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M06-12

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Intern
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Joined: 27 Mar 2017
Posts: 5

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GMAT 1: 720 Q50 V38
Re: M06-12 [#permalink]

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New post 18 May 2017, 02:01
It is always so time taking to arrive at the integer values of the equations like this by picking numbers! Is there any material on how we can quicken this step?
Thank you!

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Re: M06-12 [#permalink]

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New post 30 Oct 2017, 06:14
(5,25) clearly violates the condition that the total number of heads and feet between the cows and chickens equals 100, so C cannot be a possible answer. (5,25) = 5(3) + 25 (5)
= 15 + 125 = 140.

Sachin07 wrote:
A farm has chickens, cows and sheep. The number of chickens and cows combined is 3 times the number of 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


Let \(C\) be a number of chickens, \(W\) - cows, and \(S\) - sheep.

Chickens and cows together have 100 feet and heads: \(100=1C+1W+4W+2C\), assuming that 1 is the number of heads per either unit and 4 and 2 are the number of legs per \(C\) and \(W\) accordingly. Therefore: \(100 = 3C + 5W\).

It is given that \(W \gt C\), and we also see that \(C\) should be a multiple of 5 for \(3C + 5W\) to add up to 100. Picking a few numbers, we get two pairs: (5, 17) and (10, 14).

\(S = \frac{(C+W)}{3}\); \(5 + 17 = 22\) which is not divisible by 3, but \(10 + 14 = 24\) which is divisible by 3, therefore only the second pair will work. So, \(S = 8\).


Answer: B



another pair of (5,25) is possible and as per this the answer can be C.10
[/quote]

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Re: M06-12 [#permalink]

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New post 17 Nov 2017, 17:58
First post here, so sorry if it is put in the wrong place.

This is a solution to - M06-12

Chicken + Cows = 3 times Sheep
=> Chicken + Cows + Sheep = 3 Sheep + Sheep = 4 Sheep

Therefore, the answer need to be a multiple of 4. Only one choice i.e. 8.

Thanks!

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Re: M06-12   [#permalink] 17 Nov 2017, 17:58

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