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.

\(A=P\left(1+\frac{r}{m}\right)^{n \times m}\)

\(A=15,000\left(1+\frac{0.12}{3}\right)^{3 \times 3}\)

\(A=15,000(1+0.04)^{9}\)

\(A=15,000(1.04)^{9}\)

\(A=15,000(1.4230)\)

\(A=\$ 21,345\)

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