Foundation November 2017 Paper 1 Q28
28 Here is a rectangle.
All measurements are in centimetres.
The area of the rectangle is 48 cm2.
Show that \(y = 3\) (4)
| Answer | Mark | Notes | ||
|---|---|---|---|---|
| Shows reasoning to reach \(y = 3\) | M1 | forms equation eg \(2x + 6 = 5x - 9\) | \(48 \div 3\ (= 16)\) | \(3(2x + 6) = 48\) or \(3(5x - 9) = 48\), condone missing bracket |
| M1 | isolates \(x\) and number terms \(3x = 15\) | forms equation \(2x + 6 = \text{``}16\text{''}\) or \(5x - 9 = \text{``}16\text{''}\) | Isolates \(x\) and number terms \(6x = \text{``}30\text{''}\) or \(15x = \text{``}75\text{''}\) | |
| M1 | substitutes “5” into side length eg \(2 \times 5 + 6\ (= 16)\) | isolates \(x\) and number terms \(2x = \text{``}10\text{''}\) or \(5x = \text{``}25\text{''}\) | forms the second equation | |
| A1 | \(48 \div 16 = 3\) or \(16 \times 3 = 48\) | shows \(x = 5\) for both solutions | \(x = 5\) from 2 different equations. | |