A. 1110110 = ………….(10)
Jawab : 1024 512 256 128 64 32 16 8 4 2 1
1 1 1 0 1 1 0
= 64+32+16+4+2
= 118..(10)
B. 10111011 = …………..(10)
Jawab : 1024 512 256 128 64 32 16 8 4 2 1
1 0 1 1 1 0 1 1
= 128+32+16+8+2+1
= 187..(10)
C. 110010 1 =………….(10)
Jawab : 1024 512 256 128 64 32 16 8 4 2 1
1 1 0 0 1 0 1
= 64+32+4+1
= 101..(10)
D. 215…(8) = …………….(10)
Jawab :
2 1 5
010 001 101
= 128 + 8 +4+1
= 141 …(10)
E. 765…(8) =…………(16)
Jawab : 1024 512 256 128 64 32 16 8 4 2 1
0 1 0 1
1 1 1 1
0 0 0 1
7 6 5 = 111110101
= 4 + 1 = 5
111 110 101
= 8+4+2+1
= 15 = 1
765…(8)= 1F5……….(16)
F. 457…(8) = ……………..(2)
Jawab :
4 5 7
011 101 111
= 011101111….(2)
G. AD8….(16)=…….(10)
Jawab : A D 8
1010 1110 1000
=
2048 1024 512 256 128 64 32 16 8 4 2 1
1 0 1 0 1 1 1 0 1 0 0 0
= 2776 (10)
H. F4AB … (16)=….. (10) 16 8 4 2 1
Jawab : 0 1 0 1 1
65536 32768 16384 8192 4096 2048 1024 512 256 128 64 32 1 1 1 1 0 1 0 0 1 0 1
1111010010101011….(10)
= 62635…(10)
I. BDF…(16)=….. (2)
Jawab : B D F
11 13 15
= 111315… (10)
DIBAGI 2 HINGGA NILAI DESIMAL HABIS
SEHINGGA NILAI YANG DIDAPAT MENJADI NILAI BINNER
= 101111011111…(2)
J. CD8…. (16)= ….(2)
Jawab : C D 8
12 13 8
= 12138…(10)
DIBAGI 2 HINGGA NILAI DESIMAL HABIS
SEHINGGA NILAI YANG DIDAPAT MENJADI NILAI BINNER
= 110011011000
2.
a. 192.168.1.64/24
jawab:
SM= 255.255.255.0
IPN=192.168.1.0
IPB=192.168.1.255
RI= 192.168.1.1 S/D 192.168.1.254
B. . 192.168.1.80 /24
jawab:
SM= 255.255.255.0
IPN=192.168.1.0
IPB=192.168.1.255
RI= 192.168.1.1 S/D 192.168.1.254
C. 192.168.1.212 /24
jawab:
SM= 255.255.255.0
IPN=192.168.1.0
IPB=192.168.1.255
RI= 192.168.1.1 S/D 192.168.1.254
D. 192.168.1.128/24
jawab:
SM= 255.255.255.0
IPN=192.168.1.0
IPB=192.168.1.255
RI= 192.168.1.1 S/D 192.168.1.254
. 192.168.1.192/24
jawab:
SM= 255.255.255.0
IPN=192.168.1.0
IPB=192.168.1.255
RI= 192.168.1.1 S/D 192.168.1.254
