Answer :
Alright, let's break down the problem and find the solution step-by-step.
### Step 1: Calculate Loan Amount for Each Option
- Option A:
- Discount on car: \[tex]$3,600 - Price after discount: \$[/tex]65,200 - \[tex]$3,600 = \$[/tex]61,600
- Down payment: \[tex]$7,500 - Loan amount \(P_A\): \$[/tex]61,600 - \[tex]$7,500 = \$[/tex]54,100
- Option B:
- No discount, so the price remains: \[tex]$65,200 - Down payment: \$[/tex]7,500
- Loan amount [tex]\(P_B\)[/tex]: \[tex]$65,200 - \$[/tex]7,500 = \[tex]$57,700 ### Step 2: Define the Payment Formula The loan payment formula for fixed installment loans is: \[ PMT = \frac{P \left(\frac{r}{n}\right)}{\left[1 - \left(1 + \frac{r}{n}\right)^{-nt}\right]} \] Where: - \(P\) is the loan amount - \(r\) is the annual interest rate - \(n\) is the number of payments per year (monthly payments mean \(n = 12\)) - \(t\) is the loan term in years ### Step 3: Calculate Monthly Payments for Each Option #### Option A: With Interest \( P_A = \$[/tex]54,100 \)
[tex]\( r_A = 5.39\% = 0.0539 \)[/tex]
[tex]\( n = 12 \)[/tex]
[tex]\( t = 7 \)[/tex] years
Substitute these values into the formula:
[tex]\[ PMT_A = \frac{5400 \left(\frac{0.0539}{12}\right)}{\left[1 - \left(1 + \frac{0.0539}{12}\right)^{-12 \times 7}\right]} \][/tex]
First, calculate the monthly interest rate:
[tex]\[ \frac{0.0539}{12} = 0.004491667 \][/tex]
Next, calculate the denominator:
[tex]\[ 1 + 0.004491667 = 1.004491667 \][/tex]
[tex]\[ 1.004491667^{-84} \approx 0.665662374 \][/tex]
[tex]\[ 1 - 0.665662374 = 0.3343376264 \][/tex]
Therefore:
[tex]\[ PMT_A = \frac{5400 \times 0.004491667}{0.3343376264} = \frac{242.943867}{0.3343376264} \approx 726.65 \][/tex]
#### Option B: No Interest
[tex]\( P_B = \$57,700 \)[/tex]
[tex]\( r_B = 0.00 \)[/tex]
[tex]\( n = 12 \)[/tex]
[tex]\( t = 7 \)[/tex]
Substitute these values into the formula:
[tex]\[ PMT_B = \frac{57700 \left(\frac{0}{12}\right)}{\left[1 - \left(1 + \frac{0}{12}\right)^{-12 \times 7}\right]} = \frac{57700 \times 0}{\left[1 - 1\right]} = \frac{0}{0} \approx 57700 / 84 \approx 686.90 \][/tex]
### Step 4: Find the Difference in Monthly Payments
Subtract the monthly payments:
[tex]\[ \text{Difference} = PMT_A - PMT_B = 726.65 - 686.90 = 39.75 \][/tex]
The difference in the monthly payments is [tex]$39.75 ### Step 5: Determine the Most Economical Option #### Total Cost of Option A: \[ \text{Total Cost}_A = (726.65 \times 12 \times 7) + 7500 = \approx 72776.60 + 7500 = 80276.60 \] #### Total Cost of Option B: \[ \text{Total Cost}_B = (686.90 \times 12 \times 7) + 7500 = \approx 69157.6 + 7500 = 76657.60 \] Comparing the total costs: \[ \text{Option A Total Cost} = \$[/tex] 80276.60 \]
[tex]\[ \text{Option B Total Cost} = \$ 76657.60 \][/tex]
Option B is the most economical.
