Cisco Certified Network Associate 200-301 CCNA Exam Questions and Answers - Page 6 - Page 2 of 10

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 511

A device has an address of 192.168.144.21 and a mask of 255.255.255.240.
What will be the broadcast address for the subnet to which this device is attached?

A. 192.168.144.23
B. 192.168.144.28
C. 192.168.144.31
D. 192.168.144.32
Correct Answer:
C. 192.168.144.31
Answer Description:
The broadcast address for the subnet to which this device is attached will be 192.168.144.31.

To determine the broadcast address of a network where a specific address resides, you must first determine the network ID of the subnetwork where the address resides. The network ID can be obtained by determining the interval between subnet IDs. With a 28-bit mask, the decimal equivalent of the mask will be 255.255.255.240. The interval between subnets can be derived by subtracting the value of the last octet of the mask from 256. In this case, that operation would be 256 – 240. Therefore, the interval is 16.

The first network ID will always be the classful network you started with (in this case 192.168.144.0). Then each subnetwork ID in this network will fall at 16-bit intervals as follows:

192.168.144.0
192.168.144.16
192.168.144.32
192.168.144.48

At 192.168.144.48 we can stop, because the address that we are given as a guide is in the network with a subnet ID of 192.168.144.16. Therefore, since the broadcast address for this network will be 1 less than the next subnet ID (192.168.144.32), the broadcast address for the subnet to which this device is attached is 192.168.144.31.

All the other options are incorrect because none of these will be the broadcast address for the subnet to which this device is attached.

Exam Question 512

You have been asked to troubleshoot the NTP configuration of a router named R70. After executing the show run command, you receive the following partial output of the command that shows the configuration relevant to NTP:
clock timezone PST -8 clock summer-time PDT recurring ntp update-calendar ntp server 192.168.13.57 ntp server 192.168.11.58 interface Ethernet 0/0 ntp broadcast
Based on this output, which of the following statements is true?

A. the time zone is set to 8 hours less than Pacific Standard time
B. the router will listen for NTP broadcasts on interface E0/0
C. the router will send NTP broadcasts on interface E0/0
D. the router will periodically update its software clock
Correct Answer:
C. the router will send NTP broadcasts on interface E0/0
Answer Description:
The router will send NTP broadcast on its E0/0 interface. The command ntp broadcast, when executed under an interface, instructs the router to send NTP broadcast packets on the interface. Any devices on the network that are set with the ntp broadcast client command on any interface will be listening for these NTP broadcasts. While the clients will not respond in any way, they will use the information in the NTP broadcast packets to synchronize their clocks with the information.

The time zone is not set to 8 hours less than Pacific Standard Time. The value -8 in the command clock timezone PST -8 represents the number of hours of offset from UTC time, not from the time zone stated in the clock timezone command.

The router will not listen for NTP broadcasts on the interface E0/0. The ntp broadcast command, when executed under an interface, instructs the router to send NTP broadcast packets on the interface. To set the interface to listen and use NTP broadcasts, you would execute the ntp broadcast client command on the interface.

The router will not periodically update its software clock. The command ntp update-calendar configures the system to update its hardware clock from the software clock at periodic intervals.

Exam Question 513

What will an EIGRP router do if the successor route fails and there is no feasible successor?

A. EIGRP will mark the route as passive until a new successor route is determined.
B. EIGRP will redistribute routes into RIP or OSPF.
C. EIGRP will query neighboring routers until a new successor route is determined.
D. EIGRP will forward traffic to the neighbor with the lowest administrative distance.
Correct Answer:
C. EIGRP will query neighboring routers until a new successor route is determined.
Answer Description:
Feasible successors are backup routes for the successor (active) route to a remote network. If a successor route fails, and a feasible successor is available, the feasible successor will immediately become the successor and be installed in the routing table. This provides EIGRP with virtually instantaneous convergence. If no feasible successor is available, then the router must send out query packets to neighboring EIGRP routers to find an alternate path to the remote network.

EIGRP routes are marked as active when the network is converging. Passive routes are stable, converged routes.

EIGRP will not redistribute routes into RIP or OSPF. Redistribution allows information learned from one routing protocol to be converted into routes for injection into the autonomous system of another routing protocol. This allows networks learned via EIGRP, for example, to be visible and reachable from hosts in a RIP routing domain. Redistribution has nothing to do with EIGRP convergence or with the determination of a new successor route.

Administrative distance is used to determine which source of routing information is considered more trustworthy when multiple routing protocols have been implemented. Administrative distance has no effect on EIGRP convergence or the determination of a new successor route.

