Hodgson Wagner
06/08/2023 · Junior High School

Lindsey bought a new car for \( \$ 29,000 \). She paid a \( 20 \% \) down payment and financed the remaining balance for 48 months with an APR of \( 4.3 \% \). Assuming she made monthly payments, determine the total cost of Lindsey's car. Round your answer to the nearest cent, if necessary.

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To determine the total cost of Lindsey's car, we need to follow these steps: 1. **Calculate the down payment**: \[ \text{Down payment} = 20\% \text{ of } 29000 = 0.20 \times 29000 = 5800 \] 2. **Calculate the financed amount**: \[ \text{Financed amount} = \text{Total cost} - \text{Down payment} = 29000 - 5800 = 23200 \] 3. **Calculate the monthly interest rate**: The APR is \( 4.3\% \), so the monthly interest rate is: \[ \text{Monthly interest rate} = \frac{4.3\%}{12} = \frac{0.043}{12} \approx 0.00358333 \] 4. **Calculate the monthly payment using the formula for an installment loan**: The formula for the monthly payment \( M \) is given by: \[ M = P \frac{r(1 + r)^n}{(1 + r)^n - 1} \] where: - \( P \) is the loan amount (financed amount), - \( r \) is the monthly interest rate, - \( n \) is the number of payments (months). Plugging in the values: - \( P = 23200 \) - \( r \approx 0.00358333 \) - \( n = 48 \) \[ M = 23200 \frac{0.00358333(1 + 0.00358333)^{48}}{(1 + 0.00358333)^{48} - 1} \] First, calculate \( (1 + r)^{48} \): \[ (1 + 0.00358333)^{48} \approx 1.183682 \] Now substitute this back into the formula: \[ M = 23200 \frac{0.00358333 \times 1.183682}{1.183682 - 1} \] \[ M = 23200 \frac{0.004247}{0.183682} \approx 23200 \times 0.023141 \approx 536.27 \] 5. **Calculate the total payment over 48 months**: \[ \text{Total payment} = M \times n = 536.27 \times 48 \approx 25701.00 \] 6. **Calculate the total cost of the car**: \[ \text{Total cost} = \text{Down payment} + \text{Total payment} = 5800 + 25701.00 = 31501.00 \] Thus, the total cost of Lindsey's car is approximately \( \boxed{31501.00} \).

Quick Answer

The total cost of Lindsey's car is approximately \( \$31,501.00 \).
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