Mathematics • Year 10 • Unit 1 • Lesson 6
Compound Interest
Build fluency with the compound interest formula A = P(1 + R)ⁿ, one step at a time, from a fully worked example through guided practice to independent problems with annual, quarterly and monthly compounding.
1. I do, fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. Ava invests $8,000 at 4.5% p.a. compounded annually for 3 years. Find the total amount and the compound interest earned.
Step 1, Identify each variable.
P = 8,000 R = 0.045 (per year) n = 3 (years)
Reason: compounded annually means R is the annual rate as a decimal, and n is the number of years.
Step 2, Substitute into A = P(1 + R)ⁿ.
A = 8,000 × (1 + 0.045)³ = 8,000 × (1.045)³
Reason: put the values straight in. Do the bracket first.
Step 3, Evaluate the power.
(1.045)³ = 1.141166 (to 6 d.p.)
Reason: keep at least 6 decimal places here so the final dollar answer is accurate.
Step 4, Multiply by the principal.
A = 8,000 × 1.141166 = 9,129.33 (to 2 d.p.)
Reason: money is always rounded to the nearest cent at the very end.
Step 5, Find the interest by subtracting.
Compound interest = A − P = 9,129.33 − 8,000 = 1,129.33
Reason: the formula gives the total amount; interest = total − principal.
Answer: Total amount = $9,129.33; compound interest = $1,129.33.
2. We do, fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank. 4 marks
Problem. $5,000 is invested at 6% p.a. compounded quarterly for 2 years. Find the total amount.
Step 1, Adjust R and n for quarterly compounding.
R per quarter = 6% ÷ ____ = ______ (as a decimal)
n = 2 years × ____ quarters/year = ______ quarters
Step 2, Substitute into A = P(1 + R)ⁿ:
A = 5,000 × (1 + ______)____
Step 3, Evaluate the bracket and power:
(1.015)⁸ = ______________ (to 6 d.p.)
Step 4, Multiply and round to the nearest cent:
A = $ ______________
3. You do, independent practice
Show your working. The first four are foundation (annual compounding only). The middle two are standard (non-annual compounding). The last two are extension (compare or work backwards).
Foundation, annual compounding
3.1 $2,000 is invested at 5% p.a. compounded annually for 3 years. Find the total amount. 1 mark
3.2 $6,000 is invested at 4% p.a. compounded annually for 4 years. Find the compound interest earned (not the total). 2 marks
3.3 $12,500 is invested at 3.5% p.a. compounded annually for 5 years. Find the total amount, to the nearest cent. 2 marks
3.4 $1,000 is invested at 8% p.a. compounded annually. How much is in the account after 10 years? 2 marks
Standard, non-annual compounding
3.5 $15,000 is invested at 5.2% p.a. compounded quarterly for 3 years. Find the total amount. 3 marks
3.6 $3,000 is deposited in an ING Savings Maximiser at 4.8% p.a. compounded monthly for 2 years. Find the total amount and the interest earned. 3 marks
Extension, push your thinking
3.7 $20,000 is invested at 6% p.a. for 10 years. Compare simple interest with annually compounded interest, how much more does compound interest earn? 3 marks
3.8 An account is opened with $4,000 at 5% p.a. compounded annually. After how many whole years will the balance first exceed $6,000? (Try n = 1, 2, 3... and stop at the first one that works.) 2 marks
Section 2, We do (quarterly $5,000 at 6% for 2 years)
Step 1: R per quarter = 6% ÷ 4 = 0.015; n = 2 × 4 = 8 quarters.
Step 2: A = 5,000 × (1 + 0.015)8.
Step 3: (1.015)⁸ = 1.126493.
Step 4: A = 5,000 × 1.126493 = $5,632.46.
3.1, $2,000 at 5% annually for 3 years
A = 2,000 × (1.05)³ = 2,000 × 1.157625 = $2,315.25.
3.2, $6,000 at 4% annually for 4 years (interest only)
A = 6,000 × (1.04)⁴ = 6,000 × 1.169859 = $7,019.15. Compound interest = A − P = 7,019.15 − 6,000 = $1,019.15.
3.3, $12,500 at 3.5% annually for 5 years
A = 12,500 × (1.035)⁵ = 12,500 × 1.187686 = $14,846.07.
3.4, $1,000 at 8% annually for 10 years
A = 1,000 × (1.08)¹⁰ = 1,000 × 2.158925 = $2,158.92. The money more than doubles.
3.5, $15,000 at 5.2% quarterly for 3 years
R per quarter = 0.052 ÷ 4 = 0.013; n = 3 × 4 = 12.
A = 15,000 × (1.013)¹² = 15,000 × 1.167302 = $17,509.53.
3.6, $3,000 at 4.8% monthly for 2 years
R per month = 0.048 ÷ 12 = 0.004; n = 2 × 12 = 24.
A = 3,000 × (1.004)²⁴ = 3,000 × 1.100553 = $3,301.66.
Compound interest = 3,301.66 − 3,000 = $301.66.
3.7, Simple vs compound on $20,000 at 6% for 10 years
Simple: I = P × R × T = 20,000 × 0.06 × 10 = $12,000 → Total = $32,000.
Compound (annual): A = 20,000 × (1.06)¹⁰ = 20,000 × 1.790847 = $35,816.95 → Interest = $15,816.95.
Difference: 15,816.95 − 12,000 = $3,816.95 more from compounding. Interest on interest pulls ahead more every year.
3.8, Years for $4,000 at 5% to exceed $6,000
n = 1: 4,000 × 1.05 = $4,200.
n = 5: 4,000 × (1.05)⁵ = $5,105.13.
n = 8: 4,000 × (1.05)⁸ = $5,909.82, still under.
n = 9: 4,000 × (1.05)⁹ = $6,205.31, over!
Answer: 9 years.