Mr. Rychlý and Mr. Louda set out on the same hiking tour at the same time, but Mr. Rychlý started from the mountain hut and Mr. Louda from the bus at the bottom of the town, heading up to the hut. They met on the trail at 10 o'clock. Mr. Rychlý was in a hurry and reached his destination by 12 o'clock. On the other hand, Mr. Louda moved slowly and did not arrive at the hut until 18 o'clock. At what time did the gentlemen set out on their journey, knowing that each of them walked at a constant speed the whole time?

(M. Volfová)
Let's denote the speed (in $\mathrm{km} / \mathrm{h}$) of Mr. Rychlý as $v_{R}$ and Mr. Louda as $v_{L}$. The time (in hours) from their departure until their meeting is denoted as $x$. Up to the meeting, Mr. Rychlý walked $x \cdot v_{R}(\mathrm{~km})$ from the cabin, and Mr. Louda walked $x \cdot v_{L}(\mathrm{~km})$ from the bus. After the meeting, Mr. Rychlý walked for another $2 \mathrm{~h}$ to the bus, covering $2 \cdot v_{R}(\mathrm{~km})$, while Mr. Louda walked for another $8 \mathrm{~h}$ to the cabin, covering $8 \cdot v_{L}(\mathrm{~km})$.

By comparing the corresponding distances, we obtain two equations: the distance from the meeting point to the bus is $x \cdot v_{L}=2 \cdot v_{R}(\mathrm{~km})$, and the distance to the cabin is $8 \cdot v_{L}=x \cdot v_{R}(\mathrm{~km})$. From this, we can express

$$
\frac{v_{L}}{v_{R}}=\frac{2}{x}=\frac{x}{8}
$$

thus $x^{2}=16$ and $x=4$. From departure to the meeting, both gentlemen walked for 4 hours, and the total journey time was $10 \mathrm{~h}$. Therefore, they set out at 6 o'clock.