1D0-510 · Question #58
Which IP address class uses 8 bits for the network portion of the address and 24 bits for the host portion of the address?
The correct answer is A. Class A. See the full explanation below for the reasoning.
Question
Options
- AClass A
- BClass B
- CClass C
- DClass D
How the community answered
(23 responses)- A74% (17)
- B13% (3)
- C4% (1)
- D9% (2)
Community Discussion
7
Lena V.Feb 8, 2026Class A is correct. IPv4 divides addresses into classes based on how the 32 bits are split between network and host. Class A reserves the first 8 bits for the network ID (with the leading bit always 0, giving you network values 1-126) and leaves the remaining 24 bits for hosts, which means each Class A network can support over 16 million hosts. Class B splits it 16/16 and Class C splits it 24/8, so if your exam question gives you 8 network bits and 24 host bits, there is only one answer.
Prof. SaraFeb 11, 2026Lena nailed the core logic, but flag for your notes that 127.x.x.x is technically in the Class A range yet is reserved for loopback, so the usable Class A network values run 1-126, not 1-127, and that single excluded block trips up a surprising number of candidates on exam day.
Imani T.Feb 13, 2026Nailed this one fast, Class A owns that 8-bit network, 24-bit host split.
Prof. SaraFeb 17, 2026Class A is correct. The first octet (8 bits) identifies the network, leaving the remaining three octets (24 bits) for host addresses, which is why Class A supports over 16 million hosts per network, a range confirmed by its 1.0.0.0 to 126.255.255.255 span.
Grace U.Feb 18, 2026Just worth flagging for test day that 127.x.x.x is reserved for loopback and lives outside the usable Class A range, so if a question folds it into the host count or the network span, you already know to exclude it.
Grace U.Mar 5, 2026I kept second-guessing myself and almost picked Class C because that one feels like the "default" network class people talk about, but then I reminded myself that Class C actually flips it, giving you only 8 bits for hosts and 24 for the network. Once I drew that out, it was clear that Class A is the one handing you all that host space, 24 full bits worth, which is why those addresses can support such enormous numbers of hosts on a single network.
Prof. SaraMar 6, 2026Grace, you nailed the core flip, and here is the one anchor that makes it stick permanently: Class A gives you that 24-bit host field because it only needs 7 bits (not 8) for the network portion, with the leading zero of the first octet serving as the class identifier, so the math is really 7 net plus 1 implicit bit plus 24 host, not a clean 8-and-24 split the way the shorthand suggests.