When y is proportional to a power of x, write y = kxⁿ, substitute one known pair of values to find the constant k, then use y = kxⁿ to answer the rest. The most common cases at GCSE Higher are y ∝ x², y ∝ x³, y ∝ √x and y ∝ 1/x².

What does y ∝ xⁿ mean?

The symbol ∝ means "is proportional to." If y ∝ xⁿ, then doubling x multiplies y by 2ⁿ — not by 2. This is what distinguishes power proportion from simple direct proportion.

Relationship Equation form Doubling x multiplies y by
y ∝ x (direct) y = kx 2
y ∝ x² y = kx² 4
y ∝ x³ y = kx³ 8
y ∝ √x y = k√x √2 ≈ 1.41
y ∝ 1/x (inverse) y = k/x ½
y ∝ 1/x² (inverse square) y = k/x² ¼

How do you find the constant of proportionality k?

Method — three steps:

  1. Write down the equation form (e.g. y = kx²).
  2. Substitute the known pair of values (x₀, y₀) and solve for k.
  3. Rewrite the equation with k inserted, then use it to find any unknown.

Worked example 1 — y ∝ x²:

y is proportional to x². When x = 3, y = 45. Find y when x = 5.

  1. y = kx²
  2. 45 = k × 3² = 9k → k = 5
  3. Equation: y = 5x². When x = 5: y = 5 × 25 = 125

Check: 3→45, 5→125. Ratio of x values = 5/3. Ratio of y values = 125/45 = 25/9 = (5/3)². ✓

Worked example 2 — y ∝ √x:

y is proportional to the square root of x. When x = 16, y = 20. Find x when y = 35.

  1. y = k√x
  2. 20 = k√16 = 4k → k = 5
  3. Equation: y = 5√x. When y = 35: 35 = 5√x → √x = 7 → x = 49

Check: √49 = 7; y = 5 × 7 = 35 ✓

How do you handle inverse square proportion?

Inverse square proportion (y ∝ 1/x²) appears in physics-style GCSE maths problems — the intensity of light or sound decreasing with distance.

Worked example — y ∝ 1/x²:

y is inversely proportional to the square of x. When x = 2, y = 18.

  1. y = k/x²
  2. 18 = k/4 → k = 72
  3. Equation: y = 72/x².

Find y when x = 6: y = 72/36 = 2

What happens to y when x is tripled? y = 72/(3x)² = 72/9x² = (1/9) × y. So y is divided by 9. ✓ (inverting 3² = 9)

What if the problem gives you two unknowns?

Sometimes a question gives neither k explicitly nor a complete pair (x, y) — instead it gives a proportional relationship between two situations. You do not need to find k.

Worked example — ratio method:

y ∝ x³. When x doubles, what happens to y?

y₁ = kx₁³ and y₂ = k(2x₁)³ = 8kx₁³.

So y₂/y₁ = 8. y multiplies by 8.

This works for any power proportion without needing a specific value of k.

What are common mistakes in power proportion questions?

Mistake Example of error Correct approach
Using direct proportion rule y ∝ x² but writing y₂/y₁ = x₂/x₁ y₂/y₁ = (x₂/x₁)²
Forgetting to square k's calculation 45 = k × 3 instead of k × 9 Always square (or cube, or root) x first
Sign error in inverse square k/x² but writing kx² Write the equation clearly before substituting
Not checking the answer No verification step Substitute your answer back into y = kxⁿ

Frequently asked questions

How do I know which power to use?

The question tells you: "y is proportional to the square of x" means y ∝ x². "y is inversely proportional to the cube of x" means y ∝ 1/x³. Read the wording carefully and write the equation form before doing any calculation.

Can I use a table to find k?

Yes. If you have a table of x and y values, compute y/x² (for y ∝ x²) for each row. If the relationship holds, all entries in the y/x² column will be equal — that constant value is k. This is a useful check and can reveal the relationship type even when it is not stated.

Does power proportion appear on GCSE Foundation or Higher only?

Simple direct and inverse proportion (y ∝ x, y ∝ 1/x) appear on both tiers. Power proportion (y ∝ x², y ∝ x³, y ∝ √x, y ∝ 1/x²) is a Higher-tier topic. You will not see y ∝ x² on a Foundation-only paper, but it is a standard question type on Higher papers.

What happens to y when x is halved (for y ∝ x²)?

Halving x multiplies x by ½, so x² is multiplied by (½)² = ¼. Therefore y is multiplied by ¼ — it becomes one-quarter of its original value. The general rule: if x is multiplied by a factor f, then y is multiplied by fⁿ for y ∝ xⁿ.


Need step-by-step guidance on proportion problems? Professor Pi is ready to help at aitutors.me.