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: 2506
Given Kudos: 1055
GPA: 3.39
At the beginning of year 1, an investor puts [m]p[/m] dollars into an
[#permalink]
24 Aug 2021, 05:27
24 Aug 2021, 05:27
1
Expert Reply
11
Bookmarks
00:00
A
B
C
D
E
Question Stats:
60% (02:25) correct
40% (02:41) wrong based on 15 sessions
Show timer Statistics
At the beginning of year 1, an investor puts p dollars into an investment whose value increases at a variable rate of xn per year, where n is an integer ranging from 1 to 3 indicating the year. If 85<xn<110 for all n between 1 and 3, inclusive, then at the end of 3 years, the value of the investment must be between
A. $p and $2p
B. $2p and $5p
C. $5p and $10p
D. $10p and $25p
E. $25p and $75p
A. $p and $2p
B. $2p and $5p
C. $5p and $10p
D. $10p and $25p
E. $25p and $75p
Re: At the beginning of year 1, an investor puts [m]p[/m] dollars into an
[#permalink]
27 Aug 2021, 10:14
27 Aug 2021, 10:14
1
Expert Reply
The quick solution to this problem is to pick a convenient number in the allowed range of growth rates. If xnxn is always between 85 and 110, then the most convenient growth rate to pick is 100. An annual growth rate of 100% is exactly equivalent to doubling one's money. Doubling one's money three times is equivalent to multiplying one's investment by a factor of 8 (=23=23). The only range that includes $8p8p is the third range ($5p5p to $10p10p).
Computing the exact outer limits of the allowed range is much more cumbersome. We would have to cube 1.85 for the lower limit and 2.1 for the upper limit. The cube of 1.85 is 6.331625, and the cube of 2.1 is 9.261. These values fall between 5 and 10.
Answer: C.
Computing the exact outer limits of the allowed range is much more cumbersome. We would have to cube 1.85 for the lower limit and 2.1 for the upper limit. The cube of 1.85 is 6.331625, and the cube of 2.1 is 9.261. These values fall between 5 and 10.
Answer: C.
Re: At the beginning of year 1, an investor puts [m]p[/m] dollars into an
[#permalink]
25 Dec 2023, 19:01
25 Dec 2023, 19:01
Hello from the GRE Prep Club BumpBot!
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
