Mathematics Keynotes 2
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Transcript of Mathematics Keynotes 2
1449/1 – 1 hour 15 minutes40 objective questions
1449/2 – subjectiveSection A - 11 compulsory questionsSection B - 4 out of 5 questions 2 Hours 30 minutes
SCIENTIFIC CALCULATORGEOMETRICAL SET
3 marks
Example
Shade the set
1. Label each part with a Roman number
I
VIVIII IVII
C
B
A
ξ
CBA ∪∩ )(
2. Identify the shaded region
(A B) U C
I
VIVIII IVII
C
B
A
ξ
( II, III, IV U IV, V
U
III, IV, V, VI )
U
3. Identify the shaded region
III, IV, V
4. Shade the region mark with III, IV, V
( II, III, IV III, IV, V, VI ) U IV, V
III, U IV, VIV
(A B) U C
U
U
(b) 'RQP ∪∩(a) 'RP ∩
R
QP
Shade the Region
PR
Q
SETS
IntersectionUnion
Compliment of
'RP ∩
'RQP ∪∩
Linear inequalitiesShade the region that satisfies the inequalities
Know how to sketch a straight line
Understand the inequality sign < , > for dashed line
and
Application of y-intercept
≥≤ , for solid line
Shade the region that satisfies the three inequalities 8,82 <≥+≤ yandxyxy
xy =
82 += xy
0
8
82 += xyFrom , , y-intercept = 8
8<y
Full marks
Know how to sketch the straight line 8=y
82 += xy
xy =
8
82 += xy
xy =
less marks
4 marks
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
Use 7
22=π
Try this A hemisphere PQR has taken out from the
cylinder. Find the volume of the remaining solid.
32
3
2rhr ππ −
P
Q
R
10 cm
8 cm
Cylinder - hemisphere
5557
22
3
2855
7
22 ×××−×××
3
2366
The diagram shows a solid cone with radius 9 cmand 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 theremaining solid.
Use 7
22=π
Diagram 3 shows a solid cone with radius 9 cm andheight 14 cm. A cylinder with radius 3 cm and height 7cm is taken out of the solid.
14 cm
9 cm
7 cm
3 cm
2
3
1rVcone π=
hrVcylinder2π=
Write the formula first
( ) ( )
2
2
1
31 22
9 143 7
1188
coneV r hπ=
= =
( ) ( )
2
2223 7
7
198
cylinderV r hπ=
= =
Remaining solid
= 1188 – 198
= 990
Stress on correct values when substituting
Stress on correct values when substituting
Lines and Planes in 3D
4 marks
Lines and Planes in 3D
IMPORTANT NOTES :
3. SKETCH the right- angled triangle
2. Identify the angle and name it
A n g le b e t w e e n a lin e a n d a p la n e A n g le b e t w e e n t w o p la n e s
Look at c, choose WhichOne is the Nearest to C (slashed alphabet)
Look at the diagram
Look at C, Choose W O N(Non-slashed alphabets)
Arrange the line and plane in two rows
Write C in the first box
Find out the same alphabet
C
Draw 3 boxes
A D
CB
6
G
H
F
E6
6
LINES AND PLANES IN 3 DIMENSIONName the angle between the line CE and the plane EFGH
C E
E F G H
E Gθ C
E G3636 +
θ
Look at the line and the plane in the diagram
W O N
Tan θ= 6 √72
θ= 35.26°
K 2
Write K in the first box
Look at K, choose W O N Slashed- alphabet
Look at the diagram
Look at K, choose W O N Non-slashed alphabets
Slash the same alphabet
Look at the diagram
W O N
D G K
D E F G
W O N
Look at the diagramD
H
F
E
K
G
6
M
N
12
8
N
K
Mθ
Tan θ= 6 12
θ= 26.57°
Name the angle between the plane DGK and the base DEFG
K
K N M
DK = GK So, choose the midpoint of DG
EK = FKchoose M
θ
8
6
10
D
V
C
θ
W O NSlash alphabet
W O NNon-slash alphabet
A BCalculate the angle between the plane AVC and the
plane BVCA V C
B V C
AC
A
BθBC θ=
Tan θ = 10 8
51.34°
A
E
GH
D C
B
F
Name the angle between the plane ACGE with the plane
DCGH
A C G E
D C G H
A C D
between A & E, choose either one and writethe alphabet in the first box ( for rectangle only)
θ
W O NW O N
Try this..
