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DS-Prob 2

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DS-Prob 2

by er.twi.fb » Sat Jul 31, 2010 11:31 pm
Q->Of the 60 animals,2/3 are either pigs or cows. How many animals are cows?

1>The farm has twice as many cows as it has pigs
2>The farm has more then 12 pigs

OA-C

My OA-"A"
Proof- 2/3 *60 =40.
So P+C=40
1> P=2C, Substitute and you are done. Suff
2>NS


Any Suggestions guys?
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Source: — Data Sufficiency |
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by kvcpk » Sun Aug 01, 2010 12:57 am
er.twi.fb wrote:Q->Of the 60 animals,2/3 are either pigs or cows. How many animals are cows?
1>The farm has twice as many cows as it has pigs
2>The farm has more then 12 pigs
OA-C
My OA-"A"
Proof- 2/3 *60 =40.
So P+C=40
1> P=2C, Substitute and you are done. Suff
2>NS
Any Suggestions guys?
As you have written the question, your answer looks good.

But I think there is something missing in statement1.
I remeber looking at this problem earlier.
It was some thing like farm has more than twice as many cows as it has pigs.
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by er.twi.fb » Sun Aug 01, 2010 12:59 am
Oh yeah, you are right. Sorry, my bad..
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by er.twi.fb » Sun Aug 01, 2010 1:03 am
So P>2C and P>12, which tells us 2C=12,C=6.

Hence C.

Is that right Kvcpk
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by kvcpk » Sun Aug 01, 2010 1:19 am
er.twi.fb wrote:So P>2C and P>12, which tells us 2C=12,C=6.

Hence C.

Is that right Kvcpk
farm has more than twice as many cows as it has pigs means
C>2P
Add P on both sides -> C+P>3P
we know that C+P = 40
3P<40
P <13.33
and from 2 we get
p>12
Hence P has to be 13. (Number of pigs cant be fraction)
Hence C = 27.

SUFF

Hope this helps!!
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