Regarding the Que pertaing to vlan….deleting the previos Vlan will avoid any configuratinal issues araining in the future….After all the switch has to match the VTP domain name!

During the exam, you have to choose a answer even you are not sure about it, otherwise you can not go forward. Also, you can not go back to recheck your answer.

TY,

George

Today I have passed the CCNA. (860/825)
50 questions 3 labs (VTP, EIGRP, ACL). 35 from 9tut.

Also thanks a lot Brar and Sekhar (still valid from examcollection)

Ty again 9tut

from here none

Thank you in advance

Thank you in advance,

IN THE CCNA EXAM, CAN WE SKIP ANY UNKNOWN ANSWER ? OR DO WE HAVE TO ANSWER ALL FROM 1 TO 50 QUESTIONS CONTINUOSLY WITHOUT SKIP ?

THANKS

ABOUT MY EXAM:
I had total of 50 questions….
Had three simulations….ACL2, same as here at 9tut but modified as the third table of modifications above and i was dealing with host B plus different IP addresses, EIGRP almost the same as here, had to change the AS in router Home 2 from somethng 21 to 221 and add the appropriate networks also i had to delete and add some networks from router Home 1 (central router) and finaly the VTP…same as here just different question to follow the same procedures in resolving
NB: i ddnt see anythng that i had not come across before

for any other enquiries find me at ropamson@gmail.com …..best of luck for everybody working on CCNA

48 ques for exams including 3 simulation, I had EIGRP, Acesslist2 and VTP. Make sure the practice the simulation, use packet tracer or gns3. Best wishes to all!

Q5

Site-local address FEC0::/10

I think its FD00::/8
FC00::/7 – 1111:1100:x:x:x:x:x:x but the last bit of the 2nd octet must be 1 since 0 is for future use, therefore from FC00::/7 it is now FD00::/8 >>>>> from jeremy ciora of CBT NUGGETS CCNA tutorial

thanks :-)

subnet mask is 255.255.255.224

possible subnets:
.0 ~ .31
.32 ~ .63
.64 ~ .95
.96 ~ .127
increases by 32 as you can see

A – 15.234.118.63
broadcast address, not suitable for a host

B – 92.11.178.93
usable IP address on the .64 ~ .95 subnet

C – 134.178.18.56
usable IP address on the .32 ~ .63 subnet

D – 192.168.16.87
usable IP address on the .64 ~ .95 subnet

I am taking the exam on April 8, 2012. Can someone send me the latest dumps to my email rshiva62@gmail.com.

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

1st we should subtract 224 from 256….then we get 32..so the increament is 32.

32+32=64-1 to get the broadcast address.Hence the 1st network address is 32..1st address is 31 and the broadcast address is 63(64-1)..the assignable host address are therefore..31 to 62… for the 1st network that is 32..
32+32=64
64+32=96

32 (network address) 64 (network address)
31 (1st assignable address) 65 (1st assignable address)
62 (last assignable address) 94 (last assignable address)
63 (broadcast address) 95 (broadcast address)

therefore the answers are within these ranges of the two networks…

i p address 92.11.178.93 is in the 92.11.178.64 network….

i p address 134.178.18.56 is in the 134.178.18.32 network

i p address 192.168.16.87 is in the 192.168.16.32 network

Meaning B, C and D are correct.l
Thanks to all.
olesimbe@yahoo.com

Thanks Guys!

chaudaryimran@hotmail.com

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 10.1.3.0/24 network when different routing protocols are deployed? (Choose three)

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

Why and how B and D are correct answers???

what does this mean?
the command *ping 127.0.0.1* tests if the device has the TCP/IP stack installed correctly. that’s it. only local importance.

regarding your second question:
*ping 127.0.0.1* works <- device has TCP/IP stack installed
*ping 10.0.0.35* works <- same as pinging 127.0.0.1 because this is the IP of the computer on the network
*ping 10.0.0.1* fails <- the problem is on the LAN
*ping 10.5.75.250* fails <- the LAN problem is preventing packets to go to/get back from the remote address

options:
A. A remote physical layer problem exists. <- ruled out because the default gateway is unreachable
B. The host NIC is not functioning. <- it is working as it is able to perform a loopback test
C. TCP/IP has not been correctly installed on the host. <- false, the successful loopback test means the TCP/IP stack is installed correctly.
D. A local physical layer problem exists. <- only option left standing.

