Problems on Trains

1.

A train crosses a platform of 120m in 15sec, same train crosses another platform of length 180m in 18sec. then find the length of the train?

   A.) 175m
   B.) 180m
   C.) 185m
   D.) 170m

Answer: Option 'B'

Length of the train be ‘X’ 
X + 120/15 = X + 180/18 
6X + 720 = 5X + 900 
X = 180m

2.

A man can row 30 km upstream and 44 km downstream in 10 hrs. It is also known that he can row 40 km upstream and 55 km downstream in 13 hrs.Find the speed of the man in still water

   A.) 8 km/hr
   B.) 6 km/hr
   C.) 12 km/hr
   D.) 4 km/hr

Answer: Option 'A'

Let the speed of the man in still water be x kmph and speed of the stream be y kmph
Downstream speed = x+y kmph and
Upstream Speed = x-y kmph
=>30 / (x-y) + 44 / (x+y) = 10 and
40 / (x-y) + 55 / (x+y) = 13
Let 1/ (x+y) = u and 1/ (x-y) = v
----> 30 u + 44 v = 10 ---> multiply this eqn by 4
----> 40 u + 55 v = 13 ---> multiply this eqn by3
Solving these two linear equation we get u = 1/5 and v = 1/11
----> x - y = 5 and x+y = 11
Solving these two linear equation we get
-----> x = 8 and y = 3
So the speed of the man in still water = x = 8 km/ph

3.

The length of the bridge , which a train 130 m long and travelling at 45 km/hr can cross in 30 seconds, is:

   A.) 250 m
   B.) 245 m
   C.) 200 m
   D.) 225 m

Answer: Option 'B'

245 m

4.

Two trains are running at 60 Kmph and 42 Kmph respectively, in same direction. Fast train completely passes a man sitting in the slower train in 30sec. what is the length of the faster train?

   A.) 100m
   B.) 125m
   C.) 150m
   D.) 175m

Answer: Option 'C'

L = S × T 
L = 18 × 5/18 × 30 
(in same direction Speed = first train - second train) 
= 150m

5.

A 400m long train is running at 72 Kmph. how much time it will take to cross an electric pole? 

   A.) 15sec
   B.) 20sec
   C.) 19sec
   D.) 21sec

Answer: Option 'B'

Formula: Distance = Speed × Time
If we convert Kmph in to Mpsec multiply by 5/18,
400 = 72 KmpH × time
400 = 72×5/18 × time
Time = 20Sec

6.

A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr ,what is the length of the platform? 

   A.) 180 m
   B.) 200 m
   C.) 240 m
   D.) 300 m

Answer: Option 'C'

Given, time taken by train to pass the man = 20 sec
Speed of the train = 54 km/hr
---> Converting into meter/sec
=> Speed of train =54 * (5/18) m/sec = 15 m/sec
W.K.T: Speed of train = Length of train/ Time taken by train to pass the man
=> 15 =Length of train / 20
=>Length of train = 15 * 20
=>Length of train = 300 metre.
Given,A train passes a station platform in 36 seconds.
=>Speed of train = (Length of train + Length of platform) / Time taken by train to pass the platform
=> 15 = (300 +Length of platform) / 36
=> 15 * 36 = (300+Length of platform)
=> 540 =300+Length of platform
=> 540 - 300 =Length of platform
Thus,Length of platform = 240 meter.

7.

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is: 

   A.) 45 km/hr
   B.) 50 km/hr
   C.) 54 km/hr
   D.) 55 km/hr

Answer: Option 'B'

Given, Length of the train = 125 meter
Time taken by the train to pass the man = 10 sec
Speed of man =5 km/hr
Speed of the train relative to man = (125/10) m/sec
= (25/2)m/sec
=(25/2) x (18/5) km/hr
= 45 km/hr
Let the speed of the train be x km/hr.
Since Man and Train are moving in same direction, speeds must be subtracted
Then, relative speed = Speed of train -Speed of man
=> x - 5 = 45
=> x = 50 km/hr

8.

