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Paper 1

1. m = 3k − 2m + 2 = 3k

k = m + 2––––––3

Answer: D

2. p2 = k + 5–––––2

2p2 = k + 5 2p2 − 5 = k k = 2p2 − 5

Answer: A

3. 3ABBBBh − 1 = k (3ABBBBh − 1 )2 = k2

9(h − 1) = k2

h − 1 = k2

9

h = k2

9 + 1

Answer: D

4. (x − y)3 = 8 3ABBBBB(x − y)3 = 3AB8 x − y = 2 x = 2 + y

Answer: B

5. h − k(4 + h) = k h − (4k + hk) = k h − 4k − hk = k h − hk = k + 4k h(1 − k) = 5k

h = 5k

1 − kAnswer: C

6. 2u(w − 1) + 5w = 4u 2uw − 2u + 5w = 4u 2uw + 5w = 4u + 2u w(2u + 5) = 6u

w = 6u2u + 5

Answer: C

7. h = m + 23m

3mh = m + 2 3mh − m = 2 m(3h − 1) = 2 m = 2

3h – 1Answer: B

8. p – 2

q = 23

3(p − 2) = 2q 3p − 6 = 2q 3p = 2q + 6

p = 2q + 63

Answer: A

9. x = 4(x – y)y

xy = 4(x − y) xy = 4x − 4y xy + 4y = 4x y(x + 4) = 4x

y = 4xx + 4

Answer: C

10. gh–––2 = 2g − 1

gh = 2(2g – 1) gh = 4g − 2 2 = 4g – gh 2 = g(4 − h) g = 2

4 – hAnswer: C

2 Algebraic Formulae

2

Mathematics SPM Chapter 2

© Penerbitan Pelangi Sdn. Bhd.

Paper 1

1. E = 8 ABBBBB1F – G

E 2 = 38 ABBBBB1F – G 4

2

E 2 = 64 1F – G

F – G = 64E2

F = 64E2 + G

Answer: A

2. p = 5q2 – 2 p + 2 = 5q2

p + 2–––––

5 = q2

q = ABBBBBp + 2–––––5

Answer: C

3. 2e = 7f – g 2e + g = 7f 7f = 2e + g

f = 2e + g7

Answer: B

4. 2h = 4 + hk – 1

2h(k – 1) = 4 + h 2hk – 2h = 4 + h 2hk – 2h – h = 4 2hk – 3h = 4 h(2k – 3) = 4 h = 4

2k – 3Answer: B

5. h = 2k

1 + k h(1 + k) = 2k h + hk = 2k h = 2k − hk h = k(2 − h)

k = h

2 – hAnswer: D

6. T = 14

– W3

T × 12 = 14

– W32 × 12

12T = 3 – 4W 4W = 3 – 12T

W = 3 – 12T4

Answer: C

7. 6h + k

2 = hk − 3k

6h + k = 2(hk − 3k) 6h + k = 2hk − 6k k + 6k = 2hk − 6h 7k = h(2k − 6)

h = 7k

2k – 6Answer: A

Paper 1

1. p2 − q = pq

p—2 = pq + q

p = 2(pq + q) p = 2pq + 2q p − 2pq = 2q p(1 − 2q) = 2q

p = 2q

1 – 2qAnswer: B

3

Mathematics SPM Chapter 2

© Penerbitan Pelangi Sdn. Bhd.

2. mk = 5 − mk

mk × k = 5 × k − mk

× k

mk2 = 5k − m mk2 + m = 5k m(k2 + 1) = 5k

m = 5kk2 + 1

Answer: D

3. V = ABBBBB1 – t4

V 2 = ABBBBB1 – t4

2

V 2 = 1 – t4

4V 2 = 1 − t 4V 2 + t = 1 t = 1 − 4V 2

Answer: B

4. 3h − 4k = 2kh

h(3h − 4k) = 2k 3h2 − 4hk = 2k 3h2 = 2k + 4hk 3h2 = k(2 + 4h)

k = 3h2

2 + 4hAnswer: B

5. m = n(m – 1)

1 – n m(1 − n) = n(m − 1) m − mn = mn − n m = mn − n + mn m = 2mn − n m = n(2m − 1)

n = m

2m – 1Answer: A

6. u = 1 + 2t1 – 2t

u(1 − 2t) = 1 + 2t u − 2ut = 1 + 2t u − 1 = 2t + 2ut u − 1 = t(2 + 2u)

t = u – 12 + 2u

= u – 12(u + 1)

Answer: C

7. w = v(2 – w)––––––––

v – 1 w(v − 1) = v(2 − w) vw − w = 2v − vw vw − 2v + vw = w 2vw − 2v = w v(2w − 2) = w v =

w2w – 2

Answer: B

8. p + q = 4p + 3

q q(p + q) = 4p + 3 pq + q2 = 4p + 3 pq − 4p = 3 − q2

p(q − 4) = 3 − q2

p = 3 – q2

q – 4Answer: C

9. h = 2k + 1k – 5

h(k − 5) = 2k + 1 hk − 5h = 2k + 1 hk − 2k = 1 + 5h k(h − 2) = 1 + 5h

k = 1 + 5hh – 2

Answer: A

10. nm − 1

2m = 3n

nm × 2m − 12m

× 2m = 3n × 2m

2n − 1 = 6mn (6n)m = 2n − 1

m = 2n – 1

6nAnswer: B

11. 3m − 2k = k − 1

3m = k − 1 + 2k

3m = 3k − 1

3 = m(3k − 1)

m = 3

3k – 1

Answer: B

4

Mathematics SPM Chapter 2

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12. 2p3 = 2ABq − 1

2p = 3(2ABq − 1) 2p = 6ABq − 3 2p + 3 = 6ABq

ABq = 2p + 3

6

q = (2p + 3)2

36Answer: B

13. e––2

= 4e – 7ABf

eABf = 2(4e − 7)

ABf = 2(4e – 7)e

f = 2(4e – 7)e  2

= 4(4e – 7)2

e2

Answer: C

14. 4h––––––ABBBBk − 1

= 3

4h = 3ABBBBk − 1 (4h)2 = (3ABBBBk − 1 )2

16h2 = 9(k − 1)

16h2––––

9 = k − 1

k = 16h2––––

9 + 1

Answer: B

15. p – 2q

= r—3

3(p − 2) = qr 3p − 6 = qr 3p = qr + 6

p = qr + 6

3Answer: A