One-Day Deep Reasoning Revision Guide

What this guide is for

This guide is for you if you remember maths best when the method makes sense.

The aim is not to memorise disconnected tricks. The aim is to recognise the structure of a question, choose the right rule, and carry it out calmly.

Paper 1 is non-calculator, so it rewards:

• exact arithmetic
• algebraic control
• fraction, ratio and percentage fluency
• clear working
• spotting hidden relationships

Edexcel Higher Paper 1 is 1 hour 30 minutes and 80 marks. It is one of three equally weighted GCSE maths papers. Any topic from the specification can appear, so this guide prioritises the ideas that connect many topics together.

For the 2026 assessment window, Pearson Edexcel says a Higher tier formula sheet/exam aid will be provided with the question papers. This helps with formula recall, but you still need to know when and how to use each formula.

One-Day Revision Plan

Morning: Core Mechanics

Spend most of the morning on the things that make other topics easier:

1. Fractions, percentages, ratio and proportion
2. Indices, standard form and surds
3. Algebra: simplifying, expanding, factorising and solving

These topics appear directly, but they also sit underneath geometry, graphs, probability and problem solving.

Afternoon: Higher Skills

Focus on the topics where exam questions often combine several ideas:

1. Quadratics and simultaneous equations
2. Graphs and functions
3. Angles, Pythagoras, trigonometry and area/volume
4. Probability and statistics

Final Hour: Exam Practice

Do a short mixed set of questions. For every mistake, write down:

• What topic was it?
• What clue should I have noticed?
• What rule did I need?
• Was the error arithmetic, algebraic, or understanding?

The point is to improve recognition, not just collect marks.

Ultimate Last-Minute Strategy

Last-minute revision is not about learning the whole course from scratch. It is about converting the most likely knowledge into marks.

For Edexcel Higher, remember the mark balance:

• about 40% standard techniques
• about 30% reasoning and communication
• about 30% problem solving

So revision must include more than routine questions. A strong final-day plan needs:

• quick recall of methods
• mixed exam questions
• explanation questions
• multi-step problems
• careful checking habits

The 80/20 revision rule

Spend most time on topics that unlock lots of other topics:

1. Fractions, decimals, percentages and ratio
2. Algebraic manipulation and solving
3. Graphs and sequences
4. Angles, Pythagoras, trigonometry and similarity
5. Probability trees, Venn diagrams and averages

These topics are high value because they appear directly and inside harder problem-solving questions.

The final-day revision cycle

Use this cycle instead of rereading notes:

Recall -> Practise -> Mark -> Fix -> Retry

1. Recall

Close the guide and write down everything remembered about one topic.

Example:

Reverse percentages:
original = final divided by multiplier
increase by 20% means multiplier 1.2
decrease by 20% means multiplier 0.8

2. Practise

Do 3 to 6 questions on that topic.

3. Mark

Check using answers or a mark scheme.

4. Fix

Write the correction in one short sentence:

I forgot that reverse percentage divides by the multiplier.

5. Retry

Do a similar question immediately. This step matters because it proves the correction has stuck.

The three-pass exam method

Paper 1 is 90 minutes for 80 marks. That is a little over 1 minute per mark, but some marks need thinking time. Do not let one hard question eat the paper.

Yes: a sensible exam strategy is to collect the easier marks first, then come back to the hard questions. GCSE papers are not designed so that you must finish question 1 before question 2. You are allowed to move on and return.

Pass 1: collect secure marks

Do the questions you recognise quickly.

If stuck after about 60 to 90 seconds, put a star next to it and move on.

Aim:

Protect the marks you know how to get.

This does not mean rushing. It means not sacrificing five easy marks later in the paper because one difficult early question stole too much time.

Pass 2: attack the starred questions

Return to questions that looked possible but needed more thought.

Write something useful:

• a formula
• a diagram
• an equation
• a ratio table
• a first calculation
• a labelled triangle

Method marks often come from visible working, even if the final answer is wrong.

Pass 3: check and rescue

Use the final minutes to check:

• Did I answer the actual question?
• Did I include units?
• Did I round only at the end?
• Did I show working on multi-mark questions?
• Is my answer reasonable?
• Did I accidentally leave a question blank?

The non-calculator fluency list

Before Paper 1, these should feel automatic:

Times tables

Know multiplication facts up to:

12 × 12

They appear inside factorising, fractions, ratio, area, probability and surds.

Square numbers

Know at least:

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100
11² = 121
12² = 144
13² = 169
14² = 196
15² = 225

These help with Pythagoras, surds, indices, area and quadratics.

Cube numbers

Know:

1³ = 1
2³ = 8
3³ = 27
4³ = 64
5³ = 125

These help with volume, indices and bounds.

Fraction, decimal and percentage equivalents

Know these cold:

These save time and make estimates easier.

Paper 1 mental arithmetic tricks

Build percentages from 10%

Example:

Find 35% of 240.

10% = 24
30% = 3 × 10% = 3 × 24 = 72
5% = 12
35% = 84

Divide by fractions using reciprocals

6 divided by 3/4 = 6 × 4/3 = 8

Use factor pairs

If multiplying:

25 × 36

Think:

25 × 4 × 9 = 100 × 9 = 900

Factor pairs are two numbers that multiply to make a target number.

Example:

factor pairs of 36:
1 × 36
2 × 18
3 × 12
4 × 9
6 × 6

How to find them:

Start at 1 and test small numbers in order. If a number divides exactly, write the pair.

Factor pairs help with:

• simplifying fractions
• factorising quadratics
• mental multiplication
• finding square factors in surds

Estimate before calculating

Before doing a calculation, ask:

What size should the answer be?

This catches place-value errors.

What to do when completely stuck

Do not freeze. Start with one of these:

Write down what you know.
Draw or label a diagram.
Define x.
Make a table.
Write a formula.
Try a simpler number.
Look for a ratio, difference, total or multiplier.

For wordy questions, translate one sentence at a time.

For proof questions, start with algebra:

even = 2n
odd = 2n + 1
consecutive = n, n + 1

For geometry, mark the diagram with every known angle or length.

The night-before plan

Do not stay up late trying to learn obscure topics. The best final evening is:

1. Review the memory hooks.
2. Do 20 to 30 minutes of mixed questions.
3. Correct mistakes carefully.
4. Pack equipment.
5. Sleep.

For Paper 1, equipment should include:

• black pen
• pencil
• ruler
• protractor
• pair of compasses
• eraser

A clean formula sheet/exam aid will be provided in the exam for 2026. Do not take your own copy into the exam.

The brain does maths better when it is rested.

Emergency Two-Hour Plan

Use this if there is very little time left.

First 20 minutes: rescue facts

Review:

• square numbers to 15 squared
• cube numbers to 5 cubed
• fraction, decimal and percentage equivalents
• index laws
• Pythagoras and trig ratios
• percentage multipliers

Do not spend this time reading long explanations. This is memory activation.

Next 45 minutes: high-yield method practice

Do short questions on:

• fractions and percentages
• ratio
• algebra solving
• expanding and factorising
• Pythagoras and trigonometry
• probability trees

Mark immediately. Fix mistakes immediately.

Next 35 minutes: mixed exam questions

Do mixed questions, not topic-by-topic. The exam will not tell you the topic, so practise recognising it.

For each question, write one of these beside it:

N = number
A = algebra
R = ratio/proportion
G = geometry
P = probability
S = statistics

This trains structure recognition.

Final 20 minutes: error review

Look only at mistakes and memory hooks.

Ask:

• Which mistakes keep repeating?
• Which topics are worth one more example?
• Which formulas must I keep in mind?

Then stop. A tired brain starts losing easy marks.

1. What Paper 1 Is Really Testing

The four big skills

1. Recognising structure

Most questions are asking:

What kind of relationship is this?

Examples:

• repeated multiplication means indices
• equal parts means fractions
• relative amounts means ratio
• constant rate means linear graph
• changing rate means quadratic or curve
• unknown value means algebra
• same shape means similarity or trigonometry

2. Using rules accurately

Once the structure is known, the method follows. A lot of marks come from carrying out ordinary steps cleanly.

3. Preserving equality

In algebra, equations must stay balanced. If you do something to one side, you must do the equivalent to the other side.

4. Linking ideas

Higher questions often combine topics.

For example, a geometry question might involve:

• angle facts
• forming an algebraic equation
• solving the equation
• checking the answer fits the diagram

The survival principle

Mathematics is usually about preserving or undoing relationships.

Important pairs:

• expanding and factorising
• squaring and square rooting
• multiplying and dividing
• increasing by a percentage and reversing a percentage
• drawing a graph and reading solutions from it
• a function and its inverse

When stuck, ask:

What operation has been done, and how can I undo it?

2. Non-Calculator Number Skills

Fractions

A fraction is division:

a / b = a divided by b

The numerator is the amount being divided. The denominator tells you how many equal parts the whole has been split into.

In:

3/4

the top number is the numerator:

3 = how many parts we have

the bottom number is the denominator:

4 = how many equal parts the whole is split into

Think:

numerator
-----------
denominator

So 3/4 means 3 out of 4 equal parts.

Why equivalent fractions work

1/2 = 2/4 = 50/100

Multiplying the top and bottom by the same number does not change the value because the size of each part and the number of parts change together.

Example:

1/2 = one out of two equal parts
2/4 = two out of four equal parts

The second fraction has twice as many parts, but the parts are half as big. The overall amount is unchanged.

Algebraically, this works because multiplying by k/k is multiplying by 1:

(a × k) / (b × k) = a / b

since:

k / k = 1

Adding and subtracting fractions

You need a common denominator because the pieces must be the same size.

1/3 + 1/4

Thirds and quarters are different units. Convert both into twelfths:

1/3 = 4/12
1/4 = 3/12

So:

1/3 + 1/4 = 7/12

Multiplying fractions

2/3 × 4/5 = 8/15

Multiplying by a fraction means taking a fraction of something. Fractions act like scaling instructions.

Dividing fractions

2/3 divided by 4/5

This means:

How many lots of 4/5 fit into 2/3?

Dividing by a fraction is the same as multiplying by its reciprocal:

2/3 × 5/4 = 10/12 = 5/6

Why "flip and multiply" works:

Dividing by a number means "how many of that thing fit into the first thing?" Dividing by 4/5 asks: how many chunks of size 4/5 fit into 2/3?

A chunk of 4/5 is 4/5 of a whole. If we multiply our amount by 5/4, we are scaling it up so that 4/5-sized chunks become whole units. After that scaling, the question "how many chunks fit?" turns into ordinary multiplication.

A reciprocal is the number that multiplies with the original number to make 1.

Examples:

reciprocal of 4 is 1/4 because 4 × 1/4 = 1
reciprocal of 2/3 is 3/2 because 2/3 × 3/2 = 1

Why it helps:

Dividing by 4/5 means asking how many 4/5 chunks fit. Multiplying by 5/4 reverses the 4/5 scaling, so it turns the question back into an ordinary multiplication.

the top and bottom have a common factor

Simplify 36/48.

36/48 = 3/4

Simplify:
1. 18/24
2. 45/60
3. 56/64

1. 3/4
2. 3/4
3. 7/8

fractions are being added or subtracted

2/3 + 5/6

2/3 = 4/6
4/6 + 5/6 = 9/6 = 3/2 = 1 1/2

2/3 + 5/6 is not 7/9

1. 1/4 + 2/3
2. 5/6 - 1/3
3. 3/5 + 7/10

1. 11/12
2. 1/2
3. 13/10 or 1 3/10

the question says "of" or asks for a fraction of a quantity

you are finding how many fractional parts fit into something

multiply: straight across
divide: keep, change, flip

3/4 × 8/9
= 24/36
= 2/3

5/6 divided by 2/3
= 5/6 × 3/2
= 15/12
= 5/4

1. 2/5 × 15/4
2. 7/8 divided by 1/4
3. 3/10 of 50

1. 3/2
2. 7/2
3. 15

Decimals and place value

Decimals are fractions written in base 10.

