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Question

A train P starts from station X for station Y and another train Q starts from station Y for station X at the same time. After passing each other, P takes 4 hours to reach Y and Q takes 1 hour to reach X. If the average speed of P is 60 km/h, then what is the average speed of Q ?

This question was previously asked in
CDS 1 2026 Maths Question Paper (12-Apr-2026)
The correct answer is
120 km/h

Solving the Train Speed Problem

This problem involves calculating the average speed of a train (Q) given information about another train (P) that starts simultaneously from the opposite station. We are given the speed of train P and the time each train takes to reach its destination *after* passing each other.

Key Concept: Train Speeds After Meeting

For two trains starting at the same time from points A and B towards each other, if they meet at point C, and after meeting, train 1 takes time '\(t_1\)' to reach B and train 2 takes time '\(t_2\)' to reach A, then the ratio of their speeds is given by:

\(\frac{\text{Speed of Train 1}}{\text{Speed of Train 2}} = \sqrt{\frac{\text{Time taken by Train 2}}{\text{Time taken by Train 1}}}\)

In our case:

  • Train P starts from X towards Y. Let its speed be '\(S_P\)'.
  • Train Q starts from Y towards X. Let its speed be '\(S_Q\)'.
  • Time taken by P to reach Y after meeting = '\(T_{P_{after}}\)' = 4 hours.
  • Time taken by Q to reach X after meeting = '\(T_{Q_{after}}\)' = 1 hour.
  • Given Speed of P, '\(S_P\)' = 60 km/h.

Applying the Train Speed Formula

Using the formula:

\(\frac{S_P}{S_Q} = \sqrt{\frac{T_{Q_{after}}}{T_{P_{after}}}}\)

Calculation Steps

  1. Substitute the given values into the formula:

    \(\frac{60 \text{ km/h}}{S_Q} = \sqrt{\frac{1 \text{ hour}}{4 \text{ hours}}}\)

  2. Simplify the square root:

    \(\frac{60}{S_Q} = \sqrt{\frac{1}{4}}\)

    \(\frac{60}{S_Q} = \frac{1}{2}\)

  3. Solve for '\(S_Q\)':

    Cross-multiply to find '\(S_Q\)':

    \(S_Q = 60 \times 2\)

    \(S_Q = 120 \text{ km/h}\)

Therefore, the average speed of train Q is 120 km/h.

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