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AP Networking IP Addressing and Subnetting Practice Test

Work through 75 IP addressing and subnetting questions with subnet boundaries, host ranges, VLSM, masks, gateway checks, overlapping networks, and IPv6 scope.

Independent practice · Published and updated October 1, 2026 · Free for students and teachers
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What this test measures

This test checks prefix interpretation, address roles, capacity, VLSM allocation, local-versus-remote decisions, and IPv6 scope. Each difficulty has 25 questions; complete all three for 75. Easy emphasizes direct application, Medium combines evidence, and Hard asks you to choose among plausible diagnoses or evaluate what a result actually proves.

These are independent instructional questions, not released College Board items or a replica of Bluebook. The difficulty labels are local teaching choices; a score here does not predict an AP score.

For explanation before testing, use the ip addressing and subnetting guide. Return to AP Networking Practice Tests by Topic to compare results across topics instead of repeating only your strongest subject.

Interactive testWorked reasoning75 explained questionsNext study step

Take the 25-question test for your selected level

Choose one answer to receive immediate feedback. Your first selection is scored; read the explanation before moving on. Use keyboard Tab to reach controls and arrow keys within the answer group. You can use these as untimed diagnostic practice, pausing to write your reasoning.

Changing difficulty or restarting clears the current attempt. Download your log first if you want to keep it.

Read the boundary before naming the fault

Representative reasoning

192.168.10.77/26 belongs to the block beginning at .64. A /26 has 64 addresses, so the block ends at .127. The conventional usable range is .65 through .126. Writing the three boundaries before selecting an answer prevents confusing the last host with the broadcast address.

Find the containing block before naming the address role. A last octet of .0 or .255 is not enough: the prefix determines which bits are host bits.

A tempting shortcut to avoid

Two addresses sharing their first three octets need not be local under /25, /26, or /27. Conversely, 192.168.0.255/23 can be a usable host because the containing subnet ends at 192.168.1.255. Read the actual prefix rather than importing a /24 assumption.

A worked decision

A /24 divided among 100, 50, and 20-host LANs can use .0/25, .128/26, and .192/27. These provide 126, 62, and 30 conventional usable addresses. Place the largest block first, check alignment, and compare full address spans for overlap. Counting demand alone does not prove an allocation fits.

For a second representation of the same skill, review the related topic explanation. Compare its evidence with your calculation or packet-path sketch rather than memorizing the answer letter.

How to read the explained question library

All 75 items are available below as ordinary HTML. Open a question to view its choices, answer, and explanation. For retrieval practice, read the prompt, write a decision without opening it, then compare your reasoning. This library also gives teachers a stable question ID for discussion.

The six groups contain two halves of each difficulty, with 12 and 13 questions respectively. The interactive version uses the same questions and answer keys.

Easy explained questions · 1–12

Start by identifying the function or applying one explicit rule. These 12 questions are 1–12 of the easy set.

addressing-easy-1 · How many bits are in an IPv4 address?
  1. 32
  2. 128
  3. 64
  4. 16

Correct answer: A. 32

IPv4 uses four 8-bit octets, totaling 32 bits. IPv6 addresses contain 128 bits.

addressing-easy-2 · Which mask corresponds to /24?
  1. 255.255.255.128
  2. 255.255.0.0
  3. 255.255.255.0
  4. 255.255.255.252

Correct answer: C. 255.255.255.0

A /24 has 24 leading one bits: three full octets followed by eight zero bits.

addressing-easy-3 · What does an IPv4 subnet mask separate?
  1. DNS names from passwords
  2. TCP from UDP ports
  3. Network bits from host bits
  4. MAC vendor bits from VLAN IDs

Correct answer: C. Network bits from host bits

The mask tells the host which address bits define its subnet. It does not encode service ports.

addressing-easy-4 · Which address is private under RFC 1918?
  1. 192.169.1.1
  2. 8.8.8.8
  3. 10.4.5.6
  4. 172.32.1.1

Correct answer: C. 10.4.5.6

10.0.0.0/8 is private. The private 172 range ends at 172.31.255.255, and the private 192 range is 192.168.0.0/16.

