1956 AHSME Problems/Problem 13

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

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


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.

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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