Chapter 1 Practice Paper: Primes, HCF and LCM — mark scheme
Secondary 1 · Lower Secondary Mathematics (G2 / G3)
Mark scheme
Primes, HCF and LCM
Chapter 1 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 | B59 has no factors other than 1 and itself. 51 = 3 × 17, and 63 and 81 are divisible by 3. | 1 |
| 2 | CA prime number has exactly two factors. The number 1 has only one factor, so it is neither prime nor composite. | 1 |
| 3 | ASplitting 60 completely into primes gives 2 × 2 × 3 × 5 = 2² × 3 × 5. The other options still contain composite factors. | 1 |
| 4 | D18 = 2 × 3² and 24 = 2³ × 3. The HCF takes the lowest power of each common prime: 2 × 3 = 6. | 1 |
| 5 | B6 = 2 × 3 and 8 = 2³. The LCM takes the highest power of each prime that appears: 2³ × 3 = 24. | 1 |
| 6 | C40 = 2 × 2 × 2 × 5 = 2³ × 5. The other options include the composite numbers 4 and 10. | 1 |
| 7 | D81 = 9 × 9, and in prime factors 3 appears four times — an even power, which every perfect square has. | 1 |
| 8 | A64 = 4 × 4 × 4 = 4³, so the cube root of 64 is 4. | 1 |
| 9 | BThe common primes are 2 and 3. Take the lowest power of each: 2² × 3 = 12. | 1 |
| 10 | CThe next common multiple of 8 and 12 is the LCM, 24 minutes. So the bells ring together again at 09:24. | 1 |
Section B: Structured questions[20 marks]
| Question | Answer and marking points | Marks |
|---|---|---|
| 1(a) | 84 = 2 × 2 × 3 × 7 = 2² × 3 × 7
The factor tree or ladder method must end in primes. Accept an un-ordered product such as 3 × 2² × 7. | 1 |
| 1(b) | 126 = 2 × 3 × 3 × 7 = 2 × 3² × 7
| 1 |
| 1(c) | 2 × 3 × 7 = 42
Accept 42 with or without the working line, but the factor product must be visible. | 1 |
| 1(d) | 2² × 3² × 7 = 252
A common error is to multiply 84 × 126. Award no marks for 10584. | 2 |
| Total for Question 1 | 5 | |
| 2(a) | 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³
| 1 |
| 2(b) | Group the prime factors into three identical sets: 216 = (2 × 3) × (2 × 3) × (2 × 3) = 6 × 6 × 6. Taking one number from each set gives the cube root, so the cube root of 216 is 6.
| 2 |
| 2(c) | 3 × 7 = 21
| 2 |
| Total for Question 2 | 5 | |
| 3(a) | The length must be a common factor of 96 and 120, and the longest such length is the HCF. Listing factors: 96 = 1 × 96, 2 × 48, 3 × 32, 4 × 24, 6 × 16, 8 × 12, and 120 = 24 × 5, so 24 is a factor of both. No larger common factor works, so each piece is 24 cm long. Check: 96 ÷ 24 = 4 and 120 ÷ 24 = 5, with no remainders.
Accept any valid method: listing factors, a factor tree, or the ladder method. The recognition of HCF earns the first mark even if the arithmetic slips. | 3 |
| 3(b) | 96 ÷ 24 + 120 ÷ 24 = 4 + 5 = 9
| 2 |
| Total for Question 3 | 5 | |
| 4(a) | 15 = 3 × 5 and 20 = 2² × 5. The LCM is 2² × 3 × 5 = 60, so both ferries leave together again after 60 minutes.
| 2 |
| 4(b) | 09:45
| 1 |
| 4(c) | 30 is not a common multiple of 15 and 20. Although 30 is a multiple of 15, 30 ÷ 20 = 1.5, so the 20-minute ferry does not leave at 09:15. Both intervals must divide the time exactly.
Do not accept "30 is too small" without a reason tied to the intervals. | 2 |
| Total for Question 4 | 5 |
— End of mark scheme —