June 2019 Paper 2 Q15

15 Rearrange \(\quad m = \dfrac{2p + 1}{p} + \dfrac{p + 5}{3p} \quad\) to make \(p\) the subject. [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(3mp = 3(2p + 1) + p + 5\) or \((m =)\ \dfrac{3(2p + 1)}{3p} + \dfrac{p + 5}{3p}\) or \((m =)\ \dfrac{6p + 3 + p + 5}{3p}\)	M1	oe fractions eliminated or common denominator eg \((m =)\ \dfrac{3p(2p + 1)}{3p^2} + \dfrac{p(p + 5)}{3p^2}\) or \((m =)\ \dfrac{6p^2 + 3p + p^2 + 5p}{3p^2}\)
\(3mp = 6p + 3 + p + 5\) or \(3mp = 7p + 8\)	M1dep	oe brackets expanded and fractions eliminated eg \(3mp^2 = 7p^2 + 8p\) implies M2
\(3mp - 7p = 8\) or \(\dfrac{8}{3m - 7}\) or \(\dfrac{-8}{7 - 3m}\)	M1dep	oe terms collected eg \(p(3m - 7) = 8\) or \(7p - 3mp = -8\) implies M3
\(p = \dfrac{8}{3m - 7}\) or \(p = \dfrac{-8}{7 - 3m}\)	A1	oe eg \(\dfrac{8}{3m - 7} = p\)
Alternative method 2
\((m =)\ \dfrac{3(2p + 1)}{3p} + \dfrac{p + 5}{3p}\) or \((m =)\ \dfrac{6p + 3 + p + 5}{3p}\)	M1	oe common denominator eg \((m =)\ \dfrac{3p(2p + 1)}{3p^2} + \dfrac{p(p + 5)}{3p^2}\) or \((m =)\ \dfrac{6p^2 + 3p + p^2 + 5p}{3p^2}\)
\(m = \dfrac{7p + 8}{3p}\) and \(m = \dfrac{7}{3} + \dfrac{8}{3p}\) and \(m - \dfrac{7}{3} = \dfrac{8}{3p}\)	M1dep	simplifies numerator and isolates term in \(p\) eg \(m = \dfrac{7p^2 + 8p}{3p^2}\) and \(m = \dfrac{7}{3} + \dfrac{8}{3p}\) and \(m - \dfrac{7}{3} = \dfrac{8}{3p}\) implies M2
\(\dfrac{3m - 7}{3} = \dfrac{8}{3p}\)	M1dep	converts \(m - \dfrac{7}{3}\) to a single fraction implies M3
\(p = \dfrac{8}{3m - 7}\) or \(p = \dfrac{-8}{7 - 3m}\)	A1	oe eg \(\dfrac{8}{3m - 7} = p\)

\(p = \dfrac{8}{3m - 7}\) in working but \(\dfrac{8}{3m - 7}\) on answer line	M3A1
Allow recovery of missing brackets
\(p = \dfrac{8}{3m - 7}\) followed by incorrect further work	M3A0
Allow equivalences for A1 eg \(p = \dfrac{\frac{8}{3}}{\frac{3m - 7}{3}}\)	M3A1
Do not regard eg \(3m(p) = 7p + 8\) as having unexpanded brackets	M1M1dep