Pembahasan Soal UN Matematika SMP Tahun 2014 Paket 1 www.olimattohir.blogspot.com

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+ −  + −

= + −

= −

= = 8

36  8 x  24

=

x = ×

x =

x = 12 buah

= = =

:  × : = × : =

= × = 2

:  : =

= × = 2 atau



×

=

(i) Besar bunga = 920.000 – 800.000 = 120.000 (ii) Besar bunga = � × 9% × 800.000

Sehingga: 120.000 = � × × 800.000  120 = � × 3 × 8

 bulan =

× = 20 bulan

b = −

− =

−

= = 5

U2 = a + (2-1)5 6 = a + 5 a = 1

U40 = a + (40 – 1)5 U40 = 1 + (39)5 U40 = 1 + 195 U40 =196

b = −

− =

−

= = 5

U3 = a + (3-1)5 18 = a + 10 a = 8

S24 = (2×8 + (24 – 1)5)

S24 = 12(16 + (23)5)

S24 = 12(16 + 115)

S24 = 12(131)

S24 =1572

Barisan gaji: 3.000.000, 3.500.000, 4.000.000, 4.500.000,....

S10 = (2×3.000.000 + (9)500.000)

S10 = 5(6.000.000 + 4.500.000)

S10 = 5(10.500.000)

S10 = 52.500.000

(i) 15x2y – 20xy2 = 5xy(3x – 4y) (ii) p2– 16 = (p + 4)(p– 4)

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3x + 6 = 5x + 20 6 – 20 = 5x – 3x

– 14 = 2x

– 7 = x x = – 7

K = 2(p + l) 44 = 2(p + l) 22 = (p + l)

22 = (3x + 4) + (2x + 3) 22 = 5x + 7

22 – 7 = 5x 15 = 5x 3 = x x = 3

p = 3x + 4 = 3(3) + 4 = 9 + 4 = 13 l = 2x + 3 = 2(3) + 3 = 6 + 3 = 9

Banyak anggota D = 6

Banyak himpunan bagian dari D adalah 26 = 64

P C 40

12

23

-1

2

=

1

1

40 – (11 + 12) = 40 – 23 = 17 Menulis cerpen = 17 + 12 = 29

f(x) = 8 – 2x f(k) = 8 – 2k

–10 = 8 – 2k

–18 = –2k

 9 = k

 k = 9

Jadi, x + 12 = – 7 + 12 = 5

m1 = −₊ = – 5 m1×m2 = –1

–5 × m2 = –1 m2 = m = 2

Memotong sumbu-y di titik (0, 2)

m = −

Memotong sumbu-y bukan di titik (0, 2) m = = 2

Memotong sumbu-y bukan di titik (0, 2)

m = = 1

Memotong sumbu-y bukan di titik (0, 2)

m = = 2

Memotong sumbu-y di titik (0, 2)

A. m2 = 5 B. m2 = C. m2 = – 5 D. m2 = – 5

m1 = − −_{− −} =−₋ = 2

y –y1= m(x–x1), (ambil x1 = 3 dan y1 = 1)

y – 1 = 2(x– 3)

y – 1 = 2x– 6

y = 2x– 5

y = 2x– 5 , titik A (10, p)

p = 2(10) – 5

p = 20 – 5 p = 15

x – 3y– 5 = 0  ×2  2x – 6y– 10 = 0 2x – 5y– 9 = 0  ×1  2x – 5y– 9 = 0

–y– 1 = 0 y = –1 sehingga x – 3y– 5 = 0

x – 3(–1) – 5 = 0

x + 3 – 5 = 0

x – 2 = 0

x = 2

Nilai 3x + 2y = 3(2) + 2(–1) = 6 – 2

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5a + 3j = 79.000  ×2  10a + 6j = 158.000 3a + 2j = 49.000  ×3  9a + 6j = 147.000

a = 11.000

Panjang tali = + = × =

Panjang perkiraan dengan , adalah 150 × 1,41 = 211,5  212

K L

N M

B C D

A

K L

N M

B C D

A 8cm

cm

8

cm

10

cm

10

cm

10

Luas daerah yang diarsir = × Luas persegi KLMN = × (8)2

= 16 cm2

(4 + 4 + 6 + 6) × 2 = (20) × 2 = 40 cm

4 cm

6 cm 6 cm

atau

1 2 4 3 (i) 1 dengan 3

(ii) 2 dengan 4 (iii) 12 dengan 34

(iv) 14 dengan 23 ada 4 pasang

x



Q R

P

S

T R

10 cm 15 cm

6 cm

=



=

 TR = ×

 TR = 4 cm Jadi, PT = PR – TR

= 15 – 4 PT = 11 cm

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D A

E

B _C

E

x  x 

= =

A + B = 1800 (2x + 30) + (5x + 10) = 180

7x + 40 = 180 7x = 140 x = 20

K L

M

4 1 2 3

Jadi, B = (5x + 10) = 5(20) + 10 = 100 + 10

B = (5x + 10) = 1100

K L

M

3

(i)

K L

M

2

(ii)

K L

M

1

(iii)

K _L

M

4

(iv)

