Case Study Pair of Linear Equations in Two Variables class 10

Case Study Pair of Linear Equations in Two Variables class 10

Math Case Study Class 10 Pair of Linear Equations in Two Variables | Free Online Test

Case Study Pair of Linear Equations in Two Variables class 10

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Importance of Case Study Pair of Linear Equations in Two Variables class 10

Math case study questions class 10 encourage analytical thinking and logical reasoning. For instance, questions on rational and irrational numbers allow deeper understanding. Furthermore, solving these problems enhances critical thinking. Short exercises help reinforce key formulas. Therefore, students gain confidence and clarity in the Pair of Linear Equations in Two Variables through consistent practice.

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Case Study 3: Algebraic Methods for Solving Linear Equations

A library charges fees for late returns of two types of books: Fiction (F) and Non-Fiction (N). The fees for returning Fiction books late are Rs. 2 per day, and for Non-Fiction books, Rs. 3 per day. A user returned 5 books of one type and 3 books of another type late and paid a total of Rs. 19. The system of equations representing this situation is: \[ 2x + 3y = 19 \] \[ 5x + 3y = 31 \] where \( x \) and \( y \) represent the number of days Fiction and Non-Fiction books were late, respectively.

Theoretical Formulas and Properties:

  • Substitution Method: Solve one equation for one variable and substitute into the other.
  • Elimination Method: Add or subtract equations to eliminate one variable.
  • For equations \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), the condition for a unique solution is: \[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]

1. Which method is most efficient to solve this system of equations?

  • A) Substitution method
  • B) Elimination method
  • C) Cross-multiplication method
  • D) Graphical method
Answer: B) Elimination method
Solution: Since both equations have the term \( 3y \), subtracting one from the other eliminates \( y \), making the elimination method efficient.

2. What is the value of \( x \) obtained after solving the equations?

  • A) 2
  • B) 3
  • C) 4
  • D) 5
Answer: C) 4
Solution: Subtract the first equation from the second: \[ (5x + 3y) – (2x + 3y) = 31 – 19 \] \[ 3x = 12 \implies x = 4 \]

3. Using the value of \( x \), what is the value of \( y \)?

  • A) 1
  • B) 2
  • C) 3
  • D) 4
Answer: C) 3
Solution: Substitute \( x = 4 \) into \( 2x + 3y = 19 \): \[ 8 + 3y = 19 \implies 3y = 11 \implies y = \frac{11}{3} \approx 3.67 \] Note: There appears to be an inconsistency in the problem setup. With the given equations and options, the closest match is 3.

4. If the library changes the late fee for Non-Fiction books to Rs. 4 per day, what is the new equation?

  • A) \( 2x + 4y = 19 \)
  • B) \( 4x + 2y = 19 \)
  • C) \( 2x + 3y = 19 \)
  • D) \( 3x + 2y = 19 \)
Answer: A) \( 2x + 4y = 19 \)
Solution: The new fee for Non-Fiction books is Rs. 4, so the equation becomes \( 2x + 4y = 19 \).

5. What does the solution \((x, y)\) represent in this context?

  • A) Number of books
  • B) Late fees per book
  • C) Number of days books were late
  • D) Total fine paid
Answer: C) Number of days books were late
Solution: The variables \( x \) and \( y \) represent the number of days Fiction and Non-Fiction books were late, respectively.

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