olimpiade primagama madura mencari juara 2015 matematika smp kode soal 15333 (babak penyisihan)

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SD.A 2 SMPN 1 Tambelangan

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PEMBAHASAN

OLIMPIADE PRIMAGAMA MADURA MENCARI JUARA 2015

MATEMATIKA SMP KODE SOAL 15333 (BABAK PENYISIHAN NO.1-30)

( versi penulis : www.siap-osn.blogspot.com )

1. Jawaban : 𝐶. 8 4 − 𝜋 𝑐𝑚2

Pembahasan :

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑚𝑏𝑎𝑟 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶

𝐾𝑎𝑟𝑒𝑛𝑎 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑂𝑃𝑄 𝑚𝑒𝑟𝑢𝑝𝑎𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑑𝑎𝑛 ∠𝑃𝑂𝑄 = 45𝑜 𝑚𝑎𝑘𝑎 ∶

𝑃𝑄 = 𝑂𝑃 = 8 𝑐𝑚

𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑂𝑃𝑄 =1

2 .𝑂𝑃 .𝑃𝑄

=1

2 .8 .8

= 32

𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝑃𝑂𝑅 =∠𝑃𝑂𝑄

360𝑜 .𝜋 .𝑂𝑃2

=45𝑜

360𝑜 .𝜋 . 82

=1

8 .𝜋 . 82

= 8 𝜋

𝐿𝑎𝑟𝑠𝑖𝑟𝑎𝑛 = 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑂𝑃𝑄 − 𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝑃𝑂𝑅

= 32 − 8 𝜋

= 8 4 − 𝜋 𝑐𝑚2 𝐶

2. Jawaban : 𝐵. 30

Pembahasan :

22020−22016−90

22015−3=

25+2015 −21+2015−90

22015−3

=25 .22015 −21 .22015−90

22015−3

=32 .22015 −2 .22015−90

22015 −3

=30 .22015 −90

22015 −3

=30 . 22015 −3

22015 −3

= 30 𝐵

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3. Jawaban : 𝐶. 729

Pembahasan :

𝑥1

6 =7

3− 2− 2

𝑥1

6 =7

3− 2 .

3+ 2

3+ 2− 2

𝑥1

6 =7 . 3+ 2

32− 2 2 − 2

𝑥1

6 =7 . 3+ 2

9−2− 2

𝑥1

6 =7 . 3+ 2

7− 2

𝑥1

6 = 3 + 2 − 2

𝑥1

6 = 3

𝑥 = 36

𝑥 = 729 𝐶

4. Jawaban : 𝐵. 72 𝑘𝑚 𝑗𝑎𝑚

Pembahasan :

𝑉1 = 60

𝑠1

𝑡1= 60

𝑠

60= 𝑡1 → 𝑡1 =

𝑠

60

𝑉2 = 90

𝑠2

𝑡2= 90

𝑠

90= 𝑡2 → 𝑡2 =

𝑠

90

𝑉𝑟𝑎𝑡𝑎 −𝑟𝑎𝑡𝑎 =𝑠1+𝑠2

𝑡1+𝑡2

=𝑠+𝑠𝑠

60+

𝑠

90

=2𝑠

3𝑠+2𝑠

180

=2𝑠5𝑠

180

= 2𝑠 .180

5𝑠

=360𝑠

5𝑠

= 72 𝑘𝑚 𝑗𝑎𝑚 𝐵

5. Jawaban : 𝑁. 16

Pembahasan :

𝐵𝑎𝑛𝑦𝑎𝑘 𝑓𝑎𝑘𝑡𝑜𝑟 𝑑𝑎𝑟𝑖 𝑕𝑎𝑠𝑖𝑙 𝑝𝑒𝑟𝑘𝑎𝑙𝑖𝑎𝑛 𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 𝑏𝑒𝑟𝑏𝑒𝑑𝑎 = 2𝑛





𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘 𝑓𝑎𝑘𝑡𝑜𝑟 𝑑𝑎𝑟𝑖 𝑕𝑎𝑠𝑖𝑙 𝑝𝑒𝑟𝑘𝑎𝑙𝑖𝑎𝑛 4 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑟𝑖𝑚𝑎 𝑏𝑒𝑟𝑏𝑒𝑑𝑎 = 24

= 16 𝑁

6. Jawaban : 𝐴. 5

18

Pembahasan :

𝑀𝑒𝑟𝑎𝑕 = 3

𝑃𝑢𝑡𝑖𝑕 = 4

𝐵𝑖𝑟𝑢 = 2

𝑇𝑜𝑡𝑎𝑙 = 3 + 4 + 2 = 9

𝐶𝑎𝑟𝑎 𝐼 ∶

𝑃 2 𝑘𝑒𝑙𝑒𝑟𝑒𝑛𝑔 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑠𝑎𝑚𝑎 = 𝑃 2 𝑀𝑒𝑟𝑎𝑕 + 𝑃 2 𝑃𝑢𝑡𝑖𝑕 + 𝑃 2 𝐵𝑖𝑟𝑢

=3

9 .

2

8+

4

9 .

3

8+

2

9 .