0.7 = 7/10
0.07 = 7/100
0.007 = 7/1000

For non-calculator arithmetic, line up place values carefully. A common mistake is treating 0.7 and 0.07 as close in size. They are ten times different.

Percentages

Percent means "per hundred".

35% = 35/100 = 0.35

Percentages are also scaling instructions.

Percentage increase

Increase by 20%:

Original = 100%
Increase = 20%
New amount = 120%
Multiplier = 1.2

So increasing 60 by 20%:

60 × 1.2 = 72

Percentage decrease

Decrease by 20%:

Original = 100%
Decrease = 20%
Remaining = 80%
Multiplier = 0.8

Reverse percentages

If the final amount after a 20% increase is 72:

72 = 120%

So:

original = 72 / 1.2 = 60

The key idea:

Do not subtract 20% from 72. The 20% was based on the original amount, not the final amount.

the question asks for a percentage of a number

Find 35% of 240.

10% = 24
30% = 3 × 10% = 3 × 24 = 72
5% = 12
35% = 84

1. 15% of 80
2. 35% of 200
3. 12.5% of 64

1. 12
2. 70
3. 8

an amount is increased or decreased by a percentage

Increase 70 by 15%.

100% + 15% = 115%
115% = 1.15
70 × 1.15 = 80.5

10% of 70 = 7
5% of 70 = 3.5
15% of 70 = 10.5
70 + 10.5 = 80.5

Find 20% of 150.

10% = 15
20% = 2 × 15 = 30

Increase 80 by 15%.

10% = 8
5% = 4
15% = 12
80 + 12 = 92

A price is increased by 25% to 75 pounds.
Find the original price.

final = 125%
75 = 125%
1% = 75 / 125
100% = 75 / 1.25 = 60

60 pounds

1. Increase 50 by 20%.
2. Decrease 90 by 30%.
3. Increase 160 by 12.5%.

1. 60
2. 63
3. 180

the final amount is known and the original is required

After a 20% increase, a price is 72. Find the original price.

original = 100%
final = 120%
120% = 1.2
72 / 1.2 = 60

1. After a 25% increase, the value is 100. Find the original.
2. After a 10% decrease, the value is 45. Find the original.
3. After a 40% increase, the value is 84. Find the original.

1. 80
2. 50
3. 60

Ratio

A ratio compares relative sizes.

2 : 3

means:

For every 2 parts of one quantity, there are 3 parts of the other.

It is multiplicative, not additive.

Sharing in a ratio

Share 60 in the ratio 2 : 3.

Total parts:

2 + 3 = 5

One part:

60 / 5 = 12

So:

2 parts = 24
3 parts = 36

Proportion

Direct proportion

If y is directly proportional to x, then as x doubles, y doubles.

y = kx

k is the constant multiplier.

Inverse proportion

If y is inversely proportional to x, then as x doubles, y halves.

y = k/x

The product xy stays constant.

there is a total amount and a ratio

Share 84 in the ratio 3 : 4.

3 + 4 = 7 parts
84 / 7 = 12
3 parts = 36
4 parts = 48

36 : 48

1. Share 60 in the ratio 2 : 3.
2. Share 96 in the ratio 5 : 7.
3. Share 120 in the ratio 1 : 2 : 3.

1. 24 and 36
2. 40 and 56
3. 20, 40 and 60

the question gives how much bigger one part is than another

The ratio of boys to girls is 5 : 3.
There are 18 more boys than girls.
How many children are there altogether?

5 - 3 = 2 parts

2 parts = 18
1 part = 9

5 + 3 = 8
8 parts = 8 × 9 = 72

1. Ratio A : B is 7 : 4. A is 21 more than B. Find A + B.
2. Ratio red : blue is 2 : 5. There are 24 more blue than red. Find the total.

1. 77
2. 56

one quantity scales in the same direction as another

y is directly proportional to x.
When x = 4, y = 18.
Find y when x = 10.

y = kx
18 = 4k
k = 4.5
y = 4.5x
when x = 10, y = 45

1. y is directly proportional to x. When x = 6, y = 30. Find y when x = 8.
2. y is directly proportional to x. When x = 5, y = 12. Find x when y = 36.

1. 40
2. 15

one quantity increases while the other decreases proportionally

y is inversely proportional to x.
When x = 5, y = 12.
Find y when x = 20.

y = k/x
12 = k/5
k = 60
y = 60/x
when x = 20, y = 3

1. y is inversely proportional to x. When x = 4, y = 15. Find y when x = 10.
2. y is inversely proportional to x. When x = 8, y = 6. Find x when y = 12.

1. 6
2. 4

3. Indices, Standard Form and Surds

Indices

An index means repeated multiplication:

a⁴ = a × a × a × a

In normal maths writing, this is:

a⁴ = a × a × a × a

The small raised number tells you how many copies of the base are being multiplied.

base = a
index/power = 4

Multiplication rule

a^m × a^n = am+n

Reason:

You are counting total factors of a.

Example:

a³ × a²
= (a × a × a) × (a × a)
= a × a × a × a × a
= a⁵

So:

a³ × a² = a3+2 = a⁵

Division rule

a^m / a^n = am-n

Reason:

Shared factors cancel.

Example:

a⁵ / a²
= (a × a × a × a × a) / (a × a)

Two a factors cancel from the top and bottom:

= a × a × a
= a³

So:

a⁵ / a² = a5-2 = a³

Important:

Only use these rules when the base is the same.

2³ x 2⁴ = 2⁷

but:

2³ x 3⁴

cannot be combined into one power.

Zero powers

a⁰ = 1

because:

a³ / a³ = 1

and also:

a3-3 = a⁰

Negative powers

a^(-1) = 1/a
a^(-n) = 1 / a^n

Why this works:

The division rule says a^m / a^n = a^(m-n). Look at what happens when the bottom power is bigger than the top one:

a² / a⁵ = a^(2 - 5) = a^(-3)

but you can also work it out directly:
a² / a⁵ = (a × a) / (a × a × a × a × a) = 1 / (a × a × a) = 1/a³

Both results have to be the same number, so a^(-3) = 1/a³. A negative power is the rule's way of saying "this is on the bottom of a fraction".

Fractional powers

a1/2 = square root of a

because squaring it gets back to a.

Standard form

Standard form writes very large or very small numbers as:

a x 10^n

where:

1 <= a < 10

Examples:

45000 = 4.5 × 10⁴
0.0032 = 3.2 × 10−³

The power of 10 records how many times the decimal point has moved.

Example:

45000 -> 4.5

The decimal point moved 4 places left, so:

45000 = 4.5 × 10⁴

For a small number:

0.0032 -> 3.2

The decimal point moved 3 places right, so:

0.0032 = 3.2 × 10−³

Positive powers mean large numbers. Negative powers mean small numbers.

Surds

A surd is a square root that cannot be simplified to a whole number.

Examples:

√9 = 3

This is not a surd after simplifying because the answer is a whole number.

√2, √3, √5, √7

These are surds because their decimal answers go on forever without repeating. GCSE maths often keeps them in exact root form instead of rounding.

√12

The problem is that √12 is not a whole number. So we look for a square number that is hiding inside 12.

Square numbers are useful because their square roots are clean:

√4 = 2
√9 = 3
√16 = 4

Now split 12 into:

12 = 4 × 3

This is useful because 4 is a square number.

√12 = √(4 × 3) = 2√3

Why this works:

√12 = √(4 × 3)

The square root sign applies to the whole multiplication. You can split it:

√(4 × 3) = √4 × √3

Now simplify the part that has a clean square root:

√4 = 2

So:

√4 × √3 = 2 × √3

Mathematicians write 2 × √3 as:

2√3

So:

√12 = 2√3

Check that √12 and 2√3 really are the same number

It is fair to be suspicious. We started with √12, did some moves, and ended up with 2√3. They look completely different. Are they really equal?

Yes — and we can check it two ways. Decimals first, then with proper reasoning.

Check 1: turn both into decimals

If √12 and 2√3 really are the same number, they must give the same decimal.

√12  ≈ 3.4641016...

√3   ≈ 1.7320508...
2√3  = 2 × √3
     = 2 × 1.7320508...
     ≈ 3.4641016...

Same decimal. So they are the same number, just written differently.

Check 2: square both of them

The square of √12 is, by the very definition of "square root":

(√12)² = 12

Why: the square root of a number is the number that squares to give it. So squaring undoes the square root — the two operations are inverses of each other, just like adding 5 and subtracting 5 cancel out.

Now square 2√3:

(2√3)²
= 2√3 × 2√3
= 2 × 2 × √3 × √3
= 4 × 3
= 12

Both expressions square to 12. Both are positive. So they must be the same number.

Why exam mark schemes prefer 2√3:

• The number under the root is as small as possible, so the value is easier to picture.
• It makes adding and subtracting surds possible. For instance 2√3 + 5√3 = 7√3 — you could not collect terms from √12 + √75 without simplifying first.
• It is the form mark schemes expect on Paper 1.

The aim is to find the biggest square number hidden inside the root.

Useful square numbers:

4, 9, 16, 25, 36, 49, 64, 81, 100

Rationalising a denominator

Rationalising means removing a surd from the denominator of a fraction.

The denominator is the bottom of the fraction. In GCSE exact answers, examiners usually prefer no square root left on the bottom.

The key idea is:

√5 × √5 = 5

because a square root multiplied by itself gives the number inside the root.

Example:

3 / √5

We want to remove √5 from the bottom. Multiply the bottom by √5:

√5 × √5 = 5

But to keep the fraction the same value, whatever we do to the bottom we must also do to the top.

So multiply by:

√5 / √5

This is allowed because:

√5 / √5 = 1

Now:

3 / √5
= 3 / √5 × √5 / √5
= 3√5 / 5

The bottom is now rational because it is 5, not a square root.

terms have the same base

multiply same base: add powers
divide same base: subtract powers
power of a power: multiply powers
negative power: reciprocal
zero power: 1

x⁵ × x³ / x²
= x5+3-2
= x⁶

1. a⁴ × a⁷
2. y⁹ / y³
3. (x²)^5
4. 5⁰

1. a¹¹
2. y⁶
3. x¹⁰
4. 1

numbers are very large or very small

0.00072 = 7.2 × 10−⁴

1. Write 56000 in standard form.
2. Write 0.0034 in standard form.
3. Work out (3 × 10⁵)(4 × 10²).

1. 5.6 × 10⁴
2. 3.4 × 10−³
3. 1.2 × 10⁸

a square root cannot be written as a whole number

√72
= √(36 × 2)
= 6√2

Simplify √20.

20 = 4 × 5
√20 = √4 × √5 = 2√5

Simplify 3√18.

18 = 9 × 2
√18 = √9 × √2 = 3√2

3√18 = 3 × 3√2 = 9√2

Simplify √50 + √8.

√50 = √(25 × 2) = 5√2
√8 = √(4 × 2) = 2√2

5√2 + 2√2 = 7√2

multiply top and bottom by the surd in the denominator

5/√3
= 5√3/3

Rationalise 1/√2.

1/√2 × √2/√2
= √2/2

Rationalise 6/√3.

