Form 4 – Chapter 1 Standard Form
Passport To Success
(Fully–worked Solutions)
Paper 1
1 27.035 = 27.0 (3 sig. fig.)
3⬍5
Answer: B
2 4.23 × 10–4 = 0.0004.23
= 0.000423
Answer: B
429 000
4.29 × 105
3 ————–
= —————
–2
1.5 × 10
1.5 × 10–2
4.29
105
= —— × ——
10–2
1.5
= 2.86 × 105 – (–2)
= 2.86 × 107
Answer: B
4 2.35 × 108 – 2.48 × 107
= 2.35 × 108 – 0.248 × 101 × 107
= 2.35 × 108 – 0.248 × 108
= (2.35 – 0.248) × 108
= 2.102 × 108
Answer: D
4
Weblink
Suc Math SPM (Passport).indd 4
10/7/2008 3:17:36 PM
Form 4 – Chapter 2 Quadratic Expressions and Equations
Passport To Success
(Fully–worked Solutions)
Paper 1
1 3h(1 – h) + (h – 1)2
= 3h – 3h2 + h2 – 2h + 1
= –2h2 + h + 1
Answer: D
3
(9p – 1)2
81p – 18p + 1
72p2 – 18p + 1
(12p – 1)(6p – 1)
2
9p2
9p2
0
0
1
1
p = —– or —
12
6
=
=
=
=
Paper 2
2
8 – 10
3m = —
m
8 – 10m
3m = ————–
m
3m2 = 8 – 10m
3m2 + 10m – 8 = 0
(3m – 2)(m + 4) = 0
2 or –4
m=—
3
Weblink
Suc Math SPM (Passport).indd 5
5
10/7/2008 3:17:44 PM
Form 4 – Chapter 3 Sets
Passport To Success
(Fully–worked Solutions)
Paper 1
1 A = {4, 9}
Set A has 2n = 22 = 4 subsets.
The subsets are {4}, {9}, {4, 9}, { }.
Answer: D
2
ξ
Answer: B
T
K
5
15
4 ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20}
P = {6, 11, 16}
Q = {1, 2, 3, 4, 6, 8, 12}
P ’ 艚 Q = {1, 2, 3, 4, 8, 12}
n(P ’ 艚 Q) = 6
Paper 2
5 (a)
30
P
Q
20
R
It is given that
n(K 艚 T ) = 5.
n(K ) – n(K 艚 T )
= 20 – 5
= 15
n(T ) – n(K 艚 T )
= 35 – 5
= 30
(P 艚 Q) 艛 R
(b)
Q
P
n(ξ) – n(K 艛 T )
= 70 – (15 + 5 + 30)
= 20
R
Set (P 艚 Q)’
Answer: A
Union
3
ξ
with
Q
P
P
R
R
Set R
Q
In the above diagram,
(a) the shaded region
represents the set
(Q 艛 R)’, and
(b) the shaded region
represents the set
P.
The intersection of (a) and (b) is the set
that is required by the shaded region of the
question i.e. (Q 艛 R)’ 艚 P.
Uniting (P 艚 Q)’ and R, we have
(P 艚 Q)’ 艛 R, as shown in the following
Venn diagram.
Q
P
R
Answer: C
6
Weblink
Suc Math SPM (Passport).indd 6
10/7/2008 3:17:52 PM
Form 4 – Chapter 4 Mathematical Reasoning
Passport To Success
(Fully–worked Solutions)
Paper 2
1 (a) 3m2 + 5m – 2 = 0 is not a statement.
3 (a)
(i) 15 ÷ 3 = 5 and 72 = 14 is false.
‘15 ÷ 3 = 5’ is true.
‘72 = 14’ is false.
‘true’ and ‘false’ is ‘false’.
This is because we cannot
determine its truth value.
(b) Premise 1: All sets which contain n
elements have 2n subsets.
Premise 2: Set A contains 3 elements.
Conclusion: Set A has 23 subsets.
The given argument is a type 1 argument.
Premise 1: All P is Q.
Premise 2: R is P.
