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HKDSE Maths Compulsory Paper 2 Mock Paper

This 45-question multiple-choice sample allows 1 hour 15 minutes. Section A carries 30 marks and Section B carries 15 marks. It covers algebra, geometry, trigonometry, probability and data handling. The model answers identify the correct choices and explain the reasoning.

45 questions45 marks1 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 A30 questions · 30 marks
Section B15 questions · 15 marks

Topics in this sample

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

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Hong Kong Diploma of Secondary Education Examination Mock Paper

Mathematics (Compulsory Part)

Paper 2: Multiple Choice

Time Allowed: 1 hr 15 minTotal Marks: 45

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

A

Question 1

Solve

x^2-5x+6=0

.(1 mark)

x=1

x=6

x=2

x=3

x=-2

x=-3

x=2

only

Question 2

The function

f

is defined by

f(x)=2x-3

. Find

f(4)

.(1 mark)

-5

Question 3

Solve

3x+7<16

.(1 mark)

x<3

x>3

\displaystyle x<\frac{23}{3}

x\leq 3

Page2

Question 4

Simplify

\displaystyle \dfrac{2^3\times 2^4}{2^2}

.(1 mark)

2^5

2^9

2^3

2^2

Question 5

The first three terms of an arithmetic sequence are

7, 11, 15

. Find its 10th term.(1 mark)

Question 6

A quantity

y

varies directly as

x

. When

x=4

y=18

. Find

y

when

x=10

.(1 mark)

7.2

Page3

Question 7

The points

A(1,2)

and

B(5,10)

lie on a straight line. Find its gradient.(1 mark)

\displaystyle \frac{1}{2}

-2

Question 8

A circle has centre

O

and radius 5 cm. A chord is 8 cm long. What is the perpendicular distance from

O

to the chord?(1 mark)

2 cm

3 cm

4 cm

\sqrt{41}

Question 9

In a right-angled triangle, the acute angle

\theta

has opposite side 6 and hypotenuse 10. Find

\sin\theta

.(1 mark)

\displaystyle \frac{3}{5}

\displaystyle \frac{4}{5}

\displaystyle \frac{5}{3}

\displaystyle \frac{3}{4}

Page4

Question 10

How many different arrangements can be made using all the letters of the word MATH?(1 mark)

Question 11

A fair six-sided die is rolled once. What is the probability of obtaining an even number?(1 mark)

\displaystyle \frac{1}{6}

\displaystyle \frac{1}{3}

\displaystyle \frac{1}{2}

\displaystyle \frac{2}{3}

Question 12

The values

2,4,4,6

have what population standard deviation?(1 mark)

\sqrt{2}

\sqrt{8}

Page5

Question 13

Which is the equation of a line with gradient

-3

and

y

-intercept 2?(1 mark)

y=3x+2

y=-3x+2

y=-3x-2

y=2x-3

Question 14

f(x)=x^2+1

, which expression is

f(x-2)

?(1 mark)

x^2-3

(x-2)^2+1

(x+2)^2+1

x^2-2+1

Question 15

A fair coin is tossed twice. What is the probability of getting exactly one head?(1 mark)

\displaystyle \frac{1}{4}

\displaystyle \frac{1}{2}

\displaystyle \frac{3}{4}

Page6

Question 16

The data are

3,5,7,9

. What is the range?(1 mark)

Question 17

Solve

\log_{10}x=2

.(1 mark)

x=20

x=100

x=10

x=0.01

Question 18

Which is a factorisation of

x^2-9

?(1 mark)

(x-3)^2

(x+3)^2

(x-3)(x+3)

(x-9)(x+1)

Page7

Question 19

The line

y=2x+1

is translated 3 units upwards. What is the equation of its image?(1 mark)

y=2x+4

y=2x-2

y=5x+1

y=2(x+3)+1

Question 20

The two roots of

x^2-7x+10=0

are

\alpha

and

\beta

. Find

\alpha+\beta

.(1 mark)

