Percubaan Addmath Kelantan 2009 Paper 2

10 Solution by scale drawing is not accepted. Penyelesaian secara lukisan berskala tidak diterima.

Diagram 10 shows a triangle PQR. Point S lies on PQ.

Rajah 10 menunjukkan segitiga PQR. Titik S terletak pada garis PQ.

f(x)

P (5, k)

S(2, 3)

O R(4,0) x

Q

Diagram 10 Rajah 10

(a) A point M moves such that its distance from point S is always 5 units.

Find the equation of the locus of M. [3 marks]

Suatu titik M bergerak dengan keadaan jaraknya dari titik S adalah sentiasa 5 unit.

Cari persamaan lokus M. [3 markah]

(b) It is given that point P and Q lie on the locus of M. Diberi bahawa titik P dan titik Q terletak pada lokus M.

Calculate Hitung

(i) the value of k, nilai k,

(ii) the coordinates of Q.

koordinat Q. [5 marks] [5 markah]

(c) Hence, find the area, in unit2 , of triangle PQR. [2 marks] Seterusnya, cari luas, dalam unit2, bagi segitiga PQR. [2 markah]

Percubaan Addmath Kelantan 2009 Paper 2

11 (a) A student is considered pass the test whenever six of ten questions are being answered correctly. If 8 students are chosen at random, calculate the probability that

Seorang pelajar dianggap berjaya dalam ujian jika dapat menjawab enam dari sepuluh soalan yang diberi. Jika 8 orang pelajar dipilih secara rawak, hitung

kebarangkalian

(i) exactly 2 students pass,

tepat 2 orang pelajar berjaya,

(ii) more than 1 student pass the test.

lebih daripada seorang berjaya dalam ujian tersebut.

[5 marks] [5 markah]

(b) The heights of students in School A are normally distributed with a mean of 150 cm and a standard deviation of 10 cm.

Ketinggian pelajar di Sekolah A adalah bertaburan secara normal dengan minnya 150 cm dan sisihan piawai 10 cm.

(i) Height exceeds 170 cm are classified as tall. It is found that 57 students are

tall. Find the total enrolment of the school.

Ketinggian melebihi 170 cm dikatakan tinggi di sekolah itu. Didapati 57 pelajar adalah tinggi. Cari enrolmen sekolah itu.

(ii) If 30% of the students are categorized as short. What is the height which is still short?

Jika 30% daripada pelajar itu dikategorikan rendah. Berapakah ukuran ketinggian yang masih dikatakan rendah?

[5 marks] [5 markah]

Percubaan Addmath Kelantan 2009 Paper 2

Section C Bahagian C

[20 marks]

[20 markah]

Answer any two questions from this section. Jawab mana-mana dua soalan daripada bahagian ini.

12 A particle moves along a straight line and passes through a fixed point O. Its velocity, v m s-1, is given by v = 2t2 – 7t + 3, where t is the time in seconds after leaving points O. Satu zarah bergerak di sepanjang suatu garis lurus dan melalui satu titik tetap O. Halajunya, v m s-1, diberi oleh v = 2t2 – 7t + 3, dengan t ialah masa dalam saat selepas melalui titik O.

Find Cari

(a) (i) the time interval during which the particle moves towards the left,

julat masa apabila zarah itu bergerak arah ke kiri,

(ii) the acceleration of the particle when t = 2 seconds,

pecutan zarah itu apabila t = 2 saat,

(iii) the time when the velocity is constant.

masa apabila halajunya malar.

[5 marks] [5 markah]

(b) Sketch the velocity-time graph of the motion of the particle for O ≤ t ≤ 3.

Lakarkan graf halaju-masa bagi pergerakan zarah itu untuk O ≤ t ≤ 3.

[2 marks] [2 markah]

(c) Calculate the total distance, in m, traveled during the first 3 seconds after leaving point O.

Hitung jumlah jarak, dalam m, yang dilalui dalam 3 saat yang pertama selepas melalui titik O.

[3 marks] [3 markah]

Percubaan Addmath Kelantan 2009 Paper 2

13 Table 13 shows the prices, the price indices and percentage of usage of four items A, B, C and D, which are the main ingredients in the manufacturing of a types of biscuits.

Jadual 13 menunjukkan harga, indeks harga dan peratus penggunaan empat barangan A, B, C dan D, yang merupakan bahan utama dalam penghasilan sejenis biskut.

