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1 Penerbitan Pelangi Sdn. Bhd.

Paper 1

1. m= 3k2

m+ 2 = 3k

k=m+ 2

3

Answer: D

2. p2=k+ 5

2

2p2= k+ 5

2p25 = k

k= 2p25

Answer: A

3. 3h1 = k(3h1 )2= k2

9(h1) = k2

h1 =k2

9

h=k2

9+ 1

Answer: D

4. (xy)3= 8

3(xy)3= 38 xy= 2

x= 2 +y

Answer: B

5. hk(4 + h) = kh(4k+ hk) = k

h 4khk= k

hhk= k + 4k

h(1 k) = 5k

h=5k

1 k

Answer: C

6. 2u(w1) + 5w = 4u

2uw2u+ 5w = 4u

2uw+ 5w = 4u+ 2u

w(2u+ 5) = 6u

w =6u

2u+ 5

Answer: C

7. h=m+ 2

3m

3mh= m+ 2

3mhm= 2

m(3h1) = 2

m=2

3h 1

Answer: B

8.p 2

q =

2

3

3(p2) = 2q

3p

6 = 2q 3p = 2q+ 6

p =2q+ 6

3

Answer: A

9. x=4(x y)

y xy= 4(xy)

xy= 4x4y

xy+ 4y= 4x

y(x+ 4) = 4x

y=

4x

x+ 4

Answer: C

10.gh

2

= 2g1

gh= 2(2g 1)

gh= 4g2

2 = 4g gh

2 = g(4 h)

g=2

4 h

Answer: C

CHAPTER

2 Algebraic Formulae

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2

Mathematics SPM Chapter 2

Penerbitan Pelangi Sdn. Bhd.

Paper 1

1. E = 8 1F G

E2 = 8 1F G

2

E 2 = 641

F G

F G =64

E2

F =64

E2+ G

Answer: A

2. p= 5q2 2

p+ 2 = 5q2

p+ 2

5= q2

q= p+ 25

Answer: C

3. 2e= 7f g

2e+ g= 7f

7f= 2e+ g

f=2e+ g

7

Answer: B

4. 2h=4 + h

k 1

2h(k 1) = 4 + h

2hk 2h= 4 + h

2hk 2h h= 4

2hk 3h= 4 h(2k 3) = 4

h=4

2k 3

Answer: B

5. 1 + 3w2= 3(4w2 e)

1 + 3w2= 12w2 3e

1 + 3e= 12w2 3w2

1 + 3e= 9w2

w2=1 + 3e

9

w= 1 + 3e9 w=

1 + 3e

3

Answer: D

6. T=1

4

W

3

T12 =1

4

W

312

12T= 3 4W

4W= 3 12T

W=3 12T

4

Answer: C

7.6h+ k

2= hk3k

6h+ k= 2(hk3k)

6h+ k= 2hk6k

k+ 6k= 2hk6h

7k= h(2k6)

h=7k

2k 6Answer: A

Paper 1

1.p

2q=pq

p2 =pq+ q

p= 2(pq+ q)

p= 2pq+ 2q

p 2pq= 2q

p(1 2q) = 2q

p=2q

1 2q

Answer: B

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3

Mathematics SPM Chapter 2

Penerbitan Pelangi Sdn. Bhd.

2. mk= 5 m

k

mkk= 5 km

kk

mk2= 5km

mk2+ m= 5k

m(k2+ 1) = 5k

m= 5kk2+ 1

Answer: D

3. V= 1 t4

V 2= 1 t4

2

V 2=1 t

4

4V 2= 1 t

4V

2

+ t= 1 t= 1 4V 2

Answer: B

4. 3h4k=2k

h

h(3h4k) = 2k

3h24hk= 2k

3h2= 2k+ 4hk

3h2= k(2 + 4h)

k=3h2

2 + 4h

Answer: B

5. m=n(m 1)

1 n

m(1 n) = n(m1)

mmn= mnn

m= mnn+ mn

m= 2mnn

m= n(2m1)

n=m

2m 1

Answer: A

6. u=1 + 2t

1 2t

u(1 2t) = 1 + 2t

u2ut= 1 + 2t

u1 = 2t+ 2ut

u1 = t(2 + 2u)

t=u 1

2 + 2u

=u 1

2(u+ 1)

Answer: C

7. w=v(2 w)

v 1

w(v1) = v(2 w)

vww= 2vvw

vw2v+ vw= w

2vw2v= w

v(2w2) = w

v= w2w 2

Answer: B

8. p+ q=4p+ 3

q

q(p+ q) = 4p+ 3

pq+ q2= 4p+ 3

pq4p= 3 q2

p(q4) = 3 q2

p=3q2

q 4

Answer: C

9. h=2k+ 1

k 5

h(k5) = 2k+ 1

hk5h= 2k+ 1

hk2k= 1 + 5h

k(h2) = 1 + 5h

k=1 + 5h

h 2

Answer: A

10.nm

1

2m= 3n

nm

2m1

2m2m= 3n2m

2n1 = 6mn

(6n)m= 2n1

m=2n 1

6n

Answer: B

11.3

m2k= k1

3m

= k1 + 2k

3m

= 3k1

3 = m(3k1)

m=3

3k 1

Answer: B

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Mathematics SPM Chapter 2

Penerbitan Pelangi Sdn. Bhd.

12.2p

3= 2q 1

2p= 3(2q 1) 2p= 6q 32p+ 3 = 6q

q =2p+ 3

6

q=(2p+ 3)2

36

Answer: B

13.e

2

=4e 7

fef= 2(4e 7)

f =2(4e 7)

e

f=2(4e 7)

e 2

=4(4e 7)2

e2

Answer: C

14.4h

k1

= 3

4h= 3k1 (4h)2= (3k1 )2

16h2= 9(k1)

16h2

9 = k1

k=16h2

9

+1

Answer: B

15.p 2

3q+ 1 = 4r 1

p2 = (4r 1)(3q+ 1)

p2 = 4r(3q ) + 4r 3q 1

p 2 4r+ 1 = 4r(3q ) 3q

p

4r 1 = 3

q(4r 1)

3q =p 4r 1

4r 1

q = p 4r 14r 1 3

Answer: D

16. x= t2 5 y= t+ 4 From , t =y 4 t2 = (y 4)2

Substitute into ,x= t

2 5

x= (y 4)2 5

=y2 8y+ 16 5

=y2 8y+ 11

Answer: D