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CCNA Addressing Subnet Formats Questions

43 questions · Addressing Subnet Formats topic · All types, answers revealed

1
MCQmedium

A network administrator is configuring a point-to-point WAN link between two Cisco routers. To conserve IPv4 address space, the administrator decides to use a /30 subnet mask on the serial interface. Which of the following best describes the number of usable host addresses provided by this subnet mask?

A.1 usable host address
B.2 usable host addresses
C.4 usable host addresses
D.6 usable host addresses
AnswerB

A /30 subnet mask leaves 2 bits for host addresses, yielding 2^2 - 2 = 2 usable addresses. This is ideal for point-to-point links because it minimizes address waste, providing exactly one address for each end of the link while reserving the network and broadcast addresses.

Why this answer

A /30 subnet mask provides 2 usable host addresses because it allocates 2 host bits, resulting in 4 total addresses minus the network and broadcast addresses. This is the standard choice for point-to-point WAN links, conserving address space while ensuring both routers have a unique IP.

Exam trap

The trap here is forgetting to subtract the network and broadcast addresses when calculating usable hosts, leading to an answer of 4 instead of 2.

2
MCQhard

Refer to the exhibit. An administrator is configuring subnets on a Cisco router. Based on the output, what is the broadcast address for the subnet connected to Gig0/0?

A.172.16.5.16
B.172.16.5.15
C.172.16.5.31
D.172.16.5.255
AnswerB

The /28 mask indicates a block size of 16. The subnet range for 172.16.5.0/28 starts at 172.16.5.0 and ends at 172.16.5.15. The address 172.16.5.15 serves as the broadcast address, which is the final address in the specific subnet block reserved for the interface.

Why this answer

The interface 172.16.5.1/28 belongs to the subnet starting at 172.16.5.0. Since the /28 mask provides 16 addresses (block size of 16), the range is 172.16.5.0 to 172.16.5.15. The broadcast address is the last address in the subnet, which is 172.16.5.15.

Identifying broadcast addresses is essential for troubleshooting traffic propagation and verifying proper segmentation in multi-interface router environments.

Exam trap

Candidates often confuse the network ID or the first usable host address with the broadcast address, failing to calculate the absolute final address of the block.

3
MCQhard

A network administrator is configuring a point-to-point link between two routers and wants to use the most efficient subnet mask that allows for only two usable host addresses. However, the administrator is told to use a /28 mask instead for future expansion. How many usable host addresses are provided by a /28 mask?

A.2 usable host addresses
B.6 usable host addresses
C.14 usable host addresses
D.30 usable host addresses
AnswerC

A /28 subnet mask uses 28 bits for the network portion, leaving exactly 4 bits for host identification. Calculating two to the power of four results in 16 total addresses, but after subtracting the mandatory network and broadcast addresses, 14 usable IPs remain for assignment to networking equipment.

Why this answer

Determining the number of usable hosts requires identifying the total number of bits remaining for the host portion after the subnet mask is applied. For a /28 prefix, 4 bits are available for hosts, resulting in 16 total addresses. Subtracting the network ID and the broadcast address leaves 14 addresses that can be assigned to devices.

Exam trap

Candidates frequently forget to subtract two for the network ID and broadcast address, mistakenly stating that a /28 mask provides 16 usable host addresses instead of 14.

4
MCQeasy

Which of the following is a Class C IP address?

A.10.5.5.5
B.172.16.1.1
C.192.168.1.1
D.224.1.1.1
AnswerC

Class C addresses begin with a first octet in the range 192-223. Since 192 falls within this range, 192.168.1.1 is correctly identified as a Class C address. This classification was historically important for defining the default subnet mask as /24, which is still the most commonly used default setting today.

Why this answer

Class C IP addresses have a first octet in the range 192–223, and 192.168.1.1 falls within that range. Class C addresses use a default /24 subnet mask (255.255.255.0) and were historically assigned to small networks. This makes 192.168.1.1 the correct answer.

Exam trap

100-150 often tests memorization of classful first-octet ranges — candidates misremember Class B as 172–191 or confuse Class D multicast (224–239) with Class C, especially when the address looks like a normal host address.

How to eliminate wrong answers

Option A is wrong because 10.5.5.5 falls in the 10.0.0.0/8 range, which is Class A (first octet 1–126). Option B is wrong because 172.16.1.1 falls in the 172.16.0.0–172.31.255.255 range, which is Class B (first octet 128–191). Option D is wrong because 224.1.1.1 falls in the 224–239 range, which is Class D, reserved for multicast, not a unicast Class C address.

5
MCQeasy

What is the decimal equivalent of the subnet mask 255.255.255.224?

A./25
B./26
C./27
D./28
AnswerC

A /27 mask means 27 bits are set to one. The first three octets are 24 bits. The last octet has 3 bits set (11100000), resulting in 224. This matches the provided decimal mask, confirming that /27 is the correct CIDR prefix for this network segment configuration.

Why this answer

The subnet mask 255.255.255.224 corresponds to a /27 CIDR notation. The first three octets are all ones, and the last octet consists of three ones followed by five zeros (11100000), which equals 224 in decimal. Identifying CIDR masks from decimal notation is a core competency for Cisco technicians, as it is required for rapid network design and troubleshooting in real-world infrastructure environments.

Exam trap

Candidates often guess the CIDR prefix by miscounting the number of active bits in the fourth octet, confusing 224 with masks like 192 or 240.

6
MCQmedium

In an IPv6 environment, which address type is automatically configured on every IPv6-enabled interface and is used for neighbor discovery and router solicitation?

A.Unique Local Address
B.Global Unicast Address
D.Anycast Address
AnswerC

Link-local addresses are automatically created as soon as IPv6 is enabled on an interface. They use the FE80::/10 prefix and provide the necessary connectivity for nodes to discover each other on the local wire, even before any global or local routing has been established by an administrator.

Why this answer

A Link-Local Address (LLA) is automatically configured on every IPv6-enabled interface, even before any router advertisement is received. It is formed using the fe80::/10 prefix and is required for Neighbor Discovery Protocol (NDP) operations including router solicitation, router advertisement, and address resolution.

Exam trap

The trap is confusing which IPv6 address type is automatically configured — candidates often pick Global Unicast or Unique Local because they assume 'default' means internet-routable, missing that link-local is the mandatory auto-configured type.

