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CCNA Practice Test - 11
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QuestionClick to flip
Given that the IP address and subnet mask is 172.16.134.64 and255.255.255.224 respectively, which of the following options would describe this address?
AnswerClick to flip back
A
This is a network address.
💡 Explanation:
With IP address 172.16.134.64 and mask 255.255.255.224 (/27), we calculate the network address. The block size for /27 is 32 (2^(32-27) = 32). Network addresses are multiples of 32: 172.16.134.0, 172.16.134.32, 172.16.134.64, etc. The address 172.16.134.64 falls exactly on a network boundary (0, 32, 64, 96...), making it a network address, not a usable host address. The broadcast address would be 172.16.134.95 (network address + block size - 1). Usable host addresses are 172.16.134.65-94.