03[a math cd]

1 © Penerbitan Pelangi Sdn. Bhd.

Paper 1

1. (e3)2 = e3 × 2

= e6

Answer: D

2. (h1—2 )8 = h

1—2

× 8

= h4

Answer: B

3. 5−3 = 153

Answer: B

4. t−

4—3 = 1

t4—3

= 1––––3ABt4

Answer: D

5. (9n4)1—2 = (9)

1—2 (n4)

1—2

= AB9 (n4 ×

1—2 )

= 3n2

Answer: C

6. (8x3)2—3 = (8)

2—3 (x3)

2—3

= (3AB8 )2(x3 ×

2—3 )

= (2)2(x2) = 4x2

Answer: D

7. p4

–––9

3—2

= (p4)

3—2

–––––(9)

3—2

= p

4 × 3—2

––––––(AB9 )3

= p6

––––(3)3

= p6

–––27

Answer: D

8. (m2)4 × m−2 = m2 × 4 × m−2

= m8 + (−2)

= m6

Answer: A

9. (p3)4 ÷ p5 = p3 × 4 ÷ p5

= p12 − 5

= p7

Answer: C

10. (x4)2 = x2n

x4 × 2 = x2n

x8 = x2n

∴ 8 = 2n n = 4

Answer: B

11. 3−2 × 27 = 3m

3−2 × 33 = 3m

3−2 + 3 = 3m

31 = 3m

∴ m = 1

Answer: C

CHAPTER

3 IndicesCHAPTER

2

Mathematics SPM Chapter 3

© Penerbitan Pelangi Sdn. Bhd.

12. (2q3)2 × 3q−4 = (2q3)(2q3) × 3q−4

= 2 × 2 × 3 × q3 + 3 + (−4)

= 12q2

Answer: C

13. (2e2)3 ÷ 4e = (2)3(e2 × 3) ÷ 4e

= 8e6–––4e1

= 2e6 − 1

= 2e5

Answer: D

14. (5−1 × 31—2 )2 = (5−1)2 × (3

1—2

× 2)

= 5−2 × 31

= 1–––52

× 3

= 3–––25

Answer: C

Paper 1

1. 1t n = 6

–

1—2

t –n = 6–

1—2

∴ t = 6, n = 12

Answer: B

2. (h4k3)–1

h2k –5

= h– 4k –3

h2k –5

= h– 4 − 2k −3 − (−5)

= h–6k2

Answer: D

3. 203—5 = x 20y

203—5 = 20

y—x

∴ x = 5, y = 3

Answer: A

4. (16m4)1—4

––––––––2n

× m3n4

= 24 ×

1—4 m

4 ×

1—4

2n

× m3n4

= 2m2n

× m3n4

= m1 + 3 × n4 – 1

= m4n3

Answer: C

5. p3

4q3 = p3q–3

4

Answer: C

6. n × n

5—4

–––––n

1—4

= n1—2 × n

5—4 × n

–

1—4

= n1—2

+

5—4

+ (–

1—4

)

= n3—2

Answer: C

7. 1 23 2–5

= 1 23 2–5 ×

1—2

= 1 23 2

–

5—2

= 1 32 2

5—2

Answer: C

8. (2a3 × 3b5—2 ) 2–––––––––––

a2b3

= 22a3 × 2 × 32b5—2 ×

2

––––––––––––––a2b3

= 4a6 × 9b5

a2b3

= 36a6 – 2b5 – 3

= 36a4b2

Answer: C

3



9. 1 23 2–3

= 1

1 23 23

= 1827

= 1 ÷ 827

= 1 × 278

= 278

Answer: B

10. 2x = 16–––23x

2x = 24–––23x

2x = 24 − 3x

∴ x = 4 − 3x x + 3x = 4 4x = 4 x = 1

Answer: B

11. 6–

1—2

= 1––––6

1—2

Answer: D

12. 42 × 31—2

–––––––62

4

= 42 × 4 × 31—2

× 4

–––––––––––62 × 4

= 48 × 32––––––

68

= (22)8 × 32––––––––(2 × 3)8

= 216 × 32–––––––28 × 38

= 216 − 8 × 32 − 8

= 28 × 3−6

Answer: B

Paper 1

1. h5—4 × (h

1—4 × k

4—3 )3

= h5—4 × (h

1—4

× 3k

4—3

× 3)

= h5—4 × h

3—4 k4

= h5—4

+

3—4 × k4

= h2k4

Answer: D

2. (2t−3m)2 × 5t2

= (22t−3 × 2m1 × 2) × 5t2

= 4t−6m2 × 5t2

= 20t−6 + 2m2

= 20t−4m2

Answer: A

3. (3h−2k)3 × 1—3 h2k−1

= (33h−2 × 3k1 × 3) × 3−1h2k−1

= 33 + (−1)h−6k3 × h2k−1

= 32h−6 + 2k3 + (−1)

= 9h−4k2

Answer: D

4. 2m3n––––––––

(3mn−2)2

= 2m3n–––––––––––32m1 × 2n−2 × 2

= 2m3n––––––9m2n−4

= 2m3 − 2n1 − (−4)–––––––––––

9

= 2mn5–––––

9Answer: C

5. (ef 4)2 ÷ e−3f 9= (e1 × 2f 4 × 2) ÷ e−3f 9= e2f 8 ÷ e−3f 9= e2 − (−3)f 8 − 9

= e5f −1

Answer: A

4



6. (x−2y3)3 × x7y2

= (x−2 × 3y3 × 3) × x7y2

= x−6y9 × x7y2

= x−6 + 7y9 + 2

= xy11

Answer: C

7. (2p2q1—3 )3 × p3q−4

= (23p2 × 3q1—3

× 3) × p3q−4

= 8p6q1 × p3q−4

= 8p6 + 3q1 + (−4)

