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SULIT 2 1449 / 1
1449 / 1 2008 Hak Cipta Bahagian Pendidikan & Latihan (Menengah) MARA SULIT
M AT H E M AT I C A L F O R MU L A ERUMUS MATE MATIK
The following formulae may be helpful in answering the questions. The symbols given are the onescommonly used.Ru m u s -ru m u s b e ri k u t bol e h m e m ban t u anda m e njawab s oalan . S i m bol- s i m bol yang dib e ri adalah bia s a diguna k an .
R E L AT I O N SP E R K A I TA N
1 n m n m a a a 10 Pythagoras TheoremT e or e m Pi t hagora s
222 b a c
2 n m n m a a a
3 n m n m a a )( 11S n An
AP
4a c b d
b c ad A 11 12 AP AP 1'
5 Distance / jara k 1312
12x x y y
m
212
212 y y x x
6 Midpoint / Ti t i k t e ngah 14interceptintercept
x y m
2
,
2
, 2121 y y x x y x
x a s an in t p y a s an in t p m
7takentimetravelleddistance
speedAverage
dia m bi l yang m a s a dilalui yang jara k
laju Pura t a
8dataof number
dataof sumMean
da t a bilangan
da t a nilai t a m bah ha s il Min
9sfrequencieof sum
frequencyclassmark of SumMean
keke rapan t a m bah ha s il
keke rapa n k e la s t e ngah t i t i k nilai t a m bah ha s il Min
Mathematics 1 dan 2 MRSM trial SPM 2008
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SHAPES AND SPA C EBE NTUK DAN RUANG
1 Area of trapezium =2
1 × sum of parallel sides × height
Lua s t rap e ziu m = 2
1
× ha s il t a m bah dua s i s i s e lari × t inggi
2 Circumference of circle = d = 2 r Lili t an bula t an = d = 2 j
3 Area of circle = r 2 Lua s bula t an = j 2
4 Curved surface area of cylinder = 2 rh Lua s p e r m u k aan m e l e ng k ung s ilind e r = 2 j t
5 Surface area of sphere = 4 r 2
Lua s p e r m u k aan sf e ra = 4 j 2
6 Volume of right prism = cross sectional area lengthI s ipadu pri s m a t e ga k = lua s ke raain r e n t a s panjang
7 Volume of cylinder = r 2h I s ipadu s ilind e r = j 2t
8 Volume of cone = h r 23
1
I s ipadu k on = t j 231
9 Volume of sphere = 3
34 r
I s ipadu sf e ra = 3
34 j
10 Volume of right pyramid =3
1 base area height
I s ipadu pyra m id t e ga k =3
1 lua s t apa k t inggi
11 Sum of interior angles of a polygonHa s il t a m bah s udu t p e dala m an polygon
= 1802n
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12360
centreatsubtendedanglecircleof ncecircumfere
lengtharc
360
pu s a t s udu t
bula t an lili t an l e ng k u k panjang
13360
centreatsubtendedanglecircleof areasector of area
360tan
pu s a t s udu t
bula lua s
s ek t or lua s
14 Scale factor,PAPA
k '
Fa c t or s k ala , PAPAk
'
15 Area of image = k 2 area of objectLua s i m e j = k 2 lua s obj ek
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An s w e r al l qu e st ion s .Jawab s e mua s oalan .
1 Round off 69919 correct to three significant figures.
Bundar k an 69919 b e t ul ke pada t iga ang k a b e r e r t i .
A 700B 7000
C 69900
D 69920
2 Express 0.0196 in standard form.
Ung k ap k an 0.0196 dala m b e n t u k piawai .
A 1.96 × 10-3
B 1.96 × 10 -2
C 1.96 × 210
D 1.96 × 310
3 11023
0015.0=
A 4.6875 × 104
B 2.005 × 10 -1
C 4.6875 × 10 -4
D 3.0 × 10 -5
4 Given that x is a number in base 2 such that 31 8 < x < 11110 2, the possible value of x is
Dib e ri bahawa x ial ah no m bor dala m a s a s 2 d e ngan ke adaan 318 < x < 11110 2, nilai yang
m ung k in bagi x ialah
A 10111 2
B 11000 2
C 11011 2
D 11111 2
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5 Express 3 × 53 + 2 as a number in base five.
Ung k ap k an 3 × 53 + 2 s e bagai no m bor dala m a s a s li m a .
