A shopkeeper first raises the price of Jewellery by x% then he decreases the new price by x%. After such up down cycle, the price of a Jewellery decreased by Rs. 21025. After a second up down cycle the Jewellery was sold for Rs. 484416. What was the original price of the jewellery.2

Explanation:-

$$\eqalign{ & {\text{Let the initial price}} \cr & = Rs.\,10000p \cr & {\text{price after first increment}} \cr & = 10000p + 100xp \cr & {\text{price after first decrement}} \cr & = 10000p + 100xp - \left( {100px + p{x^2}} \right) \cr & = 10000p - p{x^2} \cr & {\text{Now,}}\,{\text{total}}\,{\text{decrement}}, \cr & p{x^2} = 21025\,.............\left( 1 \right) \cr & \cr & {\text{price after second increment}}, \cr & 10000p - p{x^2} + 100xp - \frac{{p{x^3}}}{{100}} \cr & {\text{price after second increment}}, \cr & 10000p - {p^2} + 100xp - \frac{{{p^3}}}{{100}} - 100xp + \frac{{p{x^3}}}{{100}} - p{x^2} + \frac{{p{x^4}}}{{10000}} \cr & = 10000p - 2p{x^2} + \frac{{p{x^2}}}{{10000}} \cr & = 484416\,................\left( 2 \right) \cr & \cr & {\text{on solving equation equn }}\left( {\text{1}} \right){\text{ and }}\left( {\text{2}} \right){\text{, We get}} \cr & x = 20 \cr & {\text{substituting back we get,}} \cr & p = 525625 \cr}$$