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Financial Maths, STD1 F1 2021 HSC 19 v1

Maya purchased a motorbike for $12 000. The value of the motorbike decreases according to a linear model. The graph shows the value of the motorbike, $\(V\), against the time, \(t\) months, since it was purchased.
 

  1. By how much does the value of the motorbike decrease every 10 months?   (1 mark)

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  2. Find the value of the motorbike after 4 years.   (1 mark)

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  3. Identify ONE problem with using this model to determine the value of Maya's motorbike over time.   (1 mark)

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Show Answers Only

a.     \(\$1000\)

b.     \(\$7200\)

c.     \(\text{Motorbike will have a negative value after 120 months.}\)

Show Worked Solution

a.    \(\text{Total decrease over 100 months}= 12\,000-2000= \$10\,000\)

\(\text{Decrease per 10 months}= \dfrac{10\,000}{10}= \$1000\)
 

b.    \(\text{4 years} = 4 \times 12 = 48\ \text{months}\)

\(\text{Depreciation rate}= \$100\ \text{per month}\)

\(V = 12\,000-(100 \times 48)=\$7200\)
 

c.    \(\text{Model limitations:}\)

\(\text{The linear model will eventually predict a value of \$0 (at 120 months) and negative}\)

\(\text{values beyond that, which is unrealistic.}\)

Filed Under: Simple Interest and S/L Depreciation (Std 1-X) Tagged With: Band 4, Band 6, smc-6965-20-Straight-line Depreciation

Financial Maths, STD1 F1 2023 HSC 6 MC v1

A courier van was valued at $48 000 when new. The value of the van depreciates at a rate of 18 cents per kilometre travelled.

What is the value of the van after it has travelled a total distance of 95 400 km?

  1. $17 172
  2. $30 828
  3. $37 440
  4. $65 172
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Depreciation} = 95\,400\times\dfrac{18}{100}=\$17\,172\)

\(\text{Value} = 48\,000-17\,172=\$30\,828\)

\(\Rightarrow B\)

Filed Under: Simple Interest and S/L Depreciation (Std 1-X) Tagged With: Band 4, smc-6965-20-Straight-line Depreciation

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