DS question - Any easy method?

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DS question - Any easy method?

by singh21 » Sun Nov 09, 2008 7:34 pm
Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows?

(1) The farm has more than as many cows as it has pigs.

(2) The farm has more than 12 pigs.

The answer is (C)
Source: — Data Sufficiency |

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Re: DS question - Any easy method?

by logitech » Sun Nov 09, 2008 8:33 pm
singh21 wrote:Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows?

(1) The farm has more than as many cows as it has pigs.

(2) The farm has more than 12 pigs.

The answer is (C)
I don't like the language of the problem and I doubt that the answer is C.

How about E ? 13 pigs 17 cows vs 14 pigs 16 cows and etc
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by singh21 » Sun Nov 09, 2008 8:58 pm
This is an old GMAT real question and the answer is indeed (C). I know it is confusing ....and couldn't make sense out of it either :(

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by tnis0612 » Mon Nov 10, 2008 8:57 am
Yeah sure looks like E to me. All that is saying is that cows+pigs = 40.

more cows than pigs. 21 cows 19 pigs, 22 cows 18 pigs, etc.
more than 12 pigs. 21 cows 19 pigs, 22 cows 18 pigs...27 cows 13 pigs.

Unless there is something major left out of this question I really dont see how it could possibly be C.

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Re: DS question - Any easy method?

by wdouglasf » Mon Nov 10, 2008 11:59 am
singh21 wrote:Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows?

(1) The farm has more than as many cows as it has pigs.

(2) The farm has more than 12 pigs.

The answer is (C)
Are you sure that Statement 1 is written correctly here? Should it read "the farm has more than twice as many cows as it has pigs"?

If so, then the statements together are, in fact, sufficient. If 2/3 of the 60 animals are pigs and cows, then P + C = 40.

From the first statement we would know that C > 2P.

From the second statement we know that P > 12.

So when we combine them, we see that P has to be at least 13, and that C has to be more than twice this number, or more than 26. Therefore, if we plug the least possible value for P, which is 13, into the first statement, we see that C would equal 27. 27 is greater than 2P, so it works.

However, if P equals 14, then C would equal 26. 26 is less than 2P (which would be 28), so this will not hold true using both statements. Therefore, the only values that work with both statements are P=13 and C=27, so together they are SUFFICIENT.

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by tnis0612 » Wed Nov 12, 2008 5:41 am
Good catch. I'd say that has to be what the problem is. As it is the wording really woudlnt be right anyway. I dont think they would say "has more than as many"...more than twice as many certainly sounds more likely.

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by logitech » Wed Nov 12, 2008 8:53 am
I hate typos!
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