Mathematics Standard • Year 12 • Module 7 • Lesson 2
Compound Interest — Problem Set
Apply compound interest to realistic Australian savings, term-deposit and credit-card scenarios — and quantify the cost of more frequent compounding.
Problem 1 — Term deposit comparison (compound vs simple)
Mei deposits $20,000 into a 5-year term deposit at her local bank, which pays 4.6% p.a. compounded annually. Her brother places $20,000 with a different institution paying 4.6% p.a. simple interest for the same period.
Set up: What are we solving for?
(i) Calculate the final amount in Mei's account after 5 years. 2 marks
(ii) Calculate the final amount in her brother's account after 5 years. 1 mark
(iii) Calculate how much more Mei earns than her brother, and state which mathematical effect is responsible. 2 marks
Stuck? Revisit lesson § Compound Interest — interest on interest earns extra over multiple years.Problem 2 — Choosing a compounding frequency
A savings account pays 6% p.a. but the customer can choose how often interest is compounded. Asha plans to deposit $10,000 and leave it untouched for 10 years.
Option 1: 6% p.a. compounded annually (k = 1).
Option 2: 6% p.a. compounded quarterly (k = 4).
Option 3: 6% p.a. compounded monthly (k = 12).
Set up: What are we solving for?
(i) Calculate A under Option 1 after 10 years. 1 mark
(ii) Calculate A under Option 3 after 10 years. 2 marks
(iii) Recommend the best option and state the dollar gap between Option 1 and Option 3 over the 10 years. 2 marks
Stuck? Revisit lesson § Compounding Frequency — more frequent compounding gives a higher effective rate.Problem 3 — Which credit card is more expensive?
Two credit cards advertise different nominal rates.
Card P: 19.9% p.a. compounded daily (k = 365).
Card Q: 20.5% p.a. compounded monthly (k = 12).
Set up: What are we solving for?
(i) Calculate the effective annual rate of Card P, to 2 decimal places of a percent. 2 marks
(ii) Calculate the effective annual rate of Card Q, to 2 decimal places of a percent. 2 marks
(iii) State which card is the more expensive for a customer who carries a balance, with a one-sentence justification. 1 mark
Stuck? Revisit lesson § Effective Annual Rate — always compare effective rates, never nominal.Problem 4 — One-off super contribution growth
Liam makes a one-off $25,000 voluntary contribution to his superannuation at age 30. The fund earns 7.2% p.a. compounded quarterly, and Liam does not touch the contribution until age 60 (a period of 30 years).
Set up: What are we solving for?
(i) Calculate the total in this part of his fund at age 60. 2 marks
(ii) Calculate the total interest earned. 1 mark
(iii) If Liam had instead made the same contribution at age 40 (so it grew for only 20 years), how much smaller would the age-60 balance be? 2 marks
Stuck? Revisit lesson § Compound Interest — the same contribution invested 10 years earlier grows substantially larger.Problem 5 — Solving for an unknown principal
A retirement fund needs to grow from an initial deposit to exactly $50,000 in 8 years. The fund earns 5.4% p.a. compounded monthly.
Set up: What are we solving for?
(i) Write A = P(1 + r/k)^(kn) and identify r/k and kn for this scenario. 1 mark
(ii) Rearrange to make P the subject and calculate the required initial deposit, to the nearest dollar. 3 marks
(iii) State how much interest the account will earn over the 8 years. 1 mark
Stuck? Revisit lesson § Compound Interest — rearrange A = P(1 + r/k)^(kn) by dividing both sides by (1 + r/k)^(kn).How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Compound vs simple term deposits
Set up. Find Mei's compound final amount, her brother's simple final amount, then the difference.
(i) Mei: A = 20,000 × (1.046)^5 = 20,000 × 1.252198 = $25,043.96.
(ii) Brother: A = 20,000 × (1 + 0.046 × 5) = 20,000 × 1.23 = $24,600.00.
(iii) Difference = $25,043.96 − $24,600.00 = $443.96. The extra comes from interest earning interest in Mei's account — each year's interest is added to the principal and itself earns 4.6% in subsequent years.
Problem 2 — Compounding frequency on $10,000 at 6% for 10 years
Set up. Apply A = P(1 + r/k)^(kn) for k = 1 and k = 12 over 10 years and compare.
(i) k = 1: A = 10,000 × (1.06)^10 = 10,000 × 1.790847 = $17,908.48.
(ii) k = 12: r/k = 0.005, kn = 120. A = 10,000 × (1.005)^120 = 10,000 × 1.819397 = $18,193.97.
(iii) Option 3 (monthly) is best, $18,193.97 vs $17,908.48 — a gap of $285.49 over 10 years. More frequent compounding gives a higher effective annual rate (6.17% vs 6%).
Problem 3 — Credit card comparison
Set up. Compute the effective annual rate for each card; whichever is higher costs more.
(i) Card P: r/k = 0.199/365 ≈ 0.000545. Effective = (1.000545)^365 − 1 = 1.22010 − 1 ≈ 22.01% p.a.
(ii) Card Q: r/k = 0.205/12 ≈ 0.017083. Effective = (1.017083)^12 − 1 = 1.22538 − 1 ≈ 22.54% p.a.
(iii) Card Q is more expensive (22.54% effective vs 22.01%), despite Card P having more frequent compounding — the higher nominal rate on Q wins out.
Problem 4 — $25,000 super for 30 years vs 20 years at 7.2% quarterly
Set up. Compute the age-60 amount for n = 30, then n = 20, then subtract.
(i) r/k = 0.018, kn = 120. A = 25,000 × (1.018)^120 = 25,000 × 8.46415 = $211,603.69.
(ii) Interest = $211,603.69 − $25,000 = $186,603.69.
(iii) If contributed at 40: kn = 80. A = 25,000 × (1.018)^80 = 25,000 × 4.17118 = $104,279.45. Shortfall = $211,603.69 − $104,279.45 = $107,324.24. (Investing 10 years earlier roughly doubles the final balance.)
Problem 5 — Find the principal for $50,000 in 8 years at 5.4% monthly
Set up. Identify r/k and kn, rearrange A = P(1 + r/k)^(kn) to P = A / (1 + r/k)^(kn), substitute, then find interest.
(i) r/k = 0.054 / 12 = 0.0045. kn = 12 × 8 = 96.
(ii) P = 50,000 / (1.0045)^96 = 50,000 / 1.538559 = $32,498 (to the nearest dollar).
(iii) Interest = $50,000 − $32,498 = $17,502.