H
E
UG
T
L M
N
R
F
P
Name the angle between the plane LUM with the plane LPNM
Answer : URT
Find the angle between the plane JFE and the plane DEF.
L
E
DF
J
M5
513
5
M
L
E
D
135
5
F
J
5
Identify the angle JMD∠
22 513
5
−=tan JMD
∠ JMD'3722
62.22o
o
atau
=
5 M A R K S
MATHEMATICAL REASONING
9229 −=+
Is the following sentence a statement ? Give your reason.
Statement Not acceptedyes
A statement.It is a false statement.
answer
Make a conclusion for the number sequence below
712
312
112
012
3
2
1
0
=−=−=−=−
...,3,2,1,0,12 =− nn
3 d o t s
F u ll m a r k s
Ies s marks
2n ─ 1
2n ─ 1 , n = 0 , 1 , 2 , 3 , ..
2n ─ 1 , n = 0 , 1 , 2 , 3 ,
2n ─ 1 , n = 0 , 1 , 2 ,
12553 == pifonlyandifp
Write two implications from this compound statement
If , then p = 125
If p = 125 , then
53 =p
53 =p F u ll m a r k s
, then p = 125
If p = 125 ,
53 =p
53 =p n o m a r k
Complete the following argument
164 ≠x
Premise1 : If 4x = 16 , then x = 4
Premise 2 :
Conclusion :
4≠x
Elimination methodElimination method
SIMULTANEOUS LINEAR EQUATIONS
Substitution method
Matrix method
4 M A R K S
923
1 =− qp
965 −=+ qp
Solve the simultaneous linear equations
When there is a fraction, you must have the same denominator first
93
6 =− qp
276 =− qp
Elimination methodElimination method
93
6min92
3
1 =−=− qpatordenosameqp
965 −=+ qp
1
2
1 2+
3
186
=∴=p
p
4
2763
−=∴=−⇒
q
q
1276 =−⇒ qp
C o r r e c t o p e r a t io n
Substitution methodSubstitution method
qpqp 627276 +=∴=−⇒
965 −=+ qp1
2
4
9630135
96)627(5
−=∴−=++−=++
q
3
)4(627
=∴−+=⇒
p
p1 2substitute into
93
6min92
3
1 =−=− qpatordenosameqp
C o r r e c t o p e r a t io n
Matrix methodMatrix method
4,3 −==∴ qp
−
=
−9
9
65
23
1
q
p
( )
−
−−×−×=
9
9
3
15
26
25631
1
q
p
−
=
4
3
q
p
C o r r e c t m a t r ix f o r m
THE STRAIGHT LINE – 6 MARKS
REMEMBER :
• Gradient
• Equation of a line21
21
xx
yym
−−=
cmxy +=1=+
b
y
a
x
R E M E M B E R :
3. Parallel lines , same gradient
4. Perpendicular lines , the product of their
gradients =
21 mm =
1−
121 −=mm
THE STRAIGHT LINE
R E M E M B E R :
5. x-intercept , substitute
6. y-intercept , substitute
0=y
0=x
THE STRAIGHT LINE
Important notes
12−=x-intercept
12−=x
( )0,12−x-intercept is
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
x
P
O
S
R
T (2,-5)•
••
•
x + 2y = 14
U•
(b) Find the equation of the straight line ST and hence, state its x-intercept.
(a) Find the value of its
y-intercept from x + 2y = 14.
(a) PU is parallel to the x-axis.
Find the value of its y-intercept from x + 2y = 14.y
x
P
O
S
R
T (2,-5)•
••
•
x + 2y = 14
U• 2 14
14
2 2
72
y x
xy
xy
= − +
= − +
= − +
y-intercept!
y-intercept =7
Find the equation of the straight line ST and hence, state its x-intercept.
