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Pembahasan osn matematika smp 2014 tingkat kabupaten (bagian a pilihan ganda) 2

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Pembahasan osn matematika smp 2014 tingkat kabupaten (bagian a pilihan ganda) 2

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Pembahasan osn matematika smp 2014 tingkat kabupaten (bagian a pilihan ganda) 2

  1. 1. www.siap-osn.blogspot.com @ Maret 2014 SD.A 2 N.H Pembahasan OSN Matematika SMP 2014 / Page 1 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” PEMBAHASAN OSN MATEMATIKA SMP 2014 TINGKAT KABUPATEN BAGIAN A : PILIHAN GANDA BAGIAN A : PILIHAN GANDA 1-10 Sudah dipostingkan di www.siap-osn.blogspot.com 11. Jawaban : 𝐵. 1 Pembahasan : 𝑥2 = 𝑦2 + 100 𝑥2 − 𝑦2 = 100 𝑥 − 𝑦 . 𝑥 + 𝑦 = 100 𝐾𝑎𝑟𝑒𝑛𝑎 𝑥 𝑑𝑎𝑛 𝑦 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑠𝑙𝑖, 𝑚𝑎𝑘𝑎 𝑑𝑎𝑝𝑎𝑡 𝑑𝑖𝑏𝑢𝑎𝑡 𝑏𝑒𝑏𝑒𝑟𝑎𝑝𝑎 𝑘𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛 𝑑𝑎𝑟𝑖 𝑥 − 𝑦 . 𝑥 + 𝑦 = 100 𝑦𝑎𝑖𝑡𝑢 ∶ 𝑥 − 𝑦 . 𝑥 + 𝑦 = 1 .100 → 𝑥 − 𝑦 = 1 𝑥 + 𝑦 = 100 2𝑥 = 101 𝑥 = 101 2 𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢ℎ𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑥 𝑏𝑢𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑠𝑙𝑖 𝑥 − 𝑦 . 𝑥 + 𝑦 = 2 .50 → 𝑥 − 𝑦 = 2 𝑥 + 𝑦 = 50 2𝑥 = 52 𝑥 = 52 2 𝑥 = 26 → 𝑥 + 𝑦 = 50 26 + 𝑦 = 50 𝑦 = 50 − 26 𝑦 = 24 → 𝑚𝑒𝑚𝑒𝑛𝑢ℎ𝑖 26,24 𝑥 − 𝑦 . 𝑥 + 𝑦 = 4 .25 → 𝑥 − 𝑦 = 4 𝑥 + 𝑦 = 25 2𝑥 = 29 𝑥 = 29 2 𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢ℎ𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑥 𝑏𝑢𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑠𝑙𝑖 𝑥 − 𝑦 . 