Question

The price of an article was increased by \[P\% \]. Later the new price was decreased by \[P\% \]. If the latest price was \[Rs.1\], then the original price was?
A. \[Rs.5\]
B. \[Rs.\dfrac{{1 - {p^2}}}{{100}}\]
C. \[Rs.\dfrac{{10,000}}{{10,000 - {p^2}}}\]
D. \[Rs.\sqrt {\dfrac{{1 - {p^2}}}{{100}}} \]

Answer 1

Hint: Here we use the concept of percentage and assuming the original price as a variable we add and subtract the percentage for the two situations, then we equate the latest price with the equation formed.
* If a value \[x\] is increased by \[P\% \] then the new value $ = x(1 + \dfrac{P}{{100}})$
* If a value \[y\] is decreased by \[P\% \] then the new value $ = y(1 - \dfrac{P}{{100}})$

Complete step-by-step answer:
Let the Initial price of the article was \[Rs.x\]
First we find the price of the article after increasing by \[P\% \]
We know \[P\% \] of \[x = \dfrac{P}{{100}}x\]
Then the new price $ = x + \dfrac{P}{{100}}x$
                                    $ = x(1 + \dfrac{P}{{100}})$
Now we find the expression of the price obtained after decreasing the new price by \[P\% \]
 The changed price will be obtained by decreasing $x(1 + \dfrac{P}{{100}})$ by \[P\% \]
We know \[P\% \] of $x(1 + \dfrac{P}{{100}}) = \dfrac{P}{{100}}\left\{ {x\left( {1 + \dfrac{P}{{100}}} \right)} \right\}$
Therefore, decreasing by \[P\% \]
Then the changed price will be the price after increasing the percentage minus the percentage of that price.
  \[ = x + \dfrac{P}{{100}}x - \dfrac{P}{{100}}(x + \dfrac{P}{{100}}x)\]
Take common \[(x + \dfrac{P}{{100}}x)\].
\[ = (x + \dfrac{P}{{100}}x)(1 - \dfrac{P}{{100}})\]
Take common \[x\]
\[
   = x(1 + \dfrac{P}{{100}})[1 - \dfrac{P}{{100}}] \\
   = x(1 - \dfrac{{{P^2}}}{{{{100}^2}}}) \\
 \] {since, \[(a - b)(a + b) = {a^2} - {b^2}\]}
Equate that changed price with the price given in the question
$x(1 - \dfrac{{{P^2}}}{{10000}}) = 1$
Taking LCM on LHS of the equation
$x(\dfrac{{10000 - {P^2}}}{{10000}}) = 1$
Multiply both sides of the equation by \[10000\]
$x(\dfrac{{10000 - {P^2}}}{{10000}}) \times 10000 = 1 \times 10000$
Cancel out the same terms from numerator and denominator
$x(10000 - {P^2}) = 10000$
Divide both sides by $(10000 - {P^2})$
$\dfrac{{x(10000 - {P^2})}}{{(10000 - {P^2})}} = \dfrac{{10000}}{{(10000 - {P^2})}}$
Cancel the same terms from denominator and numerator.
$x = \dfrac{{10000}}{{(10000 - {P^2})}}$
The original price $x = \dfrac{{10000}}{{(10000 - {P^2})}}$
So the original price is $Rs.\dfrac{{10000}}{{(10000 - {P^2})}}$

So, the correct answer is “Option C”.

Note: Students are likely to make mistakes while calculating the decreased percentage part as they decrease the percentage from initial price which is wrong, they have to take the increased value and then apply the decreasing part on that value. Also, always write the final answer along with the unit.