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.