Marking Scheme(a)

8/6/2019 Marking Scheme(a)

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MARKING SCHEME

MATHEMATICSPMR



20 Subjective Questions ( 60 marks )

Time : 1 hour 45 minutes



Popular Topics

1. Fractions - ( 2 marks)2. Decimals ± (2 marks)

3. Integer s/ Dir ected Number s

± (2 marks)

4. Squar e ,Squar e Roots,Cubes, Cubes Roots

( 3marks )

5. Algebraic Expr essions ± 3

to 4 questions- ( 7 m )6. Linear Equations ± (3m)

7. Algebraic Formulae ± (3 m)

8. Geometr ical,constr uctions ± (6m)

9. Loci in Two Dimensions ±

(5 m)

10. Transformations ± (4m)

11.Statistics - (4 to 7m)

12.Indices ± (5m)

13.Linear Inequalities-(4m)

14.Graphs of Functions(4 m)

15. Tr igonometry ± (2m)



1 Calculate the value of 3

2)6.0(14 z

. [ 2 marks ] Answer:

10

914

10

914

2

3

10

614

3

2

10

614

!

!

v!

z!

K1

N1




2)6.0(14 z


20

1814

20

1814

2

3

10

614

3

2

10

614

!

!

v!

z!

K1

N0




2)6.0(14 z


10

914

201814

20

1814

2

3

10

614

32

10614

!

!

!

v!

z!

K1

N1




2)6.0(14 z


9.14

9.014

10

9

14

2

3

10

614

3

2

10

614

!

!

!

v!

z!

K1

N1



2 Calculate the value of ¹ º

¸©ª

¨v

3

2

5

4

8

11

and expr ess the answer as a f raction in its lowest term.

[ 2 marks ] Answer:

203

15

2

8

915

1012

8

9

!

v!

¹

º

¸©

ª

¨

v!

K1

N1




¸©ª

¨v

3

2

5

4

8

11


[ 2 marks ] Answer:

12018

15

2

8

915

1012

8

9

!

v!

¹

º

¸©

ª

¨

v!

K1

N0




¸©ª

¨v

3

2

5

4

8

11


[ 2 marks ] Answer:

12018

15

2

8

9

15

1012

8

9

!

v!

¹ º

¸©ª

¨

v!

K1

N1

20

3!



3 (a) Find the value of 49.0

(b) Calculate the value of 3

116

25¹¹ º

¸©©ª

¨

[ 3 marks ]

= 0.7 P1

Answer:

(a)

(b)

64

1

4

1

1

4

5

3

3

!

¹ º

¸©ª

¨!

¹

º

¸©

ª

¨!

K1

N1



3 (a) Find the value of 49.0

(b) Calculate the value of 3

116

25¹¹ º

¸©©ª

¨

[ 3 marks ]

= 0.7 P1

Answer:

(a)

(b)

64

1

8

1

1

4

5

2

3

!

¹ º

¸©ª

¨!

¹

º

¸©

ª

¨!

K0

N0



y

x

R

R'

2 4 6 8

2

4

8

6

4. In Diagram 1, R ' is the image of R under transformation W.

Descr ibe in full transformation W

Answer:

Translation ¹¹

º

¸©©

ª

¨

4

6

translation 6 units to the r ight, 4 units down.

¹¹ º

¸©©ª

¨

4

6.

Translation or

P2Or

awar d P1 -



y

x

R

R'

2 4 6 8

2

4

8

6



Answer:

Transilation ¹¹

º

¸©©

ª

¨

4

6

¹ º

¸

©ª

¨

4

6.

Translation

Or

P1

awar d P1 -



y

x

R

R'

2 4 6 8

2

4

8

6



Answer:

Translation 4

6

)4,6(

.

Transilation

Or

P1

P0



5 In Diagram 2, PQR is a straight line and PT = 3

1RS .

= 15

9

=

5

3

K1

N1

P R

T

5 c m 3 c m

S

y

Q

13 c m

Find the value of cos y º.

Answer :

(2 marks)




1RS .

