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hscscience Maths Adv · Y12
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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

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

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 calculating which deal leaves you with more money? Make a prediction and explain your reasoning.

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02
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.

Comparison table of simple vs compound interest formulas
$A = P(1+r)^n$ 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.
03
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 $n$
04
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.

Value money across time

See why a dollar today is not a dollar next year.

05
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.

Work the examples

Follow three worked problems moving money forwards and backwards in time.

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.

1
$I = Prn = 5{,}000 \times 0.04 \times 3$
Identify $P = 5000$, $r = 0.04$ (4% as decimal), $n = 3$ years.
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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.

1
$A = P(1 + r)^n = 5{,}000 \times (1.04)^3$
$P = 5000$, $r = 0.04$, $n = 3$. Use the compound interest formula.
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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?

1
$r_{\text{monthly}} = \dfrac{0.06}{12} = 0.005$
The rate is 6% per year, but interest compounds monthly. Divide by 12 to get the rate per period.
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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?

Compounding Period Conversion Reference Frequency Periods/year Periodic rate r Total periods n (t years) Annually 1 r = R n = t Semi-annually 2 r = R/2 n = 2t Quarterly 4 r = R/4 n = 4t Monthly 12 r = R/12 n = 12t Daily 365 r = R/365 n = 365t Critical: R is the annual rate (as a decimal). r and n must always use the same time unit.

Avoid the traps

Fix the annual-rate-with-monthly-periods slip that costs marks.

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 $n = 2 \times$ $= 8$.

Drill it and revisit

Run the drill, then return to your opening answer and name what has changed.

Quick-fire practice · 5 calculations

1

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

2

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

3

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

4

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

5

$P = \$20{,}000$, $r = 5\%$ 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?

10
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.

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Practise the time value of money

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

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 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)

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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)

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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)

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📖 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].