pembahasan soal un matematika smp tahun 2014 paket 1 (m2suidhat.blogspot.com)
TRANSCRIPT
http://m2suidhat.blogspot.com/
http://m2suidhat.blogspot.com/
Mathematics Sport
πππ
π+ π
π
π β π
π
π
ππ
π+
π
π β
ππ
π
= πππ
ππ+
ππ
ππ β
πππ
ππ
= πππ
ππ β
πππ
ππ
= πππ
ππ
= 8ππ
ππ
36 8
x 24
π
ππ=
π
ππ x = ππ Γ
π
ππ x =
ππ
π
x = 12 buah
πππππ = ππ
ππ = ππ = ππ
ππ : π ππΓ π : π
= ππΓ π : π
= ππ
= π Γ π
= 2 π
ππ : π ππ:π
= ππ
= π Γ π
= 2 π
atau
Paket 1
http://m2suidhat.blogspot.com/
π
π
π
π Γ
π
π =
π π
π
(i) Besar bunga = 920.000 β 800.000 = 120.000
(ii) Besar bunga = πππππ
ππΓ 9% Γ 800.000
Sehingga: 120.000 = πππππ
ππ Γ
π
πππ Γ 800.000 120 =
πππππ
πΓ 3 Γ 8
bulan = πππ
πΓπ = 20 bulan
b = πΌπβπΌπ
πβπ =
ππβπ
π =
ππ
π = 5
U2 = a + (2-1)5
6 = a + 5
a = 1
U40 = a + (40 β 1)5
U40 = 1 + (39)5
U40 = 1 + 195
U40 =196
b = πΌπβπΌπ
πβπ =
ππβππ
π =
ππ
π = 5
U3 = a + (3-1)5
18 = a + 10
a = 8
S24 = ππ
π(2Γ8 + (24 β 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 = ππ
π(2Γ3.000.000 + (9)500.000)
S10 = 5(6.000.000 + 4.500.000)
S10 = 5(10.500.000)
S10 = 52.500.000
(i) 15x2y β 20xy
2 = 5xy(3x β 4y)
(ii) p2 β 16 = (p + 4)(p β 4)
(iii) 3a2 + 8a β 3 = (3a β 1)(a + 3)
Paket 1
http://m2suidhat.blogspot.com/
3x + 6 = 5x + 20
6 β 20 = 5x β 3x
β 14 = 2x
β 7 = x
x = β 7
K = 2(p + l)
44 = 2(p + l)
22 = (p + l)
22 = (3x + 4) + (2x + 3)
22 = 5x + 7
22 β 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
-12
= 1
1
40 β (11 + 12) = 40 β 23 = 17
Menulis cerpen = 17 + 12 = 29
f(x) = 8 β 2x f(k) = 8 β 2k
β10 = 8 β 2k
β18 = β2k
9 = k
k = 9
Jadi, x + 12 = β 7 + 12 = 5
Paket 1
http://m2suidhat.blogspot.com/
m1 = πβπ
π+π = β 5
m1Γm2 = β1
β5 Γ m2 = β1
m2 = π
π
m = 2
Memotong sumbu-y di titik (0, 2)
m = βπ
π
Memotong sumbu-y bukan di titik (0, 2) m =
π
π = 2
Memotong sumbu-y bukan di titik (0, 2)
m = π
π = 1
Memotong sumbu-y bukan di titik (0, 2)
m = π
π = 2
Memotong sumbu-y di titik (0, 2)
A. m2 = 5
B. m2 = π
π
C. m2 = β 5
D. m2 = β 5
m1 = βππβπ
βπβπ =
βππ
βπ = 2
y β y1 = m(x β x1), (ambil x1 = 3 dan y1 = 1)
y β 1 = 2(x β 3)
y β 1 = 2x β 6
y = 2x β 5
y = 2x β 5 , titik A (10, p)
p = 2(10) β 5
p = 20 β 5
p = 15
x β 3y β 5 = 0 Γ2 2x β 6y β 10 = 0
2x β 5y β 9 = 0 Γ1 2x β 5y β 9 = 0
βy β 1 = 0
y = β1 sehingga x β 3y β 5 = 0
x β 3(β1) β 5 = 0
x + 3 β 5 = 0
x β 2 = 0
x = 2
Nilai 3x + 2y = 3(2) + 2(β1)
