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 available free from your board.

Related: Inequalities revision notes

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Section A

1. State the one rule that distinguishes solving an inequality from solving an equation. [2]

2. List the integer values of n such that −3 < n ≤ 2. [2]

Section B

3. Solve:

(a) 5x − 7 > 2x + 8 [3] (b) 4 − 3x ≥ 19 [3] (c) −2 < (3x + 1)/2 ≤ 5 [4]

4. Represent the solution to 5x − 7 > 2x + 8 on a number line, explaining the convention used. [3]

5. A school orders x tables and y chairs. Each table costs $40 and each chair costs $15. The school has a maximum budget of $600, and needs at least twice as many chairs as tables.

(a) Write down two inequalities in x and y representing these conditions. [3] (b) Determine, showing your reasoning, whether x = 5, y = 12 satisfies both inequalities. [2]

6. Solve the inequality (7 − 2x)/3 ≥ 1, and state the greatest integer value of x satisfying it. [4]

7. A region is defined by y < 2x + 1, y ≥ −1 and x ≤ 4.

(a) Explain how you would show this region on a graph. [4] (b) State whether the point (2, 3) lies inside the region, showing your working. [3]

8. Represent x < 1 and y ≥ 1 on the same diagram, explaining the shading convention used. [4]

9. A diagram shows a shaded region bounded by a solid horizontal line at y = 2 and a broken vertical line at x = 3, with the unshaded region above y = 2 and to the left of x = 3. State the two inequalities defining the unshaded region. [3]

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Answers

1. When both sides are multiplied or divided by a negative number, the direction of the inequality sign must be reversed [1] [1].

2. −2, −1, 0, 1, 2 [1] [1].

3. (a) 3x > 15 [1] [1]; x > 5 [1]. (b) −3x ≥ 15 [1]; dividing by −3 reverses the sign [1]; x ≤ −5 [1]. (c) Multiply throughout by 2: −4 < 3x + 1 ≤ 10 [1]; subtract 1: −5 < 3x ≤ 9 [1]; divide by 3 [1]; −5/3 < x ≤ 3 [1].

4. Draw a number line and mark 5 [1]. Use an open (unfilled) circle at 5, because the inequality is strict and 5 is not included [1]. Draw an arrow to the right from the circle, indicating all values greater than 5 [1]. (A filled circle would be used for ≥ or ≤.)

5. (a) Cost constraint: 40x + 15y ≤ 600 [1] [1]; at-least-twice-as-many-chairs constraint: y ≥ 2x [1]. (b) Substituting x = 5, y = 12: 40(5) + 15(12) = 200 + 180 = 380, and 380 ≤ 600 ✓ [1]; y ≥ 2x → 12 ≥ 10 ✓, so both inequalities are satisfied [1].

6. Multiply both sides by 3: 7 − 2x ≥ 3 [1]; subtract 7 from both sides: −2x ≥ −4 [1]; divide by −2, reversing the sign: x ≤ 2 [1]. The greatest integer value satisfying this is x = 2 [1].

7. (a) Draw the line y = 2x + 1 as a dashed line, since the inequality is strict and points on it are excluded [1]; draw y = −1 and x = 4 as solid lines, since those inequalities include equality [1]. Shade the unwanted regions — above y = 2x + 1, below y = −1 and to the right of x = 4 — leaving the required region (below y = 2x + 1, above y = −1 and to the left of x = 4) unshaded [1]; test a point such as the origin in each inequality to confirm which side to shade [1]. (b) Check each: y < 2x + 1 → 3 < 5 ✓ [1]; y ≥ −1 → 3 ≥ −1 ✓ [1]; x ≤ 4 → 2 ≤ 4 ✓. All three are satisfied, so (2, 3) lies inside the region [1].

8. Draw a broken vertical line at x = 1, since the inequality is strict [1], shading the unwanted side (x ≥ 1, to the right) [1]. Draw a solid horizontal line at y = 1, since the inequality is inclusive [1], shading the unwanted side (y < 1, below it) [1]. The convention is to shade the region not wanted, leaving the required region unshaded and visually clear.

9. The line y = 2 is solid, so the boundary is inclusive: y ≥ 2 [1] [1]. The line x = 3 is broken, so the boundary is strict, and the unshaded region lies to the left: x < 3 [1]. (Note: optimising a quantity subject to a system of inequalities — linear programming — is not part of this syllabus; the skill required stops at representing, solving and reading off regions.)

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Where marks are usually lost

• Forgetting to reverse the sign when dividing by a negative.
• Writing a worded constraint the wrong way round (e.g. reading “at least” as ≤ instead of ≥).
• Using a solid line for a strict inequality.
• Treating a double inequality one side at a time and losing a bound.
• Shading the wanted region instead of the unwanted region (the convention is the opposite of what many students expect).