages of 6 men for 14 days be 84 dollars, what will be the wages of 9 men for 16 days?' Ans. $144. 10. If 100 men in 40 days of 10 hours each, build a wall 30 feet long, 8 feet high, and 2 feet thick, how many men must be employed to build a wall 40 feet in length, 6 feet high, and 4 feet thick, in 20 days, by working 8 hours each day? Ans. 500 men. 11. In how many days, working 9 hours a day, will 24 men dig a trench 420 yards long, 5 yards wide, and 3 yards deep, if 248 men, working 11 hours a day, in 5 days, dig a trench 230 yards long, 3 yards wide, and 2 yards deep? Ans. 288 for days. 12. Suppose that 50 men, by working 5 hours each day, can dig, in 54 days, 24 cellars, which are each 36 feet long, 21 feet wide, and 10 feet deep, how many men would be required to dig, in 27 days, 18 cellars, which are each 48 feet long, 28 feet wide, and 9 feet deep, provided they work only 3 hours each day ? Ans.200 men.

PRACTIC E.

** 111. PRACTICE is a short method of finding the answer to such questions in the Rule of Three as have a unit for their first term. So named, because in the ordinary practical business of life very frequent use is made of it. As an example, suppose one bushel of apples to be worth 50 cents, what is the value of 18; bushels? Had the apples been worth $1 per bushel, it is plain that 18% bushels would have been worth $18}, that is, $1850. Now since 50 cents is just half of one dollar, they must have been worth half of $18.50=$9.25. In order to work by this rule, we must make use of aliquot parts. An aliquot part of any thing is an exact part. In the above example, 50 cents is an aliquot part of $1, since it is exactly half of $1. We will give some aliquot parts which are in frequent use, in the following

TA B L E OF A LIQ U OT PARTS.

cts. & limo. yr. &. 36 | d. &e 50 = } | 6 = } 10 =} |6 = } 33} = } || 4 = } 6 8d. =#| |4 = } 25 = } | 3 = } 5 =#| |3 = } 20 = } || 2 = } 4 =} |2 = } 164- $ | 1 = + 3 4d. =#|1} = } 12}= # 15da. = }of Imo. 2 6d. =#| |1} = } 10 = 1, 10 = } “ |2 = |*|1} = # 8}= th 6 – # &g 1 8d. =# 1 - * d. 6}= I's 5 - # 44 1 4d. =# 2far - } of 1. 5 = #| 3 = % “ | 1 =#| || = + “

what is Practice? What is an aliquot part of anything 4 Repeat all the aliquot parts of a dollar as given in the above table. Repeat in the same way all the other aliquot parts of the table.

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