1MA0 Higher June 2012 Paper 2 Q20
20 Make \(t\) the subject of the formula \(2(d - t) = 4t + 7\) (3)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(2d - 2t = 4t + 7\) \(2d - 7 = 4t + 2t\) \(2d - 7 = 6t\) \(\dfrac{2d - 7}{6}\) | \(\dfrac{2d - 7}{6}\) | 3 | B1 for \(2d - 2t\) or \(2t + \dfrac{7}{2}\) oe M1 for rearranging 4 terms correctly to isolate terms in \(t\) e.g. ‘\(2d\)’ – 7 = \(4t\) + ‘\(2t\)’ or \(2d - 7 = 6t\) or \(-6t = 7 - 2d\) seen A1 for \(\dfrac{2d - 7}{6}\) oe |