Higher November 2023 Paper 2 Q1
1
(a) Expand and simplify \(3(2y - 5) + 7(y + 2)\) (2)
(b) Factorise fully \(6x^2 + 15x\) (2)
(c) Make \(g\) the subject of the formula \(f = 3g + 11\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(13y - 1\) | M1 | for method to expand one bracket or collect like terms, eg \(3 \times 2y - 3 \times 5\ (= 6y - 15)\) or \(7 \times y + 7 \times 2\ (= 7y + 14)\) or \(3 \times 2y + 7 \times y\ (= 6y + 7y)\) or \(3 \times {-5} + 7 \times 2\ (= -15 + 14)\) |
| A1 | oe |
Additional guidance
May be implied by \(13y\) or \(-1\)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(3x(2x + 5)\) | B2 | oe |
| (B1 | for \(3(2x^2 + 5x)\) or \(x(6x + 15)\) or \(3x(ax + b)\)) |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(g = \dfrac{f - 11}{3}\) | M1 | for correct first step to rearrange, eg \(f - 11 = 3g + 11 - 11\) or \(f - 11 = 3g\) or \(\dfrac{f}{3} = \dfrac{3g}{3} + \dfrac{11}{3}\) or \(-3g = 11 - f\) or answer ambiguously shown, eg \(g = f - 11 \div 3\) or given as \(\dfrac{f - 11}{3}\) |
| A1 | oe |
Additional guidance
May be seen in different equivalent forms