Cisco 100-105 Free Practice Questions

Answer: B 

Explanation: 

With the 224 there are 8 networks with increments of 32 One of these is 32 33 62 63 where 63 is broadcast so 62 is last valid host out of given choices. 

Answer: D 

Explanation: 

IPv6 has three types of addresses, which can be categorized by type and scope: 

Unicast addresses. A packet is delivered to one interface. 

Multicast addresses. A packet is delivered to multiple interfaces. 

Anycast addresses. A packet is delivered to the nearest of multiple interfaces (in terms of 

routing distance). 

IPv6 does not use broadcast messages. 

Unicast and anycast addresses in IPv6 have the following scopes (for multicast addresses, 

the scope are built into the address structure): 

Link-local. The scope is the local link (nodes on the same subnet). 

Site-local. The scope is the organization (private site addressing). 

Global. The scope is global (IPv6 Internet addresses). 

In addition, IPv6 has special addresses such as the loopback address. The scope of a 

special address depends on the type of special address. 

Much of the IPv6 address space is unassigned. 

http://technet.microsoft.com/en-us/library/cc757359(v=ws.10).aspx 

Answer: D 

Explanation: 

The “show ip ospf database” command displays the link states. Here is an example: 

Here is the lsa database on R2. 

R2#show ip ospf database 

OSPF Router with ID (2.2.2.2) (Process ID 1) 

Router Link States (Area 0) 

Link ID ADV Router Age Seq# Checksum Link count 

2.2.2.2 2.2.2.2 793 0x80000003 0x004F85 2 

10.4.4.4 10.4.4.4 776 0x80000004 0x005643 1 

111.111.111.111 111.111.111.111 755 0x80000005 0x0059CA 2 

133.133.133.133 133.133.133.133 775 0x80000005 0x00B5B1 2 Net Link States (Area 0) Link ID ADV Router Age Seq# Checksum10.1.1.1 111.111.111.111 794 0x80000001 0x001E8B 

10.2.2.3 133.133.133.133 812 0x80000001 0x004BA9 

10.4.4.1 111.111.111.111 755 0x80000001 0x007F16 

10.4.4.3 133.133.133.133 775 0x80000001 0x00C31F 

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 relationship—an adjacency—with 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 

Answer: B 

Explanation: 

Initially the MAC address had to be resolved, but once it is resolved and is in the ARP table of the router, pings go through immediately. 

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 

Answer: A 

Explanation: 

Your CCNA certification exam is likely going to contain questions about Telnet, an application-level protocol that allows remote communication between two networking devices. With Telnet use being as common as it is, you had better know the details of how to configure it in order to pass your CCNA exam and to work in real-world networks. 

The basic concept is pretty simple - we want to configure R1, but we're at R2. If we telnet successfully to R1, we will be able to configure R1 if we've been given the proper permission levels. In this CCNA case study, R2 has an IP address of 172.12.123.2 and R1 an address of 172.12.123.1. Let's try to telnet from R2 to R1. 

R2#telnet 172.12.123.1 Trying 172.12.123.1 ... Open Password required, but none set [Connection to 172.12.123.1 closed by foreign host] 

This seems like a problem, but it's a problem we're happy to have. A Cisco router will not let any user telnet to it by default. That's a good thing, because we don't want just anyone connecting to our router! The "password required" message means that no password has been set on the VTY lines on R1. Let's do so now. 

R1(config)#line vty 0 4 

R1(config-line)#password baseball 

A password of "baseball" has been set on the VTY lines, so we shouldn't have any trouble using Telnet to get from R2 to R1. Let's try that now. 

R2#telnet 172.12.123.1 

Trying 172.12.123.1 ... Open 

User Access Verification 

Password: 

R1> 

We're in, and placed into user exec mode. 

Reference: 

http://www.mcmcse.com/cisco/guides/telnet_passwords_and_privilege_levels.shtml 

Answer: B 

Explanation: 

Certain types of passwords, such as Line passwords, by default appear in clear text in the configuration file. You can use the service password-encryption command to make them more secure. Once this command is entered, each password configured is automatically encrypted and thus rendered illegible inside the configuration file (much as the Enable/Enable Secret passwords are). Securing Line passwords is doubly important in networks on which TFTP servers are used, because TFTP backup entails routinely moving config files across networks—and config files, of course, contain Line passwords. 

Answer: B 

Explanation: 

Hubs create single collision and broadcast domains, so in this case there will be a single collision domain for each of the two hubs. 

Answer: B,D,F 

Explanation: 

The router will forward the frames out the interface toward the destination – B is correct. Since the router will has the end station already in it’s MAC table as see by the “show arp” command, it will replace the destination MAC address to that of PC2 – D is correct. The router will then replace the source IP address to 172.16.40.1 – E is correct. 

Answer: C 

Explanation: 

SStatic EEGP DEIGRP RRIP OOSPF 

Default Administrative distance of EIGRP protocol is 90 then answer is C 

Default Distance Value TableThis table lists the administrative distance default values of the protocols that Cisco supports: 

Route Source Default Distance Values 

Connected interface 0 Static route 1 Enhanced Interior Gateway Routing Protocol (EIGRP) summary route 5 External Border Gateway Protocol (BGP) 20 Internal EIGRP 90 IGRP 100 OSPF 110 Intermediate System-to-Intermediate System (IS-IS) 115 Routing Information Protocol (RIP) 120 Exterior Gateway Protocol (EGP) 140 On Demand Routing (ODR) 160 External EIGRP 170 Internal BGP 200 Unknown* 255 

Answer: C 

Explanation: 

Administrative distance is the feature that routers use in order to select the best path when there are two or more different routes to the same destination from two different routing protocols. Administrative distance defines the reliability of a routing protocol. Each routing protocol is prioritized in order of most to least reliable (believable) with the help of an administrative distance value. 

Default Distance Value Table 

This table lists the administrative distance default values of the protocols that Cisco supports: 

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. 

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 

Answer: B