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Kaedah-kaedah yangbetul

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1. Using matrices, calculate

value of x and y

3x 4y= -15x 6y= 2

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3x 4y= -15x 6y= 2

15x 20 = -515x 18y= 6

- 2y= -11

y= 112

3x 4(5.5) = -1x= 7

Mark = 0

Penyelesaian tidakmenggunakankaedah matriks

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3x 4y= -15x 6y= 2 Matrices form

2

1

65

43

y

x

Inverse matrices

2

1

35

46

5)4()6(3

1

y

x

Answers

x =7y = 11

2

Formula

Mark = 1

Marks = 2

Mark = 1

2315

2416

2

1

y

x

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3x 4y= -15x 6y= 2

Inverse matrices

2

1

35

46

)5)(4()6(3

1

y

x

3 marks

Answers

x =7

y = 11

2

Tidak menulisbentuk matriks

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3x 4y= -15x 6y= 2

2

1

65

43

y

x

Matrices form

3 marks

Answers

x =7

y = 11

2

2

1

35

46

5)4()6(3

1

y

x

Wrong method

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3x 4y= -15x 6y= 2

2

1

65

43

y

x

Matrices form

1 mark

Answers

x =7

y = 11

2

35

46

2

1

5)4()6(3

1

y

x

Wrong method

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3x 4y= -15x 6y= 2

Inverse matrices

2

1

35

46

)5(43)6(

1

y

x

2

1

65

43

y

x

Matrices form

1 markAnswers

x =7

y = 11

2

wrongmethod

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JAWAPAN AKHIR TIDAK

LENGKAP

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3x 4y= -15x 6y= 2

Inverse matrices

2

1

35

46

)4(5)6(3

1

y

x

2

1

65

43

y

x

Matrices form

Answers

2

117

y

x

3 marks

Jawapan akhirtidak lengkap

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Mengenal pasti pemberian

markah pada langkah-langkahtertentu

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21

52equationquadratictheSolve2.

2

m

mm
http://../teknik%20menjawab/quadratic%20equation.gsp

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21

52 2

m

mm

)1(252

2

mmm2252 2 mmm

0232 2 mm0)2)(12( mm

2

1m 2m

Mark = 1

012 m 02 m

02252 2 mmm

Mark = 1

Mark = 1

Mark = 1

12 m

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21

52 2

m

mm

)1(252

2

mmm

2252 2 mmm

02322

mm

2

1m

2m

Mark = 2

Tidakmenggunakankaedahpemfaktoran ataukaedah rumus

2

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21

52 2

m

mm

)1(252

2

mmm2252 2 mmm

0232 2 mm

0)2)(2

1( mm

2

1m 2m

02

1 m 02 m

02252 2 mmm

Marks= 3

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Lakukan semakan semula

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243

1322

1

qp

qp

3. Calculate the value of pand qthat

satisfy the following simultaneouslinear equations:
http://../teknik%20menjawab/simultanous%20equation.gsp

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243

1322

1

qp

qp

282 p

243

264

qp

qp

14p

26414 q

10q

Mark = 1

14264 q

404 q

82

)10(4)14(3

243 qp

semakan

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243

1322

1

qp

qp

244 p

243

264

qp

qp

6p

2646 q

5q

Markah penuh = 4

1 mark

1 mark

1 mark

1 mark

6264 q

204 q

420

q

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243

1322

1

qp

qp

6p

5q

Markah = 0

2)5(4)6(3

13)5(2)6(2

1

Tidak bolehmenggunakan kaedah

cuba jaya

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Berhati-hati denganjawapan anda

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Express

mn

n

m 15

75

5

1 as a single fraction in its simplest form.

mnn3

12

mn

n

1545

mnn3

12 mn

n15

110

A

D

C

B

4.

