A trader bought a number of articles for ₹ 1200. Ten were damaged and he sold each of the remaining articles at ₹ 2 more than what he paid for it, thus getting a profit of ₹ 60 on the whole transaction.

Taking the number of articles he bought as x, form an equation in x and solve it.

Let no. of articles bought be x,

Cost price of each article = ₹ $\dfrac{1200}{x}$

S.P of each article = ₹ $\Big(\dfrac{1200}{x} + 2\Big)$

S.P. of (x - 10) articles = (x - 10)$\Big(\dfrac{1200}{x} + 2\Big)$

Profit = ₹ 60

$\therefore (x - 10)\Big(\dfrac{1200}{x} + 2\Big) - 1200 = 60 \\[1em] \Rightarrow 1200 + 2x - \dfrac{12000}{x} - 20 - 1200 = 60 \\[1em] \Rightarrow 2x - \dfrac{12000}{x} = 80 \\[1em] \Rightarrow \dfrac{2x^2 - 12000}{x} = 80 \\[1em] \Rightarrow 2x^2 - 12000 = 80x \\[1em] \Rightarrow 2x^2 - 12000 - 80x = 0 \\[1em] \Rightarrow 2(x^2 - 6000 - 40x = 0) \\[1em] \Rightarrow x^2 - 40x - 6000 = 0 ……..(i) \\[1em] \Rightarrow x^2 - 100x + 60x - 6000 = 0 \\[1em] \Rightarrow x(x - 100) + 60(x - 100) = 0 \\[1em] \Rightarrow (x + 60)(x - 100) = 0 \\[1em] \Rightarrow x = -60 \text{ or } x = 100.$

Since, no. of articles cannot be negative,

∴ x ≠ -60.

From (i) quadratic equation = x² - 40x - 6000 = 0.

Hence, no. of articles bought = 100 and quadratic equation = x² - 40x - 6000 = 0.