Question

In a Flight of 600 km and the aircraft was float down due to bad weather its average speed for the trip was reduce by 200 km per hour and a time of flight increased by 30 minutes

Answer 1

Let duration of the flight was X hours.
Increased time = 30 minutes = 1/2 hours.
Now,
(600 /X ) - {600 /(X + (1/2))} = 200.

X*(2X +1) = 3
2X^2 +X = 3
2X^2 + X - 3 = 0

(2X + 3) (X-1) = 0
On Solving, We get,
X = - 3 , 1.
X is the time so it can not be negative.
X = 1 hours.

Hope this helps.....
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Answer 2

original time =xkm/hr
increased time=x+1/2km/hr
distance =600km

600/x-600/x+1/2=200
600(1/x-1/x+1/2)=200
1/x-2/2x+1=200/600
2x+1-2x/2x²+x=1/3
1/2x²+x=1/3
3=2x²+x
2x²+x-3=0
2x²+3x-2x-3=0
x(2x+3)-1(2x+3)=0
(x-1)(2x+3)=0

x=1
duration of the flight is 1hour.