www.ketam.pja.my -nota 2 matematik tingkatan 4 dan 5 spm
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Form 4 – Chapter 1 Standard FormPassport To Success
(Fully–worked Solutions)
Paper 1 1 27.035 = 27.0 (3 sig. fig.)
Answer: B
2 4.23 × 10–4 = 0.0004.23 = 0.000423
Answer: B
3 429 000————–1.5 × 10–2
= 4.29 × 105—————1.5 × 10–2
= 4.29——1.5
× 105——10–2
= 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
3 � 5
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Form 4 – Chapter 2 Quadratic Expressions and EquationsPassport To Success
(Fully–worked Solutions)
Paper 1 1 3h(1 – h) + (h – 1)2
= 3h – 3h2 + h2 – 2h + 1= –2h2 + h + 1
Answer: D
Paper 2
2 3m = 8—m
– 10
3m = 8 – 10m————–m
3m2 = 8 – 10m 3m2 + 10m – 8 = 0(3m – 2)(m + 4) = 0
m = 2—
3 or –4
3 (9p – 1)2 = 9p2
81p2 – 18p + 1 = 9p2
72p2 – 18p + 1 = 0(12p – 1)(6p – 1) = 0
p = 1—–12
or 1—6
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P
Q
R
Set R
Set (P � Q)’
Union with
PQ
R
PQ
R
PQ
R
Form 4 – Chapter 3 SetsPassport 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
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 ’ � Q = {1, 2, 3, 4, 8, 12}n(P ’ � Q) = 6
Answer: B
Paper 2 5 (a)
(P � Q) � R
(b)
Uniting (P � Q)’ and R, we have (P � Q)’ � R, as shown in the following Venn diagram.
Answer: A
3
In the above diagram,(a) the shaded region represents the set
(Q � 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 � R)’ � P.
Answer: C
ξP
Q
R
ξK T
15 5 30
20
n(K ) – n(K � T )= 20 – 5= 15
It is given that n(K � T ) = 5.
n(T ) – n(K � T )= 35 – 5= 30
n(ξ) – n(K � T )= 70 – (15 + 5 + 30)= 20
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Form 4 – Chapter 4 Mathematical ReasoningPassport To Success
(Fully–worked Solutions)
Paper 2 1 (a) 3m2 + 5m – 2 = 0 is not a statement.
(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.
(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.’
(b) If x � 3, then x < 8.
The converse of the above statement is ‘If x � 8, then x � 3.’
The converse is false.
(c) Premise 1: If set M is a subset of set N, then M � N = M.
Premise 2: M � N ≠ M Conclusion: Set M is not a subset
of set N.
3 (a) (i) 15 ÷ 3 = 5 and 72 = 14 is false.
(ii) 24 is a multiple of 6 or 1—7
> 1—5
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.
(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.
This is because we cannot determine its truth value.
The given argument is a type 1 argument.
Premise 1: All P is Q.Premise 2: R is P.Conclusion: R is Q.
whereP : ‘3 elements’Q : ‘have 23 subsets’R : ‘Set A’
A quadratic equation may have ‘two distinct roots’, ‘two equal roots’ or ‘no roots’.
When x � 8, x = 7, 6, 5, 4,… but x = 7, 6, 5 and 4 is not less than 3.
The given argument is a type 3 argument.
Premise 1: If p, then q.Premise 2: Not q.Conclusion: Not p.
wherep : ‘set M is a subset of set N ’q : ‘M � N = M ’
‘15 ÷ 3 = 5’ is true.‘72 = 14’ is false.‘true’ and ‘false’ is ‘false’.
‘24 is a multiple of 6’ is true.
‘ 1—7
� 1—5
’ is false.
‘true’ or ‘false’ is ‘true’.
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.’
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Form 4 – Chapter 5 The Straight LinePassport To Success
(Fully–worked Solutions)
Paper 1 1 2x + 5y = 7 5y = –2x + 7
y = – 2—5
x + 7—5
∴ m = – 2—5
Answer: A
2 3x + 6y + 5 = 0 6y = –3x – 5
y = – 1—2
x – 5—6
∴ c = – 5—6
Answer: B
Paper 2 3
(a) mDE
= mGF
= – 1—3
The equation of DE is y = mx + c, i.e.
y = – 1—3
x + c
Since DE passes through point D(–2, 3), x = –2 and y = 3.
