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6.03 Analysing depreciation

Interactive practice questions

A 2012 Holden Commodore is priced at $\$33000$$33000 and depreciates by approximately $\$4000$$4000 per year.

a

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 $\editable{}$
b

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

Yes

A

No

B
c

This depreciation method is known as:

Straight Line Depreciation

A

Constant Change Depreciation

B

Declining Balance Depreciation

C

Zero Return Depreciation

D
Easy
2min

The spectator attendance at an annual sporting event was recorded for four consecutive years from its first year of running: $44500$44500, $43800$43800, $43100$43100, $42400$42400

Easy
1min

It is estimated that a house purchased for $\$254500$$254500 will depreciate by an average of $\$9400$$9400 each year.

Easy
1min

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

Easy
3min
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Outcomes

3.3.1.4

use arithmetic sequences to model and analyse practical situations involving linear growth or decay, such as analysing a simple interest loan or investment, calculating a taxi fare based on the flag fall and the charge per kilometre, or calculating the value of an office photocopier at the end of each year using the straight-line method or the unit cost method of depreciation

3.3.2.4

use geometric sequences to model and analyse (numerically or graphically only) practical problems involving geometric growth and decay (logarithmic solutions not required), such as analysing a compound interest loan or investment, the growth of a bacterial population that doubles in size each hour or the decreasing height of the bounce of a ball at each bounce; or calculating the value of office furniture at the end of each year using the declining (reducing) balance method to depreciate

4.1.1.3

solve problems involving compound interest loans or investments, e.g. determining the future value of a loan, the number of compounding periods for an investment to exceed a given value, the interest rate needed for an investment to exceed a given value

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