Analysis on the HKCEE/HKDSE questions

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Analysis on the HKCEE/HKDSE questions

2003 2004 2005 2006 2007

1. Percentages ^{11, 12} ¹² ¹² ^{10, 11} ^{10, 11}

2. Numbers and Estimation ⁹ ¹²

3. Polynomials 2, 6, 37, 38, 39 10, 37, 38, 40 4, 10, 38, 40 3, 4, 6, 40 2, 3, 40

4. Formulas 3 2 2 2

5. Equations ^{5, 7, 41} 6, 7, 8, 41, 42 5, 7, 8 8, 9 4, 7, 42

6. Functions and Graphs ^{1, 42} ^{3, 5} ^{3, 6} ^{5, 7} ^{5, 8, 38}

7. Rate, Ratio and Variations ^{14, 15} 13, 14, 15, 16 13, 14 13, 14, 15 13, 14

8. Exponential and Logarithmic

Functions ^{4, 13, 40} ^{1, 4, 39} ^{1, 37, 39} 1, 37, 38, 39 1, 37, 39, 41

9. Geometry (1) – Straight Lines

and Rectilinear Figures 17, 18, 27, 28, 53 17, 18, 27, 28 26, 27, 28, 29, 30,

43, 52 26, 48, 49 19, 27, 28, 50

10. Geometry (2) – Circles 25, 50, 51, 52 23, 24, 25, 26, 50, 51 24, 25, 49, 50, 51 19, 46, 47 48, 49

11. Transformation and Symmetry ²⁵ ^{25, 26}

12. Coordinate Geometry (1) ³¹ ³¹ ³² ^{27, 31} ^{29, 30}

13. Coordinate Geometry (2) ^{29, 30, 54} 29, 30, 52, 53 31, 33, 53, 54 28, 29, 30, 50, 51 31, 32, 51, 52

14. Sequence ¹⁰ ^{11, 44} ^{11, 42} ^{12, 42, 43} ^{9, 44, 45}

15. Mensuration 19, 20, 21, 44 45 16, 17, 18, 19 17, 18, 20 16, 17, 18

16. Trigonometry ^{22, 45, 46} ^{20, 46, 47} 20, 44, 45, 46 21, 22, 44 20, 21, 22, 46

17. Applications of Trigonometry 16, 23, 24, 26, 48, 49 19, 21, 22, 48, 49 15, 21, 22, 23, 47, 48 16, 23, 24, 45 15, 23, 24, 47

18. Inequalities and Linear

Programming ^{8, 43} ^{9, 43} ^{9, 41} ⁴¹ ^{6, 43}

19. Permutation and Combination

20. Probability ^{34, 35} ^{33, 34} ^{35, 36} 32, 33, 52, 53 33, 53, 54

21. Statistics ^{32, 33, 36} 32, 35, 36, 54 34 34, 35, 36, 54 34, 35, 36

Note: 1. The numbers listed above refer to the question numbers in the HKCEE/HKDSE mathematics Paper II that year.

2. ‘Permutation and Combination’ is a new topic in the HKDSE syllabus.

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2008 2009 2010 2011 HKDSE 2012 HKDSE 2013 HKDSE 2014

12 10, 11 13, 14 10, 11 8 10, 11 9, 10

16, 17 15 11, 17 14 13, 14, 35 4, 36 11, 36

3, 4, 5 3, 4, 5, 41 3, 4, 5, 41 3, 40 2, 3, 4, 31 3, 8, 9, 31 2, 3, 31

2 2 1 2 2

8, 41 8 7, 8 6 5 6, 35 4, 8

6, 9, 10, 37 6, 37 6, 9, 37 7, 8, 37 6, 34, 38 7 5, 35, 38

13, 14, 15 13, 14 15, 16 12, 13 9, 10, 11 12, 13 13

1, 38, 39, 40 1, 38, 39, 40 2, 38, 39, 40 1, 38, 39, 41 1, 32, 33 1, 32, 33, 34 1, 32, 33, 34

