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Knewton Problem

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by akhilsuhag » Wed Jul 27, 2011 12:42 pm
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 2/3 full, the number of sheep in the pen will decrease by 1/4.

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

The answer is A only and the explanation provided is:

" Statement 1 tells us that if 8 sheep are removed from the pen when it is 2/3 full, then the number of sheep in the pen will decrease by 1/4. This means that 8 is equal to 1/4 of 2/3 of the pen's maximum capacity."

I was hoping if someone could explain how to get to this. I am gettin lost and confused in the words.


Thanks.
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Source: — Data Sufficiency |

by clock60 » Wed Jul 27, 2011 1:23 pm
hi, i`ll try to translate in math terms
let it be p the number of sheeps in the pen
(2/3)*p-8=(2/3)*p-(2/3)*p*1/4
(2/3)*p-8=(2/3)*p(1-1/4)
(2/3)*p-8=(2/3)*p*(3/4)
(2/3)*p-8=(1/2)*p
p*(2/3-1/2)=8
p*(1/6)=8
p=48
so suff
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by Asher » Tue Aug 02, 2011 8:00 pm
Hi clock60,

I may sound like a nitwit, but I dint quite understand the explanation.

how did you arrive at the equation : (2/3)*p-8=(2/3)*p-(2/3)*p*1/4

:oops: :oops:
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by gmatboost » Tue Aug 02, 2011 8:52 pm
Hi,

I'll try to rephrase the explanation a bit.

Statement 1 says: If 8 sheep are removed from the pen when it is 2/3 full, the number of sheep in the pen will decrease by 1/4.

Forget for a moment that the pen is 2/3 full. Focus on the fact that if 8 are removed, the number in the pen will decrease by 1/4. That means that these 8 removed sheep make up 1/4 of what is currently in the pen.

So if x = currently in the pen, then x * 1/4 = 8, or x = 32.

Now, we go back to the fact that the pen is 2/3 full. So, 32 sheep represents 2/3 of a full pen.
Since this is a DS question, DON'T CALCULATE the answer. You should feel confident that if you know that 2/3 * n = 32, you can find n. So Statement 1 is sufficient.

You could even take this reasoning one step further:
From the start, you know that some OBTAINABLE fraction of the pen is equal to 8 sheep. I say it is obtainable because we know both how full the pen is (2/3) and what fraction of the sheep in the pen are removed (1/4). So, we could set up the "equation"
OBTAINABLE fraction * full pen = 8.
Since we COULD calculate OBTAINABLE fraction, we can conclude that Statement 1 is sufficient without doing any more math. It requires a fair amount of confidence to draw that conclusion early on, but doing so can save you precious time on the GMAT, so it's worth considering.

Let me know what you think.
Greg Michnikov, Founder of GMAT Boost

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by Asher » Tue Aug 02, 2011 9:21 pm
Thank you gmatboost. Got it :) Kudos!
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by tpr-becky » Wed Aug 03, 2011 11:13 am
another way to decode the algebra for statement 1 is to read it 1 phrase at a time - using S for the max number of sheep.

if 8 sheep are removed (means - 8), when the pen is 2/3 full (means 2/3(s)) is thus translated to:

2/3(S) - 8

the number of sheep will decrease (this means to subtract from the original number of sheep):

2/3(s) - (----)

by 1/4 (you need to find 1/4 of 2/3(S)) of means multiply so 1/4(2/3)(S)

thus the full formula is 2/3(S)- 8 = 2/3(S) - 1/4(2/3)(S)

you can see that with only one variable, and no exponents, you could solve this problem (no need to actually do it)

The second statement only gives you a current number but no relationship to the entire population.
Becky
Master GMAT Instructor
The Princeton Review
Irvine, CA
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by akhilsuhag » Thu Aug 04, 2011 1:12 am
Thanks guys!!
Made the picture so much more clearer!!
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