An article is marked at ₹ 2,250. By selling it at a discount of 12%, the dealer makes a profit of 10%. Find :

(i) the selling price of the article.

(ii) the cost price of the article for the dealer.

(i) Given:

M.P. of an article = ₹ 2,250

Discount % = 12 %

$\text{Discount\%} = \dfrac{\text{Discount}}{\text{M.P.}} \times 100\\[1em] \Rightarrow 12 = \dfrac{\text{Discount}}{2250} \times 100\\[1em] \Rightarrow \text{Discount} = \dfrac{12 \times 2250}{100}\\[1em] = \dfrac{27,000}{100}\\[1em] = 270$

$\text{Discount = M.P. - S.P.}\\[1em] \Rightarrow 270 = 2,250 - \text{S.P.}\\[1em] \Rightarrow \text{S.P.} = 2,250 - 270\\[1em] \Rightarrow \text{S.P.} = 1,980$

The S.P. of the article = ₹ 1,980.

(ii) S.P. of the article = ₹ 1,980

Profit of the article = 10%

Let the C.P. of the article be ₹ $x$.

$\text{Profit\%} = \dfrac{\text{Profit}}{\text{C.P.}}\times 100\\[1em] \Rightarrow 10 = \dfrac{\text{Profit}}{x}\times 100\\[1em] \Rightarrow \text{Profit} = \dfrac{10 \times x}{100}\\[1em] = \dfrac{10x}{100}\\[1em] = \dfrac{x}{10}$

and,

$\text{Profit = S.P. - C.P.}\\[1em] \Rightarrow \dfrac{x}{10} = 1980 - x\\[1em] \Rightarrow \dfrac{x}{10} + x = 1980 \\[1em] \Rightarrow \dfrac{x}{10} + \dfrac{10x}{10} = 1980 \\[1em] \Rightarrow \dfrac{(x + 10x)}{10} = 1980 \\[1em] \Rightarrow \dfrac{11x}{10} = 1980 \\[1em] \Rightarrow x = \dfrac{1980 \times 10}{11} \\[1em] \Rightarrow x = \dfrac{19800}{11} \\[1em] \Rightarrow x = 1800$

The cost price of the article = ₹ 1,800