Session 1 · Questions 1–5: connect familiar methods
Allow 30–40 minutes, then 10 minutes to mark. Attempt without a hint first. If stuck, reveal one starting hint; return later with the solution covered. Write a reason beside each important step.
Free GCSE maths · online and printable
Grade 7–9 GCSE maths challenge pack
Twenty substantial Higher-tier questions linking algebra, geometry, graphs and probability. Try a hint when you need a first step, then compare your reasoning with a detailed worked solution.
This is a selected practice resource, not a full specification, mock paper or grade prediction. Completing it cannot guarantee a grade. Ask your teacher which tier and topics are appropriate.
Robinson Tuition · www.robinsontuition.com/gcse-maths-grade-7-9
Use the pack on paper
Choose the sections you need. Your browser can print or save them as a PDF. Questions include working space; worked answers begin separately.
Choose A4, standard margins and 100% scale. Switch off browser headers and footers if you prefer. No answers or notes are sent anywhere.
How to use this pack
- Try the questions on paper first. Follow each calculator label and keep your working.
- Open the worked answer when you are ready. Compare the method as well as the result; equivalent correct methods count.
- Record where you worked independently, needed a hint, made an error or skipped. Use the topic map to choose what to practise next.
Marks are suggested maximums for these original practice questions, not an exam-board mark scheme. The worked steps show the method; ask a teacher if you are unsure about partial credit. No score from this page changes Maths Streak or tracker progress.
Open your four-session plan
Four focused sessions
Session 2 · Questions 6–10: algebra, functions and chance
Allow 30–40 minutes, then 10 minutes to mark. Attempt without a hint first. If stuck, reveal one starting hint; return later with the solution covered. Write a reason beside each important step.
Session 3 · Questions 11–15: geometry and reasoning
Allow 30–40 minutes, then 10 minutes to mark. Attempt without a hint first. If stuck, reveal one starting hint; return later with the solution covered. Write a reason beside each important step.
Session 4 · Questions 16–20: combine ideas
Allow 30–40 minutes, then 10 minutes to mark. Attempt without a hint first. If stuck, reveal one starting hint; return later with the solution covered. Write a reason beside each important step.
The order moves towards longer linked arguments. These are practice labels, not official grades for individual questions. Skip a question and return when needed.
Higher-tier challenges
20 original questions · 87 suggested marks · 30–40 minutes per group of five
Work on paper. Write a reason or method, even when the final answer is short. Leave a question and return later if needed.
Higher-tier challenges · Question 1
Expand and simplify (4 − √3)2
Give your answer in the form a + b√3, where a and b are integers.
Working
Show a starting hint
Expand all four products. Keep the square of the surd and combine the two middle terms.
Higher-tier challenges · Question 2
- Work out the value of 64−23
- Work out the value of (8116)−34
Working
Show a starting hint
A negative power gives a reciprocal. Deal with the root before taking the remaining positive power.
Higher-tier challenges · Question 3
Solve 3x2 - 11x + 6 = 0
Working
Show a starting hint
Look for two brackets whose first terms multiply to 3x² and whose constant terms multiply to 6.
Higher-tier challenges · Question 4
Two solid trophies are mathematically similar and are made from the same metal.
The small trophy has height 6 cm and mass 120 g.
The large trophy has height 9 cm.
Work out the mass of the large trophy.
Working
Show a starting hint
The linear scale factor is 9/6. With the same material, mass scales like volume.
Higher-tier challenges · Question 5
A, B and C are points on a circle with centre O.
C lies on the major arc AB.
Angle OAB = 28°.
Work out the size of angle ACB. Give a reason for each stage of your working.
Working
Show a starting hint
Join O to B. Use the equal radii to find the central angle before using the angle-at-the-centre theorem.
Higher-tier challenges · Question 6
Simplify fully
[x2 + 7x + 10x2 − 4] ÷ [x2 + 10x + 253x − 6]
Working
Show a starting hint
Factorise each polynomial. Dividing by a fraction means multiplying by its reciprocal; cancel factors, not terms. Retain the original excluded x-values.
Higher-tier challenges · Question 7
a is directly proportional to the square of b.
When b = 2, a = 12.
b is directly proportional to the cube of c.
When c = 2, b = 16.
Find a formula for a in terms of c.
Working
Show a starting hint
Write both proportional relationships with constants. Find each constant before substituting one formula into the other.
Higher-tier challenges · Question 8
- Using the discriminant, explain why the equation x2 + 3x + 7 = 0 has no real solutions.
- Write down the number of real solutions of the equation x2 - 6x + 9 = 0. Give a reason for your answer.
Working
Show a starting hint
Evaluate b² − 4ac. Consider what happens to the square root in the quadratic formula when this is negative or zero.
Higher-tier challenges · Question 9
f(x) = x + 23 g(x) = x2 − 1
Find fg(3)
Find f−1(x)
Working
Show a starting hint
fg(3) means f(g(3)). For the inverse, set y = f(x), then rearrange to make x the subject.
