O Level Maths 4024 — every topic, worked properly. Free.

A complete study guide for Cambridge O Level Mathematics (Syllabus D) 4024, mapped one-to-one to all 68 sub-topics of the official syllabus for exams in 2025–2030. Every unit gives you the method, a fully worked example, the mistakes examiners see most, and a skill check with a complete solution.

How to use it: maths is not learned by reading. For each unit — read the method, cover the worked example and re-do it yourself, then attempt the skill check before opening the solution. If you get it wrong, the fix is not to read again; it's to do three more of the same type from a past paper.

📄 9 plain-English chapter handouts →✎ Practice & self-test →

CAIE 4024 · exams 2025–203068 syllabus unitsPaper 1 non-calculatorPaper 2 calculatorFree & shareable

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The two papers

Both papers are 2 hours, 100 marks, worth 50% each — and both can test any topic. The only structural difference is the calculator, and that difference changes how you must prepare.

	P1 Paper 1 — Non-calculator	P2 Paper 2 — Calculator
Time / marks	2 hours · 100 marks	2 hours · 100 marks
Calculator	Not allowed	Scientific calculator required (no graphical/algebraic)
Content	Any topic except 1.14 Using a calculator	Any topic
Answers	Usually exact: fractions, surds, in terms of π	3 significant figures (1 d.p. for angles) unless told otherwise
Weighting	50%	50%

You also need compasses, a protractor and a ruler for both papers. Tracing paper can be requested in the exam (useful for transformations).

Assessment objectives: AO1 (knowledge and technique) is 40–50%; AO2 (analyse, interpret and communicate) is 50–60% — over half the marks are for choosing a strategy, connecting topics and communicating clearly, not just executing a procedure. That is why method marks matter so much.

Write every line of working. Marks are awarded for method (M) and accuracy (A) separately — a wrong final answer with correct working usually still scores most of the marks, while a bare correct answer to a multi-step question can lose them if working was required.

Rounding too early on Paper 2. Keep full accuracy in the calculator (use ANS or memory) and round only the final answer. If a later part uses your answer, use the unrounded value.

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Formula sheet: what's given vs what you must memorise

Page 2 of both papers gives you a short list of formulas. Everything else you must know. Students lose easy marks by memorising the given list and forgetting the rest — this table fixes that.

Given to you in the exam

Quantity	Formula
Area of triangle	A = ½bh
Area of circle	A = πr²
Circumference of circle	C = 2πr
Curved surface area of cylinder	A = 2πrh
Curved surface area of cone	A = πrl
Surface area of sphere	A = 4πr²
Volume of prism	V = Al
Volume of pyramid	V = ⅓Ah
Volume of cylinder	V = πr²h
Volume of cone	V = ⅓πr²h
Volume of sphere	V = 4/3 πr³
Quadratic formula	x = (−b ± √(b² − 4ac)) / 2a

NOT given — you must know these

Topic	Formula you must recall
Speed	speed = distance ÷ time
Density / pressure	density = mass ÷ volume
Pythagoras	a² + b² = c²
Trig ratios (right-angled)	sin = O/H, cos = A/H, tan = O/A
Sine rule	a/sin A = b/sin B = c/sin C
Cosine rule	a² = b² + c² − 2bc·cos A
Area of any triangle	A = ½ab·sin C
Arc length / sector area	arc = (θ/360)×2πr · sector = (θ/360)×πr²
Compound interest	A = P(1 + r/100)ⁿ
Exponential growth/decay	A = P(1 ± r/100)ⁿ
Gradient	m = (y₂ − y₁)/(x₂ − x₁)
Midpoint	((x₁+x₂)/2, (y₁+y₂)/2)
Length of a line segment	√((x₂−x₁)² + (y₂−y₁)²)
Equation of a line	y = mx + c
Perpendicular gradients	m₁ × m₂ = −1
Magnitude of a vector	\|(x y)\| = √(x² + y²)
Mean from frequency table	mean = Σfx ÷ Σf
Histogram	frequency density = frequency ÷ class width
Similar shapes	areas ∝ k², volumes ∝ k³

Copy the "NOT given" table onto one page and test yourself on it weekly. Roughly a fifth of all marks in the paper start with recalling one of these.

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Study planner & progress

All 68 syllabus units. Tick one when you can do a past-paper question on it unaided — not when you've read it. Your ticks are saved on this device only — nothing is sent anywhere, and there is no account to create.

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Topic 1 · 18 units

Number

The biggest topic in the syllabus and the foundation of Paper 1. Number work is where non-calculator fluency is built — fractions, indices, surds and percentages appear inside almost every other topic, so weakness here costs marks everywhere else.

1.1Types of number

Know and identify: natural numbers (1, 2, 3, …), integers (…−2, −1, 0, 1, 2…), prime numbers (exactly two factors — note 1 is not prime, 2 is the only even prime), square numbers (1, 4, 9, 16…), cube numbers (1, 8, 27, 64…), common factors, common multiples, rational numbers (can be written as a fraction of integers) and irrational numbers (√2, π), and reciprocals (the reciprocal of a/b is b/a).

Method — HCF and LCM by prime factorsWrite each number as a product of primes. HCF: multiply the primes common to both, using the lowest power. LCM: multiply every prime that appears, using the highest power.

Worked example

Find the HCF and LCM of 36 and 84.

36 = 2² × 3²
84 = 2² × 3 × 7
HCF: common primes 2 and 3, lowest powers → 2² × 3 = 12
LCM: all primes, highest powers → 2² × 3² × 7 = 252

Check: HCF × LCM = 12 × 252 = 3024 = 36 × 84 ✓ (always true for two numbers)

Skill check: Express 72 as a product of its prime factors, and find the LCM of 72 and 120.

Solution: 72 = 2³ × 3². 120 = 2³ × 3 × 5. LCM takes every prime at its highest power: 2³ × 3² × 5 = 8 × 9 × 5 = 360. (HCF would be 2³ × 3 = 24.)

1.2Sets

Set notation you must know:

n(A)	number of elements in A	∈ / ∉	is / is not an element of
A′	complement of A (everything not in A)	∅	the empty set
ℰ	the universal set	A ⊆ B	A is a subset of B
A ∪ B	union — in A or B (or both)	A ∩ B	intersection — in A and B

Venn diagrams are limited to two or three sets.

Method — two-set Venn problemsAlways fill the intersection first, then subtract it from each set total to get the "only" regions, then the outside region last.

Worked example

In a class of 30, 18 study Physics, 15 study Chemistry, 5 study neither. How many study both?

Studying at least one = 30 − 5 = 25
Let both = x. Then (18 − x) + x + (15 − x) = 25
33 − x = 25 → x = 8

So Physics only = 10, both = 8, Chemistry only = 7, neither = 5. Total 30 ✓

Skill check: ℰ = {1,2,…,12}, A = {multiples of 3}, B = {even numbers}. List A ∩ B and find n(A ∪ B)′.

Solution: A = {3,6,9,12}, B = {2,4,6,8,10,12}. A ∩ B = {6, 12}. A ∪ B = {2,3,4,6,8,9,10,12} → n(A ∪ B) = 8, so n(A ∪ B)′ = 12 − 8 = 4 (the elements 1, 5, 7, 11).

1.3Powers and roots

You are expected to recall squares and square roots from 1 to 15, and cubes and cube roots of 1, 2, 3, 4, 5 and 10.

n	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
n²	1	4	9	16	25	36	49	64	81	100	121	144	169	196	225

Cubes to recall: 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125, 10³ = 1000.

Worked example

Without a calculator, work out 5 × ∛8 + √169.

∛8 = 2, so 5 × 2 = 10
√169 = 13
Total = 23

Skill check: Work out √196 − ∛125.

Solution: √196 = 14, ∛125 = 5, so 14 − 5 = 9.

1.4Fractions, decimals and percentages

Convert freely between proper fractions, improper fractions, mixed numbers, decimals and percentages. Always give fractions in simplest form. Recurring decimal notation: 0.17 = 0.1777…, 0.123 = 0.123123…

Method — recurring decimal to fractionLet x = the decimal. Multiply by a power of 10 to move past the non-recurring part, and by another to move a whole recurring block. Subtract to eliminate the recurring tail, then solve.

Worked example

Write 0.17 (= 0.1777…) as a fraction in its lowest terms.

Let x = 0.1777…
10x = 1.777… (past the non-recurring "1")
100x = 17.777…
Subtract: 100x − 10x = 17.777… − 1.777… → 90x = 16
x = 16/90 = 8/45

Skill check: Write 0.36 (= 0.3636…) as a fraction in its simplest form.

Solution: Let x = 0.3636…; 100x = 36.3636…; subtract: 99x = 36; x = 36/99 = 4/11 (divide top and bottom by 9).

1.5Ordering 1.6The four operations

1.5 Order quantities by size using =, ≠, >, <, ⩾, ⩽. To compare fractions, convert to a common denominator or to decimals.

