Edexcel International GCSE Maths Changing the Subject Past Paper Questions and Solutions (4MA1 Higher)

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

Formulae and Changing the Subject: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every formulae and changing the subject question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 3 questions worth 21 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.

18 marksHigherJanuary 2020 · Paper 1HR (4MA1/1HR) · Q3

(a) Make a the subject of d = g + 2ac (2)

(b) Factorise fully 9ef − 12f (2)

(c) Expand and simplify (x + 2)(x − 5) (2)

(d) Simplify fully n4 × n7⁄n5 (2)

Show solution

Part (a)

  1. Subtract g from both sides: d − g = 2ac
  2. Divide by 2c.

Answer: a = d − g⁄2c

Part (b)

  1. The HCF of the two terms is 3f.

Answer: 3f(3e − 4)

Part (c)

  1. x2 − 5x + 2x − 10

Answer: x2 − 3x − 10

Part (d)

  1. Top: n4+7 = n11. Then divide: n11−5

Answer: n6

25 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q6

(a) Make c the subject of A = c⁄y − 5z (2)

(b) Write down the value of g0 (1)

(c) Factorise x2 − 11x + 24 (2)

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Part (a)

  1. Add 5z to both sides: A + 5z = c⁄y
  2. Multiply by y.

Answer: c = y(A + 5z)

Part (b)

  1. Any non-zero number to the power 0 is 1.

Answer: 1

Part (c)

  1. Two numbers that multiply to +24 and add to −11: −3 and −8.

Answer: (x − 3)(x − 8)

38 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q17

(a) Factorise 6y2 − y − 5 (2)

(b) Make f the subject of w = 2f + 3⁄8 − f (3)

(c) Express 4x2 − 8x + 7 in the form a(x + b)2 + c where a, b and c are integers. (3)

Show solution

Part (a)

  1. 6 × (−5) = −30. Two numbers that multiply to −30 and add to −1: −6 and +5.
  2. 6y2 − 6y + 5y − 5 = 6y(y − 1) + 5(y − 1)

Answer: (6y + 5)(y − 1)

Part (b)

  1. Multiply both sides by (8 − f): w(8 − f) = 2f + 3, so 8w − wf = 2f + 3
  2. Get the f terms on one side: 8w − 3 = 2f + wf
  3. Factorise: 8w − 3 = f(2 + w)

Answer: f = 8w − 3⁄w + 2

Part (c)

  1. Take out 4 from the x terms: 4(x2 − 2x) + 7
  2. x2 − 2x = (x − 1)2 − 1, so 4[(x − 1)2 − 1] + 7 = 4(x − 1)2 − 4 + 7

Answer: 4(x − 1)2 + 3

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