13.117. A cyclist traveled 60 km from point $A$ to point $B$. On the return trip, he rode the first hour at the same speed, then stopped for 20 minutes. Resuming his journey, he increased his speed by 4 km/h and therefore spent as much time traveling from $B$ to $A$ as he did traveling from $A$ to $B$. Determine the cyclist's speed on the trip from $A$ to $B$.
## Solution.

Let $x$ km/h be the speed of the cyclist on the way from $A$ to $B$. The time spent on the way from $A$ to $B$ is $\frac{60}{x}$ hours; the time spent on the return trip is $-1+\frac{1}{3}+\frac{60-x}{x+4}$. According to the condition, $\frac{60}{x}=1+\frac{1}{3}+\frac{60-x}{x+4}$, from which $x=20$ km/h.

Answer: 20 km/h.