Higher-tier challenges · Question 10
A bag contains 7 red beads and 5 blue beads.
Two beads are taken from the bag at random, without replacement.
Work out the probability that exactly one of the beads is red.
Working
Show a starting hint
Exactly one red can occur in either order. Find both paths; the second denominator is one smaller.
Higher-tier challenges · Question 11
B is due east of A. AB = 250 m.
The bearing of a mast C from A is 040°.
The bearing of C from B is 300°.
Work out the distance AC. Give your answer correct to 3 significant figures.
Working
Show a starting hint
Sketch the east–west line and north lines. Convert the bearings into the triangle’s internal angles, then use the sine rule.
Higher-tier challenges · Question 12
In triangle ABC, AB = 8 cm, BC = 12 cm and AC = 15 cm.
- Work out the size of angle ABC. Give your answer correct to 1 decimal place.
- Work out the area of the triangle. Give your answer correct to 3 significant figures.
Working
Show a starting hint
Find the angle between the 8 cm and 12 cm sides using the cosine rule, then use ½ab sin C for area.
Higher-tier challenges · Question 13
ABCDEFGH is a cuboid with base ABCD and top face EFGH, where E is directly above A, F is directly above B, G is directly above C and H is directly above D.
AB = 5 cm, BC = 12 cm and CG = 9 cm.
Calculate the size of the angle between the line AG and the plane BCGF.
Give your answer correct to 1 decimal place.
Working
Show a starting hint
Identify the perpendicular projection of A onto the face. Find a face diagonal first, then use the right-angled triangle containing AG.
Higher-tier challenges · Question 14
Solve 2x2 + 5x − 3 ≤ 0
Working
Show a starting hint
Factorise or solve the corresponding quadratic equation. Decide where the upward parabola is at or below zero.
Higher-tier challenges · Question 15
A, B and P are points on a circle with centre O.
O lies inside triangle ABP.
Prove that angle AOB = 2 × angle APB.
Working
Show a starting hint
Join P to O and continue the line beyond O. The two triangles made with radii are isosceles; use exterior angles.
Higher-tier challenges · Question 16
The curve C has equation y = x2 − 4x − 5
Write x2 − 4x − 5 in the form (x − a)2 + b
Sketch C, showing clearly the coordinates of the turning point and the coordinates of the points where C crosses the axes.
Working
Show a starting hint
Complete the square for the turning point. Set y = 0 for x-intercepts and x = 0 for the y-intercept.
Higher-tier challenges · Question 17
The circle C has equation x2 + y2 = 10
The line L has equation y = x − 2
Find the coordinates of the points where L intersects C.
Working
Show a starting hint
Replace y in the circle equation by the expression from the line. Find both x-values and pair each with its own y-value.
Higher-tier challenges · Question 18
The nth term of a quadratic sequence is an2 + bn, where a and b are constants.
The 2nd term of the sequence is 10.
The 4th term of the sequence is 36.
- Find an expression, in terms of n, for the nth term of the sequence.
- Find the value of n for which the nth term of the sequence is 105.
Working
Show a starting hint
Substitute n = 2 and n = 4 to form simultaneous equations for a and b. A term number must be a positive integer.
Higher-tier challenges · Question 19
A bag contains x black counters and 6 white counters.
Two counters are taken at random, without replacement.
The probability that the two counters are different colours is 12.
Find the possible values of x.
Working
Show a starting hint
Different colours can occur in two orders. Write both probabilities using the total x + 6, and solve the resulting equation.
Higher-tier challenges · Question 20
The points A and B have position vectors
OA = 4a + 3b and OB = 8a + 11b.
- Find vector AB in terms of a and b.
- The point P lies on AB such that AP : PB = 3 : 1. Work out vector OP in terms of a and b.
- The point D has position vector OD = 14a + 18b. Prove that O, P and D lie on the same straight line.
Working
Show a starting hint
Subtract position vectors to find AB. Use the fraction AP/AB, then compare OP with OD and identify their common starting point.
Turn your answers into priorities
Record “independent”, “partly”, “revisit” or “skipped” beside each question. A single answer is only one piece of evidence. Start with two or three topics where the method was unclear, and retry a different question after practice. If you are unsure about partial marks, ask a teacher to look at your working.
