Class 8 Math Case Study on Linear Equations in One Variable | CBSE Guide

Class 8 Math Case Study on Linear Equations in One Variable

A class 8 math case study on linear equations in one variable helps students understand how algebra connects to real-life situations. It focuses on solving equations that contain only one unknown variable, improving logical and analytical skills. Moreover, students learn structured problem-solving through math case study questions.

Importance of Linear Equations in Real Life

Linear equations in one variable are essential for future topics like algebra and geometry. They prepare students for Case Study math questions for class 9 and math case study questions class 9. Additionally, these problems encourage accuracy, reasoning, and conceptual clarity through simple yet practical examples.

Practice and Application

To master math case study questions, students should solve NCERT exercises regularly. Furthermore, consistent practice builds confidence and makes learning enjoyable while strengthening mathematical foundations.

Case Study 2: Mixture Problem — Preparing a Cost-Effective Rice Blend

A small grocery shop wants to prepare a 100 kilogram mixture of two varieties of rice. The first variety (Type A) costs Rs. 36 per kg and the second variety (Type B) costs Rs. 60 per kg. The shopkeeper aims to sell the final 100 kg mixture at Rs. 48 per kg so that the selling price equals the weighted average cost. To maintain consistency and simplicity in accounting, he decides to mix an integral number of kilograms of each type. Using the idea of forming linear equations in one variable, the shopkeeper models the situation by taking the quantity of Type A rice as the unknown variable and expressing the total cost of the mixture in terms of that variable. The problem requires forming the correct linear equation, simplifying it using transposition (balancing method), solving for the unknown, and interpreting the solution practically (i.e., how many kilograms of each rice type are used). While solving, one must check arithmetic carefully and ensure the computed quantities are realistic (non-negative and integral).

1. If we let \( x \) denote the kilograms of Type A rice used in the mixture, then the kilograms of Type B rice will be:
Solution:

Total mixture is 100 kg. If \(x\) kg is Type A, the remaining mass for Type B is \(100-x\).

2. The correct linear equation representing the total cost of the mixture (so that average cost is Rs. 48/kg) is:
Solution:

Total cost = cost of Type A \(+\) cost of Type B. The required total cost equals \(48\) rupees per kg times \(100\) kg, hence \(36x + 60(100-x)=4800\).

3. Simplify the equation \(36x + 60(100 – x) = 4800\).
Solution:

Expanding gives \(36x + 6000 – 60x = 4800\). Combine like terms: \(-24x + 6000 = 4800\). Subtract 6000 from both sides: \(-24x = -1200\).

4. What is the value of \( x \)?
Solution:

From \(-24x = -1200\), divide both sides by \(-24\): \(x = \dfrac{-1200}{-24} = 50\). Thus Type A rice used is \(50\) kg.

5. How many kilograms of Type B rice are used in the mixture?
Solution:

Since total is 100 kg and Type A is \(50\) kg, Type B \(=100-50=50\) kg.

Review your answers and solutions below:

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