Conclusion: R is Q.
where
P : ‘3 elements’
Q : ‘have 23 subsets’
R : ‘Set A’
(c) 1 = 2(1)3 – 1
15 = 2(2)3 – 1
53 = 2(3)3 – 1
127 = 2(4)3 – 1
The nth term is 2n3 – 1,
n = 1, 2, 3, 4, …
2 (a) ‘Some quadratic equations have two
distinct roots.’
A quadratic equation may have ‘two distinct
roots’, ‘two equal roots’ or ‘no roots’.
If x ⬍ 3, then x < 8.
(b)
The converse of the above statement is
‘If x ⬍ 8, then x ⬍ 3.’
1 >—
1 is
(ii) 24 is a multiple of 6 or —
7
5
true.
‘24 is a multiple of 6’ is true.
1
1
‘ — ⬎ — ’ is false.
7
5
‘true’ or ‘false’ is ‘true’.
(b) Premise 1: If the side a rhombus is
5 cm, then its perimeter is
20 cm.
Premise 2: The side of rhombus P is
5 cm.
Conclusion: The perimeter of
rhombus P is 20 cm.
The given argument is a type 2 argument.
Premise 1: If p, then q.
Premise 2: p is true.
Conclusion: q is true.
where
p : ‘The side of rhombus P is 5 cm.’
q : ‘The perimeter of rhombus P is 20 cm.’
(c) 5x ⬎ 10 if and only if x ⬎ 2.
Implication 1: If 5x ⬎ 10, then x ⬎ 2.
Implication 2: If x ⬎ 2, then 5x ⬎ 10.
The converse is false.
When x ⬍ 8, x = 7, 6, 5, 4,… but x = 7, 6,
5 and 4 is not less than 3.
(c) Premise 1: If set M is a subset of set N,
then M 艚 N = M.
Premise 2: M 艚 N ≠ M
Conclusion: Set M is not a subset
of set N.
The given argument is a type 3 argument.
Premise 1: If p, then q.
Premise 2: Not q.
Conclusion: Not p.
where
p : ‘set M is a subset of set N ’
q : ‘M 艚 N = M ’
Weblink
Suc Math SPM (Passport).indd 7
7
10/7/2008 3:18:06 PM
Form 4 – Chapter 5 The Straight Line
Passport To Success
(Fully–worked Solutions)
Paper 1
1 2x + 5y = 7
5y = –2x + 7
2x+ 7
y = –—
—
5
5
2
∴ m = –—
5
Since DE passes through point D(–2, 3),
x = –2 and y = 3.
1 (–2) + c
3 = –—
3
2 = 7
c=3–—
—
3
3
Hence, the equation of DE is
1x+ 7.
y = –—
—
3
3
At the x-axis, y = 0.
1x+ 7
0 = –—
—
3
3
0 = –x + 7
x =7
∴ x-intercept = 7
Answer: A
2 3x + 6y + 5 = 0
6y = –3x – 5
1x– 5
y = –—
—
2
6
5
∴ c = –—
6
Answer: B
Paper 2
3
(b) G is point (–2, 0).
The equation of GF is y = mx + c, i.e.
1x+c
y = –—
3
y
D(–2, 3)
E
G(–2, 0)
O
x
F
1
(a) mDE = mGF = – —
3
The equation of DE is y = mx + c, i.e.
1x+c
y = –—
3
8
Since GF passes through point G(–2, 0),
x = –2 and y = 3.
1 (–2) + c
0 = –—
3
2
c = –—
3
Hence, the equation of GF is
1x – 2
y = –—
—
3
3
3y = –x – 2
Weblink
Suc Math SPM (Passport).indd 8
10/7/2008 3:18:16 PM
Form 4 – Chapter 6 Statistics III
Passport To Success
(Fully–worked Solutions)
Paper 2
1 (a)
Distance (km)
Midpoint (x)
f
fx
Class boundaries
21 – 30
25.5
Tally
2
51.0
20.5 – 30.5
31 – 40
35.5
4
142.0
30.5 – 40.5
41 – 50
45.5
11
500.5
40.5 – 50.5
51 – 60
55.5
10
555.0
50.5 – 60.5
61 – 70
65.5
8
524.0
60.5 – 70.5
71 – 80
75.5
4
302.0
70.5 – 80.5
81 – 90
85.5
1
85.5
80.5 – 90.5
Σf = 40
Σfx = 2160
Σfx
2160
(b) x = —–– = —––– = 54 km
40
Σf
40
(c) (i), (ii)
Cumulative frequency
35
12
Frequency
10
8
30
25
20
15
6
10
4
5
2
0
20.5
30.5
40.5
50.5
60.5
70.5
80.5
90.5
0
29.5
39.5
49.5
59.5
f
Cumulative
frequency
29.5
0
0
30 – 39
39.5
4
4
40 – 49
49.5
5
9
50 – 59
59.5
7
16
60 – 69
69.5
10
26
70 – 79
79.5
7
33
80 – 89
89.5
5
38
90 – 99
99.5
2
40
Marks
Upper
boundary
20 – 29
Tally
79.5
89.5
99.5
Marks
75
Distance (km)
2
69.5
(c)
(i) Q3 = 75
3 of the
(ii) The third quartile means —
4
students (i.e. 30 students) have
marks of 75 and below.