addressing-easy-5 · What is the network address of 192.168.7.44/24?
  1. 192.168.7.255
  2. 192.168.7.44
  3. 192.168.7.1
  4. 192.168.7.0

Correct answer: D. 192.168.7.0

The /24 mask keeps the first three octets and clears the host octet. The host's own address is not the network identifier.

addressing-easy-6 · What is the broadcast address of 192.168.7.0/24?
  1. 192.168.7.255
  2. 192.168.8.0
  3. 192.168.7.0
  4. 192.168.7.1

Correct answer: A. 192.168.7.255

All eight host bits become one, yielding .255. The network identifier has all host bits zero.

addressing-easy-7 · How many conventional usable host addresses are in an IPv4 /24 LAN?
  1. 255
  2. 254
  3. 128
  4. 256

Correct answer: B. 254

There are 256 addresses. A conventional LAN reserves the network and broadcast addresses, leaving 254 host addresses.

addressing-easy-8 · Which is a usable host on 192.168.8.0/24?
  1. 192.168.8.10
  2. 192.168.8.0
  3. 192.168.8.255
  4. 192.168.9.10

Correct answer: A. 192.168.8.10

The usable range is .1 through .254. The other choices are the network identifier, broadcast, or another subnet.

addressing-easy-9 · A host is 10.2.3.4/24. Which destination is considered local?
  1. 10.2.3.99
  2. 11.2.3.99
  3. 10.3.3.99
  4. 10.2.4.99

Correct answer: A. 10.2.3.99

Both addresses must share the first 24 bits. Similar-looking octets do not compensate for a different third octet.

addressing-easy-10 · Which address is IPv4 loopback?
  1. 0.0.0.0
  2. 192.168.1.1
  3. 127.0.0.1
  4. 169.254.1.1

Correct answer: C. 127.0.0.1

127.0.0.0/8 is reserved for loopback. Traffic to loopback tests the local host rather than the external LAN.

addressing-easy-11 · A Windows client self-assigns 169.254.20.8. What is the strongest initial clue?
  1. Normal DHCP IPv4 assignment did not complete
  2. Its Internet route is guaranteed
  3. The server offered that exact lease
  4. DNS is definitely broken

Correct answer: A. Normal DHCP IPv4 assignment did not complete

An IPv4 link-local address suggests automatic DHCP assignment failed on this client. The address alone does not locate the underlying cause.

addressing-easy-12 · What does CIDR /26 mean?
  1. 26 usable hosts
  2. 26 network-prefix bits
  3. 26 VLANs
  4. 26 address octets

Correct answer: B. 26 network-prefix bits

The slash value counts prefix bits. Host count must be calculated from the six remaining bits.

Easy explained questions · 13–25

Start by identifying the function or applying one explicit rule. These 13 questions are 13–25 of the easy set.

addressing-easy-13 · Which mask corresponds to /25?
  1. 255.255.255.192
  2. 255.255.255.0
  3. 255.255.255.224
  4. 255.255.255.128

Correct answer: D. 255.255.255.128

The fourth octet begins with one network bit: binary 10000000 equals 128.

addressing-easy-14 · Which mask corresponds to /26?
  1. 255.255.255.240
  2. 255.255.0.0
  3. 255.255.255.192
  4. 255.255.255.128

Correct answer: C. 255.255.255.192

Two leading one bits in the fourth octet give binary 11000000, or 192.

addressing-easy-15 · What is the total address count in an IPv4 /27?
  1. 32
  2. 27
  3. 64
  4. 30

Correct answer: A. 32

Five host bits remain, so 2 to the fifth is 32 total addresses. Conventional usable count is 30.

addressing-easy-16 · Which value is a valid IPv4 octet?
  1. 256
  2. -1
  3. 300
  4. 254

Correct answer: D. 254

Each IPv4 octet represents eight bits and ranges from 0 through 255.

addressing-easy-17 · Which is a private IPv4 address?
  1. 172.20.5.4
  2. 172.40.5.4
  3. 192.170.1.2
  4. 203.0.114.8

Correct answer: A. 172.20.5.4

172.16.0.0/12 spans second-octet values 16 through 31. Not every address beginning with 172 is private.

