The price of an article was increased $p\%$. Later the new price was decreased $p\%$. If the last price was one dollar, the original price was: 
$\textbf{(A)}\ \frac{1-p^2}{200}\qquad \textbf{(B)}\ \frac{\sqrt{1-p^2}}{100}\qquad \textbf{(C)}\ \text{one dollar}\qquad\\ \textbf{(D)}\ 1-\frac{p^2}{10000-p^2}\qquad \textbf{(E)}\ \frac{10000}{10000-p^2}$
Converting the word problem in to an equation, we take x as the initial amount, we get \[x(1+\dfrac{p}{100})(1-\dfrac{p}{100}) = 1\] Using, $(a+b)(a-b) = a^{2} - b^{2}$. We now simplify it to, \[x(1-\dfrac{p^{2}}{10000})=1\] Simplifying to, \[x(\dfrac{10000-p^{2}}{10000}) = 1\] Dividing both sides by $\dfrac{10000-p^{2}}{10000}$, we get the value of x. $\boxed{\textbf{(E)}\dfrac{10000}{10000-p^{2}}}$