Post on 16-Jul-2015
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 1
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
PEMBAHASAN
OSN MATEMATIKA SMP 2015 TINGKAT KABUPATEN
BAGIAN B : ISIAN SINGKAT
BAGIAN B : ISIAN SINGKAT
1. Jawaban : 𝑥 = −4,−1
Pembahasan :
𝑥 𝑎𝑑𝑎𝑙𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡
𝑥2 + 5𝑥 + 6 𝑎𝑑𝑎𝑙𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎
𝑥2 + 5𝑥 + 6 = 𝑥 + 2 . (𝑥 + 3)
𝐾𝑎𝑟𝑒𝑛𝑎 𝑥 + 2 < 𝑥 + 3 𝑚𝑎𝑘𝑎 𝑢𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 = 2, 3, 5, 7, 11,… 𝑏𝑒𝑟𝑙𝑎𝑘𝑢 ∶
𝑈𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 = 2
𝑥2 + 5𝑥 + 6 = 𝑥 + 2 . 𝑥 + 3 = 2 = 1 .2 = −2 . −1
𝐼𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑤𝑎 ∶
𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = 1 .2
𝑥 + 2 = 1 𝑑𝑎𝑛 𝑥 + 3 = 2
𝑥 = 1 − 2 𝑑𝑎𝑛 𝑥 = 2 − 3
𝑥 = −1 𝑑𝑎𝑛 𝑥 = −1 (𝑚𝑒𝑚𝑒𝑛𝑢𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑠𝑎𝑚𝑎)
𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = −2 . −1
𝑥 + 2 = −2 𝑑𝑎𝑛 𝑥 + 3 = −1
𝑥 = −2 − 2 𝑑𝑎𝑛 𝑥 = −1 − 3
𝑥 = −4 𝑑𝑎𝑛 𝑥 = −4 (𝑚𝑒𝑚𝑒𝑛𝑢𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑠𝑎𝑚𝑎)
𝑈𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 = 3
𝑥2 + 5𝑥 + 6 = 𝑥 + 2 . 𝑥 + 3 = 3 = 1 .3 = −3 . −1
𝐼𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑤𝑎 ∶
𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = 1 .3
𝑥 + 2 = 1 𝑑𝑎𝑛 𝑥 + 3 = 3
𝑥 = 1 − 2 𝑑𝑎𝑛 𝑥 = 3 − 3
𝑥 = −1 𝑑𝑎𝑛 𝑥 = 0 (𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎)
𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = −3 . −1
𝑥 + 2 = −3 𝑑𝑎𝑛 𝑥 + 3 = −1
𝑥 = −3 − 2 𝑑𝑎𝑛 𝑥 = −1 − 3
𝑥 = −5 𝑑𝑎𝑛 𝑥 = −4 (𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎)
𝑈𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 = 5
𝑥2 + 5𝑥 + 6 = 𝑥 + 2 . 𝑥 + 3 = 5 = 1 .5 = −5 . −1
𝐼𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑤𝑎 ∶
𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = 1 .5
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 2
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
𝑥 + 2 = 1 𝑑𝑎𝑛 𝑥 + 3 = 5
𝑥 = 1 − 2 𝑑𝑎𝑛 𝑥 = 5 − 3
𝑥 = −1 𝑑𝑎𝑛 𝑥 = 2 (𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎)
𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = −5 . −1
𝑥 + 2 = −5 𝑑𝑎𝑛 𝑥 + 3 = −1
𝑥 = −5 − 2 𝑑𝑎𝑛 𝑥 = −1 − 3
𝑥 = −7 𝑑𝑎𝑛 𝑥 = −4 (𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎)
𝐷𝑎𝑟𝑖 𝑝𝑒𝑟𝑖𝑡𝑢𝑛𝑔𝑎𝑛 𝑑𝑖𝑎𝑡𝑎𝑠 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑝𝑜𝑙𝑎 𝑏𝑎𝑤𝑎 𝑢𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 ≥ 3
𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑑𝑖𝑝𝑒𝑟𝑜𝑙𝑒 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑦𝑎𝑛𝑔 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎
𝐽𝑎𝑑𝑖 𝑛𝑖𝑙𝑎𝑖 𝑥 = −4,−1
2. Jawaban : 12
Pembahasan :
𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 𝑚𝑒𝑙𝑎𝑙𝑢𝑖 𝑡𝑖𝑡𝑖𝑘 (−2, 6)
𝑠𝑢𝑚𝑏𝑢 𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑛𝑦𝑎 𝑥 = −1
𝑎, 𝑏, 𝑑𝑎𝑛 𝑐 𝑚𝑒𝑟𝑢𝑝𝑎𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑔𝑒𝑛𝑎𝑝 𝑝𝑜𝑠𝑖𝑡𝑖𝑓 𝑏𝑒𝑟𝑢𝑟𝑢𝑡𝑎𝑛
−2 𝑥
, 6 𝑦
→ 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐
6 = 𝑎 . −2 2 + 𝑏 . −2 + 𝑐
6 = 4𝑎 − 2𝑏 + 𝑐
4𝑎 − 2𝑏 + 𝑐 = 6 … 1
𝑆𝑢𝑚𝑏𝑢 𝑠𝑖𝑚𝑒𝑡𝑟𝑖 𝑥 = −1 → 𝑥 = −𝑏
2𝑎
−1 = −𝑏
2𝑎
−2𝑎 = −𝑏
𝑏 = 2𝑎 … 2
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑠𝑖𝑘𝑎𝑛 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 2 𝑘𝑒 1 ∶
4𝑎 − 2𝑏 + 𝑐 = 6
4𝑎 − 2 . 2𝑎 + 𝑐 = 6
4𝑎 − 4𝑎 + 𝑐 = 6
𝑐 = 6
𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑎 𝑐 = 6 , 𝑏 = 2𝑎 𝑑𝑎𝑛 𝑘𝑎𝑟𝑒𝑛𝑎 𝑎, 𝑏,𝑑𝑎𝑛 𝑐 𝑚𝑒𝑟𝑢𝑝𝑎𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑔𝑒𝑛𝑎𝑝 𝑝𝑜𝑠𝑖𝑡𝑖𝑓 𝑏𝑒𝑟𝑢𝑟𝑢𝑡𝑎𝑛
𝑚𝑎𝑘𝑎 𝑏𝑒𝑟𝑙𝑎𝑘𝑢 ∶
2 𝑎
, 4 𝑏=2𝑎
, 6
𝐽𝑎𝑑𝑖 𝑎 + 𝑏 + 𝑐 = 2 + 4 + 6 = 12
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 3
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
3. Jawaban : 7 3 + 12 𝑐𝑚2
Pembahasan :
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝑃,𝑄 𝑑𝑎𝑛 𝑅 𝑎𝑑𝑎𝑙𝑎 𝑡𝑖𝑡𝑖𝑘 𝑠𝑖𝑛𝑔𝑔𝑢𝑛𝑔 𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛 𝑝𝑎𝑑𝑎 𝑠𝑖𝑠𝑖 − 𝑠𝑖𝑠𝑖 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐷, 𝑠𝑒𝑖𝑛𝑔𝑔𝑎 ∶
∠𝐴𝑃𝐷 = ∠𝐶𝑃𝐷 = ∠𝐷𝑅𝑆 = ∠𝐶𝑅𝑆 = ∠𝐴𝑄𝑆 = ∠𝐷𝑄𝑆 = 90𝑜
𝑆𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 → 𝐴𝐵 = 𝐴𝐶
𝑆𝑅 = 𝑆𝑄 = 𝑃𝑆 = 1
𝑅𝐷 = 3
3 𝑐𝑚
∠𝑆𝐷𝑅 = 60𝑜
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 ∶
∠𝐷𝑆𝑅 = 180𝑜 − 90𝑜 − 60𝑜 = 30𝑜
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑄𝑆 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶
∠𝑄𝐷𝑆 = ∠𝑆𝐷𝑅 = 60𝑜
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 ∶
∠𝐷𝐶𝑃 = 180𝑜 − 90𝑜 − 60𝑜 = 30𝑜
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 4
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 ∶
∠𝐷𝐴𝑃 = 180𝑜 − 90𝑜 − 60𝑜 = 30𝑜
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 ∶
∠𝐴𝐵𝐶 = ∠𝐴𝐶𝐵 = 30𝑜
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐷 ∶
∠𝐴𝐷𝐵 = 180𝑜 − 60𝑜 − 60𝑜 = 60𝑜
∠𝐵𝐴𝐷 = 180𝑜 − 60𝑜 − 30𝑜 = 90𝑜
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 ∶
𝐷𝑆 = 𝑅𝐷2 + 𝑆𝑅2 = 3
3
2
+ 12 = 3
9+ 1 =
3
9+
9
9=
12
9=
12
9=
4 .3
9=
2 3
3
𝐷𝑃 = 𝑃𝑆 + 𝐷𝑆 = 1 +2 3
3
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶
𝐶𝑃
𝑆𝑅=
𝐷𝑃
𝑅𝐷
𝐶𝑃
1=
1+2 3
3
3
3
𝐶𝑃 = 1 +2 3
3 .
