soal-dan-pembahasan-un-matematika-smp-2014-paket-3

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

Sosuke D. Aizen 2 SMPN 1 Tambelangan

Soal dan Pembahasan UN MATEMATIKA SMP 2014 /

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PAKET 3

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6

2=

6

2 . 2

2=

6 2

2= 3 2

24 ∶ 3 = 24: 3 = 8 = 4 .2 = 4 . 2 = 2 2

8134 = 34

34 = 34 .

34 = 33 = 27

𝑝𝑒𝑘𝑒𝑟𝑗𝑎 = 55 → 𝑠𝑒𝑙𝑒𝑠𝑎𝑖 = 130 𝑕𝑎𝑟𝑖

𝑝𝑒𝑘𝑒𝑟𝑗𝑎 = 55 + 10 = 65 → 𝑠𝑒𝑙𝑒𝑠𝑎𝑖 =55

65 .130 = 110 𝑕𝑎𝑟𝑖

81

2+ 1

1

4− 7

3

5= 8 + 1 − 7 +

1

2+

1

4−

3

5 = 2 +

10

20+

5

20−

12

20 = 2 +

3

20= 2

3

20







920000 = 800000 + 9% .𝑛 .800000

920000 − 800000 =9

100 . 𝑛 .800000

120000 = 72000 .𝑛

120000

72000= 𝑛

5

3= 𝑛

𝑛 =5

3 𝑡𝑎𝑕𝑢𝑛

𝑛 =5

3 .12 𝑏𝑢𝑙𝑎𝑛

𝑛 = 20 𝑏𝑢𝑙𝑎𝑛

𝑏 =𝑈5 − 𝑈3

5 − 3=

31 − 17

2=

14

2= 7

𝑈20 = 𝑈5 + 20 − 5 . 𝑏 = 31 + 15 . 7 = 31 + 105 = 136

𝑏 =𝑈7 − 𝑈3

7 − 3=

38 − 18

4=

20

4= 5

𝑎 = 𝑈1 = 𝑈3 + 1 − 3 .𝑏 = 18 − 2 . 5 = 18 − 10 = 8

𝑆𝑛 =𝑛

2 . 2𝑎 + 𝑛 − 1 .𝑏

𝑆24 =24

2 . 2 . 8 + 24 − 1 . 5 = 12 . 16 + 115 = 12 .131 = 1572

𝑎 = 15

𝑈5 = 23

𝑆𝑛 =𝑛

2 . 𝑎 + 𝑈𝑛

𝑆5 =5

2 . 15 + 𝑈5 =

5

2 . 15 + 23 =

5

2 .38 = 95

2 . 𝑝 + 𝑙 = 72 2 . 5𝑥 − 1 + 2𝑥 + 2 = 72

2 . 7𝑥 + 1 = 72 14𝑥 + 2 = 72 14𝑥 = 72 − 2

𝑥 =70

14= 5

𝑝 = 5𝑥 − 1 = 5 .5 − 1 = 25 − 1 = 24 𝑙 = 2𝑥 + 2 = 2 .5 + 2 = 12

2𝑦 − 3 = 12 − 𝑦 2𝑦 + 𝑦 = 12 + 3 3𝑦 = 15

𝑦 =15

3

𝑦 = 5 → 𝑦 − 2 = 5 − 2 = 3







1 𝑥2 + 𝑥𝑦 − 2𝑦2 = 𝑥 − 𝑦 . 𝑥 + 2𝑦

2 16𝑥2 − 25𝑦2 = 42𝑥2 − 52𝑦2 = 4𝑥 + 5𝑦 . 4𝑥 − 5𝑦

3 3𝑥2 + 4𝑥 − 4 = 𝑥 + 2 . 3𝑥 − 2

𝑛 𝑃 = 5

𝐵𝑎𝑛𝑦𝑎𝑘 𝑕𝑖𝑚𝑝𝑢𝑛𝑎𝑛 𝑏𝑎𝑔𝑖𝑎𝑛 𝑑𝑎𝑟𝑖 𝑃 = 2𝑛 𝑃 = 25 = 32

𝑛 𝑃 ∪ 𝐶 = 40 ; 𝑛 𝑃 = 23 ; 𝑛 𝑃 ∩ 𝐶 = 12

𝑛 𝑃 ∪ 𝐶 = 𝑛 𝑃 + 𝑛 𝐶 − 𝑛 𝑃 ∩ 𝐶 40 = 23 + 𝑛 𝐶 − 12 40 = 11 + 𝑛 𝐶 40 − 11 = 𝑛 𝐶 29 = 𝑛 𝐶 𝑛 𝐶 = 29

