Content
Analysis on the HKCEE/HKDSE questions
1. Percentages 2
2. Numbers and Estimation 13
3. Polynomials 22
4. Formulas 32
5. Equations 39
6. Functions and Graphs 51
7. Rate, Ratio and Variations 65
8. Exponential and Logarithmic Functions 75
9. Geometry (1) – Straight Lines and Rectilinear Figures 85
10. Geometry (2) – Circles 100
11. Transformation and Symmetry 114
12. Coordinate Geometry (1) 125
13. Coordinate Geometry (2) 137
14. Sequence 150
15. Mensuration 162
16. Trigonometry 174
17. Applications of Trigonometry 185
18. Inequalities and Linear Programming 196
19. Permutation and Combination 210
20. Probability 220
21. Statistics 235
Sample Sample
Analysis on the HKCEE/HKDSE Paper 1 Questions
2003 2004 2005 2006 2007
1. Percentages 5 3 6, 16(a) 6 6
2. Numbers and Estimation 11 10
3. Polynomials 3 6 3, 10(b) 3 3, 14(a)
4. Formulas 1 2 1 1
5. Equations 6 7 15(b) 5, 7
6. Functions and Graphs 4 10
7. Rate, Ratio and Variations 10(a) 10(a) 5, 10(a) 15(a) 14(b)
8. Exponential and Logarithmic
Functions 4 1 2, 16(b) 1 2
9. Geometry (1) – Straight Lines
and Rectilinear Figures 8 12 8 5 8
10. Geometry (2) – Circles 17(a) 16(a)(b) 17(a) 16(a) 17(a)
11. Transformation and Symmetry
12. Coordinate Geometry (1) 7
13. Coordinate Geometry (2) 12, 17(b) 13, 16(c) 13, 17(b) 12, 16(b) 13, 17(b)
14. Sequence 7, 15 15 7
15. Mensuration 13 9, 14(a)(c) 9, 12 4, 13 9, 11
16. Trigonometry
17. Applications of Trigonometry 9, 14 5, 17 14 17 16
18. Inequalities and Linear
Programming 2, 10(b) 10(b) 4 2
19. Permutation and Combination
20. Probability 16 8 11, 15(c) 8, 14(b) 12(c), 15
21. Statistics 11 11 15(a)(b) 9, 14(a) 4, 12(a)(b)(d)
Note: 1. The numbers listed above refer to the question numbers in the HKCEE Mathematics/HKDSE Mathematics (Compulsory Part) Paper 1 that year.
2. ‘Permutation and Combination’ is a new topic in the HKDSE curriculum.
Sample
Analysis on the HKCEE/HKDSE Paper 1 Questions
2008 2009 2010 2011 HKDSE 2012 HKDSE 2013 HKDSE 2014
9, 16(b) 7 7 7 4 6
8 4 8 4 8 3
3 3 3 3 3, 12 2, 7
1 5 1 2 2 5
3, 6 6 6 5 4
11 12, 15 16 11, 12(b) 13 17 13(b)
10 6 11 11 10, 13(a)
1, 16(c) 2 1 2 1, 19 1 1, 15
9 11 9 9, 12(a) 7 9
17(a) 8
8, 9 8 6 8
12, 17(b) 12 16 14, 17 14 12
16(a) 4, 17 15 19 19 16
13 13 13 13 9, 12 13 14
4, 15 17 15 17 18 18 17
2 16 2 6 5 18
5, 14(a) 5, 14 14 14 16 10(b)(ii), 16 19
10, 14(b) 10 11 5, 10 7, 10, 15 9, 10(a)(b)(i), 15 4, 11
Sample Sample
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 × (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
1Let’s Try
2Let’s Try
3Find 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%.
2
Sample
1
Percentages5
Exercise
Conventional Questions Section A(1)
1. Michael’s monthly salary is $18 000, which is 20% higher than that of Sally and 10% less than that of Anne. What percentage is the monthly salary of Sally less than that of Anne? (3 marks)
2. There are 44 students in a group. The students are assigned to sit in two rooms A and B, such that the number of students in room A is 20% more than that in room B.
(a) Find the number of students in each room.
(b) Ken claims that if 4 students from room A are assigned to room B, the number of students in the two rooms will be the same. Do you agree? Explain your answer.
Reference:HKDSE 12Q4 (5 marks) 3. There are 240 children in a swimming pool.
(a) If the number of adults is 60% more than that of children, find the total number of swimmers in the swimming pool.
(b) It is known that 420 swimmers are female. If 70% of children are female, what percentage of adults are female?
Reference:HKCEE 08Q8 (4 marks) 4. The marked price of a statue was $500 last year. This year, the marked price is increased by 25%.
(a) Find the marked price of the statue this year.
(b) If there is a discount of 25% on the price of the statue, find the selling price.
