OCR GCSE Mathematics: Inequalities and the language of functions (J560) – 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 sections 6.04 (Algebraic inequalities) and 6.05 (Language of functions) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). Questions 1 to 5 are suitable for both tiers. Questions 6 to 12 are on content the specification gives to Higher tier learners only, and are labelled (Higher tier only).

Revise the content in the study guide and the revision notes. Course hub: OCR GCSE Mathematics. Checklist: OCR GCSE Mathematics checklist. Not sure where to start? Try the free diagnostics.

Questions

1. On a number line, a line joins a closed circle at −3 to an open circle at 4.

(a) Write down the inequality shown. [2] (b) Write down all the integers that satisfy it. [1]

2. Solve 6 − 5x > 21. Show your answer on a number line. [4]

3. Solve −4 ≤ 3x + 2 < 20, and write down the integer solutions. [4]

4. A hall costs £12 to hire plus £2.75 for each ticket sold to cover costs. The organiser can spend at most £50. Write an inequality, solve it, and state the greatest number of tickets. [4]

5. A machine is y = 4x − 5.

(a) Find the output when the input is 7. [1] (b) Find the input that gives an output of 31. [2] (c) Draw a machine diagram for y = 3(x − 2). [1]

6. (Higher tier only) Solve x² + 5x − 36 < 0. Give your answer in set notation. [4]

7. (Higher tier only) Solve 2x² − 9x − 5 ≥ 0. [4]

8. (Higher tier only)

(a) Solve x² − 10x + 6 ≤ 0, leaving your answer with a surd. [3] (b) Write down the integers that satisfy it. [2]

9. (Higher tier only) A triangle has corners at (1, 2), (9, 2) and (1, 6). Write down three inequalities whose solution set is the region inside and on the triangle. [3]

10. (Higher tier only) The region R satisfies x ≥ 0, y > 1 and 3x + 2y ≤ 12.

(a) Draw the boundary lines and say which are dashed. [3] (b) How many points with whole-number coordinates lie in R? [2]

11. (Higher tier only)

(a) Machine A multiplies by 4 and then subtracts 3. The output is 29. Find the input. [2] (b) A machine is y = (2x + 5)/3. Describe its inverse as a machine, and write x in terms of y. [3]

12. (Higher tier only) Machine P adds 2 and then squares. Machine Q multiplies by 3 and then subtracts 1.

(a) Write an expression for P then Q, for input x. [2] (b) Write an expression for Q then P, for input x. [2] (c) Find the values of x for which P then Q gives 47. [3]

Answers

1. (a) The left end is included and the right end is not. −3 ≤ x < 4 [2] (b) −3, −2, −1, 0, 1, 2, 3 [1] Examiner insight: Match each circle to its symbol before writing: closed goes with ≤ and open goes with <, and both ends need the correct x position.

2. −5x > 15 [1] Divide by −5 and reverse the sign: x < −3 [1] Open circle at −3 [1] Arrow pointing left [1] Examiner insight: An open circle at −3 and an arrow to the left is the pair that matches x < −3; a correct circle with the arrow pointing right shows the sign was not reversed.

3. −6 ≤ 3x < 18 [1] −2 ≤ x < 6 [2] Integers: −2, −1, 0, 1, 2, 3, 4, 5 [1] Examiner insight: Do the same step to every part, and check the strict end: 6 is not allowed, so the list stops at 5.

4. 12 + 2.75n ≤ 50 [1] 2.75n ≤ 38 [1] n ≤ 13.8… [1] n must be whole: 13 tickets [1] Examiner insight: Rounding 13.8 up to 14 would make the cost exceed £50; in a context, state the whole-number answer that satisfies the inequality.

5. (a) 4 × 7 − 5 = 23 [1] (b) 4x − 5 = 31 [1]; 4x = 36, so x = 9 [1] (c) x → [ − 2 ] → [ × 3 ] → y [1] Examiner insight: In (c) the bracket means subtract first and multiply second; drawing the steps in the other order is a common slip.

