13.257. A courier on a moped left city $A$ for city $B$, which are 120 km apart. One hour after this, a second courier on a motorcycle left $A$ and, catching up with the first, immediately handed him an errand and then turned back, returning to $A$ at the moment the first courier reached $B$. What is the speed of the first courier if the speed of the second is 50 km/h?
Solution.

Let $v$ km/h be the speed of the first courier; he traveled for $t$ hours to the meeting point. According to the problem, $v t=50(t-1)$. The first courier arrived in $B$ after $\frac{120}{v}$ hours, and the second courier returned to $A$ after $\left(\frac{2 v t}{50}+1\right)$ hours.

$$
\text { According to the problem } \frac{2 v t+1}{50}+1=\frac{120}{v} \text {. }
$$

Solving the system $\left\{\begin{array}{l}v t=50(t-1), \\ \frac{2 v t}{50}+1=\frac{120}{v}\end{array}\right.$,

we find $v=30$ km/h.

Answer: 30 km/h.