Linear Equations in One Variable Case Study Class 8 PDF Download

Linear Equations in One Variable Case Study Class 8 PDF

A linear equations in one variable case study class 8 PDF helps students apply mathematical concepts in real-life situations. It includes math case study questions that improve logical reasoning and understanding of algebra. Moreover, it strengthens problem-solving skills essential for exams.

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Practicing linear equations in one variable case study questions builds a solid foundation for higher classes. It prepares students for Case Study math questions for class 9 and math case study questions class 9. Additionally, it enhances analytical thinking and clarity in equations.

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Case Study 4: Dividing Prize Money with Fractional Relationships

A mathematics club organized a competitive quiz and awarded a total cash prize of Rs. 960 to be shared among three winners: Asha, Bhavik and Chirag. The club announced that Asha should receive twice the amount Bhavik receives because Asha answered more high-value questions correctly. Additionally, Chirag is to receive one quarter of the combined amount received by Asha and Bhavik together, since he contributed in tie-break rounds. The organizers prefer to set up a single linear equation in one variable to determine each person’s share, using Bhavik’s share as the unknown. This case requires forming a precise linear equation from the word description, performing transposition and balancing to isolate the variable, checking that all shares are non-negative integers, and interpreting the result in context. Students must also verify the solution by summing the shares to confirm they equal the total prize. The problem illustrates how fractional relationships and proportional statements reduce to simple linear equations in one variable, and highlights the importance of correct arithmetic during simplification.

1. If Bhavik’s share is \(x\) rupees, what is Asha’s share in terms of \(x\)?
Solution:

It is given that Asha receives twice Bhavik’s amount, so Asha’s share \(=2x\).

2. If Asha and Bhavik together receive \(3x\), what is Chirag’s share according to the statement “Chirag receives one quarter of the combined amount of Asha and Bhavik”?
Solution:

Combined amount of Asha and Bhavik \(=2x+x=3x\). One quarter of this is \( \dfrac{1}{4}\times 3x = \dfrac{3x}{4}\).

3. Which linear equation correctly represents the total prize distribution?
Solution:

Asha \(=2x\), Bhavik \(=x\), Chirag \(=\dfrac{3x}{4}\). Sum must equal Rs. 960, hence \(2x + x + \dfrac{3x}{4} = 960\).

4. Solve the equation \(2x + x + \dfrac{3x}{4} = 960\) to find \(x\).
Solution:

Combine terms: \(3x + \dfrac{3x}{4} = \dfrac{12x + 3x}{4} = \dfrac{15x}{4}\). So \(\dfrac{15x}{4} = 960\). Multiply both sides by 4: \(15x = 3840\). Divide by 15: \(x = \dfrac{3840}{15} = 256\).

5. Using \(x = 256\), what are the actual shares of Asha, Bhavik and Chirag respectively?
Solution:

Bhavik \(=x=256\). Asha \(=2x=512\). Chirag \(=\dfrac{3x}{4}=\dfrac{3\times 256}{4}=3\times 64=192\). Check sum: \(512+256+192=960\), which matches the total prize.

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