Free notes · Secondary 3 and 4
O-Level E-Math notes
Free O-Level E-Math notes on one page, for the 2026 O-Level (4052) and the 2027 SEC G3 Mathematics (K310). Full notes for 12 of the 18 syllabus chapters, each with the key idea, the facts and formulas to know, a worked exam question, the common mistakes and exam tips. The other chapters have a checklist of what the syllabus asks for, so the page maps the whole syllabus.
Free to read, print or save as a PDF, with no sign-up. The print layout keeps every chapter and drops the menus. After each chapter, practise on the O-Level E-Math prelim papers with answers.
- 18 chapters
- Free to print
How to use these notes
- Pick one chapter. Read the key idea and the must-know points.
- Close the page and write down what you remember, then check and mark what you missed.
- Try the worked exam question before reading its steps, and read the common mistakes.
- Do questions on that chapter from a school paper and mark them. Redo the wrong ones a few days later.
Syllabus edition
These notes follow SEAB's O-Level Mathematics syllabus 4052, examined in 2026, and SEC G3 Mathematics K310, examined from 2027. The two documents list the same 18 topics in three strands and the same two papers. Sources: SEAB, O-Level Mathematics syllabus 4052 (2026), SEAB, SEC G3 Mathematics syllabus K310 (2027).
Exam format
| Paper | What it is | Time | Marks | Weighting |
|---|---|---|---|---|
| Paper 1 | About 26 short-answer questions. Answer all. | 2 h 15 min | 90 | 50% |
| Paper 2 | 9 to 10 questions of different lengths. The last applies maths to a real-world scenario. Answer all. | 2 h 15 min | 90 | 50% |
- Relevant formulae are printed. An approved calculator may be used in both papers. Bring geometrical instruments.
- Missing essential working loses marks. Give non-exact answers to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.
The whole syllabus, chapter by chapter
All 18 chapters of the syllabus in order. 12 have full notes on this page. The rest have a checklist of what the syllabus asks you to do.
| Chapter | What it covers | On this page |
|---|---|---|
| 1. Numbers and their Operations | Primes, HCF and LCM, types of numbers, rounding, standard form and indices. | Full notes, plus checklist |
| 2. Ratio and Proportion | Ratios, map scales, and direct and inverse proportion. | Checklist |
| 3. Percentage | Percentages of quantities, increases, decreases and reverse percentages. | Checklist |
| 4. Rate and Speed | Average rate, average speed and unit conversion. | Checklist |
| 5. Algebraic Expressions and Formulae | Expanding, factorising, algebraic fractions and changing the subject. | Full notes |
| 6. Functions and Graphs | Linear, quadratic, power and exponential graphs, and gradients of curves. | Full notes |
| 7. Equations and Inequalities | Linear, fractional, simultaneous and quadratic equations, and linear inequalities. | Quadratic Equations, Simultaneous Equations, plus checklist |
| 8. Set Language and Notation | Set notation and Venn diagrams. | Checklist |
| 9. Matrices | Displaying data in matrices and matrix arithmetic. | Checklist |
| 10. Angles, Triangles and Polygons | Angle facts, parallel lines, polygons and constructions. | Checklist |
| 11. Congruence and Similarity | Congruent and similar figures, scale drawings, and area and volume ratios. | Full notes |
| 12. Properties of Circles | Symmetry and angle properties of circles. | Full notes |
| 13. Pythagoras' Theorem and Trigonometry | Right-angled triangles, sine and cosine rules, bearings, and 3D problems. | Full notes |
| 14. Mensuration | Areas, volumes, surface areas, arcs, sectors and radians. | Full notes |
| 15. Coordinate Geometry | Gradient, length and the equation of a straight line. | Full notes |
| 16. Vectors in Two Dimensions | Vector notation, magnitude, position vectors and vector geometry. | Full notes |
| 17. Data Handling and Analysis | Statistical diagrams, averages, quartiles and standard deviation. | Full notes |
| 18. Probability | Single and combined events, possibility and tree diagrams. | Full notes |
O-Level E-Math at a glance
| Chapter | The one thing to remember |
|---|---|
| Indices and Standard Form | aᵐ × aⁿ = a^(m + n), aᵐ ÷ aⁿ = a^(m − n), (aᵐ)ⁿ = a^(mn) |
| Algebraic Expressions and Formulae | (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b². |
| Functions and Graphs | y = mx + c: gradient m, y-intercept c. |
| Quadratic Equations | Factorise: if (x − p)(x − q) = 0, then x = p or x = q. |
| Simultaneous Equations | Elimination: same signs subtract, different signs add. |
| Congruence and Similarity | Congruence tests: SSS, SAS, ASA or AAS, and RHS. AAA and SSA are not tests. |
| Angle Properties of Circles | Angle at centre = 2 × angle at circumference, standing on the same arc. |
| Trigonometry | SOH CAH TOA: sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj. |
| Mensuration | π radians = 180°, so multiply degrees by π/180 to get radians. |
| Coordinate Geometry | Gradient m = (y₂ − y₁)/(x₂ − x₁), the change in y over the change in x. |
| Vectors | Magnitude of the column vector (x, y) = √(x² + y²) |
| Statistics and Probability | Estimated mean of grouped data = Σfx ÷ Σf, using class mid-values as x. |
Notes 1 of 12
Indices and Standard Form
Key idea. Indices show repeated multiplication, so aⁿ means n factors of a. With the same base, add indices when multiplying, subtract them when dividing and multiply them for a power of a power. A zero index gives 1, a negative index gives a reciprocal and a fractional index gives a root. Standard form writes a number as A × 10ⁿ, with 1 ≤ A < 10 and n an integer.
Must know
- aᵐ × aⁿ = a^(m + n), aᵐ ÷ aⁿ = a^(m − n), (aᵐ)ⁿ = a^(mn)
- a⁰ = 1, a⁻ⁿ = 1/aⁿ, and a^(m/n) is the nth root of a, raised to the power m.
- Standard form: A × 10ⁿ with 1 ≤ A < 10, so 0.000 304 = 3.04 × 10⁻⁴.
What the syllabus asks you to do
- Use the laws of indices
- Use positive, negative, zero and fractional indices
- Use standard form A × 10ⁿ, where n is an integer and 1 ≤ A < 10
- Round answers to a required number of decimal places or significant figures
The laws of indices, for a non-zero base a
| Law | Rule | Example |
|---|---|---|
| Multiply, same base | aᵐ × aⁿ = a^(m + n) | 2³ × 2⁴ = 2⁷ = 128 |
| Divide, same base | aᵐ ÷ aⁿ = a^(m − n) | 12x⁷ ÷ 4x² = 3x⁵ |
| Power of a power | (aᵐ)ⁿ = a^(mn) | (x³)² = x⁶ |
| Power of a product | (ab)ⁿ = aⁿbⁿ | (2x³)⁴ = 16x¹² |
| Zero index | a⁰ = 1 | 7⁰ = 1 |
| Negative index | a⁻ⁿ = 1/aⁿ | 5⁻² = 1/25 |
| Fractional index | a^(m/n) = (nth root of a)ᵐ | 27^(2/3) = 3² = 9 |
Worked exam question
Simplify (2x³y⁻²)³ ÷ (4x⁵y⁻⁴), giving your answer with positive indices.
- Raise each factor in the first bracket to the power 3: 2³ = 8, (x³)³ = x⁹ and (y⁻²)³ = y⁻⁶, so the first part is 8x⁹y⁻⁶.
