Bunuel wrote:
Company X’s stock price drops by p%, and then drops by p% again, where 0<p<100. In terms of p, by what percent would the stock price need to increase in order to return to its initial value?
A. \(2p\%\)
B. \(p^2\%\)
C. \(\frac{200p}{100−2p}\%\)
D. \(\frac{100p^2}{10000−p^2}\%\)
E. \(\frac{100p(200−p)}{(p−100)^2}\%\)
Let, Initial stock price of \(X\) is \(100\).
After dropping \(p\%\), the value will be \(100-p\).
Again after dropping \(p\%\), the value will be \((100-p)(1-\frac{p}{100})\)
\((100-p)(1-\frac{p}{100})\)
\(=100-p-p+\frac{p^2}{100}\)
\(=100-2p+\frac{p^2}{100}\)
\(=\frac{(10000-200p+p^2)}{100}\)
\(=\frac{(100-p)^2}{100}\)
\(=\frac{(p-100)^2}{100}\)
∴ Net decrease in value is \(100-\frac{(p-100)^2}{100} = \frac{10000-(p-100)^2}{100}\)
Now the question becomes \(\frac{10000-(p-100)^2}{100}\) is what percent of \(\frac{(p-100)^2}{100}\)?
\(\frac{\frac{10000-(p-100)^2}{100}}{\frac{(p-100)^2}{100}}*100\)
\(=\frac{10000-(p-100)^2}{100}*\frac{100}{(p-100)^2}*100\)
\(=\frac{(10000-(p-100)^2)*100}{(p-100)^2}\)
\(=\frac{(10000-(p^2-2p.100+100^2))*100}{(p-100)^2}\)
\(=\frac{(10000-p^2+2p.100-10000)*100}{(p-100)^2}\)
\(=\frac{(200p-p^2)*100}{(p-100)^2}\)
\(=\frac{100p(200-p)}{(p-100)^2}\)
Answer: E. \(\frac{100p(200-p)}{(p-100)^2}\%\)
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