2010 PAT T4 MATEMATIK SKEMA

JABATAN PELAJARAN MELAKA PEPERIKSAAN SELARAS AKHIR TAHUN TINGKATAN 4, 2010

MATHEMATICS

SKEMA PEMARKAHAN

MATHEMATICS PAPER 1

1

A

11

B

21

A

31

B

2

C

12

B

22

D

32

D

3

A

13

D

23

C

33

D

4

B

14

B

24

B

34

B

5

A

15

D

25

C

35

C

6

A

16

C

26

C

36

B

7

D

17

D

27

A

37

A

8

A

18

B

28

C

38

D

9

C

19

D

29

B

39

D

10

C

20

C

30

D

40

C

MATHEMATICS PAPER 2 No 1

Peraturan Pemarkahan

Markah

(a) Q R P 1

(b)

Q R P 2

3

2

4 x  8 y  16 or equivalent

K1 3 y  3 or equivalent

K1

OR x

8  4y or x  4  2 y or 2

3 y  3

or equivalent

equivalent

K1 K1

OR   5 4  8  1    (2)(5)  (4)(4)   4 2 13 

K2

x2

N1

y  1

N1 4

No 3

Peraturan Pemarkahan

Markah

< MUN

1

tan < MUN =

2

< MUN = 15.640 or 150 39’

1

Note: NU = 25 seen award 1 mark. 4

4(a)

False

1

4(b)

If x5 = -32 then x = -2 If x = -2 then x5 = -32

1 1

4(c) 1 450

1 5

5(a)

or

1

5(b)

Some

1

5(c)

5 is greater than zero

1

n = 0, 1, 2, …

1 1

5(d)

5

6

x2 – 3x = 0 x(x – 3) = 0 x=0 x=3

1 1 1 1 4

No

Peraturan Pemarkahan

7(a)

1

1

k=7 7(b)

Markah

mPQ = 3 = (-2) + c

1 1 1

y= x+4 7(c)

0=

+4

x-intercept = -8 8

(a) x  50 120 (b) 2 y  60

1 1 7 P1

y

(c)

K1 N1

QRS  180  60 or 120

K1 z  360  120  120  40

K1

z  80

N1 6

9

1  12  9  10 2 22  22  7 7

K1

K1

1 22 2  12  9  10  2 7 2 7

K1

452

N1 4

No 10

Peraturan Pemarkahan

(a)

K1

n( B ) 2  72 3

48 (b)

N1

n( R )  20, n( B )  46

P( R) 

Markah

20 66

or

10 33

K1 N1 4

11

(a)

120 22 90 22  2   7 or  2   12 360 7 360 7 120 22 90 22  2 7   2   12  7  12  5 360 7 360 7 57

(b)

11  57.52 21

120 22 90 22 1   7  7 or   12  12 or  7  12 360 7 360 7 2 120 22 90 22 1  77    12  12   7  12 360 7 360 7 2 122

10  122.48 21

K1

K1

N1

K1 K1

N1

6

No 12

Peraturan Pemarkahan

Markah

(a) 57

P2

Note: 55 – 59 , give P1 (b) Frequency Mass (g)

Upper Boundary

i

Cumulative Frequency

I

40 – 44

12

44.5

12

II

45 – 49

20

49.5

32

III

50 – 54

32

54.5

64

55 – 59

48

59.5

112

60 – 64

44

64.5

156

65 – 69

26

69.5

182

70 – 74

14

74.5

196

75 – 79

4

79.5

200

IV V VI VII VIII

Upper boundary: ( I until VIII )

P1

Cumulative frequency: ( I until VIII )

P2

(c) Graph: Uniform scale and correct axis

P1

Plot all 9 points correctly

K2

Smooth curve

N1

(d) (i) 166 (ii) 63.5 -52.5 = 11 Note: Do not accept answers without an ogive.

P1 K1 N1 12

No 13

Peraturan Pemarkahan

Markah

(a) Mass (kg)

Frequency

Midpoint

Jisim (kg)

Kekerapan

Titik Tengah

0

37

5

42

I II

(b)

35 – 39 40 – 44

III

45 – 49

7

47

IV

50 – 54

18

52

V

55 – 59

13

57

VI

60 – 64

4

62

VII

65 – 69

2

67

VIII

70 – 74

1

72

IX

75 – 79

0

77

Class Intervals: ( III until IX )

P1

Frequency:

( I until IX )

P2

Midpoint:

( I until IX )

P1

(42  5)  (47  7)  (52  18)  (57  13)  (62  4)  (67  2)  (72  1) 50

=

K2

2670 50

N1 = 53.4

No

Peraturan Pemarkahan

Markah

(c) Graph: Uniform scale and correct axis,

P1

Plot all 9 points correctly using midpoint or class interval as x – axis

K2

Correct polygon.

N1

(d) Modal class is 50 – 54

P1

OR Any correct information from the frequency polygon 12

Note: Do not accept answer without a frequency polygon. 14(a)

14(b)

14(c)

(i) 4n(2m + 3)

1

(ii) (k – 5)(k – 2)

1

(i) {23, 29, 31}

1

(ii) {30, 31, 35} Note: = {20, 23, 25,29, 30, 31, 35} seen award 1 mark

2

(iii) 2 Note:

2 = {23, 29} seen award 1 mark 2

(i) (ii) mPQ = -3

1

6 = -3(1) + c

1

y = -3x + 9

1 12

No

Peraturan Pemarkahan

Markah 1

15(a) 1

2 15(b)

4 tan 260 or 1.951

1 2

tan x =

1 0

0

x = 65.92 or 65 55’ 15(c)

1