skemajawapan 2009

7/30/2019 SkemaJawapan 2009

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ANSWER TO SCOREPENILAIAN MENENGAH RENDAH 2009

MATHEMATICSANSWERS FOR PAPER 1 (SET 1)

1. D 21. A2. A 22. D3. C 23. B4. D 24. C5. C 25. B6. A 26. B7. B 27. D

8. C 28. B9. C 29. A

10. B 30. C11. B 31. B12. B 32. A13. C 33. C14. B 34. B15. D 35. C16. D 36. B17. D 37. A18. B 38. C19. B 39. A20. A 40. D

MARKING SCHEME

1. The number 345,678 is written as 350,000 when it is rounded off to theA nearest ten = 345 680B nearest hundred = 345 700C nearest thousand = 346 000

D nearest ten thousand = 350 000

175


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2. Which of the following fractions has value greater than4

3?

A 87

0.875168

147

8

7=

B3

2 0.667 or

168

112

3

2=

C2

1 0.5168

84

2

1=

D7

5 0.714168

120

7

5=

3. 33100000

168= 3 +

100000

168

= 3 + 0.00168

= 3.00168

4. Town H = 8C (Coolest on a particular day)

5. m = 23n = 29m + n = 23 + 29

= 52

6. Factors of 54 = 1, 2, 3, 6, 9, 18, 27, 29Answer: A = 2, 3, 18

7. Broken eggs perday = 9000100

1 = 90

Number of broken eggs produced in a week = 7 90

= 630

8. Total length of road constructed = 5.54 km + 6.78 km= 12.32 km

= 12 km 320 m

9. x = 180 55 (180 125)= 180 55 55= 180 110= 70

10

.

PTS = 180 140 = 140

p + q + r+s + PTS = (5 2) X 180p + q + r+s + 140 = 540

176


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p + q + r+s = 540 140 = 400

11.

3x + 2x + 3x + 115 + 122 + 75 = (6 2) X 1808x + 312 = 720

x =o720 312

8

o

= 51

12.

Interior angle of pentagon = ( )o

5 - 2 180

5= 108

Interior angle of hexagon = ( )o6 - 2 180

6= 120

The value ofm is = 360 - 108 - 120 = 132

13.

Perimeter is the total length of the boundaries.= PQ+QW+WV+VU+UR+RP

1. each side of square QRVW and equilateral triangle RVU is 12cm.2. PQR is a right angled triangle, therefore

PR = 5 + 12

= 25 + 144= 169= 13cm

3. Perimeter of the whole diagram = 5 + 12 + 12 + 12 + 12 + 13= 66cm

14. 1. Area of XYZ = 2

1x B x H

= x 14 x 15= 105cm

2. Area of VWZ =2

1x B x H

= x 10 x 6= 30cm

3. Area of shaded reqion = Area 1 Area 2= 105 30= 75 cm

15 Area of whole diagram177


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. = (10 cm 15 cm) (5 cm 4 cm)= 150 cm - 20 cm= 130 cm

16.

Which of the following geometric solid is a cone?

A

Cube

B

Cylinder

C

Sphere

D

cone

17. Volume of the cone =

hr2

3

1

= 21431 2

= 21163

1

= 16 7= 112 cm

Volume of the cylinder = Volume of the cone

Height of the cylinder = 4 h = 112

16h = 112

178


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h =

16

112

= 7 cm

18.

Length of base = 36= 6 cm

Total surface area of pyramid = (4 2

1 6 cm 5 cm) + 36 cm

= 60 cm + 36 cm= 96 cm

19.

PQ = 5 3= 25 9= 16

PQ = 16= 4 cm

Total surface area of the prism= 2 (

2

1 3 cm 4 cm) + (3 cm 12 cm) + (5 cm 12 cm) + (4 cm 12 cm)

= 12 cm + 36 cm + 60 cm + 48 cm= 156 cm

20.

PSX = RSX

21. x = 2

130360 00

179

Locus X

P

S

Q

R

Locus X

P

S

Q

R


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=2

2300

= 115

22.

