$1 \cdot 127$ If the length of a rectangle increases by $p \%$, to keep the area of the rectangle unchanged, the width of the rectangle should decrease by
(A) $p \%$.
(B) $(1-p) \%$.
(C) $\frac{p}{1+p} \%$.
(D) $\frac{100 p}{100+p} \%$.
(China Beijing Junior High School Mathematics Competition, 1993)
[Solution] Let the original length and width of the rectangle be $a$ and $b$, respectively, and let the width of the rectangle decrease by $x \%$. According to the problem, we have
$$
[a(1+p \%)][b(1-x \%)]=a b,
$$

then
$$
\begin{aligned}
x \% & =1-\frac{1}{1+p \%}=\frac{p \%}{1+p \%} \\
& =\frac{p}{100+p} \cdot 100 \% \\
& =\frac{100 p}{100+p} \% .
\end{aligned}
$$

Therefore, the answer is $(D)$.