A 235
B 325
C 302 5
D 3002 5
6 In Diagram 1, ABCD is a quadrilateral and AB is parallel to CE .
Dala m Rajah 1 , ABCD ialah s e buah s i s i e m pa t dan AB adalah s e lari d e ngan CE .
The value of x is
Nilai x ialah
A 65
B 70
C 80
D 120
B
D
C
A
E
x o
75o
115 o
145 o
Diagram 1Rajah 1
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7 In Diagram 2, PQR S is a rhombus.
Dala m Rajah 2 , PQR S adalah s e buah ro m bu s .
The value of x is
Nilai x ialah
A 15
B 20
C 25
D 35
S P
R
T
25o
Q
x o
70o
65o
Diagram 2Rajah 2
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8 In Diagram 3, AB C and CDE are tangents to the circle with centre O , at B and D respectively.
Dala m Rajah 3 , ABC dan CD E adalah t ang e n un t u k bula t an b e rpu s a t O , m a s ing- m a s ing
di t i t i k B dan D .
The value of x is
Nilai bagi x ialah
A 30
B 32
C 58D 62
B
A
O
E
D
C F
28o32o
x o
Diagram 3Rajah 3
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9 In Diagram 4, P ́ is the image of P under a certain translation. Q ́ is the image of Q under thesame translation.
Dala m Rajah 4 , P ́ ialah i m e j bagi P di bawah s a t u t ran s la s i t e r t e n t u . Q ́ ialah i m e j bagi Q di bawah t ran s la s i yang s a m a .
Find the coordinates of Q .
Cari k an k oordina t Q
A ( 3 , 10)
B (13 , 0)C ( 5 , 12)
D (10 , 13)
0
2
4
x 104 146
.
8 12
6
8
2
y
Q ́
P
P ́
Diagram 4Rajah 4
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10 Diagram 5 shows the point X which is the image of point Y under a reflection.
Rajah 5 m e nunju kk an t i t i k X adalah i m e j bagi t i t i k Y di bawah s a t u pan t ulan .
Diagram 5Rajah 5
Determine the axis of the reflection.
T e n t u k an pa k s i pan t ulan t e r s e bu t .
A TV
B S U
C S V
D TU
. . .
. . . U
V X
T
Y S
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12 In Diagram 7, PTR is a straight line. Given that c o s x = 13
5 and s in y =
5
3.
Dala m Rajah 7 , PTR ialah gari s luru s . Dib e ri bahawa k o s x =
13
5dan
s in y = 5
3
Find the length, in cm, of PT .
Cari k an panjang , dala m c m , bagi PT .
A 10
B 22
C 23
D 26
R
P S
T 5 cm
10 cmQ
Diagram 7Rajah 7
y o
x o
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15 In Diagram 9, MN is a vertical television antenna on top of a building. Given that the anglesof elevation of M and N from the point Q are 31.3° and 28.1° respectively.
Dala m Rajah 9 , MN ialah an t e na t e l e vi s y e n yang t e ga k di pun c a k s e buah bangunan . Dib e ri s udu t donga k an M dan N dari t i t i k Q m a s ing- m a s ing ialah 31.3° dan 28.1° .
M
P 50m Q
Calculate the height, in m, of the television antenna.
Hi t ung k an t inggi , dala m m , an t e na t e l e vi s y e n t e r s e bu t .
A 2.42
B 2.80
C 3.70
D 11.40
N
Diagram 9Rajah 9
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17 An aeroplane flew 1320 nautical miles from P ( 80° N , 10° E ) to Q via the North Pole.Find the position of Q .
S e buah p e s awa t t e rbang 1320 ba t u nau t i k a dari P ( 80° U , 10° T ) ke Q m e lalui Ku t ub
U t ara . Cari ke dudu k an Q .
A ( 80° N , 10° E )
( 80° U , 10° T )
B ( 80° N , 170° W )
( 80° U , 170° B )
C ( 78° N , 30° E )( 78° U , 30° T )
D ( 78° N , 170° W )
( 78° U , 170° B )
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18 In Diagram 11, NO S is the polar axis of the Earth. Given that PON = 50° andQOR = 120°.
Dala m Rajah 11 , NO S ialah pa k s i bu m i . Dib e ri PON = 50° dan QOR = 120°.
Diagram 11Rajah 11
Find the position of P .
Cari k an ke dudu k an P .
