Edexcel International GCSE Mathematics A 4MA1: Proportion – Revision Notes

These notes cover 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 it. The whole section is Higher tier only, so it is tested on papers 1H and 2H, where a calculator may be used. For full explanations and longer worked examples, read the study guide first.

When you have finished these notes, try the practice questions. The course hub links the rest of 4MA1, and the printable checklist lets you tick off each statement. The algebra you need for rearranging is revised in the algebraic manipulation revision notes.

What the specification asks (2.5A)

• Set up problems involving direct or inverse proportion.
• Relate algebraic solutions to graphical representation of the equations.
• Only these forms are included: y ∝ x, y ∝ 1/x, y ∝ x², y ∝ 1/x², y ∝ x³, y ∝ 1/x³, y ∝ √x, y ∝ 1/√x.

Foundation statement 1.7D (unknown quantities in direct proportion, missing values in a table) is assumed knowledge.

Key words

• ∝ means “is proportional to”.
• Directly proportional / varies directly as / varies as: y = k × (power of x).
• Inversely proportional / varies inversely as: y = k ÷ (power of x).
• Constant of proportionality: the number k. It is the same for every pair of values.

Formulae

In words	Symbols	Formula	Constant check
y is directly proportional to x	y ∝ x	y = kx	y/x = k
y is proportional to the square of x	y ∝ x²	y = kx²	y/x² = k
y is proportional to the cube of x	y ∝ x³	y = kx³	y/x³ = k
y is proportional to the square root of x	y ∝ √x	y = k√x	y/√x = k
y is inversely proportional to x	y ∝ 1/x	y = k/x	xy = k
y is inversely proportional to the square of x	y ∝ 1/x²	y = k/x²	x²y = k
y is inversely proportional to the cube of x	y ∝ 1/x³	y = k/x³	x³y = k
y is inversely proportional to the square root of x	y ∝ 1/√x	y = k/√x	y√x = k

None of these is on the Higher tier formulae sheet.

Method in steps

Finding and using a proportion formula

1. Write the formula with k (pick the power or root from the wording).
2. Substitute the given pair of values.
3. Solve for k, then rewrite the formula with the number in place of k.
4. Substitute the new value, or rearrange to find the unknown.

Undoing the power when finding x

• y = kx²: divide by k, then square-root.
• y = kx³: divide by k, then cube-root.
• y = k√x: divide by k, then square.
• y = k/x²: multiply by x², divide by y, then square-root.

Worked reminder

P is inversely proportional to √Q. P = 6 when Q = 25. Find P when Q = 36.

P = k/√Q
6 = k/5   so   k = 30
P = 30/√Q = 30/6 = 5

Worked reminder: from a graph

A graph of y against x has the shape of y ∝ x² and passes through the point (2, 12). Find y when x = 4.

y = kx²
12 = k × 2² = 4k   so   k = 3
y = 3x² = 3 × 16 = 48

Pick a point where both coordinates can be read exactly, ideally one marked on the graph.

Worked reminder: no k needed

y is inversely proportional to x. When x = 3, y = 20. Find y when x = 12, without finding k.

x is multiplied by 12 ÷ 3 = 4
y ∝ 1/x, so y is divided by 4
y = 20 ÷ 4 = 5

Finding k (k = 60, y = 60/12 = 5) gives the same answer. Use whichever you are less likely to get wrong.

Scale factors

If y ∝ xⁿ and x is multiplied by c, y is multiplied by cⁿ. If y ∝ 1/xⁿ, y is divided by cⁿ.

Form	Power n	x × 2	x × 10	x × 1.1 (10% rise)
y ∝ x	1	y × 2	y × 10	y × 1.1
y ∝ x²	2	y × 4	y × 100	y × 1.21
y ∝ x³	3	y × 8	y × 1000	y × 1.331
y ∝ √x	½	y × √2	y × √10	y × √1.1
y ∝ 1/x²	−2	y ÷ 4	y ÷ 100	y ÷ 1.21

Percentage change: convert to a multiplier, raise it to the power, then convert back. A 10% rise in x with y ∝ x³ gives y × 1.331, a 33.1% rise.

