13.436 On the river section from $A$ to $B$, the current is so weak that it can be neglected; on the section from $B$ to $C$, the current affects the boat's movement. The boat covers the distance downstream from $A$ to $C$ in 6 hours, and upstream from $C$ to $\boldsymbol{A}$ in 7 hours. If the current on the section from $\boldsymbol{A}$ to $\boldsymbol{B}$ were the same as on the section from $B$ to $C$, the entire journey from $A$ to $C$ would take 5.5 hours. How long would it take the same boat to move upstream from $C$ to $A$ in this case? The boat's own speed is constant in all cases.
Solution. Let $x$ be the speed of the boat relative to the shore downstream, and $y$ be the speed of the boat relative to the shore upstream. Then $\frac{x+y}{2}$ is the boat's own speed. Let $z$ be the time it takes for the boat to travel the segment from $A$ to $B$ (or from $B$ to $A$), where there is no current. Then the segment from $B$ to $C$ is traveled in $(6-z)$ hours, and the segment from $C$ to $B$ in $(7-z)$ hours. Therefore, $\frac{6-z}{7-z}=\frac{y}{x}$.

If the current appeared on the segment from $\boldsymbol{A}$ to $\boldsymbol{B}$, the boat would travel the segment from $A$ to $B$ in $(z-0.5)$ hours (by the condition). Therefore, $\frac{z-0.5}{z}=\frac{\frac{x+y}{2}}{x}$.

$$
\text { Thus, }\left\{\begin{array}{l}
\frac{6-z}{7-z}=\frac{y}{x} \\
\frac{2 z-1}{z}=\frac{x+y}{2}
\end{array} \Rightarrow \frac{6-z}{7-z}=\frac{z-1}{z} \Rightarrow z=\frac{7}{2} \Rightarrow \frac{y}{x}=\frac{5}{7}\right.
$$

Therefore, the segment from $C$ to $A$ would be traveled by the boat $7 / 5$ times slower than the segment from $A$ to $C$, i.e., in $5.5 \cdot \frac{7}{5}=7.7$ hours.

Answer: 7.7 hours.