Worked Solutions
Algebra — Worked Solutions (Preliminary Maths Standard)
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Worked examples for Preliminary Maths Standard algebra. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Substituting into a formula
Question
The stopping distance of a car is given by $d = 0.4v + \dfrac{v^2}{20}$, where $d$ is the distance in metres and $v$ is the speed in metres per second. Find the stopping distance when the car is travelling at $v = 30$ m/s.
Solution
Substitute $v = 30$ straight into the formula and evaluate each term.
$d = 0.4(30) + \dfrac{30^2}{20}$.
First term: $0.4 \times 30 = 12$. Second term: $\dfrac{900}{20} = 45$.
$d = 12 + 45 = 57$ m.
Write the units. A stopping distance with no metres attached loses the final mark.
The formula is already done for us — our job is just to put the right number in the right place. We're told $v = 30$, so everywhere we see $v$ we write $30$.
$d = 0.4(30) + \dfrac{30^2}{20}$.
Take it one term at a time so nothing slips. The first term is $0.4 \times 30 = 12$. For the second, square first ($30^2 = 900$) then divide ($900 \div 20 = 45$).
Adding them: $d = 12 + 45 = 57$ metres.
Doing it term-by-term and keeping the units the whole way through is what stops careless slips on a question that's really just careful arithmetic.
Substitute $v = 30$.
- $d = 0.4(30) + \dfrac{30^2}{20}$
- $0.4 \times 30 = 12$
- $\dfrac{900}{20} = 45$
- $d = 12 + 45 = 57$ m
Where the marks go
- 1 mark: Correct substitution of $v = 30$ into the formula
- 1 mark: Correct evaluation of both terms ($12$ and $45$)
- 1 mark: Correct answer with units ($57$ m)
Key idea
Substitution means replacing the variable with its value, then evaluating each term carefully — and always include units in the final answer.
Example 2 — Linear equation from a model
Question
A plumber charges a $\$80$ call-out fee plus $\$65$ per hour. The total cost $C$ for a job of $h$ hours is $C = 80 + 65h$. A customer is charged $\$340$. Set up and solve a linear equation to find how many hours the plumber worked.
Solution
The total is $\$340$, so set the cost formula equal to $340$ and solve for $h$.
$80 + 65h = 340$.
Subtract the call-out fee: $65h = 260$.
Divide: $h = \dfrac{260}{65} = 4$.
The plumber worked $4$ hours. Undo the constant first, then the coefficient — that order keeps the algebra clean.
We want to find $h$, the number of hours. The clue is that the total charge equals $\$340$, so we set the whole cost expression equal to that total.
$80 + 65h = 340$.
To get $h$ on its own, we peel away everything attached to it. First the $\$80$ call-out fee, which is added on, so we subtract it from both sides: $65h = 340 - 80 = 260$.
Now $h$ is only multiplied by $65$, so we divide both sides by $65$: $h = \dfrac{260}{65} = 4$.
So the plumber worked $4$ hours. The reason we subtract before we divide is that we reverse the operations in the opposite order they were applied — like taking off your shoes before your socks.
Set cost $= 340$.
- $80 + 65h = 340$
- $65h = 260$
- $h = \dfrac{260}{65} = 4$
Plumber worked $4$ hours.
Where the marks go
- 1 mark: Sets up the correct equation $80 + 65h = 340$
- 1 mark: Subtracts the call-out fee to get $65h = 260$
- 1 mark: Divides to solve $h = 4$
- 1 mark: States the answer in context (4 hours)
Key idea
To solve a linear model, set the formula equal to the given value, then undo the operations in reverse — subtract the constant, then divide by the coefficient.