1

8

=6

72+

12

72+

2

72

=20

72

=5

18 𝐴

𝐶𝑎𝑟𝑎 𝐼𝐼 ∶

𝑃 2 𝑘𝑒𝑙𝑒𝑟𝑒𝑛𝑔 𝑏𝑒𝑟𝑤𝑎𝑟𝑛𝑎 𝑠𝑎𝑚𝑎 = 𝑃 2 𝑀𝑒𝑟𝑎𝑕 + 𝑃 2 𝑃𝑢𝑡𝑖𝑕 + 𝑃 2 𝐵𝑖𝑟𝑢

=𝐶3 2

𝐶9 2+

𝐶4 2

𝐶9 2+

𝐶2 2

𝐶9 2

=3

36+

6

36+

1

36

=10

36

=5

18 𝐴

7. Jawaban : 𝐶. 110

Pembahasan :

4𝑥 + 4−𝑥 = 23

22 𝑥 + 22 −𝑥 = 23

22𝑥 + 2−2𝑥 = 23

2𝑥 + 2−𝑥 2 = 2𝑥 + 2−𝑥 . 2𝑥 + 2−𝑥

= 2𝑥 . 2𝑥 + 2𝑥 . 2−𝑥 + 2−𝑥 . 2𝑥 + 2−𝑥 . 2−𝑥

= 2𝑥+𝑥 + 2𝑥+ −𝑥 + 2−𝑥+𝑥 + 2−𝑥+ −𝑥

= 22𝑥 + 20 + 20 + 2−2𝑥

= 22𝑥 + 1 + 1 + 2−2𝑥

= 22𝑥 + 2−2𝑥 + 2

= 23 + 2

= 25

2𝑥 + 2−𝑥 = 25





2𝑥 + 2−𝑥 = 5

22𝑥 + 2−2𝑥 . 2𝑥 + 2−𝑥 = 23 .5

22𝑥 . 2𝑥 + 22𝑥 . 2−𝑥 + 2−2𝑥 . 2𝑥 + 2−2𝑥 . 2−𝑥 = 115

22𝑥+𝑥 + 22𝑥+ −𝑥 + 2−2𝑥+𝑥 + 2−2𝑥+ −𝑥 = 115

23𝑥 + 2𝑥 + 2−𝑥 + 2−3𝑥 = 115

23𝑥 + 5 + 2−3𝑥 = 115

23𝑥 + 2−3𝑥 + 5 = 115

23 𝑥 + 23 −𝑥 = 115 − 5

8𝑥 + 8−𝑥 = 110 𝐶


6𝑚−3

Pembahasan :

log 124 = 𝑚 +1

2

log 3 .44 = 𝑚 +1

2

log 34 + log 44 = 𝑚 +1

2

log 322+ 1 = 𝑚 +

1

2

1

2 . log 32 = 𝑚 +

1

2− 1

1

2 . log 32 = 𝑚 −

1

2

log 32 = 2 . 𝑚 −1

2

log 32 = 2𝑚 − 1

log 23 =1

2𝑚−1

log 1627 = log 2433

=4

3 . log 23

=4

3 .

1

2𝑚−1

=4

6𝑚−3 𝐴

9. Jawaban : 𝐷. 𝑆𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖

Pembahasan :

𝐴 2 𝑥1

, 3 𝑦1

𝑑𝑎𝑛 𝐵 8 𝑥2

, 5 𝑦2

𝐴𝐵 = 𝑥2 − 𝑥1 2 + 𝑦2 − 𝑦1

2

= 8 − 2 2 + 5 − 3 2

= 62 + 22

= 36 + 4

= 40





𝐵 8 𝑥1

, 5 𝑦1

𝑑𝑎𝑛 𝐶 6𝑥2

, 11 𝑦2

𝐵𝐶 = 𝑥2 − 𝑥1 2 + 𝑦2 − 𝑦1

2

= 6 − 8 2 + 11 − 5 2

= −2 2 + 62

= 4 + 36

= 40

𝐴 2 𝑥1

, 3 𝑦1

𝑑𝑎𝑛 𝐶 6𝑥2

, 11 𝑦2

𝐴𝐶 = 𝑥2 − 𝑥1 2 + 𝑦2 − 𝑦1

2

= 6 − 2 2 + 11 − 3 2

= 42 + 82

= 16 + 64

= 80

𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 𝐴𝐵 = 𝐵𝐶 = 40 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝑑𝑎𝑛 𝐴𝐶 = 80 , 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 ∶

𝐴𝐶 = 𝐴𝐵2 + 𝐵𝐶2

= 40 2

+ 40 2

= 40 + 40

= 80 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢

𝐽𝑎𝑑𝑖 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐵𝐶 𝑎𝑑𝑎𝑙𝑎𝑕 𝑆𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑠𝑎𝑚𝑎 𝑘𝑎𝑘𝑖 𝐷

10. Jawaban : 𝐷. 24𝜋 + 18 3 𝑐𝑚2

Pembahasan :


𝐴𝐶 = 6 𝑐𝑚

𝐴𝐵 = 6 𝑐𝑚

𝐴𝐷 =1

2 .𝐴𝐵 =

1

2 .6 = 3 𝑐𝑚

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝐶𝐷 ∶

𝐶𝐷 = 𝐴𝐶2 − 𝐴𝐷2

= 62 − 32





= 36 − 9

= 27

= 9 .3

= 3 3

𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 ∶ 𝐴𝐶 = 6 𝑐𝑚 ,𝐴𝐷 = 3 𝑐𝑚 ,𝐶𝐷 = 3 3 𝑐𝑚 , 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 𝑑𝑒𝑛𝑔𝑎𝑛 𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑝𝑒𝑟𝑏𝑎𝑛𝑑𝑖𝑛𝑔𝑎𝑛

𝑠𝑖𝑠𝑖 𝑝𝑎𝑑𝑎 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑚𝑎𝑘𝑎 𝑑𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 𝑏𝑒𝑠𝑎𝑟 𝑠𝑢𝑑𝑢𝑡 𝑠𝑒𝑏𝑎𝑔𝑎𝑖 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶

𝐴𝐶 ∶ 𝐴𝐷 ∶ 𝐶𝐷 = 6 ∶ 3 ∶ 3 3 = 2 ∶ 1 ∶ 3 → ∠𝐴𝐷𝐶 ∶ ∠𝐴𝐶𝐷 ∶ ∠𝐶𝐴𝐷 = 90𝑜 ∶ 30𝑜 ∶ 60𝑜

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐷 𝑑𝑎𝑛 𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 ∶

