Mathematics Standard • Year 11 • Module 3 • Lesson 11

Simple Interest

Build fluency with I = Prn: substitute correctly, rearrange to find P, r or n, and convert between annual and monthly time periods cleanly.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Write the simple interest formula and the total-amount formula.

I = ____________ A = ____________

Q1.2 Convert each rate to a decimal: 5% = ________, 4.5% = ________, 6.25% = ________.

Q1.3 Rearrange I = Prn for each unknown.

P = ____________ r = ____________ n = ____________

Stuck? Revisit lesson § Finding the Unknown Variable — Rearranging I = Prn.

2. Worked example — total amount after a fixed term

Follow each line of working. Every step has a reason on the right.

Problem. Mei invests $8,400 in a 4-year term deposit at 4.6% per annum simple interest. Calculate (a) the interest earned and (b) the total amount at maturity.

Step 1 — Identify P, r, n.

P = $8,400 r = 0.046 (per annum) n = 4 (years)

Reason: convert percent to decimal — 4.6 ÷ 100 = 0.046. Both r and n are annual.

Step 2 — Apply I = Prn.

I = $8,400 × 0.046 × 4 = $1,545.60

Step 3 — Find total amount.

A = P + I = $8,400 + $1,545.60 = $9,945.60

Conclusion. Mei earns $1,545.60 in interest; the total amount at maturity is $9,945.60.

3. Faded example — finding the rate

A 3-year loan of $6,500 charges simple interest. Over the 3 years the borrower repays $7,397.50. Find the annual simple interest rate (%). Fill the blanks. 4 marks

Step 1 — Find the interest:

I = A − P = $7,397.50 − $ ________ = $ ____________

Step 2 — Identify variables for r: P = $ ________, n = ________ years.

Step 3 — Rearrange and substitute:

r = I ÷ (P × n) = $ ________ ÷ ($ ________ × ____) = ____________

Step 4 — Convert to percentage: r = ________ × 100 = ____________%.

Check. I = $6,500 × ________ × 3 = $ ____________ ✓

Stuck? Revisit lesson § Worked Example 2 — Finding the Interest Rate.

4. Graduated practice — Simple interest calculations

Show your working below each part. Convert percentages to decimals before substituting.

Foundation — single-step substitution (4 questions)

Q	Problem	Answer
4.1 1	Find I: P = $4,000, r = 5% p.a., n = 3 years.
4.2 1	Find A: P = $5,000, r = 4% p.a., n = 2 years.
4.3 1	Convert 4.8% per annum to a monthly rate (as a decimal).
4.4 1	How many years is 30 months?

Standard — typical HSC difficulty (6 questions)

Show at least one line of substitution and clearly label your final answer with units.

4.5 Calculate the simple interest on $9,600 at 5.4% per annum for 5 years. 2 marks

4.6 Find the total amount after $7,200 is invested at 4.2% per annum simple interest for 4 years. 2 marks

4.7 Lena borrows $8,400 at 6.5% per annum simple interest. How much interest is charged after 18 months? 2 marks

4.8 A 4-year investment of $6,500 earns $1,170 in simple interest. Find the annual interest rate. 2 marks

4.9 A loan of $12,000 at 7% per annum simple interest accrues $2,520 in interest. For how many years was the loan held? 2 marks

4.10 A 3-year simple-interest investment grows from $4,800 to $5,520. Find the annual interest rate. 2 marks

Extension — multi-step with unit conversion (2 questions)

4.11 An investment of $7,500 at 4.8% per annum simple interest is expected to grow to $8,400. How many months will this take? 3 marks

4.12 Compare two simple-interest investments: Option X — $5,000 at 5% p.a. for 4 years; Option Y — $5,000 at 4.5% p.a. for 5 years. Which earns more interest, and by how much? 3 marks

Stuck on 4.11? Find I needed = $900, then convert the rate to monthly (0.048 ÷ 12 = 0.004) and solve n = I ÷ (P × r) for months.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I used 0.05 (not 5) and got $600 — not $60,000.

For 4.2 I added I to P at the end ($5,000 + $400 = $5,400) — not multiplied.

For 4.3 I divided 0.048 by 12 to get the monthly rate 0.004 — and remembered the unit must match n.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Formulas

I = Prn. A = P + I (or equivalently A = P(1 + rn)).

Q1.2 — Percent → decimal

5% = 0.05. 4.5% = 0.045. 6.25% = 0.0625.

Q1.3 — Rearrangements

P = I ÷ (rn). r = I ÷ (Pn). n = I ÷ (Pr).

Q3 — Faded example (3-year loan, A = $7,397.50)

Step 1: I = $7,397.50 − $6,500 = $897.50.
Step 2: P = $6,500, n = 3 years.
Step 3: r = $897.50 ÷ ($6,500 × 3) = $897.50 ÷ $19,500 = 0.046.
Step 4: r = 0.046 × 100 = 4.6% per annum.
Check: I = $6,500 × 0.046 × 3 = $897.00 (≈ $897.50 with small rounding) ✓

Q4.1 — I from P, r, n

I = $4,000 × 0.05 × 3 = $600.00.

Q4.2 — A from P, r, n

I = $5,000 × 0.04 × 2 = $400. A = $5,000 + $400 = $5,400.00.

Q4.3 — Annual → monthly rate

r = 0.048 ÷ 12 = 0.004 per month (= 0.4% per month).

Q4.4 — Months to years

30 ÷ 12 = 2.5 years.

Q4.5 — Interest on $9,600 at 5.4% for 5 years

I = $9,600 × 0.054 × 5 = $2,592.00.

Q4.6 — Total on $7,200 at 4.2% for 4 years

I = $7,200 × 0.042 × 4 = $1,209.60. A = $7,200 + $1,209.60 = $8,409.60.

Q4.7 — Interest on $8,400 loan over 18 months at 6.5%

n = 18 ÷ 12 = 1.5 years. I = $8,400 × 0.065 × 1.5 = $819.00.

Q4.8 — Rate from $1,170 interest over 4 years on $6,500

r = $1,170 ÷ ($6,500 × 4) = $1,170 ÷ $26,000 = 0.045 → 4.5% per annum.

Q4.9 — Time from $2,520 interest at 7% on $12,000

n = $2,520 ÷ ($12,000 × 0.07) = $2,520 ÷ $840 = 3 years.

Q4.10 — Rate from $4,800 → $5,520 over 3 years

I = $5,520 − $4,800 = $720. r = $720 ÷ ($4,800 × 3) = $720 ÷ $14,400 = 0.05 → 5% per annum.

Q4.11 — Months to reach $8,400 from $7,500 at 4.8% p.a.

I needed = $8,400 − $7,500 = $900.
Monthly rate = 0.048 ÷ 12 = 0.004.
n = $900 ÷ ($7,500 × 0.004) = $900 ÷ $30 = 30 months.

Q4.12 — Compare X and Y

Option X: I = $5,000 × 0.05 × 4 = $1,000.
Option Y: I = $5,000 × 0.045 × 5 = $1,125.
Option Y earns $125 more interest than Option X over its term. (Despite the lower rate, the extra year of interest more than makes up the difference.)