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 Rs90, find the number of articles produced and the cost of each article.​

Answer 1

Given:

The total cost of production = Rs. 90

Now:

Let the number of pottery articles produced be 'x'.

Therefore:

Production of each article = Rs. (2x + 3)      

A.T.Q:

$\implies$ $\boxed{\sf{2x^{2}+3x-90=0}}$

So:

$\implies$ 2x² + 3x – 90 = 0

$\implies$ 2x² + 15x – 12x – 90 = 0

$\implies$ x(2x + 15) – 6(2x + 15) = 0

$\implies$ (2x + 15)(x – 6) = 0

Now:

$\implies$ 2x + 15 = 0 

$\implies$ 2x = -15

$\implies$ x = -15/2

Or:

$\implies$ x – 6 = 0

$\implies$ x = 6

So:

$\implies$ x = -15/2 or x = 6

Note: As the number of articles produced can only be a +ve integer, therefore, 'x' can only be 6.

Hence:

Number of articles produced = 6

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

Answer 2

Answer:-

Number of article produced = 6

Cost of each article = ₹ 15

Given :-

Total cost of production = ₹90

To find :-

The number of article produced and the cost of each article.

Solution:-

Let the number of each article produced be x and cost of each article be y.

A/Q

The cost of article is 3 more than 2times the number of article produced.

→$y = 2x + 3$

Total cost of production = Cost of each particle × Number of article produced.

→$xy = 90$

• Put the value of y.

→$x ( 2x +3) = 90$

→$2x^2 +3x = 90$

→$2x^2 +3x -90 = 0$

• Solving this quadratic equation by middle splitting term.

→$2x^2 -12x +15 x -90 = 0$

→$2x(x-6) +15(x-6) =0$

→$(2x+15) (x-6) = 0$

→$2x +15 = 0 \\ x -6 = 0$

→$x = \dfrac{-15}{2}\\ x = 6$

• Neglect negative value.

→y = 2x + 3

→y = 2 × 6 + 3

→y = 12 + 3

→y = 15

hence,

The number of article is 6 and cost of production is ₹ 15.