P = 180 - (50 + 70)= 180 - 120= 60

x = 180 - (90 + 60)= 180 - 110= 30

or

x = 180 - (110 + 40)= 180 - 150= 30

or

x = 70 - 40= 30

23. TOU = 40UOV = 60WOP = 50ROS = 30

Answer = B

24. Area of the shaded part =

27

7

22-

25.3

7

22

=

49722

-

25.12722

= 154 cm - 38.5 cm= 115.5 cm

25.

PT = 20 - 16= 400 256= 144

PT = 144

= 12 cm

QR = 12 + 5180


7/37

= 144 + 25= 169

QR = 169= 13 cm

26. CDBE

= ACAB

6

BE=9

6

BE =9

36

= 4 cm

27.

x + 3y = 9 .. i2x 3y = 6 .ii

i + ii; 3x = 15

x =3

15

= 5

28. 4

3y


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p = 9

30.

OP = 22 68 + = 3664 += 100= 10 units

OQ = 22 310 +

= 9100 += 109OQ 10

OR = 22 29 += 481+= 85

OR 10

OS = 22 68 +

= 3664 += 100= 10 units

182


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OT = 22 511 += 25121+= 146

OT 10

OP = OS

31.

Perimeter of the shaded region

= ( 147

222

4

1 ) + ( 77

222

4

1 ) + 7 cm + 21 cm

= 22 cm + 11 cm + 7 cm + 21 cm= 61 cm

32.

Mimi : RM 150 + ( 8 X RM 35 + 3 X RM 40 + 3 X RM 38 )= RM 664

Roger : RM 150 + ( 7 X RM 35 + 2 X RM 40 + 3 X RM 38 )= RM 589

Difference = RM 664 - RM 589= RM 75

33.

A : B B : C7 (3) 4 (3) 6 (2) 5 (2)

21 : 12 12 : 10

A : B : C = 21 : 12 : 10

34.

** Convert to 24Hours system

0740 + 0630 + 0020 = 1430

Time Daud arrive = 2.30 p.m.

35.

Total students = 160

x =160

32 360 = 72

y =160

36 360 = 81

153

183


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36.

0

5

10

15

20

25

30

A B C

BOYS

GIRLS

Diagram 24

The bar chart shows the nmber of students in three classes. Which of the following statementis true?A The total number of students in three classes is 125

A = 45, B = 40, C = 45A + B + C = 120

B The total number of girls in Class A and Class B are the same.

C The number of boys in this school is greater than the number of girls.Boys = 20 + 15 + 20

= 55

Girls = 25 + 25 + 15

= 65The number of boys in this school is less than the number of girls.

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D The number of girls in class C is greater than the number of girls in class B.

The number of girls in class B = 25The number of girls in class C = 15The number of girls in class C is less than the number of girls in class B.

37

.

Mode = B = 13

Choose the answer which is less than 13.Suitable answer is A = 12

38.

Total frequency = 30

Median =2

1615 datadata thth +

=2

33+

= 2

6

= 3

39.

y = 3x2 2xSubstitute x = -2Then, y = 3x2 2x

y = 3(-2)2 2(-2)y = 3 (4) + 4

y = 12 + 4y = 16

40.

y = x + 6The graph intersect y-axis at 6.(-6, 0) and (0, 6) certify the functiony =x + 6

185


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ANSWER TO SCOREPENILAIAN MENENGAH RENDAH 2009

MATHEMATICS

ANSWERS FOR PAPER 1 (SET 2)

1. B 21. C2. D 22. A3. C 23. B4. A 24. A5. C 25. B6. D 26. A7. B 27. C

8. D 28. C9. D 29. D10. B 30. C11. A 31. B12. B 32. C13. C 33. D14. D 34. C15. B 35. B16. A 36. C17. B 37. B18. D 38. C

19. C 39. D20. A 40. B

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MARKING SCHEME

1. Which of the following numbers when rounded off to the nearest hundred thousand becomes8700000 ?

A 8754321 = 8800000B 8654321 = 8700000 C 8643210 = 8600000D 8554321 = 8600000

2. 250 (4.5 30) (2.5 19) (0.5 18) = 58.50

3.(6

1 300) 20 = 30. Percentage durians left =

300

30 100% = 10%

4. P = -12 6= -18

Q = 18

5. Original price = RM 24.00

Selling price =11

2412

RM

= 22RM

Swee Lan uses2

1 of her savings.

