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Question Solution and Marking Scheme Sub Marks 9

(a) (i) Median = = 52.7 (ii) Mean = [6(14.5) + 8(24.5) + 11(34.5) + 17(44.5) + 25(54.5) + 14(64.5) + 12(74.5) + 7(84.5)] = 51.7

(iii) Standard deviation =

301905100 51.72

= 18.61 *For 301905 or [6(14.5)2 + 8(24.5)2 + 11(34.5)2 + 17(44.5)2 + 25(54.5)2 + 14(64.5)2 + 12(74.5)2 + 7(84.5)2] and follow through for the mean (b) Axes with uniform scale and correct labelling for the boundaries or midpoints. All eight(8) bars correct. Correct method for determining the mode. 53.0 < mode < 54.0

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K1

1002 42

25 1049.5

1100

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K1

www.banksoalanspm.com

Question 9


Question Solution and Marking Scheme Sub Marks

10 (a) p = 0.65 and q = 0.35 or seen P(X 3) = 1 - P(X 3) or 1 - [P(X = 0) + P(X = 1) + P(X = 2)] or P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) 1 - 10C0(0.65)0(0.35)10 - 10C1(0.65)1(0.35)9 - 10C2(0.65)2(0.35)8 or 10C3(0.65)3(0.35)7 + 10C4(0.65)4(0.35)6 + 10C5(0.65)5(0.35)5 + 10C6(0.65)6(0.35)4 + 10C7(0.65)7(0.35)3 + 10 (0.35)

C8(0.65)8

+

0.9

2 0 10C9(0.65)9(0.35)1 + 10C10(0.65)10(0.35)

952

) ( = 1.1

1 - [P(Z

0.7675

(b i) z (ii) x = 44.7 kg (iii)

> 1.1) + P(Z > 1.3)]

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x 5510 1.03

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4210 55

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11 (a) (i) 7.1375 or 7.138 cm2 (ii) 43.565 or 43.57 cm (iii) * 46.4125 or 46.41 cm (b) (i) or 1.2 rad (ii) (5)2(1.2*) - BR2(1.2*) = 12.6 QR = 2*(1.2*) 2.4 cm Note: 1.2* follow through on and 2* follow through on BR

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K112 5

22 900sin

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K152 54 5

12 5

2 12 5

2 12 5

2 7.1375

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65 N1

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K112

12

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Full Marks

12

(a) (i) RM2.00 (ii) 137.5 (iii) RM2.40 (b) 131.17 (c) * RM3279.25 Note: 131.17* follow through on the composite index. (d) or 100.27

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125100 12580 137.550 15060 12030100 80 50 60 30

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P2006

2500 100 131.17 K1

N1

0.600.55 100 2.45

2.40K1100

10280 102100 10260 0.600.55 10050 2.45

2.40 10030

100 80 50 60 30K1

N1



Full Marks

13

(a) AE2 = 202 + AE2 - 2(AE)(20)cos 30o 11.55 cm (b) (20)(11.55*)sin 30o 2(CE)(10)sin 60o = (20)(11.55*)sin 30o 6.668 cm Note: 11.55* follow through on AE (c) CD2 = 6.668*2 + 102 - 2(6.668*)(10)cos 60o 8.819 cm Note: 6.668* follow through on CE (d) (i) 40.90o (ii)

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12 K1

12

12

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sinCDE6.668* 60o

8.819*sin

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C

EF

D

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Full Marks

14

(a) 6t - 8 -8 ms-2 (b) tA = 1 s or tB = s, 2 s 1 s t 2 s (c) for the correct limits 8 m (d) SA = 10t - 5t2 or SB = t3 - 4t2 + 4t t3 - 4t2 + 4t = 10t - 5t 2 s

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23

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1

3P1

10t 10t2

2 K1

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Full Marks

15

(a) I : 2x + 3y 90 II : 3x + 2y 180 III : 2x - y 40

(b) Draw correctly at least one (1) straight line of (a) with correct axes and uniform scale. All three (3) lines of (a) correctly drawn. Region R correctly shaded and labelled. (c) (i) (ii) 3x + 12y = any constant or its line representation in the graph 3(37*) + 12(34*) RM519 Note: 37* and 34* follow through from any clear indication or reading of a maximum point using any line of 3x + 12y in the region of R in the graph.