addressing-easy-18 · How many bits are in an IPv6 address?
  1. 64
  2. 128
  3. 32
  4. 48

Correct answer: B. 128

IPv6 uses 128-bit addresses. A common subnet prefix is /64, but that is not the full address length.

addressing-easy-19 · In IPv6 notation, what does :: replace?
  1. A decimal subnet mask
  2. One consecutive run of zero groups
  3. Any missing nonzero digits
  4. Every repeated hexadecimal group

Correct answer: B. One consecutive run of zero groups

Double-colon compression represents consecutive all-zero 16-bit groups and may appear only once in one address.

addressing-easy-20 · Which IPv6 prefix identifies link-local unicast space?
  1. ff00::/8
  2. ::1/128
  3. fe80::/10
  4. 2000::/3

Correct answer: C. fe80::/10

fe80::/10 is link-local. ff00::/8 is multicast; ::1 is loopback.

addressing-easy-21 · Which IPv6 address is loopback?
  1. fe80::1
  2. 2001:db8::1
  3. ff02::1
  4. ::1

Correct answer: D. ::1

::1 refers to the host itself. A link-local address identifies an interface on a link instead.

addressing-easy-22 · Which statement about IPv6 broadcast is correct?
  1. Every subnet ends in a broadcast .255
  2. Only routers can receive multicast
  3. Broadcast requires a /24
  4. IPv6 uses multicast instead of broadcast

Correct answer: D. IPv6 uses multicast instead of broadcast

IPv6 does not use IPv4-style broadcast addresses. Relevant one-to-many discovery uses multicast.

addressing-easy-23 · What often happens when two LAN hosts use the same IPv4 address?
  1. Their bandwidth doubles
  2. Traffic may reach the wrong host or become intermittent
  3. DNS automatically repairs both
  4. Their MAC addresses become identical

Correct answer: B. Traffic may reach the wrong host or become intermittent

Conflicting IP-to-MAC mappings can disrupt delivery. Unique MAC addresses do not make a duplicate IP safe.

addressing-easy-24 · What does a DHCP reservation usually provide?
  1. A consistent lease address for an identified client
  2. A faster Ethernet clock
  3. A replacement subnet mask standard
  4. A public DNS zone

Correct answer: A. A consistent lease address for an identified client

A reservation ties address allocation to a client identifier, commonly a MAC address. The client still uses DHCP.

addressing-easy-25 · On an ordinary IPv4 subnet, what address should a directly reachable default gateway have?
  1. Any arbitrary remote address
  2. A usable address in the host's local subnet
  3. Always 127.0.0.1
  4. The subnet's broadcast address

Correct answer: B. A usable address in the host's local subnet

The host normally needs to reach its next hop on the local link. A gateway outside that subnet requires special configuration beyond this ordinary case.

Medium explained questions · 1–12

Combine the configuration with the symptom before selecting a decision. These 12 questions are 1–12 of the medium set.

addressing-medium-1 · Find the network containing 192.168.10.77/26.
  1. 192.168.10.0/26
  2. 192.168.10.77/26
  3. 192.168.10.128/26
  4. 192.168.10.64/26

Correct answer: D. 192.168.10.64/26

A /26 uses blocks of 64: 0, 64, 128, 192. Address 77 lies in the 64–127 block.

addressing-medium-2 · What is the broadcast of 192.168.10.64/26?
  1. 192.168.10.127
  2. 192.168.10.128
  3. 192.168.10.126
  4. 192.168.10.255

Correct answer: A. 192.168.10.127

The 64-address block starts at 64 and ends at 127. .126 is the last conventional usable host.

addressing-medium-3 · What is the usable host range of 10.0.0.96/27?
  1. 10.0.0.98–10.0.0.128
  2. 10.0.0.97–10.0.0.126
  3. 10.0.0.1–10.0.0.30
  4. 10.0.0.96–10.0.0.127

Correct answer: B. 10.0.0.97–10.0.0.126

The /27 block ends at 127. Reserve .96 for the network and .127 for broadcast.

