hong kong standard assessment103.253.147.205/math eng sample.pdf · hkdse standard assessment 2013...

SAMPLE

@香港標準評核保留版權 Hong Kong Standard Assessment All Right Reserved 2012

香港標準評核

HONG KONG STANDARD ASSESSMENT

香港中學文憑全港標準模擬考試 2013

HKDSE STANDARD ASSESSMENT 2013

數學科 (必修部分)

試卷一及二

MATHEMATICS (COMPULSORY PART)

PAPER 1 & 2

學習套件

(連參考答案)

LEARNING KIT

(WITH SUGGESTED SOLUTION)

英文版

ENGLISH VERSION

本學習套件由香港標準評核專為本試卷而編寫，只供參與學校教師及學生參考之用。

This set of Learning Kit has been prepared by the Hong Kong Standard Assessment for participated

school teachers’ and students’ reference only.

@ 香港標準評核保留版權 Hong Kong Standard Assessment All Right Reserved 2012

2

SAMPLE

Law of Indices1

pm

p 42

19=

+−

)2(419 mpp +=−

pmpp 4819 +=−

pmpp 4189 =−−

p

pm4

1−=

2

22 494 ba −

( )( )baba 7272 −+=

22 49472 baba +−+

( )22 49472 baba −−+=

( )( )bababa 727272 −+−+=

( )( )baba 72172 +−+=

( )( )bababa −+=− 22

( )222 2 bababa −=+−

( )222 2 bababa +=++

Factorization

Expansion

)1( baaba −=−

3

Concept Review

Concept Review

yxyx

2

59− yxx

y29

5

=

29

15

+

−

=xy

11

4

xy

=

nmnm aaa +=⋅

nmn

m

aaa −=

n

mnnm

ba

ba

=

m

n

n

m

ab

ba

=−

−

PAPER I


PAP

ER I

3

SAMPLE

1 - 5

%80500%)251( ×=+x

320=x

Concept Review

sellingprice cost

profit

dis-count

markedprice

price sellingprofit) (1cost =+× %

price selling) discount (1price marked =−× %

4a

b

80$=

5.73$=

... which one, the jacket or the shirt, has a higher profit ?

From the question:

REMARK

profitprofit cost =× %

profitcost price selling =−

5

2518

123

50

125357

=

=

=+

×

x

x

xx baba =:

Concept Review

Ratio

25182×=

kg) 1.44 ( kg 2536

=

Let $x be the cost of the jacket. Then, we have

Thus, the cost of the jacket is $ 320.

profit %profit ≠

The profit of the jacket %25320×=

The profit of the shirt 5.262%80420 −×=

Thus, the jacket has a higher profit

Let x be the weight of a Chinese book

Then, the weight of a Mathematics book is x35

The weight of 2 Chinese books


4

SAMPLE

6a

7ai

902

75120

=

+=

x

x

Concept Review

ii

b

205

250455=

−=

REMARK

Since the given inter-quartile range, i.e. 120, in (a)(ii) isonly an assumption for that part of the question, it is regarded as invalid in this part and cannot be used.

205185

350535

<=

−=

( )

291217

341731

<−>+

−>+

xxx

xx

2332

825

−≥

−≥≤−

x

xx

223

<≤−∴ x

b

and

ans: 1

Concept Review

There is a simple rule to follow: If we multiply or divide an inequality by a NEGATIVE number, we must REVERSE the inequality (flip the inequality sign).

Inequality

Stem-and-leaf diagram

Stem (10) Leaf (1)

5

6

7

1 3 7

2 2

8

51 , 53 , 57 , 62 , 62 , 78

Q1 Q3Q2median

2

6257 +=

E.g.

lowerlimit

upperlimit

Q1 Q3Q2

Box-and-whisker diagram

lowerlimit

upperlimit

median

345$2

350340 median The

=

+=

2

2902802

420300 range quartile-inter The +−

++=

x

REMARK

xx NOT , +300

A company in ondistributi theof range The

B company in ondistributi theof range The

Thus, the distribution of the daily wages in company B is not more dispersed than that of the company A.

