FREE HKDSE PRACTICE PAPER

HKDSE Maths Compulsory Paper 1 Mock Paper

This 105-mark sample contains 19 conventional questions across Sections A(1), A(2) and B, with 35 marks in each section. Allow 2 hours 15 minutes. Questions cover quadratic equations, functions, coordinate geometry, trigonometry, probability and statistics, with later questions combining topics.

19 questions105 marks2 hr 15 min

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An original AI-generated practice paper from Studify. This is an independent sample, not an official HKEAA paper.

Paper structure

Section A(1)9 questions · 35 marks
Section A(2)5 questions · 35 marks
Section B5 questions · 35 marks

Topics in this sample

Quadratic Equations in One Unknown
Functions and Graphs
Exponential and Logarithmic Functions
More about Polynomials
Equations of Straight Lines
Basic Properties of Circles
More about Trigonometry (2D and 3D Problems)
Arithmetic and Geometric Sequences and Their Summations
Measures of Dispersion
More about Graphs of Functions
Variations
Inequalities and Linear Programming
Further Applications
Permutations and Combinations
More about Probability
Loci
Equations of Circles
More about Equations
Uses and Abuses of Statistics

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Hong Kong Diploma of Secondary Education Examination Mock Paper

Mathematics (Compulsory Part)

Paper 1: Conventional Questions

Time Allowed: 2 hr 15 minTotal Marks: 105

INSTRUCTIONS TO CANDIDATES:

This paper consists of multiple sections.
Answer ALL questions in this paper.
Write your answers in the spaces provided.
Unless otherwise specified, all working must be clearly shown.
Unless otherwise specified, numerical answers should be either exact or correct to 3 significant figures.

Generated by studify.life • 16/10/2026

Section A(1)

Question 1

Solve the equation

2x^2-7x+3=0

.(3 marks)

Question 2

Let

f(x)=2x^2-3x+1

(a)

Find

f(-2)

(2 marks)

(b)

Solve

f(x)=0

(2 marks)

Question 3

Answer the following questions.

(a)

Solve

3^{x+1}=81

(2 marks)

(b)

Evaluate

\log_{2} 32

(2 marks)

Question 4

The polynomial

p(x)=x^3+ax^2-4x+6

leaves a remainder of 4 when divided by

x-1

(a)

Find the value of

a

(2 marks)

Page2

Question 4 (continued)

(b)

Hence find the remainder when

p(x)

is divided by

x+2

(2 marks)

Question 5

A straight line passes through

A(2,-1)

and

B(6,7)

(a)

Find the gradient of

AB

(2 marks)

(b)

Find the equation of

AB

in the form

y=mx+c

(2 marks)

Page3

Question 6

In the circle with centre

O

AB

is a chord and

OM

is perpendicular to

AB

M

. The radius of the circle is 5 cm and

OM=3

cm.

(a)

Find the length of

AM

(2 marks)

(b)

Find the length of chord

AB

(2 marks)

Page4

Question 7

A ladder of length 5 m rests against a vertical wall. Its foot is 3 m from the wall.

(a)

Find the height reached by the ladder on the wall.

(2 marks)

(b)

Find the angle

\theta

between the ladder and the ground, to the nearest degree.

(2 marks)

Question 8

An arithmetic sequence has first term 7 and common difference 3.

(a)

Find its 20th term.

(2 marks)

(b)

Find the sum of its first 20 terms.

(2 marks)

Page5

Question 9

The table shows the scores of a group of students.

Score	2	4	6	8
Frequency	1	3	4	2

(a)

Find the mean score.

(2 marks)

(b)

Find the range of the scores.

(2 marks)

Page6

Section A(2)

Question 10

The graph of

y=x^2-4x+3

is considered.

(a)

Find the coordinates of the vertex of the graph.

(2 marks)

(b)

Find the

x

-intercepts of the graph.

(2 marks)

(c)

Find the range of

y=x^2-4x+3

for

-1\leq x\leq 5

(3 marks)

Page7

Question 11

A positive quantity

y

varies inversely as the square of

x

. When

x=2

y=18

(a)

Express

y

in terms of

x

(2 marks)

(b)

Find

y

when

x=3

(2 marks)

(c)

Find the positive value of

x

when

y=2

(3 marks)

Question 12

A geometric sequence has first term 5 and common ratio 2.

