N10-009 · Networking Concepts · Updated July 29, 2026
Subnetting Practice Problems with Worked Answers
Nearly every subnetting question on the Network+ exam reduces to three numbers: the block size for the prefix, the network address below the host, and the broadcast address one below the next network. Get those, and host ranges, same-subnet checks, and subnet counts fall out for free. The ten problems below run from a straight /27 split to mask-mismatch troubleshooting, each solved in full so you can check your method as well as your answer.
The only reference table you need
Block size is 256 minus the mask value in the octet the prefix ends in. Every network address is a multiple of that block size.
| Prefix | Dotted-decimal mask | Block size | Total addresses | Usable hosts |
|---|---|---|---|---|
| /22 | 255.255.252.0 | 4 (3rd octet) | 1024 | 1022 |
| /23 | 255.255.254.0 | 2 (3rd octet) | 512 | 510 |
| /24 | 255.255.255.0 | 1 (3rd octet) | 256 | 254 |
| /25 | 255.255.255.128 | 128 | 128 | 126 |
| /26 | 255.255.255.192 | 64 | 64 | 62 |
| /27 | 255.255.255.224 | 32 | 32 | 30 |
| /28 | 255.255.255.240 | 16 | 16 | 14 |
| /30 | 255.255.255.252 | 4 | 4 | 2 |
Problem 1 — Splitting a /24 into /27s
Q: You subnet 192.168.44.0/24 into /27s. How many subnets, how many usable hosts each, and what are the network addresses?
You moved the prefix from /24 to /27, borrowing 3 host bits. Borrowed bits produce 2^3 = 8 subnets. Host bits remaining: 32 − 27 = 5, so usable hosts = 2^5 − 2 = 30 each. Block size is 256 − 224 = 32, so networks step by 32:
192.168.44.0 .32 .64 .96 .128 .160 .192 .224The last subnet is 192.168.44.224/27, broadcast 192.168.44.255, usable 192.168.44.225–192.168.44.254.
Problem 2 — Network and broadcast for a given host
Q: A host is 10.14.7.203/26. Give its network address, broadcast address, and usable range.
/26 puts the boundary in the fourth octet with block size 64. The multiple of 64 at or below 203 is 192 (64 × 3 = 192), so the network is 10.14.7.192. The next network would be 10.14.7.256, which rolls into 10.14.8.0, so the broadcast is one below: 10.14.7.255. In binary:
host 203 = 1100 1011
mask 192 = 1100 0000
AND = 1100 0000 = 192 -> network 10.14.7.192
OR hosts = 1111 1111 = 255 -> broadcast 10.14.7.255Usable: 10.14.7.193 – 10.14.7.254, 62 addresses.
Problem 3 — Same subnet or not? (/25)
Q: 172.22.9.100/25 and 172.22.9.130/25 — same subnet?
Block size 128, so a /25 splits the fourth octet into 0–127 and 128–255. The first host sits in 172.22.9.0/25 (broadcast 172.22.9.127). The second sits in 172.22.9.128/25 (broadcast 172.22.9.255). Different subnets — they need a router between them even if they share a switch port VLAN. The clue is that 100 and 130 straddle 128.
Problem 4 — Same subnet or not? (/23)
Q: 10.55.14.200/23 and 10.55.15.30/23 — same subnet?
A /23 borrows into the third octet, block size 2. Third octet 14 is an even multiple of 2, so the network is 10.55.14.0 and it spans two full third-octet values: 10.55.14.0 through 10.55.15.255, broadcast 10.55.15.255. Both addresses fall inside, so yes, same subnet, usable range 10.55.14.1–10.55.15.254. A differing third octet does not by itself mean different subnets once the prefix is shorter than /24.
Problem 5 — Counting subnets from borrowed bits
Q: A /20 is subnetted by borrowing 6 more bits. How many equal subnets? And what happens if someone asks you to borrow 5 bits from a /28?
Borrowing b bits always yields 2^b subnets, regardless of the starting prefix. Here b = 6, so 64 subnets, each a /26 with 62 usable hosts. The second half is a trap: a /28 has only 32 − 28 = 4 host bits left. Borrowing 5 would produce a /33, and IPv4 prefixes cannot exceed /32 — there is no such subnet. When the arithmetic pushes past /32, the correct answer is that the operation is impossible, not a host count.
Problem 6 — Same subnet or not? (/22, the hard one)
Q: 192.168.19.250/22 and 192.168.20.5/22 — same subnet?
Block size for /22 is 256 − 252 = 4, counted in the third octet, so valid networks are third octets 0, 4, 8, 12, 16, 20, 24… The multiple of 4 at or below 19 is 16: first host is in 192.168.16.0/22, broadcast 192.168.19.255. For the second, 20 is itself a multiple of 4: 192.168.20.0/22, broadcast 192.168.23.255. Different subnets — even though the two addresses are only 11 apart numerically, the boundary at 192.168.20.0 falls between them. This is the classic near-miss.
