Question

A man sold a chair and a table together for Rs 760, thereby marking a profit of 25% on chair and 10% on table. By selling them together for Rs 767.50, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.

Answer 1

Let the cost price of the chair be Rs. x and the cost price of the table be Rs. y.
Then, we have:
If the profit is 25%, then the selling price of the chair = Rs. $\left(x+25\%\mathrm{of}x\right)$ = Rs. $\left(x+\frac{25x}{100}\right)$ = Rs. $\left(\frac{125}{100}x\right)$
If the profit is 10%, then the selling price of the table = Rs. $\left(y+10\%\mathrm{of}y\right)$ = Rs. $\left(y+\frac{10y}{100}\right)$ = Rs. $\left(\frac{110}{100}y\right)$
Given:
Selling price of the chair and the table = Rs. 760
⇒ Rs. $\left(\frac{125}{100}x\right)$ + Rs. $\left(\frac{110}{100}y\right)$ = Rs. 760
⇒ $\frac{25}{20}x+\frac{22}{20}y=760$
⇒ 25x + 22y = 15200 .....(i)

Again, we have:
If the profit is 10%, then the selling price of the chair = Rs. $\left(x+10\%\mathrm{of}x\right)$ = Rs. $\left(x+\frac{10x}{100}\right)$ = Rs. $\left(\frac{110}{100}x\right)$
If the profit is 25%, then the selling price of the table = Rs. $\left(y+25\%\mathrm{of}y\right)$ = Rs. $\left(y+\frac{25y}{100}\right)$ = Rs. $\left(\frac{125}{100}y\right)$
Selling price of the chair and the table = Rs. 767.50
⇒ Rs. $\left(\frac{110}{100}x\right)$ + Rs. $\left(\frac{125}{100}y\right)$ = Rs. 767.50
⇒ $\frac{22}{20}x+\frac{25}{20}y=767.50$
⇒ 22x + 25y = 15350 .....(ii)

On adding (i) and (ii), we get:
47x + 47y = 30550
⇒ 47(x + y) = 30550
⇒ (x + y) = 650 ....(iii)
On subtracting (ii) from (i), we get:
3x − 3y = −150
⇒ 3(x − y) = −150
⇒ (x − y) = −50 ....(iv)
On adding (iii) and (iv), we get:
2x = (650 − 50) = 600
⇒ x = 300
On substituting x = 300 in (iii), we get:
300 + y = 650
⇒ y = (650 − 300) = 350
Hence, the cost price of the chair is Rs. 300 and the cost price of the table is Rs. 350.