Problem 27

Question

Comparing loan options. Kathy plans to finance \(\$ 12,000\) for a new car through an amortized loan. The lender offers two options: (1) a 5-yr term at an annual interest rate of \(5.2 \%,\) compounded monthly, and (2) a 6-yr term at an annual interest rate of \(5 \%,\) compounded monthly. a) Find the monthly payments for options 1 and 2 . b) Assume that Kathy makes every monthly payment. Find her total payments for options 1 and 2 . c) Assume that Kathy intends to make every monthly payment. Which option will result in less interest paid, and how much less?

Step-by-Step Solution

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Answer
Option 1 costs less interest by \$194.40.
1Step 1: Calculate Monthly Interest Rate for Option 1
For the first option, we have an annual interest rate of 5.2%, compounded monthly. The monthly interest rate is \(\frac{5.2}{12} = 0.4333\%\), or \(\frac{5.2}{100 \times 12} = 0.004333\) as a decimal.
2Step 2: Calculate Monthly Payment for Option 1
Use the formula for the monthly payment of an amortized loan:\[M = P \frac{r(1 + r)^n}{(1 + r)^n - 1}\]where \(P = 12000\), \(r = 0.004333\), and \(n = 60\). Substituting these values gives:\[M = 12000 \frac{0.004333(1 + 0.004333)^{60}}{(1 + 0.004333)^{60} - 1}\]Upon calculation, \(M \approx \$228.60\).
3Step 3: Calculate Monthly Interest Rate for Option 2
For the second option, the annual interest rate is 5%, compounded monthly. The monthly interest rate is \(\frac{5}{12} = 0.4167\%\), or \(\frac{5}{100 \times 12} = 0.004167\) as a decimal.
4Step 4: Calculate Monthly Payment for Option 2
Using the same formula for the monthly payment:\[M = P \frac{r(1 + r)^n}{(1 + r)^n - 1}\]where \(P = 12000\), \(r = 0.004167\), and \(n = 72\). Substituting these values gives:\[M = 12000 \frac{0.004167(1 + 0.004167)^{72}}{(1 + 0.004167)^{72} - 1}\]Upon calculation, \(M \approx \$193.20\).
5Step 5: Calculate Total Payment for Option 1
Multiply the monthly payment by the number of payments:\[\text{Total Payment} = 228.60 \times 60 = \$13,716\].
6Step 6: Calculate Total Payment for Option 2
Multiply the monthly payment by the number of payments:\[\text{Total Payment} = 193.20 \times 72 = \$13,910.40\].
7Step 7: Calculate Total Interest Paid for Both Options
For option 1, the total interest is the total payment minus the loan amount:\[13716 - 12000 = \\(1716\].For option 2:\[13910.40 - 12000 = \\)1910.40\].
8Step 8: Compare Interest Payments of Both Options
Subtract the total interest of option 1 from option 2:\[1910.40 - 1716 = \\(194.40\].Option 1 results in \\)194.40 less interest paid compared to option 2.

Key Concepts

Monthly Payments CalculationInterest Rate ComparisonTotal Interest Paid
Monthly Payments Calculation
To figure out the monthly payment amount for an amortized loan, understanding the formula is key. An amortized loan is repaid over regular installments, usually monthly, that cover both principal and interest. This calculation can be daunting, but broken down, it's more manageable.

The fundamental formula used is:
  • \(M = P \frac{r(1 + r)^n}{(1 + r)^n - 1}\)
where:
  • \(M\) is the monthly payment.
  • \(P\) is the principal amount of the loan, in Kathy's case, this is \(\\(12,000\).
  • \(r\) is the monthly interest rate, found by dividing the annual rate by 12.
  • \(n\) is the total number of payments (months).
For instance, for option 1, with a 5.2% annual interest rate and a 5-year term:
  • First, convert the annual interest rate to a monthly rate: \(5.2\% \div 12 = 0.4333\%\). As a decimal, that's \(0.004333\).
  • There are 60 months (5 years) of payments.
  • Inserting these values into the formula yields a monthly payment of approximately \(\\)228.60\).
Similarly, for option 2, with a 5% annual interest rate and a 6-year term:
  • The monthly interest rate is \(5\% \div 12 = 0.4167\%\), or \(0.004167\) as a decimal.
  • Here, there will be 72 months of payments.
  • This setup results in a monthly payment of about \(\$193.20\).
These mathematical steps can help anyone calculate their own loan repayments.
Interest Rate Comparison
Understanding the difference in interest rates and how they affect your payments is crucial when choosing between loan options. Even seemingly small differences in rates can make quite an impact over time.

Interest, in this context, is the cost of borrowing money. It gets added onto the principal you owe. Monthly compounding means the interest is recalculated each month on the current amount you owe. Thus, a lower interest rate can save you money.

In the given exercise:
  • Option 1 offers a 5.2% annual interest rate.
  • Option 2 gives a 5% annual interest rate.
While option 2 has a lower annual interest rate, the total number of payments also needs to be considered. Lower interest might sound appealing, but with more payments (72 months versus 60), the overall interest paid could still be more. The choice depends on which factor, lower monthly payments or total interest paid, better serves your financial goals.

See how the different rates affect total costs by comparing the calculations of total interest below.
Total Interest Paid
Knowing how much extra you'll pay in interest over the life of a loan is just as important as the monthly payment alternative you select. This total is what you'll spend simply due to borrowing money, beyond repaying the initial borrowed amount.

To find out how much interest is paid:
  • Determine the total amount paid over the life of the loan.
  • Subtract the principal from this total amount.
For option 1:
  • The total paid over 60 months is \(\\(13,716\).
  • By subtracting the initial \(\\)12,000\), the interest paid is \(\\(1,716\).
And for option 2:
  • Total payments over 72 months amount to \(\\)13,910.40\).
  • After deducting the loan principal, the total interest cost comes out to \(\\(1,910.40\).
By comparing the two, option 1 saves \(\\)194.40\) in interest compared to option 2. Thus, if minimizing interest expense is a priority, option 1 could be the better choice.

Understanding these principles helps in making informed financial decisions, ensuring that you can balance monthly budget constraints with long-term cost savings.