Higher June 2025 Paper 1R Q6
6
(a) Write down the inequality shown on the number line. (2)
(b) Solve the inequality \(7a - 5 \leqslant 3a + 28\)
Show clear algebraic working. (2)
Show clear algebraic working. (2)
| Scheme | Marks |
|---|---|
| \(-2 \lt x \leqslant 1\) | B2 |
| (2) |
Notes
B2: accept \(1 \geqslant x \gt -2\) or \(x \gt -2, x \leqslant 1\)
if not B2 then B1 for \(-2 \lt x\) or \(x \leqslant 1\) or
\(-2 \leqslant x \lt 1\) or \(-2 \leqslant x \leqslant 1\) or \(-2 \lt x \lt 1\)
Condone use of a variable other than \(x\) but not 0
| Scheme | Marks |
|---|---|
| \(7a - 3a \leqslant 28 + 5\) or \(4a \leqslant 33\) or \(-5 - 28 \leqslant 3a - 7a\) or \(-33 \leqslant -4a\) | M1 |
| Working required Answer: \(a \leqslant 8.25\) | A1 |
| (2) | |
| (4 marks) |
Notes
M1: for \(a\) terms on one side and numbers on the other.
Condone = rather than \(\leqslant\) or any other sign for this mark.
A1: (dep on M1) oe eg \(a \leqslant \dfrac{33}{4}\) or \(a \leqslant 8\dfrac{1}{4}\) or \(8.25 \geqslant a\)
must have correct sign on answer line
(sight of correct answer in working space and just 8.25 on answer line gains M1 only).