Pembahasan osn matematika smp 2015 tingkat kabupaten (bagian b isian singkat)

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Sosuke D. Aizen 2 SMPN 1 Tambelangan
Pembahasan OSN Matematika SMP 2015 / Page 1
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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 𝑏𝑎𝑕𝑤𝑎 ∶
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 (𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 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 (𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑠𝑎𝑚𝑎)
𝑥2
+ 5𝑥 + 6 = 𝑥 + 2 . 𝑥 + 3 = 3 = 1 .3 = −3 . −1
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 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 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 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎)
𝑥2
+ 5𝑥 + 6 = 𝑥 + 2 . 𝑥 + 3 = 5 = 1 .5 = −5 . −1
𝑈𝑛𝑡𝑢𝑘 ∶ 𝑥 + 2 . 𝑥 + 3 = 1 .5

𝑥 + 2 = 1 𝑑𝑎𝑛 𝑥 + 3 = 5
𝑥 = 1 − 2 𝑑𝑎𝑛 𝑥 = 5 − 3
𝑥 = −1 𝑑𝑎𝑛 𝑥 = 2 (𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 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 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 𝑘𝑎𝑟𝑒𝑛𝑎 𝑛𝑖𝑙𝑎𝑖 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑠𝑎𝑚𝑎)
𝐷𝑎𝑟𝑖 𝑝𝑒𝑟𝑕𝑖𝑡𝑢𝑛𝑔𝑎𝑛 𝑑𝑖𝑎𝑡𝑎𝑠 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑝𝑜𝑙𝑎 𝑏𝑎𝑕𝑤𝑎 𝑢𝑛𝑡𝑢𝑘 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 ≥ 3
𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢𝑕𝑖 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

3. Jawaban : 7 3 + 12 𝑐𝑚2
Pembahasan :
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝑃, 𝑄 𝑑𝑎𝑛 𝑅 𝑎𝑑𝑎𝑙𝑎𝑕 𝑡𝑖𝑡𝑖𝑘 𝑠𝑖𝑛𝑔𝑔𝑢𝑛𝑔 𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛 𝑝𝑎𝑑𝑎 𝑠𝑖𝑠𝑖 − 𝑠𝑖𝑠𝑖 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐷, 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 ∶
∠𝐴𝑃𝐷 = ∠𝐶𝑃𝐷 = ∠𝐷𝑅𝑆 = ∠𝐶𝑅𝑆 = ∠𝐴𝑄𝑆 = ∠𝐷𝑄𝑆 = 90 𝑜
𝑆𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎𝑕 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 → 𝐴𝐵 = 𝐴𝐶
𝑆𝑅 = 𝑆𝑄 = 𝑃𝑆 = 1
𝑅𝐷 =
3
3
𝑐𝑚
∠𝑆𝐷𝑅 = 60 𝑜
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 ∶
∠𝐷𝑆𝑅 = 180 𝑜
− 90 𝑜
− 60 𝑜
= 30 𝑜
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑄𝑆 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶
∠𝑄𝐷𝑆 = ∠𝑆𝐷𝑅 = 60 𝑜
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 ∶
∠𝐷𝐶𝑃 = 180 𝑜
− 90 𝑜
− 60 𝑜
= 30 𝑜

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 ∶
∠𝐷𝐴𝑃 = 180 𝑜
− 90 𝑜
− 60 𝑜
= 30 𝑜
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 ∶
∠𝐴𝐵𝐶 = ∠𝐴𝐶𝐵 = 30 𝑜
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐷 ∶
∠𝐴𝐷𝐵 = 180 𝑜
− 60 𝑜
− 60 𝑜
= 60 𝑜
∠𝐵𝐴𝐷 = 180 𝑜
− 60 𝑜
− 30 𝑜
= 90 𝑜
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 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
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 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
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶
𝐴𝑃 = 𝐶𝑃 = 3 + 2
𝐴𝐵 = 𝐴𝐶 = 𝐴𝑃 + 𝐶𝑃 = 3 + 2 + 3 + 2 = 2 3 + 4
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 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

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 ∶
𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 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
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶
𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 = 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐶𝑃𝐷
=
7 3
6
+ 2
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐵𝐴𝐷 ∶
𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 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
𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 = 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 + 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 + 𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 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

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
𝐴 𝐼+𝐼𝐼
=
55𝑉
104
153 𝑉
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 − 𝑎2
𝑏2
= 209 − 𝑎2
+ 4𝑎𝑏 + 4𝑏2
− 209 − 𝑎2
− 4𝑎𝑏 + 4𝑏2
209 − 𝑎2
𝑏2
= 209 − 𝑎2
− 4𝑎𝑏 − 4𝑏2
− 209 − 𝑎2
+ 4𝑎𝑏 − 4𝑏2
209 − 𝑎2
𝑏2
= 209 − 𝑎2
− 4𝑎𝑏 − 4𝑏2
− 209 + 𝑎2
− 4𝑎𝑏 + 4𝑏2
209 − 𝑎2
𝑏2
= −8𝑎𝑏
209 = 𝑎2
𝑏2
− 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

6. Jawaban : 10080
Pembahasan :
𝑈1 + 𝑈2 + 𝑈3 + 𝑈4 = 70 → 𝑆4 = 70
𝑈5 + 𝑈6 + ⋯ + 𝑈16 = 690
𝑈1 + 𝑈2 + ⋯ + 𝑈16 = 760
𝑆16 = 760
𝑆 𝑛 =
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)
𝑆 𝑛 =
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

