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At a satellite office, the migration lead is configuring a Cisco interface in a /29 IPv4 network. Which dotted-decimal mask matches the given prefix?
Prefix /29 sets 29 leading one bits, producing dotted mask 255.255.255.248. The remaining 3 bits define host numbering.
Mask 255.255.240.0 represents /20, not /29. Counting contiguous one bits gives the distinction.
Mask 255.255.255.128 represents /25, not /29. Counting contiguous one bits gives the distinction.
Mask 255.255.254.0 represents /23, not /29. Counting contiguous one bits gives the distinction.
At a nonprofit organization, the infrastructure lead needs a single IPv4 subnet for 5 devices with no extra reserve requirement. Which mask is the smallest address block that can hold those devices?
A /29 block offers 6 usable addresses, meeting the 5-host requirement with the smallest fitting normal IPv4 LAN block. It leaves 1 addresses for immediate growth.
A /28 block is larger than necessary for 5 endpoints; it consumes extra addresses even though a /29 block meets the stated limit.
A /30 block offers only 2 usable hosts, less than the 5 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.
At a regional office, the migration lead must validate address pairs on LAN 192.168.195.128/25 before assigning endpoints. Which statement is correct?
192.168.195.129 and 192.168.195.254 are both strictly within 192.168.195.128/25, so both are usable and on the same on-link prefix. Their computed masked network addresses match 192.168.195.128.
192.168.196.0 is in the next subnet after 192.168.195.255. A shared first three octets does not prove same-subnet membership when the mask is /25.
192.168.195.128 has all host bits zero and is the subnet identifier. It is not ordinarily assigned to a host interface.
192.168.195.255 has all host bits one and addresses the subnet broadcast. A host normally uses an interior address.
At a biotech lab, the branch support lead must document the network address containing host 10.101.16.191/25. Which address is that network ID?
Masking 10.101.16.191 with /25 yields network 10.101.16.128; the subnet block contains 128 addresses through 10.101.16.255. That is the correct network ID for an SVI or static route.
This is the next subnet after 10.101.16.128/25; the given host 10.101.16.191 resides in the preceding block. The prefix boundary is fixed by 255.255.255.128.
10.101.16.255 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 /25 network containing 10.101.16.191. A network ID is obtained by zeroing all 7 host bits.
At a warehouse, the migration lead reviews host 10.56.84.130/23. Which IPv4 address is the directed broadcast of that subnet?
For 10.56.84.0/23, all 9 host bits set to one yield broadcast 10.56.85.255. The final usable host, if any, is one address before it.
10.56.84.0 has host bits set to zero, so it is the network ID rather than broadcast. The block ends at 10.56.85.255.
10.56.86.0 begins the next subnet block. It cannot be the broadcast for the subnet containing 10.56.84.130.
10.56.85.254 is the highest usable host in the /23 block, not the directed broadcast. It lacks the all-ones host pattern.
At a retail branch, the security and networking lead is planning an ordinary IPv4 LAN with mask /28. How many addresses can be assigned to hosts, excluding subnet and directed broadcast?
A /28 subnet has 4 host bits, or 16 total IPv4 addresses; excluding network and broadcast leaves 14 usable host addresses. This is the traditional LAN-host count.
8 corresponds to splitting address space rather than counting both usable host endpoints in a /28 subnet. The actual block has 16 total addresses.
16 includes the network and directed broadcast addresses. Ordinary IPv4 LAN host assignment excludes both endpoints.
This count does not follow the /28 host-bit calculation; the correct operation is 2^(32-28) - 2 = 14. Subnet capacity should be validated against required growth.
At a manufacturing plant, the infrastructure lead needs a valid unicast gateway address in subnet 172.20.185.144/29. Which proposed IPv4 address is usable without using its network or broadcast address?
172.20.185.145 falls strictly between 172.20.185.144 and 172.20.185.151; it is a valid host inside 172.20.185.144/29. That meets the requested gateway address constraint.
172.20.185.152 is beyond the upper boundary of 172.20.185.144/29. Even though it looks adjacent, it belongs to the next subnet.
172.20.185.151 is the all-one host-bit directed broadcast for 172.20.185.144/29; using it as the gateway would conflict with broadcast behavior.
172.20.185.144 is the all-zero host-bit network address for 172.20.185.144/29, not a normal interface IP on this LAN.
At a payment processor, a network operations engineer is configuring a Cisco interface in a /27 IPv4 network. Which dotted-decimal mask matches the given prefix?
Prefix /27 sets 27 leading one bits, producing dotted mask 255.255.255.224. The remaining 5 bits define host numbering.
Mask 255.255.255.128 represents /25, not /27. Counting contiguous one bits gives the distinction.
Mask 255.255.254.0 represents /23, not /27. Counting contiguous one bits gives the distinction.
Mask 255.255.255.252 represents /30, not /27. Counting contiguous one bits gives the distinction.
At an insurance company, a network administrator needs a single IPv4 subnet for 124 devices with no extra reserve requirement. Which mask is the smallest address block that can hold those devices?
A /25 block offers 126 usable addresses, meeting the 124-host requirement with the smallest fitting normal IPv4 LAN block. It leaves 2 addresses for immediate growth.
A /24 block is larger than necessary for 124 endpoints; it consumes extra addresses even though a /25 block meets the stated limit.
A /26 block offers only 62 usable hosts, less than the 124 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.
At a bank branch, the campus network lead must validate address pairs on LAN 192.168.192.192/27 before assigning endpoints. Which statement is correct?
192.168.192.193 and 192.168.192.222 are both strictly within 192.168.192.192/27, so both are usable and on the same on-link prefix. Their computed masked network addresses match 192.168.192.192.
192.168.192.192 has all host bits zero and is the subnet identifier. It is not ordinarily assigned to a host interface.
192.168.192.224 is in the next subnet after 192.168.192.223. A shared first three octets does not prove same-subnet membership when the mask is /27.
192.168.192.223 has all host bits one and addresses the subnet broadcast. A host normally uses an interior address.
At a clinic, the branch support lead must document the network address containing host 10.17.122.46/24. Which address is that network ID?
Masking 10.17.122.46 with /24 yields network 10.17.122.0; the subnet block contains 256 addresses through 10.17.122.255. That is the correct network ID for an SVI or static route.
This is the next subnet after 10.17.122.0/24; the given host 10.17.122.46 resides in the preceding block. The prefix boundary is fixed by 255.255.255.0.
10.17.122.255 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 /24 network containing 10.17.122.46. A network ID is obtained by zeroing all 8 host bits.
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