Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3

Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

Let the length of the train be L metres

Speed of first train =  3600 18 x L m/hour

Speed of secxond train =  3600 12 x L m/hour

When running in opposite directions, relative speed = 200 L + 300 L m/hour

Distance to be covered = L + L = 2L metre

Time taken =  2 L 500 L x 3600 sec

                 =14.4 sec

Total distance covered = 7 2 + 1 4miles = 15 4miles  

Time taken =  15 4 * 75hrs =  1 20 hrs = 1 20 * 60 m i n = 3 min

Let the length of the train = L mts

Relative speed of train and man = 74 - 8 = 66 kmph = 66 x 5/18 m/s

=> 66 x 5/18 = L/9

=> L = 165 mts.

Speed of train 1 =  120 10 m / s e c = 12 m/sec 

Speed of train 2 = 120 15 m / s e c  = 8 m/sec 

if they travel in opposite direction, relative speed = 12 + 8 = 20 m/sec

distance covered = 120 + 120 = 240 m

time = distance/speed = 240/20 = 12 sec

Relative speed = (120 + 80) km/hr

=(200*5/18)m/s = (500/9)m/s

Let the length of the other train be x metres.

Then, x+270/9 = 500/9 

=>  x + 270 = 500

=>  x = 230.

Let the distance be 'd' kms.

According to the given data,

$$\begin{gathered} d 30 - d 50 = 40 60 hrs \\ = > 6 d = 300 \\ = > d = 50 kms . \\ \end{gathered}$$

Given speed of the first train = 60 km/hr = 60 x 5/18 = 50/3 m/s

Let the speed of the second train = x m/s

Then, the difference in the speed is given by

50 3 - x = 120 18

=> x = 10 m/s

=> 10 x 18/5 = 36 km/hr

Relative speed = (60 + 40) km/hr =[ 100 x ( 5 / 18 ) ]m/sec = ( 250 /9 ) m/sec.
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = [ 300 x ( 9/250 ) ] sec = ( 54/ 5 )sec = 10.8 sec.