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(x^{2} + 400x)

1920000 = 2x^{2} + 800x

2x^{2} + 800x - 1920000 = 0

2(x^{2} + 400x - 960000) = 0

x^{2} + 400x - 960000 = 0

x^{2} - 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