Linear inequalities & number lines; quadratic inequalities – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator4 marks

The inequality shown on the number line is for values of $x$.

(a) Write down the inequality. [2]

(b) $x$ is an integer. Write down all the possible values of $x$. [2]

(a) for one end correct, e.g. $x > -2$ or $x \le 3$ (or for $-2 \le x \le 3$ / $-2 < x < 3$)B1
(a) (total B2) for $-2 < x \le 3$ oe, e.g. $3 \ge x > -2$ (accept $x > -2$ and $x \le 3$)B1
(b) for at least 4 correct values with no more than one incorrect valueB1
(b) (total B2) for exactly $-1, 0, 1, 2, 3$ and no othersB1
Answer: (a) $-2 < x \le 3$ (b) $-1, 0, 1, 2, 3$
Examiner tip: Open circle means 'not included' ($<$), filled circle means 'included' ($\le$) – so $-2$ is NOT in the list but 3 is.

2Grade 4–5Non-calculator3 marks

(a) Solve $5x - 3 > 2x + 9$ [2]

(b) Write down the smallest integer that satisfies $5x - 3 > 2x + 9$ [1]

(a) for a correct first step collecting terms, e.g. $3x > 12$ (allow $=$ or wrong inequality sign for this mark)M1
$x > 4$ (must have the correct inequality sign)A1
(b) $5$ (ft from their inequality in (a) if of the form $x > a$)B1
Answer: (a) $x > 4$ (b) $5$
Examiner tip: In (b) 4 does not work – $x$ must be STRICTLY greater than 4, so the smallest integer is 5.

3Grade 4–5Non-calculator3 marks

(a) Solve $-7 < 3x + 2 \le 11$ [2]

(b) $n$ is an integer. Write down all the values of $n$ that satisfy $-7 < 3n + 2 \le 11$ [1]

(a) for subtracting 2 from all three parts, $-9 < 3x \le 9$, or for one end correct, e.g. $x > -3$ or $x \le 3$M1
$-3 < x \le 3$A1
(b) $-2, -1, 0, 1, 2, 3$ (ft their (a))B1
Answer: (a) $-3 < x \le 3$ (b) $-2, -1, 0, 1, 2, 3$
Examiner tip: Whatever you do to one part of a double inequality you must do to all THREE parts.

4Grade 6–7HigherNon-calculator3 marks

Solve $\dfrac{4 - 3x}{2} \ge x - 8$ [3]

for multiplying both sides by 2 correctly: $4 - 3x \ge 2x - 16$M1
for collecting terms correctly, e.g. $20 \ge 5x$ or $-5x \ge -20$M1
$x \le 4$ oe, e.g. $4 \ge x$A1
Answer: $x \le 4$
Examiner tip: If you divide by $-5$ you must reverse the inequality sign – $x \ge 4$ is the classic error.

5Grade 6–7HigherNon-calculator3 marks

Solve $x^2 - 2x - 15 < 0$ [3]

for factorising: $(x - 5)(x + 3)$, or a correct method to solve $x^2 - 2x - 15 = 0$M1
for the critical values $-3$ and $5$M1
$-3 < x < 5$ (accept $x > -3$ and $x < 5$, or $\{x : -3 < x < 5\}$)A1
Answer: $-3 < x < 5$
Examiner tip: Sketch the U-shaped graph: '$< 0$' means BELOW the $x$-axis, which is the region BETWEEN the roots.

6Grade 8–9HigherNon-calculator3 marks

Solve $2x^2 + x \ge 6$
Write your answer in set notation. [3]

for rearranging to $2x^2 + x - 6 \ge 0$ (or $=0$) and a correct method to find the critical values, e.g. factorising to $(2x - 3)(x + 2)$ or correct substitution into the quadratic formulaM1
for the critical values $-2$ and $\frac{3}{2}$ (accept 1.5)A1
$\{x : x \le -2\} \cup \{x : x \ge \frac{3}{2}\}$ oe, e.g. $\{x : x \le -2 \text{ or } x \ge 1.5\}$ (correct inequalities not in set notation, e.g. $x \le -2$ or $x \ge 1.5$, score M1A1A0; strict inequalities, $-2 \le x \le 1.5$, $\{x : -2 \le x \le 1.5\}$ or $-2 \ge x \ge 1.5$ score A0)A1
Answer: $\{x : x \le -2\} \cup \{x : x \ge 1.5\}$
Examiner tip: Sketch $y = 2x^2 + x - 6$: '$\ge 0$' means the parts of the curve ON or ABOVE the $x$-axis, i.e. the two OUTSIDE regions; write each region as its own set and join them with $\cup$, e.g. $\{x : x \le -2\} \cup \{x : x \ge 1.5\}$, never as $-2 \ge x \ge 1.5$.

