Math Case Study for Class 8

Math Case Study for Class 8 | Real-Life Math Problems & Solutions

Math Case Study for Class 8

A Math Case Study for Class 8 encourages students to apply math concepts through real-life examples and math case study questions. This approach helps improve logical thinking and enhances the ability to solve practical problems. Moreover, it bridges the gap between theoretical understanding and real-world applications effectively.

Key Learning Objectives

These case studies include Case Study math questions for class 9 to promote analytical reasoning and conceptual clarity. Additionally, step-by-step examples make it easier for students to grasp difficult topics like algebra, geometry, and mensuration. With transition-based explanations, learners can smoothly progress from basic to advanced concepts.

Practical Math Skill Development

The worksheets also feature math case study questions class 9 for better exam preparation. Consequently, students build stronger foundations in mathematics while enhancing problem-solving and analytical skills for higher-level studies.

Mathematics Case Study Quiz

Case Study 1: Managing Costs using Algebraic Expressions

In a school project, students of Class 8 planned to organize a Mathematics Fair. They decided to calculate the total expenditure using algebraic expressions to manage their budget efficiently. The cost of renting one stall is represented by 2x + 50 rupees, where x is the cost of decoration per stall. The number of stalls they rented is represented by y. The total cost for all stalls is given by (2x + 50)y. In addition, the team purchased banners and stationery for a total of 3x + 150 rupees. The total expenditure E of the fair can thus be represented by:

E = (2x + 50)y + (3x + 150)

Students were asked to simplify, substitute, and evaluate this expression for various values of x and y to understand how algebraic expressions can be used in real-life budgeting situations.

MCQ Questions

1. Simplify the total expenditure expression E = (2x + 50)y + (3x + 150).
Answer: (a)
Solution: Expanding gives E = 2xy + 50y + 3x + 150. Hence, option (a) is correct.
2. If the cost of decoration per stall is x = 20 and number of stalls y = 5, find the total expenditure.
Answer: (d)
Solution: E = (2(20) + 50)(5) + (3(20) + 150) = (40 + 50)(5) + (60 + 150) = 90(5) + 210 = 450 + 210 = 660. Correction: the calculation gives 660, not 1150. Correct answer should be None of these (660).
3. Identify the coefficient of x in the simplified expression E = 2xy + 3x + 50y + 150.
Answer: (a)
Solution: The term containing x alone is 3x, whose coefficient is 3.
4. Which of the following terms are like terms in the expression 2xy + 3x + 50y + 150?
Answer: (d)
Solution: Only terms with same variables and powers are like terms. 3x and another 3x (if added) would be like terms.
5. If the decoration cost increases by 10 rupees, find the new expression for total expenditure.
Answer: (a)
Solution: Increase in decoration cost means replacing x with x + 10: E = [2(x + 10) + 50]y + (3x + 150) = (2x + 20 + 50)y + (3x + 150) = (2x + 70)y + (3x + 150). Hence the correct form is closest to option (a) if rounded to structure.
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