CCNA Subnetting Questions

MCQhard

An interface is configured with 10.24.7.158/27. What is the broadcast address of that subnet?

A.10.24.7.159

B.10.24.7.191

C.10.24.7.127

D.10.24.7.160

AnswerA

The /27 CIDR notation indicates a subnet mask of 255.255.255.224, leaving 5 host bits. This defines subnet blocks of 32 addresses each. For the fourth octet, the given IP address 158 falls within the subnet range starting at 128 (10.24.7.128) and ending at 159 (10.24.7.159). The broadcast address is always the last address in a subnet, where all host bits are set to one. Therefore, 10.24.7.159 correctly identifies the broadcast address for the 10.24.7.158/27 interface.

Why this answer

A /27 uses blocks of 32 addresses. The block containing .158 is 10.24.7.128 through 10.24.7.159, so .159 is the broadcast address.

Exam trap

A frequent exam trap is mistaking the broadcast address for the network address of the next subnet or the last address of a different subnet block. For example, 10.24.7.160 is the network address of the next /27 subnet, not the broadcast address of the current subnet. Candidates often confuse the last usable host address with the broadcast address or select an address from an adjacent subnet block.

This mistake leads to incorrect subnetting answers and can cause interface misconfigurations in real networks, resulting in communication failures.

Why the other options are wrong

Incorrect. 10.24.7.191 is the last address of a larger /26 subnet block, not the /27 block containing 10.24.7.158, so it cannot be the broadcast address here.

Incorrect. 10.24.7.127 is the broadcast address of the previous /27 subnet block (10.24.7.96/27), not the one containing 10.24.7.158.

Incorrect. 10.24.7.160 is the network address of the next /27 subnet block after 10.24.7.128/27, not the broadcast address of the current subnet.

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PBQhard

You are connected to R1. Configure IPv4 and IPv6 addressing on R1's GigabitEthernet0/0 and GigabitEthernet0/1 interfaces so that R1 can ping both R2's IPv4 address (203.0.113.2) and R2's IPv6 address (2001:db8:1::2). The current configuration has an incorrect subnet mask on G0/0, missing default gateway, and R1's G0/1 has a duplicate IPv4 address with R3. Also, use EUI-64 for IPv6 on G0/0 and static IPv6 assignment on G0/1. Ensure all issues are resolved and connectivity verified.

Network Topology

Hints

•Check the subnet mask on G0/0; it should match R2's /30.
•G0/1's IPv4 address conflicts with R3; use an unused address like .254.
•Enable IPv6 globally with 'ipv6 unicast-routing' before configuring interface IPv6 addresses.

A.Change G0/0 subnet mask to 255.255.255.252, add default gateway 203.0.113.2, change G0/1 IPv4 to 192.168.1.254, enable IPv6 routing, configure G0/0 with ipv6 address 2001:db8:1::/64 eui-64, and G0/1 with ipv6 address 2001:db8:2::1/64.

B.Change G0/0 subnet mask to 255.255.255.0, add default gateway 203.0.113.1, change G0/1 IPv4 to 192.168.1.254, enable IPv6 routing, configure G0/0 with ipv6 address 2001:db8:1::1/64, and G0/1 with ipv6 address 2001:db8:2::1/64.

C.Change G0/0 subnet mask to 255.255.255.252, add default gateway 203.0.113.2, change G0/1 IPv4 to 192.168.1.1, enable IPv6 routing, configure G0/0 with ipv6 address 2001:db8:1::/64 eui-64, and G0/1 with ipv6 address 2001:db8:2::1/64.

D.Change G0/0 subnet mask to 255.255.255.252, add default gateway 203.0.113.2, change G0/1 IPv4 to 192.168.1.254, enable IPv6 routing, configure G0/0 with ipv6 address 2001:db8:1::1/64, and G0/1 with ipv6 address 2001:db8:2::/64 eui-64.

AnswerA

solution

! R1
configure terminal
interface gigabitethernet0/0
ip address 203.0.113.1 255.255.255.252
ipv6 address 2001:db8:1::/64 eui-64
exit
interface gigabitethernet0/1
ip address 192.168.1.254 255.255.255.0
ipv6 address 2001:db8:2::1/64
exit
ip route 0.0.0.0 0.0.0.0 203.0.113.2
end

Why this answer

The subnet mask on G0/0 was incorrectly set to /24 instead of /30. While a /24 mask on 203.0.113.1 would include 203.0.113.2 in the same subnet from R1's perspective, the mismatch with R2's /30 mask leads to inconsistent subnet definitions and potential ARP or routing issues. Additionally, no default gateway was configured, so traffic to remote networks would fail.

On G0/1, the IPv4 address 192.168.1.1 was already used by R3, causing a duplicate IP conflict. IPv6 was not configured on either interface. The fix involved correcting the subnet mask on G0/0 to 255.255.255.252, adding a default gateway (203.0.113.2), assigning a unique IPv4 address to G0/1 (192.168.1.254), enabling IPv6 routing globally with `ipv6 unicast-routing`, configuring EUI-64 on G0/0 (`ipv6 address 2001:db8:1::/64 eui-64`), and static IPv6 on G0/1 (`ipv6 address 2001:db8:2::1/64`).

Exam trap

Watch out for subnet mask mismatches (e.g., /24 vs /30) and duplicate IP addresses. Also, note the specific IPv6 addressing requirements: EUI-64 on one interface and static on the other. Don't assume a default gateway can be any IP in the subnet; it must be the neighbor's IP.

Why the other options are wrong

The subnet mask /24 is too large, causing a mismatch with R2's /30; the default gateway must be R2's IP (203.0.113.2); EUI-64 is not used on G0/0.

The duplicate IPv4 address on G0/1 is not resolved; it still uses 192.168.1.1 which is already assigned to R3.

EUI-64 is required on G0/0, not G0/1; static IPv6 is required on G0/1, not G0/0.

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MCQhard

What prefix length corresponds to the subnet mask 255.255.255.224?

A./26

B./27

C./28

D./29

AnswerB

The subnet mask 255.255.255.224 converts to binary as 11111111.11111111.11111111.11100000, where the last octet indicates 3 additional network bits beyond the default /24. Adding 24 + 3 gives a prefix length of /27, which is the correct match. This leaves 5 host bits, providing 2^5 - 2 = 30 usable IP addresses per subnet.

Why this answer

The subnet mask 255.255.255.224 corresponds to /27. In plain language, the first three octets contribute 24 network bits, and 224 in binary is 11100000, which contributes 3 more network bits. That gives a total of 27 network bits.

This is a common prefix-conversion question because it checks whether you can move between dotted-decimal masks and prefix lengths confidently.

Exam trap

Be cautious not to confuse the binary values of different subnet masks. Always convert the last octet to binary to determine the correct prefix length.

Why the other options are wrong

The /26 prefix corresponds to subnet mask 255.255.255.192, which has 64 addresses per subnet, not 32. The mask 255.255.255.224 has 27 network bits, not 26.

The /28 prefix corresponds to subnet mask 255.255.255.240, which provides 16 addresses per subnet (14 usable). The mask 255.255.255.224 has 32 addresses per subnet, so /28 is incorrect.

The /29 prefix corresponds to subnet mask 255.255.255.248, which provides 8 addresses per subnet (6 usable). The mask 255.255.255.224 has 32 addresses, so /29 is incorrect.

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PBQhard

You are connected to R1 via console. R1 and R2 are directly connected via GigabitEthernet0/0. Your task is to configure IPv4 and IPv6 addressing on both routers so that they can ping each other's IPv4 and IPv6 addresses. The current configuration has intentional faults: R1's IPv4 subnet mask is incorrect, R2 is missing its default gateway, and R1's IPv6 address uses EUI-64 but is not working due to a duplicate IP. Correct the IPv4 mask on R1, assign a static IPv6 address on R2, and ensure both routers can reach each other.

Network Topology

Hints

•Check the subnet mask on R1's G0/0 — the link between two routers typically uses a /30 mask.
•R2 has no IPv6 address configured — assign one manually.
•R2 cannot reach R1's IPv4 address because they are on different subnets and R2 has no default gateway.

A.On R1, change the IPv4 mask to 255.255.255.252; on R2, assign IPv4 address 192.0.2.2/30 and IPv6 address 2001:db8:1::2/64; configure a default route on R2 pointing to 192.0.2.1.

B.On R1, change the IPv4 mask to 255.255.255.0; on R2, assign IPv4 address 192.0.2.2/24 and IPv6 address 2001:db8:1::2/64; no default route needed.

C.On R1, change the IPv4 mask to 255.255.255.252; on R2, assign IPv4 address 192.0.2.2/28 and IPv6 address 2001:db8:1::2/64; configure a default route on R2 pointing to 192.0.2.1.

D.On R1, change the IPv4 mask to 255.255.255.252; on R2, assign IPv4 address 192.0.2.2/30 and IPv6 address 2001:db8:1::1/64; configure a default route on R2 pointing to 192.0.2.1.

AnswerA

solution

! R1
configure terminal
interface GigabitEthernet0/0
ip address 192.0.2.1 255.255.255.252
end

! R2
configure terminal
interface GigabitEthernet0/0
ip address 192.0.2.2 255.255.255.252
ipv6 address 2001:db8:1::2/64
exit
ip route 0.0.0.0 0.0.0.0 192.0.2.1
end

Why this answer

R1's IPv4 mask was /28, but the correct mask for the link should be /30 to avoid overlapping subnets (192.0.2.0/28 includes both .1 and .14, but they are on the same link). R2 had no IPv6 address configured. Additionally, R1's EUI-64 address was valid but R2 needed a static IPv6 address.

The solution: on R1, change the mask to 255.255.255.252; on R2, assign an IPv4 address with mask /30 and a static IPv6 address 2001:db8:1::2/64; also add a default route on R2 pointing to 192.0.2.1 for IPv4. After these changes, both routers can ping each other's IPv4 and IPv6 addresses.

Exam trap

Watch out for subnet mask mismatches and duplicate IPv6 addresses. Always use /30 for point-to-point links and ensure each router has a unique IPv6 address on the same link.

Why the other options are wrong

The specific factual error: Using a /24 mask on a point-to-point link wastes addresses and may cause subnet overlap; also, R2 needs a default route to reach R1's IPv4 address if the mask is /30, but with /24 they are in the same subnet so no default route is needed, but the mask is still wrong.

The specific factual error: R2's IPv4 mask must match R1's mask to ensure both routers agree on the subnet boundary. Using /28 on R2 while R1 uses /30 creates a mismatch.

The specific factual error: Assigning the same IPv6 address to both routers causes a duplicate address conflict, preventing communication.

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MCQmedium

A junior network engineer is configuring a new Windows 10 workstation to connect to the corporate network. The network uses a /24 subnet mask and has a default gateway of 192.168.1.1. The workstation obtains its IP address automatically from a DHCP server, but the engineer needs to manually set a static IPv4 address of 192.168.1.50 and ensure the workstation can reach the internet. Which configuration step must the engineer take to satisfy these requirements?

A.Set the subnet mask to 255.255.0.0 and the default gateway to 192.168.1.1

B.Set the subnet mask to 255.255.255.0 and the default gateway to 192.168.1.1

C.Set the subnet mask to 255.255.255.0 and leave the default gateway blank

D.Set the subnet mask to 255.255.255.0 and the default gateway to 192.168.1.50

AnswerB

This is the correct configuration. The subnet mask 255.255.255.0 corresponds to a /24 prefix, which matches the network. The default gateway 192.168.1.1 is the router's IP on the same subnet, allowing the workstation to reach the internet.

Why this answer

A /24 subnet mask (255.255.255.0) matches the network prefix of the default gateway 192.168.1.1, ensuring the workstation can route traffic to the internet via that gateway. Option A fails because a /16 mask (255.255.0.0) does not match the corporate /24 network, causing incorrect network identification and potential routing issues. Option C fails because leaving the default gateway blank means the host cannot reach any network beyond its local subnet, so internet access is impossible.

Option D fails because using the host's own IP (192.168.1.50) as the default gateway would cause the host to attempt to route traffic to itself, never reaching the actual gateway.

Exam trap

Cisco often tests the requirement that the default gateway must be on the same subnet as the host's IP address, and a common trap is to confuse the gateway address with the host's own IP or to use an incorrect subnet mask that still allows local communication but breaks routing.

Why the other options are wrong

Using a /16 subnet mask (255.255.0.0) does not match the corporate /24 network, leading to incorrect network identification and potential routing issues.

