A train leaves a station A at 7 am and reaches another station B at 11 am. Another train leaves B at 8 am and reaches A at 11.30 am. The two trains cross one another at

Time taken by 1st train to travel from A to B = 11-7 = 4 hours

Time taken by 2nd train to travel from B to A = 11:30-8 = 3.5 hours

=> ratio of time taken by 1st train to 2nd train = $$4 : \frac{7}{2}$$ = 8 : 7

Since, speed is inversely proportion to time

=> Ratio of speeds of 1st train to 2nd train = 7 : 8

Let the speed of 1st train = $$7x$$ and 2nd train = $$8x$$ km/hr

Distance between the two stations = time * speed = $$7x * 4 = 28x$$ km

We know that, 1st train starts one hour early, thus it will cover $$7x$$ distance till the time 2nd train starts.

So, at 8.00 a.m., remaining distance between two trains = $$28x-7x = 21x$$ km

Also, the two trains are moving in opposite directions, =>relative speed of two trains = $$8x+7x = 15$$ km/hr

Now, time taken to meet = $$\frac{21x}{15x} = \frac{7}{5}$$ hours

=> $$\frac{7}{5} * 60$$ = 84 minutes after 8.00 a.m.

=> Time when they meet = 8.00 a.m. + 84 min = 9:24 a.m.