About this A-Level Math practice test

These practice questions mirror the roughly two-thirds Pure and one-third Applied split tested across AQA, Edexcel, and OCR specifications, covering core Pure topics like algebra, trigonometry, and calculus alongside Statistics and Mechanics. Each worked solution lays out the full line-by-line method, since that's exactly what earns the method marks A-Level examiners are trained to award.

A-Level Mathematics Practice Test sample questions

These starter questions help you launch a math mock test quickly. Swap them with your own worksheet, notebook, or textbook questions any time.

1. 1. Algebra and functions: The function f is defined by f(x) = (2x - 1)/(x + 3), x ∈ ℝ, x ≠ -3. Find f⁻¹(x), stating its domain.

2. 2. Algebra and functions: Given that p(x) = 2x³ - x² - 7x + 2, show that (x - 2) is a factor of p(x), and hence solve the equation p(x) = 0 completely, giving your answers in exact (surd) form where necessary.

3. 3. Algebra and functions: Solve the inequality |2x - 3| < |x + 1|.

4. 4. Coordinate geometry: A circle C has equation x² + y² - 6x + 4y - 12 = 0. Find the coordinates of the centre and the radius of C, and determine whether the point (7, 1) lies inside, outside, or on the circle.

5. 5. Coordinate geometry: Points A(-2, 5) and B(4, -1) are the endpoints of a line segment. Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.

6. 6. Coordinate geometry: The line y = x + k is a tangent to the curve y = x² - 4x + 10. Find the value of k.

7. 7. Sequences and series: The 5th term of an arithmetic sequence is 23 and the 12th term is 58. Find the first term and the common difference, and hence find the sum of the first 20 terms of the sequence.

8. 8. Sequences and series: A geometric series has first term 8 and common ratio 3/4. Find the sum to infinity of the series, and find the smallest number of terms, n, for which the sum of the first n terms exceeds 99% of the sum to infinity.

9. 9. Sequences and series: Find the value of Σ (r=1 to 15) (3r - 2).

10. 10. Trigonometry: Solve, for 0 ≤ x < 360°, the equation 5 sin x - 3 cos x = 2, giving your answers to 1 decimal place.

11. 11. Trigonometry: Prove the identity (1 - cos 2θ)/(sin 2θ) ≡ tan θ.

12. 12. Trigonometry: Solve, for 0 ≤ x < 2π, the equation sin 2x = cos x, giving your answers as exact multiples of π.

13. 13. Exponentials and logarithms: Solve the equation 5^(2x-1) = 3^(x+2), giving your answer to 3 significant figures.

14. 14. Exponentials and logarithms: Solve the equation log₂(x + 3) - log₂(x - 1) = 2.

15. 15. Exponentials and logarithms: The number of bacteria in a culture is modelled by N = N₀e^(kt), where t is the time in hours. Initially there are 200 bacteria, and after 3 hours there are 650 bacteria. Find the value of k to 3 significant figures, and hence find the time taken for the population to reach 5000.

16. 16. Differentiation and integration: Find dy/dx in simplified form, given that y = x² ln(3x), and find the exact gradient of the curve at x = 1.

17. 17. Differentiation and integration: Using the substitution u = x² + 1, find the exact value of ∫ (from 0 to 2) x(x² + 1)³ dx.

18. 18. Differentiation and integration: A curve satisfies the differential equation dy/dx = 2y/(x + 1), for x > -1, and passes through the point (1, 4). Find y in terms of x.

19. 19. Vectors: Points A and B have position vectors a = 2i - 3j + k and b = 5i + j - 2k relative to origin O. Find the vector AB, its magnitude, and a unit vector in the direction of AB.

20. 20. Vectors: Points C and D have position vectors c = 4i + 2j - k and d = i - 2j + 2k. Find the angle between OC and OD, giving your answer in degrees to 1 decimal place.

21. 21. Statistical sampling and data: A researcher wants a sample of 40 students from a school roll of 850 students, listed alphabetically. Describe how systematic sampling could be used to select the sample, and state one advantage of this method over simple random sampling in this context.

