Numeracy Intermediate November 2024 Unit 2 Q6
6. 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 |