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A farm has chickens, cows and sheep. There are three times

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A farm has chickens, cows and sheep. There are three times  [#permalink]

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New post 12 Jan 2012, 08:00
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A farm has chickens, cows and sheep. There are three times the number of chickens and cows than sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

A. 5
B. 8
C. 10
D. 14
E. 17

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Re: a farm has chickens, cows, sheeps  [#permalink]

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New post 19 Jan 2012, 05:02
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1
manalq8 wrote:
A farm has chickens, cows and sheep. There are three times the number of chickens and cows than sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

5
8
10
14
17

what is the quickest way for this? help please


Divisibility (properties of a multiple) approach can also give an answer quite easily:

"There are three times the number of chickens and cows than sheep" --> {cows}+{chickens}=3*{sheep} --> the {cows}+{chickens} is a multiple of 3.

"Together, cows and chickens have a total of 100 feet and heads" --> 5*{cows}+3*{chickens}=100 --> now, as both 5*{cows} and 100 are multiples of 5 then {chickens} must also be a multiple of 5:
{chickens}=5 --> {cows}=17 --> {cows}+{chickens}=22 which is NOT a multiple of 3;
{chickens}=10 --> {cows}=14 --> {cows}+{chickens}=24 which is a multiple of 3;
{chickens}=15 --> {cows}=11 --> not a valid solution as given that {chickens}<{cows}.

{cows}+{chickens}=10+24=32=3*{sheep} --> {sheep}=8.

Answer: B.
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Re: a farm has chickens, cows, sheeps  [#permalink]

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New post 19 Jan 2012, 00:57
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2
My answer is 8

Chicken - ch
Cows - C
Sheep -S

ch+C=3S
C> ch and c>s
Each cow has 4 legs and 1 head
Each chicken has 2 legs and 1 head
So 5c+3ch=100 (sum of legs and head)
There are 2 possible solutions to this equation
c=11 and ch=9 or
c=14 and ch=10

Since from first equation where ch+c=3s

The sum of ch and c should be divisbile by 3. 20 is not so the only possible solution is c=14 and ch=10. So s=8
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Re: a farm has chickens, cows, sheeps  [#permalink]

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New post 19 Jan 2012, 02:16
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manalq8 wrote:
A farm has chickens, cows and sheep. There are three times the number of chickens and cows than sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

5
8
10
14
17

what is the quickest way for this? help please


Using options, in my opinion!

When I read the question, I note the following:

chickens + cows = 3 * sheep

Chickens have 2 legs and one head. Cows have 4 legs and one head.
Total legs + heads = 100
If chicken = cows, average number of legs would be 3 for every head. So out of 100, we would have 75 legs and 25 heads (i.e. 25 = chickens + cows)
But we know cows > chickens. So average number of legs must be more than 3. Then, we must have fewer than 25 animals (chickens + cows).

If number of sheep = 10/14/17, three times would be 30/42/51 but this cannot be the number of (chickens + cows).
So only possible options are 5 and 8.
If number of sheep = 5, number of (chickens + cows) = 15. Even if they are all cows, they would have only 60 legs and total legs + heads = 15+60 = 75. We cannot make 100 in this case.

So answer has to be 8.
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Re: A farm has chickens, cows and sheep. There are three times  [#permalink]

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New post 22 May 2013, 04:35
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manalq8 wrote:
A farm has chickens, cows and sheep. There are three times the number of chickens and cows than sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

A. 5
B. 8
C. 10
D. 14
E. 17


Just another approach :

Cows+Chicken = 3*Sheep.

Also, Cows>Chicken and Cows>Sheep.

Also, Cows+Chicken+4*Cows+2*Chicken = 100

or 5Cows+3Chicken = 100 --> 2Cows + 3*(Cows+Chicken) = 100 --> 2Cows + 9*Sheep = 100. Replace the options for Sheep,

A= odd , thus discard
C = # of Cows<# of Sheep, thus discard
D and E = # of Cows coming as negative,discard

B.
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Re: A farm has chickens, cows and sheep. There are three times  [#permalink]

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New post 24 Feb 2016, 19:47
I was confused because I thought that there were 100 feet and (100) heads. Is the real GMAT this ambiguous in the wording?
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Re: A farm has chickens, cows and sheep. There are three times  [#permalink]

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New post 24 Feb 2016, 22:55
DJ1986 wrote:
I was confused because I thought that there were 100 feet and (100) heads. Is the real GMAT this ambiguous in the wording?


In my opinion, the wording is not ambiguous. It says there are a total of 100 feet and heads.
If there were 100 feet and 100 heads, it would have been written as such. Also, then the animals would not have been cows and chickens since per head there need to be at least 2 feet. We certainly cannot have equal number of heads and feet until and unless we are talking about molluscs.
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Re: A farm has chickens, cows and sheep. There are three times  [#permalink]

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New post 25 Feb 2016, 12:48
VeritasPrepKarishma wrote:
DJ1986 wrote:
I was confused because I thought that there were 100 feet and (100) heads. Is the real GMAT this ambiguous in the wording?


In my opinion, the wording is not ambiguous. It says there are a total of 100 feet and heads.
If there were 100 feet and 100 heads, it would have been written as such. Also, then the animals would not have been cows and chickens since per head there need to be at least 2 feet. We certainly cannot have equal number of heads and feet until and unless we are talking about molluscs.


Now that you mention it, what I was thinking really doesn't make sense. I was imagining fried chicken and someone ate all the legs!
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Re: A farm has chickens, cows and sheep. There are three times  [#permalink]

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New post 25 Feb 2016, 21:59
DJ1986 wrote:
VeritasPrepKarishma wrote:
DJ1986 wrote:
I was confused because I thought that there were 100 feet and (100) heads. Is the real GMAT this ambiguous in the wording?


In my opinion, the wording is not ambiguous. It says there are a total of 100 feet and heads.
If there were 100 feet and 100 heads, it would have been written as such. Also, then the animals would not have been cows and chickens since per head there need to be at least 2 feet. We certainly cannot have equal number of heads and feet until and unless we are talking about molluscs.


Now that you mention it, what I was thinking really doesn't make sense. I was imagining fried chicken and someone ate all the legs!


:) Don't study on an empty stomach!
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Re: A farm has chickens, cows and sheep. There are three times  [#permalink]

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New post 25 Feb 2016, 23:31
A farm has chickens, cows and sheep. There are three times the number of chickens and cows than sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

A. 5
B. 8
C. 10
D. 14
E. 17


Solving this question with a common sense approach,

considering 100 feet and the fact cows>Chicken( 20 cows and 10 chickens which exactly adds upto 100 feet and 30 heads)

Pluging in the answers, when sheep's 5.. the cow+chicken= 15(doesnt make sense)

Sheep's 8 Cow+chicken=24(makes sense because the numbers possible)

Sheep's 10 cow +chicken=30( Values much higher than derived) and hence no need to check for other answers
and the answer is B)8.
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Re: A farm has chickens, cows and sheep. There are three times  [#permalink]

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New post 15 Feb 2018, 13:48
2
chickens - a, cows - b , sheep - c

a + b = 3c
3a + 3b = 9c ----(1)

3a + 5b = 100 (heads and feet)
9c - 3b + 5b = 100
b = 50 - (9c/2) => to make b integer, c must be even

given cows > sheeps => b > c => 50 - (9c/2) > c
= (11c/2) < 50
= c < (100/11) = c <= 9, even number less than 9 in the options is 8 (B)
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Re: A farm has chickens, cows and sheep. There are three times  [#permalink]

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Re: A farm has chickens, cows and sheep. There are three times   [#permalink] 20 Oct 2019, 10:14
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