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mn

n

15

45

mn

nn

15

573

mn

n

mn

n

15

75

15

3

mn

n

m 15

75

5

1

(3n)

(3n)

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mn

n

3

12

mn

n

15

)12(5

mn

n

15

510

mn

nn

15

573

mn

n

mn

n

15

75

15

3

mn

n

m 15

75

5

1

(3n)

(3n)

1

3

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Penggunaan rumus-rumus yang

bersesuaian

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4 cm

The diameter of cylinder is 7 cm and the volume ofthe solid is 231 cm2. Calculate the height of the cone,in cm.

5.

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4 cm

The volume of cylinder:

hr2

The diameter ofcylinder is 7 cm andthe volume of thesolid is 231 cm2.Calculate the height

of the cone, in cm.

4

2

7

7

222

1 mark

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4 cm

The volume of cone

The diameter of cylinderis 7 cm and the volume ofthe solid is 231 cm2.Calculate the height ofthe cone, in cm.

hr2

3

1

h

2

2

7

7

22

3

1

1 mark

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The volume of solid:

42

7

7

22 2

h

2

2

7

7

22

3

1=231

The height of cone:

h = 61 mark

1 mark

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Langkah-langkah kerja yang

lengkap dan betul

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427

722

2

h

2

2

7

7

22

3

1

The volume of cylinder:

The volume of cone :

The height of cone:

h = 6

Markah = 2

Tinggalkan langkahmengira isipadu pepejal

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Bundarkan jawapankepada sekurang-

kurangnya 2 tempatperpuluhan atau

4 angka bererti

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35

H

E

GF

I

O

22

7

In Diagram , OFIE is a quadrant of a circle with centre O

and GH is an arc of another circle with centre O. OFG andOIH are straight lines. OF = FG = 14 cm, and

Using

calculatea) the perimeter, in cm, of the whole diagramb) the area, in cm2, of the shaded region.

35GOH

6.

H

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35

H

E

GF

I

O

227

In Diagram , OFIE is a quadrant of a circle withcentre O and GH is an arc of another circle withcentre O. OFG and OIH are straight linesOF = FG = 14 cm, and

Using

calculate a) the perimeter, in cm, of the whole diagram

35GOH

Perimeter = EI + HG + OE + IH + GO

360

5514

7

222

360

3528

7

222 + 14 + 14 + 28

= 86.6

2 marks

0 mark

H

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35

H

E

GF

I

O

227


Using


35GOH


360

5514

7

222

360

3528

7

222 + 14 + 14 + 28

= 86.56

2 marks

1 mark

H

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35

H

E

GF

I

O

227


Using


35GOH


360

5514

7

222

360

3528

7

222 + 14 + 14 + 28

2 marks

1 mark9586

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Satu jawapan sahajasebagai jawapan akhir

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35

H

E

GF

I

O

227

In Diagram , OFIE is a quadrant of a circle withcentre O and GH is an arc of another circle withcentre O. OFG and OIH are straight lines. OF = FG= 14 cm, and

Using

calculateb) the area, in cm2, of the shaded region.

35GOH

The area of the shaded region= OEI + OHG OIF

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35

H

E

GF

I

O

227


Using


35GOH

Area of the shaded region = OEI + OHG OIF

360

5514

7

22 2

360

3528

7

22 2 360

3514

7

22 2

= 273

2 marks

0 mark

97273

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35

H

E

GF

I

O

227


Using


35GOH


360

5514

7

22 2

360

3528

7

22 2 360

3514

7

22 2

2 marks

1 mark9

7273

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35

H

E

GF

I

O

227


Using


35GOH


360

5514

7

22 2

360

3528

7

22 2 360

3514

7

22 2

= 273.78

2 marks

1 mark

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Jawapan akhir bentuk

pecahan hendaklah dalambentuk termudah

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0 10 15 27

16

Speed (ms-1

)

Time (s)

7. Calculate the rate of change of speed, in ms-2 , of the particlein the last 12 seconds

1527

160

1216

1 mark

0 mark

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0 10 15 27

16

Speed (ms-1

)

Time (s)

