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# Of the 60 animals on a certain farm, 2/3 are either pigs or

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Of the 60 animals on a certain farm, 2/3 are either pigs or [#permalink]

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22 Jan 2006, 23:59
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Question Stats:

57% (02:29) correct 43% (01:36) wrong based on 209 sessions

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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 twice as many cows at it has pigs.

(2) The farm has more than 12 pigs.

OPEN DISCUSSION OF THIS QUESTION IS HERE: of-the-60-animals-on-a-certain-farm-2-3-are-either-cows-or-92062.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Jan 2014, 06:00, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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23 Jan 2006, 00:03
If it is E I will Explain
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23 Jan 2006, 00:14
The ans is C. This is a tricky one.

Questions stem gives us P+C=40.

1) says C=2P+r, so basically P=(40-r)/3, multiple value possible.
2) says P>12, again multiple values possible.

Combining 1) and 2), min P=13. If P is more than 13, C is more than 28 and the sum exceeds 40. So P=13, C=27.

Ans = (C).
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23 Jan 2006, 02:41
Total pigs and cows = 40

St1: Could be 1, 39 or 2,38 etc etc.... INSUFF

St2: Could be 13, 27 or 14, 26 etc etc.... INSUFF

Combined:

Only case that satisfies both the statements is 13,27.
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23 Jan 2006, 10:27
ps_dahiya wrote:
Total pigs and cows = 40

St1: Could be 1, 39 or 2,38 etc etc.... INSUFF

St2: Could be 13, 27 or 14, 26 etc etc.... INSUFF

Combined:

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
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23 Jan 2006, 10:47
Here is the blunder I did. The question says "2/3 are either pigs or cows" and I thought 40 are either pigs or cows not both. How stupid?
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23 Jan 2006, 11:01
Bhai wrote:
Here is the blunder I did. The question says "2/3 are either pigs or cows" and I thought 40 are either pigs or cows not both. How stupid?

The true is born in dispute))
Bhai wish I were so stupid as you are)))
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23 Jan 2006, 12:51
C...

C+P = 2/3(60) = 40.

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.

Thus C = 27 and P = 13.
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23 Jan 2006, 13:05
Yurik79 wrote:
ps_dahiya wrote:
Total pigs and cows = 40

St1: Could be 1, 39 or 2,38 etc etc.... INSUFF

St2: Could be 13, 27 or 14, 26 etc etc.... INSUFF

Combined:

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.
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23 Jan 2006, 14:29
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23 Jan 2006, 23:06
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watever wrote:
The ans is C. This is a tricky one.

Questions stem gives us P+C=40.

1) says C=2P+r, so basically P=(40-r)/3, multiple value possible.
2) says P>12, again multiple values possible.

Combining 1) and 2), min P=13. If P is more than 13, C is more than 28 and the sum exceeds 40. So P=13, C=27.

Ans = (C).

i missed C=2P+r --> I overlook things easily.. thats y i scored less last time.

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vazlkaiye porkalam vazltuthan parkanum.... porkalam maralam porkalthan maruma

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23 Jan 2006, 23:40
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
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26 Jun 2006, 17:01

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 ...
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Re: Of the 60 animals on a certain farm, 2/3 are either pigs or [#permalink]

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04 Jan 2014, 13:45
Hi

Can someone please explain why Pig+cow=40. why don't we take either pig=40 or cow=40.
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Re: Of the 60 animals on a certain farm, 2/3 are either pigs or [#permalink]

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05 Jan 2014, 06:01
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vikrantgulia wrote:
Hi

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: of-the-60-animals-on-a-certain-farm-2-3-are-either-cows-or-92062.html
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Re: Of the 60 animals on a certain farm, 2/3 are either pigs or   [#permalink] 05 Jan 2014, 06:01
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