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Admissions tests / ESAT / Maths 1 / Algebra

Demanding. 15 questions, 15 marks, about 28 minutes.

ESAT Mathematics 1: Algebra, set 3

Algebraic notation and manipulation, expanding and factorising, formulae, linear and quadratic equations, simultaneous equations, inequalities, sequences and graphs.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    Factorise fully: 4x^2 - 4x - 15.

    1. A (4x - 5)(x + 3)
    2. B (2x - 5)(2x + 3)
    3. C (2x + 5)(2x - 3)
    4. D (x - 3)(4x + 5)
  2. 21 mark

    Solve 2x^2 - 6x + 1 = 0, giving your answer in exact surd form.

    1. A x = (3 +/- sqrt(11))/2
    2. B x = 3 +/- sqrt(7)
    3. C x = (-3 +/- sqrt(7))/2
    4. D x = (3 +/- sqrt(7))/2
  3. 31 mark

    Make x the subject of the formula y = (3 - 2x)/(x + 4).

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

    Find the equation of the perpendicular bisector of the line segment joining A(-3, 1) and B(3, 5), giving your answer in the form y = mx + c.

    1. A y = (2/3)x + 3
    2. B y = (3/2)x + 3
    3. C y = -(3/2)x + 3
    4. D y = -(3/2)x + 2
  5. 51 mark

    By completing the square, find the coordinates of the turning point of y = 2x^2 - 12x + 7.

    1. A (-3, -11)
    2. B (3, -11)
    3. C (-11, 3)
    4. D (3, -2)
  6. 61 mark

    A graph is undefined at x = 0 and has two separate branches that both approach y = 0 as x tends to +infinity or -infinity. For x > 0 the branch lies entirely above the x-axis, and for x < 0 the branch lies entirely below the x-axis. Which of these could be the equation of the graph?

    1. A y = 1/x
    2. B y = -1/x
    3. C y = 1/x^2
    4. D y = x
  7. 71 mark

    A velocity-time graph shows an object decelerating uniformly from 20 m/s to 4 m/s over the first 8 seconds, then travelling at a constant 4 m/s for the next 5 seconds, before decelerating uniformly to rest over the final 2 seconds. Find the total distance travelled.

    1. A 116 m
    2. B 184 m
    3. C 124 m
    4. D 120 m
  8. 81 mark

    Solve the inequality 3(2x - 1) < 5(x + 1) - 4, giving your answer in the form x < k or x > k.

    1. A x < 4
    2. B x < 8
    3. C x < 2
    4. D x < -2
  9. 91 mark

    A sequence is defined by T(1) = 3 and T(n+1) = 2 x T(n) - 5 for n >= 1. Find T(4).

    1. A 1
    2. B 59
    3. C -11
    4. D -3
  10. 101 mark

    Simplify fully, giving your answer with positive indices: (x^(-2) y^3)^2 / (x^3 y^(-1)).

    1. A y^4/x^5
    2. B y^5/x
    3. C x y^7
    4. D y^7/x^7
  11. 111 mark

    Tickets to a school concert cost c pounds for a child and a pounds for an adult. Three child tickets and five adult tickets cost 41 pounds in total; two child tickets and three adult tickets cost 25 pounds in total. Find the cost of one adult ticket.

    1. A 7 pounds
    2. B 2 pounds
    3. C 6 pounds
    4. D 16 pounds
  12. 121 mark

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

    1. A x^2 - x - 6
    2. B x^3 - 7x - 6
    3. C x^3 + 2x^2 - 5x - 6
    4. D x^3 + 2x^2 - 7x - 6
  13. 131 mark

    The kinetic energy of an object is given by E = (1/2)mv^2. A second object has half the mass and three times the speed of the first. Find the ratio of the second object's kinetic energy to the first object's kinetic energy, E2 : E1.

    1. A 3 : 2
    2. B 9 : 1
    3. C 9 : 4
    4. D 9 : 2
  14. 141 mark

    A line has equation 3x + 4y = 12. Find the equation, in the form y = mx + c, of the line parallel to this one that passes through the point (8, -1).