Calculate the rate of change of speed, in ms-2 , of the particle inthe last 12 seconds

1527

160

34

1 mark

1 mark

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0 10 15 27

16

Speed (ms-1

)

Time (s)


1527

160

33.1

1 mark

1 mark

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0 10 15 27

16

Speed (ms-1

)

Time (s)


1527

160

3.1

1 mark

0 mark

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0 10 15 27

16

Speed (ms-1

)

Time (s)


1527

016

34

0 mark

0 mark

Diagram shows two boxes, P and Q. Box P contains four cards labelled8

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List all the possible outcomes,

Sample space, S ={

A

B

C

D

3 4 5

(A,4)(A,3) (A,5)

Box P

Box Q

1 mark

Diagram shows two boxes, P and Q. Box P contains four cards labelledwith letters and box Q contains three cards labelled with numbers.

A B C DBox P

3 4 5Box Q

(B,4)(B,3) (B,5)

(C,4)(C,3) (C,5)

(D,4)(D,3) (D,5) }

Two cards are pickedat random, a card from

box P and another cardfrom box Q.

8.

Diagram shows two boxes, P and Q. Box P contains four cards labelled

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Sample space, S ={ (A,4)(A,3) (A,5)

Diagram shows two boxes, P and Q. Box P contains four cards labelledwith letters and box Q contains three cards labelled with numbers.

A B C DBox P

3 4 5Box Q

(B,4)(B,3) (B,5)

(C,4)(C,3) (C,5)

(D,4)(D,3) (D,5) }

(b)Hence, find the probability that

(i) a card labelled with letter B and a card labelled with an odd number.

{ (B,3) (B,5) }

Probability =212

(ii) a card labelled with letter B or a card labelled with an odd number.

(B,4){(B,3) (B,5) (A,3) (A,5) (C,3) (C,5) (D,3) (D,5) }

Probability =912

Two cards are pickedat random, a card from

box P and another cardfrom box Q.

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Jawapan yang tepat danjelas

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a) Write down two implications based on the following sentence

P q > 0if and only if p > q

9.

b) Make a general conclusion

7 = 3 (2)1 + 1

14 = 3 (2)2 + 1

27 = 3 (2)3 + 1

=

2 marks

2 marks

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p q > 0if and only if p > q

SPM 2004

i) If p

q > 0the p > q

ii) Ifp > q the p q > 00 mark

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p q > 0if and only if p > q

SPM 2004

i) p

q > 0if and only if p > q

ii) p > q If and only if p q > 0

0 mark

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7 = 3 (2)1 + 1

14 = 3 (2)2

+ 127 = 3 (2)3 + 1

=

3 (2)n + 1 , n = 1, 2, 3.

1 mark

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7 = 3 (2)1 + 1

14 = 3 (2)2

+ 127 = 3 (2)3 + 1

=

3 (2)n+ 1 , n = 1, 2, 3,

2 marks

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Jawab semua soalan

10.

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KLRM is the image of ABCD under the combined transformations WU.Describe in full, the transformation U and the transformation W.

U is rotationW is enlargement 2 marks

KLRM is the image of ABCD under the combined transformations

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gWU.Describe in full, the transformation U and the transformation W

U is rotation at (0,1)W is enlargement of scale factor 3

4 marks


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4 marks


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4 marks

10.

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KLRM is the image of ABCD under the combined transformations WU.Describe in full, the transformation U and the transformation W.

U is rotation of 90 clockwise at (0,1)W is enlargement of scale factor 3 at (2,3)

6 marks
http://f/ceramah/ceramah%20okt%202010/transformation.gsp

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Lukisan bukan lakaran dan

gunakan peralatan yangbersesuaian

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11.