22. 22. Statistical sampling and data: The diameters (mm) of a sample of 8 ball bearings are: 12.1, 12.3, 11.9, 12.0, 12.4, 12.2, 11.8, 12.3. Calculate the mean and the standard deviation of this sample.

23. 23. Probability distributions: A biased coin has P(Head) = 0.3 and is tossed 10 times. Find the probability of obtaining exactly 4 heads, and the probability of obtaining at least 2 heads. Give your answers to 3 significant figures.

24. 24. Probability distributions: The heights of adult males in a population are modelled by X ~ N(178, 6.5²) (in cm). Find P(X > 185), and find the height, to 1 decimal place, that is exceeded by only 10% of adult males.

25. 25. Hypothesis testing: A machine dispenses liquid into bottles with amounts normally distributed with standard deviation 8 ml and mean 500 ml. A manager suspects the machine is over-filling and tests, at the 5% significance level, whether the mean amount dispensed has increased, using a random sample of 25 bottles with sample mean 503.2 ml. Carry out the hypothesis test, stating your hypotheses and conclusion clearly.

26. 26. Kinematics: A particle moving in a straight line with constant acceleration passes through point A with speed 4 m/s. Six seconds later it passes through point B with speed 19 m/s. Find the acceleration of the particle and the distance AB.

27. 27. Kinematics: A ball is thrown vertically upwards from ground level with initial speed 21 m/s. Using g = 9.8 m/s² and ignoring air resistance, find the greatest height reached by the ball and the total time taken for it to return to the ground.

28. 28. Forces and Newton's laws: A box of mass 15 kg rests on a rough horizontal floor, where the coefficient of friction between the box and the floor is 0.4. A horizontal force of 70 N is applied to the box. Show that the box moves, and find its acceleration. (Take g = 9.8 m/s².)

29. 29. Forces and Newton's laws: Particles P (mass 5 kg) and Q (mass 3 kg) are connected by a light inextensible string passing over a smooth fixed pulley at the edge of a smooth horizontal table. P lies on the table and Q hangs freely below the pulley. Find the acceleration of the system and the tension in the string. (Take g = 9.8 m/s².)

30. 30. Moments: A uniform beam AB has length 4 m and weight 200 N. It rests horizontally on two supports, one at end A and one at point C which is 3 m from A. A load of 150 N is placed at end B. Find the reaction forces at the two supports.

Answer key & quick explanations

Short answers for the sample questions above. Use this to self-check before generating a fresh AI-built mock test.

1. 1. f⁻¹(x) for f(x)=(2x-1)/(x+3)

   f⁻¹(x) = (1 + 3x)/(2 - x), domain x ≠ 2

   Let y = (2x-1)/(x+3) and multiply through: y(x+3) = 2x-1, so yx + 3y = 2x - 1. Collect x terms: x(y-2) = -1-3y, giving x = (1+3y)/(2-y). Swapping x and y gives f⁻¹(x) = (1+3x)/(2-x); the domain excludes x=2 since this is the value f(x) can never take (a horizontal asymptote).

2. 2. Factorise p(x)=2x³-x²-7x+2

   x = 2, x = (-3+√17)/4, x = (-3-√17)/4

   Substituting x=2 gives p(2) = 16 - 4 - 14 + 2 = 0, confirming (x-2) is a factor by the factor theorem. Dividing p(x) by (x-2) gives the quotient 2x²+3x-1, so p(x) = (x-2)(2x²+3x-1). Solving 2x²+3x-1=0 by the quadratic formula gives x = (-3±√17)/4, since the discriminant is 9+8=17.

3. 3. Solve |2x-3| < |x+1|

   2/3 < x < 4

   Since both sides are non-negative, squaring preserves the inequality: (2x-3)² < (x+1)². Expanding gives 4x²-12x+9 < x²+2x+1, i.e. 3x²-14x+8 < 0, which factorises as (x-4)(3x-2) < 0. This is a positive quadratic that is negative between its roots, so 2/3 < x < 4.

