Question: Britta has been accepted into a 2-year Medical Assistant program at a career school. Suppose that Britta decided to take out a private loan for $6000 for which loan payments start as soon as the loan amount is deposited in her account and continue for 10 years. The interest rate is 6.1%. a. Determine her monthly payment. b. What is the total amount she will pay back? c. What is the total interest amount?
1. **State the problem:**
Britta takes a loan of $6000 with an interest rate of 6.1% per year, to be paid monthly over 10 years. We need to find:
a. Monthly payment
b. Total amount paid back
c. Total interest paid
2. **Formula used:**
For a loan with principal $P$, annual interest rate $r$, number of payments per year $n$, and total years $t$, the monthly payment $M$ is given by the amortization formula:
$$M = P \times \frac{\frac{r}{n}}{1 - (1 + \frac{r}{n})^{-nt}}$$
Where:
- $P = 6000$
- $r = 0.061$ (6.1% as decimal)
- $n = 12$ (monthly payments)
- $t = 10$
3. **Calculate monthly interest rate:**
$$\frac{r}{n} = \frac{0.061}{12} = 0.0050833$$
4. **Calculate total number of payments:**
$$nt = 12 \times 10 = 120$$
5. **Calculate denominator:**
$$1 - (1 + 0.0050833)^{-120} = 1 - (1.0050833)^{-120}$$
Calculate $(1.0050833)^{-120}$:
$$ (1.0050833)^{120} \approx e^{120 \times \ln(1.0050833)} \approx e^{120 \times 0.00507} = e^{0.6084} \approx 1.837$$
So,
$$ (1.0050833)^{-120} = \frac{1}{1.837} \approx 0.544$$
Therefore,
$$1 - 0.544 = 0.456$$
6. **Calculate monthly payment $M$:**
$$M = 6000 \times \frac{0.0050833}{0.456} = 6000 \times 0.01115 = 66.9$$
7. **Calculate total amount paid back:**
$$\text{Total paid} = M \times nt = 66.9 \times 120 = 8028$$
8. **Calculate total interest paid:**
$$\text{Interest} = \text{Total paid} - P = 8028 - 6000 = 2028$$
**Final answers:**
- a. Monthly payment = $66.90$
- b. Total amount paid back = $8028$
- c. Total interest paid = $2028$