1449-1&2 skema mat pat f4 2013

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013

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1449/2 © 2013 Hak Cipta BPSBPSK [Lihat Halaman Sebelah

SULIT

1449/1

1449/2

Peraturan

Pemarkahan

Matematik

Kertas 1 & 2Oktober

2013

BAHAGIAN PENGURUSAN

SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN

KEMENTERIAN PENDIDIKAN MALAYSIA

PENTAKSIRAN DIGNOSTIK SBP 2013

PEPERIKSAAN AKHIR TAHUN TINGKATAN 4

MATEMATIK

Kertas 1 & Kertas 2

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan Pemarkahan ini mengandungi 9 halaman bercetak

AMARAN

Peraturan Pemarkahan ini SULIT dan Hak Cipta Sekolah Berasrama

Penuh. Kegunaannya khusus untuk pemeriksa yang berkenaan sahaja.

Sebarang maklumat dalam peraturan pemarkahan ini tidak boleh dimaklumkan

kepada sesiapa. Peraturan Pemarkahan ini juga tidak boleh dikeluarkan dalam

apa jua bentuk penulisan dan percetakan.

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013


SULIT 1449/1

[Lihat Halaman Sebelah

SULIT

PAPER 1

QUESTION ANSWER QUESTION ANSWER

1 B 21 D

2 C 22 C

3 D 23 D

4 C 24 C

5 B 25 A

6 C 26 B

7 C 27 B

8 B 28 B

9 A 29 A

10 D 30 D

11 B 31 D

12 B 32 C

13 A 33 D

14 B 34 B

15 C 35 C

16 D 36 B

17 B 37 A

18 A 38 C

19 B 39 D

20 A 40 B

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013


SULIT 1449/2

Question Solution and Mark Scheme Marks

1449/2 © 2013 Hak Cipta BPSBPSK [ Lihat Halaman Sebelah

SULIT

1 (a)

(b)

P1

P2

3

212 p

2+ p – 6 = 0

( 3 p – 2 )( 4 p + 3) = 0

p =3

2, -

4

3

K1

K1

N1,N1 4

3 4m + 6n = -12 or equivalent

-7n = 28 or equivalent

n = -4, m = 3

K1

K1

N1, N1 4

4 WHV seen

HV = 22 912

15

tan WHV =15

12

38o40’/38.7

0

P1

K1

K1

N1 4

RP

Q

P

Q

R

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013


SULIT 1449/2



SULIT

5 12 × 8 × 8

3

1×

7

22 × 3.5

2 × 12

(12 × 8 × 8) – (3

1 ×7

22 × 3.52 × 12)

614

K1

K1

K1

N1 4

6 (a)False P1

(b)If B B A , then A B A

If A B A , then B B A

P1

P1

(c) n + 3n,

n = 1, 2, 3, 4…… K1 N1 5

7(a)

5

7

EF m

c )5(5

72

55

7 x y

P1

K1

N1

(b) When x = 0, y-intercept = 5

When y = 0, x-intercept =7

25

P1

P1 5

8

(a) P(digit 5 ) = 16

3

P1

(b) B = { a multiple of 5 }

= { 35, 40, 45, 50}

P( B ) =4

1

K1

N1

(c) C = {36, 41, 46 }

P ( C ) =16

3

K1

N1 5

9

(a)

147

22

2360

45

or 147

22

2360

60

or

147

222

360

90

22143

2141411

K1

K1

N1

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013


SULIT 1449/2



SULIT

Perimeter = 67.75/3

275 cm

(b) 14147

22

360

45 or 1414

7

22

360

60 or

14147

22

360

90 or 1414

1541963

210277 or 42

3

210277

Area of whole diagram = 67.221/3

2221 cm²

K1

K1

N1

6

10(a) 3

8

312

k

k = 5

K1

N1

(b)5

3ORm P1

(c) c )8(5

312

5

36c

5

36

5

3 x y or 3635 x y

P1

K1

N1

6

11. (a) (i) Some

(ii) All(b) Converse: If p > 5, then p > 10. False

(c) 5 is not a factor of 81

P1

P1

P1P1P2 6

12. (a) (i) A = {34,40,41,42,43,44,45}

B = {31,37,41,43}

(ii) {32,34,36,40,41,42,43,44,45}

(iii) 4

P1

P1

K1N1

KIN1

(b) (i) n (G’) = 40 – 15

= 25

(ii) n (G H) = 15 + 21 – 6

= 30

K1

N1

K1

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013


SULIT 1449/2



SULIT

(iii) number of students = 15 – 6

= 9

N1

K1 N1 12

13. (a) (i) (7,2)

(ii) (3,1)(iii) ( - 5 ,3)

N1

N1 N2

(b) Reflection in the line x = - 1 P3

(c) (i) Scale factor, k = 3

(ii) Area of EFGH = 32 x Area of ABCD

207 = 9 x Area of ABCD

23 = Area of ABCD

Area of AEHGFBCD = 207 - 23

= 184 cm2

K1

K1

N1

K1

N1 12

14. Class

interval

Midpoint Frequency

25 – 29 27 7

30 - 34 32 9

35 – 39 37 7

40 – 44 42 6

45 – 49 47 1

50 – 54 52 2

Class interval : I – IV

Midpoint : I – VIFrequency : I – VI

P1

P1

P2

(b)(i) 30 – 34 P1

(ii) Mean =

216797

252147642737932727

= 35.59 / 35.6

K2

N1

(c)

Axes drawn in the correct direction, the uniformscale in the range given

6 bars drawn correctly

Complete histogram with scale given

P1

K2

N1 12

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013


SULIT 1449/2



SULIT

15 . (a)Estimated mean =

100

)58(5)53(14)48(26)43(28)38(17)33(10

= 44.6

Upper Boundary

Sempadan atas

Cumulative

frequency

Kekerapan

longgokan

30.5 0

35.5 10

40.5 27

45.5 55

50.5 81

55.5 9560.5 100

Column I

Column II

Axes drawn in the correct direction, the uniform scale

for 30.5 ≤ x ≤ 60.5 and 0 ≤ y ≤ 100

Horizontal axis labelled using upper boundary or usethe values of upper boundary for plotting

8 points plotted correctly or the ogive passed through

them

The ogive completed and passed through 8 points

correctly

Any correct information

25% 0f 100 teachers =

100

25× 100

= 25= 100 – 25 = 75

x = 49 (value read from the graph)

K2

N1

P1

P2

K1

K2

N1

K1

N1

12

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013


Graph for Question 14

27 32 37 42 47 52

1

2

3

4

5

6

7

8

9

10

Frequency

Books

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013


Graph for Question 15

10

40

30

90

20

70

60

50

100

80

30.5 40.535.5 55.550.5 60.545.5

X

X

X

X

X

X

X

Cumulative frequency

Age

( years)

8/17/2019 1449-1&2 SKEMA MAT PAT F4 2013


1449-1&2 skema mat pat f4 2013

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