Here are the key steps to solve quadratic equations: 1. Factorize the quadratic expression if possible. This allows using the zero product property. 2. Use the quadratic formula if factorizing is not possible: x = (-b ± √(b^2 - 4ac)) / 2a 3. Solve for the roots. The roots are the values of x that make the quadratic equation equal to 0. 4. Check your solutions in the original equation to verify they are correct roots. 5. Determine the nature of the roots: - If the discriminant (b^2 - 4ac) is greater than 0, there are two real distinct roots. - If the discriminant

1. 1449/1 – 1 hour 15 minutes
40 objective questions
1449/2 – subjective
Section A - 11 compulsory
questions
Section B - 4 out of 5
questions
2 Hours 30 minutes
SCIENTIFIC CALCULATOR
GEOMETRIC AL SET

2. 3 marks

3. Example
ξ B
A C
II III IV V VI
I
Shade the set ( A ∩ B ) ∪ C
1. Label each part with a Roman number

4. ξ B
A C
II III IV V VI
I
2. Identify the shaded region
U
(A B) U C
U
( II, III, IV III, IV, V, VI ) U IV, V

5. 3. Identify the shaded region
U
( II, III, IV III, IV, V, VI ) U IV, V
III, IV U IV, V
III, IV, V
4. Shade the region mark with III, IV, V
U
(A B) U C

6. SETS
Shade the Region
(a) P ∩ R ' (b) P ∩ Q ∪ R'
Q
Q P
P
R
Intersection R
Union
Com plim of
ent

7. P ∩ R'
P ∩ Q ∪ R'

8. Linear inequalities
Shade the region that satisfies the inequalities
Know how to sketch a straight line
Application of y-intercept
Understand the inequality sign
< , > for dashed line
and
≤ , ≥ for solid line

9. Shade the region that satisfies the three
inequalities y ≤ 2 x + 8 , y ≥ x and y < 8
y=x
0
y = 2x + 8

10. Know how to sketch the straight line y =8
From , y = 2 x + 8 , y-intercept = 8
y = 2x + 8
y=x
8
y<8
Full marks

11. y = 2x + 8
y=x
8
less marks

12. 4 marks

13. Solid Geometry & Volume
Combination of two
solids
1. Determine the two solids involved
2.Choose the operations + or -
3.Write the correct formulae
4.Substitute the values of r, h, d
22
Use π=
7

14. Try this
A hemisphere PQR has taken out from the
cylinder.
Find the volume of the remaining solid.
P R Cylinder - hemisphere
2 3
πr h − πr
2
3
Q 8 cm 22 2 22
×5× 5×8 − × 5× 5× 5
7 3 7
2
366
10 cm 3

15. The diagram shows a solid cone with radius 9 cm
and height 14 cm. A cylinder with radius 3 cm
and height 7 cm is taken out of the solid.
Calculate the volume,
in cm3 , of the
remaining solid.
22
Use π =
7

16. Diagram 3 shows a solid cone with radius 9 cm and
height 14 cm. A cylinder with radius 3 cm and height 7
cm is taken out of the solid.
14 cm
Write the formula first
7 cm 1 2
Vcone = πr
3 cm
3
9 cm
Vcylinder = π r h
2

17. 1
Vcone = πr 2 h
3 Stress on correct values when substituting
1  22 
( 9 ) ( 14 )
2
= 
3 7 
Remaining solid
=1188
= 1188 – 198
Vcylinder =πr h 2
= 990
 22  Stress on correct values
( 3 ) ( 7 )
2
= when substituting
 7 
=198

18. Lines and Planes
in 3D
4 marks

19. Lines and Planes in 3D
IMPORTANT NOTES :
3. SKETC therig a le
H ht- ng d
tria le
ng
2 Id ntify thea lea na eit
. e ng nd m
n g le b e t w e e n a lin e a n d a p la
A n g le b e t w e e n t w o p la n e s

20. LINES AND PLANES IN 3 DIMENSION
Name the angle between the line CE and the plane EFGH
Arrange the line and plane in two rows Find out the same alphabet
Look at the line and the plane in the diagram
C Write C in the first box
B
Look at c, choose Which
One is the Nearest to C
A D
(slashed alphabet)
6 C E
Look at C,
WON E F G
Choose W O N
H
(Non-slashed alphabets)
F G
Look at the diagram
θ 6
C
E C E G
6 H
Tan θ= 6 Draw 3 boxes
K2
√72 θ
E 36 + 36 G
θ= 35.26°

