13.398 Two friends decided to go hunting. One of them lives 46 km from the hunting base, the other, who has a car, lives 30 km from the base between the base and his friend's house. They set off at the same time, with the car owner driving towards his friend who was walking. Upon meeting, they drove together to the base and arrived there one hour after leaving their homes. If the pedestrian had left his house 2 hours and 40 minutes earlier than the car owner, they would have met 11 km from the pedestrian's house. What is the speed of the car?
Solution. Let $x$ and $y$ be the speeds (km/h) of the pedestrian and the car. Then $y \cdot 1$ is the total distance the car has traveled, from which $\frac{y-30}{2}$ is the distance the car traveled before meeting the pedestrian, $16-\frac{y-30}{2}=\frac{62-y}{2}$ is the distance the pedestrian traveled before the meeting. Therefore, $\frac{y-30}{2 y}=\frac{62-y}{2 x}$. According to the second condition of the problem, $\frac{5}{y}+2 \frac{2}{3}=\frac{11}{x}$. Thus,

$\left\{\begin{array}{l}\frac{5}{y}+\frac{8}{3}=\frac{11}{x}, \\ \frac{y-30}{y(62-y)}=\frac{1}{x}\end{array} \Rightarrow y=60\right.$.

Answer: 60 km/h.