Your weak spots

Insights load after your first practice round.

Module 7 · L1 of 20 ~40 min ⚡ +95 XP available

Introduction to Financial Mathematics

Every dollar you earn today is worth more than a dollar tomorrow. Banks, super funds, and lenders all use the same mathematical machinery to turn time into money, or money into debt. In this lesson you'll learn the two foundational models: simple interest and compound interest, and discover why the difference between them grows explosively over time.

Today's hook, A friend offers you $10,000 at 8% simple interest or $10,000 at 6% compound interest for 10 years. The higher rate doesn't guarantee the bigger payout. Why?

0/5QUESTS

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

Recall, your gut answer first

+5 XP warm-up

A friend offers you two deals on $10,000:

Deal A: Simple interest at 8% per year for 10 years.

Deal B: Compound interest at 6% per year for 10 years.

Without calculatingwhich deal leaves you with more money? Make a prediction and explain your reasoning.

auto-saved

The two formulas you need to own

+5 XP to read

Financial maths starts with two core formulas. One produces a straight line; the other an explosion. Lock these in, everything in Module 7 builds from them.

Simple interest charges the same dollar amount every period, interest on principal only. Compound interest charges interest on the growing total, interest on principal and accumulated interest.

$A = P(1+r)^n$ grows exponentially

Simple interest, linear

Same interest every period. Graph is a straight line. Formula: $I = Prn$ and $A = P(1+rn)$.

Compound interest, exponential

Interest earns interest. Graph curves upward. Formula: $A = P(1+r)^n$.

Time value of money

$10,000 today grows to $12,762.82 at 5% over 5 years. Money has a time dimension.

What you'll master

Know

Key facts

Simple interest: $I = Prn$ and $A = P(1+rn)$
Compound interest: $A = P(1+r)^n$
The difference between nominal rate and periodic rate
How to convert years ↔ months for $n$

Understand

Concepts

Why compound interest grows exponentially while simple grows linearly
The time value of money: today's dollar beats tomorrow's
How compounding frequency affects total return

Can do

Skills

Calculate simple and compound interest for any $P$, $r$, $n$
Convert between different compounding periods
Compare two financial products using total return
Transpose formulas to find $P$, $r$, or $n$

Key terms

Principal ($P$)The initial amount of money invested or borrowed.

Interest ($I$)The cost of borrowing or the reward for investing, as a percentage of the principal.

Simple interestInterest calculated only on the original principal for the entire term.

Compound interestInterest calculated on the principal plus all accumulated interest from previous periods.

Compounding periodThe interval at which interest is calculated and added (e.g. annually, monthly, daily).

Nominal rateThe advertised annual interest rate before adjusting for compounding frequency.

The time value of money

core concept

Would you rather have $10,000 today or $10,000 in five years? Almost everyone chooses today, and mathematics explains exactly why.

If you invest $10,000 today at 5% p.a. compound interest, it grows to:

$$A = 10{,}000 \times (1.05)^5 = \$12{,}762.82$$

So $10,000 today is equivalent to $12,762.82 in five years at that rate. This is the foundation of all financial mathematics: money has a time dimension. Every loan, investment, and savings account is just a rearrangement of this idea.

Real-world anchor, Lottery payouts. When someone wins $50 million in Oz Lotto, they're often offered a choice: a smaller lump sum now, or the full amount spread over 20 years. The lump sum is the present value of that future stream, compound interest in reverse.

The time value of money: $1 today is worth more than $1 in the future because it can be invested.; Simple interest: $I = Prn$ (interest only); $A = P(1+rn)$ (total amount)

Pause, copy the time-value principle ($1 today is worth more than $1 in the future) and the simple interest formulas $I = Prn$ and $A = P(1+rn)$ into your book.

Did you get this? True or false: simple interest grows linearly, while compound interest grows exponentially.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · SIMPLE INTEREST

You invest $5,000 at 4% p.a. simple interest for 3 years. Find the total interest earned and the final amount.

$I = Prn = 5{,}000 \times 0.04 \times 3$

Identify $P = 5000$, $r = 0.04$ (4% as decimal), $n = 3$ years.

PROBLEM 2 · COMPOUND INTEREST

You invest $5,000 at 4% p.a. compound interest for 3 years. Find the final amount and compare with simple interest.

$A = P(1 + r)^n = 5{,}000 \times (1.04)^3$

$P = 5000$, $r = 0.04$, $n = 3$. Use the compound interest formula.

PROBLEM 3 · MONTHLY COMPOUNDING

Maya invests $8,000 in an account paying 6% p.a. compounded monthly. How much will she have after 4 years?

$r_{\text{monthly}} = \dfrac{0.06}{12} = 0.005$

The rate is 6% per year, but interest compounds monthly. Divide by 12 to get the rate per period.

Quick check: An account earns 4.8% p.a. compounded monthly. What is the correct periodic rate $r$ and number of periods $n$ for 2 years?

