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Q71. - (Topic 3) 

Which statements are TRUE regarding Internet Protocol version 6 (IPv6) addresses? (Choose three.) 

A. An IPv6 address is divided into eight 16-bit groups. 

B. A double colon (::) can only be used once in a single IPv6 address. 

C. IPv6 addresses are 196 bits in length. 

D. Leading zeros cannot be omitted in an IPv6 address. 

E. Groups with a value of 0 can be represented with a single 0 in IPv6 address. 

Answer: A,B,E 

Explanation: 

IPv6 addresses are divided into eight 16-bit groups, a double colon (::) can only be used 

once in an IPv6 address, and groups with a value of 0 can be represented with a single 0 in 

an IPv6 address. 

The following statements are also true regarding IPv6 address: 

IPv6 addresses are 128 bits in length. 

Eight 16-bit groups are divided by a colon (:). 

Multiple groups of 16-bit 0s can be represented with double colon (::). 

Double colons (::) represent only 0s. 

Leading zeros can be omitted in an IPv6 address. 

The option stating that IPv6 addresses are 196 bits in length is incorrect. IPv6 addresses 

are 128 bits in length. 

The option stating that leading zeros cannot be omitted in an IPv6 address is incorrect. 

Leading zeros can be omitted in an IPv6 address. 


Q72. - (Topic 3) 

What is the subnet address of 172.16.159.159/22? 

A. 172.16.0.0 

B. 172.16.128.0 

C. 172.16.156.0 

D. 172.16.159.0 

E. 172.16.159.128 

F. 172.16.192.0 

Answer:

Explanation: 

Converting to binary format it comes to 11111111.11111111.11111100.00000000 or 

255.255.252.0 Starting with 172.16.0.0 and having increment of 4 we get. 


Q73. - (Topic 5) 

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:

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. 


Q74. - (Topic 3) 

Which of the following describe the process identifier that is used to run OSPF on a router? (Choose two) 

A. It is locally significant. 

B. It is globally significant. 

C. It is needed to identify a unique instance of an OSPF database. 

D. It is an optional parameter required only if multiple OSPF processes are running on the router. 

E. All routers in the same OSPF area must have the same process ID if they are to exchange routing information. 

Answer: A,C 

Explanation: 

https://learningnetwork.cisco.com/thread/6248 They are locally significant only, and have no bearing on the structure of any OSPF packet or LSA update. So you can have a separate process-id on every single router in your network if you so desire. 


Q75. - (Topic 5) 

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:

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. 


Q76. - (Topic 7) 

Which component of a routing table entry represents the subnet mask? 

A. routing protocol code 

B. prefix 

C. metric 

D. network mask 

Answer:

Explanation: 

IP Routing Table Entry TypesAn entry in the IP routing table contains the following information in the order presented: Network ID. The network ID or destination corresponding to the route. The network ID can be class-based, subnet, or supernet network ID, or an IP address for a host route. Network Mask. The mask that is used to match a destination IP address to the network ID. Next Hop. The IP address of the next hop. Interface. An indication of which network interface is used to forward the IP packet. Metric. A number used to indicate the cost of the route so the best route among possible multiple routes to the same destination can be selected. A common use of the metric is to indicate the number of hops (routers crossed) to the network ID. Routing table entries can be used to store the following types of routes: Directly Attached Network IDs. Routes for network IDs that are directly attached. For directly attached networks, the Next Hop field can be blank or contain the IP address of the interface on that network. Remote Network IDs. Routes for network IDs that are not directly attached but are available across other routers. For remote networks, the Next Hop field is the IP address of a local router in between the forwarding node and the remote network. Host Routes. A route to a specific IP address. Host routes allow routing to occur on a per-IP address basis. For host routes, the network ID is the IP address of the specified host and the network mask is 255.255.255.255. Default Route. The default route is designed to be used when a more specific network ID or host route is not found. The default route network ID is 0.0.0.0 with the network mask of 

0.0.0.0. 


Q77. - (Topic 3) 

What is the network address for the host with IP address 192.168.23.61/28? 

A. 192.168.23.0 

B. 192.168.23.32 

C. 192.168.23.48 

D. 192.168.23.56 

E. 192.168.23.60 

Answer:

Explanation: 

Convert bit-length prefix to quad-dotted decimal representation, then from it find the number of bits used for subnetting you can find previously calculated number of subnets by separating subnets each having value of last bit used for subnet masking Find that your IP address is in which subnet, that subnet's first address is network address and last address is broadcast address. Based on above steps the answer is option C 


Q78. CORRECT TEXT - (Topic 6) 