Higher-tier challenges: question-to-topic map
Q1: Surds
Evidence / next step: ____________________
Practise this skill (Higher)
Q2: Fractional and Negative Indices
Evidence / next step: ____________________
Practise this skill (Higher)
Q3: Factorising Harder Quadratics
Evidence / next step: ____________________
Practise this skill (Higher)
Q4: Similar Shapes (Area and Volume)
Evidence / next step: ____________________
Practise this skill (Higher)
Q5: Circle Theorems
Evidence / next step: ____________________
Practise this skill (Higher)
Q6: Algebraic Fractions
Evidence / next step: ____________________
Practise this skill (Higher)
Q7: Direct and Inverse Proportion (Algebraic)
Evidence / next step: ____________________
Practise this skill (Higher)
Q8: Quadratic Formula
Evidence / next step: ____________________
Practise this skill (Higher)
Q9: Inverse and Composite Functions
Evidence / next step: ____________________
Practise this skill (Higher)
Q10: Conditional Probability
Evidence / next step: ____________________
Practise this skill (Higher)
Q11: The Sine Rule
Evidence / next step: ____________________
Practise this skill (Higher)
Q12: The Cosine Rule
Evidence / next step: ____________________
Practise this skill (Higher)
Q13: 3D Pythagoras and Trigonometry
Evidence / next step: ____________________
Practise this skill (Higher)
Q14: Quadratic Inequalities
Evidence / next step: ____________________
Practise this skill (Higher)
Q15: Proof of the Circle Theorems
Evidence / next step: ____________________
Practise this skill (Higher)
Q16: Completing the Square
Evidence / next step: ____________________
Practise this skill (Higher)
Q17: Quadratic Simultaneous Equations
Evidence / next step: ____________________
Practise this skill (Higher)
Q18: The Nth Term of a Quadratic Sequence
Evidence / next step: ____________________
Practise this skill (Higher)
Q19: Probability Equation Questions
Evidence / next step: ____________________
Practise this skill (Higher)
Q20: Vectors Proof Questions
Evidence / next step: ____________________
Practise this skill (Higher)
For a guided summary, try the short topic check. Neither this pack nor that check predicts a GCSE grade or assesses the whole course.
Mistake-review sheet
Choose useful errors, not just the hardest questions. Cover the answer and redo the question after writing the corrected method.
Review 1
Question / topic: __________________________________________
Error: knowledge / method / accuracy / timing / skipped
What happened? __________________________________________
Corrected method: ________________________________________
________________________________________________________
Next practice and date: ____________________________________
Review 2
Question / topic: __________________________________________
Error: knowledge / method / accuracy / timing / skipped
What happened? __________________________________________
Corrected method: ________________________________________
________________________________________________________
Next practice and date: ____________________________________
Review 3
Question / topic: __________________________________________
Error: knowledge / method / accuracy / timing / skipped
What happened? __________________________________________
Corrected method: ________________________________________
________________________________________________________
Next practice and date: ____________________________________
Review 4
Question / topic: __________________________________________
Error: knowledge / method / accuracy / timing / skipped
What happened? __________________________________________
Corrected method: ________________________________________
________________________________________________________
Next practice and date: ____________________________________
Worked answers
Keep this section covered until you have tried the questions. A matching final answer does not establish that all the reasoning is correct.
Credit an equivalent correct method. A correct method can earn its checkpoint even if later arithmetic is wrong; only award an accuracy checkpoint when its stated result is correct. Do not award the same checkpoint twice. For a proof, the reasons are essential. This original guidance is not an exam-board scheme.
Higher-tier challenges · Answer 1: Surds
Higher-tier challenges · Answer 1
3 suggested marks · compare each step, not just the final answer.
(4 − √3)² = 16 − 4√3 − 4√3 + (√3)².
(√3)² = 3, so this is 16 − 8√3 + 3.
Collect terms: 19 − 8√3, so a = 19 and b = −8.
19 - 8√3
Suggested marking checkpoints
- 1 mark: Multiply all four products when expanding.
- 1 mark: Use (√3)² = 3.
- 1 mark: 19 − 8√3, so a = 19 and b = −8.
Next practice: Surds workout · Learn this method
Higher-tier challenges · Answer 2: Fractional and Negative Indices
Higher-tier challenges · Answer 2
4 suggested marks · compare each step, not just the final answer.
(a) 6423 = (cube root of 64)² = 4² = 16, so 64−23 = 116.
(b) The negative index turns the fraction over: (8116)−34 = (1681)34.
1634 = (fourth root of 16)³ = 2³ = 8 and 8134 = 3³ = 27.
So the answer is 827.
(a) 116; (b) 827
Suggested marking checkpoints
- 1 mark: (a) cube root of 64 = 4, then square = 16.
- 1 mark: Reciprocal gives 1/16.
- 1 mark: (b) Turn the fraction over and correctly take fourth roots.
- 1 mark: Cube to obtain 8/27.
Next practice: Fractional and Negative Indices workout · Learn this method
Higher-tier challenges · Answer 3: Factorising Harder Quadratics
Higher-tier challenges · Answer 3
3 suggested marks · compare each step, not just the final answer.
Look for two numbers that multiply to 3 × 6 = 18 and add to -11: they are -9 and -2.
Split the middle term: 3x2 - 9x - 2x + 6 = 3x(x - 3) - 2(x - 3) = (3x - 2)(x - 3).
