12.05 Applications of simple and compound interest
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.
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%$2.1% in the last quarter, or perhaps $0.8%$0.8% over the last year.
So if a house is valued by a real estate agent to be worth $\$580000$$580000, what would the house be worth one year later, if house prices appreciated by $0.8%$0.8% over the last year?
This final value of the house can be calculated using the compound interest formula.
$A=P\left(1+r\right)^n$A=P(1+r)n
where:
$A$A is the final amount of money (principal and interest together)
$P$P is the principal (the initial amount of money invested)
$r$r is the interest rate expressed as a decimal
$n$n is the number of time periods
So in our example we would have:
| Value after one year: | $A$A | $=$= | $580000\times1.008^1$580000×1.0081 |
| $=$= | $\$584640$$584640 |
When depreciating an item, perhaps a car, you would be subtracting the interest rate.
Appreciation uses a positive rate, $+r$+r.
Depreciation uses a negative rate, $-r$−r
QUESTION 1
If a piece of land appreciates at an average rate of $3.7%$3.7% per annum and its current value is $\$430000$$430000, calculate its value in $3$3 years. Give your answer to the nearest dollar.
QUESTION 2
A vintage collectors item that costs $\$6000$$6000, appreciates at approximately $6.6%$6.6% p.a.
After how many full years, $n$n, will the value of the vintage collectors item be over $\$15000$$15000?
Items can also appreciate or depreciate using straight line or flat rate method. The straight line method assumes the value of depreciation is constant per period.
The straight line graph assumes a $\$20000$$20000 initial value with a depreciation of $\$3000$$3000 p.a.
Practice question
Question 3
A 2012 Holden Commodore is priced at $\$33000$$33000 and depreciates by approximately $\$4000$$4000 per year.
Complete the following table:
Year Price (dollars) $0$0 $33000$33000 $1$1 $\editable{}$ $2$2 $\editable{}$ $3$3 $\editable{}$ $4$4 $\editable{}$ $5$5 $\editable{}$ By this calculation method, will the car ever be worth nothing?
Yes
ANo
BThis depreciation method is known as:
Straight Line Depreciation
AConstant Change Depreciation
BDeclining Balance Depreciation
CZero Return Depreciation
D
Question 4
The graph shows the depreciation of a car's value over 4 years.
What is the initial value of the car?
By how much did the car depreciate each year ?
After how many years will the car be worth $\$14400$$14400 ?
What is the value of the car after 4 years ?
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.
Practice questions
Question 5
A one year sports club membership currently costs $\$332$$332. Calculate the cost in $6$6 years’ time if the inflation rate is on average $2.6%$2.6% per annum. Give your answer correct to the nearest dollar.