addressing-medium-4 · A conventional LAN needs 50 hosts. What is the smallest suitable prefix?
  1. /27
  2. /29
  3. /28
  4. /26

Correct answer: D. /26

A /26 provides 62 usable addresses; /27 provides only 30. Smaller prefix numbers represent larger address blocks.

addressing-medium-5 · A conventional LAN needs 12 hosts. What is the smallest suitable prefix?
  1. /30
  2. /28
  3. /31
  4. /29

Correct answer: B. /28

A /28 has 16 total and 14 conventional usable addresses. A /29 has only six usable.

addressing-medium-6 · How many /26 subnets fit inside one /24?
  1. 8
  2. 16
  3. 4
  4. 2

Correct answer: C. 4

Two additional network bits create 2 squared, or four, equal subnets.

addressing-medium-7 · Host A is 192.168.1.30/27; B is 192.168.1.40/27. What must be recognized?
  1. Both have duplicate addresses
  2. They are local because the first three octets match
  3. They are in different subnets
  4. Both are broadcast addresses

Correct answer: C. They are in different subnets

A lies in .0–.31 and B in .32–.63. The /27 boundary matters more than visual similarity.

addressing-medium-8 · What is the network of 10.3.19.200/20?
  1. 10.3.16.0/20
  2. 10.3.0.0/20
  3. 10.3.19.0/20
  4. 10.3.20.0/20

Correct answer: A. 10.3.16.0/20

A /20 advances by 16 in the third octet. Third-octet value 19 lies in the 16–31 block.

addressing-medium-9 · What is the broadcast of 172.16.8.0/23?
  1. 172.16.9.255
  2. 172.16.10.0
  3. 172.16.9.254
  4. 172.16.8.255

Correct answer: A. 172.16.9.255

A /23 covers two consecutive third-octet values. This block spans 172.16.8.0 through 172.16.9.255.

addressing-medium-10 · Host 192.168.4.130/25 has gateway 192.168.4.1. What is wrong in an ordinary LAN design?
  1. The prefix has no usable addresses
  2. The gateway is outside the host's /25 subnet
  3. The gateway is broadcast
  4. The host is a network address

Correct answer: B. The gateway is outside the host's /25 subnet

The host belongs to .128–.255; .1 belongs to .0–.127. Sharing the first three octets is insufficient.

addressing-medium-11 · Which address is valid on 192.168.4.128/25?
  1. 192.168.4.255
  2. 192.168.4.127
  3. 192.168.4.128
  4. 192.168.4.129

Correct answer: D. 192.168.4.129

The usable range is .129–.254. .128 is network and .255 broadcast for this subnet.

addressing-medium-12 · A /24 is split into /28 blocks. How many blocks result?
  1. 8
  2. 16
  3. 4
  4. 32

Correct answer: B. 16

Four prefix bits are added, yielding 2 to the fourth, or 16 subnets.

Medium explained questions · 13–25

Combine the configuration with the symptom before selecting a decision. These 13 questions are 13–25 of the medium set.

addressing-medium-13 · What is the network containing 10.0.0.14/30?
  1. 10.0.0.12/30
  2. 10.0.0.16/30
  3. 10.0.0.8/30
  4. 10.0.0.14/30

Correct answer: A. 10.0.0.12/30

A /30 advances by four. The block is .12–.15, with usable hosts .13 and .14.

addressing-medium-14 · A conventional /30 link uses 10.0.0.12/30. Which pair is usable?
  1. 10.0.0.13 and 10.0.0.14
  2. 10.0.0.12 and 10.0.0.13
  3. 10.0.0.14 and 10.0.0.15
  4. 10.0.0.15 and 10.0.0.16

Correct answer: A. 10.0.0.13 and 10.0.0.14

The network is .12 and broadcast is .15. The two interior addresses are usable.

addressing-medium-15 · Which summary exactly covers 192.168.8.0/24 and 192.168.9.0/24?
  1. 192.168.0.0/16
  2. 192.168.8.0/25
  3. 192.168.8.0/23
  4. 192.168.9.0/23

Correct answer: C. 192.168.8.0/23

The two adjacent /24s begin on an even third-octet /23 boundary. A /16 includes many additional networks.