𝑥 + 𝑦 = 5 .20 → 𝑥 − 𝑦 = 5 𝑥 + 𝑦 = 20 2𝑥 = 25 𝑥 = 25 2 𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢ℎ𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑥 𝑏𝑢𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑠𝑙𝑖 𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘 𝑝𝑎𝑠𝑎𝑛𝑔𝑎𝑛 𝑥, 𝑦 𝑦𝑎𝑛𝑔 𝑚𝑒𝑚𝑒𝑛𝑢ℎ𝑖 𝑎𝑑𝑎𝑙𝑎ℎ 1 𝐵
  2. 2. www.siap-osn.blogspot.com @ Maret 2014 SD.A 2 N.H Pembahasan OSN Matematika SMP 2014 / Page 2 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 12. Jawaban : 𝐶. 𝐻𝑖𝑚𝑝𝑢𝑛𝑎𝑛 𝐴 𝑡𝑒𝑟𝑡𝑢𝑡𝑢𝑝 𝑡𝑒𝑟ℎ𝑎𝑑𝑎𝑝 𝑜𝑝𝑒𝑟𝑎𝑠𝑖 𝑝𝑒𝑛𝑗𝑢𝑚𝑙𝑎ℎ𝑎𝑛 𝑑𝑎𝑛 𝑝𝑒𝑟𝑘𝑎𝑙𝑖𝑎𝑛 Pembahasan : 𝐴 = 0,2,4, 6,… 𝐴 + 𝐴 = 0,2, 4, 6,… + 0,2, 4,6, … = 0,2, 4,6, … → 𝐴 + 𝐴 = 𝐴 (𝑡𝑒𝑟𝑡𝑢𝑡𝑢𝑝) 𝐴 . 𝐴 = 0, 2,4, 6,… . 0,2, 4, 6,… = 0, 4,8, 12,… → 𝐴 . 𝐴 ⊂ 𝐴 𝑡𝑒𝑟𝑡𝑢𝑡𝑢𝑝 𝐴 − 𝐴 = 0, 2, 4, 6, … − 0, 2, 4, 6, … = … , −6, −4, −2,0, 2, 4,6, … → 𝐴 − 𝐴 ≠ 𝐴 (𝑡𝑖𝑑𝑎𝑘 𝑡𝑒𝑟𝑡𝑢𝑡𝑢𝑝 ) 𝐽𝑎𝑑𝑖 𝑝𝑒𝑟𝑛𝑦𝑎𝑡𝑎𝑎𝑛 𝑦𝑎𝑛𝑔 𝑏𝑒𝑛𝑎𝑟 𝑎𝑑𝑎𝑙𝑎ℎ 𝐻𝑖𝑚𝑝𝑢𝑛𝑎𝑛 𝐴 𝑡𝑒𝑟𝑡𝑢𝑡𝑢𝑝 𝑡𝑒𝑟ℎ𝑎𝑑𝑎𝑝 𝑜𝑝𝑒𝑟𝑎𝑠𝑖 𝑝𝑒𝑛𝑗𝑢𝑚𝑙𝑎ℎ𝑎𝑛 𝑑𝑎𝑛 𝑝𝑒𝑟𝑘𝑎𝑙𝑖𝑎𝑛 𝐶 13. Jawaban : 𝐶. 4 3 3 Pembahasan : 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐴𝐵 = 𝐵𝐶 = 𝐴𝐶 = 2 𝑡 = 𝐵𝐶2 − 1 2 . 𝐴𝐵 2 = 22 − 1 2 .2 2 = 4 − 12 = 4 − 1 = 3 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 ∶ 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 = 1 2 . 𝐴𝐵 . 𝑡 = 1 2 .2 . 3 = 3 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 𝑑𝑎𝑛 𝐷𝐸𝐹 ∶ 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐸𝐹 = 1 4 . 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 = 1 4 . 