How to eliminate wrong answers

Option A is wrong because Unique Local Addresses (fc00::/7) are manually or DHCPv6-assigned for private internal communication — they are not automatically configured on every interface. Option B is wrong because Global Unicast Addresses (2000::/3) require router advertisement or manual/DHCPv6 configuration and are used for internet-routable communication. Option D is wrong because Anycast addresses are assigned to multiple interfaces and routed to the nearest one — they are not automatically configured and serve a different purpose (e.g., DNS root servers).

7
MCQmedium

A network technician is configuring a small branch office network. The ISP has assigned a single public IPv4 address that must be shared by all internal devices. The technician needs a solution that allows multiple internal hosts to communicate on the internet simultaneously without requiring a one-to-one mapping of internal to external addresses. Which technology should the technician implement?

A.Port Address Translation (PAT)
B.Static NAT with a one-to-one mapping
C.Dynamic NAT with a pool of public addresses
D.Proxy ARP
AnswerA

PAT, also called NAT overload, allows many internal hosts to share a single public IPv4 address by translating both IP addresses and source port numbers. Each session is uniquely identified by the translated port, so thousands of internal devices can communicate simultaneously using one public address. This directly matches the requirement of sharing a single ISP-assigned address.

Why this answer

Port Address Translation (PAT) is the correct choice because it enables many-to-one translation, allowing multiple internal hosts to share a single public IPv4 address by multiplexing sessions using source port numbers. Dynamic NAT and static NAT both require multiple public addresses or one-to-one mappings, which are not available. Proxy ARP does not perform translation and is irrelevant to internet sharing.

Exam trap

The trap here is confusing dynamic NAT with PAT, assuming that any form of NAT can share a single public address among many hosts.

8
MCQmedium

An engineer needs to subnet a 192.168.10.0/24 network to provide at least 5 usable subnets, with each subnet accommodating a minimum of 25 host devices. Which subnet mask meets these specific design requirements?

A.255.255.255.192
B.255.255.255.224
C.255.255.255.240
D.255.255.255.248
AnswerB

A /27 mask creates 8 subnets by borrowing 3 host bits, satisfying the minimum of 5 subnets. Each subnet contains 32 total addresses, yielding 30 usable IP addresses which satisfies the requirement for at least 25 hosts.

Why this answer

The subnet mask /27 provides 32 total IP addresses per subnet, leaving 30 usable hosts after excluding the network and broadcast addresses. This satisfies the requirement for 25 hosts while yielding 8 total subnets from the original /24 network, easily meeting the minimum threshold of 5 subnets.

Exam trap

Candidates often confuse the number of required usable subnets with host requirements, selecting a mask that satisfies total hosts but provides too few subnets, or vice versa.

9
MCQeasy

A host with the IP address 192.168.1.50 and a subnet mask of 255.255.255.0 is attempting to send data to a server at 10.1.1.5. Which address must be configured on the host to allow the data to exit the local subnet?

A.DNS Server Address
D.Broadcast Address
AnswerB

The default gateway is the IP address of the router interface connected to the local subnet. When a host determines that a destination IP is on a remote network, it forwards the packet to the gateway's MAC address, allowing the router to handle the transition to other network segments.

Why this answer

The host 192.168.1.50/24 and the server 10.1.1.5 are on different subnets, so the host cannot ARP for the server directly. It must forward the frame to its default gateway, which is the router interface on the local subnet that knows how to reach remote networks. Without a default gateway, the host can only communicate with devices in 192.168.1.0/24.

Exam trap

The trap is that candidates confuse DNS with routing; DNS resolves names but does not move packets between subnets, so the default gateway is the only correct answer for off-subnet communication.

How to eliminate wrong answers

Option A is wrong because DNS resolves hostnames to IP addresses; it does not route packets off the local subnet. Option C is wrong because the loopback address (127.0.0.1) is a virtual interface for testing the local TCP/IP stack and never carries traffic to remote hosts. Option D is wrong because the broadcast address (192.168.1.255) is used to reach all hosts on the local subnet, not to reach remote networks.

10
MCQhard

A network administrator is planning to deploy IPv6 on a Cisco router. The administrator needs to assign a global unicast address to the GigabitEthernet0/0 interface. Which of the following addresses is a valid global unicast IPv6 address that can be manually configured on the interface?

A.FE80::1
B.FF02::1
C.2001:DB8:ACAD:1::1
D.::1
AnswerC

2001:DB8::/32 is the documentation prefix reserved for examples, but within a private lab or documentation context, it is a valid global unicast address. In a real deployment, a global unicast address from the provider's allocated range would be used. This address is in the 2000::/3 range, which is designated for global unicast, and can be manually configured on the interface.

Why this answer

Global unicast IPv6 addresses are in the range 2000::/3, which includes 2001:DB8::/32. The address 2001:DB8:ACAD:1::1 falls within this range and can be manually configured on an interface. Link-local addresses (FE80::/10), multicast addresses (FF00::/8), and the loopback address (::1) are not global unicast and serve different purposes.

Therefore, the correct choice is the address in the 2001:DB8::/32 range.

Exam trap

The trap here is confusing link-local addresses like FE80::1 with global unicast addresses, as both can be assigned to interfaces but serve different scopes.

11
MCQeasy

A small branch office has been assigned the network 192.168.50.0/24. The network administrator must divide it into four equal-sized subnets to match four separate departments. Which subnet mask should be configured on the router interfaces to create exactly four usable subnets?

A.255.255.255.240
B.255.255.255.224
C.255.255.255.192
D.255.255.255.128
AnswerC

Borrowing two host bits from the /24 prefix yields a /26 mask, which is 255.255.255.192. This creates four subnets (00, 01, 10, 11) each containing 64 addresses, or 62 usable hosts after removing the network and broadcast addresses. This matches the requirement of four equal departments exactly.

Why this answer

To create four equal subnets from a single /24 block, two host bits must be borrowed, converting the prefix to /26. The resulting mask 255.255.255.192 divides the 192.168.50.0 range into four blocks of 64 addresses each, providing 62 usable host addresses per department. Masks that borrow fewer bits yield too few subnets, while borrowing more bits creates excess subnets with reduced host capacity.

Exam trap

The trap here is assuming that any mask larger than /24 automatically satisfies a subnet-count requirement, when the number of borrowed bits must match the exact number of required subnets.

12
MCQeasy

Which IPv6 address type is represented by the prefix fe80::/10?

A.Global Unicast
B.Unique Local
D.Multicast
AnswerC

Link-Local addresses are automatically assigned to interfaces for link-specific communication. The range fe80::/10 is specifically reserved for this purpose, ensuring that devices can communicate with neighbors even without a DHCPv6 server or manual configuration, which is a mandatory requirement for IPv6 operational stability.

