Cisco Certified Network Associate 200-301 CCNA Exam Questions and Answers – Page 2

The latest Cisco Certified Network Associate 200-301 CCNA certification actual real practice exam question and answer (Q&A) dumps are available free, which are helpful for you to pass the Cisco Certified Network Associate 200-301 CCNA exam and earn Cisco Certified Network Associate 200-301 CCNA certification.

Exam Question 161

Which Cisco IOS command allows you to change the setting of the configuration register?
A. boot config
B. configuration-register edit
C. config-register
D. edit configuration-register

Correct Answer:
C. config-register
Answer Description:
The config-register command is used to change the setting of the configuration register. The configuration register has the boot field setting, which specifies the order in which the router should look for bootstrap information. The router contains a 16-bit software register, which is stored in the non-volatile random access memory (NVRAM). The config-register command is used to modify the default configuration register. The most common use of changing this register is to instruct the router to ignore the stored configuration file and boot as a new router with no configuration. This process is normally used when a router has a password that is not known and must be reset. For security purposes, this procedure can only be performed from the console connection, which means it requires physical access to the router.

Normally the setting of this register is 0x2102, which tells the router to look for a configuration file. If the file exists, it will use it. If none exists, the router will boot into ROM and present the user with a menu-based setup. This would be the default behavior for a new router as well.

To view the value of the configuration register, use the show version command as displayed below. The register setting can be seen at the bottom of the output in bold.

The register setting can be seen at the bottom of the output in bold.
The register setting can be seen at the bottom of the output in bold.

Configuration register is 0x2102

To change this setting would require issuing these commands, followed by a restart:

Router#configure terminal
Enter configuration commands, one per line. End with CNTL/Z.
Router(config)#config
Router(config)#config-register 0x2142

By setting register to 0x2142, the router will ignore a configuration file at reboot if it exists. The router will then enter setup mode and prompt for you to enter initial system configuration information, as would happen with a new router. This enables the user to bypass an unknown password, since the password is contained in the file.

The boot config command is incorrect because this command is used to set the device where the configuration file is located (flash, slot, etc.) and file name for the configuration file, which helps the router to configure itself during startup.

The configuration-register edit command and the edit configuration-register commands are incorrect because they are not valid Cisco IOS commands.

Objective: Infrastructure Management
Sub-Objective: Perform device maintenance

Exam Question 162

Which Cisco Internetwork Operating System (IOS) command can be used to configure the location of the configuration file?
A. boot buffersize
B. configure
C. boot config
D. service config

Correct Answer:
C. boot config
Answer Description:
The boot config command will configure the location of the configuration file. It must be followed by the copy run start command to be effective at next reboot. The syntax of the command is as follows:

boot config device:filename

The parameters of the command are as follows:

  • Device: Specifies the device that contains the configuration file.
  • Filename: Specifies the name of the configuration file.

The boot buffersize command is incorrect because this command is used to modify the buffer size used to load the IOS image. Moreover, this command no longer functions in IOS 12.4.

The configure command is incorrect because this command is used to enter the global configuration mode.

The service config command is incorrect because this command is used to enable autoloading of configuration files from a network server.

Objective: Infrastructure Management
Sub-Objective: Perform device maintenance

Exam Question 163

Refer to the following configuration on a Cisco router to allow Telnet access to remote users:

Router(config)#line vty 0 2
Router(config-line)#login
Router(config-line)#password guest

How many users can Telnet into this router at the same time?
A. 0
B. 1
C. 2
D. 3
E. 5

Correct Answer:
D. 3
Answer Description:
The given configuration will allow three users to Telnet into the router at the same time. The line vty 0 2 command specifies a range from 0 to 2; therefore, three simultaneous Telnet sessions are allowed on this Cisco router. The commands in the exhibit can be explained as follows:

Router(config)#line vty 0 2 (determines which of the five possible terminal lines are being configured. In this case, they are lines 0 through 2. It also determines the number of lines available, in that any line with no password configured will be unusable.)
Router(config-line)#login (specifies that a password will be required)
Router(config-line)#password guest (specifies the password)

The default configuration allows five simultaneous Telnet sessions on the Cisco router. For the default configuration, you would issue the vty 0 4 command in global configuration mode.

