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S4 Maths: Inequalities and Linear Programming (Paper 2 (intermediate))
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 2 (intermediate))
Paper 2 (intermediate): 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 (2/3)x − 1 < 5.
- FoundationSolve the inequality 0.5x + 2.4 ≥ 3.9.
- FoundationSolve the inequality 7 − 4x > 3 − 2x.
- IntermediateSolve the inequality 3x − 4 ≤ 5x + 8.
- IntermediateSolve −4 ≤ 2x − 6 < 8 and count the integer solutions.
- IntermediateFor integer x, find the smallest solution of (x + 3)/2 > 4.
- ChallengeSolve 5(x − 1) ≤ 2(x + 7) and give the greatest integer solution.
- ChallengeIf 2x + 1 > 7 and 3x − 2 < 16 hold together, list all integer solutions.
- ChallengeThe vertices of a feasible region are (0,0), (4,0) and (0,3). Find the maximum of P = 3x + 2y.
- Error redo(Error clinic) Mei solved x/3 + 2 ≥ 5 as x ≥ 3. What did she miss, and what is the correct answer?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | x < 9 | Add 1 to get (2/3)x < 6, then multiply by 3/2 to get x < 9. |
| 2 | x ≥ 3 | Subtract 2.4 to get 0.5x ≥ 1.5, then divide by 0.5 to get x ≥ 3. |
| 3 | x < 2 | Add 4x and subtract 3 to get 4 > 2x, then divide by 2 to get x < 2. |
| 4 | x ≥ −6 | Rearrange to −2x ≤ 12, then divide by −2 and reverse the sign to get x ≥ −6. |
| 5 | 1 ≤ x < 7, giving 6 integer solutions | Add 6 to all parts to get 2 ≤ 2x < 14, then halve to get 1 ≤ x < 7; the integers are 1 to 6. |
| 6 | 6 | Multiply by 2 to get x + 3 > 8, so x > 5 and the smallest integer is 6. |
| 7 | x ≤ 19/3, about 6.333, so the greatest integer is 6 | Expand to 5x − 5 ≤ 2x + 14, move terms to get 3x ≤ 19, and the greatest integer is 6. |
| 8 | x = 4 and x = 5 | The first gives x > 3 and the second x < 6, so only 4 and 5 satisfy both. |
| 9 | 12 | The vertices give 0, 12 and 6, so the maximum is 12, taken at (4,0). |
| 10 | The correct answer is x ≥ 9; she forgot that 2 must also be multiplied by 3 | Subtract 2 to get x/3 ≥ 3, then multiply by 3 to get x ≥ 9. |
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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