3 = – 1—3
(–2) + c
c = 3 – 2—3
= 7—3
Hence, the equation of DE is
y = – 1—3
x + 7—3
.
At the x-axis, y = 0.
0 = – 1—3
x + 7—3
0 = –x + 7 x = 7 ∴ x-intercept = 7
(b) G is point (–2, 0). The equation of GF is y = mx + c, i.e.
y = – 1—3
x + c
Since GF passes through point G(–2, 0), x = –2 and y = 3.
0 = – 1—3
(–2) + c
c = – 2—3
Hence, the equation of GF is
y = – 1—3
x – 2—3
3y = –x – 2
y
O x
D(–2, 3)
G(–2, 0)E
F
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Form 4 – Chapter 6 Statistics IIIPassport To Success
(Fully–worked Solutions)
Paper 2
1 (a) Distance (km) Midpoint (x) Tally f fx Class boundaries
21 – 30 25.5 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
(b) x = Σfx
—––Σf
= 2160—–––40
= 54 km
(c) (i), (ii)
(c) (i) Q3 = 75
(ii) The third quartile means 3—4
of the
students (i.e. 30 students) have marks of 75 and below.
2 MarksUpper
boundaryTally f
Cumulative frequency
20 – 29 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
20.5
12
10
8
6
4
2
0
Fre
quen
cy
30.5 40.5 50.5 60.5 70.5 80.5 90.5
Distance (km)
40
35
30
25
20
15
10
5
029.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5
Cum
ulat
ive
freq
uenc
y
Marks75
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Bab 7 tidak ada
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Form 4 – Chapter 8 Circles IIIPassport To Success
(Fully–worked Solutions)
Paper 1 1
∠ABD = ∠ADE = 54°
∠CBD = 180° – ∠ABD = 180° – 54° = 126°
∠BDC = 180° – ∠DBC——————–2
= 180° – 126°—————–2
= 27°
t° = ∠BDC = 27°
Answer: A
2
∠PQB = ∠PBA = 50°
∠PQO + ∠OQB = 50° 20° + ∠OQB = 50° ∠OQB = 30°
∠OBQ = ∠OQB = 30°
∠BOQ = 180° – ∠OBQ – ∠OQB = 180° – 30° – 30° = 120°
y° = ∠BOQ———–
2 = 120°
——2
= 60°
x° = ∠QPB = 60°
∴ x° + y° = 60° + 60° = 120°
Answer: D
A
Angle in the alternate segment
Angles on a straight line
BD = BC
Angle in the alternate segment
B
C
D
E
126°27°
54°
27°
t° 54°Angle in the alternate segment
OB = OQ
Angles in a triangle
A B C
P
Q
y°20°
30°120°
30°50° x°
O
The angle subtended by an arc at the centre of a circle is twice the angle at the circumference.
Angle in the alternate segment
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Form 4 – Chapter 9 Trigonometry IIPassport To Success
(Fully–worked Solutions)
Paper 1 1
sin x° = 5—9
sin ∠RSQ = 5—9
RQ——SQ
= 5—9
10——SQ
= 5—9
5SQ = 90 SQ = 18
∴ TQ = SQ – ST = 18 – 13 = 5 cm
In �UTQ, based on the Pythagorean triples, TU = 12 cm.
tan y° = –tan ∠TQU = – 12—–5
Answer: D
2
∠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.
P
QR
S
T
U
x°
13 cm
12 cm
13 cm5 cm
10 cm
V
y°
cos θ = –cos ∠PQR = – 8—–10
= – 4—5
Answer: A
3 The graph of y = cos x for 0° � x � 180° is as shown below.
Answer: B
4 cos θ = –0.4226 Basic ∠ = 65°
∴ θ1 = 180° – 65° = 115° ∴ θ2 = 180° + 65° = 245°
Answer: A
∠POR is the basic angle which corresponds to θ. cos θ is negative because θ is an angle in the third quadrant.
y
O R x
θ
10 6
8
T
P(8, 6)
∠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.
y
1
090°
180°x
–1
cos θ is negative in the second and third quadrants.
