Question

an aeroplane left 30 minutes later than IIT secluded time and in order to reach is destination 1500 km away in time it has to increase its speed by 250 km per hour from its usual speed determine its usual speed

Answer 1

Let the 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 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