Mathematics Standard • Year 11 • Module 1 • Lesson 12

Direct Variation and Proportional Relationships

Build fluency recognising direct variation y = kx, finding the constant k = y/x, and telling proportional relationships apart from other linear ones.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Write the general form of a direct variation equation.

y = ______________

Q1.2 Write the formula for the constant of variation k.

k = ______________

Q1.3 A direct variation graph always passes through which point?

( ___ , ___ )

Q1.4 Circle the equations that show direct variation.

(a) y = 6x (b) y = 6x + 1 (c) C = 4k (d) C = 12 + 4k (e) d = 60t

Stuck? Revisit lesson § Key Formula — Direct Variation. The "+ something" makes it not direct variation.

2. Worked example — find the constant of variation

Follow each line of working. Every step has a reason on the right.

Problem. A car travels 180 km in 3 hours at a constant speed. Write a direct variation equation for distance d in terms of time t, then use it to find d when t = 5.

Step 1 — Check the relationship is direct variation.

No fixed starting distance ⇒ use d = kt

Reason: at time t = 0 the car has covered 0 km, so the graph passes through the origin.

Step 2 — Find the constant k using k = d/t.

k = 180 / 3 = 60

Reason: divide a matching pair where the input is not zero. The units of k here are km/h.

Step 3 — Write the equation.

d = 60t

Reason: substitute k = 60 into d = kt.

Step 4 — Substitute t = 5 to predict the new distance.

d = 60(5) = 300

Reason: brackets keep the multiplication explicit.

Conclusion. The car would travel 300 km in 5 hours at the same speed.

3. Faded example — fill in the missing steps

Apples cost $4 per kilogram with no fixed fee. Write a direct variation equation for cost C in terms of kilograms k, then find the cost of 3.5 kg. Fill in each blank. 4 marks

Step 1 — Check there is no fixed fee:

There is no flat charge before any apples are bought, so the equation has the form C = ______________.

Step 2 — Identify the constant k:

k = ____________ ($ per kg)

Step 3 — Write the equation:

C = ____________

Step 4 — Substitute k = 3.5 with brackets:

C = 4(____) = ____________ dollars

Stuck? Revisit lesson § Worked Example 1 — Write a direct variation equation.

4. Graduated practice — direct variation

Show your working in the space below each part. Where units are given, include them in the answer.

Foundation — find k from a single pair (4 questions)

Q	Problem	Answer
4.1 1	Find k for y = kx when x = 4 and y = 20.
4.2 1	Find k for y = kx when x = 3 and y = 21.
4.3 1	For y = 5x, find y when x = 8.
4.4 1	For C = 4k, find C when k = 7.

Standard — write the equation from a context (6 questions)

Decide whether it's direct variation, find k, write y = kx, and substitute.

4.5 A recipe uses 250 g of flour for each cake. Write a formula for flour F in terms of cakes c, then find F when c = 6. 2 marks

4.6 A worker earns $28 per hour with no allowance. Write a direct variation formula for pay P in terms of hours h. 2 marks

4.7 A table has points (0,0), (2,14), (5,35). Find k and write y = kx. Use the equation to predict y when x = 8. 2 marks

4.8 Decide whether y = 7x + 3 is a direct variation relationship. Justify your answer in one sentence. 2 marks

4.9 Apples are $4 per kilogram. Find the cost of (a) 2.5 kg, (b) 6 kg. Use C = 4k. 2 marks

4.10 A printer prints 12 pages per minute. Write a direct variation formula for pages p in terms of minutes m, and use it to find p when m = 7.5. 2 marks

Extension — proportional reasoning (2 questions)

4.11 If y varies directly with x and y = 45 when x = 9, (a) find k, (b) write the equation, (c) find y when x = 14, and (d) find x when y = 100. 3 marks

4.12 A car uses 6 L of fuel for every 80 km driven (direct variation). (a) Find k. (b) Write the equation for fuel F in terms of distance d. (c) Predict the fuel needed for 350 km. 3 marks

Stuck on 4.12? k = 6/80 = 0.075 L per km.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I divided y by x to get k = 20/4 = 5.

For 4.3 I substituted x = 8 to get y = 5(8) = 40 (no "+ something").

For 4.4 I substituted k = 7 to get C = 4(7) = $28.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Direct variation general form

y = kx.

Q1.2 — Constant of variation

k = y/x (use a matching pair where x ≠ 0).

Q1.3 — Special point

Through the origin (0, 0).

Q1.4 — Spot direct variation

Direct variation: (a) y = 6x, (c) C = 4k, (e) d = 60t. Not direct variation: (b) and (d) — both have a starting value added on.

Q3 — Faded apples example

Step 1: C = kk (no constant added; the form is C = constant × k).
Step 2: k = 4.
Step 3: C = 4k.
Step 4: C = 4(3.5) = $14.

Q4.1 — Find k (x = 4, y = 20)

k = 20 / 4 = 5.

Q4.2 — Find k (x = 3, y = 21)

k = 21 / 3 = 7.

Q4.3 — y = 5x, x = 8

y = 5(8) = 40.

Q4.4 — C = 4k, k = 7

C = 4(7) = $28.

Q4.5 — Flour recipe

F = 250c. For c = 6: F = 250(6) = 1500 g (= 1.5 kg).

Q4.6 — Worker pay

P = 28h (where P is pay in dollars and h is hours worked).

Q4.7 — Points (0,0), (2,14), (5,35)

k = 14/2 = 7 (and 35/5 = 7 confirms it). Equation: y = 7x. At x = 8: y = 7(8) = 56.

Q4.8 — Is y = 7x + 3 direct variation?

No. The +3 means the line does not pass through the origin — at x = 0, y = 3 (not 0), so this is linear but not direct variation.

Q4.9 — Apple cost

(a) C = 4(2.5) = $10.00. (b) C = 4(6) = $24.00.

Q4.10 — Printer pages

p = 12m. At m = 7.5: p = 12(7.5) = 90 pages.

Q4.11 — y varies directly with x

(a) k = 45/9 = 5. (b) y = 5x. (c) y = 5(14) = 70. (d) 100 = 5x ⇒ x = 20.

Q4.12 — Car fuel consumption

(a) k = 6/80 = 0.075 L/km. (b) F = 0.075d. (c) F = 0.075(350) = 26.25 L.