Network+ Subnetting Questions Explained

Subnetting questions appear on the Network+ exam and require the same core skills as CCNA subnetting but with less complexity. The exam tests CIDR notation, converting between prefix and dotted decimal, calculating usable hosts, and identifying subnet boundaries.

CIDR Notation and Subnet Masks

CIDR (Classless Inter-Domain Routing) notation represents a subnet mask as a prefix length — the number of bits set to 1 in the mask.

Common conversions:

The host count formula: 2^(32 - prefix) - 2

For /26: 2^(32-26) - 2 = 2^6 - 2 = 64 - 2 = 62 usable hosts.

The Block Size Method

The block size is the number of addresses in each subnet. For subnets in the last octet: block size = 256 - (interesting octet value in the mask).

For /26 (mask .192): block size = 256 - 192 = 64.

Subnets start at multiples of the block size: 0, 64, 128, 192.

Identifying Which Subnet an Address Belongs To

Given address 192.168.1.100/26, which subnet?

Block size = 64. Subnets: 0, 64, 128, 192. 100 is between 64 and 128, so the subnet is 192.168.1.64/26. Broadcast = 192.168.1.127. Last host = .126. First host = .65.

Method: divide the host octet by the block size → 100 / 64 = 1.5 → round down to 1 → 1 × 64 = 64.

Working Backwards from Required Hosts

Network+ questions often ask: "You need to accommodate 50 hosts. What is the smallest subnet?"

Work backwards:

• /27 gives 30 hosts — not enough
• /26 gives 62 hosts — enough
• Answer: /26

The smallest subnet that fits N hosts: find the smallest power of 2 greater than N + 2, then determine the prefix.

50 hosts + 2 (network + broadcast) = 52. Next power of 2 above 52 = 64. 64 = 2^6, so 6 host bits, prefix = 32 - 6 = /26.

Subnetting with Third-Octet Variation

For prefixes like /20 or /22, the subnet boundary is in the third octet.

For /22 (mask 255.255.252.0): block size in the third octet = 256 - 252 = 4.

Subnets in the third octet: 0, 4, 8, 12...

Given 172.16.14.50/22:

• Third octet = 14
• 14 / 4 = 3.5 → round down to 3 → 3 × 4 = 12
• Subnet: 172.16.12.0/22
• Broadcast: 172.16.15.255 (next boundary at .16.0, so broadcast is .15.255)
• Usable range: 172.16.12.1 to 172.16.15.254

Private vs Public IP Ranges

Network+ also tests knowledge of private IP ranges (RFC 1918):

"Which address range is used by APIPA?" — 169.254.0.0/16.

"Which of the following is not a private IP address?" — Any address outside the three RFC 1918 ranges above.

Practice Network+ subnetting questions with timed practice to build calculation speed for exam day.

The Block Size Method — Full Reference Table

For Network+ N10-009, subnetting questions focus on the /24 through /30 range. Memorise this table and you can answer the majority of subnetting questions without binary conversion:

Block size = 256 minus the last non-255 octet of the subnet mask. The block size tells you how far apart subnet boundaries are and how many total addresses are in each subnet (usable = block size minus 2).

For subnetting in the third octet (/17 through /23):

VLSM Worked Example — From Scratch

Given the address block 192.168.50.0/24 and these requirements:

• Site A: 60 hosts
• Site B: 25 hosts
• Site C: 12 hosts
• Site D: 10 hosts
• Point-to-point link 1: 2 hosts
• Point-to-point link 2: 2 hosts

Sort largest first, then allocate:

Site A — 60 hosts: Need /26 (62 usable). Assign 192.168.50.0/26.

• Network: 192.168.50.0
• Usable: 192.168.50.1 – 192.168.50.62
• Broadcast: 192.168.50.63

Site B — 25 hosts: Need /27 (30 usable). Assign 192.168.50.64/27.

• Network: 192.168.50.64
• Usable: 192.168.50.65 – 192.168.50.94
• Broadcast: 192.168.50.95

Site C — 12 hosts: Need /28 (14 usable). Assign 192.168.50.96/28.

• Network: 192.168.50.96
• Usable: 192.168.50.97 – 192.168.50.110
• Broadcast: 192.168.50.111

Site D — 10 hosts: Need /28 (14 usable — /29 only gives 6, not enough). Assign 192.168.50.112/28.

• Network: 192.168.50.112
• Usable: 192.168.50.113 – 192.168.50.126
• Broadcast: 192.168.50.127

Point-to-point link 1: Need /30 (2 usable). Assign 192.168.50.128/30.