-10

-7

Question 21

y

varies inversely as

x

, and

y=6

when

x=2

, find

y

when

x=3

.(1 mark)

Page8

Question 22

A sector has radius 6 cm and central angle

60^\circ

. Find its arc length.(1 mark)

2\pi

6\pi

\pi

12\pi

Question 23

Find the 5th term of the geometric sequence

3,6,12,\ldots

.(1 mark)

Question 24

Solve

2(x-1)=3x+4

.(1 mark)

x=-6

x=2

x=-2

x=6

Page9

Question 25

The circle has equation

(x-2)^2+(y+1)^2=16

. What are its centre and radius?(1 mark)

Centre

(2,-1)

, radius 4

Centre

(-2,1)

, radius 4

Centre

(2,-1)

, radius 16

Centre

(2,1)

, radius 16

Question 26

In the diagram,

AB

is a diameter of the circle and

C

lies on the circumference. Find

\angle ACB

.(1 mark)

45^\circ

90^\circ

180^\circ

60^\circ

Page10

Question 27

Which value is the median of

1, 3, 4, 8, 10

?(1 mark)

5.2

Question 28

\tan\theta=1

and

0^\circ<\theta<90^\circ

, find

\theta

.(1 mark)

30^\circ

45^\circ

60^\circ

90^\circ

Question 29

Which point lies on the graph

y=x^2

?(1 mark)

(2,4)

(2,2)

(4,2)

(-2,-4)

Page11

Question 30

A linear programming feasible region is bounded by

x\geq0

y\geq0

x+y\leq6

and

x\leq4

. What is the maximum value of

P=2x+y

?(1 mark)

Page12

B

Question 31

The roots of

x^2-4x+8=0

are

\alpha

and

\beta

. Find

\alpha^2+\beta^2

.(1 mark)

-8

Question 32

Solve

\log_2(x-1)+\log_2(x-3)=3

.(1 mark)

x=5

x=-1

x=4

x=3

Question 33

A polynomial

p(x)

leaves remainder 5 when divided by

x-2

, and remainder

-1

when divided by

x+1

. What is the remainder when

p(x)

is divided by

(x-2)(x+1)

?(1 mark)

2x+1

-2x+1

2x-1

x+3

Page13

Question 34

Solve

\displaystyle \dfrac{1}{x-1}+\dfrac{1}{x+1}=1

.(1 mark)

x=1\pm\sqrt{2}

x=\pm\sqrt{2}

x=2

only

x=0

Question 35

The sum of the first 20 terms of an arithmetic sequence is 610. Its first term is 2. Find its common difference.(1 mark)

\displaystyle \frac{61}{10}

\displaystyle \frac{59}{19}

Question 36

A positive quantity

y

varies directly as

x^2

and inversely as

\sqrt z

. When

x=2

and

z=9

y=8

. Find

y

when

x=3

and

z=4

.(1 mark)

\displaystyle \frac{27}{2}

Page14

Question 37

A circular sector has radius 10 cm and arc length 6 cm. Find its area.(1 mark)

30 cm

^2

60 cm

^2

16 cm

^2

3 cm

^2

Question 38

A vertical tower

AB

stands on horizontal ground. From point

C

, 20 m from its foot

A

, the angle of elevation of its top

B

35^\circ

. Find the height of the tower, to the nearest metre.(1 mark)

11 m

14 m

29 m

16 m

Page15

Question 39

From 7 distinct people, a committee of 3 is selected. Two particular people, P and Q, must not both be on the committee. How many committees are possible?(1 mark)

Question 40

A bag contains 4 red and 3 blue counters. Two counters are drawn at random without replacement. Find the probability that they are of different colours.(1 mark)

\displaystyle \frac{4}{7}

\displaystyle \frac{3}{7}

\displaystyle \frac{1}{2}

\displaystyle \frac{2}{7}

Question 41

The scores of a group have mean 50 and population standard deviation 8. Each score

x

is transformed to

y=3x-10

. What are the mean and population standard deviation of the transformed scores?(1 mark)