6/√3 × √3/√3
= 6√3/3
= 2√3

Rationalise 4/(2√5).

4/(2√5) = 2/√5

2/√5 × √5/√5
= 2√5/5

√(9 + 16) is not √9 + √16

1. Simplify √48.
2. Simplify √75.
3. Rationalise 4/√7.

1. 4√3
2. 5√3
3. 4√7/7

4. Algebra

What algebra is

Algebra is a way of describing number patterns when the exact value can vary.

A letter is not just an unknown. It can represent a changing value, a general value, or a quantity we need to find.

One useful analogy:

Algebra is like using a labelled box for a number.

x = the number in the box

If the question says:

3x + 5

that means:

3 lots of the box, then add 5

Algebra feels abstract, but most of it is just careful bookkeeping. You are keeping track of unknown quantities without needing to know their values yet.

Another analogy:

An equation is like a balance scale.

left side = right side

The two sides may look different, but they have the same value. Solving means doing legal moves that keep the balance equal until the letter is on its own.

Simplifying expressions

3x + 5x = 8x

Reason:

3 groups of x plus 5 groups of x makes 8 groups of x.

You can only combine like terms:

3x + 5y

cannot become 8xy, because x and y are different objects.

Why this rule exists:

Each letter stands for a different number that we do not yet know. 3x means "three copies of whatever x is", and 5y means "five copies of whatever y is". Until we know what x and y are, we cannot combine them — they are different things.

So:

3x + 5x = 8x
3x + 5y stays as 3x + 5y

Expanding brackets

3(x + 4)

means 3 groups of (x + 4), so:

3x + 12

This is the distributive law:

a(b + c) = ab + ac

Double brackets

(x + 2)(x + 5)

Every term in the first bracket multiplies every term in the second:

Each cell is one of the four products. Add them up: x² + 5x + 2x + 10 = x² + 7x + 10.

So the four multiplications are:

x times x  = x² × times 5  = 5x
2 times x  = 2x
2 times 5  = 10

Arrow version:

Four arrows, four products. The teal arrows show what the x pairs with; the red arrows show what the 2 pairs with. Add the products to finish:

(x + 2)(x + 5)
= x·x + x·5 + 2·x + 2·5
= x² + 5x + 2x + 10
= x² + 7x + 10

Then collect the middle terms:

x² + 5x + 2x + 10
= x² + 7x + 10

Factorising

Factorising is the reverse of expanding.

Example:

6x + 12

Both terms contain a factor of 6:

6(x + 2)

Quadratic factorising

For:

x² + 7x + 12

find two numbers that:

• multiply to 12
• add to 7

The numbers are 3 and 4, so:

x² + 7x + 12 = (x + 3)(x + 4)

Why this works:

When expanded:

(x + 3)(x + 4) = x² + 4x + 3x + 12

The middle term comes from adding the two cross-products.

Solving equations

An equation is balanced.

That means the left side and right side have the same value:

left side = right side

Solving means finding the value of the letter that keeps the balance true.

The important rule is:

Whatever you do to one side, you must do to the other side.

Why?

Because if two sides are equal, changing only one side would break the equality.

Think of a balance scale:

If you remove 5 from the left pan, you must remove 5 from the right pan too, otherwise the scale is no longer balanced.

Example:

3x + 5 = 20

The left side means:

3 lots of x, plus 5

We want x on its own eventually, so first remove the +5.

The opposite of adding 5 is subtracting 5.

But we must subtract 5 from both sides:

3x + 5 - 5 = 20 - 5
3x = 15

Now the equation says:

3x = 15

This means:

3 lots of × make 15

So one lot of x is found by splitting 15 into 3 equal groups:

3x / 3 = 15 / 3
x = 5

So "undo" means use the opposite operation:

• undo adding by subtracting
• undo subtracting by adding
• undo multiplying by dividing
• undo dividing by multiplying

And always do the same thing to both sides.

Inequalities

Inequalities use symbols such as:

<, >, <=, >=

An inequality is like an equation, but instead of saying two things are equal, it says one side is bigger or smaller.

x > 4

means:

x is greater than 4

On a number line, that means values to the right of 4:

The open circle at 4 means "4 is not included" (because the inequality is strict). For x ≥ 4 we would use a filled circle.

The open circle means 4 itself is not included.

Most solving steps work like equations:

x + 3 < 10

Subtract 3 from both sides:

x < 7

The key extra rule:

When multiplying or dividing by a negative number, reverse the inequality sign.

Why this happens:

Multiplying by a negative number flips the number line. The positives swap places with the negatives, so things that were further to the right end up further to the left. Anything that was bigger than becomes smaller than, and vice versa.

You can test this with any pair of numbers:

Example:

3 > 1

This is true. But multiply both numbers by -1:

-3 and -1

Now:

-3 < -1

The direction has reversed.

Example:

-2x > 8

To get x on its own, divide by -2.

Because we divide by a negative, reverse the sign:

x < -4

Check with a value less than -4, such as x = -5:

-2(-5) = 10
10 > 8

It works.

Simultaneous equations

Simultaneous equations are two conditions that must both be true.

They are called simultaneous because both equations are true at the same time for the same values of the letters.

Think of each equation as a clue. One clue is usually not enough to find two unknowns, but two good clues can pin down one pair of values.

Example:

x + y = 10
x - y = 2

Add the equations:

2x = 12
x = 6

Then:

y = 4

The reason elimination works is that you combine equations in a way that removes one variable while preserving the shared solution.

Step-by-step reasoning:

   x + y = 10
   x - y = 2

Add the left sides and add the right sides:

(x + y) + (x - y) = 10 + 2

The +y and -y cancel because together they make zero:

x + x = 12

This is the same as:

2x = 12

There are two equal x parts. To find one x, split 12 into 2 equal parts:

12 / 2 = 6

So:

x = 6

Now put x = 6 back into one of the original equations:

x + y = 10
6 + y = 10
y = 4

Check:

6 + 4 = 10
6 - 4 = 2

Both equations work, so the solution is correct.

a term is multiplying a bracket

-3(2x - 5)
= -6x + 15

1. 4(x + 7)
2. -2(3x - 6)
3. 5(2a - 3b)

1. 4x + 28
2. -6x + 12
3. 10a - 15b

two brackets are multiplied

(x + 6)(x - 2)
= x² - 2x + 6x - 12
= x² + 4x - 12

x² - 2x + 6x - 12 = x² + 4x - 12

(x + 3)(x + 4)

x² + 4x + 3x + 12
= x² + 7x + 12

(x - 5)(x + 2)

x² + 2x - 5x - 10
= x² - 3x - 10

(2x + 1)(x - 3)

2x times x = 2x²
2x times -3 = -6x
1 times x = x
1 times -3 = -3

2x² - 6x + x - 3
= 2x² - 5x - 3

1. (x + 3)(x + 4)
2. (x - 5)(x + 2)
3. (2x + 1)(x - 3)

1. x² + 7x + 12
2. x² - 3x - 10
3. 2x² - 5x - 3

terms share a common factor

12x + 18
= 6(2x + 3)

1. 8x + 20
2. 15a - 25
3. 6x² + 9x

1. 4(2x + 5)
2. 5(3a - 5)
3. 3x(2x + 3)

the expression looks like x² + bx + c

x² - x - 12

-4 and 3

x² - x - 12 = (x - 4)(x + 3)

1. x² + 9x + 20
2. x² - 7x + 12
3. x² + 2x - 15

1. (x + 4)(x + 5)
2. (x - 3)(x - 4)
3. (x + 5)(x - 3)

two square terms are subtracted

a² - b² = (a - b)(a + b)

x² - 49
= (x - 7)(x + 7)

1. x² - 25
2. 4x² - 9
3. 100 - y²

1. (x - 5)(x + 5)
2. (2x - 3)(2x + 3)
3. (10 - y)(10 + y)

there is one unknown and an equals sign

4x - 7 = 2x + 9

2x - 7 = 9
2x = 16
x = 8

3x + 4 = 19

3x = 15
x = 5

5x - 6 = 2x + 12

3x - 6 = 12
3x = 18
x = 6

2(x + 3) = 3x - 5

2x + 6 = 3x - 5

6 = x - 5

x = 11

1. 3x + 4 = 19
2. 5x - 6 = 2x + 12
3. 2(x + 3) = 18

1. x = 5
2. x = 6
3. x = 6

x appears in a fraction or the equation contains fractional terms

x/3 + 2 = 5

x + 6 = 15
x = 9

1. x/4 + 3 = 8
2. x/5 - 2 = 4
3. x/2 + x/3 = 10

1. x = 20
2. x = 30
3. x = 12

a quadratic equation can be factorised

x² + 5x + 6 = 0

(x + 2)(x + 3) = 0

x + 2 = 0 or x + 3 = 0
x = -2 or x = -3

1. x² - 9x + 20 = 0
2. x² + x - 12 = 0
3. x² - 16 = 0

1. x = 4 or x = 5
2. x = 3 or x = -4
3. x = 4 or x = -4

there are two equations with two unknowns

2x + y = 13
x - y = 2

3x = 15
x = 5

5 - y = 2
y = 3

x + y = 9
x - y = 3

2x = 12
x = 6

6 + y = 9
y = 3

2x + y = 11
x + y = 7

(2x + y) - (x + y) = 11 - 7
x = 4

4 + y = 7
y = 3

3x + 2y = 16
x + 2y = 8

(3x + 2y) - (x + 2y) = 16 - 8
2x = 8
x = 4

4 + 2y = 8
2y = 4
y = 2

1. x + y = 9, x - y = 3
2. 2x + y = 11, x + y = 7
3. 3x - y = 10, x + y = 6

1. x = 6, y = 3
2. x = 4, y = 3
3. x = 4, y = 2

5. Graphs and Functions

Linear graphs

y = mx + c

means:

• m is the gradient
• c is the y-intercept

The gradient is:

change in y / change in x

It measures rate of change.

Visual idea:

This line rises as it moves to the right, so it has a positive gradient.

For:

y = 2x + 1

the +1 tells you the line crosses the y-axis at 1.

To draw a straight line, make a small table of values:

Plot the points and join them with a straight line.

Parallel lines

Parallel lines have the same gradient because they rise at the same rate.

Perpendicular lines

For perpendicular lines:

m1 × m2 = -1

Their gradients are negative reciprocals.

Example:

If one gradient is 2, the perpendicular gradient is:

-1/2

Quadratic graphs

y = x²

is curved because the rate of change is not constant.

As x increases, the outputs increase faster.

Roots

Roots are the x values where:

y = 0

On a graph, these are the points where the curve crosses the x-axis.

In algebra, roots are solutions to an equation.

Visual idea:

A root is where the curve meets the x-axis. At that point y = 0.

At the roots, the graph touches or crosses the x-axis, so y = 0.

Functions

A function is a rule that turns an input into an output.

Example:

f(x) = 2x + 3

If x = 5:

f(5) = 2(5) + 3 = 13

Inverse functions

An inverse function undoes the original function.

If:

f(x) = 2x + 3

then the inverse reverses the steps:

1. subtract 3
2. divide by 2

So:

f−¹(x) = (x - 3) / 2

the graph or equation is linear

Draw y = 2x + 1.

c = 1, so start at (0, 1)
m = 2, so go up 2 for every 1 right
points: (0, 1), (1, 3), (2, 5)

1. State the gradient and y-intercept of y = 3x - 4.
2. State the gradient and y-intercept of y = -2x + 5.
3. Find the gradient between (1, 4) and (3, 10).

1. gradient 3, y-intercept -4
2. gradient -2, y-intercept 5
3. 3

the sequence has a constant difference

Find the nth term of 5, 8, 11, 14, ...

