A fruit-seller bought x apples for ₹ 1,200.

(i) Write the cost price of each apple in terms of x.

(ii) If 10 of the apples were rotten and he sold each of the rest at ₹ 3 more than the cost price of each, write the selling price of (x − 10) apples.

(iii) If he made a profit of ₹ 60 in this transaction, form an equation in x and solve it to evaluate x.

(i) Given,

Cost of x apples = ₹ 1,200

Cost of one apple = ₹ $\dfrac{1200}{x}$

Hence, cost of each apple in terms of x = ₹ $\dfrac{1200}{x}$.

(ii) Given,

Fruit-seller sells each apple at ₹ 3 more than the cost price of each.

Selling price of each apple = ₹ $\dfrac{1200}{x}$ + 3

Selling price of (x − 10) apples = ₹ $(x - 10)\Big(\dfrac{1200}{x} + 3 \Big)$

Hence, total selling price = ₹ $(x - 10)\Big(\dfrac{1200}{x} + 3 \Big)$.

(iii) Cost price = ₹ 1,200

Profit = ₹ 60

Selling price = Cost price + Profit = 1200 + 60 = ₹ 1,260

$\Rightarrow 1260 = (x - 10)\Big(\dfrac{1200}{x} + 3 \Big) \\[1em] \Rightarrow 1260 = (x - 10)\Big(\dfrac{1200 + 3x}{x} \Big) \\[1em] \Rightarrow 1260x = (x - 10)(1200 + 3x) \\[1em] \Rightarrow 1260x = 1200x + 3x^2 - 12000 - 30x \\[1em] \Rightarrow 0 = 1200x - 1260x - 30x + 3x^2 - 12000 \\[1em] \Rightarrow 3x^2 - 90x - 12000 = 0 \\[1em] \Rightarrow 3(x^2 - 30x - 4000) = 0 \\[1em] \Rightarrow x^2 - 30x - 4000 = 0 \\[1em] \Rightarrow x^2 - 80x + 50x - 4000 = 0 \\[1em] \Rightarrow x(x - 80) + 50(x - 80) = 0 \\[1em] \Rightarrow (x + 50) = 0 \text{ or } (x - 80) = 0 \text{…..[Using zero-product rule]} \\[1em] \Rightarrow x = -50 \text{ or } x = 80.$

Since,

Number of apples cannot be negative.

Thus, x = 80

Hence, obtained equation x² - 30x - 4000 = 0 and the value x = 80.