Foundation November 2018 Paper 1 Q5
5 Work out \(\quad 83 \times 26\) [3 marks]
| Answer | Mark | Comments | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Alternative method 1 | |||||||||||
 | M1 | at least one row correct, with the 0 correct for multiplication by the multiple of 10 you may see the rows of working switched | |||||||||
| their 498 + their 1660 or their 78 + their 2080 | M1dep | ||||||||||
| 2158 | A1 | ||||||||||
| Alternative method 2 | |||||||||||
| M1 | at least three of the calculated values correct may be seen as 4 calculations, not in a grid | |||||||||
| their 1600 + their 480 + their 60 + their 18 | M1dep | ||||||||||
| 2158 | A1 | ||||||||||
| Alternative method 3 | |||||||||||
 | M1 | at least three of the calculated values correct | |||||||||
| Total calculated for each diagonal with at least one correct carrying figure | M1dep | clear attempt to add each diagonal | |||||||||
| 2158 | A1 | ||||||||||
Additional guidance
| \(20 \times 80 + 6 \times 3 \quad (= 1618)\) | M0A0 |
| Alternative method 1: if the place holder 0 is missing or misaligned, allow this to be evidenced by their 8 as the units value in the answer, or an ‘x’ in place of the 0 | |
| Alternative method 2: if numbers are broken down further they must have at least 8 of the calculated values correct in example below (oe) eg 40 40 3 and 10 10 6 (ie a maximum of one error) | |
| Alternative method 3: diagonals must slope the correct way for M1 (unless recovered) | |
| Diagonal lines not present is M0 unless this is recovered by seeing correct totals around the grid | |
Example of alternate method 3 with carrying completed once | M1M1depA0 |