addressing-medium-16 · Host 10.1.2.20/16 sends to 10.1.99.30. Which next-hop behavior is expected without a more specific route?
  1. Treat the destination as on-link and resolve its link-layer address
  2. Always use the default gateway
  3. Send to the DNS server first
  4. Convert the packet to IPv6

Correct answer: A. Treat the destination as on-link and resolve its link-layer address

Both share 10.1 under /16. The connected subnet normally causes local neighbor resolution rather than default-route use.

addressing-medium-17 · What is the mask for /22?
  1. 255.255.254.0
  2. 255.255.248.0
  3. 255.255.252.0
  4. 255.255.255.252

Correct answer: C. 255.255.252.0

Six network bits in the third octet equal 252. The last octet contains eight host bits.

addressing-medium-18 · What is the conventional usable-host count of /22?
  1. 1024
  2. 254
  3. 1022
  4. 510

Correct answer: C. 1022

Ten host bits give 1024 addresses; subtract the conventional network and broadcast reservations.

addressing-medium-19 · What is the fully expanded IPv6 form of 2001:db8::5?
  1. 2001:db8:5
  2. 2001:0db8:0000:0000:0000:0005
  3. 2001:0db8:0000:0000:0000:0000:0000:0005
  4. 2001:0db8:0005:0000:0000:0000:0000:0000

Correct answer: C. 2001:0db8:0000:0000:0000:0000:0000:0005

An IPv6 address contains eight 16-bit groups. The compression here supplies five zero groups between db8 and 5.

addressing-medium-20 · Which pair belongs to the same /64 IPv6 subnet?
  1. 2001:db8:1:2::10 and 2001:db8:1:2::20
  2. 2001:db8:1:2::10 and 2001:db8:1:3::20
  3. 2001:db8:1:2::10 and 2001:db9:1:2::20
  4. 2001:db8:1:2::10 and 2001:db8:2:2::20

Correct answer: A. 2001:db8:1:2::10 and 2001:db8:1:2::20

A /64 fixes the first four 16-bit groups. Addresses differing after those groups can share the subnet.

addressing-medium-21 · A DHCP pool includes a printer's manually assigned IP. What should be changed?
  1. Remove the subnet mask
  2. Give every client the printer MAC
  3. Use the broadcast as gateway
  4. Exclude or reserve the printer address appropriately

Correct answer: D. Exclude or reserve the printer address appropriately

Overlapping static and dynamic allocation risks duplicate addresses. Coordination prevents the conflict without changing the whole subnet.

addressing-medium-22 · What is the network of 172.20.6.65/26?
  1. 172.20.6.64/26
  2. 172.20.6.65/26
  3. 172.20.6.0/26
  4. 172.20.6.128/26

Correct answer: A. 172.20.6.64/26

Blocks advance by 64. .65 is the first conventional usable host in the .64 block.

addressing-medium-23 · Which conventional subnet provides at least 100 usable addresses with least waste?
  1. /28
  2. /26
  3. /25
  4. /27

Correct answer: C. /25

A /25 offers 126 usable addresses. A /26 offers only 62 and cannot meet the requirement.

addressing-medium-24 · A host has 192.168.2.64/26. Why is this unsuitable as a conventional host address?
  1. It is the subnet identifier
  2. The first octet is invalid
  3. It is the broadcast
  4. It is always private multicast

Correct answer: A. It is the subnet identifier

The .64/26 boundary begins this subnet. The host range starts at .65.

addressing-medium-25 · A client has valid IP and mask but no gateway. What can still work in an ordinary LAN?
  1. Every remote subnet
  2. All public websites by IP
  3. Any routed VLAN
  4. Same-subnet communication

Correct answer: D. Same-subnet communication

Local destinations do not require a router. Off-subnet destinations normally need an appropriate route or gateway.

