Pembahasan Soal UN Matematika SMP Tahun

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Paket 1
Mathematics Sport
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

−

+

=
=
=

+
+

−

−

−

=8
36  8
x  24

=

x=

=

=

:


=

=
=
=2

 x = 36 × 3

×

 x = 108 buah

=

×

×

×

:

:

:


=

atau

=
=2

×

:

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Paket 1



×

=

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

Sehingga: 120.000 =

b=

−

=

−

=

−

U 2 = a + (2-1)5
6=a+5
a=1
b=

−

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

−

−

=

=

×

=5

=5

Barisan gaji: 3.000.000,

× 800.000  120 =

�

 bulan =

U 40 = a + (40 – 1)5

U 40 = 1 + (39)5
U 40 = 1 + 195
U 40 =196

×3×8
×

= 20

(2×8 + (24 – 1)5)
S 24 = 12(16 + (23)5)
S 24 = 12(16 + 115)
S 24 = 12(131)
S 24 =1572
S 24 =

3.500.000, 4.000.000, 4.500.000,....

S 10 =

(2×3.000.000 + (9)500.000)
S 10 = 5(6.000.000 + 4.500.000)
S 10 = 5(10.500.000)
S 10 = 52.500.000

(i) 15x2y – 20xy2 = 5xy(3x – 4y)
(ii) p2 – 16 = (p + 4)(p – 4)
(iii) 3a2 + 8a – 3 = (3a – 1)(a + 3)

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Paket 1
3x + 6 = 5x + 20
6 – 20 = 5x – 3x
– 14 = 2x
–7=x
x=–7

Jadi, x + 12 = – 7 + 12 = 5

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
23 -12 = 11

40

C
12

f(x) = 8 – 2x  f(k) = 8 – 2k
 –10 = 8 – 2k
 –18 = –2k
 9=k
 k=9

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

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Paket 1
Memotong sumbu-y di titik (0, 2)
m=2
m= −

m=

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

=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)

−

=–5
m1×m2 = –1
–5 × m2 = –1
m2 =
m1 =

−

−

+

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

sehingga x – 3y – 5 = 0
x – 3(–1) – 5 = 0
x+3–5=0
x–2=0
x=2

–y – 1 = 0
y = –1

y = 2x – 5 , titik A (10, p)
p = 2(10) – 5
p = 20 – 5
p = 15

Nilai 3x + 2y = 3(2) + 2(–1)
=6–2
=4

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Paket 1

5a + 3j = 79.000
3a + 2j = 49.000

 ×2  10a + 6j = 158.000
 ×3  9a + 6j = 147.000
a = 11.000

Panjang tali =
+
Panjang perkiraan dengan

=

,

×

=
adalah 150 × 1,41 = 211,5  212

D
D

10 cm

10 cm

C

C

10 cm
A

M

N

M

N

B

A

B

K
L

K

Luas daerah yang diarsir =

× Luas persegi KLMN

= × (8)2
= 16 cm2

8 cm

8 cm
L

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Paket 1

4 cm
4 cm

6 cm

atau

6 cm

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

4
(i) 1 dengan 3
(ii) 2 dengan 4
(iii) 12 dengan 34
(iv) 14 dengan 23

3
2
1

ada 4 pasang

P
x

15 cm

S


Q

10 cm

=



=

 TR =

6 cm
R T

×

 TR = 4 cm
Jadi, PT = PR – TR
= 15 – 4
PT = 11 cm

R

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Paket 1
E
E

x

D



A

B



x

=

A + B
(2x + 30) + (5x + 10)
7x + 40
7x

= 1800
= 180
= 180
= 140
x = 20 M

=

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

M
M

C

M

3

3
K

2
1
K

2
4

K
K

(i)

K

L

Panjang busur AB =

1

L

(ii)

L

�

�

× 2 r

=
=

(iv)
(iii)

×2×

× 44 × 2

=
= 17, 6 cm

L

4

× 14

L

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Paket 1

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
+
=
×
= (30 × 5) × 20
= 150 × 20
V = 3000

× 20

Dengan menggunakan teorema pythagoras
didapat panajang TE = 10 cm
8 cm
O

12 cm

E

L = luas alas + 4 × luas sisi tegak
= 122 + 4 × × 12 × 10
= 144 + 240
L = 384 cm2

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Paket 1

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

180

130 525

640 360

 30
200  2035

Rata-rata =

×

+

+ +

+

=

= 67, 83

Banyak siswa yang nilainya lebih dari rata-rata = 30 – 9 = 21

= 130

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Paket 1

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
P=

�

��

�

� � ��

 ada 4

=

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/