CCNA Network Infrastructure and Connectivity Practice Test 10 — 15 Questions

Question 1

A network engineer must summarize the following routes before advertising them upstream:

172.16.32.0/24
172.16.33.0/24
172.16.34.0/24
172.16.35.0/24

Which summary route should be used?

Accepted Answer

172.16.32.0/22. The correct summary is 172.16.32.0/22 because a /22 covers exactly four consecutive /24 networks when the starting boundary is aligned correctly. This is the part many people miss: summarization is not only about how many networks fit into a block, but also where that block starts. Here the four /24 networks begin neatly at 172.16.32.0 and continue through 172.16.35.255, which is the exact range a /22 covers. A /23 would be too small, while a /21 would be unnecessarily broad and could advertise addresses you do not intend to include. The /22 beginning at 172.16.34.0 is not on a valid /22 boundary, so that option is misaligned.

Answer

172.16.32.0/23

Answer

172.16.32.0/21

Answer

172.16.34.0/22

Question 2

A switch port was configured for sticky MAC learning. Which two statements accurately describe how the feature behaves?

Accepted Answer

The switch can dynamically learn MAC addresses and add them to the running configuration as secure MAC addresses.. Sticky MAC learning is Cisco’s way of letting a port learn device MAC addresses automatically, while still treating them as secure addresses under port security. In everyday language, it saves the administrator from typing each allowed MAC address by hand. As devices connect, the switch can learn their MAC addresses and place them into the running configuration as sticky secure MACs. If the administrator later saves the configuration, those learned entries can also be written into startup-config and survive a reboot. The feature does not replace port security; it works as part of port security. It also does not change the port into a trunk or disable the maximum address count. So the two correct ideas are dynamic secure learning and the ability to preserve those learned MACs by saving the configuration.

Answer

Sticky learning removes the need to enable port security on the interface.

Answer

Sticky learning automatically converts the interface into a trunk port.

Answer

Sticky learning prevents the maximum secure MAC limit from being enforced.

Question 3

A switch unexpectedly blocks a link toward the distribution layer. Gi1/0/24 shows a path cost of 4 while Gi1/0/23 shows a path cost of 19. Why did interface Gi1/0/24 become the root port instead of Gi1/0/23?

Accepted Answer

has a lower root path cost to the root bridge. Spanning Tree chooses a root port by looking for the best path toward the root bridge. In this case, Gi1/0/24 shows a cost of 4, while Gi1/0/23 shows a cost of 19. Lower cost is better, so Gi1/0/24 is selected as the root port and moves into forwarding. Gi1/0/23 becomes an alternate port and is placed into a blocking state to prevent a loop. STP compares root path cost first; only if the cost is tied does it move on to tie-breakers like sender bridge ID and port ID. The lower cost on Gi1/0/24 explains why that port won the root-port election.

Answer

Gi1/0/24 has a lower port number, so STP always prefers it first.

Answer

STP prefers interfaces with the highest path cost to reduce loops.

Answer

Gi1/0/23 is blocked because alternate ports are always chosen over root ports.

Question 4

Match the data format to its most accurate characteristic.

Accepted Answer

JSON: Lightweight, key-value based data format. JSON is lightweight and key-value based, XML uses tags and attributes, YAML relies on indentation, CSV is for tabular data, Protobuf is a binary schema-driven format, and HTML is for web page structure.

Answer

XML: Binary schema-driven format

Answer

YAML: Uses tags and attributes

Answer

CSV: For web page structure

Question 5

Match the security feature to its main purpose.

Accepted Answer

Firewall: Filters traffic based on security rules. ACLs are correct because they use permit and deny statements to filter traffic based on source/destination IP, protocol, or port. DHCP Snooping is correct because it identifies trusted ports and builds a DHCP binding table to block rogue DHCP servers and prevent spoofed DHCP messages. DAI is correct because it leverages the DHCP Snooping binding table to validate ARP packets, dropping those that do not match trusted bindings and thus preventing ARP spoofing attacks. Port Security is correct because it restricts the number and specific MAC addresses learned on a switch port, mitigating MAC flooding and unauthorized device access.

