Foundation June 2022 Paper 2 Q21
21 Show that 2125 can be written as
a cube number multiplied by a prime number between 10 and 20 [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 125 and 17 or \(5^3\) and 17 or 5 and 5 and 5 and 17 | B2 | together in any order eg \(125 \times 17\) or \(17 \times 5^3\) or 5, 5, 5, 17 or \(2125 \div 17 = 125\) or \(2125 \div 125 = 17\) B1 at least three of 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197 etc (allow \(2^3, 3^3, 4^3\) etc) or all four of 11, 13, 17, 19 (ignore any numbers not between 10 and 20) or (cube number \(\gt 1\)) \(\times\) (prime number between 10 and 20) or \(2125 \div\) (cube number \(\gt 1\)) or \(2125 \div\) (prime number between 10 and 20) |
Additional guidance
| B1 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts | |
| B2 responses may be seen on a factor tree | |
| B1 for three cube numbers given in index form – evaluations can be ignored eg \(4^3\ \ 5^3\ \ 6^3\) scores B1 with no evaluations or with incorrect evaluations | |
| B1 for multiplications or divisions – evaluation can be ignored eg1 \(2^3 \times 13\) scores B1 with no evaluation or evaluated incorrectly eg2 \(2125 \div 27\) scores B1 with no evaluation or evaluated incorrectly eg3 \(2125 \div 11\) scores B1 with no evaluation or evaluated incorrectly | |
| 125 and 17 seen in multiple attempts is B2 if 2125 included eg \(125 \times 17 = 2125\) or \(2125 \div 17 = 125\) or \(2125 \div 125 = 17\) seen amongst multiple attempts | B2 |
| 125 and 17 seen in multiple attempts is B1 if 2125 not included eg \(125 \times 17\) seen amongst multiple attempts | B1 |
| 11 13 15 17 19 does not score B1 unless 11 13 17 19 selected | |
| Incomplete list eg 11 13 19 does not score B1 |