Soal dan Pembahasan UN MATEMATIKA SMP 2014 Page 1

PAKET 1

5

5

=

5

5

.

5

5

=

5

5

5

=

5

60

∶

5 =

60: 5 =

12 =

4 .3 =

4 .

3 =

2

3

64

56

=

2

6 5 6

= 2

6 .

5

6

= 2

5

=

32

𝑎𝑦𝑎𝑚

= 60

→

𝑝𝑒𝑟𝑠𝑒𝑑𝑖𝑎𝑎𝑛

𝑚𝑎𝑘𝑎𝑛𝑎𝑛

= 24

𝑕𝑎𝑟𝑖

𝑎𝑦𝑎𝑚

= 60

−

15 = 45

→

𝑝𝑒𝑟𝑠𝑒𝑑𝑖𝑎𝑎𝑛

𝑚𝑎𝑘𝑎𝑛𝑎𝑛

=

60

45

.24 =

32

𝑕𝑎𝑟𝑖

2

1

3

+ 5

1

4

−

1

1

2

=

7

3

+

21

4

−

3

2

=

28

12

+

63

12

−

18

12

=

73

12

=

6

1

12

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𝑏

=

𝑈

9

− 𝑈

5

9

−

5

=

20

−

8

4

=

12

4

= 3

𝑈

10

=

𝑈

9

+

10

−

9

.

𝑏

= 20 + 1 . 3 = 20 + 3 =

23

𝑏=𝑈7− 𝑈3 7−3 =

38−18

4 =

20 4 = 5

𝑎=𝑈1=𝑈3+ 1−3.𝑏= 18−2 . 5 = 18−10 = 8

𝑆𝑛=𝑛₂ . 2𝑎+ 𝑛 −1.𝑏

𝑆24= 24

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

𝐵𝑎𝑟𝑖𝑠𝑎𝑛𝑘𝑢𝑟𝑠𝑖 ∶20 , 23 ,…,𝑈20

𝑎= 20

𝑏= 3

𝑆𝑛=𝑛

2 . 2𝑎+ 𝑛 −1 .𝑏

𝑆20= 20

2 . 2 . 20 + 20−1 . 3 = 10 . 40 + 57 = 10 .97 =970 𝑘𝑢𝑟𝑠𝑖

920000 = 800000 + 9% .𝑛 .800000

920000−800000 = 9

100 .𝑛 .800000 120000 = 72000 .𝑛

120000 72000 =𝑛

5 3=𝑛

𝑛=5

3𝑡𝑎𝑕𝑢𝑛

𝑛=5

3 .12 𝑏𝑢𝑙𝑎𝑛 𝑛=20 𝑏𝑢𝑙𝑎𝑛

2 . 𝑝+𝑙 = 144 2 . 3𝑥+ 10 + 𝑥+ 10 = 144 2 . 4𝑥+ 20 = 144 8𝑥+ 40 = 144 8𝑥= 144−40

𝑥=104 8 = 13

𝑝= 3𝑥+ 10

= 3 .13 + 10 = 39 + 10 =49 𝑐𝑚

𝑙 =𝑥+ 10

= 13 + 10 =23 _𝑐𝑚

5

𝑥 −

3

𝑥

= 12 + 8

2

𝑥

= 20

𝑥

=

20

2

𝑥

= 10

→

𝑥

+ 3 = 10 + 3 =

13

𝑖

9

𝑎𝑏

+ 21

𝑎𝑐

= 3

𝑎

.

3

𝑏

+ 7

𝑐

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 𝑜𝑟𝑎𝑛𝑔

𝑛 𝑃

= 3

𝐵𝑎𝑛𝑦𝑎𝑘

𝑕𝑖𝑚𝑝𝑢𝑛𝑎𝑛

𝑏𝑎𝑔𝑖𝑎𝑛

𝑑𝑎𝑟𝑖

𝑃

= 2

𝑛 𝑃

= 2

3

=

8

3𝑥+ 4𝑦= 17 → 3𝑥+ 4𝑦= 17 4𝑥 −2𝑦= 8 → 8𝑥 −4𝑦= 16 11𝑥= 33

𝑥=33 11= 3

𝑥= 3 → 3𝑥+ 4𝑦= 17 3 . 3 + 4𝑦= 17

9 + 4𝑦= 17 4𝑦= 17−9

𝑦=8 4= 2

2𝑥+ 3𝑦= 2 . 3 + 3 . 2 = 6 + 6 =12

3𝐴+ 5𝐵= 39000 → 3𝐴+ 5𝐵= 39000 𝐴+𝐵= 11000 → 3𝐴+ 3𝐵= 33000 2𝐵= 6000

_𝐵=6000 2 = 3000 𝐵= 3000 → 𝐴+𝐵= 11000

𝐴+ 3000 = 11000

𝐴= 11000−3000 𝐴= 8000

4𝐴+ 2𝐵= 4 . 8000 + 2 . 3000 = 32000 + 6000 =38000

𝑓 𝑥

= 3

𝑥

+ 5

𝑓 𝑎

= 3

𝑎

+ 5 =

−

7

3

𝑎

=

−

7

−

5

𝑥

= 0

→

𝑦

= 2 .