6. x² + 5x − 36 = (x + 9)(x − 4) [1] Critical values −9 and 4 [1] U-shaped curve below the axis between the roots: −9 < x < 4 [1] {x | −9 < x < 4} [1] Examiner insight: Write the critical values and a quick sketch; if the final range slips, they still show your method. “Less than 0” means between the roots.

7. 2x² − 9x − 5 = (2x + 1)(x − 5) [1] Critical values −1/2 and 5 [1] U-shaped curve: the values at or above zero lie beyond both roots [1] x ≤ −1/2 or x ≥ 5 [1] Examiner insight: “Greater than or equal to 0” needs two separate parts joined by or; one chain such as 5 ≤ x ≤ −1/2 describes no numbers.

8. (a) (x − 5)² − 19 ≤ 0 [1] Critical values 5 ± √19 [1] Between the roots: 5 − √19 ≤ x ≤ 5 + √19 [1] (b) √19 is about 4.36, so the range is about 0.64 to 9.36 [1] Integers: 1, 2, 3, 4, 5, 6, 7, 8, 9 [1] Examiner insight: Keep the exact roots for (a). Using decimal estimates only for (b) lets you check which whole numbers sit inside the range.

9. The left side is x = 1, so x ≥ 1 [1] The bottom side is y = 2, so y ≥ 2 [1] The slanted side has gradient (6 − 2)/(1 − 9) = −1/2 and passes through (9, 2), so x + 2y = 13: x + 2y ≤ 13 [1] Examiner insight: Test one point inside, such as (2, 3), in each inequality: 2 ≥ 1, 3 ≥ 2 and 2 + 6 ≤ 13 all hold, which confirms the symbols.

10. (a) x = 0 (solid) [1] y = 1 (dashed, because y > 1 is strict) [1] 3x + 2y = 12 (solid) through (4, 0) and (0, 6) [1] (b) y = 2: x = 0, 1, 2. y = 3: x = 0, 1, 2. y = 4: x = 0, 1. y = 5: x = 0. y = 6: x = 0. [1] Total 10 points [1] Examiner insight: Work row by row and test the inequality for each y. Points on y = 1 are excluded, and points on the solid lines are included.

11. (a) Reverse: add 3 [1], then divide by 4: (29 + 3)/4 = 8 [1] (b) The machine multiplies by 2, adds 5, then divides by 3. The inverse multiplies by 3, subtracts 5, then divides by 2. [2] x = (3y − 5)/2 [1] Examiner insight: An inverse needs the reversed order as well as opposite steps; check by feeding an output back in, such as y = 11 giving x = 14.

12. (a) P gives (x + 2)² [1]; then Q: 3(x + 2)² − 1 [1] (b) Q gives 3x − 1 [1]; then P: (3x + 1)² [1] (c) 3(x + 2)² − 1 = 47 [1] (x + 2)² = 16, so x + 2 = 4 or −4 [1] x = 2 or x = −6 [1] Examiner insight: In (b), 3x − 1 plus 2 is 3x + 1 before squaring, and in (c) a square root gives both signs, so stopping at x = 2 loses a solution.

Where marks are usually lost

• Dividing by a negative number and keeping the original inequality sign.
• Showing the right circle on a number line but an arrow in the wrong direction.
• Forgetting that a strict inequality leaves out its boundary integer.
• Rounding up in a context question when the inequality needs the largest whole number that works.
• Stopping at the critical values of a quadratic without stating the range.
• Choosing between the roots when the inequality needs outside the roots, or the reverse.
• Writing one chain for a two-part answer instead of two inequalities joined by or.
• Using the wrong style of line (solid or dashed) for the boundary of a region.
• Reversing the steps of an inverse but keeping the original order.
• Missing the negative solution after taking a square root in a composite.

Next steps

• Condense the methods: revision notes
• Re-learn anything you missed: study guide
• Course hub: OCR GCSE Mathematics
• Checklist: OCR GCSE Mathematics checklist
• Find weak spots with the free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR GCSE (9-1) Mathematics (J560) specification, Version 2.1 (August 2026), first assessment 2017, published by OCR (Oxford Cambridge and RSA) - sections 6.04 Algebraic inequalities and 6.05 Language of functions.