The price of an article was increased by r %. Later the new price was

1.	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\(\big(\frac{1-r^2}{100}\big)\)(c) Rs \(\frac{\sqrt{1-r^2}}{100}\)(d) Rs \(\frac{10000}{10000-r^2}\)
Answer» (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 = \(\frac{10000}{10000-r^2}\)

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\(\big(\frac{1-r^2}{100}\big)\)(c) Rs \(\frac{\sqrt{1-r^2}}{100}\)(d) Rs \(\frac{10000}{10000-r^2}\)

Answer»

(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 = \(\frac{10000}{10000-r^2}\)