A shopkeeper sold a table and a chair for ₹ 1,050, thereby making a profit of 10% on the table and 25% on the chair. If he had taken a profit of 25% on the table and 10% on the chair, then he would have got ₹ 1,065. What is the cost price of 1 table and 1 chair.

Let cost price of the table be ₹ x and cost price of the chair be ₹ y.

According to case 1 :

⇒ Profit on table = 10%

Selling Price of table = Cost price (1 + Profit%) = $x \Big(1 + \dfrac{10}{100}\Big)$ = ₹ x × 1.10

⇒ Profit on chair = 25%

Selling Price of chair = Cost price (1 + Profit%) = $y \Big(1 + \dfrac{25}{100}\Big)$ = ₹ y × 1.25

⇒ 1.10x + 1.25y = 1050

Multiply the equation by 100,

⇒ 100(1.10x + 1.25y) = 100 × 1050

⇒ 110x + 125y = 105000

⇒ 5(22x + 25y) = 5 × 21000

⇒ 22x + 25y = 21000

⇒ 22x = 21000 - 25y

⇒ x = $\dfrac{21000 - 25y}{22}$     ……(1)

According to case 2 :

⇒ Profit on table = 25%

Selling Price of table = Cost price (1 + Profit%) = $x \Big(1 + \dfrac{25}{100}\Big)$ = ₹ x × 1.25

⇒ Profit on chair = 10%

Selling Price of chair = Cost price (1 + Profit%) = $y \Big(1 + \dfrac{10}{100}\Big)$ = ₹ y × 1.10

⇒ 1.25x + 1.10y = 1065

Multiply the equation by 100,

⇒ 100(1.25x + 1.10y) = 1065 × 100

⇒ 125x + 110y = 106500

⇒ 5(25x + 22y) = 5 × 21300

⇒ 25x + 22y = 21300     …….(2)

Substituting value of x from equation 1 in (2), we get :

$\Rightarrow 25 \Big(\dfrac{21000 - 25y}{22}\Big) + 22y = 21300 \\[1em] \Rightarrow \dfrac{25(21000 - 25y) + 484y}{22} = 21300 \\[1em] \Rightarrow 525000 - 625y + 484y = 468600 \\[1em] \Rightarrow -141y = 468600 - 525000 \\[1em] \Rightarrow -141y = -56400 \\[1em] \Rightarrow y = \dfrac{-56400}{-141} = 400.$

Substituting value of y in equation 1, we get :

$\Rightarrow x = \dfrac{21000 - 25y}{22} \\[1em] \Rightarrow x = \dfrac{21000 - 25 \times 400}{22} \\[1em] \Rightarrow x = \dfrac{21000 - 10000}{22} \\[1em] \Rightarrow x = \dfrac{11000}{22} = 500.$

Hence, cost Price of Table = ₹ 500 and cost Price of Chair = ₹ 400.