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New Cisco 100-105 Exam Dumps Collection (Question 12 – Question 21)

Question No: 12

Refer to the exhibit.

An administrator replaced the 10/100 Mb NIC in a desktop PC with a 1 Gb NIC and now the PC will not connect to the network. The administrator began troubleshooting on the switch. Using the switch output shown, what is the cause of the problem?

A. Speed is set to 100Mb/s.

B. Input flow control is off.

C. Encapsulation is set to ARPA.

D. The port is administratively down.

E. The counters have never been cleared.

Answer: A

Explanation:

For PC to switch connectivity, the speed settings must match. In this case, the 1 Gb NIC will not be able to communicate with a 100Mb fast Ethernet interface, unless the 1Gb NIC can be configured to connect at 100Mb.

Question No: 13

Which command can you use to manually assign a static IPV6 address to a router interface?

A. ipv6 address PREFIX_1::1/64

B. ipv6 autoconfig 2001:db8:2222:7272::72/64

C. ipv6 autoconfig

D. ipv6 address 2001:db8:2222:7272::72/64

Answer: D

Explanation:

An example of configuring IPv6 on an interface is shown below: Router(config)# interface fastethernet 0/1

Router(config-if)# ipv6 address 3000::2222:1/64

Question No: 14

What information can be used by a router running a link-state protocol to build and maintain its topological database? (Choose two.)

A. hello packets

B. SAP messages sent by other routers

C. LSAs from other routers

D. beacons received on point-to-point links

E. routing tables received from other link-state routers

F. TTL packets from designated routers

Answer: A,C

Explanation: Reference 1:

http://www.ciscopress.com/articles/article.asp?p=24090&seqNum=4

Link state protocols, sometimes called shortest path first or distributed database protocols, are built around a well-known algorithm from graph theory, E. W. Dijkstra'a shortest path algorithm. Examples of link state routing protocols are:

Open Shortest Path First (OSPF) for IP

The ISO's Intermediate System to Intermediate System (IS-IS) for CLNS and IP DEC's DNA Phase V

Novell's NetWare Link Services Protocol (NLSP)

Although link state protocols are rightly considered more complex than distance vector protocols, the basic functionality is not complex at all:

1. Each router establishes a relationshipu2014an adjacencyu2014with each of its neighbors.

2. Each router sends link state advertisements (LSAs), some

3. Each router stores a copy of all the LSAs it has seen in a database. If all works well, the databases in all routers should be identical.

4. The completed topological database, also called the link state database, describes a graph of the internetwork. Using the Dijkstra algorithm, each router calculates the shortest path to each network and enters this information into the route table.

OSPF Tutorial

Question No: 15

A network administrator is connecting PC hosts A and B directly through their Ethernet interfaces as shown in the graphic. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts? (Choose two.)

A. A crossover cable should be used in place of the straight-through cable.

B. A rollover cable should be used in place of the straight-through cable.

C. The subnet masks should be set to 255.255.255.192

D. A default gateway needs to be set on each host.

E. The hosts must be reconfigured to use private IP addresses for direct connections of this type.

F. The subnet masks should be set to 255.255.255.0

Answer: A,F

Explanation:

If you need to connect two computers but you don't have access to a network and can't set up an ad hoc network, you can use an Ethernet crossover cable to create a direct cable connection.

Generally speaking, a crossover cable is constructed by reversing (or crossing over) the order of the wires inside so that it can connect two computers directly. A crossover cable looks almost exactly like a regular Ethernet cable (a straight-through cable), so make sure you have a crossover cable before following these steps.

Both devices need to be on the same subnet, and since one PC is using 192.1.1.20 and the other is using 192.1.1.201, the subnet mask should be changed to 255.255.255.0.

Question No: 16

If an Ethernet port on a router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?

