Foundation June 2023 Paper 2 Q15
15. A shop sells storage boxes.
A stack of four of these boxes is shown below.
Diagram not drawn to scale
The height of a single box is 35·2 cm.
The height of four of the same boxes when stacked is 52 cm.
A different stack of these boxes has a height of 85·6 cm.
How many boxes are in this stack? [5]
| Scheme | Marks | Comments |
|---|---|---|
| (52 – 35.2) ÷ 3 | M1 | May be seen in stages |
| = 5.6 (cm) | A1 | |
| (85.6 – 35.2) ÷ 5.6 or 35.2 + 9 × 5.6 oe | m1 | FT ‘their (52 – 35.2) ÷ 3’ Note: If a candidate is awarded M1 A0 and attempts repeated additions with their incorrect 5.6, there needs to be enough additions to get to a value just below or just above 85.6. They may use a mixture of + ‘their 5.6’ (1box) and + ‘their 16.8’ (3boxes) within their additions. |
| 10 (boxes) | A2 | Not from incorrect working A1 for 9 (boxes) If no marks, from a method starting with (52 – 35.2) ÷ 4 = 4.2 award SC2 for 12 boxes from (85.6 – 52) ÷ 4.2 + 4 or (85.6 – 35.2) ÷ 4.2 SC1 for 8 boxes from (85.6 – 52) ÷ 4.2 |
| (5) |
Alternative
Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| (52 – 35.2) ÷ 3 | M1 | |
| = 5.6 (cm) | A1 | |
| (85.6 – 52) ÷ 5.6 or 52 + 5.6 + 5.6 + 5.6 + 5.6 + 5.6 + 5.6 oe or | m1 | FT ‘their (52 – 35.2) ÷ 3’ Note: If a candidate is awarded M1 A0 and attempts repeated additions with their incorrect 5.6, there needs to be enough additions to get to a value just below or just above 85.6. They may use a mixture of + ‘their 5.6’ (1box) and + ‘their 16.8’ (3boxes) within their additions. |
| 10 (boxes) | A2 | Not from incorrect working A1 for 6 (boxes) |
Alternative method 2
| Scheme | Marks | Comments |
|---|---|---|
| 52 – 35.2 = 16.8 with an attempt at repeated additions to get to 85.6. At least 1 addition must be attempted. e.g. 35.2 + 16.8 + ……. or 52 + 16.8 + ……. | M2 | |
| 35.2 + 16.8 + 16.8 + 16.8 ( = 85.6) or (85.6 – 35.2) ÷ 16.8 OR 52 + 16.8 + 16.8 or (85.6 – 52) ÷ 16.8 | m1 | Note: If the answer to 52 – 35.2 is incorrect and they have attempted repeated additions with their incorrect 16.8 onto either 35.2 or 52. Marks can only be awarded if there are enough additions to get to a value just below or just above 85.6. In this case award M1 m1. Note: (85.6 – 35.2) ÷ 16.8 or (85.6 – 52) ÷ 16.8 implies M2. If the answer to 52 – 35.2 is incorrect but the correct divisions are shown with their incorrect 16.8 then award M1 m1. |
| 10 (boxes) | A2 | Not from incorrect working A1 for 1 + 3 + 3 + 3 or 4 + 3 + 3 or 9 boxes or 6 boxes (as appropriate for their method) |