Simple interest and compound interest do not only apply to bank loans and investments. Concepts such as appreciation & depreciation as well as inflation are examples of practical applications of the compound and simple interest.

Ideas

Appreciation and depreciation
Straight line method
Inflation

Appreciation and depreciation

The compound formula is just one of a variety of methods for calculating the value of appreciating or depreciating assets.

When an object or investment is said to appreciate, this means it increases in value by a given average percentage.

When an object or investment is said to depreciate, this means it decreases in value by a given average percentage.

For example, property prices in a given suburb might have appreciated by 2.1\% in the last quarter, or perhaps 0.8\% over the last year. So if a house is valued by a real estate agent to be worth \$580\,000, what would the house be worth one year later, if house prices appreciated by 0.8\% over the last year?

This final value of the house can be calculated using the compound interest formula: A=P(1+r)^n where A is the final amount of money, P is the principal, r is the interest rate per year, expressed as a decimal and n is the number of time periods.

So in our example we would have:

\displaystyle A	\displaystyle =	\displaystyle 580\,000 \times 1.008^1
	\displaystyle =	\displaystyle \$584\,640

When depreciating an item, perhaps a car, you would be subtracting the interest rate.

Appreciation uses a positive rate, + r. Depreciation uses a negative rate, - r.

Examples

Example 1

If a piece of land appreciates at an average rate of 3.7\% per annum and its current value is \$430\,000, calculate its value in 3 years. Give your answer to the nearest dollar.

Worked Solution

Create a strategy

We can use the formula: A=P(1 + r)^n

Apply the idea

We are given: P=430\,000,\,r=0.037,\,n=3.

\displaystyle \text{Future value}	\displaystyle =	\displaystyle 430\,000 (1 + 0.037)^{3}	Substitute the values
	\displaystyle =	\displaystyle 430\,000 \times 1.0037^3	Evaluate the sum inside the brackets
	\displaystyle =	\displaystyle \$479\,518	Evaluate and round to nearest dollar

Example 2

A vintage collectors item that costs \$6000,appreciates at approximately 6.6\% p.a. After how many full years, will the value of the vintage collectors item be over \$15\,000?

Worked Solution

Create a strategy

We can use the formula: A=P(1 + r)^n

Apply the idea

We are given: A=15\,000,\,P=6000,\,r=0.066.

\displaystyle 15\,000	\displaystyle =	\displaystyle 6000 (1 + 0.066)^{n}	Substitute the values
\displaystyle 15\,000	\displaystyle =	\displaystyle 6000 \times 1.066^{n}	Evaluate the sum inside the brackets
\displaystyle \dfrac{15\,000}{6000}	\displaystyle =	\displaystyle \dfrac{6000}{6000} \times 1.066^{n}	Divide both sides by 6000
\displaystyle 2.5	\displaystyle =	\displaystyle 1.066^{n}	Evaluate

To find the value of n we need to substitute a number that will make the 1.006^{n} greater than 2.5.

\displaystyle 1.066^{15}	\displaystyle >	\displaystyle 2.5	Substitute the value of n
\displaystyle 2.6	\displaystyle >	\displaystyle 2.5	Evaluate

So the correct value of n is 15.

Idea summary

Compound interest formula:

\displaystyle A=P(1+ r)^n

\bm{A}

is the final amount of money

\bm{P}

is the principal

\bm{r}

is the interest rate expressed as a decimal

\bm{n}

is the number of time periods

Straight line method

Items can also appreciate or depreciate using straight line or flat rate method. The straight line method assumes the value of depreciation is constant per period.

\text{Number of periods (years)}

5000

10000

15000

20000

\text{Values (\$)}

The straight line graph assumes a \$20\,000 initial value with a depreciation of \$3000 p.a.

Examples

Example 3

A 2012 Holden Commodore is priced at \$33\,000 and depreciates by approximately \$4000 per year.

Complete the following table:

Year	Price (dollars)
0	33\,000
1
2
3
4
5

Worked Solution

Create a strategy

Subtract the depreciation amount from the original value for the first year, and then subtract the amount from the remaining value of the car every year.

Apply the idea

Year	Price (dollars)
0	33\,000
1	29\,000
2	25\,000
3	21\,000
4	17\,000
5	13\,000

By this calculation method, will the car ever be worth nothing?

