Exam Code: 100-105 (Practice Exam Latest Test Questions VCE PDF)
Exam Name: Cisco Interconnecting Cisco Networking Devices Part 1 (ICND1 v3.0)
Certification Provider: Cisco
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Q1. - (Topic 3)
Refer to the topology. Your company has decided to connect the main office with three other remote branch offices using point-to-point serial links.
You are required to troubleshoot and resolve OSPF neighbor adjacency issues between the main office and the routers located in the remote branch offices.
R1 does not form an OSPF neighbor adjacency with R2. Which option would fix the issue?
A. R1 ethernetO/1 is shutdown. Configure no shutdown command.
B. R1 ethernetO/1 configured with a non-default OSPF hello interval of 25: configure no ip ospf hello-interval 25
C. R2 ethernetO/1 and R3 ethernetO/O are configured with a non-default OSPF hello interval of 25; configure no ip ospf hello-interval 25
D. Enable OSPF for R1 ethernetO/1; configure ip ospf 1 area 0 command under ethernetO/1
Looking at the configuration of R1, we see that R1 is configured with a hello interval of 25 on interface Ethernet 0/1 while R2 is left with the default of 10 (not configured).
Q2. CORRECT TEXT - (Topic 6)
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 220.127.116.11 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: 18.104.22.168 Broadcast address: 22.214.171.124 (because 126.96.36.199 is the second subnetwork, so the previous IP - 188.8.131.52 - is the broadcast address of the first subnet). -> The second assignable host address of this subnetwork is 184.108.40.206/27 Assign the second assignable host address to Fa0/0 interface of Apopka router: Apopka(config)#interface Fa0/0 Apopka(config-if)#ip address 220.127.116.11 255.255.255.224 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 18.104.22.168 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
Q3. - (Topic 3)
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?
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
Q4. - (Topic 3)
Refer to the exhibit.
A network associate has configured OSPF with the command:
City(config-router)# network 192.168.12.64 0.0.0.63 area 0
After completing the configuration, the associate discovers that not all the interfaces are participating in OSPF. Which three of the interfaces shown in the exhibit will participate in OSPF according to this configuration statement? (Choose three.)
A. FastEthernet0 /0
B. FastEthernet0 /1
The “network 192.168.12.64 0.0.0.63 equals to network 192.168.12.64/26. This network has:
Increment: 64 (/26= 1111 1111.1111 1111.1111 1111.1100 0000)
Network address: 192.168.12.64
Broadcast address: 192.168.12.127
Therefore all interface in the range of this network will join OSPF.
Q5. - (Topic 5)
The hosts in the LAN are not able to connect to the Internet. Which commands will correct this issue?
A. Option A
B. Option B
C. Option C
D. Option D
E. Option E
Do a “show ip int brief” and you will see that Fa0/1 has an IP address assigned, but it is shut down.
Improved 100-105 pdf exam:
Q6. - (Topic 3)
R1 is configured with the default configuration of OSPF. From the following list of IP addresses configured on R1, which address will the OSPF process select as the router ID?
The Router ID (RID) is an IP address used to identify the router and is chosen using the following sequencE.
The highest IP address assigned to a loopback (logical) interface. + If a loopback interface is not defined, the highest IP address of all active router's physical interfaces will be chosen.
The router ID can be manually assigned In this case, because a loopback interface is not configured so the highest active IP address 192.168.0.1 is chosen as the router ID.
Q7. - (Topic 5)
If a host experiences intermittent issues that relate to congestion within a network while remaining connected, what could cause congestion on this LAN?
A. half-duplex operation
B. broadcast storms
C. network segmentation
A broadcast storm can consume sufficient network resources so as to render the network unable to transport normal traffic.
Topic 6, Simulation
Q8. - (Topic 1)
Refer to the exhibit.
The host in Kiev sends a request for an HTML document to the server in Minsk. What will be the source IP address of the packet as it leaves the Kiev router?
Although the source and destination MAC address will change as a packet traverses a network, the source and destination IP address will not unless network address translation (NAT) is being done, which is not the case here.
Q9. - (Topic 5)
An administrator has connected devices to a switch and, for security reasons, wants the dynamically learned MAC addresses from the address table added to the running configuration.
What must be done to accomplish this?
A. Enable port security and use the keyword sticky.
B. Set the switchport mode to trunk and save the running configuration.
C. Use the switchport protected command to have the MAC addresses added to the configuration.
D. Use the no switchport port-security command to allow MAC addresses to be added to the configuration.
One can configure MAC addresses to be sticky. These can be dynamically learned or manually configured, stored in the address table, and added to the running configuration. If these addresses are saved in the configuration file, the interface does not need to dynamically relearn them when the switch restarts, hence enabling security as desired.
Q10. - (Topic 3)
Refer to the exhibit.
The two routers have had their startup configurations cleared and have been restarted. At a minimum, what must the administrator do to enable CDP to exchange information between R1 and R2?
A. Configure the router with the cdp enable command.
B. Enter no shutdown commands on the R1 and R2 fa0/1 interfaces.
C. Configure IP addressing and no shutdown commands on both the R1 and R2 fa0/1 interfaces.
D. Configure IP addressing and no shutdown commands on either of the R1 or R2 fa0/1 interfaces.
If the no shut down commands are not entered, then CDP can exchange information between the two routers. By default, all Cisco device interfaces and ports are shut down and need to be manually enabled.