1.6 Use the four operations with integers, fractions and decimals — including negatives, improper fractions and mixed numbers — with correct order of operations (BIDMAS: Brackets, Indices, Division/Multiplication, Addition/Subtraction) and brackets.

Worked example

Work out 2⅓ − 1½ × ⅔, giving your answer as a fraction in its lowest terms.

Multiplication first: 1½ × ⅔ = 3/2 × 2/3 = 6/6 = 1
2⅓ − 1 = 1⅓ (or 4/3)

Working left to right instead of using BIDMAS. In 2 + 3 × 4 the answer is 14, not 20. On Paper 1 this single slip is worth several marks a year.

Skill check: Work out −5 + 12 ÷ (−3) − (−4).

Solution: Division first: 12 ÷ (−3) = −4. So −5 + (−4) − (−4) = −5 − 4 + 4 = −5.

1.7Indices I

Method — the index laws

a^m × aⁿ = a^m+n · a^m ÷ aⁿ = a^m−n · (a^m)ⁿ = a^mn · a⁰ = 1 · a⁻ⁿ = 1/aⁿ · a^1/n = ⁿ√a · a^m/n = (ⁿ√a)^m

Worked example

Without a calculator find the value of (a) 7⁻², (b) 81^1/2, (c) 8^−2/3.

(a) 7⁻² = 1/7² = 1/49
(b) 81^1/2 = √81 = 9
(c) Deal with the root first: 8^1/3 = 2. Then square: 2² = 4. Then the minus sign inverts: ¼

Thinking a negative index makes the answer negative. It does not — it makes it a reciprocal. 2⁻³ = ⅛, never −8.

Skill check: Simplify 2⁻³ × 2⁴ and evaluate 16^3/4.

Solution: 2⁻³ × 2⁴ = 2⁻³⁺⁴ = 2¹ = 2. For 16^3/4: fourth root of 16 is 2, then 2³ = 8.

1.8Standard form

Standard form is A × 10ⁿ where 1 ⩽ A < 10 and n is an integer. Large numbers give positive n; small numbers give negative n.

Worked example

Work out (3.2 × 10⁵) × (4 × 10⁻⁸), giving your answer in standard form.

Numbers: 3.2 × 4 = 12.8
Powers: 10⁵ × 10⁻⁸ = 10⁻³
12.8 × 10⁻³ is not standard form (12.8 ⩾ 10). Adjust: 1.28 × 10¹ × 10⁻³ = 1.28 × 10⁻²

Leaving an answer like 12.8 × 10⁻³ or 0.45 × 10⁶. The number in front must be at least 1 and less than 10 — always check before writing your final answer.

Skill check: Work out (6 × 10⁷) ÷ (1.5 × 10⁻²) in standard form.

Solution: 6 ÷ 1.5 = 4; 10⁷ ÷ 10⁻² = 10⁷⁻⁽⁻²⁾ = 10⁹. Answer: 4 × 10⁹.

1.9Estimation

Round values to a stated accuracy (decimal places or significant figures), estimate calculations by rounding each number to 1 significant figure, and round final answers sensibly for the context.

Worked example

By writing each number correct to 1 significant figure, estimate 41.3 ÷ (9.79 × 0.765).

41.3 → 40, 9.79 → 10, 0.765 → 0.8
10 × 0.8 = 8
40 ÷ 8 = 5

(True value ≈ 5.51 — an estimate should be close, not exact.)

Estimation questions demand you show the rounded values. Writing only the final answer, even a correct one, loses the method mark.

Skill check: Estimate (19.6 × 4.87) ÷ 0.51 using 1 significant figure values.

Solution: 19.6 → 20, 4.87 → 5, 0.51 → 0.5. So (20 × 5) ÷ 0.5 = 100 ÷ 0.5 = 200. (True value ≈ 187.2.)

1.10Limits of accuracy

Method — boundsA value rounded to a given accuracy lies within half a unit either side. Measured to the nearest cm: 12 cm → lower bound 11.5, upper bound 12.5. For calculations: to get the largest result use the largest values (but for a subtraction or division, use the largest top and the smallest bottom).

Worked example

A rectangle measures 12 cm by 8 cm, each to the nearest centimetre. Find the upper bound of its area, and the lower bound of its perimeter.

Bounds: length 11.5 ⩽ l < 12.5; width 7.5 ⩽ w < 8.5
Upper bound of area = 12.5 × 8.5 = 106.25 cm²
Lower bound of perimeter = 2(11.5 + 7.5) = 38 cm

Worked example — division

A car travels 150 m (to the nearest 10 m) in 12 s (to the nearest second). Find the lower bound of its speed.

Distance: 145 ⩽ d < 155. Time: 11.5 ⩽ t < 12.5
Slowest = smallest distance ÷ largest time = 145 ÷ 12.5
= 11.6 m/s

Skill check: x = 4.2 and y = 2.5, both correct to 1 decimal place. Find the upper bound of x − y.

Solution: Bounds: 4.15 ⩽ x < 4.25 and 2.45 ⩽ y < 2.55. For the largest difference take the largest x and the smallest y: 4.25 − 2.45 = 1.8.

1.11Ratio and proportion

Method — sharing in a ratioAdd the parts to get the total number of shares. Divide the amount by the total shares to find one share. Multiply back up for each portion.

Worked example

Share $540 between three people in the ratio 2 : 3 : 4.

Total shares = 2 + 3 + 4 = 9
One share = 540 ÷ 9 = 60
Amounts = 2×60, 3×60, 4×60 = $120, $180, $240 (check: they sum to 540 ✓)

Watch for questions giving the difference between two portions rather than the total, e.g. "the largest share is $120 more than the smallest". Here the difference is 4 − 2 = 2 shares, so one share = $60.

Skill check: A recipe for 4 people needs 300 g of rice. How much is needed for 10 people?

Solution: Direct proportion. For 1 person: 300 ÷ 4 = 75 g. For 10: 75 × 10 = 750 g.

1.12Rates

Common rates: pay per hour, exchange rates, flow rates, fuel consumption. Other measures: pressure, density, population density — formulas for these will be given in the question. But speed = distance ÷ time must be recalled.

Worked example

A cyclist travels 45 km in 3 hours 45 minutes. Find the average speed in km/h.

Convert time to hours: 45 min = 45/60 = 0.75 h, so time = 3.75 h
Speed = 45 ÷ 3.75 = 12 km/h

Entering 3 hours 45 minutes as 3.45 on the calculator. Minutes must be converted to a decimal fraction of an hour: 3 h 45 min = 3.75 h.

Skill check: A metal block has mass 480 g and volume 60 cm³. Find its density in g/cm³.

Solution: density = mass ÷ volume = 480 ÷ 60 = 8 g/cm³.

1.13Percentages

Method — multipliersIncrease by r%: multiply by (1 + r/100). Decrease by r%: multiply by (1 − r/100). Reverse percentage: divide by the multiplier. Compound interest: A = P(1 + r/100)ⁿ — this formula is not given.

Worked example — reverse percentage

A shop sells a jacket for $84 after adding 20% profit to the cost price. Find the cost price.

Selling price = cost × 1.2
Cost = 84 ÷ 1.2 = $70

Check: 70 × 1.2 = 84 ✓ (Subtracting 20% of 84 gives $67.20 — wrong, and the classic trap.)

Worked example — compound interest

$5000 is invested at 6% per year compound interest. Find the value after 3 years, and the interest earned.

A = 5000 × (1.06)³
= 5000 × 1.191016 = $5955.08 (to the nearest cent)
Interest = 5955.08 − 5000 = $955.08

Answering the wrong question: many marks are lost by giving the final amount when the question asked for the interest (or vice versa). Underline what is being asked.

Skill check: After a 15% discount a phone costs $459. What was the original price?

Solution: Reverse percentage: the multiplier is 0.85, so original = 459 ÷ 0.85 = $540. Check: 540 × 0.85 = 459 ✓

1.14Using a calculator P2 only

This is the one sub-topic that cannot be examined on Paper 1. You need to:

Use the calculator efficiently — never round mid-calculation; use ANS or memory to carry full accuracy forward.
Enter values properly: 2 hours 30 minutes as 2.5 or with the degrees-minutes-seconds key.
Interpret the display in context: in money, 4.8 means $4.80; in time, 3.25 hours means 3 hours 15 minutes.

On Paper 2, use π from the calculator (or 3.142) and give non-exact answers to 3 significant figures — 1 decimal place for angles — unless told otherwise.

1.15Time 1.16Money

1.15 Calculate with seconds, minutes, hours, days, weeks, months and years (take 1 year = 365 days). Convert between the 24-hour clock (03 15 and 15 15) and the 12-hour clock (3.15 a.m. and 3.15 p.m.). Read clocks and timetables, including time zones and time differences.

