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Practice Materials

Mathematics Resources

Exam-style questions with full worked solutions, written by Dr Krishan P. Chauhan. Select your level and topic below. Tap any question to reveal the solution.

GCSE Foundation Tier

Grades 1–5 · AQA, Edexcel, OCR · Click any question to see the solution

Number & Fractions

2 marks
Work out ¾ + ⅖. Give your answer as a fraction in its simplest form.
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Worked Solution

1Convert to a common denominator: ¾ = 15/20 and ⅖ = 8/20
2Add the numerators: 15 + 8 = 23
3Answer: 23/20 (already in simplest form as 23 is prime)
Examiner tip: Always find the lowest common multiple of the denominators first. Show each conversion step for full marks.
3 marks
The price of a coat is reduced by 15% in a sale. The sale price is £85. Work out the original price.
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Worked Solution

1The sale price represents 100% − 15% = 85% of the original price
2Set up the equation: 85% × original = £85
3Divide: £85 ÷ 0.85 = £100
Examiner tip: Many students multiply 85 by 0.85 — this is wrong. If you are going backwards from a percentage reduction, you must divide, not multiply.
2 marks
Write 0.000456 in standard form.
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Worked Solution

1Identify the first significant figure: 4
2Count how many places the decimal point moves: 4 places to the right
3Answer: 4.56 × 10−&sup4;
Examiner tip: The power is negative because the number is less than 1. The number before × must always be between 1 and 10.

Algebra

2 marks
Expand and simplify: 3(2x + 4) + 2(x − 3)
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Worked Solution

1Expand the first bracket: 6x + 12
2Expand the second bracket: 2x − 6
3Collect like terms: 6x + 2x = 8x; 12 − 6 = 6
4Answer: 8x + 6
Examiner tip: Always expand before you collect like terms. Do not try to do both in one step under exam pressure.
2 marks
Solve: 3x + 7 = 22
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Worked Solution

1Subtract 7 from both sides: 3x = 15
2Divide both sides by 3: x = 5
Examiner tip: Show every step. A fully correct method with an arithmetic error still earns the method mark.
3 marks
The nth term of a sequence is 4n − 1. (a) Find the 10th term. (b) Is 75 a term in this sequence? Show your reasoning.
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Worked Solution

1(a) Substitute n = 10: 4(10) − 1 = 40 − 1 = 39
2(b) Set 4n − 1 = 75: 4n = 76, n = 19
3Since n = 19 is a positive whole number, 75 is the 19th term in the sequence.
Examiner tip: Part (b) always requires you to show working. Stating ‘yes’ without showing n is a whole number will not earn full marks.

Geometry & Measures

3 marks
A train travels 240 miles in 3 hours 20 minutes. Calculate the average speed in mph.
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Worked Solution

1Convert time to hours: 3 hours 20 minutes = 3 + 20/60 = 3⅓ hours
2Use Speed = Distance ÷ Time
3240 ÷ (10/3) = 240 × 3/10 = 72 mph
Examiner tip: Convert units before you apply the formula. Leaving time in mixed hours and minutes is the most common error on this type of question.
4 marks
The diagram shows a rectangle with length (2x + 3) cm and width (x − 1) cm. The perimeter is 34 cm. Find the value of x and the area of the rectangle.
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Worked Solution

1Perimeter = 2(length + width): 2(2x + 3 + x − 1) = 34
2Simplify: 2(3x + 2) = 34 ⇒ 6x + 4 = 34 ⇒ 6x = 30 ⇒ x = 5
3Length = 2(5) + 3 = 13 cm; Width = 5 − 1 = 4 cm
4Area = 13 × 4 = 52 cm²
Examiner tip: Label your answer. ‘52’ on its own loses the units mark. Write cm².

