Olimpiade primagama madura mencari juara 2015 matematika smp kode soal 15333 (babak penyisihan)

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www.siap-osn.blogspot.com Pembahasan Olimpiade PMMJ 2015 Matematika SMP (Babak Penyisihan) / Page 1
Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat”
PEMBAHASAN
OLIMPIADE PRIMAGAMA MADURA MENCARI JUARA 2015
MATEMATIKA SMP KODE SOAL 15333 (BABAK PENYISIHAN NO.1-30)
( versi penulis : www.siap-osn.blogspot.com )
1. Jawaban : 𝐶. 8 4 − 𝜋 𝑐𝑚2
Pembahasan :
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐾𝑎𝑟𝑒𝑛𝑎 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑂𝑃𝑄 𝑚𝑒𝑟𝑢𝑝𝑎𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑑𝑎𝑛 ∠𝑃𝑂𝑄 = 45 𝑜
𝑚𝑎𝑘𝑎 ∶
𝑃𝑄 = 𝑂𝑃 = 8 𝑐𝑚
𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑂𝑃𝑄 =
1
2
. 𝑂𝑃 . 𝑃𝑄
=
1
2
.8 .8
= 32
𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝑃𝑂𝑅 =
∠𝑃𝑂𝑄
360 𝑜 . 𝜋 . 𝑂𝑃2
=
45 𝑜
360 𝑜 . 𝜋 . 82
=
1
8
. 𝜋 . 82
= 8 𝜋
𝐿 𝑎𝑟𝑠𝑖𝑟𝑎𝑛 = 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑂𝑃𝑄 − 𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝑃𝑂𝑅
= 32 − 8 𝜋
= 8 4 − 𝜋 𝑐𝑚2
𝐶
2. Jawaban : 𝐵. 30
Pembahasan :
22020 −22016 −90
22015 −3
=
25+2015 −21+2015 −90
22015 −3
=
25 .22015 −21 .22015 −90
22015 −3
=
32 .22015 −2 .22015 −90
22015 −3
=
30 .22015 −90
22015 −3
=
30 . 22015 −3
22015 −3
= 30 𝐵

3. Jawaban : 𝐶. 729
Pembahasan :
𝑥
1
6 =
7
3− 2
− 2
𝑥
1
6 =
7
3− 2
.
3+ 2
3+ 2
− 2
𝑥
1
6 =
7 . 3+ 2
32− 2
2 − 2
𝑥
1
6 =
7 . 3+ 2
9−2
− 2
𝑥
1
6 =
7 . 3+ 2
7
− 2
𝑥
1
6 = 3 + 2 − 2
𝑥
1
6 = 3
𝑥 = 36
𝑥 = 729 𝐶
4. Jawaban : 𝐵. 72 𝑘𝑚 𝑗𝑎𝑚
Pembahasan :
𝑉1 = 60
𝑠1
𝑡1
= 60
𝑠
60
= 𝑡1 → 𝑡1 =
𝑠
60
𝑉2 = 90
𝑠2
𝑡2
= 90
𝑠
90
= 𝑡2 → 𝑡2 =
𝑠
90
𝑉𝑟𝑎𝑡𝑎 −𝑟𝑎𝑡𝑎 =
𝑠1+𝑠2
𝑡1+𝑡2
=
𝑠+𝑠
𝑠
60
+
𝑠
90
=
2𝑠
3𝑠+2𝑠
180
=
2𝑠
5𝑠
180
= 2𝑠 .
180
5𝑠
=
360𝑠
5𝑠
= 72 𝑘𝑚 𝑗𝑎𝑚 𝐵
5. Jawaban : 𝑁. 16
Pembahasan :
𝐵𝑎𝑛𝑦𝑎𝑘 𝑓𝑎𝑘𝑡𝑜𝑟 𝑑𝑎𝑟𝑖 𝑕𝑎𝑠𝑖𝑙 𝑝𝑒𝑟𝑘𝑎𝑙𝑖𝑎𝑛 𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 𝑏𝑒𝑟𝑏𝑒𝑑𝑎 = 2 𝑛

𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘 𝑓𝑎𝑘𝑡𝑜𝑟 𝑑𝑎𝑟𝑖 𝑕𝑎𝑠𝑖𝑙 𝑝𝑒𝑟𝑘𝑎𝑙𝑖𝑎𝑛 4 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 𝑏𝑒𝑟𝑏𝑒𝑑𝑎 = 24
= 16 𝑁
6. Jawaban : 𝐴.
5
18
Pembahasan :
𝑀𝑒𝑟𝑎𝑕 = 3
𝑃𝑢𝑡𝑖𝑕 = 4
𝐵𝑖𝑟𝑢 = 2
𝑇𝑜𝑡𝑎𝑙 = 3 + 4 + 2 = 9
𝐶𝑎𝑟𝑎 𝐼 ∶
𝑃 2 𝑘𝑒𝑙𝑒𝑟𝑒𝑛𝑔 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑠𝑎𝑚𝑎 = 𝑃 2 𝑀𝑒𝑟𝑎𝑕 + 𝑃 2 𝑃𝑢𝑡𝑖𝑕 + 𝑃 2 𝐵𝑖𝑟𝑢
=
3
9
.
2
8
+
4
9
.
3
8
+
2
9
.
1
8
=
6
72
+
12
72
+
2
72
=
20
72
=
5
18
𝐴
𝐶𝑎𝑟𝑎 𝐼𝐼 ∶
𝑃 2 𝑘𝑒𝑙𝑒𝑟𝑒𝑛𝑔 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑠𝑎𝑚𝑎 = 𝑃 2 𝑀𝑒𝑟𝑎𝑕 + 𝑃 2 𝑃𝑢𝑡𝑖𝑕 + 𝑃 2 𝐵𝑖𝑟𝑢
=
𝐶3 2
𝐶9 2
+
𝐶4 2
𝐶9 2
+
𝐶2 2
𝐶9 2
=
3
36
+
6
36
+
1
36
=
10
36
=
5
18
𝐴
7. Jawaban : 𝐶. 110
Pembahasan :
4 𝑥
+ 4−𝑥
= 23
22 𝑥
+ 22 −𝑥
= 23
22𝑥
+ 2−2𝑥
= 23
2 𝑥
+ 2−𝑥 2
= 2 𝑥
+ 2−𝑥
. 2 𝑥
+ 2−𝑥
= 2 𝑥
. 2 𝑥
+ 2 𝑥
. 2−𝑥
+ 2−𝑥
. 2 𝑥
+ 2−𝑥
. 2−𝑥
= 2 𝑥+𝑥
+ 2 𝑥+ −𝑥
+ 2−𝑥+𝑥
+ 2−𝑥+ −𝑥
= 22𝑥
+ 20
+ 20
+ 2−2𝑥
= 22𝑥
+ 1 + 1 + 2−2𝑥
= 22𝑥
+ 2−2𝑥
+ 2
= 23 + 2
= 25
2 𝑥
+ 2−𝑥
= 25

