CCNA Subnetting Practice Questions:20+ with Answer

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Subnetting is just splitting one IP network into smaller portions called subnets. And honestly? It’s one of the most tested and most  important topics on the CCNA 200-301 exam.

 

Network engineer do subnetting in daily practice specially when they plan to design a LAN, assign VLANs to switches,setting  ospf protocol ,writing ACLs. Do subnetting fast and right way ,other you will struggle .

 

Subnetting fixes three big problems:

 

  1. Saves IP addresses (IPv4 only has ~4.3 billion total)

 

  1. Split traffic – Better security and control

 

  1. Shrinks broadcast domains – less noise, better speed.

 

  1. Key Terms You must know

 

Here’s a quick table. Memorize this stuff.

 

Entity Definition Example
IP address 32‑bit address which is label  for any device     192.168.10.50
Subnet mask Separates network from host    255.255.255.0
CIDR prefix Number of network bits/Slash notation       /24
Network address   is the   First address, all host bits zero 192.168.10.0  
Broadcast address Last address 192.168.10.255
Usable host range IPs you can actually allot 1 to 254
Wildcard mask Inverse of subnet mask 0.0.0.255

 

VLSM    Different prefix lengths in one network               /26, /28, /30

Subnets created               2^S  S means =bits borrowed                  2^3=  8 subnets

Usable hosts      2^h – 2 (h = host bits left)             2^6 – 2 = 62

IPv4 address classes  (still on the test)

 

Class First octet Default mask      Max hosts            Use
A 1–126        /8 (255.0.0.0)        16.7 million Big companies
B 128–191            /16 (255.255.0.0)            65,534 Mid‑size
C 192–223         /24(255.255.255.0)             254               Small LANs
D 224–239                 –               – Multicast
E 240–255                 –               – Research

 

127.0.0.0  is basically loopback  address, which is  not usable. When DHCP fails APIPA

  1. The Binary Stuff – Yes, You Need This

 

You cannot skip binary. The CCNA will check if you can convert decimal to binary and CIDR fast.

 

Octet bit values: 128, 64, 32, 16, 8, 4, 2, 1.

 

Core formulas

 

Number of subnets = 2^S (s = bits borrowed)

 

Usable hosts = 2^h – 2 (h = host bits left)

 

Block size = 256 minus the mask octet

 

Wildcard = 255 minus mask octet

 

Subnet mask cheat sheet (from /24 to /32)

 

CIDR      Mask     Block     Usable hosts      Subnets from /24

/24         255.255.255.0                   256        254            1

/25         255.255.255.128                128         126         2

/26         255.255.255.192                64           62           4

/27         255.255.255.224                32           30           8

/28         255.255.255.240                16           14           16

/29         255.255.255.248                8              6              32

/30         255.255.255.252                4              2              64

/31         255.255.255.254                2              2*           128

/32         255.255.255.255                1              1              256

*/31 is for point‑to‑point links – no wasted network/broadcast addresses.

 

  1. Beginner Practice Questions

 

Try these first. Answers below.

 

Q1 – What’s the default subnet mask for a Class B address?

  1. 255.0.0.0
  2. 255.255.0.0
  3. 255.255.255.0
  4. 255.255.255.128

 

Explanation: Class B = 128–191 first octet, default /16 = 255.255.0.0.

 

Q2 – How many usable hosts in a /28 subnet?

  1. 14
  2. 16
  3. 30
  4. 32

 

Explanation: 32-28=4 host bits → 2^4=16 total – 2 = 14 usable.

 

Q3 – Broadcast address for 192.168.1.64/26?

  1. 192.168.1.64
  2. 192.168.1.127
  3. 192.168.1.128
  4. 192.168.1.255

 

Explanation: /26 block size = 64. Subnet .64, next .128 → broadcast = .127.

 

Q4 – Host 10.0.0.1/8 – what’s the network address?

  1. 10.0.0.0
  2. 10.255.255.255
  3. 10.0.0.1
  4. 0.0.0.1

 

Explanation: /8 means first octet only. Set host bits to zero → 10.0.0.0.

 

Q5 – Wildcard mask for /24?

  1. 255.255.255.0
  2. 0.0.0.255
  3. 0.255.255.255
  4. 255.0.0.0

 

Explanation: 255.255.255.255 minus 255.255.255.0 = 0.0.0.255.

 

  1. Intermediate Questions

 

Q6 – Need at least 50 hosts per subnet from 172.16.0.0/16. Which mask?

  1. /25 (126 hosts)
  2. /26 (62 hosts)
  3. /27 (30 hosts)
  4. /28 (14 hosts)

 

Explanation: /26 gives 62 usable – enough. /27 only 30. Always pick the smallest that fits.

