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Admissions tests / ESAT / Maths 1 / Algebra

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

ESAT Mathematics 1: Algebra, set 2

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

    Which expression correctly represents 'the sum of a number x and 7, then multiplied by 3'?

    1. A 3x + 7
    2. B 3(x + 7)
    3. C x + 21
    4. D x + 10
  2. 21 mark

    Simplify fully: (x^3)^-2 x x^7.

    1. A x
    2. B x^13
    3. C x^8
    4. D 1/x^13
  3. 31 mark

    Given that y = 3x^2 - 2x + 1, find the value of y when x = -2.

    1. A 9
    2. B 41
    3. C -7
    4. D 17
  4. 41 mark

    Expand and simplify (2x - 3)^2.

    1. A 4x^2 - 9
    2. B 4x^2 + 12x + 9
    3. C 4x^2 - 12x + 9
    4. D 2x^2 - 12x + 9
  5. 51 mark

    Factorise fully: 4x^2 - 25.

    1. A (4x - 5)(4x + 5)
    2. B (2x - 5)(2x + 5)
    3. C (2x + 5)(2x + 5)
    4. D 2x(2x - 25)
  6. 61 mark

    Simplify fully: (x^2 - 9)/(2x) x (4x)/(x - 3).

    1. A 2x - 6
    2. B 4x + 12
    3. C x + 3
    4. D 2x + 6
  7. 71 mark

    Make x the subject of the formula: ax + b = cx + d.

    1. A x = (d - b)/(a - c)
    2. B x = (d - b)/(a + c)
    3. C x = (b - d)/(a - c)
    4. D x = (d + b)/(c - a)
  8. 81 mark

    The expression 4(x + a) - 8 is identical to 4x + 12 for all values of x. Find the value of a.

    1. A a = 1
    2. B a = 3
    3. C a = 5
    4. D a = 20
  9. 91 mark

    A line has equation y = -3x + 4. Find the equation of the line that is parallel to it and passes through the point (2, 1).

    1. A y = (1/3)x + 1/3
    2. B y = -3x + 7
    3. C y = -3x + 1
    4. D y = 3x - 5
  10. 101 mark

    The quadratic y = x^2 - 4x - 12 crosses the x-axis at two points. Find the x-coordinates of these points by factorising.

    1. A x = 6 or x = -2
    2. B x = 6 or x = 2
    3. C x = -3 or x = 4
    4. D x = -6 or x = 2
  11. 111 mark

    A function has the following outputs: when x = 1, y = 6; when x = 2, y = 3; when x = 3, y = 2; when x = 6, y = 1. Which type of function is this?

    1. A Linear function
    2. B Quadratic function
    3. C Exponential function
    4. D Reciprocal function
  12. 121 mark

    A hiker's distance-time graph shows they cover a total of 20 km while walking, taking 4 hours of actual walking time spread across a journey that lasts 5 hours in total, including rest stops. What is the hiker's average speed for the whole 5-hour journey?

    1. A 5 km/h
    2. B 1/4 km/h
    3. C 4 km/h
    4. D 20 km/h
  13. 131 mark

    Find the two values of x that satisfy the simultaneous equations y = x^2 - 2x and y = 3x - 6.

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

    Solve x^2 + 4x + 1 = 0 by completing the square, giving your answer in the form x = -2 +/- sqrt(k). Find the value of k.

    1. A k = 5
    2. B k = 3
    3. C k = 12
    4. D k = 4
  15. 151 mark

    A sequence is defined by the term-to-term rule: each term after the first is found by multiplying the previous term by 2 and then subtracting 3. The first term is 5. Find the third term.

    1. A 7
    2. B 29
    3. C 3
    4. D 11

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. The instruction is: first add x and 7 to get the sum (x + 7), then multiply that whole sum by 3.
    2. Multiplying a sum by a number means multiplying the entire bracket, not just one term inside it.
    3. So the expression that represents this is 3(x + 7), option B.
    • Why not A: Multiplies only the x-term by 3 and leaves the 7 unscaled, as if the instruction were '3 times x, then add 7' rather than '3 times the whole sum'.
    • Why not C: Multiplies only the 7 by 3 to get 21 but leaves the x term unscaled, giving x + 21 instead of scaling the entire sum by 3.
    • Why not D: Misreads 'multiplied by 3' as 'add 3', combining 7 and 3 into 10 instead of multiplying the whole sum by 3.

