The passenger train takes 5 hrs less for a journey of 360km . If its speed is increased by 50 kmph from its normal speed . The normal speed( to the nearest kmph) is
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Let the normal speed be X
Distance = 360 km
Speed = distance/time
X = 360/time
Time = 360/X
Increased speed = x+50
Time = 360/x+50
As per question
360/X - 360/X+50 = 5
360(X + 50 - X)/ X^2+ 50X = 5
X^2 +50X - 3600 =0
X = -d +- root of b^2 -4ac / 2a
X = -50 +- 130 /2
X = 40 or -140
Speed can't be in negative so
Original speed of the train was 40km/h
Hope it helped you
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