$A$ 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.

(N/A) Let the number of articles produced be $x$.
Therefore,the cost of production of each article $= (2x + 3)$ rupees.
It is given that the total cost of production is ₹ $90$.
Thus,$x(2x + 3) = 90$.
Expanding the equation: $2x^2 + 3x - 90 = 0$.
Factoring the quadratic equation: $2x^2 + 15x - 12x - 90 = 0$.
$x(2x + 15) - 6(2x + 15) = 0$.
$(2x + 15)(x - 6) = 0$.
This gives $x = -15/2$ or $x = 6$.
Since the number of articles produced must be a positive integer,we have $x = 6$.
Therefore,the number of articles produced is $6$,and the cost of each article is $2(6) + 3 = 15$ rupees.