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Elementary Algebra: Paper 1 Challenge

Ten harder Paper 1 questions that state a condition rather than name a method: an unknown index, a letter constant that can leave an equation with no solution, integer solutions read from a factorization, conjugate surds, and claims to test. This is the Challenge tier; work the standard Paper 1 set first. Detailed solutions can be requested by email.

10 questions · 48 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 2021 P1 Q1 · indices: unknown index; testing a claim

(a) For all non-zero values of $a$ and $b$ , $\frac{a ^{3} b ^{- 2}}{( a ^{- 1} b ^{k} ) ^{2}} = \frac{a ^{m}}{b ^{8}}$ , where $k$ and $m$ are constants. Find $k$ and $m$ .

(b) Someone claims that $\frac{8 ^{n}}{4 ^{n - 1}}$ is an even integer for every integer $n$ . Is the claim correct? Explain your answer.

(5 marks)

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(a) $k = 3$ , $m = 5$ (b) Not correct. The expression equals $2^{n + 2}$ ; for $n = - 2$ it is 1, which is odd.

“Every integer” includes negative integers. The claim holds only for n ≥ −1.

Question 2 · Modelled on 2019 P2 Q5 · subject appears twice; an impossible value

Consider the formula $y = \frac{2 x + 3}{x - 1}$ , where $x \neq = 1$ .

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

(b) Explain why $y$ can never be equal to 2.

(4 marks)

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(a) $x = \frac{y + 3}{y - 2}$ (b) $y = 2$ would give $2 x - 2 = 2 x + 3$ , i.e. $- 2 = 3$ , which is impossible.

In (a) the step x(y − 2) = y + 3 already shows the problem: dividing by y − 2 needs y ≠ 2.

Question 3 · Modelled on 2026 P2 Q6 · linear equation with a letter constant

Consider the equation $a (x - 2) = 3 x + 4$ , where $a$ is a constant.

(a) Show that the equation has no solution when $a = 3$ .

(b) For $a \neq = 3$ , express the solution $x$ in terms of $a$ .

(c) How many integers $a$ are there such that the solution $x$ is an integer?

(5 marks)

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(b) $x = \frac{2 a + 4}{a - 3}$ (c) 8

In (c), write the solution as 2 + 10/(a − 3). It is an integer exactly when a − 3 divides 10.

Question 4 · Extension · no direct past-paper model · directed numbers: deducing signs; testing a claim

Let $a$ , $b$ and $c$ be non-zero numbers such that $ab < 0$ , $a - b > 0$ and $b c > 0$ .

(a) Determine whether each of $a$ , $b$ and $c$ is positive or negative.

(b) Someone claims that $(a - c) (b - c) > 0$ must be true. Is the claim correct? Explain your answer.

(4 marks)

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(a) $a$ positive, $b$ negative, $c$ negative (b) Not correct. $a = 5, b = - 3, c = - 1$ satisfy the conditions but give $(6) (- 2) = - 12$ .

a − c is always positive, but b − c can have either sign, because b and c are both negative with no order given.

Question 5 · Modelled on 2019 P2 Q4 · identities: completing a form; an identity that cannot exist

(a) Find the constants $A$ and $B$ such that $x^{2} + 4 x + 7 \equiv (x + A)^{2} + B$ .

(b) Someone claims that there is a constant $k$ such that $(x + k)^{2} - (x - k)^{2} \equiv 12 x + k$ . Is the claim correct? Explain your answer.

(4 marks)

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(a) $A = 2$ , $B = 3$ (b) Not correct. $4 k x \equiv 12 x + k$ needs both $4 k = 12$ and $k = 0$ .

An identity must match every coefficient. Matching the x-terms gives k = 3, but the constant terms then fail.

Question 6 · Modelled on 2022 P1 Q4 · perfect square, difference of squares; integer solutions

(a) Factorize $x^{2} - y^{2} - 6 x + 9$ .

(b) Find all pairs of positive integers $(x, y)$ such that $x^{2} - y^{2} - 6 x + 9 = 7$ .

(5 marks)

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(a) $(x - y - 3) (x + y - 3)$ (b) $(7, 3)$ only

The two factors are integers whose product is 7, and the second exceeds the first by 2y > 0. That leaves 1 × 7 or (−7) × (−1).

Question 7 · Extension · no direct past-paper model · surds: conjugates; a product that is rational

Let $a = 5 + 2$ and $b = 5 - 2$ .

(a) Find $ab$ and $a - b$ .

(b) Hence, without expanding $a^{2}$ or $b^{2}$ , find the value of $a^{2} + b^{2}$ .

(c) Someone claims that the product of two irrational numbers must be irrational. Is the claim correct? Explain your answer.

(5 marks)

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(a) $ab = 1$ , $a - b = 4$ (b) 18 (c) Not correct. $a$ and $b$ are irrational but $ab = 1$ .

In (b), use a² + b² = (a − b)² + 2ab = 16 + 2.

Question 8 · Modelled on 2015 P2 Q3 · simultaneous equations with no solution or infinitely many

Consider the simultaneous equations ${2 x + k y = 5 4 x + 6 y = c$ , where $k$ and $c$ are constants.

(a) It is given that $x = 1, y = 1$ is a solution. Find $k$ and $c$ .

(b) Using the values in (a), someone claims that $x = 1, y = 1$ is the only solution. Is the claim correct? Explain your answer.

(c) Find all values of $k$ and $c$ for which the simultaneous equations have no solution.

(5 marks)

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(a) $k = 3$ , $c = 10$ (b) Not correct; the second equation is twice the first, e.g. $x = 4, y = - 1$ also works. (c) $k = 3$ and $c \neq = 10$

When the second equation is a multiple of the first, the two lines coincide. When only the x and y parts are multiples, the lines are parallel.

Question 9 · Modelled on 2014 P1 Q5 · formula with reciprocals; testing a claim about change

For a lens, the distances $u$ cm and $v$ cm and the focal length $f$ cm satisfy $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ , where $u > f > 0$ .

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

(b) It is given that $f = 10$ . Find $v$ when $u = 15$ , and when $u = 20$ .

(c) Someone claims that for a fixed $f$ , the larger $u$ is, the larger $v$ is. Is the claim correct? Explain your answer.

(5 marks)

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(a) $v = \frac{u f}{u - f}$ (b) 30 and 20 (c) Not correct. By (b), $u$ rises from 15 to 20 while $v$ falls from 30 to 20.

Combine the two fractions on one side before inverting: 1/v = (u − f)/(uf).

Question 10 · Modelled on 2021 P1 Q5 · simultaneous equations; parity; least total

A box contains only $2 coins and $5 coins, with a total value of $137.

(a) If there are 40 coins in the box, find the number of each kind of coin.

(b) Now the number of coins is not given. Prove that the number of $5 coins in the box must be odd.

(c) Find the least possible number of coins in the box.

(6 marks)

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(a) 21 $2 coins and 19 $5 coins (c) 28 (one $2 coin and twenty-seven $5 coins)

In (c), using more $5 coins means fewer coins in total, so take as many $5 coins as possible while 137 − 5y stays a positive even number.

Continue practising this topic: Try the Paper 1 long questions first if you have not, 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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