while boarding an aeroplane, a passenger got hurt. The pilot, showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 min. To reach the destination, 1500 km away, in time, the pilot increased the speed by 100km/hr. Find the original speed /hr of the plane.
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let the speed be x km/hr
distance given = 1500 km
time= distance/speed
T1 (time 1) =1500/x
T2 (time 2) = 1500/x+100
acc to the ques
T1-(30/60)hr (late by 30 mins) = T2
=1500/x -1/2 = 1500/x+100
=1500/x - 1500/x+100 = 1/2
=1500x + 1,50,000 -1500x = 1/2
x²+100x
=1,50,000= 1/2 X ( x²+ 100x )
2 X 1,50,000 = x² + 100x
3,00,000 = x² + 100x
x² + 100x - 3,00,000 = 0
splitting the middle term
x² + 600x -500x -3,00,000 = 0
x ( x + 600 ) - 500 ( x + 600 ) = 0
( x - 500 ) ( x + 600 ) = 0
x = 500 or x = - 600
since speed cannot be negative hence we consider x = 500
therefore x = original speed = 500 km/hr
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