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Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
11 Jul 2014, 17:20

Expert's post

rbaudoin10 wrote:

What about the 20 extra animals ? Why can't they be cows or pigs as well within that set ? Nothing is mentioned about these 20 animals. We just know that 40 animals are either pigs or cows. Does anyone get my point ?

Having 13 pigs, 27 cows, and 20 other animals is a possiblity.

Having 14 pigs, 30 cows and 16 other animals is another possibility.

Following that logic, E is the right answer.

To add to what Bunuel said:

It is similar to: Of the 100 people in a room, 40 are men. What does this mean to you? It means that 60 are not men, right?

Similarly, Of the 60 animals on a certain farm, 40 are either pigs or cows. This means that rest of the 20 are neither pigs nor cows!

When you are given concrete numbers, it implies that the group has been considered. If only some people were considered, you would have been given "Of the 100 people in a room, at least 40 are men." _________________

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
30 Aug 2014, 10:31

Bunuel wrote:

shrivastavarohit wrote:

Guys I don't know how long this post has been here however I thought I would add my 2 cents and see if what I feel should be the answer.

Question says something either a cow or a pig form the ratio of 2/3. I don't feel 60 number is of much significance since this is DS.

1) says the mix is 1 to 2. So if by simple math I find what's remaining out of 2/3 the ratio comes out to be 1/3. At this moment for me the information in the statement 1 becomes helpful.

Since 2wice of 1/3 is 2/3 and we know that 2wice as many cows hence this information is sufficient.

Posted from my mobile device

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 as it has pigs --> so we have is c>2p and not c=2p --> as c+p=40 (p=40-c and 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.

Answer: C.

One more thing: if the ratio indeed were c=2p, then the question would be flawed as solving c=2p and c+p=40 gives p=13.3, but # of pigs can not be a fraction it MUST be an integer.

Hi Bunuel,

Two questions:

-How do you know that Cmin is 27? I can se how Pmax is 13, but I don't really see how that translates onto Cmin? -If I were to add Stm 1 and Stm 2, I get c-p>12. That, algebraically, doesn't help solve the problem Am I correct?

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
31 Aug 2014, 23:48

Expert's post

russ9 wrote:

Bunuel wrote:

shrivastavarohit wrote:

Guys I don't know how long this post has been here however I thought I would add my 2 cents and see if what I feel should be the answer.

Question says something either a cow or a pig form the ratio of 2/3. I don't feel 60 number is of much significance since this is DS.

1) says the mix is 1 to 2. So if by simple math I find what's remaining out of 2/3 the ratio comes out to be 1/3. At this moment for me the information in the statement 1 becomes helpful.

Since 2wice of 1/3 is 2/3 and we know that 2wice as many cows hence this information is sufficient.

Posted from my mobile device

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 as it has pigs --> so we have is c>2p and not c=2p --> as c+p=40 (p=40-c and 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.

Answer: C.

One more thing: if the ratio indeed were c=2p, then the question would be flawed as solving c=2p and c+p=40 gives p=13.3, but # of pigs can not be a fraction it MUST be an integer.

Hi Bunuel,

Two questions:

-How do you know that Cmin is 27? I can se how Pmax is 13, but I don't really see how that translates onto Cmin? -If I were to add Stm 1 and Stm 2, I get c-p>12. That, algebraically, doesn't help solve the problem Am I correct?

There are total of 40 pigs and cows. We know that there are at most 13 pigs (13 or less). Thus there are at least 27 cows.

Or: p + c = 40 and p <= 13 --> (40 - c) <= 13 --> c >= 27.

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
29 Sep 2014, 06:58

1

This post received KUDOS

Bunuel wrote:

ngoctraiden1905 wrote:

From your ans I think the stem means 2/3 of 60 animals are both pigs and cows, not either pigs or cows. I think there is some problem with stem here?

How it's possible for an animal to be BOTH a pig and a cow? 2/3 of 60 animals are either pigs or cows. So there are total of 40 cows and pigs. Yansta8's solution is correct.

C it is.

Maybe they are half cow, half bear, half pig.... Cowbearpig _________________

What we think, we become

gmatclubot

Re: Of the 60 animals on a certain farm, 2/3 are either cows or
[#permalink]
29 Sep 2014, 06:58