## Task 4 - 150734

A train departs from station B exactly 15 minutes later than the timetable specifies. Therefore, it travels at $120\%$ of the usual average speed on this route until the delay is caught up.

After how many minutes (calculated from the actual departure time of the train) is this the case?
If $v$ is the usual average speed on the route, then the train travels at the speed $\frac{120}{100} v=\frac{6}{5} v$.

If $s$ is the length of the route from $B$ to the point where the delay is caught up, and if $t$ is the travel time of the train from $B$ to this point, then on the one hand $s=\frac{6}{5} v \cdot t$, and on the other hand the usual travel time (in minutes) for the mentioned route is $t+15$, so $s=v \cdot(t+15)$.

This leads to $\frac{6}{5} v t=v t+15 v$, thus $\frac{1}{5} v t=15 v$, hence $t=5 \cdot 15=75$. The delay is therefore caught up in 75 minutes.