 

Q7 – Subnet 192.168.5.0/24 with /27 mask – how many subnets?

  1. 4
  2. 6
  3. 8
  4. 16

 

Explanation: Borrow 3 bits (27-24=3) → 2^3=8 subnets.

 

Q8 – Router IP 172.20.15.193/27 – what’s the network address?

  1. 172.20.15.160
  2. 172.20.15.192
  3. 172.20.15.224
  4. 172.20.15.128

 

Explanation: /27 block=32. Subnets: .0, .32, .64, .96, .128, .160, .192, .224. .193 falls in .192.

 

Q9 – Which private IP range is Class A per RFC 1918?

  1. 192.168.0.0/16
  2. 172.16.0.0/12
  3. 10.0.0.0/8
  4. 169.254.0.0/16

 

Explanation: RFC 1918: Class A = 10.0.0.0/8, Class B = 172.16–31, Class C = 192.168.

 

Q10 – Max usable hosts in /29?

  1. 8
  2. 6
  3. 14
  4. 30

 

Explanation: 32-29=3 host bits → 2^3=8 total – 2 = 6.

 

  1. Advanced Questions

 

Q11 – VLSM: network 192.168.100.0/24. Need subnets: 60 hosts, 28 hosts, 12 hosts. Mask for the 60‑host subnet?

  1. /25 (126 hosts)
  2. /26 (62 hosts)
  3. /27 (30 hosts)
  4. /28 (14 hosts)

 

Explanation: 60 hosts needs at least 62. /26 fits. /25 wastes space. Always smallest that works.

 

Q12 – Given 203.0.113.0/24. Need 100 hosts, 50, 25. Third subnet’s network address?

  1. 203.0.113.0
  2. 203.0.113.128
  3. 203.0.113.192
  4. 203.0.113.224

 

Explanation: Largest first: 100 hosts → /25 (0–127). 50 hosts → /26 (128–191). 25 hosts → /27 (192–223). Third subnet = 203.0.113.192.

 

Q13 – ACL entry: permit ip 10.10.20.0 0.0.0.63. What CIDR?

  1. /24
  2. /25
  3. /26
  4. /27

 

Explanation: Wildcard 0.0.0.63 means 6 host bits (64-1). Prefix = 32-6 = /26.

 

Q14 – Borrow bits from /24 to get at least 12 subnets, maximize hosts per subnet?

  1. 3
  2. 4
  3. 5
  4. 6

 

Explanation: 2^4=16 subnets (≥12). 2^3=8 not enough. Borrow 4 leaves 4 host bits → 14 hosts each.

 

Q15 – Host 172.31.200.130/22 belongs to which subnet?

  1. 172.31.198.0/22
  2. 172.31.200.0/22
  3. 172.31.196.0/22
  4. 172.31.202.0/22

 

Explanation: /22 block in third octet = 4. Subnets: .196, .200, .204. .200.130 in .200 block.

 

  1. VLSM Design Walkthrough (Real Scenario)

 

You got to know VLSM for the CCNA. Golden rule: biggest subnet first.

 

Scenario: 192.168.50.0/24. Need:

Site A = 100 hosts, Site B = 50, Site C = 25, Site D = 10, two WAN links with 2 hosts each.

 

Site Mask Network Usable range Broadcast
A (100) /25 192.168.50.0 1–126  127
B (50) /26 192.168.50.128 129–190 191
C (25) /27 192.168.50.192 193–222 223
D (10) /28 192.168.50.224 225–238 239
WAN1 /30 192.168.50.240 241–242 243
WAN2 /30 192.168.50.244 245–246 247

 

Unused: .248–255 (8 addresses). Efficiency ~97%. Way better than classful.

More Practice Questions

Q16 – What is 255.255.255.255?

  1. Network address
  2. Loopback
  3. Limited broadcast
  4. Multicast

 

Q17 – How many host bits in /20?

  1. 20
  2. 12
  3. 8
  4. 4

 

Q18 – Most IP‑efficient WAN link subnet for two routers?

  1. /28
  2. /29
  3. /30
  4. /31

 

Q19 – Two addresses in same /26 subnet? (choose pair)

  1. 172.16.5.63 and .65
  2. 10.0.0.190 and .193
  3. 192.168.1.128 and .191
  4. 203.0.113.64 and .127

 

Q20 – Most appropriate mask for serial point‑to‑point OSPF link?

  1. /24
  2. /28
  3. /30
  4. /32

 

CCNA Subnetting Practice Questions:20+ with Answer

 

 

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