An aeroplane take 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.
Solution
Let the usual speed of aero plane be x km/hr. Then,
Increased speed of the aero plane = (x + 100)km/hr
Time taken by the aero plane under usual speed to cover 1200 km = `1200/x`hr
Time taken by the aero plane under increased speed to cover 1200 km = `1200/(x+100)hr`
Therefore,
`1200/x-1200/(x+100)=1`
`(1200(x+100)-1200x)/(x(x+100x))=1`
`(1200x+120000-1200x)/(x^2+100x)=1`
`12000/(x^2+100x)=1`
120000 = x2 + 100x
x2 + 100x - 120000 = 0
x2 - 300x + 400x - 120000 = 0
x(x - 300) + 400(x - 300) = 0
(x - 300)(x + 400) = 0
So, either
x - 300 = 0
x = 300
Or
x + 400 = 0
x = -400
But, the speed of the aero plane can never be negative.
Hence, the usual speed of train is x = 300 km/hr
