$2 \cdot 2$ An item's price increases by $p \%$ and then the new price decreases by $p \%$. If the final price is one dollar, then the original price was
(A) $\frac{1-p^{2}}{200}$.
(B) $\frac{\sqrt{1-p^{2}}}{100}$.
(C) one dollar.
(D) $1-\frac{p^{2}}{10000-p^{2}}$.
(E) $\frac{10000}{10000-p^{2}}$.
(3rd American High School Mathematics Examination, 1952)
［Solution］Let the original price of the item be $x$ dollars. According to the problem, we have
$$
x(1+p \%)(1-p \%)=1 \text {, }
$$

which gives $x=\frac{10000}{10000-p^{2}}$.
Therefore, the correct choice is $(E)$.