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.)