Edexcel International GCSE Maths Algebraic Indices Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Equations, Formulae and Identities

Algebraic Notation and Indices: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every algebraic notation and indices question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 5 questions worth 22 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

16 marksHigherJanuary 2022 · Paper 1H (4MA1/1H) · Q1

(a) Simplify a7 × a4 (1)

(b) Simplify w15 ÷ w3 (1)

(c) Simplify (8x5y3)2 (2)

(d) Make t the subject of c = t3 − 8v (2)

Show solution

Part (a)

  1. Add the indices: 7 + 4 = 11

Answer: a11

Part (b)

  1. Subtract the indices: 15 − 3 = 12

Answer: w12

Part (c)

  1. 82 = 64, (x5)2 = x10, (y3)2 = y6

Answer: 64x10y6

Part (d)

  1. Add 8v to both sides: t3 = c + 8v
  2. Take the cube root of both sides.

Answer: t = ∛(c + 8v)

24 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q4

x5 × x7 = xm

(a) Find the value of m. (1)

y8 ÷ y3 = yn

(b) Find the value of n. (1)

(c) Simplify fully (5a4r2)3 (2)

Show solution

Part (a)

  1. Multiplying powers of the same letter: add the indices. 5 + 7 = 12

Answer: m = 12

Part (b)

  1. Dividing powers of the same letter: subtract the indices. 8 − 3 = 5

Answer: n = 5

Part (c)

  1. Cube every part of the bracket: 53 = 125, (a4)3 = a12, (r2)3 = r6

Answer: 125a12r6

34 marksHigherJune 2023 · Paper 1HR (4MA1/1HR) · Q6

(a) Simplify m10 ÷ m3 (1)

kn × k4 = k12

(b) Write down the value of n. (1)

(c) Simplify (3x6y8)2 (2)

Show solution

Part (a)

  1. Subtract the indices when dividing: 10 − 3 = 7

Answer: m7

Part (b)

  1. Indices add when multiplying, so n + 4 = 12

Answer: n = 8

Part (c)

  1. Square each part: 32 = 9, (x6)2 = x12, (y8)2 = y16

Answer: 9x12y16

44 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q9

(a) Simplify 8 × (4t)0 (1)

x6 ÷ x−5 = xp

(b) Find the value of p. (1)

(c) Simplify fully (2k2m4)3 (2)

Show solution

Part (a)

  1. (4t)0 = 1, so 8 × 1

Answer: 8

Part (b)

  1. Subtract the indices: 6 − (−5) = 11

Answer: p = 11

Part (c)

  1. 23 = 8, (k2)3 = k6, (m4)3 = m12

Answer: 8k6m12

54 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q10

(a) Simplify (2p)0 where p > 0 (1)

y9 × y−3 = yn

(b) Find the value of n. (1)

(c) Simplify fully (5a4c2)3 (2)

Show solution

Part (a)

  1. Anything (except 0) to the power 0 is 1.

Answer: 1

Part (b)

  1. When multiplying powers of the same letter, add the indices: 9 + (−3) = 6

Answer: n = 6

Part (c)

  1. The power 3 applies to every part inside the bracket.
  2. 53 = 125
  3. (a4)3 = a4×3 = a12 (power of a power: multiply the indices)
  4. (c2)3 = c2×3 = c6

Answer: 125a12c6

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