A train leaves Delhi at 5:00 a.m. and reaches Kanpur at 10:00 a.m. Another leaves Kanpur at 7:00 a.m. and reaches Delhi at 2:00 p.m. When do they meet?

Introduction / Context:
Opposite-end departures with different start times again reduce to a single linear equation by expressing each distance at meet in terms of t hours after the first start.

Given Data / Assumptions:

• Delhi→Kanpur: 5:00 to 10:00 → 5 h.
• Kanpur→Delhi: 7:00 to 14:00 → 7 h.
• Constant speeds; same route distance D.

Concept / Approach:
Speeds: v1 = D/5, v2 = D/7. If they meet t hours after 5:00, the second has traveled for (t − 2) hours. Sum of distances equals D.

Step-by-Step Solution:
1 = t/5 + (t − 2)/7 (after dividing by D).Multiply by 35: 35 = 7t + 5(t − 2) = 12t − 10 → 12t = 45 → t = 15/4 h = 3 h 45 min.Meet time = 5:00 + 3:45 = 8:45 a.m.

Verification / Alternative check:
Check proportions of remaining times to endpoints; they should match the computed speeds.

Why Other Options Are Wrong:
6:45 a.m. ignores the 7:00 a.m. start of the second train; 3:45 p.m. is after both arrive.

Common Pitfalls:
Using average start times or assuming equal speeds.

Final Answer:
8:45 a.m.