A train departs from the station according to the schedule. After traveling 8 kilometers, the driver looks at his watch and sees that the hour and minute hands are exactly overlapping. In the first 8 kilometers, the train traveled at an average speed of 33 kilometers per hour. When did the train depart from the station?
Solution. The hour and minute hands first overlap at 12 o'clock. (It is sufficient to examine a 12-hour period.) Let the time elapsed until the next meeting be $t$ (minutes). The minute hand turns $360^{\circ}$ in 60 minutes, or 6 degrees per minute. The hour hand turns $30^{\circ}$ in 60 minutes, or 0.5 degrees per minute.

The two hands will overlap again when the difference in their angles is a multiple of $360^{\circ}$, that is,

$$
6 t-0.5 t=5.5 t=k \cdot 360^{\circ}, \quad k=1,2,3, \ldots
$$

From this:

$$
t=k \cdot \frac{360}{5.5}=k \cdot 65 \frac{5}{11} \quad(\text { minutes })
$$

The train traveled $8 \text{ km}$ at a speed of $33 \text{ km/h}$; the time elapsed since departure is:

$$
t^{\prime}=\frac{480}{33}=14 \frac{6}{11} \quad(\text { minutes })
$$

If the locomotive driver looked at his watch exactly at noon, then the train

$$
12 \text{ o'clock } 00 \text{ minutes }-14 \frac{6}{11} \text{ minutes }=11 \text{ o'clock } 45 \frac{5}{11}
$$

minutes departed.

The clock hands will next overlap $65 \frac{5}{11}$ minutes later, in this case, the train would have to depart at

$$
11 \text{ o'clock } 45 \frac{5}{11} \text{ minutes }+65 \frac{5}{11} \text{ minutes }=12 \text{ o'clock } 50 \frac{10}{11}
$$

minutes, etc.

If we calculate the values corresponding to $k=2,3, \ldots$ from formula (1) - and assume that trains only depart at whole minute times - we get two suitable values, and these are: 22:40, and 10:40.