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hscscience Maths Std · Y11
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Module 3 · L11 of 14 ~50 min MS-F1 ⚡ +90 XP available

Simple Interest

Simple interest is calculated only on the original principal — the same dollar amount is added every period. Master $I = Prn$, learn to rearrange for any unknown, convert time units, and compare investments. The formula $I = Prn$ is the engine of this entire lesson.

Today's hook — If you put $5,000 in a bank account that pays 4% interest per year, how much would you have after 3 years? Most people guess $5,600 — but is that right? Who benefits more from simple interest: the borrower or the lender?
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Worksheets

Practise this lesson

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

01
Recall — your gut answer first
+5 XP warm-up

If you put $5,000 in a bank account that pays 4% interest per year, how much would you have after 3 years? Simple interest assumes the bank pays you the same dollar amount every single year, based only on your original deposit. It never grows on itself. Who do you think benefits more from simple interest — the borrower or the lender?

Without calculating — write your gut feeling. We'll revisit this at the end of the lesson.

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02
The key formulas you need to own
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Simple interest in Maths Standard centres on three formulas. Lock these in before the worked examples.

$I = Prn$ gives the interest earned or charged. $A = P + I$ gives the total amount. The rearrangements let you solve for any unknown when the other three values are known.

INTEREST I = Prn TOTAL AMOUNT A = P + I REARRANGED P=I/rn · r=I/Pn · n=I/Pr P = principal ($) r = rate (decimal) n = periods r and n must match After finding r, × 100 for %
$r$ and $n$ must use the same time unit before substituting
Rate as decimal first
4.5% per annum means $r = 0.045$, not 4.5. Substituting 4.5 gives an answer 100 times too large.
Match the time units
If $r$ is annual, $n$ must be in years. 18 months = 1.5 years. Or convert $r$ to a monthly rate (annual rate ÷ 12).
Linear growth
Simple interest grows by the same dollar amount each period — it is straight-line, not exponential. A graph would be a straight line.
03
What you'll master
Know

Key facts

  • The simple interest formula $I = Prn$
  • The total amount formula $A = P + I$
  • The rearrangements for $P$, $r$, and $n$
  • That simple interest grows linearly
  • That $r$ and $n$ must use matching time units
Understand

Concepts

  • Why simple interest is only calculated on the original principal
  • Why mismatched time units give wrong answers
  • How to compare two investments on the same basis
  • The difference between simple and compound interest (preview)
Can do

Skills

  • Calculate $I$ and $A$ for any principal, rate, and time
  • Rearrange to find unknown $P$, $r$, or $n$
  • Convert months to years and vice versa for $n$
  • Compare two investment options and state the better one with the dollar difference
04
Key terms
Simple InterestInterest calculated only on the original principal: $I = P \times r \times n$.
Principal ($P$)The initial amount of money invested or borrowed.
Rate ($r$)The interest rate per period expressed as a decimal (e.g. 4% = 0.04).
Term ($n$)The number of time periods — must use the same unit as $r$.
Interest Earned ($I$)The total dollar amount of interest over the full term.
Total Amount ($A$)The final value: principal plus all interest earned, $A = P + I$.
05
The Simple Interest Formula — $I = Prn$
core concept

Simple interest is calculated only on the original principal — it does not grow on previously earned interest, making it predictable and linear over time.

In the formula $I = Prn$: $P$ is the principal (the original amount invested or borrowed), $r$ is the interest rate expressed as a decimal per time period (so 6% per annum = 0.06), and $n$ is the number of time periods. The formula gives $I$, the total interest earned or charged. To find the total amount, use $A = P + I$.

