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An investment club has had an average rate of return of 15%
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26 Apr 2017, 18:13
26 Apr 2017, 18:13
Expert Reply
00:00
Question Stats:
90% (00:53) correct
9% (01:23) wrong based on 33 sessions
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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. \(1000(5)^{0.15}\)
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
Question: 7
Page: 322
Re: An investment club has had an average rate of return of 15%
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04 May 2017, 01:03
04 May 2017, 01:03
Expert Reply
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.
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.
Re: An investment club has had an average rate of return of 15%
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16 Jan 2019, 11:08
16 Jan 2019, 11:08
\(1000(1+15/100)^5\)
\(1000(1.15)^5\)
Answer B
\(1000(1.15)^5\)
Answer B
