70(1004). From two settlements $A$ and $B$, two tourists set out towards each other at the same time. Upon meeting, it turns out that the tourist who left from $A$ has walked 2 km more than the second tourist. Continuing their movement at the same speed, the first tourist arrives in $B$ after 1 hour 36 minutes, and the second tourist arrives in $A$ after 2 hours 30 minutes after the meeting. Find the distance $A B$ and the speed of each tourist.

$\mathrm{Pe}$ she n ie. Let the first tourist (who left from $A$) have walked $s$ kilometers by the time they met, then the second tourist walked $(s-2)$ kilometers by the time they met. Since after the meeting, the first tourist walked $(s-2)$ kilometers in 1.6 hours, his speed is $\frac{s-2}{96}$ kilometers per minute, and the speed of the second tourist is $\frac{s}{150}$ kilometers per minute. Since before the meeting, the tourists walked the same amount of time, we have the equation

$$
s: \frac{s-2}{96}=(s-2): \frac{s}{150}, \text { i.e. } 9 s^{2}-100 s+100=0
$$

from which $s_{1}=10$ (km), $s_{2}=\frac{10}{9}$ (km). The condition of the problem $(s>2)$ is satisfied only by the first root.

Thus, the distance $A B$ is 18 km $(10+(10-2))$, the speed of the first tourist is $-\frac{8}{96}=\frac{1}{12}$ km/min, or 5 km/h, and the speed of the second tourist is $-\frac{10}{150}=\frac{1}{15}$ km/min, or 4 km/h.

Answer: 18 km, 5 km/h, 4 km/h.

$71(1005)$. Two pedestrians set out towards each other from points $A$ and $B$ at the same time and met after 2 hours. How much time did each pedestrian spend on the journey $A B$ if the first, who left from $A$, arrived at point $B$ 1 hour 40 minutes later than the second arrived at point $A$?
Solution. Method I. Let the speed of the first pedestrian (who left from $A$) be $v_{1}$ kilometers per hour, and the speed of the second pedestrian be $v_{2}$ kilometers per hour. Then, before the meeting, the first pedestrian walked $2 v_{1}$ kilometers, and the second pedestrian walked $2 v_{2}$ kilometers, and after the meeting, it was the opposite. Since after the meeting, the first pedestrian spent 1 hour and 40 minutes more than the second, we have the equation

$$
\frac{2 v_{2}}{v_{1}}-\frac{2 v_{1}}{v_{2}}=\frac{5}{3}
$$

To solve this equation, let $\frac{v_{2}}{v_{1}}=t$. Then

$$
\begin{gathered}
2 t-2 \cdot \frac{1}{t}=\frac{5}{3}, 6 t^{2}-5 t-6=0 \\
t_{1}=\frac{3}{2}, t_{2}=-\frac{2}{3}
\end{gathered}
$$

Only the positive root $t=\frac{3}{2}$ satisfies the condition of the problem.

We have $\frac{v_{2}}{v_{1}}=\frac{3}{2}, 2 v_{2}=3 v_{1}$. Since $A B=2 v_{1}+2 v_{2}=5 v_{1}=$ $=\frac{10}{3} v_{2}$, the first tourist spent 5 hours on the entire journey $\left(5 v_{1}: v_{1}=5\right)$, and the second 3 hours and 20 minutes $\left(\frac{10}{3} v_{2}: v_{2}=3 \frac{1}{3}\right)$.

Method II. Let the second pedestrian spent $t$ hours on the entire journey, then the first pedestrian spent $\left(t+1 \frac{2}{3}\right)$ hours. If we denote the distance $A B$ as $s$ kilometers, then the speed of the first pedestrian is $\frac{s}{t+1 \frac{2}{3}}$ kilometers per hour, and the speed of the second pedestrian is $\frac{s}{t}$ kilometers per hour.

In 2 hours, the pedestrians walked $\left(\frac{2 s}{t+1 \frac{2}{3}}+\frac{2 s}{t}\right)$ kilometers, or $s$ kilometers.

## We have the equation

$$
\frac{2 s}{t+\frac{5}{3}}+\frac{2 s}{t}=s
$$

Dividing term by term by $s \neq 0$ and simplifying, we get:

$$
3 t^{2}-7 t-10=0
$$

from which $t_{1}=3 \frac{1}{3}, t_{2}0)$ only the root $v=4$ satisfies.

Answer: 4 km/h.