Networking Fundamentals Practice Question 6: Subnet 26 Host Range
Networking Fundamentals · Test 1 · question 6 of 10
A server is configured as 192.168.10.77/26. A technician needs to give a new printer a static address in the same subnet so the two can talk directly without a router. Which address is valid for the printer?
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Answer:
A /26 leaves 32 − 26 = 6 host bits, so each block has 2^6 = 64 addresses. The mask is 255.255.255.192, and the /26 blocks in the last octet start at 0, 64, 128 and 192. 77 falls in the 64-127 block:
- Network address: 192.168.10.64 - Broadcast address: 192.168.10.127 - Usable host range: 192.168.10.65 to 192.168.10.126 (64 − 2 = 62 hosts)
192.168.10.100 is inside the usable range, so D is correct.
- A. 192.168.10.64 is the network address, where the host bits are all 0. It identifies the subnet itself and can't be assigned to a device. - B. 192.168.10.127 is the subnet's directed broadcast address, where the host bits are all 1. It can't be assigned to a device either. - C. 192.168.10.130 is a valid host address, but it sits in the next block (192.168.10.128/26). The printer would be on a different subnet and could reach the server only through a router.
References: RFC 4632 - Classless Inter-domain Routing (CIDR), section 3.1; RFC 1122 - Requirements for Internet Hosts, section 3.2.1.3