Pembahasan Soal UN Matematika SMP Tahun 2014 Paket 4 www.olimattohir.blogspot.com

Paket 4

+ −  + −

= + − = − =

= 1

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24 orang  132 hari 32 orang x hari

=

x = 132 ×

x = 99 hari

= = =

:  ∶

=

=

(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 = � −�

− =

−

=

= 2

U5 = a + (4-1)2

7 = a + 6

a = 1

U20 = a + (19 – 1)2 U20 = 1 + (18)2 U20 = 1 + 36

U20 =37

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 =1.572

(i) x2+ xy– 2y2 = (x –y)(x + 2y)

(ii) 16x2– 25y2 = (4x + 5y)(4x – 5y) (iii) 3x2 + 4x– 4 = (x + 2)(3x – 2)

9x + 5 = 2x– 9 9x – 2x = – 9 – 5 7x = –14

x = –2

Jadi, x + 11 = –2 + 11 = 9

K = 2(p + l) 144 = 2(p + l) 72 = (p + l)

72 = (3x + 10) + (x + 10) 72 = 4x + 20

72 – 20 = 4x 52 = 4x 13 = x x = 13

p = 3x + 10 = 3(13) + 10 = 39 + 10 = 49

l = x + 10 = (13) + 10 = 13 + 10 = 23

Deret banyak tiang beton: 12 + 20 + 28 + ..., S11

Sn = (2.a + (n– 1)b)

S11 = (2.12 + (11 – 1)8)

S11 = 11(52)

Paket 4

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) = –2x + 7 f(k) = –2k + 7

 11 = –2k + 7

 11 – 7 = –2k

 4 = –2k

 –2 = k

m = –2

Memotong sumbu-y di titik (0, 6)

m = − = −

Memotong sumbu-y di titik (0, 6)

m = − = −

Memotong sumbu-y bukan di titik (0, 6)

m = =

Memotong sumbu-y bukan di titik (0, 6)

m = =

Memotong sumbu-y bukan di titik (0, 6)

m1 =

-- -- = − = −

m1×m2 = –1 − × m2 = –1

m2 = 2

A. m2 = 2

B. m2 = – 2

C. m2 =

D. m2 = −

atau

titik (3,2)

y–y1 = m2(x –x1) y– 2 = 2(x– 3)

y = 2x– 1

m1 = −₋ =− = –2

y –y1= m(x–x1), (ambil x1 = 1 dan y1 = 2)

y – 2 = –2(x– 1)

y – 2 = –2x + 2

y = –2x + 4

y = –2x + 4 , titik C (3, p)

p = –2(3) + 4

p = –6 + 4

Paket 4

3x + 4y = 17  ×1  3x + 4y = 17 4x – 2y = 8  ×2  8x – 4y = 16

11x = 33 x = 3

sehingga y = 2

Nilai 2x + 3y = 2(3) + 3(2) = 6 + 6 = 12

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

Luas daerah yang diarsir = × Luas persegi KLMN

= × (6)2 = 9 cm2

K L

N M

B C D

A

K L

N M

B C D

A cm

6

cm

8

cm

8

cm

(10 + 12 + 4)× 2 = (26) × 2 = 52 cm

1 ₂

4 3

(i) 1 dengan 4 (ii) 2 dengan 3

(iii) 12 dengan 34 _{ada 3 pasang}

=



=

 AC = 4 ×

 AC = 6 cm

 

