Edexcel International GCSE Mathematics A 4MA1: Proportion – Study Guide

This study guide covers section 2.5 Proportion of Topic 2 (Equations, formulae and identities) in the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on that specification. Section 2.5 is Higher tier only: it appears only in the Higher tier content, so it is examined on papers 1H and 2H. Each Higher paper is 2 hours and 100 marks, and a calculator may be used on both.

The page teaches one specification statement in depth: setting up direct and inverse proportion problems, and linking the algebra to the graphs. When you have worked through it, test your recall with the revision notes and then try the practice questions. The rest of Topic 2 is on the existing pages for use of symbols and algebraic manipulation, with revision notes and practice. The Edexcel IGCSE Mathematics course hub links the whole course, and the printable checklist lists every statement.

What this section covers

Specification	What you must be able to do	Tier
2.5A	Set up problems involving direct or inverse proportion	Higher tier only
2.5A	Relate algebraic solutions to the graphs of the equations	Higher tier only
2.5A (notes)	Work only with y ∝ x, y ∝ 1/x, y ∝ x², y ∝ 1/x², y ∝ x³, y ∝ 1/x³, y ∝ √x, y ∝ 1/√x	Higher tier only

The specification notes limit the forms to those eight. You will not be asked about other powers, such as y ∝ x⁴ or a cube root.

Higher questions also assume the Foundation content. Foundation statement 1.7D asks you to “calculate an unknown quantity from quantities that vary in direct proportion” and find a missing value in a table. Section 2.5 extends that idea to powers, roots and inverse relationships, written as formulae.

The language of proportion

The symbol ∝ means “is proportional to”. Exam questions use several phrasings for the same idea:

• “y is directly proportional to x”, “y varies directly as x” and “y varies as x” all mean y ∝ x.
• “y is inversely proportional to x” and “y varies inversely as x” mean y ∝ 1/x.
• “y is proportional to the square of x” means y ∝ x².
• “y is inversely proportional to the square root of x” means y ∝ 1/√x.

Every proportion statement becomes an equation by replacing ∝ with “= k ×”, where k is a constant called the constant of proportionality:

Statement	Formula
y ∝ x	y = kx
y ∝ x²	y = kx²
y ∝ x³	y = kx³
y ∝ √x	y = k√x
y ∝ 1/x	y = k/x
y ∝ 1/x²	y = k/x²
y ∝ 1/x³	y = k/x³
y ∝ 1/√x	y = k/√x

These formulae are not on the Higher tier formulae sheet, so learn the method rather than looking for them.

The four-step method

Almost every proportion question follows the same four steps.

1. Write the formula with k, choosing the right power or root.
2. Substitute the pair of values you are given.
3. Solve for k and write the full formula with k replaced by its value.
4. Use the formula to find the unknown, either by substituting or by rearranging.

Step 3 matters. Many questions say “find a formula for y in terms of x”, and the mark for that needs the formula with a number in place of k, not just the value of k on its own.

Direct proportion

When y is directly proportional to a power or root of x, y increases as x increases (for positive k and x).

Worked example 1: proportional to a square

E is directly proportional to the square of v. When v = 3, E = 36.

(a) Find a formula for E in terms of v. (b) Find E when v = 5. (c) Find v when E = 196, given v > 0.

(a)  E = kv²
     36 = k × 3²   so   36 = 9k   so   k = 4
     E = 4v²
(b)  E = 4 × 5² = 4 × 25 = 100
(c)  196 = 4v²   so   v² = 49   so   v = 7

In (c) you take a square root. v = −7 also satisfies v² = 49, but the question says v > 0. Always read the context: lengths, speeds and times are positive.

Worked example 2: proportional to a square root

y is directly proportional to √x. When x = 25, y = 20.

(a) Find y when x = 64. (b) Find x when y = 6.

     y = k√x
     20 = k × √25 = 5k   so   k = 4
     y = 4√x
(a)  y = 4 × √64 = 4 × 8 = 32
(b)  6 = 4√x   so   √x = 1.5   so   x = 1.5² = 2.25

To undo a square root, square both sides. To undo a cube, take the cube root.

Inverse proportion

When y is inversely proportional to a power or root of x, y decreases as x increases. The product of y and the power of x stays constant: for y ∝ 1/x, the value of xy is always k.

Worked example 3: inverse square

F is inversely proportional to the square of r. When r = 3, F = 20.

(a) Find a formula for F in terms of r. (b) Find F when r = 6. (c) Find r when F = 0.8, given r > 0.

(a)  F = k/r²
     20 = k/3² = k/9   so   k = 180
     F = 180/r²
(b)  F = 180/6² = 180/36 = 5
(c)  0.8 = 180/r²   so   r² = 180/0.8 = 225   so   r = 15

In (c), multiply both sides by r² first, then divide by 0.8. Doing it in one line on a calculator is fine, but write the rearranged equation so a method mark can be given.

