70(1186). From points $M$ and $K$, which are 70 km apart, a bus and a cyclist set off towards each other simultaneously. They met after 1 hour and 24 minutes. Continuing their journey at the same speed, the bus arrived in $K$ and, after a 20-minute stop, set off back. Find the speeds of the bus and the cyclist, given that the bus overtook the cyclist 2 hours and 41 minutes after their first meeting.
Solution. Let the speed of the bus be $v_{1}$ km/h, and the speed of the cyclist be $v_{2}$ km/h. The bus and the cyclist met after 1 hour and 24 minutes, i.e., after $\frac{7}{5}$ hours, so we have the equation $\frac{7}{5} v_{1}+\frac{7}{5} v_{2}=70$.

Since the cyclist had been riding for $4 \frac{1}{12}$ hours (1 hour 24 minutes + 2 hours 41 minutes = 4 hours 5 minutes) before being overtaken by the bus, and the bus had been traveling for 20 minutes less, i.e., $3 \frac{3}{4}$ hours, and considering that the bus traveled 70 km more than the cyclist during this time, we have the second equation: $\frac{15}{4} v_{1}-\frac{49}{12} v_{2}=70$.

We obtain the system of equations:

$$
\left\{\begin{array}{l}
\frac{7}{5} v_{1}+\frac{7}{5} v_{2}=70 \\
\frac{15}{4} v_{1}-\frac{49}{12} v_{2}=70
\end{array}\right.
$$

After simplification, we have:

$$
\left\{\begin{array}{l}
v_{1}+v_{2}=50 \\
45 v_{1}-49 v_{2}=840
\end{array}\right.
$$

From this, $v_{1}=35$ km/h, $v_{2}=15$ km/h.

Answer: the speed of the bus is 35 km/h, the speed of the cyclist is $15 \mathrm{km} /$ h.