13.122. Two brothers had tickets to a stadium located 20 km from their home. To get to the stadium, they decided to use their bicycle and agreed that they would leave simultaneously, one on the bicycle and the other on foot; after covering part of the distance, the first would leave the bicycle, and the second, upon reaching the place where the bicycle was left, would continue on it and catch up with the first at the entrance to the stadium. Where should the first brother leave the bicycle and how much time will the journey take if each of the brothers will walk uniformly at a speed of 4 km/h, and ride 5 times faster?
Solution.

Let $x$ km be the distance the second brother walks. He will spend $\frac{x}{4}$ hours on this path. The remaining $20-x$ km he will travel in $\frac{20-x}{4 \cdot 5}$ hours. The total time the second brother spends traveling is $\frac{x}{4}+\frac{20-x}{20}$ hours. The first brother will travel $x$ km in $\frac{x}{20}$ hours and walk $(20-x)$ km in $\frac{20-x}{4}$ hours. He will be on the road for $\frac{x}{20}+\frac{20-x}{4}$ hours. According to the condition, $\frac{x}{4}+\frac{20-x}{20}=\frac{x}{20}+\frac{20-x}{4}$, from which $x=10$ km. The total time for the entire journey is $\frac{x}{20}+\frac{20-x}{4}=\frac{10}{20}+\frac{20-10}{4}=3$ hours.

Answer: at the midpoint of the journey; 3 hours.