    1. A y = (4/3)x - 35/3
    2. B y = -(4/3)x + 29/3
    3. C y = -(3/4)x + 5
    4. D y = -(3/4)x - 7
  15. 151 mark

    A distance-time graph is a straight line passing through the points (2, 15) and (6, 35), where distance d is measured in km and time t in hours. Find the equation of the line in the form d = mt + c.

    1. A d = 5t + 5
    2. B d = -5t + 25
    3. C d = 5t + 15
    4. D d = 5t + 35

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. For 4x^2 - 4x - 15, look for factors of the form (2x + p)(2x + q) where p and q multiply to -15 and combine so the outer and inner products sum to -4x.
    2. Try p = -5 and q = 3: (2x - 5)(2x + 3) expands to 4x^2 + 6x - 10x - 15 = 4x^2 - 4x - 15, which matches.
    3. So 4x^2 - 4x - 15 factorises as (2x - 5)(2x + 3), option B.
    • Why not A: Splits the x^2 coefficient as 4 x 1 instead of 2 x 2, pairing 4x with -5 and x with +3; expanding gives 4x^2 + 7x - 15, with the wrong middle term.
    • Why not C: Keeps the correct 2 x 2 split but assigns the minus sign to the wrong bracket, giving (2x + 5)(2x - 3); expanding gives 4x^2 + 4x - 15, the wrong sign on the middle term.
    • Why not D: Splits the x^2 coefficient as 1 x 4 the other way round and swaps which bracket is negative, giving (x - 3)(4x + 5); expanding gives 4x^2 - 7x - 15, still the wrong middle term.

  2. Question 2Answer: D

    1. For 2x^2 - 6x + 1 = 0, use the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / (2a) with a = 2, b = -6, c = 1.
    2. Compute the discriminant: b^2 - 4ac = 36 - 4(2)(1) = 36 - 8 = 28, so sqrt(28) = sqrt(4 x 7) = 2sqrt(7).
    3. Substitute: x = (6 +/- 2sqrt(7)) / 4. Dividing numerator and denominator by 2 gives x = (3 +/- sqrt(7))/2, option D.
    • Why not A: Adds 4ac inside the discriminant instead of subtracting it, computing b^2 + 4ac = 36 + 8 = 44 instead of b^2 - 4ac = 28, giving sqrt(44) = 2sqrt(11) and the wrong answer (3 +/- sqrt(11))/2.
    • Why not B: Uses the coefficient a alone as the denominator instead of 2a, dividing (6 +/- 2sqrt(7)) by 2 instead of 4, missing the extra factor of 2.
    • Why not C: Keeps b = -6 instead of using -b = 6 in the numerator of the quadratic formula, giving (-6 +/- 2sqrt(7))/4 = (-3 +/- sqrt(7))/2.

  3. Question 3Answer: A

    1. Start with y = (3 - 2x)/(x + 4) and multiply both sides by (x + 4): y(x + 4) = 3 - 2x.
    2. Expand the left-hand side: yx + 4y = 3 - 2x.
    3. Collect the x terms on one side: yx + 2x = 3 - 4y, so x(y + 2) = 3 - 4y.
    4. Divide both sides by (y + 2): x = (3 - 4y)/(y + 2), option A.
    • Why not B: Makes a sign error moving the 4y term across the equation, adding it to the right-hand side instead of subtracting, giving x(y + 2) = 3 + 4y instead of 3 - 4y.
    • Why not C: Forgets to multiply the 4 inside the bracket by y when expanding y(x + 4), keeping it as +4 instead of +4y, which collapses most of the equation and leaves x = -1/(y + 2).
    • Why not D: Makes a sign error collecting the x terms, subtracting 2x from both sides instead of adding it, giving x(y - 2) = 3 - 4y instead of x(y + 2) = 3 - 4y.

