Mathematics Advanced • Year 12 • Module 7 • Lesson 12

Superannuation Modelling

Practise HSC-style writing on super projections and an extended response on the impact of fees.

Master · Past-Paper Style

1. Short-answer questions

1.1 A worker has $30,000 in super, salary $85,000, employer contribution 11.5%, fund return 7% p.a., fees 1.2% p.a.
(a) Find the annual contribution C and the net return rnet.
(b) Using the closed form, project the balance after 30 years to the nearest dollar.    3 marks    Band 3

1.2 Two super funds both quote a 7% gross return. Fund A's fees are 1.5% p.a.; Fund B's fees are 0.5% p.a. A 25-year-old with A0 = $0, salary $70,000 and 11.5% employer contribution invests for 40 years.
(a) Calculate the projected balance in each fund.
(b) State the dollar gap and express it as a percentage of Fund B's balance.    3 marks    Band 3-4

1.3 A super recurrence is given by An+1 = 1.058An + 9,200 with A0 = $20,000.
(a) Iterate to find A1 and A2.
(b) Verify A2 using the closed form, and explain why the two methods have to give the same value.
(c) Interpret the values of rnet = 0.058 and C = $9,200: what salary and which combination of return/fees could have produced these numbers?    4 marks    Band 4

Stuck on 1.3(c)? Solve C = salary × 0.115 for the salary, then pick any (return, fee) pair that differs by 5.8%.

2. Extended response

2.1 A 25-year-old has $0 in super. Salary $80,000. Employer contribution 11.5%. Three funds are available, all advertised as "7% p.a. growth", but they charge different fees.

Fund Low: 7% return, 0.5% fees.

Fund Mid: 7% return, 1.0% fees.

Fund High: 7% return, 1.5% fees.

(a) Calculate each fund's net return rnet and the projected balance at age 65 (40 years), showing the closed-form substitution.
(b) Rank the funds from highest to lowest balance, and quantify the dollar penalty of each 0.5% fee step (Low→Mid, Mid→High).
(c) Explain in 2-3 sentences why a small annual fee difference results in a six-figure final difference, referencing how fees enter the recurrence relation and compound over time.    8 marks    Band 5-6

Explicit marking criteria

Part (a), 4 marks

1 mark correct C = 80,000 × 0.115 = $9,200 stated explicitly.

1 mark correct three rnet values (0.065, 0.060, 0.055).

1 mark at least two projected balances correct to the nearest thousand dollars.

1 mark all three projected balances correct to the nearest thousand dollars.

Part (b), 2 marks

1 mark correct ranking Low > Mid > High.

1 mark correct dollar gaps for both steps.

Part (c), 2 marks

1 mark explicitly states that fees compound against the balance just as returns compound for it (or equivalent phrasing).

1 mark links the recurrence An+1 = (1 + r − f)An + C to the long term: small differences in r − f raise (1 + r − f)n by a large factor over decades.

Your response:

Stuck on (c)? Reference the lesson's "Hidden Cost of Fees" table: a 1% fee gap can cost over $600k over 40 years.

How did this worksheet feel?

What I'll revisit before next class:

Answers, sample responses + marking notes

1.1, $30,000 + 11.5% on $85,000, return 7%, fees 1.2%, n = 30 (3 marks)

Sample response. (a) C = 85,000 × 0.115 = $9,775. rnet = 0.07 − 0.012 = 0.058. (b) (1.058)³⁰ ≈ 5.404. A30 = 30,000(5.404) + 9,775 × (4.404 ÷ 0.058) = 162,120 + 9,775 × 75.931 = 162,120 + 742,222 ≈ $904,342.

Marking notes. 1 mark, correct C and rnet. 1 mark, correct substitution into closed form. 1 mark, correct numerical answer (accept ±$2,000 for rounding). A bald final answer with no working scores 1/3.

