
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)?
\[
\left( A \right)\quad 2756.25 \\
\left( B \right)\quad 2256.25 \\
\left( C \right)\quad 2500 \\
\left( D \right)\quad 2000 \\
\]
Answer
592.8k+ views
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.
Complete step-by-step answer:
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 $P$.
Complete step-by-step answer:
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 $P$.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What steps did the French revolutionaries take to create class 11 social science CBSE

The transition element that has lowest enthalpy of class 11 chemistry CBSE

Can anyone list 10 advantages and disadvantages of friction

