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Admissions tests / ESAT / Maths 1 / Algebra

Stretch. 15 questions, 15 marks, about 32 minutes.

ESAT Mathematics 1: Algebra, set 4

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

    Simplify fully: (x^4 - 16)/(x^2 - 4).

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

    Simplify fully, giving your answer with positive indices: (8x^(-3))^(2/3) x (x^5)^(-1).

    1. A 64/x^7
    2. B 4/x^3
    3. C 4x^7
    4. D 4/x^7
  3. 31 mark

    The period T (in seconds) of a pendulum of length L (in metres) is given by T = 2 x pi x sqrt(L/g), where g = 10 m/s^2. A pendulum has period T = pi seconds. Find its length L.

    1. A 2.5 m
    2. B 5 m
    3. C 10 m
    4. D 0.25 m
  4. 41 mark

    The expression (x + a)(x - 4) + b is identical to x^2 - x - 10 for all values of x. Find the value of a - b.

    1. A 13
    2. B -1
    3. C 1
    4. D -5
  5. 51 mark

    The parabola y = x^2 - 2x - 15 crosses the x-axis at two points. Its turning point is reflected in the x-axis to give point C. Find the coordinates of C.

    1. A (1, -16)
    2. B (1, 16)
    3. C (-1, 16)
    4. D (2, 15)
  6. 61 mark

    The equation sin x = 0.5 has solutions x = 30 degrees and x = 150 degrees in the range 0 <= x <= 360 degrees. Using the periodicity of the sine graph, how many solutions does sin x = 0.5 have in the range 0 <= x <= 1080 degrees?

    1. A 6
    2. B 3
    3. C 8
    4. D 4
  7. 71 mark

    A mobile phone tariff charges a fixed monthly fee plus a constant rate per minute used, so that cost = fixed fee + rate x minutes. In one month, 100 minutes of calls cost 25 pounds; in another month, 180 minutes of calls cost 37 pounds. Find the fixed monthly fee.

    1. A 0.15 pounds
    2. B 22 pounds
    3. C 23.5 pounds
    4. D 10 pounds
  8. 81 mark

    A quadratic sequence has nth term T(n) = n^2 - 9n + 20. For how many positive integer values of n is T(n) negative?

    1. A Infinitely many
    2. B 0
    3. C 2
    4. D 1
  9. 91 mark

    The line y = x + k is tangent to the curve y = x^2 - 3x + 7 (it touches the curve at exactly one point). Find the value of k.

    1. A 4.75
    2. B 11
    3. C 3
    4. D -3
  10. 101 mark

    Which of the following points satisfies both of the inequalities y > 2x - 3 and x + y <= 4?

    1. A (-2, 3)
    2. B (3, 4)
    3. C (1, -2)
    4. D (4, 1)
  11. 111 mark

    Exactly one of the following statements is a genuine identity, true for every value of x. Which one?

    1. A (x + 3)^2 = x^2 + 9
    2. B 2(x + 3) = 2x + 3
    3. C (x + 1)^2 = x^2 + 2x - 1
    4. D (x + 3)(x - 3) = x^2 - 9
  12. 121 mark

    A sequence is defined by u(1) = 3 and u(n+1) = (u(n))^2 for n >= 1. Express u(4) as a power of 3.

    1. A 3^2
    2. B 3^4
    3. C 3^8
    4. D 3^16
  13. 131 mark

    Simplify fully as a single fraction: 3/(x - 2) - 2/(x + 1).

    1. A (x - 1)/((x - 2)(x + 1))
    2. B (x + 7)/((x - 2)(x + 1))
    3. C (7 - 2x)/((x - 2)(x + 1))
    4. D (x + 7)/(x - 2)
  14. 141 mark

    The quadratic equation 2x^2 - 7x + 4 = 0 has roots alpha and beta. Without finding alpha and beta individually, find the value of alpha^2 + beta^2.