3
3
𝐶𝑃 =3
3+ 2
𝐶𝑃 =3
3 . 3
3+ 2
𝐶𝑃 = 3 + 2
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶
𝐴𝑃 = 𝐶𝑃 = 3 + 2
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 ∶
𝐴𝐵 = 𝐴𝐶 = 𝐴𝑃 + 𝐶𝑃 = 3 + 2 + 3 + 2 = 2 3 + 4
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐵𝐴𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶
𝐴𝐷
𝑅𝐷=
𝐴𝐵
𝑆𝑅
𝐴𝐷
3
3
=2 3+4
1
𝐴𝐷 = 2 3 + 4 . 3
3
𝐴𝐷 = 2 +4 3
3
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 5
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 ∶
𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 =1
2 .𝐷𝑃 .𝐶𝑃
=1
2 . 1 +
2 3
3 . 3 + 2
=1
2 . 3 + 2 + 2 +
4 3
3
=1
2 .
3 3
3+ 4 +
4 3
3
=1
2 .
7 3
3+ 4
=7 3
6+ 2
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶
𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 = 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐶𝑃𝐷
=7 3
6+ 2
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐵𝐴𝐷 ∶
𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐵𝐴𝐷 =1
2 .𝐴𝐵 .𝐴𝐷
=1
2 . 2 3 + 4 . 2 +
4 3
3
=1
2 . 4 3 + 8 + 8 +
16 3
3
=1
2 .
12 3
3+ 16 +
16 3
3
=1
2 .
28 3
3+ 16
=14 3
3+ 8
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 ∶
𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 = 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 + 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 + 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐵𝐴𝐷
= 7 3
6+ 2 +
7 3
6+ 2 +
14 3
3+ 8
=14 3
6+ 12 +
14 3
3
=7 3
3+ 12 +
14 3
3
=21 3
3+ 12
= 7 3 + 12
𝐽𝑎𝑑𝑖 𝑙𝑢𝑎𝑠 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎 7 3 + 12 𝑐𝑚2
4. Jawaban : 55 ∶ 153
Pembahasan :
𝐵𝑜𝑡𝑜𝑙 𝐼 → 𝐺𝐼 ∶ 𝐴𝐼 = 2 ∶ 11
𝐵𝑜𝑡𝑜𝑙 𝐼𝐼 → 𝐺𝐼𝐼 ∶ 𝐴𝐼𝐼 = 3 ∶ 5
𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶
𝑉𝑏𝑜𝑡𝑜𝑙 = 𝑉
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 6
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
𝑉𝑜𝑙𝑢𝑚𝑒 𝐺𝑢𝑙𝑎 𝑑𝑎𝑛 𝐴𝑖𝑟 𝑑𝑎𝑙𝑎𝑚 𝑏𝑜𝑡𝑜𝑙 𝐼 𝑑𝑎𝑛 𝑏𝑜𝑡𝑜𝑙 𝐼𝐼 𝑠𝑒𝑏𝑒𝑙𝑢𝑚 𝑑𝑖𝑐𝑎𝑚𝑝𝑢𝑟 ∶
𝐺𝐼 =2
2+11 .𝑉 =
2𝑉
13
𝐴𝐼 =11
2+11 .𝑉 =
11𝑉
13
𝐺𝐼𝐼 =3
3+5 .𝑉 =
3𝑉
8
𝐴𝐼𝐼 =5
3+5 .𝑉 =
5𝑉
8
𝑉𝑜𝑙𝑢𝑚𝑒 𝐺𝑢𝑙𝑎 𝑑𝑎𝑛 𝐴𝑖𝑟 𝑠𝑒𝑡𝑒𝑙𝑎 𝑏𝑜𝑡𝑜𝑙 𝐼 𝑑𝑎𝑛 𝑏𝑜𝑡𝑜𝑙 𝐼𝐼 𝑑𝑖𝑐𝑎𝑚𝑝𝑢𝑟 ∶
𝐺𝐼+𝐼𝐼 = 𝐺𝐼 + 𝐺𝐼𝐼 =2𝑉
13+
3𝑉
8=
16𝑉
104+
39𝑉
104=
55𝑉
104
𝐴𝐼+𝐼𝐼 = 𝐴𝐼 + 𝐴𝐼𝐼 =11𝑉
13+
5𝑉
8=
88𝑉
104+
65𝑉
104=
153𝑉
104
𝐺𝐼+𝐼𝐼 ∶ 𝐴𝐼+𝐼𝐼 =𝐺𝐼+𝐼𝐼
𝐴𝐼+𝐼𝐼=
55𝑉
104153𝑉
104
=55𝑉
104 .
