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.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
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.
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.
Key facts
- Simple interest formula $I = Prn$
- Total amount formula $A = P(1+rn)$
- Definitions of $P$, $r$, and $n$
Concepts
- Why interest is proportional to time
- The difference between interest and total amount
- When simple interest produces linear growth
Skills
- Calculate simple interest
- Find the total amount after any period
- Rearrange to find $P$, $r$, or $n$
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:
$$A = P + I = P(1 + rn)$$$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.
What to write in your book
- $I = Prn$ — interest = principal × rate × time. Rate must be a decimal.
- $A = P(1 + rn)$ — total amount including principal.
- Simple interest grows linearly — same interest earned every period.
Quick check: $3000 is invested at 5% p.a. simple interest for 4 years. What is the interest earned?
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) |
$P = \dfrac{I}{rn} = \dfrac{720}{0.06 \times 4} = \dfrac{720}{0.24} = \$3000$
What to write in your book
- $P = \dfrac{I}{rn}$, $\quad r = \dfrac{I}{Pn}$, $\quad n = \dfrac{I}{Pr}$
- When given $A$ and not $I$: find $I = A - P$ first, then rearrange.
- Rate answer must be converted back to percentage: multiply by 100 and add % symbol.
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
A loan of $8000 at 7.5% p.a. simple interest. Find the total repayment after 5 years.
Find the principal that earns $450 interest at 5% p.a. over 3 years.
Quick-fire practice · 2 activities
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.
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?
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
What to write in your book
- Rate must always be a decimal — divide percentage by 100.
- $I$ = interest only; $A$ = total amount. Check what the question asks.
- Units of $r$ and $n$ must match — both per year, or both per month.
Match each variable to its description:
Top 3 list: Name THREE real-world situations where simple interest is used. For each, identify who pays the interest and who receives it.
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?
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?
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)
SA 2. A loan of $P$ at 6% p.a. simple interest amounts to $6440 after 4 years. Find $P$. (2 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)
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.
Five timed questions on simple interest, rearranging formulas, and linear growth. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering questions on simple interest. Pool: lesson 1.
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