A network administrator has been assigned the 192.168.2.64/26 network and assigns the following computer's IP addresses: 192.168.2.70 192.168.2.100 192.168.2.125 192.168.2.130 When the administrator…

Question

Options

• AChange the IP of 192.168.2.130 to 192.168.1.128.
• BChange the IP of 192.168.2.130 to 192.168.2.126.
• CChange the default gateway of 192.168.2.130 accordingly.
• DReconfigure the switch for the right VLAN.

• A
  4% (1)
• B
  75% (21)
• C
  14% (4)
• D
  7% (2)

Explanation

The assigned network is 192.168.2.64/26. A /26 prefix gives 64 addresses per subnet, so the valid range is 192.168.2.64 (network) through 192.168.2.127 (broadcast), with usable host addresses from .65 to .126. The IPs .70, .100, and .125 all fall within this range. However, 192.168.2.130 falls in the next subnet (192.168.2.128/26, usable range .129–.190), placing it on a different network. Since it is on a different subnet, it cannot communicate directly with the other three hosts. The fix is to change .130 to a valid address within the original /26 subnet. Option B (192.168.2.126) is the last usable address in the correct subnet and resolves the problem. Option A is on a completely different network (192.168.1.x), which is wrong. Option C (changing the default gateway) would not fix an incorrect IP assignment. Option D (VLAN reconfiguration) is not the issue here.

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