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Let's break down the problem step by step.

### Given:
- **Ananya's Flight (Flight 1)** travels from Point A to Point C (distance: 1200 km).
  - Due to bad weather, the speed of the flight is reduced by **200 km/h**, and the time increases by **30 minutes**.
  
- **Madhuri's Flight (Flight 2)** travels from Point B to Point C (distance: 1800 km).
  - The average speed of Flight 2 is **720 km/h**.
  
- The question asks us to find the **actual speed** of Flight 1 in **good weather** (i.e., its normal speed).

### Approach:
Let the actual (good weather) speed of Flight 1 be denoted as **S** (in km/h).

1. **Time taken by Flight 1 under normal conditions**:
   \[
   \text{Time}_\text{normal} = \frac{1200}{S}
   \]
   
2. **Time taken by Flight 1 in bad weather**:
   Since the speed is reduced by 200 km/h, the bad weather speed is **S - 200 km/h**. So, the time taken in bad weather:
   \[
   \text{Time}_\text{bad} = \frac{1200}{S - 200}
   \]
   
3. **Time difference**:
   The time difference between normal and bad weather is **30 minutes**, which is **0.5 hours**. Therefore:
   \[
   \text{Time}_\text{bad} - \text{Time}_\text{normal} = 0.5
   \]
   Substituting the expressions for time, we get:
   \[
   \frac{1200}{S - 200} - \frac{1200}{S} = 0.5
   \]

### Solve for \( S \):
We can solve this equation algebraically.

Multiply both sides by \( S(S - 200) \) to eliminate the denominators:
\[
1200S - 1200(S - 200) = 0.5 \cdot S(S - 200)
\]
Simplify:
\[
1200S - 1200S + 240000 = 0.5(S^2 - 200S)
\]
\[
240000 = 0.5(S^2 - 200S)
\]
Multiply by 2 to clear the fraction:
\[
480000 = S^2 - 200S
\]
Rearrange into a quadratic equation:
\[
S^2 - 200S - 480000 = 0
\]

### Solving the quadratic equation:
Using the quadratic formula:
\[
S = \frac{-(-200) \pm \sqrt{(-200)^2 - 4(1)(-480000)}}{2(1)}
\]
\[
S = \frac{200 \pm \sqrt{40000 + 1920000}}{2}
\]
\[
S = \frac{200 \pm \sqrt{1960000}}{2}
\]
\[
S = \frac{200 \pm 1400}{2}
\]

So, we have two possible solutions:
\[
S = \frac{200 + 1400}{2} = 800 \quad \text{or} \quad S = \frac{200 - 1400}{2} = -600
\]

Since speed cannot be negative, we have:
\[
S = 800 \, \text{km/h}
\]

### Conclusion:
The actual speed of Flight 1 (in good weather) is **800 km/h**.

Would you like further details, or do you have any specific questions?

---

Here are five related questions to think about:
1. What would happen if the time delay was different (e.g., 45 minutes instead of 30)?
2. How would the equation change if the distance between A and C were longer or shorter?
3. How does Madhuri's flight speed relate to Ananya's if both flights faced weather issues?
4. What would happen if both flights were affected by the weather?
5. How could you calculate the total time difference for both flights under the same weather conditions?

**Tip:** Always verify your units (km/h, hours, etc.) to ensure you're solving the problem correctly.

Consider three Airports A, B, and C. Two friends Ananya and Madhuri want to meet at Airport C. Ananya boarded Flight 1 from Point A to C which is 1200 km. Due to bad weather, Flight 1 slowed down, and the average speed was reduced by 200 km/h and the time increased by 30 minutes. Madhuri boarded Flight 2 from Point B to C, which is 1800 km, the average speed of Flight 2 is 720 km/h. What is the speed of flight 1 in km/hr if weather is good (Actual Speed of flight 1)?

This problem asks to find the actual speed of Flight 1 under good weather conditions. The solution involves quadratic equations and speed-time relations, with a step-by-step breakdown for high school students. It includes solving for time differences caused by weather conditions.

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