  4. Question 4Answer: C

    1. Find the gradient of AB: m = (5 - 1)/(3 - (-3)) = 4/6 = 2/3.
    2. The perpendicular bisector has gradient equal to the negative reciprocal: -1/(2/3) = -3/2.
    3. Find the midpoint of AB: ((-3 + 3)/2, (1 + 5)/2) = (0, 3).
    4. Use the midpoint and the perpendicular gradient in y - 3 = -(3/2)(x - 0), which rearranges to y = -(3/2)x + 3, option C.
    • Why not A: Uses the gradient of AB itself (2/3) instead of finding the perpendicular gradient, keeping the correct midpoint but the wrong slope.
    • Why not B: Takes the reciprocal of AB's gradient (2/3) to get 3/2, but forgets to also flip its sign, using +3/2 instead of the correct -3/2.
    • Why not D: Finds the correct perpendicular gradient but makes a sign error averaging the y-coordinates, computing (5 - 1)/2 = 2 instead of (5 + 1)/2 = 3 for the midpoint's y-value.

  5. Question 5Answer: B

    1. Factor out the coefficient of x^2 from the first two terms: y = 2(x^2 - 6x) + 7.
    2. Complete the square inside the bracket: x^2 - 6x = (x - 3)^2 - 9, so y = 2[(x - 3)^2 - 9] + 7 = 2(x - 3)^2 - 18 + 7 = 2(x - 3)^2 - 11.
    3. Since the coefficient of the squared bracket is positive, this is a minimum, occurring where the bracket is zero, i.e. at x = 3, giving y = -11.
    4. The turning point is (3, -11), option B.
    • Why not A: Writes the completed square as 2(x + 3)^2 - 11 instead of 2(x - 3)^2 - 11, flipping the sign of the x-coordinate of the turning point.
    • Why not C: Reads the coordinates out of the completed-square form 2(x - 3)^2 - 11 in the wrong order, writing the turning point as (-11, 3) instead of (3, -11).
    • Why not D: Forgets to multiply the correction term by the factored-out 2, writing 2(x - 3)^2 - 9 + 7 instead of 2(x - 3)^2 - 18 + 7, giving a y-coordinate of -2 instead of -11.

  6. Question 6Answer: A

    1. A graph undefined at x = 0 with two separate branches approaching y = 0 as x tends to +/- infinity, positive for x > 0 and negative for x < 0, is the standard shape of a reciprocal function.
    2. y = 1/x has exactly this behaviour: for x > 0, y = 1/x > 0, and for x < 0, y = 1/x < 0, with a vertical asymptote at x = 0 and a horizontal asymptote at y = 0.
    3. So the correct function is y = 1/x, option A.
    • Why not B: Reflects the graph in the x-axis, giving negative y for positive x and positive y for negative x, the opposite of the branches described.
    • Why not C: Squares the denominator, which makes y positive for every nonzero x, so both branches lie above the x-axis instead of one above and one below.
    • Why not D: Has no asymptote and is defined at x = 0, so it cannot produce two separate branches either side of a vertical asymptote there.

  7. Question 7Answer: D

    1. Phase 1 (0 to 8 s): the object decelerates uniformly from 20 m/s to 4 m/s, so this section is a trapezium of area (1/2)(20 + 4)(8) = (1/2)(24)(8) = 96 m.
    2. Phase 2 (8 to 13 s): constant speed 4 m/s for 5 s, a rectangle of area 4 x 5 = 20 m.
    3. Phase 3 (13 to 15 s): uniform deceleration from 4 m/s to rest, a triangle of area (1/2)(4)(2) = 4 m.
    4. Total distance = 96 + 20 + 4 = 120 m, option D.
    • Why not A: Correctly calculates the first two phases (96 m and 20 m) but forgets to include the distance covered during the final 2 seconds of deceleration, omitting the last triangle entirely.
    • Why not B: Treats the first phase as travelling at the constant initial speed of 20 m/s for the whole 8 seconds (a rectangle of area 160 m) instead of a trapezium, ignoring that the object was decelerating throughout that phase.
    • Why not C: Treats the final phase as travelling at the constant speed of 4 m/s for the full 2 seconds (a rectangle of area 8 m) instead of decelerating uniformly to rest, which should give a triangle of area 4 m.

