$3 \cdot 22$ A person rows 15 miles, if he rows at his usual speed, then the time taken downstream is 5 hours less than that taken upstream; if he rows at twice his usual speed, then the time taken downstream is only 1 hour less than that taken upstream. In miles per hour, the speed of the current is
(A) 2 .
(B) $\frac{5}{2}$.
(C) 3 .
(D) $\frac{7}{2}$.
(E) 4 .
(17th American High School Mathematics Examination, 1966)
[Solution] Let $c$ be the speed of the current, and $m$ be the speed of the boat in still water. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{15}{m+c}=\frac{15}{m-c}-5, \\
\frac{15}{2 m+c}=\frac{15}{2 m-c}-1 .
\end{array}\right.
$$

Simplifying, we get
$$
\left\{\begin{array}{l}
5 m^{2}-5 c^{2}=30 c, \\
4 m^{2}-c^{2}=30 c .
\end{array}\right.
$$
Subtracting (4) from (3) gives $m^{2}-4 c^{2}=0$, which yields $m=2 c$ (the negative value is discarded). Substituting $m=2c$ into (4) and simplifying, we get $15 c-30=0$, which gives $c=2$. Therefore, the answer is $(A)$.