Stations A and B are 110 km apart. Train T1 starts from A at 7:00 a.m. at 20 km/h toward B. Train T2 starts from B at 8:00 a.m. at 25 km/h toward A. At what time do they meet?

Introduction / Context:Here the trains do not start simultaneously: one starts an hour earlier. We first adjust the separation at the moment the second train starts, then use relative speed to find the meeting time.

Given Data / Assumptions:

• A–B distance = 110 km.
• T1: 20 km/h from 7:00 a.m.
• T2: 25 km/h from 8:00 a.m.

Concept / Approach:By 8:00 a.m., T1 has covered 20 km; remaining separation is 110 − 20 = 90 km. With both now moving toward each other, their relative speed is 20 + 25 = 45 km/h.

Step-by-Step Solution:Remaining separation at 8:00 a.m. = 90 km.Time after 8:00 a.m. to meet = 90 / 45 = 2 h.Meeting time = 8:00 a.m. + 2 h = 10:00 a.m.

Verification / Alternative check:T1 distance by 10:00 a.m. = 3 h * 20 = 60 km; T2 = 2 h * 25 = 50 km; total = 110 km.

Why Other Options Are Wrong:9:00 a.m. or 11:00 a.m. do not satisfy the summed distances equal to 110 km under the given speeds.

Common Pitfalls:Using total 110 km directly with sum speeds without accounting for the first hour of T1.

Final Answer:10:00 a.m.