Edexcel International GCSE Mathematics A 4MA1: Inequalities – 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 2.8 Inequalities in Topic 2 (Equations, formulae and identities) of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Questions 1–7 use Foundation statements 2.8A–E and are for both tiers. Questions 8–12 are marked (Higher): they use the Higher tier only statements on quadratic inequalities and harder regions. A calculator may be used on every 4MA1 paper, but show your working, because most marks here are for method.

Learn the content first in the inequalities study guide and the inequalities revision notes. For factorising practice, which questions 9–11 rely on, use the algebraic manipulation practice questions. The course hub is Edexcel IGCSE Mathematics and the printable checklist lists every statement.

Questions

1. (Both tiers)

(a) n is an integer and −2 < n ≤ 4. List all the possible values of n. [2] (b) A number line shows an open circle at −1 and a closed circle at 3, joined by a line. Write down the inequality shown. [2]

2. (Both tiers)

(a) Solve 7x − 4 ≤ 3x + 10. [2] (b) Solve 11 − 4x > 23. [2] (c) Describe how your answer to (b) is shown on a number line. [1]

3. (Both tiers)

(a) Solve −5 < 3x + 4 ≤ 13. [3] (b) Write down the integers that satisfy −5 < 3x + 4 ≤ 13. [1]

4. (Both tiers)

(a) Solve (2x − 1)/5 > (x + 3)/4. [3] (b) Write down the smallest integer that satisfies the inequality. [1]

5. (Both tiers) A rectangle has length (x + 3) cm and width (2x − 1) cm. Its perimeter is at most 40 cm.

(a) Show that 6x + 4 ≤ 40. [2] (b) The width must be greater than 0. Find the range of possible values of x, giving your answer as a double-ended inequality. [3]

6. (Both tiers) Use a coordinate grid with x and y from 0 to 8.

(a) Draw the lines x = 2, y = 1 and x + y = 7. [2] (b) Show the region R that satisfies x ≥ 2, y ≥ 1 and x + y ≤ 7. Label it R. [1] (c) Write down the coordinates of the vertices of R. [2] (d) Decide whether each of the points (3, 3) and (4, 4) lies in R. Show your working. [2]

7. (Both tiers) A triangular region R has vertices (1, 1), (5, 1) and (5, 5). Its sides lie on the lines y = 1, x = 5 and y = x.

(a) Write down the three inequalities that define R. [3] (b) Explain why the point (3, 4) does not lie in R. [1]

8. (Higher)

(a) Solve x² ≤ 49. [2] (b) Solve 4x² > 81. [2]

9. (Higher)

(a) Solve 3x² ≥ 10x + 8. [4] (b) Describe how your answer to (a) is shown on a number line. [1]

10. (Higher)

(a) Solve x² − 6x + 4 < 0, giving your answer in exact form. [3] (b) Write down all the integers that satisfy x² − 6x + 4 < 0. [2]

11. (Higher) A rectangle has length (x + 5) cm and width x cm. Its area is greater than 24 cm² and its perimeter is less than 30 cm.

(a) Show that x² + 5x − 24 > 0. [1] (b) Solve x² + 5x − 24 > 0. [3] (c) Use the perimeter to write down and solve a second inequality in x. [2] (d) Hence find the range of possible values of x. [1]

12. (Higher) The region R is defined by the inequalities y < 2x + 2, x + 2y ≤ 14 and y ≥ 1.

(a) Find the coordinates of the three vertices of R. [4] (b) Calculate the area of R. [2] (c) Decide whether each of the points (6, 4) and (0, 3) lies in R. Give a reason for each. [2]

Answers

1. (a) At least four correct values, with at most one wrong value [1]; −1, 0, 1, 2, 3, 4 [1] (b) Limits −1 and 3 [1]; −1 < x ≤ 3 [1] Examiner insight: In (a), including −2 or leaving out 0 costs the accuracy mark; a list with one wrong value usually keeps only the method mark.

2. (a) 4x ≤ 14 [1]; x ≤ 3.5 [1] (b) −4x > 12 [1]; x < −3 [1] (dividing by −4 reverses the sign) (c) Open circle at −3 with an arrow pointing left [1] Examiner insight: An answer of x > −3 in (b) loses the accuracy mark but keeps the method mark for −4x > 12; follow-through is then allowed in (c) only for the circle type and direction matching that answer.

3. (a) Subtract 4 from all three parts: −9 < 3x ≤ 9 [1]; divide all three parts by 3 [1]; −3 < x ≤ 3 [1] (b) −2, −1, 0, 1, 2, 3 [1] Examiner insight: Two separate answers, x > −3 and x ≤ 3, are acceptable for the method marks, but a single wrong symbol (≤ in place of <) loses the final accuracy mark.

4. (a) Multiply both sides by 20: 4(2x − 1) > 5(x + 3) [1]; 8x − 4 > 5x + 15 [1]; x > 19/3 (or x > 6.33…) [1] (b) 7 [1] Examiner insight: The first method mark needs both numerators multiplied correctly; writing 2x − 1 > x + 3 after “cross-multiplying” earns nothing.

5. (a) Perimeter = 2(x + 3) + 2(2x − 1) [1] = 2x + 6 + 4x − 2 = 6x + 4, so 6x + 4 ≤ 40 [1] (b) 6x ≤ 36, so x ≤ 6 [1]; 2x − 1 > 0, so x > 1/2 [1]; 1/2 < x ≤ 6 [1] Examiner insight: On a “show that” the expanded perimeter 6x + 4 must appear before the given inequality; quoting 6x + 4 ≤ 40 with no expression for the perimeter scores 0.