∠𝐴𝐶𝐵 = 2 .∠𝐴𝐶𝐷 = 2 . 30𝑜 = 60𝑜

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 ∶

𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 =∠𝐴𝐶𝐵

360𝑜 .𝜋 .𝐴𝐶2

=60𝑜

360𝑜 .𝜋 . 62

=1

6 .𝜋 . 62

= 6𝜋

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝐶𝐷 𝑑𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐵 ∶

𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐵 = 2 . 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐷

= 2 .1

2 .𝐶𝐷 .𝐴𝐷

= 2 .1

2 .3 3 .3

= 9 3

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑒𝑚𝑏𝑒𝑟𝑒𝑛𝑔 𝐴𝐵 ∶

𝐿𝑡𝑒𝑚𝑏𝑒𝑟𝑒𝑛𝑔 𝐴𝐵 = 𝐿𝑗𝑢𝑟𝑖𝑛𝑔 𝐴𝐶𝐵 − 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐴𝐶𝐵

= 6𝜋 − 9 3

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛 𝑑𝑒𝑛𝑔𝑎𝑛 𝑡𝑖𝑡𝑖𝑘 𝑝𝑢𝑠𝑎𝑡 𝐶 ∶

𝐿𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛 𝐶 = 𝜋 .𝐴𝐶2

= 𝜋 . 62

= 36𝜋

𝐿𝑎𝑟𝑠𝑖𝑟𝑎𝑛 = 𝐿𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛 𝐶 − 2 . 𝐿𝑡𝑒𝑚𝑏𝑒𝑟𝑒𝑛𝑔 𝐴𝐵

= 36𝜋 − 2 . 6𝜋 − 9 3

= 36𝜋 − 12𝜋 + 18 3

= 24𝜋 + 18 3 𝑐𝑚2 𝐷





11. Jawaban : 𝐷. ± 24

Pembahasan :

𝑃𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 𝑔𝑎𝑟𝑖𝑠 ∶ 3𝑥 + 4𝑦 = 𝑝

𝑇𝑖𝑡𝑖𝑘 𝑝𝑜𝑡𝑜𝑛𝑔 𝑑𝑒𝑛𝑔𝑎𝑛 𝑠𝑢𝑚𝑏𝑢 𝑥 ∶

𝑦 = 0 → 3𝑥 + 4𝑦 = 𝑝

3𝑥 + 4 . 0 = 𝑝

3𝑥 = 𝑝

𝑥 =𝑝

3 →

𝑝

3, 0

𝑇𝑖𝑡𝑖𝑘 𝑝𝑜𝑡𝑜𝑛𝑔 𝑑𝑒𝑛𝑔𝑎𝑛 𝑠𝑢𝑚𝑏𝑢 𝑦 ∶

𝑥 = 0 → 3𝑥 + 4𝑦 = 𝑝

3 . 0 + 4𝑦 = 𝑝

4𝑦 = 𝑝

𝑦 =𝑝

4 → 0,

𝑝

4

𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 =1

2 .𝑎 . 𝑡

24 =1

2 .𝑝

3 .𝑝

4

24 =𝑝2

24

24 .24 = 𝑝2

± 24 .24 = 𝑝

±24 = 𝑝

𝑝 = ±24 𝐷

12. Jawaban : 𝑁. 4 3

Pembahasan :


𝐾𝑂 = 6

𝑁𝑂 = 5

𝑀𝑂 = 37

𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶

𝐷𝑁 = 𝐵𝐾 = 𝐴𝑂 = 𝑎

𝐴𝑁 = 𝐶𝑀 = 𝐷𝑂 = 𝑏

𝐴𝐾 = 𝐶𝐿 = 𝐵𝑂 = 𝑐

𝐵𝐿 = 𝐷𝑀 = 𝐶𝑂 = 𝑑





𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐷𝑁 ∶

𝐷𝑁2 + 𝐷𝑂2 = 𝑁𝑂2

𝑎2 + 𝑏2 = 52

𝑎2 + 𝑏2 = 25 … (1)

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐴𝐾 ∶

𝐴𝑂2 + 𝐴𝐾2 = 𝐾𝑂2

𝑎2 + 𝑐2 = 62

𝑎2 + 𝑐2 = 36 … (2)

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐵𝐿 ∶

𝐵𝑂2 + 𝐵𝐿2 = 𝐿𝑂2

𝑐2 + 𝑑2 = 𝐿𝑂2

𝑐2 + 𝑑2 = 𝐿𝑂2 … (3)

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝑂𝐶𝑀 ∶

𝐶𝑀2 + 𝐶𝑂2 = 𝑀𝑂2

𝑏2 + 𝑑2 = 37 2

𝑏2 + 𝑑2 = 37 … (4)

𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 1 𝑑𝑎𝑛 2 ∶

𝑎2 + 𝑏2 = 25

𝑎2 + 𝑐2 = 36

𝑏2 − 𝑐2 = −11 … (5)

𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 (4) 𝑑𝑎𝑛 (5) ∶

𝑏2 + 𝑑2 = 37

𝑏2 − 𝑐2 = −11

𝑑2 + 𝑐2 = 48 … (6)


𝑑2 + 𝑐2 = 48

𝑐2 + 𝑑2 = 𝐿𝑂2

0 = 48 − 𝐿𝑂2

𝐿𝑂2 = 48

𝐿𝑂 = 48

𝐿𝑂 = 16 .3

𝐿𝑂 = 4 3 𝑁





13. Jawaban : 𝐶. 𝑡 30

Pembahasan :

𝑝 = 5𝑡

𝑙 = 2𝑡

𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑟𝑢𝑎𝑛𝑔 𝑏𝑎𝑙𝑜𝑘 = 𝑝2 + 𝑙2 + 𝑡2

= 5𝑡 2 + 2𝑡 2 + 𝑡2

= 25𝑡2 + 4𝑡2 + 𝑡2

= 30𝑡2

= 𝑡 30 𝐶

14. Jawaban : 𝐵. 𝑥 = 5 𝑎𝑡𝑎𝑢 𝑥 =1

2

Pembahasan :


