Mat Dasar Kode 323 jogjastudent.com

323
PEMBAHASAN SELEKSI NASIONAL MASUK PERGURUAN TINGGI NEGERI (SNMPTN)
Mata Pelajaran : Matematika Dasar
Tanggal
: 12 Juni 2012
Kode Soal
: 323

1. Jawab: D
Eksponen
a b = 2020 − 219 = 219 2 − 1 = 219
a=2
b = 19
a + b = 21
2. Jawab: A
Sistem Persamaan Linear
2x + y = 3
2y + z = 5
2z + x = 7 +

5. Jawab: E

Fungsi Kuadrat
2

Puncak (α β ) → y − β = a ( x − α )
2

y + 1 = a ( x − 2)

Puncak ( 2 −1)

2

−5 + 1 = a ( 0 − 2 )

Melalui ( 0 −5)

−4 = 4a
a = −1
2


y + 1 = −1 ( x − 2 )

y + 1 = − ( x2 − 4 x + 4 )

3x + 3y + 3z = 15

y = −x2 + 4 x − 5

x+y+z =5

f x = − x2 + 4 x − 5
f 5 = −25 + 20 − 5 = −10

3. Jawab: D
Persamaan Kuadrat
x1 = p − 1
x2 = p + 1
x2 − 4 x + a = 0
b
x1 + x2 = −

a
2p = 4 → p = 2
x1 = 1
x2 = 3
x1 x2 = a
1⋅3 = a → a = 3
4. Jawab: B
Logaritma
3
a = 2 → a = 32 = 9

(

2

a )( a

b) = 2

b = 2 → b = 22 = 4

a + b = 13

6. Jawab: E
Matriks
a b 1 2
1 1
=
c d 1 5
3 2

a b
1 1
1
=
1⋅ 5 − 2⋅1 3 2
c d
=

5 −2
−1 1


1 5 − 1 −2 + 1
3 15 − 2 −6 + 2

4
3
=
13
3

1
3
4

3


a+b+c+d =

2


7. Jawab: A
Statistika
fk 7 = 19
fk 8 − fk 7
⋅ 100
fk

4 1 13 4
− +
− =4
3 3 3 3

fk 8 = 22

fk = 25

3
⋅ 100
25

= 12%
=

Halaman 1

323
8. Jawab: C
Deret Aritmatika
a = 1001
b = 997 − 1001 = −4
Un < 0

11. Jawab: B
Statistika
n1 = 9
n2 = 1

x 1 = 61
x2 =


a + ( n − 1) b < 0

x = 63 5

1001 + ( n − 1)( −4 ) < 0

x=

1001 − 4n + 4 < 0
−4n < −1005
4n > 1005
n > 251 14
n = 252

9 ⋅ 61 + x 2
10
635 = 549 + x2
63 5 =

x2 = 86


9. Jawab: E
Fungsi
f x = 2x + b

12. Jawab: D
Aritmetika Sosial
Sewa = 15.000.000
Banyak barang = 3.000
Bahan baku = 2.000
Biaya bahan baku total = 3.000 x 2.000
= 6.000.000
Modal = biaya + bahan baku
= 15.000.000 + 6.000.000
= 21.000.000
Harga penjualan minimum agar tidak rugi
21 000 000
=
= 7 000
3 000


f −1 ( 9 ) = 3

maka f(3) = 3
6+b=9
b=3
f x = 2x + 3
f (f 3

n1 x 1 + n2 x 2
n1 + n2

) = f ( 9 ) = 21

10. Jawab: D
Program Linear
y

13. Jawab: A
Deret Geometri

Sn = 3 ( 2 − 21−n )

8

a = S1 = 3 ( 2 − 20 ) = 3

6

1
2
a + 2r = 3 + 1 = 4
r = 2−1 =

4

3

6

12

f ( x y ) = 3x + 4 y
f ( 3 6 ) = 9 + 24 = 33
f ( 6 4 ) = 18 + 16 = 34
f

= 33

Halaman 2

323
14. Jawab: B
Bidang Datar

L=

1
5

1
1
→y=
5
5x
2y = 1
2
2
=1→ x =
5x
5
xy =

15. Jawab: C
Pertidaksamaan
( x + 5)( x − 3) < x − 3
x 2 + 2 x − 15 < x − 3
x 2 + x − 12 < 0
( x + 4 )( x − 3) < 0
+



+

3
−4
−4 < x < 3

Halaman 3