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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

7. Two train 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 45 kmph respectively. In how much time will they cross each other, if they are running in the same direction?
 
  • A. 80 sec
  • B. 252 sec
  • C. 320 sec
  • D. 330 sec
Ans: B.
Sol.
Relative speed = (45 - 40) kmph = 5 kmph = [5 * 5/18] m/sec = 25/18 m/sec.
Total distance covered = Sum of lengths of trains = 350 m.
∴ Time taken = [350 * 18/25] sec = 252 sec.
 
8. A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?
 
  • A. 4
  • B. 5
  • C. 6
  • D. 9
Ans: C.
Sol.
Speed of train relative to man = (60 + 6) km/hr = 66 km/hr
= [66 * 5/18] m/sec = [55/3] m/sec.
∴ Time taken to pass the man = [110 * 3/55] sec = 6 sec.
 
 
9. A jogger running at 9 kmph alongside a railway track is 40 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
 
  • A. 36 sec
  • B. 38 sec
  • C. 44 sec
  • D. 60 sec
Ans: A.
Sol.
Speed of train relative to jogger = (45-9) km/hr = 36 km/hr
= [36 * 5/18] m/sec = 10 m/sec.
Distance to be covered = (240 + 120) m =360 m.
∴ Time taken = [360/10] sec = 36 sec.