Question

A plane left 30 minutes late than it's scheduled time and in order to reach it's destination 1500 km away in time it has to increase it's speed by 100km/hr from it's usual speed. Find it's usual speed.

Answer 1

Let the usual time taken by the aeroplane = x km/hr

Distance to the destination = 1500 km

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

 

------------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 = 100 km/hr

 

[1500 / (x - 1/2)] - [1500 / x] = 100

150((10/x-1/2)-10/x0=100

15 /x -1/2- 15/x =1  

30x -30x +15=2x^2-x

15=2x^2-x

2x^2-x-15=0

2x^2 - 6x+5x-15=0

2x(x-3)+5(x-3)

2x+5=0  or x-3 =0

x=-5/2   or x=3

Since, the time can not be negative,

The usual time taken by the aeroplane = 3hrs

and the usual speed = (1500 / 3) = 500km/hr

hope this helped

#bebrainly

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