Hard explained questions · 1–12

State what the evidence proves and reject conclusions that reach beyond it. These 12 questions are 1–12 of the hard set.

addressing-hard-1 · Allocate a /24 to LANs needing 100, 50, and 20 hosts. Which sequence fits without overlap?
  1. /25 at .0, /26 at .128, /27 at .192
  2. /26 at .0, /26 at .64, /27 at .128
  3. /25 at .0, /25 at .128, /27 at .192
  4. /25 at .0, /26 at .64, /27 at .96

Correct answer: A. /25 at .0, /26 at .128, /27 at .192

The chosen blocks provide 126, 62, and 30 usable addresses and do not overlap. The other plans lack capacity or allocate inside existing blocks.

addressing-hard-2 · A is 192.168.10.70/26, B is 192.168.10.130/24; both share a Layer 2 segment. What inconsistency can cause asymmetric behavior?
  1. Both agree they are local
  2. Their MAC addresses must match
  3. A treats B as remote while B treats A as local
  4. Both agree they are remote

Correct answer: C. A treats B as remote while B treats A as local

A's subnet is .64–.127, excluding B. B's /24 includes A. Different on-link decisions can send traffic along different paths.

addressing-hard-3 · A client uses 10.0.5.40/24 and gateway 10.0.5.1. A remote VPN needs 10.0.5.80. Why may delivery fail?
  1. The overlapping local subnet can cause local ARP instead of VPN routing
  2. DNS necessarily changes the mask
  3. The address is broadcast
  4. A /24 has no host .80

Correct answer: A. The overlapping local subnet can cause local ARP instead of VPN routing

The connected local route competes with the remote addressing plan. Inspect actual routes and avoid overlapping spaces or provide an intentional specific-route design.

addressing-hard-4 · Which single CIDR block exactly covers 10.0.8.0 through 10.0.11.255?
  1. 10.0.8.0/23
  2. 10.0.0.0/21
  3. 10.0.10.0/22
  4. 10.0.8.0/22

Correct answer: D. 10.0.8.0/22

A /22 spans four third-octet values on a multiple-of-four boundary. Values 8 through 11 meet both conditions.

addressing-hard-5 · Two LANs are configured as 192.168.1.0/25 and 192.168.1.64/26. What must be fixed?
  1. Both are public ranges
  2. They need longer DNS names
  3. The second subnet overlaps the first
  4. Only the MAC table can resolve the design

Correct answer: C. The second subnet overlaps the first

The first block covers .0–.127; the second covers .64–.127 inside it. Routed LANs need an intentional nonoverlapping address plan.

addressing-hard-6 · A /27 LAN has 28 clients and needs a gateway plus two new printers. Will conventional usable capacity suffice?
  1. Yes: the gateway does not need an IP
  2. Yes: all 32 addresses are hosts
  3. No: /27 provides only 16 addresses
  4. No: 31 addresses are needed but only 30 are usable

Correct answer: D. No: 31 addresses are needed but only 30 are usable

Count every addressed interface: 28 + 1 + 2 = 31. Conventional /27 capacity is 30, so expand or redesign.

addressing-hard-7 · A host is 172.16.31.200/20. What are its network and broadcast?
  1. 172.16.0.0 and 172.16.15.255
  2. 172.16.32.0 and 172.16.47.255
  3. 172.16.16.0 and 172.16.31.255
  4. 172.16.31.0 and 172.16.31.255

Correct answer: C. 172.16.16.0 and 172.16.31.255

/20 blocks advance by 16 in the third octet. Third-octet value 31 is the last value in the 16–31 block.

addressing-hard-8 · A proposed /23 is written as 192.168.5.0/23. What is the actual containing network?
  1. 192.168.5.0/23 as a boundary
  2. 192.168.6.0/23
  3. 192.168.0.0/23
  4. 192.168.4.0/23

Correct answer: D. 192.168.4.0/23

A /23 aligns on even third-octet values. Applying the mask to .5 yields .4, covering .4 and .5.

addressing-hard-9 · A routed interface uses 10.10.10.1/24; its DHCP scope sends clients /16 masks. What symptom is plausible?
  1. DNS answers cannot contain A records
  2. Clients ARP for some destinations that the router considers remote
  3. Clients always refuse every address
  4. IPv4 becomes IPv6

Correct answer: B. Clients ARP for some destinations that the router considers remote

