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The price of an article was increased by $p\%$, later the new price was decreased by $p\%$. If the last price was Re. $1$ then the original price was:

- $\dfrac{1-p^{2}}{200}\\$
- $\dfrac{\sqrt{1-p^{2}}}{100} \\$
- $1-\dfrac{p^{2}}{10,000-p^{2}} \\$
- $\dfrac{10,000}{10,000-p^{2}}$

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Ans is option (D)

Let the original price be $x$ Rs. Price after $p\%$ increase: $x(1+\frac{p}{100})$ Rs.

Price after $p\%$ decrease: $x(1+\frac{p}{100})-[\frac{p}{100}\times x(1+\frac{p}{100})]=1$ (given in question)

$\therefore$ $x(1+\frac{p}{100})(1-\frac{p}{100})=1$ $\Rightarrow$ $x=\frac{10000}{10000-p^{2}}$ Rs.

Let the original price be $x$ Rs. Price after $p\%$ increase: $x(1+\frac{p}{100})$ Rs.

Price after $p\%$ decrease: $x(1+\frac{p}{100})-[\frac{p}{100}\times x(1+\frac{p}{100})]=1$ (given in question)

$\therefore$ $x(1+\frac{p}{100})(1-\frac{p}{100})=1$ $\Rightarrow$ $x=\frac{10000}{10000-p^{2}}$ Rs.