Download Cisco.PracticeDumps.200-125.2017-09-17.1e.346q.vcex

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Exam	CCNA Cisco Certified Network Associate CCNA (v3.0)
Number	200-125
File Name	Cisco.PracticeDumps.200-125.2017-09-17.1e.346q.vcex
Size	14.66 Mb
Posted	September 17, 2017
Downloads	154

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Demo Questions

Question 1

Refer to the exhibit. What will Router1 do when it receives the data frame shown? (Choose three.)

A: Router1 will strip off the source MAC address and replace it with the MAC address 0000.0c36.6965.
B: Router1 will strip off the source IP address and replace it with the IP address 192.168.40.1.
C: Router1 will strip off the destination MAC address and replace it with the MAC address 0000.0c07.4320.
D: Router1 will strip off the destination IP address and replace it with the IP address of 192.168.40.1.
E: Router1 will forward the data packet out interface FastEthernet0/1.
F: Router1 will forward the data packet out interface FastEthernet0/2.

Correct Answer: ACF

The router will decapsulate the frame, thus removing the source and destination MAC addresses. It will consult its routing table to work out on which interface to send the packet which is going to 192.168.40.5: interface FastEthernet0/2. The router will then encapsulate the packet into a new frame with its own MAC address as the source MAC address and the MAC address of 192.168.40.5 as the destination MAC address.
Looking at the ARP table, the MAC address for 192.168.40.5 is 0000.0c07.4320 (the one which has been dynamically learned and has age 9 minutes). So what is the router’s MAC address on this link? The ARP table shows a static entry (no age) for 192.168.40.1 which is 0000.0c36.6965. As it is static, we can conclude that these are the router’s IP and MAC address.

Question 2

Refer to the exhibit. Which three statements correctly describe Network Device A? (Choose three.)

A: With a network wide mask of 255.255.255.128, each interface does not require an IP address.
B: With a network wide mask of 255.255.255.128, each interface does require an IP address on a unique IP subnet.
C: With a network wide mask of 255.255.255.0, it must be a Layer 2 device for the PCs to communicate with each other.
D: With a network wide mask of 255.255.255.0, it must be a Layer 3 device for the PCs to communicate with each other.
E: With a network wide mask of 255.255.254.0, each interface does not require an IP address.

Correct Answer: BDE

If the device is a Layer 2 transparent device (hub, bridge, switch) then the two PCs must be on the same subnet. The smallest subnet which would contain both of the IP addresses is 10.1.0.0/23 with the range 10.1.0.0 up to 10.1.1.255. The mask for this is 255.255.254.0. On the other hand, if the device is a router, then we would need a netmask smaller than 255.255.254.0 and we would have two subnets.
The masks of 255.255.255.128 and 255.255.255.0 are both smaller than 255.255.254.0, so this rules out A and C as the answer and confirms B and D as the answer. E is the situation where we have the smallest subnet that contains both IP addresses, so we can have a Layer 2 device which does not require an IP address.

Question 3

Which layer in the OSI reference model is responsible for determining the availability of the receiving program and checking to see if enough resources exist for that communication?

A: transport
B: network
C: presentation
D: session
E: application

Correct Answer: E

Programs live at the top end of the OSI model, so this rules out the network (delivery of packets across multiple links) and transport (end-to-end reliability and ports). The presentation layer is concerned with the representation of data. The session layer provides the mechanism for opening, closing and managing a session between end-user application processes, i.e., a semi-permanent dialogue (Wikipedia).
Wikipedia says: In the OSI model, the definition of its application layer is narrower in scope. The OSI model defines the application layer as being the user interface. The OSI application layer is responsible for displaying data and images to the user in a human-recognizable format. So the best answer is the application layer, as this will see if there is a program (user interface) available to receive communications.

Question 4

Which of the following describes the roles of devices in a WAN? (Choose three.)

A: A CSU/DSU terminates a digital local loop.
B: A modem terminates a digital local loop.
C: A CSU/DSU terminates an analog local loop.
D: A modem terminates an analog local loop.
E: A router is commonly considered a DTE device.
F: A router is commonly considered a DCE device.

Correct Answer: ADE

Modems and CSU/DSU are considered DCE: data communications equipment, parts of the WAN itself. Routers are considered DTE: data terminating equipment, right on the edge of the WAN and not really part of it. Modems are considered to be analog devices and terminate an analog local loop. CSU/DSU devices are considered to be digital and terminate a digital local loop.

Question 5

Refer to the exhibit.

Host A pings interface S0/0 on router 3. What is the TTL value for that ping?

A: 252
B: 253
C: 254
D: 255

Correct Answer: B

The default TTL is 255, and as a packet passes through a router the TTL is decremented. By the time the ping packet reaches S0/0 on Router 3, it has passed through two routers and the TTL is now 253. Note that the S0/0 interface responds to the ping: the packet is not routed by Router 3 and so the TTL in the ping packet is not decremented to 252.

Question 6

A network administrator is verifying the configuration of a newly installed host by establishing an FTP connection to a remote server.
What is the highest layer of the protocol stack that the network administrator is using for this operation?

A: application
B: presentation
C: session
D: transport
E: internet
F: data link

Correct Answer: A

FTP is an application layer protocol, so this is the highest OSI layer involved in the operation.

Question 7

Refer to the exhibit. After HostA pings HostB, which entry will be in the ARP cache of HostA to support this transmission?

Correct Answer: D

Hosts A and B are on different subnets and are separated by routers, so the MAC address of HostB will never get back to HostA: Router2 will decapsulate the frames that HostB sends. Router1 will re-encapsulate the ping packet sent by HostB and put its own MAC address in the frame. Therefore, HostA will put 192.168.6.1 and 000f.2480.8916 into its ARP table as a dynamic entry.

Question 8

A network interface port has collision detection and carrier sensing enabled on a shared twisted pair network. From this statement, what is known about the network interface port?

A: This is a 10 Mb/s switch port.
B: This is a 100 Mb/s switch port.
C: This is an Ethernet port operating at half duplex.
D: This is an Ethernet port operating at full duplex.
E: This is a port on a network interface card in a PC.

Correct Answer: C

The only time that CSMA/CD (carrier sense medium access with collision detection) is ever needed is when the medium is shared by all devices. In this situation, devices can never send and receive at the same time; they must operate in half-duplex mode. The bitrate is irrelevant.

Question 9

A receiving host computes the checksum on a frame and determines that the frame is damaged. The frame is then discarded. At which OSI layer did this happen?

A: physical
B: session
C: data link
D: transport
E: network

Correct Answer: C

Operations on frames occur at the data link layer.

Question 10

Which of the following correctly describe steps in the OSI data encapsulation process? (Choose two)

A: The transport layer divides a data stream into segments and may add reliability and flow control information.
B: The data link layer adds physical source and destination addresses and an FCS to the segment.
C: Packets are created when the network layer encapsulates a frame with source and destination host addresses and protocol-related control information.
D: Packets are created when the network layer adds Layer 3 addresses and control information to a segment.
E: The presentation layer translates bits into voltages for transmission across the physical link.

Correct Answer: AD

The key word here is encapsulation. The transport layer breaks the stream into segments, with ports, reliability and flow control fields added (TCP) (A). The network layer encapsulates the segment (not a frame) into packets by adding layer 3 addresses (D). The data link layer adds physical source and destination addresses and an FCS to a packet, not a segment.

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