104
153𝑉=
55
153= 55 ∶ 153
𝐽𝑎𝑑𝑖 𝑟𝑎𝑠𝑖𝑜 𝑘𝑎𝑛𝑑𝑢𝑛𝑔𝑎𝑛 𝑔𝑢𝑙𝑎 𝑑𝑎𝑛 𝑎𝑖𝑟 𝑎𝑠𝑖𝑙 𝑐𝑎𝑚𝑝𝑢𝑟𝑎𝑛𝑛𝑦𝑎 𝑎𝑑𝑎𝑙𝑎 55 ∶ 153
5. Jawaban : 19
Pembahasan :
𝑓 𝑥 = 209 − 𝑥2
𝑓 𝑎𝑏 = 𝑓 𝑎 + 2𝑏 − 𝑓 𝑎 − 2𝑏 𝑑𝑖𝑚𝑎𝑛𝑎 ∶ 𝑎 , 𝑏 𝑎𝑑𝑎𝑙𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑝𝑜𝑠𝑖𝑡𝑖𝑓 𝑑𝑎𝑛 𝑎 < 𝑏
𝑓 𝑎𝑏 = 𝑓 𝑎 + 2𝑏 − 𝑓 𝑎 − 2𝑏
209 − 𝑎𝑏 2 = 209 − 𝑎 + 2𝑏 2 − 209 − 𝑎 − 2𝑏 2
209 − 𝑎2𝑏2 = 209 − 𝑎2 + 4𝑎𝑏 + 4𝑏2 − 209 − 𝑎2 − 4𝑎𝑏 + 4𝑏2
209 − 𝑎2𝑏2 = 209 − 𝑎2 − 4𝑎𝑏 − 4𝑏2 − 209 − 𝑎2 + 4𝑎𝑏 − 4𝑏2
209 − 𝑎2𝑏2 = 209 − 𝑎2 − 4𝑎𝑏 − 4𝑏2 − 209 + 𝑎2 − 4𝑎𝑏 + 4𝑏2
209 − 𝑎2𝑏2 = −8𝑎𝑏
209 = 𝑎2𝑏2 − 8𝑎𝑏
19 .11 = 𝑎𝑏 . 𝑎𝑏 − 8 → 𝑎𝑏 = 19
𝑎𝑏 − 8 = 11
𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑎 𝑎𝑏 = 19 𝑦𝑎𝑛𝑔 𝑚𝑒𝑟𝑢𝑝𝑎𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 ,𝑑𝑎𝑛 𝑘𝑎𝑟𝑒𝑛𝑎 𝑎 < 𝑏 𝑚𝑎𝑘𝑎 𝑏𝑒𝑟𝑙𝑎𝑘𝑢 ∶
𝑎𝑏 = 19
𝑎𝑏 = 1 . 19 → 𝑎 = 1
𝑏 = 19
𝐽𝑎𝑑𝑖 𝑛𝑖𝑙𝑎𝑖 𝑏
𝑎=
19
1= 19
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 7
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
6. Jawaban : 10080
Pembahasan :
𝑈1 + 𝑈2 + 𝑈3 + 𝑈4 = 70 → 𝑆4 = 70
𝑈5 + 𝑈6 + ⋯+ 𝑈16 = 690
𝑈1 + 𝑈2 + ⋯+ 𝑈16 = 760
𝑆16 = 760
𝑆𝑛 =𝑛
2 . 2𝑎 + 𝑛 − 1 . 𝑏
𝑆4 =4
2 . 2𝑎 + 4 − 1 . 𝑏 = 70
2 . 2𝑎 + 3𝑏 = 70
2𝑎 + 3𝑏 =70
2
2𝑎 + 3𝑏 = 35 … (1)
𝑆𝑛 =𝑛
2 . 2𝑎 + 𝑛 − 1 . 𝑏
𝑆16 =16
2 . 2𝑎 + 16 − 1 . 𝑏 = 760
8 . 2𝑎 + 15𝑏 = 760
2𝑎 + 15𝑏 =760
8
2𝑎 + 15𝑏 = 95 … (2)
𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 2 𝑑𝑎𝑛 (1):
2𝑎 + 15𝑏 = 95
2𝑎 + 3𝑏 = 35
12𝑏 = 60
𝑏 =60
12
𝑏 = 5 → 1 :
