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Elementary Algebra: Paper 1 Long Questions

Twelve Paper 1 questions on the junior algebra that opens every HKDSE paper: laws of indices, change of subject, factorization by the cross-method and by grouping, identities, surds, and linear equations in one and two unknowns. The set starts gently and ends with questions that ask you to test a claim. Detailed solutions can be requested by email.

12 questions · 47 marks · about 60 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 2025 P1 Q1 · laws of integral indices

Simplify $\frac{( m ^{- 2} n ^{3} ) ^{4}}{m ^{5} n ^{- 2}}$ and express your answer with positive indices.

(3 marks)

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$\frac{n ^{14}}{m ^{13}}$

Raise every index inside the bracket to the 4th power first. Dividing by n⁻² adds 2 to the index of n.

Question 2 · Modelled on 2024 P1 Q2 · change of subject

Make $b$ the subject of the formula $\frac{3 a + b}{4} = \frac{b - 2 c}{a}$ .

(3 marks)

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$b = \frac{3 a ^{2} + 8 c}{4 - a}$

Clear the fractions, move every term containing b to one side, then take b out as a common factor.

Question 3 · Modelled on 2024 P1 Q3 · cross-method, then grouping

Factorize

(a) $3 x^{2} - 10 x y - 8 y^{2}$ ,

(b) $3 x^{2} - 10 x y - 8 y^{2} - 6 x + 24 y$ .

(3 marks)

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(a) $(3 x + 2 y) (x - 4 y)$ (b) $(x - 4 y) (3 x + 2 y - 6)$

In (b), −6x + 24y = −6(x − 4y), which shares the factor x − 4y with (a).

Question 4 · Junior skill · no direct past-paper model · linear equation with fractions

Solve the equation $\frac{2 x - 1}{3} - \frac{x + 4}{4} = \frac{x - 2}{6}$ .

(3 marks)

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$x = 4$

Multiply every term by the L.C.M. 12, and keep the brackets: −3(x + 4) = −3x − 12.

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

Show that the value of $\frac{8 ^{n + 1} \cdot 4 ^{- n}}{2 ^{n - 3}}$ does not depend on $n$ , and find this value.

(3 marks)

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The expression equals $2^{6} = 64$ for every $n$ .

Write 8 and 4 as powers of 2. The indices then add to 6, with no n left.

Question 6 · Modelled on 2013 P1 Q3 · difference of squares; evaluating with factors

(a) Factorize $9 a^{2} - 16 b^{2}$ .

(b) Hence factorize $9 a^{2} - 16 b^{2} - 12 a + 16 b$ .

(c) It is given that $3 a - 4 b = 5$ and $3 a + 4 b = 7$ . Using (b), find the value of $9 a^{2} - 16 b^{2} - 12 a + 16 b$ .

(4 marks)

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(a) $(3 a - 4 b) (3 a + 4 b)$ (b) $(3 a - 4 b) (3 a + 4 b - 4)$ (c) 15

In (c) there is no need to find a and b: substitute 5 and 7 straight into the factors, 5 × (7 − 4) = 15.

Question 7 · Modelled on 2013 P1 Q4 · simultaneous linear equations from a context

A cinema sells adult tickets and student tickets. The total price of 3 adult tickets and 5 student tickets is $555, while the total price of 2 adult tickets and 7 student tickets is $579.

(a) Find the price of an adult ticket and the price of a student ticket.

(b) A teacher buys 4 adult tickets and some student tickets for $1 044. How many student tickets does the teacher buy?

(4 marks)

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(a) Adult $90, student $57 (b) 12

Eliminate one unknown: multiply the first equation by 2 and the second by 3, then subtract.

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

It is given that $5 x^{2} - 7 x + 6 \equiv A (x - 1)^{2} + B (x - 1) + C$ , where $A$ , $B$ and $C$ are constants.

(a) Find $A$ , $B$ and $C$ .

(b) Hence, without using a calculator, find the value of $5 x^{2} - 7 x + 6$ when $x = 1.1$ .

(4 marks)

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(a) $A = 5$ , $B = 3$ , $C = 4$ (b) 4.35

Expand the right-hand side and compare coefficients. In (b), x − 1 = 0.1 makes every term easy to work out by hand.

Question 9 · Junior skill · no direct past-paper model · rationalizing a denominator; rational or not

Let $k = \frac{6}{3 - 3}$ .

(a) Express $k$ in the form $p + q 3$ , where $p$ and $q$ are integers.

(b) Is $k^{2} - 6 k$ a rational number? Explain your answer.

(4 marks)

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(a) $k = 3 + 3$ (b) Yes. $k^{2} - 6 k = - 6$ , an integer.

Multiply top and bottom by 3 + √3. In (b) the √3 terms cancel: 12 + 6√3 − 18 − 6√3 = −6.

Question 10 · Modelled on 2014 P1 Q5 · change of subject; substitution; a value that fits both scales

A temperature of $C$ degrees Celsius is equal to $F$ degrees Fahrenheit, where $F = \frac{9 C}{5} + 32$ .

(a) Make $C$ the subject of the formula.

(b) A thermometer reads 95 °F. Find the temperature in degrees Celsius.

(c) Is there a temperature at which the reading in °C is equal to the reading in °F? Explain your answer.

(5 marks)

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(a) $C = \frac{5 ( F - 32 )}{9}$ (b) 35 °C (c) Yes, at −40 (−40 °C = −40 °F).

For (c), put F = C into the formula and solve the linear equation. A negative answer is a real temperature.

Question 11 · Modelled on 2026 P1 Q5 · simultaneous equations; whole-number solutions

In a shop, the price of a pen is $$ x$ and the price of a notebook is $$ y$ . The total price of 4 pens and 3 notebooks is $66, while the total price of 3 pens and 5 notebooks is $77.

(a) Find $x$ and $y$ .

(b) Mary spends exactly $150 on some pens and some notebooks from the shop, buying at least one of each. Someone claims that Mary buys 15 items in total. Is the claim correct? Explain your answer.

(6 marks)

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(a) $x = 9$ , $y = 10$ (b) Not correct. The only possibility is 10 pens and 6 notebooks, which is 16 items.

In (b), 9p + 10n = 150 forces 9p to end in 0, so p is a multiple of 10. With at least one notebook, p = 10.

Question 12 · Modelled on 2024 P2 Q1 · proving an identity; using it to compute and to test a claim

(a) Prove that $(a + b)^{2} - (a - b)^{2} \equiv 4 ab$ .

(b) Hence, without using a calculator, find the value of $202 7^{2} - 202 5^{2}$ .

(c) Someone claims that if $(a + b)^{2} = (a - b)^{2}$ , then $a = 0$ . Is the claim correct? Explain your answer.

(5 marks)

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(b) 8 104 (c) Not correct. $4 ab = 0$ means $a = 0$ or $b = 0$ ; for example $a = 3, b = 0$ .

In (b), choose a = 2 026 and b = 1, so that a + b = 2 027 and a − b = 2 025.

Continue practising this topic: Try the Paper 1 Challenge for harder variants, or move to Paper 2 multiple-choice — also available as Paper 2 Challenge MC.

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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