13.267. Two motorcyclists started racing in the same direction from the same starting point simultaneously: one at a speed of 80 km/h, the other at a speed of 60 km/h. Half an hour later, a third racer started from the same starting point in the same direction. Find his speed, given that he caught up with the first racer 1 hour and 15 minutes later than the second.
Solution.

Let $x$ km/h be the speed of the third racer. He caught up with the first racer after $t_{1}$ hours, and the second racer after $t_{2}$ hours. According to the problem,

$$
\begin{aligned}
& \left\{\begin{array}{l}
80 t_{1}=x\left(t_{1}-0.5\right), \\
60 t_{2}=x\left(t_{2}-0.5\right) \\
t_{1}-t_{2}=1.25,
\end{array}\right. \\
& \text { from which } x=100 \text { km/h. }
\end{aligned}
$$

Answer: 100 km/h.