Her savings was 222 RM= 44RM

6. A 238 = remainder 2

158 = remainder 3

B 339 =

159 = remainder 4C 8325 = remainder 1

187


14/37

5525 =

D 15345 =

9545 =

Common multiple of 3 and 5 is 45

7.5 rolls of rope x

13

2m = 17.5 m

5 rolls of rope x 1.69m = 8.45mLength of remaining rope = 17.5m 8.45m = 9.05m

8. 2400 + 0445 05152330

9. x +x + 35 + 62 + (180 - 147) = 180

2 x + 130 = 1802 x = 180 - 1302 x = 50x = 25

10.

Scale factor, k=PQ

Q'P'

k= 22

4=

11. Interior angle of regular hexagon = 6

180)26(

=

1206

720=

The value ofy =2

120180

=

302

60=

12.

3x +x = 180

x = 4180

= 45

Number of sides =

45

360

= 8 sides

188


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13.

1 m = RM 13.60

The Perimeter of Rectangular Compound

= 2 (25m) + 2 (75m)

= 50m + 150m

= 200m

Cost of fencing

= 200 RM 13.60

= RM 2720

14

. 2

1 14 h = 168

7h = 168

h =7

168

= 24 cm

15. 2

1 3 EF = 6

EF = 6 3

2

= 4 cm

Area of the whole diagram

= 6 cm 2 + (12 cm 4 cm)

= 6 cm 2 + 48 cm 2

= 54 cm 2

16.

Right prism

17.

Volume of cuboid: Volume of right prism = base area x height= 21 6 11 66 h = 1386

= 1386 h =66

1386

h = 21

189


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18. Aminah ages = 4

34

Bala ages =12

1013

12

19

12

94

12

19

4

34 =+=+

Chee Meng ages =

24

1146

24

1115

12

223

2

5

12

718

2

5

12

1013

12

94

2

5

12

1013

4

34

2

12 ==

=

=

+=

+

19.

J = K = 70

M = 90

JLM = 180 - (90 + 70)

= 180 - 160

= 20

20.

E = 100

x = 180 - (100 + 30)

= 180 - 130

= 50

21. Area of the shaded region=2

1 8

=2

1 64

= 32

22.

PQ = 10 - 8= 100 - 64= 36

PQ = 36= 6 cm

Area of the coloured region

= (5 ) (2

1 8 cm 6 cm)

= 25 - 24

23.

x = 180 - 120= 60

190


17/37

or

x =2

120

= 60

24. Height of triangle =22

1517 = 225289= 64= 8 m

Area of triangle =2

1 15 m 8 m

= 60 m

Number of paint tins = 60 m 5 m= 12

25.

180 - 70 = 110

Line XY non-parallel to line PQ.

26.

2x 3y = -17

2 (-4) 3y = -17

-8 3y = -17

3y = 17 8

= 9

y =3

9

= 3

27.

x -3 (a)x < 2 . (b)-3 x < 2

28.

C (-7, 8) and E (5, -2)

Let thex-coordinate of D bex : 2 57 + =x

191

70O

70O

X

Y

P

Q

70O

70O

X

Y

P

Q

70O

70O

X

Y

P

Q


18/37

2

2=x

Thus, x = -1

Let they-coordinate of D bey :2

)2(8 +=y

26

=y

Thus, y = 3

Then, the coordinate of D is ( -1, 3 )

29.

OJ = 5 units OM = 6 units

OK = 3 + 4 OL = 8 + 6= 25 = 100

OK = 5 units OL = 10 units

Hence, distance of OJ = distance of OK

30.

