Subnetting MCQ 60 Practice Tests With Answers (2026)
PerfectNotes TeamUpdated: June 2026~25 min practice 60 Questions · 3 Levels 60 Questions · 3 Levels
Subnetting MCQ practice questions are essential for preparing for competitive exams, certifications (CompTIA Network+, CCNA, CCNP), and technical interviews. This comprehensive MCQ platform provides 60 carefully curated practice questions covering IP address structure, CIDR notation, subnet masks, host calculations, VLSM design, and route summarization.
These questions are organized into three progressive difficulty levels of 20 questions each: Basics (covering IP address structure, address classes, CIDR fundamentals, default masks, host counting formula), Concepts (covering block size calculations, network/broadcast identification, mask conversions, subnet planning for specific host counts), and Advanced (covering VLSM multi-department design, supernetting and route summarization, RFC 3021, IPv6 /64 networks, overlapping subnet detection). Each question includes a verified, in-depth explanation.
Practice in Study Mode to reveal answers and detailed explanations instantly, or use Exam Mode for timed testing and real-time scoring to simulate CompTIA Network+, CCNA, or CCNP certification exam conditions.
Contents
- 1.Basics (20 Questions)Address classes · default masks · CIDR notation · host calculations
- 2.Concepts (20 Questions)Block sizes · host capacity · mask conversions · network addresses
- 3.Advanced (20 Questions)VLSM · supernetting · summarization · RFC standards
- 4.Conclusionsummary · next steps · study tips
- 5.Key Takeawaysquick-fire bullet recap of essential facts
- 6.Quick Review Summaryconcept · definition · key fact table
- 7.FAQcommon questions answered
Subnetting MCQ 60 Practice Tests With Answers (2026) — Basics
1What is the total number of bits in an IPv4 address?
CorrectC: 32 bits
An IPv4 address consists of 32 bits, typically divided into four octets (8 bits each), each representing a decimal number 0-255.
IncorrectC: 32 bits
An IPv4 address consists of 32 bits, typically divided into four octets (8 bits each), each representing a decimal number 0-255.
2What is the default subnet mask for a Class C network?
CorrectA: 255.255.255.0
Class C networks use a default subnet mask of 255.255.255.0 (/24), reserving the first three octets for the network and the last octet for hosts.
IncorrectA: 255.255.255.0
Class C networks use a default subnet mask of 255.255.255.0 (/24), reserving the first three octets for the network and the last octet for hosts.
3Which of the following is a private IP address range exclusively for Class B networks?
CorrectD: 172.16.0.0 – 172.31.255.255
RFC 1918 designates 172.16.0.0 – 172.31.255.255 as private Class B addresses; 10.0.0.0/8 is Class A private; 192.168.0.0/16 is Class C private.
IncorrectD: 172.16.0.0 – 172.31.255.255
RFC 1918 designates 172.16.0.0 – 172.31.255.255 as private Class B addresses; 10.0.0.0/8 is Class A private; 192.168.0.0/16 is Class C private.
4What does CIDR stand for in networking?
CorrectB: Classless Inter-Domain Routing
CIDR (Classless Inter-Domain Routing) allows flexible IP address allocation without adhering to rigid class boundaries, introduced in RFC 1519.
IncorrectB: Classless Inter-Domain Routing
CIDR (Classless Inter-Domain Routing) allows flexible IP address allocation without adhering to rigid class boundaries, introduced in RFC 1519.
5In the CIDR notation 192.168.1.0/24, what does the /24 explicitly represent?
CorrectB: The number of bits used for the network portion
The /24 in CIDR notation indicates 24 bits are allocated to the network portion, leaving 8 bits for hosts.
IncorrectB: The number of bits used for the network portion
The /24 in CIDR notation indicates 24 bits are allocated to the network portion, leaving 8 bits for hosts.
6Which portion of an IP address identifies the specific device on a local network?
CorrectC: Host ID
The Host ID (or host bits) uniquely identifies individual devices within a specific subnet or network segment.