Set each factor to zero: 3x - 2 = 0 gives x = 23, and x - 3 = 0 gives x = 3.
Check x = 3: 3(9) - 33 + 6 = 27 - 27 = 0, correct.
x = 23, x = 3
Suggested marking checkpoints
- 1 mark: Correct factorisation (3x − 2)(x − 3), or correct quadratic-formula substitution.
- 1 mark: One correct root.
- 1 mark: Both roots 2/3 and 3 from valid working.
Next practice: Factorising Harder Quadratics workout · Learn this method
Higher-tier challenges · Answer 4: Similar Shapes (Area and Volume)
Higher-tier challenges · Answer 4
4 suggested marks · compare each step, not just the final answer.
Length scale factor = 9 ÷ 6 = 1.5.
Mass is proportional to volume, so the mass scale factor = 1.53 = 3.375.
Mass of the large trophy = 120 × 3.375 = 405 g.
405 g
Suggested marking checkpoints
- 1 mark: Length scale factor 1.5.
- 1 mark: Recognise, explicitly or through the calculation, that the same metal makes mass proportional to volume.
- 1 mark: Volume/mass scale factor 1.5³ = 3.375.
- 1 mark: 405 g.
Next practice: Similar Shapes (Area and Volume) workout · Learn this method
Higher-tier challenges · Answer 5: Circle Theorems
Higher-tier challenges · Answer 5
4 suggested marks · compare each step, not just the final answer.
OA = OB because both are radii, so triangle OAB is isosceles and angle OBA = angle OAB = 28°.
Angles in a triangle add up to 180°, so angle AOB = 180° - 28° - 28° = 124°.
The angle at the centre is twice the angle at the circumference standing on the same arc.
Angle ACB = 124° ÷ 2 = 62°.
62°
Suggested marking checkpoints
- 1 mark: OA = OB radii, hence both base angles 28°.
- 1 mark: Central angle 124° using triangle sum.
- 1 mark: State angle at centre is twice angle at circumference on same arc.
- 1 mark: 62°.
Next practice: Circle Theorems workout · Learn this method
Higher-tier challenges · Answer 6: Algebraic Fractions
Higher-tier challenges · Answer 6
4 suggested marks · compare each step, not just the final answer.
Factorise each expression: x^2 + 7x + 10 = (x + 2)(x + 5), x^2 - 4 = (x - 2)(x + 2), x^2 + 10x + 25 = (x + 5)^2, 3x - 6 = 3(x - 2).
Dividing by a fraction is multiplying by its reciprocal: [(x + 2)(x + 5)(x - 2)(x + 2)] x [3(x - 2)(x + 5)^2].
Cancel (x + 2): gives x + 5x - 2 x 3(x - 2)(x + 5)^2.
Cancel (x - 2) and one factor of (x + 5) to leave 3x + 5.
3x+5
Domain check: the original expression is undefined at x = −2 or 2 because a denominator is zero, and at x = −5 because the divisor is zero. The simplified form 3/(x + 5) is valid only with all three original exclusions retained.
Suggested marking checkpoints
- 1 mark: Factorise the original expressions correctly.
- 1 mark: Invert the divisor and multiply.
- 1 mark: Cancel common factors rather than terms.
- 1 mark: 3/(x + 5). Original exclusions −2, 2, −5 are a useful domain check, but their omission does not lose this mark: the question only asks to simplify.
Next practice: Algebraic Fractions workout · Learn this method
Higher-tier challenges · Answer 7: Direct and Inverse Proportion (Algebraic)
Higher-tier challenges · Answer 7
4 suggested marks · compare each step, not just the final answer.
From the first relation: a = kb2 and 12 = k × 4, so k = 3 and a = 3b2.
From the second relation: b = mc3 and 16 = m × 8, so m = 2 and b = 2c3.
Substitute: a = 3(2c3)2 = 3 × 4c6 = 12c6.
Check with c = 2: b = 16 and a = 3 × 256 = 768 = 12 × 26.
a = 12c^6
Suggested marking checkpoints
- 1 mark: a = kb² and k = 3.
- 1 mark: b = mc³ and m = 2.
- 1 mark: Substitute a = 3(2c³)².
- 1 mark: a = 12c⁶.
Next practice: Direct and Inverse Proportion (Algebraic) workout · Learn this method
Higher-tier challenges · Answer 8: Quadratic Formula
Higher-tier challenges · Answer 8
4 suggested marks · compare each step, not just the final answer.
The discriminant of ax2 + bx + c = 0 is b2 - 4ac.
For x2 + 3x + 7 = 0: discriminant = 9 - 28 = -19. A negative number has no real square root, so the formula gives no real solutions.
For x2 - 6x + 9 = 0: discriminant = 36 - 36 = 0, so there is exactly one repeated real solution.
In fact x2 - 6x + 9 = (x - 3)2, so the repeated solution is x = 3.