GCSE Higher Tier

Grades 4–9 · AQA, Edexcel, OCR · Click any question to see the solution

Algebra & Proof

3 marks
Prove that the product of two consecutive even numbers is always divisible by 8.
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Worked Solution

1Let the two consecutive even numbers be 2n and 2n + 2 (where n is an integer)
2Product = 2n(2n + 2) = 4n(n + 1)
3Since n and n+1 are consecutive integers, one must be even, so n(n+1) is divisible by 2
4Therefore 4n(n+1) is divisible by 4 × 2 = 8 □
Examiner tip: Proof questions require fully general algebraic arguments. Never use specific numbers in a proof.
3 marks
Solve: x² + 5x − 6 = 0
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Worked Solution

1Factorise: find two numbers that multiply to −6 and add to +5 ⇒ +6 and −1
2Write: (x + 6)(x − 1) = 0
3Either x + 6 = 0 or x − 1 = 0
4Solutions: x = −6 or x = 1
Examiner tip: If the quadratic does not factorise, use the formula x = (−b ± √(b²−4ac)) / 2a. Always check both solutions.
4 marks
A circle has equation x² + y² = 25. Find the equation of the tangent at the point (3, 4).
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Worked Solution

1The centre of the circle is (0, 0)
2Gradient of radius to (3, 4): m = 4/3 = 4/3
3The tangent is perpendicular to the radius, so its gradient = −3/4
4Equation of tangent: y − 4 = −3/4(x − 3) ⇒ 4y − 16 = −3x + 9
5Answer: 3x + 4y = 25
Examiner tip: The radius to the tangent point is always perpendicular to the tangent. Use m⊂1; × m⊂2; = −1.

Probability & Statistics

3 marks
A bag contains 3 red, 5 blue, and 2 green balls. Two balls are drawn without replacement. Find P(both red).
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Worked Solution

1P(first ball red) = 3/10
2P(second ball red | first was red) = 2/9 (one less red, one less total)
3P(both red) = 3/10 × 2/9 = 6/90 = 1/15
Examiner tip: ‘Without replacement’ means the denominator changes for the second event. This is the most commonly missed distinction in conditional probability.
4 marks
The mean of 5 numbers is 12. Four of the numbers are 8, 15, 10, and 14. Find the fifth number.
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Worked Solution

1Total sum = mean × number of values = 12 × 5 = 60
2Sum of known four: 8 + 15 + 10 + 14 = 47
3Fifth number = 60 − 47 = 13
Examiner tip: Always use Total = Mean × n as your starting point for reverse mean problems.

Geometry & Trigonometry

2 marks
A sphere has radius 5 cm. Calculate the volume. Give your answer in terms of π.
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Worked Solution

1Formula: V = &frac43;πr³
2V = &frac43; × π × 5³ = &frac43; × π × 125 = 500π/3 cm³
Examiner tip: You must learn the volume formulas for sphere (4/3πr³), cone (1/3πr²h), and pyramid (1/3 × base area × h). These are not given in the formula sheet in all specifications.
4 marks
ABC is a triangle. AB = 7 cm, BC = 9 cm, angle ABC = 65°. Find the area of triangle ABC.
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Worked Solution

1Area of triangle = ½ × a × b × sin(C)
2Here a = 7, b = 9, C = 65°
3Area = ½ × 7 × 9 × sin(65°)
4= ½ × 63 × 0.9063 = 28.5 cm² (3 s.f.)
Examiner tip: Use the sine area formula when you have two sides and the included angle. Make sure your calculator is in degrees mode.

A-Level Pure Mathematics

Year 1 & Year 2 · AQA, Edexcel, OCR · Click any question to see the solution

Calculus

2 marks
Differentiate: y = 3x&sup5; − 4x³ + 7x − 2
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Worked Solution

1Apply the power rule to each term: d/dx(ax⊃) = nax⊃−¹
2dy/dx = 15x&sup4; − 12x² + 7
3(The constant −2 differentiates to 0)
Examiner tip: Differentiate term by term. The constant always disappears. Do not forget to reduce the power by 1.
3 marks
Find the equation of the tangent to y = x³ − 2x at the point where x = 2.
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Worked Solution

