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Australia
Year 10

4.01 Compound interest

Lesson

Introduction

An investment with compound interest differs from one with simple interest in the way that it always calculates interest based on the current total amount. In other words, any interest that is earned will be taken into account for the next time interest is calculated.

Compound interest

Since compound interest always calculates the interest based on the current total, we can think of it as a continuous application of simple interest, one period at a time, where we update the principal amount to match the new total after each application.

We can calculate compound interest in this way using a table like the one below. In this example, the starting principal amount is \$1000 with a compounding interest rate of 5\% per annum.

YearPrincipal usedCompound interest earnedNew total
1\$1000\$50\$1050
2\$1050\$52.50\$1102.50
3\$1102.50\$55.13\$1157.63

If we check, we will see that the new total at the end of a year is always equal to 1.05 times the principal amount used, which was equal to the total at the end of the previous year.

With each earning of interest the total increases, increasing the principal amount used which in turn increases the amount of interest earned each year. What this tells us is that compound interest results in a total that increases at an increasing rate.

When calculating compound interest using repeated applications of simple interest, we notice that the new total at the end of each year is simply a percentage increase of the previous year's total.

Using the example with \$1000 invested at an interest rate of 5\% p.a., compounding annually, our calculations take the form:

  • \text{End of 1st year} = 1000 \times 1.05

  • \text{End of 2nd year} = 1000 \times 1.05 \times 1.05

  • \text{End of 3rd year} = 1000 \times 1.05 \times 1.05 \times 1.05

If we don't evaluate each year's total, we can see a pattern form. Since each percentage increase will be the same, we can just apply it as many times as there are periods of interest.

It is also worth noting that since we are multiplying by the same amount for each percentage increase when calculating compound interest, we could also compress these terms using indices, like so:

  • \text{End of 1st year} = 1000 \times 1.05

  • \text{End of 2nd year} = 1000 \times {1.05}^2

  • \text{End of 3rd year} = 1000 \times {1.05}^3

Examples

Example 1

\$3200 is invested for three years at a rate of 6\% p.a., compounding annually.

a

Complete the table row for the third year.

Balance + interestTotal balanceInterest earned
First year-\$3200\$192
Second year\$3200 + \$192\$3392\$203.52
Third year\$3392 + \$ \,⬚\$\,⬚\$ \, ⬚
Fourth year\$\, ⬚\$ \, ⬚-
Worked Solution
Create a strategy

To find the interest, multiply the balance by the interest rate.

Apply the idea

We add the interest and balance from the second year to fill in the first 2 boxes:

Balance + interestTotal balanceInterest earned
First year-\$3200\$192
Second year\$3200 + \$192\$3392\$203.52
Third year\$3392 + \$203.523595.52\$
Fourth year\$\, ⬚\$ \, ⬚-
\displaystyle \text{Interest earned}\displaystyle =\displaystyle \$3595.52 \times 0.06Multiply the balance by the interest rate
\displaystyle =\displaystyle \$215.73Evaluate
Balance + interestTotal balanceInterest earned
First year-\$3200\$192
Second year\$3200 + \$192\$3392\$203.52
Third year\$3392 + \$203.523595.52\$215.73
Fourth year\$\, ⬚\$ \, ⬚-
b

Complete the table row for the Fourth Year, to determine the final value of the investment.

Worked Solution
Create a strategy

Complete the fourth row using the same method used in part (a).

Apply the idea

We add the interest and balance from the third year to fill in the first 2 boxes:

Balance + interestTotal balanceInterest earned
First year-\$3200\$192
Second year\$3200 + \$192\$3392\$203.52
Third year\$3392 + \$203.523595.52\$215.73
Fourth year\$3595.52+\$215.73\$3811.25-
c

Calculate the total interest earned over the three years.

Worked Solution
Create a strategy

Add the values in the "interest earned" column from the table in part (b).

Apply the idea
\displaystyle \text{Total interest}\displaystyle =\displaystyle \$192 + \$203.52 + \$215.73Add the interest
\displaystyle =\displaystyle \$611.25

Example 2

Bill invests \$15\,000 at an interest rate of 2.8 \% p.a., compounding annually. After how many years will Bill's investment be greater than \$17\,500?

A
7 years
B
12 years
C
4 years
D
6 years
Worked Solution
Create a strategy

Find the total amount at the end of each year by increasing the value by the interest rate.

Apply the idea

Start with the initial investment amount and multiply it by 1+0.028=1.028 to increase it.

\displaystyle \text{End of 1st year}\displaystyle =\displaystyle \$15\,000 \times 1.028Increase by the interest rate
\displaystyle =\displaystyle \$15\,420Evaluate

Use the answer of the previous year to find the amount at the end of the next year.

\displaystyle \text{End of 2nd year}\displaystyle =\displaystyle \$15\,420 \times 1.028Increase by the interest rate
\displaystyle =\displaystyle \$15\,851.76Evaluate
\displaystyle \text{End of 3rd year}\displaystyle =\displaystyle \$15\,851.76 \times 1.028Increase by the interest rate
\displaystyle =\displaystyle \$16\,295.61Evaluate
\displaystyle \text{End of 4th year}\displaystyle =\displaystyle \$16\,295.61 \times 1.028Increase by the interest rate
\displaystyle =\displaystyle \$16\,751.89Evaluate
\displaystyle \text{End of 5th year}\displaystyle =\displaystyle \$16\,751.89 \times 1.028Increase by the interest rate
\displaystyle =\displaystyle \$17\,220.94Evaluate
\displaystyle \text{End of 6th year}\displaystyle =\displaystyle \$17\,220.94 \times 1.028Increase by the interest rate
\displaystyle =\displaystyle \$17\,703.13Evaluate

After 6 years the value is greater than \$17\,500. So the correct answer is option D.