Why this answer

The fe80::/10 prefix is reserved for Link-Local addresses in IPv6. These addresses are automatically configured on all IPv6-enabled interfaces and are essential for neighbor discovery and routing protocol communication on a single local link. Unlike global unicast addresses, these are not routable outside the local segment, making them fundamental for basic connectivity and troubleshooting in IPv6 environments.

Exam trap

Candidates frequently confuse IPv6 link-local addresses with global unicast or unique local addresses, assuming they are routable across multiple router hops on the wider network.

13
MCQhard

Refer to the exhibit. An administrator is reviewing the interface configuration. Which statement regarding these subnets is true?

A.The subnets overlap, causing a routing conflict.
B.The subnets are contiguous and correctly separated by 16 addresses each.
C.The configuration will fail because the IP addresses are not on the same network.
D.The mask /28 is invalid for these specific IP addresses.
AnswerB

Each /28 subnet consumes 16 addresses (2^4). Starting at .0, the next block begins at .16, and the following at .32. This represents a perfect, non-overlapping contiguous addressing scheme, which is standard practice for hierarchical network design and ensures efficient routing table management on the Cisco router.

Why this answer

The exhibit shows three interfaces configured with /28 masks. A /28 mask provides 16 IP addresses per block. The first subnet is 172.16.5.0, the second is 172.16.5.16, and the third is 172.16.5.32.

Because each interface is assigned the first host address of its respective block, the configuration is valid and follows classless inter-domain routing principles for non-overlapping subnets.

Exam trap

Candidates often miscalculate block boundaries when looking at exhibits, mistakenly thinking that subnets overlap or are improperly spaced when they are actually contiguous and correctly separated by 16 addresses each.

14
Multi-Selectmedium

Which TWO of the following statements correctly describe the characteristics of IPv4 address classes?

Select 2 answers
A.Class A addresses begin with a binary 0 in the first octet.
B.Class B addresses use the first 24 bits for the network portion.
C.Class C addresses begin with a binary 110 in the first octet.
D.Class D addresses are reserved for private internal usage.
E.Class E addresses are used for public internet routing.
AnswersA, C

The first bit of a Class A address is always 0. This allows the first octet to range from 1 to 126, providing a massive number of host addresses per network. This binary structure is the foundational rule for determining address class ranges in IPv4 networks.

Why this answer

Understanding classful boundaries is historical but necessary for identifying legacy network structures and subnetting fundamentals. Class A uses the first octet for network identification with a leading 0, while Class C uses three octets for the network portion with a leading 110. This knowledge assists in recognizing address ranges and identifying potential overlaps in misconfigured legacy network environments.

Exam trap

Candidates often confuse the leading binary bits for Class A, B, and C addresses because they memorize decimal ranges instead of binary prefixes.

15
MCQhard

An organization is using Variable Length Subnet Masking (VLSM) to optimize their 10.0.0.0/24 space. They need one subnet for 100 hosts and two subnets for 50 hosts each. Which of the following set of subnets correctly fulfills this requirement without overlapping?

A.10.0.0.0/25, 10.0.0.64/26, 10.0.0.128/26
B.10.0.0.0/25, 10.0.0.128/26, 10.0.0.192/26
C.10.0.0.0/24, 10.0.0.128/26, 10.0.0.192/26
D.10.0.0.0/26, 10.0.0.64/26, 10.0.0.128/25
AnswerB

This set of subnets perfectly utilizes the /24 block. The /25 mask provides 126 usable hosts for the first group. The remaining space is then split into two /26 blocks, each providing 62 usable hosts. There is no overlap, and all host requirements are met with maximum efficiency.

Why this answer

VLSM allows for efficient use of IP space by using different masks for different subnets. To accommodate 100 hosts, a /25 (128 addresses) is needed. To accommodate 50 hosts, /26 subnets (64 addresses) are needed. 10.0.0.0/25 covers 0-127.

Then, 10.0.0.128/26 covers 128-191, and 10.0.0.192/26 covers 192-255.

Exam trap

Candidates often miscalculate block sizes or create overlapping ranges by failing to account for binary boundary rules when allocating smaller subnets after a large /25 block.

16
MCQmedium

Which of the following is the correct representation of the IPv6 address 2001:0DB8:0000:0000:0008:0800:200C:417A in compressed format?

A.2001:DB8::8:800:200C:417A
B.2001:DB8:0:0:8:800:200C:417A
C.2001:DB8::0000::8:800:200C:417A
D.2001:DB8:0000:0000:8:800:200C:417A
AnswerA

This is the correct compressed form. The leading zeros in 0DB8 were removed, and the two blocks of zeros were successfully represented by the double-colon. Additionally, the leading zeros in 0800 were removed, resulting in 800, which is the correct and most efficient way to express this IPv6 address.

Why this answer

The correct compressed format is 2001:DB8::8:800:200C:417A. IPv6 compression rules (RFC 5952) allow leading zeros in each group to be omitted and a single contiguous sequence of all-zero groups to be replaced with '::'. Here, the two consecutive zero groups (0000:0000) are replaced by '::', and leading zeros in other groups are dropped (0DB8→DB8, 0008→8, 0800→800).

Exam trap

The trap is using '::' more than once or failing to compress the longest zero run — candidates often pick an option with double '::' or leave leading zeros in place, not realizing RFC 5952 mandates a single canonical form.

How to eliminate wrong answers

Option B is wrong because it does not use the '::' notation to compress the two consecutive zero groups — while technically valid, it is not the fully compressed form. Option C is wrong because it uses '::' twice, which is invalid — the double-colon can only appear once in an IPv6 address to avoid ambiguity. Option D is wrong because it leaves the leading zeros in 0000:0000 uncompressed and only partially compresses other groups, which is not the correct fully compressed representation.

17
Multi-Selectmedium

Which THREE fields are found in the standard IPv4 header and are directly related to the delivery or identification of the packet across subnets?

Select 3 answers
A.Source IP Address
B.Destination IP Address
C.Time to Live (TTL)
E.Sequence Number
AnswersA, B, C

This 32-bit field identifies the original sender of the packet. It is essential for the destination host to know where to send return traffic. Routers also use this field in conjunction with security features like Access Control Lists to permit or deny traffic based on the source network.

