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A trader bought a number of articles for Rs. 1,200. Ten were damaged and he sold each of the remaining articles at Rs. 2 more than what he paid for it

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Question

A trader bought a number of articles for Rs. 1,200. Ten were damaged and he sold each of the remaining articles at Rs. 2 more than what he paid for it, thus getting a profit of Rs. 60 on the whole transaction. Taking the number of articles he bought as x, form an equation in x and solve it.

Sum
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Solution

Number of articles bought by the trader = x

It is given that the trader bought the articles for Rs. 1200

So, the cost of one article = Rs. `1200/x`

Ten articles were damaged. So, the number of articles left = x – 10

Selling price of each of (x – 10) articles = Rs. `(x - 10) (1200/x + 2)`

Profit = Rs. 60

`∴ (x - 10)(1200/x + 2) - 1200 = 60`

`1200 + 2x - 12000/x - 20 - 1200 = 60`

`2x - 12000/x - 80 = 0`

2x2 – 80x – 12000 = 0 

x2 – 40x – 6000 = 0

x2 – 100x + 60x – 6000 = 0

x(x – 100) + 60(x – 100) = 0

(x – 100)(x + 60) = 0

x = 100, – 60

Number of articles cannot be negative.

So, x = 100.

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Miscellaneous Problems
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Chapter 6: Solving (simple) Problems (Based on Quadratic Equations) - Exercise 6 (D) [Page 78]

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Selina Concise Mathematics [English] Class 10 ICSE
Chapter 6 Solving (simple) Problems (Based on Quadratic Equations)
Exercise 6 (D) | Q 11. | Page 78
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