    1. A 33/4
    2. B 49/4
    3. C 41/4
    4. D 17/4
  15. 151 mark

    The value V (in pounds) of a car t complete years after purchase is modeled by V = 12500 x (0.8)^t. After how many complete years does the value first fall below 6000 pounds?

    1. A 3
    2. B 5
    3. C 4
    4. D 2

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. Recognise x^4 - 16 as a difference of two squares: x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4).
    2. The denominator is x^2 - 4, which is exactly one of the two factors of the numerator.
    3. Cancel the common factor (x^2 - 4): (x^2 - 4)(x^2 + 4)/(x^2 - 4) = x^2 + 4.
    4. So the expression simplifies fully to x^2 + 4, option B.
    • Why not A: Incorrectly splits the fraction term by term, treating (x^4 - 16)/(x^2 - 4) as x^4/x^2 - 16/4 instead of factorising first; this gives x^2 - 4, which does not equal the original expression (at x = 0 the original is (-16)/(-4) = 4, but x^2 - 4 gives -4).
    • Why not C: Factorises fully as (x - 2)(x + 2)(x^2 + 4) over (x - 2)(x + 2), cancels only the (x + 2) factor, then mistakenly treats the denominator as used up entirely, forgetting the remaining (x - 2) still cancels too.
    • Why not D: Correctly reaches (x^2 - 4)(x^2 + 4)/(x^2 - 4) = x^2 + 4, but then over-cancels, also removing the constant +4, reasoning (wrongly) that the 16 has already been accounted for.

  2. Question 2Answer: D

    1. Apply the power of a power rule: (8x^(-3))^(2/3) = 8^(2/3) x x^(-3 x 2/3) = 8^(2/3) x x^-2.
    2. Since 8^(1/3) = 2, 8^(2/3) = 2^2 = 4, so this part is 4x^-2.
    3. (x^5)^(-1) = x^-5.
    4. Multiply: 4x^-2 x x^-5 = 4x^(-2-5) = 4x^-7 = 4/x^7, option D.
    • Why not A: Computes 8^(2/3) as 8^2 = 64, forgetting to also take the cube root; it should be (8^(1/3))^2 = 2^2 = 4.
    • Why not B: Multiplies the exponent -3 by 2/3 and drops the negative sign, getting x^2 instead of x^-2, so the final power of x is -3 instead of -7.
    • Why not C: Correctly reaches 4x^-7, but converts to positive indices by only flipping the sign of the exponent, writing 4x^7 instead of taking the reciprocal to get 4/x^7.

  3. Question 3Answer: A

    1. Divide both sides by 2 x pi: T/(2 x pi) = sqrt(L/g).
    2. Square both sides: (T/(2 x pi))^2 = L/g, so L = g x T^2/(4 x pi^2).
    3. Substitute T = pi and g = 10: L = 10 x pi^2/(4 x pi^2). The pi^2 terms cancel.
    4. L = 10/4 = 2.5 m, option A.
    • Why not B: Forgets to square both sides when rearranging, using L = g x T/(2 x pi) directly instead of L = g x T^2/(2 x pi)^2, giving L = 10 x pi/(2 x pi) = 5.
    • Why not C: Squares T correctly but forgets to also square the 2 in the denominator 2 x pi, using pi^2 instead of 4 x pi^2, giving L = 10 x pi^2/pi^2 = 10.
    • Why not D: Correctly isolates T^2/(4 x pi^2) but forgets to multiply back by g at the end, giving L = pi^2/(4 x pi^2) = 0.25.

