A. I and iii
B. Ii and iv
C. I, ii and iv
D. Ii, iii and iv
A. 14
B. 15
C. 16
D. 30
Explanation: A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets, each with 30 hosts. Does it matter if this mask is used with a Class A, B, or C network address? Not at all. The number of host bits would never change.
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.255.240
D. 255.255.255.248
Explanation: You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts which is less than 15, so this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts so this may work. The mask 255.255.255.192 provides 4 subnets, each with 60 hosts so this may work. Comparing both the possible masks, 255.255.255.224 provides the best answer.
A. 2
B. 3
C. 4
D. 5
Explanation: A 240 mask is 4 subnet bits and provides 16 subnets, each with 14 hosts. We need more subnets, so letβs add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 host bits (6 hosts per subnet). This is the best answer.
A. 172.16.112.0
B. 172.16.0.0
C. 172.16.96.0
D. 172.16.255.0
Explanation: A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on-which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet
A. 6
B. 8
C. 30
D. 32
Explanation: A /29 (255.255.255.248), regardless of the class of address, has only 3 host bits. Six hosts are the maximum number of hosts on this LAN, including the router interface. Out of the 8 addresses possible with the host bits, the first and the last address are for the subnet id and broadcast address respectively.
A. 172.16.36.0
B. 172.16.48.0
C. 172.16.64.0
D. 172.16.0.0
Explanation: A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.
A. 7 subnets, 30 hosts each
B. 8 subnets, 8,190 hosts each
C. 8 subnets, 2,046 hosts each
D. 7 subnets, 2,046 hosts each
Explanation: A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.