2𝑥 − 5𝑦 = 18 → 10𝑥 − 25𝑦 = 90

5𝑥 − 3𝑦 = 26 → 10𝑥 − 6𝑦 = 52

−19𝑦 = 38

𝑦 =38

−19= −2

𝑦 = −2 → 2𝑥 − 5𝑦 = 18

2𝑥 − 5 . −2 = 18

2𝑥 + 10 = 18

2𝑥 = 18 − 10

𝑥 =8

2= 4

4𝑥 − 7𝑦 = 4 . 4 − 7 . −2

= 16 + 14

= 30

3𝐺 + 4𝐸 = 95000 → 3𝐺 + 4𝐸 = 95000

5𝐺 + 2𝐸 = 65000 → 10𝐺 + 4𝐸 = 130000

−7𝐺 = −35000

𝐺 =−35000

−7= 5000

𝐺 = 5000 → 5𝐺 + 2𝐸 = 65000

5 . 5000 + 2𝐸 = 65000

25000 + 2𝐸 = 65000

2𝐸 = 65000 − 25000

𝐸 =40000

2= 20000

4𝐺 + 5𝐸 = 4 . 5000 + 5 . 20000 = 20000 + 100000

= 120000

𝑓 𝑥 = 5𝑥 − 2 𝑓 𝑘 = 5𝑘 − 2 = 18 5𝑘 = 18 + 2

𝑘 =20

5

𝑘 = 4







𝑥 = 0 → 𝑦 = 0 − 1

𝑦 = −1 0, −1 𝑦𝑎𝑛𝑔 𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖 𝑔𝑎𝑚𝑏𝑎𝑟 𝑝𝑎𝑑𝑎 𝑝𝑖𝑙𝑖 𝑕𝑎𝑛 𝐴 𝑑𝑎𝑛 𝐵

𝑦 = 0 → 0 = 𝑥 − 1

1 = 𝑥

𝑥 = 1 1,0 𝐽𝑎𝑑𝑖 𝑦𝑎𝑛𝑔 𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖 𝑔𝑎𝑚𝑏𝑎𝑟 𝑝𝑎𝑑𝑎 𝑝𝑖𝑙𝑖 𝑕𝑎𝑛 𝐴

