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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 1

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 or a circle, sketch the configuration yourself if it helps.
  1. 11 mark

    Factorise the expression x^2 + 2x - 63 fully.

    1. A (x + 9)(x - 7)
    2. B (x - 9)(x + 7)
    3. C (x - 7)(x - 9)
    4. D (x + 7)(x + 9)
  2. 21 mark

    Simplify fully: (x^2 - 16) / (x^2 + x - 12).

    1. A (x + 4)/(x + 3)
    2. B (x - 4)/(x - 3)
    3. C (x - 4)/(x + 3)
    4. D (x + 4)/(x - 3)
  3. 31 mark

    Make x the subject of the formula y = (3x + 2)/(x - 1).

    1. A x = 3/(y - 3)
    2. B x = y - 1
    3. C x = (y + 2)/(y - 3)
    4. D x = (2 - y)/(y - 3)
  4. 41 mark

    One solution of the equation 2x^2 - 5x - 3 = 0 is negative. Find the other, positive, solution.

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

    Line L has equation 2x + 3y = 12. What is the gradient of a line that is perpendicular to L?

    1. A -2/3
    2. B 3/2
    3. C -3/2
    4. D 2/3
  6. 61 mark

    A sequence begins 5, 12, 23, 38, ... . Find an expression for the nth term.

    1. A 2n^2 + n + 2
    2. B 2n^2 + n + 3
    3. C 4n^2 + n + 2
    4. D 2n^2 + 3n
  7. 71 mark

    A cuboid has edges of length 3 cm, 4 cm and 12 cm. Find the length of its longest diagonal, in cm.

    1. A 19
    2. B 5
    3. C 17
    4. D 13
  8. 81 mark

    PQ is a tangent to a circle at the point T. TR is a chord of the circle, and the angle between the tangent PQ and the chord TR is 52 degrees. Find the angle subtended by TR at a point on the major arc (the angle in the alternate segment).

    1. A 128
    2. B 38
    3. C 52
    4. D 26
  9. 91 mark

    A sector of a circle has radius 6 cm and an angle of 60 degrees at the centre. Find the exact area of the sector, in cm^2.

    1. A 36*pi
    2. B 6*pi
    3. C 12*pi
    4. D 2*pi
  10. 101 mark

    A right-angled triangle has a hypotenuse of length 8 cm and one angle of 30 degrees. Find the length of the side opposite the 30 degree angle, in cm.

    1. A 4*sqrt(3)
    2. B 16
    3. C (8*sqrt(3))/3
    4. D 4
  11. 111 mark

    OABC is a parallelogram with vertices in that order, and O is the origin. Vector OA = a and vector OC = c. M is the midpoint of AB. Find the vector OM in terms of a and c.

    1. A a + (1/2)c
    2. B (1/2)a + c
    3. C (1/2)(a + c)
    4. D a + c
  12. 121 mark

    Shape S has area 45 cm^2. Shape T is an enlargement of S with scale factor 2/3. Find the area of T, in cm^2.

    1. A 30
    2. B 67.5
    3. C 20
    4. D 101.25

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: A

    1. To factorise x^2 + 2x - 63, find two numbers that multiply to give -63 (the constant term) and add to give 2 (the coefficient of x).
    2. Since the product is negative, the two numbers have opposite signs; since the sum is positive, the number with the larger size is positive.
    3. 9 and -7 work: 9 * (-7) = -63 and 9 + (-7) = 2.
    4. So x^2 + 2x - 63 = (x + 9)(x - 7).
    5. The answer is A.
    • Why not B: Swaps which factor is positive and which is negative, giving (x - 9)(x + 7), which expands to x^2 - 2x - 63: the constant term is right but the sign of the x term is wrong.
    • Why not C: Takes both factors negative, as though the constant term were +63 rather than -63, giving (x - 7)(x - 9), which expands to x^2 - 16x + 63.
    • Why not D: Takes both factors positive, ignoring that the constant term is negative, giving (x + 7)(x + 9), which expands to x^2 + 16x + 63.

  2. Question 2Answer: B

    1. Factorise the numerator: x^2 - 16 is a difference of two squares, so it equals (x - 4)(x + 4).
    2. Factorise the denominator: two numbers multiplying to -12 and adding to 1 are 4 and -3, so x^2 + x - 12 = (x + 4)(x - 3).
    3. The factor (x + 4) is common to numerator and denominator, so it cancels (for x != -4).
    4. What remains is (x - 4)/(x - 3).
    5. The answer is B.
    • Why not A: Mis-factorises the denominator as (x - 4)(x + 3) instead of (x + 4)(x - 3) (a sign slip splitting the +1 coefficient of x into two numbers), then cancels the shared (x - 4) factor, leaving (x + 4)/(x + 3).
    • Why not C: Factorises and cancels the shared (x + 4) factor correctly, but mis-copies the remaining denominator factor as (x + 3) instead of (x - 3).
    • Why not D: Factorises and cancels correctly, but mis-copies the remaining numerator factor as (x + 4) instead of (x - 4).

