1986 AHSME Problems/Problem 20

Problem

Suppose $x$ and $y$ are inversely proportional and positive. If $x$ increases by $p\%$ , then $y$ decreases by

$\textbf{(A)}\ p\%\qquad \textbf{(B)}\ \frac{p}{1+p}\%\qquad \textbf{(C)}\ \frac{100}{p}\%\qquad \textbf{(D)}\ \frac{p}{100+p}\%\qquad \textbf{(E)}\ \frac{100p}{100+p}\%$

Solution

We see that $x$ is multiplied by $\frac{100+p}{100}$ when it is increased by $p$ %. Therefore, $y$ is multiplied by $\frac{100}{100+p}$ , and so it is decreased by $\frac{p}{100+p}$ times itself. Therefore, it is decreased by $\frac{100p}{100+p}$ %, and so the answer is $\boxed{E}$ .

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1986 AHSME Problems/Problem 20

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