public assessment of the hkdse mathematics examination

Public Assessment of the HKDSE Mathematics Examination1.ExamFormatPaper1(Conventionalquestions)Duration:2hours15minutes;Percentage:65%

Section Marks Number of Questions Other Details

A(1) 35 8 – 11Elementary questions on the Foundation Topics of the Compulsory Part together with the Foundation Part of the Secondary 1–3 Mathematics Curriculum.

A(2) 35 4 – 7Harder questions on the Foundation Topics of the Compulsory Part together with the Foundation Part of the Secondary 1–3 Mathematics Curriculum.

B 35 4 – 7Questions on the Compulsory Part together with the Foundation Part and the Non-Foundation Part of the Secondary 1–3 Mathematics Curriculum.

Paper2(Multiple-choicequestionswith4options)Duration:1hour15minutes;Percentage:35%

Section Number of Questions Other Details

A 30Questions on the Foundation Topics of the Compulsory Part together with the Foundation Part of the Secondary 1–3 Mathematics Curriculum.

B 15Questions on the Compulsory Part together with the Foundation Part and the Non-Foundation Part of the Secondary 1–3 Mathematics Curriculum.

2.Standards-referencedReportingThe HKDSE makes use of standards-referenced reporting, which means candidates’ levels of performance will be reported with reference to a set of standards as defined by cut scores on the variable or scale for a given subject. The following diagram represents the set of standards for a given subject:

Cut scores

U 1 2 3 4 5

Variable/scale

Within the context of the HKDSE there will be five cut scores, which will be used to distinguish five levels of performance (1–5), with 5 being the highest. The Level 5 candidates with the best performance will have their results annotated with the symbols ∗∗ and the next top group with the symbol ∗. A performance below the threshold cut score for Level 1 will be labelled as ‘Unclassified’ (U).

– II –

Exam StrategiesA. TimeAllocation

Paper1

Section Suggested Time Allocation Approximate Time per Question

A(1) 40 minutes 3 – 5 minutes

A(2) 40 minutes 5 – 10 minutes

B 50 minutes 5 – 15 minutes

– In general, spend 5 minutes for every 4 marks.– Allow 5 minutes for final checking.

Paper2

Section Suggested Time Allocation Approximate Time per Question

A 45 minutes 1.5 minutes

B 20 – 25 minutes 1.5 – 2 minutes

– Allow 5 – 10 minutes for final checking.

B. AnsweringSkills

ForPaper1:

• Skip the questions that you do not have confidence on. Go back to the skipped questions after you have finished the others.

• Show your formulas and steps rather than just writing down the answers. In case you do not get the correct answer, you can get marks for the correct methods used.

• Make sure your numerical answers are either exact or correct to 3 significant figures, unless otherwise specified.

• Give intermediate results correct to more than 3 significant figures to avoid accumulated errors in the final answers.

• Make sure you give a unit, if any, to each answer.

• In answering questions involving drawing,– Always use a pencil.– Draw exact and precise figures with tools like rulers and protractors.– Choose an appropriate scale.– Label significant items like graph titles and axis names.

• In answering questions which involve sketching, show rough figures of the correct shapes.

• Pay attention to question wordings.– ‘Write down’ – You can just write down the answer without showing your working steps.– ‘Hence’ – You have to use the results obtained earlier to get the answer.

– III –

Distribution of Exam QuestionsPaper1

Topics HKCEE 06 HKCEE 07 HKCEE 08 HKCEE 09 HKCEE 10 HKDSESample

Number Systems / / / / / /Percentages 6 6 8, 16(b) 7 7 4Rate, Ratio, Variations and Estimation

9(b) & (c), 11, 15(a) & (b)(ii)

10, 12(b), 14(b)

7 4 8, 10 11, 12

Polynomials 3 3, 14(a)(ii) / 3 3 3, 10(a) & (b)(i)

Indices, Surds, Exponential Functions and Logarithmic Functions

1 2 1 2 1 1, 17

Sequences / / 16(a) & (c) 15(c) 4(a), 17(b)(ii) 15Equations 9(a),

10(a)(ii) & (b), 15(b)(iii)