  8. Question 8Answer: A

    1. Expand both sides: 3(2x - 1) = 6x - 3, and 5(x + 1) - 4 = 5x + 5 - 4 = 5x + 1.
    2. So the inequality becomes 6x - 3 < 5x + 1.
    3. Subtract 5x from both sides: x - 3 < 1. Add 3 to both sides: x < 4.
    4. The solution is x < 4, option A.
    • Why not B: Expands 5(x + 1) - 4 by forgetting to subtract the 4, treating the right-hand side as just 5x + 5 instead of 5x + 1, giving 6x - 3 < 5x + 5 and so x < 8.
    • Why not C: Distributes 3(2x - 1) only over the first term inside the bracket, treating it as 6x - 1 instead of 6x - 3, giving 6x - 1 < 5x + 1 and so x < 2.
    • Why not D: Makes a sign error collecting the constant terms, computing 1 - 3 = -2 instead of 1 + 3 = 4 when moving the -3 across the inequality, giving x < -2.

  9. Question 9Answer: C

    1. T(1) = 3.
    2. T(2) = 2 x T(1) - 5 = 2(3) - 5 = 1.
    3. T(3) = 2 x T(2) - 5 = 2(1) - 5 = -3.
    4. T(4) = 2 x T(3) - 5 = 2(-3) - 5 = -11, option C.
    • Why not A: Misreads the recurrence as applying the rule once directly to the first term, computing T(4) = 2 x T(1) - 5 = 1, instead of stepping through T(2) and T(3) first.
    • Why not B: Uses the rule T(n+1) = 2 x T(n) + 5 instead of 2 x T(n) - 5, adding 5 at each step instead of subtracting it, which compounds into a much larger final value.
    • Why not D: Stops one step too early, reporting T(3) instead of continuing the recurrence one more step to reach T(4).

  10. Question 10Answer: D

    1. Square everything inside the bracket: (x^(-2) y^3)^2 = x^(-2 x 2) y^(3 x 2) = x^(-4) y^6.
    2. Divide by x^3 y^(-1) using the index law for division: x^(-4-3) y^(6-(-1)) = x^(-7) y^7.
    3. Write with positive indices: x^(-7) y^7 = y^7/x^7, option D.
    • Why not A: Forgets to apply the outer power of 2 to the bracket (x^(-2) y^3)^2, using x^(-2) y^3 unchanged before dividing, giving x^(-2-3) y^(3+1) = y^4/x^5.
    • Why not B: Correctly squares the bracket to get x^(-4) y^6, but then adds the exponents when dividing instead of subtracting, giving x^(-4+3) y^(6-1) = y^5/x.
    • Why not C: Drops the negative sign inside the bracket before squaring, treating x^(-2) as x^2, so (x^2 y^3)^2 = x^4 y^6; dividing by x^3 y^(-1) then gives x^(4-3) y^(6+1) = x y^7.

  11. Question 11Answer: A

    1. Let c = the cost of one child ticket and a = the cost of one adult ticket. The two equations are 3c + 5a = 41 and 2c + 3a = 25.
    2. Multiply the first equation by 2 and the second by 3 to match the coefficient of c: 6c + 10a = 82 and 6c + 9a = 75.
    3. Subtract the second from the first: 10a - 9a = 82 - 75, so a = 7.
    4. Substituting a = 7 into 2c + 3a = 25 gives 2c + 21 = 25, so c = 2, confirming the system; the adult ticket costs 7 pounds, option A.
    • Why not B: Solves the simultaneous equations correctly but reports the cost of a child ticket (2 pounds) instead of the adult ticket the question asks for.
    • Why not C: Makes an arithmetic slip in the elimination step, computing 82 - 75 as 6 instead of 7.
    • Why not D: Subtracts the two original equations directly without first scaling them to match a coefficient, then misreads the resulting equation c + 2a = 16 as if it already gave the value of a alone.

