Question

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 Rs.90, find the number of articles produced and the cost of each article.

Answer 1

Hint: First of all this is a very simple and a very easy problem. This problem deals with basic concepts in algebraic mathematics. In order to solve this problem we should be able to write mathematical expressions from the given data, and also we should be able to equate these mathematical expressions to the given relatable data, and obtain correct results. We should be familiar in solving quadratic expressions as well.

Complete step-by-step solution:
Given that there is a cottage industry and a certain number of pottery articles in one day.
Also given that the cost of the production for each article for a particular day is 3 more than twice the number of articles produced on that particular day.
Given that the total cost of production on that particular day is Rs.90
We have to find the total number of articles produced on that day.
Also we have to find the cost of each article produced on that particular day.
Let the number of pottery articles produced in one day = $x$
Given that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day, is expressed mathematically as given below:
$ \Rightarrow 3 + 2(x)$
$\therefore $The cost of production of each article is : $3 + 2x$
Given that total cost of production on that day in rupees = 90
We know that the total cost of production of articles is given by the product of total number of articles produced and the cost of each article, is expressed mathematically as given below:
$ \Rightarrow $Total number of articles produced = $x$
$ \Rightarrow $The cost of each article produced = $3 + 2x$
Now the total cost of production of the articles is given by : $x(3 + 2x)$
Now already given that the total cost of production on that day was Rs.90, hence equating the above obtained expression to Rs.90, in order to get the value of $x$, which is the total no. of articles produced, as given below:
$ \Rightarrow x(3 + 2x) = 90$
$ \Rightarrow 3x + 2{x^2} = 90$
 Now we have obtained a quadratic expression, solve this quadratic expression to get the value of $x$, as given below:
$ \Rightarrow 2{x^2} + 3x - 90 = 0$
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {{{(3)}^2} - 4(2)( - 90)} }}{{2(2)}}$
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {9 + 4(2)(90)} }}{4}$
Simplifying the above expression to get the value of $x$, as given below:
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {729} }}{4}$
$ \Rightarrow x = \dfrac{{ - 3 \pm 27}}{4}$
As the number of articles cannot be a negative number, hence considering only:
$ \Rightarrow x = \dfrac{{ - 3 + 27}}{4}$
$ \Rightarrow x = 6$
$\therefore $The total no. of articles produced in a day = 6
The cost of each article is given by $3 + 2x$, by substituting the value of $x$, as given below:
$ \Rightarrow 3 + 2(6)$
$ \Rightarrow 3 + 12 = 15$
$\therefore $The cost of each article is Rs.15.

The number of articles produced and the cost of each article is 6 and Rs.15 respectively.

Note: While solving such problems we should understand that, there is a quadratic expression, and to solve any quadratic expression which is in the form of $a{x^2} + bx + c = 0$, the roots of the quadratic expression are given by the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where here $a$ is the coefficient of the term ${x^2}$, and $b$ is the coefficient of the term $x$, and $c$ is the constant in the quadratic expression.