trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]

7
PEPERIKSAAN PERCUBAAN SPM MATHEMATICS FORM 5 SGI 2013 Kertas 1 Total = Paper 1 + Paper 2 = 40 + 100 = 140 1 C 11 D 21 B 31 B 2 B 12 C 22 A 32 A 3 C 13 D 23 C 33 D 4 D 14 B 24 B 34 A 5 C 15 C 25 B 35 D 6 D 16 D 26 D 36 A 7 A 17 B 27 D 37 B 8 A 18 C 28 B 38 C 9 C 19 D 29 C 39 D 10 B 20 B 30 B 40 A Kertas 2 Q Solution Submarks Full marks 1 Dotted line y = 3 drawn and Correct shaded region Note : 1. Solid line y = 2 drawn (1 mark) 2. Give 1 mark if shaded region between 2 inequalities only 3M 3 2 (a) 6 40 = 240 (b) 0 5 0 40 or equivalent = 8 ms 1 (units not important) (c) 1 1 ( 40 5) (240) 40 ( 11) 2 2 t t t ) 11 ( 40 2 1 ) 240 ( ) 5 40 2 1 ( = 28 or equivalent t = 15 s (units not important) 1M 1M 1M 1M 1M 1M 1M 7 3 3 1 4 22 21 2 3 7 2 or 5 . 10 5 . 10 5 . 10 7 22 3 4 2 1 or 2 22 14 35 7 2 or 35 7 7 7 22 Volume = 1 4 22 10.5 10.5 10.5 2 3 7 + 22 7 7 35 7 = 7815.5 or 2 1 7815 or 15631 2 cm 3 (units not important) 1M 1M 1M 3 x + y = 8 y = 8 2x y = 2 8 2 x 0 8 4 y

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Trial SPM 2013 / Kertas Percubaan SPM 2013 SMK St George Taiping

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Page 1: Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]

PEPERIKSAAN PERCUBAAN SPM MATHEMATICS FORM 5 SGI 2013

Kertas 1

Total = Paper 1 + Paper 2

= 40 + 100

= 140

1 C 11 D 21 B 31 B

2 B 12 C 22 A 32 A

3 C 13 D 23 C 33 D

4 D 14 B 24 B 34 A

5 C 15 C 25 B 35 D

6 D 16 D 26 D 36 A

7 A 17 B 27 D 37 B

8 A 18 C 28 B 38 C

9 C 19 D 29 C 39 D

10 B 20 B 30 B 40 A

Kertas 2

Q Solution Submarks Full

marks

1

Dotted line y = 3 drawn and Correct shaded region

Note : 1. Solid line y = 2 drawn (– 1 mark)

2. Give 1 mark if shaded region between 2 inequalities only

3M

3

2 (a) 640

= 240

(b)05

040

or equivalent

= 8 ms1(units not important)

(c)1 1

( 40 5) (240) 40 ( 11)2 2

t

t

t )11(402

1)240()540

2

1(

= 28 or equivalent

t = 15 s (units not important)

1M

1M

1M

1M

1M

1M

1M

7

3 31 4 22 21

2 3 7 2

or 5.105.105.10

7

22

3

4

2

1 or

222 14

357 2

or 35777

22

Volume =1 4 22

10.5 10.5 10.52 3 7

+

227 7 35

7

= 7815.5 or 2

17815 or

15631

2 cm

3 (units not important)

1M

1M

1M

3

x + y = 8

y = 8 – 2x

y = 2

8

2

x 0 8 4

y

Page 2: Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]

4

(a)(i) False/ Palsu

(ii) True/ Benar

(b) k + 1 is not an odd integer/ k + 1 bukan satu integer ganjil

(c) 4n – 3, n = 2, 3, 4, ... or equivalent

Note : 4n – 3 only (Give 1 mark)

1M

1M

2M

2M

6

5 LNM

tan LNM = 8

6 or equivalent

36.87o

or 36o 52

1M

1M

1M

3

6 3x2 − 5x − 2 = 0

(3x + 1) (x − 2) = 0

x = 3

1 , 2

1M

1M

1M,1M

4

7 4m – n = 26

4m + 3n = 2

–4n = 24 or

12 ( 6) 13

2m or 4m – (-6) = 26 or

or 4m + 3(-6) = 2

m = 5, n = 6

Note : 5

6

m

n

only award 1 mark

1M

1M

1M,1M

4

8

(a) M PR = 2 and c = 4

y = 2x + 4

(b) Coordinate T : (2, 0) or x = 2

y = 2(2) + 4 or equivalent

R(2, 8)

1M

1M

1M

1M

1M

5

9

(a) Sampel space/ Ruang sampel = {(S,C), (S,O), (S,R), (S,E),

(C,S), (C,O), (C,R), (C,E),

(O,S), (O,C), (O,R), (O,E),

(R,S), (R,C), (R,O), (R,E),

(E,S), (E,C), (E,O), (E,R)}

(b)(i) {(C,S), (C,O), (C,R), (C,E)}

4

20 or

1

5

(ii){(S,C), (S,R), (C,S), (C,R), (O,E), (R,S), (R,C), (E,O)}

8

20 or

2

5

Note : Accept correct answer only if list out

1M

1M

1M

1M

1M

5

10 (a)

60 222 14

360 7 or equivalent

Perimeter = 60 22

2 14360 7

+14 + 14 + 7 + 7 or equivalent

= 56.67 or 3

256 or

170

3 cm (units not important)

1M

1M

1M

1132

22

4 3

11331

21 2

2(3) 4( ) 4 22

5

6

m

n

m

n

Page 3: Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]

