If a $3,000 deposit is made into a savings account that pays 6 percent interest, compounded monthly, and there are no

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If a $3,000 deposit is made into a savings account that pays 6 percent interest, compounded monthly, and there are no other deposits or withdrawals from the account, how much money, rounded to the nearest dollar, is in the account at the end of one year?

A. $2,160
B. $3,180
C. $3,185
D. $5,160
E. $6,037

Answer: C

Source: Princeton Review

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$$Compound\ interest\ =\ p\left(1+\frac{r\%}{n}\right)^{nt}$$
$$where\ p=principal\ =\ $3000$$
$$r=rate\ =\ 6\%$$
$$n=\ number\ of\ times\ interest\ is\ compounded\ =\ 12$$
$$t=time\ /year\ =\ 1$$
$$compound\ interest\ =\ 3000\left(1+\frac{6}{100}\cdot\frac{1}{12}\right)^{12}$$
$$\ \ 3000\left(1+\frac{6}{1200}\right)^{12}$$
$$\ \ 3000\cdot1.005^{12}$$
$$\ \ 3000\cdot1.0617$$
$$$3185.1$$
$$Answer\ =\ C$$

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M7MBA wrote:
Wed Sep 09, 2020 6:53 am
If a $3,000 deposit is made into a savings account that pays 6 percent interest, compounded monthly, and there are no other deposits or withdrawals from the account, how much money, rounded to the nearest dollar, is in the account at the end of one year?

A. $2,160
B. $3,180
C. $3,185
D. $5,160
E. $6,037

Answer: C

Solution:

We use the compound interest formula A = P(1 + r/n)^nt, where A is the ending amount, P is the starting amount (P = $3,000), r is the interest rate (r = 0.06), n is the number of times the interest is compounded per year (n = 12), and t is the number of years (t = 1).

The amount of money in the account, rounded to the nearest dollar, at the end of one year is:

A = 3000(1 + 0.06/12)^12 = 3000(1.005)^12 = $3,185

Answer: C

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