3 = 3 4 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐸𝐹 𝑑𝑎𝑛 𝐺𝐻𝐼 ∶ 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐺𝐻𝐼 = 1 4 . 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐸𝐹 = 1 4 . 3 4 = 3 16 𝐿 𝑠𝑒𝑙𝑢𝑟𝑢 ℎ 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 = 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 + 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐸𝐹 + 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐺𝐻𝐼 + ⋯ = 3 𝑈1 + 3 4 𝑈2 + 3 16 𝑈3 + ⋯ 𝑑𝑒𝑟𝑒𝑡 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖 𝑡𝑎𝑘 ℎ𝑖𝑛𝑔𝑔𝑎 → 𝑎 = 𝑈1 = 3 𝑟 = 𝑈2 𝑈1 = 3 4 3 = 1 4 = 𝑎 1−𝑟 = 3 1− 1 4 = 3 3 4 = 4 3 3 𝐽𝑎𝑑𝑖 𝑙𝑢𝑎𝑠 𝑠𝑒𝑙𝑢𝑟𝑢ℎ 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑎𝑑𝑎𝑙𝑎ℎ 4 3 3 𝐶
  3. 3. www.siap-osn.blogspot.com @ Maret 2014 SD.A 2 N.H Pembahasan OSN Matematika SMP 2014 / Page 3 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 14. Jawaban : 𝐶. 7 Pembahasan : 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝑀𝑢𝑙𝑎 𝑚𝑢𝑙𝑎 𝑘𝑎𝑡𝑎𝑘 𝑏𝑒𝑟𝑎𝑑𝑎 𝑝𝑎𝑑𝑎 𝑝𝑜𝑠𝑖𝑠𝑖 𝑛𝑜𝑚𝑜𝑟1 𝐾𝑎𝑡𝑎𝑘 𝑚𝑒𝑙𝑜𝑚𝑝𝑎𝑡1 𝑠𝑎𝑡𝑢𝑎𝑛 𝑠𝑒𝑎𝑟𝑎ℎ 𝑗𝑎𝑟𝑢𝑚 𝑗𝑎𝑚 𝑗𝑖𝑘𝑎 𝑏𝑒𝑟𝑎𝑑𝑎 𝑝𝑎𝑑𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 𝐾𝑎𝑡𝑎𝑘 𝑚𝑒𝑙𝑜𝑚𝑝𝑎𝑡3 𝑠𝑎𝑡𝑢𝑎𝑛 𝑠𝑒𝑎𝑟𝑎ℎ 𝑗𝑎𝑟𝑢𝑚 𝑗𝑎𝑚 𝑗𝑖𝑘𝑎 𝑏𝑒𝑟𝑎𝑑𝑎 𝑝𝑎𝑑𝑎 𝑏𝑢𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 𝑃𝑜𝑙𝑎 𝑙𝑜𝑛𝑐𝑎𝑡𝑎𝑛 𝑘𝑎𝑡𝑎𝑘 ∶ 4 , 7 , 8,1 4 𝑏𝑒𝑟𝑢𝑙𝑎𝑛𝑔 ,4 ,7 ,8, 1 4 𝑏𝑒𝑟𝑢𝑙𝑎𝑛𝑔 , 4 , 7 ,8,1 4 𝑏𝑒𝑟𝑢𝑙𝑎𝑛𝑔 ,…. 2014 = 4 .503 + 2 𝐽𝑎𝑑𝑖 𝑝𝑜𝑠𝑖𝑠𝑖 𝑘𝑎𝑡𝑎𝑘 𝑠𝑒𝑡𝑒𝑙𝑎ℎ 𝑚𝑒𝑙𝑜𝑚𝑝𝑎𝑡 2014 𝑘𝑎𝑙𝑖 𝑎𝑑𝑎𝑙𝑎ℎ 7 𝐶 15. Jawaban : 𝐷. 50 Pembahasan : 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝑥 = 180 𝑜 − 110 𝑜 − 35 𝑜 = 35 𝑜 2𝑥 = 2 . 35 𝑜 = 70 𝑜 𝑦 + 𝑥 + 60 = 180 𝑜 𝑦 + 𝑥 = 180 𝑜 − 60 𝑜 𝑦 + 𝑥 = 120 𝑜 𝑦 + 𝑥 − 2𝑥 = 120 𝑜 − 70 𝑜 𝑦 − 𝑥 = 50 𝑜 𝐽𝑎𝑑𝑖 𝑏𝑒𝑠𝑎𝑟 𝑠𝑢𝑑𝑢𝑡 𝑦 − 𝑥 𝑎𝑑𝑎𝑙𝑎ℎ 50 𝑜 𝐷 1 2 3 4
