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Source: — Data Sufficiency |

by bww » Sat May 12, 2007 6:39 pm
Of 60 animals, 40 are either pigs or cows.

I) We have more than twice as many cows as pigs. Good to keep in mind, but not sufficient.

II) There are more than 12 pigs on the farm. Again, not sufficient.

By combining the two statements, we know that there are more than 12 pigs and more than double that number of cows. Let us assume there are 13 pigs. That means there must be 27 cows. 27 is more than twice 13. If we try 14 pigs, that means we have 26 cows, which is not even twice 14.

Hence, C.
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by mikeclarke44 » Sun Jun 10, 2007 5:57 pm
Yes,
but if you had a 2 to 1 ratio of cows to pigs and you knew there were 40 pigs and cows
wouldn't you have 2/3 * 40 cows
and 1/3 x 40 pigs????
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by jayhawk2001 » Sun Jun 10, 2007 8:48 pm
mikeclarke44 wrote:Yes,
but if you had a 2 to 1 ratio of cows to pigs and you knew there were 40 pigs and cows
wouldn't you have 2/3 * 40 cows
and 1/3 x 40 pigs????
Nope. The question says "more than twice as many" not "twice as many"
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by mikeclarke44 » Mon Jun 11, 2007 1:21 pm
Ah yes, tricky. I'd better be wary of the details!
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by anujgoyal11 » Tue Jul 03, 2007 11:27 pm
Hi Friends,
It's a simple mathemetical equation question.

X+Y=40
Condition
1) X>2*Y
2) Y>12
So by this we are come up that
if Y=13(By using Condition 2) then
By Condition 1 ) X>26 and
By X+Y=40 it is X = 27 so if we join both then we come up with one solution that is
X + 27 and Y = 13
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Additionally....

by GMATpaduan » Tue Jul 10, 2007 10:01 am
Mike - sometimes the math gets in the way of logic. You can't have 2/3 * 40 cows b/c then you would have a piece of a cow....which doesn't make sense unless you're cooking steak
GOOD LUCK ALL!
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