Soal dan pembahasan un matematika smp 2014 paket 1

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Soal dan Pembahasan UN MATEMATIKA SMP 2014 / Page 1
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PAKET 1
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5
5
=
5
5
.
5
5
=
5 5
5
= 5
60 ∶ 5 = 60: 5 = 12 = 4 .3 = 4 . 3 = 2 3
64
5
6 = 26
5
6 = 26 .
5
6 = 25
= 32
𝑎𝑦𝑎𝑚 = 60 → 𝑝𝑒𝑟𝑠𝑒𝑑𝑖𝑎𝑎𝑛 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 = 24 𝑕𝑎𝑟𝑖
𝑎𝑦𝑎𝑚 = 60 − 15 = 45 → 𝑝𝑒𝑟𝑠𝑒𝑑𝑖𝑎𝑎𝑛 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 =
60
45
.24 = 32 𝑕𝑎𝑟𝑖
2
1
3
+ 5
1
4
− 1
1
2
=
7
3
+
21
4
−
3
2
=
28
12
+
63
12
−
18
12
=
73
12
= 6
1
12

𝑏 =
𝑈9 − 𝑈5
9 − 5
=
20 − 8
4
=
12
4
= 3
𝑈10 = 𝑈9 + 10 − 9 . 𝑏 = 20 + 1 . 3 = 20 + 3 = 23
𝑏 =
𝑈7 − 𝑈3
7 − 3
=
38 − 18
4
=
20
4
= 5
𝑎 = 𝑈1 = 𝑈3 + 1 − 3 . 𝑏 = 18 − 2 . 5 = 18 − 10 = 8
𝑆𝑛 =
𝑛
2
. 2𝑎 + 𝑛 − 1 . 𝑏
𝑆24 =
24
2
. 2 . 8 + 24 − 1 . 5 = 12 . 16 + 115 = 12 .131 = 1572
𝐵𝑎𝑟𝑖𝑠𝑎𝑛 𝑘𝑢𝑟𝑠𝑖 ∶ 20 , 23 , … , 𝑈20
𝑎 = 20
𝑏 = 3
𝑆 𝑛 =
𝑛
2
. 2𝑎 + 𝑛 − 1 . 𝑏
𝑆20 =
20
2
. 2 . 20 + 20 − 1 . 3 = 10 . 40 + 57 = 10 .97 = 970 𝑘𝑢𝑟𝑠𝑖
920000 = 800000 + 9% . 𝑛 .800000
920000 − 800000 =
9
100
. 𝑛 .800000
120000 = 72000 . 𝑛
120000
72000
= 𝑛
5
3
= 𝑛
𝑛 =
5
3
𝑡𝑎𝑕𝑢𝑛
𝑛 =
5
3
.12 𝑏𝑢𝑙𝑎𝑛
𝑛 = 20 𝑏𝑢𝑙𝑎𝑛
2 . 𝑝 + 𝑙 = 144
2 . 3𝑥 + 10 + 𝑥 + 10 = 144
2 . 4𝑥 + 20 = 144
8𝑥 + 40 = 144
8𝑥 = 144 − 40
𝑥 =
104
8
= 13
𝑝 = 3𝑥 + 10
= 3 .13 + 10
= 39 + 10
= 49 𝑐𝑚
𝑙 = 𝑥 + 10
= 13 + 10
= 23 𝑐𝑚
5𝑥 − 3𝑥 = 12 + 8
2𝑥 = 20
𝑥 =
20
2
𝑥 = 10 → 𝑥 + 3 = 10 + 3 = 13
𝑖 9𝑎𝑏 + 21𝑎𝑐 = 3𝑎 . 3𝑏 + 7𝑐
𝑖𝑖 𝑥2
− 9 = 𝑥2
− 32
= 𝑥 − 3 . 𝑥 + 3
𝑖𝑖𝑖 3𝑝2
− 𝑝 − 2 = 3𝑝 + 2 . 𝑝 − 1

