Pembahasan Mtk UN 2013 paket 02
SMP NEGERI 3 KALIBAGOR
PAKET 2
Alfa Kristanti
1
SMP NEGERI 3 KALIBAGOR
Alfa Kristanti
2
SMP NEGERI 3 KALIBAGOR
Pelangi Matematika Viandre
Al4-kristanty.blogspot.com
2−3 + 3−2 =
7 7 ×
1
1
1 1
9
8
17
+
=
+
=
+
=
23 32
8 9
72 72
72
14 = 7 7 × 14 = 7 7 × 7 × 2 = 7 49 × 2
= 7 7 2 = 49 2
2
2
1
23 ∶ 13 − 45 =
8
3
5
8
∶3−
5
5
= −
21
21
5
8
3
=3×5−
13
= −
5
4−3
Selisih = 4 + 3 × 560.000 =
21
5
= −2
1
7
3
5
× 560.000
= 80.000
9
Bunga 9 bln = 12 × 12% ×
= 9%
Tabungan setelah 9 bln = 100% T + 9% T = 109% T
109% T = 3.815.000 T =
3.815.000
109%
=
3.815.000
= 3.815.000 ×
109
100
100
109
= 3.500.000
Alfa Kristanti
3
SMP NEGERI 3 KALIBAGOR
b=
b=
7−
7−4
4
=
37−19
3
18
=6
3
Sn = 2 ( 2a + (n – 1)b )
U4 = a + 3b
19 = a + 3(6)
19 = a + 18 a = 19 – 18
a=1
24
2
S24 =
( 2(1) + (24 – 1) 6) = 12 (2 + (23) 6) = 12 (2 + 138)
= 12 (140) = 1.680
Un = a + (n – 1)b
a = U1 = 7
U52 = 7 + (52 – 1) 5 = 7 + (51) 5
b = U2 – U1 = 12 – 7 = 5
= 7 + 255 = 262
r=
2
1
=
9
27
=
1
3
Un = a × r n – 1 = 27 ×
Un = 33 × 3−
a = U1 = 27
+1
−1
1
3
= 33 −
+1
= 33 × 3−1
= 34−
−1
3x – 2 < 8x + 13
3x – 8x < 13 + 2
5x < 15
x> 3
×
×
m=
I. 4x2 – 10x = 2x (2x – 5)
II. 7x2 – 49 = 7(x2 – 7)
III. x2 – 3x – 18 = (x – 6)(x + 3)
IV. x2 + 5x – 36 = (x + 9)(x – 4)
2− 1
2− 1
=
−3 − (−6)
−4 − −1
y – y1 = m (x – x1)
=
y – (1) = 1 (x – (–6))
y + 1 = 1 (x + 6)
y+ 1 = – x 6
−3+6
−4+1
=
3
−3
= −1
y+x=–6–1
x+y=–7
Alfa Kristanti
4
SMP NEGERI 3 KALIBAGOR
ax + by + c = 0
a=8;b=4
m=
−
=
8
−4
= −2
Jumlah bilangan terbesar dan terkecil
2
= × jml ketiga bilangan
3
=
2
3
× 162 = 108
M = { 1, 3, 5, 7 }
N = { 5, 7, 11 }
M N = { 1, 3, 5, 7, 11 }
f(3) = 15
3a + b = 15
3a + b = 15
2a + b = 0
5a = 15
a=3
f(2) = 0
2a + b = 0
4p + 2q = 18.000
4p + 2(1.000) = 18.000
4p + 2.000 = 18.000
4p = 16.000
p = 4.000
20p + 30q = 20(4000)+30(1000)
= 80.000 + 30.000
= 110.000
Misal :
Parkir mobil = p
Parkir motor = q
N
12 12
O
M
1
2
LN =
M
A
o
80
80o
C
38
o
o
62
384
12
B
K
= 32 OL =
2
=
32
2
= 16
2+
2 =
KL =
122 + 162
KL = 144 + 256 = 400 = 20
Keliling = 4× KL = 4 × 20 = 80
× 24 × LN = 384
12 × LN = 384
L
f(5) = 3(5) + ( 6)
= 15 6
= 21
3p + 5q = 17.000 × 4 12p + 20q = 68.000
