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.
Correct Answer: Rs. 5,25,625
Let the initial price = Rs. 10000p
Price after first increment = 10000p + 100xp
Price after first decrement
= 10000p + 100xp - (100px + px2)
= 10000p - px2
Now, total decrement, px2 = 21025 . . . . . (1)
Price after second increment,
= 10000p - px2 + 100xp - $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$
Price after second increment,
= 10000p - p2 + 100xp - $$\frac{{{{\text{p}}^3}}}{{100}}$$ - 100xp + $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$ - px2 + $$\frac{{{\text{p}}{{\text{x}}^4}}}{{10000}}$$
= 10000p - 2px2 + $$\frac{{{\text{p}}{{\text{x}}^2}}}{{10000}}$$
= 484416 . . . . . . (2)
On solving equation (1) and (2), We get
x = 20
Substituting back we get,
p = 5,25,625
Price after first increment = 10000p + 100xp
Price after first decrement
= 10000p + 100xp - (100px + px2)
= 10000p - px2
Now, total decrement, px2 = 21025 . . . . . (1)
Price after second increment,
= 10000p - px2 + 100xp - $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$
Price after second increment,
= 10000p - p2 + 100xp - $$\frac{{{{\text{p}}^3}}}{{100}}$$ - 100xp + $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$ - px2 + $$\frac{{{\text{p}}{{\text{x}}^4}}}{{10000}}$$
= 10000p - 2px2 + $$\frac{{{\text{p}}{{\text{x}}^2}}}{{10000}}$$
= 484416 . . . . . . (2)
On solving equation (1) and (2), We get
x = 20
Substituting back we get,
p = 5,25,625