2 𝑥
+ 2−𝑥
= 5
22𝑥
+ 2−2𝑥
. 2 𝑥
+ 2−𝑥
= 23 .5
22𝑥
. 2 𝑥
+ 22𝑥
. 2−𝑥
+ 2−2𝑥
. 2 𝑥
+ 2−2𝑥
. 2−𝑥
= 115
22𝑥+𝑥
+ 22𝑥+ −𝑥
+ 2−2𝑥+𝑥
+ 2−2𝑥+ −𝑥
= 115
23𝑥
+ 2 𝑥
+ 2−𝑥
+ 2−3𝑥
= 115
23𝑥
+ 5 + 2−3𝑥
= 115
23𝑥
+ 2−3𝑥
+ 5 = 115
23 𝑥
+ 23 −𝑥
= 115 − 5
8 𝑥
+ 8−𝑥
= 110 𝐶
8. Jawaban : 𝐴.
4
6𝑚−3
Pembahasan :
log 124
= 𝑚 +
1
2
log 3 .44
= 𝑚 +
1
2
log 34
+ log 44
= 𝑚 +
1
2
log 322
+ 1 = 𝑚 +
1
2
1
2
. log 32
= 𝑚 +
1
2
− 1
1
2
. log 32
= 𝑚 −
1
2
log 32
= 2 . 𝑚 −
1
2
log 32
= 2𝑚 − 1
log 23
=
1
2𝑚−1
log 1627
= log2433
=
4
3
. log 23
=
4
3
.
1
2𝑚−1
=
4
6𝑚−3
𝐴
9. Jawaban : 𝐷. 𝑆𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖
Pembahasan :
𝐴 2
𝑥1
, 3
𝑦1
𝑑𝑎𝑛 𝐵 8
𝑥2
, 5
𝑦2
𝐴𝐵 = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
= 8 − 2 2 + 5 − 3 2
= 62 + 22
= 36 + 4
= 40

𝐵 8
𝑥1
, 5
𝑦1
𝑑𝑎𝑛 𝐶 6
𝑥2
, 11
𝑦2
𝐵𝐶 = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
= 6 − 8 2 + 11 − 5 2
= −2 2 + 62
= 4 + 36
= 40
𝐴 2
𝑥1
, 3
𝑦1
𝑑𝑎𝑛 𝐶 6
𝑥2
, 11
𝑦2
𝐴𝐶 = 𝑥2 − 𝑥1
2 + 𝑦2 − 𝑦1
2
= 6 − 2 2 + 11 − 3 2
= 42 + 82
= 16 + 64
= 80
𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 𝐴𝐵 = 𝐵𝐶 = 40 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝑑𝑎𝑛 𝐴𝐶 = 80 , 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 ∶
𝐴𝐶 = 𝐴𝐵2 + 𝐵𝐶2
= 40
2
+ 40
2
= 40 + 40
= 80 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢
𝐽𝑎𝑑𝑖 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎𝑕 𝑆𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐷
10. Jawaban : 𝐷. 24𝜋 + 18 3 𝑐𝑚2
Pembahasan :
𝐴𝐶 = 6 𝑐𝑚
𝐴𝐵 = 6 𝑐𝑚
𝐴𝐷 =
1
2
. 𝐴𝐵 =
1
2
.6 = 3 𝑐𝑚
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝐶𝐷 ∶
𝐶𝐷 = 𝐴𝐶2 − 𝐴𝐷2
= 62 − 32