Answer: Router>enable 

Router#config terminal Router(config)#hostname Apopka 2) Enable-secret password (cisco10): Apopka(config)#enable secret cisco10 3) Set the console password to RouterPass: Apopka(config)#line console 0 Apopka(config-line)#password RouterPass Apopka(config-line)#login Apopka(config-line)#exit 4) Set the Telnet password to scan90: Apopka(config)#line vty 0 4 Apopka(config-line)#password scan90 Apopka(config-line)#login Apopka(config-line)#exit 5) Configure Ethernet interface (on the right) of router Apopka: The subnet mask of the Ethernet network 209.165.201.0 is 27. From this subnet mask, we can find out the increment by converting it into binary form, that is /27 = 1111 1111.1111 1111.1111 1111.1110 0000. Pay more attention to the last bit 1 because it tells us the increment, using the formula: Increment = 2place of the last bit 1 (starts counting from 0,from right to left), in this case increment = 25 = 32. Therefore: Increment: 32 Network address: 209.165.201.0 Broadcast address: 209.165.201.31 (because 209.165.201.32 is the second subnetwork, so the previous IP - 209.165.201.31 - is the broadcast address of the first subnet). -> The second assignable host address of this subnetwork is 209.165.201.2/27 Assign the second assignable host address to Fa0/0 interface of Apopka router: Apopka(config)#interface Fa0/0 Apopka(config-if)#ip address 209.165.201.2 255.255.255.224 Apopka(config-if)#no shutdown Apopka(config-if)#exit 6) Configure Serial interface (on the left) of router Apopka: Using the same method to find out the increment of the Serial network: Serial network 192.0.2.128/28: Increment: 16 (/28 = 1111 1111.1111 1111.1111 1111.1111 0000) Network address: 192.0.2.128 (because 8 * 16 = 128 so 192.0.2.128 is also the network address of this subnet) Broadcast address: 192.0.2.143 -> The last assignable host address in this subnet is 192.0.2.142/28. Assign the last assignable host address to S0/0/0 interface of Apopka router: Apopka(config)#interface S0/0/0 (or use interface S0/0 if not successful) Apopka(config-if)#ip address 192.0.2.142 255.255.255.240 Apopka(config-if)#no shutdown Apopka(config-if)#exit 7) Configure RIP v2 routing protocol: Apopka(config)#router rip Apopka(config-router)#version 2 Apopka(config-router)#network 209.165.201.0 Apopka(config-router)#network 192.0.2.128 Apopka(config-router)#end Save the configuration: Apopka#copy running-config startup-config Finally, you should use the ping command to verify all are working properly! 

Topic 7, Mix Questions 


Q79. - (Topic 3) 

An administrator must assign static IP addresses to the servers in a network. For network 192.168.20.24/29, the router is assigned the first usable host address while the sales server is given the last usable host address. 

Which of the following should be entered into the IP properties box for the sales server? 

A. IP address: 192.168.20.14 Subnet Mask: 255.255.255.248 Default Gateway: 192.168.20.9 

B. IP address: 192.168.20.254 Subnet Mask: 255.255.255.0 Default Gateway: 192.168.20.1 

C. IP address: 192.168.20.30 Subnet Mask: 255.255.255.248 Default Gateway: 192.168.20.25 

D. IP address: 192.168.20.30 Subnet Mask: 255.255.255.240 Default Gateway: 192.168.20.17 

E. IP address: 192.168.20.30 Subnet Mask: 255.255.255.240 Default Gateway: 192.168.20.25 

Answer:

Explanation: 

With network 192.168.20.24/29 we have: 

Increment: 8 (/29 = 255.255.255.248 = 11111000 for the last octet) 

Network address: 192.168.20.24 (because 24 = 8 * 3) 

Broadcast address: 192.168.20.31 (because 31 = 24 + 8 – 1) 

Therefore the first usable IP address is 192.168.20.25 (assigned to the router) and the last usable IP address is 192.168.20.30 (assigned to the sales server). The IP address of the router is also the default gateway of the sales server. 


Q80. - (Topic 3) 

Which two commands will display the current IP address and basic Layer 1 and 2 status of an interface? (Choose two.) 

A. router#show version 

B. router#show ip interface 

C. router#show protocols 

D. router#show controllers 

E. router#show running-config 

Answer: B,C 

Explanation: 

The outputs of “show protocols” and “show ip interface” are shown below: 

Global values:Internet Protocol routing is enabledSerial0/0 is up, line protocol is downInternet address is 10.1.1.1/30Serial0/1 is up, line protocol is downInternet address is 209.65.200.225/30Serial0/2 is up, line protocol is downSerial0/3 is up, line protocol is downNVI0 is up, line protocol is upInterface is unnumbered. Using address of NVI0 (0.0.0.0)Loopback0 is up, line protocol is upInternet address is 10.1.10.1/32Loopback1 is up, line protocol is upInternet address is 10.1.2.1/27Loopback6 is up, line protocol is up 

Serial0/0 is up, line protocol is downInternet address is 10.1.1.1/30Broadcast address is 255.255.255.255Address determined by non-volatile memoryMTU is 1500 bytesHelper address is not setDirected broadcast forwarding is disabledMulticast reserved groups joined: 224.0.0.5Outgoing access list is not setInbound access list is not setProxy ARP is enabledLocal Proxy ARP is disabledSecurity level is defaultSplit horizon is disabledICMP redirects are always sentICMP unreachables are always sentICMP mask replies are never sentIP fast switching is enabledIP fast switching on the same interface is enabledIP Flow switching is disabledIP CEF switching is disabledIP Feature Fast switching turbo vectorIP multicast fast switching is enabledIP multicast distributed fast switching is disabledIP route-cache flags are FastRouter Discovery is disabledIP output packet accounting is disabledIP access violation accounting is disabledTCP/IP header compression is disabledRTP/IP header compression is disabledPolicy routing is disabledNetwork address translation is enabled, interface in domain insideBGP Policy Mapping is disabledWCCP Redirect outbound is disabledWCCP Redirect inbound is disabledWCCP Redirect exclude is disabled