A ( 40° N , 20° W )
( 40° U , 20° B )
B ( 40° N , 120° W )
( 40° U , 120° B )
C ( 50° N , 20° W )
( 50° U , 20° B )
D ( 50° N , 120° W )
( 50° U , 120° B )
O
N
S
P
Q
R
100o
E
0o
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22 Given that 5(8 m ) =2
5 m 5 , find the value of m
.
Dib e ri 5(8 m ) =2
5 m 5 , c ar i k an nilai m .
A 3
B 3
25
C 6
D 32
8
23 Expressm
m
44
m
m 52as a single fraction in its simplest form.
Ung k ap k an m
m
44
m
m 52 s e bagai s a t u p ec ahan t unggal dala m b e n t u k t e r m udah .
A m
m
4247
B m
m 4
716
C m
m
41
D m
m
4167
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24 32
23
3)3(
n m m n
m n =
A 2
22
m
n
B m
n 22
C m n 53
D m n 23
25 Given that 2 3x = 32(2 x ), find the value of x .
Dib e ri 23x = 32(2 x ), c ari nilai x .
A 4
3
B 1
C 3
5
D 45
26 Given that h is an integer, find all the values of h that satisfy both inequalities
4
h < 2 and 28 7h 3.
Dib e ri h ialah in t e g e r , c ar i k an s e m ua nilai h yang m e m ua s k an ke dua-dua ke t a k s a m aan
4
h < 2 dan 28 7h 3.
A 5, 6, 7
B 4, 5, 6, 7
C 5, 6, 7, 8
D 4, 5, 6, 7, 8
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27 Table 1 shows the distribution of marks for 50 students.
Jadual 1 m e nunju kk an t aburan m ar k ah bagi 50 orang p e lajar .
MarksMar k ah
Number of studentsBilangan p e lajar
6 - 10 7
11 - 15 11
16 - 20 8
21 - 25 12
26 - 30 9
31 - 35 3
Calculate the mean mark .
Hi t ung k an m in m ar k ah .
A 17.4
B 19.4
C 21.4
D 37.6
Table 1Jadual 1
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28 Histogram in Diagram 12 shows the weight, in kg, of 30 students.
Hi st ogra m dala m Rajah 12 m e nunju kk an b e ra t , dala m k g , bagi 30 orang p e lajar .
Number of students
Bilangan P e lajar
10
8
6
4
2
35 - 39 40 - 44 45 - 49 50 - 54 55 59 Weight / B e ra t (kg)
Diagram 12Rajah 12
Calculate the percentage of students whose weight is more than 44 kg.
Hi t ung k an p e ra t u s p e lajar yang m e m punyai b e ra t l e bih daripada 44 k g .
A 26.67
B 33.33
C 53.33
D 86.67
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32 Diagram 14 is a Venn diagram sho set R and set S .
Rajah 14 ialah s e buah ga m barajah V e nn yang m e nunju kk an s e t s e t R dan s e t S .
Diagram 14
Rajah 14
Which of the following represents the shaded region?
Yang m ana k ah m e wa k ili ran t au yang b e rlor ek ?
A R S ́
B R S ́
C S R´
D S R´
R
S
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33 In Diagram 15, PQ is parallel to R S . The equation of the straight line R S is y = 2x 7.The point F lies on the x axis.
Dala m Rajah 15 , PQ dan R S adalah s e lari . P e r s a m aan bagi gari s luru s R S ialah y = 2x 7 . Ti t i k F b e rada di a t a s pa k s i x .
Diagram 15Rajah 15
Find the value of t .
Cari nilai t .
A 20
B 10
C 5D 2.5
34 Find the equation of the straight line with the y intercept of 5 and passes through P (6 , 3).
Cari k an p e r s a m aan gari s luru s d e ngan pin t a s an y ialah 5 dan m e lalui P (6 , 3).
A 534 x y
B 534 x y
C 543 x y
D 543 x y
10
F (t , 0)
y = 2x 7
Q
R
S
x
y
P
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38 Table 3 shows the values of the variables v u , and w where u varies directly as the squareof v and inversely as w .
Jadual 3 m e nunju kk an nilai bagi p e m bol e h ubah v u , dan w d e ngan ke adaan u b e rubah s ec ara lang s ung d e ngan k ua s a dua v dan b e rubah s ec ara s ong s ang d e ngan w
.
u v w
40 4 2
r 6 4
Table 3Jadual 3
Calculate the value of r .
