GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Oct 2019, 12:40 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Company X’s stock price drops by p%, and then drops by p% again, where

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Company X’s stock price drops by p%, and then drops by p% again, where  [#permalink]

Show Tags 00:00

Difficulty:   55% (hard)

Question Stats: 63% (03:22) correct 37% (03:13) wrong based on 85 sessions

HideShow timer Statistics

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. $$\frac{100p(200−p)}{(p−100)^2}\%$$

_________________
examPAL Representative P
Joined: 07 Dec 2017
Posts: 1153
Company X’s stock price drops by p%, and then drops by p% again, where  [#permalink]

Show Tags

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.
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.
_________________
Retired Moderator P
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 484
Company X’s stock price drops by p%, and then drops by p% again, where  [#permalink]

Show Tags

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. $$\frac{100p(200-p)}{(p-100)^2}\%$$
_________________
Hasan Mahmud
Math Expert V
Joined: 02 Aug 2009
Posts: 7978
Re: Company X’s stock price drops by p%, and then drops by p% again, where  [#permalink]

Show Tags

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
_________________
Manager  B
Joined: 08 Sep 2016
Posts: 102
Re: Company X’s stock price drops by p%, and then drops by p% again, where  [#permalink]

Show Tags

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%.

Target Test Prep Representative D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8104
Location: United States (CA)
Re: Company X’s stock price drops by p%, and then drops by p% again, where  [#permalink]

Show Tags

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

_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: Company X’s stock price drops by p%, and then drops by p% again, where   [#permalink] 18 Jan 2018, 08:47
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  