### Final Answers:
- Difference in monthly payments: [tex]\(\$ 39.75\)[/tex]
- Most economical option: [tex]\(\text{Option B}\)[/tex]
### Step 1: Calculate Loan Amount for Each Option
- Option A:
- Discount on car: \[tex]$3,600 - Price after discount: \$[/tex]65,200 - \[tex]$3,600 = \$[/tex]61,600
- Down payment: \[tex]$7,500 - Loan amount \(P_A\): \$[/tex]61,600 - \[tex]$7,500 = \$[/tex]54,100
- Option B:
- No discount, so the price remains: \[tex]$65,200 - Down payment: \$[/tex]7,500
- Loan amount [tex]\(P_B\)[/tex]: \[tex]$65,200 - \$[/tex]7,500 = \[tex]$57,700 ### Step 2: Define the Payment Formula The loan payment formula for fixed installment loans is: \[ PMT = \frac{P \left(\frac{r}{n}\right)}{\left[1 - \left(1 + \frac{r}{n}\right)^{-nt}\right]} \] Where: - \(P\) is the loan amount - \(r\) is the annual interest rate - \(n\) is the number of payments per year (monthly payments mean \(n = 12\)) - \(t\) is the loan term in years ### Step 3: Calculate Monthly Payments for Each Option #### Option A: With Interest \( P_A = \$[/tex]54,100 \)
[tex]\( r_A = 5.39\% = 0.0539 \)[/tex]
[tex]\( n = 12 \)[/tex]
[tex]\( t = 7 \)[/tex] years
Substitute these values into the formula:
[tex]\[ PMT_A = \frac{5400 \left(\frac{0.0539}{12}\right)}{\left[1 - \left(1 + \frac{0.0539}{12}\right)^{-12 \times 7}\right]} \][/tex]
First, calculate the monthly interest rate:
[tex]\[ \frac{0.0539}{12} = 0.004491667 \][/tex]
Next, calculate the denominator:
[tex]\[ 1 + 0.004491667 = 1.004491667 \][/tex]
[tex]\[ 1.004491667^{-84} \approx 0.665662374 \][/tex]
[tex]\[ 1 - 0.665662374 = 0.3343376264 \][/tex]
Therefore:
[tex]\[ PMT_A = \frac{5400 \times 0.004491667}{0.3343376264} = \frac{242.943867}{0.3343376264} \approx 726.65 \][/tex]
#### Option B: No Interest
[tex]\( P_B = \$57,700 \)[/tex]
[tex]\( r_B = 0.00 \)[/tex]
[tex]\( n = 12 \)[/tex]
[tex]\( t = 7 \)[/tex]
Substitute these values into the formula:
[tex]\[ PMT_B = \frac{57700 \left(\frac{0}{12}\right)}{\left[1 - \left(1 + \frac{0}{12}\right)^{-12 \times 7}\right]} = \frac{57700 \times 0}{\left[1 - 1\right]} = \frac{0}{0} \approx 57700 / 84 \approx 686.90 \][/tex]
### Step 4: Find the Difference in Monthly Payments
Subtract the monthly payments:
[tex]\[ \text{Difference} = PMT_A - PMT_B = 726.65 - 686.90 = 39.75 \][/tex]
The difference in the monthly payments is [tex]$39.75 ### Step 5: Determine the Most Economical Option #### Total Cost of Option A: \[ \text{Total Cost}_A = (726.65 \times 12 \times 7) + 7500 = \approx 72776.60 + 7500 = 80276.60 \] #### Total Cost of Option B: \[ \text{Total Cost}_B = (686.90 \times 12 \times 7) + 7500 = \approx 69157.6 + 7500 = 76657.60 \] Comparing the total costs: \[ \text{Option A Total Cost} = \$[/tex] 80276.60 \]
[tex]\[ \text{Option B Total Cost} = \$ 76657.60 \][/tex]
Option B is the most economical.
### Final Answers:
- Difference in monthly payments: [tex]\(\$ 39.75\)[/tex]
- Most economical option: [tex]\(\text{Option B}\)[/tex]