In: Finance
(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:
Account Holder |
Amount Deposited |
Annual Interest Rate |
Compounding Periods per Year (M) |
Compounding Periods (Years) |
Theodore Logan TIT |
$ 1,000 |
10% |
1 |
10 |
Vernell Coles |
95,000 |
12 |
12 |
1 |
Tina Elliott |
8,000 |
12 |
6 |
2 |
Wayne Robinson |
120,000 |
8 |
4 |
2 |
Eunice Chung |
30,000 |
10 |
2 |
4 |
Kelly Cravens |
15,000 |
12 |
3 |
3 |
Given data:
Account Holder: Theodore Logan III
Principal Amount \(P=\$ 1,000\)
Rate of Interest \(r=10 \%\)
Compounding Period per year \(m=1\) (number of years compounding)
Period of deposit \(n=10\) years
Step 1:
Let us assume that the amount to be received by Theodore Logan III is \(A\).
\(A=P\left(1+\frac{r}{m}\right)^{n \times m}\)
\(A=1,000\left(1+\frac{0.10}{1}\right)^{10 \times 1}\)
\(A=1,000(1.10)^{10}\)
\(A=1,000(2.594)\)
\(A=\$ 2,594\)
Account Holder: Vernell Coles
Principal Amount \(P=\$ 95,000\)
Rate of Interest \(r=12 \%\)
Compounding Period per year \(m=12\) (number of years compounding)
Period of deposit \(n=1\) year
Step 1 :
Let us assume that the amount to be received by Vernell Coles is \(A\).
\(A=P\left(1+\frac{r}{m}\right)^{n \times m}\)
\(A=95,000\left(1+\frac{0.12}{12}\right)^{1 \times 12}\)
\(A=95,000(1+0.01)^{12}\)
\(A=95,000(1.01)^{12}\)
\(A=95,000(1.127)\)
\(A=\$ 1,07,065\)
Given data:
Account Holder: Tina Elliott
Principal Amount \(P=\$ 8,000\)
Rate of Interest \(r=12 \%\)
Compounding Period per year \(m=6\) (number of years compounding)
Period of deposit \(n=2\) years
Step 1:
Let us assume that the amount to be received by Tina Elliott is \(A\).
\(A=P\left(1+\frac{r}{m}\right)^{n \times m}\)
\(A=8,000\left(1+\frac{0.12}{6}\right)^{2 \times 6}\)
\(A=8,000(1+0.02)^{12}\)
\(A=8,000(1.02)^{12}\)
\(A=8,000(1.268)\)
\(A=\$ 10,144\)
Account Holder: Wayne Robinson
Principal Amount \(=\$ 120,000\)
Rate of Interest \(=8 \%\)
Compounding Period per year \(m=4\)
Period of deposit \(\mathrm{n}=2\) years
Step 1 :
Let us assume that the amount to be received by Wayne Robinson is A.
\(A=P\left(1+\frac{r}{m}\right)^{n \times m}\)
\(A=120,000\left(1+\frac{0.08}{4}\right)^{2 \times 4}\)
\(A=120,000(1+0.02)^{8}\)
\(A=120,000(1.02)^{8}\)
\(A=120,000(1.1720)\)
\(A=\$ 1,40,640\)
Account Holder: Eunice Chung
Principal Amount \(=\$ 30,000\)
Rate of Interest \(=10 \%\)
Compounding Period per year \(m=2\)
Period of deposit \(\mathrm{n}=4\) years
Step 1
Let us assume that the amount to be received by Eunice Chung is A.
\(A=P\left(1+\frac{r}{m}\right)^{n \times m}\)
\(A=30,000\left(1+\frac{0.10}{2}\right)^{4 \times 2}\)
\(A=30,000(1+0.05)^{8}\)
\(A=30,000(1.05)^{8}\)
\(A=30,000(1.4770)\)
\(A=\$ 44,310\)
Account Holder: Kelly Cravens
Principal Amount \(=\$ 15,000\)
Rate of Interest \(=12 \%\)
Compounding Period per year \(m=3\)
Period of deposit \(n=3\) years
Step 1
Let us assume that the amount to be received by Kelly Cravens is A.
\(A=P\left(1+\frac{r}{m}\right)^{n \times m}\)
\(A=15,000\left(1+\frac{0.12}{3}\right)^{3 \times 3}\)
\(A=15,000(1+0.04)^{9}\)
\(A=15,000(1.04)^{9}\)
\(A=15,000(1.4230)\)
\(A=\$ 21,345\)