Question

An aeroplane was late by 30 minutes due to some unavoidable circumstances and in order to reach the destination 1500km away in time the pilot increased the speed of the plane by 250km/hr.find the original speed of the aeroplane

Answer 1

Let the usual time taken by the aeroplane = x km/hr
Distance to the destination = 1500 km

Case (i)
Speed = Distance / Time = (1500 / x) Hrs

Case (ii)
Time taken by the aeroplane = (x - 1/2) Hrs

Speed = Distance / Time = 1500 / (x - 1/2) Hrs

Increased speed = 250 km/hr

⇒ [1500 / (x - 1/2)] - [1500 / x] = 250

⇒ 1/(2x2 - x) = 1/6

⇒ 2x2 - x = 6

⇒ (x - 2)(2x + 3) = 0

⇒ x = 2 or -3/2

Since, the time can not be negative,

The usual time taken by the aeroplane = 2 hrs

and the usual speed = (1500 / 2) = 750 km/hr.

Answer 2

Solution :-

Let the original speed of train be x km/hr

New speed = (x + 250) km/hr

We know that,

Time = Distance / Speed

Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.

According to the question,

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

=> (1500x + 37500 - 1500x)/x(x + 250) = 1/2

=> 2(37500) = x(x + 250)

=> 75000 = x² + 250x

=> x² + 250x - 75000 = 0

=> x² + 1000x - 750x - 75000 = 0

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

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

=> x = 750 or x = - 1000

∴ x ≠ - 1000 (Because speed can't be negative)

Hence,

Its usual speed = 750 km/hr