7. Jawaban : 1 ∶ 2
Pembahasan :
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐴𝐵 𝑠𝑒𝑗𝑎𝑗𝑎𝑟 𝐸𝐹
𝐴𝐸 = 𝐵𝐹
𝐴𝐵 = 2 . 𝐸𝐹
𝐴𝑃 = 𝑃𝐵 = 𝐷𝑄 = 𝑄𝐶 = 𝐸𝐹
𝐴𝐷 ⊥ 𝐴𝐵 𝑑𝑎𝑛 𝐸𝐻 ⊥ 𝐸𝐹
𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶
𝑡𝑡𝑟𝑎𝑝𝑒𝑠𝑖𝑢𝑚 𝐴𝐵𝐹𝐸 = 𝑡 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝑃𝐸 = 𝑡𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 𝑃𝐵𝐹𝐸 = 𝐸𝑅
𝑉𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸.𝐷𝑄𝐻 ∶ 𝑉𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸.𝑄𝐶𝐺𝐻 =
𝑉 𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸 .𝐷𝑄𝐻
𝑉 𝑝𝑟𝑖𝑠𝑚𝑎 𝑃𝐵𝐹𝐸 .𝑄𝐶𝐺𝐻
=
𝐿 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝑃𝐸 . 𝑡 𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝑃𝐸 .𝐷𝑄𝐻
𝐿 𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 𝑃𝐵𝐹𝐸 . 𝑡 𝑝𝑟𝑖𝑠𝑚𝑎 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
𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐾𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛 𝑝𝑒𝑚𝑖𝑙𝑖𝑕𝑎𝑛 𝐵𝑎𝑛𝑦𝑎𝑘 𝑐𝑎𝑟𝑎
𝑝𝑒𝑛𝑦𝑢𝑠𝑢𝑛𝑎𝑛𝑀𝑎𝑡𝑒𝑚𝑎𝑡𝑖𝑘𝑎
𝐴 , 𝐵 , 𝐶 , 𝐷
𝐼𝑃𝐴
𝐴 , 𝐵 , 𝐶 , 𝐸
𝐼𝑃𝑆
𝐴 , 𝐷 , 𝐸 , 𝐹
𝐴
𝐶 𝐷 , 𝐸 , 𝐹 3
𝐸 𝐷 , 𝐹 2

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 𝑡𝑎𝑕𝑢𝑛 𝑖𝑛𝑖 𝑎𝑑𝑎
𝑠𝑒𝑏𝑎𝑛𝑦𝑎𝑘 28
9. Jawaban : 𝐴 −8, 6 , 𝐵 −8, 10 , 𝐶 −4, 6
Pembahasan :
∆𝐴𝐵𝐶 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑡𝑒𝑟𝑕𝑎𝑑𝑎𝑝 𝑠𝑢𝑚𝑏𝑢 𝑌, 𝑘𝑒𝑚𝑢𝑑𝑖𝑎𝑛 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑙𝑎𝑔𝑖 𝑡𝑒𝑟𝑕𝑎𝑑𝑎𝑝 𝑔𝑎𝑟𝑖𝑠 𝑦 = 3 ,
𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 𝑕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 𝑡𝑒𝑟𝑕𝑎𝑑𝑎𝑝 𝑔𝑎𝑟𝑖𝑠 𝑦 = 3 , 𝑘𝑒𝑚𝑢𝑑𝑖𝑎𝑛 𝑑𝑖𝑐𝑒𝑟𝑚𝑖𝑛𝑘𝑎𝑛 𝑡𝑒𝑟𝑕𝑎𝑑𝑎𝑝 𝑠𝑢𝑚𝑏𝑢 𝑌
𝐽𝑎𝑑𝑖 𝑘𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡 𝑡𝑖𝑡𝑖𝑘 − 𝑡𝑖𝑡𝑖𝑘 ∆𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎𝑕 𝐴 −8, 6 , 𝐵 −8, 10 , 𝐶 −4, 6

10. Jawaban : 61600
Pembahasan :
𝑃𝑢𝑡𝑖𝑕 = 3
𝑀𝑒𝑟𝑎𝑕 = 3
𝐾𝑢𝑛𝑖𝑛𝑔 = 3
𝐻𝑖𝑗𝑎𝑢 = 3
𝐵𝑖𝑟𝑢 = 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) , 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 𝑢𝑛𝑡𝑢𝑘 𝑚𝑒𝑚𝑝𝑒𝑟𝑚𝑢𝑑𝑎𝑕 𝑝𝑒𝑟𝑕𝑖𝑡𝑢𝑛𝑔𝑎𝑛, 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 𝑝𝑒𝑟𝑕𝑖𝑡𝑢𝑛𝑔𝑎𝑛 𝑠𝑒𝑏𝑎𝑔𝑎𝑖 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶
𝐵𝑎𝑛𝑦𝑎𝑘𝑛𝑦𝑎 𝑠𝑢𝑠𝑢𝑛𝑎𝑛 𝑔𝑒𝑙𝑎𝑛𝑔 𝑦𝑎𝑛𝑔 𝑚𝑢𝑛𝑔𝑘𝑖𝑛 𝑑𝑖𝑏𝑢𝑎𝑡 =
12!
3! .3! .3! .3! . 3!
𝑏𝑎𝑛𝑦𝑎𝑘 𝑝𝑒𝑚𝑢𝑡𝑎𝑟𝑎𝑛 / 𝑏𝑎𝑙𝑖𝑘𝑎𝑛
= 61600

Pembahasan osn matematika smp 2015 tingkat kabupaten (bagian b isian singkat)

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Pembahasan osn matematika smp 2015 tingkat kabupaten (bagian b isian singkat)