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Admissions tests / ESAT / Maths 1 / Algebra

Test standard. 15 questions, 15 marks, about 22 minutes.

ESAT Mathematics 1: Algebra, set 1

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

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

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

    Simplify fully: 5(2x - 3) + 4x.

    1. A 30x - 15
    2. B 14x + 15
    3. C 14x - 15
    4. D 14x - 3
  2. 21 mark

    Simplify fully: (2x^3)^2 / x^4.

    1. A 2x^2
    2. B 4x^2
    3. C x^2
    4. D 4x^10
  3. 31 mark

    A formula used in mechanics is s = ut + (1/2)at^2, where s is distance, u is initial speed, a is acceleration and t is time. Find s when u = 3, a = 4 and t = 2.

    1. A 22
    2. B 10
    3. C 20
    4. D 14
  4. 41 mark

    Expand and simplify (3x - 2)(2x + 5).

    1. A 6x^2 + 11x - 10
    2. B 6x^2 - 11x - 10
    3. C 6x^2 + 19x - 10
    4. D 6x^2 + 11x + 10
  5. 51 mark

    Factorise fully: 2x^2 + 7x + 3.

    1. A (2x + 1)(x + 3)
    2. B (2x + 3)(x + 1)
    3. C (2x - 1)(x - 3)
    4. D (x + 1)(x + 3)
  6. 61 mark

    Simplify fully: (x^2 - 9) / (x^2 + x - 6).

    1. A (x + 3)/(x + 2)
    2. B (x - 3)/(x - 2)
    3. C (x - 3)/(x + 2)
    4. D x - 3
  7. 71 mark

    Make x the subject of the formula y = (3x + 2)/5.

    1. A x = (y - 2)/3
    2. B x = (5y - 2)/3
    3. C x = (5y + 2)/3
    4. D x = 5y/3 - 2
  8. 81 mark

    For which value of k is 4(x + 3) - 2x = 2x + k true for every value of x?

    1. A k = 6
    2. B k = 10
    3. C k = 12
    4. D k = 24
  9. 91 mark

    A line passes through the points (-2, 5) and (4, -7). Find the equation of the line that is perpendicular to this line and passes through (4, -7).

    1. A y = (1/2)x - 9
    2. B y = -2x + 1
    3. C y = (1/2)x - 11
    4. D y = -(1/2)x - 5
  10. 101 mark

    By completing the square, find the roots of x^2 - 6x + 5 = 0.

    1. A x = 7 or x = -1
    2. B No real roots
    3. C x = -1 or x = -5
    4. D x = 1 or x = 5
  11. 111 mark

    A quantity of a radioactive substance decreases over time such that it halves every fixed period. Which type of function best models the mass of the substance remaining, m, as a function of time, t?

    1. A An exponential function, m = m0 x k^t, with 0 < k < 1
    2. B A linear function, m = m0 - ct
    3. C A reciprocal function, m = k/t
    4. D A quadratic function, m = m0 - ct^2
  12. 121 mark

    A speed-time graph shows an object accelerating uniformly from 0 m/s to 12 m/s over the first 4 seconds, then travelling at a constant 12 m/s for the next 6 seconds. What total distance does the object travel in these 10 seconds?

    1. A 72 m
    2. B 120 m
    3. C 60 m
    4. D 96 m
  13. 131 mark

    Find the two values of x that satisfy the simultaneous equations y = x + 1 and y = x^2 - 5.

    1. A x = 2 or x = -3
    2. B x = 6 or x = -1
    3. C x = 3 or x = -2
    4. D x = -6 or x = 1
  14. 141 mark

    Solve the inequality 5 - 2x >= 1, giving your answer in the form x <= k or x >= k.

    1. A x <= 2
    2. B x >= -3
    3. C x >= 2
    4. D x <= -3
  15. 151 mark

    A sequence begins 4, 9, 16, 25, 36, ... Find an expression for the nth term.

    1. A n^2 + 2n
    2. B 2n^2 + 2n + 1
    3. C n^2 + 5n - 2
    4. D n^2 + 2n + 1

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: C

    1. Distribute the 5 across both terms in the bracket: 5(2x - 3) = 5 x 2x - 5 x 3 = 10x - 15.
    2. Add the remaining 4x term: 10x - 15 + 4x = 14x - 15.
    3. So 5(2x - 3) + 4x simplifies to 14x - 15, which is option C.
    • Why not A: Adds the 4x inside the bracket before distributing, effectively computing 5(2x - 3 + 4x) instead of 5(2x - 3) + 4x.
    • Why not B: Makes a sign error when distributing, treating 5 x (-3) as +15 instead of -15.
    • Why not D: Distributes the 5 only over the first term inside the bracket (5 x 2x) and forgets to multiply the -3 term by 5 as well.