0

−

1

𝑦

=

−

1

0,

−

1

𝑦𝑎𝑛𝑔𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖𝑔𝑎𝑚𝑏𝑎𝑟 𝑝𝑎𝑑𝑎𝑝𝑖𝑙𝑖 𝑕𝑎𝑛𝐴𝑑𝑎𝑛𝐵

𝑥

= 2

→

𝑦

= 2 .

2

−

1

𝑦

= 4

−

1

𝑦

= 3

2,3

𝐽𝑎𝑑𝑖 𝑦𝑎𝑛𝑔 𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖𝑔𝑎𝑚𝑏𝑎𝑟 𝑝𝑎𝑑𝑎𝑝𝑖𝑙𝑖 𝑕𝑎𝑛𝐴

𝑃 −3

𝑥1

, 8

𝑦1

𝑑𝑎𝑛 𝑄 2

𝑥2

, 5

𝑦2

𝑚𝑃𝑄=𝑦_𝑥2− 𝑦1

2− 𝑥1

= 5−8 2−(−3)=

−3

2 + 3=− 3 5

𝑆𝑦𝑎𝑟𝑎𝑡𝑡𝑒𝑔𝑎𝑘𝑙𝑢𝑟𝑢𝑠 ∶ 𝑚 .𝑚_𝑃𝑄=−1 𝑚 . −3

5 =−1

𝑚=−1 . −5 3

𝑚=5 3

𝐴. 3𝑥 −5𝑦 −14 = 0 → 𝑚=3 5 𝐵. 3𝑥+ 5𝑦+ 14 = 0 → 𝑚=−3

5 𝐶. 5𝑥+ 3𝑦 −42 = 0 → 𝑚=−5 3

𝐴 2

𝑥1

, 7

𝑦1

; 𝐵 −3

𝑥2

,−3

𝑦2

; 𝐶 3

𝑥

,𝑎

𝑦 𝑦 − 𝑦1

𝑦2− 𝑦1

= 𝑥 − 𝑥1

𝑥2− 𝑥1

𝑎 −7

−3−7=

3−2

−3−2

𝑎 −7

−10 =

1

−5

𝑎 −7 = 1

−5 . −10 𝑎 −7 = 2

𝑎= 2 + 7

𝑎= 9

𝑃𝑎𝑛𝑗𝑎𝑛𝑔𝑡𝑎𝑙𝑖 𝑛𝑖𝑙𝑎𝑖𝑎𝑠𝑙𝑖 = 1502_{+ 150}2_{= 150}2 _{.2 = 150 2}

𝑃𝑎𝑛𝑗𝑎𝑛𝑔𝑡𝑎𝑙𝑖 𝑝𝑒𝑛𝑑𝑒𝑘𝑎𝑡𝑎𝑛 = 1502_{+ 150}2_{= 22500 + 22500 = 45000≈ 44944 =212 𝑚}

𝐶𝐸

𝐴𝐶

=

𝐷𝐸

𝐴𝐵

𝐶𝐸

15

=

8

12

𝐶𝐸

=

8

12

.15

𝐿

𝑎𝑟𝑠𝑖𝑟𝑎𝑛

=

1

4

.

𝐿

𝑉𝑊𝑋𝑌

=

1

4

. 10

2

=

1

4

.100

=

25

𝑐𝑚

2

𝐴𝐵

𝐶𝐷

=

𝐴𝐸

𝐸𝐶

=

𝐾𝑏𝑎𝑛𝑔𝑢𝑛

= 2 .

