**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.**

**Solution:**

Let x be the number of articles produced.

Therefore, cost of production of each article = Rs (2*x* + 3)

Given the total cost of production is Rs. 90

∴ *x*(2*x* + 3) = 90

⇒ 2*x*^{2 }+ 3*x* – 90 = 0

⇒ 2*x*^{2 }+ 15*x* -12*x* – 90 = 0

⇒ *x*(2*x* + 15) -6(2*x* + 15) = 0

⇒ (2*x* + 15)(*x* – 6) = 0

Thus, either 2*x* + 15 = 0 or *x* – 6 = 0

⇒ *x* = -15/2 or *x* = 6

As the number of articles produced can only be a positive integer, x can only be 6.

Hence, the number of articles produced = 6

Cost of each article = 2 × 6 + 3 = Rs. 15