## Problems on Trains Questions and Answers with Detailed Solution:

__Question No. 01__**Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:**

(A) 50 m

(B) 72 m

(C) 80 m

(D) 82 m

Answer: Option
A

__Explanation:__
Let the
length of each train be

*x*meters.
Then,
distance covered = 2

*x*meters.
Relative
speed = (46 - 36) km/hr

= [10 × (5/18)]
m/sec

= (25/9) m/sec

∴2

*x*/36 = 25/9
=> 2

*x*= 100
=>

*x*= 50.

__Question No. 02__**Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?**

(A) 9 a.m.

(B) 10 a.m.

(C) 10.30
a.m.

(D) 11 a.m.

Answer:
Option B

__Explanation:__
Suppose they
meet

*x*hours after 7 a.m.
Distance
covered by A in

*x*hours = 20*x*km.
Distance
covered by B in (

*x*- 1) hours = 25(*x*- 1) km.
∴ 20

*x*+ 25(*x*- 1) = 110
=> 45

*x*= 135
=>

*x*= 3
So, they
meet at 10 a.m.

__Question No. 03__**A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:**

(A) 45 m

(B) 50 m

(C) 54 m

(D) 72 m

Answer:
Option B

__Explanation:__
2 kmph = [2
× (5/18)] m/sec = (5/9) m/sec

4 kmph = [4
× (5/18)] m/sec = (10/9) m/sec

Let the
length of the train be

*x*meters and its speed by*y*m/sec.
Then, [

*x*/{*y*- (5/9)}] = 9 and [*x*/{*y*- (10/9)}] = 10
∴ 9

*y*- 5 =*x*and 10(9*y*- 10) = 9*x*
=> 9

*y*-*x*= 5 and 90*y*- 9*x*= 100
On solving,
we get:

*x*= 50
∴ Length
of the train is 50 m.

__Question No. 04__**A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?**

(A) 69.5
km/hr

(B) 70 km/hr

(C) 79 km/hr

(D) 79.2
km/hr

Answer:
Option D

__Explanation:__
Let the
length of the train be

*x*meters and its speed by*y*m/sec.
Then,

*x*/*y*= 8
=>

*x*= 8*y*
Now, [(

*x*+ 264)/20] =*y*
=> 8

*y*+ 264 = 20*y*
=>

*y*= 22
∴ Speed
= 22 m/sec = [22 × (18/5)] km/hr = 79.2 km/hr.

__Question No. 05__**A train traveling at 48 kmph completely crosses another train having half its length and traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is:**

(A) 400 m

(B) 450 m

(C) 560 m

(D) 600 m

Answer:
Option A

__Explanation:__
Let the
length of the first train be

*x*meters.
Then, the
length of the second train is (

*x*/2) meters
Relative
speed = (48 + 42) kmph = [90 × (5/18)] m/sec = 25 m/sec

∴[

*x*+ (*x*/2)]/25 = 12
Or, (3

*x*/2) = 300
Or,

*x*= 200
∴ Length
of first train = 200 m.

Let the
length of platform be

*y*meters.
Speed of the
first train = [48 × (5/18)] m/sec = (40/3) m/sec

∴ [(200
+

*y*) × (3/40)] = 45
=> 600 + 3

*y*= 1800
=>

*y*= 400 m.

__Question No. 06__**Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.**

(A) 12 sec

(B) 24 sec

(C) 48 sec

(D) 60 sec

Answer:
Option B

__Explanation:__
Relative speed = (45 + 30) km/hr = [75 ×
(5/18)] m/sec = (125/6) m/sec.

We have to find the time
taken by the slower train to pass the DRIVER of the faster train and not the
complete train.

So, distance covered =
Length of the slower train.

Therefore, Distance
covered = 500 m.

∴ Required time = [500 × (6/125)] = 24 sec.

__Question No. 07__**Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:**

(A) 30 km/hr

(B) 45 km/hr

(C) 60 km/hr

(D) 75 km/hr

Answer:
Option C

__Explanation:__
Let the
speed of the slower train be

*x*m/sec.
Then, speed
of the faster train = 2

*x*m/sec.
Relative
speed = (

*x*+ 2*x*) m/sec = 3*x*m/sec.
∴
(100 + 100)/8 = 3

*x*
=> 24

*x*= 200
=>

*x*= 25/3
So, speed of
the faster train = (50/3) m/sec = [(50/3) × (18/5)] km/hr = 60 km/hr.

**Trains Aptitude: Next Tests**