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Olimpiade primagama madura mencari juara 2015 matematika smp kode
soal 15333 (babak penyisihan)
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Sosuke Aizen, Working at LOS ESPADA
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Published on Dec 25, 2014
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Olimpiade primagama madura mencari juara 2015 matematika smp kode soal 15333 (babak penyisihan)
1. 1. www.siap-osn.blogspot.com @ Desember 2014 SD.A 2 SMPN 1 Tambelangan 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 : 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
Matematika SMP (Babak Penyisihan) / Page 2 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 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 𝑝𝑒𝑟𝑘𝑎𝑙𝑖𝑎𝑛 𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 𝑏𝑒𝑟𝑏𝑒𝑑𝑎 = 2 𝑛
Matematika SMP (Babak Penyisihan) / Page 3 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘
𝑓𝑎𝑘𝑡𝑜𝑟 𝑑𝑎𝑟𝑖 𝑝𝑒𝑟𝑘𝑎𝑙𝑖𝑎𝑛 4 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 𝑏𝑒𝑟𝑏𝑒𝑑𝑎 = 24 = 16 𝑁 6. Jawaban : 𝐴. 5 18 Pembahasan : = 3 = 4 𝐵𝑖𝑟𝑢 = 2 𝑇𝑜𝑡𝑎𝑙 =
3 + 4 + 2 = 9 𝐶𝑎𝑟𝑎 𝐼 ∶ 𝑃 2 𝑘𝑒𝑙𝑒𝑟𝑒𝑛𝑔 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑠𝑎𝑚𝑎 = 𝑃 2 + 𝑃 2 + 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 + 𝑃 2 + 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
Matematika SMP (Babak Penyisihan) / Page 4 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 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
Matematika SMP (Babak Penyisihan) / Page 5 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 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 𝐴𝐵 = 𝐵𝐶 = 40 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝑑𝑎𝑛 𝐴𝐶 = 80 , ∶ 𝐴𝐶 = 𝐴𝐵2 +
𝐵𝐶2 = 40 2 + 40 2 = 40 + 40 = 80 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐽𝑎𝑑𝑖 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 𝑆𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐷 10. Jawaban : 𝐷. 24𝜋
+ 18 3 𝑐𝑚2 Pembahasan : 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐴𝐶 = 6 𝑐𝑚 𝐴𝐵 = 6 𝑐𝑚 𝐴𝐷 = 1 2 . 𝐴𝐵 = 1 2 .6 = 3 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −
𝑠𝑖𝑘𝑢 𝐴𝐶𝐷 ∶ 𝐶𝐷 = 𝐴𝐶2 − 𝐴𝐷2 = 62 − 32
Matematika SMP (Babak Penyisihan) / Page 6 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” = 36 − 9 = 27 = 9
.3 = 3 3 ∶ 𝐴𝐶 = 6 𝑐𝑚 , 𝐴𝐷 = 3 𝑐𝑚 , 𝐶𝐷 = 3 3 𝑐𝑚 , 𝑑𝑒𝑛𝑔𝑎𝑛 𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑝𝑒𝑟𝑏𝑎𝑛𝑑𝑖𝑛𝑔𝑎𝑛 𝑠𝑖𝑠𝑖 𝑝𝑎𝑑𝑎 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢
𝑚𝑎𝑘𝑎 𝑏𝑒𝑠𝑎𝑟 𝑠𝑢𝑑𝑢𝑡 𝑠𝑒𝑏𝑎𝑔𝑎𝑖 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐴𝐶 ∶ 𝐴𝐷 ∶ 𝐶𝐷 = 6 ∶ 3 ∶ 3 3 = 2 ∶ 1 ∶ 3 → ∠𝐴𝐷𝐶 ∶ ∠𝐴𝐶𝐷 ∶ ∠𝐶𝐴𝐷 = 90 𝑜 ∶ 30 𝑜 ∶ 60 𝑜
𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐷 𝑑𝑎𝑛 𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 ∶ ∠𝐴𝐶𝐵 = 2 . ∠𝐴𝐶𝐷 = 2 . 30 𝑜 = 60 𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 ∶ 𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 = ∠𝐴𝐶𝐵 360 𝑜 . 𝜋 .
