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PROBLEMS ON TRAINS

Aptitude

PROBLEMS ON TRAINS -> IMPORTANT FORMULAE

1. a km/hr = [a * 5/18]m/s.
2. a m/s = [a * 18/5] km/hr.
3. Time taken by a trian 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 l metres to pass a stationary object of length b metres is the time taken by the train to cover (l + 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 is = (u + v) m/s
7. If two trains of length a metres and b metres are moving in opposite directions at u
8. If two trains of length a metres and b metres are moving in the same direciton at u m/s and v m/s, then the time taken by the faster train to cross the
slower train = (a + b)/(u - v) sec.
9. If tow 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) = (√b : √a).

PROBLEMS ON TRAINS -> SOLVED EXAMPLES

1. Two tain 100 metres and 120 metres long are running in the same direction with speeds of 72 km/hr and 54 km/hr. In how much time will the first train cross the second?
  Sol. Relative speed of the train = (72 - 54) km/hr = 18 km/hr
= [18 * 5/18] m/sec = 5 m/sec.
Time taken by the trains to cross each other
= Time taken to cover (100 + 120) m at 5 m/sec = [220/5]sec = 44 sec.
2. A train 220 m long is running with a speed of 59 kmph. In what time will it pas a man who is running at 7 kmph in the direction opposite to that in which the tain is going?
  Sol. Speed of the train relative to man = (59 + 7) kmph
= [66 * 5/18] m/sec = [55/3] m/sec.
Time taken by the train to cross the man
= Time taken by it to cover 220m at [55/3] m/sec
= [220 * 3/55] sec = 12 sec.
3. A man siting in a trian which is travelling at 50 kmph observes that a goods trian, travelling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.
  Sol.
Relative speed = [280/9] m/sec = [280/9 * 18/5] kmph = 112 kmph.
∴ Speed of goods train = (112 - 50) kmph = 62 kmph.

PROBLEMS ON TRAINS -> Exercise

4. A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
 
  • A. 190
  • B. 200
  • C. 225
  • D. 230
Ans: D.
Sol.
Relative speed = (120 + 80) km/hr [200 * 5/18] m/sec = [500/9] m/sec.
Let the length of the other tain be x metres.
Then, x + 270 / 9 = 500/9 ⇔ x + 270 = 500 ⇔ x = 230.
 
5. 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. 45
  • B. 50
  • C. 60
  • D. 80
Ans: B.
Sol.
Let the length of each tain be x metres.
Then, distance covered = 2x metres.
Relaive speed = (46 - 36) km/hr = [10 * 5/18] m/sec = [25/9] m/sec.
∴ 2x/36 = 25/9 ⇔ 2x = 100 ⇔ x = 50.
 
 
6. A train 125 m long passes a man, running at 5 kmph in the same direction in which the train is going, in 10 seconds. The speed of the train is:
 
  • A. 38 km/hr
  • B. 40 km/hr
  • C. 44 km/hr
  • D. 50 km/hr
Ans: D.
Sol.
Speed of the train relative to man = [125/10] m/sec = [25/2] m/sec.
= [25/2 * 18/5] km/hr = 45 km/hr.
Let the speed of the train be x kmph. Then, relative speed = (x - 5) kmph.
∴ x - 5 = 45 or x = 50 kmph.