= 36 − 9
= 27
= 9 .3
= 3 3
𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 ∶ 𝐴𝐶 = 6 𝑐𝑚 , 𝐴𝐷 = 3 𝑐𝑚 , 𝐶𝐷 = 3 3 𝑐𝑚 , 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 𝑑𝑒𝑛𝑔𝑎𝑛 𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑝𝑒𝑟𝑏𝑎𝑛𝑑𝑖𝑛𝑔𝑎𝑛
𝑠𝑖𝑠𝑖 𝑝𝑎𝑑𝑎 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑚𝑎𝑘𝑎 𝑑𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 𝑏𝑒𝑠𝑎𝑟 𝑠𝑢𝑑𝑢𝑡 𝑠𝑒𝑏𝑎𝑔𝑎𝑖 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐴𝐶 ∶ 𝐴𝐷 ∶ 𝐶𝐷 = 6 ∶ 3 ∶ 3 3 = 2 ∶ 1 ∶ 3 → ∠𝐴𝐷𝐶 ∶ ∠𝐴𝐶𝐷 ∶ ∠𝐶𝐴𝐷 = 90 𝑜
∶ 30 𝑜
∶ 60 𝑜
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐷 𝑑𝑎𝑛 𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 ∶
∠𝐴𝐶𝐵 = 2 . ∠𝐴𝐶𝐷 = 2 . 30 𝑜
= 60 𝑜
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 ∶
𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 =
∠𝐴𝐶𝐵
360 𝑜 . 𝜋 . 𝐴𝐶2
=
60 𝑜
360 𝑜 . 𝜋 . 62
=
1
6
. 𝜋 . 62
= 6𝜋
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝐶𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐵 ∶
𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐵 = 2 . 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐷
= 2 .
1
2
. 𝐶𝐷 . 𝐴𝐷
= 2 .
1
2
.3 3 .3
= 9 3
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑒𝑚𝑏𝑒𝑟𝑒𝑛𝑔 𝐴𝐵 ∶
𝐿𝑡𝑒𝑚𝑏𝑒𝑟𝑒𝑛𝑔 𝐴𝐵 = 𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 − 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐵
= 6𝜋 − 9 3
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛 𝑑𝑒𝑛𝑔𝑎𝑛 𝑡𝑖𝑡𝑖𝑘 𝑝𝑢𝑠𝑎𝑡 𝐶 ∶
𝐿𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛 𝐶 = 𝜋 . 𝐴𝐶2
= 𝜋 . 62
= 36𝜋
𝐿 𝑎𝑟𝑠𝑖𝑟𝑎𝑛 = 𝐿𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛 𝐶 − 2 . 𝐿𝑡𝑒𝑚𝑏𝑒𝑟𝑒𝑛𝑔 𝐴𝐵
= 36𝜋 − 2 . 6𝜋 − 9 3
= 36𝜋 − 12𝜋 + 18 3
= 24𝜋 + 18 3 𝑐𝑚2
𝐷

11. Jawaban : 𝐷. ± 24
Pembahasan :
𝑃𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 𝑔𝑎𝑟𝑖𝑠 ∶ 3𝑥 + 4𝑦 = 𝑝
𝑇𝑖𝑡𝑖𝑘 𝑝𝑜𝑡𝑜𝑛𝑔 𝑑𝑒𝑛𝑔𝑎𝑛 𝑠𝑢𝑚𝑏𝑢 𝑥 ∶
𝑦 = 0 → 3𝑥 + 4𝑦 = 𝑝
3𝑥 + 4 . 0 = 𝑝
3𝑥 = 𝑝
𝑥 =
𝑝
3
→
𝑝
3
, 0
𝑇𝑖𝑡𝑖𝑘 𝑝𝑜𝑡𝑜𝑛𝑔 𝑑𝑒𝑛𝑔𝑎𝑛 𝑠𝑢𝑚𝑏𝑢 𝑦 ∶
𝑥 = 0 → 3𝑥 + 4𝑦 = 𝑝
3 . 0 + 4𝑦 = 𝑝
4𝑦 = 𝑝
𝑦 =
𝑝
4
→ 0,
𝑝
4
𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 =
1
2
. 𝑎 . 𝑡
24 =
1
2
.
𝑝
3
.
𝑝
4
24 =
𝑝2
24
24 .24 = 𝑝2
± 24 .24 = 𝑝
±24 = 𝑝
𝑝 = ±24 𝐷
12. Jawaban : 𝑁. 4 3
Pembahasan :
𝐾𝑂 = 6
𝑁𝑂 = 5
𝑀𝑂 = 37
𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶
𝐷𝑁 = 𝐵𝐾 = 𝐴𝑂 = 𝑎
𝐴𝑁 = 𝐶𝑀 = 𝐷𝑂 = 𝑏
𝐴𝐾 = 𝐶𝐿 = 𝐵𝑂 = 𝑐
𝐵𝐿 = 𝐷𝑀 = 𝐶𝑂 = 𝑑

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐷𝑁 ∶
𝐷𝑁2
+ 𝐷𝑂2
= 𝑁𝑂2
𝑎2
+ 𝑏2
= 52
𝑎2
+ 𝑏2
= 25 … (1)
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐴𝐾 ∶
𝐴𝑂2
+ 𝐴𝐾2
= 𝐾𝑂2
𝑎2
+ 𝑐2
= 62
𝑎2
+ 𝑐2
= 36 … (2)
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐵𝐿 ∶
𝐵𝑂2
+ 𝐵𝐿2
= 𝐿𝑂2
𝑐2
+ 𝑑2
= 𝐿𝑂2
𝑐2
+ 𝑑2
= 𝐿𝑂2
… (3)
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐶𝑀 ∶
𝐶𝑀2
+ 𝐶𝑂2
= 𝑀𝑂2
𝑏2
+ 𝑑2
= 37
2
𝑏2
+ 𝑑2
= 37 … (4)
𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 1 𝑑𝑎𝑛 2 ∶
𝑎2
+ 𝑏2
= 25
𝑎2
+ 𝑐2
= 36
𝑏2
− 𝑐2
= −11 … (5)
𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 (4) 𝑑𝑎𝑛 (5) ∶
𝑏2
+ 𝑑2
= 37
𝑏2
− 𝑐2
= −11
𝑑2
+ 𝑐2
= 48 … (6)
𝑑2
+ 𝑐2
= 48
𝑐2
+ 𝑑2
= 𝐿𝑂2
0 = 48 − 𝐿𝑂2
𝐿𝑂2
= 48
𝐿𝑂 = 48
𝐿𝑂 = 16 .3
𝐿𝑂 = 4 3 𝑁