Question 6

Which two statements accurately describe UDP compared with TCP?

Accepted Answer

UDP is connectionless.. UDP is designed for simplicity and speed rather than built-in reliability. In plain terms, it sends data without creating a formal conversation first. That is why it is called connectionless. Because it does not perform the same reliability features as TCP, its header is smaller and the protocol adds less overhead. This makes UDP a good fit for applications that care more about speed or low delay than guaranteed delivery at the transport layer.

UDP does not perform a three-way handshake, and it does not guarantee delivery. It is also used by many real services, including DNS in common query scenarios.

Answer

UDP always guarantees delivery and sequencing.

Answer

UDP requires a three-way handshake before application data can be sent.

Answer

UDP cannot be used by DNS.

Question 7

Which two statements accurately compare IPv4 private addresses and public addresses?

Accepted Answer

Private IPv4 addresses are not directly Internet-routable.. Private IPv4 addresses are intended for internal use and are not directly routable on the public Internet. In plain language, they are designed for use inside organizations, homes, and other local environments without consuming globally unique public space. Public addresses, by contrast, are intended to be unique and routable across the Internet. This is one of the main reasons NAT became so common in IPv4 environments.

CCNA questions often test this distinction because learners sometimes confuse “valid inside a LAN” with “routable everywhere.” Private addressing is extremely useful, but it does not eliminate the need for translation or public addressing when reaching the Internet. The two correct statements are the ones that preserve that basic separation between internal-use ranges and globally routable address space.

Answer

Private IPv4 addresses always require OSPF to function inside a LAN.

Answer

Public IPv4 addresses cannot exist on Internet-facing devices.

Answer

Private and public IPv4 addresses are both automatically translated by ARP.

Question 8

Which two statements accurately describe ARP in an IPv4 Ethernet network?

Accepted Answer

ARP resolves a known IPv4 address to a MAC address on the local segment.. ARP is the mechanism used to map a known IPv4 address to a Layer 2 MAC address on the local network segment. In plain language, if a device knows the IP address it wants to reach on the same LAN, ARP helps it discover the correct Ethernet destination MAC address to use in the frame. That is why ARP is so important for local delivery in IPv4 Ethernet environments. Without it, devices would know where they want to send traffic logically, but not how to address the actual frame on the local link.

ARP does not cross routers in the usual way, and it is not a routing protocol. It does not determine best paths to remote networks. It simply helps with local resolution of IPv4-to-MAC information. This distinction matters a lot on CCNA questions because many wrong answers try to blur the line between local neighbor resolution and routing behavior.

Answer

ARP is used to choose the best Layer 3 path across multiple routers.

Answer

ARP can normally resolve the MAC address of a host located across a routed network.

Answer

ARP replaces the need for a default gateway.

Question 9

A host with address 192.168.1.130/26 needs to identify its local subnet. Which subnet does it belong to?

Accepted Answer

192.168.1.128/26. A /26 creates subnets in blocks of 64 addresses each. In plain language, that means the fourth octet ranges are 0–63, 64–127, 128–191, and 192–255. Since 192.168.1.130 falls within the 128–191 block, the host belongs to the 192.168.1.128/26 subnet. That is the local network boundary the host will use to decide what is on-link and what requires the default gateway.

This kind of question is a very common CCNA subnetting task. The main challenge is not the arithmetic itself but recognizing the block size and locating the destination inside the correct range. Once you know a /26 moves in increments of 64, the right subnet becomes much easier to see.

Answer

192.168.1.0/26

Answer

192.168.1.64/26

Answer

192.168.1.192/26

Question 10

A subnet uses the mask 255.255.255.224. How many usable host addresses does it provide?

Accepted Answer

30. A mask of 255.255.255.224 corresponds to a /27 prefix. In plain language, that leaves 5 host bits available in the address, which creates 32 total addresses in each subnet. Two of those are reserved: one for the network address and one for the broadcast address. That leaves 30 usable host addresses.