  4. Question 4Answer: C

    1. Expand the left-hand side: (x + a)(x - 4) + b = x^2 - 4x + ax - 4a + b = x^2 + (a - 4)x + (b - 4a).
    2. For this to be identical to x^2 - x - 10 for every x, match coefficients: a - 4 = -1, so a = 3.
    3. Match the constant terms: b - 4a = -10, so b - 12 = -10, giving b = 2.
    4. a - b = 3 - 2 = 1, option C.
    • Why not A: Correctly finds a = 3 from the x-coefficient, but forgets that -4a also contributes to the constant term, so sets b = -10 directly instead of solving b - 4a = -10; this gives a - b = 3 - (-10) = 13.
    • Why not B: Correctly finds a = 3 and b = 2, but then computes b - a instead of the a - b the question asks for, giving 2 - 3 = -1.
    • Why not D: Makes a sign error expanding, using the x-coefficient as -(a - 4) instead of (a - 4), so -(a-4) = -1 gives a = 5; substituting into b - 4a = -10 then gives b = 10, so a - b = 5 - 10 = -5.

  5. Question 5Answer: B

    1. Factorise x^2 - 2x - 15 = (x - 5)(x + 3), so the parabola crosses the x-axis at x = 5 and x = -3.
    2. A parabola is symmetric about the vertical line through the midpoint of its roots: x = (5 + (-3))/2 = 1.
    3. Substitute x = 1 into y = x^2 - 2x - 15: y = 1 - 2 - 15 = -16, so the turning point is (1, -16).
    4. Reflecting in the x-axis negates the y-coordinate: C = (1, 16), option B.
    • Why not A: Correctly finds the turning point (1, -16) but forgets to reflect it in the x-axis, giving the turning point itself instead of its reflection.
    • Why not C: Reflects in the y-axis instead of the x-axis, negating the x-coordinate rather than the y-coordinate.
    • Why not D: Finds the vertex's x-coordinate as the sum of the roots (5 + (-3) = 2) rather than their average, forgetting to divide by 2; substituting x = 2 into the original equation then also shifts the y-value, and reflecting gives (2, 15).

  6. Question 6Answer: A

    1. The sine graph repeats every 360 degrees, and within each period sin x = 0.5 has exactly two solutions.
    2. The range 0 to 1080 degrees contains exactly 1080/360 = 3 complete periods.
    3. So the total number of solutions is 3 x 2 = 6: at x = 30, 150, 390, 510, 750, 870 degrees.
    4. That gives 6 solutions, option A.
    • Why not B: Assumes there is only one solution per 360-degree period (forgetting the sine graph crosses a given value twice per cycle), giving 3 periods x 1 = 3.
    • Why not C: Miscounts the number of complete 360-degree periods in 1080 degrees as 4 instead of 3, giving 4 x 2 = 8.
    • Why not D: Miscounts the number of complete 360-degree periods in 1080 degrees as 2 instead of 3, giving 2 x 2 = 4.

  7. Question 7Answer: D

    1. The rate per minute is the gradient between the two data points: (37 - 25)/(180 - 100) = 12/80 = 0.15 pounds per minute.
    2. Use the pair (100 minutes, 25 pounds): 25 = fixed fee + 0.15 x 100 = fixed fee + 15.
    3. So the fixed fee = 25 - 15 = 10 pounds, option D.
    • Why not A: Correctly calculates the per-minute rate (0.15 pounds), but gives that instead of the fixed fee the question actually asks for.
    • Why not B: Finds the correct rate of 0.15 pounds per minute, but substitutes it into the wrong data pair, computing fee = 37 - 0.15 x 100 = 22 instead of using the matching pair 25 and 100.
    • Why not C: Finds the correct rate of 0.15 pounds per minute, but makes a decimal-place slip computing 0.15 x 100 as 1.5 instead of 15, giving fee = 25 - 1.5 = 23.5.