- Divide the numbers: 8 ÷ 4 = 2.
- Subtract the indices of x: x⁹ ÷ x⁵ = x⁴.
- Subtract the indices of y: y⁻⁶ ÷ y⁻⁴ = y^(−6 − (−4)) = y⁻².
- Combine to get 2x⁴y⁻², then use y⁻² = 1/y² to remove the negative index.
Answer: 2x⁴/y²
Common mistakes
- Writing 2³ × 2⁴ = 4⁷.Keep the base and add only the indices: 2³ × 2⁴ = 2⁷ = 128. Check with 8 × 16 = 128.
- Writing (x³)² = x⁵.Write it out once, (x³)² = x³ × x³ = x⁶, and remember that a power outside a bracket multiplies the indices.
- Writing 7⁰ = 0 or (3x)⁰ = 3.Remember a⁰ = 1 for any non-zero a, because aⁿ ÷ aⁿ = 1. So 7⁰ = 1 and (3x)⁰ = 1.
- Writing 2x⁻³ = 1/(2x³).Look for a bracket. Without one, only x is affected: 2x⁻³ = 2/x³. With one, (2x)⁻³ = 1/(8x³).
- Writing 16^(−1/2) = −4.Treat the minus as 'one over' and the fraction as a root: 16^(−1/2) = 1/√16 = 1/4.
- Leaving an answer as 28.8 × 10² or 0.8 × 10⁸.Check that the front number is at least 1 and less than 10, then adjust: 2.88 × 10³ and 8 × 10⁷.
Exam tips
- Write every term as a power of a prime base before simplifying expressions such as 4ⁿ × 8 ÷ 2ⁿ. Once everything is a power of 2, the laws apply directly.
- For fractional indices without a calculator, take the root before the power. 64^(2/3) is (∛64)² = 4² = 16, which is far easier than finding 64² first.
- When a question says give your answer in standard form, check that the number in front is at least 1 and less than 10, and that you have written × 10ⁿ rather than copying a calculator display.
Notes 2 of 12
Algebraic Expressions and Formulae
Key idea. Algebraic manipulation means rewriting expressions without changing their value. Expand by multiplying every term in one bracket by every term in the other. Factorise by taking out the highest common factor first, then using grouping, a quadratic pattern or a difference of two squares. Simplify algebraic fractions by factorising and cancelling factors, and change the subject by undoing operations in reverse order.
Must know
- (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².
- a² − b² = (a − b)(a + b), the difference of two squares.
- In algebraic fractions, cancel only common factors, never single terms.
- If the new subject appears twice, collect those terms on one side and factorise.
What the syllabus asks you to do
- Expand the product of algebraic expressions
- Use (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b)
- Extract common factors and factorise linear expressions of the form ax + bx + kay + kby
- Factorise quadratic expressions ax² + bx + c
- Multiply and divide simple algebraic fractions, and add and subtract algebraic fractions with linear or quadratic denominators
- Change the subject of a formula
- Evaluate algebraic expressions and formulae, and find the value of an unknown quantity in a given formula
Factorising methods, with examples not used elsewhere on this page
| Type | Example | Factorised form |
|---|---|---|
| Common factor | 6x²y − 9xy² | 3xy(2x − 3y) |
| Grouping | x³ + 2x² + 3x + 6 | (x² + 3)(x + 2) |
| Quadratic with a = 1 | x² + x − 12 | (x + 4)(x − 3) |
| Quadratic with a ≠ 1 | 3x² − 10x − 8 | (3x + 2)(x − 4) |
| Difference of two squares | 25 − 4y² | (5 − 2y)(5 + 2y) |
| Perfect square | 9x² + 30x + 25 | (3x + 5)² |
Clear the fraction
Multiply both sides by (x − 1) to get y(x − 1) = 3x + 2.
Expand the bracket
xy − y = 3x + 2
Collect the x terms
xy − 3x = y + 2
Factorise out x
x(y − 3) = y + 2
Divide to finish
x = (y + 2)/(y − 3)
Worked exam question
Make x the subject of the formula y = (3x + 2)/(x − 1). Hence find x when y = 5.
- Multiply both sides by (x − 1) to clear the fraction: y(x − 1) = 3x + 2.
- Expand the bracket: xy − y = 3x + 2.
- Collect the x terms on one side and the other terms on the other side: xy − 3x = y + 2.
- Factorise out x: x(y − 3) = y + 2.
- Divide by (y − 3): x = (y + 2)/(y − 3).
- Substitute y = 5: x = 7/2 = 3.5.
- Check in the original formula: (3(3.5) + 2)/(3.5 − 1) = 12.5/2.5 = 5, which is correct.
Answer: x = (y + 2)/(y − 3), and x = 3.5 when y = 5
Common mistakes
- Expanding (x + 5)² as x² + 25.Write the bracket twice, (x + 5)(x + 5), and multiply out to find the middle term: x² + 10x + 25.
- Cancelling the x in (x + 6)/(x + 3), or turning 6/3 into 2.Factorise first, then cancel only a bracket or number that multiplies the entire numerator and the entire denominator.
- Writing 3x + 3 − 2x − 4 for 3(x + 1) − 2(x − 2).Keep the second numerator in a bracket and expand carefully: −2(x − 2) = −2x + 4.
- Stopping at 2(x² − 25) when asked to factorise 2x² − 50 completely.After each step, check every bracket for a difference of two squares or a quadratic: 2(x − 5)(x + 5).
- Ending with x = (y + 2 + 3x)/y when changing the subject.Collect every x term on one side, factorise out x, then divide, as in x(y − 3) = y + 2.
- Square-rooting x² = a² + b² to get x = a + b.Root the whole side: x = √(a² + b²). Test with a = 3 and b = 4: √25 = 5, not 7.
Exam tips
- After factorising, expand your answer in your head to check that it gives back the original expression. This takes seconds and catches sign errors.
- When asked to simplify an algebraic fraction, your first move should almost always be to factorise top and bottom. A factor that appears in both is a strong hint that you are on the right track.
- In change-the-subject questions, write one operation per line. Short lines make it easier for you to spot where a sign went wrong and for a marker to follow your method.
Notes 3 of 12
Functions and Graphs
Key idea. Functions and graphs is about drawing and reading graphs of linear, quadratic, cubic, reciprocal and exponential functions. Plot points from a table of values and join them with a smooth curve. To solve a related equation, rearrange it so one side matches the drawn curve and draw the other side as a straight line. Estimate the gradient at a point by drawing a tangent.
Must know
- y = mx + c: gradient m, y-intercept c.
- y = (x − h)² + k has turning point (h, k), which is a minimum.
- Solutions are the x-coordinates where the curve meets the line you draw.
- Gradient of a curve at a point = gradient of the tangent at that point.
