Foundation June 2024 Paper 2R Q17
17 Write 1400 as a product of powers of its prime factors.
Show your working clearly.
(3)
| Scheme | Marks | ||||||
|---|---|---|---|---|---|---|---|
| eg \(2 \times 2 \times 350\) or \(2 \times 7 \times 100\) or \(2 \times 5 \times 140\) or \(5 \times 7 \times 40\) or \(5 \times 5 \times 56\) or (\(14 \times 100 =\)) \(2 \times 7 \times 100\) or (\(28 \times 50 = 4 \times 7 \times 50 =\)) \(2 \times 2 \times 7 \times 50 =\) or
| M1 | ||||||
eg 2, 2, 2, 5, 5, 7 | M1 | ||||||
| Working required Answer: \(2^3 \times 5^2 \times 7\) | A1 | ||||||
| (3) | |||||||
| (3 marks) |
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, 350 (see LHS for examples of the amount of work needed for the award of this mark)
Example of 3 stages with 1 incorrect stage: \(1400 = 10 \times 14 = 2 \times 5 \times 2 \times 7\)
M1: dep on M1 for \(2 \times 2 \times 2 \times 5 \times 5 \times 7\) or \(2^3, 5^2, 7\) or \(2^3 + 5^2 + 7\) (Ignore 1’s) (may be seen in a fully correct factor tree or ladder)
A1: dep on M2 (do not allow 1 in the final answer)
Can be in any order (allow \(2^3 . 5^2 . 7\)) but must be in index form as asked for.