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Note for Problems on Trains - PT by Placement Factory

  • Problems on Trains - PT
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  • Quantitative Aptitude
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4. Self Analysis Time …………………………………………………………………..4 Problems On Trains Introduction Page 5 of 18

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In the discussion till now (as in the diginotes of the Speed Time and Distance); we never included the length of the moving object in the distance covered. The problem based on trains is a special case of speed time and distance in which the length of the moving objects becomes the distance covered. How we would be able to identify whether the length of the moving objects should considered as the distance covered? It will be told by the “Time of Crossing”. That will tell us whether we should consider the length of the moving object as distance or not as illustrated with the help of the following discussion. Self Analysis Time 1. A train running at the speed of 60 km/hr. crosses a pole in 9 seconds. What is the length of the train? Page 6 of 18

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D  ST 5   L   60   m / s  9 s 18    L  150 m 2. A train 125 m long passes a man, running at 5 km/hr. in the same direction in which the train is going, in 10 seconds. Find the speed of the train. D  ST 5   125m  S Train  5   m / s  10 s 18    S Train  50 km / hr. 3. Find the length of the bridge, which a train 130 meters long and travelling at 45 km/hr. can cross in 30 seconds. D  ST 5   L  130   45  m / s  30s 18    L  245m 4. Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. Find the ratio of their speeds. Page 7 of 18

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D1  S1T1 D2  S 2T2  L1  S1  27 s  L2  S 2  17 s  L1  27 S1  L2  17 S 2 D  ST  L1  L1  S1  S1   23s  27 S1  17 S 2  23S1  23S 2  4 S1  6 S 2  S1:S 2  6:4  S1:S 2  3:2 5. A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr., what is the length of the platform? D  ST 5   LT   54  m/s  20s 18    LT  300m Now D  ST 5   LT  LP   54  m/s  36s 18    300m  LP  540m  LP  240m 6. A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long? Page 8 of 18

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