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Applications of Compound Interest - Depreciation


All the applications of simple interest and compound interest that we have looked at so far look at situations where amounts appreciate, or increase in value. However, items may also decrease in value- this is known as depreciation. Cars, machinery, computers and electronics are examples of things that depreciate. In other words, they're worth less over time than when you bought them.

Straight-line depreciation

Think back to simple interest. Do you remember how the interest was only calculated on the principal amount so each year you earned the same amount of interest as the previous year? Well, straight-line depreciation is similar except that an item loses a fixed amount each year.

Reducing-Balance Depreciation

The more common form of depreciation is reducing-balance depreciation. The rules for calculating this kind of depreciation are similar to calculating compound interest. The formula is just slightly different- see if you can spot the change. The depreciation is:


Did you notice that instead of $1+r$1+r, the formula is now $1-r$1r?



Question 1

Kathleen deposited $\$6500$$6500 into a new superannuation account. This amount decreased by $2%$2% each year for $3$3 consecutive years.

What was the value of her superannuation after $3$3 years? Give your answer to $2$2 decimal places if necessary.

Think: How do we substitute these values into the depreciation formula?


$A$A $=$= $P\left(1-r\right)^n$P(1r)n
  $=$= $6500\times\left(1-0.02\right)^3$6500×(10.02)3
  $=$= $\$6117.75$$6117.75


Question 2

A car salesman received a total commission of $\$4461$$4461 at the beginning of the month. He expects that since fewer cars are being purchased, his commission will decrease by $2%$2% each month over the next few months.

A)According to his prediction, what will be his commission in $13$13 months time? Give your answer to the nearest dollar.

Think: His commission is depreciating, so let's sub the values into the depreciation formula.

$Commission$Commission $=$= $P\left(1-r\right)^n$P(1r)n
  $=$= $4461\times\left(1-0.02\right)^{13}$4461×(10.02)13
  $=$= $4198.65$4198.65...
  $=$= $\$4199$$4199


B) Using the rounded value from the previous part, what is the predicted decrease in the salesman's monthly commission from this month to in $13$13 months' time?

Think: What is the difference between his two commission amounts?

Do: $4461-4199=\$262$44614199=$262


Question 3

A microwave that costs $\$700$$700, depreciates at $10%$10% p.a.

  1. After how many full years will the value of the microwave be under $525?

  2. After how many full years will the value of the microwave be under $350?

  3. After how many full years will the value of the microwave be under $70?


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