13. Jawaban : 𝐶. 𝑡 30
Pembahasan :
𝑝 = 5𝑡
𝑙 = 2𝑡
𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑟𝑢𝑎𝑛𝑔 𝑏𝑎𝑙𝑜𝑘 = 𝑝2 + 𝑙2 + 𝑡2
= 5𝑡 2 + 2𝑡 2 + 𝑡2
= 25𝑡2 + 4𝑡2 + 𝑡2
= 30𝑡2
= 𝑡 30 𝐶
14. Jawaban : 𝐵. 𝑥 = 5 𝑎𝑡𝑎𝑢 𝑥 =
1
2
Pembahasan :
𝑥 = 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑦𝑎𝑛𝑔 𝑑𝑖𝑚𝑎𝑘𝑠𝑢𝑑
2𝑥 +
5
𝑥
= 11
𝑥 . 2𝑥 +
5
𝑥
= 𝑥 . 11 𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 𝑥
2𝑥2
+ 5 = 11𝑥
2𝑥2
− 11𝑥 + 5 = 0
1
2
. 2𝑥 − 10 . 2𝑥 − 1 = 0
1
2
. 2 . 𝑥 − 5 . 2𝑥 − 1 = 0
𝑥 − 5 . 2𝑥 − 1 = 0
𝑥 = 5 𝑎𝑡𝑎𝑢 𝑥 =
1
2
𝐵
15. Jawaban : 𝐴. 𝑅𝑝. 61.000, −
Pembahasan :
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐺𝑎𝑗𝑖 𝑎𝑤𝑎𝑙 𝐵𝑒𝑠𝑎𝑟 𝑘𝑒𝑛𝑎𝑖𝑘𝑎𝑛 𝑔𝑎𝑗𝑖
260000
15
100
. 260000 = 39000
370000
15
100
. 370000 = 55500
470000
15
100
. 470000 = 70500
650000
10
100
. 650000 = 65000
750000
10
100
. 750000 = 75000

𝑅𝑎𝑡𝑎 − 𝑟𝑎𝑡𝑎 𝑏𝑒𝑠𝑎𝑟 𝑘𝑒𝑛𝑎𝑖𝑘𝑎𝑛 𝑔𝑎𝑗𝑖 =
39000+55500+70500+65000+75000
5
=
305000
5
= 61000 𝐴
16. Jawaban : 𝑁. 868
Pembahasan :
𝐽𝑢𝑚𝑙𝑎𝑕 𝑠𝑒𝑚𝑢𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑛𝑡𝑎𝑟𝑎 1 − 100 𝑦𝑎𝑛𝑔 𝑕𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 4 ∶
4 + 8 + 12 + ⋯ + 100
𝑑𝑒𝑟𝑒𝑡 𝑎𝑟𝑖𝑡𝑚𝑎𝑡𝑖𝑘𝑎 ∶
𝑛=
100
4
=25
𝑎=4
𝑈25=100
=
𝑛
2
. 𝑎 + 𝑈𝑛 =
25
2
. 4 + 𝑈25 =
25
2
. 4 + 100 =
25
2
. 104 = 1300
𝐽𝑢𝑚𝑙𝑎𝑕 𝑠𝑒𝑚𝑢𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑛𝑡𝑎𝑟𝑎 1 − 100 𝑦𝑎𝑛𝑔 𝑕𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 3 𝑑𝑎𝑛 4 ∶
12 + 24 + 36 + ⋯ + 96
𝑑𝑒𝑟𝑒𝑡 𝑎𝑟𝑖𝑡𝑚𝑎𝑡𝑖𝑘𝑎 ∶
𝑛=
96
12
=8
𝑎=12
𝑈8=96
=
𝑛
2
. 𝑎 + 𝑈𝑛 =
8
2
. 12 + 𝑈8 =
8
2
. 12 + 96 =
8
2
. 108 = 432
𝐽𝑎𝑑𝑖 𝐽𝑢𝑚𝑙𝑎𝑕 𝑠𝑒𝑚𝑢𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑛𝑡𝑎𝑟𝑎 1 − 100 𝑦𝑎𝑛𝑔 𝑕𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 4 𝑡𝑒𝑡𝑎𝑝𝑖 𝑡𝑖𝑑𝑎𝑘 𝑕𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 3
𝑎𝑑𝑎𝑙𝑎𝑕 1300 − 432 = 868 𝑁
17. Jawaban : 𝐴. 915
Pembahasan :
𝑎𝑏𝑐 = 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑦𝑎𝑛𝑔 𝑑𝑖𝑚𝑎𝑘𝑠𝑢𝑑
𝑎 + 𝑏 + 𝑐 = 15 … 1
𝑐𝑏𝑎 = 𝑎𝑏𝑐 − 396
100𝑐 + 10𝑏 + 𝑎 = 100𝑎 + 10𝑏 + 𝑐 − 396
100𝑐 + 10𝑏 + 𝑎 − 100𝑎 − 10𝑏 − 𝑐 = −396
99𝑐 − 99𝑎 = −396
𝑐 − 𝑎 = −4 … 2 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑏𝑎𝑔𝑖 99)
𝑎𝑐𝑏 = 𝑎𝑏𝑐 + 36
100𝑎 + 10𝑐 + 𝑏 = 100𝑎 + 10𝑏 + 𝑐 + 36
100𝑎 + 10𝑐 + 𝑏 − 100𝑎 − 10𝑏 − 𝑐 = 36
−9𝑏 + 9𝑐 = 36
−𝑏 + 𝑐 = 4 … 3 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑏𝑎𝑔𝑖 9)
𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 1 , 2 𝑑𝑎𝑛 3 ∶
𝑎 + 𝑏 + 𝑐 = 15
𝑐 − 𝑎 = −4
−𝑏 + 𝑐 = 4
3𝑐 = 15

