Practise 36 exam-style questions on Inequalities for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
36 questions
1–5 marks each
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20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Adding \(2x\) to both sides keeps the \(x\) term positive and avoids dividing by a negative number. If you divide both sides by \(-2\) instead, you must reverse the inequality sign: \(-2x \le 6\) gives \(x \ge -3\).
Mark scheme
1 mark
M1 for \(-2x \le 6\) or \(5 - 11 \le 2x\)
oe; correct answer with no working scores full marks
1 mark
A1 for \(x \ge -3\)
oe, e.g. \(-3 \le x\); do not accept \(x \le -3\)
Question 2
2 marks
Ben is solving the inequality \(6 - 3x < 12\) Here is his working. \(-3x < 6\) \(x < -2\)
Ben's answer is wrong. Explain his mistake and give the correct answer.
Model answer
Ben divided both sides by \(-3\) without reversing the inequality sign. \(-3x < 6\) \(x > -2\)
Explanation
Multiplying or dividing an inequality by a negative number reverses its direction. Check with a value: \(x = 0\) gives \(6 < 12\), which is true, and 0 satisfies \(x > -2\) but not Ben's \(x < -2\).
Mark scheme
1 mark
C1 for explaining that when you divide by a negative number (\(-3\)) the inequality sign must be reversed
Rearrange so that one side is 0 before factorising; you cannot use \(x \le 12\) or \(x - 4 \le 12\). The quadratic is negative or zero between its roots, and both critical values are included because of ≤.
Mark scheme
1 mark
M1 for rearranging to \(x^2 - 4x - 12 \le 0\)
correct answer with no working scores full marks
1 mark
M1 for factorising, \((x - 6)(x + 2)\), giving critical values 6 and \(-2\)
1 mark
A1 for \(-2 \le x \le 6\)
oe
Question 4
4 marks
The region R is defined by the inequalities \(x \ge 0\qquad y \ge x\qquad x + y \le 6\)
Work out the area of R.
Model answer
The boundaries are \(x = 0\) (the \(y\)-axis), \(y = x\) and \(x + y = 6\). \(y = x\) and \(x + y = 6\) meet where \(2x = 6\), at \((3, 3)\). \(y = x\) meets \(x = 0\) at \((0, 0)\); \(x + y = 6\) meets \(x = 0\) at \((0, 6)\). R is the triangle with vertices \((0, 0)\), \((0, 6)\) and \((3, 3)\). Base along the \(y\)-axis \(= 6\), perpendicular height \(= 3\) Area \(= \dfrac{1}{2} \times 6 \times 3 = 9\)
Explanation
Find the corners of the region by solving pairs of boundary equations simultaneously. A quick sketch shows that R is a triangle with one side on the \(y\)-axis, so use that side as the base and the horizontal distance to \((3, 3)\) as the height.
Mark scheme
1 mark
P1 for identifying the boundary lines \(x = 0\), \(y = x\) and \(x + y = 6\), and that R is a triangle
correct answer with no working scores full marks
1 mark
P1 for finding the vertex where \(y = x\) meets \(x + y = 6\), \((3, 3)\)
1 mark
P1 for the other vertices \((0, 0)\) and \((0, 6)\) and a correct method for the area, e.g. \(\dfrac{1}{2} \times 6 \times 3\)
1 mark
A1 for 9 (square units)
cao
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