An aeroplane lefts 30 minute later than it's shedule time and in order to reach its destination 1500 km away in time. It had to increase its speed by 250 km/hr. from its usual speed. Determine its usual time.

Let the usual speed of the plane be 'x' hours

Time taken to cover 1500km with usual speed = 1500/x hrs

time taken to cover 1500km with speed of (x+250) = 1500/(x+250) hrs

1500/x = 1500/(x+250) + 1/2

1500/x - 1500/(x+250) = 1/2

1500x + 1500*250 - 1500x / (x*(x+250)) = 1/2

1500*250 / x²+250x = 1/2

750000 = x² + 250x

x²+250x - 750000 = 0

x² + 1000x - 750x - 750000 = 0

x ( x+1000) - 750 (x+1000) =0

(x+1000)(x-750) = 0

which means x = -1000 , which is not possible , as speed cannot be negative

or x = 750

so the usual speed of the plane is 750 kmph

usual time = distance / speed = 1500 / 750 = 150/75 = 2 hours !

Please approve !

Approved