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Admissions tests / TMUA / Paper 1 / GCSE algebra and geometry

Test standard. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: GCSE algebra and geometry, set 2

Higher-tier GCSE algebra and geometry: algebraic manipulation, sequences, graphs, mensuration, circle theorems, Pythagoras, trigonometry and vectors.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator is allowed.
  • Each question has exactly one correct answer.
  • No diagrams are provided. Where a question describes a shape, graph or configuration, sketch it yourself if it helps.
  1. 11 mark

    Evaluate 8^(-2/3), giving your answer as a fraction.

    1. A 4
    2. B 1/4
    3. C -4
    4. D 1/64
  2. 21 mark

    A quantity is given by the formula s = ut + (1/2)at^2. Given that u = 3, a = -2 and t = 4, find the value of s.

    1. A 28
    2. B 4
    3. C -20
    4. D -4
  3. 31 mark

    Expand and simplify fully: (2x - 3)(3x + 5) - (x - 1)^2.

    1. A 5x^2 + 3x - 16
    2. B 5x^2 - x - 14
    3. C 5x^2 + x - 16
    4. D 5x^2 + x - 14
  4. 41 mark

    The curve y = x^2 - 4x and the line y = 2x - 5 intersect at two points. Find the product of the x-coordinates of the two points of intersection.

    1. A 6
    2. B -15
    3. C 5
    4. D -5
  5. 51 mark

    Write y = x^2 - 6x + 5 in the form (x - p)^2 - q by completing the square, and hence state the coordinates of the turning point of the curve.

    1. A (-3, -4)
    2. B (3, 5)
    3. C (3, 4)
    4. D (3, -4)
  6. 61 mark

    A particle's speed-time graph is a straight line from (0, 4) to (10, 24), where speed is measured in metres per second and time in seconds. Find the distance travelled in the first 10 seconds, in metres.

    1. A 280
    2. B 140
    3. C 120
    4. D 240
  7. 71 mark

    Solve the inequality 5 - 3x <= 2x + 20.

    1. A x <= -3
    2. B x >= 3
    3. C x >= -3
    4. D x <= 3
  8. 81 mark

    The exterior angle of a regular polygon is 24 degrees. Find the sum of its interior angles, in degrees.

    1. A 2340
    2. B 156
    3. C 3960
    4. D 2700
  9. 91 mark

    ABCD is a cyclic quadrilateral. Angle A = (3x + 10) degrees and angle C = (2x + 30) degrees are opposite angles of the quadrilateral. Find the value of x.

    1. A 64
    2. B 28
    3. C 36
    4. D 20
  10. 101 mark

    In triangle ABC, point D lies on AB and point E lies on AC, with DE parallel to BC. AD = 6 cm, DB = 4 cm and DE = 9 cm. Find the length of BC, in cm.

    1. A 5.4
    2. B 13
    3. C 6
    4. D 15
  11. 111 mark

    A cone has radius 6 cm and height 10 cm. Find its volume, in cm^3, in terms of pi.

    1. A 120*pi
    2. B 480*pi
    3. C 200*pi
    4. D 360*pi
  12. 121 mark

    The bearing of point Q from point P is 072 degrees. Find the bearing of point P from Q, in degrees.

    1. A 108
    2. B 162
    3. C 252
    4. D 72

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: B

    1. A fractional exponent m/n means: take the nth root, then raise to the power m (or the other order, which gives the same result).
    2. 8^(1/3) = 2, since 2^3 = 8.
    3. So 8^(2/3) = (8^(1/3))^2 = 2^2 = 4.
    4. The negative sign in the exponent means take the reciprocal: 8^(-2/3) = 1/(8^(2/3)) = 1/4.
    5. The answer is B.
    • Why not A: Drops the negative sign in the exponent entirely, computing 8^(2/3) instead of 8^(-2/3).
    • Why not C: Treats the negative sign as making the whole result negative rather than as an instruction to take a reciprocal, giving -(8^(2/3)) = -4.
    • Why not D: Applies the -2 part of the exponent (8^(-2) = 1/64) but forgets to also apply the /3 part, so never takes the cube root.

  2. Question 2Answer: D

    1. Substitute u = 3, a = -2 and t = 4 into s = ut + (1/2)at^2.
    2. ut = 3 * 4 = 12.
    3. t^2 = 4^2 = 16, so (1/2)at^2 = 0.5 * (-2) * 16 = -16.
    4. s = 12 + (-16) = -4.
    5. The answer is D.
    • Why not A: Forgets the negative sign on a, substituting a = 2 instead of a = -2, giving s = 12 + 0.5(2)(16) = 28.
    • Why not B: Mistakes t^2 for 2t, using 2(4) = 8 in place of 4^2 = 16, giving s = 12 + 0.5(-2)(8) = 4.
    • Why not C: Omits the 1/2 coefficient on the second term, giving s = 12 + (-2)(16) = -20.

