13.098. A pedestrian and a cyclist set off simultaneously towards each other from cities $A$ and $B$, the distance between which is 40 km, and meet 2 hours after departure. Then they continue their journey, with the cyclist arriving in $A$ 7 hours and 30 minutes earlier than the pedestrian in $B$. Find the speeds of the pedestrian and the cyclist, assuming that they remained constant throughout.
Solution.

We will fill in the table of speed, distance, and time values in the order indicated by the numbers (1), (2), ..., (12):

| Tourist | Before Meeting |  |  | After Meeting |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | speed, km $/ \mathbf{4}$ | time, h | distance, | speed,  km $/ \mathbf{4}$ | time, | distance,  km |
| Pedestrian | (1) $x$ | (3) 2 | (5) $2 x$ | (7) $x$ | (11)  $40-2 x$ | (9)  $40-2 x$ |
| Cyclist | (2)  $20-x$ | (4) 2 | (6)  $40-2 x$ | (8)  $20-x$ | (12)  $\frac{2 x}{20-x}$ | (10)  $2 x$ |

According to the condition $\frac{40-2 x}{x}-\frac{2 x}{20-x}=\frac{15}{2}$, from which $x=4$ (km/h).

Answer: 4 and 16 km/h.