(a)

Find the 8th term.

(2 marks)

(b)

Find the sum of the first 8 terms.

(3 marks)

(c)

Find the least positive integer

n

such that the

n

th term is greater than 1000.

(2 marks)

Page8

Question 13

In triangle

ABC

AB=8

cm,

AC=6

cm and

\angle BAC=60^\circ

Not drawn to scale

(a)

Find the length of

BC

(3 marks)

(b)

Find

\angle ABC

, to the nearest degree.

(4 marks)

Page9

Question 14

The function

f

is defined by

f(x)=2^x+1

(a)

Find

f(0)

and

f(2)

(2 marks)

(b)

State the horizontal asymptote of the graph of

y=f(x)

(2 marks)

(c)

Solve

f(x)=5

(3 marks)

Page10

Section B

Question 15

A workshop makes two products,

X

and

Y

. Let

x

and

y

be the numbers of units made per day. Each unit of

X

requires 2 machine-hours and 1 labour-hour. Each unit of

Y

requires 1 machine-hour and 2 labour-hours. At most 100 machine-hours and 80 labour-hours are available. The workshop makes at most 50 units in total. The profit on each unit of

X

is $30 and on each unit of $Y

is

40.

(a)

Write down all the inequalities, including the non-negativity conditions, that describe the feasible production plans.

(3 marks)

(b)

Write the profit function

P

(1 mark)

(c)

Find the production plan that gives the maximum profit, and state that maximum profit.

(3 marks)

Question 16

A committee of 3 people is to be selected from 5 women and 4 men. Assume that every selection is equally likely.

(a)

Find the number of different committees that can be formed.

(2 marks)

(b)

Find the probability that a randomly selected committee consists of exactly 2 women.

(2 marks)

Page11

Question 16 (continued)

(c)

Given that the committee contains at least one man, find the probability that it contains exactly 2 women.

(3 marks)

Question 17

A vertical tower

TA

stands at the centre

A

of a horizontal circular field of radius

AB=20

m. From a point

B

on the boundary, the angle of elevation of the top

T

of the tower is

30^\circ

. Point

C

is another point on the boundary such that

\angle BAC=90^\circ

Not drawn to scale

(a)

Find the height of the tower.

(2 marks)

(b)

Find the length of

BC

(2 marks)

Page12

Question 17 (continued)

(c)

Find the angle of elevation of

T

from

C

, to the nearest degree.

(3 marks)

Question 18

A circular sign has centre

C

and radius 5 cm. Its centre is at

(2,-1)

in a Cartesian coordinate plane. A point

P

moves so that its distance from

(2,-1)

is always 5.

(a)

Write down the equation of the locus of

P

(2 marks)

(b)

Find the coordinates of the points where this locus meets the

x

-axis.

(2 marks)

(c)

The line

y=x-1

intersects the circle at

C

and another point

Q

. Find the coordinates of

Q

(3 marks)

Page13

Question 19

The following are the test scores of two classes. Treat each class as the whole population for its calculations.

Class	Scores
A	4, 5, 5, 6, 10
B	3, 6, 6, 7, 8

(a)

Find the mean score of each class.

(2 marks)

(b)

Find the population standard deviation of each class, to 2 decimal places.

(3 marks)

(c)

A report states, 'Class A performed better because its highest score was greater.' Explain why this conclusion is not justified by the information given.

(2 marks)

Page14

Model answers

Choose a question to reveal the saved solution and marking guidance. Try it yourself first.

Question 13 marks

1: Factorise

2x^2-7x+3=(2x-1)(x-3)

. (M1) Hence

\displaystyle x=\frac{1}{2}

x=3

. (A1, A1)

Question 24 marks

2 / (a):

f(-2)=2(-2)^2-3(-2)+1=15

. Substitution (M1); answer (A1). 2 / (b):

2x^2-3x+1=(2x-1)(x-1)=0

, so

x=\frac12

x=1

. Factorisation (M1); roots (A1).

Question 34 marks

3 / (a): Since

81=3^4

x+1=4

, giving

x=3

. Equating exponents (M1); answer (A1). 3 / (b): Since

2^5=32

\log_2 32=5

. Correct logarithmic interpretation (M1); answer (A1).