The too-wide client mask classifies other 10.10 networks as on-link. Clients may ARP locally instead of sending through the router.

addressing-hard-10 · You split 192.168.50.0/24 into eight equal conventional LANs. What prefix and usable count apply to each?
  1. /26 with 62 usable
  2. /27 with 30 usable
  3. /25 with 126 usable
  4. /28 with 14 usable

Correct answer: B. /27 with 30 usable

Three borrowed bits create eight subnets; /24 + 3 = /27. Five host bits leave 32 total and 30 conventional usable.

addressing-hard-11 · A point-to-point IPv4 link supports RFC 3021 /31 addressing. How many endpoint addresses can that /31 provide?
  1. 30
  2. 0
  3. 1
  4. 2

Correct answer: D. 2

The /31 point-to-point exception uses both addresses as endpoints. Do not apply the ordinary LAN subtraction rule indiscriminately.

addressing-hard-12 · A routed host route is 192.0.2.9/32. What does its prefix identify?
  1. One exact IPv4 address
  2. A LAN with 30 hosts
  3. An invalid prefix in every context
  4. A broadcast to 32 hosts

Correct answer: A. One exact IPv4 address

A /32 fixes all 32 bits and matches one address. It is useful for host routes even though it is not a conventional LAN block.

Hard explained questions · 13–25

State what the evidence proves and reject conclusions that reach beyond it. These 13 questions are 13–25 of the hard set.

addressing-hard-13 · Two devices alternate replies for 10.5.5.20, and ARP captures show different MACs advertising it. What is best supported?
  1. A wider subnet is definitely required
  2. An address conflict or identity impersonation deserves investigation
  3. Every ARP change is harmless
  4. The gateway is definitely down

Correct answer: B. An address conflict or identity impersonation deserves investigation

Competing advertisements for one IP explain unstable delivery. Confirm authorized assignment and endpoint identity before declaring a benign change.

addressing-hard-14 · A network needs 500 client IPs plus one gateway. What is the smallest conventional subnet that fits?
  1. /26
  2. /25
  3. /23
  4. /24

Correct answer: C. /23

A /23 provides 510 usable addresses, enough for 501 interfaces. A /24 offers only 254.

addressing-hard-15 · A server is 10.8.2.254/23. Which proposed gateway is in the same subnet?
  1. 10.9.2.1
  2. 10.8.1.1
  3. 10.8.4.1
  4. 10.8.3.1

Correct answer: D. 10.8.3.1

The /23 beginning at 10.8.2.0 spans third-octet values 2 and 3. The .3.1 address is therefore local.

addressing-hard-16 · Four /26 subnets fill a /24. Which broadcast belongs to the third subnet in ascending order?
  1. .192
  2. .127
  3. .255
  4. .191

Correct answer: D. .191

Subnet starts are .0, .64, .128, and .192. The third block ends at .191.

addressing-hard-17 · You allocate 10.0.0.0/26 and 10.0.0.64/27. What is the earliest aligned free /27 afterward?
  1. 10.0.0.96/27
  2. 10.0.0.80/27
  3. 10.0.0.95/27
  4. 10.0.0.64/27

Correct answer: A. 10.0.0.96/27

The allocated spans end at .63 and .95. The next /27-aligned boundary is .96, not .80 or .95.

addressing-hard-18 · A host is 192.168.6.200/26. Which gateway proposal is unusable because it is broadcast?
  1. 192.168.6.194
  2. 192.168.6.193
  3. 192.168.6.255
  4. 192.168.6.254

Correct answer: C. 192.168.6.255

The host's subnet is .192–.255, with broadcast .255. The other listed addresses fall in the usable range.

addressing-hard-19 · A static client uses 192.168.9.10/24; all DHCP clients use 192.168.9.128/25. Why does the static address not belong to their IP subnet?
  1. The static client must have a duplicate MAC
  2. Private IPs cannot mix
  3. The hostname is too short
  4. Their /25 begins at .128 and excludes .10

Correct answer: D. Their /25 begins at .128 and excludes .10

VLAN co-location does not create identical IP subnet membership. Correct the address plan and mask together.

