NZ Level 6 (NZC) Level 1 (NCEA)
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Comparing Loans
Lesson

Loans come with different interest rates, with various methods of calculating the interest and the repayment amounts, and with different durations over which the loans are to be repaid. There may be extra charges that affect the total amount that the borrower will pay back.

When borrowing money, it is wise to make sure that the terms of the loan are clearly understood and that the agreement the borrower enters into is the least expensive one available that meets his or her requirements. However, it is not always easy to compare the loans offered by several lenders because of the variety of different options that are offered.

Finance for cars and other consumer items is often quoted with a fixed rate of simple interest and a repayment schedule of monthly amounts spread over an agreed number of months. There may be extra fees added in and sometimes there is an initial period during which no payment or reduced payments are made. Whatever the details of the arrangement, what really matters is how much in total will be paid to the lender over the whole period of the loan.

If we know how much is to be borrowed initially and how much will be repaid, we can work out an effective simple interest rate for the loan. This will make it possible to compare different loans.

Example 1

To enable me to buy a car, a car dealer offers to lend me $\$9000$$9000 at $4.5%$4.5% per annum simple interest. There will be a $\$25$$25 establishment fee and an annual administration fee of $\$20$$20 charged five times during the loan, which will run for $60$60 months. The interest rate seems reasonable, at first, but I decide to work out the effective rate, given the extra charges.

The interest over the $60$60 months, which I convert to $5$5 years because the interest is given as an annual rate, is $\frac{4.5}{100}\times9000\times5=\$2025$4.5100×9000×5=$2025. Over the period of the loan, I must repay the interest and the principle and the extra fees. This comes to $2025+9000+25+\left(5\times20\right)=\$11150$2025+9000+25+(5×20)=$11150.

The cost of the loan is, therefore, $11150-9000=\$2150$111509000=$2150 or $2150\div5=\$430$2150÷5=$430 per year. Thus, the effective interest rate is $\frac{430}{9000}\times100\approx4.78%$4309000×1004.78%. This effective rate is also called a comparison rate. As expected, it is slightly higher than the advertised rate. To be precise, it is $4.78-4.5=0.38%$4.784.5=0.38% higher.

 

Comparisons between this and other types of loans can be made if enough information is available to make it possible to know the full cost of the loan. Some financial institutions offer a 'reducing balance' type of loan so that the interest is charged on a decreasing amount owed as the loan progresses. 

The interest rate quoted for this type of loan would normally be higher than that for a simple interest loan but the actual amount paid could turn out to be less, particularly if the borrower makes bigger repayments than the minimum necessary. The calculation for this type of loan is more complicated to begin with and matters are made worse by the possibility that the interest rate can vary during the course of the loan.

Housing loans are of this kind. The financial instrument through which money is borrowed to purchase a house is called a mortgage. Lenders typically offer two different mortgage interest rates initially, depending on whether the rate is to be fixed for the whole of the repayment period or whether it is to be allowed to vary in line with changes to the central bank rates that lenders themselves have to pay on their borrowings.

A variable rate is usually lower than the prevailing fixed rate because it involves less risk to the lender. But the variable rate is risky for the borrower because if the rate moves significantly higher, which sometimes happens, the interest rises and borrowers may have to increase their repayments or extend the duration of the loan. 

Example 2

Still wanting to arrange finance to purchase the car in Example $1$1, I discuss my options with my credit union. I am told that I can borrow the $\$9000$$9000 at $8.25%$8.25% per annum interest calculated monthly on the balance outstanding, with no extra charges. (This is a reducing balance loan.) There would be $60$60 monthly installments in the repayment schedule, each of $\$184$$184. How does this compare with the loan offered by the car dealer?

Since there are $60$60 payments of $\$184$$184, I will repay $60\times184=\$11040$60×184=$11040 in total. The interest component of this amount is $11040-9000=\$2040$110409000=$2040. This is $2040\div5=\$408$2040÷5=$408 per year. The effective annual interest rate, then, is $\frac{408}{9000}\times100=4.53%$4089000×100=4.53%. This is better than the dealer's finance by a small margin of $4.78-4.53=0.25%$4.784.53=0.25%

 

Example 3

The table presents the details of three types of home loans.

Loan Interest Rate (p.a.) Legal Fees Application Fee Service Fee (p.a.)
A: Fixed for 5 years, then variable $5.99%$5.99% $\$160$$160 $\$550$$550 $\$20$$20
B: Variable $5.74%$5.74% $\$100$$100 $\$375$$375 $\$25$$25
C: Fixed $6.03%$6.03% $\$160$$160 $\$775$$775 $\$15$$15
  1. Which loan option currently has the lowest annual interest rate?

    A

    A

    B

    B

    C

    C

    A

    A

    B

    B

    C

    C
  2. Calculate the total fees that loan A will incur in the first year.

  3. If Loan C is taken out on a $30$30 year term, calculate the total fees incurred over the term of the loan.

  4. The table shows the expected change in the variable rate for the next 4 years.

    Year Rate change
    1 +$0.3%$0.3%
    2 -$0.18%$0.18%
    3 -$0.04%$0.04%
    4 +$0.32%$0.32%

    In 4 years, by what percentage will the variable rate of Loan B exceed the fixed rate of Loan C?

  5. Home loan rates are expected to fluctuate significantly for the next decade, and as the rate changes, loan repayments change accordingly. If Maria wants to know her exact repayments over the 7 year term of her loan, which should she choose?

    A

    A

    B

    B

    C

    C

    A

    A

    B

    B

    C

    C

Example 4

Comparison rates allow for potential loan recipients to quickly compare different loans, accounting for interest rates, fees and other charges.

Loan Interest Rate Upfront Establishment Fee Service Fee
Standard Variable Rate A $5.34%$5.34% $\$350$$350 $\$10$$10
Standard Variable Rate B $5.34%$5.34% $\$250$$250 $\$10$$10
  1. Which loan's comparison rate would you expect to be the higher of the two?

    Standard Variable Rate A

    A

    Standard Variable Rate B

    B

    Standard Variable Rate A

    A

    Standard Variable Rate B

    B
  2. The comparison rates of each of the loans is shown in the table below.

    Loan Interest Rate (p.a.) Comparison Rate (p.a.)
    Standard Variable Rate A $5.34%$5.34% $5.5%$5.5%
    Standard Variable Rate B $5.34%$5.34% $5.53%$5.53%

    What is the difference between the Interest rate and Comparison rate for Variable Rate Loan B? Express your answer as a percentage.

Example 5

Luke is aiming to secure a loan worth $\$400000$$400000 for a new house.

The terms of the loan are such that he will have to make monthly repayments of $\$2300$$2300 for $25$25 years.

  1. How much will Luke have to pay on the loan in the first year?

  2. Determine the total amount Luke will repay over $25$25 years.

  3. Calulate the total interest payment.

  4. The annual nominal rate for the loan is $2.9%$2.9%.

    Calculate the effective interest rate if interest is compounded monthly. Give your answer as a percentage correct to two decimal places.

  5. If interest were compounded fortnightly at a nominal rate of $2.9%$2.9% p.a., instead of monthly, which of the following scenarios would occur?

    You would pay more interest over the lifetime of the loan.

    A

    You would pay less interest over the lifetime of the loan.

    B

    You would pay more interest over the lifetime of the loan.

    A

    You would pay less interest over the lifetime of the loan.

    B

Outcomes

NA6-3

Apply everyday compounding rates

91026

Apply numeric reasoning in solving problems

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