21. Name the angle between the plane DGK and the base DEFG
WON
DK = GKW O N
Slash the same alphabet
Write K in the first box
So, choose the midpoint of DG
H Look at K, choose W O N
Slashed- alphabet
K
D G K
8 G
D E F G
N
θ 6 Look at the diagram
D F Look at the diagram
Look at the diagram
12 M
K N M
E Look at K, choose W O N K
Tan θ= 6 EK = FK Non-slashed alphabets
choose M
12
θ= 26.57° θ
N M

22. V
WON
WON
Non-slash alphabet
Slash alphabet
6
D
C
θ
8
A 10 B
Calculate the angle between the plane AVC and the
plane BVC
A V C
A
Tan θ = 10
B V C
8
A C B θ= 51.34° θ
C B

23. A B
Name the angle between the
WON
E
F
plane ACGE with the plane
WON DCGH
A C G E
D θ C
D C G H
H
G
A C D
between A & E, choose either one and write
the alphabet in the first box ( for rectangle only)

24. Try this..
H U
G
E F
T
P N
L M
R
Name the angle between the plane LUM
with the plane LPNM
Answer : URT

25. Find the angle between the plane JFE
and the plane DEF.
L J
5
F D
5
M 13
5
E

26. L J
5
F D
5
M 13
5
E
Identify the angle ∠ JMD

27. 5
tan JMD =
13 − 5
2 2
∠ JMD = 22.62 o
o '
atau 22 37

28. 5M A R K S

29. MATHEMATICAL REASONING
Is the following sentence a statement ?
Give your reason.
9 + 2 = 2 −9
Statement yes Not accepted
answer
A statement.
It is a false statement.

30. Make a conclusion for the number sequence
below
2 −1= 0
0
F u ll m a r k s
2 −1=1
1
3 d o ts
2 −1= 3
2
2 −1= 7
3
2 − 1, n = 0 ,1, 2 , 3 , ...
n

31. n
2 ─ 1 , n = 0 , 1 , 2 , 3 , ..
n
2 ─ 1 , n =0,1,2,3,
n
2 ─ 1 , n =0,1,2,
n
2 ─ 1
Ie s s marks

32. Write two implications from this
compound statement
3 p = 5 if and only if p = 125
If 3 p = 5 , then p = 125
F u ll m a r k
If p = 125 , then 3 p =5
3 p = 5 , then p = 125
n o m a rk
If p = 125 , 3 p =5

33. Complete the following argument
Premise1 : If 4x = 16 , then x = 4
Premise 2 : x ≠ 4
Conclusion : 4 x ≠ 16

34. SIMULTANEOUS LINEAR
EQUATIONS
Elimination method
Substitution method
Matrix method
4 M AR KS

35. Solve the simultaneous linear equations
1
p − 2q = 9
3
5 p + 6q = −9
When there is a fraction, you must have the
same denominator first
p − 6q
=9
3
p − 6q = 27

36. Elimination method
1 p − 6q
p − 2q = 9 same deno min ator =9
3 3
⇒ p −6q = 27 1
5 p + 6q = −9 2
1 +2 6 p = 18
r r e c t o p e r a t io n ∴ p=3
⇒ 3 − 6q = 27
∴ q = −4

37. Substitution method
1 p − 6q
p − 2q = 9 same deno min ator =9
3 3
⇒ p −6q = 27 ∴ p = 27 +6q
C o r r e c t o p e r a t io n
1
5 p + 6q = −9 2
substitute 1 into 2
5 ( 27 + 6q ) + 6q = − 9 ⇒ p = 27 + 6 (− 4 )
135 + 30q + 6q = −9 ∴ p=3
∴ q = −4

38. Matrix method 1  p  9 
 − 2
 = 
 3  q   − 9
 5 6    
C o r r e c t m a t r ix f o r m
 p 1  6 2 9 
 =   
1  
q  1 − 5 − 9 
  × 6 − 5 × ( − 2)  3
3
 p  3 
 =   ∴ p = 3 , q = −4
 q   −4
   

