Question

I don't know what I'm doing wrong. I triple checked with the 2nd question: Calculators online, my scientific calculator, everything comes up with the same number but per system, my answer is wrong. Please help

(Related to Checkpoint 5.3) (Compound interest with non-annual periods) Calculate the amount of money that will be in each of the following accounts at the end of the given deposit period: Compounding Periods Per Year (M) Account Holder Theodore Logan III Vernell Coles Tina Elliot Wayne Robinson Eunice Chung Kelly Cravens Amount Deposited $ 1,000 96,000 9,000 118,000 30,000 17,000 Annual Interest Rate 12% 10 12 8 18 10 2 3 6 12 Compounding Periods (Years) 10 3 6 4 4 3 1 a. The amount of money in Theodore Logan III's account at the end of 10 years will be $ 3.262.04. (Round to the nearest cent.) b. The amount of money in Vernell Coles' account at the end of 3 year(s) will be $ 127776.00. (Round to the nearest cent.)

Answer 1

Calculating the Future Value of Deposit at the end of different periods:-

a).

$Future Value = Invested Amount(1+\frac{r}{m})^{n*m}$

Where, Invested Amount = $1000

r = Periodic Interest rate = 12%

n= no of periods = 10

m = no of times compounding in a year = 4

FutureValue 1000 *(1 + 10*4 0.12 4

Future Value = $3262.04

b).

$Future Value = Invested Amount(1+\frac{r}{m})^{n*m}$

Where, Invested Amount = $96,000

r = Periodic Interest rate = 10%

n= no of periods = 3

m = no of times compounding in a year = 2

$Future Value =96000*(1+\frac{0.10}{2})^{2*3}$

FutureValue 96000 *(1 + 0.05)

Future Value = $128,649.18

c).

$Future Value = Invested Amount(1+\frac{r}{m})^{n*m}$

Where, Invested Amount = $9,000

r = Periodic Interest rate = 12%

n= no of periods = 6

m = no of times compounding in a year = 3

$Future Value =9000*(1+\frac{0.12}{3})^{3*6}$

Future Value = $18,232.35

d).

$Future Value = Invested Amount(1+\frac{r}{m})^{n*m}$

Where, Invested Amount = $118,000

r = Periodic Interest rate = 8%

n= no of periods = 4

m = no of times compounding in a year = 6

$Future Value =118000*(1+\frac{0.08}{6})^{6*4}$

Future Value = $162,157.82

e).

$Future Value = Invested Amount(1+\frac{r}{m})^{n*m}$

Where, Invested Amount = $30,000

r = Periodic Interest rate = 18%

n= no of periods = 4

m = no of times compounding in a year = 12

$Future Value =30000*(1+\frac{0.18}{12})^{4*12}$

Future Value = $61,304.35

f).

$Future Value = Invested Amount(1+\frac{r}{m})^{n*m}$

Where, Invested Amount = $17,000

r = Periodic Interest rate = 10%

n= no of periods = 3

m = no of times compounding in a year = 1

$Future Value =17000*(1+\frac{0.10}{1})^{3*1}$

Future Value = $22,627

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