Mathematics Standard • Year 11 • Module 1 • Lesson 12
Direct Variation and Proportional Relationships
Practise HSC-style writing on direct variation — three multi-mark short answers and one extended response with marking criteria.
1. Short-answer questions
1.1 A worker is paid $32 per hour with no allowance. Let P be the pay in dollars after h hours.
(a) Write a direct variation equation for P in terms of h.
(b) Calculate the pay for a 6.5-hour shift. 3 marks Band 3
1.2 A table contains the points (0, 0), (3, 21) and (5, 35).
(a) Show that the data fits a direct variation relationship.
(b) Find the constant k and write the equation y = kx. 3 marks Band 3-4
1.3 Explain why C = 10 + 4k is not a direct variation relationship even though it is linear. Use the words intercept, origin and fixed in your explanation. 3 marks Band 4
Stuck on 1.3? Substitute k = 0 and see what C is — for direct variation C should also be 0.2. Extended response
2.1 A petrol station offers a fuel-loyalty discount. The discounted price per litre is constant. The table shows the cost C (dollars) for L litres of fuel filled.
L 0 10 20 30 50
C 0 18 36 54 90
(a) Show, using two ratios from the table, that C varies directly with L.
(b) Find the constant of variation k. State its units and explain what k means in this context.
(c) Write the direct variation equation linking C and L.
(d) A driver puts in 42.5 L. Use the equation to find the total cost.
(e) A driver pays $63 at the bowser. Use the equation to find how many litres they bought. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — computes at least two ratios (e.g. 18/10 = 1.8 and 36/20 = 1.8) and notes they are equal.
Part (b) — 2 marks
• 1 mark — k = 1.8 with units dollars per litre ($/L).
• 1 mark — explains k = $1.80 is the discounted price per litre of fuel.
Part (c) — 1 mark
• 1 mark — writes C = 1.8L (with variables defined or clearly named).
Part (d) — 1 mark
• 1 mark — substitutes and finds C = $76.50.
Part (e) — 2 marks
• 1 mark — rearranges 1.8L = 63 to L = 63/1.8.
• 1 mark — final answer L = 35 litres.
Your response:
Stuck on (a)? Direct variation needs constant C/L for all pairs where L ≠ 0. Compute two ratios and compare.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Hourly pay (3 marks)
Sample response.
(a) P = 32h (where P is pay in dollars and h is hours worked).
(b) P = 32(6.5) = $208.
Marking notes. 1 mark — correct direct variation equation (no constant added). 1 mark — correct substitution shown. 1 mark — final answer $208.
1.2 — Table direct variation (3 marks)
Sample response.
(a) The table includes (0, 0). For the non-zero pairs, y/x = 21/3 = 7 and 35/5 = 7. The ratio is constant, so it is direct variation.
(b) k = 7. Equation: y = 7x.
Marking notes. 1 mark — shows both ratios are 7 (or notes constant y/x). 1 mark — states k = 7. 1 mark — correct equation y = 7x.
1.3 — Why C = 10 + 4k is not direct variation (3 marks)
Sample response. Direct variation has the form y = kx and passes through the origin. In C = 10 + 4k, the y-intercept is 10, not 0 — at k = 0 the cost is $10 because of a fixed starting charge. Since the line does not pass through the origin, the relationship is linear but not direct variation.
Marking notes. 1 mark — uses the word origin correctly. 1 mark — identifies the intercept 10 as a fixed/starting cost. 1 mark — clear conclusion that it's linear but not direct variation. Missing one key term: deduct 1 mark.
2.1 — Fuel loyalty (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Showing direct variation.
Compute the ratio C/L for two non-zero pairs:
C/L at L = 10: 18/10 = 1.8. C/L at L = 30: 54/30 = 1.8. (Also 36/20 = 1.8 and 90/50 = 1.8.)
The ratio is constant, so C varies directly with L. [1 mark.]
(b) Constant of variation.
k = 1.8 dollars per litre ($/L). [1 mark — value + units.]
This means each litre of fuel costs $1.80 at the loyalty rate. [1 mark — meaning explained.]
(c) Equation.
C = 1.8L (where C is the total cost in dollars and L is litres of fuel). [1 mark.]
(d) 42.5 L fill-up.
C = 1.8(42.5) = $76.50. [1 mark.]
(e) Working backwards from $63.
1.8L = 63 ⇒ L = 63/1.8. [1 mark — correct rearrangement.]
L = 35 litres. [1 mark.]
Total: 7/7.
Band descriptors for marker.
Band 3: Computes one ratio correctly and possibly substitutes into a wrong equation. ≈ 2-3 marks.
Band 4: Shows direct variation and finds k = 1.8, but loses units or fails to interpret what k means. ≈ 4-5 marks.
Band 5: Equation correct and forward calculation correct; reverse calculation (e) attempted but loses one mark for arithmetic. ≈ 5-6 marks.
Band 6: All parts complete: ratios shown, k with units, equation, both forward and reverse substitutions. 7/7.