[Free] 2017(July) Ensurepass Testking Cisco 200-125 Latest Dumps 241-250

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CCNA Routing and Switching v3.0

QUESTION 241

Refer to the exhibit. The Bigtime router is unable to authenticate to the Littletime router. What is the cause of the problem?

 

clip_image002

 

A.

The usernames are incorrectly configured on the two routers.

B.

The passwords do not match on the two routers.

C.

CHAP authentication cannot be used on a serial interface.

D.

The routers cannot be connected from interface S0/0 to interface S0/0.

E.

With CHAP authentication, one router must authenticate to another router. The routers cannot be configured to authenticate to each other.

 

Correct Answer: B

Explanation:

With CHAP authentication, the configured passwords must be identical on each router. Here, it is configured as little123 on one side and big123 on the other.

 

 

QUESTION 242

Which statement describes VRRP object tracking?

 

A.

It monitors traffic flow and link utilization.

B.

It ensures the best VRRP router is the virtual router master for the group.

C.

It causes traffic to dynamically move to higher bandwidth links.

D.

It thwarts man-in-the-middle attacks.

 

Correct Answer: B

Explanation:

Object tracking is the process of tracking the state of a configured object and uses that state to determine the priority of the VRRP router in a VRRP group.

 

 

QUESTION 243

What are two benefits of using NAT? (Choose two.)

 

A.

NAT facilitates end-to-end communication when IPsec is enabled.

B.

NAT eliminates the need to re-address all hosts that require external access.

C.

NAT conserves addresses through host MAC-level multiplexing.

D.

Dynamic NAT facilitates connections from the outside of the network.

E.

NAT accelerates the routing process because no modifications are made on the packets.

F.

NAT protects network security because private networks are not advertised.

 

Correct Answer: BF

Explanation:

By not revealing the internal IP addresses, NAT adds some security to the inside network – > F is correct.

NAT has to modify the source IP addresses in the packets -> E is not correct.

Connection from the outside of the network through a “NAT” network is more difficult than a more network because IP addresses of inside hosts are hidden -> C is not correct.

In order for IPsec to work with NAT we need to allow additional protocols, including Internet Key Exchange (IKE), Encapsulating Security Payload (ESP) and Authentication Header (AH) -> more complex -> A is not correct.

By allocating specific public IP addresses to inside hosts, NAT eliminates the need to re- address the inside hosts -> B is correct.

NAT does conserve addresses but not through host MAC-level multiplexing. It conserves addresses by allowing many private IP addresses to use the same public IP address to go to the Internet -> C is not correct.

 

 

QUESTION 244

Given an IP address 172.16.28.252 with a subnet mask of 255.255.240.0, what is the correct network address?

 

A.

172.16.16.0

B.

172.16.0.0

C.

172.16.24.0

D.

172.16.28.0

 

Correct Answer: A

Explanation:

For this example, the network range is 172.16.16.1 – 172.16.31.254, the network address is 172.16.16.0 and the broadcast IP address is 172.16.31.255.

 

 

QUESTION 245

Which two statements describe characteristics of IPv6 unicast addressing? (Choose two.)

 

A.

Global addresses start with 2000::/3.

B.

Link-local addresses start with FE00:/12.

C.

Link-local addresses start with FF00::/10.

D.

There is only one loopback address and it is ::1.

E.

If a global address is assigned to an interface, then that is the only allowable address for the interface.

 

Correct Answer: AD

Explanation:

Below is the list of common kinds of IPv6 addresses:

 

clip_image003

 

 

QUESTION 246

What is the alternative notation for the IPv6 address B514:82C3:0000:0000:0029:EC7A:0000:EC72?

 

A.

B514 : 82C3 : 0029 : EC7A : EC72

B.

B514 : 82C3 :: 0029 : EC7A : EC72

C.

B514 : 82C3 : 0029 :: EC7A : 0000 : EC72

D.

