An article is marked $$x$$​% (0 < $$x$$​ < 40) above its cost price. It is sold by giving $$\frac{x}{2}$$​% discount on its marked price. If there is a profit of $$10\frac{1}{2}$$​%, then what is the value of $$x$$​ ?

Correct option is D

Given:
The article is marked x% above its cost price, and a discount of $$\frac{x}{2}\%$$​ is given on the marked price. The profit earned is $$10 \frac{1}{2}\%$$​or 10.5%.

Formula Used:

Selling Price = Marked Price $$\times \left(1 - \frac{\text{Discount Percentage}}{100}\right)$$​

Profit Percentage = $$\frac{\text{Selling Price} - \text{Cost Price}}{\text{Cost Price}} \times 100$$​

Solution:
Let the cost price be C.

Marked Price = $$C \times \left(1 + \frac{x}{100}\right)$$​

Selling Price = Marked Price $$\times \left(1 - \frac{x}{200}\right)$$​

Profit Percentage = 10.5%, so

​$$\frac{\text{Selling Price} - C}{C} \times 100 = 10.5$$​

Substitute Selling Price:

​$$\frac{C \times \left(1 + \frac{x}{100}\right) \times \left(1 - \frac{x}{200}\right) - C}{C} \times 100 = 10.5$$​

​$$\left(1 + \frac{x}{100}\right) \times \left(1 - \frac{x}{200}\right) = 1.105$$​

​$$1 + \frac{x}{100} - \frac{x}{200} - \frac{x^2}{20000} = 1.105$$​

​$$1 + \frac{x}{200} - \frac{x^2}{20000} = 1.105$$​

​$$\frac{x}{200} - \frac{x^2}{20000} = 0.105$$​

​$$100x - x^2 = 2100$$​

​$$x^2 - 100x + 2100 = 0$$​

x = $$\frac{-(-100) \pm \sqrt{(-100)^2 - 4(1)(2100)}}{2(1)}$$

x =$$\frac{100 \pm \sqrt{10000 - 8400}}{2}$$​

x = $$\frac{100 \pm \sqrt{1600}}{2}$$​

x = $$\frac{100 \pm 40}{2}$$​

x = $$\frac{100 + 40}{2} = 70 \quad \text{or} \quad x = \frac{100 - 40}{2} = 30$$​

Since  x < 40,So the value of x is 30.

Thuus, the correct option is (d) 30

Alternate Method:

Let C.P. = 100%

Profit = 10.5%

SP = CP + Profit = 100% + 10.5% = 110.5%
By checking options:

Option (d): x = 30 %

so, MP = 100% + 30% = 130%

Discount = 15% = $$-\frac{3}{20} = \frac{17}{20}$$​​

SP = 130% $$\times \frac{17}{20}$$

SP = 110.5% (Satisfied)