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hscscience Maths Std · Y12
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Module 7 · L1 of 12 ~25 min MST-12-S2-02 ⚡ +50 XP available

Simple Interest

You deposit $\$1000$ in a savings account that pays 5% per year simple interest. After one year you earn $\$50$. After two years, $\$100$. After ten years, $\$500$. The interest never compounds, it is always calculated on the original principal. Simple interest is the most straightforward way to calculate the cost of borrowing or the return on lending, and it forms the foundation for understanding every other interest calculation in this module.

Today's hook, You lend $2000 to a friend at 6% per year simple interest. They repay after 3 years. How much interest do they owe? How much total do they repay? Predict before reading on.
0/5QUESTS

Get oriented

Recall what you already know, meet the key ideas and settle the terms.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall, your gut answer first
+5 XP warm-up

You lend $2000 to a friend at 6% per year simple interest. They repay after 3 years. How much interest do they owe? How much total do they repay?

Before reading on write your gut feeling. We will revisit this at the end of the lesson.

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Key ideas for this lesson
reference

Simple interest is the foundation of all financial mathematics in this module. Two formulas are essential.

Simple Interest: $I = P \times r \times n$, where $P$ is the principal, $r$ is the rate per period as a decimal, and $n$ is the number of periods.

Total Amount: $A = P + I = P(1 + rn)$, the total you receive or repay including the original principal.

The simple-interest formula and its variables.
Interest is always calculated on the original principal it never earns further interest.
Rate as decimal
Always convert the percentage rate before substituting. 5% p.a. becomes $r = 0.05$. Never use 5 directly in the formula.
Linear growth
$A = P + Prn$ is a linear function of $n$. The graph of amount vs time is a straight line, the same interest is added each period.
Units must match
If $r$ is per year, then $n$ must be in years. If $r$ is per month, $n$ must be in months. Mixing units is the most common error.
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What you will master
Know

Key facts

  • Simple interest formula $I = Prn$
  • Total amount formula $A = P(1+rn)$
  • Definitions of $P$, $r$, and $n$
Understand

Concepts

  • Why interest is proportional to time
  • The difference between interest and total amount
  • When simple interest produces linear growth
Can do

Skills

  • Calculate simple interest
  • Find the total amount after any period
  • Rearrange to find $P$, $r$, or $n$
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Key terms
Principal ($P$)The initial amount borrowed or invested, the base on which all interest is calculated.
Interest Rate ($r$)The percentage charged or earned per period, expressed as a decimal in calculations.
Time ($n$)The number of time periods over which interest accrues. Units must match the rate.
Simple Interest ($I$)Interest calculated only on the original principal: $I = Prn$.
Total Amount ($A$)Principal plus interest: $A = P + I = P(1+rn)$.
Linear GrowthGrowth by a constant amount each period, producing a straight-line graph.

See money grow in a straight line

Work out why simple interest is linear, then check you can read I = Prn.

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The linear growth of money
core concept

Simple interest is calculated as a fixed percentage of the principal for each time period. Unlike compound interest, the interest itself does not earn further interest, the same dollar amount is added every period.

$$I = P \times r \times n$$

Where $P$ is the principal (initial amount), $r$ is the interest rate per period as a decimal, and $n$ is the number of periods.

The total amount (principal plus interest) is:

$$A = P + I = P(1 + rn)$$
Worked example: $5000 invested at 4% p.a. simple interest for 3 years.
$I = 5000 \times 0.04 \times 3 = \$600$
$A = 5000 + 600 = \$5600$

Because $A = P + Prn$, this is a linear function of $n$. Graphing amount vs time gives a straight line with slope $Pr$ and y-intercept $P$. The same amount of interest ($Pr$ dollars) is added every year.

Real-world use: Simple interest appears in short-term personal loans, some car loans, government bonds, and store credit arrangements. Its predictability makes both borrower and lender comfortable for short durations.

Simple interest grows linearly: the total amount A = P + Prn = P(1 + rn). A graph of A against n is a straight line with slope Pr and y-intercept P. The interest per year is constant regardless of balance.

