Foundation January 2023 Paper 2R Q14
14
(a) Expand \(x(10 - x)\) (1)
(b) Factorise \(6y + 27\) (1)
(c) Make \(m\) the subject of the formula \(h = \dfrac{m}{2} + 4\) (2)
(d) Solve \(7g + 3 = 2g - 5\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
| Scheme | Marks |
|---|---|
| \(10x - x^2\) | B1 |
| (1) |
Notes
B1: oe eg \(-x^2 + 10x\)
| Scheme | Marks |
|---|---|
| \(3(2y + 9)\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| eg \(h - 4 = \dfrac{m}{2}\) or \(2h = m + 8\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(m = 2(h - 4)\) | A1 |
| (2) |
Notes
M1: for a correct first step
A1: oe eg \(m = 2h - 8\)
SC award M1 for \(m = 2h - 4\) or \(m = h - 8\)
| Scheme | Marks |
|---|---|
| eg \(7g - 2g + 3 = -5\) or \(5g + 3 = -5\) or \(7g = 2g - 5 - 3\) or \(7g = 2g - 8\) | M1 |
| eg \(7g - 2g = -5 - 3\) or \(5g = -8\) | M1 |
Working required Answer: \(-\dfrac{8}{5}\) | A1 |
| (3) | |
| (7 marks) |
Notes
M1: for correctly collecting the terms in \(g\) on one side or the numbers on one side
M1: for a correct rearrangement with terms in \(g\) on one side and numbers on the other. Award of this mark implies the first M1
A1: (dep on M1) oe eg \(-1\dfrac{3}{5}\) or –1.6