Q6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that 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 Rs. 90, find the number of articles produced and the cost of each article.

Let the number of articles produced be \(x\).

Thus, the cost of production of each article can be expressed as \(2x + 3\).

As the total cost of production on that day is 90 Rs.,

\(x(2x + 3) = 90\)

\(2x^2 + 3x - 90 = 0\)

(Splitting * 3x * as * -12x + 15x *)

\(2x^2 + -12x +15x - 90 = 0\)

\(2x(x - 6) + 15(x - 6) = 0\)

\((2x + 15)(x - 6) = 0\)

The roots of this equation are the values of x for which \((2x + 15)(x - 6) = 0 \)

which are,

\(2x + 15 = 0 \) or \(x - 6 = 0\)

Thus, \(x = \frac{-15}{2}\) or \(x = 6\)

Since the number of articles produced cannot be a negative number, \(x = \frac{-15}{2}\) is rejected.

** Thus, the number of articles produced is 6 and the cost of each article is \(2*6 + 3\) = 15Rs.
**