Draw the elevation of the

solid as viewed from P
http://f/ceramah/ceramah%20okt%202010/Plan%20&%20elevation2.gsp

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8 cm

5 cm

4 cm

3 cm

1 mark

sketch

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4 marks(full mark)

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5 cm

3 cm

8 cm 0 mark

Songsang sisi

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8 cm

5 cm

4 cm

3 cm

0 mark

SPM 2007

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SPM 2007

Draw the elevation of thesolid as viewed from Q

4 cm
http://f/ceramah/ceramah%20okt%202010/Plan%20&%20elevation2.gsp

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5 marks

7 cm

7 cm

4 cm

1 cm

4 cm

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1 mark

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No 2

M k F C l i

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Mark Frequency Cumulativefrequency

11 15 1

16 20 3

21 25 6

26 30 10

31 35 11

36 40 7

41 45 2

The table shows the frequency distribution of mark of 40 students in ageography test.

(a) Complete the table(b) Using a scale of 2 cm to 5 marks on x-axis and 2 cm to 5 students on y-axis,

draw an ogive to illustrate the data.(c) Use your ogive to estimate the

(i) upper quartile,(ii) hence, explain briefly the meaning of the upper quartile.

M k F C l ti

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11 15 1

16 20 3

21 25 6

26 30 10

31 35 11

36 40 7

41 45 2

1

4

10

20

31

38

40

+ 3

+ 6

+ 10

+ 11

+ 7

+ 2

The table shows the frequency distribution of mark of 40 students in ageography test.

(a) Complete the table(b) Using a scale of 2 cm to 5 marks on x-axis and 2 cm to 5 students on y-axis,

draw an ogive to illustrate the data.(c) Use your ogive to estimate the

(i) upper quartile,(ii) hence, explain briefly the meaning of the upper quartile.

Mark Frequency Cumulativefrequency 1 mark

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11 15 1 1

16 20 3 4

21 25 6 10

26 30 10 20

31 35 11 31

36 40 7 38

41 45 2 40

10.5 15.5 20.5 25.5 30.5 35.5 40.5 45.5

Mark

C

umulativefrequency

0

5

10

15

20

25

30

35

40

2 marks

1 mark

c) Use your ogive to estimate the(i) upper quartile

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10.5 15.5 20.5 25.5 30.5 35.5 40.5 45.5

Mark

C

umulativefrequency

0

5

10

15

20

25

30

35

40

(i) upper quartile,(ii) hence, explain briefly the

meaning of the upper quartile.

(i) Upper quartile

30404

3

Answer

35(ii)

30 students scored lessthan 35 marks.

or

10 students scoredmore than 35 marks.


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10

5

0

15

20

25

30

35

40

10.5 15.5 20.5 25.5 30.5 35.5 40.5 45.5

Lengkung tidak licin

11 15 1 1

16 20 3 4

21 25 6 10

26 30 10 2031 35 11 31

36 40 7 38

41 45 2 40

C

umulativefrequency

Mark

(a) Using a scale of 2 cm to 5 markson x-axis and 2 cm to 5 studentson y-axis, draw an ogive toillustrate the data.


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10

5

0

15

20

25

30

35

40

10.5 15.5 20.5 25.5 30.5 35.5 40.5 45.5

Lengkung tidak melalui titik

11 15 1 1

16 20 3 4

21 25 6 10

26 30 10 2031 35 11 31

36 40 7 38

41 45 2 40

C

umulativefrequency

Mark

(b) Using a scale of 2 cm to 5 markson x-axis and 2 cm to 5 studentson y-axis, draw an ogive toillustrate the data.