𝑥 = 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑦𝑎𝑛𝑔 𝑑𝑖𝑚𝑎𝑘𝑠𝑢𝑑

2𝑥 +5

𝑥= 11

𝑥 . 2𝑥 +5

𝑥 = 𝑥 . 11 𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 𝑥

2𝑥2 + 5 = 11𝑥

2𝑥2 − 11𝑥 + 5 = 0

1

2 . 2𝑥 − 10 . 2𝑥 − 1 = 0

1

2 . 2 . 𝑥 − 5 . 2𝑥 − 1 = 0

𝑥 − 5 . 2𝑥 − 1 = 0

𝑥 = 5 𝑎𝑡𝑎𝑢 𝑥 =1

2 𝐵

15. Jawaban : 𝐴. 𝑅𝑝. 61.000,−

Pembahasan :

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶

𝐺𝑎𝑗𝑖 𝑎𝑤𝑎𝑙 𝐵𝑒𝑠𝑎𝑟 𝑘𝑒𝑛𝑎𝑖𝑘𝑎𝑛 𝑔𝑎𝑗𝑖

260000 15

100 . 260000 = 39000

370000 15

100 . 370000 = 55500

470000 15

100 . 470000 = 70500

650000 10

100 . 650000 = 65000

750000 10

100 . 750000 = 75000





𝑅𝑎𝑡𝑎 − 𝑟𝑎𝑡𝑎 𝑏𝑒𝑠𝑎𝑟 𝑘𝑒𝑛𝑎𝑖𝑘𝑎𝑛 𝑔𝑎𝑗𝑖 =39000 +55500 +70500 +65000 +75000

5=

305000

5= 61000 𝐴

16. Jawaban : 𝑁. 868

Pembahasan :

𝐽𝑢𝑚𝑙𝑎𝑕 𝑠𝑒𝑚𝑢𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑛𝑡𝑎𝑟𝑎 1 − 100 𝑦𝑎𝑛𝑔 𝑕𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 4 ∶

4 + 8 + 12 + ⋯+ 100 𝑑𝑒𝑟𝑒𝑡 𝑎𝑟𝑖𝑡𝑚𝑎𝑡𝑖𝑘𝑎 ∶

𝑛=100

4=25

𝑎=4𝑈25 =100

=𝑛

2 . 𝑎 + 𝑈𝑛 =

25

2 . 4 + 𝑈25 =

25

2 . 4 + 100 =

25

2 . 104 = 1300

𝐽𝑢𝑚𝑙𝑎𝑕 𝑠𝑒𝑚𝑢𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑛𝑡𝑎𝑟𝑎 1 − 100 𝑦𝑎𝑛𝑔 𝑕𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 3 𝑑𝑎𝑛 4 ∶

12 + 24 + 36 + ⋯+ 96 𝑑𝑒𝑟𝑒𝑡 𝑎𝑟𝑖𝑡𝑚𝑎𝑡𝑖𝑘𝑎 ∶

𝑛=96

12=8

𝑎=12𝑈8=96

=𝑛

2 . 𝑎 + 𝑈𝑛 =

8

2 . 12 + 𝑈8 =

8

2 . 12 + 96 =

8

2 . 108 = 432

𝐽𝑎𝑑𝑖 𝐽𝑢𝑚𝑙𝑎𝑕 𝑠𝑒𝑚𝑢𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑛𝑡𝑎𝑟𝑎 1 − 100 𝑦𝑎𝑛𝑔 𝑕𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 4 𝑡𝑒𝑡𝑎𝑝𝑖 𝑡𝑖𝑑𝑎𝑘 𝑕𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 3

𝑎𝑑𝑎𝑙𝑎𝑕 1300 − 432 = 868 𝑁


Pembahasan :


𝑎𝑏𝑐 = 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑦𝑎𝑛𝑔 𝑑𝑖𝑚𝑎𝑘𝑠𝑢𝑑

𝑎 + 𝑏 + 𝑐 = 15 … 1

𝑐𝑏𝑎 = 𝑎𝑏𝑐 − 396

100𝑐 + 10𝑏 + 𝑎 = 100𝑎 + 10𝑏 + 𝑐 − 396

100𝑐 + 10𝑏 + 𝑎 − 100𝑎 − 10𝑏 − 𝑐 = −396

99𝑐 − 99𝑎 = −396

𝑐 − 𝑎 = −4 … 2 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑏𝑎𝑔𝑖 99)

𝑎𝑐𝑏 = 𝑎𝑏𝑐 + 36

100𝑎 + 10𝑐 + 𝑏 = 100𝑎 + 10𝑏 + 𝑐 + 36

100𝑎 + 10𝑐 + 𝑏 − 100𝑎 − 10𝑏 − 𝑐 = 36

−9𝑏 + 9𝑐 = 36

−𝑏 + 𝑐 = 4 … 3 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑏𝑎𝑔𝑖 9)

𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 1 , 2 𝑑𝑎𝑛 3 ∶

𝑎 + 𝑏 + 𝑐 = 15

𝑐 − 𝑎 = −4

−𝑏 + 𝑐 = 4

3𝑐 = 15





𝑐 =15

3

𝑐 = 5

𝑐 = 5 → 𝑐 − 𝑎 = −4 … 2

5 − 𝑎 = −4

5 + 4 = 𝑎

9 = 𝑎

𝑎 = 9

𝑐 = 5 → −𝑏 + 𝑐 = 4 … 3

−𝑏 + 5 = 4

5 − 4 = 𝑏

1 = 𝑏

𝑏 = 1

𝐽𝑎𝑑𝑖 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎𝑕 𝑎𝑏𝑐 = 915 𝐴

18. Jawaban : 𝐶. 1 𝑎𝑡𝑎𝑢 5

Pembahasan :