This is a classic subnetting question because it tests whether you can move from mask to prefix idea to host count without getting lost. Many learners remember the block size but forget to subtract the network and broadcast entries. The correct answer comes from that full logic chain: /27 means 32 total, and therefore 30 usable.

Answer

14

Answer

62

Answer

126

Question 11

A host has the address 10.10.10.94/27. Which subnet contains that host?

Accepted Answer

10.10.10.64/27. A /27 mask creates subnets in blocks of 32 addresses. In plain language, that means the fourth-octet ranges are 0–31, 32–63, 64–95, 96–127, and so on. Since the host address ends in 94, it falls inside the 64–95 block. That means the subnet is 10.10.10.64/27.

This is a classic subnetting task because it checks whether you can move from prefix length to block size and then locate the host inside the correct range. The key skill is recognizing the increment boundary and not guessing based only on the nearest familiar address.

Answer

10.10.10.32/27

Answer

10.10.10.96/27

Answer

10.10.10.0/27

Question 12

A subnet must support 14 usable hosts. Which prefix is the smallest that meets the requirement?

Accepted Answer

/28. To support 14 usable hosts, the subnet must have 16 total addresses, because two are reserved for the network and broadcast addresses. In plain language, you need enough address slots so that after those two reserved entries are removed, 14 remain. A /28 provides exactly that: 16 total addresses and 14 usable addresses.

This is a classic minimum-subnet-size question because it tests whether you can work backward from host requirement to prefix length. A /29 would be too small, while a /27 would work but waste more addresses than necessary.

Answer

/29

Answer

/27

Answer

/26

Question 13

A subnet must support at least 62 usable hosts. Which prefix will create the smallest subnet that meets the requirement?

Accepted Answer

/26. To support at least 62 usable hosts, the subnet needs 64 total addresses, because two of those will be reserved for the network and broadcast addresses. In plain language, the target is not 62 total addresses; it is 62 usable ones after the two reserved values are taken away. A /26 provides exactly 64 total addresses and therefore 62 usable host addresses.

This is a classic minimum-prefix question because it checks whether you can convert a host requirement into the correct power-of-two subnet size without over-allocating unnecessarily. A /27 would be too small, while /25 would work but would waste more addresses than needed. The smallest valid prefix is /26.

Answer

/27

Answer

/25

Answer

/24

Question 14

Which two statements accurately describe IPv6 link-local addresses?

Accepted Answer

They are used for communication on the local segment only.. IPv6 link-local addresses are designed for communication on the local segment only. In plain language, they allow devices to talk to nearby neighbors without needing globally routable addresses. They play an important role in IPv6 functions such as Neighbor Discovery and are commonly used when hosts communicate with the default gateway on the same link. These addresses are normal and expected in IPv6 environments.

They are not globally Internet-routable, and they are not just emergency fallbacks for DHCPv6 failure. The two correct answers are the ones that preserve their local-link purpose and their importance in standard IPv6 behavior rather than treating them as optional or globally reachable.

Answer

They are globally routable across the Internet.

Answer

They exist only when DHCPv6 fails.

Answer

They replace the need for any default gateway logic.

Question 15

A subnet uses the prefix /29. How many usable host addresses are available in each subnet?

Accepted Answer

6. A /29 prefix leaves 3 host bits, which creates 8 total addresses per subnet. In plain language, two of those addresses cannot be assigned to hosts because one identifies the subnet itself and one is reserved as the broadcast address. That leaves 6 usable host addresses. This is a standard CCNA calculation because it checks whether you understand both the total address count and the subtraction of the reserved addresses.

Many candidates remember powers of two but forget to account for the network and broadcast addresses when the question asks for usable hosts. The safest process is to calculate the total size first and then reduce it by two. That is how you arrive at 6 usable addresses for a /29.

Answer

2

Answer

14

Answer

30