Intern
Joined: 07 Jan 2017
Posts: 19
Location: India
Concentration: Other, Other
WE:Medicine and Health (Pharmaceuticals and Biotech)
M06-12
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11 Feb 2017, 20:01
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[/quote]
another pair of (5,25) is possible and as per this the answer can be C.10