B514 : 82C3 :: 00
29 : EC7A : 0 : EC72

 

Correct Answer: D

Explanation:

There are two ways that an IPv6 address can be additionally compressed: compressing leading zeros and substituting a group of consecutive zeros with a single double colon (::). Both of these can be used in any number of combinations to notate the same address. It is important to note that the double colon (::) can only be used once within a single IPv6 address notation. So, the extra 0’s can only be compressed once.

 

 

QUESTION 247

What are three components that comprise the SNMP framework? (Choose three.)

 

A.

MIB

B.

agent

C.

set

D.

AES

E.

supervisor

F.

manager

 

Correct Answer: ABF

Explanation:

The SNMP framework consists of three parts:

 

clip_image005An SNMP manager — The system used to control and monitor the activities of network devices using SNMP.

clip_image005[1]An SNMP agent — The software component within the managed device that maintains the data for the device and reports these data, as needed, to managing systems. Cisco Nexus 1000V supports the agent and MIB. To enable the SNMP agent, you must define the relationship between the manager and the agent.

clip_image005[2]A managed information base (MIB) — The collection of managed objects on the SNMP agent.

 

SNMP is defined in RFCs 3411 to 3418.

 

Reference: http://www.cisco.com/c/en/us/td/docs/switches/datacenter/nexus1000/sw/4_0_4_s_v_1_3/system_management/configuration/guide/n1000v_system/n1000v_system_10snmp.html

 

 

QUESTION 248

What SNMP message alerts the manager to a condition on the network?

 

A.

response

B.

get

C.

trap

D.

capture

 

Correct Answer: C

Explanation:

An agent can send unsolicited traps to the manager. Traps are messages alerting the SNMP manager to a condition on the network. Traps can mean improper user authentication, restarts, link status (up or down), MAC address tracking, closing of a TCP connection, loss of connection to a neighbor, or other significant events.

Reference: http://www.cisco.com/c/en/us/td/docs/switches/lan/catalyst2950/software/release/12-1_9_ea1/configuration/guide/scg/swsnmp.html

 

 

QUESTION 249

Which two are features of IPv6? (Choose two.)

 

A.

anycast

B.

broadcast

C.

multicast

D.

podcast

E.

allcast

 

Correct Answer: AC

Explanation:

IPv6 addresses are classified by the primary addressing and routing methodologies common in networking: unicast addressing, anycast addressing, and multicast addressing. A unicast address identifies a single network interface. The Internet Protocol delivers packets sent to a unicast address to that specific interface. An anycast address is assigned to a group of interfaces, usually belonging to different nodes. A packet sent to an anycast address is delivered to just one of the member interfaces, typically the nearest host, according to the routing protocol’s definition of distance. Anycast addresses cannot be identified easily, they have the same format as unicast addresses, and differ only by their presence in the network at multiple points. Almost any unicast address can be employed as an anycast address.

A multicast address is also used by multiple hosts, which acquire the multicast address destination by participating in the multicast distribution protocol among the network routers. A packet that is sent to a multicast address is delivered to all interfaces that have joined the corresponding multicast group.

 

 

QUESTION 250

The network administrator is asked to configure 113 point-to-point links. Which IP addressing scheme defines the address range and subnet mask that meet the requirement and waste the fewest subnet and host addresses?

 

A.

10.10.0.0/16 subnetted with mask 255.255.255.252

B.

10.10.0.0/18 subnetted with mask 255.255.255.252

C.

10.10.1.0/24 subnetted with mask 255.255.255.252

D.

10.10.0.0/23 subnetted with mask 255.255.255.252

E.

10.10.1.0/25 subnetted with mask 255.255.255.252

 

Correct Answer: D

Explanation:

We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits (because 2^7 = 128).

The network used for point-to-point connection should be /30.

So our initial network should be 30 – 7 = 23.

So 10.10.0.0/23 is the correct answer.

You can understand it more clearly when writing it in binary form:

/23 = 1111 1111.1111 1110.0000 0000

/30 = 1111 1111.1111 1111.1111 1100 (borrow 7 bits)

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