In a flight of 600kman aircraft …

Let the original speed of the aircraft be x km/hr.

Then, new speed = (x- 200) km/hr.

Duration of flight at original speed = {tex}{600 \over x}hrs{/tex}

Duration of flight at reduced speed = {tex}{600 \over x-200}hrs{/tex}

According to the question

{tex}{600 \over x-200} - {600 \over x} = {1 \over 2}{/tex}

{tex}\implies {600x - 600x +120000 \over x(x-200)} = {1 \over 2}{/tex}

{tex}\implies {120000 \over x^2 -200x} = {1 \over 2}{/tex}

{tex}\implies {/tex}x² - 200x -240000 = 0

{tex}\implies {/tex}x² - 600 x + 400 x - 240000 = 0

{tex}\implies {/tex}x (x - 600) + 400 (x - 600) = 0

{tex}\implies {/tex}(x- 600) (x + 400) = 0

Either x - 600 =0 or x + 400 = 0

{tex}\implies {/tex}x = 600, -400

since Speed cannot be negative. So x = 600

So, original speed of the aircraft was 600 km /hr.

Hence, duration of flight = {tex}{600 \over x}hrs = {600 \over 600}hrs = 1 hr{/tex}