Higher June 2024 Paper 2R Q6
6
Show clear algebraic working. (3)
| Scheme | Marks |
|---|---|
eg \(2f = 12f - 51\) or \(\dfrac{f}{3} = \dfrac{4}{2}f - \dfrac{17}{2}\) or \(\dfrac{f}{3} = 2f - \dfrac{17}{2}\) or \(0.3f = 2f - 8.5\) or \(f = 6f - \dfrac{51}{2}\) or \(f = 6f - 25.5\) or \(17 = 4f - \dfrac{2}{3}f\) or \(17 = 4f - 0.6f\) or \(17 = 4f - 0.7f\) or \(\dfrac{2}{3}f - 4f = -17\) or \(0.6f - 4f = -17\) or \(0.7f - 4f = -17\) | M1 |
eg \(-10f = -51\) or \(10f = 51\) or \(\dfrac{5f}{3} = \dfrac{17}{2}\) or \(5f = \dfrac{51}{2}\) or \(17 = \dfrac{10f}{3}\) or \(3.3f = 17\) or \(-\dfrac{10f}{3} = -17\) or \(-3.3f = -17\) | M1 |
Working required Answer: \(\dfrac{51}{10}\) | A1 |
| (3) |
Notes
M1: for a correct first step – multiplying both sides by 3 correctly and expanding to find \(2f = 12f - 51\) or \(2f = -51 + 12f\)
or
writing the RHS as 2 terms each over 2
(Allow decimals to 1dp or better – rounded or truncated)
M1: for a correct 2 term equation in the form \(af = b\)
ft the following equations only
\(2f = 12f - 17\) oe
\(2f = 4f - 51\) oe
\(6f = 12f - 51\) oe
(Allow decimals to 1dp or better – rounded or truncated)
A1: (dep on at least M1) oe
| Scheme | Marks |
|---|---|
| 1 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(3a^3h^4\) | B2 |
| (2) |
Notes
| Scheme | Marks |
|---|---|
| \(4x^3y(5x^2 + 3y^3)\) | B2 |
| (2) | |
| (8 marks) |
Notes
B2: for \(4x^3y(5x^2 + 3y^3)\)
B1 for any correct factorisation with at least a 2 term factor outside the bracket
eg \(2x^3y(10x^2 + 6y^3)\) or \(x^3y(20x^2 + 12y^3)\) or
\(2x(10x^4y + 6x^2y^4)\) or \(4y(5x^5 + 3x^3y^3)\) or
\(4x^3(5x^2y + 3y^4)\) etc
or the correct highest common factor and a 2 term expression with at most one incorrect term
eg \(4x^3y(5x^2 + \ldots)\) or \(4x^3y(\ldots + 3y^3)\)