Reference:HKCEE 01Q8 (3 marks) 5. Wendy’s salary had been reduced by 5% last year and then increased by 5% this year. Wendy claims that her
salary this year is the same as the salary before any change was made. Is Wendy correct? Explain your answer.
Reference:HKCEE 06Q6 (3 marks) 6. A spherical balloon has a radius of 4 cm. Its radius is now increased by 25%.
(a) Find, in terms of π, the original surface area and the new surface area of the balloon.
(b) Find the percentage change in the surface area of the balloon.
Reference:HKCEE 02Q6 (4 marks) 7. In the figure, the height and the base of a triangle is h cm and b cm respectively.
If the base is increased by 60% while the area remains unchanged, what is the
percentage change in the height? (4 marks)
b cm h cm
Sample Sample
9
1
PercentagesMultiple-choice Questions Level 1
For questions 36 – 37 and 50 – 51, refer to the table below.
Tax rate Rates 5% p.a.
Property Tax 15% p.a.
Profits Tax Corporations – 16.5%
Unincorporated businesses – 15%
Salaries Tax
On the first $40 000 – 2%
On the next $40 000 – 7%
On the next $40 000 – 12%
Remainder – 17%
22. If A is 10% greater than B and B is 10% less than C, then A is
A. 1% less than C.
B. 10% less than C.
C. 1% greater than C.
D. 10% greater than C.
23. Mr. Chan bought 20 eggs at $2.5 each. If they are sold at $60, what is the profit percentage?
A. 10%
B. 20%
C. 30%
D. 40%
24. The cost and the marked price of a bicycle are $600 and $800 respectively. If the bicycle is sold at a discount of $10%, find the profit.
A. $120 B. $140 C. $160 D. $180
25. If the length and the width of a rectangle are both decreased by x% so that the area is decreased by 19%, then x =
A. 5.
B. 10.
C. 15.
D. 20.
26. The height of a triangle is decreased by 20% while the area remains unchanged. Find the percentage change in the base.
A. Increased by 20%
B. Decreased by 20%
C. Increased by 25%
D. Decreased by 25%
27. A number is first increased by 4% and then decreased by x%. If the final value is 70% of the original value, find x.
A. 32 913 B. 41 5
12 C. 54 3 11 D. 64 1
4
28. The marked prices of computers A and B are $3000 and $4000 respectively. Computer A is sold with 10% discount and computer B is sold with r%
discount. If the overall selling price is $5700, then r =
A. 10.
B. 15.
C. 20.
D. 25.
Reference:
HKDSE 14Q10
Reference:
HKCEE 11Q11
Sample Sample
Sample
15
3
Polynomials3 Polynomials
Instant Drill
1. A
(x + 2)(2x2 – 3x + 1)
= 2x3 – 3x2 + x + 4x2 – 6x + 2
= 2x3 + x2 – 5x + 2 2. C
Ax2 + Bx + 1 (x + 1)2 x2 + 2x + 1
` A = 1, B = 2 Guidelines
Expand the expression and then compare the coeffcients of the like terms.
3. D
4x2 – y2 = (2x)2 – y2 = (2x – y)(2x + y) 4. C
Let f(x) = 2x3 – x2 + x – 2.
Remainder = f(1)
= 2(1)3 – (1)2 + (1) – 2
= 2 – 1 + 1 – 2
= 0 5. C
2xyz = (2x)(yz) 4x2 = (2x)2
` G.C.D. = 2x 6. A
2 12 1 1 x+ + x
–
= 2
2 1 1 1 1
1 2 1 2 1
x x
x x x
x
+ × + × +
+ –– –
= 2 2 2 1 2 1 1
x x
x x
( − )+( + )
( + )( − )
= 4 1 2 1x 1
x x
− ( + ) −( )
Guidelines
Find the L.C.M. of the denominators first.
Conventional Questions
Section A(1)
1. (a) (2x – 3y + z) – (x + 2y – z) + (y + 2 – z)
= 2x – 3y + z – x – 2y + z + y + 2 – z
= x – 4y + z + 2 [1A]
(b) (x – y)(x + y) – (2x + 3y)(x – 2y)
= (x2 – y2) – (2x2 – 4xy + 3xy – 6y2) [1M]
= x2 – y2 – 2x2 + xy + 6y2
= –x2 + 5y2 + xy [1A]
2. Ax(x – 2) + B(x + 1) 2x2 + x + 5 Ax2 – 2Ax + Bx + B 2x2 + x + 5
Ax2 + (B – 2A)x + B 2x2 + x + 5 [1M]
By comparing the coefficients, we have
A = 2 [1A]
B = 5 [1A]
3. (a) x2 + xy – 2x – 2y = x(x + y) – 2(x + y)
= (x – 2)(x + y) [1A]
(b) 2y3 – 18y(x – z)2
= 2y[y2 – 9(x – z)2] [1M]
= 2y[y + 3(x – z)][y – 3(x – z)]
= 2y(y + 3x – 3z)(y – 3x + 3z) [1A]
4. (a) x2 + 4xy + 4y2 = x2 + 2(x)(2y) + (2y)2
= (x + 2y)2 [1A]
(b) x2 + 4xy + 4y2 – 9z2
= (x + 2y)2 – (3z)2 [1M]
= (x + 2y – 3z)(x + 2y + 3z) [1A]
Guidelines
Try to use the result in (a) to factorize the expression.
