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John invests $10,000 at the monthly constant compounded rate of annual

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Math Revolution GMAT Instructor
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John invests $10,000 at the monthly constant compounded rate of annual  [#permalink]

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New post 04 Jan 2017, 06:11
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

58% (01:00) correct 42% (01:00) wrong based on 91 sessions

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John invests $10,000 at the monthly constant compounded rate of annually 11 percent. After t years, what is the amount including interest?

A. \($10,000(1+0.11/12)^t\)
B. \($10,000(1+0.11/12)^1^2^t\)
C. \($10,000(1+0.11)^1^2^t\)
D. \($10,000(1+0.11)^t\)
E. \($10,000(0.11/12)^1^2^t\)

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Re: John invests $10,000 at the monthly constant compounded rate of annual  [#permalink]

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New post 06 Jan 2017, 01:43
1
==> \($10,000(1+0.11/12)^1^2^t\) becomes the answer, because 11% of yearly compound interest rate is divided by month, and the amount including the interest in t years.

Hence, the answer is B.
Answer: B
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John invests $10,000 at the monthly constant compounded rate of annual  [#permalink]

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New post 10 Sep 2018, 14:01
MathRevolution wrote:
John invests $10,000 at the monthly constant compounded rate of annually 11 percent. After t years, what is the amount including interest?

A. \($10,000(1+0.11/12)^t\)
B. \($10,000(1+0.11/12)^1^2^t\)
C. \($10,000(1+0.11)^1^2^t\)
D. \($10,000(1+0.11)^t\)
E. \($10,000(0.11/12)^1^2^t\)

This question asks for the formula for compound interest.

\(A = P(1+\frac{r}{n})^{nt}\)
\(A\) = amount (or future value) including principal and interest
\(P\) = principal ($10,000)
\(r\) = yearly interest rate in decimal form (0.11)
\(n\) = number of times per year that interest is compounded (monthly = 12)
\(t\) = time in years

After \(t\) years, John's $10,000 investment at an annual interest rate of 11%, compounded monthly, is given by

\(A=$10,000(1+\frac{0.11}{12})^{12t}\)

ANSWER B

\(\frac{r}{n}\) sometimes confuses people. Divide \(r\) by \(n\) first, then add to 1 and raise to the power of (n*t). Some problems require the division. For example, if the rate were 8% annually, compounded quarterly (4 times a year), after t=2 years:
\(A=P(1+\frac{0.08}{4})^{nt}\)
\(A=P(1 + 0.02)^{n*t}\)
\(A=P(1.02)^{(4*2)}\)
\(A = P(1.02)^{8}\)
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John invests $10,000 at the monthly constant compounded rate of annual &nbs [#permalink] 10 Sep 2018, 14:01
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John invests $10,000 at the monthly constant compounded rate of annual

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