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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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[Lihat s e be lah1449 / 1 2008 Hak Cipta Bahagian Pendidikan & Latihan (Menengah) MARA SULIT

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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[Lihat s e be lah1449 / 2 2008 Hak Cipta Bahagian Pendidikan & Latihan (Menengah) MARA S U L I T

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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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

For

U s e

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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

For

U s e

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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

For

U s e

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Answer / Jawapan :

(a)

(b) Premise 2 / Pr e m i s 2 :

(c)

For

U s e

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Answer / Jawapan :

(a)

(b)

(c)

For

U s e

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Answer / Jawapan :

(a)

(b)

For

U s e

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Answer / Jawapan :

(a)

(b)

(c)

For

U s e

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Answer/ Jawapan :

For

U s e

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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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G raph for Que s tion 16Graf un t uk Soalan 16

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