Pause, copy A = P(1 + rn) with all variable definitions, the slope (Pr per year), the y-intercept (P at n = 0), and note this gives a straight-line graph of A versus n into your book.

Quick check: $3000 is invested at 5% p.a. simple interest for 4 years. What is the interest earned?

Rearrange the formula

Make P, r or n the subject, then test yourself on the rearrangement.

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Rearranging the formula
core concept

Simple interest grows linearly: A = P(1 + rn) is a straight line with y-intercept P and slope Pr when graphed against n. When the question gives you A and asks for P, r, or n, rearrange the formula: P = A / (1 + rn), r = (A/P − 1) / n, and n = (A/P − 1) / r, always convert percentage rates to decimals first.

Given any three of $I$, $P$, $r$, $n$, you can rearrange $I = Prn$ to find the fourth variable.

Find Rearrangement In words
$P$$P = \dfrac{I}{rn}$Interest divided by (rate times time)
$r$$r = \dfrac{I}{Pn}$Interest divided by (principal times time)
$n$$n = \dfrac{I}{Pr}$Interest divided by (principal times rate)
Worked example, find P: An investment earns $720 interest at 6% p.a. over 4 years. What was the principal?
$P = \dfrac{I}{rn} = \dfrac{720}{0.06 \times 4} = \dfrac{720}{0.24} = \$3000$
Using total amount: If given $A$ instead of $I$, find interest first: $I = A - P$. Or rearrange directly: $P = \dfrac{A}{1+rn}$.

Rearranging A = P(1 + rn): to find P = A/(1 + rn), rate r = (A/P − 1)/n, time n = (A/P − 1)/r. Always convert percentage rates to decimals before calculating; convert decimal answers back to percentages if required.

Pause, copy the three rearrangements from A = P(1 + rn): P = A/(1 + rn), r = (A/P − 1)/n, n = (A/P − 1)/r, remembering to convert % to decimal and convert decimal r answers back to % into your book.

True or false: To find the interest rate, you divide the interest by the product of the principal and the number of periods.

Worked examples · reveal each step

Work the examples and drill

Follow two worked problems, then run the quick-fire practice.

PROBLEM 1 · CALCULATE INTEREST AND TOTAL AMOUNT

A loan of $8000 at 7.5% p.a. simple interest. Find the total repayment after 5 years.

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$P = 8000,\quad r = 0.075,\quad n = 5$
Identify the variables and convert rate to decimal (7.5% = 0.075)
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PROBLEM 2 · FIND THE PRINCIPAL

Find the principal that earns $450 interest at 5% p.a. over 3 years.

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$I = 450,\quad r = 0.05,\quad n = 3$
Write down what we know
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Quick-fire practice · 2 activities

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Calculate, 5 questions. (a) Calculate simple interest on $\$3500$ at 4.5% p.a. for 4 years. (b) Find the total amount after 6 years if $\$6000$ is invested at 3.2% p.a. simple interest. (c) What rate is needed for $\$4000$ to earn $\$960$ interest over 5 years? (d) How long will $\$2500$ take to grow to $\$3250$ at 6% p.a. simple interest? (e) Find the principal that earns $540 at 4.5% p.a. over 4 years.

2

Analyse and connect. (a) Why does simple interest produce linear growth? Sketch the shape of the graph of $A$ vs $n$ for $P = \$1000$, $r = 5\%$. (b) A store offers "buy now, pay later" with 0% interest for 12 months, then 18% p.a. simple interest. Explain the trap. (c) Compare simple interest on $10,000 at 5% for 10 years vs 10% for 5 years. Which earns more?

Avoid the interest traps

Fix the decimal, total-amount and unit slips that cost marks.

Fill the gap: $\$4000$ invested at 4.8% p.a. simple interest for 5 years earns interest of $\$$.

