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.