71 (1187). A tourist left from $A$ to $B$. After 1 hour and 20 minutes, a cyclist left from $A$ in the same direction and overtook the tourist after 30 minutes. Upon arriving at $B$, the cyclist, without stopping, turned back and met the tourist 1.5 hours after the first meeting. Find the speeds of the tourist and the cyclist, given that the distance $A B$ is 24 km.

72
Solution. Let the speed of the cyclist be $v_{1}$ km/h, and the speed of the tourist be $v_{2}$ km/h. Since the cyclist overtook the tourist 30 minutes $\left(\frac{1}{2}\right.$ hours) after departure and traveled $\frac{1}{2} v_{1}$ km, while the tourist walked 1 hour 20 minutes longer than the cyclist $\left(1 \frac{1}{3} \text{ hours} + \frac{1}{2} \text{ hours} = 1 \frac{5}{6} \text{ hours}\right)$ and walked a distance of $\frac{11}{6} v_{2}$ km during this time, we have the equation $\frac{1}{2} v_{1} = \frac{11}{6} v_{2}$.

Until the second meeting, the tourist walked $3 \frac{1}{3}$ hours $\left(1 \frac{5}{6} \text{ hours} + 1 \frac{1}{2} \text{ hours} = 3 \frac{1}{3} \text{ hours}\right)$, and the cyclist - 2 hours $\left(\frac{1}{2} \text{ hours} + 1 \frac{1}{2} \text{ hours} = 2 \text{ hours}\right)$. Since the sum of the distances traveled by the tourist and the cyclist until the second meeting is 48 km $(24 \cdot 2 = 48)$, we have the equation $2 v_{1} + \frac{10}{3} v_{2} = 48$.

We obtain the system of equations:

$$
\left\{\begin{array}{l}
2 v_{1} + \frac{10}{3} v_{2} = 48 \\
\frac{1}{2} v_{1} - \frac{11}{6} v_{2} = 0
\end{array}\right.
$$

Multiplying both sides of the second equation by 4, we can easily solve this system:

$$
\left\{\begin{array}{l}
2 v_{1} - \frac{22}{3} v_{2} = 0 \\
2 v_{1} + \frac{10}{3} v_{2} = 48
\end{array}\right.
$$

from which $\frac{32}{3} v_{2} = 48$, i.e., $v_{2} = 4.5$ km/h, $v_{1} = 16.5$ km/h.

Answer: the speed of the cyclist is 16.5 km/h, the speed of the tourist is 4.5 km/h.

$\triangle 72$ (1192). A train traveled from $A$ to $B$ at a speed of 60 km/h, and returned from $B$ to $A$ at a speed 20 km/h less. What is the average speed of the train?

Solution. Let the distance from $A$ to $B$ be $s$ kilometers. Then the train traveled from $A$ to $B$ in $\frac{s}{60}$ hours, and from $B$ to $A$ in $\frac{s}{40}$ hours. Since the average speed is the quotient of the distance divided by the time, we have:

$$
v_{\mathrm{cp}} = \frac{2 s}{\frac{s}{60} + \frac{s}{40}} = 48 \text{ (km/h)}
$$

Note. Draw the students' attention to the fact that the average speed is not equal to the arithmetic mean of the speeds on different segments of the journey.

Answer: 48 km/h.