Mathematics Standard • Year 12 • Module 7 • Lesson 11
Mixed Financial Problems — Skill Drill
Build fluency choosing the right tool: identify which formula a question needs (simple/compound interest, FV/PV of annuity, loan repayment, flat-vs-reducing) and apply it cleanly.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Name the formula needed for each situation.
(a) Lump sum P invested at rate r for n years, compounded: ____________________
(b) Regular contribution M each period earning rate r for n periods: ____________________
(c) Borrowed PV, repaid in equal instalments M for n periods: ____________________
Q1.2 Complete the 5-step problem-solving strategy from the lesson.
1. Identify what to find. 2. List ____________. 3. Choose the ____________. 4. Calculate. 5. ____________.
Q1.3 Two loans have the same period and principal: Loan P is 6% flat, Loan Q is 6% reducing balance. Which charges more total interest? Why in one short phrase?
____________________________________________________
2. Worked example — save vs borrow
Follow each line of working. Every step has a reason on the right.
Problem. Jordan needs $20,000 for a car. Option A: save $500/month at 4% p.a. compounded monthly until reaching $20,000. Option B: take a 5-year loan at 6.5% p.a. compounded monthly (M = $391.32). Find time to save in Option A, total cost in Option B, and the dollar difference.
Step 1 — Option A is a future-value-of-annuity problem.
FV = M × [(1+r)ⁿ − 1] / r with r = 0.04/12 ≈ 0.003333, FV = 20,000
[(1.003333)ⁿ − 1] / 0.003333 = 20,000 / 500 = 40
(1.003333)ⁿ = 1.13333 → n ≈ ln(1.13333) / ln(1.003333) ≈ 37.5 months ≈ 38 months
Reason: this is regular contributions earning interest — the FV-of-annuity formula matches the situation.
Step 2 — Option B is a loan-repayment problem.
Total cost = M × n = 391.32 × 60 = $23,479. Interest = $3,479.
Reason: equal-instalment borrowing — total cost = M × n.
Step 3 — Compare and conclude.
Saving: pay only $19,000 contributions (≈$1,000 interest earned). Borrowing: pay $23,479. Difference: $4,479 more if you borrow.
Reason: choose the right formula for each option, then compare totals on the same basis.
Conclusion. Saving takes ~38 months but costs $4,479 less; borrowing gives the car now but at a $4,479 premium.
3. Faded example — refinancing break-even
A borrower with $300,000 remaining at 5.4% p.a. (M = $1,997 over 240 months) can refinance to 4.8% (M = $1,946 over 240 months) but pays $1,200 in fees. Fill in the blanks to find the break-even month. 4 marks
Step 1 — Monthly saving:
Saving = $1,997 − $1,946 = $ ____________ per month
Step 2 — Total interest saving over 20 years:
Total saving = $ ____________ × 240 = $ ____________
Step 3 — Net saving after fees:
Net = $ ____________ − $1,200 = $ ____________
Step 4 — Break-even month (fees ÷ monthly saving):
Break-even = $1,200 / $ ____________ ≈ ____________ months
Conclusion. After about ____________ months the refinance has paid for itself; over the full 20 years the borrower saves $ ____________ net.
4. Graduated practice — choose-the-formula drills
For each question, name the formula and show your working.
Foundation — identify the right tool (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | $5,000 invested for 4 years at 5% p.a. compounded annually. Find FV. | |
| 4.2 1 | $200/month invested for 10 years at 6% p.a. compounded monthly. Find FV of annuity. | |
| 4.3 1 | $15,000 borrowed at 8% p.a. compounded monthly over 3 years. The monthly repayment formula gives M = $470.05. Find total interest. | |
| 4.4 1 | Calculate total flat interest on a $10,000 loan at 6% flat over 4 years. |
Standard — typical HSC difficulty (6 questions)
Show formulas before substituting; label final answers with units.
4.5 A car costs $25,000. Saving plan: $500/month at 4.8% p.a. compounded monthly. Use the FV-of-annuity formula with the given values to find the approximate months to reach the goal. (You may use trial and error.) 3 marks
4.6 A 5-year $25,000 loan at 7.2% p.a. compounded monthly has M = $497.65. Find the total amount paid and the total interest. 2 marks
4.7 Compare 4.5 (save) and 4.6 (borrow): how much more (or less) does the borrower pay than the saver? 1 mark
4.8 A $50,000 windfall could either pay off a $50,000 mortgage at 5.2% p.a. or be invested at expected 7% p.a. Compare nominal balances after 10 years (mortgage option treated as a guaranteed 5.2% saving). 3 marks
4.9 A 0% finance offer over 12 months on a $4,000 purchase has a $300 establishment fee. Calculate (a) total cost, (b) equivalent flat rate p.a. 2 marks
4.10 A credit card charges 19.99% p.a. on $3,000. A personal loan offers 8% p.a. compounded monthly over 3 years (M = $94.01 per $1,000). Calculate the monthly repayment and total cost for the $3,000 personal loan, and compare to keeping the debt on the card. 3 marks
Extension — break-even and integrated decisions (2 questions)
4.11 A bank offers 0.3% lower mortgage rate with a $2,000 fee. On a $300,000 loan over 25 years, monthly repayment falls by $48. Find the break-even month and the total saving over 25 years. 3 marks
4.12 Linh has $5,000 credit-card debt at 20% and a $10,000 car loan at 8% reducing balance. She can afford $400/month total. Compare (A) splitting evenly $200/$200 and (B) "avalanche" — minimum $100 to the car, $300 to the card. Explain in one sentence which is cheaper and why. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Formula naming
(a) Compound interest: A = P(1 + r)ⁿ. (b) FV of annuity: FV = M × [(1+r)ⁿ − 1] / r. (c) Loan repayment: M = PV × r / [1 − (1+r)⁻ⁿ].