4. 4. Circle x²+y²-6x+4y-12=0, point (7,1)

   Centre (3, -2), radius 5; the point (7,1) lies exactly on the circle

   Completing the square: (x-3)² - 9 + (y+2)² - 4 - 12 = 0, so (x-3)² + (y+2)² = 25, giving centre (3,-2) and radius 5. The distance from the centre to (7,1) is √[(7-3)²+(1+2)²] = √(16+9) = √25 = 5, which equals the radius, so the point lies on the circle.

5. 5. Perpendicular bisector of A(-2,5), B(4,-1)

   y = x + 1

   The midpoint of AB is ((-2+4)/2, (5-1)/2) = (1,2). The gradient of AB is (-1-5)/(4+2) = -1, so the perpendicular gradient is 1 (negative reciprocal). Using y - 2 = 1(x-1) gives y = x + 1.

6. 6. Tangent y=x+k to y=x²-4x+10

   k = 15/4

   Substituting y=x+k into the curve's equation gives x+k = x²-4x+10, i.e. x²-5x+(10-k)=0. For tangency this quadratic must have a repeated root, so the discriminant is zero: 25 - 4(10-k) = 0. Solving gives 4k = 15, so k = 15/4.

7. 7. Arithmetic sequence, a5=23, a12=58

   a = 3, d = 5; S₂₀ = 1010

   Using a+4d=23 and a+11d=58, subtracting gives 7d=35, so d=5, and then a = 23-4(5) = 3. The sum of the first 20 terms is S₂₀ = (20/2)[2(3)+19(5)] = 10(6+95) = 10 × 101 = 1010.

8. 8. Geometric series a=8, r=3/4

   S∞ = 32; smallest n = 17

   The sum to infinity is S∞ = a/(1-r) = 8/(1/4) = 32. We need Sₙ = 32(1-(3/4)ⁿ) > 0.99 × 32, i.e. (3/4)ⁿ < 0.01; taking logs gives n > ln(0.01)/ln(0.75) ≈ 16.01. Checking n=16 gives (0.75)¹⁶ ≈ 0.01002 (just above 0.01), while n=17 gives (0.75)¹⁷ ≈ 0.00752 (below 0.01), so the smallest integer value is n = 17.

9. 9. Σ(r=1 to 15)(3r-2)

   330

   Split the sum: Σ(3r-2) = 3Σr - 2(15). Using Σr = 15(16)/2 = 120, this gives 3(120) - 30 = 360 - 30 = 330.

10. 10. Solve 5sinx - 3cosx=2, 0≤x<360°

    x = 51.0°, 190.9°

    Write 5sinx - 3cosx in the form R sin(x-α), where R=√(5²+3²)=√34 and tanα=3/5, giving α≈30.96°. The equation becomes √34 sin(x-α) = 2, so sin(x-α) = 2/√34 ≈ 0.3430, giving x-α ≈ 20.06° or 159.94°. Adding α to each gives x ≈ 51.0° and x ≈ 190.9° within the given range.

11. 11. Prove (1-cos2θ)/sin2θ ≡ tanθ

    Identity proved using double-angle formulae

    Using cos2θ = 1-2sin²θ, the numerator becomes 1-(1-2sin²θ) = 2sin²θ. Using sin2θ = 2sinθcosθ for the denominator, the expression becomes 2sin²θ/(2sinθcosθ) = sinθ/cosθ = tanθ, as required.

12. 12. Solve sin2x=cosx, 0≤x<2π

    x = π/6, π/2, 5π/6, 3π/2

    Using sin2x = 2sinxcosx, the equation becomes 2sinxcosx = cosx, i.e. cosx(2sinx-1)=0. Either cosx=0, giving x=π/2 or 3π/2, or sinx=1/2, giving x=π/6 or 5π/6.

13. 13. Solve 5^(2x-1)=3^(x+2)

    x ≈ 1.80 (3 s.f.)

