$1 \cdot 124$ Suppose two positive numbers $x$ and $y$ are inversely proportional. If $x$ increases by $p \%$, then $y$ decreases by
(A) $p \%$.
(B) $\frac{p}{1+p} \%$.
(C) $\frac{100}{p} \%$.
(D) $\frac{p}{100+p} \%$.
(E) $\frac{100 p}{100+p} \%$.
(37th American High School Mathematics Examination, 1986)
[Solution] Let $y$ decrease by $q \%$. According to the problem, we have
$$
(1+p \%) x \cdot(1-q \%) y=1 \text{. }
$$

Given $x y=1$, it follows that $1-q \%= \frac{1}{1+p \%}$, thus $q \%=1-\frac{1}{1+p \%}=\frac{p}{100+p}=\frac{100 p}{100+p} \%$. Therefore, the answer is $(E)$.