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Quadratic Equations
EXERCISE 4.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ` 90, find the number of articles produced and the cost of each article.

Step-by-Step Solution

Key Concept: Translate the word problem into algebraic equations, use the relation between the number of articles and the cost per article to form a quadratic equation, and solve it using factorisation or the quadratic formula. The physically meaningful (positive integer) root gives the required number of articles; substitute back to obtain the cost per article.
1. Introduce variables
Let \(n\) be the number of articles produced in a day.
Let \(c\) be the cost (in rupees) of producing one article.

2. Write the given relations
- Cost per article is "3 more than twice the number of articles":
$$c = 2n + 3 \tag{1}$$
- Total cost of production is \(90\) rupees:
$$\text{Total cost}= n \times c = 90 \tag{2}$$

3. Substitute (1) into (2)
$$n(2n + 3) = 90$$
$$2n^{2} + 3n - 90 = 0 \tag{3}$$

4. Solve the quadratic equation
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) for \(a=2,\; b=3,\; c=-90\):
$$\Delta = b^{2} - 4ac = 3^{2} - 4(2)(-90) = 9 + 720 = 729 = 27^{2}$$
$$n = \frac{-3 \pm 27}{2\times 2} = \frac{-3 \pm 27}{4}$$
- Positive root: \(n = \frac{-3 + 27}{4} = \frac{24}{4} = 6\)
- Negative root: \(n = \frac{-3 - 27}{4} = \frac{-30}{4} = -7.5\) (reject, because number of articles cannot be negative).

5. Find the cost per article using (1):
$$c = 2n + 3 = 2(6) + 3 = 12 + 3 = 15\text{ rupees}$$

6. Verification
Total cost = \(n \times c = 6 \times 15 = 90\) rupees, which matches the given total cost.

Hence, the cottage industry produced 6 articles, each costing ₹15.

Correct Answer: Number of articles = 6; Cost per article = ₹15
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