Why this answer

The IPv4 header contains several fields essential for routing and delivery. The Source IP Address and Destination IP Address are used by routers to make forwarding decisions. The Time to Live (TTL) field prevents packets from looping infinitely by decrementing at each hop, ensuring the network remains stable during convergence.

Exam trap

Candidates often mistakenly select the Protocol or Version fields. While technically in the header, these identify the packet type or version rather than facilitating the actual routing delivery across distinct subnet boundaries.

18
MCQeasy

A network administrator is documenting the addressing scheme for a new subnet. The subnet uses the IPv4 address 192.168.100.0 with a prefix length of /26. How many usable host addresses are available in this subnet?

A.62
B.126
C.30
D.64
AnswerA

A /26 prefix leaves 6 host bits (32 - 26 = 6). The total number of addresses is 2^6 = 64. Subtracting the network address and the broadcast address yields 62 usable host addresses. This is the correct calculation for a /26 subnet.

Why this answer

A /26 subnet mask provides 6 host bits, resulting in 2^6 = 64 total addresses. After reserving the network address and the broadcast address, 62 addresses remain for host assignment. This is a fundamental subnetting calculation that every network technician must master.

Exam trap

The trap here is forgetting to subtract the two reserved addresses (network and broadcast) and mistakenly selecting the total number of addresses.

19
MCQhard

A network engineer is troubleshooting connectivity on a host configured with the address 10.10.10.130/26. The host cannot reach a server at 10.10.10.200. Assuming no ACLs or routing issues, what is the most likely explanation?

A.The server address 10.10.10.200 is the broadcast address of the host subnet and cannot be used.
B.The host address 10.10.10.130 is invalid because it falls in a reserved range within the /26 subnet.
C.The host and server are in different subnets, so the host must use its default gateway to reach the server.
D.The two devices are in the same subnet, so the server must be down or have a firewall blocking traffic.
AnswerC

With a /26 mask, 10.10.10.130 falls in the subnet 10.10.10.128-10.10.10.191, while 10.10.10.200 belongs to 10.10.10.192-10.10.10.255. Since they are on different subnets, the host must forward traffic to its default gateway. If the gateway is missing or misconfigured, connectivity fails.

Why this answer

The /26 mask splits the 10.10.10.0 range into blocks of 64 addresses. The host at 10.10.10.130 belongs to the 10.10.10.128 subnet, while the server at 10.10.10.200 belongs to the 10.10.10.192 subnet. Because they are on different subnets, the host must forward traffic through a default gateway.

If the gateway is absent or wrong, communication fails even though both addresses are individually valid.

Exam trap

The trap here is assuming that two addresses close in value are on the same subnet, when the /26 mask actually places them in separate 64-address blocks.

20
Multi-Selectmedium

Which TWO of the following statements regarding IPv4 address types are true?

Select 2 answers
A.Public IP addresses are globally routable across the internet.
B.Private IP addresses can be routed directly across the public internet.
C.The 127.0.0.0/8 range is used for private internal communication.
D.Loopback addresses are used for testing the local TCP/IP stack.
E.All IPv4 addresses are assigned automatically by the ISP.
AnswersA, D

Public IP addresses are assigned by IANA and regional registries to ensure unique, globally routable communication. These are the only addresses that can reach and be reached from the public internet directly, making them indispensable for hosting public-facing services like web servers, mail servers, and corporate network gateways.

Why this answer

Understanding the distinction between private, public, and special-use IPv4 addresses is critical for network security and traffic routing. Private addresses allow internal segmentation without internet visibility, while special addresses like the loopback range are essential for local service testing. Mastering these types prevents common errors like NAT misconfiguration or routing traffic into non-existent or internal-only address blocks across the network infrastructure.

Exam trap

Candidates often mistakenly believe that private IP addresses are routable on the internet or confuse the specific loopback range (127.0.0.0/8) with other reserved address ranges.

21
MCQhard

Refer to the exhibit. What is the maximum number of usable host IP addresses available on this subnet?

A.30
B.62
C.64
D.126
AnswerB

With a /26 prefix, 6 bits are allocated for host addressing. The calculation 2^6 - 2 = 62 provides the exact number of usable host IP addresses. This is the correct capacity for this subnet, ensuring accurate planning for the number of connected devices on the GigabitEthernet0/0 interface.

Why this answer

The subnet is defined as a /26 mask. A /26 mask uses 6 bits for hosts (32 - 26 = 6). The formula for usable hosts is 2^6 - 2.

This equals 64 minus 2, which results in 62 usable host addresses. Correctly calculating host capacity is vital for capacity planning and ensuring that subnets are sized appropriately to avoid IP address exhaustion within the enterprise network.

Exam trap

Candidates often forget to subtract the two reserved addresses (network and broadcast), leading to an answer of 64 instead of the correct 62 usable host addresses.

22
MCQhard

A network engineer is configuring a large subnet for a campus and uses the 172.16.0.0/23 prefix. What is the last usable host address available in this subnet?

A.172.16.0.254
B.172.16.1.255
C.172.16.1.254
D.172.16.2.254
AnswerC

The /23 subnet mask (255.255.254.0) creates a block of 512 total addresses. For the 172.16.0.0 network, the range spans from 172.16.0.0 to 172.16.1.255. Subtracting the network and broadcast addresses leaves 172.16.1.254 as the final IP address that can be manually or dynamically assigned to a host.

Why this answer

A /23 subnet has 9 host bits (32 - 23 = 9), giving 2^9 - 2 = 510 usable hosts. The subnet 172.16.0.0/23 spans 172.16.0.0 through 172.16.1.255, with the broadcast address at 172.16.1.255 and the last usable host at 172.16.1.254.

Exam trap

The trap is stopping at the /24 boundary (172.16.0.255) and picking 172.16.0.254 — candidates forget that a /23 borrows a bit from the third octet and extends into 172.16.1.x.

How to eliminate wrong answers

Option A is wrong because 172.16.0.254 is a valid host in the middle of the /23 range, not the last usable address — the range extends into the 172.16.1.0 octet. Option B is wrong because 172.16.1.255 is the broadcast address for the /23 subnet and cannot be assigned to a host. Option D is wrong because 172.16.2.254 falls outside the /23 range entirely — it belongs to the next subnet (172.16.2.0/23).

23
MCQmedium

What is the directed broadcast address for the subnet in which the host 172.16.50.131/29 resides?

A.172.16.50.135
B.172.16.50.159
C.172.16.50.255
D.172.16.50.133
AnswerA

In a /29 network, the subnets increment by 8. The host 172.16.50.131 falls into the 172.16.50.128 subnet. The broadcast address is always the last address in the block, which is calculated as the next network ID (128 + 8 = 136) minus one, resulting in 172.16.50.135.

