7.1. The younger brother takes 25 minutes to reach school, while the older brother takes 15 minutes to walk the same route. How many minutes after the younger brother leaves home will the older brother catch up to him if he leaves 8 minutes later?
Answer: in 17 minutes. Solution: Let $S$ be the distance from home to school. Since the younger brother covers this distance in 25 minutes, in 8 minutes he will cover the distance $\frac{8 S}{25}$. After the older brother leaves the house, the rate at which they are closing the distance between them will be $\frac{S}{15}-\frac{S}{25}=\frac{2 S}{75}$. To find the time from this moment until they meet, we divide the distance between them by the rate of closing. We get $\frac{8 S}{25}: \frac{2 S}{75}=12$ (minutes). Taking into account the 5-minute delay, we get the answer of 17 minutes (and since this is less than 25 minutes, the older brother does catch up with the younger one).