Year 10 - Compound interest - Free practice

Year 10 Compound Interest Worksheets

Growth, decay, compound interest in applied contexts.

Quick answer

How do you calculate compound interest?

Use A = P(1 + r/m)^(mn), where P is the principal, r is the annual interest rate as a decimal, m is the number of compounding periods per year, and n is the number of years. For annual compounding m = 1, so the formula simplifies to A = P(1 + r)^n. For example, 1000 dollars at 5% compounded annually for 3 years gives A = 1000 × 1.05^3 = 1157.63 dollars.

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Worked examples

Year 10 compound interest: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Calculating compound interest
Calculate the value of an investment of $8000 after 3 years at an interest rate of 5% per annum, compounded annually.
Show worked solution
Key idea: Use the compound interest formula $A = P(1 + r)^n$ , where $P$ is the principal, $r$ is the annual interest rate as a decimal, and $n$ is the number of compounding periods.
1. Identify the values: $P = 8000$ , $r = 0.05$ , $n = 3$ .
2. Substitute into the formula:
  $A = 8000(1 + 0.05)^3$
3. Simplify:
  $A = 8000 \times (1.05)^3 = 8000 \times 1.157625 = 9261.00$
4. Find the interest earned: $9261.00 - 8000 = 1261.00$ .
Answer: The investment is worth $9261.00, so the compound interest earned is $1261.00.
Example 2 · Finding the principal
After 4 years of earning 6% per annum compound interest (compounded annually), an investment is worth $15 000. What was the original principal? Give your answer to the nearest dollar.
Show worked solution
Key idea: Rearrange $A = P(1 + r)^n$ to make $P$ the subject: $P = \dfrac{A}{(1 + r)^n}$ .
1. Identify the values: $A = 15000$ , $r = 0.06$ , $n = 4$ .
2. Calculate $(1.06)^4$ :
  $(1.06)^4 = 1.26247696$
3. Divide to find $P$ :
  $P = \frac{15000}{1.26247696} \approx 11881.40$
Answer: The original principal was approximately $11 881.
Example 3 · Compound interest in context
Mia borrows $12 000 to pay for a car. The loan charges 8% per annum compound interest, compounded annually. Mia makes no repayments. After 5 years, how much more does she owe compared with what she would owe under a simple interest loan at the same rate over the same period? Give your answer to the nearest dollar.
Show worked solution
Key idea: Calculate each type of interest separately and find the difference.
1. Compound interest final amount:
  $A = 12000 \times (1.08)^5 = 12000 \times 1.46932807 \approx 17631.94$
2. Simple interest final amount:
  $A = P(1 + rn) = 12000 \times (1 + 0.08 \times 5) = 12000 \times 1.4 = 16800.00$
3. Find the difference:
  $17631.94 - 16800.00 = 831.94$
Answer: Mia owes approximately $832 more under compound interest than under simple interest.

Common mistakes

Where students usually lose marks

Using the rate as a percentage instead of a decimal

$A = 8000 \times (1 + 5)^3$

Always convert the percentage to a decimal before substituting. An interest rate of 5% becomes $r = 0.05$ , so you write $(1 + 0.05)$ , not $(1 + 5)$ .

Confusing n with the number of years

For monthly compounding over 2 years, writing $n = 2$ instead of $n = 24$ .

$n$ is the total number of compounding periods, not the number of years. Multiply the number of years by the number of compounding periods per year. Two years compounded monthly gives $n = 2 \times 12 = 24$ .

Reporting interest instead of the total amount (or vice versa)

The question asks for the total amount but the student writes only the interest earned.

Read the question carefully. $A = P(1 + r)^n$ gives the total amount. To find only the interest, subtract the principal: $I = A - P$ .

Rounding too early

$(1.05)^3 \approx 1.158$ , then $8000 \times 1.158 = 9264$ instead of $9261.

Keep all decimal places in your calculator throughout the calculation and round only your final answer. Early rounding builds up errors, especially over many compounding periods.

Try one yourself

Your turn

Anika invests $4000 at an annual interest rate of 6%, compounded quarterly, for 5 years. What is the value of the investment at the end of 5 years, correct to 2 decimal places? Give your answer in dollars.

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About this worksheet

Why we built it

Year 10 financial maths jumps from simple to compound interest, and adds general growth/decay scenarios. These worksheets give your student practice across each variation - calculating final value, finding the principal, finding time.

What's covered

Sub-skills your student will practise

✓Calculating compound interest using A = P(1 + r)^n
✓Finding the principal given the final amount
✓Comparing simple vs compound interest
✓Population growth and depreciation problems
✓Different compounding frequencies (annual, quarterly, monthly)

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Frequently asked

Questions parents ask about compound interest

What's the compound interest formula?

A = P(1 + r)^n, where A is the final amount, P is the principal, r is the rate per period (as a decimal), and n is the number of periods. For monthly compounding over a year at 12%, r = 0.01 and n = 12.

How does compound interest differ from simple interest?

Simple interest pays the same dollar amount each year (based on the original principal). Compound interest pays interest on the principal plus accumulated interest, so the dollar amounts grow each year.

What's compounding frequency?

How often interest is added to the balance. Annual (once a year), semi-annual (twice), quarterly (four times), monthly (twelve), daily (365). More frequent compounding gives a higher final amount for the same nominal rate.

Is the formula the same for depreciation?

Almost - depreciation uses A = P(1 - r)^n, with a minus instead of a plus. A car worth $30,000 depreciating 15% per year for 4 years: A = 30000 x (0.85)^4 = $15,660.

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