Question

How Do I solve Q16 ??

Answer 1

Answer:

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