4p + 2q = 18.000 × 3 12p + 6q = 54.000
14q = 14.000
q = 1.000
Lbelahketupat ABCD = 384
1
× KM × LN = 384
2
K
2a + b = 0
2(3) + b = 0
6+b=0
b=06
b =6
38o
62o
L
Pada gambar di samping:
ABC KLM maka
AB = LM, BC = KL,
AC = KM
Alfa Kristanti
5
SMP NEGERI 3 KALIBAGOR
Q
24
FC =
R
A
=
9
12
30
P
C
18
15
12
24
1
=2 ,
=
15
30
1
=2 ,
=
9
18
1
=2
B
4 × 20 + 6 × 5
4 +6
=
80 + 30
10
=
110
10
= 11
L daerah tdk diarsir
= LKLMN + LPQRS – 2 × L daerah diarsir
= 202 + 10 × 15 – 2 × 67
= 400 + 150 – 134
= 416
Banyak pohon =
=
�
2(42)
3
�
=
�
ℎ
84
3
=
2(24+ 18)
3
= 28
A
B
(3x 5) + (2x + 5) = 90
5x + 0 = 90
5x = 90
x = 18
D
C
Penyiku POR = ROQ = 2x + 5 = 2(18) + 5 = 36 + 5 = 41
Alfa Kristanti
6
SMP NEGERI 3 KALIBAGOR
AOE = 2 × ABE
= 2 × 32
= 64
Sudut keliling menghdp
busur yg sama maka
ABE = ACE = ADE
96
ABE = 3 = 32
luar = 262 − 12 − 2
676 − 100 =
=
LOBC =
=
2
=
676 − 102
576 = 24
×
120
50
× 30
= 72
Panjang rusuk 1 balok = 4 (p + l + t) = 4 (40 + 24 + 36)
= 4 (100) = 400 cm = 4 m
Kawat tersedia = 10 m
10
Banyak kerangka balok yg dibuat = 4 = 2,5 2
Garis pelukis
BC =
1
3
1
3
4
=
72
4
t = 152 − 92 =
t = 144 = 12
2
= 18
225 − 81
t
15
9
Vlimas = La × t = × 18 × 12 = 324 × 4 = 1.296
Alfa Kristanti
7
SMP NEGERI 3 KALIBAGOR
dbola = rusuk kubus
dbola = 30, maka :
30
r = 2 = 15
4
4
4
Vbola = 3 π r3 = 3 π × 153 = 3 π × 3.375
= 4 π × 1.125 = 4.500 π
Ldinding = 2 (pt + lt) = 2(8 × 5 + 6 × 5)
= 2(40 + 30) = 2(70) = 140
Biaya = 140 × Rp 50.000,- = Rp 7.000.000,-
2
8 2
s
s
Lkubus = 6 s2
=6× 8 2
= 6 × 64
= 384
s2 + s 2 = 8 2
2 s2 = 64 × 2
2 s2 = 128
s2 = 64 s = 64 = 8
d = 28 r = 14
22
Ltabung = 2πr (r + t) = 2 × 7 × 14 (14 + 26)
= 88 (40) = 3.520
Jumlah siswa = 18 + 22 = 40 anak
Jml tinggi = 18 × 156 + 22 × 152 = 2.808 + 3.344 = 6.152
Rata-rata tinggi =
��
=
6.152
40
= 153,8
Data
5 6 7 8 9
frekuensi 2 3 4 3 2
Modus
Alfa Kristanti
8
SMP NEGERI 3 KALIBAGOR
32
30
Selisih produksi pupuk Maret dan Mei = 10 – 6 = 4
Alfa Kristanti
9
SMP NEGERI 3 KALIBAGOR
n(S) = 2 × 2 × 2 = 8
3 gambar = GGG
1
P (3 gambar) = 8
6
5
5
4
3
3
2
2
Banyak kelereng = 6 + 5 + 3 + 3 + 2 + 4 + 2 + 5 = 30
Kelereng merah = 6
6
P ( 1 merah) = 30 × 100 % = 20 %
