Chapter 5 Practice Paper: Linear Equations and Simple Inequalities — mark scheme
Secondary 1 · Lower Secondary Mathematics (G2 / G3)
Mark scheme
Linear Equations and Simple Inequalities
Chapter 5 Practice Paper · Maximum mark: 30
Notes for markers
- Award one mark for each point marked [1], up to the marks shown for the part.
- Accept answers worded differently that show the same understanding.
- Units are required where the answer line shows one. ecf: allow an error carried forward.
Section A: Multiple-choice questions[10 marks]
| Question | Answer | |
|---|---|---|
| 1 | ADivide both sides by 4: x = 28 ÷ 4 = 7. | 1 |
| 2 | BSubtract 9 from both sides: x = 3 - 9 = -6. | 1 |
| 3 | CAdd 4 to both sides to get 3x = 15, then divide by 3: x = 5. | 1 |
| 4 | DMultiply both sides by 5: x = -2 × 5 = -10. | 1 |
| 5 | BExpand to 2x + 6 = 14, so 2x = 8 and x = 4. (Dividing both sides by 2 first gives x + 3 = 7, the same answer faster.) | 1 |
| 6 | CMultiply both sides by 4: 3x - 2 = 16, so 3x = 18 and x = 6. | 1 |
| 7 | ASubtract 7 from both sides: -2x = -6. Divide by -2: x = 3. Check: 7 - 2(3) = 1. | 1 |
| 8 | DAdd 6 to both sides: x > 7. The inequality sign keeps its direction when the same number is added to both sides. | 1 |
| 9 | BDividing by the negative number -3 flips the inequality: x < 12 ÷ (-3), so x < -4. | 1 |
| 10 | DTwice the number n is 2n, and 5 more gives 2n + 5 = 17, so 2n = 12 and n = 6. | 1 |
Section B: Structured questions[20 marks]
| Question | Answer and marking points | Marks |
|---|---|---|
| 1(a) | x = -2 - 7 = -9 [1] | 1 |
| 1(b) | 5x = 22 + 3 = 25, so x = 5.
| 2 |
| 1(c) | Subtract 4: -3x = 9, then divide by -3: x = -3.
Award the second mark only if the sign is handled correctly; x = 3 fails the check 4 - 3(3) = -5. | 2 |
| Total for Question 1 | 5 | |
| 2(a) | Multiply both sides by 2: x + 5 = 14, so x = 9.
| 2 |
| 2(b) | Multiply both sides by 3: 2x - 1 = 15, so 2x = 16 and x = 8.
A common error is adding 1 before clearing the fraction; award marks for correct working from wherever the student restarts. | 2 |
| 2(c) | With x = 8: (2 × 8 - 1)/3 = 15/3 = 5, which matches the right-hand side.
| 1 |
| Total for Question 2 | 5 | |
| 3(a) | Jun has y dollars and Wei has y + 12 dollars, so y + (y + 12) = 58.
| 2 |
| 3(b) | 2y + 12 = 58, so 2y = 46 and y = 23. Jun has $23.
| 2 |
| 3(c) | 23 + 12 = $35. [1] | 1 |
| Total for Question 3 | 5 | |
| 4(a) | x < 5 [1] | 1 |
| 4(b) | 3x > 12, so x > 4.
| 2 |
| 4(c) | Divide both sides by -2 and flip the inequality: x < -5. Dividing by a negative number reverses the direction of the inequality.
The answer x > -5 earns the explanation mark only if the flip is mentioned; the final answer must show the reversal. | 2 |
| Total for Question 4 | 5 |
— End of mark scheme —