  2. Question 2Answer: B

    1. Square everything inside the bracket: (2x^3)^2 = 2^2 x^(3 x 2) = 4x^6.
    2. Divide by x^4 using the index law x^a / x^b = x^(a-b): 4x^6 / x^4 = 4x^(6-4) = 4x^2.
    3. So (2x^3)^2 / x^4 simplifies to 4x^2, option B.
    • Why not A: Squares the x^3 term correctly to get x^6 but does not square the coefficient 2, carrying it over unchanged as if (2x^3)^2 = 2x^6.
    • Why not C: Ignores the coefficient 2 altogether when squaring the bracket, as if (2x^3)^2 = x^6.
    • Why not D: Adds the exponents instead of subtracting when dividing by x^4, computing 4x^(6+4) instead of 4x^(6-4).

  3. Question 3Answer: D

    1. Substitute u = 3, a = 4, t = 2 into s = ut + (1/2)at^2.
    2. Compute the first term: ut = 3 x 2 = 6.
    3. Compute the second term: (1/2)at^2 = (1/2) x 4 x 2^2 = (1/2) x 4 x 4 = 8.
    4. Add the two terms: s = 6 + 8 = 14, so the answer is D.
    • Why not A: Omits the 1/2 coefficient on the acceleration term, computing ut + at^2 = 6 + 16 = 22 instead of ut + (1/2)at^2.
    • Why not B: Uses t instead of t^2 in the second term, computing ut + (1/2)at = 6 + 4 = 10.
    • Why not C: Misreads the formula's structure and combines u with (1/2)a before multiplying by t^2, giving (u + (1/2)a) t^2 = 5 x 4 = 20 with the ut term dropped entirely.

  4. Question 4Answer: A

    1. Multiply each term in the first bracket by each term in the second: 3x x 2x = 6x^2, 3x x 5 = 15x, -2 x 2x = -4x, -2 x 5 = -10.
    2. Combine the like (x) terms: 15x - 4x = 11x.
    3. Add everything together: 6x^2 + 11x - 10, which is option A.
    • Why not B: Combines the middle terms 15x and -4x with a sign error, getting -11x instead of +11x.
    • Why not C: Treats -4x as +4x when combining middle terms, adding 15x + 4x = 19x instead of 15x - 4x = 11x.
    • Why not D: Drops the negative sign on the -2 when multiplying the constant terms, computing 2 x 5 = +10 instead of -2 x 5 = -10.

  5. Question 5Answer: A

    1. For 2x^2 + 7x + 3, look for factors of the form (2x + p)(x + q) where p and q are factors of 3, the constant term.
    2. The factor pairs of 3 are 1 and 3. Try (2x + 1)(x + 3): expanding gives 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3, which matches.
    3. So 2x^2 + 7x + 3 factorises as (2x + 1)(x + 3), option A.
    • Why not C: Chooses negative factors of 3 even though both the x-term and constant are positive, writing (2x - 1)(x - 3); expanding this gives 2x^2 - 7x + 3, with the wrong sign on the middle term.
    • Why not D: Correctly finds the factor pair (2x + 1)(x + 3) but then drops the leading coefficient 2 from the first bracket, writing (x + 1)(x + 3) instead.
    • Why not B: Pairs the factors of 3 with the wrong terms, writing (2x + 3)(x + 1); expanding this gives 2x^2 + 5x + 3, not 2x^2 + 7x + 3.

  6. Question 6Answer: B

    1. Factorise the numerator: x^2 - 9 is a difference of two squares, so x^2 - 9 = (x - 3)(x + 3).
    2. Factorise the denominator: x^2 + x - 6 = (x + 3)(x - 2), since 3 x (-2) = -6 and 3 + (-2) = 1.
    3. Cancel the common factor (x + 3) from the top and bottom: [(x - 3)(x + 3)] / [(x + 3)(x - 2)] = (x - 3)/(x - 2).
    4. So the fully simplified fraction is (x - 3)/(x - 2), option B.
    • Why not A: Factorises the denominator with the signs swapped, as (x - 3)(x + 2) instead of (x + 3)(x - 2), then cancels the resulting common (x - 3) factor, leaving (x + 3)/(x + 2).
    • Why not C: Factorises the denominator with a sign error on the second bracket, as (x + 3)(x + 2) instead of (x + 3)(x - 2), then correctly cancels the common (x + 3) factor, leaving (x - 3)/(x + 2).
    • Why not D: Cancels the whole denominator (x + 3)(x - 2) against the (x + 3) factor in the numerator, incorrectly treating the entire denominator as a single common factor and leaving just x - 3.