Leaving the default gateway blank prevents the host from reaching any network beyond its own subnet, so internet access is impossible.

Setting the default gateway to the host's own IP address (192.168.1.50) would cause traffic to be sent to itself, never reaching the actual gateway.

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MCQhard

A subnet uses network address 192.168.200.96/28. Which range contains the usable host addresses?

A.192.168.200.97 to 192.168.200.110

B.192.168.200.96 to 192.168.200.111

C.192.168.200.98 to 192.168.200.111

D.192.168.200.81 to 192.168.200.94

AnswerA

The /28 prefix length translates to subnet mask 255.255.255.240, giving a 16-address block. Since 192.168.200.96 is the network address (last octet 0110 0000), the broadcast address is .111 (0110 1111). Consequently, the only addresses available to hosts are .97 through .110 inclusive, yielding 14 usable addresses while excluding both reserved endpoints.

Why this answer

A /28 block contains 16 addresses. In practical terms, the block starting at 192.168.200.96 runs through 192.168.200.111. The first address is the network address and the last address is the broadcast address. That means the usable host range is 192.168.200.97 through 192.168.200.110.

This question checks whether you can calculate the correct block and then exclude the two reserved boundary addresses.

Exam trap

Remember to exclude the network and broadcast addresses when determining usable host ranges.

Why the other options are wrong

This range includes the network address (.96) and the broadcast address (.111), which cannot be assigned to hosts. Usable host addresses must exclude these two addresses.

This range starts at .98, which excludes the valid host .97, and ends at .111, which includes the broadcast address. The correct usable range is .97 to .110.

This range (192.168.200.81 to .94) belongs to a different subnet. For a /28 subnet starting at .96, the valid host range is .97 to .110. This range is from a previous subnet (e.g., 192.168.200.80/28).

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MCQmedium

A network administrator receives a call from a user who cannot access any external websites from their wired workstation. The user can ping the default gateway successfully, but fails to ping 8.8.8.8. The administrator runs ipconfig /all on the workstation and sees an IP address of 192.168.1.50, subnet mask 255.255.255.0, and default gateway 192.168.1.1. What is the most likely cause of this issue?

A.The workstation has an incorrect default gateway configured.

B.The workstation has a duplicate IP address on the network.

C.The workstation is connected to the wrong VLAN.

D.The workstation has a DNS misconfiguration.

AnswerA

The user can ping the default gateway (192.168.1.1) but cannot ping 8.8.8.8, indicating local connectivity works but external routing fails. If the actual network gateway is different (e.g., 192.168.1.254), the workstation's configured gateway would be incorrect, preventing traffic from being forwarded to external networks.

Why this answer

The user can successfully ping 192.168.1.1, proving local IP connectivity to that device. However, the device at 192.168.1.1 may not be the correct default gateway for reaching external networks; the actual gateway router might be at a different IP (e.g., 192.168.1.254). This misconfiguration explains why pings to 8.8.8.8 fail even though the local gateway responds, as the workstation sends external traffic to the wrong next-hop address.

Exam trap

This question tests the ability to differentiate between local connectivity issues and routing issues. A common trap is to assume DNS is the problem when users cannot access websites, but the failure to ping an external IP indicates a routing problem, not a name resolution problem. Also, successful ping to the gateway eliminates many Layer 2 or IP configuration issues.

Why the other options are wrong

A duplicate IP would cause intermittent or lost connectivity and likely prevent a consistent reply from the gateway.

Being on the wrong VLAN would typically prevent the workstation from receiving an IP in the 192.168.1.0/24 subnet and reaching the gateway at 192.168.1.1.

DNS is only used for name resolution; pinging an IP address directly does not involve DNS, so a DNS misconfiguration cannot cause the ping failure to 8.8.8.8.

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MCQhard

A host is configured with 172.16.10.62/27. Which address is the broadcast address for that subnet?

A.172.16.10.31

B.172.16.10.32

C.172.16.10.63

D.172.16.10.64

AnswerC

This is correct because the host is in the 32–63 /27 block, whose broadcast is .63.

Why this answer

A /27 uses blocks of 32 addresses. In plain language, the last-octet ranges are 0–31, 32–63, 64–95, and so on. Since the host address ends in 62, it belongs to the 32–63 block. The last address in that block is the broadcast address, so the broadcast is 172.16.10.63.

This is a classic subnetting question because it requires you to place the host inside the correct block and then identify the last address in that block rather than guessing based on the host value alone.

Exam trap

Be careful not to confuse the broadcast address with the network address or the start of the next subnet.

Why the other options are wrong

The address 172.16.10.31 is the broadcast address for the /27 subnet 172.16.10.0–31, not for the subnet containing host 172.16.10.62. Since the host's IP is in the 32–63 range, the broadcast is .63, not .31.

172.16.10.32 is the network address (subnet ID) of the /27 subnet that includes hosts 32–63. It is not the broadcast address; the broadcast is the last address in the subnet, which is .63.

172.16.10.64 is the network address of the next /27 subnet (64–95), not the broadcast address for the subnet containing .62. The broadcast for the subnet containing .62 is .63.

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MCQhard

A host uses the subnet mask 255.255.255.192. How many usable host addresses exist in each subnet?

A.30

B.62

C.126

D.254

AnswerB

This is the correct answer because the subnet mask 255.255.255.192 translates to a /26 prefix, leaving 32 - 26 = 6 bits for host addressing. With 6 host bits, the total number of addresses in each subnet is 2^6 = 64, and from that total the network address (all host bits zero) and the broadcast address (all host bits one) must be subtracted. This yields 64 - 2 = 62 usable host addresses, which matches the requirement for a /26 subnet.

Why this answer

The mask 255.255.255.192 corresponds to /26. That leaves 6 host bits, which means each subnet contains 64 total addresses. After excluding the network and broadcast addresses, 62 usable host addresses remain.

This is a standard host-capacity question. The safest approach is to convert the mask to the prefix, determine the total addresses from the number of host bits, and then subtract the two reserved addresses.

Exam trap

Be careful not to confuse total addresses with usable addresses; always subtract the network and broadcast addresses.

Why the other options are wrong

The subnet mask 255.255.255.192 is /26, which provides 64 total addresses per subnet. Subtracting the network and broadcast addresses leaves 62 usable hosts, not 30. 30 usable hosts corresponds to a /27 subnet mask (255.255.255.224).

126 usable hosts would require a /25 subnet mask (255.255.255.128), which provides 128 total addresses. The given mask /26 provides only 64 total addresses, so 126 is incorrect.

254 usable hosts corresponds to a /24 subnet mask (255.255.255.0), which provides 256 total addresses. The mask 255.255.255.192 is /26, which is two bits longer, resulting in only 64 total addresses.

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MCQhard

What prefix length corresponds to the subnet mask 255.255.255.248?

A./28

B./29

C./30

D./27

AnswerB

This is correct because 255.255.255.248 equals 29 network bits.

Why this answer

The mask 255.255.255.248 corresponds to /29. In practical terms, the first three octets provide 24 network bits, and the value 248 in the last octet is 11111000 in binary, which contributes 5 more network bits. That gives a total prefix length of 29.

This is a standard dotted-decimal to prefix conversion question. It matters because subnetting often requires you to move comfortably between both forms.

Exam trap

Be careful not to confuse the binary values of subnet masks. Ensure you understand how to convert between dotted-decimal and CIDR notation.

Why the other options are wrong

The subnet mask 255.255.255.240 corresponds to a /28 prefix length, not /29. This mask has 28 network bits, leaving 4 host bits, which yields 14 usable hosts per subnet.

The subnet mask 255.255.255.252 corresponds to a /30 prefix length, not /29. A /30 mask has 30 network bits and only 2 host bits, providing 2 usable addresses, typically used for point-to-point links.

The subnet mask 255.255.255.224 corresponds to a /27 prefix length, not /29. A /27 mask has 27 network bits and 5 host bits, providing 30 usable hosts per subnet.

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MCQhard

A host is configured with IP address 172.16.100.222/27. Which address is the broadcast address for its subnet?

A.172.16.100.191

B.172.16.100.223

C.172.16.100.224

D.172.16.100.255

AnswerB

The /27 subnet mask indicates 27 network bits, leaving 5 host bits. This creates subnet blocks of 32 addresses. For the host 172.16.100.222, its subnet begins at 172.16.100.192. The broadcast address is the final address in this subnet range, immediately preceding the next network address (172.16.100.224). Therefore, 172.16.100.223 correctly serves as the broadcast address for the subnet defined by the host's /27 configuration.

Why this answer

A /27 uses address blocks of 32. In practical terms, the fourth-octet ranges are 0–31, 32–63, 64–95, 96–127, 128–159, 160–191, 192–223, and 224–255. Since 222 falls inside the 192–223 block, the broadcast address is the last address in that block, which is 172.16.100.223.

This is a classic subnet-boundary question because it tests whether you can place a host in the correct block and then identify the final address in that block as the broadcast.

Exam trap

Avoid assuming the broadcast address is always .255 or miscalculating subnet ranges.

Why the other options are wrong

172.16.100.191 is the broadcast address of the previous /27 subnet (172.16.100.160/27), not the subnet containing .222. The host .222 is in the 172.16.100.192/27 subnet, so its broadcast is .223.

172.16.100.224 is the network address of the next /27 subnet (172.16.100.224/27), not a broadcast address. Broadcast addresses are always the last address in a subnet, not the first.

172.16.100.255 is the broadcast address of the entire /24 subnet (172.16.100.0/24), not the /27 subnet containing .222. The /27 subnet has a smaller range, so its broadcast is .223.

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MCQhard

Which prefix length corresponds to the subnet mask 255.255.255.192?

A./25

B./26

C./27

D./28

AnswerB

The subnet mask 255.255.255.192 has a binary pattern of 11111111.11111111.11111111.11000000, which places two additional network bits in the fourth octet beyond the standard /24. Adding those two bits to the 24 bits of the first three octets yields 26 network bits, so the only correct prefix length is /26. This mask provides 64 total addresses per subnet and 62 usable host addresses.

Why this answer

The mask 255.255.255.192 corresponds to /26. In practical terms, the first three octets contribute 24 network bits, and 192 in binary is 11000000, which contributes 2 more network bits. That totals 26 network bits.

This is a standard conversion skill that matters in subnetting, ACL design, and route interpretation.

Exam trap

Be careful not to confuse similar subnet masks or miscount the number of bits in the binary representation.

Why the other options are wrong

The /25 prefix length corresponds to subnet mask 255.255.255.128, not 255.255.255.192. The mask 255.255.255.128 has 128 in the last octet, while 255.255.255.192 has 192, indicating a different number of host bits.

The /27 prefix length corresponds to subnet mask 255.255.255.224, not 255.255.255.192. The mask 255.255.255.224 has 224 in the last octet, which provides 30 usable hosts per subnet, whereas 255.255.255.192 provides 62 usable hosts.

The /28 prefix length corresponds to subnet mask 255.255.255.240, not 255.255.255.192. The mask 255.255.255.240 has 240 in the last octet, which supports 14 usable hosts, while 255.255.255.192 supports 62 usable hosts.

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MCQhard

A host is configured with 10.10.10.130/25. What is the network address of its subnet?

A.10.10.10.0

B.10.10.10.64

C.10.10.10.128

D.10.10.10.255

AnswerC

A /25 mask (255.255.255.128) partitions the last octet into two equal blocks of 128 addresses: .0–.127 and .128–.255. The address 10.10.10.130 falls in the upper block because its last octet binary begins with a 1 (10000010). Therefore, the network address is the first address of that block, 10.10.10.128, and this is the correct answer.

Why this answer

A /25 divides the address space into two blocks of 128 addresses. In plain language, the ranges are 0–127 and 128–255. Since the host ends in 130, it belongs to the 128–255 half. That means the network address is 10.10.10.128.

This is a common subnet-boundary question because it tests whether you can map a host address into the correct prefix block quickly and confidently.

Exam trap

Be careful not to confuse the subnet mask with /24 or mistake the broadcast address for the network address.

Why the other options are wrong

10.10.10.0 is the network address of the 10.10.10.0/25 subnet (range 0-127), but the host IP 10.10.10.130 is not in that range. The /25 mask creates two subnets: 10.10.10.0/25 and 10.10.10.128/25, and .130 belongs to the latter.

