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A certain farm has a group of sheep, some of which are rams (males) an
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Updated on: 15 Oct 2014, 06:56
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A certain farm has a group of sheep, some of which are rams (males) and the rest ewes (females). The ratio of rams to ewes on the farm is 4 to 5. The sheep are divided into three pens, each of which contains the same number of sheep. If the ratio of rams to ewes in the first pen is 4 to 11, and if the ratio of rams to ewes in the second pen is the same as that of rams to ewes in the third, which of the following is the ratio of rams to ewes in the third pen? A. 8/7 B. 2/3 C. ½ D. 3/12 E. 1/6
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Originally posted by anceer on 15 Oct 2014, 06:36.
Last edited by Bunuel on 15 Oct 2014, 06:56, edited 1 time in total.
Edited the question.




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Re: A certain farm has a group of sheep, some of which are rams (males) an
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15 Oct 2014, 07:12
anceer wrote: A certain farm has a group of sheep, some of which are rams (males) and the rest ewes (females). The ratio of rams to ewes on the farm is 4 to 5. The sheep are divided into three pens, each of which contains the same number of sheep. If the ratio of rams to ewes in the first pen is 4 to 11, and if the ratio of rams to ewes in the second pen is the same as that of rams to ewes in the third, which of the following is the ratio of rams to ewes in the third pen?
A. 8/7 B. 2/3 C. ½ D. 3/12 E. 1/6 The ratio of rams to ewes on the farm is 4 to 5 > rams:ewes = 4x:5x > total sheep = 9x; The sheep are divided into three pens, each of which contains the same number of sheep > each pen contains 3x sheep. The ratio of rams to ewes in the first pen is 4 to 11 > 4y:11y > sheep in in the first pen = 15y. We know that each pen contains 3x sheep, thus 3x = 15y > x = 5y. From rams:ewes = 4x:5x, we have rams:ewes = 20y:25y. Together in the second and third pins there are (20y  4y) =16y rams and (25y  11y) = 14y ewes, so 8y rams and 7y ewes in each. Ratio = 8y:7y = 8:7. Answer: A. Or you can simply plug numbers, foir example, assume that there are 45 (should be a multiple of 9) sheep and proceed.
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Re: A certain farm has a group of sheep, some of which are rams (males) an
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11 Jan 2017, 20:26
Since the ratio of rams to ewes in the first pen is 4:11, let the number of rams in the first pen be 4x and the number of ewes be 11x. Let r be the number of rams in the second pen and let e be the number of ewes in the second pen. Since the number of sheep in each pen is the same, we can construct the following equation: 4x + 11x = r + e, or 15x = r + e. Since the number of sheep in each pen is the same, we know that the number of rams in the second and third pens together is 2r and the number of ewes in the second and third pens together is 2e. Therefore, the total number of rams is 4x + 2r. The total number of ewes is 11x + 2e. Since the overall ratio of rams to ewes on the farm is 4:5, we can construct and simplify the following equation: We can find the ratio of r to e by setting the equations we have equal to each other. First, though, we must multiply each one by coefficients to make them equal the same value: Since both equations now equal 120x, we can set them equal to each other and simplify: The correct answer is A.
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Re: A certain farm has a group of sheep, some of which are rams (males) an
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11 Jan 2017, 20:38
Another approach : It is important to remember that if the ratio of one group to another is x:y, the total number of objects in the two groups together must be a multiple of x + y. So since the ratio of rams to ewes on the farm is 4 to 5, the total number of sheep must be a multiple of 9 (4 parts plus 5 parts). And since the ratio of rams to ewes in the first pen is 4 to 11, the total number of sheep in the first pen must be a multiple of 15 (4 parts plus 11 parts). Since the number of sheep in each pen is the same, the total number of sheep must be a multiple of both 9 and 15. If we assume that the total number of sheep is 45 (the lowest common multiple of 9 and 15), the number of rams is 20 and the number of ewes is 25 (ratio 4:5). 45/3 = 15, so there are 15 sheep in each pen. Therefore, there are 4 rams and 11 ewes in the first pen (ratio 4:11). This leaves 20  4 = 16 rams and 25  11 = 14 ewes in the other two pens. Since the second and third pens have the same ratio of rams to ewes, they must have 16/2 = 8 rams and 14/2 = 7 ewes each, for a ratio of 8:7 or 8/7.
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A certain farm has a group of sheep, some of which are rams (males) an
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21 Jan 2018, 21:53
anceer wrote: A certain farm has a group of sheep, some of which are rams (males) and the rest ewes (females). The ratio of rams to ewes on the farm is 4 to 5. The sheep are divided into three pens, each of which contains the same number of sheep. If the ratio of rams to ewes in the first pen is 4 to 11, and if the ratio of rams to ewes in the second pen is the same as that of rams to ewes in the third, which of the following is the ratio of rams to ewes in the third pen?
A. 8/7 B. 2/3 C. ½ D. 3/12 E. 1/6 let r=rams in pen 3 e=ewes in pen 3 15s=total sheep in pen 1 r+e=15s r+e is a multiple of 15 plugging in choice A, assume r+e=8 rams+7 ewes=15 sheep pen 1 must also contain 15 sheep: 4 rams+11 ewes thus, total farm rams/ewes ratio=20/25=4/5 A



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Re: A certain farm has a group of sheep, some of which are rams (males) an
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22 Jan 2018, 00:14
anceer wrote: A certain farm has a group of sheep, some of which are rams (males) and the rest ewes (females). The ratio of rams to ewes on the farm is 4 to 5. The sheep are divided into three pens, each of which contains the same number of sheep. If the ratio of rams to ewes in the first pen is 4 to 11, and if the ratio of rams to ewes in the second pen is the same as that of rams to ewes in the third, which of the following is the ratio of rams to ewes in the third pen?
A. 8/7 B. 2/3 C. ½ D. 3/12 E. 1/6 Use weighted averages. You have a "total population" which you have split. This should remind you of weighted averages. Concentration of rams in total population = 4/9 Concentration of rams in first pen = 4/15 Concentration of rams in second and third pens combined = x (same concentration in both pens) Since number of sheep in each pen is the same, the weight of first pen:second and third combined = 1:2 1/2 = (x  4/9)/(4/9  4/15) 1/2 *(4/3)*(2/15) + 4/9 = x 8/15 = x Ratio of rams:ewes = 8:7 Answer (A)
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