As Trains are moving in same direction,

Relative Speed = 60-20 = 40 kmph

-->  40 × 5 18 = 100 9 m/sec

Length of Train= Speed * Time

Length =  100 9 × 6

$$\begin{gathered} ? 200 3 = 66 2 3 m t s \\ \end{gathered}$$ 

Speed of the Train K is given by s = d/t = 240/20 = 12 m/s
Distance covered by Train K in 50 seconds = 12 x 50 = 600 mts.
But it crosses Train L in 50 seconds
Therefore, the length of the Train L is = 600 - 240 = 360 mts.

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27 x meters, and

length of the second train = 17 y meters.

(27 x + 17 y) / (x + y) = 23

=> 27 x + 17 y = 23 x + 23 y

=> 4 x = 6 y

=> x/y = 3/2.

Let the length of the train be 'L' mts

let the speed of the train be 'S' m/s

Given it crosses a pole in 10 sec=> L/S = 10 ......(1)
Given it takes 20 sec (double of pole) to cross a platform of length 200 mts

=> (L + 200)/S = 20

=> L/S + 200/S = 20

But from (1) L/S = 10

=> 200/S = 20 - 10

=> S = 20 m/s

Then, from (1)

=> L = 10 x 20 = 200 mts.

Hence, the length of the train = 200 mts.

Relative speed = 42 + 36 = 78 km/hr = 65 3 m/s

Distance = (520 + 520) =1040 mts.

Time =  1040 × 3 65= 48 sec

It is given train X leave station A at 6:30 am, here it is asked to calculate the distance from A when the trains meet, the
Distance traveled by train left at 6:30 am upto 7:40 am i.e. in 1 hr. 10 min. or 7/6 hours = 30 x 7/6 = 35 km
So train leaving at 7:40 am will meet first train after covering a distance of 35 km. with relative speed of 40-30=10 km/hr.
Hence time taken = 35/10 = 3.5 hours or 3 hours 30 minutes
So distance from A = Distance traveled by 2nd train in 3 hr. 30 min
= 40 x 3.5 = 140 km.

Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec.
[(25/2) x (18/5)] km/hr = 45 km/hr.
Let the speed of the train be 'S' km/hr.
Then, relative speed = (S - 5) km/hr.
S - 5 = 45 => S = 50 km/hr.

Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.

[66*(5/18)] m/sec = (55/3) m/sec

Time taken to pass the man = [110*(3/55)]m/sec = 6 sec.

Given L1 = 140 m

L2 = 160 m

S1 = 60 km/hr

S2 = 80 km/hr

From the question we get,

S1 + S2 = (L1 + L2) / T

=> (60 + 80) 5/18 m/s = 140 + 160/T

=> T = 54/7 = 7.71 sec