6. (a) x = 2 and y = 1 drawn correctly [1]; x + y = 7 drawn through (0, 7) and (7, 0) [1] (b) The triangle to the right of x = 2, above y = 1 and below x + y = 7, labelled R [1] (c) Any two correct vertices [1]; (2, 1), (2, 5) and (6, 1) [1] (d) (3, 3): 3 ≥ 2, 3 ≥ 1 and 3 + 3 = 6 ≤ 7, so (3, 3) is in R [1]; (4, 4): 4 + 4 = 8, which is not ≤ 7, so (4, 4) is not in R [1] Examiner insight: The region mark in (b) needs the correct region clearly identified as the question instructs; the specification says boundary conventions (solid or dashed lines) are not required at this level.

7. (a) y ≥ 1 [1]; x ≤ 5 [1]; y ≤ x [1] (b) At (3, 4), y = 4 is greater than x = 3, so y ≤ x is not satisfied [1] Examiner insight: Each inequality earns its own mark, so one wrong symbol loses one mark, not three; check each by substituting a point inside the triangle such as (4, 2).

8. (a) Critical values ±7 [1]; −7 ≤ x ≤ 7 [1] (b) x² > 81/4, critical values ±9/2 [1]; x < −4.5 or x > 4.5 [1] Examiner insight: Writing x ≤ ±7 or x > ±4.5 shows the critical values and usually earns the method mark only; the accuracy mark needs the correct inside or outside form.

9. (a) 3x² − 10x − 8 ≥ 0 [1]; (3x + 2)(x − 4) ≥ 0 [1]; critical values −2/3 and 4 [1]; x ≤ −2/3 or x ≥ 4 [1] (b) Closed circle at −2/3 with an arrow left, and a closed circle at 4 with an arrow right [1] Examiner insight: Finding the critical values earns the method marks, but the final mark needs the correct region; writing −2/3 ≤ x ≤ 4 (inside instead of outside) loses it.

10. (a) (x − 3)² − 5 < 0, or the quadratic formula with a = 1, b = −6, c = 4 [1]; critical values x = 3 ± √5 [1]; 3 − √5 < x < 3 + √5 [1] (b) 3 − √5 ≈ 0.76 and 3 + √5 ≈ 5.24 [1]; 1, 2, 3, 4, 5 [1] Examiner insight: “Exact form” means surds; a final answer of 0.764 < x < 5.24 loses the accuracy mark in (a), although the decimals are the right tool in (b).

11. (a) x(x + 5) > 24, so x² + 5x > 24, so x² + 5x − 24 > 0 [1] (b) (x + 8)(x − 3) > 0 [1]; critical values −8 and 3 [1]; x < −8 or x > 3 [1] (c) 2(x + 5) + 2x < 30 [1]; 4x + 10 < 30, so x < 5 [1] (d) x is a length, so x > 0 and x < −8 is rejected: 3 < x < 5 [1] Examiner insight: The final mark in (d) needs the negative solution rejected using the context; carrying x < −8 into the answer loses it even if (b) and (c) are correct.

12. (a) y = 1 and y = 2x + 2: 2x + 2 = 1, so (−0.5, 1) [1]; y = 1 and x + 2y = 14: (12, 1) [1]; x + 2(2x + 2) = 14 [1]; 5x = 10, so (2, 6) [1] (b) Base = 12 − (−0.5) = 12.5 and height = 6 − 1 = 5 [1]; area = ½ × 12.5 × 5 = 31.25 square units [1] (c) (6, 4): 4 < 14, 6 + 8 = 14 ≤ 14 and 4 ≥ 1, so (6, 4) is in R (on the boundary x + 2y = 14, which is included) [1]; (0, 3): 2(0) + 2 = 2 and 3 < 2 is false, so (0, 3) is not in R [1] Examiner insight: The method mark for the third vertex is for substituting one equation into the other; a vertex read from a rough sketch with no working gets no credit if it is inaccurate.

Where marks are usually lost

• Not reversing the sign after dividing by a negative coefficient, as in 2(b).
• Applying an operation to only two of the three parts of a double-ended inequality.
• Clearing fractions by multiplying only one side, or forgetting the brackets around a numerator.
• Using the wrong circle type: closed for < or >, open for ≤ or ≥.
• Listing integers that include an excluded end value, or missing 0.
• Drawing x = 2 as a horizontal line, or x + y = 7 through the wrong intercepts.
• Leaving a “show that” without the expression it starts from.
• (Higher) Stopping at critical values, or choosing “inside” when the inequality needs “outside”.
• (Higher) Giving decimals when the question asks for exact form.
• (Higher) Keeping a negative length in a context answer.

Next steps

• Refresh the methods: inequalities revision notes
• Re-learn anything you missed: inequalities study guide
• Rest of Topic 2: use of symbols and algebraic manipulation
• Course hub: Edexcel IGCSE Mathematics
• Track coverage: printable checklist
• Check other topics: all free 10-minute diagnostics
• Want help from a teacher? Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities, section 2.8 Inequalities (Foundation tier statements 2.8A–E and Higher tier statements 2.8A–B).