13.221. Two riders set out simultaneously from points $A$ and $C$ to point $B$, and despite the fact that $C$ was 20 km farther from $B$ than $A$ was from $B$, they arrived at $B$ simultaneously. Find the distance from $C$ to $B$, if the rider who set out from $C$ traveled each kilometer 1 minute 15 seconds faster than the rider who set out from $A$, and the rider who set out from $A$ arrived at $B$ after 5 hours.
Solution.

Let $AB = x$ km, $BC = x + 20$ km.

The speed of the rider from $A - \frac{x}{5}$ km/h, the speed of the rider from $C - \frac{x + 20}{5}$ km/h.

According to the condition $\frac{5}{x} - \frac{5}{x + 20} = \frac{1}{48}$, from which $x = 60$ km.

Then $CB = x + 20 = 80$ km.

Answer: 80 km.