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

## The correct option is A Rs. 525625Let the original price be p, then the decrease in value of p after one cycle. =p(x100)2=21025……(i) Again the final value after second cycle ⇒p(1+x100)(1−x100)(1+x100)(1−x100)=484416⇒p[1−(x100)2]2=484416……(ii) Dividing equation (ii) by equation (i) [1−(x100)2]2(x100)2=48441621025=2304100⇒ 1−(x100)2(x100)=√2304100=4810 Let x100=k, then 1−k2k=4810⇒10k2+48k−10=0⇒5k2−24x−5=0⇒k=5 or k=−15(inadmissible value)So,x=20%Hence,p(x100)2=21025⇒p=525625

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