Numeracy Higher June 2022 Unit 2 Q6
6. Dewi and Cai are building a fence.
They are using fence boards that overlap each other.
The diagrams below show the first 3 boards in position, and how the boards overlap.
Plan view of how the boards overlap
Diagrams not drawn to scale
The fence needs to cover a length of 5700 mm, correct to the nearest 100 mm.
When in position, each board has a width of 100 mm, correct to the nearest 5 mm.
Each overlap is exactly 20 mm, as shown in the diagram above.
Calculate the minimum number of boards that Dewi and Cai need to buy to guarantee that they have enough to build the fence.
You must show all your working. [6]
| Scheme | Marks | Comments |
|---|---|---|
| M marks may be awarded from working with multiples of e.g. 77.5 and/or 97.5 to reach e.g. 5750 | ||
| \(\dfrac{5750}{97.5 - 20}\) or \(\dfrac{5750}{77.5}\) | M3 | M2 for \(\dfrac{\text{length}}{\text{width} - 20}\), where 5700 < length ≤ 5800 and 95 ≤ width < 100 M1 for \(\dfrac{5750}{97.5}\) |
| \(-\dfrac{97.5}{97.5 - 20}\) or \(-\dfrac{97.5}{77.5}\) | m1 | FT from M2 for ‘their 97.5’ |
| (May be seen as \(\dfrac{5750 - 97.5}{97.5 - 20}\) or \(\dfrac{5652.5}{77.5}\)) | ||
| = 72.9(3…) or 73 | A1 | FT is possible from m0 provided M3 or M2 previously awarded From M3, 5750 ÷ 77.5 = 74.19(3…) rounded down to 74 (boards) is awarded M3m0A1 unless further correct working seen |
| (Number of boards needed =) 74 (boards) | A1 | FT from M2m1A1 for a correct evaluation using their bounds, rounded up and +1 If no marks awarded, and from a misinterpretation of the question, SC4 for an answer of 69 boards from \(\dfrac{5650}{102.5 - 20} - \dfrac{102.5}{102.5 - 20} + 1\) or \(\dfrac{5547.5}{82.5} + 1\) OR SC3 for an answer of 67(.242…) or 68 from \(\dfrac{5650}{102.5 - 20} - \dfrac{102.5}{102.5 - 20}\) or \(\dfrac{5547.5}{82.5}\) OR SC2 for a correct evaluation (rounded, truncated or unrounded) of the calculation \(\dfrac{\text{length – width}}{\text{width} - 20}\), where 5600 ≤ length < 5700 and 100 < width ≤ 105 OR SC1 for an answer of 68(.484…) or 69 boards from \(\dfrac{5650}{102.5 - 20}\) or \(\dfrac{5650}{82.5}\) OR If no marks awarded, SC1 for sight of 97.5 and 5750 |
Alternative
6. Alternative method:
| Scheme | Marks | Comments |
|---|---|---|
| M marks may be awarded from working with multiples of e.g. 77.5 and/or 97.5 to reach e.g. 5750 | ||
| \(\dfrac{5750}{97.5 - 20}\) or \(\dfrac{5750}{77.5}\) | M3 | M2 for \(\dfrac{\text{length}}{\text{width} - 20}\), where 5700 < length ≤ 5800 and 95 ≤ width < 100 M1 for \(\dfrac{5750}{97.5}\) |
| \(-\dfrac{20}{97.5 - 20}\) or \(-\dfrac{20}{77.5}\) | m1 | FT from M2 for ‘their 97.5’ |
| (May be seen as \(\dfrac{5750 - 20}{97.5 - 20}\) or \(\dfrac{5730}{77.5}\)) | ||
| (Number of boards needed =) 74 (boards) | A2 | FT from M2m1 for a correct evaluation using their bounds, rounded up A1 for 73(.9354…) OR A1 on FT from M3m0 for 74.19(3…) or 75 An answer of 74.19(3…) rounded down to 74 (boards) is awarded M3m0A1 unless further correct working seen OR A1 on FT from M2m1 for an unrounded correct evaluation using their bounds If no marks awarded, and from a misinterpretation of the question, SC4 for an answer of 69 boards from \(\dfrac{5650}{102.5 - 20} - \dfrac{20}{102.5 - 20}\) or \(\dfrac{5630}{82.5}\) OR SC3 for an answer of 68(.2424…) boards from \(\dfrac{5650}{102.5 - 20} - \dfrac{20}{102.5 - 20}\) or \(\dfrac{5630}{82.5}\) OR SC2 for a correct evaluation (rounded, truncated or unrounded) of the calculation \(\dfrac{\text{length – 20}}{\text{width} - 20}\), where 5600 ≤ length < 5700 and 100 < width ≤ 105 OR SC1 for an answer of 68(.484…) or 69 boards from \(\dfrac{5650}{102.5 - 20}\) or \(\dfrac{5650}{82.5}\) OR If no marks awarded, SC1 for sight of 97.5 AND 5750 |