A. 1024

B. 2046

C. 4094

D. 4096

E. 8190

Answer: C

Explanation:

Each octet represents eight bits. The bits, in turn, represent (from left to right): 128, 64, 32 , 16 , 8, 4, 2, 1

Add them up and you get 255. Add one for the all zeros option, and the total is 256. Now, take away one of these for the network address (all zeros) and another for the broadcast address (all ones). Each octet represents 254 possible hosts. Or 254 possible

networks. Unless you have subnet zero set on your network gear, in which case you could conceivably have 255.

The CIDR addressing format (/20) tells us that 20 bits are used for the network portion, so the maximum number of networks are 2^20 minus one if you have subnet zero enabled, or minus 2 if not.

You asked about the number of hosts. That will be 32 minus the number of network bits, minus two. So calculate it as (2^(32-20))-2, or (2^12)-2 = 4094

Question No: 17

Which two statements describe the IP address 10.16.3.65/23? (Choose two.)

A. The subnet address is 10.16.3.0 255.255.254.0.

B. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.

C. The last valid host address in the subnet is 10.16.2.254 255.255.254.0

D. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.

E. The network is not subnetted.

Answer: B,D

Explanation:

The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 – 254). So this makes the subnets in 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254

Question No: 18

A router has learned three possible routes that could be used to reach a destination network. One route is from EIGRP and has a composite metric of 20514560. Another route is from OSPF with a metric of 782. The last is from RIPv2 and has a metric of 4. Which route or routes will the router install in the routing table?

A. the OSPF route

B. the EIGRP route

C. the RIPv2 route

D. all three routes

E. the OSPF and RIPv2 routes

Answer: B

Explanation:

When one route is advertised by more than one routing protocol, the router will choose to use the routing protocol which has lowest Administrative Distance. The Administrative

Distances of popular routing protocols are listed below:

Question No: 19

Refer to the exhibit.

A user cannot reach any web sites on the Internet, but others in the department are not having a problem.

What is the most likely cause of the problem?

A. IP routing is not enabled.

B. The default gateway is not in the same subnet.

C. A DNS server address is not reachable by the PC.

D. A DHCP server address is not reachable by the PC.

E. NAT has not been configured on the router that connects to the Internet.

Answer: C

Explanation:

Answer C is only answer that makes sense. IP routing does not need to be enabled on PCu2019s, this is a router function. We can see from the output that the PC and default gateway are on the same subnet. DHCP has not been enabled on this PC so it has been configured with a static address so reaching the DHCP server is not the issue. Finally, NAT must be configured correctly or the other users in the department would also be having issues.

Question No: 20

Refer to the exhibit.

The MAC address table is shown in its entirety. The Ethernet frame that is shown arrives at the switch.

What two operations will the switch perform when it receives this frame? (Choose two.)

A. The switch will not forward a frame with this destination MAC address.

B. The MAC address of 0000.00aa.aaaa will be added to the MAC Address Table.

C. The MAC address of ffff.ffff.ffff will be added to the MAC address table.

D. The frame will be forwarded out of all the active switch ports except for port fa0/0.

E. The frame will be forwarded out of fa0/0 and fa0/1 only.

F. The frame will be forwarded out of all the ports on the switch.

Answer: B,D

Explanation:

If the switch already has the MAC address in its table for the destination, it will forward the frame directly to the destination port. If it was not already in its MAC table, then they frame would have been flooded out all ports except for the port that it came from.

Question No: 21

Refer to the exhibit.

Host A can communicate with Host B but not with Hosts C or D. How can the network administrator solve this problem?

A. Configure Hosts C and D with IP addresses in the 192.168.2.0 network.

B. Install a router and configure a route to route between VLANs 2 and 3.

C. Install a second switch and put Hosts C and D on that switch while Hosts A and B remain on the original switch.

D. Enable the VLAN trunking protocol on the switch.

Answer: B

Explanation:

Two VLANs require a router in between otherwise they cannot communicate. Different VLANs and different IP subnets need a router to route between them.

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