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At a biotech lab, an implementation partner needs a single IPv4 subnet for 252 devices with no extra reserve requirement. Which mask is the smallest address block that can hold those devices?
A /24 block offers 254 usable addresses, meeting the 252-host requirement with the smallest fitting normal IPv4 LAN block. It leaves 2 addresses for immediate growth.
This option fails the smallest-fitting-block criterion or has unsuitable capacity. Compare each prefix using 2^(32-p)-2 rather than treating larger prefix numbers as larger subnets.
A /25 block offers only 126 usable hosts, less than the 252 required. Using the next longer mask would exhaust host capacity.
A /23 block is larger than necessary for 252 endpoints; it consumes extra addresses even though a /24 block meets the stated limit.
At a distribution center, the migration lead must validate address pairs on LAN 192.168.228.64/26 before assigning endpoints. Which statement is correct?
192.168.228.65 and 192.168.228.126 are both strictly within 192.168.228.64/26, so both are usable and on the same on-link prefix. Their computed masked network addresses match 192.168.228.64.
192.168.228.127 has all host bits one and addresses the subnet broadcast. A host normally uses an interior address.
192.168.228.64 has all host bits zero and is the subnet identifier. It is not ordinarily assigned to a host interface.
192.168.228.128 is in the next subnet after 192.168.228.127. A shared first three octets does not prove same-subnet membership when the mask is /26.
At an enterprise data center, the campus network lead must document the network address containing host 10.101.118.149/29. Which address is that network ID?
Masking 10.101.118.149 with /29 yields network 10.101.118.144; the subnet block contains 8 addresses through 10.101.118.151. That is the correct network ID for an SVI or static route.
10.101.118.151 is the last address in the block, not its network address. A broadcast address has host bits set to one.
This address identifies the adjacent lower block, not the /29 network containing 10.101.118.149. A network ID is obtained by zeroing all 3 host bits.
This is the next subnet after 10.101.118.144/29; the given host 10.101.118.149 resides in the preceding block. The prefix boundary is fixed by 255.255.255.248.
At a payment processor, an implementation partner reviews host 10.125.84.86/25. Which IPv4 address is the directed broadcast of that subnet?
For 10.125.84.0/25, all 7 host bits set to one yield broadcast 10.125.84.127. The final usable host, if any, is one address before it.
10.125.84.128 begins the next subnet block. It cannot be the broadcast for the subnet containing 10.125.84.86.
10.125.84.126 is the highest usable host in the /25 block, not the directed broadcast. It lacks the all-ones host pattern.
10.125.84.0 has host bits set to zero, so it is the network ID rather than broadcast. The block ends at 10.125.84.127.
At a logistics provider, the site engineer is planning an ordinary IPv4 LAN with mask /25. How many addresses can be assigned to hosts, excluding subnet and directed broadcast?
A /25 subnet has 7 host bits, or 128 total IPv4 addresses; excluding network and broadcast leaves 126 usable host addresses. This is the traditional LAN-host count.
128 includes the network and directed broadcast addresses. Ordinary IPv4 LAN host assignment excludes both endpoints.
64 corresponds to splitting address space rather than counting both usable host endpoints in a /25 subnet. The actual block has 128 total addresses.
This count does not follow the /25 host-bit calculation; the correct operation is 2^(32-25) - 2 = 126. Subnet capacity should be validated against required growth.
At an airport terminal, the campus network lead needs a valid unicast gateway address in subnet 172.20.89.192/27. Which proposed IPv4 address is usable without using its network or broadcast address?
172.20.89.218 falls strictly between 172.20.89.192 and 172.20.89.223; it is a valid host inside 172.20.89.192/27. That meets the requested gateway address constraint.
172.20.89.223 is the all-one host-bit directed broadcast for 172.20.89.192/27; using it as the gateway would conflict with broadcast behavior.
172.20.89.192 is the all-zero host-bit network address for 172.20.89.192/27, not a normal interface IP on this LAN.
172.20.89.224 is beyond the upper boundary of 172.20.89.192/27. Even though it looks adjacent, it belongs to the next subnet.
At a satellite office, the site engineer is configuring a Cisco interface in a /30 IPv4 network. Which dotted-decimal mask matches the given prefix?
Prefix /30 sets 30 leading one bits, producing dotted mask 255.255.255.252. The remaining 2 bits define host numbering.
Mask 255.255.254.0 represents /23, not /30. Counting contiguous one bits gives the distinction.
Mask 255.255.255.240 represents /28, not /30. Counting contiguous one bits gives the distinction.
Mask 255.255.255.224 represents /27, not /30. Counting contiguous one bits gives the distinction.
At a customer support center, the security and networking lead needs a single IPv4 subnet for 12 devices with no extra reserve requirement. Which mask is the smallest address block that can hold those devices?
A /28 block offers 14 usable addresses, meeting the 12-host requirement with the smallest fitting normal IPv4 LAN block. It leaves 2 addresses for immediate growth.
A /29 block offers only 6 usable hosts, less than the 12 required. Using the next longer mask would exhaust host capacity.
This option fails the smallest-fitting-block criterion or has unsuitable capacity. Compare each prefix using 2^(32-p)-2 rather than treating larger prefix numbers as larger subnets.
A /27 block is larger than necessary for 12 endpoints; it consumes extra addresses even though a /28 block meets the stated limit.
At a supply depot, the branch support lead must validate address pairs on LAN 192.168.37.0/25 before assigning endpoints. Which statement is correct?
192.168.37.1 and 192.168.37.126 are both strictly within 192.168.37.0/25, so both are usable and on the same on-link prefix. Their computed masked network addresses match 192.168.37.0.
192.168.37.0 has all host bits zero and is the subnet identifier. It is not ordinarily assigned to a host interface.
192.168.37.127 has all host bits one and addresses the subnet broadcast. A host normally uses an interior address.
192.168.37.128 is in the next subnet after 192.168.37.127. A shared first three octets does not prove same-subnet membership when the mask is /25.
At a public library, the security and networking lead must document the network address containing host 10.167.233.236/29. Which address is that network ID?
Masking 10.167.233.236 with /29 yields network 10.167.233.232; the subnet block contains 8 addresses through 10.167.233.239. That is the correct network ID for an SVI or static route.
This is the next subnet after 10.167.233.232/29; the given host 10.167.233.236 resides in the preceding block. The prefix boundary is fixed by 255.255.255.248.
This address identifies the adjacent lower block, not the /29 network containing 10.167.233.236. A network ID is obtained by zeroing all 3 host bits.
10.167.233.239 is the last address in the block, not its network address. A broadcast address has host bits set to one.
At a field-service firm, the migration lead reviews host 10.101.49.37/23. Which IPv4 address is the directed broadcast of that subnet?
For 10.101.48.0/23, all 9 host bits set to one yield broadcast 10.101.49.255. The final usable host, if any, is one address before it.
10.101.48.0 has host bits set to zero, so it is the network ID rather than broadcast. The block ends at 10.101.49.255.
10.101.49.254 is the highest usable host in the /23 block, not the directed broadcast. It lacks the all-ones host pattern.
10.101.50.0 begins the next subnet block. It cannot be the broadcast for the subnet containing 10.101.49.37.
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