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

www.siap-osn.blogspot.com @ Maret 2015

Sosuke D. Aizen 2 SMPN 1 Tambelangan

Pembahasan OSN Matematika SMP 2015 /

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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

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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 = 90𝑜

𝑆𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎𝑕 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 → 𝐴𝐵 = 𝐴𝐶

𝑆𝑅 = 𝑆𝑄 = 𝑃𝑆 = 1

𝑅𝐷 = 3

3 𝑐𝑚

∠𝑆𝐷𝑅 = 60𝑜

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 ∶

∠𝐷𝑆𝑅 = 180𝑜 − 90𝑜 − 60𝑜 = 30𝑜

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑅𝑆 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐷𝑄𝑆 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶

∠𝑄𝐷𝑆 = ∠𝑆𝐷𝑅 = 60𝑜

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 ∶

∠𝐷𝐶𝑃 = 180𝑜 − 90𝑜 − 60𝑜 = 30𝑜







𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 ∶

∠𝐷𝐴𝑃 = 180𝑜 − 90𝑜 − 60𝑜 = 30𝑜

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐴𝐵𝐶 ∶

∠𝐴𝐵𝐶 = ∠𝐴𝐶𝐵 = 30𝑜

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐷 ∶

∠𝐴𝐷𝐵 = 180𝑜 − 60𝑜 − 60𝑜 = 60𝑜

∠𝐵𝐴𝐷 = 180𝑜 − 60𝑜 − 30𝑜 = 90𝑜

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 ∶

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

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐶𝑃𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 𝑦𝑎𝑛𝑔 𝑘𝑜𝑛𝑔𝑟𝑢𝑒𝑛 ∶

𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐴𝑃𝐷 = 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 −𝑠𝑖𝑘𝑢 𝐶𝑃𝐷

=7 3

6+ 2

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 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


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







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







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







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 = 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

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶

𝐾𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛 𝑝𝑒𝑚𝑖𝑙𝑖𝑕𝑎𝑛 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







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𝑕 = 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) , 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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Transcript of Pembahasan osn matematika smp 2015 tingkat kabupaten (bagian b isian singkat)