Let the original speed = x km/hr
∴ Normal time of flight = \(\frac{2800}{x}\) hours
∴ Time of flight, when speed is reduced = \(\frac{2800}{(x-100)}\) hours
According to the given condition, \(\frac{2800}{(x-100)}-\)\(\frac{2800}{x}\) = \(\frac{30}{60}=\frac12\)
Neglecting the –ve value.
Original speed of the air craft = 800 km/hr
Original duration of flight = \(\frac{2800}{800} = 3\frac12\) hours.