Mathematics Standard • Year 11 • Module 1 • Lesson 11
Intercepts and Linear Equations
Practise HSC-style writing on intercepts and linear equations — three multi-mark short answers and one extended response with marking criteria.
1. Short-answer questions
1.1 A scooter-share company charges $4 to unlock the scooter, then $0.45 per minute of ride time. Let C be the total cost in dollars and t be the ride time in minutes.
(a) Write a linear equation for C in terms of t.
(b) Calculate the cost of a 12-minute ride. 3 marks Band 3
1.2 Pay for a casual job is given by P = 120 + 25w, where P is total pay in dollars and w is the number of weeks worked.
(a) State, in plain English, what 120 and 25 represent.
(b) Find P when w = 8. 3 marks Band 3-4
1.3 A table shows the height H (cm) of a candle after t hours of burning.
t = 0, 1, 2, 3 → H = 18, 16, 14, 12.
(a) Write a linear equation for H in terms of t.
(b) Use the equation to predict H when t = 7.
(c) Explain what the gradient represents in real life. 4 marks Band 4
2. Extended response
2.1 Two ride-share apps charge using the formulas below for a trip of k kilometres.
App A: C = 6 + 2.20k, where C is the cost in dollars.
App B: C = 3.50 + 2.70k, where C is the cost in dollars.
(a) State, in plain English, what each gradient and y-intercept represents for each app.
(b) Calculate the cost of a 5 km trip with each app, and state which is cheaper for this trip.
(c) Calculate the cost of a 10 km trip with each app.
(d) Find the number of kilometres at which the two apps charge the same. State in a conclusion sentence which app is cheaper for short trips and which is cheaper for long trips. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — y-intercepts identified as booking/start fees ($6 for A, $3.50 for B).
• 1 mark — gradients identified as cost per kilometre ($2.20 for A, $2.70 for B).
Part (b) — 1 mark
• 1 mark — both 5 km costs correct (A = $17, B = $17) and notes they are equal at 5 km.
Part (c) — 1 mark
• 1 mark — both 10 km costs correct (A = $28, B = $30.50).
Part (d) — 3 marks
• 1 mark — sets up 6 + 2.20k = 3.50 + 2.70k.
• 1 mark — correctly solves to k = 5 km.
• 1 mark — clear conclusion sentence: App B is cheaper for trips under 5 km (lower booking fee); App A is cheaper for trips over 5 km (lower per-km rate).
Your response:
Stuck on (d)? Set the two equations equal and solve, then test a value below and above to decide which app wins on each side.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Scooter share (3 marks)
Sample response.
(a) C = 4 + 0.45t (where C is cost in dollars and t is ride time in minutes).
(b) C = 4 + 0.45(12) = 4 + 5.40 = $9.40.
Marking notes. 1 mark — correct equation. 1 mark — correct substitution shown. 1 mark — final answer $9.40 with cents.
1.2 — Casual pay (3 marks)
Sample response.
(a) The 120 is a fixed starting amount/bonus already credited before any weeks are worked. The 25 is the weekly pay (dollars added per week).
(b) P = 120 + 25(8) = 120 + 200 = $320.
Marking notes. 1 mark — meaning of 120. 1 mark — meaning of 25. 1 mark — correct $320.
1.3 — Burning candle (4 marks)
Sample response.
(a) At t = 0, H = 18. The height drops by 2 cm each hour, so the gradient is −2. H = 18 − 2t.
(b) H = 18 − 2(7) = 18 − 14 = 4 cm.
(c) The gradient −2 means the candle loses 2 cm of height for every hour it burns.
Marking notes. 1 mark — correct equation (including the minus sign). 1 mark — correct intercept 18 identified. 1 mark — correct value 4 cm for t = 7. 1 mark — gradient interpreted as "2 cm lost per hour" with the negative explained.
2.1 — Ride-share apps (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Meaning of gradients and intercepts.
App A's y-intercept $6 and App B's y-intercept $3.50 are fixed booking/start fees charged before any kilometres are driven. [1 mark.]
App A's gradient $2.20/km and App B's gradient $2.70/km are the per-kilometre charges. [1 mark.]
(b) 5 km trip.
A: C = 6 + 2.20(5) = 6 + 11 = $17.00. B: C = 3.50 + 2.70(5) = 3.50 + 13.50 = $17.00. Both apps cost the same at exactly 5 km. [1 mark.]
(c) 10 km trip.
A: C = 6 + 2.20(10) = 6 + 22 = $28.00. B: C = 3.50 + 2.70(10) = 3.50 + 27 = $30.50. [1 mark.]
(d) Cross-over equation.
Set costs equal: 6 + 2.20k = 3.50 + 2.70k. [1 mark — correct equation.]
Rearrange: 6 − 3.50 = 2.70k − 2.20k ⇒ 2.50 = 0.50k ⇒ k = 5 km. [1 mark — correct cut-off.]
Conclusion: App B is cheaper for trips shorter than 5 km (lower booking fee); App A is cheaper for trips longer than 5 km (lower per-km rate). At exactly 5 km both apps cost $17. [1 mark — clear conclusion with both ranges named.]
Total: 7/7.
Band descriptors for marker.
Band 3: Identifies one intercept correctly and substitutes for one app at one distance; no comparison. ≈ 2-3 marks.
Band 4: All four 5/10 km costs correct; identifies booking fee vs per-km rate; does not attempt the equation in (d). ≈ 4-5 marks.
Band 5: Sets up and solves 6 + 2.20k = 3.50 + 2.70k correctly but conclusion is incomplete (e.g. names only the cut-off, not which app wins on each side). ≈ 5-6 marks.
Band 6: Complete, correct, conclusion sentence names both ranges (under 5 km vs over 5 km) and which app wins. 7/7.