𝑐 =
15
3
𝑐 = 5
𝑐 = 5 → 𝑐 − 𝑎 = −4 … 2
5 − 𝑎 = −4
5 + 4 = 𝑎
9 = 𝑎
𝑎 = 9
𝑐 = 5 → −𝑏 + 𝑐 = 4 … 3
−𝑏 + 5 = 4
5 − 4 = 𝑏
1 = 𝑏
𝑏 = 1
𝐽𝑎𝑑𝑖 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎𝑕 𝑎𝑏𝑐 = 915 𝐴
18. Jawaban : 𝐶. 1 𝑎𝑡𝑎𝑢 5
Pembahasan :
𝑥 − 3 2
= −2 𝑥 − 3 + 8
𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 𝑥 − 3 = 𝑝 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 ∶
𝑝2
= −2𝑝 + 8
𝑝2
+ 2𝑝 − 8 = 0
𝑝 − 2 . 𝑝 + 4 = 0
𝑝 = 2
𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖
𝑎𝑡𝑎𝑢 𝑝 = −4
𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖
𝑝 = 2 → 𝑥 − 3 = 𝑝
𝑥 − 3 = 2 → 𝑢𝑛𝑡𝑢𝑘 𝑥 − 3 = + → 𝑥 − 3 = 2
𝑥 = 2 + 3
𝑥 = 5
→ 𝑢𝑛𝑡𝑢𝑘 𝑥 − 3 = − → − 𝑥 − 3 = 2
−𝑥 + 3 = 2
3 − 2 = 𝑥
1 = 𝑥
𝑥 = 1
𝐽𝑎𝑑𝑖 𝑠𝑜𝑙𝑢𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎𝑕 1 𝑎𝑡𝑎𝑢 5 𝐶

19. Jawaban : 𝐵.
1
5
Pembahasan :
𝑛 𝑆 = 40
𝑛 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑆𝑐𝑖𝑒𝑛𝑐𝑒 = 24
𝑛 𝑀𝑎𝑡𝑕𝑒𝑚𝑎𝑡𝑖𝑐𝑠 = 13
𝑛 𝐿𝑖𝑘𝑒 𝑏𝑜𝑡𝑕 = 5
𝑛 𝐷𝑖𝑠𝑙𝑖𝑘𝑒 𝑏𝑜𝑡𝑕 = 𝑛 𝑆 − 𝑛 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑆𝑐𝑖𝑒𝑛𝑐𝑒 + 𝑛 𝑀𝑎𝑡𝑕𝑒𝑚𝑎𝑡𝑖𝑐𝑠 − 𝑛 𝐿𝑖𝑘𝑒 𝑏𝑜𝑡𝑕
= 40 − 24 + 13 − 5
= 40 − 32
= 8
𝑃 𝐷𝑖𝑠𝑙𝑖𝑘𝑒 𝑏𝑜𝑡𝑕 =
𝑛 𝐷𝑖𝑠𝑙𝑖𝑘𝑒 𝑏𝑜𝑡 𝑕
𝑛 𝑆
=
8
40
=
1
5
𝐵
Pembahasan :
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐵𝑖𝑙𝑎𝑛𝑔𝑎𝑛 ∶
301 − 999
𝑅𝑎𝑡𝑢𝑠𝑎𝑛 𝑃𝑢𝑙𝑢𝑕𝑎𝑛 𝑆𝑎𝑡𝑢𝑎𝑛
𝐾𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛
𝑎𝑛𝑔𝑘𝑎 𝑦𝑎𝑛𝑔 𝑑𝑖𝑝𝑖𝑙𝑖𝑕
3 2 3
5 3 5
6 5 7
7 6 9
9 7
9
𝐵𝑎𝑛𝑦𝑎𝑘
𝑎𝑛𝑔𝑘𝑎 𝑦𝑎𝑛𝑔 𝑑𝑖𝑝𝑖𝑙𝑖𝑕
5 − 1
𝑡𝑖𝑑𝑎𝑘 𝑏𝑜𝑙𝑒 𝑕 𝑏𝑒𝑟𝑢𝑙𝑎𝑛𝑔
= 4 6 − 2
𝑡𝑖𝑑𝑎𝑘 𝑏𝑜𝑙𝑒 𝑕 𝑏𝑒𝑟𝑢𝑙𝑎𝑛𝑔
= 4
4
𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘 𝑐𝑎𝑟𝑎 𝑚𝑒𝑛𝑦𝑢𝑠𝑢𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎𝑕 4 .4 .4 = 64 𝐵
21. Jawaban : 𝐷. 20,25 𝑐𝑚2
Pembahasan :
𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷 = 𝐷𝐸 = 𝑝𝑎𝑛𝑗𝑎𝑛𝑔 𝑠𝑖𝑠𝑖 𝑝𝑒𝑟𝑠𝑒𝑔𝑖 = 9 𝑐𝑚