  7. Question 7Answer: B

    1. Start with y = (3x + 2)/5 and multiply both sides by 5 to clear the fraction: 5y = 3x + 2.
    2. Subtract 2 from both sides: 5y - 2 = 3x.
    3. Divide both sides by 3: x = (5y - 2)/3, which is option B.
    • Why not C: Makes a sign error rearranging 5y = 3x + 2, adding 2 to both sides instead of subtracting it, giving 3x = 5y + 2.
    • Why not D: Divides only the 5y term by 3 and forgets to divide the -2 as well, giving x = 5y/3 - 2 instead of x = (5y - 2)/3.
    • Why not A: Forgets to multiply both sides by 5 first, treating y = (3x + 2)/5 as if it were simply y = 3x + 2, giving x = (y - 2)/3.

  8. Question 8Answer: C

    1. Simplify the left-hand side: 4(x + 3) - 2x = 4x + 12 - 2x = 2x + 12.
    2. For 2x + 12 = 2x + k to hold for every value of x, this is an identity, so the constant terms must match once the x terms agree.
    3. So k must equal 12, option C.
    • Why not A: Divides the constant term by 2 by mistake, perhaps trying to also halve the constant the way the x-terms combine, giving k = 6 instead of k = 12.
    • Why not D: Forgets to combine 4x and -2x into 2x on the left-hand side first, and compares 4x + 12 to 2x + k term by term, doubling the constant to k = 24.
    • Why not B: Makes an arithmetic slip expanding the bracket, computing 4 x 3 as 10 instead of 12.

  9. Question 9Answer: A

    1. Find the gradient of the original line through (-2, 5) and (4, -7): m = (-7 - 5)/(4 - (-2)) = -12/6 = -2.
    2. A line perpendicular to it has gradient equal to the negative reciprocal: m = -1/(-2) = 1/2.
    3. Use the point (4, -7) with this gradient in y - y1 = m(x - x1): y - (-7) = (1/2)(x - 4), so y + 7 = (1/2)x - 2.
    4. Rearranging gives y = (1/2)x - 9, option A.
    • Why not B: Uses the original line's gradient of -2 directly instead of finding the negative reciprocal needed for a perpendicular line.
    • Why not D: Takes the reciprocal of the original gradient but forgets to flip its sign, using -1/2 instead of the correct perpendicular gradient of 1/2.
    • Why not C: Uses the correct perpendicular gradient of 1/2 but fails to multiply the 4 inside the bracket by 1/2, keeping it as -4 instead of -2.

  10. Question 10Answer: D

    1. Complete the square: x^2 - 6x + 5 = (x - 3)^2 - 9 + 5 = (x - 3)^2 - 4.
    2. Set the expression equal to zero: (x - 3)^2 - 4 = 0, so (x - 3)^2 = 4.
    3. Take the square root of both sides: x - 3 = 2 or x - 3 = -2.
    4. Solve each case: x = 5 or x = 1, so the roots are x = 1 and x = 5, option D.
    • Why not A: Forgets to take the square root of 4 when solving (x - 3)^2 = 4, and instead sets x - 3 equal to 4 or -4 directly, giving x = 7 or x = -1.
    • Why not B: Forgets to subtract 9 when completing the square, incorrectly writing x^2 - 6x + 5 as (x - 3)^2 + 5, which gives (x - 3)^2 = -5 and wrongly concludes there are no real roots.
    • Why not C: Writes the completed square form with the wrong sign inside the bracket, as (x + 3)^2 - 4 instead of (x - 3)^2 - 4, and so solves x + 3 = 2 or x + 3 = -2 to get x = -1 or x = -5.

  11. Question 11Answer: A

    1. Halving over equal fixed periods means the mass is multiplied by the same factor, 1/2, each period, not reduced by the same fixed amount.
    2. A quantity that is repeatedly multiplied by a constant factor over equal time intervals is modelled by an exponential function, m = m0 x k^t, where here k = 1/2.
    3. So the correct model is an exponential function, option A.
    • Why not C: Mistakes 'the mass keeps halving' for inverse proportion between mass and time, which would make the mass fall in a different pattern to repeated halving over equal time periods.
    • Why not D: Mistakes the shape of a decaying exponential curve for part of a downward parabola, which would mean the rate of decrease keeps changing in a way that does not match repeated halving over equal time periods.
    • Why not B: Models the decrease as happening at a constant rate (a straight line), which would mean the mass drops by the same fixed amount each period rather than by the same fixed fraction.