Weblink
Suc Math SPM (Passport).indd 9
9
10/7/2008 3:18:25 PM
Bab 7 tidak ada
Form 4 – Chapter 8 Circles III
Passport To Success
(Fully–worked Solutions)
Paper 1
1
2
C
Q
P
126°
27°
B
y°
54°
50°
27°
A
D
t°
C
B
Angle in the
alternate segment
∠PQB = ∠PBA = 50°
E
∠ABD = ∠ADE = 54°
Angle in the
alternate segment
∠CBD = 180° – ∠ABD
= 180° – 54°
= 126°
Angles on a
straight line
180° – ∠DBC
∠BDC = ——————–
2
180°
–
126°
= —————–
2
= 27°
Answer: A
30°
x°
54°
A
t° = ∠BDC = 27°
20°
30°
O 120°
BD = BC
∠PQO + ∠OQB = 50°
20° + ∠OQB = 50°
∠OQB = 30°
∠OBQ = ∠OQB = 30°
OB = OQ
∠BOQ = 180° – ∠OBQ – ∠OQB
= 180° – 30° – 30°
Angles in a triangle
= 120°
∠BOQ
120°
y° = ———– = —— = 60°
2
2
Angle in the
alternate segment
The angle subtended by an arc at the centre of a
circle is twice the angle at the circumference.
x° = ∠QPB = 60°
Angle in the
alternate segment
∴ x° + y° = 60° + 60° = 120°
Answer: D
10
Weblink
Suc Math SPM (Passport).indd 10
10/7/2008 3:18:33 PM
Form 4 – Chapter 9 Trigonometry II
Passport To Success
(Fully–worked Solutions)
Paper 1
1
8
4
cos θ = –cos ∠PQR = – —– = – —
10
5
V
S
x°
∠POR is the basic angle
which corresponds to θ.
cos θ is negative because θ is
an angle in the third quadrant.
U
13 cm
12 cm
T
R
Q
10 cm
Answer: A
13 cm
5 cm
3 The graph of y = cos x for 0° ⱕ x ⱕ 180° is
as shown below.
y°
P
y
5
sin x° = —
9
5
sin ∠RSQ = —
9
RQ
—— =
SQ
10
—— =
SQ
5SQ =
SQ =
5
—
9
5
—
9
90
18
1
∠RSQ is the basic angle
which corresponds to the
obtuse ∠VSQ (x °). sin x °
is positive because x ° is an
angle in the second quadrant.
180°
0
x
90°
–1
Answer: B
4
cos θ = –0.4226
Basic ∠ = 65°
∴ TQ = SQ – ST = 18 – 13 = 5 cm
In ⌬UTQ, based on the Pythagorean triples,
TU = 12 cm.
θ2
65°
θ1
65°
12
tan y° = –tan ∠TQU = – —–
5
∠TQU is the basic angle which corresponds to the
obtuse ∠PQU (y °). tan y ° is negative because y ° is
an angle in the second quadrant.