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐴𝐻 𝑑𝑎𝑛 𝐷𝐵𝑂 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶
𝐵𝑂
𝐴𝐻
=
𝐵𝐷
𝐴𝐷
𝐵𝑂
9
=
9+9
9+9+9
𝐵𝑂
9
=
18
27
𝐵𝑂 =
18
27
.9
𝐵𝑂 = 6
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐴𝐻 𝑑𝑎𝑛 𝐷𝐶𝑃 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶
𝐶𝑃
𝐴𝐻
=
𝐶𝐷
𝐴𝐷
𝐶𝑃
9
=
9
9+9+9
𝐶𝑃
9
=
9
27
𝐶𝑃 =
9
27
.9
𝐶𝑃 = 3
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐺𝐸𝐷 𝑑𝑎𝑛 𝐺𝐹𝑄 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶
𝐹𝑄
𝐷𝐸
=
𝐹𝐺
𝐸𝐺
𝐹𝑄
9
=
9
9+9
𝐹𝑄
9
=
9
18
𝐹𝑄 =
9
18
.9
𝐹𝑄 =
9
2
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑟𝑖𝑠 𝐵𝐺 ∶
𝑂𝐺 = 𝐵𝐺 − 𝐵𝑂 = 9 − 6 = 3
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑟𝑖𝑠 𝐶𝐹 ∶
𝑃𝑄 = 𝐶𝐹 − 𝐶𝑃 − 𝐹𝑄 = 9 − 3 −
9
2
=
3
2
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝑂𝐺 𝑑𝑎𝑛 𝐷𝑃𝑄 ∶
𝐿 𝑎𝑟𝑠𝑖𝑟𝑎𝑛 = 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝑂𝐺 − 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝑃𝑄
=
1
2
. 𝑂𝐺 . 𝐵𝐷 −
1
2
. 𝑃𝑄 . 𝐶𝐷
=
1
2
.3 . 9 + 9 −
1
2
.
3
2
.9
=
1
2
.3 .18 −
1
2
.
3
2
.9
=
54
2
−
27
4
=
108
4
−
27
4
=
81
4
= 20,25 𝑐𝑚2
𝐷

22. Jawaban : 𝐷. 60 7
Pembahasan :
𝐴𝐵 = 4 𝑐𝑚
𝐵𝐶 = 8 𝑐𝑚
𝐴𝐶 = 4 2 𝑐𝑚
𝐴𝐷 = 15 𝑐𝑚
𝐵𝑇 = 𝑥
𝐶𝑇 = 𝐵𝐶 − 𝐵𝑇 = 8 − 𝑥
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑇𝐵 ∶
𝐴𝑇2
= 𝐴𝐵2
− 𝐵𝑇2
𝐴𝑇2
= 42
− 𝑥2
𝐴𝑇2
= 16 − 𝑥2
𝐴𝑇2
= 16 − 𝑥2
… (1)
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑇𝐶 ∶
𝐴𝑇2
= 𝐴𝐶2
− 𝐶𝑇2
𝐴𝑇2
= 4 2
2
− 8 − 𝑥 2
𝐴𝑇2
= 16 .2 − 64 − 16𝑥 + 𝑥2
𝐴𝑇2
= 32 − 64 + 16𝑥 − 𝑥2
𝐴𝑇2
= −32 + 16𝑥 − 𝑥2
… (2)
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑠𝑖𝑘𝑎𝑛 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 2 𝑘𝑒 1 ∶
𝐴𝑇2
= 16 − 𝑥2
−32 + 16𝑥 − 𝑥2
= 16 − 𝑥2
16𝑥 = 16 − 𝑥2
+ 32 + 𝑥2
16𝑥 = 48
𝑥 =
48
16
𝑥 = 3
𝑥 = 3 → 𝐴𝑇2
= 16 − 𝑥2
… 1
𝐴𝑇2
= 16 − 32
𝐴𝑇2
= 16 − 9

𝐴𝑇2
= 7
𝐴𝑇 = 7
𝐿 𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝐵𝐶.𝐷𝐸𝐹 = 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 . 𝑡 𝑝𝑟𝑖𝑠𝑚𝑎
=
1
2
. 𝐵𝐶 . 𝐴𝑇 . 𝐴𝐷
=
1
2
.8 . 7 .15
= 60 7 𝐷
23. Jawaban : 𝐴. 52 𝑚 (𝑏𝑢𝑘𝑎𝑛 𝑐𝑚)
Pembahasan :
𝑀𝐴 = 𝐵𝐶 − 𝐷𝑀 = 26 − 12 = 14
𝐴𝐸 = 𝐶𝐷 − 𝐵𝐸 = 36 − 24 = 12
𝑀𝐿 + 𝐾𝐽 + 𝐼𝐻 + 𝐺𝐹 = 𝐴𝐸 = 12
𝐸𝐹 + 𝐺𝐻 + 𝐼𝐽 + 𝐾𝐿 = 𝑀𝐴 = 14
𝐾𝑟𝑢𝑚𝑎 𝑕 = 𝐴𝑀 + 𝐴𝐸 + 𝑀𝐿 + 𝐾𝐽 + 𝐼𝐻 + 𝐺𝐹 + 𝐸𝐹 + 𝐺𝐻 + 𝐼𝐽 + 𝐾𝐿
= 14 + 12 + 12 + 14
= 52 𝑚 (𝑏𝑢𝑘𝑎𝑛 𝑐𝑚) 𝐴
24. Jawaban : 𝐷. 2
1
2
Pembahasan :
2𝑥 − 𝑎 >
𝑥−1
2
+
𝑎𝑥
3
𝑚𝑒𝑚𝑖𝑙𝑖𝑘𝑖 𝑠𝑜𝑙𝑢𝑠𝑖 𝑥 > 3
2𝑥 − 𝑎 >
𝑥−1
2
+
𝑎𝑥
3
2𝑥 − 𝑎 >
3 . 𝑥−1
6
+
2 .𝑎𝑥
6
2𝑥 − 𝑎 >
3𝑥−3
6
+
2𝑎𝑥
6
2𝑥 − 𝑎 >
3𝑥−3+2𝑎𝑥
6
6 . 2𝑥 − 𝑎 > 3𝑥 − 3 + 2𝑎𝑥
12𝑥 − 6𝑎 > 3𝑥 − 3 + 2𝑎𝑥
12𝑥 − 3𝑥 − 2𝑎𝑥 > −3 + 6𝑎

