20. Let two positive numbers $x$ and $y$ be 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) $\frac{100 p}{100+p} \%$
26. When $x$ and $y$ are inversely proportional, it means that if $x$ is multiplied by $k$, then $y$ should be divided by $k$. Let $x^{\prime}$ and $y^{\prime}$ be the new values of $x$ and $y$ after $x$ increases by $p \%$, then
$$
\begin{aligned}
x^{\prime} & =\left(1+\frac{p}{100}\right) x, \\
y^{\prime} & =\frac{y}{1+\frac{p}{100}}=\frac{100 y}{100+p} .
\end{aligned}
$$

By definition, the percentage decrease in $y$ is
$$
\begin{array}{l}
1\left(10\left(-\frac{y-y^{\prime}}{y}\right)=100\left(1-\frac{100}{100+p}\right)\right. \\
=\frac{100}{100+p} .
\end{array}
$$