A trader buys x articles for a total cost of ₹600.

(i) Write down the cost of one article in terms of x.

If the cost per article were ₹5 more, the number of articles that can be bought for ₹600 would be four less.

(ii) Write down the equation in x for the above situation and solve it to find x.

(i) Total cost = ₹ 600

No. of articles = x

Cost of each article = ₹ $\dfrac{600}{x}$

Hence, cost of each article = ₹ $\dfrac{600}{x}$.

(ii) According to question,

New cost per article = ₹ $\dfrac{600}{x}$ + 5

New number of articles bought for ₹ 600 = $\dfrac{600}{\dfrac{600}{x} + 5}$

Given, new number of articles is four less than original number of articles,

$\therefore \dfrac{600}{\dfrac{600}{x} + 5} = x - 4 \\[1em] \Rightarrow \dfrac{600}{x - 4} = \dfrac{600}{x} + 5 \\[1em] \Rightarrow \dfrac{600}{x - 4} - \dfrac{600}{x} = 5 ……(i) \\[1em] \Rightarrow \dfrac{600x - 600(x - 4)}{x(x - 4)} = 5 \\[1em] \Rightarrow \dfrac{600x - 600x + 2400}{x^2 - 4x} = 5 \\[1em] \Rightarrow \dfrac{2400}{x^2 - 4x} = 5 \\[1em] \Rightarrow 2400 = 5(x^2 - 4x) \\[1em] \Rightarrow x^2 - 4x = 480 \\[1em] \Rightarrow x^2 - 4x - 480 = 0 \\[1em] \Rightarrow x^2 - 24x + 20x - 480 = 0 \\[1em] \Rightarrow x(x - 24) + 20(x - 24) = 0 \\[1em] \Rightarrow (x + 20)(x - 24) = 0 \\[1em] \Rightarrow x + 20 = 0 \text{ or } x - 24 = 0 \\[1em] \Rightarrow x = -20, 24.$

Since no. of articles cannot be negative,

∴ x ≠ -20.

From (i) we get quadratic equation,

$\dfrac{600}{x - 4} - \dfrac{600}{x} = 5.$

and x = 24.