45 (1287). A rider and a pedestrian simultaneously set off from point $A$ to point $B$. The rider, arriving in $B$ 50 minutes earlier than the pedestrian, returned back to $A$. On the return trip, he met the pedestrian 2 kilometers from $B$. The rider spent 1 hour and 40 minutes on the entire journey. Find the distance from $A$ to $B$ and the speed of the rider and the pedestrian.
S o l u t i o n. Since the rider spent 1 hour 40 minutes on the entire journey, he spent 50 minutes on the journey from $A$ to $B$ (1 hour 40 minutes: $2=50$ minutes). Then the pedestrian spent 1 hour 40 minutes on the journey from $A$ to $B$, as he arrived in $B$ 50 minutes later than the rider. Therefore, the pedestrian's speed is half the rider's speed. This means that when the rider was in $B$, the pedestrian was halfway through the journey. Since they met 2 km from $B$ on the return journey, the pedestrian had traveled 1 km from the halfway point (2 km $: 2=1$ km). Therefore, the distance from the halfway point to $B$ is 3 km, and the total distance between $A$ and $B$ is 6 km. Since the rider covered this distance in 50 minutes $\left(\frac{5}{6}\right.$ hours $)$, his speed is 7.2 km/h $\left(6: \frac{5}{6}=7.2\right)$; the pedestrian's speed is half of that: $7.2: 2=3.6$ km/h.

A n s w e r: 6 km; 7.2 km/h; 3.6 km/h.