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10

5

0

15

20

25

30

35

40

10.5 15.5 20.5 25.5 30.5 35.5 40.5 45.5

Lengkung mempunyaidua garis

C

umulativefrequency

11 15 1 1

16 20 3 4

21 25 6 10

26 30 10 2031 35 11 31

36 40 7 38

41 45 2 40

Mark



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10

5

0

15

20

25

30

35

40

10.5 15.5 20.5 25.5 30.5 35.5 40.5 45.5

Tidak boleh melukis garis lurus

C

umulativefrequency

11 15 1 1

16 20 3 4

21 25 6 10

26

30 10 2031 35 11 31

36 40 7 38

41 45 2 40

Mark



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10

5

0

15

20

25

30

35

40

8.0 13.0 18.0 23.0 28.0 33.0 38.0 43.0Tidak boleh menanda titik tengah

C

umulativefrequency

11 15 1 1

16 20 3 4

21 25 6 10

26 30 10 20

31 35 11 31

36 40 7 38

41 45 2 40

Mark


a) Complete Table 2 in the answer space for the equation y= -2x + 5x+ 8

No.15
http://statistics.gsp/

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a) Co p ete ab e t e a s e space o t e equat o y 5 8by writing down the values of ywhen x= -2 and x= 3 [2 marks]

Table 2

x -3.5 -3 -2 -1 0 1 2 3 4

y -34 -25 1 8 11 10 -4

b) By using a scale 2 cm to 1 unit on the x axis and 2cm to 5 unit on theyaxis , draw the graph of y= -2x + 5x+8 for -3.5 x 4 [4 marks]

c) From the graph, find(i) the value of ywhen x= 3.3(ii) the value of xwhen y= -16 [2 marks]

d) Find and draw a suitable straight line on your graph to determine the valuesof x which satisfy the equation 2x 2x = -10 for -3.5 x 4. State thevalues of x [4 marks]

a) Complete Table 2 in the answer space for the equation y= -2x + 5x+ 8

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by writing down the values of ywhen x= -2 and x= 3 [2 marks]

Table 2

x -3.5 -3 -2 -1 0 1 2 3 4

y -34 -25 1 8 11 10 -4

y= -2x + 5x+ 8

y= -2(-2) + 5(-2)+ 8

= -10

y= -2(3) + 5(3)+ 8

= 5

-10 5

y10b) By using a scale 2

cm to 1 unit on the

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-1-3 -2 10 2 3 4-4x

-35

-30

-25

-20

-15

-10

-5

5

cm to 1 unit on thex axis and 2 cm to5 unit on the y axis, draw the graph of

y= -2x + 5x+8 for-3.5 x 4[4 marks]

y10x -3.5 -3 -2 -1 0 1 2 3 4

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-1-3 -2 10 2 3 4-4x

-35

-30

-25

-20

-15

-10

-5

5y -34 -25 -10 1 8 11 10 5 -4

y10c) From the graph, find

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-1-3 -2 10 2 3 4-4x

-35

-30

-25

-20

-15

-10

-5

5(i) the value of y when x = 3.3

(ii) the value of xwhen y= -16

[2 marks]

(i) When x=3.3, y = 2.5

(ii) When y=-16, x = -2.5

Onthe graph, shade the region which satisfy the three

i liti 6 2 4 d 1

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inequalities y - x + 6 , y 2x - 4 and x > 1

y= - x + 6

y=2x- 4

y

x

6

60

-4

2

3 marks


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x = 1


y= - x + 6

y=2x- 4

y

x

6

60

-4

2

1 mark


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y= - x + 6

y=2x- 4

y

x

6

60

-4

2

1 mark


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y= - x + 6

y=2x- 4

y

x

6

60

-4

2

1 mark


inequalities y x + 6 y 2x 4 and x > 1

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y= - x + 6

y=2x- 4

y

x

6

60

-4

2

3 marks

No.3
http://f/ceramah/ceramah%20okt%202010/linear%20inequalities.gsp

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Shade the P Q

SPM 2006

SPM 2006
http://../teknik%20menjawab/set.gsp

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Shade the ( P U Q) R
http://../teknik%20menjawab/set.gsp

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2. In Diagram , O is the origin, point R lies on the x-axis and point P lies onthe y-axis. Straight line PU is parallel to x-axis and the straight line PR is

parallel to straight line ST. The equation of straight line PR is x + 2y = 14.

(a) State the equation of the straight line PU.

(b) Find the equation of the straight line ST and

hence, state its x-intercept.