limit lowerlimit upper ondistributiof range The −=

13 QQ −= range quartile-inter The


PAP

ER I

5

SAMPLE

6 - 9

8a ( ) πππ 90)3(3

34

21 23 =+

h

872990918

===+

hhh

πππππ

( )( ) ( )( ) ( )2

22

cm 75

38323421

π

πππ

=

++=

b

r

Concpet Review

Volume ofa sphere

3

34 rπ= Surface area

of a sphere 24 rπ=

r

Volume ofa cylinder hr π2=

h

πr2

h

h

πr2

Curved surface Area

hrπ2=

9 °=∠=∠ 77DEBACB

°=∠=∠ 77ACBABC

°=°−°−°=∠−∠−°=∠

267777180

180 ABCACBCAB

( )°=°=

∠=∠

52262

2 CABBOC

77A

O

B

C

D

E

77A

O

B

C

D

E

a

b

c

a + b + c =180°77

AO

B

C

D

E

(∠ sum of △)(∠at center twice ∠at circumference)

(base ∠s of isos △)

（ext.∠, cyclic quad.）

The total surface area of the solid


6

SAMPLE

17ai

==

3110418000

4abab

18000311043 =b

ii ( )122.115000×=

133742=

iii ( )

( )

146235.5923log2.1log

9232.1

91412.1

1400002.0

12.118000

7000000012.1

12.118000500

<

<

<

<−

<−

<−

−×

x

x

x

x

x

x

MnM n loglog =

Logarithmic Functions

1)( −= narnT

Concept Review

=a =n =r

( ) ( )1

1−−

=rranS

n

Sum of geometric sequence

Geometric progression/ Geometric sequence

firstterm

no. ofterms

( )252.1 4 <b

( )

)(

)(25

)(2.1

)4(4

mQ

mP

mP

mP

=

<

=

+∴

2001

2002

2003

2004

2005

2006

abT =)1(

babT ⋅=)2(

2)3( babT ⋅=

3)4( babT ⋅=

4)5( babT ⋅=

...

REMARK

a (15000) is NOT the first term

ab (18000) is the first term

So, we have

Solving, we have b = 1.2 and a = 15000

The population in 2012

but

commonratio

Thus, the money will be insufficient to give allowance

to all citizens in 2006

Note that

Thus, after 2005, the population of City B is higher than that of City A in each year.


PAP

ER I

7

SAMPLE

17 - 18

18a OB = OC

OCBOBC ∠=∠

BP = CP

PCBPBC ∠=∠

PCOPCBOCBPBCOBCOBP

∠=∠+∠=∠+∠=∠

°=∠ 90OBP

°=∠ 90PCO

(radii)

(base ∠s isos. △)

(radii)


OB

D

rr

°=∠+∠ 180OCPOBPbi

−++

=2

)4(0,2

60

ii

)2,3( −=

( ) ( )22 0203 −−+−= 13=

( ) ( ) 1323 22 =++− yx

( ) ( ) 222 rkyhx =−+−

( )kh,= r=

( )11 , yx

( )22 , yx

A

B

( ) ( )212

212 yyxxAB −+−=

Distance between two points

a

b

P

QR

S

°=+ 180ba

SRQP ,,,If

then are concyclic.(opp. ∠s, supp.)

Concept Review

°=∠ 90PCO

(Converse of ∠ in a semicircle)

REMARK

∵

∴

(tangent ⊥ radius)

O

A BT

If OT passes through the centerand AB touches the circle at T,

ABOT ⊥then (tangent ⊥ radius)

tangent ⊥ radius

O

A BT

ABOT ⊥then AB is the tangent to the circle at T.

If OT is a radius and ,

(converse of tangent ⊥ radius)

Converse of tangent ⊥ radius


∴CP is the tangent to the circle ACB at C (converse of tangent ⊥ radius)

Hence, O, B, P and C are concyclic. (opp. ∠s, supp.)

