Compound-interest revision | Year 12

01 / LEARN

Understand the idea.

The balance grows by a multiplier

Compound interest earns interest on earlier interest as well as on the starting principal. At a periodic rate $i$ , each closing balance becomes the next opening balance, so every period multiplies the account by $1 + i$ .

Use $i$ as a decimal and count matching periods: annual compounding for three years means $n = 3$ . The model $B_{n} = P {(1 + i)}^{n}$ assumes a constant rate and no deposits, withdrawals or fees; keep calculator precision until the final amount.

B_{n} = P {(1 + i)}^{n}

Read the result correctly

The balance includes the money originally invested. To isolate the interest earned, subtract the principal: $I = B_{n} - P$ . To recover a starting principal from a future balance, reverse the growth by dividing: $P = \frac{B_{n}}{{(1 + i)}^{n}}$ .

Depreciation uses the fraction remaining each period. A loss rate $d$ gives multiplier $1 - d$ , applied repeatedly to the current value. For nominal annual interest compounded monthly, first use $i = \frac{j}{12}$ and count months, rather than combining annual rates with monthly counts.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Build the model

AUD 1,200 earns 4% interest annually for 3 years. Find the final balance and total interest.

Worked solution

1
Identify the annual rate and period count.
$P = 1200, i = 0.04, n = 3$
Annual crediting gives one interest period per year.
2
Grow the principal for three periods.
$\begin{aligned} B_{3} & = 1200 {(1.04)}^{3} \\ = 1349.8368 \end{aligned}$
Each annual multiplier acts on the whole preceding balance.
3
Separate the interest from the final balance.
$\begin{aligned} I & = 1349.8368 - 1200 \\ = 149.8368 \end{aligned}$
The original AUD 1,200 is part of the balance, not interest.
4
Round both money amounts to cents.
$B_{3} \approx AUD 1349.84, I \approx AUD 149.84$
Rounding at the end preserves the compound calculation.

Answer

AUD 1,349.84; interest AUD 149.84

EXAMPLE 02

Use the model to make a decision

A tool costing AUD 2,500 loses 12% of its value each year. Estimate its value after 4 years.

Worked solution

1
Find the proportion of value retained each year.
$1 - 0.12 = 0.88$
A 12% loss leaves 88% of the current value.
2
Apply the retained-value multiplier four times.
$\begin{aligned} V_{4} & = 2500 {(0.88)}^{4} \\ = 1499.2384 \end{aligned}$
Each loss is calculated on the reduced value from the previous year.
3
State the estimated value to cents.
$V_{4} \approx AUD 1499.24$
The calculated amount is the remaining value after four years.

Answer

AUD 1,499.24

03 / YOUR TURN

Try it yourself.

Check your understanding

AUD 900 grows to AUD 1,050. What interest has been earned?

Show answer & reasoning

AUD 150.00

The AUD 1,050 balance includes the AUD 900 originally invested. Interest is the increase, $1050 - 900 = 150$ , so AUD 150.00 has been earned.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Find the balance of AUD 3,400 after 5 years at 3.5% compounded annually.
3 marks
AUD 900 grows to AUD 1,050. What interest has been earned?
3 marks
An account reaches AUD 8,000 after 6 years at 5% compounded annually. Find its original principal.
3 marks
AUD 2,000 earns a nominal annual rate of 6%, compounded monthly, for 2 years. Find the balance.
3 marks
A model uses $B = 6000 {(1.02)}^{8}$ . State the periodic rate and the number of periods.
3 marks
Compare AUD 5,000 invested for 4 years at 4% simple interest and at 4% annual compound interest.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the balance of AUD 3,400 after 5 years at 3.5% compounded annually.
3 marks
Use $i = 0.035$ and $n = 5$ .
$3400 {(1.035)}^{5} =$ AUD 4,038.13.
AUD 4,038.13
AUD 900 grows to AUD 1,050. What interest has been earned?
3 marks
$Interest = final balance - principal$ .
$1050 - 900 = 150$ .
AUD 150.00
An account reaches AUD 8,000 after 6 years at 5% compounded annually. Find its original principal.
3 marks
Rearrange $P = \frac{B}{{(1 + i)}^{n}}$ .
$P = \frac{8000}{{(1.05)}^{6}} =$ AUD 5,969.72.
AUD 5,969.72
AUD 2,000 earns a nominal annual rate of 6%, compounded monthly, for 2 years. Find the balance.
3 marks
The monthly rate is $\frac{0.06}{12} = 0.005$ and there are 24 months.
$2000 {(1.005)}^{24} =$ AUD 2,254.32.
AUD 2,254.32
A model uses $B = 6000 {(1.02)}^{8}$ . State the periodic rate and the number of periods.
3 marks
The multiplier 1.02 equals $1 + i$ , so $i = 0.02$ .
The exponent counts 8 periods. The formula alone does not identify whether the periods are months or years.
2% per period; 8 periods.
Compare AUD 5,000 invested for 4 years at 4% simple interest and at 4% annual compound interest.
3 marks
Simple balance $= 5000 (1 + 0.04 \times 4) = 5800$ .
Compound balance $= 5000 {(1.04)}^{4} =$ AUD 5,849.29.
Subtract the simple balance from the compound balance.
Simple: AUD 5,800.00; compound: AUD 5,849.29; compound is AUD 49.29 higher.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics financial models: investments and annuities, superannuation, income streams and borrowing. Original practice uses supplied hypothetical rates, taxes and fees; statutory or commercial conditions are not inferred.

Official curriculum source ↗

In this lesson

Understand the idea.

The balance grows by a multiplier

Read the result correctly

See the method in action.

Build the model

Identify the annual rate and period count.

Grow the principal for three periods.

Separate the interest from the final balance.

Round both money amounts to cents.

Use the model to make a decision

Find the proportion of value retained each year.

Apply the retained-value multiplier four times.

State the estimated value to cents.

Try it yourself.

Check your understanding

Keep practising