=

15

9

=

RS

QR

K1N0

P R

T

5 c m 3 c m

S

y

Q

13 c m


Answer :

(2 marks)

cos y º




1RS .

=

13

3

=

RS

QR

K0N0

If label QR = 9 cm or

label RS = 15 cm

Awar d B1

P R

T

5 c m 3 c m

S

y

Q

13 c m


Answer :

(2 marks)

cos y º

9 cm

15 cm



6 Diagram 3 shows a r ight pr ism with a squar e base.

3 cm

4 cm

Net of prism must be correct

The drawing of the net for the

Prism must be in actual

measurement or accor dingto actual scale

Accurate drawing

K1

K1

N1




3 cm

4 cm





Accurate drawing

K1

K1

N0




3 cm

4 cm





Accurate drawing

K0

K0

N0



7. Diagram 4 is drawn on a squar e gr id paper. On the squar e gr id, draw

the image of tr iangle JKL under a 90º anticlockwise r otation

about the point P .

J

K L

P

J '

K '

L'

P2





about the point P .

J

K L

P

'

P2





about the point P .

J

K L

P

J '

K '

L'

P0



8. Diagram 5 is a pie char t which shows the number of pupils

in five gr oups who complete an assignment dur ing a

motivational camp. ( 5 marks)

Tuah

10Jebat

x

Kasturi

2x

Lekiu

12

Lekir 8

It is given that the total number of pupils whoComplete the assignment is 36.

(a) Find the value of x

(b) Calculate the angle of the sector representing

the Tuah group. Show your working.

(C) State the mode of your data

Answer :

X = 2

r!

rv!

100

360

36

10

Mode = Lekiu

P2

P1

K1

N1






Tuah

10Jebat

x

Kasturi

2x

Lekiu

12

Lekir 8






Answer :

4

r!

rv!

100

1010

Mode = Lekiu , 12 P0

CJP1

K1

N1

2






Tuah

10Jebat

x

Kasturi

2x

Lekiu

12

Lekir 8






Answer :

X = 2

r!

rv!

100

360

36

10

Mode = Lekiu (12) P2

P1

K1

N1






Tuah

10Jebat

x

Kasturi

2x

Lekiu

12

Lekir 8






Answer :

X = 2

r!

rv!

100

360

36

10

Mode = Lekiu

Fr equency = 12

P2

P1

K1

N1



2

2

9 Simplif y (2 x ± 3 ) ± (6 + 7 x )

Answer :

(2 marks)

= 4 x -12 x + 9 - 6 ± 7 x

± 19 x + 3

2

= 4 x 2

K1

N1



2

2

9 Simplif y (2 x ± 3 ) ± (6 + 7 x )

Answer :

(2 marks)

= 4 x -12 x + 9 - 6 ± 7 x

± 19 x + 3= 4 x

K1

N0



2

2

9 Simplif y (2 x ± 3 ) ± (6 + 7 x )

Answer :

(2 marks)

= 4 x -12 x + 9 - 6 + 7 x

± 5 x + 3

2

= 4 x 2

K1

N0



10 (a) Solve the inequality 2 + x � 7.

(b) List all the integer value of x which satisf y both

the inequalites 1 ± x < 2 and 212

1e x [ 4 marks ]

Answer : (a) 5u x

(b)1- x < 2

-x < 2 - 1

-x < 1

x > -1

x = 0, 1, 2,3, 4, 5, 6

P1

K1K1

N1

6

23

122

1

212

1

e

ve

e

e

x

x

x

x






1e x [ 4 marks ]

Answer : (a) 5u x

(b)

x = 0, 1, 2, 3, 4, 5, 6

P1

K1

K0

N0

6e x

x < -1






1e x [ 4 marks ]

Answer : (a) 5u x

(b) x > -1

x = -1, 0, 1, 2, 3, 4, 5, 6

P1

K1

K1N0

6e x






1e x [ 4 marks ]

Answer : (a) 5u x

(b)

61 e x

x = 0, 1, 2, 3, 4, 5, 6

Note :

P1

N1

K1K1

6e x

x < -1

Marking Scheme(a)

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