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

qq

qq

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