Free Year 11 Simple and Compound Interest
An HSC/VCE-style study guide comparing simple interest (I = Prn) with compound interest (A = P(1 + r)ⁿ), including compounding more often than once a year. Worked examples find the final amount, the interest earned, the principal needed to reach a target, the time for money to double and the rate from a known growth, with a six-question self-check and answers.
Instant download · No sign-up · Free for personal, classroom and homeschool use
- Year
- Year 11
- Subject
- Mathematics
- Topic
- Financial Mathematics
- Difficulty
- Level 2 · Standard
- Estimated time
- 25 minutes
- Curriculum
- Australian Curriculum
- Answers
- Not applicable
- Format
- PDF (A4) + print
Students will practise
- calculating simple interest and the final amount with I = Prn
- calculating compound interest with A = P(1 + r)ⁿ, matching the rate to the compounding period
- finding the principal or the time needed to reach a target amount
- comparing simple and compound interest over the same period
Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.
What's next?
Completed: Simple and Compound Interest
Ready for more? Move on to Year 12 Mathematics.
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 11 students (typically ages 16–17) working on financial mathematics. It is pitched at level 2 · standard.
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 25 minutes. Short, regular sessions work best: two or three a week beats one long one.
Do I need to sign up to download?
No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.
What should we do next?
Explore more Year 11 mathematics resources.
Preview
Year 11 · Mathematics · Financial Mathematics
Simple and Compound Interest
Student name:
Date:
What you need to know
Simple interest is paid on the original amount (the principal) only, so the same dollar amount is added every period: I = Prn, where P is the principal, r is the interest rate per period as a decimal and n is the number of periods. The final amount is A = P + I.
Compound interest is paid on the principal and on interest already earned, so the balance grows faster each period: A = P(1 + r)[sup n], and the interest earned is I = A − P. The key skill is matching r and n to the compounding period: for 6% p.a. compounded monthly, r = 0.06 ÷ 12 = 0.005 per month and n counts months.
| Nominal rate | Compounded | r per period | n for 2 years |
|---|---|---|---|
| 6% p.a. | annually | 0.06 | 2 |
| 6% p.a. | half-yearly | 0.03 | 4 |
| 6% p.a. | quarterly | 0.015 | 8 |
| 6% p.a. | monthly | 0.005 | 24 |
Simple interest: $5000 at 4% p.a. simple interest for 3 years
- I = Prn = 5000 × 0.04 × 3 = $600
- A = 5000 + 600 = $5600
Compound interest: $5000 at 4% p.a. compounded annually for 3 years
- A = P(1 + r)n = 5000 × 1.043 = 5000 × 1.124864 = $5624.32
- Interest earned: 5624.32 − 5000 = $624.32, which is $24.32 more than simple interest over the same 3 years. The gap grows quickly with more years.
Monthly compounding: $8000 at 6% p.a. compounded monthly for 2 years
- r = 0.06 ÷ 12 = 0.005 per month; n = 2 × 12 = 24 months
- A = 8000 × 1.00524 = 8000 × 1.12716… = $9017.28
- Interest: 9017.28 − 8000 = $1017.28
Find the principal: how much must be invested now at 5% p.a. compounded annually to have $20 000 in 5 years?
- 20 000 = P × 1.055
- P = = 20 = $15 670.52
- This is sometimes called the present value of $20 000.
Find the time: how long does $1000 take to double at 7% p.a. compounded annually?
- 1000 × 1.07n = 2000, so 1.07n = 2
- Take logs: n ln 1.07 = ln 2, so n = = = 10.24…
- Interest is only added at the end of each year, so it takes 11 years to be at least double. (You can also find this by trial and error with the table function on a calculator.)
Find the rate: $3000 grows to $3600 in 4 years with annual compounding. What is the rate?
- 3000(1 + r)4 = 3600, so (1 + r)4 = 1.2
- 1 + r = 1.21/4 = 1.0466…
- r = 0.0466…, so the rate is 4.66% p.a. (to 2 decimal places)
Check yourself
- Find the simple interest on $2000 at 5% p.a. for 4 years.
- Find the final amount when $2000 is invested at 5% p.a. compounded annually for 4 years.
- Find the final amount when $10 000 is invested at 3% p.a. compounded monthly for 1 year.
- How much must be invested now at 6% p.a. compounded annually to have $5000 in 3 years?
- How many whole years does it take money to at least double at 10% p.a. compounded annually?
- Find the interest earned on $1500 at 8% p.a. compounded quarterly for 2 years.
Answers: 1. 2000 × 0.05 × 4 = $400 2. 2000 × 1.054 = $2431.01 3. 10 000 × 1.002512 = $10 304.16 4. 5000 ÷ 1.063 = $4198.10 5. n = ln 2 ÷ ln 1.1 = 7.27…, so 8 years 6. 1500 × 1.028 = $1757.49, so interest = $257.49
Preview — the PDF contains the full resource.
Financial Mathematics in other year levels
- Annuities, Loans and Depreciation
Study Guide · Year 12 · Financial Mathematics · Level 3 · Challenge
About this resource
- Created by
- Success Tutoring
- Last reviewed
- 1 October 2026
- How it was made
- Written by our team
Free to print and share for personal, classroom and homeschool use. Please don't resell. Spotted an error? Let us know.