Let the speed of the second train be x km/hr.

Relative speed	= (x + 50) km/hr
	= (x + 50) x 5 m/sec 18
	= 250 + 5x m/sec. 18

Distance covered = (108 + 112) = 220 m.

	220	= 6
	250 + 5x 18

 250 + 5x = 660

 x = 82 km/hr.

2 kmph =		2 x	5		m/sec =	5	m/sec.
			18			9

4 kmph =		4 x	5		m/sec =	10	m/sec.
			18			9

Let the length of the train be x metres and its speed by y m/sec.

Then,		x		= 9 and		x		= 10.
		y - 5 9				y - 10 9

 9y - 5 = x and 10(9y - 10) = 9x

 9y - x = 5 and 90y - 9x = 100.

On solving, we get: x = 50.

 Length of the train is 50 m.

Let the length of the first train be x metres.

Then, the length of the second train is		x		metres.
		2

Relative speed = (48 + 42) kmph =		90 x	5		m/sec = 25 m/sec.
			18

	[x + (x/2)]	= 12 or	3x	= 300     or     x = 200.
	25		2

 Length of first train = 200 m.

Let the length of platform be y metres.

Speed of the first train =		48 x	5		m/sec =	40	m/sec.
			18			3

(200 + y) x	3	= 45
	40

 600 + 3y = 1800

 y = 400 m.

Suppose they meet x hours after 7 a.m.

Distance covered by A in x hours = 20x km.

Distance covered by B in (x - 1) hours = 25(x - 1) km.

 20x + 25(x - 1) = 110

 45x = 135

 x = 3.

So, they meet at 10 a.m.