Question:

A train 'X' leaves station 'A' at 3 p.m. and reaches station 'B' at 4:30 p.m., while another train 'Y' leaves station 'B' at 3:00 p.m. and reaches station 'A' at 4:00 p.m. These two trains cross each other at:

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When two objects move towards each other, add their speeds to find the relative speed, then divide the total distance by this to find the meeting time.
Updated On: Aug 14, 2025
  • 3:30 p.m.
  • 3:20 p.m.
  • 3:36 p.m.
  • 3:40 p.m.
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The Correct Option is C

Solution and Explanation

Step 1: Understand the travel times.
Train X takes 1.5 hours (from 3:00 p.m. to 4:30 p.m.) to travel from A to B.
Train Y takes 1 hour (from 3:00 p.m. to 4:00 p.m.) to travel from B to A.
Step 2: Relative speeds in terms of distance.
Let the total distance between A and B be \(D\) km.
Speed of X = \(D / 1.5 = \frac{2D}{3}\) km/hr.
Speed of Y = \(D / 1 = D\) km/hr.
Step 3: Timeline alignment.
Both trains start at 3:00 p.m., but X starts from A and Y starts from B.
We want the time when their combined distances equal \(D\).
Step 4: Combined speed.
Combined speed = \(\frac{2D}{3} + D = \frac{5D}{3}\) km/hr.
Step 5: Time taken to meet.
Time = Distance / Speed = \(D / (\frac{5D}{3}) = \frac{3}{5}\) hours = 36 minutes.
Step 6: Meeting time.
Meeting time = 3:00 p.m. + 36 minutes = 3:36 p.m.
Thus, the answer is \(\boxed{3:36 \ \text{p.m.}}\).
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