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

Revision as of 16:22, 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

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

1956 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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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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 ==Solution 1==
-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.
+Let there be <math>x</math> cows and <math>y</math> chickens. Then, there are <math>4x+2y</math> legs and <math>x+y</math> heads. Solving: <cmath>(4x+2y)=14+2(x+y)</cmath> <cmath>4x+2y=14+2x+2y</cmath> <cmath>2x=14</cmath> <cmath>x=\boxed{\textbf{(C) }7}</cmath>
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Difference between revisions of "1956 AHSME Problems/Problem 6"

Revision as of 16:22, 14 March 2023

Problem 6

Solution 1

See Also