An article is bought for ₹ x and is sold at the profit of x %. If its selling price is ₹ 56, the cost price is :

1. ₹ 60

2. ₹ 50

3. ₹ 40

4. ₹ 28

Given,

C.P. = ₹ x

Profit = x %

S.P. = C.P. + Profit

$\Rightarrow 56 = x + \dfrac{x}{100} \times x \\[1em] \Rightarrow 56 = x + \dfrac{x^2}{100} \\[1em] \Rightarrow \dfrac{x^2 + 100x}{100} = 56 \\[1em] \Rightarrow x^2 + 100x = 5600 \\[1em] \Rightarrow x^2 + 100x - 5600 = 0\\[1em] \Rightarrow x^2 + 140x - 40x - 5600 = 0 \\[1em] \Rightarrow x(x + 140) - 40(x + 140) = 0 \\[1em] \Rightarrow (x - 40)(x + 140) = 0 \\[1em] \Rightarrow x - 40 = 0 \text{ or }x + 140 =0 \\[1em] \Rightarrow x = 40 \text{ or } x = -140.$

Since, cost cannot be negative.

∴ x ≠ -140.

Hence, Option 3 is the correct option.