39. THE STRAIGHT LINE – 6 MARKS
REMEMBER :
y1 − y2
• Gradient m =
x1 − x2
• Equation of a line
y = mx + c
x y
+ = 1
a b

40. THE STRAIGHT LINE
REMEMB ER :
3. Parallel lines , same gradient m1 = m2
4. Perpendicular lines , the product of their
gradients = − 1
m1m2 = −1

41. THE STRAIGHT LINE
REMEMB ER :
5. x-intercept , substitute y=0
6. y-intercept , substitute x=0

42. Important notes
x-intercept = − 12
x = − 12
x-intercept is ( − 12 , 0 )

43. In Diagram 2, O is the origin, point R lies on the x-
axis and point P lies on the y-axis. Straight line PU
is parallel to the x-axis and straight line PR is
parallel to the straight line ST. The equation of
straight line PR is x + 2y = 14.
y (a) Find the value of its
• • U
P
y-intercept from x + 2y = 14.
x + 2y = 14
S
• (b) Find the equation of the
O •
R
x straight line ST and hence,
state its x-intercept.
• T (2,-5)

44. (a) PU is parallel to the x-axis.
Find the value of its y-intercept from x + 2y = 14.
y
P
• • U 2 y = − x + 14
x 14
x + 2y = 14 y=− +
S 2 2
•
• x
O R
x
y = − +7
2
• T (2,-5)
y-intercept!
y-intercept =7

45. Find the equation of the straight line ST and
hence, state its x-intercept.
y From part (a), we have
• • U x
P
2 y = − x + 14 → y = − + 7
2
1
x + 2y = 14 Therefore, gradient ST , m = −
S 2
• 1
• x Substitute m = − and po int (2, −5) in
O R 2
y = mx + c,
 1
−5 =  −  (2) + c
• T (2,-5)
 2
c = −5 + 1 = −4
Thus, equation ST is
1
y = − x−4
2

46. hence, state its x-intercept.
1
y = − x−4
2
x − int ercept ∴ y = 0
1
4=− x
2
x = −8
x-intercept = – 8

47. QUADRATIC EQUATIONS
4 MARKS

48. QUADRATIC EQUATION
REARRANGE TO GENERAL FORM OF
QUADRATIC EQUATION
ax + bx + c = 0
2
Factorise ( )( )=0
State the values of x

49. Solve the quadratic equation
3n + 6n = 7 ( 1 − 2n )
2
3n + 6n = 7 (1 − 2n ) ⇒ 3n + 6n − 7 + 14n = 0
2 2
∴ 3n + 20n − 7 = 0
2 Fa c to rs
mus t be
g iv e n b y
u s in g w h o le
( 3n − 1 ) ( n + 7 ) = 0
n u m b e rs
1
∴ n= , −7
3

50. 2k − 5
2
Solve the quadratic equation = 3k
3
2k − 5 = 9k
2
2k − 9k − 5 = 0
2
( 2k + 1) ( k − 5) = 0
1
k =− , k =5
2

51. MATRICES
NOTES
1. When the matrix has no inverse
ad − bc = 0
2. MATRIX FORM
3. Formula of the inverse matrix
4. State the value of x and of y

52.  5 − 9
The inverse of matrix 
 2 − 3  is

 
1  − 3 9

 q 5 
p 
Find the values of p and q

53. Use the inverse formula
1  − 3 9
inverse =  
 − 2 5
5 × (−3) − (−9) × 2  
1  − 3 9 Compare with the
=   − 2 5  given inverse matrix
3 
1  − 3 9 ∴ p= 3

 q 5 
p  q= −2

54. Calculate the value of x and the value of y by
using matrix method
2 x − 3 y = −14
− x + 3 y = 13
Form a matrix equation
 2 − 3  x   − 14 

 − 1 3   y  =  13 
   
    

55. Write the inverse formula IN FRONT
1  d − b
 
− c a 
F u ll m a r k s
ad − bc  
 x  1  3 3   − 14 
 =  
 y  3  1 2   13  
    
 x   − 1 ∴ x = −1
 =  
 y  4 
    y=4

56.  2 − 3  x   − 14 
   = 
 −1 3   y   13  
    
 x  1  3 3   − 14 
 = 
 y  3  1 2   13 
 
    
 x   − 1
 =  
 y  4 
   
le s s m a r k s

57. Wrong arrangement
 x   2 − 3  − 14 
 
 y   − 1 3  =  13 
  
    
 x  1  2 3  − 14 
 =  
 y  3  1 3  13  
    
 x  1  3 − 3   − 14 
 = 
 y  3  − 1 2   13 
  n o m a rk
    
x  − 14  1  3 3 
 = 
 y  13  3  1 2 
  
     