  2. Question 2Answer: A

    1. Apply the power of a power rule to the bracket: (x^3)^-2 = x^(3 x -2) = x^-6.
    2. Multiply by x^7 using the rule x^a x x^b = x^(a+b): x^-6 x x^7 = x^(-6+7) = x^1.
    3. So (x^3)^-2 x x^7 simplifies to x, option A.
    • Why not B: Forgets the negative sign on the outer power, treating (x^3)^-2 as x^6 (as if the power were +2), then adds exponents: x^6 x x^7 = x^13.
    • Why not C: Applies the power of a power rule as addition instead of multiplication, treating (x^3)^-2 as x^(3-2) = x^1, then adds exponents: x^1 x x^7 = x^8.
    • Why not D: Correctly finds (x^3)^-2 = x^-6, but then subtracts the exponents instead of adding when multiplying by x^7, giving x^(-6-7) = x^-13 = 1/x^13.

  3. Question 3Answer: D

    1. Substitute x = -2 into each term of y = 3x^2 - 2x + 1 separately.
    2. First term: 3x^2 = 3 x (-2)^2 = 3 x 4 = 12.
    3. Second term: -2x = -2 x (-2) = 4.
    4. Add all three terms with the +1: 12 + 4 + 1 = 17, so y = 17, option D.
    • Why not A: Makes a sign error on the middle term, treating -2x at x = -2 as -2 x 2 = -4 instead of -2 x (-2) = +4, giving 12 - 4 + 1 = 9.
    • Why not B: Squares the whole term 3x instead of just x, treating 3x^2 as (3x)^2 = 36 at x = -2, giving 36 + 4 + 1 = 41.
    • Why not C: Squares -2 as if a negative squared stays negative, using (-2)^2 = -4 instead of +4, giving 3 x (-4) + 4 + 1 = -7.

  4. Question 4Answer: C

    1. Write (2x - 3)^2 as (2x - 3)(2x - 3) and expand fully.
    2. Multiply each term: 2x x 2x = 4x^2, 2x x (-3) = -6x, (-3) x 2x = -6x, (-3) x (-3) = 9.
    3. Combine the two middle terms: -6x - 6x = -12x.
    4. Add everything together: 4x^2 - 12x + 9, option C.
    • Why not A: Treats (2x - 3)^2 as a difference of two squares, squaring each term separately and dropping the middle term entirely, giving 4x^2 - 9 instead of expanding the bracket properly.
    • Why not B: Squares each term correctly but makes a sign error on the cross term, treating (-3)(2x) as positive, giving +12x instead of -12x.
    • Why not D: Forgets to square the coefficient 2 inside the bracket, using 2x^2 instead of (2x)^2 = 4x^2, while getting the rest of the expansion right.

  5. Question 5Answer: B

    1. 4x^2 - 25 is a difference of two squares, since 4x^2 = (2x)^2 and 25 = 5^2.
    2. A difference of two squares a^2 - b^2 factorises as (a - b)(a + b), so here a = 2x and b = 5.
    3. This gives (2x - 5)(2x + 5), option B. Expanding checks: (2x)(2x) + (2x)(5) - (5)(2x) - (5)(5) = 4x^2 - 25.
    • Why not A: Takes the square root of the x^2 term but forgets to also take the square root of the coefficient 4, using 4x instead of 2x in each bracket.
    • Why not C: Uses two positive signs, forgetting that a difference of two squares needs one positive and one negative factor, so this expands to a perfect square trinomial rather than back to 4x^2 - 25.
    • Why not D: Factors out only the term 2x as a common factor without recognising the difference-of-two-squares structure; this is not even a valid factorisation, since 2x does not divide -25.

  6. Question 6Answer: D

    1. Factorise x^2 - 9 as a difference of two squares: x^2 - 9 = (x - 3)(x + 3).
    2. The expression becomes [(x - 3)(x + 3)/(2x)] x [4x/(x - 3)].
    3. Cancel the common factor (x - 3), and cancel x from 4x and 2x, leaving (x + 3) x 2.
    4. Multiply out: 2(x + 3) = 2x + 6, option D.
    • Why not A: Cancels the wrong bracket, treating (x + 3) as the common factor to cancel instead of (x - 3), leaving (x - 3) uncancelled and giving 2(x - 3) = 2x - 6.
    • Why not B: Cancels the (x - 3) factor correctly but simplifies 4x/(2x) as 4 instead of 2, forgetting that both the 4 and the x share a common factor with the 2x, giving 4(x + 3) = 4x + 12.
    • Why not C: Simplifies 4x/(2x) as 1 instead of 2, cancelling the coefficient entirely rather than reducing it, leaving just (x + 3).

