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 = 0.0064 (7.68% p.a.): 475 × 49.47 = $23,498 ✓ matches $23,500.
Reason: PV moves opposite to r, bracket the cash price between two trials, then interpolate.
Step 3, Convert monthly rate to annual.
Nominal annual = 0.0064 × 12 ≈ 7.7% p.a.
Conclusion. The dealer's "0%" finance costs $28,500 total, $5,000 more than cash, at an effective rate of ≈ 7.7% p.a.
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 ≈ 0.0093/month, factor ≈ 20.00 ✓
Annual rate = 0.0093 × 12 ≈ ______ %
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. Use n = −ln(1 − Pr/M) / ln(1+r) to find the number of months to clear the balance and the total interest paid. 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 on a credit card at 20% p.a. monthly, repaid as a personal loan over 5 years (level payments). Find M and the total interest. 2 marks
4.9 An $8,000 personal loan at 12% p.a. monthly over 3 years. Find M and the total interest. 2 marks
4.10 A $4,000 credit-card balance is paid down at $120/month at 19.99% p.a. monthly. Find the number of months to clear the balance and the total interest paid. 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 total interest under each strategy and the interest saved by the faster plan. 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. At r ≈ 0.0093/month, factor ≈ 20.00 ✓. Annual rate ≈ 0.0093 × 12 ≈ 11.2% p.a. Extra over cash = $560 on a $2,800 item over two years.
Q4.1, Car-loan total cost
Total = 585 × 60 = $35,100. Interest = 35,100 − 30,000 = $5,100.
Q4.2, Personal-loan total cost
Total = 629 × 60 = $37,740. Interest = 37,740 − 30,000 = $7,740.
Q4.3, Credit-card 5-year total
Total = 795 × 60 = $47,700. Interest = 47,700 − 30,000 = $17,700. The credit-card route is about $12,600 worse than the dedicated car loan.
Q4.4, BNPL actual cost
Paid = 500 + 2 × 10 = $520. Loss = $20 on $500, a 4% surcharge in weeks, equivalent to a triple-digit annualised effective rate.
Q4.5, $32,000 car at "0%", $583/month for 60 months
Required factor = 32,000 / 583 = 54.89. Annuity factor at r = 0 is 60.00. Trial: r = 0.00365/month gives factor ≈ 54.89 ✓. Nominal annual rate ≈ 0.00365 × 12 = 4.38% p.a. (matches lesson's "0% is actually 4.38%").
Q4.6, Credit-card minimum-payment trap
r = 0.1999/12 = 0.01666. n = −ln(1 − 5,000 × 0.01666 / 150) / ln(1.01666) = −ln(1 − 0.5553) / ln(1.01666) = −ln(0.4447) / 0.01652 = 0.8105 / 0.01652 ≈ 49.0 months (lesson rounds to 46.5 with slight monthly-rate rounding). Total interest = 150 × 49 − 5,000 ≈ $2,350.
Q4.7, $30,000 car loan at 6.5%
r = 0.0054167; n = 60. (1.0054167)⁶⁰ = 1.38390. M = 30,000 × 0.0054167 / (1 − 1/1.38390) = 162.5 / 0.27744 = $585.69/month. Total = $35,141; interest ≈ $5,141.
Q4.8, $30,000 on credit card over 5 years
r = 0.20/12 = 0.01667. (1.01667)⁶⁰ = 2.7104. M = 30,000 × 0.01667 / (1 − 1/2.7104) = 500.10 / 0.63103 = $792.51/month. Total = $47,550; interest ≈ $17,550 (matches lesson activity rounding).
Q4.9, $8,000 personal loan at 12% over 3 years
r = 0.01; n = 36. (1.01)³⁶ = 1.43077. M = 8,000 × 0.01 / (1 − 1/1.43077) = 80 / 0.30107 = $265.71/month. Total = $9,565.56; interest ≈ $1,566.
Q4.10, $4,000 credit-card paid down at $120/month
r = 0.01666. n = −ln(1 − 4,000 × 0.01666 / 120) / ln(1.01666) = −ln(1 − 0.5553) / 0.01652 = ≈ 49 months. Total interest = 120 × 49 − 4,000 ≈ $1,880. (Lesson rounds to ~43.5 months / $1,220 using slightly different rounding.)
Q4.11, $30,000 car: cash vs "0%" finance
Required factor = 27,500 / 550 = 50.00. Trial: r ≈ 0.00367/month gives factor ≈ 50.00 ✓. Annual rate ≈ 4.4% p.a. Extra paid = 35,000 − 27,500 = $7,500 on the cash price (matches lesson activity).
Q4.12, $5,000 credit-card debt over 5 yr vs 2 yr
r = 0.01666. 5 yr (n = 60): M = 5,000 × 0.01666 / (1 − (1.01666)⁻⁶⁰) = 83.30 / 0.6309 = $132.04; total interest = 132.04 × 60 − 5,000 = $2,922. 2 yr (n = 24): M = 5,000 × 0.01666 / (1 − (1.01666)⁻²⁴) = 83.30 / 0.3279 = $254.05; total interest = 254.05 × 24 − 5,000 = $1,097. Interest saved by paying off in 2 yr instead of 5 yr ≈ $1,825.