Foundation November 2019 Paper 1 Q5
5 Work out \(\quad 76 \times 24\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – traditional method | ||
| 304 or 1520 with the 0 correct for the multiplication by 20 or 144 or 1680 with the 0 correct for the multiplication by 70 | M1 | values may be seen separately or in rows if 1520 or 1680 incorrect, placeholder 0, or equivalent must be present |
| their 304 + their 1520 or their 144 + their 1680 | M1dep | |
| 1824 | A1 | |
| Alternative method 2 – grid method | ||
| At least three of 1400, 280, 120 and 24 | M1 | may not be in a grid |
| their 1400 + their 280 + their 120 + their 24 | M1dep | |
| 1824 | A1 | |
| Alternative method 3 – Napier’s bones | ||
 | M1 | oe at least three of the calculated values correct |
| Attempt to total correctly four diagonals for their table with carrying figure seen | M1dep | |
| 1824 | A1 | |
| Alternative method 4 – breaking calculation down | ||
| Calculation broken down correctly with a maximum of one calculation error | M1 | eg \(76 \times 10 \times 2\) (+) \(70 \times 4\) (+) \(6 \times 4\) with at least two of 1520, 280 and 24 correct |
| Addition of their parts | M1dep | eg 1520 + 280 + 24 |
| 1824 | A1 | |
Additional guidance
| \(70 \times 20 + 6 \times 4\) (= 1424) | M0M0A0 |
| Alt 1 304 + 152 = 456 | M0M0A0 |
| Alt 1 If the 0 is missing, allow 0 to be replaced by \(x\) or a placeholder space (may be implied by their 4 in units column of their final answer) | |
| Alt 3 Diagonal lines must slope consistently for M1 unless recovered | |
| Alt 3 Diagonal lines missing is M0 unless recovered | |
| Alt 3 For M1M1dep, a carrying figure must be seen or implied | |
| Alt 3 Answer must be clearly stated and not left “around” the grid |