Answer :
Let's go through each step carefully to determine the monthly payments and the total interest for each loan option, and then identify the lower-cost option and the amount of savings.
### Loan A:
- Principal Amount (P): [tex]$6000 - Annual Interest Rate (r\_A): 9.2% or 0.092 - Duration (t\_A): 4 years - Number of Payments per Year (n): 12 (monthly payments) To determine the monthly payment, we use the loan payment formula: \[ PMT = \frac{P \left( \frac{r}{n} \right)}{1 - \left(1 + \frac{r}{n}\right)^{-nt}} \] First, we calculate the monthly interest rate: \[ \frac{r\_A}{n} = \frac{0.092}{12} = 0.0076667 \] Next, we calculate the number of monthly payments: \[ n \times t\_A = 12 \times 4 = 48 \text{ months} \] Substituting these values into the formula, we get: \[ PMT\_A = \frac{6000 \left( 0.0076667 \right)}{1 - \left(1 + 0.0076667\right)^{-48}} \] After performing the calculations, the monthly payment for Loan A is: \[ PMT\_A \approx \$[/tex]149.88
\]
To find the total interest paid over the life of Loan A:
[tex]\[ \text{Total Payment} = PMT\_A \times 48 = 149.88 \times 48 = 7194.24 \][/tex]
[tex]\[ \text{Total Interest} = \text{Total Payment} - Principal = 7194.24 - 6000 = 1194.24 \][/tex]
### Loan B:
- Principal Amount (P): [tex]$6000 - Annual Interest Rate (r\_B): 9.1% or 0.091 - Duration (t\_B): 3 years - Number of Payments per Year (n): 12 (monthly payments) We use the same loan payment formula as above. First, we calculate the monthly interest rate for Loan B: \[ \frac{r\_B}{n} = \frac{0.091}{12} = 0.0075833 \] Next, we calculate the number of monthly payments: \[ n \times t\_B = 12 \times 3 = 36 \text{ months} \] Substituting these values into the formula, we get: \[ PMT\_B = \frac{6000 \left( 0.0075833 \right)}{1 - \left(1 + 0.0075833\right)^{-36}} \] After performing the calculations, the monthly payment for Loan B is: \[ PMT\_B \approx \$[/tex]191.08
\]
To find the total interest paid over the life of Loan B:
[tex]\[ \text{Total Payment} = PMT\_B \times 36 = 191.08 \times 36 = 6878.88 \][/tex]
[tex]\[ \text{Total Interest} = \text{Total Payment} - Principal = 6878.88 - 6000 = 878.88 \][/tex]
### Comparison of Loan A and Loan B:
- Total Interest for Loan A: [tex]$1194.24 - Total Interest for Loan B: $[/tex]878.88
Between the two loans, Loan B has a lower total interest cost, making it the lower-cost option.
### Amount of Savings:
Diana will save the difference in total interest if she chooses Loan B over Loan A:
[tex]\[ \text{Savings} = 1194.24 - 878.88 = 315.36 \][/tex]
### Conclusion:
- The monthly payment for Loan A is \[tex]$149.88 - The total interest paid for Loan A is \$[/tex]1194.24
- The monthly payment for Loan B is \[tex]$191.08 - The total interest paid for Loan B is \$[/tex]878.88
- The lower-cost option is Loan B
- The amount of savings will be \$315.36
### Loan A:
- Principal Amount (P): [tex]$6000 - Annual Interest Rate (r\_A): 9.2% or 0.092 - Duration (t\_A): 4 years - Number of Payments per Year (n): 12 (monthly payments) To determine the monthly payment, we use the loan payment formula: \[ PMT = \frac{P \left( \frac{r}{n} \right)}{1 - \left(1 + \frac{r}{n}\right)^{-nt}} \] First, we calculate the monthly interest rate: \[ \frac{r\_A}{n} = \frac{0.092}{12} = 0.0076667 \] Next, we calculate the number of monthly payments: \[ n \times t\_A = 12 \times 4 = 48 \text{ months} \] Substituting these values into the formula, we get: \[ PMT\_A = \frac{6000 \left( 0.0076667 \right)}{1 - \left(1 + 0.0076667\right)^{-48}} \] After performing the calculations, the monthly payment for Loan A is: \[ PMT\_A \approx \$[/tex]149.88
\]
To find the total interest paid over the life of Loan A:
[tex]\[ \text{Total Payment} = PMT\_A \times 48 = 149.88 \times 48 = 7194.24 \][/tex]
[tex]\[ \text{Total Interest} = \text{Total Payment} - Principal = 7194.24 - 6000 = 1194.24 \][/tex]
### Loan B:
- Principal Amount (P): [tex]$6000 - Annual Interest Rate (r\_B): 9.1% or 0.091 - Duration (t\_B): 3 years - Number of Payments per Year (n): 12 (monthly payments) We use the same loan payment formula as above. First, we calculate the monthly interest rate for Loan B: \[ \frac{r\_B}{n} = \frac{0.091}{12} = 0.0075833 \] Next, we calculate the number of monthly payments: \[ n \times t\_B = 12 \times 3 = 36 \text{ months} \] Substituting these values into the formula, we get: \[ PMT\_B = \frac{6000 \left( 0.0075833 \right)}{1 - \left(1 + 0.0075833\right)^{-36}} \] After performing the calculations, the monthly payment for Loan B is: \[ PMT\_B \approx \$[/tex]191.08
\]
To find the total interest paid over the life of Loan B:
[tex]\[ \text{Total Payment} = PMT\_B \times 36 = 191.08 \times 36 = 6878.88 \][/tex]
[tex]\[ \text{Total Interest} = \text{Total Payment} - Principal = 6878.88 - 6000 = 878.88 \][/tex]
### Comparison of Loan A and Loan B:
- Total Interest for Loan A: [tex]$1194.24 - Total Interest for Loan B: $[/tex]878.88
Between the two loans, Loan B has a lower total interest cost, making it the lower-cost option.
### Amount of Savings:
Diana will save the difference in total interest if she chooses Loan B over Loan A:
[tex]\[ \text{Savings} = 1194.24 - 878.88 = 315.36 \][/tex]
### Conclusion:
- The monthly payment for Loan A is \[tex]$149.88 - The total interest paid for Loan A is \$[/tex]1194.24
- The monthly payment for Loan B is \[tex]$191.08 - The total interest paid for Loan B is \$[/tex]878.88
- The lower-cost option is Loan B
- The amount of savings will be \$315.36
