3. Suppose that you decide to borrow $$ \$ 25,000 $$ for a new car, You can seled 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.

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**a. Loan A (Three-year loan at $$1.9\%$$)**

We use the annuity formula for the monthly payment:
$$
A = P \cdot \frac{i(1+i)^n}{(1+i)^n - 1},
$$
where  
$$P = 25000$$ (the principal),  
$$i$$ is the monthly interest rate, and  
$$n$$ is the total number of monthly payments.

1. **Determine the monthly interest rate and number of payments.**  
   The annual rate is $$1.9\%$$, so
   $$
   i = \frac{0.019}{12} \approx 0.00158333.
   $$
   For a three‐year loan,
   $$
   n = 3 \times 12 = 36.
   $$

2. **Compute $$(1+i)^n$$.**  
   $$
   (1+i)^n = (1.00158333)^{36}.
   $$
   Evaluating (using logarithms or a calculator) gives approximately:
   $$
   (1+i)^n \approx 1.0586.
   $$

3. **Substitute into the payment formula.**  
   Using the form
   $$
   A = P \cdot \frac{i}{1 - (1+i)^{-n}},
   $$
   first compute:
   $$
   (1+i)^{-n} = \frac{1}{1.0586} \approx 0.9450.
   $$
   Then the denominator is:
   $$
   1 - (1+i)^{-n} \approx 1 - 0.9450 = 0.0550.
   $$
   Now, the numerator is:
   $$
   P \cdot i = 25000 \times 0.00158333 \approx 39.5833.
   $$
   Thus, the monthly payment is:
   $$
   A \approx \frac{39.5833}{0.0550} \approx 719.70.
   $$
   (Due to rounding at intermediate stages, one may also obtain a value near $$\$714$$ using an alternative form. Using a high‐precision calculation, a monthly payment of about $$\$715$$ is typical. For our purposes, we can state that the monthly payment is approximately $$\$715$$.)

4. **Total Interest for Loan A.**  
   The total amount paid over 36 months is:
   $$
   \text{Total Payment} \approx 715 \times 36 \approx 25740.
   $$
   The total interest paid is then:
   $$
   \text{Interest} \approx 25740 - 25000 = 740.
   $$

---

**b. Loan B (Five-year loan at $$2.4\%$$)**

1. **Determine the monthly interest rate and number of payments.**  
   The annual rate is $$2.4\%$$, so
   $$
   i = \frac{0.024}{12} = 0.002.
   $$
   For a five‐year loan,
   $$
   n = 5 \times 12 = 60.
   $$

2. **Compute $$(1+i)^n$$.**  
   $$
   (1+i)^n = (1.002)^{60}.
   $$
   Evaluating gives approximately:
   $$
   (1+i)^n \approx 1.1276.
   $$

3. **Substitute into the payment formula.**  
   Again using
   $$
   A = P \cdot \frac{i}{1 - (1+i)^{-n}},
   $$
   first compute:
   $$
   (1+i)^{-n} = \frac{1}{1.1276} \approx 0.8868.
   $$
   The denominator is:
   $$
   1 - (1+i)^{-n} \approx 1 - 0.8868 = 0.1132.
   $$
   The numerator is:
   $$
   P \cdot i = 25000 \times 0.002 = 50.
   $$
   Thus, the monthly payment is:
   $$
   A \approx \frac{50}{0.1132} \approx 441.5.
   $$

4. **Total Interest for Loan B.**  
   The total amount paid over 60 months is:
   $$
   \text{Total Payment} \approx 441.5 \times 60 \approx 26490.
   $$
   The total interest paid is then:
   $$
   \text{Interest} \approx 26490 - 25000 = 1490.
   $$

---

**c. Comparison**

- **Monthly Payments:**  
  - Loan A (3-year): Approximately $$\$715$$ per month.  
  - Loan B (5-year): Approximately $$\$442$$ per month.

- **Total Interest Paid:**  
  - Loan A: Approximately $$\$740$$ in interest.  
  - Loan B: Approximately $$\$1490$$ in interest.

**Conclusion:**  
Loan A has a higher monthly payment but results in considerably less total interest paid compared to Loan B, which offers lower monthly payments but a higher overall cost in interest.
**a. Loan A (3-year loan at 1.9%)** - **Monthly Payment:** Approximately \$715 - **Total Interest:** \$740 **b. Loan B (5-year loan at 2.4%)** - **Monthly Payment:** Approximately \$442 - **Total Interest:** \$1,490 **c. Comparison** - **Monthly Payments:** Loan A is higher (\$715) than Loan B (\$442). - **Total Interest:** Loan A has less interest (\$740) compared to Loan B (\$1,490). **Summary:** Loan A offers higher monthly payments but less total interest, while Loan B provides lower monthly payments with more total interest over the loan period.

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Beyond the Answer

For Loan A (3-year loan at 1.9%):

For Loan B (5-year loan at 2.4%):

Summary:

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