Two trains Ajanta express and Taj express simultaneously started on two parallel tracks from Meerut to Nagpur, which are 390 km a part. The ratio of the speed of Ajanta express and Taj express is 6 : 7. After how long (in kms) travelling. Taj express exchanges the speed with Ajanta express so that both the trains reach at their destination simultaneously?

A's speed / B's speed = ?(Time taken by B to reach X) / (Time taken by A to reach Y)
? 45/B's speed = ?(10/3) x (5/24) = 5/6
? B's speed = 45 x (6/5) km/hr = 54 km/hr.

Speed of first train = (100/10) = 10 m/sec.
Let the speed of second train be x m/sec
? 200/ 10 + x = 8
? 200 = 80 + 8x
? x = 15
? Speed of second train = 15 m/sec
= (15 x 18/5)
= 54 km/hr.

Relative speed = (84 + 60) km/h
= 144 km/h = 144 x (5/18) m/s
= 8 x 5 = 40 m/s

Distance covered in passing each other
= 512 + 528 = 1040 m

? Required time = 1040/40 = 26 s

Length of the train = 280 m
Length of the platform = 280 x 3 = 840 m
Taken time = 6 min 40 s
= (360 + 40) = 400 s
? Required speed = Distance / Time
= (840 + 280)/400
= 1120/400
= 28 m/s

Speed of the train relative to man
= (240 + 24) km/h = 264 km/h
= 264 x (5/18) m/s = 220/3 m/s

Distance covered in passing the man = 440 m
? Time taken = (440/220) x 3 = 6 s

Let the distance between Mokama and Hazipur be D km
Time taken by train P to cover D km = 1 h
Time taken by Q to cover D km = 3/2 h
? Speed of train P = D km/h
Speed of train Q = 2D/3 km/h .
Let they cross each other T h after 5 : 00 pm.
Then, DT + 2DT/ 3 = D
? T + 2T/3 = 1
? T(1 + 2/3) = 1
? T = 3/5h = (3/5) x 60 = 3 x 12 = 36 min.

Hence, the two trains meet 36 min after 5 : 00 pm.
? Meeting time = 5:36 pm

Let speed in the return journey = V
? Speed in onward journey = (125/100) x V = 5V/4 km/h
Average speed = (2x 5V/4 x V) / (5V/4 + V) = 10V/9 km/h
? 1600 x (9/10V) = 30 [? halts for 2 h ]
? V = 1600 x 9/30 x 10 = 48 km/h
? Speed in onward journey = 5V/4 = 5/4 x 48 = 60 km/h

Speed of the person = 4 km/h = 4.5 x (5/18) = m/s = 5/4 m/s = 1.25 m/s
Speed of 2nd person = 5.4 km/h = 5.4 x (5/18) m/s = 3/2 m/s = 1.5 m/s

Let speed of train be x m/s.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5
? 8.4x - 10.5 = 8.5x - 12.75
? 0.1x = 2.25
? x = 22.5
? Speed of the train = 225 x (18/5 ) = 81 km/h .

Given, a = 45 km/h, y = ?, t₁ = 4 h 48 min and t₂ = 3 h 20 min
Using y = a?t₁ / t₂= 45?4 h and 48 min /3 h and 20 min
= 45?(24/5h) / (10/3 h) = 45?24 x 3/5 x 10
= 45 x ?1.44 = 45 x 12 = 54 km/h

Let the length of slower train be L meters and the
length of faster train be (L/2) meters
Their relative speed = (36 + 54) km/hr
= 90 x (5/18)
= 25 m/sec

? 3L / (2 x 25) = 12
? 3L = 600
? L = 200
? Length of slower train = 200 meters
Let the length of platform by M meters
Then, 200 + M/[36 x (5/18)] = 90 sec.
? 200 + M = 900
? M = 700 meters
Length of platform = 700 meters.