21, 27, 28, 52 26, 27, 28, 51 25, 26, 27 23, 27, 28 17, 22 21 16, 22

50, 51 48, 49, 50 49, 50 48, 49 20, 41 19, 41 20, 21, 41

25, 26 29 23, 24 25, 26 15 23

29, 30, 32 30, 33, 52 30 29, 30 23 43

31, 53 31, 32, 53 29, 31, 32, 51, 52 31, 32, 51 24, 25, 26, 42 14, 24, 25, 42 24, 25, 26, 42

11, 43, 44 12, 42, 43 12, 43, 44 9, 44, 45 12, 37 38 14, 37

7, 18, 19, 20 7, 17, 18, 19, 20 18, 19, 20 16, 17, 18, 19 15, 16, 21 16, 17, 18 15, 17

23, 24, 45, 46, 47 21, 24, 25, 45, 46 22, 45, 46 20, 21, 22, 46 18, 19, 39 22, 23, 39 18, 19, 39

22, 48, 49 16, 22, 23, 47 21, 28, 47, 48 15, 24, 47, 50 40 20, 40 40

42 9, 44 10, 42 4, 5, 42, 43 7, 36 5, 37 6, 7, 12

43 44 43

33 34 33, 53 33, 52 27, 28, 44 26 27

34, 35, 36, 54 35, 36, 54 34, 35, 36, 54 34, 35, 36, 53, 54 29, 30, 45 27, 28, 29, 30, 45 28, 29, 30, 44, 45

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Paper II of the HKDSE mathematics examination consists of two sections: Sections A and B. There are 30 questions in Section A and 15 questions in Section B. As the time allowed is 75 minutes, students are required to answer one question in about 1.5 minutes on average. In order to spend time in the most effective way, it is essential for candidates to learn some useful strategies in answering multiple-choice questions.

Strategy 1 Direct computation

In questions such as finding interest, area of plane figures or solving equations with quadratic formulas, candidates can apply the related formulas and calculate the answers directly. However, as not all the formulas are provided in the examination paper, candidates should be familiar with some useful formulas.

Example 1

A sum of $30 000 is deposited for 1 year at 6% p.a., compounded monthly. Find the interest correct to the nearest dollar.

A. $1750 B. $1800 C. $1850 D. $1900

Solution:

Interest = $30 000 × 1

12

 +

 



6%

12 – $30 000 = $1850 (cor. to the nearest dollar) The answer is C.

Compound interest I = P(1 + r%)ⁿ – P

Example 2

In the figure, the solid consists of a hemisphere and a cylinder. If the base radius and the height of the cylinder are 6 cm and 5 cm respectively, find the volume of the solid.

A. 204π B. 324π C. 396π D. 468π

12 cm Reference: HKDSE 13 Q11

Reference: HKDSE 13 Q17

Multiple-choice question tackling strategy

6 cm

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Percentages

1

B. Profit , Loss and Discount

1. Profit percentage（盈利百分數）= Profit

Cost × 100% = Selling price Cost Cost

− × 100%

Loss percentage（虧蝕百分數）= Loss

Cost × 100% = Cost Selling price Cost

− × 100%

2. Selling price = Cost × (1 + Profit percentage) or

= Cost × (1 – Loss percentage)

For example, if Peter bought a computer for $4000 and then sold it at a loss of 20%, then

selling price = $4000(1 – 20%) = $3200

3. Discount percentage（折扣百分數）= Marked price Selling price Marked price

− × 100%

Selling price = Marked price × (1 – Discount percentage)

Key Points

A. Percentage Change

Percentage change（百分增減）= New value Original value Original value

−

New value = Original value × (1 + Percentage change)

For example, if a number is changed from 12 000 to 9000, then percentage change = 9000 12 000

12 000

− × 100% = –25%

Note: A positive percentage change means a percentage increase while a negative percentage change means a percentage decrease.

Percentages

Let’s Try Ans: 1. (a) 156 (b) 117 2. Profit percentage = 15% 3. $210

Let’s Try

¹

Let’s Try

²

Let’s Try

³

Find the new value if 130 is (a) increased by 20%;

(b) decreased by 10%.

John bought a phone at $3200 and then sold it at $3680. Find the profit or loss percentage.