58. CIRCLES : Perimeter and Area
1. Use the correct formulae
3. Substitute with the correct values.
22
π =
7

59. P
S 9 cm
7 cm W U
6 cm
T
30O
R
Q 6 cm V 6 cm
a) Find the perimeter of the shaded region
b) Calculate the area of the shaded region

60. P
S 9 cm
7 cm W U
6 cm
T
30O
R
Q 6 cm V 6 cm
Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT
Area = area of triangle – area of hemisphere –
area of the sector

61. (a).
Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT
 30 22   180 22  7  
= 9+ × 2 × × 6  + 6 + ( 9 − 7) + 
 360 × 2 × 7 ×  2  

 360 7    
218
=
7
1
31
7 31.14

62. (b).
Area
1    180 22  7  2  30 22 2  
=  × 12 × 9  − × ×  − × × 6 
2   360 7  2   360 7
 
709
=
28
9
25
28 25.32

63. 1    180 22  7  2  30 22 2  
a) =  × 12 × 9  −  × ×  − × × 6 
2   360 7  2   360 7 
709
= A n s w e r s a r e in t h e w ro n g
28 a ns w e r s pa c e s
b)
 30 22   180 22  7  
= 9+ × 2 × × 6  + 6 + ( 9 − 7) + 
 360 × 2 × 7 ×  2  

 360 7    
218
=
7

64. Diagram 4 shows two sectors ORST and OUV with the same
centre O. RWO is a semicircle with diameter RO and RO=2OV.
ROV and OUT are straight lines. OV=7cm and angle UOV= 60˚
Calculate
(a) perimeter of the whole diagram ,
(b) area of the shaded region.
S
T
7 c U
W
m
o
6 0V
R O
14 c 7 c
m m

65. a ) perimeter = RO + OV + RST + TU + UV
 120 22 
=14 + 7 +  × 2 × ×14  + 7
 360 7
Stress on the correct values when substituting

 60 22 
+ × 2× × 7
 360 7 
2
= 64
3

66. b) area = OUV + ORST − OWR
 60 22 2   120 22 2
= × ×7  +  × × 14 
 360 7   360 7 
Stress on the correct values when substituting
 180 22 2 
− × ×7 
 360 7 
= 154

67. 4 marks

68. PROBABILITY II
n( A )
P( A) =
n( S )

69. A group of 5 boys and 4 girls take part in a study on the
type of plants found in a reserved forest area. Each
day, two pupils are chosen at random to write a
report.
• Calculate the probability that both pupils chosen to
write the report on the first day are boys.
(b) Two boys has written the report on the first day.
They are then exempted from writing the report on
the second day. Calculate the probability that
both pupils chosen to write the report on the second
day are the same gender.

70. To choose a boy, the probability,
5 4
P (1 boy ) =
st
P (2 boy ) =
nd
9 8
5 4 5
P (both boys ) = × =
9 8 18

71. Two boys then exempted from writing the report on the
second day
Both boys-2 boys Both girls-2 girls
n(all girls )
n(all boys ) P (one girl ) =
P (one boy ) = n(all pupils )
n(all pupils )
4
3 P (1 girl ) =
st
P (1 boy ) =
st
7
7
3
2 P (2nd girl ) =
P (2nd boy ) = 6
6
4 3 2
3 2 1 Thus, P(2boys ) = × =
Thus, P(2boys ) = × = 7 6 7
7 6 7
1 2 3
P(both pupils) = + =
7 7 7

72. Society Number of Student
boy girl
Science 3 5 4
Consumer 6 7
a. If two students were chosen at random from the science
society, calculate the probability that both are girls
GxG 5 x 4 = 5
8 7 14
b. If two students were chosen at random from the group of
boys, calculate the probability that both boys came from the
same society
S x S or C x C 3/9 x2/8 + 6/9 x 5/8 =1/2

73. 5 marks

74. GRADIENT AND AREA UNDER A
GRAPH
Distance
Speed Constant/
Uniform speed Object stops
Time Time
m = rate of change of speed m = Rate of change of distance
= speed/time = distance/ time
= acceleration / deceleration = speed
Area under the graph is the
distance