Idea summary

We can use repeated applications of simple interest to find the compound interest of an investment or loan.

Appreciation and depreciation

Appreciation is when an item's value increases over time, usually by some fixed percentage of its current value. An example of this is how antiques or collectables become rarer and more expensive over time.

Depreciation is when an item's value decreases over time, again by some fixed percentage of its current value. An example of this is how a new car will lose value over time as it gets older and more out of date.

Since we often calculate appreciation and depreciation in terms of annual percentage increases or decreases, we can model their change in value the same way that we do with compound interest.

Looking at it one year at a time, appreciation is simply a repeated application of percentage increases, exactly like compound interest.

Examples

Example 3

A new book depreciates in value by \dfrac{33}{5} \, \% every month. The book is currently valued at \$60.

a

How much will the book be valued at in one month's time?

Worked Solution
Create a strategy

Decrease the value by the percentage rate.

Apply the idea

If the book depreciates by \dfrac{33}{5} \%, the remaining percentage will be (100 - \dfrac{33}{5}) \, \%= 93.4 \%.

\displaystyle \text{Value in one month}\displaystyle =\displaystyle \$60 \times 0.934Multiply by 0.934
\displaystyle =\displaystyle \$56.04Evaluate
b

How much will the book be valued at in two months' time?

Worked Solution
Create a strategy

Decrease the value after 1 month by the percentage rate again.

Apply the idea
\displaystyle \text{Value}\displaystyle =\displaystyle \$56.04 \times 0.934Multiply by 0.934
\displaystyle =\displaystyle \$52.34Evaluate
Reflect and check

We could also have multiplied the original value by 0.934 twice, one for each month:

\displaystyle \text{Value}\displaystyle =\displaystyle \$60 \times 0.934 \times 0.934Multiply by 0.934 twice
\displaystyle =\displaystyle \$52.34Evaluate
c

Which of the following expressions can be used to calculate the value of the book in 4 months' time? Select all that apply.

A
60 \times 0.9934 \times 4
B
60 - 60 \times (0.066)^4
C
60 \times (0.934)^4
D
60 \times 0.934 \times 0.934 \times 0.934 \times 0.934
Worked Solution
Create a strategy

Use the pattern from parts (a) and (b).

Apply the idea

In part (a), we multiplied \$60 by 0.934 one time to find the book's value after one month. In part (b), we multiplied \$60 by 0.934 two times to find the book's value after two months.

If we want to find the value after 4 months, we will have to multiply the book's original value by 0.934 four times.

\displaystyle \text{Value}\displaystyle =\displaystyle 60 \times 0.934 \times 0.934 \times 0.934 \times 0.934Multiply by 0.934 four times
\displaystyle =\displaystyle 60 \times (0.934)^4Write in index form

So the correct answers are options C and D.

d

Using either of the expressions found in the previous part, calculate the value of the book in 4 months' time.

Worked Solution
Create a strategy

Use the expression found in part (c).

Apply the idea
\displaystyle \text{Value}\displaystyle =\displaystyle \$60 \times (0.934)^4Use the compressed expression
\displaystyle =\displaystyle \$45.66Evaluate
Idea summary

Appreciation is when an item's value increases over time, usually by a fixed percentage. Appreciation requires repeated applications of percentage increase.

Depreciation is when an item's value decreases over time, usually by a fixed percentage. Depreciation requires repeated applications of percentage decrease.

Different periods

So far, the examples we have looked at have all had interest calculated annually. However, interest can be calculated over any period and the interest rate per period will need to match it.

To match the interest rate to the period duration, we can use interest rate conversions.

It is also important to match the total number of periods to the duration of the investment. To do this, we can simply divide the duration of the investment by the duration of the period.

Examples

Example 4

Victoria borrows \$35\,000 at a rate of 4.8\% p.a, compounding monthly.

a

After 4 months, Victoria repays the loan all at once. How much money does she pay back in total?

Worked Solution
Create a strategy

Since interest is compounding monthly, interest will be calculated 12 times a year. So we need to divide the rate by 12.

Apply the idea
\displaystyle \text{Monthly rate}\displaystyle =\displaystyle \dfrac{4.8}{12} \, \%Divide by 12
\displaystyle =\displaystyle 0.4 \%Evaluate

To find the value after 4 months, we need to increase the value by 0.4\% four times.

\displaystyle \text{Amount to be repaid}\displaystyle =\displaystyle \$35\,000 \times 1.004\times 1.004\times 1.004\times 1.004Increase by the rate 4 times
\displaystyle =\displaystyle \$35 \, 563.37Evaluate
b

How much interest was generated on the loan over the four months?

Worked Solution
Create a strategy

Subtract the initial amount borrowed from the final amount repaid.

Apply the idea
\displaystyle \text{Interest}\displaystyle =\displaystyle \$35\,563.37 - \$35\, 000Subtract initial amount
\displaystyle =\displaystyle \$563.37
Idea summary

To match the interest rate to the period duration, we can use interest rate conversions.

Compounded rateDivide annual rate by
\text{Daily}365
\text{Weekly}52
\text{Monthly}12
\text{Quarterly}4
\text{Half-yealy}2

Outcomes

ACMNA229

Connect the compound interest formula to repeated applications of simple interest using appropriate digital technologies

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