𝐴𝐶2 = 60 𝑜 360 𝑜 . 𝜋 . 62 = 1 6 . 𝜋 . 62 = 6 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 𝑡𝑒𝑚𝑏𝑒𝑟𝑒𝑛𝑔 𝐴𝐵 ∶ 𝐿𝑡𝑒𝑚𝑏𝑒𝑟𝑒𝑛𝑔 𝐴𝐵 = 𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 − 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐵 = 6𝜋 − 9 3
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
Matematika SMP (Babak Penyisihan) / Page 7 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 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 : 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐾𝑂 = 6 𝑁𝑂 = 5 𝑀𝑂 = 37 𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 𝐷𝑁 = 𝐵𝐾 = 𝐴𝑂
= 𝑎 𝐴𝑁 = 𝐶𝑀 = 𝐷𝑂 = 𝑏 𝐴𝐾 = 𝐶𝐿 = 𝐵𝑂 = 𝑐 𝐵𝐿 = 𝐷𝑀 = 𝐶𝑂 = 𝑑
Matematika SMP (Babak Penyisihan) / Page 8 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 𝑃𝑒𝑟
𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐷𝑁 ∶ 𝐷𝑁2 + 𝐷𝑂2 = 𝑁𝑂2 𝑎2 + 𝑏2 = 52 𝑎2 + 𝑏2 = 25 … (1) 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐴𝐾 ∶ 𝐴𝑂2 + 𝐴𝐾2 = 𝐾𝑂2 𝑎2 +
𝑐2 = 62 𝑎2 + 𝑐2 = 36 … (2) 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐵𝐿 ∶ 𝐵𝑂2 + 𝐵𝐿2 = 𝐿𝑂2 𝑐2 + 𝑑2 = 𝐿𝑂2 𝑐2 + 𝑑2 = 𝐿𝑂2 … (3)
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) 𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 (6) 𝑑𝑎𝑛 (3) ∶ 𝑑2 + 𝑐2 = 48 𝑐2 + 𝑑2
= 𝐿𝑂2 0 = 48 − 𝐿𝑂2 𝐿𝑂2 = 48 𝐿𝑂 = 48 𝐿𝑂 = 16 .3 𝐿𝑂 = 4 3 𝑁
Matematika SMP (Babak Penyisihan) / Page 9 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 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 : 𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 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 :
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
Matematika SMP (Babak Penyisihan) / Page 10 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 𝑅𝑎𝑡𝑎 − 𝑟𝑎𝑡𝑎
𝑏𝑒𝑠𝑎𝑟 𝑘𝑒𝑛𝑎𝑖𝑘𝑎𝑛 𝑔𝑎𝑗𝑖 = 39000+55500+70500+65000+75000 5 = 305000 5 = 61000 𝐴 16. Jawaban : 𝑁. 868 Pembahasan : 𝑠𝑒𝑚𝑢𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛
𝑎𝑛𝑡𝑎𝑟𝑎 1 − 100 𝑦𝑎𝑛𝑔 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 𝑠𝑒𝑚𝑢𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑛𝑡𝑎𝑟𝑎 1 − 100 𝑦𝑎𝑛𝑔 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 𝑠𝑒𝑚𝑢𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑛𝑡𝑎𝑟𝑎 1 − 100 𝑦𝑎𝑛𝑔
𝑑𝑖𝑏𝑎𝑔𝑖 4 𝑡𝑒𝑡𝑎𝑝𝑖 𝑡𝑖𝑑𝑎𝑘 𝑑𝑖𝑏𝑎𝑔𝑖 3 1300 − 432 = 868 𝑁 17. Jawaban : 𝐴. 915 Pembahasan : 𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 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
Matematika SMP (Babak Penyisihan) / Page 11 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 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 𝑎𝑏𝑐 =
915 𝐴 18. Jawaban : 𝐶. 1 𝑎𝑡𝑎𝑢 5 Pembahasan : 𝑥 − 3 2 = −2 𝑥 − 3 + 8 𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 𝑥 − 3 = 𝑝 ∶ 𝑝2 = −2𝑝 + 8 𝑝2 + 2𝑝 − 8 = 0 𝑝 − 2 . 𝑝 + 4 = 0 𝑝
= 2 𝑚𝑒𝑚𝑒𝑛𝑢 𝑝 = −4 𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢 = 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 1 𝑎𝑡𝑎𝑢 5 𝐶
Matematika SMP (Babak Penyisihan) / Page 12 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 19. Jawaban : 𝐵.

