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

 It is currently 19 Oct 2018, 21:22

### 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

# At the beginning of year 1, an investor puts p dollars into an investm

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50002
At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

### Show Tags

17 Nov 2014, 12:48
00:00

Difficulty:

75% (hard)

Question Stats:

57% (02:34) correct 43% (02:51) wrong based on 158 sessions

### HideShow timer Statistics

Tough and Tricky questions: Percents and Interest Problems.

At the beginning of year 1, an investor puts $$p$$ dollars into an investment whose value increases at a variable rate of $$x_n%$$ per year, where $$n$$ is an integer ranging from 1 to 3 indicating the year. If $$85 \lt x_n \lt 110$$ 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$$$75p$$

Kudos for a correct solution.

_________________
Manager
Joined: 21 Jan 2014
Posts: 62
WE: General Management (Non-Profit and Government)
Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

### Show Tags

17 Nov 2014, 23:43
1
for finding the range :

Max limit = p(1+110/100)^3=10p
Min limit = p(1+85/100)^3= 5p

Current Student
Joined: 02 Sep 2014
Posts: 89
Location: United States
Schools: Haas EWMBA '20
GMAT 1: 770 Q50 V44
GPA: 3.97
Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

### Show Tags

17 Nov 2014, 23:52
1
2
Bunuel wrote:

Tough and Tricky questions: Percents and Interest Problems.

At the beginning of year 1, an investor puts $$p$$ dollars into an investment whose value increases at a variable rate of $$x_n%$$ per year, where $$n$$ is an integer ranging from 1 to 3 indicating the year. If $$85 \lt x_n \lt 110$$ 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$$$75p$$
Kudos for a correct solution.

Nice problem! I think the whole xn business is just a tricky way of saying this is a compound interest problem.

Since the min rate of interest is 86% the min value of investment will be (1+0.86)^3*p = (1.86^3)*p. Similarly max value will be (2.1)^3*p.

2.1^3 is 9.5 or you can guess little over 8 so 10p may be good estimate.
1.86^3 is trickier to estimate. I knew root(3) = 1.73. So 1.86^2 is close to 3 and 3*1.8 is 5.4. So 5p is a close lower bound.

So range is 5p to 10p.
Intern
Joined: 04 Jul 2011
Posts: 9
Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

### Show Tags

18 Nov 2014, 00:04
2
At the beginning of year 1, an investor puts p dollars into an investment whose value increases at a variable rate of x_n% per year, where n is an integer ranging from 1 to 3 indicating the year. If 85 \lt x_n \lt 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

The min. value for Xn for 1, 2 & 3 year can be 86%, 87% & 88% ( They can be as close to 85 as possible but for simplicity i am taking these rate and carefull about variable so cannot take all value same)
similarly Max value can be 109%, 108% & 107%
Now consider p 100
than solve by increasing it by saubsequent rate for both max and min for
for min rate it will be around 546 and for max it will be around 816 that is the range of final value and by puting p=100 in option u will get option C as the range.
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1829
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

### Show Tags

18 Nov 2014, 01:43
2
Answer = C = $5p and$10p

Min limit $$= (1 + \frac{86}{100})^3 = (1 + 0.86)^3 = 1.86^3 =$$ Below 8

Max limit $$= (1 + \frac{109}{100})^3 = (1 + 1.09)^3 = 2.09^3 =$$ Above 8
_________________

Kindly press "+1 Kudos" to appreciate

Math Expert
Joined: 02 Sep 2009
Posts: 50002
Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

### Show Tags

18 Nov 2014, 08:08
3
1
Official Solution:

At the beginning of year 1, an investor puts $$p$$ dollars into an investment whose value increases at a variable rate of $$x_n%$$ per year, where $$n$$ is an integer ranging from 1 to 3 indicating the year. If $$85 \lt x_n \lt 110$$ 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$$$75p$$

The quick solution to this problem is to pick a convenient number in the allowed range of growth rates. If $$x_n$$ 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 ($$= 2^3$$). The only range that includes $$$8p$$ is the third range ($$$5p$$ to $$$10p$$). 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. _________________ Senior Manager Joined: 03 Apr 2013 Posts: 280 Location: India Concentration: Marketing, Finance GMAT 1: 740 Q50 V41 GPA: 3 Re: At the beginning of year 1, an investor puts p dollars into an investm [#permalink] ### Show Tags 07 Nov 2016, 23:49 Bunuel wrote: Tough and Tricky questions: Percents and Interest Problems. At the beginning of year 1, an investor puts $$p$$ dollars into an investment whose value increases at a variable rate of $$x_n%$$ per year, where $$n$$ is an integer ranging from 1 to 3 indicating the year. If $$85 \lt x_n \lt 110$$ 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 \$$$75p$$
Kudos for a correct solution.

One question...How do we know that they are talking about compound interest and not simple interest?
_________________

Spread some love..Like = +1 Kudos

Manager
Joined: 15 Apr 2016
Posts: 76
Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

### Show Tags

12 Nov 2016, 22:27
Even i have the same query as ShashankDave

Bunuel ,can you please clarify the below query

ShashankDave wrote:
One question...How do we know that they are talking about compound interest and not simple interest?

_________________

Cheers,
Shri
-------------------------------
GMAT is not an Exam... it is a war .. Let's Conquer !!!

Non-Human User
Joined: 09 Sep 2013
Posts: 8459
Re: At the beginning of year 1, an investor puts p dollars into an investm  [#permalink]

### Show Tags

19 Apr 2018, 23:26
Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.
_________________
Re: At the beginning of year 1, an investor puts p dollars into an investm &nbs [#permalink] 19 Apr 2018, 23:26
Display posts from previous: Sort by