Question

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

• A. Rs. $1$
• B. Rs. $\left(\frac{1-r^2}{100}\right)$
• C. Rs. $\frac{\sqrt{1-r^2}}{100}$
• D. Rs. $\frac{10000}{10000-r^2}$

Answer 1

The correct option is D Rs. $\frac{10000}{10000-r^2}$

Let the original price of the article be Rs. $x$

Then Increased price of the article $=\left(\frac{100+r}{100}\right)\times x$
Decreased price of the article after $\%$ decrease
$=\left(\frac{100+r}{100}\right)\left(\frac{100-r}{100}\right)\times x=\left(\frac{10000-r^2}{10000}\right)\times x$
Given, $\frac{10000-r^2}{10000}\times x=1$

$\Rightarrow x=\frac{10000}{10000-r^2}$