AQA GCSE Mathematics 8300: Notation, vocabulary and manipulation – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.2.1 Notation, vocabulary and manipulation (A1 to A7) of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). Questions and parts on content from the specification’s “Higher content only” column are labelled (Higher); everything else suits both tiers. Each question says whether it is non-calculator (Paper 1 style) or calculator (Papers 2 and 3 style).

Learn the content first in the study guide for this section and the revision notes for this section. For mixed questions across the whole of Topic 2, use the Algebra practice questions. The course hub is AQA GCSE Mathematics and the printable checklist lists every statement.

Questions

1. (non-calculator) Write each of these in its simplest algebraic form.

(a) a × 7 × b [1] (b) n × n × n × p [1] (c) (x + 3) ÷ 4 [1] (d) 0.25 × y [1]

2. (calculator)

(a) The time, T seconds, for one swing of a pendulum is given by T = 2π√(l/g). Calculate T when l = 0.8 and g = 9.8. Give your answer to 3 significant figures. [2] (b) The formula F = 9C/5 + 32 converts a temperature C in °C to F in °F. Work out F when C = −15. [2]

3. (non-calculator)

(a) Here are four lines of algebra.

A: 7 − 2w B: 7 − 2w = 1 C: 7 − 2w < 1 D: 2(w − 3) ≡ 2w − 6

Write down which one is an expression, which is an equation, which is an inequality and which is an identity. [2] (b) Show that 3 is a factor of 6w − 9. [1]

4. (non-calculator) Simplify fully.

(a) 5p − 3q − 2p + 7q [1] (b) 2a⁵ × 6a³ [1] (c) 24m⁶n⁴ ÷ 8m²n [2] (d) (k⁴)³ [1]

5. (non-calculator) Expand and simplify.

(a) 4(2x − 3) − 3(x − 5) [2] (b) (x − 6)(x + 2) [2] (c) (3x + 1)² [2]

6. (non-calculator) Factorise fully.

(a) 8x²y − 20xy² [2] (b) x² − 11x + 28 [2] (c) w² − 121 [1]

7. (non-calculator) Expand and simplify (5 − √2)(3 + √2). Give your answer in the form a + b√2, where a and b are integers. [3]

8. (calculator)

(a) The area of a trapezium is A = ½(a + b)h. Make h the subject. [2] (b) The volume of a cone is V = ⅓πr²h. Make r the subject. Hence find the radius of a cone with volume 500 cm³ and height 12 cm, to 3 significant figures. [4]

9. (non-calculator)

(a) Show that 5(x + 2) − 2(x − 1) ≡ 3x + 12. [2] (b) Is 4(x − 1) = 3x + 2 an equation or an identity? Give a reason for your answer. [2]

10. (non-calculator) (Higher)

(a) Prove that (n + 5)² − (n + 1)² is a multiple of 8 for every integer n. [4] (b) Prove that the sum of any three consecutive even numbers is a multiple of 6. [3]

11. (non-calculator) (Higher)

(a) Factorise 3x² − 5x − 12. [2] (b) Hence simplify (3x² − 5x − 12)/(x² − 9). [2] (c) Write 4/(x − 3) − 1/(x + 3) as a single fraction in its simplest form. [3] (d) Expand and simplify (x + 1)(x − 2)(2x + 3). [3]

12. (non-calculator)

(a) A function machine multiplies the input by 4 and then subtracts 3. Find the output when the input is −2. [1]

The function machine in part (a) is the function f(x) = 4x − 3. Also, g(x) = x² + 2.

(b) (Higher) Find f⁻¹(x). [2] (c) (Higher) Find fg(x). Give your answer in its simplest form. [2] (d) (Higher) Work out gf(2). [2] (e) (Higher) Explain why fg(x) can never be less than 5. [1]

Answers

1. (a) 7ab [1] (b) n³p [1] (c) (x + 3)/4 [1] (d) ¼y or y/4 [1] Examiner insight: On (d), 0.25y is not accepted as simplest form, because the specification asks for coefficients written as fractions rather than decimals; on (a), b7a or 7 × a × b would also lose the mark.

2. (a) T = 2π√(0.8 ÷ 9.8) = 1.7951… [1] T = 1.80 (seconds) [1] (b) F = 9(−15)/5 + 32 [1] = −27 + 32 = 5 [1] Examiner insight: Write down the unrounded calculator value (1.7951…) before rounding: if the rounding is wrong, the first line still earns the method mark, and 1.8 alone is not 3 significant figures.

3. (a) Two or three correctly named [1]; all four correct: A expression, B equation, C inequality, D identity [1] (b) 6w − 9 = 3(2w − 3), so 3 divides exactly into it [1] Examiner insight: For (b), saying “6 and 9 are both in the 3 times table” without the factorised form is weaker; writing 3(2w − 3) shows the factor directly.

4. (a) 3p + 4q [1] (b) 12a⁸ [1] (c) Two of the three parts correct (3, m⁴, n³) [1]; 3m⁴n³ [1] (d) k¹² [1] Examiner insight: (k⁴)³ = k⁷ is a common wrong answer that scores zero: raising a power to a power multiplies the indices, so there is no method mark for adding them.

