Pembahasan SNMPTN 2012 Mat Dasar Kode 322 jogjastudent.com
322
PEMBAHASAN SELEKSI NASIONAL MASUK PERGURUAN TINGGI NEGERI (SNMPTN)
Mata Pelajaran : Matematika Dasar
Tanggal
: 12 juni 2012
Kode Soal
: 322
1. Jawab: D
Bilangan Real
a b = 220 − 219 = 219 (2 − 1) = 219
a = 2 dan b = 19 , maka a = b = 21
2. Jawab: E
Statistika
f X +f X
X= 1 1 2 2
f1 + f2
X 1 = 7,0
f1 = 4
28 + 4 X 2
8
64 = 28 + 4 X 2
8=
X = 8,0
f1 + f2 = 8
X2 = 9
X 2 − X 1 = 2,0
3. Jawab: D
Persamaan Kuadrat
x2 − 4x + a = 0
x1 = p + 1 ; x 2 = p − 1
x1 + x2 = 2 p = 4
p=2
x1 ⋅ x 2 = a
(3)(1) = a
(p + 1)(p − 1) = a
a=3
6. Jawab: C
Persamaan Linier
Modal A = x rupiah
Modal B = 10.000.000 – x
Keuntungan A + Keuntungan B = 904.000
0,08 x + 0,10 (10.000.000 − x ) = 904.000
0,08 x + 10.000.000 − 0,10 x = 904.000
0,02 x = 96.000
x = 4.800.000
7. Jawab: A
Statistika
fi = 25
f8 = 21 − 19 = 3
3
Persentase =
(100% ) = 12%
25
8. Jawab: B
Program Linier
f (x , y ) = 4 x + 3y
2 x + 3y ≤ 18 ; x ≥ 3 ; y ≥ 2
y
6
4. Jawab:
2x − y = 6
C
2y + 3z = 4
3x − z = 8
5x + y + 2z = 18
+
5. Jawab: D
Barisan dan Deret
Tiga bilangan barisan aritmetika dengan beda 6
Misal ( x − 6 ), (x ), (x + 6 )
(x − 6 ), (x ), (x + 6 + 12) barisan geometri
2
x = (x − 6 )(x + 18 )
x 2 = x 2 + 12 x − 6 (18 )
12x = 6 (18 )
x =9
Jumlah ketiga bilangan = 3x = 27
2
A
B
x
0
3
9
A (3, 2) → f (3, 2) = 12
B (6, 2) → f (6, 2) = 30
C (3, 4 ) → f (3, 4 ) = 24
Nilai maksimum = 30
Halaman 1
322
9. Jawab: A
Fungsi
f ( x ) = ax + 3
Luas tiap persegi panjang =
AB =
f ( f (x )) = 4 x + 9
f (ax + 3) = 4 x + 9
a 2 x + 3a + 3 = 4 x + 9
a 2 + 3a + 3 = 4 + 9 = 13
[
log 14 = 3 log (7)(2) = 3 log 7+ 3 log 2
14. Jawab: E
Matriks dan Determinan
2 0
; det( A ) = 2
A⋅B =
0 2
det( A ) ⋅ det(B ) = (2)(2) − (0 )(0 )
1 xy + 1
=y+ =
x
xy
11. Jawab: E
Fungsi Kuadrat
Fungsi kuadrat melalui (− 1, 0 ) ; (2, 0 )
y = a ( x + 1)(x − 2)
dan (0, 2)
2 = a (0 + 1)(0 − 2)
melalui (0, 2)
2 = −2a
a = −1
y = f (x ) = −( x + 1)(x − 2)
f (7) = −(7 + 1)(7 − 2) = −40
2 det(B ) = 4
(
det B ⋅ A
1
4
−1
det(B ) = 2
) = det(B)⋅ det(A )
−1
= det(B )
(x + 1)(x + 2) − (x + 2) ≥ 0
(x + 2)[(x + 1) − 1] ≥ 0
(x + 2)(x ) ≥ 0
1
2
+
−
−2
A
1
1
=2
=1
det( A )
2
15. Jawab: C
Pertidaksamaan
(x + 1)(x + 2) ≥ (x + 2)
12. Jawab: B
Bidang Datar
4
5
]
= 5(5)2 − 6(5) − 5(4 )2 − 6(4 )
= 95 − 56
= 39
10. Jawab: D
Logaritma
2
log 3 = x ; 3 log 7 = y
1
4
1
2
13. Jawab: C
Barisan dan Deret
S n = S n 2 − 6n
U5 = S 5 − S 4
a (ax + 3) + 3 = 4 x + 9
3
1
5
B
+
0