7, 12(a), 14(a)(i)

3 6 6, 16 5, 9(a), 10(b)(ii)

Formulas, Functions and Graphs

10(a)(i), 15(b)(i) & (iv)

1 6, 11 1, 12(a), 15(a) & (b)

5, 16(a) & (b)(ii)

2

Inequalities and Linear Programming

2 5 2 16(a)(ii) & (b) 2 /

Mensuration 4, 13 9, 11, 13(c)(ii), 16(a)

13 12(b)(ii), 13 12(d), 13 6

Deductive Geometry / 17(a) / 11 9(b) /Angles, Straight Lines and Rectilinear Figures

5 8 9 / 9(a) 9(b), 19(a)(ii)

Circles 16(a) / 17(a) / / 7, 13, 19(a)(i)Transformation and Symmetry

/ / / 9 / /

Coordinate Geometry 7, 12, 16(b)(i) – (ii), 16(b)(iii)

13(a) – (c)(i), 17(b)

12, 17(b)(i) – (ii), 17(b)(iii)

8, 12(b)(i), 16(a)(i)

12(a) – (c), 17(a)

8, 19(b)

Trigonometry 17 16(b), 16(c) 4, 15 17 4(b), 15, 17(b)(i)

18

Permutation and Combination

/ / / / / /

Probability 8(b), 14(b) 12(c), 15(a), 15(b)

5, 14(a)(i), 14(a)(ii) & (b)

5, 14(b)(i), 14(b)(ii) & (iii)

14(a),14(b)(iii)

16

Statistics 8(a), 14(a) 4, 12(d) 10 10, 14(a) 11, 14(b)(i) 14

* Non-foundation questions are in bold.– VI –

Useful FormulasJuniorSecondary1. Estimation,ApproximationandErrors

(a) Absolute error = estimated value - exact value

(b) Maximum absolute error= largest possible uncertainty of an estimation or a measurement

(c) Relative error = Maximum absolute error

Measured value or

= Absolute error

Exact value(d) Percentage error = Relative error × 100%

2. Percentages(a) Percentage change

= New value - Original value

Original value × 100%

(b) (i) New value= Original value × (1 + Percentage increase)

(ii) New value= Original value × (1 - Percentage decrease)

(c) Profit and loss Percentage change

= Selling price - Cost price

Cost price × 100%

If the percentage change > 0, then there is a profit.If the percentage change < 0, then there is a loss.

(d) Selling price = Cost price × (1 + Profit percentage) or = Cost price × (1 - Loss percentage)

(e) Discount percentage

= Marked price - Selling price

Marked price × 100%

(f) Selling price= Marked price × (1 - Discount percentage)

(g) Let P be the prinicipal, r% be the interest rate per period, n be the number of periods and A be the total amount.

(i) Simple interest

(1) Interest, I = P × r% × n

(2) Total amount, A = P + I

(ii) Compound interest

(1) Total amount, A = P × (1 + r%)n

(2) Compound interest, I = P × (1 + r%)n - P

(h) Let n be the number of periods

(i) Growth New value= Original value × (1 + Growth rate)n

(ii) Depreciation New value= Original value × (1 - Depreciation rate)n

3. Polynomials,IndicesandSurds(a) Division algorithm

Dividend = Divisor × Quotient + Remainder,where the remainder is zero or a polynomial with degree less than that of the divisor.

(b) Properties of Surds

(i) ab a b=

(ii) ab

ab

=

4. AlgebraicRelations(a) (a + b)2 ≡ a2 + 2ab + b2

(b) (a - b)2 ≡ a2 - 2ab + b2

(c) a2 - b2 ≡ (a + b)(a - b)

(d) a3 + b3 ≡ (a + b)(a2 - ab + b2)

(e) a3 - b3 ≡ (a - b)(a2 + ab + b2)

5. CoordinateGeometry(a) The distance between A(x1, y1) and B(x2, y2) is

AB x x y y= − + −( ) ( )1 22

1 22 .