  12. Question 12Answer: B

    1. First expand (x + 2)(x - 3): x^2 - 3x + 2x - 6 = x^2 - x - 6.
    2. Multiply this by the remaining bracket (x + 1): (x^2 - x - 6)(x + 1) = x^3 + x^2 - x^2 - x - 6x - 6.
    3. Combine like terms: the x^2 and -x^2 terms cancel, and -x - 6x = -7x, leaving x^3 - 7x - 6, option B.
    • Why not A: Expands the first two brackets correctly to x^2 - x - 6, but forgets to multiply by the third bracket (x + 1) at all, leaving the answer as a quadratic.
    • Why not C: Makes a sign error expanding the first two brackets, combining -3x and +2x as +x instead of -x, giving x^2 + x - 6, and then correctly multiplies this (wrong) quadratic by (x + 1).
    • Why not D: Correctly expands (x + 2)(x - 3) as x^2 - x - 6, but when multiplying by (x + 1) incorrectly combines the two x^2 terms as +2x^2 instead of letting them cancel to zero.

  13. Question 13Answer: D

    1. E = (1/2)mv^2, so for the second object with mass m/2 and speed 3v: E2 = (1/2)(m/2)(3v)^2 = (1/2)(m/2)(9v^2).
    2. Multiply the numerical factors: (1/2)(1/2)(9) = 9/4, so E2 = (9/4)mv^2.
    3. Since E1 = (1/2)mv^2 = (2/4)mv^2, the ratio E2 : E1 = (9/4) : (2/4) = 9 : 2, option D.
    • Why not A: Forgets to square the speed factor, using the tripled speed as a single multiplier (3) rather than 3^2, combined with the halved-mass factor: (1/2) x 3 = 3/2.
    • Why not B: Correctly squares the speed factor (3^2 = 9) but forgets to also apply the halved-mass factor, ignoring that the second object's mass is only half as large.
    • Why not C: Mistakenly squares the mass factor as well as the speed factor, computing (1/2)^2 x 3^2 = (1/4) x 9 = 9/4, when only the speed is squared in the formula, not the mass.

  14. Question 14Answer: C

    1. Rearrange 3x + 4y = 12 into gradient form: 4y = 12 - 3x, so y = 3 - (3/4)x. The gradient is -3/4.
    2. A line parallel to this one has the same gradient, -3/4.
    3. Using the point (8, -1): y - (-1) = -(3/4)(x - 8), so y + 1 = -(3/4)x + 6.
    4. Rearranging gives y = -(3/4)x + 5, option C.
    • Why not A: Uses the perpendicular gradient (4/3) instead of the parallel gradient (-3/4), since a parallel line must share the same gradient as the original.
    • Why not B: Misreads the gradient of the original line by inverting which coefficient goes on top, using -4/3 instead of -3/4 when rearranging 3x + 4y = 12.
    • Why not D: Correctly finds the parallel gradient (-3/4), but substitutes the point with the wrong sign, using (x + 8) instead of (x - 8) in the point-gradient form.

  15. Question 15Answer: A

    1. Find the gradient: m = (35 - 15)/(6 - 2) = 20/4 = 5.
    2. Use the point (2, 15) to find the intercept: 15 = 5(2) + c, so c = 15 - 10 = 5.
    3. The equation of the line is d = 5t + 5, option A.
    • Why not B: Computes the gradient with the coordinates swapped, using (15 - 35)/(6 - 2) instead of (35 - 15)/(6 - 2), getting -5 instead of 5, then finds the intercept using this wrong gradient.
    • Why not C: Correctly finds the gradient (5) but then uses the d-value at t = 2 directly as the y-intercept, forgetting to subtract the m x t term (5 x 2 = 10) first.
    • Why not D: Correctly finds the gradient (5) but misreads the point (6, 35) as directly giving the y-intercept, using c = 35 instead of solving 35 = 5(6) + c for c.

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