Mathematics Standard • Year 11 • Module 1 • Lesson 3
Equations from Worded Problems
Practise HSC-style writing on worded equations — three multi-mark short answers and one extended response with marking criteria.
1. Short-answer questions
1.1 Movie tickets cost $14 each plus a $5 booking fee. The total is $61. By forming and solving an equation, find the number of tickets bought. State your final answer in a sentence. 3 marks Band 3
1.2 A van hire costs $45 plus $20 per hour. The total cost is $145. Define your variable, write the equation, solve it, and interpret the answer. 4 marks Band 3-4
1.3 A student writes "$30 + 5 = x$" to model "a $30 starting amount plus $5 each week becomes $80".
(a) Explain in one sentence what is wrong with the student's equation.
(b) Write the correct equation (define your variable).
(c) Solve the correct equation and state, in a sentence, how many weeks it takes. 4 marks Band 4
2. Extended response
2.1 A school is choosing between two excursion bus quotes for a class of students.
Bus A: A flat hire fee of $480 plus $9 per student.
Bus B: A flat hire fee of $300 plus $13 per student.
(a) Let n be the number of students. Write a cost equation for each bus.
(b) The school's budget is $1,020. For each bus, write and solve the equation that finds the maximum number of students that can be carried within this budget. State, in a sentence, the maximum whole number of students each bus allows.
(c) For what number of students do the two buses cost the same? Form and solve the relevant equation.
(d) Write a conclusion sentence recommending which bus the school should book if they expect close to 45 students. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — both cost equations correct: A: C = 480 + 9n; B: C = 300 + 13n.
Part (b) — 3 marks
• 1 mark — sets up 480 + 9n = 1020 and 300 + 13n = 1020.
• 1 mark — Bus A: n = 60 (exact); Bus B: n = 720/13 ≈ 55.4 → max 55 students.
• 1 mark — both answers stated in a sentence with the rounding-down reason for Bus B.
Part (c) — 2 marks
• 1 mark — sets up 480 + 9n = 300 + 13n.
• 1 mark — solves correctly: 180 = 4n, n = 45.
Part (d) — 1 mark
• 1 mark — clear recommendation sentence that uses the n = 45 break-even (e.g. at exactly 45 students costs are equal; for <45 Bus B is cheaper, for >45 Bus A is cheaper).
Your response:
Stuck on (c)? Set the two cost expressions equal: 480 + 9n = 300 + 13n. Move all n's to one side and constants to the other.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Movie tickets (3 marks)
Sample response.
Let t = the number of tickets bought. 5 + 14t = 61 ⇒ 14t = 56 ⇒ t = 4.
4 tickets were bought.
Marking notes. 1 mark — correct equation (variable defined). 1 mark — correct algebra to t = 4. 1 mark — interpretation sentence with the number. A bare "t = 4" without context loses the interpretation mark.
1.2 — Van hire (4 marks)
Sample response.
Let h = the number of hours of van hire. 45 + 20h = 145 ⇒ 20h = 100 ⇒ h = 5.
The van was hired for 5 hours.
Marking notes. 1 mark — variable defined. 1 mark — correct equation. 1 mark — correct algebra. 1 mark — interpretation sentence.
1.3 — Identify the error and fix it (4 marks)
(a) Sample response. The student wrote "5" instead of "5w" — they forgot to multiply the weekly $5 by the number of weeks, and they left the target $80 out of the equation entirely.
(b) Sample response. Let w = the number of weeks. 30 + 5w = 80.
(c) Sample response. 5w = 50 ⇒ w = 10. It takes 10 weeks to reach $80.
Marking notes. 1 mark — identifies that "5" should be "5w" (the weekly rate must multiply weeks). 1 mark — correct equation 30 + 5w = 80. 1 mark — correct solving to w = 10. 1 mark — interpretation sentence. Note: noting that the student also omitted the target $80 strengthens (a) but is not required for the mark.
2.1 — Bus quote comparison (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Cost equations.
Bus A: C = 480 + 9n. Bus B: C = 300 + 13n. [1 mark — both correct.]
(b) Within $1,020 budget.
Bus A: 480 + 9n = 1020 ⇒ 9n = 540 ⇒ n = 60.
Bus B: 300 + 13n = 1020 ⇒ 13n = 720 ⇒ n = 720/13 ≈ 55.38, so the maximum whole number is 55 students. [1 mark — equations set up. 1 mark — both n values correct (60 and 55.38). 1 mark — Bus B rounded DOWN to 55 with reason that 56 would exceed the budget.]
Bus A can take up to 60 students within $1,020; Bus B can take up to 55 students within the same budget.
(c) Break-even.
Set costs equal: 480 + 9n = 300 + 13n. [1 mark — break-even equation.]
Subtract 9n and 300: 180 = 4n ⇒ n = 45. [1 mark — solves correctly.]
(d) Recommendation.
Both buses cost exactly the same at 45 students. Above 45 students Bus A becomes cheaper (its lower per-student rate of $9 dominates); below 45 students Bus B is cheaper (its lower hire fee dominates). Since the school expects "close to 45", the price difference is small either way — Bus A is the safer choice if final numbers may climb above 45. [1 mark — clear, context-sensitive recommendation.]
Total: 7/7.
Band descriptors for marker.
Band 3: Writes one cost equation; finds max students for one bus. ≈ 2-3 marks.
Band 4: Both equations and both maximums correct; no break-even calculation. ≈ 4 marks.
Band 5: Break-even n = 45 found, but recommendation in (d) is incomplete (does not name which bus to choose). ≈ 5-6 marks.
Band 6: Complete, all working correct, Bus B rounded DOWN to 55 with reason, AND a recommendation sentence that references the n = 45 break-even. 7/7.