1. At 9:00, a pedestrian set off on a journey. An hour later, a cyclist set off after him from the same starting point. At 10:30, the cyclist caught up with the pedestrian and continued on, but after some time, the bicycle broke down. After $3a$ minutes of repair, the cyclist resumed his journey, following the pedestrian, and caught up with him again at 13:00. How many minutes did the repair take? (The pedestrian's speed is constant, and he moved without stopping. The cyclist's speed is also constant, and he moved with only one break for repair.) (I. Rubanov)
Answer: 100 minutes. Solution: The cyclist caught up with the pedestrian half an hour after his start and one and a half hours after the pedestrian's start. This means he is moving three times faster than the pedestrian. By the time of the second meeting with the cyclist, the pedestrian had been walking for 4 hours = 240 minutes. If the cyclist had been riding without stopping for repairs, he would have reached this place in $240: 3=80$ minutes. In reality, he spent 3 hours = 180 minutes on this. Therefore, he spent $180-80=100$ minutes on repairs.