Mathematics Standard • Year 11 • Module 2 • Lesson 1
Working With Formulas and Units
Apply unit conversions and formula substitution to realistic measurement scenarios — decompose, calculate, conclude with units.
Problem 1 — Mixed units in a rectangle (tiling)
A bathroom floor is rectangular, 3.5 m long and 80 cm wide. You want to tile the whole floor.
Set up: What are we solving for?
(i) Why can you NOT just multiply 3.5 × 80 to get the area? Explain in one sentence. 1 mark
(ii) Convert 80 cm to metres, then use A = ℓ × w to calculate the floor area in m². 2 marks
(iii) Tiles cost $48 per m². Find the total cost to tile the floor. 2 marks
Stuck? Revisit lesson § Common Mistakes — "Mixing units in a formula gives nonsense." Convert to the same unit first.Problem 2 — Pool volume, capacity and water cost
A rectangular backyard swimming pool is 8 m long, 4 m wide and 1.5 m deep.
Set up: What are we solving for?
(i) Use V = ℓ × w × h to find the volume of the pool in m³. 1 mark
(ii) Convert the volume to litres. (Hint: 1 m³ = 1000 L.) 1 mark
(iii) Water is sold at $2.30 per kilolitre. Find the cost to fill the pool from empty. 2 marks
Stuck? Revisit lesson § Worked Example 4 — Volume & Capacity. 1 m³ = 1000 L = 1 kL.Problem 3 — Paddock area in hectares (rural)
A rectangular paddock is 320 m long and 250 m wide.
Set up: What are we solving for?
(i) Calculate the area of the paddock in m². 1 mark
(ii) Convert this area to hectares. (Recall: 1 ha = 10 000 m².) 1 mark
(iii) The farmer plans 1.2 sheep per hectare. How many sheep can the paddock support? State your final answer as a whole number with a one-sentence justification of the rounding direction. 3 marks
Stuck? Revisit lesson § Area Units — hectares sit between m² and km².Problem 4 — Circular fountain (substituting into A = πr²)
A circular fountain in a park has a radius of 2.5 m.
Set up: What are we solving for?
(i) Find the area of the fountain in m², correct to 2 decimal places. Use the π button on your calculator. 2 marks
(ii) Convert the area to cm². 1 mark
(iii) The council wants to fence the perimeter (circumference = 2πr). Find the circumference to 2 d.p. and the fencing cost at $42/m. 2 marks
Stuck? Revisit lesson § Card 1 — substitute into a formula, evaluate, write the unit. Never use 3.14 for π.Problem 5 — Find the unit-conversion error (payroll calc)
A student attempts to calculate the area of a rectangular advertising board that is 4 m long and 50 cm wide. Their working is shown.
Length = 4 m, Width = 50 cm
A = 4 × 50 = 200 m²
Conclusion: the board is 200 m² in area.
Set up: What are we solving for?
(i) Identify the error in the student's working in one sentence. 1 mark
(ii) Write the correct calculation, showing the unit conversion line clearly. Give the answer in m². 2 marks
(iii) A 200 m² board would be the size of a small house block. Briefly explain why the student's "sense check" should have made them suspicious. 2 marks
Stuck? Revisit lesson § Common Mistakes — units MUST match before substituting into a formula.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Mixed-unit tiling
Set up. We are finding the floor area (in m²) and the tile cost, but the dimensions are given in two different units — we must convert before multiplying.
(i) Because 3.5 m and 80 cm are in different units. Multiplying gives "m × cm", which is not a recognised area unit and gives the wrong numerical value.
(ii) Convert: 80 cm = 80 ÷ 100 = 0.8 m. A = ℓ × w = 3.5 × 0.8 = 2.8 m².
(iii) Cost = 2.8 × $48 = $134.40.
Problem 2 — Backyard pool
Set up. We are finding the pool's volume, converting it into a capacity in kilolitres, then multiplying by the water price.
(i) V = 8 × 4 × 1.5 = 48 m³.
(ii) 48 m³ × 1000 = 48 000 L.
(iii) 48 000 L ÷ 1000 = 48 kL. Cost = 48 × $2.30 = $110.40.
Problem 3 — Paddock area in hectares
Set up. We are finding the paddock area in m², converting to hectares, then using the stocking rate to find a whole number of sheep.
(i) A = 320 × 250 = 80 000 m².
(ii) 80 000 ÷ 10 000 = 8 ha.
(iii) Number of sheep = 8 × 1.2 = 9.6 → round DOWN to 9 sheep. Justification: you cannot have 0.6 of a sheep, and rounding up would exceed the safe stocking rate (the paddock cannot support 10 full sheep). Common error: rounding up to 10 — that overgrazes the paddock.
Problem 4 — Circular fountain
Set up. We are substituting r = 2.5 into A = πr² and C = 2πr, then converting units and finding fencing cost.
(i) A = π × (2.5)² = π × 6.25 = 19.635... ≈ 19.63 m² (to 2 d.p.).
(ii) 19.63 × 10 000 = 196 300 cm² (using the rounded value; exact: 196 349.5 cm²).
(iii) C = 2π × 2.5 = 5π = 15.708... ≈ 15.71 m. Cost = 15.71 × $42 = $659.82.
Problem 5 — Find the error
Set up. We are auditing a student's working for a unit-mismatch bug, then correcting it.
(i) The student multiplied 4 m by 50 cm without first converting to the same unit, so the "200" is not in m² (it is a meaningless m·cm product, and the value is also far too large).
(ii) Convert 50 cm = 0.5 m. A = 4 × 0.5 = 2 m².
(iii) 200 m² is roughly the footprint of a small house — clearly impossible for an advertising board half a metre wide. A quick sense check ("would this fit on a wall?") would have flagged the bug. Marking note: an answer of "$2 m²" without identifying the unit-mismatch cause earns 1/3.