𝐴 3 𝑥1

, 2 𝑦1

𝑑𝑎𝑛 𝐵 −1 𝑥2

, 4 𝑦2

𝑚𝐴𝐵 =𝑦2 − 𝑦1

𝑥2 − 𝑥1

=4 − 2

−1 − 3=

2

−4= −

1

2

𝑆𝑦𝑎𝑟𝑎𝑡 𝑡𝑒𝑔𝑎𝑘 𝑙𝑢𝑟𝑢𝑠 ∶ 𝑚 .𝑚𝐴𝐵 = −1

𝑚 . −1

2 = −1

𝑚 = −1 . −2 𝑚 = 2

𝐴. 𝑦 = 2𝑥 − 1 → 𝑚 = 2 𝐵. 𝑦 = −2𝑥 + 7 → 𝑚 = −2

𝐶. 𝑦 =1

2𝑥 + 2𝑦

𝑦 − 2𝑦 =1

2𝑥

−𝑦 =1

2𝑥

𝑦 = −1

2𝑥 → 𝑚 = −

1

2

𝐷. 𝑦 = −1

2𝑥 + 4 → 𝑚 = −

1

2







𝐾 −2 𝑥1

, 3 𝑦1

; 𝐿 1 𝑥2

, −3 𝑦2

; 𝑀 4 𝑥

, 𝑎 𝑦

𝑦 − 𝑦1

𝑦2 − 𝑦1

=𝑥 − 𝑥1

𝑥2 − 𝑥1

𝑎 − 3

−3 − 3=

4 − −2

1 − −2

𝑎 − 3

−3 − 3=

4 + 2

1 + 2

𝑎 − 3

−6=

6

3

𝑎 − 3 =6

3 . −6

𝑎 − 3 = −12

𝑎 = −12 + 3

𝑎 = −9

𝑃𝑎𝑛𝑗𝑎𝑛𝑔 𝑡𝑎𝑙𝑖 𝑛𝑖𝑙𝑎𝑖 𝑎𝑠𝑙𝑖 = 1502 + 1502 = 1502 .2 = 150 2

𝑃𝑎𝑛𝑗𝑎𝑛𝑔 𝑡𝑎𝑙𝑖 𝑝𝑒𝑛𝑑𝑒𝑘𝑎𝑡𝑎𝑛 = 1502 + 1502 = 22500 + 22500 = 45000 ≈ 44944 = 212 𝑚







𝑃𝑇

𝑃𝑅=

𝑆𝑇

𝑄𝑅

𝑃𝑇

15=

6

10

𝑃𝑇 =6

10 .15

𝐶𝐸 = 9

𝑀𝑁

𝐾𝐿=

𝑁𝑃

𝑃𝐿

𝐿𝑎𝑟𝑠𝑖𝑟𝑎𝑛 =1

4 . 𝐿𝐸𝐹𝐺𝐻

=1

4 . 82

=1

4 .64

= 16 𝑐𝑚2







𝐶𝐷 + 𝐻𝐼 = 𝐴𝐵 = 18 𝑐𝑚 𝐷𝐸 + 𝐹𝐺 + 𝐼𝐴 = 𝐵𝐶 = 20 𝑐𝑚 𝐸𝐹 = 𝐺𝐻 = 4 𝑐𝑚 𝐾𝑏𝑎𝑛𝑔𝑢𝑛 = 2 . 𝐴𝐵 + 2 . 𝐵𝐶 + 2 . 𝐸𝐹

= 2 . 18 + 2 . 20 + 2 . 4 = 36 + 40 + 8 = 84 𝑐𝑚

𝑃𝑎𝑛𝑗𝑎𝑔 𝑏𝑢𝑠𝑢𝑟 𝐴𝐵 =∠𝐴𝑂𝐵

360𝑜 . 𝐾𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛

=72𝑜

360𝑜 .2 .𝜋 . 𝑟

=72𝑜

360𝑜 .2 .

22

7 .35

= 44

𝐴𝐵 = 𝑃𝐿2 + 𝑅 − 𝑟 2

= 162 + 17 − 5 2 = 162 + 122 = 256 + 144 = 400 = 20

∠𝐴 + ∠𝐵 = 90𝑜

4𝑥 − 13 𝑜 + 3𝑥 + 5 𝑜 = 90𝑜

4𝑥𝑜 − 13𝑜 + 3𝑥𝑜 + 5𝑜 = 90𝑜

7𝑥𝑜 − 8𝑜 = 90𝑜

7𝑥𝑜 = 90𝑜 + 8𝑜

𝑥𝑜 =98𝑜

7

𝑥𝑜 = 14𝑜

∠𝐵 = 3𝑥 + 5 𝑜

= 3 .14 + 5 𝑜

= 42 + 5 𝑜

= 47𝑜







𝑉𝑝𝑟𝑖𝑠𝑚𝑎 = 𝐿𝑏𝑒𝑙𝑎 𝑕 𝑘𝑒𝑡𝑢𝑝𝑎𝑡 . 𝑡𝑝𝑟𝑖𝑠𝑚𝑎

= 1

2 . 𝑑1 .𝑑2 . 𝑡𝑝𝑟𝑖𝑠𝑚𝑎

= 1

2 .16 .20 . 24

= 160 . 24

= 3840 𝑐𝑚3

𝐿𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑙𝑖𝑚𝑎𝑠 = 𝐿𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝐴𝐵𝐶𝐷 + 4 .𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐵𝐶𝑇

