Mathematics Standard • Year 11 • Module 1 • Lesson 1
Algebraic Language, Variables and Substitution
Apply substitution to real Australian contexts — delivery costs, room areas, fuel calculations and trip planning.
Problem 1 — Delivery cost formula (substitution into a real rule)
A Sydney bike-delivery company charges using the rule C = 8 + 1.50k, where C is the cost in dollars and k is the number of kilometres travelled.
Set up: What are we solving for?
(i) Calculate the cost for a 4 km delivery. 1 mark
(ii) Calculate the cost for a 14 km delivery. 1 mark
(iii) Explain in one sentence what the 8 and the 1.50 represent in real life. 2 marks
Stuck? Revisit lesson § Variables Let a Rule Work More Than Once — the constant is a fixed fee, the coefficient is the per-km charge.Problem 2 — Carpeting a rectangular room
Mia is buying carpet for a rectangular living room that measures 5.4 m long and 3.5 m wide. The carpet costs $48 per square metre. Use A = lw for the area.
Set up: What are we solving for?
(i) Calculate the floor area of the room. 1 mark
(ii) Calculate the cost of carpet, ignoring any wastage. 2 marks
(iii) The supplier insists on rounding the area UP to the next whole square metre. Recalculate the cost. 1 mark
Stuck? Revisit lesson § Worked Example 2 — Substitute into a perimeter formula (same idea for area).Problem 3 — Sydney to Newcastle drive
A car travels at an average speed of 95 km/h on the M1. Use d = st to model the distance covered.
Set up: What are we solving for?
(i) Calculate the distance covered in 1.75 hours. 1 mark
(ii) Sydney to Newcastle is approximately 162 km. Using d = st, calculate the time it would take at 95 km/h. Give your answer in hours and minutes. 2 marks
(iii) Explain why your time in (ii) is the minimum driving time, not necessarily the actual time. 1 mark
Stuck on (ii)? Rearrange d = st to t = d/s, substitute, then convert the decimal hours to minutes (e.g. 0.7 h × 60 = 42 min).Problem 4 — Australian holiday temperature check
Tom is checking the temperature for a US trip. The conversion formula from Fahrenheit to Celsius is C = (F − 32) / 1.8, where C is in °C and F is in °F.
Set up: What are we solving for?
(i) Convert 86 °F to Celsius. Use brackets in your substitution. 2 marks
(ii) Convert 14 °F to Celsius. Show that the answer is negative. 2 marks
(iii) Explain in one sentence why brackets around (F − 32) are essential in this formula. 1 mark
Stuck? Revisit lesson § Brackets Matter With Negative Values.Problem 5 — Fuel cost on a road trip
A car uses fuel at the rate of 8.2 L per 100 km. Petrol costs $1.95 per litre. The fuel cost formula for a trip of d kilometres is:
F = 1.95 × (8.2/100) × d (where F is the fuel cost in dollars, d is distance in km)
Set up: What are we solving for?
(i) Calculate the fuel cost for a 250 km trip. 2 marks
(ii) Calculate the fuel cost for a 612 km Sydney-to-Melbourne leg. 2 marks
(iii) Explain in one sentence what the 8.2/100 in the formula means in plain English. 1 mark
Stuck? 8.2/100 = 0.082 L per km, so the formula is "price per litre × litres per km × kilometres".How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Delivery cost
Set up. Substitute the value of k into C = 8 + 1.50k and interpret what each number means.
(i) C = 8 + 1.50(4) = 8 + 6 = $14.00.
(ii) C = 8 + 1.50(14) = 8 + 21 = $29.00.
(iii) The 8 is the fixed booking fee charged regardless of distance. The 1.50 is the cost per kilometre. (Conclusion: as k grows, only the variable part grows; the booking fee is paid every time.)
Problem 2 — Carpeting the room
Set up. Use A = lw to find the floor area, then multiply by the price per m².
(i) A = 5.4 × 3.5 = 18.9 m².
(ii) Cost = 18.9 × $48 = $907.20.
(iii) Rounded up: 19 m² × $48 = $912.00 (an extra $4.80 because the supplier won't cut a partial square metre).
Problem 3 — M1 driving
Set up. Substitute into d = st for distance; rearrange to t = d/s for time.
(i) d = 95 × 1.75 = 166.25 km.
(ii) t = 162 / 95 = 1.7053... hours. 0.7053 h × 60 ≈ 42.3 min. Time ≈ 1 hour 42 minutes.
(iii) The 95 km/h is an average speed; real driving includes slowing for traffic, lights and roadworks, so the actual trip is usually longer than the calculated minimum.
Problem 4 — Temperature conversion
Set up. Substitute F into C = (F − 32)/1.8, keeping brackets so the subtraction happens before the division.
(i) C = (86 − 32) / 1.8 = 54 / 1.8 = 30 °C.
(ii) C = (14 − 32) / 1.8 = (−18) / 1.8 = −10 °C. The negative comes from F < 32.
(iii) Without brackets, "F − 32 / 1.8" would divide only the 32 by 1.8, then subtract from F — producing a completely different (wrong) value. The brackets force the subtraction to happen first.
Problem 5 — Fuel cost
Set up. Substitute distance d into F = 1.95 × (8.2/100) × d. Each factor has a real meaning — price/L, L/km, total km.
(i) F = 1.95 × 0.082 × 250 = 0.1599 × 250 = $39.98 (rounded to nearest cent).
(ii) F = 1.95 × 0.082 × 612 = 0.1599 × 612 = $97.86 (to nearest cent).
(iii) 8.2/100 means the car uses 8.2 litres of fuel for every 100 km driven, i.e. 0.082 L per km — the fuel-economy figure quoted on the car's sticker.