Two stations A and B are 110 km apart on a straight line. One train starts…

Concept: In a two-body meeting problem, two objects move toward each other along a fixed distance and meet at the exact instant the sum of the distances each has individually covered equals that total distance. When one body starts later than the other, measure both distances using a common "meeting time" variable, but reduce the late starter's travel time by its delay.

1. Let the trains meet x hours after 7 a.m. (Train A's start time).

2. Train A travels for x hours at 20 kmph, covering 20x km.

3. Train B starts one hour later, at 8 a.m., so it travels for (x − 1) hours at 25 kmph, covering 25(x − 1) km.

4. Since the two stations are 110 km apart, at the meeting instant the two distances together equal 110 km: 20x + 25(x − 1) = 110.

5. Expand and simplify: 20x + 25x − 25 = 110, so 45x = 135.

6. Solve for x: x = 3 hours.

7. So the meeting time is 3 hours after 7 a.m., i.e. 10 a.m.

Cross-check: In 3 hours, Train A covers 20 × 3 = 60 km. From 8 a.m. to 10 a.m. is 2 hours, so Train B covers 25 × 2 = 50 km. Together, 60 + 50 = 110 km, exactly the distance between the stations — confirming the meeting time found above.