GSEC · Question #25
In a /24 subnet, which of the following is a valid broadcast address?
The correct answer is D. 192.10.10.255. In a /24 subnet, the broadcast address sets all 8 host bits to 1, producing 255 in the last octet.
Question
In a /24 subnet, which of the following is a valid broadcast address?
Options
- A200.11.11.1
- B221.10.10.10
- C245.20.30.254
- D192.10.10.255
How the community answered
(41 responses)- A2% (1)
- B7% (3)
- C12% (5)
- D78% (32)
Why each option
In a /24 subnet, the broadcast address sets all 8 host bits to 1, producing 255 in the last octet.
200.11.11.1 has a last octet of 1, which represents the first usable host address in a /24 subnet, not the broadcast address.
221.10.10.10 has a last octet of 10, which is a valid host address within the /24 host range, not the broadcast address.
245.20.30.254 has a last octet of 254, which is the last usable host address in a /24 subnet - one value below the broadcast address of 255.
A /24 subnet allocates 24 bits to the network portion and 8 bits to hosts. Setting all 8 host bits to 1 yields a decimal value of 255 in the last octet. 192.10.10.255 satisfies this requirement exactly, making it the valid directed broadcast address for any /24 network.
Concept tested: IPv4 /24 subnet broadcast address identification
Source: https://learn.microsoft.com/en-us/troubleshoot/windows-client/networking/tcpip-addressing-and-subnetting
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