𝑥 − 3 2 = −2 𝑥 − 3 + 8

𝑀𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ 𝑥 − 3 = 𝑝 𝑠𝑒𝑕𝑖𝑛𝑔𝑔𝑎 ∶

𝑝2 = −2𝑝 + 8

𝑝2 + 2𝑝 − 8 = 0

𝑝 − 2 . 𝑝 + 4 = 0

𝑝 = 2 𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖

𝑎𝑡𝑎𝑢 𝑝 = −4 𝑡𝑖𝑑𝑎𝑘 𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖

𝑝 = 2 → 𝑥 − 3 = 𝑝

𝑥 − 3 = 2 → 𝑢𝑛𝑡𝑢𝑘 𝑥 − 3 = + → 𝑥 − 3 = 2

𝑥 = 2 + 3

𝑥 = 5

→ 𝑢𝑛𝑡𝑢𝑘 𝑥 − 3 = − → − 𝑥 − 3 = 2

−𝑥 + 3 = 2

3 − 2 = 𝑥

1 = 𝑥

𝑥 = 1

𝐽𝑎𝑑𝑖 𝑠𝑜𝑙𝑢𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎𝑕 1 𝑎𝑡𝑎𝑢 5 𝐶






5

Pembahasan :

𝑛 𝑆 = 40

𝑛 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑆𝑐𝑖𝑒𝑛𝑐𝑒 = 24

𝑛 𝑀𝑎𝑡𝑕𝑒𝑚𝑎𝑡𝑖𝑐𝑠 = 13

𝑛 𝐿𝑖𝑘𝑒 𝑏𝑜𝑡𝑕 = 5

𝑛 𝐷𝑖𝑠𝑙𝑖𝑘𝑒 𝑏𝑜𝑡𝑕 = 𝑛 𝑆 − 𝑛 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑆𝑐𝑖𝑒𝑛𝑐𝑒 + 𝑛 𝑀𝑎𝑡𝑕𝑒𝑚𝑎𝑡𝑖𝑐𝑠 − 𝑛 𝐿𝑖𝑘𝑒 𝑏𝑜𝑡𝑕

= 40 − 24 + 13 − 5

= 40 − 32

= 8

𝑃 𝐷𝑖𝑠𝑙𝑖𝑘𝑒 𝑏𝑜𝑡𝑕 =𝑛 𝐷𝑖𝑠𝑙𝑖𝑘𝑒 𝑏𝑜𝑡𝑕

𝑛 𝑆

=8

40

=1

5 𝐵


Pembahasan :

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑡𝑎𝑏𝑒𝑙 𝑏𝑒𝑟𝑖𝑘𝑢𝑡 ∶

𝐵𝑖𝑙𝑎𝑛𝑔𝑎𝑛 ∶

301 − 999 𝑅𝑎𝑡𝑢𝑠𝑎𝑛 𝑃𝑢𝑙𝑢𝑕𝑎𝑛 𝑆𝑎𝑡𝑢𝑎𝑛

𝐾𝑒𝑚𝑢𝑛𝑔𝑘𝑖𝑛𝑎𝑛

𝑎𝑛𝑔𝑘𝑎 𝑦𝑎𝑛𝑔 𝑑𝑖𝑝𝑖𝑙𝑖𝑕

3 2 3

5 3 5

6 5 7

7 6 9

9 7

9

𝐵𝑎𝑛𝑦𝑎𝑘

𝑎𝑛𝑔𝑘𝑎 𝑦𝑎𝑛𝑔 𝑑𝑖𝑝𝑖𝑙𝑖𝑕

5 − 1 𝑡𝑖𝑑𝑎𝑘 𝑏𝑜𝑙𝑒 𝑕 𝑏𝑒𝑟𝑢𝑙𝑎𝑛𝑔

= 4 6 − 2 𝑡𝑖𝑑𝑎𝑘 𝑏𝑜𝑙𝑒 𝑕 𝑏𝑒𝑟𝑢𝑙𝑎𝑛𝑔

= 4 4

𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘 𝑐𝑎𝑟𝑎 𝑚𝑒𝑛𝑦𝑢𝑠𝑢𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑡𝑒𝑟𝑠𝑒𝑏𝑢𝑡 𝑎𝑑𝑎𝑙𝑎𝑕 4 .4 .4 = 64 𝐵

21. Jawaban : 𝐷. 20,25 𝑐𝑚2

Pembahasan :


𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷 = 𝐷𝐸 = 𝑝𝑎𝑛𝑗𝑎𝑛𝑔 𝑠𝑖𝑠𝑖 𝑝𝑒𝑟𝑠𝑒𝑔𝑖 = 9 𝑐𝑚





𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐴𝐻 𝑑𝑎𝑛 𝐷𝐵𝑂 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶

𝐵𝑂

𝐴𝐻=

𝐵𝐷

𝐴𝐷

𝐵𝑂

9=

9+9

9+9+9

𝐵𝑂

9=

18

27

𝐵𝑂 =18

27 .9

𝐵𝑂 = 6

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝐴𝐻 𝑑𝑎𝑛 𝐷𝐶𝑃 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶

𝐶𝑃

𝐴𝐻=

𝐶𝐷

𝐴𝐷

𝐶𝑃

9=

9

9+9+9

𝐶𝑃

9=

9

27

𝐶𝑃 =9

27 .9

𝐶𝑃 = 3

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐺𝐸𝐷 𝑑𝑎𝑛 𝐺𝐹𝑄 𝑦𝑎𝑛𝑔 𝑠𝑒𝑏𝑎𝑛𝑔𝑢𝑛 ∶

𝐹𝑄

𝐷𝐸=

𝐹𝐺

𝐸𝐺

𝐹𝑄

9=

9

9+9

𝐹𝑄

9=

9

18

𝐹𝑄 =9

18 .9

𝐹𝑄 =9

2

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑟𝑖𝑠 𝐵𝐺 ∶

𝑂𝐺 = 𝐵𝐺 − 𝐵𝑂 = 9 − 6 = 3

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑔𝑎𝑟𝑖𝑠 𝐶𝐹 ∶

𝑃𝑄 = 𝐶𝐹 − 𝐶𝑃 − 𝐹𝑄 = 9 − 3 −9

2=

3

2

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝑂𝐺 𝑑𝑎𝑛 𝐷𝑃𝑄 ∶

𝐿𝑎𝑟𝑠𝑖𝑟𝑎𝑛 = 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝑂𝐺 − 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐷𝑃𝑄

= 1

2 .𝑂𝐺 .𝐵𝐷 −

1

2 .𝑃𝑄 .𝐶𝐷

= 1

2 .3 . 9 + 9 −

1

2 .

