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Bunuel
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Company X’s stock price drops by p%, and then drops by p% again, where 10 Jan 2018, 20:50
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E
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65% (hard)

Question Stats:

68% (02:59) correct 32% (02:53) wrong based on 444 sessions

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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. \(\frac{100p(200−p)}{(p−100)^2}\%\)
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E
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Company X’s stock price drops by p%, and then drops by p% again, where 11 Jan 2018, 02:55
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Bunuel
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. \(\frac{100p(200−p)}{(p−100)^2}\%\)
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Instead of creating an equation, we'll try easy numbers.
This is an Alternative approach.

Say there were 100 units of stock which decreased by p=10% twice.
Then we now have 100 - 10 - 9 = 81 units of stock.
So we need to increase by 19/81 = a bit less than 1/4 = 25% to get back to the original.
Plugging p=10% into our answers:
A. 20% No!
B. 100%. No!
C. 2000/80 = 200/8 = 100/4 = 25%. No!
D. 10,000/9,900. No!
E. 1000*190/(-90)^2 = 190,000/8,100 = about 190/8 = a bit less than 25 (because 8*25 = 200). Yes!

(E) is our answer.
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Company X’s stock price drops by p%, and then drops by p% again, where 13 Jan 2018, 00:24
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Bunuel
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. \(\frac{100p(200−p)}{(p−100)^2}\%\)
Show more

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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Company X’s stock price drops by p%, and then drops by p% again, where 13 Jan 2018, 02:33
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Bunuel
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. \(\frac{100p(200−p)}{(p−100)^2}\%\)
Show more


Try substitution with smart numbers...
Here it would be 100 initial price and reduction of 20 each time..
So 100*80/100*80/100=64...
Total decrease =100-64=36 and percentage=3600/64.. You can leave it here as our choices are in fraction
E is
\(\frac{100p(200−p)}{(p−100)^2}\%\) ...
\(\frac{100*20(200−20)}{(20−100)^2}\%=100*20*180/80*80=20*180/64=3600/64\)
Exactly what we are looking for..
E
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Company X’s stock price drops by p%, and then drops by p% again, where 14 Jan 2018, 13:46
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Plugged in values

P= 50%
original value =100.
First drop = 100*(50/100) = 50
Second drop = 50*(50/100) = 25

For the stock to return to original value, I used the following equation: New-Old/New

(25 -100)/25 = 300%

I then plugged in P=50 in answer E and got the answer matching 300%.

Answer E
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Company X’s stock price drops by p%, and then drops by p% again, where 18 Jan 2018, 08:47
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Bunuel
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. \(\frac{100p(200−p)}{(p−100)^2}\%\)
Show more

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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Company X’s stock price drops by p%, and then drops by p% again, where 12 Jan 2026, 08:27
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Bunuel Is there any way to solve this question faster. I am taking over 3 mins using variable substitution.
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