1.16 Calculate with money and convert between currencies.

Worked example

A flight leaves Karachi at 23 40 and takes 7 hours 50 minutes. Dubai time is 1 hour behind Karachi. At what local time does it land?

Arrival in Karachi time: 23 40 + 7 h 50 min. Add hours: 23 40 → 06 40 (next day). Add 50 min: 06 40 → 07 30.
Dubai is 1 hour behind: 07 30 − 1 h = 06 30 local time

Worked example — currency

$1 = 278 rupees. Convert (a) $45 to rupees, (b) 100 000 rupees to dollars.

(a) 45 × 278 = 12 510 rupees
(b) 100 000 ÷ 278 = $359.71 (2 d.p.)

Skill check: A train leaves at 14 55 and arrives at 18 20. How long is the journey?

Solution: 14 55 → 18 55 would be 4 hours, which overshoots by 35 minutes. So the journey is 4 h − 35 min = 3 hours 25 minutes. (Or: 14 55 → 15 00 is 5 min; 15 00 → 18 20 is 3 h 20 min; total 3 h 25 min.)

1.17Exponential growth and decay

Same multiplier idea as compound interest, applied to depreciation, population change and similar. Knowledge of e is not required.

value after n periods = P × (1 ± r/100)ⁿ

Worked example

A car bought for $25 000 depreciates by 15% each year. Find its value after 4 years.

Multiplier for a 15% decrease = 0.85
Value = 25 000 × 0.85⁴
0.85⁴ = 0.52200625, so value = $13 050.16 (to the nearest cent)

Note it is not 25 000 − 4 × 15% = $10 000; each year's loss is smaller than the last.

Skill check: A town's population of 40 000 grows by 3% per year. Find the population after 5 years, to the nearest hundred.

Solution: 40 000 × 1.03⁵ = 40 000 × 1.159274… = 46 370.9…, so 46 400 to the nearest hundred.

1.18Surds

Method — simplifying surdsFind the largest square factor inside the root and take its root outside: √ab = √a × √b. To rationalise a denominator, multiply top and bottom by the surd (or by the conjugate if the denominator has two terms — change the sign between them).

Worked example — simplifying

Simplify √200 − √32.

√200 = √(100 × 2) = 10√2
√32 = √(16 × 2) = 4√2
10√2 − 4√2 = 6√2

Worked example — rationalising with a conjugate

Rationalise the denominator of 1/(−1 + √3).

Conjugate of (−1 + √3) is (1 + √3) — multiply top and bottom by it
Denominator: (−1 + √3)(1 + √3) = −1 − √3 + √3 + 3 = 2
Answer: (1 + √3)/2

Skill check: Simplify 10/√5, and write √75 + √12 in the form a√b.

Solution: 10/√5 = (10 × √5)/(√5 × √5) = 10√5/5 = 2√5. And √75 = 5√3, √12 = 2√3, so the sum is 7√3.

Topic 2 · 12 units

Algebra and graphs

The largest source of marks after Number, and the topic that most separates grades. Almost everything here rests on two skills: manipulating expressions without sign errors, and knowing which method a question is asking for.

2.1Introduction to algebra

Letters represent generalised numbers. You must substitute numbers into expressions and formulas confidently — including negatives.

Worked example

Find the value of 3x² − 2y when x = −2 and y = 5.

3x² means 3 × (x²), not (3x)². So x² = (−2)² = 4, giving 3 × 4 = 12
2y = 10
12 − 10 = 2

Squaring a negative wrongly: (−2)² = +4, but −2² = −4. Always bracket a negative before squaring it.

2.2Algebraic manipulation

Collect like terms, expand products (including three brackets), and factorise fully — by common factors, by grouping, using the difference of two squares, and quadratics.

Method — key factorising patterns

Common factor: 9x² + 15xy = 3x(3x + 5y)
Difference of two squares: a² − b² = (a − b)(a + b) — e.g. 9x² − 16 = (3x − 4)(3x + 4)
Quadratic (a = 1): find two numbers that multiply to c and add to b
Grouping (4 terms): ax + ay + bx + by = a(x+y) + b(x+y) = (a+b)(x+y)

Worked example — expanding

Expand and simplify (3x + y)(x − 4y).

3x × x = 3x²; 3x × (−4y) = −12xy
y × x = xy; y × (−4y) = −4y²
Collect: 3x² − 11xy − 4y²

Worked example — three brackets

Expand (x − 2)(x + 3)(2x + 1).

First two: (x − 2)(x + 3) = x² + x − 6
Multiply by (2x + 1): 2x(x² + x − 6) = 2x³ + 2x² − 12x; 1(x² + x − 6) = x² + x − 6
Add: 2x³ + 3x² − 11x − 6

Skill check: Factorise fully (a) x² − 5x − 14, (b) 4x² − 36.

Solution: (a) Two numbers multiplying to −14 and adding to −5: −7 and +2 → (x − 7)(x + 2). (b) "Fully" means take the common factor first: 4(x² − 9) = 4(x − 3)(x + 3).

2.3Algebraic fractions

MethodTo simplify: factorise top and bottom fully, then cancel common factors. To add or subtract: use a common denominator (usually the product of the two denominators), combine numerators, then simplify.

Worked example — simplifying

Simplify (x² − 9)/(x² + 7x + 12).

Top: difference of two squares → (x − 3)(x + 3)
Bottom: factors of 12 adding to 7 → (x + 3)(x + 4)
Cancel (x + 3): (x − 3)/(x + 4)

Worked example — adding

Write 2/(x + 2) + 3/(2x − 1) as a single fraction.

Common denominator (x + 2)(2x − 1)
Numerator: 2(2x − 1) + 3(x + 2) = 4x − 2 + 3x + 6 = 7x + 4
Answer: (7x + 4)/[(x + 2)(2x − 1)]

Cancelling terms instead of factors. In (x + 3)/(x + 4) you cannot cancel the x's — cancelling is only allowed between complete bracketed factors.

2.4Indices II

The same index laws as 1.7, now applied to algebra and to equations with unknown powers. Logarithms are not required — instead, write both sides with the same base and equate the powers.

Worked example — equation

Solve 5^x+1 = 25^x.

Write 25 as 5²: 25^x = (5²)^x = 5^2x
Same base, so equate powers: x + 1 = 2x
x = 1

Worked example — simplifying

Simplify (2x⁵/3)³.

Cube everything: 2³ = 8, (x⁵)³ = x¹⁵, 3³ = 27
= 8x¹⁵/27

Skill check: Simplify 15x⁵ ÷ 3x², and solve 2^3x = 32.

Solution: 15 ÷ 3 = 5 and x⁵⁻² = x³, so 5x³. For the equation: 32 = 2⁵, so 3x = 5 and x = 5/3.

2.5Equations

You must construct expressions and equations, then solve: linear equations, fractional equations, simultaneous linear equations, quadratics (three methods), and rearrange formulas — including when the subject appears twice or under a power/root.

Worked example — linear

Solve 5 − 2x = 3(x + 7).

Expand: 5 − 2x = 3x + 21
Collect: 5 − 21 = 3x + 2x → −16 = 5x
x = −16/5 = −3.2

Worked example — simultaneous

Solve 3x + 2y = 16 and x − y = 2.

From the second: x = y + 2
Substitute: 3(y + 2) + 2y = 16 → 5y + 6 = 16 → y = 2
x = 2 + 2 = 4. So x = 4, y = 2 (check in the first equation: 12 + 4 = 16 ✓)

Worked example — fractional equation

Solve 2/(x + 2) + 3/(2x − 1) = 1.

Multiply every term by (x + 2)(2x − 1): 2(2x − 1) + 3(x + 2) = (x + 2)(2x − 1)
7x + 4 = 2x² + 3x − 2
0 = 2x² − 4x − 6, i.e. x² − 2x − 3 = 0
(x − 3)(x + 1) = 0 → x = 3 or x = −1

Worked example — quadratic three ways

(a) Factorise: 2x² + x − 6 = 0 → (2x − 3)(x + 2) = 0 → x = 1.5 or x = −2

(b) Completing the square: x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7. (Halve the x-coefficient, square it, subtract it back.)

Worked example — subject appears twice

Make x the subject of y = (x + 3)/(x − 2).

Multiply out: y(x − 2) = x + 3 → xy − 2y = x + 3
Gather x-terms on one side: xy − x = 2y + 3
Factorise: x(y − 1) = 2y + 3
x = (2y + 3)/(y − 1)

If a quadratic does not factorise easily, use the formula (it is given on page 2). If the question says "give your answer in surd form" or "leave in terms of √", do not convert to a decimal.

Skill check: Solve x² − 6x + 4 = 0, giving answers in surd form.

Solution: x = (6 ± √(36 − 16))/2 = (6 ± √20)/2 = (6 ± 2√5)/2 = 3 ± √5.