Worked example 4: inverse square root

a is inversely proportional to √b. When b = 16, a = 3.

(a) Find a when b = 9. (b) Find b when a = 2.

     a = k/√b
     3 = k/√16 = k/4   so   k = 12
     a = 12/√b
(a)  a = 12/√9 = 12/3 = 4
(b)  2 = 12/√b   so   √b = 6   so   b = 36

Worked example 5: inverse cube

y is inversely proportional to x³. When x = 2, y = 500. Find y when x = 5.

     y = k/x³
     500 = k/8   so   k = 4000
     y = 4000/5³ = 4000/125 = 32

Scale factors: what happens when x changes

Some questions give no numbers for k at all. They ask what happens to y when x is multiplied by some factor. If y = kxⁿ and x is multiplied by c, then y is multiplied by cⁿ. For inverse forms, y is divided by cⁿ.

Relationship	x multiplied by 2	x multiplied by 3
y ∝ x	y × 2	y × 3
y ∝ x²	y × 4	y × 9
y ∝ x³	y × 8	y × 27
y ∝ 1/x	y ÷ 2	y ÷ 3
y ∝ 1/x²	y ÷ 4	y ÷ 9

Worked example 6: percentage change

y is directly proportional to x³. x is increased by 20%. Find the percentage increase in y.

x is multiplied by 1.2
y is multiplied by 1.2³ = 1.728
percentage increase in y = (1.728 − 1) × 100 = 72.8%

A common wrong answer is 60% (three lots of 20%). Powers multiply; they do not add percentages. This links to repeated percentage change in Higher statement 1.6A.

Recognising proportion from a table

To test a table of values, work out the quantity that should stay constant.

• For y ∝ x, check y/x.
• For y ∝ x², check y/x².
• For y ∝ 1/x, check xy.
• For y ∝ 1/x², check x²y.

Worked example 7: identify the relationship

x	2	4	5	8
y	40	20	16	10

As x doubles from 2 to 4, y halves, which suggests inverse proportion. Check xy: 2 × 40 = 80, 4 × 20 = 80, 5 × 16 = 80, 8 × 10 = 80. The product is constant, so y is inversely proportional to x and y = 80/x. When x = 32, y = 80/32 = 2.5.

Graphs of proportion

The specification asks you to relate the algebra to the graphs. For positive k, and x ≥ 0 (or x > 0 for inverse forms), each graph has a recognisable shape.

Relationship	Shape of the graph of y against x
y ∝ x	Straight line through the origin; the gradient is k
y ∝ x²	Curve through the origin, getting steeper (half of a parabola)
y ∝ x³	Curve through the origin; with the same k, flatter than y = kx² between 0 and 1 and steeper after x = 1
y ∝ √x	Curve through the origin, getting less steep
y ∝ 1/x, 1/x², 1/x³, 1/√x	Curve that falls as x increases, getting close to both axes but never touching them

Two facts link graphs to the algebra:

• Only direct proportion to x gives a straight line. A straight line that does not pass through the origin is not proportion, because when x = 0, y = k × 0 = 0.
• Every point on the graph satisfies the formula. If you can read one point from the graph, substitute it to find k.

A further idea: if y = kx², then plotting y against x² (instead of against x) gives a straight line through the origin with gradient k. The same works for any of the eight forms, using the right power of x on the horizontal axis.

Worked example 8: formula from a graph

A graph of y against x is a straight line through the origin and the point (4, 10). Find y when x = 7.

straight line through the origin, so y = kx
10 = k × 4   so   k = 2.5
y = 2.5x,  so when x = 7, y = 2.5 × 7 = 17.5

Common errors

• Writing y = x² + k instead of y = kx². The constant multiplies; it is never added.
• Using the wrong power: “inversely proportional to the square of x” is y = k/x², not y = k/x or y = 1/(kx²).
• Giving k but not the formula when the question asks for “a formula for y in terms of x”.
• Forgetting to square-root (or cube-root) at the end when finding x from y = kx² (or kx³).
• Forgetting to square at the end when solving y = k√x for x.
• Rounding k too early. If k is a long decimal, keep it as a fraction or in the calculator memory.
• Adding percentages for powers: a 20% rise in x gives a 44% rise in y when y ∝ x², not 40%.
• Assuming any rising curve is y ∝ x². Check where the graph passes through the origin and whether it gets steeper or less steep.

Where to go next

Fix the formulae and methods in the revision notes, then test yourself with the practice questions, which have full mark schemes. To find other gaps across the course, try the free diagnostics.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities, section 2.5 Proportion (Higher tier only).