Soal dan pembahasan un matematika smp 2014 paket 1

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PAKET 1

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5

5=

5

5 . 5

5=

5 5

5= 5

60 ∶ 5 = 60: 5 = 12 = 4 .3 = 4 . 3 = 2 3

6456 = 26

56 = 26 .

56 = 25 = 32

𝑎𝑦𝑎𝑚 = 60 → 𝑝𝑒𝑟𝑠𝑒𝑑𝑖𝑎𝑎𝑛 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 = 24 𝑕𝑎𝑟𝑖

𝑎𝑦𝑎𝑚 = 60 − 15 = 45 → 𝑝𝑒𝑟𝑠𝑒𝑑𝑖𝑎𝑎𝑛 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 =60

45 .24 = 32 𝑕𝑎𝑟𝑖

21

3+ 5

1

4− 1

1

2=

7

3+

21

4−

3

2=

28

12+

63

12−

18

12=

73

12= 6

1

12







𝑏 =𝑈9 − 𝑈5

9 − 5=

20 − 8

4=

12

4= 3

𝑈10 = 𝑈9 + 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 =𝑛

2 . 2𝑎 + 𝑛 − 1 .𝑏

𝑆24 =24

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

𝐵𝑎𝑟𝑖𝑠𝑎𝑛 𝑘𝑢𝑟𝑠𝑖 ∶ 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

3 𝑡𝑎𝑕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

5𝑥 − 3𝑥 = 12 + 8 2𝑥 = 20

𝑥 =20

2

𝑥 = 10 → 𝑥 + 3 = 10 + 3 = 13

𝑖 9𝑎𝑏 + 21𝑎𝑐 = 3𝑎 . 3𝑏 + 7𝑐

𝑖𝑖 𝑥2 − 9 = 𝑥2 − 32 = 𝑥 − 3 . 𝑥 + 3

𝑖𝑖𝑖 3𝑝2 − 𝑝 − 2 = 3𝑝 + 2 . 𝑝 − 1







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 = 2𝑛 𝑃 = 23 = 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 = 3𝑥 + 5 𝑓 𝑎 = 3𝑎 + 5 = −7 3𝑎 = −7 − 5

𝑎 =−12

3

𝑎 = −4







𝑥 = 0 → 𝑦 = 2 . 0 − 1

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

𝑥 = 2 → 𝑦 = 2 . 2 − 1

𝑦 = 4 − 1

𝑦 = 3 2,3 𝐽𝑎𝑑𝑖 𝑦𝑎𝑛𝑔 𝑚𝑒𝑚𝑒𝑛𝑢 𝑕𝑖 𝑔𝑎𝑚𝑏𝑎𝑟 𝑝𝑎𝑑𝑎 𝑝𝑖𝑙𝑖 𝑕𝑎𝑛 𝐴

𝑃 −3 𝑥1

, 8 𝑦1

𝑑𝑎𝑛 𝑄 2 𝑥2

, 5 𝑦2

𝑚𝑃𝑄 =𝑦2 − 𝑦1

𝑥2 − 𝑥1

=5 − 8

2 − (−3)=

−3

2 + 3= −

3

5

𝑆𝑦𝑎𝑟𝑎𝑡 𝑡𝑒𝑔𝑎𝑘 𝑙𝑢𝑟𝑢𝑠 ∶ 𝑚 .𝑚𝑃𝑄 = −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