IncorrectC: Host ID
The Host ID (or host bits) uniquely identifies individual devices within a specific subnet or network segment.
7What is the valid first-octet decimal range for a traditional Class A IP address?
CorrectA: 1–126
Class A addresses begin with first octet 1–126 (127 is reserved for loopback); the leading bit is always 0.
IncorrectA: 1–126
Class A addresses begin with first octet 1–126 (127 is reserved for loopback); the leading bit is always 0.
8Which IP address block is reserved specifically for loopback testing (localhost)?
CorrectD: 127.0.0.0/8
The 127.0.0.0/8 block is reserved for loopback addresses; 127.0.0.1 specifically represents localhost used for internal testing.
IncorrectD: 127.0.0.0/8
The 127.0.0.0/8 block is reserved for loopback addresses; 127.0.0.1 specifically represents localhost used for internal testing.
9How many usable host addresses are physically available in a standard /24 network?
CorrectA: 254
A /24 network contains 256 total addresses (2^8), but the first (network) and last (broadcast) are reserved, leaving 254 usable hosts.
IncorrectA: 254
A /24 network contains 256 total addresses (2^8), but the first (network) and last (broadcast) are reserved, leaving 254 usable hosts.
10What is the mathematical formula used to calculate the number of usable hosts in a given subnet?
CorrectC: 2^n - 2
The formula 2^n - 2 calculates usable hosts (where n = number of host bits), subtracting the network and broadcast addresses.
IncorrectC: 2^n - 2
The formula 2^n - 2 calculates usable hosts (where n = number of host bits), subtracting the network and broadcast addresses.
11Why do network engineers subtract 2 when calculating the number of usable hosts in a subnet?
CorrectD: To reserve the Network address and the Broadcast address
The first address is the network address (all host bits 0) and the last is the broadcast address (all host bits 1), neither usable for hosts.
IncorrectD: To reserve the Network address and the Broadcast address
The first address is the network address (all host bits 0) and the last is the broadcast address (all host bits 1), neither usable for hosts.
12What is the primary purpose of a broadcast address within a subnet?
CorrectB: To send a single packet to all hosts located on that specific subnet
The broadcast address (all host bits set to 1) sends a single packet to all hosts within that specific subnet simultaneously.
IncorrectB: To send a single packet to all hosts located on that specific subnet
The broadcast address (all host bits set to 1) sends a single packet to all hosts within that specific subnet simultaneously.
13Which traditional IP address class is officially reserved for Multicast groups?
CorrectC: Class D
Class D (224.0.0.0 – 239.255.255.255) is designated for multicast; Class E (240.0.0.0 – 255.255.255.255) is reserved for future use.
IncorrectC: Class D
Class D (224.0.0.0 – 239.255.255.255) is designated for multicast; Class E (240.0.0.0 – 255.255.255.255) is reserved for future use.
14What is the process of borrowing bits from the host portion of an IP address to create smaller networks called?
CorrectA: Subnetting
Subnetting divides a large network into smaller subnetworks by borrowing bits from the host portion and designating them for the subnet.
IncorrectA: Subnetting
Subnetting divides a large network into smaller subnetworks by borrowing bits from the host portion and designating them for the subnet.
15If a subnet mask is 255.255.255.0, how many bits are permanently assigned to the network portion?
CorrectB: 24
The subnet mask 255.255.255.0 has 24 bits set to 1 (network portion) and 8 bits set to 0 (host portion), equivalent to /24.
IncorrectB: 24
The subnet mask 255.255.255.0 has 24 bits set to 1 (network portion) and 8 bits set to 0 (host portion), equivalent to /24.
16Which of the following is an example of a Class C private IP address?
CorrectD: 192.168.10.50
RFC 1918 reserves 192.168.0.0/16 (Class C range) for private use; 10.1.1.1 is Class A private; 172.16.5.5 is Class B private.
