Higher November 2018 Paper 4 Q8
8
(a) Two numbers, \(P\) and \(Q\), are written as products of their prime factors.\[P = 2^5 \times 3^2 \times 5^3 \times 11 \qquad Q = 2^4 \times 3 \times 5^4 \times 7\]
(i) Find the lowest common multiple (LCM) of \(P\) and \(Q\). [2]
(ii) The number \(C\) is written as the product of its prime factors.
\(C = 2^3 \times 3 \times 5^2\)
Work out \(P \div C\), leaving your answer as a product of powers of prime numbers. [2]
\(C = 2^3 \times 3 \times 5^2\)
Work out \(P \div C\), leaving your answer as a product of powers of prime numbers. [2]
(b)
(i) Write 450 as a product of its prime factors. [3]
(ii) Find the highest common factor (HCF) of 270 and 450. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 13 860 000 oe | 2 | M1 for \(2^5 \times 3^2 \times 5^4 \times 11 \times 7\) with at most one error | condone \(2^5 \times 3^2 \times 5^4 \times 11 \times 7\) for 2 marks |
| (ii) \(2^2 \times 3 \times 5 \times 11\) isw | 2 | M1 for answer one step away | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(2 \times 3^2 \times 5^2\) | 3 | B2 for answer one step away or a correct diagram e.g. factor tree or B1 for 2, 3 and 5 identified e.g. could be in a factor tree | |
| (ii) 90 | 3 | M2 for [270 =] 2 × 3 × 3 × 3 × 5 oe or M1 for 2, 3 and 5 as factors of 270 or for an answer of 2, 3, 5, 6, 9, 10, 15, 18, 30 or 45 | Accept in factor tree or a division spine, allow M1 if one step away |