S
T
A
C
65°
65°
θ2
θ1
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Form 4 – Chapter 10 Angles of Elevation and DepressionPassport To Success
(Fully–worked Solutions)
Paper 1 1
In �RTP,
tan 40° = x—–
15
x = 15 × tan 40° = 12.586 m ∴ RS = 2x = 2 × 12.586 = 25.17 m
Answer: C
2
In �PQR,
tan 40° = x—–
18
x = 18 × tan 40° = 15.10 m
∴ Height of tree = 15.10 + (3 – 1) = 17.10 m
Answer: B
P
Q
R
S
T40°
40°
x m
15 m
15 m
Angle of elevation
Angle of depression
P
3 m
Q
R40°
x m
1 m
18 m
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Form 4 – Chapter 11 Lines and Planes in 3–DimensionsPassport To Success
(Fully–worked Solutions)
Paper 1 1
The line is KN.The normal is KJ.The orthogonal projection is JN.
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
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.
Hence, the angle between the planes NCM and QBC is ∠NMQ.
Answer: A
Paper 2 3
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.
Let N be the midpoint of AD. In �ANM,
tan ∠MAN = 8—5
∠MAN = 57.99° (or ∠MAD = 57.99°)
J
K L
M
N
Orthogonal projection
Normal
A B
CD
P Q
M
N
A
B C
D
EF
G H
M
N
10 cm
8 cm
12 cm
5 cm 5 cm
A
M
N5 cm
8 cm
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53 52 51 50
2 3 4 05
∴ 23405 = 2 × 53 + 3 × 52 + 4 × 51 + 0 × 50
But it is given that: 23405 = 2 × 53 + 3 × 52 + y × 51 + 0 × 50
Hence, by comparison, y = 4.
Answer: D
4
∴ 15710 = 2358
But it is given that 15710 = 2k58. Hence, by comparison, k = 3.
Answer: C
8 157
8 19 –5
8 2 –3
0 –2
Form 5 – Chapter 1 Number BasesPassport To Success
(Fully–worked Solutions)
Paper 1 1
Answer: A
2
Answer: D
3
1 1
1 1 0 0 12
+ 1 1 1 0 12
1 1 0 1 1 02
12 + 12 = 10212 + 12 + 12 = 112
100 000 1112
421 421 421
4 0 78
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Form 5 – Chapter 2 Graphs of Functions IIPassport To Success
(Fully–worked Solutions)
Paper 1 1 The general shape of the cubic graph
y = –2x3 – 9 is
The y–intercept of y = –2x3 – 9 is –9. Hence, the graph of y = –2x3 – 9 is as
shown below.
Answer: D
Paper 2 2 For y � –2x + 4, shade above the straight
line y = –2x + 4 and it should be a solid line.
For y � 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.
3 (a) When x = –1, y = 8 – (–1)3 = 9 When x = 1.5, y = 8 – (1.5)3 = 4.625(b)
(c) From the graph, (i) when x = 1.25, y = 6 (ii) when y = 8.4, x = –0.75
(d) y = 8 – x3
+ 0 = x3 – x – 8 y = –x
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.
y
O x
–9
6
4
2
O
–2
–4
y
x–2 2 4
y = x
+ 1
y = –2x + 4
10
8
6
4
2
8.4
–1.0–0.75
–0.5 0.5 1.0
1.25
1.5 2.0 2.5
2.15x
y
y = 8 – x 3
y = –x–2
–4
–6
–8
0
Graph drawn
Given equation
This is the equation of the straight line which has to be drawn.
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Form 5 – Chapter 3 Transformations IIIPassport To Success
(Fully–worked Solutions)
Paper 2 1
(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).
(b) (i) A(1, 3) ⎯→ A’(5, 0) ⎯→ A’’(0, 5)
(ii) A(1, 3) ⎯→ A’(3, 1) ⎯→ A’’(7, –2)
(c)
(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.
