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

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Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink] New post 11 Oct 2009, 23:59
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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] New post 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.

ANSWER = C.
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Re: Word problem from GMATPrep (2) [#permalink] New post 12 Oct 2009, 06:08
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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] New post 12 Oct 2009, 11:28
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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.
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Re: Word problem from GMATPrep (2) [#permalink] New post 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.
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Re: Word problem from GMATPrep (2) [#permalink] New post 12 Oct 2009, 17:50
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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.
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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] New post 12 Oct 2009, 18:02
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 write-up was pointing towards anything else. I do appreciate all the club members help.

:oops:
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Re: Word problem from GMATPrep (2) [#permalink] New post 12 Oct 2009, 21:24
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 write-up was pointing towards anything else. I do appreciate all the club members help.

:oops:


No offence taking. We're all here to help each other! :)
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Re: Word problem from GMATPrep (2) [#permalink] New post 13 Oct 2009, 04:45
I do appreciate your understanding! :-D
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Re: cow and pig problem [#permalink] New post 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.
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Re: DS-Pigs and Cows [#permalink] New post 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.
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Re: DS-Pigs and Cows [#permalink] New post 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.

Together 13<P>12 so P=13, C=27
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Re: cow and pig problem [#permalink] New post 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.

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

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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 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.
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Re: Pigs and cows [#permalink] New post 16 Sep 2010, 15:24
Geronimo 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 it has pigs.
(2) The farm has more than 12 pigs

Why isn't statement (1) sufficient (i.e. cows = 40) ?

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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Re: cow and pig problem [#permalink] New post 16 Sep 2010, 16:16
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] New post 18 Sep 2010, 13:50
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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.

Last edited by LJ on 18 Sep 2010, 14:25, edited 1 time in total.
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Re: Word problem from GMATPrep (2) [#permalink] New post 16 Mar 2011, 00:20
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 animals to be BOTH pigs and cows. 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.


I guess that ngoctraiden thought 2/3 of 60 = 40 animals are all pigs or 40animals are all cows. btw, i've learnt an intersting difference in languages between English and some other languages. This stem should be translated as people here already said: 40 animals include pigs and cows: p+c = 40
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Re: cow and pig problem [#permalink] New post 16 Mar 2011, 03:49
So 40 are either P or C

C = 2P + k, where k is an integer

So P = 5, then C = 35 which is 10 + 25

P = 4, then C = 36 which is 8 + 28

So (1) is insufficient

From (2) we have p > 13, which is not sufficient as p can be 14, 15 etc. and C can be 26, 25, so (2) is insufficient

From (1) and (2), if P = 13, then C = 27 which is = 2*13 + 1, hence the answer is (C).
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Re: Pig and Cows [#permalink] New post 03 May 2011, 08:55
udaymathapati wrote:
Attachment:
M-Q30.JPG


Farm has 60 animals.
2/3 are either cows or pigs i.e. = 2/3*60=40

St 1--> Farm has more than twice as many cows as it has pigs.
i.e. C > 2p

if p = 1, C is 39
if p = 10, C is 30
You need a number to restrict P; not sufficient

St 2--> The farm has more than 12 pigs
P could be any number and therefore C could be any.

Combining both statements, P is restricted i.e. if you take P as 13, C is 27 (and satisfied C>2P)

if you take P as 14, C is 26 (but doesnt satisfy C > 2p)

i.e. you have P = 13 and C=27. Answer is C.
Re: Pig and Cows   [#permalink] 03 May 2011, 08:55
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