Exam Question 514

Examine the output of the show ip route command below:

Which of the following statements is FALSE?

A. The route to 30.30.30.30 uses a cost of 21
B. The command ip route 192.168.2.0 255.255.255.0 172.16.14.2 200 will replace the current route to 192.168.2.0/24
C. The route to 192.168.2.0/24 uses the default administrative distance
D. Traffic will be load balanced across two routes to 30.30.30.30
Correct Answer:
B. The command ip route 192.168.2.0 255.255.255.0 172.16.14.2 200 will replace the current route to 192.168.2.0/24
Answer Description:
The command ip route 10.10.10.0 255.255.255.0 172.16.14.2 200 will NOT replace the current route to 10.0.0.0/24.

When you execute the ip route command to enter a static route, the administrative distance can be altered by adding the desired distance value to the end of the command. In this scenario, the administrative distance value was set to 200. The route to the 10.10.10.0/24 network that is currently in the table was learned by OSPF and is using the default administrative distance of 110. Since 110 is lower than 200, the new static route will not be added to the routing table UNLESS the current route becomes unavailable.

The route to 30.30.30.30 does uses a cost of 21, as is indicated by the value on the right side of the forward slash within the brackets found in the route entry, [50/21].

The route to 192.168.2.0/24 uses the default administrative distance. It was learned from OSPF, which has a default distance of 110. Its administrative distance is indicated by the value on the left side of the forward slash within the brackets found in the route entry, [110/20].

Traffic will be load balanced across two routes to 30.30.30.30 because they have equal cost of 21. This cost is indicated by the value on the right side of the forward slash within the brackets found in the route entry, [50/21].

Exam Question 515

What is the purpose of using the show arp command?

A. To view the ARP statistics only for a particular interface
B. To view details regarding neighboring devices discovered by ARP
C. To view global ARP information such as timer and hold time
D. To view the Address Resolution Protocol (ARP) cache
Correct Answer:
D. To view the Address Resolution Protocol (ARP) cache
Answer Description:
The show arp command is used to view the Address Resolution Protocol (ARP) cache. ARP is used by the Internet Protocol (IP) to find the Media Access Control (MAC) address or the hardware address of a host. The main function of ARP is to translate IP addresses to MAC addresses. The process of obtaining the address of a computer in the network is known as address resolution. This process is accomplished by sending an ARP packet from a source to a destination host. The destination host responds to the ARP packet by replying back to the source and including its own MAC address. Once the source host receives the reply, it will update its ARP cache with the new MAC address.

The following is a brief description of the parameters used with this command:

ip-address: An optional parameter that displays specific ARP entries.
locationnode-id: An optional parameter that displays the ARP entry for a specific location. The method for entering the node-id argument is rack/slot/module notation.
hardware-address: An optional parameter that displays ARP entries that match the 48-bit MAC address.
traffic: An optional parameter that displays ARP traffic statistics.
interface instance: Either a physical interface instance or a virtual interface instance:
Physical interface instance: the naming notation is rack/slot/module/port and a slash mark between values is required as part of the notation where:
rack refers to the chassis number of the rack.
slot refers to the physical slot number of the line card.
module refers to the module number. A physical layer interface module (PLIM) is always 0.
port refers to the physical port number of the interface.
Virtual interface instance: the number range is variable depending on the type of interface.
trace: An optional parameter that displays the ARP entries in the buffer.
error: An optional parameter that displays the ARP error logs.
dev: An optional parameter that displays the ARP internal logs.
events: An optional parameter that displays the ARP events logs.
table: An optional parameter that displays the ARP cache logs.
packets: An optional parameter that displays the ARP packet receive and reply logs.
type instance: An optional parameter that specifies the interface for which you want to view the ARP cache.

An example of the output of the show arp command is shown below along with a diagram of the network in which the router resides.

R1#show arp Protocol Address Age (min) Hardware Addr Type Interface Internet 192.0.5.1 120 0000.a710.4baf ARPA FastEthernet 0/1 Internet 192.0.5.6 105 0000.a710.859b ARPA FastEthernet 0/1 Internet 58.59.6.3 42 0000.a710.68cd ARPA FastEthernet 0/0 Internet 58.59.6.4 59 0000.0c01.7bbd ARPA FastEthernet 0/0

From the information above, we can make the following conclusions about the actions R1 will take when it receives data from PC1 destined for the Web server:

The data frames will be forwarded out the Fa0/0 interface of R1
R1 will place the MAC address of the Web Server (0000.0c01.7bbd) in the destination MAC address of the frames
R1 will put the MAC address if the forwarding Fa0/0 interface (0000.a710.68cd) in the place of the source MAC address

The option stating that the show arp command is used to view the ARP statistics only for a particular interface is incorrect because this command is used to view the ARP cache. You can also view the information for a particular interface with the help of the interface instance parameter.