Centre of rotation
G H
H Gy
6
4
2
O 2x
A
B C
P Q
R
4 6
8
6
4
2
O 2 4 6 8x
K L
P Q
R
M
y
y = 6
Q ′
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Form 5 – Chapter 4 MatricesPassport To Success
(Fully–worked Solutions)
Paper 1
1 –2 h� � 3 5 – 2 0 2� � –3 k
= –2 0� � 9 7
–2 h� � 3 5 – 0 4� � –6 2k
= –2 0� � 9 7
–2 h – 4� � 9 5 – 2k = –2 0� � 9 7
h – 4 = 0 h = 4
∴ h + k = 4 – 1 = 3
Answer: C
5 – 2k = 7 5 – 7 = 2k –2 = 2k k = –1
2 2A – 1 0� � –1 4 = 5 2� � 9 8
2A = 5 2� � 9 8 + 1 0� � –1 4
2A = 6 2� � 8 12
A = 1—2
6 2� � 8 12
A = 3 1� � 4 6
Answer: D
3 (k 4) 2 0� � –k 7 = (14 28)
(k(2) + 4(–k) k(0) + 4(7)) = (14 28) (–2k 28) = (14 28) –2k = 14 k = –7
Answer: D
Paper 2
4 (a) PQ = 1 0� � 0 1
PP –1 = 1 0� � 0 1
∴ Q = P –1
= 1—————–—3(3) – (–4)(2)
3 4� � –2 3
= 1—–
17 3 4� � –2 3
But it is given that Q = 1—k
3 h� � –2 3.
Hence, by comparison, k = 17 and h = 4.
(b) 3x – 4y = –5 2x + 3y = 8
The matrix equation is
3 –4� � 2 3 x� � y
= –5� � 8
x� � y =
1—–17
3 4� � –2 3 –5� � 8
x � � y =
1—–17
17� � 34
x� � y = 1� � 2
∴ x = 1, y = 2
5 (a) Let A = 2 –1� � –6 4
A–1 = 1—————–—–2(4) – (–1)(–6)
4 1� � 6 2
= 1—2
4 1� � 6 2
= 2 1—
2� � 3 1
But it is given that A–1 = 2 h� � 3 1.
Hence, by comparison, h = 1—2
.
(b) 2m – n = 6 –6m + 4n = –20
The matrix equation is
2 –1� � –6 4 m� � n
= 6� � –20
m� � n =
2 1—2� � 3 1
6� � –20
m� � n =
2 × 6 – 1—2 (20)
3 × 6 – 1(20)� � m� � n
= 2� � –2
∴ m = 2, n = –2
Px� �y =
–5� �8
P –1Px� �y = P –1 –5� �8
Ix� �y = P –1 –5� �8
x� �y = P –1 –5� �8
Am� �n =
6� �–20
A–1Am� �n = A–1 6� �–20
Im� �n = A–1 6� �–20
m� �n = A–1 6� �–20
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Form 5 – Chapter 5 VariationsPassport 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
k = 9—–
27
k = 1—3
∴ y = 1—3
x3
When x = k and y = 8—3
,
8—3
= 1—3
k3
k3 = 8 k = 2
Answer: A 2 Q ∝
1——3 R
Q = k——R
1—3
, where k is a constant
Q = kR 1–— 3
Answer: B
3 s ∝ m—n
s = km—–n
, where k is a constant
When m = 2 and n = 8, s = 1—2
1—2
= k(2)—––
8 4k = 8 k = 2
∴ s = 2m—–n
When s = 25 and m = 50,
25 = 2(50)—–––
n
n = 100—––25
n = 4
Answer: C
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Chapter 6 Gradient and Area Under a GraphPassport To Success
(Fully–worked Solutions)
Paper 2 1
(a) Rate of change of speed from nth s to
18th s = – 5—2
m s–2
–10 – 0�———–�18 – n
= – 5—2
5(18 – n) = 20 90 – 5n = 20 5n = 90 – 20 5n = 70
n = 70—–5
n = 14
(b) (i) Length of time the particle travels at a uniform speed
= n – 7 = 14 – 7 = 7 s
(ii) Average speed in the first 7 s
= Total distance——————–
Total time =
Area P + Area Q——————–—
Total time
1—2
(8 + 16)(4) + 1—2
(16 + 10)(3) = —————————————— 7 =
87—–7
= 12 3—7
m s–1
Negative gradient
Speed (m s–1)
16
108
Time (s)O4 7 n 18
P Q
Horizontal part of the graph
Distance (m)
d
400
O5 11 22
Time (min)
2
(a) The length of time Normala stops for a rest
