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Elementary Algebra: Paper 2 Challenge MC

Twelve harder Paper 2 questions. Each has a distractor that is the right answer to the wrong question: an odd negative power treated as even, a constant that empties the right-hand side instead of the coefficient, parallel lines mistaken for the same line. This is the Challenge tier; work the standard Paper 2 set first. Detailed solutions can be requested by email.

12 questions · about 25 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 · Modelled on 2015 P2 Q2 · negative index on a negative base

$(- 2 a^{- 1})^{- 3} =$

$\frac{a ^{3}}{8}$
$- 8 a^{3}$
$- \frac{a ^{3}}{8}$
$- \frac{1}{8 a ^{3}}$

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C. $(- 2)^{- 3} = \frac{1}{( - 2 ) ^{3}} = - \frac{1}{8}$ and $(a^{- 1})^{- 3} = a^{3}$ .

Question 2 · Modelled on 2026 P2 Q2 · factorizing powers before cancelling

$\frac{2 ^{2027} - 2 ^{2025}}{2 ^{2026} + 2 ^{2024}} =$

$\frac{6}{5}$
$\frac{3}{5}$
$2$
$0$

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A. Take out $2^{2024}$ : $\frac{2 ^{2024} ( 8 - 2 )}{2 ^{2024} ( 4 + 1 )} = \frac{6}{5}$ .

Question 3 · Extension · no direct past-paper model · order of negative numbers, reciprocals and products

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

I. $a^{2} > b^{2}$
II. $\frac{1}{a} > \frac{1}{b}$
III. $ab > b^{2}$

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

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D. With $a = - 3, b = - 1$ : $9 > 1$ , $- \frac{1}{3} > - 1$ , $3 > 1$ . In general each statement follows from $a < b < 0$ .

Question 4 · Modelled on 2019 P2 Q5 · subject appears twice

If $y = \frac{a x + 1}{x - a}$ , where $a$ is a constant, then $x =$

$\frac{a y - 1}{y - a}$
$\frac{a y + 1}{y - a}$
$\frac{a y + 1}{y + a}$
$\frac{a y + 1}{a - y}$

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B. $y (x - a) = a x + 1$ gives $x y - a x = a y + 1$ , so $x (y - a) = a y + 1$ .

Question 5 · Modelled on 2014 P1 Q2 · factorizing four terms by grouping

$a^{3} - a^{2} b - a b^{2} + b^{3} =$

$(a + b)^{2} (a - b)$
$(a - b) (a^{2} + b^{2})$
$(a - b)^{3}$
$(a - b)^{2} (a + b)$

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D. $a^{2} (a - b) - b^{2} (a - b) = (a - b) (a^{2} - b^{2}) = (a - b)^{2} (a + b)$ .

Question 6 · Modelled on 2019 P2 Q4 · which equations are identities

Which of the following are identities?

I. $(x - 3)^{2} = (3 - x)^{2}$
II. $(x - y)^{3} = (y - x)^{3}$
III. $(x + 1)^{2} - x^{2} = 2 x + 1$

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

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C. I and III hold for every value; II fails at $x = 1, y = 0$ (1 ≠ −1).

Question 7 · Modelled on 2026 P2 Q6 · linear equation with no solution

The equation $k (x + 1) = 2 x + 3$ has no solution. Find the constant $k$ .

$2$
$3$
$- 2$
$2$ or $3$

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A. $(k - 2) x = 3 - k$ . No solution needs $k - 2 = 0$ while $3 - k \neq = 0$ : $k = 2$ .

Question 8 · Modelled on 2015 P2 Q3 · simultaneous equations with infinitely many solutions

The simultaneous equations ${2 x - k y = 3 6 x - 9 y = c$ have infinitely many solutions. Which of the following is true?

$k = 3$ and $c \neq = 9$
$k = - 3$ and $c = 9$
$k = 3$ and $c = 3$
$k = 3$ and $c = 9$

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D. The second equation must be 3 times the first: $3 (- k) = - 9$ and $3 \times 3 = c$ .

Question 9 · Extension · no direct past-paper model · rationalizing two denominators

$\frac{1}{3 - 2} - \frac{1}{3 + 2} =$

$23$
$0$
$22$
$\frac{2 2}{5}$

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C. Each denominator rationalizes to $(3)^{2} - (2)^{2} = 1$ , so the value is $(3 + 2) - (3 - 2)$ .

Question 10 · Extension · no direct past-paper model · what must be irrational

Let $p$ and $q$ be irrational numbers. Which of the following must be irrational?

I. $p + q$
II. $pq$
III. $p + 1$

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

Show answer

B. $2 + (- 2) = 0$ and $2 \times 2 = 2$ are rational. If $p + 1$ were rational, $p = (p + 1) - 1$ would be rational too.

Question 11 · Modelled on 2024 P2 Q5 · setting up equations: a two-digit number

The sum of the digits of a two-digit number is 11. When the two digits are interchanged, the new number is 27 greater than the original number. The original number is

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A. With digits $a$ and $b$ : $a + b = 11$ and $(10 b + a) - (10 a + b) = 27$ , i.e. $b - a = 3$ . So $a = 4, b = 7$ .

Question 12 · Section B · Modelled on 2024 P2 Q32 · hexadecimal place value

$3 \times 1 6^{5} + 10 \times 1 6^{2} + 1 =$

$30 A 00 1_{16}$
$300 A 0 1_{16}$
$3000 A 1_{16}$
$3 A 000 1_{16}$

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B. Place values from $1 6^{5}$ down to $1 6^{0}$ hold the digits 3, 0, 0, A, 0, 1.

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

Elementary Algebra is the junior secondary groundwork that the HKDSE tests at the start of both papers: a three-mark index or change-of-subject question and a factorization in Paper 1 Section A(1), and several quick items near the start of Paper 2. Most lost marks come from signs: a negative index in the denominator, a minus sign in front of a bracket, and the sign of a denominator after changing the subject. The remainder and factor theorems, H.C.F. and L.C.M., and algebraic fractions are practised in the Polynomials, Remainder and Factor Theorems sets.

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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