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.

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1956 AHSME Problems/Problem 6

Problem 6

Solution 1

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