Pembahasan Soal UN Matematika SMP Tahun 2014 Paket 2 www.olimattohir.blogspot.com

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Paket 2
Mathematics Sport
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

−

+

+
+

=

−

=
=

=8

1,2 jam  80 km/jam
1 jam
 x km/jam

=
=

,

 x = 80 × 1,2

 x = 96 km/jam

=

=


:

=
=
=

×

×

:
:

×

×

×

=



×

=

=3

−

−

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Paket 2
(i) Besar bunga = 920.000 – 800.000 = 120.000
�
(ii) Besar bunga =
× 9% × 800.000

�

Sehingga: 120.000 =

b=

� −�

=

−

U 5 = a + (4-1)3
8=a+9
a = –1
b=

� −�
−

=

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

−

=

−

�

 bulan =

U 10 = a + (9 – 1)3
U 10 = –1 + (8)3
U 10 = –1 + 24
U 10 =23

=3

=

Barisan kursi: 23,

× 800.000  120 =

×

=5

×3×8
×

= 20 bulan

U n = U n-1 + b
U = U9 + 3

atau 10
U 10 = 20 + 3
U 10 =23

(2×8 + (24 – 1)5)
S24 = 12(16 + (23)5)
S24 = 12(16 + 115)
S24 = 12(131)
S 24 =1572
S24 =

25, 27, 29,....

S15 =

(2×23 + (14)2)
S15 = 15(23 + 14)
S15 = 15(37)
S15 = 555

(i) 14x2 + 7y = 7(2x2 + y)
(ii) x2 – 25 = (x + 5)(p – 5)
(iii) 3x2 + 5x – 12 = (3x – 4)(x + 3)

5x + 8 = 3x + 16
5x – 3x = 16 – 8
2x = 8
x=4

Jadi, x + 11 = 4 + 11 = 15

K = 2(p + l)
72 = 2(p + l)
36 = (p + l)
36 = (5x – 1) + (2x + 2)
36 = 7x + 1

36 – 1 = 7x
35 = 7x
5=x

x=5

p = 5x – 1 = 5(5) – 1 = 25 – 1 = 24
l = 2x + 2 = 2(5) + 2 = 10 + 2 = 12

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Paket 2
Banyak anggota Q = 6
Banyak himpunan bagian dari Q adalah 26 = 64

P
23 -12 = 11

40

C

12

40 – (11 + 12) = 40 – 23 = 17

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

Menulis cerpen = 17 + 12 = 29

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Paket 2
Memotong sumbu-y di titik (0, –2)
m=3
m=

=3

Memotong sumbu-y di titik (0, –2)

m=

=

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

m=

−

= –2

Memotong sumbu-y di titik (0, –2)

m=

−

= −

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

m1 =

−
+

m1 = m2 =
m2 =

=

=

A. m2 = 3
B. m2 =
C. m2 = −
D. m2 = – 3

atau

y – y1 = m2(x – x1)
y+0=
y= x+2

(x + 6)

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Paket 2

− −

−

m1 =
= =2
− −
−
y – y1 = m(x – x1), (ambil x1 = 1 dan y1 = 5)
y – 5 = 2(x – 1)
y – 5 = 2x – 2
y = 2x + 3

y = 2x + 3 , titik R (3, a)
a = 2(3) + 3
a=6+3
a=9

2x – 5y = 18  ×3  6x – 15y = 54
5x – 3y = 26  ×5  25x – 15y = 130
–19x
sehingga y = – 2

Nilai 4x – 7y = 4(4) – 7(–2)
= 16 + 14
= 30

= –76
x =4

2k + 4g = 7.000  ×7  14k + 28g = 49.000
5k + 7g = 13.000  ×4  20k + 28g = 52.000
– 6k

= 3.000
k = 500
1 lusin Kelereng = 12 × 500 = 6000

Panjang tali =
+
Panjang perkiraan dengan

=

,

×

=
adalah 150 × 1,41 = 211,5  212

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Paket 2
D
D

10 cm

10 cm

C

C

10 cm
A

M

N

M

N

B

A

B

K
K

L

Luas daerah yang diarsir =

× Luas persegi KLMN

= × (8)2
= 16 cm2
33

(32 + 33 + 11) × 2 = (76) × 2 = 152 cm

8 cm

8 cm
L

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Paket 2

3

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

2

4
1

180o – 115o = 65o

ada 4 pasang

C
E
10 cm



A


=



65o
6 + 14 = 20 cm

=

 BE =

o
8 cm 65

B

×

 BE = 16 cm

P
=

P

M

x



N

K

x



L

� �
=
�
�

D



B

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Paket 2

P + Q
(x + 17) + (3x – 7)
4x + 10
4x

Jadi, Q = (3x – 7) = 3(20) – 7
= 60 – 7
Q = (3x – 7) = 530

= 900
= 90
= 90
= 80
x = 20

C
C

C

4

C
A

2

1

A

B
A

C
BA

B

B
A

3
B

Panjang busur AB =

�

�

× 2 r

=
=

×2×

× 44 × 2

=
= 17, 6 cm
d2 = p2 – (r1 + r2)2

 p2 = d2 + (r1 + r2)2
= 242 + (7 + 3)2
= 576 + (10)2
= 576 + 100
= 676
 p = 26

Rusuk = 8 × 3 = 24
Sisi = 8 + 2 = 10

× 14

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Paket 2

V =

� �

×t

×

× 24
=
= (16 × 10) × 24
V = 3.840

TO=16 cm

O

24 cm

E

Dengan menggunakan teorema pythagoras didapat panajang TE = 20 cm
L = luas alas + 4 × luas sisi tegak
= 242 + 4 × × 24 × 20
= 576 + 960
L = 1.536 cm2

s
24
LP = Luas tabung tanpa tutup + luas selimut kerucut
= (πr2 + 2πrt) + πrs
= ( ×72 + 2× ×7×12) + ×7×25
= (22×7 + 2×22×12) + 22×25
= (154 + 528) + 550
= (682) + 550
LP = 1.232 cm2

7

s=
s=
s = 25

+

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Paket 2

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 =

4

10

42 84

48 18

 30
 206

Rata-rata =

×

+

+ +

+

= 6, 87

Banyak siswa yang nilainya lebih dari rata-rata = 30 – 10 = 20

=

= 130

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Paket 2

Pukul 16.00
Pukul 17.00
Pukul 18.00
Pukul 19.00
Pukul 20.00

 38oC
 38,20oC
 38,10oC
 39oC
 38,20oC

Nomor bola ganjil  1, 3, 5, 7, 9
P=

�

�

�

Jadi, Pukul 18.30  38,50oC

 ada 5

=

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/