3n

3, 6, 9, 12

nth term = 3n + 2

1. 4, 7, 10, 13, ...
2. 9, 14, 19, 24, ...
3. 20, 17, 14, 11, ...

1. 3n + 1
2. 5n + 4
3. 23 - 3n

function notation such as f(x) appears

f(x) = 2x² - 5
Find f(4).

f(4) = 2(4²) - 5
= 32 - 5
= 27

1. f(x) = 3x + 7. Find f(5).
2. g(x) = x² - 4. Find g(-3).
3. h(x) = 10 - 2x. Find h(6).

1. 22
2. 5
3. -2

6. Geometry and Measures

Angle facts

Straight line

Angles on a straight line add to:

180 degrees

This is half a full turn.

Around a point

Angles around a point add to:

360 degrees

This is one full turn.

Triangle

Angles in a triangle add to:

180 degrees

A useful way to remember this is that the triangle's angles can be rearranged to form a straight line.

Quadrilateral

Angles in a quadrilateral add to:

360 degrees

because a quadrilateral can be split into two triangles.

Parallel line angles

A transversal is a line that crosses two parallel lines. The crossings make eight angles. Because the parallel lines point in the same direction, the angles at the top crossing are identical copies of the angles at the bottom crossing — just slid along the transversal.

Three rules to recognise:

• Corresponding angles are equal — they sit in the same position at both crossings.
• Alternate angles are equal — these form a Z-shape across the transversal.
• Co-interior angles add to 180° — these sit on the same side of the transversal, between the parallel lines, forming a C-shape. They are supplementary because together they fill a half-turn formed by the transversal.

Pythagoras

For a right-angled triangle:

a² + b² = c²

where c is the hypotenuse, the longest side opposite the right angle.

Side c (the hypotenuse) is the sloping side opposite the right angle. It is always the longest side.

Why it works:

Pythagoras compares the areas of squares built on the sides, not just the side lengths. That is why the lengths are squared.

The two smaller squares (one on each shorter side) together have the same total area as the square on the hypotenuse. That is the geometric meaning of a² + b² = c².

Trigonometry

Trig ratios compare sides in right-angled triangles.

sin(theta) = opposite / hypotenuse
cos(theta) = adjacent / hypotenuse
tan(theta) = opposite / adjacent

Side labels depend on which angle you are using. The labels are not fixed to the triangle — they are fixed to the angle θ.

From the angle θ (theta), look at each side and ask "where is this in relation to theta?"

Same physical triangle. θ has moved to a different corner. Now the side that was "opposite" is "adjacent", and what was "adjacent" is "opposite". Only the hypotenuse keeps its label — because it is defined by the right angle, not by θ.

SOH CAH TOA reminds you which ratio to choose once the sides are labelled:

SOH:  sin θ = Opposite / Hypotenuse
CAH:  cos θ = Adjacent / Hypotenuse
TOA:  tan θ = Opposite / Adjacent

Where the exact trig values come from

Exact values are often taught as pure memory facts. They are easier to remember if you have seen where they come from. There are only two special triangles to know.

The 30°–60°–90° triangle (from an equilateral triangle)

Take an equilateral triangle with sides of length 2. All its angles are 60°. Cut it down the middle. The cut bisects the top angle, so the new triangle has angles 30°, 60° and 90°. The base is now half — length 1. The slanted side stays length 2. By Pythagoras, the height is √(2² − 1²) = √3.

Now read the trig ratios off this triangle:

For the 30° angle (top corner of the isolated triangle):
  opposite to 30° = 1      adjacent = √3      hypotenuse = 2
  sin 30° = 1/2            cos 30° = √3/2     tan 30° = 1/√3

For the 60° angle (bottom-left corner):
  opposite to 60° = √3     adjacent = 1       hypotenuse = 2
  sin 60° = √3/2           cos 60° = 1/2      tan 60° = √3

Same triangle, two angles, six values. Once you have drawn this once, you can re-derive any of them in seconds.

The 45°–45°–90° triangle (from a unit square)

Take a square with sides of length 1. Cut it along its diagonal. The diagonal splits the corner angles in half, so each acute angle is 45°. By Pythagoras the diagonal has length √(1² + 1²) = √2.

Both acute angles are 45°, so the opposite and adjacent sides are equal:

sin 45° = 1/√2 = √2/2
cos 45° = 1/√2 = √2/2
tan 45° = 1/1   = 1

And the boundary cases (0° and 90°) come from imagining the triangle being squashed flat:

At 0°:  the "opposite" side shrinks to zero  → sin 0° = 0,   cos 0° = 1,   tan 0° = 0
At 90°: the "opposite" side equals the hypotenuse → sin 90° = 1, cos 90° = 0, tan 90° = undefined

So all the exact values come from just two pictures — the half-equilateral and the half-square.

Area and volume

Area measures two-dimensional space, so units are squared:

cm², m²

Volume measures three-dimensional space, so units are cubed:

cm³, m³

If a length scale factor is k:

• area scale factor is k²
• volume scale factor is k³

This is because area has two dimensions and volume has three.

Similar shapes

Similar shapes have the same shape but may be different sizes.

Corresponding lengths are multiplied by the same scale factor.

If one length doubles, every length doubles. The angles stay the same.

the question gives a diagram with unknown angles

Two angles on a straight line are 3x and 60 degrees.
Find x.

3x + 60 = 180
3x = 120
x = 40

1. A triangle has angles 50, 70 and x. Find x.
2. Angles around a point are 90, 140 and x. Find x.
3. A regular pentagon has exterior angle x. Find x.

1. 60 degrees
2. 130 degrees
3. 72 degrees

there is a right-angled triangle and a missing side

A right-angled triangle has shorter sides 6 cm and 8 cm.
Find the hypotenuse.

c² = 6² + 8²
c² = 36 + 64
c² = 100
c = 10

Short sides are 5 cm and 12 cm. Find the hypotenuse.

c² = 5² + 12²
c² = 25 + 144 = 169
c = 13

Hypotenuse is 13 cm. One shorter side is 5 cm.
Find the other shorter side.

missing² = 13² - 5²
missing² = 169 - 25 = 144
missing = 12

A ladder is 10 m long and reaches 8 m up a wall.
How far is the foot of the ladder from the wall?

distance² = 10² - 8²
distance² = 100 - 64 = 36
distance = 6

short² = hypotenuse² - other short²

1. Short sides 5 and 12. Find hypotenuse.
2. Hypotenuse 13, one short side 5. Find the other short side.
3. Short sides 9 and 12. Find hypotenuse.

1. 13
2. 12
3. 15

there is a right-angled triangle involving an angle and sides

In a right-angled triangle, angle theta is 30 degrees.
The hypotenuse is 10 cm.
Find the opposite side.

sin(theta) = opposite / hypotenuse
sin(30) = opposite / 10
1/2 = opposite / 10
opposite = 5

sin 30 = x/12. Find x.

sin 30 = 1/2
1/2 = x/12
x = 6

cos 60 = x/20. Find x.

cos 60 = 1/2
1/2 = x/20
x = 10

tan 45 = x/7. Find x.

tan 45 = 1
1 = x/7
x = 7

1. sin 30 = x/12. Find x.
2. cos 60 = x/20. Find x.
3. tan 45 = x/7. Find x.

1. 6
2. 10
3. 7

two shapes are the same shape but different sizes

Two similar shapes have matching lengths 4 cm and 10 cm.
The smaller area is 12 cm².
Find the larger area.

10 / 4 = 2.5

2.5² = 6.25

12 × 6.25 = 75 cm²

1. Length scale factor is 3. Area scale factor?
2. Length scale factor is 2. Volume scale factor?
3. Small volume 20 cm³, length scale factor 4. Large volume?

1. 9
2. 8
3. 1280 cm³

7. Probability

Core idea

Probability measures likelihood on a scale from 0 to 1.

0 <= P <= 1

• 0 means impossible
• 1 means certain

Mutually exclusive events

Mutually exclusive events cannot happen at the same time.

Example:

Rolling a 2 and rolling a 5 on one die roll.

Add probabilities:

P(A or B) = P(A) + P(B)

Reason:

If two outcomes cannot overlap, their probabilities sit in separate boxes. To find the chance of either one happening, add the boxes.

So:

P(2 or 5) = P(2) + P(5) = 1/6 + 1/6 = 2/6 = 1/3

Independent events

Independent events do not affect each other.

Example:

Flipping a coin and rolling a die.

Multiply probabilities:

P(A and B) = P(A) × P(B)

Reason:

Each event filters the possible outcomes.

Example:

P(heads and rolling a 6)
= P(heads) × P(6)
= 1/2 × 1/6
= 1/12

The coin does not change the die. The die does not change the coin. That is what independent means.

Tree diagrams

Multiply along branches.

Add separate routes that lead to the same desired outcome.

Visual idea for two coin flips:

Multiply along the branches. Add across different routes if more than one route reaches the wanted outcome.

To find one head and one tail:

HT or TH
= 1/4 + 1/4
= 1/2

Example idea:

P(two heads) = 1/2 × 1/2 = 1/4

there are two or more events happening in sequence

A bag has 3 red counters and 2 blue counters.
One counter is taken and replaced.
Then another is taken.
Find P(two red).

P(red first) = 3/5
P(red second) = 3/5
P(two red) = 3/5 × 3/5 = 9/25

P(two red) = 3/5 × 2/4 = 6/20 = 3/10

A fair coin is flipped twice. Find P(two heads).

P(HH) = 1/2 × 1/2 = 1/4

A bag has 4 red and 2 blue counters.
Two counters are chosen with replacement.
Find P(two red).

P(red then red) = 4/6 × 4/6 = 16/36 = 4/9

A bag has 4 red and 2 blue counters.
Two counters are chosen without replacement.
Find P(two red).

P(red then red) = 4/6 × 3/5 = 12/30 = 2/5

A bag has 4 green and 1 yellow counter.
Two counters are chosen without replacement.
1. Find P(two green).
2. Find P(green then yellow).
3. Find P(one green and one yellow in any order).

1. 4/5 × 3/4 = 3/5
2. 4/5 × 1/4 = 1/5
3. 1/5 + 1/5 = 2/5

items are sorted into overlapping groups

In a class of 30, 18 study French, 12 study Spanish, and 5 study both.
Find how many study neither.

18 - 5 = 13

12 - 5 = 7

13 + 5 + 7 = 25

30 - 25 = 5

In a group of 40, 22 like tea, 18 like coffee, 7 like both.
1. How many like tea only?
2. How many like coffee only?
3. How many like neither?

1. 15
2. 11
3. 7

8. Statistics

Mean

The mean is the balancing point.

mean = total / number of values

If the mean of 5 numbers is 8, their total is:

5 × 8 = 40

This reverse idea is often tested.

Median

The median is the middle value when data is ordered.

It is about position, not total.

Mode

The mode is the most common value.

Range

range = largest - smallest

It measures spread, but only using the two extreme values.

Frequency tables

For grouped or repeated data, remember:

total = value × frequency

Visual example:

The frequency tells you how many times each score appears.

