13.399 The train was delayed at the departure station for 1 hour 42 minutes. Upon receiving the departure signal, the driver followed this schedule: on the section constituting 0.9 of the entire route from the departure station to the destination station, he maintained a speed 20% higher than the usual speed, and on 0.1 of the route, he drove the train at a speed 25% higher than the usual speed. As a result, the train arrived at the destination station without being late. What is the duration of the train's journey between the stations at the usual speed?
Solution. Let $z$ be the distance between stations, $x$ be the usual speed of the train.

$$
\begin{aligned}
& \text { Then } \frac{0.9 z}{1.2 x}+\frac{0.1 z}{1.25 x}-\text { is the travel time of the train, hence } \\
& \frac{0.9 z}{1.2 x}+\frac{0.1 z}{1.25 x}=\frac{z}{x}-1 \frac{42}{60} \Rightarrow \frac{z}{x}=10 .
\end{aligned}
$$

Answer: 10 hours.