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A leaves station P at 8:00 a.m. and reaches station Q at 11:00 a.m. B leaves station Q at 9:00 a.m. and reaches station P at

11:00 a.m. At what time will they cross each other?

This questions was previously asked in

SSC MTS 15th July 2022 Shift-1

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10:00 a.m.

10:05 a.m.

9:48 a.m.

9:52 a.m.

Explanation:

- A travels from P to Q in 3 hours, leaving at 8:00 a.m. and arriving at 11:00 a.m. This gives A a speed, which we'll assume is constant, covering the distance PQ in 3 hours.

- B travels from Q to P in 2 hours, leaving at 9:00 a.m. and arriving at 11:00 a.m. B travels the same distance in 2 hours, implying B is faster than A.

- Both start moving towards each other when B leaves at 9:00 a.m.

- When they meet, the time taken by both is equal. Let t be the time in hours after 9:00 a.m. when they meet.

By setting the distances covered by both equal, we can set up the equation:

(A's speed) x t = (Total distance)/(3/5) x (2 - t),

where Total distance is the same in both cases.

Solving gives t = 48/60 = 0.8 hours.

- So, they cross each other at 9:48 a.m.

Option 3: 9:48 a.m. is the correct answer.