A bag of marked price $280 is sold at a discount of 25%. Find the selling price.

1

Instant Drill 1

Peter’s monthly salary is changed from $10 000 to $12 000. The percentage change in his salary is

A. 5%. B. 10%. C. 15%. D. 20%.

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HKDSE Exam Series – Mathematics Multiple-choice Questions (Compulsory Part) (Latest Upgraded Edition)

28

7. (2a – b)(2a + b – 1) = A. 2a² – b² – a – b B. 2a² – b² – a + b C. 4a² – b² – 2a – b D. 4a² – b² – 2a + b

8. (3x – 1)(x² – 2x + 1) = A. 3x³ – 7x² + 5x – 1 B. 3x³ – 7x² + 5x + 1 C. 3x³ – 7x² + x – 1 D. 3x³ – 7x² + x + 1

9. (2x² + 4x – 1)(2 – x) = A. 2x³ – 9x + 2 B. –2x³ + 9x – 2 C. x³ + 9x² – 2 D. x³ – 9x² – 2

10. (3x – 2y)² – (3x + 2y)² = A. –4xy

B. –24xy C. –32xy D. –64y

11. Find the quotient when 2x³ + 7x² + 7x + 2 is divided by x + 1.

A. 2x² + 5x + 2 B. 2x² + 9x + 2 C. 2x² + 5x + 12 D. 2x² + 9x + 12

Reference:

HKCEE 11Q3

Reference:

HKCEE 05Q4

Reference:

HKDSE 12Q2

Basic Exercise

A. Polynomials and Their Operations

1. What is the constant term of 2x³ + 4x² – 5x + 7?

A. 2 B. 4 C. 5 D. 7

2. What is the coefficient of x² in x³ – 3x² + x – 2?

A. 2 B. 3 C. –2 D. –3

3. What is the degree of the polynomial 2m – 3n + 3mn² + 2m²?

A. 1 B. 2 C. 3 D. 4

4. (6x³ – 2x² + x – 3) + (1 – x – x²) = A. 6x³ – 3x² – 2

B. 6x³ – x² + 2x – 2 C. 6x³ – 3x² + 2x – 2 D. 6x³ + x² + 2x – 2 5. (x² – 1) – (2x³ – 3x² + 2) =

A. –2x³ – 2x² + 1 B. 2x³ – 2x² + 1 C. –2x³ + 4x² – 3 D. –2x³ + 4x² – 1

6. Find the coefficient of x³ in the result of (x³ + 3x² + 2x + 1) – (x³ – 2x + 1).

A. 0 B. 1 C. 2 D. 3

Reference:

HKCEE 08Q4

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96

Revision Test

Level 1

1. (− )

 

 =

3 1

3

2017

2018

A. –1 3 B. –3 C.

1 3 D. 3

2. (3a³)³ 3a⁴ = A. 3a² B. 3a⁵ C. 9a² D. 9a⁵

3. (4^{n + 1} · 8²ⁿ)² = A. 2^{8n + 1} B. 2^{8n + 2} C. 2^{16n + 4} D. 2^{16n + 6}

4. x y x y

3 2

1 2 3 2 2

5

−

=

( )

A. ( )xy

2 10

B. ( )xy

3 10

C. x⁻ y

19 10

11 10

D. x y⁻

3 2

Reference:

HKCEE 11Q1

Reference:

HKDSE 12Q1

Reference:

HKDSE 14Q1

Reference:

HKDSE 13Q1

NF

5. If 1 49

1 343

2 3



 

 =  



x y

and x, y are non-zero integers, then x : y =

A. 1 : 2 . B. 4 : 3 . C. 9 : 4.

D. 16 : 9.

6. If a > 0, then 81a− 49a− 4a= A. 0.

B. 2 a. C. 14 a . D. 28 a. 7. If a > 0, then 2

16 2 a

a a + a = A. a.