𝐷. 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

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

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

𝐶𝐸

𝐴𝐶=

𝐷𝐸

𝐴𝐵

𝐶𝐸

15=

8

12

𝐶𝐸 =8

12 .15

𝐶𝐸 = 10 𝑐𝑚







𝐿𝑎𝑟𝑠𝑖𝑟𝑎𝑛 =1

4 . 𝐿𝑉𝑊𝑋𝑌

=1

4 . 102

=1

4 .100

= 25 𝑐𝑚2

𝐴𝐵

𝐶𝐷=

𝐴𝐸

𝐸𝐶=

𝐵𝐸

𝐸𝐷







𝐾𝑏𝑎𝑛𝑔𝑢𝑛 = 2 . 17 + 8 + 5 + 6 + 5 + 4

= 2 . 45 = 90 𝑐𝑚

𝑃𝑎𝑛𝑗𝑎𝑔 𝑏𝑢𝑠𝑢𝑟 𝐴𝐵 =∠𝐴𝑂𝐵

360𝑜 . 𝐾𝑙𝑖𝑛𝑔𝑘𝑎𝑟𝑎𝑛

=60𝑜

360𝑜 .2 .𝜋 . 𝑟

=60𝑜

360𝑜 .2 . 3,14 .10

= 10,466 = 10,47 𝑐𝑚

∠𝐴 + ∠𝐵 = 180𝑜

5𝑦 − 16 𝑜 + 2𝑦𝑜 = 180𝑜

5𝑦𝑜 − 16𝑜 + 2𝑦𝑜 = 180𝑜

7𝑦𝑜 − 16𝑜 = 180𝑜

7𝑦𝑜 = 180𝑜 + 16𝑜

7𝑦𝑜 = 196𝑜

𝑦𝑜 =196𝑜

7

𝑦𝑜 = 28𝑜

∠𝐴 = 5𝑦 − 16 𝑜

= 5 .28 − 16 𝑜

= 140 − 16 𝑜

= 124𝑜

𝐴𝐵 = 𝑃𝐿2 + 𝑅 − 𝑟 2

= 242 + 12 − 5 2 = 242 + 72 = 576 + 49 = 625 = 25 𝑐𝑚







𝑉𝑝𝑟𝑖𝑠𝑚𝑎 = 𝐿𝑡𝑟𝑎𝑝𝑒𝑠𝑖𝑢𝑚 . 𝑡𝑝𝑟𝑖𝑠𝑚𝑎

= 1

2 . 𝐽𝑢𝑚𝑙𝑎𝑕 𝑠𝑖𝑠𝑖 𝑠𝑒𝑗𝑎𝑗𝑎𝑟 . 𝑡𝑡𝑟𝑎𝑝𝑒𝑠𝑖𝑢𝑚 . 𝑡𝑝𝑟𝑖𝑠𝑚𝑎

= 1

2 . 8 + 12 .5 . 10

= 50 . 10

= 500 𝑐𝑚3

𝐿𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑙𝑖𝑚𝑎𝑠 = 𝐿𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝐴𝐵𝐶𝐷 + 4 .𝐿𝑠𝑒𝑔𝑖𝑡𝑖𝑔𝑎 𝐵𝐶𝑇

= 𝐴𝐵2 + 4 .1

2 . 𝐵𝐶 . 𝑇𝑃

= 162 + 4 .1

2 .16 .17

= 256 + 544

= 800 𝑐𝑚2

𝐴𝐵 = 𝐵𝐶 =𝐾𝑝𝑒𝑟𝑠𝑒𝑔𝑖 𝐴𝐵𝐶𝐷

4=

64

4= 16

𝑂𝑃 =1

2 .𝐴𝐵 =

1

2 .16 = 8

𝑇𝑃 = 𝑂𝑇2 + 𝑂𝑃2 = 152 + 82

= 225 + 64 = 289 = 17

𝑂𝐷 = 𝑟 =1

2 . 𝐴𝐵 =

1

2 .14 = 7

𝑂𝑃 = 36 − 𝐴𝐷 = 36 − 12 = 24

𝐷𝑃 = 𝑠 = 𝑂𝐷2 + 𝑂𝑃2 = 72 + 242

= 49 + 576 = 625 = 25

𝐿𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑏𝑎𝑛𝑔𝑢𝑛 = 𝐿𝑝𝑒𝑟𝑚𝑢𝑘𝑎𝑎𝑛 𝑡𝑎𝑏𝑢𝑛𝑔 𝑡𝑎𝑛𝑝𝑎 𝑡𝑢𝑡𝑢𝑝 + 𝐿𝑠𝑒𝑙𝑖𝑚𝑢 𝑡 𝑘𝑒𝑟𝑢𝑐𝑢𝑡

= 𝜋𝑟2 + 2𝜋𝑟𝑡 + 𝜋𝑟𝑠

= 22

7 . 72 + 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

𝑆𝑖𝑠𝑖 = 𝐴𝐵𝐶𝐷𝐸𝐹, 𝐴𝐵𝑇, 𝐵𝐶𝑇, 𝐶𝐷𝑇, 𝐷𝐸𝑇, 𝐸𝐹𝑇, 𝐹𝐴𝑇 → 7







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

𝑈𝑟𝑢𝑡𝑎𝑛 𝑑𝑎𝑡𝑎 ∶ 165, 166, 168, 168, 170, 171, 171 𝑀𝑒𝑑𝑖𝑎𝑛

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

(𝐵𝑒𝑙𝑢𝑚 𝑡𝑒𝑛𝑡𝑢 𝑥 𝑔𝑎𝑏𝑢𝑛𝑔𝑎𝑛 = 130)



(𝑃𝑎𝑠𝑡𝑖 𝑥 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 =

2

8



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Soal dan pembahasan un matematika smp 2014 paket 1