7Grade 6–7HigherNon-calculator3 marks

$n$ is an integer such that

$$3n - 1 > 2 \quad \text{and} \quad n^2 < 20$$

Find all the possible values of $n$. [3]

for solving $3n - 1 > 2$ to get $n > 1$M1
for $-\sqrt{20} < n < \sqrt{20}$ (or $-4.47… < n < 4.47…$), or for listing integers $-4$ to $4$ for $n^2 < 20$M1
$2, 3, 4$ and no othersA1
Answer: $n = 2, 3, 4$
Examiner tip: $n^2 < 20$ also allows negative values, but these are ruled out by $n > 1$ – the answer must satisfy BOTH.

8Grade 1–3FoundationNon-calculator4 marks

(a) On the number line, show the inequality $-1 \le x < 4$ [2]

(b) $n$ is a whole number. Write down the largest value of $n$ that satisfies $-1 \le n < 4$ [1]

(a) a line drawn between $-1$ and $4$ with a circle at one end correct (filled circle at $-1$ or open circle at $4$), or both circles correct with no line / an incorrect lineB1
(a) (total B2) fully correct: filled (solid) circle at $-1$, open (hollow) circle at $4$, joined by a line. Allow arrows on the line provided circles are correct; do not accept the line extending beyond either circleB1
(b) $3$ cao (not 4)B1
(c) any correct inequality, e.g. $x < 6$, $x \le 5$, $x < 5.5$, $4 < x \le 5$ oe (must include 5 and exclude 6)B1
Answer: (a) filled circle at $-1$, open circle at $4$, joined by a line (b) $3$ (c) e.g. $x < 6$
Examiner tip: An open circle means the end value is NOT included, so 4 is not allowed – the largest whole number is 3.

9Grade 4–5Calculator3 marks

A taxi company charges a fixed fee of £3 plus £2.40 for each mile travelled.

Priya has £25 to spend on a taxi journey.

(a) Using $m$ for the number of miles, write down an inequality to show this information. [1]

(b) The taxi only charges for whole numbers of miles.
Work out the greatest number of miles Priya can travel. [2]

(a) $3 + 2.4m \le 25$ oe, e.g. $2.40m + 3 \le 25$ or $300 + 240m \le 2500$ ($3 + 2.4m < 25$ scores B0, as Priya can spend exactly £25)B1
(b) for $(25 - 3) \div 2.4$ or $22 \div 2.4$ oe, or $m \le 9.16...$, or a build-up showing £24.60 for 9 miles and £27 for 10 milesM1
(b) $9$ cao (answers of 9.16..., 9.2 or 10 score A0)A1
Answer: (a) $3 + 2.4m \le 25$ (b) 9 miles
Examiner tip: Do not round 9.16... to the nearest whole number – the answer must be rounded DOWN, because 10 miles would cost £27, which is more than £25.

10Grade 4–5Non-calculator4 marks

Kai is solving the inequality $11 - 3x < 2$

Here is his working.

$$\begin{aligned} 11 - 3x &< 2 \\ -3x &< -9 \\ x &< 3 \end{aligned}$$

(a) Explain the mistake Kai has made. [1]

(b) Write down the correct solution to $11 - 3x < 2$ [1]

(a) explanation that when he divided both sides by $-3$ (a negative number) the inequality sign should have been reversed, e.g. 'the $<$ should become $>$ when dividing by $-3$'C1
(b) $x > 3$ oe, e.g. $3 < x$B1
(c) for a correct first step collecting $x$ terms on one side and numbers on the other, e.g. $18 \ge 6x$ or $-6x \ge -18$M1
(c) $x \le 3$ oe, e.g. $3 \ge x$ (answer $x \ge 3$ or $x = 3$ scores A0)A1
Answer: (a) He divided by $-3$ without reversing the inequality sign (b) $x > 3$ (c) $x \le 3$
Examiner tip: To avoid dividing by a negative altogether, move the $x$ terms to the side where they stay positive: $5 + 13 \ge 2x + 4x$ gives $18 \ge 6x$.

11Grade 4–5Non-calculator5 marks

(a) Solve $4(x - 1) < x + 11$ [2]

(b) Solve $6 - x \le 3x + 14$ [2]

(a) for expanding and a correct step towards isolating $x$, e.g. $4x - 4 < x + 11$ leading to $3x < 15$M1
(a) $x < 5$ oeA1
(b) for a correct step collecting terms, e.g. $-8 \le 4x$ or $-4x \le 8$M1
(b) $x \ge -2$ oe, e.g. $-2 \le x$A1
(c) filled circle at $-2$ and open circle at $5$ joined by a line (ft their two inequalities from (a) and (b) provided they overlap and the circle types match their signs)B1ft
Answer: (a) $x < 5$ (b) $x \ge -2$ (c) filled circle at $-2$, open circle at $5$, joined by a line ($-2 \le x < 5$)
Examiner tip: In (c) only the overlap of the two solutions is shown, so the line runs from $-2$ (included, filled) to $5$ (not included, open) – not two separate arrows.

