Moderate -0.5 This is a straightforward algebraic manipulation question requiring students to rearrange a formula to make 'a' the subject. It involves basic algebraic operations (multiplying out, collecting like terms, factorising) but is simpler than a typical A-level question as it's a single-step problem with no conceptual depth or problem-solving required.
for multiplying up correctly (may also expand brackets); annotate if partially correct
\(a(2c-5)-2a=3c\) or \(2ac-7a=3c\) or ft
M1
for collecting \(a\) terms on one side, remaining terms on other; ft only if two or more \(a\) terms
\(a(2c-7)=3c\) or ft
M1
for factorising \(a\) terms; ft only if two or more \(a\) terms needing factorising; may be earned before 2nd M1
\([a=]\frac{3c}{2c-7}\) or simplified equivalent or ft as final answer
M1
for division by their two-term factor; for all 4 marks work must be fully correct and simplified; candidates expressing \(c\) in terms of \(a\): treat as MR after first common M, applying MR\(-1\) [4]
# Question 4:
$a(2c-5)=3c+2a$ or $2ac-5a=3c+2a$ | M1 | for multiplying up correctly (may also expand brackets); annotate if partially correct
$a(2c-5)-2a=3c$ or $2ac-7a=3c$ or ft | M1 | for collecting $a$ terms on one side, remaining terms on other; ft only if two or more $a$ terms
$a(2c-7)=3c$ or ft | M1 | for factorising $a$ terms; ft only if two or more $a$ terms needing factorising; may be earned before 2nd M1
$[a=]\frac{3c}{2c-7}$ or simplified equivalent or ft as final answer | M1 | for division by their two-term factor; for all 4 marks work must be fully correct and simplified; candidates expressing $c$ in terms of $a$: treat as MR after first common M, applying MR$-1$ [4]
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