1956 AHSME Problems/Problem 13

Given two positive integers $x$ and $y$ with $x < y$ . The percent that $x$ is less than $y$ is:

$\textbf{(A)}\ \frac{100(y-x)}{x}\qquad \textbf{(B)}\ \frac{100(x-y)}{x}\qquad \textbf{(C)}\ \frac{100(y-x)}{y}\qquad \\ \textbf{(D)}\ 100(y-x)\qquad \textbf{(E)}\ 100(x - y)$

Solution

Suppose that $x$ is $p$ percent less than $y$ . Then $x = \frac{100 - p}{100}y$ , so that $y - x = \frac{p}{100}y$ . Solving for $p$ , we get $p = \frac{100(y-x)}{y}$ , so our answer is $\boxed{\textbf{(C)}}$ and we are done.

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1956 AHSME Problems/Problem 13

Solution

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