2 14 72
1, ,
21
int (2, 5)2
,
15 (2)
2
5 1 4
,
14
2
xy x y
Therefore gradient ST m
Substitute m and po in
y mx c
c
c
Thus equation ST is
y x
= − + → = − +
= −
= − −
= +
− = − + = − + = −
= − −
y
x
P
O
S
R
T (2,-5)•
••
•
x + 2y = 14
U•From part (a), we have
hence, state its x-intercept.
82
14
0int
42
1
−=
−=
=∴−
−−=
x
x
yerceptx
xy
x-intercept = – 8
QUADRATIC EQUATIONS
4 MARKS
QUADRATIC EQUATION
REARRANGE TO GENERAL FORM OF QUADRATIC EQUATION
02 =++ cbxax
Factorise ( ) ( ) 0=
State the values of x
( ) ( )7,
3
1
0713
−=∴
=+−
n
nn
07203
014763)21(7632
22
=−+∴=+−+⇒−=+
nn
nnnnnn
Solve the quadratic equation
( )nnn 21763 2 −=+
F a c t o r s m u s t b e g iv e n b y
u s in g w h o le n u m b e r s
Solve the quadratic equation
22 53
3
kk
− =
( ) ( )5,
2
1
0512
0592
9522
2
=−=
=−+=−−
=−
kk
kk
kk
kk
MATRICESNOTES1. When the matrix has no inverse
2. MATRIX FORM
3. Formula of the inverse matrix
4. State the value of x and of y
0=− bcad
Find the values of p and q
−5
931
qp
The inverse of matrix is
−−
32
95
−−
=
−−
×−−−×=
52
93
3
1
52
93
2)9()3(5
1inverse
Compare with the given inverse matrix
2
3
−==∴
q
p
Use the inverse formula
−5
931
qp
Calculate the value of x and the value of y by using matrix method
−=
−
−13
14
31
32
y
x
133
1432
=+−−=−
yx
yx
Form a matrix equation
−
=
13
14
21
33
3
1
y
x
−=
4
1
y
x
4
1
=−=∴
y
x
Write the inverse formula IN FRONT
−
−− ac
bd
bcad
1
F u ll m a r k s
−
=
13
14
21
33
3
1
y
x
−=
−
−13
14
31
32
y
x
−=
4
1
y
x
le s s m a r k s
Wrong arrangement
−=
−
−
−=
−
=
21
33
3
1
13
14
13
14
21
33
3
1
13
14
31
32
3
1
y
x
y
x
y
x
−=
−
−
13
14
31
32
y
x
n o m a r k
CIRCLES : Perimeter and Area
7
22=π
1. Use the correct formulae
3. Substitute with the correct values.
V
S
P
R6 cm
T
6 cm
9 cm
30O
7 cm
6 cm
U
Q
W
a) Find the perimeter of the shaded region
b) Calculate the area of the shaded region
V
S
P
R6 cm
T
6 cm
9 cm
30O
7 cm
6 cm
U
Q
W
Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT
Area = area of triangle – area of hemisphere – area of the sector
(a).
Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT
( )
×××+−++
×××+=
2
7
7
222
360
1807966
7
222
360
309
7
218=
7
131 14.31
( )
×××+−++
×××+=
2
7
7
222
360
1807966
7
222
360
309
7
218=
(b).
××−
××−
××= 2
2
67
22
360
30
2
7
7
22
360
180912
2
1
28
709=
28
925 32.25
Area
b)
a)
××−
××−
××= 2
2
67
22
360
30
2
7
7
22
360
180912
2
1
28
709=
( )
×××+−++
×××+=
2
7
7
222
360
1807966
7
222
360
309
7
218=
A n s w e r s a r e in t h e w r o n g a n s w e r s p a c e s
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.
U
T
S
W
OR V7 cm
14 cm
6 0 o
7 cm
3
264
77
222
360
60
7147
222
360
120714
)
=
×××+
+
×××++=
++++= UVTURSTOVROperimetera
Stress on the correct values when substituting
154
77
22
360
180
147
22
360
1207
7
22
360
60
)
2
22
=
××−
××+
××=
−+= OWRORSTOUVareab
Stress on the correct values when substituting
4 marks
PROBABILITY II
)(
)()(
Sn
AnAP =
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.