Problem 7 — Maximum subnets with a host-count floor
Q: Subdivide 172.24.0.0/22 into the maximum number of equal subnets, each supporting at least 50 usable hosts. How many subnets, and what prefix?
Work from the host requirement, not the subnet count. You need 2^h − 2 ≥ 50. With h = 5 you get 30 — short. With h = 6 you get 62, which clears it, and h = 6 means prefix 32 − 6 = /26. Going any longer (/27) would break the 50-host floor, so /26 is the maximum split. Subnet count = 2^(26 − 22) = 16 subnets. They run 172.24.0.0/26, 172.24.0.64/26, 172.24.0.128/26 … up to 172.24.3.192/26, which ends exactly on the parent’s last address, 172.24.3.255.
Problem 8 — Same drill, different numbers
Q: Subdivide 10.5.0.0/21 into the maximum number of equal subnets, each supporting at least 100 usable hosts.
2^6 − 2 = 62 is too small; 2^7 − 2 = 126 works, so h = 7 and the prefix is /25. Subnets = 2^(25 − 21) = 16, running from 10.5.0.0/25 through 10.5.7.128/25 (the parent /21 spans 10.5.0.0–10.5.7.255).
Try the same method on 192.168.0.0/20 with a 200-host floor: 2^7 − 2 = 126 fails, 2^8 − 2 = 254 passes, so /24 — and 2^(24 − 20) = 16 subnets, 192.168.0.0/24 through 192.168.15.0/24.
Problem 9 — Mask mismatch on one switch
Q: A workstation at 10.12.40.20 with mask 255.255.254.0 (/23) cannot reliably reach a file server at 10.12.41.200 with mask 255.255.255.128 (/25). The design calls for /25 everywhere on this segment. What is the most likely cause?
Compute what each device believes:
workstation /23 : 10.12.40.0 - 10.12.41.255 -> server looks LOCAL
server /25 : 10.12.41.128 - 10.12.41.255 -> workstation looks REMOTEThe workstation ARPs for the server and sends frames directly; the server considers the workstation off-subnet and hands replies to its default gateway. The result is asymmetric, unreliable, or entirely failed communication. The cause is the workstation’s subnet mask — it should be /25, which would place it in 10.12.40.0/25 and route both directions through the gateway consistently. Notice that only the host deviating from the documented design is wrong; the server matches the standard.
Problem 10 — Reading a 169.254 address
Q: A DHCP client shows 169.254.118.7 with mask 255.255.0.0 and no default gateway. It can ping another workstation on the same switch but nothing else. What happened?
The client got no DHCP response and self-assigned from 169.254.0.0/16 using Automatic Private IP Addressing (APIPA), the link-local mechanism described in RFC 3927 (Request for Comments 3927). Two facts explain the symptoms: the range is link-local only, so routers do not forward it, and APIPA supplies no default gateway and no DNS server. Peer pings inside the same broadcast domain succeed because ARP resolves locally; anything past the segment fails. Treat a 169.254 address plus an empty gateway field as proof of DHCP failure, not as an addressing design.
How the N10-009 exam tests this
- Two addresses, one prefix, “same subnet?” — the pair is chosen to straddle a boundary. Always compute both network addresses; never judge by how close the addresses look.
- Host-count floors. You’re asked for the maximum number of subnets that still meet a minimum usable-host requirement. Solve 2^h − 2 first, then convert to a prefix, then count subnets.
- Impossible prefixes. Borrowed-bit questions occasionally push past /32; the answer is that no such subnet exists.
- Broken connectivity between two hosts on one switch. Compare each host’s computed subnet against the stated design standard — the misconfigured mask is the one that disagrees with the design.
When these ten feel easy, move to full-length practice exams and hold the same accuracy under time pressure.
Quick reference
- Block size = 256 − (mask value in the octet where the prefix ends); network addresses are multiples of it.
- Broadcast = next network address − 1; first usable = network + 1; last usable = broadcast − 1.
- Usable hosts = 2^(32 − prefix) − 2, with /31 (2 usable, RFC 3021) and /32 (1 address) as the exceptions.
- Borrowing b bits creates 2^b subnets; the count of subnets between prefixes is 2^(new − old).
- For prefixes shorter than /24, do the arithmetic in the third octet: /23 steps by 2, /22 by 4, /21 by 8, /20 by 16.
- A mismatched mask makes one host treat a peer as local while the peer treats it as remote — the classic one-way-traffic symptom.
- 169.254.x.x /16 with no gateway is APIPA after a DHCP failure; it never routes off the local segment.
- For the underlying theory see subnetting explained and network and broadcast addresses; for unequal-sized allocations see VLSM.