Graphs: must-know shapes

With k > 0 and x > 0:

• y ∝ x: straight line through the origin, gradient k.
• y ∝ x² and y ∝ x³: curves through the origin that get steeper.
• y ∝ √x: curve through the origin that gets less steep.
• All four inverse forms: curves that decrease, approaching both axes without touching them.

Reading a graph: any point on it satisfies the formula, so substitute one clear point to find k.

Straight-line test: y ∝ x² gives a straight line through the origin if you plot y against x² (gradient k). The same holds for each form with the matching power on the x-axis.

Must-know distinctions

	Direct proportion	Inverse proportion
Formula	y = k × xⁿ	y = k ÷ xⁿ
As x increases (k > 0)	y increases	y decreases
What stays constant	y ÷ xⁿ	y × xⁿ
Graph through origin?	Yes	No, the graph never meets either axis

• “Proportional to the square of x” (y = kx²) is not the same as “the square of y is proportional to x” (y² = kx). The specification only lists forms with y on its own, so read which quantity is squared.
• A straight line not through the origin (such as y = 3x + 2) is a linear relationship but not direct proportion.
• k is a constant: if your value of k changes between two pairs of data, the relationship is not the one you assumed.

Quick self-test

1. y is directly proportional to x. y = 14 when x = 4. Find y when x = 10.
2. y ∝ x². y = 48 when x = 4. Find y when x = 5.
3. y is inversely proportional to x. y = 6 when x = 5. Find y when x = 12.
4. y ∝ 1/x². y = 10 when x = 2. Find x when y = 0.4, given x > 0.
5. y is proportional to the cube of x. y = 54 when x = 3. Find y when x = 5.
6. y ∝ √x. y = 10 when x = 4. Find x when y = 35.
7. y ∝ 1/x³. y = 2 when x = 2. Find y when x = 4.
8. y ∝ x². x is doubled. What happens to y?
9. y ∝ 1/√x. x is multiplied by 9. What happens to y?
10. Which of the eight forms has a straight-line graph through the origin when y is plotted against x?
11. y ∝ x³. x is increased by 10%. Find the percentage increase in y.
12. The table shows x = 1, 2, 3 with y = 4, 16, 36. Write a formula for y in terms of x.

Answers

1. k = 14/4 = 3.5, y = 3.5x, so y = 35.
2. k = 48/16 = 3, y = 3x², so y = 75.
3. k = 6 × 5 = 30, y = 30/x, so y = 2.5.
4. k = 10 × 4 = 40, so 0.4 = 40/x², x² = 100, x = 10.
5. k = 54/27 = 2, y = 2x³, so y = 250.
6. k = 10/2 = 5, so 35 = 5√x, √x = 7, x = 49.
7. k = 2 × 8 = 16, y = 16/x³, so y = 16/64 = 0.25.
8. y is multiplied by 4 (2² = 4).
9. y is divided by 3 (√9 = 3).
10. y ∝ x only.
11. 1.1³ = 1.331, so a 33.1% increase.
12. y/x² = 4 every time, so y = 4x².

Where marks are usually lost

• Writing y = x² + k, or y = kx + c, instead of y = kx². In proportion k multiplies and there is no added constant.
• Choosing y = k/x when the question says “inversely proportional to the square of x”.
• Stopping at “k = 3” when the question asks for a formula for y in terms of x; write y = 3x².
• Leaving x² = 100 as the answer, or forgetting to square √x = 7 to get 49.
• Giving both ±10 when the context (length, time, speed) or the question says the value is positive.
• Adding percentages for a power: a 10% rise in x with y ∝ x² is a 21% rise in y, not 20%.
• Rounding k early, so later answers are slightly out. Keep exact values or use the calculator memory.
• Calling any curve through the origin “y ∝ x²” without checking whether it gets steeper (x², x³) or less steep (√x).
• Saying a straight line with a y-intercept shows direct proportion.

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).