Foundation June 2023 Paper 1 Q23
23
Show your working clearly. (2)
\(A = 2 \times 2 \times 2 \times 3 \times 3 \times 5\)
\(B = 2 \times 2 \times 3 \times 3 \times 3 \times 5\)
Show your working clearly. (2)
| Scheme | Marks | ||||||
|---|---|---|---|---|---|---|---|
| eg \(2 \times 2 \times 75\) or \(3 \times 5 \times 20\) or \(2 \times 3 \times 50\) or \(5^2 \times 12\) or
| M1 | ||||||
| Working required Answer: \(2 \times 2 \times 3 \times 5 \times 5\) | A1 | ||||||
| (2) |
Notes
M1: for 2 correct stages in prime factorisation with 0 incorrect stages
or at least 3 stages in prime factorisation with no more than 1 incorrect stage.
Each stage gives 2 factors – may be in a factor tree or a table or listed eg 2, 2, 75 (see LHS for examples of the amount of work needed for the award of this mark). Example of 3 stages with 1 incorrect stage:
\(300 = 100 \times 30 = 2 \times 50 \times 5 \times 6\)
A1: dep on M1, oe eg \(2^2 \times 3 \times 5^2\)
| Scheme | Marks |
|---|---|
| (\(5A =\)) \(2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5\) oe (= 1800) or (\(5A =\)) \(2^3 \times 3^2 \times 5^2\) (= 1800) or (\(7B =\)) \(2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 7\) oe (= 3780) or (\(7B =\)) \(2^2 \times 3^3 \times 5 \times 7\) (= 3780) | M1 |
| Working required Answer: 37800 | A1 |
| (2) | |
| (4 marks) |
Notes
M1: for method to find \(5A\) or \(7B\) as prime factors (may be seen in factor tree, table or Venn diagram) or as an integer
or for listing at least 3 multiples of each number eg 1800, 3600, 5400... and 3780, 7560, 11340...
or for an answer of 1080 oe eg \(2^3 \times 3^3 \times 5\)
A1: dep on M1, oe eg \(2^3 \times 3^3 \times 5^2 \times 7\)