Mean 140, standard deviation 24

Mean 140, standard deviation 14

Mean 40, standard deviation 24

Mean 150, standard deviation 8

Page16

Question 42

For

x\ne0

, the graph of

\displaystyle y=\dfrac{2}{x-1}+3

has which vertical and horizontal asymptotes?(1 mark)

x=1

and

y=3

x=-1

and

y=3

x=1

and

y=2

x=3

and

y=1

Question 43

A circle has centre

(1,2)

and passes through

(4,6)

. Which is its equation?(1 mark)

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

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

(x-1)^2+(y-2)^2=5

(x-4)^2+(y-6)^2=25

Page17

Question 44

The locus of a point

P

satisfying

PA=PB

, where

A(-2,0)

and

B(4,0)

, is which line?(1 mark)

x=1

y=0

x=2

y=1

Question 45

A rectangular garden has perimeter 40 m. Its length is

x

m and width is

(20-x)

m. Which value of

x

gives the maximum area?(1 mark)

Page18

Model answers

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

Question 11 mark

Correct option: B

The factorisation is

x^2-5x+6=(x-2)(x-3)

. Hence

x=2

x=3

(A1). Option B is correct. A and C do not satisfy the equation; D omits one root.

Question 21 mark

Correct option: A

Substitute

x=4

f(4)=2(4)-3=5

(A1). Thus A. B omits the subtraction, C adds 3, and D has an incorrect sign.

Question 31 mark

Correct option: A

Subtract 7 to obtain

3x<9

; dividing by positive 3 gives

x<3

(A1). Thus A. B reverses the direction, C fails to subtract 7 correctly, and D incorrectly includes equality.

Question 41 mark

Correct option: A

Using index laws, add exponents when multiplying and subtract when dividing:

2^{3+4-2}=2^5

(A1). A is correct. B adds rather than subtracting the denominator exponent; C and D use incorrect exponent arithmetic.

Question 51 mark

Correct option: B

The common difference is 4. The 10th term is

7+9(4)=43

(A1). B is correct. A uses only eight increments, C uses ten increments, and D uses seven increments.

Question 61 mark

Correct option: B

Since

y=kx

k=18/4=4.5

. At

x=10

y=4.5(10)=45

(A1). B is correct. A applies inverse scaling, while C and D use incorrect scale factors.

Question 71 mark

Correct option: A

The gradient is

(10-2)/(5-1)=8/4=2

(A1). Therefore A. B inverts the ratio, C uses only the vertical change, and D gives the wrong sign.

Question 81 mark

Correct option: B

The perpendicular from the centre bisects the chord, giving a right triangle with hypotenuse 5 and one leg 4. Thus the distance is

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

cm (A1). B is correct. A confuses a radius difference, C uses the half-chord, and D adds rather than subtracts squares.

Question 91 mark

Correct option: A

By definition,

\sin\theta=\text{opposite}/\text{hypotenuse}=6/10=3/5

(A1). A is correct. B is the adjacent-to-hypotenuse ratio, C is greater than 1, and D uses the wrong denominator.

Question 101 mark

Correct option: C

The four letters are distinct, so the number of arrangements is

4!=24

(A1). C is correct. A omits a factor, B squares 4, and D counts only the choice of a first letter.

Question 111 mark

Correct option: C

The even outcomes are 2, 4 and 6, so the probability is

3/6=1/2

(A1). C is correct. A counts only one outcome; B and D use an incorrect number of favourable outcomes.

Question 121 mark

Correct option: A

The mean is 4. The population variance is

[(2-4)^2+0^2+0^2+(6-4)^2]/4=8/4=2

, so the standard deviation is

\sqrt{2}

(A1). A is correct. B reports the range, C omits division by the population size, and D is not the square root of the variance.

Question 131 mark

Correct option: B

In slope-intercept form

y=mx+c

m=-3

and

c=2

, hence

y=-3x+2

(A1). B is correct. A has the wrong gradient, C the wrong intercept, and D interchanges the two constants.

