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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

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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 _________________

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
25 Nov 2014, 03:17

I guess we cannot solve it using Matrix method as , Both cannot be Cow and Pig .

Is there any alternative method to solve this .

Thanks

Bunuel wrote:

Syed wrote:

yangsta8 wrote:

Question stem says that 2/3 of 60 are pigs or cows. That means 40 animals are pigs or cows. So all we need to have sufficiency is either number of pigs or number of cows.

1) Farm has more than 2 cows for 1 pig (at least 26.6 animals of the 40 are cows). This just tells us that the farm has at least 27 pigs and that the max number of cows is 13. For example the farm could have 30 pigs and 10 cows. Not a definitive number. Insufficient.

2) Farm has more than 12 pigs. Again not enough info. Insuff

1+2) Statements together tell us: 12 < number of pigs is <=13 Which means number pigs = 13 Number of cows = 27.

ANSWER = C.

I do agree with 'yangsta8' that the answer is 'C' but (if I am NOT wrong), Stmnt#1 says the number of Cows are more than twice than the pigs. Thus, the cows could be 27 and Pigs could be 13; and For example, the farm could have 30 cows and 10 pigs. ------------------------------- Hi Bunuel and Yangsta8 - Please correct me if I am wrong (which is very much possible)!

c+p=40

(1) c>2p --> min # of cows is 27 and max # pigs is 13, so there can be any combination not violating this and totaling 40. Not sufficient (2) p>12 Not sufficient

(1)+(2) p>12 but max of p is 13, hence p=13 --> c=27

You are right there can be 27 cows (min) and 13 pigs (max) or 30 cows and 10 pigs.

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
14 Mar 2015, 06:34

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.

Sorry Bunuel/Experts, i could not get why is A not sufficient .

statement A : C > 2P then must be C>P for all positive integer values of C and P . now we know Cows are more in number , so from question stem 2/3 are cows and hence 1/3 will be pigs . where am i missing( i know i am falling in a intended trap of this question ) ? _________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
14 Mar 2015, 06:48

Expert's post

Lucky2783 wrote:

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.

Sorry Bunuel/Experts, i could not get why is A not sufficient .

statement A : C > 2P then must be C>P for all positive integer values of C and P . now we know Cows are more in number , so from question stem 2/3 are cows and hence 1/3 will be pigs . where am i missing( i know i am falling in a intended trap of this question ) ?

Your doubt is answered on previous pages.

Of the 60 animals on a certain farm, 2/3 are either pigs or cows. Hence 40 animals are either pigs or cows, meaning that out of 40 animals some are pigs and the rest are cows: p+c=40.

(1) The farm has more than twice as many cows as pigs --> c > 2p --> min # of cows is 27 and max # pigs is 13, so there can be any combination not violating this and totaling 40. Not sufficient _________________

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
14 Mar 2015, 06:51

Lucky2783 wrote:

Sorry Bunuel/Experts, i could not get why is A not sufficient .

statement A : C > 2P then must be C>P for all positive integer values of C and P . now we know Cows are more in number , so from question stem 2/3 are cows and hence 1/3 will be pigs . where am i missing( i know i am falling in a intended trap of this question ) ?

hi Lucky2783, what are you are missing is the meaning of question stem... it says 2/3 are either pigs or cows.... this means that out of 60, 2/3rd that is 40 are either pigs or cows and rest can be any other animal other than this two... it does not say that 2/3 are one of the two and 1/3 remaining are the other of the two...

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 pigs

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
14 Mar 2015, 08:21

chetan2u wrote:

Lucky2783 wrote:

Sorry Bunuel/Experts, i could not get why is A not sufficient .

statement A : C > 2P then must be C>P for all positive integer values of C and P . now we know Cows are more in number , so from question stem 2/3 are cows and hence 1/3 will be pigs . where am i missing( i know i am falling in a intended trap of this question ) ?

hi Lucky2783, what are you are missing is the meaning of question stem... it says 2/3 are either pigs or cows.... this means that out of 60, 2/3rd that is 40 are either pigs or cows and rest can be any other animal other than this two... it does not say that 2/3 are one of the two and 1/3 remaining are the other of the two...

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 pigs

(2) The farm has more than 12 pigs

ohhh god !! thats a very callous mistake at my end . thanks chetan. _________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

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