$3 \cdot 21$ A ship sails from port $A$ to port $B$ downstream in 6 hours, and from port $B$ to port $A$ upstream in 8 hours. If in still water, the time required to sail from port $A$ to port $B$ is
(A) 7 hours.
(B) $6 \frac{6}{7}$ hours.
(C) $7 \frac{1}{2}$ hours.
(D) $6 \frac{1}{2}$ hours.
(China Guangzhou, Wuhan, Fuzhou and other five cities junior high school mathematics league, 1990)
[Solution] Let the distance from $A$ to $B$ be $S$, the speed of the boat in still water be $x$, and the speed of the current be $y$. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{S}{x+y}=6, \\
\frac{S}{x-y}=8.
\end{array} \text { Then } \left\{\begin{array}{l}
x+y=\frac{S}{6}, \\
x-y=\frac{S}{8}.
\end{array} \text { Solving, we get } x=\frac{7 S}{48}\right.\right. \text { . }
$$

Thus, the time required for the boat to travel from $A$ to $B$ in still water is $S \div \frac{7}{48} S=6 \frac{6}{7}$ (hours). Therefore, the answer is $(B)$.