AC = 12 + 9= 225

AC = 15 = radius

Length of arc AD = Angle at the centre circumference360

= 150 2r

360= 150 2

7

22 15

360= 39.29 cm

Thus, the perimeter = 39.29 + 15 + 9 + 12= 75.29 cm.

31.

x : 624 = 12 : 13

624

x

= 1312

X = 12/13 624

= RM 576

32. 6

3=42

21

33

.

30 % of the total number of computer sold over the four month period is 1950

192


19/37

34.

10 square grids = 20 0001 square grid = 2 000

So , for February, 6 square grids = 12 000

35.

2y + 5y = 140

7y = 140

y =7

140

y = 20

Mangosteens = 5y

5y = 5 20

= 100

Number of mangosteens

= 252360

100

= 70

36.

Total income = RM1 400 + RM1 800 + RM1 350 + RM1 700= RM6 250

37

.

Min =5

46235 ++++

=5

20

= 4

38.

y = x + 2 because (0, 2) and (-2, 0) satisfy the equation and the graph intersect y-axis at 2.

39.

y = -2x2 + 1

Substitute x = -1Then, y = - 2x2 + 1

y = - 2(-1)2 + 1y = - 2 (1) + 1y = -2 + 1y = -1 so s = -1

193


20/37

Substitute x = 2Then, y = - 2x2 + 1

y = - 2(2)2 + 1y = - 2 (4) + 1y = -8 + 1y = -7 so t = -7

s + t = (-1 ) + (- 7)= 1 7= 8

40.

B (intersect at point G)

2009 FORECAST PAPER 2SET 1

Answer Scheme

1. 42

6 10= 7 10

194

B C

A D

E

FG

H


21/37

= 17

[ 2 marks]

2.

=4.0

44.110

10OR =

4

4.26

=4

4.14 = 3.6

= 3.6

[ 2 marks]

3. a)3

33

6

1

216

1

=

=6

1

b) = ( ) 21327+

= ( )214

= 196

[ 3 marks]

4. a) 2 ( r 2 ) = 10 5r2r 4 = 10 5r

7r = 14r = 2

b) s + 6 = 4s

2s + 6 = 8s

195


22/37

s 8s = 6 7s = 6

s =7

6

[ 3 marks]

5.

Image correctly drawn - 2 marks[ 2 marks]

6.TransformationPis a translation

6

8.

[2 marks]

7. a) x8 xy4 = )2(4 yxx

b) k6 183 + k = 3 (2k + k 6) = 3 (2k 3) (k+ 2)

[ 2 marks]

196

L

M

A B

C

D

C

D

B

A

L

M

A B

C

D

C

D

B

A


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8. a) 8x + 6

b) ( ) ( )443 xx

164

4123

+++

x

xx

[ 3 marks]

9. (3) 5(2 )

5

n n

mn

=

3 10 5

5

n n

mn

+=

8 10

5

n

mn

=

[ 3 marks]

10. a) x < 14 - 8 x < 6

b) - 3x < 11 - 2 and -3

1x 7 - 8

- 3x < 9 -3

1x - 1

- x < 3 - x - 3 x > - 3 x 3

x = - 2, - 1, 0, 1, 2, 3

[ 3 marks]

11. km 452 =+54

2 = km54 = km

[ 3 marks]

12. a) 2412 qpqp

= 2142 + qp

= 36qp

= 63

p

q

b) = 3 2

197


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Number of cars

= 9 [ 3 marks]

13.

Key :

WomenMan

Uniform scale 1 markEach set of bars 1 mark

[ 4 marks]

14. a)

Number Frequency

2 43 34 25 36 2

All the frequency correct 2 marks3 or 4 correct 1 mark

b) 5.32

43=

+

[3 marks]

15.= 634

2

1

= 36 cm [2 marks]

16. a) PR ( 2 marks )

198

Locus X

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

Kancil

Persona

Waja

Types of cars

Q

S


25/37

b) i. locus x correctly drawn ii. locus y correctly drawn c) intersection point correctly marked

[5 marks]

199

OP R

P

Locus y

Intersectionpoint


26/37

17.

a) uniform correct scale on y-axis 1 mark

b) 9 points correctly marked - 2 marksc) graph correctly drawn - 1 mark[4 marks]

200

y35

30

25

20

15

10

5

0

-5

-10

-15

-3 -2 -1 1 2 3 4 x


27/37

T

Q

18.

