Mathematics Advanced • Year 12 • Module 7 • Lesson 1
Introduction to Financial Mathematics
Apply simple and compound interest to real saving, borrowing and product-comparison scenarios.
Problem 1 — Two deals from a friend (the "Think First" scenario)
A friend offers two deals on $10,000:
Deal A: 8% p.a. simple interest for 10 years.
Deal B: 6% p.a. compound interest (annually) for 10 years.
Set up: What are we solving for?
(i) Find the final amount under each deal, to the nearest cent. 3 marks
(ii) State which deal pays more after 10 years and by how many dollars. 1 mark
(iii) Determine, by trial, the first whole year in which Deal B overtakes Deal A. Explain in one sentence why this happens eventually for any pair of "lower-rate compound vs higher-rate simple". 3 marks
Stuck? Revisit lesson § Think First and § Compound Interest.Problem 2 — Holiday-fund term deposit
A family deposits $25,000 into a term deposit paying 4.8% p.a. compounded monthly. The deposit runs for 3 years before being used for an overseas holiday.
Set up: What are we solving for?
(i) State the monthly rate and the total number of compounding periods. 1 mark
(ii) Find the value of the deposit at maturity, to the nearest cent. 2 marks
(iii) A competing bank offers 5.0% p.a. compounded annually for the same term. Which product pays more, and by how much? 3 marks
Problem 3 — Car loan (borrowing side)
Tom borrows $12,000 from a personal lender at 5.4% p.a. simple interest. He repays the loan in one lump sum after 3.5 years.
Set up: What are we solving for?
(i) Find the total amount Tom must repay. 1 mark
(ii) A different lender offers the same loan at 5.4% p.a. compounded annually. Find the total repayment and the dollar difference between the two products. 3 marks
(iii) Explain in one sentence why a borrower would prefer the simple-interest product and a lender would prefer the compound-interest product. 1 mark
Stuck? Revisit lesson § Compounding Periods — Critical Rule.Problem 4 — Long-horizon savings (the "3% is not low" question)
An aunt argues that a 3% p.a. savings rate "is too low to bother with." She is choosing where to park $50,000 for 30 years.
Set up: What are we solving for?
(i) Calculate the final balance at 3% p.a. compounded annually over 30 years. 2 marks
(ii) Compare with 3% p.a. simple interest over 30 years. State the dollar difference. 2 marks
(iii) Write a one-line counter-argument to the aunt: use the final compound balance to make a quantitative point about long horizons and exponential growth. 2 marks
Problem 5 — Reading the marketing carefully
A bank advertises a savings account at "4.5% p.a. compounded monthly." A customer assumes that "4.5% over one year" means the final balance after one year is exactly $10,450 on a $10,000 deposit.
Set up: What are we solving for?
(i) Compute the actual balance after one year, to the nearest cent. 2 marks
(ii) Express the actual annual return as a percentage, to two decimal places, and explain in one sentence why it exceeds 4.5%. 2 marks
(iii) If the same nominal rate compounded daily (n = 365), recompute the actual return for one year. By how many extra dollars does daily compounding beat monthly on a $10,000 deposit over one year? 2 marks
Stuck? Revisit lesson § Compounding Periods table.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Two deals on $10,000
Set up. We are computing the final amount under each interest model and identifying which is larger and when their orderings swap.
(i) Deal A (simple 8% for 10 yr): A = 10,000(1 + 0.08 × 10) = $18,000.00. Deal B (compound 6% for 10 yr): A = 10,000(1.06)¹⁰ = 10,000 × 1.790847 = $17,908.48.
(ii) Deal A pays more by 18,000 − 17,908.48 = $91.52.
(iii) At year 13: A = 10,000(1.06)¹³ ≈ $21,329 versus simple $20,400. Year 13 is the first year Deal B overtakes Deal A. Why: simple interest is linear in n while compound interest is exponential in n; exponential growth eventually overtakes any straight-line schedule.
Problem 2 — Term deposit
Set up. We are valuing a monthly-compounded deposit at maturity and comparing it against an annually-compounded competitor at a higher headline rate.
(i) r = 0.048/12 = 0.004 per month; n = 3 × 12 = 36 periods.
(ii) A = 25,000(1.004)³⁶ = 25,000 × 1.154594 = $28,864.85.
(iii) Competitor: A = 25,000(1.05)³ = 25,000 × 1.157625 = $28,940.63. The annual-compounding 5.0% product pays $28,940.63 − $28,864.85 = $75.78 more, even though monthly compounding is "more frequent" — because the headline rate is higher.
Problem 3 — Car loan
Set up. We are computing total repayment under two interest models and explaining the borrower/lender preference.
(i) A = 12,000(1 + 0.054 × 3.5) = 12,000 × 1.189 = $14,268.00.
(ii) Compound: A = 12,000(1.054)3.5. Compute (1.054)3.5 = e3.5 ln 1.054 = e3.5 × 0.052585 = e0.184048 = 1.20208. A = 12,000 × 1.20208 = $14,425.00 (to nearest cent). Difference = 14,425.00 − 14,268.00 = $157.00.
(iii) Borrowers want simple interest because it does not compound on accrued interest — they pay less. Lenders want compound interest because the interest itself earns interest — they receive more.
Problem 4 — 3% over 30 years on $50,000
Set up. We are projecting a long-horizon balance under compound and simple regimes to refute the claim that 3% is "not worth bothering with".
(i) A = 50,000(1.03)³⁰ = 50,000 × 2.42726 = $121,363.12.
(ii) Simple: A = 50,000(1 + 0.03 × 30) = 50,000 × 1.90 = $95,000.00. Difference = 121,363.12 − 95,000 = $26,363.12. Compound earns over $26,000 more than simple on the same rate.
(iii) Sample line: "At only 3% p.a., $50,000 grows to over $121,000 in 30 years — more than doubling — because exponential growth at any positive rate beats your intuition over long horizons."
Problem 5 — "4.5% p.a. monthly"
Set up. We are converting a monthly-compounded nominal rate into the actual one-year return and comparing it against daily compounding.
(i) r = 0.045/12 = 0.00375; n = 12. A = 10,000(1.00375)¹² = 10,000 × 1.045938 = $10,459.38.
(ii) Actual annual return = (10,459.38 − 10,000) ÷ 10,000 × 100 = 4.59%. It exceeds 4.5% because each month's interest is added to the balance and itself earns interest in the following months — exactly the (1 + r/n)ⁿ − 1 effect.
(iii) Daily: r = 0.045/365; n = 365. A = 10,000(1 + 0.045/365)³⁶⁵ = 10,000 × 1.046028 = $10,460.28. Daily beats monthly by 10,460.28 − 10,459.38 = $0.90 per year on a $10,000 deposit — small at one year, larger over decades.