# mathematics paper 2(set1)

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• 8/9/2019 Mathematics Paper 2(SET1)

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MATHEMATICAL FORMULAE RUMUS MATEMATIK

The following formulae may be helpful in answering the questions. The symbols given are

the ones commonly used. Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang diberiadalah yang biasa digunakan.

RELATIONS PERKAITAN

1 a m a n = a m + n

2 a m a n = a m n

3 (am)n = a m n

4 Distance / jarak = 2122

12 )()( y y x x +

5 Mid Point / Titik Tengah

(x , y) =2

21 x x + ,2

21 y y +

6 Average Speed =

Purata laju =

7 Mean =Min =

8 Pythagoras Theorem / Teorem Pithagoras

c2 = a 2 + b2

jarak yang dilalui masamasa yang diambil

distance travelledtime taken

sum of datanumber of data

hasil tambah nilai databilangan data

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SHAPES AND SPACE BENTUK DAN RUANG

1 Area of rectangle = length width Luas segiempat tepat = panjang lebar

2 Area of triangle =2

1 base height

Luas segitiga =2

1 tapak tinggi

3 Area of parallelogram = base heightLuas segiempat selari = tapak tinggi

4 Area of trapezium =2

1 sum of parallel side height

Luas trapezium =2

1 hasil tambah dus sisi selari tinggi

5 Circumference of circle = d = 2 r Lilitan bulatan = d = 2 j

6 Area of circle = r 2

Luas bulatan = j2

7 Curved surface area of cylinder = 2 rh

Luas permukaan melengkung silinder = 2

jt

8 Surface area of sphere = 4 r 2

Luas permukaan sfera = 4 j2

9 Volume of right prism = cross sectional area length Isipadu prisma tegak = luas keratan rentas panjang

10 Volume of cuboid = length width height Isipadu kuboid = panjang lebar tinggi

11 Volume of cylinder = r 2

12 Volume of cone =3

1r 2h

1 j2t

13 Volume of sphere = 34

r 3

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4 j3

14 Volume of right pyramid =3

1 base area height

luas tapak tinggi

15 Sum of interior angles of a polygon = ( n 2) 180 Hasil tambah sudut pedalaman poligon = (n 2) 180

16 =

=

17. =

=

18. Scale factor / Faktor skala, k =

19. Area of image = k 2 area of object Luas imej = k 2 luas objek

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Area of sector Area of circle

Arc lengthCircumference

Angle subtended at centre360

PA PA

Angle subtended at centre360

Panjang lengkok Lilitan bulatan

Sudut pusat

360

Luas sektor Luas bulatan

Sudut pusat

360

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Answer all questions.Jawab semua soalan For

ExaminersUse

1 Calculate the value of 5

23)

3

2+

5

31( and express the answer as a fraction in

its lowest term .

Hitung nilai bagi 52

3)32

+53

1( dan nyatakan jawapan itu sebagai satu

pecahan dalam bentuk teringkas. [2 marks ]

2 Calculate the value of 501

-103

(+6.0 ) and express the answer as decimal.

Hitung nilai bagi 501

-103

(+6.0 ) dan nyatakan jawapan itu sebagai nombor

perpuluhan.

[2 marks ] [2 markah ]

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2

1

2

2

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3 a) Find the value of 3 000125.0 Cari nlai bagi 3 000125.0

b) Calculate the value of )3-144(+4 23

Kira nlai bagi [3 marks ] [3 markah ]

4 a) Solve each of the following equations:Selesaikan tiap-tiap persamaan berikut

a) 2p = -49 5 p

b) 7=5

2+3 x

[3 marks ] [3 markah

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3

4

3

3

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5. Diagram 1 in the answer space shows that B is the image of A under areflection in the line MN..

Rajah 1 di ruang jawapan menunjukkan B adalah imej bagi A di bawah satu pantulan pada garis MN.

Draw the line MN Lukiskan garis MN tersebut

[2 marks ]

[2markah

Diagram 1Rajah 1

6. Diagram 2 shows two polygons , P and P , drawn on a grid of equal squareswith sides of 1 unit.

Rajah 2 menunjukkan dua poligon, M dan M , yang dilukis pada grid segiempat sama bersisi 1 unit.

For Examiners

Use

B

A

2

5

3

5

2

8

P

P

Diagram 2 Rajah 2 3

7

2

6

gredE 3

12

4

12

4

11

3

13

its tsuni

aj 2

17

3

16

For Examiners

Use-4 -3 -2 -1 0 1 2 xy 3

20

6

15

P Q

R

S T

U U

5

18

O 2

19

3

9

2

10

P8 cm

Number of student P RQ 6.0 cmDiagram 6

Rajah 6[6 marks ][6 markah ] R

P

ForEaminersUse

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MARKING SCHEME

ITEM MARKING SCHEME SUB-MARKS

MARKS

1.

3175

15

34

=3

2

K1

N1 2

2.0.6 +(

50

14) or 0.6 + 0.28 or 0.6 +

25

7

= 0.88

K1

N1

2

3.a) 0.05

b) 64 + ( 12 -9) or 64 + 3

67

P1

K1

N1

3

4. a) p = -7

b ) 3x = 33

x = 11

N1

K1

N1

3

5. Line Of reflection MN correct

Note:- i) Dotted line drawn award P1P2

2

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6.

Translation ( )2-5

Note:- i) Translation - award P1

ii) ( )2-

5- award P1

iii) Do not accept wrong spelling, or (5, 2), or

( )2-

5

P2

2

7.a) xs( 7x + s)

b) cd( 1 - d)

= cd( 1 d) (1 +d)

N1

K1

N1

3

8.22 6y-3xy+2xy- x

= 22 6y-xy+ x K1

N1

2

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17 cm

9 cmR

QS

x

Diagram 3 Rajah 3

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9 X 3 or x 6

3 x 6

3, 4, 5, 6 ( list the integers)

K1

K1

N1

3

10PR = 10 ( Theorem pythagoras)

Sin x =3

2K1

N12

11

Note:- i) 3 or 4 points marked correctly award K1ii) Do not accept dotted line, free hand drawing.

iii) Drawing starts from 0 award N 0

K1Uniform

scale

K25 pointsmarked

correctly.

N1Straightline

drawn to join each

point.

4

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12. Number of students taking tuition classes

Bilangan pelajar yang mengambil kelas tusyen

Mathematicmatematik

ScienceSains

English Bahasa Inggeris

represent/ mewakili 40 student/pelajar.

3

132-2 2

3

2

1

2

1

4 pq pq

2-223+

21

2 pq22 q

K1

K1

N1

3

14- Construct 30- Extend line 30- Line segment 6.0 cm- Complete triangle PQR - Construct angle bisector

(a) Angle Q = 6.3

K2K1K1K1K1

6

15 Scale correctAll point correctly plottedIf 2 point incorrect award - 1 mark Smooth line.

K1K2K1

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16

N3

17 22 2=18 T m F

m F T 29=

K1 N1 2

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18a)Circle with centre O and radius of 2 units

- Award 1 mark if the circle drawn

(i) locus of Y (ii) Locus o f Z

(a) Intersection locus of Y and Z

K1K1

N1

P2

5

19

2

3-=

4=4

64

1=4

3-2

2

n

n

n

K1 N1

2

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P Q

R

S T

U

Locus Y

Locus Z

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20 (a) Score Frequency0 111 13

2 103 8

Mode = 1 N2

N1

3

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