13.222. The distance between stations $A$ and $B$ is 103 km. A train left $A$ for $B$ and, after traveling a certain distance, was delayed, and therefore the remaining distance to $B$ was traveled at a speed 4 km/h greater than the initial speed. Find the original speed of the train, given that the remaining distance to $B$ was 23 km longer than the distance traveled before the delay, and that the time taken to travel the remaining distance after the delay was 15 minutes more than the time taken to travel the distance before the delay.
## Solution.

Let $x$ km/h be the original speed of the train, $t$ h be the time until its stop.

Then $\left\{\begin{array}{l}x \cdot t=40, \\ (x+4)\left(t+\frac{1}{4}\right)=63,\end{array}\right.$ from which $x=80$ km/h.

Answer: $80 \mathrm{km} /$ h.