What the syllabus asks you to do
- Draw the graph of a set of ordered pairs as a representation of a relationship between two variables
- Describe the properties of quadratic graphs: the sign of the coefficient of x², maximum and minimum points, and symmetry
- Draw graphs of power functions y = axⁿ, where n = −2, −1, 0, 1, 2 or 3, and simple sums of not more than three of these
- Draw graphs of exponential functions y = kaˣ, where a is a positive integer
- Solve quadratic equations by the graphical method
- Estimate the gradient of a curve by drawing a tangent
Graph shapes to recognise and sketch
| Equation | Shape | Key features |
|---|---|---|
| y = mx + c | Straight line | Gradient m, crosses the y-axis at c |
| y = ax² + bx + c | Parabola | U-shaped if a > 0, upside-down U if a < 0, symmetric about the turning point |
| y = ax³ | Cubic curve | Passes through the origin, rises from left to right if a > 0 |
| y = a/x | Two separate branches | Never touches either axis, branches in the first and third quadrants if a > 0 |
| y = a/x² | Two separate branches | Never touches the axes, both branches above the x-axis if a > 0 |
| y = kaˣ with k > 0, a > 1 | Exponential growth curve | Passes through (0, k), approaches the x-axis on the left |
A U-shaped curve for x from 0 to 5, starting at (0, 1), falling to its minimum point (2, −3) and rising to (5, 6). A straight line y = x − 1 runs from (0, −1) to (5, 4). The line cuts the curve twice, at x ≈ 0.4 and x ≈ 4.6, which are the solutions of x² − 5x + 2 = 0.
Worked exam question
(a) Calculate the values of y = x² − 4x + 1 for x = 0, 1, 2, 3, 4 and 5, and draw the graph. (b) Write down the minimum point and the equation of the line of symmetry. (c) By drawing a suitable straight line, solve x² − 5x + 2 = 0.
- Substitute each x value. For example, x = 1 gives 1 − 4 + 1 = −2. The y values for x = 0 to 5 are 1, −2, −3, −2, 1 and 6.
- Plot the six points and join them with a smooth U-shaped curve.
- The lowest point is (2, −3), and the curve is symmetrical about the vertical line through it, x = 2.
- For part (c), subtract the equations: (x² − 4x + 1) − (x² − 5x + 2) = x − 1.
- So x² − 5x + 2 = 0 is the same as x² − 4x + 1 = x − 1. Draw the line y = x − 1, for example through (0, −1) and (5, 4).
- Read the x-coordinates of the two intersection points: x ≈ 0.4 and x ≈ 4.6.
- Check with the formula: x = (5 ± √17)/2 gives 0.44 and 4.56, which agrees with the graph.
Answer: (b) Minimum point (2, −3), line of symmetry x = 2. (c) Draw y = x − 1, giving x ≈ 0.4 or x ≈ 4.6.
Common mistakes
- Joining the plotted points of a quadratic or reciprocal graph with a ruler.Sketch the shape lightly first, then draw one smooth freehand curve through every point. Use a ruler only for straight lines and tangents.
- Typing −2² for x = −2 and getting −4 instead of 4.Put negative values in brackets every time: (−2)² = 4.
- Drawing y = −3 instead of y = 3 to solve x² − 3x − 3 = 0 from the curve y = x² − 3x.Rearrange so one side is exactly the curve's expression, then expand back to check that you get the original equation.
- Joining the two branches of y = 12/x across the y-axis.Remember y = a/x is not defined at x = 0. Draw each branch separately, approaching the axes without touching them.
- Counting squares for a gradient when 1 square is 0.5 on the x-axis and 2 on the y-axis.Read actual coordinates from the scales, then calculate change in y ÷ change in x.
- Drawing a tangent that cuts through the curve near the point.Adjust the ruler until the line touches the curve at the point and the curve stays on one side of it nearby.
Exam tips
- Plot points with small, neat crosses. If one point sits clearly off the smooth curve, recheck the calculation for that point before you draw.
- When reading solutions from a graph, draw faint vertical lines from the intersection points down to the x-axis, and give values to the accuracy of the scale.
- For a tangent, make the line as long as the grid allows and pick two points on it with easy coordinates. A larger triangle gives a more accurate gradient.
Notes 4 of 12
Quadratic Equations
Key idea. A quadratic equation has the form ax² + bx + c = 0 with a ≠ 0. Rearrange so one side is zero, then factorise if you can. If it does not factorise, complete the square or use the formula x = (−b ± √(b² − 4ac)) / (2a). The solutions are the x-intercepts of the graph y = ax² + bx + c, and there are at most two of them.
Must know
- Factorise: if (x − p)(x − q) = 0, then x = p or x = q.
- Formula: x = (−b ± √(b² − 4ac)) / (2a), substituting a, b and c with their signs.
- Completed square: x² + bx + c = (x + b/2)² − (b/2)² + c.
- y = (x − h)² + k has minimum point (h, k) and line of symmetry x = h.
What the syllabus asks you to do
- Solve quadratic equations in one unknown by factorisation
- Solve quadratic equations using the formula
- Solve quadratic equations by completing the square for y = x² + px + q
- Solve quadratic equations by the graphical method
- Solve fractional equations that can be reduced to quadratic equations
- Formulate quadratic equations to solve problems
- Sketch quadratic graphs given in the form y = (x − p)² + q or y = −(x − p)² + q, showing the maximum or minimum point and symmetry
Choosing a method for a quadratic equation
| Method | Use it when | Example |
|---|---|---|
| Factorisation | The expression splits into brackets with whole numbers | x² − 7x + 12 = 0 gives (x − 3)(x − 4) = 0 |
| Difference of two squares | There is no x term and both terms are squares | 4x² − 9 = 0 gives (2x − 3)(2x + 3) = 0 |
| Common factor | There is no constant term | 3x² − 12x = 0 gives 3x(x − 4) = 0 |
| Completing the square | The question asks for the form (x + p)² + q or a turning point | x² + 6x + 2 = (x + 3)² − 7 |
| Quadratic formula | Answers are needed to 2 decimal places or 3 significant figures | 2x² − 5x − 10 = 0 gives x = (5 ± √105)/4 |
A U-shaped parabola drawn for x from 0 to 8. It starts at (0, 5), falls to its lowest point (4, −11) and rises symmetrically back to (8, 5). It crosses the x-axis twice, near x = 0.68 and x = 7.32, which are the solutions of x² − 8x + 5 = 0. The line of symmetry is x = 4.
Worked exam question
Solve the equation (2x + 1)(x − 3) = 7, giving your answers correct to 2 decimal places.
- Expand the left-hand side: 2x² − 6x + x − 3 = 2x² − 5x − 3.
- Make one side zero: 2x² − 5x − 3 − 7 = 0, so 2x² − 5x − 10 = 0.
- Identify a = 2, b = −5 and c = −10.
- Work out b² − 4ac = (−5)² − 4(2)(−10) = 25 + 80 = 105.
- Substitute into the formula: x = (5 ± √105) / 4, where √105 = 10.247 to 3 decimal places.
- So x = (5 + 10.247) / 4 = 3.8117 or x = (5 − 10.247) / 4 = −1.3117.
- Check by substitution: 2(3.8117)² − 5(3.8117) − 10 is approximately 0, as required.
Answer: x = 3.81 or x = −1.31, correct to 2 decimal places
Common mistakes
- Writing x − 2 = 4 or x + 1 = 4 from (x − 2)(x + 1) = 4.Expand and collect terms first: x² − x − 6 = 0, so (x − 3)(x + 2) = 0, giving x = 3 or x = −2.
- Dividing x² = 5x by x and answering only x = 5.Rearrange to x² − 5x = 0 and factorise to x(x − 5) = 0, giving x = 0 or x = 5.
- Substituting b = −5 into the formula as −b = −5 or b² = −25.Write a, b and c with their signs first, then substitute in brackets: −(−5) = 5 and (−5)² = 25.
