A person starts walking from a point P at 2 a.m. and reaches Q at 5 a.m. on the same day. Another person starts walking from Q at 4 a.m. and reaches P at 9 a.m. on the same day. They will cross each other at

It is clear that the first person (say A) takes 3 hours to cover the distance between P and Q, while the second person (say B) takes 5 hours to cover the same distance. Hence, we can assume their speeds to be 5x kmph and 3x kmph (the speeds are in the reverse ratio of time taken when the distance travelled is same).

Hence, the distance is 15x km (speed$$\times$$time).

Now, until B did not start (4 a.m.), A would have covered 10x km in 2 hrs (from 2 a.m.). The distance left is 5x km between A and B. 

Now, B will start moving and we shall use the concept of relative speed on the remaining 5x km. Since A and B are moving towards each other, their speed will be added. Hence, the resultant speed is 8x kmph. 

They will meet at t= 5x/8x, which is 5/8 hours. This is 37 minutes and 30 seconds, which will be counted from when B started (4 a.m.). Hence, they will meet at 4:37:30 a.m.