69 (1185). From $A$ to $B$, the distance between which is 37 km, two buses with the same speed left at 7:18 AM and 7:48 AM, respectively. A cyclist, who left $B$ for $A$ at 7:28 AM, met the first bus at 7:58 AM and the second bus at 8:19 AM. Find the speeds of the cyclist and the buses.
Solution. Let the speed of the buses be $v_{1}$ km/min, and the speed of the cyclist be $v_{2}$ km/min. Since the first bus traveled for 40 minutes (7:58 AM - 7:18 AM = 40 minutes) before meeting the cyclist, the second bus for 31 minutes (8:19 AM - 7:48 AM = 31 minutes), and the cyclist spent 30 minutes (7:58 AM - 7:28 AM = 30 minutes) and 51 minutes (8:19 AM - 7:28 AM = 51 minutes) before meeting the buses, we have the system of equations:

$$
\left\{\begin{array}{l}
40 v_{1}+30 v_{2}=37 \\
31 v_{1}+51 v_{2}=37
\end{array}\right.
$$

from which $v_{1}=0.7$ km/min, or 42 km/h; $v_{2}=0.3$ km/min, or 18 km/h.

Remark. If the speed of the buses $v_{1}$ and the speed of the cyclist $v_{2}$ are expressed in kilometers per hour, the solution of the problem reduces to the system:

$$
\left\{\begin{array}{l}
\frac{2}{3} v_{1}+\frac{1}{2} v_{2}=37 \\
\frac{31}{60} v_{1}+\frac{51}{60} v_{2}=37
\end{array}\right.
$$

Obviously, the numerical data of this system complicate its solution.

Answer: the speed of the bus is 42 km/h, the speed of the cyclist is 18 km/h.