Panjang busur AB = �_� × 2 r = × 2 × × 14 = × 44 × 2

= = 17, 6 cm

d2 = p2– (r1 + r2)2 p2 = d2 + (r1 + r2)2 = 162 + (8 + 4)2

= 256 + (12)2

= 256 + 144 = 400

p = 20

Ada 18 (6×3) rusuk dan 8 (6 + 2) sisi

V = La × t

= + × × 20 = (30 × 5) × 20

= 150 × 20 V = 3000

Dengan menggunakan teorema pythagoras didapat panajang TE = 10 cm

L = luas alas + 4 × luas sisi tegak = 122 + 4 × × 12 × 10 = 144 + 240

L = 384 cm2

12 cm 8 cm

E O

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LP = Luas tabung tanpa tutup + luas setengah bola = (πr2+ 2πrt) + 2πr2

= ( ×72 + 2× ×7×20) + 2× ×72 = (22×7 + 2×22×20) + 2×22×7 LP = 1.342

4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10

Median = + = 6,5

Bisa jadi, rata-ratanya = 130

Bisa jadi, rata-ratanya = 130 Bisa jadi, rata-ratanya = 130

Bisa dipastikan, rata-ratanya = 130

Rata-rata = × + +

+ + = = 130

180 420 525 640 360 200  2325

 30

Rata-rata = = 77, 50

Banyak siswa yang nilainya lebih dari rata-rata = 30 – 16 = 14

Senin – Selasa dan Kamis –Jum’at  penurunan 1000 Selasa – Rabu  kenaikan 2000

Rabu –Kamis dan Jum’at – Sabtu  uang saku tetap Senin – Selasa  penurunan 1000

Nomor bola kurang dari 5  1, 2, 3, 4  ada 4

P = � �� � � ��

� =

Disusun oleh : Mohammad Tohir Jika ada saran, kritik maupun masukan silahkan kirim ke- My email: mohammadtohir@yahoo.com

Terima kasih. My blog : http://m2suidhat.blogspot.com/

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5a + 3j = 79.000  ×2  10a + 6j = 158.000 3a + 2j = 49.000  ×3  9a + 6j = 147.000

a = 11.000

Panjang tali = + = × =

Panjang perkiraan dengan , adalah 150 × 1,41 = 211,5  212

K L

N M

B C D

A

K L

N M

B C D

A 8cm

cm

8

cm

10

cm

10

cm

10

Luas daerah yang diarsir = × Luas persegi KLMN = × (8)2

= 16 cm2

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(4 + 4 + 6 + 6) × 2 = (20) × 2 = 40 cm

4 cm 4 cm

6 cm 6 cm

atau

1 2 4 3 (i) 1 dengan 3

(ii) 2 dengan 4 (iii) 12 dengan 34

(iv) 14 dengan 23 ada 4 pasang

x



Q R

P

S

T R

10 cm 15 cm

6 cm

=



=

 TR = ×

 TR = 4 cm Jadi, PT = PR – TR

= 15 – 4

PT = 11 cm

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D A

E

B _C

E

x  x 

= =

A + B = 1800 (2x + 30) + (5x + 10) = 180

7x + 40 = 180 7x = 140 x = 20

K L

M

4

1 2

3

Jadi, B = (5x + 10) = 5(20) + 10 = 100 + 10

B = (5x + 10) = 1100

K L

M

3

(i)

K L

M

2

(ii)

K L

M

1

(iii)

K _L

M

4

(iv)

Panjang busur AB = �_� × 2 r = × 2 × × 14 = × 44 × 2

=

= 17, 6 cm

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d2 = p2– (r1 + r2)2 p2 = d2 + (r1 + r2)2 = 162 + (8 + 4)2

= 256 + (12)2

= 256 + 144 = 400

p = 20

Ada 18 (6×3) rusuk dan 8 (6 + 2) sisi

V = La × t

= + × × 20 = (30 × 5) × 20

= 150 × 20

V = 3000

Dengan menggunakan teorema pythagoras didapat panajang TE = 10 cm

L = luas alas + 4 × luas sisi tegak = 122 + 4 × × 12 × 10 = 144 + 240

L = 384 cm2

12 cm 8 cm

E O

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LP = Luas tabung tanpa tutup + luas setengah bola = (πr2+ 2πrt) + 2πr2

= ( ×72 + 2× ×7×20) + 2× ×72 = (22×7 + 2×22×20) + 2×22×7

LP = 1.342

4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10

Median = + = 6,5

Bisa jadi, rata-ratanya = 130

Bisa jadi, rata-ratanya = 130 Bisa jadi, rata-ratanya = 130

Bisa dipastikan, rata-ratanya = 130

Rata-rata = × + +

+ + = = 130

180 420 525 640 360 200  2325

 30

Rata-rata = = 77, 50

Banyak siswa yang nilainya lebih dari rata-rata = 30 – 16 = 14

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Senin – Selasa dan Kamis –Jum’at  penurunan 1000

Selasa – Rabu  kenaikan 2000

Rabu –Kamis dan Jum’at – Sabtu  uang saku tetap Senin – Selasa  penurunan 1000

Nomor bola kurang dari 5  1, 2, 3, 4  ada 4

P = � �� � � ��

� =

Disusun oleh : Mohammad Tohir Jika ada saran, kritik maupun masukan silahkan kirim ke- My email: mohammadtohir@yahoo.com

Terima kasih. My blog : http://m2suidhat.blogspot.com/

http://matematohir.wordpress.com/