1449-1&2 SKEMA MAT PAT F4 2013
Transcript of 1449-1&2 SKEMA MAT PAT F4 2013
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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.
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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
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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
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Question Solution and Mark Scheme Marks
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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
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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
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(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
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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
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Graph for Question 14
27 32 37 42 47 52
1
2
3
4
5
6
7
8
9
10
Frequency
Books
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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)
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