= 8p9q−3

Answer: B

8. (27m3n9)1—3 ÷ 9m2n−1

= (271—3 m

3 ×

1—3 n

9 ×

1—3 ) ÷ 9m2n−1

= 3m1n3 ÷ 9m2n−1

= 3m1 − 2n3 − (−1)

–––––––––––9

= m−1n4–––––

3Answer: C

9. 3g × (16h6)1—2

–––––––––––(g9h–3)

1—3

= 3g × 16

1—2 h

6 × 1—2

––––––––––––g

9 × 1—3 h

−3 ×

1—3

= 3g × 4h3––––––––

g3h–1

= 12g1 − 3h3 − (−1)

= 12g−2h4

= 12h4–––––

g2

Answer: D

10. (x3y1—3 )2 ÷ (x6y5)

1—3

= (x3 × 2y1—3

× 2) ÷ (x

6 ×

1—3 y

5 ×

1—3 )

= x6y2—3 ÷ x2y

5—3

= x6 − 2y2—3

−

5—3

= x4y−1

Answer: C

11. [2p × (3p)1—2 ]4 ÷ (p–12)

2—3

= [(2p)4 × (3p)2] ÷ p–8

= (2p)4 × (3p)2

p–8

= 144p14

Answer: B

12. (16 × 3−4)1—2 ÷ (23 × 3−5)

= (161—2 × 3

−4 ×

1—2 ) ÷ (23 × 3−5)

= (22 × 3−2) ÷ (23 × 3−5)= 22 − 3 × 3−2 − (−5)

= 2−1 × 33

= 33

21

= 272

Answer: A

13. (27 × 2−6)1—3 × (2−4 × 3)

= (271—3 × 2

−6 ×

1—3 ) × 2−4 × 31

= 31 × 2−2 × 2−4 × 31

= 31 + 1 × 2−2 + (−4)

= 32 × 2−6

= 32–––26

= 9–––64

Answer: B

14. (41—3 × 5)3 ÷ 56

–––24

= (41—3

× 3 × 53) ÷ 56

–––24

= (22 × 53) ÷ (56 × 2−4)= 22 − (−4) × 53 − 6

= 26 × 5−3

= 26–––53

= 64––––125

Answer: C

5



15. (35—2 × 54)

1—2 × 3

3—4 × 5−3

= (35—2

×

1—2 × 5

4 ×

1—2 ) × 3

3—4 × 5−3

= 35—4 × 52 × 3

3—4 × 5−3

= 35—4

+

3—4 × 52 + (−3)

= 32 × 5−1

= 32–––51

= 9––5

Answer: A

16. (9 × 2−6)1—2 ÷ (2−5 × 34)

= (91—2 × 2

−6 ×

1—2 ) ÷ (2−5 × 34)

= (31 × 2−3) ÷ (2−5 × 34)= 31 − 4 × 2−3 − (−5)

= 3−3 × 22

= 22–––33

= 4–––27

Answer: C

17. (641—3 × 5−1)2 ÷ (5−4 × 23)

= (4 × 5−1)2 ÷ (5−4 × 23)= (22 × 2 × 5−1 × 2) ÷ (5−4 × 23)= (24 × 5−2) ÷ (5−4 × 23)= 24 − 3 × 5−2 − (−4)

= 21 × 52

= 2 × 25= 50Answer: A

18. 813—4 × (32)3

––––––––––96

= (34)

3—4 × 32 × 3

–––––––––––(32)6

= 33 × 36––––––

312

= 33 + 6 − 12

= 3−3

= 1–––33

= 1–––27

Answer: C

19. 125−

2—3 × (34)

1—2

––––––––––––25–2

= (53)−

2—3 × 3

4 ×

1—2

––––––––––––(52)−2

= 5−2 × 32–––––––

5−4

= 5−2 − (−4) × 32

= 52 × 32

= 25 × 9= 225

Answer: D

20. 324—5

–––––––––––(212 × 7−4)

1—2

= (25)4—5

––––––––––––2

12 ×

1—2 × 7

−4 ×

1—2

= 24–––––––26 × 7−2

= 24 − 6 × 72

= 2−2 × 72

= 72–––22

= 49–––4

Answer: C

21. 27−

5—3

––––––––––9−4 × (4

1—3 )6

= (33)−

5—3

––––––––––––(32)−4 × 4

1—3

× 6

= 3−5–––––––3−8 × 42

= 3−5 − (−8) × 4−2

= 33 × 4−2

= 33–––42

= 27–––16

Answer: B

6



22. 8–––27

4—3 ÷ (5

−

1—2 × 4

1—2 )4

= 23–––33

4—3 ÷ (5

−

1—2

× 4 × 4

1—2

× 4)

= 2—3

4 ÷ (5−2 × 42)

= 24–––34

÷ (5−2 × 24)

= 24 – 4––––

34 × 52

= 20×52

––––––34

= 25–––81

Answer: A

23. (36 × 7−4)1—3 ÷ (7

2—3 × 9−1)

= (36 ×

1—3 × 7

−4 ×

1—3 ) ÷ (7

2—3 × 3−2)

= (32 × 7−

4—3 ) ÷ (7

2—3 × 3−2)

= 32 − (−2) × 7− 4—

3 −

2—3

= 34 × 7−2

= 34–––72

= 8149

Answer: D

24. 2p – 1 + 2p + 1 – 2p

= 2p(2 –1 + 2 – 1)

= 2p 12

+ 1

= 2p × 32

= 3(2p – 1)

Answer: D

03[a math cd]

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