41(1283). The path from $A$ to $B$ goes 3 km uphill, 6 km downhill, and 12 km on flat ground. The motorcyclist covered this path in 1 hour and 7 minutes, and the return path in 1 hour and 16 minutes. Find the motorcyclist's speed uphill and downhill, if his speed on flat ground was 18 km/h. (Note that the motorcyclist traveled at constant speeds both uphill and downhill, both on the path $A B$ and on the return path.)
Solution. I method. On a flat surface in one direction, the motorcyclist traveled $\frac{2}{3}$ hours $\left(12: 18=\frac{2}{3}\right)$, or 40 minutes. Then, 3 km uphill and 6 km downhill, the motorcyclist traveled 27 minutes, and 6 km uphill and 3 km downhill, the motorcyclist traveled 36 minutes. If we denote the motorcyclist's speed uphill and downhill as $v_{1}$ (km/min) and $v_{2}$ (km/min) respectively, we will have the system of equations:

$$
\left\{\begin{array}{l}
\frac{3}{v_{1}}+\frac{6}{v_{2}}=27 \\
\frac{3}{v_{2}}+\frac{6}{v_{1}}=36
\end{array}\right.
$$

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from which $v_{1}=\frac{1}{5}$ km/min, or 12 km/h, $v_{2}=\frac{1}{2}$ km/min, or 30 km/h.

II method. The problem can also be solved without setting up a system of equations.

Indeed, if 3 km uphill and 6 km downhill took the motorcyclist 27 minutes, then by doubling the distance, we find that for a distance of 6 km uphill and 12 km downhill, the motorcyclist would need 54 minutes. Since the motorcyclist, according to the problem, spent 36 minutes on 6 km uphill and 3 km downhill, it would take 18 minutes $(54-36=18)$ for 9 km $(12-3=9)$ downhill. Therefore, the motorcyclist's speed downhill is $\frac{9}{18}=\frac{1}{2}$ km/min, or 30 km/h.

Similarly, we find that for a distance of 12 km uphill and 6 km downhill, the motorcyclist would need 72 minutes $(36 \cdot 2=72)$. Then, for a distance of 9 km uphill, it would take 45 minutes $(72-27=45)$. Therefore, the motorcyclist's speed uphill is $\frac{9}{45}=$ $=\frac{1}{5}$ km/min, or 12 km/h.

Answer: 12 km/h, 30 km/h.