(a) b^2 - 4ac = 3^2 - 4(1)(7) = -19, which is negative, so there are no real solutions; (b) One repeated real solution, because b^2 - 4ac = 36 - 36 = 0
Suggested marking checkpoints
- 1 mark: (a) Discriminant −19.
- 1 mark: Explain a negative discriminant has no real square root, hence no real solutions.
- 1 mark: (b) Discriminant 0.
- 1 mark: One repeated real solution because both ± branches coincide.
Next practice: Quadratic Formula workout · Learn this method
Higher-tier challenges · Answer 9: Inverse and Composite Functions
Higher-tier challenges · Answer 9
4 suggested marks · compare each step, not just the final answer.
a) g(3) = 9 - 1 = 8, then f(8) = 8 + 23 = 103.
b) Write y = x + 23, so 3y = x + 2 and x = 3y - 2, giving f^-1(x) = 3x - 2.
b) Check: f(3x - 2) = 3x - 2 + 23 = x.
(a) 103; (b) f^-1(x) = 3x - 2
Suggested marking checkpoints
- 1 mark: (a) g(3) = 8.
- 1 mark: f(8) = 10/3.
- 1 mark: (b) Rearrange y = (x + 2)/3 to x = 3y − 2.
- 1 mark: Inverse f⁻¹(x) = 3x − 2.
Next practice: Inverse and Composite Functions workout · Learn this method
Higher-tier challenges · Answer 10: Conditional Probability
Higher-tier challenges · Answer 10
4 suggested marks · compare each step, not just the final answer.
P(red then blue) = 712 × 511 = 35132.
P(blue then red) = 512 × 711 = 35132.
P(exactly one red) = 35132 + 35132 = 70132 = 3566
3566
Suggested marking checkpoints
- 1 mark: Identify both red-blue orders without replacement.
- 1 mark: Correct probability 35/132 for one order.
- 1 mark: Correct probability 35/132 for the other order.
- 1 mark: Add: 35/66.
Next practice: Conditional Probability workout · Learn this method
Higher-tier challenges · Answer 11: The Sine Rule
Higher-tier challenges · Answer 11
4 suggested marks · compare each step, not just the final answer.
At A: the bearing of C is 040° and B is due east (bearing 090°), so angle CAB = 90° - 40° = 50°.
At B: A is due west (bearing 270°) and the bearing of C is 300°, so angle CBA = 300° - 270° = 30°.
Angle ACB = 180° - 50° - 30° = 100°.
Sine rule: ACsin(CBA) = ABsin(ACB), so AC = 250 × sin 30° ÷ sin 100°.
AC = 125 ÷ 0.98480... = 126.928... = 127 m (3 sf).
127 m
Suggested marking checkpoints
- 1 mark: Internal angles at A and B are 50° and 30°.
- 1 mark: Angle at C is 100°.
- 1 mark: Correct sine-rule equation AC/sin30° = 250/sin100°.
- 1 mark: 127 m (3 significant figures).
Next practice: The Sine Rule workout · Learn this method
Higher-tier challenges · Answer 12: The Cosine Rule
Higher-tier challenges · Answer 12
5 suggested marks · compare each step, not just the final answer.
a) Angle ABC is opposite AC. Cosine rule: cos B = AB2 + BC2 - AC22 × AB × BC.
cos B = 64 + 144 - 2252 × 8 × 12 = -17192 = -0.08854...
B = cos-1(-0.08854...) = 95.079...° = 95.1° (1 dp).
b) Area = 12 × AB × BC × sin B = 12 × 8 × 12 × sin 95.079...°.
Area = 48 × 0.99607... = 47.811... = 47.8 cm2 (3 sf).
(a) 95.1°; (b) 47.8 cm²
Suggested marking checkpoints
- 1 mark: (a) Correct cosine-rule rearrangement for angle B.
- 1 mark: cos B = −17/192.
- 1 mark: 95.1° (1 decimal place).
- 1 mark: (b) Use ½ × 8 × 12 × sin B with unrounded B.
- 1 mark: 47.8 cm² (3 significant figures).
Next practice: The Cosine Rule workout · Learn this method
Higher-tier challenges · Answer 13: 3D Pythagoras and Trigonometry
Higher-tier challenges · Answer 13
4 suggested marks · compare each step, not just the final answer.
AB is perpendicular to the plane BCGF, because AB is perpendicular to both BC and BF.
So the projection of A onto the plane BCGF is B, the projection of AG onto the plane is BG, and the required angle is angle AGB.
BG is a diagonal of the face BCGF: BG² = BC² + CG² = 12² + 9² = 144 + 81 = 225, so BG = 15 cm.
Triangle ABG has a right angle at B because AB is perpendicular to the plane BCGF.
tan(angle AGB) = AB ÷ BG = 5 ÷ 15 = 0.3333...
angle AGB = tan−1(13) = 18.43...°
The angle is 18.4° (1 decimal place).