2.6Inequalities

Method & conventions

Solve like an equation — but reverse the inequality sign whenever you multiply or divide by a negative.
Number lines: open circle ○ for strict (<, >), closed circle ● for inclusive (⩽, ⩾).
Graphs: broken line for strict, solid line for inclusive; shade the unwanted region unless told otherwise.

Worked example — double inequality

Solve −3 ⩽ 3x − 2 < 7.

Add 2 throughout: −1 ⩽ 3x < 9
Divide by 3 throughout: −⅓ ⩽ x < 3
Number line: closed circle at −⅓, open circle at 3, line joining them.

Forgetting to flip the sign: from −2x > 6, dividing by −2 gives x < −3, not x > −3.

Skill check: Solve 3x < 2x + 4 and list the integers satisfying both this and x ⩾ 1.

Solution: 3x − 2x < 4 → x < 4. Combined with x ⩾ 1: 1 ⩽ x < 4, so the integers are 1, 2, 3.

2.7Sequences

Continue sequences, describe the term-to-term rule, and find the nth term. Sequences may be linear, quadratic, cubic or exponential, and simple combinations. Subscript notation (T_n) may be used.

Method — identifying the type

First differences constant → linear: nth term = (difference)n + (term before the first).
Second differences constant → quadratic: the n² coefficient is half the second difference.
Constant ratio → exponential: nth term = a × rⁿ⁻¹.

Worked example — linear

Find the nth term of 7, 11, 15, 19, …

First difference = 4 → term contains 4n
4n gives 4, 8, 12, 16 — each is 3 less than the sequence
T_n = 4n + 3

Worked example — quadratic

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

First differences: 5, 7, 9, 11. Second differences: 2 (constant) → quadratic
Coefficient of n² = 2 ÷ 2 = 1, so start with n²: 1, 4, 9, 16, 25
Subtract from the sequence: 2, 4, 6, 8, 10 → that is 2n
T_n = n² + 2n (check n = 4: 16 + 8 = 24 ✓)

Skill check: Find the nth term of 2, 6, 18, 54, … and hence the 6th term.

Solution: Constant ratio 3 → exponential. T_n = 2 × 3ⁿ⁻¹. 6th term = 2 × 3⁵ = 2 × 243 = 486.

2.8Proportion

Express direct and inverse proportion algebraically, including square, square root, cube and cube root relationships. You must know the symbol ∝.

Method — three steps, always1. Write the relationship with a constant: y ∝ x² means y = kx². (Inverse: y ∝ 1/x means y = k/x.) 2. Substitute the given pair to find k. 3. Use the completed formula.

Worked example

y is directly proportional to x². When x = 3, y = 18. Find y when x = 5.

y = kx²
18 = k × 9 → k = 2, so y = 2x²
x = 5: y = 2 × 25 = 50

Skill check: p is inversely proportional to √q. When q = 16, p = 3. Find p when q = 4.

Solution: p = k/√q. Then 3 = k/4 → k = 12, so p = 12/√q. When q = 4: p = 12/2 = 6.

2.9Graphs in practical situations

Interpret travel graphs and conversion graphs, draw graphs from data, apply rate of change to distance–time and speed–time graphs (acceleration and deceleration), and find distance as the area under a speed–time graph (linear sections only). Includes estimating a gradient by drawing a tangent.

Graph type	Gradient means	Area under means
Distance–time	Speed	(no meaning)
Speed–time	Acceleration (negative = deceleration)	Distance travelled

Speed–time graph: gradient gives acceleration, the shaded area gives total distance.

Worked example

A car accelerates uniformly from rest to 20 m/s in 8 s, travels at 20 m/s for 12 s, then decelerates uniformly to rest in 5 s. Find (a) the acceleration in the first stage, (b) the total distance.

(a) Acceleration = change in speed ÷ time = 20 ÷ 8 = 2.5 m/s²
(b) Split the area into a triangle, rectangle and triangle:
Triangle 1 = ½ × 8 × 20 = 80 m; Rectangle = 12 × 20 = 240 m; Triangle 2 = ½ × 5 × 20 = 50 m
Total = 370 m

Skill check: On a distance–time graph, a straight line goes from (0 s, 0 m) to (40 s, 300 m), then is horizontal until 60 s. Describe the motion and find the speed in the first stage.

Solution: Constant speed of 300 ÷ 40 = 7.5 m/s for the first 40 s; then the horizontal line means the gradient is zero — the object is stationary for the last 20 s.

2.10Graphs of functions

Construct tables of values and draw graphs of functions of the form axⁿ (sums of up to three terms, with n = −2, −1, −½, 0, ½, 1, 2, 3) and ab^x + c. Solve equations graphically (including line-meets-curve), interpret roots, draw exponential growth/decay graphs, and estimate gradients of curves by drawing tangents.

Method — solving graphicallyThe roots of f(x) = 0 are where the curve crosses the x-axis. To solve f(x) = g(x), draw both and read off the x-coordinates of the intersections. To solve a different equation from your existing curve, rearrange it into "your curve = a straight line", then draw that line.

Worked example

The graph of y = x³ + x − 4 has been drawn. Explain how to use it to solve x³ + x − 6 = 0.

Rearrange so one side matches the drawn curve: x³ + x − 6 = 0 → x³ + x − 4 = 2
So draw the horizontal line y = 2
The x-coordinate(s) where the line meets the curve give the solution(s) — here x ≈ 1.6

When plotting, use a sharp pencil, plot every point from your table, and join with a smooth curve — never a series of straight segments. A table with one wrong value is the most common cause of a lost curve mark, so re-check any point that breaks the pattern.

2.11Sketching curves

Recognise and sketch linear, quadratic, cubic, reciprocal and exponential graphs. Knowledge of turning points, roots, symmetry and asymptotes is required, and you must find turning points of quadratics by completing the square.

Function	Shape and features
ax + by = c	Straight line
y = ax² + bx + c	Parabola: U-shape if a > 0, ∩-shape if a < 0; symmetric about the turning point
y = ax³ + b or ax³ + bx² + cx	Cubic: up to two turning points; opposite ends go opposite ways
y = a/x + b	Reciprocal: two branches; asymptotes x = 0 and y = b
y = ar^x + b	Exponential: rapid growth (or decay); horizontal asymptote y = b

Worked example

Find the turning point and roots of y = x² − 4x + 3, and sketch it.

Complete the square: x² − 4x + 3 = (x − 2)² − 4 + 3 = (x − 2)² − 1
Turning point: (2, −1), a minimum since the x² coefficient is positive
Roots: factorise (x − 1)(x − 3) = 0 → crosses at x = 1 and x = 3
y-intercept: x = 0 gives y = 3. Sketch a U-shape through (1,0), (3,0), (0,3), lowest at (2,−1)

Skill check: Write y = x² + 6x + 5 in completed square form and state its minimum point.

Solution: (x + 3)² − 9 + 5 = (x + 3)² − 4. Minimum at (−3, −4).

2.12Functions

Use function notation, domain and range; find inverse functions f⁻¹(x); form composite functions where gf(x) means g(f(x)). Mapping diagrams may appear. You are not expected to find domains/ranges of composite functions.

Method — inverse functionWrite y = f(x), swap x and y, then make y the subject. Replace y with f⁻¹(x).

Worked example

f(x) = 3x − 5 and g(x) = x². Find (a) f⁻¹(x), (b) fg(x), (c) gf(2).

(a) y = 3x − 5 → swap: x = 3y − 5 → y = (x + 5)/3, so f⁻¹(x) = (x + 5)/3
(b) fg(x) = f(g(x)) = f(x²) = 3x² − 5
(c) gf(2): first f(2) = 1, then g(1) = 1

Applying composite functions in the wrong order. In gf(x) the function nearest the x acts first — f, then g. And fg(x) is almost never the same as gf(x).

Skill check: f(x) = 3/(x + 2), g(x) = (3x + 5)². Find fg(x).

Solution: fg(x) = f((3x + 5)²) = 3/[(3x + 5)² + 2] — substitute g into the x of f, and leave it as a single fraction.

Topic 3 · 7 units

Coordinate geometry

Short, formula-driven and highly predictable — one of the best marks-per-hour topics in the syllabus. None of these formulas are given on the exam formula sheet, so all four must be memorised.

3.1Coordinates 3.2Drawing linear graphs

3.1 Use and interpret Cartesian coordinates in two dimensions — points are written (x, y), across then up.

3.2 Draw a straight-line graph from its equation. Quickest reliable method: build a small table of three values of x (use the third as a check), plot, and join with a ruled line extended across the given grid.

For a line like 2x + 3y = 12, the fastest sketch uses intercepts: put x = 0 → y = 4, and y = 0 → x = 6. Plot (0,4) and (6,0) and rule the line.

3.3Gradient 3.4Length and midpoint

The three formulas — memorise (not given)

gradient m = (y₂ − y₁) ÷ (x₂ − x₁)

midpoint = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

length = √[ (x₂ − x₁)² + (y₂ − y₁)² ] (Pythagoras in disguise)

Worked example

A is (−1, 2) and B is (5, 10). Find the gradient of AB, the midpoint of AB, and the length AB.

