Page 8 - Grab A+ Kertas Model SPM Matematik

JAWAPAN / ANSWERS

Kertas Model 1/ Model Paper 1 y° = 180° – 108° – 56°
= 16°
Kertas 1/ Paper 1 x + y = 52 + 16
= 68
1. A 0.000867 – 3.2 × 10 –5
= 8.67 × 10 – 0.32 × 10 –4
–4
= (8.67 – 0.32) × 10 –4 7. C Sudut pedalaman bagi heksagon sekata
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= 8.35 × 10 –4 Interior angle of regular hexagon
(6 – 2) × 180°
= —–—————–
2. B 52.10528 720° 6
= —–––
6
Tambah 1 kepada digit 0 kerana digit seterusnya, = 120°
5, adalah sama dengan 5. Sudut pedalaman bagi poligon P
Add 1 to the digit 0 since the next digit, 5, is Interior angle of polygon P
equal to 5. = 360° – 120° – 90°

= 52.11 (4 angka bererti/ significant figures) = 150°
Bilangan sisi bagi poligon P
Number of sides for polygon P
3. D Jejari/ Radius = 0.7 m 360°
= 70 cm = —–————–
180° – 150°
Isi padu silinder/ Volume of cylinder = 12
= πr h
2
22
2
= —– × 35 × 120 8. B ∠POB = 180° – 28° – 90°
7
= 462 000 = 62°
= 4.62 × 10 5 x + y = 62
2
2
2
4. B 3(5 ) + 5 + 19 9. C QS = 32.5 – 12.5 2
4
= 3(625) + 25 + 19 = 900
900
= 1919 QS = ABBB
10
= 30 cm
8 1919 Baki/ Remainder kos x°/ cos x° = – —–––
30
8 239 ... 7 32.5
12
8 29 ... 7 = – —–
13
8 3 ... 5
0 ... 3 10. C sin 45° = kos 45° / cos 45°

sin 225° = kos 225° / cos 225
∴ 3(5 ) + 5 + 19 = 3577
2
4
8 ∴ x = 45°, 225°
5. D 110111 + 101011 = p + 132
3
a b
3 –2
3 –9
–2 4
2 2 10 8 11. D ABBBB × 3a b ÷ 6a b
p = 110111 + 101011 – 132 8 — 1
10
2
2
–2 4
3 –9 3
3 –2
p = (1 × 2 + 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 + 1 × 2 ) = (a b ) × 3a b ÷ 6a b
0
3
4
2
1
5
10
3 –2
–2 4
–3
+ (1 × 2 + 0 × 2 + 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 ) = ab × 3a b ÷ 6a b
2
3
1
4
5
0
3
– (1 × 8 + 3 × 8 + 2 × 8 ) = —(a 1 + (–2) – 3 )(b –3 + 4 – (–2) )
0
2
1
p = 8 10 6
1
10
–4
3
∴ p = 8 = —(a )(b )
2
b
3
= ——
6. A Sudut pedalaman bagi pentagon 2a 4
Interior angle of pentagon 256
x
(5 – 2) × 180° 12. B 32 = ——
= —–—————– 8 x
5 32 × 8 = 256
x
x
= 108° 256 = 256
x
∠ABF = 360° – (2 × 108° + 88°) x = 1
= 56°
x° = 108° – 56° 5p 2p – 4
3
= 52° 13. A —– + 3 , ——––
5
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