IncorrectD: 192.168.10.50
RFC 1918 reserves 192.168.0.0/16 (Class C range) for private use; 10.1.1.1 is Class A private; 172.16.5.5 is Class B private.
17What is the exact binary representation of the decimal number 255?
CorrectA: 11111111
255 in binary is 11111111 (8 bits all set to 1), a common value in subnet masks and broadcast addresses.
IncorrectA: 11111111
255 in binary is 11111111 (8 bits all set to 1), a common value in subnet masks and broadcast addresses.
18An IP address containing all host bits set to 0 typically represents what network component?
CorrectC: The network address
The network address has all host bits set to 0, identifying the network itself; the first usable host has the last bit set to 1.
IncorrectC: The network address
The network address has all host bits set to 0, identifying the network itself; the first usable host has the last bit set to 1.
19What is the default subnet mask applied to a Class B network?
CorrectD: 255.255.0.0
Class B networks use the default subnet mask 255.255.0.0 (/16), reserving the first two octets for network and last two for hosts.
IncorrectD: 255.255.0.0
Class B networks use the default subnet mask 255.255.0.0 (/16), reserving the first two octets for network and last two for hosts.
20Which IP address block is traditionally utilized as the Automatic Private IP Addressing (APIPA) fallback range?
CorrectB: 169.254.0.0/16
APIPA (Automatic Private IP Addressing) uses 169.254.0.0/16 as a fallback when DHCP is unavailable, allowing hosts to self-assign addresses.
IncorrectB: 169.254.0.0/16
APIPA (Automatic Private IP Addressing) uses 169.254.0.0/16 as a fallback when DHCP is unavailable, allowing hosts to self-assign addresses.
Subnetting MCQ 60 Practice Tests With Answers (2026) — Concepts
1What is the corresponding decimal subnet mask for a /26 prefix?
CorrectB: 255.255.255.192
/26 means 26 bits for network; the last octet is 11000000 in binary = 192. Each subnet has 2^(32-26) = 2^6 = 64 addresses.
IncorrectB: 255.255.255.192
/26 means 26 bits for network; the last octet is 11000000 in binary = 192. Each subnet has 2^(32-26) = 2^6 = 64 addresses.
2How many usable hosts can be dynamically assigned in a /28 subnet?
CorrectD: 14
A /28 subnet has 2^(32-28) = 2^4 = 16 total addresses. Subtracting network and broadcast: 16 - 2 = 14 usable hosts.
IncorrectD: 14
A /28 subnet has 2^(32-28) = 2^4 = 16 total addresses. Subtracting network and broadcast: 16 - 2 = 14 usable hosts.
3You have a Class C network (192.168.1.0/24). You need to create subnets that can support up to 25 hosts each. Which subnet mask should you use?
CorrectA: 255.255.255.224 (/27)
/26 provides 62 hosts (too many); /27 provides 30 usable hosts (sufficient for 25); /28 provides 14 (insufficient). /27 is optimal.
IncorrectA: 255.255.255.224 (/27)
/26 provides 62 hosts (too many); /27 provides 30 usable hosts (sufficient for 25); /28 provides 14 (insufficient). /27 is optimal.
4What is the true Network Address for the IP 192.168.1.50/27?
CorrectC: 192.168.1.32
/27 has block size 32. 50 ÷ 32 = 1 remainder 18; network address: 1 × 32 = 32. Valid range: 192.168.1.32–63.
IncorrectC: 192.168.1.32
/27 has block size 32. 50 ÷ 32 = 1 remainder 18; network address: 1 × 32 = 32. Valid range: 192.168.1.32–63.
5What is the resulting Broadcast Address for the IP 192.168.1.50/27?
CorrectC: 192.168.1.63
For the network 192.168.1.32/27 (block size 32), broadcast is 192.168.1.32 + 31 = 192.168.1.63.
IncorrectC: 192.168.1.63
For the network 192.168.1.32/27 (block size 32), broadcast is 192.168.1.32 + 31 = 192.168.1.63.
