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At the beginning of a year ,the owner of a jewel shop raised the price of all the jewels in his shop by x% and lowered them by x%. The price of one jewel after this up and down cycle reduced by Rs. 100. The owner carried out the same procedure after a month. After this second up-down cycle,the price of that jewel was Rs. 2304. Find the original price of that jewel(in Rs.)
Answer & Solution
Correct Answer:
Option
A
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{original}}\,{\text{price}} = y, \cr
& {\text{After}}\,{\text{first}}\,{\text{change,}}\,{\text{it}}\,{\text{becomes}}, \cr
& y \times \left( {1 + {\frac{x}{{100}}} } \right) \cr
& {\text{After}}\,{\text{second}}\,{\text{change,}}\,{\text{it}}\,{\text{becomes}} \cr
& y \times \left( {1 + {\frac{x}{{100}}} } \right)\left( {1 - {\frac{x}{{100}}} } \right) \cr
& = y\left( {1 - {{\left( {\frac{x}{{100}}} \right)}^2}} \right) \cr
& {\text{Thus}}, \cr
& {x^2} \times y = {10^6} - - - - \left( 1 \right) \cr
& {x^2} = \frac{{{{10}^6}}}{y} \cr
& {\text{Now}}, \cr
& y{\left( {1 - {\frac{{{{10}^6}}}{{10000y}}} } \right)^2} \cr
& = 2304\left( {{\text{similar}}\,{\text{to}}\,{\text{above}}} \right) \cr
& y{\left( {1 - \frac{{100}}{y}} \right)^2} = 2304 \cr
& y = 2500 \cr} $$
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LoginSo, 100>>>130>>>91
Here the change is (100-91)=9
Which is always X^2/100 percent
Here, (30)^2/100 percent.
So the relation will always be X^2/100 percent.
Here in the math we can say that
2500*(X^2/100 )=100
So X^2=4,
So the loss is 4%.
Back calculating it from 2500>>2400>>2304.
So 2500 is the right ans.
And y(1-100/y)2 = 2304
y = 2500. How is it possible??