A network administrator has subnetted the 172.16.0.0 network using a subnet mask of 255.255.255.192. A duplicate IP address of 172.16.2.120 has accidentally been configured on a workstation in the…

Question

Options

• A172.16.1.80
• B172.16.2.80
• C172.16.1.64
• D172.16.2.64
• E172.16.2.127
• F172.16.2.128

• A
  7% (3)
• B
  85% (39)
• C
  2% (1)
• D
  4% (2)
• E
  2% (1)

Explanation

With a /26 mask (255.255.255.192), the block size is 64 (256 − 192), so subnets in the third octet .2 run as: .2.0–.2.63, 172.16.2.64–172.16.2.127, .2.128–.2.191, etc. Since 172.16.2.120 lands in the 172.16.2.64–.127 subnet, any replacement address must also fall in that range - making 172.16.2.80 the only valid host address among the choices.

Why the distractors fail:

• A (172.16.1.80) and C (172.16.1.64) are in the .1.x range - a completely different third-octet subnet.
• D (172.16.2.64) is the network address of the subnet (the first address), which is never assignable to a host.
• E (172.16.2.127) is the broadcast address of the subnet (the last address), also unassignable.
• F (172.16.2.128) is the network address of the next subnet, placing it outside the target subnet entirely.

Memory tip: With /26 subnets, remember the magic number is 64 - subnet boundaries always land on multiples of 64 (0, 64, 128, 192). To identify valid hosts, find which boundary pair your IP falls between, then eliminate the first address (network) and last address (broadcast) of that range.

No community discussion yet for this question.