Answer: D
T
C
∴ θ1 = 180° – 65° = 115°
∴ θ2 = 180° + 65° = 245°
cos θ is negative in the
second and third quadrants.
y
2
A
S
P(8, 6)
10
Answer: A
6
θ
O
8
R
x
T
Weblink
Suc Math SPM (Passport).indd 11
11
10/7/2008 3:18:40 PM
Form 4 – Chapter 10 Angles of Elevation and Depression
Passport To Success
(Fully–worked Solutions)
Paper 1
1
2
Angle of depression
Q
R
xm
40°
Angle of elevation
P
xm
40°
15 m
P
40°
3m
T
R
1m
18 m
Q
15 m
S
In ⌬RTP,
12
In ⌬PQR,
x
tan 40° = —–
18
x = 18 × tan 40°
= 15.10 m
x
tan 40° = —–
15
x = 15 × tan 40°
= 12.586 m
∴ RS = 2x = 2 × 12.586 = 25.17 m
∴ Height of tree = 15.10 + (3 – 1)
= 17.10 m
Answer: C
Answer: B
Weblink
Suc Math SPM (Passport).indd 12
10/7/2008 3:18:50 PM
Form 4 – Chapter 11 Lines and Planes in 3–Dimensions
Passport To Success
(Fully–worked Solutions)
Paper 1
1
Paper 2
3
N
J
8 cm
Normal
N
D
G
H
L
12 cm
The line is KN.
The normal is KJ.
The orthogonal projection is JN.
F 5 cm M 5 cm E
The angle between the line KN and the
plane NMJ is the angle between the line
KN and its orthogonal projection (JN), i.e.
∠KNJ.
Answer: A
2
C
M
A
K
10 cm
B
Orthogonal
projection
The line of intersection of the planes ABM
and ABCD is AB.
∠BAM is a right angle on the plane ABM.
∠BAD is a right angle on the plane ABCD.
Hence, the angle between the planes ABM
and ABCD is ∠MAD.
N
P
Q
A
N
C
D
8 cm
M
A
5 cm
B
M
The line of intersection of the planes NCM
and QBC is MC.
∠NMC is a right angle on the plane NCM.
∠QMC is a right angle on the plane QBC.
Let N be the midpoint of AD.
In ⌬ANM,
Hence, the angle between the planes NCM
and QBC is ∠NMQ.
8
tan ∠MAN = —
5
Answer: A
∠MAN = 57.99° (or ∠MAD = 57.99°)
Weblink
Suc Math SPM (Passport).indd 13
13
10/7/2008 3:18:58 PM
Form 5 – Chapter 1 Number Bases
Passport To Success
(Fully–worked Solutions)
Paper 1
1
1
∴ 23405 = 2 × 53 + 3 × 52 + 4 × 51
+ 0 × 50
1
1 1 0 0 12
1 1 1 0 12
1 1 0 1 1 02
+
But it is given that:
23405 = 2 × 53 + 3 × 52 + y × 51 + 0 × 50
Hence, by comparison, y = 4.
12 + 12 = 102
12 + 12 + 12 = 112
Answer: D
Answer: A
4
2
000 1112
8 19 –5
421
421
421
8 2
–3
4
0
78
0
–2
Answer: D
3
14
8 157
100
53
52
51
50
∴ 15710 = 2358
But it is given that 15710 = 2k58.
Hence, by comparison, k = 3.
2
3
4
05
Answer: C
Weblink
Suc Math SPM (Passport).indd 14
10/7/2008 3:19:05 PM
Form 5 – Chapter 2 Graphs of Functions II
Passport To Success
(Fully–worked Solutions)
Paper 1
1 The general shape of the cubic graph
y = –2x3 – 9 is
3 (a) When x = –1, y = 8 – (–1)3 = 9
When x = 1.5, y = 8 – (1.5)3 = 4.625
(b)
y
10
8.4
8
The y–intercept of y = –2x3 – 9 is –9.
Hence, the graph of y = –2x3 – 9 is as
shown below.
6
y = 8 – x3
4
2
y
1.25
–1.0 –0.5
–0.75
x
O
0
–2
0.5
1.0
2.15
1.5 2.0
x
2.5
y = –x
–4
–9
–6
–8
Answer: D
Paper 2
2 For y ⱖ –2x + 4, shade above the straight
line y = –2x + 4 and it should be a solid
line.
For y ⱕ x + 1, shade below the straight line
y = x + 1 and it should be a solid line.
For x ⬍ 4, shade to the left of the straight
line x = 4 and it should be a dashed line.