2𝑎 + 3𝑏 = 35
2𝑎 + 3 . 5 = 35
2𝑎 + 15 = 35
2𝑎 = 35 − 15
2𝑎 = 20
𝑎 =20
2
𝑎 = 10
𝑈𝑛 = 𝑎 + 𝑛 − 1 . 𝑏
𝑈2015 = 10 + 2015 − 1 .5
= 10 + 2014 .5
= 10 + 10070
= 10080
𝐽𝑎𝑑𝑖 𝑠𝑢𝑘𝑢 𝑘𝑒 − 2015 𝑏𝑎𝑟𝑖𝑠𝑎𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎 10080
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 8
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
7. Jawaban : 1 ∶ 2
Pembahasan :
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐴𝐵 𝑠𝑒𝑗𝑎𝑗𝑎𝑟 𝐸𝐹
𝐴𝐸 = 𝐵𝐹
𝐴𝐵 = 2 .𝐸𝐹
𝐴𝑃 = 𝑃𝐵 = 𝐷𝑄 = 𝑄𝐶 = 𝐸𝐹
𝐴𝐷 ⊥ 𝐴𝐵 𝑑𝑎𝑛 𝐸𝐻 ⊥ 𝐸𝐹
𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶
𝑡𝑡𝑟𝑎𝑝𝑒𝑠𝑖𝑢𝑚 𝐴𝐵𝐹𝐸 = 𝑡𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝑃𝐸 = 𝑡𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 𝑃𝐵𝐹𝐸 = 𝐸𝑅
𝑉𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸 .𝐷𝑄𝐻 ∶ 𝑉𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸 .𝑄𝐶𝐺𝐻 =𝑉𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸 .𝐷𝑄𝐻
𝑉𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸 .𝑄𝐶𝐺𝐻
=𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝑃𝐸 . 𝑡𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸 .𝐷𝑄𝐻
𝐿𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 𝑃𝐵𝐹𝐸 . 𝑡𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸 .𝑄𝐶𝐺𝐻
=1
2 . 𝐴𝑃 . 𝐸𝑅 . 𝐸𝐻
𝑃𝐵 . 𝐸𝑅 . 𝐸𝐻
=1
2 . 𝐸𝐹 . 𝐸𝑅 . 𝐸𝐻
𝐸𝐹 . 𝐸𝑅 . 𝐸𝐻
=1
2
= 1 ∶ 2