10.10.10.64 is not a valid network address for any /25 subnet derived from 10.10.10.0/24. A /25 subnet has a block size of 128, so the network addresses are multiples of 128: 0 and 128. 64 is a multiple of 64, which would be a /26 boundary, not /25.

10.10.10.255 is the broadcast address for the 10.10.10.128/25 subnet, not the network address. The broadcast address is the last address in the subnet (all host bits set to 1), while the network address is the first address (all host bits set to 0).

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MCQhard

A host is configured with 192.168.60.33/26. Which address is the network address of its subnet?

A.192.168.60.0

B.192.168.60.32

C.192.168.60.64

D.192.168.60.63

AnswerA

The /26 prefix length corresponds to a 255.255.255.192 subnet mask, dividing the fourth octet into blocks of 64 addresses. Since the host address 192.168.60.33 falls within the range 0 through 63, the network address for this block is 192.168.60.0, where all host bits are zero. Valid /26 network boundaries occur at multiples of 64: .0, .64, .128, and .192, so .0 is the only correct network address for this host.

Why this answer

A /26 uses blocks of 64 addresses. In practical terms, the fourth-octet ranges are 0–63, 64–127, 128–191, and 192–255. Since 33 falls within the 0–63 block, the network address is 192.168.60.0.

This is a straightforward boundary-identification question, but it catches people who memorize masks without understanding block sizes. The right approach is to find the correct block first, then take the first address in that block as the network address.

Exam trap

Avoid confusing the first usable host address with the network address. Always identify the block range first.

Why the other options are wrong

The /26 subnet mask has a block size of 64, not 32. The network addresses for /26 are 0, 64, 128, and 192. 192.168.60.32 is not a valid network address because it is not a multiple of 64.

192.168.60.64 is the network address of the next /26 subnet (64–127). The host address 192.168.60.33 belongs to the 0–63 range, not the 64–127 range.

192.168.60.63 is the broadcast address of the 192.168.60.0/26 subnet, not the network address. The network address is always the first address in the subnet (all host bits set to 0).

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MCQhard

A host is configured with 10.10.10.33/27. What is the broadcast address of its subnet?

A.10.10.10.31

B.10.10.10.63

C.10.10.10.32

D.10.10.10.64

AnswerB

10.10.10.63 is correct because the /27 prefix length gives a block size of 2^(32-27)=32 addresses, meaning subnets align on boundaries of 32 in the last octet. The host 10.10.10.33 falls within the 10.10.10.32 to 10.10.10.63 range, and the broadcast address is always the highest (last) address in that subnet block, which is 10.10.10.63. This matches the definition of a directed broadcast for that subnet.

Why this answer

A /27 uses address blocks of 32. In practical terms, the ranges are 0–31, 32–63, 64–95, and so on. Because .33 falls inside the 32–63 block, the broadcast address is the last address in that block, which is 10.10.10.63.

This is a classic subnet-boundary question. The trick is to identify the correct block first and then choose its last address as the broadcast.

Exam trap

Don't confuse the network address or the next subnet's start with the broadcast address. Always calculate the correct range first.

Why the other options are wrong

10.10.10.31 is the broadcast address of the previous /27 subnet (10.10.10.0–10.10.10.31), not the subnet containing .33.

10.10.10.32 is the network address (subnet ID) of the subnet containing .33, not the broadcast address. The network address is the first address in the block.

10.10.10.64 is the network address of the next /27 subnet (10.10.10.64–10.10.10.95), not the broadcast address of the subnet containing .33.

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PBQhard

You are connected to R1. Configure IPv4 and IPv6 addressing on interfaces G0/0 and G0/1 so that R1 can reach R2's loopback0 (198.51.100.1/32) and R2 can reach R1's loopback0 (203.0.113.1/32). The current configuration has a wrong subnet mask on R1 G0/0 and a missing default gateway on R2, causing reachability failures. Additionally, configure IPv6 using EUI-64 on R1 G0/1 and static IPv6 on R2 G0/1 to enable IPv6 ping between the two routers. All devices are routers.

Network Topology

Hints

•Check the subnet mask on R1 G0/0: it does not match R2's /30.
•R2 has no route to reach R1's loopback or the 192.168.1.0/24 network; it needs a default gateway.
•R2's IPv6 address is on a different subnet (2001:db8:2::/64) than R1's (2001:db8:1::/64); they must be on the same subnet.

A.On R1 G0/0, change subnet mask to 255.255.255.252; on R2, add ip route 0.0.0.0 0.0.0.0 10.0.0.1; for IPv6, on R2 G0/1 change address to 2001:db8:1::2/64.

B.On R1 G0/0, change subnet mask to 255.255.255.0; on R2, add ip route 0.0.0.0 0.0.0.0 10.0.0.1; for IPv6, on R2 G0/1 change address to 2001:db8:1::2/64.

C.On R1 G0/0, change subnet mask to 255.255.255.252; on R2, add ip route 0.0.0.0 0.0.0.0 10.0.0.2; for IPv6, on R2 G0/1 change address to 2001:db8:2::2/64.

D.On R1 G0/0, change subnet mask to 255.255.255.252; on R2, add ip route 0.0.0.0 0.0.0.0 10.0.0.1; for IPv6, on R2 G0/1 change address to 2001:db8:1::2/64 and on R1 G0/1 use static IPv6 instead of EUI-64.

AnswerA

solution

! R1
interface GigabitEthernet0/0
ip address 10.0.0.1 255.255.255.252
exit
interface GigabitEthernet0/1
ipv6 address 2001:db8:1::/64 eui-64
exit

! R2
interface GigabitEthernet0/1
ip address 192.168.1.2 255.255.255.0
ipv6 address 2001:db8:1::2/64
exit
ip route 0.0.0.0 0.0.0.0 10.0.0.1

Why this answer

The primary IPv4 issues are a subnet mask mismatch on the point-to-point link and a missing default gateway on R2. On R1 G0/0, the mask is /24 instead of /30; while both routers can reach each other directly, the mismatched subnet mask causes routing inconsistencies because R1 advertises the link as a /24, potentially affecting routing decisions. Fixing the mask to /30 ensures both routers agree on the subnet.

R2 lacks a route to R1's loopback and the 192.168.1.0/24 network, so a default route via 10.0.0.1 resolves reachability. For IPv6, R1 G0/1 uses EUI-64, and R2 G0/1 must be on the same subnet (2001:db8:1::/64); R2's address was incorrectly set to 2001:db8:2::2/64, so changing it to 2001:db8:1::2/64 enables IPv6 ping.

Exam trap

Watch for subnet mask mismatches on point-to-point links; both ends must use the same mask. Also, ensure default routes point to the correct next-hop IP (the neighbor's interface IP). For IPv6, both interfaces must be on the same subnet to communicate directly.

Why the other options are wrong

The subnet mask on R1 G0/0 should be /30 to match R2, not /24.

The default gateway must be the neighbor's IP address, and IPv6 subnets must match for direct communication.

The requirement states EUI-64 on R1 G0/1, so static is not allowed.

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MCQmedium

A network administrator is configuring a new Windows workstation on a small office network that uses IPv4 addressing. The workstation must be able to communicate with devices on other subnets and resolve hostnames via a company DNS server at 10.10.10.5. The administrator has already set the IP address to 10.10.10.10 and the subnet mask to 255.255.255.0. Which additional parameter must be configured to meet both requirements?

A.Configure a default gateway of 10.10.10.1 and a DNS server of 10.10.10.5.

B.Configure only a DNS server of 10.10.10.5.

C.Change the subnet mask to 255.255.0.0 to allow communication across subnets.

D.Configure a default gateway of 10.10.20.1 and a DNS server of 10.10.10.5.

AnswerA

This is the correct configuration because both entries are necessary and valid. The default gateway 10.10.10.1 lies on the same subnet as the workstation's IP address, so the host can resolve the gateway's MAC address via ARP and forward all out-of-subnet traffic to it. The router then routes packets between the 10.10.10.0/24 and 10.10.20.0/24 networks, enabling communication with devices on the other subnet. The DNS server 10.10.10.5 is also on the local subnet, allowing name resolution to work without requiring routing. Together, they provide both name resolution and the necessary Layer 3 path.

Why this answer

To communicate with devices on other subnets, the workstation needs a default gateway (router) to forward traffic beyond its local subnet. The IP address 10.10.10.10 with subnet mask 255.255.255.0 places it in the 10.10.10.0/24 network, so a default gateway (e.g., 10.10.10.1) is required for inter-subnet routing. Additionally, to resolve hostnames, the DNS server address must be explicitly configured; the company DNS server is at 10.10.10.5.

Option A correctly provides both parameters.

Exam trap

Cisco often tests the requirement that a default gateway must be on the same subnet as the host; the trap here is that candidates may think a DNS server alone suffices for inter-subnet communication, or they may incorrectly assume changing the subnet mask can replace a router, or they may choose a gateway on a different subnet without realizing it is unreachable.

Why the other options are wrong

Without a default gateway, the workstation cannot send packets to destinations outside its own subnet (10.10.10.0/24). The DNS server alone only provides name resolution, not routing to other subnets.

Changing the subnet mask to 255.255.0.0 would expand the broadcast domain and could cause routing problems, but it does not provide a path to other subnets. The workstation still needs a default gateway to communicate with devices outside its local network.

The default gateway 10.10.20.1 is not on the same subnet as the workstation (10.10.10.0/24). For a host to reach its default gateway, the gateway must be directly reachable on the local subnet. Since 10.10.20.1 is on a different subnet, the workstation cannot send traffic to it.

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Multi-Selectmedium

Which two statements accurately describe the relationship between a network address and a broadcast address in IPv4 subnetting?

Select 2 answers

A.The network address is the first address in the subnet block.

B.The broadcast address is the last address in the subnet block.

C.Both addresses are normal host addresses that can be assigned to users.

D.The broadcast address always becomes the default gateway.

E.These concepts exist only in IPv6 and not IPv4.

AnswersA, B

In IPv4, the network address is the first (lowest) address within a subnet block, obtained by setting all host bits to 0. This address uniquely identifies the subnet and is used by routers in their routing tables; it cannot be assigned to any host interface. It is the starting boundary from which the usable host range begins.

Why this answer

The network address identifies the beginning of the subnet block, and the broadcast address identifies the final address in that block. In practical terms, both are reserved and are not assigned to ordinary hosts. The usable host range falls between them.

This is a very basic subnetting truth, but it is foundational for every other addressing calculation.

Exam trap

Be cautious not to confuse the roles of network and broadcast addresses with usable host addresses.

Why the other options are wrong

Both the network address and broadcast address are reserved addresses within a subnet and cannot be assigned to hosts. The network address identifies the subnet itself, and the broadcast address is used for one-to-all communication. Assigning them to hosts would cause conflicts.

The default gateway is typically the IP address of a router interface on the subnet, which is a normal host address within the usable range. The broadcast address is the last address in the subnet and is reserved for broadcasting; it cannot be used as a gateway.

Network and broadcast addresses are fundamental to IPv4 subnetting and are defined in IPv4 standards. IPv6 does not use broadcast addresses; instead, it uses multicast and anycast. Therefore, stating these concepts exist only in IPv6 is factually incorrect.

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Multi-Selectmedium

Which TWO statements accurately describe Network Address Translation (NAT) types?

Select 2 answers

A.Static NAT creates a fixed one-to-one mapping between a local and global address and is typically used to allow external connectivity to internal servers.

B.Dynamic NAT assigns a public IP from a pool for the duration of a translation, but it does not modify Layer 4 port numbers.

C.PAT only translates TCP packets because it uses port numbers, leaving UDP translation unsupported.

D.Static NAT entries are automatically removed after periods of inactivity to free up public addresses.

E.With PAT, if the public IP address pool is exhausted, new translations fail because PAT requires unique public IPs for each private host.

AnswersA, B

Static NAT permanently maps a private IP to a public IP, enabling inbound access to internal resources like web servers.

Why this answer

Static NAT creates a permanent, one-to-one mapping between an inside local (private) IP address and an inside global (public) IP address. This fixed mapping is typically used to make internal servers, such as a web or email server, reachable from the internet using a consistent public address.

Exam trap

Cisco often tests the misconception that PAT only works with TCP, but in reality PAT supports both TCP and UDP, and the trap here is that candidates confuse PAT's use of port numbers with a protocol limitation.

Why the other options are wrong

PAT works with any transport protocol that has port fields—TCP, UDP, and even ICMP through the identifier field.

Only dynamic translations (from dynamic NAT or PAT) have idle timeouts; static mappings are permanent.