The region which satisfies all the given
inequalities is as follows.
y
6
4
y
=
x
+
(c) From the graph,
(i) when x = 1.25, y = 6
(ii) when y = 8.4, x = –0.75
Graph drawn
(d) y = 8 – x3
+ 0 = x3 – x – 8
Given equation
y = –x
This is the equation of the straight
line which has to be drawn.
From the graph, the value of x which
satisfies the equation x3 – x – 8 = 0
is the x-coordinate of the point of
intersection of the curve y = 8 – x3 and
the straight line y = –x, i.e. x = 2.15.
1
2
–2
O
–4
4
+4
–2x
y=
–2
x
2
Weblink
Suc Math SPM (Passport).indd 15
15
10/7/2008 3:19:14 PM
Form 5 – Chapter 3 Transformations III
Passport To Success
(Fully–worked Solutions)
Paper 2
1
(b)
y
Centre of rotation
H
G
(ii) A(1, 3) ⎯→
A’(3, 1) ⎯→
A’’(7, –2)
R
6
(c)
4
A
y
Q′
K
Q
P
2
L
8
B
O
G
H
(i) A(1, 3) ⎯→
A’(5, 0) ⎯→
A’’(0, 5)
C
2
4
6
x
6
R
y=6
4
(a) Draw a line segment to join the points
A and P. Construct a perpendicular
bisector of the line segment AP. …➀
Draw a line segment to join the points
BQ. Construct a perpendicular bisector
of the line segment BQ. …➁
(i) The centre of rotation is the point
of intersection of the perpendicular
bisectors of ➀ and ➁, i.e. (3, 5).
(ii) The angle of rotation is 90°
(anticlockwise).
16
P
Q
M
2
O
2
4
6
8
x
(i) ⌬PQR is transformed to ⌬KQ’R
under transformation V, i.e.
reflection in the straight line
y = 6.
(ii) ⌬KQ’R is transformed to ⌬KLM
under transformation W, i.e.
enlargement with centre (5, 9)
and a scale factor of 2.
Weblink
Suc Math SPM (Passport).indd 16
10/7/2008 3:19:22 PM
Form 5 – Chapter 4 Matrices
Passport To Success
(Fully–worked Solutions)
Paper 1
1 –2 h – 2 0 2 = –2
3 5
–3 k
9
–2 h – 0 4 = –2
3 5
–6 2k
9
–2 h – 4 = –2
9 5 – 2k
9
0
7
0
7
0
7
5 – 2k
5–7
–2
k
h–4=0
h =4
∴h+k=4–1=3
Hence, by comparison, k = 17 and
h = 4.
=
=
=
=
(b) 3x – 4y = –5
2x + 3y = 8
The matrix equation is
3 –4 x = –5
2 3 y
8
1
x = —– 3 4 –5
17 –2 3
y
8
1
x = —– 17
–5
x
17 34 P y = 8 y
–5
x
x = 1
P –1P = P –1 y
8
y
2
x
–5
–1
7
2k
2k
–1
Answer: C
2 2A –
–11 40 = 95
2A = 5
9
2A = 6
8
1
A =—
2
A = 3
4
2
8
2 + 1
8
–1
2
12
6 2
8 12
1
6
(k
(k(2) + 4(–k)
4) 2 0 =
–k 7
k(0) + 4(7)) =
(–2k 28) =
–2k =
k=
0
4
5 (a) Let A =
28)
(14
(14
14
–7
28)
28)
4
3
mn = –22 ∴ Q = P –1
4
3
The matrix equation is
2 –1 m = 6
–6 4 n
–20
1
6
m = 2 —
2
n
3 1 –20
1 (20)
m = 2×6–—
2
n
3 × 6 – 1(20)
1
= —– 3
17 –2
1
A–1 = —————–—– 4 1
2(4) – (–1)(–6) 6 2
1 4 1
=—
2 6 2
1
2 —
2
=
3 1
But it is given that A–1 = 2 h .
3 1
1.
Hence, by comparison, h = —
2
(b) 2m – n = 6
–6m + 4n = –20
(14
1
= —————–— 3
3(3) – (–4)(2) –2
–1
–62 –14 Paper 2
0
1
0
1
y =P 8
x
y = P –58 I
Answer: D
4 (a) PQ = 1
0
PP –1 = 1
0
∴ x = 1, y = 2
Answer: D
3
∴ m = 2, n = –2
n = –20
6
m
A = A n
–20
m
6
I = A n
–20
6
mn = A –20
A
m
A–1
–1
–1
–1
1 3
But it is given that Q = —
k –2
h .