𝐽𝑎𝑑𝑖 𝑝𝑒𝑟𝑏𝑎𝑛𝑑𝑖𝑛𝑔𝑎𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸.𝐷𝑄𝐻 𝑑𝑎𝑛 𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸.𝑄𝐶𝐺𝐻 𝑎𝑑𝑎𝑙𝑎 1 ∶ 2
8. Jawaban : 28
Pembahasan :
𝑀𝑎𝑡𝑒𝑚𝑎𝑡𝑖𝑘𝑎 → 𝐴 ,𝐵 ,𝐶 ,𝐷
𝐼𝑃𝐴 → 𝐴 ,𝐵 ,𝐶 ,𝐸
𝐼𝑃𝑆 → 𝐴 ,𝐷 ,𝐸 ,𝐹
𝐴 𝑑𝑎𝑛 𝐵 𝑏𝑒𝑟𝑠𝑎𝑢𝑑𝑎𝑟𝑎 , 𝑗𝑎𝑑𝑖 𝑗𝑖𝑘𝑎 𝐴 𝑡𝑒𝑟𝑝𝑖𝑙𝑖 𝑚𝑎𝑘𝑎 𝐵 𝑡𝑖𝑑𝑎𝑘 𝑡𝑒𝑟𝑝𝑖𝑙𝑖, 𝑏𝑒𝑔𝑖𝑡𝑢 𝑝𝑢𝑙𝑎 𝑠𝑒𝑏𝑎𝑙𝑖𝑘𝑛𝑦𝑎
𝑃𝑒𝑟𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐾𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛 𝑝𝑒𝑚𝑖𝑙𝑖𝑎𝑛 𝐵𝑎𝑛𝑦𝑎𝑘 𝑐𝑎𝑟𝑎
𝑝𝑒𝑛𝑦𝑢𝑠𝑢𝑛𝑎𝑛 𝑀𝑎𝑡𝑒𝑚𝑎𝑡𝑖𝑘𝑎 𝐴 ,𝐵 ,𝐶 ,𝐷
𝐼𝑃𝐴 𝐴 ,𝐵 ,𝐶 ,𝐸
𝐼𝑃𝑆 𝐴 ,𝐷 ,𝐸 ,𝐹
𝐴 𝐶 𝐷 ,𝐸 ,𝐹 3
𝐸 𝐷 ,𝐹 2
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 9
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
𝐵 𝐶 𝐷 ,𝐸 ,𝐹 3
𝐸 𝐷 ,𝐹 2
𝐶
𝐴 𝐷 ,𝐸 ,𝐹 3
𝐵 𝐷 ,𝐸 ,𝐹 3
𝐸 𝐴 ,𝐷 ,𝐹 3
𝐷
𝐴 𝐸 ,𝐹 2
𝐵 𝐸 ,𝐹 2
𝐶 𝐴 ,𝐸 ,𝐹 3
𝐸 𝐴 ,𝐹 2
𝑇𝑜𝑡𝑎𝑙 𝑏𝑎𝑛𝑦𝑎𝑘 𝑐𝑎𝑟𝑎 𝑝𝑒𝑛𝑦𝑢𝑠𝑢𝑛𝑎𝑛 28
𝐽𝑎𝑑𝑖 𝑐𝑎𝑟𝑎 𝑦𝑎𝑛𝑔 𝑚𝑢𝑛𝑔𝑘𝑖𝑛 𝑢𝑛𝑡𝑢𝑘 𝑚𝑒𝑚𝑖𝑙𝑖 𝑤𝑎𝑘𝑖𝑙 𝑠𝑒𝑘𝑜𝑙𝑎 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑘𝑒 𝑂𝑆𝑁 𝑆𝑀𝑃 𝑡𝑎𝑢𝑛 𝑖𝑛𝑖 𝑎𝑑𝑎
𝑠𝑒𝑏𝑎𝑛𝑦𝑎𝑘 28
9. Jawaban : 𝐴 −8, 6 ,𝐵 −8, 10 ,𝐶 −4, 6
Pembahasan :
∆𝐴𝐵𝐶 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑡𝑒𝑟𝑎𝑑𝑎𝑝 𝑠𝑢𝑚𝑏𝑢 𝑌,𝑘𝑒𝑚𝑢𝑑𝑖𝑎𝑛 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑙𝑎𝑔𝑖 𝑡𝑒𝑟𝑎𝑑𝑎𝑝 𝑔𝑎𝑟𝑖𝑠 𝑦 = 3 ,
𝑠𝑒𝑖𝑛𝑔𝑔𝑎 𝑎𝑠𝑖𝑙 𝑝𝑒𝑛𝑐𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑛𝑦𝑎 𝑎𝑑𝑎𝑙𝑎 ∆𝐴′𝐵′𝐶′ , 𝑦𝑎𝑖𝑡𝑢 𝐴′ 8, 0 ,𝐵′ 8,−4 ,𝐶′ 4, 0
𝐷𝑒𝑛𝑔𝑎𝑛 𝑑𝑒𝑚𝑖𝑘𝑖𝑎𝑛 𝑢𝑛𝑡𝑢𝑘 𝑚𝑒𝑛𝑑𝑎𝑝𝑎𝑡𝑘𝑎𝑛 𝑘𝑒𝑚𝑏𝑎𝑙𝑖 ∆𝐴𝐵𝐶, 𝑙𝑎𝑘𝑢𝑘𝑎𝑛 𝑙𝑎𝑛𝑔𝑘𝑎 𝑚𝑢𝑛𝑑𝑢𝑟 𝑦𝑎𝑖𝑡𝑢 ∶
∆𝐴′𝐵′𝐶′ 𝑎𝑟𝑢𝑠 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑡𝑒𝑟𝑎𝑑𝑎𝑝 𝑔𝑎𝑟𝑖𝑠 𝑦 = 3 ,𝑘𝑒𝑚𝑢𝑑𝑖𝑎𝑛 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑡𝑒𝑟𝑎𝑑𝑎𝑝 𝑠𝑢𝑚𝑏𝑢 𝑌
