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Let the usual speed of the train be x km/hr. Usual time = Distance / Speed = 180/x hr If the speed is increased by 2 km/hr, that is (x + 2) km/hr. The new time is = 180/(x+z) hr 180/x - 180/(x+z) = 3 180x + 360 -180 = x (3x + 6) x (3x + 6) = 360 x(x+2) = 120 x ² + 2x = 120 x ² + 2x - 120 = 0 x = 10, x = -12 Therefore, usual speed of the passenger train is 10 km/hr.
Alternate method: Let speed of train is x kmph. Distance = (Gap of time× Product of speed ) / (Gap of Speed) 180 = (3hrs× x× (x+2)) /2 360 = 3hrs× x× (x+2) 120 = x× (x+2) So x = 10 kmph
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