9𝑥 − 2𝑎𝑥 > −3 + 6𝑎
9 − 2𝑎 𝑥 > −3 + 6𝑎
𝑥 >
−3+6𝑎
9−2𝑎
𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 𝑥 >
−3+6𝑎
9−2𝑎
𝑑𝑎𝑛 𝑥 > 3 , 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑕𝑤𝑎 ∶
−3+6𝑎
9−2𝑎
= 3
−3 + 6𝑎 = 3 . 9 − 2𝑎
−3 + 6𝑎 = 27 − 6𝑎
6𝑎 + 6𝑎 = 27 + 3
12𝑎 = 30
𝑎 =
30
12
𝑎 =
5
2
𝑎 = 2
1
2
𝐷
Pembahasan :
𝑥2
− 10𝑥 + 2015 = 0 𝑚𝑒𝑚𝑖𝑙𝑖𝑘𝑖 𝑠𝑜𝑙𝑢𝑠𝑖 𝑝 𝑑𝑎𝑛 𝑞
𝑥2
− 10𝑥 + 2015
𝑎=1
𝑏=−10
𝑐=2015
= 0 → 𝑝 + 𝑞 =
−𝑏
𝑎
=
− −10
1
= 10
𝑥 = 𝑝 → 𝑥2
− 10𝑥 + 2015 = 0
𝑝2
− 10𝑝 + 2015 = 0
𝑝2
= 10𝑝 − 2015
𝑝2
+ 10𝑞 + 2014 = 10𝑝 − 2015 + 10𝑞 + 2014
= 10𝑝 + 10𝑞 − 1
= 10 . 𝑝 + 𝑞 − 1
= 10 . 10 − 1
= 100 − 1
= 99 𝐵
26. Jawaban : 𝐶. 2017
Pembahasan :
𝑀𝑒𝑛𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑦𝑎𝑛𝑔 𝑙𝑒𝑏𝑖𝑕 𝑏𝑒𝑠𝑎𝑟 1 𝑠𝑎𝑡𝑢𝑎𝑛 𝑑𝑎𝑟𝑖 𝑥2 − 2𝑥 + 9 ∶
𝑥2 − 2𝑥 + 9 < 𝑥2
𝑥2 − 2𝑥 + 9 < 𝑥
𝑢𝑛𝑡𝑢𝑘 𝑥 = 2015 → 𝑥2 − 2𝑥 + 9 < 𝑥
𝑥2 − 2𝑥 + 9 < 2015

𝑀𝑒𝑛𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑦𝑎𝑛𝑔 𝑙𝑒𝑏𝑖𝑕 𝑘𝑒𝑐𝑖𝑙 1 𝑠𝑎𝑡𝑢𝑎𝑛 𝑑𝑎𝑟𝑖 4𝑥2 + 4𝑥 + 9 ∶
4𝑥2 + 4𝑥 + 1 < 4𝑥2 + 4𝑥 + 9
2𝑥 + 1 2 < 4𝑥2 + 4𝑥 + 9
2𝑥 + 1 < 4𝑥2 + 4𝑥 + 9
𝑢𝑛𝑡𝑢𝑘 𝑥 = 2015 → 2𝑥 + 1 < 4𝑥2 + 4𝑥 + 9
2 .2015 + 1 < 4𝑥2 + 4𝑥 + 9
4030 + 1 < 4𝑥2 + 4𝑥 + 9
4031 < 4𝑥2 + 4𝑥 + 9
𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 𝑥2 − 2𝑥 + 9 < 2015 𝑑𝑎𝑛 4031 < 4𝑥2 + 4𝑥 + 9 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑕𝑤𝑎 ∶
𝑥2 − 2𝑥 + 9 < 2015 ≤ ⋯ ≤ 4031 < 4𝑥2 + 4𝑥 + 9
𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘𝑛𝑦𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑦𝑎𝑛𝑔 𝑏𝑒𝑟𝑎𝑑𝑎 𝑑𝑖𝑎𝑛𝑡𝑎𝑟𝑎 𝑥2 − 2𝑥 + 9 𝑑𝑎𝑛 4𝑥2 + 4𝑥 + 9
𝑎𝑑𝑎𝑙𝑎𝑕 4031 − 2015 + 1 = 2017 𝐶
27. Jawaban : 𝐶. 94,5
Pembahasan :
𝐴𝐷 = 𝐴𝐵 = 4 𝑐𝑚
𝐴𝐹 = 3,5 𝑐𝑚
𝐿 𝑎𝑟𝑠𝑖𝑟𝑎𝑛 = 4 . 𝐿 𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝑝𝑎𝑛𝑗𝑎𝑛𝑔 𝐴𝐷𝐸𝐹 + 𝐿𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛
= 4 . 𝐴𝐷 . 𝐴𝐹 + 𝜋 . 𝐴𝐹2
= 4 . 4 .3,5 +
22
7
. 3,52
= 56 + 38,5
= 94,5
𝐽𝑎𝑑𝑖 𝑙𝑢𝑎𝑠 𝑑𝑎𝑒𝑟𝑎𝑕 𝑦𝑎𝑛𝑔 𝑑𝑖𝑏𝑎𝑡𝑎𝑠𝑖 𝑜𝑙𝑒𝑕 𝑙𝑖𝑛𝑡𝑎𝑠𝑎𝑛 𝑘𝑢𝑐𝑖𝑛𝑔 𝑎𝑑𝑎𝑙𝑎𝑕 94,5 𝐶
28. Jawaban : 𝐴. 12
Pembahasan :
𝑓 𝑥 +
1
𝑥
. 𝑓 −𝑥 = 𝑥 + 6

