The price of an article was increased by r %. | vedclass

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(D) Let the original price be $P$.According to the problem,the price is first increased by $r \%$ and then decreased by $r \%$.The final price is give

Vedclass · Accepted Answer

$₹ \left(\frac{10000}{10000-r^{2}}\right)$

Vedclass · Answer

(D) Let the original price be $P$.According to the problem,the price is first increased by $r \%$ and then decreased by $r \%$.The final price is given as $₹ 1$.So,$P 	imes (1 + \frac{r}{100}) 	imes (1 - \frac{r}{100}) = 1$.Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$,we get:$P 	imes (1 - (\frac{r}{100})^2) = 1$.$P 	imes (1 - \frac{r^2}{10000}) = 1$.$P 	imes (\frac{10000 - r^2}{10000}) = 1$.Therefore,$P = \frac{10000}{10000 - r^2}$.

The price of an article was increased by $r \%$. Later,the new price was decreased by $r \%$. If the latest price was $₹ 1$,then the original price was:

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