Solution

If the speed of train A is 10 m/s less than that of train B. Let’s assume the speed of train A is ‘y’ m/s. y = (speed of train B) - 10 speed of train B = (y+10) m/s Train B whose length is (l+40) metre can cross a pole in 20 seconds. (y+10) = (l+40)/20 20(y+10) = (l+40) 20y+200 = (l+40) 20y = (l+40)-200 20y = (l-160) y = (l-160)/20    Eq.(i) The length of train A is 80 metre less than that of train B. length of train A = (l+40)-80 = (l-40) metre Train A can cross 168 metre long platform in 44 seconds. [(l-40)+168]/44 = y    Eq.(ii) Eq.(i) = Eq.(ii) (l-160)/20 = [(l-40)+168]/44 (l-160)/5 = [(l-40)+168]/11 11l-1760 = 5l-200+840 11l-1760 = 5l+640 11l-5l = 640+1760 6l = 2400 Value of ‘l’ = 400