13.355. A car, having traveled a distance from $A$ to $B$, equal to 300 km, turned back and after 1 hour 12 minutes from leaving $B$, increased its speed by 16 km/h. As a result, it spent 48 minutes less on the return trip than on the trip from $A$ to $B$. Find the original speed of the car.
Solution.

Let $V$ be the original speed of the car. According to the condition, we have

$$
\frac{300}{V}-\left(1 \frac{12}{60}+\frac{300-1 \frac{12}{60} V}{V+16}\right)=\frac{48}{60}, \quad \text { from which } V=60 \text { (km/h) }
$$

Answer: 60 km/h.