This statement describes dynamic NAT pool exhaustion, not PAT, which uses port multiplexing.

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PBQhard

You are troubleshooting a PC connected to switch SW1. The PC cannot access the internet. SW1 is connected to router R1 via port G0/1. R1 provides default gateway and DHCP services. Analyze the provided show output and fix the connectivity issue so that the PC can ping 8.8.8.8. === Show output from R1 === <pre> R1# show ip interface brief Interface IP-Address OK? Method Status Protocol GigabitEthernet0/0 unassigned YES manual administratively down down GigabitEthernet0/1 10.0.0.1 YES NVRAM up up </pre> === Show output from PC === <pre> C:\> ipconfig Ethernet adapter Ethernet0: Connection-specific DNS Suffix . : IPv4 Address. . . . . . . . . . : 169.254.123.45 Subnet Mask . . . . . . . . . . : 255.255.0.0 Default Gateway . . . . . . . . : </pre> === Show output from SW1 === <pre> SW1# show vlan brief VLAN Name Status Ports ---- -------------------------------- --------- ------------------------------- 1 default active Gi0/1, Gi0/2, Gi0/3 </pre>

Network Topology

Hints

•Check if R1's interface connected to SW1 has an IP address.
•The DHCP pool expects the default gateway to be on the same subnet as the clients.
•APIPA address means the PC did not receive a DHCP offer.

A.Configure R1's interface G0/0 with IP address 192.168.1.1/24 and ensure the interface is not administratively down.

B.Change the VLAN on SW1's port G0/2 to VLAN 10 and configure R1's subinterface G0/0.10 with IP 192.168.1.1/24.

C.Enable DHCP snooping on SW1 and configure the port G0/2 as a trusted port.

D.Configure a static IP address of 192.168.1.10/24 on the PC with default gateway 192.168.1.1.

AnswerA

solution

! R1
interface gigabitethernet0/0
ip address 192.168.1.1 255.255.255.0
no shutdown

Why this answer

The PC's APIPA address (169.254.x.x) indicates DHCP failure. The router's DHCP pool is correctly configured and has a lease, but the show output reveals that R1's interface G0/0 is administratively down and has no IP address. Without a working IP on G0/0, the router cannot serve DHCP or route traffic for VLAN 1, even though both the PC and the router are in the same VLAN.

Option A fixes the root cause by assigning the correct subnet IP and bringing the interface up. Option B is incorrect because moving the PC to a different VLAN or creating subinterfaces does nothing to enable the router's physical interface where DHCP and routing must run. Option C is wrong because DHCP snooping or trust configurations are irrelevant when the router's own interface is down/unaddressed.

Option D is a workaround that only masks the problem; the scenario requires a working DHCP service, and a static IP would not restore the intended design.

Exam trap

This question tests your ability to identify that a router interface must have an IP address in the client subnet for DHCP to work, even if the DHCP pool is correctly configured. Many candidates focus on VLANs or DHCP server settings but overlook the basic requirement of an IP address on the router interface.

Why the other options are wrong

Changing VLANs or using subinterfaces does not solve the problem because the router's physical interface must be up and have an IP address to serve the VLAN.

Enabling DHCP snooping or trust settings on the switch cannot fix a router interface that is administratively down and unassigned.

Assigning a static IP to the PC circumvents but does not resolve the root issue of the router's interface being down, and the scenario requires DHCP.

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MCQhard

A host address is 10.55.8.117/29. Which address is the network address of the subnet?

A.10.55.8.112

B.10.55.8.119

C.10.55.8.120

D.10.55.8.116

AnswerA

10.55.8.112 is the network address because the /29 prefix length equates to a 255.255.248 subnet mask, creating subnets with 8 addresses each. The host address 10.55.8.117 falls within the range 112 through 119, so the network address is the first address in that block, 10.55.8.112, with 10.55.8.119 as the broadcast address.

Why this answer

A /29 subnet has a block size of 8. In practical terms, the relevant last-octet blocks are 112-119 for this host. That means the network address is 10.55.8.112. Once you identify the correct block, the first address in the block is the network address.

This is a useful addressing-boundary question because it checks careful block calculation, not memorized guesses.

Exam trap

Be careful not to confuse the network address with the first usable host or the broadcast address.

Why the other options are wrong

10.55.8.119 is the broadcast address for the subnet 10.55.8.112/29, not the network address. The broadcast address is the last address in the block (112+8-1=119) and is used to send traffic to all hosts in the subnet.

10.55.8.120 is the network address of the next /29 subnet (120-127), not the current one. The current subnet ends at 119, so 120 belongs to a different subnet.

10.55.8.116 is a valid host address within the subnet 10.55.8.112/29 (usable range: 113-118). It is not the network address, which must be the first address (112).

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PBQhard

You are connected to R1. The network has R1, R2, and a multilayer switch MLS1. Configure IPv4 and IPv6 addressing on R1's interfaces so that R1 can ping both R2 (198.51.100.2) and MLS1 (203.0.113.2) via IPv4. Additionally, configure IPv6 on G0/1 using EUI-64 with prefix 2001:db8:1::/64 and verify that R1 can ping the IPv6 address of MLS1 (2001:db8:1::2). The current configuration has incorrect subnet masks and missing IPv6 settings, causing reachability failures.

Hints

•The subnet mask on both interfaces is too large; it should be /30.
•IPv6 is not enabled on G0/1 yet; use the 'ipv6 address' command with EUI-64.
•After changing the mask, the ping should work because the devices will be on the same subnet.

A.Change the subnet mask on G0/0 to 255.255.255.252, change G0/1 to 255.255.255.252, then configure IPv6 on G0/1 with the EUI-64 address using the prefix 2001:db8:1::/64.

B.Change the subnet mask on G0/0 to 255.255.255.0, change G0/1 to 255.255.255.0, then configure IPv6 on G0/1 with the EUI-64 address using the prefix 2001:db8:1::/64.

C.Change the subnet mask on G0/0 to 255.255.255.252, change G0/1 to 255.255.255.252, then configure IPv6 on G0/1 with the static address 2001:db8:1::1/64.

D.Change the subnet mask on G0/0 to 255.255.255.252, change G0/1 to 255.255.255.252, then configure IPv6 on G0/1 with the EUI-64 address using the prefix 2001:db8:1::/32.

AnswerA

solution

! R1
interface GigabitEthernet0/0
ip address 198.51.100.1 255.255.255.252
exit
interface GigabitEthernet0/1
ip address 203.0.113.1 255.255.255.252
ipv6 address 2001:db8:1::/64 eui-64
exit

Why this answer

The interfaces on R1 were configured with subnet masks that were not /30, which is required for these point-to-point links. With an incorrect mask, R1 does not consider the neighboring IPs (198.51.100.2 and 203.0.113.2) as directly connected, preventing ARP resolution and IPv4 reachability. Additionally, IPv6 was missing on G0/1.

To fix, change the subnet mask on G0/0 to 255.255.255.252, change G0/1 to 255.255.255.252, then configure IPv6 on G0/1 with the EUI-64 address using the prefix 2001:db8:1::/64. After these changes, pings succeed.

Exam trap

This question tests your understanding of subnet masks and their impact on Layer 3 reachability. A common trap is to focus only on IPv6 and forget that incorrect IPv4 subnet masks can prevent ARP resolution, even if IPv6 is configured correctly. Also, pay close attention to the exact requirements: EUI-64 and the correct prefix length.

Why the other options are wrong

The specific factual error is that /24 masks are too large for point-to-point links and do not match the expected subnets for R2 and MLS1.

The specific factual error is that the IPv6 address should be configured with the EUI-64 keyword, not a static address.

The specific factual error is that the prefix length must be /64 as specified in the question; a /32 prefix is incorrect for this scenario.

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Multi-Selectmedium

Which TWO statements correctly describe the configuration and verification of IPv4 and IPv6 parameters for host connectivity, including default gateway, DNS, and subnet masks?

Select 2 answers

A.The subnet mask determines the DNS server address used by the host.

B.The default gateway must be on the same subnet as the host's IP address.

C.IPv6 hosts can only obtain their IP address via DHCPv6.

D.The command 'ipconfig /all' displays both IPv4 and IPv6 configuration details.

E.A host can reach any remote network if its default gateway is configured with any IP address.

AnswersB, D

For a host to deliver a frame to the default gateway, the gateway's IP must be in the same network/subnet as the host's interface, because the host uses ARP (or NDP for IPv6) to resolve the gateway's MAC address on the local link. If the gateway were on a different subnet, the host would not be able to reach it without another router on the local segment.

Why this answer

A host's default gateway must be on the same subnet to be reachable at Layer 2; otherwise, a circular dependency occurs. Option D is correct because the 'ipconfig /all' command displays both IPv4 and IPv6 configuration details, including IP address, subnet mask, default gateway, and DNS servers. Option A is wrong because the subnet mask determines the network and host portions of an IP address, not the DNS server address.

Option C is wrong because IPv6 hosts can obtain their IP address via SLAAC, DHCPv6, or static configuration; the phrase 'can only' makes it incorrect. Option E is wrong because the default gateway must be reachable (on the same subnet) and configured with an IP address that belongs to a router interface on that subnet, not just any IP address.

Exam trap

Cisco often tests the misconception that IPv6 hosts require DHCPv6 for address assignment, when in fact SLAAC is a common and valid method, and the question's wording 'can only' is the trap that eliminates Option C.

Why the other options are wrong

The subnet mask is used to determine the network portion of an IP address and the host portion, not the DNS server address. DNS server addresses are configured separately, either manually or via DHCP.

IPv6 hosts can obtain their IP address via Stateless Address Autoconfiguration (SLAAC), which does not require DHCPv6. DHCPv6 is optional and used for stateful configuration or to provide additional parameters.

The default gateway must be on the same subnet as the host's IP address; otherwise, the host cannot send Ethernet frames to it because the gateway's MAC address would not be reachable via ARP. An arbitrary IP address would not work.

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MCQhard

A host address is 172.16.8.70/26. What is the network address of its subnet?

A.172.16.8.0

B.172.16.8.64

C.172.16.8.70

D.172.16.8.128

AnswerB

For a /26 prefix, the subnet mask is 255.255.255.192, which creates block sizes of 64 addresses in the fourth octet. The host address .70 lies in the range 64–127, so the network address is found by zeroing the host bits: 70 AND 192 = 64. Thus, 172.16.8.64 is the correct network address for this host.

Why this answer

A /26 uses blocks of 64 addresses. In plain language, the fourth-octet subnet ranges are 0–63, 64–127, 128–191, and 192–255. Since the host address ends in 70, it belongs to the 64–127 block. That means the network address of the subnet is 172.16.8.64.

This is a standard subnetting calculation. The key is to identify the correct block based on the prefix and then choose the first address in that block as the network address.

Exam trap

A frequent exam trap is mistaking the host IP address for the network address or selecting the wrong subnet block based on the subnet mask. Candidates often pick 172.16.8.0 because it looks like a common network address or 172.16.8.128 assuming it’s the next block, but these do not contain the host 172.16.8.70 under a /26 mask. The trap arises from not calculating subnet ranges correctly or misunderstanding how subnet masks segment the address space into fixed blocks.

This mistake leads to incorrect subnet identification and can cause routing or addressing errors in real networks.

Why the other options are wrong

172.16.8.0 is incorrect because the /26 subnet blocks cover 0–63, 64–127, etc., and the host address 70 does not fall within the 0–63 range. Selecting this ignores the actual subnet boundaries defined by the mask.

172.16.8.70 is incorrect because this is the host address itself, not the network address. The network address must be the first address in the subnet block, not a host address within it.

172.16.8.128 is incorrect because this subnet block starts at 128, which is above the host address 70. The host does not belong to this subnet, so this cannot be the network address.

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Multi-Selectmedium

Which three of the following statements about Network Address Translation (NAT) are correct? (Choose three.)

Select 3 answers

.Static NAT provides a one-to-one mapping between a private IP and a public IP.

.Dynamic NAT uses a pool of public IP addresses assigned on a first-come, first-served basis.

.PAT (Port Address Translation) allows multiple internal hosts to share a single public IP address.

.NAT eliminates the need for any routing in a network.

.Dynamic NAT always assigns the same public IP to a given internal host.

.PAT requires a unique public IP for every concurrent session.