Common errors · the 3 traps that cost marks

Trap 01

Using the annual rate with monthly periods

If interest compounds monthly, you must divide the annual rate by 12. Using $r = 0.06$ and $n = 24$ months gives a wildly wrong answer. Always check: rate and periods share the same time unit.

Trap 02

Confusing I with A

$I = Prn$ gives only the interest earned. The question usually asks for the final amount: $A = P + I = P(1+rn)$. Check whether the question says "interest" or "total amount".

Trap 03

Thinking compound always wins in the short run

For $n = 1$, simple and compound interest are identical. The compound advantage only becomes meaningful over multiple periods. For $n < 1$ (partial year), simple interest can actually appear higher if rates aren't converted.

Fill the gap: A $12,000 investment earns 6% p.a. compounded quarterly for 2 years. The periodic rate is $r = 0.06 \div$ and the total number of periods is $n = 2 \times$ $= 8$.

Quick-fire practice · 5 calculations

Find $A$ using simple interest: $P = \$6{,}000$, $r = 5\%$ p.a., $n = 4$ years.

Find $A$ using compound interest: $P = \$6{,}000$, $r = 5\%$ p.a., $n = 4$ years.

$P = \$15{,}000$, 3.6% p.a. compounded monthly, 3 years. Find $A$.

Tom borrows $12,000 at 5.4% p.a. simple interest for 3.5 years. What total amount must he repay?

$P = \$20{,}000$, $r = 5\%$ p.a. compound, $n = 15$ years. By how much does compound beat simple?

Odd one out: Three of these are correct statements about compound interest. Which one is wrong?

Revisit your thinking

Earlier you predicted which deal was better. Let's check:

Deal A (simple 8%, 10 years): $A = 10{,}000 \times (1 + 0.08 \times 10) = \$18{,}000$

Deal B (compound 6%, 10 years): $A = 10{,}000 \times (1.06)^{10} = \$17{,}908.48$

Despite the lower rate, compound interest nearly catches up to the higher simple rate over 10 years. Extend to 15 years: Deal A gives $\$22{,}000$ while Deal B gives $\$23{,}965.58$, compound takes the lead. Exponential growth always wins in the end.

auto-saved

Multiple choice

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

Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 43 marks

Q1. A sum of $6,500 is invested at 4.8% p.a. for 4 years. Calculate the final amount under (a) simple interest and (b) compound interest (annual). (c) Find the difference. (3 marks)

auto-saved

ApplyBand 43 marks

Q2. $15,000 is invested at 7.2% p.a. compounded monthly for 3 years. Find (a) the periodic rate, (b) the total number of periods, and (c) the final amount. (3 marks)

auto-saved

AnalyseBand 54 marks

Q3. $P = \$50{,}000$ is invested at $r = 6\%$ p.a. simple interest. A second investor places $\$50{,}000$ at $r = 6\%$ p.a. compound interest. (a) After how many years does the compound investor's balance exceed the simple investor's balance by more than $\$10{,}000$? (b) Explain why the gap continues to grow. (4 marks)

auto-saved

📖 Comprehensive answers (click to reveal)

Drill 1: $A = 6000(1 + 0.05 \times 4) = \$7{,}200$ · 2: $A = 6000(1.05)^4 = \$7{,}293.04$ · 3: $r = 0.003$, $n = 36$, $A = 15000(1.003)^{36} = \$16{,}662.09$ · 4: $A = 12000(1 + 0.054 \times 3.5) = \$14{,}268.00$ · 5: Simple $= \$35{,}000$; Compound $= \$41{,}578.56$; Difference $= \$6{,}578.56$

Q1 (3 marks): (a) $A = 6500(1 + 0.048 \times 4) = \$7{,}748.00$ [1]. (b) $A = 6500(1.048)^4 = \$7{,}940.47$ [1]. (c) Difference $= 7940.47 - 7748.00 = \$192.47$ [1].

Q2 (3 marks): (a) $r = 0.072 / 12 = 0.006$ [1]. (b) $n = 3 \times 12 = 36$ [1]. (c) $A = 15000(1.006)^{36} = \$18{,}698.47$ [1].

Q3 (4 marks): (a) At year 14: Simple $= 50000 \times (1 + 0.06 \times 14) = \$92{,}000$; Compound $= 50000 \times (1.06)^{14} = \$112{,}738$; gap $= \$20{,}738 > \$10{,}000$. At year 10: Compound $= \$89{,}542$; Simple $= \$80{,}000$; gap $= \$9{,}542$ (not yet). So gap exceeds $10{,}000 between year 10 and 11 [2]. (b) Compound interest applies interest to an ever-growing base; the additional interest earned each year grows every period, so the gap between linear and exponential growth widens continuously, it never shrinks [2].

Boss battle · The Banker

earn bronze · silver · gold

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

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering simple and compound interest questions. Pool: lesson 1.

Mark lesson as complete

Tick when you've finished the practice and review.

← Module 6 · Lesson 15 Lesson 2 · Compound Interest in Depth →

Module overview · Maths Advanced · Checkpoint 1