18.4°
Suggested marking checkpoints
- 1 mark: Identify BG as projection and AGB as the required angle, because AB is perpendicular to the face.
- 1 mark: BG = √(12² + 9²) = 15 cm.
- 1 mark: tan AGB = 5/15.
- 1 mark: 18.4° (1 decimal place).
Next practice: 3D Pythagoras and Trigonometry workout · Learn this method
Higher-tier challenges · Answer 14: Quadratic Inequalities
Higher-tier challenges · Answer 14
5 suggested marks · compare each step, not just the final answer.
Factorise: 2x^2 + 5x - 3 = (2x - 1)(x + 3).
Check: (2x - 1)(x + 3) = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3.
The roots are x = 12 and x = -3, and the parabola is U-shaped.
The expression is less than or equal to zero between the roots.
So -3 ≤ x ≤ 12.
-3 ≤ x ≤ 12
Suggested marking checkpoints
- 1 mark: Factorise (2x − 1)(x + 3).
- 1 mark: Root −3.
- 1 mark: Root 1/2.
- 1 mark: Select the interval between the roots, with a sign test or upward-parabola reason.
- 1 mark: −3 ≤ x ≤ 1/2, including both endpoints.
Next practice: Quadratic Inequalities workout · Learn this method
Higher-tier challenges · Answer 15: Proof of the Circle Theorems
Higher-tier challenges · Answer 15
4 suggested marks · compare each step, not just the final answer.
Join P to O and extend the line to a point D on the other side of O.
OA = OP because both are radii, so triangle OAP is isosceles and angle OPA = angle OAP. Call this angle x.
The exterior angle of a triangle equals the sum of the two interior opposite angles, so angle AOD = angle OPA + angle OAP = 2x.
OB = OP because both are radii, so triangle OBP is isosceles and angle OPB = angle OBP. Call this angle y.
By the same exterior angle argument, angle BOD = 2y.
Angle AOB = angle AOD + angle DOB = 2x + 2y = 2(x + y).
Angle APB = angle APD + angle DPB = x + y.
Therefore angle AOB = 2 × angle APB, which proves that the angle at the centre is twice the angle at the circumference standing on the same arc.
Extending PO to D and using the exterior angles of the isosceles triangles OAP and OBP gives angle AOB = 2x + 2y = 2 angle APB
Suggested marking checkpoints
- 1 mark: Join PO; use equal radii to justify the equal base angles x and y. Extend PO if using exterior angles.
- 1 mark: Obtain angle AOD = 2x by exterior angles, or angle AOP = 180° − 2x by the triangle sum.
- 1 mark: Similarly obtain BOD = 2y, or BOP = 180° − 2y.
- 1 mark: Combine the exterior angles, or use the 360° sum around O, to obtain AOB = 2(x + y) = 2APB.
Next practice: Proof of the Circle Theorems workout · Learn this method
Higher-tier challenges · Answer 16: Completing the Square
Higher-tier challenges · Answer 16
5 suggested marks · compare each step, not just the final answer.
a) x^2 - 4x - 5 = (x - 2)^2 - 4 - 5 = (x - 2)^2 - 9.
a) Check by expanding: x^2 - 4x + 4 - 9 = x^2 - 4x - 5.
b) The turning point is (2, -9).
b) For the x-intercepts, factorise: x^2 - 4x - 5 = (x - 5)(x + 1), so the curve crosses at (5, 0) and (-1, 0).
b) When x = 0, y = -5, so the y-intercept is (0, -5).
b) Draw a U-shaped parabola through these points with its minimum at (2, -9).
(a) (x - 2)^2 - 9; (b) U-shaped parabola, turning point (2, -9), crossing the x-axis at (-1, 0) and (5, 0) and the y-axis at (0, -5)
Suggested marking checkpoints
- 1 mark: (a) Correct bracket (x − 2)².
- 1 mark: Correct adjustment gives (x − 2)² − 9.
- 1 mark: (b) Turning point (2, −9).
- 1 mark: Both x-intercepts (−1, 0), (5, 0) and y-intercept (0, −5).
- 1 mark: Upward curve through these points, symmetric about x = 2.
Next practice: Completing the Square workout · Learn this method
Higher-tier challenges · Answer 17: Quadratic Simultaneous Equations
Higher-tier challenges · Answer 17
5 suggested marks · compare each step, not just the final answer.
Substitute y = x - 2 into the circle equation: x^2 + (x - 2)^2 = 10.
Expand: x^2 + x^2 - 4x + 4 = 10, so 2x^2 - 4x - 6 = 0.
Divide by 2: x^2 - 2x - 3 = 0, so (x - 3)(x + 1) = 0 and x = 3 or x = -1.
When x = 3, y = 1. When x = -1, y = -3.
Check: 9 + 1 = 10 and 1 + 9 = 10.
