**Q) In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100 km/h and by doing so, the time of flight is increased by 30 minutes. Find the original duration of the flight.**

**Ans: **

Let’s consider original Speed of the flight be S and original duration of flight be T hours

We know that Speed = Distance / Time

Distance = Speed x Time

∴ 2800 = S x T

∴ S = ………… (i)

Next, by given condition, if speed is reduced by 100 km/h, the flight time is increased by 30 minutes (= hours),

∴ 2800 = (S – 100) x ( T + )

∴ 2800 = ( – 100) ( T + )

∴ 28 = ( – 1)(T + )

∴ 28 =

∴ 56 T = (28 – T) (2 T + 1)

∴ 56 T = 56 T – 2 T^{2} + 28 – T

∴ 2 T^{2} + T – 28 = 0

∴ 2 T^{2 }+ 8 T – 7 T – 28 = 0

∴ 2 T(T + 4) – 7 (T + 4) = 0

∴ (T + 4) (2 T – 7) = 0

∴ T = -4, T =

Here we reject T = – 4 because time can not be negative, and accept T =

**Therefore the original duration of the fight is hours or 3 hours and 30 minutes.**

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