23(1099). A bus left city $M$ for city $N$ at a speed of 40 km/h. A quarter of an hour later, it met a car traveling from city $N$. This car reached city $M$, 15 minutes later it set off back to city $N$ and overtook the bus 20 km from city $N$. Find the distance between cities $M$ and $N$, if the speed of the car is 50 km/h.
Solution. Since the bus traveled 10 km in a quarter of an hour $(40: 4=10)$, the bus met the car 10 km from city $M$. If the distance between the cities is $s$ kilometers, then by the time the car overtook the bus, the bus had traveled $(s-30)$ kilometers from the moment of the meeting, spending $\frac{s-30}{40}$ hours on this. By that time, the car had traveled 20 km more, i.e., $(s-10)$ kilometers, spending $\frac{s-10}{50}$ hours on the entire journey (from the moment of the meeting). Considering that the car stood in city $M$ for 15 minutes $\left(\frac{1}{4} \text{ hours}\right)$, we get the equation

$$
\frac{s-10}{50}+\frac{1}{4}=\frac{s-30}{40}
$$

from which $s=160$ km.

Answer: 160 km.

$24(1100)$. Two boys started on a running track 50 m long with a 1-second interval. The boy who started second caught up with the first 10 m from the starting line, ran to the end of the track, and then ran back at the same speed. At what distance from the end of the track did he meet the first boy, if it is known that this meeting occurred 10 seconds after the first boy started?

Solution. Let the speed of the boy who started first be $v_{1}$ meters per second, and the speed of the boy who started second be $v_{2}$ meters per second $\left(v_{2}>v_{1}\right)$. The distance of 10 m, each of the starters ran in $\frac{10}{v_{1}}$ seconds and $\frac{10}{v_{2}}$ seconds, respectively. Since one boy caught up with the other 10 m from the starting line, and they started with a 1-second interval, we have the equation

$$
\frac{10}{v_{1}}-\frac{10}{v_{2}}=1
$$

Since, according to the problem, the boys met 10 seconds after the first boy started (9 seconds after the second boy started) and by the time of the meeting, both had run a distance equal to twice the length of the running track ( $50 \cdot 2=100$ ), we have the equation

$$
10 v_{1}+9 v_{2}=100
$$

Thus, we have a system of two equations with two unknowns:

$$
\left\{\begin{array}{l}
\frac{10}{v_{1}}-\frac{10}{v_{2}}=1 \\
10 v_{1}+9 v_{2}=100
\end{array}\right.
$$

From the first equation, express one of the unknowns (for example, $v_{2}$) and substitute its value into the second equation. We have $v_{2}=\frac{10 v_{1}}{10-v_{1}}, v_{1}^{2}-29 v_{1}+100=0$, from which $v_{1}^{\prime}=4 \text{ m/s}, v_{1}^{\prime \prime}=25 \text{ m/s}$.

Obviously, the condition of the problem is satisfied by one root $v_{1}=$ $=4 \text{ m/s}$.

Thus, the speed of the first boy is $4 \text{ m/s}$. The second boy met the first at a distance of 10 m from the end of the track $(10=50-10 \cdot 4)$.

Remark. To verify, it is advisable to find the speed of the second boy $\left(v_{2}=\frac{10 v_{1}}{10-v_{1}}=\frac{10 \cdot 4}{10-4}=6 \frac{2}{3} \text{ m/s}\right)$ and calculate the required distance in another way:

$$
9 v_{2}-50=9 \cdot \frac{20}{3}-50=10 \text{ m}
$$

Answer: 10 m.