#### Question

A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.

#### Solution

et the cost price of article be Rs. *x*.

Then, gain percent = *x*

Therefore, the selling price of article

`=(x+x/100xx x)`

`=(x^2+100x)/100`

It is given that

`(x^2+100x)/100=24`

x^{2} + 100x = 2400

x^{2} + 100x - 2400 = 0

x^{2} + 120x - 20x - 2400 = 0

x(x + 120) - 20(x + 120) = 0

(x + 120)(x - 20) = 0

x + 120 = 0

x = -120

Or

x - 20 = 0

x = 20

Because *x *cannot be negative.

Thus,* x = 20 *is the require solution.

Therefore, the cost price of article be x = Rs. 20

Is there an error in this question or solution?

#### APPEARS IN

Solution A Dealer Sells an Article for Rs. 24 and Gains as Much Percent as the Cost Price of the Article. Find the Cost Price of the Article. Concept: Solutions of Quadratic Equations by Factorization.