Mathematics Advanced • Year 12 • Module 7 • Lesson 17
Car Loans, Personal Loans and Credit Cards
Build fluency with total-cost, effective-rate and minimum-payment calculations for consumer loans.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the formulas:
Total loan cost (level monthly payments): Total = ____________________
Implicit rate (loan PV identity): P = M × ____________________
Q1.2 A car loan of $30,000 at 6.5% p.a. compounded monthly is repaid over 5 years. State the periodic rate r and the number of periods n.
r = ____________ n = ____________
Q1.3 In one sentence, state why "0% finance" on a car is rarely a true 0% loan.
2. Worked example, the lesson's 0%-finance car
Follow every line. Each step has a short reason.
Problem. A $25,000 car has cash price $23,500. Dealer finance: $0 down, $475/month for 60 months at "0%". Find (a) the total cost of finance and (b) the effective annual interest rate.
Step 1, Compute total finance cost.
Total = M × n = 475 × 60 = $28,500
Reason: with level monthly payments, the total handed over is just payment × periods.
Step 2, Solve the PV identity for r.
23,500 = 475 × [1 − (1 + r)⁻⁶⁰] / r
Trial r = 0.005 (6.0% p.a.): 475 × 51.73 = $24,572, too high a PV → r too low.
Trial r = 0.0075 (9.0% p.a.): 475 × 48.17 = $22,881, PV too low → r too high.
Trial r = 0.00655577/month: the present value of 60 end-of-month payments of $475 is $23,500. Nominal annual rate = 12r ≈ 7.87%; effective annual rate = (1+r)¹² − 1 ≈ 8.16%.
Reason: PV moves opposite to r, bracket the cash price between two trials, then interpolate.
Step 3, Convert monthly rate to annual.
Nominal annual = 12r ≈ 7.87% p.a.; effective annual = (1+r)¹² − 1 ≈ 8.16% p.a.
Conclusion. The 60 payments total $28,500, which is $5,000 above the $23,500 cash price. The implied nominal annual rate is about 7.87% and effective annual rate about 8.16%.
3. Faded example, fill in the missing steps
A $3,000 laptop. Cash price $2,800. Finance offer: $140/month for 24 months. 4 marks
Step 1, Total finance cost:
Total = 140 × ______ = $______________
Step 2, PV identity:
2,800 = 140 × [1 − (1 + r)⁻²⁴] / r
Required annuity factor = 2,800 / 140 = ______
Step 3, Trial r:
At r ≈ 0.01513084/month, the factor is 20.00 ✓
Nominal annual rate = 12r ≈ ______ %; effective annual rate = (1+r)¹² − 1 ≈ ______ %
Conclusion. Total finance = $______________, extra over cash = $______________, effective rate ≈ ______ % p.a.
4. Graduated practice, total cost and effective rate
Show the substitution and the final amount for each. Use the same time units for r and n.
Foundation, single-step total cost (4 questions)
| Q | Scenario | Working & answer |
|---|---|---|
| 4.1 1 | Car loan: $585/month for 60 months. State total cost. | |
| 4.2 1 | Personal loan: $629/month for 60 months. State total cost. | |
| 4.3 1 | Credit-card monthly repayment $795 over 60 months. State total cost and interest paid on a $30,000 balance. | |
| 4.4 1 | BNPL: $500 purchase, two missed payments at $10 late fee each. State actual cost. |
Standard, typical HSC difficulty (6 questions)
Show working in the space below each part, at least one substitution line and one evaluation line.
4.5 $32,000 car. Dealer offers 0% finance: $583/month for 60 months. Find the effective monthly rate r and quote it as a nominal annual rate. 2 marks
4.6 A $5,000 credit-card balance is repaid at the minimum $150/month at 19.99% p.a. compounded monthly. Assume interest is applied monthly before each end-of-month payment. Find the continuous estimate of months from the formula, then use a monthly balance recurrence to determine the scheduled payment count, final partial payment and total interest. 3 marks
4.7 $30,000 car at 6.5% p.a. monthly over 5 years. Use M = Pr / [1 − (1+r)⁻ⁿ] to find the monthly repayment and the total interest. 2 marks
4.8 The same $30,000 borrowed at 20% p.a. compounded monthly and repaid by 60 equal end-of-month payments. Find M and formula total interest, rounding currency only after evaluating the full expression. 2 marks
4.9 An $8,000 personal loan at 12% p.a. monthly over 3 years. Find M and total interest using the formula, rounding the final currency values only. 2 marks
4.10 A $4,000 credit-card balance is paid down at $120/month at 19.99% p.a. monthly. Assume interest is applied monthly before each end-of-month payment. Find the continuous estimate of months from the formula, then use a monthly balance recurrence to determine the scheduled payment count, final partial payment and total interest. 2 marks
Extension, combine concepts (2 questions)
4.11 A dealer offers $30,000 car as either (a) cash at $27,500 or (b) $0 down, $550/month for 60 months at "0%". Find the effective annual rate of (b) and the extra dollars paid versus cash. 3 marks
4.12 A $5,000 credit-card debt at 19.99% p.a. is paid down either (a) over 5 years (≈ $132/month) or (b) over 2 years (≈ $254/month). Find the formula payment and total interest under each strategy, then the difference in total interest. State that formula totals assume unrounded level payments. 3 marks
5. Self-check the easy 3
Tick the first three once you have checked the method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, Formulas
Total = M × n. P = M × [1 − (1 + r)⁻ⁿ] / r.
