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

Step 1: Define variables and formulate equations

We need to find the original duration of the flight. Let's denote the original speed of the aircraft as $v$ km/hr and the original time taken as $t$ hours. The total distance of the flight is given as 600 km.

Step 2: Set up equations based on given information

The fundamental relationship between distance, speed, and time is $\text{Distance} = \text{Speed} \times \text{Time}$. So, for the original flight, we have $v \cdot t = 600$. Due to bad weather, the speed was reduced by 200 km/hr, making the new speed $v - 200$. The time of flight increased by 30 minutes, which is 0.5 hours, so the new time is $t + 0.5$. The distance remains 600 km, leading to the second equation: $(v - 200)(t + 0.5) = 600$.

Step 3: Substitute and simplify the equations

From equation (1), we can express $v$ as $v = \frac{600}{t}$. Substitute this expression for $v$ into equation (2). This will give us an equation solely in terms of $t$. Expand the equation and simplify.

Step 4: Form a quadratic equation in terms of t

Continue simplifying the equation. Multiply the entire equation by $t$ to eliminate the fraction, and then rearrange the terms to form a standard quadratic equation of the form $at^2 + bt + c = 0$. Divide by -100 to simplify the coefficients.

Step 5: Solve the quadratic equation for t

We have a quadratic equation $2t^2 + t - 3 = 0$. We can solve this using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=2$, $b=1$, and $c=-3$. Calculate the two possible values for $t$.

Step 6: Determine the valid duration

We get two possible values for $t$: $1$ hour and $-1.5$ hours. Since time cannot be negative, we discard the negative solution. Therefore, the original duration of the flight is 1 hour.

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