# NIELIT 2019 Feb Scientist C - Section D: 28

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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:

1. $\dfrac{1-p^{2}}{200}\\$
2. $\dfrac{\sqrt{1-p^{2}}}{100} \\$
3. $1-\dfrac{p^{2}}{10,000-p^{2}} \\$
4. $\dfrac{10,000}{10,000-p^{2}}$

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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.
238 points 2 2 4

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