Practice › Number and Algebra › Linear Inequalities in One Unknown

Linear Inequalities in One Unknown: Paper 2 Challenge Multiple-Choice

Ten harder Paper 2 questions, each with a distractor that is the right number attached to the wrong end: a boundary included when it should not be, or a sign that was never reversed. This is the Challenge tier; work the standard Paper 2 set first. Detailed solutions can be requested by email.

10 questions · about 20 minutes · answers below each question

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

Question 1 · No direct past-paper model · endpoints that swap

If $- 3 < x \leq 4$ , then the range of values of $5 - 2 x$ is

$- 11 < 5 - 2 x \leq 3$ .
$- 11 \leq 5 - 2 x < 3$ .
$- 3 < 5 - 2 x \leq 11$ .
$- 3 \leq 5 - 2 x < 11$ .

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D. Multiplying by $- 2$ reverses both signs: $6 > - 2 x \geq - 8$ , so $- 3 \leq 5 - 2 x < 11$ . Option C has the right numbers but keeps the hollow end on the left; it is the included end $x = 4$ that produces $- 3$ .

Question 2 · Modelled on 2020 P2 Q13 · runs it backwards

Let $k$ be a constant. If the solution of $2 (x - k) < 5 x + 8$ is $x > - 6$ , then $k =$

$- 13$ .
$- 5$ .
$5$ .
$13$ .

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C. The inequality gives $x > - \frac{2 k + 8}{3}$ . Setting $- \frac{2 k + 8}{3} = - 6$ gives $k = 5$ . Option B comes from expanding $2 (x - k)$ as $2 x + 2 k$ .

Question 3 · No direct past-paper model · states a condition

Let $a$ be a constant. If no real number satisfies the compound inequality $x > 2 a - 3$ and $x \leq a + 1$ , then the range of values of $a$ is

$a < 4$ .
$a \leq 4$ .
$a > 4$ .
$a \geq 4$ .

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D. There is no solution when the lower boundary is not below the upper one: $2 a - 3 \geq a + 1$ , so $a \geq 4$ . At $a = 4$ the pair reads $x > 5$ and $x \leq 5$ , which still has no solution, so option C wrongly leaves out $4$ .

Question 4 · Modelled on 2020 P1 Q6 · inverts what is asked

How many integers do not satisfy the compound inequality $2 x + 1 \leq - 7$ or $3 x - 4 > 11$ ?

$8$
$9$
$10$
$11$

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B. The solution is $x \leq - 4$ or $x > 5$ , so the numbers left out are $- 4 < x \leq 5$ : the integers $- 3$ to $5$ , which is $9$ . Note that $- 4$ satisfies the inequality but $5$ does not.

Question 5 · Modelled on 2014 P2 Q6 · harder statements

If $a > b > 0$ and $c < 0$ , which of the following must be true?

I. $\frac{c}{a} > \frac{c}{b}$
II. $a^{2} c < b^{2} c$
III. $a + c > b - c$

I and II only
I and III only
II and III only
I, II and III

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A. Since $a > b > 0$ , $\frac{1}{a} < \frac{1}{b}$ and $a^{2} > b^{2}$ ; multiplying each by the negative $c$ reverses the sign, so I and II are true. III fails for $a = 2$ , $b = 1$ , $c = - 5$ : it reads $- 3 > 6$ .

Question 6 · No direct past-paper model · a trap

Let $m$ be a negative constant. The solution of $m (x - 2) > 3 m$ is

$x < 5$ .
$x > 5$ .
$x < - 1$ .
$x > - 1$ .

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A. Dividing both sides by $m$ , which is negative, reverses the sign: $x - 2 < 3$ , so $x < 5$ . Option B is what you get by cancelling $m$ as if it were positive.

Question 7 · Modelled on 2025 P1 Q6 · two endpoint decisions

Let $p$ be the greatest integer satisfying $3 x + 2 < 20$ and let $q$ be the least integer satisfying $7 - 2 x \leq 15$ . Then $p + q =$

$0$ .
$1$ .
$2$ .
$3$ .

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B. $x < 6$ gives $p = 5$ , because 6 is not included. $x \geq - 4$ gives $q = - 4$ , because $- 4$ is included. So $p + q = 1$ . Taking $p = 6$ gives 2.

Question 8 · No direct past-paper model · word problem, two conditions

The length and the width of a rectangle are $(2 x + 3)$ cm and $5$ cm respectively. If the perimeter of the rectangle does not exceed $40$ cm and its area is greater than $45$ cm², find the range of values of $x$ .

$x > 3$
$x \leq 6$
$3 < x \leq 6$
$3 \leq x < 6$

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C. Perimeter: $2 (2 x + 3 + 5) \leq 40$ gives $x \leq 6$ . Area: $5 (2 x + 3) > 45$ gives $x > 3$ . Both must hold, so $3 < x \leq 6$ . “Does not exceed” allows equality; “greater than” does not.

Question 9 · Modelled on 2023 P1 Q4 · runs it backwards

Let $k$ be a constant. If exactly $4$ positive integers satisfy the inequality $3 x - k \leq 0$ , then the range of values of $k$ is

$12 < k < 15$ .
$12 < k \leq 15$ .
$12 \leq k < 15$ .
$12 \leq k \leq 15$ .

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C. The inequality gives $x \leq \frac{k}{3}$ . The positive integers $1, 2, 3, 4$ satisfy it and $5$ does not when $4 \leq \frac{k}{3} < 5$ , so $12 \leq k < 15$ . At $k = 12$ , $x \leq 4$ still includes 4; at $k = 15$ , $x \leq 5$ lets in a fifth.

Question 10 · Modelled on 2016 P1 Q6 · inverts what is asked

Which of the following compound inequalities are satisfied by all real numbers $x$ ?

I. $x > 3$ or $x < 5$
II. $x > 5$ and $x < 3$
III. $x \geq 4$ or $x \leq 4$

I and II only
I and III only
II and III only
I, II and III

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B. I: any number that is not greater than 3 is certainly less than 5, so every number satisfies one part. III: every number is at least 4 or at most 4. II is the opposite extreme: no number is both greater than 5 and less than 3.

Continue practising this topic: Try the Paper 2 multiple-choice first if you have not, or move to Paper 1 long questions — also available as Paper 1 Challenge.

Where this sits in the syllabus

Linear Inequalities in One Unknown is junior secondary content that the HKDSE keeps examining: a four-mark compound inequality in Paper 1 Section A(1) in most years, and usually one question in Paper 2 Section A. Marks are lost in three places: not reversing the sign when dividing by a negative number, treating “and” as “or”, and miscounting the integers at an end that is or is not included.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

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