  3. Question 3Answer: C

    1. Multiply both sides by (x - 1): y(x - 1) = 3x + 2.
    2. Expand the left-hand side: yx - y = 3x + 2.
    3. Collect the terms in x on one side: yx - 3x = y + 2.
    4. Factorise out x: x(y - 3) = y + 2.
    5. Divide both sides by (y - 3): x = (y + 2)/(y - 3), so the answer is C.
    • Why not A: Expands y(x - 1) as yx - 1 instead of yx - y, losing the factor of y on the constant term, and so arrives at x = 3/(y - 3).
    • Why not B: Correctly reaches x(y - 3) = y + 2, but then tries to isolate x by subtracting 3 rather than dividing by (y - 3), which is not a valid inverse operation, giving x = y - 1.
    • Why not D: Makes a sign slip moving -y to the other side of the equation, writing 2 - y instead of y + 2, giving x = (2 - y)/(y - 3).

  4. Question 4Answer: D

    1. Compare 2x^2 - 5x - 3 = 0 with ax^2 + bx + c = 0: a = 2, b = -5, c = -3.
    2. To factorise, find two numbers with product a*c = 2 * (-3) = -6 and sum b = -5: these are -6 and 1.
    3. Rewrite: 2x^2 - 6x + x - 3 = 0, so 2x(x - 3) + (x - 3) = 0, giving (2x + 1)(x - 3) = 0.
    4. So x = -1/2 or x = 3.
    5. The positive solution is 3, so the answer is D.
    • Why not A: Factorises as (2x - 1)(x + 3) = 0, which expands to 2x^2 + 5x - 3 (the wrong sign on the x term), and gives the positive root of that equation, 1/2.
    • Why not B: Factorises as (2x - 3)(x + 1) = 0, which expands to 2x^2 - x - 3 (the wrong coefficient of x), and gives the positive root of that equation, 3/2.
    • Why not C: Applies the quadratic formula correctly as far as x = (5 +/- 7)/(2a), but divides by a rather than 2a, computing (5 + 7)/2 = 6 instead of (5 + 7)/4 = 3.

  5. Question 5Answer: B

    1. Rearrange 2x + 3y = 12 into the form y = mx + c: 3y = -2x + 12, so y = (-2/3)x + 4.
    2. The gradient of L is therefore -2/3.
    3. Two gradients m1 and m2 are perpendicular exactly when m1 * m2 = -1, so the perpendicular gradient is the negative reciprocal of -2/3.
    4. The negative reciprocal of -2/3 is 3/2, since (-2/3) * (3/2) = -1.
    5. The answer is B.
    • Why not A: Gives the gradient of L itself, -2/3, rather than applying the perpendicular condition, which would instead describe a line parallel to L.
    • Why not C: Takes the reciprocal of the gradient of L, 1/(-2/3) = -3/2, but forgets that a perpendicular gradient must also change sign.
    • Why not D: Rearranges 2x + 3y = 12 with a sign slip to get y = (2/3)x + 4, and gives this (wrong) gradient of L directly as the answer, without applying the perpendicular condition at all.

  6. Question 6Answer: A

    1. Find the first differences of the sequence: 12 - 5 = 7, 23 - 12 = 11, 38 - 23 = 15.
    2. Find the second differences: 11 - 7 = 4 and 15 - 11 = 4. A constant second difference confirms a quadratic sequence.
    3. The coefficient of n^2 is half the second difference: a = 4/2 = 2.
    4. Subtracting 2n^2 from each term leaves 3, 4, 5, 6 for n = 1, 2, 3, 4, which is the linear sequence n + 2.
    5. So the nth term is 2n^2 + n + 2, and the answer is A.
    • Why not B: Sets up the simultaneous equations for the coefficients correctly, but makes an arithmetic slip solving them, getting the constant term c = 3 instead of 2.
    • Why not C: Uses the second difference itself, 4, as the coefficient of n^2, instead of halving it, giving a = 4 rather than a = 2.
    • Why not D: Makes an arithmetic slip finding the coefficient of n, getting b = 3 instead of 1, and omits the constant term entirely.

