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Paper 1

1. 135

3 significant figures

Answer: C

2. 20074 significant figures

Answer: D

3. 0.045 2 significant figures

Answer: B

4. 6000 1 significant figure

Answer: B

5. 42 = 40 (1 significant figure)

Answer: C

6. 26. = 2.6 × 101

1 Answer: B

7. 10. = 1.0 × 101

1

Answer: B

8. 100. = 1.0 × 102

2 1

Answer: B

9. 5 = 5 × 1 = 5 × 100

Hence, n = 0

Answer: A

10. 0.1 = 1 × 10−1

1

Answer: D

11. 0.002 = 2 × 10−3

2 31

Hence, n = −3

Answer: D

Paper 1

1. 7.048 = 7.0 (2 significant figures)

Answer: A

2. 0.000429

3 41 2

= 4.29 × 10–4

Answer: B

3. 5.0678 = 5.07 (3 significant figures)

Answer: A

4. 6 3 2 1 . 7

= 6.3217 × 103

Answer: C

3 2 1

CHAPTER

10 Standard FormCHAPTER

2

Mathematics SPM Chapter 10

© Penerbitan Pelangi Sdn. Bhd.

5. 465.2 × 10–4

10–6

= 4.652 × 102 × 10–4

10–6

= 4.652 × 102 + (–4) – (–6)

= 4.652 × 104

Answer: B

6. 4.08 × 106

= 4 080 000 5 64321

= 4 080 000

Answer: B

7. (5 – 1.056) ÷ 50= 3.944 ÷ 50= 0.07888= 0.0789 (3 significant figures)

Answer: D

8. Volume of water

= 60100

× 160 × 100 × 120

= 1 152 000= 1.152 × 106 cm3

Answer: D

9. Volume of rod

= 227

× 0.0052 × 7

= 0.00055 m3

Density = MassVolume

7800 = Mass of rod0.00055

Mass of rod = 7800 × 0.00055 = 4.29 kg

Number of solid spheres

= 4.295 × 10 –3

= 8.58 × 102

Answer: C

10. 0.0003007

3 421

= 3.007 × 10−4

Answer: D

11. 5.31 × 10−6 − 1.8 × 10−7

= 5.31 × 10−6 − 0.18 × 101 × 10−7

= 5.31 × 10−6 − 0.18 × 10−6

= (5.31 − 0.18) × 10−6

= 5.13 × 10−6

Answer: A

12. 4.016 = 4.02 (3 significant figures)

Answer: D

13. 0.00000086 = 8.6 × 10–7

Compare with p × 10n

Hence, p = 8.6 and n = –7

Answer: C

14. 0.0641.6 × 108

= 6.4 × 10–2

1.6 × 108

= 6.41.6

× 10–2 – 8

= 4 × 10–10

Answer: A

15. Number of defective bulbs

= 125.8 × 106 × 0.02100

= 125.8 × 106 × 0.02 × 10–2

= 2.516 × 104

Answer: C

16. 80 526 = 80 500 (3 significant figures)

Answer: A

17. 6.415 × 10−6

= 0.000006415

2 13456

= 0.000006415

Answer: C

3

Mathematics SPM Chapter 10

© Penerbitan Pelangi Sdn. Bhd.

18. 0.04–––––––––8 000 000

= 40 × 10–3–––––––––8 × 106

= 40–––8 × 10−3 − 6

= 5 × 10−9

Answer: D

19. 0.0112(2 × 104)3

= 0.011223 × 1012

= 0.01128 × 10−12

= 0.0014 × 10−12

= 1.4 × 10–3 × 10–12 = 1.4 × 10–15

Answer: D

20. 0.0062 − 4 × 10−6

= 6.2 × 10−3 − 0.004 × 103 × 10−6

= 6.2 × 10−3 − 0.004 × 10−3

= (6.2 − 0.004) × 10−3

= 6.196 × 10−3

Answer: A

21. 0.0648 = 0.065 (2 significant figures)

Answer: B

22. Total volume of water in 40 cylindrical containers

= 40 × 227

× 1002 × 350

= 4.4 × 108 cm3

Volume of water in each cuboid container

= 4.4 × 108

50= 0.088 × 108

= 8.8 × 10–2 × 108

= 8.8 × 106 cm3

Answer: D

23. 60 781 = 60 800 (3 significant figures)