This table represents:

2, 2, 2, 4, 4, 4, 4, 4, 6, 6

The mean from a frequency table is:

sum of (value × frequency) / sum of frequencies

values repeat and frequencies are given

total score = 2x3 + 4x5 + 6x2
= 6 + 20 + 12
= 38
total frequency = 3 + 5 + 2 = 10
mean = 38/10 = 3.8

(1x2 + 3x4 + 5x4) / 10
= 34/10
= 3.4

data is grouped into intervals

5, 15, 25

estimated total = 5x4 + 15x6 + 25x2
= 20 + 90 + 50
= 160
total frequency = 12
estimated mean = 160/12 = 13.3...

midpoints 5, 15, 25
(5x3 + 15x5 + 25x2) / 10
= 140/10
= 14

the question asks you to compare two sets of data

Class A has a higher median, so its typical score is higher.
Class B has a smaller interquartile range, so its scores are more consistent.

9. Common Higher Paper 1 Problem Types

Forming equations

When a question gives relationships in words, translate them into algebra.

Example:

"Three more than twice a number is 17"

becomes:

2x + 3 = 17

Proof

A proof must work for all possible cases, not just one example.

Example:

Prove the sum of two consecutive integers is odd.

Let the first integer be:

n

The next integer is:

n + 1

Sum:

n + (n + 1) = 2n + 1

This is odd because it is one more than an even number.

the question asks prove, show that, or explain why something is always true

Prove that the sum of two consecutive odd numbers is even.

2n + 1

2n + 3

(2n + 1) + (2n + 3)
= 4n + 4
= 2(2n + 2)

1. Prove that the sum of two consecutive integers is odd.
2. Prove that the square of an even number is even.

1. n + (n + 1) = 2n + 1, which is odd.
2. Let the even number be 2n. Its square is (2n)^2 = 4n² = 2(2n²), which is even.

Bounds

Bounds are about the range of possible values after rounding.

If a length is 8.4 cm to 1 decimal place:

lower bound = 8.35
upper bound = 8.45

The true value could be 8.35 but not 8.45.

For maximum or minimum calculations:

• to maximise a fraction, use biggest numerator and smallest denominator
• to minimise a fraction, use smallest numerator and biggest denominator

values have been rounded

8.4 cm rounded to 1 decimal place

lower bound = 8.35
upper bound = 8.45

8.35 <= x < 8.45

1. Find bounds for 12 cm rounded to the nearest cm.
2. Find bounds for 3.6 kg rounded to 1 decimal place.
3. Find bounds for 250 rounded to the nearest 10.

1. 11.5 <= x < 12.5
2. 3.55 <= x < 3.65
3. 245 <= x < 255

Rearranging formulae

Treat the formula like an equation. The target letter must be isolated.

Example:

y = 3x + 4

Make x the subject:

y - 4 = 3x
x = (y - 4) / 3

The logic is the same as solving equations.

10. How To Remember Methods

The best memory tricks in maths are not random slogans. They work because they attach a method to a reason.

When revising, do not only ask:

What do I do?

Also ask:

Why does that method fit this situation?

That is what makes the method easier to remember in the exam.

The master memory sentence

Use this for almost every GCSE question:

Structure first. Rule second. Steps third. Check last.

Meaning:

1. Work out what type of relationship the question contains.
2. Choose the rule that belongs to that relationship.
3. Carry out the method one line at a time.
4. Check if the answer makes sense.

The three exam questions to ask yourself

Before starting any question, pause for five seconds and ask:

What have they given me?
What do they want?
What connects the two?

The third question is where the method usually appears.

Example:

If they give a right-angled triangle and want a missing side, the connection is probably Pythagoras or trigonometry.

If they give a final price and want the original, the connection is reverse percentage.

Trigger words and what they usually mean

These clues are not magic, but they are very good starting points.

The "undo" memory trick

Many methods are just undoing operations in reverse order.

Example:

3x + 5 = 20

The x was multiplied by 3, then 5 was added.

Undo in reverse:

subtract 5
divide by 3

Number methods

Fractions

Memory hook:

Same bottoms to add. Straight across to multiply. Flip to divide.

Why:

• adding needs same-sized pieces
• multiplying fractions means scaling
• dividing asks how many groups fit, so the reciprocal reverses the divisor

Percentages

Memory hook:

Tell the 100% story.

Always start from:

original = 100%

Then:

• increase by 20% means 120% = 1.2
• decrease by 20% means 80% = 0.8
• reverse percentage means divide by the multiplier

Ratio

Memory hook:

Parts first, value second.

For a ratio problem:

1. Add the ratio parts.
2. Find one part.
3. Scale to the required parts.

Ratio is about relative size, so never treat 2 : 3 as a difference of 1. It means 2 parts compared with 3 parts.

Standard form

Memory hook:

One digit before the decimal.

The number at the front must satisfy:

1 <= a < 10

The power of 10 records the movement of the decimal point.

Surds

Memory hook:

Find the hidden square.

Example:

√72 = √(36 × 2) = 6√2

Surd simplification is mostly about spotting square factors:

4, 9, 16, 25, 36, 49, 64, 81, 100

Algebra methods

Simplifying

Memory hook:

Only same species combine.

3x + 5x = 8x

but:

3x + 5y

cannot be combined.

Expanding

Memory hook:

Everything inside gets touched.

For:

3(x + 4)

the 3 multiplies both terms:

3x + 12

Double brackets

Memory hook:

Each term shakes hands with each term.

For:

(x + 2)(x + 5)

multiply every pairing:

x times x
x times 5
2 times x
2 times 5

Factorising

Memory hook:

Find what they all share.

For:

6x + 12

both terms share 6:

6(x + 2)

Quadratic factorising

Memory hook:

Product and sum.

For:

x² + 7x + 12

ask:

What multiplies to 12 and adds to 7?

Answer:

3 and 4

So:

(x + 3)(x + 4)

Solving equations

Memory hook:

Balance, opposite, reverse order.

• Balance: do equivalent operations to both sides.
• Opposite: use the inverse operation.
• Reverse order: undo the outside operation before the inside operation.

Rearranging formulae

Memory hook:

Same as solving, but the answer is a letter.

Do not be frightened by several letters. Treat every other letter as if it were a number.

Graph methods

Straight lines

Memory hook:

y = mx + c means movement and crossing.

• m is the gradient, the movement rate.
• c is where the graph crosses the y-axis.

Gradient

Memory hook:

Up or down over across.

gradient = change in y / change in x

Roots

Memory hook:

Roots are where the graph hits the ground.

The "ground" is the x-axis, where:

y = 0

Geometry methods

Angles

Memory hook:

Turns explain angles.

• straight line = half turn = 180 degrees
• around a point = full turn = 360 degrees
• triangle = half turn = 180 degrees
• quadrilateral = two triangles = 360 degrees

Pythagoras

Memory hook:

Short squared plus short squared equals long squared.

The hypotenuse is always the longest side and is always opposite the right angle.

Trigonometry

Memory hook:

Angle first. Labels second. Ratio third.

Steps:

1. Mark the chosen angle.
2. Label opposite, adjacent and hypotenuse.
3. Choose sine, cosine or tangent.

SOHCAHTOA is useful only after the sides are labelled:

SOH: sin = opposite / hypotenuse
CAH: cos = adjacent / hypotenuse
TOA: tan = opposite / adjacent

Similar shapes

Memory hook:

Same shape, same multiplier.

All matching lengths use the same scale factor.

If the length scale factor is k, then:

area scale factor = k²
volume scale factor = k³

Probability methods

Memory hook:

AND means multiply. OR means add routes.

• A and B: multiply along branches.
• A or B: add separate routes.

For tree diagrams:

multiply along
add across

Statistics methods

Mean

Memory hook:

Mean is fair share.

mean = total / number

Reverse version:

total = mean × number

Median

Memory hook:

Median is middle after ordering.

Always put the data in order first.

Range

Memory hook:

Range is big minus small.

Proof methods

Memory hook:

One example shows. Algebra proves.

A proof must work for every possible number, not just a few examples.

Useful algebra forms:

even number = 2n
odd number = 2n + 1
consecutive integers = n, n + 1
multiple of 5 = 5n

The mistake-fixing method

When a question goes wrong, label the type of mistake:

• arithmetic mistake
• sign mistake
• fraction mistake
• copied the question wrongly
• chose the wrong method
• stopped before answering the question
• rounded too early
• forgot units

This is powerful because different mistakes need different fixes.

If the mistake was "wrong method", revise recognition clues.

If the mistake was "sign error", slow down around negatives.

If the mistake was "stopped too early", underline the final instruction in the question.

The mini whiteboard drill

For fast one-day revision, make tiny method cards. Each card should have three lines:

Topic:
Clue:
Method:

Examples:

Topic: Reverse percentage
Clue: final amount after increase/decrease
Method: divide by multiplier

Topic: Pythagoras
Clue: right-angled triangle, missing side
Method: short² + short² = long²

Topic: Quadratic factorising
Clue: x² + bx + c
Method: find two numbers with product c and sum b

This turns revision into recognition practice, which is exactly what helps under exam conditions.

11. Mini Paper 1 Mixed Mock

This is a short mixed practice set. Try it without notes first.

Suggested time:

35 to 40 minutes

Show working. Some questions are easy marks, some are designed to make you think.

Questions

1. Fractions

Work out:

3/4 + 5/6

Give your answer as a mixed number.

2. Percentages

A jacket is reduced by 20% in a sale. The sale price is 48 pounds.

Find the original price.

3. Ratio

The ratio of adults to children at an event is 5 : 3.

There are 24 more adults than children.

How many people are at the event altogether?

4. Indices

Simplify:

x⁷ × x³ / x⁴

5. Surds

Simplify:

√98

6. Expanding

Expand and simplify:

(x - 4)(x + 7)

7. Factorising

Factorise fully:

x² - 8x + 15

8. Solving

Solve:

3(x - 2) = 2x + 9

9. Simultaneous equations

Solve:

x + y = 11
x - y = 3

10. Sequence

Find the nth term of:

7, 11, 15, 19, ...

11. Straight line graph

Find the gradient of the line through:

(2, 5) and (6, 17)

12. Pythagoras

A right-angled triangle has hypotenuse 17 cm and one shorter side 8 cm.

Find the other shorter side.

13. Trigonometry

In a right-angled triangle:

sin 30 = x / 14

Find x.

14. Similarity

Two similar shapes have length scale factor 3.

The smaller shape has area 8 cm².

Find the area of the larger shape.

15. Probability

A bag contains 4 red counters and 2 blue counters.

Two counters are chosen without replacement.

Find the probability that both counters are red.

16. Venn diagrams

In a group of 50 people:

• 28 study History
• 24 study Geography
• 10 study both

How many study neither History nor Geography?

17. Mean from frequency table

Find the mean score.

18. Bounds

A length is given as 6.8 cm to 1 decimal place.

Write the lower and upper bounds.

19. Proof

Prove that the sum of two consecutive integers is odd.

20. Multi-step problem

The price of a bike increases by 10% and is then reduced by 10%.

The original price was 200 pounds.

Is the final price 200 pounds? Show working.

3/4 = 9/12
5/6 = 10/12
9/12 + 10/12 = 19/12 = 1 7/12

48 = 80%
original = 48 / 0.8 = 60

5 - 3 = 2
2 parts = 24
1 part = 12
total parts = 8
8 parts = 96

x7+3-4 = x⁶

√98 = √(49 × 2) = 7√2

(x - 4)(x + 7)
= x² + 7x - 4x - 28
= x² + 3x - 28

-3 and -5

x² - 8x + 15 = (x - 3)(x - 5)

3x - 6 = 2x + 9
x - 6 = 9
x = 15

2x = 14
x = 7

7 + y = 11
y = 4

4n gives 4, 8, 12, 16
actual sequence is 3 higher
nth term = 4n + 3

gradient = change in y / change in x
= (17 - 5) / (6 - 2)
= 12/4
= 3

other side squared = 17² - 8²
= 289 - 64
= 225
other side = 15

sin 30 = 1/2
1/2 = x/14
x = 7

3² = 9

8 × 9 = 72 cm²

P(red then red) = 4/6 × 3/5
= 12/30
= 2/5

28 - 10 = 18

24 - 10 = 14

18 + 10 + 14 = 42

50 - 42 = 8

total score = 2x4 + 5x3 + 7x3
= 8 + 15 + 21
= 44
total frequency = 10
mean = 44/10 = 4.4

6.75 <= x < 6.85

n and n + 1

n + (n + 1) = 2n + 1

200 × 1.1 = 220

220 × 0.9 = 198

Paper 1 Higher Extras: Prioritise If Time

These topics can appear on Higher Paper 1, but they are not all equally urgent. Start with Priority A if time is limited. Treat Priority C as stretch material once the core methods are secure.