6Which of the following represents a valid usable host IP within the 10.0.0.0/29 network?
CorrectA: 10.0.0.6
/29 has block size 8: 10.0.0.0 (network), 10.0.0.1–6 (usable), 10.0.0.7 (broadcast). 10.0.0.6 is the last usable host.
IncorrectA: 10.0.0.6
/29 has block size 8: 10.0.0.0 (network), 10.0.0.1–6 (usable), 10.0.0.7 (broadcast). 10.0.0.6 is the last usable host.
7What is the block size (or magic number) for a subnet mask of 255.255.255.240?
CorrectD: 16
255.255.255.240 (/28) means last octet is 11110000 in binary. Block size = 256 - 240 = 16.
IncorrectD: 16
255.255.255.240 (/28) means last octet is 11110000 in binary. Block size = 256 - 240 = 16.
8If you borrow 3 bits from the host portion of a standard Class C network, how many subnets do you effectively create?
CorrectB: 8
Borrowing 3 bits creates 2^3 = 8 subnets. Original /24 becomes /27 (24 + 3), with each subnet having 32 addresses (2^(32-27)).
IncorrectB: 8
Borrowing 3 bits creates 2^3 = 8 subnets. Original /24 becomes /27 (24 + 3), with each subnet having 32 addresses (2^(32-27)).
9What is the correct decimal subnet mask for a /22 prefix?
CorrectD: 255.255.252.0
/22 = 22 network bits. Last two octets: 11111100 00000000 = 252.0. Subnet mask: 255.255.252.0.
IncorrectD: 255.255.252.0
/22 = 22 network bits. Last two octets: 11111100 00000000 = 252.0. Subnet mask: 255.255.252.0.
10How many usable host addresses are available within a /22 network?
CorrectB: 1022
/22 has 2^(32-22) = 2^10 = 1024 total addresses. Usable hosts: 1024 - 2 = 1022.
IncorrectB: 1022
/22 has 2^(32-22) = 2^10 = 1024 total addresses. Usable hosts: 1024 - 2 = 1022.
11You are given the network 192.168.10.64/26. What is the network address of the next consecutive subnet?
CorrectC: 192.168.10.128
/26 has block size 64. Current network: 192.168.10.64–127. Next network: 192.168.10.128.
IncorrectC: 192.168.10.128
/26 has block size 64. Current network: 192.168.10.64–127. Next network: 192.168.10.128.
12What is the broadcast address of the network 10.1.1.200/25?
CorrectA: 10.1.1.255
/25 has block size 128. 200 ÷ 128 = 1 remainder 72; network: 10.1.1.128. Broadcast: 10.1.1.128 + 127 = 10.1.1.255.
IncorrectA: 10.1.1.255
/25 has block size 128. 200 ÷ 128 = 1 remainder 72; network: 10.1.1.128. Broadcast: 10.1.1.128 + 127 = 10.1.1.255.
13Which subnet mask provides exactly 2 usable hosts, commonly utilized for point-to-point WAN links?
CorrectA: /30
/30 has 4 total addresses (2^2); with network/broadcast removed: 4 - 2 = 2 usable hosts, ideal for point-to-point links.
IncorrectA: /30
/30 has 4 total addresses (2^2); with network/broadcast removed: 4 - 2 = 2 usable hosts, ideal for point-to-point links.
14What is the calculated Network Address of a host assigned the IP 172.16.5.100/21?
CorrectC: 172.16.0.0
/21 = 3 borrowed bits from Class B (which is normally /16). Block size = 256 - 248 = 8 (in 3rd octet). 5 ÷ 8 = 0 remainder 5; network starts at 172.16.0.0.
IncorrectC: 172.16.0.0
/21 = 3 borrowed bits from Class B (which is normally /16). Block size = 256 - 248 = 8 (in 3rd octet). 5 ÷ 8 = 0 remainder 5; network starts at 172.16.0.0.
15What is the designated Broadcast Address of the 172.16.5.100/21 network?
