Question

An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.

Answer 1

Let the usual speed of aeroplane be x km/hr. Then,

Increased speed of the aeroplane = (x + 250) km/hr

Time taken by the aeroplane under usual speed to cover 1250 km = `1250/x` hr

Time taken by the aeroplane under increased speed to cover 1250 km = `1250/(x +250)` hr

Therefore,

`1250/x-1250/(x+250)=50/60`

`(1250(x+250)-1250x)/(x(x+250))=5/6`

`(1250x+312500-1250x)/(x^2+250x)=5/6`

`312500/(x^2+250x)=5/6`

312500(6) = 5(x² + 250x)

1875000 = 5x² + 1250x

5x² + 1250x - 1875000 = 0

5(x² + 250x - 375000) = 0

x² + 250x - 375000 = 0

x² - 500x + 750x - 375000 = 0

x(x - 500) + 750(x - 500) = 0

(x - 500)(x + 750) = 0

So, either 

x - 500 = 0

x = 500

Or

x + 750 = 0

x = -750

But, the speed of the aeroplane can never be negative.

Hence, the usual speed of train is x = 500 km/hr.