$3 \cdot 46$ When one ounce of water is added to an acid and water mixture, the new mixture is $20 \%$ acid. When one ounce of acid is added to the new mixture, the result is $33 \frac{1}{3} \%$ acid. The percent of acid in the original mixture is
(A) $22 \%$.
(B) $24 \%$.
(C) $25 \%$
(D) $30 \%$
(E) $33 \frac{1}{3} \%$.
(24th American High School Mathematics Examination, 1973)
［Solution］Let $(x, y)$ represent the ounces of (water, acid) in the original mixture.
After adding $\mathrm{loz}$ (ounces) of water and $1 \mathrm{oz}$ of acid, in the total mixture volumes $(x+y+1)$ and $(x+y+2)_{\mathrm{Oz}}$, we have acid $y \mathrm{Oz}$ and $(y+1)_{\mathrm{Oz}}$, respectively, such that
$$
\frac{y}{x+y+1}=\frac{1}{5}, \frac{y+1}{x+y+2}=\frac{1}{3} .
$$

Thus, we get $(x, y)=(3,1)$. From this, the percentage of acid in the original mixture is
$$
\frac{y}{x+y}=\frac{1}{3+1}=\frac{1}{4}=25 \% .
$$

Therefore, the answer is $(C)$.