1956 AHSME Problems/Problem 6

Problem 6

In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 14$


Solution 1

Suppose that there are $o$ cows and $h$ chickens. Then there are $4o + 2h$ legs and $o + h$ heads. Then we have $(4o + 2h) = 14 + 2(o + h)$. This expands to $4o + 2h = 14 + 2o + 2h$. Canceling $2o + 2h$ from both sides, we get $2o = 14$, implying that $o = 7$. Therefore, the answer is $\boxed{\textbf{(C)}}$, and we are done.

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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