5. (a) x2 + 5xy – 6y2
= (x + 6y)(x – y) [1A]
(b) x2 + 5xy – 6y2 + 5x + 30y
= (x + 6y)(x – y) + 5(x + 6y) (By (a)) [1M]
= (x + 6y)(x – y + 5) [1A]
6. (a) 9x2 – 16y2
= (3x – 4y)(3x + 4y) [1A]
(b) 9x2 – 16y2 + 6x – 8y
= (3x – 4y)(3x + 4y) + 2(3x – 4y) [1M]
= (3x – 4y)(3x + 4y + 2) [1A]
x +6y x –y +6xy –xy = 5xy
Sample Sample
16
HKDSE Exam Series – Intensive Practice (CQ & MCQ) for Mathematics (Compulsory Part) Solution Guide
7. (a)
x y
x
y x y
3 3
5
2 3
- -
^ h ^ - h
=
x y
x
x y
y
6 3
10
6 3
3
- +
^ h ^ - h
= 3
x y
x y
6 10 3
- +
^ h [1A]
(b)
x x
x 2 1
1
2 1 + -
- =
x x
x
x x
x x 2 1 2 1
2 1
2 1 2 1 2 1
+ -
- -
+ -
] ] ] +
] ] ]
g g g
g g g [1M]
=
2x 1 2x 1
x x x
2 1 2 2
+ -
- - -
] ]g g
=
x x
x x
2 1 2 1
2 2 1
+ -
- + -
] ]g g [1A]
Guidelines
Express each fraction with the same denominator, (2x + 1)(2x – 1).
8.
g
002x2 – 5x + 9 2x + 5 4x3 + 0x2 – 7x + 9 4x3 + 10x2
–10x2 – 7x –10x2 – 25x 18x + 9 18x + 45 –36
[1M]
` Quotient = 2x2 – 5x + 9 [1A]
Remainder = –36 [1A]
Guidelines
Add zeros to the missing terms to avoid making mistakes.
9.
g
004x – 12 8x3 – 4x2 + 9x – 11 8x3 + 20x2 – 4x –24x2 + 13x – 11 –24x2 – 60x + 12 73x – 23 2x2 + 5x – 1
[1M]
` Quotient = 4x – 12 [1A]
Remainder = 73x – 23 [1A]
10. (a) Remainder = 2(–2)3 + 3(–2)2 – 6(–2) + 5 [1M]
= 13 [1A]
(b) Remainder = 2 1
2 1
2
2 3 6 1 5
3 2
+ - +
c m c m c m [1M]
= 3 [1A]
11. p(1) = k
5(1)3 + k(1)2 – (k – 1)(1) + 2 = k [1M]
5 + k – k + 1 + 2 = k
k = 8 [2A]
12. Let f(x) = 2x3 – 5x2 – 23x – 10.
f 2 -1 c m = 2
2
1 5
2 1 23
2 1 10
3 2
- - - -
c m c m c m [1M]
= 1 4 5
2 23 10 -4- + -
= 0 [1A]
` By the factor theorem, 2x + 1 is a factor of f(x). [1A]
13. (a) Let f(x) = x3 + 6x2 + 3x – 10.
f(1) = (1)3 + 6(1)2 + 3(1) – 10 [1M]
= 1 + 6 + 3 – 10
= 0
` By the factor theorem, x – 1 is a factor of f(x).
[1A]
(b)
g
00x2 + 7x + 10 x – 1 x3 + 6x2 + 3x – 10 x3 – x2
7x2 + 3x 7x2 – 7x 10x – 10 10x – 10
[1M]
f(x) = (x – 1)(x2 + 7x + 10)
= (x – 1)(x + 5)(x + 2) [1A]
14. (a) f(3) = 2(3)3 – 9(3)2 + 7(3) + 6 [1M]
= 54 – 81 + 21 + 6
= 0 [1A]
(b) a f(3) = 0
` x – 3 is a factor of f(x).
By long division,
g
002x2 – 3x – 2 x – 3 2x3 – 9x2 + 7x + 6 2x3 – 6x2
–3x2 + 7x –3x2 + 9x –2x + 6
–2x + 6 [1M]
f(x) = (x – 3)(2x2 – 3x – 2)
= (x – 3)(2x + 1)(x – 2) [1A]
Common mistakes
Candidates may apply the long division directly without stating that x – 3 is a factor of f(x).
