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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

13. A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?
 
  • A. 200
  • B. 250
  • C. 260
  • D. 270
Ans: D.
Sol.
Speed = [72 * 5/18] m/sec = 20 m/sec; Time = 26 sec.
Let the length of the train be x metres.
Then, x + 250 / 26 = 20 ⇔ x + 250 = 520 ⇔ x = 270.
 
14. A train covers a distance of 12 km in 10 minutes. If it takes 6 seconds to pass a telegraph post, then the length of the train is :
 
  • A. 120 m
  • B. 140 m
  • C. 240 m
  • D. 300 m
Ans: A.
Sol.
Speed = [12/10 * 60] km/hr = [72 * 5/18] m/sec = 20 m/sec.
Length of the train = (speed * Time) = (20 * 6) m = 120 m.
 
 
15. A train 132 m long pass a telegraph pole in 6 seconds. Find the speed of the train.
 
  • A. 66 km/hr
  • B. 68.4 km/hr
  • C. 72 km/hr
  • D. 79.2 km/hr
Ans: D.
Sol.
Speed = [132/6] m/sec = [22 * 18/5] km/hr = 79.2 km/hr.