Worked Solutions
Algebra — Worked Solutions (HSC Maths Standard 2)
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Worked examples for HSC Maths Standard 2 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 — Linear and non-linear modelling
Question
A drone hire company charges a fixed booking fee plus a constant rate per hour. Hiring the drone for 2 hours costs \$95, and hiring it for 5 hours costs \$200.
(a) Write a linear equation for the total cost $C$ dollars in terms of the number of hours $h$.
(b) The company also offers a maintenance plan whose cost follows $M = 4h^2 + 20$. For how many whole hours is the hire cost $C$ cheaper than the maintenance plan $M$?
Solution
Find the rate first. Cost rises by $200 - 95 = 105$ over $5 - 2 = 3$ hours, so the rate is $\dfrac{105}{3} = 35$ per hour.
Back-substitute to get the fixed fee: at $h = 2$, $95 = 35(2) + b$, so $b = 25$. Hence $C = 35h + 25$.
(b) Want $C < M$: $35h + 25 < 4h^2 + 20$, i.e. $4h^2 - 35h - 5 > 0$. The positive root is $h = \dfrac{35 + \sqrt{1225 + 80}}{8} = \dfrac{35 + \sqrt{1305}}{8} \approx 8.86$.
So $C < M$ once $h > 8.86$, meaning from 9 whole hours onward. Read the inequality direction carefully — that's where marks are lost.
Let's build the linear model. A linear cost has the form $C = (\text{rate})h + (\text{fixed fee})$. The rate is how fast the cost changes, which is the change in cost divided by the change in hours: $\dfrac{200 - 95}{5 - 2} = \dfrac{105}{3} = 35$ dollars per hour.
Now we need the fixed fee. Use a known point — at $h = 2$ the cost is \$95: $95 = 35 \times 2 + b = 70 + b$, so $b = 25$. That gives $C = 35h + 25$.
(b) "$C$ cheaper than $M$" means $C < M$, so $35h + 25 < 4h^2 + 20$. Rearranging into the usual form: $4h^2 - 35h - 5 > 0$. Solving the quadratic, the positive crossing point is $h = \dfrac{35 + \sqrt{1305}}{8} \approx 8.86$. The $h^2$ term grows fastest, so beyond that point $M$ overtakes $C$ — the hire is cheaper from 9 hours onward.
(a) Linear model $C = mh + b$.
- Rate: $m = \dfrac{200 - 95}{5 - 2} = 35$
- Fee: $95 = 35(2) + b \Rightarrow b = 25$
- $C = 35h + 25$
(b) $C < M$:
- $35h + 25 < 4h^2 + 20$
- $4h^2 - 35h - 5 > 0$
- Positive root $h = \dfrac{35 + \sqrt{1305}}{8} \approx 8.86$
- Cheaper from $h = 9$ hours onward
Where the marks go
- 1 mark: Correct rate of \$35 per hour
- 1 mark: Correct linear equation $C = 35h + 25$
- 1 mark: Solves $C < M$ and states 9 whole hours onward
Key idea
A linear model is rate × quantity + fixed amount; comparing it with a non-linear model means solving an inequality and checking which side satisfies it.
Example 2 — Simultaneous equations
Question
A school canteen sells two snack packs. On Monday it sells 12 fruit packs and 8 muesli packs for total takings of \$76. On Tuesday it sells 9 fruit packs and 14 muesli packs for total takings of \$83.
Let $f$ be the price of a fruit pack and $m$ the price of a muesli pack. Set up and solve a pair of simultaneous equations to find the price of each pack.
Solution
Set up the two equations straight from the takings:
$12f + 8m = 76$ … (1) $9f + 14m = 83$ … (2)
Use elimination. Multiply (1) by 3 and (2) by 4 to match the $f$ terms: $36f + 24m = 228$ and $36f + 56m = 332$. Subtract: $32m = 104$, so $m = 3.25$.
Back-substitute into (1): $12f + 8(3.25) = 76 \Rightarrow 12f = 50 \Rightarrow f = 4.166\ldots$ — check that: $12f = 76 - 26 = 50$, $f \approx 4.17$.
So a fruit pack is about \$4.17 and a muesli pack \$3.25. Always verify both numbers in the other equation before you commit.
First, turn the words into equations. Each day's takings are (number of fruit packs)$\times f$ plus (number of muesli packs)$\times m$:
$12f + 8m = 76$ … (1) $9f + 14m = 83$ … (2)
To solve them together we eliminate one variable. I'll line up the $f$ terms: multiply (1) by 3 to get $36f + 24m = 228$, and (2) by 4 to get $36f + 56m = 332$. Subtracting removes $f$: $32m = 104$, so $m = 3.25$.
Now substitute back to find $f$. Using (1): $12f + 8(3.25) = 76$, so $12f + 26 = 76$, giving $12f = 50$ and $f \approx 4.17$. The reason elimination works is that making the coefficients equal lets us cancel that variable cleanly. So fruit packs are \$4.17 and muesli packs \$3.25.
Equations:
- $12f + 8m = 76$ … (1)
- $9f + 14m = 83$ … (2)
Eliminate $f$:
- (1)$\times 3$: $36f + 24m = 228$
- (2)$\times 4$: $36f + 56m = 332$
- Subtract: $32m = 104 \Rightarrow m = 3.25$
Back-substitute:
- $12f + 8(3.25) = 76 \Rightarrow 12f = 50 \Rightarrow f \approx 4.17$
Fruit \$4.17, muesli \$3.25.
Where the marks go
- 1 mark: Correct pair of simultaneous equations from the context
- 1 mark: Valid elimination/substitution step matching coefficients
- 1 mark: Correct muesli price $m = 3.25$
- 1 mark: Correct fruit price $f \approx 4.17$
Key idea
Two unknowns need two equations; matching coefficients lets you eliminate one variable, then back-substitute for the other.