Foundation June 2018 Paper 1 Q7
7 A rectangular carpet measures 8 m by 6 m
Part of the carpet is covered by a square rug of length 2 m
Not drawn accurately
Show that \(\quad \dfrac{1}{12} \quad\) of the carpet is covered by the rug. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(6 \times 8\) or 48 or \(2^2\) or \(2 \times 2\) or 4 | M1 | may be on diagram |
| \(48 \div 4 = 12\) or \(48 \div 12 = 4\) or \(4 \times 12 = 48\) or \(\dfrac{4}{48}\ (=)\ \dfrac{1}{12}\) | A1 | oe eg \(48 \div 2 = 24\) and \(24 \div 2 = 12\) |
| Alternative method 2 | ||
| \(6 \div 2\) or \(2 \div 6\) or \(8 \div 2\) or \(2 \div 8\) | M1 | |
| \(3 \times 4 = 12\) or \(\dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{12}\) with full working seen | A1 | Need to justify where this product comes from with M1 work seen |
| Alternative method 3 | ||
| One row of 4 squares drawn or one column of 3 squares drawn | M1 | Mark intention, not accuracy of drawing, 2m labels not required |
| Rectangle split into 4 columns and 3 rows | A1 | |
Additional guidance
| (\(2 \times 2 = 4\), \(6 \times 8 = 48\) and) 4 is \(\dfrac{1}{12}\) of 48 | M1A1 |
| 4 12s are 48 | M1A1 |
| \(8 \times 6 = 48\), \(12 \div 48 = 4\) (cannot condone incorrect order as ‘show that’) | M1A0 |
| \(\dfrac{4}{48}\) so correct | M1A0 |
| Beware 4 (or 12) arising from incorrect working eg \(2 + 2 = 4\), \(8 + 6 = 14\), \(14 - 2 = 12\) | M0A0 |
| \(2 \times 2 + 2 \times 2 = 8\) (misconception on area of rug) cannot score for \(2 \times 2\) | M0A0 |
| \(6 \times 8 = 48\) and \(48 \times 2 = 96\) (ignore additional ‘method’ and give M1 for 48) \(6 \times 8 = 48\) and \(48 \div 2 = 24\) (ignore additional ‘method’ and give M1 for 48) \(6 \times 8 \times 2\) (ignore additional ‘method’ and give M1 for \(6 \times 8\)) | M1A0 |
| \(6 \times 8 = 48\) and \(48 \div 2 \div 2 = 12\) (equivalent to dividing by 4) | M1A1 |
| Ignore references to perimeter or units if it is clear they are working out area |