x ≤ −2 atau x ≥ 0
1
5
1
Halaman 2
PEMBAHASAN SELEKSI NASIONAL MASUK PERGURUAN TINGGI NEGERI (SNMPTN)
Mata Pelajaran : Matematika Dasar
Tanggal
: 12 juni 2012
Kode Soal
: 322
1. Jawab: D
Bilangan Real
a b = 220 − 219 = 219 (2 − 1) = 219
a = 2 dan b = 19 , maka a = b = 21
2. Jawab: E
Statistika
f X +f X
X= 1 1 2 2
f1 + f2
X 1 = 7,0
f1 = 4
28 + 4 X 2
8
64 = 28 + 4 X 2
8=
X = 8,0
f1 + f2 = 8
X2 = 9
X 2 − X 1 = 2,0
3. Jawab: D
Persamaan Kuadrat
x2 − 4x + a = 0
x1 = p + 1 ; x 2 = p − 1
x1 + x2 = 2 p = 4
p=2
x1 ⋅ x 2 = a
(3)(1) = a
(p + 1)(p − 1) = a
a=3
6. Jawab: C
Persamaan Linier
Modal A = x rupiah
Modal B = 10.000.000 – x
Keuntungan A + Keuntungan B = 904.000
0,08 x + 0,10 (10.000.000 − x ) = 904.000
0,08 x + 10.000.000 − 0,10 x = 904.000
0,02 x = 96.000
x = 4.800.000
7. Jawab: A
Statistika
fi = 25
f8 = 21 − 19 = 3
3
Persentase =
(100% ) = 12%
25
8. Jawab: B
Program Linier
f (x , y ) = 4 x + 3y
2 x + 3y ≤ 18 ; x ≥ 3 ; y ≥ 2
y
6
4. Jawab:
2x − y = 6
C
2y + 3z = 4
3x − z = 8
5x + y + 2z = 18
+
5. Jawab: D
Barisan dan Deret
Tiga bilangan barisan aritmetika dengan beda 6
Misal ( x − 6 ), (x ), (x + 6 )
(x − 6 ), (x ), (x + 6 + 12) barisan geometri
2
x = (x − 6 )(x + 18 )
x 2 = x 2 + 12 x − 6 (18 )
12x = 6 (18 )
x =9
Jumlah ketiga bilangan = 3x = 27
2
A
B
x
0
3
9
A (3, 2) → f (3, 2) = 12
B (6, 2) → f (6, 2) = 30
C (3, 4 ) → f (3, 4 ) = 24
Nilai maksimum = 30
Halaman 1
322
9. Jawab: A
Fungsi
f ( x ) = ax + 3
Luas tiap persegi panjang =
AB =
f ( f (x )) = 4 x + 9
f (ax + 3) = 4 x + 9
a 2 x + 3a + 3 = 4 x + 9
a 2 + 3a + 3 = 4 + 9 = 13
[
log 14 = 3 log (7)(2) = 3 log 7+ 3 log 2
14. Jawab: E
Matriks dan Determinan
2 0
; det( A ) = 2
A⋅B =
0 2
det( A ) ⋅ det(B ) = (2)(2) − (0 )(0 )
1 xy + 1
=y+ =
x
xy
11. Jawab: E
Fungsi Kuadrat
Fungsi kuadrat melalui (− 1, 0 ) ; (2, 0 )
y = a ( x + 1)(x − 2)
dan (0, 2)
2 = a (0 + 1)(0 − 2)
melalui (0, 2)
2 = −2a
a = −1
y = f (x ) = −( x + 1)(x − 2)
f (7) = −(7 + 1)(7 − 2) = −40
2 det(B ) = 4
(
det B ⋅ A
1
4
−1
det(B ) = 2
) = det(B)⋅ det(A )
−1
= det(B )
(x + 1)(x + 2) − (x + 2) ≥ 0
(x + 2)[(x + 1) − 1] ≥ 0
(x + 2)(x ) ≥ 0
1
2
+
−
−2
A
1
1
=2
=1
det( A )
2
15. Jawab: C
Pertidaksamaan
(x + 1)(x + 2) ≥ (x + 2)
12. Jawab: B
Bidang Datar
4
5
]
= 5(5)2 − 6(5) − 5(4 )2 − 6(4 )
= 95 − 56
= 39
10. Jawab: D
Logaritma
2
log 3 = x ; 3 log 7 = y
1
4
1
2
13. Jawab: C
Barisan dan Deret
S n = S n 2 − 6n
U5 = S 5 − S 4
a (ax + 3) + 3 = 4 x + 9
3
1
5
B
+
0
x ≤ −2 atau x ≥ 0
1
5
1
Halaman 2