Solution. Let the speed of the pedestrian be $v_{1}$ meters per minute, the speed of the cyclist be $v_{2}$ meters per minute, and $t$ minutes be the time the pedestrian spent from the moment of the second meeting with the cyclist until arriving at point $B$.

According to the problem, we form a system of equations:

$$
\left\{\begin{array}{l}
\frac{20 v_{1}+10 v_{1}+t v_{1}+10 v_{1}+t v_{1}}{v_{2}}=20 \\
\frac{20 v_{1} \cdot 2+10 v_{1}}{v_{2}}=10
\end{array}\right.
$$

156

$$
\left\{\begin{array}{l}
20 v_{2}=40 v_{1}+2 t v_{1} \\
10 v_{2}=50 v_{1}
\end{array}\right.
$$

from which $t=30$.

Therefore, the required time (the time it takes for the pedestrian to travel from $A$ to $B$) is 1 hour $(20+10+30=60$ (min)).

Answer: 1 hour.
145(1173). A pedestrian and a cyclist set off from point $A$ to point $B$ simultaneously. At point $B$, the cyclist turns back and meets the pedestrian 20 minutes after the start of the journey. Without stopping, the cyclist continues to point $A$, turns back, and catches up with the pedestrian 10 minutes after the first meeting. How long will it take the pedestrian to walk from $A$ to $B$?