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

Answer 1

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

• Cost of production of each article (in rupees) was 3 more than twice the number of articles.
• Total cost of production on that day was ₹90

Let the Number of the articles be x

again, Cost of each articles be ₹ (2x + 3)

$\\ \circ \ {\pmb{\underline{\sf{ According \ to \ Question: }}}} \\ \\ \colon\implies{\sf{ x(2x+3) = 90 }} \\ \\ \colon\implies{\sf{ 2x^2 + 3x = 90 }} \\ \\ \colon\implies{\sf{ 2x^2 + 3x - 90 = 0 }}$

Now, It's time to Find the roots of Equation as that:-

$\\ \colon\Rightarrow{\underline{\boxed{\sf{ x = \dfrac{ -b \pm \sqrt{ b^2 - 4ac} }{2a} }}}} \\ \\ \\ \colon\implies{\sf{ x = \dfrac{ -3 \pm \sqrt{ 3^2 - 4 \times 2 \times (-90)} }{2 \times 2} }} \\ \\ \\ \colon\implies{\sf{ x = \dfrac{ -3 \pm \sqrt{ 9 +720} }{4} }} \\ \\ \\ \colon\implies{\sf{ x = \dfrac{ -3 \pm \sqrt{ 729} }{4} }} \\ \\ \\ \colon\implies{\sf{ x = \dfrac{ -3 \pm 27 }{4} }} \\ \\ \\ \colon\implies{\sf{ x = \dfrac{ -3 + 27 }{4} \ and \ \dfrac{-3-27}{4} }} \\ \\ \\ \colon\implies{\sf{ x = \dfrac{ 24 }{4} \ and \ \dfrac{-30}{4} }} \\ \\ \\ \colon\implies{\underline{\boxed{\sf\red{ x = 6 \ and \ -7.5 }}}}$

Neglecting Negative value (—ve).

Hence,

• Number of the articles be x = 6
• Cost of each articles be ₹ (2x + 3) = ₹(2×6+3) = ₹ 15

Answer 2

Answer:

It is the correct answer.

Step-by-step explanation:

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