= 6 β 2
= 4
Paket 1
http://m2suidhat.blogspot.com/
5a + 3j = 79.000 Γ2 10a + 6j = 158.000
3a + 2j = 49.000 Γ3 9a + 6j = 147.000
a = 11.000
Panjang tali = ππππ + ππππ = π Γ ππππ =πππ π
Panjang perkiraan dengan π π,ππ adalah 150 Γ 1,41 = 211,5 212
K L
N M
B
C
D
A
K L
N M
B
CD
A cm8
cm8
cm10
cm10
cm10
Luas daerah yang diarsir = π
π Γ Luas persegi KLMN
= π
π Γ (8)
2
= 16 cm2
Paket 1
http://m2suidhat.blogspot. com/
(4 + 4 + 6 + 6) Γ 2 = (20) Γ 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
π»πΉ
πΈπΉ=
πΉπΊ
π·πΉ
π»πΉ
ππ=
π
ππ
TR = ππΓπ
ππ
TR = 4 cm
Jadi, PT = PR β TR
= 15 β 4
PT = 11 cm
Paket 1
http://m2suidhat.blogspot.com/
D
A
E
B
C
E
x
x
π¨π«
π©πͺ=π¨π¬
π¬πͺ=π«π¬
π©π¬
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 = πππ¨
ππππ¨ Γ 2 r =
π
π Γ 2 Γ
ππ
π Γ 14
= π
π Γ 44 Γ 2
= ππ
π
= 17, 6 cm
Paket 1
http://m2suidhat.blogspot.com/
d2 = p
2 β (r1 + r2)
2 p
2 = d
2 + (r1 + r2)
2
= 162 + (8 + 4)
2
= 256 + (12)2
= 256 + 144
= 400
p = 20
Ada 18 (6Γ3) rusuk dan 8 (6 + 2) sisi
V = La Γ t
= ππ+ππ
πΓ ππ Γ 20
= (30 Γ 5) Γ 20
= 150 Γ 20
V = 3000
Dengan menggunakan teorema pythagoras
didapat panajang TE = 10 cm
L = luas alas + 4 Γ luas sisi tegak
= 122 + 4 Γ
π
π Γ 12 Γ 10
= 144 + 240
L = 384 cm2
12 cm
8 cm
E
O
Paket 1
http://m2suidhat.blogspot.com/
LP = Luas tabung tanpa tutup + luas setengah bola
= (Οr2 + 2Οrt) + 2Οr
2
= (ππ
πΓ7
2 + 2Γ
ππ
πΓ7Γ20) + 2Γ
ππ
πΓ7
2
= (22Γ7 + 2Γ22Γ20) + 2Γ22Γ7
LP = 1.342
4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10
Median = π+π
π = 6,5
Bisa jadi, rata-ratanya = 130
Bisa jadi, rata-ratanya = 130
Bisa jadi, rata-ratanya = 130
Bisa dipastikan, rata-ratanya = 130
Rata-rata = ππΓπππ+πππ+πππ
ππ+π+π =
ππππ
ππ = 130
180 130 525 640 360 200 2035
30 Rata-rata =
ππππ
ππ = 67, 83
Banyak siswa yang nilainya lebih dari rata-rata = 30 β 9 = 21
Paket 1
http://m2suidhat.blogspot.com/
Senin β Selasa dan Kamis β Jumβat penurunan 1000
Selasa β Rabu kenaikan 2000
Rabu β Kamis dan Jumβat β Sabtu uang saku tetap
Senin β Selasa penurunan 1000
Nomor bola kurang dari 5 1, 2, 3, 4 ada 4
P = πππππ ππππ ππππππ π πππ π
ππππππ πππππ ππππ =
π
π
Disusun oleh : Mohammad Tohir
Jika ada saran, kritik maupun masukan
silahkan kirim ke- My email: [email protected]
Terima kasih.
My blog : http://m2suidhat.blogspot.com/
http://matematohir.wordpress.com/
Paket 1