Q1.2, Monthly rate and periods
r = 0.065 ÷ 12 = 0.005417. n = 5 × 12 = 60.
Q1.3, The 0% trick
The dealer inflates the sticker price (e.g. $35,000 finance vs $32,000 cash). The "extra" $3,000 is interest baked into the price; the rate is hidden, not absent.
Q3, Faded example: $3,000 laptop
Total = 140 × 24 = $3,360. Required annuity factor = 2,800 / 140 = 20.00. Solving [1−(1+r)⁻²⁴]/r = 20 gives r ≈ 0.01513084/month. Nominal annual rate = 12r ≈ 18.16%; effective annual rate = (1+r)¹²−1 ≈ 19.75%. Difference from the $2,800 cash price = $560.
Q4.1, Car-loan total cost
Total = $585 × 60 = $35,100.
Q4.2, Personal-loan total cost
Total = $629 × 60 = $37,740.
Q4.3, Credit-card 5-year total
Total = $795 × 60 = $47,700. Interest = 47,700 − 30,000 = $17,700.
Q4.4, BNPL actual cost
Paid = 500 + 2 × 10 = $520. The two late fees add a total surcharge of $20, or 4% of the $500 cash price. The supplied figures do not determine an effective annual rate without a timing model.
Q4.5, $32,000 car at "0%", $583/month for 60 months
Required factor = 32,000 / 583 = 54.8885. Solving [1−(1+r)⁻⁶⁰]/r = 54.8885 gives r ≈ 0.00296689/month. Nominal annual rate = 12r ≈ 3.56% p.a.; effective annual rate = (1+r)¹²−1 ≈ 3.62% p.a..
Q4.6, Credit-card minimum-payment trap
r = 0.1999/12 = 0.016658333… per month. The formula gives n = −ln(1 − 5,000r/150)/ln(1+r) ≈ 49.0467 months, a continuous estimate. With interest applied monthly before each end-of-month payment, 49 payments of $150 leave a final payment of about $7.06 in month 50; simulated interest is $2,357.06.
Q4.7, $30,000 car loan at 6.5%
r = 0.065/12; n = 60. M = 30,000r/[1−(1+r)⁻⁶⁰] = $586.98/month (unrounded M = $586.984446…). Using the unrounded payment, total paid is $35,219.07 and interest is $5,219.07.
Q4.8, $30,000 on credit card over 5 years
r = 0.20/12; n = 60. M = 30,000r/[1−(1+r)⁻⁶⁰] = $794.82/month (unrounded M = $794.816511…). Using unrounded M, total paid is $47,688.99 and interest is $17,688.99.
Q4.9, $8,000 personal loan at 12% over 3 years
r = 0.12/12; n = 36. M = 8,000r/[1−(1+r)⁻³⁶] = $265.71/month (unrounded M = $265.714479…). Using unrounded M, total paid is $9,565.72 and interest is $1,565.72.
Q4.10, $4,000 credit-card paid down at $120/month
r = 0.1999/12 = 0.016658333… per month. The formula gives n = −ln(1 − 4,000r/120)/ln(1+r) ≈ 49.0467 months, a continuous estimate. With interest applied monthly before each end-of-month payment, 49 payments of $120 leave a final payment of about $5.65 in month 50; simulated interest is $1,885.65.
Q4.11, $30,000 car: cash vs "0%" finance
For a $27,500 amount financed with 60 end-of-month payments of $550, solve 27,500 = 550[1−(1+r)⁻⁶⁰]/r. This gives r ≈ 0.00618341/month, nominal annual rate 7.42% and effective annual rate 7.68%. The payments total $33,000, a $5,500 difference from the cash price.
Q4.12, $5,000 credit-card debt over 5 yr vs 2 yr
r = 0.1999/12 = 0.016658333… per month. For 60 payments, M = 5,000r/[1−(1+r)⁻⁶⁰] = $132.441599…; unrounded-formula total interest is $2,946.50. For 24 payments, M = $254.454588…; unrounded-formula total interest is $1,106.91. These are formula totals using unrounded payments; if payments are rounded to cents and the final payment is adjusted, the monthly recurrence gives about $2,946.56 and $1,106.93 respectively.