3
Weblink
Suc Math SPM (Passport).indd 17
6
17
10/7/2008 3:19:35 PM
Form 5 – Chapter 5 Variations
Passport To Success
(Fully–worked Solutions)
Paper 1
1 y ∝ x3
y = kx3, where k is a constant
When x = 3, y = 9,
9 = k(3)3
9
k = —–
27
1
k=—
3
1 x3
∴y=—
3
8,
When x = k and y = —
3
8 =—
1 k3
—
3
3
k3 = 8
k =2
Answer: A
1
2 Q ∝ ——
3
R
k
Q = ——
, where k is a constant
1
—
R3
Q = kR
m
3 s∝—
n
km
s = —– , where k is a constant
n
1
When m = 2 and n = 8, s = —
2
k(2)
1 = —––
—
8
2
4k = 8
k =2
2m
∴ s = —–
n
When s = 25 and m = 50,
2(50)
25 = —–––
n
100
n = —––
25
n =4
Answer: C
1
–—
3
Answer: B
18
Weblink
Suc Math SPM (Passport).indd 18
10/7/2008 3:19:42 PM
Chapter 6 Gradient and Area Under a Graph
Passport To Success
(Fully–worked Solutions)
Paper 2
1
2
Distance (m)
d
Speed (m s–1)
16
400
10
8
Q
P
O
O
7
4
n
18
Time (s)
(a) Rate of change of speed from nth s to
5 m s–2
18th s = – —
2
冢
冣
10 – 0
5
– ———– = – —
18 – n
2
5(18 – n)
90 – 5n
5n
5n
Negative gradient
20
20
90 – 20
70
70
n = —–
5
n = 14
(b)
=
=
=
=
5
11
22
Time (min)
(a) The length of time Normala stops for a
rest
= 11 – 5
Horizontal part of
= 6 minutes
the graph
(b) Speed in the first 5 minutes
400
Gradient
= —––
5
= 80 m min–1
(c) Average speed
Total distance
——————–
Total time
d
—–
22
d
= 30 m min–1
= 30
= 30
= 660
(i) Length of time the particle travels
at a uniform speed
=n–7
Horizontal part of
= 14 – 7
the graph
=7s
(ii) Average speed in the first 7 s
Total distance
= ——————–
Total time
Area P + Area Q
= ——————–—
Total time
1
1 (16 + 10)(3)
— (8 + 16)(4) + —
2
2
= ——————————————
7
87
= —–
7
3 m s–1
= 12 —
7
Weblink
Suc Math SPM (Passport).indd 19
19
10/7/2008 3:19:50 PM
Form 5 – Chapter 7 Probability I & II
Passport To Success
(Fully–worked Solutions)
Paper 1
1 Let S – Sample space
A – Event that the card drawn is a factor
of 48
A = {6, 12, 24, 16, 3, 4, 8}
= {3, 4, 6, 8, 12, 16, 24}
n(A) = 7
7
n(A)
∴ P(A) = —––– = —–
12
n(S)
Answer: D
2 Let M – Event that a male fish is chosen
F – Event that a female fish is chosen
S – Sample space
5
7
P(F) = 1 – —– = —–
12
12
n(F)
7
—––– = —–
12
n(S)
35
7
—––– = —–
12
n(S)
12
n(S) = —– × 35
7
= 60
Answer: A
3 Let R
B
H
S
–
–
–
–
P(B) =
n(H) =
=
=
Event of drawing a red pen
Event of drawing a blue pen
Event of drawing a black pen
Sample space
2
— × 50 = 20
5
n(S) – n(R) – n(B)
50 – 18 – 20
12
Answer: A
20
4 Let M – Event that a male student is chosen
F – Event that a female student is chosen
H – Event that a student carrying a
handphone is chosen
S – Sample space
n(S) = n(M) + n(F) = 24 + 16 = 40
5 × 40 = 25
n(H) = P(H) × n(S) = —
8
Hence, the number of male students who
carry handphones