𝐽𝑎𝑑𝑖 𝑘𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡 𝑡𝑖𝑡𝑖𝑘 − 𝑡𝑖𝑡𝑖𝑘 ∆𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎 𝐴 −8, 6 ,𝐵 −8, 10 ,𝐶 −4, 6
www.siap-osn.blogspot.com @ Maret 2015
Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 10
Download Soal dan Pembahasan OSN Matematika SMP Lainnya di “ www.siap-osn.blogspot.com ”
10. Jawaban : 61600
Pembahasan :
𝑃𝑢𝑡𝑖 = 3
𝑀𝑒𝑟𝑎 = 3𝐾𝑢𝑛𝑖𝑛𝑔 = 3𝐻𝑖𝑗𝑎𝑢 = 3𝐵𝑖𝑟𝑢 = 3
𝐽𝑢𝑚𝑙𝑎 = 12
𝑀𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑝𝑎𝑑𝑎 𝑔𝑒𝑙𝑎𝑛𝑔 𝑑𝑖𝑠𝑢𝑠𝑢𝑛 𝑑𝑒𝑛𝑔𝑎𝑛 𝑎𝑡𝑢𝑟𝑎𝑛 𝑑𝑖𝑎𝑛𝑡𝑎𝑟𝑎 2 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑝𝑢𝑡𝑖
𝑠𝑒𝑙𝑎𝑙𝑢 𝑡𝑒𝑟𝑑𝑎𝑝𝑎𝑡 4 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑠𝑒𝑙𝑎𝑖𝑛 𝑝𝑢𝑡𝑖
𝐾𝑎𝑟𝑒𝑛𝑎 𝑀𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑑𝑖𝑠𝑢𝑠𝑢𝑛 𝑑𝑒𝑛𝑔𝑎𝑛 𝑎𝑡𝑢𝑟𝑎𝑛 𝑑𝑖𝑎𝑛𝑡𝑎𝑟𝑎 2 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑝𝑢𝑡𝑖
𝑠𝑒𝑙𝑎𝑙𝑢 𝑡𝑒𝑟𝑑𝑎𝑝𝑎𝑡 4 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑠𝑒𝑙𝑎𝑖𝑛 𝑝𝑢𝑡𝑖,𝑚𝑎𝑘𝑎 𝑚𝑎𝑛𝑖𝑘 −𝑚𝑎𝑛𝑖𝑘 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡
𝑏𝑒𝑟𝑝𝑜𝑙𝑎 ∶
𝐷𝑒𝑛𝑔𝑎𝑛 𝑑𝑒𝑚𝑖𝑘𝑖𝑎𝑛 𝑝𝑜𝑠𝑖𝑠𝑖 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑝𝑢𝑡𝑖 𝑡𝑒𝑡𝑎𝑝, 𝑡𝑒𝑡𝑎𝑝𝑖 𝑢𝑛𝑡𝑢𝑘 𝑚𝑎𝑛𝑖𝑘 −𝑚𝑎𝑛𝑖𝑘 𝑚𝑒𝑟𝑎,𝑘𝑢𝑛𝑖𝑛𝑔,
𝑖𝑗𝑎𝑢 𝑑𝑎𝑛 𝑏𝑖𝑟𝑢 𝑑𝑎𝑝𝑎𝑡 𝑑𝑖𝑠𝑢𝑠𝑢𝑛 𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑠𝑖 𝑑𝑎𝑟𝑖 𝑢𝑛𝑠𝑢𝑟 𝑦𝑎𝑛𝑔 𝑠𝑎𝑚𝑎 ∶
𝐵𝑎𝑛𝑦𝑎𝑘 𝑝𝑒𝑛𝑦𝑢𝑠𝑢𝑛𝑎𝑛 𝑚𝑎𝑛𝑖𝑘 −𝑚𝑎𝑛𝑖𝑘 𝑚𝑒𝑟𝑎, 𝑏𝑖𝑟𝑢,𝑘𝑢𝑛𝑖𝑛𝑔 𝑑𝑎𝑛 𝑖𝑗𝑎𝑢 =12!