You must configure a password when enabling a router for Telnet access. Without a password, the login access to the router will be disabled and you will receive the following error message if you try to Telnet to the router:

router# telnet 10.10.10.1
Trying 10.10.10.1 ... Open
Password required, but none set
[Connection to 10.10.10.1 closed by foreign host]

Objective: Infrastructure Management
Sub-Objective: Configure and verify device management

Exam Question 164

Which of the following are characteristics of Enhanced Interior Gateway Routing Protocol (EIGRP)? (Choose all that apply.)
A. Requires a hierarchical physical topology
B. Does not require a hierarchical physical topology
C. Uses Diffusing Update Algorithm (DUAL) to provide loop prevention
D. Uses Bellman-Ford algorithm to provide loop prevention
E. Supports Message-Digest Algorithm 5 (MD5) authentication
F. Does not support Message-Digest Algorithm 5 (MD5) authentication
G. Can differentiate between internal and external routes
H. Uses a 32-bit metric

Correct Answer:
B. Does not require a hierarchical physical topology
C. Uses Diffusing Update Algorithm (DUAL) to provide loop prevention
E. Supports Message-Digest Algorithm 5 (MD5) authentication
G. Can differentiate between internal and external routes
H. Uses a 32-bit metric
Answer Description:
EIGRP does not require a hierarchical physical topology. It uses Diffusing Update Algorithm (DUAL) to provide loop prevention, and it supports Message-Digest Algorithm 5 (MD5) authentication. It can differentiate between internal and external routes, and uses a 32-bit metric.

EIGRP is a classless protocol that allows the use of variable length subnet masks (VLSM) and supports classless interdomain routing (CIDR) for allocation of IP addresses. The following are characteristics of EIGRP:

  • Supports large networks due to high scalability
  • Provides fast convergence using the Diffusing Update Algorithm (DUAL)
  • Performs equal and unequal load balancing by default
  • Supports variable length subnet masks (VLSM) and classless interdomain routing (CIDR)
  • Is a hybrid routing protocol (distance-vector protocol) that also provides link-state protocol characteristics
  • Is a classless protocol
  • Sends partial route updates only when there are changes, reducing bandwidth usage for routing updates
  • Has an administrative distance of 90 for EIGRP internal routes, 170 for EIGRP external routes, and 5 for EIGRP summary routes
  • Is used only with Cisco platforms
  • Provides support for IP IPX and AppleTalk protocols
  • Can differentiate between internal and external routes
  • Uses a 32-bit metric

EIGRP can load-balance up to four unequal cost paths. To do so, use the variance n command to instruct the router to include routes with a metric of less than n times the minimum metric route for that destination.

The variable n can take a value between 1 and 128. The default is 1, which means equal cost load balancing.

The option stating that EIGRP requires a hierarchical physical topology is incorrect because EIGRP does not require or support a hierarchical routing topology.

The option stating that EIGRP uses Bellman-Ford algorithm to provide loop prevention is incorrect. EIGRP uses DUAL to provide loop prevention.

Objective: Routing Fundamentals
Sub-Objective: Configure, verify, and troubleshoot EIGRP for IPv4 (excluding authentication, filtering, manual summarization, redistribution, stub)

Exam Question 165

You have implemented SNMP v3 in your network. After making the configuration changes, you find that technicians in the TECHS group cannot access the MIB. You execute the show run command and receive the following output that relates to SNMP:

You execute the show run command and receive the following output that relates to SNMP.
You execute the show run command and receive the following output that relates to SNMP.