= Total number of students who carry
handphones – Number of female students
who carry handphones
= 25 – 7
= 18
Answer: D
Paper 2
5 (a) Let R – Event that a red cube is drawn
Y – Event that a yellow cube is drawn
P(RR or YY)
2
4
3
6
= — × —– + — × —–
5
10
5
10
13
Outcomes
= —–
5
—
25
R
RR
11
5R
2
(b)
—
冢
冣 冢
6Y
6
—
Bowl 11
4
—
11
4R
Y
7Y
7
—
Bowl 11
R
5
2R
3Y
Jar
冣
3
—
5
Y
RY
R
YR
Y
YY
P(RR or YY) = P(RR) + P(YY)
2
5
3
7
= — × —– + — × —–
5
11
5
11
31
= —–
55
冢
冣 冢
冣
Weblink
Suc Math SPM (Passport).indd 20
10/7/2008 3:19:59 PM
Form 5 – Chapter 8 Bearing
Passport To Success
(Fully–worked Solutions)
Paper 1
1
Bearing of R from T
2
N2
N1
N2
T
V
42°
N1
Q
50°
20°
42°
S
Bearing of P
from R
42°
R
P
∠TRS = ∠TSR = 42°
TS = TR
∠VTR = ∠TRS = 42°
RS//VT and alternate
angles are equal.
∴ Bearing of R from T
= 180° + 42°
= 222°
Answer: C
130°
R
∠N1QR = 50°
It is given that the bearing of
R from Q is 050°.
∠N2RQ = 180° – 50° = 130°
QN1//RN2 and the sum of
interior angles is 180°.
∴ Bearing of P from R
= 360° – (130° + 20°)
= 210°
Answer: C
Weblink
Suc Math SPM (Passport).indd 21
21
10/7/2008 3:20:06 PM
Form 5 – Chapter 9 Earth as a Sphere
Passport To Success
(Fully–worked Solutions)
Paper 1
1
Paper 2
3
N
N
Q
O
F
65°
G
0°
P
42°
J
O
35°
H
S
T
M
35°E
Since the longitude of point H is 35°E,
∠GOH = 35°.
Since the difference in longitude between
point F and point H is 100°, ∠FOH = 100°.
∴ ∠GOF = 100° – 35° = 65°
Therefore, the longitude of point F is 65°W.
Hence, the longitude of point J is
(180 – 65)°E = 115°E
55°W
S
42°S
10°E
(a) Longitude of point P = (180 – 55)°E
= 125°E
(b) Distance of MT
= (55 + 10) × 60 × cos 42°
= 2898.3 n.m.
(c) Distance of MQ = 4740 n.m.
∠MOQ × 60 = 4740
4740
∠MOQ = ——— = 79°
60
Hence, the latitude of point Q
= (79 – 42)°N
= 37°N
Answer: B
2
N
D
50°N
H
A
40°
O
40°
B
F 40°S
S
Distance of MNP
(d) Time = ——————–—–
Speed
∠MOP = 180°
because MP is the
diameter of the earth.
180 × 60
= ————–
660
= 16.36 hours
= 16 hours 22 minutes
0.36 hours
= 0.36 × 60
= 22 minutes
Since the difference in latitude between
point D and point F is 90°, then
∠BOF = 90° – 50° = 40°
Therefore, ∠AOH = 40° because FOH is the
diameter of the earth.
Hence, the latitude of point H is 40°N.
Answer: A
22
Weblink
Suc Math SPM (Passport).indd 22
10/7/2008 3:20:14 PM
Form 5 – Chapter 10 Plans and Elevations
Passport To Success
(Fully–worked Solutions)
Paper 2
1 (a)
(ii)
M/N 3 cm J/R
T/S 2 cm
L
3 cm
B/A
4 cm
V/U
J/M
1 cm
N/T
5 cm
7 cm
Q/P
U/P
C
R/S
Elevation as viewed from Y
L/V
K/Q
Plan
(b)
(i)
L/A/M 3 cm B/J
U/T 2 cm 1 cm
V/N
6 cm
5 cm
P/S
Q/C/R
Elevation as viewed from X
Weblink
Suc Math SPM (Passport).indd 23
23
10/7/2008 3:20:21 PM