In a flight of 3000 km, an aircraft was slowed down by bad weather. Its average speed for the trip was reduced by 100 km/hour and the time increased by one hour. Find the original duration of the flight.

Let initial speed be $$x$$ km/hr, => New speed = $$(x-100)$$ km/hr

Let original duration of flight = $$t$$ hours and actual time taken = $$(t+1)$$ hours

Total distance = $$xt=3000$$ ------------(i)

Speed is inversely proportional to time,

=> $$\frac{x}{x-100}=\frac{t+1}{t}$$

=> $$xt=xt+x-100t-100$$

=> $$x-100t=100$$

=> $$\frac{3000}{t}-100t=100$$

=> $$-100t^2-100t+3000=0$$

=> $$t^2+t-30=0$$

=> $$(t+6)(t-5)=0$$

=> $$t=5,-6$$

$$\because$$ Time cannot be negative, => Original duration = 5 hours

=> Ans - (A)