75. GRADIENT AND AREA UNDER A GRAPH
REMEMBER :
2. Le ng th o f time is to tal time take n
2. AREA o f trape zium
3. Dis tanc e -time g raph , g radie nt = s pe e d
e quivale nt to the rate o f c hang e o f
dis tanc e
4. S pe e d -time g raph , g radie nt =
ac c e le ratio n e quivale nt to the rate o f
c hang e o f s pe e d

76. Speed (ms-1)
21 (t ,21)
9
1 (t ,0)
O 5 12 t Time (s)
(a) State the length of time, in s, that the particle
moves with uniform speed.
(b) Calculate the rate of change of speed, in
ms-1 , in the first 5 seconds.

77. State the length of time, in s, that the particle moves
with uniform speed.
12 − 5 = 7
Speed (ms-1)
21 (t ,21)
Time (s)
9
1 (t ,0)
O 5 12 t

78. -1
(b) Calculate the rate of change of speed, in ms , in the
first 5 seconds.
Speed (ms-1)
21 (t ,21)
9
1 (t ,0)
O 5 12 t Time (s)
9 −1 8
gradient of the green straight line = =
12 − 5 7

79. Calculate the value of t, if the total distance
travelled for t seconds is 148 metres.
Speed (ms-1)
21 (t ,21)
9
1 (t ,0)
O 5 12 t
Total distance= area under the graph
Stress on the correct values when substituting
1 1
148 = (1 + 9 ) 5 + ( 7 × 9 ) + ( 9 + 21) ( t −12 )
2 2
10t = 180
t = 18

80. SUGGESTED TIME
15 – 20 MINUTES
PER QUESTION

81. 12 (a) Complete table 1 in the answer space for the
equation y =2x2-x-3. ( 2 marks)
x -2 -1 -0.5 1 2 3 4 4.5 5
y 7 -2 -2 3 12 33 42
12 (a) In the table 1 , find the value of m and the value of n
for the equation y = 2 x 3 − 12 x + 3 ( 2 marks)
x -3 -2 -1 0 1 2
y -15 m 13 3 -7 n

82. 12 (a) Complete table 1 in the answer space for the
equation y =2x2 – x - 3. ( 2 marks)
x -2 -1 -0.5 1 2 3 4 4.5 5
y 7 -2 -2 3 12 33 42
0 25

83. 12
GRAPHS OF FUNCTIONS
1. Fill in the blanks in the table
y = 2 x − 12 x + 3
3
-2 2
x
m n
y
11 −5

84. 2. Draw the graphs of functions
Scales and the range of x are given
−3 ≤ x ≤ 5
Plot the points accurately
Can use flexible curve

85. through
the curve must be smooth, passing
through each point

86. losing marks
Using your own scale
12 marks – 1 mark

87. -10
-15
-20

91. e12c. e a n s w e r in t h e a n s w e r s p a c e s p r
th
x =1.5

92. 12c. The first equation from (a)
y = x − 6x + 5
2
The second equation from (c)
x = 7x −4
2
eliminate variables that have indices ,
2 1 3
x ,x ,
x

93. 12c. The first equation from (a)
y = x − 6x + 5
2
The second equation from (c)
x = 7x −4
2
eliminate variables that have indices
y − x = x − 6 x + 5 − (7 x − 4)
2 2
y = x − x − 6x − 7x + 5 + 4
2 2
y = −13 x + 9

94. 12d.

95. TRANSFORMATIONS III
A Combined Transformation RS means
transformation S followed by transformation R.
- Use the right terminologies
- Start the answer with the right transformation
- No short form
- Describe in full the transformation – with the
correct properties.

