Foundation November 2021 Paper 2 Q12
12 \(T = 6p - 4d\)
(a) Work out the value of \(T\) when \(p = 8\) and \(d = 3\) (2)
\(T = 6p - 4d\)
(b) Work out the value of \(p\) when \(T = -41\) and \(d = 5\) (3)
(c) Solve \(\quad 4(x - 3) = 7x + 15\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
| Scheme | Marks |
|---|---|
| \(6 \times 8 - 4 \times 3\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 36 | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(-41 = 6 \times p - 4 \times 5\) or \(6p = T + 4d\) or \(6p = -41 + 4 \times 5\) | M1 |
\(6p = -41 + 20\) or \(6p = -21\) \(-6p = 41 - 20\) or \(-6p = 21\) \(p = \dfrac{-41 + 20}{6}\) or \(p = \dfrac{41 - 20}{-6}\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(-\dfrac{7}{2}\) | A1 |
| (3) |
Notes
M1: for correct substitution into the correct formula or a correct rearrangement for \(6p\)
A1: Oe
If no marks awarded SCB1 for –266
| Scheme | Marks |
|---|---|
| \(4x - 12\) or \(x - 3 = \dfrac{7x}{4} + \dfrac{15}{4}\) oe | M1 |
| \(4x - 7x = 15 + 12\) or \(-12 - 15 = 7x - 4x\) or \(-3x = 27\) or \(-27 = 3x\) | M1 |
| Working required Answer: \(-9\) | A1 |
| (3) | |
| (8 marks) |
Notes
M1: for a correct expansion of bracket or division of all terms in a correct equation by 4
M1: for a correct rearrangement within a correct equation with \(x\) terms on one side and the numbers on the other side
A1: dep on M1
(SCB1 for an answer of \(x = -6\) with working shown from \(4x - 3 = 7x + 15\))