CCNA Practice Test 12 — 15 Questions

Question 1

You are connected to the console of R1. The network uses IPv6 with EUI-64. R1's GigabitEthernet0/0 interface has MAC address 0011.2233.4455. You must configure the interface to generate an IPv6 link-local address using the 'ipv6 enable' command, and also assign a global unicast address 2001:db8:1::/64 using EUI-64. The interface is currently administratively down.

Accepted Answer

R1(config-if)# ipv6 enable
R1(config-if)# ipv6 address 2001:db8:1::/64 eui-64
R1(config-if)# no shutdown. The ipv6 enable command explicitly creates a link-local address as required by the scenario. The global unicast address with the eui-64 keyword automatically derives the interface ID from the MAC address. Option B is incorrect because it tries to manually configure a link-local address with eui-64, which is unnecessary and invalid. Option C omits the ipv6 enable command, failing the explicit requirement. Option D assigns a static host portion instead of using eui-64.

Answer

R1(config-if)# ipv6 address fe80::/10 eui-64
R1(config-if)# ipv6 address 2001:db8:1::/64 eui-64
R1(config-if)# no shutdown

Answer

R1(config-if)# ipv6 address 2001:db8:1::/64 eui-64
R1(config-if)# no shutdown

Answer

R1(config-if)# ipv6 enable
R1(config-if)# ipv6 address 2001:db8:1::1/64
R1(config-if)# no shutdown

Question 2

A host address is 192.168.22.145/28. Which subnet contains that host?

Accepted Answer

192.168.22.144/28. A /28 prefix creates address blocks of 16. In practical terms, the fourth-octet ranges are 0–15, 16–31, 32–47, and so on. Because 145 falls inside the 144–159 range, the network address of the containing subnet is 192.168.22.144/28.

This type of question checks whether you can move from prefix length to block size and then place a host into the correct interval. The common mistake is choosing a nearby familiar number instead of calculating the actual block boundary.

Answer

192.168.22.128/28

Answer

192.168.22.160/28

Answer

192.168.22.148/28

Question 3

A branch router is configured for NAT overload. The inside interface Gi0/0 is correctly marked ip nat inside, and the outside interface Gi0/1 is ip nat outside. The NAT statement uses access-list 1 permit 10.1.1.0 0.0.0.255 with ip nat inside source list 1 interface Gi0/1 overload. Inside hosts are in the 192.168.1.0/24 subnet and still reach the ISP with their private addresses. What is the most likely reason?

Accepted Answer

The ACL used for NAT does not match the inside client subnet.. The ACL matches the wrong inside subnet. NAT overload will only translate traffic that matches the source list or route map tied to the NAT statement. The interfaces are marked inside and outside correctly, so the bad match criteria is the most likely failure point.

Answer

GigabitEthernet0/0 should be configured as ip nat inside.

Answer

PAT cannot use an interface address as the translated source.

Answer

The router must run OSPF before NAT overload can function.

Question 4

You are connected to R1. The network uses a router-on-a-stick design with a single switch (SW1) and two VLANs (10 and 20). Currently, hosts in VLAN 10 cannot ping hosts in VLAN 20, and the trunk between R1 and SW1 shows a native VLAN mismatch. Examine the provided configuration and output, then apply the necessary corrections to R1 so that inter-VLAN routing works correctly.

Accepted Answer

Change the native VLAN on R1's trunk interface to 99, correct the encapsulation on G0/0.20 to dot1Q 20, and enable IP routing globally.. The configuration has three issues: 1) The native VLAN on the trunk is VLAN 1, but the switch expects VLAN 99 (common mismatch scenario). 2) Subinterface G0/0.20 uses encapsulation dot1Q 10 instead of 20, causing VLAN 20 traffic to be mis-tagged. 3) The 'ip routing' command is missing globally, so R1 cannot route between subinterfaces. To fix, configure the main interface G0/0 with 'encapsulation dot1Q 99 native' to set the native VLAN to 99, correct the encapsulation on G0/0.20 to 'dot1Q 20', and enable IP routing with 'ip routing'.

Answer

Change the native VLAN on R1's trunk interface to 99, correct the encapsulation on G0/0.20 to dot1Q 20, but do not enable IP routing because it is enabled by default.

Answer

Change the native VLAN on R1's trunk interface to 1, correct the encapsulation on G0/0.20 to dot1Q 20, and enable IP routing globally.

Answer

Change the native VLAN on R1's trunk interface to 99, correct the encapsulation on G0/0.10 to dot1Q 10, and enable IP routing globally.

Question 5

Drag and drop the following steps into the correct order to configure HSRP on a router and verify the active/standby election process.

Accepted Answer

1. Enter interface configuration mode on the desired interface. 2. Configure the HSRP group number and virtual IP address using the 'standby group-number ip virtual-ip' command. 3. Set the HSRP priority using the 'standby group-number priority priority-value' command. 4. Enable preemption using the 'standby group-number preempt' command. 5. Verify the HSRP state using 'show standby' or 'show standby brief'. 6. Test failover by shutting down the active router's interface and observing the state change.. The order begins with interface and HSRP group configuration, then priority and preempt. Verification shows the current state, and failover testing validates the election process works correctly.

Question 6

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

Accepted Answer

10.24.7.159. 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.

Answer

10.24.7.191

Answer

10.24.7.127

Answer

10.24.7.160

Question 7

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

Accepted Answer

192.168.30.63. 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.

Answer

192.168.30.64

Answer

192.168.30.255

Answer

192.168.30.31

Question 8

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

Accepted Answer

192.168.200.97 to 192.168.200.110. 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.

