A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.
Solution
Let the usual speed of plane be x km/hr. Then,
Increased speed of the plane = (x + 400)km/hr
Time taken by the plane under usual speed to cover 1600km = `1600/x`hr
Time taken by the plane under increased speed to cover 1600km = `1600/(x+400)`hr
Therefore,
`1600/x-1600/(x+400)=40/60`
`(1600(x+400)-1600x)/(x(x+400))=2/3`
`(1600x+640000-1600x)/(x^2+400x)=2/3`
`640000/(x^2+400x)=2/3`
640000(3) = 2(x2 + 400x)
1920000 = 2x2 + 800x
2x2 + 800x - 1920000 = 0
2(x2 + 400x - 960000) = 0
x2 + 400x - 960000 = 0
x2 - 800x + 1200x - 960000 = 0
x(x - 800) + 1200(x - 800) = 0
(x - 800)(x + 1200) = 0
So, either
x - 800 = 0
x = 800
Or
x + 1200 = 0
x = -1200
But, the speed of the plane can never be negative.
Hence, the usual speed of train is x = 800 km/hr
