The distance between stations A and B is 220 km. A train leaves A for B at 80 km/h. After 30 minutes, another train leaves B for A at 100 km/h. At what distance from A will the trains meet?

Introduction / Context:
This is a head-on meeting problem with a delayed start. Use the lead distance gained by the first train before the second starts, then use relative speed after both are moving to find the meeting time and position.

Given Data / Assumptions:

• Total separation AB = 220 km.
• Train A speed = 80 km/h; Train B speed = 100 km/h.
• Train B starts 0.5 h later.

Concept / Approach:
First, distance covered by Train A in 0.5 h: d_lead = 80 * 0.5 = 40 km. Remaining separation when B starts: 220 − 40 = 180 km. After B starts, closing speed v_rel = 80 + 100 = 180 km/h; time to meet thereafter = remaining / v_rel = 180/180 = 1 h. Meeting distance from A = 80 * (0.5 + 1) = 120 km.

Step-by-Step Solution:

Verification / Alternative check:
Positions after 1 h from B’s start: A covered 80 km more (total 120), B covered 100 km; 120 + 100 = 220 km—consistent.

Why Other Options Are Wrong:
130, 140, 150, 110 km do not satisfy the distance-time constraints with the delay.

Common Pitfalls:
Ignoring the 30-minute head start; using difference of speeds instead of sum for opposite directions.

Final Answer:
120 km