R xO

T(2,-5)

S

U

x + 2y = 14

y


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the y axis. Straight line PU is parallel to x axis and the straight line PR isparallel to straight line ST. The equation of straight line PR is x + 2y = 14.

(a) State the equation of the straight line PU.

RxO

T(2,-5)

S

U

x + 2y = 14

y

Substitute x=0 into x + 2y = 14

x + 2y = 14

0 + 2y = 14

y = 7


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t e y a s St a g t e U s pa a e to a s a d t e st a g t e sparallel to straight line ST. The equation of straight line PR is x + 2y = 14.

(b) Find the equation of the straight line ST and hence, state its x-intercept.

Gradient of ST= gradient of PR

R xO

T(2,-5)

S

U

x + 2y = 14

y

P

x + 2y = 14

72

x

y

2

1ST ofgradient

1ST ofgradient

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c 2

2

5

2fg

cmxy

4c

42

xyisSTofequationthe

cxy 2

4x

yisSTofequationthe

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8x

42

yisSTofequationthe

42

0 x

42

into0y xysubstitute

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b) Given the matrix equations

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b) Given the matrix equations

1

4

85

67

y

x 8 41

5 7 1

x k

y h

and

i) Find the value of h and k

ii) Hence, find the value of xand y.

i) Find the value of h and k

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1

4

85

67

y

x

1

4

75

68

)5)(6()8(7

1

y

x

1

4

75

68

26

1

1

4

75

81 k

h

h = 26 , k = 6

ii) Hence, find the value of xand y.

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1

4

75

68

26

1

y

x

)1(7)4(5

)1(6)4(8

26

1

13

26

26

1

2

1

1

y

x

6

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EKLMN is the image of EFGHJ under transformation V and EKLMN is the image of

KPQRS under transformation W. Describe in full, the transformation V and thetransformation W.Given that the area of EFGHJ is 26.8cm2 , calculate the area of shaded region

6

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EKLMN is the imageof EFGHJ under

transformation V andEKLMN is the imageof KPQRS undertransformation W.Describe in full, thetransformation V andthe transformation W.

V = enlargement of scale factor 2 at ( 7, 2 )

W = rotation of 90 clockwise at ( 1, 2 )

6

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EKLMN is the image ofEFGHJ under transformation

V and EKLMN is the image ofKPQRS under transformationW. Given that the area ofEFGHJ is 26.8cm2 , calculatethe area of shaded region

Area of image = k2 x area of object

EKLMN = 22 x EFGHJ

EFGHJ + x= 4 x EFGHJ

26.8 + x= 4 x 26.8

x= 80.4

Let the area of the shaded region = x

T 7 I th di b l KLM i

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LK

T

N

M

78035 0x

7. In the diagram below, KLM is

a tangent to the circle with centre

O, at L. TNM is a straight line,

find the value ofx.

43

35102180102

78180

78

x

LNM

TLKTNL

K 8. In the diagram below , KLM is a

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4 cmx

U

LM 8 cm

right-angled triangle.Given thatKUL is a straight line and

, find the value ofcos x 5

4

tan MKL

6

10KL

5

48

5

4tan

UL

KL

MKL

5

3

10

6

MULcosxcos

1086 22

MU

10. In the diagram below,NOS

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g

is the axis of the earth.A,B and C

are three points on the surface of

the earth.AB is the diameter of theearth. The longitude of point Cis
http://earth.gsp/

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SOALAN-SOALANBENGKEL

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4. The diagram shows a speed-time graph of thet f ti l f i d f 35

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movement of a particle for a period of 35 s.

10 35Time (s)

Speed (m s-1)

20

12

t0

(a) State the uniform speed,in ms-1, of the particle.

(b) Given the distance travelled bythe particle at a uniform speed is250m,

calculate(i) the value of t,(ii) the average of speed, in ms-1,of the particle for the period of 35seconds.

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10 35Time (s)

Speed (m s-1)

20

12

t0

(a) State the uniform speed,in ms-1, of the particle.