addressing-hard-20 · Which summary exactly covers 10.4.12.0/24 through 10.4.15.0/24?
  1. 10.4.12.0/22
  2. 10.4.0.0/20
  3. 10.4.14.0/22
  4. 10.4.12.0/23

Correct answer: A. 10.4.12.0/22

Four aligned /24s form one /22. The /20 overincludes additional networks, and .14 is not the correct /22 boundary.

addressing-hard-21 · A technician claims every address ending in .255 is broadcast. Which counterexample refutes that?
  1. 255.255.255.255 is an ordinary host
  2. 192.168.1.255/23 is always usable
  3. 192.168.0.255/23 is a usable host address
  4. 192.168.0.255/24 is usable

Correct answer: C. 192.168.0.255/23 is a usable host address

In 192.168.0.0/23, broadcast is 192.168.1.255. The prefix, not the final octet alone, determines address role.

addressing-hard-22 · A technician claims every address ending in .0 is a network identifier. Which is a usable counterexample?
  1. 192.168.0.0/23
  2. 192.168.1.0/23
  3. 192.168.2.0/23
  4. 10.0.0.0/24

Correct answer: B. 192.168.1.0/23

192.168.1.0 lies inside 192.168.0.0/23 and is neither its network nor broadcast. Host bits span more than one octet.

addressing-hard-23 · An IPv6 link-local ping works only when an interface zone is specified. Why can that be necessary?
  1. Link-local addresses are globally unique routes
  2. DNS converts the zone to a VLAN
  3. The same link-local scope can exist on multiple interfaces
  4. IPv6 has no interfaces

Correct answer: C. The same link-local scope can exist on multiple interfaces

A zone identifier tells the host which local link to use. Link-local scope is not a globally routable destination identity.

addressing-hard-24 · A dual-stack client reaches a service over IPv4 but fails over IPv6. What should be tested?
  1. Only IPv4 DHCP lease time
  2. Only Ethernet category
  3. IPv6 address, neighbor, route, and service-policy dependencies separately
  4. Whether all IPv4 addresses are public

Correct answer: C. IPv6 address, neighbor, route, and service-policy dependencies separately

Successful IPv4 proves that stack's tested path, not IPv6 routing or filtering. Compare both families explicitly.

addressing-hard-25 · A /24 must hold LANs with 120 and 120 hosts plus a conventional /30 routed link. Does it fit?
  1. No: two /25 blocks consume the entire /24 before the link
  2. Yes: 242 hosts always fit in 254 usable
  3. Yes: the /30 can overlap one LAN
  4. No: a /25 holds only 62 hosts

Correct answer: A. No: two /25 blocks consume the entire /24 before the link

Each 120-host LAN needs a 128-address /25. Block allocation, not merely summing usable host demand, determines whether space remains.

After address-boundary calculations, examine assignment ownership and the lease lifecycle in AP Networking DHCP Leases and Relays Practice Test. Its original Easy, Medium and Hard scenarios add 75 explained items.

After prefix and scope practice, investigate router advertisements and local adjacency in AP Networking IPv6 and Neighbor Discovery Practice Test. Its 75 explained questions add a separate Easy, Medium, and Hard diagnostic for that next reasoning step.

Use the result to choose your next study step

Explain a mask error through forwarding behavior. A host configured with /16 where /24 is intended may ARP for another 10.x destination locally instead of using the gateway. Show the host's network calculation, the destination classification, and the observed ARP or route evidence before changing DHCP or static settings.

Record whether each miss concerned boundary calculation, address planning, scope. Copy one decisive observation and one rejected inference into your error log. Then review the focused supporting resource, explain a missed item without its options, and retry the relevant level after a separate study session.

For course-level transfer, move to Unit 1 practice, Unit 2 practice, Unit 3 practice. When the topic labels no longer guide your answer, use full practice exams and artifact analysis to combine selection, evidence, and explanation.

Teacher use

Assign one level as a diagnostic, collect explanations for two missed items, and group students by the missing step rather than total score alone. Ask pairs to compare a correct conclusion with a tempting overclaim. Do not treat repeated exposure to the same answer as independent evidence of mastery.