Worked Solution

Apply the idea

If we keep subtracting the same amount annually beyond 5 years, the car will be worth nothing.

This depreciation method is known as:

Straight line depreciation

Constant change depreciation

Declining balance depreciation

Zero return depreciation

Worked Solution

Create a strategy

As we can see in part (a), there is a constant value of depreciation every year.

Apply the idea

The correct answer is option A: Straight line depreciation.

Example 4

The graph shows the depreciation of a car's value over 4 years.

\text{Age (years)}

9000

18000

27000

36000

\text{Value } (\$)

What is the initial value of the car?

Worked Solution

Create a strategy

Find the value at year 0.

Apply the idea

The y-intercept on the graph is (0,36\,000). So at year 0 the value was \$36\,000.

The initial value is \$36\,000.

By how much did the car depreciate each year?

Worked Solution

Create a strategy

Subtract the value of the car after one year from its initial value.

Apply the idea

First we need to know how much each interval on the vertical axis is increasing by.

\text{Age (years)}

9000

18000

27000

36000

\text{Value } (\$)

Each \$9000 on the vertical axis is divided into 5 intervals. So each interval represents 9000\div 5=\$1800.

After 1 year, the value of the car is 1 interval above \$27\,000. So the value is 27\,000+1800=\$28\,800.

Now we are ready to subtract the value of the car after one year from its initial value.

\displaystyle \text{Depreciation}	\displaystyle =	\displaystyle 36\,000 - 28\,800	Subtract the values
	\displaystyle =	\displaystyle \$7200	Evaluate

After how many years will the car be worth \$14\,400?

Worked Solution

Create a strategy

Divide the amount of depreciation that we want by the depreciation rate per year.

Apply the idea

The depreciation amount can be found by subtracting \$14\,400 from the initial value:

\displaystyle \text{Depreciation}	\displaystyle =	\displaystyle 36\,000-14\,400	Subtract 14\,400 from the initial value
	\displaystyle =	\displaystyle \$21\,600	Evaluate

To find how many years this will take, divide the depreciation by the depreciation rate of \$7200 per year.

\displaystyle \text{Years}	\displaystyle =	\displaystyle \dfrac{21\,600}{7200}	Divide by the depreciation per year
	\displaystyle =	\displaystyle 3	Evaluate

The car will be worth \$14\,400 after 3 years.

What is the value of the car after 4 years ?

Worked Solution

Create a strategy

In the graph, find the y-value of the point in the depreciation line which corresponds to 4 years.

Apply the idea

\text{Age (years)}

9000

18000

27000

36000

\text{Value } (\$)

The value that corresponds to the 4 years hits the vertical axis 1 interval below 9000.

We learnt from part (b) that each interval represents \$1800.

So the corresponding value is 9000-1800=\$7200.

The car will be worth \$7200 after 4 years.

Idea summary

The straight line method assumes the value of depreciation is constant per period.

Inflation

Inflation is a very similar concept to appreciation, but instead of looking at the increase in value of an investment, we instead examine the increase in the prices of goods and services in an economy over time.

The rate of inflation is expressed as a percentage. As inflation causes an increase in prices, the compound interest formula can be used again.

Often, the rate of inflation in a particular country is reported as the average annual inflation rate.

Examples

Example 5

A one year sports club membership currently costs \$332. Calculate the cost in 6 years time if the inflation rate is 2.6\% per annum. Round your answer to the nearest dollar.

Worked Solution

Create a strategy

We can use the formula: A=P(1+r)^{n}

Apply the idea

We are given: P=332,\, r=0.026,\, n= 6

\displaystyle \text{Future cost}	\displaystyle =	\displaystyle 332 (1+ 0.026)^{6}	Substitute the values
	\displaystyle =	\displaystyle 332 \times 1.026 ^{6}	Evaluate the sum inside the brackets
	\displaystyle =	\displaystyle \$387	Evaluate

Idea summary

Inflation examine the increase in the prices of goods and services in an economy over time.

Outcomes

U2.AoS4.8

scientific notation, exact and approximate answers, significant figures and rounding

4.08 Interest applications

Introduction

Ideas

Appreciation and depreciation

Examples

Example 1

Example 2

Straight line method

Examples

Example 3

Example 4

Inflation

Examples

Example 5

Outcomes

U2.AoS4.8

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