What is preventing the TECHS group from viewing the MIB?
A. The presence of the keyword priv in the command creating the RESTRICTED group
B. A mismatch between the authentication mechanism and the encryption type in the command creating the TECHS user
C. The absence of an access list defining the stations that can used by the TECHS group
D. The presence of the keyword auth in the command creating the TECHS user

Correct Answer:
C. The absence of an access list defining the stations that can used by the TECHS group
Answer Description:
The command that creates the TECHS group ends with the parameter access 99:

snmp-server group TECHS v3 priv read TECHS access 99

This indicates that the access list number 99 is specifying the IP addresses of the stations allowed to connect to the MIB for the group. Since the access list is missing from the configuration, no IP addresses will be allowed, and no connections can be made by the group.

The presence of the keyword priv in the command creating the TECHS group is not causing the issue. This keyword indicates that encryption (privacy) and authentication should both be used on all transmissions by the group.

In SMNPv3, there are three combinations of security that can be used:

  • noAuthNoPriv- no authentication and no encryption; includes the noauth keyword in the configuration
  • AuthNoPriv – messages are authenticated but not encrypted; includes the auth keyword in the configuration
  • AuthPriv – messages are authenticated and encrypted; includes the priv keyword in the configuration

There is no mismatch between the authentication mechanism and the encryption type in the command creating the TECHS user.

snmp-server user TECHS TECHS v3 auth sha CISCO priv des56 CISCO

In the preceding command, the section auth sha CISCO specified that messages are authenticated using SHA with a key of CISCO. It does not need to the match the section priv des56 CISCO, which indicates that encryption (priv) will be provided using DES56 with a key of CISCO.

The presence of the keyword auth in the command creating the TECHS user is not causing the issue. This line indicates that that messages are authenticated using SHA with a key of CISCO.

Objective: Infrastructure Management
Sub-Objective: Configure and verify device-monitoring protocols

Exam Question 166

Your company’s network must make the most efficient use of the IP address space. In the following diagram, the circles define separate network segments. The requirements of each network segment are given in the diagram. (Click the Exhibit(s) button.)

The requirements of each network segment are given in the diagram.
The requirements of each network segment are given in the diagram.

Users complain of connectivity issues. You need to discover the problems with the network configuration.
What are the three problems with the network diagram? (Choose three.)
A. The 172.16.1.0/30 segment requires more user address space.
B. The 172.16.2.0/26 segment requires more user address space.
C. The 172.16.3.0/25 segment requires more user address space.
D. The 172.16.2.64/26 segment requires more user address space.
E. Interface fa0/2 has an IP address that belongs to the 172.16.2.64/26 segment.
F. Interface fa0/4 has an IP address that belongs to the 172.16.2.0/26 segment.
G. Interface fa0/3 has an IP address outside the 172.16.3.0/25 segment.

Correct Answer:
A. The 172.16.1.0/30 segment requires more user address space.
F. Interface fa0/4 has an IP address that belongs to the 172.16.2.0/26 segment.
G. Interface fa0/3 has an IP address outside the 172.16.3.0/25 segment.
Answer Description:
The given exhibit has three problems:

  • The 172.16.1.0/30 segment requires more user address space.
  • Interface Fa0/4 has an IP address that belongs to the 172.16.2.0/26 segment.
  • Interface Fa0/3 has an IP address outside the 172.16.3.0/25 segment.

The 172.16.1.0/30 segment, as configured, will only support two hosts. This segment needs to support three hosts, the two servers, and the Fa0/1 interface. The number of hosts that a subnet is capable of supporting is a function of the number of host bits in the subnet mask. When that has been determined, the following formula can be used to determine the number of hosts yielded by the mask:

2n – 2 = X (where n = the number of host bits in the mask and X = the number of hosts supported)

In this example with a 30-bit mask, 2 host bits are left in the mask. When that is plugged into the formula, it yields only two usable addresses. The -2 in the formula represents the two addresses in each subnet that cannot be assigned to hosts, the network ID and the broadcast address. Therefore, the segment should be configured with the 172.16.1.0/29 address range, which supports up to six hosts.