20 m s-1

Speed (m s-1) (b) Given the distance travelledby the particle at a uniform speed

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10 35Time (s)

20

12

t0

y p pis 250m,

t 10 x 20 = 250

t 10 = 12.5

t = 22.5

calculate(i) the value of t

Speed (m s-1) (b) Given the distance travelled by the particleat a uniform speed is 250m,

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10 35Time (s)

20

12

t0

Calculate(ii) the average of speed, in ms-1,of the particle for

the period of 35 seconds.

takentime

travelleddistancespeedAverage

35

205.122

125010)2012(

2

1

29.15

a) Calculate the angleNo.17

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S R

QP

C

B

D

A

D C

RS

B

between the plane BRS

and the plane CDRS

b) Calculate the

angle between the

line BS and the

plane PQRS

4 cm

5 cm

12 cm

9. In the diagram below, EF is avertical flag pole FG is horizontal.
http://angle%202%20planes.gsp/

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E

F G59 m

The angle of depression of G fromE is 32 0. The height, in m, of the

flag pole is

87.36EF

5932tanEF

5932tan

EF

320

320

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Polygons I and II

TOPIK-TOPIK TINGKATAN 1 - 3

Paper 1

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Polygons I and II

Transformations I and II

Algebraic Expressions II and III

Algebraic FormulaeLinear Equations I and II

Indices

Linear Inequalities

Statistics I and II

Circles I and II

Solid Geometry I and III

Paper 1

Paper1Paper1 and 2

Paper1

Paper1and 2

Paper1

Paper1

Paper 1

Paper1

Paper2

TOPIK-TOPIK TINGKATAN 4

St d d F Paper 1

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Standard Form

Sets

The straight line

Probability 1

Circles III

Trigonometry II

Angles of Elevation and Depression

Lines and Planes in 3-Dimension

Paper 1

Paper 1 and 2

Paper1 and 2

Paper1

Paper1

Paper1

Paper1

Paper1 and 2

Quadratics Expression & Equations Paper 1 and 2

Mathematical Reasoning

Statistics Paper2

Paper2

TOPIK-TOPIK TINGKATAN 5

Number Bases Paper 1

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Graphs of Function II

Matrices

Variations

Bearing

Earth As A Sphere

Gradient & AreaUnder A Graph

Probability II

Transformations III

Paper1 and 2

p

Paper2

Paper1 and 2

Paper1

Paper2

Paper1 and 2

Paper1

Paper1 and 2

Plan & Elevation Paper2

No.5

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5 marks

First box Second box

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List all the possible outcomes,

Sample space, S ={

A

2

B

C

D F3 E 4 G

(A,3)(A,D) (A,G)(A,E) (A,4) (A,F)

(C,3)(C,D) (C,G) }(C,E) (C,4) (C,F)

(2,3)(2,D) (2,G)(2,E) (2,4) (2,F)

(B,3)(B,D) (B,G)(B,E) (B,4) (B,F)

Firstbox

Second box

1 mark

Sample space, S ={ (A,3)(A,D) (A,G)(A,E) (A,4) (A,F)

(2,3)(2,D) (2,G)(2,E) (2,4) (2,F)

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A

2

B

C

D F3 E 4 G

(C,3)(C,D) (C,G)(C,E) (C,4) (C,F)

(B,3)(B,D) (B,G)(B,E) (B,4) (B,F)

Firstbox

Second box

By listing the outcomes, find theprobability that(a) Both cards are labelled with anumber

The outcomes is (2,3), (2,4).

The probability is24

2

1 mark

1 mark

1 mark

Sample space, S ={ (A,3)(A,D) (A,G)(A,E) (A,4) (A,F)

(2,3)(2,D) (2,G)(2,E) (2,4) (2,F)

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A

2

B

C

D F3 E 4 G

(C,3)(C,D) (C,G)(C,E) (C,4) (C,F)

(B,3)(B,D) (B,G)(B,E) (B,4) (B,F)

Firstbox

Second box

By listing the outcomes, find theprobability that(b) One card is labelled with anumber and the other card islabelled with a letter.

