13.131. The beacon keeper, inspecting his river section in an ordinary rowboat, traveled 12.5 km upstream and then returned to the starting point along the same route. In this voyage, he overcame every 3 km against the current and every 5 km with the current in the same average time intervals, and was on the way for exactly 8 hours in total. Find the speed of the current and the time of the beacon keeper's voyage upstream.
## Solution.

Let $v$ km/h be the own speed of the boat, $v_{\mathrm{T}}$ km/h be the speed of the current. According to the condition, $\left\{\begin{array}{l}\frac{3}{v-v_{T}}=\frac{5}{v+v_{\mathrm{T}}} \\ \frac{12.5}{v-v_{T}}+\frac{12.5}{v+v_{T}}=8,\end{array}\right.$ from which $v_{\mathrm{T}}=\frac{5}{6}$ km/h, $v=\frac{20}{6}$ km/h. The time of the buoy tender's voyage upstream is $-\frac{12.5}{\frac{20}{6}-\frac{5}{6}}=5$ hours. Answer: $\frac{5}{6}$ km/h; 5 hours.