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Elementary Algebra: Paper 2 Multiple-Choice

Fifteen Paper 2 questions in the style of the first section of the paper: directed numbers, indices, expansion and factorization, identities, change of subject, simultaneous equations, rational and irrational numbers, and one Section B question on binary numbers. Detailed solutions can be requested by email.

15 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 · Extension · no direct past-paper model · directed numbers and fractions without a calculator

$(- \frac{2}{3})^{2} \div (- \frac{1}{6}) - (- 2)^{3} =$

$\frac{16}{3}$
$- \frac{32}{3}$
$\frac{32}{3}$
$\frac{10}{3}$

Show answer

A. $\frac{4}{9} \times (- 6) = - \frac{8}{3}$ , and $- (- 8) = + 8$ , so the value is $- \frac{8}{3} + 8 = \frac{16}{3}$ .

Question 2 · Extension · no direct past-paper model · directed numbers: which expressions must be positive

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

I. $b - a$
II. $ab$
III. $a^{2} b$

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

Show answer

C. $b - a = b + (- a)$ is a sum of two positives; $ab < 0$ ; $a^{2} > 0$ so $a^{2} b > 0$ .

Question 3 · Modelled on 2026 P2 Q3 · expanding squares

$(2 a - b)^{2} - (a - 2 b)^{2} =$

$3 a^{2} - 8 ab - 3 b^{2}$
$3 a^{2} - 3 b^{2}$
$5 a^{2} - 8 ab + 5 b^{2}$
$3 a^{2} + 3 b^{2}$

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B. $(4 a^{2} - 4 ab + b^{2}) - (a^{2} - 4 ab + 4 b^{2}) = 3 a^{2} - 3 b^{2}$ ; the $ab$ terms cancel.

Question 4 · Modelled on 2015 P2 Q2 · laws of integral indices

$\frac{( 3 x ^{- 2} y ) ^{3}}{9 x ^{4} y ^{- 1}} =$

$\frac{3 y ^{2}}{x ^{10}}$
$\frac{y ^{4}}{3 x ^{10}}$
$\frac{3 y ^{4}}{x ^{2}}$
$\frac{3 y ^{4}}{x ^{10}}$

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D. $\frac{27 x ^{- 6} y ^{3}}{9 x ^{4} y ^{- 1}} = 3 x^{- 10} y^{4}$ .

Question 5 · Modelled on 2026 P2 Q2 · indices with a letter in the index

$\frac{2 5 ^{n + 1}}{5 ^{2 n - 1}} =$

$5$
$25$
$125$
$5^{4 n + 1}$

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C. $2 5^{n + 1} = 5^{2 n + 2}$ , and $5^{(2 n + 2) - (2 n - 1)} = 5^{3}$ .

Question 6 · Modelled on 2019 P2 Q5 · change of subject with fractions

If $\frac{1}{a} - \frac{2}{b} = 3$ , then $b =$

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

Show answer

A. $\frac{2}{b} = \frac{1}{a} - 3 = \frac{1 - 3 a}{a}$ , so $b = \frac{2 a}{1 - 3 a}$ .

Question 7 · Modelled on 2013 P1 Q3 · difference of squares, then grouping

$x^{2} - 4 y^{2} - x - 2 y =$

$(x - 2 y) (x + 2 y - 1)$
$(x + 2 y) (x - 2 y + 1)$
$(x - 2 y) (x + 2 y + 1)$
$(x + 2 y) (x - 2 y - 1)$

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D. $x^{2} - 4 y^{2} = (x + 2 y) (x - 2 y)$ and $- x - 2 y = - (x + 2 y)$ ; take out $x + 2 y$ .

Question 8 · Modelled on 2019 P2 Q4 · identity: comparing coefficients

If $a$ and $b$ are constants such that $(x + a) (x - 3) + b \equiv x^{2} + 2 x + 1$ , then

$a = 5$ and $b = - 14$ .
$a = 5$ and $b = 16$ .
$a = - 1$ and $b = - 2$ .
$a = - 5$ and $b = - 14$ .

Show answer

B. Compare coefficients: $a - 3 = 2$ gives $a = 5$ ; $- 3 a + b = 1$ gives $b = 16$ .

Question 9 · Junior skill · no direct past-paper model · linear equation with fractions

The solution of $\frac{x + 2}{3} - \frac{x - 1}{2} = 1$ is

$x = - 5$ .
$x = 13$ .
$x = 1$ .
$x = - 1$ .

Show answer

C. Multiply by 6: $2 x + 4 - 3 x + 3 = 6$ , so $- x = - 1$ and $x = 1$ .

Question 10 · Modelled on 2015 P2 Q11 · algebraic expressions from a given ratio

If $\frac{x}{y} = \frac{3}{4}$ , then $\frac{2 x + y}{3 x - y} =$

$2$ .
$\frac{10}{13}$ .
$\frac{1}{2}$ .
$\frac{11}{9}$ .

Show answer

A. Put $x = 3 k, y = 4 k$ : $\frac{6 k + 4 k}{9 k - 4 k} = \frac{10 k}{5 k} = 2$ .

Question 11 · Junior skill · no direct past-paper model · identity for a sum of squares

If $m + n = 7$ and $mn = 10$ , then $m^{2} + n^{2} =$

$39$
$49$
$69$
$29$

Show answer

D. $m^{2} + n^{2} = (m + n)^{2} - 2 mn = 49 - 20$ .

Question 12 · Modelled on 2015 P2 Q3 · simultaneous linear equations

If $3 x - 2 y = 13$ and $x + 4 y = - 5$ , then $x + y =$

$5$
$1$
$3$
$- 11$

Show answer

B. From the second equation $x = - 5 - 4 y$ ; then $3 (- 5 - 4 y) - 2 y = 13$ gives $y = - 2$ , $x = 3$ .

Question 13 · Modelled on 2024 P2 Q5 · simultaneous equations from a context

The total price of 3 kg of apples and 2 kg of pears is $96. The price of 1 kg of pears is $8 more than the price of 1 kg of apples. The price of 1 kg of pears is

$16.
$19.2.
$24.
$25.6.

Show answer

C. Let apples be $$ a$ per kg: $3 a + 2 (a + 8) = 96$ , so $a = 16$ and pears cost $24.

Question 14 · Junior skill · no direct past-paper model · rational and irrational numbers

Which of the following are irrational numbers?

I. $2.25$
II. $18 - 8$
III. $(3 + 1) (3 - 1)$

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

Show answer

B. $2.25 = 1.5$ ; $18 - 8 = 32 - 22 = 2$ ; $(3 + 1) (3 - 1) = 2$ .

Question 15 · Section B · Modelled on 2026 P2 Q31 · binary numbers

$101100000001 1_{2} =$

$11 \times 2^{9} + 3$
$11 \times 2^{10} + 3$
$13 \times 2^{9} + 3$
$11 \times 2^{9} + 6$

Show answer

A. The number has 13 digits. The block $101 1_{2} = 11$ is followed by 9 digits, so it contributes $11 \times 2^{9}$ ; the last digits $1 1_{2} = 3$ .

Continue practising this topic: Try the Paper 2 Challenge MC for harder variants, 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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