Why this answer

For 172.16.50.131/29, the subnet mask is 255.255.255.248, meaning each subnet has 8 addresses (2^3). The subnet containing .131 starts at .128 (network address) and ends at .135 (broadcast address). The directed broadcast address is the last address in the subnet, which is 172.16.50.135.

Exam trap

The trap is miscalculating the subnet block size for /29 or confusing the broadcast address with the next subnet's network address; candidates must remember that /29 yields 8-address blocks.

How to eliminate wrong answers

Option B is wrong because 172.16.50.159 is the broadcast address for a /27 subnet (mask 255.255.255.224), not /29. Option C is wrong because 172.16.50.255 is the broadcast address for a /24 subnet, which is much larger than /29. Option D is wrong because 172.16.50.133 is a usable host address within the /29 subnet, not the broadcast address.

24
MCQhard

A network administrator is subnetting the 172.16.0.0/22 network into smaller subnets to support point-to-point WAN links between routers. Each WAN link requires exactly two usable IP addresses. Which subnet mask should the administrator use to maximize the number of subnets while meeting the requirement?

A.255.255.255.240
B.255.255.255.252
C.255.255.255.254
D.255.255.255.248
AnswerB

A /30 subnet mask (255.255.255.252) provides 4 total addresses, with 2 usable host addresses after reserving the network and broadcast addresses. This is the most efficient mask for point-to-point links, maximizing the number of subnets. It perfectly meets the requirement of exactly two usable IP addresses.

Why this answer

For point-to-point WAN links requiring exactly two usable IP addresses, a /30 subnet mask (255.255.255.252) is the standard choice. It provides two usable addresses and minimizes address waste. While /31 can also work on supported devices, /30 is universally compatible and is the most common recommendation in Cisco curricula.

Exam trap

The trap here is selecting a /31 mask due to its efficiency, overlooking that many devices do not support RFC 3021 point-to-point links.

25
MCQmedium

A company uses the 172.16.0.0/16 private address block. The network architect needs to create subnets that each support at least 500 hosts while maximizing the number of subnets available. Which subnet mask should be chosen?

A.255.255.252.0
B.255.255.254.0
C.255.255.248.0
D.255.255.255.0
AnswerB

A /23 mask provides 9 host bits, yielding 510 usable hosts per subnet, which satisfies the 500-host minimum. It also creates 128 subnets from the /16 block, maximizing subnet count while meeting the host requirement. This is the most efficient choice because a /22 would provide more hosts than necessary and reduce the subnet total.

Why this answer

The design requires at least 500 hosts per subnet while maximizing subnet count from 172.16.0.0/16. A /23 mask provides 510 usable hosts, just above the threshold, and yields 128 subnets. A /22 would provide more hosts but fewer subnets, while a /24 would not meet the host requirement.

Therefore, 255.255.254.0 is the optimal balance between host capacity and subnet quantity.

Exam trap

The trap here is choosing a mask that comfortably exceeds the host requirement without checking whether a smaller subnet could still meet it and produce more subnets.

26
MCQmedium

A network technician is tasked with dividing the 192.168.10.0/24 network into four equal-sized subnets to accommodate different departments. Which subnet mask should be applied to achieve this requirement, and how many usable host addresses will be available in each subnet?

A.255.255.255.128 with 126 hosts
B.255.255.255.224 with 30 hosts
C.255.255.255.192 with 62 hosts
D.255.255.255.240 with 14 hosts
AnswerC

Borrowing two bits from the host portion of a /24 network creates four subnets with 62 usable addresses each. The mask 255.255.255.192 correctly identifies the boundary for 64-address blocks, ensuring that the technician can adequately support four separate departments while maximizing the available host capacity for each.

Why this answer

To divide a /24 network into four equal subnets, you must borrow two bits from the host portion, changing the prefix from /24 to /26. A /26 prefix corresponds to a subnet mask of 255.255.255.192. Each resulting subnet contains 64 total addresses, but subtracting the network ID and broadcast address leaves 62 usable host addresses for devices.

Exam trap

Candidates often select the wrong subnet mask or miscalculate usable hosts by forgetting to borrow two bits for four subnets, confusing /24 with /26 boundaries.

27
MCQmedium

An administrator needs to subnet the 192.168.10.0/24 network into subnets that support at least 25 hosts each. Which CIDR mask provides the most efficient use of addresses while meeting the requirement?

A./25
B./26
C./27
D./28
AnswerC

A /27 mask provides 30 usable host addresses. This is the smallest possible subnet size that accommodates 25 hosts, adhering to the requirement for maximum efficiency. It leaves only 5 addresses wasted per subnet, which is the optimal mathematical balance for this specific host count requirement.

Why this answer

To support 25 hosts, the subnetwork must provide at least 27 usable IP addresses, accounting for network and broadcast IDs. The formula 2^n - 2 >= 25 indicates n=5 host bits are required, leaving 27 bits for the network portion. Using a /27 mask results in 30 usable hosts per subnet, which is the most efficient configuration compared to larger allocations that would waste addresses.

Exam trap

Candidates often select /26 instead of /27 because they miscalculate the number of host bits needed, forgetting to account for the network ID and broadcast address in their math.

28
MCQhard

A network engineer is configuring IPv6 on a Cisco router interface. The engineer assigns the address 2001:db8:acad:1::1/64 to GigabitEthernet0/0. Which of the following statements is true regarding the interface's IPv6 configuration?

A.The interface will use the global unicast address as its link-local address.
B.The interface will automatically generate a link-local address using the EUI-64 format.
C.The interface will require manual configuration of a link-local address to enable IPv6.
D.The interface will not have a link-local address because only global unicast addresses are configured.
AnswerB

When an IPv6 address is configured on a Cisco router interface, the interface automatically generates a link-local address (starting with fe80::) using the EUI-64 format by default, unless a link-local address is manually configured. This link-local address is used for neighbor discovery and routing protocols.

Why this answer

When an IPv6 global unicast address is configured on a Cisco router interface, the interface automatically creates a link-local address using the EUI-64 format. This link-local address is mandatory for IPv6 operations, including neighbor discovery and routing protocols, and is generated without additional configuration.

Exam trap

The trap here is assuming that configuring a global unicast address eliminates the need for a link-local address, or that manual configuration is required for link-local.