- Writing only x + 3 = √7 from (x + 3)² = 7.Write ± every time you square-root both sides: x + 3 = ±√7, which gives two answers.
- Rounding √105 to 10 before finishing, which gives 3.75 instead of 3.81.Keep at least 4 or 5 significant figures, such as 10.247, and round only in the final line.
- Giving x = −9 as the width of a rectangle.Check each root against the situation, reject impossible values with a short reason, and answer in a sentence with units.
Exam tips
- If the question says give your answers to 2 decimal places, go straight to the formula rather than spending time hunting for factors that do not exist.
- Write the formula out in full once, then substitute using brackets, such as (−5)² and 4(2)(−10). This makes your working easy to follow if you need to find a slip.
- In word problems, show the step where you reject a solution and give the reason in a few words. The final answer should be a sentence with units.
- Check solutions by substituting back into the original equation, not the rearranged one, so that you also catch errors in your expansion.
Notes 5 of 12
Simultaneous Equations
Key idea. Simultaneous equations are two equations in two unknowns that must both be true at once. To solve a linear pair, make the coefficients of one unknown equal and add or subtract to eliminate it, or substitute an expression for one unknown into the other equation. The solution is the point where the two straight-line graphs cross. Always check it in both original equations.
Must know
- Elimination: same signs subtract, different signs add.
- Substitution: replace the unknown with its expression in brackets, such as 3(x − 4).
- The solution (x, y) is the intersection point of the two lines.
- Parallel lines give no solution, and identical lines give infinitely many.
What the syllabus asks you to do
- Solve simultaneous linear equations in two variables by the elimination method
- Solve simultaneous linear equations in two variables by the substitution method
- Solve simultaneous linear equations in two variables by the graphical method
- Formulate simultaneous linear equations to solve problems
Elimination or substitution: choosing the first step
| Situation | Best method | First step |
|---|---|---|
| One unknown is already the subject, such as y = x − 4 | Substitution | Replace y in the other equation with (x − 4) |
| Matching coefficients with different signs, such as +2y and −2y | Elimination | Add the equations to remove y |
| Matching coefficients with the same sign, such as 3x and 3x | Elimination | Subtract one equation from the other |
| No matching coefficients, such as 3x + 2y = 16 and 5x − 3y = 14 | Elimination | Multiply to make 6y and −6y, then add |
| An unknown with coefficient 1, such as x + 2y = 7 | Substitution | Rearrange to x = 7 − 2y and substitute |
Two straight lines on axes with x from 0 to 6. The line 3x + 2y = 16 slopes down from (0, 8) to (6, −1). The line 5x − 3y = 14 slopes up from about (0, −4.7) to about (6, 5.3). They cross at exactly one point, (4, 2), so x = 4 and y = 2 is the only solution.
Worked exam question
Solve the simultaneous equations 3x + 2y = 16 and 5x − 3y = 14.
- Label the equations: 3x + 2y = 16 is (1) and 5x − 3y = 14 is (2).
- Multiply (1) by 3 to get 9x + 6y = 48, which is (3).
- Multiply (2) by 2 to get 10x − 6y = 28, which is (4).
- The y terms have different signs, so add (3) and (4): 19x = 76, which gives x = 4.
- Substitute x = 4 into (1): 12 + 2y = 16, so 2y = 4 and y = 2.
- Check in (2): 5(4) − 3(2) = 20 − 6 = 14, which is correct.
Answer: x = 4 and y = 2
Common mistakes
- Multiplying 3x + 2y = 16 by 3 and writing 9x + 6y = 16.Multiply every term on both sides, and check the new right-hand side: 3 × 16 = 48.
- Adding two equations that both contain +6y and getting 12y.Circle the matching terms before combining. Same signs subtract, different signs add.
- Subtracting 2x − 3y = 5 and writing −3y instead of +3y.Put the subtracted equation in brackets, or multiply it by −1 and add, so every sign changes.
- Substituting y = x − 4 into 2x + 3y = 13 as 2x + 3x − 4.Always write the substituted expression in brackets: 2x + 3(x − 4) = 13.
- Stopping at c = 5 without finding a, or without saying what the numbers mean.Find both values and answer in context, for example 'an adult ticket costs $12 and a child ticket costs $5'.
- Checking the answer in the equation just used to find the second unknown.Check in the other original equation, which tests both values independently.
Exam tips
- Before multiplying, decide which unknown needs the smaller multipliers. Eliminating the unknown whose coefficients are already factors of each other saves arithmetic.
- If the question asks you to solve graphically, read the intersection point carefully from the grid and give values to the accuracy the scale allows, usually to the nearest tenth of a unit.
- When a word problem gives two different totals, you almost always need two unknowns and two equations. Write a 'let' statement with units first, so your equations are clear to you and to anyone marking them.
Notes 6 of 12
Congruence and Similarity
Key idea. Congruent figures have the same shape and size, and similar figures have the same shape with all lengths in the same ratio. Triangles are congruent by SSS, SAS, ASA or AAS, or RHS, and similar when two pairs of angles match or the sides are in proportion. If lengths are in the ratio k, areas are in the ratio k² and volumes in the ratio k³.
Must know
- Congruence tests: SSS, SAS, ASA or AAS, and RHS. AAA and SSA are not tests.
- Similar figures have equal angles and corresponding sides in the same ratio.
- Length ratio k gives area ratio k² and volume ratio k³.
- On a 1 : n map, 1 cm represents n cm, so 1 cm² represents n² cm².
What the syllabus asks you to do
- Determine whether two triangles are congruent or similar
- Use the properties of similar triangles and polygons: corresponding angles are equal and corresponding sides are proportional
- Use the ratio of areas of similar plane figures and the ratio of volumes of similar solids
- Use map scales for distance and area
- Solve simple problems involving similarity and congruence
Tests for congruent and similar triangles
| Information | Congruent triangles | Similar triangles |
|---|---|---|
| Angles | ASA or AAS: two pairs of equal angles and a corresponding side | Two pairs of equal angles |
| Three sides | SSS: three pairs of equal sides | Three pairs of sides in the same ratio |
| Two sides and an angle | SAS: two pairs of equal sides and equal included angles | Two pairs of sides in the same ratio and equal included angles |
| Right-angled triangles | RHS: right angle, equal hypotenuses and one other pair of equal sides | One more pair of equal angles is enough |
| Not valid | AAA and SSA | SSA |
Worked exam question
In triangle ABC, D lies on AB and E lies on AC such that DE is parallel to BC. AD = 4 cm, DB = 6 cm and BC = 15 cm. (a) Show that triangle ADE is similar to triangle ABC. (b) Find DE. (c) Given that the area of triangle ADE is 12 cm², find the area of the trapezium DBCE.
- Angle DAE = angle BAC, since the angle at A is common to both triangles.
- Angle ADE = angle ABC, since they are corresponding angles with DE parallel to BC. So triangle ADE is similar to triangle ABC, as two pairs of angles are equal.
- AB = AD + DB = 4 + 6 = 10 cm, so the length ratio is AD/AB = 4/10 = 2/5.
- DE/BC = 2/5, so DE = (2/5) × 15 = 6 cm.
- The area ratio is (2/5)² = 4/25, so the area of triangle ABC = 12 × 25/4 = 75 cm².
- Area of DBCE = 75 − 12 = 63 cm².
Answer: (b) DE = 6 cm (c) Area of DBCE = 63 cm²
Common mistakes
- Using AD/DB = 4/6 instead of AD/AB = 4/10 when DE is parallel to BC.Redraw triangles ADE and ABC separately, label the full side lengths, then match corresponding sides.