CorrectB: 172.16.15.255
/21 network starting at 172.16.0.0 has block size 8 (3rd octet). Broadcast: 172.16.7.255.
IncorrectB: 172.16.15.255
/21 network starting at 172.16.0.0 has block size 8 (3rd octet). Broadcast: 172.16.7.255.
16What is the first usable host address belonging to the subnet 192.168.100.128/25?
CorrectB: 192.168.100.129
/25 network 192.168.100.128 has network address ending in .128; first usable host is .129.
IncorrectB: 192.168.100.129
/25 network 192.168.100.128 has network address ending in .128; first usable host is .129.
17What is the last usable host address belonging to the subnet 192.168.100.128/25?
CorrectB: 192.168.100.254
/25 block size 128. Broadcast: 192.168.100.128 + 127 = 192.168.100.255. Last usable: .254.
IncorrectB: 192.168.100.254
/25 block size 128. Broadcast: 192.168.100.128 + 127 = 192.168.100.255. Last usable: .254.
18Which of the following IP addresses falls identically into the same subnet as 10.5.5.50/28?
CorrectD: 10.5.5.55
/28 block size 16: 50 ÷ 16 = 3 remainder 2; network = 3 × 16 = 48. Range: 10.5.5.48–63. Only 10.5.5.55 falls in this range.
IncorrectD: 10.5.5.55
/28 block size 16: 50 ÷ 16 = 3 remainder 2; network = 3 × 16 = 48. Range: 10.5.5.48–63. Only 10.5.5.55 falls in this range.
19You need to create 12 subnets from a single Class C network. What is the absolute minimum number of bits you must borrow?
CorrectC: 4
2^n subnets: 2^2 = 4 (insufficient); 2^3 = 8 (insufficient); 2^4 = 16 (sufficient). Must borrow minimum 4 bits.
IncorrectC: 4
2^n subnets: 2^2 = 4 (insufficient); 2^3 = 8 (insufficient); 2^4 = 16 (sufficient). Must borrow minimum 4 bits.
20Which decimal subnet mask is mathematically equivalent to the prefix /29?
CorrectA: 255.255.255.248
/29 = 29 network bits; last octet: 11111000 = 248. Subnet mask: 255.255.255.248.
IncorrectA: 255.255.255.248
/29 = 29 network bits; last octet: 11111000 = 248. Subnet mask: 255.255.255.248.
Subnetting MCQ 60 Practice Tests With Answers (2026) — Advanced
1What does the networking acronym VLSM formally stand for?
CorrectD: Variable Length Subnet Masking
VLSM (Variable Length Subnet Masking) allows different subnet sizes within the same network, optimizing address allocation efficiency.
IncorrectD: Variable Length Subnet Masking
VLSM (Variable Length Subnet Masking) allows different subnet sizes within the same network, optimizing address allocation efficiency.
2What is the primary architectural benefit of deploying VLSM in network design?
CorrectC: It allows subnets of different sizes, drastically reducing wasted IP addresses
VLSM enables tailored subnet sizing for different requirements, minimizing address waste compared to fixed-size subnets.
IncorrectC: It allows subnets of different sizes, drastically reducing wasted IP addresses
VLSM enables tailored subnet sizing for different requirements, minimizing address waste compared to fixed-size subnets.
3You want to summarize four continuous routes into a single supernet route: 192.168.0.0/24, 192.168.1.0/24, 192.168.2.0/24, and 192.168.3.0/24. What is the summarized route?
CorrectB: 192.168.0.0/22
4 networks = 2^2; need 2 borrowed bits back. /24 - 2 = /22. Summarized: 192.168.0.0/22 covers all four networks.
IncorrectB: 192.168.0.0/22
4 networks = 2^2; need 2 borrowed bits back. /24 - 2 = /22. Summarized: 192.168.0.0/22 covers all four networks.
4What is the most efficient summarized route for strictly encompassing 10.1.4.0/24 and 10.1.5.0/24?