The options stating that the show arp command is used to view the details of neighboring devices discovered by the ARP or to view global ARP information, such as hold time and timer, are both incorrect because these are both Cisco Discovery protocol (CDP) functions, not ARP functions. The show cdp neighbors detail command is used to display details regarding the neighboring devices that are discovered by CDP, and the show cdp command displays global CDP information, such as timer and hold-time.

Exam Question 516

What are the three types of Internet Protocol version 6 (IPv6) addresses? (Choose three.)

A. Unicast
B. Broadcast
C. Dual-cast
D. Anycast
E. Multicast
Correct Answer:
A. Unicast
D. Anycast
E. Multicast
Answer Description:
Unicast, multicast, and anycast are types of IPv6 addresses.

The following are the IPv6 address types:

Unicast address: These types of addresses are used to define a single destination interface. A packet sent to a unicast address is delivered to the specific interface.
Multicast address: These types of addresses are used to define a group of hosts. When a packet is sent to a multicast address, it is delivered to all the hosts identified by that address. Multicast addresses begin with the prefix FF00::/8 and the second octet identifies the range over which the multicast address is propagated. Some special case IPv6 multicast addresses:
– FF01:0:0:0:0:0:0:1: Indicates all-nodes address for interface-local scope.
– FF02:0:0:0:0:0:0:2: Indicates all-routers address for link-local.
Anycast address: These types of addresses are used to identify a set of devices. These addresses are also assigned to more than one interface belonging to different nodes. A packet sent to an anycast address is delivered to just one of the interfaces, based on which one is closest. For example, if an anycast address is assigned to a set of routers, one in India and another in the U.S., the users in the U.S. will be routed to U.S. routers and the users in India will be routed to a server located in India.

The broadcast option is incorrect because these types of addresses are not supported by IPv6. Broadcast functionality is provided by multicast addressing.

The dual-cast option is incorrect because this is not a valid Cisco address type.

Exam Question 517

Which media access control method is used by Ethernet technology to minimize collisions in the network?

A. CSMA/CD
B. token passing
C. back-on algorithm
D. full-duplex
Correct Answer:
A. CSMA/CD
Answer Description:
Carrier Sense Multiple Access – Collision Detection (CSMA/CD) is used by Ethernet technology to minimize collisions in the network. The CSMA/CD method uses a back-off algorithm to calculate random time for retransmission after a collision. When two stations start transmitting at the same time, their signals will collide. The CSMA/CD method detects the collision, and both stations hold the retransmission for a certain amount of time that is determined by the back-off algorithm. This is an effort to help ensure that the retransmitted frames do not collide.

Token passing is used by the token-ring network topology to control communication on the network.

Full-duplex is the Ethernet communication mode that allows workstation to send and receive simultaneously. With the use of full-duplex, the bandwidth of the station can effectively be doubled. Hubs are not capable of handling full-duplex communication. You need dedicated switch ports to allow full-duplex communication.

The back-on algorithm is an invalid option. There is no such contention method.

Exam Question 518

How many IP addresses are available for hosts in the 192.168.16.64 /26 subnet?

A. 14
B. 30
C. 62
D. 126
Correct Answer:
C. 62
Answer Description:
There are 62 IP addresses available for hosts in the 192.168.16.64 /26 subnet.

The number of host addresses is calculated as 2n – 2, where n is the number of host bits and 2 is subtracted to exclude the network address and the broadcast address.

An IP address has 32 available bits divided into four octets. In the 192.168.16.66 /26 address, the /26 indicates that there are 26 masking bits, or that 26 bits are reserved for the network portion of the address.
This leaves 6 bits for the host addresses (32 – 26 = 6).

The following formula is used to calculate the number of IP addresses available for hosts:

Network address: 192.168.16.0
Subnet mask in decimal: 255.255.255.192
Subnet mask in binary: 11111111.11111111.1111111.11000000
Number of bits used for masking = 2⁶
Number of hosts bits in the address = 6

Using the formula for calculating the number of hosts per subnet, we find:

Hosts formula: 2^{number-of-host-bits – 2}
Hosts: 2⁶ – 2 = 62

For subnet 192.168.16.64, the valid host range starts from 192.168.16.65 and runs to 192.168.16.126. For subnet 192.168.16.128, the valid host range starts from 192.168.16.129 and runs to 192.168.16.190.