3! .3! .3! .3!
𝑇𝑒𝑡𝑎𝑝𝑖 , 𝑠𝑒𝑏𝑢𝑎 𝑔𝑒𝑙𝑎𝑛𝑔 𝑏𝑖𝑠𝑎 𝑑𝑖 𝑝𝑢𝑡𝑎𝑟 𝑑𝑎𝑛 𝑑𝑖𝑏𝑎𝑙𝑖𝑘 𝑘𝑎𝑛𝑎𝑛 𝑘𝑖𝑟𝑖𝑛𝑦𝑎 (𝑦𝑎𝑛𝑔 𝑘𝑎𝑛𝑎𝑛 𝑏𝑖𝑠𝑎 𝑗𝑎𝑑𝑖 𝑘𝑖𝑟𝑖, 𝑏𝑒𝑔𝑖𝑡𝑢
𝑗𝑢𝑔𝑎 𝑠𝑒𝑏𝑎𝑙𝑖𝑘𝑛𝑦𝑎) , 𝑠𝑒𝑖𝑛𝑔𝑔𝑎 𝑢𝑛𝑡𝑢𝑘 𝑚𝑒𝑚𝑝𝑒𝑟𝑚𝑢𝑑𝑎 𝑝𝑒𝑟𝑖𝑡𝑢𝑛𝑔𝑎𝑛,𝑔𝑒𝑙𝑎𝑛𝑔 𝑑𝑖𝑘𝑒𝑙𝑜𝑚𝑝𝑜𝑘𝑘𝑎𝑛 𝑗𝑎𝑑𝑖
3 𝑏𝑎𝑔𝑖𝑎𝑛 𝑑𝑒𝑛𝑔𝑎𝑛 𝑎𝑐𝑢𝑎𝑛 𝑚𝑎𝑛𝑖𝑘 − 𝑚𝑎𝑛𝑖𝑘 𝑝𝑢𝑡𝑖, 𝑠𝑒𝑏𝑎𝑔𝑎𝑖 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐵𝑎𝑛𝑦𝑎𝑘 𝑘𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛 𝑔𝑒𝑙𝑎𝑛𝑔 𝑑𝑖𝑝𝑢𝑡𝑎𝑟 𝑎𝑡𝑎𝑢 𝑑𝑖𝑏𝑜𝑙𝑎𝑘 − 𝑏𝑎𝑙𝑖𝑘 = 3!
𝐴𝑔𝑎𝑟 𝑔𝑒𝑙𝑎𝑛𝑔 𝑦𝑎𝑛𝑔 𝑑𝑖𝑠𝑢𝑠𝑢𝑛 𝑡𝑖𝑑𝑎𝑘 𝑎𝑑𝑎 𝑦𝑎𝑛𝑔 𝑙𝑒𝑏𝑖 𝑑𝑎𝑟𝑖 𝑠𝑎𝑡𝑢 𝑝𝑜𝑙𝑎 𝑔𝑒𝑙𝑎𝑛𝑔 (𝑎𝑘𝑖𝑏𝑎𝑡 𝑝𝑒𝑚𝑢𝑡𝑎𝑟𝑎𝑛 𝑎𝑡𝑎𝑢
𝑑𝑖𝑏𝑜𝑙𝑎𝑘 − 𝑏𝑎𝑙𝑖𝑘) 𝑚𝑎𝑘𝑎 𝑎𝑟𝑢𝑠 𝑑𝑖𝑙𝑎𝑘𝑢𝑘𝑎𝑛 𝑝𝑒𝑟𝑖𝑡𝑢𝑛𝑔𝑎𝑛 𝑠𝑒𝑏𝑎𝑔𝑎𝑖 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐵𝑎𝑛𝑦𝑎𝑘𝑛𝑦𝑎 𝑠𝑢𝑠𝑢𝑛𝑎𝑛 𝑔𝑒𝑙𝑎𝑛𝑔 𝑦𝑎𝑛𝑔 𝑚𝑢𝑛𝑔𝑘𝑖𝑛 𝑑𝑖𝑏𝑢𝑎𝑡 =12!
3! .3! .3! .3! . 3! 𝑏𝑎𝑛𝑦𝑎𝑘 𝑝𝑒𝑚𝑢𝑡𝑎𝑟𝑎𝑛 / 𝑏𝑎𝑙𝑖𝑘𝑎𝑛
= 61600