TNPSC Problems on Trains Questions & Answers Pdf - 1
Question: 1
Two trains 200 metres and 150 metres long are running on parallel rails in the same direction at speeds of 40 km/hr and 45 km/hr respectively. Time taken by the faster train to cross the slowed train will be
(A) 72 seconds
(B) 132 seconds
(C) 252 seconds
(D) 272 seconds
Ans: C
Relative speed = (45 - 40) km/hr = 5 km/hr
= $$(5 × {5}/{18})$$ m/sec = $$({25}/{18})$$m/sec.
Total distance covered = Sum of lengths of trains = (200 + 150) = 350 m.
∴ Time taken = $$(350 × {18}/{25})$$ sec = 252 sec.
Question: 2
A man sitting in a train which is travelling at 50 kmph observes that a good train, travelling in opposite direction, takes 9 seconds to pass him. If the good train is 280 m long, find its speed.
(A) 62 kmph
(B) 64 kmph
(C) 68 kmph
(D) 69 kmph
Ans: A
Relative speed = $$({280} / {9})$$m/sec = $$({280} /{ 9} × {18} / {5})$$kmph = 112 kmph.
∴ Speed of goods train = (112 - 50)kmph = 62 kmph.
Question: 3
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) 40 km/hr
(B) 48 km/hr
(C) 50 km/hr
(D) 70 km/hr
Ans: C
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.
Question: 4
A train 280 m long, running with a speed of 63 km/hr will pass a tree in
(A) 14 sec
(B) 15 sec
(C) 16 sec
(D) 18 sec
Ans: C
Speed = $$(63 × {5}/{18})$$ m/sec = $${35}/{2}$$ m/sec.
Time taken = $$(280 × {2}/{35})$$ sec = 16 sec.
Question: 5
Two trains A and B start running together from the same point in the same direction, at the speeds of 60 kmph and 72 kmph respectively. If the length of each of the train is 240 metres, how long will it take for the train B to cross train A?
(A) 1 min 12 secs
(B) 1 min 24 secs
(C) 2 min 12 secs
(D) 2 min 24 secs
Ans: D
Relative speed = (72 - 60) km/hr = 12 km/hr.
= $$(12 × {5}/{18})$$ m/sec = $$({10}/{3})$$m/sec. Total distance covered = Sum of lengths of trains = (240 + 240) m = 480 m.
∴ Time taken = $$(480 × {3}/{10})$$sec = 144 sec = 2 min 24 sec.
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