Gradient = (10 − 2)/(5 − (−1)) = 8/6 = 4/3
Midpoint = ((−1 + 5)/2, (2 + 10)/2) = (2, 6)
Length = √(6² + 8²) = √(36 + 64) = √100 = 10

Subtracting in inconsistent orders for the gradient — if you do y₂ − y₁ on top you must do x₂ − x₁ on the bottom. Reversing one gives the wrong sign.

Skill check: P is (3, −4) and Q is (−3, 4). Find the length PQ and the midpoint.

Solution: Length = √((−3−3)² + (4−(−4))²) = √(36 + 64) = 10. Midpoint = ((3 + −3)/2, (−4 + 4)/2) = (0, 0).

3.5Equations of linear graphs 3.6Parallel lines 3.7Perpendicular lines

Every straight line can be written y = mx + c, where m is the gradient and c the y-intercept.

Method — the two rules

Parallel lines have equal gradients: m₁ = m₂.
Perpendicular lines have gradients whose product is −1: m₁ × m₂ = −1. In practice: flip the fraction and change the sign (the "negative reciprocal").

Worked example

Find the equation of the line through (2, 7) that is perpendicular to y = ½x + 4.

Given gradient = ½, so the perpendicular gradient = −2 (negative reciprocal)
Use y = mx + c with the point: 7 = −2(2) + c → 7 = −4 + c → c = 11
y = −2x + 11

Skill check: Find the equation of the line parallel to y = 3x − 1 passing through (−2, 5).

Solution: Parallel → same gradient, m = 3. Then 5 = 3(−2) + c → 5 = −6 + c → c = 11. So y = 3x + 11.

Topic 4 · 8 units

Geometry

Angle reasoning is where "give a reason" marks live — and where they are most often thrown away. Every angle you state must come with the correct named property, in the examiner's language.

4.1Geometrical terms

General: point, vertex, line, plane, parallel, perpendicular, perpendicular bisector, bearing, right/acute/obtuse/reflex angles, interior and exterior angles, similar, congruent, scale factor. (You are not asked to prove two shapes congruent.)

Triangles: equilateral, isosceles, scalene, right-angled. Quadrilaterals: square, rectangle, kite, rhombus, parallelogram, trapezium. Polygons: regular/irregular, pentagon, hexagon, octagon, decagon. Solids: cube, cuboid, prism, cylinder, pyramid, cone, sphere, hemisphere, frustum, plus face, surface and edge.

Circle vocabulary: centre, radius (radii), diameter, circumference, semicircle, chord, tangent, major and minor arc, sector, segment.

Mixing up arc (part of the circumference), sector (the "pizza slice" bounded by two radii) and segment (the region cut off by a chord). Questions use these words precisely and so must you.

4.2Geometrical constructions 4.3Scale drawings

4.2 Measure and draw lines and angles; construct a triangle given all three sides using ruler and compasses only — and leave your construction arcs visible, they carry marks. Draw, use and interpret nets (cubes, cuboids, prisms, pyramids), including using net measurements to find surface areas and volumes. Note: perpendicular bisector and angle bisector constructions are not required on this syllabus.

4.3 Draw and interpret scale drawings, and use three-figure bearings — always measured clockwise from north, written with three digits (e.g. 025°, 170°, 305°).

Method — back bearingsTo find the bearing of A from B when you know the bearing of B from A: add 180° if the original is less than 180°, subtract 180° if it is 180° or more.

Worked example

The bearing of B from A is 025°. Find the bearing of A from B.

025° is less than 180°, so add: 025° + 180° = 205°

Skill check: The bearing of Q from P is 310°. Find the bearing of P from Q.

Solution: 310° ⩾ 180°, so subtract: 310° − 180° = 130°.

4.4Similarity

Method — the scale factor rulesIf two shapes are similar with length scale factor k:

lengths × k · areas × k² · volumes × k³

Triangles are similar if their angles are equal (equiangular) — you may be asked to show this with geometric reasons.

Worked example

Two similar cones have heights 6 cm and 9 cm. The smaller has volume 40 cm³ and surface area 30 cm². Find the volume and surface area of the larger.

Length scale factor k = 9/6 = 1.5
Volume scale factor = k³ = 3.375 → volume = 40 × 3.375 = 135 cm³
Area scale factor = k² = 2.25 → surface area = 30 × 2.25 = 67.5 cm²

Using the length scale factor for area or volume. If lengths double, areas become four times bigger and volumes eight times bigger.

Skill check: Two similar jugs have volumes 250 ml and 2000 ml. The smaller is 10 cm tall. Find the height of the larger.

Solution: Volume factor = 2000/250 = 8, so k³ = 8 → k = 2. Height = 10 × 2 = 20 cm.

4.5Symmetry

Recognise line symmetry and order of rotational symmetry in 2D, including the symmetry properties of triangles, quadrilaterals and polygons; and recognise symmetry properties of prisms, cylinders, pyramids and cones (planes and axes of symmetry).

Shape	Lines of symmetry	Rotational symmetry order
Square	4	4
Rectangle	2	2
Rhombus	2	2
Parallelogram	0	2
Kite	1	1
Equilateral triangle	3	3
Isosceles triangle	1	1
Regular n-gon	n	n

Saying a parallelogram has 2 lines of symmetry — it has none (its diagonals are not mirror lines), though it does have rotational symmetry of order 2.

4.6Angles

The properties — and the words to use

Angles at a point sum to 360°; angles on a straight line sum to 180°.
Vertically opposite angles are equal.
Angle sum of a triangle = 180°; of a quadrilateral = 360°.
In parallel lines: corresponding angles are equal (F-shape), alternate angles are equal (Z-shape), co-interior angles sum to 180° (C-shape, also called supplementary).

Polygon formulas

sum of interior angles = (n − 2) × 180° · each exterior angle of a regular polygon = 360° ÷ n · interior + exterior = 180°

Worked example

A regular polygon has interior angles of 156°. How many sides does it have?

Exterior angle = 180° − 156° = 24°
n = 360° ÷ 24° = 15 sides

Three-letter angle notation is required: "angle ABC" means the angle at vertex B — the middle letter is always the vertex. When a question says "give a reason", one correct named property earns the mark; vague answers like "because it looks equal" earn nothing.

Skill check: Find the sum of the interior angles of a decagon, and the size of each interior angle if it is regular.

Solution: A decagon has n = 10. Sum = (10 − 2) × 180° = 1440°. If regular, each interior angle = 1440 ÷ 10 = 144° (check: exterior = 360/10 = 36°, and 144 + 36 = 180 ✓).

4.7Circle theorems I

The four most-tested circle theorems. Also required: tangent ⊥ radius, and the alternate segment theorem.

The six theorems — learn these exact phrases

The angle in a semicircle is 90°.
The angle between a tangent and a radius is 90°.
The angle at the centre is twice the angle at the circumference (same arc).
Angles in the same segment are equal.
Opposite angles of a cyclic quadrilateral sum to 180° (supplementary).
Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.

Worked example

A, B, C, D lie on a circle. Angle ABC = 95° and angle BAD = 70°. Find angles ADC and BCD.

ABCD is a cyclic quadrilateral, so opposite angles sum to 180°
ADC = 180° − 95° = 85° (opposite to ABC)
BCD = 180° − 70° = 110° (opposite to BAD)
Check: 95 + 70 + 85 + 110 = 360° ✓

Every circle-theorem answer needs its reason named in full — "angle at the centre is twice the angle at the circumference", not "circle theorem" or "double rule". The reason is usually worth as much as the number.

Skill check: O is the centre of a circle. Points A and B are on the circumference and angle AOB = 130°. Point C is on the major arc. Find angle ACB.

Solution: Angle at the centre is twice the angle at the circumference on the same arc, so angle ACB = 130 ÷ 2 = 65°.

4.8Circle theorems II — symmetry properties

The three symmetry properties

Equal chords are equidistant from the centre (and chords equidistant from the centre are equal).
The perpendicular bisector of a chord passes through the centre — so a radius drawn perpendicular to a chord bisects it.
Tangents from an external point are equal in length (creating an isosceles triangle with the two radii).

Worked example

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find its perpendicular distance from the centre.

The perpendicular from the centre bisects the chord → half-chord = 8 cm
Right-angled triangle with hypotenuse = radius = 10, one leg = 8
Distance = √(10² − 8²) = √36 = 6 cm

Skill check: Two tangents from point P touch a circle centre O at A and B. Angle APB = 48°. Find angle AOB.

Solution: Tangent ⊥ radius gives angles OAP = OBP = 90°. OAPB is a quadrilateral, so angles sum to 360°: angle AOB = 360 − 90 − 90 − 48 = 132°.

Topic 5 · 5 units

Mensuration

Area, volume and surface area. The formula sheet gives you the solids — but not the flat shapes (except triangles), and not the circle's flat end faces when you build total surface area. That gap is where the marks go.