    Taking natural logs of both sides gives (2x-1)ln5 = (x+2)ln3. Expanding and collecting x terms: x(2ln5 - ln3) = 2ln3 + ln5. Substituting ln5≈1.6094 and ln3≈1.0986 gives x = 3.8067/2.1203 ≈ 1.80 to 3 significant figures.

14. 14. Solve log2(x+3)-log2(x-1)=2

    x = 7/3

    Using the quotient law, log₂[(x+3)/(x-1)] = 2, so (x+3)/(x-1) = 2² = 4. This gives x+3 = 4x-4, so 3x = 7, x = 7/3, which satisfies x>1 so is a valid solution.

15. 15. Bacteria growth N=N0 e^(kt)

    k ≈ 0.393 per hour; t ≈ 8.19 hours

    Using N(3)=650=200e^(3k) gives e^(3k)=3.25, so 3k=ln3.25 and k≈0.393 (3 s.f.). To reach 5000: 5000=200e^(kt) gives e^(kt)=25, so kt=ln25≈3.2189, and t = 3.2189/0.392885 ≈ 8.19 hours.

16. 16. Differentiate y=x²ln(3x)

    dy/dx = 2x ln(3x) + x; gradient at x=1 is 1+2ln3 ≈ 3.20

    Using the product rule with u=x², v=ln(3x): du/dx=2x and dv/dx=1/x (since d/dx[ln(3x)] = 1/x by the chain rule). So dy/dx = 2x ln(3x) + x²(1/x) = 2x ln(3x) + x. At x=1, dy/dx = 2ln3 + 1 ≈ 3.20.

17. 17. ∫₀² x(x²+1)³ dx by substitution

    78

    With u=x²+1, du=2x dx, so x dx = du/2; the limits become u=1 (x=0) and u=5 (x=2). The integral becomes (1/2)∫₁⁵ u³ du = (1/2)[u⁴/4] from 1 to 5 = (1/8)(625-1) = 78.

18. 18. Solve dy/dx=2y/(x+1), through (1,4)

    y = (x+1)²

    Separating variables: (1/y)dy = 2/(x+1) dx. Integrating both sides gives ln|y| = 2ln|x+1| + C, so y = A(x+1)² for constant A. Substituting (1,4) gives 4=4A, so A=1, giving y=(x+1)².

19. 19. Vector AB, a=2i-3j+k, b=5i+j-2k

    AB = 3i+4j-3k; |AB| = √34; unit vector = (3i+4j-3k)/√34

    AB = b - a = (5-2)i + (1-(-3))j + (-2-1)k = 3i+4j-3k. Its magnitude is √(3²+4²+(-3)²) = √34. Dividing the vector by its magnitude gives the unit vector (3i+4j-3k)/√34.

20. 20. Angle between OC, OD

    θ ≈ 98.4°

    The scalar product is c·d = (4)(1)+(2)(-2)+(-1)(2) = 4-4-2 = -2. The magnitudes are |c|=√21 and |d|=√9=3, so cosθ = -2/(3√21) ≈ -0.1455. Taking arccos gives θ ≈ 98.4°; the negative cosine confirms the angle is obtuse.

21. 21. Systematic sampling, 40 from 850

    Sampling interval k = 850/40 ≈ 21 (round to nearest whole number)

    The sampling interval is k = 850/40 = 21.25, rounded to 21; a random starting point between 1 and 21 is chosen from the list, and then every 21st student thereafter is selected until 40 are obtained. This method is quicker and simpler to carry out than simple random sampling while still spreading the sample evenly across the whole population list.

22. 22. Mean and sd of ball bearing diameters

    Mean = 12.125 mm; standard deviation ≈ 0.198 mm

    The mean is found from Σx/n = 97.0/8 = 12.125 mm. The sum of squared deviations from the mean is 0.315, so the variance is 0.315/8 = 0.039375, and the standard deviation is √0.039375 ≈ 0.198 mm.