(b) Slope of a straight line joining A(x1, y1) and B(x2, y2)

is, m y yx x

=−−

=2 1

2 1tanθ ,

where q is the inclination of the straight line.

(c) If P(x, y) is the mid-point of A(x1, y1) and B(x2, y2),

then xx x

=+1 22

and yy y

=+1 22

.

(d) If P(x, y) is a point dividing the line segment joining A(x1, y1) and B(x2, y2) internally in the ratio r : s, that is, AP : PB = r : s, then

x sx rxr s

=++

1 2 and y sy ryr s

=++

1 2 .

– VIII –

– 2 –

Answers written in the margins will not be marked.

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Page total© 2011 Hong Kong Educational Publishing Co.

SECTION A(1) (35 marks)

1. Simplify ( )x yxy

2 3

6 and express the answer with positive indices. (3 marks)

2. Find the remainder when x5 - 3x2 + 2x - 1 is divided by x + 1. (2 marks)

CP MOCK 1 PAPER 1-2


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Please stick the barcode label here.

Go on to the next page

SECTION A(2) (35 marks)

11. In a bag, there are x cards marked ‘10’ to ‘19’, y cards marked ‘20’ to ‘29’ and z cards marked ‘30’ to ‘39’. A card is drawn randomly from the bag. After the number is recorded, the card is put back into the bag. The records of 30 draws are shown in the stem-and-leaf diagram.

Stem (10) Leaf (1)

1 2 3 3 5 6 8 9 9 0

2 1 3 4 6 8 9

3 1 1 1 1 2 2 5 5 6 6 6 7 7 8 9

(a) State one mistake in the stem-and-leaf diagram. (1 mark)

(b) Using the stem-and-leaf diagram, estimate

(i) x : y : z,

(ii) the mean and the standard deviation of the cards’ numbers.(3 marks)

CP MOCK 2 PAPER 1-7


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SECTION B (35 marks)

17. On the Richter scale, the relationship between the magnitude R of an earthquake and the energy E (in J) released by the earthquake is E = a(10)bR, where a and b are constants. The following table shows the magnitudes and energies released in two earthquakes.

Earthquake Magnitude of the earthquake Energy released

Sendai earthquake 9.0 2 × 1018 J

Kobe earthquake 7.2 4 × 1015 J

Find the values of a and b. (4 marks)

CP MOCK 2 PAPER 1-13

– 166 – © 2011 Hong Kong Educational Publishing Co.

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.

Section A

1. If n is a positive integer, which of the following must be even?

I. 3(n + 2)

II. (3n + 2)2

III. 33n + 2

A. I only

B. II only

C. III only

D. None of the above

2. If A is greater than B by 10% and B is 90% of C, then

A. A = 0.99C.

B. A = C.

C. C = 0.99A.

D. C = 1.01A.

3. When x5 + 2x + 1 is divided by x2 + 1, the quotient and the remainder are x3 - x and R respectively. What are the degree of R and the coefficient of x in R?

Degree Coefficient of x in R

A. 1 3

B. 1 -3

C. 2 3

D. 2 -3

CP MOCK 5 PAPER 2-2


Section B

31. 1

3 21

3 2+−

−=

i iA. -4i

B. - 47i

C. 2 3i

D. 2 37i

32. log3

3

43 8a bb

=

A. log32ab

B. log3(-2ab)

C. log3 2ab

D. log log

log3 3

3

2 ab

33. The graph of y = x2 + kx + 3 intersects with the line y = x - 1 at P(x1, y1) and Q(x2, y2). y1 + y2 =

A. -1 - k.

B. -k.

C. 1 - k.

D. 2.

34. Two terms, a and b, are inserted between 2 and 4 2 such that the four consecutive terms form a geometric sequence. If a, c and b form an arithmetic sequence, then c =

A. 2 2 .

B. 2 2+ .

C. 2 2- .

D. 4.

CP MOCK 5 PAPER 2-10

Mathematics: Mock Exam Papers Compulsory Part Solution Guide


(b)

The shape of the conical container and the shape of the water in the container are similar.