OP is a diameter OP is a diameter

Centre of circle OBPC

Radius of circle OBPC

Equation of the circle OBPC:

Equations of Cricles

Standard Form:

Center Radius


8

SAMPLE

PAPER II

( )( )bababa −+=− 22

( )222 2 bababa −=+−

( )222 2 bababa +=++

Factorization

Expansion

)1( baaba −=−

Concept Review

Law of Indices

Concept Review1 B ( )

4

106

10

6

10

323

10

32

32322642

424

x

xxxxx

xx

=

=

=

=

−

×

2 A ( )( ) ( )( )

( ) ( )( )

124124

12

)1(2121212

22

22

22

−+−=

+−−=

−−=

−−−+=+−−+

bbabba

ba

babababa

3 A

23

1211

2131

212

21

−=

−

−+

−

−

+

−=

−= f

The Remainder Remainder theorem

If is a polynomial,then

)(xP

Concept Review

4 C ( ) 16 2 −++= mx

mxxmxx

+++=

−+++=

351213612

2

2

RHS

mn

m+=

=35122

41,6 ==∴ nm

( )

=−÷

abPbaxxP )(Remainder of

By comparing the coefficient of x and the constant term,

we have

nmnm aaa +=⋅

nmn

m

aaa −=

n

mnnm

ba

ba

=

m

n

n

m

ab

ba

=−

−


PAP

ER II

9

SAMPLE

SAMPLE

1 - 9

5 D

=−=+

⇒

=+−=+

)...(1423)...(1735

173231735

iiyxiyx

yxyx

x = 4

6 B

133523

≥≥≥+

xx

x

1≥x∴

or

2147

1715

≥−≤−

≤−

xxx

7 D

4

2

843256

2

2

=

=

=+=−

yx

yx

yxyxyx

1:4: 22 =yx∴

baba =:

Concept Review

Ratio

8 B

x

xxxx

=

×

−−++

%3143

%1541

1211

41

121

9 A Concept Review

speed =distance

time

Concept Review

(1) Method of Substitution

(2) Method of Elimination

(3) Using Calculator Formula (can be easily found by searching " simultaneous Linear equations in 2 unknowns + (your calculator model no.)" on the internet.)

Simultaneous Equations

Concept Review

There is a simple rule to follow: If we multiply or divide an inequality by a NEGATIVE number, we must REVERSE the inequality (flip the inequality sign).

Inequality

Let the saving of Sophia be x

Solve (i) and (ii),

speed original

distancetime original =

( )

time originalspeed

distancespeed

distancetime new

×=

=

+×=

8.0

8.0

%)251

Hence, the time it takes to travel the same distance is reduced by 20 %


10

SAMPLE

22 C

△ AEH, △ DEP, △ DBG, △ PFG, △ KFC, △ KPH

Concept Review

23 D ( )

( )8108018027203601802

=°=°×−°=°−°×−

nn

n

a

d

b cAB

DC

°= 360

24 C ( ) 190210cos2 −=°−°=x

( ) 390210sin2 =°−°=y

∴P0 = )3,1(−

y

xO

210

2

2

90

P0 (x, y)

P

25 AConcept Review

26 C

31628

26 22

=−

−+

=Radius

ba =

db =

°=+ 180ca(corr. s, AB // CD)∠

(alt. s, AB // CD)∠

(int. s, AB // CD)∠

Angles and Parallel Lines

Condition of Similar △s

1. AAA2. 3 sides proportional3. 2 sides proportional, int. ∠ equal

triangles similar to ∆ ABC

Let the polygon has n sides

(n ─ 2)×180

(4 ─ 2)×180 = 360(3 ─ 2)×180 = 180

(5 ─ 2)×180 = 540

n

∠ sum of polygon

Sum of the exterior anglesof convex polygon

Let the rectangular coordinates of P0 be (x, y)

y-intercept of L1 = 4 ∴The y-intercept of L2 = 4

Slope of L1 = m

∴The slope of L2 = m1

−

Hence, the equation of L2 is y =

m1

− x + 4 If , then .Conversely, if , then .

31 LL ⊥ 131 −=×mm

31 LL ⊥ 131 −=×mm

Slopes of Perpendicular lines:

O

y

x

c slope = m

Slope-intercept form

y = mx+c

Concept Review

( ) ( ) 222 rkyhx =−+−

Concept Review

022 =+−++ FEyDxyx

FED

−

+

=

22

22

−−=

2,

2ED

General Form:


Radius

Center

Standard Form:

Center ( )kh,= Radius r=

Centre = (–3, 4) , ∴ the center is at quadrant II

∵ The x coordinate of the center = –3 and the radius = 3,

∴ the graph touches the y-axis.