To choose a boy, the probability,
4(2 )
8ndP boy =
9
5)1( =boyP st
18
5
8
4
9
5)( =×=boysbothP
( )( )
( )
3(1 )
72
(2 )6
3 2 1, (2 )
7 6 7
st
nd
n all boysP one boy
n all pupils
P boy
P boy
Thus P boys
=
=
=
= × =
( )( )
( )
4(1 )
73
(2 )6
4 3 2, (2 )
7 6 7
st
nd
n all girlsP one girl
n all pupils
P girl
P girl
Thus P boys
=
=
=
= × =
1
7
Both girls-2 girls
2
7
Two boys then exempted from writing the report on the second day
Both boys-2 boys
P(both pupils) = =+7
3
76Consumer
53Science
girlboy
Number of StudentSociety
b. If two students were chosen at random from the group of boys, calculate the probability that both boys came from the same society
5
8 x 4
7= 5 14
S x S or C x C 3/9 x2/8 + 6/9 x 5/8 =1/2
4
G x G
a. If two students were chosen at random from the science society, calculate the probability that both are girls
5 marks
GRADIENT AND AREA UNDER A GRAPH
Speed
Time
m = rate of change of speed
= speed/time
= acceleration / deceleration
Area under the graph is the distance
Distance
Time
m = Rate of change of distance
= distance/ time
= speed
Constant/Uniform speed Object stops
:REMEMBER2. Length of time is total time taken
2. AREA of trapezium
3. - , = Dis tance time graph gradient s peed equivalent to the rate of change of
dis tance
4. - , = Speed time graph gradient acce leration equivalent to the rate of
change of s peed
GRADIENT AND AREA UNDER A GRAPH
(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.
Speed (ms-1)
21
9
O 5 12 t
(t ,21)
(t ,0)1
Time (s)
State the length of time, in s, that the particle moves with uniform speed.
Speed (ms-1)
21
9
O 5 12 t
(t ,21)
(t ,0)1
Time (s)
7512 =−
(b) Calculate the rate of change of speed, in ms-1 , in the first 5 seconds.
Speed (ms-1)
21
9
O 5 12 t
(t ,21)
(t ,0)1
Time (s)
7
8
512
19 =−−=gradient of the green straight line
Calculate the value of t, if the total distance travelled for t seconds is 148 metres.
Total distance= area under the graph
Speed (ms-1)
21
9
O 5 12 t
(t ,21)
(t ,0)1
( ) ( ) ( ) ( )
18
18010
122192
197591
2
1148
==
−++×++=
t
t
tStress on the correct values when substituting
SUGGESTED TIME
15 – 20 MINUTES PER QUESTION
12 (a) Complete table 1 in the answer space for the equation y =2x2-x-3. ( 2 marks)
4233123-2-27y
54.54321-0.5-1-2x
3122 3 +−= xxy12 (a) In the table 1 , find the value of m and the value of n for the equation ( 2 marks)
n-7313m-15y
210-1-2-3x
12 (a) Complete table 1 in the answer space for the equation y =2x2 – x - 3. ( 2 marks)
4233123-2-27y
54.54321-0.5-1-2x
0 25
1. Fill in the blanks in the table
GRAPHS OF FUNCTIONS12
3122 3 +−= xxy
2-2
mxy n
11 5−
53 ≤≤− xScales and the range of x are given
Plot the points accurately
Can use flexible curve
2. Draw the graphs of functions
the curve must be smooth, passing through each point
through
losing marks
Using your own scale 12 marks – 1 mark
-10-15-20
12c.
x =1.5
W r it e t h 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 o v id e d
12c. The first equation from (a)
eliminate variables that have indices ,
562 +−= xxy
472 −= xxThe second equation from (c)
xxx
1,, 32
12c. The first equation from (a)
eliminate variables that have indices
562 +−= xxy
472 −= xxThe second equation from (c)
913
4576
)47(5622
22
+−=++−−−=−−+−=−
xy
xxxxy
xxxxy
12d.
