Pembahasan mtk un 2013 paket 04

10
SMP NEGERI 3 KALIBAGOR Alfa Kristanti 1 PAKET 4

Transcript of Pembahasan mtk un 2013 paket 04

Page 1: Pembahasan mtk un 2013 paket 04

SMP NEGERI 3 KALIBAGOR

Alfa Kristanti

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PAKET 4

Page 2: Pembahasan mtk un 2013 paket 04

SMP NEGERI 3 KALIBAGOR

Alfa Kristanti

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Page 3: Pembahasan mtk un 2013 paket 04

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Alfa Kristanti

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Pelangi Matematika Viandre Al4-kristanty.blogspot.com

2−1 + 3−1 = 1

21+

1

31=

1

2+

1

3=

3

6+

2

6=

5

6

7 7 × 14 = 7 7 × 14 = 7 7 × 7 × 2 = 7 49 × 2

= 7 7 2 = 49 2

31

2+ 2

2

5∶ 1

1

5= 3

1

2+

12

5∶

6

5= 3

1

2+

12

5 ×

5

6

= 31

2+

2

1=

7

2+

4

2=

11

2

Selisih = 5 − 3

5 + 3 × 24 =

2

8 × 24 = 6

Bunga 9 bln = 9

12 × 12% × 𝑇 = 9% 𝑇

Tabungan setelah 9 bln = 100% T + 9% T = 109% T

109% T = 3.815.000 T = 3.815.000

109% =

3.815.000109

100

= 3.815.000 × 100

109

= 3.500.000

Page 4: Pembahasan mtk un 2013 paket 04

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b = 𝑈7− 𝑈3

7− 3=

22 −10

4

b = 12

4= 3

U3 = a + 2b

10 = a + 2(3)

10 = a + 6 a = 10 – 6

a = 4

Sn = 𝑛

2 ( 2a + (n – 1)b )

S30 = 30

2 ( 2(4) + (30 – 1) 3) = 15 (8 + (29) 3) = 15 (8 + 87)

= 15 (95) = 1.425

a = U1 = 3

b = U2 – U1 = 7 – 3 = 4

Un = a + (n – 1)b

U45 = 3 + (45 – 1) 4 = 3 + (44) 4

= 3 + 176 = 179

r = 𝑈2

𝑈1=

32

64=

1

2

a = U1 = 64

Un = a × r n – 1

= 64 × 1

2 𝑛−1

= 26 × 2 𝑛−1

Un = 26 × 2−𝑛 + 1 = 26 − 𝑛 + 1 = 27−𝑛

x – 3 ≤ 5 – 3x

x + 3x ≤ 5 + 3

4x ≤ 8

x ≤ 2

I. 2x2 + x – 3 = (2x + 3)(x – 1)

II. x2 + x – 6 = (x + 3)(x – 2)

III. 4x2 – 9 = (2x)

2 – 3

2 = (2x + 3)(2x – 3)

IV. 6x2 – 15x = 3x(2x – 5)

× ×

m = 𝑦2− 𝑦1

𝑥2 − 𝑥1=

2 − −5

3− 1=

2 + 5

2=

7

2

y – y1 = m (x – x1)

y – (5) = 7

2 (x – 1)

y + 5 = 7

2 (x – 1)

2(y + 5) = 7(x – 1)

2y + 10 = 7x – 7

2y – 7x = – 7 – 10

– 7x + 2y = – 17

7x – 2y = 17

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ax + by + c = 0

a = 6 ; b = 3

m = − 𝑎

𝑏=

− − 6

3=

6

3= 2

Jumlah bilangan terbesar dan terkecil

= 2

3 × jml ketiga bilangan

= 2

3 × 84 = 56

M = { 1, 3, 5, 7 }

N = { 5, 7, 11 }

M N = { 1, 3, 5, 7, 11 }

f(5) = 15

5a + b = 15

f(5) = 5

5a + b = 5

5a + b = 15

5a + b = 5

10a = 20

a = 2

5a + b = 5

5( 2) + b = 5

10 + b = 5

b = 5 + 10

b = 5

F(1) = 2 (1) + 5

= 2 + 5

= 3

Misal :

Harga pulpen = p

Harga pensil = q

5p + 4q = 30.000 × 3 15p + 12q = 90.000

2p + 6q = 23.000 × 2 4p + 12q = 46.000

11p = 44.000

p = 4.000

2p + 6q = 23.000

2(4000) + 6q = 23.000

8.000 + 6q = 23.000

6q = 15.000

q = 2.500

3p + 2q = 3(4000)+2(2500)

