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Quizzes > Computers & Technology > Computers & IT

Master Subnetting: Take the Practice Quiz Now!

Think you can ace these subnetting example questions? Start now!

Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
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Use this subnetting quiz to practice IP addressing and subnet masks with quick, scored questions. Build speed and accuracy for exams or on the job as you work out prefix length, network ID, and broadcast address. For extra practice, check more questions or try the mask quiz .

What is the default subnet mask for a Class C IPv4 network?
255.255.255.0
255.0.0.0
255.255.255.255
255.255.0.0
Class C networks use a 24-bit default mask of 255.255.255.0, meaning the first three octets define the network. This mask supports up to 256 total addresses per network. It is part of the original classful addressing scheme.
How many usable host addresses are available in a /24 subnet?
252
255
256
254
A /24 subnet has 2^8 = 256 total addresses, but two addresses are reserved for the network ID and the broadcast address. Therefore, the total usable hosts are 256 - 2 = 254.
What is the network address of 192.168.1.130/25?
192.168.1.64
192.168.1.192
192.168.1.0
192.168.1.128
A /25 mask is 255.255.255.128, creating subnets of 0 - 127 and 128 - 255 in the last octet. The address 192.168.1.130 falls in the second block, so its network ID is 192.168.1.128.
What is the broadcast address of the 10.0.0.0/30 network?
10.0.0.2
10.0.0.4
10.0.0.3
10.0.0.1
A /30 network mask is 255.255.255.252, providing four addresses. The highest address in the block is the broadcast address. For 10.0.0.0/30, the addresses are 10.0.0.0 - 10.0.0.3, making .3 the broadcast.
Which subnet mask corresponds to a /26 prefix?
255.255.255.224
255.255.255.192
255.255.255.128
255.255.255.240
A /26 prefix corresponds to a 26-bit mask: 11111111.11111111.11111111.11000000. In decimal notation, that is 255.255.255.192.
How many /26 subnets can be created from a single /24 network?
8
4
16
2
Moving from a /24 to a /26 borrows 2 bits for subnetting, yielding 2^2 = 4 subnets. Each /26 subnet contains 64 addresses.
How many bits are used for host addresses in a /29 subnet?
2
6
5
3
A /29 mask uses 29 bits for the network portion and leaves 3 bits for hosts (32 - 29 = 3). That provides 2^3 = 8 total addresses per subnet.
What is the network ID for 172.16.5.20/16?
172.16.255.255
172.16.5.0
172.16.0.0
172.0.0.0
A /16 mask covers the first two octets, so the network portion is 172.16.x.x. All host bits are zeroed out, giving a network ID of 172.16.0.0.
How many usable host addresses are available in a /26 subnet?
62
58
60
64
A /26 subnet has 2^6 = 64 total addresses, with two reserved addresses for network and broadcast. Thus, 64 - 2 = 62 usable hosts.
Given the network 192.168.10.0/28, what is the network address of the fourth subnet?
192.168.10.48
192.168.10.32
192.168.10.16
192.168.10.64
A /28 subnet has 16 addresses per block. The subnets start at .0, .16, .32, .48, so the fourth (starting count at one) network ID is 192.168.10.48.
What is the wildcard mask corresponding to 255.255.255.248?
0.0.0.8
0.0.0.7
0.0.0.1
0.0.0.15
A wildcard mask is the bitwise inverse of a subnet mask. Inverting 255.255.255.248 yields 0.0.0.7. Wildcard masks are used in ACLs on Cisco devices.
If you need at least 500 hosts in a single subnet, which subnet mask should you choose?
255.255.254.0
255.255.255.0
255.255.0.0
255.255.252.0
To support 500 hosts, you need at least 512 addresses (2^9). A /23 mask (255.255.254.0) provides 512 total addresses, 510 usable.
What is the CIDR notation for the subnet mask 255.255.252.0?
/23
/24
/21
/22
255.255.252.0 in binary is 11111111.11111111.11111100.00000000, which has 22 leading ones. Therefore, it corresponds to /22.
How many /22 subnets can be created from a /20 network?
16
2
8
4
Going from /20 to /22 borrows 2 bits for subnetting, resulting in 2^2 = 4 subnets of size /22.
What is the broadcast address for the 192.168.2.64/26 subnet?
192.168.2.191
192.168.2.65
192.168.2.127
192.168.2.63
A /26 subnet spans 64 addresses; 192.168.2.64 - 192.168.2.127. The highest address, .127, is the broadcast address.
How many bits must be borrowed from the host portion to create 30 subnets from a /24 network?
4
6
3
5
To create at least 30 subnets, you need to borrow 5 bits (2^5 = 32 subnets). Borrowing 4 bits would yield only 16 subnets.
You have the network 10.10.0.0/16 and need at least 1000 hosts per subnet. What is the smallest subnet mask you can use?
/22
/23
/20
/21
To accommodate 1000 hosts, you need at least 1024 addresses (2^10). A /22 mask leaves 10 bits for hosts (2^10 = 1024).
Which summarized network would cover both 192.168.4.0/24 and 192.168.5.0/24?
192.168.5.0/23
192.168.4.0/24
192.168.4.0/23
192.168.4.0/22
Aggregating two /24 networks that differ in the last octet by 1 requires a /23 mask, combining them into 192.168.4.0 - 192.168.5.255.
What is the subnet increment (block size) for subnets on a /29 network?
16
32
8
4
A /29 mask is 255.255.255.248, which increments subnets by 8 in the last octet (256 - 248 = 8).
Which IP address falls within the subnet 192.168.100.64/26?
192.168.100.128
192.168.100.70
192.168.100.63
192.168.100.64
The /26 subnet 192.168.100.64/26 spans addresses .64 - .127. The first address (.64) is the network ID, so .70 is a valid host.
How many usable host addresses are there in a /21 subnet?
2044
2048
4094
2046
A /21 subnet provides 2^(32-21) = 2^11 = 2048 total addresses. Subtracting 2 for network and broadcast leaves 2046 usable hosts.
How many /26 subnets can you create from a /16 network?
64
256
4096
1024
A /16 to /26 requires borrowing 10 bits (26 - 16), yielding 2^10 = 1024 subnets. Each subnet has 64 addresses.
In a variable-length subnet mask (VLSM) environment, which mask provides at least 30 host addresses?
/26
/28
/25
/27
A /27 subnet has 2^(32-27) = 2^5 = 32 total addresses, 30 usable after excluding network and broadcast. This meets the requirement for 30 hosts.
Which summarized network covers the address range from 192.168.0.0 through 192.168.3.255?
192.168.0.0/23
192.168.0.0/21
192.168.0.0/22
192.168.0.0/24
Aggregating four contiguous /24 networks (0.0 - 3.255) results in a /22 summarize route (255.255.252.0), covering 192.168.0.0 - 192.168.3.255.
What is the summarized route for networks 172.16.0.0/23 and 172.16.2.0/23?
172.16.0.0/22
172.16.0.0/23
172.16.0.0/21
172.16.2.0/22
Two /23 networks starting at .0 and .2 combine into a /22 prefix (255.255.252.0) covering 172.16.0.0 - 172.16.3.255. Summarization reduces route entries.
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Study Outcomes