  3. Question 3Answer: A

    1. Expand (2x - 3)(3x + 5): 2x * 3x + 2x * 5 - 3 * 3x - 3 * 5 = 6x^2 + 10x - 9x - 15 = 6x^2 + x - 15.
    2. Expand (x - 1)^2 = x^2 - 2x + 1.
    3. Subtract: 6x^2 + x - 15 - (x^2 - 2x + 1) = 6x^2 + x - 15 - x^2 + 2x - 1.
    4. Collect like terms: (6x^2 - x^2) + (x + 2x) + (-15 - 1) = 5x^2 + 3x - 16.
    5. The answer is A.
    • Why not B: Distributes the minus sign onto only the first term of the expansion of (x - 1)^2, leaving the -2x and +1 terms with their original signs instead of flipping all three, so subtracts x^2 + 2x - 1 rather than x^2 - 2x + 1.
    • Why not C: Makes a sign slip expanding (2x - 3)(3x + 5), combining the cross terms as -10x + 9x = -x instead of 10x - 9x = x, giving 6x^2 - x - 15 before the subtraction.
    • Why not D: Treats (x - 1)^2 as a difference of two squares, x^2 - 1, omitting the middle term -2x entirely.

  4. Question 4Answer: C

    1. At an intersection, x^2 - 4x = 2x - 5.
    2. Rearranging: x^2 - 4x - 2x + 5 = 0, so x^2 - 6x + 5 = 0.
    3. This factorises as (x - 1)(x - 5) = 0, so the two x-coordinates are 1 and 5.
    4. Alternatively, for ax^2 + bx + c = 0 the product of the roots is c/a = 5/1 = 5.
    5. The product of the x-coordinates is 1 * 5 = 5, so the answer is C.
    • Why not A: Finds the sum of the two x-coordinates (which is 6, from -b/a) rather than their product.
    • Why not B: Finds the product of the corresponding y-coordinates instead of the x-coordinates: the two points are (1, -3) and (5, 5), and -3 * 5 = -15.
    • Why not D: Makes a sign error transposing the constant, writing x^2 - 6x - 5 = 0 instead of x^2 - 6x + 5 = 0, which would give a product of roots of -5.

  5. Question 5Answer: D

    1. x^2 - 6x + 5 = (x - 3)^2 - 3^2 + 5, since (x - 3)^2 = x^2 - 6x + 9.
    2. So x^2 - 6x + 5 = (x - 3)^2 - 9 + 5 = (x - 3)^2 - 4.
    3. This is of the form (x - p)^2 - q with p = 3 and q = 4.
    4. The turning point of y = (x - p)^2 - q is at (p, -q), so the turning point is (3, -4).
    5. The answer is D.
    • Why not A: Mishandles the sign when completing the square, taking the vertex x-coordinate directly as half of -6 without the sign flip that the (x - p)^2 form requires, giving p = -3 instead of p = 3.
    • Why not B: Correctly finds p = 3, but forgets to subtract the extra square term (9) that completing the square introduces, simply replacing x^2 - 6x with (x - 3)^2 and leaving the original +5 unchanged.
    • Why not C: Correctly finds p = 3 and the adjustment -9 + 5 = -4, but drops the negative sign when stating the y-coordinate of the turning point.

  6. Question 6Answer: B

    1. Distance travelled is the area under a speed-time graph.
    2. The region between t = 0 and t = 10 is a trapezium with parallel sides of length 4 and 24 (the speeds at the two ends) and width 10 (the time).
    3. Area of a trapezium = (1/2)(sum of parallel sides)(width) = (1/2)(4 + 24)(10).
    4. = (1/2)(28)(10) = 140.
    5. The answer is B.
    • Why not A: Adds the two speeds and multiplies by the time, (4 + 24) * 10 = 280, forgetting to halve the sum as the trapezium area formula requires.
    • Why not C: Assumes the particle starts at rest, treating the graph as a straight line from (0, 0) to (10, 24) and computing (1/2)(24)(10) = 120, which ignores the initial speed of 4.
    • Why not D: Treats the final speed as constant throughout, computing 24 * 10 = 240 instead of accounting for the steady increase from 4 to 24.