Key rule: The time period for $r$ and $n$ must match. If $r$ is a per annum rate, $n$ must be in years. If $r$ is a monthly rate, $n$ must be in months. Mixing annual rates with monthly periods is the most common error in simple interest questions.
1 2 3 4 5 6 7 Years Amount ($) P Simple interest — straight line (linear growth) Simple Interest: Same Dollar Amount Added Each Period
Insight: Simple interest is most commonly used for short-term loans and some government bonds. For longer-term investments, compound interest is almost always used in practice — which is why Lesson 12 matters more for real-world financial decisions.
What to write in your book
  • $I = Prn$ — convert rate to decimal first; match time units for $r$ and $n$.
  • $A = P + I$ — always add interest to principal for the total amount.
  • Rearrangements: $P = I \div (rn)$; $r = I \div (Pn)$; $n = I \div (Pr)$.
  • After finding $r$, multiply by 100 and state the unit (% per annum, etc.).

Quick check: An investment earns simple interest at 5% per annum. What value of $r$ should be substituted into $I = Prn$?

06
Finding the Unknown — Rearranging $I = Prn$
core concept

HSC questions don't always ask you to find $I$ — they may give you the interest and ask for the principal, rate, or time instead, requiring you to rearrange the formula.

The formula $I = Prn$ can be rearranged to find any one variable if the other three are known:

  • Finding $P$: $P = I \div (rn)$
  • Finding $r$: $r = I \div (Pn)$ — then multiply by 100 to express as a percentage
  • Finding $n$: $n = I \div (Pr)$ — the answer is in the same time unit as $r$
Strategy: A reliable method is to substitute the known values into $I = Prn$ first, then solve for the unknown algebraically. Always check your answer by substituting all values back into $I = Prn$ to verify.
Common error — finding $n$ in months: When solving for $n$, the unit of $n$ matches the unit of $r$. If $r$ was given as a monthly rate, $n$ comes out in months. If the question asks for the time in years, divide by 12.
What to write in your book
  • To find $r$: rearrange to $r = I \div (Pn)$, solve, then multiply by 100 for percentage.
  • To find $n$: rearrange to $n = I \div (Pr)$. Units of $n$ match units of $r$.
  • For "how long to reach $X$": find the required interest first $I = A - P$, then use $n = I \div (Pr)$.

True or false: To find the interest rate from $I = Prn$, rearrange to get $r = I \div (Pn)$, then multiply the result by 100 to convert to a percentage.

PROBLEM 1 · FINDING TOTAL INTEREST AND AMOUNT

Yuki invests $12,500 at a simple interest rate of 5.2% per annum for 3 years. Calculate: (a) the interest earned, and (b) the total amount at the end of 3 years.

1
Identify variables
$P = \$12{,}500$, $r = 0.052$, $n = 3$
Convert 5.2% per annum to a decimal: 5.2 ÷ 100 = 0.052. Rate and time are both annual, so the periods match.
PROBLEM 2 · FINDING THE INTEREST RATE

Lena borrows $8,400 and repays $9,576 after 2 years under a simple interest arrangement. What was the annual interest rate?

1
Find $I$
$I = A - P = \$9{,}576 - \$8{,}400 = \$1{,}176.00$
The interest is the difference between what Lena repaid and what she borrowed.
PROBLEM 3 · FINDING TIME — TARGET AMOUNT

How many months will it take for an investment of $6,000 at 4.8% per annum simple interest to grow to $6,840?

1
Find required interest
$I = \$6{,}840 - \$6{,}000 = \$840$
Find the required interest — the difference between target amount and principal.
PROBLEM 4 · COMPARING TWO INVESTMENTS

Option A invests $8,000 at 4.6% per annum simple interest for 5 years. Option B invests $8,000 at 4.2% per annum simple interest for 6 years. Which option gives the greater total amount, and by how much?

1
Option A
$I_A = \$8{,}000 \times 0.046 \times 5 = \$1{,}840$
$A_A = \$8{,}000 + \$1{,}840 = \$9{,}840$
Calculate the simple interest first, then add it to the principal.
What to write in your book
  • For "time to reach target" questions: find $I = A - P$ first, then $n = I \div (Pr)$.
  • Comparison answers must name the better option AND give the dollar difference.
  • Don't confuse $A = P(1+rn)$ (simple, linear) with $A = P(1+r)^n$ (compound, exponential).