B. a. C. 2 a. D. 3 a. 8. 1110111(2) =

A. 2⁵ + 2⁴ + 2³ + 2² + 2 + 1 B. 2⁵ + 2⁴ + 2³ + 2³ + 2² + 2 C. 2⁶ + 2⁵ + 2⁴ + 2² + 2 + 1 D. 2⁶ + 2⁵ + 2⁴ + 2³ + 2² + 2 9. 11(16) × A3(16) =

A. B4₍₁₆₎ B. F5₍₁₆₎ C. 62B(16)

D. AD3(16)

10. Convert 2010(10) into a hexadecimal number.

A. 5F8₍₁₆₎ B. 7DA₍₁₆₎ C. D50(16)

D. 10B6(16)

NF

Reference:

HKCEE 07Q37

NF

Reference:

HKCEE 11Q41

NF

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Not to be taken away before the end of the examination session Pan Lloyds Publishers Ltd

MATHEMATICS Compulsory Part MOCK EXAM 1

PAPER 2

(1¼ hours)

INSTRUCTIONS

1. Read carefully the instructions on the Answer Sheet. After the announcement of the start of the examination, you should first stick a barcode label and insert the information required in the spaces provided. No extra time will be given for sticking on the barcode label after the ‘Time is up’

announcement.

2. When told to open this book, you should check that all the questions are there. Look for the words ‘END OF PAPER’ after the last question.

3. All questions carry equal marks.

4. ANSWER ALL QUESTIONS. You are advised to use an HB pencil to mark all the answers on the Answer Sheet, so that wrong marks can be completely erased with a clean rubber. You must mark the answers clearly; otherwise you will lose marks if the answers cannot be captured.

5. You should mark only ONE answer for each question. If you mark more than one answer, you will receive NO MARKS for that question.

6. No marks will be deducted for wrong answers.

MC

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290

5. Let f(x) = 2x³ – 3x² – x + 1. When f(x) is divided by x + 1, the remainder is

A. –6.

B. –3.

C. 4.

D. 7.

6. Solve (x + 2a)² = 9a², where a is a constant.

A. x = a

B. x = –3a or x = 3a C. x = –5a or x = a D. x = –5a or x = 3a

7. Find the range of values of k such that the quadratic equation 2x² + x + 2k = k – x has real roots.

A. k≤1 4 B. k≤1 2 C. k≥1 4 D. k≥1 2

8. In the figure, the graph of y = x² + kx – 3 passes through the vertex (a, –4). Find the equation of the axis of symmetry of the graph.

A. x = –2 B. x = –1 C. x = 1 D. x = 2

x y

0 3

(a, –4)

Section A 1. (16 ⋅ 8^{n – 1})² =

A. 2^{6n – 1} B. 2^{6n + 2} C. 2^{6n + 3} D. 2^{7n + 1} 2. If 2

3 4 x y

− = , then y = A. 2x – 3.

B. 2 – 3x.

C. 3(x + 4).

D. 6(x – 2).

3. If x + 3y + 2 = y – x = –3, then y = A. 2.

B. 1.

C. 0.

D. –2.

4. ru + 2su – rv – 2sv + rt + 2st = A. (r + 2s)(u – v + t)

B. (r – 2s)(u – v + t) C. (r + 2s)(u + v + t) D. (r – 2s)(u – v – t)

There are 30 questions in Section A and 15 questions in Section B.

The diagrams in this paper are not necessarily drawn to scale.

Choose the best answer for each question.

y = x² + kx – 3

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1 Solution Guide

3. A

Percentage increase = (110 0) cm 00 cm

3 3

− 10

1 × 100%

= 10%

4. B

Percentage change = (40 50) kg 50 kg

− × 100%

= –20%

5. A 6. B

Let $x be the sales amount last month.

x(1 + 10%) = 60 500 1.1x = 60 500 x = 55 000

The sales amount last month was $55 000.

7. A

Let l and w be the original length and width of the rectangle respectively.

