Numeracy Higher November 2022 Unit 1 Q6
6. Luke wants to lay wooden flooring in the living room of his house.
The plan view of his living room floor is shown below.
Diagram not drawn to scale
All the measurements shown on the diagram are correct to the nearest 0·2 m.
Luke knows that:
- each pack of wooden flooring covers exactly 3 m2
- 10% needs to be added to the area of the floor that is to be covered, to ensure he has enough wooden flooring.
Calculate the minimum number of packs needed to guarantee that Luke has enough wooden flooring to cover his living room floor.
You must show all your working. [6]
Number of packs is ……………
| Scheme | Marks | Comments |
|---|---|---|
| (Area of floor =) length × width + ½ × base × height | M1 | Must be from working with bounds If lower bounds used, they must be in a similar range to those shown for M1 below Use of any formula to calculate the complete area of the floor |
| 4.5 × 8.1 + \(\dfrac{5.1 \times (5.5 - 4.5)}{2}\) (36.45) (2.55) | M2 | M1 for ‘their 4.5 × 8.1 + \(\dfrac{5.1 \times (5.5 - 4.5)}{2}\)’ where 4.4 < ‘their 4.5’ \(\leqslant\) 4.6 and 8 < ‘their 8.1’ \(\leqslant\) 8.2 and 5(.0) < ‘their 5.1’ \(\leqslant\) 5.2 and 5.4 < ‘their 5.5’ \(\leqslant\) 5.6 If using ½ × base × height, allow 0.8 \(\leqslant\) height \(\leqslant\) 1.2 |
| = 39 (m2) | A1 | CAO |
| (Amount of flooring needed =) 39 + 0.1 × 39 or equivalent | M1 | FT ‘their 39’ provided 1 mark previously awarded Accept 39/3 + 0.1×39/3 |
| = 42.9 (m2) AND 15 packs needed | A1 | Accept 43 (m2) AND 15 packs needed Accept 14.3 AND 15 packs needed On FT, accept the amount of flooring correct to 1 d.p. or rounded up to the nearest whole number for ‘their 39 + 0.1 × 39’ Allow M1A1 for an answer of 15 packs from an area of 39 (m2) provided no incorrect work seen FT ‘their 39’ If no marks awarded, SC1 for sight of 4.5, 8.1, and 5.1 |
Alternative method for first 4 marks:
| Scheme | Marks | Comments |
|---|---|---|
| (Area of floor =) length × width + ½×(sum of parallel sides)×height | M1 | Must be from working with bounds If lower bounds used, they must be in a similar range to those shown for M1 below Use of any formula to calculate the complete area of the floor |
| 4.5 × (8.1 – 5.1) + \(\dfrac{(5.5 + 4.5) \times 5.1}{2}\) (13.5) (25.5) | M2 | M1 for ‘their 4.5 × (8.1−5.1) + \(\dfrac{(5.5 + 4.5) \times 5.1}{2}\)’ where 4.4 < ‘their 4.5’ \(\leqslant\) 4.6 and 8 < ‘their 8.1’ \(\leqslant\) 8.2 and 5(.0) < ‘their 5.1’ \(\leqslant\) 5.2 and 5.4 < ‘their 5.5’ \(\leqslant\) 5.6 Allow width of rectangle to be 2.8 \(\leqslant\) width \(\leqslant\) 3.2 |
| = 39 (m2) | A1 | CAO |