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Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
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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 as pigs (2) The farm has more than 12 pigs
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Re: Word problem from GMATPrep (2) [#permalink]
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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.
ANSWER = C.



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Re: Word problem from GMATPrep (2) [#permalink]
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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?



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Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
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Re: Word problem from GMATPrep (2) [#permalink]
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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)!



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Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
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12 Oct 2009, 18:50
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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.
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Re: Word problem from GMATPrep (2) [#permalink]
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Guys! I am a novive. My point was (dfinitely) NOT to point a mistake rather I was trying to understand the process and clarifying my understaning. Hence, please accept my apology if in anyway my writeup was pointing towards anything else. I do appreciate all the club members help.



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Re: Word problem from GMATPrep (2) [#permalink]
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Syed wrote: Guys! I am a novive. My point was (dfinitely) NOT to point a mistake rather I was trying to understand the process and clarifying my understaning. Hence, please accept my apology if in anyway my writeup was pointing towards anything else. I do appreciate all the club members help. No offence taking. We're all here to help each other!



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Re: cow and pig problem [#permalink]
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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.



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Re: DSPigs and Cows [#permalink]
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11 May 2010, 14:27
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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.
Together 13<P>12 so P=13, C=27



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Re: cow and pig problem [#permalink]
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12 May 2010, 10:37
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
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Re: cow and pig problem [#permalink]
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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=40c and c=40p) > \(40p>2p\), \(13.3>p\), \(p_{max}=13\) 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.
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Re: cow and pig problem [#permalink]
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i think the confusion for a few people (me included!) was the statement: "2/3 are either pigs or cows" i incorrectly started the problem by interpreting the above as either one of two cases are possible: 1) 40 pigs or 2) 40 cows after working the problem, i realized that there's no way this problem is so easy. then i managed to work out that the above simply meant: "40 = p + c" lol! too much verbal review, i guess, and neglecting quant.
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Re: Pigs or Cows? [#permalink]
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Given:
X = total # of animals = 60 (2/3)x = P + C = 40
Question: C = ?
(1): C > 2P
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.
Originally posted by LJ on 18 Sep 2010, 14:50.
Last edited by LJ on 18 Sep 2010, 15:25, edited 1 time in total.



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Re: cow and pig problem [#permalink]
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40 = P + C C > 2P C > 2(40  C) => 3C > 80 => C > 26 Not Sufficient 2) P > 12 Not sufficient (1) + (2) C> 26 and P > 12 So C = 27, P = 13 Answer  C
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Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
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08 Feb 2013, 07:01
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...
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Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
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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)
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Re: cow and pig problem [#permalink]
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09 Nov 2013, 11:31
subhashghosh wrote: 40 = P + C
C > 2P
C > 2(40  C)
=> 3C > 80
=> C > 26
Not Sufficient
2)
P > 12 Not sufficient (1) + (2) C> 26 and P > 12 So C = 27, P = 13 Answer  C Hi Guys, I thought this to be a very interesting approach  even better than picking numbers. But something is quite confusing when one tries to develop statement (1) and (2) together to find the solution. There isn't an algebraically proof? Is it pick numbers the only method? Here is my puzzle:  Statement (1): Not sufficient C > 26  Statement (2): Not sufficient P > 12 Or, substituting variables: 40  C >12 C < 38  Statement (1) and (2) together: 26 < C < 38 > ??? Why is this not the right approach?



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Re: cow and pig problem [#permalink]
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11 Nov 2013, 01:33
nechets wrote: subhashghosh wrote: 40 = P + C
C > 2P
C > 2(40  C)
=> 3C > 80
=> C > 26
Not Sufficient
2)
P > 12 Not sufficient (1) + (2) C> 26 and P > 12 So C = 27, P = 13 Answer  C Hi Guys, I thought this to be a very interesting approach  even better than picking numbers. But something is quite confusing when one tries to develop statement (1) and (2) together to find the solution. There isn't an algebraically proof? Is it pick numbers the only method? Here is my puzzle:  Statement (1): Not sufficient C > 26  Statement (2): Not sufficient P > 12 Or, substituting variables: 40  C >12 C < 38 Statement (1) and (2) together: 26 < C < 38 > ??? Why is this not the right approach? From \(40  c > 12\) you get \(c < 28\) not \(c < 38\). Thus when we combine we get \(26<c<28\) > \(c=27\). Hope it helps.
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Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]
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19 Mar 2014, 20:33
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




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