Math Problem Statement

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)?

Solution

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: Timenormal=1200S\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: Timebad=1200S200\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: TimebadTimenormal=0.5\text{Time}_\text{bad} - \text{Time}_\text{normal} = 0.5 Substituting the expressions for time, we get: 1200S2001200S=0.5\frac{1200}{S - 200} - \frac{1200}{S} = 0.5

Solve for SS:

We can solve this equation algebraically.

Multiply both sides by S(S200)S(S - 200) to eliminate the denominators: 1200S1200(S200)=0.5S(S200)1200S - 1200(S - 200) = 0.5 \cdot S(S - 200) Simplify: 1200S1200S+240000=0.5(S2200S)1200S - 1200S + 240000 = 0.5(S^2 - 200S) 240000=0.5(S2200S)240000 = 0.5(S^2 - 200S) Multiply by 2 to clear the fraction: 480000=S2200S480000 = S^2 - 200S Rearrange into a quadratic equation: S2200S480000=0S^2 - 200S - 480000 = 0

Solving the quadratic equation:

Using the quadratic formula: S=(200)±(200)24(1)(480000)2(1)S = \frac{-(-200) \pm \sqrt{(-200)^2 - 4(1)(-480000)}}{2(1)} S=200±40000+19200002S = \frac{200 \pm \sqrt{40000 + 1920000}}{2} S=200±19600002S = \frac{200 \pm \sqrt{1960000}}{2} S=200±14002S = \frac{200 \pm 1400}{2}

So, we have two possible solutions: S=200+14002=800orS=20014002=600S = \frac{200 + 1400}{2} = 800 \quad \text{or} \quad S = \frac{200 - 1400}{2} = -600

Since speed cannot be negative, we have: S=800km/hS = 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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Time and Speed Relations

Formulas

Time = Distance / Speed
Quadratic equation formula ax^2 + bx + c = 0

Theorems

Quadratic formula

Suitable Grade Level

Grades 9-12