𝑛 𝑃 ∪ 𝐶 = 40 ; 𝑛 𝑃 = 23 ; 𝑛 𝑃 ∩ 𝐶 = 12
𝑛 𝑃 ∪ 𝐶 = 𝑛 𝑃 + 𝑛 𝐶 − 𝑛 𝑃 ∩ 𝐶
40 = 23 + 𝑛 𝐶 − 12
40 = 11 + 𝑛 𝐶
40 − 11 = 𝑛 𝐶
29 = 𝑛 𝐶
𝑛 𝐶 = 29 𝑜𝑟𝑎𝑛𝑔
𝑛 𝑃 = 3
𝐵𝑎𝑛𝑦𝑎𝑘 𝑕𝑖𝑚𝑝𝑢𝑛𝑎𝑛 𝑏𝑎𝑔𝑖𝑎𝑛 𝑑𝑎𝑟𝑖 𝑃 = 2 𝑛 𝑃
= 23
= 8
3𝑥 + 4𝑦 = 17 → 3𝑥 + 4𝑦 = 17
4𝑥 − 2𝑦 = 8 → 8𝑥 − 4𝑦 = 16
11𝑥 = 33
𝑥 =
33
11
= 3
𝑥 = 3 → 3𝑥 + 4𝑦 = 17
3 . 3 + 4𝑦 = 17
9 + 4𝑦 = 17
4𝑦 = 17 − 9
𝑦 =
8
4
= 2
2𝑥 + 3𝑦 = 2 . 3 + 3 . 2
= 6 + 6
= 12
3𝐴 + 5𝐵 = 39000 → 3𝐴 + 5𝐵 = 39000
𝐴 + 𝐵 = 11000 → 3𝐴 + 3𝐵 = 33000
2𝐵 = 6000
𝐵 =
6000
2
= 3000
𝐵 = 3000 → 𝐴 + 𝐵 = 11000
𝐴 + 3000 = 11000
𝐴 = 11000 − 3000
𝐴 = 8000
4𝐴 + 2𝐵 = 4 . 8000 + 2 . 3000
= 32000 + 6000
= 38000
𝑓 𝑥 = 3𝑥 + 5
𝑓 𝑎 = 3𝑎 + 5 = −7
3𝑎 = −7 − 5
𝑎 =
−12
3
𝑎 = −4

𝑥 = 0 → 𝑦 = 2 . 0 − 1
𝑦 = −1 0, −1
𝑦𝑎𝑛𝑔 𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖 𝑔𝑎𝑚𝑏𝑎𝑟 𝑝𝑎𝑑𝑎 𝑝𝑖𝑙𝑖 𝑕𝑎𝑛 𝐴 𝑑𝑎𝑛 𝐵
𝑥 = 2 → 𝑦 = 2 . 2 − 1
𝑦 = 4 − 1
𝑦 = 3 2,3
𝐽𝑎𝑑𝑖 𝑦𝑎𝑛𝑔 𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖 𝑔𝑎𝑚𝑏𝑎𝑟 𝑝𝑎𝑑𝑎 𝑝𝑖𝑙𝑖 𝑕𝑎𝑛 𝐴
𝑃 −3
𝑥1
, 8
𝑦1
𝑑𝑎𝑛 𝑄 2
𝑥2
, 5
𝑦2
𝑚 𝑃𝑄 =
𝑦2 − 𝑦1
𝑥2 − 𝑥1
=
5 − 8
2 − (−3)
=
−3
2 + 3
= −
3
5
𝑆𝑦𝑎𝑟𝑎𝑡 𝑡𝑒𝑔𝑎𝑘 𝑙𝑢𝑟𝑢𝑠 ∶ 𝑚 . 𝑚 𝑃𝑄 = −1
𝑚 . −
3
5
= −1
𝑚 = −1 . −
5
3
𝑚 =
5
3
𝐴. 3𝑥 − 5𝑦 − 14 = 0 → 𝑚 =
3
5
𝐵. 3𝑥 + 5𝑦 + 14 = 0 → 𝑚 = −
3
5
𝐶. 5𝑥 + 3𝑦 − 42 = 0 → 𝑚 = −
5
3
𝐷. 5𝑥 − 3𝑦 − 42 = 0 → 𝑚 =
5
3

