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Re: Word problem from GMATPrep (2) [#permalink]
12 Oct 2009, 02:01

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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.

Re: cow and pig problem [#permalink]
12 May 2010, 13:49

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Expert's post

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. _________________

Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
12 Oct 2009, 17:50

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Syed wrote:

yangsta8 wrote:

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

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.

Think there was simple typo from yangsta8. _________________

Re: DS-Pigs and Cows [#permalink]
11 May 2010, 13:27

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I like this question. If you read it too fast (as I often do), it’s super confusing.

IMO the answer has to be C.

From the stem, we know that there are 40 cows and pigs, or C+P=40.

St 1: Tells us that the most pigs we can have is 13, or P<13. Any more than that, then their won’t be twice as many cows. But there could be less than 13 pigs. Insuf.

St 2: Now tells us there has to be at least 13 pigs, or P>12 Alone this is not very helpful, but combined with one, we can get the answer.

Insufficient. We are told there are more than twice as many cows as there are pigs, but we are given no indication as to how many more than twice as many.

(2): P > 12 Insufficient. We have no idea how many pigs or cows there are, just that there are at least 13 pigs.

(1) & (2)

Sufficient. Given P > 12 and C > 2P, we need to have at least 26 cows to satisfy statement 2. Since 13 + 26 = 39, it is not possible to have more than 13 pigs on the farm and still have C + P < 41.

Last edited by LJ on 18 Sep 2010, 14:25, edited 1 time in total.

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
09 Feb 2013, 23:05

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Archit143 wrote:

HI all I am facing little difficulty in interpretation of the question, My specifc doubts are:- 1. Here either or means C+p = 40 , since a pig cannot be a cow at the same time. What if the question had a scenario, where a case of both was possible.....should we consider c+p- both = 40 in that case. I am haveing a doubt with either or statement 2. In case of question stating "What was the number of cows", what should we infer that it requires number of only cow ie cow - both or only cow + both, i am facing difficulty...... 3."I. The farm has more than twice as many cows as it has pigs " Can we interpret this as 'For every pig there were more than twice cow" ie c/p>2/1

Pls help me in clearing my doubts...

Regards Archit

Responding to a pm:

Either A or B means either A or B or both. If they mean to say that both should not be included then they will say 'Either A or B but not both'

What is the number of A? implies all A (including those who can be B too) What is the number of only A ? implies those A who are B too are not to be counted.

3."I. The farm has more than twice as many cows as it has pigs " Can we interpret this as 'For every pig there were more than twice cow" ie c/p>2/1

Yes, that's correct. You can write it as c/p > 2 or as c > 2p (same thing) _________________

Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
19 Mar 2014, 20:34

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Expert's post

TooLong150 wrote:

Could we add across inequalities as I did below? If not, when is it allowed to add across inequalities? (1) c > 2p NS

(2) p > 12 NS

(1) + (2) c + p > 2p + 12 40 > 2p + 12 2p < 28 p < 14 Because (2) gives p > 12, with the above statement, p = 13 and we can solve for c. S

You can add inequalities as long as both the inequalities have the same sign. a > b c > d gives a+c > b + d Think logically: a is greater than b and c is greater than d so a+c will be greater than b+d because you are adding the larger numbers together. So what you have done above is correct.

If the inequality signs are different, you cannot add them.

Also, you can subtract inequalities when they have opposite signs but prefer not to do that to avoid confusion. Just flip the sign of one inequality by multiplying it by -1 and then add them up. _________________

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

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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

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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 _________________

Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
12 Oct 2009, 11:28

Expert's post

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.

Re: Word problem from GMATPrep (2) [#permalink]
12 Oct 2009, 17:13

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)!

Re: cow and pig problem [#permalink]
30 Mar 2010, 16:01

The key word in statement 1 is "more" than twice as many. If "more" wasn't there, you would be right. From the stem we know that #pigs + #cows = 40

Statement 1: If you had 1 pig and 39 cows, you have more than twice as many cows as pigs (a lot more than twice the number actually). If you had 2 pigs and 38 cows, the statement still holds true. Therefore, insufficient by itself.

Statement 2: If you had 13 pigs and 27 cows, the statement holds true. If you had 14 pigs and 26 cows, the statement still holds true. Therefore, insufficient by itself.

Together: Start with the first posible option according to statement 2: 13 pigs and 27 cows works for both statements. The next option (14 pigs and 26 cows) violates statement 1 because you don't have more than twice as many cows as pigs. Every option after this will have the same problem. Therefore you only have one option - sufficient to determine the answer. C.

Re: DS-Pigs and Cows [#permalink]
11 May 2010, 13:17

2/3 of 60 animals are pigs or cows. so 40 of the animals are pigs or cows. We are not told what the rest are. The question is how many of the animals are cows?

Statement one says there are more than twice as many cows as pigs we would write 2P < C. This is not sufficient to give the answer.

Statement two says the farm has 12 pigs. Since we know that there are 40 cows and pigs we can simply subtract the 12 from the total pigs and cows and get that there are 28 cows. Of course we don't have to complete the math to say that number two is sufficient.

Re: cow and pig problem [#permalink]
12 May 2010, 09:37

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 _________________

------------------------------------------------------------------------- Ros. Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people. -------------------------------------------------------------------------

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. (2) The farm has more than 12 pigs

1. so we have either 40 cows OR pigs. so 1 is saying C> 2P so we could have P =10 and C=30 which is twice as P. OR we could have P =14 and C = 26 which is more. notice C changes! INSUFF

2. so we know there are more than 12 pigs... which tells us nothing about cows so INSUFF

C. ok. so we have C > 2*12 = C>24 since we know P =12 we can figure out cows is 28 _________________

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