Lakshman Patel RJIT
asked
in Quantitative Aptitude
Apr 1, 2020
recategorized
Nov 8, 2020
by Krithiga2101

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0 votes

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.