Difference between revisions of "1956 AHSME Problems/Problem 6"

Revision as of 17:14, 14 March 2023

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 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
-{{AHSME box|year=1956|num-b=5|num-a=7}}
+{{AHSME 50p box|year=1956|num-b=5|num-a=7}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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Difference between revisions of "1956 AHSME Problems/Problem 6"

Revision as of 17:14, 14 March 2023

Problem 6

Solution 1

See Also