Price per unit (RM) Harga per unit (RM)

Item Item 2005 2007

Price index for the year 2007 based on the year 2005

Indeks harga pada tahun 2007 berasas tahun 2005

Percentage of usage (%)

Peratus penggunaan (%)

A p 45 125 7m

B 55 q 120 8m

C 40 42 105 28

D 50 47 r 9m

Table 13 Jadual 13

(a) Find the value of p, of q and of r. [3 marks]

Cari nilai p, nilai q dan nilai r. [3 markah]

(b) State the value of m. Hence, calculate the composite index for the cost of manufacturing the biscuits in the year 2007 based on the year 2005. [3 marks]

Nyatakan nilai m. Seterusnya, hitung nombor indeks gubahan bagi kos penghasilan biskut itu pada tahun 2007 berasaskan tahun 2005. [3 markah]

(c) Calculate the price of a box of biscuits in the year 2005 if the corresponding price in the year 2007 is RM 25.70

[2 marks] Hitung harga sekotak biskut yang sepadan pada tahun 2005 jika harganya pada tahun 2007 ialah RM 25.70 [2 markah]

(d) The coast of manufacturing the biscuits is expected to increase by 20% from the year 2007 to the year 2009. Find the expected composite index for the year 2009

based on the year 2005. [2 marks]

Kos penghasilan biskut ini dijangka meningkat sebanyak 20% dari tahun 2007 ke tahun 2009. Cari nombor indeks gubahan yang dijangkakan pada tahun 2009 berasaskan tahun 2005. [2 markah]

Percubaan Addmath Kelantan 2009 Paper 2

14 Diagram 14 shows two triangle ABC and ACD. ∠ABC is obtuse and sin ∠ABC = 53 .

Given that AB = 7 cm, BC = 5 cm and CD = 9 cm.

Rajah 14 menunjukkan dua segitiga ABC dan ACD. ∠ ABC adalah cakah dan sin ABC = ∠53

.

Diberi bahawa AB = 7 cm, BC = 5 cm and CD = 9 cm.

Diagram 14 Rajah 14 Calculate Hitung

(a) the length , in cm, of AC, [4 marks] panjang, dalam cm, bagi AC, [4 markah]

(b) ADC, [2 marks] ∠

[2 markah]

(c) the area, in cm2 , of the ABCD. [4 marks] luas, dalam cm2, ABCD. [4 markah]

15 Use graph paper to answer this question.

40o

D

A

B

C

Percubaan Addmath Kelantan 2009 Paper 2

Gunakan kertas graf untuk menjawab soalan ini.

The English Language society sells x packets of snack A and y packets of snack B at

Inggeris menjual x bungkus snek A dan y bungkus snek B pada hari

snacks is based on the following constraints :

: The number of packets of snack B is at least 250.

ber of snacks A and B is at least 500 but not more than 800.

(a) Write three inequalities, other than x

a school funfair.

Persatuan Bahasa

pasaria sekolah. The number of packets of

Bilangan bungkus snek berdasarkan pada kekangan berikut :

I : The number of packets of snack A is at least 150.

Bilangan bungkus snek A sekurang-kurangnya 150. II

Bilangan bungkus snek B sekurang-kurangnya 250.

III : The total num

Jumlah bilangan snek A dan B sekurang-kurangnya 500 tetapi tidak lebih dari 800.

0 and y 0 , which satisfy all the 3 marks]

above constraints [

Tulis tiga ketaksamaan selain dari x 0 and y 0 , yang memenuhi semua

(b) By using a scale of 4 cm to 200 packets of a snacks on both axes, traints. [3 marks]

[3 markah]

(c) The profit of a packet of snack A is RM0.15 and the profit of snack B is RM0.40

lah RM0.15 dan keuntungan bagi snek B ialah

s A and B,

and B.

