Foundation November 2024 Paper 1 Q15
15
(a) Factorise \(6a + 15\) (1)
(b) Solve \(4(3y + 1) = 28\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(3(2a + 5)\) | B1 |
| Answer | Mark | Mark scheme |
|---|---|---|
| 2 | M1 | for correct expansion of brackets, ie \(12y + 4\) or dividing throughout by 4 as a first step to solve equation, eg \(3y + 1 = 28 \div 4\) |
| M1 | for isolating terms in \(y\), eg \(12y = 28 - 4\) or \(3y = 7 - 1\) | |
| A1 | cao |
Additional guidance
For M marks step must be carried out not just intention shown.
For example, if you see
\(\begin{array}{rcl} 4(3y + 1) &=& 28 \\ \div 4 && \div 4 \end{array}\)
Award M1 for:
\(3y + 1 = k\) with \(k \neq 28\) or 112
ft their equation of the form \(ay \pm b = c\)
For example, if you see
\(\begin{array}{rcl} 12y + 4 &=& 28 \\ -4 && -4 \end{array}\)
Award M1 for:
\(12y = k\) with \(k \neq 28\) or 32