A certain farm has a group of sheep, some of which are rams (males) and the rest ewes (females). The ratio of rams to

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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


OA A

Source: Princeton Review

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BTGmoderatorDC wrote:
Wed Mar 22, 2023 2:03 pm
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


OA A

Source: Princeton Review
We can let the total number of rams and ewes in the farm be 4n and 5n, respectively. We can let the number of rams and ewes in the first pen be 4m and 11m, respectively. Thus we have 4n - 4m rams and 5n - 11m ewes left for the remaining two pens. Since the ratio of rams to ewes is equal in the second and third pen, each pen must have:

(4n - 4m)/2 = 2n - 2m rams

and

(5n - 11m)/2 = 2.5n - 5.5m ewes.

Since all 3 pens have an equal number of sheep and the total number of sheep is 9n, then each pen has 3n sheep.

However, the first pen has 4m + 11m = 15m sheep; thus:

3n = 15m

n = 5m

Substituting n = 5m in 2n - 2m and 2.5n - 5.5m, we have:

2(5m) - 2m = 8m rams

and

2.5(5m) - 5.5m = 7m ewes in the third pen.

Therefore, the ratio of rams to ewes in the third pen is 8m/7m = 8/7.

Alternate Solution:

Let’s let the total number of rams on this farm be 40 and the total number of ewes on this farm to be 50 (note the ratio of rams to ewes is 40/50 = 4/5). Then, in total, there are 40 + 50 = 90 sheep on this farm, and since each pen has an equal number of sheep, there are 90/3 = 30 sheep in each pen.

Since the ratio of rams to ewes in the first pen is 4:11, we can let the number of rams be 4x and the number of ewes be 11x where x is some integer. Then, there are 4x + 11x = 15x sheep in the first pen, which we know to be equal to 30. Thus, 15x = 30 and x = 2. In the first pen, there are 4*2 = 8 rams and 11*2 = 22 ewes.

Then, in pens 2 and 3 combined, there are 40 - 8 = 32 rams and 50 - 22 = 28 ewes. Since the ratio of rams to ewes in the second and third pens is the same, so is the ratio of rams to ewes in second and third pens combined. Thus, the ratio of rams to ewes in the third pen (also the ratio of rams to ewes in the second pen and also the ratio of rams to ewes in the second and third pen combined) is 32/28 = 8/7.

Answer: A

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