The distance between points $A$ and $B$ is 40 km. A pedestrian left $A$ at 4 o'clock. When he had walked half the distance, he was caught up by a cyclist who had left $A$ at 7:20. An hour after this, the pedestrian met another cyclist who had left $B$ at 8:30. The speeds of the cyclists are the same. Determine the speed of the pedestrian.
The second cyclist was on the road for $11 / 6$ hours less than the first (by the time of the second cyclist's meeting with the pedestrian). If he had set out simultaneously with the first cyclist, he would have been halfway through his journey one hour earlier than the moment of his meeting with the pedestrian. Therefore, it took him $1 / 6$ of an hour to travel the distance from the meeting point with the pedestrian to the halfway point. This same distance the pedestrian covers in one hour. Therefore, the speed of the cyclist is 6 times the speed of the pedestrian.

The pedestrian spent $7 \frac{1}{3}-4=10 / 3$ hours more on 20 km than the first cyclist. Thus, he covered 20 km in $6 / 5.10 / 3=4$ hours.

## Answer

5 km/h.