Question

In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:

Answer 1

The duration of the flight is 1 hourLet the duration of the flight be x hours.Then,600/x-600/x + (1/2)= 200600/x-12002x + 1/= 200 x(2x + 1) = 3 2x2 + x - 3 = 0 (2x + 3)(x - 1) = 0 x = 1 hr. 

Answer 2

Explanation:

distance= 600km

let time= x

delayed time =x+1/2

                     =$\frac{2x+1}{2}$

original speed= 600/x

new speed=600÷(2x+1/2)= 1200/2x+1

600/x - 1200/2x+1 = 200

600 (1/x -2/2x+1)=200       [600 and 200 cancels]

3  ( $\frac{2x+1-2x}{2x^{2}+x }$ )  =1                 [2x and -2x cancels]

3 = $2x^{2} +x$

2x^2 +x -3=0

Solving this eq. we get solutions as 1 and -3/2

since, time cannot be -ve

duration= 1hr.