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S4 Maths: Inequalities and Linear Programming (Paper 4 (error clinic))
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 4 (error clinic))
Paper 4 (error clinic): 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 − 8 < 7.
- FoundationSolve the inequality 9 − 2x ≥ 5.
- FoundationSolve the inequality x/5 + 1 > 3.
- IntermediateSolve the inequality 6x + 5 ≥ 4x − 3.
- IntermediateSolve the inequality 2(x + 4) > 5x − 1.
- IntermediateSolve −1 < 3x + 2 ≤ 11 and count the integer solutions.
- ChallengeSolve the inequality 4 − 3x ≤ x + 8.
- ChallengeFor positive integer x, solve 5x − 2 < 18 and list the solutions.
- ChallengeA feasible region has vertices (0,0), (6,0), (2,4) and (0,3). Find the maximum of P = x + 4y.
- Error redo(Error clinic) Ling solved 5 − x > 2 as x > 3. Where is the mistake, and what is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | x < 5 | Add 8 to get 3x < 15, then divide by 3 to get x < 5. |
| 2 | x ≤ 2 | Rearrange to −2x ≥ −4, then divide by −2 and reverse the sign to get x ≤ 2. |
| 3 | x > 10 | Subtract 1 to get x/5 > 2, then multiply by 5 to get x > 10. |
| 4 | x ≥ −4 | Subtract 4x and 5 to get 2x ≥ −8, then divide by 2 to get x ≥ −4. |
| 5 | x < 3 | Expand to 2x + 8 > 5x − 1, move terms to get 9 > 3x, so x < 3. |
| 6 | −1 < x ≤ 3, giving 4 integer solutions | Subtract 2 to get −3 < 3x ≤ 9, then divide by 3 to get −1 < x ≤ 3; the integers are 0 to 3. |
| 7 | x ≥ −1 | Add 3x and subtract 8 to get −4 ≤ 4x, so x ≥ −1. |
| 8 | x < 4, so x = 1, 2 or 3 | Rearrange to 5x < 20, that is x < 4, so the positive integers are 1 to 3. |
| 9 | 18, taken at (2,4) | Substituting each vertex gives 0, 6, 18 and 12, so the maximum is 18. |
| 10 | The correct answer is x < 3; −x > −3 becomes x < 3 after multiplying by −1 | From 5 − x > 2 we get −x > −3, and multiplying by −1 gives x < 3. |
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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