Can anyone help I have seen a questionon p4sure about dada transfer being slow betwean source and destination the QOS requested by the tarnsport layer is not being mantained at which layer should the trouble shooting begin? I thought transport, as this is where windowing etc is but they have the answer down as network, now I just thought network was responsible for logical addressing can anyone explain the answer?

TIA

Odoyl wrote on January 28th, 2011

Mr. Hassan,
If you are borrowing 6 bits, why is your mask reflecting that you are borrowing 10 bits?
You’re default mask is 255.255.0.0 and you are showing 255.255.255.192 as your answer.

—————

Sorry. I was away for too long. Odoyl I sincerely hope that you would have figured it out by now and have become certified as well but for newcomers that could get stuck at same place, I’ll explain it anyways.

While solving for 50 hosts, I’m NOT BORROWING 6 BITS, I’m RESERVING 6 BITS FOR HOSTS.

Cheers

@ doda_zeta@yahoo.com

can sm1 send me the latest dump on phathu6@webmail.co.za.

am planning 2 take my exam in 2 weeks tym.
tanks

But in the explanation part,
I think the most recent RFC revision says that site-local addresses start with FC00::/7 (FD00::/8 to be used while FC00::/8 to be used by “authorities”) and not FEC0::/10 anymore.

FYI
:)

quoting wikipedia
“In 1995, RFC 1884 reserved the block fec0::/10 for site-local addresses, that could be used within a “site” for private IPv6 networks. However, insufficient definition of the term site lead to confusion over the resulting routing rules. RFC 3879 (September 2004) deprecated this address range, and postulated solutions to its problems. In October 2005, RFC 4193 was published, reserving the address block fc00::/7 for use in private IPv6 networks, and defining the associated term unique local addresses. ”
http://en.wikipedia.org/wiki/Unique_local_address

Now write new subnet mask with only 6 bits for hosts and assign rest to the network:
11111111.11111111.11111111.11000000
i.e., 255.255.255.192 (Option E)

Option D is incorrect because if maximum required hosts are 50, there is no harm in having some extra hosts available to you i.e., 62 but solving for 30 hosts DOES NOT FULFILL the requirement. You MUST always solve for networks and hosts demanded in the question. You can always have more but you can never ever provide less than what was required. Hope that explains

ip address 192.168.16.87 is in the 192.168.16.64 network

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

1st we should subtract 224 from 256….then we get 32..so the increament is 32.

32+32=64…….therefore….1st network address is 32..1st address is 31 and the broadcast address is 63(64-1)..the assignable host address are therefore..31 to 62… for the 1st network that is 32..
32+32=64
64+32=96

32 (network address) 64 (network address)
31 (1st assignable address) 65 (1st assignable address)
62 (last assignable address) 94 (last assignable address)
63 (broadcast address) 95 (broadcast address)

therefore the answers are within these ranges of the two networks…

ip address 92.11.178.93 is in the 92.11.178.64 network….

ip address 134.178.18.56 is in the 134.178.18.32 network

ip address 192.168.16.87 is in the 192.168.16.32 network

Answer: B C D

I am new to 9tut and want to participate in its forum,

I have to understand one question, but i couldnt find where i can post my question.

please guide me

if any one need help from me pls email to

hasy_001@yahoo.com

Address is of class B, so default mask is:
255.255.0.0 or /16. This is fixed. We can not go back (I mean lower than 16: 15, 14… etc) in the network bits.

Default mask:
255.255.0.0
11111111.11111111.00000000.00000000

Solving for Subnets
——————-

Subnets required: 300
So network bits we need to borrow from host portion = n = ?
300 can be achieved using what power of 2?
Answer: 9
So 2^n = 2^9 = 512
So we need to borrow 9 more bits from host portion (Zeroes) and include them in network part (Ones)
New subnet mask:
11111111.11111111.11111111.10000000
which is 255.255.255.128 (Option B)

Solving for Hosts
—————–

Hosts required: 50
So how many bits do we need to reserver for hosts = h = ?
50 can be achieved using what power of 2?
Answer: 6
Total IP address = (2^h) = (2^6) = 64
Nodes assignable addresses = (2^h)-2 = 62 (Subtracted 2 for network and broadcast ID)

Now write new subnet mask with only 6 bits for hosts and assign rest to the network:
11111111.11111111.11111111.11000000
i.e., 255.255.255.192 (Option E)

Option D is incorrect because if maximum required hosts are 50, there is no harm in having some extra hosts available to you i.e., 62 but solving for 30 hosts DOES NOT FULFILL the requirement. You MUST always solve for networks and hosts demanded in the question. You can always have more but you can never ever provide less than what was required. Hope that explains.

Cheers