Pembahasan Soal Mtk UN 2013 Paket 02

download Pembahasan Soal Mtk UN 2013 Paket 02

of 10

  • date post

    27-Dec-2015
  • Category

    Documents

  • view

    73
  • download

    1

Embed Size (px)

description

Pembahasan Soal Ujian Nasional Mata pelajaran Matematika SMP Tahun Pelajaran 2013/2014

Transcript of Pembahasan Soal Mtk UN 2013 Paket 02

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    1

    PAKET 2

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    2

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    3

    Pelangi Matematika Viandre Al4-kristanty.blogspot.com

    23 + 32 = 1

    23+

    1

    32=

    1

    8+

    1

    9=

    9

    72+

    8

    72=

    17

    72

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

    = 7 7 2 = 49 2

    22

    3 1

    2

    3 4

    1

    5=

    8

    3

    5

    3

    21

    5=

    8

    3

    3

    5

    21

    5

    = 8

    5

    21

    5=

    13

    5= 2

    3

    5

    Selisih = 4 3

    4 + 3 560.000 =

    1

    7 560.000

    = 80.000

    Bunga 9 bln = 9

    12 12% = 9%

    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

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    4

    b = 18

    3= 6

    b = 7 4

    74=

    3719

    3

    U4 = a + 3b

    19 = a + 3(6)

    19 = a + 18 a = 19 18 a = 1

    Sn =

    2 ( 2a + (n 1)b )

    S24 = 24

    2 ( 2(1) + (24 1) 6) = 12 (2 + (23) 6) = 12 (2 + 138)

    = 12 (140) = 1.680

    a = U1 = 7

    b = U2 U1 = 12 7 = 5

    Un = a + (n 1)b

    U52 = 7 + (52 1) 5 = 7 + (51) 5

    = 7 + 255 = 262

    r = 21

    = 9

    27=

    1

    3

    a = U1 = 27

    Un = a r n 1

    = 27 1

    3 1

    = 33 31 1

    Un = 33 3 + 1 = 33 + 1 = 34

    3x 2 < 8x + 13 3x 8x < 13 + 2

    5x < 15

    x > 3

    I. 4x2 10x = 2x (2x 5)

    II. 7x2 49 = 7(x2 7)

    III. x2 3x 18 = (x 6)(x + 3)

    IV. x2 + 5x 36 = (x + 9)(x 4)

    m = 2 12 1

    = 3 (6)

    4 1 =

    3+6

    4+1=

    3

    3= 1

    y y1 = m (x x1)

    y (1) = 1 (x (6))

    y + 1 = 1 (x + 6)

    y + 1 = x 6

    y + x = 6 1

    x + y = 7

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    5

    ax + by + c = 0

    a = 8 ; b = 4

    m =

    =

    8

    4= 2

    M = { 1, 3, 5, 7 }

    N = { 5, 7, 11 }

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

    f(3) = 15

    3a + b = 15

    f(2) = 0

    2a + b = 0

    3a + b = 15

    2a + b = 0

    5a = 15 a = 3

    2a + b = 0

    2(3) + b = 0

    6 + b = 0

    b = 0 6

    b = 6

    f(5) = 3(5) + ( 6)

    = 15 6

    = 21

    Misal :

    Parkir mobil = p

    Parkir motor = q

    12 12 K

    L

    M

    N

    O

    Lbelahketupat ABCD = 384

    1

    2 KM LN = 384

    1

    2 24 LN = 384

    12 LN = 384

    4p + 2q = 18.000

    4p + 2(1.000) = 18.000

    4p + 2.000 = 18.000

    4p = 16.000

    p = 4.000

    20p + 30q = 20(4000)+30(1000)

    = 80.000 + 30.000

    = 110.000

    3p + 5q = 17.000 4 12p + 20q = 68.000

    4p + 2q = 18.000 3 12p + 6q = 54.000 14q = 14.000

    q = 1.000

    LN = 384

    12 = 32 OL =

    2=

    32

    2 = 16

    KL = 2 + 2 = 122 + 162

    KL = 144 + 256 = 400 = 20 Keliling = 4 KL = 4 20 = 80

    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

    Jumlah bilangan terbesar dan terkecil

    = 2

    3 jml ketiga bilangan

    = 2

    3 162 = 108

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    6

    Q

    30

    24

    18

    P R B

    C

    A

    12 9

    15

    =

    12

    24=

    1

    2 ,

    =

    15

    30=

    1

    2 ,

    =

    9

    18=

    1

    2

    FC = 4 20 + 6 5

    4 + 6

    = 80 + 30

    10=

    110

    10 = 11

    L daerah tdk diarsir = LKLMN + LPQRS 2 L daerah diarsir = 20

    2 + 10 15 2 67

    = 400 + 150 134 = 416

    Banyak pohon =

    =

    2(24+ 18)

    3

    = 2(42)

    3=

    84

    3 = 28

    B

    A

    C D

    (3x 5) + (2x + 5) = 90 5x + 0 = 90

    5x = 90

    x = 18

    Penyiku POR = ROQ = 2x + 5 = 2(18) + 5 = 36 + 5 = 41

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    7

    Sudut keliling menghdp

    busur yg sama maka

    ABE = ACE = ADE

    ABE = 96

    3 = 32

    AOE = 2 ABE = 2 32

    = 64

    = 676 100 = 576 = 24

    luar = 262 12 2 2 = 676 102

    LOBC =

    = 120

    50 30

    = 72

    Panjang rusuk 1 balok = 4 (p + l + t) = 4 (40 + 24 + 36)

    = 4 (100) = 400 cm = 4 m

    Kawat tersedia = 10 m

    Banyak kerangka balok yg dibuat = 10

    4 = 2,5 2

    Garis pelukis

    BC =

    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

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    8

    dbola = rusuk kubus

    dbola = 30, maka :

    r = 30

    2 = 15

    Vbola = 4

    3 r3 =

    4

    3 153 =

    4

    3 3.375

    = 4 1.125 = 4.500

    Ldinding = 2 (pt + lt) = 2(8 5 + 6 5)

    = 2(40 + 30) = 2(70) = 140

    Biaya = 140 Rp 50.000,- = Rp 7.000.000,-

    s2 + s

    2 = 8 2

    2

    2 s2 = 64 2

    2 s2 = 128

    s2 = 64 s = 64 = 8

    8 2 s

    s

    Lkubus = 6 s2

    = 6 8 2 = 6 64

    = 384

    d = 28 r = 14

    Ltabung = 2r (r + t) = 2 22

    7 14 (14 + 26)

    = 88 (40) = 3.520

    Jumlah siswa = 18 + 22 = 40 anak

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

    Rata-rata tinggi =

    =

    6.152

    40 = 153,8

    Data 5 6 7 8 9

    frekuensi 2 3 4 3 2

    Modus

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    9

    32 30

    Selisih produksi pupuk Maret dan Mei = 10 6 = 4

  • SMP NEGERI 3 KALIBAGOR

    Alfa Kristanti

    10

    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 %