= 11 – 5 = 6 minutes
(b) Speed in the first 5 minutes
= 400—––
5 = 80 m min–1
(c) Average speed = 30 m min–1
Total distance——————–
Total time = 30
d—–22
= 30
d = 660
Horizontal part of the graph
Gradient
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Form 5 – Chapter 7 Probability I & IIPassport 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
∴ P(A) = n(A)
—–––n(S)
= 7—–12
Answer: D
2 Let M – Event that a male fish is chosen F – Event that a female fish is chosen S – Sample space
P(F) = 1 – 5—–
12 =
7—–12
n(F)
—–––n(S)
= 7—–
12
35
—–––n(S)
= 7—–
12
n(S) = 12—–7
× 35
= 60
Answer: A
3 Let R – Event of drawing a red pen B – Event of drawing a blue pen H – Event of drawing a black pen S – Sample space
P(B) = 2—5
× 50 = 20
n(H) = n(S) – n(R) – n(B) = 50 – 18 – 20 = 12
Answer: A
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
n(H) = P(H) × n(S) = 5—8
× 40 = 25
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—5
× 4—–10 � + � 3—
5 ×
6—–10 �
= 13—–25
(b)
P(RR or YY) = P(RR) + P(YY)
= � 2—5
× 5—–11 � + � 3—
5 ×
7—–11 �
= 31—–55
2R3Y
Jar
5R6Y
Bowl
4R7Y
Bowl
R
Y
2—5
3—5
R
Y
R
Y
RR
RY
YR
YY
5—11
6—114—11
7—11
Outcomes
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Form 5 – Chapter 8 BearingPassport To Success
(Fully–worked Solutions)
Paper 1 1
∠TRS = ∠TSR = 42°
∠VTR = ∠TRS = 42°
∴ Bearing of R from T = 180° + 42° = 222°
Answer: C
2
∠N1QR = 50°
∠N2RQ = 180° – 50° = 130°
∴ Bearing of P from R = 360° – (130° + 20°) = 210°
Answer: C
TS = TR
RS //VT and alternate angles are equal.
It is given that the bearing of R from Q is 050°.
QN1//RN2 and the sum of interior angles is 180°.
P
QR
N1
N2
50°20°
130°
Bearing of P from R
N2
T
V
N1
SR
42°
42°42°
Bearing of R from T
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Weblink22
0.36 hours= 0.36 × 60= 22 minutes
∠MOP = 180° because MP is the diameter of the earth.
N
O
S
PQ
M
42°
55°W10°E
42°ST
Form 5 – Chapter 9 Earth as a SpherePassport To Success
(Fully–worked Solutions)
Paper 1 1
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
Answer: B
2
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
Paper 2 3
(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
∠MOQ =
4740———
60 = 79°
Hence, the latitude of point Q = (79 – 42)°N = 37°N
(d) Time = Distance of MNP——————–—–
Speed
=
180 × 60————–
660
= 16.36 hours = 16 hours 22 minutes
N
S
F
G
O
H
J65° 35°
0° 35°E
N
S
A B
H
F
D
O
50°N
40°40°
40°S
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Paper 2 1 (a)
(b) (i)
Form 5 – Chapter 10 Plans and ElevationsPassport To Success
(Fully–worked Solutions)
U/T
(ii)
T/SM/N J/R
2 cm3 cm
7 cm
U/P L/V K/QPlan
L/A/M
V/N
B/J
2 cm
3 cm
5 cm6 cm
P/S Q/C/R
Elevation as viewed from X
L B/A J/M3 cm 4 cm
V/U N/T1 cm
5 cm
Q/P C R/S
Elevation as viewed from Y
1 cm
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