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
@ 香港標準評核 保留版權 Hong Kong Standard Assessment All Right Reserved 2012
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
@ 香港標準評核 保留版權 Hong Kong Standard Assessment All Right Reserved 2012
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
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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
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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.
@ 香港標準評核 保留版權 Hong Kong Standard Assessment All Right Reserved 2012
PAP
ER I
7
SAMPLE
17 - 18
18a OB = OC
OCBOBC ∠=∠
BP = CP
PCBPBC ∠=∠
PCOPCBOCBPBCOBCOBP
∠=∠+∠=∠+∠=∠
°=∠ 90OBP
°=∠ 90PCO
(radii)
(base ∠s isos. △)
(radii)
(base ∠s isos. △)
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
(base ∠s isos. △)
∴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
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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
=−
−
@ 香港標準評核 保留版權 Hong Kong Standard Assessment All Right Reserved 2012
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 %
@ 香港標準評核 保留版權 Hong Kong Standard Assessment All Right Reserved 2012
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:
Equations of Cricles
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
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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
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12
SAMPLE
( ) ( ) 222 rkyhx =−+−
Concept Review
Equations of Cricles
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=
@ 香港標準評核 保留版權 Hong Kong Standard Assessment All Right Reserved 2012
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
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