1.2, Fund A vs Fund B at 7%/1.5% vs 7%/0.5%, 40 years (3 marks)

Sample response. C = 70,000 × 0.115 = $8,050. (a) Fund A (rnet = 0.055): A40 = 8,050 × [(1.055)⁴⁰ − 1] ÷ 0.055 = 8,050 × 136.605 ≈ $1,099,673. Fund B (rnet = 0.065): A40 = 8,050 × [(1.065)⁴⁰ − 1] ÷ 0.065 = 8,050 × 175.633 ≈ $1,413,847. (b) Gap ≈ $314,174, about 22% of Fund B's balance.

Marking notes. 1 mark, correct C. 1 mark, both projections within ±$5,000. 1 mark, correct gap and percentage. Students who use the wrong (gross) rate score 0/2 on the calculation.

1.3, Recurrence An+1 = 1.058An + 9,200, A0 = 20,000 (4 marks)

Sample response. (a) A1 = 1.058(20,000) + 9,200 = 21,160 + 9,200 = $30,360. A2 = 1.058(30,360) + 9,200 = 32,120.88 + 9,200 = $41,320.88. (b) Closed form: A2 = 20,000(1.058)² + 9,200 × ((1.058)² − 1) ÷ 0.058 = 20,000(1.119364) + 9,200 × (0.119364 ÷ 0.058) = 22,387.28 + 18,933.60 = $41,320.88. That is the recurrence answer to the cent, and it has to be: the closed form is the recurrence unrolled, so the two can never disagree. Any gap you get is rounding you introduced yourself. (c) C = 9,200 = salary × 0.115 implies salary = $80,000. rnet = 5.8% could come from return 7%, fees 1.2%; or return 6.5%, fees 0.7%, many pairs work, since only the difference matters.

Marking notes. 1 mark, correct A1 and A2. 1 mark, correct closed-form A2. 1 mark, explicit acknowledgement of the rounding gap (no need to compute exactly). 1 mark, correct salary $80,000 and a valid return/fee pair.

2.1, Three funds at 7% with different fees, 40 years (8 marks): sample Band-6 response

Sample Band-6 response.

C = 80,000 × 0.115 = $9,200 per year. The three net returns are:

Fund Low: rnet = 0.07 − 0.005 = 0.065

Fund Mid: rnet = 0.07 − 0.010 = 0.060

Fund High: rnet = 0.07 − 0.015 = 0.055

[2 marks: C correct, all three rnet correct]

(a) Projected balances at age 65 (n = 40, A0 = 0).

Fund Low: A40 = 9,200 × [(1.065)⁴⁰ − 1] ÷ 0.065 = 9,200 × 175.633 ≈ $1,615,824.

Fund Mid: A40 = 9,200 × [(1.060)⁴⁰ − 1] ÷ 0.060 = 9,200 × 154.762 ≈ $1,423,810.

Fund High: A40 = 9,200 × [(1.055)⁴⁰ − 1] ÷ 0.055 = 9,200 × 136.605 ≈ $1,256,766.

[2 marks: all three balances correct]

(b) Ranking and penalties. Low > Mid > High. Low → Mid penalty = 1,615,824 − 1,423,810 ≈ $192,000. Mid → High penalty = 1,423,810 − 1,256,766 ≈ $167,000. [2 marks]

(c) Why a 0.5% fee step costs six figures. Each year's fee reduces the growth factor in the recurrence An+1 = (1 + r − f)An + C, so fees compound against the balance just as returns compound for it. Over 40 years, a 0.5% gap in rnet raises the growth factor (1 + rnet)⁴⁰ by a factor of roughly 1.13–1.14, multiplying through the entire balance. Because the balance reaches seven figures, a 13% multiplicative change is worth roughly $170,000, for the cost of choosing the next-cheaper fund. [2 marks]

Total: 8/8.

Band descriptors for marker.

Band 3: Two of three balances correct, ranking present but no quantification of fee penalty. ≈ 4 marks.

Band 4: All three balances correct, ranking with one penalty quantified. Part (c) restates "fees reduce return" without compound reasoning. ≈ 5-6 marks.

Band 5: Correct calculations and ranking, both penalty values, structural explanation of compounding in (c). ≈ 6-7 marks.

Band 6: Full calculations to within a few thousand dollars, ranking with both penalties, and part (c) explicitly references the recurrence form and the compounding multiplier (1 + r − f)n. 8/8.