Question

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

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

Answer 1

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

Let the original price of product be 'P'

Price increased by r%

$\therefore$ Increased price$=P(1+\frac{r}{100})$

New price decreased by r%

$\therefore$ Latest price$=\left[P(1+\frac{r}{100})\right](1-\frac{r}{100})$

$=P(1+\frac{r^2}{{100}^2})$

Given latest price=Re. $1$

$\therefore P(1+\frac{r^2}{{100}^2})=1$

$P=$Rs. $\left(\frac{10000}{10000-r^2}\right)$