(d) \(\frac{10000}{10000-r^2}\)
Let the original price of the article be Rs x. Then,
Increased price of the article =\(\big(\frac{100+r}{100}\big)\) x x
Decreased price of the article after r % decrease
= \(\big(\frac{100-r}{100}\big)\)\(\big(\frac{100-r}{100}\big)\)x x
Given\(\frac{10000-r^2}{10000}\) x x = 1
\(\Rightarrow\) x = \(\frac{10000}{10000-r^2}\)