29
Multi-Selecthard

A network administrator is documenting the IPv6 addressing scheme for a new campus. The administrator must identify which statements about IPv6 address structure and notation are accurate. (Choose two.)

Select 2 answers
A.An IPv6 address can contain multiple double colon sequences as long as each replaces a different zero run.
B.The IPv6 prefix length is expressed in dotted-decimal notation following a space, such as 2001:db8::1 255.255.255.0.
C.Leading zeros within each 16-bit group of an IPv6 address can be omitted to shorten the notation.
D.Trailing zeros at the end of a hextet may be removed to shorten the address, so 2001:0db0::1 can be written as 2001:0db::1.
E.A single contiguous sequence of zero groups within an IPv6 address may be compressed once using a double colon.
AnswersC, E

Each hextet can drop leading zeros, so 0db8 becomes db8 and 0001 becomes 1. This reduces visual clutter without changing the address value. Trailing zeros within a group, however, cannot be dropped, because doing so would alter the numeric value of that hextet and produce a different address.

Why this answer

IPv6 text representation permits two key shortcuts: suppressing leading zeros inside each hextet and replacing one contiguous run of zero hextets with a double colon. The double colon may appear only once to keep the address unambiguous. Dotted-decimal masks and removal of trailing zeros are not valid, since they would either mix IPv4 conventions or alter the numeric value of the address.

Exam trap

The trap here is confusing leading-zero suppression with trailing-zero removal, and assuming the double colon can be used more than once in a single address.

30
MCQmedium

A network engineer needs to allocate a subnet for a link connecting two routers. Which prefix provides the most efficient addressing for a point-to-point connection?

A./24
B./28
C./29
D./30
AnswerD

A /30 mask provides exactly two usable host addresses, which is the perfect amount for a point-to-point router connection. It eliminates address wastage and is the industry-standard prefix for serial or Ethernet links connecting two devices directly, making it the most efficient choice for this specific networking scenario.

Why this answer

A /30 mask leaves 2 host bits, yielding exactly 2 usable addresses — one for each router interface on the point-to-point link. That is the minimum needed for a functional link, so /30 is the most efficient allocation without wasting addresses.

Exam trap

The trap is forgetting that a point-to-point link still needs two usable addresses plus network and broadcast, so candidates pick /31 or /32 thinking 'smaller is always more efficient' — /31 is only valid under RFC 3021 and /32 is a host route, not a link.

How to eliminate wrong answers

Option A is wrong because /24 provides 254 usable hosts, which is massive overkill for a two-router link and wastes 252 addresses. Option B is wrong because /28 provides 14 usable hosts, still far more than the two endpoints require. Option C is wrong because /29 provides 6 usable hosts, which is more than needed for a point-to-point connection and therefore less efficient than /30.

31
MCQmedium

An administrator needs to subnet the 192.168.10.0/24 network into subnets that can support at least 25 hosts each. Which subnet mask provides the maximum number of subnets while meeting the host requirement?

A.255.255.255.248
B.255.255.255.240
C.255.255.255.224
D.255.255.255.192
AnswerC

A /27 mask provides 30 usable host addresses, which satisfies the 25-host requirement while maximizing subnet density. This prefix length is the most efficient choice because any larger mask would reduce host availability below the threshold, and any smaller mask would unnecessarily waste IP address space.

Why this answer

To support at least 25 hosts, the subnet must have at least 5 host bits (2^5 - 2 = 30 usable hosts). A /27 mask (255.255.255.224) leaves 5 host bits, providing 30 usable addresses per subnet and yielding 8 subnets from a /24. This maximizes subnet count while still meeting the 25-host requirement.

Exam trap

The trap is that candidates pick the mask with the most hosts (the /26) instead of the mask that maximizes subnet count while still meeting the minimum host requirement — the question wording 'maximum number of subnets' is the key.

How to eliminate wrong answers

Option A (/29, 255.255.255.248) leaves only 3 host bits, providing just 6 usable hosts — far short of 25. Option B (/28, 255.255.255.240) leaves 4 host bits, providing 14 usable hosts — still short of 25. Option D (/26, 255.255.255.192) leaves 6 host bits, providing 62 usable hosts, which meets the requirement but yields only 4 subnets, fewer than the /27's 8 subnets.

32
Multi-Selectmedium

Which TWO of the following statements regarding IPv6 address representation are accurate?

Select 2 answers
A.The double-colon (::) can be used multiple times in a single IPv6 address to shorten it.
B.Leading zeros in a 16-bit field can be omitted.
C.An IPv6 address must always be written in full 32-character hexadecimal format.
D.The double-colon (::) represents one or more groups of 16-bit fields containing all zeros.
E.Trailing zeros in a 16-bit field can be omitted.
AnswersB, D

The IPv6 specification allows for the omission of leading zeros within each hexadecimal quartette. For example, 0db8 becomes db8. This convention significantly reduces the length of written addresses, making them easier for network administrators to manage, read, and type into device configuration files without introducing errors.

Why this answer

Option B is correct because RFC 4291 allows leading zeros within any 16-bit hexadecimal field (hextet) to be dropped, so 0db8 can be written as db8. Option D is correct because the double-colon (::) is the compression notation that stands in for one or more consecutive 16-bit groups that are all zeros, and it may appear only once in an address. Option A is wrong because :: can be used only a single time per address; using it twice would make the number of omitted zero groups ambiguous.

Option C is wrong because IPv6 addresses are commonly written in compressed form, not always as a full 32-character hexadecimal string. Option E is wrong because only leading zeros may be omitted; trailing zeros are significant and must be retained.

Exam trap

The trap is confusing leading and trailing zero omission rules and believing :: can be used multiple times; the exam tests exact knowledge of RFC 4291 compression rules.

33
MCQmedium

What is the broadcast address for the network 172.16.20.0/24?

A.172.16.20.0
B.172.16.20.1
C.172.16.20.254
D.172.16.20.255
AnswerD

In a /24 network, the broadcast address is calculated by setting all host bits (the last 8 bits) to one. 11111111 in binary is 255. Therefore, 172.16.20.255 is the broadcast address, which is the correct identifier for sending traffic to all hosts residing on this specific local subnet.

Why this answer

In a /24 subnet (255.255.255.0), the network occupies the first three octets, and the last octet is available for hosts. The broadcast address is the very last address in the subnet range, where all host bits are set to one. For the 172.16.20.0 network, the last address is 172.16.20.255.

This calculation is essential for identifying network boundaries and verifying traffic scope in Cisco configurations.