Alfa Kristanti
10
PAKET 2
Alfa Kristanti
1
SMP NEGERI 3 KALIBAGOR
Alfa Kristanti
2
SMP NEGERI 3 KALIBAGOR
Pelangi Matematika Viandre
Al4-kristanty.blogspot.com
2−3 + 3−2 =
7 7 ×
1
1
1 1
9
8
17
+
=
+
=
+
=
23 32
8 9
72 72
72
14 = 7 7 × 14 = 7 7 × 7 × 2 = 7 49 × 2
= 7 7 2 = 49 2
2
2
1
23 ∶ 13 − 45 =
8
3
5
8
∶3−
5
5
= −
21
21
5
8
3
=3×5−
13
= −
5
4−3
Selisih = 4 + 3 × 560.000 =
21
5
= −2
1
7
3
5
× 560.000
= 80.000
9
Bunga 9 bln = 12 × 12% ×
= 9%
Tabungan setelah 9 bln = 100% T + 9% T = 109% T
109% T = 3.815.000 T =
3.815.000
109%
=
3.815.000
= 3.815.000 ×
109
100
100
109
= 3.500.000
Alfa Kristanti
3
SMP NEGERI 3 KALIBAGOR
b=
b=
7−
7−4
4
=
37−19
3
18
=6
3
Sn = 2 ( 2a + (n – 1)b )
U4 = a + 3b
19 = a + 3(6)
19 = a + 18 a = 19 – 18
a=1
24
2
S24 =
( 2(1) + (24 – 1) 6) = 12 (2 + (23) 6) = 12 (2 + 138)
= 12 (140) = 1.680
Un = a + (n – 1)b
a = U1 = 7
U52 = 7 + (52 – 1) 5 = 7 + (51) 5
b = U2 – U1 = 12 – 7 = 5
= 7 + 255 = 262
r=
2
1
=
9
27
=
1
3
Un = a × r n – 1 = 27 ×
Un = 33 × 3−
a = U1 = 27
+1
−1
1
3
= 33 −
+1
= 33 × 3−1
= 34−
−1
3x – 2 < 8x + 13
3x – 8x < 13 + 2
5x < 15
x> 3
×
×
m=
I. 4x2 – 10x = 2x (2x – 5)
II. 7x2 – 49 = 7(x2 – 7)
III. x2 – 3x – 18 = (x – 6)(x + 3)
IV. x2 + 5x – 36 = (x + 9)(x – 4)
2− 1
2− 1
=
−3 − (−6)
−4 − −1
y – y1 = m (x – x1)
=
y – (1) = 1 (x – (–6))
y + 1 = 1 (x + 6)
y+ 1 = – x 6
−3+6
−4+1
=
3
−3
= −1
y+x=–6–1
x+y=–7
Alfa Kristanti
4
SMP NEGERI 3 KALIBAGOR
ax + by + c = 0
a=8;b=4
m=
−
=
8
−4
= −2
Jumlah bilangan terbesar dan terkecil
2
= × jml ketiga bilangan
3
=
2
3
× 162 = 108
M = { 1, 3, 5, 7 }
N = { 5, 7, 11 }
M N = { 1, 3, 5, 7, 11 }
f(3) = 15
3a + b = 15
3a + b = 15
2a + b = 0
5a = 15
a=3
f(2) = 0
2a + b = 0
4p + 2q = 18.000
4p + 2(1.000) = 18.000
4p + 2.000 = 18.000
4p = 16.000
p = 4.000
20p + 30q = 20(4000)+30(1000)
= 80.000 + 30.000
= 110.000
Misal :
Parkir mobil = p
Parkir motor = q
N
12 12
O
M
1
2
LN =
M
A
o
80
80o
C
38
o
o
62
384
12
B
K
= 32 OL =
2
=
32
2
= 16
2+
2 =
KL =
122 + 162
KL = 144 + 256 = 400 = 20
Keliling = 4× KL = 4 × 20 = 80
× 24 × LN = 384
12 × LN = 384
L
f(5) = 3(5) + ( 6)
= 15 6
= 21
3p + 5q = 17.000 × 4 12p + 20q = 68.000