CorrectA: 10.1.4.0/23
Two consecutive /24 networks (4 and 5) combine into a single /23: 10.1.4.0/23 (covers 10.1.4.0–10.1.5.255).
IncorrectA: 10.1.4.0/23
Two consecutive /24 networks (4 and 5) combine into a single /23: 10.1.4.0/23 (covers 10.1.4.0–10.1.5.255).
5You are given the network 10.0.0.0/16. You need to create subnets that can support exactly 500 hosts each. What is the most efficient subnet mask to deploy?
CorrectC: /23
/22 provides 1022 hosts (wasteful); /23 provides 510 hosts (sufficient for 500, minimal waste); /24 provides 254 (insufficient).
IncorrectC: /23
/22 provides 1022 hosts (wasteful); /23 provides 510 hosts (sufficient for 500, minimal waste); /24 provides 254 (insufficient).
6How many separate /24 subnets can you successfully carve out of a single /20 network block?
CorrectB: 16
/20 to /24 = 4 bits difference; 2^4 = 16 subnets of /24 fit within a /20 block.
IncorrectB: 16
/20 to /24 = 4 bits difference; 2^4 = 16 subnets of /24 fit within a /20 block.
7What is the valid host range for the specific network containing the IP address 10.25.25.25/20?
CorrectA: 10.25.16.1 – 10.25.31.254
/20 = 4 bits in 3rd octet. 25 ÷ 16 = 1 remainder 9; network: 10.25.16.0. Range: 10.25.16.1–31.254.
IncorrectA: 10.25.16.1 – 10.25.31.254
/20 = 4 bits in 3rd octet. 25 ÷ 16 = 1 remainder 9; network: 10.25.16.0. Range: 10.25.16.1–31.254.
8Which IP address naturally overlaps (belongs to the exact same subnet) with 172.16.32.0/19?
CorrectD: 172.16.47.255
/19 block size 32 (3rd octet). Network: 32; range: 172.16.32.0–63.255. Only 172.16.47.255 falls in this range.
IncorrectD: 172.16.47.255
/19 block size 32 (3rd octet). Network: 32; range: 172.16.32.0–63.255. Only 172.16.47.255 falls in this range.
9In IPv6, what is the default prefix length universally utilized for a standard local area network (LAN) segment?
CorrectA: /64
IPv6 LAN segments use /64 by default, providing 2^64 addresses per subnet; /48 is for organizations, /32 for ISPs.
IncorrectA: /64
IPv6 LAN segments use /64 by default, providing 2^64 addresses per subnet; /48 is for organizations, /32 for ISPs.
10In a standard IPv6 /64 subnet, how many bits are implicitly reserved for the Interface ID (host portion)?
CorrectD: 64 bits
IPv6 /64 allocates 64 bits to network prefix and 64 bits to Interface ID (host portion).
IncorrectD: 64 bits
IPv6 /64 allocates 64 bits to network prefix and 64 bits to Interface ID (host portion).
11What is the maximum number of point-to-point /30 subnets you can safely carve out of a single /24 network?
CorrectC: 64
/24 to /30 = 6 bits difference; 2^6 = 64 separate /30 subnets fit within a /24.
IncorrectC: 64
/24 to /30 = 6 bits difference; 2^6 = 64 separate /30 subnets fit within a /24.
12What is the corresponding OSPF wildcard mask for a subnet mask of 255.255.248.0?
CorrectB: 0.0.7.255
Wildcard mask = 255.255.255.255 - subnet mask. 255.255.255.255 - 255.255.248.0 = 0.0.7.255.
IncorrectB: 0.0.7.255
Wildcard mask = 255.255.255.255 - subnet mask. 255.255.255.255 - 255.255.248.0 = 0.0.7.255.
13Per RFC 3021, which subnet mask is officially designated as usable for point-to-point links to save IP addresses, yielding exactly two addresses by removing the network/broadcast limitations?
