18. A group of tourists was supposed to arrive at the station at 5 o'clock. By this time, a bus was supposed to arrive from the tourist base to pick them up. However, arriving at the station at 3:15 PM, the tourists, not waiting for the bus, started walking to the tourist base. Meeting the bus on the way, they got in and arrived at the tourist base 15 minutes earlier than the scheduled time. At what speed were the tourists walking before they met the bus, if the bus's speed was 60 km/h?
18. Since the bus arrived at the tourist base 15 minutes earlier, it should have traveled from the meeting point with the tourists to the station for $\frac{1}{8}$ hours (15 minutes / 2 = $=7.5$ minutes $=\frac{1}{8}$ hours). In this time, it would travel 7.5 km $(60: 8=7.5)$, which means the tourists walked 7.5 km on foot. They covered this distance in 1.5 hours $(5-3.25=$ $=1.75 ; 1.75-0.25=1.5)$. Therefore, their speed was 5 km/h $(7.5: 1.5=5)$.

150