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# The owner of an art shop conducts his business in the following manner. Every once in a while he raises his prices by $X\%$. Then a while later he reduces all the new prices by $X\%$. After one such up-down cycle, the price of painting decreased by $Rs.441$. After a second up-down cycle, the painting was sold for $Rs.1944.811$. What was the original price of the painting (in Rs)?$\left( A \right)\quad 2756.25 \\ \left( B \right)\quad 2256.25 \\ \left( C \right)\quad 2500 \\ \left( D \right)\quad 2000 \\$

Last updated date: 22nd Jun 2024
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Hint: In this question we have two unknowns, the original price of the painting and the $X$. Therefore, we need at least two equations to find out the exact value of these two unknowns. We will be obtaining these equations from the conditions and data given to us in the question.

Let us assume that the original price of the painting is $P$rupees.
Now, the owner increased the price by $X\%$. Therefore, the new price will be
$\Rightarrow price\;1 = P + \left( {X\% \;of\;P} \right) \\ \Rightarrow price\;1 = P + \left( {\dfrac{X}{{100}} \times P} \right) \\ \Rightarrow price\;1 = \left( {\dfrac{{100 + X}}{{100}}} \right)P \\$
At this point of time, the owner reduced the price by $X\%$. Therefore, the new price of the painting will be
$\Rightarrow price\;2 = price\;1 - \left( {X\% \;of\;price\;1} \right) \\ \Rightarrow price\;2 = \left( {\dfrac{{100 - X}}{{100}}} \right)(price\;1) \\$
Now substituting the value of $price\;1$in it, we will get
$\Rightarrow price\;2 = \left( {\dfrac{{100 - X}}{{100}}} \right)\left( {\dfrac{{100 + X}}{{100}}} \right)P$
Now the owner again raised the price by $X\%$. Therefore the new price will be
$\Rightarrow price\;3 = price\;2 + \left( {X\% \;of\;price\;2} \right) \\ \Rightarrow price\;3 = \left( {\dfrac{{100 + X}}{{100}}} \right)(price\;2) \\$
Now substituting the value of $price\;2$ in it, we get
$\Rightarrow price\;3 = \left( {\dfrac{{100 + X}}{{100}}} \right)\left( {\dfrac{{100 - X}}{{100}}} \right)\left( {\dfrac{{100 + X}}{{100}}} \right)P$
At last, the owner decreased the price by $X\%$. Therefore, the final price is
$\Rightarrow price\;4 = price\;3 - \left( {X\% \;of\;price\;3} \right) \\ \Rightarrow price\;4 = \left( {\dfrac{{100 - X}}{{100}}} \right)(price\;3) \\$
Now substituting the value of $price\;3$in it , we get
$\Rightarrow price\;4 = \left( {\dfrac{{100 - X}}{{100}}} \right)\left( {\dfrac{{100 + X}}{{100}}} \right)\left( {\dfrac{{100 - X}}{{100}}} \right)\left( {\dfrac{{100 + X}}{{100}}} \right)P$
In the question , we are told that
$\Rightarrow P - price\;2 = 441$ and $Final\;price = price\;4 = 1944.811$
Using the above information, we will form the final two required equations.
Part $1$:
$\Rightarrow P - price\;2 = 441$
Substituting value from above available data, we get
$\Rightarrow P - price\;2 = 441 \\ \Rightarrow P - \left( {\dfrac{{100 - X}}{{100}}} \right)\left( {\dfrac{{100 + X}}{{100}}} \right)P = 441 \\$
Which on simplifying gives us
$\Rightarrow P = {\left( {\dfrac{{2100}}{X}} \right)^2}$
Part $2$:
$\Rightarrow price\;4 = 1944.811 \\ \Rightarrow price\;4 = \left( {\dfrac{{100 - X}}{{100}}} \right)\left( {\dfrac{{100 + X}}{{100}}} \right)\left( {\dfrac{{100 - X}}{{100}}} \right)\left( {\dfrac{{100 + X}}{{100}}} \right)P = 1944.811 \\$
Which on simplifying gives us
$\Rightarrow {\left( {\dfrac{{{{10}^4} - {X^2}}}{{{{10}^4}}}} \right)^2}P = 1944.811$
Now, at last we have two very simplified equations from part $1$ and part $2$. Since, we are only interested in finding the painting price $P$. Therefore, we will substitute value of ${X^2}$ from part $1$ to part $2$
$\Rightarrow {\left( {\dfrac{{{{10}^4} - {X^2}}}{{{{10}^4}}}} \right)^2}P = 1944.811 \\ \Rightarrow {\left( {\dfrac{{{{10}^4} - \left( {\dfrac{{441 \times {{10}^4}}}{P}} \right)}}{{{{10}^4}}}} \right)^2}P = 1944.811 \\$
Which on further simplifying becomes
$\Rightarrow {P^2} - 2826.811P + 194481 = 0$
On solving this quadratic equation, we get
$P = 70.55$ or $P = 2756.25$
Hence, the correct answer is $\left( A \right)\;P = 2756.25$

Note: We must take care which unknown variable we need to eliminate. If, we have eliminated $P$ instead of $X$, then the question would have further calculations like substituting the calculated value of $X$to find the Value of $P$.