NIELIT 2019 Feb Scientist C - Section D: 28

Question

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:

• A. $\dfrac{1-p^{2}}{200}\\$
• B. $\dfrac{\sqrt{1-p^{2}}}{100} \\$
• C. $1-\dfrac{p^{2}}{10,000-p^{2}} \\$
• D. $\dfrac{10,000}{10,000-p^{2}}$

Answer 1

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.