soal dan pembahasan un matematika smp 2014 paket 1
DESCRIPTION
www.smpn1gantung.sch.idTRANSCRIPT
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 1
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 2
PAKET 1
5
5=
5
5 . 5
5=
5 5
5= 5
60 βΆ 5 = 60: 5 = 12 = 4 .3 = 4 . 3 = 2 3
6456 = 26
56 = 26 .
56 = 25 = 32
ππ¦ππ = 60 β ππππ ππππππ πππππππ = 24 ππππ
ππ¦ππ = 60 β 15 = 45 β ππππ ππππππ πππππππ =60
45 .24 = 32 ππππ
21
3+ 5
1
4β 1
1
2=
7
3+
21
4β
3
2=
28
12+
63
12β
18
12=
73
12= 6
1
12
www.smpn1gantung.sch.id
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 3
π =π9 β π5
9 β 5=
20 β 8
4=
12
4= 3
π10 = π9 + 10 β 9 . π = 20 + 1 . 3 = 20 + 3 = 23
π =π7 β π3
7 β 3=
38 β 18
4=
20
4= 5
π = π1 = π3 + 1 β 3 .π = 18 β 2 . 5 = 18 β 10 = 8
ππ =π
2 . 2π + π β 1 .π
π24 =24
2 . 2 . 8 + 24 β 1 . 5 = 12 . 16 + 115 = 12 .131 = 1572
π΅ππππ ππ ππ’ππ π βΆ 20 , 23 ,β¦ , π20
π = 20
π = 3
ππ =π
2 . 2π + π β 1 .π
π20 =20
2 . 2 . 20 + 20 β 1 . 3 = 10 . 40 + 57 = 10 .97 = 970 ππ’ππ π
920000 = 800000 + 9% .π .800000
920000 β 800000 =9
100 . π .800000
120000 = 72000 .π
120000
72000= π
5
3= π
π =5
3 π‘πππ’π
π =5
3 .12 ππ’πππ
π = 20 ππ’πππ
2 . π + π = 144
2 . 3π₯ + 10 + π₯ + 10 = 144
2 . 4π₯ + 20 = 144 8π₯ + 40 = 144 8π₯ = 144 β 40
π₯ =104
8= 13
π = 3π₯ + 10 = 3 .13 + 10 = 39 + 10 = 49 ππ π = π₯ + 10 = 13 + 10 = 23 ππ
5π₯ β 3π₯ = 12 + 8 2π₯ = 20
π₯ =20
2
π₯ = 10 β π₯ + 3 = 10 + 3 = 13
π 9ππ + 21ππ = 3π . 3π + 7π
ππ π₯2 β 9 = π₯2 β 32 = π₯ β 3 . π₯ + 3
πππ 3π2 β π β 2 = 3π + 2 . π β 1
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 4
π π βͺ πΆ = 40 ; π π = 23 ; π π β© πΆ = 12
π π βͺ πΆ = π π + π πΆ β π π β© πΆ 40 = 23 + π πΆ β 12 40 = 11 + π πΆ 40 β 11 = π πΆ 29 = π πΆ π πΆ = 29 πππππ
π π = 3
π΅πππ¦ππ πππππ’πππ ππππππ ππππ π = 2π π = 23 = 8
3π₯ + 4π¦ = 17 β 3π₯ + 4π¦ = 17
4π₯ β 2π¦ = 8 β 8π₯ β 4π¦ = 16
11π₯ = 33
π₯ =33
11= 3
π₯ = 3 β 3π₯ + 4π¦ = 17
3 . 3 + 4π¦ = 17
9 + 4π¦ = 17
4π¦ = 17 β 9
π¦ =8
4= 2
2π₯ + 3π¦ = 2 . 3 + 3 . 2
= 6 + 6
= 12
3π΄ + 5π΅ = 39000 β 3π΄ + 5π΅ = 39000
π΄ + π΅ = 11000 β 3π΄ + 3π΅ = 33000
2π΅ = 6000
π΅ =6000
2= 3000
π΅ = 3000 β π΄ + π΅ = 11000
π΄ + 3000 = 11000
π΄ = 11000 β 3000
π΄ = 8000
4π΄ + 2π΅ = 4 . 8000 + 2 . 3000 = 32000 + 6000
= 38000
π π₯ = 3π₯ + 5 π π = 3π + 5 = β7 3π = β7 β 5
π =β12
3
π = β4
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 5
π₯ = 0 β π¦ = 2 . 0 β 1
π¦ = β1 0, β1 π¦πππ ππππππ’ ππ ππππππ ππππ ππππ πππ π΄ πππ π΅
π₯ = 2 β π¦ = 2 . 2 β 1
π¦ = 4 β 1
π¦ = 3 2,3 π½πππ π¦πππ ππππππ’ ππ ππππππ ππππ ππππ πππ π΄
π β3 π₯1
, 8 π¦1
πππ π 2 π₯2
, 5 π¦2
πππ =π¦2 β π¦1
π₯2 β π₯1
=5 β 8
2 β (β3)=
β3
2 + 3= β
3
5
ππ¦ππππ‘ π‘ππππ ππ’ππ’π βΆ π .πππ = β1
π . β3
5 = β1
π = β1 . β5
3
π =5
3
π΄. 3π₯ β 5π¦ β 14 = 0 β π =3
5
π΅. 3π₯ + 5π¦ + 14 = 0 β π = β3
5
πΆ. 5π₯ + 3π¦ β 42 = 0 β π = β5
3
π·. 5π₯ β 3π¦ β 42 = 0 β π =5