  8. Question 8Answer: B

    1. Factorise T(n) = n^2 - 9n + 20 = (n - 4)(n - 5).
    2. Since the coefficient of n^2 is positive, this parabola opens upwards and is negative only strictly between its roots, i.e. for 4 < n < 5.
    3. There is no positive integer strictly between 4 and 5: n = 4 gives T(4) = 0 and n = 5 gives T(5) = 0, not negative.
    4. So there are 0 positive integer values of n for which T(n) is negative, option B.
    • Why not A: Assumes a quadratic with a minimum must eventually take negative values for large n, without checking where its actual roots are; since both roots are close together and positive, the sequence is never negative at an integer value of n.
    • Why not C: Believes n = 4 and n = 5 give T(n) < 0, when direct substitution shows T(4) = T(5) = 0 exactly, since these are the roots, not values where the sequence is negative.
    • Why not D: Assumes exactly one integer lies strictly between the roots 4 and 5, without checking that no integer actually lies in that open interval.

  9. Question 9Answer: C

    1. Set the line equal to the curve: x + k = x^2 - 3x + 7, which rearranges to x^2 - 4x + (7 - k) = 0.
    2. A line is tangent to a curve exactly when this equation has a repeated root, i.e. when its discriminant is zero.
    3. Discriminant: (-4)^2 - 4(1)(7 - k) = 16 - 28 + 4k = 4k - 12. Set this equal to zero: 4k = 12.
    4. So k = 3, option C.
    • Why not A: Forgets that moving the x term from the right-hand side also changes the coefficient of x on the left, keeping it as x^2 - 3x + (7-k) = 0 instead of x^2 - 4x + (7-k) = 0, giving discriminant 9 - 4(7-k) = 0 and k = 19/4 = 4.75.
    • Why not B: Uses the discriminant formula as b^2 + 4ac instead of b^2 - 4ac on the correct equation x^2 - 4x + (7-k) = 0, giving 16 + 4(7-k) = 0 and k = 11.
    • Why not D: Makes a sign error moving k across, using x^2 - 4x + (7+k) = 0 instead of x^2 - 4x + (7-k) = 0, giving discriminant 16 - 4(7+k) = 0 and k = -3.

  10. Question 10Answer: A

    1. Check each point against both inequalities.
    2. For (-2, 3): y > 2x - 3 gives 3 > 2(-2) - 3 = -7, which is true. x + y <= 4 gives -2 + 3 = 1 <= 4, which is true. Both hold.
    3. The other three points each fail at least one inequality, as detailed above.
    4. So (-2, 3), option A, is the only point satisfying both inequalities.
    • Why not B: Satisfies y > 2x - 3 (4 > 3), but fails x + y <= 4, since 3 + 4 = 7 is greater than 4.
    • Why not C: Satisfies x + y <= 4 (-1 <= 4), but fails y > 2x - 3, since -2 is not greater than 2(1) - 3 = -1.
    • Why not D: Fails both inequalities: y = 1 is not greater than 2(4) - 3 = 5, and x + y = 5 is greater than 4.

  11. Question 11Answer: D

    1. An identity must hold for every value of x, so check each option by expanding its left-hand side.
    2. (x + 3)(x - 3) is a difference of two squares and expands exactly to x^2 - 9 for every value of x, matching the right-hand side.
    3. The other three options each expand to something that differs from their stated right-hand side by a fixed nonzero amount, so none of them hold for every x.
    4. So (x + 3)(x - 3) = x^2 - 9, option D, is the genuine identity.
    • Why not A: (x + 3)^2 actually expands to x^2 + 6x + 9, which has an extra 6x term absent from x^2 + 9; the two sides are equal only at the single value x = 0, so this is an equation, not an identity.
    • Why not B: Distributing the 2 across the bracket correctly gives 2x + 6, not 2x + 3; the two sides differ by a constant 3 for every x, so this is never true, let alone an identity.
    • Why not C: (x + 1)^2 expands correctly to x^2 + 2x + 1, but the right-hand side has -1 instead of +1; the two sides differ by a constant 2 for every x, so this is never true either.

