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animals [#permalink] New post 08 Feb 2009, 23:05
Pls explain ur answer
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Re: animals [#permalink] New post 09 Feb 2009, 01:50
I'd say the answer is C.
From the question, we get that the total # of Pigs and Cows are 2/3*(60)=40
1) 2C>P
insuff. (P=1, C=39 or P=2, C=38)
2) P>12
insuff. (P=13, C=27 or P=14, C=26)

Together:
If P=13, then C=27. Both 1)+2) cond. are fulfilled
If P=14, then C=26, 2C<P, thus, condi. 1 is not fulfilled. So P=13, C=27 is the only scenario that works.
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Re: animals [#permalink] New post 09 Feb 2009, 04:37
total = 60
2/3 of it is either cows or pigs
So 40 = cows + pigs

Stmt1: more than twice as many cows as it has pigs ==>C >2P
P=1, C = 39
.
.
.
p=13, c= 27 --- till here it satisfies the condtion C>2P

Still we cannot say the value of C (39......27)

Stmt2: says p>12.. we cannot conclude the value of C coz C can be from 27..to ...1

Combining 1,2 gives p=13 and C=27

C

Algebric way of solving this problem,

from Q, P+C = 40 ---1)
Stmt 1: C>2P --- 2)
sub 2) in 1) 3P<40 --- So P<13.3333

Not SUFF. since P can vary b/w 1 to 13

Stmt2: p>12 we cannot conclude the value of C coz C can be from 27..to ...1

Combining both
12<p<13.33

So absolute value is 13

hence C
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Re: animals [#permalink] New post 09 Feb 2009, 04:53
each statement not sufficient indiviually

combined
C+P = 2/3*60 = 40

P >12 assume p=13
C>2P C an be 27,28,29..
but only 27 will fit..to satisfy P=13 and C+p=40

Sufficient

C is the answer
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Re: animals [#permalink] New post 09 Feb 2009, 05:03
C is the answer! What is the OA?
Re: animals   [#permalink] 09 Feb 2009, 05:03
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