Foundation November 2024 Paper 2 Q17
17 Anika has a shelf 79.6 cm long.
She has many books, each of width 3.4 cm.
Anika puts two paperweights, each of width 5 cm, and the maximum possible number of books on the shelf.
Work out the amount of space on the shelf that is left over.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1.6 with correct working | 5 | M2 for \(\dfrac{79.6 - 2 \times 5}{3.4}\) oe | “Correct working” requires evidence of at least M2 For M2 allow for use of trials to try to make 69.6 e.g. \(3.4 \times 20\) [= 68] with \(79.6 - 2 \times 5\) oe seen Accept attempt with repeated addition/subtraction of 3.4 for M2 isw using estimation only after correct values shown for M2 |
| or M1 for \(79.6 - 2 \times 5\) oe implied by 69.6 AND | |||
| B1 for 20 [. …] or [total length of books = ] 68 [cm] AND | 20[. …] seen or used as max number of books For B1 ignore remainders with 20 68 must be their total length of books and not just a value in working | ||
| M1 for \(79.6 - (\textit{their}\ 20 \times 3.4 + 2 \times 5)\) | their 20 must be written and working shown for M1. M1 Dep on answer < 3.4 Could be implied by e.g. repeated subtraction | ||
| If 0 or 1 scored, instead award SC2 for answer 1.6 with no working or insufficient working | Alternative method: M2B1M1 or M2B0M1 depending on “20” for \(\left(\dfrac{79.6 - 2 \times 5}{3.4} - \textit{their}\ 20\right) \times 3.4\) Alt scheme if consistent omission of two paperweights or one paperweight with no adjustment later M1 for \(\dfrac{79.6[-5]}{3.4}\) oe B2 for answer 11.6 or B1 for 68 [cm] OR M1 for \(\dfrac{79.6[-5]}{3.4}\) oe B1 for 23.[…] ignore remainders or for 78.2 or 21.[…] ignore remainders or for 71.4 A1 for answer 1.4 or 3.2 | ||