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Jill invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. Approximately how much

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Jill invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. Approximately how much

by AAPL » Mon Feb 01, 2021 4:47 am

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Magoosh

Jill invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. Approximately how much does she have in her account after two years?

A. $10079.44
B. $10815.83
C. $12652.61
D. $14232.14
E. $20598.11

OA B
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Re: Jill invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. Approximately how mu

by deloitte247 » Thu Feb 04, 2021 7:43 am
$$U\sin g\ compound\ interest\ formular\ =>$$
$$Amount\ \left(A\right)\ =\ P\left(1+\frac{r}{n}\right)^{nt}$$
$$where\ A\ =\ amount\ =\ ??$$
$$\ p\ =\ principal\ =\ $10,000$$
$$\ r\ =\ rate=\ 3.96\%\ =\ 0.0396$$
$$n\ =\ number\ of\ times\ interest\ is\ compounded\ for\ this\ period$$
$$=2\cdot2=4\ \left(twice\ a\ year\ for\ 2\ years\right)$$
$$t\ =\ time\ =\ 2\ years$$
$$A=10000\left(1+\frac{0.0396}{4}\right)^{2\cdot4}$$
$$A=10000\left(\frac{4+0.0396}{4}\right)^8$$
$$A=10000\left(1.0099\right)^8$$
$$A=10819.99$$
$$which\ is\ approximately\ $10,820$$
$$The\ only\ option\ that\ suits\ this\ approximation\ is\ B$$
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Re: Jill invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. Approximately how mu

by Scott@TargetTestPrep » Fri Feb 19, 2021 6:42 am
AAPL wrote: ↑
Mon Feb 01, 2021 4:47 am
 Magoosh

Jill invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. Approximately how much does she have in her account after two years?

A. $10079.44
B. $10815.83
C. $12652.61
D. $14232.14
E. $20598.11

OA B
Solution:

We’ll use the compound interest formula, A = P (1 + r/n)^nt, where P = principal, r = annual interest rate, n = the number of compounding periods in 1 year, and t = the number of years. Thus, the amount, A, Jill will have after two years is:

A = 10,000(1 + 0.0396/2)^(2 x 2) ≈ $10,815.83

Answer: B

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