Fill the gap: An investment of $7,200 earns simple interest at 4.5% per annum for 18 months. The interest earned is $ (18 months = 1.5 years; $I = 7200 \times 0.045 \times 1.5$).

Trap 01
Not converting rate to a decimal
Using $r = 4.5$ instead of $r = 0.045$ gives an answer 100 times too large. Always divide the percentage by 100 before substituting.
Trap 02
Mixing time units
If $r$ is an annual rate and $n$ is given in months, you must convert — either $n$ to years (÷ 12) or $r$ to a monthly rate (÷ 12). Never mix annual and monthly directly.
Trap 03
Writing $A$ instead of $I$ as the answer
If the question asks for interest earned, the answer is $I$, not $A$. Read carefully: "interest" means $I$; "total amount" or "final value" means $A = P + I$.
What to write in your book
  • Rate always as decimal before substituting: 6% → 0.06.
  • 18 months at annual rate: convert to 1.5 years or convert rate to monthly (÷ 12).
  • "Interest" means $I$; "total amount" means $A = P + I$. Read the question carefully.

Match each setup:

1

Calculate the simple interest on $5,000 at 3.6% per annum for 4 years.

2

Find the total amount for $9,000 at 5% per annum simple interest for 3 years.

3

An account earns $1,020 simple interest in 4 years at 3.4% per annum. Find the principal.

4

A loan of $4,800 earns simple interest of $432 over 18 months. What annual simple interest rate was charged?

5

An investment of $5,500 earns simple interest at 6% per annum. How many years until the total amount reaches $7,150?

Top 3 list: Name THREE things you must check before substituting values into $I = Prn$.

09
Revisit your thinking

Look back at what you wrote in the Think First section. $5,000 at 4% for 3 years: $I = 5000 \times 0.04 \times 3 = \$600$, so $A = \$5,600$ — the guess was right. Simple interest benefits borrowers (they never pay interest on accumulated interest) but is less favourable for investors than compound interest over long periods.

What has changed? What did you get right? What surprised you?

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01
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. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 32 marks

SA 1. Find the simple interest earned on $7,200 invested at 4.5% per annum for 18 months. (2 marks)

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ApplyBand 32 marks

SA 2. An account earns $1,020 simple interest in 4 years at 3.4% per annum. Find the principal. (2 marks)

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AnalyseBand 43 marks

SA 3. Compare two investments: Option A is $9,500 at 4.1% simple interest for 3 years; Option B is $9,500 at 3.8% simple interest for 4 years. Which gives the greater total amount, and by how much? (3 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: $I = 5000 \times 0.036 \times 4 = \$720$ · 2: $A = 9000 + 9000 \times 0.05 \times 3 = \$10{,}350$ · 3: $P = 1020 \div (0.034 \times 4) = \$7{,}500$ · 4: 18 months = 1.5 years; $r = 432 \div (4800 \times 1.5) = 0.06 = 6\%$ p.a. · 5: $I = 7150 - 5500 = 1650$; $n = 1650 \div (5500 \times 0.06) = 5$ years

SA 1 (2 marks): Convert: 18 months = 1.5 years [1]. $I = 7200 \times 0.045 \times 1.5 = \$486.00$ [1].

SA 2 (2 marks): $P = I \div (rn)$ [1]. $P = 1020 \div (0.034 \times 4) = 1020 \div 0.136 = \$7{,}500.00$ [1].

SA 3 (3 marks): Option A: $I = 9500 \times 0.041 \times 3 = \$1{,}168.50$; $A = \$10{,}668.50$ [1]. Option B: $I = 9500 \times 0.038 \times 4 = \$1{,}444.00$; $A = \$10{,}944.00$ [1]. Option B is greater by $\$10{,}944.00 - \$10{,}668.50 = \$275.50$ [1].

01
Boss battle · The Interest Inspector
earn bronze · silver · gold

Five timed questions on simple interest. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering questions on simple interest. Pool: lessons 1–11.

Mark lesson as complete

Tick when you've finished the practice and review.