Foundation June 2017 Paper 1 Q5
5 Work out \(\quad 58 \times 73\) [3 marks]
| Answer | Mark | Comments | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Alternative method 1 | |||||||||||
| \[\begin{array}{r} 73 \\ \times \;\; 58 \\ \hline 584 \\ 3650 \end{array}\]or\[\begin{array}{r} 58 \\ \times \;\; 73 \\ \hline 174 \\ 4060 \end{array}\] | 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 174 + their 4060 or their 584 + their 3650 | M1dep | ||||||||||
| 4234 | A1 | ||||||||||
| Alternative method 2 | |||||||||||
| M1 | At least three correct values | |||||||||
| their 3500 + their 560 + their 150 + their 24 | M1dep | ||||||||||
| 4234 | A1 | ||||||||||
| Alternative method 3 | |||||||||||
![]() | M1 | At least three of the 2-digit numbers correct | |||||||||
| Total calculated for each diagonal with at least one correct carrying figure | M1dep | Clear attempt to add each diagonal | |||||||||
| 4234 | A1 | ||||||||||
Additional guidance
| \(50 \times 70 + 8 \times 3 \quad (= 3524)\) | M0M0A0 |
| Alternative method 1 – if the place holder 0 is missing or mis-aligned, allow this to be evidenced by their 4 as the units value in their answer | |
| For alternative method 3, diagonals must slope the correct way | |
| Diagonal lines not present is M0 unless this is recovered by seeing correct totals around the grid | |
Example of alternative method 3 with carrying completed once![]() | M1M1depA0 |