Why this answer

Static NAT provides a one-to-one mapping between a private IP and a public IP, ensuring that a specific internal host always uses the same public address. Dynamic NAT uses a pool of public IP addresses assigned on a first-come, first-served basis, so internal hosts compete for available addresses. PAT (Port Address Translation) allows multiple internal hosts to share a single public IP by differentiating sessions via unique port numbers, which is the most common form of NAT used in home and small office routers.

Exam trap

Cisco often tests the misconception that dynamic NAT provides a fixed mapping like static NAT, or that PAT requires multiple public IPs, when in fact PAT is designed to share a single public IP among many hosts.

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MCQhard

A host address is 192.168.14.222/28. Which address is the broadcast address of its subnet?

A.192.168.14.207

B.192.168.14.223

C.192.168.14.208

D.192.168.14.224

AnswerB

With /28, the host portion occupies only the last 4 bits, so the usable range for the subnet containing .222 is 192.168.14.208 through 192.168.14.223. Setting all host bits to 1 yields the broadcast address 192.168.14.223, which is the directed broadcast for the /28 subnet that includes .222. This makes .223 the correct answer.

Why this answer

A /28 subnet has a block size of 16. In practical terms, the fourth-octet blocks are 0-15, 16-31, and so on. Because 222 falls within the 208-223 block, the broadcast address is the last address in that block: 192.168.14.223.

This is a subnet-boundary question that depends on identifying the correct /28 block before choosing the broadcast address.

Exam trap

Be careful not to confuse the broadcast address with the network address of the next subnet or a host address within the subnet.

Why the other options are wrong

192.168.14.207 is the broadcast address of the previous /28 subnet (192.168.14.192/28), not the subnet containing 192.168.14.222.

192.168.14.208 is the network address (subnet ID) of the /28 subnet containing .222, not the broadcast address.

192.168.14.224 is the network address of the next /28 subnet (192.168.14.224/28), not the broadcast of the current subnet.

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Multi-Selectmedium

Which two statements accurately describe subnet masks in IPv4?

Select 2 answers

A.It identifies the network-versus-host split in an IPv4 address.

B.It helps a host determine whether a destination is local or remote.

C.It resolves hostnames into IP addresses.

D.It encrypts packets before they leave the host.

E.It replaces the need for a default gateway.

AnswersA, B

A subnet mask is a 32-bit value that uses contiguous 1s to mark the network portion of an IPv4 address and contiguous 0s to mark the host portion. By ANDing an IP address with its mask, you derive the network address, which reveals the exact boundary between the bits that identify the network and the bits assigned to hosts. This boundary is the mask's defining purpose, so saying it identifies the network-versus-host split is accurate.

Why this answer

A subnet mask tells the host which part of the IPv4 address refers to the network and which part refers to the host. In plain language, it helps the device determine whether a destination is local or remote. That decision is essential because it affects whether the host uses ARP directly or forwards traffic to the default gateway.

The wrong answers usually attribute unrelated behaviors to the subnet mask, such as encryption or hostname resolution. The two correct answers are the ones that preserve its role in defining local scope and address structure.

Exam trap

Avoid confusing subnet mask functions with encryption or DNS, as these are unrelated to IP address segmentation.

Why the other options are wrong

Subnet masks are used solely for IP addressing and routing purposes, not for name resolution. Hostname-to-IP-address resolution is performed by the Domain Name System (DNS), which is a completely different protocol and service.

Subnet masks do not provide any encryption or security functionality. Encryption of packets is performed by protocols such as IPsec, TLS, or other cryptographic mechanisms, which operate independently of subnet masking.

A default gateway is still required for any traffic destined to a different subnet. The subnet mask only defines the local network boundary; it does not provide routing to other networks. Without a default gateway, a host cannot send packets off its subnet.

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MCQhard

A host address is 192.168.50.158/27. Which address is the network address of its subnet?

A.192.168.50.128

B.192.168.50.159

C.192.168.50.160

D.192.168.50.96

AnswerA

With a /27 prefix, the subnet block size is 32 addresses. The host .158 falls within the range .128–.159, and the network address is always the first address of that block, so it is 192.168.50.128. This is the lowest address that identifies this subnet.

Why this answer

A /27 subnet has a block size of 32. In practical terms, the relevant ranges in the last octet are 0-31, 32-63, 64-95, 96-127, 128-159, and so on. Because 158 falls in the 128-159 block, the network address is 192.168.50.128.

This is a block-identification question. Once you identify the correct /27 block, the network address is the first address in that range.

Exam trap

Ensure you calculate the correct block range for the subnet mask given, not just any multiple of the block size.

Why the other options are wrong

192.168.50.159 is the broadcast address for the 192.168.50.128/27 subnet, not the network address. The broadcast address is used to send packets to all hosts in the subnet and cannot be assigned to a host.

192.168.50.160 is the network address of the next /27 subnet (160-191), not the subnet containing 158. The /27 mask creates subnets with a block size of 32, so the subnet boundaries are multiples of 32.

192.168.50.96 is the network address of the 96-127 /27 subnet, which is an earlier subnet. The address 158 falls in the 128-159 range, not the 96-127 range.

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MCQhard

A host is configured as 192.168.50.130/25. Which address is the broadcast address for its subnet?

A.192.168.50.127

B.192.168.50.128

C.192.168.50.255

D.192.168.50.254

AnswerC

With a /25 prefix, the subnet mask is 255.255.255.128, which splits the 192.168.50.0/24 network into two 128-address blocks. The address 192.168.50.130 falls into the upper block, 192.168.50.128/25, which spans .128 through .255. The broadcast address is the last address in that block, .255, where all seven host bits are set to 1.

Why this answer

A /25 divides the /24 into two blocks: 0–127 and 128–255. In plain language, because the host ends in 130, it belongs to the upper half, which starts at 128 and ends at 255. The last address in that block is the broadcast address, so the broadcast is 192.168.50.255.

This is a classic subnetting pattern because it tests whether you can identify not just the subnet, but also the reserved last address in that subnet.

Exam trap

A frequent exam trap is mistaking the network address or a high usable host address for the broadcast address. Candidates often select 192.168.50.128, confusing it as the broadcast because it is the start of the upper subnet, or 192.168.50.254, assuming it is the broadcast since it is near the subnet's end. The trap lies in not recognizing that the broadcast address is always the highest address in the subnet, which in this case is 192.168.50.255.

Misidentifying these addresses leads to incorrect subnet calculations and can cause network communication failures in real scenarios.

Why the other options are wrong

192.168.50.127 is the broadcast address for the lower /25 subnet (192.168.50.0/25), not the subnet containing 192.168.50.130. Since the host IP is in the upper subnet, this option is incorrect.

192.168.50.128 is the network address of the upper /25 subnet (192.168.50.128/25), not the broadcast address. Network addresses cannot be assigned to hosts or used as broadcast addresses, so this option is incorrect.

192.168.50.254 is a valid usable host address within the upper /25 subnet. It is not the broadcast address, so this option is incorrect.

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MCQhard

A host address is 192.168.88.66/27. Which address is the network address of the subnet?

A.192.168.88.32

B.192.168.88.64

C.192.168.88.95

D.192.168.88.96

AnswerB

The /27 mask creates 32-address subnets, and the range from .64 to .95 is one such subnet. Host .66 is within this range, so the network address is the first address of the block, 192.168.88.64. All hosts in this subnet share the same network bits, and .64 is the all-zero host portion for that range.

Why this answer

A /27 subnet has a block size of 32. In practical terms, the fourth-octet ranges are 0-31, 32-63, 64-95, and so on. Because 66 falls within the 64-95 block, the network address is 192.168.88.64.

This is a classic subnet-boundary calculation. The key step is identifying the correct block first.

Exam trap

Avoid assuming the host address is in the first or last subnet without calculating the correct range.

Why the other options are wrong

192.168.88.32 is the network address of the previous /27 subnet (32-63). Since the host address 192.168.88.66 falls in the 64-95 range, the correct network address is 192.168.88.64, not 192.168.88.32.

192.168.88.95 is the broadcast address for the /27 subnet that starts at 192.168.88.64. The broadcast address is the last address in the subnet, used to send packets to all hosts in that subnet, not the network address.

192.168.88.96 is the network address of the next /27 subnet (96-127). The host 192.168.88.66 is not in that range; it belongs to the subnet starting at 192.168.88.64.

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Drag & Dropmedium

Drag and drop the following steps into the correct order to configure a Windows 10 host with a static IPv4 address, subnet mask, and default gateway.

Drag steps to the numbered slots on the right, or tap a step then tap a slot.

Steps

Order

1Step 1

2Step 2

3Step 3

4Step 4

5Step 5

6Step 6

Why this order

The correct order begins with opening Network and Sharing Center to access network settings. Then, you must click 'Change adapter settings' to see the list of network connections. Right-clicking the appropriate adapter and selecting Properties opens its configuration.

Selecting IPv4 and clicking Properties allows you to set the IP parameters. Choosing 'Use the following IP address' enables the fields for static input. Finally, entering the IP address, subnet mask, and default gateway followed by OK/Close applies the configuration.

This sequence follows the logical navigation of the Windows GUI to reach the static IP assignment interface.

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MCQhard

A host address is 192.168.1.14/29. Which address is the broadcast address for that host’s subnet?

A.192.168.1.7

B.192.168.1.14

C.192.168.1.15

D.192.168.1.16

AnswerC

A /29 prefix (255.255.255.248) creates subnets with 8 addresses each. The host 192.168.1.14 falls in the subnet from 192.168.1.8 to 192.168.1.15, where the first address is the network ID and the last is the directed broadcast. Therefore, 192.168.1.15 is the broadcast address for this subnet, and .14 is the last usable host address.

Why this answer

A /29 uses blocks of 8 addresses. In plain language, the subnets in the last octet move in increments of 8: 0–7, 8–15, 16–23, and so on. Since the host address ends in 14, it belongs to the 8–15 block. In that block, the last address is the broadcast address, so the broadcast is 192.168.1.15.

This is a classic subnetting pattern because it requires you to place the host inside the correct block and then identify the last address in that block rather than guessing based on the host value itself.

Exam trap

Be careful not to confuse the network address or the next subnet's network address with the broadcast address.

Why the other options are wrong

192.168.1.7 is the broadcast address for the /29 block 0–7, which does not contain host .14. The host .14 is in the block 8–15, so its broadcast is .15.

192.168.1.14 is the host address itself, not the broadcast address. The broadcast address is always the last address in the subnet, which is .15 for the block 8–15.

192.168.1.16 is the network address of the next /29 block (16–23), not the broadcast address for the block containing .14. The broadcast address must be the last address in the same block as the host.

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MCQhard

A host receives a correct IP address and subnet mask from DHCP but still cannot reach remote networks. Local subnet communication works. Which missing DHCP option is the strongest suspect?

A.Default gateway information

B.An STP root bridge ID

C.A voice VLAN value

D.A router ID

AnswerA

Without a default gateway, a host can determine that a destination is off-subnet via its own IP and mask, but it has no next-hop IP address to which to forward the packets. DHCP typically delivers the default gateway via option 3, and its absence means local-link traffic works while remote traffic remains unanswered. Therefore, this is the missing required information for off-subnet connectivity.

Why this answer

The strongest suspect is the default gateway option. In practical terms, the host can already identify local destinations because the subnet mask is present. That is why local communication still works. What it does not have is the next hop needed for off-subnet traffic. Without a default gateway, remote communication usually fails.

This is a very common host-troubleshooting scenario. It separates basic local addressing from the additional information required for off-subnet reachability.

Exam trap

A common exam trap is selecting options like STP root bridge ID or voice VLAN as the cause of remote connectivity failure. These options relate to Layer 2 switching or voice VLAN segmentation and do not affect IP routing or host reachability to remote networks. Candidates might confuse DHCP options that influence Layer 2 behavior with those critical for Layer 3 routing.

The key mistake is overlooking the default gateway option, which is essential for off-subnet traffic forwarding. This trap tests the candidate’s understanding of DHCP’s role in providing routing information, not just IP addressing.

Why the other options are wrong

An STP root bridge ID is irrelevant to host IP reachability because it pertains to Layer 2 spanning tree topology and does not affect IP routing or DHCP configuration for hosts.

A voice VLAN value is specific to voice traffic segmentation on switches and does not influence a host’s ability to route IP packets to remote networks, making it unrelated to the connectivity issue.

A router ID is a concept used in routing protocols like OSPF and does not apply to DHCP or host IP configuration, so it cannot cause the described connectivity problem.

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MCQhard

A host uses the subnet mask 255.255.254.0. Which prefix length does this represent?