∵ The y coordinate of the center >3 and the radius = 3,

∴ the graph does not touch the x-axis


PAP

ER II

11

SAMPLE

SAMPLE

27 A

12563

107

108

109

1010

=

×××=

Concept Review

P ( A ⋂ B ) = P(A) x P(B)

28 A

2142

=

=

A

D

D

No. of favourable outcomes= 2

No. of all possible outcomes = 4

29 D

Freq

uenc

y

Marks

Q1 Q3

(Frequency Polygon)

30 D ( ) ( ) ( )

23

321

+=

+++++=

x

xxxmx

( ) ( ) ( )

x

y

mxy

yyym

=+<+=

+++++=

22

3321 ( ) ( ) ( ) ( ) ( )

nxxxxxxxxxx nn

221

23

22

21 ... −+−++−+−+−

= −σ

( )[ ] ( )[ ] ( )[ ]32

3232221 222

=+−+++−+++−+

=xxxxxxsx

( )[ ] ( )[ ] ( )[ ]xy syyyyyys ==

+−+++−+++−+=

32

3232221 222

If A and B are two independent events with probabilities P(A) and P(B) respectively, then the probability that A and B will both happen is:

Multiplication Law of Probability

The probability that the password is formed by different digits

The probability that A sits next to D

Concept Review

No. of favourable outcomes

No. of all possible outcomesP(E) =

The probability of an event E is a measure of the chance that the event will happen, and it is usually denoted as P(E).

Probability

(Box-and-whisker diagram)

lowerlimit

upperlimit

median

Concept Review

Standard Deviation

22 - 30


12

SAMPLE

( ) ( ) 222 rkyhx =−+−

Concept Review


Standard Form:

Center ( )kh,= Radius r=

42 B

x

y

O R

QP

13

3 3

2 2

2 (3, – 2)

43 B

Centre of the circle = (3, – 2) and radius of the circle = 13

Hence, the coordinates of P, Q and R are (0, – 4), (6, – 4) and (6, 0) respectively.

Area of the circle = ( ) ππ 13132=

Area of the rectangle = 2446 =×

Hence, the area of the shaded region = ( )2413 −π sq. unit.

Since Michael must be included in the list, Amy has to invite 9 friends out of 17 friends.

If Peter and Jenny attend the party, Amy has to invite 7 friends out of 15 friends.

Number of ways Amy has to choose her friends to attend her Christmas party = 15

7C= If Peter and Jenny will not attend the party, Amy has to invite 9 friends out of 15 friends.

Number of ways Amy has to choose her friends to attend her Christmas party = 15

9C=

Hence, number of ways Amy has to choose her friends to attend her Christmas party

11440

159

157

=+= CC

Concept Review

Combination

The no. of combinations of r objects from n distinct objects without repetition

nrC=


PAP

ER II

13

SAMPLE

SAMPLE

45 A

Concept Review

If the salary of all the workers in a factory increase by $300,

I. Mean of salary will increase by $300

II. Median of salary will increase by $300

III. Standard deviation unchanged

Add a common constant (c) to each item of the set of data

each item xi mean median standard deviation

range inter-quartile range

xi + c + c + c unchanged unchanged unchanged

@香港標準評核保留版權 Hong Kong Standard Assessment All Right Reserved 2012

Complementary Events

1)'()( =+ EPEP

)(1)'( EPEP −=

or

Concept Review

P ( A ⋂ B ) = P(A) x P(B)

If A and B are two independent events with probabilities P(A) and P(B) respectively, then the probability that A and B will both happen is:

Multiplication Law of Probability44 C

case 1

case 5

case 4

case 3

case 2

REMARK

5

The probability that Peter hits all the targets

24332

32

32

32

32

32

=

=

The probability that Peter hits 4 targets out of 5 fires

24380

5321

32

32

32

32

=

×

−

=

∴The required probability

243112

24380

24332

=

+=

42 - 45

hong kong standard assessment103.253.147.205/math eng sample.pdf · hkdse standard assessment 2013...

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