$2 \cdot 76$ A river has a constant current of 3 miles per hour. A motorboat travels 4 miles downstream and then returns to the starting point at a constant speed, taking a total of 1 hour (ignoring the time to turn the boat around). The ratio of the speed downstream to the speed upstream is
(A) $4: 3$.
(B) $3: 2$.
(C) $5: 3$.
(D) $2: 1$.
(E) $5: 2$.
(22nd American High School Mathematics Examination, 1971)
$\cdot$[Solution] Let $v$ be the speed of the steamboat in still water (in miles per hour). According to the problem, we have
$$
1=\frac{4}{v+3}+\frac{4}{v-3},
$$

which simplifies to
$$
v^{2}-9=4(v-3)+4(v+3),
$$

or
$$
\begin{array}{l}
v^{2}-8 v-9=0, \\
(v-9)(v+1)=0 .
\end{array}
$$

Solving this, we get
$v=9$. (The negative value has been discarded)
$\therefore \quad$ The downstream speed $=\frac{v+3}{v-3}=\frac{12}{6}=\frac{2}{1}$. Thus, the ratio is $2: 1$. Therefore, the answer is $(D)$.