Answer: D

24. 825 000 = 8.25 × 105

2 1345

Answer: D

25. 0.0000036 − 1.5 × 10−7

= 3.6 × 10−6 − 0.15 × 101 × 10−7

= 3.6 × 10−6 − 0.15 × 10−6

= (3.6 − 0.15) × 10−6

= 3.45 × 10−6

Answer: C

26. Mass of sugar in each bag

= 320 × 103 × 0.85

= 3.2 × 102 × 103 × 8 × 10–1

5= 5.12 × 102 + 3 – 1

= 5.12 × 104

Answer: B

Paper 1

1. 318 = 300 (1 significant figure)

Answer: C

2. 466 = 470 (2 significant figures)

Answer: A

3. 1.5276 = 1.53 (3 significant figures)

Answer: A

4. 0.1064 = 0.11 (2 significant figures)

Answer: A

5. 1.00764 = 1.008 (4 significant figures)

Answer: A

6. 200 765 = 200 800 (4 significant figures)

Answer: D

7. 21.048 = 21.0 (3 significant figures)

Answer: B

8. 0.6852––––––0.043

= 10 706.25= 10 710 (4 significant figures)

Answer: A

4

Mathematics SPM Chapter 10

© Penerbitan Pelangi Sdn. Bhd.

9. RM84 695 = RM85 000 (2 significant figures)

Answer: D

10. 943 560 = 944 000 (3 significant figures)

Answer: D

11. Breadth of the rectangle

= 336–––––20.5= 16.3902= 16.4 (3 significant figures)

Answer: A

12. Amount received by each person

= 2.56 × 106–––––––––10

= 256 000= 260 000 (2 significant figures)

Answer: D

13. 2730 = 2.73 × 103

123

Answer: C

14. 0.0834 = 8.34 × 10−2

21

Answer: B

15. 2.7 × 106 + 590 000= 2.7 × 106 + 0.59 × 106

= (2.7 + 0.59) × 106

= 3.29 × 106

Answer: C

16. 780 000 − 7.8 × 104

= 7.8 × 105 − 0.78 × 101 × 104

= 7.8 × 105 − 0.78 × 105

= (7.8 − 0.78) × 105

= 7.02 × 105

Answer: B

17. 8.5 × 10−4 − 2.32 × 10−5

= 8.5 × 10−4 − 0.232 × 101 × 10−5

= 8.5 × 10−4 − 0.232 × 10−4

= (8.5 − 0.232) × 10−4

= 8.268 × 10−4

Answer: A

18. 0.000024 + 1.53 × 10−5

= 2.4 × 10−5 + 1.53 × 10−5

= (2.4 + 1.53) × 10−5

= 3.93 × 10−5

Answer: C

19. 44 000–––––––2 × 1010

= 4.4 × 104–––––––––2 × 1010

= 4.4–––2 × 104 − 10

= 2.2 × 10−6

Answer: B

20. 24 000––––––––6 × 10–5

= 24 × 103––––––––6 × 10–5

= 24–––6 × 103 − (−5)

= 4 × 108

Answer: C

21. 85 000–––––––––(2 × 105)2

= 8.5 × 104–––––––––22 × 105 × 2

= 8.5 × 104–––––––––4 × 1010

= 8.5––––4 × 104 − 10

= 2.125 × 10−6

Answer: A

22. 0.0064–––––––––(4 × 10–6)2

= 64 × 10–4––––––––––42 × 10–6 × 2

= 64 × 10–4–––––––––16 × 10–12

= 64–––16 × 10−4 − (−12)

= 4 × 108

Answer: B

23. 8.4 × 109 − 6.2 × 108

= 8.4 × 109 − 0.62 × 101 × 108

= 8.4 × 109 − 0.62 × 109

= (8.4 − 0.62) × 109

= 7.78 × 109

Answer: D

5

Mathematics SPM Chapter 10

© Penerbitan Pelangi Sdn. Bhd.

24. 8.2 × 10−6 + 7.0 × 10−8

= 8.2 × 10−6 + 0.07 × 102 × 10−8

= 8.2 × 10−6 + 0.07 × 10−6

= (8.2 + 0.07) × 10−6

= 8.27 × 10−6

Answer: D

25. Area of the rectangle PQRS= 2.5 × 102 × 1.2 × 102

= 3 × 104 cm2

Answer: C

26. Distance travelled

= 660 × 103 × 15–––60= 165 000= 1.65 × 105 m

Answer: C

27. Mass of 100 atoms= 6.02 × 10−29 × 103 × 102

= 6.02 × 10−29 + 3 + 2

= 6.02 × 10−24

Answer: B

28.

0.62 m

0.9 m

Area of the triangle

= 1—2 × (0.9 × 103) × (0.62 × 103)

= 2.79 × 105 mm2

Answer: D

29. Area of land that each person gets

= 5 × 104–––––––8

m2

= 6.25 × 103 m2

Answer: D