# 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 are running at 60 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 6 seconds. What is the length of the fast train ?

 A) Option A B) Option B C) Option C D) Option D

Answer & Explanation Answer: A) Option A

Explanation:

As Trains are moving in same direction,

Relative Speed = 60-20 = 40 kmph

--> $40×518=1009$ m/sec

Length of Train= Speed * Time

Length = $1009×6$

13 11557
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

Answer & Explanation Answer: B) 40 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.

6 11530
Q:

A train for Fathehpur leaves for every 2 hrs 30 min from Agra station. An announcement was made that train left 37 mins ago and next train comes at 17:00hrs. At what time was the announcement made ?

 A) 15:07 hrs B) 15:20 hrs C) 15:05 hrs D) 15:00 hrs

Answer & Explanation Answer: A) 15:07 hrs

Explanation:

Next train comes at 17:00 hrs.
So, last train will be = 17:00hrs - 2:30hrs
= 14:30hrs
Announcement made after 37 min of the last train.
S0, 14:30hrs + 00:37 min
= 15:07 hrs.

13 11165
Q:

A train travelling with a speed of 60 km/hr catches another train travelling in the same direction and then leaves it 120m behind in 18 seconds. The speed of the second train is

 A) 42 kmph B) 72 kmph C) 36 kmph D) 44 kmph

Answer & Explanation Answer: C) 36 kmph

Explanation:

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

=> x = 10 m/s

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

25 11066
Q:

An Engine length 1000 m moving at 10 m/s. A bird is flying from engine to end with 'x' m/s and coming back at '2x' m/s. Take total time of bird travelling as 187.5 s. Find x and 2x in km/hr ?

 A) 21.4 and 42.8 B) 25.2 and 50.4 C) 31.4208 and 62.8416 D) 33.12 and 66.24

Answer & Explanation Answer: C) 31.4208 and 62.8416

Explanation:

Birds speeds in mtrs/sec is 'x' and '2x' .
While flying from Engine to end, relative speed = (x+10) m/sec
from end to engine, flying speed = (2x - 10) mtr/sec
so
1000/(x+10) + 1000/(2x-10) = 187.5 secs
solving it, we get
so x = 8.728 m/sec and 2x= 17.456 m/sec

x = 31.4208 km/hr and 2x = 62.8416 km/hr.

13 10909
Q:

Two trains of equal length are running on parallel lines in the same direction at 36 km/hr and 26 km/hr. The faster train passes the slower train in 36 sec. The length of each train is ?

 A) 28 mts B) 54 mts C) 24 mts D) 50 mts

Answer & Explanation Answer: D) 50 mts

Explanation:

Let the length of each train be x mts.

Then, distance covered = 2x mts.

Relative speed = 36 - 26 = 10 km/hr.

= 10 x 5/18 = 25/9 m/sec.

2x/36 = 25/9 => x = 50 mts.

11 10692
Q:

A train moving at 2/3 rd of its normal speed reaches its destination 20 minutes late. Find the normal time taken ?

 A) 4/3 hrs B) 2/3 hrs C) 3/2 hrs D) 1/4 hrs

Answer & Explanation Answer: B) 2/3 hrs

Explanation:

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

5 10478
Q:

A train is traveling at 48 kmph . It crosses another train having half of its length , traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform ?

 A) 500 B) 400 C) 360 D) 480

Explanation:

Speed of train 1 = 48 kmph
Let the length of train 1 = 2x meter

Speed of train 2 = 42 kmph
Length of train 2 = x meter (because it is half of train 1's length)

Distance = 2x + x = 3x
Relative speed= 48+42 = 90 kmph = (90*5/18)  m/s = 25 m/s

Time = 12 s

Distance/time = speed

=>3x/12 = 25 (25*12)/3

Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s

Speed of train 1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y     [where y is the length of the platform]

x =100m => 200+y = 45* 40/3
y = 400m