3

2 .9

= 1

2 .3 .18 −

1

2 .

3

2 .9

=54

2−

27

4

=108

4−

27

4

=81

4

= 20,25 𝑐𝑚2 𝐷





22. Jawaban : 𝐷. 60 7

Pembahasan :


𝐴𝐵 = 4 𝑐𝑚

𝐵𝐶 = 8 𝑐𝑚

𝐴𝐶 = 4 2 𝑐𝑚

𝐴𝐷 = 15 𝑐𝑚


𝐵𝑇 = 𝑥

𝐶𝑇 = 𝐵𝐶 − 𝐵𝑇 = 8 − 𝑥

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑇𝐵 ∶

𝐴𝑇2 = 𝐴𝐵2 − 𝐵𝑇2

𝐴𝑇2 = 42 − 𝑥2

𝐴𝑇2 = 16 − 𝑥2

𝐴𝑇2 = 16 − 𝑥2 … (1)

𝑃𝑒𝑟𝑕𝑎𝑡𝑖𝑘𝑎𝑛 𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝑠𝑖𝑘𝑢 − 𝑠𝑖𝑘𝑢 𝐴𝑇𝐶 ∶

𝐴𝑇2 = 𝐴𝐶2 − 𝐶𝑇2

𝐴𝑇2 = 4 2 2− 8 − 𝑥 2

𝐴𝑇2 = 16 .2 − 64 − 16𝑥 + 𝑥2

𝐴𝑇2 = 32 − 64 + 16𝑥 − 𝑥2

𝐴𝑇2 = −32 + 16𝑥 − 𝑥2 … (2)

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑠𝑖𝑘𝑎𝑛 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 2 𝑘𝑒 1 ∶

𝐴𝑇2 = 16 − 𝑥2

−32 + 16𝑥 − 𝑥2 = 16 − 𝑥2

16𝑥 = 16 − 𝑥2 + 32 + 𝑥2

16𝑥 = 48

𝑥 =48

16

𝑥 = 3

𝑥 = 3 → 𝐴𝑇2 = 16 − 𝑥2 … 1

𝐴𝑇2 = 16 − 32

𝐴𝑇2 = 16 − 9





𝐴𝑇2 = 7

𝐴𝑇 = 7

𝐿𝑝𝑟𝑖𝑠𝑚𝑎 𝐴𝐵𝐶 .𝐷𝐸𝐹 = 𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 . 𝑡𝑝𝑟𝑖𝑠𝑚𝑎

= 1

2 .𝐵𝐶 .𝐴𝑇 .𝐴𝐷

= 1

2 .8 . 7 .15

= 60 7 𝐷

23. Jawaban : 𝐴. 52 𝑚 (𝑏𝑢𝑘𝑎𝑛 𝑐𝑚)

Pembahasan :


𝑀𝐴 = 𝐵𝐶 − 𝐷𝑀 = 26 − 12 = 14

𝐴𝐸 = 𝐶𝐷 − 𝐵𝐸 = 36 − 24 = 12

𝑀𝐿 + 𝐾𝐽 + 𝐼𝐻 + 𝐺𝐹 = 𝐴𝐸 = 12

𝐸𝐹 + 𝐺𝐻 + 𝐼𝐽 + 𝐾𝐿 = 𝑀𝐴 = 14

𝐾𝑟𝑢𝑚𝑎 𝑕 = 𝐴𝑀 + 𝐴𝐸 + 𝑀𝐿 + 𝐾𝐽 + 𝐼𝐻 + 𝐺𝐹 + 𝐸𝐹 + 𝐺𝐻 + 𝐼𝐽 + 𝐾𝐿

= 14 + 12 + 12 + 14

= 52 𝑚 (𝑏𝑢𝑘𝑎𝑛 𝑐𝑚) 𝐴

24. Jawaban : 𝐷. 21

2

Pembahasan :

2𝑥 − 𝑎 >𝑥−1

2+

𝑎𝑥

3 𝑚𝑒𝑚𝑖𝑙𝑖𝑘𝑖 𝑠𝑜𝑙𝑢𝑠𝑖 𝑥 > 3

2𝑥 − 𝑎 >𝑥−1

2+

𝑎𝑥

3

2𝑥 − 𝑎 >3 . 𝑥−1

6+

2 .𝑎𝑥

6

2𝑥 − 𝑎 >3𝑥−3

6+

2𝑎𝑥

6

2𝑥 − 𝑎 >3𝑥−3+2𝑎𝑥

6

6 . 2𝑥 − 𝑎 > 3𝑥 − 3 + 2𝑎𝑥

12𝑥 − 6𝑎 > 3𝑥 − 3 + 2𝑎𝑥

12𝑥 − 3𝑥 − 2𝑎𝑥 > −3 + 6𝑎





9𝑥 − 2𝑎𝑥 > −3 + 6𝑎

9 − 2𝑎 𝑥 > −3 + 6𝑎

𝑥 >−3+6𝑎

9−2𝑎

𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 𝑥 >−3+6𝑎

9−2𝑎 𝑑𝑎𝑛 𝑥 > 3 , 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑕𝑤𝑎 ∶

−3+6𝑎

9−2𝑎= 3

−3 + 6𝑎 = 3 . 9 − 2𝑎

−3 + 6𝑎 = 27 − 6𝑎

6𝑎 + 6𝑎 = 27 + 3

12𝑎 = 30

𝑎 =30

12

𝑎 =5

2

𝑎 = 21

2 𝐷


Pembahasan :

