Higher November 2020 Paper 1R Q3
3
(a) Simplify \(\;h^7 \times h^2\) (1)
\(G = c^2 - 4c\)
(b) Find the value of \(G\) when \(c = -5\) (2)
(c) Solve \(\;\dfrac{5x - 3}{4} = 2x + 3\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
| Scheme | Marks |
|---|---|
| \(h^9\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| (−5)² − 4 × −5 oe e.g. 25 + 20 | M1 |
| 45 | A1 |
| (2) |
Notes
M1: for a correct substitution
| Scheme | Marks |
|---|---|
| \(5x - 3 = 4(2x + 3)\) oe or \(\dfrac{5x}{4} - \dfrac{3}{4} = 2x + 3\) oe | M1 |
e.g. \(5x - 8x = 12 + 3\) or \(-3x = 12 + 3\) or \(8x - 5x = -12 - 3\) or \(3x = -12 - 3\) or \(-\dfrac{3}{4} - 3 = 2x - \dfrac{5x}{4}\) or \(-\dfrac{15}{4} = \dfrac{3x}{4}\) | M1 |
| Working required Answer: −5 | A1 |
| (3) | |
| (6 marks) |
Notes
M1: for correctly removing the denominator, condone missing brackets
M1: for a correct rearrangement with terms in \(x\) on one side and numbers on the other,
allow correct rearrangement of their equation in the form \(ax + b = cx + d\)
A1: dep on at least M1
SCB2 for an answer of
\(x = -2\) coming from \(5x - 3 = 8x + 3\)
or \(x = 5\) coming from \(5x - 3 = 2x + 12\)