17 + 8 + 5 + 6 + 5 + 4

= 2 . 45

=

90

𝑐𝑚

𝑃𝑎𝑛𝑗𝑎𝑔𝑏𝑢𝑠𝑢𝑟𝐴𝐵=∠𝐴𝑂𝐵

360𝑜 .𝐾𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛

= 60

𝑜

360𝑜 .2 .𝜋 .𝑟

= 60

𝑜

360𝑜 .2 . 3,14 .10 = 10,466 =10,47 𝑐𝑚

∠𝐴

+

∠𝐵

= 180

𝑜

5

𝑦 −

16

𝑜

+ 2

𝑦

𝑜

= 180

𝑜

5

𝑦

𝑜

−

16

𝑜

+ 2

𝑦

𝑜

= 180

𝑜

7

𝑦

𝑜

−

16

𝑜

= 180

𝑜

7

𝑦

𝑜

= 180

𝑜

+ 16

𝑜

7

𝑦

𝑜

= 196

𝑜

𝑦

𝑜

=

196

𝑜

7

𝑦

𝑜

= 28

𝑜

∠𝐴

=

5

𝑦 −

16

𝑜

=

5 .28

−

16

𝑜

=

140

−

16

𝑜

=

124

𝑜

𝐴𝐵

=

𝑃𝐿

2

+

𝑅 − 𝑟

2

𝑉𝑝𝑟𝑖𝑠𝑚𝑎 =𝐿𝑡𝑟𝑎𝑝𝑒𝑠𝑖𝑢𝑚 .𝑡𝑝𝑟𝑖𝑠𝑚𝑎

= 1

2 .𝐽𝑢𝑚𝑙𝑎𝑕𝑠𝑖𝑠𝑖𝑠𝑒𝑗𝑎𝑗𝑎𝑟 .𝑡𝑡𝑟𝑎𝑝𝑒𝑠𝑖𝑢𝑚 .𝑡𝑝𝑟𝑖𝑠𝑚𝑎

= 1

2 . 8 + 12 .5 . 10 = 50 . 10

=500 𝑐𝑚3

𝐿𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑙𝑖𝑚𝑎𝑠=𝐿𝑝𝑒𝑟𝑠𝑒𝑔𝑖𝐴𝐵𝐶𝐷+ 4 .𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎𝐵𝐶𝑇

=𝐴𝐵2_{+ 4 .}1 2 .𝐵𝐶 .𝑇𝑃

= 162_{+ 4 .}1 2 .16 .17 = 256 + 544 =800 𝑐𝑚2

𝐴𝐵=𝐵𝐶=𝐾𝑝𝑒𝑟𝑠𝑒𝑔𝑖𝐴𝐵𝐶𝐷

4 =

64 4 = 16

𝑂𝑃=1

2 .𝐴𝐵= 1 2 .16 = 8

𝑇𝑃= 𝑂𝑇2_+𝑂𝑃2_{= 15}2_{+ 8}2

= 225 + 64 = 289 = 17

𝑂𝐷=𝑟=1

2 .𝐴𝐵= 1 2 .14 = 7 𝑂𝑃= 36− 𝐴𝐷= 36−12 = 24

𝐷𝑃=𝑠= 𝑂𝐷2_+𝑂𝑃2_{= 7}2_{+ 24}2

= 49 + 576 = 625 = 25

𝐿𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑏𝑎𝑛𝑔𝑢𝑛 =𝐿𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑡𝑎𝑏𝑢𝑛𝑔𝑡𝑎𝑛𝑝𝑎𝑡𝑢𝑡𝑢𝑝+𝐿𝑠𝑒𝑙𝑖𝑚𝑢 𝑡𝑘𝑒𝑟𝑢𝑐𝑢𝑡

= 𝜋𝑟2_{+ 2𝜋𝑟𝑡 +𝜋𝑟𝑠}

= 22 7 . 7

2_{+ 2 .}22 7 .7 .12 +

22 7 .7 .25 = 154 + 528 + 550

= 682 + 550 =1232 𝑐𝑚2

𝑅𝑢𝑠𝑢𝑘

=

𝐴𝐵

,

𝐵𝐶

,

𝐶𝐷

,

𝐷𝐸

,

𝐸𝐹

,

𝐹𝐴

,

𝐴𝑇

,

𝐵𝑇

,

𝐶𝑇

,

𝐷𝑇

,

𝐸𝑇

,

𝐹𝑇

→

12

𝑥

=

3 .3 + 4 .5 + 5 .12 + 6 .7 + 7 .6 + 8 .4 + 9 .3

3 + 5 + 12 + 7 + 6 + 4 + 3

=

232

40

= 5,8

𝐵𝑎𝑛𝑦𝑎𝑘

𝑠𝑖𝑠𝑤𝑎

𝑦𝑎𝑛𝑔

𝑙𝑢𝑙𝑢𝑠

= 7 + 6 + 4 + 3 =

20

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

𝑈𝑟𝑢𝑡𝑎𝑛𝑑𝑎𝑡𝑎 ∶165, 166, 168, 168, 170, 171, 171 𝑀𝑒𝑑𝑖𝑎𝑛

, 172, 173, 173, 175, 178, 182

(𝐵𝑒𝑙𝑢𝑚𝑡𝑒𝑛𝑡𝑢𝑥 _{𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛} = 130)

(𝐵𝑒𝑙𝑢𝑚𝑡𝑒𝑛𝑡𝑢𝑥 _{𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛} = 130)

(𝐵𝑒𝑙𝑢𝑚𝑡𝑒𝑛𝑡𝑢𝑥 _{𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛} = 130)

150 + 250 =

400

𝑜𝑟𝑎𝑛𝑔

𝑛 𝑏𝑜𝑙𝑎

𝑏𝑒𝑟𝑛𝑜𝑚𝑜𝑟

𝑙𝑒𝑏𝑖𝑕

𝑑𝑎𝑟𝑖

6

= 2

𝑛 𝑆

= 8

𝑃 𝑏𝑜𝑙𝑎

𝑏𝑒𝑟𝑛𝑜𝑚𝑜𝑟

𝑙𝑒𝑏𝑖𝑕

𝑑𝑎𝑟𝑖

6

=

𝑛 𝑏𝑜𝑙𝑎

𝑏𝑒𝑟𝑛𝑜𝑚𝑜𝑟

𝑙𝑒𝑏𝑖𝑕

𝑑𝑎𝑟𝑖

6

𝑛 𝑆

=