Let f (r) = r3 − r2 − 180. f (6) = 216 − 36 − 180 = 0∴ r − 6 is a factor of f (r). r3 − r2 − 180 = 0 (r − 6)(r2 + 5r + 30) = 0 1M∴ r = 6 or r2 + 5r + 30 = 0 (rejected) 1ALet h cm be the height of the conical container.

h

h6 1

186

15−

=

=∴ Height of the conical container = 15 cm

(4)

16. (a)

1A (1)

(b)

As shown in the figure,β = 180° − (60° + 50°) (int. ∠s, // lines) 1M = 70°∴ The bearing of R from Q = 070° 1A

(2)

1M

1A

(c)

If Chris walks in the direction perpendicular to PR, then the walking distance is the shortest.

Let x be the distance between P and R.

∴ θ = 10° and α = 50° (alt. ∠s, // lines)∠PRQ = 60°∠QPR = α + 10° = 60° 1M∠RQP = β − θ = 60°∴ ∆PQR is an equilateral triangle.Note that they will meet at the mid-point of PR.

∴ Walking distance of Martin = x2

Walking distance of Chris= The distance from Q to the meeting position= °

=

x

x

sin6032∴

The walking time of Chris and Martin are the same.

∴ Average walking speed of Chris : Average walking speed of Martin

= Walking distance of Chris : Walking distance of Martin

=

=

32 23 1

x x:

: (4)

Average speed =Distance

Time

Section B

17. When E = 2 × 1018, R = 9.0,2 × 1018 = a(10)9.0b............(1)When E = 4 × 1015, R = 7.2,4 × 1015 = a(10)7.2b............(2)

( )( )12

:2 104 10

1010

500 101 8 500

18

15

9 0

7 2

1 8

××

=

==

.

.

.

. log

b

b

b

b

b ==

=

log.

.

5001 8

1 50 (cor. to 3 sig. fig.)

1M

1A

1M

1A

1M

1A

Mock Exam 6


18.

If α and β are the equal roots of the quadratic equation, ∆ = b2 − 4ac = 0.

�

∴ b2 = 4ac∴ ∆ = b2 − 4ac = 0The quadratic equation has double roots.� α = β∴

�

log ( )log ( ) log

log

log

2

2 2

2

2

12

2

α βα βα β

α

+ −= + −

= +

=

1212

2 2

2

2

12

2

(log log )

log

log ( )log

α β

αβ

αβα

+

=

==

log ( ) (log log )2 2 2

2

1 12

4

α β α β

α β

+ − = +

=

=b ac

The candidate confuses the condition with the result which needs to be proved. Although he/she does not have any mistakes in the operations, only 1 mark is obtained due to the problem with the logic.

1M

∴

∴∴

Candidates should understand the given condition and the steps for the proof.

log ( ) (log log )

log ( ) log (l

2 2 2

2 2

1 12

2 12

α β α β

α β

+ − = +

+ − = oog log )

log log

log log ( )

2 2

2 2

2 2

12

12

2

α β

α β αβ

α β αβ

+

+ =

+ = 22

12

2

2 2

2

2

22

42

α β αβ

α β αβ

α αβ β αβ

α αβ

+ =

+

=

+ + =

+ +

( )

ββ αβ

α αβ β

α βα β

α β

2

2 2

2

42 0

00

=

− + =

− =− =

=

( )

∴� ∆ of the equation = 0,� b2 − 4ac = 0 b2 = 4ac 1A (4)

The candidate successfully starts the proof from the condition and there are no operational mistakes.

1A

1M

1M

19. The number of flowers in each row are:4, 5, 8, 8, 16, 11, 32, 14, ... Total number of flowers

= + + + + + + + +4 5 8 8 16 11 32 14 �� 20 terms

== + + + + + + + + +( ) (4 8 16 32 5 8 11 14�� 10 terms

))

( ) [ ( ) (

10104 2 1

2 1102

2 5 10

terms� ��

= −−

+ + − 11 3

4

)( )]

= 277 (4)

1A

1A + 1A

1M

public assessment of the hkdse mathematics examination

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