5. (a) 8x − 12 − 3x + 15 [1] = 5x + 3 [1] (b) x² + 2x − 6x − 12 [1] = x² − 4x − 12 [1] (c) (3x + 1)(3x + 1) = 9x² + 3x + 3x + 1 [1] = 9x² + 6x + 1 [1] Examiner insight: In (a), −3 × −5 = +15 is where the sign slips; one correct expansion of all four terms usually earns the method mark even if the final collection is wrong.

6. (a) A partial factorisation such as 4x(2xy − 5y²) or 2xy(4x − 10y) [1]; 4xy(2x − 5y) [1] (b) Numbers −4 and −7 identified (product 28, sum −11) [1]; (x − 4)(x − 7) [1] (c) (w + 11)(w − 11) [1] Examiner insight: “Factorise fully” means the highest common factor; a correct but incomplete factorisation in (a) earns only the first mark.

7. 15 + 5√2 − 3√2 − √2 × √2 [1]; √2 × √2 = 2, so 15 + 2√2 − 2 [1] = 13 + 2√2 [1] Examiner insight: Four terms must appear in the expansion; giving 15 − 2 = 13 and forgetting the surd terms, or writing √2 × √2 = 4, loses the accuracy marks.

8. (a) 2A = (a + b)h [1]; h = 2A/(a + b) [1] (b) 3V = πr²h, so r² = 3V/(πh) [1]; r = √(3V/(πh)) [1] r = √(3 × 500 ÷ (π × 12)) = √39.788… [1] = 6.3078… so r = 6.31 cm [1] Examiner insight: A follow-through mark is usually available for substituting into your own rearranged formula, but only if the rearrangement is written down.

9. (a) 5x + 10 − 2x + 2 [1] = 3x + 12, which is the right-hand side for every x ✓ [1] (b) It is an equation [1], because it is true only for one value: 4x − 4 = 3x + 2 gives x = 6 [1] Examiner insight: On a “show that”, the answer is given, so both marks depend on the expansion line being written out, including −2 × −1 = +2.

10. (a) (n + 5)² = n² + 10n + 25 and (n + 1)² = n² + 2n + 1 [1] Difference = n² + 10n + 25 − n² − 2n − 1 [1] = 8n + 24 [1] = 8(n + 3), and n + 3 is an integer, so the expression is a multiple of 8 [1] (b) Let the numbers be 2n, 2n + 2 and 2n + 4 [1] Sum = 6n + 6 [1] = 6(n + 1), which is a multiple of 6 [1] Examiner insight: The final mark in each proof is for the concluding statement linked to the factorised form; 8n + 24 on its own, or checking n = 1, 2, 3, does not complete the proof.

11. (a) ac = −36; use −9 and 4: 3x² − 9x + 4x − 12 [1] = (3x + 4)(x − 3) [1] (b) x² − 9 = (x + 3)(x − 3) [1]; cancelling (x − 3) gives (3x + 4)/(x + 3) [1] (c) [4(x + 3) − (x − 3)] / ((x − 3)(x + 3)) [1] = (4x + 12 − x + 3) / ((x − 3)(x + 3)) [1] = (3x + 15)/((x − 3)(x + 3)), or 3(x + 5)/(x² − 9) [1] (d) (x + 1)(x − 2) = x² − x − 2 [1] (x² − x − 2)(2x + 3) = 2x³ + 3x² − 2x² − 3x − 4x − 6 [1] = 2x³ + x² − 7x − 6 [1] Examiner insight: “Hence” in (b) means the factorisation from (a) must be used; in (c), the most common lost mark is writing − x − 3 instead of − x + 3 when subtracting the second numerator.

12. (a) 4 × (−2) − 3 = −11 [1] (b) y = 4x − 3, so x = (y + 3)/4 [1]; f⁻¹(x) = (x + 3)/4 [1] (c) fg(x) = 4(x² + 2) − 3 [1] = 4x² + 5 [1] (d) f(2) = 5 [1]; gf(2) = g(5) = 25 + 2 = 27 [1] (e) x² ≥ 0 for every x, so 4x² + 5 ≥ 5 [1] Examiner insight: For fg(x), the function written next to x acts first; working out g(f(x)) instead gives 16x² − 24x + 11 and usually scores nothing for (c).

Where marks are usually lost

• Using decimal coefficients such as 0.25y, or leaving × signs in an answer.
• Squaring a negative without brackets during substitution.
• Adding indices when one power is raised to another.
• Expanding (3x + 1)² as 9x² + 1 with no middle term.
• Stopping at a partial factorisation when the question says “factorise fully”.
• In a surd expansion, forgetting that √2 × √2 = 2 and not 4, or dropping the surd terms.
• Rearranging a formula but not writing the rearranged version before substituting, which loses follow-through credit.
• Subtracting a bracketed numerator without changing every sign inside it (Higher).
• Missing the concluding sentence in a proof, or testing numbers instead of using n (Higher).
• Composing functions in the wrong order (Higher).

Next steps

• Revision notes for this section
• Study guide for this section
• Algebra study guide (all of Topic 2)
• AQA GCSE Mathematics hub
• Printable AQA GCSE Mathematics checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Mathematics (8300) specification, for teaching from September 2015, exams from May/June 2017, version 1.0, published by AQA – section 3.2.1 Notation, vocabulary and manipulation (A1 to A7) and Appendix: mathematical formulae.