23. 23. X~B(10,0.3): P(X=4), P(X≥2)

    P(X=4) ≈ 0.200; P(X≥2) ≈ 0.851

    Using P(X=4) = C(10,4)(0.3)⁴(0.7)⁶ = 210 × 0.0081 × 0.117649 ≈ 0.200. For at least 2 heads, use the complement: P(X≥2) = 1 - P(X=0) - P(X=1) = 1 - (0.7)¹⁰ - 10(0.3)(0.7)⁹ ≈ 1 - 0.0282 - 0.1211 = 0.851.

24. 24. X~N(178,6.5²): P(X>185), 10% threshold

    P(X>185) ≈ 0.141; height ≈ 186.3 cm

    Standardising, z = (185-178)/6.5 ≈ 1.077, and using the standard normal distribution, P(Z>1.077) ≈ 0.141. For the top 10%, we need z such that P(Z>z)=0.10, which is z=1.2816 from tables; the corresponding height is 178+1.2816(6.5) ≈ 186.3 cm.

25. 25. Hypothesis test, mean fill 503.2 ml

    Reject H0; there is significant evidence the mean has increased

    The hypotheses are H0: μ=500 against H1: μ>500 (one-tailed test). The test statistic is z = (503.2-500)/(8/√25) = 3.2/1.6 = 2.0, which exceeds the critical value of 1.6449 at the 5% significance level. Since 2.0 > 1.6449, we reject H0 and conclude there is sufficient evidence that the mean amount dispensed has increased.

26. 26. SUVAT: u=4, v=19, t=6

    a = 2.5 m/s²; AB = 69 m

    Using v=u+at: 19=4+6a, giving a=2.5 m/s². The distance is found using s=((u+v)/2)t = ((4+19)/2)(6) = 11.5×6 = 69 m.

27. 27. Ball thrown up at 21 m/s

    Greatest height = 22.5 m; total time = 4.29 s

    At maximum height v=0, so using v²=u²-2gh: 0=21²-2(9.8)h gives h=441/19.6=22.5 m. The total time to return to the ground (by symmetry, or using v=u-gt with v=-21 at landing) is t=2u/g=2(21)/9.8≈4.29 s.

28. 28. Box, μ=0.4, mass 15kg, force 70N

    Box moves; a ≈ 0.747 m/s²

    The normal reaction is R=mg=15(9.8)=147 N, so the maximum friction force is F=μR=0.4(147)=58.8 N. Since the applied force 70 N exceeds this, the box moves, and Newton's second law gives ma=70-58.8=11.2, so a=11.2/15≈0.747 m/s².

29. 29. Connected particles P(5kg), Q(3kg), pulley

    a = 3.675 m/s²; T = 18.375 N

    For Q (hanging): 3g - T = 3a. For P (on the smooth table): T = 5a. Adding these equations to eliminate T: 3g = 8a, so a = 3(9.8)/8 = 3.675 m/s², and T = 5(3.675) = 18.375 N.

30. 30. Beam moments: RA, RC

    R_A ≈ 16.7 N; R_C ≈ 333.3 N

    Taking moments about A: R_C(3) = 200(2) + 150(4), since the weight acts at the midpoint (2 m from A) and the load acts at B (4 m from A). This gives R_C = 1000/3 ≈ 333.3 N. Resolving vertically, R_A + R_C = 350, so R_A = 350 - 1000/3 = 50/3 ≈ 16.7 N.

Curriculum Mapping & Learning Guide

Use this breakdown to identify which skills each question tests and guide post-test review.

A-Level Math units covered

1. Chapter 1: Pure: Algebra and functions
2. Chapter 2: Pure: Coordinate geometry
3. Chapter 3: Pure: Sequences and series
4. Chapter 4: Pure: Trigonometry
5. Chapter 5: Pure: Exponentials and logarithms
6. Chapter 6: Pure: Differentiation and integration
7. Chapter 7: Pure: Vectors
8. Chapter 8: Statistics: Statistical sampling and data
9. Chapter 9: Statistics: Probability distributions (Binomial, Normal)
10. Chapter 10: Statistics: Hypothesis testing
11. Chapter 11: Mechanics: Kinematics
12. Chapter 12: Mechanics: Forces and Newton's Laws
13. Chapter 13: Mechanics: Moments

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