Interface fa0/4, as configured, has an IP address that belongs to the 172.16.2.0/26 segment. With a 26-bit mask and the chosen class B address, the following network IDs are created:
172.16.0.0
172.16.0.64
172.16.1.128
172.16.1.192
172.16.2.0
172.16.2.64
172.16.2.128
172.16.2.192
172.16.2.0
172.16.2.64
172.16.2.128
172.16.2.192
…and so on, incrementing each time by 64 in the last octet

The 172.16.2.0/26 segment is allocated host addresses in the 172.16.2.1 through 172.16.2.62 range (the last address, 172.16.2.63, is the broadcast address and cannot be assigned). Interface fa0/4 should be assigned an IP address in the 172.16.2.64/26 range, which includes host addresses in the 172.16.2.65 through 172.16.2.126 range.

Interface Fa0/3, as configured, has an IP address outside the 172.16.3.0/25 segment. With a 25-bit mask and the chosen class B address, the following network IDs are created:
172.16.0.0
172.16.0.128
172.16.1.0
172.16.1.128
172.16.2.0
172.16.2.128
172.16.3.0
172.16.3.128
…and so on, incrementing each time by 128 in the last octet

Interface Fa0/3 should be allocated an IP address in the 172.16.3.1 through 172.16.3.126 range.

The 172.16.2.0/26 segment does not require more user address space. With a 26-bit mask, 6 bits are left for hosts, and by using the above formula it can be determined that it will yield 62 hosts. It requires 32.

The 172.16.2.64/26 segment does not require more user address space. With a 26-bit mask, 6 bits are left for hosts, and by using the above formula it can be determined that it will yield 62 hosts. It requires 32.

Interface Fa0/2 does not have an IP address that belongs to the 172.16.2.64/26 segment. The 172.16.2.64/26 segment includes addresses 172.16.2.65-172.16.5.126. Because its address is 172.16.2.1, it belongs in the 172.16.2.0/26 network (from 172.16.2.1-172.16.2.62), so it is correctly configured.

The network should be configured as shown in the following image:

The network should be configured as shown in the following image.
The network should be configured as shown in the following image.

Objective: Network Fundamentals
Sub-Objective: Configure, verify, and troubleshoot IPv4 addressing and subnetting

Exam Question 167

Which of the following are classless routing protocols? (Choose four.)
A. Open Shortest Path First (OSPF)
B. Enhanced Interior Gateway Routing Protocol (EIGRP)
C. Interior Gateway Routing Protocol (IGRP)
D. Routing Information Protocol version 1 (RIPv1)
E. Border Gateway Protocol (BGP)
F. Routing Information Protocol version 2 (RIPv2)

Correct Answer:
A. Open Shortest Path First (OSPF)
B. Enhanced Interior Gateway Routing Protocol (EIGRP)
E. Border Gateway Protocol (BGP)
F. Routing Information Protocol version 2 (RIPv2)
Answer Description:
Open Shortest Path First (OSPF), Enhanced Interior Gateway Routing Protocol (EIGRP), Border Gateway Protocol (BGP), and Routing Information Protocol version 2 (RIPv2) are classless routing protocols.

Intermediate-System-to-Intermediate System (IS-IS) is also a classless routing protocol.

The options IGRP and RIPv1 are incorrect because these are classful routing protocols.

The following are characteristics of classless routing protocols:

  • The subnet mask is advertised with each route by using classless routing protocols.
  • Flexible route summarization and supernetting (CIDR) are allowed in classless routing protocols.
  • Classless routing protocols support variable length subnet masks (VLSM), which allow different subnets of a given IP network to be configured with different subnet masks.

One of the main advantages of using a classless routing protocol is its ability to minimize the effects of discontiguous networks. When subnets of the same classful network are separated by another classful network, the networks are called discontiguous. Examine the diagram below:

When subnets of the same classful network are separated by another classful network, the networks are called discontiguous.
When subnets of the same classful network are separated by another classful network, the networks are called discontiguous.

