Free Year 12 Annuities, Loans and Depreciation
An HSC/VCE-style study guide to the financial models met in senior maths: straight-line and declining-balance depreciation, the future and present value of an annuity, calculating loan repayments and total interest, and recurrence relations for a reducing-balance loan. Worked examples use realistic car, savings and home-loan figures, with a six-question self-check and answers.
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- Year
- Year 12
- Subject
- Mathematics
- Topic
- Financial Mathematics
- Difficulty
- Level 3 · Challenge
- Estimated time
- 30 minutes
- Curriculum
- Australian Curriculum
- Answers
- Not applicable
- Format
- PDF (A4) + print
Students will practise
- calculating the value of an asset under straight-line and declining-balance depreciation
- finding the future value and present value of an annuity
- calculating the regular repayment and total interest on a reducing-balance loan
- using a recurrence relation to track a loan or investment balance period by period
Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.
What's next?
Completed: Annuities, Loans and Depreciation
More resources for this topic are coming soon.
How to use this study guide
- Read it together first, pausing at each worked example to try the step before reading the answer.
- Attempt the "Check yourself" questions at the end without looking back.
- Then practise with a worksheet from the pathway above and finish with the topic test.
Common questions
Who is this study guide for?
Year 12 students (typically ages 17–18) working on financial mathematics. It is pitched at level 3 · challenge.
Are the answers included?
This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.
How long does it take?
About 30 minutes. Short, regular sessions work best: two or three a week beats one long one.
Do I need to sign up to download?
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Year 12 · Mathematics · Financial Mathematics
Annuities, Loans and Depreciation
Student name:
Date:
What you need to know
Depreciation is the loss in value of an asset over time. With straight-line (flat-rate) depreciation the same amount D is lost each period: V = V₀ − Dn. With declining-balance (reducing-balance) depreciation the same percentage r is lost each period: V = V₀(1 − r)n, so the value falls quickly at first and never quite reaches zero.
An annuity is a series of equal payments M made at regular intervals with compound interest at rate r per period. Its future value is what the payments grow to (a savings plan); its present value is the single amount today that is worth the same as all the payments (a loan). A reducing-balance loan is an annuity in reverse: each period, interest is added to the balance, then the repayment is subtracted. The same step written as a recurrence relation is Vₙ₊₁ = Vₙ(1 + r) − M.
| Model | Formula | Notes |
|---|---|---|
| Straight-line depreciation | V = V₀ − Dn | D = amount lost per period |
| Declining-balance depreciation | V = V₀(1 − r)n | r = rate per period as a decimal |
| Future value of an annuity | FV = M × | payments at the end of each period |
| Present value of an annuity | PV = M × | also the amount that can be borrowed |
| Loan repayment | M = | rearranged from the PV formula |
| Recurrence (loan) | Vₙ₊₁ = Vₙ(1 + r) − M, V₀ = amount borrowed | interest first, then repayment |
A $30 000 car depreciates by $4500 per year (straight line). Find its value after 4 years, and when it is written off.
- V = 30 000 − 4500 × 4 = 30 000 − 18 000 = $12 000
- Written off when V = 0: 30 000 ÷ 4500 = 6.67, so during the 7th year (after 6 years it is still worth $3000).
The same $30 000 car depreciates at 20% p.a. (declining balance). Find its value after 4 years.
- V = 30 000 × (1 − 0.2)4 = 30 000 × 0.84 = 30 000 × 0.4096 = $12 288
- Compare: straight-line gave $12 000 after 4 years, but declining balance loses more early on ($6000 in year 1 versus $4500) and less later.
Savings plan: $500 is deposited at the end of every month for 10 years at 6% p.a. compounded monthly. Find the future value.
- r = 0.06 ÷ 12 = 0.005; n = 10 × 12 = 120
- FV = 500 × = 500 × = 500 × 163.879…
- FV = $81 939.67. The deposits total $60 000, so $21 939.67 is interest.
Home loan: $400 000 is borrowed at 6% p.a. compounded monthly over 30 years. Find the monthly repayment and the total interest.
- r = 0.005; n = 360
- M = = = = $2398.20 per month
- Total repaid = 2398.20 × 360 = $863 352
- Total interest = 863 352 − 400 000 = $463 352 — more than the amount borrowed.
Recurrence relation: $20 000 is borrowed at 12% p.a. compounded monthly, with repayments of $600 per month. Find the balance after 3 months.
- r = 0.12 ÷ 12 = 0.01, so Vₙ₊₁ = 1.01Vₙ − 600 with V₀ = 20 000.
- V₁ = 1.01 × 20 000 − 600 = 20 200 − 600 = 19 600
- V₂ = 1.01 × 19 600 − 600 = 19 796 − 600 = 19 196
- V₃ = 1.01 × 19 196 − 600 = 19 387.96 − 600 = $18 787.96
- Interest in month 1 was $200 and the principal fell by $400; each month slightly more of the $600 goes to principal.
Present value: how much must be invested now at 5% p.a. compounded annually to provide $2000 at the end of each year for 5 years?
- PV = 2000 × = 2000 × = 2000 × 4.3295…
- PV = $8658.95. The payments total $10 000; the rest comes from interest earned on the money still invested.
Check yourself
- A $12 000 machine depreciates by $1500 per year. Find its value after 5 years.
- A $12 000 machine depreciates at 15% p.a. (declining balance). Find its value after 3 years.
- V₀ = 5000 and Vₙ₊₁ = 1.02Vₙ − 300. Find V₁ and V₂.
- Find the future value of $1000 deposited at the end of each year for 3 years at 4% p.a.
- $10 000 is borrowed at 12% p.a. compounded monthly over 12 months. Find the monthly repayment.
- Find the total interest paid on the loan in question 5.
Answers: 1. 12 000 − 7500 = $4500 2. 12 000 × 0.853 = $7369.50 3. V₁ = 5100 − 300 = 4800; V₂ = 4896 − 300 = 4596 4. 1000 × = $3121.60 5. M = = $888.49 6. 888.49 × 12 − 10 000 = $661.88
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Financial Mathematics in other year levels
- Simple and Compound Interest
Study Guide · Year 11 · Financial Mathematics · Level 2 · Standard
About this resource
- Created by
- Success Tutoring
- Last reviewed
- 1 October 2026
- How it was made
- Written by our team
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