25(1101). The distance between the ports $A$ and $B$ is covered by the steamboat downstream in 5 hours, and upstream in 6 hours. How many hours will it take for a raft to drift downstream this distance?
Solution. I method. To answer the question of the problem, the distance $AB$ needs to be divided by the speed of the current. Let $AB = s$ kilometers. Then the speed of the steamer downstream is $\frac{s}{5}$ kilometers per hour, the speed upstream is $-\frac{s}{6}$ kilometers per hour, the speed of the current is $-\left(\frac{s}{5}-\frac{s}{6}\right): 2=\frac{s}{60}$ kilometers per hour. The required time is $-s: \frac{s}{60}=60$ hours.

II method. Let the speed of the steamer in still water be $v_{1}$ kilometers per hour, and the speed of the current be $v_{2}$ kilometers per hour. Then, according to the condition, the distance $AB$ is equal to $5\left(v_{1}+v_{2}\right)$ kilometers, or $6\left(v_{1}-v_{2}\right)$ kilometers. From the equation $5\left(v_{1}+v_{2}\right)=6\left(v_{1}-v_{2}\right)$, we find $v_{1}=11 v_{2}$. Then the distance $AB$ is equal to $60 v_{2}$ kilometers, and the required time is $-60 v_{2}: v_{2}=60$ hours.

Answer: 60 hours.

$\square$ 26(1102). A motorboat traveled 90 km downstream in a certain time. In the same time, it would have traveled 70 km upstream. What distance would a log float in this time?

I method. Let $t$ hours be the time it takes for the motorboat to travel 90 km downstream (and also 70 km upstream). Then the speed of the motorboat downstream is $\frac{90}{t}$ kilometers per hour, the speed upstream is $\frac{70}{t}$ kilometers per hour, the speed of the current is $\left(\frac{90}{t}-\frac{70}{t}\right): 2=\frac{10}{t}$ kilometers per hour, and the required distance is $\frac{10}{t} \cdot t=10$ km.

II method. Let $v$ kilometers per hour be the motorboat's own speed, and $v_{1}$ kilometers per hour be the speed of the current. Then $\left(v+v_{1}\right)$ kilometers per hour is the speed of the motorboat downstream, and $\left(v-v_{1}\right)$ kilometers per hour is the speed of the motorboat upstream.

Downstream, the motorboat covers a distance of 90 km in $\frac{90}{v+v_{1}}$ hours. To cover a distance of 70 km upstream, the motorboat needs $\frac{70}{v-v_{1}}$ hours.

According to the condition, we have $\frac{90}{v+v_{1}}=\frac{70}{v-v_{1}}$, from which $v=8 v_{1}$.

Thus, the speed of the current is 9 times less than the speed of the motorboat downstream $\left(v+v_{1}=8 v_{1}+v_{1}=9 v_{1}\right)$. Therefore, in the same time, a log will float a distance 9 times less than the motorboat downstream. Thus, the required distance is 10 km $(90: 9=10)$.

Answer: 10 km.

$\square 27(1103)$. From points $A$ and $B$, two cyclists simultaneously set off towards each other and met 30 km from $B$.

Upon arriving at $A$ and $B$, they turned back. The second meeting occurred 18 km from $A$. Find the distance between $A$ and $B$.

Solution. I method. It is not difficult to understand that from the start of the movement to the first meeting, both cyclists together traveled a distance equal to $AB$, and by the time of the second meeting, they traveled three times the distance ( $3AB$ ). Thus, each of them traveled three times more by the second meeting than by the first. The cyclist who set off from $B$ traveled 30 km to the first meeting. Therefore, by the second meeting, he traveled 90 km ( $30 \cdot 3=90$ ). Therefore, the distance from $A$ to $B$ is 72 km $(90-18=72)$.

II method. Let $AB = s$ (in km). Since the speeds of both cyclists are constant, the ratios of the distances traveled by the cyclists to the first and second meetings are equal. Thus, we have the equation

$$
\frac{s-30}{30}=\frac{2s-18}{s+18}
$$

from which $s=72$ km.

Answer: 72 km.