  7. Question 7Answer: A

    1. Start with ax + b = cx + d and collect all the x-terms on one side: subtract cx from both sides to get ax - cx + b = d.
    2. Subtract b from both sides: ax - cx = d - b.
    3. Factor out x on the left: x(a - c) = d - b.
    4. Divide both sides by (a - c): x = (d - b)/(a - c), option A.
    • Why not B: Collects the x-terms on one side correctly in principle but combines a and c by adding rather than subtracting when factoring x out, giving (a + c)x instead of (a - c)x.
    • Why not C: Moves the constant terms to the wrong side, subtracting d and adding b in the wrong order, giving b - d in the numerator instead of d - b.
    • Why not D: Makes sign errors on both sides at once: adds b instead of subtracting it, and swaps the order in the denominator to c - a instead of a - c.

  8. Question 8Answer: C

    1. Expand the left-hand side: 4(x + a) - 8 = 4x + 4a - 8.
    2. Compare the constant terms with the right-hand side, since the 4x terms already match: 4a - 8 = 12.
    3. Add 8 to both sides: 4a = 20.
    4. Divide by 4: a = 5, option C.
    • Why not A: Moves the -8 to the other side with the wrong sign, computing 4a = 12 - 8 = 4 instead of 4a = 12 + 8 = 20, giving a = 1.
    • Why not B: Ignores the -8 term completely when comparing constants, as if 4(x + a) - 8 simplified to just 4x + 4a, giving 4a = 12 and a = 3.
    • Why not D: Forgets that a is multiplied by 4 inside the bracket, and instead treats the equation as a - 8 = 12, giving a = 20 directly without dividing by 4.

  9. Question 9Answer: B

    1. Parallel lines have the same gradient, so the new line also has gradient -3.
    2. Use the point (2, 1) in y = -3x + c to find c: 1 = -3(2) + c, so 1 = -6 + c.
    3. Solve for c: c = 1 + 6 = 7.
    4. The equation of the parallel line is y = -3x + 7, option B.
    • Why not A: Uses the negative reciprocal of the gradient, 1/3, which is the gradient needed for a perpendicular line, instead of keeping the same gradient -3 that a parallel line requires, then solves 1 = (1/3)(2) + c for c, giving c = 1/3.
    • Why not C: Substitutes the point's y-coordinate directly as the new y-intercept without doing the algebra, using c = 1 instead of solving 1 = -3(2) + c for c.
    • Why not D: Drops the negative sign from the gradient throughout, using +3 instead of -3, so the line produced is not parallel to the original at all; solving 1 = 3(2) + c for c then gives c = -5.

  10. Question 10Answer: A

    1. To factorise x^2 - 4x - 12, find two numbers that multiply to -12 and add to -4.
    2. Those numbers are -6 and 2, since -6 x 2 = -12 and -6 + 2 = -4.
    3. So x^2 - 4x - 12 = (x - 6)(x + 2).
    4. Setting each factor to zero gives x = 6 or x = -2, option A.
    • Why not B: Finds the correct pair of factors but drops the negative sign on the second root, giving x = 2 instead of x = -2.
    • Why not C: Chooses a factor pair of -12 (namely -3 and 4) that multiplies correctly but does not check that the pair also adds to -4, since -3 + 4 = 1, not -4.
    • Why not D: Flips the sign of both roots, effectively factorising as (x + 6)(x - 2) instead of (x - 6)(x + 2), which expands to x^2 + 4x - 12, not the original expression.

  11. Question 11Answer: D

    1. Check the product of each pair of x and y values: 1 x 6 = 6, 2 x 3 = 6, 3 x 2 = 6, 6 x 1 = 6.
    2. Since x times y is always equal to the same constant, 6, this means y = 6/x, which is a reciprocal function.
    3. So the correct answer is D, a reciprocal function.
    • Why not A: Notices that y decreases steadily as x increases and assumes a constant rate of change (a straight line), without checking whether the product xy stays constant rather than the difference between values.
    • Why not B: Sees the values curving downward and assumes this must be part of a downward parabola, without checking that the relationship is multiplicative (xy = 6) rather than following a squared pattern.
    • Why not C: Sees a rapidly shrinking sequence of y-values and assumes exponential decay, without checking that x and y multiply to the same constant each time rather than y being repeatedly scaled by a fixed factor.

