Prefix Mask (last octet) Block Usable hosts /24 0 (255.255.255.0) 256 254 /25 128 128 126 /26 192 64 62 /27 224 32 30 /28 240 16 14 /29 248 8 6 /30 252 4 2
Q1 — Network & broadcast of 192.168.10.100/27
Find: the network address, broadcast address, and usable host range.
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Work: /27 = mask 255.255.255.224, block size 256 − 224 = 32. Step the fourth octet in 32s: .0, .32, .64, .96, .128. The host .100 lands in the .96 block (covers 96–127).
Answer: network 192.168.10.96, broadcast 192.168.10.127, usable .97 – .126 (30 hosts).
Q2 — Usable hosts in a /26
Find: how many usable host addresses a /26 provides.
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Work: host bits = 32 − 26 = 6, so 2^6 = 64 total addresses. Subtract 2 (network + broadcast).
Answer: 64 − 2 = 62 usable hosts.
Q3 — Subnet mask for a /29
Find: the /29 mask in dotted-decimal, and hosts per subnet.
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Work: /29 sets 29 bits, leaving 3 host bits. The fourth octet is 11111000 = 248. Block size 256 − 248 = 8, so 2^3 − 2 = 6 usable hosts.
Answer: 255.255.255.248, 6 usable hosts per subnet.
Q4 — Which subnet does 172.16.20.75/28 belong to?
Find: the subnet (network + broadcast) containing 172.16.20.75.
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Work: /28 = mask 255.255.255.240, block size 16. Step in 16s: .0, .16, .32, .48, .64, .80. The host .75 is in the .64 block (64–79).
Answer: network 172.16.20.64, broadcast 172.16.20.79, usable .65 – .78.
Q5 — First and last usable host in 192.168.1.192/26
Find: the first and last assignable host addresses.
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Work: /26 block size 64. The .192 block covers 192–255, so the broadcast is .255. First usable = network + 1, last usable = broadcast − 1.
Answer: first 192.168.1.193, last 192.168.1.254.
Q6 — Borrowing 3 bits from a /24
Find: how many subnets you get, and the new prefix.
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Work: subnets = 2^(borrowed bits) = 2^3 = 8. The prefix grows by 3: /24 → /27.
Answer: 8 subnets, each a /27 with 30 usable hosts.
Q7 — Pick a subnet for 50 hosts (VLSM sizing)
Find: the smallest subnet that holds at least 50 hosts.
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Work: need host bits h where 2^h − 2 ≥ 50. 2^5 − 2 = 30 (too small), 2^6 − 2 = 62 (fits). 6 host bits → prefix 32 − 6 = 26.
Answer: a /26 (62 usable hosts). This is the core move in VLSM — see what is VLSM.
Q8 — Convert /22 to a mask and count hosts
Find: the /22 mask in dotted-decimal and usable hosts.
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Work: 22 bits = 255.255.11111100.0 = 255.255.252.0. Host bits = 32 − 22 = 10, so 2^10 − 2 = 1022.
Answer: mask 255.255.252.0, 1022 usable hosts. (Note the subnetting happens in the third octet — block size 256 − 252 = 4 there.)
Q9 — Wildcard mask for 255.255.255.240
Find: the wildcard mask (for an ACL or OSPF statement).
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Work: a wildcard is the inverse of the mask — subtract each octet from 255. 255−255, 255−255, 255−255, 255−240.
Answer: 0.0.0.15. More on this in what is a wildcard mask.
Q10 — Are 192.168.5.130 and 192.168.5.160 in the same /26?
Find: whether both hosts share one /26 subnet.
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Work: /26 block size 64: subnets .0–.63, .64–.127, .128–.191, .192–.255. Both .130 and .160 fall in .128–.191.
Answer: yes — both are in network 192.168.5.128/26 (broadcast .191), so they can talk without a router.
Q11 — Usable hosts on a /30
Find: usable hosts, and why it's used for router links.
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Work: 2 host bits → 2^2 − 2 = 2.
Answer: 2 usable hosts — exactly the two ends of a point-to-point router link, which is why /30 is the classic WAN-link prefix.
Q12 — Magic-number method on 200.100.50.181/26
Find: the network and broadcast for this host.
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Work: /26 block size 64. Steps: .0, .64, .128, .192. Host .181 sits in the .128 block (128–191).
Answer: network 200.100.50.128, broadcast 200.100.50.191, usable .129 – .190.