1986 AHSME Problems/Problem 20

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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 $\boxed{E}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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