Mathematics Standard • Year 11 • Module 3 • Lesson 12
Compound Interest
Apply the compound interest toolkit to realistic investment, loan and comparison scenarios — decompose, calculate, conclude.
Problem 1 — Term deposit with monthly compounding (graduate's savings)
Hana deposits $11,500 from her first-year salary into a 4-year term deposit. The bank advertises a rate of 5.4% per annum compounding monthly. She makes no further deposits or withdrawals.
Set up: What are we solving for?
(i) State the adjusted values of r (per month) and n (months) you will use in A = P(1 + r)n. 1 mark
(ii) Calculate the value of Hana's investment after 4 years. Show (1 + r)n as a separate line. 2 marks
(iii) Calculate the interest earned over the 4 years. 1 mark
Stuck? Revisit lesson § A Reliable Compound Interest Workflow — adjust r AND n before substituting.Problem 2 — Simple vs compound interest (credit union)
A credit union offers two 5-year products on a $20,000 deposit.
Product A: Simple interest at 4.5% per annum.
Product B: Compound interest at 4.5% per annum, compounding annually.
Set up: What are we solving for?
(i) Calculate the interest earned under Product A. 1 mark
(ii) Calculate the final amount and the interest earned under Product B. 2 marks
(iii) State which product earns more interest over 5 years, and by how much. Write a clear conclusion sentence. 2 marks
Stuck? Revisit lesson § Comparing Simple and Compound Interest — same P, same rate, same time.Problem 3 — Credit card debt allowed to grow (no repayments)
Liam owes $4,200 on a credit card at 19.2% per annum compounding monthly. He makes no repayments for 18 months.
Set up: What are we solving for?
(i) State the adjusted values of r and n. 1 mark
(ii) Calculate the amount Liam owes after 18 months. 2 marks
(iii) Calculate the total interest charged, then explain in one sentence why compound interest on a loan is described as "debt that grows by itself". 2 marks
Stuck? Revisit lesson § Worked Example 3 — even with no further spending, the balance grows because interest is added every month.Problem 4 — Choosing compounding frequency (the same nominal rate)
A bank offers a $25,000 investment at a nominal rate of 6% per annum for 5 years. The customer chooses the compounding frequency: annually, quarterly, or monthly.
Set up: What are we solving for?
(i) Calculate the final amount for annual compounding. 1 mark
(ii) Calculate the final amount for quarterly compounding. 1 mark
(iii) Calculate the final amount for monthly compounding. Then rank the three options and state the difference between the best and worst. 3 marks
Stuck? Revisit lesson § Compounding Periods — more frequent compounding = more interest, but the increases get smaller.Problem 5 — Working backwards from a target amount (saving for a deposit)
Mei wants $50,000 in 5 years' time to use as a house deposit. A high-interest savings account compounds annually at 6.2% per annum.
Set up: What are we solving for?
(i) Suppose Mei deposits $35,000 today. Calculate the final amount after 5 years and decide whether she meets her $50,000 target. 2 marks
(ii) Suppose instead she deposits $37,500 today. Calculate the final amount after 5 years and decide whether this is enough. 2 marks
(iii) Looking at parts (i) and (ii), state in one sentence why depositing $35,000 falls short while $37,500 succeeds — referring to the growth multiplier (1.062)5. 2 marks
Stuck? The multiplier (1.062)5 ≈ 1.3509, so the final amount is roughly the deposit × 1.35.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Hana's term deposit
Set up. We are using compound interest with monthly compounding to find the future value of $11,500 after 4 years, then the interest earned.
(i) r = 0.054 ÷ 12 = 0.0045 per month; n = 4 × 12 = 48 months.
(ii) (1.0045)48 = 1.240618…
A = 11,500 × 1.240618 = $14,267.11.
(iii) I = $14,267.11 − $11,500 = $2,767.11. (Slips to watch: forgetting to subtract P; rounding (1.0045)48 to 1.24 before multiplying — that gives $14,260 and a small but avoidable mark loss.)
Problem 2 — Simple vs compound on $20,000 at 4.5%, 5 years
Set up. We calculate the interest under both products at the same P, r and time, then state which is higher and by how much.
(i) IA = Prn = 20,000 × 0.045 × 5 = $4,500.00.
(ii) (1.045)5 = 1.246182…; AB = 20,000 × 1.246182 = $24,923.64; IB = $24,923.64 − $20,000 = $4,923.64.
(iii) Difference = $4,923.64 − $4,500.00 = $423.64. Product B (compound) earns more by $423.64 over 5 years. (A bare "$423.64" with no naming of which product loses the conclusion mark.)
Problem 3 — Liam's unpaid credit card
Set up. We find the amount owed after 18 months of monthly compounding with no repayments, then the interest charged.
(i) r = 0.192 ÷ 12 = 0.016 per month; n = 18 months.
(ii) (1.016)18 = 1.331697…; A = 4,200 × 1.331697 = $5,593.13.
(iii) Interest = $5,593.13 − $4,200 = $1,393.13. The debt is "growing by itself" because each month's interest is added to the balance and the next month's interest is calculated on the new, larger balance — interest is earning interest, so the balance grows exponentially even without further spending.
Problem 4 — Comparing compounding frequencies on $25,000 at 6% p.a., 5 years
Set up. Calculate the final amount for each compounding frequency at the same nominal rate, then rank.
(i) Annual: (1.06)5 = 1.338226; A = 25,000 × 1.338226 = $33,455.64.
(ii) Quarterly: r = 0.015; n = 20. (1.015)20 = 1.346855; A = 25,000 × 1.346855 = $33,671.38.
(iii) Monthly: r = 0.005; n = 60. (1.005)60 = 1.348850; A = 25,000 × 1.348850 = $33,721.25.
Ranking (highest to lowest): monthly > quarterly > annual. Best − worst = $33,721.25 − $33,455.64 = $265.61. (Note: the gap shrinks each time you increase the frequency — diminishing returns.)
Problem 5 — Working backwards to a $50,000 target
Set up. Test two deposit amounts at 6.2% p.a. compounding annually for 5 years to see which clears $50,000.
(i) (1.062)5 = 1.350859…; A = 35,000 × 1.350859 = $47,280.06. This is $2,719.94 short of $50,000 — does NOT meet the target.
(ii) A = 37,500 × 1.350859 = $50,657.21. This exceeds the $50,000 target by $657.21 — meets the target.
(iii) Each deposit is multiplied by the same growth factor (1.062)5 ≈ 1.3509. The minimum deposit needed = $50,000 ÷ 1.3509 ≈ $37,012, so $35,000 falls short but $37,500 clears the bar with about $657 to spare.