Exam trap

Candidates often select the network ID (172.16.20.0) or the last usable host address (172.16.20.254) instead of the absolute final address in the subnet range for the broadcast.

34
MCQmedium

A network engineer is configuring IPv6 on a Cisco router interface. The engineer wants to ensure that the interface can automatically generate a link-local address and participate in neighbor discovery without manual configuration. Which IPv6 address type is automatically generated on the interface when IPv6 is enabled?

A.Unique local address
C.Multicast address
D.Global unicast address
AnswerB

Enabling IPv6 on an interface, with 'ipv6 enable' or by configuring any IPv6 address, automatically creates a link-local address in FE80::/10, on Cisco routers usually from the MAC address using modified EUI-64. Link-local addresses are used for neighbour discovery, router advertisements and routing protocol neighbours, and are never routed. Many engineers set them manually, such as FE80::1, to make gateways easier to read.

Why this answer

Enabling IPv6 on an interface automatically creates a link-local address, which is essential for neighbor discovery and local link communication. Global unicast and unique local addresses require additional configuration or router advertisements. Multicast addresses are not unicast addresses assigned to the interface.

Exam trap

The trap here is assuming that a global unicast address is automatically configured, when in fact only the link-local address is generated without external input.

35
MCQeasy

A small office technician is assigning IPv4 addresses to devices on a network that will not connect to the internet and uses the 192.168.50.0/24 range. The technician needs to pick an address from the valid host range for this network. Which address should the technician choose?

A.192.168.50.1
B.192.168.50.255
C.192.168.50.0
D.192.168.51.1
AnswerA

In the 192.168.50.0/24 network, the first usable host address is 192.168.50.1, since .0 is the network address and .255 is the broadcast address. This address falls within the valid host range of .1 through .254 and can be assigned to a device interface without conflict.

Why this answer

For any subnet, the network address and the broadcast address are reserved and cannot be assigned to hosts. In 192.168.50.0/24, the usable host range is 192.168.50.1 through 192.168.50.254, so the only valid host address listed is 192.168.50.1. The address ending in .0 is the network, .255 is the broadcast, and 192.168.51.1 is in a different subnet.

Exam trap

The trap here is assuming that any address in a private range is automatically a valid host address, ignoring the reserved network and broadcast values within the specific subnet.

36
Multi-Selectmedium

Which TWO statements correctly describe valid rules for compressing or representing IPv6 addresses?

Select 2 answers
A.Leading zeros in a quartet can be omitted.
B.Consecutive fields of zeros can be replaced by '::' only once.
C.Trailing zeros in a quartet can be omitted to save space.
D.A single quartet of '0000' must always be compressed to '::'.
E.IPv6 addresses use dots to separate quartets for better readability.
AnswersA, B

Removing leading zeros is a standard compression method that simplifies the appearance of an address without changing its value. For example, the quartet '0db8' can be written as 'db8'. This reduces the character count and makes the address easier for network technicians to read and document.

Why this answer

Option A is correct because RFC 4291 permits omitting leading zeros within any 16-bit hexadecimal quartet, so '0db8' can be written as 'db8' and '0001' as '1'. Option B is correct because the '::' shorthand may be used only once in an address; it represents one or more consecutive groups of zeros, and using it twice would make the address ambiguous. Option C is wrong because only leading zeros may be dropped, not trailing zeros, since removing trailing zeros would change the value of the quartet.

Option D is wrong because compression is optional, not mandatory, and a single '0000' quartet may be written as '0' or left uncompressed. Option E is wrong because IPv6 uses colons to separate quartets, while dots appear only in mixed IPv4-embedded notation such as '::ffff:192.168.0.1'.

Exam trap

The trap is mixing up leading and trailing zero rules and assuming :: can be used multiple times; the exam tests precise recall of RFC 4291 compression constraints.

37
Multi-Selectmedium

Which THREE of the following are valid private IPv4 address ranges as defined in RFC 1918?

Select 3 answers
A.10.0.0.0/8
B.172.16.0.0/12
C.192.168.0.0/16
D.127.0.0.0/8
E.224.0.0.0/4
AnswersA, B, C

The 10.0.0.0/8 block is the largest private address range, spanning from 10.0.0.0 to 10.255.255.255. It is commonly used in enterprise environments to facilitate internal communication across large-scale distributed networks without the need for public IP address registration for every internal host on the network.

Why this answer

RFC 1918 defines specific address blocks reserved for private networks. These addresses are not routable on the public internet and are used internally to conserve public IPv4 address space. Technicians must be able to identify these ranges to configure NAT, DHCP, and internal routing policies accurately, as misusing public addresses in private ranges can lead to connectivity failures or security issues.

Exam trap

Candidates often misidentify the 172.16.0.0/12 range, erroneously including the entire 172.0.0.0/8 block instead of the specific RFC 1918 reserved private address range.

38
Multi-Selecthard

Which TWO of the following statements are true regarding IPv4 and IPv6 coexistence?

Select 2 answers
A.Dual-stack enables a device to run IPv4 and IPv6 protocol stacks simultaneously.
B.IPv6 addresses are directly compatible with IPv4 routing tables.
C.Tunneling encapsulates IPv6 packets inside IPv4 headers.
D.IPv4 is natively supported by all IPv6-only devices.
E.Broadcast traffic is still required for IPv6 address resolution.
AnswersA, C

Dual-stack is the most common transition mechanism, allowing interfaces to support both protocol versions at the same time. This ensures compatibility with legacy IPv4-only hosts while enabling native IPv6 traffic for modern devices, providing a seamless operational transition within the enterprise network architecture without requiring an immediate, complete overhaul.

Why this answer

As the industry transitions from IPv4 to IPv6, coexistence mechanisms are required. Dual-stack allows devices to run both protocols simultaneously, while tunneling encapsulates IPv6 packets within IPv4 headers. These methods ensure that legacy systems can still communicate while modern IPv6-enabled infrastructure is deployed, which is a critical skill for network administrators managing hybrid, multi-protocol corporate environments during the long-term industry shift.

Exam trap

Candidates often confuse tunneling with translation, incorrectly assuming that tunneling replaces IPv6 addresses with IPv4 addresses rather than simply encapsulating the IPv6 packet within an IPv4 payload.

39
MCQhard

Refer to the exhibit. Based on the IPv6 configuration of the GigabitEthernet0/0 interface, which statement is true regarding the connectivity of this router?

