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# S95-37

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Math Expert
Joined: 02 Sep 2009
Posts: 52154

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16 Sep 2014, 00:50
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Difficulty:

35% (medium)

Question Stats:

73% (01:25) correct 27% (01:22) wrong based on 33 sessions

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What is the maximum number of sheep that Ruben's pen will hold?

(1) If $$8$$ sheep are removed from the pen when it is $$\frac{2}{3}$$ full, the number of sheep in the pen will decrease by $$\frac{1}{4}$$.

(2) Currently, there are 12 sheep in the pen.

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Joined: 02 Sep 2009
Posts: 52154

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16 Sep 2014, 00:50
Official Solution:

The question asks us to find the maximum number of sheep that Ruben's pen can hold.

Statement 1 tells us that if 8 sheep are removed from the pen when it is $$\frac{2}{3}$$ full, then the number of sheep in the pen will decrease by $$\frac{1}{4}$$. This means that 8 is equal to $$\frac{1}{4}$$ of $$\frac{2}{3}$$ of the pen's maximum capacity. Let us call $$m$$ the maximum capacity of Ruben's pen. Then $$\frac{2}{3}m$$ is the number of sheep in a pen that is $$\frac{2}{3}$$ full. Now we set up the equation $$8 = \frac{1}{4} \times \frac{2}{3}m$$. (Remember that "of" means multiplication in math problems.) We can solve for the single variable $$m$$ and thus for the maximum capacity of the pen. Statement 1 is sufficient to answer the question. Eliminate answer choices B, C, and E. The correct answer choice is either A or D.

Statement 2 tells us that the pen currently contains 12 sheep. This statement tells us that the pen can hold at least 12 sheep, but it gives us no clues as to what the maximum capacity of the pen is. Statement 2 alone is NOT sufficient to answer the question. Eliminate answer choice D.

Though we do not need to solve for $$m$$ explicitly, we will for illustration. From statement 1, we have the equation $$8 = \frac{1}{4} \times \frac{2}{3}m$$. Multiplying the fractions on the right side together, we get $$8 = \frac{2}{12}m = \frac{1}{6}m$$. Multiplying both sides by 6, we find that $$m = 48$$ is the maximum capacity of Ruben's pen.

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Joined: 07 Jul 2017
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03 Jan 2018, 06:22
Bunuel wrote:
Official Solution:

The question asks us to find the maximum number of sheep that Ruben's pen can hold.

Statement 1 tells us that if 8 sheep are removed from the pen when it is $$\frac{2}{3}$$ full, then the number of sheep in the pen will decrease by $$\frac{1}{4}$$. This means that 8 is equal to $$\frac{1}{4}$$ of $$\frac{2}{3}$$ of the pen's maximum capacity. Let us call $$m$$ the maximum capacity of Ruben's pen. Then $$\frac{2}{3}m$$ is the number of sheep in a pen that is $$\frac{2}{3}$$ full. Now we set up the equation $$8 = \frac{1}{4} \times \frac{2}{3}m$$. (Remember that "of" means multiplication in math problems.) We can solve for the single variable $$m$$ and thus for the maximum capacity of the pen. Statement 1 is sufficient to answer the question. Eliminate answer choices B, C, and E. The correct answer choice is either A or D.

Statement 2 tells us that the pen currently contains 12 sheep. This statement tells us that the pen can hold at least 12 sheep, but it gives us no clues as to what the maximum capacity of the pen is. Statement 2 alone is NOT sufficient to answer the question. Eliminate answer choice D.

Though we do not need to solve for $$m$$ explicitly, we will for illustration. From statement 1, we have the equation $$8 = \frac{1}{4} \times \frac{2}{3}m$$. Multiplying the fractions on the right side together, we get $$8 = \frac{2}{12}m = \frac{1}{6}m$$. Multiplying both sides by 6, we find that $$m = 48$$ is the maximum capacity of Ruben's pen.

This question seems easy..... and in the form, it is mentioned as 25% i.e medium diff............ but in the GMAT free test provided by the club its a 600 level question? ..... which is the legit evaluation of the difficulty of the question?
Math Expert
Joined: 02 Sep 2009
Posts: 52154

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03 Jan 2018, 06:26
ChrisChris wrote:
Bunuel wrote:
Official Solution:

The question asks us to find the maximum number of sheep that Ruben's pen can hold.

Statement 1 tells us that if 8 sheep are removed from the pen when it is $$\frac{2}{3}$$ full, then the number of sheep in the pen will decrease by $$\frac{1}{4}$$. This means that 8 is equal to $$\frac{1}{4}$$ of $$\frac{2}{3}$$ of the pen's maximum capacity. Let us call $$m$$ the maximum capacity of Ruben's pen. Then $$\frac{2}{3}m$$ is the number of sheep in a pen that is $$\frac{2}{3}$$ full. Now we set up the equation $$8 = \frac{1}{4} \times \frac{2}{3}m$$. (Remember that "of" means multiplication in math problems.) We can solve for the single variable $$m$$ and thus for the maximum capacity of the pen. Statement 1 is sufficient to answer the question. Eliminate answer choices B, C, and E. The correct answer choice is either A or D.

Statement 2 tells us that the pen currently contains 12 sheep. This statement tells us that the pen can hold at least 12 sheep, but it gives us no clues as to what the maximum capacity of the pen is. Statement 2 alone is NOT sufficient to answer the question. Eliminate answer choice D.

Though we do not need to solve for $$m$$ explicitly, we will for illustration. From statement 1, we have the equation $$8 = \frac{1}{4} \times \frac{2}{3}m$$. Multiplying the fractions on the right side together, we get $$8 = \frac{2}{12}m = \frac{1}{6}m$$. Multiplying both sides by 6, we find that $$m = 48$$ is the maximum capacity of Ruben's pen.

This question seems easy..... and in the form, it is mentioned as 25% i.e medium diff............ but in the GMAT free test provided by the club its a 600 level question? ..... which is the legit evaluation of the difficulty of the question?

600-level and medium is the same difficulty and it seems right to me.
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Re: S95-37 &nbs [#permalink] 03 Jan 2018, 06:26
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# S95-37

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