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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

22. Two tains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?
 
  • A. 12
  • B. 14
  • C. 16
  • D. 20
Ans: A.
Sol.
Speed of the first train = [120 / 10] m/sec = 12 m/sec.
Speed of the second train = [120 / 15] m/sec = 8 m/sec.
Relative speed = (12 + 8) = m/sec = 20 m/sec.
∴ Required time = (120 + 120) / 20 secc = 12 sec.
 
23. Two tains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is:
 
  • A. 12
  • B. 24
  • C. 36
  • D. 48
Ans: C.
Sol.
Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (120 + 120)/12 ⇔ 2x = 20 ⇔ x = 10.
∴ Speed of each train = 10 m/sec = [10 * 18/5] km/hr = 36 km/hr.
 
 
24. Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is :
 
  • A. 48
  • B. 52
  • C. 58
  • D. 66
Ans: A.
Sol.
Relative sped = (60 + 90) km/hr
= [150 * 5/18] m/sec = [125 / 3] m/sec.
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
Required time = [2000 * 3/125] sec = 48 sec.