General Mathematics · Past QCAA questions All simple familiar questions
2021 · Paper 1 · Simple familiar

Question 20

[4 marks] Scientific calculator
Topic guidesPremiumTopic guide 5 · Where the first term sits
Game plansPremiumGame plan 3 · Growth and decay in sequences
The question

A farmer bought a tractor for $45 100 at the start of 2012. It depreciates by $2700 each year.

Identify and use a mathematical model to determine the value of the tractor at the start of 2021.

Watch the situation first

Ten values, so only nine drops

Each bar is the tractor’s value at the start of a year. Watch the same amount come off every year, then count how many times it comes off before 2021.

Start of 2012: the tractor is worth $45 100 Every year the value drops by the same $2700 Number the terms: 2012 is t₁, so 2021 is t₁₀ A constant difference: an arithmetic sequence Check: a linear function gives the same value 123456789 45 100 42 400 39 700 37 000 34 300 31 600 28 900 26 200 23 500 20 800 45 100 $20 800 2012 2013 2014 2015 2016 2017 2018 2019 2020 2021 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10 t1 = 45 100, d = −2700, n = 10 t10 = 45 100 + (10 − 1)(−2700) = 45 100 − 24 300 = $20 800 y = −2700x + 45 100 x = 9 years after 2012 y = −2700 × 9 + 45 100 −2700 each year: 45 100, 42 400, 39 700, … From t₁ to t₁₀ there are 9 drops, not 10 Formula book: tn = t1 + (n − 1)d Check: −24 300 + 45 100 = 20 800, the same value

The first value is the purchase price, before any depreciation. That is why the tenth value has only nine drops in it.

Now for the mathematics

Match each year to its term, then test the working

On the left, slide through the years and watch the term number and the years since purchase move together, always one apart. On the right, pick the model, the working and the final line. Each wrong one is an easy slip to make.

45 100 42 400 39 700 37 000 34 300 31 600 28 900 26 200 23 500 20 800 2012201320142015201620172018201920202021 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10
Arithmetic sequence, term number
Linear function, years after purchase

Check an easy slip

Pick one. The feedback says exactly what the marker would see.

QCAA marking guide · 4 marks

Two marks before you touch the calculator

Try each step yourself before you reveal it. The guide gives a mark for naming the model and one for its parameters, then one for the substitution and one for the value with its units.

Step 11 mark

The same amount comes off every year, so the values at the start of each year have a common difference. Say so, and write the rule from the formula book.

Arithmetic sequence:

$$t_n=t_1+(n-1)d$$

The guide also accepts a linear function, $y=mx+c$, or a recurrence relation that takes 2700 off nine times. Any of the three works, as long as you name it.

Marker: correctly identifies the model

Step 21 mark

The purchase price is the first term. The value falls, so the common difference is negative. Count the terms: the start of 2012 is term 1, 2013 is term 2, and so on up to 2021.

$$t_1=45\,100,\quad d=-2700,\quad n=10$$

For the linear function the parameters are $c=45\,100$, $m=-2700$ and $x=9$, the number of years after the start of 2012. This mark can be implied by your substitution.

Marker: correctly identifies the parameters $t_1$, $d$ and $n$

Step 31 mark

Substitute all three parameters into the rule.

$$t_{10}=45\,100+(10-1)(-2700)$$

The guide writes this as $45\,100-2700(10-1)$, which is the same thing. Follow-through applies, so if your $n$ was wrong you can still earn this mark and the next by substituting and evaluating correctly.

Marker: substitutes values into appropriate model

Step 41 mark

Evaluate, then answer in context with the units.

$$\begin{aligned}t_{10}&=45\,100-24\,300\\&=20\,800\end{aligned}$$

The tractor will be worth $20 800 at the start of 2021.

As a check, the linear model gives $y=-2700\times9+45\,100=20\,800$, the same value.

Marker: determines value of tractor, including units

Putting it all together

Each phrase of the question gave you something

“bought a tractor for $45 100 at the start of 2012”
The first term, $t_1=45\,100$. The start of 2012 is term 1.
“depreciates by $2700 each year”
The same amount every year, so a common difference, $d=-2700$. Negative, because the value falls.
“Identify … a mathematical model”
Name it in words. “Arithmetic sequence” is worth a mark on its own.
“at the start of 2021”
Term 10. From the start of 2012 that is nine years, so nine drops.
“the value of the tractor”
A dollar amount in a sentence: $20 800.
What makes this simple familiar

One rule from the formula book, with the first term and the common difference handed to you in the first sentence. The only real decision is which term the start of 2021 is. The 2021 subject report asked for more practice at defining variables and explaining with mathematical reasoning, and that is what the first two marks reward here.

If I had to guess where you might lose marks, it would be the term number. From 2012 to 2021 is nine years, so it is tempting to put $n=9$ into the rule. That gives $23 500, the value at the start of 2020. Taking off ten lots of 2700 gives $18 100, the value a year too late. Either way, follow-through can still save two marks. The last easy loss is leaving off the dollar sign, because the fourth mark asks for units.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2021 QCAA General Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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