Background Info: Tom finds a second personal loan option. This loan would also require him to repay the principal in one lump sum after three years.

Loan Option B
- Principal: [tex]\$9,000[/tex]
- Type of Interest: Compound Interest
- Interest Rate: [tex]8\%[/tex]
- Rate of Accrual: Once per year

Use the formula for annual compound interest:
[tex]
A = P \left(1 + \frac{r}{n}\right)^{nt}
[/tex]

Where:
- [tex]A[/tex] is the total amount owed
- [tex]P[/tex] is the principal amount ([tex]\$9,000[/tex])
- [tex]r[/tex] is the annual interest rate ([tex]0.08[/tex])
- [tex]n[/tex] is the number of times the interest is compounded per year ([tex]1[/tex])
- [tex]t[/tex] is the number of years ([tex]3[/tex])

Calculate the total amount that Tom would repay.

A. [tex]\$10,337[/tex]
B. [tex]\$11,337[/tex]
C. [tex]\[tex]$12,337[/tex]
D. [tex]\$[/tex]13,337[/tex]



Answer :

Sure, let's break down the problem step by step using the compound interest formula given:

[tex]\[ A=P\left(1+\frac{r}{n}\right)^{n t} \][/tex]

Here:
- [tex]\( P \)[/tex] is the principal amount (initial amount of the loan),
- [tex]\( r \)[/tex] is the annual interest rate,
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year,
- [tex]\( t \)[/tex] is the number of years the money is borrowed for,
- [tex]\( A \)[/tex] is the final amount that needs to be repaid.

Given the details for Loan Option B:
- Principal, [tex]\( P = \$9,000 \)[/tex]
- Annual interest rate, [tex]\( r = 8\% = 0.08 \)[/tex]
- Compounded annually hence, [tex]\( n = 1 \)[/tex]
- Time, [tex]\( t = 3 \)[/tex] years

Substituting these values into the compound interest formula:

[tex]\[ A = 9000 \left( 1 + \frac{0.08}{1} \right)^{1 \cdot 3} \][/tex]

Simplifying inside the parentheses:

[tex]\[ A = 9000 \left( 1 + 0.08 \right)^3 \][/tex]

[tex]\[ A = 9000 \left( 1.08 \right)^3 \][/tex]

[tex]\[ A = 9000 \times 1.259712 \][/tex]

[tex]\[ A \approx 11337.408 \][/tex]

Therefore, the total amount that Tom would repay after 3 years is approximately:

[tex]\[ \boxed{11337.41} \][/tex]

Among the given options, [tex]\( \$ 11,337 \)[/tex] is the closest and correct amount.