Question

A dealer bought two tables for Rs. 3120. He sold one of them at a loss of $15\%$ and other at a gain of $36\%$. Then he found that each table was sold for the same price, find the cost price of each table.

Answer 1

The cost price of two tables $=Rs.3120$

Let the $C{P}_1$ of $1$st table be $Rs.x$

And, the $C{P}_2$ of the $2$nd table $=Rs.(3120-x)$

$1$st table is sold at a loss of $15\%$

$\therefore S{P}_1=\frac{100-loss\%}{100}\times C{P}_1$

$=\frac{100-15}{100}\times x$

$=\frac{85x}{100}=\frac{17x}{20}$

$2$nd table is sold at a gain of $36\%$

$\therefore S{P}_2=\frac{100+gain\%}{100}\times C{P}_2$

$=\frac{100+36}{100}\times (3120-x)$

$=\frac{136(3120-x)}{100}$

$=\frac{34(3120-x)}{25}$

It is given that $S{P}_1=S{P}_2$

$\frac{17x}{20}=\frac{34(3120-x)}{25}$

$\Rightarrow \frac{x}{4}=\frac{2(3120-x)}{5}$

$\Rightarrow 5x=8(3120-x)$

$\Rightarrow 5x=8\times 3120-8x$

$\Rightarrow 5x+8x=8\times 3120$

$\Rightarrow x=\frac{8\times 3120}{13}$

$\therefore x=1920$

Hence, the cost price of $1$st table is $Rs.1920$ and Cost price of $2$nd table is $(3120-1920=RS.1200$)