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In a flight of $2800 \text{ km}$, an aircraft was slowed down due to bad weather. Its average speed is reduced by $100 \text{ km/h}$ and by doing so, the time of flight is increased by $30$ minutes. Find the original duration of the flight.
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Let original speed of aircraft be $x \text{ km/hr}$.
A.T.Q.
$\frac{2800}{x-100} - \frac{2800}{x} = \frac{1}{2}$
$\Rightarrow x^2 - 100x - 560000 = 0$
$\Rightarrow (x-800)(x + 700) = 0$
$x \neq -700$ So, $x = 800$
Original Duration $\frac{2800}{800} = \frac{7}{2}$ hrs or $3$ hrs $30$ min.
A.T.Q.
$\frac{2800}{x-100} - \frac{2800}{x} = \frac{1}{2}$
$\Rightarrow x^2 - 100x - 560000 = 0$
$\Rightarrow (x-800)(x + 700) = 0$
$x \neq -700$ So, $x = 800$
Original Duration $\frac{2800}{800} = \frac{7}{2}$ hrs or $3$ hrs $30$ min.