Mathematics Standard • Year 11 • Module 1 • Lesson 8
Formula and Equation Synthesis
Practise HSC-style writing on mixed algebra strategy — three multi-mark short answers and one extended response with marking criteria.
1. Short-answer questions
1.1 A hire company charges $35 plus $18 per hour. The total cost is $143. Find the hire time. Explain in one sentence which strategy you chose and why. 4 marks Band 3-4
1.2 Use d = st to find the time when d = 210 km and s = 70 km/h. You must rearrange the formula before substituting. 3 marks Band 3
1.3 A table has outputs 14, 20, 26, 32 for inputs 0, 1, 2, 3.
(a) Write a formula linking output y to input x.
(b) Test your formula using input 3.
(c) Predict the output for input 9. 4 marks Band 4
2. Extended response
2.1 A small landscaping business uses three different cost rules.
Lawn mowing: M = 25 + 18a, where M is cost in dollars and a is lawn area in 10 m² units.
Hedge trimming: T = 40h, where T is cost in dollars and h is hours worked.
Garden table: for new clients, area covered (m²) at hours 0, 1, 2, 3 is 0, 20, 40, 60.
(a) Calculate M for a lawn with a = 4 (i.e. 40 m²).
(b) A trimming job costs $180. Use T = 40h to find the hours worked, showing your equation working.
(c) From the garden table, write a formula linking area A (m²) to hours h, then predict A at h = 5.5 h.
(d) The owner quotes a customer a total of $313 for a single job that includes a lawn mowing (a = 5) AND some hedge trimming. Write down the cost equation, solve to find how many hours of trimming were billed, then state the answer as a customer-facing sentence. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — M = 25 + 18(4) = $97.
Part (b) — 1 mark
• 1 mark — 40h = 180 → h = 4.5 hours.
Part (c) — 2 marks
• 1 mark — A = 20h.
• 1 mark — A(5.5) = 110 m².
Part (d) — 3 marks
• 1 mark — writes total = M + T equation: 313 = (25 + 18 × 5) + 40h, i.e. 313 = 115 + 40h.
• 1 mark — solves to h = 4.95, accept h ≈ 4.95 h or 5 hours rounded.
• 1 mark — customer-facing conclusion sentence (e.g. "The $313 quote covers a 50 m² lawn mow plus about 5 hours of hedge trimming").
Your response:
Stuck on (d)? Compute the lawn cost first (M at a = 5 is $115). Subtract from $313 to leave the trimming portion, then divide by $40/h.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Hire company (4 marks)
Sample response.
Strategy: Solve an equation — the total is known and the hire time is unknown.
Let h = hours. 35 + 18h = 143 → 18h = 108 → h = 6 hours.
Check: 35 + 18(6) = 143 ✓.
Marking notes. 1 mark — names strategy correctly. 1 mark — sets up the equation. 1 mark — solves to h = 6. 1 mark — reasonableness check or units stated.
1.2 — Rearrange d = st (3 marks)
Sample response.
Rearrange d = st to t = d/s.
t = 210 / 70 = 3 hours.
Marking notes. 1 mark — correct rearrangement. 1 mark — correct substitution. 1 mark — answer with units.
1.3 — Build, test and predict (4 marks)
(a) Sample response. Start = 14 (at x = 0); rate = +6 per step. Formula: y = 14 + 6x.
(b) Sample response. Test x = 3: y = 14 + 6(3) = 14 + 18 = 32 ✓ matches the table.
(c) Sample response. Predict x = 9: y = 14 + 6(9) = 14 + 54 = 68.
Marking notes. 1 mark — identifies start value 14. 1 mark — identifies rate +6. 1 mark — correct formula. 1 mark — correct prediction at x = 9.
2.1 — Landscaping business (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Lawn mowing at a = 4.
M = 25 + 18(4) = 25 + 72 = $97. [1 mark.]
(b) Trimming hours.
40h = 180 → h = 180/40 = 4.5 hours. [1 mark.]
(c) Garden table formula and prediction.
At h = 0, A = 0; rate = +20 m² per hour. Formula: A = 20h. [1 mark.]
A(5.5) = 20(5.5) = 110 m². [1 mark.]
(d) Combined-job equation.
Lawn cost at a = 5: M = 25 + 18(5) = $115.
Total cost equation: 115 + 40h = 313. [1 mark — equation set up.]
40h = 198 → h = 4.95 h. [1 mark — solves correctly.]
Conclusion: The $313 quote covers a 50 m² lawn mow ($115) plus about 5 hours of hedge trimming ($198). [1 mark — customer-facing summary.]
Total: 7/7.
Band descriptors for marker.
Band 3: Substitutes for (a) and solves (b), but does not link the two formulas in (d). ≈ 3 marks.
Band 4: (a)-(c) all correct, but (d) only computes the lawn cost without setting up the combined equation. ≈ 4-5 marks.
Band 5: (d) equation set up correctly but arithmetic slip in solving, or conclusion missing the dollar split. ≈ 5-6 marks.
Band 6: Complete and correct, with a customer-facing sentence that names both the lawn and the trimming hours. 7/7.