  4. 4. www.siap-osn.blogspot.com @ Maret 2014 SD.A 2 N.H Pembahasan OSN Matematika SMP 2014 / Page 4 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 16. Jawaban : 𝐵. 0,30 Pembahasan : 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐽𝑢𝑚𝑙𝑎ℎ 𝑠𝑖𝑠𝑤𝑎 = 100 𝑃𝑒𝑙𝑢𝑎𝑛𝑔 𝑠𝑖𝑠𝑤𝑎 𝑚𝑒𝑛𝑔𝑖𝑟𝑖𝑚 𝑠𝑚𝑠 𝑡𝑖𝑑𝑎𝑘 𝑙𝑒𝑏𝑖ℎ 𝑑𝑎𝑟𝑖 30 𝑘𝑎𝑙𝑖 = 5% + 10% + 15% = 30% = 30 100 = 0,30 𝐽𝑎𝑑𝑖 𝑝𝑒𝑙𝑢𝑎𝑛𝑔 𝑠𝑖𝑠𝑤𝑎 𝑚𝑒𝑛𝑔𝑖𝑟𝑖𝑚 𝑠𝑚𝑠 𝑡𝑖𝑑𝑎𝑘 𝑙𝑒𝑏𝑖ℎ 𝑑𝑎𝑟𝑖 30 𝑘𝑎𝑙𝑖 𝑎𝑑𝑎𝑙𝑎ℎ 0,30 𝐵 17. Jawaban : 𝐶. 12 Pembahasan : 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐷 𝑑𝑎𝑛 𝐷𝐹𝐺 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛∶ 𝐵𝐷 𝐷𝐺 = 𝐴𝐵 𝐹𝐺 8+𝑥 8 = 7 4 8 + 𝑥 = 7 4 .8 8 + 𝑥 = 14 𝑥 = 14 − 8 𝑥 = 6 → 𝐵𝐷 = 8 + 𝑥 𝐵𝐷 = 8 + 6 𝐵𝐷 = 14 𝐽𝑢𝑚𝑙𝑎ℎ 𝑠𝑚𝑠 𝑃𝑒𝑟𝑠𝑒𝑛𝑡𝑎𝑠𝑒 1 − 10 5% 11 − 20 10% 21 − 30 15% 31 − 40 20% 41 𝑎𝑡𝑎𝑢 𝑙𝑒𝑏𝑖ℎ 25% 𝑇𝑖𝑑𝑎𝑘 𝑚𝑒𝑛𝑔𝑖𝑟𝑖𝑚 𝑠𝑚𝑠 25%
  5. 5. www.siap-osn.blogspot.com @ Maret 2014 SD.A 2 N.H Pembahasan OSN Matematika SMP 2014 / Page 5 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐵𝐶𝐷 𝑑𝑎𝑛 𝐵𝐸𝐺 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛∶ 𝑥 = 6 𝐵𝐷 = 14 𝐸𝐹 𝐶𝐷 = 𝐵𝐹 𝐵𝐷 𝑦 14 = 6 14 𝑦 = 6 14 .14 𝑦 = 6 𝑥 + 𝑦 = 6 + 6 = 12 𝐽𝑎𝑑𝑖 𝑛𝑖𝑙𝑎𝑖 𝑥 + 𝑦 𝑎𝑑𝑎𝑙𝑎ℎ 12 𝐶 18. Jawaban : 𝐵. 3 Pembahasan : 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐾𝑒𝑡𝑒𝑟𝑎𝑛𝑔𝑎𝑛 𝑈𝑚𝑢𝑟 ≤ 25 𝑈𝑚𝑢𝑟 > 25 𝐽𝑢𝑚𝑙𝑎ℎ 𝑜𝑟𝑎𝑛𝑔 = 75 75 − 50 = 25 50 𝑆𝑢𝑘𝑎 𝑝𝑒𝑑𝑎𝑠 = 27 7 27 − 7 = 20 𝑆𝑢𝑘𝑎 𝑚𝑎𝑛𝑖𝑠= 28 28 − 25 = 3 25 𝑇𝑖𝑑𝑎𝑘 𝑠𝑢𝑘𝑎 𝑝𝑒𝑑𝑎𝑠 𝑑𝑎𝑛 𝑚𝑎𝑛𝑖𝑠 = 25 25 − 7 = 18 7 𝑆𝑢𝑘𝑎 𝑝𝑒𝑑𝑎𝑠 𝑑𝑎𝑛 