CorrectB: /31
RFC 3021 designates /31 as usable for point-to-point links, providing exactly 2 addresses without network/broadcast overhead.
IncorrectB: /31
RFC 3021 designates /31 as usable for point-to-point links, providing exactly 2 addresses without network/broadcast overhead.
14Host A has the IP 192.168.1.14/28. Host B has the IP 192.168.1.20/28. If Host A tries to ping Host B, where will the packet be sent first?
CorrectC: To the default gateway (Router)
/28 block size 16. Host A (14) is in 192.168.1.0/28; Host B (20) is in 192.168.1.16/28. Different subnets require routing.
IncorrectC: To the default gateway (Router)
/28 block size 16. Host A (14) is in 192.168.1.0/28; Host B (20) is in 192.168.1.16/28. Different subnets require routing.
15In a /23 subnet, how many *total* IP addresses mathematically exist (including the network and broadcast addresses)?
CorrectD: 512
/23 has 2^(32-23) = 2^9 = 512 total addresses.
IncorrectD: 512
/23 has 2^(32-23) = 2^9 = 512 total addresses.
16You are designing a network that requires 2,000 hosts in a single broadcast domain. What is the most efficient subnet mask to apply?
CorrectB: /20
/20 = 4096 hosts; /21 = 2048 hosts; /22 = 1024 hosts; /23 = 512 hosts. /20 is most efficient for 2000 hosts.
IncorrectB: /20
/20 = 4096 hosts; /21 = 2048 hosts; /22 = 1024 hosts; /23 = 512 hosts. /20 is most efficient for 2000 hosts.
17You are given 192.168.100.0/24. You must design a VLSM scheme for four departments needing 60, 30, 14, and 14 hosts respectively. Which sequence of masks correctly satisfies this with the least waste?
CorrectD: /26, /27, /28, /28
/26 (62 hosts), /27 (30 hosts), /28 (14 hosts), /28 (14 hosts) = 60+30+14+14 = 118. Total capacity: 62+30+14+14 = 120. Minimal waste.
IncorrectD: /26, /27, /28, /28
/26 (62 hosts), /27 (30 hosts), /28 (14 hosts), /28 (14 hosts) = 60+30+14+14 = 118. Total capacity: 62+30+14+14 = 120. Minimal waste.
18What is the definitively calculated Network address for 10.200.150.150/13?
CorrectC: 10.192.0.0
/13 = 3 bits in 2nd octet. 256 - (2^(8-3)) = 256 - 32 = 224. 200 ÷ 224 = 0 remainder 200, but 200 > 192; network: 10.192.0.0.
IncorrectC: 10.192.0.0
/13 = 3 bits in 2nd octet. 256 - (2^(8-3)) = 256 - 32 = 224. 200 ÷ 224 = 0 remainder 200, but 200 > 192; network: 10.192.0.0.
19If you attempt to summarize 172.20.16.0/20 and 172.20.32.0/20 into the smallest possible single supernet, what mask is strictly required?
CorrectB: /18
Both /20 networks; 16 and 32 in 3rd octet differ at bit 5; need /18 to encompass both: 172.20.0.0/18.
IncorrectB: /18
Both /20 networks; 16 and 32 in 3rd octet differ at bit 5; need /18 to encompass both: 172.20.0.0/18.
20Which of the following statements about legacy IPv4 classful networking is historically true?
CorrectC: The leading bit sequence for a Class B network is 10
Class B networks have leading bits 10 (binary). Class A: leading bit 0 (/8); Class D: 1110 (multicast, not workstations).
IncorrectC: The leading bit sequence for a Class B network is 10
Class B networks have leading bits 10 (binary). Class A: leading bit 0 (/8); Class D: 1110 (multicast, not workstations).
Conclusion: Mastering Subnetting
These 60 MCQs cover the complete spectrum of subnetting knowledge — from recognizing Class C default masks and calculating the /24 formula through advanced VLSM multi-department allocation, route summarization across autonomous boundaries, and RFC 3021 point-to-point optimization.
