Mathematics Standard • Year 11 • Module 3 • Lesson 12
Compound Interest
Build fluency with the compound interest formula A = P(1 + r)n — including adjusting r and n for compounding frequency and separating amount from interest.
1. Quick recall
Answer each part in the space provided. 1 mark each
Q1.1 Write the compound interest formula and name each variable.
A = ____________ P = ____________ r = ____________ n = ____________
Q1.2 For each compounding frequency, write the number of periods per year:
Annually: ____ Semi-annually: ____ Quarterly: ____ Monthly: ____
Q1.3 A loan compounds quarterly at 8% per annum for 3 years. State the adjusted values: r = ____________ n = ____________
2. Worked example — compound interest with quarterly compounding
Follow each line. Every step has a reason on the right.
Problem. $14,000 is invested at 6% per annum compounding quarterly for 4 years. Find (a) the final amount, (b) the interest earned.
Step 1 — Adjust r and n for quarterly compounding.
r = 0.06 ÷ 4 = 0.015 per quarter
n = 4 × 4 = 16 quarters
Reason: divide the annual rate by 4, multiply the years by 4 to get total quarters.
Step 2 — Substitute into A = P(1 + r)n.
A = 14,000 × (1.015)16
(1.015)16 = 1.26899 (calculator)
Reason: show (1 + r)n on its own line — markers award a method mark for this even if the final arithmetic slips.
Step 3 — Multiply by the principal.
A = 14,000 × 1.26899 = $17,765.86
Reason: this is the future value (amount), not the interest.
Step 4 — Subtract P to find interest earned.
I = A − P = $17,765.86 − $14,000 = $3,765.86
Reason: the formula gives the total amount, not the interest. The interest question always needs the subtraction.
Conclusion. (a) A = $17,765.86; (b) I = $3,765.86.
3. Faded example — fill in the missing steps
$7,500 is borrowed at 9% per annum compounding monthly. The borrower makes no repayments. Find the amount owed after 2 years. Fill in each blank. 4 marks
Step 1 — Adjust r and n for monthly compounding:
r = 0.09 ÷ ____ = ____________ per month
n = 2 × ____ = ____________ months
Step 2 — Substitute: A = 7,500 × (1 + ____)____
Step 3 — Calculate the bracket: (1.0075)24 = ____________
Step 4 — Multiply by P: A = 7,500 × ____________ = $ ____________
Conclusion. The borrower owes $ ____________ after 2 years.
4. Graduated practice — compound interest calculations
Show one line of substitution and clearly label your final answer with units. Keep dollar amounts to 2 decimal places.
Foundation — annual compounding only (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | Find A for P = $5,000, r = 0.04, n = 3 (annual). | |
| 4.2 1 | Find A for P = $10,000, r = 0.05, n = 6 (annual). | |
| 4.3 1 | For an investment with A = $13,400 and P = $10,000, calculate the interest earned (I = A − P). | |
| 4.4 1 | An annual rate of 7.2% compounding monthly. State r per period. |
Standard — typical HSC difficulty (6 questions)
Show at least one line of substitution and the value of (1 + r)n as a separate line.
4.5 $9,500 is invested at 4.8% per annum compounding annually for 6 years. Calculate the final amount. 2 marks
4.6 $8,000 is invested at 5% per annum compounding semi-annually for 3 years. Calculate the final amount. 2 marks
4.7 A loan of $6,800 compounds monthly at 8.4% per annum for 18 months. Calculate the amount owed. 2 marks
4.8 $15,000 compounds annually at 5% for 4 years. Find the interest earned. 2 marks
4.9 $12,000 is invested at 6% per annum compounding quarterly for 2 years. Find the final amount. 2 marks
4.10 Use the Rule of 72 to estimate how many years it takes for money to double at 6% per annum compound interest, then check by calculating A for P = $1,000, r = 0.06, n = 12. 2 marks
Extension — comparison (2 questions)
4.11 $18,000 is invested for 4 years at 5.5% per annum. Compare the interest earned under (a) simple interest, and (b) compound interest compounding annually. State which earns more and by how much. 3 marks
4.12 $10,000 is invested for 3 years at 7.2% per annum. Compare the final amount with annual compounding versus monthly compounding. State which gives more and by how much. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Variables
A = P(1 + r)n. A = final amount; P = principal; r = rate per compounding period (decimal); n = number of compounding periods.
