[ [motion problems ]

A cyclist left point $A$ for point $B$. At the same time, a pedestrian left point $B$ for point $A$ to meet the cyclist. After they met, the cyclist turned back, while the pedestrian continued his journey. It is known that the cyclist returned to point $A$ 30 minutes earlier than the pedestrian, and his speed was 5 times the speed of the pedestrian. How much time did the pedestrian spend on the journey from $A$ to $B$?
Before the meeting, the pedestrian had walked a distance five times less than the cyclist had ridden, that is, $1 / 6$ of the entire path. By the time the cyclist returned to point $A$, the pedestrian had walked the same distance again, meaning he still had $2 / 3$ of the path left. For this, he spent 30 minutes. Consequently, for the entire distance, he would need one and a half times more time.

## Answer

45 minutes.