Numeracy Foundation November 2024 Unit 2 Q10
10. Alfred has been given the job of varnishing a floor.
A plan of the floor is shown in the diagram below.
Diagram not drawn to scale
One tin of varnish contains enough to cover an area of 1·6 m2.
Alfred says,
| I only need 3 tins of varnish to do this job. |
Is Alfred correct?
Yes ☐ No ☐
You must show all your working. [4]
| Scheme | Marks | Comments |
|---|---|---|
| (Area of floor) ½ × 4.3 × 2.7 5.8 (m2) or 5.805 (m2) or 5.81 (m2) | M1 A1 | Accept 6 (m2) provided correct working seen |
| (3 tins covers) 3 × 1.6 (= 4.8) OR (Number of tins of varnish) 5.805 ÷ 1.6 (= 3.628…) OR (Area per tin if only 3 tins) 5.805 ÷ 3 (= 1.935) | M1 | Use of information given in the question OR FT ‘their derived area of floor’, including if not dimensionally correct, e.g. allow FT if ‘their derived area’ = 2.7 + 4.3 = 7 or 4.3 × 2.7 = 11.61 or \(\sqrt{4.3^2 + 2.7^2}\) = 5.077… or \(\sqrt{4.3^2 + 2.7^2}\) + 2.7 + 4.3 = 12.077… |
| States or unambiguously implies ‘No’ AND a suitable correctly evaluated calculation, e.g.
| A1 | FT ‘their derived area of floor’, provided their area calculation involves 4.3 × 2.7 and is dimensionally correct Ignore additional spurious statements or calculations provided a correct comparison with ‘their number of tins’ with 3, or ‘their area’ with 4.8(m2) is given |