𝐴 2
𝑥1
, 7
𝑦1
; 𝐵 −3
𝑥2
, −3
𝑦2
; 𝐶 3
𝑥
, 𝑎
𝑦
𝑦 − 𝑦1
𝑦2 − 𝑦1
=
𝑥 − 𝑥1
𝑥2 − 𝑥1
𝑎 − 7
−3 − 7
=
3 − 2
−3 − 2
𝑎 − 7
−10
=
1
−5
𝑎 − 7 =
1
−5
. −10
𝑎 − 7 = 2
𝑎 = 2 + 7
𝑎 = 9
𝑃𝑎𝑛𝑗𝑎𝑛𝑔 𝑡𝑎𝑙𝑖 𝑛𝑖𝑙𝑎𝑖 𝑎𝑠𝑙𝑖 = 1502 + 1502 = 1502 .2 = 150 2
𝑃𝑎𝑛𝑗𝑎𝑛𝑔 𝑡𝑎𝑙𝑖 𝑝𝑒𝑛𝑑𝑒𝑘𝑎𝑡𝑎𝑛 = 1502 + 1502 = 22500 + 22500 = 45000 ≈ 44944 = 212 𝑚
𝐶𝐸
𝐴𝐶
=
𝐷𝐸
𝐴𝐵
𝐶𝐸
15
=
8
12
𝐶𝐸 =
8
12
.15
𝐶𝐸 = 10 𝑐𝑚

𝐿 𝑎𝑟𝑠𝑖𝑟𝑎𝑛 =
1
4
. 𝐿 𝑉𝑊𝑋𝑌
=
1
4
. 102
=
1
4
.100
= 25 𝑐𝑚2
𝐴𝐵
𝐶𝐷
=
𝐴𝐸
𝐸𝐶
=
𝐵𝐸
𝐸𝐷

𝐾𝑏𝑎𝑛𝑔𝑢𝑛 = 2 . 17 + 8 + 5 + 6 + 5 + 4
= 2 . 45
= 90 𝑐𝑚
𝑃𝑎𝑛𝑗𝑎𝑔 𝑏𝑢𝑠𝑢𝑟 𝐴𝐵 =
∠𝐴𝑂𝐵
360 𝑜
. 𝐾𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛
=
60 𝑜
360 𝑜 .2 . 𝜋 . 𝑟
=
60 𝑜
360 𝑜
.2 . 3,14 .10
= 10,466 = 10,47 𝑐𝑚
∠𝐴 + ∠𝐵 = 180 𝑜
5𝑦 − 16 𝑜
+ 2𝑦 𝑜
= 180 𝑜
5𝑦 𝑜
− 16 𝑜
+ 2𝑦 𝑜
= 180 𝑜
7𝑦 𝑜
− 16 𝑜
= 180 𝑜
7𝑦 𝑜
= 180 𝑜
+ 16 𝑜
7𝑦 𝑜
= 196 𝑜
𝑦 𝑜
=
196 𝑜
7
𝑦 𝑜
= 28 𝑜
∠𝐴 = 5𝑦 − 16 𝑜
= 5 .28 − 16 𝑜
= 140 − 16 𝑜
= 124 𝑜
𝐴𝐵 = 𝑃𝐿2 + 𝑅 − 𝑟 2
= 242 + 12 − 5 2 = 242 + 72 = 576 + 49 = 625 = 25 𝑐𝑚