  1. Understand Subnetting Fundamentals -

    Develop a clear understanding of IP addressing concepts, including network and host portions, subnet masks, and CIDR notation.

  2. Calculate Subnet Masks and IP Ranges -

    Perform precise calculations to determine subnet masks, network addresses, broadcast addresses, and usable host ranges.

  3. Design Custom Subnetting Schemes -

    Apply subnetting principles to create efficient network designs that meet specific size and segmentation requirements.

  4. Differentiate Addressing Methods -

    Distinguish between classful and classless IP addressing, understanding when and how to use CIDR for flexible subnetting.

  5. Evaluate Your Proficiency -

    Assess your skills by completing the scored subnetting practice questions quiz, identifying areas of strength and opportunities for improvement.

Cheat Sheet

  1. IP Address Classes and Default Masks -

    Review the five IPv4 address classes (A - E) and their default subnet masks (e.g., Class A: /8, Class B: /16, Class C: /24) to quickly distinguish network and host portions when tackling subnetting example questions. Use a simple mnemonic like "Aunt Betty Cooks" for /8, /16, /24, and verify against IANA's IPv4 Address Space Registry.

  2. Converting Masks to Prefix Lengths -

    Practice converting dotted-decimal masks (e.g., 255.255.254.0) into CIDR notation (/23) by counting the number of "1" bits in the binary representation, a skill often tested in subnetting practice questions. Refer to Cisco's binary mask tables for quick validation and reinforce your learning with a subnetting practice questions quiz.

  3. Calculating Block Size and Subnet Increments -

    Use the "magic number" method (256 minus the mask octet) to determine block size and identify subnet increments (e.g., 256 - 192=64 for a /26 mask), a trick featured in many sample subnetting questions. Apply this across various octets to master boundary calculations and cross-check with reputable networking tutorials.

  4. Identifying Network, Broadcast, and Host Ranges -

    Perform a bitwise AND operation between an IP address and its mask to find the network address, then add block size minus one for the broadcast address; valid host IPs fall between these values. This core technique appears in every subnetting quiz and is rigorously defined in RFC 4632 for accurate ip address subnetting practice.

  5. Implementing VLSM and CIDR Summarization -

    Apply Variable Length Subnet Masking (VLSM) to allocate subnets of different sizes based on host requirements, then summarize multiple routes using CIDR aggregation to streamline routing tables. Hone this skill through subnetting example questions drawn from CCNA and Juniper Networks certification guides.

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