Question 9

Leigh borrows $35 000 to complete home renovations. Compound interest will be charged at the rate of 4.6% per annum, compounding monthly.

If Leigh makes monthly payments of $1529.24, the amount of interest he has paid after one year is closest to:

$1253

B. $1610

C. $16 649

D. $17 098

E. $18 350

Question 10

Henri is considering a 5-year interest-only loan for a car valued at $30,000. The loan has a fixed interest rate of 8% per year compounded monthly. After 3 years, how much would Henri still owe on his loan?




A. $12,000




В. $13449.66




C. $2264.50




D. $25,521.26




E. $30,000


### Question 9

To determine the amount of interest Leigh has paid after one year, we need to break down the total amount paid and the principal repaid.

Given:
- Principal $P = \$35,000$
- Annual interest rate $r = 4.6\%$
- Monthly payment $R = \$1529.24$
- Loan period for interest calculation = 1 year

**Step 1: Convert annual interest rate to monthly rate**
\[
r_{\text{monthly}} = \frac{4.6\%}{12} = \frac{0.046}{12} \approx 0.003833
\]

**Step 2: Calculate total payments made in one year**
\[
\text{Total payments} = 1529.24 \times 12 = 18,350.88
\]

**Step 3: Calculate the balance after one year**
Using the formula for the remaining balance on an amortizing loan:
\[
B = P \times (1 + r_{\text{monthly}})^{12} - \frac{R}{r_{\text{monthly}}} \times \left[(1 + r_{\text{monthly}})^{12} - 1\right]
\]

\[
B = 35000 \times (1 + 0.003833)^{12} - \frac{1529.24}{0.003833} \times \left[(1 + 0.003833)^{12} - 1\right]
\]

First, calculate $(1 + 0.003833)^{12}$:
\[
(1 + 0.003833)^{12} \approx 1.047
\]

Now, calculate the balance $B$:
\[
B = 35000 \times 1.047 - \frac{1529.24}{0.003833} \times (1.047 - 1)
\]

\[
B = 36645 - \frac{1529.24}{0.003833} \times 0.047
\]

\[
B = 36645 - 398868.81 \times 0.047 \approx 36645 - 18746.64 = 17898.36
\]

**Step 4: Calculate the total principal repaid in one year**
\[
\text{Principal repaid} = 35000 - 17898.36 = 17101.64
\]

**Step 5: Calculate the total interest paid in one year**
\[
\text{Total interest paid} = \text{Total payments} - \text{Principal repaid}
\]

\[
\text{Total interest paid} = 18350.88 - 17101.64 = 1249.24
\]

Thus, the closest answer is:
\[
\boxed{A. \, 1253}
\]

### Question 10

Henri is considering a 5-year interest-only loan for a car valued at $30,000 with a fixed interest rate of 8% per year compounded monthly. 

Given:
- Principal $P = \$30,000$
- Annual interest rate $r = 8\%$
- Loan term = 5 years
- Time passed = 3 years

For an interest-only loan, the principal does not reduce over the life of the loan. The borrower only pays the interest periodically.

**Step 1: Convert annual interest rate to monthly rate**
\[
r_{\text{monthly}} = \frac{8\%}{12} = \frac{0.08}{12} \approx 0.006667
\]

**Step 2: Calculate the monthly interest payment**
\[
\text{Monthly interest payment} = P \times r_{\text{monthly}} = 30000 \times 0.006667 = 200
\]

**Step 3: Calculate the total interest paid after 3 years**
\[
\text{Total interest paid in 3 years} = 200 \times 12 \times 3 = 7200
\]

Since this is an interest-only loan, the principal remains unchanged, so Henri would still owe the original loan amount:
\[
\boxed{E. \, 30000}
\]

Would you like more details or have any questions?

Here are 8 related questions:
1. What is the difference between an interest-only loan and an amortizing loan?
2. How does compounding frequency affect the total interest paid on a loan?
3. What would be the monthly payments for a fully amortizing loan of $30,000 at 8% over 5 years?
4. How do you calculate the remaining balance on an amortizing loan?
5. How can you determine the total interest paid over the life of a loan?
6. What are the advantages and disadvantages of an interest-only loan?
7. How does the interest rate impact the monthly payment on a loan?
8. Can you explain how the formula for the annuity payment $R$ is derived?

**Tip:** When dealing with interest-only loans, remember that the principal remains the same throughout the loan period, and only the interest is paid periodically.

Learn how to calculate compound interest and remaining loan balances with detailed step-by-step solutions. Solve problems involving compound interest rates, amortization schedules, and interest-only loans. Suitable for advanced high school and college-level mathematics.

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