Class 8 Linear Equations Case Study Questions with Answers

Class 8 Linear Equations Case Study Questions with Answers | Solved Worksheet

Class 8 Linear Equations Case Study Questions with Answers

A class 8 linear equations case study with answers helps students relate mathematical ideas to real-life situations. These math case study questions focus on equations in one variable and strengthen analytical thinking. Moreover, such practice improves accuracy and conceptual clarity.

Importance of Practicing Linear Equations

Solving linear equations case study questions builds logical reasoning and prepares students for Case Study math questions for class 9 and math case study questions class 9. Additionally, understanding solved examples enhances confidence and exam performance.

How to Prepare Effectively

To master math case study questions, students should practice NCERT-based worksheets regularly. Furthermore, solving daily practice papers and revising solved answers ensures consistent improvement in problem-solving skills.

Case Study 5: Discounted Sales and Profit Calculation at a Stationery Shop

Mr. Verma runs a neighbourhood stationery shop. He purchases a batch of 120 identical notebooks at a cost price of Rs. 45 each. For display, he marks each notebook at Rs. 60. During a festive sale he decided to sell some notebooks at the full marked price and the remaining notebooks at a discount of 20% on the marked price to attract more buyers. By the end of the day all 120 notebooks were sold and the total revenue collected was Rs. 6480. Mr. Verma wants to know how many notebooks were sold at the full marked price and how many at the discounted price. He also wants to calculate his total profit for the day. To solve this, students should assign a suitable variable, form a single linear equation based on the total number of notebooks and total revenue, use the transposition (balancing) method to isolate the variable, and then interpret the solution practically. While solving, it is important to check that the number of notebooks found is a non-negative integer and that the computed profit is consistent with the cost and revenue data.

1. What is the selling price of a notebook sold at 20% discount on the marked price of Rs. 60?
Solution:

A 20% discount on Rs. 60 reduces the price by \(0.20\times 60 = 12\). Thus discounted price \(=60-12=48\) rupees.

2. If \(x\) denotes the number of notebooks sold at full marked price, which equation correctly represents the total revenue being Rs. 6480?
Solution:

Revenue \(=\) (full-price sold)\(\times 60\) \(+\) (discounted sold)\(\times 48\). Since discounted sold \(=120-x\), equation is \(60x + 48(120-x)=6480\).

3. Simplify the equation \(60x + 48(120 – x) = 6480\) and find the value of \(x\).
Solution:

Expand: \(60x + 5760 – 48x = 6480\). Combine like terms: \(12x + 5760 = 6480\). Subtract 5760: \(12x = 720\). Divide by 12: \(x = 60\).

4. How many notebooks were sold at the discounted price?
Solution:

Discounted sold \(=120-x = 120-60 = 60\).

5. What is Mr. Verma’s total profit for the day?
Solution:

Total cost \(=120\times 45 = 5400\). Total revenue \(=6480\). Profit \(=6480-5400 = 1080\) rupees.

Review your answers and solutions below:

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