Question 141 mark

Correct option: B

Replace every input

x

x-2

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

(A1). B is correct. A and D incorrectly expand or substitute only part of the expression; C shifts in the opposite direction.

Question 151 mark

Correct option: B

The four equally likely outcomes are HH, HT, TH and TT. Exactly one head occurs in HT and TH, so the probability is

2/4=1/2

(A1). B is correct. The other options miscount favourable outcomes or assert certainty.

Question 161 mark

Correct option: B

Range = maximum minus minimum

=9-3=6

(A1). B is correct. A is half the range, C is the maximum, and D is the sum of the extremes.

Question 171 mark

Correct option: B

By the definition of logarithm,

x=10^2=100

(A1). B is correct. A multiplies instead of exponentiating, C uses exponent 1, and D uses exponent

-2

Question 181 mark

Correct option: C

Use the difference of two squares:

x^2-9=x^2-3^2=(x-3)(x+3)

(A1). C is correct. A and B expand to trinomials with middle terms; D expands to a different polynomial.

Question 191 mark

Correct option: A

A vertical translation upwards by 3 adds 3 to every

y

-value:

y=2x+1+3=2x+4

(A1). A is correct. B shifts down, C changes the gradient, and D represents a horizontal shift.

Question 201 mark

Correct option: C

For

ax^2+bx+c=0

, the sum of roots is

-b/a

. Here it is

-(-7)/1=7

(A1). C is correct. A and D confuse product and sum; B uses the coefficient with the wrong sign.

Question 211 mark

Correct option: A

For inverse variation,

xy=k

. From the given values

k=12

, so when

x=3

y=12/3=4

(A1). A is correct. B applies direct variation, while C and D do not preserve the product.

Question 221 mark

Correct option: A

Arc length

=(60/360)(2\pi\times6)=2\pi

cm (A1). A is correct. B is a semicircle length, C is half the correct arc length, and D is the full circumference.

Question 231 mark

Correct option: C

The common ratio is 2. The terms are

3,6,12,24,48

, so the 5th term is 48 (A1). C is correct. A is the 4th term, B does not follow the ratio, and D applies an extra doubling.

Question 241 mark

Correct option: A

Expand and collect:

2x-2=3x+4

, hence

-6=x

(A1). A is correct. The other options result from sign or transposition errors.

Question 251 mark

Correct option: A

Compare with

(x-h)^2+(y-k)^2=r^2

. The centre is

(2,-1)

and

r=\sqrt{16}=4

(A1). A is correct. B reverses both centre signs, C mistakes

r^2

for

r

, and D has the wrong

y

-coordinate and radius.

Question 261 mark

Correct option: B

The angle subtended by a diameter at the circumference is a right angle, so

\angle ACB=90^\circ

(A1). B is correct. The other choices contradict the angle-in-a-semicircle theorem.

Question 271 mark

Correct option: B

The ordered data set has five values; its middle (third) value is 4 (A1). B is correct. A and D are adjacent order statistics, while C is the mean rather than the median.

Question 281 mark

Correct option: B

In the acute range,

\tan45^\circ=1

, hence

\theta=45^\circ

(A1). B is correct. The tangent values at 30° and 60° differ from 1, and tangent is undefined at 90°.

Question 291 mark

Correct option: A

For

x=2

y=x^2=4

, so

(2,4)

lies on the graph (A1). A is correct. B and C do not satisfy

y=x^2

, and the square of

-2

is positive, not

-4

Question 301 mark

Correct option: B

The vertices are

(0,0),(4,0),(4,2),(0,6)

. The corresponding

P

values are

0,8,10,6

. Thus the maximum is 10 at

(4,2)

(A1). B is correct; the other options are values at other vertices or an incorrect evaluation.

Question 311 mark

Correct option: D

By Vieta's relations,

\alpha+\beta=4

and

\alpha\beta=8

. Hence

\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=16-16=0

(A1). D is correct. A gives the product, B omits the product correction, and C incorrectly negates the final result.

