Higher November 2018 Paper 1 Q19
19 In a chess club, there are \(x\) boys and \(y\) girls.
(a) If 5 more boys and 8 more girls join, there would be half as many boys as girls.
Show that \(\quad y = 2x + 2\) [2 marks]
(b) If instead,
10 more boys and 1 more girl join, there would be the same number of boys and girls.
Work out \(x\) and \(y\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2(x + 5) = y + 8\) or \(2x + 10 = y + 8\) | M1 | oe eg \(\ \dfrac{x + 5}{y + 8} = \dfrac{1}{2}\) or \(\dfrac{y + 8}{x + 5} = 2\) |
| \(2x + 10 = y + 8\) and \(y = 2x + 2\) | A1 |
| Answer | Mark | Comments |
|---|---|---|
| \(x + 10 = y + 1\) | M1 | oe |
| Eliminates \(x\) or \(y\) from their \((x + 10) = y + 1\) and \(y = 2x + 2\) | M1 | their \((x + 10) = y + 1\) must be an equation in \(x\) and \(y\) eg \(x + 10 = y - 1\) (and \(y = 2x + 2\)) followed by \(x + 11 = 2x + 2\) |
| \(x = 7\) and \(y = 16\) | A1 |
Additional guidance
| \(x = 7\) or \(y = 16\) with no value or an incorrect value for the other unknown and no working worth M marks | M0M0A0 |