The key to subnetting mastery is developing speed and accuracythrough the "block size" (magic number) method. Once you can instantly calculate subnet boundaries, identify overlapping networks, and design optimal VLSM schemes, subnetting becomes second nature. Track your performance across difficulty levels and focus on weak areas until your accuracy reaches 95%+.
After completing this MCQ set, deepen your implementation knowledge with the full Subnetting theory notes, practice with Computer Networks interview questions, and explore related MCQ pages on IPv4 vs IPv6, OSI Model, and Routing Protocols to see subnetting applied across modern network architectures.
📌 Key Takeaways — Subnetting
- 32-bit Address Structure: IPv4 addresses contain 32 bits divided into network and host portions by a subnet mask.
- Address Classes: Class A (/8), Class B (/16), Class C (/24), Class D (multicast), Class E (reserved).
- Private Ranges (RFC 1918): 10.0.0.0/8, 172.16.0.0/12, 192.168.0.0/16 for internal networks only.
- Host Calculation Formula: Usable hosts = 2^n - 2 (where n = number of host bits). Subtract 2 for network and broadcast addresses.
- Magic Number (Block Size): Calculate as 256 - (subnet mask octet value). Use this to determine subnet increments quickly.
- Network Address: All host bits set to 0; identifies the subnet itself in routing tables.
- Broadcast Address: All host bits set to 1; sends a single packet to all hosts on that subnet.
- Subnetting: Borrowing bits from host portion to create smaller networks; increases number of subnets, decreases hosts per subnet.
- VLSM (Variable Length Subnet Masking): Allows different subnet sizes within the same network for optimal address allocation and minimal waste.
- Supernetting / Route Summarization: Combining multiple consecutive subnets into a single larger network to reduce routing table entries.
- /30 for Point-to-Point: 4 addresses (2 usable) ideal for router-to-router WAN connections.
- RFC 3021 (/31): Designates /31 (2 addresses total, both usable) for point-to-point links, eliminating network/broadcast overhead.
- IPv6 /64 LAN Default: IPv6 standard LAN segment prefix length is /64 (64-bit network, 64-bit Interface ID).
- Wildcard Mask: OSPF and ACLs use inverted subnet masks: 255.255.255.255 - subnet mask = wildcard mask.
Quick Review & Summary
Use this reference table to quickly recall common subnetting formulas and prefix equivalents.
| Prefix | Subnet Mask | Block Size | Total Addresses | Usable Hosts |
|---|---|---|---|---|
| /24 | 255.255.255.0 | 256 | 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 |
| /29 | 255.255.255.248 | 8 | 8 | 6 |
| /30 | 255.255.255.252 | 4 | 4 | 2 |
| /31 (RFC 3021) | 255.255.255.254 | 2 | 2 | 2* |
| /22 | 255.255.252.0 | 4 (3rd octet) | 1024 | 1022 |
| /21 | 255.255.248.0 | 8 (3rd octet) | 2048 | 2046 |
| /20 | 255.255.240.0 | 16 (3rd octet) | 4096 | 4094 |
| /16 | 255.255.0.0 | 256 (3rd octet) | 65536 | 65534 |
| /8 | 255.0.0.0 | 256 (2nd octet) | 16777216 | 16777214 |
*RFC 3021: /31 yields 2 usable addresses for point-to-point links (no network/broadcast overhead)
Frequently Asked Questions
Q. How many Subnetting MCQs are available on this page?
Q. What topics do these Subnetting MCQs cover?
Q. Are these MCQs suitable for CompTIA Network+ or CCNA exam preparation?
Q. Why must 2 addresses be subtracted when calculating usable hosts?
Q. What is the "block size" or "magic number" and how is it calculated?
Q. What is VLSM and why is it important?
Q. How do you summarize (supernet) multiple routes?
Q. What is the /31 subnet and why is RFC 3021 important?
Struggling with some questions? Re-read the full Theory Guide: Subnetting