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Manhattan GMAT challenge question

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by gabriel » Fri Nov 23, 2007 11:00 pm
The question can be found on www.manhattangmat.com

Some of the people in Town X are left-handed, some are tall, some are both, and some are neither. In Town Y, three times as many people are left-handed as are left-handed in Town X, three times as many people are tall as are tall in Town X, three times as many people are both as are both in Town X, but no one is neither. If the total number of people in Town X is four times greater than the total number of people in Town Y, which of the following could be the number of people in Town X who are neither left-handed nor tall?

(A) 23
(B) 39
(C) 72
(D) 143
(E) 199
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Re: Manhattan GMAT challenge question

by projase » Sat Nov 24, 2007 5:44 am
The equations are the followings and we need N.

(1) X = L+T-B+N =4Y
(2) Y=3L+3T-3B=3(L+T-B) =>Y/3 =(L+T-B)

Putting (2) in (1): Y/3+N = 4Y => 11/3*Y=N

The only alternative that is a multiple of 11 is (D).
Where there's a will, there's a way...
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by cris » Sat Nov 24, 2007 8:51 pm
i think that the correct answer is C (not D)
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by gabriel » Sat Nov 24, 2007 9:32 pm
cris wrote:i think that the correct answer is C (not D)
Care to explain cris ..
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by cris » Sun Nov 25, 2007 10:13 am
Gabriel, not sure 100%, but I think the answer needs to be a multiple of 12...and only C is.

Why 12? 3(that repets in L, T and N)x4
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by gabriel » Sun Nov 25, 2007 10:42 pm
cris wrote:Gabriel, not sure 100%, but I think the answer needs to be a multiple of 12...and only C is.

Why 12? 3(that repets in L, T and N)x4
Well, you have used the correct logic but seems like you have missed a step or 2, my answer is same as that of projase... D .. as the number should be a multiple of 11 ..

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by samirpandeyit62 » Mon Nov 26, 2007 12:10 am
IMO D
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