Task B-2.6. Two cyclists started simultaneously from two places $A$ and $B$ towards each other. After one hour, the first cyclist had traveled $10 \mathrm{~km}$ more than the second cyclist. The first cyclist arrived 50 minutes earlier at place $B$ than the second cyclist at place $A$. What is the distance between places $A$ and $B$?
## Solution.

If the distance $|A B|=2 x$,

and the cyclists meet at point $C$, then

$$
\begin{aligned}
& |A C|=x+5 \\
& |B C|=x-5
\end{aligned}
$$

The speed (in km/h) of the first cyclist is $\frac{x+5}{1}$, and the second cyclist's speed is $\frac{x-5}{1}$.

The first cyclist will cover the distance $|A B|$ in time $\frac{2 x}{x+5}$, and the second cyclist will take $\frac{2 x}{x-5}$.

Thus, we have the equation $\frac{2 x}{x+5}=\frac{2 x}{x-5}-\frac{5}{6}$, or

$$
x^{2}-24 x-25=0
$$

The solutions are $x_{1}=25 \mathrm{~km}$ and $x_{2}<0$.

The distance is $|A B|=50 \mathrm{~km}$.