GRE Prep 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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
The value of an investment increases by x% during January and decrease
[#permalink]
31 Mar 2021, 06:00
31 Mar 2021, 06:00
1
1
Expert Reply
1
Bookmarks
00:00
Question Stats:
100% (02:53) correct
0% (00:00) wrong based on 4 sessions
Hide Show timer Statistics
The value of an investment increases by x% during January and decreases by y% during February. If the value of the investment is the same at the end of February as at the beginning of January, what is y in terms of x ?
A. (200x)\(100 + 2x)
B. x(2 + x)\(1 + x)^2
C. 2x\1 + 2x
D. x(200 + x)\10,000
E. 100 –( 10,000 \ 100 + x)
A. (200x)\(100 + 2x)
B. x(2 + x)\(1 + x)^2
C. 2x\1 + 2x
D. x(200 + x)\10,000
E. 100 –( 10,000 \ 100 + x)
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: The value of an investment increases by x% during January and decrease
[#permalink]
31 Mar 2021, 06:11
31 Mar 2021, 06:11
1
GeminiHeat wrote:
The value of an investment increases by x% during January and decreases by y% during February. If the value of the investment is the same at the end of February as at the beginning of January, what is y in terms of x ?
A. (200x)\(100 + 2x)
B. x(2 + x)\(1 + x)^2
C. 2x\1 + 2x
D. x(200 + x)\10,000
E. 100 –( 10,000 \ 100 + x)
A. (200x)\(100 + 2x)
B. x(2 + x)\(1 + x)^2
C. 2x\1 + 2x
D. x(200 + x)\10,000
E. 100 –( 10,000 \ 100 + x)
Let the investment at the beginning of January be \(I\)
Value at the end of January \(= (1 + \frac{x}{100})I\)
Value at the end of February \(= (1 - \frac{y}{100)}(1 + \frac{x}{100})I\)
Now, \(I = (1 - \frac{y}{100)}(1 + \frac{x}{100})I\)
\(1 = \frac{(100 - y)}{100}\frac{(100 + x)}{100}\)
\(10000 = (100 - y))(100 + x)\)
\(\frac{10000}{(100 + x)} = 100 - y\)
\(y = 100 - \frac{10000}{(100 + x)}\)
Hence, option E
gmatclubot
Re: The value of an investment increases by x% during January and decrease [#permalink]
31 Mar 2021, 06:11
Moderators:
