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S4 Maths: Inequalities and Linear Programming (Paper 1 (basics))
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 1 (basics))
Paper 1 (basics): 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 2x + 3 > 11.
- FoundationSolve the inequality 5x − 7 ≤ 18.
- FoundationSolve the inequality −3x < 12.
- IntermediateSolve the inequality 4x − 5 ≥ 2x + 7.
- IntermediateSolve the inequality 3(x − 2) < 12.
- IntermediateSolve the inequality x/2 + 3 > 7.
- ChallengeSolve the compound inequality −2 < x + 1 ≤ 5 and count the integer solutions.
- ChallengeFind the greatest integer solution of 4x + 1 ≤ 3x + 6.
- ChallengeSolve 6 − 2x ≥ 0 and list its positive integer solutions.
- Error redo(Error clinic) Ming solved −2x ≥ 8 as x ≥ −4. What went wrong, and what is the correct answer?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | x > 4 | Subtract 3 to get 2x > 8, then divide by 2; the sign stays the same. |
| 2 | x ≤ 5 | Add 7 to get 5x ≤ 25, then divide by 5 to get x ≤ 5. |
| 3 | x > −4 | Dividing by the negative number −3 reverses the sign from less than to greater than. |
| 4 | x ≥ 6 | Subtract 2x to get 2x − 5 ≥ 7, then add 5 to get 2x ≥ 12. |
| 5 | x < 6 | Expand to 3x − 6 < 12, then add 6 and divide by 3. |
| 6 | x > 8 | Subtract 3 to get x/2 > 4, then multiply both sides by 2. |
| 7 | −3 < x ≤ 4, giving 7 integers | Subtract 1 from all three parts to get −3 < x ≤ 4; the integers run from −2 to 4, which is 7 of them. |
| 8 | 5 | Move terms to get x ≤ 5, so the greatest integer solution is 5. |
| 9 | x ≤ 3, so the positive integers are 1, 2 and 3 | Rearrange to −2x ≥ −6, then divide by −2 and reverse the sign to get x ≤ 3. |
| 10 | The correct answer is x ≤ −4; dividing by a negative number reverses the sign | Dividing −2x ≥ 8 by −2 gives −4 and turns ≥ into ≤. |
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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