5.1Units of measure

Use metric units of mass, length, area, volume and capacity, and convert between them: mm/cm/m/km, mm²/cm²/m²/km², mm³/cm³/m³, ml/l, g/kg.

Method — area and volume conversionsSquare or cube the length conversion factor:

1 cm = 10 mm → 1 cm² = 100 mm² → 1 cm³ = 1000 mm³

1 m = 100 cm → 1 m² = 10 000 cm² → 1 m³ = 1 000 000 cm³

1 cm³ = 1 ml · 1000 cm³ = 1 litre · 1 m³ = 1000 litres

Dividing by 100 to convert cm² to m². It is 10 000, because area involves two dimensions. This single error costs marks every series.

Skill check: A tank holds 2.5 m³ of water. How many litres is this, and how many cm³?

Solution: 1 m³ = 1000 litres, so 2.5 m³ = 2500 litres. And 1 m³ = 1 000 000 cm³, so 2.5 m³ = 2 500 000 cm³.

5.2Area and perimeter

Perimeter and area of rectangle, triangle, parallelogram and trapezium. Only the triangle formula is given in the exam — the rest must be recalled.

Formulas to memorise

rectangle: A = lw · parallelogram: A = bh · trapezium: A = ½(a + b)h · triangle: A = ½bh (given)

In every case h is the perpendicular height, not a slant side.

Worked example

A trapezium has parallel sides 8 cm and 14 cm, and perpendicular height 5 cm. Find its area.

A = ½(8 + 14) × 5
= ½ × 22 × 5 = 55 cm²

Skill check: A parallelogram has base 12 cm, slant side 7 cm and perpendicular height 6 cm. Find its area.

Solution: Use the perpendicular height, not the slant side: A = 12 × 6 = 72 cm². (The 7 cm is a distractor.)

5.3Circles, arcs and sectors

Circumference (C = 2πr) and area (A = πr²) are given. Arc length and sector area are not — but both are just fractions of the whole circle.

Method — fraction of a circle

arc length = (θ/360) × 2πr · sector area = (θ/360) × πr²

Perimeter of a sector = arc + two radii (a common omission).

Worked example

A sector has radius 10 cm and angle 72°. Find (a) the arc length, (b) the sector area, (c) the perimeter of the sector. Give exact answers in terms of π and then to 3 s.f.

Fraction = 72/360 = 1/5
(a) arc = ⅕ × 2π × 10 = 4π cm ≈ 12.6 cm
(b) area = ⅕ × π × 10² = 20π cm² ≈ 62.8 cm²
(c) perimeter = arc + 2 radii = 4π + 20 = 32.6 cm (3 s.f.)

If the question says "give your answer in terms of π", leave it as 4π — converting to 12.566 loses the mark. If it doesn't, give 3 significant figures.

Skill check: A sector of a circle of radius 6 cm has area 12π cm². Find its angle.

Solution: (θ/360) × π × 36 = 12π → θ/360 = 12/36 = ⅓ → θ = 120°.

5.4Surface area and volume

Cuboid, prism, cylinder, sphere, pyramid and cone. Note "prism" means any solid with a uniform cross-section — so volume of a prism = cross-sectional area × length.

Solid	Volume	Surface area
Cuboid	lwh (not given)	2(lw + lh + wh) (not given)
Prism	Al (given)	2 × cross-section + perimeter × length
Cylinder	πr²h (given)	curved 2πrh (given); total = 2πrh + 2πr²
Cone	⅓πr²h (given)	curved πrl (given); total = πrl + πr²
Sphere	4/3 πr³ (given)	4πr² (given)
Pyramid	⅓Ah (given)	base + triangular faces

Worked example

A closed cylinder has radius 5 cm and height 12 cm. Find its volume and total surface area, in terms of π.

Volume = π × 5² × 12 = 300π cm³ (≈ 942 cm³)
Curved surface = 2π × 5 × 12 = 120π
Two circular ends = 2 × π × 5² = 50π
Total surface area = 170π cm² (≈ 534 cm²)

Worked example — cone with slant height

A cone has radius 6 cm and slant height 10 cm. Find its volume.

The volume formula needs the vertical height, so use Pythagoras: h = √(10² − 6²) = √64 = 8 cm
Volume = ⅓ × π × 36 × 8 = 96π cm³ ≈ 302 cm³

Using slant height l in the volume formula (it needs the perpendicular height h), or forgetting the flat circular faces when a question asks for total surface area of a cylinder or cone — the formula sheet only gives you the curved part.

Skill check: A sphere has radius 3 cm. Find its volume and surface area in terms of π.

Solution: V = 4/3 × π × 27 = 36π cm³. SA = 4 × π × 9 = 36π cm². (A numerical coincidence at r = 3 — the units differ.)

5.5Compound shapes and parts of shapes

Perimeters and areas of compound shapes and parts of shapes; surface areas and volumes of compound solids and parts of solids — including frustums (a cone with its top cut off).

Method — compound solidsSplit into standard solids and add volumes. For surface area, add only the surfaces that are actually exposed — the joining faces are hidden and must not be counted. For a frustum, work out the whole cone minus the small cone removed (similar shapes, 4.4, gives you the small cone's dimensions).

Worked example

A solid is a cylinder of radius 4 cm and height 10 cm with a hemisphere of radius 4 cm on top. Find its total volume and total surface area, in terms of π.

Cylinder volume = π × 16 × 10 = 160π
Hemisphere volume = ½ × 4/3 × π × 4³ = ½ × 256π/3 = 128π/3
Total volume = 160π + 128π/3 = 608π/3 cm³ ≈ 637 cm³
Surfaces exposed: curved cylinder 2π(4)(10) = 80π; base circle π(16) = 16π; curved hemisphere ½ × 4π(16) = 32π
Total surface area = 80π + 16π + 32π = 128π cm² ≈ 402 cm² (the flat top of the cylinder is not exposed)

Skill check: A shape is a semicircle of radius 7 cm sitting on the top edge of a rectangle 14 cm by 5 cm. Find the total area in terms of π.

Solution: Rectangle = 14 × 5 = 70 cm². Semicircle = ½ × π × 7² = 24.5π cm². Total = (70 + 24.5π) cm² ≈ 147 cm².

Topic 6 · 4 units

Trigonometry

Reliable, heavily-tested marks. The whole topic reduces to one decision: is the triangle right-angled (Pythagoras/SOHCAHTOA) or not (sine rule/cosine rule)? Get that right and the rest is arithmetic.

6.1Pythagoras' theorem

a² + b² = c² where c is the hypotenuse (opposite the right angle) — not given in the exam

Worked example

A ladder 15 m long leans against a wall with its foot 9 m from the base. How far up the wall does it reach?

The ladder is the hypotenuse: 9² + h² = 15²
81 + h² = 225 → h² = 144
h = 12 m

Adding when you should subtract. To find a shorter side, subtract: leg² = hyp² − other leg². Adding gives an answer bigger than the hypotenuse, which is impossible — always sanity-check.

6.2Right-angled triangles

SOHCAHTOA — not given in the exam

sin θ = Opposite/Hypotenuse · cos θ = Adjacent/Hypotenuse · tan θ = Opposite/Adjacent

Label the triangle from the angle you are using: H is always opposite the right angle; O is opposite your angle; A is the remaining side. To find an angle, use the inverse (sin⁻¹, cos⁻¹, tan⁻¹).

Also required: the perpendicular distance from a point to a line is the shortest distance, and calculations with angles of elevation and depression (both measured from the horizontal). Bearings may be combined with this. Angles are given and answered in degrees, to 1 decimal place.

Worked example — elevation

From a point 50 m from the foot of a tower, the angle of elevation of the top is 32°. Find the height of the tower.

Known: adjacent = 50, want: opposite = h → use tan
tan 32° = h/50
h = 50 × tan 32° = 50 × 0.62487 = 31.2 m (3 s.f.)

Worked example — finding an angle

A right-angled triangle has hypotenuse 13 cm and opposite side 5 cm. Find the angle.

sin θ = 5/13 = 0.3846…
θ = sin⁻¹(0.3846…) = 22.6° (1 d.p.)

The angle of depression from the top of a cliff down to a boat equals the angle of elevation from the boat up to the cliff top (alternate angles between horizontals). Drawing both horizontals makes almost every such question straightforward.

Skill check: A kite string 40 m long makes an angle of 55° with the horizontal ground. How high is the kite?

Solution: The string is the hypotenuse, height is opposite → sin 55° = h/40 → h = 40 × sin 55° = 40 × 0.81915 = 32.8 m (3 s.f.).

6.3Non-right-angled triangles

The sine rule, cosine rule and the area formula are given on the formula sheet in this syllabus — but knowing which to choose is the examined skill. Includes obtuse angles and the ambiguous case.

