CAT 2017 Quantitative Ability

Question : 32 of 34

Marks: +1, -0

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 ₹ 21025. After a second up down cycle the jewellery was sold for ₹ 484416. What was the original price of the jewellery.

₹ 500000
₹ 600625
₹ 525625
₹ 526000

Validate

Solution:

Let the original price P, then the decrease in value of P afterone cycle

=p(\;\;{x}/{100})^{2}=21025

Again the final value after second cycle

P(1+\;\;{x}/{100})(1-\;\;{x}/{100})(1+\;\;{x}/{100})(1-\;\;{x}/{100})=484416

P[1-(\;\;{x}/{100})^{2}]=484416

Dividing Eq. (ii) by Eq (i), we get

\;\;{[1-(\;\;{x}/{100})^{2}]^{2}}/{(\;\;{x}/{100})^{2}}=\;\;{484416}/{21025}=\;\;{2304}/{100}

\;\;{1-(\;\;{x}/{100})^{2}}/{(\;\;{x}/{100})^{2}}=√{\;\;{2304}/{100}}=\;\;{48}/{10}

Let

\;\;{x}/{100}=k,\;\text"then"\;\;\;{1-k^{2}}/{k}=\;\;{48}/{10}

10k^{2}+48k-10=0

5k^{2}-24x-5=0

k=5\;\text"or"\;k=-\;\;{1}/{5}

x = 20%

Hence,

P(\;\;{x}/{100})^{2}=21025

P = 525625

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