Answer-A passenger train takes 3 hours less for a journey of 180 km, if its speed is increased by 2 km/hour...

Q: Answer-A passenger train takes 3 hours less for a journey of 180 km, if its speed is increased by 2 km/hour...

Let the usual speed of the train be x kmhr. Usual time Distance Speed 180x hr If the speed is increased by 2 kmhr, that is x 2 kmhr. The new time is 180xz hr 180x - 180xz 3 180x 360 -180 x 3x 6 x 3x 6 360 xx2 120 x sup2 2x 120 x sup2 2x - 120 0 x 10, x -12 Therefore, usual speed of the passenger train is 10 kmhr. Alternate method Let speed of train is x kmph. Distance Gap of time times Product of speed Gap of Speed 180 3hrs times x times x2 2 360 3hrs times x times x2 120 x times x2 So x 10 kmph

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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

A passenger train takes 3 hours less for a journey of

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