A shopkeeper sells each item for \(x\) rupees and sells a total of \(80-x\) items. If his total revenue is 1500 rupees, what is the smaller value of \(x\)?
Answer and explanation
Correct answer: 30
Total revenue equals price per item multiplied by the number of items, so \(x(80-x)=1500\). Rearranging gives \(x^2-80x+1500=0\), or \((x-30)(x-50)=0\). Thus, \(x=30\) or \(x=50\); the smaller value is 30. In such problems, first form the revenue equation and then compare both roots.
Frequently asked questions
What is the correct answer to this question?
30
Why is this the correct answer?
Total revenue equals price per item multiplied by the number of items, so \(x(80-x)=1500\). Rearranging gives \(x^2-80x+1500=0\), or \((x-30)(x-50)=0\). Thus, \(x=30\) or \(x=50\); the smaller value is 30. In such problems, first form the revenue equation and then compare both roots.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Word Problems and Applications.
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