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.

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Solution

Let the number of articles produced be $x$ .
Therefore, cost of production of each article = $₹ (2 x + 3)$
It is given that the total production is $₹ 90$ .
$∴ x (2 x + 3) = 90$
$\Rightarrow 2 x^{2} + 3 x - 90 = 0$
$\Rightarrow 2 x^{2} + 15 x - 12 x - 90 = 0$
(Splitting the middle term 3x as 15x-12x because $15 x \times - 12 x = - 180 x^{2}$ )
$\Rightarrow x (2 x + 15) - 6 (2 x + 15) = 0$
$\Rightarrow (2 x + 15) (x - 6) = 0$
Either, $2 x + 15 = 0 o r x - 6 = 0$
$\Rightarrow x = - \frac{15}{2} o r x = 6$
As the number of articles produced can only be a positive integer, so $x$ can only be $6$ .
Hence, number of articles produced = $6$
Cost of each article = $2 \times 6 + 3 = ₹ 15$ Let the number of articles produced be $x$ .
Therefore, cost of production of each article = $₹ (2 x + 3)$
It is given that the total production is $₹ 90$ .
$∴ x (2 x + 3) = 90$
$\Rightarrow 2 x^{2} + 3 x - 90 = 0$
$\Rightarrow 2 x^{2} + 15 x - 12 x - 90 = 0$
(Splitting the middle term 3x as 15x-12x because $15 x \times - 12 x = - 180 x^{2}$ )
$\Rightarrow x (2 x + 15) - 6 (2 x + 15) = 0$
$\Rightarrow (2 x + 15) (x - 6) = 0$
Either, $2 x + 15 = 0 o r x - 6 = 0$
$\Rightarrow x = - \frac{15}{2} o r x = 6$
As the number of articles produced can only be a positive integer, so $x$ can only be $6$ .
Hence, number of articles produced = $6$
Cost of each article = $2 \times 6 + 3 = ₹ 15$

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