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New CCNA – General Questions.

Most of these questions appeared earlier this years (2013). Questions can sometime be twisted just to test your CCNA ability. Try as much as possible to read about topics provided through the links.

Here are some of the basic questions and answers to common CCNA. Also provided are brief explanations and links for more or in-depth explanation.

Question 1:

An administrator issues the command “ping 127.0.0.1″ from the command line prompt on a PC host named PC1. If an ICMP reply is received, what does this confirm?

A – The PC host PC1 has connectivity with a local host
B – The PC host PC1 has connectivity with a Layer 3 device
C – The PC host PC1 has a default gateway correctly configured
D – The PC host PC1 has connectivity up to Layer 5 of the OSI model
E – The PC host PC1 has the TCP/IP protocol stack correctly installed

Answer: E

Explanation:

The “ping 127.0.0.1″ command is used to troubleshoot the local network card and the TCP/IP stack to make sure is working correctly. If an ICMP echo reply is received, that shows there are connectivity to a particular host in the network.

Question 2:

You want to ping the loopback address of your local host. What will you type? (choose two)

A.    Ping 127.0.0.1

B.     Ping 0.0.0.0

C.     Ping ::1

D.    Trace 0.0.::1

Answer: A, C

Explanation:

The loopback address with IPv4 is 127.0.0.1. With IPv6, the loopback address is ::1 or 0:0:0:0:0:0:0:1

Question 3:

Your company wants to reconfigure a Catalyst 2950. Which actions must be taken to erase the old configuration? (Choose three)

A – Erase flash
B – Restart the switch
C – Delete the VLAN database
D – Erase the startup configuration

Answer: B C D

Read more on VLAN and how to configure and delete VLAN database.

 
 
Question 4:

If the subnet mask is 255.255.255.224, which of the following addresses can be assigned to network hosts? (Choose three)

A – 15.234.118.63
B – 92.11.178.93
C – 134.178.18.56
D – 192.168.16.87

Answer: B C D

Explanation:

Lets use the cram VLSM Subnet Cram table:

Bit Value

128

64

32

16

8

4

2

1

Bit Borrowed

1

2

3

4

5

6

7

8

Usable host address

126

62

30

14

6

2



Subnet Mask

128

192

224

240

248

252

255

256

Subnet Prefix/CIDR

/25

/26

/27

/28

/29

/30



As you can see that mask 224 means 32 bits value and 2 bits where borrowed and 30 usable ip addresses.

So the network range will be: 0, 32, 64, 96, 128, 160, 192, 224

Broadcast range: 31, 63, 95, 127, 159, 191, 223.

So the host addresses that satisfy these conditions will be those of B, C, D because, they don’t belong network or broadcast addresses in this range

 Question 5:

Study the exhibit carefully, can you tell which three description are correct about the ways used by the router R1 to choose a path to the 10.10.5.0/24 network when different routing protocols are deployed? (Choose three)

 
orbitco-ccna-pastquestions.com

A – When RIPv2 is the routing protocol, only the path R1-R4 is to be installed into the routing  table by default
B – When RIPv2 is the routing protocol, the equal cost paths R1-R3-R4 and R1-R2-R4 are to be installed in the routing table
C – If both EIGRP and OSPF are working on the network with their default configurations, the EIGRP paths will be installed in the routing table
D – By default, if EIGRP is the routing protocol, the equal cost paths R1-R3-R4 and R1-R2-R4 will be installed in the routing table

Answer: A C D

Explanation:

RIP uses hop count as the metric for path selection so only the path R1-R4 (with only 2 hops) will be installed into the routing table.

If EIGRP and OSPF were configured on the router, the EIGRP paths will be installed in the routing table because the default administrative distance of EIGRP is 90 over OSPF with AD of 110. The network routers will likely choose EIGRP over OSPF because EIGRP has lower administrative distance value.


 
 
 


 
 
 
 
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