Q1.2 — Strategy
1. Identify. 2. List given information. 3. Choose the right formula(s). 4. Calculate. 5. Compare and conclude.
Q1.3 — Flat vs reducing
Loan P (flat) charges more total interest because flat charges interest on the original principal for the whole term, while reducing-balance only charges on what's still owed.
Q3 — Faded example (refinancing break-even)
Monthly saving = 1,997 − 1,946 = $51. Total over 240 months = $51 × 240 = $12,240. Net after $1,200 fees = $12,240 − $1,200 = $11,040. Break-even = 1,200 / 51 ≈ 23.5 months. Conclusion: after about 24 months the refinance has paid for itself; lifetime net saving ≈ $11,040.
Q4.1 — Compound FV
FV = 5,000 × (1.05)⁴ ≈ 5,000 × 1.2155 ≈ $6,078.
Q4.2 — FV of annuity
r = 0.005, n = 120. FV = 200 × [(1.005)¹²⁰ − 1] / 0.005 = 200 × [1.8194 − 1] / 0.005 = 200 × 163.88 ≈ $32,776.
Q4.3 — Total interest on a loan
Total = 470.05 × 36 = $16,922 (to nearest dollar). Interest = 16,922 − 15,000 = $1,922.
Q4.4 — Flat interest
I = 10,000 × 0.06 × 4 = $2,400.
Q4.5 — Months to save $25,000
r = 0.04/12 ≈ 0.003333. [(1.003333)ⁿ − 1] / 0.003333 = 25,000 / 500 = 50. (1.003333)ⁿ = 1.1667 → n ≈ ln 1.1667 / ln 1.003333 ≈ 46.4 → about 47 months (just under 4 years).
Q4.6 — $25,000 loan total cost
Total = 497.65 × 60 = $29,859 (to nearest dollar). Interest = 29,859 − 25,000 = $4,859.
Q4.7 — Save vs borrow gap
Saving: contributes 500 × 47 = $23,500 (plus interest earned) — pays exactly $25,000 in dollars contributed including interest, but interest is received, not paid. Borrowing pays $29,859. Borrower pays about $4,859 more than the saver in pure costs.
Q4.8 — $50,000 windfall over 10 years
Mortgage paydown (5.2%): equivalent saving = 50,000 × (1.052)¹⁰ ≈ 50,000 × 1.6602 ≈ $83,011. Invest at 7%: FV = 50,000 × (1.07)¹⁰ ≈ 50,000 × 1.9672 ≈ $98,358. Investing wins by ~$15,347 in nominal terms (but with risk; mortgage is guaranteed).
Q4.9 — "0% finance" with fee
(a) Total = $4,000 + $300 = $4,300. (b) Extra paid = $300 over 1 year on $4,000. r_flat = 300 / (4,000 × 1) = 7.5% p.a. flat.
Q4.10 — Card vs personal loan on $3,000
Personal loan M = 3 × $94.01 = $282.03/month. Total = 282.03 × 36 = $10,153. Interest = $381 (to nearest dollar) on $3,000? Recompute: 282.03 × 36 = $10,153 — that's the wrong scale. Correctly: M_per1000 = $94.01 was the loan-formula calculation that should produce M ≈ $31.34 for $1,000? Reread: actually M = $94.01 per $1,000 over 36 months at 8% looks too high — re-verify. Re-derive: r = 0.00667, n = 36, M = 1,000 × 0.00667 / [1 − (1.00667)⁻³⁶] ≈ 6.67 / 0.21295 ≈ $31.34 per $1,000. So for $3,000: M ≈ $94.02/month, Total ≈ $3,384.72, Interest ≈ $384.72. Compared to the credit card at 19.99% p.a. on $3,000 — even paying $94/month, the card would take ~3.5 years and cost ~$960 in interest. Personal loan is far cheaper — saves about $575 in interest.
Q4.11 — Refinance break-even
Break-even = 2,000 / 48 ≈ 42 months (3.5 years). Total saving over 25 years = $48 × 300 − $2,000 = $14,400 − $2,000 = $12,400 net.
Q4.12 — Debt-avalanche strategy
Strategy B (avalanche) is cheaper. Every $1 paid to the 20% card saves 20¢/year of interest; the same $1 to the 8% car loan saves only 8¢/year — so prioritising the highest-rate debt clears total interest fastest.