[4 marks]

END OF QUESTION PAPER

kekangan di atas. [3 markah]

construct and shade the region R which satisfies all the above cons

Dengan menggunakan skala 4 cm kepada 200 bungkus snek pada kedua-dua paksi,

bina dan lorek rantau R yang memenuhi semua kekangan di atas.

by using the graph from (b), find

Keuntungan bagi sebungkus snek A ia RM0.40, dengan menggunakan graf dari (b), cari

(i) the maximum profit from the sale of snack keuntungan maksimum jualan snek A dan B,

(ii) the minimum profit from the sale of snacks A keuntungan minimum jualan snek A dan B

[4 markah]

KERTAS SOALAN TAMAT

ADDITIONAL MATHEMATICS PEPERIKSAAN PERCUBAAN SPM 2009(KELANTAN)

MARKING SCHEME PAPER 2

No Solution Marks Total

1

n = 12

m – 1 @ 2n = m – 2 @ m = 2n + 2

m2 – 9 = 2( 12

m – 1) @ m2 – 9 = m – 2 (2n + 2)2 – 9 = 2n

m2 – m – 7 = 0 4n2 + 6n – 5 = 0

2( 1) ( 1) 4(1)( 7)2(1)

m− − ± − − −

= @ 26 (6) 4(4)( 5)

2(4)n

− ± − −=

m = 3.193, – 2.193 n = 0.596, – 2.096

n = 0.597, – 2.097 m = 3.192, – 2.192

P1

K1

K1

N1(both) N1(both)

5

2(a)

(b)

(c)

, ∴k = 2

When x = 2, 2

2 3

161 3(2

d ydx

= + = > 0)

so, the turning point (2,6) is a minimum point

2

8y x dxx

= −∫

2 8

2xy c

x= + +

Therefore,

K1 N1

K1

N1

K1

K1

N1

7

3(a)

(b)

Let x is diameter of smallest circle first term a = x and common difference d = 2 [2( x) + 4(2 )] = 60 x = 8 a = 8 d = 4.5

[ ]2(8 ) ( 1)2 2602n nπ π π+ − =

n2 + 7n − 260 = 0 (n + 20)(n − 13) = 0 n = 13

P1 K1

N1

P1

K1

K1 N1

7

4(a)

(b)

|sin x| =

π2x

−|sin x| = −π2x

y = −|sin x| + 1

= −π2x + 1

The number of solution is 4

P1 P1 P1 P1

K1

K1(Graph)

N1

7

1

2л л 3л 2

0 л 2

y = −|sin x| + 1

× y = −sin x

y = −|sin x| y = −|sin x| + 1

5(a)

(b)

(c)

4 12

Times taken

x f fx fx2

10 – 12 11 4 44 484 13 – 15 14 12 168 2352 16 – 18 17 15 255 4335 19 – 21 20 8 160 3200 22 – 24 23 6 138 3174

Σfx = 765

Σfx2 =13545

σ = 213545 765

45 45⎛ ⎞− ⎜ ⎟⎝ ⎠

= 3.464

P1

P1

K1

K1 K1

N1

6

6(a)(i)

(ii)

(b)

RP RQ QP= +

uuur uuur uuur

= 6x + 5 y

OQ OP PQ= +uuur uuur uuur

= 2x – 5y

OS = h OQ = h ( 2x – 5 y ) = 2hx – 5hy PS = k PR = k (–6x –5 y ) = –6kx – 5ky OS = OP + PS = 2 x – 6kx – 5ky = (2 – 6k)x – 5ky 2hx – 5hy = (2 – 6k)x – 5ky 2h = 2 – 6k or 5h = 5k

N1

N1

P1

P1

K1

K1

7

when h = k 2h = 2 – 6h h = ¼ and k = ¼

N1

7(a)

(b)

(c)

y² = 4 – (y + 2)

(y + 2)(y – 1) = 0

y = – 2 , y = 1

when y = 1, x = 1 + 2 ; P(3, 1)

Area =

=

=

= ]

= unit2

Volume = =

= =

=

K1

N1

K1

K1 K1

N1

K1

K1

K1

N1

10

9(a)

(b)

(c)

(d)

OP = 6 cm cos = 0.6 = 0.9274 PN = 8 cm MN = 10 x 0.9274 = 9.274 Perimeter = 8 + 9.274 + 4 = 21.274 cm The area = (0.9274) - x 6 x 8 = 22.37 cm2

P1

K1 N1 P1 K1 N1

N1

K1 K1 N1

10

10(a)

(b)(i)

(ii)

(c)

= 0

= 7 k > 0, k = 7

Midpoint of PQ = (2, 3)

or = 3

Therefore, Q (-1,-1)

= 17 unit2

K1 K1

N1

K1

K1

N1

K1

N1

K1

N1

10

11(a)(i)

(ii)

(b)(i)

p = 0.6 q = 0.4 p(X = 2) = 8 C (0.6) (0.4) 2

2 6

= 0.04129 p(X > 1) = 1 − p(X = 0) − p(X= 1)

