Higher June 2025 Paper 1 Q7
7
| Answer | Mark | Mark scheme |
|---|---|---|
| \(6 - 3m\) | B1 | for \(6 - 3m\) oe |
Additional guidance
Accept \(3(2 - m)\)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(x \geqslant 10\) | M1 | for a correct first step working with an inequality or an equation, eg \(7 + x + 8 \leqslant \dfrac{5x}{2} - 8 + 8\) or \(15 + x \leqslant \dfrac{5x}{2}\) or \(7 + x - x \leqslant \dfrac{5x}{2} - 8 - x\) or \(7 \leqslant \dfrac{3x}{2} - 8\) or \(7 \times 2 + x \times 2 \leqslant \dfrac{5x}{2} \times 2 - 8 \times 2\) or \(14 + 2x \leqslant 5x - 16\) |
| M1 | (dep M1) for a correct second step, eg subtracts \(x\) from both sides or adds 8 to both sides or subtracts \(2x\) from both sides or multiplies both sides by 2 or gives the critical value of 10 | |
| A1 | for \(x \geqslant 10\) as final answer |
Additional guidance
Can work with an equation or incorrect inequality symbol for both M marks
For M marks step must be carried out not just intention shown.
For example, if you see \[\begin{array}{ccc} 7 + x &\leqslant& \dfrac{5x}{2} - 8 \\ +8 && +8 \end{array}\] award M1 for \(k + x \leqslant \dfrac{5x}{2}\) where \(k \gt 7\)
or indicating \(-x\) and reaching \(7 \leqslant kx - 8\) where \(k \lt \dfrac{5}{2}\)
or indicating multiplying by 2 and obtaining an equation or inequality with no more than one term incorrect and no term unchanged.
The first 2 marks can be awarded for critical value of 10, eg \(x = 10\)
Accept \(10 \leqslant x\)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(2.5 \lt y \lt 4\) | M1 | for a correct first step, eg \(9 - 4 \lt 2y \lt 12 - 4\) or \(5 \lt 2y \lt 8\) or \(9 \div 2 \lt y + 2 \lt 12 \div 2\) or \(4.5 \lt y + 2 \lt 6\) or showing 2.5 and 4 as the critical values |
| A1 | for \(2.5 \lt y \lt 4\) oe as final answer |
Additional guidance
For M mark condone use of “=” and incorrect inequality signs
For M mark step must be carried out not just intention shown.
For example, if you see \[\begin{array}{ccccc} 9 &\lt & 2y + 4 &\lt & 12 \\ -4 && -4 && -4 \end{array}\] award M1 for \(a \lt 2y \lt b\) where \(a \lt 9\) and \(b \lt 12\)
or if you see \[\begin{array}{ccccc} 9 &\lt & 2y + 4 &\lt & 12 \\ \div 2 && \div 2 && \div 2 \end{array}\] award M1 for \(a \lt y + 2 \lt b\) where \(a \lt 9\) and \(b \lt 12\)
Accept \(y \gt 2.5\) and \(y \lt 4\) oe