lollar. 3. Suppose that you decide to borrow \( \$ 25,000 \) for a new car. You can select one of the following loans, each requiring regular monthly payments: Installment Loan A: three-year loan at \( 1.9 \% \) Installment Loan B: five-year loan at \( 2.4 \% \), a. Find the monthly payments and the total interest for Loan A. b. Find the monthly payments and the total interest for Loan B. c. Compare the monthly payments and the total interest for the two loans.

**Loan A (3-year loan at \(1.9\%\) per year):** 1. The monthly interest rate is \[ r_A=\frac{0.019}{12}\approx0.00158333. \] 2. The total number of monthly payments is \[ n_A=3\times 12=36. \] 3. The monthly payment is found using the formula \[ m_A=\frac{r_A \cdot P}{1-(1+r_A)^{-n_A}}, \] where \(P=\$25,\!000\). Substituting the values gives \[ m_A=\frac{0.00158333\times25000}{1-(1.00158333)^{-36}}. \] 4. Compute the numerator: \[ 0.00158333\times25000\approx39.58325. \] 5. Estimate the denominator: - First, note that \[ (1.00158333)^{36}\approx e^{36\times0.00158333}\approx e^{0.057}\approx1.0587. \] - Therefore, \[ (1.00158333)^{-36}\approx\frac{1}{1.0587}\approx0.94465. \] - So, \[ 1-(1.00158333)^{-36}\approx1-0.94465=0.05535. \] 6. The monthly payment is then \[ m_A\approx\frac{39.58325}{0.05535}\approx715.3. \] Thus, the monthly payment is approximately \(\$715.30\). 7. The total amount paid over 36 months is \[ 36\times715.30\approx25750.8. \] 8. The total interest paid is \[ 25750.8-25000\approx750.8. \] So, the total interest is approximately \(\$750.80\). --- **Loan B (5-year loan at \(2.4\%\) per year):** 1. The monthly interest rate is \[ r_B=\frac{0.024}{12}=0.002. \] 2. The total number of monthly payments is \[ n_B=5\times 12=60. \] 3. The monthly payment is given by \[ m_B=\frac{r_B\cdot P}{1-(1+r_B)^{-n_B}}, \] with \(P=\$25,\!000\). Substituting in the values yields \[ m_B=\frac{0.002\times25000}{1-(1.002)^{-60}}. \] 4. Compute the numerator: \[ 0.002\times25000=50. \] 5. Estimate the denominator: - Compute \[ (1.002)^{60}\approx e^{60\times\ln(1.002)}. \] Since \(\ln(1.002)\approx0.002\) (more precisely about 0.001998), then \[ 60\times0.002\approx0.12, \] and \[ e^{0.12}\approx1.1275. \] - Therefore, \[ (1.002)^{-60}\approx\frac{1}{1.1275}\approx0.8865. \] - This gives \[ 1-(1.002)^{-60}\approx1-0.8865=0.1135. \] 6. The monthly payment is then \[ m_B\approx\frac{50}{0.1135}\approx440.7. \] So, the monthly payment is approximately \(\$440.70\). 7. The total amount paid over 60 months is \[ 60\times440.7\approx26442. \] 8. The total interest paid is \[ 26442-25000\approx1442. \] Thus, the total interest is approximately \(\$1,442.00\). --- **Comparison:** - **Monthly Payment:** \(\textbf{Loan A} \approx \$715.30\) per month vs. \(\textbf{Loan B} \approx \$440.70\) per month. - **Total Interest:** \(\textbf{Loan A} \approx \$750.80\) total interest vs. \(\textbf{Loan B} \approx \$1,442.00\) total interest. Even though Loan B offers a lower monthly payment, it results in paying nearly twice the total interest compared to Loan A.