Original area = lw

New area = (1 + 10%)l × (1 – 15%)w = 0.935lw

Percentage change = 0 935. lw lw lw

− × 100%

= –6.5%

1 Percentages

Instant Drill

1. D

Percentage change = $ 12 000 10 000

$10 000

( − )× 100%

= 20%

2. A

Selling price = $200 × (1 – 10%) = $200 × 0.9 = $180 3. D

Amount = $500 × (1 + 4%)² = $500 × 1.04² = $540.8

Compound interest = $540.8 – $500

= $40.8

4. A

Value = $2 000 000 × (1 + 3%)⁴ = $2 000 000 × 1.03⁴

= $2 250 000 (cor. to 3 sig. fig.) 5. C

Annual rent of the flat = $8000 × 12

= $96 000 Property tax = $96 000 × 0.8 × 15%

= $11 520

Basic Exercise

1. D

Number of candies in bag B

= 500 × (1 + 20%)

= 600

Let x be the number of candies in bag C.

x × (1 + 25%) = 600 x = 600 ÷ 1.25

= 480

The number of candies in bag C is 480.

2. D

Percentage of candidates who passed in both tests

= 70% × 80%

= 56%

Common mistakes

Remember that the new value is 40 kg instead of 50 kg.

8. B

Let r be the original radius of the circle.

New radius of the circle = (1 – 20%)r

= 0.8r

Percentage change

= (0.8 )r r r

2 2

2

− π π

π × 100%

= −0 36 ²

2

. r

r π

π × 100%

= – 36%

Guidelines

Express the new value in terms of the original value. Then we can find the percentage change with the formula.

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17 Solution Guide

14. C

For A: (x + y)³= x³ + 3x²y + 3xy²+ y³

≠ x³ + y³ For B:

(4x + 1)(4x²– 4x + 1) = 16x³ – 16x²+ 4x + 4x² – 4x+ 1

= 16x³ – 12x² + 1

≠ 4x³ + 1 For C: 27x³– y³=(3x)³ – y³

=(3x – y)(9x²+ 3xy+ y²) For D: 8x³– 1 = (2x)³– 1³

= (2x – 1)(4x²+ 2x+ 1)

≠ (2x – 1)(4x² + 4x + 1)

13. B

For A: (x + 3)²= x² + 6x + 9

≠ x² + 9 For B: 4x(x + 5)= 4x² + 20x

For C: –(x + 1)(3x – 2) = –3x²– x+ 2

≠ 3x² + x – 2 For D: x²– 4 = (x – 2)(x + 2)

≠ (x – 4)(x + 4) Guidelines

When performing long division of polynomials, add ‘0’s to the missing terms of the dividend can avoid mistakes.

Guidelines

The difference of squares and the sum and difference of cubes are very common in public examination. Candidates should familiarize with these identities.

9. B

(2x² + 4x – 1)(2 – x)

= 4x² + 8x – 2 – 2x³ – 4x²+ x

= –2x³ + 9x – 2 10. B

(3x – 2y)² – (3x + 2y)²

= [(3x – 2y) – (3x + 2y)][(3x – 2y) + (3x + 2y)]

= (–4y)(6x)

= –24xy 11. A

2 5 2

2 2

5 5

7 5

2 2

2 7 7 2 1

2

2 2 2 3

2

3 + +

+ + + + +

+ + +

+ x x

x x x x

12. C

15. A

(x – 3)²= x² – 6x + 9 ≠ x² – 9 16. D

L.H.S. = x² – 4ax + 2 R.H.S.= (x – 2a)²+ 2b = x² – 4ax + 4a² + 2b

By comparing the like terms, we have 4a² + 2b = 2

b = 1 – 2a² 17. C

L.H.S. = x² + 2a² – b² R.H.S.= 2(x – a)²– (x + b)²

= 2x² – 4ax + 2a² – x² – 2bx – b² = x² – (4a + 2b)x + 2a² – b² By comparing the like terms, we have 4a + 2b = 0

b = – 2a 18. A

L.H.S. = (ax + b)(x – 1) = ax² – (a – b)x – b R.H.S. = 3x² – x – 2

By comparing the like terms, we have a = 3, b = 2.

19. C

L.H.S. = Ax³ + Bx² – 3x – 2 R.H.S. = (2x² + 1)(–3x – 2) = –6x³– 4x² – 3x – 2

By comparing the like terms, we have A = –6, B = –4.