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

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


𝑥 − 𝑦 . 𝑥 + 𝑦 = 5 .20 → 𝑥 − 𝑦 = 5

𝑥 + 𝑦 = 20

2𝑥 = 25

𝑥 =25


𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘 𝑝𝑎𝑠𝑎𝑛𝑔𝑎𝑛 𝑥,𝑦 𝑦𝑎𝑛𝑔 𝑚𝑒𝑚𝑒𝑛𝑢ℎ𝑖 𝑎𝑑𝑎𝑙𝑎ℎ 1 𝐵

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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 ∶

𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 =1

2 .𝐴𝐵 . 𝑡 =

1

2 .2 . 3 = 3

𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 𝑑𝑎𝑛 𝐷𝐸𝐹 ∶

𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐸𝐹 =1

4 .𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 =

1

4 . 3 =

3

4

𝑃𝑒𝑟ℎ𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐸𝐹 𝑑𝑎𝑛 𝐺𝐻𝐼 ∶

𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐺𝐻𝐼 =1

4 .𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐸𝐹 =

1

4 . 3

4=

3

16

𝐿𝑠𝑒𝑙𝑢𝑟𝑢 ℎ 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 = 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 + 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐸𝐹 + 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 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





14. Jawaban : 𝐶 . 7

Pembahasan :


𝑀𝑢𝑙𝑎 𝑚𝑢𝑙𝑎 𝑘𝑎𝑡𝑎𝑘 𝑏𝑒𝑟𝑎𝑑𝑎 𝑝𝑎𝑑𝑎 𝑝𝑜𝑠𝑖𝑠𝑖 𝑛𝑜𝑚𝑜𝑟 1

𝐾𝑎𝑡𝑎𝑘 𝑚𝑒𝑙𝑜𝑚𝑝𝑎𝑡 1 𝑠𝑎𝑡𝑢𝑎𝑛 𝑠𝑒𝑎𝑟𝑎ℎ 𝑗𝑎𝑟𝑢𝑚 𝑗𝑎𝑚 𝑗𝑖𝑘𝑎 𝑏𝑒𝑟𝑎𝑑𝑎 𝑝𝑎𝑑𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎

𝐾𝑎𝑡𝑎𝑘 𝑚𝑒𝑙𝑜𝑚𝑝𝑎𝑡 3 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 :


𝑥 = 180𝑜 − 110𝑜 − 35𝑜 = 35𝑜

2𝑥 = 2 . 35𝑜 = 70𝑜

𝑦+ 𝑥 + 60 = 180𝑜

𝑦 + 𝑥 = 180𝑜 − 60𝑜

𝑦 + 𝑥 = 120𝑜

𝑦+ 𝑥 − 2𝑥 = 120𝑜 − 70𝑜

𝑦 − 𝑥 = 50𝑜

𝐽𝑎𝑑𝑖 𝑏𝑒𝑠𝑎𝑟 𝑠𝑢𝑑𝑢𝑡 𝑦− 𝑥 𝑎𝑑𝑎𝑙𝑎ℎ 50𝑜 𝐷

1 2 3 4





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 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%





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


Pembahasan :


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 :


𝐿𝑢𝑎𝑠 𝑠𝑎𝑡𝑢 𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝑘𝑒𝑐𝑖𝑙 = 4 𝑚2

𝐵𝑖𝑠𝑎 𝑑𝑖𝑙𝑖ℎ𝑎𝑡 𝑏𝑎ℎ𝑤𝑎 𝑏𝑎𝑛𝑔𝑢𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑡𝑒𝑟𝑑𝑖𝑟𝑖 𝑑𝑎𝑟𝑖 4 𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 𝑦𝑎𝑛𝑔 𝑏𝑒𝑟𝑢𝑘𝑢𝑟𝑎𝑛 𝑠𝑎𝑚𝑎





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

𝐿𝑏𝑎𝑛𝑔𝑢𝑛 𝑑𝑎𝑡𝑎𝑟 = 4 .𝐿𝑗𝑎𝑗𝑎𝑟 𝑔𝑒𝑛𝑗𝑎𝑛𝑔 = 4 .36 = 144

𝐽𝑎𝑑𝑖 𝑙𝑢𝑎𝑠 𝑏𝑎𝑛𝑔𝑢𝑛 𝑑𝑎𝑡𝑎𝑟 𝑝𝑎𝑑𝑎 𝑔𝑎𝑚𝑏𝑎𝑟 𝑎𝑑𝑎𝑙𝑎ℎ 144 𝐶


Pembahasan :


𝐾𝑎𝑛𝑡𝑜𝑛𝑔 𝐼 𝐾𝑎𝑛𝑡𝑜𝑛𝑔 𝐼𝐼 𝐾𝑎𝑛𝑡𝑜𝑛𝑔 𝐼𝐼𝐼 𝐵𝑎𝑛𝑦𝑎𝑘 𝑝𝑎𝑠𝑎𝑛𝑔𝑎𝑛 𝑤𝑎𝑟𝑛𝑎

𝑀 𝐾

3 𝑀 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



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

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