Suppose $x$ and $y$ are inversely proportional and positive. If $x$ increases by $p\%$, then $y$ decreases by 
$\textbf{(A)}\ p\%\qquad \textbf{(B)}\ \frac{p}{1+p}\%\qquad \textbf{(C)}\ \frac{100}{p}\%\qquad \textbf{(D)}\ \frac{p}{100+p}\%\qquad \textbf{(E)}\ \frac{100p}{100+p}\%$
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 $\boxed{E}$.