  12. Question 12Answer: C

    1. Average speed over the whole journey is found using total distance divided by total time, including any rest stops.
    2. The total distance travelled is 20 km, and the total time for the whole journey is 5 hours.
    3. Average speed = 20 km / 5 hours = 4 km/h, option C.
    • Why not A: Divides the total distance by only the walking time (4 hours), ignoring the rest stops, instead of dividing by the whole 5-hour journey.
    • Why not B: Inverts the calculation, dividing the total time by the total distance (5/20) instead of the distance by the time, giving a speed with the units the wrong way round.
    • Why not D: Works out the length of the rest period as 5 - 4 = 1 hour, then mistakenly divides the total distance by this 1 hour instead of by the whole 5-hour journey.

  13. Question 13Answer: A

    1. Set the two expressions for y equal to each other: x^2 - 2x = 3x - 6.
    2. Rearrange so everything is on one side: x^2 - 2x - 3x + 6 = 0, which simplifies to x^2 - 5x + 6 = 0.
    3. Factorise: we need two numbers that multiply to 6 and add to -5, which are -2 and -3, so x^2 - 5x + 6 = (x - 2)(x - 3).
    4. Setting each factor to zero gives x = 2 or x = 3, option A.
    • Why not B: Combines the two x-terms with a sign error, treating -2x - 3x as -2x + 3x = x instead of -5x, which leads to solving x^2 + x - 6 = 0 rather than x^2 - 5x + 6 = 0.
    • Why not C: Moves the constant term to the wrong side, keeping it as -6 instead of flipping its sign to +6, which leads to solving x^2 - 5x - 6 = 0 rather than x^2 - 5x + 6 = 0.
    • Why not D: Flips the sign of both correct roots throughout, as if the factorised form were (x + 2)(x + 3) instead of (x - 2)(x - 3).

  14. Question 14Answer: B

    1. Complete the square: x^2 + 4x + 1 = (x + 2)^2 - 4 + 1 = (x + 2)^2 - 3.
    2. Set this equal to zero: (x + 2)^2 - 3 = 0, so (x + 2)^2 = 3.
    3. Take the square root of both sides: x + 2 = sqrt(3) or x + 2 = -sqrt(3), so x = -2 +/- sqrt(3).
    4. Comparing with the form x = -2 +/- sqrt(k), the value of k is 3, option B.
    • Why not A: Uses b^2 + 4ac inside the calculation instead of b^2 - 4ac, effectively adding the constant term's contribution instead of subtracting it, which corresponds to a discriminant of 20 rather than 12 and gives k = 5 instead of k = 3.
    • Why not C: Correctly reaches a discriminant of 12 but stops at the unsimplified sqrt(12) instead of simplifying it, reading off k = 12 instead of recognising that sqrt(12) = 2 x sqrt(3), which corresponds to k = 3 in the required form.
    • Why not D: Forgets to add back the original constant term (+1) after completing the square, leaving (x + 2)^2 - 4 = 0 instead of (x + 2)^2 - 3 = 0, giving k = 4 instead of k = 3.

  15. Question 15Answer: D

    1. The rule is: multiply the previous term by 2, then subtract 3, to get the next term.
    2. Starting from the first term, 5, find the second term: 2 x 5 - 3 = 7.
    3. Find the third term by applying the rule again to the second term: 2 x 7 - 3 = 11.
    4. So the third term is 11, option D.
    • Why not A: Applies the rule only once from the first term and reports the resulting second term, 7, as if it were the third term, instead of applying the rule twice.
    • Why not B: Adds 3 instead of subtracting it at each step: the second term becomes 2 x 5 + 3 = 13, and the third term becomes 2 x 13 + 3 = 29.
    • Why not C: Misapplies the term-to-term rule as if it were a position-to-term rule, substituting the position number 3 directly into 2n - 3 to get 2 x 3 - 3 = 3, instead of applying the rule step by step from the first term.

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