96. y E H
4
F
y=3
D G
2 Rotation
C
180o
A B o
centre ( 0,3 )
-4 -2 2 4
Describe in full the transformation PQ

97. Enlargement
centre( 2,6 )
6
E H Scale factor 3
4
J M
2 G
K
2 4 6 8
-2
L

98. losing marks
Please use the right term

99. rotation
enlargement correct direction
correct centre
k : sf : ratio ( m: n)
Reflection
at a point A
enlargement
correct scale factor
reflection
enlargement rotation
correct centre
enlargement
rotation
angle translation

100. STATISTICS spm
2006 FREQUENCY
POLYGON
2005 HISTOGRAM
2004 HISTOGRAM
2003 OGIVE

101. STATISTICS
Mean , median , modal class
2.Frequency Polygon
3.Histogram
4.Ogive
5.Information of the graph

102. Marks Midpoint Frequency
20 – 24 22 5
25 - 29 27 7
30 - 34 32 8
35 - 39 37 10
40 - 44 42 6
45 - 49 47 4
50 - 54 52 2

103. MEAN TABLE
x fx
92 92 x 4 = 368 10257
97 97 x 10 = 970
mean =
96
102 102 x 26 = 2652
107 107 x 24 = 2568
112 112 x 17 = 1904 = 106.84
117 117 x 9 = 1053
122 122 x 4 = 488
127 127 x 2 = 254
10257

104. These are students’ answers
Mean =
0 x 38 + 4 x 43 + 6 x 48 +
12 x 53 + 9 x 58 + 5 x 63 + 6 x 68 + 8 x 73
0 + 4 + 6 + 12 + 9 + 5 + 6 + 8
= 58.5

105. Frequency polygon table
frequency midpoint
height f x
90 - 94 4 92
95 - 99 10 97
100 - 104 26 102
105 - 109 24 107
110 - 114 17 112
115 - 119 9 117
120 - 124 4 122
125 - 129 2 127
JUMLAH 96

106. FREQUENCY POLYGON
POLIGON KEKERAPAN
F 30
R
F u ll m a r k s
25
KEKERAPAN
E
20
Q
U 15
E
10
N
C 5
Y 0
87 92 97 102 107 112 117 122 127 132
MIDPOINT
TITIK TENGAH

107. POLIGON KEKERAPAN
FREQUENCY POLYGON
F
30
R
E 25 le s s m a r k s
KEKERAPAN
Q 20
U
15
E
N 10
C 5
Y
0
7
7
2
2
2
2
7
87
92
97
12
13
10
10
11
11
12
MIDPOINT
MARKAH

108. TABLE FOR HISTOGRAM
frequency midpoint Upper boundary
CLASS f x
36 – 40 0 38 40.5
41 – 45 4 43 45.5
46 – 50 6 48 50.5
51 – 55 12 53 55.5
56 – 60 10 58 60.5
61 – 65 5 63 65.5
66 – 70 5 68 70.5
71 – 75 8 73 75.5
JUMLAH 50

109. F u ll m a r k

110. 40.5 45.5 50.5 55.5 60.5 65.5 70.5 75.5
F u ll m a r k s

111. 41 - 45
46 - 50
51 - 55
56 - 60
61 - 65
61 - 70
71 - 75
F u ll m a r k

112. Le s s m a r
o u ld h a v e g a p

113. 41 - 45 46 - 50 51 - 55 61 - 65 71 - 75
56 - 60 61 - 70
no mark

114. Table for OGIVE
frequency Cumulative Upper boundary
CLASS f frequency
36 – 40 0 0 40.5
41 – 45 4 4 45.5
46 – 50 6 10 50.5
51 – 55 12 22 55.5
56 – 60 10 32 60.5
61 – 65 5 37 65.5
66 – 70 5 42 70.5
71 – 75 8 50 75.5
∑ 50

115. Cumulative F u ll m a r k
120
frequency OGIF
100
LONGGOKAN
KEKERAPAN
80
60
40
20
0 5
5
5
5
5
5
.5
.5
.5
.5
4.
9.
4.
9.
4.
9.
84
89
94
99
10
10
11
11
12
12
SEMPADAN ATAS
Upper boundary

116. Cumulative
frequency
120 OGIF
100
LONGGOKAN
KEKERAPAN
80
60 le s s m a r k
40
20
0
5
5
5
5
5
5
.5
.5
.5
.5
4.
9.
4.
9.
4.
9.
84
89
94
99
10
11
11
12
10
12
SEMPADAN ATAS
Upper boundary

117. Cumulative OGIF
frequency
N o m a rk
110
100
LONGGOKAN
KEKERAPAN
90
80
70
60
50
40
30
20
10
0
0
0
0
0
5
5
5
80
90
85
95
10
10
11
11
12
12
13
limit MARKAH