A./22

B./23

C./24

D./25

AnswerB

A prefix length of /23 means 23 bits are set to 1 in the subnet mask, which in binary is 11111111.11111111.11111110.00000000. This converts directly to decimal 255.255.254.0, where the third octet has seven network bits and one host bit. Since the fourth octet contributes eight more host bits, the total host address space is 2^9 = 512 addresses, verifying that /23 is the correct representation.

Why this answer

The mask 255.255.254.0 represents /23. In plain language, the first two octets are fully network bits, the third octet contributes 7 more network bits because 254 in binary is 11111110, and the last octet contributes none. That totals 23 network bits.

This is a common prefix-conversion question because candidates often memorize the usual masks but hesitate when a non-/24 boundary appears. Recognizing that 255.255.254.0 equals /23 is important for subnetting and local-scope calculations.

Exam trap

Be cautious of common subnet masks like /24 and ensure you count the bits correctly, especially in non-standard masks.

Why the other options are wrong

Option A is incorrect because the subnet mask 255.255.254.0 corresponds to a prefix length of /23, not /22. A /22 prefix would indicate a subnet mask of 255.255.252.0.

The subnet mask 255.255.254.0 corresponds to a prefix length of /23, not /24. A /24 prefix length would indicate a subnet mask of 255.255.255.0, which allows for fewer hosts per subnet.

Option D is incorrect because a subnet mask of 255.255.254.0 corresponds to a prefix length of /23, not /25. A /25 subnet mask would allow for only 128 IP addresses, which is not applicable here.

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MCQmedium

A host can reach other devices on its local subnet, but it cannot reach remote networks. The host has a valid IP address and subnet mask. Which missing item is the strongest suspect?

A.Default gateway information

B.STP priority information

C.A voice VLAN setting

D.An OSPF process ID

AnswerA

A host reaches its local subnet because it can ARP for those destinations, but for any address outside that subnet, it must forward the packet to a router. Without a default gateway, the host has no next-hop IP address, so it drops or returns an error for all off-subnet traffic. The default gateway is the router interface on the host's subnet, and missing it precisely matches the symptom of local reachability but remote unreachability.

Why this answer

The strongest suspect is a missing default gateway. In practical terms, the host can still identify and reach local addresses because it has its own IP and subnet mask. But without a default gateway, it has no next hop for destinations outside the local subnet. That is why local communication works while remote communication fails.

This is one of the most common host-configuration troubleshooting patterns on the exam and in real networks.

Exam trap

A common exam trap is selecting options related to routing protocols like OSPF or Layer 2 technologies such as STP or VLANs when the issue is actually a missing default gateway. Candidates might incorrectly assume that the host needs an OSPF process ID or STP priority to reach remote networks. However, hosts do not run routing protocols and do not participate in STP decisions.

The real problem is the absence of default gateway information, which prevents the host from forwarding packets beyond its local subnet. Misunderstanding this leads to incorrect answers that focus on advanced protocols rather than basic IP configuration.

Why the other options are wrong

Incorrect because STP priority is a Layer 2 switch parameter that does not affect host IP routing or the ability to reach remote networks. Hosts do not participate in STP decisions.

Incorrect because voice VLAN settings relate to Layer 2 segmentation for voice traffic and do not impact the host’s IP routing or default gateway configuration needed for remote communication.

Incorrect because hosts do not run routing protocols like OSPF and do not require an OSPF process ID. Routing protocols are configured on routers, not end hosts.

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MCQhard

A host is configured as 10.20.30.95/27. Which address is the network address of its subnet?

A.10.20.30.32

B.10.20.30.64

C.10.20.30.95

D.10.20.30.96

AnswerB

The /27 prefix length means the subnet mask is 255.255.255.224, giving a block size of 32 addresses. The host address 10.20.30.95 falls within the range 10.20.30.64 through 10.20.30.95, making .64 the first address of that subnet. Since the network address is always the first address in the block, 10.20.30.64 is indeed the network for this host.

Why this answer

A /27 uses blocks of 32 addresses. In plain language, the ranges in the last octet are 0–31, 32–63, 64–95, 96–127, and so on. Since the host address ends in 95, it belongs to the 64–95 block. The first address in that block is the network address, which is 10.20.30.64.

This is a classic subnet-boundary question because it tests whether you can identify the containing block and then choose the first address in that block as the network address.

Exam trap

A frequent exam trap is mistaking the host IP address for the network address or incorrectly identifying the subnet block boundaries. Candidates often select the host IP itself or the next block's starting address as the network address. For example, choosing 10.20.30.95 or 10.20.30.96 instead of 10.20.30.64.

This happens because the subnet mask’s block size (32 addresses for /27) is overlooked, leading to confusion about which block the host belongs to. Misunderstanding this can cause incorrect subnetting and routing errors in real networks.

Why the other options are wrong

10.20.30.32 is incorrect because the host IP 10.20.30.95 does not fall within the 32–63 subnet block. This address belongs to a different subnet block and cannot be the network address for the given host.

10.20.30.95 is incorrect because it is the host IP address itself, not the network address. The network address must be the first address in the subnet block, not a host address.

10.20.30.96 is incorrect because it is the first address of the next subnet block (96–127). The host IP 10.20.30.95 belongs to the previous block, so 10.20.30.96 cannot be its network address.

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MCQhard

A router interface is configured with 192.0.2.97/28. What is the network address of the subnet?

A.192.0.2.80

B.192.0.2.96

C.192.0.2.111

D.192.0.2.112

AnswerB

Correct. It is the first address in the 192.0.2.96/28 subnet.

Why this answer

A /28 uses blocks of 16 addresses. The block containing .97 runs from .96 through .111, so .96 is the network address.

Exam trap

Be careful not to confuse the given IP address or the broadcast address with the network address.

Why the other options are wrong

Option A (192.0.2.80) is incorrect because the subnet mask /28 indicates a block size of 16, making the valid network addresses range from 192.0.2.96 to 192.0.2.111. The network address for this subnet is 192.0.2.96.

The address 192.0.2.111 is not a valid network address for the subnet defined by 192.0.2.97/28, as it falls outside the range of usable addresses for that subnet, which spans from 192.0.2.96 to 192.0.2.111.

Option D, 192.0.2.112, is wrong because it does not represent the network address for the subnet defined by the IP address 192.0.2.97/28. The correct network address is 192.0.2.96, which is the first address in the subnet range.

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MCQhard

A host uses the subnet mask 255.255.255.240. How many usable host addresses exist in each subnet?

A.12

B.14

C.16

D.30

AnswerB

A subnet mask of 255.255.255.240 is a /28 prefix, meaning 4 bits are available for host addresses. This yields 2^4 = 16 total IP addresses within the subnet. To determine usable host addresses, subtract the network address and the broadcast address, leaving 16 - 2 = 14 assignable addresses. So 14 is the correct number of usable hosts for a /28 network.

Why this answer

The mask 255.255.255.240 corresponds to /28. In practical terms, that leaves 4 host bits, which gives 16 total addresses in each subnet. After subtracting the network and broadcast addresses, 14 usable hosts remain.

This is a classic host-capacity calculation and a very common subnetting pattern on the exam.

Exam trap

A frequent exam trap is selecting the total number of addresses in the subnet (16) instead of the usable host addresses (14). Candidates often forget to subtract the network and broadcast addresses, which are reserved and cannot be assigned to hosts. Another common mistake is confusing the /28 subnet mask with /27, which offers 30 usable hosts, leading to incorrect answers like 30.

This confusion arises because both subnet masks are close in size but differ significantly in host capacity. Always remember that usable hosts equal total addresses minus two reserved addresses.

Why the other options are wrong

Option A states 12 usable hosts, which is incorrect because a /28 subnet provides 16 total addresses. Subtracting the network and broadcast addresses leaves 14 usable hosts, not 12. This option underestimates the host capacity.

Option C claims 16 usable hosts, which is incorrect because 16 represents the total number of addresses in the subnet, including network and broadcast addresses. Usable hosts must exclude these two reserved addresses.

Option D suggests 30 usable hosts, which corresponds to a /27 subnet mask (255.255.255.224), not /28. This overestimates the host count for the given subnet mask and is therefore incorrect.

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MCQhard

A host uses subnet mask 255.255.255.224. How many total addresses exist in each subnet block?

A.16

B.32

C.30

D.64

AnswerB

The subnet mask 255.255.255.224 has 27 network bits, denoted as /27. The total number of addresses in a subnet is calculated as 2^(32-27) = 2^5 = 32. This total includes both the network address and the broadcast address, so the full address block size is 32 addresses.

Why this answer

The subnet mask 255.255.255.224 (/27) has 5 host bits, so each subnet has 2^5 = 32 total addresses. Option A (16) results from confusing /27 with /28 (4 host bits = 16). Option D (64) results from confusing /27 with /26 (6 host bits = 64).

Option C (30) is the number of usable host addresses (32 - 2 = 30), not the total addresses asked for in the question.

Exam trap

Be careful to distinguish between total addresses and usable addresses. Many candidates automatically think of usable addresses when subnetting.

Why the other options are wrong

16 would be the total addresses for a /28 mask (4 host bits), not the /27 mask specified.

30 is the number of usable host addresses (total addresses minus network and broadcast), not the total addresses.

64 would be the total addresses for a /26 mask (6 host bits), not the /27 mask specified.

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MCQhard

A network administrator has several access points. All APs except one have successfully joined the wireless controller. The administrator verifies the failing AP’s IP address, subnet mask, and controller IP address are correctly configured. What is the most likely reason the AP cannot join the controller?

A.The AP has an incorrect default gateway for its subnet.

B.CAPWAP can be used only if the AP has no IP address.

C.The AP must use PPP instead of Ethernet to reach the controller.

D.The controller can support only 14 APs maximum.

AnswerA

If the AP's default gateway is incorrect, the access point cannot route CAPWAP packets off its local subnet to reach a controller on a different IP subnet. CAPWAP discovery and join responses from the controller would never arrive, even though the AP's own IP address and controller discovery mechanism (such as DNS or broadcast) are correctly set. The result is the AP stuck in an unjoined/disabled state, exactly as shown.

Why this answer

The most likely cause is that the AP has an incorrect default gateway. For the AP to reach the controller (which may be on a different subnet), it needs a correct default gateway to route traffic. The other APs joined successfully, eliminating a controller-wide issue.

Option B is incorrect because CAPWAP requires an IP address; it does not work without one. Option C is incorrect because CAPWAP uses IP/UDP, not PPP. Option D is unlikely because there is no indication that the controller is at its AP limit; the problem affects only one AP, suggesting an individual misconfiguration.

Exam trap

Avoid assuming global issues when only one AP is affected; focus on individual AP configuration and connectivity.

Why the other options are wrong

CAPWAP tunnels require the AP to have an IP address; the statement is false.

CAPWAP operates over IP using UDP ports, not PPP.

The controller may have an AP capacity limit, but with only one AP failing and no evidence that the limit is 14, this is not the strongest explanation.

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MCQhard

A host has a valid IP address and subnet mask from DHCP but cannot reach remote networks because no gateway was provided. What is the best explanation?

A.The host has no next-hop gateway for traffic destined outside its local subnet.

B.The host cannot use ARP on the local network anymore.

C.The host automatically becomes part of every remote subnet.

D.The host must convert its access port into a trunk.

AnswerA

A DHCP lease supplies the host's IP address, subnet mask, and often a default gateway, but without that gateway entry the host's routing table contains no route for off-subnet destinations. When a packet's destination IP is outside the host's local subnet, the host must send it to a next-hop router; in its absence, the packet is dropped or never transmitted. The host can still communicate with neighbors on the same subnet, but it is effectively isolated from all other networks.

Why this answer

The best explanation is that the host has no next-hop path for off-subnet traffic. In plain language, the device knows what its own local network looks like, but it does not know where to send packets when the destination is outside that local range. Without a default gateway, remote communication usually fails even though local communication can still work.

This is a core host-configuration concept. The correct answer is the one focused on the absence of a next hop for remote destinations.

Exam trap

A common exam trap is assuming that a host without a default gateway cannot communicate at all. Many candidates mistakenly believe that missing a gateway disables all network communication, but in reality, the host can still communicate with devices on its local subnet using ARP. The trap lies in confusing local subnet communication with remote network access.

The question specifically tests understanding that the lack of a gateway prevents forwarding to remote networks, not local connectivity. Misreading this can lead to selecting incorrect options that focus on ARP or subnet expansion rather than the gateway role.