Hi t ung k an nilai bagi r .
A 12
B 30
C 45
D 180
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1449 / 2 2008 Hak Cipta Bahagian Pendidikan & Latihan (Menengah) MARA S U L I T
M AT H E M AT I C A L F O R MU L A ERUMUS MATE MATIK
The following formulae may be helpful in answering the questions. The symbols given are theones commonly used.Ru m u s -ru m u s b e ri k u t bol e h m e m ban t u anda m e njawab s oalan . S i m bol- s i m bol yang dib e ri adalah bia s a diguna k an .
R E L AT I O N SPERKAITAN
1 n m n m a a a 10 Pythagoras TheoremT e or e m Pi t hagora s
222 b a c
2 n m n m a a a
3 n m n m a a )( 11S n An
AP
4 a c
b d b c ad A
1112 AP AP 1
'
5 Distance / Jara k 1312
12x x y y
m
212
212 y y x x
6 Midpoint / Ti t i k t e ngah 14interceptintercept
x y
m
2,
2, 2121 y y x x y x
x a s an in t p y a s an in t p
m
7takentimetravelleddistance
speedAverage
dia m bi l yang m a s a dilalui yang jara k
laju Pura t a
8dataof number
dataof sumMean
da t a bilangan da t a nilai t a m bah ha s il Min
9sfrequencieof sum
frequencyclassmark of sumMean
keke rapan t a m bah ha s il k e k e rapa n k e la s t e ngah t i t i k nilai t a m bah ha s il
Min
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SHAPES AND SPA C E
BE NTUK DAN RUANG
1 Area of trapezium =2
1 × sum of parallel sides × height
Lua s t rap e ziu m =2
1 × ha s il t a m bah dua s i s i s e lari × t inggi
2 Circumference of circle = d = 2 r Lili t an bula t an = d = 2 j
3 Area of circle = r 2 Lua s bula t an = j 2
4 Curved surface area of cylinder = 2 rh Lua s p e r m u k aan m e l e ng k ung s ilind e r = 2 j t
5 Surface area of sphere = 4 r 2
Lua s p e r m u k aan sf e ra = 4 j 2
6 Volume of right prism = cross sectional area lengthI s ipadu pri s m a t e ga k = lua s ke raain r e n t a s panjang
7 Volume of cylinder = r 2h I s ipadu s ilind e r = j 2t
8 Volume of cone = h r 2
31
I s ipadu k on = t j 231
9 Volume of sphere = 3
34 r
I s ipadu sf e ra = 3
34 j
10 Volume of right pyramid = 3
1
base area height
I s ipadu pyra m id t e ga k =3
1 lua s t apa k t inggi
11 Sum of interior angles of a polygonHa s il t a m bah s udu t p e dala m an polygon
= 1802n
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SULIT 4 1449 / 2
1449 / 2 2008 Hak Cipta Bahagian Pendidikan & Latihan (Menengah) MARA S U L I T
12360
centreatsubtendedanglecircleof ncecircumfere
lengtharc
360
pu s a t s udu t bula t an lili t an
l e ng k u k panjang
13360
centreatsubtendedanglecircleof areasector of area
360tanpu s a t s udu t
bula lua s
s ek t or lua s
14 Scale factor,PAPAk
'
Fa c t or s k ala ,PAPAk
'
15 Area of image = k 2 area of objectLua s i m e j = k 2 lua s obj ek
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Sec tion ABahagian A
[ 52 m ar k s ][ 52 m ar k ah ]
1 On the graph provided, shade the region which satisfies the three inequalities 4 3 2x and y < 5.
Pada gra f yang di s e dia k an , lor ekk an ran t au yang m e m ua s k an ke t iga- t iga ke t a k s a m aan 4 3 2x dan y < 5.
[ 3 m ar k s ][ 3 m ar k ah ]
Answer / Jawapan :
y = 3 2x y = x 4 y
x O
For
U s e
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2 Calculate the value of v and of w that satisfy the following simultaneous linear equations.
Hi t ung k an nilai v dan nilai w yang m e m ua s k an p e r s a m aan s e r e n t a k b e ri k u t .