B C

A

4 cm



B E

D

9 cm



3 + 9 = 12 cm



Paket 4

A B

E

C _D

E

x  x 

= =

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

7x + 40 = 180 7x = 140

x = 20

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

B = (5x + 10) = 1100

Q P

R

A

B

Q P

R

A

B 1

Q P

R 3

Q P

R

4

Q P

R

Panjang busur PQ = �_� ×  d = × × 42 = × 22 × 6 = 3 × 11 × 3 = 99 cm

d2 = p2– (r1 + r2)2 p2 = d2 + (r1 + r2)2

= 202 + (10 + 5)2

= 400 + (15)2

= 625

p = 25 Rusuk = 7 × 2 = 14

Sisi = 7 + 1 = 8

V = La × t

= + × × 10

= (10 × 5) × 10 = 50 × 10 V = 500

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

Paket 4

LP = Luas tabung tanpa tutup + luas setengah bola

= (πr2_{+ 2πrt) + 2πr}2

= ( ×72 + 2× ×7×10) + 2× ×72 = (22×7 + 2×22×10) + 2×22×7

LP = 902

3, 4, 4, 5, 5, 6, 7, 7, 7, 8, 9, 9

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

20 30 49 104 54 50  307

 40 Rata-rata = = 7, 675

Banyak siswa yang nilainya kurang dari rata-rata = 40 – 24 = 16

atau

Banyak siswa yang nilainya kurang dari rata-rata = 4 + 5 + 7 = 16

Pukul 14.00  38,00oC

Pukul 15.00  37,50oC

Pukul 16.00  38,00oC Pukul 17.00  38,00oC Pukul 18.00  39,00oC Pukul 19.00  38,00oC

Pukul 20.00  37,50oC

Pukul 21.00  37,00oC

Nomor bola lebih dari 6 adalah 7, 8  ada 2

P = � �

� =

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3x + 4y = 17  ×1  3x + 4y = 17 4x – 2y = 8  ×2  8x – 4y = 16

11x = 33 x = 3 sehingga y = 2

Nilai 2x + 3y = 2(3) + 3(2) = 6 + 6 = 12

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

Luas daerah yang diarsir = × Luas persegi KLMN = × (6)2

= 9 cm2

K L

N M

B C D

A

K L

N M

B C D

A cm

6

cm

8

cm

8

cm

(10 + 12 + 4)× 2 = (26) × 2 = 52 cm

1 ₂ 4 3

(i) 1 dengan 4 (ii) 2 dengan 3

(iii) 12 dengan 34 _{ada 3 pasang}

=



=

 AC = 4 ×  AC = 6 cm

 

B C

A

4 cm



B E

D

9 cm 

3 + 9 = 12 cm



A B E

C _D

E

x  x 

= =

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

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

Jadi, B = (5x + 10) = 5(20) + 10 = 100 + 10 B = (5x + 10) = 1100

Q P

R

A

B

Q P

R

A

B 1

Q P

R 3

Q P R

4

Q P

R

Panjang busur PQ = _� ×  d = × × 42 = × 22 × 6 = 3 × 11 × 3 = 99 cm

d2 = p2– (r1 + r2)2 p2 = d2 + (r1 + r2)2

= 202 + (10 + 5)2

= 400 + (15)2

= 625 p = 25 Rusuk = 7 × 2 = 14

Sisi = 7 + 1 = 8

V = La × t

= + × × 10 = (10 × 5) × 10 = 50 × 10 V = 500

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

LP = Luas tabung tanpa tutup + luas setengah bola

= (πr2_{+ 2πrt) + 2πr}2

= ( ×72 + 2× ×7×10) + 2× ×72 = (22×7 + 2×22×10) + 2×22×7

LP = 902

3, 4, 4, 5, 5, 6, 7, 7, 7, 8, 9, 9

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 20 30 49 104 54 50  307

 40 Rata-rata = = 7, 675 Banyak siswa yang nilainya kurang dari rata-rata = 40 – 24 = 16

atau

Banyak siswa yang nilainya kurang dari rata-rata = 4 + 5 + 7 = 16

Pukul 14.00  38,00oC

Pukul 15.00  37,50oC

Pukul 16.00  38,00oC Pukul 17.00  38,00oC Pukul 18.00  39,00oC Pukul 19.00  38,00oC

Pukul 20.00  37,50oC

Pukul 21.00  37,00oC

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/