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Question: The owner of an art shop conducts his business in the following manner. Every once in a while he rai...

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

(A)2756.25 (B)2256.25 (C)2500 (D)2000  \left( A \right)\quad 2756.25 \\\ \left( B \right)\quad 2256.25 \\\ \left( C \right)\quad 2500 \\\ \left( D \right)\quad 2000 \\\
Explanation

Solution

In this question we have two unknowns, the original price of the painting and the XX. 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 PPrupees.
Now, the owner increased the price by X%X\% . Therefore, the new price will be
price  1=P+(X%  of  P) price  1=P+(X100×P) price  1=(100+X100)P  \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%X\% . Therefore, the new price of the painting will be
price  2=price  1(X%  of  price  1) price  2=(100X100)(price  1)  \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  1price\;1in it, we will get
price  2=(100X100)(100+X100)P\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%X\% . Therefore the new price will be
price  3=price  2+(X%  of  price  2) price  3=(100+X100)(price  2)  \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  2price\;2 in it, we get
price  3=(100+X100)(100X100)(100+X100)P\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%X\% . Therefore, the final price is
price  4=price  3(X%  of  price  3) price  4=(100X100)(price  3)  \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  3price\;3in it , we get
price  4=(100X100)(100+X100)(100X100)(100+X100)P\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
Pprice  2=441\Rightarrow P - price\;2 = 441 and Final  price=price  4=1944.811Final\;price = price\;4 = 1944.811
Using the above information, we will form the final two required equations.
Part 11:
Pprice  2=441\Rightarrow P - price\;2 = 441
Substituting value from above available data, we get
Pprice  2=441 P(100X100)(100+X100)P=441  \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
P=(2100X)2\Rightarrow P = {\left( {\dfrac{{2100}}{X}} \right)^2}
Part 22:

price  4=1944.811 price  4=(100X100)(100+X100)(100X100)(100+X100)P=1944.811  \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
(104X2104)2P=1944.811\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 11 and part 22. Since, we are only interested in finding the painting price PP. Therefore, we will substitute value of X2{X^2} from part 11 to part 22
(104X2104)2P=1944.811 (104(441×104P)104)2P=1944.811  \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
P22826.811P+194481=0\Rightarrow {P^2} - 2826.811P + 194481 = 0
On solving this quadratic equation, we get
P=70.55P = 70.55 or P=2756.25P = 2756.25
Hence, the correct answer is (A)  P=2756.25\left( A \right)\;P = 2756.25

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