The LAN networks extending from Router A and Router B are derived from the same Class C network, 192.168.1.0/24. A classful routing protocol such as RIP v1 would not be able to determine the direction to send the packets, but since classless protocols include the subnet mask in advertisements, they would not suffer the same problem. Whenever networks with non-default subnet masks are used, a classless routing protocol will be required.

Below are some examples of networks that do not have default masks. You can recognize them by the fact that they are not /8, /16, or /24.

192.168.10.0/27
10.5.6.0/22
172.68.0.0/18

All of the classless protocols discussed here are interior routing protocols with the exception of Border Gateway Protocol (BGP), which is an external routing protocol used to connect different autonomous systems. For example, BGP would be used to connect two OSPF autonomous systems (AS).

Objective: Routing Fundamentals
Sub-Objective: Interpret the components of routing table

Exam Question 168

You are configuring a serial link between a Cisco router and a router produced by another vendor. What would be the advantages of using Point to Point Protocol (PPP) over High Level Data Link Control (HDLC) in this scenario?
A. HDLC has a proprietary “type” field that may be incompatible with equipment from other vendors.
B. HDLC is not available on non-Cisco routers.
C. PPP is faster.
D. PPP performs error checking.

Correct Answer:
A. HDLC has a proprietary “type” field that may be incompatible with equipment from other vendors.
Answer Description:
High Level Data Link Control (HDLC) has a proprietary “type” field that may be incompatible with equipment from other vendors. It is recommended that PPP always be used when combining equipment from multiple vendors because this Data Link layer WAN protocol is an industry standard. PPP is implemented in the same manner on all PPP-capable equipment.

HDLC is available on non-Cisco routers. However, the Cisco implementation has a “type” field that may prevent the connection from working.

PPP is not faster than HDLC.

PPP performs error checking, but so does HDLC.

Objective: WAN Technologies
Sub-Objective: Describe WAN access connectivity options

Exam Question 169

Which of the following are NOT valid IPv6 addresses? (Choose all that apply.)
A. 225.1.4.2
B. ::FFFF:10.2.4.1
C. ::
D. 2001:0:42:3:ff::1
E. fe80:2030:31:24
F. 2001:42:4:0:0:1:34:0
G. 2003:dead:bef:4dad:ab33:46:abab:62

Correct Answer:
A. 225.1.4.2
F. 2001:42:4:0:0:1:34:0
Answer Description:
The addresses 255.1.4.2 and fe80:2030:31:24 are not valid IPv6 addresses.

225.1.4.2 is incorrect because it is an IPv4 multicast address. The address fe80:2030:31:24 is incorrect because it does not represent a 16-byte IPv6 address, with colons separating each 2-byte segment.

IPv6 addresses are 16 bytes, or 128 bits in length. The following are valid IPv6 addresses.

  • ::FFFF:10.2.4.1 is an example of an IPv4-compatible IPv6 address, where the first 10 bytes (80 bits) of the address are set to 0 the next 2 bytes (16 bits) are set to FFFF and the last 32 bits are the IPv4 address
  • :: is the IPv6 “unspecified address.” It is a unicast address not assigned to any interface, and is used by a DHCP-dependent host prior to allocating a real IPv6 address.
  • 2001:0:42:3:ff::1 is a valid IP address, with the :: representing two segments (4 bytes) of compressed zeros.
  • 2001:42:4:0:0:1:34:0 is a valid IP address, with only the leading zeros of each segment truncated.
  • 2003:dead:beef:4dad:ab33:46:abab:62 has 16 bytes, is divided correctly by colons into eight sections, utilizes the dropping of leading zeros in each section correctly, and uses the letters a-f in the three section that spell out dead beef 4 dad.

Objective: Network Fundamentals
Sub-Objective: Compare and contrast IPv6 address types

Exam Question 170

Given the following output, which statements can be determined to be true? (Choose three.)

