To solve for the monthly payments and total amounts paid for both financing durations using the given formula, we'll break down the problem step-by-step.
Given Information:
- Amount to finance (A) = $33,000
- Annual interest rate (r) = 3.5% = 0.035 (as a decimal)
- For monthly payments, the number of times interest is applied per year (n) = 12
Formula:
The formula provided is: \[ P = \frac{A \cdot \left( \frac{r}{n} \right) \cdot (1 + \frac{r}{n})^{nt}}{(1 + \frac{r}{n})^{nt} - 1} \]
1. Calculating for a 5-Year Loan (t = 5):
With the provided values:
- A = $33,000
- r = 0.035
- n = 12
- t = 5
Calculate monthly payment (M):
\[ M = \frac{33000 \cdot \left( \frac{0.035}{12} \right) \cdot \left( 1 + \frac{0.035}{12} \right)^{12 \times 5}}{\left( 1 + \frac{0.035}{12} \right)^{12 \times 5} - 1} \]
Calculate \( \frac{r}{n} \): \[ \frac{0.035}{12} \approx 0.00291667 \]
Calculate \( nt \): \[ nt = 12 \cdot 5 = 60 \]
Now calculate \( (1 + \frac{0.035}{12})^{60} \): \[ (1 + 0.00291667)^{60} \approx (1.00291667)^{60} \approx 1.185178 \]
Putting it into the formula: \[ M = \frac{33000 \cdot 0.00291667 \cdot 1.185178}{1.185178 - 1} \] \[ M = \frac{33000 \cdot 0.00291667 \cdot 1.185178}{0.185178} \] \[ M \approx \frac{33000 \cdot 0.00345069}{0.185178} \approx \frac{113.86}{0.185178} \approx 614.56 \]
Thus, the monthly payment for the 5-year loan is approximately $614.56.
Calculate Total Loan Amount Paid:
Total amount paid over 5 years: \[ \text{Total Paid} = M \times (\text{Number of payments}) \] Number of payments = 12 payments/year \(\times\) 5 years = 60 payments.
\[ \text{Total Paid} = 614.56 \times 60 \approx 36873.60 \]
2. Calculating for a 6-Year Loan (t = 6):
Now, we do the same for a 6-Year loan (t = 6):
- t = 6
Calculate monthly payment (M):
\[ M = \frac{33000 \cdot \left( \frac{0.035}{12} \right) \cdot \left( 1 + \frac{0.035}{12} \right)^{12 \times 6}}{\left( 1 + \frac{0.035}{12} \right)^{12 \times 6} - 1} \]
Here we need to calculate \( nt \) again: \[ nt = 12 \cdot 6 = 72 \]
Calculate \( (1 + \frac{0.035}{12})^{72} \): \[ (1 + 0.00291667)^{72} \approx (1.00291667)^{72} \approx 1.219202 \]
Putting it into the formula: \[ M = \frac{33000 \cdot 0.00291667 \cdot 1.219202}{1.219202 - 1} \] \[ M = \frac{33000 \cdot 0.00291667 \cdot 1.219202}{0.219202} \] \[ M \approx \frac{33000 \cdot 0.00355774}{0.219202} \approx \frac{117.36}{0.219202} \approx 535.72 \]
Thus, the monthly payment for the 6-year loan is approximately $535.72.
Calculate Total Loan Amount Paid:
Total amount paid over 6 years: \[ \text{Total Paid} = 535.72 \times 72 \approx 38511.84 \]
Summary of Results:
5-Year Loan:
- Monthly payment (M) = $614.56
- Total loan amount (Principal + Interest) = $36,873.60
6-Year Loan:
- Monthly payment (M) = $535.72
- Total loan amount (Principal + Interest) = $38,511.84