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# An investment club has had an average rate of return of 15%

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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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An investment club has had an average rate of return of 15% [#permalink]  26 Apr 2017, 18:13
Expert's post
00:00

Question Stats:

90% (00:36) correct 10% (00:57) wrong based on 10 sessions
An investment club has had an average rate of return of 15% per year for the past 6 years. If Teresa invests \$1000 today and neither adds nor subtracts money from the club, how much will Teresa have invested after 5 years assuming that the rate of return does not change?

A. $$1000 + 1.15^5$$
B. $$1000(1.15)^5$$
C. $$1000 + 0.15$$
D. $$1000(0.15)^5$$
E. $$1000(5)^{0.15}$$

Drill 4
Question: 7
Page: 322
[Reveal] Spoiler: OA

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Sandy
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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 106

Kudos [?]: 1820 [0], given: 397

Re: An investment club has had an average rate of return of 15% [#permalink]  04 May 2017, 01:03
Expert's post
Explanation

To solve this question, remember that the formula for finding the result of periodic increases at a certain rate is

$$(Original_Amount)(1 + rate)^{number_of_periods}$$.

In this case, the final amount would be $$1000(1.15)^5$$.

The correct answer is choice (B).

If you forgot the formula, you could calculate the final amount after 5 years, and then calculate all the answers for a match.
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Sandy
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Re: An investment club has had an average rate of return of 15% [#permalink]  16 Jan 2019, 11:08
$$1000(1+15/100)^5$$
$$1000(1.15)^5$$

Manager
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Re: An investment club has had an average rate of return of 15% [#permalink]  12 Feb 2019, 21:28
Expert's post
Just remember the compound interest formula

Interest= Principal(1+rate)^(time period)
or
I=P(1+r)^t

Just list each variable

I = unknown
P=1000
r=0.15
t=5

thus...
I=1000(1+0.15)^5
I=1000(1.15)^5

Re: An investment club has had an average rate of return of 15%   [#permalink] 12 Feb 2019, 21:28
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