2w 3v = 4
w v 3
1= 5
[ 4 m ar k s ][ 4 m ar k ah ]
Answer / Jawapan :
3 Solve the following quadratic equation:
S e l e s ai k an p e r s a m aan k uadra t i k b e ri k u t :
7x = ( 2x + 3 )( 1 2x )
[ 4 m ar k s ][ 4 m ar k ah ]
Answer / Jawapan :
For
U s e
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4 Diagram 1 shows a right prism. The base S TU is a right angled triangle. The triangle S TU is the uniform cross section of the prism. W is the midpoint of TU .
Rajah 1 m e nunju kk an s e buah pri s m a t e ga k . Tapa k pri s m a adalah s e gi t iga S TU yang b e r s udu t t e ga k
. S e gi t iga S TU adalah ke ra t an r e n t a s s e raga m pri s m a i t u . W adalah t i t i k t e
ngah gari s
TU .
Calculate the angle between the line PW and the base S TU .
Hi t ung k an s udu t an t ara gari s PW dan s a t ah S TU [ 4 m ar k s ] [ 4 m ar k ah ]
Answer / Jawapan :
13 cm
12 cm
W10 cm
P
S U
Q
R
Diagram 1Rajah 1
T
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5 Diagram 2 shows a trapezium PQR S . The equation of the straight line R S is y 2x = 10 and equation of the straight line P S is y = 4x + 40.
Rajah 2 m e nunju kk an s e buah t rap e ziu m PQR S . P e r s a m aan gari s luru s R S ialah y 2x = 10 dan p e r s a m aan gari s luru s P S ialah y = 4x + 40.
Find
Cari
(a) the x intercept of line S R .
pin t a s an x bagi gari s luru s S R .
(b) the equation of the straight line PQ .
p e r s a m aan gari s luru s PQ .
(c) the coordinates of point S .
k oordina t t i t i k S .
[ 5 m ar k s ][ 5 m ar k ah ]
P
Q ( 10 , 45 )
R
S
y
x
y 2x = 10
Diagram 2Rajah 2
y = 4x + 40
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6 Diagram 3 shows a semicircle ORT with centre S . OPQ is a sector of a circlewith centre O .
Rajah 3 m e nunju kk an s e m ibula t an ORT b e rpu s a t S . OPQ adalah s ek t or bula t an
b e rpu s a t O .
Given that PT = T S = 6 cm .
Di b e ri PT = T S = 6 c m .
[ Use / Guna =7
22 ]
Calculate
Hi t ung k an
(a) the perimeter, in cm, of the whole diagram.
p e ri m e t e r , dala m c m , s e luruh rajah i t u .
(b) the area, in cm 2, of the shaded region .
lua s , dala m c m 2
, k awa s an yang b e rlor ek .
[ 6 m ar k s ][ 6 m ar k ah ]
Q
45
O S T
R
P
Diagram 3Rajah 3
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Answer / Jawapan :
(a)
(b) Premise 2 / Pr e m i s 2 :
(c)
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Answer / Jawapan :
(a)
(b)
(c)
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Answer / Jawapan :
(a)
(b)
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Answer / Jawapan :
(a)
(b)
(c)
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Answer/ Jawapan :
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Sec tion BBahagian B
[ 48 m ar k s ][ 48 m ar k ah ]
Answer any four questions from this section.Jawab m ana- m ana e mpat s oalan daripada bahagian ini .
12 (a) Complete Table 2 in the answer space for the equation y = x ³ + 8x 6 by writingdown the values of y when x = 4 and x = 3.
Le ng k ap k an Jadual 2 di ruang jawapan bagi p e r s a m aan y = x ⇡ + 8x 6 d e ngan m e nuli s nilai-nila i y apabila x = 4 dan x = 3 .
[ 2 m ar k s ] [ 2 m ar k ah ]
(b) For this part of the question, use the graph paper provided on page 25.You may use a flexible curve ruler.
Un t u k ce raian s oalan ini , guna k an ke r t a s gra f yang di s e dia k an pada hala m an 25. Anda bol e h m e ngguna k an p e m bari s f l ek s ib e l .
By using a scale of 2 cm to 1 unit on the x axis and 2 cm to 5 units on the y axis,draw the graph of y = x ³ + 8x 6 for x .
D e ngan m e ngguna k an s k ala 2 c m ke pada 1 uni t pada pa k s i x dan 2 c m ke pada 5 uni t pada pa k s i y , lu k i s k an gra f y = x ³ + 8x 6 bagi 4 x 3.5.
