4. The train is traveling at a speed of 60 kilometers per hour, making stops every 48 kilometers. The duration of each stop, except the fifth, is 10 minutes, and the fifth stop is half an hour. How far has the train traveled if it departed at noon on September 29 and arrived at its destination on October 1 at 10:00 PM?
Solution: The train was on the way for 58 hours.

The train covers a section of 48 kilometers in $\frac{4}{5}$ of an hour.

Let the train make $N$ stops during its journey, then the time of its movement will be $\left(\frac{4}{5}+\frac{1}{6}\right)(N-1)+\left(\frac{4}{5}+\frac{1}{2}\right)+t=58$, where $t$ is the time to cover the last section of the path, which is more than 0 km but no more than 48 km, so $t \in\left(0 ; \frac{4}{5}\right]$. $\left(\frac{4}{5}+\frac{1}{6}\right)(N-1)+\left(\frac{4}{5}+\frac{1}{2}\right)+t=58$ $\frac{29}{30} N+\frac{1}{3}+t=58$ $t=57 \frac{2}{3}-\frac{29}{30} N$ $0<57 \frac{2}{3}-\frac{29}{30} N \leq \frac{4}{5}$ $56 \frac{13}{15} \leq \frac{29}{30} N<57 \frac{2}{3}$ $58 \frac{24}{29} \leq N<59 \frac{19}{29}$

We get that $\mathrm{N}=59, t=57 \frac{2}{3}-\frac{29}{30} \cdot 59=\frac{19}{30}$.

The distance covered by the train, $N \cdot 48+t \cdot 60=59 \cdot 48+\frac{19}{30} \cdot 60=2870$.

Answer: 2870 km.