𝑥2 − 10𝑥 + 2015 = 0 𝑚𝑒𝑚𝑖𝑙𝑖𝑘𝑖 𝑠𝑜𝑙𝑢𝑠𝑖 𝑝 𝑑𝑎𝑛 𝑞

𝑥2 − 10𝑥 + 2015 𝑎=1

𝑏=−10𝑐=2015

= 0 → 𝑝 + 𝑞 =−𝑏

𝑎=

− −10

1= 10

𝑥 = 𝑝 → 𝑥2 − 10𝑥 + 2015 = 0

𝑝2 − 10𝑝 + 2015 = 0

𝑝2 = 10𝑝 − 2015

𝑝2 + 10𝑞 + 2014 = 10𝑝 − 2015 + 10𝑞 + 2014

= 10𝑝 + 10𝑞 − 1

= 10 . 𝑝 + 𝑞 − 1

= 10 . 10 − 1

= 100 − 1

= 99 𝐵

26. Jawaban : 𝐶. 2017

Pembahasan :

𝑀𝑒𝑛𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑦𝑎𝑛𝑔 𝑙𝑒𝑏𝑖𝑕 𝑏𝑒𝑠𝑎𝑟 1 𝑠𝑎𝑡𝑢𝑎𝑛 𝑑𝑎𝑟𝑖 𝑥2 − 2𝑥 + 9 ∶

𝑥2 − 2𝑥 + 9 < 𝑥2

𝑥2 − 2𝑥 + 9 < 𝑥

𝑢𝑛𝑡𝑢𝑘 𝑥 = 2015 → 𝑥2 − 2𝑥 + 9 < 𝑥

𝑥2 − 2𝑥 + 9 < 2015





𝑀𝑒𝑛𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑦𝑎𝑛𝑔 𝑙𝑒𝑏𝑖𝑕 𝑘𝑒𝑐𝑖𝑙 1 𝑠𝑎𝑡𝑢𝑎𝑛 𝑑𝑎𝑟𝑖 4𝑥2 + 4𝑥 + 9 ∶

4𝑥2 + 4𝑥 + 1 < 4𝑥2 + 4𝑥 + 9

2𝑥 + 1 2 < 4𝑥2 + 4𝑥 + 9

2𝑥 + 1 < 4𝑥2 + 4𝑥 + 9

𝑢𝑛𝑡𝑢𝑘 𝑥 = 2015 → 2𝑥 + 1 < 4𝑥2 + 4𝑥 + 9

2 .2015 + 1 < 4𝑥2 + 4𝑥 + 9

4030 + 1 < 4𝑥2 + 4𝑥 + 9

4031 < 4𝑥2 + 4𝑥 + 9

𝐷𝑖𝑝𝑒𝑟𝑜𝑙𝑒𝑕 𝑥2 − 2𝑥 + 9 < 2015 𝑑𝑎𝑛 4031 < 4𝑥2 + 4𝑥 + 9 𝑖𝑛𝑖 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎𝑕𝑤𝑎 ∶

𝑥2 − 2𝑥 + 9 < 2015 ≤ ⋯ ≤ 4031 < 4𝑥2 + 4𝑥 + 9

𝐽𝑎𝑑𝑖 𝑏𝑎𝑛𝑦𝑎𝑘𝑛𝑦𝑎 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑢𝑙𝑎𝑡 𝑦𝑎𝑛𝑔 𝑏𝑒𝑟𝑎𝑑𝑎 𝑑𝑖𝑎𝑛𝑡𝑎𝑟𝑎 𝑥2 − 2𝑥 + 9 𝑑𝑎𝑛 4𝑥2 + 4𝑥 + 9

𝑎𝑑𝑎𝑙𝑎𝑕 4031 − 2015 + 1 = 2017 𝐶

27. Jawaban : 𝐶. 94,5

Pembahasan :


𝐴𝐷 = 𝐴𝐵 = 4 𝑐𝑚

𝐴𝐹 = 3,5 𝑐𝑚

𝐿𝑎𝑟𝑠𝑖𝑟𝑎𝑛 = 4 . 𝐿𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝑝𝑎𝑛𝑗𝑎𝑛𝑔 𝐴𝐷𝐸𝐹 + 𝐿𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛

= 4 .𝐴𝐷 .𝐴𝐹 + 𝜋 .𝐴𝐹2

= 4 . 4 .3,5 +22

7 . 3,52

= 56 + 38,5

= 94,5

𝐽𝑎𝑑𝑖 𝑙𝑢𝑎𝑠 𝑑𝑎𝑒𝑟𝑎𝑕 𝑦𝑎𝑛𝑔 𝑑𝑖𝑏𝑎𝑡𝑎𝑠𝑖 𝑜𝑙𝑒𝑕 𝑙𝑖𝑛𝑡𝑎𝑠𝑎𝑛 𝑘𝑢𝑐𝑖𝑛𝑔 𝑎𝑑𝑎𝑙𝑎𝑕 94,5 𝐶


Pembahasan :

𝑓 𝑥 +1

𝑥 .𝑓 −𝑥 = 𝑥 + 6





𝑥 = 2 → 𝑓 𝑥 +1

𝑥 .𝑓 −𝑥 = 𝑥 + 6

𝑓 2 +1

2 .𝑓 −2 = 2 + 6

𝑓 2 +1

2 .𝑓 −2 = 8

2 . 𝑓 2 +1

2 .𝑓 −2 = 2 . 8 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 2)

2 . 𝑓 2 + 𝑓 −2 = 16 … (1)