A Combined Transformation RS means
transformation S followed by transformation R.
TRANSFORMATIONS III
- Use the right terminologies- Start the answer with the right transformation- No short form- Describe in full the transformation – with the correct properties.
-2-4
2
2 4
A B
D
C
y
4
E H
F
G
y = 3
o
Rotation
180o
centre ( 0,3 )
Describe in full the transformation PQ
2 4 6 8
2
4
6
K
J
G
-2
M
L
E H
Enlargement
centre( 2,6 )
Scale factor 3
losing marks
Please use the right term
enlargement
translation
rotation
reflection
enlargementcorrect centre
rotation angle
enlargementcorrect scale factor
rotationcorrect directionenlargement
correct centrek : sf : ratio ( m: n)
Reflection at a point A
STATISTICS spm
2006 FREQUENCY POLYGON2005 HISTOGRAM2004 HISTOGRAM2003 OGIVE
Mean , median , modal class2.Frequency Polygon3.Histogram4.Ogive5.Information of the graph
STATISTICS
25250 - 54
44745 - 49
64240 - 44
103735 - 39
83230 - 34
72725 - 29
52220 – 24
FrequencyMidpointMarks
10257
127 x 2 = 254127
122 x 4 = 488122
117 x 9 = 1053117
112 x 17 = 1904112
107 x 24 = 2568107
102 x 26 = 2652102
97 x 10 = 97097
92 x 4 = 36892
f x xMEAN TABLE
9610257=mean
84.106=
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
5.58=
Mean =
These are students’ answers
96JUMLAH
2125 - 129
4120 - 124
9115 - 119
17110 - 114
24105 - 109
26100 - 104
1095 - 99
490 - 94
frequency f height
Frequency polygon table
127
122
117
112
107
102
97
92
midpoint x
POLIGON KEKERAPAN
0
5
10
15
20
25
30
87 92 97 102 107 112 117 122 127 132
TITIK TENGAH
KE
KE
RA
PA
N
MIDPOINT
F
R
E
Q
U
E
N
C
Y
FREQUENCY POLYGON
MIDPOINT
F
R
E
Q
U
E
N
C
Y
FREQUENCY POLYGON
F u ll m a r k s
POLIGON KEKERAPAN
0
5
10
15
20
25
30
87 92 97 102
107
112
117
122
127
132
MARKAH
KE
KE
RA
PA
N
MIDPOINT
F
R
E
Q
U
E
N
C
Y
FREQUENCY POLYGON
le s s m a r k s
TABLE FOR HISTOGRAM
JUMLAH
71 – 75
66 – 70
61 – 65
56 – 60
51 – 55
46 – 50
41 – 45
36 – 40
CLASS
50
Upper boundarymidpoint x
frequency f
8
5
5
10
12
6
4
0
73
68
63
58
53
48
43
38
75.5
70.5
65.5
60.5
55.5
50.5
45.5
40.5
F u ll m a r k
40.5 45.5 50.5 55.5 60.5 65.5 70.5 75.5
F u ll m a r k s
41 -
45
46 -
50
51 -
55
56 -
60
61 -
65
61 -
70
71 -
75
F u ll m a r k s
L e s s m a r k
S h o u ld h a v e g a p
41 - 45 46 - 50 51 - 5556 - 60
61 - 6561 - 70
71 - 75
no mark
Table for OGIVE
71 – 75
66 – 70
61 – 65
56 – 60
51 – 55
46 – 50
41 – 45
36 – 40
CLASS
50
Upper boundaryCumulativefrequency
frequency f
8
5
5
10
12
6
4
0
50
42
37
32
22
10
4
0
75.5
70.5
65.5
60.5
55.5
50.5
45.5
40.5
∑
OGIF
0
20
40
60
80
100
120
84.5
89.5
94.5
99.5
104.5
109.5
114.5
119.5
124.5
129.5
SEMPADAN ATAS
KE
KE
RA
PA
N
LO
NG
GO
KA
N
Upper boundary
Cumulative frequency
F u ll m a r k s
OGIF
0
20
40
60
80
100
120
84.5
89.5
94.5
99.5
104.5
109.5
114.5
119.5
124.5
129.5
SEMPADAN ATAS
KE
KE
RA
PA
N
LO
NG
GO
KA
N
Upper boundary
Cumulative frequency
le s s m a r k s
OGIF
0102030405060708090
100110
80 85 90 95 100
105
110
115
120
125
130
MARKAH
KE
KE
RA
PA
N
LO
NG
GO
KA
N
limit
Cumulative frequency
N o m a r k
2
1
4
1
4
3
Median( Second Quartile)
First Quartile
Third Quartile
Ogive Of Time Taken For 100 Students To Complete Their Compositions
