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

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

Inequalities: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every inequalities question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 11 questions worth 43 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.

13 marksHigherJanuary 2022 · Paper 1HR (4MA1/1HR) · Q1

n is an integer.

(a) Write down all the values of n such that −2 ≤ n < 3 (2)

(b) On the number line, represent the inequality y ≤ 1 (1)

Show solution

Part (a)

  1. −2 is included (≤) and 3 is not (<).

Answer: −2, −1, 0, 1, 2

Part (b)

  1. Filled circle at 1 (included), with an arrow pointing left for all the smaller values.
−3−2−10123y

Answer: Filled circle at 1 with an arrow to the left

24 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q3

(a) On the number line, show the inequality −2 ≤ y < 1 (2)

n is an integer.

(b) Write down all the values of n that satisfy −3.4 < n ≤ 2 (2)

Show solution

Part (a)

  1. −2 is included (≤), so use a filled circle. 1 is not included (<), so use an open circle. Join them.
−3−2−10123y

Answer: Filled circle at −2, open circle at 1, joined by a line

Part (b)

  1. Integers greater than −3.4 start at −3. 2 is included.

Answer: −3, −2, −1, 0, 1, 2

35 marksHigherJanuary 2022 · Paper 1H (4MA1/1H) · Q6

(a) Solve the inequality 5x − 7 ≤ 2 (2)

(b)(i) Factorise y2 − 2y − 35 (2)

(ii) Hence, solve y2 − 2y − 35 = 0 (1)

Show solution

Part (a)

  1. Add 7: 5x ≤ 9

Answer: x ≤ 1.8

Part (b)(i)

  1. Two numbers that multiply to −35 and add to −2: −7 and +5.

Answer: (y − 7)(y + 5)

Part (b)(ii)

  1. Set each bracket to zero.

Answer: y = 7 or y = −5

44 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q6
−3−2−10123x

(a) Write down the inequality shown on the number line. (2)

(b) Solve the inequality 7a − 5 ≤ 3a + 28. Show clear algebraic working. (2)

Show solution

Part (a)

  1. An open circle means the number is not included (<). A filled circle means it is included (≤).
  2. x is bigger than −2 (not equal) and less than or equal to 1.

Answer: −2 < x ≤ 1

Part (b)

  1. Subtract 3a from both sides: 4a − 5 ≤ 28
  2. Add 5 to both sides: 4a ≤ 33
  3. Divide by 4: a ≤ 8.25

Answer: a ≤ 8.25

54 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q7

−4 ≤ 2y < 6 and y is an integer.

(a) Write down all the possible values of y. (2)

(b) Solve the inequality 7t − 3 ≤ 2t + 31. Show your working clearly. (2)

Show solution

Part (a)

  1. Divide every part by 2: −2 ≤ y < 3
  2. −2 is included (≤) but 3 is not (<).

Answer: −2, −1, 0, 1, 2

Part (b)

  1. Subtract 2t from both sides: 5t − 3 ≤ 31
  2. Add 3: 5t ≤ 34

Answer: t ≤ 6.8

64 marksHigherJune 2023 · Paper 1H (4MA1/1H) · Q7

The region R, shown shaded in the diagram, is bounded by three straight lines. Find the inequalities that define R.

−4−3−2−1012345−3−2−1123456789xyR
Show solution
  1. Find the equation of each boundary line.
  2. The vertical line passes through x = 1. R is to the left of it, so x ≤ 1.
  3. The horizontal line is y = −2. R is above it, so y ≥ −2.
  4. The sloping line passes through (−2, 0) and (0, 4): gradient = 4 ÷ 2 = 2 and y-intercept 4, so it is y = 2x + 4.
  5. Test the point (0, 0), which is in R: 0 ≤ 2(0) + 4 is true, so y ≤ 2x + 4.

Answer: x ≤ 1, y ≥ −2, y ≤ 2x + 4

75 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q9

(i) Solve the inequalities −7 ≤ 2x − 3 < 5 (3)

(ii) On the number line, represent the solution set to part (i). (2)

Show solution

Part (i)

  1. Add 3 to every part: −4 ≤ 2x < 8
  2. Divide every part by 2: −2 ≤ x < 4

Answer: −2 ≤ x < 4

Part (ii)

  1. A filled circle at −2 (included, ≤) and an open circle at 4 (not included, <), joined by a line.
−6−4−20246

Answer: Filled circle at −2, open circle at 4, line joining them

82 marksHigherJune 2022 · Paper 1HR (4MA1/1HR) · Q9

Solve the inequality 3 − 4x ≤ 11

Show solution
  1. Subtract 3 from both sides: −4x ≤ 8
  2. Divide both sides by −4. Dividing an inequality by a negative number reverses the sign.

Answer: x ≥ −2

Tip: Alternative that avoids dividing by a negative: add 4x and subtract 11 to get −8 ≤ 4x, so −2 ≤ x.
93 marksHigherJanuary 2020 · Paper 1H (4MA1/1H) · Q10

The shaded region in the diagram is bounded by three lines. The equation of one of the lines is y = 1⁄3x − 2. Write down the three inequalities that define the shaded region.

−6−4−20246−6−4−2246xyy = x/3 − 2
Show solution
  1. The vertical line is x = −1, and the region is to its right: x ≥ −1
  2. The region is above the given line: y ≥ 1⁄3x − 2
  3. The third line passes through (0, 4) and (4, 0): gradient −1, intercept 4, so y = 4 − x (or x + y = 4). The region is below it: y ≤ 4 − x

Answer: x ≥ −1, y ≥ 1⁄3x − 2, x + y ≤ 4

106 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q10

(a) Factorise y2 − 2y − 48 (2)

(b) Write down the inequality shown on the number line (a filled circle at 3 with an arrow going left). (1)

(c) Solve the inequality 7w + 6 > 12w + 14 (3)

Show solution

Part (a)

  1. Two numbers that multiply to −48 and add to −2: −8 and +6.

Answer: (y − 8)(y + 6)

Part (b)

  1. The filled circle means 3 is included, and the arrow goes towards smaller numbers.

Answer: x ≤ 3

Part (c)

  1. Subtract 7w and 14 from both sides: −8 > 5w
  2. Divide by 5: −1.6 > w

Answer: w < −1.6

113 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q11

The region R, shown shaded in the diagram, is bounded by the straight lines with equations 2x + y = 6, 2y = 5x + 1 and 3y + 2x = 4. Write down the three inequalities that define R.

−3−2−10123456−4−3−2−112345678xyR
Show solution
  1. Pick a point inside R, for example (2, 1), and test it in each equation to see which side of each line R is on.
  2. 2x + y = 4 + 1 = 5, which is less than 6, so 2x + y ≤ 6
  3. 2y = 2 and 5x + 1 = 11, so 2y ≤ 5x + 1
  4. 3y + 2x = 3 + 4 = 7, which is more than 4, so 3y + 2x ≥ 4

Answer: 2x + y ≤ 6, 2y ≤ 5x + 1, 3y + 2x ≥ 4

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