Question

a plane left 30 minutes late than its scheduled time and in order ti reach the distination 1500 km away in time it had to increase its speed by 100 km hr from itsusual 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 + 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