(3, 1) and (-1, -3)
Suggested marking checkpoints
- 1 mark: Substitute x² + (x − 2)² = 10.
- 1 mark: Simplify to x² − 2x − 3 = 0.
- 1 mark: Both roots x = 3, −1.
- 1 mark: Find each corresponding y from y = x − 2.
- 1 mark: Correctly paired points (3,1) and (−1,−3).
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Higher-tier challenges · Answer 18: The Nth Term of a Quadratic Sequence
Higher-tier challenges · Answer 18
6 suggested marks · compare each step, not just the final answer.
a) n = 2 gives 4a + 2b = 10, so 2a + b = 5. n = 4 gives 16a + 4b = 36, so 4a + b = 9.
Subtracting: 2a = 4, so a = 2, and then b = 5 - 4 = 1.
The nth term is 2n^2 + n.
Check: 2nd term = 8 + 2 = 10; 4th term = 32 + 4 = 36.
b) Solve 2n^2 + n = 105, so 2n^2 + n - 105 = 0.
Factorise: (2n + 15)(n - 7) = 2n^2 - 14n + 15n - 105 = 2n^2 + n - 105, so n = 7 or n = -152.
n must be a positive integer, so n = 7.
Check: 2(49) + 7 = 98 + 7 = 105.
(a) 2n^2 + n; (b) 7
Suggested marking checkpoints
- 1 mark: (a) Equations 4a + 2b = 10 and 16a + 4b = 36.
- 1 mark: Solve for a = 2 and b = 1.
- 1 mark: nth term 2n² + n.
- 1 mark: (b) Form 2n² + n − 105 = 0.
- 1 mark: Solve to obtain 7 and −15/2.
- 1 mark: Select n = 7 and reject the other root because a term position is a positive integer.
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Higher-tier challenges · Answer 19: Probability Equation Questions
Higher-tier challenges · Answer 19
6 suggested marks · compare each step, not just the final answer.
The bag holds x + 6 counters.
P(black then white) = xx + 6 × 6x + 5 and P(white then black) = 6x + 6 × xx + 5.
P(different colours) = 12x(x + 6)(x + 5) = 12.
24x = (x + 6)(x + 5) = x² + 11x + 30.
x² - 13x + 30 = 0, so (x - 3)(x - 10) = 0.
x = 3 or x = 10.
Check x = 3: 2 × 3 × 69 × 8 = 3672 = 12. Check x = 10: 2 × 10 × 616 × 15 = 120240 = 12.
x = 3 or x = 10
Suggested marking checkpoints
- 1 mark: Use total x + 6 then x + 5 after removal.
- 1 mark: Write both different-colour path probabilities.
- 1 mark: Equation 12x/((x + 6)(x + 5)) = 1/2.
- 1 mark: Rearrange to x² − 13x + 30 = 0.
- 1 mark: Factorise (x − 3)(x − 10) = 0.
- 1 mark: Both possible counts x = 3 and x = 10, verified as valid in the original probability.
Next practice: Probability Equation Questions workout · Learn this method
Higher-tier challenges · Answer 20: Vectors Proof Questions
Higher-tier challenges · Answer 20
5 suggested marks · compare each step, not just the final answer.
a) Vector AB = OB − OA = (8a + 11b) − (4a + 3b) = 4a + 8b.
b) AP : PB = 3 : 1, so AP = 34AB = 34(4a + 8b) = 3a + 6b.
Vector OP = OA + AP = 4a + 3b + 3a + 6b = 7a + 9b.
c) OD = 14a + 18b = 2(7a + 9b) = 2 × OP.
So OD is a scalar multiple of OP, which means OD and OP are parallel.
OD and OP both pass through the point O.
Therefore O, P and D lie on the same straight line, with P the midpoint of OD.
(a) AB = 4a + 8b; (b) OP = 7a + 9b; (c) OD = 14a + 18b = 2(7a + 9b) = 2OP, so OD is a scalar multiple of OP; both pass through O, so O, P and D are collinear
Suggested marking checkpoints
- 1 mark: (a) AB = OB − OA = 4a + 8b.
- 1 mark: (b) AP = 3/4 AB = 3a + 6b.
- 1 mark: OP = OA + AP = 7a + 9b.
- 1 mark: (c) Establish OD = 2OP.
- 1 mark: Explain the vectors are parallel and share O, so O, P and D are collinear.
Next practice: Vectors Proof Questions workout · Learn this method
Learn the methods
Surds
A surd is an exact root left unevaluated. To simplify a square root, take out square factors: √80 = √(16 × 5) = 4√5. Only like surds combine, just as like algebraic terms do. When squaring a bracket, multiply every term by every term; the middle terms do not disappear.
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Fractional and Negative Indices
A negative power means a reciprocal: a⁻ⁿ = 1/aⁿ for a ≠ 0. A fractional power a^(m/n) means take the nth root then raise to power m (use positive bases here). Thus 64^(−2/3) = 1/(cube root of 64)² = 1/16. The negative sign does not make the answer negative.
