112(981). A pedestrian left point $A$ for point $B$. After 1 hour and 24 minutes, a cyclist left point $A$ in the same direction. After one hour, the cyclist was 1 km behind the pedestrian, and another hour later, the distance remaining for the cyclist to $B$ was half the distance remaining for the pedestrian. Find the speeds of the pedestrian and the cyclist, given that the distance $A B$ is 27 km.
Solution. Let the speed of the pedestrian be $v_{1}$ kilometers per hour, and the speed of the cyclist be $v_{2}$ kilometers per hour. Since in 2 hours and 24 minutes (i.e., $\frac{12}{5}$ hours), the pedestrian walked 1 km more than the cyclist traveled in 1 hour, we have the equation

$$
\frac{12}{5} v_{1}-v_{2}=1
$$

Since after 2 hours of travel, the cyclist had a remaining distance to point $B$ of $\left(27-2 v_{2}\right)$ kilometers, which is half the distance the pedestrian had left to point $B$ $\left(27-\frac{17}{5} v_{1}\right)$, we have the equation

$$
27-\frac{17}{5} v_{1}=2\left(27-2 v_{2}\right), \text { or } 4 v_{2}=27+\frac{17}{5} v_{1} \text {. }
$$

Thus, we have the system of equations:

$$
\left\{\begin{array}{l}
\frac{12}{5} v_{1}-v_{2}=1 \\
4 v_{2}=27+\frac{17}{5} v_{1}
\end{array}\right.
$$

from which we find $v_{1}=5, v_{2}=11$.

Answer: the speed of the pedestrian is 5 km/h, the speed of the cyclist is $11 \mathrm{km} / \mathrm{h}$.