Question

A plane left 30min later then its schsduled tym and in order to reach the destination 1500km away in time it had to increase its speed by 100km/hr. Find d usual speed

Answer 1

Sol: 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 Distance to the destination = 1500 km 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 + 100) 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 + 100) = 1/2

=> (1500x + 15000 - 1500x)/x(x + 100) = 1/2

=> 2(15000) = x(x + 100)

=> 30000 = x² + 100x

=> x² + 100x - 30000 = 0

=> x² + 600x - 500x - 30000 = 0

=> x(x + 600) - 500(x + 600) = 0

=> (x - 500) (x + 600) = 0

=> x = 500 or x = - 600

∴ x ≠ - 600 (Because speed can't be negative)

Hence,

Its usual speed = 500 km/hr