Construction of point S 1 markConstruction of rectangle PQRS 1 markConstruction of perpendicular line 1 markStraight line drawn from R 1 markLabel T correctly marked 1 mark

[5 marks]

19.

201

P

QR

S

Q

P

P

R

R

O


28/37

Image correctly drawn 2 marks[2 marks]

20.a) tan xo =

6

8

=3

4

b) length ofMPQ = 8+9 = 17 cm

[3 marks]

202


29/37

Set 2 Paper 2 ( Forecast Question )Answer Scheme

1.=2

3

15

10

15

21

=2

3

15

11

=10

11or 1

10

1

[2 marks]

2.= 9.1

5

13

= 9.1

135

= 3.5

[3 marks]

203


30/37

3.

Diagram 3Tcorrectly marked

[2 marks]

4.

Diagram 4

Quadrilateral WXYZ correctly drawn [2 marks]

5.

204

- 6 - 4 - 2 0 2 4 6 8

8

6

4

2

-2

x

y

P

Q R

Q R

P

T

- 6 - 4 - 2 0 2 4 6 8

8

6

4

2

-2

x

y

P

Q R

Q R

P

T

P

Q

R

S

W

X

Z

Y

P

Q

R

S

W

X

Z

Y


31/37

Diagram 5

Image correctly drawn [2 marks]

6. kp 53 = 35

+= kp

[2 marks]

7. a) 4 6x = 14 6x = 18

x = 3

b) 3m 10 = 8m 3m8m = 10

5m = 10m = 2

[4 marks]

8. = abbbab 8104736 ++ = 14ab 7b 3

[2 marks]

9.

205

L M

K

K

L

M

L M

K

K

L

M


32/37

O

S R

QP

Any 2 lines correctly drawn 1 markAny 3 lines correctly drawn 2 marksAll the line correctly drawn 3 marks

[3 marks]

10.

intersection of locus y and z

Diagram 10

206

Locus y

Locus z

Locus x


33/37

Locus x correctly drawn 1 markLocus y correctly drawn 1 markLocus z correctly drawn 1 markIntersection point correctly marked 1 mark

[4 marks]

11. a) = _2x 6 + 5x + 10x= 3x + 4

b) = n mmnmn 33 + = n mmn 34 +

[3 marks]

12.=

2

32 +a3

4 a

=6

)4(2)32(3 + aa

=6

8296 ++ aa

=6

174 +a

[3 marks]

13.a) JLM

b) KNLM = JN

JM

6

LM=3

5

LM =3

5x 6

= 10cm [3 marks]

14. 2k + 2 - 42k - 6

k - 3 4 - k > 0

207


34/37

- k > - 4k < 4

k = - 3, - 2, - 1, 0, 1, 2, 3

[3 marks]

15. a) May

(b)1

3502

175= [3 marks]

208


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16.Answer :

-2 -1 0 1 2 3 4 x

y- axis correctly marked 1 mark7 points correctly marked 2 marks ( 6 points 1 mark )All the 7 points joined 1 mark

Smooth curve - 1 mark[5 marks]

209

Y

10

8

6

4

2

-2


36/37

17.

90 correctly constructed at B 1 markBC correctly constructed 1 mark60 correctly constructed at A 1 markAD correctly constructed 1 markCD correctly joined 1 mark [5 marks]

18. = 2844

3 mmh37 mh

= 3287443 + mh = 9381 mh

=3

981

h

m

[3 marks]

19. = 23

2

3

2

1

2

1

2323 = 22 23 = 9 4= 36

[3 marks]

210

B

C

A

6 cm

3 cm

D

60


37/37

20.tan xo=

3

4

tan x =9

QR

9

QR=3

4

QR =3

4x 9

= 12

PR = 9 2 + 12 2 = 225= 15cm

[3 marks]

skemajawapan 2009

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