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File Edit Format View Help Jennifer Kaufman bought a used Toyota Prius for \( \$ 18,500 \). She made a \( \$ 3,500 \) down payment and is financing the balance at a particular bank over a 3 year period at \( 12 \% \) interest. As her banker, calculate what equal monthly payments (in \( \$ \) ) will be required by Jennifer to amortize the car loan. (Round your answer to the nearest cent.)

Ask by Pollard Reeves. in the United States
Mar 21,2025

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Jennifer's monthly payment is approximately \$498.50.

Solución

We start by noting that Jennifer is financing \[ P = \$18\,500 - \$3\,500 = \$15\,000. \] The annual interest rate is \( 12\% \), so the monthly interest rate is \[ r = \frac{0.12}{12} = 0.01. \] The loan term is 3 years, or \[ n = 3 \times 12 = 36 \text{ months}. \] The formula for the monthly payment \( M \) to fully amortize a loan is \[ M = P \times \frac{r(1+r)^n}{(1+r)^n-1}. \] Substitute the known values: \[ M = 15000 \times \frac{0.01(1.01)^{36}}{(1.01)^{36}-1}. \] First, we calculate \((1.01)^{36}\). Using a calculator, we find \[ (1.01)^{36} \approx 1.430768. \] Now compute the numerator: \[ 0.01 \times (1.01)^{36} \approx 0.01 \times 1.430768 = 0.01430768, \] \[ \text{Numerator} = 15000 \times 0.01430768 \approx 214.6152. \] Next, calculate the denominator: \[ (1.01)^{36} - 1 \approx 1.430768 - 1 = 0.430768. \] Thus, the monthly payment is \[ M \approx \frac{214.6152}{0.430768} \approx 498.50. \] Jennifer's equal monthly payment is approximately \(\$498.50\).

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Bonus Knowledge

To calculate Jennifer's equal monthly payments, we first need to determine the loan amount. She bought the Prius for \( \$ 18,500 \) and made a down payment of \( \$ 3,500 \), so: \[ \text{Loan Amount} = 18500 - 3500 = 15000 \] Next, we use the loan amortization formula, which is: \[ M = P \frac{r(1 + r)^n}{(1 + r)^n - 1} \] Where: - \( M \) is the monthly payment - \( P \) is the loan amount (\$ 15,000) - \( r \) is the monthly interest rate (annual rate / 12 months) - \( n \) is the total number of payments (loan term in months) The annual interest rate is \( 12\% \), so the monthly rate \( r \) is: \[ r = \frac{12\%}{12} = 0.01 \] Since she is financing the loan for 3 years, the total number of payments \( n \) is: \[ n = 3 \times 12 = 36 \] Now we can plug these numbers into the formula: \[ M = 15000 \frac{0.01(1 + 0.01)^{36}}{(1 + 0.01)^{36} - 1} \] Calculating \( (1 + 0.01)^{36} \): \[ (1.01)^{36} \approx 1.430768 \] Now substitute this value back into the formula: \[ M = 15000 \frac{0.01 \times 1.430768}{1.430768 - 1} \] \[ = 15000 \frac{0.01430768}{0.430768} \] \[ = 15000 \times 0.033227644 \] \[ \approx 498.41466 \] Rounding this to the nearest cent, the monthly payments Jennifer needs to make are approximately: \[ \text{Monthly Payment} \approx \$ 498.41 \]

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