Method — choosing the rule

What you know	Use
Two sides and the angle between them (SAS) → find third side	Cosine rule
All three sides (SSS) → find an angle	Cosine rule (rearranged)
An angle and its opposite side, plus one more (ASA/AAS/SSA)	Sine rule
Two sides and the included angle → find area	Area = ½ab sin C

Worked example — cosine rule

In triangle ABC, b = 7 cm, c = 9 cm and angle A = 60°. Find a.

a² = b² + c² − 2bc cos A = 49 + 81 − 2(7)(9)(cos 60°)
= 130 − 126 × 0.5 = 130 − 63 = 67
a = √67 = 8.19 cm (3 s.f.)

Worked example — area

Find the area of that same triangle.

Area = ½ × 7 × 9 × sin 60° = 31.5 × 0.86603 = 27.3 cm² (3 s.f.)

Worked example — sine rule and the ambiguous case

In triangle ABC, a = 8, b = 11 and angle A = 40°. Find angle B.

sin B/11 = sin 40°/8 → sin B = 11 × sin 40° ÷ 8 = 0.88378
B = sin⁻¹(0.88378) = 62.1°
But sin is also positive for obtuse angles: B = 180° − 62.1° = 117.9° is a second valid answer here (the ambiguous case) — check which fits the diagram or the question's description.

Skill check: A triangle has sides 5 cm, 6 cm and 9 cm. Find its largest angle.

Solution: The largest angle is opposite the longest side (9). Cosine rule rearranged: cos θ = (5² + 6² − 9²)/(2 × 5 × 6) = (25 + 36 − 81)/60 = −20/60 = −0.3333. θ = cos⁻¹(−0.3333) = 109.5° (1 d.p.). The negative cosine correctly signals an obtuse angle.

6.4Pythagoras and trigonometry in 3D

Solve problems in three dimensions, including the angle between a line and a plane.

Method — always find a right-angled triangle1. Sketch the solid and mark the line you want. 2. Drop a perpendicular from the top of the line to the plane. 3. Join the foot of that perpendicular to the line's start point — this is the projection of the line on the plane. 4. You now have a right-angled triangle; the required angle sits between the line and its projection.

Worked example

A cuboid measures 8 cm by 6 cm by 5 cm. Find (a) the length of the space diagonal, (b) the angle this diagonal makes with the base.

(a) Base diagonal first: √(8² + 6²) = √100 = 10 cm
Space diagonal = √(10² + 5²) = √125 = 11.2 cm (3 s.f.)
(b) The projection on the base is that 10 cm diagonal; the vertical rise is 5 cm
tan θ = 5/10 = 0.5 → θ = 26.6° (1 d.p.)

In 3D questions, always draw the 2D triangle you are actually using, separately and to a sensible size, labelling all three parts. Working straight from the 3D picture is where sides get mixed up.

Skill check: A square-based pyramid has base edge 10 cm and vertical height 12 cm. Find the length of an edge from the apex to a base corner.

Solution: Half the base diagonal = ½ × √(10² + 10²) = ½ × √200 = 7.071 cm. Edge = √(12² + 7.071²) = √(144 + 50) = √194 = 13.9 cm (3 s.f.).

Topic 7 · 4 units

Transformations and vectors

Transformation questions are marked strictly: to "describe fully" you must give every required detail. Vectors then formalise translation into arithmetic you can do algebraically.

7.1Transformations

Method — "describe fully" checklist

Transformation	You MUST state
Reflection	the word "reflection" + the equation of the mirror line (e.g. y = x)
Rotation	"rotation" + centre, angle (multiples of 90°) and direction (clockwise/anticlockwise)
Enlargement	"enlargement" + centre and scale factor (may be fractional or negative)
Translation	"translation" + the column vector (x above y)

Naming two transformations (e.g. "reflection and translation") scores zero — one transformation only.

Scale factors: a factor between 0 and 1 makes the image smaller; a negative factor puts the image on the opposite side of the centre and turns it upside down.

Worked example

Triangle A has vertices (1,1), (3,1), (1,4). It is enlarged by scale factor −2 about the origin. Find the image vertices.

For a centre at the origin, multiply each coordinate by the scale factor
(1,1) → (−2,−2); (3,1) → (−6,−2); (1,4) → (−2,−8)
The image is twice as large, on the opposite side of the origin and rotated 180° in appearance.

Skill check: A shape is reflected so that the point (2, 5) maps to (5, 2). What is the mirror line?

Solution: The coordinates have swapped, which is the signature of reflection in y = x. (Swapping and negating both would be y = −x.)

7.2Vectors in two dimensions 7.3Magnitude

Vectors are written as a column, as AB (with an arrow above), or as a bold letter a. Add and subtract vectors, and multiply by a scalar.

Method

Add/subtract component-wise: (3, −4) + (1, 6) = (4, 2)

Scalar multiple: 3(2, −1) = (6, −3) — same direction, three times as long

Magnitude: |(x, y)| = √(x² + y²) (not given)

Worked example

a = (3, −4) and b = (−1, 2). Find (a) 2a + 3b, (b) |a|.

(a) 2a = (6, −8); 3b = (−3, 6); sum = (3, −2)
(b) |a| = √(3² + (−4)²) = √25 = 5

Writing a magnitude as a vector or with a negative value. Magnitude is a length — always a single positive number.

7.4Vector geometry

Represent vectors by directed line segments, use position vectors, and express given vectors in terms of two known vectors — the classic "prove these points are collinear" question.

Method — routes and position vectors

Any journey can be broken into steps: AB = AO + OB = −OA + OB = b − a (position vectors from origin O).
Reversing a vector reverses its sign: BA = −AB.
If XY = k × XZ for a scalar k, the points are collinear (they lie on the same straight line, since the vectors are parallel and share point X).

Worked example

OACB is a parallelogram with OA = a and OB = b. M is the midpoint of AC. Express OM in terms of a and b.

In the parallelogram, AC is parallel and equal to OB, so AC = b
M is the midpoint of AC, so AM = ½b
OM = OA + AM = a + ½b

Skill check: OP = p, OQ = q. R lies on PQ with PR : RQ = 1 : 2. Express OR in terms of p and q.

Solution: PQ = q − p. R is ⅓ of the way along, so PR = ⅓(q − p). Then OR = OP + PR = p + ⅓(q − p) = ⅔p + ⅓q.

Topic 8 · 3 units

Probability

Small topic, generous marks. Nearly every harder question is a tree diagram — and the single most important question to ask is whether the item is replaced.

8.1Introduction to probability

Probability runs from 0 (impossible) to 1 (certain), and may be given as a fraction, decimal or percentage — but never as a ratio or "1 in 4". Notation: P(A) is the probability of A; P(A′) is the probability of not A.

P(A′) = 1 − P(A)

Probabilities of a single event may need to be read from tables, graphs or Venn diagrams.

Worked example

P(B) = 0.8. Find P(B′).

P(B′) = 1 − 0.8 = 0.2

Skill check: A bag has 4 red, 5 green and 6 yellow balls. One is taken at random. Find P(not green).

Solution: Total = 15. P(green) = 5/15 = ⅓, so P(not green) = 1 − ⅓ = ⅔ (or count directly: 10/15 = ⅔).

8.2Relative and expected frequencies

Method

relative frequency = number of successes ÷ number of trials (an estimate of probability)

expected frequency = probability × number of trials

Know the terms fair (all outcomes equally likely), bias (they are not) and random. The more trials, the closer relative frequency gets to the true probability.

Worked example

A spinner is spun 200 times and lands on red 70 times. (a) Estimate P(red). (b) If it is spun 500 more times, how many reds are expected?

(a) Relative frequency = 70/200 = 0.35
(b) Expected = 0.35 × 500 = 175

Skill check: A fair six-sided die is rolled 300 times. How many times would you expect a number greater than 4?

Solution: Outcomes greater than 4 are 5 and 6, so P = 2/6 = ⅓. Expected = ⅓ × 300 = 100.

8.3Probability of combined events

Use sample space diagrams, Venn diagrams (notation P(A ∩ B) and P(A ∪ B)) and tree diagrams. Combined events may be with or without replacement. On tree diagrams, write outcomes at the ends of branches and probabilities beside them.

Method — the two rules

AND → multiply (along the branches of a tree).
OR → add (across different complete branches).
Without replacement: on the second pick, both the numerator and the total go down by one. With replacement, nothing changes.
"At least one" is nearly always fastest as 1 − P(none).

Worked example — without replacement

A bag contains 5 red and 3 blue counters. Two are taken at random without replacement. Find (a) P(both red), (b) P(one of each colour), (c) P(at least one red).

(a) P(RR) = 5/8 × 4/7 = 20/56 = 5/14
(b) Two ways: red then blue, or blue then red
= (5/8 × 3/7) + (3/8 × 5/7) = 15/56 + 15/56 = 30/56 = 15/28
(c) P(at least one red) = 1 − P(no reds) = 1 − (3/8 × 2/7) = 1 − 6/56 = 50/56 = 25/28

Forgetting to reduce the denominator on the second pick in a "without replacement" question — the total must drop from 8 to 7. Read the question twice for the words "replaced" or "not replaced".

