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?

Let's assume the distance between stations P and Q is '6y' km.

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

Total time taken by A to reach from station P to Q = 3 hours

$$speed=\frac{distance}{time}$$

$$speed=\frac{6y}{3}$$

= 2y km/h
B leaves station Q at 9:00 a.m. and reaches station P at 11:00 a.m.

Total time taken by B to reach from station Q to P = 2 hours

$$speed=\frac{distance}{time}$$

$$speed=\frac{6y}{2}$$

= 3y km/h

As given in the question that A started one hour earlier than B. So the distance covered by A in 1 hour is '2y'. Now remaining distance will be covered by both of them together to for crossing each other.

The remaining distance between A and B at 9:00 a.m. = 6y-2y = 4y km

Time taken by both of them to cross each other (after 9:00 a.m.) = $$\frac{distance}{relative\ speed}$$

= $$\frac{4y}{2y+3y}$$

= $$\frac{4y}{5y}$$

= 0.8 hour

= 48 minutes [1 hour = 60 minutes. then 0.8 hour = $$0.8\times60$$ = 48 minutes.]

total time taken by both of them to cross each other = 1 hour 48 minutes.

A started at  8:00 a.m.. So they will meet with each other = 8:00 a.m. + 1 hour 48 minutes

= 9:48 a.m.