𝑥 = −2 → 𝑓 𝑥 +1

𝑥 . 𝑓 −𝑥 = 𝑥 + 6

𝑓 −2 +1

−2 .𝑓 − −2 = −2 + 6

𝑓 −2 −1

2 .𝑓 2 = 4

2 . 𝑓 −2 −1

2 .𝑓 2 = 2 . 4 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 2)

2 .𝑓 −2 − 𝑓 2 = 8

−𝑓 2 + 2 .𝑓 −2 = 8 … (2)

𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑠𝑖 𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 1 𝑑𝑎𝑛 2 ∶

2 .𝑓 2 + 𝑓 −2 = 16 → 4 .𝑓 2 + 2 .𝑓 −2 = 32 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 2)

−𝑓 2 + 2 .𝑓 −2 = 8 → −𝑓 2 + 2 .𝑓 −2 = 8

5 . 𝑓 2 = 24

𝑓 2 =24

5

𝑥 = 3 → 𝑓 𝑥 +1

𝑥 .𝑓 −𝑥 = 𝑥 + 6

𝑓 3 +1

3 .𝑓 −3 = 3 + 6

𝑓 3 +1

3 .𝑓 −3 = 9

3 . 𝑓 3 +1

3 .𝑓 −3 = 3 . 9 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 3)

3 . 𝑓 3 + 𝑓 −3 = 27 … (3)

𝑥 = −3 → 𝑓 𝑥 +1

𝑥 . 𝑓 −𝑥 = 𝑥 + 6

𝑓 −3 +1

−3 .𝑓 − −3 = −3 + 6

𝑓 −3 −1

3 .𝑓 3 = 3

3 . 𝑓 −3 −1

3 .𝑓 3 = 3 . 3 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 3)

3 .𝑓 −3 − 𝑓 3 = 9

−𝑓 3 + 3 .𝑓 −3 = 9 … (4)


3 .𝑓 3 + 𝑓 −3 = 27 → 9 .𝑓 3 + 3 .𝑓 −3 = 81 (𝑟𝑢𝑎𝑠 𝑘𝑖𝑟𝑖 𝑑𝑎𝑛 𝑘𝑎𝑛𝑎𝑛 𝑑𝑖𝑘𝑎𝑙𝑖𝑘𝑎𝑛 3)

−𝑓 3 + 3 .𝑓 −3 = 9 → −𝑓 3 + 3 .𝑓 −3 = 9

10 . 𝑓 3 = 72

𝑓 3 =72

10





𝑓 3 =36

5

𝑓 2 + 𝑓 3 =24

5+

36

5

=60

5

= 12 𝐴

29. Jawaban : 𝐵. 160 𝜋

Pembahasan :


𝑟 = 8 𝑐𝑚

𝐿𝑎𝑟𝑠𝑖𝑟𝑎𝑛 =3 .360−180

360 .𝜋 . 𝑟2

=1080−180

360 .𝜋 . 82

=900

360 .𝜋 .64

= 160 𝜋 𝐵

30. Jawaban : 𝐷. −1

2 2 𝑎𝑡𝑎𝑢 2

Pembahasan :


1 +1

𝑥 1 +

1

𝑥 1 +

1

𝑥 1 +

1

𝑥 … = 𝑝

1 +1

𝑥 1 +

1

𝑥 1 +

1

𝑥 1 +

1

𝑥 … = 𝑝

1 +1

𝑥 1 +

1

𝑥 1 +

1

𝑥 1 +

1

𝑥 … = 𝑝2

1

𝑥 1 +

1

𝑥 1 +

1

𝑥 1 +

1

𝑥 … = 𝑝2 − 1





1 +1

𝑥 1 +

1

𝑥 1 +

1

𝑥 … = 𝑥 . 𝑝2 − 1

𝑝 = 𝑥 . 𝑝2 − 1

𝑝 = 𝑥 .𝑝2 − 𝑥

0 = 𝑥 .𝑝2 − 𝑝 − 𝑥

𝑥 .𝑝2 − 𝑝 − 𝑥 = 0

𝑥 = 2 → 𝑥 .𝑝2 − 𝑝 − 𝑥 = 0

2 .𝑝2 − 𝑝 − 2 𝑎= 2𝑏=−1

𝑐=− 2

= 0

𝑝1,2 =−𝑏± 𝑏2−4𝑎𝑐

2𝑎

𝑝1,2 =− −1 ± −1 2−4 . 2 . − 2

2 . 2

𝑝1,2 =1± 1+8

2 . 2

𝑝1,2 =1± 9

2 . 2

𝑝1,2 =1±3

2 . 2

𝑝 =1−3

2 . 2 𝑎𝑡𝑎𝑢 𝑝 =

1+3

2 . 2

𝑝 =−2

2 . 2 𝑎𝑡𝑎𝑢 𝑝 =

4

2 . 2

𝑝 =−2

2 . 2 . 2

2 𝑎𝑡𝑎𝑢 𝑝 =

4

2 . 2 . 2

2

𝑝 =−2 2

2 .2 𝑎𝑡𝑎𝑢 𝑝 =

4 2

2 .2

𝑝 = −1

2 2 𝑎𝑡𝑎𝑢 𝑝 = 2

𝐽𝑎𝑑𝑖 𝑛𝑖𝑙𝑎𝑖 𝑑𝑎𝑟𝑖 1 +1

𝑥 1 +

1

𝑥 1 +

1

𝑥 1 +

1

𝑥 … 𝑎𝑑𝑎𝑙𝑎𝑕 −

1

2 2 𝑎𝑡𝑎𝑢 2 𝐷

"𝑆𝑒𝑚𝑜𝑔𝑎 𝑏𝑒𝑟𝑚𝑎𝑛𝑓𝑎𝑎𝑡"

"𝑆𝐷.𝐴"


olimpiade primagama madura mencari juara 2015 matematika smp kode soal 15333 (babak penyisihan)

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