= 1 − 8C (0.6)0 (0.4) − 08 8C1 (0.6)1 (0.4) 7

= 0.9915

P(X > 170) = P(Z > 170 15010− )

= P(Z > 2) = 0.0228

P1 K1 N1

K1

N1

K1

Area PQR

(ii)

0228.057 = 2500

P(X < k) = 0.3

P(Z < 15010

k − ) = 0.3

10150−k = −0.524

k = 144.76 cm

N1

K1 N1

N1

12(a)(i)

(ii)

(iii)

(b)

2t2 – 7t + 3 < O (2t – 1)(t – 3) < O

1 32

t< <

dvadt

=

= 4t – 7 = 4(2) – 7 = 1 m s-1 4t – 7 = O

t = 74

when t = O, v = 3 when v = O, 2t2 – 7t + 3 = O (2t – 1)(t – 3) = O

t = 12

or 3

when t = 74

,

v = 2( 74

)2 -7( 74

) + 3 = – 3 18

K1

N1

K1

N1

N1

(c)

s = 2(2 7 3)t t d− +∫ t

= 3 22 7 33 2

t t t− + + c

When t = O, s = O, c = O

= 3 22 7 33 2

t t t− + + c

When t = 12

, s = 3 22 1 7 1 1( ) ( ) 3( )3 2 2 2 2

− + = 1724

When t = 3, s = 3 22 7(3) (3) 3(3) 43 2

− + = −12

Total distance traveled = 1724

+ 1724

+ 142

= 5 1112

m

K1

v (m s-1)

3

N1 N1 [any 2 points

include (74

,– 3 18

)] t(s)

– 3 18

74

0 1

2 3

t = 3

t = O

142

−

1

O

K1

N1

1724

2t =

13(a)

(b)

(c)

(d)

45 100p

= ×125 or 120 or 10055q

= ×47 10050

r = ×

p = 36 q = 66 r = 94

7m + 8m + 28 + 9m = 100 m = 3

125(21) 120(24) 105(28) 94(27)100

I + + +=

= 109.83

2005

2005

25.70 100 109.83

23.40P

P RM

× =

=

2009

2009

100 12025.70

30.8430.84 10023.40131.79

P

P

I

× =

=

= ×

=

K1

N2, 1, 0

P1

K1

N1

K1

N1

K1

N1

10

14(a)

(b)

(c)

sin ABC = 3/5 = 36o 52’ // 36.87o

∠ ABC = 180o - 36o 52’ = 143o 8’ AC2 = 72 + 52 – 2 x 7 x 5 cos 143o 8’ = 11.40 cm

11.40

SinADC = 940Sin

sin ADC = 0.8142 ∠ ADC = 54o 30’ // 54.51o ∆ ABC = ½ x 7 x 5 x Sin 143o 8’ = 10.499 cm2 ∠ACD = 180 o – 40o – 54o 30’ = 85o 30’ AADC = ½ x 9 x 11.40 x Sin 85 o 30 ’

= 51.142 cm2 Area of ABCD = 10.499 + 51.142 = 61 .641 cm2

P1

P1 K1 N1

K1

N1

K1

P1

K1

N1

10

120 109.83 1000 100

×OR 10

I = ×

= 131.79

No. 8

y2 x 1.0 1.5 2.0 2.5 3.0 3.5

y2 32.49 42.25 53.29 64.00 75.69 86.49

Correct at least 2 decimal places

x 0.5 1.0 1.5 4.0 3.5 3.0 2.5 2.0

40

30

20

10

0

50

60

70

80

X

X

X

(a) ~ Correct axes, uniform scale and one point plotted correctly K1 ~ All 6 points plotted correctly N1 ~ Line of best fit N1 y2 = p2 x + p2 q P1

(b) (i) gradient, p2 = 64 212.5 0.5

−−

K1

= 21.5 p = 4.637 N1 (ii) y-intercept, p2 q = 10 K1 q = 0.465 N1 (iii) y = 7 , y2 = 49 x = 1.8 N1

X

90

X

X

N1

No.15

a) i)x ≥ 150

ii)y 250 ≥iii) 500≤ x + y≤ 800

c) max, Profit

=0.15(150) + 0.4(650) = RM282.50 min, Profit =0.15(250) + 0.4(250) = RM137.50

R