Skill check: A box has 4 faulty and 16 working bulbs. Two are chosen at random without replacement. Find the probability that exactly one is faulty.

Solution: Faulty-then-working: 4/20 × 16/19 = 64/380. Working-then-faulty: 16/20 × 4/19 = 64/380. Total = 128/380 = 32/95 ≈ 0.337.

Topic 9 · 7 units

Statistics

Mostly method-following, with two topics that reliably separate grades: cumulative frequency (reading quartiles correctly) and histograms (frequency density, not frequency, on the vertical axis).

9.1Classifying data 9.2Interpreting data

9.1 Classify and tabulate statistical data — tally tables and two-way tables.

9.2 Read, interpret and draw inferences from tables and diagrams; compare two data sets using averages and measures of spread; and appreciate the restrictions on drawing conclusions from data.

When asked to compare two data sets you must comment on both an average and a measure of spread, in context. "Class A's median score (62) is higher than Class B's (55), so Class A performed better on average, but Class A's range (40) is larger, so their results were less consistent" earns full marks; quoting numbers alone does not.

9.3Averages and measures of spread

Measure	How	Best used when
Mean	total ÷ number of values	Data is fairly symmetric; uses every value
Median	middle value when ordered	There are extreme values (outliers) — it is not distorted by them
Mode	most common value	Data is categorical (e.g. favourite colour, shoe size sold)
Range	largest − smallest	Measuring spread/consistency (but sensitive to outliers)

Method — estimated mean from grouped dataUse the midpoint of each class as x, multiply by frequency f, then:

estimated mean = Σfx ÷ Σf

It is an estimate because the original values within each class are unknown. The modal class is simply the class with the highest frequency.

Worked example

Estimate the mean of this grouped data and state the modal class.

Time t (min)	0 < t ⩽ 10	10 < t ⩽ 20	20 < t ⩽ 30	30 < t ⩽ 40
Frequency	4	11	15	10

Midpoints: 5, 15, 25, 35
fx: 4×5 = 20; 11×15 = 165; 15×25 = 375; 10×35 = 350
Σf = 40, Σfx = 910
Estimated mean = 910 ÷ 40 = 22.75 minutes
Modal class = highest frequency (15) → 20 < t ⩽ 30

Dividing by the number of classes (4) instead of the total frequency (40) — always divide by Σf.

Skill check: Find the median and range of 7, 3, 9, 4, 12, 3, 8.

Solution: Order first: 3, 3, 4, 7, 8, 9, 12. There are 7 values, so the median is the 4th = 7. Range = 12 − 3 = 9. (The mode is 3.)

9.4Statistical charts and diagrams

Draw and interpret bar charts (including composite/stacked and dual/side-by-side), pie charts, pictograms and simple frequency distributions.

Method — pie charts

angle for a category = (frequency ÷ total frequency) × 360°

To work backwards: frequency = (angle ÷ 360°) × total.

Worked example

In a survey of 90 students, 25 chose cricket. Find the angle for cricket on a pie chart.

(25 ÷ 90) × 360° = 25 × 4 = 100°

Skill check: On a pie chart of 72 people, one sector has an angle of 65°. How many people does it represent?

Solution: (65 ÷ 360) × 72 = 65 ÷ 5 = 13 people.

9.5Scatter diagrams

Draw and interpret scatter diagrams (plot points as small crosses ×), understand positive, negative and zero correlation, and draw and use a line of best fit.

Method — line of best fit rulesA single ruled line drawn by inspection, extending across the full data set, with roughly equal numbers of points either side along its whole length. It need not pass through any actual point.

Describing correlation properly: state the type and its meaning in context — "strong positive correlation: as revision time increases, test score tends to increase."

Beware extrapolation. Using the line of best fit far outside the plotted data is unreliable, and questions often award a mark for saying exactly that. Also remember correlation does not prove that one variable causes the other.

9.6Cumulative frequency diagrams

Method

Cumulative frequency = running total of frequencies.
Plot each cumulative total against the upper class boundary (not the midpoint), mark points with crosses, and join with a smooth curve.
Read off with n = total frequency:

median at n/2 · lower quartile at n/4 · upper quartile at 3n/4 · IQR = UQ − LQ

Worked example

120 students' marks are shown on a cumulative frequency curve. Explain how to find the median and interquartile range.

n = 120, so median is read at cumulative frequency 60 — go across from 60 to the curve, then down to the mark axis.
Lower quartile at 120 ÷ 4 = 30; upper quartile at 3 × 120 ÷ 4 = 90.
IQR = UQ − LQ. If UQ = 68 and LQ = 44, then IQR = 24 marks.
To find "how many scored more than 70", read the cumulative frequency at 70 (say 96) and subtract from the total: 120 − 96 = 24 students.

Using n/2 on a cumulative frequency curve is correct — but plotting against class midpoints instead of upper boundaries is not. Cumulative frequency always means "this value or less", so it belongs at the top of the class.

Skill check: A cumulative frequency curve covers 200 people. At what cumulative frequencies do you read the lower quartile, median and upper quartile?

Solution: LQ at 200 ÷ 4 = 50; median at 200 ÷ 2 = 100; UQ at 3 × 200 ÷ 4 = 150.

9.7Histograms

Histograms have unequal class widths, so the vertical axis is labelled Frequency density, never frequency. The area of each bar represents the frequency.

frequency density = frequency ÷ class width ⟺ frequency = frequency density × class width

Worked example

Complete the frequency densities for this data.

Class	Frequency	Class width	Frequency density
0 < x ⩽ 10	15	10	15 ÷ 10 = 1.5
10 < x ⩽ 20	24	10	24 ÷ 10 = 2.4
20 < x ⩽ 50	36	30	36 ÷ 30 = 1.2
50 < x ⩽ 90	20	40	20 ÷ 40 = 0.5

Note the third class has the highest frequency but not the highest bar — that is exactly what frequency density corrects for.

Worked example — reading backwards

A histogram bar covers 30 < x ⩽ 45 with a frequency density of 2.4. Find the frequency.

Class width = 45 − 30 = 15
Frequency = 2.4 × 15 = 36

Reading the bar height as the frequency. On a histogram the height is density — you must multiply by the class width to recover the frequency.

Skill check: A class 60 < x ⩽ 100 contains 28 values. Find the frequency density.

Solution: Class width = 40, so frequency density = 28 ÷ 40 = 0.7.

Reference

Notation, accuracy and command words

Accuracy rules that earn or lose marks

Paper 2: give non-exact answers to 3 significant figures — except angles in degrees, which go to 1 decimal place — unless the question specifies otherwise.
Never round part-way through. Carry the full calculator value forward; if a later part uses your answer, use the unrounded version.
Use π from the calculator (or 3.142). If asked for an answer "in terms of π", leave the π in.
Paper 1: answers are usually meant to be exact — leave fractions, surds and π rather than attempting decimals.
Money answers normally need 2 decimal places ($4.80, not $4.8); do not round money to 3 significant figures.

Command words

Word	What it demands
Work out	Calculate — show your working
Calculate	Obtain a numerical answer, showing the relevant working
Show that	Prove the given result — you must reach the stated answer with visible steps, and cannot simply quote it
Give an exact answer	Leave as a fraction, surd or in terms of π — no decimals
Explain / Give a reason	State the mathematical property by name (especially in geometry)
Describe fully	Give every required detail (see the transformation checklist in 7.1)
Estimate	Round to 1 significant figure first, and show those rounded values
Sketch	A neat freehand shape showing key features (intercepts, turning points, asymptotes) — no plotting required
Construct	Use ruler and compasses, and leave the construction arcs visible

"Show that" questions are free marks if you work towards the answer step by step — and near-zero marks if you simply restate it. Never write the given answer and stop.

Reference

Free past papers & how to revise this subject

Official (free)

Cambridge International — 4024 subject page: syllabus, specimen papers, past papers, mark schemes and examiner reports.
Examiner reports name the exact questions candidates got wrong each series — read the one for every paper you sit.

Free archives

GCE Guide — past papers by year and variant.
PastPapers.co — another full CAIE archive.
Physics & Maths Tutor — topic-sorted question sets.

The method that actually works for maths

Learn: read one unit and re-do its worked example with the solution covered.
Drill: do 5–10 questions of that same type from topic-sorted past papers, until the method is automatic.
Correct: mark honestly with the mark scheme. Keep an error log with the cause of each mistake (sign error? wrong formula? misread?) — patterns will appear fast.
Mix: once several topics are solid, do mixed exercises. Choosing the right method is a separate skill from executing it, and it is what Paper 2 really tests.
Time: in the last six weeks, full papers under exam conditions — Paper 1 with the calculator physically out of reach.