Question 321 mark

Correct option: A

The domain requires

x>3

. Combine logs:

(x-1)(x-3)=2^3=8

, so

x^2-4x-5=0

, giving

x=5

-1

. Only

x=5

satisfies the domain (A1). A is correct. B is extraneous; C and D do not satisfy the original equation.

Question 331 mark

Correct option: A

The remainder is linear, say

ax+b

. It satisfies

2a+b=5

and

-a+b=-1

. Subtracting gives

3a=6

, so

a=2

b=1

. The remainder is

2x+1

(A1). A is correct; the other options fail at least one of the two required values.

Question 341 mark

Correct option: A

For

x\ne\pm1

, combine fractions:

2x/(x^2-1)=1

, giving

x^2-2x-1=0

. Thus

x=1\pm\sqrt2

, neither excluded (A1). A is correct. B solves a different quadratic, and C and D do not satisfy the equation.

Question 351 mark

Correct option: B

Use

\displaystyle S_{20}=\frac{20}{2}[2(2)+19d]

. Thus

610=10(4+19d)

, so

19d=57

and

d=3

(A1). B is correct. A confuses first term with difference; C and D result from incorrect division or term counts.

Question 361 mark

Correct option: B

Write

y=kx^2/\sqrt z

. From the initial values,

8=k(4)/3

, so

k=6

. At

(x,z)=(3,4)

y=6(9)/2=27

(A1). B is correct. A omits the square scaling, C reverses or misapplies the square-root factor, and D uses an incorrect ratio.

Question 371 mark

Correct option: A

For a sector,

A=\frac12 r\ell

when radius is

r

and arc length is

\ell

. Hence

A=\frac12(10)(6)=30

^2

(A1). A is correct. B omits the factor one-half; C and D use unrelated combinations of the given values.

Question 381 mark

Correct option: B

Use

\tan35^\circ=h/20

, so

h=20\tan35^\circ\approx14.0

m (A1). B is correct. A uses sine, C uses the reciprocal tangent ratio, and D is an inaccurate rounding or estimate.

Question 391 mark

Correct option: B

There are

\binom73=35

committees in total. Those containing both P and Q are formed by choosing one of the other five, giving 5. Therefore

35-5=30

(A1). B is correct. A subtracts the wrong number; C ignores the restriction, and D undercounts valid committees.

Question 401 mark

Correct option: A

The probability is

P(RB)+P(BR)=(4/7)(3/6)+(3/7)(4/6)=24/42=4/7

(A1). A is correct. B counts only one order; C assumes an incorrect equally likely split; D fails to account for both orders.

Question 411 mark

Correct option: A

The mean transforms as

3(50)-10=140

. Subtracting a constant does not change dispersion, and multiplication by 3 multiplies the standard deviation by 3, giving 24 (A1). A is correct. B incorrectly subtracts 10 from the standard deviation; C transforms the mean wrongly; D ignores the scaling.

Question 421 mark

Correct option: A

The denominator is zero at

x=1

, giving the vertical asymptote; as

x

becomes large, the fraction tends to 0, giving horizontal asymptote

y=3

(A1). A is correct. The other options confuse numerator, denominator shift, or vertical translation.

Question 431 mark

Correct option: A

The squared radius is

(4-1)^2+(6-2)^2=9+16=25

. Hence

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

(A1). A is correct. B has the wrong centre, C uses the radius rather than its square, and D puts the centre at the point on the circumference.

Question 441 mark

Correct option: A

The locus of points equidistant from A and B is the perpendicular bisector of

AB

. The segment has midpoint

(({-2+4})/2,0)=(1,0)

and is horizontal, so its perpendicular bisector is

x=1

(A1). A is correct; the other lines are not the perpendicular bisector.

Question 451 mark

Correct option: B

The area is

A=x(20-x)=-x^2+20x=-(x-10)^2+100

, which is maximized when

x=10

(A1). B is correct. The remaining options do not give the vertex;

x=20

gives zero area and the choices 5 and 15 give smaller equal areas.