  7. Question 7Answer: C

    1. Start with 5 - 3x <= 2x + 20.
    2. Add 3x to both sides: 5 <= 5x + 20.
    3. Subtract 20 from both sides: -15 <= 5x.
    4. Divide both sides by 5 (a positive number, so the inequality direction is unchanged): -3 <= x, which is the same as x >= -3.
    5. The answer is C.
    • Why not A: Rearranges instead to -5x <= 15 (moving both x-terms to the left), which is valid, but then divides both sides by -5 without reversing the inequality sign, giving x <= -3 instead of x >= -3.
    • Why not B: Reaches -15 <= 5x correctly but drops the negative sign dividing -15 by 5, writing x >= 3 instead of x >= -3.
    • Why not D: Reaches -5x <= 15 correctly, by subtracting 2x from both sides and then subtracting 5, but then drops the negative sign on the x term itself, writing 5x <= 15 in place of -5x <= 15, and so gets x <= 3.

  8. Question 8Answer: A

    1. The exterior angles of a polygon sum to 360 degrees, and for a regular polygon each exterior angle is 360/n, where n is the number of sides.
    2. Here 360/n = 24, so n = 360/24 = 15.
    3. The sum of the interior angles of an n-sided polygon is (n - 2) * 180.
    4. With n = 15, this is (15 - 2) * 180 = 13 * 180 = 2340.
    5. The answer is A.
    • Why not B: Finds only the size of one interior angle, 180 - 24 = 156, rather than the sum of all the interior angles.
    • Why not C: Mistakes the given exterior angle value, 24, for the number of sides of the polygon, computing (24 - 2) * 180 = 3960.
    • Why not D: Correctly finds the number of sides as 15, but forgets to subtract 2 in the interior-angle-sum formula, computing 15 * 180 = 2700 instead of 13 * 180.

  9. Question 9Answer: B

    1. Opposite angles of a cyclic quadrilateral sum to 180 degrees.
    2. So (3x + 10) + (2x + 30) = 180.
    3. 5x + 40 = 180.
    4. 5x = 140, so x = 28.
    5. The answer is B.
    • Why not A: Uses 360 degrees instead of 180 degrees for the sum of this one pair of opposite angles, solving (3x + 10) + (2x + 30) = 360.
    • Why not C: Omits subtracting the constants 10 and 30 before dividing, solving 5x = 180 instead of 5x = 140.
    • Why not D: Treats opposite angles of the cyclic quadrilateral as equal (as in a parallelogram) rather than supplementary, solving 3x + 10 = 2x + 30.

  10. Question 10Answer: D

    1. Since DE is parallel to BC, triangle ADE is similar to triangle ABC.
    2. AB = AD + DB = 6 + 4 = 10.
    3. The scale factor from the smaller triangle to the larger one is AB/AD = 10/6 = 5/3.
    4. Corresponding sides scale by the same factor, so BC = DE * (5/3) = 9 * (5/3) = 15.
    5. The answer is D.
    • Why not A: Uses the reciprocal ratio AD/AB instead of AB/AD, scaling DE down rather than up: 9 * (6/10) = 5.4.
    • Why not B: Assumes BC exceeds DE by the length DB and simply adds the two lengths directly, 9 + 4 = 13, instead of scaling by a ratio.
    • Why not C: Uses the ratio of DB to AD instead of AB to AD, computing 9 * (4/6) = 6.

  11. Question 11Answer: A

    1. The volume of a cone is (1/3) * pi * r^2 * h.
    2. Here r = 6 and h = 10, so r^2 = 36.
    3. Volume = (1/3) * pi * 36 * 10 = (1/3) * 360 * pi = 120*pi.
    4. The answer is A.
    • Why not B: Uses the diameter (12 cm) in place of the radius, computing (1/3) * pi * 12^2 * 10 = 480*pi.
    • Why not C: Squares the height instead of the radius, computing (1/3) * pi * 6 * 10^2 = 200*pi.
    • Why not D: Uses the formula for the volume of a cylinder instead of a cone, forgetting the factor of 1/3: pi * 6^2 * 10 = 360*pi.

  12. Question 12Answer: C

    1. A back bearing (the bearing in the reverse direction) is found by adding 180 degrees if the given bearing is less than 180 degrees, or subtracting 180 degrees if it is 180 degrees or more.
    2. Here the bearing of Q from P is 072 degrees, which is less than 180.
    3. So the bearing of P from Q is 072 + 180 = 252 degrees.
    4. The answer is C.
    • Why not A: Applies the rule for when the given bearing is at least 180 degrees (subtract 180) rather than the rule for a bearing under 180 degrees (add 180), and then drops the resulting negative sign: 072 - 180 = -108, taken as 108.
    • Why not B: Adds 90 degrees instead of 180 degrees, as though reversing a direction only required a right angle: 072 + 90 = 162.
    • Why not D: Assumes the back bearing equals the original bearing, forgetting that a bearing depends on the direction of travel.

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