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Ncert solutions

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**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.**

Let us say, the number of articles produced be *x*.

Therefore, cost of production of each article = Rs (2*x* + 3)

Given, total cost of production is Rs.90

∴ *x*(2*x* + 3) = 90

⇒ 2*x*^{2 }+ 3*x* – 90 = 0

⇒ 2*x*^{2 }+ 15*x* -12*x* – 90 = 0

⇒ *x*(2*x* + 15) -6(2*x* + 15) = 0

⇒ (2*x* + 15)(*x* – 6) = 0

Thus, either 2*x* + 15 = 0 or *x* – 6 = 0

⇒ *x* = -15/2 or *x* = 6

As the number of articles produced can only be a positive integer, therefore, x can only be 6.

Hence, number of articles produced = 6

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

Let us say, the number of articles produced be *x*.

Therefore, cost of production of each article = Rs (2*x* + 3)

Given, total cost of production is Rs.90

∴ *x*(2*x* + 3) = 90

⇒ 2*x*^{2 }+ 3*x* – 90 = 0

⇒ 2*x*^{2 }+ 15*x* -12*x* – 90 = 0

⇒ *x*(2*x* + 15) -6(2*x* + 15) = 0

⇒ (2*x* + 15)(*x* – 6) = 0

Thus, either 2*x* + 15 = 0 or *x* – 6 = 0

⇒ *x* = -15/2 or *x* = 6

As the number of articles produced can only be a positive integer, therefore, x can only be 6.

Hence, number of articles produced = 6

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

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