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S4 Maths: Inequalities and Linear Programming (Paper 3 (challenge))
Tutor Prep Guide · Solving linear inequalities in one variable, when the sign reverses, the solution set of a compound inequality, counting integer solutions, and the optimum of a linear objective at a vertex of the feasible region
S4 Maths: Inequalities and Linear Programming (Paper 3 (challenge))
Paper 3 (challenge): ten questions on inequalities and linear programming with answers and one-line notes.
Direct answers Level: S4 maths; format: linear and compound inequalities, integer solutions and a linear-programming optimum.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S4 Maths: Inequalities and Linear Programming |
| Level | S4 |
02 Questions (10 items)
- FoundationSolve the inequality (3x − 1)/2 ≤ (x + 4)/3.
- FoundationSolve the compound inequality −5 ≤ 2(x − 1) < 9.
- FoundationSolve the inequality 2x + 7 < 3x − 1.
- IntermediateThe sum of three consecutive integers is at most 42. Find the largest possible set.
- IntermediateThree times a number minus 5 is not less than twice the number plus 7. Find the smallest integer.
- IntermediateA rectangle is 5 cm longer than it is wide and its perimeter is at most 50 cm. Find the greatest width.
- ChallengeGiven x + y ≥ 4, x ≤ 3, y ≤ 5, x ≥ 0 and y ≥ 0, find the maximum of P = 2x + 3y.
- ChallengeSolve the inequality (2x − 3)/4 ≥ (x + 1)/6.
- ChallengeFor integer x with x² ≤ 50 and x > 6, how many values are possible?
- Error redo(Error clinic) Keung solved −x/4 ≥ 3 as x ≥ −12. Where is the mistake, and what is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | x ≤ 11/7, about 1.571 | Multiply both sides by 6 to get 3(3x − 1) ≤ 2(x + 4), which reduces to 7x ≤ 11. |
| 2 | −1.5 ≤ x < 5.5 | Add 2 to all parts to get −3 ≤ 2x < 11, then halve to get −1.5 ≤ x < 5.5. |
| 3 | x > 8 | Subtract 2x and add 1 to get 8 < x, that is x > 8. |
| 4 | 13, 14 and 15 | With smallest n, 3n + 3 ≤ 42 gives n ≤ 13, so the largest set is 13 to 15. |
| 5 | 12 | Let the number be x, then 3x − 5 ≥ 2x + 7, which gives x ≥ 12. |
| 6 | 10 cm | With width w, 2(2w + 5) ≤ 50 gives 4w ≤ 40, so w ≤ 10. |
| 7 | 21, taken at (3,5) | The vertices are (0,5), (3,1) and (3,5), giving 15, 9 and 21, so the maximum is 21. |
| 8 | x ≥ 11/4, that is 2.75 | Multiply both sides by 12 to get 3(2x − 3) ≥ 2(x + 1), which reduces to 4x ≥ 11. |
| 9 | 1, namely x = 7 | Seven squared is 49 which fits, while eight squared is 64 and too large, so only 7 works. |
| 10 | The correct answer is x ≤ −12; multiplying by a negative reverses the sign | Multiply by −4: 3 × (−4) = −12, and ≥ must become ≤. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Check the answers, copy the working of any wrong item, then try another paper of this chapter.Tutor Registration →
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