Year 11 · Mathematics · Number
Compound interest and depreciation
Use the compound interest formula to calculate the growth of savings and loans and the depreciation of assets, and compare compound with simple interest.
Awaiting educator review. This material was drafted by our content team with AI assistance and is waiting for review by a qualified educator. Please check it suits your child before using it. Created 10 October 2026.
These lessons were written around Australian Curriculum strands and matched to New Zealand year levels by age (NZ Year 1 is the first year of school). They are not yet mapped to The New Zealand Curriculum, so check them against your own learning programme.
Learning goals
Learners will
- Explain the difference between simple and compound interest.
- Use A = P(1 + r)^n to calculate the final amount of an investment.
- Adjust the rate and number of periods when interest is compounded more often than yearly.
- Use A = P(1 − r)^n to model depreciation.
Success looks like
- I can calculate the final amount and the interest earned for a compound interest investment.
- I can find the rate per period and the number of periods for monthly or quarterly compounding.
- I can calculate the value of an item after it depreciates for several years.
The big idea
With simple interest, interest is earned only on the original amount. $2000 at 5% p.a. earns $100 every year, so after 3 years the total interest is $300.
With compound interest, each year's interest is added to the balance and then earns interest itself. Year 1: $2000 × 1.05 = $2100. Year 2: $2100 × 1.05 = $2205. Year 3: $2205 × 1.05 = $2315.25. The interest is $315.25, which is more than the simple interest of $300.
Multiplying by 1.05 three times is the same as multiplying by 1.05^3, which leads to the formula A = P(1 + r)^n, where A is the final amount, P is the principal, r is the interest rate per period written as a decimal, and n is the number of compounding periods.
The interest earned is the final amount minus the principal: I = A − P.
Interest rates are quoted per year, but interest is often compounded more frequently. Divide the yearly rate by the number of periods in a year and multiply the number of years by the same number. For 6% p.a. compounded monthly for 2 years: r = 0.06 ÷ 12 = 0.005 and n = 2 × 12 = 24.
Depreciation works the same way, except the value goes down. If a car loses 15% of its value each year, it keeps 85% of its value, so its value after n years is A = P(1 − r)^n.
Round money to the nearest cent only at the final step. Rounding during the calculation can make the answer a few cents out.
Worked examples
Compounded annually
$2000 is invested at 5% p.a., compounded annually, for 3 years. Find the final amount and the interest earned.
- P = 2000, r = 0.05, n = 3.
- A = 2000 × (1.05)^3 = 2000 × 1.157625 = 2315.25.
- Interest = 2315.25 − 2000 = 315.25.
Answer: Final amount $2315.25; interest $315.25.
Compounded monthly
$5000 is invested at 6% p.a., compounded monthly, for 2 years. Find the final amount.
- Rate per month: r = 0.06 ÷ 12 = 0.005.
- Number of months: n = 2 × 12 = 24.
- A = 5000 × (1.005)^24 ≈ 5000 × 1.127160 ≈ 5635.80.
Answer: $5635.80 (to the nearest cent).
Depreciation
A car bought for $30 000 depreciates at 15% p.a. Find its value after 2 years.
- Each year it keeps 100% − 15% = 85% of its value, so multiply by 0.85.
- A = 30 000 × (0.85)^2 = 30 000 × 0.7225 = 21 675.
Answer: $21 675
Practice check
Have a go, then check your answers. Each answer comes with an explanation.
Answer guide for parents
Which formula gives the final amount A when a principal P earns compound interest at rate r per period for n periods?
A = P(1 + r)^n — Each period multiplies the balance by (1 + r), so after n periods A = P(1 + r)^n. The formula with (1 − r) models depreciation.
$1000 is invested at 10% p.a., compounded annually, for 2 years. What is the final amount?
$1210 — A = 1000 × 1.1^2 = 1000 × 1.21 = $1210.
An account pays 8% p.a., compounded quarterly. What is the interest rate per quarter?
2% — There are 4 quarters in a year, so the rate per quarter is 8% ÷ 4 = 2%.
The same principal is invested for 5 years at the same yearly rate. Which earns more interest?
Compound interest — After the first year, compound interest also earns interest on previous interest, so it grows faster than simple interest.
Interest is compounded monthly for 3 years. How many compounding periods is that?
36 — There are 12 months in each year, so n = 3 × 12 = 36.
$500 is invested at 4% p.a., compounded annually, for 2 years. Find the final amount in dollars.
$540.80 — A = 500 × 1.04^2 = 500 × 1.0816 = $540.80.
$4000 is invested at 5% p.a., compounded annually, for 2 years. How much interest is earned, in dollars?
$410 — A = 4000 × 1.05^2 = 4000 × 1.1025 = $4410. Interest = 4410 − 4000 = $410.
A laptop costs $1500 and depreciates at 20% p.a. What is its value after 2 years, in dollars?
$960 — A = 1500 × 0.8^2 = 1500 × 0.64 = $960.
Watch out for
- Using the percentage as a whole number, such as 1 + 5 instead of 1 + 0.05. Convert the rate to a decimal first.
- Using the yearly rate with monthly periods. Both r and n must match the compounding period.
- Giving the final amount when the question asks for the interest. Re-read the question and subtract the principal if needed.
- Thinking that 15% depreciation for 2 years is a 30% loss. Each year's 15% is taken from a smaller value, so the total loss is 27.75%.
Tips for parents
- Build a year-by-year table together for the first example. Seeing the balance grow step by step makes the formula meaningful.
- Look at a real savings account or loan advertisement and discuss what the quoted rate and compounding period mean.
- Talk about how compound interest works against you on credit cards and loans, not just in your favour on savings.
Go further
- Use repeated calculation to find how many years it takes $1000 to double at 6% p.a. compounded annually. Compare with the estimate 72 ÷ 6 = 12 years.
- Compare $10 000 invested for 5 years at 5% p.a. compounded yearly, quarterly and monthly. How much difference does the compounding period make?
- Graph the value of a $30 000 car depreciating at 15% p.a. over 10 years. Describe the shape of the graph.