Quadratic Equations Case Study Class 10

Quadratic Equations Case Study Class 10

Quadratic Equations Case Study Class 10

Understanding Quadratic Equations Case Study Class 10 is essential for board exam preparation. These case study questions class 10 maths quadratic equations help students connect concepts with real-life problems. Since Class 10 is crucial, solving case study class 10 builds confidence and strengthens problem-solving skills. Moreover, these questions are part of class 10 maths chapter 4 case study questions.

Importance of Quadratic Equations Case Based Questions

Practicing case based questions improves logical thinking. Consequently, students learn to apply formulas beyond textbooks. Teachers often recommend cbse maths quadratic equations case study for revision before exams. In addition, such practice helps identify weak areas. By solving case studies, students enhance their ability to analyze situations effectively.

Preparation with Online Tests

Our free online tests on case study class 10 include detailed explanations. Hence, students can self-assess regularly. The case study questions class 10 maths provided cover diverse scenarios. Therefore, learners can boost their confidence and ensure accuracy in exams.

Case Study: Revenue and Profit using Quadratic Equations

Case Study 1

Rohit runs a small business that manufactures and sells handcrafted candles. He notices that the profit he earns depends on the number of candles he produces each day. After collecting data for a month, he figures out that the profit \( P \) (in rupees) he earns by selling \( x \) candles can be modeled by the quadratic equation: \[ P(x) = -2x^2 + 100x – 800 \] This equation helps him understand how to optimize production. The quadratic nature of this equation indicates that there is a maximum profit he can achieve, which corresponds to the vertex of the parabola. He also realizes that if he produces either too few or too many candles, the business will run at a loss. With this model, he wants to understand the most profitable number of candles to produce, break-even points, and interpret real-life decisions based on this mathematical model.

Important Formulas and Concepts:

  • Standard form of a quadratic equation: \( ax^2 + bx + c = 0 \)
  • Vertex of a parabola: \( x = \frac{-b}{2a} \)
  • Discriminant: \( D = b^2 – 4ac \)
  • Roots of the quadratic equation: \( x = \frac{-b \pm \sqrt{D}}{2a} \)

1. What is the number of candles Rohit should produce daily to earn maximum profit?

  • A) 10
  • B) 15
  • C) 25
  • D) 30
Answer: C) 25
Solution: For maximum profit, use vertex formula: \[ x = \frac{-b}{2a} = \frac{-100}{2 \times (-2)} = \frac{-100}{-4} = 25 \]

2. What is the maximum profit Rohit can earn?

  • A) Rs.450
  • B) Rs.600
  • C) Rs.425
  • D) Rs.500
Answer: A) Rs.450
Solution: Substitute \( x = 25 \) in the equation: \[ P(25) = -2(25)^2 + 100(25) – 800 = -1250 + 2500 – 800 = 450 \]

3. What are the break-even points where profit is zero?

  • A) 10, 40
  • B) 20, 30
  • C) 5, 80
  • D) 25, 45
Answer: A) 10, 40
Solution: Solve: \[ -2x^2 + 100x – 800 = 0 \Rightarrow x^2 – 50x + 400 = 0 \] \[ x = \frac{50 \pm \sqrt{(-50)^2 – 4 \cdot 1 \cdot 400}}{2} = \frac{50 \pm \sqrt{2500 – 1600}}{2} = \frac{50 \pm \sqrt{900}}{2} \] \[ x = \frac{50 \pm 30}{2} \Rightarrow x = 10 \text{ or } 40 \]

4. What is the nature of roots of the equation \( -2x^2 + 100x – 800 = 0 \)?

  • A) Real and equal
  • B) Real and distinct
  • C) Imaginary
  • D) Cannot be determined
Answer: B) Real and distinct
Solution: Discriminant \( D = 100^2 – 4 \cdot (-2) \cdot (-800) = 10000 – 6400 = 3600 > 0 \). Hence, roots are real and distinct.

5. Which of the following quadratic equations has equal roots?

  • A) \( x^2 + 4x + 5 = 0 \)
  • B) \( x^2 + 6x + 9 = 0 \)
  • C) \( x^2 + 2x + 3 = 0 \)
  • D) \( x^2 – 7x + 10 = 0 \)
Answer: B) \( x^2 + 6x + 9 = 0 \)
Solution: For equal roots, the discriminant \( D = b^2 – 4ac = 0 \)
(a) \( D = 4^2 – 4(1)(5) = 16 – 20 = -4 \) (not equal roots)
(b) \( D = 6^2 – 4(1)(9) = 36 – 36 = 0 \) equal roots
(c) \( D = 2^2 – 4(1)(3) = 4 – 12 = -8 \)
(d) \( D = (-7)^2 – 4(1)(10) = 49 – 40 = 9 \)

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