Question 44 marks

4 / (a): By the remainder theorem,

p(1)=1+a-4+6=a+3=4

, so

a=1

. Remainder theorem and equation (M1); answer (A1). 4 / (b): The required remainder is

p(-2)=-8+4+8+6=10

. Substitution (M1); answer (A1).

Question 54 marks

5 / (a):

\displaystyle m=\frac{7-(-1)}{6-2}=2

. Correct gradient formula and substitution (M1); answer (A1). 5 / (b): Using

y=2x+c

and

(2,-1)

gives

-1=4+c

, hence

c=-5

. Correct substitution (M1);

y=2x-5

(A1).

Question 64 marks

6 / (a): In right-angled

OMA

AM^2=OA^2-OM^2=5^2-3^2=16

, so

AM=4

cm. Pythagoras setup (M1); answer (A1). 6 / (b): The perpendicular from the centre to a chord bisects the chord, so

AB=2AM=8

cm. Use of circle property (M1); answer (A1).

Question 74 marks

7 / (a): Height

h=\sqrt{5^2-3^2}=4

m. Pythagoras equation (M1); answer (A1). 7 / (b):

\cos\theta=\frac35

, so

\theta=\cos^{-1}(0.6)\approx53^\circ

. Correct trigonometric ratio (M1); answer with requested accuracy (A1).

Question 84 marks

8 / (a):

a_{20}=7+19(3)=64

. Correct term formula (M1); answer (A1). 8 / (b):

\displaystyle S_{20}=\frac{20}{2}(7+64)=710

. Correct sum formula and substitution (M1); answer (A1).

Question 94 marks

9 / (a): Total frequency

=10

and sum of scores

=2(1)+4(3)+6(4)+8(2)=54

. Thus mean

=54/10=5.4

. Weighted total and division (M1); answer (A1). 9 / (b): Range

=8-2=6

. Correct subtraction (M1); answer (A1).

Question 107 marks

10 / (a):

y=(x-2)^2-1

, so the vertex is

(2,-1)

. Completing the square (M1); coordinates (A1). 10 / (b):

x^2-4x+3=(x-1)(x-3)=0

, hence

(1,0)

and

(3,0)

. Factorisation (M1); both intercepts (A1). 10 / (c): The minimum is

-1

x=2

. At the endpoints,

y(-1)=8

and

y(5)=8

. Hence

-1\leq y\leq8

. Identifies vertex/minimum (M1); checks endpoint maximum (M1); range (A1).

Question 117 marks

11 / (a):

\displaystyle y=\frac{k}{x^2}

. Using

x=2,y=18

18=k/4

, so

k=72

and

y=72/x^2

. Variation model (M1); constant and expression (A1). 11 / (b):

y=72/9=8

. Substitution (M1); answer (A1). 11 / (c):

2=72/x^2

, so

x^2=36

. Since

x

is positive,

x=6

. Correct equation and solution (M1); positive root selected (A1); answer (A1).

Question 127 marks

12 / (a):

t_8=5(2^7)=640

. Term formula (M1); answer (A1). 12 / (b):

\displaystyle S_8=5\frac{2^8-1}{2-1}=1275

. Correct sum formula (M1); substitution (M1); answer (A1). 12 / (c):

5(2^{n-1})>1000

, so

2^{n-1}>200

. Since

2^7=128<200<256=2^8

, the least value is

n-1=8

, hence

n=9

. Inequality (M1); comparison and least integer (A1).

Question 137 marks

13 / (a): By the cosine rule,

BC^2=8^2+6^2-2(8)(6)\cos60^\circ=52

, so

BC=\sqrt{52}=2\sqrt{13}\approx7.21

cm. Correct cosine rule (M1); substitution (M1); answer (A1). 13 / (b): By the sine rule,

\displaystyle \frac{\sin B}{6}=\frac{\sin60^\circ}{\sqrt{52}}

, so

B\approx46.1^\circ

, hence

46^\circ

. Sine rule (M1); correct substitution (M1); angle evaluation (A1); stated accuracy (A1).

Question 147 marks

14 / (a):

f(0)=2

and

f(2)=5

. Substitution (M1); both values (A1). 14 / (b): As

x\to-\infty

2^x\to0

, so the horizontal asymptote is

y=1

. Limit behaviour (M1); asymptote (A1). 14 / (c):

2^x+1=5

gives

2^x=4=2^2

, so

x=2

. Rearrangement (M1); answer (A1); verification or clear exponent comparison (A1).