1 5 Pembahasan : 𝑛 𝑆 = 40 𝑛 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑆𝑐𝑖𝑒𝑛𝑐𝑒 = 24 𝑛 = 13 𝑛 𝐿𝑖𝑘𝑒 = 5 𝑛 𝐷𝑖𝑠𝑙𝑖𝑘𝑒 = 𝑛 𝑆 − 𝑛 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑆𝑐𝑖𝑒𝑛𝑐𝑒 + 𝑛
− 𝑛 𝐿𝑖𝑘𝑒 = 40 − 24 + 13 − 5 = 40 − 32 = 8 𝑃 𝐷𝑖𝑠𝑙𝑖𝑘𝑒 = 𝑛 𝐷𝑖𝑠𝑙𝑖𝑘𝑒 𝑏𝑜𝑡 𝑆 = 8 40 = 1 5 𝐵 20. Jawaban : 𝐵. 64
Pembahasan : 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐵𝑖𝑙𝑎𝑛𝑔𝑎𝑛 ∶ 301 − 999 𝑅𝑎𝑡𝑢𝑠𝑎𝑛 𝑆𝑎𝑡𝑢𝑎𝑛 𝐾𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛 𝑎𝑛𝑔𝑘𝑎 𝑦𝑎𝑛𝑔 3 2 3 5
3 5 6 5 7 7 6 9 9 7 9 𝐵𝑎𝑛𝑦𝑎𝑘 𝑎𝑛𝑔𝑘𝑎 𝑦𝑎𝑛𝑔 5 − 1 𝑡𝑖𝑑𝑎𝑘 𝑏𝑜𝑙𝑒 𝑏𝑒𝑟𝑢𝑙𝑎𝑛𝑔 = 4 6 − 2 𝑡𝑖𝑑𝑎𝑘 𝑏𝑜𝑙𝑒 𝑏𝑒𝑟𝑢𝑙𝑎𝑛𝑔 = 4 4 𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘 𝑐𝑎𝑟𝑎
𝑚𝑒𝑛𝑦𝑢𝑠𝑢𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 4 .4 .4 = 64 𝐵 21. Jawaban : 𝐷. 20,25 𝑐𝑚2 Pembahasan : 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷 =
𝐷𝐸 = 𝑝𝑎𝑛𝑗𝑎𝑛𝑔 𝑠𝑖𝑠𝑖 𝑝𝑒𝑟𝑠𝑒𝑔𝑖 = 9 𝑐𝑚
Matematika SMP (Babak Penyisihan) / Page 13 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat”
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
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 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 𝑔𝑎𝑟𝑖𝑠 𝐵𝐺 ∶ 𝑂𝐺 = 𝐵𝐺 − 𝐵𝑂 = 9 − 6 = 3
𝑔𝑎𝑟𝑖𝑠 𝐶𝐹 ∶ 𝑃𝑄 = 𝐶𝐹 − 𝐶𝑃 − 𝐹𝑄 = 9 − 3 − 9 2 = 3 2 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
Matematika SMP (Babak Penyisihan) / Page 14 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 22. Jawaban : 𝐷.