Answer

192.168.200.96 to 192.168.200.111

Answer

192.168.200.98 to 192.168.200.111

Answer

192.168.200.81 to 192.168.200.94

Question 9

A network administrator is troubleshooting an issue where internal hosts can ping the company's web server by IP address (192.0.2.10) but cannot access it using the fully qualified domain name www.example.com. The DNS server (192.0.2.5) is reachable and responds to queries. The administrator runs nslookup www.example.com from a host and receives the following output:

C:\> nslookup www.example.com
Server:  UnKnown
Address:  192.0.2.5

Name:    www.example.com
Address:  192.0.2.20

Based on the output, what is the most likely cause of the problem?

Accepted Answer

The DNS A record for www.example.com is incorrect; update it to point to 192.0.2.10.. The nslookup output shows that www.example.com resolves to 192.0.2.20, but the web server is at 192.0.2.10. This indicates the DNS A record is incorrect and must be updated to point to the correct IP. Pinging by IP works because it bypasses DNS, confirming network connectivity. The host's DNS cache is not the issue because nslookup queries the server directly and still returns the wrong address. The firewall is not involved since pinging by IP succeeds, and the DNS server is authoritative (the response is received).

Answer

The host's DNS cache is corrupted; flush it using ipconfig /flushdns.

Answer

The web server's firewall is blocking traffic from the host; add an allow rule.

Answer

The DNS server is not authoritative for the example.com zone; delegate the zone to a different server.

Question 10

Which TWO statements correctly describe OSPFv2 router-id selection and verification in a single-area configuration?

Accepted Answer

When no 'router-id' is configured, a loopback interface with the highest IP address is preferred over a physical interface for the router-id.. OSPFv2 selects the router-ID based on the highest IP address of any loopback interface when no explicit 'router-id' is configured, making D correct. The 'show ip ospf' command displays the current router-ID, verifying choice E. Option A is incorrect because the router-ID is derived from IP addresses, not MAC addresses. Option B fails because changing the router-ID requires a reload or clearing the OSPF process to take effect. Option C is wrong because each router must have a unique router-ID; they do not need to match across the area.

Answer

The OSPF router-id is automatically derived from the MAC address of the first Ethernet interface.

Answer

If the router-id is changed using the 'router-id' command, the change takes effect immediately without any additional action.

Answer

The router-id must be the same on all routers in a single OSPF area.

Question 11

Exhibit: OSPF neighbors are not reaching FULL state on an Ethernet segment with multiple routers. The output of show ip ospf neighbor on R2 shows a neighbor in the 2WAY/DROTHER state. What is the most likely reason?

Accepted Answer

The 2WAY state with another DROTHER on a broadcast segment is normal. On a broadcast Ethernet network, two routers normally become fully adjacent through the DR or BDR. If the local router is stuck in 2WAY with another DROTHER, that is normal behavior. It is not a fault by itself.

Answer

Authentication mismatch between R2 and 3.3.3.3

Answer

R2 must be configured as a point-to-point network type

Answer

R2 has a duplicate router ID

Question 12

Which TWO statements accurately describe the characteristics and deployment considerations for fiber optic cabling in a modern enterprise network?

Accepted Answer

When using a 10GBASE-SR SFP+ transceiver over OM3 multimode fiber, the maximum supported distance is 300 meters.. The correct statements are that 10GBASE-SR over OM3 multimode fiber supports up to 300 meters and that fiber optic cabling is immune to electromagnetic interference (EMI), making it ideal for electrically noisy environments. Single-mode fiber actually has a smaller core diameter than multimode, making option A incorrect. Multimode fiber is designed for shorter links, so option B is wrong. Standard 1000BASE-LX SFP transceivers over single-mode fiber are limited to 5 km, not 10 km, so option C is inaccurate.

Answer

Single-mode fiber (SMF) typically uses a larger core diameter than multimode fiber (MMF).

Answer

Multimode fiber (MMF) is generally preferred for longer-distance links, such as between buildings on a campus network.

Answer

A 1000BASE-LX SFP transceiver operating over single-mode fiber can support distances up to 10 km.

Question 13

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

Accepted Answer

10.1.1.35. 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.

Answer

10.1.1.31

Answer

10.1.1.32

Answer

10.1.1.36

Question 14

Which of the following statements about VLAN configuration and trunking on a Cisco switch are correct? (Choose all that apply.)

Accepted Answer

The native VLAN on a trunk link is used for untagged traffic and should match on both ends of the link.. The native VLAN on a trunk carries untagged traffic, and mismatched native VLANs can cause spanning-tree issues, VLAN hopping, and misdirected control traffic, so both ends must match. VLAN 1 is a system-defined VLAN that cannot be deleted or shut down, making it always available. Dynamic Trunking Protocol (DTP) is a Cisco proprietary protocol that can automatically negotiate trunk links when interfaces are set to desirable or auto. The 'switchport trunk allowed vlan' command prunes VLANs on a trunk, restricting which VLANs are permitted. A trunk link forwards traffic only for VLANs that exist in the local VLAN database, so the statement that it can only carry globally created VLANs is correct. The incorrect option is that an access port can carry multiple VLANs; in reality, an access port is assigned to a single VLAN and cannot carry traffic for multiple VLANs like a trunk.

Answer

A trunk link can only carry VLANs that are globally created on the switch.

Answer

A switchport configured as an access port can carry multiple VLANs.

Question 15

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

Accepted Answer

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

Answer

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

Answer

The broadcast address always becomes the default gateway.

Answer

These concepts exist only in IPv6 and not IPv4.