A man spent ₹2800 on buying a number of plants priced at ₹x each. Because of the number involved, the supplier reduced the price of each plant by one rupee. The man finally paid ₹2730 and received 10 more plants . Find x.

In first case,

Amount spent = ₹2800

Price of each plant = ₹x

No. of plants = $\dfrac{2800}{x}$

In second case,

Amount spent = ₹2730

Price of plant is reduced by ₹1 , so new price = ₹(x - 1)

Given, in this case 10 more plants were received hence, no. of plants = $\dfrac{2800}{x} + 10$

According to question,

$\Rightarrow \big(\dfrac{2800}{x} + 10\big)(x - 1) = 2730 \\[1em] \Rightarrow \big(\dfrac{2800 + 10x}{x}\big)(x - 1) = 2730 \\[1em] \Rightarrow \dfrac{(2800 + 10x)(x - 1)}{x} = 2730 \\[1em] \Rightarrow \dfrac{2800x - 2800 + 10x^2 - 10x}{x} = 2730 \\[1em] \Rightarrow 10x^2 + 2790x - 2800 = 2730x \text{ (On cross multiplication) } \\[1em] \Rightarrow 10x^2 + 60x - 2800 = 0 \\[1em] \Rightarrow x^2 + 6x - 280 = 0 \text{ (On dividing by 10) } \\[1em] \Rightarrow x^2 + 20x - 14x - 280 = 0 \\[1em] \Rightarrow x(x + 20) - 14(x + 20) = 0 \\[1em] \Rightarrow (x - 14)(x + 20) = 0 \\[1em] x = 14 \text{ or } x = -20$

Since price cannot be negative hence , x ≠ -20

Hence, the value of x is 14.