= 𝐴𝐵2 + 4 .1

2 . 𝐵𝐶 . 𝑇𝑃

= 102 + 4 .1

2 .10 .13

= 100 + 260

= 360 𝑐𝑚2

𝐴𝐵 = 𝐵𝐶 =𝐾𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝐴𝐵𝐶𝐷

4=

40

4= 10

𝑂𝑃 =1

2 .𝐴𝐵 =

1

2 .10 = 5

𝑇𝑃 = 𝑂𝑇2 + 𝑂𝑃2 = 122 + 52

= 144 + 25 = 169 = 13

𝑂𝐷 = 𝑟 =1

2 . 𝐴𝐵 =

1

2 .1,4 = 0,7

𝑂𝑃 = 2,4

𝐷𝑃 = 𝑠 = 𝑂𝐷2 + 𝑂𝑃2 = 0,72 + 2,42

= 0,49 + 5,76 = 6,25 = 2,5

𝐿𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑏𝑎𝑛𝑔𝑢𝑛 = 𝐿𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑡𝑎𝑏𝑢𝑛𝑔 𝑡𝑎𝑛𝑝𝑎 𝑡𝑢𝑡𝑢𝑝 + 𝐿𝑠𝑒𝑙𝑖𝑚𝑢𝑡 𝑘𝑒𝑟𝑢𝑐𝑢𝑡

= 𝜋𝑟2 + 2𝜋𝑟𝑡 + 𝜋𝑟𝑠

= 22

7 . 0,72 + 2 .

22

7 .0,7 .5 +

22

7 .0,7 .2,5

= 1,54 + 22 + 5,5

= 23,54 + 5,5

= 29,04 𝑐𝑚2

𝑅𝑢𝑠𝑢𝑘 𝑙𝑖𝑚𝑎𝑠 𝑠𝑒𝑔𝑖 − 𝑛 = 2 . 𝑛

𝑅𝑢𝑠𝑢𝑘 𝑙𝑖𝑚𝑎𝑠 𝑠𝑒𝑔𝑖 − 7 = 2 .7 = 14

𝑆𝑖𝑠𝑖 𝑙𝑖𝑚𝑎𝑠 𝑠𝑒𝑔𝑖 − 𝑛 = 𝑛 + 1

𝑆𝑖𝑠𝑖 𝑙𝑖𝑚𝑎𝑠 𝑠𝑒𝑔𝑖 − 7 = 7 + 1 = 8







𝑥 =5 .4 + 6 .5 + 7 .7 + 8 .13 + 9 .6 + 10 .5

4 + 5 + 7 + 13 + 6 + 5=

307

40= 7,675

𝐵𝑎𝑛𝑦𝑎𝑘 𝑠𝑖𝑠𝑤𝑎 𝑦𝑎𝑛𝑔 𝑖𝑘𝑢𝑡 𝑟𝑒𝑚𝑒𝑑𝑖𝑎𝑙 = 4 + 5 + 7 = 16 𝑜𝑟𝑎𝑛𝑔

𝑥 𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛 =𝑛𝑝 .𝑥 𝑝 + 133 + 127

𝑛𝑝 + 1 + 1

=23 .130 + 133 + 127

23 + 1 + 1

=23 .130 + 260

25

=23 .130 + 2 .130

25

=130 . 23 + 2

25

=130 . 25

25

= 130

(𝐵𝑒𝑙𝑢𝑚 𝑡𝑒𝑛𝑡𝑢 𝑥 𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛 = 130)



(𝑃𝑎𝑠𝑡𝑖 𝑥 𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛 = 130)

𝑈𝑟𝑢𝑡𝑎𝑛 𝑑𝑎𝑡𝑎 ∶ 45, 55, 55, 60, 65, 70, 70, 75, 85, 85, 85, 85, 90, 90, 95, 100

𝑀𝑒𝑑𝑖𝑎𝑛 =75 + 85

2=

160

2= 80







𝑅𝑝200.000.000,00 − 𝑅𝑝192.500.000,00 = 𝑅𝑝7.500.000,00

𝑛 𝑏𝑜𝑙𝑎 𝑏𝑒𝑟𝑛𝑜𝑚𝑜𝑟 𝑔𝑎𝑛𝑗𝑖𝑙 = 5

𝑛 𝑆 = 9

𝑃 𝑏𝑜𝑙𝑎 𝑏𝑒𝑟𝑛𝑜𝑚𝑜𝑟 𝑔𝑎𝑛𝑗𝑖𝑙 =𝑛 𝑏𝑜𝑙𝑎 𝑏𝑒𝑟𝑛𝑜𝑚𝑜𝑟 𝑔𝑎𝑛𝑗𝑖𝑙

𝑛 𝑆 =

5

9



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