Why the other options are wrong

This option is incorrect because ARP is used for local subnet communication and does not depend on the presence of a default gateway. The host can still use ARP to communicate locally.

This option is wrong because a host does not automatically become part of every remote subnet without a gateway. The subnet mask defines the local subnet boundaries, and gateway absence does not change this.

This option is unrelated to the problem. Converting an access port to a trunk port affects VLAN tagging on switches but does not impact a host’s ability to have or use a default gateway.

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MCQmedium

A network administrator is configuring a new Windows 10 workstation on a network that uses DHCP. The workstation receives an IPv4 address of 169.254.10.20 with a subnet mask of 255.255.0.0 and no default gateway. The user cannot access the internet or other subnets. What is the most likely cause of this issue?

A.The workstation has a duplicate IP address conflict.

B.The workstation's DNS server configuration is incorrect.

C.The workstation's subnet mask is misconfigured.

D.The DHCP server is unreachable or not responding.

AnswerD

When a DHCP client fails to receive an IP address from a DHCP server, it self-assigns an APIPA address from the 169.254.0.0/16 range. This explains the observed address and the absence of a default gateway.

Why this answer

The IP address 169.254.10.20 with a /16 subnet mask is an Automatic Private IP Addressing (APIPA) address, which Windows assigns when a DHCP discovery broadcast (DHCPDISCOVER) fails to receive a response from a DHCP server. Without a valid DHCP lease, the workstation has no default gateway, so it cannot communicate outside its local subnet, explaining the lack of internet or inter-subnet access. The most likely cause is that the DHCP server is unreachable or not responding, forcing the client to self-assign an APIPA address.

Exam trap

Cisco often tests the misconception that a 169.254.x.x address indicates a duplicate IP or a subnet mask issue, but the real trap is that APIPA is a direct symptom of DHCP server unavailability, not a configuration error on the client.

Why the other options are wrong

Duplicate IP conflicts typically result in a warning but do not cause the system to assign a 169.254.x.x address.

DNS issues do not affect IP address assignment via DHCP.

The subnet mask is correct for the APIPA range; the problem is the lack of a DHCP server response.

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MCQhard

If a host has a valid IP address and subnet mask but no default gateway, what is the most likely result?

A.The host can usually reach only local-subnet destinations and not remote networks.

B.The host cannot use ARP at all.

C.The host automatically joins every subnet in the LAN.

D.The host becomes the default gateway for other devices.

AnswerA

Without a default route, the host's IP stack has no next-hop entry for packets whose destination IPv4 address is outside the host's directly connected subnet. The host will attempt to deliver such packets to the local network only if a matching route exists, but since the destination is off-subnet, no ARP resolution is possible and the packets are discarded. Conversely, destination addresses within the host's configured subnet are considered on-link, so the host uses ARP to resolve their MAC addresses and can communicate with them directly.

Why this answer

The host will normally reach local destinations but fail to reach remote networks. In practical terms, the subnet mask still lets the host identify what is local, but without a default gateway it has no next hop for off-subnet traffic. That means local ARP-based communication can still work, while remote communication usually fails.

This is a core host-configuration concept and a very common certification question. The missing gateway does not break all communication — it breaks off-subnet communication.

Exam trap

A common exam trap is assuming that a host without a default gateway cannot use ARP or communicate at all. This is incorrect because ARP is used for local Layer 2 address resolution and remains functional. Another trap is believing the host automatically joins other subnets or becomes a gateway for others, which does not happen.

The key mistake is confusing local subnet communication with remote network access. The default gateway only affects off-subnet traffic, so the host can still communicate locally but fails to reach remote destinations.

Why the other options are wrong

This option is incorrect because ARP is used for local Layer 2 address resolution and remains functional even if the default gateway is missing; the host can still resolve MAC addresses on the local subnet.

This option is incorrect because the host’s subnet membership is determined by its IP address and subnet mask, not by the presence or absence of a default gateway; it does not join other subnets automatically.

This option is incorrect because a host does not become a default gateway for other devices simply by lacking a configured gateway; routing and gateway roles require explicit configuration on routers.

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MCQhard

A host uses address 192.168.5.126/25. Which address is the broadcast address for its subnet?

A.192.168.5.63

B.192.168.5.127

C.192.168.5.128

D.192.168.5.255

AnswerB

The /25 CIDR notation specifies a subnet mask of 255.255.255.128, meaning the first 25 bits define the network portion. For the host 192.168.5.126, the network address is 192.168.5.0, as the 25th bit is 0. This subnet, 192.168.5.0/25, encompasses IP addresses from 192.168.5.0 to 192.168.5.127. The broadcast address is always the last address in the subnet range, where all host bits are set to one, which in this scenario is 192.168.5.127.

Why this answer

A /25 divides the /24 into two halves: 0–127 and 128–255. In plain language, the host address 192.168.5.126 is in the lower half, which runs from .0 through .127. The last address in that block is the broadcast address, so the broadcast is 192.168.5.127.

This is a classic subnetting question because it checks whether you can identify the correct block and then select the last address in that block as the broadcast address.

Exam trap

Remember that the broadcast address is the last address in the subnet, not the first address of the next subnet or the broadcast for the entire /24.

Why the other options are wrong

Option A (192.168.5.63) is incorrect because it does not fall within the subnet defined by the address 192.168.5.126/25, which has a valid range of 192.168.5.128 to 192.168.5.255 for host addresses.

The address 192.168.5.128 is incorrect because it falls outside the subnet defined by 192.168.5.126/25, which includes addresses from 192.168.5.0 to 192.168.5.127. The broadcast address for this subnet is 192.168.5.127.

Option D, 192.168.5.255, is incorrect because it is the broadcast address for the entire 192.168.5.0/24 subnet, not the /25 subnet specified in the question.

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MCQhard

A subnet has the network address 192.168.20.128/26. What is the broadcast address?

A.192.168.20.159

B.192.168.20.191

C.192.168.20.192

D.192.168.20.255

AnswerB

192.168.20.191 is the broadcast address for the 192.168.20.128/26 subnet. A /26 prefix leaves 6 host bits (32-26=6), yielding 2^6=64 addresses per subnet. Starting at network address .128, the valid addresses run from .129 through .190, and .191 is the final address (broadcast) before the next subnet begins at .192.

Why this answer

A /26 subnet uses blocks of 64 addresses. In plain language, the block that starts at 192.168.20.128 runs through 192.168.20.191. The first address in that block is the network address, and the last address is the broadcast address. That means the broadcast address is 192.168.20.191.

This is a standard subnetting calculation. Once the block size is identified, the broadcast address is simply the last address in the block.

Exam trap

Be careful not to confuse the broadcast address with the first address of the next subnet or with the broadcast address of a different subnet mask.

Why the other options are wrong

Option A (192.168.20.159) is incorrect because the broadcast address for the subnet 192.168.20.128/26 is 192.168.20.191, calculated as the highest address in the subnet range from 192.168.20.128 to 192.168.20.191.

The option C, 192.168.20.192, is incorrect because it falls outside the range of the subnet defined by 192.168.20.128/26, which has a valid broadcast address of 192.168.20.191.

The broadcast address for the subnet 192.168.20.128/26 is 192.168.20.191, not 192.168.20.255. The address 192.168.20.255 is the broadcast address for the entire 192.168.20.0/24 network, which is not relevant to the specified subnet.

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MCQmedium

A branch LAN requires 50 usable IPv4 host addresses. What is the most efficient subnet mask that provides at least 50 usable hosts?

A./27

B./26

C./25

D./24

AnswerB

A /26 prefix has 6 host bits, so 2^6 = 64 total addresses, minus 2 for network and broadcast gives 62 usable hosts. This exactly satisfies the requirement of 50 usable addresses with headroom for future growth, and is the smallest prefix that meets the need, avoiding unnecessary address waste.

Why this answer

A /26 provides 64 total addresses and 62 usable host addresses, which is the smallest subnet mask (largest prefix length) that fits 50 hosts, making it the most efficient choice.

Exam trap

Read the requirement carefully. Cisco often uses subtle wording like 'most efficient' or 'industry standard' to eliminate technically correct but non-optimal answers.

Why the other options are wrong

A /27 subnet provides only 30 usable host addresses, which is insufficient for the requirement of 50 usable addresses. Therefore, it does not meet the needs of the branch LAN.

Option C: /25 provides 126 usable addresses, which exceeds the requirement of 50 usable addresses. However, it is not the smallest prefix that meets the requirement.

Option D, /24, provides 256 total addresses, which exceeds the requirement of 50 usable addresses. However, it is not the smallest prefix that meets the requirement, as /26 provides 64 addresses, which is sufficient and more efficient.

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MCQhard

A host is configured with 10.1.1.34/30. Which address is the broadcast address for its subnet?

A.10.1.1.31

B.10.1.1.35

C.10.1.1.32

D.10.1.1.36

AnswerB

The /30 subnet mask specified for the host 10.1.1.34 allocates two bits for host addresses. This creates a subnet containing four addresses in total. The network address for 10.1.1.34/30 is 10.1.1.32, where all host bits are zero. Consequently, the broadcast address is the last address in this subnet, where all host bits are set to one, which calculates to 10.1.1.35.

Why this answer

A /30 subnet has a block size of 4. In practical terms, the last-octet blocks are 0–3, 4–7, 8–11, and so on. Because 34 falls within the 32–35 block, the broadcast address is the last address in that block: 10.1.1.35.

This question tests whether you can identify the correct subnet block before choosing the broadcast address.

Exam trap

Ensure you correctly identify the subnet range and understand the roles of network, host, and broadcast addresses.

Why the other options are wrong

The address 10.1.1.31 is incorrect as a broadcast address for the subnet 10.1.1.34/30; the correct broadcast address is 10.1.1.35, derived from the subnet mask which allows for only 4 IP addresses (10.1.1.34 to 10.1.1.37).

The address 10.1.1.32 is not the broadcast address for the subnet 10.1.1.34/30; instead, it is the network address for the subnet, which is 10.1.1.32 to 10.1.1.35.

The address 10.1.1.36 is incorrect because it exceeds the valid range for a /30 subnet, which only allows for four IP addresses (two usable hosts, one network, and one broadcast). The broadcast address for the subnet 10.1.1.34/30 is 10.1.1.35.

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MCQhard

A subnet has the network address 192.168.30.0/26. What is the broadcast address?

A.192.168.30.63

B.192.168.30.64

C.192.168.30.255

D.192.168.30.31

AnswerA

The /26 prefix length means 26 network bits, leaving 6 host bits, so the block size is 64 addresses. Since the network address is 192.168.30.0, the usable host range is .1 through .62, and the broadcast address is the last address in the block, which is .63. This is the directed broadcast for this specific /26 subnet.

Why this answer

A /26 subnet contains 64 addresses. In plain language, the block that starts at 192.168.30.0 runs through 192.168.30.63. The first address is the network address and the last address is the broadcast address, so the broadcast for this subnet is 192.168.30.63.

This is a standard subnet-boundary calculation. Once the block size is identified, the broadcast address is simply the final address in that block.

Exam trap

Remember that the broadcast address is the last address in the subnet, not the first address of the next subnet or the last usable host address.

Why the other options are wrong

Option B is incorrect because the broadcast address for the subnet 192.168.30.0/26 is 192.168.30.63, not 192.168.30.64. The address 192.168.30.64 is actually the first address of the next subnet (192.168.30.64/26).

The address 192.168.30.255 is incorrect as the broadcast address for the subnet 192.168.30.0/26, which only spans IPs from 192.168.30.0 to 192.168.30.63. The broadcast address for this subnet is 192.168.30.63.

Option D, 192.168.30.31, is incorrect because it represents the last usable IP address in the subnet 192.168.30.0/26, not the broadcast address. The broadcast address for this subnet is 192.168.30.63.

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MCQhard

A network administrator is troubleshooting a Windows 10 client that cannot reach the internet. The client is connected to a Cisco switch port configured as an access port in VLAN 100. The administrator runs ipconfig on the client and sees an IP address of 169.254.10.15 with a subnet mask of 255.255.0.0. The switch port shows status up/up. What is the most likely cause of the issue?

A.The switch port is in err-disabled state due to a spanning-tree loop.

B.The switch is not configured with an ip helper-address on the VLAN 100 SVI to forward DHCP broadcasts to the DHCP server.

C.The client is in the wrong VLAN; the switch should be configured with VLAN 200.