RouterA2# show ip ospf neighbor
Neighbor ID Pri State Dead Time Address Interface
192.168.23.2 1 FULL/BDR 00:00:29 10.24.4.2 FastEthernet1/0
192.168.45.2 2 FULL/BDR 00:00:24 10.1.0.5 FastEthernet0/0
192.168.85.1 1 FULL/- 00:00:33 10.6.4.10 Serial0/1
192.168.90.3 1 FULL/DR 00:00:32 10.5.5.2 FastEthernet0/1
192.168.67.3 1 FULL/DR 00:00:20 10.4.9.20 FastEthernet0/2
192.168.90.1 1 FULL/BDR 00:00:23 10.5.5.4 FastEthernet0/1
<Output omitted>

A. This router is the DR for subnet 10.1.0.0.
B. The DR for the network connected to Fa0/0 has an interface priority greater than 2.
C. The DR for the network connected to Fa0/1 has a router ID of 10.5.5.2.
D. The DR for the serial subnet is 192.168.85.1.
E. This router is neither the DR nor the BDR for the Fa0/1 subnet.
F. RouterA2 is connected to more than one multi-access network.

Correct Answer:
B. The DR for the network connected to Fa0/0 has an interface priority greater than 2.
E. This router is neither the DR nor the BDR for the Fa0/1 subnet.
F. RouterA2 is connected to more than one multi-access network.
Answer Description:
The show ip ospf neighbor command displays a list of all OSPF routers with which you have established a neighbor relationship. The following describes the command output:

  • Neighbor ID: the Router ID (RID) of the neighboring router
  • Pri: the interface priority of the neighboring router, which is used to determine which router should serve the function of a Designated Router (DR)
  • State: the functional state of the neighboring router
  • Dead Time: the period that the router will wait to hear a Hello packet from this neighbor before declaring the neighbor down
  • Address: the IP address of the neighboring router on this subnet
  • Interface: the local interface over which the neighbor relationship (adjacency) was formed

The output for neighbor 192.168.45.2 is as follows:

192.168.45.2 2 FULL/BDR 00:00:24 10.1.0.5 FastEthernet0/0

This indicates that the interface priority of neighbor 192.168.45.2 is 2. The default OSPF interface priority is 1, and the highest interface priority determines the designated router (DR) for a subnet. This same line reveals that this neighbor is currently the backup designated router (BDR) for this segment, which indicates that another router became the DR. It can be then be assumed that the DR router has an interface priority higher than 2. (The router serving the DR function is not present in the truncated sample output.)

The output for the two neighbors discovered on F0/1 is as follows:

192.168.90.3 1 FULL/DR 00:00:32 10.5.5.2 FastEthernet0/1
192.168.90.1 1 FULL/BDR 00:00:23 10.5.5.4 FastEthernet0/1

This output indicates that router 192.168.90.3 is the DR, and router 192.168.90.1 is the BDR for this network. Since there can only be one DR and BDR per segment, this indicates that the local router is neither the DR nor the BDR. (OSPF considers these DROther routers.)

The fact that multiple DRs are listed in this output indicates that RouterA2 is connected to more than one multi-access segment, since each segment will elect a DR.

It cannot be determined if this router is the DR for subnet 10.1.0.0. The output indicates that router 192.168.45.2 is the BDR for this network, but with the truncated output, it cannot be determined if this router is the DR.

The DR for the network connected to Fa0/1 does not have a router ID of 10.5.5.2. The Address field of the show ip ospf neighbor command indicates the IP address of the neighbor’s interface, not the router ID of the neighbor.

The DR for the serial subnet is not 192.168.85.1, since point-to-point serial interfaces do not elect DRs and BDRs. This is indicated by the output below:

192.168.85.1 1 FULL/- 00:00:33 10.6.4.10 Serial0/1

Objective: Routing Fundamentals
Sub-Objective: Configure, verify, and troubleshoot single area and multi-area OSPFv2 for IPv4 (excluding authentication, filtering, manual summarization, redistribution, stub, virtual-link, and LSAs)