118. Third Quartile
3
4
1
Median( Second Quartile)
2
1
First Quartile
4

119. Ogive Of Time Taken For 100 Students To Complete
Their Compositions
105
100
3 95
100× 90
4 85
80
Finding the third quartile
Cumulative Frequency
75
70
65
100 60
55 Finding the median
2 50
(second quartile)
45
40
35
100 30
Finding the first quartile
25
4 20
15
10
5
0
0 20 40 60 80 100
Time ( minutes )

120. INTERQUARTILE
Ogive Of Time Taken For 100 Students To Complete RANGE
Their Compositions
105
100
Q3 − Q1
3
Q3
95
100× 90
4 85
80 Finding the third quartile
Cumulative Frequency
75
70
65
100 60
Finding the median
2 55
50 (second quartile)
45
40
35 Q1
100 30 Finding the first quartile
4 25
20
15
10
5
0
0 20 40 60 80 100
Time ( minutes )

121. Information of the graph
50 students took 60 minutes to complete
their composition.
Interquartile range is 20 minutes
Median is 60 minutes
n f o r m a t io n m u s t c o m e f r o m t h e g

122. PLANS AND ELEVATIONS
CORRECT SHAPE
Satisfy the given CONDITIONS
MEASUREMENT MUST BE
ACCURATE
LATERAL INVERSION is not accepted
(SONGSANG SISI TIDAK DITERIMA)

123. 15b(ii).
H id d e n lin e

124. losing marks

125. 15
Case 1:
Double line
Bold line

126. 15a.
Case 2 :
Sizes – Bigger or Smaller

127. Case 3:
extension
Case 4 :
gap

128. 15
Case 5:
Not a right angles

129. Draw a full scale
i. The plan of the solid
ii. the elevation of the solid as viewed from Y
iii. the elevation of the solid as viewed from X
X
Y

130. F u ll m a r k

131. N o m a rk

132. EARTH
AS A SPHERE

133. Longitude
N
MG 0°
40°
20°
40° E
S
60° E

134. TWO meridians form a GREAT
circle
N
30 w 150 E
S

135. U
135 W 45 E
S

136. P(60 N, 30 W ) and Q are two points on the surface of the earth
where PQ is the diameter of the parallel latitude of P and Q.
The position of point Q is
A. ( 60 N, 150 W) C. (60 S, 150 E )
B. ( 60 N 150 E) D. (60 S, 150 W )
N
P Q 150 E
30 W
60 60
S

137. J( 30 S, 80 E ) and K are two points on the earth where
JK is the diameter of the earth. The location of K is
A. ( 30 S, 100E) C. ( 30 N, 100W )
B. ( 30 S, 80 E) D. ( 30 N, 80 W)
N
80° E
100° W K
300
300
J
S

138. a)P is a point on the surface of the earth such
that JP is the diameter of the earth. State the
position of P.
b) Calculate the value of x, if the distance from J
to K measured along the meridian is 4200
nautical miles.
c)Calculate the value of y, if the distance from J
due west to L measured along the common
parallel of latitude and then due south to M.
f) If the average speed for the whole flight
is 600 knots, calculate the time taken for the
whole flight.

139. 16. Interpretation of question – sketch the
earth
o
40 W o
J K 140 E
o
50
O

140. 16b. JM = MK
∴ angle at the centre JOM = 90 o
Greenwich
o
40 W o
J o
K 140 E
40
o
50
M
o o
50 E , 50 N

141. 16c.
JK = 80 × 60
= 4800
J K
o
80
o o
50 50
O

142. 16b. JMK route
∴ angle at the centre JOM = 180 o
= 180 × 60 × cos 50 o
o
40 B o
J 180 K
Greenwich
M
6942.106
∴ average speed =
13
= 534

143. Reminders
 Calculate , find , solve ,
– all steps are clearly shown
 State – only the answer is required
 Unit / label
– must be correct if written

144. Reminders
Basic Mathematical Skills such as
addition, division, subtraction,
multiplication.
Algebraic and Trigonometric skills
Formulae and its applications,
Formulae and its substitutions.
Round off only at the last answer
line.

145. Reminders
 Allsteps must be clearly shown.
 Read the instructions and questions very
carefully .
 The answer must be in the lowest form, to
4 significant figures and to 2 decimal
places.
 Master the calculator
 Do not sleep during the exam!