[ 4 m ar k s ][ 4 m ar k ah ]
(c ) From your graph, findDar ipada gra f anda , c ar i
(i) the value of y when x = 2.5nilai y apabila x = 2.5
(ii) the value of x when y = 15nilai x apabila y = 15 [ 2 m ar k s ]
[ 2 m ar k ah ]
(d) Draw a suitable straight line on your graph to find the values of x whichsatisfy the equation x ³ = 10 x 10 for x State these values of x .
Lu k i s k an s a t u gari s luru s yang s e s uai pada gra f anda un t u k m e n c ari nilai-nilai x yang m e m ua s k an p e r s a m aan x ³ = 10 x 10 bagi x Nya t a k an nilai-nilai x i t u .
[ 4 m ar k s ][ 4 m ar k ah ]
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Answer / Jawapan :
(a)
x 4 3 2 1 0 1 2 3 3.5
y 3 14 13 6 1 2 20.9
(b) Refer graph on page 25.
Ruju k gra f di hala m an 25.
(c) ( i ) y
( ii ) x
(d) x =
Table 2Jadual 2
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G raph for Que s tion 12Graf un t uk Soalan 12
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13 You are not allowed to use graph paper to answer this question.
Anda t idak dib e nar k an m e ngguna k an ke r t a s gra f un t u k m e njawab s oalan ini .
(a) Diagram 6 (i) shows a solid in the shape of a right prism with a rectangular baseABCD on a horizontal plane. The surface ABGFE is the uniform cross-section of the solid. Rectangle FGHJ is an inclined plane and rectangle EFJK is a horizontal
plane. Edges AE , BG , C H and DK are vertical.
Rajah 6 (i) m e nunju kk an p e p e jal b e rb e n t u k pri s m a d e ngan t apa k s e gi e m pa t t e pa t ABCD t e rl e t a k di a t a s s a t ah m e ngu f u k
. P e r m u k aan ABG F E ialah ke ra t an r e n t a s s e raga m bagi p e p e jal i t u . S e gi e m pa t t e pa t FGHJ ialah s a t ah c ondong dan s e gi e m pa t t e pa t EFJK ialah s a t ah m e ngu f u k
. S i s i AE , BG , CH dan DK adalah t e ga k
.
Draw to full scale, the elevation of the solid on a vertical plane parallel to AB asviewed from X .
Lu k i s k an d e ngan s k ala p e nuh , donga k an p e p e jal i t u pada s a t ah m e n c an c ang yang s e lari d e ngan AB s e bagai m ana diliha t dari X .
[ 3 m ar k s ][ 3 m ar k ah ]
C E
B
D
A
F
6 cm
5 cm
3 cm
X
H
K J
G
6 cm2 cm
Diagram 6 (i)Rajah 6 (i)
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Answer / Jawapan :
(a)
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(b) A solid in the shape of a prism with uniform cross-section LMP and LD = MN = PC is joined to the prism in Diagram 6 (i) to form a combined solid as shown inDiagram 6 (ii). Both of the prisms are joined at the plane CDKJH . Plane ABCN ishorizontal, ADN and LKD are straight lines.
S e buah p
e p
e jal pri
s m
a d e ngan
ke ra
t an r
e n
t a
s s e
raga m
LMP dan LD = MN = PC digabung d e ngan pri s m a dala m Rajah 6 (i) bagi m e m b e n t u k p e p e jal s e p e r t i di t unju kk an dala m Rajah 6 (ii) . K e dua-dua p e p e jal t e r s e bu t digabung k an pada s a t ah CD KJH . S a t ah ABCN adalah m e ngu f u k
, ADN dan LK D ialah gari s luru s .
Draw to full scale,
Lu k i s k an d e ngan s k ala p e nuh ,
(i) the plan of the combined solid,
p e lan gabungan p e p e jal i t u . [ 4 m ar k s ][ 4 m ar k ah ]
E
B
D
A
F
6 cm
5 cm
3 cm Y
H
K J 6 cm2 cm
Diagram 6 (ii)Rajah 6 (ii)
L
M
P
C
G N
3 cm
4 cm
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(ii) the elevation of the combined solid on a vertical plane parallel to BC asviewed from Y .
Donga k an gabungan p e p e jal i t u pada s a t ah m e n c an c ang yang s e lari d e ngan BC s e bagai m ana diliha t dari Y .
