13.272. From post office $A$, a car departed in the direction of post office $B$. After 20 minutes, a motorcyclist set off after it at a speed of 60 km/h. Catching up with the car, the motorcyclist handed a package to the driver and immediately turned back. The car arrived at $B$ at the moment when the motorcyclist was halfway between the meeting point and $A$. Determine the speed of the car, if the distance from $A$ to $B$ is 82.5 km.
Solution.

Let $t$ be the time it takes for the motorcyclist to catch up with the car, and $v$ be the speed of the car. According to the problem, we have the system

$\left\{\begin{array}{l}v\left(t+\frac{20}{60}\right)=60 \cdot t, \\ \frac{82.5-60 \cdot t}{v}=\frac{t}{2},\end{array}\right.$

from which $v=45(\mathrm{km} / \mathrm{h})$.

Answer: 45 km $/$ h.