  12. Question 12Answer: C

    1. u(1) = 3 = 3^1.
    2. u(2) = (u(1))^2 = (3^1)^2 = 3^2.
    3. u(3) = (u(2))^2 = (3^2)^2 = 3^4.
    4. u(4) = (u(3))^2 = (3^4)^2 = 3^8, option C. Each squaring doubles the exponent: 1, 2, 4, 8.
    • Why not A: Stops after only one squaring step, treating u(2) = 3^2 as the answer instead of continuing the recurrence two more times to reach u(4).
    • Why not B: Stops after two squaring steps, treating u(3) = 3^4 as the answer instead of continuing the recurrence one more time to reach u(4).
    • Why not D: Applies one squaring step too many, computing u(5) = (3^8)^2 = 3^16 instead of stopping at u(4).

  13. Question 13Answer: B

    1. The common denominator is (x - 2)(x + 1).
    2. Rewrite each fraction over this denominator: 3/(x - 2) = 3(x + 1)/((x - 2)(x + 1)) and 2/(x + 1) = 2(x - 2)/((x - 2)(x + 1)).
    3. Subtract the numerators: 3(x + 1) - 2(x - 2) = 3x + 3 - 2x + 4 = x + 7.
    4. So the combined fraction is (x + 7)/((x - 2)(x + 1)), option B.
    • Why not A: Makes a sign error expanding -2(x - 2), treating it as -2x - 4 instead of -2x + 4, giving numerator 3x + 3 - 2x - 4 = x - 1.
    • Why not C: Forgets to scale the first fraction's numerator by (x + 1), using the numerator 3 - 2(x - 2) = 7 - 2x instead of 3(x + 1) - 2(x - 2) = x + 7.
    • Why not D: Correctly finds the numerator x + 7, but drops the (x + 1) factor from the common denominator, leaving only (x - 2).

  14. Question 14Answer: A

    1. Divide the equation by 2: x^2 - (7/2)x + 2 = 0, so alpha + beta = 7/2 and alpha x beta = 2, by comparing to x^2 - (sum)x + (product) = 0.
    2. Use the identity alpha^2 + beta^2 = (alpha + beta)^2 - 2 x alpha x beta.
    3. Substitute: (7/2)^2 - 2(2) = 49/4 - 4 = 49/4 - 16/4 = 33/4.
    4. So alpha^2 + beta^2 = 33/4, option A. This avoids solving the quadratic for its (irrational) roots directly.
    • Why not B: Computes only (alpha + beta)^2 and forgets to subtract 2 x alpha x beta at all, giving 49/4.
    • Why not C: Uses the formula (alpha + beta)^2 - alpha x beta instead of (alpha + beta)^2 - 2 x alpha x beta, subtracting the product once instead of twice: 49/4 - 2 = 41/4.
    • Why not D: Confuses the identity with (alpha - beta)^2 = (alpha + beta)^2 - 4 x alpha x beta, subtracting 4 times the product instead of 2 times: 49/4 - 8 = 17/4.

  15. Question 15Answer: C

    1. Multiply by 0.8 for each complete year: t = 1: 12500 x 0.8 = 10000. t = 2: 10000 x 0.8 = 8000. t = 3: 8000 x 0.8 = 6400. t = 4: 6400 x 0.8 = 5120.
    2. At t = 3, the value is 6400 pounds, still above 6000.
    3. At t = 4, the value is 5120 pounds, which is below 6000.
    4. So the value first falls below 6000 pounds after 4 complete years, option C.
    • Why not A: Misjudges the value at t = 3 (12500 x 0.8^3 = 6400 pounds) as already below 6000 pounds, when 6400 is still above the threshold.
    • Why not B: Makes an arithmetic slip computing the value at t = 4, concluding it is still above 6000 pounds, and so checks one year further than necessary.
    • Why not D: Wrongly assumes the value halves every year (a multiplier of 0.5 instead of 0.8), giving 12500, 6250, 3125, ... which first drops below 6000 pounds at t = 2.

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