Ex.4.2 Q6 Quadratic Equations Solutions - NCERT Maths Class 10

A cottage industry produces 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 cost of the production on that day is ₹ \(90,\) find the number of articles produced and the cost of each article.

Text Solution

What is known?

(i) On a particular day that cost of production of each article (in rupees) was \(3\) more than twice the no. of articles produced on that day.

(ii) The total cost of the production is ₹ \(90.\)

What is Unknown?

Number of articles produced and cost of each article.

Reasoning:

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

Therefore, the cost (in rupees) of each article will be \((3 + 2x)\)

Total cost of production \(=\) Cost of each article \(\times\) Total number of articles

\[90{\rm{ }} = {\rm{ }}\left( {3{\rm{ }} + {\rm{ }}2x} \right)\left( x \right)\]

Steps:

\[\begin{align}90 &= \left( {3 + 2x} \right)\left( x \right)\\\left( {3 + 2x} \right)\left( x \right) &= 90\\
3x + 2x^2 &= 90 \end{align}\]

\[\begin{align} 2x^2 + 3x-90 &= 0\\  2x^2 \!\! + \!\! 15x \!- \! 12x \!-90  &= 0\\ \left(\! {2x + 15} 
 \! \right) \! - \! 6 \! \left( 2x \! +\! 15 \right)  & = 0\\\left( {2x + 15} \right)\left( {x - 6} \right) &= 0\end{align}\]

\[ \begin{align} 2x + 15 = 0 &\qquad x-6 = 0\\2x = - 15 &\qquad x = 6\\x = - \left(\frac{15}{2}\right)&\qquad x = 6\end{align}\]

Number of articles cannot be a negative number.

\[\therefore \,\,x = 6\]

Cost of each article

\[\begin{align}& = {\rm{ }}3{\rm{ }} + {\rm{ }}2x\\& = {\rm{ }}3{\rm{ }} + {\rm{ }}2{\rm{ }}\left( 6 \right)\\& = {\rm{ Rs}}{\rm{. }}\,15 \end{align}\]

Cost of each article is \(\rm{Rs.}\,15.\)

Number of articles produced is \(6.\)