The quadratic formula

Use when:

a quadratic ax² + bx + c = 0 will not factorise neatly

Formula:

x = ( -b ± √(b² - 4ac) ) / (2a)

Why it works:

The formula is what you get when you complete the square on the general quadratic ax² + bx + c = 0 once and for all. Someone did the completing-the-square algebra one time, kept the letters a, b and c in symbolic form, and ended up with the formula. So instead of completing the square every time, you can just substitute the numbers in.

This also explains the discriminant b² − 4ac: it is the bit that sits under the square root. If it is positive, the square root is a real number → two solutions. If it is zero, the square root is zero → one repeated solution. If it is negative, you would be square-rooting a negative number → no real solutions (the parabola never touches the x-axis).

Worked example:

Solve x² + 5x + 3 = 0.

a = 1, b = 5, c = 3.

b² - 4ac = 25 - 12 = 13

x = ( -5 ± √13 ) / 2

Leave the answer in surd form on Paper 1.

Completing the square

Use when:

you need the turning point of a quadratic
or you need to solve when factorising fails
or the question says "in the form (x + p)² + q"

Method for x² + bx + c:

1. Halve the coefficient of x. Call it p.
2. Write (x + p)².
3. (x + p)² expands to x² + 2px + p², so it has an extra p².
4. Subtract p² and add the original c.

Worked example:

Write x² + 6x + 11 in the form (x + p)² + q.

half of 6 is 3
(x + 3)² = x² + 6x + 9
we need + 11, but already have + 9
so add 2:
x² + 6x + 11 = (x + 3)² + 2

The turning point is at (-3, 2). The graph of y = x² + 6x + 11 is a U-shape with its lowest point there.

Algebraic fractions

The same rules as numerical fractions, just with letters.

Simplifying

Factorise the top and the bottom, then cancel any common factor.

(x² - 9) / (x² + 5x + 6)
= (x - 3)(x + 3) / ((x + 2)(x + 3))
= (x - 3) / (x + 2)

Adding and subtracting

Make a common denominator first.

1/x + 2/(x + 1)
= (x + 1)/(x(x + 1)) + 2x/(x(x + 1))
= (x + 1 + 2x) / (x(x + 1))
= (3x + 1) / (x(x + 1))

Multiplying and dividing

Multiply: top times top, bottom times bottom, then simplify.

Divide: keep the first, change ÷ to ×, flip the second.

Recurring decimals to fractions

Use when:

a decimal has a repeating block, e.g. 0.272727... or 0.16666...

Method:

1. Let x equal the recurring decimal.
2. Multiply by 10, 100, or 1000 so that the recurring blocks line up.
3. Subtract the original from the new equation. The recurring tails cancel.
4. Solve for x.

Worked example:

Convert 0.27 (with 27 recurring) to a fraction.

Let x = 0.272727...
Then 100x = 27.272727...

Subtract:
100x - x = 27.272727... - 0.272727...
99x = 27
x = 27/99 = 3/11

Worked example with a non-recurring start:

Convert 0.16 (with 6 recurring) to a fraction.

Let x = 0.16666...
Then 10x  = 1.6666...
Then 100x = 16.6666...

Subtract:
100x - 10x = 16.6666... - 1.6666...
90x = 15
x = 15/90 = 1/6

Compound measures: speed, density, pressure

speed    = distance / time
density  = mass / volume
pressure = force / area

Three quantities, three formulas, all the same shape: the middle one divided by the bottom one.

Rearrange when needed:

distance = speed × time
time     = distance / speed

Worked example:

A car travels 90 miles in 1 hour 30 minutes. Find the speed in mph.

time = 1.5 hours
speed = 90 / 1.5 = 60 mph

Compound interest, growth and decay

Compound interest means interest is added on top of interest. Each year you multiply by the same multiplier.

final = original × (multiplier)^n

Worked example:

£1000 invested at 10% compound interest for 2 years.

multiplier = 1.10
final = 1000 × 1.10²

1.10² = 1.21
final = 1000 × 1.21 = £1210

Decay works the same way with a multiplier less than 1.

A car worth £8000 loses 15% of its value each year.

multiplier = 0.85
after 2 years: 8000 × 0.85² = 8000 × 0.7225 = £5780

Quadratic sequences

A linear sequence has a constant first difference. A quadratic sequence has a constant second difference.

sequence:    3,  8, 15, 24, 35, ...
1st diff:      5,  7,  9, 11
2nd diff:        2,  2,  2

The coefficient of n² in the nth term is half the second difference.

Method:

1. Find the second difference. Halve it. That is the coefficient of n².
2. Write down the values of that an² sequence.
3. Subtract from the original sequence to get a remainder sequence.
4. The remainder is a linear sequence. Find its nth term.
5. Add it on.

Worked example:

Find the nth term of 3, 8, 15, 24, 35, ...

second difference = 2
coefficient of n² = 1, so try n²: 1, 4, 9, 16, 25
remainder: 2, 4, 6, 8, 10
remainder nth term: 2n
nth term = n² + 2n

Geometric and Fibonacci-type sequences

A geometric sequence multiplies by the same number each step.

2, 6, 18, 54, ...     common ratio = 3

Each term is the previous term multiplied by the ratio.

A Fibonacci-type sequence adds the previous two terms.

1, 1, 2, 3, 5, 8, 13, ...

Higher exam questions sometimes give the rule and ask you to find missing terms by working forwards or backwards. Substitute carefully and form simple equations if needed.

Composite functions

A composite function applies one function and then another.

fg(x) means do g first, then f.

Worked example:

f(x) = 2x + 1
g(x) = x²

fg(x) = f(g(x)) = f(x²) = 2x² + 1
gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)²

Inverse functions: the full method

An inverse function reverses what the original function does.

Method:

1. Write y = f(x).
2. Swap x and y.
3. Rearrange to make y the subject.
4. Replace y with f^(-1)(x).

Worked example:

f(x) = (3x - 4) / 5

y = (3x - 4) / 5
swap:  x = (3y - 4) / 5
solve: 5x = 3y - 4
       5x + 4 = 3y
       y = (5x + 4) / 3

f^(-1)(x) = (5x + 4) / 3

Iteration Stretch / skim

Iteration uses a formula again and again, feeding the answer back in.

Notation:

x_(n+1) = formula in x_n

To find x_2, put x_1 into the right-hand side. To find x_3, put x_2 in. And so on.

Worked example:

x_(n+1) = √(2x_n + 5),  x_1 = 2

x_2 = √(2(2) + 5) = √9 = 3
x_3 = √(2(3) + 5) = √11
x_4 = √(2(√11) + 5)

Sketching quadratics

To sketch y = ax² + bx + c, find:

1. shape: positive a means U-shape, negative a means n-shape
2. y-intercept: c (set x = 0)
3. x-intercepts (roots): factorise or use the quadratic formula and set y = 0
4. turning point: complete the square, or use symmetry between the roots

Other graph shapes to recognise

Cubics have an S-shape (going from bottom left to top right). The basic reciprocal y = 1/x has two branches and never touches the axes. Exponential graphs y = aˣ with a > 1 grow rapidly and pass through (0, 1).

Trigonometric graphs Stretch / skim

Both sine and cosine wave between -1 and +1. sin x starts at 0; cos x starts at 1. They have the same shape but cosine is shifted 90° to the left of sine.

Useful symmetries (Higher):

sin(180 - x) = sin x
cos(180 - x) = -cos x
sin(-x) = -sin x
cos(-x) = cos x

You can see these in the sine and cosine graphs above. The first one — sin(180 - x) = sin x — is the key fact for finding the second solution to a trig equation in 0° ≤ x ≤ 360°.

Sine rule, cosine rule and area formula If time

These formulae are for triangles that are not right-angled. They can appear on Higher Paper 1, especially when the values are exact-friendly, such as angles of 30°, 45°, 60° or 90°. If the arithmetic looks ugly, check whether the question expects an exact surd form rather than a decimal.

Sine rule

  a        b        c
─────  =  ─────  =  ─────
sin A    sin B    sin C

Each side is divided by the sine of the angle opposite that side. The "opposite" pairing is what makes the rule work: angle A sits opposite side a, and so on.

Worked example (Paper 1 style — angle of 30°)

In triangle ABC, angle A = 30°, angle B = 90°, side b = 8 cm.
Find side a.

Use the sine rule with the two sides we know about (a and b):
  a / sin A = b / sin B
  a / sin 30° = 8 / sin 90°
  a / (1/2)   = 8 / 1            ← exact values, no calculator
  a / (1/2)   = 8
  a           = 8 × (1/2)        ← multiply both sides by 1/2
  a           = 4 cm

When the sine rule gives two answers

The sine rule sometimes has two valid answers because sin(180° − x) = sin x. If you find sin x = 1/2, that means x could be 30° or 150° — both have the same sine. The diagram in the question usually tells you which one is geometrically possible.

Cosine rule

a² = b² + c² - 2bc cos A

Use it when you know two sides (b and c) and the angle between them (A) and want the third side (a).

Worked example (Paper 1 style — angle of 60°)

Triangle has b = 5 cm, c = 8 cm, and the angle between them A = 60°.
Find the third side a.

a² = b² + c² - 2bc cos A
a² = 5² + 8² - 2(5)(8) cos 60°
a² = 25 + 64 - 80 × (1/2)         ← cos 60° = 1/2 (exact value)
a² = 89 - 40
a² = 49
a  = 7 cm

Rearranged version to find an angle when you know all three sides:

cos A = (b² + c² - a²) / (2bc)

Area of a triangle = ½ab sin C

Use it when you know two sides and the angle between them and want the area.

area = ½ × a × b × sin C

Here a and b are the two sides you know and C is the angle between them.

Worked example (June 2022 Paper 1 style — angle of 60°)

Triangle has two sides of length 4 cm with a 60° angle between them.
Find the area, giving your answer in surd form.

area = ½ × 4 × 4 × sin 60°
     = ½ × 16 × √3/2              ← sin 60° = √3/2 (exact value)
     = 8 × √3/2
     = 4√3 cm²

Solving trig equations using the graph Stretch / skim

When a question asks to solve sin x = (some value) for 0° ≤ x ≤ 360°, there are usually two answers. You use the sine curve's symmetry to find them both.

Method:

1. Find the reference angle from the exact-value table.
2. Use the symmetry rule sin(180° − x) = sin x to find the second.
3. Check both answers are inside the given range.

Worked example (May 2024 Q22(b) style)

Solve  2 sin x° = -1  for 0 ≤ x ≤ 360.