05
101520253035404550556065707580859095
100105
0 20 40 60 80 100
Time ( minutes )
Cu
mu
lati
ve F
req
uen
cy
Finding the median
(second quartile)
Finding the first quartile
Finding the third quartile
2
100
4
3100×
4
100
Ogive Of Time Taken For 100 Students To Complete Their Compositions
05
101520253035404550556065707580859095
100105
0 20 40 60 80 100
Time ( minutes )
Cu
mu
lati
ve F
req
uen
cy
Finding the median
(second quartile)
Finding the first quartile
Finding the third quartile
2
100
4
3100×
4
100
INTERQUARTILERANGE
3Q
1Q
13 QQ −
Information of the graph
Interquartile range is 20 minutes
50 students took 60 minutes to complete their composition.
Median is 60 minutes
T h e in f o r m a t io n m u s t c o m e f r o m t h e g r a p h
PLANS AND ELEVATIONS
CORRECT SHAPE
Satisfy the given CONDITIONS
MEASUREMENT MUST BE ACCURATE
LATERAL INVERSION is not accepted(SONGSANG SISI TIDAK DITERIMA)
15b(ii).
H id d e n lin e
losing marks
15
Case 1:Double line
Bold line
15a.
Case 2 :Sizes – Bigger or Smaller
Case 3:extension
Case 4 :gap
15
Case 5:Not a right angles
Y
X
Draw a full scale i. The plan of the solidii. the elevation of the solid as viewed from Yiii. the elevation of the solid as viewed from X
F u ll m a r k
N o m a r k
EARTH AS A SPHERE
LongitudeN
S
MG 0°
40°
40° E
20°
60° E
TWO meridians form a GREAT circle
N
S
30 w 150 E
U
S
45 E135 W
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
S
30 W30 W6060
P 150 EQ
60
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
S
80° E 100° W
300
J
K
300
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.
16. Interpretation of question – sketch the earth
J
O
K
50o
40oW 140oE
16b. JM = MKoJOMcentretheatangle 90=∴
J K
M
40oW 140oE
Greenwich
40o
50o
50oE , 50oN
16c.
J
O
K
50o50o
80o
4800
6080
=×=JK
16b. JMK route
J K
M
40oB
Greenwich
180o
o50cos60180 ××=
13
106.6942=∴ speedaverage 534=
oJOMcentretheatangle 180=∴
Calculate , find , solveCalculate , find , solve , , – – all steps are clearly shownall steps are clearly shown
State State – only the answer is required– only the answer is required
Unit / labelUnit / label – – must be correct if writtenmust be correct if written
Reminders
RemindersRemindersBasic Mathematical Skills such as Basic Mathematical Skills such as
addition, division, subtraction, addition, division, subtraction, multiplicationmultiplication..
AlgebraicAlgebraic and T and Trigonometric rigonometric skillsskillsFormulae and its Formulae and its applicationsapplications, ,
Formulae and its Formulae and its substitutionssubstitutions. . Round offRound off only at the last answer only at the last answer
line.line.
All All stepssteps must be clearly shown. must be clearly shown.Read the instructions and questions very Read the instructions and questions very
carefully .carefully .The answer must be in the The answer must be in the lowest form, to lowest form, to
44 significant figures and to significant figures and to 22 decimal decimal places.places.
Master the Master the calculatorcalculatorDo not Do not sleepsleep during the exam! during the exam!
RemindersReminders