IPv4 Addressing

Correct Answer: 10.0.5.25

The subnet 10.0.4.0/22 has a subnet mask of 255.255.252.0. This means the first 22 bits are fixed and are used for network addressing, while the last 10 bits are used for hosts within that subnet. The /22 translates to 4 contiguous 1's in the subnet mask's 3rd octet which allows for the 3rd octet ranges from 4 to 7 (4 + 0 to 3, since 2^2=4 and 0-3 is a set of 4). Thus, valid addresses within this subnet range from 10.0.4.1 to 10.0.7.254. The IP address 10.0.5.25 falls within this range and is therefore a correct choice. The other IP addresses given are not within the correct subnet: - 10.0.8.5 is out of range as the third octet, 8, is higher than the 4-7 range allowed by the subnet mask. - 10.0.3.255 is below the starting address of the subnet 10.0.4.0. - 10.0.11.10 is well beyond the range of the 3rd octet permitted by the /22 subnet mask.

Correct Answer: 255.255.255.192

Given a /24 subnet, breaking down into four equal subnets means the subnet will be broken down to /26. Each of these four subnet has 64 addresses. In decimal, the new subnet mask for a /26 network is 255.255.255.192. Next, the subnet mask 255.255.255.224 represents a /27 network (32 addresses). The subnet mask 255.255.255.240 represents a /28 network (16 addresses) which makes for a much smaller network than the required. Lastly, the subnet mask 255.255.255.128 represents a /25 network (128 addresses) which is only half of the /24 network but not broken down into four. Therefore, the correct choice is 255.255.255.192 which correctly subnets /24 into four /26 equal subnets.

Correct Answer: 192.168.0.127

The network 192.168.0.0 with a subnet mask of /25, or 255.255.255.128, has the following characteristics: The subnet mask indicates that the first 25 bits are the network bits and the remaining 7 bits are used for host addresses. Thus, the network can support 2^7 - 2 = 126 usable IP addresses (subtracting 2 for the network and broadcast addresses). The network address is 192.168.0.0 and it ends with the broadcast address before the next subnet begins. Since the subnet has 128 addresses total (0 to 127), the broadcast address for this subnet is the last address in the range, which is 192.168.0.127. The other answers are incorrect as they do not align with this subnet's address range: 192.168.0.128 is the first address of the next subnet, 192.168.0.255 is the last address of an entire /24 subnet, and 192.168.0.64 is an address within the range of the given /25 subnet, not a broadcast address.