Question
Train A takes 24 seconds to cross a pole and 20sec to
cross a man walking at the speed of 5.6 meter per second towards it. Train B travels 1 meter per second slower than train A and takes 25%. Less time than train need to cross a pole. If the speed of train B had been 7 meter per second less than the time taken by it to cross 114-meter-long platform is ‘5a’. Find the value of ‘a’.Solution
Let the speed of the train A = y m/s Then the length of train ‘A’= 24*y = 24y mt. Relative speed of train A wrt man = (y+5.6) m/s Also length of the train A = (y+5.6)*20 = (20y+112) mt. So, 24y = 20y+112 Y=28 Speed of train A is 28 m/s. Speed of train B = 28-1 = 27m/s. Time taken by Train ‘B’ to cross pole = 24*0.75 = 18 sec So length of train ‘B’ = 18*27 =486mt. New speed of train ‘B’ = 27-7 = 20m/s Required time taken ‘B’ = (486+114)/20 = 30 seconds 5a=30sec a=6
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