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

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16 Sep 2010, 16:24
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]

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16 Sep 2010, 17:16
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: Word problem from GMATPrep (2) [#permalink]

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16 Mar 2011, 01:20
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]

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16 Mar 2011, 04:49
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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03 May 2011, 09:55
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.
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Re: cow and pig problem [#permalink]

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04 May 2011, 03:38
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

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Re: 60 animals in a farm [#permalink]

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07 May 2011, 15:13
Read the word 'more' in the 1st statement.

so there are total 40 animals that are either cow or pigs.
st - 1==> you can have 8 pigs and 32 cows OR 10 pigs and 30 cows OR 12 pigs and 28 cows OR 13 pigs and 27 cows ... so not sufficient.
st - 2 not sufficient ==> if pigs are more than 12 it could be that pigs are 13 or 14 or whatsoever ... and remaining cows.

if you consider together... basically you noticed that in st - 1, you need something to restrict # of pigs which is provided by st - 2.
so considering both, you can just have 1 case - pigs 13 and cows 27.
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Re: cow and pig problem [#permalink]

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08 May 2011, 00:49
p+c = 40

a c> 2p means p = 1 ; c = 39 or p = 4 ; c = 36. Not sufficient.

b p>12 means p = 13 gives c = 27 or p = 14 gives c = 26. Not sufficient.

a+b only p = 13 and c = 27 fits for p+c = 40

hence C
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Re: cow and pig problem [#permalink]

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08 May 2011, 13:25
p +c =40

c?

1. Not sufficient

c>2p
we dont know much about p.

2. Not sufficient

p>12
we could be any of the following

p c
13 27
14 26
15 25

Together , Its sufficient

as p = 13 and c=27 is the only combination that satisfies the given condition.

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08 Jan 2012, 02:53
2/3 of the animals are cow and pigs = 40 animals = Pigs + Cows

according to 1) C > 2 P insuff
2) P > 12 insuff

from 1) & 2)
C > 24
i.e the possible combinations
C P
25 15
26 14
27 13 ==> this one satisfys both 1 & 2

Hence C
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09 Jan 2012, 12:12
Rephrase: C + P = 40

1. C > 2P . Insuff
2. P > 12. insuff

Together, the only (C, P) combination that meets the sum is (13, 27). 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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15 Jul 2012, 08:43
The question tells us that:
$$p + c = 40$$

Restrictions:
$$p$$ and $$c$$ must be non-negative integers. (We can't have -2 pigs or 1.5 cows.)

(1)
$$c > 2p$$

Let's say that $$c = 2p$$. Then:
$$p + c = 40$$
$$p + (2p) = 40$$
$$p = 13.33$$

If $$p$$ were an integer greater than 13.33, then $$p + c$$ would be greater than 40 (for example, if $$p = 14$$, then $$p + c > (14) + 2(14)$$, $$p + c > 42$$). But $$p + c$$ cannot be greater than 40, since the question says that $$p + c = 40$$. Therefore:
$$p < =13.33$$

Insufficient because $$p$$ could be one of a number of non-negative integers less than 13.33 (for example, 13 or 2), and thus $$c$$ could be one of a number of non-negative integers ($$c$$ = 40 minus whatever $$p$$ is).

(2)
$$p > 12$$

Insufficient because $$p$$ could be one of a number of integers (for example, 13 or 39), and thus $$c$$ could be one of a number of integers ($$c$$ = 40 minus whatever $$p$$ is).

(1) and (2) together are sufficient because:
(1) $$p <= 13.33$$
(2) $$p > 12$$
Therefore:
$$12 < p <= 13.33$$
$$p$$ must equal 13 since 13 is the only integer that is greater than 12 and less than or equal to 13.33.
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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...

Regards
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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

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

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.
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Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]

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20 Mar 2014, 03:00
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

Hope this helps.
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Re: Word problem from GMATPrep (2) [#permalink]

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29 Apr 2014, 03:03
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.

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.

Hi bunnel,

I need clarification here

2/3 are either pigs or cows

i am understanding (2/3)*60 = 40 (cows or pigs) how it can be c+p = 40 question is saying either cow or pig so this can be c or p but not C+P

Thanks.
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Re: Of the 60 animals on a certain farm, 2/3 are either cows or [#permalink]

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29 Apr 2014, 03:43
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

From question, it's given that

p+c = (2*60)/3 = 40

Statement 1

p>2c

c could 10, 8...
not sufficient

Statement 2

p>12
not sufficient

Combining both Statements
p>2c
and
p>12

We need to have at least 26 cows
And it's not possible.

Hence C.
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Re: Word problem from GMATPrep (2) [#permalink]

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29 Apr 2014, 03:52
PathFinder007 wrote:
Hi bunnel,

I need clarification here

2/3 are either pigs or cows

i am understanding (2/3)*60 = 40 (cows or pigs) how it can be c+p = 40 question is saying either cow or pig so this can be c or p but not C+P

Thanks.

40 animals are either pigs or cows, so out of 40 animals some are pigs and the rest are cows: p+c=40. How else?
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Re: Word problem from GMATPrep (2)   [#permalink] 29 Apr 2014, 03:52

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