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Factorising Harder Quadratics
For ax² + bx + c, find two numbers whose product is ac and sum is b. Split bx using those numbers, then factorise in pairs. If a product equals zero, at least one factor is zero, so solve both linear equations. Check by expanding; the quadratic formula is also a valid solution method unless factorising is specifically requested.
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Similar Shapes (Area and Volume)
Similar shapes have the same shape and one constant length scale factor k. Area scales by k² because two lengths are multiplied; volume scales by k³ because three are multiplied. Take a square root of an area ratio to recover k. Mass scales like volume only when the material has the same density.
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Circle Theorems
First identify the chord or arc used by both angles. Radii of the same circle are equal, giving isosceles triangles. Angles in the same segment are equal; opposite angles in a cyclic quadrilateral total 180°; the angle at the centre is twice the angle at the circumference on the same arc. State the relevant fact at each step.
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Algebraic Fractions
Factorise numerators and denominators before cancelling. Cancellation removes a common non-zero factor of a product, not a term in a sum. Division by a fraction is multiplication by its reciprocal. Keep every excluded value from the original denominators and exclude values that make the original divisor zero.
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Direct and Inverse Proportion (Algebraic)
Replace a proportionality statement with an equation containing a constant. Direct proportion to b² gives a = kb²; inverse proportion to d² gives F = k/d². Use a known pair to find the constant, then substitute the new value. When combining rules, substitute the whole expression in brackets before taking a power.
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Inverse and Composite Functions
A function is a rule that takes an input to an output. fg(x) means f(g(x)): apply g first, then use its output as f’s input. An inverse undoes a function; f⁻¹ does not mean 1/f. Write y = f(x), undo the operations to make x the subject, then rename the input for the inverse.
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Conditional Probability
A conditional probability uses the group that is still possible after the first event. Without replacement, both the total and the relevant colour count may change. Multiply along one path; add different mutually exclusive paths. “Exactly one” often needs two paths, such as red then blue and blue then red.
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The Sine Rule
Bearings are measured clockwise from north. Convert them to internal triangle angles first. The sine rule pairs each side with its opposite angle: a/sin A = b/sin B. Choose a known opposite pair and the required side’s opposite angle. Use degrees and retain calculator precision until the final answer.
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The Cosine Rule
Use the cosine rule when three sides are known or two sides and their included angle are known: a² = b² + c² − 2bc cos A. For an angle, rearrange to cos A = (b² + c² − a²)/(2bc). Here a is opposite A. The area formula ½bc sin A uses the angle between b and c, and works for obtuse angles too.
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3D Pythagoras and Trigonometry
The angle between a line and a plane is measured to the line’s perpendicular projection onto the plane. Drop a perpendicular from the off-plane endpoint. This makes a right triangle containing the original line and its projection. Find any needed face diagonal with Pythagoras before using trigonometry in that triangle.
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Quadratic Inequalities
Solve the corresponding equation to locate the boundary roots. The sign of the quadratic is constant between successive roots. An upward parabola is negative between two distinct roots and positive outside; test a value to check. Include boundary values for ≤ or ≥, and exclude them for < or >.
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Proof of the Circle Theorems
A proof must work for any diagram meeting the stated conditions. Add a useful construction, give unknown angles letters, and justify each equality using radii, isosceles triangles and triangle angle facts. Combine the resulting equalities to reach the exact statement; measuring one picture is not proof.
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Completing the Square
Half the coefficient of x to form a bracket. Expanding (x + p)² gives x² + 2px + p², so subtract the extra p² to keep equality. In y = (x − a)² + b, the square is smallest at x = a, giving turning point (a,b). Solving a positive square requires both the positive and negative square roots.
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Quadratic Simultaneous Equations
At an intersection, both equations hold for the same x and y. Substitute the linear expression for y into the other equation using brackets. Solve the resulting quadratic, then find y separately for each x. Keep the coordinates paired and check each pair in both original equations.
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The Nth Term of a Quadratic Sequence
Substitute known positions into the proposed nth-term expression to form equations for its coefficients. Solve those equations and check the known terms. To find a position, set the expression equal to the target value. Reject any solution that is not a positive integer because positions are 1, 2, 3, ….
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Probability Equation Questions
Represent the unknown count with a letter and build probabilities from favourable counts divided by current totals. Add the mutually exclusive orders for different colours. Set this probability equal to the given value, clear denominators and solve. Finally check each solution is a valid whole-number count and satisfies the original probability.
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Vectors Proof Questions
A position vector starts at O. To travel from A to B, subtract OA from OB. A ratio AP:PB = 3:1 means AP is three of the four equal parts of AB. Scalar-multiple vectors are parallel; to prove points lie on one line, also establish that the relevant vectors share a point.
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