= 12.000 + 5.000

= 17.000

26 26

26 26

10 10 A

B

C

D

E

S = 𝐾

4=

104

4= 26

DE = 𝐴𝐷2 − 𝐴𝐸2 = 262 − 102 = 676 − 100 = 576 = 24

BD = 2 × DE = 2 × 24 = 48

Lbelahketupat ABCD = 1

2 × d1 × d2 =

1

2 × AC × BD = =

1

2 × 20 × 48 = 480

Pada gambar di samping:

ABC KLM maka

AB = LM, BC = KL,

AC = KM L

M

K 62

o

80o

B

A

C 62

o

80o

38o

38o

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P

16

12 10

Q R A

B

C

6 5

8

𝐴𝐶

𝑄𝑅=

8

16=

1

2 ,

𝐴𝐵

𝑃𝑅=

5

10=

1

2 ,

𝐵𝐶

𝑃𝑄=

6

12=

1

2

KL = 3 × 12 + 2 × 27

3 + 2

= 36 + 54

5=

90

5 = 18

L daerah tdk diarsir

= LEFGH + LABCD – 2 × L daerah diarsir

= 42 + 12 × 8 – 2 × 8

= 16 + 96 – 16

= 96

Banyak pohon cemara = 𝐾𝑒𝑙𝑖𝑙𝑖𝑛𝑔 𝑡𝑎𝑚𝑎𝑛

𝑗𝑎𝑟𝑎𝑘 𝑡𝑎𝑛𝑎𝑚=

2(48+ 32)

4

= 2(80)

4 = 40

Q

P

R

(3x 6) + (4x + 25) = 180

7x + 19 = 180

7x = 180 – 19

7x = 161

x = 23

Pelurus AOC = COB = 4x + 25

= 4(23) + 25 = 92 + 25 = 117

Page 7: Pembahasan mtk un 2013 paket 04

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Sudut keliling menghdp

busur yg sama maka

ABE = ACE =ADE

ABE = 96

3 = 32

AOE = 2 × ABE

= 2 × 32

= 64

= 289 − 64 = 225 = 15

dalam = 172 − 5 + 3 2 = 289 − 82

LAOD = 𝐴𝑂𝐷

𝐵𝑂𝐶 × 𝐿𝐵𝑂𝐶

= 40

100 × 50

= 20

Panjang kawat minimal = Panjang rusuk balok

= 4 (p + l + t) = 4 (40 + 40 + 35)

= 4 (115)

= 460

Garis pelukis

BC = 𝐾𝐴𝐵𝐶𝐷

4=

72

4 = 18

t

9

15 t = 152 − 92 = 225 − 81

t = 144 = 12

Vlimas = 1

3 La × t =

1

3 × 18

2 × 12 = 324 × 4 = 1.296

Page 8: Pembahasan mtk un 2013 paket 04

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dbola = rusuk kubus

dbola = 24, maka :

r = 24

2 = 12

Vbola = 4

3 π r

3 =

4

3 π × 12

3 =

4

3 π × 1.728

= 4 π × 576 = 2.304 π

Ldinding = 2 (pt + lt) = 2(6 × 4 + 8 × 4)

= 2(24 + 32) = 2(56) = 112

Biaya = 112 × Rp 50.000,- = Rp 5.600.000,-

s2 + s

2 = 10

2

2 s2 = 100

s2 = 50

s = 50 = 25 × 2 = 5 2

10 s

s

Lkubus = 6 s2

= 6 × 5 2 2

= 6 × 50

= 300

Ltabung = 2πr (r + t) = 2 × 22

7 × 35 (35 + 10)

= 220 (45) = 9.900

Jumlah siswa = 18 + 22 = 40 anak

Jml tinggi = 18 × 156 + 22 × 152 = 2.808 + 3.344 = 6.152

Rata-rata tinggi = 𝑗𝑚𝑙 𝑡𝑖𝑛𝑔𝑔𝑖

𝑏𝑎𝑛𝑦𝑎𝑘 𝑠𝑖𝑠𝑤𝑎=

6.152

40 = 153,8

Data 5 6 7 8 9

frekuensi 3 2 3 4 2

Modus

Page 9: Pembahasan mtk un 2013 paket 04

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32 30

Banyak siswa dapat nilai kurang dari 8 = 1 + 4 + 5 = 10

Page 10: Pembahasan mtk un 2013 paket 04

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n(S) = 2 × 2 × 2 = 8

3 gambar = GGG

P (3 gambar) = 1

8

6

5 5

2 2

4

3 3

Banyak kelereng = 6 + 5 + 3 + 3 + 2 + 4 + 2 + 5 = 30

Kelereng merah = 6

P ( 1 merah) = 6

30 × 100 % = 20 %