(b) 260 2214

360 7 or

277

22

360

60 or 290 22

7360 7

or equivalent

Area of the shaded region/ Luas kawasan berlorek

= 2 2 260 22 60 22 90 2214 7 14 7 7

360 7 360 7 360 7

or equivalent

= 136.5 or 1

1362

or 273

2 cm

2 (units not important)

1M

1M

1M

6

11

(a) Q = )4(3)5(2

1

53

42

=

53

42

2

1 or

1 2

3 52 2

(b)

8

13

23

45

y

x

8

13

53

42

2

1

y

x

2

1,3 yx

Note : 3

12

x

y

only award 1 mark

1M

1M

1M

1M

1M,1M

6

12

(a) x = –1, y = 12

x = 3, y = – 4

(b) Graph

All axis drawn with correct direction and uniform scale

7 points and 2 points* have been plot correctly

(7 or 8 points have been plot correctly get 1 mark)

Smooth curve and continue without straight line and passing through the

correct 8 points for –3 x 4.

(c) (i) 10 9y

(ii) 0.5 0.4x

(d) Identify equation y = 5x – 5

Straight line y = 5x – 5 correctly drawn

x = 0.4 ≤ x ≤ 0.5 , x = 3.75 x 3.85

1M

1M

1M

2M

1M

1M

1M

1M

1M

1M, 1M

12

13 (a) (i) (4, 9)

(ii) (4, 9) → (4, 3)

(b)(i)(a) U = Rotation, 900 clockwise, about centre (2, 4)

Note : 1. Rotation, 900 clockwise, get 2 marks

2. Rotation, about centre (2, 4), get 2 marks

3. Rotation, get 1 mark

(b) W = Enlargement, with centre (7, 4) , of scale factor 2

Note : 1. Enlargement, with centre (7, 4), get 2 marks

2. Enlargement, of scale factor 2 , get 2 marks

3. Enlargement, get 1 mark

(ii)22 A

22 120A A or equivalent

A = 40 m2

(units not important)

1M

1M, 1M

3M

3M

1M

1M

1M

12

Page 4: Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]

14 (a) (i) 12

(ii) I II III

Column I (All Correct)

Column II (All Correct)

Column III (All Correct)

(iii) (3 112 12 117 26 122 24 127 12 132 8 137 6 142 3 147)

94

= 126.95

(b) All axis in correct direction, uniform scale, x-axis label as upper boundry

All 8 points* correctly plot.

(6 or 7 points* correctly plot get 1 mark)

Ogive pass through point (109.5, 0)

Smooth curve without any straight line and pass all the 8 points correctly

(c) Median = 125.5 0.5

Height

(cm)

Tinggi

(cm)

Mid

point

Titik

Tengah

Frequency

Kekerapan

Cumulative

Frequency

Kekerapan

Longgokan

Upper

Boundaries

Sempadan

atas

105 109 107 0 0 109.5

110 – 114 112 3 3 114.5

115 – 119 117 12 15 119.5

120 – 124 122 26 41 124.5

125 – 129 127 24 65 129.5

130 – 134 132 x = 12 77 134.5

135 – 139 137 8 85 139.5

140 – 144 142 6 91 144.5

145 - 149 147 3 94 149.5

1M

1M

1M

1M

1M

1M

1M

2M

1M

1M

1M

12

15 (a) Plan

or

The shape must be correct in quadrilateral form. All lines must be drawn in full.

JG = NK > GN = KJ > JI = QK

The measurement is accurate to 0.2 cm. (one way ) and the angles at all

verticals of the rectangle are 90 1

(b) (i) Viewed from A

The shape must be correct. All lines must be drawn in full.

JV > VF > IH > GH > FG, UV > VZ = YU = FG

Curve must be smooth

The measurement is accurate to 0.2 cm. (one way ) and the angles at all

verticals of the rectangle are 90 1

1M

1M

1M

1M

1M

1M

1M

IH JE GF

2 cm 4 cm

5 cm

KL NM QP 5 cm

2 cm

4 cm

IH

JE

GF

KL

NM

QP

TCW

JK

Y ZSR

UX VEL

IQ

FM

HP GN

2 cm

5 cm

4 cm

3 cm

4 cm 6 cm

Page 5: Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]

(i) Viewed from B

The shape must be correct in quadrilateral form. All lines must be drawn in full.

ICGF and QWNM are two parallel straight lines

Note: Ignore straight line CW

C and W are joined dotted lines to shape a rectangular CFMW

VM = MQ > QI = TV > IC > CT > CG = WN

The measurement is accurate to 0.2 cm. (one way ) and the angles at all

verticals of the rectangle are 90 1

1M

1M

1M

1M

1M

12

16 (a) R = 70o N

(b) Q(70o S, 50

o W)

(c) 60)9070(

= 9600 n.m.

(d)(i) 600 2 =1200 n.m.

(ii) 70cos60

2600

= 58.48o or 58

o 29

58.48o + 50

o = 108.48

oW or 108

o 29

’W

2M

2M

2M

1M

1M

2M

1M

1M

12

3 cm

2 cm

3 cm

3 cm 5 cm

T

IJ

C

ZY

VU FE

QK

MLX

NPR

W

GHS

Page 6: Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]

Graph For Question 12

-1 -2 -3 -4

y

x 1 2 3

5

10

15

20

25

-5

-10

0

(-2, 16)

(-1, 12)

(0, 2)

(1, -8)

(2, -12)

(3.5, 4.4)

(4, 22)

(3, -4)

(-3, 8)

Page 7: Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]

Graph For Question 14

Graph for Question 14

(-3, -29)

50

10

60

40

20

30

70

80

90

100

0

Upper

Boundary

124.5 129.5 139.5 144.5 134.5 119.5 114.5 109.5 149.5

Cumulative

Frequency