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CCNA Subnetting Practice Questions: 10 Worked Examples

Subnetting is the skill CCNA candidates fear most and use most — on the exam and on the job. Every question below is original and fully worked using the magic-number method: find the interesting octet, compute the block size (256 − mask value), and count. Cover the answer, work each one yourself, then check the walkthrough.

The magic-number method in 20 seconds: (1) Find the interesting octet — the first octet, left to right, where the mask is neither 255 nor 0. (2) Magic number = 256 − (mask value in that octet). (3) Subnet boundaries are multiples of the magic number. Network address = the multiple at or below your IP; broadcast = one less than the next multiple.

CIDRMaskMagic #Usable hosts
/24255.255.255.0256254
/25255.255.255.128128126
/26255.255.255.1926462
/27255.255.255.2243230
/28255.255.255.2401614
/29255.255.255.24886
/30255.255.255.25242

The Questions

1. Host 192.168.44.77/26. What is the network address, broadcast address, and usable host range?

Correct answer: A. Explanation: /26 → mask 255.255.255.192. Interesting octet is the 4th; magic number = 256 − 192 = 64. Subnet boundaries: 0, 64, 128, 192. The host .77 falls in the 64-block, so network = 192.168.44.64, broadcast = one below the next boundary = .127, usable hosts = .65 through .126. B describes the 0-block (wrong subnet); C's broadcast is off by one; D mistakes the host address for the network address.

2. You must subnet 10.0.0.0/16 to support at least 200 subnets with at least 200 usable hosts per subnet. Which is the most efficient mask?

Correct answer: C. Explanation: Always check both constraints. Subnets: from a /16, /23 borrows 7 bits → 128 subnets (too few), /24 borrows 8 bits → 256 subnets (enough). Hosts: /24 leaves 8 host bits → 254 usable (enough); /25 leaves 7 host bits → 126 usable (too few). So /24 is the tightest mask satisfying both. /22 and /23 fail the subnet count; /25 fails the host count. The exam loves this double-constraint format — verify subnets first, then hosts.

3. Which of these addresses is a valid usable host address?

Correct answer: D. Explanation: /26 → magic number 64. Subnet boundaries: 0, 64, 128, 192. A (.63) is the broadcast address of the 0-block; B (.64) is the network address of the 64-block; C (.127) is the broadcast address of the 64-block. Network and broadcast addresses are never assignable to hosts, leaving D (.100) — inside the 64-block's usable range of .65–.126 — as the only valid host. The exam tests exactly this recognition at speed: find the boundaries first, then eliminate.

4. A branch office needs 60 usable host addresses on one subnet. What is the most efficient mask?

Correct answer: B. Explanation: Need 2n − 2 ≥ 60 → 2n ≥ 62 → n = 6 host bits (64 addresses, 62 usable). 32 − 6 = /26. "Most efficient" means the smallest subnet that still fits — /25 (126 usable) wastes 66 addresses; /27 (30 usable) doesn't fit at all. The exam constantly asks for the smallest sufficient subnet, so always test the tighter mask first.

5. Two routers are connected by a point-to-point serial link. The engineer assigns 192.168.99.0/31 to the link. Which statement is true?

Correct answer: B. Explanation: RFC 3021 defines /31 subnets for point-to-point links: with exactly 2 addresses and no broadcast domain (only two devices that already know each other), the network and broadcast addresses are unnecessary — both addresses are usable. This saves address space versus a /30 (which wastes 2 of 4). The classic exam trap is A: /30 is the traditional answer, but /31 is valid and preferred on modern equipment.

6. An ACL must match the entire 192.168.16.0/20 network. What is the correct wildcard mask?

Correct answer: A. Explanation: Wildcard = 255.255.255.255 − subnet mask. /20 → mask 255.255.240.0 → wildcard 0.0.15.255. The fast method: in the interesting octet, wildcard value = magic number − 1 = 16 − 1 = 15. C is the subnet mask itself (the #1 wildcard trap); B inverts the wrong octet; D only matches a /24.

7. Which single summary route covers 192.168.32.0/24 through 192.168.39.0/24 with no extra networks?

Correct answer: B. Explanation: The range spans third-octet values 32–39: that's 8 subnets. 8 = 23, so borrow 3 fewer bits than /24 → /21. Check boundaries: /21 → magic number 8 in the third octet → blocks are 0, 8, 16, 24, 32, 40… The 32-block covers 32–39 exactly — no more, no less. /20 (magic 16) would also cover 32–47, leaking 8 extra subnets (40–47) into the summary; /22 only covers 32–35.

8. Host 10.200.75.14 with mask 255.255.240.0. What is the broadcast address of its subnet?

Correct answer: B. Explanation: Mask 255.255.240.0 = /20. Interesting octet is the 3rd; magic number = 256 − 240 = 16. Boundaries: …, 64, 80, 96. The host's third octet (75) falls in the 64-block → network 10.200.64.0, broadcast = one below the next boundary = 10.200.79.255. A assumes a /24; C is the next network address (a favorite trap); D uses magic 32 (that's /19).

9. How many total subnets and usable hosts per subnet does 172.16.0.0/19 provide?

Correct answer: A. Explanation: 172.16.0.0 is Class B → default /16. Borrowing 3 bits for /19 → 23 = 8 subnets. Host bits = 32 − 19 = 13 → 213 − 2 = 8,192 − 2 = 8,190 usable hosts. The exam wants both numbers from one mask — compute borrowed bits and host bits separately, and never forget the −2 on hosts.

10. An IPv6 host has address 2001:db8:acad::50/64. Which statement is correct?

Correct answer: A. Explanation: In IPv6 the /64 prefix is the standard subnet size: first 64 bits = network, last 64 = interface ID (usually derived via EUI-64 or random). B is wrong — IPv6 has no broadcast concept and no −2 subtraction; effectively the whole 264 space is usable. C is wrong — :: may appear only once per address (otherwise the decompression is ambiguous). D confuses the host address with the network address (2001:db8:acad::).

Quick self-check

If you worked all 10 without peeking, you're in good shape. If questions 2 or 6 tripped you up, re-read them — they test the two most-missed mechanics on the exam: verifying both constraints when sizing a subnet (subnets AND hosts), and deriving wildcard masks from the magic number instead of guessing. On exam day, the magic-number method should take you under 30 seconds per question.

How fast should I be at subnetting for the CCNA?
Aim for 30–60 seconds per question. The exam gives you roughly 100 minutes for ~100 questions, and subnetting questions are designed to be quick if your method is automatic — slow down only on VLSM and summarization items, which reward careful boundary checks.
Do I need to memorize the subnet mask table?
Memorize the /24–/30 row cold (masks, magic numbers, usable hosts) and be able to derive the rest with the magic-number method. Rote-memorizing every CIDR value is unnecessary; fast derivation beats recall under pressure.

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