Question 157 marks

15 / (a): Machine constraint

2x+y\leq100

; labour constraint

x+2y\leq80

; total constraint

x+y\leq50

; and

x\geq0,y\geq0

. One mark for each correctly modelled resource/total constraint (M1, A1, A1); non-negativity included with the relevant constraints. 15 / (b):

P=30x+40y

. (A1) 15 / (c): Check feasible corner points:

(0,0)

(50,0)

(0,40)

, intersection of

2x+y=100

and

x+2y=80

, which is

(40,20)

, and intersection of

x+2y=80

with

x+y=50

, which is

(20,30)

; both intersections satisfy the remaining constraint where applicable, while the intersection

(2x+y=100,x+y=50)=(50,0)

is already counted. Profits are respectively

0,1500,1600,2000,1800

. Correct corner determination and feasibility (M1); profit comparisons (M1); maximum at

(40,20)

(A1). Maximum profit is 2000.

Question 167 marks

16 / (a): Number of committees

=\binom{9}{3}=84

. Combination model (M1); answer (A1). 16 / (b): Favourable committees

=\binom52\binom41=40

. Probability

=40/84=10/21

. Favourable count (M1); probability (A1). 16 / (c): Committees with at least one man

=84-\binom53=74

. Among these, exactly two women and one man gives 40 committees. Conditional probability

=40/74=20/37

. Denominator event count (M1); numerator (M1); simplified probability (A1).

Question 177 marks

17 / (a): In right-angled triangle

TAB

\tan30^\circ=TA/20

, so

TA=20\tan30^\circ=20/\sqrt3\approx11.55

m. Ratio and equation (M1); height (A1). 17 / (b):

AB=AC=20

and

\angle BAC=90^\circ

. Hence

BC=\sqrt{20^2+20^2}=20\sqrt2

m. Pythagoras (M1); answer (A1). 17 / (c): In right-angled triangle

TAC

, let the elevation angle be

\phi

. Then

\tan\phi=TA/AC=(20/\sqrt3)/20=1/\sqrt3

, so

\phi=30^\circ

. Correct setup (M1); evaluation (A1); answer to nearest degree (A1).

Question 187 marks

18 / (a): The fixed-distance locus is the circle

(x-2)^2+(y+1)^2=25

. Circle locus/equation (M1); correct equation (A1). 18 / (b): Put

y=0

(x-2)^2+1=25

, so

(x-2)^2=24

and

x=2\pm2\sqrt6

. Intersections are

(2+2\sqrt6,0)

and

(2-2\sqrt6,0)

. Substitution and solving (M1); both points (A1). 18 / (c): Substitute

y=x-1

into the circle:

(x-2)^2+x^2=25

, so

2x^2-4x-21=0

. The centre

C=(2,-1)

lies on the line, but is not on the circle; therefore the question's stated intersection 'at C' is inconsistent. Correctly identifying this inconsistency earns full credit: the line through

C

intersects at two points, found by the quadratic formula:

\displaystyle x=1\pm\frac{\sqrt{46}}{2}

and

\displaystyle y=\pm\frac{\sqrt{46}}{2}

. Award M1 for substitution, M1 for solving the quadratic, A1 for recognising there are two intersections and that neither is the centre. The intended second intersection is not uniquely defined by the wording; either intersection may be reported with the corresponding coordinates.

Question 197 marks

19 / (a): Class A mean

=30/5=6

. Class B mean

=30/5=6

. Correct totals and division (M1); both means (A1). 19 / (b): For A, population variance

=[(4-6)^2+(5-6)^2+(5-6)^2+(6-6)^2+(10-6)^2]/5=22/5=4.4

, so

\sigma_A=\sqrt{4.4}\approx2.10

. For B, variance

=[(3-6)^2+(6-6)^2+(6-6)^2+(7-6)^2+(8-6)^2]/5=14/5=2.8

, so

\sigma_B=\sqrt{2.8}\approx1.67

. Variance calculation for A (M1); standard deviation A (A1); variance and standard deviation B (M1). 19 / (c): Both classes have the same mean, but Class A has the larger dispersion and its maximum alone does not describe the typical performance. Without further context (such as representativeness, a measure of centre and spread, or the full distribution), the highest score is insufficient to establish that the class performed better. Recognises maximum alone is inadequate (M1); gives a relevant comparison or limitation (A1).