Foundation June 2022 Paper 1 Q4
4 Jamie has some empty boxes.
Each box can hold 73 pencils.
Jamie has 590 pencils.
Jamie says that eight boxes are needed to hold all of the pencils.
Is Jamie correct?
You must show your working. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| No, they need 9 boxes oe with correct working | 2 | M1 for \(590 \div 73\), \(590 \div 8\) or \(73 \times 8\) If 0 scored, SC1 for 8.08[2…], 8.1, 73.7[5], 73.8, 584 or 6 | Correct working requires M1 Allow M1 for repeated addition/subtraction if method shown. If only numbers listed addition must reach 584, subtraction must reach 6 See Exemplars |
Appendix
Exemplar responses for Q4
| Response | Mark |
|---|---|
| \(73 \times 8\) No 8 boxes hold 584 pencils | 2 |
| No 8 boxes hold 584 pencils | SC1 |
| \(73 \times 8\) No Jamie needs an extra box for the other 6 pencils | 2 |
| There are 6 remaining pencils | SC1 |
| \(590 \div 73 = 8.08\) you can’t have a fraction of a box (doesn’t say whether Jamie is correct) | M1 |
| Jamie is incorrect 8 boxes will only hold 584 pencils | SC1 |
| Jamie is correct \(590 \div 73 = 8\) | M1 |
| No, they need an extra box | 0 |