The outcomes is (A,3),(A,4),(2,D),(2,E),(2,F),(2,G)(B,3),(B,4),(C,3),(C,4).

The probability is

24

10

1 mark

1 mark

1 mark

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5 marks

242

62

41 a)

b)

2410

62

43

64

41

2 marks

2 marks

1Simplify

No.9

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22

1

1 )4( nm Simplify

22

1

1 )4( nm Simplify

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22

1

214

nm

)4( nmSimplify

nm 24

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The data in Diagram 1 shows the marks obtained by 42 studentsin a Mathematics final exam.

51 20 45 31 26 40 30

No.13

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51 20 45 31 26 40 3048 32 37 41 25 36 3846 38 28 37 39 23 39

33 35 42 29 38 31 4942 34 26 35 43 42 2226 47 40 48 44 34 54

Diagram 1

(a) Based on the data in Diagram 1 and using a class interval 5 marks,complete Table 1 in the answer space. [4 marks]

(b) From the table in a),(i) state the modal class,(ii) calculate the mean mark for the Mathematics final exam and give

your answer correct to 2 decimal places. [4 marks](c) By using a scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1student on the vertical axis, draw a histogram based on the data. [4 marks]

Marks Frequency Midpoint

20-24

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0

25-29

No.14
http://f/ceramah/ceramah%202010/statistics.gsp

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Draw the plan of the solid
http://plan%20%26%20elevation2.gsp/

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3 marks(full mark)

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2 marks

extension

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0 mark

Extension is > 0.4 cm

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2 marks

Small gap

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2 marks

> 91

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0 mark

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1 mark

Soalan 4 SPM 2002

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DA

C(6, -5)

B(9, 10)

y

O x

Find the gradient of the straight of BC

B(9, 10)

y

yy

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DA

C(6, -5)

O x

12

12

xx

yy

69

)5(10

5

Jawapan

Markah penuh = 2


B(9, 10)

y

yy

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DA

C(6, -5)

O x

12

12

xx

yy

69

)5(10

1

5

Jawapan

Markah = 1

Tidak meringkaskanjawapan akhir


Soalan 6 SPM 2004y

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R(4, 12)

x

Q

P(3, -6)

O

OPQR is the parallelogram and O is the origin, find the y-interceptof the straight line PQ

R(4, 12)

y

m =04012

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x

Q

P(3, -6)

O

cmxy

c

3

04

0126

15c

Markah penuh = 3

OPQR is the parallelogram and O is theorigin, find the y-intercept of the straightline PQ

R(4, 12)

y

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x

Q

P(3, -6)

O

cmxy

c

3

04

0126

15c

Markah = 2

)15-,0(isintercept-ythe

Kesalahan keranajawapan akhir yang salah

OPQR is the parallelogram and O isthe origin, find the y-intercept of thestraight line PQ

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Indices

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Inverse Matrices

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Gradient of the straight line

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7 cm

4 cm

6 cm

3 cm

8 cm

5 cm

Q

4 cm

Jawapan akhir mestilah dalambentuk teringkas
http://../teknik%20menjawab/Plan%20&%20elevation.gsp

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be tu te g as

Salah Betul

3-2

_ 32

31

3

1

2

2

4

434

=34

10Jawapan yang sama

Jawapan akhir mestilah betul kepada sekurang-kurang dua tempat perpuluhan atau empat angka

bererti

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bererti

Salah Betul

=13

21 21.33=13

21 21.3

=213.19 213.2 (4 s.f.)=213.19 213 (3 s.f.)

Jawapan akhir dalam darjah danminit

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=213.670 213.70(1 t.p.)

=21031.3 21.31 (minit yang hampir)

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Kalkulator saintifik

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SMK KG RAJA 2011