In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 `(km)/(hr)` and the time of flight increased by 30 minutes. What is the duration of the flight?

Options

A. 2 hours
B. `1 1/2 hours`
C. `1/2 hours`
D. 1 hour

Answer

To solve the problem, we need to find the duration of the flight given the distance, the reduction in speed, and the increase in time.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let the original time of flight be \( t \) hours.
- The distance of the flight is given as \( 600 \) km.
- The original speed of the aircraft can be expressed as \( \frac{600}{t} \) km/h.

2. **Calculate the Reduced Speed:**
- The average speed is reduced by \( 200 \) km/h due to bad weather. Therefore, the new speed is:
\[ \text{New Speed} = \frac{600}{t} - 200 \]

3. **Determine the Increased Time:**
- The time of flight increased by \( 30 \) minutes, which is \( \frac{1}{2} \) hour. Thus, the new time of flight is:
\[ \text{New Time} = t + \frac{1}{2} \]

4. **Set Up the Equation:**
- The distance remains the same (600 km), so we can set up the equation using the new speed and new time:
\[ \text{Distance} = \text{New Speed} \times \text{New Time} \]
Substituting the values:
\[ 600 = \left( \frac{600}{t} - 200 \right) \left( t + \frac{1}{2} \right) \]

5. **Expand the Equation:**
- Expanding the right side:
\[ 600 = \left( \frac{600}{t} - 200 \right) \left( t + \frac{1}{2} \right) \]
\[ = \frac{600(t + \frac{1}{2})}{t} - 200(t + \frac{1}{2}) \]
\[ = \frac{600t + 300}{t} - 200t - 100 \]
\[ = 600 + \frac{300}{t} - 200t - 100 \]
\[ = 500 + \frac{300}{t} - 200t \]

6. **Rearranging the Equation:**
- Now we can rearrange the equation:
\[ 600 = 500 + \frac{300}{t} - 200t \]
\[ 100 = \frac{300}{t} - 200t \]
\[ 200t^2 + 100t - 300 = 0 \]

7. **Simplifying the Quadratic Equation:**
- Dividing the entire equation by \( 100 \):
\[ 2t^2 + t - 3 = 0 \]

8. **Using the Quadratic Formula:**
- The quadratic formula is given by \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = 1, c = -3 \):
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} \]
\[ = \frac{-1 \pm \sqrt{1 + 24}}{4} \]
\[ = \frac{-1 \pm 5}{4} \]
This gives us two possible solutions:
\[ t = 1 \quad \text{(taking the positive root)} \]
\[ t = -\frac{3}{2} \quad \text{(not valid as time cannot be negative)} \]

9. **Conclusion:**
- The original duration of the flight is \( 1 \) hour. Therefore, the duration of the flight is:
\[ \text{Duration of the flight} = 1 \text{ hour} \]