4MA0 Higher June 2013 Paper 3R Q4
4
(a) Expand \(6(3a - 2b + c)\) (1)
(b) Factorise \(t^2 - 10t\) (2)
(c) Solve \(x = \dfrac{7 - 2x}{3}\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
| Scheme | Marks |
|---|---|
| \(18a - 12b + 6c\) (oe) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(t(t - 10)\) | B2 |
| (2) |
Notes
B2: also accept \((t \pm 0)(t - 10)\) for B2
B1 for factors which, when expanded and simplified, give only two terms, one of which is correct.
SC B1 for \(t(t - 10t)\)
| Scheme | Marks |
|---|---|
| \(3x = 7 - 2x\) | M1 |
| \(5x = 7\) or \(5x - 7 = 0\) | M1 |
| 1.4oe | A1 |
| (3) |
Notes
M1: or \(x = \dfrac{7}{3} - \dfrac{2x}{3}\)
M1: or \(\dfrac{5x}{3} = \dfrac{7}{3}\) or \(x + \dfrac{2x}{3} = \dfrac{7}{3}\)
A1: Answer dependent on at least M1