The only common factor is 5, so 5x − 20 = 5(x − 4).
(b) [2]
Group the terms in pairs: 3ax² − 3bx − ax + b = 3x(ax − b) − (ax − b). Both groups now share (ax − b), so the answer is (ax − b)(3x − 1). Check by expanding: 3ax² − ax − 3bx + b ✓
Q2Evaluate (9/4)3/2. [2 marks]
Take the power in two steps — the half is a square root, the 3 is a cube. √(9/4) = 3/2, then (3/2)³ = 27/8. So (9/4)3/2 = 27/8 (= 3.375).
Q3Simplify x²(y⁵)²x⁻³y³. [2 marks]
Deal with the bracket first: (y⁵)² = y¹⁰. The numerator is x²y¹⁰. For the division, subtract the indices: x2 − (−3) = x⁵ and y10 − 3 = y⁷. So the answer is x⁵y⁷. (A negative index in the denominator becomes a positive index on top.)
Q4Solve 6x² + 5x − 6 = 0. [3 marks]
Factorise: 6x² + 5x − 6 = (3x − 2)(2x + 3). (Check the middle term: 9x − 4x = 5x ✓) So 3x − 2 = 0 or 2x + 3 = 0, giving x = 2/3 or x = −3/2. (Or use the formula: x = [−5 ± √(25 + 144)]/12 = (−5 ± 13)/12.)
Q5Express 1/(x² − 9) − 2/(x + 3) as a single fraction in its simplest form. [2 marks]
Factorise the first denominator: x² − 9 = (x − 3)(x + 3). Put both fractions over that denominator:1(x−3)(x+3) − 2(x − 3)(x−3)(x+3) = 1 − 2x + 6(x−3)(x+3) = 7 − 2x(x−3)(x+3)
Q6Solve: (a) 5x = 1/25 (b) 33x+5 = 9x−1[3 marks]
(a) [1]
Write both sides with base 5: 1/25 = 5⁻², so x = −2. x = −2
(b) [2]
Write both sides with base 3: 9x−1 = (3²)x−1 = 32x−2. So 3x + 5 = 2x − 2, giving x = −7.
Q7Shade one square of the 4 × 4 grid so that (a) the shape has exactly one line of symmetry [1], (b) the shape has rotational symmetry of order 2 [1]. The squares shaded on the paper are the five grey ones.[2 marks]
(a) [1]
The printed shape has no symmetry of its own. Shading row 1, column 2 makes the whole top row shaded and gives a single line of symmetry: the vertical middle line (colours pair up as 1↔4 and 2↔3, and the two bottom corners are already a pair). Other answers that also score: row 2 column 4 (diagonal through the top-left corner), row 3 column 1 (the other diagonal), row 4 column 3 (the horizontal middle line) — each of those gives exactly one line, never two.
(b) [1]
Shade row 4, column 2. A half-turn about the centre then maps the shape onto itself: top-left ↔ bottom-right, top-third ↔ bottom-second, top-fourth ↔ bottom-left. That square is the only one that works.
Take out the square factors: √75 = √(25×3) = 5√3 and √48 = √(16×3) = 4√3. So 5√3 − 4√3 = √3 (= 1.73 to 3 s.f.).
(b) [3]
Multiply top and bottom by the conjugate of the denominator, (3 − √7): top: (2 − √7)(3 − √7) = 6 − 2√7 − 3√7 + 7 = 13 − 5√7; bottom: (3 + √7)(3 − √7) = 9 − 7 = 2. So the answer is (13 − 5√7)/2 (= 6.5 − 2.5√7).
Q9Evaluate, leaving your answer in standard form: (a) 1.07 × 10⁵ − 4.7 × 10⁴ (b) (6 × 10⁻³)² [4 marks]
(a) [2]
Write both in ordinary numbers: 107 000 − 47 000 = 60 000. In standard form that is 6 × 10⁴.
(b) [2]
Square both parts: 6² = 36 and (10⁻³)² = 10⁻⁶. So 36 × 10⁻⁶ = 3.6 × 10⁻⁵ (the 36 is not standard form, so it becomes 3.6 and the index rises by one).
Q10Rearrange 5bx = 2 − 3x/(b + 2) to make x the subject. [4 marks]
Multiply every term by (b + 2) to clear the fraction: 5bx(b + 2) = 2(b + 2) − 3x. Expand the left: 5b²x + 10bx = 2b + 4 − 3x. Bring the x-terms together: 5b²x + 10bx + 3x = 2b + 4. Factorise the left: x(5b² + 10b + 3) = 2(b + 2). So x = 2(b + 2)5b² + 10b + 3.
Rounding: the paper asks for non-exact answers to 3 significant figures, which is what the answers above use. Marks: the brackets shown are the paper's own. These solutions are worked, then recomputed in code (78 checks) before this page is built.