A.The router can communicate with any host on the internet.
B.The router can only communicate with other hosts on the local link.
C.The interface is disabled because it lacks a global address.
D.The router will use the FF02::1 address to route external traffic.
AnswerB

Link-local addresses starting with FE80 are automatically generated or manually assigned to facilitate communication within a single local segment. Because no global unicast address is present, the router cannot forward packets from this interface to remote networks or receive traffic originating from outside the local broadcast domain.

Why this answer

The exhibit shows a GigabitEthernet0/0 interface configured with only a link-local IPv6 address (FE80::/10 range, typically starting with FE80::). A link-local address is automatically generated and is only valid on the local segment; it is never routed beyond the local link. Therefore the router can communicate only with directly connected IPv6 hosts on that link, not with the internet or remote networks.

Exam trap

The trap is that candidates see an IPv6 address on the interface and assume full connectivity, forgetting that link-local (FE80::/10) addresses are non-routable and confined to the local segment.

How to eliminate wrong answers

Option A is wrong because internet reachability requires a globally routable IPv6 address (2000::/3) plus a default route; a link-local-only interface cannot reach the internet. Option C is wrong because the interface is not disabled; link-local addresses are fully functional for on-link communication and the interface remains up/up. Option D is wrong because FF02::1 is the all-nodes multicast address used for neighbor discovery, not a routing address for external traffic.

40
MCQeasy

Which component of an IPv4 address does the subnet mask identify?

A.The MAC address of the destination device.
B.The network and host portions of the address.
C.The default gateway for the host.
D.The TCP port number for the application.
AnswerB

The primary role of the subnet mask is to identify which bits of an IPv4 address belong to the network prefix and which bits belong to the individual host. This binary comparison allows the device to route traffic to the correct local or remote network segment accurately.

Why this answer

The subnet mask is a 32-bit value that, when ANDed with an IPv4 address, separates the network portion (bits set to 1 in the mask) from the host portion (bits set to 0). It tells the device which part of the address identifies the network and which part identifies the specific host on that network.

Exam trap

The trap is confusing Layer 2 (MAC), Layer 3 (IP/mask), and Layer 4 (ports) concepts — candidates pick the MAC address option because both are 'addresses,' but the mask is purely a Layer 3 network/host boundary tool.

How to eliminate wrong answers

Option A is wrong because MAC addresses are Layer 2 hardware addresses resolved via ARP, not derived from the subnet mask. Option C is wrong because the default gateway is a router IP configured separately; the mask only helps the host decide whether a destination is local or remote, not what the gateway is. Option D is wrong because TCP/UDP port numbers are Layer 4 identifiers and have no relationship to the subnet mask.

41
MCQmedium

A network technician is configuring a point-to-point serial link between two Cisco routers. To conserve address space, the engineer wants the smallest possible subnet that still provides two usable host addresses. Which subnet mask should be applied to the serial interfaces?

A.255.255.255.252
B.255.255.255.254
C.255.255.255.128
D.255.255.255.248
AnswerA

A /30 mask uses 30 network bits, leaving 2 host bits, which yields exactly 4 addresses: two usable hosts plus network and broadcast. This is the traditional smallest subnet for point-to-point links and conserves address space efficiently. Cisco serial interfaces commonly use this mask for WAN connections between routers.

Why this answer

A point-to-point serial link requires only two host addresses, one for each router interface. The /30 mask 255.255.255.252 provides exactly four addresses, of which two are usable hosts, making it the smallest conventional subnet for this purpose. Larger masks waste addresses, and while /31 is an option in some modern implementations, the /30 remains the standard expected configuration for conserving address space on WAN links.

Exam trap

The trap here is overlooking that two usable hosts require exactly two host bits, and choosing a mask that provides more addresses than necessary instead of the minimal /30.

42
Multi-Selecthard

A network designer is planning to use the 10.50.0.0/16 address space. Which TWO of the following represent valid network IDs for subnets created by further dividing this space into /24 segments?

Select 2 answers
A.10.50.1.0
B.10.50.255.0
C.10.51.0.0
D.10.50.0.255
E.10.50.1.1
AnswersA, B

In a /24 subnetting scheme derived from a /16, each increment of 1 in the third octet represents a new network. 10.50.1.0 is the network ID for the second available /24 subnet within the 10.50.0.0/16 block, providing 254 usable host addresses for that specific segment.

Why this answer

Options A (10.50.1.0) and B (10.50.255.0) are correct because when 10.50.0.0/16 is divided into /24 subnets, the third octet becomes the subnet identifier and the fourth octet must be 0, giving valid network IDs such as 10.50.0.0, 10.50.1.0, and 10.50.255.0, all of which fall within the 10.50.0.0–10.50.255.0 range. Option C (10.51.0.0) is outside the 10.50.0.0/16 block, since 10.51.x.x belongs to a different /16. Option D (10.50.0.255) is a host address within 10.50.0.0/24, not a network ID, because the host portion is non-zero.

Option E (10.50.1.1) is also a host address in the 10.50.1.0/24 subnet, not a valid network ID.

Exam trap

The trap is picking addresses that look like network IDs but are actually host or broadcast addresses (ending in .255 or .1), or picking a subnet outside the parent block — candidates must verify both the parent range and the .0 network-ID rule.

43
MCQmedium

A network administrator is documenting a subnet used for a point-to-point WAN link between two Cisco routers. The link is configured with the address 10.10.10.1/30. What is the network address and the usable host range for this subnet?

A.Network 10.10.10.0, usable hosts 10.10.10.1 through 10.10.10.4
B.Network 10.10.10.4, usable hosts 10.10.10.5 through 10.10.10.6
C.Network 10.10.10.1, usable hosts 10.10.10.2 through 10.10.10.3
D.Network 10.10.10.0, usable hosts 10.10.10.1 through 10.10.10.2
AnswerD

A /30 mask leaves only two host bits, so each subnet has four addresses: network, two usable hosts, and broadcast. For 10.10.10.1/30, the network is 10.10.10.0, usable hosts are .1 and .2, and the broadcast is .3. This is the standard addressing for point-to-point links.

Why this answer

A /30 prefix provides two host bits, yielding exactly four addresses per subnet: one network, two usable hosts, and one broadcast. For the address 10.10.10.1/30, the subnet boundaries are multiples of four, so the network is 10.10.10.0, usable hosts are .1 and .2, and the broadcast is .3. This design is common on point-to-point links because it conserves address space.

Exam trap

The trap here is treating the configured interface address as the network address and then counting usable hosts from the next number, rather than calculating the subnet boundary from the mask.

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