60 7 Pembahasan : 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐴𝐵 = 4 𝑐𝑚 𝐵𝐶 = 8 𝑐𝑚 𝐴𝐶 = 4 2 𝑐𝑚 𝐴𝐷 = 15 𝑐𝑚 𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 𝐵𝑇 = 𝑥 𝐶𝑇 = 𝐵𝐶 − 𝐵𝑇 = 8 − 𝑥
𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑇𝐵 ∶ 𝐴𝑇2 = 𝐴𝐵2 − 𝐵𝑇2 𝐴𝑇2 = 42 − 𝑥2 𝐴𝑇2 = 16 − 𝑥2 𝐴𝑇2 = 16 − 𝑥2 … (1) 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
Matematika SMP (Babak Penyisihan) / Page 15 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 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 :
𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 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
Matematika SMP (Babak Penyisihan) / Page 16 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 9𝑥 − 2𝑎𝑥 > −3 +
6𝑎 9 − 2𝑎 𝑥 > −3 + 6𝑎 𝑥 > −3+6 9−2 𝑥 > −3+6𝑎 9−2𝑎 𝑑𝑎𝑛 𝑥 > 3 , 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 ∶ −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 25. Jawaban : 𝐵. 99 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 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
Matematika SMP (Babak Penyisihan) / Page 17 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 𝑀𝑒𝑛𝑒𝑛𝑡𝑢𝑘𝑎𝑛
𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑦𝑎𝑛𝑔 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 𝑥2 − 2𝑥 + 9 < 2015 𝑑𝑎𝑛 4031 <
4𝑥2 + 4𝑥 + 9 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 ∶ 𝑥2 − 2𝑥 + 9 < 2015 ≤ ⋯ ≤ 4031 < 4𝑥2 + 4𝑥 + 9 𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘𝑛𝑦𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑦𝑎𝑛𝑔 𝑏𝑒𝑟𝑎𝑑𝑎 𝑑𝑖𝑎𝑛𝑡𝑎𝑟𝑎 𝑥2 −
2𝑥 + 9 𝑑𝑎𝑛 4𝑥2 + 4𝑥 + 9 4031 − 2015 + 1 = 2017 𝐶 27. Jawaban : 𝐶. 94,5 Pembahasan : 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 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
𝑦𝑎𝑛𝑔 𝑑𝑖𝑏𝑎𝑡𝑎𝑠𝑖 𝑙𝑖𝑛𝑡𝑎𝑠𝑎𝑛 𝑘𝑢𝑐𝑖𝑛𝑔 94,5 𝐶 28. Jawaban : 𝐴. 12 Pembahasan : 𝑓 𝑥 + 1 𝑥 . 𝑓 −𝑥 = 𝑥 + 6
Matematika SMP (Babak Penyisihan) / Page 18 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 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) 𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 (3) 𝑑𝑎𝑛 (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
Matematika SMP (Babak Penyisihan) / Page 19 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 𝑓 3 = 36 5 𝑓 2 +
𝑓 3 = 24 5 + 36 5 = 60 5 = 12 𝐴 29. Jawaban : 𝐵. 160 𝜋 Pembahasan : 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 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 : 𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 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
Matematika SMP (Babak Penyisihan) / Page 20 Blog tentang : “Soal Matematika dan Pembahasannya untuk SD, SMP, SMA atau Sederajat” 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 𝑎𝑡𝑎𝑢 𝑝
= 4 2 . 2 𝑝 = −2 2 . 2 . 2 2 𝑎𝑡𝑎𝑢 𝑝 = 4 2 . 2 . 2 2 𝑝 = −2 2 2 .2 𝑎𝑡𝑎𝑢 𝑝 = 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 …
− 1 2 2 𝑎𝑡𝑎𝑢 2 𝐷 "𝑆𝑒𝑚𝑜𝑔𝑎 𝑏𝑒𝑟𝑚𝑎𝑛𝑓𝑎𝑎𝑡" "𝑆𝐷. 𝐴"
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