Question

a plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time ,it increase its speed by 250km/hr from its usual speed .find its usual speed?

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
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 + 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