  7. Question 7Answer: D

    1. For a cuboid with edges a, b, c, the longest diagonal d satisfies d^2 = a^2 + b^2 + c^2, by Pythagoras' theorem applied in three dimensions.
    2. Here a = 3, b = 4, c = 12, so d^2 = 9 + 16 + 144 = 169.
    3. d = sqrt(169) = 13.
    4. The answer is D.
    • Why not A: Adds the three edge lengths directly, 3 + 4 + 12 = 19, instead of applying Pythagoras' theorem.
    • Why not B: Applies Pythagoras' theorem only to the 3 cm and 4 cm edges, finding the diagonal of one face (5 cm), and forgets the cuboid has a third dimension.
    • Why not C: Correctly finds the diagonal of the 3 cm by 4 cm face as 5 cm, but then adds the 12 cm edge directly, 5 + 12 = 17, instead of combining all three edges under one square root.

  8. Question 8Answer: C

    1. The alternate segment theorem states that the angle between a tangent and a chord drawn from the point of contact equals the angle subtended by that chord in the alternate segment.
    2. Here the tangent-chord angle at T is 52 degrees.
    3. So the angle subtended by TR at a point on the major arc is also 52 degrees.
    4. The answer is C.
    • Why not A: Assumes the tangent-chord angle and the angle in the alternate segment are supplementary, as with opposite angles of a cyclic quadrilateral, giving 180 - 52 = 128.
    • Why not B: Assumes the tangent-radius right angle applies here and subtracts the given angle from 90 degrees, giving 90 - 52 = 38.
    • Why not D: Confuses this with the rule relating the angle at the centre to the angle at the circumference and halves the given angle, giving 52 / 2 = 26.

  9. Question 9Answer: B

    1. The area of a sector is (theta/360) * pi * r^2, where theta is the angle at the centre in degrees.
    2. Here theta = 60 and r = 6, so r^2 = 36.
    3. The fraction of the circle is 60/360 = 1/6.
    4. Area = (1/6) * pi * 36 = 6*pi.
    5. The answer is B.
    • Why not A: Calculates the area of the whole circle, pi * 6^2 = 36*pi, and forgets to scale down by the fraction of the circle the sector occupies.
    • Why not C: Divides the angle by 180 instead of 360, as though a full turn were 180 degrees, giving (60/180) * pi * 36 = 12*pi.
    • Why not D: Uses the arc length formula, (60/360) * 2 * pi * 6, instead of the sector area formula, giving 2*pi.

  10. Question 10Answer: D

    1. In a right-angled triangle, sin(angle) = opposite/hypotenuse.
    2. Here the angle is 30 degrees and the hypotenuse is 8 cm, so the opposite side = 8 * sin(30).
    3. sin(30) = 1/2 exactly, one of the standard angles that must be known without a calculator.
    4. opposite = 8 * (1/2) = 4.
    5. The answer is D.
    • Why not A: Uses cos(30) instead of sin(30), finding the side adjacent to the 30 degree angle rather than the side opposite it: 8 * cos(30) = 8 * (sqrt(3)/2) = 4*sqrt(3).
    • Why not B: Divides by sin(30) instead of multiplying by it: 8 / sin(30) = 8 / (1/2) = 16.
    • Why not C: Uses tan(30) as though it related the opposite side to the hypotenuse, computing 8 * tan(30) = 8 * (1/sqrt(3)) = (8*sqrt(3))/3.

  11. Question 11Answer: A

    1. In parallelogram OABC (vertices in order), side AB is parallel and equal to side OC, so vector AB = c.
    2. The position vector of B is therefore OA + AB = a + c.
    3. M is the midpoint of AB, so OM = OA + (1/2) * AB = a + (1/2)c.
    4. The answer is A.
    • Why not B: Halves a instead of c, as though the midpoint condition applied to side OA rather than to side AB.
    • Why not C: Finds the midpoint of the diagonal OB instead of the midpoint of the side AB, giving (1/2)(a + c).
    • Why not D: Gives the position vector of B itself, a + c, and forgets to take the midpoint at all.

  12. Question 12Answer: C

    1. When a shape is enlarged by a linear scale factor k, its area is multiplied by k^2, not by k.
    2. Here k = 2/3, so the area scale factor is (2/3)^2 = 4/9.
    3. Area of T = 45 * 4/9 = 20.
    4. The answer is C.
    • Why not A: Applies the linear scale factor directly to the area instead of squaring it first: 45 * (2/3) = 30.
    • Why not B: Uses the reciprocal of the scale factor directly, without squaring it: 45 * (3/2) = 67.5.
    • Why not D: Uses the reciprocal of the scale factor, squared, as if going from T back to S: 45 * (3/2)^2 = 45 * (9/4) = 101.25.

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