Answer
414.9k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)