𝑥 = 2 → 𝑓 𝑥 +
1
𝑥
. 𝑓 −𝑥 = 𝑥 + 6
𝑓 2 +
1
2
. 𝑓 −2 = 2 + 6
𝑓 2 +
1
2
. 𝑓 −2 = 8
2 . 𝑓 2 +
1
2
. 𝑓 −2 = 2 . 8 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 2)
2 . 𝑓 2 + 𝑓 −2 = 16 … (1)
𝑥 = −2 → 𝑓 𝑥 +
1
𝑥
. 𝑓 −𝑥 = 𝑥 + 6
𝑓 −2 +
1
−2
. 𝑓 − −2 = −2 + 6
𝑓 −2 −
1
2
. 𝑓 2 = 4
2 . 𝑓 −2 −
1
2
. 𝑓 2 = 2 . 4 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 2)
2 . 𝑓 −2 − 𝑓 2 = 8
−𝑓 2 + 2 . 𝑓 −2 = 8 … (2)
𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 1 𝑑𝑎𝑛 2 ∶
2 . 𝑓 2 + 𝑓 −2 = 16 → 4 . 𝑓 2 + 2 . 𝑓 −2 = 32 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 2)
−𝑓 2 + 2 . 𝑓 −2 = 8 → −𝑓 2 + 2 . 𝑓 −2 = 8
5 . 𝑓 2 = 24
𝑓 2 =
24
5
𝑥 = 3 → 𝑓 𝑥 +
1
𝑥
. 𝑓 −𝑥 = 𝑥 + 6
𝑓 3 +
1
3
. 𝑓 −3 = 3 + 6
𝑓 3 +
1
3
. 𝑓 −3 = 9
3 . 𝑓 3 +
1
3
. 𝑓 −3 = 3 . 9 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 3)
3 . 𝑓 3 + 𝑓 −3 = 27 … (3)
𝑥 = −3 → 𝑓 𝑥 +
1
𝑥
. 𝑓 −𝑥 = 𝑥 + 6
𝑓 −3 +
1
−3
. 𝑓 − −3 = −3 + 6
𝑓 −3 −
1
3
. 𝑓 3 = 3
3 . 𝑓 −3 −
1
3
. 𝑓 3 = 3 . 3 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 3)
3 . 𝑓 −3 − 𝑓 3 = 9
−𝑓 3 + 3 . 𝑓 −3 = 9 … (4)
3 . 𝑓 3 + 𝑓 −3 = 27 → 9 . 𝑓 3 + 3 . 𝑓 −3 = 81 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 3)
−𝑓 3 + 3 . 𝑓 −3 = 9 → −𝑓 3 + 3 . 𝑓 −3 = 9
10 . 𝑓 3 = 72
𝑓 3 =
72
10

𝑓 3 =
36
5
𝑓 2 + 𝑓 3 =
24
5
+
36
5
=
60
5
= 12 𝐴
29. Jawaban : 𝐵. 160 𝜋
Pembahasan :
𝑟 = 8 𝑐𝑚
𝐿 𝑎𝑟𝑠𝑖𝑟𝑎𝑛 =
3 .360−180
360
. 𝜋 . 𝑟2
=
1080−180
360
. 𝜋 . 82
=
900
360
. 𝜋 .64
= 160 𝜋 𝐵
30. Jawaban : 𝐷. −
1
2
2 𝑎𝑡𝑎𝑢 2
Pembahasan :
1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
… = 𝑝
1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
… = 𝑝
1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
… = 𝑝2
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
… = 𝑝2
− 1

1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
… = 𝑥 . 𝑝2
− 1
𝑝 = 𝑥 . 𝑝2
− 1
𝑝 = 𝑥 . 𝑝2
− 𝑥
0 = 𝑥 . 𝑝2
− 𝑝 − 𝑥
𝑥 . 𝑝2
− 𝑝 − 𝑥 = 0
𝑥 = 2 → 𝑥 . 𝑝2
− 𝑝 − 𝑥 = 0
2 . 𝑝2
− 𝑝 − 2
𝑎= 2
𝑏=−1
𝑐=− 2
= 0
𝑝1,2 =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
𝑝1,2 =
− −1 ± −1 2−4 . 2 . − 2
2 . 2
𝑝1,2 =
1± 1+8
2 . 2
𝑝1,2 =
1± 9
2 . 2
𝑝1,2 =
1±3
2 . 2
𝑝 =
1−3
2 . 2
𝑎𝑡𝑎𝑢 𝑝 =
1+3
2 . 2
𝑝 =
−2
2 . 2
4
2 . 2
𝑝 =
−2
2 . 2
.
2
2
4
2 . 2
.
2
2
𝑝 =
−2 2
2 .2
4 2
2 .2
𝑝 = −
1
2
2 𝑎𝑡𝑎𝑢 𝑝 = 2
𝐽𝑎𝑑𝑖 𝑛𝑖𝑙𝑎𝑖 𝑑𝑎𝑟𝑖 1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
1 +
1
𝑥
… 𝑎𝑑𝑎𝑙𝑎𝑕 −
1
2
2 𝑎𝑡𝑎𝑢 2 𝐷
"𝑆𝑒𝑚𝑜𝑔𝑎 𝑏𝑒𝑟𝑚𝑎𝑛𝑓𝑎𝑎𝑡"
"𝑆𝐷. 𝐴"

Olimpiade primagama madura mencari juara 2015 matematika smp kode soal 15333 (babak penyisihan)

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Olimpiade primagama madura mencari juara 2015 matematika smp kode soal 15333 (babak penyisihan)