1Find y when x = 2: y = 8 − 4 = 4. Point is (2, 4)
2Differentiate: dy/dx = 3x² − 2
3Gradient at x = 2: m = 3(4) − 2 = 10
4Tangent: y − 4 = 10(x − 2) ⇒ y = 10x − 16
Examiner tip: Always find the point, then the gradient, then the equation — in that order.
3 marks
Find: ∫(4x³ − 6x + 3) dx
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Worked Solution

1Integrate term by term: ∫ax⊃ dx = ax⊃⁺¹/(n+1)
2= 4x&sup4;/4 − 6x²/2 + 3x + C
3= x&sup4; − 3x² + 3x + C
Examiner tip: Never forget + C for indefinite integration. It costs you a mark every time.
5 marks
Find the area enclosed between y = x² and y = x + 2.
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Worked Solution

1Find intersections: x² = x + 2 ⇒ x² − x − 2 = 0 ⇒ (x+1)(x−2) = 0, so x = −1 and x = 2
2Area = ∫⊃−¹⊃² [(x+2) − x²] dx (upper minus lower curve)
3= ∫⊃−¹⊃² [−x² + x + 2] dx
4= [−x³/3 + x²/2 + 2x]⊃−¹⊃²
5= (−8/3 + 2 + 4) − (1/3 + 1/2 − 2) = 9/2 square units
Examiner tip: Always sketch the curves. The area between curves is always ∫(top − bottom). Check which curve is on top in the interval.

Trigonometry

3 marks
Solve 2sin(x) + 1 = 0 for 0° ≤ x ≤ 360°.
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Worked Solution

1Rearrange: sin(x) = −1/2
2Principal value: x = sin−¹(−1/2) = −30° (not in range)
3sin is negative in 3rd and 4th quadrants: x = 180 + 30 = 210° and x = 360 − 30 = 330°
Examiner tip: Draw the CAST diagram or sketch y = sin(x) to identify which quadrants give a negative result. Always check whether your answers are in the given range.
4 marks
Prove that (1 − cos²θ)/sinθ ≡ sinθ.
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Worked Solution

1Start with the left-hand side
2Use the identity sin²θ + cos²θ = 1, so 1 − cos²θ = sin²θ
3LHS = sin²θ / sinθ
4= sinθ = RHS □
Examiner tip: In proof questions, start from one side and reach the other. Never manipulate both sides simultaneously.

Series & Sequences

2 marks
A geometric series has first term 12 and common ratio 2/3. Find the sum to infinity.
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Worked Solution

1Check: |r| = 2/3 < 1, so S∞ exists
2S∞ = a/(1−r) = 12/(1−2/3) = 12 ÷ (1/3) = 36
Examiner tip: The sum to infinity only exists when |r| < 1. Always state this condition in your solution.
4 marks
Find the first three terms of the binomial expansion of (1 + 3x)&sup5;.
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Worked Solution

1Use (1 + x)⊃ = 1 + nx + n(n−1)x²/2! + ...
2Here n = 5, replace x with 3x:
3Term 1: 1
4Term 2: 5(3x) = 15x
5Term 3: 5×4/2 × (3x)² = 10 × 9x² = 90x²
6First three terms: 1 + 15x + 90x²
Examiner tip: Substitute (3x) carefully — the coefficient 3 must be raised to the same power as x.

A-Level Applied Mathematics

Statistics & Mechanics · AQA, Edexcel, OCR · Click any question to see the solution

Statistics

3 marks
X ~ B(10, 0.3). Find P(X = 4).
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Worked Solution

1Use the binomial formula: P(X = r) = &binom;n r p⊃(1−p)⊃−⊃
2P(X = 4) = &binom;10 4 (0.3)&sup4;(0.7)&sup6;
3= 210 × 0.0081 × 0.117649
4= 0.2001 (4 d.p.)
Examiner tip: Memorise the binomial formula. Show the substitution clearly before evaluating on your calculator.
4 marks
A Normal distribution has mean 60 and standard deviation 8. Find P(50 < X < 72).
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Worked Solution

1Standardise both values: Z = (X − μ)/σ
2For X = 50: Z = (50−60)/8 = −1.25
3For X = 72: Z = (72−60)/8 = 1.5
4P(−1.25 < Z < 1.5) = Φ(1.5) − Φ(−1.25) = 0.9332 − (1−0.8944) = 0.9332 − 0.1056 = 0.8276
Examiner tip: Use Φ(−z) = 1 − Φ(z) for negative z values. Draw a sketch of the normal curve to check your answer makes sense.