- Saying shapes with lengths in the ratio 2 : 5 have areas in the ratio 2 : 5.Write all three ratios: lengths 2 : 5, areas 4 : 25, volumes 8 : 125. Pick the one that matches the quantity asked for.
- Using 25 parts for the trapezium DBCE when the area ratio ADE : ABC is 4 : 25.Note what each ratio compares. The trapezium is the whole minus the small triangle: 25 − 4 = 21 parts.
- Quoting AAA or SSA as a reason for congruence.Learn the four tests, SSS, SAS, ASA or AAS, and RHS. AAA proves similarity only, and SSA proves neither.
- Writing triangle ABC ≡ triangle QPR when A matches P and B matches Q.Match the vertices with equal angles first, then write the names in that order: triangle ABC ≡ triangle PQR.
- Giving 'parallel lines' or 'similar' as a reason.Name the property and the lines, such as 'corresponding angles, DE parallel to BC' or 'common angle'.
Exam tips
- Redraw the two triangles side by side in the same orientation and mark the equal angles. Matching corresponding sides becomes much easier.
- In a proof, write one statement per line with its reason, and finish with the test used, such as 'so triangle ABC ≡ triangle DEF (SAS)'.
- Before using a ratio, write a line such as 'length ratio 2 : 5, area ratio 4 : 25'. Seeing both ratios together stops you using the wrong one.
Notes 7 of 12
Angle Properties of Circles
Key idea. The angle properties of circles are rules for the angles made by chords, radii and tangents. The angle at the centre is twice the angle at the circumference, the angle in a semicircle is 90°, angles in the same segment are equal, and opposite angles of a cyclic quadrilateral add up to 180°. A tangent is perpendicular to the radius at the point of contact.
Must know
- Angle at centre = 2 × angle at circumference, standing on the same arc.
- The angle in a semicircle is 90°.
- Opposite angles of a cyclic quadrilateral add up to 180°.
- A tangent meets the radius at 90°, and tangents from an external point are equal.
What the syllabus asks you to do
- Use the symmetry properties of circles: equal chords are equidistant from the centre, and the perpendicular bisector of a chord passes through the centre
- Use the tangent properties: tangents from an external point are equal in length, and the line joining the external point to the centre bisects the angle between the tangents
- Use the facts that the angle in a semicircle is a right angle and that the angle between a tangent and a radius is a right angle
- Use the fact that the angle at the centre is twice the angle at the circumference
- Use the facts that angles in the same segment are equal and angles in opposite segments are supplementary
Circle properties and the reason to write for each
| Property | What it says | Reason to write |
|---|---|---|
| Angle at the centre | The angle at the centre is twice the angle at the circumference on the same arc | angle at centre = 2 × angle at circumference |
| Angle in a semicircle | A diameter subtends 90° at the circumference | angle in a semicircle |
| Same segment | Angles on the same arc, on the same side of the chord, are equal | angles in the same segment |
| Cyclic quadrilateral | Opposite angles add up to 180° | angles in opposite segments |
| Tangent and radius | They meet at 90° at the point of contact | tangent perpendicular to radius |
| Two tangents | Tangents from an external point are equal in length | tangents from an external point |
| Chord and centre | The perpendicular from the centre to a chord bisects the chord | perpendicular from centre bisects chord |
Worked exam question
A, B, C and D lie on a circle with centre O. B lies on the major arc AC and D lies on the minor arc AC, so ABCD is a cyclic quadrilateral. Angle AOC = 140°, where this is the angle facing the minor arc. Find (a) angle ABC, (b) angle ADC, (c) angle OAC, giving reasons.
- Angle ABC stands on the minor arc AC, which subtends 140° at the centre, so angle ABC = 140° ÷ 2 = 70° (angle at centre = 2 × angle at circumference).
- ABCD is a cyclic quadrilateral, so angle ADC = 180° − 70° = 110° (angles in opposite segments).
- Check with the reflex angle: reflex angle AOC = 360° − 140° = 220°, and angle ADC = 220° ÷ 2 = 110°, which agrees.
- Triangle AOC is isosceles because OA = OC (radii).
- So angle OAC = (180° − 140°) ÷ 2 = 20° (base angles of an isosceles triangle).
Answer: (a) 70° (b) 110° (c) 20°
Common mistakes
- Halving angle AOC = 140° to get 70° for a point D on the minor arc AC.Trace the arc each angle stands on. For a point on the minor arc, halve the reflex angle: 220° ÷ 2 = 110°.
- Marking 90° in a triangle drawn on a chord that only looks like a diameter.Use the semicircle property only when the question states a diameter or the chord clearly passes through O.
- Adding the opposite angles of OABC to 180°, where O is the centre.Check that all four vertices lie on the circle before using 'angles in opposite segments'.
- Missing that OA = OB makes triangle OAB isosceles.Mark every radius as soon as you see it, then look for equal base angles.
- Writing 'circle property' as the reason for an angle.Use the exact phrase, such as 'angles in the same segment' or 'tangent perpendicular to radius'.
- Marking 90° between a tangent and a radius drawn to a different point.Mark the right angle only at the point of contact, between the tangent and the radius to that point.
Exam tips
- Mark every radius on the diagram as you notice it. Equal radii reveal isosceles triangles, which are often the missing link.
- Write each angle with its reason on the same line, such as 'angle ADC = 110° (angles in opposite segments)'. The reasons are part of the answer in these questions.
- If you are stuck, look for a diameter, a tangent, or two angles standing on the same arc. Spotting these three features is the way into most circle questions.
Notes 8 of 12
Trigonometry
Key idea. Trigonometry connects the sides and angles of triangles. In a right-angled triangle, use Pythagoras' theorem and SOH CAH TOA. In any triangle, use the sine rule when a side and its opposite angle are known, the cosine rule for two sides and the included angle or for three sides, and ½ab sin C for the area. For obtuse angles, sin(180° − θ) = sin θ and cos(180° − θ) = −cos θ.
Must know
- SOH CAH TOA: sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj.
- Sine rule: a/sin A = b/sin B = c/sin C
- Cosine rule: a² = b² + c² − 2bc cos A
- Area of any triangle = ½ab sin C
What the syllabus asks you to do
- Use Pythagoras' theorem
- Use the sine, cosine and tangent of acute angles to calculate unknown sides and angles in right-angled triangles
- Extend sine and cosine to obtuse angles
- Use the sine rule and the cosine rule for any triangle, and the formula ½ab sin C for the area of a triangle
- Solve problems in two and three dimensions, including angles of elevation and depression and bearings
Which rule to use in a triangle
| What you know | Rule | What it finds |
|---|---|---|
| A right angle and two sides | Pythagoras' theorem | The third side |
| A right angle, one side and one angle | SOH CAH TOA | Another side or angle |
| Two sides and the angle between them | Cosine rule | The third side |
| All three sides | Cosine rule, cos A = (b² + c² − a²)/(2bc) | Any angle |
| A side, its opposite angle and one more side or angle | Sine rule | A missing side or angle |
| Two sides and the angle between them | ½ab sin C | The area |
Two curves on axes from 0° to 180°. The sine curve rises from 0 at 0° to 1 at 90° and falls back to 0 at 180°, symmetric about 90°, so sin 150° = sin 30° = 0.5. The cosine curve falls from 1 at 0° through 0 at 90° to −1 at 180°, so cos 120° = −cos 60° = −0.5.
