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SEKTOR SEKOLAH BERASRAMA PENUH
BAHAGIAN SEKOLAH
KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN AKHIR TAHUN
TINGKATAN 4, 2006
MARKING SCHEME
MATHEMATICS
PAPER 1 & PAPER 2
NOTE
This marking scheme serves as a guide only. Other method which are equivalent and
suitable may also be adapted and used in the awarding of marks.
Paper 1: 40 marks
x
Paper 2: × 60
100
Total Mark: P1 + P2
THANK YOU
This marking scheme consists of 7 printed pages
PEPERIKSAAN AKHIR TAHUN INGKATAN 4, SBP 2006
SKEMA JAWAPAN MATEMATIK KERTAS 1

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Q Answer Q Answer
1. B 21. D
2. B 22. D
3. C 23. D
4. A 24. A
5. A 25. B
6. C 26. C
7. A 27. B
8. A 28. D
9. A 29. D
10. B 30. C
11. C 31. D
12. C 32. C
13. D 33. C
14. A 34. C
15. B 35. A
16. C 36. C
17. D 37. B
18. C 38. B
19. D 39. A
20. C 40. A

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PEPERIKSAAN PERTENGAHAN TINGKATAN 4, SBP 2006
SKEMA JAWAPAN MATEMATIK KERTAS 2
Sub Full
Q Marking Scheme
mark mark
1 4p2 – 3p – 10 = 0 P1
(4p + 5)(p – 2) = 0 K1
5
p= − ,2 N1N1 4
4
2 p + 4q = 26 or equivalent K1
4p = 24 or equivalent K1
p=6 N1
q=5
N1 4
3 ∠PTQ or ∠QTP P2
9
tan ∠PTQ = K1
12
∠PTQ = 36 52’ or 36.9
0 0 N1 4
4 (a) –2 P1
(b) -2 = -2(4) + c K1
c=6 K1
y = – 2x + 6 N1
y-intecept = 6 P1 5
5 (a) 15 P1
or equivalent
50
(b) 1
× 50
5 K1
10 N1
(c) 10 + 10
60 K1
20
or equivalent N1 5
60
6 1 22
V= × × 8 × 12 × 15 K1
3 7
V = 480 N1
22 2
× 7 × h = 480 K1
7
240 4
h= or 3.12 N1
77

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Sub Full
Q Marking Scheme
mark mark
7 (a) 90 22 60 22
A1 = × × 14 × 14 and A2 = × ×7×7 K1K1
360 7 360 7
A1 – A2 K1
1
128 N1
3
(b) 90 22 180 22
P1 = × 2 × × 14 or P2 = ×2× ×7 K1
360 7 360 7
P1 + P2 + 14 K1
58 N1 7
8 (a) Statement. False statement P1P1
(b) 3 is not a factor of 15. False P1P1
(c) Some triangles have a right angle P1 5
P P1
9 (a) Q
R
(b) P P2
Q
R
3
10 (a) 110 P1
(b) 20 P2
(c) 40 P2 5
11 (a) (i) – 0.91 P1
(ii) –0.42 – 1 K1
–1.42 N1
(c) (i) 20 N1
(ii) 5
13 N2 6
12 (a) (i) x
y= −2 K1
2
N1
y-intercept = – 2
(ii) 2m = 8 – 4 K1
m=2 N1
(iii) 1
mPQ= mRS = P1
2
1 K1
y – 5 = (x – 4)
2
2y = x + 6 N1

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Sub Full
Q Marking Scheme
mark mark
12 (b) (i) y – 10 = 2(x – 4) or y = 2x – 8 + 10 K1
y = 2x + 2 N1
(ii) (0, –8) P1
(iii) 10 + 8 K1
4−0 N1
9
12
2
13 (a) (i) 10 P1
(ii) Identify ∠EPF P1
7
Tan ∠EPF = K1
10
0
34 59’ N1
(iii) ∠UQV P2
5 K1
(b) (i) Tan ∠PKM =
10
∠PKM = 26.60 or 26057’ N1
(ii) 10 tan 38o K1
h = 5 + 10 tan 38o K1K1
12.81 N1
14 (a) (i) (a) or P1
(b) or P1
(ii) 5∉ A ∪ B. N2
(iii) 3n 2 + 1, n = 2, 3, 4, 5, … K1N1
(b) (i) 3 P1
(ii) 11 P1
(iii) 13 P2
(iv) 9 K1N1

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Sub Full
Q Marking Scheme
mark mark
15 (a) Distance Midpoint Frequency
(km)
11 – 20 15.5 5
21 – 30 25.5 3
31 – 40 35.5 3
41 – 50 45.5 4
51 – 60 55.5 10
61 – 70 65.5 8
71 – 80 75.5 4
81 – 90 85.5 1
91 – 100 95.5 2
Column 1 – P1
Column 2 – P1
Column 3 – P2
(b) 81 N1
15.5(5) + 25.5(3) + 35.5(3) + 45.5(4) + 55.5(10) + 65.5(8) + 75.5(4) + 85.5(1) + 95.5(2)
(c)
40
2100
or K2
40
52.5
N1
(d) Refer to the graph given
9 plotted points P2
Joining points P1
A straight line from point (5.5, 0) to point (15.5, 5) and 12
(95.5, 1) and (105.5, 0) P1

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Sub Full
Q Marking Scheme
mark mark
16 (a) 72 N1
(b) 60 – 69 N1
(c) Marks Upper Cumulative
boundary frequency
20 – 29 29.5 0
30 – 39 39.5 4
40 – 49 49.5 12
50 – 59 59.5 24
60 – 69 69.5 44
70 – 79 79.5 60
80 – 89 89.5 70
90 – 99 99.5 72
Column 1 – P1
Column 2 – P1
Column 3 – P1
(d) Refer to the graph given
Uniform scale K1
8 plotted points P2
Smooth curve N1
(e) (i) 62.5 N1
(ii) 72 – 9 K1
63 N1 12
Remark:

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The graph for 15(d)
Frequency
10 ×
9
8 ×
7
6
5 ×
4 × ×
3 × ×
2 ×
1 ×
5.5
× 15.5 55.5 65.5 85.5
×
105.5
25.5 35.5 45.5 75.5 95.5
Distance (km)

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The graph for 16(b)
Cumulative frequency
80
×
70 ×
60 ×
50
×
40
30
×
20
10
×
×
× 69.5
29.5 39.5 49.5 59.5 79.5 89.5 99.5
Upper boundary

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