Step 1 — rearrange:
  sin x° = -1/2

Step 2 — find the first solution.
  We know sin 30° = +1/2.
  The sine curve dips to -1/2 in the third and fourth quadrants.
  By symmetry of sin around the x-axis below 180°:
    first solution  = 180° + 30° = 210°
    second solution = 360° - 30° = 330°

Step 3 — both 210° and 330° lie in 0° ≤ x ≤ 360°, so:
  x = 210°  or  x = 330°

Cosine version: For cos x = (value), use the symmetry cos(360° − x) = cos x. If cos x = 1/2, the solutions in [0°, 360°] are x = 60° and x = 300°.

Graph transformations of y = f(x) Stretch / skim

Why inside-the-bracket is the opposite: consider y = f(x + 2). To find the y-value at x = 0, the formula tells you to evaluate f(0 + 2) = f(2). So whatever shape f had at x = 2 now appears at x = 0. The whole graph has shifted two units to the left, not right. Adding inside the bracket effectively asks the graph "what would you be doing two steps ahead?", which pulls the graph backwards.

3D Pythagoras

For a cuboid with sides a, b and c, the longest internal diagonal d satisfies:

d² = a² + b² + c²

Why: apply Pythagoras twice. First find the diagonal of the base: p² = a² + b². Then make a right-angled triangle with that diagonal and the vertical side: d² = p² + c² = a² + b² + c².

Circle theorems

Each theorem is a fact you can quote in an angle-chasing question. The exam often asks you to give the reason in words next to the value.

1. Angle at the centre is twice the angle at the circumference

2. Angle in a semicircle is 90°

3. Angles in the same segment are equal

4. Cyclic quadrilateral: opposite angles add to 180°

5. Tangent meets radius at 90°

6. Two tangents from an external point are equal

7. Perpendicular from centre bisects a chord

8. Alternate segment theorem

Vectors

A vector has size and direction. Write it as a column:

(  3 )       3 right
(  2 )       2 up

Add by adding components. Subtract by subtracting components. Multiply by a scalar by multiplying each component.

Two vectors are parallel if one is a scalar multiple of the other:

If a = (2, 3) and b = (4, 6), then b = 2a, so a and b are parallel.

Position vectors

The position vector of a point A is the vector from the origin to A, written OA or a.

Key rule:

AB = b - a    (end position minus start position)
midpoint M of AB has position vector (a + b)/2

If a question asks "show that two lines are parallel", show that one vector is a scalar multiple of the other.

Equation of a circle (centre at origin)

x² + y² = r²

This is just Pythagoras applied to every point on the circle: the distance from the origin (0, 0) to any point (x, y) on the circle is r.

Tangent to a circle at a given point

Method:

1. Find the gradient of the radius from the origin to the point of contact.
2. Take the negative reciprocal — that is the gradient of the tangent.
3. Use y - y1 = m(x - x1) to write the tangent equation.

Worked example:

x² + y² = 25 has the point (3, 4) on it.

gradient of radius = 4/3
gradient of tangent = -3/4
tangent: y - 4 = -3/4 (x - 3)
        4y - 16 = -3x + 9
        3x + 4y = 25

Histograms

A histogram shows grouped data using frequency density on the y-axis, not frequency.

frequency density = frequency / class width

The area of each bar gives the frequency of that group.

frequency = frequency density × class width

Cumulative frequency and box plots

Cumulative frequency is a running total of frequency. Plot the cumulative frequency at the upper end of each class. Join with a smooth curve.

From a cumulative frequency curve you can read off:

• median: at n/2 on the y-axis
• lower quartile (Q1): at n/4
• upper quartile (Q3): at 3n/4
• interquartile range: Q3 - Q1

A box plot summarises five values: minimum, Q1, median, Q3, maximum. The box covers Q1 to Q3 with a line at the median; the whiskers reach the minimum and maximum.

Conditional probability

"P(A given B)" means the probability of A happening, knowing that B has happened.

P(A | B) = P(A and B) / P(B)

In a Venn diagram, you are restricting attention to the part of the universal set that belongs to B, then asking what fraction of that also lies in A.

Worked example:

50 people. 30 study French, 20 study Spanish, 8 study both.

P(studies Spanish given they study French)
= P(both) / P(French)
= 8/50 / 30/50
= 8/30
= 4/15

In tree diagram language, conditional probability is what sits on the second-level branches when the first event has already happened (the "without replacement" case).

Quadratic inequalities

Method:

1. Move everything to one side so it equals zero.
2. Find the roots of the corresponding quadratic.
3. Sketch the parabola.
4. Read off where the curve is above or below the x-axis, depending on the inequality.

Worked example:

Solve x² - x - 6 > 0.

Factorise: (x - 3)(x + 2) > 0
Roots: x = 3 and x = -2

The graph of y = x² - x - 6 is a U-shape that crosses the x-axis at -2 and 3.
It is above the x-axis (positive) outside the roots.

So x < -2 or x > 3.

Bounds for compound calculations

To maximise a calculation:

• Use the upper bound for any quantity being added or multiplied.
• Use the lower bound for any quantity being subtracted or divided by.

To minimise a calculation, do the opposite.

Worked example:

A rectangle has length 8.4 cm and width 5.2 cm, both to 1 d.p.
Find the upper bound of the area.

length:  8.35 ≤ L < 8.45
width:   5.15 ≤ W < 5.25

upper bound of area = 8.45 × 5.25 = 44.3625 cm²

Error intervals and truncation

An error interval is just bounds written as an inequality:

x = 8.4 to 1 d.p.   means   8.35 ≤ x < 8.45

A truncated value (cut off, not rounded) gives a different interval:

x = 8.4 truncated to 1 d.p.   means   8.4 ≤ x < 8.5

Exam-Style Walkthroughs (paper-tested topics)

These are full worked examples in the shape that Paper 1 actually uses: recognise the question type, set up the method, show working for every method mark, then check. Every example here is modelled on a real Edexcel 1MA1/1H question.

How to win method marks

Mark schemes award marks in stages — usually labelled P1 (process), M1 (method) and A1 (accuracy). You can pick up most of the marks on a 4-mark question even with a wrong final answer, provided your working is visible.

Compound measures: density, speed, pressure

density  = mass / volume
speed    = distance / time
pressure = force / area

Exam clue: the question gives two of the three quantities and asks for the third — or asks for a ratio of densities or speeds.

Worked example (May 2022 Paper 1, Q7 style)

Cube A has side 3 cm and mass 81 g.
Cube B has side 4 cm and mass 128 g.
Find the ratio  density of A : density of B  in integer form.

Method (mark each line):

volume of A = 3³ = 27 cm³                                  ← M1 (volume A)
volume of B = 4³ = 64 cm³                                  ← M1 (volume B)
density of A = 81/27 = 3 g/cm³                             ← M1 (one density)
density of B = 128/64 = 2 g/cm³
ratio = 3 : 2                                              ← A1 (final ratio)

Unit reminders (non-calculator):

1 m  = 100 cm = 1000 mm
1 km = 1000 m
1 m² = 10 000 cm²       (because (100)² = 10 000)
1 m³ = 1 000 000 cm³    (because (100)³ = 1 000 000)
1 kg = 1000 g
1 hour = 60 min = 3600 seconds
m/s to km/h: multiply by 3.6
km/h to m/s: divide by 3.6

Cumulative frequency: the full pipeline

Exam clue: a grouped frequency table and a graph grid, asking for median, quartiles, IQR, or "how many were less than X".

Method:

1. Cumulative frequency is a running total. Add each row's frequency to the previous total.
2. Plot each cumulative frequency against the upper class boundary of its group (not the midpoint).
3. Join with a smooth curve (or a series of straight line segments — both are accepted).
4. To read the median, go from n/2 on the y-axis across to the curve, then down to the x-axis.
5. Q1 is at n/4. Q3 is at 3n/4. IQR = Q3 − Q1.

Worked example (May 2022 Q10 style)

Profit (£x)      Frequency      Cumulative frequency
0  ≤ x < 50         10                10
50 ≤ x < 100        15                25
100 ≤ x < 150       25                50
150 ≤ x < 200       30                80
200 ≤ x < 250       5                 85
250 ≤ x < 300       15                100

Total n = 100, so:
  median at y = 50 → read curve → x ≈ £150
  Q1     at y = 25 → x ≈ £100
  Q3     at y = 75 → x ≈ £188
  IQR    ≈ 188 − 100 = £88

Box plots

A box plot summarises five numbers: minimum, lower quartile (Q1), median, upper quartile (Q3), maximum. The box covers Q1 to Q3 with a line at the median, and whiskers reach out to the min and max.

To compare two box plots, say one thing about average (compare medians) and one thing about spread (compare IQRs or ranges), always in the context of the question.

Histograms with unequal class widths

A histogram looks like a bar chart but uses frequency density on the y-axis. The area of each bar is the frequency.

frequency density = frequency / class width
frequency         = frequency density × class width

Worked example

Time t (min)   Frequency    Class width    Frequency density
0 < t ≤ 10        8            10                0.8
10 < t ≤ 20      18            10                1.8
20 < t ≤ 40      24            20                1.2
40 < t ≤ 70      15            30                0.5

To find the frequency for a partial bar (e.g. 15 < t ≤ 35), use area = height × width across each piece of the partial range, then add.

Rate graphs: speed-time, distance-time, volume-time

On any graph where the y-axis is a rate and the x-axis is time:

gradient at a point  =  the rate of change of the quantity
                        (e.g. acceleration on a speed-time graph)
area under the graph =  total quantity accumulated
                        (e.g. distance on a speed-time graph)

Exam clue: "estimate the gradient at t = …" → draw a tangent by eye and read off two points.

"What does the area under the graph represent?" → on a speed-time graph it represents distance; on a volume-time graph it represents total volume; on a flow-rate graph it represents the total amount that has flowed.

Worked example (May 2022 Q14 style)

A speed-time graph has speed in m/s on the y-axis and time in seconds on the x-axis.
Estimate the gradient at t = 2 seconds.

Method:
1. Draw a tangent to the curve at t = 2 — a straight line that just touches the curve.
2. Pick two points on the tangent, e.g. (0, 1) and (4, 4.6).
3. gradient = (4.6 − 1) / (4 − 0) = 3.6 / 4 = 0.9 m/s²

The units of the gradient are m/s ÷ s = m/s² (acceleration).

"What does the gradient represent?" Answer in words tied to the graph's quantities — e.g. on a volume-time graph it represents the rate of change of volume (litres per second).

Combined direct and inverse proportion

Higher Paper 1 sometimes chains two proportion statements: y ∝ √t and t ∝ 1/x³. The method is the same as one proportion, just done twice.

Worked example (May 2022 Q17)

y is directly proportional to √t.   y = 15 when t = 9.
t is inversely proportional to x³.   t = 8 when x = 2.
Find a formula for y in terms of x.

Method:

Step 1 — first relationship
y = k√t        ← P1 (set up)
15 = k × √9 = 3k
k = 5
y = 5√t

Step 2 — second relationship
t = K/x³       ← P1 (set up)
8 = K/2³ = K/8
K = 64
t = 64/x³

Step 3 — substitute t into the y formula  ← P1 (combine)
y = 5 × √(64/x³)
  = 5 × 8/x^(3/2)
  = 40/x^(3/2)        ← A1 (final form)

Other proportional forms to recognise:

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

Equation of a tangent to a circle

This is one of the most consistently asked Paper 1 questions and follows the same five-step recipe every time.

Method (memorise this order):

1. Find the gradient of the radius from the centre to the point of contact.
2. Take the negative reciprocal — that is the gradient of the tangent.
3. Use y − y₁ = m(x − x₁) with the point of contact and the tangent gradient.
4. Rearrange to the required form (often ax + by + c = 0).
5. Check by substituting the point of contact back into your equation.

