100-105 Exam
Point Checklist: icnd1 100 105
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Q121. - (Topic 5)
Which of the following is true regarding the configuration of SwitchA?
A. only 5 simultaneous remote connections are possible
B. remote connections using ssh will require a username and password
C. only connections from the local network will be possible
D. console access to SwitchA requires a password
Answer: B
Q122. - (Topic 7)
Which dynamic routing protocol uses only the hop count to determine the best path to a destination?
A. IGRP
B. RIP
C. EIGRP
D. OSPF
Answer: C
Q123. - (Topic 4)
In the configuration of NAT, what does the keyword overload signify?
A. When bandwidth is insufficient, some hosts will not be allowed to access network translation.
B. The pool of IP addresses has been exhausted.
C. Multiple internal hosts will use one IP address to access external network resources.
D. If the number of available IP addresses is exceeded, excess traffic will use the specified address pool.
Answer: C
Explanation:
The keyword overload.used in the ip nat inside source list 1 pool ovrld overload example command allows NAT to translate multiple inside devices to the single address in the pool. The types of NAT include: Static address translation (static NAT)—Allows one-to-one mapping between local and global addresses. Dynamic address translation (dynamic NAT)—Maps unregistered IP addresses to registered IP addresses from a pool of registered IP addresses. Overloading—Maps multiple unregistered IP addresses to a single registered IP address (many to one) using different ports. This method is also known as Port Address Translation (PAT). By using overloading, thousands of users can be connected to the Internet by using only one real global IP address.
Q124. - (Topic 7)
Which statement about a router on a stick is true?
A. Its date plane router traffic for a single VI AN over two or more switches.
B. It uses multiple subinterfaces of a single interface to encapsulate traffic for different VLANs on the same subnet.
C. It requires the native VLAN to be disabled.
D. It uses multiple subinterfaces of a single interface to encapsulate traffic for different VLANs.
Answer: D
Explanation:
https://www.freeccnaworkbook.com/workbooks/ccna/configuring-inter-vlan-routing-router-on-a-stick
Q125. - (Topic 3)
Refer to the exhibit.
The enterprise has decided to use the network address 172.16.0.0. The network administrator needs to design a classful addressing scheme to accommodate the three subnets, with 30, 40, and 50 hosts, as shown. What subnet mask would accommodate this network?
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.255.240
D. 255.255.255.248
E. 255.255.255.252
Answer: A
Explanation:
Subnet mask A i.e. 255.255.255.192 with CIDR of /26 which means 64 hosts per subnet which are sufficient to accommodate even the largest subnet of 50 hosts.
Q126. - (Topic 3)
On a live network, which commands will verify the operational status of router interfaces? (Choose two.)
A. Router# show interfaces
B. Router# show ip protocols
C. Router# debug interface
D. Router# show ip interface brief
E. Router# show start
Answer: A,D
Explanation:
Only two commands “show interfaces” and “show ip interface brief” reveal the status of router interfaces (up/up, for example).
Q127. - (Topic 5)
Refer to the exhibit.
After configuring two interfaces on the HQ router, the network administrator notices an error message. What must be done to fix this error?
A. The serial interface must be configured first.
B. The serial interface must use the address 192.168.1.2
C. The subnet mask of the serial interface should be changed to 255.255.255.0
D. The subnet mask of the FastEthernet interface should be changed to 255.255.255.240
E. The address of the FastEthernet interface should be changed to 192.168.1.66
Answer: D
Explanation:
The IP address 192.168.1.17 255.255.255.0 specifies that the address is part of the 192.168.1.0/24 subnet
24 mask bits = 255.255.255.0 28 mask bits = 255.255.255.240 192.168.1.0/24 subnet has a host range of 192.168.1.1 to 192.168.1.254 (0 being network and 255 being broadcoast) 192.168.1.17/28 subnet has a host range of 192.168.1.17 to 192.168.1.30 (16 being network and 31 being broadcast) 192.168.1.65/28 subnet has a host range of 192.168.1.65 - 192.168.1.78 (64 being network and 79 being broadcast)
if fa0/0 was left as /24, you can see that the host range includes the host range of 192.168.1.64/28 which conflicts. Simply speaking, you can't overlap the subnets. By changing the subnet mask of fa0/0 to 255.255.255.240, these networks would no longer overlap.
Q128. DRAG DROP - (Topic 4)
Various protocols are listed on the left. On the right are applications for the use of those protocols. Drag the protocol on the left to an associated function for that protocol on the right. (Not all options are used.)
Answer:
Q129. - (Topic 1)
Which protocol uses a connection-oriented service to deliver files between end systems?
A. TFTP
B. DNS
C. FTP
D. SNMP
E. RIP
Answer: C
Explanation:
TCP is an example of a connection-oriented protocol. It requires a logical connection to be established between the two processes before data is exchanged. The connection must be maintained during the entire time that communication is taking place, then released afterwards. The process is much like a telephone call, where a virtual circuit is established--the caller must know the person's telephone number and the phone must be answered--before the message can be delivered. TCP/IP is also a connection-oriented transport with orderly release. With orderly release, any data remaining in the buffer is sent before the connection is terminated. The release is accomplished in a three-way handshake between client and server processes. The connection-oriented protocols in the OSI protocol suite, on the other hand, do not support orderly release. Applications perform any handshake necessary for ensuring orderly release. Examples of services that use connection-oriented transport services are telnet, rlogin, and ftp.
Q130. - (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: C
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.