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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

34. A train 360 m long is running at a speed of 45 km / hr. In what time will be pass a bridge 140 m long?
 
  • A. 40 sec
  • B. 42 sec
  • C. 45 sec
  • D. 48 sec
Ans: A.
Sol.
Speed = [45x5/18] m/sec = 25/2 m/sec.
Total distance covered = (360 + 140) m = 500 m.
∴ Required time = (500 x 2/25)sec = 40 sec.
 
35. Two good 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
Ans: C.
Sol.
Relative Speed = (45 + 30) km / hr
= (75 x 5 / 18) m/sec
= (125/6) m/sec.
Distance covered = (500 + 500) m = 1000 m.
Required time = (1000 x 6 / 125) sec = 48 sec.
 
 
36. A train 280 m long, running with a speed of 63 km / hr will pass a tree in
 
  • A. 12 sec
  • B. 15 sec
  • C. 16 sec
  • D. 20 sec
Ans: C.
Sol.
Speed = (63 x 5/18) m/sec.
          = 35 /2 m / sec.
Time taken = (280 x 2/35) sec = 16 sec.