Worked example (May 2022 Q20)

A circle has centre (−1, 3). The point A(6, 8) lies on the circle.
Find an equation of the tangent to the circle at A
in the form ax + by + c = 0.

Step 1 — gradient of radius                       ← P1
m_radius = (8 − 3) / (6 − (−1))
        = 5 / 7

Step 2 — gradient of tangent (negative reciprocal) ← P1
m_tangent = −7/5

Step 3 — substitute into y − y₁ = m(x − x₁)        ← P1
y − 8 = −7/5 (x − 6)

Step 4 — rearrange                                 ← A1
5(y − 8) = −7(x − 6)
5y − 40 = −7x + 42
7x + 5y − 82 = 0

Step 5 — check
7(6) + 5(8) − 82 = 42 + 40 − 82 = 0  ✓

If the circle is x² + y² = r² and the point is (p, q):

centre is (0, 0)
m_radius  = q/p
m_tangent = −p/q
tangent:    y − q = (−p/q)(x − p)

Solving an algebraic-fractions equation

Exam clue: two fractions added together equal a number, with x in each denominator. Answer often required in (p ± √q) / r form.

Worked example (May 2022 Q19)

Solve:    1/(2x − 1)  +  3/(x − 1)  =  1
Give your answer in the form  (p ± √q) / 2.

Step 1 — common denominator                       ← M1
1(x − 1) + 3(2x − 1)
─────────────────────  = 1
   (2x − 1)(x − 1)

Step 2 — multiply both sides by the denominator and expand   ← M1
(x − 1) + 3(2x − 1) = (2x − 1)(x − 1)
x − 1 + 6x − 3      = 2x² − 3x + 1
7x − 4              = 2x² − 3x + 1

Step 3 — rearrange to a three-term quadratic
0 = 2x² − 10x + 5

Step 4 — use the quadratic formula              ← M1 (dep)
a = 2, b = −10, c = 5
b² − 4ac = 100 − 40 = 60

x = (10 ± √60) / 4
  = (10 ± 2√15) / 4
  = (5 ± √15) / 2                                ← A1

Rationalising a denominator with two terms

Exam clue: the denominator has a surd added to or subtracted from a number, e.g. 2 + √3 or √5 − 1.

Method: multiply top and bottom by the conjugate (the same expression with the sign flipped). This uses the difference of two squares to clear the surd.

Worked example

Rationalise   1 / (2 + √3).

Multiply top and bottom by (2 − √3):

  1        (2 − √3)
─────  ×  ─────────
2 + √3    (2 − √3)

denominator: (2 + √3)(2 − √3) = 2² − (√3)² = 4 − 3 = 1

numerator:   1 × (2 − √3) = 2 − √3

Answer: (2 − √3) / 1 = 2 − √3

Worked example with a number on top

Rationalise   6 / (√5 − 1).

Multiply by (√5 + 1)/(√5 + 1):

denominator: (√5 − 1)(√5 + 1) = 5 − 1 = 4
numerator:   6(√5 + 1) = 6√5 + 6

Answer: (6√5 + 6) / 4 = (3√5 + 3) / 2

Similarity: area and volume ratio chains

If two similar shapes have length scale factor k, then:

area scale factor   = k²
volume scale factor = k³

Working backwards from an area or volume ratio:

given area ratio a : b      → length ratio is √a : √b
given volume ratio a : b    → length ratio is ∛a : ∛b

Worked example

Two similar cones have volumes in the ratio 27 : 64.
Find the ratio of their surface areas.

Step 1 — length scale factor
volume ratio 27 : 64
length ratio ∛27 : ∛64 = 3 : 4

Step 2 — area scale factor
area ratio = 3² : 4² = 9 : 16

Circle theorem questions: writing the reason

Examiners award a separate mark for the reason. Use exact wording:

13. Exam Technique

The five-step method

For every question:

1. Identify the structure.
2. Identify exactly what is being asked.
3. Choose the rule or relationship.
4. Work one clear line at a time.
5. Check whether the answer is reasonable.

Command words

• Calculate: find a numerical answer.
• Solve: find the value or values that make the equation true.
• Simplify: write in a cleaner equivalent form.
• Expand: remove brackets.
• Factorise: put brackets back in.
• Show that: prove the given result through working.
• Explain: give a reason, not just an answer.
• Hence: use the previous part.

Common traps

Sign errors

Be careful when subtracting negatives:

5 - (-3) = 8

Fraction errors

Do not add denominators:

1/3 + 1/4 is not 2/7

Rounding too early

Keep exact values as long as possible.

Units

Check whether the question uses mm, cm, m, kg, grams, minutes or hours.

Diagrams

If the paper says "not drawn accurately", do not rely on appearance. Use angle facts, lengths and algebra.

14. Confidence Checklist

Use this after revising the topics. Tick each statement if it feels secure. Circle any that need one more example.

Number

• I can simplify fractions.
• I can add and subtract fractions using a common denominator.
• I can multiply and divide fractions.
• I can find percentages of amounts without a calculator.
• I can use percentage multipliers.
• I can do reverse percentage questions.
• I know key fraction, decimal and percentage equivalents.
• I can use index laws.
• I can write numbers in standard form.
• I can simplify surds.
• I can rationalise a simple surd denominator.
• I can find upper and lower bounds after rounding.

Ratio and Proportion

• I can share an amount in a ratio.
• I can solve harder ratio questions involving a difference.
• I can use direct proportion with y = kx.
• I can use inverse proportion with y = k/x.
• I can spot when a problem is multiplicative rather than additive.

Algebra

• I can simplify like terms.
• I can expand single brackets.
• I can expand double brackets.
• I can factorise using a common factor.
• I can factorise quadratics.
• I can use difference of two squares.
• I can solve linear equations.
• I can solve equations with brackets.
• I can solve equations with fractions.
• I can solve inequalities and reverse the sign when needed.
• I can solve simultaneous equations.
• I can solve quadratic equations by factorising.
• I can rearrange a formula.
• I can write algebra for a proof.

Graphs and Functions

• I understand y = mx + c.
• I can find a gradient from two points.
• I know roots are where y = 0.
• I can find the nth term of a linear sequence.
• I can substitute into function notation.
• I can find a simple inverse function.

Geometry and Measures

• I know angles on a line, around a point, in a triangle and in a quadrilateral.
• I can use parallel line angle facts.
• I can use Pythagoras to find a hypotenuse.
• I can use Pythagoras to find a shorter side.
• I can label opposite, adjacent and hypotenuse.
• I can use SOHCAHTOA.
• I can use exact trig values for 30, 45 and 60 degrees.
• I can use length, area and volume scale factors.
• I can describe transformations fully.

Probability and Statistics

• I can calculate basic probability.
• I can use tree diagrams.
• I remember probabilities change without replacement.
• I can use Venn diagrams.
• I can calculate a mean from a frequency table.
• I can estimate a mean from grouped data.
• I can compare distributions using average and spread.
• I can calculate conditional probability.
• I can read frequency from a histogram (using area).
• I can read median and quartiles from a cumulative frequency graph.
• I can construct and interpret a box plot.

Paper 1 Higher Extras Checklist

• I can use the quadratic formula and leave answers in surd form.
• I can complete the square and find the turning point.
• I can simplify, add and subtract algebraic fractions.
• I can convert a recurring decimal into a fraction.
• I know speed/density/pressure formulas and can rearrange them.
• I can apply compound interest with a multiplier raised to a power.
• I can find the nth term of a quadratic sequence.
• I can evaluate composite functions like fg(x).
• I can find an inverse function by swapping x and y.
• I can carry out an iteration to find an approximate root.
• I can sketch quadratics showing roots, y-intercept and turning point.
• I recognise cubic, reciprocal and exponential graph shapes.
• I know the shape of sin x and cos x and their key values.
• I know the exact values of sin/cos/tan at 0, 30, 45, 60, 90.
• I can describe and apply transformations of f(x).
• I can use 3D Pythagoras for the diagonal of a cuboid.
• I can quote and apply all 8 circle theorems with reasons.
• I can add, subtract and find magnitudes of vectors.
• I can prove three points are collinear or that lines are parallel using vectors.
• I know x² + y² = r² for a circle and can find a tangent equation.
• I can solve quadratic inequalities by sketching.
• I can find upper and lower bounds for a compound calculation.

15. Final Rescue And Common Traps

Use this section at the end of revision, not at the beginning. It is a recap sheet for the final few minutes.

Exam-room strategy

Collect easy marks first. Star hard questions. Come back.

Do not let one difficult question steal time from questions you can do.

Method triggers

Formula and fact reminders

These are here as a final recap. The 2026 exam provides a formula sheet/exam aid, but you still need to know how to use the formulae.

gradient = change in y / change in x

a^m × a^n = a^(m+n)
a^m / a^n = a^(m-n)
(a^m)^n = a^(mn)
a⁰ = 1
a⁻¹ = 1/a

a² + b² = c²

sin = opposite / hypotenuse
cos = adjacent / hypotenuse
tan = opposite / adjacent

area scale factor = length scale factor squared
volume scale factor = length scale factor cubed

mean = total / number
total = mean × number

Exact values to remember

sin 0°  = 0          cos 0°  = 1          tan 0°  = 0
sin 30° = 1/2        cos 30° = √3/2       tan 30° = 1/√3 = √3/3
sin 45° = √2/2       cos 45° = √2/2       tan 45° = 1
sin 60° = √3/2       cos 60° = 1/2        tan 60° = √3
sin 90° = 1          cos 90° = 0          tan 90° = undefined

1/2 = 0.5  = 50%
1/4 = 0.25 = 25%
3/4 = 0.75 = 75%
1/5 = 0.2  = 20%
1/8 = 0.125 = 12.5%
3/8 = 0.375 = 37.5%
5/8 = 0.625 = 62.5%

High-impact mistake list

Algebra

(x + y)² is not x² + y²

Correct expansion:

(x + y)² = x² + 2xy + y²

Negatives outside brackets

-3(x - 4) = -3x + 12

The negative multiplies everything inside the bracket.

Pythagoras

After finding c², remember to square root if the question asks for a length. Also check: the hypotenuse must be the longest side.

Bounds

If a value is rounded to 1 decimal place, the boundary is halfway to the next tenth:

8.4 to 1 decimal place means 8.35 ≤ x < 8.45

Transformations

A full description usually needs: reflection (mirror line), rotation (centre, angle and direction), enlargement (centre and scale factor), translation (vector).

Probability without replacement

If something is not replaced, the denominator changes on the second branch.

Last checks

• Have I answered the actual question?
• Did I show working for multi-mark questions?
• Did I include units?
• Did I round only at the end?
• Is the answer sensible?
• Did I go back to starred questions?

16. Quick Recognition Table

17. Final Priority List

Must revise

• fractions
• percentages and reverse percentages
• ratio and proportion
• indices
• algebraic simplifying
• expanding and factorising
• solving equations
• angles
• linear graphs
• probability

High-value Higher topics

• quadratics (factorising, formula, completing the square)
• simultaneous equations
• surds (simplifying and rationalising)
• standard form
• exact trig values for 0, 30, 45, 60 and 90
• functions (composite and inverse)
• circle theorems
• vectors
• algebraic fractions
• graph transformations
• bounds and error intervals
• similarity (with area and volume scale factors)
• proof
• rearranging formulae

Final memory principle

Maths is not a pile of random rules. It is built from equivalence, balance, reversal, scaling, structure and logical steps. When a question feels unfamiliar, ask:

What relationship is this question built on?

Once the relationship is identified, the method usually becomes much easier to choose.