Mechanics

4 marks
A particle accelerates uniformly from rest. After 6 seconds its velocity is 18 m/s. Find (a) the acceleration and (b) the distance travelled.
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Worked Solution

1(a) Use v = u + at: 18 = 0 + a(6) ⇒ a = 3 m/s²
2(b) Use s = ut + ½at²: s = 0(6) + ½(3)(36) = 54 m
Examiner tip: Use the SUVAT equations. List your known and unknown quantities first: s=?, u=0, v=18, a=?, t=6.

IB Mathematics

AA SL/HL & AI SL/HL · IB Diploma Programme · Click any question to see the solution

IB Exam technique note: The IB Mathematics papers require clear communication of method, not just correct answers. “Hence” questions require you to use the result of the previous part. “Show that” questions require full working — the answer is given. Marks are awarded for the journey, not the destination.

AA SL — Functions & Calculus

6 marks (AA SL)
Let f(x) = 2x³ − 3x² − 12x + 4. (a) Find f′(x). (b) Find the x-coordinates of the stationary points. (c) Classify each stationary point.
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Worked Solution

1(a) f′(x) = 6x² − 6x − 12
2(b) Set f′(x) = 0: 6x² − 6x − 12 = 0 ⇒ x² − x − 2 = 0 ⇒ (x−2)(x+1) = 0
3Stationary points at x = 2 and x = −1
4(c) f′′(x) = 12x − 6
5At x = 2: f′′(2) = 18 > 0 ⇒ Local minimum
6At x = −1: f′′(−1) = −18 < 0 ⇒ Local maximum
Examiner tip: The second derivative test: f′′ > 0 means minimum, f′′ < 0 means maximum, f′′ = 0 is inconclusive (use first derivative test).
4 marks (AA HL)
Show that d/dx[ln(sin x)] = cot x.
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Worked Solution

1Use the chain rule: d/dx[ln(u)] = 1/u × du/dx
2Here u = sin x, so du/dx = cos x
3d/dx[ln(sin x)] = (1/sin x) × cos x
4= cos x / sin x = cot x □
Examiner tip: ‘Show that’ requires every step written out. Do not skip from ln(sin x) straight to cot x.

Key Stage 3 Mathematics

Years 7–9 · Foundation for GCSE · Click any question to see the solution

A note for KS3 students: These questions are designed to build genuine understanding, not just exam familiarity. Work through every step. If you cannot do a question, identify specifically which concept you are missing — then ask your tutor to address that concept directly.
2 marks (Year 7)
Work out 3/8 of 560.
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Worked Solution

1Divide 560 by 8 first: 560 ÷ 8 = 70
2Multiply by 3: 70 × 3 = 210
Examiner tip: Always divide by the denominator first, then multiply by the numerator. This avoids working with large numbers.
3 marks (Year 8)
Simplify: 4a² × 3a³
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Worked Solution

1Multiply the coefficients: 4 × 3 = 12
2Add the powers of a: a² × a³ = a²⊃⁺³ = a&sup5;
3Answer: 12a&sup5;
Examiner tip: When multiplying powers of the same base, add the exponents. When dividing, subtract them.
3 marks (Year 9)
A recipe for 6 people uses 450g of flour. How much flour is needed for 10 people?
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Worked Solution

1Find the amount per person: 450 ÷ 6 = 75g per person
2Multiply by 10: 75 × 10 = 750g
Examiner tip: This is a direct proportion problem. Always find the unit rate first, then scale up.

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