Worked exam question
In triangle ABC, AB = 8 cm, AC = 11 cm and angle BAC = 50°. Find (a) the length of BC, (b) the area of triangle ABC, (c) angle ABC.
- Two sides and the included angle are known, so use the cosine rule: BC² = 8² + 11² − 2(8)(11) cos 50°.
- BC² = 64 + 121 − 176 cos 50° = 185 − 113.131 = 71.869.
- BC = √71.869 = 8.4776 cm, which is 8.48 cm to 3 significant figures.
- Area = ½ × 8 × 11 × sin 50° = 44 × 0.76604 = 33.706 cm².
- For angle ABC, use the cosine rule with all three sides: cos B = (8² + 8.4776² − 11²)/(2 × 8 × 8.4776) = 14.870/135.64 = 0.10963.
- B = cos⁻¹(0.10963) = 83.7°. Using the cosine rule here avoids having to decide between an acute and an obtuse answer.
Answer: (a) BC = 8.48 cm (b) Area = 33.7 cm² (c) Angle ABC = 83.7°
Common mistakes
- Getting sin 30° = −0.988 and carrying on.Check the mode at the start of every paper: sin 30° should give 0.5.
- Using sin⁻¹ for an angle that should be 110° and getting 70°.Use the cosine rule for an angle that might be obtuse, or check whether it is opposite the longest side.
- Working out (b² + c² − 2bc) first, then multiplying by cos A.Calculate 2bc cos A as one product and subtract it from b² + c², or key in the whole formula with brackets.
- Writing a direction 40° east of north as 40° or 320°.Measure clockwise from north and always write three figures, so this direction is 040°.
- Measuring an angle of elevation from the vertical pole instead of from the ground.Draw the horizontal line from the observer first, then mark the angle between it and the line of sight.
- Rounding BC to 8.5 before finding angle B, which gives 83.6° instead of 83.7°.Keep 4 or 5 significant figures in working, or use the calculator memory, and round only the final answer.
Exam tips
- Before choosing a rule, mark on your sketch which sides and angles you know. Two sides and the angle between them points to the cosine rule, and a complete side and opposite angle pair points to the sine rule.
- In bearings questions, draw a north line at every point and use the parallel north lines to find angles inside the triangle, using alternate or co-interior angles.
- Give angles to 1 decimal place and lengths to 3 significant figures unless the question says otherwise, but keep more figures in your working.
Notes 9 of 12
Mensuration
Key idea. Mensuration is the calculation of lengths, areas and volumes. For circles, use arc length s = rθ and sector area ½r²θ with θ in radians, and find a segment by subtracting the triangle area ½r² sin θ from the sector. For solids, use the volume and surface area formulas for prisms, cylinders, pyramids, cones and spheres, keeping all units the same.
Must know
- π radians = 180°, so multiply degrees by π/180 to get radians.
- Arc length s = rθ and sector area = ½r²θ, with θ in radians.
- Segment area = ½r²θ − ½r² sin θ
- Cone volume = ⅓πr²h, sphere volume = (4/3)πr³, sphere surface area = 4πr².
What the syllabus asks you to do
- Use radian measure of angle, including conversion between radians and degrees
- Find arc length, sector area and the area of a segment of a circle
- Find the volume and surface area of a cube, cuboid, prism, cylinder, pyramid, cone and sphere
- Solve problems involving the volume and surface area of composite solids
- Convert between cm² and m², and between cm³ and m³
Volume and surface area formulas for O-Level solids
| Solid | Volume | Surface area |
|---|---|---|
| Prism | Area of cross-section × length | Sum of the areas of all the faces |
| Cylinder | πr²h | Curved 2πrh, total 2πrh + 2πr² if closed |
| Pyramid | ⅓ × base area × perpendicular height | Base plus the triangular faces |
| Cone | ⅓πr²h | Curved πrl, where l² = r² + h² |
| Sphere | (4/3)πr³ | 4πr² |
| Hemisphere | (2/3)πr³ | Curved 2πr², plus πr² for the flat face if solid |
Check the angle unit
θ = 1.2 radians, so set the calculator to radian mode.
Sector area
½r²θ = ½ × 100 × 1.2 = 60 cm²
Triangle area
½r² sin θ = 50 sin 1.2 = 46.602 cm²
Segment area
60 − 46.602 = 13.4 cm² to 3 significant figures
Segment perimeter
Arc 12 cm plus chord 11.293 cm gives 23.3 cm
Worked exam question
A circle has centre O and radius 10 cm. The chord AB subtends an angle of 1.2 radians at O. Find (a) the length of the minor arc AB, (b) the area of the minor segment cut off by AB, (c) the perimeter of that segment.
- Arc length s = rθ = 10 × 1.2 = 12 cm.
- Sector area = ½r²θ = ½ × 100 × 1.2 = 60 cm².
- Area of triangle AOB = ½r² sin θ = ½ × 100 × sin 1.2 = 50 × 0.93204 = 46.602 cm², with the calculator in radian mode.
- Segment area = 60 − 46.602 = 13.398 cm², which is 13.4 cm² to 3 significant figures.
- For the chord, split triangle AOB into two right-angled triangles with angle 0.6 radians at O: AB = 2 × 10 × sin 0.6 = 20 × 0.56464 = 11.293 cm.
- Perimeter of the segment = arc + chord = 12 + 11.293 = 23.293 cm.
Answer: (a) 12 cm (b) 13.4 cm² (c) 23.3 cm
Common mistakes
- Working out sin 1.2 in degree mode and getting 0.0209 instead of 0.932.Switch to radian mode for radian angles, or convert first: 1.2 radians = 68.75°.
- Using s = rθ with θ = 60 for a 60° sector.Use s = rθ only with radians. For degrees, use (θ/360°) × 2πr, or convert 60° to π/3 first.
- Giving only the arc length as the perimeter of a sector.Trace the whole boundary with a finger: arc plus two radii, so perimeter = rθ + 2r.
- Using the slant height in ⅓πr²h for a cone.Volume uses the perpendicular height h, curved area uses the slant height l. Find the missing one with l² = r² + h².
- Adding the circle where a hemisphere joins a cylinder into the surface area.List only the faces you could touch from outside before adding them up.
- Using a radius of 0.5 m with a height of 120 cm in one formula.Convert everything to one unit first, here 50 cm and 120 cm, and remember 1 m³ = 1 000 000 cm³.
Exam tips
- Write down the formula you are using before substituting, so it is clear which height or radius each number stands for.
- Keep π as a symbol until the last line, especially in melting and recasting questions, where π often cancels and leaves a clean answer.
- Check whether the question wants an exact answer in terms of π, an answer to 3 significant figures, or an answer in a different unit such as litres.
Notes 10 of 12
Coordinate Geometry
Key idea. Coordinate geometry uses coordinates to describe straight lines. The gradient is (y₂ − y₁)/(x₂ − x₁), the length of a segment is √((x₂ − x₁)² + (y₂ − y₁)²) by Pythagoras' theorem, and every non-vertical line has an equation y = mx + c, where m is the gradient and c is the y-intercept. Parallel lines have equal gradients.
Must know
- Gradient m = (y₂ − y₁)/(x₂ − x₁), the change in y over the change in x.
- Length of a segment = √((x₂ − x₁)² + (y₂ − y₁)²)
- Equation of a line: y = mx + c, with c the y-intercept.
