Mathematics Standard • Year 12 • Module 7 • Lesson 12
Module Review — Problem Set
Apply every lesson in the module to one realistic, integrated scenario — short-term saving, long-term investing, borrowing, debt strategies and inflation in Australian contexts.
Problem 1 — Comparing four ways to grow $10,000
Tara has $10,000 to invest for 6 years. She is considering four options:
Option A: 5% p.a. simple interest, paid annually.
Option B: 5% p.a. compounded annually.
Option C: 5% p.a. compounded quarterly.
Option D: 5% p.a. compounded monthly.
Set up: What are we solving for?
(i) Calculate the FV (or final amount) of each option to the nearest dollar. 2 marks
(ii) State the dollar difference between Option A and Option D. 1 mark
(iii) Calculate the effective annual rate of Option D and explain in one sentence why more frequent compounding gives a higher final amount even though the headline rate is the same. 2 marks
Stuck? Revisit lesson § Module Formula Summary — simple interest, compound interest, effective rate.Problem 2 — Saving $400/month into super from age 30 to 65
A 30-year-old contributes $400/month into super for 35 years at an expected 7% p.a. compounded monthly. (Use r = 0.07/12, n = 420.)
Set up: What are we solving for?
(i) Calculate the projected balance at age 65 using FV of annuity. 2 marks
(ii) Calculate the total amount the person actually contributed over the 35 years, and the dollar interest earned. 2 marks
(iii) Comment in one sentence on what this tells you about starting super contributions early. 2 marks
Stuck? Revisit lesson § Activity 3 Q2 — the age-30-to-65 super calculation.Problem 3 — Refinancing a $450,000 mortgage
A couple borrow $450,000 at 5.2% p.a. compounded monthly over 30 years; their monthly repayment is $2,470.21. After 5 years they refinance the remaining balance ($419,300) to 4.5% over the remaining 25 years; the new monthly is $2,328.43.
Set up: What are we solving for?
(i) Calculate the total amount paid in the first 5 years. 1 mark
(ii) Calculate the total amount paid over the remaining 25 years after refinancing. 1 mark
(iii) Calculate the total interest saved by refinancing, compared to keeping the original 30-year schedule at 5.2% (total = $2,470.21 × 360). State the saving in one summary sentence. 3 marks
Stuck? Revisit lesson § Activity 3 Q1 — refinancing a $450,000 mortgage.Problem 4 — Store's "0% finance" with a fee
A retailer offers a $4,000 fridge on the following terms: "0% finance over 12 months" with a $400 establishment fee added at the start. The customer pays in 12 equal monthly instalments.
Set up: What are we solving for?
(i) Calculate the total cost and equal monthly instalment. 1 mark
(ii) Calculate the equivalent flat rate per annum for the $400 extra paid. 1 mark
(iii) Approximate the equivalent reducing-balance annual rate using r_red ≈ 2·n·r_flat / (n + 1) with n = 12. Comment in one sentence on whether "0% finance" is an honest description. 2 marks
Stuck? Revisit lesson § Activity 3 Q3 — 0% finance with a $400 fee.Problem 5 — Inflation-adjusted FV of a 20-year investment
An investor places $50,000 into a balanced fund expected to return 6.5% p.a. compounded annually for 20 years. Inflation is expected to run at 2.5% p.a.
Set up: What are we solving for?
(i) Calculate the nominal FV after 20 years. 1 mark
(ii) Calculate the real return rate and the purchasing power after 20 years in today's dollars. 2 marks
(iii) In one or two sentences, explain why ignoring inflation can lead to a misleading sense of how much your investment has grown in real terms. 2 marks
Stuck? Revisit Lesson 10 § Comparing Investments — Real return = Nominal − Inflation.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Four ways to grow $10,000
Set up. Apply each formula to the same P, r, n.
(i) A: A = 10,000 × (1 + 0.05 × 6) = $13,000. B: A = 10,000 × (1.05)⁶ ≈ $13,401. C: A = 10,000 × (1 + 0.05/4)²⁴ ≈ 10,000 × 1.3474 ≈ $13,474. D: A = 10,000 × (1 + 0.05/12)⁷² ≈ 10,000 × 1.3489 ≈ $13,489.
(ii) D − A = 13,489 − 13,000 = $489.
(iii) r_eff(D) = (1 + 0.05/12)¹² − 1 ≈ 5.12%. More frequent compounding adds interest on interest sooner, so the realised annual rate creeps above the nominal 5%.
Problem 2 — Super from age 30
Set up. FV-of-annuity over 35 years with monthly contributions.
(i) r = 0.07/12 ≈ 0.005833, n = 420. FV = 400 × [(1.005833)⁴²⁰ − 1] / 0.005833 ≈ 400 × 1,766.8 ≈ $706,720.
(ii) Contributed = 400 × 420 = $168,000. Interest earned = 706,720 − 168,000 = $538,720.
(iii) The investor contributes only $168,000 but ends with about $707,000 — over $0.75 of every final dollar comes from compound interest. Starting early is what makes the time-driven compounding so powerful.
Problem 3 — Refinance saving on $450,000
Set up. Compute first-5-year total + remaining-25-year total; compare to original 30-year total.
(i) First 5 yr = 2,470.21 × 60 = $148,213.
(ii) Remaining 25 yr = 2,328.43 × 300 = $698,529.
(iii) Actual total = 148,213 + 698,529 = $846,742. Original 30-yr total at 5.2% = 2,470.21 × 360 = $889,276. Interest saved ≈ $42,534 from a single mid-life refinance.
Problem 4 — "0% finance" with fee
Set up. Total cost via fee; back-out flat and reducing-balance equivalents.
(i) Total = $4,000 + $400 = $4,400. Monthly = 4,400 / 12 = $366.67.
(ii) r_flat = 400 / (4,000 × 1) = 10% p.a. flat.
(iii) r_red ≈ 2 × 12 × 0.10 / 13 = 2.4 / 13 ≈ 18.5% reducing balance. "0% finance" is misleading — the fee structure is mathematically equivalent to a ~10% flat / ~18.5% reducing-balance loan, comparable to a credit card.
Problem 5 — Inflation adjustment
Set up. Compute nominal FV; compute real-return rate; compute purchasing power.
(i) FV = 50,000 × (1.065)²⁰ ≈ 50,000 × 3.5236 ≈ $176,182.
(ii) Real return = 6.5 − 2.5 = 4% p.a. PP = 50,000 × (1.04)²⁰ ≈ 50,000 × 2.1911 ≈ $109,556 in today's dollars.
(iii) The $176,182 nominal figure looks like a 3.5× increase but in today's dollars the investor's actual purchasing power is closer to 2.2×. Ignoring inflation overstates real wealth growth by roughly $66,000 in this scenario.