Higher June 2025 Paper 5 Q16
16 Rearrange this formula to make \(a\) the subject.
\(p = \dfrac{a + 2b}{3 - a}\) [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(a = \frac{3p - 2b}{1 + p}\) final answer | 4 | For 4 marks accept \(a = \frac{2b - 3p}{-p - 1}\) as final answer If answer \(a\) = 3\(p\) – 2\(b\)/1 + \(p\) check working and if seen written correctly allow 4 marks | |
| M1 for \(p(3 - a) = a + 2b\) or better | |||
| M1 for expanding bracket | Each method step FT previous step \(3p - pa = a + 2b\) if correct, implies M1M1 | ||
| M1 for isolating terms in \(a\) | \(3p - 2b = a + pa\) if correct Dep on at least two terms in \(a\) when fraction removed | ||
| M1 for removing \(a\) as a factor and dividing by bracket to the answer | If answer incorrect then max 3 marks | ||