Question

A shopkeeper sells a table at $8\%$ a profit and a chair at a $10\%$ discount, thereby getting$₹1008$. If he had sold the table at $10\%$ profit and the chair at $8\%$ discount, he would have got $₹20$ more. Find the cost price of the table and the list price of the chair.

Answer 1

Step 1: Framing the equation using first condition

$\Rightarrow$Let the Cost Price of the table be $₹\mathrm{x}$

Let the Cost Price of the chair be $₹\mathrm{y}$

According to the first condition given in the question:

A shopkeeper sells a table at '$8\%$' profit and a chair at a $10\%$ discount.

$\Rightarrow$The formula for Selling Price(SP), in terms of Cost Price(CP) and Profit%, $\mathrm{SP}=\frac{100+\mathrm{Profit}\%}{100}\times \mathrm{CP}$

$\therefore$The selling price (SP) of the table$=\frac{100+8}{100}\mathrm{x}=\frac{108\mathrm{x}}{100}$………($1$)

$\therefore$The Selling Price(SP) of the chair$=\mathrm{y}-\frac{10\mathrm{y}}{100}=\frac{90\mathrm{y}}{100}$……….($2$)

Now by adding both the equations [$1$] and [$2$], we get.

$\Rightarrow \frac{108\mathrm{x}}{100}+\frac{90\mathrm{y}}{100}=1008$

$108\mathrm{x}+90\mathrm{y}=100800$……………………………………………($3$)

Step 2: Framing the equation using second condition

According to the second condition given in the question:

A shopkeeper sells a table at 10% profit and a chair at an 8% discount, And he gets $₹20$ more, i.e, $₹1028$

$\therefore$The selling price (SP) of the table$=\frac{100+10}{100}\mathrm{x}=\frac{110\mathrm{x}}{100}$……………….($4$)

$\therefore$The selling price (SP) of the chair$=\mathrm{y}-\frac{8\mathrm{y}}{100}=\frac{92\mathrm{y}}{100}$…………………($5$)

Now by adding both the equations –[$4$] and [$5$], we get,

$\Rightarrow$$\frac{110\mathrm{x}}{100}+\frac{92\mathrm{y}}{100}=1028$

$110\mathrm{x}+92\mathrm{y}=102800$…………………………………………….($6$)

Now by subtracting equation [$3$] from equation [$6$], we get,

$\Rightarrow 2\mathrm{x}+2\mathrm{y}=2000$

$\mathrm{x}+\mathrm{y}=1000$

$\mathrm{y}=1000-\mathrm{x}$………………………………………………….($7$)

Step 3: Finding the cost price of the table and the list price of the chair.

Now by substituting the value of $\mathrm{y}$ from equation ($7$) in equation ($3$), we get,

$\begin{array}{rcl}& \Rightarrow & 108\mathrm{x}+90\left(1000-\mathrm{x}\right)=100800\\ 108\mathrm{x}-90\mathrm{x}& =& 100800-90000\\ 18\mathrm{x}& =& 10800\\ \mathrm{x}& =& ₹600\end{array}$

Substituting the value of $\mathrm{x}$ in equation($7$), we get,

$\begin{array}{rcl}& \Rightarrow & \mathrm{y}=1000-600\\ \mathrm{y}& =& ₹400\end{array}$

Therefore, the cost price of the table is $₹600$ and the cost price of the chair is $₹400$