𝑉𝑝𝑟𝑖𝑠𝑚𝑎 = 𝐿𝑡𝑟𝑎𝑝𝑒𝑠𝑖𝑢𝑚 . 𝑡 𝑝𝑟𝑖𝑠𝑚𝑎
=
1
2
. 𝐽𝑢𝑚𝑙𝑎𝑕 𝑠𝑖𝑠𝑖 𝑠𝑒𝑗𝑎𝑗𝑎𝑟 . 𝑡𝑡𝑟𝑎𝑝𝑒𝑠𝑖𝑢𝑚 . 𝑡 𝑝𝑟𝑖𝑠𝑚𝑎
=
1
2
. 8 + 12 .5 . 10
= 50 . 10
= 500 𝑐𝑚3
𝐿 𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑙𝑖𝑚𝑎𝑠 = 𝐿 𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝐴𝐵𝐶𝐷 + 4 . 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐵𝐶𝑇
= 𝐴𝐵2
+ 4 .
1
2
. 𝐵𝐶 . 𝑇𝑃
= 162
+ 4 .
1
2
.16 .17
= 256 + 544
= 800 𝑐𝑚2
𝐴𝐵 = 𝐵𝐶 =
𝐾𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝐴𝐵𝐶𝐷
4
=
64
4
= 16
𝑂𝑃 =
1
2
. 𝐴𝐵 =
1
2
.16 = 8
𝑇𝑃 = 𝑂𝑇2 + 𝑂𝑃2 = 152 + 82
= 225 + 64 = 289 = 17
𝑂𝐷 = 𝑟 =
1
2
. 𝐴𝐵 =
1
2
.14 = 7
𝑂𝑃 = 36 − 𝐴𝐷 = 36 − 12 = 24
𝐷𝑃 = 𝑠 = 𝑂𝐷2 + 𝑂𝑃2 = 72 + 242
= 49 + 576 = 625 = 25
𝐿 𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑏𝑎𝑛𝑔𝑢𝑛 = 𝐿 𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑡𝑎𝑏𝑢𝑛𝑔 𝑡𝑎𝑛𝑝𝑎 𝑡𝑢𝑡𝑢𝑝 + 𝐿 𝑠𝑒𝑙𝑖𝑚𝑢 𝑡 𝑘𝑒𝑟𝑢𝑐𝑢𝑡
= 𝜋𝑟2
+ 2𝜋𝑟𝑡 + 𝜋𝑟𝑠
=
22
7
. 72
+ 2 .
22
7
.7 .12 +
22
7
.7 .25
= 154 + 528 + 550
= 682 + 550
= 1232 𝑐𝑚2
𝑅𝑢𝑠𝑢𝑘 = 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, 𝐷𝐸, 𝐸𝐹, 𝐹𝐴, 𝐴𝑇, 𝐵𝑇, 𝐶𝑇, 𝐷𝑇, 𝐸𝑇, 𝐹𝑇 → 12
𝑆𝑖𝑠𝑖 = 𝐴𝐵𝐶𝐷𝐸𝐹, 𝐴𝐵𝑇, 𝐵𝐶𝑇, 𝐶𝐷𝑇, 𝐷𝐸𝑇, 𝐸𝐹𝑇, 𝐹𝐴𝑇 → 7

𝑥 =
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
𝐵𝑎𝑛𝑦𝑎𝑘 𝑠𝑖𝑠𝑤𝑎 𝑦𝑎𝑛𝑔 𝑙𝑢𝑙𝑢𝑠 = 7 + 6 + 4 + 3 = 20 𝑜𝑟𝑎𝑛𝑔
𝑥 𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛 =
𝑛 𝑝 . 𝑥 𝑝 + 133 + 127
𝑛 𝑝 + 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
𝑈𝑟𝑢𝑡𝑎𝑛 𝑑𝑎𝑡𝑎 ∶ 165, 166,168, 168,170, 171, 171
𝑀𝑒𝑑𝑖𝑎𝑛
, 172, 173, 173, 175, 178, 182
(𝐵𝑒𝑙𝑢𝑚 𝑡𝑒𝑛𝑡𝑢 𝑥 𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛 = 130)
(𝑃𝑎𝑠𝑡𝑖 𝑥 𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛 = 130)

150 + 250 = 400 𝑜𝑟𝑎𝑛𝑔
𝑛 𝑏𝑜𝑙𝑎 𝑏𝑒𝑟𝑛𝑜𝑚𝑜𝑟 𝑙𝑒𝑏𝑖𝑕 𝑑𝑎𝑟𝑖 6 = 2
𝑛 𝑆 = 8
𝑃 𝑏𝑜𝑙𝑎 𝑏𝑒𝑟𝑛𝑜𝑚𝑜𝑟 𝑙𝑒𝑏𝑖𝑕 𝑑𝑎𝑟𝑖 6 =
𝑛 𝑏𝑜𝑙𝑎 𝑏𝑒𝑟𝑛𝑜𝑚𝑜𝑟 𝑙𝑒𝑏𝑖𝑕 𝑑𝑎𝑟𝑖 6
𝑛 𝑆
=
2
8

Soal dan pembahasan un matematika smp 2014 paket 1

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