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. \(\frac{100p(200−p)}{(p−100)^2}\%\)

If we let the original price = m, the,n after dropping by p% twice, we have:

(100 - p)/100 x (100 - p)/100 x m = new price

Let the percent increase needed to return to its original value be n percent. We can create the following equation:

(100 - p)/100 x (100 - p)/100 x m x (100 + n)/100 = m

Dividing both sides by m, we have:

(100 - p)/100 x (100 - p)/100 x (100 + n)/100 = 1

(100 - p)^2(100 + n)/100^3 = 1

Multiplying both sides by 100^3 and dividing both sides by (100 - p)^2, we have

100 + n = 100^3/(100 - p)^2

n = 100^3/(100 - p)^2 - 100

We obtain the common denominator of (100 - p)^2 on the right side of the equation:

n = 100^3/(100 - p)^2 - 100(100 - p)^2/(100 - p)^2

We factor the common 100 from both terms of the right side of the equation:

n = 100/(100 - p)^2 x (100^2 - (100 - p)^2)

Recognizing that (100^2 - (100 - p)^2) is a difference of squares, we factor and combine like terms:

n = 100/(100 - p)^2 x [(100 - (100 - p))(100 + (100 - p))]

n = 100/(100 - p)^2 x [p(200 - p)]

n = 100p(200 - p)/(100 - p)^2

Answer: E

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Scott Woodbury-Stewart

Founder and CEO

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