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

Let there be $x$ cows and $y$ chickens. Then, there are $4x+2y$ legs and $x+y$ heads. Writing the equation: $4x+2y=14+2(x+y)$ $4x+2y=14+2x+2y$ $2x=14$ $x=\boxed{\textbf{(C) }7}$

1956 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1956 AHSME Problems/Problem 6

Problem 6

Solution 1

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