The options 14, 30, and 126 are incorrect because 62 IP addresses are available for hosts in the 192.168.16.64/26 subnet.

The correct mask for the size network desired is critical to proper network function. For example, assume a router has an interface Fa0/0 hosting a LAN with 20 computers configured as shown in the following output of show interfaces command:

Router# show interfaces Fastethernet0 is up, line protocol is up Hardware address is 000b.12bb.4587 Internet address 192.168.10.30/30

In this example, the computers will not be able to access anything beyond the LAN because the mask /30 only allows for 2 addresses when 21 (including the router interface) are required.

Exam Question 519

Which of the following statements are TRUE regarding the following output? (Choose all that apply.)

A. There are four default routes on this router.
B. There are four physically connected interfaces on this router.
C. This router is running EIGRP.
D. The metric for the routes learned via a routing protocol is 90.
E. A packet for the 192.168.52.0 network will be load-balanced across two paths.
Correct Answer:
C. This router is running EIGRP.
E. A packet for the 192.168.52.0 network will be load-balanced across two paths.
Answer Description:
This router is running EIGRP and a packet for the 192.168.52.0 network will be load-balanced across two paths.

EIGRP routes display with a D code in the leftmost column of the show ip route command. The D stands for Diffusing Update Algorithm (DUAL), which is the algorithm used by EIGRP to determine the best and potential backup paths to each remote network. There are four EIGRP-learned routes in this exhibit.

When two routes with equal metrics exist in the routing table, EIGRP will send packets using both paths. In the output there are two routes listed for the 192.168.52.0 network. Both have the same metric value (2172416). Therefore, packets will be sent to that network via the Serial 0/1/0 interface to the neighbor at 192.168.15.254 and via the Serial 0/0/0 interface to the neighbor at 192.168.15.5. Both paths, either directly or indirectly, lead to the 192.168.52.0 network, and both paths have the same cost.

There are not four default routes on this router. The D represents EIGRP-learned routes, not default routes. There is one default route, as indicated by the line of output that says Gateway of last resort is 192.168.15.1 to network 0.0.0.0. Because Serial0/1/0 is directly connected to the 192.168.15.0 network, packets that are destined for networks not found in the routing table will be sent out on that interface.

The C in the leftmost column of the show ip route command represents directly connected networks, of which there are four in the exhibit. Closer examination, however, reveals that one of these entries (for network 192.168.33.0) is connected to a loopback interface (Loopback1), as opposed to a physical interface:
C 192.168.33.0 is directly connected, Loopback1

Loopback interfaces are virtual, software interfaces that appear in the routing table, but do not represent a physical interface on the router. Therefore, there are three physically connected interfaces on this router, not four.

The metric for the routes learned via a routing protocol is not 90. The 90 in the scenario output is the administrative distance (AD) of the route, and the 2196545 is the metric value (see below):
D 192.168.25.0 [90/2196545] via 192.168.20.254, 0:01:20, Serial0/0/1

Exam Question 520

You are purchasing a device to upgrade your network. You need to determine the type of device required, as well as the number and type of required interfaces. The device will host three LAN subnets and a T1 Internet connection.

Which of the following device and interface combinations will support this requirement without providing any unnecessary interfaces or using subinterfaces?

A. a switch with one Ethernet interface and three serial interfaces
B. a router with one serial interface and three Ethernet interfaces
C. a router with one serial interface and one Ethernet interface
D. a switch with one modem and three serial interfaces
Correct Answer:
B. a router with one serial interface and three Ethernet interfaces
Answer Description:
This deployment will require a router with one serial interface and three Ethernet interfaces. When LAN subnets and the Internet must be connected, you must deploy a device that can make decisions based on IP addresses. This is the function of a router. Each LAN subnet will require a separate Ethernet interface, and the T1 connection requires a serial interface, so the router must have one serial interface and three Ethernet interfaces.

A switch cannot be used to connect separate subnets and the Internet. This requires a router. Switches make forwarding decisions based on MAC addresses. In this deployment, decisions must be made on the basis of IP addresses. Moreover, switches only have Ethernet interfaces, so a switch could not handle the T1 connection.

A router with one serial and one Ethernet interface will not be sufficient. Each LAN subnet will require a separate Ethernet interface.