𝑚𝑎𝑛𝑖𝑠 = 5 7 + 3 + 18 − 25 = 3 20 + 25 + 7 − 50 = 2 𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘 𝑜𝑟𝑎𝑛𝑔 𝑦𝑎𝑛𝑔 𝑏𝑒𝑟𝑢𝑚𝑢𝑟 𝑡𝑖𝑑𝑎𝑘 𝑙𝑒𝑏𝑖ℎ 𝑑𝑎𝑟𝑖 25 𝑡𝑎ℎ𝑢𝑛 𝑦𝑎𝑛𝑔 𝑚𝑒𝑛𝑦𝑢𝑘𝑎𝑖 𝑚𝑎𝑠𝑎𝑘𝑎𝑛 𝑝𝑒𝑑𝑎𝑠 𝑑𝑎𝑛 𝑗𝑢𝑔𝑎 𝑚𝑎𝑠𝑎𝑘𝑎𝑛 𝑚𝑎𝑛𝑖𝑠 𝑎𝑑𝑎𝑙𝑎ℎ 3 𝐵 19. Jawaban : 𝐶. 144 Pembahasan : 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐿𝑢𝑎𝑠 𝑠𝑎𝑡𝑢 𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝑘𝑒𝑐𝑖𝑙 = 4 𝑚2 𝐵𝑖𝑠𝑎 𝑑𝑖𝑙𝑖ℎ𝑎𝑡 𝑏𝑎ℎ𝑤𝑎 𝑏𝑎𝑛𝑔𝑢𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑡𝑒𝑟𝑑𝑖𝑟𝑖 𝑑𝑎𝑟𝑖 4 𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 𝑦𝑎𝑛𝑔 𝑏𝑒𝑟𝑢𝑘𝑢𝑟𝑎𝑛 𝑠𝑎𝑚𝑎
  6. 6. www.siap-osn.blogspot.com @ Maret 2014 SD.A 2 N.H Pembahasan OSN Matematika SMP 2014 / Page 6 Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ” 𝐿𝑢𝑎𝑠 𝑠𝑎𝑡𝑢 𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝑘𝑒𝑐𝑖𝑙 = 4 𝑠2 = 4 𝑠 = 4 𝑠 = 2 𝑎 = 4,5 . 𝑠 = 4,5 .2 = 9 𝑡 = 2 . 𝑠 = 2 .2 = 4 𝐿𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 = 𝑎 . 𝑡 = 9 .4 = 36 𝐿 𝑏𝑎𝑛𝑔𝑢𝑛 𝑑𝑎𝑡𝑎𝑟 = 4 . 𝐿𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 = 4 .36 = 144 𝐽𝑎𝑑𝑖 𝑙𝑢𝑎𝑠 𝑏𝑎𝑛𝑔𝑢𝑛 𝑑𝑎𝑡𝑎𝑟 𝑝𝑎𝑑𝑎 𝑔𝑎𝑚𝑏𝑎𝑟 𝑎𝑑𝑎𝑙𝑎ℎ 144 𝐶 20. Jawaban : 𝐵. 18 Pembahasan : 𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶ 𝐾𝑎𝑛𝑡𝑜𝑛𝑔 𝐼 𝐾𝑎𝑛𝑡𝑜𝑛𝑔 𝐼𝐼 𝐾𝑎𝑛𝑡𝑜𝑛𝑔 𝐼𝐼𝐼 𝐵𝑎𝑛𝑦𝑎𝑘 𝑝𝑎𝑠𝑎𝑛𝑔𝑎𝑛 𝑤𝑎𝑟𝑛𝑎 𝑀 𝐾 3𝑀 𝐻 𝐾 𝐻 𝑊𝑎𝑟𝑛𝑎 1 𝑊𝑎𝑟𝑛𝑎 2 𝐵𝑎𝑛𝑦𝑎𝑘 𝑝𝑒𝑛𝑦𝑢𝑠𝑢𝑛𝑎𝑛 6 1 6 5 2 4 3 3 4 2 5 1 6 𝐽𝑎𝑑𝑖 𝑚𝑎𝑘𝑠𝑖𝑚𝑎𝑙 𝑏𝑎𝑛𝑦𝑎𝑘 𝑠𝑖𝑠𝑤𝑎 𝑦𝑎𝑛𝑔 𝑎𝑑𝑎 𝑑𝑖 𝑘𝑒𝑙𝑎𝑠 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎ℎ 3 .6 = 18 𝐵 BAGIAN B : ISIAN SINGKAT pada posting berikutnya di www.siap-osn.blogspot.com

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