• Usable: 192.168.50.129 – 192.168.50.130

Point-to-point link 2: Need /30. Assign 192.168.50.132/30.

• Usable: 192.168.50.133 – 192.168.50.134

Total used: .0 through .135 (136 addresses). Remaining: .136 through .255 (120 addresses available for future use). The /24 is not exhausted.

Supernetting / Route Summarization — The Compression Direction

Supernetting combines multiple subnets into a single summary route, reducing the routing table. This is the opposite of subnetting.

Example: You have four contiguous /24 networks:

• 192.168.4.0/24
• 192.168.5.0/24
• 192.168.6.0/24
• 192.168.7.0/24

To summarize: find the common bits. In binary, the third octet:

• 4 = 00000100
• 5 = 00000101
• 6 = 00000110
• 7 = 00000111

Common bits: 000001 (first 6 bits match). Add to the /16 for the first two octets: /16 + 6 = /22. The summary route is 192.168.4.0/22.

The summary route covers .4.0 through .7.255 — verify by checking: block size for /22 is 4 (in the third octet), starting at .4, next boundary at .8. That covers exactly .4, .5, .6, and .7.

Exam question: "Which summary route covers 192.168.4.0/24 through 192.168.7.0/24?" → 192.168.4.0/22.

The requirement for summarization: the networks must be contiguous (no gaps) and start on a boundary aligned with the summary prefix. 192.168.4.0/22 is valid because 4 is a multiple of 4 (the block size for /22). 192.168.5.0/22 would not be valid because 5 is not a multiple of 4.

IPv6 Addressing Basics — Network+ Now Tests This

Network+ N10-009 includes IPv6 addressing questions. The key facts:

Address structure: 128 bits, written as 8 groups of 4 hex digits separated by colons. :: represents one or more consecutive groups of all zeros (used once per address). Example: 2001:db8::1 is short for 2001:0db8:0000:0000:0000:0000:0000:0001.

Subnet size: always /64 for user-facing subnets. The first 64 bits are the network prefix, the last 64 bits are the interface identifier. You will not subnet smaller than /64 in normal practice — the interface ID needs all 64 bits for EUI-64 or random generation.

EUI-64: a method for generating the 64-bit interface identifier from the device's 48-bit MAC address. The MAC is split in half, FF:FE is inserted in the middle, and the 7th bit is flipped. This creates a globally unique interface ID. Not all systems use EUI-64 — modern OS versions use random IDs for privacy.

Special address ranges:

• fe80::/10 — link-local addresses (like 169.254.x.x in IPv4). Auto-configured, not routable. Every IPv6 interface always has a link-local address.
• ff00::/8 — multicast. IPv6 has no broadcast — broadcast functions are replaced by multicast.
• ::1 — loopback (equivalent to 127.0.0.1)
• 2001:db8::/32 — documentation and example addresses (not routable)

Address assignment: SLAAC (Stateless Address Autoconfiguration) — the client uses the router's advertised prefix + its own interface ID. No DHCP server needed. Stateless DHCPv6 — provides DNS and other options but not the IP address (SLAAC handles the IP). Stateful DHCPv6 — provides the full IP address and options, like DHCPv4.

The Exam's Trick Subnetting Questions

"How many usable host addresses does a /27 provide?" Block size = 32, usable = 32 - 2 = 30. Wrong answer trap: 32.

"How many /26 subnets can be created from a single /24?" /24 block size = 256, /26 block size = 64. 256 ÷ 64 = 4 subnets. Wrong answer trap: 2 (candidates confuse with /25 which gives 2).

"What is the broadcast address of the subnet containing 10.10.10.65/26?" /26 block size = 64. Boundaries at 0, 64, 128. 65 falls in the .64 subnet. Next boundary: 128. Broadcast = 128 - 1 = 10.10.10.127. Wrong answer traps: .127 (correct), .63 (wrong subnet), .191 (next subnet's broadcast).

"A host has IP 172.16.45.200 with subnet mask 255.255.240.0. What is the network address?" /20, interesting in third octet, block size = 256 - 240 = 16. Third octet boundaries: 0, 16, 32, 48 — 45 falls in the 32 block. Network: 172.16.32.0. Broadcast: 172.16.47.255.

"A /29 subnet is used for a point-to-point link. How many addresses are wasted?" /29 has 8 total, 6 usable. A point-to-point link needs 2 usable. Wasted = 6 - 2 = 4 usable addresses (plus 2 for network/broadcast = 6 total addresses consumed, 4 host addresses unused). A /30 would be more efficient.

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