Foundation June 2025 Paper 1R Q23
23
(a) Expand \(x(x - 3)\) (1)
(b) Make \(t\) the subject of \(m = \dfrac{t + 4}{5}\) (2)
(c) Simplify \(a^6 \times a^{10}\) (1)
(d) Simplify \(c^{30} \div c^{12}\) (1)
(e)
(i) Factorise \(y^2 - 10y + 21\) (2)
(ii) Hence, solve \(y^2 - 10y + 21 = 0\) (1)
| Scheme | Marks |
|---|---|
| \(x^2 - 3x\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
eg \(5m = t + 4\) or \(m = \dfrac{t}{5} + \dfrac{4}{5}\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(t = 5m - 4\) | A1 |
| (2) |
Notes
M1: for a correct first step
A1: oe eg \(t = 5\left(m - \dfrac{4}{5}\right)\) or \(t = -4 + 5m\)
\(5m - 4\) only on answer line scores M1 unless \(t = 5m - 4\) is seen in the working then score M1A1
| Scheme | Marks |
|---|---|
| \(a^{16}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(c^{18}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| (i) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \((y - 3)(y - 7)\) | A1 |
| (2) | |
| (ii) Answer: 3, 7 | B1 |
| (1) | |
| (8 marks) |
Notes
M1: for \((y \pm 3)(y \pm 7)\)
or for \((y \pm a)(y \pm b)\) with \(ab = 21\) or \(a + b = -10\)
A1: for correct factors
B1: ft dep on factorising in the form \((y \pm p)(y \pm q)\)