- Horizontal line y = k has gradient 0, and a vertical line x = k has no defined gradient.
What the syllabus asks you to do
- Find the gradient of a straight line given the coordinates of two points on it
- Find the length of a line segment given the coordinates of its end points
- Interpret and find the equation of a straight line graph in the form y = mx + c
- Solve geometric problems involving the use of coordinates
Gradient, length and equation for A(1, 2) and B(7, 10)
| Quantity | Formula or method | For A(1, 2) and B(7, 10) |
|---|---|---|
| Gradient | (y₂ − y₁)/(x₂ − x₁) | (10 − 2)/(7 − 1) = 8/6 = 4/3 |
| Length | √((x₂ − x₁)² + (y₂ − y₁)²) | √(6² + 8²) = √100 = 10 units |
| Equation | y = mx + c, substitute a point to find c | y = (4/3)x + 2/3, or 3y = 4x + 2 |
| y-intercept | Put x = 0 in the equation | (0, 2/3) |
| x-intercept | Put y = 0 in the equation | 0 = 4x + 2, so (−1/2, 0) |
A straight line segment joins A(1, 2) to B(7, 10). A dashed right-angled triangle is drawn beneath it, with a horizontal side of 6 units from (1, 2) to (7, 2) and a vertical side of 8 units up to B. The gradient is 8/6 = 4/3 and, by Pythagoras' theorem, the length of AB is 10 units.
Worked exam question
The points A(1, 2) and B(7, 10) are given. Find (a) the gradient of AB, (b) the length of AB, (c) the equation of the line AB, (d) the coordinates of the point where the line AB meets the y-axis.
- Gradient = (10 − 2)/(7 − 1) = 8/6 = 4/3.
- Length = √((7 − 1)² + (10 − 2)²) = √(36 + 64) = √100 = 10 units.
- Use y = mx + c with m = 4/3 and the point B(7, 10): 10 = (4/3)(7) + c = 28/3 + c, so c = 2/3.
- The equation is y = (4/3)x + 2/3, which can also be written as 3y = 4x + 2.
- Check with A: (4/3)(1) + 2/3 = 6/3 = 2, which is correct.
- The line meets the y-axis where x = 0, so y = 2/3.
Answer: (a) 4/3 (b) 10 units (c) 3y = 4x + 2 (d) (0, 2/3)
Common mistakes
- Working out (10 − 2)/(1 − 7) for A(1, 2) and B(7, 10), and getting −4/3 instead of 4/3.Choose one point to go first and use it first on both the top and the bottom: (10 − 2)/(7 − 1).
- Calculating change in x over change in y, which gives 3/4 instead of 4/3.Say 'rise over run' and write the y-values on top every time.
- Saying the gradient of 2x + 3y = 12 is 2.Rearrange to y = mx + c first: y = −(2/3)x + 4, so the gradient is −2/3.
- From P(−2, 1) to Q(4, 9), writing the run as 4 − 2 = 2.Write brackets round every coordinate you subtract: 4 − (−2) = 6.
- Writing x = 5 for the horizontal line through (3, 5).On a horizontal line y never changes, so its equation is y = 5. The vertical line through (3, 5) is x = 3.
Exam tips
- Draw a quick sketch with the points roughly in place. It tells you whether the gradient should be positive or negative and whether a length is reasonable.
- Leave gradients as fractions such as 4/3 rather than 1.33, so the equation of the line stays exact and later substitutions give exact answers.
- Check every equation you find by substituting the coordinates of a second point that should lie on the line.
Notes 11 of 12
Vectors
Key idea. A vector has size and direction. In two dimensions it can be written as a column vector, and its magnitude is √(x² + y²). Add vectors component by component, multiply by a scalar to stretch or reverse them, and find the vector between two points as AB = OB − OA. Two vectors are parallel when one is a scalar multiple of the other.
Must know
- Magnitude of the column vector (x, y) = √(x² + y²)
- AB = OB − OA, end minus start.
- a and b are parallel if a = kb for some scalar k.
- Collinear points: AC = kAB, and the two vectors share the point A.
What the syllabus asks you to do
- Use vector notation, including column vectors, AB, a and the magnitudes of AB and a, and represent a vector as a directed line segment
- Find the magnitude of a column vector (x, y) as √(x² + y²)
- Use the sum and difference of two vectors to express given vectors in terms of two coplanar vectors
- Multiply a vector by a scalar
- Use position vectors
- Solve geometric problems involving the use of vectors
Vector skills with new examples, column vectors written as (x, y)
| Skill | Rule | Example |
|---|---|---|
| Magnitude | √(x² + y²) | The magnitude of (5, −12) is 13 |
| Addition | Add the top numbers and add the bottom numbers | (−1, 4) + (3, 2) = (2, 6) |
| Scalar multiple | Multiply both components by k | −2(3, −1) = (−6, 2) |
| Vector between points | AB = OB − OA | A(−1, 3) and B(4, 5) give AB = (5, 2) |
| Parallel test | One vector is a scalar multiple of the other | (6, 9) = 3(2, 3) |
| Point dividing a line | AP : PB = m : n gives AP = (m/(m + n))AB | AP : PB = 3 : 2 gives AP = (3/5)AB |
Axes with the origin O. A dashed line runs from O to A(2, 1) and another dashed line from O to B(5, 7). A solid line joins A to B. Going from A to B is the same as going back from A to O and then out to B, so AB = OB − OA = (3, 6), with magnitude √45, about 6.71 units.
Worked exam question
In triangle OAB, OA = a and OB = b. M is the midpoint of OB, and P lies on AB such that AP : PB = 2 : 1. (a) Express AB, OP and MP in terms of a and b. (b) The point C is such that OC = 2a + b. Show that MP is parallel to OC and find the ratio MP : OC.
- AB = OB − OA = b − a.
- AP = (2/3)AB = (2/3)(b − a), because AP is 2 parts out of 3 of AB.
- OP = OA + AP = a + (2/3)b − (2/3)a = (1/3)a + (2/3)b.
- OM = (1/2)b, since M is the midpoint of OB.
- MP = OP − OM = (1/3)a + (2/3)b − (1/2)b = (1/3)a + (1/6)b.
- Factorise: MP = (1/6)(2a + b) = (1/6)OC. Since MP is a scalar multiple of OC, MP is parallel to OC.
- The scalar 1/6 gives the ratio of lengths, so MP : OC = 1 : 6.
Answer: (a) AB = b − a, OP = (1/3)a + (2/3)b, MP = (1/3)a + (1/6)b. (b) MP = (1/6)OC, so MP is parallel to OC and MP : OC = 1 : 6.
Common mistakes
- Writing AB = OA − OB.Say the route aloud: from A back to O, then out to B, so AB = −OA + OB = OB − OA.
- Taking AP = (1/2)AB or 2AB when AP : PB = 2 : 1.Add the parts first, 2 + 1 = 3, so AP is 2 of the 3 parts: AP = (2/3)AB.
- Saying the magnitude of a + b equals the magnitude of a plus the magnitude of b.Find the column vector of a + b first, then its magnitude. For (3, 0) + (0, 4) the magnitude is 5, not 7.
- Claiming X, Y and Z are collinear only because XY is parallel to YZ.State the shared point as well: XY and YZ are parallel and share the point Y, so X, Y and Z are collinear.
