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Only case that satisfies both the statements is 13,27.

I guess we are both wrong C should be the answer
The farm has more than twice as many cows as it has pigs. Hence there could be only 13 pigs and 27 cows
_________________

I. C > 2P gives 40 = ( greater than 2P ) + P....INSUFF.
II. P > 12 . If P = 13, then C could be 27. If P =14, then C could be 26.INSUFF.

I + II gives

40 = greater than 2P + P ( which is greater than 12 )/
Let P =12, then 40 = (>2P) + 13...the number of cows has to be 27.
Let P =13, then 40 = (>2P) + 14..the number of cows need to be greater than 28, but then the total is greater than 40...INVALID.

Only case that satisfies both the statements is 13,27.

I guess we are both wrong C should be the answer The farm has more than twice as many cows as it has pigs. Hence there could be only 13 pigs and 27 cows

I replied C already. There is only one case when both the statements are taken together. So it means, its SUFF.
_________________

2/3 are either pigs or cows means 40
a) Pigs =x Cows >2x Insuff
b)>12 pigs Insuff
Combine
and 3x<=40=>x<=13 with B we know only x=13 stands out , so remaining are cows

I did the same mistake as Bhai. I thought that we had only cow and pigs in the farm, and that we had either 40 pigs + 20 cows, or 40 cows + 20 pigs...
I guess I have to work my verbal a bit tougher then ...

Can someone please explain why Pig+cow=40. why don't we take either pig=40 or cow=40.

Of the 60 animals on a certain farm, 2/3 are either pigs or cows. How many of the animals are cows?

2/3 of 60 animals are either pigs or cows. So there are total of 40 cows and pigs.

(1) The farm has more than twice as many cows as it has pigs --> \(c>2p\) --> as \(c+p=40\) then \(c=40-p\) --> \(40-p>2p\) --> \(13.3>p\) --> \(p_{max}=13\) and \(c_{min}=27\). Many combinations are possible: (27,13), (28, 12), ... Not sufficient.

(2) \(p>12\). Not sufficient

(1)+(2) \(p>12\) but \(p_{max}=13\), hence \(p=13\) --> \(c=27\). Sufficient.