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34 N1

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Question 15 (b)


(SKEMA PERMARKAHAN MATEMATIK TAMBAHAN) PERCUBAAN 2007 KERTAS 2

NO.

SOALAN PENYELESAIAN PER

MARKAHAN 1 Menyatakan atau 3 2m n 7 723 6m mn

7 2

3

nm

atau

7 3

2

mn

atau setara

Menghapuskan m atau n

*2 7 3

3 62

mm m

7

atau * 2 *

7 2 7 23 6 7

3 3

n nn

Selesaikan persamaan kuadratik Menggunakan rumus atau Penyempurnaan kuasa dua m = 0.257 , 2.591 atau n = 3.114 , 7.387 ATAU n = 3.114 , 7.387 atau m = 0.257 , 2.591

P 1 P 1 K 1 K1 N 1 N 1


2

NO. SOALAN

PENYELESAIAN PER MARKAHAN

2 (a)

(b)

2( ) 2 ( 1) 1 3f x x atau setara

2( ) 2( 1) 1f x x

Nilai maksimum f(x) ialah -1

2 4 21x x 0 Memfaktorkan ungkapan kuadratik dan mencari julat dengan kaedah yang sesuai 7 3x

K1 N1 N1 K1 K1 N1

3(a)

(b)(i)

(ii)

Menggunakan rumus sudut berganda

sin 2 2sin cosx x x dan 2cos 2 1 2sinx x

memansuhkan sebutan dalam 2

2sin cos

2sin

x x

x

menyatakan cos

cotsin

xx

x

Mendapatkan tan2

p

Menggantikan ke dalam 2

2 tan2tan

1 tan

2

2tan

1

p

p

2

2sin

1

p

p

atau

2

2

1cos

1

p

p

menggantikan p = 2 ke dalam

** 2

2 2

2 12

1 1

p p

p p

24

25

P1 K1 N1 P1 N1 K1 K1 N1


3

NO. SOALAN

PENYELESAIAN PER MARKAHAN

4(a)

(b)

Mencari

dy

dx dengan betul dan

menggantikan 1x

kaMengguna n * 1m m1 2

2y x

34dy

xdx

3

4( 0.1)

4y

0.00625

1

1

1

K1 K1 N K K N1

5 (a)

(b)

Kamirkan

2(3 2 2)x x d x

3 2 2x x x c

-3) ke dalam

)

Gantikan (1,

* 3( 2 2y x x x c untuk mencari nilai c

3 2 2 3y x x x

Kamirkan

24

dyy

1

161

y

Gunakan had ke dalam *hasil kamiran

4

2

4

K1

1

1

K1 N1 K1 K N


4

6(a)

(b)(i)

(ii)

, , , ....2 16 128

xy xy xy

atau mana-mana tiga sebutan berturutan.

apatkan sekurang-kurangnya dua nisbah bagi

D

16 128

2 16

,xy xy

xy xy atau setara

1

8r

kan

Guna 1 146nar

1** 1 125600 6

4 4

n

atau Senaraikan setiap sebutan sehingga 1

46

25600, 3200, 400, 50, 146

5n

Gunakan 1

aS

r

*

*

25600, 1

11

8

r

1

292577

P1 K1

1

1

1

1

1

N K N K N


5

7 (a)

(b)

(c)

y 3.3 2.4

1.8 1.5 1.3 1.1

xy 1.65 3.60 4.50 5.25 5.85 6.05

Plo e y b n se )

Garis lurus penyuaian terbaik

t xy m lawan (paksi etul da skala ragam

6 *titik ditanda betul . xy by a

8.3a

kecerunan b 2.0b

1 b1

xy a a

ecerunan = 12

1

1

1

1

k 0.

N K1 N1 N P1 N1 K1 N P N1


6

8(a)(i)

(ii)

(b)(i)

(ii)

1 7 3 11

242 8 22 k

0

enggunakan

6 12 48k

6 1k or k

M 1 2 1m m

2

3

2m

*5y

3

4 2

x

311

2y x atau setara

Menggunakan

10AP 2 2( 1) ( 2) 10x y

2 2 2 4 5x y x y 0

Gantikan titik (4,1) ke dalam *persamaan lokus.

okus melalui titik (4,1)

K1

1

1

1

1

1

1

1

L

K N K K N1 K N K1 N


7


- drveenasforte.com · (b) draw correctly at least one (1) straight line of (a) with correct axes...

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