Problems on Trains Questions

The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.

Relative speed = 72 -36 = 36 kmph = 36 x 5/18 = 10 mps.

Time required = d/s = 500/10 = 50 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 train relative to jogger = 45 - 9 = 36 km/hr.
= 36 x 5/18 = 10 m/sec.

Distance to be covered = 260 + 140 = 400 m.

Time taken = 400/10 = 40 sec.

Let the original speed and time is S and T
then distance = S x T
Now the speed changes to 2/3S and T is T+20
As the distance is same
S x T = 2/3Sx(T+20)
solving this we get t = 40 minutes
=40/60 = 2/3 hour

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.

Speed = 78 x 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (900 + x)/60 = 65/3
x = 400 m.

Total time taken = k/40 + 2k/20 hours

= 5k/40 = k/8 hours

Average speed = 3k/(k/8) = 24 kmph.