Question

The price of an article was increased $p$%. Later the new price was decreased $p$%. If the last price was $Re.1$, the original price was

• A. $\frac{1-p^2}{200}$
• B. $\frac{\sqrt{1-p^2}}{200}$
• C. $1Re.$
• D. $1-\frac{p^2}{10,000-p^2}$
• E. $\frac{10,000}{10,000-p^2}$

Answer 1

Answer: (E)

$\frac{10,000}{10,000-p^2}$

Let the original price be $P$, Then $P_1=(1+\frac{P}{100})P$.
$P_2=(1-\frac{P}{100})P_1=(1-\frac{P}{100})(1+\frac{P}{100})P=1$;
$\therefore P=\frac{1}{1-\left(p^2/{100}^2\right)}=\frac{10,000}{10,000-p^2}$.