# Problems on Trains Questions

FACTS  AND  FORMULAE  FOR  PROBLEMS  ON  TRAINS

1. a km/hr = [a x (5/18)] m/s.

2. a m/s = [a x (18/5)] km/hr.

3. Time taken by a train of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres.

4. Time taken by a train of length 1 metres to pass a stationary object of length b metres is the time taken by the train to cover (1 + b) metres.

5. Suppose two trains or two bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relatives speed = (u - v) m/s.

6. Suppose two trains or two bodies are moving in opposite directions at u m/s and v m/s, then their relative speed = (u + v) m/s.

7. If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then time taken by the trains to cross each other = $\frac{\left(a+b\right)}{\left(u+v\right)}$sec.

8. If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then the time taken by the faster train to cross the slower train = $\frac{\left(a+b\right)}{\left(u-v\right)}$sec.

9. If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then (A's speed) : (B’s speed) = $\left(\sqrt{b}:\sqrt{a}\right)$

Q:

Two trains namely X & Y leave station 'A' at 6.30am and 7.40am and travel at 30km/hr and 40 km/hr respectively. How many kms from 'A' will the trains meet ?

 A) 140 kms B) 120 kms C) 96 kms D) 142 kms

Explanation:

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.

17 9963
Q:

Train X crosses a stationary train Y in 60 seconds and a pole in 25 seconds with the same speed. The length of the train X is 300 m. What is the length of the stationary train Y ?

 A) 360 m B) 420 m C) 460 m D) 320 m

Explanation:

Let the length of the stationary train Y be LY
Given that length of train X, LX = 300 m
Let the speed of Train X be V.
Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.
=> 300/V = 25 ---> ( 1 )
(300 + LY) / V = 60 ---> ( 2 )
From (1) V = 300/25 = 12 m/sec.
From (2) (300 + LY)/12 = 60
=> 300 + LY = 60 (12) = 720
=> LY = 720 - 300 = 420 m
Length of the stationary train = 420 m

6 9870
Q:

A train 800 metres long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 minute, then the length of the tunnel (in meters) is :

 A) 130 B) 360 C) 500 D) 540

Explanation:

Speed =[ 78 x ( 5/18 ) ] m/sec = 65/3 m/sec.

Time = 1 minute = 60 sec.

Let the length of the tunnel be x metres.

Then, [ (800 + x )/ 60 ]= 65/3

$\inline \fn_jvn \Rightarrow$3(800 + x) = 3900

$\inline \fn_jvn \Rightarrow$ x = 500.

14 9870
Q:

A jogger running at 9 km/hr along side a railway track is 260 m ahead of the engine of a 140 m long train running at 45 km/hr in the same direction. In how much time will the train pass the jogger ?

 A) 25 sec B) 40 sec C) 38 sec D) 32 sec

Explanation:

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.

5 9632
Q:

Find the length of a train if it takes 10 seconds to cross a pole and double of this time to cross a platform of length 200 mts ?

 A) 180 mts B) 190 mts C) 200 mts D) 210 mts

Explanation:

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.

12 7798
Q:

A train covered k km at 40 kmph and another 2k km at 20 kmph. Find the average speed of the train in covering the entire 3k km ?

 A) 28 kmph B) 24 kmph C) 26 kmph D) 32 kmph

Explanation:

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

= 5k/40 = k/8 hours

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

7 7656
Q:

How many seconds will a 600 m long train take to cross a man walking with a speed of 7 km/hr in the direction of the moving train if the speed of the train is 67 km/hr ?

 A) 29 sec B) 58 sec C) 36 sec D) 44 sec

Explanation:

Speed of train relative to man = 67 - 7 = 60 km/hr.
= 60 x 5/18 = 50/3 m/sec.

Time taken to pass the man = 600 x 3/50 = 36 sec.

4 7345
Q:

The length of a brigade which a train 130 m long and traveling at 45kmph can cross in 30 seconds, is

 A) 196 mts B) 220 mts C) 230 mts D) 245 mts

Explanation:

Let the length of the brigade is x mts
Then Distance = 130 + x mts
given speed = 45 kmph = 45 x5/18 m/s
Time = 30 sec
T = D/S
=> 30 = 130+x/(45x5/18)
=> x = 245 mts.