Question

A super bazar prices on items in rupees and paise. So that when $4\% $ sale tax is added, no rounding is necessary because the result is exactly in $n$ rupees, where $n$ is a positive integer. Find the smallest value of $n$

Answer 1

Hint: Let final rupees of an item be $n$
And let initial rupees before tax added be $x$
Now when sale tax is $4\% $
Then $n$ becomes sum of $x$ and $4\% $ of $x$
Hence find the value of $n$ for when $x$ is a terminating decimal.

Complete step-by-step answer:
So, according to the question, Let the initial rupees before tax added be $x$ by super bazar prices.
Now after including sales tax of $4\% $, let final price will become $n$
And here it is saying $n$ must be a positive integer.
So we know initial price be $x$
And after adding $4\% $ tax it becomes $n$. So
$
  n = x + 4\% x \\
  n = x + \dfrac{4}{{100}}x \\
  n = \dfrac{{100x + 4x}}{x} \\
 $
So we get $n$ in terms of $x$
$n = \dfrac{{104x}}{{100}} = \dfrac{{26x}}{{25}}$
So $x = \dfrac{{25n}}{{26}}$
Now as we know $x$ can be rupees and paise. So $x$ must be terminating decimal and $n$ must be a positive integer.
So $n$ is a factor of $26$
So $n$ can be \[1,2,13,26\]
So it can be represented in form of rupees and paise
But for $n = 1,2$, $x$ can’t be terminated
So $n$ must be either \[13\,\,or\,\,26\]
So question is saying for smallest \[ + n\]
So answer be $n = 13$

Note: If we include any tax on any item then its price will increase by percentage of tax of the actual price of the item
Suppose the actual price is $x$ and tax of $n\% $ is applied. So it’s price becomes $x + \dfrac{{nx}}{{100}}\,$ or $\,x + n\% \,of\,x$