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: \[4x+2y=14+2(x+y)\] \[4x+2y=14+2x+2y\] \[2x=14\] \[x=\boxed{\textbf{(B) }7}\]

See Also

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