- Writing a single-letter vector a with no underline in handwritten answers.Underline every single-letter vector and put an arrow over AB, so it cannot be read as a number or a length.
- Writing a + (2/3)(b − a) as a + (2/3)b − a.Expand every term in the bracket: a + (2/3)b − (2/3)a = (1/3)a + (2/3)b.
Exam tips
- Draw arrows on the diagram for every vector you find. A route that follows an arrow forwards gives a plus sign, and going against an arrow gives a minus sign.
- When asked to show that two vectors are parallel, factorise to make the scalar multiple visible, such as (1/3)a + (1/6)b = (1/6)(2a + b), and then state the conclusion in words.
- Keep fractions exact. Decimals such as 0.333a make it hard to see scalar multiples and ratios.
Notes 12 of 12
Statistics and Probability
Key idea. Statistics summarises data with averages, such as the mean and median, and measures of spread, such as the interquartile range and standard deviation. Read the median and quartiles from a cumulative frequency curve at n/2, n/4 and 3n/4. Probability measures how likely an event is, on a scale from 0 to 1. For combined events, multiply along tree diagram branches and add separate outcomes.
Must know
- Estimated mean of grouped data = Σfx ÷ Σf, using class mid-values as x.
- Standard deviation = √(Σfx²/Σf − (Σfx/Σf)²)
- Interquartile range = upper quartile − lower quartile.
- P(not A) = 1 − P(A). Multiply for 'and', add for 'or'.
What the syllabus asks you to do
- Use the mean, mode and median as measures of central tendency, and calculate the mean for grouped data
- Analyse and interpret cumulative frequency diagrams and box-and-whisker plots, including quartiles and percentiles
- Use the range, interquartile range and standard deviation as measures of spread
- Calculate the standard deviation for a set of data, grouped and ungrouped
- Use the mean and standard deviation to compare two sets of data
- Find the probability of single events and of simple combined events, using possibility diagrams and tree diagrams
- Add probabilities of mutually exclusive events and multiply probabilities of independent events
Averages and measures of spread
| Measure | How to find it | Best used when |
|---|---|---|
| Mean | Total ÷ number of values, or Σfx ÷ Σf | The data have no extreme values |
| Median | Middle value, or read at n/2 on a cumulative frequency curve | The data are skewed or have extreme values |
| Mode | Most common value | You need the most popular value or category |
| Range | Largest value − smallest value | A quick, rough measure of spread is enough |
| Interquartile range | Upper quartile − lower quartile | Comparing spread alongside medians |
| Standard deviation | √(Σfx²/Σf − (Σfx/Σf)²) | Comparing spread alongside means |
An S-shaped cumulative frequency graph rising from 0 at 0 marks to 80 at 70 marks, passing through 12 at 20 marks, 26 at 30, 48 at 40 and 66 at 50. Dashed lines from 40 on the vertical axis give a median of about 36 marks. Lines from 20 and 60 give quartiles of about 26 and 47 marks, so the interquartile range is about 21 marks.
Worked exam question
A bag contains 5 red balls and 3 blue balls. Two balls are taken at random, one after the other, without replacement. Find the probability that (a) both balls are red, (b) the balls are of different colours, (c) at least one ball is blue.
- First ball: P(red) = 5/8 and P(blue) = 3/8.
- Second ball after a red: 4 red and 3 blue remain, so P(red) = 4/7 and P(blue) = 3/7. After a blue: 5 red and 2 blue remain, so P(red) = 5/7 and P(blue) = 2/7.
- (a) P(red, red) = 5/8 × 4/7 = 20/56 = 5/14.
- (b) P(red, blue) + P(blue, red) = (5/8 × 3/7) + (3/8 × 5/7) = 15/56 + 15/56 = 30/56 = 15/28.
- (c) At least one blue is every outcome except red then red, so the probability is 1 − 5/14 = 9/14.
- Check: P(blue, blue) = 3/8 × 2/7 = 6/56, and 20/56 + 30/56 + 6/56 = 1, as it should be.
Answer: (a) 5/14 (b) 15/28 (c) 9/14
Common mistakes
- Plotting the cumulative frequency for 10 < x ≤ 20 at the mid-value 15.Plot at the upper class boundary, here 20, because cumulative frequency counts every value up to the end of the class.
- Dividing Σfx = 320 by 3 classes instead of by Σf = 20 values.Add up the frequency column first and divide Σfx by that total: 320 ÷ 20 = 16.
- Using 5/8 for a second red ball after one red has been taken from 5 red and 3 blue.After each draw, reduce the total and the colour taken by one, so the second red is 4/7.
- Adding 5/8 + 3/7 for 'red then blue'.Multiply along a path for 'and': 5/8 × 3/7 = 15/56. Add only separate whole paths for 'or'.
- Writing only 'Set A is higher' when comparing two sets of data.Make two comparisons, one of average and one of spread, each in context, such as 'Class A scored higher and more consistently'.
- Listing every path to find P(at least one blue).Use P(at least one) = 1 − P(none). It is faster and leaves less room to miss a path.
Exam tips
- In cumulative frequency questions, draw clear horizontal and vertical lines on the graph to show where you read each value. This shows your method and helps you read the scale accurately.
- In tree diagram questions, write the probabilities on every branch before answering any part, and check that each set of branches from one point adds up to 1.
- Leave probabilities as fractions in their simplest form unless the question asks for a decimal. Fractions avoid rounding errors when you multiply along branches.
Syllabus checklist for the rest
What the syllabus asks you to do in the chapters, and parts of chapters, that do not have full notes yet. Each point is the syllabus learning outcome in plain words. Tick each one you can do without looking.
Chapter 1. Numbers and their Operations: also in the syllabus
2 points to check
- Find primes, prime factorisation, HCF and LCM, and squares, cubes, square roots and cube roots by prime factorisation.
- Work with negative numbers, integers, rational and real numbers, and order numbers on a number line.
Chapter 2. Ratio and Proportion
1 point to check
- Write ratios in their simplest form, use map scales for distance and area, and solve direct and inverse proportion problems.
Chapter 3. Percentage
1 point to check
- Express one quantity as a percentage of another, compare by percentage, use percentages over 100%, increase or decrease by a percentage, and find reverse percentages.
Chapter 4. Rate and Speed
1 point to check
- Find average rate and average speed, and convert units such as km/h to m/s.
Chapter 7. Equations and Inequalities: also in the syllabus
1 point to check
- Solve linear inequalities in one variable and show the solution on a number line.
Chapter 8. Set Language and Notation
1 point to check
- Use set notation for union, intersection, complement, subsets, the empty set, the universal set and n(A), and draw and read Venn diagrams.
Chapter 9. Matrices
1 point to check
- Show and interpret information in a matrix, multiply a matrix by a scalar, and find the sum and product of two matrices where possible.
Chapter 10. Angles, Triangles and Polygons
3 points to check
- Use angles on a straight line, at a point and vertically opposite, and corresponding, alternate and interior angles with parallel lines.
- Use the properties of triangles, special quadrilaterals and regular polygons, and the interior and exterior angle sums of a convex polygon.
- Construct simple figures with compasses, ruler, set squares and protractor.
If your child is struggling
- Find the exact chapter and point. The checklists and common mistakes on this page show where an answer breaks down.
- Go back to that chapter's worked question and do it again without the steps.
- Ask the school teacher which chapters the class has covered and which are tested next.
- If several chapters are weak, work on one at a time, with short timed questions, rather than rereading everything.