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04 Jan 2017, 07:11
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45% (medium)

Question Stats:

58% (00:55) correct 42% (01:00) wrong based on 100 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. $$10,000(0.11/12)^1^2^t$$ _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: John invests $10,000 at the monthly constant compounded rate of annual [#permalink] ### Show Tags 06 Jan 2017, 02:43 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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John invests $10,000 at the monthly constant compounded rate of annual [#permalink] ### Show Tags 10 Sep 2018, 15:01 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

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

$$\frac{r}{n}$$ 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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