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New Cisco 200-125 Exam Dumps Collection (Question 13 - Question 22)

Q13. Which three statements about HSRP operation are true? (Choose three.)

A. The virtual IP address and virtual MA+K44C address are active on the HSRP Master router.

B. The HSRP default timers are a 3 second hello interval and a 10 second dead interval.

C. HSRP supports only clear-text authentication.

D. The HSRP virtual IP address must be on a different subnet than the routers' interfaces on the same LAN.

E. The HSRP virtual IP address must be the same as one of the router's interface addresses on the LAN.

F. HSRP supports up to 255 groups per interface, enabling an administrative form of load balancing.

Answer: A,B,F

Explanation:

The virtual MAC address of HSRP version 1 is 0000.0C07.ACxx, where xx is the HSRP

group number in hexadecimal based on the respective interface. For example, HSRP group 10 uses the HSRP virtual MAC address of 0000.0C07.AC0A. HSRP version 2 uses a virtual MAC address of 0000.0C9F.FXXX (XXX: HSRP group in hexadecimal).


Q14. Which three of these statements regarding 802.1Q trunking are correct? (Choose three.)

A. 802.1Q native VLAN frames are untagged by default.

B. 802.1Q trunking ports can also be secure ports.

C. 802.1Q trunks can use 10 Mb/s Ethernet interfaces.

D. 802.1Q trunks require full-duplex, point-to-point connectivity.

E. 802.1Q trunks should have native VLANs that are the same at both ends.

Answer: A,C,E

Explanation:

By default, 802.1Q trunk defined Native VLAN in order to forward unmarked frame. Switches can forward Layer 2 frame from Native VLAN on unmarked trunks port. Receiver

switches will transmit all unmarked packets to Native VLAN. Native VLAN is the default VLAN configuration of port. Note for the 802.1Q trunk ports between two devices, the same Native VLAN configuration is required on both sides of the link. If the Native VLAN in 802.1Q trunk ports on same trunk link is properly configured, it could lead to layer 2 loops. The 802.1Q trunk link transmits VLAN information through Ethernet.


Q15. Refer to the exhibit.

A new subnet with 60 hosts has been added to the network. Which subnet address should this network use to provide enough usable addresses while wasting the fewest addresses?

A. 192.168.1.56/26

B. 192.168.1.56/27

C. 192.168.1.64/26

D. 192.168.1.64/27

Answer: C

Explanation:

A subnet with 60 host is 2*2*2*2*2*2 = 64 -2 == 62

6 bits needed for hosts part. Therefore subnet bits are 2 bits (8-6) in fourth octet. 8bits+ 8bits+ 8bits + 2bits = /26

/26 bits subnet is 24bits + 11000000 = 24bits + 192 256 u2013 192 = 64

0 -63

64 u2013 127


Q16. Refer to the exhibit.

Based on the exhibited routing table, how will packets from a host within the 192.168.10.192/26 LAN be forwarded to 192.168.10.1?

A. The router will forward packets from R3 to R2 to R1.

B. The router will forward packets from R3 to R1 to R2.

C. The router will forward packets from R3 to R2 to R1 AND from R3 to R1.

D. The router will forward packets from R3 to R1.

Answer: C

Explanation:

From the routing table we learn that network 192.168.10.0/30 is learned via 2 equal-cost paths (192.168.10.9 &192.168.10.5) -> traffic to this network will be load-balancing.


Q17. What are three features of the IPv6 protocol? (Choose three.)

A. optional IPsec

B. autoconfiguration

C. no broadcasts

D. complicated header

E. plug-and-play

F. checksums

Answer: B,C,E

Explanation:

An important feature of IPv6 is that it allows plug and play option to the network devices by allowing them to configure themselves independently. It is possible to plug a node into an IPv6 network without requiring any human intervention. This feature was critical to allow network connectivity to an increasing number of mobile devices. This is accomplished by autoconfiguration.

IPv6 does not implement traditional IP broadcast, i.e. the transmission of a packet to all hosts on the attached link using a special broadcast address, and therefore does not define broadcast addresses. In IPv6, the same result can be achieved by sending a packet to the link-local all nodes multicast group at address ff02::1, which is analogous to IPv4 multicast to address 224.0.0.1.


Q18. Which two are advantages of static routing when compared to dynamic routing? (Choose two.)

A. Configuration complexity decreases as network size increases.

B. Security increases because only the network administrator may change the routing table.

C. Route summarization is computed automatically by the router.

D. Routing tables adapt automatically to topology changes.

E. An efficient algorithm is used to build routing tables, using automatic updates.

F. Routing updates are automatically sent to neighbors.

G. Routing traffic load is reduced when used in stub network links.

Answer: B,G

Explanation:

Since static routing is a manual process, it can be argued that it is more secure (and more prone to human errors) since the network administrator will need to make changes to the routing table directly. Also, in stub networks where there is only a single uplink connection, the load is reduced as stub routers just need a single static default route, instead of many routes that all have the same next hop IP address.


Q19. Refer to the exhibit.

Which address and mask combination represents a summary of the routes learned by EIGRP?

A. 192.168.25.0 255.255.255.240

B. 192.168.25.0 255.255.255.252

C. 192.168.25.16 255.255.255.240

D. 192.168.25.16 255.255.255.252

E. 192.168.25.28 255.255.255.240

F. 192.168.25.28 255.255.255.252

Answer: C

Explanation:

The binary version of 20 is 10100. The binary version of 16 is 10000. The binary version of 24 is 11000. The binary version of 28 is 11100.

The subnet mask is /28. The mask is 255.255.255.240.

Note:

From the output above, EIGRP learned 4 routes and we need to find out the summary of them:

+ 192.168.25.16

+ 192.168.25.20

+ 192.168.25.24

+ 192.168.25.28

-> The increment should bE. 28 u2013 16 = 12 but 12 is not an exponentiation of 2 so we must

choose 16 (24). Therefore the subnet mask is /28 (=1111 1111.1111 1111.1111

1111.11110000) = 255.255.255.240.

So, the best answer should be 192.168.25.16 255.255.255.240.


Q20. Refer to the exhibit.

Which switch provides the spanning-tree designated port role for the network segment that services the printers?

A. Switch1

B. Switch2

C. Switch3

D. Switch4

Answer: C

Explanation:

Printers are connected by hubs. Decide the switch that provides the spanning-tree designated port role between Switch3 and Switch4. They have the same priority 32768. Compare their MAC addresses. Switch3 with a smaller MAC address will provide a designated port for printers.


Q21. Refer to the exhibit.

A network administrator configures a new router and enters the copy startup-config running-config command on the router. The network administrator powers down the router and sets it up at a remote location. When the router starts, it enters the system configuration dialog as shown. What is the cause of the problem?

A. The network administrator failed to save the configuration.

B. The configuration register is set to 0x2100.

C. The boot system flash command is missing from the configuration.

D. The configuration register is set to 0x2102.

E. The router is configured with the boot system startup command.

Answer: A

Explanation:

The u201cSystem Configuration Dialogu201d appears only when no startup configuration file is found. The network administrator has made a mistake because the command u201ccopy startup-config running-configu201d will copy the startup config (which is empty) over the running config (which is configured by the administrator). So everything configured was deleted.

Note: We can tell the router to ignore the start-up configuration on the next reload by setting the register to 0u00d72142. This will make the u201cSystem Configuration Dialogu201d appear at the next reload.


Q22. 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 :: 0029 : EC7A : 0 : EC72

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 0u2019s can only be compressed once.