12Grade 6–7HigherNon-calculator3 marks

The region $\mathbf{R}$ is shaded on the grid.

Write down the three inequalities that define the region $\mathbf{R}$. [3]

$y \ge 1$ oe, e.g. $1 \le y$B1
$y \le 2x$ oe, e.g. $2x \ge y$ or $y - 2x \le 0$B1
$x + y \le 6$ oe, e.g. $y \le 6 - x$. (Lines are solid, so $\le$ / $\ge$ are expected: if all three boundaries are correct but strict signs are used throughout, award B2 in total)B1
Answer: $y \ge 1$, $\;y \le 2x$, $\;x + y \le 6$
Examiner tip: Find each line's equation first (gradient and intercept), then test a point inside R, e.g. $(2, 2)$, to decide which way each sign faces.

13Grade 8–9HigherNon-calculator4 marks

$x$ and $y$ are integers such that

$$x \ge 1, \qquad y \ge x, \qquad x + 2y < 10$$

Work out the greatest possible value of $3x + y$
You must show your working. [4]

for a correct method to identify points, e.g. drawing the lines $x = 1$, $y = x$ and $x + 2y = 10$ and indicating the correct region, or listing at least 3 correct integer pairs (e.g. $(1,1), (1,4), (2,3)$)M1
for a complete list of the 7 points $(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (3,3)$ with no incorrect points, or a correctly shaded region with these integer points markedM1
for evaluating $3x + y$ at least two correct points, including $(3,3)$ or $(2,3)$M1
$12$ cao (from $(3,3)$; answers from points outside the region, e.g. 15 from $(4,3)$ or 14 from $(4,2)$, score A0)A1
Answer: $12$ (at $x = 3$, $y = 3$)
Examiner tip: The corner of the region where $y = x$ meets $x + 2y = 10$ is $\left(\tfrac{10}{3}, \tfrac{10}{3}\right)$, which is not an integer point, so check the integer points nearby: $(4, 3)$ breaks $y \ge x$, so the best point is $(3, 3)$.

14Grade 6–7HigherNon-calculator4 marks

The graph of $y = x^2 + 2x - 8$ is drawn on the grid.

(a) Use the graph to solve $x^2 + 2x - 8 < 0$ [2]

(b) Use the graph to solve $x^2 + 2x - 8 \le 7$ [2]

(a) for identifying the critical values $-4$ and $2$ (where the curve crosses the $x$-axis)M1
(a) $-4 < x < 2$ oe, e.g. $x > -4$ and $x < 2$, or $\{x : -4 < x < 2\}$ (A0 for $x < -4$, $x > 2$, or for $\le$ signs)A1
(b) for drawing or using the line $y = 7$, or for identifying both $x = -5$ and $x = 3$M1
(b) $-5 \le x \le 3$ oe (A0 for strict signs or for two separate inequalities facing outwards)A1
Answer: (a) $-4 < x < 2$ (b) $-5 \le x \le 3$
Examiner tip: 'Less than' means the part of the curve BELOW the line – here that is between the two crossing points, so the answer is one combined inequality, not two.

15Grade 8–9HigherNon-calculator5 marks

A rectangle has length $(x + 3)$ cm and width $(x - 1)$ cm.

The area of the rectangle is less than $21\text{ cm}^2$.

Find the range of possible values of $x$.
Give your answer using set notation. [5]

for forming the inequality $(x + 3)(x - 1) < 21$ oeM1
for expanding and rearranging to $x^2 + 2x - 24 < 0$ (or $= 0$) and a correct method to find the critical values, e.g. $(x + 6)(x - 4)$M1
for the critical values $-6$ and $4$A1
for recognising that the width must be positive, so $x > 1$ (or stating $x - 1 > 0$)B1
$\{x : 1 < x < 4\}$ oe in set notation, e.g. $\{x : x > 1\} \cap \{x : x < 4\}$ (correct inequality $1 < x < 4$ not in set notation scores M1M1A1B1A0; $\{x : -6 < x < 4\}$ scores A0)A1
Answer: $\{x : 1 < x < 4\}$
Examiner tip: Most students stop at $-6 < x < 4$ – but a length cannot be negative or zero, so the width $x - 1 > 0$ cuts the range down to $1 < x < 4$.

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