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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6495
GMAT 1: 760 Q51 V42
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John invests $10,000 at the monthly constant compounded rate of annual [#permalink] ### Show Tags 04 Jan 2017, 06:11 00:00 Difficulty: 45% (medium) Question Stats: 58% (01:00) correct 42% (01:00) wrong based on 91 sessions ### HideShow timer Statistics 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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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6495 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: John invests$10,000 at the monthly constant compounded rate of annual  [#permalink]

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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.

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior SC Moderator Joined: 22 May 2016 Posts: 2082 John invests$10,000 at the monthly constant compounded rate of annual  [#permalink]

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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: $$A=P(1+\frac{0.08}{4})^{nt}$$ $$A=P(1 + 0.02)^{n*t}$$ $$A=P(1.02)^{(4*2)}$$ $$A = P(1.02)^{8}$$ John invests$10,000 at the monthly constant compounded rate of annual &nbs [#permalink] 10 Sep 2018, 14:01
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