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
(a)₹ 1
(b)$₹\left(\frac{1-r^{2}}{100}\right)$
(c)$₹ \frac{\sqrt{1-r^{2}}}{100}$
(d)$₹\left(\frac{10000}{10000-r^{2}}\right)$
Answer
Answer (as printed): D
Explanation
Let the original price be ₹ $x$. $$\begin{aligned} & \therefore \quad(100-r) \% \text { of }(100+r) \% \text { of } x=1 \\ & \Rightarrow \quad \frac{(100-r)}{100} \times \frac{(100+r)}{100} \times x=1 \\ & \Rightarrow \quad x=\frac{100 \times 100}{(100-r)(100+r)}=\frac{10000}{\left(10000-r^{2}\right)} . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.