Subnetting Practice Questions

Subnetting is the topic people struggle with most on the Network+ exam — and it's also one of the most learnable. It isn't really math; it's a handful of patterns you can recognize in seconds once you've seen them enough times. Read the quick reference below, then try the practice questions. Answers stay hidden until you click, so you can genuinely test yourself.

Why it feels harder than it is

Objective 1.7 asks you to work out subnet masks, how many hosts fit in a subnet, and where each network starts and ends — with a clock running. Here's the thing most people get wrong about that: they assume they're failing because they don't understand binary well enough.

Usually that isn't it. They understand it fine, they just haven't done it often enough to be quick. Subnetting under time pressure is a recall skill, closer to knowing your times tables than to solving an equation. The fix isn't more theory — it's more repetitions.

CIDR quick reference

Where does “usable hosts” come from? Take 2^{(host bits)} and subtract 2. You subtract two because every subnet gives up its first address to name the network itself and its last address to the broadcast — neither can be assigned to a device. A /24 has 256 addresses but only 254 usable ones for that reason.

You'll also see VLSM (Variable Length Subnet Masking) in the objectives. It sounds technical, but it just means using different sized subnets inside the same network instead of making them all the same. A link between two routers needs exactly 2 addresses; giving it 254 wastes 252 of them. VLSM is how you avoid that.

Example 1: finding the network, broadcast, and usable range

Let's take 192.168.20.50/27 and find out which network it belongs to. Three steps:

1. Find the block size. A /27 leaves 5 host bits in the last octet (8 − 3 borrowed bits = 5), so each block is 2⁵ = 32 addresses wide: .0, .32, .64, .96, .128...
2. Find which block .50 falls in. The blocks go 0–31, 32–63, 64–95... .50 falls in the 32–63 block, so the network address is 192.168.20.32.
3. Find the broadcast and usable range. The block runs .32–.63, so the broadcast address is 192.168.20.63, and the usable host range is 192.168.20.33 – 192.168.20.62 (30 usable hosts).

Example 2: VLSM — giving each subnet only what it needs

Say you have 10.10.0.0/24 and three departments to serve: one needs 100 addresses, one needs 50, and one needs 20. You could split the block into equal quarters, but then the small department sits on a pile of addresses it will never use. VLSM lets you size each piece to fit.

1. Size for 100 hosts first (largest first). You need 2ⁿ − 2 ≥ 100, so n = 7 host bits (126 usable) → /25. Assign 10.10.0.0/25 (range .0–.127).
2. Size for 50 hosts next. You need 2ⁿ − 2 ≥ 50, so n = 6 host bits (62 usable) → /26. The first /25 used .0–.127, so this subnet starts at .128: 10.10.0.128/26 (range .128–.191).
3. Size for 20 hosts last. You need 2ⁿ − 2 ≥ 20, so n = 5 host bits (30 usable) → /27. The next free block starts at .192: 10.10.0.192/27 (range .192–.223).

The order matters here, so it's worth saying plainly: always start with the biggest subnet. If you hand out the small ones first, the leftover space gets chopped up and the large subnet no longer fits anywhere — and you'll have to start over. Biggest first, every time.

Try these yourself

Work each one out before you open the answer. Getting one wrong now is a lot cheaper than getting it wrong on exam day.