A man sold a chair and a table for ₹ 2,178, thereby making a profit of 12% on the chair and 16% on the table. By selling them for ₹ 2,154, he gains 16% on the chair and 12% on the table. Find the cost price of each.

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

Given,

When sold for ₹ 2,178, he makes profit of 12% on the chair and 16% on the table.

$\therefore x + \dfrac{12}{100}x + y + \dfrac{16}{100}y = 2178 \\[1em] \Rightarrow \dfrac{100x + 12x}{100} + \dfrac{100y + 16y}{100} = 2178 \\[1em] \Rightarrow \dfrac{112x}{100} + \dfrac{116y}{100} = 2178 \\[1em] \Rightarrow 112x + 116y = 2178 \times 100 \\[1em] \Rightarrow 112x + 116y = 217800\\[1em] \Rightarrow 4(28x + 29y) = 217800 \\[1em] \Rightarrow 28x + 29y = \dfrac{217800}{4} \\[1em] \Rightarrow 28x + 29y = 54450 \text{ ….(1) }$

Given,

When sold for ₹ 2,154, he makes profit of 16% on the chair and 12% on the table.

$\therefore x + \dfrac{16}{100}x + y + \dfrac{12}{100}y = 2154 \\[1em] \Rightarrow \dfrac{100x + 16x}{100} + \dfrac{100y + 12}{100} = 2154 \\[1em] \Rightarrow \dfrac{116x}{100} + \dfrac{112y}{100} = 2154 \\[1em] \Rightarrow 116x + 112y = 2154 \times 100 \\[1em] \Rightarrow 116x + 112y = 215400 \\[1em] \Rightarrow 4(29x + 28y) = 215400 \\[1em] \Rightarrow 29x + 28y = \dfrac{215400}{4} \\[1em] \Rightarrow 29x + 28y = 53850 \text{ ……..(2) }$

Subtracting equation eqn 1 from 2, we get :

⇒ 29x + 28y - (28x + 29y) = 53850 - 54450

⇒ 29x + 28y - 28x - 29y = -600

⇒ x - y = -600

⇒ x = y - 600     ….(3)

Substituting value of x from equation (3) in equation (1), we get :

⇒ 28x + 29y = 54450

⇒ 28(y - 600) + 29y = 54450

⇒ 28y - 16800 + 29y = 54450

⇒ 57y = 54450 + 16800

⇒ 57y = 71250

⇒ y = $\dfrac{71250}{57}$

⇒ y = ₹ 1,250

Substituting value of y in equation (2), we get :

⇒ 29x + 28 × 1250 = 53850

⇒ 29x + 35000 = 53850

⇒ 29x = 53850 - 35000

⇒ 29x = 18850

⇒ x = $\dfrac{18850}{29}$

⇒ x = ₹ 650

Hence, cost price of table = ₹ 1,250 and cost price of chair = ₹ 650.