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Quadratic Equations
RD Sharma
CBSE
Grade 10

Question:

In a flight of $600\text{ km}$, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by $200\text{ km/h}$ and the time of flight increased by $30$ minutes. Find the original duration of the flight.

Step-by-Step Solution

Key Concept: Original speed $= v$. $\\dfrac{600}{v - 200} - \\dfrac{600}{v} = \\dfrac{1}{2} \Rightarrow 600\left(\dfrac{200}{v(v - 200)}\right) = \\dfrac{1}{2} \Rightarrow v^2 - 200v - 240000 = 0 \Rightarrow (v - 600)(v + 400) = 0 \Rightarrow v = 600\text{ km/h}$. Duration $= 600/600 = 1$ hour.
$\\dfrac{600}{v - 200} - \\dfrac{600}{v} = \\dfrac{1}{2}$. [1.5 Marks]
$v^2 - 200v - 240000 = 0 \Rightarrow (v - 600)(v + 400) = 0 \Rightarrow v = 600\text{ km/h}$. [2.0 Marks]
Original duration $= \dfrac{600}{600} = 1$ hour. [1.5 Marks]

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🎯 Official CBSE Marking Scheme:
Forming equation $\dfrac{600}{v - 200} - \dfrac{600}{v} = \dfrac{1}{2}$: 1.5 Marks
Solving $v = 600\text{ km/h}$: 2.0 Marks
Finding duration $= 1$ hour: 1.5 Marks

Correct Answer:
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