13.306. Two hours after departure, the train stopped for 30 minutes. On the remaining part of the route to the station, repair work was being carried out, and the train was allowed a speed that was $1 / 3$ of its initial speed, as a result of which the train arrived at the station 1 hour and 10 minutes late. The next day, the train stopped 14 km closer to the final station, and under the same conditions, the delay was reduced to 50 minutes. Determine the distance between the stations and the speed of the train.
Solution.

Let $x$ be the distance between the stations, $V$ be the speed of the train. According to the problem, we have the system

$$
\left\{\begin{array}{l}
2+\frac{30}{60}+\frac{x-2 V}{\frac{1}{3} V}=\frac{x}{V}+1+\frac{10}{60} \\
\frac{2 V+14}{V}+\frac{30}{60}+\frac{x-(2 V+14)}{\frac{1}{4} V}=\frac{x}{V}+\frac{50}{60}
\end{array}\right.
$$

from which $x=196$ (km), $V=84$ (km/h).

Answer: 196 km; 84 km/h.