3
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 6
π΄ 2 π₯1
, 7 π¦1
; π΅ β3 π₯2
, β3 π¦2
; πΆ 3 π₯
, π π¦
π¦ β π¦1
π¦2 β π¦1
=π₯ β π₯1
π₯2 β π₯1
π β 7
β3 β 7=
3 β 2
β3 β 2
π β 7
β10=
1
β5
π β 7 =1
β5 . β10
π β 7 = 2
π = 2 + 7
π = 9
πππππππ π‘πππ πππππ ππ ππ = 1502 + 1502 = 1502 .2 = 150 2
πππππππ π‘πππ ππππππππ‘ππ = 1502 + 1502 = 22500 + 22500 = 45000 β 44944 = 212 π
πΆπΈ
π΄πΆ=
π·πΈ
π΄π΅
πΆπΈ
15=
8
12
πΆπΈ =8
12 .15
πΆπΈ = 10 ππ
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 7
πΏπππ ππππ =1
4 . πΏππππ
=1
4 . 102
=1
4 .100
= 25 ππ2
π΄π΅
πΆπ·=
π΄πΈ
πΈπΆ=
π΅πΈ
πΈπ·
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 8
πΎπππππ’π = 2 . 17 + 8 + 5 + 6 + 5 + 4
= 2 . 45 = 90 ππ
ππππππ ππ’π π’π π΄π΅ =β π΄ππ΅
360π . πΎπππππππππ
=60π
360π .2 .π . π
=60π
360π .2 . 3,14 .10
= 10,466 = 10,47 ππ
β π΄ + β π΅ = 180π
5π¦ β 16 π + 2π¦π = 180π
5π¦π β 16π + 2π¦π = 180π
7π¦π β 16π = 180π
7π¦π = 180π + 16π
7π¦π = 196π
π¦π =196π
7
π¦π = 28π
β π΄ = 5π¦ β 16 π
= 5 .28 β 16 π
= 140 β 16 π
= 124π
π΄π΅ = ππΏ2 + π β π 2
= 242 + 12 β 5 2 = 242 + 72 = 576 + 49 = 625 = 25 ππ
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 9
πππππ ππ = πΏπ‘πππππ ππ’π . π‘ππππ ππ
= 1
2 . π½π’ππππ π ππ π π ππππππ . π‘π‘πππππ ππ’π . π‘ππππ ππ
= 1
2 . 8 + 12 .5 . 10
= 50 . 10
= 500 ππ3
πΏπππππ’ππππ πππππ = πΏππππ πππ π΄π΅πΆπ· + 4 .πΏπ ππππ‘πππ π΅πΆπ
= π΄π΅2 + 4 .1
2 . π΅πΆ . ππ
= 162 + 4 .1
2 .16 .17
= 256 + 544
= 800 ππ2
π΄π΅ = π΅πΆ =πΎππππ πππ π΄π΅πΆπ·
4=
64
4= 16
ππ =1
2 .π΄π΅ =
1
2 .16 = 8
ππ = ππ2 + ππ2 = 152 + 82
= 225 + 64 = 289 = 17
ππ· = π =1
2 . π΄π΅ =
1
2 .14 = 7
ππ = 36 β π΄π· = 36 β 12 = 24
π·π = π = ππ·2 + ππ2 = 72 + 242
= 49 + 576 = 625 = 25
πΏπππππ’ππππ πππππ’π = πΏπππππ’ππππ π‘πππ’ππ π‘ππππ π‘π’π‘π’π + πΏπ πππππ’ π‘ ππππ’ππ’π‘
= ππ2 + 2πππ‘ + πππ
= 22
7 . 72 + 2 .
22
7 .7 .12 +
22
7 .7 .25
= 154 + 528 + 550
= 682 + 550
= 1232 ππ2
π π’π π’π = π΄π΅, π΅πΆ, πΆπ·, π·πΈ, πΈπΉ, πΉπ΄, π΄π, π΅π, πΆπ, π·π, πΈπ, πΉπ β 12
πππ π = π΄π΅πΆπ·πΈπΉ, π΄π΅π, π΅πΆπ, πΆπ·π, π·πΈπ, πΈπΉπ, πΉπ΄π β 7
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 10
π₯ =3 .3 + 4 .5 + 5 .12 + 6 .7 + 7 .6 + 8 .4 + 9 .3
3 + 5 + 12 + 7 + 6 + 4 + 3=
232
40= 5,8
π΅πππ¦ππ π ππ π€π π¦πππ ππ’ππ’π = 7 + 6 + 4 + 3 = 20 πππππ
π₯ ππππ’ππππ =ππ .π₯ π + 133 + 127
ππ + 1 + 1
=23 .130 + 133 + 127
23 + 1 + 1
=23 .130 + 260
25
=23 .130 + 2 .130
25
=130 . 23 + 2
25
=130 . 25
25
= 130
πππ’π‘ππ πππ‘π βΆ 165, 166, 168, 168, 170, 171, 171 ππππππ
, 172, 173, 173, 175, 178, 182
(π΅πππ’π π‘πππ‘π’ π₯ ππππ’ππππ = 130)
(π΅πππ’π π‘πππ‘π’ π₯ ππππ’ππππ = 130)
(π΅πππ’π π‘πππ‘π’ π₯ ππππ’ππππ = 130)
(πππ π‘π π₯ ππππ’ππππ = 130)
Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 11
150 + 250 = 400 πππππ
π ππππ ππππππππ πππππ ππππ 6 = 2
π π = 8
π ππππ ππππππππ πππππ ππππ 6 =π ππππ ππππππππ πππππ ππππ 6
π π =
2
8