A dealer sells a table at $20 %$ profit. Had he purchased it for $6 %$ lesser cost and sold for Rs. $50$ less he would have earned a profit of $25 %$ . Find the cost of the table.

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Solution

Let the cost of table $= R s . x$
Profit $= 20 %$
$S P = C o s t + P r o f i t = x + 20 %$ of $x$ $= x + \frac{x}{5} = \frac{6 x}{5}$
Now, New cost price $= x - \frac{6}{100} \times x$ $= \frac{94 x}{100}$
New selling price $= \frac{6 x}{5} - 50$
New Profit = New SP - New CP $= \frac{6 x}{5} - 50 - \frac{94 x}{100} = \frac{26 x}{100} - 50$
New Profit $= 25 %$ of New CP $= \frac{25}{100} \times \frac{94 x}{100} = \frac{47 x}{200}$
Hence, $\frac{47 x}{200} = \frac{26 x}{100} - 50$
$\frac{47 x - 52 x}{200} = - 50$
$5 x = 200 \times 50$
$x = 2000 R s$
Thus, Cost Price $= R s . 2000$

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