Question

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 articles (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 articles.​

Answer 1

$\huge\underline\mathbb {SOLUTION:-}$

$\mathsf {Let\:cost\:of\:articles\:be\:Rs\:x.}$

We are given total cost of production on that particular day = Rs 90.

$\therefore \mathsf {Total\:number\:of\:articles\:produced\:that\:day\: = \frac{90}{x} }$

$\underline\mathsf \red {According\:to\:given\:conditions,}$

$\mathsf {x = 2\bigg(\frac{90}{x}\bigg) + 3}$

$\implies \mathsf {x = \frac{180}{x}+ 3}$

$\implies \mathsf {x = \frac{180 + 3x}{x} }$

$\implies \mathsf {x^2 = 180 + 3x}$

$\implies \mathsf {x^2 - 3x - 180 = 0}$

$\implies \mathsf {x^2 - 15x + 12x - 180 - 0}$

$\implies \mathsf {x (x - 15) + 12(x - 15) = 0}$

$\implies \mathsf {(x - 15)\:(x + 12) = 0}$

$\implies \mathsf {x = 15,\: -12}$

$\mathsf {Cost\:cannot\:be\:in\:negative,}$

$\therefore \mathsf {We\:discard\: x = 12}$

$\therefore \mathsf {x = Rs\:15\:which\:is\:the\:cost\:of\:production\:of\:each\:article.}$

$\mathsf {Number\:of\:articles\:produced\:on\:that\:particular\:day,}$

$\implies \mathsf \blue {\frac{90}{15} = 16}$

Answer 2

Let us assume, number of pottery in a day = x

So, cost of production of each article = 2x + 3

according to question:-

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

$2x {}^{2} + 3x = 90$

$2x {}^{2} + 3x - 90 = 0$

$2x {}^{2} - 12x + 15x - 90 = 0$

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

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

$\: Hence, \red{x = - 15/2 \: and \: x = 6}$

• Now, number of articles produced = 6

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