A cottage industry produces a certain number of pottery articles in a day. The cost of production of each article is 3 rupees more than twice the number of articles produced that day. If the total cost of production is ₹90, find the number of articles produced and the cost of each article.
Number of articles = 6, Cost of each article = ₹15
Step 1: Introduce the variable. Let the number of pottery articles produced in a day be \(n\).
Step 2: Express the cost per article. According to the question, the cost of production of each article (in rupees) is 3 more than twice the number of articles produced. So,
\[\text{Cost per article} = 2n + 3\]
Step 3: Form the equation using total cost. Total cost = (number of articles) × (cost per article). The total cost is given as ₹90, so:
\[n(2n + 3) = 90\]
Step 4: Expand and rearrange.
Multiply out the brackets:
\[2n^2 + 3n = 90\]
Bring 90 to the left side to get a standard quadratic equation:
\[2n^2 + 3n - 90 = 0\]
Step 5: Factorise the quadratic. We want two numbers whose product is \(2 \times (-90) = -180\) and whose sum is \(+3\). These numbers are \(+15\) and \(-12\) because:
\[15 \times (-12) = -180, \quad 15 + (-12) = 3\]
So we split the middle term \(3n\) as \(15n - 12n\):
\[2n^2 + 15n - 12n - 90 = 0\]
Group the terms:
\[(2n^2 + 15n) + (-12n - 90) = 0\]
Factor each group:
\[n(2n + 15) - 6(2n + 15) = 0\]
Take \((2n + 15)\) common:
\[(2n + 15)(n - 6) = 0\]
Step 6: Solve for \(n\). Using the zero product rule:
\[2n + 15 = 0 \Rightarrow 2n = -15 \Rightarrow n = -\dfrac{15}{2}\]
\[n - 6 = 0 \Rightarrow n = 6\]
Step 7: Choose the meaningful solution. The number of articles produced cannot be negative or fractional, so we reject \(n = -\dfrac{15}{2}\) and accept:
\[n = 6\]
Step 8: Find the cost per article.
Cost per article = \(2n + 3 = 2(6) + 3 = 12 + 3 = 15\) rupees.
Conclusion: The cottage industry produces 6 articles in that day, and the cost of each article is ₹15.