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'

Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
Therefore 20x + 25(x - 1) = 110 => 45x = 135 => x = 3.
So, they meet at 10 a.m.

9.

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.) 10 a.m.
   B.) 11 a.m.
   C.) 8 a.m.
   D.) 9 a.m.

Answer: Option 'A'

Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
Therefore 20x + 25(x - 1) = 110
=> 45x = 135
=> x = 3.
So, they meet at 10 a.m.

10.

320m long train is running at 72 Kmph. how much time it will take to cross a platform of 180m long? 

   A.) 20sec
   B.) 25sec
   C.) 30sec
   D.) 27sec

Answer: Option 'B'

Total Length = platform Length+ Train Length 
Total Length= 500m 
D = S × T 
500 = 72 × time 
500 = 72 × 5/18 × time 
Time = 25Sec 

11.

A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is: 

   A.) 200 m
   B.) 150 m
   C.) 50 m
   D.) Data inadequate

Answer: Option 'B'

Given time taken by train to pass the pole =15 seconds
=> W.K.T: Time = Distance / speed
=> Time = Length of train / speed of train
=> 15 = Length of train/ speed of train
=>speed of train =Length of train /15 ---> eqn(1)
Given, time taken by train to pass the platform of 100 m long =25 seconds
=>Time = (Length of train + Length of the platform) / speed of train
=> 25 = (Length of train + 100) /speed of train
Substituting eqn (1) in the above eqn, we get
=> 25 =(Length of train + 100) / (Length of train /15)
=> 25 × (Length of train /15) =(Length of train + 100)
=> (5 / 3) × Length of train =(Length of train + 100)
=> 5 × Length of train = 3 (Length of train + 100)
=> 5 × Length of train =3 × Length of train + 300
=> 2 × Length of train = 300
=> Length of train = 150 meter.

12.

A train travels from City X to City Y at a speed of 44 kmph, while on its return journey the train was travelling at speed of 77 kmph. Find the average speed of the train. 

   A.) 56
   B.) 58.5
   C.) 60
   D.) 60.5

Answer: Option 'A'

Let x be the speed of the train while going from City X to Y.
Let y be the speed of the train while going from City Y to X.
Average speed of the train is given by = 2xy / (x+y) 
Average speed = (2 x 44 x 77) / (44 + 77 )
Average speed = 56

13.

A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is:

   A.) 48 km/hr
   B.) 54 km/hr
   C.) 66 km/hr
   D.) 82 km/hr

Answer: Option 'D'

Let the speed of the second train be x km/hr.
Relative speed = (x + 50) km/hr = [( x + 50 ) x 5/18 ] m/sec = [ 250 + 5x / 18 ] m/sec.
Distance covered = (108 + 112) = 220 m. 220/ ( 250 + 5x /18 ) = 6 => 250 + 5x = 660 => x = 82 km/hr.

14.

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.) 55 m
   C.) 50 m
   D.) 65 m

Answer: Option 'C'

2 kmph = (2 x 5/18) m/sec = 5/9 m/sec.
4 kmph = (4 x 5/18 m/sec = 10/9 m/sec.
Let the length of the train be x metres and its speed by y m/sec.
Then, ( x/(y - 5) = 9 and ( x/(y - 10) = 10.
Therefore 9y - 5 = x and 10(9y - 10) = 9x
=> 9y - x = 5 and 90y - 9x = 100.
On solving, we get: x = 50.

15.

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is: 

   A.) 54 km/hr
   B.) 50 km/hr
   C.) 55 km/hr
   D.) 45 km/hr

Answer: Option 'B'

Given, Length of the train = 125 meter
Time taken by the train to pass the man = 10 sec
Speed of man =5 km/hr
Speed of the train relative to man = (125/10) m/sec
= (25/2)m/sec
=(25/2) x (18/5) km/hr
= 45 km/hr
Let the speed of the train be x km/hr.
Since Man and Train are moving in same direction, speeds must be subtracted
Then, relative speed = Speed of train -Speed of man
=> x - 5 = 45
=> x = 50 km/hr

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