Q1.2 — Periods per year
Annually: 1; Semi-annually: 2; Quarterly: 4; Monthly: 12.
Q1.3 — Adjusted values
r = 0.08 ÷ 4 = 0.02 per quarter; n = 3 × 4 = 12 quarters.
Q3 — Faded example (monthly compounding loan)
Step 1: r = 0.09 ÷ 12 = 0.0075 per month; n = 2 × 12 = 24 months.
Step 2: A = 7,500 × (1 + 0.0075)24.
Step 3: (1.0075)24 = 1.19641…
Step 4: A = 7,500 × 1.19641 = $8,973.09.
Conclusion: The borrower owes $8,973.09 after 2 years.
Q4.1 — P = $5,000, r = 0.04, n = 3
A = 5,000 × (1.04)3 = 5,000 × 1.124864 = $5,624.32.
Q4.2 — P = $10,000, r = 0.05, n = 6
A = 10,000 × (1.05)6 = 10,000 × 1.340096 = $13,400.96.
Q4.3 — Interest from A − P
I = $13,400 − $10,000 = $3,400.00.
Q4.4 — Monthly rate from 7.2% p.a.
r = 0.072 ÷ 12 = 0.006 per month (or 0.6% per month).
Q4.5 — $9,500 at 4.8% p.a. annual for 6 years
(1.048)6 = 1.32738… A = 9,500 × 1.32738 = $12,610.09.
Q4.6 — $8,000 at 5% p.a. semi-annual for 3 years
r = 0.05 ÷ 2 = 0.025; n = 3 × 2 = 6. (1.025)6 = 1.15969… A = 8,000 × 1.15969 = $9,277.55.
Q4.7 — Loan $6,800 at 8.4% p.a. monthly for 18 months
r = 0.084 ÷ 12 = 0.007; n = 18. (1.007)18 = 1.13342… A = 6,800 × 1.13342 = $7,707.26.
Q4.8 — $15,000 at 5% p.a. annual for 4 years, interest only
A = 15,000 × (1.05)4 = 15,000 × 1.21551 = $18,232.59. I = $18,232.59 − $15,000 = $3,232.59.
Q4.9 — $12,000 at 6% p.a. quarterly for 2 years
r = 0.06 ÷ 4 = 0.015; n = 2 × 4 = 8. (1.015)8 = 1.12649… A = 12,000 × 1.12649 = $13,517.91.
Q4.10 — Rule of 72 check
Rule of 72: 72 ÷ 6 = 12 years to double. Check: A = 1,000 × (1.06)12 = 1,000 × 2.012196 = $2,012.20 — slightly more than double, confirming 12 years is about right.
Q4.11 — Simple vs compound on $18,000 at 5.5% for 4 years
Simple: I = 18,000 × 0.055 × 4 = $3,960.00.
Compound: A = 18,000 × (1.055)4 = 18,000 × 1.23882 = $22,298.85; I = $22,298.85 − $18,000 = $4,298.85.
Compound earns more by $4,298.85 − $3,960.00 = $338.85.
Q4.12 — Annual vs monthly compounding on $10,000 at 7.2% for 3 years
Annual: (1.072)3 = 1.231925; A = $12,319.25.
Monthly: r = 0.072 ÷ 12 = 0.006; n = 36. (1.006)36 = 1.240122; A = $12,401.22.
Monthly compounding gives more by $12,401.22 − $12,319.25 = $81.97.