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# Company X’s stock price drops by p%, and then drops by p% again, where

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Math Expert
Joined: 02 Sep 2009
Posts: 46251
Company X’s stock price drops by p%, and then drops by p% again, where [#permalink]

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10 Jan 2018, 20:50
00:00

Difficulty:

65% (hard)

Question Stats:

62% (01:56) correct 38% (02:27) wrong based on 69 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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Company X’s stock price drops by p%, and then drops by p% again, where [#permalink]

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11 Jan 2018, 02:55
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. $$\frac{100p(200−p)}{(p−100)^2}\%$$

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.
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!

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Company X’s stock price drops by p%, and then drops by p% again, where [#permalink]

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13 Jan 2018, 00:24
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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Hasan Mahmud

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Joined: 02 Aug 2009
Posts: 5911
Re: Company X’s stock price drops by p%, and then drops by p% again, where [#permalink]

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13 Jan 2018, 02:33
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. $$\frac{100p(200−p)}{(p−100)^2}\%$$

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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Re: Company X’s stock price drops by p%, and then drops by p% again, where [#permalink]

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14 Jan 2018, 13:46
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%.

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Re: Company X’s stock price drops by p%, and then drops by p% again, where [#permalink]

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18 Jan 2018, 08:47
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

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Re: Company X’s stock price drops by p%, and then drops by p% again, where   [#permalink] 18 Jan 2018, 08:47
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