Common errors · the 3 traps that cost marks

Trap 01
Using the rate as a percentage, not a decimal
Writing $I = 4000 \times 5 \times 3 = 60\,000$ instead of $I = 4000 \times 0.05 \times 3 = 600$. Always divide the percentage by 100 before substituting into the formula.
Trap 02
Confusing interest with total amount
Questions often ask for the total amount, not just the interest. $A = P + I$, not just $I$. Re-read the question carefully to check which value is required.
Trap 03
Mismatched units for rate and time
If the rate is per annum but time is in months, convert months to years (divide by 12) before applying the formula. The units of $r$ and $n$ must always match.

Match each variable to its description:

  • P
  • r
  • n
  • A
  • Number of time periods
  • Principal plus interest
  • Initial amount borrowed or invested
  • Interest rate per period as a decimal

Revisit your thinking

Return to your opening answer and name what has changed.

Top 3 list: Name THREE real-world situations where simple interest is used. For each, identify who pays the interest and who receives it.

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Revisit your thinking

The answer: $I = 2000 \times 0.06 \times 3 = \$360$. Total repayment $= 2000 + 360 = \$2360$.

Simple interest is straightforward because the interest is always calculated on the original $2000, not on the growing balance. This makes it predictable, but far less powerful for investors than compound interest, the topic of the next lesson.

What has changed in your understanding? What did you get right? What surprised you?

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Practise simple interest

Answer the question bank, then write full short-answer responses.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next.

Q1. $6000 is invested at 3.5% p.a. simple interest for 4 years. What is the total amount?

Q2. Simple interest on $P$ at rate $r$ for $n$ periods is $I$. Which expression gives the principal?

Q3. The graph of total amount $A$ against time $n$ under simple interest is:

Q4. A loan of $P$ at 6% p.a. simple interest amounts to $\$6440$ after 4 years. What is $P$?

Q5. What annual interest rate gives $\$960$ simple interest on $\$4000$ over 5 years?

02
Short answer
ApplyBand 42 marks

SA 1. (a) Calculate the simple interest on $4500 invested at 3.8% p.a. for 5 years. (b) Find the total amount. (c) How much more would be earned if the rate were 4.5%? (2 marks)

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ApplyBand 42 marks

SA 2. A loan of $P$ at 6% p.a. simple interest amounts to $\$6440$ after 4 years. Find $P$. (2 marks)

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AnalyseBand 53 marks

SA 3. (a) Prove that the total amount $A$ under simple interest grows linearly with time. (b) Two investments both start at $\$5000$. Investment X earns 8% p.a. simple interest; Investment Y earns 6% p.a. simple interest. After how many years will X be worth exactly $\$1000$ more than Y? (c) Explain why simple interest is rare for long-term investments. (3 marks)

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Comprehensive answers (click to reveal)

MC 1, B: $I = 6000 \times 0.035 \times 4 = \$840$. $A = 6000 + 840 = \$6840$.

MC 2, C: Rearranging $I = Prn$ gives $P = I \div (rn)$.

MC 3, A: $A = P + Prn$ is linear in $n$, a straight line with slope $Pr$ and y-intercept $P$.

MC 4, D: $6440 = P(1 + 0.06 \times 4) = 1.24P$. $P = 6440 \div 1.24 = \$5200$.

MC 5, B: $r = I \div (Pn) = 960 \div (4000 \times 5) = 960 \div 20\,000 = 0.048 = 4.8\%$.

SA 1 (2 marks): (a) $I = 4500 \times 0.038 \times 5 = \$855$ [0.5]. (b) $A = \$5355$ [0.5]. (c) At 4.5%: $I = \$1012.50$. Extra $= \$157.50$ [1].

SA 2 (2 marks): $P = \$5200$ [2].

SA 3 (3 marks): (a) $A = P + Prn$, linear in $n$ [1]. (b) $n = 10$ years [1]. (c) Simple interest never earns interest on interest, over decades compound interest dramatically outperforms [1].

Drill 1: (a) $\$630$. (b) $\$7152$. (c) $4.8\%$. (d) $5$ years. (e) $\$3000$.

Drill 2: (a) Linear because $n$ appears to the first power only, constant increment $Pr$ each year. (b) When the 0% period ends, the entire outstanding balance starts accruing 18% p.a., not just new purchases. (c) Both earn $5000 interest, exactly equal.