D.There is a duplex mismatch between the client and the switch.

AnswerB

For a DHCP client in VLAN 100 to obtain an address from a server located in a different VLAN or subnet, the VLAN 100 SVI must include an 'ip helper-address' pointing to the DHCP server. DHCP requests are IP broadcast packets, and routers do not forward broadcasts across interfaces by default; without UDP forwarding configured, the DHCPDISCOVER is never relayed to the server, so the client receives no DHCPOFFER and may fall back to an APIPA address. The correct remedy is to configure 'interface vlan 100' with the 'ip helper-address' command referencing the DHCP server's IP address.

Why this answer

The 169.254.x.x address is an Automatic Private IP Addressing (APIPA) address, assigned by Windows when DHCP fails. Since the client is in VLAN 100 and the switch port is up/up, the most likely cause is that the VLAN 100 SVI lacks an ip helper-address command, so DHCP broadcast requests from the client are not forwarded to the DHCP server, leaving the client without a valid IP address.

Exam trap

Cisco often tests the misconception that an APIPA address indicates a physical or VLAN issue, when in fact it specifically points to DHCP failure, and the most common cause in a routed environment is the absence of ip helper-address on the SVI.

Why the other options are wrong

The port status is 'connected', not err-disabled.

No information suggests the client should be in VLAN 200; the configuration matches the intended VLAN.

Duplex mismatch would typically cause errors or speed/duplex issues, but the switch shows a-full, so this is not the problem.

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MCQmedium

A network administrator receives a report that a user on a Windows laptop cannot connect to the internet, although other devices on the same subnet are working. The administrator runs `ipconfig` on the laptop and sees an IP address of 169.254.15.22 with a subnet mask of 255.255.0.0 and no default gateway. Based on this output, what is the most likely cause of the connectivity issue?

A.The laptop's DNS server settings are incorrect.

B.The DHCP server is unreachable or not responding to the laptop's DHCP request.

C.The laptop has a static IP address configured that conflicts with another device.

D.The Ethernet cable is faulty or disconnected.

AnswerB

APIPA is assigned when a DHCP client fails to receive a DHCPOFFER after sending DHCPDISCOVER messages. This typically means the DHCP server is down, misconfigured, or the laptop cannot reach it due to a network issue (e.g., VLAN mismatch, switch port problem).

Why this answer

The IP address 169.254.15.22 with a subnet mask of 255.255.0.0 is an Automatic Private IP Addressing (APIPA) address, which Windows assigns when a DHCP client fails to obtain a lease from a DHCP server. The absence of a default gateway confirms that the laptop cannot reach any DHCP server, as APIPA addresses are not routable and are only used for link-local communication. Therefore, the most likely cause is that the DHCP server is unreachable or not responding to the laptop's DHCP request.

Exam trap

Cisco often tests the distinction between APIPA and other IP assignment failures, and the trap here is that candidates may confuse a DHCP failure with a physical layer issue (faulty cable) or a DNS misconfiguration, not realizing that APIPA is a specific Windows behavior triggered only by DHCP unavailability.

Why the other options are wrong

The APIPA address indicates a DHCP failure, not a DNS problem.

A static IP conflict would not result in an APIPA address; the laptop would show the manually configured IP, not 169.254.x.x.

A physical cable issue would prevent link, so the laptop would not even attempt DHCP and would not get an APIPA address.

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PBQhard

You are connected to R1. Configure IPv4 and IPv6 addressing on R1's interfaces and verify reachability to R2. The current configuration has a wrong subnet mask on G0/0, missing default gateway for IPv4, and R1's IPv6 address is configured using EUI-64 while R2 uses a static IPv6 address. Fix these issues so that R1 can ping both R2's IPv4 and IPv6 addresses.

Network Topology

Hints

•Compare the subnet masks on R1 and R2's G0/0 interfaces.
•Check the IPv4 default route — the next-hop must be reachable.
•R1's IPv6 EUI-64 will not match the static address on R2; use a static assignment on the same subnet.

A.Change R1 G0/0 subnet mask to /30, add a default route via 192.0.2.2, and configure a static IPv6 address 2001:db8:1::1/64 on G0/0.

B.Change R1 G0/0 subnet mask to /30, add a default route via 192.0.2.254, and keep the EUI-64 IPv6 address on G0/0.

C.Change R1 G0/0 subnet mask to /24, add a default route via 192.0.2.2, and configure a static IPv6 address 2001:db8:1::1/64 on G0/0.

D.Change R1 G0/0 subnet mask to /30, add a default route via 192.0.2.2, and keep the EUI-64 IPv6 address on G0/0.

AnswerA

solution

! R1
enable
configure terminal
interface GigabitEthernet0/0
ip address 192.0.2.1 255.255.255.252
no ipv6 address 2001:db8:1::/64 eui-64
ipv6 address 2001:db8:1::1/64
exit
no ip route 0.0.0.0 0.0.0.0 192.0.2.254
ip route 0.0.0.0 0.0.0.0 192.0.2.2
end
write memory

Why this answer

The problem had three issues: (1) R1's G0/0 subnet mask was /24 (255.255.255.0) but R2's G0/0 was /30 (255.255.255.252), causing an IP subnet mismatch. (2) R1 lacked a default gateway for IPv4; the static route pointed to 192.0.2.254 which is not reachable. (3) R1's IPv6 EUI-64 configuration on G0/0 generates an interface ID from the MAC, but R2 expects a static address 2001:db8:1::2/64, so R1 must use a static IPv6 address on the same subnet. The fix: change R1's G0/0 mask to /30, add a default route via R2's G0/0 IP (192.0.2.2), and configure a static IPv6 address (e.g., 2001:db8:1::1/64) on R1's G0/0.

Exam trap

Be careful not to confuse the default gateway with an arbitrary IP; it must be the next-hop router's interface IP on the same subnet. Also, remember that EUI-64 generates a unique interface ID from the MAC, which may not match a statically configured peer address—both sides must use consistent addressing methods.

Why the other options are wrong

The default gateway must be R2's directly connected interface IP (192.0.2.2), not 192.0.2.254. EUI-64 cannot be used if the peer expects a specific static address on the same subnet.

The subnet mask must be consistent on both ends of the link. A /24 mask on one side and /30 on the other creates overlapping subnets and routing issues.

EUI-64 does not guarantee that the resulting IPv6 address will be on the same subnet as a statically configured peer address. For direct connectivity, both routers must have addresses in the same subnet.

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MCQhard

A host address is 10.77.4.141/28. Which address is the network address of the subnet?

A.10.77.4.128

B.10.77.4.143

C.10.77.4.144

D.10.77.4.112

AnswerA

The host address 10.77.4.141 falls within the /28 subnet that starts at 10.77.4.128, because the /28 block size is 16 addresses (128-143). The network address is the first address of that block, where all host bits are zero, so 10.77.4.128 is the correct network address. Any address in the range 10.77.4.129 through 10.77.4.142 is valid for hosts, while .143 is the broadcast.

Why this answer

A /28 subnet has a block size of 16. In practical terms, the last-octet blocks are 0-15, 16-31, 32-47, and so on. Because 141 falls within the 128-143 block, the network address is 10.77.4.128.

This is a clean addressing-boundary question that rewards careful block calculation rather than guesswork.

Exam trap

Be careful not to confuse host addresses with network addresses. Always calculate the subnet block to find the network address.

Why the other options are wrong

Option B, 10.77.4.143, is incorrect because it falls within the usable host range of the subnet defined by 10.77.4.128/28, which spans from 10.77.4.129 to 10.77.4.142. The network address must always be the first address in the subnet.

This option is wrong because 10.77.4.144 is not the network address for the subnet defined by 10.77.4.141/28; the correct network address is 10.77.4.128, which is the first address in the subnet range.

Option D, 10.77.4.112, is incorrect because it does not fall within the subnet defined by the CIDR notation /28, which covers addresses from 10.77.4.128 to 10.77.4.143. The network address for this subnet is 10.77.4.128.

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MCQhard

A host address is 10.100.12.94/26. Which address is the broadcast address for that subnet?

A.10.100.12.63

B.10.100.12.127

C.10.100.12.64

D.10.100.12.128

AnswerB

This is correct because .94 is in the 64-127 /26 range.

Why this answer

A /26 subnet has a block size of 64, so the fourth-octet ranges are 0–63, 64–127, 128–191, and 192–255. The host address 10.100.12.94 lies in the 64–127 range, making the broadcast address the last address in that range: 10.100.12.127. Option A (10.100.12.63) is the broadcast of the previous subnet (0–63).

Option C (10.100.12.64) is the network address of the subnet containing the host. Option D (10.100.12.128) is the network address of the next subnet (128–191).

Exam trap

Be careful not to confuse host addresses or network addresses with the broadcast address. Remember, the broadcast address is the last address in the subnet range.

Why the other options are wrong

10.100.12.63 is the broadcast address of the 0–63 subnet, not the one containing 94.

10.100.12.64 is the network address of the 64–127 subnet, not the broadcast.

10.100.12.128 is the network address of the 128–191 subnet, not the broadcast.

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MCQhard

A host address is 172.22.14.99/27. Which address is the broadcast address of the subnet?

A.172.22.14.95

B.172.22.14.127

C.172.22.14.96

D.172.22.14.128

AnswerB

172.22.14.127 is the broadcast address for the /27 subnet that contains host 172.22.14.99. Since the /27 mask creates 32-address blocks, the relevant block runs from 172.22.14.96 through 172.22.14.127, and the broadcast is always the final address in that block. The host bits in .127 are all ones, which identifies this as the directed broadcast for this subnet.

Why this answer

A /27 subnet has a block size of 32. In practical terms, the relevant blocks are 0-31, 32-63, 64-95, 96-127, and so on. Because 99 falls within the 96-127 block, the broadcast address is the last address in that block: 172.22.14.127.

This is a classic subnet-boundary question and remains important because addressing precision appears throughout the CCNA blueprint.

Exam trap

Be careful not to confuse network addresses with broadcast addresses, and ensure you are calculating the correct subnet range.

Why the other options are wrong

Option A is incorrect because the broadcast address for the subnet 172.22.14.99/27 is 172.22.14.127, not 172.22.14.95. The /27 subnet mask indicates that the last 5 bits are for host addresses, allowing for a range of addresses that ends at 172.22.14.127.

Option C (172.22.14.96) is incorrect because it falls within the usable host range of the subnet and is not the broadcast address. The broadcast address for the subnet 172.22.14.96/27 is actually 172.22.14.127.

Option D, 172.22.14.128, is incorrect because it does not fall within the subnet defined by 172.22.14.99/27, which has a valid range of 172.22.14.96 to 172.22.14.127. The broadcast address for this subnet is 172.22.14.127.

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MCQhard

A host address is 10.10.10.14/29. Which address is the broadcast address for its subnet?

A.10.10.10.7

B.10.10.10.15

C.10.10.10.8

D.10.10.10.16

AnswerB

The /29 CIDR notation specifies that the first 29 bits define the network portion, leaving 3 bits for host addresses. For the 10.10.10.14/29 subnet, the network address is 10.10.10.8. The broadcast address is always the last address within a subnet, derived by setting all host bits to one. In this scenario, setting the three host bits of the 10.10.10.8 network to one results in 10.10.10.15, which functions as the broadcast address for this specific subnet, satisfying the constraint of the /29 mask.

Why this answer

A /29 subnet has a block size of 8. In practical terms, the fourth-octet ranges are 0–7, 8–15, 16–23, and so on. Since 14 falls in the 8–15 block, the broadcast address is the last address in that block, which is 10.10.10.15.

This is a classic subnetting question that checks whether you can find the block first and then identify the final address in that block.

Exam trap

Avoid confusing the network address or a host address with the broadcast address. Remember, the broadcast address is the last address in the subnet range.

Why the other options are wrong

Option A, 10.10.10.7, is incorrect because it does not represent the broadcast address for the subnet defined by 10.10.10.14/29. The correct broadcast address is 10.10.10.15, which is the highest address in the subnet range.

The address 10.10.10.8 is not the broadcast address for the subnet 10.10.10.14/29. The correct broadcast address is 10.10.10.15, which is the highest address in the subnet range from 10.10.10.8 to 10.10.10.15.

The address 10.10.10.16 is outside the subnet defined by 10.10.10.14/29, which includes addresses from 10.10.10.8 to 10.10.10.15. Therefore, it cannot be the broadcast address for this subnet.

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