Mathematics Standard • Year 11 • Module 1 • Lesson 12
Direct Variation and Proportional Relationships
Apply direct variation y = kx to real Australian contexts — fruit prices, fuel use, hourly pay, currency conversion and printing tasks.
Problem 1 — Buying apples at the market
Apples at a Sydney market cost $4 per kilogram with no fixed charge. Let C be the cost in dollars for k kilograms of apples.
Set up: What are we solving for?
(i) Write the direct variation equation for C in terms of k. State the constant of variation and its units. 2 marks
(ii) Calculate the cost of 2.5 kg and the cost of 7.2 kg. 2 marks
(iii) A customer is charged $26. How many kilograms did they buy? 1 mark
Stuck on (iii)? Solve 4k = 26 for k.Problem 2 — Car fuel for a country drive
A car uses 6 L of petrol for every 80 km driven at constant highway speed. Assume fuel use varies directly with distance.
Set up: What are we solving for?
(i) Find the constant of variation k (litres per km). 1 mark
(ii) Write the equation for fuel F (litres) in terms of distance d (km), then calculate the fuel needed for a 320 km Sydney-to-Canberra drive. 2 marks
(iii) The tank holds 45 L. Using your equation, find the maximum distance the car can travel on a full tank. 2 marks
Stuck? Revisit lesson § Worked Example 2 — Find the constant of variation.Problem 3 — Retail casual pay
A casual retail employee earns $28 per hour, with no shift allowance.
Set up: What are we solving for?
(i) Write a direct variation equation for pay P in terms of hours worked h. 1 mark
(ii) Calculate the pay for a 5.5-hour shift. 1 mark
(iii) The employee wants to earn $420 this fortnight. How many hours must they work? 2 marks
(iv) The next month they receive a $40 weekly travel allowance on top of hourly pay. Explain in one sentence whether their total weekly pay is still a direct variation relationship. 1 mark
Stuck on (iv)? A fixed allowance added on top means the graph no longer passes through the origin.Problem 4 — Converting AUD to USD
On a particular day, 1 AUD converts to 0.66 USD (no commission). The conversion is a direct variation between U (USD) and A (AUD).
Set up: What are we solving for?
(i) State the constant of variation k and write the equation for U in terms of A. 2 marks
(ii) Convert $500 AUD to USD. 1 mark
(iii) A traveller exchanges money and receives $264 USD. How many AUD did they hand over? 2 marks
Stuck on (iii)? Rearrange U = 0.66A to A = U/0.66.Problem 5 — A library printing job
A school library printer prints 12 pages per minute, with no warm-up delay. Assume pages printed varies directly with minutes running.
Set up: What are we solving for?
(i) Write the direct variation equation for pages p in terms of minutes m. 1 mark
(ii) Calculate the number of pages printed in 15 minutes and in 7.5 minutes. 2 marks
(iii) A teacher needs to print a 450-page booklet. How long will the print job take? Give your answer in minutes and seconds. 2 marks
Stuck on (iii)? 450 / 12 = 37.5 min ⇒ 37 min 30 s.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Apples at the market
Set up. Build the direct variation C = 4k, then substitute/rearrange.
(i) C = 4k. The constant of variation is k = 4 dollars per kilogram.
(ii) 2.5 kg: C = 4(2.5) = $10.00. 7.2 kg: C = 4(7.2) = $28.80.
(iii) 4k = 26 ⇒ k = 26/4 = 6.5 kg.
Problem 2 — Car fuel
Set up. Find k = fuel per km, then F = kd.
(i) k = 6/80 = 0.075 L/km.
(ii) F = 0.075d. For 320 km: F = 0.075(320) = 24 L.
(iii) 0.075d = 45 ⇒ d = 45/0.075 = 600 km on a full tank.
Problem 3 — Retail pay
Set up. Build P = 28h, then deal with the new allowance.
(i) P = 28h.
(ii) P = 28(5.5) = $154.00.
(iii) 28h = 420 ⇒ h = 420/28 = 15 hours.
(iv) No. Adding a fixed $40 allowance makes the new formula P = 40 + 28h, which is linear but not direct variation (the graph no longer passes through the origin — at h = 0, pay = $40, not $0).
Problem 4 — AUD to USD
Set up. Build the direct variation U = 0.66A.
(i) k = 0.66 (USD per AUD). Equation: U = 0.66A.
(ii) U = 0.66(500) = $330 USD.
(iii) 0.66A = 264 ⇒ A = 264/0.66 = $400 AUD.
Problem 5 — Library printer
Set up. Build p = 12m.
(i) p = 12m.
(ii) 15 min: p = 12(15) = 180 pages. 7.5 min: p = 12(7.5) = 90 pages.
(iii) 12m = 450 ⇒ m = 450/12 = 37.5 minutes = 37 min 30 s.