  12. Question 12Answer: D

    1. During the first 4 seconds the object accelerates uniformly from 0 to 12 m/s, so this section of the graph is a triangle with area (1/2) x 4 x 12 = 24 m.
    2. During the next 6 seconds the speed is constant at 12 m/s, so this section is a rectangle with area 6 x 12 = 72 m.
    3. The total distance is the sum of the two areas: 24 + 72 = 96 m, option D.
    • Why not A: Only calculates the area of the constant-speed rectangle (6 s x 12 m/s = 72 m) and forgets to include the distance covered during the first 4 seconds of acceleration.
    • Why not B: Treats the whole 10 seconds as if the object travelled at the constant final speed of 12 m/s throughout, computing 10 x 12 = 120 m, which ignores that the object started from rest.
    • Why not C: Applies the average speed of the first phase (6 m/s, halfway between 0 and 12) to the entire 10 second interval, computing 6 x 10 = 60 m, instead of only to the first 4 seconds.

  13. Question 13Answer: C

    1. Set the two expressions for y equal to each other: x + 1 = x^2 - 5.
    2. Rearrange so everything is on one side: 0 = x^2 - x - 6, i.e. x^2 - x - 6 = 0.
    3. Factorise: we need two numbers that multiply to -6 and add to -1, which are -3 and 2, so x^2 - x - 6 = (x - 3)(x + 2).
    4. Setting each factor to zero gives x = 3 or x = -2, option C.
    • Why not B: Picks a factor pair of -6 (namely 6 and -1) that gives the right product but does not check that the pair also sums to -1, the coefficient of x; 6 and -1 sum to 5, not -1.
    • Why not D: Picks the factor pair -6 and 1, which also gives the right product but sums to -5, not -1, so does not actually satisfy the equation.
    • Why not A: Rearranges x + 1 = x^2 - 5 with a sign error on the linear term, getting x^2 + x - 6 = 0 instead of x^2 - x - 6 = 0, which factorises as (x + 3)(x - 2) giving x = -3 or x = 2.

  14. Question 14Answer: A

    1. Start with 5 - 2x >= 1 and subtract 5 from both sides: -2x >= 1 - 5, so -2x >= -4.
    2. Divide both sides by -2. Dividing an inequality by a negative number reverses its direction, so x <= 2.
    3. The solution is x <= 2, option A.
    • Why not C: Divides both sides by -2 correctly in every other respect but forgets to reverse the inequality sign when dividing by a negative number, giving x >= 2 instead of x <= 2.
    • Why not D: Makes a sign error moving the 5 across the inequality, computing -2x >= 1 + 5 = 6 instead of -2x >= 1 - 5 = -4, though the inequality is correctly reversed when dividing by -2.
    • Why not B: Combines both errors: moves the 5 with the wrong sign (getting -2x >= 6) and then also forgets to reverse the inequality sign when dividing by -2, giving x >= -3.

  15. Question 15Answer: D

    1. Find the first differences of 4, 9, 16, 25, 36: 5, 7, 9, 11.
    2. Find the second differences of those: 2, 2, 2, which is constant, confirming this is a quadratic sequence with n^2 coefficient equal to half the second difference, so a = 2/2 = 1.
    3. Subtract n^2 from each term to find what remains: 4 - 1 = 3, 9 - 4 = 5, 16 - 9 = 7, 25 - 16 = 9, 36 - 25 = 11, which is the linear sequence 3, 5, 7, 9, 11 with nth term 2n + 1.
    4. Combine the two parts: the nth term of the original sequence is n^2 + 2n + 1, option D.
    • Why not A: Correctly finds the n^2 coefficient and the linear coefficient but forgets to include the constant term, giving n^2 + 2n instead of n^2 + 2n + 1.
    • Why not C: Uses the sequence's first term and first common difference directly as the constant and linear coefficients, without first subtracting n^2 from each term to find the remaining linear sequence.
    • Why not B: Uses the full second difference (2) as the coefficient of n^2 instead of half of it, forgetting that the coefficient of n^2 in a quadratic sequence is always half the constant second difference, while still keeping the correctly found linear part 2n + 1 unchanged, giving 2n^2 + 2n + 1.

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