Question

In the Garbar Jhala, Aminabad a shopkeeper first raises the price of a Jewellery by x% then he decreases the new price by x%. After one 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?

• A. Rs. 5,00,000
• B. Rs. 6,00,625
• C. Rs. 525625
• D. Rs. 5,26,000

Answer 1

The correct option is C Rs. 525625

Let the original price be p, then the decrease in value of p after one cycle.
$=p{\left(\frac{x}{100}\right)}^2=21025\dots \dots \left(i\right)$
Again the final value after second cycle
$\Rightarrow p(1+\frac{x}{100})(1-\frac{x}{100})(1+\frac{x}{100})(1-\frac{x}{100})=484416\Rightarrow p{[1-{\left(\frac{x}{100}\right)}^2]}^2=484416\dots \dots \left(ii\right)$
Dividing equation (ii) by equation (i)
$\frac{{[1-{\left(\frac{x}{100}\right)}^2]}^2}{{\left(\frac{x}{100}\right)}^2}=\frac{484416}{21025}=\frac{2304}{100}\Rightarrow \frac{1-{\left(\frac{x}{100}\right)}^2}{\left(\frac{x}{100}\right)}=\sqrt{\frac{2304}{100}}=\frac{48}{10}$
Let $\frac{x}{100}=k$, then
$\begin{array}{ccc}& \frac{1-k^2}{k}=\frac{48}{10}& \\ \Rightarrow & 10k^2+48k-10=0& \\ \Rightarrow & 5k^2-24x-5=0& \\ \Rightarrow & k=5ork=-\frac{1}{5}& \left(inadmissiblevalue\right)\\ So,& x=20\%& \\ Hence,& p{\left(\frac{x}{100}\right)}^2=21025& \\ \Rightarrow & p=525625& \end{array}$