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

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Suppose that there are <math>o</math> cows and <math>h</math> chickens. Then there are <math>4o + 2h</math> legs and <math>o + h</math> heads. Then we have <math>(4o + 2h) = 14 + 2(o + h)</math>. This expands to <math>4o + 2h = 14 + 2o + 2h</math>. Canceling <math>2o + 2h</math> from both sides, we get <math>2o = 14</math>, implying that <math>o = 7</math>. Therefore, the answer is <math>\boxed{\textbf{(C)}}</math>, and we are done.
 
Suppose that there are <math>o</math> cows and <math>h</math> chickens. Then there are <math>4o + 2h</math> legs and <math>o + h</math> heads. Then we have <math>(4o + 2h) = 14 + 2(o + h)</math>. This expands to <math>4o + 2h = 14 + 2o + 2h</math>. Canceling <math>2o + 2h</math> from both sides, we get <math>2o = 14</math>, implying that <math>o = 7</math>. Therefore, the answer is <math>\boxed{\textbf{(C)}}</math>, and we are done.
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==See Also==
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{{AHSME box|year=1956|num-b=5|num-a=7}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 20:21, 12 February 2021

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 (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
All AHSME Problems and Solutions

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