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VARIATION · LESSON 09

Variation Final Checkpoint

Bring together direct, inverse, joint and partial variation in one final assessment.

⏱️ 35–45 min 📘 Checkpoint ✏️ Worked examples included

🏗️ Start With a Building Project

A building project combines wages, workers, days, floor dimensions and delivery charges. Use the clues to identify each type of variation before solving.

Observe the quantities→Describe the change→Choose the model

🎉 Variation Final Checkpoint

This checkpoint combines direct, inverse, joint and partial variation. Attempt it without opening earlier lessons first.

Show the equation before substituting values. This makes your reasoning clear and helps you earn method marks.

Confidence Checklist

Section A: Core Skills

  1. Write an equation for y varying directly as x².
  2. Write an equation for p varying inversely as q.
  3. Write an equation for V varying jointly as l, w and h.
  4. Write an equation for C varying partly as d and partly as a constant.
  5. State the meaning of k in d = kt when d is distance and t is time.

Section B: Direct and Inverse Variation

  1. y ∝ x² and y = 45 when x = 3. Find y when x = 7.
  2. y ∝ 1/x and y = 18 when x = 4. Find x when y = 6.
  3. A car uses fuel directly proportional to distance. It uses 9 litres for 135 km. Find the fuel for 420 km.
  4. Twenty-four workers complete a task in 15 days. How many workers are needed to complete it in 9 days?

Section C: Joint and Partial Variation

  1. z ∝ xy². If z = 96 when x = 3 and y = 4, find z when x = 5 and y = 2.
  2. y = kx + c. When x = 4, y = 19 and when x = 9, y = 34. Find y when x = 15.
  3. The cost of hiring equipment contains a fixed fee and a daily charge. Three days cost US$41 and eight days cost US$86. Find the cost for twelve days.

Section D: Extension Thinking

  1. t varies inversely as n². If t = 20 when n = 3, find t when n = 5.
  2. P varies jointly as a and b and inversely as c. Write the equation.
  3. P = 48 when a = 4, b = 6 and c = 3. Find P when a = 5, b = 9 and c = 2.
  4. Explain why a delivery cost of C = 2d + 5 cannot represent direct variation between cost and distance.

✅ Check Your Work

Show checkpoint answers

Section A

  1. y = kx²
  2. p = k/q
  3. V = klwh
  4. C = kd + c
  5. k is the constant speed.

Section B

  1. 245.
    y = kx²
    45 = 9k, so k = 5.
    y = 5(7²) = 245.
  2. x = 12.
    k = xy = 18×4 = 72.
    6 = 72/x, so x = 12.
  3. 28 litres.
  4. 40 workers.

Section C

  1. 40.
    96 = k(3)(16), so k = 2.
    z = 2(5)(4) = 40.
  2. 52.
    Gradient k = (34−19)/(9−4) = 3.
    19 = 3(4) + c, so c = 7.
    y = 3(15) + 7 = 52.
  3. US$122.
    41 = 3k + c
    86 = 8k + c
    45 = 5k, so k = 9.
    c = 14.
    Cost = 9(12) + 14 = 122.

Section D

  1. 7.2.
    t = k/n²
    20 = k/9, so k = 180.
    t = 180/25 = 7.2.
  2. P = kab/c
  3. 135.
    48 = k(4)(6)/3 = 8k, so k = 6.
    P = 6(5)(9)/2 = 135.
  4. Because cost is US$5 when distance is zero. A direct variation graph must pass through the origin and have no fixed term.

Celebration Message

You can now recognise and solve several kinds of variation. More importantly, you can explain what the relationship means rather than memorising a formula without context.

Revision Guidance and Readiness