[ 5 m ar k s ]
[ 5m
ar k ah ]Answer / Jawapan :
(b) (i) , (ii)
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14 K ( 50 S , 30 E ) , L ( 50 S , 90 W ) and M are three points on the surface of the earth.KM is the diameter of its parallel of latitude.
K ( 50 S , 30 T ) , L ( 50 S , 90 B ) dan M adalah t iga t i t i k yang b e rada di p e r m u k aan bu m i . KM ia lah dia m e t e r s e larian la t i t ud .
(a) (i) State the longitude of M .
Nya t a k an longi t ud bagi M .
(ii) Find the ratio of the distance from K to L to the distance from L to M measured along the common parallel of latitude.
Cari ni s bah jara k dari K ke L ke pada jara k dari L ke M diu k ur s e panjang s e larian la t i t ud s e punya .
(iii) Calculate the shortest distance, in nautical miles, from K to M measuredalong the surface of the earth.
Hi t ung k an jara k t e rd ek a t , dala m ba t u nau t i k a , dari K ke M diu k ur s e panjang p e r m u k aan bu m i .
(iv) Calculate the distance in nautical miles from M to the east to L measuredalong the common parallel of latitude.
Hi t ung k an jara k , dala m ba t u nau t i k a , dari M ke t i m ur ke L diu k ur
s e panjang s e larian la t i t ud s e punya .
[ 8 m ar k s ][ 8 m ar k ah ]
(b) An aeroplane with speed 540 knots flew from K due west to point L. The planethen flew due north to the point T . T is 2400 nautical miles north of L.
S e buah k apal t e rbang d e ngan ke lajuan 540 k no t t e rbang dari K ke arah bara t hingga s a m pai di L. K e m udian k apal t e rbang i t u t e rbang ke arah u t ara hingga s a m pai ke T . T t e rl e t a k 2400 ba t u nau t i k a ke u t ara L.
Calculate
Hi t ung k an
(i) latitude of T .
la t i t ud bagi T .
(ii) time taken, in hours, for the whole journey
Ma s a dia m bil , dala m ja m , ke s e luruhan p e n e rbangan i t u .
[ 4 m ar k s ][ 4 m ar k ah ]
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Answer / Jawapan :
(a) (i)
(ii)
(iii)
(iv)
(b) (i)
(ii)
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15 (a) Transformation P is a reflection in the line x = 1 and transformation R isa rotation of 90 clockwise about point ( 1 , 2 )
P e nj
e l m
aan P adalah pan t ulan pada gari
s luru
s x = 1 dan p
e nj
e l m
aan Radalah pu t aran 90 i k u t arah ja m pada pu s a t ( 1 , 2 ).
(i) State the coordinates of the image of point M ( 2 , 5 ) under thecombined transformations RP
Nya t a k an k oordina t i m e j bagi t i t i k M ( 2 , 5 ) di bawah gabungan p e nj e l m aan RP.
(ii) If the point S ( 2 , 1 ) is the image of point T ( x , y ) under thetransformation PR, state the value of x and of y .
Ji k a t i t i k S ( 2 , 1 ) adalah i m e j t i t i k T ( x , y ) di bawah p e nj e l m aan PR,nya t a k an nilai x dan nilai y .
[ 4 m ar k s ][ 4 m ar k ah ]
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(ii) The ratio of the area of triangle F GH to the area of the square EFHJ is 1 : 2.Given the area of ABCD is 36 cm 2, calculate the area, in cm 2, of the triangleF GH .
Ni s bah lua s s e gi t iga FGH ke pada lua s s e gi e m pa t s a m a EFHJ adalah 1 : 2.
Dib e ri lua
s ABCD ialah 36 cm
2
, hi t ung
k an lua
s , dala
m
cm2
,s e
gi t iga F GH .[ 8 m ar k s ][ 8 m ar k ah ]
Answer / Jawapan :
(a) (i)
(ii)
(b) (i) a)
b)
(ii)
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Answer / Jawapan :
(a) Complete the table below .Le ng k ap k an jadual di bawah .
MarksMar k ah
FrequencyK eke rapan
Cumulative frequencyK eke rapan longgo k an
Upper boundaryS e m padan a t a s
30 34 0 0
35 39 2
40 44 9
45 49 13
50 54 16
55 59 12
60 64
65 69
(b) Refer graph on page 39.Ruju k gra f di hala m an 39.
(c) (i)
(ii)
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