Suppose you take out a 60 -month installment loan to finance one year of tuition for \( \$ 13,100 \). The payments are \( \$ 327.50 \) per month and the total fir charge is \( \$ 6,550 \). Using Table \( 13-1 \), calculate APR? Select one: a. \( 18 \% \) b. \( 17.25 \% \) c. \( 10 \% \) d. \( 6 \% \)

We start with the annuity‐formula for the monthly payment on an installment loan. If the principal is \( P \), the monthly interest rate is \( r \), and the number of payments is \( n \), then the monthly payment \( \text{PMT} \) is given by \[ \text{PMT} = P \cdot \frac{r(1+r)^n}{(1+r)^n-1}. \] In our problem we have: - \( P = 13\,100 \), - \( \text{PMT} = 327.50 \), - \( n = 60 \) months. Thus, the equation is \[ 327.50 = 13\,100 \cdot \frac{r(1+r)^{60}}{(1+r)^{60}-1}. \] Our task is to solve for \( r \) (the monthly rate) and then multiply by 12 to find the nominal annual percentage rate (APR). Because this equation cannot be solved for \( r \) in closed‐form, we use trial and error (or interpolation) to approximate \( r \). 1. Suppose \( r = 0.01 \) (1% per month). Compute \( (1+0.01)^{60} \approx 1.816 \). Then \[ \frac{r(1+r)^{60}}{(1+r)^{60}-1} \approx \frac{0.01 \times 1.816}{1.816-1} \approx \frac{0.01816}{0.816} \approx 0.02227. \] Payment would be \[ 13\,100 \times 0.02227 \approx 291.7, \] which is below \(327.50\). So \( r \) must be larger. 2. Suppose \( r = 0.015 \) (1.5% per month). Compute \( (1+0.015)^{60} \). Using the approximation \[ (1.015)^{60} \approx e^{60\ln(1.015)} \quad \text{with} \quad \ln(1.015) \approx 0.014888, \] we have \[ (1.015)^{60} \approx e^{60 \times 0.014888} \approx e^{0.8933} \approx 2.444. \] Then \[ \frac{0.015 \times 2.444}{2.444-1} = \frac{0.03666}{1.444} \approx 0.02538. \] Payment becomes \[ 13\,100 \times 0.02538 \approx 332, \] which is slightly above \(327.50\). 3. Now, try a slightly lower \( r \), say \( r \approx 0.0145 \) (1.45% per month). First, estimate \[ (1.0145)^{60} \approx e^{60\ln(1.0145)}. \] Using \(\ln(1.0145) \approx 0.014404\), we get \[ (1.0145)^{60} \approx e^{60 \times 0.014404} \approx e^{0.86424} \approx 2.373. \] Then \[ \frac{0.0145 \times 2.373}{2.373-1} = \frac{0.03443}{1.373} \approx 0.02508. \] Multiplying by the principal gives \[ 13\,100 \times 0.02508 \approx 328.5, \] which is very close to \(327.50\). A slight adjustment would yield nearly the desired payment. Since the monthly rate is approximately \( 1.45\% \), the nominal APR is \[ \text{APR} \approx 12 \times 1.45\% \approx 17.4\%. \] Among the options provided, the closest answer is \[ 17.25\%. \] Thus, the answer is \( \textbf{(b)}\;17.25\% \).