Mathematics Standard • Year 12 • Module 7 • Lesson 11

Mixed Financial Problems — Past-Paper Style

Practise HSC Mathematics Standard 2-style writing on integrated financial decisions — three short-answer questions and one longer scenario with marking criteria.

Master · Past-Paper Style

1. Short-answer questions

1.1 A family needs $15,000 for home renovations.
(a) Calculate the monthly repayment on a $15,000 personal loan at 9.6% p.a. compounded monthly over 3 years (use M = PV × r / [1 − (1+r)⁻ⁿ]).
(b) Calculate the total interest payable.    2 marks    Band 3

1.2 A bank offers a 0.3% lower mortgage rate but charges a $2,000 fee on a $300,000 mortgage over 25 years. The monthly repayment drops by $48.
(a) Calculate the break-even month (fees ÷ monthly saving).
(b) Calculate the total dollar saving over 25 years, net of fees.    3 marks    Band 3-4

1.3 Aiden has a $5,000 credit-card balance at 20% p.a. (monthly r = 0.20/12) and a $12,000 car loan at 8% p.a. compounded monthly (monthly r ≈ 0.00667). He can pay $400/month in total.
(a) Calculate the Month-1 interest on each debt.
(b) State whether he should split payments evenly or pay the credit card down first, and justify mathematically in one sentence using your figures from (a).    4 marks    Band 4

Stuck on 1.3(b)? Compare interest charged per dollar paid against each debt — the higher-rate debt earns the bigger interest saving per $1 paid down.

2. Extended response

2.1 Bianca needs $25,000 for a car. She is comparing three options.

Option A: Save $600/month at 4.8% p.a. compounded monthly until she has $25,000 (use FV-of-annuity formula).

Option B: Reducing-balance loan: $25,000 at 7.2% p.a. compounded monthly over 5 years, M = $497.65.

Option C: Dealer flat-rate finance: $25,000 at 4% flat over 5 years (equal monthly instalments).

(a) Approximately how many months does Option A take? (Solve 600 × [(1.004)ⁿ − 1] / 0.004 = 25,000 by trial — give n to the nearest month.)
(b) Calculate the total amount paid under Option B and the total interest.
(c) Calculate the total amount paid under Option C and the equal monthly instalment.
(d) Calculate the approximate equivalent reducing-balance rate for Option C using r_red ≈ 2·n·r_flat / (n + 1).
(e) Recommend the cheapest option, naming the option and the total dollar saving compared to Option C, and state one practical trade-off (e.g. waiting for the car, monthly cashflow).    7 marks    Band 5-6

Explicit marking criteria

Part (a) — 1 mark

1 mark — correct n to the nearest month using trial in the FV-of-annuity formula.

Part (b) — 1 mark

1 mark — correct total and interest using M × 60 − $25,000.

Part (c) — 1 mark

1 mark — correct total (I = P × r_flat × n; Total = P + I) and monthly instalment = Total / 60.

Part (d) — 1 mark

1 mark — correct approximate reducing-balance equivalent of Option C.

Part (e) — 3 marks

1 mark — comparison of all three on the same "total dollar cost" basis.

1 mark — recommendation names the cheapest option AND quotes the dollar saving vs Option C.

1 mark — names one realistic practical trade-off.

Your response:

Stuck on (e)? Identify the smallest total-cost figure, name that option, quote the dollar gap vs Option C, AND add a realistic trade-off — usually "waiting for the car" if A wins.

How did this worksheet feel?

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — Personal loan repayment (2 marks)

(a) Sample response. r = 0.096/12 = 0.008, n = 36. M = 15,000 × 0.008 / [1 − (1.008)⁻³⁶] = 120 / 0.24959 ≈ $480.76/month.

(b) Sample response. Total = 480.76 × 36 = $17,307. Interest = $17,307 − $15,000 = $2,307.

Marking notes. 1 mark — correct M with substitution shown. 1 mark — correct total interest. A bare "$480" without working scores 1/2.

1.2 — Refinance break-even (3 marks)

(a) Sample response. Break-even = 2,000 / 48 ≈ 41.7 months — i.e. about 42 months (3.5 years).

(b) Sample response. Total saving = $48 × 300 = $14,400 over 25 years. Net of fees = $14,400 − $2,000 = $12,400 saved.

Marking notes. (a) 1 mark — correct break-even calculation. (b) 1 mark — gross saving. 1 mark — net of fees.

1.3 — Debt avalanche (4 marks)

(a) Sample response. Card I₁ = 5,000 × 0.20/12 ≈ $83.33. Car I₁ = 12,000 × 0.08/12 ≈ $80.00.

(b) Sample response. Aiden should pay the credit card down first. Every extra dollar paid against the card reduces a 20%-p.a. interest charge; the same dollar paid against the car loan only saves 8%-p.a. interest, so directing all spare funds to the highest-rate debt minimises total interest paid.

Marking notes. (a) 1 mark — card interest correct. 1 mark — car interest correct. (b) 1 mark — names "card first" / "avalanche". 1 mark — justification references the rate differential / "more interest saved per $1 paid".

2.1 — Bianca's $25,000 car (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

(a) Option A — saving plan.

r = 0.048/12 = 0.004. Solve 600 × [(1.004)ⁿ − 1] / 0.004 = 25,000 → (1.004)ⁿ ≈ 1.16667 → n ≈ ln 1.16667 / ln 1.004 ≈ 38.6 → 39 months. [1 mark — correct n to the nearest month using trial in the FV-of-annuity formula.]

(b) Option B — reducing-balance loan.

Total = 497.65 × 60 = $29,859. Interest = $29,859 − $25,000 = $4,859. [1 mark — correct total and interest.]

(c) Option C — dealer flat-rate finance.

I_flat = 25,000 × 0.04 × 5 = $5,000. Total = $30,000. M = 30,000 / 60 = $500/month. [1 mark — correct total and monthly instalment.]

(d) True rate of Option C.

r_red ≈ 2 × 60 × 0.04 / 61 = 4.8 / 61 ≈ 7.9% reducing balance. [1 mark — correct approximation.]

(e) Comparison and recommendation.

Total cost ranking: A ≈ $25,000 paid in contributions + ≈ $400 interest earned (a net cost close to $24,600 in real terms), B = $29,859, C = $30,000. [1 mark — comparison of all three on the same total-cost basis.]

Conclusion: Bianca should take Option A (save), paying about $25,000 total versus Option C's $30,000 — a saving of about $5,000. [1 mark — names Option A and quotes the dollar saving vs Option C.] The practical trade-off is the 39-month wait before she actually has the car; if she needs reliable transport for work right now, Option B (loan) at $29,859 is the next-cheapest immediate option. [1 mark — one realistic practical trade-off named.]

Total: 7/7.

Band descriptors for marker.

Band 3: Computes Option B and Option C totals correctly; Option A months omitted or true-rate missing. ≈ 3 marks.

Band 4: All four numerical parts correct, but no recommendation sentence or no dollar gap stated. ≈ 4-5 marks.

Band 5: Full numerical solution and recommendation that names Option A with the dollar saving, but no trade-off discussion. ≈ 6 marks.

Band 6: Complete: four correct numerical parts, recommendation with dollar saving vs Option C, AND a realistic practical trade-off sentence. 7/7.