4p + 2q = 18.000 × 3 12p + 6q = 54.000
14q = 14.000
q = 1.000
Lbelahketupat ABCD = 384
1
× KM × LN = 384
2
K
2a + b = 0
2(3) + b = 0
6+b=0
b=06
b =6
38o
62o
L
Pada gambar di samping:
ABC KLM maka
AB = LM, BC = KL,
AC = KM
Alfa Kristanti
5
SMP NEGERI 3 KALIBAGOR
Q
24
FC =
R
A
=
9
12
30
P
C
18
15
12
24
1
=2 ,
=
15
30
1
=2 ,
=
9
18
1
=2
B
4 × 20 + 6 × 5
4 +6
=
80 + 30
10
=
110
10
= 11
L daerah tdk diarsir
= LKLMN + LPQRS – 2 × L daerah diarsir
= 202 + 10 × 15 – 2 × 67
= 400 + 150 – 134
= 416
Banyak pohon =
=
�
2(42)
3
�
=
�
ℎ
84
3
=
2(24+ 18)
3
= 28
A
B
(3x 5) + (2x + 5) = 90
5x + 0 = 90
5x = 90
x = 18
D
C
Penyiku POR = ROQ = 2x + 5 = 2(18) + 5 = 36 + 5 = 41
Alfa Kristanti
6
SMP NEGERI 3 KALIBAGOR
AOE = 2 × ABE
= 2 × 32
= 64
Sudut keliling menghdp
busur yg sama maka
ABE = ACE = ADE
96
ABE = 3 = 32
luar = 262 − 12 − 2
676 − 100 =
=
LOBC =
=
2
=
676 − 102
576 = 24
×
120
50
× 30
= 72
Panjang rusuk 1 balok = 4 (p + l + t) = 4 (40 + 24 + 36)
= 4 (100) = 400 cm = 4 m
Kawat tersedia = 10 m
10
Banyak kerangka balok yg dibuat = 4 = 2,5 2
Garis pelukis
BC =
1
3
1
3
4
=
72
4
t = 152 − 92 =
t = 144 = 12
2
= 18
225 − 81
t
15
9
Vlimas = La × t = × 18 × 12 = 324 × 4 = 1.296
Alfa Kristanti
7
SMP NEGERI 3 KALIBAGOR
dbola = rusuk kubus
dbola = 30, maka :
30
r = 2 = 15
4
4
4
Vbola = 3 π r3 = 3 π × 153 = 3 π × 3.375
= 4 π × 1.125 = 4.500 π
Ldinding = 2 (pt + lt) = 2(8 × 5 + 6 × 5)
= 2(40 + 30) = 2(70) = 140
Biaya = 140 × Rp 50.000,- = Rp 7.000.000,-
2
8 2
s
s
Lkubus = 6 s2
=6× 8 2
= 6 × 64
= 384
s2 + s 2 = 8 2
2 s2 = 64 × 2
2 s2 = 128
s2 = 64 s = 64 = 8
d = 28 r = 14
22
Ltabung = 2πr (r + t) = 2 × 7 × 14 (14 + 26)
= 88 (40) = 3.520
Jumlah siswa = 18 + 22 = 40 anak
Jml tinggi = 18 × 156 + 22 × 152 = 2.808 + 3.344 = 6.152
Rata-rata tinggi =
��
=
6.152
40
= 153,8
Data
5 6 7 8 9
frekuensi 2 3 4 3 2
Modus
Alfa Kristanti
8
SMP NEGERI 3 KALIBAGOR
32
30
Selisih produksi pupuk Maret dan Mei = 10 – 6 = 4
Alfa Kristanti
9
SMP NEGERI 3 KALIBAGOR
n(S) = 2 × 2 × 2 = 8
3 gambar = GGG
1
P (3 gambar) = 8
6
5
5
4
3
3
2
2
Banyak kelereng = 6 + 5 + 3 + 3 + 2 + 4 + 2 + 5 = 30
Kelereng merah = 6
6
P ( 1 merah) = 30 × 100 % = 20 %
Alfa Kristanti
10