8.1. The older brother's journey to school takes 12 minutes, while the younger brother's journey (along the same route) takes 20 minutes. How many minutes will pass from the moment the younger brother leaves home until he is caught up by the older brother, if the older brother left 5 minutes after the younger brother?
Answer: In 12.5 minutes.

Solution: Let $S$ be the distance from home to school. Since the younger brother covers this distance in 20 minutes, in 5 minutes he will cover the distance $\frac{5 S}{20}=\frac{S}{4}$. After the older brother leaves the house, the rate at which the distance between them decreases will be $\frac{S}{12}-\frac{S}{20}=\frac{S}{30}$. To find the time from this moment until they meet, we divide the distance between them by the rate of decrease. We get $\frac{S}{4}: \frac{S}{30}=7.5$ (minutes). Taking into account the 5-minute delay, we get the answer of 12.5 minutes (and since this is less than 20 minutes, the older brother catches up with the younger one).