Foundation November 2022 Paper 1 Q15
15
(a) Solve \(\quad 11x - 3 = 6x + 1\) [3 marks]
(b) Solve \(\quad \dfrac{2x}{5} = 14\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(11x - 6x\) or \(6x - 11x\) or \(\pm 5x\) or \((+)1 + 3\) or \(-3 - 1\) or \(\pm 4\) | M1 | oe terms in \(x\) or constant terms collected |
| \(5x = 4\) or \(-5x = -4\) | A1 | may be implied eg \(4 \div 5\) or \(-4 \div -5\) or \(\dfrac{-4}{-5}\) |
| \(\dfrac{4}{5}\) or 0.8 | A1ft | oe ft any equation of the form \(5x = a\) or \(-5x = a\) or \(bx = 4\) or \(bx = -4\) |
Additional guidance
| Ignore attempt to convert or simplify after correct answer seen | |
| Trial and improvement scores 3 or 0 | |
| \(5x - 4\ (= 0)\) with no further work | M1A0A0 |
| \(\dfrac{4}{5}\) and \(5x = 4\) on answer line | M1A1A1 |
| Embedded answer eg \(11 \times 0.8 - 3 = 6 \times 0.8 + 1\) | M1A1A0 |
| ft answers must be exact or rounded to 2 dp or better eg \(17x = 4\), answer \(\dfrac{4}{17}\) eg \(17x = -4\), answer \(-0.24\) | M1A0A1ft M1A0A1ft |
| \(5x + 4\) or \(5x + 4 = 0\) or \(17x - 4\) or \(17x - 4 = 0\) etc with no further work | M1 |
| \(\pm 5x\) or \(\pm 4\) must not have come from incorrect working |
| Answer | Mark | Comments |
|---|---|---|
| \(2x = 14 \times 5\) or \(2x = 70\) or \(\dfrac{x}{5} = 14 \div 2\) or \(\dfrac{x}{5} = 7\) or \(14 \times 5 \div 2\) or \(70 \div 2\) | M1 | oe eg \(14 \div 0.4\) |
| 35 | A1 |
Additional guidance
| Trial and improvement scores 2 or 0 | |
| Embedded answer eg \(\dfrac{2 \times 35}{5}\) | M1A0 |
| \(\dfrac{2x}{5} = \dfrac{14 \times 5}{5}\) | M1 |